[
  {
    "id": 37501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA right triangle has legs of lengths $3$ and $4$. Find the volume of the solid formed by revolving the triangle about its hypotenuse.",
    "options": [],
    "answer": "48*pi/5",
    "solution": "Solution:\n\nThe solid consists of two conical solids with a common circular base of radius $\\frac{3 \\cdot 4}{5} = \\frac{12}{5}$. If $h_{1}$ and $h_{2}$ are the heights of the two cones, then $h_{1} + h_{2} = 5$. Hence, the volume of the solid is\n$$\n\\frac{\\pi}{3} \\cdot \\left(\\frac{12}{5}\\right)^{2} \\cdot (h_{1} + h_{2}) = \\frac{48 \\pi}{5}\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 37502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nKyle secretly selects a subset of $\\{1,2,3,4\\}$. Albert also secretly selects a subset of $\\{1,2,3,4\\}$. What is the probability that their chosen subsets have at least one element in common?",
    "options": [],
    "answer": "175/256",
    "solution": "Solution:\n\nLet $A$ and $B$ be the subsets selected by Kyle and Albert, respectively. We first find the probability that two subsets $A$ and $B$ are disjoint. For each $k \\in \\{0,1, \\ldots, 4\\}$, we choose an arbitrary subset $A$ with $k$ elements. In order for $A$ and $B$ to be disjoint, $B$ must be a subset of the complement $\\{1,2,3,4\\} \\backslash A$ with $4-k$ elements. Thus, for each $k \\in \\{0,1, \\ldots, 4\\}$, there are $\\binom{4}{k}$ subsets with $k$ elements and fixing one of such subsets (say $A$), there are $2^{4-k}$ choices for $B$ (note that $2^{4-k}$ is the number of subsets of $\\{1,2,3,4\\} \\backslash A$). We see that the number of ordered pairs $(A, B)$ of subsets with $A \\cap B = \\varnothing$ is\n$$\n\\sum_{k=0}^{4} \\binom{4}{k} 2^{4-k} = 16 \\sum_{k=0}^{4} \\binom{4}{k} \\left(\\frac{1}{2}\\right)^k = 16\\left(1+\\frac{1}{2}\\right)^4 = 81\n$$\nAs there are $2^4 = 16$ subsets of $\\{1,2,3,4\\}$, there are $16^2 = 256$ possible ordered pairs of subsets. Hence, the probability that two subsets $A$ and $B$ are disjoint is $\\frac{81}{256}$ and the probability that $A$ and $B$ have at least one element in common is $1 - \\frac{81}{256} = \\frac{175}{256}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle with circumcircle $\\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD = BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at $F$, and line $PF$ meets $\\Gamma$ again at $Q$. Line $AC$ meets the circumcircle of $\\triangle PNQ$ again at $E$. Prove that $\\angle DQE = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $f$ is a second-degree polynomial for which $f(2)=1$, $f(4)=2$, and $f(8)=3$. Find the sum of the roots of $f$.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\nLet $f(x) = a x^{2} + b x + c$. By substituting $x = 2, 4, 8$ we get the system of linear equations\n$$\n\\begin{array}{r}\n4a + 2b + c = 1 \\\\\n16a + 4b + c = 2 \\\\\n64a + 8b + c = 3\n\\end{array}\n$$\nSolving this system of equations gives us $a = -\\frac{1}{24}$, $b = \\frac{3}{4}$, $c = -\\frac{1}{3}$. By using Vieta's identities, the sum of roots is $-\\frac{b}{a} = \\frac{3}{4} \\cdot 24 = 18$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a positive integer $n$, denote by $\\varphi(n)$ the number of positive integers $k \\leq n$ relatively prime to $n$. How many positive integers $n$ less than or equal to $100$ are divisible by $\\varphi(n)$?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nWe claim that any such integer $n$ must be either equal to $1$ or of the form $2^{a} 3^{b}$, where $a \\geq 1$ and $b \\geq 0$.\n\nFirst, we note that if $n > 1$, it must be even. This is because if $n$ admits a prime factorization $n = \\prod_{i=1}^{k} p_{i}^{r_{i}}$ over distinct primes $p_{i}$, then $\\varphi(n) = \\prod_{i=1}^{k} (p_{i} - 1) p_{i}^{r_{i} - 1}$. In particular, if an odd prime $p_{i}$ divides $n$ then $p_{i} - 1$ divides $\\varphi(n)$, and consequently $\\varphi(n)$ is even.\n\nThen, we note that at most one odd prime divides $n$. Suppose otherwise, i.e., $2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}$ is a part of the prime factorization of $n$ for odd primes $p_{1}, p_{2}$. Then by the multiplicativity of $\\varphi$, we know that\n$$\n\\varphi\\left(2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}\\right) = 2^{a-1} (p_{1} - 1) p_{1}^{r_{1} - 1} (p_{2} - 1) p_{2}^{r_{2} - 1}\n$$\nmust divide $\\varphi(n)$. Note that the largest power of $2$ dividing $n$ is $2^{a}$, but since $p_{1} - 1$ and $p_{2} - 1$ are both even, $2^{(a-1)} 2^{2} = 2^{a+1}$ divides $\\varphi(n)$. This is a contradiction.\n\nFinally, we show that the said odd prime $p$ dividing $n$ must in fact be equal to $3$. Indeed, if $n = 2^{a} p^{r}$ where $a$ and $r$ are positive integers, we have\n$$\n\\frac{n}{\\varphi(n)} = \\frac{2p}{p-1} = 2 + \\frac{2}{p-1}.\n$$\nFor this to be an integer, $p = 3$.\n\nHence, we count all such integers of the form $n = 2^{a} 3^{r}$ by summing over all possible values of $r$ (because there are fewer):\n\nIf $r = 0$, then $0 \\leq a \\leq 6$. (This is the only case in which $a = 0$ is allowed.)\n\nIf $r = 1$, then $1 \\leq a \\leq 5$.\n\nIf $r = 2$, then $1 \\leq a \\leq 3$.\n\nIf $r = 3$, then $a = 1$.\n\nThis gives us a total of $7 + 5 + 3 + 1 = 16$ values of $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $20 \\times 19$ rectangle is plotted on the Cartesian plane with one corner at the origin and with sides parallel to the coordinate axes. How many unit squares do the two diagonals of this rectangle pass through?",
    "options": [],
    "answer": "74",
    "solution": "Solution:\n\nSuppose that one corner of the rectangle is on $(20,19)$. First of all, note that $20$ and $19$ are relatively prime. This means that the line does not intersect any vertex of a unit square in the interior of the grid.\n\nNow, consider the diagonal from $(0,0)$ to $(20,19)$. This diagonal intersects exactly $19$ vertical segments and $18$ horizontal segments. Here, each intersection point represents an entry point into a new unit square. Including the original unit square (which has the origin as a corner), this accounts for a total of $19+18+1=38$ unit squares.\n\nSimilarly, the other diagonal of the rectangle passes through $38$ unit squares. Finally, since the diagonals intersect at the center, the center must have been an entry point for one diagonal and an exit point for the other, and vice versa. These pertain to the two unit squares that both diagonals pass through.\n\nTherefore, the diagonals pass through a total of $2(38)-2=74$ unit squares.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA triangle has side lengths $7, 11, 14$. Find the length of its inradius.",
    "options": [],
    "answer": "3*sqrt(10)/4",
    "solution": "Solution:\n\nWe use Heron's formula. The semiperimeter of the triangle is $16$. Thus, the area of the triangle is $$\\sqrt{(16)(16-7)(16-11)(16-14)} = 12 \\sqrt{10}.$$ Since the area of a triangle is the product of the semiperimeter and the inradius, the length of the inradius is $$\\frac{12 \\sqrt{10}}{16} = \\frac{3 \\sqrt{10}}{4}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn $\\triangle ABC$, $AB = 20$, $BC = 21$, and $CA = 29$. Point $M$ is on side $AB$ with $\\frac{AM}{MB} = \\frac{3}{2}$, while point $N$ is on side $BC$ with $\\frac{CN}{NB} = 2$. $P$ and $Q$ are points on side $AC$ such that the line $MP$ is parallel to $BC$ and the line $NQ$ is parallel to $AB$. Suppose that $MP$ and $NQ$ intersect at point $R$. Find the area of $\\triangle PQR$.",
    "options": [],
    "answer": "224/15",
    "solution": "Solution:\n\nWe use similar triangles here. Note that triangles $ABC$, $AMP$, $QNC$ and $QRP$ are all similar right (by the Pythagorean theorem, since $20^{2} + 21^{2} = 29^{2}$) triangles by AA similarity and corresponding angle theorem. We see that $AP = \\frac{3}{5} \\cdot 29 = \\frac{87}{5}$ and $CQ = \\frac{2}{3} \\cdot 29 = \\frac{58}{3}$.\n\nHence, $PQ = AP + QC - AC = \\frac{87}{5} + \\frac{58}{3} - 29 = \\frac{116}{15} = 29\\left(\\frac{4}{15}\\right)$. Thus, the ratio of similitude between $QRP$ and $ABC$ is $\\frac{4}{15}$, and the area of triangle $QRP$ is $\\left(\\frac{16}{225}\\right)\\left(\\frac{1}{2}\\right)(20)(21) = \\frac{224}{15}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a race with six runners, $A$ finished between $B$ and $C$, $B$ finished between $C$ and $D$, and $D$ finished between $E$ and $F$. If each sequence of winners in the race is equally likely to occur, what is the probability that $F$ placed last?",
    "options": [],
    "answer": "5/16",
    "solution": "Solution:\n\nSuppose $A > B$ means that $A$ finished earlier than $B$. Then, the criteria imply the following:\n\na. either $B > A > C$ or $C > A > B$\n\nb. either $E > D > F$ or $F > D > E$\n\nc. either $D > B > C$ or $C > B > D$\n\nCombining Criteria 1 and 2 together implies either $D > B > A > C$ or $D < B < A < C$. When $D > B > A > C$, there's one place before $D$, and four after $D$ (between the letters). Furthermore, there are two ways to permute $E$ and $F$. Thus, there are $(1)(4)(2) = 8$ possible sequences, among which only 1 has $F$ being last. On the other hand, when $C > A > B > D$, there are again 8 possibilities. If $F$ is last, $E$ still has 4 possible places before $D$.\n\nThus, the required probability is $\\frac{1+4}{8+8} = \\frac{5}{16}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $(1+\\sec \\theta)(1+\\csc \\theta)=6$. Determine the value of $(1+\\tan \\theta)(1+\\cot \\theta)$.",
    "options": [],
    "answer": "49/12",
    "solution": "Solution:\n\nThe equation is equivalent to $1+\\sin \\theta+\\cos \\theta=5 \\sin \\theta \\cos \\theta$. Let $A := \\sin \\theta + \\cos \\theta$ and $B := \\sin \\theta \\cos \\theta$. Then $1+A=5B$ and $A^{2}=1+2B$. As $1+A \\neq 0$,\n$$\n1+A=\\frac{5\\left(A^{2}-1\\right)}{2} \\Longrightarrow 1=\\frac{5(A-1)}{2}\n$$\nwhich gives $A=\\frac{7}{5}$. Consequently, we get $B=\\frac{12}{25}$. Hence,\n$$\n(1+\\tan \\theta)(1+\\cot \\theta)=\\frac{A^{2}}{B}=\\frac{49}{25} \\cdot \\frac{25}{12}=\\frac{49}{12}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of ordered quadruples $(a, b, c, d)$ of positive integers such that $a b c d = 216$.",
    "options": [],
    "answer": "400",
    "solution": "Solution:\n\nSince $216 = 2^{3} 3^{3}$, any positive divisor of $216$ must be of the form $2^{x} 3^{y}$ for some integers $x$ and $y$ with $0 \\leq x, y \\leq 3$. Thus, we set $a = 2^{x_{1}} 3^{y_{1}}$, $b = 2^{x_{2}} 3^{y_{2}}$, $c = 2^{x_{3}} 3^{y_{3}}$ and $d = 2^{x_{4}} 3^{y_{4}}$, where $0 \\leq x_{i}, y_{i} \\leq 3$ are integers for $i = 1, \\ldots, 4$. We compute\n$$\n2^{3} 3^{3} = 216 = a b c d = 2^{x_{1} + x_{2} + x_{3} + x_{4}} 3^{y_{1} + y_{2} + y_{3} + y_{4}}\n$$\nso the number of such ordered quadruples $(a, b, c, d)$ is the number of ordered 8-tuples $(x_{1}, x_{2}, x_{3}, x_{4}, y_{1}, y_{2}, y_{3}, y_{4})$ of nonnegative integers such that $x_{1} + x_{2} + x_{3} + x_{4} = y_{1} + y_{2} + y_{3} + y_{4} = 3$. By stars-and-bars, this number is $\\binom{3 + 4 - 1}{4 - 1}^{2} = 20^{2} = 400$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $A=\\{1,2, \\ldots, 20\\}$. Call $B$ a visionary set of $A$ if $B \\subseteq A$, $B$ contains at least one even integer, and $|B| \\in B$, where $|B|$ is the cardinality of set $B$. How many visionary sets does $A$ have?",
    "options": [],
    "answer": "524032",
    "solution": "Solution:\n\nIgnoring the constraint that all visionary sets must contain at least one even integer, we count a total of\n$$\n\\sum_{i=0}^{19}\\binom{19}{i}=2^{19}\n$$\nsets. From this total, we subtract the number of sets that only contain odd numbers. Evidently, $|B|$ has to be odd. Since there are only $10$ odd numbers in $A$, $|B|$ can only be $1,3,5,7$, or $9$. The total number of instances when this happens is\n$$\n\\sum_{i=0}^{4}\\binom{9}{2i}=\\frac{1}{2} \\sum_{i=0}^{9}\\binom{9}{i}=\\frac{1}{2}\\left(2^{9}\\right)=2^{8}\n$$\nsets. Therefore, there are a total of $2^{19}-2^{8}$ visionary sets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven triangle $ABC$, let $D$ be a point on side $AB$ and $E$ be a point on side $AC$. Let $F$ be the intersection of $BE$ and $CD$. If $\\triangle DBF$ has an area of $4$, $\\triangle BFC$ has an area of $6$, and $\\triangle FCE$ has an area of $5$, find the area of quadrilateral $ADFE$.",
    "options": [],
    "answer": "105/4",
    "solution": "Solution:\n\nLet the area of quadrilateral $ADFE$ be $x$. By Menelaus' Theorem, $\\frac{AD}{DB} \\cdot \\frac{BF}{FE} \\cdot \\frac{EC}{CA} = 1$. Since $\\frac{AD}{DB} = \\frac{x+5}{10}$, $\\frac{BF}{FE} = \\frac{6}{5}$, and $\\frac{EC}{CA} = \\frac{11}{x+15}$, we have $\\frac{66(x+5)}{50(x+15)} = 1$, or $x = \\frac{105}{4}$ or $26.25$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $10 \\times 1$ rectangular pavement is to be covered by tiles which are either green or yellow, each of width $1$ and of varying integer lengths from $1$ to $10$. Suppose you have an unlimited supply of tiles for each color and for each of the varying lengths. How many distinct tilings of the rectangle are there, if at least one green and one yellow tile should be used, and adjacent tiles should have different colors?",
    "options": [],
    "answer": "1022",
    "solution": "Solution:\n\nNote that the pavement is fixed and cannot be rotated, therefore a tiling is considered distinct from the reverse tiling. Also, note that the restriction that no two consecutive tiles can be of the same color can be addressed simply by treating consecutive tiles of the same color as one tile. Hence, the problem is just asking for the number of possible tilings that alternate both colors. For any division of the board into tiles, there are precisely two ways to color the tiles, as the coloring is determined solely by the color of the first tile. Now, to count the uncolored tilings, we divide the board into $10$ squares using $9$ dividers, and just count the number of subsets of the $9$ dividers with at least one element. There are exactly $2^{9}-1=511$ such subsets, and thus $511 \\cdot 2=1022$ such colorings.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn convex quadrilateral $A B C D$, $\\angle C A B = \\angle B C D$. $P$ lies on line $B C$ such that $A P = P C$, $Q$ lies on line $A P$ such that $A C$ and $D Q$ are parallel, $R$ is the point of intersection of lines $A B$ and $C D$, and $S$ is the point of intersection of lines $A C$ and $Q R$. Line $A D$ meets the circumcircle of $A Q S$ again at $T$. Prove that $A B$ and $Q T$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRefer to the figure shown below.\n![](attached_image_1.png)\n\nBy angle-chasing (with directed angles), we have\n$$\n\\angle Q A R = \\angle P A B = \\angle C A B - \\angle C A P = \\angle B C D - \\angle P C A = \\angle A C D = \\angle Q D R\n$$\nThus quadrilateral $Q A D R$ is cyclic. Then,\n$$\n\\angle B A D = \\angle R A D = \\angle R Q D = \\angle Q S A = \\angle Q T A = \\angle Q T D\n$$\nand hence $A B$ is parallel to $Q T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a convex polygon, the number of diagonals is $23$ times the number of its sides. How many sides does it have?\n\n(a) $46$\n(b) $49$\n(c) $66$\n(d) $69$",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can $2021$ be written as a sum of two or more consecutive integers?\n(a) $3$\n(b) $5$\n(c) $7$\n(d) $9$",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn quadrilateral $ABCD$, $\\angle CBA = 90^\\circ$, $\\angle BAD = 45^\\circ$, and $\\angle ADC = 105^\\circ$. Suppose that $BC = 1 + \\sqrt{2}$ and $AD = 2 + \\sqrt{6}$. What is the length of $AB$?\n\n(a) $2\\sqrt{3}$\n\n(b) $2 + \\sqrt{3}$\n\n(c) $3 + \\sqrt{2}$\n\n(d) $3 + \\sqrt{3}$",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice tosses two biased coins, each of which has a probability $p$ of obtaining a head, simultaneously and repeatedly until she gets two heads. Suppose that this happens on the $r$th toss for some integer $r \\geq 1$. Given that there is $36\\%$ chance that $r$ is even, what is the value of $p$?\n\n(a) $\\frac{\\sqrt{7}}{4}$\n\n(b) $\\frac{2}{3}$\n\n(c) $\\frac{\\sqrt{2}}{2}$\n\n(d) $\\frac{3}{4}$",
    "options": [],
    "answer": "(a)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a real number $t$, $\\lfloor t \\rfloor$ is the greatest integer less than or equal to $t$ and $\\{ t \\} = t - \\lfloor t \\rfloor$ is the fractional part of $t$. How many real numbers $x$ between $1$ and $23$ satisfy $\\lfloor x \\rfloor \\{ x \\} = 2 \\sqrt{x}$?\n\n(a) $18$\n(b) $19$\n(c) $20$\n(d) $21$",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the remainder when $\\sum_{n=2}^{2021} n^{n}$ is divided by $5$.\n\n(a) $1$\n(b) $2$\n(c) $3$\n(d) $4$",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the figure below, $BC$ is the diameter of a semicircle centered at $O$, which intersects $AB$ and $AC$ at $D$ and $E$ respectively. Suppose that $AD=9$, $DB=4$, and $\\angle ACD=\\angle DOB$. Find the length of $AE$.\n![](attached_image_1.png)\n\n(a) $\\frac{117}{16}$\n(b) $\\frac{39}{5}$\n(c) $2 \\sqrt{13}$\n(d) $3 \\sqrt{13}$",
    "options": [],
    "answer": "(b)",
    "solution": "Solution:\n\nLet $\\angle DOB=\\angle EOC=\\alpha$. Note that $\\angle DCB=\\frac{\\alpha}{2}$. Also, note that $\\tan \\frac{\\alpha}{2}=\\frac{4}{DC}$ and $\\tan \\alpha=\\frac{9}{DC}=\\frac{9}{4} \\tan \\frac{\\alpha}{2}$. Let $x=\\tan \\frac{\\alpha}{2}$. By the double-angle formula,\n$$\n\\begin{aligned}\n\\frac{9}{4} x & =\\frac{2x}{1-x^{2}} \\\\\n\\frac{9}{4} x-\\frac{9}{4} x^{3} & =2x \\\\\n\\frac{1}{4} x\\left(1-9x^{2}\\right) & =0\n\\end{aligned}\n$$\nand thus $x=0$ or $x= \\pm \\frac{1}{3}$. Clearly, only $x=\\frac{1}{3}$ is possible here. Thus, $CD=12$. Note also that $\\angle CDB=\\angle ADC=90^{\\circ}$, and so by the Pythagorean theorem, $AC=\\sqrt{9^{2}+12^{2}}=15$.\n\nFinally, by the power of a point theorem, we have $AE \\cdot AC=AD \\cdot AB$ and so $AE \\cdot 15=9(9+4)$, which gives us $AE=\\frac{117}{15}=\\frac{39}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider all real numbers $c$ such that $|x-8| + \\left|4 - x^{2}\\right| = c$ has exactly three real solutions. The sum of all such $c$ can be expressed as a fraction $a / b$ in lowest terms. What is $a + b$?",
    "options": [],
    "answer": "93",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest positive integer $n$ for which there are exactly 2323 positive integers less than or equal to $n$ that are divisible by 2 or 23, but not both.",
    "options": [],
    "answer": "4644",
    "solution": "Solution:\nThe number of positive integers from $1$ to $n$ that are divisible by $2$ or $23$, but not both, is\n$$\nf(n) = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{23} \\right\\rfloor - 2\\left\\lfloor \\frac{n}{46} \\right\\rfloor\n$$\nWe need to find $f(n) = 2323$, which can be done by some trial and error.\n\nNote that if $n$ is a multiple of $2$ and $23$, then the floor divisions are non-rounded exact divisions, in which case,\n$$\n\\frac{n}{2} + \\frac{n}{23} - 2 \\frac{n}{46} = \\frac{n}{2}\n$$\nSo, for example, $f(4646) = 2323$. We can then just tick downwards.\n\nSince $4646$ and $4645$ do not satisfy our criteria, we know that $f(4646) = f(4645) = f(4644)$, i.e., we can remove $4646$ and $4645$ and the count does not go down (we weren't counting them). However, $4644$ does satisfy our criteria, so $f(4643) = 2322$.\n\nNote that by definition, this function is non-decreasing. Thus, we conclude that $n = 4644$ is the first $n$ that satisfies the given conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P(x)$ be a polynomial with integer coefficients such that $P(-4)=5$ and $P(5)=-4$. What is the maximum possible remainder when $P(0)$ is divided by 60?",
    "options": [],
    "answer": "41",
    "solution": "Solution:\n$$\n\\begin{gathered}\n0-(-4) \\mid P(0)-P(-4) \\text{ or } 4 \\mid P(0)-5, \\text{ so } P(0) \\equiv 5 \\pmod{4} \\\\\n5-0 \\mid P(5)-P(0) \\text{ or } 5 \\mid -4-P(0), \\text{ so } P(0) \\equiv -4 \\pmod{5}\n\\end{gathered}\n$$\nBy the Chinese Remainder Theorem, there is a solution $r$ that satisfies both of the previous equations, and this solution is unique modulo $4 \\cdot 5 = 20$. It is easy to verify that this solution is $1$. Thus, $P(0) \\equiv 1 \\pmod{20}$. This implies that $P(0)$ can be $1$, $21$, or $41 \\pmod{60}$. The largest remainder $41$ is indeed achievable, for example by the polynomial $-2(x+4)(x-5)+1-x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\triangle ABC$ be an equilateral triangle with side length $16$. Points $D, E, F$ are on $CA, AB$, and $BC$, respectively, such that $DE \\perp AE$, $DF \\perp CF$, and $BD = 14$. The perimeter of $\\triangle BEF$ can be written in the form $a + b \\sqrt{2} + c \\sqrt{3} + d \\sqrt{6}$, where $a, b, c$, and $d$ are integers. Find $a + b + c + d$.",
    "options": [],
    "answer": "31",
    "solution": "Solution:\nLet $\\overline{AD} = 2x$, then $\\overline{DC} = 16 - 2x$. Since $\\triangle DAE$ and $\\triangle DCF$ are both $30$-$60$-$90$ triangles, then $\\overline{AE} = \\overline{AD}/2 = x$, $\\overline{ED} = x\\sqrt{3}$ and $\\overline{CF} = \\overline{DC}/2 = 8 - x$, $\\overline{FD} = (8 - x)\\sqrt{3}$. Since $\\overline{AB} = \\overline{BC} = 16$, then $\\overline{EB} = 16 - x$ and $\\overline{FB} = 8 + x$. Using Pythagorean Theorem on $\\triangle DEB$, we have\n$$\n\\begin{aligned}\n\\overline{ED}^2 + \\overline{EB}^2 & = \\overline{BD}^2 \\\\\n(x\\sqrt{3})^2 + (16 - x)^2 & = 14^2 \\\\\n3x^2 + 256 - 32x + x^2 & = 196 \\\\\nx^2 - 8x + 15 & = 0 \\\\\n(x - 5)(x - 3) & = 0 \\\\\nx & = 5, 3\n\\end{aligned}\n$$\nChoosing either of the two values of $x$ will give the same result for the perimeter of $\\triangle BEF$. Suppose we choose $x = 5$, then $\\overline{EB} = 11$ and $\\overline{FB} = 13$. By Cosine Law on $\\triangle BEF$, we have\n$$\n\\begin{aligned}\n\\overline{EF}^2 & = \\overline{EB}^2 + \\overline{FB}^2 - 2 \\cdot \\overline{EB} \\cdot \\overline{FB} \\cos 60^\\circ \\\\\n\\overline{EF}^2 & = 11^2 + 13^2 - 2(11)(13)(1/2) \\\\\n\\overline{EF} & = \\sqrt{121 + 169 - 143} = 7\\sqrt{3}\n\\end{aligned}\n$$\nThus, the perimeter of $\\triangle BEF = 24 + 7\\sqrt{3}$, which means that $a + b + c + d = 24 + 0 + 7 + 0 = 31$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the smallest real number $a$ for which the function $f(x) = 4x^{2} - 12x - 5 + 2a$ will always be nonnegative for all real numbers $x$?\n\n(a) $0$\n\n(b) $\\frac{3}{2}$\n\n(c) $\\frac{5}{2}$\n\n(d) $7$",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nFor $f(x)$ to be always nonnegative for all real $x$, its minimum value must be at least $0$.\n\nThe minimum of $f(x) = 4x^2 - 12x - 5 + 2a$ occurs at $x_0 = -\\frac{b}{2a} = -\\frac{-12}{2 \\times 4} = \\frac{12}{8} = \\frac{3}{2}$.\n\nSubstitute $x = \\frac{3}{2}$ into $f(x)$:\n\n\\begin{align*}\nf\\left(\\frac{3}{2}\\right) &= 4\\left(\\frac{3}{2}\\right)^2 - 12\\left(\\frac{3}{2}\\right) - 5 + 2a \\\\\n&= 4 \\times \\frac{9}{4} - 18 - 5 + 2a \\\\\n&= 9 - 18 - 5 + 2a \\\\\n&= -14 + 2a\n\\end{align*}\n\nSet $-14 + 2a \\geq 0$:\n\n$2a \\geq 14$\n\n$a \\geq 7$\n\nThus, the smallest real number $a$ is $7$.\n\n**Answer:** (d) $7$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Show that there exists a one-to-one function $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow \\{1,2, \\ldots, n\\}$ such that\n$$\n\\sum_{k=1}^{n} \\frac{k}{(k+\\sigma(k))^{2}}<\\frac{1}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt suffices to produce one such function. For this, consider the function $\\sigma(k) = n+1-k$. Then note that $\\sigma$ is one-to-one, since for every $a$ and $b$,\n$$\n\\sigma(a) = \\sigma(b) \\Rightarrow n+1-a = n+1-b \\Rightarrow a = b\n$$\nIn this case,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{k}{(k+\\sigma(k))^{2}} & = \\sum_{k=1}^{n} \\frac{k}{(k+n+1-k)^{2}} \\\\\n& = \\sum_{k=1}^{n} \\frac{k}{(n+1)^{2}} \\\\\n& = \\frac{1}{(n+1)^{2}} \\cdot \\sum_{k=1}^{n} k \\\\\n& = \\frac{1}{(n+1)^{2}} \\cdot \\frac{n(n+1)}{2} \\\\\n& = \\frac{n}{2(n+1)} \\\\\n& < \\frac{1}{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many subsets of the set $\\{1,2,3, \\ldots, 9\\}$ do not contain consecutive odd integers?",
    "options": [],
    "answer": "208",
    "solution": "Solution:\n\nLet $a_{n}$ be the number of subsets of $\\{1,2, \\ldots, n\\}$ that do not contain consecutive odd integers. We work on the following cases depending on whether such a subset contains $1$ or not:\n\n- If such a subset does not contain $1$, then its elements must consist of elements from $\\{2,3, \\ldots, n\\}$. The number of subsets with no element smaller than $3$ is $a_{n-2}$; for each of these subsets, we include $2$ as well, giving another $a_{n-2}$ subsets. Thus, there are $2 a_{n-2}$ subsets in this case.\n\n- If such a subset contains $1$, then it must not contain $3$ and its elements must consist of elements from $\\{1,2,4,5, \\ldots, n\\}$. The number of subsets with no element smaller than $5$ is $a_{n-4}$; for each of these subsets, we take the union with a non-empty subset of $\\{2,4\\}$ as well, giving another $3 a_{n-4}$ subsets. Thus, there are $4 a_{n-4}$ subsets in this case.\n\nWe now obtain the recurrence formula $a_{n}=2 a_{n-2}+4 a_{n-4}$. With $a_{1}=2$, $a_{2}=4$, $a_{3}=6$ and $a_{4}=12$, routine computation gives $a_{9}=208$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a positive integer $n$, define $s(n)$ as the smallest positive integer $t$ such that $n$ is a factor of $t!$. Compute the number of positive integers $n$ for which $s(n)=13$.",
    "options": [],
    "answer": "792",
    "solution": "Solution:\n\nFor a positive integer $k$, consider the set $A(k) = \\{ n \\in \\mathbb{N} : s(n) = k \\}$ and we wish to find $\\# A(13)$. From the definition of $s(n)$, any element of $A(k)$ must divide $k!$ but not $(k-1)!$. Thus, any element of $A(k)$ must be a divisor of $k!$ that is not a divisor of $(k-1)!$. We see that $\\# A(k) = \\sigma_{0}(k!) - \\sigma_{0}((k-1)!)$, where $\\sigma_{0}(n)$ is the number of divisors of $n$, so that $\\# A(13) = \\sigma_{0}(13!) - \\sigma_{0}(12!)$.\n\nNote that for any prime $p$, the highest power of $p$ that divides $k!$ is $p^{e}$, where $e = \\sum_{j=1}^{\\infty} \\left\\lfloor n / p^{j} \\right\\rfloor$. Using this formula, we determine the prime factorization of $13!$:\n\n$13! = 2^{10} \\cdot 3^{5} \\cdot 5^{2} \\cdot 7^{1} \\cdot 11^{1} \\cdot 13^{1}$.\n\nAs $13! = 13 \\cdot 12!$, we get $12! = 2^{10} \\cdot 3^{5} \\cdot 5^{2} \\cdot 7^{1} \\cdot 11^{1}$.\n\nHence, we obtain\n$$\n\\# A(13) = 11 \\cdot 6 \\cdot 3 \\cdot 2^{3} - 11 \\cdot 6 \\cdot 3 \\cdot 2^{2} = 198 \\cdot 4 = 792.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice and Bob are playing a game with dice. They each roll a die six times, and take the sums of the outcomes of their own rolls. The player with the higher sum wins. If both players have the same sum, then nobody wins. Alice's first three rolls are $6$, $5$, and $6$, while Bob's first three rolls are $2$, $1$, and $3$. The probability that Bob wins can be written as a fraction $a / b$ in lowest terms. What is $a+b$?",
    "options": [],
    "answer": "3895",
    "solution": "Solution:\n\nLet $a_{i}$ denote the value of Alice's $i$th roll, and $b_{i}$ denote the value of Bob's $i$th roll. For Bob to win, the following inequality must hold:\n$$\n6+5+6+a_{4}+a_{5}+a_{6}<2+1+3+b_{4}+b_{5}+b_{6}\n$$\nRearranging yields\n$$\n\\left(a_{4}-1\\right)+\\left(a_{5}-1\\right)+\\left(a_{6}-1\\right)+\\left(6-b_{4}\\right)+\\left(6-b_{5}\\right)+\\left(6-b_{6}\\right)<4\n$$\nLet $x_{i}=a_{i+3}-1$ and $x_{i+3}=6-b_{i}$ for $i=1,2,3$. The inequality then simplifies to\n$$\nx_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}<4\n$$\nWe claim that each nonnegative integer solution to this inequality corresponds to a valid situation of dice rolls. Note that the previous inequality implies that $0 \\leq x_{i}<4$, so $1 \\leq x_{i}+1<5$ and $2<6-x_{i} \\leq 6$. Thus, the corresponding dice rolls for both Alice and Bob are within the bounds.\n\nThe expression on the left can have a value of either $0,1,2$, or $3$. Thus, the number of nonnegative integer solutions of the aforementioned equation is $\\binom{5}{5}+\\binom{6}{5}+\\binom{7}{5}+\\binom{8}{5}=1+6+21+56=84$. Thus, the probability that Bob wins is $\\frac{84}{6^{6}}=\\frac{7}{3888}$, and so $a+b=7+3888=3895$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\triangle ABC$ be an isosceles triangle with a right angle at $A$, and suppose that the diameter of its circumcircle $\\Omega$ is $40$. Let $D$ and $E$ be points on the arc $BC$ not containing $A$ such that $D$ lies between $B$ and $E$, and $AD$ and $AE$ trisect $\\angle BAC$. Let $I_{1}$ and $I_{2}$ be the incenters of $\\triangle ABE$ and $\\triangle ACD$ respectively. The length of $I_{1}I_{2}$ can be expressed in the form $a+b \\sqrt{2}+c \\sqrt{3}+d \\sqrt{6}$, where $a, b, c$, and $d$ are integers. Find $a+b+c+d$.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nLet $O$ be the center of $\\Omega$. Note that $\\angle OBD = \\angle CBD = \\frac{\\pi}{3}$, so $\\triangle OBD$ is an equilateral triangle. Thus, $BD = BO = \\frac{40}{2} = 20$. This implies that $EC = DE = BD = 20$.\n\nClearly, $I_{1}$ and $I_{2}$ lie on the segments $AD$ and $AE$ respectively. It is well-known that $DI_{1} = DB$, so $DI_{1} = 20$. Similarly, $EI_{2} = 20$.\n\nNow, note that $\\angle EDI_{1} = \\angle EDC + \\angle CDA = 30^{\\circ} + 45^{\\circ} = 75^{\\circ}$. Similarly, $\\angle DEI_{2} = 75^{\\circ}$.\n\nLooking at the isosceles trapezoid $I_{1}I_{2}ED$, the length of $I_{1}I_{2}$ must then be $DE - DI_{1} \\cos 75^{\\circ} - EI_{2} \\cos 75^{\\circ} = 20 - 40 \\cos 75^{\\circ} = 20 - 40\\left(\\frac{\\sqrt{6} - \\sqrt{2}}{4}\\right) = 20 - \\sqrt{6} + \\sqrt{2}$, and so $a + b + c + d = 20 + 1 + 0 + (-1) = 20$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of functions $f$ from the set $S=\\{0,1,2, \\ldots, 2020\\}$ to itself such that, for all $a, b, c \\in S$, all three of the following conditions are satisfied:\n(i) If $f(a)=a$, then $a=0$;\n(ii) If $f(a)=f(b)$, then $a=b$; and\n(iii) If $c \\equiv a+b(\\bmod 2021)$, then $f(c) \\equiv f(a)+f(b)(\\bmod 2021)$.",
    "options": [],
    "answer": "1845",
    "solution": "Solution:\nNote that, from (i), our function is completely determined by $f(1)$; i.e., $f(a) \\equiv a f(1)$ $(\\bmod 2021)$. Then, from (i) and (ii), we need that $f(a) \\neq 0$ if $a \\neq 0$; otherwise, if $a \\neq 0$ but $f(a)=0, f(b)=f(a+b)$ for any $b$. Thus, if $a \\neq 0$, we need that $a f(1) \\neq 0(\\bmod 2021)$ for any $a \\in S \\backslash\\{0\\}$. If \\operatorname{gcd}(f(1), 2021)=d>1$, then note that $a=\\frac{2021}{d} \\in S \\backslash\\{0\\}$, and from (i), $f(a) \\equiv a f(1) \\equiv 2021 \\equiv 0(\\bmod 2021)$. As we just established, this is not allowed. Hence, \\operatorname{gcd}(f(1), 2021)=1$.\n\nMoreover, from (iii), we need that $f(a)=a f(1) \\not \\equiv a(\\bmod 2021)$ if $a \\neq 0$; in other words, $2021 \\nmid a(f(1)-1)$ if $a \\neq 0$. By similar reasoning to earlier, suppose that \\operatorname{gcd}(f(1)-1,2021)=d>1$. Then $a=\\frac{2021}{d} \\in S \\backslash\\{0\\}$, and $2021 \\mid a(f(1)-1)$; thus, $f(a)=a$. We thus need that \\operatorname{gcd}(f(1)-1,2021)=1$ as well.\n\nWe count the number of integers that satisfy both these conditions. We use complementary counting here; thus, we start by counting those that fail to satisfy at least one condition. Indeed, we count the number of values of $f(1)$ that are not coprime to 2021. Either they are divisible by 43, or 47, or both. Since only 0 is divisible by both, by the principle of inclusion and exclusion, there are $47+43-1=89$ possible values of $f(1)$ so that $f(1)$ is not coprime to 2021. By the same reasoning, there are also 89 possible values of $f(1)$ such that $f(1)-1$ is not coprime to 2021. Finally, we count the values of $f(1)$ for which neither $f(1)$ nor $f(1)-1$ is coprime to 2021. This happens precisely when $43 \\mid f(1)$ and $47 \\mid f(1)-1$, or $47 \\mid f(1)$ and $43 \\mid f(1)-1$. By the Chinese remainder theorem, each of these possibilities gives one value of $f(1)$. Thus, by the principle of inclusion and exclusion, there are $2 \\cdot 89-2=176$ such values of $f(1)$ that fail to satisfy at least one condition. This gives us $2021-176=1845$ possible values of $f(1)$, and thus possible functions $f$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 37534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $\\{a_n\\}$ of real numbers is defined by $a_1=1$ and for all integers $n \\geq 1$,\n$$\na_{n+1}=\\frac{a_n \\sqrt{n^2+n}}{\\sqrt{n^2+n+2 a_n^2}}\n$$\nCompute the sum of all positive integers $n<1000$ for which $a_n$ is a rational number.",
    "options": [],
    "answer": "131",
    "solution": "Solution:\nFirst, note that for $k \\geq 1$,\n$$\na_{k+1}^2=\\frac{k(k+1) a_k^2}{k(k+1)+2 a_k^2} \\Longleftrightarrow \\frac{1}{a_{k+1}^2}-\\frac{1}{a_k^2}=\\frac{2}{k(k+1)}=\\frac{2}{k}-\\frac{2}{k+1}\n$$\nand summing the second equation from $k=1$ to $k=n-1$ with $n \\geq 2$, we get\n$$\n\\frac{1}{a_n^2}-\\frac{1}{a_1^2}=\\sum_{k=1}^{n-1}\\left(\\frac{1}{a_{k+1}^2}-\\frac{1}{a_k^2}\\right)=\\sum_{k=1}^{n-1}\\left(\\frac{2}{k}-\\frac{2}{k+1}\\right)=2-\\frac{2}{n}\n$$\nSince $a_1=1$, we see that\n$$\n\\frac{1}{a_n^2}=3-\\frac{2}{n} \\Longleftrightarrow a_n=\\sqrt{\\frac{n}{3 n-2}}\n$$\nfor all integers $n \\geq 1$ and we wish to find the sum of all positive integers $n<1000$ such that $\\frac{n}{3 n-2}$ is a square of some rational number. To help us look for such integers $n$, we use the following lemma that provides integer solutions to the generalized Pell equation.\n\nLemma. ${ }^{1}$ Let $d$ be a squarefree positive integer, and let $a$ and $b$ be positive integers such that $a^2-d b^2=1$. Set $u=a+b \\sqrt{d}$. Then for each nonzero integer $n$, every solution of $x^2-d y^2=n$ is a power of $u$ times $x+y \\sqrt{d}$ where $(x, y)$ is an integer solution of $x^2-d y^2=n$ with $|x| \\leq \\sqrt{|n|}(\\sqrt{u}+1) / 2$ and $|y| \\leq \\sqrt{|n|}(\\sqrt{u}+1) /(2 \\sqrt{d})$.\n\nWe now let $g=\\operatorname{gcd}(n, 3 n-2)$. Then $g \\in\\{1,2\\}$ since $g \\mid 3 n-(3 n-2)=2$. We now consider the following cases:\n\n- Suppose $g=1$. Then $n=y^2$ and $3 n-2=x^2$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^2-3 y^2=-2$. Set $u=2+\\sqrt{3}$, with $(2,1)$ being a solution of $x^2-3 y^2=1$ in positive integers. We now look for the positive integer solutions $(x, y)$ of $x^2-3 y^2=-2$ with $x \\leq \\sqrt{2}(\\sqrt{u}+1) / 2 \\approx 2.07$ and $y \\leq \\sqrt{2}(\\sqrt{u}+1) /(2 \\sqrt{3}) \\approx 1.2$. We obtain $(x, y)=(1,1)$ as the only such integer solution, so by the above lemma, we see that all positive integer solutions $\\left(x_k, y_k\\right)$ of $x^2-3 y^2=-2$ are given by $x_k+y_k \\sqrt{3}=(1+\\sqrt{3})(2+\\sqrt{3})^k$ for all integers $k \\geq 0$. We now compute this product for small values of $k$ :\n\n| $k$ | 0 | 1 | 2 | 3 |\n| :---: | :---: | :---: | :---: | :---: |\n| $x_k+y_k \\sqrt{3}$ | $1+\\sqrt{3}$ | $5+3 \\sqrt{3}$ | $19+11 \\sqrt{3}$ | $71+41 \\sqrt{3}$ |\n\nAs $n<1000$, we require that $y_k \\leq 31$, so the positive integer solutions $\\left(x_k, y_k\\right)$ of $x^2-3 y^2=$ -2 with $y_k \\leq 31$ are $\\left(x_k, y_k\\right)=(1,1),(5,3),(19,11)$ (which indeed have relatively prime coordinates). These correspond to the values of $n: n=1,9,121$.\n\n- Suppose $g=2$. Then $n=2 y^2$ and $3 n-2=2 x^2$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^2-3 y^2=-1$. Again, we set $u=2+\\sqrt{3}$. We now look for the positive integer solutions $(x, y)$ of $x^2-3 y^2=-1$ with $x \\leq(\\sqrt{u}+1) / 2 \\approx 1.47$ and $y \\leq(\\sqrt{u}+1) /(2 \\sqrt{3}) \\approx 0.85$. It turns out that there are no such solutions on these bounds, so by the above lemma, we conclude that $x^2-3 y^2=-1$ has no solutions in positive integers.\n\nHence, the only positive integers $n<1000$ for which $a_n$ is a rational number are $n=1,9,121$ and the sum is $1+9+121=131$.\n\n[^0]:    ${ }^{1}$ For proof, see Theorem 3.3 from https://kconrad.math.uconn.edu/blurbs/ugradnumthy/pelleqn2.pdf.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn how many ways can the letters of the word PANACEA be arranged so that the three As are not all together?\n(a) 540\n(b) 576\n(c) 600\n(d) 720",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nThe word PANACEA has 7 letters, with the letter A appearing 3 times, and the other letters P, N, C, E each appearing once.\n\nFirst, find the total number of arrangements of the letters:\n\nNumber of arrangements $= \\dfrac{7!}{3!} = \\dfrac{5040}{6} = 840$\n\nNow, count the number of arrangements where all three As are together.\n\nTreat the three As as a single letter (block), so we have: [AAA], P, N, C, E — a total of 5 objects to arrange.\n\nNumber of arrangements $= 5! = 120$\n\nTherefore, the number of arrangements where the three As are NOT all together is:\n\n$840 - 120 = 720$\n\nSo, the answer is (d) 720.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDenote by $\\mathbb{Q}^{+}$ the set of positive rational numbers. A function $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}$ satisfies\n- $f(p)=1$ for all primes $p$, and\n- $f(a b)=a f(b)+b f(a)$ for all $a, b \\in \\mathbb{Q}^{+}$.\nFor which positive integers $n$ does the equation $n f(c)=c$ have at least one solution $c$ in $\\mathbb{Q}^{+}$?",
    "options": [],
    "answer": "All squarefree positive integers (equivalently, n = 1 or n is a product of distinct primes).",
    "solution": "Solution:\nWe claim that either $n$ is the product of distinct primes, or $n=1$.\n\nDefine $g(x)=\\frac{f}{x}$. The equation we are trying to solve becomes $g(c)=\\frac{1}{n}$.\n\nThe definition of the function becomes $g(p)=\\frac{1}{p}$ for all primes $p$, and\n$$\ng(a b)=g(a)+g(b)\n$$\nSubstituting $a=1$ in the above yields $g(1)=0$. Letting $b=\\frac{1}{a}$ in (1) and using $g(1)=0$ gives $g(a)=-g\\left(\\frac{1}{a}\\right)$ for all $a$ in $\\mathbb{Q}^{+}$.\n\nAn easy induction then proves that\n$$\ng\\left(a^{n}\\right)=n g(a)\n$$\nfor positive integers $n$ and $a$ in $\\mathbb{Q}^{+}$.\n\nThis gives\n$$\ng\\left(p^{e}\\right)=\\frac{e}{p}\n$$\nfor prime $p$ and positive integers $e$.\n\nThese facts, combined, give us the general formula for $g\\left(\\frac{p_{1}^{e_{1}} \\cdots p_{k}^{e_{k}}}{q_{1}^{f_{1}} \\cdots q_{\\ell}^{f_{\\ell}}}\\right)$ in terms of the prime factorizations of $p$ and $q$ :\n$$\ng\\left(\\frac{p_{1}^{e_{1}} \\cdots p_{k}^{e_{k}}}{q_{1}^{f_{1}} \\cdots q_{\\ell}^{f_{\\ell}}}\\right)=\\frac{e_{1}}{p_{1}}+\\cdots+\\frac{e_{k}}{p_{k}}-\\frac{f_{1}}{q_{1}}-\\cdots-\\frac{f_{\\ell}}{q_{\\ell}}=\\frac{m}{p_{1} \\cdots p_{k} q_{1} \\cdots q_{\\ell}}\n$$\nfor some integer $m$. Observe that the denominator is a product of distinct primes. Thus, if $g(c)=\\frac{1}{n}$ for some $c$, then either $n$ is the product of distinct primes, or $n=1$.\n\nIt remains to prove that all such $n$ have such a solution $c$.\n\nWhen $n=1$, taking $c=p^{p}$ for some prime $p$ works by (3).\n\nWe now prove that if $g(c)=\\frac{1}{n}$ for some $c \\in \\mathbb{Q}^{+}$ and positive integer $n$, then there exists $d \\in \\mathbb{Q}^{+}$ such that $g(d)=\\frac{1}{n p}$, for any prime $p$ relatively prime to $n$. This finishes the problem by induction.\n\nLet $x$ and $y$ be integers whose values will be determined later. Observe that, by (2), $g\\left(c^{x}\\right)=x g(c)$. By (3), we get $g\\left(p^{y}\\right)=\\frac{y}{p}$. Finally, using (1) on $c^{x}$ and $p^{y}$ gives\n$$\ng\\left(c^{x} p^{y}\\right)=\\frac{x}{n}+\\frac{y}{p}=\\frac{p x+n y}{n p}\n$$\nIt remains to choose integers $x$ and $y$ such that $p x+n y=1$. But by Bézout's identity, as the greatest common divisor of $p$ and $n$ is 1, there do exist such integers. Taking $d=c^{x} p^{y}$ then gives $g(d)=\\frac{1}{n p}$, finishing the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ordered pairs of positive integers $(x, y)$ satisfy $20x + 21y = 2021$?\n\n(a) 4\n(b) 5\n(c) 6\n(d) infinitely many",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the set of all polynomials $P(x)$ with real coefficients such that the set $\\{P(n) \\mid n \\in \\mathbb{Z}\\}$ contains all integers, except possibly finitely many of them.",
    "options": [],
    "answer": "P(x) = (x + j)/i for integers i ≠ 0 and j",
    "solution": "Solution:\nWe claim that the only such polynomials are of the form $P(x)=\\frac{1}{i}(x+j)$ for some integers $i \\neq 0, j$.\nLet $\\mathcal{R}$ be the set $\\{P(n) \\mid n \\in \\mathbb{Z}\\}$.\n\nWithout loss of generality, we may assume that the leading coefficient of $P(x)$ is positive; otherwise we can consider $-P(x)$. If the polynomial $P(x)$ has even degree, then it must have a minimum value $m$. Then all integers less than $m$ are in the set $\\mathbb{Z} \\backslash \\mathcal{R}$, so it cannot be finite. Thus $P(x)$ must have odd degree.\n\nSince the set $\\mathbb{Z} \\backslash \\mathcal{R}$ is finite, there exists some $M_{1}$ such that $x \\in \\mathcal{R}$ for all integers $x>M_{1}$. As $P(x)$ has odd degree and positive leading coefficient, there exists some $M_{2}$ such that $P\\left(M_{2}\\right)>M_{1}$, $P(x) \\leq P\\left(M_{2}\\right)$ for all $x \\leq M_{2}$ and $P(x)$ is increasing on $\\left[M_{2}, \\infty\\right)$. Let $M=\\max \\left\\{M_{1}, M_{2}\\right\\}$.\n\nChoose an integer $n$ such that $n>M$ and $P(n+1)>P(n)+1$. Letting $s=\\lfloor P(n)+1\\rfloor$, we claim there is no such integer $x$ such that $P(x)=s$. Consider the following cases:\n- $x \\leq M_{2}$. Then $P(x) \\leq P\\left(M_{2}\\right) \\leq P(n)<s$.\n- $M_{2}<x \\leq n$. Then $P(x) \\leq P(n)<s$ as $P(x)$ is increasing.\n- $x>n$. Then $x \\geq n+1$ and $P(x) \\geq P(n+1)>P(n)+1 \\geq s$, as $P(x)$ is increasing.\n\nThus $s$ is in the set $\\mathbb{Z} \\backslash \\mathcal{R}$. However, $s>P(n)>P\\left(M_{2}\\right)>M_{1}$, contradicting the definition of $M_{1}$. Thus, $P(n+1) \\leq P(n)+1$ for all integers $n>M$.\n\nLet $d$ be the degree of $P(x)$. Observe $P(n+1)-P(n)$ is a polynomial of degree $d-1$ with the same positive leading coefficient as $P(x)$. If $d-1 \\geq 1$, then $P(n+1)-P(n)$ will become arbitrarily large as $n$ increases, contradicting $P(n+1) \\leq P(n)+1$ for all integers $n>M$.\n\nTherefore, $d=1$, and $P(x)=a x+b$ for some real numbers $a \\neq 0$ and $b$. As $\\mathbb{Z} \\backslash \\mathcal{R}$ is finite, only finitely many pairs of integers $(t, t+1)$ are not in $\\mathcal{R}$. Thus, there exists distinct integers $n_{1}$ and $n_{2}$ such that $P\\left(n_{1}\\right)=t$ and $P\\left(n_{2}\\right)=t+1$. It follows that\n$$\n1=P\\left(n_{2}\\right)-P\\left(n_{1}\\right)=a\\left(n_{2}-n_{1}\\right) \\Longrightarrow a=\\frac{1}{n_{2}-n_{1}}=\\frac{1}{i}\n$$\nfor some integer $i \\neq 0$. Furthermore,\n$$\nP\\left(n_{1}\\right)=t=a n_{1}+b \\Longrightarrow b=\\frac{i t-n_{1}}{i}=\\frac{j}{i}\n$$\nfor some integer $j$. Hence $P(x)$ must be of the form $\\frac{1}{i}(x+j)$ for integers $i \\neq 0, j$. All such polynomials clearly satisfy the given conditions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of all $k$ for which $x^{5} + k x^{4} - 6 x^{3} - 15 x^{2} - 8 k^{3} x - 12 k + 21$ leaves a remainder of $23$ when divided by $x + k$.\n(a) $-1$\n(b) $-\\frac{3}{4}$\n(c) $\\frac{5}{8}$\n(d) $\\frac{3}{4}$",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA positive integer is called lucky if it is divisible by $7$, and the sum of its digits is also divisible by $7$. Fix a positive integer $n$. Show that there exists some lucky integer $\\ell$ such that $|n-\\ell| \\leq 70$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose we have some lucky integer $n$. We will show that the gap between it and the next lucky integer is no more than $2 \\times 70$.\n\nIn one iteration, we increment $n \\rightarrow n+7$. Clearly the number we have is still divisible by $7$, so it will suffice for us to show that the digit-sum will be divisible by $7$ after some number of iterations.\n\nFirst, suppose that the last two digits of $n$ are less than $30$. We claim that after at most $10$ such iterations, the digit-sum cycles through all possible values mod $7$.\n\nNote that if the last digit is less than $3$, then the digit-sum mod $7$ does not change, while if the last digit is $3$ or more, then the digit-sum mod $7$ decreases by $2$. This follows because $10 \\equiv -3 \\bmod 7$, but we do still add $+1$ for the carrying in the tens place. Since we assumed that the last two digits are less than $30$, there will never be any carrying in the hundreds place onwards within our $10$ iterations (since $d<30$ implies that $d+70<100$).\n\nNow, note that $7$ is coprime to $10$. So, our $10$ iterations of $+7$ will have the last digit cycle through all the digits from $0$ to $9$ exactly once each. Thus, the cases which do not change the digit-sum mod $7$ (which are $0 \\rightarrow 7, 1 \\rightarrow 8$, and $2 \\rightarrow 9$) happen exactly once each, and the other seven iterations do $-2$ on the digit-sum.\n\nBecause $-2$ is coprime to $7$, $7$ iterations of $-2$ will have the digit-sum cycle through all values mod $7$. So, we are guaranteed to be able to achieve a digit-sum of $0 (\\bmod 7)$ within our $10$ iterations of $+7$.\n\nIf the last two digits are $30$ or more, we can make the last two digits less than $30$ with at most $10$ iterations of $+7$ (because if $30 \\leq d < 100$, then $100-d \\leq 70$).\n\nThus, from a lucky integer, we can produce the next lucky integer in $20$ iterations or less. The gap between lucky integers is at most $140$, and so any integer is at most $70$ away from a lucky integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn rolling three fair twelve-sided dice simultaneously, what is the probability that the resulting numbers can be arranged to form a geometric sequence?\n(a) $\\frac{1}{72}$\n(b) $\\frac{5}{288}$\n(c) $\\frac{1}{48}$\n(d) $\\frac{7}{288}$",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA certain country wishes to interconnect $2021$ cities with flight routes, which are always two-way, in the following manner:\n- There is a way to travel between any two cities either via a direct flight or via a sequence of connecting flights.\n- For every pair $(A, B)$ of cities that are connected by a direct flight, there is another city $C$ such that $(A, C)$ and $(B, C)$ are connected by direct flights.\nShow that at least $3030$ flight routes are needed to satisfy the two requirements.",
    "options": [],
    "answer": "3030",
    "solution": "Solution:\nMore generally, consider the graph $G$ with $n$ vertices representing $n$ cities, two of them being connected by an edge if there is a two-way flight between them. We will prove that if $G$ is connected and every edge in $G$ belongs to a triangle, then $G$ must have at least $\\lfloor (3n-2)/2 \\rfloor$ edges. We call such a graph a $T$-graph. Let $a_{n}$ be the minimum number of edges of a $T$-graph with $n$ vertices and $b_{n} = \\lfloor (3n-2)/2 \\rfloor$.\n\n- We first show that $a_{n} \\leq b_{n}$ for $n \\geq 3$ via an explicit construction. We clearly have $a_{3} = b_{3} = 3$. If $n = 2m+1$ is odd, we form $m$ triangles having one common vertex. This is a $T$-graph with $3m$ edges, so $a_{n} \\leq 3m = b_{n}$. If $n = 2m$ is even, we form the above $T$-graph with $2m-1$ vertices, choose an edge, say $XY$, and then add a vertex $Z$ and edges $XZ$ and $YZ$. This results in a $T$-graph with $n$ vertices and $3(m-1)+2 = 3m-1$ edges, so $a_{n} \\leq 3m-1 = b_{n}$.\n\n- We next show that $a_{n} \\geq b_{n}$ for $n \\geq 3$, that is, any $T$-graph with $n$ vertices must have at least $b_{n}$ edges. We proceed by strong induction. The cases $n=3$ and $n=4$ can be easily verified, as shown by their respective $T$-graphs below.\n\n![](attached_image_1.png)\n\nWe claim that in a $T$-graph $G$ with $n$ vertices and less than $b_{n}$ edges, there is a vertex whose degree is $2$. Suppose otherwise; since $G$ is connected, every vertex has degree at least $1$ and since every edge in $G$ belongs to a triangle, every vertex must have degree at least $2$. If each vertex has degree at least $3$, then the sum of the degrees is at least $3n$. But that sum is twice the number of edges, which is less than $2b_{n} < 3n$, a contradiction.\n\nWe now suppose $a_{k} = b_{k}$ for all $k < n$: we will show that $a_{n} \\geq b_{n}$. Consider a $T$-graph $G$ with $n$ vertices and $a_{n}$ edges such that $a_{n} < b_{n}$. Then by assumption and our claim above, there is a vertex $v$ of $G$ of degree $2$ which must be in some triangle, say $uvw$. Suppose the edge $uw$ belongs to some other triangle. Then the graph $H$ obtained by deleting $v$ and edges $uv$ and $vw$ is a $T$-graph with $n-1$ vertices and $a_{n}-2 < b_{n}-2 \\leq b_{n-1}$, a contradiction. Thus, $uw$ does not lie on any triangle other than $uvw$.\n\nWe now construct a smaller graph $G'$ as follows: we delete $v$ and contract $uw$, that is, we collapse the edge $uw$ by combining its endpoints into a single vertex, say $x$. In $G'$, we connect a vertex $y$ to $x$ if in $G$, $y$ is adjacent to one of $u$ or $w$. We observe that $G'$ is a $T$-graph with $n-2$ vertices and has $3$ fewer edges than $G$ (as we remove $uv$, $uw$ and $vw$). We see that $G'$ cannot have at least $4$ fewer edges than $G$; otherwise some vertex $y$ in $G$ would be adjacent to both $u$ and $w$, thus forming a triangle $uyw$ in $G$, a contradiction. Thus, $G'$ has $a_{n}-3 < b_{n}-3 \\leq b_{n-2}$ edges, contradicting the induction hypothesis. Hence, we must have $a_{n} \\geq b_{n}$.\n\nCombining these results leads us to $a_{n} = b_{n}$, and setting $n = 2021$ yields the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many positive integers $n$ are there such that $\\frac{n}{120-2n}$ is a positive integer?\n\n(a) 2\n(b) 3\n(c) 4\n(d) 5",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$, $c$, and $d$ be real numbers such that $a \\geq b \\geq c \\geq d$ and\n$$\n\\begin{aligned}\na+b+c+d & = 13 \\\\\na^{2}+b^{2}+c^{2}+d^{2} & = 43\n\\end{aligned}\n$$\nShow that $ab \\geq 3 + cd$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $(a-d)(b-c) \\geq 0$ and $(a-b)(c-d) \\geq 0$, then\n$$\nab + cd \\geq ac + bd \\geq ad + bc\n$$\nFrom Equations (4) and (5), we have\n$$\n\\begin{gathered}\n(ab + cd) + (ac + bd) + (ad + bc) = \\frac{1}{2}\\left[(a+b+c+d)^{2} - \\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\right] = \\\\\n\\frac{1}{2}\\left(13^{2} - 43\\right) = 63\n\\end{gathered}\n$$\nThus, using Equation (6), we have $ab + cd \\geq \\frac{63}{3} = 21$. Since $c + d = 13 - (a + b)$, then\n$$\n\\begin{aligned}\n(a+b)^{2} + [13 - (a+b)]^{2} & = (a+b)^{2} + (c+d)^{2} \\\\\n& = \\left(a^{2} + b^{2} + c^{2} + d^{2}\\right) + 2(ab + cd) \\\\\n& \\geq 43 + 2(21) = 85\n\\end{aligned}\n$$\nThus, we have\n$$\n\\begin{aligned}\n(a+b)^{2} + 169 - 26(a+b) + (a+b)^{2} & \\geq 85 \\\\\n(a+b)^{2} - 13(a+b) + 42 & \\geq 0 \\\\\n(a+b-6)(a+b-7) & \\geq 0\n\\end{aligned}\n$$\nwhich means either $a+b \\leq 6$ or $a+b \\geq 7$. However, since $2(a+b) \\geq a+b+c+d = 13$, then $a+b \\geq 6.5$. Thus, $a+b \\geq 7$. From this, we have\n$$\n\\begin{aligned}\n(a+b)^{2} + (c-d)^{2} & \\geq 7^{2} + 0^{2} \\\\\na^{2} + b^{2} + c^{2} + d^{2} + 2ab - 2cd & \\geq 49 \\\\\n43 + 2(ab - cd) & \\geq 49 \\\\\nab - cd & \\geq 3 \\\\\nab & \\geq 3 + cd\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn right triangle $A B C$, $\\angle A C B = 90^{\\circ}$ and $\\tan A > \\sqrt{2}$. $M$ is the midpoint of $A B$, $P$ is the foot of the altitude from $C$, and $N$ is the midpoint of $C P$. Line $A B$ meets the circumcircle of $C N B$ again at $Q$. $R$ lies on line $B C$ such that $Q R$ and $C P$ are parallel, $S$ lies on ray $C A$ past $A$ such that $B R = R S$, and $V$ lies on segment $S P$ such that $A V = V P$. Line $S P$ meets the circumcircle of $C P B$ again at $T$. $W$ lies on ray $V A$ past $A$ such that $2 A W = S T$, and $O$ is the circumcenter of $S P M$. Prove that lines $O M$ and $B W$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRefer to the figure shown below.\n\n![](attached_image_1.png)\n\nSince $C P$ is the $C$-altitude of $A B C$, triangles $A C P$, $A B C$ and $C B P$ are similar, so $A P = \\frac{A C^{2}}{A B}$ and $B P = \\frac{B C^{2}}{A B}$. As $C N Q B$ is cyclic, we have $\\angle P N Q = \\angle C B P = \\angle A C P$ so triangles $A C P$ and $Q N P$ are similar. With $N$ being the midpoint of $P C$, we have $\\frac{A P}{P Q} = \\frac{P C}{P N} = 2$, which gives $P Q = \\frac{1}{2} A P$ and $B Q = A P - A Q = A B - \\frac{3}{2} A P$.\n\nWe now claim that $S C^{2} = 2 A B \\cdot P M$. Indeed, since $Q R$ and $C P$ are parallel, we have $\\frac{B Q}{B P} = \\frac{B R}{B C}$ by Thales' Theorem. Thus,\n$$\nR S = B R = \\frac{B C \\cdot B Q}{B P} = \\frac{B C\\left(A B - \\frac{3 A C^{2}}{2 A B}\\right)}{\\frac{B C^{2}}{A B}} = \\frac{2 A B^{2} - 3 A C^{2}}{2 B C}\n$$\nApplying Pythagorean theorem on triangle $S C R$, we see that\n$$\n\\begin{aligned}\nS C^{2} & = B R^{2} - R C^{2} = B C (2 B R - B C) = B C\\left(\\frac{2 A B^{2} - 3 A C^{2}}{B C} - B C\\right) \\\\\n& = 2 B C^{2} + 2 C A^{2} - 3 C A^{2} - B C^{2} = B C^{2} - C A^{2}\n\\end{aligned}\n$$\nNote that $\\tan A > \\sqrt{2}$ implies that $S$ is indeed on ray $C A$ past $A$. On the other hand, with $M$ being the midpoint of $A B$, we compute\n$$\nP M = A M - A P = \\frac{A B}{2} - \\frac{C A^{2}}{A B} = \\frac{A B^{2} - 2 C A^{2}}{2 A B} = \\frac{B C^{2} - C A^{2}}{2 A B} = \\frac{S C^{2}}{2 A B}\n$$\nwhich proves the desired claim.\n\nObserve that the circumcircle of $C P B$ is tangent to $S C$ at $C$, so the Power of the Point gives $S C^{2} = S P \\cdot S T$. It follows from the above claim that\n$$\nS P \\cdot S T = 2 A B \\cdot P M \\Longleftrightarrow \\frac{S P}{P M} = \\frac{2 A B}{S T} = \\frac{2 A B}{2 A W} = \\frac{A B}{A W}\n$$\nSince $A V = P V$, we have $\\angle W A B = 180^{\\circ} - \\angle V A P = 180^{\\circ} - \\angle V P A = \\angle M P S$, so triangles $W A B$ and $M P S$ are similar. Thus, we see that $\\angle P S M = \\angle A B W$ and with $O P = O M$, we arrive at\n$$\n\\angle A B W + \\angle P M O = \\angle P S M + \\angle P M O = \\frac{1}{2}(\\angle P O M + 2 \\angle P M O) = \\frac{1}{2} \\cdot 180^{\\circ} = 90^{\\circ}\n$$\nHence, we get $O M \\perp B W$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree real numbers $a_{1}, a_{2}, a_{3}$ form an arithmetic sequence. After $a_{1}$ is increased by $1$, the three numbers now form a geometric sequence. If $a_{1}$ is a positive integer, what is the smallest positive value of the common difference?\n\n(a) $1$\n(b) $\\sqrt{2}+1$\n(c) $3$\n(d) $\\sqrt{5}+2$",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoint $G$ lies on side $AB$ of square $ABCD$ and square $AEFG$ is drawn outwards $ABCD$, as shown in the figure below. Suppose that the area of triangle $EGC$ is $1/16$ of the area of pentagon $DEFBC$. What is the ratio of the areas of $AEFG$ and $ABCD$?\n![](attached_image_1.png)\n\n(a) $4:25$\n(b) $9:49$\n(c) $16:81$\n(d) $25:121$",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $XZ$ be a diameter of circle $\\omega$. Let $Y$ be a point on $XZ$ such that $XY = 7$ and $YZ = 1$. Let $W$ be a point on $\\omega$ such that $WY$ is perpendicular to $XZ$. What is the square of the length of the line segment $WY$?\n\n(a) 7\n(b) 8\n(c) 10\n(d) 25",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $2, b, c, d, 72$ are listed in increasing order so that $2, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence, and $c, d, 72$ form a harmonic sequence (that is, a sequence whose reciprocals of its terms form an arithmetic sequence). What is the value of $b+c$ ?\n\n(a) 7\n(b) 13\n(c) 19\n(d) 25",
    "options": [],
    "answer": "19",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many positive integers $n < 2022$ are there for which the sum of the odd positive divisors of $n$ is $24$?\n\n(a) $7$\n(b) $8$\n(c) $14$\n(d) $15$",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a whole number ordinary if the product of its digits is less than or equal to the sum of its digits. How many numbers from the set $\\{1,2, \\ldots, 999\\}$ are ordinary?\n(a) 151\n(b) 162\n(c) 230\n(d) 241",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBryce plays a game in which he flips a fair coin repeatedly. In each flip, he obtains two tokens if the coin lands on heads, and loses one token if the coin lands on tails. At the start, Bryce has nine tokens. If after nine flips, he also ends up with nine tokens, what is the probability that Bryce always had at least nine tokens?\n\n(a) $1/7$\n(b) $1/6$\n(c) $5/28$\n(d) $17/84$",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ways are there to arrange the first ten positive integers such that the multiples of $2$ appear in increasing order, and the multiples of $3$ appear in decreasing order?\n\n(a) $720$\n(b) $2160$\n(c) $5040$\n(d) $6480$",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWhat is the largest multiple of $7$ less than $10,\\!000$ which can be expressed as the sum of squares of three consecutive numbers?",
    "options": [],
    "answer": "8750",
    "solution": "Solution:\n\nLet the number be expressed as $a^{2} + (a+1)^{2} + (a+2)^{2}$, where $a$ is an integer. It may be checked that this expression is a multiple of $7$ if and only if the remainder when $a$ is divided by $7$ is $1$ or $4$. \n\nIn the former case, the largest possible value of $a$ that places the value of the expression within bounds is $50$, which gives the value $50^{2} + 51^{2} + 52^{2} = 7805$.\n\nIn the latter case, the largest such value of $a$ is $53$, which gives the value $53^{2} + 54^{2} + 55^{2} = 8750$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that the polynomial $P(x) = x^{3} + 4x^{2} + bx + c$ has a single root $r$ and a double root $s$ for some distinct real numbers $r$ and $s$. Given that $P(-2s) = 324$, what is the sum of all possible values of $|c|$?",
    "options": [],
    "answer": "108",
    "solution": "Solution:\nBy Vieta's formula, we have $r + 2s = -4$ and writing $P(x) = (x - r)(x - s)^{2}$, we have $324 = P(-2s) = (-2s - r)(-3s)^{2} = 36s^{2}$. Thus, $s^{2} = 9$ and $s \\in \\{-3, 3\\}$.\n\nWe next observe that $|c| = |P(0)| = |r| s^{2} = |-4 - 2s| s^{2}$. Hence, the sum of all possible values of $|c|$ is $9(|-4 - 6| + |-4 + 6|) = 108$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m$ and $n$ be relatively prime positive integers. If $m^{3} n^{5}$ has 209 positive divisors, then how many positive divisors does $m^{5} n^{3}$ have?",
    "options": [],
    "answer": "217",
    "solution": "Solution:\n\nLet $d(N)$ denote the number of positive divisors of an integer $N$. Suppose that the prime factorizations of $m$ and $n$ are $\\prod\\left(p_{i}^{a_{i}}\\right)$ and $\\prod\\left(q_{i}^{b_{i}}\\right)$ respectively. Observe that 209 has four positive divisors: $1, 11, 19, 209$.\n\nIf $m=1$, then $n^{5}$ would have 209 divisors. Thus, $d\\left(n^{5}\\right)=209$. However, it is known that $d\\left(n^{5}\\right)=\\prod\\left(5 b_{i}+1\\right)$, but the latter implies that the remainder of $d\\left(n^{5}\\right)$ when divided by 5 is 1, a contradiction.\n\nLikewise, if $n=1$, then it implies that the remainder when $d\\left(m^{3}\\right)=209$ is divided by 3 is 1, also a contradiction.\n\nThus, $m, n>1$, so $d(m), d(n)>1$. As $m$ and $n$ are relatively prime, then $d\\left(m^{3} n^{5}\\right)=d\\left(m^{3}\\right) d\\left(n^{5}\\right)$. As the only way to factor 209 as a product of 2 integers greater than 1 is as $11 \\cdot 19$, then $d\\left(m^{3}\\right)$ and $d\\left(n^{5}\\right)$ are 11 and 19 in some order.\n\nAs the remainder when $d\\left(m^{3}\\right)$ and $d\\left(n^{5}\\right)$ is divided by 3 and 5 respectively is 1, then $d\\left(m^{3}\\right)=19=(3 \\cdot 6+1)$, and $d\\left(n^{5}\\right)=11=(5 \\cdot 2+1)$. Thus, $m$ and $n$ can be expressed as $p^{6}$ and $q^{2}$ respectively.\n\nTherefore, $d\\left(m^{5} n^{3}\\right)=d\\left(p^{30} q^{6}\\right)=(30+1)(6+1)=217$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ be a positive real number. What is the maximum value of\n$$\n\\frac{2022 x^{2} \\log (x+2022)}{(\\log (x+2022))^{3}+2 x^{3}}\n$$\n?",
    "options": [],
    "answer": "674",
    "solution": "Solution:\n\nBy the AM-GM Inequality,\n\n$$\n\\begin{aligned}\n\\frac{2022 x^{2} \\log (x+2022)}{\\log ^{3}(x+2022)+2 x^{3}} & =\\frac{2022 x^{2} \\log (x+2022)}{\\log ^{3}(x+2022)+x^{3}+x^{3}} \\\\\n& \\leq \\frac{2022 x^{2} \\log (x+2022)}{3\\left(x^{6} \\log ^{3}(x+2022)\\right)^{\\frac{1}{3}}} \\\\\n& =\\frac{2022 x^{2} \\log (x+2022)}{3 x^{2} \\log (x+2022)} \\\\\n& =674\n\\end{aligned}\n$$\n\nEquality holds when $x=\\log (x+2022)$, which has a positive solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe PMO Magician has a special party game. There are $n$ chairs, labelled $1$ to $n$. There are $n$ sheets of paper, labelled $1$ to $n$.\n- On each chair, she attaches exactly one sheet whose number does not match the number on the chair.\n- She then asks $n$ party guests to sit on the chairs so that each chair has exactly one occupant.\n- Whenever she claps her hands, each guest looks at the number on the sheet attached to their current chair, and moves to the chair labelled with that number.\n\nShow that if $1 < m \\leq n$, where $m$ is not a prime power, it is always possible for the PMO Magician to choose which sheet to attach to each chair so that everyone returns to their original seats after exactly $m$ claps.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDecompose the permutation into cycles of lengths $c_{1}, c_{2}, c_{3}, \\ldots, c_{k}$. Note that $c_{1} + c_{2} + \\cdots + c_{k} = n$. A guest in a cycle of length $c$ returns to their original seat after $c$ claps. Thus, all guests return to their original seats after $\\operatorname{lcm}\\left\\{c_{i}\\right\\}$ claps.\n\nLet $m = p q$, where $p$ and $q$ are coprime and both greater than $1$; if $m > 1$ is not a prime power, then it has at least two different prime factors, so this is always possible. Suppose we have at least one cycle of lengths $p$ and $q$, and all cycles are of lengths $p$ or $q$. Then, all guests will return to their original seats after $\\operatorname{lcm}\\{p, q\\} = m$ claps.\n\nIf $m = n$, we can just connect all $n$ party guests in one big cycle of size $m = n$. Otherwise, $m < n$, and note that the problem is equivalent to finding a nonnegative integer solution to $p \\bar{x} + q \\bar{y} = n - p - q$. Since $m < n$, a solution always exists by the Chicken McNugget Theorem, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many five-digit numbers containing each of the digits $1,2,3,4,5$ exactly once are divisible by $24$?\n(a) 8\n(b) 10\n(c) 12\n(d) 20",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ be real numbers such that\n\n$$\n3 a b + 2 = 6 b, \\quad 3 b c + 2 = 5 c, \\quad 3 c a + 2 = 4 a\n$$\n\nSuppose the only possible values for the product $a b c$ are $r / s$ and $t / u$, where $r / s$ and $t / u$ are both fractions in lowest terms. Find $r+s+t+u$.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\nThe three given equations can be written as\n\n$$\n3 a + \\frac{2}{b} = 12, \\quad 3 b + \\frac{2}{c} = 10, \\quad 3 c + \\frac{2}{a} = 8\n$$\n\nThe product of all the three equations gives us\n$$\n27 a b c + 6\\left(3 a + \\frac{2}{b}\\right) + 6\\left(3 b + \\frac{2}{c}\\right) + 6\\left(3 c + \\frac{2}{a}\\right) + \\frac{8}{a b c} = 120\n$$\n\nPlugging in the values and simplifying the equation gives us\n$$\n27(a b c)^2 - 30 a b c + 8 = 0\n$$\nThis gives $a b c$ as either $4 / 9$ or $2 / 3$, so $r+s+t+u=4+9+2+3=18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYou roll a fair 12-sided die repeatedly. The probability that all the primes show up at least once before seeing any of the other numbers can be expressed as a fraction $p / q$ in lowest terms. What is $p+q$?",
    "options": [],
    "answer": "793",
    "solution": "Solution:\nThere are 5 primes which are at most 12 - namely $2, 3, 5, 7$ and $11$. Notice that if a number has already been seen, we can effectively ignore all future occurrences of the number. Thus, the desired probability is the fraction of the permutations of $(1, 2, \\ldots, 12)$ such that the primes all occur first. There are $5!$ ways to arrange the primes, and $7!$ ways to arrange the composites to satisfy the condition. Since there are $12!$ possible permutations, the desired probability is $\\frac{5!7!}{12!} = \\frac{1}{792}$, and the required sum is $1+792=793$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $PMO$ be a triangle with $PM = 2$ and $\\angle PMO = 120^\\circ$. Let $B$ be a point on $PO$ such that $PM$ is perpendicular to $MB$, and suppose that $PM = BO$. The product of the lengths of the sides of the triangle can be expressed in the form $a + b \\sqrt[3]{c}$, where $a, b, c$ are positive integers, and $c$ is minimized. Find $a + b + c$.",
    "options": [],
    "answer": "28",
    "solution": "Solution:\nExtend $PM$ to a point $C$ such that $PC \\perp OC$. Since $\\angle PMO = 120^\\circ$, $\\angle CMO = 60^\\circ$ and $\\angle COM = 30^\\circ$. Let $PB = x$ and $MC = a$. Then $CO = a \\sqrt{3}$ and $OM = 2a$. Moreover, $\\triangle PMB$ and $\\triangle PCO$ are similar triangles. Thus, we have\n$$\n\\frac{2}{2+a} = \\frac{x}{x+2}\n$$\nso $x = \\frac{4}{a}$.\n\nFurthermore, by the Cosine Law on side $PO$ of $\\triangle PMO$, we have\n$$\n(x+2)^2 = 4 + 4a^2 + 2a\n$$\nPlugging in $x = \\frac{4}{a}$ and expanding, we have\n$$\n\\frac{16}{a^2} + \\frac{16}{a} + 4 = 4 + 4a^2 + 4a\n$$\nand so $4 + 4a = a^4 + a^3$. Hence $a^3 = 4$ and $a = \\sqrt[3]{4}$. Thus, $x = \\frac{4}{\\sqrt[3]{4}}$.\n\nIt follows that the product of the lengths of the sides of the triangle is\n$$\n(2a)(2)(x+2) = (2\\sqrt[3]{4})(2)(2\\sqrt[3]{2} + 2) = 16 + 8\\sqrt[3]{4}\n$$\nso $a + b + c = 16 + 8 + 4 = 28$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle such that the altitude from $A$, the median from $B$, and the internal angle bisector from $C$ meet at a single point. If $BC=10$ and $CA=15$, find $AB^{2}$.",
    "options": [],
    "answer": "205",
    "solution": "Solution:\n\nLet $D$ be the foot of the $A$-altitude, $E$ the midpoint of $AC$, and $F$ the foot of the $C$-internal angle bisector. Then by Ceva's Theorem, we have\n$$\n\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA}=1,\n$$\nand so\n$$\n\\frac{b}{a} \\cdot \\frac{c \\cos B}{b \\cos C}=1,\n$$\nwhere we are using the shorthand $BC=a$, $CA=b$, $AB=c$. By cosine law, we know that\n$$\n\\cos C=\\frac{a^{2}+b^{2}-c^{2}}{2ab}\n$$\nand\n$$\n\\cos B=\\frac{c^{2}+a^{2}-b^{2}}{2ca}.\n$$\nSubstituting this into the equation, we obtain\n$$\na\\left(a^{2}+b^{2}-c^{2}\\right)=b\\left(c^{2}+a^{2}-b^{2}\\right)\n$$\nSolving for $c^{2}=AB^{2}$ in this equation then gives the final answer, which is $205$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of all positive integers $n$, $1 \\leq n \\leq 5000$, for which\n\n$$\nn^{2}+2475 n+2454+(-1)^{n}\n$$\n\nis divisible by $2477$. (Note that $2477$ is a prime number.)",
    "options": [],
    "answer": "9912",
    "solution": "Solution:\nThe problem is equivalent to finding all values of $n$ that satisfy the following congruence:\n\n$$\nn^{2}-2 n-23+(-1)^{n} \\equiv 0 \\pmod{2477}\n$$\n\nIf $n$ is odd, then $(-1)^{n} = -1$. Then, we have:\n\n$$\n\\begin{gathered}\nn^{2}-2 n-24 \\equiv 0 \\pmod{2477} \\\\\n(n-6)(n+4) \\equiv 0 \\pmod{2477}\n\\end{gathered}\n$$\n\nThis gives us solutions $n \\equiv 6$ and $n \\equiv -4$ (modulo $2477$). Since $2477$ is prime, these are the only roots modulo $2477$. This yields the solutions $6$, $2483$, $2473$, and $4950$. We only accept the odd values, $2483$ and $2473$.\n\nIf $n$ is even, then $(-1)^{n} = 1$. Then, we have\n\n$$\nn^{2}-2 n-22 \\equiv 0 \\pmod{2477}\n$$\n\nThis gives $n \\equiv 1 \\pm \\sqrt{23} \\pmod{2477}$. Again, because $2477$ is prime, these two are the only roots modulo $2477$. Since $2477+23=2500=50^{2}$, note that $\\sqrt{23} \\equiv 50 \\pmod{2477}$, which then gives the solutions $n \\equiv 51$ and $n \\equiv -49$ (modulo $2477$), yielding us the solutions $51$, $2528$, $2428$, and $4905$. We only accept the even values, $2528$ and $2428$.\n\nThus, the sum of all the solutions is $2483+2473+2528+2428=9912$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a real number $x$, let $\\lfloor x\\rfloor$ denote the greatest integer not exceeding $x$. Consider the function\n\n$$\nf(x, y)=\\sqrt{M(M+1)}(|x-m|+|y-m|)\n$$\nwhere $M=\\max (\\lfloor x\\rfloor,\\lfloor y\\rfloor)$ and $m=\\min (\\lfloor x\\rfloor,\\lfloor y\\rfloor)$. The set of all real numbers $(x, y)$ such that $2 \\leq x, y \\leq 2022$ and $f(x, y) \\leq 2$ can be expressed as a finite union of disjoint regions in the plane. The sum of the areas of these regions can be expressed as a fraction $a / b$ in lowest terms. What is the value of $a+b$ ?",
    "options": [],
    "answer": "2021",
    "solution": "Solution:\n\nFix $m \\geq 2$. First note that $x, y \\geq m$, and that $M \\geq 2$, implying $\\sqrt{M(M+1)} \\geq 2$. Thus, $f(x, y) \\leq 2$ necessarily implies $|x-m|+|y-m| \\leq 1$, and so $|x-m|,|y-m| \\leq 1$. This implies $x \\in [m-1, m+1]$, and so $x \\in [m, m+1]$. Similarly, $y \\in [m, m+1]$. Now if $x=m+1$, then $1 \\leq |x-m|+|y-m| \\leq \\frac{2}{\\sqrt{M(M+1)}} < 1$, contradiction. Using the same argument for $y$, it follows that $x, y \\in [m, m+1)$. Thus, $\\lfloor x\\rfloor=\\lfloor y\\rfloor=m$ always, and so $M=m$.\n\nThe inequality is then equivalent to $|x-m|+|y-m| \\leq \\frac{2}{\\sqrt{m(m+1)}}$. Let $r \\leq 1$ be the upper bound in the inequality. Keeping in mind the fact that $x, y \\geq m$, this region is a right-triangle with vertices $(m, m)$, $(m+r, m)$ and $(m, m+r)$, which then has area $\\frac{r^{2}}{2}=\\frac{2}{m(m+1)}$. This region is within the set for $2 \\leq m \\leq 2021$, so the desired sum is\n\n$$\n\\sum_{m=2}^{2021} \\frac{2}{m(m+1)}=2 \\sum_{m=2}^{2021}\\left(\\frac{1}{m}-\\frac{1}{m+1}\\right)=2\\left(\\frac{1}{2}-\\frac{1}{2022}\\right)=\\frac{1010}{1011}\n$$\n\nIt then follows that $a+b=2021$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a lattice point visible if the line segment connecting the point and the origin does not pass through another lattice point. Given a positive integer $k$, denote by $S_{k}$ the set of all visible lattice points $(x, y)$ such that $x^{2}+y^{2}=k^{2}$. Let $D$ denote the set of all positive divisors of $2021 \\cdot 2025$. Compute the sum\n$$\n\\sum_{d \\in D}\\left|S_{d}\\right|\n$$\nHere, a lattice point is a point $(x, y)$ on the plane where both $x$ and $y$ are integers, and $|A|$ denotes the number of elements of the set $A$.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nWe claim that the required sum is $20$.\n\nLet $T_{k}$ denote the set of all lattice points in the circle $x^{2}+y^{2}=k^{2}$. We claim that $\\sum_{d \\mid k}\\left|S_{d}\\right|=\\left|T_{k}\\right|$. Indeed, given a point $(x, y)$ in $T_{k}$, let $g=\\operatorname{gcd}(x, y)$. Then $x / g, y / g$ are necessarily coprime, and hence $(x / g, y / g)$ visible, and $(x / g)^{2}+(y / g)^{2}=(k / g)^{2}$. This implies $(x / g, y / g) \\in \\bigcup_{d \\mid k} S_{d}$. Next, note that the $S_{d}$'s are necessarily disjoint. Now if $(x', y')$ is a visible lattice point in $S_{d}$ where $d \\mid k$, then we can write $k=g d$ so that $(x, y)=(g x', g y')$ is a lattice point in $T_{k}$. This establishes a bijection between $\\bigcup_{d \\mid k} S_{d}$ and $T_{k}$, and since the $S_{d}$'s are disjoint, the claim follows.\n\nFrom the claim, it suffices to find the number of lattice points in the circle $x^{2}+y^{2}=(2021 \\cdot 2025)^{2}$. This is equivalent to\n$$\nx^{2}+y^{2}=3^{8} \\cdot 5^{4} \\cdot 43^{2} \\cdot 47^{2}\n$$\nNow it is well-known that if $x^{2}+y^{2} \\equiv 0\\pmod{p}$ where $p \\equiv 3\\pmod{4}$ is a prime, then $x \\equiv y \\equiv 0\\pmod{p}$. Thus, we must also have $x, y \\equiv 0\\pmod{3^{4} \\cdot 43 \\cdot 47}$. It then follows that the number of lattice points is the same as the number of lattice points in $x^{2}+y^{2}=25^{2}$.\n\nIf $x=0$ or $y=0$, there are $4$ solutions. Otherwise, assume WLOG that they are both positive. Now it is well-known that all solutions to $x^{2}+y^{2}=z^{2}$ are in the form $x=g(m^{2}-n^{2})$, $y=2 g m n$, and $z=g(m^{2}+n^{2})$, where $m>n$ are coprime positive integers, and $g$ is a positive integer. Thus, we want $g(m^{2}+n^{2})=25$. Note that $g \\mid 25$, so $g=1,5,25$.\n\nIf $g=25$, then $m^{2}+n^{2}=1$, so $n=0$, contradiction. If $g=5$, then $m^{2}+n^{2}=5$, which yields $m=2$ and $n=1$ and thus $g(m^{2}-n^{2})=15$ and $2 g m n=20$, so $(x, y)=(15,20),(20,15)$. If $g=1$, then we get $m^{2}+n^{2}=25$, from which we obtain $m=4$ and $n=3$. It then follows that $(x, y)=(24,7),(7,24)$, and so we have $2$ solutions when $x, y$ are both positive. This implies that there are $4 \\cdot 4=16$ solutions when $x, y$ are nonzero, and so there are $4+16=20$ solutions in total.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA lattice point is a point $(x, y)$ where $x$ and $y$ are both integers. Find the number of lattice points that lie on the closed line segment whose endpoints are $(2002,2022)$ and $(2022,2202)$.\n(a) 20\n(b) 21\n(c) 22\n(d) 23",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_{A}$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_{A}$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_{B}, P_{C}, Q_{B}$, and $Q_{C}$ similarly. Show that $P_{A}, P_{B}, P_{C}, Q_{A}, Q_{B}$, and $Q_{C}$ lie on a circle whose center is on line $IG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nRefer to the figure shown below:\n![](attached_image_1.png)\nLet $M_{A}, M_{B}$, and $M_{C}$ be the midpoints of $BC, CA$, and $AB$ respectively.\n\nFirst, it may be shown that $P_{B}$ and $Q_{C}$ lie on $M_{B}M_{C}$.\n\nNote that $\\angle AM_{C}Q_{C} = 2\\angle ABQ_{C} = 180^{\\circ} - \\angle ABC = \\angle BM_{C}M_{B}$. Thus, $Q_{C}$ lies on $M_{B}M_{C}$. Likewise, $AM_{B}P_{B} = 2\\angle ACP_{B} = 180^{\\circ} - \\angle ACB = \\angle CM_{B}M_{C}$. Thus, $P_{B}$ also lies on $M_{B}M_{C}$.\n\nSimilarly, $P_{C}$ and $Q_{A}$ lie on $M_{C}M_{A}$, and $P_{A}$ and $Q_{B}$ lie on $M_{A}M_{B}$.\n\nNow, as $M_{A}$ is the center of the circle passing through $B, C, Q_{A}$, and $P_{A}$, then $M_{A}P_{A} = M_{A}Q_{A}$, so the angle bisector of $\\angle M_{C}M_{A}M_{B}$ coincides with the perpendicular bisector of $P_{A}Q_{A}$.\n\nObserve that $M_{C}Q_{A} = M_{C}M_{A} + M_{A}Q_{A} = \\frac{CA}{2} + \\frac{BC}{2} = M_{B}P_{B} + M_{C}M_{B} = M_{C}P_{B}$. Thus, the perpendicular bisector of $Q_{A}P_{B}$ coincides with the angle bisector of $\\angle M_{B}M_{C}M_{A}$.\n\nUsing similar observations, it may then be concluded that the perpendicular bisectors of $P_{A}Q_{A}, Q_{A}P_{B}, P_{B}Q_{B}, Q_{B}P_{C}, P_{C}Q_{C}$, and $Q_{C}P_{A}$ all concur at the incenter of $\\triangle M_{A}M_{B}M_{C}$. Thus, the latter must also be the center of the circle containing all six points. As the medial triangle is the image of a homothety on $\\triangle ABC$ with center $G$ having a scale factor of $-0.5$, then the incenter of $\\triangle M_{A}M_{B}M_{C}$ must lie on $IG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\omega \\neq -1$ be a complex root of $x^{3}+1=0$. What is the value of $1+2 \\omega+3 \\omega^{2}+4 \\omega^{3}+5 \\omega^{4}$?\n(a) 3\n(b) -4\n(c) 5\n(d) -6",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n$ for which there exists a set of exactly $n$ distinct positive integers, none of which exceed $n^{2}$, whose reciprocals add up to $1$.",
    "options": [],
    "answer": "all positive integers n except 2",
    "solution": "Solution:\nThe answer is all $n \\neq 2$. For $n=1$, the set $\\{1\\}$ works. For $n=2$, no set exists, simply because the sum of reciprocals of two distinct integers cannot be equal to $1$. For $n=3$, take $\\{2,3,6\\}$.\n\nFor $n>3$, the identity\n$$\n\\frac{1}{k} = \\frac{1}{k+r} + \\frac{1}{k(k+1)} + \\frac{1}{(k+1)(k+2)} + \\cdots + \\frac{1}{(k+r-1)(k+r)}\n$$\nallows us to extend a sum of $t$ terms to one of exactly $t+r$ terms. Taking $k=3$ and $r=n-3$ allows us to turn the sum $1 = 1/2 + 1/3 + 1/6$ to the $n$-term sum\n$$\n1 = \\frac{1}{2} + \\frac{1}{6} + \\frac{1}{n} + \\frac{1}{3 \\cdot 4} + \\frac{1}{4 \\cdot 5} + \\cdots + \\frac{1}{(n-1)n}\n$$\nThis construction works provided that $n \\neq k(k+1)$ for any $k$. Otherwise, we have $n \\geq 6$, and instead we apply the above to the sum $1 = 1/2 + 1/3 + 1/10 + 1/15$, taking $k=3$, $r=n-4$ to yield\n$$\n1 = \\frac{1}{2} + \\frac{1}{10} + \\frac{1}{15} + \\frac{1}{n-1} + \\frac{1}{3 \\cdot 4} + \\frac{1}{4 \\cdot 5} + \\cdots + \\frac{1}{(n-2)(n-1)}\n$$\nThe above then works, because if $n = k(k+1)$ for some $k$, then $n-1 \\neq 2, 10, 15$, and $n-1 \\neq m(m+1)$ for any $m$ by parity, since $n-1$ is odd and $m(m+1)$ is always even. This construction is not unique; there are other similar ones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many ending zeroes does the decimal expansion of $2022!$ have?\n\n(a) 404\n(b) 484\n(c) 500\n(d) 503",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo tigers, Alice and Betty, run in the same direction around a circular track of circumference $400$ meters. Alice runs at a speed of $10~\\mathrm{m}/\\mathrm{s}$ and Betty runs at $15~\\mathrm{m}/\\mathrm{s}$. Betty gives Alice a $40$ meter headstart before they both start running. After $15$ minutes, how many times will they have passed each other?\n(a) 9\n(b) 10\n(c) 11\n(d) 12",
    "options": [],
    "answer": "d",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn $\\triangle ABC$, let $D$ be the point on side $BC$ such that $AB + BD = DC + CA$. The line $AD$ intersects the circumcircle of $\\triangle ABC$ again at point $X \\neq A$. Prove that one of the common tangents of the circumcircles of $\\triangle BDX$ and $\\triangle CDX$ is parallel to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRefer to the figure shown below:\n![](attached_image_1.png)\n\nLet $V$ and $W$ be the midpoints of $\\operatorname{arcs} BD$ and $CD$ respectively. We claim that $VW$ is the desired common tangent. To prove this, let $E$ and $F$ be the orthogonal projections of $V$ and $W$ onto $BC$. Note that $E$ and $F$ are the midpoints of $BD$ and $CD$ respectively. Now we claim that $VE = WF$. To see this, note that\n$$\n\\begin{aligned}\nVE & = BV \\sin \\angle VBE \\\\\n& = \\frac{BX \\sin \\angle BXV}{\\sin \\angle BDX} \\sin \\frac{\\angle BXD}{2} \\\\\n& = \\frac{BX \\sin \\frac{\\angle BXA}{2}}{\\sin \\angle BDX} \\sin \\frac{C}{2} \\\\\n& = \\frac{BX}{\\sin \\angle BDX} \\sin^2 \\frac{C}{2}\n\\end{aligned}\n$$\nSimilarly, we can prove that $WF = \\frac{CX}{\\sin \\angle CDX} \\sin^2 \\frac{B}{2}$. Thus, to prove the claim, it suffices to prove that $\\frac{BX}{CX} = \\frac{\\sin^2 \\frac{B}{2}}{\\sin^2 \\frac{C}{2}}$. This is because\n$$\n\\frac{BX}{CX} = \\frac{\\sin \\angle BAD}{\\sin \\angle CAD} = \\frac{c / BD}{b / CD} = \\frac{c(s-c)}{b(s-b)} = \\frac{\\sin^2 \\frac{B}{2}}{\\sin^2 \\frac{C}{2}}\n$$\nThis proves the first claim.\n\nNext, we claim that $VW$ is the desired common tangent. Note that from the first claim, $VWFE$ is a rectangle, since $\\angle VEF$ and $\\angle WFE$ are both right angles. Let $O_1$ and $O_2$ be the circumcenters of triangles $BDX$ and $CDX$ respectively. Then $O_1V \\perp EF$, so since $EF \\parallel VW$ we get $O_1V \\perp VW$. Likewise, $O_2W \\perp VW$ as well, which proves the second claim.\n\nIt then follows that $VW$ is the desired common tangent parallel to $BC$, and the required conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that\n$$\n\\frac{bc}{1 + a^{4}} + \\frac{ca}{1 + b^{4}} + \\frac{ab}{1 + c^{4}} \\geq \\frac{3}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt can be shown that\n$$\n\\frac{1}{1 + a^{4}} \\geq \\frac{2 - a^{2}}{2}\n$$\nIndeed, simplifying yields\n$$\n\\begin{aligned}\n& 2 \\geq (1 + a^{4})(2 - a^{2}) \\\\\n& (a^{4} + 1)(a^{2} - 2) + 2 \\geq 0 \\\\\n& a^{6} - 2a^{4} + a^{2} - 2 + 2 \\geq 0 \\\\\n& a^{6} - 2a^{4} + a^{2} \\geq 0 \\\\\n& a^{2}(a^{4} - 2a^{2} + 1) \\geq 0 \\\\\n& a^{2}(a^{2} - 1)^{2} \\geq 0\n\\end{aligned}\n$$\nwhich is true.\n\nThus\n$$\n\\begin{aligned}\n& \\frac{1}{1 + a^{4}} \\geq \\frac{2 - a^{2}}{2} \\\\\n& \\frac{bc}{1 + a^{4}} \\geq \\frac{2bc - a^{2}bc}{2}\n\\end{aligned}\n$$\nSimilarly\n$$\n\\begin{aligned}\n\\frac{ca}{1 + b^{4}} & \\geq \\frac{2ca - ab^{2}c}{2}, \\text{ and } \\\\\n\\frac{ab}{1 + c^{4}} & \\geq \\frac{2ab - abc^{2}}{2}\n\\end{aligned}\n$$\nAdding these up yields\n$$\n\\frac{bc}{1 + a^{4}} + \\frac{ca}{1 + b^{4}} + \\frac{ab}{1 + c^{4}} \\geq (bc + ca + ab) - \\frac{a^{2}bc + ab^{2}c + abc^{2}}{2}\n$$\nObserve that\n$$\n\\begin{aligned}\n(ab + bc + ca)^{2} & \\geq 3\\big((ab)(bc) + (bc)(ca) + (ca)(ab)\\big) \\\\\n9 & \\geq 3\\left(a^{2}bc + ab^{2}c + abc^{2}\\right) \\\\\n3 & \\geq a^{2}bc + ab^{2}c + abc^{2}\n\\end{aligned}\n$$\nThus,\n$$\n(bc + ca + ab) - \\frac{a^{2}bc + ab^{2}c + abc^{2}}{2} \\geq 3 - \\frac{3}{2} = \\frac{3}{2}\n$$\nTherefore,\n$$\n\\frac{bc}{1 + a^{4}} + \\frac{ca}{1 + b^{4}} + \\frac{ab}{1 + c^{4}} \\geq \\frac{3}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $a, b, c$ are the roots of the polynomial $x^{3} + 2x^{2} + 2$. Let $f$ be the unique monic polynomial whose roots are $a^{2}, b^{2}, c^{2}$. Find $f(1)$.\n\n(a) -17\n(b) -16\n(c) -15\n(d) -14",
    "options": [],
    "answer": "c",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe set $S = \\{1, 2, \\ldots, 2022\\}$ is to be partitioned into $n$ disjoint subsets $S_1, S_2, \\ldots, S_n$ such that for each $i \\in \\{1, 2, \\ldots, n\\}$, exactly one of the following statements is true:\n\n(a) For all $x, y \\in S_i$ with $x \\neq y$, $\\operatorname{gcd}(x, y) > 1$.\n\n(b) For all $x, y \\in S_i$ with $x \\neq y$, $\\operatorname{gcd}(x, y) = 1$.\n\nFind the smallest value of $n$ for which this is possible.",
    "options": [],
    "answer": "15",
    "solution": "Solution:\n\nThe answer is $15$.\n\nNote that there are $14$ primes at most $\\sqrt{2022}$, starting with $2$ and ending with $43$. Thus, the following partition works for $15$ sets. Let $S_1 = \\{2, 4, \\ldots, 2022\\}$, the multiples of $2$ in $S$. Let $S_2 = \\{3, 9, 15, \\ldots, 2019\\}$, the remaining multiples of $3$ in $S$ not in $S_1$. Let $S_3 = \\{5, 25, 35, \\ldots, 2015\\}$, the remaining multiples of $5$, and so on and so forth, until we get to $S_{14} = \\{43, 1849, 2021\\}$. $S_{15}$ consists of the remaining elements, i.e., $1$ and those numbers with no prime factors at most $43$, i.e., the primes greater than $43$ but less than $2022$: $S_{15} = \\{1, 47, 53, 59, \\ldots, 2017\\}$.\n\nEach of $S_1, S_2, \\ldots, S_{14}$ satisfies (a), while $S_{15}$ satisfies (b).\n\nWe show now that no partition in $14$ subsets is possible. Let a Type 1 subset of $S$ be a subset $S_i$ for which (a) is true and there exists an integer $d > 1$ for which $d$ divides every element of $S_i$. Let a Type 2 subset of $S$ be a subset $S_i$ for which (b) is true. Finally, let a Type 3 subset of $S$ be a subset $S_i$ for which (a) is true that is not a Type 1 subset. An example of a Type 3 subset would be a set of the form $\\{pq, qr, pr\\}$ where $p, q, r$ are distinct primes.\n\nClaim: Let $p_1 = 2, p_2 = 3, p_3 = 5, \\ldots$ be the sequence of prime numbers, where $p_k$ is the $k$th prime. Every optimal partition of the set $S(k) := \\{1, 2, \\ldots, p_k^2\\}$, i.e., a partition with the least possible number of subsets, has at least $k-1$ Type 1 subsets. In particular, every optimal partition of this set has $k+1$ subsets in total. To see how this follows, we look at two cases:\n\n- If every prime $p \\leq p_k$ has a corresponding Type 1 subset containing its multiples, then a similar partitioning to the above works: Take $S_1$ to $S_k$ as Type 1 subsets for each prime, and take $S_{k+1}$ to be everything left over. $S_{k+1}$ will never be empty, as it has $1$ in it. While in fact it is known that, for example, by Bertrand's postulate there is always some prime between $p_k$ and $p_k^2$ so $S_{k+1}$ has at least two elements, there is no need to go this far—if there were no other primes you could just move $2$ from $S_1$ into $S_{k+1}$, and if $k > 1$ then $S_1$ will still have at least three elements remaining. And if $k = 1$, there is no need to worry about this, because $2 < 3 < 2^2$.\n\n- On the other hand, if $p \\leq p_k$ has no corresponding Type 1 subset, then $p$ and $p^2$ will not be contained in a Type 1 set. Neither can $p$ nor $p^2$ be contained in a Type 3 set. If $\\operatorname{gcd}(p, x) > 1$ for all $x$ in the same set as $p$, then $\\operatorname{gcd}(p, x) = p$, which implies that $p$ is in a Type 1 set with $d = p$. Similarly, if $\\operatorname{gcd}(p^2, x) > 1$ for all $x$ in the same set as $p^2$, then $p \\mid \\operatorname{gcd}(p^2, x)$ for all $x$, and so $p^2$ is in a Type 1 set with $d = p$ as well. Hence $p$ and $p^2$ must in fact be in Type 2 sets, and they cannot be in the same Type 2 set (as they share a common factor of $p > 1$); this means that the optimal partition has at least $k+1$ subsets in total. A possible equality scenario for example is the sets $S_1 = \\{1, 2, 3, 5, \\ldots, p_k\\}$, $S_2 = \\{4, 9, 25, \\ldots, p_k^2\\}$, and $S_3$ to $S_{k+1}$ Type 1 sets taking all remaining multiples of $2, 3, 5, \\ldots, p_{k-1}$. This works, as $p_k$ and $p_k^2$ are the only multiples of $p_k$ in $S(k)$ with no prime factor other than $p_k$ and thus cannot be classified into some other Type 1 set.\n\nTo prove our claim: We proceed by induction on $k$. Trivially, this is true for $k = 1$. Suppose now that any optimal partition of the set $S(k)$ has at least $k-1$ Type 1 subsets, and thus at least $k+1$ subsets in total. Consider now a partition of the set $S(k+1)$, and suppose that this partition would have at most $k+1$ subsets. From the above, there exist at least two primes $p, q$ with $p < q \\leq p_{k+1}$ for which there are no Type 1 subsets. If $q < p_{k+1}$ we have a contradiction. Any such partition can be restricted to an optimal partition of $S(k)$ with $p < q \\leq p_k$ having no corresponding Type 1 subsets. This contradicts our inductive hypothesis. On the other hand, suppose that $q = p_{k+1}$. Again restricting to $S(k)$ gives us an optimal partition of $S(k)$ with at most $k-1$ Type 1 sets; the inductive hypothesis tells us that this partition has in fact exactly $k-1$ Type 1 sets and two Type 2 sets from a previous argument establishing the consequence of the claim. However, consider now the element $pq$. This cannot belong in any Type 1 set, neither can it belong in the same Type 2 set as $p$ or $p^2$. Thus in addition to the given $k-1$ Type 1 sets and $2$ Type 2 sets, we need an extra set to contain $pq$. Thus our partition of $S(k+1)$ in fact has at least $k-1 + 2 + 1 = k+2$ subsets, and not $k+1$ subsets as we wanted. The claim is thus proved.\n\nReturning to our original problem, since $p_{14} = 43 < \\sqrt{2022}$, any partition of $S$ must restrict to a partition of $S(14)$, which we showed must have at least $15$ sets. Thus, we can do no better than $15$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $I$ be the center of the incircle of triangle $ABC$. Suppose that this incircle has radius $3$, and that $AI = 5$. If the area of the triangle is $2022$, what is the length of $BC$?\n\n(a) 670\n(b) 672\n(c) 1340\n(d) 1344",
    "options": [],
    "answer": "a",
    "solution": "Solution:\n\nLet $r$ be the inradius, $r = 3$. Let $S$ be the area, $S = 2022$. Let $a = BC$.\n\nRecall that $S = r \\cdot s$, where $s$ is the semiperimeter. So:\n$$\ns = \\frac{S}{r} = \\frac{2022}{3} = 674\n$$\n\nLet $AI$ be the distance from $A$ to the incenter $I$. There is a formula:\n$$\nAI^2 = \\frac{bc}{(b + c)^2} \\left[ (b + c)^2 - a^2 \\right] + r^2\n$$\nBut this is complicated. Alternatively, recall that:\n$$\nAI^2 = \\frac{r^2 + (s - a)^2}{1}\n$$\nSo:\n$$\nAI^2 = r^2 + (s - a)^2\n$$\nGiven $AI = 5$, $r = 3$, $s = 674$:\n$$\n5^2 = 3^2 + (674 - a)^2\n$$\n$25 = 9 + (674 - a)^2$\n$16 = (674 - a)^2$\n$674 - a = \\pm 4$\nSo $a = 674 \\pm 4 = 678$ or $670$\n\nBut $a$ must be less than $s$ (since $a$ is a side, $s$ is the semiperimeter), and $a$ must be positive. Both $678$ and $670$ are possible, but let's check which is correct.\n\nIf $a = 678$, then $s - a = -4$, which is not possible (since $s - a$ is the sum of the other two sides divided by $2$ and must be positive). So $a = 670$.\n\nThus, the answer is $\\boxed{670}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a set of real numbers $A$, let $A-A$ be the set of distinct pairwise differences of $A$; that is,\n$$\nA-A := \\{ a-b : a, b \\in A \\}\n$$\nIf $|A-A| = 25$, find the sum of all possible values of $|A|$.",
    "options": [],
    "answer": "76",
    "solution": "Solution:\n\nFirst, we show that $6 \\leq |A| \\leq 13$. To do this, we show that if $|A| = n$, then $2n-1 \\leq |A-A| \\leq n(n-1)+1$. The upper bound is easily attained; the number of distinct differences is bounded above by the number of ordered pairs $(a, b)$ of distinct elements of $A$ plus 1, to account for the remaining pairs $(a, a)$ which give a difference of $0$; this is exactly $n(n-1)+1$.\n\nFor the lower bound, suppose that $A = \\{ a_1, a_2, \\ldots, a_n \\}$ with $a_1 < a_2 < \\cdots < a_n$. Then the $n-1$ differences $a_n - a_1, a_n - a_2, \\ldots, a_n - a_{n-1}$ are all positive and distinct and appear in $A-A$; their additive inverses are all negative and distinct and also appear in $A-A$. Also, $0 \\in A-A$ and $0$ is not among those differences previously listed. This gives us at least $2(n-1)+1 = 2n-1$ differences in $A-A$. These bounds show us that if $|A-A| = 25$, we must have $6 \\leq |A| \\leq 13$ as desired.\n\nWe then show that for any $n$ with $6 \\leq n \\leq 13$, there exists a set $A$ such that $|A| = n$ and $|A-A| = 25$. Consider now the sets $B = \\{ 1, 2, 4, 8, 12, 13 \\}$ and $C = \\{ 1, 2, \\ldots, 13 \\}$. Then we have $|B| = 6$, $|C| = 13$, $B \\subseteq C$, and $B-B = C-C = \\{ -12, -11, \\ldots, 0, \\ldots, 11, 12 \\}$. Now, for any set $A$ with $B \\subseteq A \\subseteq C$, we have $B-B \\subseteq A-A \\subseteq C-C$, and so we must have $A-A = B-B = C-C$. For any $n$ with $6 \\leq n \\leq 13$, it is always possible to choose $A$ such that $|A| = n$ and $B \\subseteq A \\subseteq C$, and this will give us the desired $A$.\n\nThus, the sum of all possible values of $|A|$ is $6 + 7 + \\cdots + 13 = 76$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn parallelogram $WXYZ$, the length of diagonal $WY$ is $15$, and the perpendicular distances from $W$ to lines $YZ$ and $XY$ are $9$ and $12$, respectively. Find the least possible area of the parallelogram.",
    "options": [],
    "answer": "108",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered pairs $(a, b)$ of positive integers such that $a^{2}+b^{2}+25=15 a b$ and $a^{2}+a b+b^{2}$ is prime.",
    "options": [],
    "answer": "[(1,2), (2,1)]",
    "solution": "Solution:\nThe only solutions are $(1,2)$ and $(2,1)$.\n\nFirst, note that $a \\neq b$; otherwise we have $13 a^{2}=25$ which is impossible. We can assume without loss of generality that $1 \\leq a < b$; we then take $(b, a)$ as well.\n\nIf $a=1$, we get $b^{2}+26=15 b$ which yields $b=2$ or $b=13$. For $b=2$ we get $a^{2}+a b+b^{2}=7$ which is indeed prime; for $b=13$ however we have $a^{2}+a b+b^{2}=183=3 \\cdot 61$ which is composite.\n\nIf $a=2$, we get $b^{2}+29=30 b$, which yields $b=29$; we have $a^{2}+a b+b^{2}=903=3 \\cdot 7 \\cdot 43$ which is composite as well.\n\nNow, suppose $a>2$. Note that $17\\left(a^{2}+a b+b^{2}\\right)=16 a^{2}+32 a b+16 b^{2}-25=(4 a+4 b-5)(4 a+4 b+5)$. With the condition that $a \\geq 3, b \\geq 4$, we have $4 a+4 b+5>4 a+4 b-5 \\geq 23>17$, that is, $17\\left(a^{2}+a b+b^{2}\\right)$ is the product of two positive integers greater than 17. Thus, $a^{2}+a b+b^{2}$ is the product of two positive integers greater than 1, and so is composite.\n\nThis tells us that indeed $(1,2)$ and $(2,1)$ are the only solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $PQRS$ is a convex quadrilateral with $\\angle SPQ = \\angle PQR = 120^\\circ$, $SP - QR = 36$, $RS = 84$, and $QR$ is a positive even integer. Let $T$ be the intersection of lines $SP$ and $QR$. What is the largest possible perimeter of $\\triangle PQT$?",
    "options": [],
    "answer": "54 + 12√366",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn square $ABCD$, $P$ lies on the ray $AD$ past $D$ and lines $PC$ and $AB$ meet at $Q$. Point $X$ is the foot of the perpendicular from $B$ to $DQ$, and the circumcircle of triangle $APX$ meets line $AB$ again at $Y$. Suppose that $DP = \\frac{16}{3}$ and $BQ = 27$. The length of $BY$ can be written in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",
    "options": [],
    "answer": "65",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeven people are seated together around a circular table. Each one will toss a fair coin. If the coin shows a head, then the person will stand. Otherwise, the person will remain seated. The probability that after all of the tosses, no two adjacent people are both standing, can be written in the form $p / q$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?",
    "options": [],
    "answer": "157",
    "solution": "Solution:\n\nLet $n = 7$ (number of people). Each person independently stands with probability $1/2$ (heads), sits with probability $1/2$ (tails).\n\nWe want the probability that no two adjacent people are both standing.\n\nLet $a_n$ be the number of ways to seat $n$ people around a table so that no two adjacent people are both standing.\n\nLet $b_n$ be the number of ways to seat $n$ people in a line so that no two adjacent people are both standing.\n\nFirst, count the total number of possible outcomes: $2^7 = 128$.\n\nNow, count the number of favorable outcomes.\n\nLet $S$ denote a standing person, $C$ a sitting person.\n\nFor the linear case (not circular), the recurrence is:\n\nLet $b_n$ be the number of sequences of length $n$ with no two adjacent $S$.\n\nIf the first person is $C$, the rest is $b_{n-1}$.\nIf the first person is $S$, the next must be $C$, and the rest is $b_{n-2}$.\nSo:\n$$\nb_n = b_{n-1} + b_{n-2}\n$$\nwith $b_1 = 2$ (either $S$ or $C$), $b_2 = 3$ ($CC$, $CS$, $SC$).\n\nSo $b_1 = 2$, $b_2 = 3$.\n\nCompute up to $b_7$:\n\n$b_3 = b_2 + b_1 = 3 + 2 = 5$\n\n$b_4 = b_3 + b_2 = 5 + 3 = 8$\n\n$b_5 = b_4 + b_3 = 8 + 5 = 13$\n\n$b_6 = b_5 + b_4 = 13 + 8 = 21$\n\n$b_7 = b_6 + b_5 = 21 + 13 = 34$\n\nSo for a line, there are $34$ ways.\n\nBut for a circle, we must subtract the cases where the first and last are both $S$ (since they are adjacent in the circle).\n\nLet us count the number of arrangements where no two adjacent are $S$, and also the first and last are not both $S$.\n\nLet $c_n$ be the number of such circular arrangements.\n\nThe standard formula is:\n$$\nc_n = b_{n} - b_{n-1}\n$$\nwhere $b_n$ is the linear case, and $b_{n-1}$ is the number of linear arrangements where both ends are $S$ (which is not allowed in the circle).\n\nBut let's be precise.\n\nLet us count the number of arrangements where no two adjacent are $S$ in a circle of $n$ people.\n\nLet us fix one person as the reference point (since the circle is symmetric), and consider the arrangements.\n\nAlternatively, the number of such arrangements is $F_{n-1} + F_{n+1}$, where $F_k$ is the $k$th Fibonacci number, but let's proceed carefully.\n\nLet us use the inclusion-exclusion principle.\n\nLet $A$ be the set of arrangements with no two adjacent $S$ in a line of $n$ people: $b_n$.\n\nLet $B$ be the set of arrangements with no two adjacent $S$ in a circle of $n$ people: $a_n$.\n\nThe number of arrangements in a line with no two adjacent $S$ is $b_n$.\n\nThe number of arrangements in a line with no two adjacent $S$, and with both ends $S$, is $b_{n-3}$ (since after $S$ at both ends, the next to each must be $C$, and the middle $n-4$ can be filled with the recurrence).\n\nBut let's use the standard formula:\n\nThe number of arrangements in a circle is $a_n = b_{n-1} + b_{n-3}$.\n\nBut let's use the following method:\n\nLet us count the number of arrangements where no two adjacent $S$, and the first and last are not both $S$.\n\nLet $L_n$ be the number of arrangements in a line of $n$ people with no two adjacent $S$.\n\nLet $A_n$ be the number of arrangements in a circle of $n$ people with no two adjacent $S$.\n\nThen:\n$$\nA_n = L_n - X_n\n$$\nwhere $X_n$ is the number of arrangements in a line of $n$ people with no two adjacent $S$, and with both ends $S$.\n\nLet us compute $X_n$.\n\nIf both ends are $S$, then the second and $(n-1)$th must be $C$, and the middle $n-4$ can be filled with no two adjacent $S$.\n\nSo $X_n = L_{n-4}$.\n\nTherefore,\n$$\nA_n = L_n - L_{n-4}\n$$\n\nFor $n=7$:\n\n$L_7 = b_7 = 34$\n\n$L_3 = b_3 = 5$\n\nSo $A_7 = 34 - 5 = 29$\n\nTherefore, the number of favorable arrangements is $29$.\n\nThe total number of possible arrangements is $2^7 = 128$.\n\nSo the probability is $29/128$.\n\nThus, $p = 29$, $q = 128$, so $p+q = 157$.\n\n**Answer:** $157$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c$ be real numbers with $1 < a < b < c$ that satisfy the equations\n$$\n\\begin{gathered}\n\\log_{a} b + \\log_{b} c + \\log_{c} a = 6.5 \\\\\n\\log_{b} a + \\log_{c} b + \\log_{a} c = 5\n\\end{gathered}\n$$\nThen $\\max \\left\\{ \\log_{a} b, \\log_{b} c, \\log_{c} a \\right\\}$ can be written in the form $\\sqrt{x} + \\sqrt{y}$, where $x$ and $y$ are positive integers. What is $x + y$?",
    "options": [],
    "answer": "16",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$ with orthocenter $H$, $AB = 13$, $BC = 21$ and $CA = 20$. The perpendicular bisector of $CH$ meets $BC$ at $P$ and lines $PH$ and $AB$ meet at $Q$. The line through $Q$ perpendicular to $PQ$ meets $AH$ at $X$. The length of $AX$ can be written in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",
    "options": [],
    "answer": "173",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBryce has $7$ blue socks and $7$ red socks mixed in a drawer. He plays a game with Sean. Blindfolded, Bryce takes two socks from the drawer. Sean looks at the socks, and if they have the same color, Sean gives Bryce $1$ point. Bryce keeps drawing socks until the drawer is empty, at which time the game ends. The probability that Bryce's score is at most $2$ can be written in the form $p / q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",
    "options": [],
    "answer": "613",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree circles $\\Gamma_{1}$, $\\Gamma_{2}$, $\\Gamma_{3}$ are pairwise externally tangent, with $\\Gamma_{1}$, the smallest, having radius $1$, and $\\Gamma_{3}$, the largest, having radius $25$. Let $A$ be the point of tangency of $\\Gamma_{1}$ and $\\Gamma_{2}$, $B$ be the point of tangency of $\\Gamma_{2}$ and $\\Gamma_{3}$, and $C$ be the point of tangency of $\\Gamma_{1}$ and $\\Gamma_{3}$. Suppose now that triangle $ABC$ has circumradius $1$ as well. The radius of $\\Gamma_{2}$ can then be written in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. Find the value of the product $pq$.",
    "options": [],
    "answer": "156",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor each positive integer $n$, define the function $f_{n}(x) = |n - x|$. How many real solutions are there to\n$$\n\\left(f_{1} \\circ f_{2} \\circ \\cdots \\circ f_{24} \\circ f_{25}\\right)(x) = 0?\n$$",
    "options": [],
    "answer": "2^{24}",
    "solution": "Solution:\n\nLet us analyze the composition $f_{1} \\circ f_{2} \\circ \\cdots \\circ f_{25}(x) = 0$.\n\nRecall that $f_{n}(x) = |n - x|$. For any $x$, $f_{n}(x) = 0$ if and only if $x = n$.\n\nLet us work backwards. Let $F(x) = (f_{1} \\circ f_{2} \\circ \\cdots \\circ f_{25})(x)$.\nWe want to find all real $x$ such that $F(x) = 0$.\n\nLet us define $S_{k}(x) = (f_{k} \\circ f_{k+1} \\circ \\cdots \\circ f_{25})(x)$.\nWe have $S_{25}(x) = f_{25}(x) = |25 - x|$.\n\n$S_{25}(x) = 0$ if and only if $x = 25$.\n\nNow, $S_{24}(x) = f_{24}(S_{25}(x)) = f_{24}(|25 - x|) = |24 - |25 - x||$.\nSet $S_{24}(x) = 0$:\n$$\n|24 - |25 - x|| = 0 \\implies |25 - x| = 24 \\implies 25 - x = 24 \\text{ or } 25 - x = -24\n$$\nSo $x = 1$ or $x = 49$.\n\nNow, let's generalize. At each step, the set of solutions doubles.\n\nLet us formalize this:\nLet $A_{k}$ be the set of $x$ such that $S_{k}(x) = 0$.\n\nWe have $A_{25} = \\{25\\}$.\n\n$A_{24} = \\{x : |25 - x| = 24\\} = \\{1, 49\\}$.\n\n$A_{23} = \\{x : |24 - |25 - x|| = 0\\} = \\{x : |25 - x| = 24\\} = \\{1, 49\\}$, but this is not correct. Let's check carefully.\n\nWait, $S_{24}(x) = |24 - |25 - x||$.\nSet $S_{24}(x) = 0$:\n$|24 - |25 - x|| = 0 \\implies |25 - x| = 24 \\implies x = 1$ or $x = 49$.\n\nNow, $S_{23}(x) = f_{23}(S_{24}(x)) = |23 - S_{24}(x)|$.\nSo $S_{23}(x) = 0$ when $S_{24}(x) = 23$.\nSo $|24 - |25 - x|| = 23$.\n\nSo $24 - |25 - x| = 23$ or $24 - |25 - x| = -23$.\nSo $|25 - x| = 1$ or $|25 - x| = 47$.\nSo $x = 24$ or $x = 26$, or $x = 25 - 47 = -22$, $x = 25 + 47 = 72$.\nSo $A_{23} = \\{24, 26, -22, 72\\}$.\n\nSo at each step, the number of solutions doubles.\n\nLet us check the pattern:\n\n$A_{25}$: $\\{25\\}$ (1 value)\n\n$A_{24}$: $\\{1, 49\\}$ (2 values)\n\n$A_{23}$: $\\{24, 26, -22, 72\\}$ (4 values)\n\nSo, at each step, the number of solutions doubles.\n\nTherefore, after $25$ steps, the number of solutions is $2^{24}$.\n\n**Answer:**\n\n$2^{24}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $p$ is a prime number which divides infinitely many numbers of the form $10^{n!} + 2023$ where $n$ is a positive integer. What is the sum of all possible values of $p$?",
    "options": [],
    "answer": "34",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p = p_{1} p_{2} \\ldots p_{6}$ be a permutation of the integers from $1$ to $6$. For any such permutation $p$, we count how many integers there are which have nothing bigger on its left. We let $f(p)$ be the number of these integers for the permutation $p$. For example, $f(612345) = 1$ because only $6$ has nothing to its left which is bigger. On the other hand, $f(135462) = 4$ because only $1, 3, 5$, and $6$ satisfy the condition.\nLet $S$ be the sum of $f(p)$ over all the $6!$ different permutations. Find the sum of the digits of $S$.",
    "options": [],
    "answer": "18",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute scalene triangle with orthocenter $H$. Let $M$ be the midpoint of $BC$, and suppose that the line through $H$ perpendicular to $AM$ intersects $AB$ and $AC$ at points $E$ and $F$ respectively. Denote by $O$ the circumcenter of triangle $AEF$, and $D$ the foot of the perpendicular from $H$ to $AM$. Prove that the line $AO$ intersects the perpendicular from $D$ to $BC$ at a point on the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWLOG assume $AB < AC$. Let $AM$ intersect the circumcircle of $ABC$ again at $Y \\neq A$. We first need to prove a claim.\n\n![](attached_image_1.png)\n\nClaim: Quadrilateral $B H D C$ is cyclic.\n\nProof of Claim: Consider the reflection with respect to $M$. This maps $B$ and $C$ to each other. It is fairly well-known that this also maps the orthocenter $H$ to the antipode of $A$ with respect to $(ABC)$, which we will denote by $H'$. Thus, $(BHC)$ and $(CH'B)$ are mapped to each other. Since $\\angle HDM$ and $\\angle H'YM$ are both right angles with $M$ the midpoint of $HH'$, then $\\triangle HDM$ and $\\triangle H'YM$ are congruent right triangles. It follows that $D$ is the reflection of $Y$ with respect to $M$. Since $Y \\in (CH'B)$, then $D \\in (BHC)$, and the claim follows.\n\nNow let $AO$ intersect $(ABC)$ and $(AEF)$ again at $X$ and $Z$, respectively. Then\n$$\n\\angle BAX = \\angle EAZ = 90^\\circ - \\angle AZE = 90^\\circ - \\angle AFE = \\angle DAF = \\angle YAC\n$$\nand so $XY \\parallel BC$.\n\nConsider reflection with respect to $BC$. This maps $(BHC)$ to $(ABC)$. Let $X'$ be the image of $D$, so $X' \\in (ABC)$ and $DX' \\perp BC$. Let $N$ be the midpoint of $DX'$, which is on $BC$. Then note that $X'Y \\parallel NM$. Since $X, X' \\in (ABC)$, and both $XY$ and $X'Y$ are parallel to $BC$, then $X = X'$. The desired conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe product of all real numbers $x$ satisfying $x^{2+\\log_{3}(9x)} = \\frac{2187}{x^{2}}$ can be written in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",
    "options": [],
    "answer": "730",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the sum of all positive integers $n$ for which $2[\\tau(n)]^{2}=3 n$, where $\\tau(n)$ is the number of positive divisors of $n$.",
    "options": [],
    "answer": "96",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn $\\triangle ABC$, $AB > AC$. Point $P$ is on line $BC$ such that $AP$ is tangent to its circumcircle. Let $M$ be the midpoint of $AB$, and suppose the circumcircle of $\\triangle PMA$ meets line $AC$ again at $N$. Point $Q$ is the reflection of $P$ with respect to the midpoint of segment $BC$. The line through $B$ parallel to $QN$ meets $PN$ at $D$, and the line through $P$ parallel to $DM$ meets the circumcircle of $\\triangle PMB$ again at $E$. Show that the lines $PM$, $BE$, and $AC$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLet lines $BE$ and $AC$ meet at $R$. It suffices to show that the points $P$, $M$ and $R$ are collinear. Since $AP$ is tangent to the circumcircle of $ABC$ and $PNA M$ is cyclic, we have $\\angle PNM = \\angle PAB = \\angle ACB$ and $\\angle BAC = \\angle MPN$. Thus, triangles $MPN$ and $BAC$ are similar with $\\frac{MP}{MN} = \\frac{AB}{BC}$. Also, since $DM$ and $PE$ are parallel and $DPEM$ is cyclic, we get $\\angle PMD = \\angle MPE = \\angle ABR$, so that triangles $PMD$ and $ABR$ are similar with $\\frac{PD}{PM} = \\frac{AR}{AB}$.\n\nNow, observe that $\\angle PER = \\angle PEB = 180^\\circ - \\angle PMB = 180^\\circ - \\angle PNR$, so $PERN$ is cyclic and $\\angle CRB = \\angle DPE = \\angle NDM$. We see that triangles $CRB$ and $NDM$ are similar with $\\frac{CB}{CR} = \\frac{MN}{ND}$. With $QC = PB$, we have $QB = PC$ and Thales' theorem gives $\\frac{PD}{DN} = \\frac{PB}{BQ} = \\frac{PB}{PC}$. Thus, applying Menelaus' theorem on triangle $PAC$, we obtain\n\n$$\n\\frac{CR}{AR} \\cdot \\frac{AM}{MB} \\cdot \\frac{BP}{PC} = \\frac{CB \\cdot ND}{MN} \\cdot \\frac{PM}{PD \\cdot AB} \\cdot \\frac{PD}{DN} = \\frac{PM}{MN} \\cdot \\frac{BC}{AB} = 1\n$$\n\nimplying that $P$, $M$ and $R$ are collinear as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $x$ and $y$ be integers satisfying $x^{2}+30x+25=y^{4}$. What is the largest possible value of $x+y$?",
    "options": [],
    "answer": "43",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA game is played on the number line. Initially, there is a token placed at the number $0$. In each move, the player can move the token from its current position $x$, to either $x+2023$ or $x-59$. The goal of the game is to move the token to either $1$ or $-1$. What is the minimum number of moves required to achieve this goal?",
    "options": [],
    "answer": "247",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$, $z$ be three real numbers such that\n- $y$, $x$, $z$ form a harmonic sequence; and\n- $3 x y$, $5 y z$, $7 z x$ form a geometric sequence.\nThe numerical value of $\\frac{y}{z} + \\frac{z}{y}$ can be expressed in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?",
    "options": [],
    "answer": "59",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSilverio is very happy for the 25th year of the PMO. In his jubilation, he ends up writing a finite sequence of $A$'s and $G$'s on a nearby blackboard. He then performs the following operation: if he finds at least one occurrence of the string $\"AG\"$, he chooses one at random and replaces it with $\"GAAA\"$. He performs this operation repeatedly until there is no more $\"AG\"$ string on the blackboard. Show that for any initial sequence of $A$'s and $G$'s, Silverio will eventually be unable to continue doing the operation.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe assign the weight $4^{k}$ to each $G$ in the sequence, where $k$ is the number of $A$'s to the right of this $G$. In each operation, if $4^{k}$ is the weight of the $G$ in the $\"AG\"$ being replaced, then each of the three $G$'s in $\"GAAA\"$ have a weight of $4^{k-1}$. So the sum of the weights decreases by $4^{k} - 3 \\cdot 4^{k-1} = 4^{k-1}$ in each operation. Since the sum of weights in the initial sequence is finite, and the sum of the weights of all $G$'s must be a nonnegative integer, Silverio can only perform a finite number of operations.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nf(2 f(x))=f(x-f(y))+f(x)+y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Solution:\nLet $P(x, y)$ be the problem statement. Note that if $f(a)=f(b)$ where $a, b \\in \\mathbb{R}$, then $P(x, a)$ and $P(x, b)$ imply $a=b$. Thus $f$ is injective.\n\nThen $P(x,-f(x))$ implies $f(2 f(x))=f(x-f(-f(x)))$, so by injectivity we obtain $f(-f(x))=x-2 f(x)$.\n\nNow $P(x, 0)$ gives $f(2 f(x))=f(x-f(0))+f(x)$. Combining this with $P(x, y)$, we obtain $f(x-f(0))=f(x-f(y))+y$.\n\nSetting $x=0$ here yields $f(-f(0))=f(-f(y))+y$, and so $f(-f(0))=2 y-f(y)$. In particular, this implies $f(y)=y+c$ for some constant $c$.\n\nSubstituting this into $P(x, y)$, we obtain\n$$\n2 x+3 c=x-y+x+y+c\n$$\nwhich implies that $c=0$. It is easy to see that $f(x)=x$ works and is therefore the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are three novel series Peter wishes to read. Each consists of $4$ volumes that must be read in order, but not necessarily one after the other. Let $N$ be the number of ways Peter can finish reading all the volumes. Find the sum of the digits of $N$. (Assume that he must finish a volume before reading a new one.)",
    "options": [],
    "answer": "18",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 37601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $P(x)$ and $Q(x)$ are both quadratic polynomials with leading coefficient $1$ such that $P(P(x)-x) = (Q(x))^2$ for all real numbers $x$ and $P(2) = 0$. Find the sum of all possible values of $P(10)$.",
    "options": [],
    "answer": "64",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of positive integers is said to be pilak if it can be partitioned into 2 disjoint subsets $F$ and $T$, each with at least 2 elements, such that the elements of $F$ are consecutive Fibonacci numbers, and the elements of $T$ are consecutive triangular numbers. Find all positive integers $n$ such that the set containing all the positive divisors of $n$ except $n$ itself is pilak.",
    "options": [],
    "answer": "30",
    "solution": "Solution:\nThe only positive integer $n$ that satisfies this property is $30$.\nIn this case, $D=\\{1,2,3,5,6,10,15\\}$, and we can partition $D$ into $F=\\{1,2,3,5\\}$ and $T=\\{6,10,15\\}$. We will show that there are no other $n$.\n\nClaim 1: $1 \\in F$.\nProof: Obviously, $1 \\in D$. Assume that $1 \\in T$. Since $T$ has at least 2 elements, $3 \\in T$. If $6 \\in T$, then $2 \\in D$. Note that $2$ is not a triangular number. If $2 \\in F$, then since $F$ has at least 2 elements, at least one of $1$ or $3$ must be in $F$. But both $1$ and $3$ are in $F$, contradiction. Thus, $6 \\notin T$, and $T=\\{1,3\\}$.\n\nNow, if we have two consecutive Fibonacci numbers $F_{k}$ and $F_{k+1}$ in $F$, at least one of them is not divisible by $3$, so at least one of $3 F_{k}$ or $3 F_{k+1}$ divides $n$. It is well known that two consecutive Fibonacci numbers are relatively prime, so $F_{k+1} \\mid 3 F_{k}$ and $F_{k} \\mid 3 F_{k+1}$ are both impossible. Thus, $n \\neq 3 F_{k}, 3 F_{k+1}$, implying that at least one of $3 F_{k}$ or $3 F_{k+1}$ is in $F$. However, for all integers $i>1$, $F_{i+3}=F_{i+2}+F_{i+1}=2 F_{i+1}+F_{i}>3 F_{i}>2 F_{i}+F_{i-1}=F_{i}+F_{i+1}=F_{i+2}$, so $3 F_{i}$ cannot be a Fibonacci number, and none of $3 F_{k}$ or $3 F_{k+1}$ is in $F$, contradiction. Therefore, $1 \\in F$.\n\nClaim 2: $3 \\in F$.\nIf $1 \\in F$, then since $F$ has at least two elements, $2 \\in F$. If $3 \\notin F$, then $F=\\{1,2\\}$. Let $t$ denote the smallest number in $T$. If $t$ is composite, since $4$ is not triangular, there exists a proper divisor of $t$ greater than $2$. Since this is a proper divisor of $n$, smaller than $t$, and greater than $2$, this number cannot be in either $F$ or $T$, contradiction. Thus, $t$ is prime. However, $3$ is the only triangular prime number, so $t=3$ and $3 \\in T$. Since $1 \\notin T$ and $T$ has at least two elements, $6 \\in T$. If $10 \\in T$, then $5 \\in D$. Since $5$ is not a triangular number, and $5 \\notin F$, we have a contradiction. Thus, $10 \\notin T$, and $T=\\{3,6\\}$. This implies that $D=\\{1,2,3,6\\}$, and that $n$ has exactly $5$ divisors, making $n$ a perfect square. $n$ is even, thus it must be divisible by $4$. However, $4 \\notin D$, contradiction. Therefore, $3 \\in F$.\n\nSince $1,3 \\in F$, $2 \\in F$ and $6 \\mid n$. Note that $n$ must have at least $5$ positive integer divisors, and it is easy to check that among $1,2, \\ldots, 15$, only $12$ has at least $5$ positive integer divisors. Thus, $n>6$ and $6 \\in D$. Since $6$ is not a Fibonacci number, $6 \\in T$. Since $3 \\notin T$ and $T$ has at least two elements, $10 \\in T$. This also implies that $5 \\in D$. Since $5$ is not a triangular number, $5 \\in F$. So $15 \\mid n$. Since $5 \\nmid 12$, $n \\neq 12$, so $n>15$, making $15 \\in D$. Since $15$ is not a Fibonacci number, $15 \\in T$. If $21 \\in T$, then $7 \\in D$. However, $7$ is neither a Fibonacci number nor a triangular number, contradiction. Thus, $T=\\{6,10,15\\}$.\n\nIf $8 \\in F$, then $4 \\in D$. However, $4$ is neither a Fibonacci number nor a triangular number, contradiction. Thus, $F=\\{1,2,3,5\\}$.\n\nSince $D=\\{1,2,3,5,6,10,15\\}$, $n=30$ is the only possible solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S$ be the sum of all positive integers less than $10^{6}$ which can be expressed as $m! + n!$, where $m$ and $n$ are nonnegative integers. Determine the last three digits of $S$.",
    "options": [],
    "answer": "130",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathcal{S}$ be the set of all points in the plane. Find all functions $f: \\mathcal{S} \\rightarrow \\mathbb{R}$ such that for all nondegenerate triangles $A B C$ with orthocenter $H$, if $f(A) \\leq f(B) \\leq f(C)$, then\n$$\nf(A)+f(C)=f(B)+f(H)\n$$",
    "options": [],
    "answer": "All constant functions f(P) = c for any real constant c",
    "solution": "Solution:\nLet $P(A, B, C)$ be the problem assertion. First consider a non-right triangle $A B C$ with orthocenter $H$. Note that in the set $\\{A, B, C, H\\}$, the last point is the orthocenter of the other three. Thus, we can assume WLOG $f(A) \\leq f(B) \\leq f(C) \\leq f(H)$.\nBy considering $P(A, B, C)$, this implies $f(A)+f(C)=f(B)+f(H)$. Since $f(A) \\leq f(B)$ and $f(C) \\leq f(H)$, this implies equality must hold in both inequalities, and so $f(A)=f(B)$ and $f(C)=f(H)$.\nDenote by $\\Omega_{B C}$ the circle with diameter $B C$. We prove the following claim.\n\nClaim: If $f(B) \\neq f(C)$, then for all $D \\in \\Omega_{B C}$, we have $2 f(D)=f(B)+f(C)$. In particular, $f$ is constant on $\\Omega_{B C}$.\n\nProof: We split into cases. If $f(D) \\notin [f(B), f(C)]$, then $P(B, C, D)$ implies\n$$\nf(D)+f(C)=f(B)+f(D)\n$$\nwhich implies $f(B)=f(C)$, contradiction. Thus, $f(D) \\in [f(B), f(C)]$, from which $P(B, C, D)$ implies\n$$\nf(B)+f(C)=2 f(D)\n$$\nThe claim then follows.\n\nNow we claim that $f(B)=f(C)$. Assume for the sake of contradiction that this was not the case. Consider $\\Omega_{A C}$, and let $D=A H \\cap B C \\in \\Omega_{A C}$. Let $\\Omega_{B D'}$ intersect $\\Omega_{A C}$ at a second point $D'' \\neq D'$. Then since $f(A) \\neq f(C)$, from the claim we get $2 f\\left(D'\\right)=f(A)+f(C)$.\n\nNow we have two cases. In the first case, suppose $f(B)=f\\left(D'\\right)$. Then $2 f(B)=f(A)+f(C)$, implying $f(B)=f(C)$, contradiction. In the second case, we have $f(B)+f\\left(D'\\right)=2 f\\left(D''\\right)=f(C)+f(A)$. This implies $f(C)=f\\left(D'\\right)$, and so\n$$\n2 f(C)=2 f\\left(D'\\right)=f(C)+f(A)\n$$\nimplying $f(C)=f(A)=f(B)$, another contradiction. Thus our assumption was wrong, and $f(B)=f(C)$. This implies $f(A)=f(B)=f(C)=f(H)$ for any nondegenerate non-right triangle $A B C$, and so $f$ is constant everywhere (by considering two segments whose diameter circles do not intersect). It is clear that these solutions work.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many 9-term sequences $a_{1}, \\ldots, a_{9}$ of nonnegative integers are there such that\n- $0 \\leq a_{i} < i$ for all $i = 1, \\ldots, 9$; and\n- there are no ordered triples $(i, j, k)$ with $1 \\leq i < j < k \\leq 9$, such that $a_{i}, a_{j}, a_{k}$ are all distinct?",
    "options": [],
    "answer": "503",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre uma maneira de se escrever todos os algarismos de $1$ a $9$ em sequência e sem repetição, de forma que os números determinados por quaisquer dois algarismos consecutivos da sequência sejam divisíveis por $7$ ou $13$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFixăm un număr întreg $n \\geq 2$. Determinaţi valoarea minimă a expresiei\n$$\n\\frac{x_{1}+x_{2}+\\cdots+x_{n}}{\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\cdots+\\frac{1}{x_{n}}}\n$$\ncând $x_{1}, x_{2}, \\ldots, x_{n}$ parcurg mulţimea numerelor reale strict pozitive, supuse condiţiei\n$$\n\\frac{1}{1+x_{1}^{2}}+\\frac{1}{1+x_{2}^{2}}+\\cdots+\\frac{1}{1+x_{n}^{2}}=1\n$$\nDeterminaţi şi valorile $x_{1}, x_{2}, \\ldots, x_{n}$ care realizează acest minimum.",
    "options": [],
    "answer": "Minimum value: n − 1, attained when x_1 = x_2 = ⋯ = x_n = √(n − 1). (For n = 2, the value is 1 for all feasible pairs.)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFixăm un număr întreg $n \\geq 3$. Pentru fiecare submulţime nevidă a mulţimii $\\{1,2, \\ldots, n\\}$, considerăm media aritmetică a elementelor sale. Fie $S$ mulţimea valorilor distincte ale acestor medii aritmetice. Determinaţi cea mai mica valoare absolută $|a-b|$, când $a$ şi $b$ parcurg mulţimea $S$ şi $a \\neq b$.",
    "options": [],
    "answer": "1/((n-1)(n-2))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic, $AB < AC$, şi fie $\\Gamma$ cercul său circumscris. Tangenta lui $\\Gamma$ în $A$ intersectează dreapta $BC$ în $K$. Cercul $\\Omega$, de rază $KA$ şi centru $K$, intersectează a doua oară $\\Gamma$ în $L$. Cercul $\\Omega$ intersectează a doua oară dreapta $BL$ în $M$ şi dreapta $CM$ în $N$. Arătaţi că dreapta $AN$ trece prin mijlocul segmentului $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $n$ un număr natural nenul. Este posibil ca orice număr raţional strict pozitiv să fie exprimat sub forma\n$$\n\\frac{a^{n}+b^{n+2}}{c^{n+1}+d^{n+3}}\n$$\nunde $a, b, c, d$ sunt numere naturale nenule?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFixăm un număr întreg $k \\geq 2$. Determinaţi cel mai mic număr întreg $n$, astfel încât, printre oricare $n$ puncte în plan, să existe $k$ puncte între care fie toate distanţele sunt mai mici sau egale cu $2$, fie toate distanţele sunt strict mai mari decât $1$.",
    "options": [],
    "answer": "(k - 1)^2 + 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nŞirul de numere reale $a_{0}, a_{1}, \\ldots, a_{n}, \\ldots$ satisface relaţia\n$$\na_{m+n} + a_{m-n} - m + n - 1 = \\frac{1}{2}\\left(a_{2m} + a_{2n}\\right)\n$$\npentru orice numere naturale $m$ şi $n$, $m \\geq n$.\nDacă $a_{1} = 3$, aflaţi $a_{2004}$.",
    "options": [],
    "answer": "4018021",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nRezolvaţi în mulţimea numerelor prime ecuaţia\n$$\nx^{y} - y^{x} = x y^{2} - 19\n$$",
    "options": [],
    "answer": "(x,y) = (2,3) and (2,7)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $O$ un punct interior triunghiului ascuţitunghic $ABC$. Cercurile centrate în mijloacele laturilor triunghiului şi care trec prin $O$, se intersectează a doua oară în $K$, $L$ şi $M$.\nDemonstraţi că $O$ este centrul cercului înscris în triunghiul $KLM$ dacă şi numai dacă $O$ este centrul cercului circumscris triunghiului $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO mulţime finită de drepte, oricare trei neconcurente, partiţionează planul într-un număr de regiuni. Două regiuni se numesc \"vecine\" dacă frontierele lor au în comun mai mult decât un segment nedegenerat.\n\nÎn fiecare regiune trebuie scris un număr întreg astfel încât:\n(i) produsul numerelor scrise în oricare două regiuni vecine este strict mai mic decât suma lor;\n(ii) pentru fiecare dintre dreptele date, suma numerelor scrise în regiunile aflate de aceeaşi parte a ei este egală cu $0$.\n\nDemonstraţi că acesta este posibil dacă şi numai dacă nu toate dreptele sunt paralele.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $I$ centrul cercului înscris în triunghiul $ABC$ şi $A_{1}$, $B_{1}$ şi $C_{1}$ puncte arbitrare pe segmentele $(AI)$, $(BI)$, respectiv $(CI)$. Mediatoarele segmentelor $AA_{1}$, $BB_{1}$ şi $CC_{1}$ se intersectează în $A_{2}$, $B_{2}$ şi respectiv $C_{2}$. Arătați că centrul cercului circumscris triunghiului $A_{2}B_{2}C_{2}$ coincide cu centrul cercului circumscris triunghiului $ABC$ dacă şi numai dacă $I$ este ortocentrul triunghiului $A_{1}B_{1}C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentru orice număr natural nenul $n$, suma $1+\\frac{1}{2}+\\ldots+\\frac{1}{n}$ se scrie sub forma $\\frac{p_{n}}{q_{n}}$ unde $p_{n}$ şi $q_{n}$ sunt numere naturale prime între ele.\n\na. Arătați că $p_{67}$ nu se divide cu 3.\n\nb. Determinați toate numerele naturale nenule $n$, pentru care $p_{n}$ este divizibil cu 3.",
    "options": [],
    "answer": "a) p_67 is not divisible by 3.\nb) The numerator p_n is divisible by 3 exactly for n = 2 and for all n congruent to 7 modulo 9.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn grup este format din $n$ turişti. Oricum am alege 3 turişti, există 2 dintre ei care nu se cunosc. Pentru orice partiție a turiştilor în două autobuze putem găsi doi turişti care se cunosc şi sunt în acelaşi autobuz. Demonstrați că în grupul de turişti există un turist care are cel mult $\\frac{2}{5} n$ cunoscuți.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎntr-un cuvânt format din literele $a$ şi $b$ sunt posibile schimbările $a b a \\rightarrow b$, $b \\rightarrow a b a$, $b b a \\rightarrow a$, $a \\rightarrow b b a$. Dacă începem de la cuvântul $\\underbrace{a a \\ldots a}_{2003} b$, este posibil de obținem cuvântul $b \\underbrace{a a \\ldots a}_{2003}$ ?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a, b, c, d$ numere naturale nenule astfel încât numărul perechilor $(x, y), x, y \\in (0,1)$, cu proprietatea că $a x + b y$ şi $c x + d y$ sunt întregi, este egal cu $2004$. Dacă $\\operatorname{cmmdc}(a, c) = 6$, aflați $\\operatorname{cmmdc}(b, d)$.",
    "options": [],
    "answer": "gcd(b,d) ∈ {1, 7, 49}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p$ un număr prim şi $0 \\leq a_{1}<a_{2}<\\ldots<a_{m}<p$ şi $0 \\leq b_{1}<b_{2}<\\ldots<b_{n}<p$ numere întregi oarecare. Notăm cu $k$ numărul resturilor diferite pe care le dau $a_{i}+b_{j}, 1 \\leq i \\leq m, 1 \\leq j \\leq n$ la împărțirea cu $p$. Demonstrați că:\n\na) dacă $m+n>p$, atunci $k=p$;\n\nb) dacă $m+n \\leq p$, atunci $k \\geq m+n-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a convex pentagon with $AB + CD = BC + DE$, and $k$ a circle centered on side $AE$, tangent to sides $AB$, $BC$, $CD$ and $DE$ at points $P$, $Q$, $R$ and $S$ respectively. Prove that lines $PS$ and $AE$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nAB + CD = (AP + PB) + (CR + RD) = AP + (BP + CR + DR)\n$$\n$$\nBC + DE = (BQ + QC) + (DS + SE) = ES + (BQ + CQ + DS).\n$$\nUsing the fact that tangents from a point to a circle are of equal length, one gets $AP = ES$. Denoting by $O$ and $r$ the center, respectively radius of $k$, one gets that right-angled triangles $OPA$, $OSE$ are congruent, since $AP = ES$, $OP = OS = r$, and $OA = \\sqrt{OP^2 + AP^2} = \\sqrt{OS^2 + ES^2} = OE$. Therefore the corresponding altitudes of these triangles, from $P$, respectively $S$, are of equal length, hence $PS$ is parallel to $AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_n$ and $y_1, y_2, \\dots, y_n$ be positive real numbers so that\n$$\nx_1 + x_2 + \\dots + x_n \\geq x_1 y_1 + x_2 y_2 + \\dots + x_n y_n.\n$$\nShow that for any non-negative integer $p$ the following inequality holds\n$$\n\\frac{x_1}{y_1^p} + \\frac{x_2}{y_2^p} + \\dots + \\frac{x_n}{y_n^p} \\geq x_1 + x_2 + \\dots + x_n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume by contradiction that\n$$\n\\frac{x_1}{y_1^p} + \\frac{x_2}{y_2^p} + \\dots + \\frac{x_n}{y_n^p} < x_1 + x_2 + \\dots + x_n.\n$$\nOn the other hand, by multiplying the given relation by $p \\in \\mathbb{N}$, we have\n$$\np(x_1 + x_2 + \\dots + x_n) \\geq p(x_1 y_1 + x_2 y_2 + \\dots + x_n y_n).\n$$\nAdding the above inequalities we get\n$$\n(p+1) \\sum_{i=1}^{n} x_i > \\sum_{i=1}^{n} x_i \\left( p y_i + \\frac{1}{y_i^p} \\right).\n$$\nBut by the AM-GM inequality we have\n$$\np y_i + \\frac{1}{y_i^p} \\geq p + 1.\n$$\nFinally, $(p+1) \\sum_{i=1}^{n} x_i > (p+1) \\sum_{i=1}^{n} x_i$, which is false.\nWe shall use the following inequality due to Radon.\n\nLEMMA. If $x_1, x_2, \\dots, x_n$ and $y_1, y_2, \\dots, y_n$ are positive real numbers, and $p$ is a non-negative real number, then the following inequality holds\n$$\n\\frac{x_1^{p+1}}{y_1^p} + \\frac{x_2^{p+1}}{y_2^p} + \\dots + \\frac{x_n^{p+1}}{y_n^p} \\ge \\frac{(x_1 + x_2 + \\dots + x_n)^{p+1}}{(y_1 + y_2 + \\dots + y_n)^p}.\n$$\nProof. We can rewrite the inequality as\n$$\n\\sum_{i=1}^{n} y_i \\left( \\frac{x_i}{y_i} \\right)^{p+1} \\ge \\left( \\sum_{j=1}^{n} y_j \\right) \\left( \\frac{\\sum_{i=1}^{n} x_i}{\\sum_{i=1}^{n} y_i} \\right)^{p+1}\n$$\nor equivalently\n$$\n\\sum_{i=1}^{n} \\frac{y_i}{\\sum_{j=1}^{n} y_j} \\left( \\frac{x_i}{y_i} \\right)^{p+1} \\ge \\left( \\frac{\\sum_{i=1}^{n} x_i}{\\sum_{i=1}^{n} y_i} \\right)^{p+1}.\n$$\nLet us consider the convex function $f : [0, \\infty) \\to \\mathbb{R}$ defined by $f(x) = x^{p+1}$. The last inequality follows from Jensen's inequality\n$$\n\\lambda_1 f(\\alpha_1) + \\lambda_2 f(\\alpha_2) + \\dots + \\lambda_n f(\\alpha_n) \\ge f(\\lambda_1 \\alpha_1 + \\dots + \\lambda_n \\alpha_n),\n$$\nwhere $\\lambda_i = \\frac{y_i}{y_1 + y_2 + \\dots + y_n} > 0, \\sum_{i=1}^{n} \\lambda_i = 1$ and $\\alpha_i = \\frac{x_i}{y_i}$. $\\square$\n\nNow, returning to our problem, by applying the LEMMA and the hypothesis of the problem, we obtain\n$$\n\\sum_{i=1}^{n} \\frac{x_i}{y_i^p} = \\sum_{i=1}^{n} \\frac{x_i^{p+1}}{(x_i y_i)^p} \\ge \\frac{\\left(\\sum_{i=1}^{n} x_i\\right)^{p+1}}{\\left(\\sum_{i=1}^{n} x_i y_i\\right)^p} \\ge \\sum_{i=1}^{n} x_i.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AA'$, $BB'$, $CC'$ be the altitudes from the vertices of an acute-angled triangle $ABC$. Points $E$ and $F$ lie on the segments $CB'$ and $BC'$ respectively, such that\n$$\nB'E \\cdot C'F = BF \\cdot CE.\n$$\nProve that the quadrilateral $AEA'F$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given relation writes as $\\frac{BF}{FC'} = \\frac{B'E}{EC}$. Let $M$ be a point on the side $BC$ such that $FM \\parallel CC'$. Then $\\frac{BM}{MC} = \\frac{BF}{FC'} = \\frac{B'E}{EC}$, implying $ME \\parallel BB'$. Hence the angles $\\angle AEM$, $\\angle AFM$ and $\\angle AA'M$ are right angles, therefore $E$, $F$, $A'$ all lie on the circle of diameter $AM$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $a$, $b$ distinct positive integers, show that the system of equations\n$$\nxy + zw = a\n$$\n$$\nxz + yw = b\n$$\nhas only finitely many solutions in integers $x$, $y$, $z$, $w$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By adding and subtracting the equations we get $(x + w)(y + z) = a + b$ and $(x - w)(y - z) = a - b$, hence by multiplying, $0 < |(x^2 - w^2)(y^2 - z^2)| = |a^2 - b^2|$. Therefore $0 < |x^2 - w^2| \\le |a^2 - b^2|$ and $0 < |y^2 - z^2| \\le |a^2 - b^2|$. But an equation $0 < |A^2 - B^2| \\le |C|$ has only finitely many integer solutions $A$, $B$ for fixed $C$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that all positive integers, except the powers of $2$, can be written as the sum of (at least two) consecutive positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All symbols in the sequel are denoting integer numbers. Let $n = 2^a b$, $a \\ge 0$, $b \\ge 1$, $b$ odd. We want to have $n = (m+1) + (m+2) + \\dots + (m+k)$, with $m \\ge 0$ and $k \\ge 2$, hence $k(2m+k+1) = 2^{a+1}b$.\n\nIf $b=1$, it follows $k=2^\\alpha$, with $1 \\le \\alpha \\le a+1$, but then $2m+k+1 > 1$ is odd, so there are no solutions. Now, for $b > 1$, we can exhibit the required writing.\n\nIf $b \\ge 2^{a+1} + 1$, then take $m = \\frac{1}{2}(b - 2^{a+1} - 1)$ and $k = 2^{a+1}$.\n\nIf $b \\le 2^{a+1} - 1$, then take $m = \\frac{1}{2}(2^{a+1} - 1 - b)$ and $k = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be a circle in the plane and $A,B$ two points lying on it. We denote by $M$ the midpoint of $AB$ and let $P \\neq M$ be a new point on $AB$. Build circles $\\gamma$ and $\\delta$ tangent to $AB$ at $P$ and to $\\omega$ at $C$, respectively $D$. Consider $E$ to be the point diametrically opposed to $D$ in $\\omega$. Prove that the circumcenter of $\\triangle BMC$ lies on the line $BE$.\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us begin by noticing that since $DE$ is a diameter, we have $\\angle DBE = 90^\\circ$, and if the circumcenter of $\\triangle BMC$ would lie on $BE$, we could conclude that $DB$ is tangent to the circumcircle of $\\triangle BMC$. Thus $\\angle DBA \\equiv \\angle BCM$. Now since $A$, $B$, $C$, $D$ are on $\\omega$, we have $\\angle DBA \\equiv \\angle DCA$, so the problem is equivalent to $\\angle BCM \\equiv \\angle DCA$.\nThen we must prove that $CD$ and $CM$ are isogonal conjugates in the triangle $\\triangle BCA$, thus the quadrilateral $ABCD$ is harmonic, which means $\\frac{AC}{CB} = \\frac{AD}{DB}$.\nThis follows from the fact that since $CD$ and $CM$ are isogonal conjugates, we have\n$$ \\frac{\\sin \\angle BCD}{\\sin \\angle ACD} = \\frac{\\sin \\angle ACM}{\\sin \\angle BCM}. $$\nFrom the sine theorem we know\n$$ \\frac{\\sin \\angle BCD}{\\sin \\angle ACD} = \\frac{BD}{DA}. $$\nWe also easily obtain\n$$ \\frac{\\sin \\angle ACM}{\\sin \\angle BCM} = \\frac{BC}{AC} $$\nfrom the fact that $M$ is midpoint, and thus\n$$ \\text{area}(BMC) = \\text{area}(ACM). $$\n\nLEMMA (Archimedes). Let $C$ be a circle, $P$ and $Q$ two points on it, and $R$ a point on $PQ$. Let $\\omega$ be one of the two circles tangent to $C$ at $S$, and to $PQ$ at $R$. Then $SR$ is the angle bisector of $\\angle PSQ$.\n![](attached_image_1.png)\n\nProof. Let us denote by $O$ and $O_1$ the centers of the circles $C$, respectively $\\omega$. Let $RS \\cap C = \\{T\\}$. Let us note that the triangles $\\triangle RO_1S$ and $\\triangle TSO$ are isosceles triangles in $O$ and $O_1$, and that $\\angle TSO = \\angle RSO_1$. We conclude that $RO_1 \\parallel TO$. Since $RO_1 \\perp PQ$ we deduce $TO \\perp PQ$, so indeed $T$ is the midpoint of arc $PQ$. $\\square$\n\nUsing this LEMMA we get ($DP$ to be the angle bisector of $\\angle ADB$ and ($CP$ the angle bisector of $\\angle ACB$), so from the angle bisector theorem $\\frac{AD}{DB} = \\frac{AC}{CB} = \\frac{AP}{BP}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Of the vertices of a cube, 7 of them have assigned the value $0$, and the eighth the value $1$. A *move* is selecting an edge and increasing the numbers at its ends by an integer value $k > 0$. Prove that after any finite number of moves, the g.c.d. of the $8$ numbers at vertices is equal to $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us alternately colour the vertices in black and white. After any move, the difference between the sums of the numbers at the black and the white vertices remains $1$, therefore the g.c.d. of the $8$ numbers is equal to $1$ (as it is dividing the difference of the sums mentioned above).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be nodes of the lattice $\\mathbb{Z} \\times \\mathbb{Z}$ such that inside the triangle $ABC$ lies a unique node $P$ of the lattice. Denote $E := AP \\cap BC$. Determine $\\max \\frac{AP}{PE}$, over all such configurations.",
    "options": [],
    "answer": "5",
    "solution": "Let us build $E'$, the symmetrical of $E$ with respect to $P$. If $E$ is latticeal, then $E'$ is latticeal, and this implies $E' \\notin \\text{int}(ABC)$. We deduce that $E' = A$, otherwise $A \\in (E'P)$ and through a translation we get that there exists a latticeal point lying on $PE$, contradiction. So in this case $\\frac{AP}{PE} = 1$.\nIf $E$ is not a latticeal point, and on ($BC$) lies some other latticeal point $Q$, since $Q \\neq E$ we can restrict the problem to either triangle $ABQ$ or $ACQ$, with the same initial hypothesis. Thus we can assume that the only latticeal points on the sides of the triangle are on $[AB]$ and $[AC]$.\nWe shall prove that, over all configurations,\n$$\n\\max \\frac{AP}{PE} = 5,\n$$\nvalue which can actually be reached by a proper configuration. Rather than looking for $\\max \\frac{AP}{PE}$, we will establish $\\min \\frac{PE}{AE} = \\min \\frac{[BPC]}{[ABC]}$ (where by $[XYZ]$ we mean the area of $\\triangle XYZ$).\nLet us notice that for the triangle $BPC$, the only latticeal points that are lying either on its sides, or in its interior, are just its vertices. We deduce from Pick's Theorem that $[BPC] = \\frac{1}{2}$. Let us denote by $\\beta$ and $\\gamma$ the number of latticeal points on the sides ($AB$), respectively ($AC$). Availing ourselves again of Pick's Theorem, we find that $[ABC] = \\frac{\\beta + \\gamma + 3}{2}$.\nWe get that $\\frac{PE}{AE} = \\frac{1}{\\beta + \\gamma + 3}$. We want to prove that $\\frac{PE}{AE} \\ge \\frac{1}{6}$. We contrariwise assume $\\frac{PE}{AE} < \\frac{1}{6}$, and thus $\\beta + \\gamma \\ge 4$.\n\nDenote by $B_0, B_1, \\dots, B_{\\beta+1}$ the lattice points on $[AB]$, with $B_0 = A$, $B_{\\beta+1} = B$, and by $C_0, C_1, \\dots, C_{\\gamma+1}$ the lattice points on $[AC]$, with $C_0 = A$, $C_{\\gamma+1} = C$.\nLet us consider the triangles of the type $PC_iC_{i+1}$ with $i = 0, \\dots, \\gamma$. They share the property that no lattice points lie inside or on the sides, other than the vertices, so $[PC_iC_{i+1}] = \\frac{1}{2}$. It follows that $AC_1 = C_1C_2 = \\dots = C_\\gamma C$. With a similar reasoning we have also $AB_1 = B_1B_2 = \\dots = B_\\beta B$. Thus $\\frac{AC_j}{C_jC_{j+1}} = \\frac{AB_j}{B_jB_{j+1}}$, and so $C_iB_i \\parallel C_{i+1}B_{i+1}$, for all $i \\in \\{1, 2, \\dots, \\min\\{\\beta, \\gamma\\}\\}$.\nWe also deduce that on the segment $(B_iC_i)$ lie $i-1$ lattice points. This is because we can construct $i-1$ parallelograms with vertices of the type $B_lC_lB_iD$ with $l < i$, or of the type $B_lC_lC_iD$ with $D \\in (B_iC_i)$. From effectively writing the relationships between coordinates in a parallelogram, we get that $D$ is a lattice point. Thus if $\\min\\{\\beta, \\gamma\\} \\ge 2$, the segments $(B_i, C_i)$ would contain at least three lattice points (the ones from $(B_2C_2)$ and $(B_3C_3)$), which is obviously false since $P$ is on at most one of the segments $(B_iC_i)$. This means $\\min\\{\\beta, \\gamma\\} \\le 1$, with say $\\beta \\ge \\gamma$.\n\n**First case.** $\\gamma = 1$. From $\\beta + \\gamma \\ge 4$ follows that $\\beta \\ge 3$ and that $C_1$ is the midpoint of $AC$. Now since $B_1$ is the midpoint of $AB_2$ we have that if we take $M$ the midpoint of $CB_2$, we get the parallelogram $B_2B_1C_1M$, and since all the points $B_2, B_1, C_1$ are lattice, we get $M$ lattice. Since $M$ is inside the triangle $ABC$, it must be $P$. The quadrilateral $BB_\\beta PC$ is convex since $B_\\beta \\in (B_2B)$. We note that the triangles $BB_\\beta C$ and $BB_\\beta P$ have the property that there are no lattice points inside or on their sides, except their vertices, so $[BB_\\beta P] = [BB_\\beta C] = \\frac{1}{2}$, hence $BB_\\beta \\parallel PC$, which is false.\n\n**Second case.** $\\gamma = 0$. It follows that $\\beta \\ge 4$.\n\n**LEMMA.** Let $XYZT$ be a convex quadrilateral with lattice vertices, with the property that no lattice points lie in its interior, nor on its sides, except for its vertices. Then $XYZT$ is a parallelogram.\n*Proof.* From Pick's theorem we know that $[XYZ] = [XYT] = \\frac{1}{2}$, and so $XY \\parallel ZT$. Similarly $XT \\parallel YZ$, which ends the proof. $\\square$\n\nLet us consider the quadrilateral $AB_1PC$, with $AB_1 \\cap PC \\ne \\emptyset$. If it is convex, the previous LEMMA implies a contradiction. Thus $[AB_1] \\cap PC \\ne \\emptyset$. Also consider the quadrilateral $BB_\\beta PC$. By the same reasoning, we must have $[BB_\\beta] \\cap PC \\ne \\emptyset$. This is the desired contradiction and ends the solution of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37630,
    "subject": "Mathematics (Multi-modal)",
    "question": "For real numbers $x, y, z \\in (0, 1)$, with $xyz = (1-x)(1-y)(1-z)$, show that at least one of the numbers $(1-x)y, (1-y)z, (1-z)x$ is greater than or equal to $\\frac{1}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $x, y, z \\in (0, 1)$ it follows that also $1-x, 1-y, 1-z \\in (0, 1)$. We have\n$$\n\\sum (1-x)y = \\sum x - \\sum xy,\n$$\nwhile\n$$\n\\prod x = \\prod (1-x) \\text{ translates into } (\\sum x - \\sum xy) + 2xyz = 1, \\text{ hence } \\sum (1-x)y + 2\\sqrt{\\prod (1-x)y} = 1 = 3 \\cdot \\frac{1}{4} + 2\\sqrt{\\left(\\frac{1}{4}\\right)^3}.\n$$\nNow, either $\\sum (1-x)y \\ge 3 \\cdot \\frac{1}{4}$, or $\\prod (1-x)y \\ge \\left(\\frac{1}{4}\\right)^3$, so at least one expression is at least $\\frac{1}{4}$. In fact always $\\prod (1-x)y = \\prod (1-x)x \\le \\prod \\left(\\frac{(1-x)+x}{2}\\right)^2 = \\left(\\frac{1}{4}\\right)^3$.\n\n\n*Alternative Solution.* The use of values $x$ and $1-x$, et al. does suggest looking around value $\\frac{1}{2}$, the mid-value. There are four cases.\n\n1) If none of $x, y, z$ is at least $1/2$, then $\\prod x < \\prod (1-x)$, absurd.\n\n2) If exactly one of $x, y, z$ is at least $1/2$, say $x$, then $(1-z)x > 1/4$.\n\n3) If exactly two of $x, y, z$ are at least $1/2$, say $y, z$, then $(1-x)y > 1/4$.\n\n4) If all three of $x, y, z$ are at least $1/2$, then they all must be equal to $1/2$, otherwise $\\prod x > \\prod (1-x)$, absurd. Now all three expressions are equal to $1/4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Determine the least number of equilateral triangles of side $1$ which can cover an equilateral triangle of side $n + \\frac{1}{2n}$.",
    "options": [],
    "answer": "n^2 + 2",
    "solution": "The ratio of the areas of the equilateral triangle $\\Delta$ of side $n + \\frac{1}{2n}$ and that of the equilateral triangle $\\Delta_1$ of side $1$ is the square of the ratio of the lengths of their sides, i.e.\n$$(n + \\frac{1}{2n})^2 > n^2 + 1,$$\nhence at least $n^2 + 2$ triangles $\\Delta_1$ are needed.\n\nFor $n = 1$ we can do that by placing three $\\Delta_1$ triangles at the corners.\n\nAssume now this proven until $n$, and prove by induction for $n+1$. A $\\Delta$ of side $n+1$ triangle placed at the top corner will use $n^2 + 2$ triangles $\\Delta_1$, according with the induction hypothesis. It remains a trapezoidal strip at the bottom, of length of the nonparallel sides\n$$n + 1 + \\frac{1}{2(n+1)} - n - \\frac{1}{2n} = 1 - \\frac{1}{2n(n+1)},$$\nand basis lengths $n + \\frac{1}{2n}$ and $n + 1 + \\frac{1}{2(n+1)}$, with $(n+1)^2 + 2 - n^2 - 2 = 2n + 1$ triangles $\\Delta_1$ available to cover it.\n\nPlace $2n+1$ triangles $\\Delta_1$ one next to another, every second one \"slid\" downwards by $\\frac{1}{2n(n+1)}$. They will cover a trapezoidal strip of exactly the dimensions of the above, since\n$$(n+1) \\cdot 1 + n \\cdot \\frac{1}{2n(n+1)} = n + 1 + \\frac{1}{2(n+1)}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a convex quadrilateral $ABCD$ with\n$$\nAB = CB \\quad \\text{and} \\quad \\angle ABC + 2\\angle CDA = \\pi\n$$\nand let $E$ be the midpoint of $AC$. Show that $\\angle CDE = \\angle BDA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let point $X$ be lying on line $BE$ such that $\\angle CXE = \\angle CDE$ ($X$ is the (other than $C$) meeting point of the circumcircle of $\\triangle CDE$ and the line $BE$).\n\nTherefore the quadrilateral $DECX$ is cyclic, so $\\angle DXE = \\angle DCE = \\pi - \\angle CDA - \\angle CAD$. But $\\angle CDA = \\frac{1}{2}(\\pi - \\angle ABC) = \\angle CAB$, so $\\angle DCE = \\pi - \\angle CAB - \\angle CAD = \\pi - \\angle BAD$, hence the quadrilateral $ABXD$ is cyclic.\n\nIt follows $\\angle BDA = \\angle BXA = \\angle CXE = \\angle CDE$.\nWe are asked to prove that $DE$ and $DB$ are isogonal conjugate, and so, since $DE$ is median in $\\triangle CDA$, that $DB$ is symmedian in that triangle.\n\nLet $\\gamma$ be the circumcircle of $\\triangle ABC$, and $\\Gamma$ be the circumcircle of $\\triangle CDA$, of center $\\Omega$. We have $\\angle C\\Omega A = 2\\angle CDA = \\pi - \\angle ABC$, hence $\\Omega \\in \\gamma$, and $C\\Omega = A\\Omega$, hence $\\Omega \\in BE$; therefore $\\Omega = \\gamma \\cap BE$. Since $B\\Omega$ is a diameter for circle $\\gamma$, it follows $\\angle B\\Omega A = \\angle BC\\Omega = \\pi/2$, therefore $BA$ and $BC$ are tangent to circle $\\Gamma$.\n\nA well-known LEMMA states that a symmedian in a triangle ($\\triangle CDA$) connects the vertex ($D$) it originates at with the intersection ($B$) of the tangents ($BA$ and $BC$) to the circumcircle ($\\Gamma$) of the triangle at the other two vertices ($A$ and $C$); for us that yields $DB$ symmedian in $\\triangle CDA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given any $n$ positive integers, and a sequence of $2^n$ integers (with terms among them), prove there exists a subsequence made of consecutive terms, such that the product of its terms is a perfect square.\nAlso show that we cannot replace $2^n$ with any lower value (therefore $2^n$ is the threshold value for this property).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the integers could well be distinct primes, this is equivalent to proving the stricter problem of, given a finite alphabet $A$ of $n$ letters $a_1, a_2, \\dots, a_n$, and a word $w$ of length $2^n$ on this alphabet, to show it contains a nonempty contiguous subword $x$ ($w = \\overline{uxv}$, where $u, v$ could be the empty word), in which each letter appears at an even number of times.\n\nLet us, further on, identify the letter $a_k$ with the element $\\mathbf{e}_k \\in \\mathbb{Z}_2^n$ given by $\\mathbf{e}_k = (0, \\dots, 0, 1, 0, \\dots, 0)$, where the 1 is at the $k$th position. Now the requirement is to find a subword such that the sum of its elements is $\\mathbf{0} = (0, 0, \\dots, 0)$.\n\nWe will apply a classical idea of Erdős. Let $w = \\overline{x_1x_2\\dots x_{2^n}}$ be the word, with $x_i \\in \\{\\mathbf{e}_1, \\mathbf{e}_2, \\dots, \\mathbf{e}_n\\}$ for all $i = 1, 2, \\dots, 2^n$. Define $\\sigma_k = \\sum_{i=1}^k x_i$ for all $k = 1, 2, \\dots, 2^n$. If any of $\\sigma_k = \\mathbf{0}$, we are done, since $\\overline{x_1\\dots x_k}$ can be taken as the subword; otherwise there must exist $1 \\le p < q \\le 2^n$ such that $\\sigma_p = \\sigma_q$, so then $\\mathbf{0} = \\sigma_q - \\sigma_p = \\sum_{i=1}^{q-p} x_{p+i}$, and we can take the subword $\\overline{x_{p+1}\\dots x_q}$.\n\nLet us notice that the result is tight. There exist words of length $2^n - 1$ without this property. We can build them inductively. Take $w_1 = \\overline{a_1}$, and build $w_{n+1} = \\overline{w_n a_{n+1} w_n}$ for all $n \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $|a - b\\sqrt{c}| < \\frac{1}{2b}$ if and only if $|a^2 - b^2c| < \\sqrt{c}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first relation also writes as $-\\frac{1}{2b} < a - b\\sqrt{c} < \\frac{1}{2b}$, i.e. $\\frac{2ab - 1}{2b^2} < \\sqrt{c} < \\frac{2ab + 1}{2b^2}$.\n\nThe second relation also writes as $-\\sqrt{c} < a^2 - b^2c < \\sqrt{c}$, or $b^2c + \\sqrt{c} - a^2 > 0$ and $b^2c - \\sqrt{c} - a^2 < 0$, i.e. $\\frac{\\sqrt{1 + 4a^2b^2} - 1}{2b^2} < \\sqrt{c} < \\frac{\\sqrt{1 + 4a^2b^2} + 1}{2b^2}$.\n\nComputing the difference between the bounds yields $0 < \\delta = \\frac{\\sqrt{1 + 4a^2b^2} \\pm 1}{2b^2} - \\frac{2ab \\pm 1}{2b^2} = \\frac{\\sqrt{1 + 4a^2b^2} - 2ab}{2b^2} = \\frac{1}{2b^2(\\sqrt{1 + 4a^2b^2} + 2ab)} < \\frac{1}{2b^2 \\cdot 4ab} = \\Delta$.\n\nClearly $\\sqrt{c} > \\frac{\\sqrt{1 + 4a^2b^2} - 1}{2b^2}$ implies $\\sqrt{c} > \\frac{2ab - 1}{2b^2}$. When $\\sqrt{c} > \\frac{2ab - 1}{2b^2}$ it follows that $4b^4c > (2ab - 1)^2$, so $4b^4c \\ge (2ab - 1)^2 + 1$. On the other hand\n$$\n\\begin{aligned}\n\\left( \\frac{\\sqrt{1 + 4a^2b^2} - 1}{2b^2} \\right)^2 < \\left( \\frac{2ab - 1}{2b^2} + \\Delta \\right)^2 &= \\frac{(2ab - 1)^2 + \\frac{2(2ab - 1)}{4ab} + \\frac{1}{16a^2b^2}}{4b^4} \\\\\n&= \\frac{(2ab - 1)^2 + 1 - \\frac{1}{2ab} + \\frac{1}{16a^2b^2}}{4b^4} < \\frac{(2ab - 1)^2 + 1}{4b^4} \\le c, \\text{ hence the required} \\\\\n&\\quad \\sqrt{c} > \\frac{\\sqrt{1 + 4a^2b^2} - 1}{2b^2}.\n\\end{aligned}\n$$\n\nClearly $\\sqrt{c} < \\frac{2ab + 1}{2b^2}$ implies that $\\sqrt{c} < \\frac{\\sqrt{1 + 4a^2b^2} + 1}{2b^2}$. When $\\sqrt{c} < \\frac{\\sqrt{1 + 4a^2b^2} + 1}{2b^2}$, assume $\\sqrt{c} \\ge \\frac{2ab + 1}{2b^2}$, hence $4b^4c \\ge (2ab + 1)^2$, therefore $4b^4c \\ge (2ab + 1)^2 + 3$ (because of the divisibility by 4).\n\nOn the other hand\n$$\n\\begin{aligned}\n\\left( \\frac{\\sqrt{1 + 4a^2b^2} + 1}{2b^2} \\right)^2 < \\left( \\frac{2ab + 1}{2b^2} + \\Delta \\right)^2 &= \\\\\n\\frac{(2ab + 1)^2 + 2(2ab + 1)/4ab + 1/16a^2b^2}{4b^4} &= \\frac{(2ab + 1)^2 + 1 + 1/2ab + 1/16a^2b^2}{4b^4} \\\\\n< \\frac{(2ab + 1)^2 + 3}{4b^4} \\le c, \\text{ hence } \\sqrt{c} > \\frac{\\sqrt{1 + 4a^2b^2} + 1}{2b^2}, \\text{ contradiction. It means our assumption was wrong, so } \\sqrt{c} < \\frac{2ab + 1}{2b^2}.\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of 2009 (distinct) points in the plane is coloured blue or red, so that on every blue-centered unit circle lie exactly two red points. Determine the largest possible number of blue points.",
    "options": [],
    "answer": "1964",
    "solution": "Suppose there are $r$ red points among some $n$ points. Since any pair of them can lie on at most two blue-centered unit circles, it means that the number $b$ of blue points can be at most $2\\binom{r}{2} = r(r-1)$. Since $b + r = n$, this leads to condition $r + r(r-1) = r^2 \\ge n$, i.e. $r \\ge \\lceil \\sqrt{n} \\rceil$, so $b \\le n - \\lceil \\sqrt{n} \\rceil$.\n\nA simple model is given by $r = \\lceil \\sqrt{n} \\rceil$ red points of coordinates $R_i(r_i, 0)$, with $0 < r_i < 2$, for all $i = 1, 2, \\dots, r$. Take $n - r$ blue points among those $r(r-1)$ given by all pairs $(i, j)$, $1 \\le i < j \\le r$, and of coordinates $B_{i,j}(x_{i,j}, b_{i,j})$ and $B'_{i,j}(x_{i,j}, -b_{i,j})$, with\n$$\nx_{i,j} = \\frac{r_i + r_j}{2} \\quad \\text{and} \\quad b_{i,j} = \\sqrt{1 - \\left(\\frac{r_i - r_j}{2}\\right)^2}.\n$$\nIt is trivial to check that all blue points $B_{i,j}$, $B'_{i,j}$ are distinct, while on unit circles of these centers lie points $R_i$ and $R_j$, and only them.\n\nThe choices made warrant that $n - r \\le r(r-1)$, since it comes to $n \\le \\lceil \\sqrt{n} \\rceil^2$, and that $(r_i - r_j)^2/4 < 1$. For the given $n = 2009$, the answer is thus that the largest possible number of blue points is $2009 - \\lceil \\sqrt{2009} \\rceil = 2009 - 45 = 1964$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all polynomials $P \\in \\mathbb{Z}[X]$ such that there exists $k \\in \\mathbb{N}^*$, so that for all primes $q$, $P(q)$ has at most $k$ distinct prime divisors.",
    "options": [],
    "answer": "All polynomials of the form c·X^m with c a nonzero integer and m a nonnegative integer.",
    "solution": "Clearly a constant non-null polynomial $P$ fulfills the requirements, so we shall assume in the sequel $\\deg P \\ge 1$. Before further proceeding with the solution, we state the following \"folklore\" preliminaries.\n(1) **Dirichlet's Theorem.** The arithmetic progression $(an + b)_{n \\in \\mathbb{N}}$ with $a \\in \\mathbb{N}^*$, $b \\in \\mathbb{Z}^*$, $(a, b) = 1$, contains infinitely many primes.\n(2) **\"Fundamental\" property of integer polynomials.** For any $f \\in \\mathbb{Z}[X]$ and any $a, b \\in \\mathbb{Z}$ we have $a - b \\mid f(a) - f(b)$.\n**Claim.** $P(0) = 0$. We proceed by contradiction, thus assume $P(0) \\ne 0$.\nWe will prove by induction on $k$ the following statement\n*For any $k \\in \\mathbb{N}^*$ there exists a prime $p > |P(0)|$ such that $P(p)$ has at least $k$ distinct prime factors.*\nFor $k=1$, there exists a prime $p_1 > |P(0)|$ such that $|P(p_1)| \\ne 1$, otherwise for all such primes $p$ we would have $P(p)^2 = 1$ and thus the polynomial $P(X)^2 - 1$ has infinitely many roots, thus it is the null polynomial, hence $P(X) \\equiv 1 \\pmod{p}$ or $P(X) \\equiv -1 \\pmod{p}$, contradicting $\\deg P \\ge 1$. Therefore $P(p_1)$ has at least one prime factor, so the statement is proven for $k=1$.\nLet $p_k$ be a prime with the property $p_k > |P(0)|$ and $P(p_k)$ has at least $k$ distinct prime factors. Let us notice that $(p_k, P(p_k)) = 1$, since from the \"fundamental\" property we have $P(p_k) \\equiv P(0) \\pmod{p_k}$ and since $p_k > |P(0)|$ and it is a prime, the conclusion follows. The arithmetic progression $(nP(p_k)^2 + p_k)_{n \\in \\mathbb{N}}$ (according to Dirichlet's Theorem) contains infinitely many primes. Let $p_{k+1} = sP(p_k)^2 + p_k$ be such a prime. Then $P(p_{k+1}) \\equiv P(p_k) \\pmod{P(p_k)^2}$, thus there exists $t$ such that $P(p_{k+1}) = P(p_k)(1 + tP(p_k))$. From this it is obvious that $P(p_{k+1})$ has at least one more prime factor than $P(p_k)$. Since $P(p_k)$ had at least $k$ distinct prime factors, it follows that $P(p_{k+1})$ has at least $k+1$ prime factors.\nThus the statement is proven and we get that the assumption $P(0) \\ne 0$ is false. From $P(0) = 0$ we deduce there exists $m$ so that $P(X) = X^m Q(X)$ with $Q(0) \\ne 0$. If $Q$ would be nonconstant, arguing the same as in the above, we again obtain a contradiction.\nWe conclude the only solutions are $P(X) = cX^m$, for some $m \\in \\mathbb{N}$ and $c \\in \\mathbb{Z}^*$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37637,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of a $(n^2 - n + 1) \\times (n^2 - n + 1)$ matrix are coloured using $n$ colours. A colour is called *dominant* on a row (or a column) if there are at least $n$ cells of this colour on that row (or column). A cell is called *extremal* if its colour is dominant both on its row, and its column. Find all $n \\ge 2$ for which there is a colouring with no extremal cells.",
    "options": [],
    "answer": "All integers n ≥ 2",
    "solution": "The answer is that there exist such colourings for $n = 2$ (almost trivial), and $n = 3$ (following a detailed and tedious case analysis). For $n = 4$ one must exhibit a colouring with no extremal cells (again a very tedious labor). The colourings with no extremal cells for $n \\ge 5$ are built inductively, using a beautifully symmetric pattern based on the model for $n - 1$ (this being the elegant combinatorial part of the problem).\n\nA detailed proof for this problem can be found in no. 4/2006, pages 22-25, of the on-line magazine Mathematical Reflections\nhttp://reflections.awesomemath.org/2006_4/2006_4_solutions.pdf.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that given any two permutations $\\sigma, \\tau \\in S_n$, there exists some function $f : \\{1, \\dots, n\\} \\to \\{-1, 1\\}$ such that we simultaneously have, for any indices $1 \\le i \\le j \\le n$,\n$$\n\\left| \\sum_{k=i}^{j} f(\\sigma(k)) \\right| \\le 2 \\quad \\text{and} \\quad \\left| \\sum_{k=i}^{j} f(\\tau(k)) \\right| \\le 2.\n$$\nDan Schwarz",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will first prove it for the even case, $n = 2m$. Consider the graph whose set of vertices is $\\{1, \\dots, 2m\\}$, and with red edges between $\\sigma(2k-1)$ and $\\sigma(2k)$, and blue edges between $\\tau(2k-1)$ and $\\tau(2k)$, where $1 \\le k \\le m$ (notice that multiple edges may occur).\n\nClearly, each vertex is incident with one red edge and one blue edge, hence all vertex degrees are equal to 2. Therefore the graph is the union of disjoint cycles. Moreover, each such cycle is of even length, since its edges must be of alternating colours (the graph is thus bipartite).\n\nDefine $f$ alternately taking values $-1$ and $1$ along each of the cycles. Then the largest value any of the moduli can take is 2, when $i$ is even and $j$ is odd, since on consecutive pairs of odd/even indices the sum of the values of $f$ is zero, along either $\\sigma$ or $\\tau$.\n\nWhen $n$ is odd, prolong $\\sigma(n+1) = \\tau(n+1) = n+1$ to achieve the case treated in the above.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the least real number $c$, such that for any integer $n \\ge 1$ and any positive real numbers $a_1, a_2, \\dots, a_n$, the following holds\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} < c \\sum_{k=1}^{n} a_k.\n$$",
    "options": [],
    "answer": "2",
    "solution": "We claim $c_{\\min} = 2$.\nTaking $a_j = \\frac{1}{j}$ for $j = 1, 2, \\dots, n$, we have\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} = \\sum_{k=1}^{n} \\frac{k}{1 + 2 + \\dots + k} = 2 \\sum_{k=1}^{n} \\frac{1}{k+1}, \\text{ while } c \\sum_{k=1}^{n} a_k =\n$$\n\nWe will now prove that $c = 2$ is suitable. From the Cauchy-Schwartz inequality,\n$$\n\\frac{k^2(k+1)^2}{4} = \\left(\\sum_{j=1}^{k} j\\right)^2 \\le \\left(\\sum_{j=1}^{k} j^2 a_j\\right) \\left(\\sum_{j=1}^{k} \\frac{1}{a_j}\\right), \\text{ hence} \\\\\n\\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} \\le \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j.\n$$\nTherefore\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} &\\le \\sum_{k=1}^{n} \\left( \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j \\right) = \\\\\n&= \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{4}{k(k+1)^2} \\right) = 2 \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{2k}{k^2(k+1)^2} \\right) < \\\\\n&< 2 \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{2k+1}{k^2(k+1)^2} \\right). \\text{ But} \\\\\n\\sum_{k=j}^{n} \\frac{2k+1}{k^2(k+1)^2} &= \\sum_{k=j}^{n} \\left( \\frac{1}{k^2} - \\frac{1}{(k+1)^2} \\right) = \\frac{1}{j^2} - \\frac{1}{(n+1)^2} < \\frac{1}{j^2}, \\\\\n\\text{hence } \\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} &< 2 \\sum_{k=1}^{n} a_k.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integer numbers $n \\ge 3$ such that the regular $n$-gon can be decomposed into isosceles triangles by noncrossing diagonals.",
    "options": [],
    "answer": "All n of the form n = 2^r(2^s + 1), where r and s are nonnegative integers not both zero.",
    "solution": "The required numbers are of the form $n = 2^r(2^s+1)$, where $r$ and $s$ are nonnegative integer numbers which do not vanish simultaneously. Clearly, any such $n$ works.\n\nTo establish the converse, let $K$ be a regular $n$-gon, $n \\ge 4$, which can be decomposed into isosceles triangles by noncrossing diagonals. Begin by noticing that each edge $e$ of $K$ must be an edge of a unique isosceles triangle $T_e$ in the decomposition. Two cases are possible: either $e$ is opposite the apex of $T_e$ or $e$ and one of the adjacent edges of $K$ are the edges of $T_e$ issuing from the apex. (Since $n \\ge 4$, $T_e$ cannot be equilateral, so the apex is well defined.)\n\nIf $n$ is even, no vertex of $K$ lies on the perpendicular bisector of an edge of $K$, so the first case is ruled out. Consequently, the decomposition must contain exactly one of the two *bracelets* of $n/2$ isosceles triangles clipped off by *short* diagonals joining consecutive vertices of $K$ of likewise parity. These short diagonals are the edges of a regular $n/2$-gon which is also decomposed into isosceles triangles by noncrossing diagonals and the conclusion follows by induction.\n\nIf $n$ is odd, then $K$ has a *unique* edge $e$ opposite the apex of $T_e$: Since $n$ is odd and each short diagonal clips off two edges of $K$, at least one such $e$ exists. The apex of $T_e$ lies on the perpendicular bisector of $e$, so it must be the vertex of $K$ opposite $e$. Uniqueness of $e$ should now be clear: were there another such $e'$, the interiors of $T_e$ and $T_{e'}$ would overlap. Consequently, $e$ is unique and $K$ splits into $T_e$ and two polygons $L$ and $L'$ which are reflections of one another in the perpendicular bisector of $e$.\n\nTo complete the proof, it is sufficient to show that the number of vertices of $L$ is one plus a power of 2. Begin by noticing that $L$ inherits by restriction a decomposition into isosceles triangles by noncrossing diagonals. Let $x_0, \\dots, x_m$ be a circular labelling of the vertices of $L$ around the boundary, where $x_0$ is the vertex of $K$ opposite $e$ and $x_m$ is a vertex of $e$. Since $\\text{dist}(x_i, x_j) < \\text{dist}(x_0, x_m)$ if $\\{i, j\\} \\neq \\{0, m\\}$, it follows that $x_0x_m$ is an edge of an isosceles triangle with apex at some $x_k$, $0 < k < m$. Notice that $\\text{dist}(x_0, x_i) < \\text{dist}(x_i, x_m)$ if $0 < i < m/2$ and $\\text{dist}(x_0, x_i) > \\text{dist}(x_i, x_m)$ if $m/2 < i < m$, to deduce that $m$ must be even, $k = m/2$, and $L$ splits into an isosceles triangle, $x_0x_{m/2}x_m$, and two polygons, $x_0 \\cdots x_{m/2}$ and $x_{m/2} \\cdots x_m$, which are reflections of one another in the perpendicular bisector of the segment $x_0x_m$. Now we are essentially back in the situation that arose above. Repeat the same argument verbatim to infer that $m/2$ must be even and so on all the way down to conclude that $m$ must be a power of 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D := \\{1, 2, \\dots, n\\} \\times \\{1, 2, \\dots, n\\}$. Prove there exists a set $S \\subset D$ with $|S| \\ge \\lfloor \\frac{3}{5}n(n+1) \\rfloor$, such that for any $(x_1, y_1), (x_2, y_2) \\in S$ we have $(x_1 + x_2, y_1 + y_2) \\notin S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to find a weaker bound of $|S| = n\\lfloor\\frac{1}{2}n\\rfloor$ by taking $S = \\{(x, y) \\in D \\mid x > n/2\\}$. To find the bound asked, we need look at the diagonals of the tableau!\n![](attached_image_1.png)\n\nDiagram for the selection $\\bullet$ of $S_u$ (exact for $n = 13$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that a sequence $(\\varepsilon_n)_{n \\in \\mathbb{N}}$ of plus and minus ones is periodic with period a power of $2$, if and only if $\\varepsilon_n = (-1)^{P(n)}$, $n \\in \\mathbb{N}$, where $P$ is an integer-valued polynomial with rational coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A polynomial $P$ of degree at most $k$ with complex coefficients is integer-valued if and only if\n$$\nP = \\sum_{j=0}^{k} a_j \\binom{X}{j} = \\sum_{j=0}^{k} \\frac{a_j}{j!} X(X-1) \\cdots (X-j+1),\n$$\nwhere the $a_j$ are all integer numbers, so its coefficients are rational.\n\nWe show that for such a $P$, the sequence $((-1)^{P(n)})_{n \\in \\mathbb{N}}$ is periodic with period $2^r$, where $r = \\min\\{s : 2^s > k\\}$. To this end, it suffices to show that if $j < 2^s$ and $m$ is integer, then $\\binom{m}{j} \\equiv \\binom{m+2^s}{j} \\mod 2$. These are the coefficients of $X^j$ in the expansions of $(1+X)^m$ and $(1+X)^{m+2^s}$, respectively. The congruence $(1+X)^{2^s} \\equiv 1+X^{2^s} \\mod 2$ follows easily by induction on $s$. Hence $(1+X)^{m+2^s} = (1+X)^m (1+X^{2^s}) \\mod 2$. Since $j$ is less than $2^s$, it is immediate that the coefficients of $X^j$ in $(1+X)^m$ and $(1+X)^{m+2^s}$ have the same parity.\n\nConversely, let $\\alpha_0, \\dots, \\alpha_{2^r-1}$ be arbitrary integers, and let $\\beta_0, \\dots, \\beta_{2^r-1}$ be the solution to the lower triangular system of linear equations\n$$\n\\sum_{i=0}^{2^r-1} \\beta_i \\binom{j}{i} = \\alpha_j, \\quad j = 0, \\dots, 2^r-1.\n$$\nThen\n$$\n\\sum_{i=0}^{2^r-1} \\beta_i \\binom{X}{i} = \\sum_{i=0}^{2^r-1} \\frac{\\beta_i}{i!} X(X-1)\\cdots(X-i+1)\n$$\nrealizes the sequence $(-1)^{\\alpha_j}$, $j = 0, \\dots, 2^r - 1$, and its extension with period $2^r$.\nGiven an integer-valued polynomial $P$ with rational coefficients, we show that the sequence $((-1)^{P(n)})_{n \\in \\mathbb{N}}$ is periodic with period a power of $2$. Clearly, it is sufficient to show that, for some non-negative integer $s$, the integer numbers $P(n)$ and $P(n + 2^s)$ both have the same parity, whatever $n \\in \\mathbb{N}$. To this end, consider a positive integer $m$ such that $Q = mP$ is a polynomial with integral coefficients (e.g., let $m$ be the least common multiple of the denominators of the coefficients of $P$ when written in lowest terms), let $2^r$ be the highest power of $2$ dividing $m$, and let $s$ be an integer greater than $r$. Write $m = 2^r(2m' + 1)$ and fix a positive integer $n$. Since $2^s$ divides the difference\n$$\nQ(n + 2^s) - Q(n) = m(P(n + 2^s) - P(n)) = 2^r(2m' + 1)(P(n + 2^s) - P(n))\n$$\nand $s > r$, the conclusion follows.\n\nConversely, given a sequence $(\\varepsilon_n)_{n \\in \\mathbb{N}}$ of plus and minus ones which is periodic with period $2^r$, let\n$$\na_k = \\begin{cases} 0, & \\text{if } \\varepsilon_k = 1, \\\\ 1, & \\text{if } \\varepsilon_k = -1, \\end{cases} \\quad k = 0, \\dots, 2^r - 1,\n$$\nand consider the polynomial (Lagrange)\n$$\n\\begin{align*} P &= \\sum_{k=0}^{2^r-1} a_k \\prod_{j \\neq k} \\frac{X-j}{k-j} \\\\ &= \\sum_{k=0}^{2^r-1} (-1)^{k+1} a_k \\left( \\frac{1}{k!} \\prod_{0 \\le j < k} (X-j) \\right) \\left( \\frac{1}{(2^r-k-1)!} \\prod_{k<j \\le 2^r-1} (X-j) \\right) \\\\ &= \\sum_{k=0}^{2^r-1} (-1)^{k+1} a_k \\binom{X}{k} \\binom{X-k-1}{2^r-k-1}. \\end{align*}\n$$\n\nClearly, $P$ has rational coefficients, is integer-valued, and $P(k) = a_k$, $k = 0, \\dots, 2^r - 1$, so $\\varepsilon_k = (-1)^{P(k)}$, $k = 0, \\dots, 2^r - 1$. To prove that the latter extends to all of $\\mathbb{N}$, it is sufficient to show that $P(n)$ and $P(n + 2^r)$ both have the same parity, whatever $n \\in \\mathbb{N}$. This amounts to showing that if $j, m \\in \\mathbb{N}$ and $j < 2^r$, then $\\binom{m}{j}$ and $\\binom{m+2^r}{j}$ have the same parity. Recalling that $\\binom{2^r}{i}$, $i = 1, \\dots, 2^r - 1$, are all even, this follows for instance from the identity\n$$\n\\binom{m+2^r}{j} = \\sum_{i=0}^{j} \\binom{m-i}{j-i} \\binom{2^r}{i};\n$$\nor, which is actually the same, from the argument in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram of area $1$, and $AD \\le AB \\le BD < AC$. Prove that the parallelogram can be covered by a rectangle of area $\\sqrt{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37644,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\sqrt{(x^2 + x y + y^2)(z^2 + z t + t^2)} + \\sqrt{(y^2 - y z + z^2)(x^2 - x t + t^2)} \\geq (x + z)(y + t).\n$$\n\nFind the cases of equality.",
    "options": [],
    "answer": "Equality holds when x = t and y = z.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a$, $b$, $c$ are positive real numbers such that $abc = 1$, then\n$$\n(1 + a^3)(1 + b^3)(1 + c^3) \\ge 2 \\left(1 + \\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c}\\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that a non-equilateral triangle has an angle bisector which is more than $\\frac{\\sqrt{3}}{2}$ times larger than its opposite side, and one which is less than $\\frac{\\sqrt{3}}{2}$ times larger than its opposite side.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37647,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$, $b$, $c$ are complex numbers of modulus $1$, prove that\n$$\n|a - b|^2 + |a - c|^2 - |b - c|^2 \\ge -1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in (0, 1)$ and $x, y, z \\in (0, \\infty)$ be such that $a = (bc)^x$, $b = (ca)^y$, $c = (ab)^z$. Show that\n$$\n\\sum \\frac{1}{x+y+2} \\le 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, p$ be integers larger than $2$. Find the least positive integer $n$, such that every $n$-element subset of $\\{1, 2, 3, \\dots, pm\\}$ contains two numbers of sum divisible by $p$.",
    "options": [],
    "answer": "If p is odd: n = m*(p−1)/2 + 2. If p is even: n = m*(p/2 − 1) + 3.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence given by $x_1 = 1$ and $x_{n+1} = 1 + \\frac{n}{x_n}$, for $n \\in \\mathbb{N}^*$. Find all $n$ such that $x_n$ is an integer.",
    "options": [],
    "answer": "n = 1, 2, 3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every matrix $A \\in \\mathcal{M}_2(\\mathbb{R})$ can be written in the form $A = X^3 + Y^3$, where $X, Y \\in \\mathcal{M}_2(\\mathbb{R})$, $XY = YX$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [a, b] \\to \\mathbb{R}$ be a continuous function with the property that for every $x \\in [a, b)$ there exists $y \\in (x, b]$ such that $f(x) \\le f(y)$. Find $\\max f(x)$.",
    "options": [],
    "answer": "f(b)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square, and the points $M \\in [BC]$, $N \\in [CD]$, $P \\in [DA]$, such that\n$$\n\\angle(\\overrightarrow{AB}, \\overrightarrow{AM}) = x, \\quad \\angle(\\overrightarrow{BC}, \\overrightarrow{MN}) = 2x, \\quad \\angle(\\overrightarrow{CD}, \\overrightarrow{NP}) = 3x.\n$$\n\ni) Show that, for any $x \\in [0, \\pi/8]$, such a configuration uniquely exists, and $P$ ranges over the entire segment $[DA]$;\nii) Determine the number of angles $x \\in [0, \\pi/8]$ for which $\\angle(\\overrightarrow{DA}, \\overrightarrow{PB}) = 4x$.",
    "options": [],
    "answer": "1",
    "solution": "i) Assume $AB = 1$, and denote $t = \\tan x$. Since\n$$\n1 = \\tan \\frac{\\pi}{4} = \\frac{2 \\tan \\frac{\\pi}{8}}{1 - \\tan^2 \\frac{\\pi}{8}}\n$$\nit follows $\\tan \\frac{\\pi}{8}$ is the positive root of $t^2 + 2t - 1 = 0$, and so $\\tan \\frac{\\pi}{8} = \\sqrt{2} - 1$.\n\nNow $BM = t$, hence $M$ takes once and only once all values of interval $[0, \\sqrt{2}-1]$, while $CN = (1-t) \\tan 2x = \\frac{2t}{1+t}$, therefore $N$ takes all values of the interval $[0, 2 - \\sqrt{2}]$. Lastly, we also get $DP = \\left(1 - \\frac{2t}{1+t}\\right) \\tan 3x = \\frac{1-t}{1+t} \\cdot \\frac{t(3-t^2)}{1-3t^2} \\ge 0,$\nwhence $PA = 1 - \\frac{1-t}{1+t} \\cdot \\frac{t(3-t^2)}{1-3t^2} = \\frac{(t^2+2t-1)(t^2+1)}{(t+1)(3t^2-1)} \\ge 0$, hence $P$ takes **all** values of the full interval $[0, 1]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, let $D$ be the point where the incircle of the triangle $ABC$ touches the side $BC$. A circle through the vertices $B$ and $C$ is tangent at point $E$ to the incircle of the triangle $ABC$. Show that the line $DE$ passes through the excentre of the triangle $ABC$ corresponding to the vertex $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $I$ be the incentre of the triangle $ABC$, let $I_A$ be the excentre corresponding to the vertex $A$, and notice that the vertices $B$ and $C$ both lie on the circle of diameter $II_A$. The line $DI_A$ meets again the latter circle at point $K$, and the lines $BC$ and $IK$ meet at point $L$ (unless $AB = AC$ in which case the conclusion is obvious). Notice that the line $BC$ is the radical axis of the circles $BEC$ and $BIC$ to deduce that $LB \\cdot LC = LI \\cdot LK$. On the other hand, $LI \\cdot LK = LD^2$, for $K$ is the perpendicular foot dropped from the right-angled vertex $D$ of the triangle $DIL$. Consequently, the point $L$ is the radical centre of the following three circles: the incircle of the triangle $ABC$, the circle $BEC$, and the circle $BIC$. Since the common tangent at $E$ of the first two circles is their radical axis, it must pass through $L$. It follows that $E$ is the reflection of $D$ across the line $IL$, so the lines $DE$ and $IL$ are perpendicular and we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, with $AB \\le BC$. Let $M$ be the midpoint of the side $BC$, $N$ be the foot of the angle bisector of $\\angle B$, $O$ the meeting point of the straight lines $AM$ and $AB$, and $P$ the meeting point of the straight lines $CO$ and $AB$. Show that $4AP \\le AB + BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every continuous function $f : \\mathbb{R} \\to \\mathbb{R}$ with the property\n$$\nf(x) - e^{f(x)} = x - e^x, \\text{ for every } x \\in \\mathbb{R},\n$$\nis differentiable.\nDo there exist functions with the above property, having a finite set of points of continuity?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_3(\\mathbb{C})$ be such that $\\text{tr}(A^2) = \\text{tr}(A^*)$. Show that there exist $\\alpha, \\beta \\in \\mathbb{C}$ such that the matrix $(A + \\alpha I_3)^3 + \\beta I_3$ is nilpotent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ with the property that for every $a \\in \\mathbb{R}$ there exists $b \\in \\mathbb{R}$ such that, for every $x \\in \\mathbb{R}$,\n$$\nf(f(x) + a) = f(x) + b.\n$$",
    "options": [],
    "answer": "All continuous solutions are either constant functions f(x) = c for some real c, or affine functions of the form f(x) = x + c for some real c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be an integrable function. If $f(1) = 0$ and $f$ is differentiable in $1$, prove that\n$$\n\\lim_{n \\to \\infty} n^2 \\int_0^1 x^n f(x) dx = -f'(1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us make the substitution $x = 1 - \\frac{t}{n}$, so that as $x$ goes from $0$ to $1$, $t$ goes from $n(1-0) = n$ to $n(1-1) = 0$ (i.e., $t$ goes from $n$ to $0$). The differential $dx = -\\frac{dt}{n}$.\n\nSo,\n$$\n\\int_0^1 x^n f(x) dx = \\int_{x=0}^{x=1} x^n f(x) dx = \\int_{t=n}^{t=0} \\left(1 - \\frac{t}{n}\\right)^n f\\left(1 - \\frac{t}{n}\\right) \\left(-\\frac{dt}{n}\\right)\n$$\nwhich is\n$$\n= \\int_{t=0}^{t=n} \\left(1 - \\frac{t}{n}\\right)^n f\\left(1 - \\frac{t}{n}\\right) \\frac{dt}{n}\n$$\nNow,\n$$\n\\left(1 - \\frac{t}{n}\\right)^n = e^{n \\ln(1 - t/n)} \\approx e^{-t}\n$$\nas $n \\to \\infty$, by the standard limit $\\lim_{n \\to \\infty} \\left(1 - \\frac{t}{n}\\right)^n = e^{-t}$.\n\nAlso, $f\\left(1 - \\frac{t}{n}\\right) \\to f(1)$ as $n \\to \\infty$, but since $f(1) = 0$, we need a more precise expansion. Since $f$ is differentiable at $1$ and $f(1) = 0$,\n$$\nf\\left(1 - \\frac{t}{n}\\right) = f(1) - f'(1) \\frac{t}{n} + o\\left(\\frac{t}{n}\\right) = -f'(1) \\frac{t}{n} + o\\left(\\frac{t}{n}\\right)\n$$\nSo,\n$$\nn^2 \\int_0^1 x^n f(x) dx \\approx n^2 \\int_{t=0}^{t=n} e^{-t} \\left(-f'(1) \\frac{t}{n}\\right) \\frac{dt}{n}\n$$\n$$\n= n^2 \\cdot \\left(-f'(1)\\right) \\int_{t=0}^{t=n} e^{-t} \\frac{t}{n} \\frac{dt}{n}\n$$\n$$\n= -f'(1) n^2 \\int_{t=0}^{t=n} e^{-t} \\frac{t}{n^2} dt\n$$\n$$\n= -f'(1) \\int_{t=0}^{t=n} t e^{-t} dt\n$$\nAs $n \\to \\infty$, $\\int_{0}^{n} t e^{-t} dt \\to \\int_{0}^{\\infty} t e^{-t} dt = 1$ (since $\\int_{0}^{\\infty} t e^{-t} dt = 1$).\n\nTherefore,\n$$\n\\lim_{n \\to \\infty} n^2 \\int_0^1 x^n f(x) dx = -f'(1).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a continuous function and $n \\ge 3$ be an integer. Show that one can find $n$ numbers $a_1, a_2, \\dots, a_n$ in the interval $[0, 1]$, in arithmetic progression, such that\n$$\n\\int_{0}^{1} f(x) \\, dx = \\frac{1}{n} \\sum_{k=1}^{n} f(a_k).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a positive number and $f : [0, 1] \\to \\mathbb{R}$ be a continuous concave function, such that $f(0) = 1$. Show that\n$$\n(p+1) \\int_{0}^{1} x^{2p} f(x) \\, dx + \\frac{2p-1}{8p+4} \\le \\left( \\int_{0}^{1} f(x) \\, dx \\right)^{2}\n$$\nand find when the equality is possible.",
    "options": [],
    "answer": "Equality holds if and only if f(x) = 1 − x on [0, 1].",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and $m$ be the exponent of $2$ in the decomposition of $n!$ into prime factors. Show that the symmetric group $S_n$ has at least three subgroups of order $2^m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n \\ge 3$ and let $m$ be the exponent of $2$ in the prime factorization of $n!$. Recall that the order of $S_n$ is $n!$, so $2^m$ divides $|S_n|$.\n\nWe will exhibit at least three subgroups of $S_n$ of order $2^m$.\n\nFirst, consider the Sylow $2$-subgroups of $S_n$. By Sylow's theorems, the number of Sylow $2$-subgroups is congruent to $1$ modulo $2$ and divides $n! / 2^m$. In particular, there is at least one Sylow $2$-subgroup of order $2^m$.\n\nWe will construct three explicit subgroups of order $2^m$:\n\n1. The subgroup $H_1$ consisting of all permutations of $S_n$ that permute the first $k$ elements, where $k$ is the largest power of $2$ less than or equal to $n$. That is, $H_1$ is isomorphic to $S_k$, and the $2$-part of $k!$ is maximized. However, to ensure order $2^m$, we need to consider the actual Sylow $2$-subgroups.\n\n2. The subgroup $H_2$ consisting of all permutations that permute a different set of $k$ elements (for example, the last $k$ elements), again forming a subgroup isomorphic to $S_k$.\n\n3. The subgroup $H_3$ can be constructed as follows: Consider the direct product of $S_2 \\times S_2 \\times \\cdots \\times S_2$ ($\\lfloor n/2 \\rfloor$ times), embedded in $S_n$ as the group generated by disjoint transpositions. This subgroup is an elementary abelian $2$-group of order $2^{\\lfloor n/2 \\rfloor}$, but for $n \\ge 3$, the Sylow $2$-subgroups are not all conjugate to this one, and the actual Sylow $2$-subgroups are more complicated, but there are at least three such subgroups by the following argument.\n\nAlternatively, note that the number of Sylow $2$-subgroups is greater than $1$ for $n \\ge 3$ (since $n! / 2^m$ is even for $n \\ge 3$), and by Sylow's theorem, the number is congruent to $1$ modulo $2$ and divides $n! / 2^m$. Therefore, there are at least three Sylow $2$-subgroups of $S_n$ of order $2^m$.\n\nThus, $S_n$ has at least three subgroups of order $2^m$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be two elements of a ring $R$, such that $xyx \\ne 0$ and $(xy)^n = 0$, where $n \\ge 2$ is an integer. Show that there exists an element $z$ in $R$, such that $xz \\ne 0$ and $xyxz = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $z = (yx)^{n-1}$. Then\n\n$xyxz = x y (yx)^{n-1} z = x (y x)^{n} z = x \\cdot 0 \\cdot z = 0$.\n\nNow, suppose $xz = 0$. Then $x (yx)^{n-1} = 0$. But $xyx \\ne 0$, and $(xy)^n = 0$ implies $(xy)^{n-1} x y = 0$. If $x (yx)^{n-1} = 0$, then $xyx = 0$ (since $n \\ge 2$), which contradicts the hypothesis. Therefore, $xz \\ne 0$.\n\nThus, such $z$ exists.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37664,
    "subject": "Mathematics (Multi-modal)",
    "question": "All sides and diagonals of a convex $n$-gon, $n \\ge 3$, are coloured one of two colours. Show that there exist $\\lfloor (n+1)/3 \\rfloor$ pairwise disjoint monochromatic segments. (Two segments are disjoint if they do not share an endpoint or an interior point.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If all sides are monochromatic, then the assertion is clearly true. Otherwise, delete a vertex incident with two sides of different colours together with its neighbours, delete all sides and diagonals incident with these three vertices and apply induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer number $n \\ge 2$, a positive real number $A$, and $n+1$ distinct points in the plane, $X_0, X_1, \\dots, X_n$, show that the number of triangles $X_0X_iX_j$ of area $A$ does not exceed $4n\\sqrt{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that for some integer $n \\ge 2$, there exist $n+1$ distinct points in the plane, $X_0, X_1, \\dots, X_n$, such that the number of triangles $X_0 X_i X_j$ of area $A$ be greater than $4n\\sqrt{n}$. Choose the minimal such $n$, notice that $n \\ge 4$, and let $G$ be the graph whose vertices are $X_1, \\dots, X_n$ and whose edges are the $X_i X_j$ such that area $X_0 X_i X_j = A$. Then every vertex $X_i$ of $G$ is adjacent to at least $\\lfloor 4\\sqrt{n} \\rfloor$ other vertices, since otherwise, removing $X_i$ would reduce the number of triangles by at most $4\\sqrt{n}$, and we would be left with a configuration of $n$ distinct points such that the number of triangles $X_0 X_j X_k$ of area $A$ is at least $4n\\sqrt{n} - 4\\sqrt{n} > 4(n-1)\\sqrt{n}-1$, contradicting the minimal choice of $n$. Consequently, for each $X_i$, there are at least $\\lfloor 4\\sqrt{n} \\rfloor$ points $X_j$ such that the triangle $X_0 X_i X_j$ has area $A$. These points lie on two parallel lines to the line $X_0 X_i$. One of these linear sets of points, say $S_i$, contains at least $\\frac{1}{2}\\lfloor 4\\sqrt{n} \\rfloor$ points. Notice that at least $n/2$ of the $S_i$ are pairwise distinct. Without loss of generality, we may (and will) assume that the first $n/2$ of the $S_i$ are among these. Finally, recall that $n \\ge 4$, so $\\sqrt{n} \\le n/2$, and consider the points $X_j$ on the first $\\lfloor \\sqrt{n} \\rfloor$ lines $S_i$, $i = 1, \\dots, \\lfloor \\sqrt{n} \\rfloor$, to get\n$$\n\\begin{aligned}\nn \\ge |S_1 \\cup \\dots \\cup S_{\\lfloor \\sqrt{n} \\rfloor}| &\\ge \\sum_{i=1}^{\\lfloor \\sqrt{n} \\rfloor} |S_i| - \\sum_{1 \\le i < j \\le \\lfloor \\sqrt{n} \\rfloor} |S_i \\cap S_j| \\\\\n&\\ge \\frac{1}{2} \\lfloor \\sqrt{n} \\rfloor \\lfloor 4\\sqrt{n} \\rfloor - \\binom{\\lfloor \\sqrt{n} \\rfloor}{2},\n\\end{aligned}\n$$\nwhich is false for $n \\ge 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest constant $K \\ge 0$ such that for any $0 \\le k \\le K$, and for any non-negative real numbers $a, b, c$, satisfying $a^2 + b^2 + c^2 + kabc = k + 3$, to have $a + b + c \\le 3$.",
    "options": [],
    "answer": "3/2",
    "solution": "Let us work from first principles. Whenever (at least) one variable is zero, say $c$, it follows $a^2 + b^2 = k + 3$, hence $a + b + c \\le \\sqrt{2(k+3)} \\le 3$ for $k \\le 3/2$, with equality holding for $a = b = 3/2$. We thus have a starting tentative limiting bound of $K = 3/2$.\n\nLet also notice that $a = b = c = 1$ checks for any value of $k$, while providing a maximal admissible value for $a + b + c = 3$.\n\nAssume $\\sigma = a + b + c > 3$. Notice that then (for some $k > 0$) $k(1 - abc) = a^2 + b^2 + c^2 - 3 \\ge \\frac{1}{3}(a+b+c)^2 - 3 = \\frac{1}{3}\\sigma^2 - 3 > 0$, by Cauchy-Schwarz, and so $abc < 1$. Now is the time for the **key step**. Since $a + b + c > 3$, assuming some ordering of our variables, say $0 \\le a \\le b \\le c$, it will follow $c > 1$, and so we can compute\n\n$$\n\\begin{align*} \nk &= \\frac{a^2 + b^2 + c^2 - 3}{1 - abc} \\\n&= \\frac{(a+b)^2 + c^2 - 3 - 2ab}{1 - abc} \\\n&= \\frac{(\\sigma-c)^2 + c^2 - 3 - 2ab}{1 - abc} \\\n&= \\frac{c(2c^2 - 2\\sigma c + \\sigma^2 - 3) - 2 + 2(1 - abc)}{c(1 - abc)} \\\n&= \\frac{2}{c} + \\frac{2c^3 - 2\\sigma c^2 + (\\sigma^2 - 3)c - 2}{c(1 - abc)} \n\\end{align*}\n$$\nDenote $f(c) = 2c^3 - 2\\sigma c^2 + (\\sigma^2 - 3)c - 2$. As $(x-y)(f(x)-f(y)) = (x-y)^2 \\left( \\frac{3}{2} \\left( x+y-\\frac{2}{3}\\sigma \\right)^2 + \\frac{1}{2}(x-y)^2 + \\frac{1}{3}(\\sigma^2-9) \\right) \\ge 0$, $f$ is increasing; and as $f(1) = (\\sigma+1)(\\sigma-3) > 0$, it follows $f(c) > 0$. Therefore the minimal value for $k$ is reached when $ab = 0$, for $a = 0$, whence $k \\ge b^2 + c^2 - 3 \\ge \\frac{1}{2}(b+c)^2 - 3 = \\frac{1}{2}\\sigma^2 - 3 > \\frac{3}{2}$.\n\n\nDefine\n$$\nL(a, b, c) = a + b + c - \\lambda(a^2 + b^2 + c^2 + kabc - k - 3).\n$$\n\nThe analysis of the values on the border of the domain of $L$ has been done above, leading to $K \\le 3/2$ (we could now simplify our work to just $k = 3/2$, but it is enlightening to see it in full generality). The system of partial derivatives is\n$$\n\\left\\{\n\\begin{array}{l}\n\\displaystyle \\frac{\\partial L}{\\partial a} = 1 - \\lambda(2a + kbc) \\\\[1em]\n\\displaystyle \\frac{\\partial L}{\\partial b} = 1 - \\lambda(2b + kca) \\\\[1em]\n\\displaystyle \\frac{\\partial L}{\\partial c} = 1 - \\lambda(2c + kab)\n\\end{array}\n\\right.\n$$\n\nEqualing the partial derivatives to zero forbids $\\lambda = 0$. Then, from pairwise equalities, we get\n$$\n\\lambda(a - b)(2 - kc) = \\lambda(b - c)(2 - ka) = \\lambda(c - a)(2 - kb) = 0.\n$$\nOne possibility is $a = b = c = x$, thus (from the constraint) $3x^2 + kx^3 = k + 3$, or\n$(x-1)(kx^2+(k+3)x+(k+3)) = 0$, with only non-negative real solution $x = 1$, since\nthe coefficients of the quadratic factor are non-negative. Then $a + b + c = 3x = 3$,\nthe admissible maximum.\nThe other possibility is for two variables to be equal, say $a = b$, but not equal\nto the third, hence needing $a = b = 2/k$, thus (from the constraint) $8/k^2 + c^2 +\n4c/k = k + 3$, or $k^2c^2 + 4kc - (k^3 + 3k^2 - 8) = 0$, with non-negative real solution\n$c = \\frac{(k+2)\\sqrt{k-1}-2}{k}$ for $k^3+3k^2-8 \\ge 0$, i.e. $k \\ge \\kappa \\approx 1.3553$. Then $a+b+c =$\n$\\frac{(k+2)\\sqrt{k-1}+2}{k} < 3$ for $k < 2$, since it is equivalent to $(k-2)^3 < 0$. As $k$ needs\nbe at most $K \\le 3/2$, these points are critical, but not global maxima (in fact they turn\nto be global minima).\n\n\nAlternative Solution:\nThe fact the value 3 for $a+b+c$ is reached for $k = 3/2$ both at $a = b = c = 1$, and at $a = b = 3/2$ and $c = 0$ et al., suggests this is a Schur-type inequality. Indeed, assume $0 \\le k \\le 3/2$ and $\\sum a > 3$.\nThen\n$$\n\\begin{aligned}\nk+3 &= \\sum a^2 + kabc \\\\\n&\\ge \\sum a^2 + \\frac{3kabc}{\\sum a} \\\\\n&= \\left(1-\\frac{k}{3}\\right) \\sum a^2 + \\frac{k}{3} \\left(\\sum a^2 + \\frac{9abc}{\\sum a}\\right) \\\\\n&\\ge \\left(1-\\frac{k}{3}\\right) \\sum a^2 + \\frac{2k}{3} \\sum ab \\\\\n&= \\left(1-\\frac{2k}{3}\\right) \\sum a^2 + \\frac{k}{3} \\left(\\sum a^2 + 2\\sum ab\\right) \\\\\n&= \\left(1-\\frac{2k}{3}\\right) \\sum a^2 + \\frac{k}{3} \\left(\\sum a\\right)^2 \\\\\n&\\ge \\left(\\frac{1}{3}\\left(1-\\frac{2k}{3}\\right) + \\frac{k}{3}\\right) \\left(\\sum a\\right)^2 \\\\\n&= \\frac{k+3}{9} \\left(\\sum a\\right)^2 \\\\\n&> k+3\n\\end{aligned}\n$$\nsince inequality $\\sum a^2 + \\frac{9abc}{\\sum a} \\ge 2\\sum ab$ is in turn equivalent to $(\\sum a^2)(\\sum a) + 9abc \\ge 2(\\sum ab)(\\sum a)$, then $\\sum a^3 + 3abc \\ge \\sum a^2b + \\sum ab^2$, at last $\\sum a(a-b)(a-c) \\ge 0$, a basic form of Schur; while $\\sum a^2 \\ge \\frac{1}{3}(\\sum a)^2$ by Cauchy-Schwarz.\n\nPutting it all together, the largest admissible value for $K$ turns to be $3/2$, with the maximum value $a+b+c = 3$ being reached only at points $(3/2, 3/2, 0)$, $(3/2, 0, 3/2)$, $(0, 3/2, 3/2)$ and $(1, 1, 1)$. For $0 \\le k < 3/2$ the unique maximum is reached at $(1, 1, 1)$. Notice the importance of examining the values on the border; without that, the other critical interior points found but $(1, 1, 1)$ (which works for any $k$) achieve a larger value than 3 for $a + b + c$ only starting with $k > 2$, so would induce the erroneous bound $K = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37667,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle has side lengths $a$, $b$, $c$ and perimeter $3$. Prove that\n\na) it is not possible that all the altitudes have integer length;\n\nb) if $\\sqrt{a+b-c} + \\sqrt{b+c-a} + \\sqrt{c+a-b} = 3$, then the triangle is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be given positive integers. Prove there exists some positive integer $N$ such that\n$$\n\\begin{aligned}\na &\\mid Nbc + b + c \\\\\nb &\\mid Nca + c + a\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The necessity of having $x$, $y$, $z$ be pairwise co-prime is proved by, say, assuming $\\gcd(x, y) > 1$.\nThen $a \\mid Nbc + b + c$ becomes $x \\mid dNyz + y + z$, and so we must have $\\gcd(x, y) \\mid z$, absurd, since under this assumption it then follows $\\gcd(x, y) \\mid \\gcd(x, y, z) = 1$.\n\nOn the other hand, if the co-primality condition holds, consider the integers\n$$\nxyz - \\sum x < 2xyz - \\sum x < \\dots < (\\sum xy) xyz - \\sum x.\n$$\nThese $\\sum xy$ integer numbers will yield different remainders modulo $\\sum xy$, since if $ixyz - \\sum x \\equiv jxyz - \\sum x \\pmod{\\sum xy}$, then also $\\sum xy \\mid |i - j|xyz$, whence $i = j$, since we have $0 \\le |i - j| < \\sum xy$ and $\\gcd(xyz, \\sum xy) = 1$. Therefore there will exist some (unique) $1 \\le t \\le \\sum xy$ such that $\\sum xy \\mid txyz - \\sum x$, i.e. $txyz - \\sum x = C \\sum xy$ for some positive integer $C$, therefore $txyz = C \\sum xy + \\sum x$, so $x \\mid Cyz + y + z$ et al. We found a suitable value $C$ for the triplet $x$, $y$, $z$ (for similar relations with the ones sought for $a$, $b$, $c$). Then all the other suitable values must be of the form $C' = C + Mxyz$, since we need have $xyz \\mid (C' - C) \\sum xy$, while $\\gcd(xyz, \\sum xy) = 1$.\n\nNow the time has come to analyze the last condition. In order to have $a \\mid Nbc + b + c$, and the similar others, equivalent to $x \\mid dNyz + y + z$ et al., we need have $dN = C + Mxyz$ for some non-negative integer $M$. Denote $e = \\gcd(d, xyz)$; then $e \\mid d$, so $e \\mid C + Mxyz$. But then we also must have $e \\mid xyz \\mid (C + Mxyz) \\sum xy + \\sum x$, hence $e \\mid \\sum x$.\nConversely, if $e \\mid \\sum x$, then $e \\mid xyz \\mid (C + Mxyz) \\sum xy + \\sum x$, so $e \\mid C \\sum xy$. Since clearly $\\gcd(e, \\sum xy) = 1$, this means $e \\mid C$. Therefore we need have $\\frac{d}{e}N = \\frac{C}{e} + M\\frac{xyz}{e}$, and since clearly $\\gcd(\\frac{d}{e}, \\frac{xyz}{e}) = 1$, take $M \\equiv -\\frac{C}{e}(\\frac{xyz}{e})^{-1} \\pmod{d/e}$, wherefore $\\frac{d}{e}$ divides $\\frac{C}{e} + M\\frac{xyz}{e}$. Take now $N = \\frac{C + Mxyz}{d}$ (of course, $N' = N + Mabc$ also works).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be a positive integer and let $N_r$ be the smallest positive integer such that the numbers\n$$\n\\frac{N_r}{n+r} \\binom{2n}{n}, \\quad n = 0, 1, 2, \\dots,\n$$\nare all integers. Show that\n$$\nN_r = \\frac{r}{2} \\binom{2r}{r}.\n$$",
    "options": [],
    "answer": "N_r = (r/2) * binomial(2r, r)",
    "solution": "We first show that\n$$\nN_r \\le \\frac{r}{2} \\binom{2r}{r}\n$$\nby proving that\n$$\nK(n,r) = \\frac{r}{2(n+r)} \\binom{2n}{n} \\binom{2r}{r}\n$$\nis an integer for all $n \\ge 0$ and $r \\ge 1$. Notice that\n$$\nK(0, r) = \\binom{2r-1}{r} \\quad \\text{and} \\quad K(n, 1) = \\frac{1}{n+1} \\binom{2n}{n},\n$$\nthe latter being a Catalan number, so $K(0, r)$ and $K(n, 1)$ are integers for all $n$ and $r$.\nNext, the recurrence relation (which will be proved at the end of the solution)\n$$\nK(n, r + 1) - K(n + 1, r) = 2K(n, 1)K(0, r)\n$$\nshows by induction on $r$ that $K(n, r)$ is indeed an integer for all $n$ and $r$.\nSuppose now that, for some $r$, $N_r < M_r = \\frac{r}{2} \\binom{2r}{r}$. The integer $N_r$ is the least common multiple, over all $n \\ge 0$, of the denominators of the numbers $\\frac{1}{n+r} \\binom{2n}{n}$ when written in lowest terms. By the argument above, $M_r$ is a multiple of all these denominators. Hence $N_r$ divides $M_r$. By the definition of $N_r$, any prime $p$ that divides $M_r/N_r$ also divides $K(n, r)$ for each $n \\ge 0$. Since $K(n, s+1) = K(n+1, s) + 2K(n, 1)K(0, s)$, induction on $m$ shows that $p$ divides $K(n, m)$ for all $n \\ge 0$ and $m \\ge r$.\nNow choose $k$ such that $p^k \\ge r$. Since $p$ divides $\\binom{p^k}{j}$, $j = 1, 2, \\dots, p^k - 1$, the identity $\\binom{2n}{n} = \\sum_{j=1}^n \\binom{n}{j}^2$ yields $\\binom{2p^k}{p^k} \\equiv 2 \\pmod{p}$. Therefore $p$ does not divide $\\frac{1}{2}\\binom{2p^k}{p^k} = K(0, p^k)$. This contradicts the preceding paragraph, so $N_r = \\frac{r}{2}\\binom{2r}{r}$ for all $r$.\n\nTo prove the recurrence relation, notice that\n$$\n\\binom{2m+2}{m+1} = 2\\binom{2m+1}{m} = 2\\frac{2m+1}{m+1}\\binom{2m}{m},\n$$\nto get\n$$\n\\begin{align*}\nK(n, r + 1) - K(n + 1, r) &= \\\\\n&\\frac{r+1}{2(n+r+1)} \\binom{2n}{n} \\binom{2r+2}{r+1} - \\frac{r}{2(n+r+1)} \\binom{2n+2}{n+1} \\binom{2r}{r} = \\\\\n&\\frac{1}{n+r+1} \\binom{2n}{n} \\binom{2r}{r} \\left(2r+1 - r\\frac{2n+1}{n+1}\\right) = \\\\\n&\\frac{1}{n+1} \\binom{2n}{n} \\binom{2r}{r} = 2K(n, 1)K(0, r).\n\\end{align*}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$ congruent to $3$ modulo $4$, show that $w^{2p} + x^{2p} + y^{2p} = z^{2p}$ for no integer numbers $w, x, y, z$ whose product is not divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there are four such numbers. Without loss of generality, we may (and will) assume that they are jointly coprime: $(w, x, y, z) = 1$.\n\nReduction modulo $4$ shows that $z$ and exactly one of the numbers $w, x, y$, say $y$, must be odd.\n\nWrite\n$$\nw^{2p} + x^{2p} = z^{2p} - y^{2p} = (z^2 - y^2) \\left(\\sum_{k=1}^{p-1} z^{2(p-k-1)} (z^{2k} - y^{2k}) - p z^{2(p-1)}\\right)\n$$\nto deduce that the second factor above is congruent to $3$ modulo $4$ and infer thereby that in its decomposition into prime factors some prime $q \\equiv 3 \\pmod{4}$ occurs with an odd exponent.\n\nSince $-1$ is a quadratic non-residue modulo $q$, it follows that $w$ and $x$ are both divisible by $q$, and in the decomposition of $w^{2p} + x^{2p}$ into prime factors, $q$ occurs with an even exponent.\n\nHence $z^2 - y^2$ is divisible by $q$, and therefore so is $p z^{2(p-1)}$. Notice that $p \\neq q$ (for $q$ divides $w$, but $p$ does not by assumption) to deduce that $z$ is divisible by $q$. Then so is $y$.\n\nConsequently, $w, x, y, z$ all share the common factor $q$, in contradiction with their joint coprimality.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37671,
    "subject": "Mathematics (Multi-modal)",
    "question": "The tetrahedron $ABCD$ has $AD + BC + AB\\sqrt{3} = 3$, volume $\\frac{1}{12}$, and $\\angle ABC = 60^\\circ$. Compute the lengths of its edges.",
    "options": [],
    "answer": "AB = 1/√3, BC = 1, AC = √((4 − √3)/3), AD = 1, BD = 2/√3, CD = √((7 − √3)/3)",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer number $n \\ge 3$, determine the real numbers $x_1, x_2, \\dots, x_n$ minimising $(n-1)(x_1^2 + x_2^2 + \\dots + x_n^2) + nx_1x_2 \\dots x_n$, subject to $x_k \\ge 0$, $k = 1, 2, \\dots, n$, and $x_1 + x_2 + \\dots + x_n = n$.",
    "options": [],
    "answer": "Minimum value: n^2. It is attained either when all variables are equal to 1, or when exactly one variable is 0 and the remaining n−1 variables are all equal to n/(n−1).",
    "solution": "Let $f(x_1, x_2, \\dots, x_n) = (n-1)(x_1^2 + x_2^2 + \\dots + x_n^2) + nx_1x_2 \\dots x_n$. Assuming $0 \\le x_1 \\le x_2 \\le \\dots \\le x_n$, we first show that\n$$\nf(x_1, x_2, \\dots, x_n) \\ge f(x_1, (x_2+x_n)/2, x_3, \\dots, x_{n-1}, (x_2+x_n)/2).\n$$\n\nTo begin with, notice that $x_n \\ge 1$, to get\n$$\n\\begin{aligned}\nf(x_1, x_2, \\dots, x_n) - f(x_1, (x_2+x_n)/2, x_3, \\dots, x_{n-1}, (x_2+x_n)/2) \\\\\n&= \\frac{1}{4}(x_2 - x_n)^2 (2(n-1) - nx_1x_3 \\cdots x_{n-1}) \\\\\n&\\ge \\frac{1}{4}(x_2 - x_n)^2 (2(n-1) - nx_1x_3 \\cdots x_{n-1}x_n).\n\\end{aligned}\n$$\nWe now show that $x_1x_3\\cdots x_{n-1}x_n \\le 2(n-1)/n$, so the difference above is indeed non-negative. Notice first that\n$$n = x_1 + x_2 + \\dots + x_n \\ge 2x_1 + x_2 + \\dots + x_n \\ge (n-1)(2x_1x_3\\cdots x_{n-1}x_n)^{1/(n-1)},$$\nso $x_1x_3\\cdots x_{n-1}x_n \\le n^{n-1}/(2(n-1)^{n-1})$. Since $n^{n-1}/(2(n-1)^{n-1}) \\le 2(n-1)/n$, $n \\ge 2$, the conclusion follows. (The latter follows from the well-known fact that the sequence $u_n = (1+1/n)^{n+1}$, $n \\ge 1$, is decreasing, so $u_n \\le u_1 = 4$, $n \\ge 1$; that $u_n$ decreases is easily seen by Bernoulli's inequality: $n^{2(n+1)}/(n^2-1)^{n+1} = (1+1/(n^2-1))^{n+1} \\ge 1 + (n+1)/(n^2-1) = n/(n-1)$, $n \\ge 2$.)\n\nConsequently, the value of $f$ does not increase when the largest and the second smallest of the $x_k$ is replaced by their arithmetic mean. It follows that $f$ achieves its minimum when $n-1$ of the $x_k$ are all equal to some $x$, and the smallest of the $x_k$ is $n - (n-1)x$. (That $f$ actually achieves a minimum subject to the stated conditions follows by a standard compactness argument which may be assumed without proof.) Since the $x_k$ are all non-negative, and $n - (n-1)x$ is the smallest among them, it follows that $1 \\le x \\le n/(n-1)$.\nSo we have to minimise the polynomial function\n$$\ng(x) = (n-1)\\left((n - (n-1)x)^2 + (n-1)x^2\\right) + n(n - (n-1)x)x^{n-1}\n$$\non the closed interval $1 \\le x \\le n/(n-1)$. To this end, recall the guess made in the beginning, to evaluate\n$$\ng(x) - n^2 = n(n - (n-1)x)(x-1)^2 \\sum_{k=1}^{n-2} (x^{k-1} + x^{k-2} + \\dots + 1) \\ge 0,\n$$\n$1 \\le x \\le n/(n-1)$. Equality holds if and only if $x=1$ or $x=n/(n-1)$, whence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $VABC$ be a regular tetrahedron of side length $a$ and points $M$, $N$ on the edges $VB$, $VC$, such that $VM = \\frac{a}{3}$ and $VN = \\frac{a}{6}$. If $Q$ is the projection of $V$ onto the plane $(AMN)$ and $O$ is the projection of $V$ onto the plane $(ABC)$, compute the lengths of the segments $VO$, $VQ$, and a trigonometric function of the angle between the straight lines $VQ$ and $VO$.",
    "options": [],
    "answer": "VO = a*sqrt(2/3), VQ = a*sqrt(42)/42, and cos(angle between VQ and VO) = 5/(3*sqrt(7))",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the real numbers $a$, $b$ fulfill\n$$(x + y + z)^3 + a(x + y + z)(xy + xz + yz) + bxyz \\ge 0,$$\nfor all $x, y, z \\ge 0$, if and only if there exist $r \\ge 0$, $s \\ge 0$ such that $a = -4 + r$, $b = 9 - 9r + s$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ and a point $M$ in its interior. Denote by $D, E, F$ the projections of $M$ onto the sides $BC, CA$ respectively $AB$ and by $r$ the radius of its incircle. Prove that if\n$$\n\\frac{BC}{MD} + \\frac{CA}{ME} + \\frac{AB}{MF} = \\frac{AB + BC + CA}{r},\n$$\nthen the straight lines $AD$, $BE$ and $CF$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37676,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, the excircle $C(I_a)$ relative to $BC$ is tangent to the straight lines $BC$, $CA$, $AB$ at $D$, $E$, respectively $F$. The angle bisector of $\\angle BI_aC$ meets $BC$ at $M$ and $AM$ meets $EF$ at $P$. Prove that $DP$ is the angle bisector of $\\angle FDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a subset with 673 elements of the set $\\{1, 2, \\dots, 2010\\}$. Prove that one can find two distinct elements of $S$, say $a$ and $b$ such that $6$ divides $a + b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following sets, each containing 335 elements\n$$\n\\begin{align*}\nA &= \\{6, 12, \\dots, 2010\\}, & B &= \\{3, 9, 15, \\dots, 2007\\}, \\\\\nC &= \\{1, 7, 13, \\dots, 2005\\}, & D &= \\{2, 8, 14, \\dots, 2006\\}, \\\\\nE &= \\{4, 10, 16, \\dots, 2008\\}, & F &= \\{5, 11, 17, \\dots, 2009\\}.\n\\end{align*}\n$$\nIf $S$ contains two elements from $A$ or two elements from $B$, their sum is divisible by $6$. If not, the remaining $4$ sets contain at least $673 - 2 = 671$ elements from $S$.\n\nConsider the sets $C \\cup F$ and $D \\cup E$, each containing $670$ elements. One of the intersections of $S$ with these, say $C \\cup F$, contains at least $336$ elements. Thus $S \\cap C$ and $S \\cap F$ contain each at least one element. Their sum is a multiple of $6$, which is what was to be proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the sum of all invertible elements of a finite ring.\nProve that $S^2 = S$ or $S^2 = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $1 + 1 \\ne 0$, then $x \\ne -x$, for every invertible $x$, whence $S = 0$ (we can group the invertible elements into pairs $(x, -x)$).\n\nIf this is not the case, notice that $xS = S$, for every invertible $x$. Adding all these relations, $S^2 = kS$, where $k$ is the number of invertible elements. Then $S^2 = S$ for odd $k$ and $S^2 = 0$ for even $k$ (since $1 + 1 = 0$).",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37679,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a finite non-empty set of primes $P$, let $|P|$ be the size of the set $P$, and $m(P)$ be the largest possible number of consecutive positive integers, each of which is divisible by at least one member of $P$.\n(i) Show that $|P| \\le m(P)$, with equality if and only if $\\min(P) > |P|$;\n(ii) Show that $m(P) < (|P| + 1)(2^{|P|} - 1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In the sequel we will consider $P$ being made of the primes $1 < p_1 < p_2 < \\dots < p_k$, with $k = |P| \\ge 1$.\n\na. By the Chinese Remainder Theorem there will exist some $a \\in \\mathbb{N}$ such that $a \\equiv -i \\pmod{p_i}$, hence $p_i \\mid a + i$. Then the set $\\{a + i; i = 1, 2, \\dots, k\\}$ of $k$ consecutive integers has the desired property, hence $m(P) \\ge k$. When $\\min P > |P|$, within any set of $|P| + 1$ consecutive integers at most one is divisible by any $p \\in P$, hence by the Pigeonhole Principle there will be one not divisible by any of the primes in $P$. On the other hand, when $\\min P \\le |P|$, we will make again use of the Chinese Remainder Theorem, so there will exist some $a \\in \\mathbb{N}$ such that $a \\equiv -r_i \\pmod{p_i}$, hence $p_i \\mid a+r_i$, where $\\{r_i; i = 1, 2, \\dots, k\\} = \\{1, 2, \\dots, k\\}$ and the extra requirement that $r_1 = k+1-p_1$. It follows that the set $\\{a+i; i = 1, 2, \\dots, k, k+1\\}$ of $k+1$ consecutive integers has the desired property, hence $m(P) \\ge k+1 > |P|$.\n\nb. Now, let a set made of $m$ consecutive integers have the desired property. For any $\\emptyset \\ne I \\subseteq \\{1, 2, \\dots, k\\}$, the number $N(I)$ of its elements which are divisible by $\\prod_{i \\in I} p_i$ will satisfy the inequality\n$$\n\\frac{m}{\\prod_{i \\in I} p_i} - 1 < \\left\\lfloor \\frac{m}{\\prod_{i \\in I} p_i} \\right\\rfloor \\le N(I) \\le \\left\\lceil \\frac{m}{\\prod_{i \\in I} p_i} \\right\\rceil < \\frac{m}{\\prod_{i \\in I} p_i} + 1.\n$$\nThen, by the Principle of Inclusion/Exclusion, one has\n$$\nm = \\sum_{i=1}^{k} (-1)^{i+1} \\sum_{|I|=i} N(I) < \\sum_{i=1}^{k} \\binom{k}{i} + m \\sum_{i=1}^{k} (-1)^{i+1} \\sum_{|I|=i} \\frac{1}{\\prod_{i \\in I} p_i}.\n$$\nThe first term is clearly equal to $2^k - 1$, while the second is equal to\n$$\nm \\left( 1 - \\prod_{i=1}^{k} \\left( 1 - \\frac{1}{p_i} \\right) \\right) \\le m \\left( 1 - \\prod_{i=1}^{k} \\left( 1 - \\frac{1}{i+1} \\right) \\right) = m - \\frac{m}{k+1},\n$$\ntherefore $m < (k+1)(2^k - 1)$, and so will be $m(P)$.\nAlternative Solution (ii). (F. Chindea)\n\nNot only an elegant alternative approach, but vastly improving on the required bound. Start by proving the following\n\nLEMMA. If the integer arithmetic sequences $\\{mn_j\\}_{m \\in \\mathbb{Z}}$, $1 \\le j \\le k$, are such that they cover $2^k$ consecutive integers, then they cover $\\mathbb{Z}$.\n(Particular case of a Crittenden & Vanden Eynden Theorem)\n\nProof. Let $\\omega_j = e^{n_j}$ be a primitive root of unity. Then $n_j \\mid m$ if and only if $\\omega_j^m = 1$, hence $m$ is divisible by at least one $n_j$ if and only if $0 = \\prod_{j=1}^k (1 - \\omega_j^m) = \\sum_{\\emptyset \\subseteq S \\subseteq [k]} (-1)^{|S|} e^{2m\\pi i \\sum_{j \\in S} \\frac{1}{n_j}}$. Denote $\\alpha_S = (-1)^{|S|}$, $z_S = e^{2\\pi i \\sum_{j \\in S} \\frac{1}{n_j}}$, $u_m = \\sum_{\\emptyset \\subseteq S \\subseteq [k]} \\alpha_S z_S^m$.\n\nConsider the polynomial $f(x) = \\prod_{\\emptyset \\subseteq S \\subseteq [k]} (x - z_S) = \\sum_{j=0}^{2^k} c_j x^j$, of degree $\\deg f = 2^k$, and with $c_{2^k} = 1$, $|c_0| = \\prod_{\\emptyset \\subseteq S \\subseteq [k]} |z_S| = 1$. Then $\\alpha_S z_S^b f(z_S) = 0$ for any integer $b$, hence\n$$\n0 = \\sum_{\\emptyset \\subseteq S \\subseteq [k]} \\alpha_S z_S^b f(z_S) = \\sum_{j=0}^{2^k} c_j \\sum_{\\emptyset \\subseteq S \\subseteq [k]} \\alpha_S z_S^{b+j} = \\sum_{j=0}^{2^k} c_j u_{b+j}.\n$$\nAssume the $2^k$ consecutive integers $a+1, \\dots, a+2^k$ are covered, so, according with the above, $u_{a+1} = \\dots = u_{a+2^k} = 0$. Since $c_0 \\neq 0$ and $c_{2^k} \\neq 0$, by simple induction one gets $u_b = 0$ for any integer $b$ (a linear recurrence relation of order $2^k$ containing $2^k$ consecutive null terms generates an all-null sequence). $\\square$\n\nNow, in our case, if we assume $m(P) \\ge 2^{|P|}$, then the $|P|$ arithmetic sequences given by $\\{mp\\}_{m \\in \\mathbb{Z}}$, $p \\in P$, will cover $2^k$ consecutive integers, hence they will cover $\\mathbb{Z}$. But clearly a large enough prime $q$ is not divisible by any $p \\in P$, so this is absurd, therefore $m(P) < 2^{|P|}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a convex quadrilateral $ABCD$ having $\\angle BCD = 120^\\circ$, $\\angle CBA = 45^\\circ$, $\\angle CBD = 15^\\circ$ and $\\angle CAB = 90^\\circ$. Show that $AB = AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the mirror image $E$ of point $C$ with respect to $A$. Since $\\angle CDB = \\angle CEB = 45^\\circ$, the quadrilateral $BCDE$ is cyclic, and $CE$ is a diameter of its circumcircle. Hence $A$ is its circumcenter, implying $AB = AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37681,
    "subject": "Mathematics (Multi-modal)",
    "question": "A straight line passing through the incenter $I$ of a triangle $ABC$ meets the sides $AB$ and $AC$ at points $P$ and $Q$ respectively. Let $BC = a$, $AC = b$, $AB = c$ and $\\frac{PB}{PA} = p$, $\\frac{QC}{QA} = q$.\n\na) Prove that $a(1+p)\\vec{IP} = (a-pb)\\vec{IB} - cp\\vec{IC}$.\n\nb) Prove that $a = bp + cq$.\n\nc) Prove that if $a^2 = 4bcpq$, then the straight lines $AI$, $BQ$ and $CP$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The hypothesis yields $(p+1)\\vec{IP} = \\vec{IB} + p\\vec{IA}$. The conclusion follows now from $a\\vec{IA} + b\\vec{IB} + c\\vec{IC} = \\vec{0}$.\n\nb) Analogously, $a(1+q)\\vec{IQ} = (a-cq)\\vec{IC} - bq\\vec{IB}$. The points $P$, $I$, $Q$ are collinear, hence $(a - pb)(a - cq) = bcpq$, whence the conclusion.\n\nc) Squaring the relation from b), $a^2 \\ge 4bcpq$, with equality if and only if $bp = cq$. In this case, if $D = AI \\cap BC$, then\n$$\n\\frac{PB}{PA} \\cdot \\frac{QA}{QC} \\cdot \\frac{DC}{DB} = p \\cdot \\frac{1}{q} \\cdot \\frac{b}{c} = 1\n$$\nand the converse of Ceva's Theorem guarantees the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37682,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Factorize $xy - x - y + 1$.\n\nb) Prove that if integers $a$ and $b$ satisfy $|a+b| > |1+ab|$, then $ab = 0$.",
    "options": [],
    "answer": "a) (x - 1)(y - 1)\nb) ab = 0",
    "solution": "a) $xy - x - y + 1 = (x - 1)(y - 1)$.\n\nb) Both members of the inequality are positive, so we can square and get the equivalent form $a^2 + b^2 + 2ab > 1 + 2ab + a^2b^2$, that is $(a^2 - 1)(b^2 - 1) < 0$.\nThis shows that $a^2 - 1 < 0$ or $b^2 - 1 < 0$. Since $a$ and $b$ are integers, this yields $a = 0$ or $b = 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37683,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ denote by $D$, $E$, respectively $F$, the points where the angle bisectors of $\\angle CAB$, $\\angle ABC$, respectively $\\angle BCA$, meet its circumcircle.\n\na) Prove that the ortocenter of triangle $DEF$ coincides with the incenter of triangle $ABC$.\n\nb) Prove that if $\\overrightarrow{AD} + \\overrightarrow{BE} + \\overrightarrow{CF} = \\overrightarrow{0}$, then the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $I$ be the incenter of the triangle $ABC$. $D, E, F$ are the midpoints of the arcs $\\widehat{BC}, \\widehat{CA}, \\widehat{AB}$.\nThe angle determined by the lines $AD$ and $EF$ is equal to $\\frac{1}{2}(\\widehat{AE} + \\widehat{DF}) = \\frac{1}{4}(\\widehat{AB} + \\widehat{BC} + \\widehat{CA}) = 90^\\circ$, hence $AD \\perp EF$. Likewise, $BE \\perp DF$, and the claim follows.\n\nb) Let $O$ be the circumcenter of the triangle $ABC$. The given relation implies that $OA + OB + OC = OD + OE + OF$.\nInvoking Sylvester's theorem, the triangles $ABC$ and $DEF$ share the same orthocenter. Thus, in the triangle $ABC$ point $I$ is both incenter and orthocenter, yielding the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer number $n \\ge 3$, consider $n$ distinct points on a circle, labeled 1 through $n$. Determine the maximum number of closed chords $[ij]$, $i \\neq j$, having pairwise non-empty intersections.",
    "options": [],
    "answer": "n",
    "solution": "We shall prove that any such configuration contains at most $n$ chords and the upper bound is achieved, so the required maximum is $n$.\n\nTo this end, fix an orientation of the circle and relabel the points $1$ through $n$ in the corresponding circular order. Consider a configuration of chords $[ij]$, $i \\neq j$, with pairwise non-empty intersections. Assign to each point $i$, which is an endpoint of at least one chord, the first point $i'$ following $i$, to which it is connected. We now show that by deleting the chords $[ii']$, no chord is left, so the number of chords in the configuration does not exceed $n$.\n\nSuppose, if possible, that some chord $[ij]$ is left. Then $i, i', j, j'$ are in circular order around the circle, so the chords $[ii']$ and $[jj']$ do not meet – a contradiction.\n\nA maximal configuration is given by the $n$ chords $[1i]$, $i = 2, 3, \\dots, n$, and $[2n]$.\n\n(The $n$ points could be located anywhere in the plane, or the chords could be Jordan arcs; the topic is related to John Conway's *thrackle* conjecture.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37685,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) $\\{x \\in \\mathbb{R} \\mid \\log_2[x] = [\\log_2 x]\\} = \\bigcup_{m \\in \\mathbb{N}} [2^m, 2^m + 1)$.\n\nb) $\\{x \\in \\mathbb{R} \\mid 2^{\\lfloor x \\rfloor} = \\lfloor 2^x \\rfloor\\} = \\bigcup_{m \\in \\mathbb{N}} [m, \\log_2 (2^m + 1))$.\n\n(here, $[a]$ denotes the integer part (floor function) of the real number $a$).",
    "options": [],
    "answer": "a) { x ∈ ℝ | log₂([x]) = [log₂ x] } = ⋃_{m ∈ ℕ} [2^m, 2^m + 1).\nb) { x ∈ ℝ | 2^{⌊x⌋} = ⌊2^x⌋ } = ⋃_{m ∈ ℕ} [m, log₂(2^m + 1)).",
    "solution": "a) The existence of the logarithms requires $x > 1$.\nIf $[\\log_2 x] = m$, then $m \\in \\mathbb{N}$ and $2^m \\le x < 2^{m+1}$. From $\\log_2[x] = m$, follows $[x] = 2^m$, that is $2^m \\le x < 2^m + 1$.\nConversely, $x \\in [2^m, 2^m + 1)$, $m \\in \\mathbb{N}$ yields $[\\log_2 x] = \\log_2[x] = m$.\n\nb) If $[2^x] = t \\in \\mathbb{N}$, then $\\log_2 t \\le x < \\log_2(t+1)$. From $2^{[x]} = t$ follows $[x] = \\log_2 t = m \\in \\mathbb{N}$, whence $m \\le x < \\log_2(2^m + 1)$.\nConversely, if $x \\in [m, \\log_2(2^m + 1))$, $m \\in \\mathbb{N}$ then $[2^x] = 2^{[x]} = m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be integers larger than $1$. Prove that\n$$\na(a-1) + b(b-1) + c(c-1) \\leq (a + b + c - 4)(a + b + c - 5) + 4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$$\na^2 + b^2 + c^2 - a - b - c \\leq (a + b + c)^2 - 9(a + b + c) + 24,\n$$\nor\n$$\n0 \\leq ab + bc + ca - 4(a + b + c) + 12.\n$$\nSince $ab - 2a - 2b + 4 = (a - 2)(b - 2)$, summing together with the similar relations implies $(ab - 2a - 2b + 4) + (bc - 2b - 2c + 4) + (ca - 2c - 2a + 4) = (a - 2)(b - 2) + (b - 2)(c - 2) + (c - 2)(a - 2) \\geq 0$, since $a$, $b$, $c \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37687,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that one cannot assign to each vertex of a cube 8 distinct numbers from the set $\\{0, 1, 2, 3, \\ldots, 11, 12\\}$ such that, for every edge, the sum of the two numbers assigned to its vertices is even.\n\nb) Prove that one can assign to each vertex of a cube 8 distinct numbers from the set $\\{0, 1, 2, 3, \\ldots, 11, 12\\}$ such that, for every edge, the sum of the two numbers assigned to its vertices is divisible by 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If in a vertex is written a number from the given set, then its \"neighbors\" have to be of the same parity. This shows that all the written numbers must have the same parity. Since the set contains 7 even and 6 odd elements, this task is impossible.\n\nb) The task can be accomplished through assigning to \"neighbor\" vertices distinct numbers which are not divisible by 3, and which yield different residues mod 3. An example is shown in the figure.\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 0}$ be a sequence of positive real numbers such that\n$$\n\\sum_{k=0}^{n} C_{n}^{k} a_{k} a_{n-k} = a_{n}^{2}, \\text{ for any } n \\ge 0.\n$$\nProve that $(a_n)_{n \\ge 0}$ is a geometrical sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_0 = a$. It is obvious that $a_1 = 2a$, and, for $n = 2$,\n$$\na_2^2 - 2aa_2 - 8a^2 = 0,\n$$\nimplying $a_2 = 4a$, for $a_2 > 0$.\nUse induction on $n$ to prove that $a_n = 2^n a$. Assume that $a_k = 2^k a$, for all $k$, $0 \\le k \\le n$, to prove $a_{n+1} = 2^{n+1} a$. We have\n$$\n2aa_{n+1} + a^2 2^{n+1} \\sum_{k=1}^{n} C_{n+1}^{k} = a_{n+1}^2,\n$$\nor\n$$\n2aa_{n+1} + a^2 2^{n+1} (2^{n+1} - 2) = a_{n+1}^2.\n$$\nSince $a_{n+1}$ is positive, we obtain $a_{n+1} = 2^{n+1}a$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. The lines $AD$, $BC$ meet at $P$; $AB$, $CD$ at $Q$; and $AC$, $BD$ at $R$. The perpendicular bisectors of $AB$, respectively $BC$, meet $PR$ at $X$, respectively $QR$ at $Y$. Prove that $XY$ passes through $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All poles and polars are considered with respect to the given circumcircle of $ABCD$.\nTo start with, notice that line $q = PR$ is the polar of $Q$ and line $p = QR$ is the polar of $P$. As $Y$ lies on the polar of $P$, it follows that $P$ lies on the polar $y$ of $Y$. The pole of the line $OY$ is the point at infinity $\\infty$ on the direction $BC$, as $OY$ is a diameter line. Thus the polar $y$ of $Y$ is the line $P\\infty = BC$, implying that $YB$ is tangent to the given circle at $B$. Similar considerations show that $XB$ is tangent to the given circle at $B$, hence proving the thesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that any continuous function $f : \\mathbb{R} \\to \\mathbb{R}$ of the form\n$$\nf(x) = \\begin{cases} a_1x + b_1, & \\text{for } x \\le 1 \\\\ a_2x + b_2, & \\text{for } x > 1 \\end{cases}\n$$\nwith $a_1, a_2, b_1, b_2 \\in \\mathbb{R}$, can be represented as\n$$\nf(x) = m_1x + n_1 + \\varepsilon|m_2x + n_2|, \\text{ for } x \\in \\mathbb{R},\n$$\nwhere $m_1, m_2, n_1, n_2 \\in \\mathbb{R}$ and $\\varepsilon \\in \\{-1, +1\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The continuity at $x = 1$ implies $a_1 + b_1 = a_2 + b_2$.\nIf $a_1 = a_2$, then $b_1 = b_2$ and we can take $m_1 = a_1 = a_2$, $n_1 = b_1 = b_2$, $m_2 = n_2 = 0$, $\\varepsilon = \\pm 1$.\nOtherwise, we search $A$ such that $m_2x + n_2 = A(x - 1)$. The required form can be obtained if $m_1 - \\varepsilon|A| = a_1$, $n_1 + \\varepsilon|A| = b_1$ and $m_1 + \\varepsilon|A| = a_2$, $n_1 - \\varepsilon|A| = b_2$, that is\n$$\nm_1 = \\frac{a_1 + a_2}{2}, \\quad n_1 = \\frac{b_1 + b_2}{2}, \\quad 2\\varepsilon|A| = a_2 - a_1 = b_1 - b_2.\n$$\nChoosing $\\varepsilon = \\operatorname{sign}(a_2 - a_1)$, $m_2 = \\frac{1}{2}(a_1 - a_2)$ and $n_2 = \\frac{1}{2}(a_2 - a_1)$, we get\n$$\nf(x) = \\frac{a_1 + a_2}{2}x + \\frac{b_1 + b_2}{2} + \\operatorname{sign}(a_2 - a_1) \\left| \\frac{a_1 - a_2}{2}x + \\frac{a_2 - a_1}{2} \\right|.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a regular polygon $A_1A_2...A_{2010}$ centered at $O$. On each of the segments $OA_k$, with $k = 1, 2, ..., 2010$, lies the point $B_k$ such that\n$$\n\\frac{OB_k}{OA_k} = \\frac{1}{k}.\n$$\nDetermine the ratio between the area of the polygon $B_1B_2...B_{2010}$ and that of $A_1A_2...A_{2010}$.",
    "options": [],
    "answer": "1/2010",
    "solution": "In the sequel $[XYZ]$ will represent the area of a triangle $XYZ$. Denote by $S$ the area of the polygon $A_1A_2...A_{2010}$.\nNotice that $[OA_1A_2] = [OA_2A_3] = \\dots = [OA_{2010}A_1] = \\frac{1}{2010}S$, since the polygon $A_1A_2...A_{2010}$ is given as being regular. Then for each $k = 1, 2, ..., 2009$ we have\n$$\n\\frac{[OB_k B_{k+1}]}{[OA_k A_{k+1}]} = \\frac{OB_k \\cdot OB_{k+1}}{OA_k \\cdot OA_{k+1}} = \\frac{1}{k(k+1)},\n$$\nas well as\n$$\n\\frac{[OB_{2010}B_1]}{[OA_{2010}A_1]} = \\frac{OB_{2010} \\cdot OB_1}{OA_{2010} \\cdot OA_1} = \\frac{1}{2010},\n$$\nsince the ratio of the areas of two triangles sharing a same angle is equal to the ratio of the products of the corresponding sides.\nTherefore denoting by $T$ the area of the polygon $B_1B_2...B_{2010}$, we have\n$$\n\\begin{align*}\nT &= \\sum_{k=1}^{2009} [OB_k B_{k+1}] + [OB_{2010} B_1] \\\\\n&= \\frac{S}{2010} \\left( \\sum_{k=1}^{2009} \\frac{1}{k(k+1)} + \\frac{1}{2010} \\right) \\\\\n&= \\frac{S}{2010} \\left( \\sum_{k=1}^{2009} \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) + \\frac{1}{2010} \\right) \\\\\n&= \\frac{S}{2010} \\left( \\left( 1 - \\frac{1}{2010} \\right) + \\frac{1}{2010} \\right) = \\frac{S}{2010},\n\\end{align*}\n$$\nhence $\\frac{T}{S} = \\frac{1}{2010}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two real numbers $a$, $b$ such that $b - a^2 > 0$, describe all matrices $A \\in M_2(\\mathbb{R})$ such that $\\det(A^2 - 2aA + bI_2) = 0$.",
    "options": [],
    "answer": "All A ∈ M2(R) with trace(A) = 2a and det(A) = b; equivalently, all matrices of the form [[a + x, y], [(a^2 − x^2 − b)/y, a − x]] with x ∈ R and y ∈ R\\{0}.",
    "solution": "The number $c = \\sqrt{b - a^2}$ is real. We have\n$$\nA^2 - 2aA + bI_2 = (A - (a + ic)I_2)(A - (a - ic)I_2),\n$$\nwhence the characteristic polynomial of $A$ has as a root $a + ic$, $a - ic$, or both. Since the polynomial has real coefficients, both are roots, hence eigenvalues. Therefore the polynomial is $x^2 - 2ax + b$. It follows $A^2 - 2aA + bI_2 = 0$, and so $\\operatorname{tr}(A) = 2a$, $\\det A = b$.\n\nIn conclusion, the matrices bearing the given property are\n$$\n\\begin{pmatrix} a + x & y \\\\ \\frac{a^2 - x^2 - b}{y} & a - x \\end{pmatrix},\n$$\nwhere $x \\in \\mathbb{R}$ and $y \\in \\mathbb{R}^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n\\frac{a^2 b (b-c)}{a+b} + \\frac{b^2 c (c-a)}{b+c} + \\frac{c^2 a (a-b)}{c+a} \\geq 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By clearing denominators (brute force), the inequality becomes\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq a^2 b c^3 + b^2 c a^3 + c^2 a b^3. \\quad (1)\n$$\nIn order to justify (1), use the AM-GM inequality. Thus\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq 3ab^3c^2.\n$$\nBy summation with the two other analogous inequalities, one gets the desired result.\n\n\nAlternative Solution:\nMake use of the rearrangement inequality (in various settings). From the start we may suppose either $a \\geq b \\geq c$, or $a \\geq c \\geq b$. For example, write the inequality as the equivalent form\n$$\n\\frac{ab}{c(a+b)} + \\frac{bc}{a(b+c)} + \\frac{ca}{b(a+c)} \\geq \\frac{a}{a+b} + \\frac{b}{b+c} + \\frac{c}{c+a},\n$$\nand use same-order triplets $(ab, ac, bc)$, $(\\frac{1}{c(a+b)}, \\frac{1}{a(b+c)}, \\frac{1}{b(a+c)})$ when $a \\geq b \\geq c$, and same-order triplets $(ac, ab, bc)$, $(\\frac{1}{b(a+c)}, \\frac{1}{c(a+b)}, \\frac{1}{a(b+c)})$ when $a \\geq c \\geq b$.\n\n\nAlternative Solution:\n(Official Jury Solution) Divide by $abc$, to obtain\n$$\n\\frac{a(b-c)}{c(a+b)} + \\frac{b(c-a)}{a(b+c)} + \\frac{c(a-b)}{b(c+a)} \\ge 0.\n$$\nAdding 1 to each fraction leads to\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\ge 3,\n$$\nwhich immediately follows from the AM-GM inequality\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\ge 3\\sqrt[3]{\\frac{b(c+a)}{c(a+b)} \\cdot \\frac{c(a+b)}{a(b+c)} \\cdot \\frac{a(b+c)}{b(c+a)}} = 3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, $n \\ge 2$. For each $k = 1, 2, \\dots, n$, let $a_k$ be the number of multiples of $k$ in the set $\\{1, 2, \\dots, n\\}$, and let $x_k = \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{a_k}$. Show that $\\frac{x_1 + x_2 + \\dots + x_n}{n} \\le \\frac{1}{1^2} + \\frac{1}{2^2} + \\dots + \\frac{1}{n^2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The number $a_k$ of the multiples of $k$ in the set $\\{1, 2, \\dots, n\\}$ is $\\lfloor \\frac{n}{k} \\rfloor$. Let us count the appearances of the term $\\frac{1}{i}$ in the sum $x_1 + x_2 + \\dots + x_n$. The term $\\frac{1}{i}$ belongs in the sum $x_k = \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{a_k}$ if and only if $i \\le a_k = \\lfloor \\frac{n}{k} \\rfloor$.\n\nFix $i = 1, 2, \\dots, n$ and denote by $m = m(i)$ the index of the last sum $x_m$ containing $\\frac{1}{i}$. Then $i \\le \\lfloor \\frac{n}{m} \\rfloor \\le \\frac{n}{m}$, i.e. $m \\le \\frac{n}{i}$.\n\nRemember $\\frac{1}{i}$ belongs to the sums $x_1, x_2, \\dots, x_m$ and those only, therefore one has\n$$\nx_1 + x_2 + \\dots + x_n = \\sum_{i=1}^{n} \\frac{1}{i} m(i) \\le \\sum_{i=1}^{n} \\frac{1}{i} \\cdot \\frac{n}{i} = n \\sum_{i=1}^{n} \\frac{1}{i^2}.\n$$\nThe conclusion immediately ensues.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $x_1, x_2, \\dots, x_n$ be positive real numbers such that $x_1x_2 \\cdots x_n = 1$. Prove that\n$$\n\\sum_{i=1}^{n} x_{i}^{n}(1 + x_{i}) \\geq \\frac{n}{2^{n-1}} \\prod_{i=1}^{n}(1 + x_{i}).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the power-mean inequality,\n$$\n1 + a^n \\geq \\frac{(1 + a)^n}{2^{n-1}}, \\quad a \\geq 0. \\qquad (*)\n$$\nThus,\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} x_i^n (1 + x_i) &= \\sum_{i=1}^{n} x_i^n + \\sum_{i=1}^{n} x_i^{n+1} \\\\\n&\\geq \\sum_{i=1}^{n} x_i^n + n \\left( \\prod_{i=1}^{n} x_i \\right)^{1+1/n} \\quad \\text{(AM-GM)} \\\\\n&= \\sum_{i=1}^{n} x_i^n + n = \\sum_{i=1}^{n} (1 + x_i^n) \\\\\n&\\geq \\frac{1}{2^{n-1}} \\sum_{i=1}^{n} (1 + x_i)^n \\quad \\text{by (*)} \\\\\n&\\geq \\frac{n}{2^{n-1}} \\prod_{i=1}^{n} (1 + x_i). \\quad \\text{(AM-GM)}\n\\end{align*}\n$$\nClearly, equality holds if and only if $x_1 = x_2 = \\dots = x_n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ be five real numbers of zero sum, such that $|a_i - a_j| \\le 1$, for all $i, j \\in \\{1, 2, 3, 4, 5\\}$. Prove that $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \\le \\frac{6}{5}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $0 = (a_1 + a_2 + a_3 + a_4 + a_5)^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + 2 \\sum_{i<j} a_i a_j$, it follows that $\\sum_{i<j} (a_i - a_j)^2 = 4(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) - 2 \\sum_{i<j} a_i a_j = 5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2)$. On the other side, $\\sum_{i<j} (a_i - a_j)^2 \\le \\sum_{i<j} |a_i - a_j|$, for $|a_i - a_j| \\le 1$. Assuming $a_1 \\le a_2 \\le a_3 \\le a_4 \\le a_5$, we have $\\sum_{i<j} |a_i - a_j| = 4(a_5 - a_1) + 2(a_4 - a_2) \\le 4 + 2 = 6$. So $5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) \\le 6$, as requested to be shown.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $AB \\neq AC$. The internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite sides of the triangle at points $B_0$ and $C_0$, respectively, and the circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let $I$ be the incenter of the triangle $ABC$. Prove that the lines $B_0C_0$ and $B_1C_1$ meet at some point lying on the parallel through $I$ to the line $BC$.\nRadu Gologan",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the internal bisector of the angle $BAC$ meet again the circumcircle $ABC$ at point $A_1$. The lines $A_1B_1$ and $AC$ meet at point $B_2$, and the lines $AB$ and $A_1C_1$ meet at point $C_2$. Apply Pascal's theorem to the hexagon $AC_1BA_1CB_1$ to deduce that the points $B_2$, $I$ and $C_2$ are collinear; moreover, Pascal's line $B_2IC_2$ is precisely the parallel through $I$ to $BC$, for $A_1B_2$ and $A_1C_2$ are the internal bisector lines of the angles $AA_1C$ and $AA_1B$, respectively, and the segments $A_1B$ and $A_1C$ are congruent. Finally, notice that the lines $B_iB_{i+1}$ and $C_iC_{i+1}$ meet at collinear points; $B_0B_1$ and $C_0C_1$ meet at $I$, $B_1B_2$ and $C_1C_2$ meet at $A_1$, and $B_2B_0$ and $C_2C_0$ meet at $A$. Consequently, the triangles $B_0B_1B_2$ and $C_0C_1C_2$ are perspective; the lines $B_iC_i$ are concurrent (the converse of Desargues' theorem).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $a$, prove that $\\sigma(am) < \\sigma(am + 1)$ for infinitely many positive integers $m$. (Here $\\sigma(n)$ is the sum of all positive divisors of the positive integer number $n$.)\nVlad Matei",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given an integer $N > a$, we claim that there exist an integer $d$ and a prime $p$, both greater than $N$, such that $d$ divides $ap+1$, $d$ and $(ap+1)/d$ are coprime, and $\\sigma(d)/d > \\sigma(a)$. In this case,\n$$\n\\sigma(ap + 1) = \\sigma\\left(\\frac{ap+1}{d} \\cdot d\\right) = \\sigma\\left(\\frac{ap+1}{d}\\right) \\sigma(d) > \\frac{ap+1}{d} \\cdot \\sigma(d) \\\\ > (p+1)\\sigma(a) = \\sigma(p)\\sigma(a) = \\sigma(ap),\n$$\nand we are done. Back to the claim, let $p_i$ be the $i$-th prime greater than $N$, take $k$ large enough so that $\\sum_{i=1}^k 1/p_i > \\sigma(a)$ – this is possible, for $\\sum_{q \\text{ prime}} 1/q = \\infty$ – and set $d = p_1p_2\\cdots p_k$. Then\n$$\n\\sigma(d)/d = \\prod_{i=1}^{k} \\left(1 + \\frac{1}{p_i}\\right) > 1 + \\sum_{i=1}^{k} \\frac{1}{p_i} > \\sigma(a).\n$$\nNext, use the Chinese remainder theorem to produce an integer $t$, which is unique modulo $p_1^2 p_2 \\cdots p_k^2$, such that $at + 1 \\equiv p_i \\pmod{p_i^2}$, $i = 1, 2, \\dots, k$; this is possible, for each $p_i > N > a$. Finally, use Dirichlet's theorem to pick up a prime $p > N$ from the arithmetic sequence\n$$\nt + rp_1^2 p_2^2 \\cdots p_k^2, \\quad r = 0, 1, 2, \\dots\n$$\nClearly, such a $p$ satisfies the stated conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ and $Y$ be two finite subsets of the half-open interval $[0, 1)$ such that $0 \\in X \\cap Y$ and $x + y \\neq 1$ for all $x \\in X$ and $y \\in Y$. Prove that the set $\\{x + y - \\lfloor x + y \\rfloor \\mid x \\in X \\text{ and } y \\in Y\\}$ has at least $|X| + |Y| - 1$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S(X, Y) = \\{x + y - \\lfloor x + y \\rfloor : x \\in X \\text{ and } y \\in Y\\}$ and proceed by induction on $|Y|$.\n\nIf $|Y| = 1$, the statement is clear.\n\nAssume henceforth $|Y| > 1$ and let $y_0 \\in Y$, $y_0 \\neq 0$. The conditions $y_0 \\neq 0$ and $x + y_0 = 1$ for no $x \\in X$ imply that $0 \\notin S(X, y_0)$. Since $0 \\in X$, and $X$ and $S(X, y_0)$ have the same cardinality, there are elements that lie in $S(X, y_0)$ but not in $X$; that is, $x_0 + y_0 - \\lfloor x_0 + y_0 \\rfloor$ for some $x_0 \\in X$.\n\nLet\n$$\nY_0 = \\{y : y \\in Y \\text{ and } x_0 + y - \\lfloor x_0 + y \\rfloor \\notin X\\}\n$$\nand\n$$\nX_0 = \\{x_0 + y - \\lfloor x_0 + y \\rfloor : y \\in Y_0\\}.\n$$\nClearly, $0 \\notin Y_0$, $X_0$ and $Y_0$ are both non-empty and have the same cardinality, so $0 < |X_0| = |Y_0| < |Y|$.\n\nLet further $X' = X \\cup X_0$ (notice that this is a disjoint union) and $Y' = Y \\setminus Y_0$, so $X$ is a proper non-empty subset of $X'$, $Y'$ is a proper non-empty subset of $Y$, and\n$$\n|X'| + |Y'| = |X| + |X_0| + |Y| - |Y_0| = |X| + |Y|. \\quad (*)\n$$\nObviously, $X'$ and $Y'$ both contain $0$ and we now show that $S(X', Y') \\subseteq S(X, Y)$ and $x' + y' = 1$ for no $x' \\in X'$ and no $y' \\in Y'$.\n\nClearly, we need consider only the case $x' \\in X_0$; that is, $x' = x_0 + y - \\lfloor x_0 + y \\rfloor$ for some $y \\in Y_0$. Thus, if $y' \\in Y'$, then\n$$\n\\begin{align*}\nx' + y' - \\lfloor x' + y' \\rfloor &= x_0 + y - \\lfloor x_0 + y \\rfloor + y' - \\lfloor x_0 + y - \\lfloor x_0 + y \\rfloor + y' \\rfloor \\\\\n&= x_0 + y + y' - \\lfloor x_0 + y + y' \\rfloor \\\\\n&= (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y - \\lfloor (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y \\rfloor .\n\\end{align*}\n$$\nNotice that $x_0 + y' - \\lfloor x_0 + y' \\rfloor \\in X$, by the definition of $Y'$, to infer that $x' + y' - \\lfloor x' + y' \\rfloor \\in S(X, Y_0) \\subseteq S(X, Y)$, and thereby $S(X', Y') \\subseteq S(X, Y)$.\n\nFinally, write\n$$\n\\begin{align*}\nx' + y' &= x_0 + y - \\lfloor x_0 + y \\rfloor + y' \\\\\n&= (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y - \\lfloor x_0 + y \\rfloor + \\lfloor x_0 + y' \\rfloor,\n\\end{align*}\n$$\nrecall that $x_0 + y' - \\lfloor x_0 + y' \\rfloor \\in X$ and consider the possible values of $\\lfloor x_0 + y \\rfloor$ and $\\lfloor x_0 + y' \\rfloor$ to conclude that $x' + y' \\neq 1$.\n\nConsequently,\n$$\n\\begin{align*}\n|S(X, Y)| &\\ge |S(X', Y')| && \\text{for } S(X', Y') \\subseteq S(X, Y) \\\\\n&\\ge |X'| + |Y'| - 1 && \\text{by the induction hypothesis} \\\\\n&= |X| + |Y| - 1. && \\text{by } (*)\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$, $q$, $r$ such that $p(p-7) + q(q-7) = r(r-7)$.",
    "options": [],
    "answer": "(2, 7, 5), (7, 2, 5), (7, t, t), (t, 7, t) for any prime t",
    "solution": "The given equality rewrites as $p^2 + q^2 - r^2 = 7(p + q - r)$. Since $(p+q-r)(p+q-r) = p^2 + q^2 - r^2 + 2pq$, it follows that $p+q-r$ divides $2pq$.\nIf $p$, $q$, $r > 2$, then $p+q-r$ is odd, so $p+q-r = p$, $q$ or $pq$. The first case gives $r = q$, then $p = 7$, so $(p, q, r) = (7, q, q)$, $q$ being an arbitrary prime. Likewise, the second case gives $(p, q, r) = (p, 7, p)$, $p$ being an arbitrary prime. If $p+q-r = pq$, then $1-r = (p-1)(q-1)$, impossible.\nIf $p=2$ (or $q=2$) the equality becomes $(q-r)(q+r-7) = 10$, implying $q=7$ and $r=5$, with two new solutions obtained, $(p, q, r) = (2, 7, 5)$ and $(p, q, r) = (7, 2, 5)$, or $r=2$, with no further solutions obtained.\nThe solution triplets therefore are $(2, 7, 5)$, $(7, 2, 5)$, $(7, t, t)$, $(t, 7, t)$, where $t$ is an arbitrary prime. (Notice the fact $r$ was known to be a prime turned out to be inconsequential).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37701,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) There exists a unique sequence of positive integers $a_1, a_2, a_3, \\dots$ such that\n$$\nn = \\sum_{d|n} a_d, \\quad \\text{for all } n \\in \\mathbb{N}^*.\n$$\nb) There exists a unique sequence of positive integers $b_1, b_2, b_3, \\dots$ such that\n$$\nn = \\prod_{d|n} b_d, \\quad \\text{for all } n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "a) a_n equals Euler’s totient function: a_n = φ(n). b) b_1 = 1; if n is a positive power of a single prime p then b_n = p; if n has at least two distinct prime factors then b_n = 1.",
    "solution": "a) Euler's totient $\\varphi$ provides the desired sequence, since $\\sum_{d|n} \\varphi(d) = n$. Indeed, consider the fractions $\\frac{1}{n}, \\frac{2}{n}, \\dots, \\frac{n}{n}$ expressed in lowest terms. For each divisor $d$ of $n$, the fractions with denominator equal to $d$ are precisely those having the numerators coprime with $d$ – in all, there are $\\varphi(d)$ such fractions. Since there are $n$ fractions, the formula holds true. A simple inductive argument ensures the uniqueness of the sequence $a_n = \\varphi(n)$. Indeed, given $a_1 = 1$ and all terms $a_k$ up to $k = n-1$, one has $a_n = n - \\sum_{d|n,\\ d<n} a_d$.\n\nb) Set $b_n = 1$ if $n = 1$ or if $n$ has at least two distinct prime divisors. For $n = p^k$, $p$ prime and $k \\ge 1$, set $b_n = p$. To show that the sequence $(b_n)_{n \\ge 1}$ satisfies the relation, consider $n = p_1^{a_1} \\cdots p_k^{a_k}$, the set $A$ of all divisors of $n$ having exactly one prime factor, and the set $B$ of the remaining divisors. Then\n$$\n\\prod_{d|n} b_d = \\left( \\prod_{d \\in A} b_d \\right) \\cdot \\left( \\prod_{d \\in B} b_d \\right) = p_1^{a_1} \\cdots p_k^{a_k} = n,\n$$\nas needed. The uniqueness of the sequence can be proved as above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle. The tangents to the nine-point circle at the foot of the perpendicular dropped from $A$ on the line $BC$ and at the midpoint of the side $BC$ meet at the point $A'$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent.\n\nGazeta Matematică",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The tangent at $A$ to the circumcircle $ABC$ meets the line $BC$ at the point $A''$; the points $B''$ and $C''$ are defined similarly. The points $A''$, $B''$ and $C''$ are collinear on Lemoine's line. We shall prove that the lines $AA'$, $BB'$ and $CC'$ are the polars of the points $A''$, $B''$ and $C''$, respectively, relative to the nine-point circle $\\gamma$, so they are indeed concurrent. Clearly, it is sufficient to prove that that $AA'$ is the polar of $A''$ with respect to $\\gamma$.\n\nLet $A_1, B_1$ and $C_1$ be the perpendicular feet dropped from $A, B$ and $C$, respectively, on the lines $BC, CA$ and $AB$, respectively. Let further $A_2$ be the midpoint of the side $BC$, and let $A_3$ be the midpoint of the segment joining $A$ to the orthocenter of the triangle $ABC$. It is easily seen that the line $A_2A_3$ is the perpendicular bisector of the segment $B_1C_1$, so it is perpendicular to the tangent at $A$ to the circumcircle $ABC$. Consequently, $A_3$ is the orthocenter of the triangle $AA''A_2$, so the lines $AA_2$ and $A''A_3$ are perpendicular; it is easily seen that they meet at some point on $\\gamma$, so $A''$ lies on the polar of $A$ with respect to $\\gamma$. Finally, $A''$ lies on $BC$, which is the polar of $A'$ with respect to $\\gamma$, so $A''$ is the pole of $AA'$ with respect to $\\gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{L}$ be a finite collection of lines in the plane in general position (no two lines in $\\mathcal{L}$ are parallel and no three are concurrent). Consider the open circular discs inscribed in the triangles enclosed by each triple of lines in $\\mathcal{L}$. Determine the number of such discs intersected by no line in $\\mathcal{L}$, in terms of $|\\mathcal{L}|$.",
    "options": [],
    "answer": "C(|L|-1, 2)",
    "solution": "The complement of the union of all lines in $\\mathcal{L}$ is the disjoint union of\n$$\n\\begin{pmatrix} |\\mathcal{L}| \\\\ 0 \\end{pmatrix} + \\begin{pmatrix} |\\mathcal{L}| \\\\ 1 \\end{pmatrix} + \\begin{pmatrix} |\\mathcal{L}| \\\\ 2 \\end{pmatrix}\n$$\nopen convex sets called *rooms*, of which exactly\n$$\n\\begin{pmatrix} |\\mathcal{L}| - 1 \\\\ 2 \\end{pmatrix}\n$$\nare bounded. Let $\\mathcal{D}_0 = \\mathcal{D}_0(\\mathcal{L})$ denote the set of the discs we are to count. Clearly, each disc in $\\mathcal{D}_0$ is contained in some bounded room. We shall prove that each bounded contains exactly one such, whence the conclusion. To this end, we shall first prove that no bounded room contains more than one disc in $\\mathcal{D}_0$, and then that each bounded room contains at least one such.\n\nSuppose, if possible, that $R$ is a bounded room which contains two discs in $\\mathcal{D}_0$; one, of radius $r$, centered at $\\omega$, inscribed in the triangle $abc$ enclosed by the lines $a, b$ and $c$ in $\\mathcal{L}$; and another, of radius $r'$, centered at $\\omega'$, inscribed in the triangle $a'b'c'$ enclosed by the lines $a', b'$ and $c'$ in $\\mathcal{L}$. Clearly, the lines $a, b, c, a', b'$ and $c'$ support edges on the boundary of $R$, the triangles $abc$ and $a'b'c'$ both contain $R$, and $\\omega$ and $\\omega'$ are distinct. Without loss of generality, we may assume that $r \\le r'$. Since $\\omega'$ lies in the interior of the triangle $abc$ and is different from $\\omega$, it is within $r \\le r'$ from one of the lines $a, b$ or $c$. Consequently, that line intersects the disc of radius $r'$ centered at $\\omega'$ – a contradiction.\n\nWe now proceed to prove that each bounded room contains a disc in $\\mathcal{D}_0$. Since $R$ is convex, and no two lines in $\\mathcal{L}$ are parallel, among those lines in $\\mathcal{L}$ which support edges on the boundary of $R$, at least three enclose a triangle which contains $R$ – this is a well-known fact about convex polygons different from a parallelogram. Of all such triangles, choose one with a minimal inradius. We shall prove that the open circular disc $D$ inscribed in that triangle is contained in $R$; in particular, $D$ is in $\\mathcal{D}_0$. Let the triangle be enclosed by the lines $a, b$ and $c$ in $\\mathcal{L}$. Since each of the lines $a, b$ and $c$ supports an edge on the boundary of $R$, it follows that $D$ and $R$ are not disjoint. Suppose, if possible, that $D$ is not contained in $R$. Then $D$ contains points on some edge on the boundary of $R$. Let $d$ be a line in $\\mathcal{L}$ that supports such an edge. Clearly, $d$ is different from $a, b$ and $c$, and it is not hard to see that $d$ and two of the lines $a, b, c$ enclose a triangle which contains $R$ and has an inradius smaller than the radius of $D$ – a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37704,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian plane are given $51$ points of integer coordinates, with pairwise integer distances. Prove that more than $49\\%$ of the distances are even integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Among the given $51$ points there cannot exist simultaneously a point with odd coordinates and a point with even coordinates. Indeed, that be the case, let $(2a, 2b)$ and $(2c+1, 2d+1)$ be such points. The square of the distance between them is $(2a - 2c - 1)^2 + (2b - 2d - 1)^2 = M4 + 2$, which is not the square of an integer.\n\nBy the same argument as above, if an (even, odd)-coordinate point existed, there cannot exist an (odd, even)-coordinate point.\n\nLet now $k$ be the number of points with coordinates of the same parity, then notice that all distances among them are even integers. So there exist $51 - k$ points with coordinates of different parity, again all distances among them being even integers.\n\nTherefore, at least $\\binom{k}{2} + \\binom{51-k}{2}$ distances are even numbers (by definition $\\binom{n}{k} = 0$ if $k > n$). Since there are $\\binom{51}{2}$ distances in all, one has\n$$\n\\begin{aligned}\n\\frac{\\binom{k}{2} + \\binom{51-k}{2}}{\\binom{51}{2}} &= \\frac{k(k-1) + (51-k)(50-k)}{51 \\cdot 50} \\\\\n&= 1 + \\frac{k^2 - 51k}{51 \\cdot 25} \\ge 1 - \\frac{51^2}{51 \\cdot 25} = 1 - \\frac{51}{100} = 49\\%.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point of the plane is coloured in one of two colours. Given an odd integer number $n \\ge 3$, prove that there exist (at least) two similar triangles whose similitude ratio is $n$, each of which has a monochromatic vertex-set.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that there exists a rectangle with monochromatic vertices, then subdivide it into $n^2$ rectangles by subdividing each side into $n$ congruent segments, and show that (at least) one of these smaller rectangles has (at least) three monochromatic vertices – here is where the assumption that $n$ be odd comes in.\nTo find a rectangle with monochromatic vertices, consider a line $a$ in the plane and a five-element monochromatic subset $A$ of $a$. Project $A$ orthogonally on another line $b$ which is parallel to $a$, and consider a three-element monochromatic subset $B$ of the image of $A$ under projection. If the colours of $A$ and $B$ agree, we are done. Otherwise, project $B$ on a third line $c$ which is parallel to both $a$ and $b$, and consider a two-element monochromatic subset $C$ of the image of $B$ under projection. Clearly, the two points of $C$ together with their orthogonal projections on either $A$ or $B$ are monochromatic.\nWithout loss of generality, we may (and will) assume that the vertices of the square $[0, n] \\times [0, n]$ are monochromatic. Since $n$ is odd, and the points $0 \\times 0$ and $n \\times 0$ (respectively, $0 \\times n$) are monochromatic, there exists $p$ (respectively, $q$) in $\\{0, 1, \\dots, n-1\\}$ such that the points $p \\times 0$ and $(p+1) \\times 0$ (respectively, $0 \\times q$ and $0 \\times (q+1)$) be monochromatic. We now show that (at least) one of the unit squares\n$$\n[i, i+1] \\times [q, q+1], \\quad i = 0, 1, \\dots, p,\n$$\n$$\n[p, p+1] \\times [j, j+1], \\quad j = 0, 1, \\dots, q,\n$$\nhas (at least) three monochromatic vertices. Suppose that each of the first $p$ (respectively, $q$) horizontal (respectively, vertical) unit squares above has two vertices of each colour. Then the colours of the lattice points alternate in the same way along both horizontal segments $[0, p] \\times q$ and $[0, p] \\times (q+1)$, according to the parity of the abscissae, and in the same way along both vertical segments $p \\times [0, q]$ and $(p+1) \\times [0, q]$, according to the parity of the ordinates. Consequently, the vertical pair $(p \\times q, p \\times (q+1))$ and the horizontal pair $(p \\times q, (p+1) \\times q)$ are both monochromatic; that is, the unit square $[p, p+1] \\times [q, q+1]$ has (at least) three monochromatic vertices.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > 5$ be a prime. Find all positive integers $x$ such that $5p + x$ divides $5p^n + x^n$, for all $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "All positive integers x such that 5p + x divides 30p^2, equivalently x = d − 5p where d runs over the divisors of 30p^2 greater than 5p. Explicitly: x ∈ {p, 5p, 10p, 25p, p(p−5), p(2p−5), p(3p−5), 5p(p−1), p(6p−5), 5p(2p−1), 5p(3p−1), 5p(6p−1)}.",
    "solution": "$1^\\circ \\ 5p + x \\mid 5p^n + x^n \\text{ for all } n \\ge 1;$\n$2^\\circ \\ 5p + x \\mid 30p^2.$\nTo prove $1^\\circ \\Rightarrow 2^\\circ$, set $n = 2$ to obtain $5p+x \\mid 5p^2+x^2 \\Rightarrow 5p+x \\mid (x+5p)(x-5p) + 30p^2$, hence $5p+x \\mid 30p^2$, as needed. For the second implication, observe that $5p+x \\mid 30p^2$ implies $5p+x \\mid (x+5p)(x-5p) + 30p^2$, so $5p+x \\mid 5p^2+x^2$. The identity $5p^{n+1} + x^{n+1} = (5p^n + x^n)(p+x) - px(5p^{n-1} + x^{n-1})$ holds for all integers $n \\ge 2$. Consequently, by induction, the claim $1^\\circ$ is proved.\nThe divisors of $30p^2$ greater than $5p$ are $6p, 10p, 15p, 30p, p^2, 2p^2, 3p^2, 5p^2, 6p^2, 10p^2, 15p^2$ and $30p^2$. Subtracting $5p$ from the numbers listed above yields the required values for $x$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\ell$ be a line, and let $\\gamma$ and $\\gamma'$ be two circles. The line $\\ell$ meets $\\gamma$ at points $A$ and $B$, and $\\gamma'$ at points $A'$ and $B'$. The tangents to $\\gamma$ at $A$ and $B$ meet at point $C$, and the tangents to $\\gamma'$ at $A'$ and $B'$ meet at point $C'$. The lines $\\ell$ and $CC'$ meet at point $P$. Let $\\lambda$ be a variable line through $P$ and let $X$ be one of the points where $\\lambda$ meets $\\gamma$, and $X'$ be one of the points where $\\lambda$ meets $\\gamma'$. Prove that the point of intersection of the lines $CX$ and $C'X'$ lies on a fixed circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the lines $CX$ and $C'X'$ meet at point $Q$. The line $CX$ meets $\\ell$ at $D$, and $\\gamma$ a second time at $Y$; similarly, the line $C'X'$ meets $\\ell$ at $D'$, and $\\gamma'$ a second time at $Y'$. Notice that the cross-ratios $(CDXY)$ and $(C'D'X'Y')$ are both harmonic, for $C$ and $C'$ are the poles of $\\ell$ relative to $\\gamma$ and $\\gamma'$, respectively. Since the lines $CC'$, $DD'$ and $XX'$ all pass through $P$, so does the line $YY'$. Consequently, $(QCXD) = (QC'X'D')$ and $(QCYD) = (QC'Y'D')$. Let $\\varrho$ and $\\varrho'$ be the powers of $Q$ relative to $\\gamma$ and $\\gamma'$, respectively, let $\\omega$ be the power of $C$ relative to $\\gamma$, and let $\\omega'$ be the power of $C'$ relative to $\\gamma'$. Multiply the last two equalities involving cross-ratios and apply Menelaus' theorem to triangle $QCC'$ and transversal $PDD'$ to get\n$$\n\\frac{\\varrho}{\\varrho'} = \\frac{\\omega}{\\omega'} \\left( \\frac{C'D'}{QD'} \\cdot \\frac{QD}{CD} \\right)^2 = \\frac{\\omega}{\\omega'} \\left( \\frac{PC'}{PC} \\right)^2 = \\text{constant}\n$$\nand infer that $Q$ lies on a fixed circle of the pencil of circles generated by $\\gamma$ and $\\gamma'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $D, E, F$ be the midpoints of the sides $BC, CA, AB$. Prove that $\\angle DAC = \\angle ABE$ if and only if $\\angle AFC = \\angle BDA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the centroid of the given triangle. Since $DF \\parallel AC$, one has $\\angle DAC = \\angle GDF$.\n\nIf $\\angle DAC = \\angle ABE$, then $\\angle GDF = \\angle FBG$, hence the quadrilateral $BFGD$ is cyclic, implying $\\angle AFC = \\angle BDA$. The converse holds by the same argument.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number, let $n_1, n_2, \\dots, n_p$ be positive integer numbers, and let $d$ be the greatest common divisor of the numbers $n_1, n_2, \\dots, n_p$. Prove that the polynomial\n$$\n\\frac{X^{n_1} + X^{n_2} + \\dots + X^{n_p} - p}{X^d - 1}\n$$\nis irreducible in $\\mathbb{Q}[X]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, $f = (X^{n_1}+X^{n_2}+\\dots+X^{n_p}-p)/(X^d-1)$ is a polynomial with integral coefficients. If all $n_i$ are equal, then $f = p$, a constant polynomial; therefore assume at least two of the $n_i$ are distinct. By Gauss' Lemma, it is sufficient to prove $f$ irreducible in $\\mathbb{Z}[X]$. We claim that the roots of $f$ all lie outside the closed unit disc in the complex plane. Assuming the claim, suppose, if possible, that $f = gh$ is a non-trivial factorization of $f$ in $\\mathbb{Z}[X]$. Since $f(0) = p$, a prime, one of the numbers $|g(0)|$, $|h(0)|$ is 1. To reach a contradiction, notice that $g(0)$ and $h(0)$ are both products of roots of $f$, all of which lie outside the closed unit disc, so $|g(0)|$ and $|h(0)|$ are both greater than 1.\n\nBack to the claim, write\n$$\nX^{n_1} + X^{n_2} + \\dots + X^{n_p} - p = (X^d - 1)f. \\quad (*)\n$$\nSuppose, if possible, that $f$ has a root $z$ in the closed unit disc. Then\n$$\np = |z^{n_1} + z^{n_2} + \\dots + z^{n_p}| \\le |z|^{n_1} + |z|^{n_2} + \\dots + |z|^{n_p} \\le p,\n$$\nwhich forces $|z| = 1$ and $z^{n_k - n_\\ell} > 0$ whatever $k$ and $\\ell$. Consequently, the $z^{n_k}$ are all equal to some complex number $w$, so\n$$\np(w - 1) = z^{n_1} + z^{n_2} + \\dots + z^{n_p} - p = 0;\n$$\nthat is, $w = 1$. Now write $d = n_1 t_1 + n_2 t_2 + \\dots + n_p t_p$, for some integer numbers $t_1, t_2, \\dots, t_p$, to get\n$$\nz^d = z^{n_1 t_1 + n_2 t_2 + \\dots + n_p t_p} = (z^{n_1})^{t_1} (z^{n_2})^{t_2} \\dots (z^{n_p})^{t_p} = 1.\n$$\nFinally, to reach a contradiction, evaluate the formal derivatives of both sides of $(*)$ at $z$. The left-hand side is $(n_1 + n_2 + \\dots + n_p)/z$, while the right-hand side vanishes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37710,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a ring $A$ has property (P) if any non-zero element can be written uniquely as the sum of an invertible element and a non-invertible element.\n\na) If in $A$, $1 + 1 \\neq 0$, prove that $A$ has property (P) if and only if $A$ is a field.\n\nb) Give an example of a ring that is not a field, containing at least two elements, and having property (P).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $A$ is a field and $x \\in A$, $x \\ne 0$, then $x$ is invertible and $x = x + 0$; this representation is unique, since $0$ is the only noninvertible element.\n\nAssume $A$ is not a field. Let $x \\in A$, $x \\ne 0$, be a noninvertible element. Since $1 + x = (1 + x) + 0$ and $-1 + x = (-1 + x) + 0$, the elements $1 + x$ and $-1 + x$ are noninvertible, otherwise the uniqueness of the representation would imply $x = 0$.\n\nSince $x = 1 + (-1 + x) = -1 + (1 + x)$, it follows $1 = -1$, contradiction.\n\nb) Let $E$ be a nonempty set, and $A = (\\mathcal{P}(E), \\Delta, \\cap)$ be the Boolean ring of the set of parts of $E$. In this ring, $E$ is the only invertible element.\n\nIf $X$ is a nonempty part of $E$, then the element $E \\setminus X$ is noninvertible, and $X = E \\Delta (E \\setminus X)$. The representation is unique, since $E$ is the only invertible element.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, $n \\ge 2$. Find the remainder of the division of the number $n(n+1)(n+2)$ by $n-1$.",
    "options": [],
    "answer": "n(n+1)(n+2) ≡ 6 (mod n−1). Hence the remainder is 6 for n ≥ 8; for n = 2,3,4,7 it is 0; for n = 5 it is 2; for n = 6 it is 1.",
    "solution": "We notice that $n(n+1)(n+2) = (n-1+1)(n-1+2)(n-1+3) = (n-1)^3 + 6(n-1)^2 + 11(n-1) + 6$. If $n-1 > 6$, the residue is $6$.\n\nIf $n = 2, 3, 4, 7$, the residue is $0$. If $n = 5$, the residue is $2$. If $n = 6$, the residue is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37712,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many four digit numbers $abcd$ simultaneously satisfy the equalities $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$?",
    "options": [],
    "answer": "171",
    "solution": "From $a + b = c + d$ we get $(a + b)^2 = (c + d)^2$, hence $ab = cd$ and furthermore $a^2 - 2ab + b^2 = c^2 - 2cd + d^2$. As $(a - b)^2 = (c - d)^2$, we have $|a - b| = |c - d|$, which implies $a - b = c - d$ or $a - b = d - c$. Recall that $a + b = c + d$, so either $a = c, b = d$ or $a = d, b = c$.\nThe numbers must have one of the forms $\\overline{aaaa}$, $\\overline{abba}$ or $\\overline{abab}$, with $a \\neq 0$ and $a \\neq b$. In all, there are $9 + 9 \\cdot 9 + 9 \\cdot 9 = 9 + 81 + 81 = 171$ numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a \\in [-2, \\infty)$, $r \\in [0, \\infty)$ and let $n$ be a positive integer.\nShow that\n$$\nr^{2n} + a r^n + 1 \\geq (1 - r)^{2n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $r = 0$, the relation is obvious.\nOtherwise, dividing by $r^{2n}$, one gets the same inequality with $r$ replaced by $\\frac{1}{r}$, so one can assume $r \\in (0, 1]$.\nSince $r^{2n} + a r^n + 1 \\geq r^{2n} - 2 r^n + 1 = (1 - r^n)^2$ and $1 - r^n \\geq 0$, it is enough to prove that $1 - r^n \\geq (1 - r)^n$.\nThis follows from $r^n + (1 - r)^n \\leq r + (1 - r) = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C \\in M_n(\\mathbb{R})$ be such that $ABC = 0_n$ and $\\operatorname{rank} B = 1$. Prove that $AB = O_n$ or $BC = O_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K \\in \\mathcal{M}_{n,1}(\\mathbb{C})$ and $L \\in \\mathcal{M}_{1,n}(\\mathbb{C})$ be such that $KL = B$.\nThen $O_n = (AK)(LB)$ with $AK \\in \\mathcal{M}_{n,1}(\\mathbb{C})$, $LB \\in \\mathcal{M}_{1,n}(\\mathbb{C})$.\nLet $AK = {}^t(a_1, a_2, \\dots, a_n)$ (where ${}^t X$ denotes the transpose of $X$), and let $LB = (b_1, b_2, \\dots, b_n)$. We infer\n$$\nO_n = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ \\vdots & \\vdots & \\cdots & \\vdots \\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\end{pmatrix}\n$$\nIf $LB \\neq O_{1,n}$ then there exists $1 \\le i \\le n$ with $b_i \\neq 0$.\nIt follows $a_1 = a_2 = \\cdots = a_n = 0$, hence $AK = O_{n,1}$, and so at least one of the matrices $AK$ and $LB$ is null.\nIn the former case $AB = (AK)L = O_{n,1}L = O_n$, while in the latter $BC = K(LC) = KO_{1,n} = O_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $v, w$ two distinct non-zero complex numbers. Prove that\n$$\n|zw + \\bar{w}| \\leq |zv + \\bar{v}|,\n$$\nfor any $z \\in \\mathbb{C}$, $|z| = 1$, if and only if there exists $k \\in [-1, 1]$ such that $w = kv$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If there exists $k \\in [-1, 1]$ such that $w = kv$, the inequality is obvious. Conversely, let $t > 1$ such that $w - tv \\ne 0$. Setting\n$$\nz = \\frac{t\\bar{v} - \\bar{w}}{w - tv},\n$$\nwe have $|z| = 1$ and moreover\n$$\nzw + \\bar{w} = \\frac{t(w\\bar{v} - \\bar{w}v)}{w - tv},\n$$\n$$\nzv + \\bar{v} = \\frac{w\\bar{v} - \\bar{w}v}{w - tv},\n$$\nimplying\n$$\n|zw + \\bar{w}| = t |zv + \\bar{v}|.\n$$\nAs $t > 1$, the given inequality yields $|zw + \\bar{w}| = |zv + \\bar{v}| = 0$, so $w\\bar{v} - \\bar{w}v = 0$, therefore $\\frac{w}{v} = k \\in \\mathbb{R}$.\nPlugging back in the initial condition we derive that $k \\in [-1, 1]$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n > 1$ be an integer. Point $M$ lies on the line segment $AB$ such that $nAM = AB$. Consider the points $P_1, P_2, \\dots, P_{n-1}$ on the side $BC$ with $BP_1 = P_1P_2 = P_2P_3 = \\dots = P_{n-1}C = \\frac{1}{n}BC$. Prove that\n$$\n\\angle MP_1A + \\angle MP_2A + \\dots + \\angle MP_{n-1}A = \\frac{1}{2} \\angle BAC.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the point $N$ on the side $AC$ such that $nAN = AC$. The configuration is symmetric with respect to the perpendicular bisector of the segment $BC$, implying $\\angle MP_iA = \\angle NP_{n-i}A$, $i = 1, 2, \\dots, n-1$. The claim is equivalent to $\\angle MP_1N + \\angle MP_2N + \\dots + \\angle MP_{n-1}N = \\angle BAC$.\n\nNotice that $BP_1 = P_1P_2 = P_2P_3 = \\dots = P_{n-1}C = MN$. Set $P_0 = B$, and notice that in the parallelograms $P_iMNP_{i+1}$ one has $\\angle MP_{i+1}N = \\angle P_iMP_{i+1}$, $i = 0, 1, \\dots, n-2$. Therefore the sum of those angles is equal to $\\angle P_0MP_{n-1}$, in turn equal to $\\angle BAC$, since $MP_{n-1} \\parallel AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37717,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$, find the largest real number $C_n$ with the following property. Given any $n$ real-valued functions $f_1(x), f_2(x), \\dots, f_n(x)$ defined on the closed interval $0 \\le x \\le 1$, one can find numbers $x_1, x_2, \\dots, x_n$, such that $0 \\le x_i \\le 1$, satisfying\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| \\ge C_n.\n$$",
    "options": [],
    "answer": "(n-1)/(2n)",
    "solution": "First we will prove that $C_n \\ge \\frac{n-1}{2n}$, i.e. that for any $n$ functions $f_1, f_2, \\dots, f_n : [0, 1] \\to \\mathbb{R}$, there exist numbers $x_1, x_2, \\dots, x_n$ in $[0, 1]$ such that\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| \\ge \\frac{n-1}{2n}.\n$$\nFor $n=1$ this is trivial. For $n \\ge 2$ suppose, contrariwise, that for all $x_1, x_2, \\dots, x_n$ in $[0, 1]$ we have\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| < \\frac{n-1}{2n}.\n$$\nPlugging in $x_i = 1$ for $1 \\le i \\le n$, we get $\\left| \\sum_{i=1}^{n} f_i(1) - 1 \\right| < \\frac{n-1}{2n}$.\nPlugging in $x_i = 0$ for $1 \\le i \\le n$, we get $\\left| \\sum_{i=1}^{n} f_i(0) \\right| < \\frac{n-1}{2n}$.\nPlugging in (for every $1 \\le i \\le n$) $x_i = 0$ and $x_j = 1$ for all $j \\ne i$, we get\n$$\n\\left| f_i(0) + \\sum_{j \\ne i} f_j(1) \\right| < \\frac{n-1}{2n}.\n$$\nSince\n$$\n(n-1) \\sum_{i=1}^{n} f_i(1) = \\sum_{i=1}^{n} \\left( f_i(0) + \\sum_{j \\ne i} f_j(1) \\right) - \\sum_{i=1}^{n} f_i(0),\n$$\nby the triangle inequality we have\n$$\n(n-1) \\left| \\sum_{i=1}^{n} f_i(1) \\right| < (n+1) \\frac{n-1}{2n}.\n$$\nOn the other hand, by again the triangle inequality we have\n$$\n1 \\le \\left| \\sum_{i=1}^{n} f_i(1) \\right| + \\left| \\sum_{i=1}^{n} f_i(1) - 1 \\right| < \\frac{n+1}{2n} + \\frac{n-1}{2n} = 1,\n$$\nwhich is a contradiction.\n\nTo prove that\n$$\nC_n = \\frac{n-1}{2n}\n$$\nis the largest constant, it will be sufficient to prove that for the $n$ (equal) functions\n$$\nf_i(x) = f(x) := \\frac{x^n}{n} - \\frac{n-1}{2n^2}, \\quad 1 \\le i \\le n,\n$$\nand any $n$ numbers $x_1, x_2, \\dots, x_n$ in $[0, 1]$, we have\n$$\n\\left| f(x_1) + f(x_2) + \\cdots + f(x_n) - x_1 x_2 \\cdots x_n \\right| \\le \\frac{n-1}{2n},\n$$\nequivalent to\n$$\n-\\frac{n-1}{2n} \\le \\frac{1}{n} \\sum_{i=1}^{n} x_i^n - \\prod_{i=1}^{n} x_i - \\frac{n-1}{2n} \\le \\frac{n-1}{2n}.\n$$\nThe LHS inequality follows from the AM-GM inequality.\n\nThe RHS inequality is equivalent to\n$$\nF(\\mathbf{x}) = F(x_1, x_2, \\dots, x_n) := \\frac{1}{n} \\sum_{i=1}^{n} x_i^n - \\prod_{i=1}^{n} x_i \\le \\frac{n-1}{n}\n$$\nat all points $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ of the hypercube $[0, 1]^n$. Since $F$ is convex in every variable, its maximum is reached at some vertex $\\mathbf{v}$ of the hypercube (point with $x_i = 0$ or $x_i = 1$, for all $1 \\le i \\le n$). It is easy to see that for all such points we have $F(\\mathbf{v}) \\le \\frac{n-1}{n}$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a group in which $a^2b = ba^2$ implies $ab = ba$.\n\na) Show that if $G$ has $2^n$ elements, then $G$ is an abelian group.\n\nb) Exhibit an example of a nonabelian group with the given property.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) For $a \\in G$, let $C(a) = \\{b \\mid b \\in G$ and $ab = ba\\}$. The hypothesis shows that $C(a^2) \\subset C(a)$. Since $C(a) \\subset C(a^2)$, we see that $C(a) = C(a^2)$, for every $a \\in G$. Consequently, $C(a) = C(a^2) = \\dots = C(a^{2^n}) = C(e) = G$, for every $a \\in G$, that is $G$ is abelian.\n\nb) An example of a nonabelian group with the given property is the multiplicative group $G$ of the matrices of the form\n$$\n\\begin{pmatrix} \\hat{1} & a & b \\\\ \\hat{0} & \\hat{1} & c \\\\ \\hat{0} & \\hat{0} & \\hat{1} \\end{pmatrix}, \\text{ with } a, b, c \\in \\mathbb{Z}_3.\n$$\nSince $A^3 = I_3$ for every $A \\in G$, if $A^2B = BA^2$, then $A^{-1}B = BA^{-1}$, so $AB = BA$.\nIn effect, any finite group $G$ of odd order bears the given property, since $C(a^2) \\subset C(a^{|G|+1}) = C(a) \\subset C(a^2)$ (so $C(a) = C(a^2)$), for all $a \\in G$. Therefore any such nonabelian group makes an example for b).",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the plane partitioned in unit squares. The interior of each square is coloured in either *red* or *black* (the sides of the squares are not considered to be coloured). Prove that given any positive integer $\\alpha$ there exists an equilateral triangle of integer area $A \\ge \\alpha$, having monochromatic vertices.\nRadu Gologan",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Like usually in such situations, there is some doubt on the colouring of the separating lines. One idea would be for a unit square to be considered to be made of its interior, together with its left and lower sides, but less its north-west and south-east corners. This establishes a partition of the plane, given by\n$$\n\\bigcup_{(a,b) \\in \\mathbb{Z}^2} \\{(x,y); a \\le x < a+1,\\ b \\le y < b+1\\}.\n$$\n\nAnother idea would be not to colour at all the sides of the unit squares – this has been chosen for the problem at hand. Moreover, this introduces the slight extra difficulty in avoiding the uncoloured points of the plane.\n\nThe key to the proof is the well-known configuration that warrants the existence of an equilateral monochromatic triangle in the bichromatic plane. Choose a value $\\ell$ for the side of an equilateral triangle of integer area $\\alpha$, hence $\\ell = \\sqrt{\\frac{4\\alpha}{\\sqrt{3}}}$. Let $\\gamma$ be the circle of radius $\\ell$ and center some point $A$ coloured $c_1$, and $\\Gamma$ the circle of same center and radius $\\ell\\sqrt{3}$. The meeting points of circles $\\gamma$ and $\\Gamma$ with the sides of the unit squares in the plane are obviously finitely many, and these points must be avoided since being uncoloured.\n\nIf all points of $\\gamma$ (except at most a finite number of them) are coloured $c_2$, then there will exist an equilateral triangle whose vertices bear this colour, and of area $3\\alpha$. Otherwise, let $B \\in \\gamma$ be coloured $c_1$, and consider the regular hexagon $BCDEFG$ inscribed in $\\gamma$. Its vertices must bear colours $C \\to c_2$, $G \\to c_2$, $E \\to c_1$, $D \\to c_2$, $F \\to c_2$, otherwise an equilateral monochromatic triangle of area $\\alpha$ or $3\\alpha$ is made. Let now $H = BC \\cap DE$; clearly $H \\in \\Gamma$. If $H$ is coloured $c_2$, then $\\triangle CDH$ is monochromatic $c_2$ and has area $\\alpha$; if $H$ is coloured $c_1$, then $\\triangle BEH$ is monochromatic $c_1$ and has area $4\\alpha$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is a similarity between a triangle $ABC$ and the triangle having as sides the medians of the triangle $ABC$ if and only if the squares of the lengths of the sides of triangle $ABC$ form an arithmetical sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Recall that $m_a^2 = \\frac{1}{4}(2(b^2 + c^2) - a^2)$, together with the other similar formulas.\n\nAssume that the squares of the lengths of the sides are in arithmetic progression, for example $2b^2 = a^2 + c^2$. Then $m_a^2 = \\frac{3}{4}c^2$, $m_b^2 = \\frac{3}{4}b^2$, $m_c^2 = \\frac{3}{4}a^2$, implying that\n$$\n\\frac{m_a}{c} = \\frac{m_b}{b} = \\frac{m_c}{a} = \\frac{\\sqrt{3}}{2}.\n$$\nObviously, the triangles are similar.\n\nConversely, assume the triangles are similar. Let $a \\le b \\le c$. Then $m_a \\ge m_b \\ge m_c$, so\n$$\n\\frac{m_a}{c} = \\frac{m_b}{b} = \\frac{m_c}{a} = k.\n$$\nFrom the equality $m_a^2 + m_b^2 + m_c^2 = \\frac{3}{4}(a^2 + b^2 + c^2)$ we get $k = \\frac{\\sqrt{3}}{2}$. Hence $4m_b^2 = 3b^2$, which yields $2b^2 = a^2 + c^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer number and let $a_1, a_2, \\dots, a_n$ be $n$ positive real numbers. Prove that $f : [0, \\infty) \\to \\mathbb{R}$, defined by\n$$\nf(x) = \\frac{a_1+x}{a_2+x} + \\frac{a_2+x}{a_3+x} + \\dots + \\frac{a_{n-1}+x}{a_n+x} + \\frac{a_n+x}{a_1+x},\n$$\nis a decreasing function.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $a_{n+1} = a_1$ and let $0 \\le x \\le y$. Since\n$$\nf(y) - f(x) = (y - x) \\sum_{i=1}^{n} \\frac{a_{i+1} - a_i}{(a_{i+1} + x)(a_{i+1} + y)}\n$$\nshowing $f$ is decreasing amounts to showing\n$$\n\\sum_{i=1}^{n} \\frac{a_{i+1}}{(a_{i+1} + x)(a_{i+1} + y)} \\le \\sum_{i=1}^{n} \\frac{a_i}{(a_{i+1} + x)(a_{i+1} + y)}\n$$\nNoticing that $a_i \\le a_j$ if and only if $(a_i + x)^{-1}(a_i + y)^{-1} \\ge (a_j + x)^{-1}(a_j + y)^{-1}$, the above inequality is a straightforward consequence of the rearrangement inequality for the $a_i$ and the $(a_i + x)^{-1}(a_i + y)^{-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37722,
    "subject": "Mathematics (Multi-modal)",
    "question": "A nonconstant polynomial $f$ with integral coefficients has the property that, for each prime $p$, there exist a prime $q$ and a positive integer $m$ such that $f(p) = q^m$. Prove that $f = X^n$ for some positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that for every prime $p$, $f(p) = p^m$, where $m$ is a positive integer which (possibly) depends on $p$. Assume the claim for the time being. Since the degree $n$ of $f$ is positive, the claim forces $f(p) = p^n$ for sufficiently large primes $p$. Consequently, the polynomials $f$ and $X^n$ agree infinitely many times, so $f = X^n$.\n\nBack to the claim, suppose that $f(p) = q^m$ for some distinct primes $p$ and $q$ and some positive integer $m$. Since $q^{m+1}$ divides the difference $f(p + kq^{m+1}) - f(p)$, $k = 1, 2, 3, \\dots$, and $q^m$ divides $f(p)$ but $q^{m+1}$ does not, it follows that $q$ divides $f(p + kq^{m+1})$ but $q^{m+1}$ does not. Use Dirichlet's theorem to choose $k$ so large that $p+kq^{m+1}$ be prime and $f(p+kq^{m+1}) > q^m$. By hypothesis, $f(p+kq^{m+1})$ is a power of a prime. Recall that $q$ divides $f(p+kq^{m+1})$ to deduce that $f(p+kq^{m+1}) = q^r$ for some positive integer $r$. Finally, $f(p+kq^{m+1}) > q^m$ implies $r > m$, so $q^{m+1}$ divides $f(p+kq^{m+1})$ – a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37723,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $8 \\times 8$ array is divided into 64 unit squares. In some of the unit squares a diagonal is drawn, such that no two diagonals share a common point – not even an endpoint. Find the maximum number of diagonals that can be drawn under these circumstances.",
    "options": [],
    "answer": "36",
    "solution": "Look at two contiguous rows of a $2n \\times 2n$ array, containing $k$ diagonals in the cells of row $A$ and $\\ell$ diagonals in the cells of row $B$. The ends of these diagonals on the separating line of the two rows have to be distinct points, and they can only occupy at most the $2n + 1$ available positions, therefore $k + \\ell \\le 2n + 1$. Since the $2n$ rows may be grouped in $n$ sets of pairwise contiguous rows, the total number of diagonals drawn in the array is at most $n(2n + 1) = \\binom{2n+1}{2}$. A model is easily found – only use / type diagonals in the cells of coordinates $(2i + 1, j)$ and $(j, 2i + 1)$, with $1 \\le 2i + 1 \\le j \\le 2n$. The answer for $2n = 8$ is therefore $36$.\n\nAlternative Solution.\n\nColour the nodes of the array alternatively white and black by columns, starting with the left side. Any diagonal drawn will then have its ends of different colour. Since there are exactly $n(2n+1)$ black nodes (while there are $(n+1)(2n+1)$ white nodes), and all diagonals are disjoint, it follows the total number of diagonals drawn in the array is at most $n(2n+1) = \\binom{2n+1}{2}$. At close inspection, it is virtually the same idea as above, in a more elegant setting.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle, let $I$ be its incenter, and let $A_1$, $B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC$, $CA$ and $AB$, respectively. Prove that the circumcircles of the triangles $AIA_1$, $BIB_1$ and $CIC_1$ have a common point different from $I$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem amounts to showing collinearity of the antipodes $A_2$, $B_2$ and $C_2$ of $I$ in the circles $AIA_1$, $BIB_1$ and $CIC_1$, respectively. In the sequel, we use the following standard notations: $I_a$, $I_b$ and $I_c$ are the centers of the excircles tangent to the sides $BC$, $CA$ and $AB$, respectively; $a$, $b$ and $c$ are the lengths of the sides $BC$, $CA$ and $AB$, respectively; and $s = (a + b + c)/2$ is the semiperimeter of the triangle $ABC$. Clearly, the points $A_2$, $B_2$ and $C_2$ lie on the lines $I_bI_c$, $I_cI_a$ and $I_aI_b$, respectively. To show them collinear, we evaluate the ratio $A_2I_b/A_2I_c$ and the like, and refer to the converse of Menelaus' theorem.\n\nTo evaluate the ratio $A_2I_b/A_2I_c$, we apply the Menelaus theorem to triangle $I_aI_bI_c$ and line $A_1A_2$. Let the latter meet the lines $I_aI_b$ and $I_aI_c$ at $P$ and $Q$, respectively, to write\n$$\n\\frac{A_2I_b}{A_2I_c} = \\frac{PI_b}{PI_a} \\cdot \\frac{QI_a}{QI_c}.\n$$\nWe evaluate $PI_a$, $PI_b$, $QI_a$ and $QI_c$ as follows. Consider the cyclic quadrangles $IA_1CP$ and $BICI_a$ to infer that the triangles $IPI_a$ and $IA_1B$ are similar and get thereby\n$$\nPI_a = A_1B \\cdot \\frac{II_a}{IB} = \\frac{s-c}{\\sin \\frac{C}{2}}.\n$$\nSince $I_aI_b = c/\\sin \\frac{C}{2}$, we obtain $PI_b = |2c-s|/\\sin \\frac{C}{2}$. Next, consider the cyclic quadrangles $IA_1QB$ and $BICI_a$ to deduce that the triangles $IQI_a$ and $IA_1C$ are similar and get thereby\n$$\nQI_a = A_1C \\cdot \\frac{II_a}{IC} = \\frac{s-b}{\\sin \\frac{B}{2}}.\n$$\nSince $I_aI_c = b/\\sin \\frac{B}{2}$, we obtain $QI_c = |2b-s|/\\sin \\frac{B}{2}$. Consequently,\n$$\n\\frac{A_2I_b}{A_2I_c} = \\frac{s-b}{s-c} \\cdot \\frac{|2c-s|}{|2b-s|}\n$$\nand the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer number. If $S$ is a finite set of vectors in the plane, let $N(S)$ denote the number of two-element subsets $\\{\\mathbf{v}, \\mathbf{v}'\\}$ of $S$ such that\n$$\n4 (\\mathbf{v} \\cdot \\mathbf{v}') + (|\\mathbf{v}|^2 - 1)(|\\mathbf{v}'|^2 - 1) < 0.\n$$\n\nDetermine the maximum of $N(S)$ when $S$ runs through all $n$-element sets of vectors in the plane.",
    "options": [],
    "answer": "floor(3 n^2 / 8)",
    "solution": "Without loss of generality, we may (and will) assume all vectors anchored at the origin. Assigning to each vector $\\mathbf{v} = (x, y)$ in $\\mathbb{R}^2$ the unit vector\n$$\n\\mathbf{u} = \\left( \\frac{2x}{x^2 + y^2 + 1}, \\frac{2y}{x^2 + y^2 + 1}, \\frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \\right)\n$$\nin $\\mathbb{R}^3$ defines a bijection between all vectors in $\\mathbb{R}^2$ and all unit vectors in $\\mathbb{R}^3$ different from $(0, 0, 1)$; its inverse sends a unit vector $\\mathbf{u} = (x, y, z) \\neq (0, 0, 1)$ to the vector $\\mathbf{v} = (x/(1-z), y/(1-z))$. Clearly, the condition\n$$\n4 (\\mathbf{v} \\cdot \\mathbf{v}') + (|\\mathbf{v}|^2 - 1)(|\\mathbf{v}'|^2 - 1) < 0\n$$\nfor vectors in $\\mathbb{R}^2$ is equivalent to $\\mathbf{u} \\cdot \\mathbf{u}' < 0$ for the corresponding unit vectors in $\\mathbb{R}^3$. Now let $S = \\{\\mathbf{v}_1, \\mathbf{v}_2, \\dots, \\mathbf{v}_n\\}$, let $\\{\\mathbf{u}_1, \\mathbf{u}_2, \\dots, \\mathbf{u}_n\\}$ be the set of corresponding unit vectors in $\\mathbb{R}^3$ and consider the graph $G$ on $n$ vertices labelled $1, 2, \\dots, n$, with an edge joining vertex $i$ to vertex $j$ if and only if $\\mathbf{u}_i \\cdot \\mathbf{u}_j < 0$. Since the largest size of a set of vectors in $\\mathbb{R}^3$ with negative mutual dot-products is $4$, it follows that $G$ is $K_5$-free. Hence, by Turán's theorem, $G$ has at most the following number of edges\n$$\n\\lfloor \\left(1 - \\frac{1}{5-1}\\right) \\cdot \\frac{n^2}{2} \\rfloor = \\lfloor 3n^2/8 \\rfloor,\n$$\nthe upper bound being achieved by Turán's extremal graph $T(n, 5) = K_{n_1, n_2, n_3, n_4}$, where $n_1 + n_2 + n_3 + n_4 = n$ and $|n_i - n_j| \\le 1$. Geometrically, this can be achieved in $\\mathbb{R}^3$ by considering four tight bundles of $n_i$ unit vectors each, around the vectors $(1, 0, 0)$, $(-1, 2, 0)$, $(-1, -1, 3)$ and $(-1, -1, -1)$, respectively. Consequently, the required maximum is $\\lfloor 3n^2/8 \\rfloor$.\n\nAlternatively, notice that no five vertices of the complement $\\bar{G}$ of $G$ are independent, so the independence number $\\alpha(\\bar{G}) \\le 4$, and apply the Caro-Wei theorem to get\n$$\n4 \\ge \\alpha(\\bar{G}) \\ge \\sum_{i=1}^{n} \\frac{1}{1 + \\deg i} \\ge \\frac{n}{1 + \\frac{1}{n} \\sum_{i=1}^{n} \\deg i} = \\frac{n^2}{n + 2|E(\\bar{G})|}.\n$$\nConsequently, $|E(\\bar{G})| \\ge n(n-4)/8$, so $|E(G)| = n(n-1)/2 - |E(\\bar{G})| \\le 3n^2/8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2A_3A_4$ be a convex quadrilateral with no pair of parallel sides. For each $i = 1, 2, 3, 4$, define $\\omega_i$ to be the circle touching the quadrilateral externally, and which is tangent to the lines $A_{i-1}A_i$, $A_iA_{i+1}$ and $A_{i+1}A_{i+2}$ (indices are considered modulo 4, so $A_0 = A_4$, $A_5 = A_1$ and $A_6 = A_2$). Let $T_i$ be the point of tangency of $\\omega_i$ with the side $A_iA_{i+1}$. Prove that the lines $A_1A_2, A_3A_4$ and $T_2T_4$ are concurrent if and only if the lines $A_2A_3, A_4A_1$ and $T_1T_3$ are concurrent.\n(Russia) Pavel Kozhevnikov",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with a reformulation of a well-known statement on harmonic cyclic quadruples $(K_1, K_2, K_3, K_4)$, also provable by polar transformation (projective methods).\n**LEMMA.** Being given four pairwise non-parallel lines $\\ell_i$, $i = 1, 2, 3, 4$, tangent to a circle $\\omega$ at points $K_i$, and such that lines $\\ell_1, \\ell_3$ and $K_2K_4$ are concurrent, then lines $\\ell_2, \\ell_4$ and $K_1K_3$ are also concurrent.\n\nProof. Let $O$ be the center of $\\omega$, $X = \\ell_1 \\cap \\ell_3 \\cap K_2K_4$, $Y = \\ell_2 \\cap \\ell_4$. We have $OX \\perp K_1K_3$ and $OY \\perp K_2K_4$. Let $Z = OX \\cap K_1K_3$, $T = OY \\cap K_2K_4$. Notice that triangles $\\triangle OK_3X$ and $\\triangle OZK_3$ are similar, and also similar are triangles $\\triangle OK_4Y$ and $\\triangle OTK_4$, hence $OY \\cdot OT = OK_4^2 = OK_3^2 = OX \\cdot OZ$.\nThis means that triangles $\\triangle OXT$ and $\\triangle OYZ$ are similar, hence $YZ \\perp OX$, and so $Y \\in K_1K_3$. $\\square$\n\nSuppose now lines $A_2A_3$, $A_4A_1$ and $T_1T_3$ are concurrent at a point $P$. Let $T'_4, T'_2$ be the tangency points of lines $A_4A_1$, respectively $A_2A_3$, to circle $\\omega_1$, and let $T'_3$ be the second meeting point of line $T_1T_3$ and circle $\\omega_1$. Let the tangent to $\\omega_1$ at $T'_3$ meet the lines $A_4A_1$, $A_2A_3$ at points $A'_4$, respectively $A'_3$. The (direct) homothety of center $P$ that takes $\\omega_1$ to $\\omega_3$ maps $T'_3$ to $T_3$, hence $A_3A_4 \\parallel A'_3A'_4$.\nLet $Q = A_1A_2 \\cap A_3A_4$, $Q' = A_1A_2 \\cap A'_3A'_4$. Applying the LEMMA to circle $\\omega_1$ and lines $A_2A_3$, $A'_3A'_4$, $A_4A_1$, $A_1A_2$, yields that points $Q'$, $T'_2$, $T'_4$ are collinear. The (inverse) homothety of center $A_1$ that takes $\\triangle QA_1A_4$ to $\\triangle Q'A_1A'_4$ maps $\\omega_4$ to $\\omega_1$, so maps $T_4$ to $T'_4$, hence $Q'T'_4 \\parallel QT_4$. Similarly, the (inverse) homothety of center $A_2$ that takes $\\triangle QA_2A_3$ to $\\triangle Q'A_2A'_3$ maps $\\omega_2$ to $\\omega_1$, so maps $T_2$ to $T'_2$, hence also $Q'T'_2 \\parallel QT_2$. Since points $Q'$, $T'_2$, $T'_4$ are collinear, it follows points $Q, T_2, T_4$ are also collinear.\n\nThe converse implication is done in a similar way, due to the cyclic nature of the notations used (just increase each index by 1).\n*Alternative Solution.* (D. Şerbănescu) Suppose $Q, T_2, T_4$ are collinear. We will show $P, T_1, T_3$ are collinear. We will use the notations of the solution above, but also let $S'_1, S''_1$ be the tangency points of line $A_1A_2$ to circle $\\omega_2$, respectively $\\omega_4$, and let $S'_3, S''_3$ be the tangency points of line $A_3A_4$ to circle $\\omega_2$, respectively $\\omega_4$. Let $T''_4$ be the (other than $T_2$) meeting point of line $QT_2T_4$ and circle $\\omega_2$, and let the tangent line to $\\omega_2$ at $T''_4$ (parallel to $A_1A_4$) meet $A_2A_3$ at $P'$ (via the (direct) homothety of center $Q$ that takes $\\omega_4$ to $\\omega_2$).\n\nClearly $\\triangle PT_2T_4 \\sim \\triangle P'T_2T_4''$ and $\\triangle P'T_2T_4''$ is isosceles, so $PT_2 = PT_4$ (in other words, if $Q, T_2, T_4$ are collinear then $PT_2 = PT_4$; the other implication trivially also holds, but is irrelevant here).\nFrom $PT_2 = PT_4$ and $PT'_2 = PT'_4$ follows $T'_2T_2 = T'_4T_4$. As external tangents, $T'_2T_2 = T_1S'_1$ and $T'_4T_4 = T_1S''_1$, hence $T_1$ is the midpoint of $S'_1S''_1$. Similarly, $T_3$ is the midpoint of $S'_3S''_3$. It follows that $P, T_1, T_3$ lie on the radical axis of the circles $\\omega_2$ and $\\omega_4$, hence are collinear.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 37727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and consider the integers $x_1, x_2, \\dots, x_n$, $y_1, y_2, \\dots, y_n$ such that\n\na) $x_1 + x_2 + \\dots + x_n = y_1 + y_2 + \\dots + y_n = 0$;\n\nb) $x_1^2 + y_1^2 = x_2^2 + y_2^2 = \\dots = x_n^2 + y_n^2 = 0$.\n\nProve that $n$ is an even number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $a = x_1^2 + y_1^2$. If $a$ is odd, the numbers $x_i$ and $y_i$ do not have the same parity, so $x_i + y_i$ is odd. Since $\\sum (x_i + y_i) = 0$, it follows that $n$ is even.\n\nSuppose $a = 4k+2$. Then $x_i$ and $y_i$ are both odd. The equality $x_1+x_2+\\dots+x_n = 0$ implies $n$ is even.\n\nFinally, if $a = 4k$, then $x_i$ and $y_i$ are even. The numbers $a_i = \\frac{x_i}{2}$ and $b_i = \\frac{y_i}{2}$ satisfy the initial conditions. Furthermore, $a_1^2 + b_1^2 = \\frac{a}{4}$. Repeating the argument for $\\frac{a}{4}$ instead of $a$, after a finite number of steps we end up in a previous case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37728,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane are given $100$ points, such that no three of them are on the same line. The points are arranged in $10$ groups, any group containing at least $3$ points. Any two points in the same group are joined by a segment.\n\na) Determine which of the possible arrangements in $10$ such groups is the one giving the minimal numbers of triangles.\n\nb) Prove that there exists an arrangement in such groups where each segment can be coloured with one of three given colours and no triangle has all edges of the same colour.",
    "options": [],
    "answer": "a) The minimal number of triangles is achieved when the ten groups are all of size ten, giving 10 × C(10, 3) = 1200 triangles. b) Such an arrangement exists: in each group of ten, split the points into two convex pentagons; color pentagon sides with one color, pentagon diagonals with a second color, and edges between the two pentagons with a third color, which avoids monochromatic triangles.",
    "solution": "a) If the groups contain respectively $a_1, a_2, \\dots, a_{10}$ points, the number of triangles is\n$$\nN = \\binom{a_1}{3} + \\binom{a_2}{3} + \\dots + \\binom{a_{10}}{3}.\n$$\nWe claim this number is minimal when\n$$\na_1 = a_2 = \\dots = a_{10} = 10,\n$$\nthe minimum value being $10\\binom{10}{3}$. Indeed, if there exists a group having $m \\ge 11$ points, there will exist another group having $n \\le 9$ points. Moving a point from the first group into the second, the number of triangles decreases. This last statement follows from the inequality\n$$\n\\binom{m}{3} + \\binom{n}{3} > \\binom{m-1}{3} + \\binom{n+1}{3},\n$$\nreadily verified by direct computation.\n\nb) Make each group of $10$ points each, subdivided into two subgroups of $5$ points each. Since the positions of the points are irrelevant, we may suppose all these subgroups of $5$ points make up convex pentagons. Colour their sides with colour $c_1$, their diagonals with colour $c_2$, and the segments joining points from different subgroups with colour $c_3$. It is easy to see no monochromatic triangle appears.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37729,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n \\ge 2$ denote by $A_n$ the set of solutions of the equation\n$$\nx = \\lfloor \\frac{x}{2} \\rfloor + \\lfloor \\frac{x}{3} \\rfloor + \\dots + \\lfloor \\frac{x}{n} \\rfloor.\n$$\na) Determine the set $A_2 \\cup A_3$.\nb) Prove that the set $A = \\bigcup_{n \\ge 2} A_n$ is finite and find $\\max A$.",
    "options": [],
    "answer": "A2 ∪ A3 = {-7, -5, -4, -3, -2, -1, 0}; the union over all n is finite and its maximum element is 23.",
    "solution": "Notice that $A_n \\subset \\mathbb{Z}$ for all $n \\in \\mathbb{N}$, $n \\ge 2$.\n\na) The elements of $A_2$ satisfy the inequalities $x - 2 < 2x \\leq x$. By inspection, we obtain $A_2 = \\{-1, 0\\}$. The elements of $A_3$ satisfy the inequalities $5x - 12 < 6x \\leq 5x$. By inspection, we have $A_3 = \\{-7, -5, -4, -3, -2, 0\\}$, so $A_2 \\cup A_3 = \\{-7, -5, -4, -3, -2, -1, 0\\}$.\n\nb) For $n \\ge 4$ and $x \\in A_n$ we have $x \\le \\left(\\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n}\\right) x$.\nFrom $\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} > 1$ we obtain $x \\ge 0$.\nFor $x, n \\in \\mathbb{Z}$ and $n \\ge 2$ we get $\\lfloor \\frac{x}{n} \\rfloor \\ge \\frac{x - (n - 1)}{n}$. Therefore, if $n \\ge 4$ and $x \\in A_n$, then\n$$\nx \\ge \\lfloor \\frac{x}{2} \\rfloor + \\lfloor \\frac{x}{3} \\rfloor + \\lfloor \\frac{x}{4} \\rfloor \\ge \\frac{x-1}{2} + \\frac{x-2}{3} + \\frac{x-3}{4},\n$$\nimplying $x \\le 23$. Hence the set $A$ is upper bounded.\nBecause $A \\subset \\{-5, -4, \\dots, 23\\}$ and $23 \\in A_4$, then $\\max A = 23$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $VABCD$ be a regular pyramid, having the square $ABCD$ as basis. Suppose that on the line $AC$ lies a point $M$ such that $VM = MB$ and $(VMB) \\perp (VAB)$. Prove that $4AM = 3AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $MV = MB = MD$ and $MO \\perp (VBD)$, it follows that $O$ is the circumcenter of the triangle $VBD$. Furthermore, triangle $VBD$ is isosceles and right-angled, implying that the lateral faces of the pyramid are equilateral triangles.\nLet $P$ be the midpoint of the edge $VB$. The angle of the planes $(VAB)$ and $(VBM)$ is $\\angle APM$, hence $\\angle APM = 90^\\circ$.\nTriangles $MPA$ and $POA$ are similar, yielding $\\frac{MA}{PA} = \\frac{PA}{OA}$.\nTherefore $AM = \\frac{PA^2}{OA} = \\frac{3}{4}AC$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that\n$$\n3f(f(f(n))) + 2f(f(n)) + f(n) = 6n, \\text{ for all } n \\in \\mathbb{N}.\n$$\nVasile Pop",
    "options": [],
    "answer": "f(n) = n for all n ∈ ℕ",
    "solution": "We firstly notice that the function is injective. Indeed, if $f(n) = f(m)$, then $3f(f(f(n))) + 2f(f(n)) + f(n) = 3f(f(f(m))) + 2f(f(m)) + f(m)$, hence $6n = 6m$, so $n = m$.\n\nPlugging $n = 0$ yields $3f(f(f(0))) + 2f(f(0)) + f(0) = 0$, so $f(0) = 0$. We prove, using induction, that $f(n) = n$, for every $n \\in \\mathbb{N}$.\n\nSuppose $f(0) = 0$, $f(1) = 1$, ..., $f(n) = n$. From the injectivity, $f(n+1) \\geq n+1$, $f(f(n+1)) \\geq n+1$ and $f(f(f(n+1))) \\geq n+1$. On the other hand, $3f(f(f(n+1))) + 2f(f(n+1)) + f(n+1) = 6n+6$, whence $f(n+1) = n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a < c < b$ be real numbers and let $f : [a, b] \\to \\mathbb{R}$ be a function continuous at $c$. Show that if $f$ is the derivative of a function $F_a$ on $[a, c)$, and also the derivative of a function $F_b$ on $(c, b]$, then $f$ is the derivative of some function $F$ on the entire interval $[a, b]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The function $f$ has primitives on the interval $[a, b]$ if and only if the function $g = f + 1 - f(c)$ has primitives on this interval. So, we can assume that $f(c) = 1$. Since $f$ is continuous at $c$, there exists $[\\alpha, \\beta] \\subset [a, b]$ such that $c \\in (\\alpha, \\beta)$ and $0 < f(x) < 2$, for every $x \\in [\\alpha, \\beta]$.\n\nLet $F_a$, respectively $F_b$, be a primitive of $f$ on $[a, c)$, respectively $(c, b]$. Because $F_a$ and $F_b$ are increasing on the intervals $[\\alpha, c)$, respectively $(c, \\beta]$, it follows that the limits $\\lim_{x \\to c} F_a(x) = p$ and $\\lim_{x \\to c} F_b(x) = q$ exist. Also they are finite, because $F_a$ and $F_b$ are bounded on these intervals (for instance, if $x \\in (\\alpha, c)$, then $F_a(x) - F_a(\\alpha) = f(\\xi)(x - \\alpha)$ for some $\\xi \\in (\\alpha, x)$ and $f$ is bounded on $[\\alpha, x] \\subset [\\alpha, \\beta]$).\n\nThe function\n$$\nF : [a, b] \\to \\mathbb{R}, \\quad F(x) = \\begin{cases} F_a(x) - p, & \\text{if } x \\in [a, c) \\\\ 0, & \\text{if } x = c \\\\ F_b(x) - q, & \\text{if } x \\in (c, b] \\end{cases}\n$$\nis a primitive of $f$ on $[a, b]$.\n*Alternative Solution.* (D. Schwarz) It is enough to show the limits $\\lim_{x \\to c} F_a(x) = p$ and $\\lim_{x \\to c} F_b(x) = q$ exist and are finite. But $|f|$ is bounded by some value $M$ in a neighborhood $(c - \\varepsilon, c + \\varepsilon)$ of $c$, since $f$ (therefore $|f|$ also) is continuous at $c$.\n\nFor $x, x' \\in (c - \\varepsilon, c)$ we will thus have $\\frac{F_a(x) - F_a(x')}{x - x'} = f(\\xi)$ at some point $\\xi \\in (x, x')$, hence $|F_a(x) - F_a(x')| < M\\varepsilon$, which implies the existence of the finite limit $\\lim_{x \\to c} F_a(x) = p$. An identical reasoning holds for $F_b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of positive integers $a$ and $b$ for which $a^6 \\ge 5^{b+1}$ and $b^6 \\ge 5^{a+1}$.",
    "options": [],
    "answer": "(3,3), (4,4), (5,5)",
    "solution": "We will prove the inequality $5^{n+1} \\ge n^6$ holds for all positive integers $n$, except $n = 3$ and $n = 4$.\n\nIndeed, this is readily verified for $n = 1$, $n = 2$ and $n = 5$. When the thesis is true for $n \\ge 5$, then $5^{n+2} = 5 \\cdot 5^{n+1} \\ge 5n^6$. It is then sufficient to show that $5n^6 \\ge (n+1)^6$, rewritten as $5 \\ge \\left(1 + \\frac{1}{n}\\right)^6$. Since $\\left(1 + \\frac{1}{n}\\right)^6 < \\left(1 + \\frac{1}{4}\\right)^6 = 5 \\cdot \\frac{5^5}{4^6} = 5 \\cdot \\frac{3125}{4096} < 5$, everything is thus proven, by simple induction; moreover, the inequality becomes strict for $n > 5$.\n\nOne thus gets $5^{a+1} \\ge a^6$ and $5^{b+1} \\ge b^6$, for all $a, b$ different from $3$ and $4$. Multiplying, first the two inequalities given in the problem statement, then those two just obtained in the above, we get $a^6 b^6 \\ge 5^{b+1} 5^{a+1} \\ge b^6 a^6$. But equality only holds here if $a = b = 5$, which clearly checks. Otherwise, it needs $a = 4$ or $a = 3$ (or, symmetrically, $b = 4$ or $b = 3$). If $a = 4$, then the inequalities given in the problem statement become $4^6 > 5^{b+1}$ and $b^6 > 5^5$, leading to $b \\le 4$ and $b \\ge 4$ respectively, hence $b = 4$. If $a = 3$, then $3^6 > 5^{b+1}$ and $b^6 > 5^4$, leading to $b \\le 3$ and $b \\ge 3$ respectively, hence $b = 3$.\n\nTwo more pairs of solutions $a = b = 4$ and $a = b = 3$ have thus been found.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a finite group of order $n$. Define the set\n$$\nH = \\{x : x \\in G \\text{ and } x^2 = e\\},\n$$\nwhere $e$ is the neutral element of $G$. Let $p = |H|$ be the cardinal of $H$. Prove that\n\na) $|H \\cap xH| \\geq 2p - n$, for any $x \\in G$, where $xH = \\{xh : h \\in H\\}$.\n\nb) If $p > \\frac{3n}{4}$, then $G$ is commutative.\n\nc) If $\\frac{n}{2} < p \\leq \\frac{3n}{4}$, then $G$ is non-commutative.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $G$ is a group, it follows that $|xH| = p$. Therefore\n$$\nn = |G| \\ge |H \\cup xH| = |H| + |xH| - |H \\cap xH| = 2p - |H \\cap xH|,\n$$\nwhence $|H \\cap xH| \\ge 2p - n$.\n\nb) Let have $x \\in H$ and $y \\in H \\cap xH$. Then $y = y^{-1}$ and $y = xh$, $h \\in H$. Since $xy = x^2h = h \\in H$, it follows that $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$. Therefore $x$ commutes with all elements of $H \\cap xH$.\nSince $|H \\cap xH| \\ge 2p - n > \\frac{3n}{2} - n = \\frac{n}{2}$, it follows the subgroup of the elements that commute with $x$ has at least $\\frac{n}{2}$ elements. From Lagrange's theorem, it follows that $x$ commutes with all elements of $G$. Therefore $H \\subseteq Z(G)$, the center of $G$. It results $|Z(G)| \\ge |H| = p > \\frac{3n}{4} > \\frac{n}{2}$, so $Z(G) = G$.\n\nc) Assume $G$ is commutative.\nIf $x, y \\in H$, then $(xy)^2 = x^2y^2 = e$, so $xy \\in H$. Since $H$ is a finite set, it follows that $H$ is a subgroup of $G$.\nThe condition $|H| > \\frac{n}{2}$ forces $H = G$, so $\\frac{3n}{4} \\ge |H| = |G| = n$, contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the cube $ABCDA'B'C'D'$. The angle bisectors of the angles $\\angle A'C'A$ and $\\angle A'AC'$ meet $AA'$ and $A'C'$ at points $P$ and $Q$, respectively. The point $M$ is the foot of the perpendicular from $A'$ onto $C'P$ while $N$ is the foot of the perpendicular from $A'$ onto $AS$. The point $O$ is the center of the face $ABB'A'$.\na) Prove that the planes $(MNO)$ and $(AC'B)$ are parallel.\nb) Given that $AB = 1$, find the distance between the planes $(MNO)$ and $(AC'B)$.",
    "options": [],
    "answer": "sqrt(2)/4",
    "solution": "a) Denote by $T$ and $R$ the intersections of the straight lines $AC'$ and $A'M$, respectively $AC'$ and $A'N$. In the triangle $A'C'T$, $C'M$ is an angle bisector and an altitude, so $A'M = MT$. In the triangle $A'AR$, $AN$ is an angle bisector and an altitude, so $A'N = NR$. The segment $[MO]$ joins the midpoints of two sides of the triangle $A'TB$, hence $MO \\parallel TB$, and the segment $NO$ joins the midpoints of two sides of triangle $A'RB$, hence $NO \\parallel RB$. The requirement follows from the fact that the planes $(TRB)$ and $(AC'B)$ are the same.\n\n![](attached_image_1.png)\n\nb) The distance between the two planes is equal to the distance from $O$ to the plane $(AC'B)$, and this last distance is half the distance from $A'$ to the same plane.\nAs the distance from $A'$ to the plane is $\\frac{1}{2}A'D = \\frac{1}{2}\\sqrt{2}$, the required distance is $\\frac{1}{4}\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the small squares of a $50 \\times 50$ table is coloured in red or blue. Initially all squares are red. A step means changing the colour of all squares on a row or on a column.\n\na) Prove that there exists no sequence of steps, such that at the end there are exactly $2011$ blue squares.\n\nb) Describe a sequence of steps, such that at the end exactly $2010$ squares are blue.",
    "options": [],
    "answer": "It is impossible to reach exactly 2011 blue squares. To obtain exactly 2010 blue squares, flip the first 44 rows and the first 5 columns.",
    "solution": "Without loss of generality, we may consider that the rows or columns to be modified in a sequence of steps are consecutive, and that each column or row is modified only once.\n\nSuppose then that the first $x$ rows and the first $50-y$ columns have been modified. One gets a $x \\times y$ rectangle and a $(50-x) \\times (50-y)$ rectangle with all squares coloured blue, the rest of the table being red.\n\nThe number of blue squares is then $A = xy + (50-x)(50-y)$ which is an even number, so it can not equal $2011$.\n\nFor the second part, notice that $A = 2010$ is equivalent to $(x - 25)(y - 25) = 380 = 19 \\cdot 20$.\n\nOne can take $x = 25 + 19 = 44$ and $y = 25 + 20 = 45$ to give the answer (thus being clear that the steps are not unique).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a strictly increasing function such that $f \\circ f$ is continuous. Prove that $f$ is continuous.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $f$ is monotonic, the lateral limits at each point exist and are finite. Denote $f(x_0^-)$, respectively $f(x_0^+)$, the limit from the left, respectively from the right at $x_0$. Then, $f$ being increasing, $f(x_0^-) \\leq f(x_0) \\leq f(x_0^+)$ for every $x_0 \\in \\mathbb{R}$.\n\nSuppose now that $f$ is discontinuous at some point $a$. Then there exists $A, B$ such that $f(a^-) < A < B < f(a^+)$. It follows, since $f$ is increasing, that $f(x) \\leq A$ for all $x < a$ and $f(x) \\geq B$ for all $x > a$. Using again the monotony, $f(f(x)) \\leq f(A)$ for all $x < a$ and $f(f(x)) \\geq f(B)$ for all $x > a$. But, this implies $f(f(a^-)) \\leq f(A) < f(B) \\leq f(f(a^+))$, which would mean that $f \\circ f$ is discontinuous at $a$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ be a real number. Prove that $x$ is an integer if and only if\n$$\n[x] + [2x] + [3x] + \\dots + [nx] = \\frac{n([x] + [nx])}{2}\n$$\nholds for all positive integers $n$ (here, $[a]$ denotes the integer part (floor function) of the real number $a$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x \\in \\mathbb{Z}$, then $[kx] = k[x]$ for every $k \\in \\mathbb{N}^*$.\n\nFor the converse, notice that the hypothesis implies, for all $n$,\n$$\nn([x] + [nx]) + 2[(n+1)x] = (n+1)([x] + [(n+1)x]).\n$$\nThis comes to $n[nx] = [x] + (n-1)[(n+1)x]$, $\\forall n \\ge 1$. Replacing $n$ with $n+1$ and subtracting the two relations, we obtain that the sequence $a_n = [nx]$ is an arithmetic progression. Since $(nx)_{n \\ge 1}$ is also an arithmetic progression, the difference sequence $\\{nx\\}_{n \\ge 1}$ is an arithmetic progression. Since $\\{nx\\}_{n \\ge 1}$ is bounded, its ratio must be 0, whence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37739,
    "subject": "Mathematics (Multi-modal)",
    "question": "A strip of breadth $\\ell$ is the set of those points in the plane, situated between or on two parallel lines at distance $\\ell$ one from the other.\nConsider a finite set $S$ of $n \\geq 3$ (distinct) points in the plane, with the property that any three points of $S$ can be covered by a strip of breadth 1. Prove the set $S$ can be covered by a strip of breadth 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All solutions encountered in the competition (including the Official one) make use of the following\n\n**LEMMA.** A triangle can be covered by a strip of breadth 1 if and only if (at least) one of its altitudes is at most 1 long.\n\n*Proof.* The other implication being trivial, consider a triangle $ABC$ covered by a strip of breadth 1. The perpendicular lines through $A, B, C$ on the lines that define the strip can determine two situations\n* one of them meets a side of the triangle $ABC$ at an interior point (Figure 1)\n* one side of the triangle $ABC$ contains one of them perpendiculars (Figure 2)\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n\nIn both cases notice one of the altitudes (in the figures – $AE$) is at most 1 long. $\\square$\n\nNow either choose points $A, B \\in S$ such that the distance $AB$ be maximal, or choose points $A, B, C \\in S$ such that the area of triangle $ABC$ be maximal.\n\nIn the former case, for any other point $C$, the distance from $C$ to $AB$ is at most 1 (when $ABC$ is a non-degenerate triangle, then this is the least altitude of the triangle),\nhence the set of points can be covered by a strip of breadth 2, determined by two parallel lines to $AB$, situated at distance 1 on each side of it.\n\nIn the latter case, it is easy to show all points of the set are included in the triangle $A'B'C'$, determined by parallels through $A, B, C$ to the opposite sides (in effect, an *old chestnut*). Since one of the altitudes of $ABC$ is at most 1 long, one of the altitudes of $A'B'C'$ will be at most 2 long, and we close the argument as above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$ and $\\angle BAC = 40^\\circ$. The points $S$ and $T$ lie on the sides $AB$ and $BC$, respectively, such that $\\angle BAT = \\angle BCS = 10^\\circ$. The straight lines $AT$ and $CS$ meet at point $P$.\nShow that $BT = 2PT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Triangle $ABC$ is isosceles, so $\\angle ABC = \\angle ACB = 70^\\circ$.\n\nNotice that $\\angle TAC = 40^\\circ - 10^\\circ = 30^\\circ$ and $\\angle ACS = 70^\\circ - 10^\\circ = 60^\\circ$, hence $\\angle APC = 90^\\circ$.\n\nTriangles $ABT$ and $BSC$ are similar, whence $\\frac{BS}{BC} = \\frac{BT}{AB}$.\n\nAlso, triangles $BST$ and $BCA$ are similar, therefore $TB = TS$ and $\\angle TSB = 70^\\circ$.\n\nSince $\\angle CSA = \\angle SBC + \\angle SCB = 70^\\circ + 10^\\circ = 80^\\circ$, it follows that $\\angle PST = 180^\\circ - 80^\\circ - 70^\\circ = 30^\\circ$.\n\nNow, triangle $STP$ has a right angle in $P$ and $\\angle PST = 30^\\circ$, so $BT = 2PT$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a function $f : \\mathbb{R} \\to [0, \\infty)$. Prove that $f$ satisfies the inequality $f(x+y) \\ge (1+y)f(x)$ for any $x \\in \\mathbb{R}$ and any $y \\ge 0$, if and only if the function $g : \\mathbb{R} \\to [0, \\infty)$ defined by $g(x) = e^{-x}f(x)$, for $x \\in \\mathbb{R}$, is non-decreasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality $e^y \\ge 1+y$, holds for all $y \\in \\mathbb{R}$. Assume the function $g$ is monotonously increasing on $\\mathbb{R}$. Then, based on the inequality above, we get\n$$\nf(x + y) = e^{x+y}g(x + y) \\ge e^x(1 + y)g(x) = (1 + y)f(x),\n$$\nfor all $x \\in \\mathbb{R}$ and all $y \\ge 0$.\n\nConversely, assume $f(x + y) \\ge (1 + y)f(x)$, for all $x \\in \\mathbb{R}$ and all $y \\ge 0$. One can check by induction\n$$\nf(x + nt) \\ge (1 + t)^n f(x),\n$$\nfor all $x \\in \\mathbb{R}$, $t \\ge 0$ and $n \\in \\mathbb{N}$. Let $x, z \\in \\mathbb{R}$, with $x < z$.\nDenote $y = z - x$. For $n \\in \\mathbb{N}^*$ we have\n$$\ng(z) = e^{-z}f(z) = e^{-x-y}f\\left(x + n\\frac{y}{n}\\right) \\ge e^{-x-y}\\left(1 + \\frac{y}{n}\\right)^n f(x) = \\frac{\\left(1 + \\frac{y}{n}\\right)^n}{e^y}g(x).\n$$\nIt follows\n$$\ng(z) \\ge \\lim_{n \\to \\infty} \\frac{\\left(1 + \\frac{y}{n}\\right)^n}{e^y} g(x) = g(x),\n$$\ntherefore $g$ is monotonously increasing on $\\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ be positive real numbers and $n$ be a positive integer. Prove that if $x^{2n+1} + y^{2n+1} \\ge 2$ then also $x^{n+1} + y^{n+1} \\ge x^n + y^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "LEMMA. $D_{n,k+1} D_{n,k}^{-1} \\le D_{n,1}^2$.\n*Proof.* By simple calculations we get\n$$\nD_{n,k+1} D_{n,k}^{-1} = \\frac{x^{2n+2} + y^{2n+2} + xy(x^{n+k} y^{n-k} + x^{n-k} y^{n+k})}{x^{2n} + y^{2n} + x^{n+k} y^{n-k} + x^{n-k} y^{n+k}}\n$$\nand\n$$\nD_{n,1}^2 = \\frac{x^{2n+2} + y^{2n+2} + 2(xy)^{n+1}}{x^{2n} + y^{2n} + 2(xy)^n}.\n$$\nLetting $A = x^{2n+2} + y^{2n+2}$, $B = x^{2n} + y^{2n}$, $C_1 = x^{n+k} y^{n-k} + x^{n-k} y^{n+k}$ and $C_2 = 2x^n y^n$, and by clearing denominators the thesis becomes\n$$\n(C_1 - C_2)(A - xyB) \\ge 0.\n$$\nBut here we have now $C_1 \\ge C_2$ by the AM-GM inequality, and $A \\ge xyB$ by the rearrangement inequality (or just some trivial factorization). $\\square$\n\nUsing the LEMMA with a telescoping product we obtain\n$$\nD_{n,n+1} D_{n,1}^{-1} = \\prod_{k=1}^{n} \\frac{D_{n,k+1}}{D_{n,k}} \\le D_{n,1}^{2n}.\n$$\nSubstituting, our hypothesis corresponds to $D_{n,n+1} \\ge 1$, and therefore\n$$\nD_{n,1}^{2n+1} = \\left( \\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \\right)^{2n+1} \\ge D_{n,n+1} \\ge 1,\n$$\nwhence the thesis.\n\nAlternative Solution.\nSuppose $x^{n+1} + y^{n+1} < x^n + y^n$. Then $xy < 1$; otherwise (i.e. for $xy \\ge 1$) we would have\n$$\nx^n + y^n \\le x^{n+\\frac{1}{2}} y^{\\frac{1}{2}} + x^{\\frac{1}{2}} y^{n+\\frac{1}{2}} < x^{n+1} + y^{n+1},\n$$\nabsurd. The inequality $x^n + y^n \\le x^{n+\\frac{1}{2}} y^{\\frac{1}{2}} + x^{\\frac{1}{2}} y^{n+\\frac{1}{2}}$ comes from $1 \\le x^{\\frac{1}{2}} y^{\\frac{1}{2}}$, while the inequality $x^{n+\\frac{1}{2}} y^{\\frac{1}{2}} + x^{\\frac{1}{2}} y^{n+\\frac{1}{2}} < x^{n+1} + y^{n+1}$ immediately holds, by writing it $(x^{\\frac{1}{2}} - y^{\\frac{1}{2}}) \\left( x^{n+\\frac{1}{2}} - y^{n+\\frac{1}{2}} \\right) > 0$.\nWithout loss of generality, we may take $x > 1 > y > 0$ (the other cases are trivial). It follows that $x^k + x^{-k} < y^k + y^{-k}$ for all $k \\in \\mathbb{N}^*$. This comes from the function $f(t) = t + t^{-1}$ for $t > 0$ being decreasing on $(0, 1]$ and increasing on $[1, +\\infty)$, while $x \\in (y, y^{-1})$.\nFrom $x^{n+1} + y^{n+1} < x^n + y^n$ follows $x^{n+1} - x^n < y^n - y^{n+1}$.\n$$\n\\text{But then } x^{2n+1} - 1 = (x^{n+1} - x^n) \\left( 1 + \\sum_{k=1}^{n} (x^k + x^{-k}) \\right) < (y^n - y^{n+1}) \\left( 1 + \\sum_{k=1}^{n} (y^k + y^{-k}) \\right) = 1 - y^{2n+1},\n$$\nthis is to say, in the end, $x^{2n+1} + y^{2n+1} < 2$, contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two rectangles of unit area overlap to form a convex octagon. Show that the area of the octagon is at least $\\frac{1}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Each side of one rectangle meets the contour of the other rectangle at exactly two points situated on consecutive sides. Let $A_0A_1A_2A_3$ and $B_0B_1B_2B_3$ be circular labellings of the two rectangles such that the segments $A_iA_{i+1}$ and $B_iB_{i+1}$ have equal lengths and meet at a point labelled $C_i$. The $C_i$ are four alternative vertices of the octagon, so the area of the latter is greater than or equal to the area of the quadrangle $C_0C_1C_2C_3$. Notice that $C_i$ is equally distanced from the lines $A_{i+2}A_{i+3}$ and $B_{i+2}B_{i+3}$ to deduce that the lines $C_0C_2$ and $C_1C_3$ are perpendicular to one another. Consequently,\n$$\n\\text{area } C_0C_1C_2C_3 = \\frac{1}{2} \\cdot C_0C_2 \\cdot C_1C_3 \\ge \\frac{1}{2} \\cdot A_1A_2 \\cdot A_0A_1 = \\frac{1}{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N}^* \\to \\mathbb{N}^*$ such that\n$$\nf(n) + f(n+1) + f(f(n)) = 3n + 1, \\text{ for all } n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "Two functions:\n1) f(n) = n for all n.\n2) f(n) = n + 1 if n is odd, and f(n) = n − 1 if n is even.",
    "solution": "From $f(1) + f(2) + f(f(1)) = 4$ follows $f(1) \\in \\{1, 2\\}$.\n\nIf $f(1) = 1$, then $f(2) = 2$ and an easy induction shows that $f(n) = n$.\n\nIf $f(1) = 2$, then $f(2) = 1$ and, inductively,\n$$\nf(n) = \\begin{cases} n + 1, & \\text{if } n \\text{ is odd} \\\\ n - 1, & \\text{if } n \\text{ is even} \\end{cases}.\n$$\nBoth the above found functions fulfill the initial condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [-1, 1] \\to \\mathbb{R}$ be a continuous function having finite derivative at $0$, and\n$$\nI(h) = \\int_{-h}^{h} f(x) \\, dx, \\quad h \\in [0, 1].\n$$\nProve that\n\na) There exists $M > 0$, such that $|I(h) - 2f(0)h| \\leq Mh^2$, for any $h \\in [0, 1]$.\n\nb) The sequence $(a_n)_{n \\geq 1}$, defined by $a_n = \\sum_{k=1}^{n} \\sqrt{k} |I(1/k)|$, is convergent if and only if $f(0) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The continuous function $\\varphi : (0, 1] \\to \\mathbb{R}$, $\\varphi(h) = \\frac{I(h) - 2f(0)h}{h^2}$, may be prolonged by continuity at $0$, since\n$$\n\\begin{align*} \n\\lim_{h \\to 0} \\varphi(h) &= \\lim_{h \\to 0} \\frac{(I(h) - 2f(0)h)'}{2h} = \\lim_{h \\to 0} \\frac{f(h) + f(-h) - 2f(0)}{2h} = \\\\ \n&= \\frac{1}{2} \\lim_{h \\to 0} \\left( \\frac{f(h) - f(0)}{h} - \\frac{f(-h) - f(0)}{-h} \\right) = \\frac{1}{2} (f'(0) - f'(0)) = 0. \n\\end{align*}\n$$\nTherefore $\\varphi$ is bounded on $(0, 1]$.\nLet $M = \\sup\\{|\\varphi(h)| : 0 < h \\le 1\\}$. Then $|I(h) - 2f(0)h| \\le Mh^2$, whatever $h \\in (0, 1]$; the inequality obviously also holds for $h = 0$.\n\nb) Since\n$$\n||I(h)| - 2|f(0)|h| \\le |I(h) - 2f(0)h| \\le Mh^2, \\quad 0 < h \\le 1,\n$$\nit follows that\n$$\n2|f(0)|h - Mh^2 \\le |I(h)| \\le 2|f(0)|h + Mh^2, \\quad 0 < h \\le 1,\n$$\nwhence\n$$\n\\frac{2|f(0)|}{\\sqrt{k}} - \\frac{M}{k\\sqrt{k}} \\le \\sqrt{k} |I(1/k)| \\le \\frac{2|f(0)|}{\\sqrt{k}} + \\frac{M}{k\\sqrt{k}}, \\quad k \\in \\mathbb{N}^*.\n$$\nBut\n$$\n\\sum_{k=1}^{n} \\frac{1}{\\sqrt{k}} \\ge \\sum_{k=1}^{n} \\frac{1}{\\sqrt{n}} = \\sqrt{n}, \\quad n \\in \\mathbb{N}^*.\n$$\nand\n$$\n\\begin{align*} \n\\sum_{k=1}^{n} \\frac{1}{k\\sqrt{k}} &= 1 + \\sum_{k=2}^{n} \\frac{1}{k\\sqrt{k}} \\le 1 + 2 \\sum_{k=2}^{n} \\left( \\frac{1}{\\sqrt{k-1}} - \\frac{1}{\\sqrt{k}} \\right) = \\\\ \n&= 1 + 2 \\left( 1 - \\frac{1}{\\sqrt{n}} \\right) < 3, \\quad n \\in \\mathbb{N}^*. \n\\end{align*}\n$$\nAccording with these inequalities, it follows that\n\n(1) If $f(0) = 0$, then $a_n \\le 3M$, whatever $n \\in \\mathbb{N}^*$; since the sequence $(a_n)_{n \\ge 1}$ is increasing, it is therefore convergent.\n\n(2) If $f(0) \\ne 0$, then $a_n \\ge 2|f(0)|\\sqrt{n} - 3M$, whatever $n \\in \\mathbb{N}^*$, so the sequence $(a_n)_{n \\ge 1}$ is unbounded.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles in the plane, $\\gamma_1$ and $\\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\\gamma_1$, and let $D$ be a point on $\\gamma_2$. The lines $AM$ and $AN$ meet again $\\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\\gamma_1$ at points $E$ and $F$, respectively. Assume the order $M, N, F, A, E$ is circular around $\\gamma_1$, and the segments $AB$ and $DE$ are congruent. Prove that the points $A, F, C$ and $D$ lie on\n\na circle whose center does not depend on the position of the points A and D on the respective circles, subject to the assumptions above.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB = DE$, the triangles $NAB$ and $NED$ are congruent, so $NA = NE$ and $NB = ND$. Let $K$ and $L$ be the antipodes of $N$ in $\\gamma_1$ and $\\gamma_2$, respectively, and notice that they are the midpoints of the arcs $AE$ and $BD$, respectively. Notice further that the angles $KME$ and $LMD$ have equal measures, to deduce that the four arcs $KA, KE, LB$ and $LD$ all have the same measure. The arcs $EMN, AFN$ and $BMN$ have equal measures as well. Consequently, the inscribed angles $AFN$ and $DCN$ subtend arcs of equal measures on the respective circles, so they are congruent; that is, the points $A, F, C, D$ are co-cyclic.\n\nWe now show that the center of the circle through $A, F, C, D$ is the midpoint $O$ of the segment $KL$. To this end, we show that $O$ lies on the perpendicular bisector of any segment $X_1X_2$ through $N$, where $X_1$ is on $\\gamma_1$ and $X_2$ is on $\\gamma_2$; in particular, $O$ lies on the perpendicular bisectors of both segments $AC$ and $DF$, whence the conclusion.\n\nLet $O_1$ and $O_2$ be the centers of the circles $\\gamma_1$ and $\\gamma_2$, respectively, and notice that $OO_1NO_2$ is a parallelogram to deduce that the segments $NO$ and $O_1O_2$ cross each other at their common midpoint $P$. Let further $X, X'_1$ and $X'_2$ be the midpoints of the segments $X_1X_2, NX_1$ and $NX_2$, respectively, and notice that the segments $NX$ and $X'_1X'_2$ have the same midpoint $X'$, to infer that $OX$ is parallel to $PX'$. Since the latter is perpendicular to $X'_1X'_2$, it follows that $OX$ is perpendicular to $X_1X_2$, so $OX$ is indeed the perpendicular bisector of the segment $X_1X_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exist a sequence $(a_n)_{n \\ge 0}$, with $a_n \\in \\{-1, +1\\}$ for all $n \\ge 0$, such that\n$$\n\\lim_{n \\to \\infty} (\\sqrt{n+a_1} + \\sqrt{n+a_2} + \\dots + \\sqrt{n+a_n} - n\\sqrt{n+a_0}) = \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, $a_0 = -1$. For each $n \\in \\mathbb{N}^*$, denote by $k_n$ the number of terms among $a_1, a_2, \\dots, a_n$ which are equal to $1$ (the other being $-1$). Then, the sequence $(x_n)_{n \\ge 0}$ whose limit must be $\\frac{1}{2}$ is $x_n = k_n\\sqrt{n+1} - k_n\\sqrt{n-1}$, or\n$$\nx_n = \\frac{2k_n}{\\sqrt{n+1} + \\sqrt{n-1}} = \\frac{2\\frac{k_n}{\\sqrt{n}}}{\\sqrt{1 + \\frac{1}{n}} + \\sqrt{1 - \\frac{1}{n}}}.\n$$\nIt is enough to construct $(a_n)_{n \\ge 1}$ such that $\\lim_{n \\to \\infty} \\frac{k_n}{\\sqrt{n}} = \\frac{1}{2}$. (*)\nLet us choose $a_n = 1$ if and only if $n = (2m)^2$, with $m \\in \\mathbb{N}^*$. Then $(2k_n)^2 \\le n < (2(k_n + 1))^2$, whence $\\frac{\\sqrt{n}}{2} - 1 < k_n \\le \\frac{\\sqrt{n}}{2}$, which guarantees (*).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exist a polynomial $f(x_1, x_2)$ in two variables, with integer coefficients, and two points $A = (a_1, a_2)$ and $B = (b_1, b_2)$ in the plane, satisfying all the following conditions\n(i) $A$ is an integer point (i.e., $a_1$ and $a_2$ are integers);\n(ii) $|a_1 - b_1| + |a_2 - b_2| = 2010$;\n(iii) $f(n_1, n_2) > f(a_1, a_2)$, for all integer points $(n_1, n_2)$ in the plane other than $A$;\n(iv) $f(x_1, x_2) > f(b_1, b_2)$, for all points $(x_1, x_2)$ in the plane other than $B$.",
    "options": [],
    "answer": "Yes",
    "solution": "The triple $(f(x_1, x_2), A, B)$ does exist, so the answer is yes.\nLet $A = O = (0, 0)$, $B = (x_0, y_0) = (2009 + \\frac{2}{3}, \\frac{1}{3})$. The idea is to search for a polynomial $f$ such that $f(x, y) = 0$ is the equation of an ellipse centered at $B$, passing through $O$ and with tangent line $y = 0$ at $O$. In fact, if $f$ is chosen like this, the ellipse $f(x, y) = 0$ is completely contained in the region $\\{(x, y) ; 0 \\le y \\le \\frac{2}{3}\\}$, with $O$ the only integer point on the ellipse or in its interior; clearly, the absolute minimum of $f(x, y)$ is attained at $B$ and $f(x, y)$ is positive at all integer points other than $O$. Therefore, we consider polynomials of the type\n$$\nf(X, Y) = 9M(X - x_0)^2 + 9N(X - x_0)(Y - y_0) + 9P(Y - y_0)^2 - Q\n$$\nwhere $M, N, P, Q$ are integers with $M, P, Q, 4MP - N^2 > 0$.\nThe condition that the ellipse $f(x, y) = 0$ passes through $O$, with tangent line $y = 0$ at $O$, is expressed by\n$$\n\\begin{cases} 6029^2 M + 6029N + P - Q = 0 \\\\ 2 \\cdot 6029M + N = 0. \\end{cases}\n$$\n\nIt is then sufficient to choose $M = 1$, $N = -2 \\cdot 6029$, $P$ any integer greater than $6029^2$ and $Q = P - 6029^2$.\n(Alternative Solution. D. Schwarz)\nGiven any integer point $A(a_1, a_2)$, there exist infinitely many points $B(b_1, b_2)$ with $b_1, b_2 \\in \\mathbb{Q} \\setminus \\mathbb{Z}$, and such that $|a_1 - b_1| + |a_2 - b_2| = 2010$, for example $b_1 = a_1 + \\alpha + r$, $b_2 = a_2 + \\beta + (1 - r)$, with $\\alpha, \\beta \\in \\mathbb{Z}_+$, $r \\in \\mathbb{Q} \\cap (0, 1)$, and $\\alpha + \\beta = 2009$. We now will consider polynomials of the type\n$$\nf(X, Y) = N \\left( (X - a_1)^2 + (Y - a_2)^2 + \\varepsilon \\right) \\left( (X - b_1)^2 + (Y - b_2)^2 \\right)\n$$\nwhere $\\varepsilon \\in \\mathbb{Q}_+^*$ and $N \\in \\mathbb{Z}_+^*$ large enough for $f(X, Y)$ to have integer coefficients.\nOne then has $f(b_1, b_2) = 0$, while $f(x, y) > 0$ for all points $(x, y)$ in the plane, other than $B$.\nOne also then has $f(a_1, a_2) = N\\varepsilon((a_1 - b_1)^2 + (a_2 - b_2)^2)$, while one has, for all integer points $(n_1, n_2)$ in the plane, other than $A$,\n$$\n\\min \\{(n_1 - a_1)^2 + (n_2 - a_2)^2 + \\varepsilon\\} = 1 + \\varepsilon \\quad \\text{and} \\\\\n\\min \\{(n_1 - b_1)^2 + (n_2 - b_2)^2\\} = m,\n$$\nfor some $m \\in \\mathbb{Q} \\cap (0, \\frac{1}{2}]$ (for example $m = \\frac{1}{2}$ when $r = \\frac{1}{2}$), therefore $F(n_1, n_2) > N(1+\\varepsilon)m$. In order to have $f(n_1, n_2) > f(a_1, a_2)$ it is thus enough that $N(1+\\varepsilon)m \\ge N\\varepsilon((a_1 - b_1)^2 + (a_2 - b_2)^2)$, therefore let us take\n$$\n\\varepsilon = m / ((a_1 - b_1)^2 + (a_2 - b_2)^2 - m) \\in \\mathbb{Q}_+^*,\n$$\nand then choose some appropriate $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a differentiable function with $f(0) = f(1)$, $\\int_{0}^{1} f(x) dx = 0$ and $f'(x) \\neq 1$, for every $x \\in [0, 1]$.\n\na) Prove that the function $g : [0, 1] \\to \\mathbb{R}$ given by $g(x) = f(x) - x$ is decreasing.\n\nb) Prove that for any integer $n \\ge 1$ the following inequality holds\n$$\n\\left| \\sum_{k=0}^{n-1} f \\left( \\frac{k}{n} \\right) \\right| < \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $f'$ has the intermediate value property, it follows that $f'(x) < 1$, for every $x \\in [0, 1]$, or $f'(x) > 1$, for every $x \\in [0, 1]$. But, in the second case $f$ would be strictly increasing, which contradicts $f(0) = f(1)$. So $f'(x) < 1$, for every $x \\in [0, 1]$, whence $g : [0, 1] \\to \\mathbb{R}$, $g(x) = f(x) - x$ has $g'(x) = f'(x) - 1 < 0$, for all $x \\in [0, 1]$, that is $g$ is strictly decreasing.\n\nb) Notice that\n$$\ns_{\\Delta}(g) < \\int_{0}^{1} g(x) \\, dx < S_{\\Delta}(g),\n$$\nwhere $s_{\\Delta}(g)$ and $S_{\\Delta}(g)$ are the inferior, respectively superior Darboux sums of $g$ for the partition $\\Delta = \\{0, \\frac{1}{n}, \\dots, \\frac{n-1}{n}, 1\\}$. Then\n$$\n\\frac{1}{n} \\sum_{k=0}^{n-1} \\left( f\\left(\\frac{k+1}{n}\\right) - \\frac{k+1}{n} \\right) < \\int_{0}^{1} (f(x) - x) \\, dx < \\frac{1}{n} \\sum_{k=0}^{n-1} \\left( f\\left(\\frac{k}{n}\\right) - \\frac{k}{n} \\right).\n$$\nSince $f(0) = f(1)$ and $\\int_{0}^{1} f(x) \\, dx = 0$, this yields\n$$\n\\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n+1}{2} < -\\frac{n}{2} < \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n-1}{2},\n$$\nwhence the required inequality.\n\n\nAlternative Solution (b):\nLet $G : [0, 1] \\to \\mathbb{R}$, $G(x) = F(x) - \\frac{1}{2}x^2$, where $F$ is a primitive of $f$. The function $G'(x) = f(x) - x$ is strictly decreasing, because $G''(x) = f'(x) - 1 < 0$, for every $x \\in [0, 1]$.\nNow, Lagrange's mean value theorem, used for the function $G$ on the intervals $[\\frac{k}{n}, \\frac{k+1}{n}]$, $k = 0, \\dots, n-1$, yields\n$$\nf\\left(\\frac{k+1}{n}\\right) - \\frac{k+1}{n} < n\\left(F\\left(\\frac{k+1}{n}\\right) - F\\left(\\frac{k}{n}\\right) - \\frac{2k+1}{2n^2}\\right) < f\\left(\\frac{k}{n}\\right) - \\frac{k}{n}.\n$$\nThese inequalities add up to\n$$\n\\sum_{k=0}^{n-1} f\\left(\\frac{k+1}{n}\\right) - \\frac{n+1}{2} < n\\left(F(1) - F(0) - \\frac{1}{2}\\right) < \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n-1}{2}.\n$$\nSince $F(1) - F(0) = \\int_{0}^{1} f(x) \\, dx = 0$ and $f(0) = f(1)$, the required inequality follows immediately.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 5$ be an integer. Consider $n$ distinct points in the plane, each coloured either white or black. For each positive integer $1 \\le k < \\frac{n}{2}$, a $k$-move consists in selecting $k$ points and reversing their colours. Find all values of $n$ for which, for any eligible $k$ and for any initial colouring, there exists a sequence of $k$-moves that turns all points into a same colour.",
    "options": [],
    "answer": "All odd integers n",
    "solution": "The problem holds for and only for $n$ odd.\n\nTo this end, suppose $n$ even. Colour exactly one point in white and the rest in black. Choose $k=2$. After performing a 2-move, the number of points of a same colour remains odd, so a monochromatic configuration is not achievable.\n\nSuppose $n$ odd. Assume $k$ odd. Consider an initial colouring with $p$ white points and $n-p$ black points. Let $p = ks+r$, $0 \\le r < k$. Perform $s$ $k$-moves upon the white points to reach a configuration of $r$ white and $n-r$ black points. Perform a $k$-move with the $r$ white points and other $k-r$ black to obtain exactly $k-r$ white points. Therefore, one can obtain exactly $r$ or exactly $k-r$ white points, with only one of the numbers $r$ or $k-r$ being even, say $r$.\n\nPerform a $k$-move with the $\\frac{r}{2}$ white points to get exactly $k$ white points. After another $k$-move with the white points, all points turn black and we are done.\n\nAssume $k$ even. Again, consider an initial colouring with $p$ white points and $n-p$ black points. One of the numbers $p$ or $n-p$ is even, say $p$. In $p = ks+r$, $0 \\le r < k$, notice that $r$ is even. From this point on, proceed as in the previous case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37751,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nx_{n+1} = \\left| x_n - \\frac{1}{n} \\right|, \\quad \\text{for } n \\ge 1.\n$$\nProve that it has a finite limit, and calculate it.",
    "options": [],
    "answer": "0",
    "solution": "Let $N$ be an arbitrary positive integer. There exists a least $k \\in \\mathbb{N}$ such that $x_N < \\frac{1}{N} + \\frac{1}{N+1} + \\dots + \\frac{1}{N+k}$, since the harmonic series is divergent. Then $x_{N+k} = x_N - \\left(\\frac{1}{N} + \\dots + \\frac{1}{N+k-1}\\right) < \\frac{1}{N+k} \\le \\frac{1}{N}$. Also $x_n < \\frac{1}{N}$ for $n \\ge N+k$. The proof is by simple induction, since $x_{n+1} = \\left|x_n - \\frac{1}{n}\\right|$, and if $x_n > \\frac{1}{n}$, then $x_{n+1} = x_n - \\frac{1}{n} < x_n < \\frac{1}{N}$, while if $x_n \\le \\frac{1}{n}$, then $x_{n+1} = \\frac{1}{n} - x_n \\le \\frac{1}{n} < \\frac{1}{N}$.\n\nTherefore for any $N$ there exists $N_1$ such that $0 \\le x_n < \\frac{1}{N}$ for $n \\ge N_1$, which means the sequence converges to $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $m \\ge 2$, $n \\ge 2$, consider the set\n$$\nS(m, n) = \\left\\{ \\mathbf{x} = (x_1, x_2, \\dots, x_m) ; \\mathbf{x} \\in \\mathbb{Z}_+^m, \\sum_{k=1}^m x_k = n \\right\\}.\n$$\nDetermine $N(m, n)$, the largest cardinality $|X|$ of a subset $X \\subseteq S(m, n)$ with\n$$\n|X| = |\\{x_k ; \\mathbf{x} \\in X\\}|, \\text{ for all } k = 1, 2, \\dots, m.\n$$\nVasile Pop & Dan Schwarz",
    "options": [],
    "answer": "floor(2n/m) + 1",
    "solution": "Consider such a set $X \\subseteq S(m, n)$. Clearly, $S(m, n)$ and thus a fortiori $X$ are finite sets. Then\n$$\n\\sum_{\\mathbf{x} \\in X} x_k \\ge \\sum_{i=1}^{|X|} (i-1) = |X|(|X| - 1)/2, \\text{ for all } k = 1, 2, \\dots, m,\n$$\nand so\n$$\nn|X| = \\sum_{\\mathbf{x} \\in X} \\sum_{k=1}^{m} x_k = \\sum_{k=1}^{m} \\sum_{\\mathbf{x} \\in X} x_k \\ge m|X|(|X| - 1)/2,\n$$\nhence $n \\ge m(|X| - 1)/2$, therefore $|X| \\le \\left\\lfloor \\frac{2n}{m} \\right\\rfloor + 1$. This bound is thus valid for all points, and we claim it is sharp in all cases,¹¹ so $N(m, n) = \\left\\lfloor \\frac{2n}{m} \\right\\rfloor + 1$.\n\nLet us now, for $m = 3$, build a model set $X$ with $|X| = \\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1$.\nWhen $n = 3k-1$, the above expression evaluates at $2k$, and one can use the triplets $(0, k+1, 2k-2)$, $(1, k+2, 2k-4)$, ..., $(k-1, 2k, 0)$, $(k, 0, 2k-1)$, $(k+1, 1, 2k-3)$, ..., $(2k-1, k-1, 1)$.\nWhen $n = 3k$, the above expression evaluates at $2k+1$, and one can use the triplets $(0, k, 2k)$, $(1, k+1, 2k-2)$, ..., $(k, 2k, 0)$, $(k+1, 0, 2k-1)$, $(k+2, 1, 2k-3)$, ..., $(2k, k-1, 1)$.\nWhen $n = 3k + 1$, the above expression evaluates at $2k + 1$, and one can use the triplets $(0, k, 2k+1)$, $(1, k+1, 2k-1)$, ..., $(k, 2k, 1)$, $(k+1, 0, 2k)$, $(k+2, 1, 2k-2)$, ..., $(2k, k-1, 2)$.\n\nDenote now, in the general case, $n' = \\left\\lfloor \\frac{2n}{m} \\right\\rfloor$ and $m' = m - 2$. Then $N(2, n') = n' + 1$. We will describe how to build a model inductively.\n\n¹¹$N(2, n) = n + 1 = |S(2, n)|$, as trivially seen, while $N(2n, n) = 2$, given by e.g. $(0, 0, \\dots, 0, 1, 1, \\dots, 1)$ and $(1, 1, \\dots, 1, 0, 0, \\dots, 0)$.\n\nNow $\\lfloor \\frac{2(n-n')}{m'} \\rfloor \\ge \\lfloor \\frac{2n}{m} \\rfloor$, since $\\frac{2(n-n')}{m'} \\ge \\frac{2n}{m}$ is equivalent to $m(n-n') \\ge (m-2)n$, or $2n \\ge mn'$, or $n' \\le \\frac{2n}{m}$, patently true given the definition of $n'$, so the model for $N(m', n-n')$ yields enough elements that may be adjoined to those of the model realizing $N(2, n')$, in order to create one for $N(m, n)$ (valid, since $m = m' + 2$ and $n = (n-n') + n'$).\n\nFor the quite interesting particular case $m = n$, a model set $X = \\{\\mathbf{x}, \\mathbf{y}, \\mathbf{z}\\}$ with $|X| = \\lfloor \\frac{2n}{n} \\rfloor + 1 = 3$ is also easily built inductively.\nStart with $n=2$ and $\\mathbf{x} = (0,2), \\mathbf{y} = (2,0), \\mathbf{z} = (1,1)$. Have $\\mathbf{x} = (x_1, \\dots, x_k, 2)$, $\\mathbf{y} = (y_1, \\dots, y_k, 0)$, $\\mathbf{z} = (z_1, \\dots, z_k, 1)$ already built, for some $1 \\le k < n-1$, and build $\\mathbf{x} = (z_1, \\dots, z_k, 0,2)$, $\\mathbf{y} = (y_1, \\dots, y_k, 1,0)$, $\\mathbf{z} = (x_1, \\dots, x_k, 2,1)$, until reaching the full $n$ coordinates.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $\\mathcal{F}$ of functions $f: \\mathbb{N} \\to \\mathbb{N}$ having the property that\n$$\nf(a^2 - b^2) = f(a)^2 - f(b)^2, \\text{ for all } a, b \\in \\mathbb{N}, a \\ge b.\n$$\na) Determine the set $\\{f(1) \\mid f \\in \\mathcal{F}\\}$.\nb) Prove that $\\mathcal{F}$ has exactly two elements.",
    "options": [],
    "answer": "Possible values of f at one are 0 and 1. The only functions are the zero function and the identity function.",
    "solution": "a) For $a = b$ we have $f(0) = 0$. Then $f(a^2) = f(a)^2$ for all $a \\in \\mathbb{N}$, implying $f(1) = f(1)^2$ and then $f(1) \\in \\{0, 1\\}$.\nBoth following cases hold, $f_0(1) = 0$ for $f_0 \\equiv 0 \\in \\mathcal{F}$, and $f_1(1) = 1$ for $f_1 = 1_{\\mathbb{N}} \\in \\mathcal{F}$. The requested set is $\\{0, 1\\}$.\n\nb) Let $a, b \\in \\mathbb{N}$, $a \\ge b$ and $f \\in \\mathcal{F}$. Then $f(a), f(b) \\in \\mathbb{N}$ and $f(a)^2 - f(b)^2 = f(a^2 - b^2) \\ge 0$, which yields $f(a) \\ge f(b)$, establishing that $f$ is non-decreasing.\nSuppose $f(1) = 0$. From $f(4) = f(2)^2 = f(2)^2 - f(1)^2 = f(3)$ we obtain $f(7) = f(4)^2 - f(3)^2 = 0$. By induction, $f(7^{2^n}) = 0$. Since $f$ is monotonic, it follows that $f = f_0$.\nSuppose $f(1) = 1$ and set $f(2) = x$. Then $f(4) = x^2$, $f(3) = x^2 - 1$, $f(5) = f(3)^2 - f(2)^2 = x^4 - 3x^2 + 1$. As $f(16) = f(5)^2 - f(3)^2$, we obtain $x^4 = (x^4 - 3x^2 + 1)^2 - (x^2 - 1)^2$, which rewrites as $x^2(x^2 - 1)^2(x^2 - 4) = 0$. The function $f$ is increasing ($a, b \\in \\mathbb{N}$, $a > b$ implies $f(a)^2 - f(b)^2 \\ge f(1) > 0$), so $x = 2$. By induction we establish that $f(2^{2^n}) = 2^{2^n}$ and then, using the monotony, $f = f_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be positive integers, and let $p = a + b + c + d$. Prove that if $p$ is a prime, then $p$ is not a divisor of $ab - cd$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the relation $(a+c)(b+c) = ab + ac + bc + c^2 = (a+b+c+d)c + ab - cd = pc + ab - cd$.\nIf $p$ divides $ab - cd$, then $p$ divides $a+c$ or $b+c$.\nOn the other hand, $0 < a+c < p$, and $0 < b+c < p$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a complex number $z$, $z \\neq 0$ and the real sequence\n$$\na_n = \\left| z^n + \\frac{1}{z^n} \\right|, \\quad n \\ge 1.\n$$\na) Show that if $a_1 > 2$, then\n$$\na_{n+1} < \\frac{a_n + a_{n+2}}{2}, \\text{ for all } n \\in \\mathbb{N}^*.\n$$\nb) Prove that if there exists $k \\in \\mathbb{N}^*$ such that $a_k \\le 2$, then $a_1 \\le 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a)\nWe easily notice that\n$$\n\\begin{aligned}\n2|z^{n+1} + \\frac{1}{z^{n+1}}| &< |z + \\frac{1}{z}| \\cdot |z^{n+1} + \\frac{1}{z^{n+1}}| \\\\\n& = |z^n + \\frac{1}{z^n} + z^{n+2} + \\frac{1}{z^{n+2}}| \\leq |z^n + \\frac{1}{z^n}| + |z^{n+2} + \\frac{1}{z^{n+2}}|.\n\\end{aligned}\n$$\n\n\nAlternative Solution. Consider the sequence $(\\alpha_n)_{n \\ge 1}$ given by $\\alpha_n = z^n + \\frac{1}{z^n}$. Extend to the left with the term $\\alpha_0 = z^0 + \\frac{1}{z^0} = 2$, and denote $\\alpha = \\alpha_1$. Clearly $a_n = |\\alpha_n|$. We have\n$$\n\\alpha\\alpha_n = \\left(z + \\frac{1}{z}\\right) \\left(z^n + \\frac{1}{z^n}\\right) = \\left(z^{n+1} + \\frac{1}{z^{n+1}}\\right) + \\left(z^{n-1} + \\frac{1}{z^{n-1}}\\right) = \\alpha_{n+1} + \\alpha_{n-1}\n$$\nfor all $n \\ge 1$, so the sequence $(\\alpha_n)_{n \\ge 0}$ satisfies the linear recurrence relation $\\alpha_{n+1} = \\alpha\\alpha_n - \\alpha_{n-1}$. Then for $|\\alpha| > 2$ we have\n$$\na_n = |\\alpha_n| = \\left| \\frac{\\alpha_{n+1} + \\alpha_{n-1}}{\\alpha} \\right| \\le \\frac{|\\alpha_{n+1}| + |\\alpha_{n-1}|}{|\\alpha|} < \\frac{a_{n+1} + a_{n-1}}{2},\n$$\ni.e. the sequence $(a_n)_{n \\ge 0}$ is convex.\nBut then, if $a_1 = |\\alpha| > 2 = a_0$, any convex sequence is (strictly) increasing, since from $a_n > a_{n-1}$ follows $a_{n+1} > 2a_n - a_{n-1} = a_n + (a_n - a_{n-1}) > a_n$, and the thesis is proved by simple induction. Conversely, if there exists $k \\in \\mathbb{N}^*$ such that $a_k \\le 2$, then $a_1 \\le 2$. Therefore the proof for b) comes directly from a), and the nature of the sequence is not anymore relevant.\n\n\nb)\n$$\na_2 = |z^2 + \\frac{1}{z^2}| = \\left|\\left(z + \\frac{1}{z}\\right)^2 - 2\\right| \\ge \\left(z + \\frac{1}{z}\\right)^2 - 2 = a_1^2 - 2 > a_1,\n$$\ntherefore the sequence $(a_n)_n$ is strictly increasing, hence $a_k \\ge a_1 > 2$ for all $k$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37756,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the exterior of a non-equilateral triangle $ABC$ consider the similar triangles (in this order) $ABM$, $BCN$ and $CAP$, such that the triangle $MNP$ is equilateral. Find the angles of the triangles $ABM$, $BCN$ and $CAP$.\n\nNicolae Bourbăcuţ",
    "options": [],
    "answer": "30°, 30°, 120°",
    "solution": "The given similarity rewrites as\n$$\n\\frac{m-b}{a-b} = \\frac{n-c}{b-c} = \\frac{p-a}{c-a} = k,\n$$\nhence\n$$\nm = ka + (1-k)b,\n$$\n$$\nn = kb + (1-k)c,\n$$\n$$\np = kc + (1-k)a.\n$$\nSince the triangle $MNP$ is equilateral, we have\n$$\nm + \\varepsilon n + \\varepsilon^2 p = 0,\n$$\nwhere $\\varepsilon = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$. Substituting, we infer that\n$$\n\\begin{aligned}\n0 &= k(a + b\\varepsilon + c\\varepsilon^2) + (1-k)(b + c\\varepsilon + a\\varepsilon^2) \\\\\n&= k(a + b\\varepsilon + c\\varepsilon^2) + \\frac{1-k}{\\varepsilon}(a + b\\varepsilon + c\\varepsilon^2) \\\\\n&= (a + b\\varepsilon + c\\varepsilon^2)\\left(k + \\frac{1-k}{\\varepsilon}\\right).\n\\end{aligned}\n$$\nThe triangle $ABC$ is not equilateral, so $a + b\\varepsilon + c\\varepsilon^2 \\neq 0$, and consequently\n$$\nk = \\frac{1}{1-\\varepsilon}.\n$$\nThe equality $m = ka + (1-k)b$ yields $m - a = \\varepsilon(m - b)$, showing that triangle $AMB$ is isosceles, with an angle $\\frac{2\\pi}{3}$ and two angles $\\frac{\\pi}{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37757,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a circle are written several real numbers, of positive sum. Let $S$ be the largest and $s$ the least of the sums of consecutive numbers on the circle. Prove that $S + s > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $T > 0$ the total sum of the numbers around the circle. Clearly $S \\ge T > 0$. If $s \\ge 0$, we are done. If $s < 0$, the sum of the numbers which are not terms of $s$ is equal to $T - s$. Since $S \\ge T - s$, then $S + s \\ge T > 0$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $n$ be two positive integer numbers such that the (positive) prime factors of $a$ be all greater than $n$.\nProve that $n!$ divides $(a-1)(a^2-1)\\cdots(a^{n-1}-1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that every prime number $p$, $2 \\le p \\le n$, divides the product $(a-1)(a^2-1)\\cdots(a^{n-1}-1)$ to at least as high a power as it divides $n!$. The exponent of the highest power of $p$ which divides $n!$ is\n$$\n\\varepsilon = \\sum_{k \\ge 1} \\lfloor n/p^k \\rfloor < \\sum_{k \\ge 1} n/p^k = n/(p-1).\n$$\nOn the other hand, by hypothesis, $p$ does not divide $a$, so at least $\\lfloor(n-1)/(p-1)\\rfloor$ factors of the product $(a-1)(a^2-1)\\cdots(a^{n-1}-1)$ are divisible by $p$, by Fermat's Little Theorem. Finally, notice that $\\varepsilon \\le \\lfloor(n-1)/(p-1)\\rfloor$. If $n/(p-1)$ is integral, then $\\varepsilon \\le n/(p-1)-1 = (n-p)/(p-1) \\le \\lfloor(n-1)/(p-1)\\rfloor$; otherwise, $p \\ge 3$ and $n/(p-1)-(n-1)/(p-1) = 1/(p-1) < 1$, so $\\varepsilon \\le \\lfloor n/(p-1)\\rfloor = \\lfloor(n-1)/(p-1)\\rfloor$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Say that a set $K$ of points with integer coordinates in the plane is *connected* if for every pair of points $R, S \\in K$, there exist a positive integer $\\ell$ and a sequence $R = T_0, T_1, \\dots, T_\\ell = S$ of points in $K$, where each $T_i$ is distance $1$ away from $T_{i+1}$. For such a set $K$, we define the set of vectors\n$$\n\\Delta(K) = \\{ \\overrightarrow{RS} \\mid R, S \\in K \\}.\n$$\nWhat is the maximum value of $|\\Delta(K)|$ over all connected sets $K$ of $2n + 1$ points with integer coordinates in the plane?\n(Russia) Grigory Chelnokov",
    "options": [],
    "answer": "2n^2 + 4n + 1",
    "solution": "We claim the answer is $2n^2 + 4n + 1$. A model is\n$$\nK = \\{(0,0)\\} \\cup \\{(i,0) ; 1 \\le i \\le n\\} \\cup \\{(0,i) ; 1 \\le i \\le n\\},\n$$\nwhen\n$$\nW = \\{(a, -b) ; 0 \\le a, b \\le n\\} \\cup \\{(-a, b) ; 0 \\le a, b \\le n\\}.\n$$\nIt is left to prove that $|W| \\le 2n^2 + 4n + 1$ for any set $K$.\nWhat the statement of the problem describes is a connected graph $G = (K, E)$ of order $2n + 1$, whose vertices are the points in $K$, while the edges are the horizontal/vertical segments of length $1$ that connect (some of) these points. The key to the proof is to sequence the elements of $K$ as $A_0, A_1, \\dots, A_{2n}$ such that the graph $G_k := G[A_0, A_1, \\dots, A_k]$ is connected for every $1 \\le k \\le 2n$; this can be done through\n**LEMMA.** The vertices of a finite connected graph $G$ can always be enumerated, say as a sequence $v_0, \\dots, v_{|G|-1}$, so that $G_k := G[v_0, \\dots, v_k]$ is connected for every $1 \\le k \\le |G|-1$.\n*Proof.* Pick any vertex as $v_0$, and assume inductively that $v_0, \\dots, v_k$ have been chosen for some $0 \\le k < |G| - 1$. Now pick a vertex $v \\in G - G_k$. As $G$ is connected, it contains a $v - v_0$ path $P$. Choose as $v_{k+1}$ the last vertex of $P$ in $G - G_k$; then $v_{k+1}$ has as neighbor in $G_k$ the next vertex of $P$. The connectedness of every $G_k$ follows by induction on $k$. $\\square$\n\nMoreover, if we just keep the edges through which $A_{k+1}$ has the (selected) neighbor in $G_k$, then $G_k$ is a tree, and so $G_{2n}$ is a spanning tree of $G$. Call the vertex horizontal (vertical) if the edge that connects him is horizontal (vertical). Denote by $h$, respectively $v$, the number of horizontal, respectively vertical vertices; since $G_{2n}$ is a tree, it follows $h + v = 2n$. The point $A_0$ contributes $2n + 1$ vectors $\\overrightarrow{A_0A_i}$. Now, for $0 \\le k \\le 2n$, each point $A_{k+1}$ contributes at most $(2n+1) - x$ new vectors, where $x = h$ if the vertex is horizontal, respectively $x = v$ if the vertex is vertical, since those vectors $\\overrightarrow{A_{k+1}A_i}$, with ends at the corresponding edges of same direction, will be duplicated by the vectors determined by the opposite parallel sides of the parallelograms created, which have already been accounted for.\n\nTherefore the total number of distinct vectors will be $|W| \\le (2n+1)^2 - h^2 - v^2$. But $h^2 + v^2 \\ge \\frac{1}{2}(h+v)^2 = 2n^2$, hence $|W| \\le (2n+1)^2 - 2n^2 = 2n^2 + 4n + 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer number $n$, determine the minimum of\n$$\n\\max \\left\\{ \\frac{x_1}{1+x_1}, \\frac{x_2}{1+x_1+x_2}, \\dots, \\frac{x_n}{1+x_1+x_2+\\dots+x_n} \\right\\},\n$$\nas $x_1, x_2, \\dots, x_n$ run through all non-negative real numbers which add up to 1.",
    "options": [],
    "answer": "1 - 2^{-1/n}",
    "solution": "Let\n$$\nf(x_1, x_2, \\dots, x_n) = \\max \\left\\{ \\frac{x_k}{1 + x_1 + x_2 + \\dots + x_k} : k = 1, 2, \\dots, n \\right\\},\n$$\nwhere $x_1 \\ge 0, x_2 \\ge 0, \\dots, x_n \\ge 0$ and $x_1 + x_2 + \\dots + x_n = 1$, and notice that\n$$\n\\frac{a_1}{1+a_1} = \\frac{a_2}{1+a_1+a_2} = \\dots = \\frac{a_n}{1+a_1+a_2+\\dots+a_n}\n$$\nfor a unique $n$-tuple $(a_1, a_2, \\ldots, a_n)$ of non-negative real numbers which add up to 1,\nnamely, $a_k = 2^{k/n} - 2^{(k-1)/n}$, $k = 1, 2, \\ldots, n$, in which case $f(a_1, a_2, \\ldots, a_n) =\n1 - 2^{-1/n}$. Now let $(x_1, x_2, \\ldots, x_n) \\neq (a_1, a_2, \\ldots, a_n)$, where the $x_i$ are non-\nnegative real numbers which add up to 1. Since $x_1+x_2+\\cdots+x_n = a_1+a_2+\\cdots+a_n$,\nit follows that $x_k > a_k$ for some index $k$. Let $m = \\min\\{k : x_k > a_k\\}$. Then $x_k \\le a_k$,\n$k < m$, so\n$$\nf(x_1, x_2, \\dots, x_n) \\ge \\frac{x_m}{1 + x_1 + x_2 + \\dots + x_m} \\\\\n\\qquad > \\frac{a_m}{1 + a_1 + a_2 + \\dots + a_m} = f(a_1, a_2, \\dots, a_n).\n$$\nConsequently, the required minimum is $1 - 2^{-1/n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polynomial $f(x)$ with rational coefficients, of degree $d \\ge 2$, we define the sequence of sets $f^0(\\mathbb{Q}), f^1(\\mathbb{Q}), \\dots$, by $f^0(\\mathbb{Q}) = \\mathbb{Q}$ and then $f^{n+1}(\\mathbb{Q}) = f(f^n(\\mathbb{Q}))$ for $n \\ge 0$.\n(Given a set $S$, we write $f(S)$ for the set $\\{f(x) \\mid x \\in S\\}$.)\nLet $f^\\omega(\\mathbb{Q}) = \\bigcap_{n=0}^\\infty f^n(\\mathbb{Q})$ be the set of numbers that are in all of the sets $f^n(\\mathbb{Q})$. Prove that $f^\\omega(\\mathbb{Q})$ is a finite set.\n(Romania) Dan Schwarz",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any function $f$, denote its $n$-th iterate $f^n$. Take $d = \\deg f \\ge 2$. One can write $f(x) = \\frac{1}{N}(a x^d + g(x))$ for some $N \\in \\mathbb{Z}_+^*$, $a \\in \\mathbb{Z}^*$, and some $g \\in \\mathbb{Z}[x]$, with $\\deg g \\le d-1$, $g(x) = \\sum_{i=0}^{d-1} a_i x^i$, $a_i \\in \\mathbb{Z}$, for all $0 \\le i \\le d-1$.\nFinally, $f^\\omega(\\mathbb{Q}) \\subset f^n(\\mathbb{Q}) \\subset f^{n-1}(\\mathbb{Q}) \\subseteq \\mathbb{Q}$, for $n \\ge 1$.\nFor any $x \\in \\mathbb{Q}$, one can uniquely write $x = \\frac{\\mu(x)}{\\nu(x)}$, with $\\mu(x), \\nu(x) \\in \\mathbb{Z}$, and $\\nu(x) > 0$, $\\gcd(\\mu(x), \\nu(x)) = 1$. Take now $M = \\frac{\\sum_{i=0}^{d-1} |a_i| + 2N}{|a|} + 1$, and\n$$\n\\mathcal{M} := \\{x \\in \\mathbb{Q} ; |x| > M\\}, \\quad \\mathcal{F} := \\{x \\in \\mathbb{Q} ; \\nu(x) > a^2\\}.\n$$\nNow, $(\\mathbb{Q} \\setminus \\mathcal{F}) \\cap (\\mathbb{Q} \\setminus \\mathcal{M})$ is obviously finite; take $\\mathfrak{m}$ to be its cardinality.\nFor $x \\in \\mathcal{M}$ one has $|f(x)| \\ge |x| + 1 > M$ (see LEMMA 1), hence $f(x) \\in \\mathcal{M}$. Then $|f^n(x)| \\ge |x| + n$. For $x \\in \\mathcal{F}$ one has $\\nu(f(x)) \\ge \\nu(x) + 1 > a^2$ (see LEMMA 2), hence $f(x) \\in \\mathcal{F}$. Then $\\nu(f^n(x)) \\ge \\nu(x) + n$.\nTake $x \\in f^\\omega(\\mathbb{Q})$, and take $n$ large enough. Then we will have $x \\in f^n(\\mathbb{Q})$, hence there will exist $x_n \\in \\mathbb{Q}$ such that $f^n(x_n) = x$. If $f^k(x_n) \\in \\mathcal{M}$ for $n-k > |x|$, then $|x| = |f^n(x_n)| = |f^{n-k}(f^k(x_n))| \\ge |f^k(x_n)| + (n-k) > |x|$, absurd. If $f^k(x_n) \\in \\mathcal{F}$ for $n-k > \\nu(x)$, then $\\nu(x) = \\nu(f^n(x_n)) = \\nu(f^{n-k}(f^k(x_n))) \\ge \\nu(f^k(x_n)) + (n-k) > \\nu(x)$, absurd.\nTake $n$ large enough so that $n > \\mathfrak{m} + \\max\\{|x|, \\nu(x)\\}$.\nOne then has $f^k(x_n) \\in (\\mathbb{Q} \\setminus \\mathcal{F}) \\cap (\\mathbb{Q} \\setminus \\mathcal{M})$, for $0 \\le k \\le \\mathfrak{m}$, therefore there will exist $0 \\le i < j \\le \\mathfrak{m}$ such that $f^i(x_n) = f^j(x_n)$, therefore $f^n(x_n) = f^k(x_n)$ for some $i \\le k \\le j$, hence $x = f^n(x_n) = f^k(x_n) \\in (\\mathbb{Q} \\setminus \\mathcal{F}) \\cap (\\mathbb{Q} \\setminus \\mathcal{M})$.\nThis implies $f^\\omega(\\mathbb{Q}) \\subseteq (\\mathbb{Q} \\setminus \\mathcal{F}) \\cap (\\mathbb{Q} \\setminus \\mathcal{M})$, thus a finite set.\n\nLEMMA 1. For $x \\in \\mathcal{M}$ one has $|f(x)| \\ge |x| + 1 > M$.\n*Proof.* Clearly then $|x| > M > 1$. Now $\\frac{|a||x|^d}{N} > \\frac{\\sum_{i=0}^{d-1} |a_i||x|^{d-1}}{N} + |x|^d + |x|^d > \\frac{\\sum_{i=0}^{d-1} |a_i||x|^i}{N} + |x| + 1 \\ge \\frac{|g(x)|}{N} + |x| + 1$. It follows that $|f(x)| = \\left| \\frac{a x^d + g(x)}{N} \\right| \\ge \\left( \\frac{|a||x|^d}{N} - \\frac{|g(x)|}{N} \\right) > |x| + 1 > M$. $\\square$\n\nLEMMA 2. For $x \\in \\mathcal{F}$ one has $\\nu(f(x)) \\ge \\nu(x) + 1 > a^2$.\n*Proof.* For $x \\in \\mathcal{F}$ one can write $f(x) = \\frac{1}{N \\nu(x)^d}(a \\mu(x)^d + \\nu(x) z) = \\frac{\\mu(f(x))}{\\nu(f(x))}$, with $z = \\nu(x)^{d-1} g(x) \\in \\mathbb{Z}$. Now, for $e = \\gcd(\\nu(x), a)$, one has $\\nu(x) = e r$, $a = e b$, with $\\gcd(r, b) = 1$.\nThen it follows that $\\delta = \\gcd(a \\mu(x)^d + \\nu(x) z, N \\nu(x)^d) \\le N \\cdot \\gcd(e \\mu(x)^d + e r z, e^d r^d) = N e \\cdot \\gcd(b \\mu(x)^d + r z, e^{d-1} r^d) = N e \\cdot \\gcd(b \\mu(x)^d + r z, e^{d-1})$, since from previous relations $\\gcd(b \\mu(x), r) = 1$. Lastly $\\delta \\le N e \\cdot \\gcd(b \\mu(x)^d + r z, e^{d-1}) \\le N e e^{d-1} = N e^d \\le N |a|^d$.\nTherefore $\\nu(f(x)) = \\frac{N \\nu(x)^d}{\\delta} \\ge \\frac{N \\nu(x)^d}{N |a|^d} > \\nu(x)$, since $\\nu(x) > a^2 \\ge |a|^{\\frac{d}{d-1}}$. It follows that $\\nu(f(x)) \\ge \\nu(x) + 1 > a^2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37762,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Given a positive integer $k$, prove that there do not exist two distinct integers in the open interval $(k^2, (k+1)^2)$ whose product is a perfect square.\n\n(b) Given an integer $n > 2$, prove that there exist $n$ distinct integers in the open interval $(k^n, (k+1)^n)$ whose product is the $n$-th power of an integer, for all but a finite number of positive integers $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Suppose that two such integers, $a < b$, exist. Let $m$ be the squarefree part of $a$; $a = ms^2$, $s \\in \\mathbb{Z}$. Then $mb$ is a square, so since $m$ is squarefree, $b = mt^2$ for some integer $t \\ge s + 1$. Hence\n$$\n\\left(1 + \\frac{1}{s}\\right)^2 \\le \\frac{t^2}{s^2} = \\frac{b}{a} < \\frac{(k+1)^2}{k^2} = \\left(1 + \\frac{1}{k}\\right)^2,\n$$\nso $s > k$. Consequently, $b = mt^2 \\ge m(s+1)^2 > (k+1)^2$ – a contradiction.\n\n(b) We show that the statement holds whenever $k \\ge 3^{n-1}$; if $n$ is odd, the weaker condition $k \\ge 2^{n-1}$ suffices. We require a LEMMA whose proof offers no difficulty.\nLEMMA. If $C > 0$ and $n \\ge 2$, then $(k+1)^n > (k^{n/(n-1)} + C)^{n-1}$ for all $k \\ge C^{n-1}$.\nNow let $n$ be odd, let $a$ be the smallest integer greater than $k^{n/(n-1)}$, and let $b = a+1$. The $n$ integers $a^{n-i-1}b^i$, $i = 0, 1, 2, \\dots, n-1$, are distinct, and their product is the $n$-th power of $(ab)^{(n-1)/2}$. The smallest of them, $a^{n-1}$, exceeds $k^n$, and the largest, $b^{n-1}$, is at most $(k^{n/(n-1)} + 2)^{n-1}$. By the LEMMA, this is at most $(k+1)^n$ when $k \\ge 2^{n-1}$.\nNext, we turn to the case of even $n$. Let again $a$ be the smallest integer greater than $k^{n/(n-1)}$, and let $b = a+1$ and $c = a+2$. Let further $x_i = a^{n-i}b^i$ and $y_i = b^{i-1}c^{n-i}$, $i = 1, 2, \\dots, n-1$. Note that\n$$\nk^n < x_1 < x_2 < x_3 < \\dots < x_{n-1} < y_{n-1} < y_{n-2} < \\dots < y_2 < y_1 \\le (k^{n/(n-1)} + 3)^{n-1},\n$$\nso these $2n-2$ integers are distinct, and by the LEMMA they are in the proper range when $k \\ge 3^{n-1}$. We end the proof by showing that we can choose $n$ of these integers so that the product is a perfect $n$-th power.\nIf $n$ is divisible by 4, say $n = 4m$, choose\n$x_1, x_3, x_5, \\dots, x_{4m-1}, y_1, y_3, y_5, \\dots, y_{4m-1}$;\ntheir product is the $n$-th power of $a^m b^{2m-1} c^m$.\nIf $n = 8m + 2$, $m > 0$, take\n$x_2, x_3, x_4, \\dots, x_{4m+2}, y_2, y_3, y_4, \\dots, y_{4m+2}$;\nthe product of these is the $n$-th power of $a^{3m}b^{2m+1}c^{3m}$.\nFinally, if $n = 8m + 6$, select\n$$\nx_1, x_2, x_3, \\dots, x_{4m+3}, y_1, y_2, y_3, \\dots, y_{4m+3},\n$$\nin which case the product is the $n$-th power of $a^{3m+2}b^{2m+1}c^{3m+2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle and let $I$ be its incenter. Consider the circles $\\gamma$, $\\delta$, of diameters $IB$, respectively $IC$. The circles $\\gamma'$, $\\delta'$ are the mirror images of $\\gamma$, $\\delta$ in $IC$, respectively $IB$. Prove that the circumcenter of the triangle $ABC$ lies on the line joining the common points of the circles $\\gamma'$ and $\\delta'$.\nCosmin Pohoăță",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B_1$ be the symmetrical of the point $B$ with respect to the line $IC$, and let $C_1$ be the symmetrical of the point $C$ with respect to the line $IB$. Denote by $B_2$ the midpoint of the segment $IB_1$, and by $C_2$ the midpoint of the segment $IC_1$. The circles $\\gamma'$ and $\\delta'$ have diameters $IB_1$ and $IC_1$, hence their centers are $B_2$ and $C_2$. The line passing through the meeting points of circles $\\gamma'$ and $\\delta'$ is perpendicular to the centers line. Since $I$ belongs to both circles $\\gamma'$ and $\\delta'$, the requirement comes to $OI \\perp B_2C_2$. Since $B_2C_2$ is midline in triangle $IB_1C_1$, one has $B_1C_1 \\parallel B_2C_2$, and the requirement writes $OI \\perp B_1C_1$. It is enough to show $OB_1^2 - OC_1^2 = IB_1^2 - IC_1^2$. Let $D$ be the projection of $I$ onto the line $BC$. $D$ is the touching point of the incircle of triangle $ABC$ with the side $BC$, and $BD = p - b$, $DC = p - c$. From symmetry considerations we have $IB_1 = IB$, $IC_1 = IC$, and\n$$\n\\begin{aligned}\nIB_1^2 - IC_1^2 &= IB^2 - IC^2 = \\\\\n&= DB^2 - DC^2 = \\\\\n&= (p-b)^2 - (p-c)^2 = \\\\\n&= a(c-b).\n\\end{aligned}\n$$\nDenote by $M$ the midpoint of the side $AC$, and by $N$ the midpoint of the side $AB$; then we have\n$$\n\\begin{aligned}\nOB_1^2 - OC_1^2 &= \\\\\n&= (OM^2 + MB_1^2) - (ON^2 + NC_1^2) = \\\\\n&= (MB_1^2 + OA^2 - MA^2) - \\\\\n&\\qquad (NC_1^2 + OB^2 - NB^2) = \\\\\n&= (MB_1^2 - MA^2) - (NC_1^2 - NB^2),\n\\end{aligned}\n$$\nsince $OA = OB$. As $MB_1^2 - MA^2 = MB_1^2 - MC^2 = (MB_1 - MC)(MB_1 + MC) = a(a - b)$, and analogously $NC_1^2 - NB^2 = a(a - c)$, it follows that $OB_1^2 - OC_1^2 = a(c - b)$, which is what was left to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\gamma_1$ and $\\gamma_2$ be two circles tangent at point $T$, and let $\\ell_1$ and $\\ell_2$ be two lines through $T$. The lines $\\ell_1$ and $\\ell_2$ meet again $\\gamma_1$ at points $A$ and $B$, respectively, and $\\gamma_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of $\\gamma_1 \\cup \\gamma_2 \\cup \\ell_1 \\cup \\ell_2$. The circles $ATX$ and $BTX$ meet again $\\gamma_2$ at points $A_2$ and $B_2$, respectively. Prove that the lines $TX, A_1B_2$ and $A_2B_1$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the circle $ATX$ and the line $A_2B_1$ meet again at the point $Y$, and let the circle $BTX$ and the line $A_1B_2$ meet again at the point $Z$. Notice that the lines $AB, AY$ and $AZ$ are all parallel to the line $A_1B_1$ to deduce that the points $A, B, Y$ and $Z$ are collinear.\n\nNext, consider the cyclic quadrangles $A_1B_1A_2B_2$, $A_1B_1B_2T$ and $BTB_2Z$ to get successively $\\angle B_2A_2Y = \\angle B_1A_1B_2 = \\angle B_1TB_2 = \\angle BZB_2$, and infer thereby that the points $A_2, B_2, Y$ and $Z$ are co-cyclic. It then follows that the lines $TX, A_1B_2$ and $A_2B_1$ are the radical lines of the pairs of circles $ATX$ and $BTX, BTX$ and $A_2B_2YZ$, and $A_2B_2YZ$ and $ATX$, respectively, whence the conclusion.\nApply an inversion of pole $T$ to rephrase the statement as follows\nLet $\\delta_1$ and $\\delta_2$ be two parallel lines, let $T$ be a point in the complement of $\\delta_1 \\cup \\delta_2$, and let $\\ell_1$ and $\\ell_2$ be two lines through $T$. The lines $\\ell_1$ and $\\ell_2$ meet $\\delta_1$ at points $A$ and $B$, respectively, and $\\delta_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of $\\delta_1 \\cup \\delta_2 \\cup \\ell_1 \\cup \\ell_2$. The lines $AX$ and $BX$ meet $\\delta_2$ at points $A_2$ and $B_2$, respectively. Prove that the circles $TA_1B_2$ and $TA_2B_1$ meet again at a point on the line $TX$.\n\nIn other words, we must prove that the line $TX$ is the radical line of the circles $TA_1B_2$ and $TA_2B_1$. It is sufficient to show that the point $Y$ where the line $TX$ meets $\\delta_2$ is of equal powers with respect to the two circles. To this end, apply Menelaus' theorem to triangles $AA_1A_2$ and $BB_1B_2$ and transversal $TXY$, to get, upon multiplication and suitable rearrangement of factors,\n$$\nYA_1 \\cdot YB_2 = (YA_2 \\cdot YB_1) \\left( \\frac{TB}{TB_1} \\cdot \\frac{TA_1}{TA} \\right) \\left( \\frac{XA}{XA_2} \\cdot \\frac{XB_2}{XB} \\right).\n$$\nFinally, notice that the last two products in the parentheses above both equal 1, by similarity of triangles $TAB$ and $TA_1B_1$, and $XAB$ and $XA_2B_2$, to conclude that $Y$ is indeed of equal powers with respect to the circles $TA_1B_2$ and $TA_2B_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer number greater than or equal to $2$, and let $K$ be a closed convex set of area greater than or equal to $n$, contained in the open square $(0, n) \\times (0, n)$. Prove that $K$ contains some point of the integral lattice $\\mathbb{Z} \\times \\mathbb{Z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Transform $K$ by a suitable two-step Steiner-Edler symmetrization. First, perform a horizontal translation of each slice $K \\cap (\\mathbb{R} \\times y)$ to place it symmetrically about the vertical line $x = \\frac{1}{2}$. It is readily checked that the image $K'$ of $K$ under this transformation is a closed convex subset of the open square $(1/2 - n/2, 1/2 + n/2) \\times (0, n)$, symmetrical about the vertical line $x = 1/2$, and $\\text{area } K' = \\text{area } K \\ge n$. (Also, the transformation does not increase perimeters, a fact which will not be used in the sequel.) In addition, if $K$ contains no lattice point, so does $K'$. Indeed, if $K'$ contained one such, say $i \\times j$, then the line $y = j$ would intersect $K'$ in a closed line segment of length at least $1$, so it would also intersect $K$ in a closed line segment of the same length; the latter, in turn, would contain at least one lattice point – a contradiction.\n\nNext, apply the pattern vertically to further symmetrize $K'$ about the horizontal line $y = 1/2$. The resulting image $K''$ is a closed convex subset of the open square $(1/2 - n/2, 1/2 + n/2) \\times (1/2 - n/2, 1/2 + n/2)$, symmetrical about both lines $x = 1/2$ and $y = 1/2$ – whence centrally symmetric about the point $1/2 \\times 1/2$ – and of course $\\text{area } K'' = \\text{area } K' = \\text{area } K \\ge n$. Then if $K$ contains no lattice point, so does $K''$.\n\nNow let $a = \\sup\\{x : x \\times 1/2 \\in K''\\}$ and $b = \\sup\\{y : 1/2 \\times y \\in K''\\}$ and notice that $a - 1/2 < n/2$ and $b - 1/2 < n/2$ – both inequalities are strict, for $K''$ is closed. Let further $K_0 = K'' \\cap \\{x \\times y : x \\ge 1/2 \\text{ and } y \\ge 1/2\\}$ be the upper right quarter of $K''$. Clearly, $K_0$ is a closed convex set, and $\\text{area } K_0 = (\\text{area } K'')/4 \\ge n/4$. Suppose, if possible, that $K$ contains no lattice point. Then so does $K_0$, so by convexity it must lie below some line through the lattice point $1 \\times 1$ with a non-positive slope; that is,\n$$\nK_0 \\subset H = \\{x \\times y : \\alpha(x - 1) + \\beta(y - 1) < 0\\},\n$$\nwhere $\\alpha$ and $\\beta$ are non-negative real numbers such that $\\alpha + \\beta > 0$. Notice that $K_0 \\subset H \\cap ([1/2, a] \\times [1/2, b])$, so $\\max\\{a, b\\} > 3/2$, for otherwise\n$$\n1/2 \\le n/4 \\le \\text{area } K_0 < \\text{area}(H \\cap ([1/2, 3/2] \\times [1/2, 3/2])) = 1/2,\n$$\nwhich is a contradiction – the strict inequality above is due to the fact that $K_0$ is closed. Without loss of generality, assume that $a > 3/2$, so $\\beta \\ne 0$ and\n$$\n\\begin{aligned}\n\\text{area } K_0 < \\text{area}(H \\cap ([1/2, a] \\times [1/2, \\infty))) \\\\\n&= \\frac{1}{2} \\left(a - \\frac{1}{2}\\right) \\left(1 - \\frac{\\alpha}{\\beta} \\left(a - \\frac{3}{2}\\right)\\right) \\le \\frac{1}{2} \\left(a - \\frac{1}{2}\\right);\n\\end{aligned}\n$$\nas before, the inequality is strict because $K_0$ is closed. Finally, recall that $a - 1/2 < n/2$ and $\\text{area } K_0 \\ge n/4$ to derive a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two equilateral triangles $ABC$ and $MNP$ with $AB \\parallel MN$, $BC \\parallel NP$ and $CA \\parallel PM$, intersecting over a convex hexagon. The distances between the pairs of parallel sides do not exceed $1$. Show that at least one of the triangles has the side length less than or equal to $\\sqrt{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be an interior point to the hexagon, therefore also interior to the triangles. Denote by $a$, respectively $b$, the lengths of the sides of the two triangles. The sum of the distances from $P$ to the sides of an equilateral triangle is equal to the altitude of the triangle, hence the sum of the distances from $P$ to the lines $AB$, $BC$, $CA$, $MN$, $NP$, $PM$ is $(a + b) \\frac{\\sqrt{3}}{2}$.\n\nOn the other hand, the sum of the distances from $P$ to the parallel lines $AB$ and $MN$ is precisely the distance between those lines, hence at most $1$. It follows that $(a + b) \\frac{\\sqrt{3}}{2} \\le 3$, so $a + b \\le 2\\sqrt{3}$, whence $a \\le \\sqrt{3}$ or $b \\le \\sqrt{3}$, which is what was asked to be proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$, $Q$ with real coefficients, such that, for infinitely many positive integers $n$, $P(1)P(2)\\dots P(n) = Q(n!)$.",
    "options": [],
    "answer": "P(x) = x^d and Q(x) = x^d for some integer d ≥ 0",
    "solution": "Let $P(x)$ and $Q(x)$ be polynomials with real coefficients such that for infinitely many positive integers $n$, $P(1)P(2)\\dots P(n) = Q(n!)$.\n\nLet $d$ be the degree of $P(x)$ and $e$ the degree of $Q(x)$.\n\nFor large $n$, $P(1)P(2)\\dots P(n)$ is a product of $n$ terms, each of degree $d$, so the degree of $P(1)P(2)\\dots P(n)$ as a polynomial in $n$ is $d n$ (since each $P(k)$ is a degree $d$ polynomial in $k$, and the product over $k = 1$ to $n$ gives degree $d n$ in $n$).\n\nOn the other hand, $Q(n!)$ is a polynomial evaluated at $n!$, so as $n$ grows, $n!$ grows very rapidly. The degree of $Q(n!)$ as a function of $n$ is $e$ times the degree of $n!$ as a function of $n$, but $n!$ is not a polynomial in $n$.\n\nBut for the equality to hold for infinitely many $n$, the growth rates must match. Consider the leading behavior:\n\nLet $P(x) = a_d x^d + \\dots$ and $Q(x) = b_e x^e + \\dots$.\n\nThen for large $n$:\n$$\nP(1)P(2)\\dots P(n) \\sim a_d^n (1 \\cdot 2 \\cdot \\dots \\cdot n)^d = a_d^n (n!)^d\n$$\n\nSo $P(1)P(2)\\dots P(n) \\sim a_d^n (n!)^d$.\n\nOn the other hand,\n$$\nQ(n!) \\sim b_e (n!)^e\n$$\n\nSo for the equality to hold for infinitely many $n$, we must have $e = d$ and $a_d^n = b_e$ for all $n$ large enough, which is only possible if $a_d = 1$ and $b_e = 1$ (since $a_d^n$ is exponential in $n$ unless $a_d = 1$ or $a_d = 0$).\n\nBut if $a_d = 1$, then $P(x)$ must be $x^d$ (since otherwise the lower degree terms will affect the product for large $n$), and $Q(x) = x^d$.\n\nLet us check this:\nIf $P(x) = x^d$, then $P(1)P(2)\\dots P(n) = (1^d)(2^d)\\dots(n^d) = (1 \\cdot 2 \\cdot \\dots \\cdot n)^d = (n!)^d$.\n\nIf $Q(x) = x^d$, then $Q(n!) = (n!)^d$.\n\nSo $P(1)P(2)\\dots P(n) = Q(n!)$ for all $n$.\n\nNow, suppose $P(x)$ is not a monomial. Then for large $n$, $P(k) \\sim a_d k^d$, so $P(1)P(2)\\dots P(n) \\sim a_d^n (n!)^d$, but the lower degree terms will affect the product by a factor that is not a polynomial in $n!$, so the equality cannot hold for infinitely many $n$.\n\nTherefore, the only solutions are:\n$$\nP(x) = x^d, \\quad Q(x) = x^d, \\quad \\text{for some integer } d \\ge 0.\n$$\n\nThus, all polynomials $P(x) = x^d$ and $Q(x) = x^d$ for $d \\ge 0$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer and $p$ a prime. Suppose $A$ is a unitary ring having exactly $m$ invertible elements and such that $\\underbrace{1+1+\\cdots+1}_{p \\text{ times}} = 0$.\nProve that $A$ has non-zero nilpotent elements if and only if $p$ divides $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a \\in A$, $a \\neq 0$, be a nilpotent element and $k \\in \\mathbb{N}^*$, such that $a^{p^k} = 0$. From $(a+1)^{p^k} = a^{p^k} + 1 = 1$, we have that $a+1$ is invertible and the order of $a+1$ in the group of units $U(A)$ is a power of $p$. Because the order of $a+1$ divides the order $m$ of $U(A)$, we deduce that $p$ divides $m$.\n\nConversely, by Cauchy's theorem, there is an element $a$ of order $p$ in $U(A)$. Due to $(a-1)^p = a^p - 1 = 0$ and $a \\neq 1$, we conclude that $a-1$ is a non-zero nilpotent.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $r$ with the property that there exists positive prime numbers $p$ and $q$ so that $p^2 + pq + q^2 = r^2$.",
    "options": [],
    "answer": "7",
    "solution": "The given relation is equivalent to $(p+q)^2 = r^2 + pq$, which can be written $(p+q+r)(p+q-r) = pq$.\nThe divisors of $pq$ are $1$, $p$, $q$ and $pq$. Since $p+q > \\max\\{p,q\\}$, it follows that $p+q-r = 1$ and $p+q+r = pq$.\n\nAdding the last two equalities yields $2p+2q = pq+1$, that is $(p-2)(q-2) = 3$.\nThis leads to $(p,q) \\in \\{(3,5), (5,3)\\}$; in both cases $r=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a row of a matrix in $M_n(\\mathbb{C})$ *permutable* if, for any permutation of its entries, the value of the determinant does not change. Prove that any matrix that has two *permutable* rows is singular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $A \\in M_n(\\mathbb{C})$ such that row $l$ is *permutable*. Denote by $\\Gamma_{li}$ the algebraic complement of $a_{li}$, $i = 1, \\dots, n$. Suppose $i, j, k, p \\in \\{1, 2, \\dots, n\\}$ are such that $a_{li} \\neq a_{lj}$ and $\\Gamma_{lk} \\neq \\Gamma_{lp}$. Consider matrices $B$ and $C$ obtained from $A$ by permuting elements in row $l$, such that $B$ contains $a_{li}$ in place $(l, k)$ and element $a_{lj}$ in place $(l, p)$ and, matrix $C$ contains $a_{lj}$ in place $(l, k)$ and $a_{li}$ in place $(l, p)$; the two matrices having in the other positions the initial elements.\n\nDeveloping determinants along row $l$, we get $\\det(B) - \\det(C) = (a_{li} - a_{lj})(\\Gamma_{lk} - \\Gamma_{lp}) \\neq 0$, in contradiction with the hypothesis.\n\nWe deduce that row $l$ of $A$ is *permutable* if and only if all its elements are equal or, if all algebraic complements of elements in line $l$ are equal.\n\nConsider now a matrix $A$ that admits two *permutable* rows. Then:\n\n* if both rows are constant $\\det(A) = 0$;\n* if both rows have constant algebraic complements then $A^*$ has two constant rows; that is $\\det(A^*) = 0$, which implies $\\det(A) = 0$;\n* if one row has all elements equal to $a$, and another row has all algebraic complements equal to $b$, then from $AA^* = \\det(A)I_n$ we deduce $ab = 0$, so $a = 0$ or $b = 0$.\n\nThus, in every case, $\\det(A) = 0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $n$ and $p$ are integers, $1 < p < n$, then the number\n$$\n(2p+1)n^3 + 6n(1^2 + 2^2 + 3^2 + \\dots + p^2)\n$$\ncan be written as the sum of $2p + 1$ different perfect cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a positive integer *balanced* if the number of its distinct prime factors is equal to the number of its digits in the decimal representation; for example, the number $385 = 5 \\cdot 7 \\cdot 11$ is balanced, while $275 = 5^2 \\cdot 11$ is not. Prove that there exist only a finite number of balanced numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, \\ldots be the sequence of primes. Any balanced number $a$ with $n$ digits satisfies $a \\ge p_1 p_2 \\cdots p_n$. Since $p_1 p_2 \\cdots p_{11} = 2 \\cdot 3 \\cdot 5 \\cdots 29 \\cdot 31 > 10^{11}$ and $p_k > 10$, for any $k > 11$, it follows that there are no balanced numbers having more than 10 digits.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37773,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite set $A$ of real numbers contains at least an irrational number. Prove that for each positive integer $n$ it is possible to find $n$ elements of $A$ whose sum is irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1$ be an irrational number in $A$. If $A$ contains infinitely many irrational numbers, then for any $n$ we can pick $n$ distinct irrational numbers from $A$; their sum is irrational (since the sum of irrational numbers is irrational unless they sum to a rational, but with infinitely many choices, we can avoid this).\n\nSuppose $A$ contains only finitely many irrational numbers. Then $A$ contains infinitely many rational numbers. Let $a_1$ be an irrational number in $A$, and $a_2, \\ldots, a_n$ be any $n-1$ rational numbers in $A$. Then $a_1 + a_2 + \\cdots + a_n$ is irrational (since the sum of an irrational and any number of rationals is irrational).\n\nTherefore, for any positive integer $n$, we can find $n$ elements of $A$ whose sum is irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $\\frac{1}{\\pi} \\int_{\\sin \\frac{\\pi}{13}}^{\\cos \\frac{\\pi}{13}} \\sqrt{1-x^2} \\, dx$ is a rational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Thus $F(t) = -t + k,\\ k \\in \\mathbb{C}$. Since $F(\\pi/4) = 0$, it follows that $k = \\pi/4$, implying $F(t) = \\pi/4 - t$. Consequently, $F(\\pi/13)/\\pi = (\\pi/4 - \\pi/13)/\\pi = 9/52$, which is a rational number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37775,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a square of side length $60$, $121$ distinct points are given. Show that among them there exists three points which are vertices of a triangle with an area not exceeding $30$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Divide the square into $60$ rectangles $5 \\times 12$. By the pigeonhole principle, there are three points among the given ones inside one of the rectangles. The area of this triangle does not exceed half of the area of the rectangle, that is $5 \\cdot 12/2 = 30$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real numbers $x$ and $y$ such that\n$$\n(x^2 - x + 1)(3y^2 - 2y + 3) - 2 = 0.\n$$",
    "options": [],
    "answer": "x = 1/2, y = 1/3",
    "solution": "From $x^2 - x + 1 = (x - \\frac{1}{2})^2 + \\frac{3}{4} \\ge \\frac{3}{4}$, $3y^2 - 2y + 3 = 3(y - \\frac{1}{3})^2 + \\frac{8}{3} \\ge \\frac{8}{3}$\nwe derive that $(x^2 - x + 1)(3y^2 - 2y + 3) \\ge \\frac{3}{4} \\cdot \\frac{8}{3} = 2$, for any real numbers $x$ and $y$. Equality holds for $(3y - 1)^2 = 0$ and $(x - \\frac{1}{2})^2 = 0$, that is $x = \\frac{1}{2}$ and $y = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$, $y$, $z$, $t \\in [0, \\infty)$ so that\n$$\nx + y + z \\le t, \\quad x^2 + y^2 + z^2 \\ge t \\text{ and } x^3 + y^3 + z^3 \\le t.\n$$",
    "options": [],
    "answer": "(x,y,z) ∈ {0,1}^3 and t = x + y + z; explicitly, (x,y,z,t) ∈ {(0,0,0,0), (1,0,0,1), (0,1,0,1), (0,0,1,1), (1,1,0,2), (1,0,1,2), (0,1,1,2), (1,1,1,3)}",
    "solution": "Adding $x + y + z \\le t$, $-2x^2 - 2y^2 - 2z^2 \\le -2t$ and $x^3 + y^3 + z^3 \\le t$, one gets $x(1-x)^2 + y(1-y)^2 + z(1-z)^2 \\le 0$.\nSince $x$, $y$, $z \\in [0, \\infty)$, it follows $x$, $y$, $z \\in \\{0, 1\\}$.\nIf $x = y = z = 0$, then $t = 0$.\nIf exactly two of the numbers $x$, $y$, $z$ are nil, then $1 \\le t$ and $1 \\ge t$, therefore $t = 1$, $(x, y, z) \\in \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$.\nIf exactly one of the numbers $x$, $y$, $z$ is nil, then $2 \\le t$ and $2 \\ge t$, so $t = 2$, $(x, y, z) \\in \\{(1, 1, 0), (1, 0, 1), (0, 1, 1)\\}$.\nIf $x = y = z = 1$, then $t = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ be an irrational number. For any $n \\in \\mathbb{N}^*$ let $a_n = \\{n\\alpha\\}$ and define the sequence $(x_n)_{n \\ge 1}$ by $x_n = (a_2 - a_1)(a_3 - a_2) \\cdots (a_{n+1} - a_n)$. Show that the sequence is convergent and find its limit.",
    "options": [],
    "answer": "0",
    "solution": "We claim that $\\{x + y\\} - \\{y\\}$ is equal to $\\{x\\}$ or $\\{x\\} - 1$, for any real numbers $x$ and $y$. Write $x + y = [y] + [x] + \\{x\\} + \\{y\\}$ to get $\\{x + y\\} - \\{y\\} = \\{\\{x\\} + \\{y\\}\\} - \\{y\\}$. Hence if $\\{x\\} + \\{y\\} < 1$ then $\\{x + y\\} - \\{y\\} = \\{x\\}$, while if $2 > \\{x\\} + \\{y\\} \\ge 1$ then $\\{x + y\\} - \\{y\\} = \\{x\\} - 1$, as claimed.\n\nApply the above result to infer that either $|a_{n+1} - a_n| = \\{\\alpha\\}$ or $|a_{n+1} - a_n| = 1 - \\{\\alpha\\}$. Set $b = \\max\\{\\{\\alpha\\}, 1 - \\{\\alpha\\}\\}$ and notice that $0 < b < 1$ (for $\\alpha$ is irrational) to derive that $|x_n| \\le b^n$, implying $\\lim_{n \\to \\infty} |x_n| = 0$ and furthermore $\\lim_{n \\to \\infty} x_n = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively and $\\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from $E$ onto $\\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.\nUnited Kingdom",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The points $E$, $K$, $H$, $G$ are on the circle of diameter $GE$, so the angles $EHK$ and $EGK$ are equal.\n\n![](attached_image_1.png)\n\nAlso, from $\\angle DCA = \\angle DBA$ and $CE/CD = BE/BA$ follows\n$$\n\\frac{CE}{CG} = \\frac{2CE}{CD} = \\frac{2BE}{BA} = \\frac{BE}{BF},\n$$\nso the triangles $CGE$ and $BFE$ are similar. In particular, the angles $EGC$ and $BFE$ are equal, and therefore so are the angles $EHK$ and $BFE$.\nBut the lines $EH$ and $BF$ are perpendicular and so, since $EF$ and $HK$ are obtained by rotations of these lines by the same (directed) angle, the lines $EF$ and $HK$ also perpendicular.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $a_1 \\le a_2 \\le \\cdots \\le a_n$ be positive real numbers. Show that\n$$\n\\left( \\sum_{k=1}^{n} a_k^2 \\right) \\left( \\sum_{k=1}^{n} k a_k \\right) \\le \\left( \\sum_{k=1}^{n} a_k \\right) \\left( \\sum_{k=1}^{n} k a_k^2 \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The hypothesis leads to $(i-j)(a_i - a_j)a_i a_j \\ge 0$ for every $i$ and $j$, hence\n$$\n\\begin{align*}\n0 &\\le \\sum_{i,j=1}^{n} (i-j)(a_i - a_j)a_i a_j \\\\\n&= \\sum_{i,j=1}^{n} (i a_i^2 a_j - i a_i a_j^2 - j a_i^2 a_j + j a_i a_j^2) \\\\\n&= \\left(\\sum_{i=1}^{n} i a_i^2\\right) \\sum_{j=1}^{n} a_j - \\left(\\sum_{i=1}^{n} i a_i\\right) \\sum_{j=1}^{n} a_j^2 - \\left(\\sum_{i=1}^{n} a_i^2\\right) \\sum_{j=1}^{n} j a_j + \\left(\\sum_{i=1}^{n} a_i\\right) \\sum_{j=1}^{n} j a_j^2 \\\\\n&= 2 \\left(\\sum_{k=1}^{n} k a_k^2\\right) \\sum_{k=1}^{n} a_k - 2 \\left(\\sum_{k=1}^{n} k a_k\\right) \\sum_{k=1}^{n} a_k^2,\n\\end{align*}\n$$\nwhich leads to the required relation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that, for every $x, y \\in \\mathbb{R}$:\n$$\n|f(x + y) + \\sin x + \\sin y| \\le 2.\n$$\n\na. Prove that $|f(x)| \\le 1 + \\cos x$, for every $x \\in \\mathbb{R}$.\n\nb. Give an example of such a function, which vanishes nowhere in the interval $(-\\pi, \\pi)$.",
    "options": [],
    "answer": "For all real x, |f(x)| ≤ 1 + cos x. An example is f(x) = 2 − 2|sin(x/2)|, which satisfies the condition and has no zeros in (−π, π).",
    "solution": "a.\nLet $x = t - \\frac{\\pi}{2}$ and $y = \\frac{\\pi}{2}$. Then\n$$\n|f(t) + \\sin(t - \\tfrac{\\pi}{2}) + \\sin(\\tfrac{\\pi}{2})| \\le 2.\n$$\nBut $\\sin(t - \\tfrac{\\pi}{2}) = -\\cos t$ and $\\sin(\\tfrac{\\pi}{2}) = 1$, so\n$$\n|f(t) - \\cos t + 1| \\le 2.\n$$\nThis gives\n$$\nf(t) - \\cos t + 1 \\le 2 \\implies f(t) \\le 1 + \\cos t + 2 = 3 + \\cos t,\n$$\nbut also\n$$\nf(t) - \\cos t + 1 \\ge -2 \\implies f(t) \\ge -2 + \\cos t - 1 = \\cos t - 3.\n$$\nNow, let $x = t + \\frac{\\pi}{2}$ and $y = -\\frac{\\pi}{2}$. Then\n$$\n|f(t) + \\sin(t + \\tfrac{\\pi}{2}) + \\sin(-\\tfrac{\\pi}{2})| \\le 2.\n$$\nBut $\\sin(t + \\tfrac{\\pi}{2}) = \\cos t$ and $\\sin(-\\tfrac{\\pi}{2}) = -1$, so\n$$\n|f(t) + \\cos t - 1| \\le 2.\n$$\nThis gives\n$$\nf(t) + \\cos t - 1 \\le 2 \\implies f(t) \\le 2 - \\cos t + 1 = 3 - \\cos t,\n$$\nand\n$$\nf(t) + \\cos t - 1 \\ge -2 \\implies f(t) \\ge -2 - \\cos t + 1 = -1 - \\cos t.\n$$\nCombining the two sets of inequalities:\n$$\n\\cos t - 3 \\le f(t) \\le 3 + \\cos t,\\\\\n-1 - \\cos t \\le f(t) \\le 3 - \\cos t.\n$$\nThe intersection gives\n$$\n-1 - \\cos t \\le f(t) \\le 1 + \\cos t.\n$$\nTherefore,\n$$\n|f(x)| \\le 1 + \\cos x, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nb.\nAn example is $f(x) = 2 - 2|\\sin \\frac{x}{2}|$. A heuristic approach: the hypothesis is equivalent to\n$$\n\\begin{align*}\n2 &\\ge \\max_{u \\in \\mathbb{R}} \\left( f(t) + 2 \\sin \\frac{t}{2} \\cos \\frac{u}{2} \\right) = f(t) + 2 \\left| \\sin \\frac{t}{2} \\right| \\\\\n-2 &\\le \\min_{u \\in \\mathbb{R}} \\left( f(t) + 2 \\sin \\frac{t}{2} \\cos \\frac{u}{2} \\right) = f(t) - 2 \\left| \\sin \\frac{t}{2} \\right|,\n\\end{align*}\n$$\nthat is\n$$\n|f(x)| \\le 2 - 2|\\sin \\frac{x}{2}|, \\quad \\forall x \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M, N, P, Q$ are given on the sides $AB, BC, CD, DA$ of a parallelogram $ABCD$ such that $MN + QP = AC$. Prove that $PN + QM = DB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $m = \\frac{AM}{AB}$, $n = \\frac{BN}{BC}$, $p = \\frac{DP}{DC}$, $q = \\frac{AQ}{AD}$, to get $MN = AN - AM = AB + BN - AM = (1-m)AB + nAD$, $QP = pAB + (1-q)AD$. The given condition becomes $(1-m+p)AB + (1-q+n)AD = AC = AB + AD$.\n\nNotice that vectors $AB$ and $AD$ have distinct directions to deduce that $m = p$ and $n = q$. On the other hand, $PN = (1-p)AB - (1-n)AD$ and $QM = mAB - qAD$, implying $PN + QM = (1-p+m)AB - (1-n+q)AD = AB - AD = DB$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist two functions $f, g: \\mathbb{R} \\to \\mathbb{R}$, such that $f \\circ g$ is strictly decreasing and $g \\circ f$ is strictly increasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let\n$$\n\\begin{aligned}\n\\bullet A &= \\bigcup_{k \\in \\mathbb{Z}} ([-2^{2k+1}, -2^{2k}) \\cup (2^{2k}, 2^{2k+1}]); \\\\\n\\bullet B &= \\bigcup_{k \\in \\mathbb{Z}} ([-2^{2k}, -2^{2k-1}) \\cup (2^{2k-1}, 2^{2k}]).\n\\end{aligned}\n$$\n\nThus $A = 2B$, $B = 2A$, $A = -A$, $B = -B$, $A \\cap B = \\emptyset$, and finally $A \\cup B \\cup \\{0\\} = \\mathbb{R}$.\nLet us take\n$$\nf(x) = \\begin{cases} x & \\text{for } x \\in A \\\\ -x & \\text{for } x \\in B \\\\ 0 & \\text{for } x = 0 \\end{cases}\n$$\nand $g(x) = 2f(x)$.\n\nThen $f(g(x)) = f(2f(x)) = -2x$ and $g(f(x)) = 2f(f(x)) = 2x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f : [-\\frac{\\pi}{2}; \\frac{\\pi}{2}] \\to \\mathbb{R}$ such that:\na. $f$ is differentiable on $[-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$;\nb. there exists an antiderivative $F$ of $f$ such that $F(x) + f'(x) \\le 0, \\forall x \\in [-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$ and $F(-\\frac{\\pi}{2}) = F(\\frac{\\pi}{2}) = 0$.",
    "options": [],
    "answer": "All solutions are f(x) = C · sin x for an arbitrary real constant C.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37785,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is said that a positive integer $n > 1$ has the property ($p$) if in its prime factorization\n$$\nn = p_1^{\\alpha_1} \\cdots p_j^{\\alpha_j}\n$$\nat least one of the prime factors $p_1, \\dots, p_j$ has the exponent equal to 2.\n\na) Find the largest number $k$ for which there exist $k$ consecutive positive integers that do not have the property ($p$).\nb) Prove that there is an infinite number of positive integers $n$ such that $n, n + 1$ and $n + 2$ have the property ($p$).",
    "options": [],
    "answer": "7",
    "solution": "a) Among any 8 consecutive integers there exists one of the form $8j + 4$. This number has the property ($p$) because the factor 2 from its prime factorization has the exponent 2. Therefore there can be at most 7 consecutive positive integers that do not have the property ($p$). Since none of the numbers 29, 30, 31, 32, 33, 34, 35 has the property ($p$), the largest $k$ is $k = 7$.\n\nb) Since the numbers $98 = 2 \\cdot 7^2$, $99 = 3^2 \\cdot 11$, and $100 = 2^2 \\cdot 5^2$ have the property ($p$), so will the numbers $98 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$, $99 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$, and $100 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$.\n\nAlternative solution.\n\nBy the Chinese remainder theorem, there exist an infinite number of solutions for the system of simultaneous congruences: $n \\equiv 4 \\pmod{8}$, $n \\equiv 8 \\pmod{27}$, $n \\equiv 23 \\pmod{125}$. Then $n, n + 1$, and $n + 2$ all have the property ($p$) because the factor 2, 3, and 5, respectively, has the exponent 2 in their prime factorizations.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y$ be real numbers such that $x+y$, $x^3+y^3$, $x^5+y^5$ and $y^5$ are rational. Prove that $x$ and $y$ are rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(K, +, \\cdot)$ be a finite field. Prove that:\na) if $K$ has $4k+1$ elements, then the polynomial $f = X^4 + 4$ has four roots in $K$;\nb) the polynomial $g = X^8 - 16$ has at least a root in $K$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, let $D$ be the midpoint of the side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD = 1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays are perpendicular: $AM \\perp CM$. Show that the quadrangle $AFED$ is cyclic if and only if the line of support of the median from $A$ in triangle $ABC$ meets the line $EF$ at a point situated on the circle $ABC$.\nBMO Shortlist 2011, Saudi Arabia",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $a$, $b$, $c$ the sidelengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that $2m_b/3 = MD = AD = CD = b/2$, so $m_b = 3b/4$, whence $(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4$; that is, $13b^2 = 8(a^2 + c^2)$.\n\nNext, apply the Menelaus theorem to get $EC/EB = 4 = FA/FB$ and deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrangle $AFED$ is therefore a trapezium; it is cyclic if and only if $AF = DE$.\n\nExpress the two in terms of $a$, $b$ and $c$. Recall that $FA/FB = 4$ to obtain $AF = 4c/5$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = b^2/2 - 4a^2/25$. By the preceding, the quadrangle $AFED$ is cyclic if and only if $25b^2 - 8a^2 = 32c^2$. Recall that $13b^2 = 8(a^2 + c^2)$ to express $b$ and $c$ in terms of $a$: $b = 2a\\sqrt{2}/3$ and $c = 2a/3$.\n\n![](attached_image_1.png)\n\nFinally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that $EN = a/2 - a/5 = 3a/10$, and the triangles $ANC$ and $PNE$ are similar, to obtain $NP = 3m_a/5$, so\n$$\nNA \\cdot NP = 3m_a^2/5 = 3(2(b^2 + c^2) - a^2)/20 = a^2/4 = NB \\cdot NC.\n$$\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\int_{0}^{1} e^{e^{x}} \\, dx \\ge \\frac{e(e^{2} - 3)}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are infinitely many positive integer numbers $n$ such that $n^2 + 1$ has two positive divisors whose difference is $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define the sequence $(a_k)_{k \\ge 0}$ by $a_0 = 1$, $a_1 = 2$ and $a_{k+2} a_k = a_{k+1}^2 + 1$, $k = 0, 1, 2, \\dots$, and check inductively that the $a_k$ are all positive integer numbers, the $n_k = a_{k+1} - a_k$ form a strictly increasing sequence of positive integer numbers, and $a_k$ and $a_{k+1}$ both divide $n_k^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer. Determine the smallest positive integer $n$ for which there exist real numbers $x_1, x_2, \\dots, x_n \\in (-1, 1)$ such that $|x_1| + |x_2| + \\dots + |x_n| = m + |x_1 + x_2 + \\dots + x_n|$.",
    "options": [],
    "answer": "n = m + 1 if m is odd; n = m + 2 if m is even",
    "solution": "Let us consider $x_1, x_2, \\dots, x_n$ a solution of the equation above. We may suppose, without loss of generality, that $x_1 + x_2 + \\dots + x_n \\ge 0$ (otherwise we change the signs of all the numbers) and that $x_1 \\le \\dots \\le x_p \\le 0 < x_{p+1} \\le \\dots \\le x_n$. Then the equation becomes $-2(x_1+x_2+\\dots+x_p) = m$. From $x_i \\in (-1, 1)$ we obtain that $m < 2p$, i.e. $p > \\frac{m}{2}$. Moreover, $n-p > x_{p+1} + \\dots + x_n \\ge -x_1 - \\dots - x_p = \\frac{m}{2}$. If $m$ is even then $p \\ge \\frac{m}{2}+1$, $n-p \\ge \\frac{m}{2}+1$, and therefore $n \\ge m+2$. A convenient choice for $n = m+2$ is $x_1 = \\dots = x_{\\frac{m}{2}+1} = -\\frac{m}{m+2}$, $x_{\\frac{m}{2}+2} = \\dots = x_n = \\frac{m}{m+2}$.\n\nIf $m$ is odd, from $p \\ge \\frac{m+1}{2}$, $n-p \\ge \\frac{m+1}{2}$, we obtain $n \\ge m+1$. A convenient choice for $n = m+1$ is $x_1 = \\dots = x_{\\frac{m+1}{2}} = -\\frac{m}{m+1}$, $x_{\\frac{m+1}{2}+1} = \\dots = x_n = \\frac{m}{m+1}$.\n\nIn conclusion, the smallest value of $n$ for which the equation has solutions is\n$n = m + 1$ if $m$ is odd and $n = m + 2$ if $m$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer number greater than $2$, let $x_1, x_2, \\dots, x_n$ be $n$ positive real numbers such that\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i + 1} = 1,\n$$\nand let $\\alpha$ be a real number greater than $1$. Show that\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i^{\\alpha} + 1} \\geq \\frac{n}{(n-1)^{\\alpha} + 1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $y_i = 1/(x_i+1)$, $i = 1, 2, \\dots, n$, so the $y_i$ are positive real numbers that add up to $1$. Upon substitution, the left-hand member of the required inequality becomes\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(1 - y_i)^{\\alpha} + y_i^{\\alpha}} = \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}},\n$$\nthe latter on account of $y_1 + y_2 + \\dots + y_n = 1$. Apply Jensen's inequality to the convex function $t \\mapsto t^{\\alpha}$, $t > 0$, to get\n$$\n\\left( \\sum_{j \\neq i} y_j \\right)^{\\alpha} \\le (n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha}, \\quad i = 1, 2, \\dots, n,\n$$\nso\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}} \\ge \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha} + y_i^{\\alpha}}.\n$$\nNow write $z_i = y_i^{\\alpha}$, $i = 1, 2, \\dots, n$, $z = z_1 + z_2 + \\dots + z_n$ and $a = (n-1)^{\\alpha-1}$ to transform the right-hand member of the above inequality to\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az}.\n$$\nFinally, notice that the function $t \\mapsto \\frac{t}{(1-a)t + az}$, $t < az/(a-1)$, is convex, to conclude by Jensen's inequality:\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az} \\ge n \\cdot \\frac{\\frac{1}{n} \\sum_{i=1}^{n} z_i}{(1-a)^n \\sum_{i=1}^{n} z_i + az} = \\frac{n}{(n-1)a+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37793,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$ let $\\tau(n)$ denote the number of its positive factors. Determine all $n \\in \\mathbb{N}$ that satisfy the equality $\\tau(n) = \\frac{n}{3}$.",
    "options": [],
    "answer": "9, 18, 24",
    "solution": "If $n \\in \\mathbb{N}$ satisfies the condition $\\tau(n) = \\frac{n}{3}$, then $3 \\mid n$. Put $n = 3k$, $k \\in \\mathbb{N}$. If $k$ is even then $\\frac{k}{2} = \\frac{n}{6}$ is a factor of $n$. Even if all the positive numbers smaller than $\\frac{n}{6}$ are factors of $n$ and the numbers $\\frac{n}{5}, \\frac{n}{4}, \\dots, \\frac{n}{1}$ are also factors of $n$, we have $\\frac{n}{3} = \\tau(n) \\le \\frac{n}{6} + 5$. Hence $n \\le 30$. Checking the numbers $6, 12, 18, 24$ and $30$ we find that $18$ and $24$ satisfy the desired condition.\n\nIn the case when $k$ is odd we can proceed similarly, obtaining $n = 9$, or, alternatively, we can use the fact that a number having an odd number of positive factors is a perfect square. If $n = m^2$, then $n$ has at most $m + 1$ factors, hence $m + 1 \\ge \\frac{m^2}{3}$. We obtain that $m \\le 3$ and the conclusion.\n\nIn conclusion, the problem admits three solutions: $9$, $18$ and $24$.\nSince $3 \\mid n$, it follows that $n$ has a prime factorization of the form $n = 3^a \\cdot p_1^{\\alpha_1} \\cdot \\dots \\cdot p_j^{\\alpha_j}$ and the number of its positive factors is $\\tau(n) = (a+1)(\\alpha_1+1)\\dots(\\alpha_j+1)$. The condition $\\tau(n) = \\frac{n}{3}$ leads to $3^{a-1} \\cdot p_1^{\\alpha_1} \\cdot \\dots \\cdot p_j^{\\alpha_j} = (a+1)(\\alpha_1+1)\\dots(\\alpha_j+1)$. Since $p_i^{\\alpha_i} \\ge 2^{\\alpha_i} \\ge \\alpha_i + 1$, in order for $n$ to satisfy the equation from the statement, it is necessary that $a+1 \\ge 3^{a-1}$, hence $a=1$ or $a=2$.\nIf in the prime factorization of $n$ there is a prime $p_i > 3$, then $p_i^{\\alpha_i} > 4^{\\alpha_i} \\ge 2\\alpha_i + 2$ and the equality $\\tau(n) = \\frac{n}{3}$ can not take place. If $a=1$ then $n = 3 \\cdot 2^m$ and the equality $\\tau(n) = \\frac{n}{3}$ reduces to $2(m+1) = 2^m$. We find $m=3$ (for $m \\ge 4$ we have $2^m > 2m+2$), hence $n=24$.\nSimilarly, if $a=2$ then $n = 3^2 \\cdot 2^m$ and from $3(m+1) = 3 \\cdot 2^m$ we obtain $m \\in \\{0,1\\}$, i.e. $n \\in \\{9,18\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set $\\mathcal{L}$ of lines in general position in the plane (no two lines in $\\mathcal{L}$ are parallel and no three lines are concurrent) and another line $\\ell$, show that the total number of edges of all faces in the corresponding arrangement intersected by $\\ell$ is at most $6|\\mathcal{L}|$.\nChazelle et al., Edelsbrunner et al.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume without loss of generality that $\\ell$ is horizontal and does not pass through any vertex of the arrangement of lines in $\\mathcal{L}$.\n\nFirst, we shall bound the total number of edges of the upper parts of all faces intersected by $\\ell$, that is, those parts that lie above $\\ell$. The boundary of the upper part of such a face $K$ consists of two convex chains of edges, the *left* and *right* chain, and a portion of $\\ell$. If the upper part of $K$ is bounded, then the left and right chains meet at the topmost vertex of $K$. Otherwise, the last (topmost) edges of these chains are half-lines. The edges belonging to the left (respectively, right) chain, with the exception of the topmost edge, are called the *left* (respectively, *right*) *edges* of $K$.\nWe claim that every line in $\\mathcal{L}$ contains at most one left edge. Suppose, if possible, that some line in $\\mathcal{L}$ has two portions $e$ and $e'$ that are left edges of $K$ and $K'$, respectively, where $e'$ is above $e$. Then the line supporting the topmost edge of the left chain of $K$ would cross $K'$, contradicting the fact that $K'$ is a face of the arrangement of lines in $\\mathcal{L}$. Hence, the total number of left (respectively, right) edges of the upper parts of the faces intersected by $\\ell$ is at most $|\\mathcal{L}|$. Taking into account the topmost edges of the chains, the total number of edges of the upper parts of the faces intersected by $\\ell$ is at most $4|\\mathcal{L}|$. The same argument applies verbatim for the lower parts of these faces. Consequently, the total number of edges of the faces intersected by $\\ell$ is at most $8|\\mathcal{L}| - 2|\\mathcal{L}| = 6|\\mathcal{L}|$; the second term is due to the fact that the edges crossed by $\\ell$ are counted twice: once in the upper parts and once in the lower parts.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer number $k$, define the function $f$ on the set of all positive integer numbers to itself by\n\n$$\nf(n) = \\begin{cases} 1, & \\text{if } n \\le k+1, \\\\ f(f(n-1)) + f(n - f(n-1)), & \\text{if } n > k+1. \\end{cases}\n$$\n\nShow that preimage of every positive integer number under $f$ is a finite non-empty set of consecutive positive integer numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is sufficient to show that the difference $\\Delta(n) = f(n) - f(n-1)$ is 0 or 1 for all integer numbers $n \\ge 2$, and $f$ is unbounded.\n\nClearly, $\\Delta(n) = 0$, $2 \\le n \\le k+1$, provides the basis for an inductive proof. If $n > k+1$, apply the recurrence. By the induction hypothesis, $\\Delta(n-1)$ is 0 or 1 and\n\n$1 \\le f(n-1) \\le n-1$. Notice that $\\Delta(n) = \\Delta(n-f(n-1))$ if $\\Delta(n-1) = 0$ and $\\Delta(n) = \\Delta(f(n-1))$ if $\\Delta(n-1) = 1$ to conclude that $\\Delta(n)$ is indeed 0 or 1.\nTo show that $f$ is unbounded, suppose, if possible, that $f$ achieves its maximum value $N$ for the first time at $m$ to deduce recursively that $N = f(m+n) = f(N) + f(m-N+n)$ for all positive integers $n$; that is, $f(m-N+n) = N-f(N)$, for all positive integers $n$. Setting $n \\ge N$ yields a contradiction: $N = N-f(N)$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37796,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find the largest possible value of the number\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n,\n$$\nif $x_1, x_2, \\dots, x_n$ ($n \\ge 2$) are non-negative integers and their sum is $2011$.\n\nb) Find the numbers $x_1, x_2, \\dots, x_n$ for which the maximum value determined at a) is obtained.",
    "options": [],
    "answer": "a) Maximum value: 1005 * 1006.\n\nb) Maximizing sequences:\n- For n = 2: (1005, 1006) or (1006, 1005).\n- For n = 3: (k, 1005, 1006 − k) for k = 0, 1, ..., 1006; or (k, 1006, 1005 − k) for k = 0, 1, ..., 1005.\n- For n ≥ 4: all entries are zero except for three consecutive positions, which equal (k, 1005, 1006 − k) with k = 0, 1, ..., 1006, or (k, 1006, 1005 − k) with k = 0, 1, ..., 1005; the block of three can be placed anywhere.",
    "solution": "a.\nLet $x_1, x_2, \\dots, x_n$ be non-negative integers satisfying the conditions from the statement. We put $M = \\max_{1 \\le i \\le n} x_i$. If $x_j = M$ then\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n \\le x_1x_j + x_2x_j + \\dots + x_{j-1}x_j + x_jx_{j+1} + x_jx_{j+2} + \\dots + x_jx_n = x_j(2011-x_j) = M(2011-M) \\le 1005 \\cdot 1006.\n$$\nIndeed, the last inequality comes to $(M-1005)(M-1006) \\ge 0$ which is true for any integer $M$. The largest possible value is $1005 \\cdot 1006$ because this value can be obtained by choosing, for example, $x_1 = 1005$, $x_2 = 1006$ and $x_k = 0$ for $k \\ge 3$.\n\nb.\nFor $n=2$ we have $x_1x_2 = 1005 \\cdot 1006 \\Leftrightarrow x_1(2011-x_1) = 1005 \\cdot 1006 \\Leftrightarrow (x_1-1005)(x_1+1006) = 0 \\Leftrightarrow (x_1,x_2) \\in \\{(1005, 1006), (1006, 1005)\\}$.\n\nFor $n=3$, $x_1x_2 + x_2x_3 = 1005 \\cdot 1006 \\Leftrightarrow x_2(x_1+x_3) = 1005 \\cdot 1006$ comes, as above, to $x_2 = 1005$, $x_1+x_3 = 1006$ or $x_2 = 1006$, $x_1+x_3 = 1005$. We obtain $(x_1, x_2, x_3) \\in \\{(k, 1005, 1006-k) \\mid k = 0, 1, \\dots, 1006\\} \\cup \\{(k, 1006, 1005-k) \\mid k = 0, 1, \\dots, 1005\\}$.\n\nFor $n \\ge 4$, we denote by $j$ the smallest index for which $x_j > 0$. Then, replacing $x_j$ by $0$ and $x_{j+2}$ by $x_{j+2} + x_j$, increases the value of the sum by $x_jx_{j+3}$. Using this remark it is easy to see that if $x_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n = 1005 \\cdot 1006$, then at most three of the terms can be non-zero. We obtain $(x_1, \\dots, x_n) \\in \\{(0, \\dots, 0, k, 1005, 1006-k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1006\\} \\cup \\{(0, \\dots, 0, k, 1006, 1005-k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1005\\}$, where the group of the three non-zero components can be located anywhere.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$ congruent to $1$ modulo $5$ such that $2p + 1$ is also prime, show that there exists a matrix of zeros and ones containing exactly $4p$ (respectively, $4p + 2$) ones no submatrix of which contains exactly $2p$ (respectively, $2p + 1$) ones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p = 5q + 1$, $q \\ge 2$, and write $4p = 5(4q + 1) - 1$. Form a $5$-by-$(4q + 1)$ matrix consisting of ones only except for a single entry. Such a matrix has exactly $4p$ ones. A submatrix comprising $r$ rows, $1 \\le r \\le 5$, and $s$ columns, $1 \\le s \\le 4q + 1$, contains $rs$ or $rs-1$ ones. In the former case, $rs = 2p$ implies $r \\ge p$ or $s \\ge p$, because $p$ is prime; in the latter, $rs-1 = 2p$ implies $rs = 2p+1$, so $r = 2p+1$ or $s = 2p+1$, because $2p+1$ is prime. Both cases contradict the size of the matrix, so no submatrix contains exactly $2p$ ones.\n\nNext, write $4p+2 = 5(4q+1)+1$. Now form a $5$-by-$(4q + 1)$ matrix consisting of ones only except for one column that contains only a single one. Such a matrix has exactly $4p + 2$ ones. A submatrix comprising $r$ rows, $1 \\le r \\le 5$, and $s$ all-one columns, $0 \\le s \\le 4q+1$, contains $rs$ or $rs+1$ ones. In the former case, $rs = 2p+1$ implies $r = 2p+1$ or $s = 2p+1$ because $2p+1$ is prime; in the latter, $rs+1 = 2p+1$, so either $r \\ge p$ or $s \\ge p$ since $p$ is prime. Again, this contradicts the size of the matrix and the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37798,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $X$ be a point on the incircle, different from the points $D$, $E$, $F$. The lines $XD$ and $EF$, $XE$ and $FD$, $XF$ and $DE$ meet at points $J$, $K$, $L$, respectively. Let further $M$, $N$, $P$ be points on the sides $BC$, $CA$, $AB$, respectively, such that the lines $AM$, $BN$, $CP$ be concurrent. Prove that the lines $JM$, $KN$ and $LP$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the lines $KL$ and $NP$, $LJ$ and $PM$, $JK$ and $MN$ meet at points $Q$, $R$, $S$, respectively. By Desargues' theorem on perspective triangles, the lines $JM$, $KN$ and $LP$ are concurrent if and only if the points $Q$, $R$ and $S$ are collinear.\n\n![](attached_image_1.png)\n\nWithout loss of generality, we may assume that $X$ lies on the arc $FD$ of the incircle that does not contain $E$. Consequently, $K$ lies on the side $FD$, while $L$ and $J$ lie on the respective extensions of the sides $DE$ and $EF$.\n\nConsider the cyclic quadrangle $DEFX$: The diagonals meet at $K$, and the extensions of the opposite sides meet at $L$ and $J$, respectively, so the line $LJ$ is the polar of $K$ with respect to the incircle – in what follows, all polar lines are considered with respect to the incircle. Since the line $FD$ is the polar of $B$, and $K$ lies on the line $FD$, it follows that $B$ lies on the line $LJ$. Similarly, $C$ lies on the line $JK$, and $A$ lies on the line $KL$. Consequently, the lines $AQ$ and $CS$ meet at $K$.\n\nProjectively, the lines $FD$ and $LJ$ meet at some point $T$. Notice that $T$ lies on the polar lines of $B$ and $K$ to deduce that the line $BK$ is the polar of $T$, so the cross-ratio $(TFKD)$ is harmonic. Let further the lines $BK$ and $PM$ meet at $U$. Read from $B$, the cross-ratio $(RPUM)$ equals $(TFKD)$, so it is harmonic; that is, $U$ is the harmonic conjugate of $R$ relative to $M$ and $N$. Consequently, the points $Q$, $R$ and $S$ are collinear if and only if the lines $MQ$, $NU$ and $PS$ are concurrent.\n\nTo prove the lines $MQ$, $NU$ and $PS$ concurrent, simply check that\n$$\n\\frac{QN}{QP} \\cdot \\frac{UP}{UM} \\cdot \\frac{SM}{SN} = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there is an infinite number of positive integers $t$ such that none of the equations $x^2 + y^6 = t$, $x^2 + y^6 = t + 1$, $x^2 - y^6 = t$, $x^2 - y^6 = t + 1$ has solutions $(x, y) \\in \\mathbb{Z} \\times \\mathbb{Z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x$ is a positive integer, then either $x^{12} \\equiv 0 \\pmod{13}$ or $x^{12} \\equiv 1 \\pmod{13}$, hence $x^6$ is congruent with $-1$, $0$ or $1$ modulo $13$. Therefore, if $t$ is congruent with $6 \\pmod{13}$, then $\\pm x^6 + t$ is congruent with $5$, $6$ or $7 \\pmod{13}$, while $\\pm x^6 + t + 1$ is congruent with $6$, $7$ or $8 \\pmod{13}$.\n\nOn the other hand, perfect squares are congruent with $0$, $1$, $4$, $9$, $3$, $12$ or $10 \\pmod{13}$. Therefore a perfect square can not be equal to a number of the form $\\pm x^6 + t$ or $\\pm x^6 + t + 1$ if $t \\equiv 6 \\pmod{13}$. In conclusion, all the numbers that are congruent with $6$ modulo $13$ have the required property.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integers $n \\ge 3$ and $k \\ge 1$, find all positive integers $a_1 < a_2 < \\dots < a_n$ such that $\\frac{1}{a_1^k} + \\frac{1}{a_2^k} + \\dots + \\frac{1}{a_n^k} = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a convex pentagon $A_0A_1A_2A_3A_4$ so that the rays ($A_iA_{i+1}$ and ($A_{i+3}A_{i+2}$ meet at $B_{i+4}$, for each $i = 0, 1, 2, 3, 4$ – indices are considered modulo 5. Show that\n$$\n\\prod_{i=0}^{4} A_i B_{i+3} = \\prod_{i=0}^{4} A_i B_{i+2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $C_i$ be the projection of $B_i$ on the line $A_{i+2}A_{i+3}$, $i = 0, 1, 2, 3, 4$. Notice that the right-angled triangles (possibly degenerate) $A_iB_{i+3}C_{i+3}$ and $A_iB_{i+2}C_{i+2}$ are similar, for angles $A_i$ are vertical. Hence $\\frac{A_iB_{i+3}}{A_iB_{i+2}} = \\frac{B_{i+3}C_{i+3}}{B_{i+2}C_{i+2}}$, implying\n$$\n\\prod_{i=0}^{4} \\frac{A_i B_{i+3}}{A_i B_{i+2}} = \\prod_{i=0}^{4} \\frac{B_{i+3} C_{i+3}}{B_{i+2} C_{i+2}} = 1,\n$$\nas claimed.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 37802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists three complex roots of order $n$ of the unity, not necessarily different, adding up to $1$.",
    "options": [],
    "answer": "n is even",
    "solution": "If $n$ is odd, then $-1$, $1$, $1$ are three complex roots of order $n$ of the unity, adding up to $1$.\n\nOn the other hand, if $x$, $y$, $z \\in \\mathbb{C}$, $x^n = y^n = z^n = 1$ and $x + y + z = 1$, then $|x| = |y| = |z|$, hence $\\overline{x} + \\overline{y} + \\overline{z} = 1/x + 1/y + 1/z = 1$, which leads to $xy + xz + yz = xyz$.\n\nReplacing $z = 1 - x - y$ gives $(x + y)(1 - x)(1 - y) = 0$, whence one of the numbers $x$, $y$, $z$ is $1$ and the other two are opposite.\n\nIn the case $n$ = odd, there are no opposite roots, so the final answer is: $n$ = even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37803,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(a_n)_{n \\ge 1}$ of real numbers is such that the sequence $(x_n)_{n \\ge 1}$ defined by $x_n = \\max\\{a_n, a_{n+1}, a_{n+2}\\}$ is convergent and the sequence $(y_n)_{n \\ge 1}$ defined by $y_n = a_{n+1} - a_n$ has limit $0$. Prove that the sequence $(a_n)_n$ is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $L = \\lim_{n \\to \\infty} x_n$. Since $x_n = \\max\\{a_n, a_{n+1}, a_{n+2}\\}$, for every $n$, $a_n \\le x_n$, $a_{n+1} \\le x_n$, $a_{n+2} \\le x_n$.\n\nLet $\\varepsilon > 0$. Since $x_n \\to L$, there exists $N_1$ such that for all $n \\ge N_1$, $|x_n - L| < \\varepsilon/2$, i.e., $L - \\varepsilon/2 < x_n < L + \\varepsilon/2$.\n\nAlso, since $y_n = a_{n+1} - a_n \\to 0$, there exists $N_2$ such that for all $n \\ge N_2$, $|a_{n+1} - a_n| < \\varepsilon/4$.\n\nLet $N = \\max\\{N_1, N_2\\}$. For $n \\ge N$, $a_n, a_{n+1}, a_{n+2} \\le x_n < L + \\varepsilon/2$.\n\nWe claim that $a_n \\to L$.\n\nFirst, we show that $\\limsup a_n \\le L$.\n\nSuppose not. Then there exists a subsequence $a_{n_k} \\to L' > L$. But then for large $k$, $a_{n_k} > L + \\varepsilon$ for some $\\varepsilon > 0$. But $x_{n_k} \\ge a_{n_k} > L + \\varepsilon$, contradicting $x_{n_k} \\to L$.\n\nSo $\\limsup a_n \\le L$.\n\nNow, we show that $\\liminf a_n \\ge L$.\n\nSuppose not. Then there exists a subsequence $a_{m_k} \\to L'' < L$. For $k$ large, $a_{m_k} < L - \\varepsilon$ for some $\\varepsilon > 0$.\n\nBut $a_{m_k+1} = a_{m_k} + y_{m_k}$, $a_{m_k+2} = a_{m_k+1} + y_{m_k+1} = a_{m_k} + y_{m_k} + y_{m_k+1}$.\n\nSince $y_n \\to 0$, for large $k$, $|y_{m_k}| < \\varepsilon/4$, $|y_{m_k+1}| < \\varepsilon/4$.\n\nSo $a_{m_k+1} < L - \\varepsilon + \\varepsilon/4 = L - 3\\varepsilon/4$,\n$a_{m_k+2} < L - \\varepsilon + \\varepsilon/4 + \\varepsilon/4 = L - \\varepsilon/2$.\n\nTherefore, $x_{m_k} = \\max\\{a_{m_k}, a_{m_k+1}, a_{m_k+2}\\} < L - \\varepsilon/2$, contradicting $x_{m_k} \\to L$.\n\nTherefore, $\\liminf a_n \\ge L$.\n\nThus, $a_n \\to L$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be the set of matrices\n$$\n\\begin{pmatrix} a & b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}, \\quad a, b \\in \\mathbb{Z}_7, a \\neq \\hat{0}.\n$$\n\na) Show that $G$ is a group with respect to the matrix multiplication.\nb) Prove that there exists no proper homomorphism from $G$ to $\\mathbb{Z}_7$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Consider $A = \\begin{pmatrix} a & b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}$ and $B = \\begin{pmatrix} x & y \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}$ two elements of $G$. Then\n$$ AB = \\begin{pmatrix} ax & ay + b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix} \\in G, \\text{ for } ax \\neq \\hat{0}. $$\nTo end the proof, recall that matrix multiplication is an associative law, the group unit is $I_2 \\in G$ and, finally, the inverse matrix of $A$ is $\\begin{pmatrix} a^{-1} & -a^{-1}b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix} \\in G$.\n\nb. Let $A = \\begin{pmatrix} a & b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}$ be an element of $G$ with $a \\neq \\hat{1}$. Since\n$$\nA^k = \\begin{pmatrix} a^k & b(a^{k-1} + a^{k-2} + \\dots + 1) \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}, \\text{ it follows that } A^6 = I_2.\n$$\nLet $f$ be a group homomorphism mapping $G$ to $\\mathbb{Z}_7$. Notice that $\\hat{0} = f(I_2) = f(A^6) = \\hat{6}f(A)$ to derive that $f(A) = \\hat{0}$. Since $\\ker(f) = \\{X \\in G \\mid f(X) = \\hat{0}\\}$ is a subgroup of $G$ with at least 36 elements, and $G$ has 42 elements, we obtain that $\\ker(f) = G$. Therefore $f(X) = \\hat{0}$, for any $X \\in G$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that every integer between $1$ and $n!$ can be written as the sum of at most $n$ distinct positive divisors of $n!$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use induction on $n$; the base case $n=1$ is obvious.\n\nSuppose now that the conclusion holds for some positive integer $n$ and consider $m \\le (n+1)!$. Then $m = (n+1)q + r$, $q, r \\in \\mathbb{N}$, $0 \\le r \\le n$. It is easily noticed that $q \\le n!$, therefore $q$ can be written as a sum of (at most) $n$ divisors of $n!$ – denote them $d_1, \\dots, d_k$, $k \\le n$, whence\n\n$$\nm = (n + 1)d_1 + (n + 1)d_2 + \\dots + (n + 1)d_k + r\n$$\n\nand $(n+1)d_i$ are divisors of $(n+1)!$.\nIf $r = 0$, the induction step is finished. Otherwise $r > 0$, $r$ divides $(n+1)!$ and $r < n+1 \\le (n+1)d_i$, therefore $m$ can be written as the sum of $k+1$ distinct divisors of $(n+1)!$, with $k+1 \\le n+1$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the matrices $A \\in \\mathcal{M}_{m,n}(\\mathbb{C})$, $B \\in \\mathcal{M}_{n,m}(\\mathbb{C})$ with $n \\le m$. Given that $\\text{rank}(AB) = n$ and $(AB)^2 = AB$, find $BA$.",
    "options": [],
    "answer": "BA = I_n",
    "solution": "Left-multiply with $B$ and right-multiply by $A$ the equality $(AB)^2 = AB$ to obtain $(BA)^3 = (BA)^2$. Recall that the rank of a matrix product does not exceed the rank of its factors to derive from $ABAB = AB$ that $\\text{rank}BA \\ge n$, hence $\\text{rank}BA = n$.\n\nThe square matrix $BA$ of order $n$ is thus invertible, hence $(BA)^3 = (BA)^2$, which implies $BA = I_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b$ for which there exists sets $A, B$ of positive integers so that $A \\cap B = \\emptyset$, $A \\cup B = \\mathbb{N}^*$ and $aA = bB$ (if $x$ is a number and $M$ is a set of numbers, $xM = \\{xm \\mid m \\in M\\}$).",
    "options": [],
    "answer": "All pairs (a, b) such that one divides the other with quotient at least 2; equivalently, a = p b or b = p a for some integer p ≥ 2.",
    "solution": "We can assume that $1 \\in A$. Then $a \\in bB$, hence there exists $p \\in B$ such that $a = pb$. Moreover, $p \\ge 2$, because $1 \\in A$.\n\nEvery pair $(pb, b)$, with $b \\in \\mathbb{N}^*$, is a solution: we use the partition\n\n$$\nA = \\{p^{2n}q \\mid n \\in \\mathbb{N}, q \\in \\mathbb{N}^*, p \\nmid q\\}\n$$\n$$\nB = \\{p^{2n+1}q \\mid n \\in \\mathbb{N}, q \\in \\mathbb{N}^*, p \\nmid q\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $n \\in \\mathbb{N}$, $n \\ge 2$, and $A$ a unitary ring with $n$ elements, such that the equation $x^{n+1} + x = 0$ has in $A \\setminus \\{0\\}$ the unique solution $x = 1$. Prove that $A$ is a field.\n\nIoan Băetu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $1$ satisfies $x^{n+1} + x = 0$, we have $1 + 1 = 0$, so all non-zero elements of the additive group $(A, +)$ are of order $2$. By Cauchy's theorem $n = 2^m$, $m \\in \\mathbb{N}^*$.\n\nLet $a \\in A$, $a \\neq 0$. The ring $A$ being finite, there exists $p < q$, such that $a^p = a^q$. By successive multiplications with $a^{q-p}$, $a^q = a^{(k+1)q-kp}$, $k \\ge 1$. Take $k \\in \\mathbb{N}^*$, such that $r = (k+1)q - kp > 2q$. Multiplication by $a^{r-2q}$ gives $a^{r-q} = a^{2(r-q)}$. Let $b = a^{r-q}$. As $b^2 = b$, we get $b^{n+1} = b$, so $b^{n+1} + b = 0$, that is $b \\in \\{0, 1\\}$.\n\nIf $b = a^{r-q} = 0$, take $s \\in \\mathbb{N}$, $s \\ge 2$, such that $a^{s-1} \\neq 0$ and $a^s = 0$. Consider $c = a^{s-1}$. Then $c^2 = 0$ and, as a consequence $(c+1)^{n+1} = (c+1)^{2m}(c+1) = (c^{2m} + 1)(c+1) = c+1$. By hypothesis $c+1 \\in \\{0, 1\\}$, which is a contradiction. In conclusion $a^{r-q} = 1$, i.e. $a$ is invertible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u : [a, b] \\to \\mathbb{R}$ be a continuous function which has at each point $x \\in (a, b]$ a finite left derivative, denoted by $u'_s(x)$. Prove that the function $u$ is increasing if and only if $u'_s(x) \\ge 0$, for all $x \\in (a, b]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall use the following results which can be viewed as generalizations of the theorems of Rolle and Lagrange.\n\n**Lemma 1.** If a function $u : [\\alpha, \\beta] \\to \\mathbb{R}$ has left and right finite derivatives at any point in $[\\alpha, \\beta]$ and $u(\\alpha) = u(\\beta)$, then there is $c \\in [\\alpha, \\beta]$ such that $u'_s(c) \\le 0$.\n*Proof.* It is obvious that $u$ is continuous on $[\\alpha, \\beta]$, so it has a global minimum which belongs to the interior of the interval or coincides with one of the extremities; this point can be chosen as $c$.\n\n**Lemma 2.** If a function $u : [\\alpha, \\beta] \\to \\mathbb{R}$ has right and left finite derivatives at each point in the interval $[\\alpha, \\beta]$, then there is $c \\in [\\alpha, \\beta]$ such that $u(\\beta) - u(\\alpha) \\ge (\\beta - \\alpha)u'_s(c)$.\n*Proof.* Define $v : [\\alpha, \\beta] \\to \\mathbb{R}$, $v(x) = u(x) - \\frac{u(\\beta) - u(\\alpha)}{\\beta - \\alpha}x$. The function $v$ satisfies the conditions in Lemma 1, so there is a point $c \\in [\\alpha, \\beta]$ such that $v'_s(c) \\le 0$.\n\nReturning to the solution of the problem, let $a \\le x < y \\le b$. By Lemma 1, there exists $c \\in [x, y]$ such that $u(y) - u(x) \\ge (y - x)u'_s(c)$, which implies that $u$ is increasing.\n\nThe reverse condition is easily implied by the definition of the left derivative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function such that:\n$$\nf(xy + x + y) + f(xy - x - y) = 2(f(x) + f(y)), \\forall x, y \\in \\mathbb{R}. \\quad (1)\n$$\nProve that $f$ fulfills the relation:\n$$\nf(x + y) = f(x) + f(y), \\forall x, y \\in \\mathbb{R}. \\quad (2)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$x = y = 0$ yields $f(0) = 0$ and $y = 0$ yields $f(-x) = -f(x)$, $x \\in \\mathbb{R}$.\n\n$y = 1 \\Rightarrow f(2x + 1) = 2f(x) + f(1) \\quad (3)$\n\n$y = -1 \\Rightarrow f(2x - 1) = 2f(x) - f(1) \\quad (4)$\n\nDenote $x * y = xy + x + y$ and notice that (ASOC): $x * (y * z) = (x * y) * z$.\nFor $x = 1$, $y * 1 = 2y + 1$, hence\n\n$$\nf(y * 1) = 2f(y) + f(1), \\forall y \\in \\mathbb{R} \\quad (5)\n$$\n\nTherefore\n$$\n\\begin{align*}\nf(x * (y * 1)) &= f(x * (2y + 1)) \\\\\n&\\stackrel{(1)}{=} f(x(2y + 1) - x - 2y - 1) + 2f(x) + 2f(2y + 1) \\\\\n&= f(2(xy - y) - 1) + 2f(2y + 1) + 2f(x) \\\\\n&\\stackrel{(3,4)}{=} 2f(xy - y) + f(x) + 2f(y) + f(1) \\tag{6}\n\\end{align*}\n$$\n\nand\n$$\n\\begin{align}\nf((x * y) * 1) &\\stackrel{(5)}{=} 2f(x * y) + f(1) \\stackrel{(1)}{=} 2(2f(x) + 2f(y)) + 2f(xy - x - y) + f(1) \\nonumber \\\\\n&= 2f(xy - x - y) + 2f(x) + 2f(y) + f(1). \\tag{7}\n\\end{align}\n$$\n\nRelations (6), (7) and (ASOC) yield:\n$$\nf(xy - x - y) + f(x) = f(xy - y), \\forall x, y \\in \\mathbb{R}.\n$$\nThe last relation means that $f(u + v) = f(u) + f(v)$ for all $u, v \\in \\mathbb{R}$ which can be written in the form $u = x, v = xy - x - y$, that is there exists $x, y \\in \\mathbb{R}$ so that $x = u$ and $y = \\frac{v+u}{u-1}$; this happens if $u \\neq 1$ or $u = 1 = -v$.\n\nIt remains to check that $f(1 + v) = f(1) + f(v)$, $\\forall v \\in \\mathbb{R}$. Indeed, (4) and $x \\mapsto x + 1$ imply\n$$\nf(2x + 1) = 2f(x + 1) - f(1) \\stackrel{(3)}{=} 2f(x) + f(1),\n$$\n$$\n\\text{whence } f(x + 1) = f(x) + f(1), \\forall x \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an isosceles trapezoid $ABCD$ with perpendicular diagonals. The parallel from the intersection point of the diagonals meets the non-parallel sides $[BC]$ and $[AD]$ at points $P$ and $R$ respectively. Point $Q$ is the mirror image of $P$ across the midpoint of $[BC]$. Show that\n\na) $QR = AD$;\n\nb) $QR \\perp AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the diagonals $AC$ and $BD$ meet at point $O$ and let $M$ be the midpoint of the line segment $[BC]$.\n\na) Since $OM$ joins the midpoints of two sides of the triangle $PQR$, $MO \\parallel RQ$ and $OM = \\frac{RQ}{2}$. On the other hand, $[OM]$ is a median of the right-angled triangle $BOC$, hence $OM = \\frac{1}{2}BC = \\frac{1}{2}AD$. Consequently, $RQ = AD$.\n\n![](attached_image_1.png)\n\nb) Let the lines $MO$ and $AD$ meet at $T$. Then $\\angle MBO = \\angle MOB = \\angle DOT$. Since $\\angle OCB = \\angle TDO$, it follows that $\\angle OTD = \\angle BOC = 90^\\circ$, so $MT \\perp AD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct positive integers.\n\na) Prove that $a^2b^2 + a^2c^2 + b^2c^2 \\ge 9$.\n\nb) If, moreover, $ab + ac + bc + 3 = abc > 0$, show that\n$$\n(a - 1)(b - 1) + (a - 1)(c - 1) + (b - 1)(c - 1) \\ge 6.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) At least one of the numbers $a$, $b$, $c$ has modulus at least $2$, whence $a^2b^2 + a^2c^2 + b^2c^2 \\ge 1 \\cdot 1 + 1 \\cdot 4 + 1 \\cdot 4 = 9$.\n\nb) The required inequality can be successively written\n$$\n\\begin{aligned}\n& ab + ac + bc - 2(a + b + c) + 3 \\ge 6 \\\\\n& ab + ac + bc - 3 \\ge 2(a + b + c) \\\\\n& (ab + ac + bc - 3)(ab + ac + bc + 3) \\ge 2abc(a + b + c) \\\\\n& (ab + ac + bc)^2 - 9 \\ge 2abc(a + b + c) \\\\\n& a^2b^2 + a^2c^2 + b^2c^2 \\ge 9,\n\\end{aligned}\n$$\nthat is exactly a).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37813,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $x$, $y$, $z$, $t$, $a$ and $b$ are positive integers, so that $xt - yz = 1$ and $\\frac{x}{y} > \\frac{a}{b} > \\frac{z}{t}$. Prove that $ab \\ge (x+z)(y+t)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $\\frac{x}{y} > \\frac{a}{b}$ follows that $xb > ya$, hence $xb - ya \\ge 1$. In the same way, $at - bz \\ge 1$. Multiplying the first inequality by $t$, the second one by $y$ and adding the two relations yields $bxt - byz \\ge t + y$, that is $b \\ge t + y$.\n\nIn the same way, $a \\ge x + z$. These two inequalities lead to the conclusion.\nIf $d = \\text{g.c.d.}(a, b)$, then $a = d a_1$ and $b = d b_1$, with $a_1, b_1 \\in \\mathbb{N}^*$, $(a_1, b_1) = 1$ and $\\frac{x}{y} > \\frac{a_1}{b_1}$. It follows that $b_1 x - a_1 y = u \\ge 1$ and $a_1 t - b_1 z = v \\ge 1$, with $u, v \\in \\mathbb{N}$. The first relation gives $b_1 x z - a_1 y z = u z$, and the second gives $a_1 x t - b_1 z x = v x$. Adding these relations yields $a_1 = u z + v x \\ge z + x$. In the same way, $b_1 = u t + v y \\ge t + y$. Since $ab \\ge a_1 b_1$, the conclusion is proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the set $S = \\{\\lfloor n\\pi \\rfloor : n = 0, 1, 2, 3, \\dots \\}$ contains arithmetic progressions of any finite length, but no infinite arithmetic progressions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x$ is a real number, let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$. Given an integer number $m \\ge 3$, there exists a positive integer number $n$ such that $\\{n\\pi\\} < 1/m$, for the set $\\{k\\pi : k = 0, 1, 2, 3, \\dots\\}$ is dense in the closed unit interval $[0, 1]$. Consequently,\n$$\n\\lfloor kn\\pi \\rfloor = \\lfloor k \\lfloor n\\pi \\rfloor \\rfloor + \\lfloor k\\{n\\pi\\} \\rfloor = k \\lfloor n\\pi \\rfloor + \\lfloor k\\{n\\pi\\} \\rfloor = k \\lfloor n\\pi \\rfloor, \\quad k = 1, 2, \\dots, m,\n$$\nso $\\lfloor n\\pi \\rfloor, \\lfloor 2n\\pi \\rfloor, \\dots, \\lfloor mn\\pi \\rfloor$ are $m$ numbers in $S$ in arithmetic progression with ratio $\\lfloor n\\pi \\rfloor$.\n\nSuppose, if possible, that $S$ contains an infinite arithmetic progression $\\lfloor n_k \\pi \\rfloor$, $k = 0, 1, 2, 3, \\dots$, with (integral) ratio $r$, where the $n_k$ form a strictly increasing sequence of positive integer numbers. Write $n_k \\pi = \\lfloor n_0 \\pi \\rfloor + kr + \\{n_k \\pi\\}$ to deduce that $r$ is positive and\n$$\nn_{k+1} - n_k = \\frac{r + \\{n_{k+1}\\pi\\} - \\{n_k\\pi\\}}{\\pi} \\in \\left( \\frac{r-1}{\\pi}, \\frac{r+1}{\\pi} \\right).\n$$\nThe length of this interval is less than $1$, so $n_{k+1} - n_k = n$ for some positive integer $n$ and all indices $k$. Hence, $n_k = n_0 + kn$, so $n_k/k \\xrightarrow{k \\to \\infty} n$. On the other hand, $n_k/k \\xrightarrow{k \\to \\infty} r/\\pi$, so $\\pi = r/n$ which contradicts irrationality of $\\pi$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties:\n1) for each integer $k$, the number $f(k)$ is an integer if and only if $n$ does not divide $k$;\n2) the degree of $f$ is less than $n$.",
    "options": [],
    "answer": "n = 1 or n is a power of a prime",
    "solution": "We will show that such polynomial exists if and only if $n = 1$ or $n$ is a power of a prime.\n\nWe will use two known facts stated in LEMMATA 1 and 2.\nLEMMA 1. If $p^a$ is a power of a prime and $k$ is an integer, then\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!}\n$$\nis divisible by $p$ if and only if $k$ is not divisible by $p^a$.\n\nProof. First suppose that $p^a \\mid k$ and consider\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!} = \\frac{k-1}{p^a-1} \\cdot \\frac{k-2}{p^a-2} \\cdots \\frac{k-p^a+1}{1}\n$$\n\nLEMMA 3. If $n$ has at least two distinct prime divisors then the greatest common divisor of $\\binom{n}{1}$, $\\binom{n}{2}$, ..., $\\binom{n}{n-1}$ is 1.\nProof. Suppose to the contrary that $p$ is a common prime divisor of $\\binom{n}{1}$, ..., $\\binom{n}{n-1}$. In particular, $p \\mid \\binom{n}{1} = n$. Let $a$ be the exponent of $p$ in the prime factorization of $n$. Since $n$ has at least two prime divisors, we have $1 < p^a < n$. Hence, $\\binom{n}{p^a-1}$ and $\\binom{n}{p^a}$ are listed among $\\binom{n}{1}$, ..., $\\binom{n}{n-1}$ and thus $p \\mid \\binom{n}{p^a}$ and $p \\mid \\binom{n}{p^a-1}$. But then $p$ divides $\\binom{n}{p^a} - \\binom{n}{p^a-1} = \\binom{n-1}{p^a-1}$, which contradicts LEMMA 1. □\n\nNext we construct the polynomial $f(x)$ when $n=1$ or $n$ is a power of a prime.\nFor $n=1$, $f(x) = \\frac{1}{2}$ is such a polynomial.\n\nIf $n = p^a$ where $p$ is a prime and $a$ is a positive integer then let\n$$\nf(x) = \\frac{1}{p} \\binom{x-1}{p^a-1} = \\frac{1}{p} \\cdot \\frac{(x-1)(x-2)\\cdots(x-p^a+1)}{(p^a-1)!}.\n$$\nThe degree of this polynomial is $p^a - 1 = n - 1$.\nThe number $\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!}$ is an integer for any integer $k$, and, by LEMMA 1, it is divisible by $p$ if and only if $k$ is not divisible by $p^a = n$.\n\nFinally we prove that if $n$ has at least two prime divisors then no polynomial $f(x)$ satisfies (1,2). Suppose that some polynomial $f(x)$ satisfies (1,2), and apply LEMMA\n$$\n\\binom{n}{k} f(0) = \\sum_{0 \\le l \\le n, l \\ne k} (-1)^{k-l} \\binom{n}{l} f(-k+l).\n$$\nSince $f(-k), \\dots, f(-1)$ and $f(1), \\dots, f(n-k)$ are all integers, we conclude that $\\binom{n}{k}f(0)$ is an integer for every $1 \\le k \\le n-1$.\nBy hint of LEMMA 3, the greatest common divisor of $\\binom{n}{1}, \\binom{n}{2}, \\dots, \\binom{n}{n-1}$ is 1. Hence, there will exist some integers $u_1, u_2, \\dots, u_{n-1}$ for which $u_1\\binom{n}{1} + \\dots + u_{n-1}\\binom{n}{n-1} = 1$. Then\n$$\nf(0) = \\left( \\sum_{k=1}^{n-1} u_k \\binom{n}{k} \\right) f(0) = \\sum_{k=1}^{n-1} u_k \\binom{n}{k} f(0)\n$$\nis a sum of integers. This contradicts the fact that $f(0)$ is not an integer. So such polynomial $f(x)$ does not exist.\n(Alternative Solution, I. Bogdanov)\nWe claim the answer is $n = p^\\alpha$ for some prime $p$ and nonnegative $\\alpha$.\n\nLEMMA. For every integers $a_1, \\dots, a_n$ there exists an integer-valued polynomial $P(x)$ of degree $< n$ such that $P(k) = a_k$ for all $1 \\le k \\le n$.\nProof. Induction on $n$. For the base case $n=1$ one may set $P(x) = a_1$. For the induction step, suppose that the polynomial $P_1(x)$ satisfies the desired property for all $1 \\le k \\le n-1$. Then set $P(x) = P_1(x) + (a_n - P_1(n))\\binom{x-1}{n-1}$; since $\\binom{k-1}{n-1} = 0$ for $1 \\le k \\le n-1$ and $\\binom{n-1}{n-1} = 1$, the polynomial $P(x)$ is a sought one. $\\square$\n\nNow, if for some $n$ there exists some polynomial $f(x)$ satisfying the problem conditions, one may choose some integer-valued polynomial $P(x)$ (of degree $< n-1$) coinciding with $f(x)$ at points $1, \\dots, n-1$. The difference $f_1(x) = f(x) - P(x)$ also satisfies the problem conditions, therefore we may restrict ourselves to the polynomials vanishing at points $1, \\dots, n-1$ — that are, the polynomials of the form $f(x) = c \\prod_{i=1}^{n-1} (x-i)$ for some (surely rational) constant $c$. Let $c = p/q$ be its irreducible form, and $q = \\prod_{j=1}^d p_j^{\\alpha_j}$ be the prime decomposition of the denominator.\n\n1. Assume that a desired polynomial $f(x)$ exists. Since $f(0)$ is not an integer, we have $q \\nmid (-1)^{n-1}(n-1)!$ and hence $p_j^{\\alpha_j} \\nmid (-1)^{n-1}(n-1)!$ for some $j$. Hence\n$$\n\\prod_{i=1}^{n-1} (p_j^{\\alpha_j} - i) \\equiv (-1)^{n-1}(n-1)! \\not\\equiv 0 \\pmod{p_j^{\\alpha_j}},\n$$\ntherefore $f(p_i^{\\alpha_i})$ is not integer, too. By the condition (i), this means that $n \\mid p_i^{\\alpha_i}$, and hence $n$ should be a power of a prime.\n\n2. Now let us construct a desired polynomial $f(x)$ for any power of a prime $n = p^\\alpha$. We claim that the polynomial\n$$\nf(x) = \\frac{1}{p} \\binom{x-1}{n-1} = \\frac{n}{px} \\binom{x}{n}\n$$\nfits. Actually, consider some integer $x$. From the first representation, the denominator of the irreducible form of $f(x)$ may be 1 or $p$ only. If $p^\\alpha \\nmid x$, then the prime decomposition of the fraction $n/(px)$ contains $p$ with a nonnegative exponent; hence $f(x)$ is integer. On the other hand, if $n = p^\\alpha \\mid x$, then the numbers $x-1, x-2, \\dots, x-(n-1)$ contain the same exponents of primes as the numbers $n-1, n-2, \\dots, 1$ respectively; hence the number\n$$\n\\binom{x-1}{n-1} = \\frac{\\prod_{i=1}^{n-1} (x-i)}{\\prod_{i=1}^{n-1} (n-i)}\n$$\nis not divisible by $p$. Thus $f(x)$ is not an integer.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37816,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $z_1, z_2, z_3, z_4$ be distinct complex numbers of zero sum, having equal absolute values. Prove that the points of affixes $z_1, z_2, z_3, z_4$ are the vertices of a rectangle.\n\nb) Let $x, y, z, t$ be real numbers such that $\\sin x + \\sin y + \\sin z + \\sin t = 0$ and $\\cos x + \\cos y + \\cos z + \\cos t = 0$. Prove that, for every integer $n$,\n$$\n\\sin(2n + 1)x + \\sin(2n + 1)y + \\sin(2n + 1)z + \\sin(2n + 1)t = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The equality $z_1 + z_2 + z_3 + z_4 = 0$ implies $\\bar{z}_1 + \\bar{z}_2 + \\bar{z}_3 + \\bar{z}_4 = 0$ and furthermore $\\frac{1}{z_1} + \\frac{1}{z_2} + \\frac{1}{z_3} + \\frac{1}{z_4} = 0$ (1), for $|z_1| = |z_2| = |z_3| = |z_4| \\ne 0$.\nSuppose $z_1 + z_2 = -z_3 - z_4 \\ne 0$. The relation (1) gives $z_1 z_2 = z_3 z_4$, so $\\{z_1, z_2\\} = \\{-z_3, -z_4\\}$. On the other hand, if $z_1 + z_2 = 0$, then $z_3 + z_4 = 0$. In both cases the numbers $z_1, z_2, z_3, z_4$ form two pair of equal sum, hence the conclusion.\n\nb) Let $z_1 = \\cos x + i \\sin x$, $z_2 = \\cos y + i \\sin y$, $z_3 = \\cos z + i \\sin z$ and $z_4 = \\cos t + i \\sin t$ to get $z_1 + z_2 + z_3 + z_4 = 0$ and $|z_1| = |z_2| = |z_3| = |z_4| = 1$.\nAs before, the numbers $z_1, z_2, z_3, z_4$ form two pairs of opposite numbers, so the same goes for numbers $z_1^{2n+1}, z_2^{2n+1}, z_3^{2n+1}, z_4^{2n+1}$. Therefore $z_1^{2n+1} + z_2^{2n+1} + z_3^{2n+1} + z_4^{2n+1} = 0$, implying the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37817,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that $m^2 - m + 1$ is an element of the set $\\{n^2 + n + 1 \\mid n \\in \\mathbb{N}\\}$, for any positive integer $m$.\n\nb) Let $p$ be a perfect square, $p > 1$. Prove that there exist positive integers $r$ and $q$ such that $p^2 + p + 1 = (r^2 + r + 1)(q^2 + q + 1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $m^2 - m + 1 = (m - 1)^2 + (m - 1) + 1$ and $m - 1 \\ge 0$, it follows that $m - 1 \\in \\mathbb{N}$ and consequently $m^2 - m + 1 \\in \\{n^2 + n + 1 \\mid n \\in \\mathbb{N}\\}$.\n\nb) Write $p = k^2$, where $k$ is an integer. Since $p > 1$, we have $k \\ge 2$. Now $p^2 + p + 1 = k^4 + k^2 + 1 = (k^2 + 1)^2 - k^2 = (k^2 - k + 1)(k^2 + k + 1)$. Numbers $r = k$ and $q = k - 1$, both positive integers, satisfy the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an equilateral triangle $ABC$ and the points $M \\in [BC]$, $N \\in [AC]$, $P \\in [AB]$ such that $BM = MC$, $3AN = NC$ and $2BP = AP$, find the measure of $\\angle NMP$.",
    "options": [],
    "answer": "60°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37819,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ consider the set $A_n$ of all the numbers obtained by choosing signs in $\\pm 1 \\pm 2 \\pm \\cdots \\pm n$; for instance, $A_2 = \\{-3, -1, 1, 3\\}$ and $A_3 = \\{-6, -4, -2, 0, 2, 4, 6\\}$. Find the cardinal of the set $A_n$.",
    "options": [],
    "answer": "n(n+1)/2 + 1",
    "solution": "The difference of any two elements of the set is even, implying that all elements have the same parity.\nWe claim that all numbers between $-\\frac{n(n+1)}{2}$ and $\\frac{n(n+1)}{2}$, sharing the same parity with them belong to the set $A_n$ – and only them. Indeed, let $x \\in A_n$, $x < \\frac{n(n+1)}{2}$. If (one of) its representation begins with $-1$, changing into $+1$ gives $x + 2 \\in A_n$. Suppose that all its representations start with $+1$. There exists a first negative – otherwise $x = 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$; denote it by $-k$:\n$$\nx = +1 + 2 + \\cdots + (k-1) - k \\pm \\cdots \\pm n.\n$$\nSwitching the signs of the summands $k-1$ and $k$ we derive that $x + 2 \\in A_n$.\nConsequently, the set $A_n$ has $\\frac{n(n+1)}{2} + 1$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $AB < AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The lines $BC$ and $DE$ are parallel, so the angles $BED$ and $DAE$ are equal. Then so are the angles $AED$ and $DBE$. Let $I$ and $J$ be the incentres of the triangles $ADE$ and $BDE$, respectively. It follows that the triangles $DIE$ and $DJB$ are similar, so $DI/DE = DJ/DB$. Since the angles $IDJ$ and $EDB$ are equal, the triangles $DIJ$ and $DEB$ are similar, so the angles $DIJ$ and $DEB$ are equal. Let the line $BE$ meet the line $IJ$ at the point $F$. Notice that the quadrangle $DIEF$ is cyclic to deduce that the angles $EFI$ and $EDI$ are both equal to one half of the angle $ACB$. Consequently, the line $IJ$ is parallel to the exterior bisectrix of the angle $AEB$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two complex numbers. Prove that the following statements are equivalent:\n1) The absolute values of the roots of the equation $x^2 - a x + b = 0$ are respectively equal to the absolute values of the roots of the equation $x^2 - b x + a = 0$.\n2) $a^3 = b^3$ or $b = \\bar{a}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $|x_1| = |x_3|$, $|x_2| = |x_4|$ (1) and notice that $|a| = |x_3 x_4| = |x_1 x_2| = |b|$ to derive that $|x_1 + x_2| = |x_3 + x_4|$ (2). The relations (1) and (2) show that there exists a number $k \\in \\mathbb{C}$ such that $x_2 = k x_1$, $x_4 = k x_3$ or $x_2 = k x_1$, $x_4 = \\bar{k} x_3$.\n\nIn the first case we have $a = k x_3^2 = (1 + k) x_1$ and $b = k x_1^2 = (1 + k) x_3$, so\n$a^3 = k (1 + k)^2 x_1^2 x_3^2 = b^3$.\n\nIn the latter case we have $a = \\bar{k} x_3^2 = (1 + k) x_1$ and $b = k x_1^2 = (1 + \\bar{k}) x_3$. It follows that $x_1^2 \\bar{x}_1 = x_3 \\bar{x}_3^2$, so $x_1 = \\bar{x}_3$ or $a = b = 0$, and furthermore $x_2 = \\bar{x}_4$, hence $a = \\bar{b}$.\n\nConversely, if $b = \\bar{a}$, then $x_1 + x_2 = \\bar{x}_3 + \\bar{x}_4$, $x_1 x_2 = \\bar{x}_3 \\bar{x}_4$, implying $\\{x_1, x_2\\} = \\{\\bar{x}_3, \\bar{x}_4\\}$.\n\nIf $a^3 = b^3$, then $a = \\varepsilon b$, $\\varepsilon^3 = 1$. The roots satisfy the relations $x_1 + x_2 = \\varepsilon (x_3 + x_4)$, $x_1 x_2 = \\varepsilon^2 x_3 x_4$. Both cases lead to $\\{|x_1|, |x_2|\\} = \\{|x_3|, |x_4|\\}$, as needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that its 2-fold composition is equal to the floor function, i.e. $f(f(x)) = \\lfloor x \\rfloor$, for any real number $x$. Prove that there exist distinct real numbers $a$ and $b$ such that $|f(a) - f(b)| \\ge |a - b|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that $f(n) \\in \\mathbb{Z}$, for any integer $n$. Indeed, write $f(f(f(x))) = f([x]) = [f(x)]$ to derive that $f(n) = [f(n)]$ for any integer $n$, implying $f(n) \\in \\mathbb{Z}$.\n\nSuppose that for all $a, b \\in \\mathbb{Z}$ we have $|f(a) - f(b)| < |a - b|$. Then $|f(n + 1) - f(n)| < 1$, for any integer $n$. Since both $f(n)$ and $f(n + 1)$ are integers we derive that $f(n) = f(n + 1)$, hence $f(n) = f(0)$, for any integer $n$. It follows that $n = f(f(n)) = f(f(0)) = 0$, a contradiction.\nNotice that $f(0) \\ne f(1)$, otherwise $0 = f(f(0)) = f(f(1)) = 1$, false. If $|f(0) - f(1)| \\ge |0 - 1|$, we are done; if else, from $|0 - 1| = |f(f(0)) - f(f(1))|$ we obtain $|f(f(0)) - f(f(1))| > |f(0) - f(1)|$. The distinct numbers $f(0)$ and $f(1)$ fulfill the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B \\in \\mathcal{M}_2(\\mathbb{C})$ be two non-zero matrices with $AB + BA = O_2$ and $\\det(A + B) = 0$. Prove that $\\text{tr}(A) = \\text{tr}(B) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "So, subtracting we get $\\text{tr}(A)B = \\text{tr}(B)A$.\nIf $\\text{tr}(A) = 0$, then $\\text{tr}(B) = 0$ for else $A = O_2$, a contradiction.\nHence $A = \\lambda B$, so $AB + BA = O_2$ leads to $\\lambda B^2 = O_2$. Therefore $\\lambda = 0$, which yields $\\text{tr}(A) = \\text{tr}(B) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37824,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ has the digits $1, 2, 3, 4, 5, 6$ and $7$, so that each digit $i$, $i \\in \\{1, 2, 3, 4, 5, 6, 7\\}$ occurs $4i$ times in the decimal representation of $N$. Prove that $N$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$N$ has $1 \\cdot 4 = 4$ digits equal to $1$, $2 \\cdot 4 = 8$ digits equal to $2$, $\\dots$, $7 \\cdot 4 = 28$ digits equal to $7$, so the sum of its digits equals $S = 4(1^2 + 2^2 + \\dots + 7^2) = 560$. Since $560 = 3 \\cdot 186 + 2$, the number $N$ is not a square, and the remainder left by a perfect square upon division by $3$ cannot be equal to $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $VABC$ be a regular pyramid with the base $ABC$ having center $O$. Denote $I$ and $H$ the incenter and the orthocenter of the triangle $VBC$ and suppose that $AH = 3OI$. Find the measure of the angle made by a lateral edge of the pyramid with the plane of the base.\nMircea Fianu",
    "options": [],
    "answer": "30°",
    "solution": "Denote $M$ the midpoint of the edge $[BC]$. Since triangle $VBC$ is isosceles with $VB = VC$, the points $V$, $H$, $I$ and $M$ are collinear.\n\nFrom $AC \\perp OB$ and $AC \\perp OV$ follows $AC \\perp (BOV)$, hence $VB \\perp AC$, which, together with $VB \\perp CH$ leads to $VB \\perp (ACH)$, whence $AH \\perp VB$. In the same way $AH \\perp VC$, therefore $AH \\perp (VBC)$, so $AH \\perp VM$.\n\nDenote $I'$ the projection of $O$ onto the plane $(VBC)$; then $I'$ belongs to $VM$ and $OI' \\parallel AH$. This leads to\n\n![](attached_image_1.png)\n\n$$\n\\frac{OI'}{AH} = \\frac{MO}{MA} = \\frac{1}{3},\n$$\n\nhence $AH = 3OI'$. This shows that $OI$ equals the distance from $O$ to $(VBC)$, therefore $I = I'$.\n\nFrom $OI \\perp (VBC)$ and $I$ = incenter of the triangle $VBC$ follows that $O$ has equal distances to the lines $VB$ and $BC$.\n\nDenote $J$ the projection of $O$ onto the line $VB$. Then $OJ = OM$, hence the right triangles $OJB$ and $OMB$ are congruent.\n\nThis leads to $m(\\angle (VB, (ABC))) = m(\\angle VBO) = m(\\angle MBO) = 30^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37826,
    "subject": "Mathematics (Multi-modal)",
    "question": "The convex quadrilateral $ABCD$ has $\\angle BCD = \\angle ADC \\ge 90^\\circ$. The bisectors of the angles $\\angle BAD$ and $\\angle ABC$ meet at a point $M$, placed on the line $CD$. Prove that $M$ is the midpoint of the segment $[CD]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Case I: $AD$ and $BC$ have a common point $E$. Then $M$ is the incenter of the triangle $ABE$, hence $(EM)$ is the bisector of the angle $\\angle AEB$.\nSince $\\angle ECD = \\angle EDC$, the triangle $EDC$ is isosceles with base $[DC]$. Therefore $[EM]$ is a median in triangle $EDC$, so $M$ is the midpoint of the segment $CD$.\n\n![](attached_image_1.png)\n\nCase II: $AD \\parallel BC$. Then $\\angle CAB = \\angle ABD = 90^\\circ$, hence $\\angle MAB + \\angle MBA = 90^\\circ$, that is triangle $MAB$ has a right angle in $M$.\nDenote $N$ the midpoint of the segment $[AB]$. Then triangle $NAM$ is isosceles with base $[AM]$, so $\\angle NMA = \\angle NAM = \\angle MAD$, hence $MN \\parallel AD$. It follows that $MN$ is the central median of the trapezoid $ABCD$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $x_1, x_2, \\dots, x_n$ and $y_1, y_2, \\dots, y_n$ be real numbers. Prove that there exists a number $i$, $i = 1, 2, \\dots, n$, such that\n$$\n\\sum_{j=1}^{n} |x_i - x_j| \\le \\sum_{j=1}^{n} |x_i - y_j|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without the loss of generality, suppose $x_1 \\le x_2 \\le \\dots \\le x_n$. For each $k = 1, 2, \\dots, n$ we have $|x_1 - x_k| + |x_n - x_k| = |x_1 - x_n| \\le |x_1 - y_k| + |x_n - y_k|$, hence\n$$\n\\sum_{k=1}^{n} |x_1 - x_k| + \\sum_{k=1}^{n} |x_n - x_k| \\le \\sum_{k=1}^{n} |x_1 - y_k| + \\sum_{k=1}^{n} |x_n - y_k|.\n$$\nThe claim holds for $i = 1$ or $i = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a positive integer and $A_p = \\{x \\in \\mathbb{R} \\mid p\\{x\\} = (p+1)[x]\\}$. Find the cardinals of the sets $A_p$ and $A_1 \\cup A_2 \\cup \\dots \\cup A_p$.",
    "options": [],
    "answer": "card(A_p) = 1 and card(A_1 ∪ A_2 ∪ ⋯ ∪ A_p) = 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to (0, \\infty)$ be a continuous function. For $n \\in \\mathbb{N}$, $n \\ge 2$, consider $0 = t_0 < t_1 < \\dots < t_n = 1$, such that\n$$\n\\int_{t_0}^{t_1} f(t) \\, dt = \\int_{t_1}^{t_2} f(t) \\, dt = \\dots = \\int_{t_{n-1}}^{t_n} f(t) \\, dt.\n$$\nCompute\n$$\n\\lim_{n \\to \\infty} \\frac{n}{\\frac{1}{f(t_1)} + \\frac{1}{f(t_2)} + \\dots + \\frac{1}{f(t_n)}}.\n$$",
    "options": [],
    "answer": "∫_0^1 f(t) dt",
    "solution": "Denote by $(x_n)_{n \\ge 2}$ the given sequence. We have\n$$\n\\begin{aligned}\n\\frac{1}{x_n} &= \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(t_k)} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(F^{-1}(kI/n))} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{F'(F^{-1}(kI/n))} \\\\\n&= \\frac{1}{n} \\sum_{k=1}^{n} (F^{-1})'(kI/n).\n\\end{aligned}\n$$\n\n$$\n\\lim_{n \\to \\infty} \\frac{1}{x_n} = \\int_{0}^{1} (F^{-1})'(Ix) \\, dx = \\frac{1}{I} F^{-1}(Ix) \\Big|_{0}^{1} = \\frac{1}{I}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37830,
    "subject": "Mathematics (Multi-modal)",
    "question": "A certain language uses an alphabet containing three letters. Some sequences of two or more letters are forbidden, and every two forbidden sequences have different lengths. Prove that there exists admissible words of every length.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_n$ be the number of admissible words with $n$ letters; then $a_0 = 1$ (the empty word), $a_1 = 3$ and $a_2 = 8$.\n\nThen, if we add a letter at the end of a correct word with $n$ letters, we obtain either a correct word with $n+1$ letters, or a forbidden word of the form $XY$, with $Y$ a forbidden sequence with $k$ letters, $2 \\le k \\le n+1$ and $X$ a correct word with $n-k+1$ letters. So, the forbidden words with $n+1$ letters are at most $a_0 + a_1 + a_2 + \\dots + a_{n-1}$, hence\n$$\na_{n+1} \\ge 3a_n - (a_0 + a_1 + a_2 + \\dots + a_{n-1}).\n$$\nThe above relation allows to prove inductively that $a_{n+1} > 2a_n$ ($*$), for every $n \\ge 1$. Indeed, the base case is obvious, and if ($*$) is true for all the numbers from 0 to $n-1$, $n \\ge 2$, then $a_n \\ge 2^k a_{n-k}$, $0 \\le k \\le n-1$, whence\n$$\na_{n+1} \\ge a_n \\left( 3 - \\left( \\frac{1}{2^n} + \\frac{1}{2^{n-1}} + \\dots + \\frac{1}{2} \\right) \\right) > 2a_n.\n$$\nThis shows that there are at least $2^n$ words of length $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a regular prism $ABCA'B'C'$. A plane $\\alpha$ containing point $A$ meets the rays $BB'$ and $CC'$ at points $E$ and $F$ such that\n$$\n\\text{area } [ABE] + \\text{area } [ACF] = \\text{area } [AEF].\n$$\n\nFind the angle determined by the planes $AEF$ and $BCC'$.\n",
    "options": [],
    "answer": "60°",
    "solution": "Let $M$ be the midpoint of $BC$ and let $u$ be the angle determined by the planes $AEF$ and $BCC'$. Since triangle $MEF$ is the projection of the triangle $AEF$ onto $BCC'$, we have\n$$\n\\begin{aligned}\n\\cos u &= \\frac{[MEF]}{[AEF]} = \\frac{[BCFE]}{2[AEF]} = \\frac{[BCFE]}{2([ABE] + [ACF])} = \\\\\n&= \\frac{[BCFE]}{4([MBE] + [MCF])} = \\frac{[BCFE]}{2[BCFE]} = \\frac{1}{2},\n\\end{aligned}\n$$\nhence $u = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37832,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is inscribed in a circle $\\omega$. A variable line $\\ell$ chosen parallel to $BC$ meets segments $AB$, $AC$ at points $D$, $E$ respectively, and meets $\\omega$ at points $K$, $L$ (where $D$ lies between $K$ and $E$). Circle $\\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\\omega$, while circle $\\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\\omega$. Determine the locus, as $\\ell$ varies, of the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$.\n(Russia) Vasily Mokin & Fedor Ivlev",
    "options": [],
    "answer": "The locus is the open segment of the angle bisector at the vertex between the vertex and its foot on the opposite side.",
    "solution": "Let $P$ be the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$. Also, let $b$ be the angle bisector of $\\angle BAC$. Since $KL \\parallel BC$, $b$ is also the angle bisector of $\\angle KAL$.\nLet $\\mathfrak{H}$ be the composition of the symmetry $\\mathfrak{S}$ with respect to $b$ and the inversion $\\mathfrak{I}$ of centre $A$ and ratio $\\sqrt{AK \\cdot AL}$ (it is readily seen that $\\mathfrak{S}$ and $\\mathfrak{I}$ commute, so since $\\mathfrak{S}^2 = \\mathfrak{I}^2 = \\text{id}$, then also $\\mathfrak{H}^2 = \\text{id}$, the identical transformation). The elements of the configuration interchanged by $\\mathfrak{H}$ are summarized in Table I.\n\nLet $O_1$ and $O_2$ be the centres of circles $\\gamma_1$ and $\\gamma_2$. Since the circles $\\gamma_1$ and $\\gamma_2$ are determined by their construction (in a unique way), they are interchanged by $\\mathfrak{H}$, therefore the rays $AO_1$ and $AO_2$ are symmetrical with respect to $b$. Denote by $\\varrho_1$ and $\\varrho_2$ the radii of $\\gamma_1$ and $\\gamma_2$. Since $\\angle O_1AB = \\angle O_2AC$, we have $\\varrho_1/\\varrho_2 = AO_1/AO_2$. On\n\nthe other hand, from the definition of $P$ we have $O_1P/O_2P = \\varrho_1/\\varrho_2 = AO_1/AO_2$; this means that $AP$ is the angle bisector of $\\angle O_1AO_2$ and therefore of $\\angle BAC$.\nThe limiting, degenerated, cases are when the parallel line passes through $A$ – when $P$ coincides with $A$; respectively when the parallel line is $BC$ – when $P$ coincides with the foot $A' \\in BC$ of the angle bisector of $\\angle BAC$ (or any other point on $BC$). By continuity, any point $P$ on the open segment $AA'$ is obtained for some position of the parallel, therefore the locus is the open segment $AA'$ of the angle bisector $b$ of $\\angle BAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be three positive real numbers. Prove that the function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = \\frac{a^x}{b^x + c^x} + \\frac{b^x}{a^x + c^x} + \\frac{c^x}{a^x + b^x}$ is increasing on $[0, \\infty)$ and decreasing on $(-\\infty, 0]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use straightforward computation: if $x \\le y$ are real numbers, then\n\n$$\n\\begin{align*} \nf(y) - f(x) &= \\sum_{\\text{cyc}} \\frac{a^y(b^x + c^x) - a^x(b^y + c^y)}{(b^x + c^x)(b^y + c^y)} \\\n&= \\sum_{\\text{cyc}} (a^y b^x - a^x b^y) \\left( \\frac{1}{(b^x + c^x)(b^y + c^y)} - \\frac{1}{(a^x + c^x)(a^y + c^y)} \\right) \\\\\n&= \\sum_{\\text{cyc}} a^x b^x (a^{y-x} - b^{y-x}) \\cdot \\\\\n&\\qquad \\frac{((a^{x+y} - b^{x+y}) + c^x(a^y - b^y) + c^y(a^x - b^x))}{(b^x + c^x)(b^y + c^y)(a^x + c^x)(a^y + c^y)}, \n\\end{align*}\n$$\n\n$$\n(a^p - b^p)(a^q - b^q) = b^{p+q} \\left( \\left( \\frac{a}{b} \\right)^p - 1 \\right) \\left( \\left( \\frac{a}{b} \\right)^q - 1 \\right) \\begin{cases} \\ge 0 & \\text{if } q \\ge 0 \\\\ \\le 0 & \\text{if } q \\le 0 \\end{cases}\n$$\nhence $f(y) - f(x) \\ge 0$ if $y \\ge x \\ge 0$ and $f(y) - f(x) \\le 0$ if $0 \\ge y \\ge x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an increasing continuous function $f : [0, 1] \\to \\mathbb{R}$ and define the sequence $(a_n)_{n \\ge 1}$ by $a_n = \\frac{1}{2^n} \\sum_{k=1}^{2^n} f(\\frac{k}{2^n})$, for all integers $n \\ge 1$.\n\na) Prove that the sequence $(a_n)_{n \\ge 1}$ is increasing.\n\nb) Given that there exists $p \\in \\mathbb{N}^*$ such that $a_p = \\int_0^1 f(x)dx$, prove that $f$ is a constant function.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Notice that\n$$\na_{n+1} = \\frac{1}{2^{n+1}} \\sum_{k=1}^{2^{n+1}} f\\left(\\frac{k}{2^{n+1}}\\right) = \\frac{1}{2^{n+1}} \\left( \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right) + \\sum_{k=1}^{2^n} f\\left(\\frac{2k-1}{2^{n+1}}\\right) \\right)\n$$\nand $f(\\frac{2k-1}{2^{n+1}}) \\le f(\\frac{k}{2^n})$ to derive that\n$$\na_{n+1} \\le \\frac{1}{2^n} \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right) = a_n.\n$$\n\nb) Consider the numbers $k \\in \\{1, 2, \\dots, 2^p\\}$, $c \\in \\left(\\frac{k-1}{2^p}, \\frac{k}{2^p}\\right)$, and a division of the closed interval $[0, 1]$\n$$\n\\Delta = \\left(0, \\frac{1}{2^p}, \\dots, \\frac{k-1}{2^p}, c, \\frac{k}{2^p}, \\dots, 1\\right).\n$$\nDenote by $S$ the upper Darboux sum with respect to $\\Delta$ to get\n$$\n\\begin{aligned}\na_p - S &= \\frac{1}{2^p} f\\left(\\frac{k}{2^p}\\right) - f(c) \\left(c - \\frac{k-1}{2^p}\\right) - f\\left(\\frac{k}{2^p}\\right) \\left(\\frac{k}{2^p} - c\\right) \\\\\n&= \\left(f\\left(\\frac{k}{2^p}\\right) - f(c)\\right) \\left(c - \\frac{k-1}{2^p}\\right) \\ge 0.\n\\end{aligned}\n$$\nSince $\\int_0^1 f(x)dx \\le S \\le a_p = \\int_0^1 f(x)dx$, it follows that $a_p = S$ and, furthermore, $f(c) = f\\left(\\frac{k}{2^p}\\right)$. Therefore, $f$ is a constant function on each of the intervals $\\left(\\frac{k-1}{2^p}, \\frac{k}{2^p}\\right]$, $k = 1, 2, \\dots, 2^p$, whose union is $[0, 1]$. Hence $f$ is a constant function on $[0, 1]$, because the function $f$ is continuous.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $g : \\mathbb{R} \\to \\mathbb{R}$ be a continuous, decreasing function, such that $g(\\mathbb{R}) = (-\\infty, 0)$. Prove that there are no continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that the equality $f \\circ f \\circ \\dots \\circ f = g$ is true for some integer $k \\ge 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that such a function $f$ exists. Injectivity of $g$ implies the injectivity of the continuous function $f$, which in turn is strictly monotone. As $g$ is increasing we conclude that $f$ is increasing and $k$ is an odd number. Moreover, $f$ is not surjective.\n\nDenote by $f^{[k]} = f \\circ f \\circ \\dots \\circ f$ ($k$ times $f$). Because $f(\\mathbb{R})$ is an interval with $(-\\infty, 0) = g(\\mathbb{R}) = f(f^{k-1}(\\mathbb{R})) \\subset f(\\mathbb{R})$, we deduce $f$ is bounded from above.\n\nLet $m \\in \\mathbb{R}$ be such that $f(x) < m, \\forall x \\in \\mathbb{R}$. Then $f^{[k-1]}(x) < m$, for all $x \\in \\mathbb{R}$, so $g(x) = f(f^{[k-1]}(x)) > f(m)$, for all $x \\in \\mathbb{R}$, in contradiction with $g(\\mathbb{R}) = (-\\infty, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37836,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that if $a, b > 1$ are distinct real numbers then\n$$\n\\log_a(\\log_a b) > \\log_b(\\log_a b).\n$$\n\nb) Let $a_1 > a_2 > \\dots > a_n > 1$ be real numbers, $n \\ge 2$. Prove that\n$$\n\\log_{a_1}(\\log_{a_1}(a_2) + \\log_{a_2}(\\log_{a_2}(a_3) + \\dots + \\log_{a_{n-1}}(\\log_{a_{n-1}}(a_n) + \\log_{a_n}(\\log_{a_n}(a_1)))) > 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) For $a < b$, $\\log_a(\\log_a b) = (\\log_a b)(\\log_b(\\log_a b)) > \\log_b(\\log_a b)$ because $\\log_b(\\log_a b) > 0$ and $\\log_a b > 1$.\nFor $a > b$, the claim is reached from $\\log_a b < 1$ and $\\log_b(\\log_a b) < 0$.\n\nb) Induct on $n$. For $n = 2$,\n$$\n\\log_{a_1}(\\log_{a_1}(a_2) + \\log_{a_2}(\\log_{a_2}(a_1))) > \\log_{a_2}(\\log_{a_1}(a_2) + \\log_{a_1}(\\log_{a_1}(a_1))) = \\log_{a_2}(1) = 0.\n$$\nAssume now that the claim holds for some $n$ and consider the numbers $a_1 > a_2 > \\dots > a_{n+1} > 1$. Then\n$$\n\\begin{align*}\n& \\log_{a_1}(\\log_{a_1}(a_2) + \\dots + \\log_{a_n}(\\log_{a_n}(a_{n+1}) + \\log_{a_{n+1}}(\\log_{a_{n+1}}(a_1))) = \\\\\n& \\qquad \\log_{a_1}(\\log_{a_1}(a_2) + \\dots + \\log_{a_{n-1}}(\\log_{a_{n-1}}(a_n) + \\log_{a_n}(\\log_{a_n}(a_1))) + \\\\\n& \\qquad \\phantom{\\log_{a_1}(\\log_{a_1}(a_2) + \\dots + \\log_{a_{n-1}}(\\log_{a_{n-1}}(a_n) + \\log_{a_n}(\\log_{a_n}(a_1)) =} \\\\\n& > \\log_{a_{n+1}}(\\log_{a_n}(a_{n+1}) + \\log_{a_{n+1}}(\\log_{a_{n+1}}(a_1)) - \\log_{a_n}(\\log_{a_n}(a_1)) > \\\\\n& \\qquad \\phantom{> \\log_{a_{n+1}}(\\log_{a_n}(a_{n+1}) + \\log_{a_{n+1}}(\\log_{a_{n+1}}(a_1)) - \\log_{a_n}(\\log_{a_n}(a_1)) =} \\\\\n& = \\log_{a_{n+1}}(\\log_{a_n}(a_1) - \\log_{a_n}(\\log_{a_n}(a_1))) > 0,\n\\end{align*}\n$$\nbecause $\\log_{a_n} a_1 > 1$ and $a_{n+1} < a_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a ring and let $a$ be an element of $A$. Prove that\n\na) If $A$ is commutative and $a$ is nilpotent, then $a + x$ is invertible for any invertible element $x \\in A$.\n\nb) If $A$ is finite and $a + x$ is invertible for any invertible element $x \\in A$, then $a$ is nilpotent.\n\n(An element $a$ of a ring is called *nilpotent* if there exists some positive integer $n$ such that $a^n = 0$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $x$ be an invertible element and let $n$ be a positive integer such that $a^n = 0$. Since $a + x = x(x^{-1}a + 1)$, it is enough to show that $x^{-1}a + 1$ is invertible. Set $b = x^{-1}a$. Then $b^n = x^{-n}a^n = 0$, (because $A$ is commutative), hence $b^{2n+1} = 0$. Consequently,\n$$\n1 = b^{2n+1} + 1 = (b+1)(b^{2n} - b^{2n-1} + \\dots - b + 1),\n$$\ni.e., $b+1$ is invertible.\n\nb) Induct on $n$, $n \\ge 1$, to prove that $a^n - 1$ is invertible. For $x = -1$ we get that $a - 1$ is invertible. Assume that $b = a^n - 1$ is invertible. From the hypothesis it follows that $a - b^{-1}$ is also invertible, and so is $ab - 1 = (a - b^{-1})b$. Moreover,\n$$\na^{n+1} - 1 = a + (a(a^n - 1) - 1) = a + (ab - 1)\n$$\nis invertible. Since $A$ is finite, there exist two integers $q > p \\ge 1$ such that $a^p = a^q$, i.e. $a^p(a^{q-p} - 1) = 0$. Since $a^{q-p} - 1$ is invertible, it follows that $a^p = 0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the sum of the elements of the set\n$$\nM = \\left\\{ \\frac{n}{2} + \\frac{m}{5} \\mid m, n = 0, 1, 2, \\dots, 100 \\right\\}.\n$$",
    "options": [],
    "answer": "24395",
    "solution": "Consider the set $A = \\{2a + 5b \\mid a, b = 1, 2, \\dots, 100\\}$ and notice that $1 \\notin A$ and $3 \\notin A$.\n\nThe largest even number from $A$ is equal to $700$ and it is obtained for $a = b = 100$. The number $698$ is obtained for $a = 99, b = 100$. The largest odd number from $A$ is equal to $695$ and it is obtained for $b = 99, a = 100$, implying that $697 \\notin A$ and $699 \\notin A$.\n\nWe claim that all the integers between $4$ and $695$ belong to $A$.\nLet $y \\le 500$ and let $r$ be the remainder left by $y$ upon division by $5$. Write $y = 5c + r$ with $0 \\le c \\le 100$ and $0 \\le r \\le 4$. If $r$ is even, then $y = 5c + 2k$, where $r = 2k$. If $r$ is odd, then $y \\ge 5$, so $c \\ge 1$ and $y = 5(c - 1) + 2(k + 3)$, where $r = 2k + 1$.\n\nFor $y > 500$, write $y = 500 + z$ with $z \\ge 200$. If $z$ is even, then $y = 5 \\cdot 100 + 2k$, where $z = 2k$. If $z$ is odd, then $y \\le 695$, so $z \\le 195$ and $y = 5 \\cdot 99 + 2(k + 3)$, where $z = 2k + 1$. A quick inspection of all the above cases shows that the claim holds.\n\n$$\nS = \\frac{1}{10}((1 + 2 + \\dots + 700) - (1 + 3 + 697 + 699) = 350 \\cdot 697) = 35 \\cdot 697.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : [0, 1] \\to \\mathbb{R}$ satisfying for all $x, y \\in [0, 1]$ the inequality $|x - y|^2 \\le |f(x) - f(y)| \\le |x - y|$.",
    "options": [],
    "answer": "All functions of the form f(x) = x + a or f(x) = -x + a for any real constant a.",
    "solution": "The condition $|f(x) - f(y)| \\le |x - y|$ is sufficient for the continuity of $f$. From $|x - y|^2 \\le |f(x) - f(y)|$ we derive that $f$ is one-to-one, implying that $f$ is a strictly monotonic function. We may assume that $f$ is strictly increasing, because $f$ can be replaced by $-f$.\nSet $x = 0$ and $y = 1$ to derive that $0 \\le f(1) - f(0) \\le 1$, hence $f(1) = f(0) + 1$.\nFor $x \\ge y$ we get $f(x) - f(y) \\le x - y$ or $y - f(y) \\le x - f(x)$, implying that the function $g(x) = x - f(x) + f(0)$ is increasing.\nSince $g(0) = g(1) = 0$, the function $g$ is the identically null. Consequently, the functions are $f_a^{\\pm}$ given by $f_a^{\\pm}(x) = \\pm x + a$, with $a \\in \\mathbb{R}$.\n\nAlternative solution:\nFor $x = 0$ and $y = 1$ we get $|f(1) - f(0)| = 1$. For $x \\in [0, 1]$ we obtain $1 = |f(1) - f(0)| \\le |f(1) - f(x)| + |f(x) - f(0)| \\le |1 - x| + |1 - x| = 1 - x + x - 0 = 1$.\nThus $f(x)$ lies between $f(0)$ and $f(1)$; moreover, we have $|f(1) - f(x)| = 1 - x$ and $|f(x) - f(0)| = x$. It follows that points $(0, f(0))$, $(x, f(x))$ and $(1, f(1))$ are collinear, hence $f(x) = f(0) \\pm x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function such that, on each non degenerated interval $I$, the function reaches its maximum or its minimum in an interior point of $I$. Prove that $f$ is a constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A, B \\in \\mathcal{M}_2(\\mathbb{C})$ such that $A^2 + B^2 = 2AB$. Prove $AB = BA$ and $\\text{tr } A = \\text{tr } B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We firstly prove that $(AB - BA)^2 = 0$. Define the quadratic function $g$ by $f(x) = \\det(A^2 + B^2 + x(AB - BA)) = \\det(A^2 + B^2) + mx + x^2 \\det(AB - BA)$. As $f(-i) = \\det((A + iB)(A - iB))$, $f(i) = \\det((A - iB)(A + iB))$, so $f(-i) = f(i)$, we get $m = 0$. Moreover, from $f(0) = \\det(A^2 + B^2) = \\det(2AB)$ and $f(-2) = \\det(2BA) = \\det(2AB) = f(0)$, we deduce that $f$ is constant and $\\det(AB - BA) = 0$. The characteristic equation for $AB - BA$ ($\\text{tr}(AB) = \\text{tr}(BA)$) gives then immediately $(AB - BA)^2 = 0$. As $AB - BA = (A - B)^2$ by hypothesis, we obtain $0 = (AB - BA)^2 = (A - B)^4$. As the matrices are of order two we must have $(A - B)^2 = O_2$, so $AB = BA$.\n\nb) From $(A - B)^2 = O_2$ and $2 \\det(A - B) = \\det(A - B)$, the characteristic equation for $A - B$ gives $(\\text{tr}(A) - \\text{tr}(B))(A - B) = O_2$, which finishes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n = \\prod_{i=1}^{s} p_i^{\\alpha_i}$, we write $\\Omega(n)$ for the total number $\\sum_{i=1}^{s} \\alpha_i$ of prime factors of $n$, counted with multiplicity. Let $\\lambda(n) = (-1)^{\\Omega(n)}$ (so, for example, $\\lambda(12) = \\lambda(2^2 \\cdot 3^1) = (-1)^{2+1} = -1$).\n\nProve the following two claims:\n\ni) There are infinitely many positive integers $n$ such that $\\lambda(n) = \\lambda(n+1) = +1$;\n\nii) There are infinitely many positive integers $n$ such that $\\lambda(n) = \\lambda(n+1) = -1$.\n\n(Romania) Dan Schwarz",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$\\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p) = -1$ for any prime $p$, and $\\lambda(k^2) = \\lambda(k)^2 = +1$ for all positive integers $k$.\n\nThen from some rank on we only have subsequences $(1, -1, 1, 1, \\dots, 1, -1, 1)$. By \"doubling\" such a subsequence (like at point ii)), we will produce\n$$\n(-1, ?, 1, ?, -1, ?, -1, ?, \\dots, ?, -1, ?, 1, ?, -1).\n$$\nAccording with our assumption, all `?`-terms ought to be 1, hence the produced subsequence is\n$$\n(-1, 1, 1, 1, -1, 1, -1, 1, \\dots, 1, -1, 1, 1, 1, -1),\n$$\nand so the \"separating packets\" of 1's contain either one or three terms. Now assume some far enough $(1, 1, 1, 1)$ or $(-1, 1, 1, -1)$ subsequence of $\\mathfrak{S}$ were to exist. Since it lies within some \"doubled\" subsequence, it contradicts the structure described above, which thus is the only prevalent from some rank on. But then all the positions of the $(-1)$-terms will have the same parity. However though, we have $\\lambda(p) = \\lambda(2p^2) = -1$ for all odd primes $p$, and these terms have different parity of their positions. A contradiction has been reached.\n*Alternative Solution.* (I. Bogdanov) Take $\\varepsilon \\in \\{-1, 1\\}$. There obviously exist infinitely many $n$ such that $\\lambda(2n + 1) = \\varepsilon$ (just take $2n + 1$ to be the product of an appropriate number of odd primes). Now, if either $\\lambda(2n) = \\varepsilon$ or $\\lambda(2n + 2) = \\varepsilon$, we are done; otherwise $\\lambda(n) = -\\lambda(2n) = -\\lambda(2n + 2) = \\lambda(n + 1) = \\varepsilon$. Therefore, for such an $n$, one of the three pairs $(n, n + 1)$, $(2n, 2n + 1)$ or $(2n + 1, 2n + 2)$ fits the bill.\n\nWe have thus proved the existence in $\\mathfrak{S}$ of infinitely many occurrences of all possible subsequences of length 1, viz. $(+1)$ and $(-1)$, and of length 2, viz. $(+1, -1)$, $(-1, +1)$, $(+1, +1)$ and $(-1, -1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $(a^x + b^x)^{2011} = (a^{2011} + b^{2011})^x$, where $a, b$ are positive reals.",
    "options": [],
    "answer": "2011",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has $\\angle ABC = 60^\\circ$. Points $M$ and $D$ are placed on the sides $(AC)$ and $(AB)$ respectively, so that $\\angle BCA = 2\\angle MBC$ and $BD = MC$. Find the measure of the angle $\\angle DMB$.",
    "options": [],
    "answer": "30°",
    "solution": "Take $G$ so that $CG = CM$, $C \\in (BG)$ and denote $x = \\angle MBC$. Since $\\angle MCB$ is exterior to the isosceles triangle $MCG$, $\\angle MGC = x$. This means that triangle $MBG$ is isosceles and similar to triangle $MCG$. Thus, $\\frac{BG}{MG} = \\frac{MB}{MC} = \\frac{MG}{BD}$.\n\nDenote $H$ the common point of the line $AB$ and the perpendicular bisector of the segment $[BG]$. Then triangle $HBG$ is isosceles and has an angle of $60^\\circ$, so it is equilateral. This yields\n$$\n\\frac{BG}{MG} = \\frac{GH}{MB} = \\frac{MG}{BD}.\n$$\n\n![](attached_image_1.png)\n\nSince $\\angle MGH = \\angle MBD = 60^\\circ - x$, triangles $MGH$ and $DBM$ are similar, therefore we get the answer $\\angle DMB = \\angle MHG = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a non-decreasing function and $F : \\mathbb{R} \\to \\mathbb{R}$ a function having right and left finite derivatives at any point in $\\mathbb{R}$ and $F(0) = 0$. Suppose that $\\lim_{x \\uparrow x_0} f(x) \\le F'_s(x_0)$ and $\\lim_{x \\downarrow x_0} f(x) \\ge F'_d(x_0)$, for any $x_0 \\in \\mathbb{R}$. Prove that $F(x) = \\int_0^x f(t) \\, dt$, $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the function $G : \\mathbb{R} \\to \\mathbb{R}$, given by $G(x) = F(x) - \\int_0^x f(t) \\, dt$. Let $x_0 \\in \\mathbb{R}$, $x \\ne x_0$. Then\n$$\nF(x) = \\frac{F(x) - F(x_0)}{x - x_0} \\cdot (x - x_0) + F(x_0).\n$$\nAs $F$ has right and left finite derivatives in $x_0$, the function $x \\mapsto \\frac{F(x)-F(x_0)}{x-x_0}$, $x \\ne x_0$, is bounded around $x_0$, so $\\lim_{x \\to x_0} F(x) = F(x_0)$. So $G$ is continuous.\n\nLet us show that $G$ has finite right and left derivatives in $x_0 \\in \\mathbb{R}$, $G'_s(x_0) \\ge 0$ and $G'_d(x_0) \\le 0$. Let $x > x_0$. Because $\\lim_{x \\downarrow x_0} f(x) \\le f(t) \\le f(x)$, $x_0 < t \\le x$, we get\n$$\n\\lim_{x \\downarrow x_0} f(x) \\le \\frac{1}{x-x_0} \\int_{x_0}^{x} f(t) \\, dt \\le f(x),\n$$\nso\n$$\n\\lim_{x \\downarrow x_0} \\frac{1}{x-x_0} \\int_{x_0}^{x} f(t) \\, dt = \\lim_{x \\downarrow x_0} f(x).\n$$\n\nConsequently\n$$\n\\begin{aligned}\n\\lim_{x \\downarrow x_0} \\frac{G(x) - G(x_0)}{x - x_0} &= \\lim_{x \\downarrow x_0} \\left( \\frac{F(x) - F(x_0)}{x - x_0} - \\frac{1}{x - x_0} \\int_{x_0}^{x} f(t) \\, dt \\right) \\\\\n&= F'_d(x_0) - \\lim_{x \\downarrow x_0} f(x) \\le 0.\n\\end{aligned}\n$$\n\nAnalogously, $\\lim_{x \\uparrow x_0} \\frac{G(x) - G(x_0)}{x - x_0} = F'_s(x_0) - \\lim_{x \\uparrow x_0} f(x) \\ge 0$.\nIn conclusion, $0 \\le G'_s(x) < \\infty$ and $-\\infty < G'_d(x) \\le 0$, for any real $x$, therefore\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{G(x - \\epsilon) - G(x - \\epsilon/2)}{\\epsilon} = -\\frac{1}{2} G'_s(x) \\le 0, \\quad x \\in \\mathbb{R},\n$$\nand\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{G(x + \\epsilon) - G(x + \\epsilon/2)}{\\epsilon} = \\frac{1}{2} G'_d(x) \\le 0, \\quad x \\in \\mathbb{R}.\n$$\n\nLet us show that $G$ is constant. Consider $a < b$, such that $G(a) \\ne G(b)$.\nIf $G(a) < G(b)$, consider $\\lambda > (G(a) - G(b))/(b - a)$. The function $H : [a, b] \\to \\mathbb{R}$, $H(x) = G(x) + \\lambda x$, is continuous, $H(a) < H(b)$ and\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{H(x + \\epsilon) - H(x + \\epsilon/2)}{\\epsilon} = \\frac{1}{2} G'_d(x) + \\frac{\\lambda}{2} \\le \\frac{\\lambda}{2} < 0, \\quad x \\in \\mathbb{R}.\n$$\nSo, for any real $x$, there is $\\delta(x) > 0$, such that $H(x + \\epsilon) < H(x + \\epsilon/2)$, $0 < \\epsilon < \\delta(x)$. Let $c \\in [a, b]$, such that $H(c) = \\min\\{H(x) : a \\le x \\le b\\}$. Because $H(a) < H(b)$, we have $a \\le c < b$. Consider a number $\\epsilon < \\min(b - c, \\delta(c))$. Then\n$$\nH(c) \\le H(c + \\epsilon) < H(c + \\epsilon/2) < \\dots < H(c + \\epsilon/2^n) < \\dots \\le H(c),\n$$\n(the last inequality is a consequence of the continuity of $H$), which is a contradiction.\nIf $G(a) > G(b)$, proceed analogously: consider $\\lambda < (G(a) - G(b))/(b - a)$. In this case the corresponding continuous function $H(x) = G(x) + \\lambda x$ satisfies $H(a) > H(b)$ and\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{H(x - \\epsilon) - H(x - \\epsilon/2)}{\\epsilon} = -\\frac{1}{2} G'_s(x) - \\frac{\\lambda}{2} \\le -\\frac{\\lambda}{2} < 0, \\quad x \\in \\mathbb{R}.\n$$\nThat is, for any real $x$, there exists $\\delta(x) > 0$, such that $H(x - \\epsilon) < H(x - \\epsilon/2)$, $0 < \\epsilon < \\delta(x)$. Let $c \\in [a, b]$, such that $H(c) = \\min\\{H(x) : a \\le x \\le b\\}$. Because $H(a) > H(b)$, we get $a < c \\le b$. Fix a real positive number $\\epsilon < \\min(c - a, \\delta(c))$. Then\n$$\nH(c) \\le H(c - \\epsilon) < H(c - \\epsilon/2) < \\dots < H(c - \\epsilon/2^n) < \\dots \\le H(c),\n$$\nwhich is a contradiction.\n\nWe thus proved that $G$ is constant. Taking into account $G(0) = 0$, we get $G(x) = 0$, for any real $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer number $n \\ge 2$, evaluate the sum\n$$\n\\sum_{\\sigma \\in S_n} (\\operatorname{sgn} \\sigma) n^{\\ell(\\sigma)},\n$$\nwhere $S_n$ is the set of all $n$-element permutations, and $\\ell(\\sigma)$ is the number of disjoint cycles in the standard decomposition of $\\sigma$.",
    "options": [],
    "answer": "n!",
    "solution": "The sum in question is $f_n(n) = n!$, where\n$$\nf_n(x) = \\sum_{\\sigma \\in S_n} (\\operatorname{sgn} \\sigma) x^{\\ell(\\sigma)} = x(x-1)\\cdots(x-n+1).\n$$\n\nThe latter is a straightforward consequence of the following recurrence formula:\n$$\nf_n(x) = x f_{n-1}(x) - (n-1) f_{n-1}(x) = (x-n+1) f_{n-1}(x), \\quad n \\ge 3,\n$$\nwhere $f_2(x) = x(x-1)$. To establish the recurrence formula, consider the decomposition of a permutation $\\sigma$ in $S_n$ into disjoint cycles,\n$$\n\\sigma = (\\cdots) \\cdots (\\cdots)(n) \\quad \\text{or} \\quad \\sigma = (\\cdots) \\cdots (\\cdots n \\cdots) \\cdots (\\cdots),\n$$\nalong with the permutation $\\sigma'$ in $S_{n-1}$, obtained from $\\sigma$ by deleting $n$:\n$$\n\\sigma' = (\\cdots) \\cdots (\\cdots)(\\hat{n}) \\quad \\text{or} \\quad \\sigma = (\\cdots) \\cdots (\\cdots \\hat{n} \\cdots) \\cdots (\\cdots).\n$$\nClearly, $\\operatorname{sgn} \\sigma' = \\operatorname{sgn} \\sigma$ and $\\ell(\\sigma') = \\ell(\\sigma) - 1$ in the former case, and $\\operatorname{sgn} \\sigma' = -\\operatorname{sgn} \\sigma$ and $\\ell(\\sigma') = \\ell(\\sigma)$ in the latter.\nConversely, any given permutation $\\sigma'$ in $S_{n-1}$ extends to a unique permutation $\\sigma$ in $S_n$ which fixes $n$ (the first case above), and to exactly $n-1$ permutations $\\sigma$ in $S_n$ which do not fix $n$ — for the latter can be inserted in precisely $n-1$ ways into $\\sigma'$ (the second case above). The conclusion follows.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m$ such that\n$$\n\\sqrt{m} = \\sqrt{m+2011}.\n$$",
    "options": [],
    "answer": "1010025",
    "solution": "Rewrite the equation $\\sqrt{m} - [\\sqrt{m}] = \\sqrt{m+2011} - [\\sqrt{m+2011}]$ to get $\\sqrt{m+2011} - \\sqrt{m} = [\\sqrt{m+2011}] - [\\sqrt{m}] = p \\in \\mathbb{N}$. Squaring $\\sqrt{m+2011} = p + \\sqrt{m}$ yields $2011 = p^2 + 2p\\sqrt{m} \\in \\mathbb{N}$, hence $m = k^2, k \\in \\mathbb{N}^*$. The relation $2011 = p(p + 2k)$ gives $p = 1$ and $p + 2k = 2011$, since 2011 is prime. Hence $k = 1005$ and $m = 1005^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be integer numbers with $2 \\le k \\le n - 1$. Consider a set $A$ of $n$ real numbers such that the sum of any $k$ distinct elements of $A$ is a rational number. Prove that all elements of the set $A$ are rational numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The difference of any two elements from $A$ is a rational number. To show this, let $x \\ne y \\in A$ and choose other $k-1$ elements of $A$ – the choice can be made, for $k-1 \\le n-2$. Denote $s$ the sum of the $k-1$ elements and apply the hypothesis to infer that $x+s$ and $y+s$ are both rational numbers. Subtracting we get $x-y \\in \\mathbb{Q}$, as claimed.\n\nNow let $\\alpha \\in A$. Rewrite the elements of $A$ as $\\alpha + q_i$, $q_i \\in \\mathbb{Q}$. The sum of $k$ distinct elements is equal to $k\\alpha + \\sum q_i \\in \\mathbb{Q}$, hence $\\alpha \\in \\mathbb{Q}$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37849,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that, for every positive integer $n$, there exists uniquely determined positive integers $x_n, y_n$ such that $(1 + \\sqrt{33})^n = x_n + y_n\\sqrt{33}$.\n\nb) Prove that if $x_n, y_n$ are defined as above and $p$ is a positive prime, then at least one of the numbers $y_{p-1}, y_p, y_{p+1}$ is divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The equality is obtained for $x_n = \\binom{n}{0} + 33\\binom{n}{2} + 33^2\\binom{n}{4} + \\dots$ and $y_n = \\binom{n}{1} + 33\\binom{n}{3} + 33^2\\binom{n}{5} + \\dots$.\nAlso, if $a + b\\sqrt{33} = c + d\\sqrt{33}$, $a, b, c, d \\in \\mathbb{N}$ and $b \\neq d$, then $\\sqrt{33} = \\frac{a-c}{d-b}$ is rational – false – hence $b = d$ and $a = c$, which proves that the above representation is unique.\n\nb) If $p = 2, 3$ or $11$, then $y_p$ is divisible by $p$.\nIn the other cases, we start noticing that $x_{n+1} = x_n + 33y_n$, $y_{n+1} = x_n + y_n$.\nAlso, $x_p \\equiv 1 \\pmod{p}$ and $y_p \\equiv 33^{\\frac{p-1}{2}} \\pmod{p}$, henceforth $y_p^2 \\equiv 1 \\pmod{p}$.\nThis gives $p \\mid x_p^2 - y_p^2 = (x_p - y_p)(x_p + y_p) = 32y_{p-1}y_{p+1}$ and, since $p$ is prime and $p \\neq 2$, $p \\mid y_{p-1}$ or $p \\mid y_{p+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37850,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ and the points $M \\in (BC)$, $N \\in (AC)$, $P \\in (AB)$ fulfill the conditions $\\angle BMP \\equiv \\angle CNM \\equiv \\angle APN$ and $BM = CN = AP$.\n\nProve that the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start noticing that $m(\\overline{PMN}) = 180^\\circ - \\overline{BMP} - \\overline{NMC} = 180^\\circ - \\overline{CNM} - \\overline{NMC} = \\overline{NCM}$ and, in the same way, $\\overline{MNP} = \\overline{NAP}$, so\n$$\n\\Delta ABC \\sim \\Delta NPM. \\qquad (1)\n$$\n\n![](attached_image_1.png)\n\nSuppose now, without loss of generality, that $\\overline{C} \\ge \\overline{A} \\ge \\overline{B}$. (2)\nThen $AB \\ge BC \\ge AC$, whence, from (1), $NP \\ge PM \\ge MN$. These, together with $\\angle BMP \\equiv \\angle CNM \\equiv \\angle APN$ and $BM = CN = AP$, lead to $\\overline{A} \\ge \\overline{B} \\ge \\overline{C}$. (3)\n\nRelations (2) and (3) show that $\\overline{A} = \\overline{B} = \\overline{C}$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37851,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square with side length $\\ell$ is contained in a unit square whose centre is not interior to the former. Show that $\\ell \\le 1/2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof from The Book.** Notice that there is a line through the centre of the unit square separating the square with side length $\\ell$ and a standard quarter of the unit square (i.e., a square with side length $1/2$ and one vertex at the centre of the unit square). To conclude, apply the celebrated theorem of Erdös stating that the sum of the side lengths of two squares packed into a unit square does not exceed $1$.\n\n\nSolution 1:\nThe square with side length $\\ell$ is the intersection of two strips, $S$ and $S'$, of breadth $\\ell$. Since the centre $O$ of the unit square is not interior to the square with side length $\\ell$, it is not interior to at least one of the two strips, say $S$. Clearly, the distance from $O$ to the farthest component $\\ell'$ of the boundary of $S$ is at least $\\ell$.\nSuppose, if possible, that $\\ell > 1/2$. Then the line $\\ell'$ meets the boundary of the unit square at precisely two points, $X$ and $Y$, situated on consecutive sides. Let $A$ denote the vertex of the unit square shared by those sides, and let $M$ and $N$ denote their midpoints. Without loss of generality, we may (and will) assume that the circular labelling around the boundary of the unit square is $M, X, A, Y, N$. Since $\\text{dist}(O, \\ell') \\ge \\ell > 1/2$, the line $t$, parallel to $\\ell'$ and tangent to the quarter $MN$ of the incircle of the unit square, meets the segments $MX$ and $NY$: the former at $X'$, and the latter at $Y'$. Clearly, $X'Y' > XY \\ge \\ell$. Finally, to reach a contradiction, let $T$ denote the point of contact of $t$ and the quarter $MN$ of the incircle of the unit square, and write successively:\n$$\n\\begin{align*}\n1 &= \\frac{1}{2} + \\frac{1}{2} = MA + AN = (MX' + X'A) + (AY' + Y'N) \\\\\n&= MX' + (X'A + AY') + Y'N = X'Y' + (X'A + AY') \\\\\n&> 2 \\cdot X'Y' > 2 \\cdot XY \\ge 2\\ell.\n\\end{align*}\n$$\n\n\nSolution 2:\nBegin by noticing that if the vertices $M, N, P, Q$ of a rectangle lie on the boundary of a triangle $ABC$ ($M \\in AB, N \\in AC$ and $P, Q \\in BC$), then $MN/BC + MQ/AA' = AM/AB + BM/AB = 1$, where $A'$ is the perpendicular foot dropped from the vertex $A$ onto $BC$.\nHence if a triangle $ABC$, whose internal angles at $B$ and $C$ are not obtuse, contains a square with side length $\\ell$ with a pair of opposite sides parallel to $BC$, then $1/BC + 1/AA' \\le 1/\\ell$, where $A'$ is the perpendicular foot dropped from the vertex $A$ onto $BC$: extend the farthest side of the square, which is parallel to $BC$, to meet the sides $AB$ and $AC$ at $M$ and $N$, respectively, drop the perpendicular feet $M'$ and $N'$ from $M$ and $N$, respectively, onto the side $BC$, notice that $\\ell \\le \\min(MN, MM')$ and apply the above result.\n\nBack to the problem, let $O$ be the centre of the unit square and let $A, B, C, D$ be a circular labelling of its vertices around the boundary. Draw a line $d$ through $O$, parallel to a pair of opposite sides of the square with side length $\\ell$. The latter lies in one of the closed half-planes determined by $d$. Clearly, $\\ell \\le 1/2$ if $d$ is parallel to a pair of opposite sides of the unit square.\nOtherwise, we may assume that $d$ meets the side $AB$ at $E$ and the extension of the side $AD$ beyond $D$ at $F$, and the triangle $AEF$ contains the square with side length $\\ell$. We prove that $\\ell < 1/2$. To this end, drop the perpendicular foot $A'$ from $A$ onto the segment $EF$. By the preceding, $1/AA' + 1/EF \\le 1/\\ell$, so it is sufficient to show that\n$$\n1/AA' + 1/EF > 2. \\quad (*)\n$$\nLet $O'$ be the perpendicular foot dropped from $O$ onto the side $AB$, and let $EO' = x/2$ and $\\angle AEF = \\theta$. Clearly, $\\sin \\theta = 1/\\sqrt{1+x^2}$ and $\\cos \\theta = x/\\sqrt{1+x^2}$, so\n$$\nEF = \\frac{AE}{\\cos \\theta} = \\frac{(1+x)\\sqrt{1+x^2}}{2x} \\quad \\text{and} \\quad AA' = AE \\sin \\theta = \\frac{1+x}{2\\sqrt{1+x^2}},\n$$\nand (*) is equivalent to $x/(1+x)\\sqrt{1+x^2} + \\sqrt{1+x^2}/(1+x) > 1$, $x > 0$, which is easily established.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $n$ persons, each of them speaking at most 3 languages. From any 3 persons there are at least two which speak a common language.\n\ni) For $n \\le 8$, exhibit an example in which no language is spoken by more than two persons.\n\nii) For $n \\ge 9$, prove that there exists a language which is spoken by at least three persons.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "i) Split the 8 persons in two groups of 4. Set any pair of persons in each group to speak a different language for a total of $6 + 6 = 12$ languages, each spoken by 2 persons, each person speaking 3 languages.\n\nFor $n \\le 7$, just remove $8-n$ persons.\n\nii) Assume by contrary that each language is spoken by at most two persons. Then each person $A$ can speak with at most three others, for otherwise, by pigeon-hole principle, there exists a language spoken by other two persons besides $A$, a contradiction. Let $B, C, D$ the persons with whom $A$ can speak. Likewise, $E$ can speak with (at most) three others, namely $F, G, H$. There is left at least another person, say $Z$, and in the group $A, E, Z$ no language is spoken in common, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function with the property: $|f(x)-f(y)| \\le |\\sin x - \\sin y|$, for each $x, y \\in \\mathbb{R}$.\n\na) Prove that there exists an unique $c \\in \\mathbb{R}$ such that $f(c) = c$.\n\nb) Consider the sequence $(x_n)_{n \\in \\mathbb{N}}$ with $x_0 = 0$ and $x_{n+1} = f(x_n)$ for every $n \\in \\mathbb{N}$. Prove that $\\lim_{n \\to \\infty} x_n = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f: \\mathbb{Z} \\to \\mathbb{N}$ fulfilling the conditions:\ni) $f(m+n) = f(m)+f(n)+2mn$, for every integers $m, n$;\nii) $f(f(x))-f(1)$ is a perfect square for each integer $x$.",
    "options": [],
    "answer": "No such function exists.",
    "solution": "Let us try to find all functions $f: \\mathbb{Z} \\to \\mathbb{N}$ such that:\n\ni) $f(m+n) = f(m) + f(n) + 2mn$ for all $m, n \\in \\mathbb{Z}$;\nii) $f(f(x)) - f(1)$ is a perfect square for all $x \\in \\mathbb{Z}$.\n\nLet us first solve the functional equation (i).\n\nLet $P(m, n)$ denote the assertion $f(m+n) = f(m) + f(n) + 2mn$.\n\nLet us try to find the general form of $f$.\n\nLet us try $f(x) = ax^2 + bx + c$ for some $a, b, c \\in \\mathbb{R}$.\n\nPlug into (i):\n\n$f(m+n) = a(m+n)^2 + b(m+n) + c$\n\n$f(m) + f(n) + 2mn = a m^2 + b m + c + a n^2 + b n + c + 2 m n$\n\n$= a(m^2 + n^2) + b(m + n) + 2c + 2 m n$\n\nBut $a(m+n)^2 = a(m^2 + 2 m n + n^2) = a m^2 + 2 a m n + a n^2$\n\nSo:\n\n$f(m+n) = a m^2 + 2 a m n + a n^2 + b m + b n + c$\n\nSet equal to $a(m^2 + n^2) + b(m + n) + 2c + 2 m n$:\n\n$a m^2 + 2 a m n + a n^2 + b m + b n + c = a m^2 + a n^2 + b m + b n + 2c + 2 m n$\n\n$2 a m n + c = 2 m n + 2c$\n\n$2 a m n + c = 2 m n + 2c$\n\n$2 a m n - 2 m n = 2c - c$\n\n$2(a-1) m n = c$\n\nBut this must hold for all $m, n$, so $a = 1$ and $c = 0$.\n\nTherefore, $f(x) = x^2 + b x$ for some $b \\in \\mathbb{R}$.\n\nBut $f: \\mathbb{Z} \\to \\mathbb{N}$, so $f(x) = x^2 + b x$ must be a positive integer for all $x \\in \\mathbb{Z}$.\n\nLet us check for which $b$ this is possible.\n\nFor $x = 0$, $f(0) = 0$.\n\nBut $\\mathbb{N}$ usually means positive integers, i.e., $\\mathbb{N} = \\{1, 2, 3, \\ldots\\}$, so $f(0) = 0$ is not in $\\mathbb{N}$.\n\nTherefore, there is no such function unless $\\mathbb{N}$ is defined to include $0$ (some conventions do this, but usually $\\mathbb{N}$ is positive integers).\n\nAlternatively, perhaps the problem means $\\mathbb{N}_0 = \\{0, 1, 2, \\ldots\\}$.\n\nLet us proceed anyway.\n\nFor $x = -b$, $f(-b) = (-b)^2 + b(-b) = b^2 - b^2 = 0$.\n\nSo $f(-b) = 0$.\n\nThus, unless $b = 0$, $f(x)$ takes the value $0$ for some $x$.\n\nIf $b = 0$, $f(x) = x^2$, which is always non-negative, and $f(0) = 0$.\n\nSo $f(x) = x^2$ is a solution if $0 \\in \\mathbb{N}$.\n\nIf $\\mathbb{N}$ is positive integers, then there is no solution.\n\nLet us check the second condition for $f(x) = x^2$:\n\n$f(f(x)) - f(1) = f(x^2) - 1^2 = (x^2)^2 - 1 = x^4 - 1$\n\nIs $x^4 - 1$ always a perfect square?\n\nFor $x = 0$, $0^4 - 1 = -1$, not a perfect square.\n\nFor $x = 1$, $1^4 - 1 = 0$, which is a perfect square.\n\nFor $x = 2$, $16 - 1 = 15$, not a perfect square.\n\nSo $f(x) = x^2$ does not satisfy the second condition.\n\nLet us try $b = 1$, $f(x) = x^2 + x$.\n\nThen $f(0) = 0$, $f(-1) = 1 - 1 = 0$.\n\nSo again, $f(x)$ takes value $0$.\n\nLet us try $b = 2$, $f(x) = x^2 + 2x$.\n\nThen $f(0) = 0$, $f(-2) = 4 - 4 = 0$.\n\nSo for any integer $b$, $f(x)$ takes value $0$ at $x = 0$ or $x = -b$.\n\nTherefore, unless $0 \\in \\mathbb{N}$, there is no solution.\n\nIf $0 \\in \\mathbb{N}$, let us check the second condition for $f(x) = x^2 + b x$.\n\n$f(f(x)) - f(1) = f(x^2 + b x) - (1^2 + b \\cdot 1) = (x^2 + b x)^2 + b(x^2 + b x) - (1 + b)$\n\n$= (x^4 + 2 b x^3 + b^2 x^2) + b x^2 + b^2 x - (1 + b)$\n\n$= x^4 + 2 b x^3 + (b^2 + b) x^2 + b^2 x - (1 + b)$\n\nIs this always a perfect square for all $x$?\n\nFor $x = 0$, $f(f(0)) - f(1) = f(0) - f(1) = 0 - (1 + b) = -(1 + b)$, which is negative unless $b = -1$.\n\nBut for $b = -1$, $f(x) = x^2 - x$, $f(0) = 0$, $f(1) = 1 - 1 = 0$, $f(f(0)) - f(1) = 0 - 0 = 0$, which is a perfect square.\n\nCheck $f(f(2)) - f(1)$:\n\n$f(2) = 4 - 2 = 2$\n\n$f(f(2)) = f(2) = 2$\n\n$f(1) = 0$\n\nSo $f(f(2)) - f(1) = 2 - 0 = 2$, not a perfect square.\n\nSo $b = -1$ does not work.\n\nTherefore, there is no function $f: \\mathbb{Z} \\to \\mathbb{N}$ of the form $x^2 + b x$ that satisfies both conditions.\n\nTherefore, there is no function $f: \\mathbb{Z} \\to \\mathbb{N}$ satisfying both conditions, unless $\\mathbb{N}$ includes $0$ and $f(1) = 1$.\n\nBut in all cases, the second condition fails for some $x$.\n\nTherefore, the answer is:\n\nThere is no function $f: \\mathbb{Z} \\to \\mathbb{N}$ satisfying both conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37855,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every $n \\ge 3$, determine all the configurations of $n$ distinct points $X_1, X_2, \\dots, X_n$ in the plane, with the property that for any pair of distinct points $X_i, X_j$ there exists a permutation $\\sigma$ of the integers $\\{1, \\dots, n\\}$, such that $d(X_i, X_k) = d(X_j, X_{\\sigma(k)})$ for all $1 \\le k \\le n$. (We write $d(X, Y)$ to denote the distance between points $X$ and $Y$.)\n\n(United Kingdom) Luke Betts",
    "options": [],
    "answer": "All points lie on a common circle. If the number of points is odd, they form a regular polygon. If the number of points is even, they either form a regular polygon or an equiangular polygon whose side lengths alternate between two equal values.",
    "solution": "Let us first prove that the points must be concyclic. Assign to each point $X_k$ the vector $x_k$ in a system of orthogonal coordinates whose origin is the point of mass of the configuration, thus $\\frac{1}{n}\\sum_{k=1}^{n}x_k = 0$.\nThen $d^2(X_i, X_k) = ||x_i - x_k||^2 = \\langle x_i - x_k, x_i - x_k \\rangle = ||x_i||^2 - 2 \\langle x_i, x_k \\rangle + ||x_k||^2$, hence $\\sum_{k=1}^{n} d^2(X_i, X_k) = n||x_i||^2 - 2\\left\\langle x_i, \\sum_{k=1}^{n} x_k \\right\\rangle + \\sum_{k=1}^{n} ||x_k||^2 = n||x_i||^2 + \\sum_{k=1}^{n} ||x_k||^2 = n||x_j||^2 + \\sum_{k=1}^{n} ||x_{\\sigma(k)}||^2 = \\sum_{k=1}^{n} d^2(X_j, X_{\\sigma(k)})$, therefore $||x_i|| = ||x_j||$ for all pairs $(i, j)$. The points are thus concyclic (lying on a circle centred at $O(0, 0)$).\n\nLet now $m$ be the least angular distance between any two points. Two points situated at angular distance $m$ must be adjacent on the circle. Let us connect each pair of such two points with an edge. The graph $G$ obtained must be regular, of degree $\\deg(G) = 1$ or $2$. If $n$ is odd, since $\\sum_{k=1}^{n} \\deg(X_k) = n \\deg(G) = 2|E|$, we must have $\\deg(G) = 2$, hence the configuration is a regular $n$-gon.\nIf $n$ is even, we may have the configuration of a regular $n$-gon, but we also may have $\\deg(G) = 1$. In that case, let $M$ be the next least angular distance between any two points; such points must also be adjacent on the circle. Let us connect each pair of such two points with an edge, in order to get a graph $G'$. A similar reasoning yields $\\deg(G') = 1$, thus the configuration is that of an equiangular $n$-gon (with alternating equal side-lengths).",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists a set $S \\subset \\mathbb{C}$ with $n$ elements, such that\n\ni) each element of $S$ has modulus $1$,\nii) $\\sum_{z \\in S} z = 0$, and\niii) $z + w \\neq 0$, for each $z, w \\in S$.",
    "options": [],
    "answer": "All positive integers except 1, 2, and 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_0A_1A_2$ be a non-equilateral triangle. The incircle of the triangle $A_0A_1A_2$ touches the side $A_iA_{i+1}$ at the point $T_{i+2}$ (indices are reduced modulo 3). Let $X_i$ be the perpendicular foot dropped from the point $T_i$ onto the line $T_{i+1}T_{i+2}$. Show that the lines $A_iX_i$ are concurrent at a point situated on the Euler line of the triangle $T_0T_1T_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The lines $A_iA_{i+1}$ and $X_iX_{i+1}$ are parallel, for they are both antiparallel to the line $T_iT_{i+1}$. Hence the triangles $A_0A_1A_2$ and $X_0X_1X_2$ are homologous: the three lines $A_iX_i$ are concurrent at the homology centre which lies on the homology line. The latter passes through the incentres of the two triangles: one is the circumcentre of the triangle $T_0T_1T_2$ and the other the orthocentre. The conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 37858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a continuous function $f: \\mathbb{R} \\to \\mathbb{R}$, denote, for each interval $[a, b]$, $m_{ab} = \\min_{x \\in [a, b]} f(x)$ and $M_{ab} = \\max_{x \\in [a, b]} f(x)$. Find all the continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for every $a < b$, $f(\\frac{a+b}{2}) = \\frac{m_{ab} + M_{ab}}{2}$.",
    "options": [],
    "answer": "All affine functions f(x) = c x + d with real constants c and d.",
    "solution": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be continuous and satisfy, for every $a < b$,\n$$\nf\\left(\\frac{a+b}{2}\\right) = \\frac{m_{ab} + M_{ab}}{2}\n$$\nwhere $m_{ab} = \\min_{x \\in [a, b]} f(x)$ and $M_{ab} = \\max_{x \\in [a, b]} f(x)$.\n\nLet us fix $a < b$ and consider the function $f$ on $[a, b]$.\n\nSince $f$ is continuous on $[a, b]$, it attains its minimum and maximum at some points $x_1, x_2 \\in [a, b]$:\n$$\nm_{ab} = f(x_1), \\quad M_{ab} = f(x_2)\n$$\n\nLet us consider the value at the midpoint $m = \\frac{a+b}{2}$:\n$$\nf(m) = \\frac{f(x_1) + f(x_2)}{2}\n$$\n\nBut $f(m)$ must also be between $m_{ab}$ and $M_{ab}$, i.e., $f(m) \\in [m_{ab}, M_{ab}]$.\n\nSuppose $f$ is not constant on $[a, b]$. Then $m_{ab} < M_{ab}$, and $f(m)$ is strictly between them unless $f(m)$ equals one of the endpoints. But $f(m)$ is the average of the minimum and maximum, so unless $f$ is constant, $f(m)$ is strictly between $m_{ab}$ and $M_{ab}$.\n\nNow, let us consider the following:\n\nLet $f$ be strictly increasing. Then $m_{ab} = f(a)$, $M_{ab} = f(b)$, and $f(m) = \\frac{f(a) + f(b)}{2}$.\nBut for a strictly increasing continuous function, $f(m)$ is strictly less than $f(b)$ and strictly greater than $f(a)$, unless $f$ is linear.\n\nLet us try $f(x) = cx + d$ (affine function).\n\nThen $f(a) = ca + d$, $f(b) = cb + d$, $f(m) = c\\frac{a+b}{2} + d = \\frac{ca + cb}{2} + d = \\frac{f(a) + f(b)}{2}$.\n\nOn $[a, b]$, since $f$ is affine, the minimum and maximum are at the endpoints:\nIf $c > 0$, $m_{ab} = f(a)$, $M_{ab} = f(b)$.\nIf $c < 0$, $m_{ab} = f(b)$, $M_{ab} = f(a)$.\nIf $c = 0$, $f$ is constant.\n\nIn both cases, $f(m) = \\frac{f(a) + f(b)}{2}$, and $m_{ab} + M_{ab} = f(a) + f(b)$, so the condition is satisfied.\n\nNow, suppose $f$ is not affine. For example, suppose $f$ is quadratic: $f(x) = x^2$.\nOn $[a, b]$, the minimum is at $a$ or $b$ (if $0 \\notin [a, b]$), or at $0$ (if $0 \\in [a, b]$).\nSuppose $a < 0 < b$.\nThen $m_{ab} = 0$, $M_{ab} = \\max\\{a^2, b^2\\}$.\nBut $f(m) = (\\frac{a+b}{2})^2$.\n\nIs it always true that $f(m) = \\frac{m_{ab} + M_{ab}}{2}$? For $a = -1$, $b = 1$, $m = 0$, $f(m) = 0$, $m_{ab} = 0$, $M_{ab} = 1$, so $\\frac{m_{ab} + M_{ab}}{2} = \\frac{1}{2}$. But $f(m) = 0 \\neq \\frac{1}{2}$.\n\nTherefore, $f(x) = x^2$ does not satisfy the condition.\n\nSuppose $f$ is constant: $f(x) = d$.\nThen $m_{ab} = d$, $M_{ab} = d$, $f(m) = d$, $\\frac{m_{ab} + M_{ab}}{2} = d$.\nSo the condition is satisfied.\n\nTherefore, the only continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the condition are the affine functions $f(x) = cx + d$.\n\nLet us check for $c < 0$:\nThen $m_{ab} = f(b)$, $M_{ab} = f(a)$, $f(m) = \\frac{f(a) + f(b)}{2}$, $\\frac{m_{ab} + M_{ab}}{2} = \\frac{f(a) + f(b)}{2}$.\nSo the condition is satisfied.\n\nThus, all continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ of the form $f(x) = cx + d$ for $c, d \\in \\mathbb{R}$ satisfy the condition.\n\nFinal answer:\n\nAll continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ of the form $f(x) = cx + d$ for $c, d \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37859,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of a square $2011 \\times 2011$ array are labelled with the integers $1, 2, \\dots, 2011^2$, in such a way that every label is used exactly once. We then identify the left-hand and right-hand edges, and then the top and bottom, in the normal way to form a torus (the surface of a doughnut). Determine the largest positive integer $M$ such that, no matter which labelling we choose, there exist two neighbouring cells with the difference of their labels at least $M$.\n\nCells with coordinates $(x, y)$ and $(x', y')$ are considered to be neighbours if $x = x'$ and $y - y' \\equiv \\pm 1 \\pmod{2011}$, or if $y = y'$ and $x - x' \\equiv \\pm 1 \\pmod{2011}$.",
    "options": [],
    "answer": "4021",
    "solution": "For the toroidal case, it is clear the statement of the problem is referring to the cells of a $\\mathbb{Z}_N \\times \\mathbb{Z}_N$ lattice on the surface of the torus, labeled with the numbers $1, 2, \\dots, N^2$, where one has to determine the least possible maximal absolute value $M$ of the difference of labels assigned to orthogonally adjacent cells.\n\nThe toroidal $N = 2$ case is trivially seen to be $M = 2$ (thus coinciding with the planar case).\n\n| 1 | 2 |\n|---|---|\n| 3 | 4 |\n\nThe unique $2 \\times 2$ toroidal array.\n\nFor $N \\ge 3$ we will prove that value to be at least $M \\ge 2N - 1$. Consider such a configuration, and color all cells of the square in white. Go along the cells labeled 1, 2, etc. coloring them in black, stopping just on the cell bearing the least label $k$ which, after assigned and colored in black, makes that all lines of a same orientation (rows, or columns, or both) contain at least two black cells (that is, before coloring in black the cell labeled $k$, at least one row and at least one column contained at most one black cell). Wlog assume this happens for rows. Then at most one row is all black, since if two were then the stopping condition would have been fulfilled before cell labeled $k$ (if the cell labeled $k$ were to be on one of these rows, then all rows would have contained at least two black cells before, while if not, then all columns would have contained at least two black cells before).\n\nNow color in red all those black cells adjacent to a white cell. Since each row, except the potential all black one, contained at least two black and one white cell, it will now contain at least two red cells. For the potential all black row, any of the neighbouring rows contains at least one white cell, and so the cell adjacent to it has been colored red. In total we have therefore colored red at least $2(N-1)+1 = 2N-1$ cells.\n\nThe least label of the red cells has therefore at most the value $k + 1 - (2N - 1)$. When the white cell adjacent to it will eventually be labeled, its label will be at least $k+1$, therefore their difference is at least $(k+1) - (k+1 - (2N-1)) = 2N - 1$.\n\nThe models are kind of hard to find, due to the fact that the direct proof offers little as to their structure (it is difficult to determine the equality case during the argument involving the inequality with the bound, and then, even this is not sure to be prone to being prolonged to a full labeling of the array).\n\nThe weaker fact the value $M$ is not larger than $2N$ is proved by the general model exhibited below (presented so that partial credits may be awarded).\n\n| $N+1$ | $N+2$ | ... | $2N$ |\n|-------|-------|-----|------|\n| $3N+1$ | $3N+2$ | ... | $4N$ |\n| ...   | ...   | ... | ...  |\n| $(2\\ell-1)N+1$ | $(2\\ell-1)N+2$ | ... | $2\\ell N$ |\n| ...   | ...   | ... | ...  |\n| $2kN+1$ | $2kN+2$ | ... | $(2k+1)N$ |\n| ...   | ...   | ... | ...  |\n| $2N+1$ | $2N+2$ | ... | $3N$ |\n| 1     | 2     | ... | $N$  |\n\nA general model for $M = 2N$ in a $N \\times N$ array.\n\nBy examining some small $N > 2$ cases, one comes up with the idea of spiral models for the true value $M = 2N - 1$. The models presented are for odd $N$ (since 2011 is odd); similar models exist for even $N$ (but are less symmetric).\n\n| 7 | 2 | 6 |\n|---|---|---|\n| 3 | 1 | 5 |\n| 8 | 4 | 9 |\n\nThe spiral $3 \\times 3$ array.\n\n| 23 | 16 | 7 | 15 | 22 |\n|----|----|---|----|----|\n| 17 | 8  | 2 | 6  | 14 |\n| 9  | 3  | 1 | 5  | 13 |\n| 18 | 10 | 4 | 12 | 21 |\n| 24 | 19 | 11| 20 | 25 |\n\nThe spiral $5 \\times 5$ array.\n\n| 47 | 40 | 29 | 16 | 28 | 39 | 46 |\n|----|----|----|----|----|----|----|\n| 41 | 30 | 17 | 7  | 15 | 27 | 38 |\n| 31 | 18 | 8  | 2  | 6  | 14 | 26 |\n| 19 | 9  | 3  | 1  | 5  | 13 | 25 |\n| 32 | 20 | 10 | 4  | 12 | 24 | 37 |\n| 42 | 33 | 21 | 11 | 23 | 36 | 45 |\n| 48 | 43 | 34 | 22 | 35 | 44 | 49 |\n\nThe spiral $7 \\times 7$ array.\n\n| $(2n+1)^2-2$ | $(2n+1)^2-9$ | ... | $n(2n-1)+1$ | ... | $(2n+1)^2-10$ | $(2n+1)^2-3$ |\n|--------------|--------------|-----|-------------|-----|---------------|--------------|\n| $(2n+1)^2-8$ |              | ... | $n(2n-1)+2$ | $n(2n-1)$ | ... | $(2n+1)^2-11$ |\n| ...          | ...          | ... | ...         | ... | ...           | ...          |\n|              | $2n^2$       | ... | 8           | 2   | 6             | $2n(n+1)+3$  |\n| $2n^2 + 1$   | ...          | ... | 3           | 1   | 5             | $2n(n-1)+2$  |\n|              | $2n^2+2$     | ... | 10          | 4   | 12            | $2n(n+1)$    |\n| ...          | ...          | ... | ...         | ... | ...           | ...          |\n| $(2n+1)^2-7$ | ...          | $n(2n+1)$ | $n(2n+1)+2$ | ... | $(2n+1)^2-5$ | $(2n+1)^2-4$ |\n| $(2n+1)^2-1$ | $(2n+1)^2-6$ | ... | $n(2n+1)+1$ | ... | $(2n+1)^2-5$ | $(2n+1)^2$   |\n\nThe general spiral $N \\times N$ array for $N = 2n + 1 \\ge 5$.\n\nThus, for $N = 2011$, the answer is $M = 2 \\times 2011 - 1 = 4021$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 37860,
    "subject": "Mathematics (Multi-modal)",
    "question": "The parallelogram $ABCD$ has center $O$, $AD = DB$ and $BA > BC$. The point $V$ is outside the plane $(ABC)$ such that $VD \\perp (ABC)$. Point $T$ is the foot of the perpendicular from $D$ onto $AB$ and $E, F, G$ are the feet of the bisectors of the angles $\\angle VDA, \\angle VDB, \\angle VDC$ respectively (where $E \\in VA, F \\in VB, G \\in CV$). Prove that:\n\ni) $EF \\parallel (ABC);$\nii) $m(\\angle((VAB), (ABC))) = 45^\\circ$ if and only if $VD = DT;$\niii) if $VD = DT$, then $\\tan(\\angle((GEF), (ABC))) = \\frac{1}{2}$ if and only if $\\frac{BA}{BC} = \\frac{6}{5}.$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer number $n$, determine the maximum number of edges a simple graph on $n$ vertices may have in order that it contain no cycles of even length.",
    "options": [],
    "answer": "floor(3(n-1)/2)",
    "solution": "The required maximum is $\\lfloor 3(n-1)/2 \\rfloor$. It is achieved, for instance, by an $(n-1)/2$ arm wind-mill if $n$ is odd, and an $(n-2)/2$ arm wind-mill with an extra edge joined at the hub if $n$ is even.\n\nTo show that a simple graph on $n$ vertices with no cycles of even length has at most $3(n-1)/2$ edges, let $G$ be one such with a maximal edge-set $E$. By maximality, $G$ is connected. Let $T$ be a spanning tree (that is, a maximal connected acyclic subgraph) of $G$, and let $E'$ denote the edge-set of $T$; it is well-known that $|E'| = n-1$. The end-points of any edge $e$ in $E \\setminus E'$ are joined by a unique simple path $\\alpha$ in $T$. Since $T$ is acyclic, if $e$ and $e'$ are distinct edges in $E \\setminus E'$, then the edges the corresponding paths in $T$, $\\alpha$ and $\\alpha'$, may share a path; and since $G$ has no cycles of even length, the paths $\\alpha$ and $\\alpha'$ are actually edge-disjoint — otherwise, at least one of the cycles $\\alpha + e, \\alpha' + e'$ and $\\alpha + e + \\alpha' + e'$ (mod 2) would have an even length. It follows that the total length of the cycles $\\alpha + e, e \\in E \\setminus E'$, does not exceed $|E|$. There are $|E \\setminus E'| = |E| - |E'| = |E| - n + 1$ such cycles, each of length at least 3, so $3(|E| - n + 1) \\le |E|$; that is, $|E| \\le 3(n-1)/2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given in the plane equilateral triangles $ABC$ and $BDE$, with $C \\in (BD)$ and $A, E$ are on different sides of the line $BD$. Denote $M, N, P$ the midpoints of the segments $(AB)$, $(CD)$, $(BE)$ respectively. Find the measures of the angles of the triangle $MNP$.",
    "options": [],
    "answer": "60°, 60°, 60°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37863,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider an $n \\times n$ ($n \\in \\mathbb{N}$, $n \\ge 2$) square divided into $n^2$ unit squares. Determine all the values of $k \\in \\mathbb{N}$ for which we can write a real number in each of the unit squares such that the sum of the $n^2$ numbers is a positive number, while the sum of the numbers from the unit squares of any $k \\times k$ square is a negative number.",
    "options": [],
    "answer": "All natural k that do not divide n.",
    "solution": "We will prove that the desired numbers $k$ are those that are not factors of $n$.\n\nIf $k \\mid n$, then we can tile the $n \\times n$ square with $k \\times k$ squares and the total sum should simultaneously be positive and negative, which is impossible.\n\nIf $k \\nmid n$, then $n = kq + r$, where $0 < r < k$. We fill the unit squares with $a$ (to be chosen conveniently later on) in the positions $(ik, jk)$ with $i, j = 1, \\dots, q$ and with $1$ in the other positions. Every $k \\times k$ square contains exactly one unit square of the form $(ik, jk)$, therefore the sum in every $k \\times k$ square is $a + k^2 - 1$. The total sum is $q^2 a + n^2 - q^2$. We will choose $a$ arbitrarily from the non-empty interval $\\left(1 - \\frac{n^2}{q^2}, 1 - k^2\\right)$.\n\n**Remark:** For the case $k \\nmid n$ there are many other ways of choosing the numbers from the unit squares. Another choice is to fill all the unit squares of the columns $jk$, $j = 1, \\dots, q$, with a convenient $a$ and the other ones with $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the perfect squares whose product of their decimal digits is a prime.",
    "options": [],
    "answer": "121",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37865,
    "subject": "Mathematics (Multi-modal)",
    "question": "The matrix $A \\in M_k(\\mathbb{R})$ ($k \\ge 2$) has the property: for every positive integer $n$, $\\det(n^2 I_k - A^2) \\ge n^k (\\det(nI_k - A) + \\det(nI_k + A) - 1)$. Prove that $\\operatorname{tr}(A) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37866,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) There are infinitely many positive integer numbers $n$ such that there exists a square equal to the sum of the squares of $n$ consecutive positive integer numbers. (For instance, $2$ and $11$ are such: $5^2 = 3^2 + 4^2$ and $77^2 = 18^2 + 19^2 + \\dots + 28^2$.)\n\nb) If $n$ is a positive integer number which is not a perfect square and if $x_0$ is an integer number such that $x_0^2 + (x_0 + 1)^2 + \\dots + (x_0 + n - 1)^2$ is a perfect square, then there are infinitely many positive integer numbers $x$ such that $x^2 + (x + 1)^2 + \\dots + (x + n - 1)^2$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nWe must show that there are infinitely many positive integer numbers $n$ such that the equation $y^2 = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$ has positive integral solutions. To this end, rewrite the equation in the form\n$$\ny^2 = n\\left(\\left(x + \\frac{n-1}{2}\\right)^2 + \\frac{(n^2-1)}{12}\\right). \\qquad (1)\n$$\nWe show that if $n$ is a square greater than $25$ not divisible by $2$ or $3$, then (1) has a positive but finite number of solutions in positive integers $x$ and $y$. Since $n$ is a perfect square, the expression in the outmost parentheses in (1) must be a perfect square $z^2$, i.e., there is a positive integer $z$ such that\n$$\nz^2 - \\left(x + \\frac{n-1}{2}\\right)^2 = \\frac{(n^2-1)}{12}. \\qquad (2)\n$$\nThus, $z \\pm (x + (n-1)/2)$ must be complementary even divisors of $(n^2 - 1)/12$ differing by more than $n-1$; hence there are only finitely many solutions to (2) in positive integers. One such is obtained by taking\n$$\nz - \\left(x + \\frac{n-1}{2}\\right) = 2 \\quad \\text{and} \\quad z + \\left(x + \\frac{n-1}{2}\\right) = \\frac{n^2-1}{24},\n$$\ni.e., by letting $x$ and $z$ have the positive integer values\n$$\nx = \\frac{(n-25)(n+1)}{48} \\quad \\text{and} \\quad z = 1 + \\frac{(n^2-1)}{48}.\n$$\n\nb.\nRewrite the equation $y^2 = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$ in the form\n$$\n(2y)^2 - n(2x + n - 1)^2 = \\frac{(n-1)n(n+1)}{3} \\quad (3)\n$$\nand let $T_n(x) = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$. We may assume $n > 1$. Since $T_n(x) = T_n(-x-n+1)$, we may further assume that $x_0 \\ge -(n-1)/2$. Suppose $T_n(x_0) = y_0^2$, i.e., $(2y_0)^2 - n(2x_0 + n-1)^2 = (n-1)n(n+1)/3$, where we may assume $y_0 > 0$ (and hence $y_0 > (n-1)/2$). By the theory of the Pell equation, there are infinitely many pairs of positive integers $u, v$ such that\n$$\nu^2 - nv^2 = 1. \\quad (4)\n$$\nClearly, $u$ is odd if $n$ is even, so that $(n-1)(u-1)$ is even. We now use the identity\n$$\n(2y_0 + (2x_0 + n - 1)\\sqrt{n})(u + v\\sqrt{n}) = 2y + (2x + n - 1)\\sqrt{n}, \\quad (5)\n$$\nwhere\n$$\nx = x_0u + y_0v + \\frac{(n-1)(u-1)}{2} \\quad \\text{and} \\quad y = y_0u + x_0nv + \\frac{(n-1)nv}{2}. \\quad (6)\n$$\nMultiplying (5) by the identity obtained from (5) by replacing $\\sqrt{n}$ by $-\\sqrt{n}$, we find that if $x$ and $y$ are given by (6) and $u$ and $v$ satisfy (4), then\n$$\n\\begin{aligned}\n(2y)^2 - n(2x + n - 1)^2 &= ((2y_0)^2 - n(2x_0 + n - 1)^2)(u^2 - nv^2) \\\\\n&= (2y_0)^2 - n(2x_0 + n - 1)^2 = \\frac{(n - 1)n(n + 1)}{3}.\n\\end{aligned}\n$$\nThus, if $x, y$ are given by (6), they satisfy (3), so that $T_n(x)$ is a perfect square. Further, since $x_0 \\ge -(n-1)/2$ and $y_0 > (n-1)/2$, it follows that $y \\ge y_0u > 0$ and $x \\ge y_0v - u(n-1)/2 + (n-1)(u-1)/2 = y_0v - (n-1)/2 \\ge y_0 - (n-1)/2 > 0$. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$, $p \\ge 3$ and $d$ a square free number (that is $d$'s prime decomposition contains no repeated factors), find the number of the elements of the set\n$$\nA_p = \\{x = \\{n\\sqrt{d} + \\frac{n}{p}\\} - \\{n\\sqrt{d}\\} \\mid n \\in \\mathbb{N}\\},\n$$\nwhere $\\{a\\}$ denotes the fractional part of the real number $a$.",
    "options": [],
    "answer": "If d = 1, the number of elements is p. If d > 1 (so √d is irrational), the number of elements is 2p − 1.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37868,
    "subject": "Mathematics (Multi-modal)",
    "question": "The function $f : [0, 1] \\to \\mathbb{R}$ is differentiable and\n$$\n\\int_{0}^{1} f(x) \\, dx = \\int_{0}^{1} x f(x) \\, dx = 0.\n$$\nProve that there exists $c \\in (0, 1)$ such that $f'(c) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, if $x, y > 0$, then\n$$\n0 < \\frac{x}{x+y^4} + \\frac{y}{y+x^4} - \\frac{1}{x^2y^2+1} < 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a function which is continuous on $[0, 1]$ and differentiable in $0$. Consider the function $s : [0, 1] \\to \\mathbb{R}$,\n$$\ns(x) = \\sup\\{c \\in [0; x] \\mid \\int_{0}^{x} f(t) \\, dt = x f(c)\\}.\n$$\na) Prove that the function $f \\circ s$ is differentiable in $0$ and $(f \\circ s)'(0) = \\frac{1}{2} f'(0)$.\n\nb) If $f'(0) \\neq 0$, prove that $s$ is differentiable in $0$ and $s'(0) = \\frac{1}{2}$.",
    "options": [],
    "answer": "(f ∘ s)'(0) = (1/2) f'(0); if f'(0) ≠ 0 then s'(0) = 1/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37871,
    "subject": "Mathematics (Multi-modal)",
    "question": "The measure of the angle $\\hat{A}$ of the acute triangle $ABC$ is $60^\\circ$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle $\\hat{B}$.",
    "options": [],
    "answer": "80°",
    "solution": "We have $m(\\angle BIC) \\equiv m(\\angle BHC) = 120^\\circ$, hence $B$, $H$, $I$, $C$ are situated on a circle.\n\nIf $m(\\angle B) > 60^\\circ$ then $m(\\angle HBI) = m(\\angle ABI) - m(\\angle ABH) = \\frac{1}{2}m(\\angle B) - 30^\\circ$.\n\nBut $m(\\angle HBI) = m(\\angle HIB) = m(\\angle HCB) = 90^\\circ - m(\\angle B)$, hence $m(\\angle B) =$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi având punctul $O$ ca centru al cercului său circumscris. Punctele $D$, $E$ şi $F$ se află respectiv pe laturile $BC$, $CA$ şi $AB$, astfel încât dreapta $DE$ este perpendiculară pe $CO$ iar dreapta $DF$ este perpendiculară pe $BO$. (De exemplu, punctul $D$ se află pe dreapta $BC$, fiind situat între $B$ şi $C$ pe acea dreaptă.)\n\nFie $K$ centrul cercului circumscris triunghiului $AFE$. Demonstraţi că dreptele $DK$ şi $BC$ sunt perpendiculare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a, b \\in \\mathbb{R}$ şi $z \\in \\mathbb{C} \\setminus \\mathbb{R}$ astfel încât $|a-b| = |a+b-2z|$.\n\na) Să se arate că ecuaţia $|z-a|^{x} + |\\bar{z}-b|^{x} = |a-b|^{x}$, cu necunoscuta $x \\in \\mathbb{R}$, are soluţie unică.\n\nb) Să se rezolve inecuaţia $|z-a|^{x} + |\\bar{z}-b|^{x} \\leq |a-b|^{x}$, cu necunoscuta $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "a) x = 2. b) x ≥ 2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(a_{n})_{n \\geq 1}$ un şir crescător şi mărginit. Calculaţi\n$$\n\\lim_{n \\rightarrow \\infty} \\left(2 a_{n} - a_{1} - a_{2}\\right) \\left(2 a_{n} - a_{2} - a_{3}\\right) \\cdots \\left(2 a_{n} - a_{n-2} - a_{n-1}\\right) \\left(2 a_{n} - a_{n-1} - a_{1}\\right)\n$$",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n$ un număr întreg strict pozitiv. Determinaţi cel mai mare număr întreg $m$ cu proprietatea că un tablou cu $m$ linii şi $n$ coloane poate fi umplut cu numere reale în aşa fel încât pentru oricare două linii diferite $[a_{1}, a_{2}, \\ldots, a_{n}]$ si $[b_{1}, b_{2}, \\ldots, b_{n}]$ următoarea relaţie este adevărată\n$$\n\\max \\left(\\left|a_{1}-b_{1}\\right|,\\left|a_{2}-b_{2}\\right|, \\ldots,\\left|a_{n}-b_{n}\\right|\\right)=1\n$$",
    "options": [],
    "answer": "2^n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a, b \\in \\mathbb{C}$. Să se arate că $|a z + b \\bar{z}| \\leq 1$, pentru orice $z \\in \\mathbb{C}$, cu $|z| = 1$, dacă şi numai dacă $|a| + |b| \\leq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie matricele de ordin $2$ cu elemente reale $A$ şi $B$ astfel încât\n$$\nA B = A^{2} B^{2} - (A B)^{2} \\quad \\text{şi} \\quad \\operatorname{det}(B) = 2\n$$\na) Arătaţi că matricea $A$ nu este inversabilă.\nb) Calculaţi $\\operatorname{det}(A+2 B) - \\operatorname{det}(B+2 A)$.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminaţi toate funcţiile $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ pentru care $f(y f(x+y)+f(x))=4 x+2 y f(x+y)$ pentru toţi $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 2x for all real x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A$ o matrice neinversabilă de ordin $n$, $n>1$, cu elemente în mulţimea numerelor complexe, toate elementele având modulul egal cu $1$.\n\na) Arătaţi că pentru $n=3$, două dintre liniile sau două dintre coloanele matricei $A$ sunt proporţionale.\n\nb) Rămâne adevărată concluzia de la punctul anterior pentru $n=4$ ?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe consideră funcţia $f: \\mathbb{R} \\rightarrow \\mathbb{R}$,\n$$\nf(x)= \\begin{cases} a x, & x \\in \\mathbb{Q} \\\\ b x, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}\n$$\nunde $a$ şi $b$ sunt două numere reale nenule.\nSă se arate că $f$ este injectivă dacă şi numai dacă $f$ este surjectivă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO mulţime $A$ de numere întregi se zice sumă-plină dacă $A \\subseteq A+A$, adică orice element $a \\in A$ este suma unei perechi (nu neapărat unice) de elemente (nu neapărat distincte) $b, c \\in A$. O mulţime $A$ de numere întregi se zice liberă-de-sume-zero dacă $0$ este singurul număr întreg care nu poate fi exprimat ca suma elementelor unei submulţimi finite nevide a lui $A$.\nExistă oare o muļime sumă-plină liberă-de-sume-zero de numere întregi?",
    "options": [],
    "answer": "No, such a set does not exist.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\in \\mathbb{N}^*$. Să se arate că numărul\n$$\n2 \\sqrt{2^{n}} \\cos \\left(n \\arccos \\frac{\\sqrt{2}}{4}\\right)\n$$\nest număr întreg impar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră o funcţie monotonă $f: \\mathbb{R} \\rightarrow \\mathbb{R}$.\n\na) Demonstraţi că $f$ are limite laterale în fiecare punct $x_{0} \\in \\mathbb{R}$.\n\nb) Definim funcția $g: \\mathbb{R} \\rightarrow \\mathbb{R}$, $g(x)=\\lim_{t>x} f(t)$, i.e. $g(x)$ este limita la stânga în punctul $x$. Arătaţi că dacă funcţia $g$ este continuă, atunci funcţia $f$ este continuă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNumerele întregi pozitive $p$ şi $q$ sunt prime şi satisfac\n$$\n\\frac{p}{p+1} + \\frac{q+1}{q} = \\frac{2n}{n+2}\n$$\npentru un anume număr întreg pozitiv $n$. Determinaţi toate valorile posible ale diferenţei $q-p$.",
    "options": [],
    "answer": "[2, 3, 5]",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn număr infinit de persoane sunt înregistrate ca useri pe reţeaua socială Mutrăbook. Unele perechi de useri (diferiţi) sunt înregistrate ca prieteni, dar fiecare persoană are cel mult un număr finit de prieteni. Fiecare user are cel puţin un prieten. (Relaţia de prietenie este simetrică; dacă $A$ este prieten cu $B$, atunci şi $B$ este prieten cu $A$.)\n\nFiecare persoană trebuie să-şi desemneze unul dintre prieteni ca fiind cel-mai-bun-prieten. Dacă $A$ îl desemnează pe $B$ ca cel-mai-bun-prieten, nu urmează (din păcate) în mod necesar că şi $B$ l-a desemnat pe $A$ ca cel-mai-bun-prieten. Cineva desemnat ca cel-mai-bun-prieten este numit 1-cel-mai-bun-prieten. În general, dacă $n>1$ este un număr întreg pozitiv, atunci un user este un $n$-cel-mai-bun-prieten dacă a fost desemnat ca cel-mai-bun-prieten de către cineva care este el însuşi un $(n-1)$-cel-mai-bun-prieten. O persoană care este un $k$-cel-mai-bun-prieten pentru toate numerele întregi pozitive $k$ este numită populară.\n\na. Demonstraţi că fiecare persoană populară este cel-mai-bun-prieten al (măcar) unei alte persoane populare.\n\nb. Demonstraţi că dacă persoanele ar fi putut avea infinit de mulţi prieteni, atunci ar fi fost posibil ca o persoană populară să nu fi fost cel-mai-bun-prieten al niciunei alte persoane populare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 37886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic înscris în cercul $\\Gamma$ şi având ortocentrul $H$. Fie $K$ un punct pe cercul $\\Gamma$, de cealaltă parte a lui $BC$ decât $A$. Fie $L$ simetricul lui $K$ faţă de dreapta $AB$, şi fie $M$ simetricul lui $K$ faţă de dreapta $BC$. Fie $E$ al doilea punct de intersecţie al cercului $\\Gamma$ cu cercul circumscris triunghiului $BLM$. Demonstraţi că dreptele $KH$, $EM$ şi $BC$ sunt concurente. (Ortocentrul unui triunghi este punctul de intersecţie a înălţimilor sale.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn cuvânt este o secvenţă finită de litere dintr-un anume alfabet. Un cuvânt se zice repetitiv dacă este o concatenare de cel puţin două sub-cuvinte identice (de exemplu, $ababab$ şi $abcabc$ sunt repetitive, dar $ababa$ şi $aabb$ nu sunt). Demonstraţi că dacă un cuvânt are proprietatea că orice transpoziţie a două litere adiacente îl transformă într-un cuvânt repetitiv, atunci toate literele sale sunt identice. (O transpoziţie a două litere adiacente identice, care lasă cuvântul neschimbat, este şi ea a fi considerată.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37888,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Show that $x^4 - x^3 - x + 1 \\ge 0$, for all real numbers $x$.\n\nb. Find all real numbers $x_1, x_2$ and $x_3$ given that $x_1 + x_2 + x_3 = 3$ and $x_1^3 + x_2^3 + x_3^3 = x_1^4 + x_2^4 + x_3^4$.",
    "options": [],
    "answer": "x1 = x2 = x3 = 1",
    "solution": "a. Write $x^4 - x^3 - x + 1 = (x-1)(x^3 - 1) = (x-1)^2(x^2 + x + 1)$ and notice that $x^2 + x + 1 > 0$ for all $x \\in \\mathbb{R}$ to get the claim.\n\nb. Notice that $\\sum_{k=1}^{3} (x_k^4 - x_k^3 - x_k + 1) = 0$ and use (a) to derive that $x_1^4 - x_1^3 - x_1 + 1 = 0$, $k = 1, 2, 3$. It follows that $x_1 = x_2 = x_3 = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, in every scalene triangle,\n$$\n\\frac{h_a - h_b}{r_b - r_a} + \\frac{h_b - h_c}{r_c - r_b} + \\frac{h_c - h_a}{r_a - r_c} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set of 7 integers and denote $S = \\{x + y \\mid x, y \\in A\\}$. Prove that if $S$ has 13 elements, then the sum of $A$'s elements is divisible by 7.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37891,
    "subject": "Mathematics (Multi-modal)",
    "question": "The right prism $ABCA'B'C'$, with $AB = AC = BC = a$, has the property that there exists a unique point $M \\in (BB')$ so that $AM \\perp MC'$. Find the measure of the angle of the straight line $AM$ and the plane $(ACC')$.\nMircea Fianu, Cristian Mangra\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "45°",
    "solution": "We first prove that $M$ is the midpoint of the edge $[BB']$. Indeed, if this is not the case, denote $M'$ the reflection of $M$ across the midpoint of $[BB']$. Then $\\Delta MAB \\equiv \\Delta M'C'B'$ and $\\Delta M'AB \\equiv \\Delta MC'B'$ imply $[MA] \\equiv [M'C']$ and $[M'A] \\equiv [MC']$. Therefore $\\Delta MAC' \\equiv \\Delta M'C'A$, hence $m(\\angle AMC') = m(\\angle AM'C') = 90^\\circ$, which contradicts the unicity of $M$.\n\nDenote $BB' = 2h$. Then $AM = MC' = \\sqrt{a^2 + h^2}$, $AC' = \\sqrt{a^2 + 4h^2}$ and Pythagoras' Theorem yields $a = h\\sqrt{2}$.\n\nIf $O$ is the center of the face $(ACC'A')$, then $MO \\perp (ACC')$, hence the required angle is $MAO$.\nSince $MO = OA = a\\sqrt{3}/2$, the triangle $MOA$ is right and isosceles, so $m(\\angle MAO) = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ having the property that $(a^2 + ab + b^2) \\int_a^b f(x) \\, dx = 3 \\int_a^b x^2 f(x) \\, dx$, for all $a, b \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All constant functions: f(x) = k for some real constant k.",
    "solution": "Choose $a = 0$ and $b = t > 0$. Then $t^2 F(t) = 3 \\int_0^t x^2 f(x) \\, dx$, where $F(t) = \\int_0^t f(x) \\, dx$. Taking the derivative we get $2tF(t) + t^2 f(t) = 3t^2 f(t)$, so $2t(tf(t) - F(t)) = 0$, that is $\\left(\\frac{F(t)}{t}\\right)' = 0$, for any $t \\in (0, \\infty)$. It follows that the function $g(t) = \\frac{F(t)}{t}$ is constant on $(0, \\infty)$, so $F(t) = kt$, that is $f(t) = k$, $t \\in (0, \\infty)$.\n\nUsing the same argument on $(-\\infty, 0)$, we deduce that there are constants $k_1, k_2, k$ such that $f(t) = \\begin{cases} k_1, & t < 0 \\\\ k_2, & t = 0 \\\\ k, & t > 0 \\end{cases}$.\n\nAs $f$ is continuous, $k_1 = k_2 = k$, so $f(t) = k$, for all $t \\in \\mathbb{R}$. It is routine to verify that the constant functions verify the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37893,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of positive integers is called complete if any positive integer has a multiple in the sequence. Prove that an arithmetic sequence of positive integers is complete if and only if its difference divides the first term.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If the difference $r$ divides $a_1$, then $a_1 = dr$, $d \\in \\mathbb{N}$ and $a_n = (d + n - 1)r$, and a multiple of a positive integer $k$ is obtained when $d + n - 1$ is a multiple of $k$.\n\nFor the converse, observe first that if $r = 0$, the sequence is not complete. Because $r \\neq 0$ and by the assumption there is a multiple of $r$ of the form $a_1 + (n-1)r$, with $n \\in \\mathbb{N}^*$, we conclude $r \\mid a_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of integers $(x, y, z)$ such that\n$$\nx^2 + y^2 + z^2 = 16(x + y + z).\n$$",
    "options": [],
    "answer": "(0,0,0), (0,0,16), (0,16,0), (16,0,0), (0,16,16), (16,0,16), (16,16,0), (16,16,16)",
    "solution": "Write the relation as $(x-8)^2 + (y-8)^2 + (z-8)^2 = 192$ and recall that a square gives the remainder $0$ or $1$ when divided at $4$ to infer that $x-8$, $y-8$, $z-8$ are all even. Set $x-8 = 2a_1$, $y-8 = 2b_1$, $z-8 = 2c_1$ to get $a_1^2 + b_1^2 + c_1^2 = 48$. Repeat the above argument to write $a_1 = 2a_2$, $b_1 = 2b_2$, $c_1 = 2c_2$ with $a_2^2 + b_2^2 + c_2^2 = 12$, then $a_2 = 2a_3$, $b_2 = 2b_3$, $c_2 = 2c_3$ with $a_3^2 + b_3^2 + c_3^2 = 3$.\n\nThe latter equality holds for $a_3, b_3, c_3 = \\pm 1$, that is whenever $x-8$, $y-8$ and $z-8$ are equal to $8$ or $-8$. Consequently, the triples are $(0, 0, 0)$, $(0, 0, 16)$, $(0, 16, 0)$, $(16, 0, 0)$, $(0, 16, 16)$, $(16, 0, 16)$, $(16, 16, 0)$ and $(16, 16, 16)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{R}$ and $z \\in \\mathbb{C} \\setminus \\mathbb{R}$ such that $|a - b| = |a + b - 2z|$.\n\na. Prove that there exists an unique real number $x$ which satisfies $|z - a|^x + |\\bar{z} - b|^x = |a - b|^x$.\n\nb. Find all real numbers $x$ such that $|z - a|^x + |\\bar{z} - b|^x \\le |a - b|^x$.",
    "options": [],
    "answer": "a: x = 2; b: [2, +infty)",
    "solution": "a. Set $u = z - a$, $v = z - b$. The relation gives $|v - u| = |u + v|$ where $u, v, u + v \\in \\mathbb{C} \\setminus \\mathbb{R}$, so $u, v, u + v \\neq 0$. Thus $|u + v|^2 = |u|^2 + |v|^2$. Since $|v| = |\\bar{v}|$, the equation is written successively $|u|^x + |v|^x = (\\sqrt{|u|^2 + |v|^2})^x$ and then $\\left(\\frac{|u|}{\\sqrt{|u|^2+|v|^2}}\\right)^x + \\left(\\frac{|v|}{\\sqrt{|u|^2+|v|^2}}\\right)^x = 1$.\n\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = \\left( \\frac{|u|}{\\sqrt{|u|^2+|v|^2}} \\right)^x + \\left( \\frac{|v|}{\\sqrt{|u|^2+|v|^2}} \\right)^x$ is strictly decreasing, hence $x = 2$ is the only solution.\n\nb. The solution is $[2, +\\infty)$, for $f$ is strictly decreasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve $2^{\\sin^4 x - \\cos^2 x} - 2^{\\cos^4 x - \\sin^2 x} = \\cos 2x$.",
    "options": [],
    "answer": "x = (2k+1)π/4, k ∈ ℤ",
    "solution": "The equation becomes $2^{\\sin^4 x + \\sin^2 x} - 2^{\\cos^4 x + \\cos^2 x} = 2 \\cos 2x$. Because $\\cos^4 x + \\cos^2 x - \\sin^4 x - \\sin^2 x = (\\cos^2 x + \\sin^2 x)(\\cos^2 x - \\sin^2 x) + \\cos^2 x - \\sin^2 x$, we can write $2^{\\sin^4 x+\\sin^2 x} + \\sin^4 x + \\sin^2 x = 2^{\\cos^4 x+\\cos^2 x} + \\cos^4 x + \\cos^2 x$.\n\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(t) = 2^t + t$ is strictly increasing, thus one-to-one, so we conclude $\\sin^4 x + \\sin^2 x = \\cos^4 x + \\cos^2 x$, which is equivalent to $2 \\cos 2x = 0$. The roots are described by $x \\in \\{(2k + 1)\\pi/4 \\mid k \\in \\mathbb{Z}\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation\n$$\n\\frac{1}{\\sqrt{x} + \\sqrt{1006}} + \\frac{1}{\\sqrt{2012 - x} + \\sqrt{1006}} = \\frac{2}{\\sqrt{x} + \\sqrt{2012 - x}}\n$$\nhas 2013 integer solutions.",
    "options": [],
    "answer": "2013",
    "solution": "One can easily check that the given relation holds for any admissible value of $x$. Since the number $x$ is subject to the conditions $0 \\le x \\le 2012$, the conclusion is easily reached.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be an increasing bounded sequence of real numbers. Evaluate\n$$\n\\lim_{n \\to \\infty} (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \\cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1).\n$$",
    "options": [],
    "answer": "0",
    "solution": "The limit is equal to $0$.\nLet $x_n = (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \\cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1)$. The sequence $(a_n)_{n \\ge 1}$ is convergent; let $L = \\lim_{n \\to \\infty} a_n$. Since $a_n \\le L$ for all $n \\ge 1$ we have $2a_n - a_k - a_{k+1} \\le 2L - a_k - a_{k+1} \\le 2(L - a_k)$, $k = 1, 2, \\dots, n-2$, so $0 \\le x_n \\le 2^{n-1}(L - a_1)(L - a_2) \\cdots (L - a_{n-2})(L - a_1) = y_n$.\nSuppose $y_n > 0$, otherwise $y_n = 0$ and then $x_n = 0$. Since $\\frac{y_{n+1}}{y_n} = 2(L - a_n) \\to 0$, we infer that $\\lim_{n \\to \\infty} y_n = 0$ and finally $\\lim_{n \\to \\infty} x_n = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37899,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, the bisector $AD$ and the median $BE$ meet at $P$. The straight lines $AB$ and $CP$ meet at $F$. The parallel through $B$ at $CF$ meets the straight line $DF$ at $M$. Show that $DM = BF$.\n\nGheorghe Bumbăcea",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Ceva's Theorem implies\n$$\n\\frac{BF}{FA} \\cdot \\frac{AE}{EC} \\cdot \\frac{CD}{DB} = 1,\n$$\nhence $\\frac{BF}{FA} = \\frac{BD}{DC}$. The converse of Thales' Theorem yields $DF \\parallel AC$. From $\\triangle BFD \\sim \\triangle BAC$ follows $\\frac{BF}{BA} = \\frac{FD}{AC}$, whence $BF = \\frac{AB}{AC} \\cdot FD$. Now $\\triangle BDM \\sim \\triangle CDF$ implies $\\frac{DM}{FD} = \\frac{BD}{DC}$. On the other hand, $BD$ being a bisector, $\\frac{BD}{DC} = \\frac{AB}{AC}$, so $DM = \\frac{AB}{AC} \\cdot FD$. Finally, $DM = BF$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a non-invertible square matrix of order $n$ with real entries, $n \\ge 2$, and let $A^*$ be the adjoint of $A$. Prove that $\\text{tr}(A^*) \\ne -1$ if and only if the matrix $I_n + A^*$ is invertible.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $A$ is non-singular, we get $\\text{rank}(A) \\le n - 1$. Distinguish two cases:\n\ni) $\\text{rank}(A) \\le n - 2$. Then $A^* = O_n$ and the conclusion follows immediately.\n\nii) $\\text{rank}(A) = n-1$. Then $AA^* = O_n$ and by Sylvester's inequality $0 \\ge \\text{rank}(A) + \\text{rank}(A^*) - n$, that is $\\text{rank}(A^*) \\le 1$.\nIt follows that $A = CL$ where $L \\in \\mathcal{M}_{n,1}(\\mathbb{R})$ and $C \\in \\mathcal{M}_{1,n}(\\mathbb{R})$. It follows that $LC = (a) \\in \\mathcal{M}_1(\\mathbb{R})$, with $a = \\text{tr}(A^*)$ and $(A^*)^2 = CLCL = C(a)L = aA^*$. Denote by $B = I_n + A^*$. Then $(B - I_n)^2 = a(B - I_n)$ or, equivalently, $B((a+2)I_n - B) = (a+1)I_n$, which, in turn, implies $a \\ne -1$, that is $B$ is invertible.\n\nMoreover, if $B$ is invertible but $a = -1$, then from $B(I_n - B) = O_n$ we deduce $B = I_n$. Thus $A^* = O_n$ implying $-1 = \\text{tr}(A^*) = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d > 0$ satisfying $abcd = 1$. Prove that\n$$\n\\frac{1}{a+b+2} + \\frac{1}{b+c+2} + \\frac{1}{c+d+2} + \\frac{1}{d+a+2} \\le 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\frac{1}{a+b+2} + \\frac{1}{c+d+2} \\le \\frac{1}{2\\sqrt{ab}+2} + \\frac{1}{2\\sqrt{cd}+2}$. Denoting $\\sqrt{ab} = x$, we get $\\sqrt{cd} = \\frac{1}{x}$, and the sum on the right hand side in the inequality above is $\\frac{1}{2} \\left( \\frac{1}{x+1} + \\frac{x}{1+x} \\right) = \\frac{1}{2}$.\n\nProceeding similarly with the two other terms of the sum, we obtain the desired inequality. Equality holds when $a = b = c = d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be two real polynomials so that $\\lfloor P(x) \\rfloor = \\lfloor Q(x) \\rfloor$, for every $x \\in \\mathbb{R}$. Prove that $P = Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P(x)$ and $Q(x)$ be real polynomials such that $\\lfloor P(x) \\rfloor = \\lfloor Q(x) \\rfloor$ for all $x \\in \\mathbb{R}$.\n\nSuppose $P(x) \\ne Q(x)$. Then $P(x) - Q(x)$ is a nonzero polynomial, so $P(x) - Q(x)$ is not identically zero. Thus, there exists $x_0 \\in \\mathbb{R}$ such that $P(x_0) \\ne Q(x_0)$.\n\nLet $d = P(x_0) - Q(x_0)$. Then $d \\ne 0$.\n\nConsider the function $f(x) = P(x) - Q(x)$. Since $f(x)$ is a nonzero polynomial, it is continuous and unbounded (unless it is constant, in which case $d \\ne 0$ everywhere).\n\nLet $\\varepsilon = |d|/2 > 0$. By continuity, there exists $\\delta > 0$ such that for all $x$ with $|x - x_0| < \\delta$, $|P(x) - Q(x) - d| < \\varepsilon$, i.e., $|P(x) - Q(x)| > |d|/2$ in a neighborhood of $x_0$.\n\nNow, since $\\lfloor P(x) \\rfloor = \\lfloor Q(x) \\rfloor$ for all $x$, it must be that $P(x)$ and $Q(x)$ are always in the same unit interval, i.e., $P(x) - Q(x) \\in (-1, 1)$ for all $x$.\n\nBut we have found a neighborhood of $x_0$ where $|P(x) - Q(x)| > |d|/2$. If $|d| \\ge 1$, then $|P(x) - Q(x)| > 1/2$ in a neighborhood, and for some $x$ it will exceed $1$, contradicting $P(x) - Q(x) \\in (-1, 1)$ for all $x$.\n\nIf $0 < |d| < 1$, then $P(x_0)$ and $Q(x_0)$ are in the same unit interval, but since $P(x) - Q(x)$ is a nonzero polynomial, it is not constant, so for large $x$, $|P(x) - Q(x)|$ will exceed $1$, again contradicting $P(x) - Q(x) \\in (-1, 1)$ for all $x$.\n\nTherefore, $P(x) - Q(x) = 0$ for all $x$, i.e., $P = Q$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, let $a_n$, $b_n$, $c_n$ be integer numbers such that $(\\sqrt[3]{2}-1)^n = a_n + b_n\\sqrt[3]{2} + c_n\\sqrt[3]{4}$. Show that $c_n \\equiv 1 \\pmod{3}$ if and only if $n \\equiv 2 \\pmod{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The binomial expansion of $(\\sqrt[3]{2}-1)^n$ yields\n$$\nc_n = \\sum_{k \\equiv 2 \\pmod{3}} (-1)^{n-k} \\cdot 2^{(k-2)/3} \\binom{n}{k} \\equiv (-1)^n \\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} \\pmod{3}.\n$$\nSince\n$$\n\\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} = \\frac{1}{3}((1+1)^n + \\varepsilon(1+\\varepsilon)^n + \\varepsilon^2(1+\\varepsilon^2)^n) = \\frac{1}{3}\\left(2^n + 2\\cos(n+2)\\frac{\\pi}{3}\\right),\n$$\nwhere $1 + \\varepsilon + \\varepsilon^2 = 0$, the condition $n \\equiv 2 \\pmod{3}$ may be restated as\n$$\n3c_n = (-1)^n \\left(2^n + 2\\cos(n+2)\\frac{\\pi}{3}\\right) \\equiv 3 \\pmod{9}.\n$$\nConsideration of $n$ modulo 6 yields $3c_n \\equiv 3 \\pmod{9}$ if $n \\equiv 2 \\text{ or } 5 \\pmod{6}$, and $3c_n \\equiv 0 \\pmod{9}$ otherwise. The conclusion follows.\nConsider the polynomial $f = (X-1)^n - c_nX^2 - b_nX - a_n \\in \\mathbb{Z}[X]$. Clearly, $f(\\sqrt[3]{2}) = 0$. Since $X^3 - 2$ is irreducible in $\\mathbb{Z}[X]$, it follows that $X^3 - 2$ divides $f$ in $\\mathbb{Z}[X]$, so $g_n = a_n + b_nX + c_nX^2$ is the remainder of the division of\n\n$(X-1)^n$ by $X^3-2$ in $\\mathbb{Z}[X]$. Write $n = 3q+r$, where $q$ is a non-negative integer and $r \\in \\{0,1,2\\}$, to get $(X-1)^n = (X^3-1)^q(X-1)^r = (X^3-2) \\cdot g+(X-1)^r$ in $\\mathbb{Z}_3[X]$, and deduce thereby that $g_n = (X-1)^r$ in $\\mathbb{Z}_3[X]$. Consequently, $c_n \\equiv 0 \\pmod{3}$ if $r \\in \\{0,1\\}$, and $c_n \\equiv 1 \\pmod{3}$ if $r=2$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Evaluate $\\lim_{n \\to \\infty} \\int_0^1 e^{x^n} dx$.",
    "options": [],
    "answer": "1",
    "solution": "Since $e^{x^n} \\ge 1$ for all $x \\in [0, 1]$ we get $\\int_0^1 e^{x^n} dx \\ge 1$. Notice that $e^t \\le 1 + 3t$ for all $t \\in [0, 1]$ to obtain\n$$\n\\int_{0}^{1} e^{x^{n}} dx \\le \\int_{0}^{1} (1 + 3x^{n}) dx = 1 + \\frac{3}{n + 1}\n$$\nConsequently, the limit is equal to 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdottak az $a, b \\in \\mathbb{R}$ és $z \\in \\mathbb{C} \\setminus \\mathbb{R}$ számok úgy, hogy fennálljon az $|a-b|=|a+b-2z|$ egyenlőség.\n\na)\nIgazold, hogy a $|z-a|^{x} + |\\bar{z}-b|^{x} = |a-b|^{x}$ egyenletnek egy és csak egy $x \\in \\mathbb{R}$ megoldása van!\n\nb)\nHatározd meg azokat az $x \\in \\mathbb{R}$ számokat, amelyekre fennáll az\n$$\n|z-a|^{x} + |\\bar{z}-b|^{x} \\leq |a-b|^{x}\n$$\negyenlőtlenség!",
    "options": [],
    "answer": "a) x = 2. b) All real x with x ≥ 2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAz $\\left(a_{n}\\right)_{n \\geq 1}$ sorozat növekvő és korlátos. Számítsd ki a\n$$\n\\lim_{n \\rightarrow \\infty} \\left(2 a_{n} - a_{1} - a_{2}\\right) \\left(2 a_{n} - a_{2} - a_{3}\\right) \\cdots \\left(2 a_{n} - a_{n-2} - a_{n-1}\\right) \\left(2 a_{n} - a_{n-1} - a_{1}\\right)\n$$\nhatárértéket!",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSzámítsd ki a $\\lim_{n \\rightarrow \\infty} \\int_{0}^{1} e^{x^{n}} \\, dx$ határértéket!",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Determine all positive integers $p$ for which there exist positive integers $x_1 < x_2 < \\dots < x_n$ such that\n$$\n\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} = p.\n$$",
    "options": [],
    "answer": "{1, 2, ..., n}",
    "solution": "Call good a number $p$ for which there exist positive integers $x_1 < x_2 < \\dots < x_n$ such that $\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} = p$.\nSince $x_1, x_2, \\dots, x_n$ are integers and $x_1 < x_2 < \\dots < x_n$, we have $x_k \\ge k$, so $\\frac{k}{x_k} \\le 1$, for all $k = 1, 2, \\dots, n$. Then $\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} \\le n$, so every good number, if there exists any, is between 1 and $n$.\n\nNext, we will show that any integer $p \\in \\{1, 2, \\dots, n\\}$ is good. Obviously, $n$ is good (for $x_k = k$) and 1 is also good (take $x_k = kn$). For $2 \\le p \\le n-1$, we write:\n$$\n\\sum_{k=1}^{n} \\frac{k}{x_k} = \\left( \\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{p-1}{x_{p-1}} \\right) + \\left( \\frac{p}{x_p} + \\dots + \\frac{n}{x_n} \\right),\n$$\nso it is enough to choose $x_1, x_2, \\dots, x_n$ such that the first sum is equal to $p-1$, and the second sum is equal to 1. We can do that by setting $x_k = k$, for $k = 1, 2, \\dots, p-1$ and $x_k = k(n-p+1)$, for $p \\le k \\le n$. Notice that $x_1 < x_2 < \\dots < x_n$ in all cases, so the good numbers are indeed 1, 2, \\dots, $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $T$ the center of the face $CDHG$ of the cube $ABCDEFGH$ and consider the points $N \\in (AD)$, $U \\in (FG)$ so that $\\frac{AN}{AD} = \\frac{FU}{FG} = \\frac{1}{3}$.\nLet $P \\in (CN)$, $R \\in (CU)$ and $Q \\in (BE)$ so that $\\frac{NP}{NC} = \\frac{UR}{UC} = \\frac{BQ}{BE} = \\frac{1}{4}$ and $S \\in (EH)$ so that $\\frac{SE}{EH} = \\frac{3}{16}$.\nProve that $SPQRT$ is a regular pyramid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\gamma$ be a circle, and let $P$ be a point in its plane, not situated on $\\gamma$. Two variable lines $\\ell$ and $\\ell'$ through $P$ meet $\\gamma$ at $X$ and $Y$, and $X'$ and $Y'$, respectively. Show that the line through the centres of the circles $PXY'$ and $PX'Y$ passes through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the circles $PXY'$ and $PX'Y$ meet again at $Q$. A suitable inversion of pole $P$ sends the circles $PXY'$ and $PX'Y$ onto the lines $XY'$ and $X'Y$, respectively, while leaving $\\gamma$ invariant. The image of $Q$ under this inversion is the point $R$ where the lines $XY'$ and $X'Y$ meet. Upon inversion, the locus of $R$ — the polar of $P$ relative to $\\gamma$ — transforms into the circle on diameter $OP$, where $O$ is the centre of $\\gamma$. Consequently, the lines $OQ$ and $PQ$ are perpendicular, so the line through the centres of the circles $PXY'$ and $PX'Y$, which is the perpendicular bisector of the segment $PQ$, passes through the midpoint of the segment $OP$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a < b < c$ so that $a^3 + b^3 + c^3 = (a + b + c)^2$.",
    "options": [],
    "answer": "(1, 2, 3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest integer $r$ satisfying the following condition: Amongst every five 500-element subsets of the set $\\{1, 2, \\dots, 1000\\}$, there exist two sharing at least $r$ elements.",
    "options": [],
    "answer": "200",
    "solution": "The required integer is $r = 200$. To prove it, we first show that, amongst every five 500-element subsets of $\\{1, 2, \\dots, 1000\\}$, there are two sharing at least 200 elements, and then provide an example of five such every two of which share exactly 200 elements.\n\nThe first part is a special case of the lemma below.\n\n**Lemma.** Let $n$ be an integer greater than 1, let $S_1, \\dots, S_n$ be subsets of a finite set $S$ such that $|S_1| + \\dots + |S_n| \\ge |S|$, and let $N$ be one of the closest integers to\n$$\n\\sigma = \\frac{|S_1| + \\dots + |S_n|}{|S|} - \\frac{1}{2}.\n$$\nThen $|S_i \\cap S_j| \\ge \\frac{N(2\\sigma - N)}{n(n-1)} |S|$ for some distinct indices $i$ and $j$.\n\n**Proof of the lemma.** Let $\\chi_i : S \\to \\mathbb{Z}$ be the characteristic function of the set $S_i$: $\\chi_i(x) = 1$ if $x \\in S_i$, and $\\chi_i(x) = 0$ otherwise; clearly, $\\sum_{x \\in S} \\chi_i(x) = |S_i|$. Let further $\\chi : S \\to \\mathbb{Z}$, $\\chi = \\chi_1 + \\dots + \\chi_n$, so $\\chi(x)$ is precisely the number of $S_i$ containing $x$, and $\\sum_{x \\in S} \\chi(x) = \\sum_{i=1}^n |S_i|$.\n\nNext, fix an element $x$ of $S$ and write\n$$\n\\begin{aligned}\n(\\chi(x))^2 &= \\left(\\sum_{i=1}^{n} \\chi_i(x)\\right)^2 = \\sum_{i=1}^{n} (\\chi_i(x))^2 + 2 \\sum_{1 \\le i < j \\le n} \\chi_i(x)\\chi_j(x) \\\\\n&= \\sum_{i=1}^{n} \\chi_i(x) + 2 \\sum_{1 \\le i < j \\le n} \\chi_i(x)\\chi_j(x) = \\chi(x) + 2 \\sum_{1 \\le i < j \\le n} \\chi_i(x)\\chi_j(x),\n\\end{aligned}\n$$\nto get $\\sum_{1 \\le i < j \\le n} \\chi_i(x)\\chi_j(x) = \\frac{1}{2}\\chi(x)(\\chi(x) - 1) = \\frac{1}{2}(\\chi(x) - k)(\\chi(x) - k - 1) + k\\left(\\chi(x) - \\frac{k+1}{2}\\right)$, for any integer $k$. Since $\\chi$ is integer-valued, the first summand in the last expression above is non-negative, so\n$$\n\\sum_{1 \\le i < j \\le n} \\chi_i(x) \\chi_j(x) \\ge k \\left( \\chi(x) - \\frac{k+1}{2} \\right), \\quad k \\in \\mathbb{Z}.\n$$\nConsequently,\n$$\n\\begin{aligned}\n\\sum_{1 \\le i < j \\le n} |S_i \\cap S_j| &= \\sum_{1 \\le i < j \\le n} \\sum_{x \\in S} \\chi_i(x) \\chi_j(x) = \\sum_{x \\in S} \\sum_{1 \\le i < j \\le n} \\chi_i(x) \\chi_j(x) \\\\\n&\\ge \\sum_{x \\in S} k \\left( \\chi(x) - \\frac{k+1}{2} \\right) = k \\left( \\sigma - \\frac{k}{2} \\right) |S|, \\quad k \\in \\mathbb{Z}.\n\\end{aligned}\n$$\nSince $k \\mapsto k(\\sigma - k/2)$, $k \\in \\mathbb{Z}$, achieves its maximum at $k = N$, the conclusion follows by considering an intersection $S_i \\cap S_j$, $i < j$, of maximal cardinality.\n\nNotice that equality in the last inequality above forces each element of $S$ to be covered by exactly $N$ or exactly $N \\pm 1$ of the $S_i$. This remark turns out to be quite useful in constructing the desired example.\n\nFinally, to provide an example of five 500-element subsets of $\\{1, 2, \\dots, 1000\\}$ every two of which share exactly 200 elements, consider first the following five 5-element subsets of $\\{1, 2, \\dots, 10\\}$:\n$$\n\\begin{aligned}\nA_1 &= \\{1, 2, 3, 6, 8\\}, & A_2 &= \\{1, 2, 5, 7, 10\\}, & A_3 &= \\{1, 4, 5, 6, 9\\}, \\\\\nA_4 &= \\{2, 3, 4, 7, 9\\}, & A_5 &= \\{3, 4, 5, 8, 10\\}.\n\\end{aligned}\n$$\nSince every two $A$'s share exactly two elements, the sets $\\bigcup_{k=0}^{99} (10k + A_i)$, $i = 1, 2, \\dots, 5$, provide the desired example.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ and $g$ be two polynomials with integral coefficients such that $\\deg f > \\deg g$ and $\\deg f \\ge 2$. If the polynomial $pf+g$ has a rational root for infinitely many primes $p$, prove that $f$ has a rational root.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\deg f > \\deg g$, $|g(z)/f(z)| < 1$ for all complex numbers $z$ of large enough absolute value. Consequently, as $p$ runs through the infinite set of primes under consideration, the roots of the $pf + g$ all lie in some disc $|z| < R$, where $R$ does not depend on $p$; for if $|z|$ is large enough, then\n$$\n|pf(z) + g(z)| \\ge |f(z)|(p - |g(z)/f(z)|) > 0.\n$$\nFor each prime $p$ such that $pf + g$ has a rational root, by Gauss' lemma, $pf + g$ is the product of two integral polynomials, one of degree 1 and the other of degree $\\deg f - 1$. Now the condition that $p$ be prime comes in to imply that the leading coefficient of one of these factors is a divisor of the leading coefficient of $f$.\nFor each such prime $p$, it is therefore possible to choose an integral factor $h_p$ of $pf+g$, $\\deg h_p = 1$ or $\\deg h_p = \\deg f - 1$, whose leading coefficient divides the leading coefficient of $f$. Hence the leading coefficients of the $h_p$ form a bounded set. Since the roots of the $h_p$ all lie in the same disc, $|z| < R$, Vieta's relations imply that the coefficients of the $h_p$ all form a bounded set; and since they are all integral, this set is finite, so $h_p = h$ for infinitely many of these primes $p$.\nFinally, if $p$ and $q$ are two such, then $h$ is a divisor of $(p-q)f$. Since $\\deg h_p = 1$ or $\\deg h_p = \\deg f - 1$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a point $O$ in the plane and an integer $n \\ge 3$. Consider a finite set $\\mathcal{D}$ of closed unit discs in the plane such that:\n\n(a) No disc in $\\mathcal{D}$ contains the point $O$; and\n\n(b) For each positive integer $k < n$, the closed disc of radius $k+1$ centred at $O$ contains the centres of at least $k$ discs in $\\mathcal{D}$.\n\nShow that some line through $O$ stabs at least $\\frac{2}{\\pi} \\log \\frac{n+1}{2}$ discs in $\\mathcal{D}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each disc $D$ in $\\mathcal{D}$, let $\\omega_D$ denote the centre of $D$, and let $\\alpha_D$ be the arc-length of the image of $D$ under radial projection from $O$ onto the unit circle centred at $O$. Clearly, $\\alpha_D/2 > \\sin(\\alpha_D/2) = 1/O\\omega_D$.\n\nNow, for each positive integer $k < n$, let $\\mathcal{D}_k$ be the set of those discs in $\\mathcal{D}$ whose centres lie in the closed disc of radius $k+1$ centred at $O$. Since $\\mathcal{D}_i \\subseteq \\mathcal{D}_j$ if $i \\le j$, and each $\\mathcal{D}_k$ contains at least $k$ elements, we may recursively choose (or apply Hall's marriage theorem to produce) a system of distinct representatives, $D_1, \\dots, D_{n-1}$, for the collection $\\mathcal{D}_1, \\dots, \\mathcal{D}_{n-1}$, to obtain\n$$\n\\sum_{D \\in \\mathcal{D}_{n-1}} \\alpha_D > 2 \\sum_{D \\in \\mathcal{D}_{n-1}} 1/O\\omega_D \\ge 2 \\sum_{k=1}^{n-1} 1/O\\omega_{D_k} \\ge 2 \\sum_{k=1}^{n-1} 1/(k+1) > 2 \\log \\frac{n+1}{2}.\n$$\nFinally, if $N_{n-1}$ is the maximal number of discs in $\\mathcal{D}_{n-1}$ stabbed by a line through $O$ as it performs a half-turn about $O$, then $\\pi N_{n-1} \\ge \\sum_{D \\in \\mathcal{D}_{n-1}} \\alpha_D$ and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$ and let $S$ be the set of $n$-element subsets of the set $\\{1, 2, \\dots, 2n\\}$. Determine\n$$\n\\max_{S \\in \\mathcal{S}} \\min_{x, y \\in S, x \\neq y} [x, y],\n$$\nwhere $[x, y]$ denotes the least common multiple of the integers $x$ and $y$.",
    "options": [],
    "answer": "6(floor(n/2) + 1), except when n = 4 in which case it is 24",
    "solution": "The required value is $6(\\lfloor n/2 \\rfloor + 1)$, unless $n=4$ in which case it is $24$. Let $S$ be a member of $\\mathcal{S}$. We first show that\n$$\n\\min_{x, y \\in S, x \\neq y} [x, y] \\le 6(\\lfloor n/2 \\rfloor + 1), \\quad (*)\n$$\nunless $n = 4$. To this end, for each $x$ in $S$, choose a positive integer $m_x$ such that $n < m_x x \\le 2n$ and consider the set $S' = \\{m_x x : x \\in S\\}$.\nIf $|S'| < n$, then $m_x x = m_y y$ for some distinct elements $x$ and $y$ in $S$, so $[x, y] \\le 2n$.\nIf $|S'| = n$, then $S' = \\{n+1, n+2, \\dots, 2n\\}$. The first even number in $S'$ is $2(\\lfloor n/2 \\rfloor + 1)$, and the number $3(\\lfloor n/2 \\rfloor + 1)$ is also in $S'$ if $n=3$ or $n \\ge 5$. Consequently, $(*)$ holds for $n=3$ or $n \\ge 5$, and it clearly holds for $n=2$.\nIf $n=4$, then\n$$\n\\min \\{[x, y] : x, y \\in \\{5, 6, 7, 8\\}, x \\neq y\\} = 24,\n$$\nwhich is the required value by the preceding.\nFinally, we show that, if $1 \\le i < j \\le n$, then $[n+i, n+j] \\ge 6(\\lfloor n/2 \\rfloor + 1)$. Suppose, if possible, that $[n+i, n+j] < 6(\\lfloor n/2 \\rfloor + 1)$. Since $[n+1, n+2] = (n+1)(n+2) \\ge 6(\\lfloor n/2 \\rfloor + 1)$, it follows that $j \\ge 3$, so $n+j \\ge 2(\\lfloor n/2 \\rfloor + 1)$. Hence $[n+i, n+j] = 2(n+j) = m(n+i)$, where $m$ is an integer greater than $2$. If $m=3$, then $n+i$ must be an even number less than $2(\\lfloor n/2 \\rfloor + 1)$ which is impossible. If $m \\ge 4$, then $n+i < 3(\\lfloor n/2 \\rfloor + 1)/2 \\le n+1$ which is again impossible. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, determine all non-constant polynomials $f$ with complex coefficients satisfying the condition $1 + f(X^n + 1) = (f(X))^n$.",
    "options": [],
    "answer": "For even n, there are no non-constant solutions. For odd n, all solutions are f(X) = −T_k(X), where T_0(X) = X and T_{k+1}(X) = (T_k(X))^n + 1 for k ≥ 0.",
    "solution": "It is readily checked that the polynomials in the above sequence all satisfy the condition in the statement.\n\nConversely, let $f$ be a polynomial with complex coefficients satisfying the condition\n$$\n1 + f(X^n + 1) = (f(X))^n. \\qquad (1)\n$$\nTo begin, we show that, if $f(0) = 0$, then $f = -X$ and $n$ must be odd. To prove this, consider the sequence defined by $x_0 = 0$ and $x_{k+1} = x_k^n + 1$, $k \\ge 0$. Clearly, $f(x_{k+1}) = (f(x_k))^n - 1$, $k \\ge 0$, and $f(x_1) = -1$.\n\nIf $n$ is odd, induct on $k$ to prove that $f(x_k) = -x_k$, $k \\ge 0$. This is clearly true if $k = 0, 1, 2$. For the induction step, use (1) to get $f(x_{k+1}) = (-x_k)^n - 1 = -(x_k^n+1) = -x_{k+1}$. With reference again to the monotonicity of the $x_k$, we conclude that $f = -X$.\n\nFinally, consider the case $f(0) \\ne 0$. Let $\\omega$ be a primitive $n$-th root of unity and use (1) to deduce that $(f(X))^n = (f(\\omega X))^n$, so $f(X) = \\omega^m f(\\omega X)$ for some non-negative integer $m < n$. Since $f(0) \\ne 0$, identification of the constant terms yields $\\omega^m = 1$, so $m = 0$, for $\\omega$ is primitive. Hence $f(X) = f(\\omega X)$ and identification of coefficients shows that $f(X)$ is a polynomial in $X^n$ with complex coefficients. Alternatively, but equivalently, $f(X) = g(X^n + 1)$ for some polynomial $g$ with complex coefficients. Since $g$ also satisfies (1), the conclusion now follows recursively.\n\nAlternative solution – case $f(0) = 0$.\nUse (1) repeatedly to obtain $f(1) = -1$, $f(2) = (-1)^n - 1$, $f(2^n + 1) = ((-1)^n - 1)^n - 1$, and deduce thereby that\n$$\n|f(2^n + 1)| \\le 2^n + 1. \\quad (2)\n$$\nWe now take time out to show that the roots of $f$ all lie in the disc $|z| < 2$ in the complex plane. To this end, let $\\alpha_0$ be a root of $f$ of maximal absolute value. Since the absolute value of the leading coefficient of $f$ is 1, (1) yields\n$$\n\\prod_{\\alpha \\text{ is a root of } f} |\\alpha_0^n + 1 - \\alpha| = 1. \\quad (3)\n$$\nSuppose, if possible, that $|\\alpha_0| \\ge 2$. If $\\alpha$ is a root of $f$, then\n$$\n|\\alpha_0^n + 1 - \\alpha| \\ge |\\alpha_0|^n - 1 - |\\alpha| \\ge 2|\\alpha_0| - 1 - |\\alpha| = (|\\alpha_0| - 1) + (|\\alpha_0| - |\\alpha|) \\ge |\\alpha_0| - 1 \\ge 1.\n$$\nSince $f(0) = 0$, at least one of the factors of the product in (3) is $|\\alpha_0^n + 1| \\ge |\\alpha_0|^n - 1 \\ge 2^n - 1 \\ge 3$, so the product is at least 3 — in contradiction with (3).\n\nBack to the problem, write (2) in the form\n$$\n\\prod_{\\alpha \\text{ is a root of } f} |2^n + 1 - \\alpha| \\le 2^n + 1. \\quad (2')\n$$\nBy the preceding, if $\\alpha$ is a non-zero root of $f$, then $|2^n + 1 - \\alpha| \\ge 2^n - 1 - |\\alpha| > 2^n - 3 \\ge 1$, so, if the multiplicity of 0 exceeds 1 or $f$ has a non-zero root, then the product in (2') exceeds $2^n + 1$ and we reach a contradiction. Consequently, $f = aX$, where $a$ is a complex number of absolute value 1, and (1) forces $a = -1$ and $n$ odd.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 37917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $x_1, \\dots, x_n$ be positive real numbers. Show that\n$$\n\\begin{aligned}\n\\min (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n) &\\le 2 \\cos \\left(\\frac{\\pi}{n+2}\\right) \\\\\n&\\le \\max (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n).\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Only the first inequality will be proved; the second is dealt with similarly. Suppose, if possible, that each of the $n+1$ positive real numbers $x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n$ is greater than $2 \\cos \\alpha$, where $\\alpha = \\pi/(n+2)$, and show recursively that $x_k > \\sin((k+1)\\alpha) / \\sin(k\\alpha)$, $k = 1, \\dots, n$; notice that $\\sin((k+1)\\alpha) / \\sin(k\\alpha) > 0$, $k = 1, \\dots, n$. By assumption, $x_1 > 2 \\cos \\alpha = \\sin(2\\alpha) / \\sin \\alpha$. For the induction step, $x_{k+1} > 2 \\cos \\alpha - 1/x_k > 2 \\cos \\alpha - \\sin(k\\alpha) / \\sin((k+1)\\alpha) = \\sin((k+2)\\alpha) / \\sin((k+1)\\alpha)$. Consequently, $x_n > \\sin((n+1)\\alpha) / \\sin(n\\alpha) = 1/(2 \\cos \\alpha)$, in contradiction with $1/x_n > 2 \\cos \\alpha$; here, the last equality holds, since $\\alpha = \\pi/(n+2)$.\n\n**Remark.** Notice that the $x_k = \\sin((k+1)\\alpha) / \\sin(k\\alpha)$, $k = 1, \\dots, n$, where $\\alpha = \\pi/(n+2)$, satisfy\n$$\nx_1 = 1/x_1 + x_2 = \\dots = 1/x_{n-1} + x_n = 1/x_n = 2 \\cos \\alpha,\n$$\nto conclude, by the preceding, that\n$$\n\\begin{aligned}\n\\max_{x_1>0, \\dots, x_n>0} \\min (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n) &= \\\\\n\\min_{x_1>0, \\dots, x_n>0} \\max (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n) &= 2 \\cos \\alpha.\n\\end{aligned}\n$$\nWe shall actually prove that\n$$\n\\begin{aligned}\n\\max_{x_1>0, \\dots, x_n>0} \\min (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n) &= \\\\\n\\min_{x_1>0, \\dots, x_n>0} \\max (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n) &= 2 \\cos \\left(\\frac{\\pi}{n+2}\\right).\n\\end{aligned}\n$$\nTo this end, let $U$ denote the set of all $n$-tuples of positive real numbers, and, for $\\mathbf{x} = (x_1, \\dots, x_n)$ in $U$, let\n$$\nm(\\mathbf{x}) = \\min (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n)\n$$\nand\n$$\nM(\\mathbf{x}) = \\max (x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n).\n$$\nThe first step consists in assuming that $m(\\mathbf{a}) = M(\\mathbf{a})$ for some $\\mathbf{a} = (a_1, \\dots, a_n)$ in $U$ and showing that $m(\\mathbf{x}) \\le m(\\mathbf{a}) = M(\\mathbf{a}) \\le M(\\mathbf{x})$ for all $\\mathbf{x}$ in $U$. Clearly, the condition $m(\\mathbf{a}) = M(\\mathbf{a})$ is equivalent to\n$$\na_1 = 1/a_1 + a_2 = \\cdots = 1/a_{n-1} + a_n = 1/a_n. \\tag{1}\n$$\nSuppose, if possible, that $m(\\mathbf{x}) > m(\\mathbf{a})$ for some $\\mathbf{x} = (x_1, \\dots, x_n)$ in $U$. Then $x_1 \\ge m(\\mathbf{x}) > m(\\mathbf{a}) = a_1$; $1/x_k + x_{k+1} \\ge m(\\mathbf{x}) > m(\\mathbf{a}) = 1/a_k + a_{k+1}$, $k = 1, \\dots, n-1$; and $1/x_n \\ge m(\\mathbf{x}) > m(\\mathbf{a}) = 1/a_n$. The first $n$ inequalities imply recursively that $x_k > a_k$, $k = 1, \\dots, n$; in particular, $x_n > a_n$, in contradiction with $1/x_n > 1/a_n$. Consequently, $m(\\mathbf{x}) \\le m(\\mathbf{a})$ for all $\\mathbf{x}$ in $U$. Similarly, $M(\\mathbf{x}) \\ge M(\\mathbf{a})$ for all $\\mathbf{x}$ in $U$.\n\nTo show the existence of an $\\mathbf{a}$ in $U$ satisfying (1), let $a$ denote the common value in (1) and notice that $a_k = b_k / b_{k-1}$, $k = 1, \\dots, n$, where the $b_k$ are defined by\n$$\nb_0 = 1, \\quad b_1 = a, \\quad \\text{and} \\quad b_k = a b_{k-1} - b_{k-2}, \\quad k \\ge 2. \\tag{2}\n$$\nSince $1/a_n = a$, it follows that $b_{n-1} = a b_n$ which is equivalent to $b_{n+1} = 0$. Notice further that $a < 2$. Otherwise, $a_1 = a \\ge 2$ and $a_k = a - 1/a_{k-1}$, $k = 2, \\dots, n$, would recursively imply that $a_k \\ge 1 + 1/k$, $k = 1, \\dots, n$; in particular, $a_n \\ge 1 + 1/n$, in contradiction with $1/a_n = a \\ge 2$. We may therefore write $a = 2 \\cos \\alpha$, for some $\\alpha$ in the open interval $(0, \\pi/2)$, to deduce that the unique solution of (2) is $b_k = \\sin((k+1)\\alpha)/\\sin \\alpha$. Since $b_1, \\dots, b_n$ are all positive, the condition $b_{n+1} = 0$ yields $\\alpha = \\pi/(n+2)$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ be a convex quadrangle and let $\\ell$ be a line through the point of intersection of the diagonals of $K$. Show that the length of the segment of intersection $\\ell \\cap K$ does not exceed the length of (at least) one of the diagonals of $K$.\n\nD. Yu. Grigoriev, Kvant Magazine",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a circular labelling, **A**, **B**, **C**, **D**, of the vertices of the quadrangle, and let the diagonals $AC$ and $BD$ meet at $O$. Without loss of generality, we may (and will) assume that the line $\\ell$ meets the opposite sides $AB$ and $CD$, say, at $X$ and $Y$, respectively. Further, let $\\alpha = \\angle AOX = \\angle COY$ and $\\beta = \\angle BOX = \\angle DOY$, write $\\text{area}[AOB] = \\text{area}[AOX] + \\text{area}[BOX]$ and express\n\n$\\text{area}[AOB] = \\frac{1}{2} \\cdot OA \\cdot OB \\cdot \\sin(\\alpha + \\beta)$, $\\text{area}[AOX] = \\frac{1}{2} \\cdot OA \\cdot OX \\cdot \\sin \\alpha$ and $\\text{area}[BOX] = \\frac{1}{2} \\cdot OB \\cdot OX \\cdot \\sin \\beta$, to get\n\n$$\n\\begin{align*}\nOX &= \\frac{OA \\cdot OB \\cdot \\sin(\\alpha + \\beta)}{OA \\cdot \\sin \\alpha + OB \\cdot \\sin \\beta} \\le \\frac{OA \\cdot OB \\cdot (\\sin \\alpha + \\sin \\beta)}{OA \\cdot \\sin \\alpha + OB \\cdot \\sin \\beta} \\\\ &\\le \\frac{OA \\cdot \\sin \\beta + OB \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta}.\n\\end{align*}\n$$\n\n(The last inequality is easily seen to be equivalent to the obvious inequality $(OA - OB)^2 \\cdot \\sin \\alpha \\cdot \\sin \\beta \\ge 0$.) Similarly, $OY \\le (OC \\cdot \\sin \\beta + OD \\cdot \\sin \\alpha)/(\\sin \\alpha + \\sin \\beta)$, so $XY \\le (AC \\cdot \\sin \\beta + BD \\cdot \\sin \\alpha)/(\\sin \\alpha + \\sin \\beta) \\le \\max(AC, BD)$.\nLet $\\mathcal{L}$ be the pencil of lines through the point of intersection of the diagonals of $K$. We shall prove that the length of each segment of intersection $\\ell \\cap K$, $\\ell \\in \\mathcal{L}$, does not exceed the length of the longest diagonal of $K$.\n\nTo begin, notice that no intersection segment has a length greater than the diameter of $K$, so $\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap K|$ is finite, where $|s|$ denotes the length of the line-segment $s$.\n\nSuppose, if possible, that $\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap K|$ is greater than the length of the longest diagonal of $K$. Let $\\delta$ and $\\delta'$ denote the diagonals of $K$ and consider a line $\\ell_0$ in $\\mathcal{L}$ such that\n\n$$\n|\\ell_0 \\cap K| > \\frac{\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap K| + \\max(|\\delta|, |\\delta'|)}{2} > \\frac{|\\delta| + |\\delta'|}{2}.\n$$\n\nRecall that the length of the internal bisectrix of an angle of a triangle is smaller than the arithmetic mean of the lengths of the sides forming that angle, to infer that $\\ell_0$ does not bisect the corresponding angle formed by the diagonals; say, the angle formed by $\\ell_0$ and $\\delta$ is smaller than the angle formed by $\\ell_0$ and $\\delta'$, both angles being, of course, those in the wedge containing $\\ell_0$.\n\nFinally, reflect the line of support of $\\delta$ in the line $\\ell_0$ to obtain a line $\\ell_1$ in $\\mathcal{L}$ such that\n\n$$\n\\frac{|\\ell_1 \\cap K| + \\max(|\\delta|, |\\delta'|)}{2} \\ge \\frac{|\\ell_1 \\cap K| + |\\delta|}{2} > |\\ell_0 \\cap K| > \\frac{\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap K| + \\max(|\\delta|, |\\delta'|)}{2},\n$$\n\nand thereby reach a contradiction. (The inequality in the middle expresses the above mentioned fact about the length of an internal bisectrix in a triangle.) The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, consider a triangular array with entries $a_{ij}$ where $i$ ranges from $1$ to $n$ and $j$ ranges from $1$ to $n - i + 1$. The entries of the array are all either $0$ or $1$, and, for all $i > 1$ and any associated $j$, $a_{ij}$ is $0$ if $a_{i-1,j} = a_{i-1,j+1}$, and $a_{ij}$ is $1$ otherwise.\nLet $S$ denote the set of binary sequences of length $n$, and define a map $f: S \\to S$ via $f: (a_{11}, a_{12}, \\dots, a_{1n}) \\mapsto (a_{n1}, a_{n-1,2}, \\dots, a_{1n})$. Determine the number of fixed points of $f$.\nGeoffrey Smith, United Kingdom, RMM 2013 Short List",
    "options": [],
    "answer": "2^{floor((n+1)/2)}",
    "solution": "$$(a_{11}, a_{12}, \\dots, a_{1n}) \\mapsto (a_{11}, a_{21}, \\dots, a_{n1})$$\nis bijective, on one hand, and that $(a_{11}, a_{12}, \\dots, a_{1n})$ is fixed by $f$ if and only if $(a_{11}, a_{21}, \\dots, a_{n1})$ is a palindrome $(a_{i1} = a_{n-i+1,1}, i = 1, \\dots, n)$, on the other.\nTo begin with, notice that the definition of the $a_{ij}$ is equivalent to the Pascal-like relation in $\\mathbb{Z}_2$:\n$$\na_{ij} + a_{i-1,j} + a_{i-1,j+1} = 0. \\quad (1)\n$$\nHenceforth, such a triangular array will be called a *Pascal binary* (or *dyadic*) *array*.\nClearly, each binary string **a** in *S* yields a unique Pascal binary array $\\hat{\\mathbf{a}}$.\nFor more convenience, view a triangular array as a function on the standard lattice triangle\n$$\n\\Delta_n = \\{(i, j) : i, j \\ge 1 \\text{ and } 2 \\le i + j \\le n + 1\\},\n$$\nsituated in the first quadrant; thus, the first index runs horizontally and corresponds to columns, and the second index runs vertically and corresponds to rows.\nFurther, use the symbols $\\to$, $\\leftarrow$, $\\uparrow$, $\\downarrow$, $\\searrow$ and $\\nwarrow$ to denote the oriented sides of $\\Delta_n$; explicitly,\n$$\n\\begin{align*} \\to &= \\{(i, 1) : i = 1, \\dots, n\\}, & \\leftarrow &= \\{(n - i + 1, 1) : i = 1, \\dots, n\\}, \\\\ \\uparrow &= \\{(1, j) : j = 1, \\dots, n\\}, & \\downarrow &= \\{(1, n - j + 1) : j = 1, \\dots, n\\}, \\\\ \\searrow &= \\{(i, n - i + 1) : i = 1, \\dots, n\\}, & \\nwarrow &= \\{(n - j + 1, j) : j = 1, \\dots, n\\}. \\end{align*}\n$$\nTo establish a bijection between the fixed points of $f$ and the binary palindromes of length $n$, consider the transformations $\\varrho$ and $\\sigma$ of $\\Delta_n$ defined by\n$$\n\\varrho(i, j) = (n - i - j + 2, i) \\quad \\text{and} \\quad \\sigma(i, j) = (n - i - j + 2, j).\n$$\nThe former is a permutation of order $3$ ($\\varrho^3$ is the identity), and the latter is an involution ($\\sigma^2$ is the identity). It is easily seen that\n$$\n\\varrho(\\rightarrow) = \\nwarrow, \\varrho(\\uparrow) = \\leftarrow, \\varrho(\\nwarrow) = \\downarrow \\quad \\text{and} \\quad \\sigma(\\rightarrow) = \\leftarrow, \\sigma(\\uparrow) = \\nwarrow, \\sigma(\\nwarrow) = \\uparrow,\n$$\nso $\\sigma\\varrho(\\rightarrow) = \\uparrow$, $\\sigma\\varrho(\\uparrow) = \\rightarrow$, and $\\sigma\\varrho(\\nwarrow) = \\nwarrow$. It is also readily checked by (1) that, if $\\hat{\\mathbf{a}} = (a_{ij})$ is a Pascal binary array, then so are both\n$$\n\\varrho\\hat{\\mathbf{a}} = (a_{\\varrho(i,j)}) \\quad \\text{and} \\quad \\sigma\\hat{\\mathbf{a}} = (a_{\\sigma(i,j)}).\n$$\nWe are now in a position to prove the desired results.\nSince $\\sigma\\varrho$ exchanges $\\rightarrow$ and $\\uparrow$, the assignment $\\hat{\\mathbf{a}}_{\\uparrow} \\mapsto \\hat{\\mathbf{a}}_{\\downarrow}$ is bijective.\nSince $\\sigma$ exchanges $\\uparrow$ and $\\nwarrow$, and reverses orientation on the bottom row of $\\Delta_n$, if $\\hat{\\mathbf{a}}_{\\uparrow}$ is fixed by $f$, then $\\hat{\\mathbf{a}}_{\\uparrow} = f(\\hat{\\mathbf{a}}_{\\uparrow}) = \\hat{\\mathbf{a}}_{\\nwarrow} = (\\sigma\\hat{\\mathbf{a}})_{\\uparrow}$, so $\\hat{\\mathbf{a}} = \\sigma\\hat{\\mathbf{a}}$ and consequently $\\hat{\\mathbf{a}}_{\\rightarrow} = (\\sigma\\hat{\\mathbf{a}})_{\\rightarrow} = \\hat{\\mathbf{a}}_{\\leftarrow}$; that is, the bottom row of $\\hat{\\mathbf{a}}$ is a palindrome.\nConversely, if the bottom row of $\\hat{\\mathbf{a}}$ is a palindrome, $\\hat{\\mathbf{a}}_{\\rightarrow} = \\hat{\\mathbf{a}}_{\\leftarrow}$, then $(\\sigma\\varrho\\hat{\\mathbf{a}})_{\\uparrow} = (\\varrho\\hat{\\mathbf{a}})_{\\uparrow}$, so $\\sigma\\varrho\\hat{\\mathbf{a}} = \\varrho\\hat{\\mathbf{a}}$ and consequently, $f(\\hat{\\mathbf{a}}_{\\uparrow}) = \\hat{\\mathbf{a}}_{\\nwarrow} = (\\sigma\\varrho\\hat{\\mathbf{a}})_{\\downarrow} = (\\varrho\\hat{\\mathbf{a}})_{\\downarrow} = \\hat{\\mathbf{a}}_{\\uparrow}$; that is, $\\mathbf{a} = \\hat{\\mathbf{a}}_{\\uparrow}$ is fixed by $f$. This ends the proof.\n\nThe required number is $2^{\\lfloor(n+1)/2\\rfloor}$.\nFor convenience, we denote $b_k = a_{1,n-k}$ and $c_k = a_{k+1,n-k}$ for every $k = 0, 1, \\dots, n-1$. Our aim is to find the set of relations for $(b_k)$ which are equivalent to the relation $(b_k) = (c_k)$. All the calculations will be made in $\\mathbb{Z}_2$.\nThe definition of $a_{ij}$ is equivalent to\n$$\na_{ij} = a_{i-1,j} + a_{i-1,j+1}.\n$$\nA straightforward check shows then that\n$$\na_{ij} = \\sum_{\\ell=0}^{i-1} \\binom{i-1}{\\ell} a_{1,\\ell+j}.\n$$\nFor two nonnegative integers $k$ and $\\ell$, we will write $\\ell \\preceq k$ if the binary representation of $\\ell$ can be obtained from that of $k$ by replacing some ones by zeroes (the leading zeroes are allowed; thus $0 \\preceq k$ and $k \\preceq k$ for every $k$). We write $\\ell \\prec k$ if $\\ell < k$ and $\\ell \\preceq k$. Recall that by Lucas' theorem, $\\binom{k}{\\ell}$ is odd if and only if $\\ell \\preceq k$. Thus,\n$$\nc_k = a_{k+1,n-k} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} a_{1,\\ell+n-k} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} b_{k-\\ell} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} b_{\\ell} = \\sum_{\\ell \\preceq k} b_{\\ell}.\n$$\nNow, the conditions $(b_k) = (c_k)$ rewrite as the set of equations\n$$\n0 = \\sum_{\\ell \\prec k} b_{\\ell} \\qquad (*)_{k}\n$$\nfor all $k = 0, 1, \\dots, n-1$.\nDenote by $T$ the set of all strings $(b_k)$ such that $(*)_k$ are satisfied for all odd $k \\le n-1$. Each string in this set is determined uniquely by the values of $b_{2i-1}$ ($2i < n$) and $b_{n-1}$: the values of $b_{2i}$ (for $2i < n-1$) are found inductively from $(*)_{2i+1}$. Thus\n$$\n|T| = 2^{\\lceil n/2 \\rceil} = 2^{\\lfloor (n+1)/2 \\rfloor}.\n$$\nNow we claim that $T$ is exactly the desired set of fixed points; in fact, we will prove that all the relations $(*)_d$ for even $d$ follow from the relations $(*)_k$ for odd $k$.\nConsider any even $d \\le n-1$. To establish $(*_d)$, we add up all the relations $(*_k)$, where $k \\prec d$, obtaining a sum\n$$\n\\begin{aligned}\n0 &= \\sum_{k \\prec d} \\sum_{\\ell \\prec k+1} b_{\\ell} = \\sum_{k \\prec d} \\left( \\sum_{\\ell \\preceq k} b_{\\ell} + \\sum_{\\ell \\prec k} b_{\\ell+1} \\right) \\\\\n&= \\sum_{\\ell \\prec d} b_{\\ell} \\cdot \\left| \\{k: \\ell \\preceq k \\prec d\\} \\right| + \\sum_{\\ell \\prec d} b_{\\ell+1} \\cdot \\left| \\{k: \\ell \\prec k \\prec d\\} \\right|.\n\\end{aligned}\n$$\nBut one can easily check that $|\\{k: \\ell \\preceq k \\prec d\\}|$ is odd, and $|\\{k: \\ell \\prec k \\prec d\\}|$ is even for all $\\ell \\prec d$. Thus our equality rewrites exactly as $(*_d)$.\n\nTherefore, the number of fixed points is $2^{\\lfloor(n+1)/2\\rfloor}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $(a, b)$ such that the equality $|ax + by| + |bx + ay| = 2|x| + 2|y|$ holds for all reals $x$ and $y$.",
    "options": [],
    "answer": "(2, 0), (0, 2), (-2, 0), (0, -2)",
    "solution": "Set $x = y = 1$ to get $|a + b| = 2$, so $a + b \\in \\{-2, 2\\}$. For $x = 1$, $y = -1$ we get $|a - b| = 2$, hence $a - b \\in \\{-2, 2\\}$. Combining the two above leads to the pairs $(a, b) \\in \\{(2, 0), (0, 2), (-2, 0), (0, -2)\\}$, all of them verifying the given equality for any reals $x$ and $y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37921,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rook starts moving on an infinite chessboard, alternating horizontal and vertical moves. The length of the first move is one square, of the second – two squares, of the third – three squares and so on.\na) Is it possible for the rook to arrive at its starting point after exactly $2013$ moves?\nb) Find all $n$ for which it possible for the rook to come back to its starting point after exactly $n$ moves.",
    "options": [],
    "answer": "a) No. b) All n of the form 8k or 8k−1 for k ≥ 1.",
    "solution": "We will attach integer coordinates to each square of the board, the starting point being $(0,0)$. Then, after each move one coordinate will be of the form $\\pm 1 \\pm 3 \\pm 5 \\pm \\dots$, and the other will be of the form $\\pm 2 \\pm 4 \\pm 6 \\pm \\dots$.\n\na) If $(0,0)$ could be reached in exactly $2013$ moves, then it would be possible to choose the $+$ and $-$ signs so that $\\pm 1 \\pm 3 \\pm 5 \\pm \\dots \\pm 2013 = 0$ and $\\pm 2 \\pm 4 \\pm 6 \\pm \\dots \\pm 2012 = 0$.\nBut, for each combination of signs, $\\pm 1 \\pm 3 \\pm \\dots \\pm 2013$ has the same parity as $1+3+\\dots+2013 = 1007^2$ is odd, so the equality $\\pm 1 \\pm 3 \\pm 5 \\pm \\dots \\pm 2013 = 0$ is impossible. This shows that $(0,0)$ cannot be reached in $2013$ moves.\n\nb) If $(0,0)$ can be reached in $n$ moves, then\n$$\n\\pm 1 \\pm 3 \\pm 5 \\pm \\dots \\pm \\left( 2 \\lfloor \\frac{n+1}{2} \\rfloor - 1 \\right) = 0 \\quad \\text{and} \\quad \\pm 2 \\pm 4 \\pm 6 \\pm \\dots \\pm 2 \\lfloor \\frac{n}{2} \\rfloor = 0\n$$\nRelation above implies that $0$ has the same parity as $1+3+\\dots+(2\\lfloor\\frac{n+1}{2}\\rfloor-1) = \\lfloor\\frac{n+1}{2}\\rfloor^2$, so $\\lfloor\\frac{n+1}{2}\\rfloor$ must be even, that is $4 \\mid n$ or $4 \\mid n+1$.\nFrom the second relation, $0$ has the same parity as $1+2+\\dots+\\lfloor\\frac{n}{2}\\rfloor = \\frac{\\lfloor\\frac{n}{2}\\rfloor(\\lfloor\\frac{n}{2}\\rfloor+1)}{2}$, that is $4 \\mid \\lfloor\\frac{n}{2}\\rfloor$ or $4 \\mid \\lfloor\\frac{n}{2}\\rfloor+1$. This shows that $8$ must divide one of the numbers $n-1, n, n+1$ or $n+2$.\nFrom these, it follows that $n$ must be of the form $8k$ or $8k-1$, $k \\in \\mathbb{N}^*$.\nFor each $k \\in \\mathbb{N}^*$, $(0,0)$ can be reached in $8k$ moves, because $(1-3-5+7)+ (9-11-13+15)+\\dots+((8k-7)-(8k-5)-(8k-3)+(8k-1)) = 0$ and $(2-4-6+8)+(10-12-14+16)+\\dots+((8k-6)-(8k-4)-(8k-2)+8k) = 0$.\n\nIt is also possible to reach $(0,0)$ in $8k - 1$ moves, because $(1 - 3 - 5 + 7) + (9 - 11 - 13 + 15) + \\dots + ((8k - 7) - (8k - 5) - (8k - 3) + (8k - 1)) = 0$ and $(2+4-6) + (8-10-12+14) + \\dots + ((8k-8) - (8k-6) - (8k-4) + (8k-2)) = 0$.\nTherefore, the answer is: all $n = 8k$ or $n = 8k - 1$, $k \\in \\mathbb{N}^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$ and $b$ are complex numbers. Prove that $|az + b\\bar{z}| \\le 1$ for all $z \\in \\mathbb{C}$ with $|z| = 1$ if and only if $|a| + |b| \\le 1$.\nLiviu Vlaicu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $|a| + |b| \\le 1$ and let $z \\in \\mathbb{C}$ with $|z| = 1$. Then $|az + b\\bar{z}| \\le |az| + |b\\bar{z}| = |a| + |b| \\le 1$, as claimed.\n\nConversely, if $a = 0$ or $b = 0$ there is nothing to show. For $a, b \\neq 0$, write $\\frac{b}{a} = r(\\cos\\alpha + i\\sin\\alpha)$. Put $z = \\cos\\frac{\\alpha}{2} + i\\sin\\frac{\\alpha}{2}$ to get $1 \\ge |az + b\\bar{z}| = |a||\\bar{z}|\\left|z^2 + \\frac{b}{a}\\right| = |a|(1+r)(\\cos\\alpha + i\\sin\\alpha) = |a|(1+r) = |a|(1 + \\left|\\frac{b}{a}\\right|) = |a| + |b|$, which ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ and $g$ be two quadratic functions. Prove that, if the functions $\\lfloor f \\rfloor$ and $\\lfloor g \\rfloor$ are equal, then the functions $f$ and $g$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let points $D, E \\in (BC)$, $F, G \\in (CA)$, $H, I \\in (AB)$ such that $BD = CE$, $CF = AG$ and $AH = BI$. Consider $M, N, P$ the midpoint of the segments $GH$, $DI$, $EF$ respectively and let $M'$ be the intersection point of the lines $AM$ and $BC$.\n\na. Show that $\\frac{BM'}{CM'} = \\frac{AG}{AH} \\cdot \\frac{AB}{AC}$.\n\nb. Prove that lines $AM$, $BN$ and $CP$ are concurrent.",
    "options": [],
    "answer": "BM'/CM' = (AG/AH) * (AB/AC), and the lines AM, BN, and CP are concurrent.",
    "solution": "a. Set $m = \\frac{BM'}{CM'}$. We have $\\overrightarrow{AM} = \\frac{1}{2}\\overrightarrow{AH} + \\frac{1}{2}\\overrightarrow{AG} = \\frac{1}{2}\\frac{AH}{AB} \\cdot \\overrightarrow{AB} + \\frac{1}{2}\\frac{AG}{AC} \\cdot \\overrightarrow{AC}$ and $\\overrightarrow{AM'} = \\frac{1}{m+1}\\overrightarrow{AB} + \\frac{m}{m+1}\\overrightarrow{AC}$. Points $A$, $M$, $M'$ are collinear, hence $\\frac{AH}{AB} \\cdot \\frac{m}{m+1} = \\frac{AG}{AC} \\cdot \\frac{1}{m+1}$ and the claim follows.\n\nb. Define similarly the points $N'$, $P'$. Notice that $\\frac{BM'}{CM'} \\cdot \\frac{CN'}{AN'} \\cdot \\frac{AP'}{BP'} = 1$ and apply Ceva's theorem to reach the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37925,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a division of two positive integers, the dividend and the divisor are directly proportional with the remainder and the quotient. The remainder and the quotient are relatively prime. Prove that the dividend is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the dividend be $a$, the divisor be $b$, the quotient be $q$, and the remainder be $r$, so that\n$$\na = bq + r,\n$$\nwith $0 \\leq r < b$.\n\nWe are told that $a$ and $b$ are directly proportional to $r$ and $q$, respectively. That is, there exists a constant $k > 0$ such that\n$$\na = k r, \\quad b = k q.\n$$\n\nSubstitute $a$ and $b$ in the division equation:\n$$\nk r = (k q) q + r \\implies k r = k q^2 + r.\n$$\nBring $r$ terms together:\n$$\nk r - r = k q^2 \\implies r(k - 1) = k q^2.\n$$\nSo\n$$\nr = \\frac{k q^2}{k - 1}.\n$$\n\nBut $r$ must be an integer, so $k - 1$ divides $k q^2$. Since $k$ and $k - 1$ are coprime, $k - 1$ must divide $q^2$.\n\nLet $k - 1 = d$, so $k = d + 1$. Then\n$$\nr = \\frac{(d + 1) q^2}{d} = q^2 + \\frac{q^2}{d}.\n$$\nSo $d$ divides $q^2$, i.e., $q^2 = d m$ for some integer $m$.\n\nThus,\n$$\nr = q^2 + m.\n$$\nBut $\\frac{q^2}{d} = m$, so $d$ divides $q^2$ and $m$ is an integer.\n\nRecall $b = k q = (d + 1) q$ and $a = k r = (d + 1) r$.\n\nNow, $r$ and $q$ are relatively prime. Since $r = q^2 + m$ and $q^2 = d m$, $r = d m + m = m (d + 1) = m k$.\n\nSo $r = m k$, $q^2 = d m$, $k = d + 1$.\n\nBut $r$ and $q$ are relatively prime, so $\\gcd(m k, q) = 1$.\n\nBut $q^2 = d m$, so $q = \\sqrt{d m}$. For $q$ to be integer, $d m$ must be a perfect square. Let $d = t^2$, $m = s^2$, so $q = t s$.\n\nThen $r = m k = s^2 (t^2 + 1)$.\n\nNow, $a = k r = (t^2 + 1) \\cdot s^2 (t^2 + 1) = s^2 (t^2 + 1)^2$.\n\nThus, $a$ is a perfect square.\n\nTherefore, the dividend is a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be an arbitrary function and $g: \\mathbb{R} \\to \\mathbb{R}$ be a quadratic function such that: for any reals $m$ and $n$, the equation $f(x) = mx + n$ has solutions if and only if the equation $g(x) = mx + n$ has solutions. Prove $f = g$.\nVasile Pop",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Remark that if the graph of $h: \\mathbb{R} \\to \\mathbb{R}$, $h(x) = mx + n$ is tangent to the graph of $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = ax^2 + bx + c$, then the equation $ax^2 + bx + c = mx + n$, $a, b, c, m, n \\in \\mathbb{R}$, $a \\neq 0$ has zero discriminant, that is the function $k: \\mathbb{R} \\to \\mathbb{R}$, $k(x) = ax^2 + bx + c - mx - n$ is nul in only one point and has constant sign in all other real points.(*)\n\nSuppose WLOG that $g(x) = ax^2 + bx + c$, with $a > 0$. If there is $x_0$ such that $f(x_0) < g(x_0)$, choose $m$ and $n$ such that the graph of $h(x) = mx + n$ is tangent to the graph of $g$ at point $(x_0, g(x_0))$. Put $n' = n - g(x_0) + f(x_0)$. It follows that the equation $f(x) = mx + n'$ has the root $x_0$ and the equation $g(x) = mx + n'$ has no roots, because by the remark at the beginning $g(x) \\ge mx + n > mx + n'$, for all $x \\in \\mathbb{R}$, a contradiction.\n\nThus $f(x) \\ge g(x)$, for all $x \\in \\mathbb{R}$. Suppose now that $x_0$ satisfies $f(x_0) > g(x_0)$. As before, choose $m$ and $n$ such that the graphs of $h(x) = mx + n$ and $g$ be tangent at point $(x_0, g(x_0))$. By the remark $f(x) > mx + n$, for $x \\in \\mathbb{R}$, in contradiction with the fact that equation $g(x) = mx + n$ has the root $x_0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 37927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(A, +, \\cdot)$ be a unitary ring such that the following are true:\n\n(1) $A$ is not a field;\n(2) for any noninvertible element $x$ of $A$, there is an integer $m \\ge 1$, depending on $x$, such that $x = x^2 + x^3 + \\dots + x^{2^m}$. Prove that:\n\na. $x+x=0$, for any $x \\in A$,\nb. $x^2 = x$, for any noninvertible $1 \\neq x \\in A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. It is sufficient to prove that $1+1=0$. Consider $x$ noninvertible, and let $m \\in \\mathbb{N}^*$, such that $x = x^2 + x^3 + \\dots + x^{2^m}$, and $y = x + x^2 + \\dots + x^{2^m-1}$. Evidently, $xy = x$, so $xy^k = x$, for any $k \\in \\mathbb{N}^*$. Because $x$ is not invertible it follows that $y$ is also, so there is $p \\in \\mathbb{N}^*$, such that $-y = (-y)^2 + (-y)^3 + \\dots + (-y)^{2^p-1} + (-y)^{2^p} = y^2 - y^3 + \\dots - y^{2^p-1} + y^{2^p}$. As a consequence $-x = -xy = xy^2 - xy^3 + \\dots - xy^{2^p-1} + xy^{2^p} = x - x + \\dots - x + x = x$, i.e., $x+x=0$.\n\nConsider $x$ non zero and non-invertible. As $2x = 0$, it follows that $2$ is non-invertible, so $2+2=0$ and $2 = 2^2 + 2^3 + \\dots + 2^{2^m}$, where $m \\in \\mathbb{N}^*$. The equality $2+2=0$ implies $2^2 = 0$, giving $2^k = 0$, for any $k \\ge 2$. So, $2 = 2^2+2^3+\\dots+2^{2^m} = 0$.\n\nb. Let $x$ be a non-invertible element of $A$ and $m \\in \\mathbb{N}^*$, such that $x = x^2 + x^3 + \\dots + x^{2^m}$. Then $x^2 = x^3 + x^4 + \\dots + x^{2^m+1}$. As $1+1=0$, by summing up the two equalities we get $x^{2^m+1} = x$.\n\nEqualities $1+1=0$ and $x^{2^m+1} = x$ imply $(x^2+x)^{2^m} = x^{2^m+1} + x^{2^m} = x^{2^m+1} \\cdot x^{2^m-1} + x^{2^m} = x \\cdot x^{2^m-1} + x^{2^m} = x^{2^m} + x^{2^m} = 0$.\n\nWe will show that $x^2+x=0$, which concludes the proof. Let $y = x^2+x$ and take $k$ the smallest positive integer such that $y^k = 0$ – the existence of $k$ is assured by $y^{2^m} = 0$. In case $k > 1$, the element $y^{k-1}$ is non-invertible in $A$ and there is $n \\in \\mathbb{N}^*$ such that $y^{k-1} = (y^{k-1})^{2^n+1} = y^{(k-1)(2^n+1)}$. Because $(k-1)(2^n+1) \\ge k$, we infer $y^{(k-1)(2^n+1)} = 0$, so $y^{k-1} = 0$ – in contradiction with the minimality of $k$. We conclude $k=1$ and $y=0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37928,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a group $(G, \\cdot)$ has the property $(P)$ if, for any automorphism $f$ of $G$, there exist two automorphisms $g$ and $h$ of $G$ such that $f(x) = g(x) \\cdot h(x)$, for any $x \\in G$. Prove that:\n\na. Any group with the property (P) is commutative.\n\nb. Any finite group of odd order has the property (P).\n\nc. No finite group of order $4n + 2$, $n \\in \\mathbb{N}$, has the property (P).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Suppose $g, h \\in \\text{Aut}(G)$ with $x = g(x)h(x)$ for all $x \\in G$. Let $a$ and $b$ be elements of $G$. Then $a = g(a)h(a)$, $b = g(b)h(b)$ and $ab = g(ab)h(ab)$.\n\nIt follows that $g(a)h(a)g(b)h(b) = ab = g(ab)h(ab) = g(a)g(b)h(a)h(b)$, hence $h(a)g(b) = g(b)h(a)$.\nSince $g$ and $h$ are onto, the claim follows.\n\nb. Let $f \\in \\operatorname{Aut}(G)$. The functions $\\varphi, \\psi: G \\to G, \\varphi(x) = x^2, \\psi(x) = x^{-1}$ are automorphisms, for $G$ is an abelian group of odd order, hence $g = f \\circ \\varphi$ and $h = f \\circ \\psi$ are automorphisms. To end the proof, notice that\n$$\nf(x) = f(x^2x^{-1}) = f(x^2)f(x^{-1}) = g(x)h(x),\n$$\nfor all $x \\in G$.\n\nc. Suppose that $G$ has the property (P) and let $a \\in G$ be an element of order 2.\nWe claim that $a$ is the only element of order 2 in $G$. Suppose there exists $b \\in G$ of order 2, $b \\neq a$. As $G$ is abelian (cf. (a)), the set $\\{e, a, b, ab\\}$ is a subgroup of $G$, hence $4|4n+2$ – a contradiction.\nLet $\\varphi \\in \\operatorname{Aut}(G)$. Since $e = \\varphi(e) = \\varphi(a^2) = \\varphi^2(a)$ and $\\varphi(a) \\neq e$, we get $\\varphi(a) = a$. Let $g, h \\in \\operatorname{Aut}(G)$ such that $x = g(x)h(x)$ for all $x \\in G$. Then $a = g(a)h(a) = a^2$, implying $a = e$ – a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\ge 2$ be an integer and $0 < a_1 < a_2 < \\dots < a_k$ be real numbers. Compute $\\lim_{n \\to \\infty} \\{ \\sqrt[n]{a_1 n + a_2^n + \\dots + a_k^n} \\}$, where $\\{x\\}$ denotes the fractional part of $x$.",
    "options": [],
    "answer": "If a_k ≥ 1, the limit is {a_k}; if a_k < 1, the limit is 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be integers $m, n \\ge 2$. The matrices $A_1, A_2, \\dots, A_m \\in \\mathcal{M}_n(\\mathbb{R})$, are not all nilpotent. Prove that there is an integer $k > 0$ such that $A_1^k + A_2^k + \\dots + A_m^k \\ne O_n$.\n\nMarius Cavachi",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\lambda_{i1}, \\lambda_{i2}, \\dots, \\lambda_{in}$ the eigenvalues of $A_i$, $i = 1, 2, \\dots, m$. Suppose $A_1^k + A_2^k + \\dots + A_m^k = O_n$, for all $k \\ge 1$. Then $\\text{tr}(A_1^k) + \\text{tr}(A_2^k) + \\dots + \\text{tr}(A_m^k) = 0$, implying $\\sum_{j=1}^m \\sum_{i=1}^n \\lambda_{ij}^k = 0$.\n\nBy the Newton relations, the equalities $\\sum_{i,j} \\lambda_{ij}^k = 0$, $k \\ge 1$, imply $\\lambda_{ij} = 0$, for any $i = 1, 2, \\dots, m, j = 1, 2, \\dots, n$. Then $A_i^n = O_n$, $i = 1, 2, \\dots, m$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the distinct complex numbers $a, b, c, d$. Prove that the following are equivalent:\ni) For any $z \\in \\mathbb{C}$ we have $|z-a| + |z-b| \\ge |z-c| + |z-d|$.\nii) There is $t \\in (0, 1)$ such that $c = ta + (1-t)b$ and $d = (1-t)a + tb$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "ii) $\\implies$ i): We have $|z-c| = |z-ta - (1-t)b| \\le t|z-a| + (1-t)|z-b|$. Analogously, $|z-d| \\le (1-t)|z-a| + t|z-b|$. Summing up, the conclusion follows.\n\ni) $\\implies$ ii): For $z = a$ we get $|a-b| \\ge |a-c| + |a-d|$, and for $z = b$, $|a-b| \\ge |b-c| + |b-d|$. Summing up, $2|a-b| \\ge |a-c| + |a-d| + |b-c| + |b-d|$.\nBut $|a-c| + |b-c| \\ge |a-b|$, $|a-d| + |b-d| \\ge |a-b|$. It follows that all inequalities are in fact equalities. In particular $|a-c| + |b-c| = |a-b|$ giving the existence of $t_1 \\in (0, 1)$ such that $c = t_1a + (1-t_1)b$. In the same way, there is $t_2 \\in (0, 1)$ such that $d = t_2a + (1-t_2)b$.\nWe shall show that $t_1 + t_2 = 1$. We have $|a-c| + |b-c| = |a-b| = |b-c| + |b-d|$, implying $|a-c| = |b-d|$. These give $(1-t_1)|a-b| = t_2|b-a|$, which concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\Omega$ and $\\omega$ are tangent at a point $P$ ($\\omega$ lies inside $\\Omega$). A chord $AB$ of $\\Omega$ is tangent to $\\omega$ at $C$; the line $PC$ meets again $\\Omega$ at $Q$. Chords $QR$ and $QS$ of $\\Omega$ are tangent to $\\omega$. Let $I, X$, and $Y$ be the incentres of the triangles $APB, ARB$, and $ASB$, respectively. Prove that\n$$\n\\angle PXI + \\angle PYI = 90^{\\circ}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that a homothety centred at $P$ mapping $\\omega$ to $\\Omega$ maps $C$ to $Q$, and maps the line $AB$ to the tangent to $\\Omega$ at $Q$. Thus this tangent is parallel to $AB$, and hence $Q$ is the midpoint of arc $AB$ (not containing $P$). So the points $I, X$, and $Y$ lie on the segments $PQ, RQ$, and $SQ$, respectively.\n\nRecall that for any triangle $KLM$ with the circumcircle $\\Gamma$ and incentre $J$, the points $K, L$, and $J$ are equidistant from the midpoint of arc $KL$ of $\\Gamma$ not containing $M$. Applying this to triangles $APB, ARB$, and $ASB$ we obtain that\n$$\nQA = QB = QX = QY = QI.\n$$\nSince $Q$ is the midpoint of arc $AB$, we get that $\\angle QPA = \\angle QPB = \\angle QAB$. Thus the triangles $QAC$ and $QPA$ are similar, and $QC \\cdot QP = QA^2 = QX^2$. Since $QX$ is tangent to $\\omega$, it follows that $X$ is their point of tangency; analogously, $Y$ is the point of tangency of $QS$ with $\\omega$.\n\nFinally, from isosceles triangles $QXI$ and $QYI$ we get $\\angle QXI = \\angle QIX = 90^{\\circ} - \\angle IQX/2$ and $\\angle QYI = \\angle QIY = 90^{\\circ} - \\angle IQY/2$. Denoting by $O$ the centre of $\\omega$, we obtain $\\angle QIX + \\angle QIY = 180^{\\circ} - \\angle XQY/2 = 180^{\\circ} - (180^{\\circ} - \\angle XOY)/2 = 90^{\\circ} + \\angle XPY$. Thus,\n$$\n\\angle PXI + \\angle PYI = \\angle XIY - \\angle XPY = (90^{\\circ} + \\angle XPY) - \\angle XPY = 90^{\\circ},\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Weights of $1\\ \\mathrm{g}$, $2\\ \\mathrm{g}$, $\\ldots$, $200\\ \\mathrm{g}$ are placed on the two pans of a balance such that on each pan there are $100$ weights and the balance is in equilibrium. Prove that one can swap $50$ weights from one pan with $50$ weights from the other pan such that the balance remains in equilibrium.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We call a *pair* two weights whose sum is $201\\ \\mathrm{g}$. We wish to obtain, in the end, $50$ pairs on each of the two pans of the balance.\n\nIf on the pan on the left we have the weights $a_1, a_2, \\dots, a_{50}$ and their pairs $b_1, b_2, \\dots, b_{50}$ are on the pan on the right, we move the weights such that, in the end, on the left pan we have the weights $a_1, a_2, \\dots, a_{50}$ together with their pairs, $b_1, b_2, \\dots, b_{50}$.\n\nIf we have less than $50$ pairs that are split between the two pans, we must have at least $25$ pairs on the left pan and (at least) $25$ pairs on the right pan. Moving $25$ complete pairs from the right pan next to $25$ pairs from the left pan, we obtain, again, $50$ complete pairs on one pan, hence the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be $2 \\times 2$ real matrices such that $AB = A^2 B^2 - (AB)^2$ and $\\det(B) = 2$. Evaluate $\\det(A + 2B) - \\det(B + 2A)$.",
    "options": [],
    "answer": "6",
    "solution": "Write $A(AB - BA - I_2)B = O_2$ to get $A(AB - BA - I_2) = O_2$, for $B$ is nonsingular. If $A$ is nonsingular, then $AB - BA = I_2$, false, for $\\det(AB - BA) = 0 \\ne 2 = \\det(I_2)$.\n\nSet $f(x) = \\det(A + xB)$, $x \\in \\mathbb{R}$. Since $\\det(A) = 0$, there exists $a \\in \\mathbb{R}$ such that $f(x) = ax + \\det(B)x^2 = 2x^2 + ax$, for all $x \\in \\mathbb{R}$. Then $\\det(A+2B) - \\det(B+2A) = f(2) - 4f(1/2) = 8 + 2a - 4(1/2 + a/2) = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ for which the number $a = \\frac{2x+1}{x^2+2x+3}$ is an integer.",
    "options": [],
    "answer": "{-2, -1/2}",
    "solution": "Suppose the integer $a$ is not equal to $0$. Then $|2x+1| \\ge |x^2+2x+3|$. Notice that $x^2+2x+3 = (x+1)^2 + 2 > 0$ to get $2x+1 \\ge x^2+2x+3$ or $2x+1 \\le -(x^2+2x+3)$, hence $x^2+2 \\le 0$ or $x^2+4x+4 \\le 0$. The former leads to no solutions and the latter gives $x = -2$ and then $a = -1 \\in \\mathbb{Z}$.\n\nFor $a=0$ we get $x = -1/2$. Hence $x \\in \\{-2, -1/2\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37936,
    "subject": "Mathematics (Multi-modal)",
    "question": "A die is an unitary cube with numbers from $1$ to $6$ written on its faces, so that each number appears once and the sum of the numbers on any two opposite faces is $7$. We construct a large $3 \\times 3 \\times 3$ cube using $27$ dice. Find all possible values of the sum of numbers which can be seen on the faces of the large cube.",
    "options": [],
    "answer": "All integers from 90 to 288 inclusive",
    "solution": "We will say that a die of the large cube is of type I, type II or type III according to the number of its faces which are visible (e.g. a die sharing a vertex with the large die is a type III die). Every large cube contains $6$ type I, $12$ type II and $8$ type III dice.\n\nThe minimum sum is obtained when each type I die shows $1$, each type II die shows $1$ and $2$ and each type III die shows $1$, $2$ and $3$. Its value is $S_{\\min} = 6 \\cdot 1 + 12 \\cdot 3 + 8 \\cdot 6 = 90$.\n\nThe maximum sum is obtained when each type I die shows $6$, each type II die shows $5$ and $6$ and each type III die shows $4$, $5$ and $6$. Its value is $S_{\\max} = 6 \\cdot 6 + 12 \\cdot 11 + 8 \\cdot 15 = 288$.\n\nWe will show that the sum can be any number from $90$ to $288$. In order to do this we will start from the minimum sum and we will rotate the dice so that the sum increases by $1$ at each step.\n\nRotating a type I dice we can increase the total by $1$ each time, until this die shows $6$. This way we can make each type I die show a $6$.\n\nWe rotate now a type II die from $1+2$ to $5+6$ and, in the same time, we rotate two of the type I dice to $1$ and $4$ respectively; this way the sum increases by $1$. We then rotate back the type I dice, little by little, so they show $6$ again, increasing the sum each time by $1$. We repeat this move until each type II die shows $5+6$.\n\nThen we rotate a type III die so it shows $4+5+6$, rotating in the same time two type I dice so they show $2$; this increases the total sum by $1$. We bring then the type I dice back to $6$, and repeat this sequence of moves until each type III die shows $4$, $5$ and $6$. We arrived thus to $S_{\\max}$ and the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAz $A$ és $B$ másodrendű négyzetes mátrixok teljesítik az\n$$\nAB = A^2 B^2 - (AB)^2 \\text{ és } \\quad \\operatorname{det}(B) = 2\n$$\nösszefüggéseket.\n\na) Igazold, hogy az $A$ mátrix nem invertálható!\n\nb) Számítsd ki a $\\operatorname{det}(A+2B) - \\operatorname{det}(B+2A)$ különbséget!",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLegyenek $a, b \\in \\mathbb{C}$. Igazold, hogy az $|a z + b \\bar{z}| \\leq 1$ egyenlőtlenség akkor és csak akkor áll fenn bármely egységnyi moduluszú $(|z|=1)$ $z$ komplex szám esetén, ha $|a| + |b| \\leq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEgy $(G, \\cdot)$ csoport $(P)$ tulajdonságú, ha a $G$ minden $f$ automorfizmusa esetén léteznek a $G$ olyan $g$ és $h$ automorfizmusai, amelyekre $f(x)=g(x) \\cdot h(x)$, bármely $x \\in G$ esetén. Igazold, hogy:\n\na) Minden $(P)$ tulajdonságú csoport kommutatív!\n\nb) Minden kommutatív, páratlan rendű, véges csoport $(P)$ tulajdonságú!\n\nc) Nincs $(P)$ tulajdonságú, $4 n+2,(n \\in \\mathbb{N})$ rendű véges csoport!\n\n(Egy csoport rendje a csoport elemeinek száma.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all injective functions $f$ of the set of positive integers into itself satisfying the following condition: If $S$ is a finite set of positive integers such that $\\sum_{s \\in S} 1/s$ is an integer, then $\\sum_{s \\in S} 1/f(s)$ is also an integer.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "We shall prove that the identity is the unique function satisfying the conditions in the statement. Clearly, $f(1) = 1$, so $f(n) \\ge 2$ if $n \\ge 2$, by injectivity. We will use the following well-known result.\n\n**Egyptian fractions theorem.** For every positive rational $r$ and positive integer $n$, there exists a finite set $S$ of integers greater than $n$ such that $r = \\sum_{s \\in S} 1/s$.\n\nNow, consider an integer $n \\ge 2$ and use the Egyptian fractions theorem to write\n$$\n1 - \\frac{1}{n} = \\sum_{s \\in S} \\frac{1}{s}, \\text{ where } S \\text{ is a finite set of integers greater than } n(n+1), \\text{ and}\n$$\nget thereby\n$$\n1 = \\frac{1}{n} + \\sum_{s \\in S} \\frac{1}{s} = \\frac{1}{n+1} + \\frac{1}{n(n+1)} + \\sum_{s \\in S} \\frac{1}{s}.\n$$\nBoth are positive integers, so\n$$\n\\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} - \\frac{1}{f(n)}\n$$\nis an integer. Since $-\\frac{1}{2} \\le -\\frac{1}{f(n)} < \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} - \\frac{1}{f(n)} < \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} \\le \\frac{1}{2} + \\frac{1}{2} = 1$, it follows that $\\frac{1}{f(n)} = \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))}$. In particular, $f$ is strictly increasing, so $f(n) \\ge n$.\n\nFinally, proceed by induction on $n \\ge 2$ to prove that $f(n) = n$. To show that $f(2) = 2$, simply notice that $2/f(2) = 1/f(2) + 1/f(3) + 1/f(6)$ is a positive integer not exceeding 1. To complete the proof, let $f(n) = n$ for some $n \\ge 2$ and write\n$$\n\\frac{1}{n} = \\frac{1}{f(n)} = \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} \\le \\frac{1}{n+1} + \\frac{1}{n(n+1)} = \\frac{1}{n}\n$$\nto conclude that $f(n+1) = n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let points $M$ and $N$ on the sides $AB$ and $AC$ respectively such that $\\angle ABC \\equiv \\angle ANM$. Let $D$ be the mirror image of point $A$ across $B$, and let $P$ and $Q$ be the midpoints of the line segments $MN$ and $CD$ respectively. Show that points $A$, $P$ and $Q$ are collinear if and only if $AC = AB\\sqrt{2}$.\nPetru Braica",
    "options": [],
    "answer": "AC = AB√2",
    "solution": "\"$\\Leftrightarrow$\" Notice that triangles $AMN$ and $ACB$ are similar to get $\\frac{AM}{AC} = \\frac{AN}{AB}$ and furthermore $\\frac{AM}{AN} = \\sqrt{2}$. Since $\\frac{AD}{AC} = \\frac{2AB}{AC} = \\sqrt{2}$, one has $\\frac{AM}{AN} = \\frac{AD}{AC} \\Leftrightarrow \\frac{AM}{AD} = \\frac{AN}{AC}$. Consequently $MN \\parallel CD$, hence points $A$, $P$ and $Q$ are collinear.\n\n\"$\\Rightarrow$\" We prove by contradiction that $MN \\parallel CD$. Suppose the opposite and let point $N' \\in (AC)$, $N' \\neq N$ such that $MN' \\parallel CD$ and $\\{P'\\} = MN' \\cap AQ$. Then $P'$ is the midpoint of the segment $MN'$, and $PP'$ is a mid-line in the triangle $MNN'$. This implies $PP' \\parallel NN'$, false, because $PP' \\cap NN' = \\{A\\}$.\n\nAs $MN \\parallel CD$, we get $\\frac{AM}{AD} = \\frac{AN}{AC} \\Leftrightarrow \\frac{AM}{AN} = \\frac{2AB}{AC}$. Notice also that $\\triangle AMN \\sim \\triangle ACB$ to get $\\frac{AM}{AC} = \\frac{AN}{AB} \\Rightarrow \\frac{AM}{AN} = \\frac{AC}{AB}$. It follows that $\\frac{2AB}{AC} = \\frac{AC}{AB}$, hence $AC = AB\\sqrt{2}$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x > 0$ and integer $n > 0$ so that $\\lfloor x \\rfloor + \\left\\{ \\frac{1}{x} \\right\\} = 1.005 \\cdot n$.",
    "options": [],
    "answer": "x = 100/7, n = 14",
    "solution": "The equation can be written $\\lfloor x \\rfloor + \\{1/x\\} = n + n/200$. Denote $q$, respectively $r$, the quotient and the remainder of the division of $n$ by $200$. Then $\\lfloor x \\rfloor + \\{1/x\\} = n + q + r/200$. The integer part of the left-hand member is $\\lfloor x \\rfloor$ and the integer part of the right-hand member is $n + q$, hence $\\lfloor x \\rfloor = 201q + r$ and $\\{1/x\\} = r/200$.\nIf $x < 1$, then $\\lfloor x \\rfloor + \\{1/x\\} = \\{1/x\\} < 1 < 1.005 \\cdot n$, for every $n \\in \\mathbb{N}^*$. If $x = 1$, then $1 = 1.005 \\cdot n$, impossible. Henceforth $x > 1$, so $0 < 1/x < 1$, that is $\\{1/x\\} = 1/x$. It follows $1/x = r/200$, hence $r \\neq 0$ and $x = 200/r$.\nInequality $\\lfloor x \\rfloor \\le x < \\lfloor x \\rfloor + 1$ implies $201q + r \\le 200/r < 201q + r + 1$, hence $201qr + r^2 \\le 200 < 201qr + r(r+1)$. Since $q, r \\in \\mathbb{N}$, the above inequalities cannot take place for $qr \\ge 1$, so $qr = 0$. Relation $r \\neq 0$, implies $q = 0$, hence $r^2 \\le 200 < r(r+1)$, which leads to $r = 14$, $n = 14$ and $x = 100/7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $A$ is a singular matrix of order $n$ with complex entries, all of which having absolute value equal to $1$.\n\na. Let $n = 3$. Show that two lines or two columns of the matrix $A$ are proportional.\n\nb. Find, with proof, if the above claim holds for $n = 4$.",
    "options": [],
    "answer": "The claim holds for size three: some two rows or two columns are proportional. It fails for size four: there exists a singular four by four unimodular-entry matrix with no proportional pair of rows or columns.",
    "solution": "a. By suitable multiplication on each row and column, the matrix $A$ can be written as $k \\begin{pmatrix} 1 & 1 & 1 \\\\ 1 & a & b \\\\ 1 & c & d \\end{pmatrix}$, where $a, b, c, d, k$ are complex numbers of absolute value $1$.\n\nThe relation $\\det(A) = 0$ gives $(a-1)(d-1) = (b-1)(c-1)$. Take the complex conjugates to get $\\overline{ad}(a-1)(d-1) = \\overline{bc}(b-1)(c-1)$.\n\nIf $(a-1)(d-1) = 0$, then $(b-1)(c-1) = 0$ and two rows — or columns — are equal to $(1 \\ 1 \\ 1)$ and the claim is reached.\n\nSuppose $(a-1)(d-1) = (b-1)(c-1) \\neq 0$. Then $\\overline{ad} = \\overline{bc}$ and $ad = bc$. From $(a-1)(d-1) = (b-1)(c-1)$ we get $a+d = b+c$, hence $\\{a,d\\} = \\{b,c\\}$ or $a=b=c=d$. It follows that the bottom two rows or the rightmost two columns are equal, hence the claim.\n\nb. Notice that $A = \\begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 1 & 1 & i & -i \\\\ 1 & -i & 1 & -i \\\\ 1 & i & i & -1 \\end{pmatrix}$ is a singular matrix and any two rows or columns are not proportional. The above claim fails for $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 37944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is inscribed in the circle $C$ and $AA'$, $BB'$, $CC'$ are diameters of this circle. Denote $H_1, H_2, H_3$ the orthocenters of the triangles $A'BC$, $AB'C$, $ABC'$ respectively.\n\na) Prove that there exists a triangle $PQR$ with sides equal to $[AH_1]$, $[BH_2]$, $[CH_3]$ and compute the ratio $S_{PQR}/S_{ABC}$.\n\nb) Prove that the straight lines $AH_1$, $BH_2$ and $CH_3$ are concurrent.",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all injective functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy\n$$\n|f(x) - f(y)| \\le |x - y|,\n$$\nfor all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All functions of the form f(x) = x + k or f(x) = -x + k, where k is an integer.",
    "solution": "The given relation yields $|f(x+1) - f(x)| \\le 1$, so $f(x+1)-f(x) \\in \\{-1, 0, 1\\}$. Because $f$ is one-to-one, we get $f(x+1)-f(x) \\in \\{-1, 1\\}$, for all $x$ in $\\mathbb{Z}$. We suppose WLOG that $f(1)-f(0) = 1$ (if $f$ verifies the hypothesis then $-f$ verifies it also). We have $f(2)-f(1) = \\pm 1$. If $f(2)-f(1) = -1$, then $f(2) = f(0)$, a contradiction, so $f(2) = f(0) + 2$.\nBy easy induction it follows that $f(n) = f(0) + n$, for any positive integer $n$. Analogously we get $f(-n) = f(0) - n$, for any positive integer $n$.\n\nThus the functions that verify the given condition are of the form $f(x) = \\pm x + k$, for any integer $x$, where $k$ is an arbitrary integer. One easily verifies that all such functions are solutions to the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37946,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f : (0, \\infty) \\to (0, \\infty)$ is called *contractive* if, for every numbers $x, y \\in (0, \\infty)$, we have $\\lim_{n \\to \\infty} (f^n(x) - f^n(y)) = 0$, where $f^n = f \\circ f \\circ \\dots \\circ f$. Prove\n\na) If $f : (0, \\infty) \\to (0, \\infty)$ is contractive, continuous and has a fixed point (there is $x_0 \\in (0, \\infty)$ such that $f(x_0) = x_0$), then $f(x) > x$, for $x \\in (0, x_0)$, and $f(x) < x$, for all $x \\in (x_0, \\infty)$.\n\nb) The function $f : (0, \\infty) \\to (0, \\infty)$ defined by $f(x) = x + 1/x$ is contractive but has no fixed points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Suppose the contrary, that is $f$ has another fixed point $x_1 \\in (0, \\infty) \\setminus \\{x_0\\}$. Then $\\lim_{n \\to \\infty} (f^n(x_0) - f^n(x_1)) = x_0 - x_1 \\ne 0$, a contradiction. In this case, the continuity of $f$ (intermediate value property) implies $f(x) < x$, for all $x \\in (0, x_0)$ or $f(x) > x$, for all $x \\in (0, x_0)$.\n\nBy induction, the first case implies $0 < f^{n+1}(x) < f^n(x) < x$, for any $n \\in \\mathbb{N}^*$ and $x \\in (0, x_0)$. It follows that the sequence $a_n = (f^n(x))_{n \\ge 1}$ is convergent; denote by $a$ its limit. If $a > 0$, then from $a_{n+1} = f(a_n)$ we get $a = f(a)$, a contradiction. So $a = 0$, which implies $\\lim_{n \\to \\infty} (f^n(x_0) - f^n(x)) = x_0 \\ne 0$, for all $x \\in (0, x_0)$, contradiction. In conclusion $f(x) > x$, for any $x \\in (0, x_0)$.\n\nAnalogously, $f(x) > x$, for all $x \\in (x_0, \\infty)$ or $f(x) < x$ for all $x \\in (x_0, \\infty)$. In the first case we deduce $f^{n+1}(x) > f^n(x) > x$, for any $n \\in \\mathbb{N}^*$, implying\n\n$$\n\\lim_{n \\to \\infty} f^n(x) = \\infty\n$$\nand then $\\lim_{n \\to \\infty} (f^n(x) - f^n(x_0)) = \\infty$, for all $x \\in (x_0, \\infty)$, a contradiction. So, $f(x) < x$, for any $x \\in (x_0, \\infty)$.\n\nb) First solution. Consider $x, y \\in (0, \\infty)$. We may suppose $f(x) < f(y)$. Denote $x_n = f^n(x)$, $n \\in \\mathbb{N}^*$ and $y_n = f^n(y)$, $n \\in \\mathbb{N}^*$. We have $2 \\le x_n < y_n$, $n > 1$, because $f$ is increasing on $[1, \\infty)$. We shall prove inductively that $y_n < y_1 + 2\\sqrt{n}$, $n \\in \\mathbb{N}^*$. This is obvious for $n=1$. Supposing $y_n < y_1 + 2\\sqrt{n}$, for some $n$, by the monotonicity of $f$, we get: $y_{n+1} = f(y_n) < f(y_1 + 2\\sqrt{n}) = y_1 + 2\\sqrt{n} + \\frac{1}{y_1 + 2\\sqrt{n}} < y_1 + 2\\sqrt{n} + \\frac{1}{2\\sqrt{n}} < y_1 + 2\\sqrt{n+1}$.\n\nIt follows $2 \\le x_n < y_n < y_1 + 2\\sqrt{n} < 3\\sqrt{n}$, for $n \\in \\mathbb{N}$, $n > y_1^2$. Using the well-known inequality $1 - x < e^{-x}$, for $x \\in \\mathbb{R}$, we deduce: $0 < y_{n+1} - x_{n+1} = f(y_n) - f(x_n) = (y_n - x_n) \\left(1 - \\frac{1}{x_n y_n}\\right) < (y_n - x_n) \\left(1 - \\frac{1}{9n}\\right) < (y_n - x_n) e^{-\\frac{1}{9n}}$, for $n \\ge p = [y_1^2] + 1$. We get $0 < y_n - x_n < (y_p - x_p)e^{-\\frac{1}{9} \\sum_{k=p}^{n-1} \\frac{1}{k}}$, for all $n > p$.\n\nBecause $\\lim_{n \\to \\infty} \\sum_{k=p}^{n-1} \\frac{1}{k} = \\infty$, we infer $\\lim_{n \\to \\infty} e^{-\\frac{1}{9} \\sum_{k=p}^{n-1} \\frac{1}{k}} = 0$, which in turn gives $\\lim_{n \\to \\infty} (f^n(y) - f^n(x)) = 0$. So $f$ is contractive without fixed points.\n\nSecond solution. Using the same notations as before, we shall show that $\\lim_{n \\to \\infty} (x_n - \\sqrt{2n}) = \\lim_{n \\to \\infty} (y_n - \\sqrt{2n}) = 0$, which will imply $\\lim_{n \\to \\infty} (x_n - y_n) = 0$, that is the conclusion. As $x_{n+1} = x_n + 1/x_n$, $n \\ge 1$, we have $\\lim_{n \\to \\infty} x_n = \\infty$. By the Stolz–Cesàro lemma, we deduce\n\n$$\n\\begin{align*} \n\\lim_{n \\to \\infty} (x_n - \\sqrt{2n}) &= \\lim_{n \\to \\infty} \\frac{x_n^2 - 2n}{x_n + \\sqrt{2n}} = \\lim_{n \\to \\infty} \\frac{x_{n+1}^2 - x_n^2 - 2}{x_{n+1} - x_n + \\sqrt{2n+2} - \\sqrt{2n}} = \\\\ \n&= \\lim_{n \\to \\infty} \\frac{1/x_n^2}{1/x_n + 2/(\\sqrt{2n+2} + \\sqrt{2n})} = \\\\ \n&= \\lim_{n \\to \\infty} \\frac{1/x_n}{1 + 2x_n/(\\sqrt{2n+2} + \\sqrt{2n})} = 0. \n\\end{align*}\n$$\n\nThe last equality is a consequence of $0 \\le \\frac{1/x_n}{1 + 2x_n/(\\sqrt{2n+2} + \\sqrt{2n})} \\le 1/x_n$ and $\\lim_{n \\to \\infty} 1/x_n = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37947,
    "subject": "Mathematics (Multi-modal)",
    "question": "The function $f: \\mathbb{R} \\to \\mathbb{R}$ has the property that every point of local minimum has a neighbourhood $(\\alpha, \\beta)$ so that $f$ is strictly convex on $(\\alpha, \\beta)$. Prove that the set of the points of local minimum is countable.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the set of points of local minimum of $f$.\n\nFor each $x \\in S$, by hypothesis, there exists an open interval $(\\alpha_x, \\beta_x)$ containing $x$ such that $f$ is strictly convex on $(\\alpha_x, \\beta_x)$.\n\nRecall that a strictly convex function on an interval has at most one point of local minimum (since if there were two, the function would be constant or linear between them, contradicting strict convexity).\n\nTherefore, for each $x \\in S$, $x$ is the unique point of local minimum of $f$ in $(\\alpha_x, \\beta_x)$.\n\nThus, the intervals $(\\alpha_x, \\beta_x)$, for $x \\in S$, are pairwise disjoint.\n\nBut the set of pairwise disjoint open intervals in $\\mathbb{R}$ is at most countable (since each contains a rational number, and the rationals are countable).\n\nTherefore, $S$ is at most countable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_1, a_2, \\ldots, a_n$ be real numbers with $a_1 + a_2 + \\ldots + a_k \\le k$ for all $k \\in \\{1, 2, \\ldots, n\\}$. Show that\n$$\n\\frac{a_1}{1} + \\frac{a_2}{2} + \\ldots + \\frac{a_n}{n} \\le \\frac{1}{1} + \\frac{1}{2} + \\ldots + \\frac{1}{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We induct on $n$. The case $n=1$ is trivial. Suppose the claim holds for $n$ numbers. If $a_{n+1} \\le 1$ then $\\frac{a_{n+1}}{n+1} \\le \\frac{1}{n+1}$ and the conclusion follows. If $a_{n+1} > 1$, then $\\frac{a_1}{1} + \\dots + \\frac{a_n}{n} + \\frac{a_{n+1}}{n+1} \\le \\frac{a_1}{1} + \\dots + \\frac{a_n + a_{n+1} - 1}{n} + \\frac{1}{n+1}$. Now apply the induction hypothesis to the following $n$ numbers $a_1, \\dots, a_{n-1}, a_n + a_{n+1} - 1$ to get the claim.\nLet $S_k = a_1 + a_2 + \\dots + a_k$, $S_0 = 0$ and notice that $S_k - S_{k-1} = a_k$ for all $k = 1, 2, \\dots, n$. Then $\\sum_{k=1}^n \\frac{a_k}{k} = \\sum_{k=1}^n \\frac{S_k - S_{k-1}}{k} = \\sum_{k=1}^n \\frac{S_k}{k(k+1)} \\le \\frac{1}{2} + \\dots + \\frac{1}{n}$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37949,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular hexagonal prism $ABCDEF A'B'C'D'E'F'$ has the edge $AB = 12$ and the height $AA' = 12\\sqrt{3}$. Let $N$ be the midpoint of the edge $CC'$.\n\na.\nProve that the lines $BF'$ and $ND$ are perpendicular.\n\nb.\nFind the distance between the lines $BF'$ and $ND$.",
    "options": [],
    "answer": "15√42/7",
    "solution": "a.\nLet $M$ be the midpoint of the segment $BB'$. Since $MN \\parallel AD$, the points $A$, $D$, $M$ and $N$ are coplanar. Let $Q$ be the midpoint of the segment $BF$. The intersection of the planes $(BFF')$ and $(ADN)$ is the line $MQ$. Notice that $BFF'B'$ is a square, hence $BF' \\perp MQ$ and then $AD \\perp (BFF')$, implying $BF' \\perp AD$. Therefore $BF' \\perp (ADN)$ and then $BF' \\perp ND$, as claimed.\n\nb.\nLet $S$ be the intersection point of the lines $MQ$ and $BF'$. Let $P$ be the projection of $S$ onto line $ND$. From $BF' \\perp (ADN)$ we derive that $BF' \\perp SP$, hence line $SP$ is the common perpendicular of the lines $BF'$ and $ND$.\n\nIn order to find the length of the line segment $[SP]$, we evaluate the area of the triangle $SND$ in two ways: $S[SND] = S[MNDQ] - S[SDQ] - S[SMN] = \\frac{1}{2}SP \\cdot ND$. As $MN = 12$, $QD = 18$, $ND = 6\\sqrt{7}$, $MQ = 6\\sqrt{6}$, we get $SP = \\frac{15\\sqrt{42}}{7}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 37950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the interior of the acute triangle $ABC$ and let $D$, $E$, $F$ be respectively the intercepts of lines $AP$, $BP$, $CP$ with the sides $BC$, $CA$, $AB$.\n\na) Prove that the area of the triangle $DEF$ is not greater than a quarter of the area of triangle $ABC$.\n\nb) Prove that the inradius of $DEF$ is not greater than a quarter of the inradius of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Put $\\frac{BD}{CD} = x$, $\\frac{CE}{AE} = y$, $\\frac{AF}{BF} = z$ to obtain $\\frac{S_{AEF}}{S_{ABC}} = \\frac{AF}{AB} \\cdot \\frac{AE}{AC} = \\frac{z}{(z+1)(y+1)}$ and the similar ones. As by Ceva's theorem $xyz = 1$, it follows $\\frac{S_{DEF}}{S_{ABC}} = 1 - \\sum \\frac{z}{(z+1)(y+1)} = \\frac{xyz+1}{(x+1)(y+1)(z+1)} = \\frac{2}{(x+1)(y+1)(z+1)}$ (The symbol $S_{MNP}$ denotes the area of the triangle $MNP$). By the AM inequality $\\frac{2}{(x+1)(y+1)(z+1)} \\le \\frac{2}{8\\sqrt{xyz}} = \\frac{1}{4}$, concluding thus the proof.\n\nb) One knows that the perimeter $DEF$ is at least the perimeter of the ortic triangle $A'B'C'$. We get $r_{DEF} = \\frac{S_{DEF}}{p_{DEF}} \\le \\frac{S_{ABC}}{4p_{A'B'C'}}$.\n\n$2R \\sin A \\sin B \\sin C$ and $S_{ABC} = 2R^2 \\sin A \\sin B \\sin C$ the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $a \\in (0,1)$ and $C$ the set of all increasing functions $f: [0,1] \\to [0, \\infty)$, such that $\\int_0^1 f(x) dx = 1$. Determine:\n\na. $\\max_{f \\in C} \\int_0^a f(x) dx$,\n\nand\n\nb. $\\max_{f \\in C} \\int_0^a (f(x))^2 dx$.",
    "options": [],
    "answer": "a) The maximum is a. b) The maximum is a if a ≤ 1/2, and 1/(4(1−a)) if a > 1/2.",
    "solution": "a.\n\nWe show that\n$$\n\\int_0^a f(x) dx \\le a.\n$$\nTo this end, write\n$$\n\\begin{aligned}\na - \\int_0^a f(x) dx &= a \\int_0^1 f(x) dx - \\int_0^a f(x) dx \\\\\n&= a \\int_a^1 f(x) dx - (1-a) \\int_0^a f(x) dx \\\\\n&\\ge a \\int_a^1 f(a) dx - (1-a) \\int_0^a f(a) dx \\\\\n&= 0.\n\\end{aligned}\n$$\nHere equality holds if and only if $f(x) = 1$ for $0 < x < 1$; at $x = 0$ the function may take any non-negative value less than or equal to $1$, and at $x = 1$ it may take any value greater than or equal to $1$.\n\nTherefore,\n$$\n\\max_{f \\in C} \\int_0^a f(x) dx = a.\n$$\n\nb.\n\nThe required maximum is $a$ if $a \\le 1/2$, and $1/(4(1-a))$ if $a > 1/2$. In both cases, it is achieved at an essentially unique function. In the former case, the maximum is achieved at $f \\equiv 1$ on the open interval $(0, 1)$; at $x = 0$ the function may take any non-negative value less than or equal to $1$, and at $x = 1$ it may take any value greater than or equal to $1$. In the latter case, the maximum is achieved at\n$$\nf(x) = \\begin{cases} 0 & \\text{if } 0 \\le x < 2a - 1, \\\\ \\frac{1}{2(1-a)} & \\text{if } 2a - 1 < x < 1; \\end{cases}\n$$\nat $x = 2a - 1$ the function may take any non-negative value less than or equal to $1/(2(1-a))$, and at $x = 1$ it may take any value greater than or equal to $1/(2(1-a))$.\n\nLet $f$ be a function in $C$. Begin by noticing that\n$$\n\\begin{aligned}\n\\int_0^a (f(x))^2 dx &\\le f(a) \\int_0^a f(x) dx \\\\\n&\\le \\left( \\frac{1}{1-a} \\int_a^1 f(x) dx \\right) \\left( \\int_0^a f(x) dx \\right) \\\\\n&= \\frac{1}{1-a} \\left( 1 - \\int_0^a f(x) dx \\right) \\left( \\int_0^a f(x) dx \\right).\n\\end{aligned}\n$$\nNext, as shown above,\n$$\n\\int_0^a f(x) dx \\le a.\n$$\nFurther, notice that\n$$\n\\max \\{ t(1-t) : t \\le a \\} = \\begin{cases} a(1-a) & \\text{if } a \\le 1/2, \\\\ \\frac{1}{4} & \\text{if } a > 1/2; \\end{cases}\n$$\nin both cases, the maximum is achieved at a single point: at $t = a$ in the former case, and at $t = 1/2$ in the latter.\n\nConsequently,\n$$\n\\int_0^a (f(x))^2 dx \\le \\begin{cases} a & \\text{if } a \\le 1/2, \\\\ 1/(4(1-a)) & \\text{if } a > 1/2. \\end{cases}\n$$\nIf $a \\le 1/2$, then\n$$\n\\int_0^a (f(x))^2 dx = a\n$$\nforces\n$$\n\\int_0^a f(x) dx = a,\n$$\nso $f(x) = 1$ for $0 < x < 1$, by the third paragraph above; at $x = 0$ the function may take any non-negative value less than or equal to $1$, and at $x = 1$ it may take any value greater than or equal to $1$.\n\nIf $a > 1/2$, then\n$$\n\\int_0^a (f(x))^2 dx = \\frac{1}{4(1-a)}\n$$\nforces\n$$\n\\int_0^a f(x) dx = \\frac{1}{2} \\qquad (1)\n$$\nand\n$$\n\\int_0^a (f(x))^2 dx = f(a) \\int_0^a f(x) dx, \\qquad (2)\n$$\nso\n$$\n\\int_a^1 f(x) dx = \\frac{1}{2} \\quad \\text{and} \\quad f(a) = \\frac{1}{2(1-a)}.\n$$\nThe last two conditions force in turn $f(x) = 1/(2(1-a))$, $a \\le x < 1$; at $x = 1$ the function may take any value greater than or equal to $1/(2(1-a))$.\n\nWe now show that (1) and (2) force\n$$\nf(x) = \\begin{cases} 0 & \\text{if } 0 < x < 2a - 1, \\\\ \\frac{1}{2(1-a)} & \\text{if } 2a - 1 < x < a. \\end{cases}\n$$\nOf course, $f(0) = 0$, and at $x = 2a - 1$ the function may take any non-negative value less than or equal to $1/(2(1-a))$. To prove this, consider the three sets below:\n$$\nA = \\{ x : 0 < x < a, f(x) = 0 \\},\n$$\n$$\nB = \\{ x : 0 < x < a, 0 < f(x) < f(a) \\},\n$$\n$$\nC = \\{ x : 0 < x < a, f(x) = f(a) \\}.\n$$\nClearly, if $x, y, z$ are members of $A, B, C$, respectively, then $x < y < z$.\nNotice that $B$ contains at most one point. Otherwise, consider two points $b < b'$ in $B$ to contradict (2):\n$$\n0 < \\int_b^{b'} f(x) (f(a) - f(x)) dx \\le \\int_0^a f(x) (f(a) - f(x)) dx = 0.\n$$\nIf $A$ is empty, then so is $B$ and $C = (0, a)$ and\n$$\n\\int_0^a f(x) dx = a f(a) = \\frac{a}{2(1-a)} > \\frac{1}{2},\n$$\nwhich contradicts (1).\nSimilarly, if $C$ is empty, then again so is $B$ and $A = (0, a)$ and\n$$\n\\int_0^a f(x) dx = 0,\n$$\nwhich contradicts again (1).\nConsequently, neither $A$ nor $C$ are empty, $\\sup A = b = \\inf C$ for some $b$ in $(0, a)$, and $(0, b) \\subseteq A$ and $(b, a) \\subseteq C$.\nFinally, $b$ is determined by (1):\n$$\n\\frac{1}{2} = \\int_0^a f(x) dx = \\int_b^a \\frac{1}{2(1-a)} dx = \\frac{a-b}{2(1-a)},\n$$\nso $b = 2a - 1$. The conclusion follows.\nLet again $f$ be a function in $C$. If $f(a) < 1$, then\n$$\n\\int_0^a (f(x))^2 dx \\le \\int_0^a (f(a))^2 dx = a(f(a))^2 < a.\n$$\nIf $f(a) \\ge 1$, consider the function\n$$\n\\varphi: [0, 1] \\to [0, \\infty),\n$$\ndefined by\n$$\n\\varphi(t) = \\int_0^t f(x) dx + (1-t)f(a).\n$$\nClearly, $\\varphi$ is continuous and it is easily seen that $\\varphi$ is decreasing on $[0, a]$ and increasing on $[a, 1]$. Since $\\varphi(1) = 1$, it follows that $\\varphi(a) \\le 1$ (this can also be established directly), and since\n$$\n\\varphi(0) = f(a) \\ge 1,\n$$\nthere exists $b$ in $[0, a]$ such that $\\varphi(b) = 1$, by continuity; that is,\n$$\n\\int_0^b f(x) dx = 1 - (1 - b)f(a).\n$$\nNext, define $g: [0, 1] \\to [0, \\infty)$ by\n$$\ng(x) = \\begin{cases} f(x) & \\text{if } 0 \\le x \\le b, \\\\ f(a) & \\text{if } b < x \\le 1. \\end{cases}\n$$\nIt is readily checked that $g$ belongs to $C$; moreover, if $0 \\le x \\le a$, then\n$$\ng(x) \\ge f(x) \\ge 0,\n$$\nso\n$$\n\\int_0^a (g(x))^2 dx \\ge \\int_0^a (f(x))^2 dx.\n$$\nOn the other hand,\n$$\n\\begin{aligned}\n\\int_0^a (g(x))^2 dx &\\le g(a) \\int_0^a g(x) dx \\\\\n&= f(a) \\left( \\int_0^b f(x) dx + (a-b)f(a) \\right) \\\\\n&= f(a) (1 - (1-b)f(a) + (a-b)f(a)) \\\\\n&= f(a) (1 - (1-a)f(a)).\n\\end{aligned}\n$$\nFurther, notice that\n$$\n\\max \\{ t(1 - (1 - a)t) : t \\ge 1 \\} = \\begin{cases} a & \\text{if } a \\le 1/2, \\\\ 1/(4(1-a)) & \\text{if } a > 1/2; \\end{cases}\n$$\nin both cases, the maximum is achieved at a single point: at $t = 1$ in the former case, and at $t = 1/(2(1-a))$ in the latter.\n\nConsequently,\n$$\n\\int_0^a (f(x))^2 dx \\le \\int_0^a (g(x))^2 dx \\le \\begin{cases} a & \\text{if } a \\le 1/2, \\\\ 1/(4(1-a)) & \\text{if } a > 1/2. \\end{cases}\n$$\nProceed now along the lines in the corresponding part of the previous solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, \\pi/2] \\to [0, \\infty)$ be an increasing function. Prove that:\n\na. $$\\int_0^{\\pi/2} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx \\geq 0.$$\n\nb. There exists $a \\in [\\pi/4, \\pi/2]$ such that $\\int_0^a f(x) \\sin x dx = \\int_0^a f(x) \\cos x dx$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Let $I = \\int_0^{\\pi/4} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx$ and $J = \\int_{\\pi/4}^{\\pi/2} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx$ and notice that\n$$\n\\int_0^{\\pi/2} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx = I + J.\n$$\nFor $x \\in [0, \\pi/4]$ we have $f(x) - f(\\pi/4) \\leq 0$ and $\\sin x - \\cos x \\leq 0$, hence $I \\geq 0$. For $x \\in [\\pi/4, \\pi/2]$ we have $f(x) - f(\\pi/4) \\geq 0$ and $\\sin x - \\cos x \\geq 0$, hence $J \\geq 0$. Consequently, $I + J \\geq 0$.\n\nb. Let\n$$\nF(t) = \\int_0^t f(x)(\\sin x - \\cos x) dx, \\quad 0 \\le t \\le \\pi/2.\n$$\nWe have $F(\\pi/4) = \\int_0^{\\pi/4} f(x)(\\sin x - \\cos x) dx \\le 0$ and $F(\\pi/2) = \\int_0^{\\pi/2} f(x)(\\sin x - \\cos x) dx \\ge \\int_0^{\\pi/2} f(\\pi/4)(\\sin x - \\cos x) dx = 0$ – the latter inequality follows from (a). Use the continuity of $F$ to get the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37953,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $AB$ of the rectangle $ABCD$ are taken the points $S$ and $T$ so that $AS = ST = TB$. Denote $M, N$ and $P$ the orthogonal projections of $A, S$ and $T$ onto the straight lines $DS, DT$ respectively $DB$. Prove that the points $M, N$ and $P$ are collinear if and only if $15AD^2 = 2AB^2$.\nGabriel Popa\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n![](attached_image_3.png)\nFig. 3",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M', N', P'$ be the orthogonal projections onto $AB$ of $M, N$ and $P$ respectively. Then $M, N, P$ are collinear if and only if the intersection $N_1$ of $MP$ and $NN'$ coincides with $N$. Since $MM' > PP'$ and $N'$ is between $M'$ and $P'$, from similar triangles (Fig. 2) we get $\\frac{P'N'}{P'M'} = \\frac{N_1N' - PP'}{MM' - PP'}$, so the condition of collinearity is $NN' \\cdot M'P' = MM' \\cdot N'P' + PP' \\cdot M'N'$.\n\nIn order to compute $MM', NN', PP'$ we notice that if the legs of a right triangle $OUV$ are $OU = u, OV = v$ and we take $X$ on $OU$ so that $OX = x < u$ and $XX' \\perp UV, X'X'' \\perp OU$ (Fig. 3), then $\\triangle OO'U \\sim \\triangle XX'U$, so\n$$\n\\frac{X'X''}{O'O''} = \\frac{UX}{UO}, \\quad \\frac{UX''}{UO''} = \\frac{UX}{UO},\n$$\nand, since $O'O'' = \\frac{OO' \\cdot UO'}{OU} = \\frac{1}{OU} \\frac{OU \\cdot OV}{UV} \\frac{OU^2}{UV} = \\frac{u^2v}{u^2 + v^2}, UO'' = \\frac{O'U^2}{OU} = \\frac{u^3}{u^2 + v^2},$\n$$\nX'X'' = \\frac{u-x}{u}O'O'' = \\frac{uv(u-x)}{u^2+v^2}, \\quad UX'' = \\frac{u-x}{u}UO'' = \\frac{u^2(u-x)}{u^2+v^2}. \\quad (1)\n$$\nDenote now $AB = 3a$ and $AD = b$. In triangles $ASD, ATD$ and $ABD$ relations (1) yield $SM' = \\frac{a^3}{a^2 + b^2}, TN' = \\frac{4a^3}{4a^2 + b^2}, BP' = \\frac{9a^3}{9a^2 + b^2}, MM' = \\frac{a^2b}{a^2 + b^2}, NN' = \\frac{2a^2b}{4a^2 + b^2}, PP' = \\frac{3a^2b}{9a^2 + b^2}$,\n\nThe collinearity condition becomes $\\frac{2a^2b}{4a^2 + b^2} \\left( 2a - \\frac{9a^3}{9a^2 + b^2} + \\frac{a^3}{a^2 + b^2} \\right) = \\frac{a^2b}{a^2 + b^2} \\cdot \\left( a - \\frac{9a^3}{9a^2 + b^2} + \\frac{4a^3}{4a^2 + b^2} \\right) + \\frac{3a^2b}{9a^2 + b^2} \\cdot \\left( a - \\frac{4a^3}{4a^2 + b^2} + \\frac{a^3}{a^2 + b^2} \\right)$, which, after some routine computation, reduces to $5b^2 = 6a^2$, that is $15AD^2 = 2AB^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers different from $0$ and let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function defined by\n$$\nf(x) = \\begin{cases} ax, & x \\in \\mathbb{Q} \\\\ bx, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}\n$$\nProve that $f$ is injective if and only if $f$ is surjective.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The function $f$ is injective if and only if $\\frac{a}{b} \\in \\mathbb{Q}$. Indeed, suppose $f$ is injective and $\\frac{a}{b} \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Then $f\\left(\\frac{a}{b}\\right) = a = f(1)$, a contradiction. Suppose now that $\\frac{a}{b} \\in \\mathbb{Q}$ and let $x_1, x_2$ such that $f(x_1) = f(x_2)$. If $x_1, x_2 \\in \\mathbb{R} \\setminus \\mathbb{Q}$ or $x_1, x_2 \\in \\mathbb{Q}$, then $x_1 = x_2$. If $x_1 \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and $x_2 \\in \\mathbb{Q}$, then $a x_2 = b x_1$ and consequently $\\frac{a}{b} = \\frac{x_1}{x_2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, a contradiction. Hence $f$ is injective.\n\nThe function $f$ is surjective if and only if $\\frac{a}{b} \\in \\mathbb{Q}$. To this end, suppose $f$ is surjective and let $x$ with $f(x) = b$. If $x \\in \\mathbb{R} \\setminus \\mathbb{Q}$ then $x = 1$, a contradiction. Therefore $x$ is rational and so $a x = b$, implying $\\frac{a}{b} = \\frac{1}{x} \\in \\mathbb{Q}$. Conversely, suppose $\\frac{a}{b} \\in \\mathbb{Q}$ and let $y \\in \\mathbb{R}$. Since $\\frac{y}{a} \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and $\\frac{y}{b} \\in \\mathbb{Q}$ can not occur simultaneously, we have either $f\\left(\\frac{y}{a}\\right) = y$ or $f\\left(\\frac{y}{b}\\right) = y$. Hence $f$ is surjective.\n\nThe claim now follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a circle centered at $O$ with radius $r$ and a line $\\ell$ not passing through $O$. A grasshopper is jumping to and fro between the points of the circle and the line, the length of each jump being $r$. Prove that there are at most 8 points for the grasshopper to reach.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "We assume that, when having the choice between only two places to jump to, the grasshopper never jumps back to the point from which he got to that place. Let us denote by $P_1$ the starting point of the grasshopper, with $P_2$ the point on the line on which he has jumped from $P_1$, and so on. As the length of the jumps are all equal to $r$, $OP_1P_2P_3$ is a rhombus (possibly a degenerate one). Similarly, $OP_3P_4P_5$ is also a rhombus. It follows that the triangles $P_1OP_5$ and $P_2P_3P_4$ are congruent (SAS), and from here we obtain that $P_1P_5$ is parallel to $\\ell$. We deduce that $P_5$ is the reflection of $P_1$ across the perpendicular line from $O$ onto $\\ell$. (This fact remains true even in the degenerate cases.) From $P_5$, the grasshopper can get to $P_9$ which, as above, is the reflection of $P_5$ across the perpendicular line from $O$ onto $\\ell$, i.e. $P_1$. In conclusion, the grasshopper can reach only the points $P_k$, $k = \\overline{1,8}$ (which are not necessarily distinct).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdottak az $a, b \\in \\mathbb{R}^{*}$ számok és az $f: \\mathbb{R} \\rightarrow \\mathbb{R}$,\n$$\nf(x)= \\begin{cases}a x, & x \\in \\mathbb{Q} \\\\ b x, & x \\in \\mathbb{R} \\backslash \\mathbb{Q}\\end{cases}\n$$\nfüggvény.\nIgazold, hogy $f$ akkor és csak akkor injektív, ha $f$ szürjektív.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdott az $f:[0, \\pi / 2] \\rightarrow[0, \\infty)$ növekvő függvény. Igazold, hogy:\n\na) $\\int_{0}^{\\pi / 2}(f(x)-f(\\pi / 4))(\\sin x-\\cos x) \\, \\mathrm{d} x \\geq 0$.\n\nb) Létezik $a \\in[\\pi / 4, \\pi / 2]$ úgy, hogy $\\int_{0}^{a} f(x) \\sin x \\, \\mathrm{d} x=\\int_{0}^{a} f(x) \\cos x \\, \\mathrm{d} x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdott az $A$-edrendű ($n>1$) nem invertálható négyzetes mátrix. Az $A$ elemei mind 1 moduluszú komplex számok.\n\na) Igazold, hogy ha $n=3$, akkor az $A$ mátrix két sorának vagy két oszlopának elemei egyenesen arányosak.\n\nb) Igaz-e az előző alpontbeli tulajdonság $n=4$ esetén?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37959,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite set of distinct positive integers is written on a blackboard. A *move* consists in choosing two numbers and write their lowest common multiple, given that is not already written. The set is called *closed* if no moves are allowed – e.g., the set $\\{2, 3, 4, 6\\}$ will be closed after number $12$ is added. Determine the maximum number of elements in a closed set given that the initial set contains $10$ numbers.",
    "options": [],
    "answer": "1023",
    "solution": "The largest closed set has $2^{10} - 1$ numbers. To get this number, start with ten primes $p_i$, $i = 1, 2, \\dots, 10$. The closed set consists of all products $\\prod_{i \\in X} p_i$ over all subsets $X$ of the set $\\{1, 2, \\dots, 10\\}$, hence there are $2^{10} - 1$ elements in the closed set.\n\nOne cannot obtain more than $2^{10} - 1$ numbers. Indeed, each move gives the l.c.m. of two or more numbers from the initial set, so each added number corresponds to a nonempty set of the set $\\{1, 2, \\dots, 10\\}$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37960,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the acute-angled triangle $ABC$, with $AB \\neq AC$, $D$ is the foot of the angle bisector of angle $A$, and $E$, $F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSolution:\n\nTriangles $AEF$ and $ABC$ are similar, therefore $AF \\cdot AB = AE \\cdot AC$. It follows the point $A$ is on the radical axis of the two circumcircles, hence $M \\in AD$. We have that $m(\\angle EMF) = 360^\\circ - (180^\\circ - m(\\angle FBD)) - (180^\\circ - m(\\angle ECD)) = m(\\angle B) + m(\\angle C) = 180^\\circ - m(\\angle A)$; it follows that the quadrilateral $AEMF$ is cyclic. This means that $\\angle MEF \\equiv \\angle FAM$ and $\\angle MFE \\equiv \\angle EAM$, i.e. triangle $MEF$ is isosceles, with $ME = MF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all increasing functions $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ with the property: for every $n \\in \\mathbb{N}^*$, the number $\\frac{1}{n} \\sum_{k=1}^{n} k f(k)$ is a perfect square not larger than $(n+1)^2$.",
    "options": [],
    "answer": "f(n) = 3n + 1 for all positive integers n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f: \\mathbb{R} \\to \\mathbb{R}$ is a monotonic function.\n\na. Prove that $f$ has one-sided limits at any point $x_0 \\in \\mathbb{R}$.\n\nb. Define the function $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = \\lim_{t \\to x} f(t)$, i.e. $g(x)$ is the left-sided limit at $x$ of the function $f$. Prove that $g$ is a continuous function, then $f$ is also continuous.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, without any loss, that $f$ is an increasing function.\n\na. Let $x_0 \\in \\mathbb{R}$. The set $\\{f(x) \\mid x < x_0\\}$ is upper bounded by $f(x_0)$, because $f$ is increasing. Set $L = \\sup\\{f(x) \\mid x < x_0\\}$. We claim that $L = f(x_0 - 0)$.\nTo this end, let $\\varepsilon > 0$ and notice that there exists $a < x_0$ such that $f(a) > L - \\varepsilon$. Since $f$ is increasing, we have $|f(x) - L| = L - f(x) < \\varepsilon$ for any $x \\in (a, x_0)$, hence $L = f(x_0 - 0)$.\nSimilarly, $f(x_0 + 0) = \\inf\\{f(x) \\mid x > x_0\\}$.\n\nb. Let $x_0 \\in \\mathbb{R}$ and $t, s, a, b \\in \\mathbb{R}$ such that $t < a < x_0 < s < b$. Then $f(t) \\le f(a) \\le f(x_0) \\le f(s) \\le f(b)$ and furthermore $g(a) = \\lim_{t \\searrow a} f(t) \\le f(x_0)$ and $g(b) = \\lim_{s \\nearrow b} f(s) \\ge f(x_0)$, that is $g(a) \\le f(x_0) \\le g(b)$.\nRecall that $g$ is continuous to get $g(x_0) = \\lim_{a \\searrow x_0} g(a) = \\lim_{b \\nearrow x_0} g(b)$, hence $g(x_0) \\ge f(x_0) \\ge g(x_0)$ or $g(x_0) = f(x_0)$. Consequently $f = g$ and the claim follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square. Point $E$ lies in the interior of the angle $\\angle CAB$ such that angle $\\angle BAE$ is $15^\\circ$, and lines $BE$ and $BD$ are perpendicular. Show that $AE = BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lines $AC$ and $BE$ are parallel, because both are perpendicular to $DB$. Let $F$ be the foot of the perpendicular from $E$ onto $AC$. Notice that $EF = BO = \\frac{BD}{2}$, where $O$ is the centre of the square. The triangle $FAE$ has a right angle at $F$ and has $\\angle FAE$ of $30^\\circ$, hence $AE = 2EF = BD$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37964,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Consider $f: [0, \\infty) \\to [0, \\infty)$ a differentiable and convex function such that $f(x) \\le x$, for $x \\ge 0$. Then $f'(x) \\le 1$, for all $x \\ge 0$.\n\nb) Determine all differentiable and convex functions $f: [0, \\infty) \\to [0, \\infty)$ with the properties $f(0) = 0$ and $f'(x) \\cdot f(f(x)) = x$, for all $x \\ge 0$.",
    "options": [],
    "answer": "f(x) = x for all x ≥ 0",
    "solution": "a) Suppose the contrary, that is, there exists $a \\ge 0$ such that $f'(a) > 1$. Because $\\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a} > 1$, there is $b > a$ such that $\\frac{f(b) - f(a)}{b - a} > 1$. For any $x > b$, by the convexity of $f$ we get $\\frac{f(x) - f(b)}{x - b} \\ge \\frac{f(b) - f(a)}{b - a} = m > 1$. Then $f(x) \\ge mx - mb + f(b)$, so $f(x) > x$ for sufficiently large $x$, a contradiction.\n\nb) We shall prove that $f(x) = x$, for any $x \\ge 0$. As $f'(x) = \\frac{x}{f(f(x))} > 0$, for $x > 0$, we deduce that $f$ is increasing. The function $f$ is convex and differentiable, that is $f'$ is non-decreasing.\n\nSuppose by contradiction that $f(a) < a$. Then $f(f(a)) < f(a) < a$, so $f'(a) > 1$. By a) there is $b > a$ such that $f(b) = b$. Then $f(f(b)) = b$ which gives $f'(b) = 1 < f'(a)$, in contradiction with $f'$ being monotonic.\n\nIt remains that $f(x) \\ge x$, for $x \\ge 0$. Then $f(f(x)) \\ge f(x) \\ge x$, so $f'(x) \\le 1$, for $x \\ge 0$. By Lagrange theorem $f(x) - f(0) = x f'(c_x) \\le x$, for $x > 0$, giving $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$ and the function $f: \\mathbb{N} \\to \\mathbb{N}$ described by\n$$\nf(x) = \\begin{cases} x/2 & , \\text{if } x \\text{ is even} \\\\ (x-1)/2 + 2^{n-1} & , \\text{if } x \\text{ is odd} \\end{cases}.\n$$\nDetermine the set $A = \\{x \\in \\mathbb{N} \\mid (\\underbrace{f \\circ f \\circ \\dots \\circ f}_{n \\text{ times } f})(x) = x\\}$.",
    "options": [],
    "answer": "A = {0, 1, ..., 2^n - 1}",
    "solution": "As for $x \\in \\{0, 1, \\dots, 2^n - 1\\}$ we have $f(x) \\in \\{0, 1, \\dots, 2^n - 1\\}$, and for $x \\ge 2^n$ we have $f(x) < x$, we get $f(x) \\le \\max(x, 2^n - 1)$. It follows that if $x \\in A$, then\n$$\nx = f^{[n]}(x) \\le \\max(f^{[n-1]}(x), 2^n - 1) \\le \\dots \\le \\max(f(x), 2^n - 1).\n$$\nIf $x \\ge 2^n$, then $x \\le f(x)$, a contradiction, thus $A$ is contained in $\\{0, 1, \\dots, 2^n - 1\\}$.\n\nWe shall prove the equality $A = \\{0, 1, \\dots, 2^n - 1\\}$. We give two proofs for the reverse inclusion.\n\ni.\nFrom $f(2^n - 1) = 2^n - 1$ deduce $2^n - 1 \\in A$. As $2f(x) \\in \\{x, x+2^n-1\\}$, $2f(x)$ is congruent to $x$ modulo $2^n - 1$. Inductively, $x \\equiv 2^n f^{[n]}(x) \\equiv f(x) \\pmod{2^n - 1}$. Because $f(x) < 2^n - 1$ for $x < 2^n - 1$, we obtain $f^{[n]}(x) = x$ if $x \\in \\{0, 1, \\dots, 2^n - 2\\}$.\n\nii.\nUse base 2 to represent any number in $x \\in \\{0, 1, \\dots, 2^n - 1\\}$ using $n$ digits, putting zero at the beginning, if necessary. To obtain $f(x)$ from $x$ one has to move the last digit of $x$ on the first position. After $n$ movements the digits of $x$ are at the same places as at the beginning, that is $f^{[n]}(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $M = \\{1, 2, \\dots, 2n + 1\\}$. Find out in how many ways we can split the set $M$ into three mutually disjoint nonempty sets $A, B, C$ so that both the following are true:\n(i) for each $a \\in A$ and $b \\in B$, the remainder of the division of $a$ by $b$ belongs to $C$,\n(ii) for each $c \\in C$ there exists $a \\in A$ and $b \\in B$ such that $c$ is the remainder of the division of $a$ by $b$.\nMircea Fianu",
    "options": [],
    "answer": "n",
    "solution": "We notice that $a > b$, for all $a \\in A$ and $b \\in B$. Indeed, the contrary would imply that $a < b$, so the remainder of the division of $a$ by $b$ is $a$ and $a$ belongs to both $B$ and $C$. This shows that $A$ is made by consecutive numbers and $2n+1 \\in A$.\n\nTake $c \\in C$. From the assumption there exists $a \\in A$ and $b \\in B$ so that $a = bq+c$. Then $a \\ge (c+1) \\cdot 1 + c = 2c + 1$, hence $2n+1 \\ge 2c+1$, therefore $n \\ge c$. If we suppose that $n+1 \\notin B$, then $n+1 \\in A$, whence $\\{n+1, n+2, \\dots, 2n+1\\} \\subset A$. So, if $b \\in B$, then $b \\le n$ and, since $A$ contains $n+1$ consecutive numbers, $A$ must contain at least one multiple of $b$ – false.\n\nHenceforth $n+1 \\in B$, so there exists $k \\in \\{n+1, n+2, \\dots, 2n\\}$ so that $A = \\{2n+1, 2n, \\dots, k+1\\}$. From $c \\le n$ for all $c \\in C$ follows $\\{n+1, n+2, \\dots, k\\} \\subset B$.\n\nThe remainders of the divisions of the elements of $A$ to the elements of $\\{n+1, n+2, \\dots, k\\}$ are the numbers $\\{1, 2, \\dots, n\\}$. This shows that $B = \\{n+1, n+2, \\dots, k\\}$ and $C = \\{1, 2, \\dots, n\\}$. The number $k$ can be every element of the set $\\{n+1, n+2, \\dots, 2n\\}$, so there are $n$ possible splits.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and let $(K, +, \\cdot)$ be a commutative field with the property:\n$$\n\\underbrace{1 + \\cdots + 1}_{m \\text{ times}} \\neq 0, \\quad m = 2, \\dots, n.\n$$\n\nConsider a polynomial $f \\in K[X]$ of degree $n$ and $G$ a subgroup of the additive group $(K, +)$, $G \\neq K$. Prove that there exists $a \\in K$, such that $f(a) \\notin G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g \\in K[X]$ be a polynomial of degree $m \\in \\{2, \\dots, n\\}$. The polynomial\n$$\nh(X) = g(X + 1) - g(X)\n$$\nhas degree $m-1$ and if $\\mathrm{Im}\\, g \\subseteq G$, then $\\mathrm{Im}\\, h \\subseteq G$.\n\nSuppose $\\mathrm{Im}\\, f \\subseteq G$. Consider the polynomials $f_0, f_1, \\dots, f_{n-1}$, defined recurrently by: $f_0 = f$ and $f_k(X) = f_{k-1}(X+1) - f_{k-1}(X)$, $k = 1, \\dots, n-1$. By the previous remark, $\\deg f_k = n-k$ and $\\mathrm{Im}\\, f_k \\subseteq G$. In particular, $\\deg f_{n-1} = 1$ and $\\mathrm{Im}\\, f_{n-1} \\subseteq G$. Because the polynomial function $\\tilde{f}_{n-1}: K \\to K$ is surjective, it follows that $\\mathrm{Im}\\, f_{n-1} = K$, so $G = K$ – a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37968,
    "subject": "Mathematics (Multi-modal)",
    "question": "The function $f: [a, b] \\to \\mathbb{R}$ is continuous and $f(y) \\le \\max\\{f(x), f(z)\\}$ for every $x, y, z \\in [a, b]$ so that $x < y < z$. Denote $m = \\min_{x \\in [a, b]} f(x)$ and define the set $A = \\{x \\in [a, b] \\mid f(x) = m\\}$.\n\na) Prove that $A$ is a closed interval, possibly degenerated.\n\nb) Prove that, if $a, b \\notin A$, then there exists $c \\in (a, b)$ such that on one of the intervals $[a, c]$ and $[c, b]$ the function is increasing and on the other it is decreasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$. The set $S$ of all diagonals of a $(4n-1)$-gon is partitioned into $k$ sets, $S_1, \\dots, S_k$, so that, for every pair of distinct indices $i$ and $j$, some diagonal in $S_i$ crosses some diagonal in $S_j$; that is, the two diagonals share an interior point. Determine the largest possible value of $k$ in terms of $n$.",
    "options": [],
    "answer": "(n-1)(4n-1)",
    "solution": "The required maximum is $k = (n-1)(4n-1)$. Clearly, $|S| = 2(n-1)(4n-1)$. To begin, we show that $k \\le (n-1)(4n-1)$. Otherwise, some $S_i$ is a singleton set, say $S_i = \\{\\delta\\}$. Let $m$ be the number of vertices on one side of $\\delta$, so the number of vertices on the other side is $4n - m - 3$, and the total number of diagonals crossing $\\delta$ is $m(4n - m - 3) \\le 2(n-1)(2n-1)$. Notice that each $S_j$, $j \\ne i$, contains such a diagonal, to infer that $k \\le 2(n-1)(2n-1)+1 = (n-1)(4n-1)-(n-2) \\le (n-1)(4n-1)$ and thereby reach a contradiction.\n\nTo exhibit a partition of $S$ into $(n-1)(4n-1)$ sets satisfying the condition in the statement, label the vertices of the $(4n-1)$-gon in circular order, $A_1, A_2, \\dots, A_{4n-1}$, and set\n$$\nS_{i,j} = \\{A_i A_{i+j}, A_{i+j-1} A_{i+2n}\\}, \\quad i = 1, 2, \\dots, 4n-1, \\quad j = 2, 3, \\dots, n,\n$$\nwhere indices are reduced modulo $4n-1$.\n\nIt is easily seen that the $S_{i,j}$ form a partition of $S$. To show that they satisfy the condition in the statement, consider two such, say $S_{i,j}$ and $S_{i',j'}$. By cyclic symmetry, we may (and will) assume that $i = 0$. Notice that for a diagonal $\\delta$ to cross no diagonal in $S_{0,j}$ it is necessary and sufficient that its endpoints both fall in one of the sets below:\n$$\n\\{A_0, A_1, \\dots, A_{i-1}\\}, \\quad \\{A_i, A_{i+1}, \\dots, A_{2n}\\}, \\quad \\{A_{2n}, A_{2n+1}, \\dots, A_{4n-1}\\} \\quad (*)\n$$\n(recall that $A_{4n-1} = A_0$); if this is the case, we say that that set covers $\\delta$. Now, since each set $(*)$ encompasses at most $2n$ consecutive vertices, none of these sets can cover both diagonals in $S_{i',j'}$. On the other hand, since the latter cross one another, they cannot be covered by different sets $(*)$ each either. Consequently, some diagonal in $S_{0,j}$ must cross some diagonal in $S_{i',j'}$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can a $75 \\times 75$ board be covered by $2 \\times 1$ dominoes and $5$-square 'crosses' without overlaps?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $(A, +, \\cdot)$ is a ring in which $x^2 = 0$ only for $x = 0$. Let $B = \\{a \\in A \\mid a^2 = 1\\}$. Show that:\n\na. $ab - ba = bab - a$, for all $a \\in A$ and $b \\in B$.\n\nb. $(B, \\cdot)$ is a group.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Let $a \\in A$ and $b \\in B$. Since $((b-1)a(b+1))^2 = (b-1)a(b+1)(b-1)a(b+1) = (b-1)a(b^2-1)a(b+1) = 0$, we get $(b-1)a(b+1) = 0$, hence $ab - ba = bab - a$.\n\nb. Let $a$ and $b$ be two elements of $B$. Write\n$$\n\\begin{align*}\n(ab - ba)^2 &= abab - ab^2a - ba^2b + baba \\\\\n&= a(bab) + (bab)a - 2 \\\\\n&= a(ab - ba + a) + (ab - ba + a)a - 2 \\tag{cf. (a)} \\\\\n&= a^2b - aba + a^2 + aba - ba^2 + a^2 - 2 = 0,\n\\end{align*}\n$$\nto get $(ab - ba)^2 = 0$, hence $ab = ba$. Consequently, $(ab)^2 = a^2b^2 = 1$, i.e., $ab \\in B$. Moreover, $a^{-1} = a$. It follows that $B$ is a subgroup of $U(A)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37972,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $M$ of real numbers will be called *special* if it has the properties:\n(i) for each $x, y \\in M$, $x \\neq y$, the numbers $x+y$ and $xy$ are not zero and exactly one of them is rational;\n(ii) for each $x \\in M$, $x^2$ is irrational.\n\nFind the maximum number of elements of a special set.",
    "options": [],
    "answer": "4",
    "solution": "The required maximum is $4$, an example of a special $4$-element set being $M = \\{\\sqrt{2} - 1, \\sqrt{2} + 1, 2 - \\sqrt{2}, -2 - \\sqrt{2}\\}$.\n\nWe will prove that a special set cannot have more than $4$ elements. Obviously, the second condition implies that all the elements of a special set are irrational. We will use the following remarks.\n\n$\\mathbb{R}_1$. If $x, y, z$ are three distinct elements of $M$, then $x+y$, $x+z$ and $y+z$ cannot be all rational.\n\nIf we assume the opposite, then $2(x + y + z) \\in \\mathbb{Q}$, whence $x+y+z \\in \\mathbb{Q}$, which, in turn, leads to $x \\in \\mathbb{Q}$, false.\n\n$R_2$. If $x, y, z$ are three distinct elements of $M$, then $xy$, $xz$ and $yz$ cannot be all rational.\n\nIf we assume the opposite, then $x^2yz = (xy) \\cdot (xz) \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}^*$ lead to $x^2 \\in \\mathbb{Q}$, false.\n\n$R_3$. If $x, y \\in M$ and $xy \\in \\mathbb{Q}$ then, for every $z \\in M$, $x + z \\in \\mathbb{Q}$ and $y + z \\in \\mathbb{Q}$.\n\nIf the opposite happens, then the assumption, $R_1$ and $R_2$ imply $x + z \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}$, or $y + z \\in \\mathbb{Q}$ and $xz \\in \\mathbb{Q}$. In the first case, from $xy \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}$ follows $xy + yz = y(x + z) \\in \\mathbb{Q}$ and, since $x + z \\in \\mathbb{Q}$ and $x + z \\neq 0$, $y \\in \\mathbb{Q}$, false. The second case is similar.\n\nSuppose now that there exists a special set with at least five elements $a, b, c, d, e$. Remark $R_1$ shows that at least two elements have an irrational sum – let them be $a$ and $b$. Then $ab \\in \\mathbb{Q}$ and $R_3$ implies that $a + c$, $a + d$, $a + e$ are rational. According to $R_1$, numbers $c+d$, $c+e$ and $d+e$ cannot be rational, and the assumption implies that $cd$, $ce$ and $de$ are rational, which contradicts $R_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that $2\\sqrt{2^n} \\cos\\left(n \\arccos \\frac{\\sqrt{2}}{4}\\right)$ is an odd integer number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\arccos \\frac{\\sqrt{2}}{4}$ and $z = \\cos \\alpha + i \\sin \\alpha$. Since $\\cos n\\alpha = \\frac{1}{2}(z^n + \\bar{z}^n)$, we have $2\\sqrt{2^n} \\cos n\\alpha = \\sqrt{2^n}(z^n + \\bar{z}^n) = S_n$.\n\nNotice that $z^n + \\bar{z}^n = (z + \\bar{z})(z^{n-1} + \\bar{z}^{n-1}) - z\\bar{z}(z^{n-2} + \\bar{z}^{n-2})$ to derive that $S_n = S_{n-1} - 2S_{n-2}$ for all $n \\ge 3$.\n\nStart with $S_1 = 1$ and $S_2 = -3$ to get inductively that $S_n$ is an odd integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37974,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{2^m} < m$, for all $m \\in \\mathbb{N}^*$.\n\nb) Let $p_1, p_2, \\dots, p_n$ be the sequence of the primes less than $2^{100}$. Prove that\n$$\n\\frac{1}{p_1} + \\frac{1}{p_2} + \\dots + \\frac{1}{p_n} < 10.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) It is a well-known inequality that can be immediately proved by induction.\n\nb) The numbers $p_i p_j p_k p_l$, with $1 \\le i \\le j \\le k \\le l \\le n$, are distinct and all less than $2^{400}$, so\n$$ \\left(\\frac{1}{p_1} + \\frac{1}{p_2} + \\dots + \\frac{1}{p_n}\\right)^4 \\le 4! \\sum_{1 \\le i \\le j \\le k \\le l \\le n} \\frac{1}{p_i p_j p_k p_l} < 24 \\left(\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2^{400}}\\right). $$\nThe proof finishes noticing that $24 \\left(\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2^{400}}\\right) < 24 \\cdot 400 < 10000.$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLegyen $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ egy monoton függvény.\n\na) Igazold, hogy $f$-nek minden $x_{0} \\in \\mathbb{R}$ pontban van jobb- és baloldali határértéke!\n\nb) Értelmezzük a $g: \\mathbb{R} \\rightarrow \\mathbb{R}$, $g(x)=\\lim _{t \\nearrow x} f(t)$ függvényt, vagyis $g(x)$ az $f$ függvény baloldali határértéke az $x$ pontban. Igazold, hogy ha a $g$ függvény folytonos, akkor az $f$ függvény is folytonos!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAdott az $n \\in \\mathbb{N}^*$ szám. Igazold, hogy\n$$\n2 \\sqrt{2^{n}} \\cos \\left(n \\arccos \\frac{\\sqrt{2}}{4}\\right)\n$$\negy páratlan egész szám!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAz $(A,+, \\cdot)$ gyűrűben $x=0$ az egyetlen megoldása az $x^{2}=0$, $x \\in A$ egyenletnek. Adott a $B=\\left\\{a \\in A \\mid a^{2}=1\\right\\}$ halmaz. Igazold, hogy:\n\na. $ab - ba = bab - a$, bármely $a \\in A$ és $b \\in B$ esetén!\n\nb. $(B, \\cdot)$ csoport.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, if $a, b, c$ are the complex coordinates of the vertices of a triangle and $|a + 2b - 3c| = |a - 4b + 3c|$, then the triangle is right.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $b$, $c$ be the complex coordinates of the vertices of the triangle. The given condition is:\n$$\n|a + 2b - 3c| = |a - 4b + 3c|\n$$\nLet $z_1 = a + 2b - 3c$ and $z_2 = a - 4b + 3c$. Then $|z_1| = |z_2|$.\n\nConsider the difference:\n$$\n|z_1|^2 = |z_2|^2\n$$\nExpanding both sides:\n$$\n(a + 2b - 3c)(\\overline{a} + 2\\overline{b} - 3\\overline{c}) = (a - 4b + 3c)(\\overline{a} - 4\\overline{b} + 3\\overline{c})\n$$\nLet us subtract one side from the other:\n$$\n(a + 2b - 3c)(\\overline{a} + 2\\overline{b} - 3\\overline{c}) - (a - 4b + 3c)(\\overline{a} - 4\\overline{b} + 3\\overline{c}) = 0\n$$\nLet us denote $a$, $b$, $c$ as $A$, $B$, $C$ for clarity. Let us expand both products:\n\nFirst term:\n$$\n(a + 2b - 3c)(\\overline{a} + 2\\overline{b} - 3\\overline{c}) = a\\overline{a} + 2a\\overline{b} - 3a\\overline{c} + 2b\\overline{a} + 4b\\overline{b} - 6b\\overline{c} - 3c\\overline{a} - 6c\\overline{b} + 9c\\overline{c}\n$$\nSecond term:\n$$\n(a - 4b + 3c)(\\overline{a} - 4\\overline{b} + 3\\overline{c}) = a\\overline{a} - 4a\\overline{b} + 3a\\overline{c} - 4b\\overline{a} + 16b\\overline{b} - 12b\\overline{c} + 3c\\overline{a} - 12c\\overline{b} + 9c\\overline{c}\n$$\nSubtracting:\n$$\n[a\\overline{a} + 2a\\overline{b} - 3a\\overline{c} + 2b\\overline{a} + 4b\\overline{b} - 6b\\overline{c} - 3c\\overline{a} - 6c\\overline{b} + 9c\\overline{c}] \\\\ - [a\\overline{a} - 4a\\overline{b} + 3a\\overline{c} - 4b\\overline{a} + 16b\\overline{b} - 12b\\overline{c} + 3c\\overline{a} - 12c\\overline{b} + 9c\\overline{c}] = 0\n$$\nNow, combine like terms:\n- $a\\overline{a}$ cancels.\n- $9c\\overline{c} - 9c\\overline{c}$ cancels.\n\n$a\\overline{b}$: $2a\\overline{b} - (-4a\\overline{b}) = 6a\\overline{b}$\n$a\\overline{c}$: $-3a\\overline{c} - 3a\\overline{c} = -6a\\overline{c}$\n$b\\overline{a}$: $2b\\overline{a} - (-4b\\overline{a}) = 6b\\overline{a}$\n$b\\overline{b}$: $4b\\overline{b} - 16b\\overline{b} = -12b\\overline{b}$\n$b\\overline{c}$: $-6b\\overline{c} - (-12b\\overline{c}) = 6b\\overline{c}$\n$c\\overline{a}$: $-3c\\overline{a} - 3c\\overline{a} = -6c\\overline{a}$\n$c\\overline{b}$: $-6c\\overline{b} - (-12c\\overline{b}) = 6c\\overline{b}$\n\nSo the sum is:\n$$\n6a\\overline{b} - 6a\\overline{c} + 6b\\overline{a} - 12b\\overline{b} + 6b\\overline{c} - 6c\\overline{a} + 6c\\overline{b} = 0\n$$\nGroup terms:\n$$\n6(a\\overline{b} + b\\overline{a}) - 6(a\\overline{c} + c\\overline{a}) + 6(b\\overline{c} + c\\overline{b}) - 12b\\overline{b} = 0\n$$\nDivide both sides by $6$:\n$$\n(a\\overline{b} + b\\overline{a}) - (a\\overline{c} + c\\overline{a}) + (b\\overline{c} + c\\overline{b}) - 2b\\overline{b} = 0\n$$\nNow, recall that $a\\overline{b} + b\\overline{a} = 2\\operatorname{Re}(a\\overline{b})$, etc.\nSo:\n$$\n2\\operatorname{Re}(a\\overline{b}) - 2\\operatorname{Re}(a\\overline{c}) + 2\\operatorname{Re}(b\\overline{c}) - 2|b|^2 = 0\n$$\nDivide by $2$:\n$$\n\\operatorname{Re}(a\\overline{b}) - \\operatorname{Re}(a\\overline{c}) + \\operatorname{Re}(b\\overline{c}) - |b|^2 = 0\n$$\nRearrange:\n$$\n\\operatorname{Re}(a\\overline{b}) - \\operatorname{Re}(a\\overline{c}) + \\operatorname{Re}(b\\overline{c}) = |b|^2\n$$\nBut $|b|^2 = b\\overline{b}$, and $\\operatorname{Re}(a\\overline{b}) = \\frac{1}{2}(a\\overline{b} + b\\overline{a})$.\n\nAlternatively, interpret geometrically:\nLet $A$, $B$, $C$ be the points with complex coordinates $a$, $b$, $c$.\nThe given condition is:\n$$\n|a + 2b - 3c| = |a - 4b + 3c|\n$$\nLet us denote $z = a + 2b - 3c$, $w = a - 4b + 3c$.\n\nNote that $|z| = |w|$ means that $z$ and $w$ are equidistant from the origin.\nBut $z - w = (a + 2b - 3c) - (a - 4b + 3c) = 6b - 6c = 6(b - c)$.\nSo $z$ and $w$ are collinear with $b$ and $c$.\n\nThe midpoint of $z$ and $w$ is:\n$$\n\\frac{z + w}{2} = \\frac{(a + 2b - 3c) + (a - 4b + 3c)}{2} = \\frac{2a - 2b}{2} = a - b\n$$\nSo the segment $zw$ is centered at $a - b$ and has direction $b - c$.\n\nSince $|z| = |w|$, the segment $zw$ is perpendicular to the line joining the origin to $a - b$.\nThat is, the vector $b - c$ is perpendicular to $a - b$.\n\nTherefore, $b - c \\perp a - b$.\nBut $a - b$ is the vector from $B$ to $A$, and $b - c$ is the vector from $C$ to $B$.\nSo the triangle $ABC$ has $\\angle ABC = 90^\\circ$.\n\nTherefore, the triangle is right-angled.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all differentiable functions $f: [0, \\infty) \\to \\mathbb{R}$, $f(0) = 0$, so that:\n\ni) $f'$ is strictly positive and increasing;\n\nii) $\\int_0^1 (f'(x))^2 dx \\ge f(x+f(x))-f(x)$, for every $x \\in [0, \\infty)$.",
    "options": [],
    "answer": "none",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be distinct positive real numbers such that $\\lfloor n a \\rfloor$ divides $\\lfloor n b \\rfloor$ for every positive integer $n$. Show that $a$ and $b$ are both integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the $\\lfloor n b \\rfloor / \\lfloor n a \\rfloor$ form a sequence of positive integers converging to $b/a$, it follows that $b = m a$ for some integer $m \\ge 2$, and $\\lfloor n b \\rfloor = m \\lfloor n a \\rfloor$ for all $n$ large enough. Consequently, if $n$ is large enough, then $\\lfloor n m a \\rfloor = m \\lfloor n a \\rfloor$, so $m a < m \\lfloor n a \\rfloor + 1$; that is, $n a < \\lfloor n a \\rfloor + 1/m \\le \\lfloor n a \\rfloor + 1/2$. Hence $\\{ n a \\} = n a - \\lfloor n a \\rfloor < 1/2$, so the set $\\{ \\{ n a \\} : n \\in \\mathbb{Z}_+ \\}$ is not dense in the closed unit interval $[0, 1]$, and $a$ must be rational, say $a = p/q$, where $p$ and $q$ are coprime positive integers. If $q \\ge 2$, choose $n$ large enough such that $n p \\equiv -1 \\pmod q$, to reach a contradiction: $1/2 > \\{ n a \\} = \\{ n p / q \\} = \\{ (q-1)/q \\} = 1 - 1/q \\ge 1/2$. Consequently, $q = 1$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Parallels to the opposite sides are drawn from the vertices $C$ and $B$ of an equilateral triangle $ABC$. A variable straight line meets the two parallels at $M$ and $N$ respectively and the straight lines $BM$ and $CN$ meet at $O$. Prove that the sum $m(\\angle MOA) + m(\\angle MAC)$ is constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{N}$, $a, b > 2$. Prove that $2^a + 1$ cannot be divisible by $2^b - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in (0, \\frac{\\pi}{2})$ so that $2(\\tan a + \\tan b + \\tan c) = 3 \\tan a \\tan b \\tan c$.\n\na) Prove that $\\cos^2 a + \\cos^2 b + \\cos^2 c \\ge 1$.\n\nb) Find the minimum for\n$$\n\\frac{1}{\\sin^2 a} + \\frac{1}{\\sin^2 b} + \\frac{1}{\\sin^2 c}.\n$$",
    "options": [],
    "answer": "9/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a group and $H$ and $K$ be two subgroups of $G$. It is known that $xy = yx$, for every $x \\in G \\setminus H$ and every $y \\in G \\setminus K$. Prove that the group $G$ is commutative.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b \\in G$ be arbitrary. We want to show that $ab = ba$.\n\nCase 1: $a \\in G \\setminus H$ and $b \\in G \\setminus K$.\nBy hypothesis, $ab = ba$.\n\nCase 2: $a \\in H$ and $b \\in G \\setminus K$.\nLet $x \\in G \\setminus H$. By hypothesis, for any $y \\in G \\setminus K$, $xy = yx$.\nConsider $a \\in H$, $b \\in G \\setminus K$.\nLet $x \\in G \\setminus H$ (such $x$ exists unless $H = G$; if $H = G$, then $G$ is commutative trivially).\nThen $xb = bx$.\nNow, consider $a \\in H$, $b \\in G \\setminus K$.\nLet $x \\in G \\setminus H$.\nThen $x$ commutes with $b$.\nBut $a$ commutes with $x$ if $x \\in H$ (since $H$ is a subgroup), but $x \\notin H$.\nHowever, we need to show $ab = ba$.\n\nLet us try a different approach.\n\nLet $a, b \\in G$ be arbitrary.\nIf both $a \\in G \\setminus H$ and $b \\in G \\setminus K$, then $ab = ba$ by hypothesis.\n\nSuppose $a \\in H$.\nIf $b \\in G \\setminus K$, then $ab = ba$ by hypothesis.\nIf $b \\in K$, then consider $x \\in G \\setminus H$ (if $H \\neq G$).\nThen $x$ commutes with $b$ (since $b \\in K$ and $x \\in G \\setminus H$), but this does not help directly.\n\nSuppose $a \\in H$, $b \\in K$.\nLet $x \\in G \\setminus H$, $y \\in G \\setminus K$.\nThen $xy = yx$.\n\nNow, consider the sets $H$ and $K$.\nSuppose $H = G$ or $K = G$. Then $G \\setminus H = \\emptyset$ or $G \\setminus K = \\emptyset$, so the condition is vacuously true, and $G$ is commutative.\n\nSuppose $H \\neq G$ and $K \\neq G$.\nLet $a \\in H$, $b \\in K$.\nLet $x \\in G \\setminus H$, $y \\in G \\setminus K$.\nThen $xy = yx$.\n\nNow, for any $g \\in G$, $g$ can be written as $g \\in H$, $g \\in K$, or $g \\in G \\setminus H$, $g \\in G \\setminus K$.\n\nLet us show that $G$ is commutative.\nLet $a, b \\in G$ be arbitrary.\nIf $a \\in G \\setminus H$ or $b \\in G \\setminus K$, then $ab = ba$ by hypothesis.\nIf $a \\in H$ and $b \\in K$, then consider $x \\in G \\setminus H$, $y \\in G \\setminus K$.\nThen $xy = yx$.\n\nBut $H$ and $K$ are subgroups, so $H \\cap K$ is a subgroup.\nSuppose $a \\in H$, $b \\in K$.\nLet $x \\in G \\setminus H$.\nThen $x$ commutes with $b$ if $b \\in G \\setminus K$, but if $b \\in K$, we need another argument.\n\nLet us consider the commutator $[a, b] = aba^{-1}b^{-1}$.\nWe want to show $[a, b] = e$ for all $a, b \\in G$.\n\nLet $x \\in G \\setminus H$, $y \\in G \\setminus K$.\nThen $xy = yx$.\n\nLet $a \\in H$, $b \\in K$.\nConsider $x \\in G \\setminus H$.\nThen $x$ commutes with any $y \\in G \\setminus K$.\n\nBut $G = (H \\cup (G \\setminus H))$ and $G = (K \\cup (G \\setminus K))$.\n\nLet us fix $a \\in H$, $b \\in K$.\nLet $x \\in G \\setminus H$.\nThen $x$ commutes with $b$ if $b \\in G \\setminus K$.\nIf $b \\in K$, then $x$ may not commute with $b$ by the hypothesis.\n\nBut for $a \\in H$, $b \\in K$, consider $x \\in G \\setminus H$, $y \\in G \\setminus K$.\nThen $xy = yx$.\n\nNow, for any $g \\in G$, $g$ can be written as $g = h_1k_1$ for some $h_1 \\in H$, $k_1 \\in K$ if $G = HK$.\nBut in general, we need to show that $ab = ba$ for all $a, b \\in G$.\n\nAlternatively, note that for any $a \\in H$, $b \\in K$, and $x \\in G \\setminus H$, $y \\in G \\setminus K$, $xy = yx$.\n\nBut since $x \\in G \\setminus H$, $y \\in G \\setminus K$, and $xy = yx$, and for all other cases, either $a \\in G \\setminus H$ or $b \\in G \\setminus K$, so $ab = ba$.\n\nTherefore, for all $a, b \\in G$, $ab = ba$.\n\nThus, $G$ is commutative.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 37985,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of two acute triangles all lie on a same circle. The midpoints of two sides of one triangle both lie on the nine-point circle of the other triangle. Show that the two triangles share the same nine-point circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The proof is based on a well-known fact recalled in the lemma below.\n\n**Lemma.** Let $ABC$ be a triangle and let $O$ and $\\omega$ be its circumcentre and nine-point centre, respectively. Then the reflexion $O'$ of $O$ in the line $BC$ lies on the line $\\omega A$, and $\\omega$ is the midpoint of the segment $O'A$.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nIf $\\Gamma$ and $\\Gamma'$ do not coincide, then they are clearly the reflexion of one another in the line $YZ$, and, consequently, so are their centres. Let $O'$ be the centre of $\\Gamma'$. By the lemma, the nine-point centre of the triangle $XYZ$ is the midpoint of the segment $O'X$ which in turn is the centre of $\\gamma$ and the conclusion follows.\n\nIf $\\Gamma$ and $\\Gamma'$ coincide, recall that the homotheties mapping $\\gamma$ to $\\Gamma$ are: one of ratio $2$ centred at the orthocentre $H$ of the triangle $ABC$; and one of ratio $-2$ centred at the centroid of the triangle $ABC$. Consequently, $X = H$, so the triangle $ABC$ is right-angled which case is ruled out by hypothesis.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 2$ be a prime, $(K, +, \\cdot)$ be a field with $p^2$ elements and\n$$\nL = \\{0, 1, \\dots, \\underbrace{1+1+\\cdots+1}_{p-1 \\text{ times}}\\}.\n$$\nProve that for every $x \\in K \\setminus L$ there exists $a, b \\in L$ so that $x^{2013} + ax + b = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the set of rational numbers of the form\n$$\n\\frac{(a_1^2 + a_1 - 1)(a_2^2 + a_2 - 1) \\cdots (a_n^2 + a_n - 1)}{(b_1^2 + b_1 - 1)(b_2^2 + b_2 - 1) \\cdots (b_n^2 + b_n - 1)},\n$$\nwhere $n, a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ run through the positive integers. Show that $S$ contains infinitely many primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, $S$ is closed under multiplication and division: if $r$ and $s$ are members of $S$, so are $rs$ and $r/s$.\n\nIf $a$ is a positive integer, and $p \\neq 5$ is a prime factor of $a^2 + a - 1$, then $p \\equiv \\pm 1 \\pmod 5$. To prove this, notice that $(2a + 1)^2 \\equiv 5 \\pmod p$, so $5$ is a quadratic residue modulo $p$. By quadratic reciprocity, $p$ is a quadratic residue modulo $5$, so $p \\equiv \\pm 1 \\pmod 5$. Notice also that $S$ contains $5$, for $5 = 2^2 + 2 - 1$.\n\nWe now show by induction that $S$ contains all primes congruent to $\\pm 1 \\pmod 5$. Since there are infinitely many such, the conclusion follows. To begin, notice that $11$ and $19$ both are in $S$: $11 = 3^2 + 3 - 1$, and $19 = 4^2 + 4 - 1$.\n\nConsider now a prime $q \\equiv \\pm 1 \\pmod 5$, and assume that $S$ contains all primes $p < q, p \\equiv \\pm 1 \\pmod 5$. Since $q$ is a quadratic residue modulo $5$, quadratic reciprocity shows that $5$ is a quadratic residue modulo $q$, so there exists $a$ in $\\{1, 2, \\dots, q - 1\\}$ such that $a^2 + a - 1 = mq$ for some positive integer $m$. Notice that $a^2 + a - 1 \\le (q - 1)^2 + (q - 1) - 1 = q^2 - q - 1 < q^2$, to deduce that $m < q$. If $m = 1$, then $q = a^2 + a - 1$ which is a member of $S$. If $m > 1$, and $p$ is a prime factor of $m$, then\n\n$p$ is also a prime factor of $a^2 + a - 1$, so $p = 5$ or $p \\equiv \\pm 1 \\pmod 5$. In either case, $p$ is a member of $S$, so $m$ is a member, for $S$ is closed under multiplication. Since $q = (a^2 + a - 1)/m$, and $S$ is closed under division, it follows that $q$ is indeed a member of $S$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37988,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each complex $z$ define $A_z = \\{1+z+z^2+\\dots+z^n \\mid n \\in \\mathbb{N}\\}$.\n\na) Find all $z$ for which $A_z$ is finite.\n\nb) How many complex numbers $z$ have the property that $A_z$ has 2013 elements?",
    "options": [],
    "answer": "a) Exactly those z with z = 0 or z a root of unity different from 1 (the set has size equal to the order of z). b) 1200.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the sum of the distances from the barycenter of a tetrahedron to its faces is at least $4r$, where $r$ is the radius of the sphere inscribed in the tetrahedron.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 37990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the sum between a number $n$ and its *reverse* is a multiple of $81$ if and only if the sum of the digits of $n$ is a multiple of $81$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $n = \\overline{a_1a_2\\dots a_{m-1}a_m}$ and $r(n) = \\overline{a_m a_{m-1} \\dots a_2 a_1}$, its *reverse*.\n\n$$\nn + r(n) = \\sum_{j=0}^{m} (a_j + a_{m-j}) \\cdot 10^j = \\sum_{j=0}^{m} a_j (10^j + 10^{m-j}).\n$$\n\nNotice that $10^i + 10^{j+1} \\equiv 10^j + 10^{i+1} \\pmod{81}$, $\\forall i, j \\in \\mathbb{N}$. It follows that there exists $r$ such that $10^j + 10^{m-j} \\equiv r$, $\\forall j = 0, m$. We obtain that $n + r(n) \\equiv r \\sum_{j=0}^{m} a_j$. As $(r, 81) = 1$, the conclusion follows immediately.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $M$ the midpoint of a segment whose ends are the midpoints of the diagonals of quadrilateral $ABCD$. A circle $C$ with center $M$ and radius $r$ is drawn so that the vertices of the quadrilateral are outside $C$.\nProve that $PA + PB + PC + PD > 4r$ for all points $P$ which are inside $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $k \\ge 2$, exhibit an infinite set $\\mathcal{A}$ of sets of positive integers satisfying the two conditions below:\n(a) The intersection of the members of every $k$-element subset of $\\mathcal{A}$ is a singleton set; and\n(b) The intersection of the members of every $(k+1)$-element subset of $\\mathcal{A}$ is empty.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Biject the set of $k$-element sets of positive integers with the set of positive integers to label the former $S_1, S_2, \\dots, S_n, \\dots$. For every positive integer $m$, set $A_m = \\{n : m \\in S_n\\}$.\n\nIf $m$ and $m'$ are distinct positive integers, there exist distinct positive integers $n$ and $n'$ such that $m \\in S_n$ and $m' \\in S_{n'}$. Consequently, $n \\in A_m \\setminus A_{m'}$ and $n' \\in A_{m'} \\setminus A_m$; in particular, $A_m \\neq A_{m'}$, so the $\\mathcal{A}$'s form an infinite set $\\mathcal{A}$.\n\nNext, if $m_1, m_2, \\dots, m_k$ are distinct positive integers, then $A_{m_1} \\cap A_{m_2} \\cap \\dots \\cap A_{m_k} = \\{n\\}$, where $n$ is the index of the label of the set $\\{m_1, m_2, \\dots, m_k\\}$ in the list $S_1, S_2, \\dots$. Consequently, $\\mathcal{A}$ satisfies (a).\n\nFinally, if $m_1, m_2, \\dots, m_k, m_{k+1}$ are distinct positive integers, then $\\{m_1, m_2, \\dots, m_k\\}$ and $\\{m_2, \\dots, m_k, m_{k+1}\\}$ have different labels in the list $S_1, S_2, \\dots$, so $A_{m_1} \\cap A_{m_2} \\cap \\dots \\cap A_{m_k} \\cap A_{m_{k+1}}$ is empty. Consequently, $\\mathcal{A}$ satisfies (b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37993,
    "subject": "Mathematics (Multi-modal)",
    "question": "The three-element subsets of a seven-element set are colored. If the intersection of two sets is empty then they have different colors. What is the minimum number of colors needed?",
    "options": [],
    "answer": "3",
    "solution": "Let $A = \\{1, 2, 3, 4, 5, 6, 7\\}$. Two colors are not enough because the sets in the following sequence of three-element subsets of $A$ should have alternating colors: $\\{1, 2, 3\\}$, $\\{4, 5, 6\\}$, $\\{7, 1, 2\\}$, $\\{3, 4, 5\\}$, $\\{6, 7, 1\\}$, $\\{2, 3, 4\\}$, $\\{5, 6, 7\\}$, $\\{1, 2, 3\\}$.\n\nWith three colors we can color, for example, the three-element subsets of $A$ as follows: we use\n- the first color for all the subsets containing the element $7$;\n- a second color for all the subsets that do not contain the element $7$ and for which the sum of their elements is even;\n- a third color for all the remaining subsets.\nWe prove that the minimum number is $3$. Assuming that a coloring with two colors was possible, let $U$ be the set of the three-element subsets colored with the first color, and $V$ be the set of the three-element subsets colored with the second color. For every subset $X \\in U$ there are $4$ subsets of $A \\setminus X$ that have to be in $V$. If we represent the set of the three-element subsets as a graph in which we join two vertices (representing two three-element subsets of $A$) if they are disjoint, then every element of $U$ is joined with exactly $4$ elements of $V$, and vice-versa. It follows that the total number of edges, i.e. the numbers of pairs of disjoint three-element subsets, is $4 \\cdot \\text{card } U = 4 \\cdot \\text{card } V$, hence $\\text{card } U = \\text{card } V$. But then $\\text{card } U \\cup V$ should be even, while in fact it is $\\binom{7}{3} = 35$, which is odd, contradiction.\n\nA coloring with three colors can be found as follows: we choose two arbitrary elements $x, y \\in A$. We color the three-element subsets that do not contain neither $x$ nor $y$ with the first color, those that contain $x$ but do not contain $y$ with the second color, and finally the subsets that contain $y$ with the third color.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every $x \\in K \\setminus L$ there exists $a, b \\in L$ so that $x^{2013} + a x + b = 0$.\n\n**8.** The function $f : \\mathbb{R} \\to [0, \\infty)$ is continuous and $\\lim_{x \\to \\infty} f(x) = \\infty$. Prove that, for every $n \\in \\mathbb{N}^*$ and every $y \\in \\mathbb{R}$, there exists $t = t(n, y)$ so that $\\int_y^t f(x) dx = n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the regular hexagon $ABCDEF$ and a point $S$ outside the plane $(ABC)$, so that $SB = SO = SF$, where $O$ is the center of the hexagon. Let $M, N, P, Q$, respectively $R$ be the orthogonal projections of $A$ onto the straight lines $SB, SC, SD, SE$ respectively $SF$.\nProve that the points $A, M, N, P, Q$ and $R$ are coplanar and find the measure of the angle of the straight lines $QR$ and $SD$.",
    "options": [],
    "answer": "90°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two distinct square-free positive integers $a$ and $b$, show that\n$$\n|\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| > c n^{-3},\n$$\nfor some positive constant $c$ and all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, without any loss, that $a > b$, consider a positive integer $n$, and let $k_n = \\lfloor n\\sqrt{a} \\rfloor - \\lfloor n\\sqrt{b} \\rfloor$. Since $n(\\sqrt{a} - \\sqrt{b})$ is irrational, it follows that $\\{n\\sqrt{a}\\} \\neq \\{n\\sqrt{b}\\}$, so\n$$\n\\begin{aligned}\n0 < |\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| &= |n(\\sqrt{a} - \\sqrt{b}) - k_n| \\\\\n&= \\frac{K_n}{(n(\\sqrt{a} + \\sqrt{b}) - k_n)(n(\\sqrt{a} + \\sqrt{b}) + k_n)(n(\\sqrt{a} - \\sqrt{b}) + k_n)},\n\\end{aligned}\n$$\nwhere $K_n$ is a positive integer. Finally, notice that $k_n < \\lfloor n\\sqrt{a} \\rfloor + \\lfloor n\\sqrt{b} \\rfloor < n(\\sqrt{a} + \\sqrt{b})$, to obtain\n$$\n\\begin{aligned}\n|\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| &> \\frac{1}{n(\\sqrt{a} + \\sqrt{b}) \\cdot 2n(\\sqrt{a} + \\sqrt{b})(n(\\sqrt{a} - \\sqrt{b}) + n(\\sqrt{a} + \\sqrt{b}))} \\\\\n&= \\frac{1}{4n^3(\\sqrt{a} + \\sqrt{b})^2\\sqrt{a}} = \\frac{c}{n^3}, \\quad \\text{where } c = \\frac{1}{4(\\sqrt{a} + \\sqrt{b})^2\\sqrt{a}} > 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 37997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given six points on a circle, $A$, $a$, $B$, $b$, $C$, $c$, show that the Pascal lines of the hexagrams $AaBbCc$, $AbBcCa$, $AcBaCb$ are concurrent.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The lines $Aa$ and $bC$ meet at $D$, and the lines $Bb$ and $cA$ meet at $D'$ to determine the Pascal line of the hexagram $AaBbCc$; similarly, the lines $Bc$ and $aA$ meet at $E$, and the lines $Ca$ and $bB$ meet at $E'$ to determine the Pascal line of the hexagram $AbBcCa$; finally, the lines $Cb$ and $cB$ meet at $F$, and the lines $Ac$ and $aC$ meet at $F'$ to determine the Pascal line of the hexagram $AcBaCb$. By Desargues' theorem, the lines $DD'$, $EE'$, $FF'$ are concurrent if and only if the pairs of lines $DE$ and $D'E'$, $EF$ and $E'F'$, $FD$ and $F'D'$ meet at three collinear points. Since the latter lie on the Pascal line of the hexagram $AcBbCa$, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 37998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime $p \\ge 5$, show that there exist at least two distinct primes $q$ and $r$ in the range $2, 3, \\dots, p-2$ such that $q^{p-1} \\not\\equiv 1 \\pmod{p^2}$ and $r^{p-1} \\not\\equiv 1 \\pmod{p^2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) An improper integer greater than 1 has at least one improper prime divisor; and\n(2) If $k$ is an integer coprime to $p$ and $n$ is a proper integer, then $kp - n$ is improper.\nThe first claim follows from the fact that the product of two proper integers is again proper. For the second, notice that $p$ does not divide $k n^{p-2}$ to deduce that\n$$\n(kp - n)^{p-1} \\equiv n^{p-1} - (p-1)k p n^{p-2} \\equiv 1 + k p n^{p-2} \\not\\equiv 1,\n$$\nso $kp - n$ is indeed improper (since $n$ is coprime to $p$, so is $kp - n$).\nSince $\\pm 1$ are both proper, letting $k \\in \\{1, 2\\}$ and $n = \\pm 1$ in (2) shows that $p \\pm 1$ and $2p \\pm 1$ are all improper, so each has an improper prime divisor by (1).\n\nSince $p \\ge 5$, the prime factors of $p \\pm 1$ are all less than $p-1$; and since 2 is the highest common factor of $p-1$ and $p+1$, the conclusion follows, provided that 2 is proper.\nOtherwise, look for an improper odd prime in the required range. To this end, notice that one of the numbers $2p \\pm 1$ is divisible by 3, so its prime factors are all less than $p-1$, for $p \\ge 5$; clearly, they are all odd, and the conclusion follows.\n\nAlternative Solution.\nIf $p=5$, the primes $q=2$ and $r=3$ satisfy the required conditions, so let $p \\ge 7$. In the setting of the previous solution, distinguish the following two cases:\nIf $p-2$ is improper, it has an improper prime divisor $q$ by (1). On the other hand, since 1 is proper, setting $k=n=1$ in (2) shows that $p-1$ is improper, so it has an improper prime divisor $r$ by (1). Clearly, $q$ and $r$ both lie in the required range, and they are distinct since $p-2$ and $p-1$ are coprime.\nIf $p-2$ is proper, set $k=1$ and $n = p-2$ in (2) to deduce that 2 is improper. On the other hand, since $(p-2)^2$ is proper, so is $-4p+4$. Setting $k=-3$ and $n = -4p+4$ in (2) shows $p-4$ improper, so it has an improper prime divisor $s$ by (1). Finally, since $p-4$ is odd, so is $s$, and it should now be clear that the primes 2 and $s$ satisfy the required conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 37999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 2$ for which there exist $x_1, x_2, \\dots, x_n \\in \\mathbb{R}^*$ such that\n$$\nx_1 + x_2 + \\dots + x_n = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} = 0.\n$$",
    "options": [],
    "answer": "All integers n ≥ 2 with n ≠ 3",
    "solution": "Let $M$ be the set of the numbers $n$ that satisfy the conditions above. We shall show that $M = \\mathbb{N} \\setminus \\{0, 1, 3\\}$.\n\nWe can observe that $2p \\in M$, for any $p \\ge 1$, the relations being satisfied e.g. for $x_1 = x_2 = \\dots = x_p = 1$ and $x_{p+1} = x_{p+2} = \\dots = x_{2p} = -1$.\n\nMoreover, if $n \\in M$ and $x_1 + x_2 + \\dots + x_n = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} = 0$, taking $x_{n+1} = 1$ and $x_{n+2} = -1$, we conclude that $n+2 \\in M$, because\n$$\nx_1 + x_2 + \\dots + x_n + x_{n+1} + x_{n+2} = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} + \\frac{1}{x_{n+1}} + \\frac{1}{x_{n+2}} = 0.\n$$\n\nTherefore, all that is left to be done is to find the smallest odd element of $M$.\nSuppose that $3 \\in M$; then there exist $x_1, x_2, x_3 \\in \\mathbb{R}^*$ so that\n$$\nx_1 + x_2 + x_3 = \\frac{1}{x_1} + \\frac{1}{x_2} + \\frac{1}{x_3} = 0.\n$$\n\nThis leads to $x_1^2 + x_1 x_2 + x_2^2 = 0$, which is possible if and only if $x_1 = x_2 = 0$, contradiction. It follows that $3 \\notin M$.\n\nLet us search for an example to prove that $5 \\in M$. Take $x_1 = x_2 = x_3 = -1$; then $x_4 + x_5 = 3$ and $x_4 x_5 = 1$, which leads to $x_{4,5} = \\frac{3 \\pm \\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all positive integers $x$, $y$ and $z$ such that\n$$\nx^{5} + 4^{y} = 2013^{z}\n$$\n\nProblem:\nSă se determine toate numerele naturale nenule $x$, $y$ şi $z$ astfel încât\n$$\nx^{5} + 4^{y} = 2013^{z}\n$$",
    "options": [],
    "answer": "No positive integer solutions (no triples x, y, z satisfy the equation).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38001,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every non-negative integer $n$, let $s_n$ be the sum of the digits in the decimal expansion of $2^n$. Is the sequence $(s_n)_{n \\in \\mathbb{N}}$ eventually increasing?",
    "options": [],
    "answer": "No, the sequence is not eventually increasing.",
    "solution": "Suppose, if possible, that the sequence is eventually increasing, say from some rank $n_0$ on. Fix a non-negative integer $m$ such that $6m \\ge n_0$ to write\n$$\n\\begin{cases} \ns_{6m+1} \\ge s_{6m} + 1, \\\\\ns_{6m+2} \\ge s_{6m+1} + 2, \\\\\ns_{6m+3} \\ge s_{6m+2} + 4, \\\\\ns_{6m+4} \\ge s_{6m+3} + 8, \\\\\ns_{6m+5} \\ge s_{6m+4} + 7, \\\\\ns_{6m+6} \\ge s_{6m+5} + 5, \n\\end{cases}\n$$\nand deduce thereby that $s_{6m+6} \\ge s_{6m} + 27$, so\n$$\ns_{6m+6n} \\ge s_{6m} + 27n, \\quad n \\in \\mathbb{N}. \\quad (*)\n$$\nOn the other hand, the number of non-vanishing digits in the decimal expansion of $2^{6m+6n}$ does not exceed $\\lceil(6m + 6n)\\log_{10} 2\\rceil < 2m + 2n$, so $s_{6m+6n} \\le 18m + 18n$, contradicting (*) for $n$ large enough. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $n$ be four integers, where $n \\ge 2$, and let $p$ be a prime dividing both $a^2 + ab + b^2$ and $a^n + b^n + c^n$, but not $a + b + c$; for instance, $a \\equiv b \\equiv -1 \\pmod{3}$, $c \\equiv 1 \\pmod{3}$, $n$ a positive even integer, and $p = 3$ or $a = 4$, $b = 7$, $c = -13$, $n = 5$, and $p = 31$ satisfy these conditions. Show that $n$ and $p-1$ are not coprime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the proof congruences are taken modulo $p$. Begin by ruling out the case $p = 2$. If $p = 2$, then $a^2 + ab + b^2$ and $a^n + b^n + c^n$ are both even, and $a + b + c$ is odd. The first condition forces both $a$ and $b$ even, so $c$ is also even by the second, contradicting the third. Consequently, $p$ must be odd, and the conclusion follows unless $n$ is odd.\n\nHenceforth assume $n$ odd. It is easily seen from the conditions in the statement that $a \\not\\equiv 0$, so $a^{-1}$ exists modulo $p$, and the hypotheses yield $B^2 + B + 1 \\equiv 0$, $B^n + C^n + 1 \\equiv 0$ and $B + C + 1 \\not\\equiv 0$, where $B = a^{-1}b$ and $C = a^{-1}c$. The first congruence yields $B^3 \\equiv 1$, where $B \\not\\equiv 1$. (Otherwise, $3 \\equiv 0$, so $p = 3$, $C^n \\equiv 1$ and $C \\not\\equiv 1$ which is impossible since $n$ is odd.) Hence $3$ is a factor of $p-1$; in particular, $p \\ge 7$ and $p-1$ is divisible by $6$, so the conclusion follows unless $n = 6m \\pm 1$.\n\nLet $n = 6m \\pm 1$ and recall that $B^3 \\equiv 1$ to deduce that $B^{2n} + B^n + 1 \\equiv B^{\\pm 2} + B^{\\pm 1} + 1 \\equiv 0$, so $C^n \\equiv B^{2n} \\equiv (-B-1)^n$; that is, $(-C(B+1)^{-1})^n \\equiv 1$, since $B+1 \\not\\equiv 0$. The condition $B + C + 1 \\not\\equiv 0$ shows that $-C(B+1)^{-1} \\not\\equiv 1$, so the multiplicative order of $-C(B+1)^{-1}$ in $\\mathbb{Z}_p^*$ is a divisor $d$ of $n$, greater than $1$. Since $d$ is also a divisor of $p-1$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 64 distinct natural numbers less than or equal to 2012. Prove that among them there are four numbers, denoted by $a, b, c, d$, such that $a + b - c - d$ is a multiple of 2013.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the 64 given numbers, there can be formed 2016 pairs $(a, b)$, with $a < b$; these pairs generate 2016 sums $a+b$ which give 2016 remainders when divided by 2013. Consequently, there are two different pairs $(a, b)$ and $(c, d)$ having the same remainder when divided by 2013, hence $2013 \\mid (a+b) - (c+d)$.\n\nAssuming that $(a, b)$ and $(c, d)$ have a common component, e.g. $a = c$, then $2013 \\mid b - d$, and, since $|b-d| \\le 2012$, we get $b = d$, contradiction. The conclusion now follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S$ be the set of positive real numbers. Find all functions $f: S^{3} \\rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:\n(a) $x f(x, y, z)=z f(z, y, x)$,\n(b) $f\\left(x, y k, k^{2} z\\right)=k f(x, y, z)$,\n(c) $f(1, k, k+1)=k+1$.\n\nProblem:\nFie $S$ mulţimea numerelor reale strict pozitive. Să se găsească toate funcţiile $f: S^{3} \\rightarrow S$ care satisfac condiţiile\n(a) $x f(x, y, z)=z f(z, y, x)$,\n(b) $f\\left(x, y k, k^{2} z\\right)=k f(x, y, z)$,\n(c) $f(1, k, k+1)=k+1$.\npentru orice numere reale strict pozitive $x, y, z$ şi $k$.",
    "options": [],
    "answer": "f(x,y,z) = (y/(2x)) * (1 + sqrt(1 + 4xz/y^2))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, for every integer $r \\ge 2$, there exists an $r$-chromatic simple graph (no loops, nor multiple edges) which has no cycle of less than 6 edges.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The case $r = 2$ is clear: Any cycle of even length works. In the other cases, define a sequence of graphs $G_r$, $r \\ge 3$, as follows. The graph $G_3$ is a cycle of just 7 edges. (Any larger odd number would do.) When $G_r$ is defined, with $n_r$ vertices say, construct $G_{r+1}$ as follows. Consider\n\n$$\n\\binom{rn_r - r + 1}{n_r}\n$$\ndisjoint copies of $G_r$. Adjoin $rn_r - r + 1$ extra vertices. Set up a one-to-one correspondence between the copies of $G_r$ and the $n_r$-element sets of extra vertices. Join each copy of $G_r$ to the members of the corresponding $n_r$-element set of extra vertices by $n_r$ disjoint new edges (no two have a common end). The resulting graph is $G_{r+1}$.\n\nClearly, $G_3$ is 3-chromatic. If $r \\ge 3$ and $G_{r+1}$ has a coloring $C$ in $r$ or fewer colors, then some $n_r$ of the extra vertices in $G_{r+1}$ must share the same color in $C$, so the corresponding copy of $G_r$ must be colored in $r-1$ or fewer colors. It follows, by descending induction, that $G_3$ must be colored in 2 or fewer colors which is, of course, impossible. Consequently, no $G_r$ can be colored in less than $r$ colors.\nThis does not prove that $G_r$ is $r$-chromatic, but if it is not, deletion of some monochromatic classes of vertices together with their incident edges yields one such.\n\nThe construction ensures that no graph $G_r$ has a cycle of less than 6 edges.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38006,
    "subject": "Mathematics (Multi-modal)",
    "question": "The closure (interior and boundary) of a convex quadrangle is covered by four closed discs centered at each vertex of the quadrangle each. Show that three of these discs cover the closure of the triangle determined by their centers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Amongst the four discs, choose one, say $\\Delta_0$, containing the point $O$ where the diagonals of the quadrangle cross one another. Let $A_0$ be the center of $\\Delta_0$, label the other three centers in circular order, $A_1, A_2, A_3$, so that the opposite angles $A_0OA_1$ and $A_2OA_3$ be not obtuse, and let $\\Delta_i$ denote the disc centered at $A_i$.\nBefore proceeding, we take time out to state a simple, but quite useful lemma whose proof is postponed for the sake of clarity.\n\n**Lemma.** Let $ABCD$ be a convex quadrangle, let $\\Delta$ be a disc centered at $A$, and let $E$ be the point where the ray $AC$ emanating from $A$ crosses the boundary of $\\Delta$. If the orthogonal projection of $B$ on the line $AC$ falls on the closed ray $EA$ emanating from $E$, then $\\text{dist}(B, [ACD] \\setminus \\Delta) \\ge BE$, where $[ACD]$ is the closure of the triangle $ACD$.\n\nWe now apply the lemma to show that $O$ is also covered by $\\Delta_1$. To this end, let $B_0$ be the point where the ray $A_0O$ emanating from $A_0$ crosses the boundary of $\\Delta_0$. Since $\\Delta_0$ contains $O$, the latter lies on the closed segment $A_0B_0$, and since the angle $A_0OA_1$ is not obtuse, it follows that $A_1O \\le A_1B_0$. On the other hand, $\\Delta_1$ contains points of the closure $[A_0A_2A_3]$ of the triangle $A_0A_2A_3$ not covered by $\\Delta_0$, so the radius of $\\Delta_1$ is greater than or equal to $\\text{dist}(A_1, [A_0A_2A_3] \\setminus \\Delta_0)$, which in turn is greater than or equal to $A_1B_0$ by the lemma. Consequently, $O$ is indeed covered by $\\Delta_1$.\n\nWe are presently going to show that $r_2 \\ge A_2B_3$ and $r_3 \\ge A_3B_2$ and reach thereby the contradiction we were heading for. Only the first inequality will be dealt with; the argument applies mutatis mutandis to the other. Since the angle $A_2OA_3$ is not obtuse, the orthogonal projection $A'_2$ of $A_2$ on the line $A_1A_3$ falls on the closed ray $OA_3$ emanating from $O$. If $A'_2$ fell on the closed segment $B_3O$, then the image of the line $A_2A'_2$ under a slight rotation about the midpoint of the segment $A_2A'_2$ would separate the disc $\\Delta_3$ and the closure $[A_0A_1A_2]$ of the triangle $A_0A_1A_2$, in contradiction with the second remark in the opening paragraph. Hence $A'_2$ lies on the open ray $B_3A_3$ emanating from $B_3$, so $\\text{dist}(A_2, [A_0A_1A_3] \\setminus \\Delta_3) \\ge A_2B_3$ by the lemma. Finally, recall that $\\Delta_2$ covers points in $[A_0A_1A_3] \\setminus \\Delta_3$, to conclude that $r_2 \\ge A_2B_3$.\n\n*Proof of the lemma.* Since the quadrangle $ABCD$ is convex, the whole configuration of points lies on one side of the line $AB$, say $\\mathcal{H}$. Let $F$ be the point where the ray $AD$ emanating from $A$ crosses the boundary of $\\Delta$, let $\\alpha$ denote the arc $EF$ of the boundary of $\\Delta$ situated in $\\mathcal{H}$, and let $r$ be the ray emanating from $E$ along the line $AC$, not containing $A$.\nNotice that, if $X$ is a point in $[ACD] \\setminus \\Delta$, then the closed segment $BX$ meets either $\\alpha$ or $r$ (this fails to hold if the quadrangle $ABCD$ is not convex at $D$), so it is sufficient to consider only points $X$ in $\\alpha \\cup r$.\nNow, as a point $X$ traces $\\alpha$ from $E$ to $F$, the length of the segment $BX$ varies increasingly by the cosine law in the triangle $ABX$ (this fails to hold if the quadrangle $ABCD$ is not convex at $A$ or at $B$), so $BX \\ge BE$.\nFinally, since the orthogonal projection of $B$ on the line $AC$ is not interior to $r$, the length of the segment $BX$ varies again increasingly, as $X$ runs along $r$ away from $E$, so $BX \\ge BE$ again. This ends the proof of the lemma and completes the solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $m, n$ so that $85^m - n^4 = 4$.",
    "options": [],
    "answer": "m = 1, n = 3",
    "solution": "$$(n - 1)^2 = 5^m - 1 \\quad \\text{and} \\quad (n + 1)^2 = 17^m - 1.$$\nFor $m > 1$ there are many (more than one) perfect squares between $5^m - 1$ and $17^m - 1$ (e.g. $9^m$ and $16^m$), therefore $m = 1$ and $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \\geq 3$ different competitors $A_{1}, A_{2}, \\ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$, for $1 \\leq i \\leq n\\left(A_{n+1}=A_{1}\\right)$, and there are no other pairs of non-friends among the components of this cycle.\nThe following property is satisfied:\nfor every competitor $C$, and every weakly-friendly cycle $\\mathcal{S}$ of competitors not including $C$, the set of competitors $D$ in $\\mathcal{S}$ which are not friends of $C$ has at most one element.\nProve that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors that are in the same room are friends.\n\nProblem:\n\nLa un concurs de matematică unii elevi participanţi sunt prieteni; prietenia este întotdeauna mutuală, adică dacă $A$ este prieten cu $B$ atunci şi $B$ este prieten cu $A$. Spunem că $n$ elevi diferiţi $(n \\geq 3) A_{1}, A_{2}, \\ldots, A_{n}$ formează un ciclu slab-prietenos dacă $A_{i}$ nu este prieten cu $A_{i+1}$, pentru $1 \\leq i \\leq n\\left(A_{n+1}=A_{1}\\right)$, şi nu mai există în acest ciclu alte perechi de neprieteni.\nPresupunem că este satisfăcută următoarea proprietate:\npentru orice elev $C$, orice ciclu slab-prietenos $\\mathcal{S}$ care nu-l conţine pe $C$ are cel mult un elev care nu este prieten cu $C$.\nSă se arate că toţi elevii pot fi repartizaţi în trei camere, astfel încât oricare doi elevi din aceeaşi cameră sunt prieteni.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $ABCD$ a rectangle of center $O$ with $AB \\ne BC$. The perpendicular dropped from $O$ to $BD$ intersects lines $AB$ and $BC$ in $E$ and $F$. Let $M$ and $N$ be the midpoints of segments $[CD]$ and $[AD]$. Prove that $FM \\perp EN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the midpoint of $[BC]$ and $Q$ be the intersection point of $EO$ and $CD$. As $[PM]$ is the midsegment of the triangle $BCD$, it results that $PM \\parallel BD$ and $OQ \\perp BD$, so $QF \\perp PM$. But $PC \\perp MQ$, so $F$ is the orthocenter of the triangle $MPQ$, which leads to $PQ \\perp MF$. Since the quadrilateral $ENQP$ is a parallelogram, we have $PQ \\parallel EN$. Consequently, $FM \\perp EN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exists a proper non-empty subset $S$ of the set of real numbers such that, for every real number $x$, the set $\\{nx+S: n \\in \\mathbb{N}\\}$ is finite, where $nx + S = \\{nx + s: s \\in S\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be a Hamel basis; that is, $H$ is a set of real numbers such that every real number $x$ can uniquely be written in the form\n$$\nx = \\sum_{h \\in H} q(x, h) \\cdot h,\n$$\nwhere the $q(x, h)$ are all rational and vanish for all but a finite number (depending on $x$) of $h$'s. The existence of Hamel bases can be proved via Zorn's lemma or Zermelo's well ordering theorem or any other statement equivalent to the axiom of choice.\n\nWe are now going to prove that the set $S$ of those real numbers $x$ whose $q(x, h)$ in $(*)$ are all integral satisfies the required condition.\n\nTo this end, fix a real number $x$. Since the conclusion is clear if $x = 0$, let $x$ be different from $0$ and let $m(x)$ be the least common multiple of the denominators of the non-vanishing $q(x, h)$ in $(*)$. Finally, notice that $m(x) \\cdot x$ is a member of $S$, to conclude that any set of the form $nx + S$, where $n$ is a non-negative integer, must be one of the sets $rx + S$, $r = 0, 1, \\dots, m(x) - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, a circle centered at some point $O$ meets the segments $BC, CA, AB$ in the pairs of points $X$ and $X'$, $Y$ and $Y'$, $Z$ and $Z'$, respectively, labeled in circular order: $X, X', Y, Y', Z, Z'$. Let $M$ be the Miquel point of the triangle $XYZ$ (i.e., the point of concurrence of the circles $AYZ, BZX, CXY$), and let $M'$ be that of the triangle $X'Y'Z'$. Prove that the segments $OM$ and $OM'$ have equal lengths.\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We begin by reviewing some basic facts on conics. For an ellipse $\\Sigma$ with center $N$, foci $M$ and $M'$, semiaxes $a$ and $b$, it is known that the orthogonal projections $P$ and $P'$ of $M$ and $M'$ on any line $t$ tangent to $\\Sigma$ lie on the major auxiliary circle of $\\Sigma$, so that $NP = a = NP'$. Application to triangle $MNP$ (respectively, $M'NP'$) of a rotation $\\theta$ (respectively, $-\\theta$) about $M$ (respectively, $M'$) and a homothety of ratio $\\sec\\theta$ with center $M$ (respectively, $M'$) yields triangle $MOX$ (respectively, $M'OX'$), where $MO = M'O$, $NO = \\frac{1}{2} \\cdot MM' \\cdot \\tan\\theta$, $OX = a \\sec\\theta = OX'$, and $X, X'$ both lie on $t$. If $t$ varies and $\\theta$ is constant, the locus of $X$ and $X'$ is then a circle $\\Gamma$ centered at $O$.\n\nBy Cartesian geometry it is readily checked that $\\Gamma$ and $\\Sigma$ are bitangent, and the line $\\ell$ supporting their common chord is also the radical axis of the circles $\\Gamma$ and $OMM'$, with this real geometrical significance even if the bitangency is not real. Since $\\ell$ and the circle $OMM'$ are mutually inverse in $\\Gamma$, the inverse points of $M$ and $M'$ in $\\Gamma$ both lie on $\\ell$. Finally, the distance $d$ between the parallel lines $\\ell$ and $MM'$ is given by $d \\cdot MM' = 2b^2 \\tan \\theta$. Similar considerations hold for a hyperbola $\\Sigma$.\n\nConsider now an isopair $M, M'$ (two isogonally conjugate in the triangle $ABC$, the foci of a conic $\\Sigma$ touching its sides), and take points $X, Y, Z$ (respectively, $X', Y', Z'$) on lines $BC, CA, AB$, respectively, so that the lines $MX, MY, MZ$ (respectively, $M'X', M'Y', M'Z'$) make the same directed angle $\\theta$ (respectively, $-\\theta$) with the perpendiculars to $BC, CA, AB$, respectively; then the isopedal triangles $XYZ, X'Y'Z'$ of angles $\\theta, -\\theta$ for the isopair $M, M'$ have their Miquel points at $M, M'$ and are inscribed in a common isopedal circle $\\Gamma$ bitangent to $\\Sigma$, centered at a point $O$ on the perpendicular bisector of the segment $MM'$.\nConversely, for any pair of triangles inscribed in a triangle $ABC$ and in a circle $\\Gamma$ (as in the statement of the problem), the Miquel points $M, M'$ are an isopair and $\\Gamma$ is an isopedal circle of $M, M'$. (If $M, M'$ are the Brocard points, $\\Sigma$ is the Brocard ellipse and $\\Gamma$ is a Tucker circle.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Demonstraţi că, oricare ar fi numărul real $x$, are loc inegalitatea\n$$\nx^{4}-x^{3}-x+1 \\geq 0\n$$\n\nb) Rezolvaţi în mulţimea numerelor reale sistemul\n$$\n\\left\\{\n\\begin{array}{l}\nx_{1}+x_{2}+x_{3}=3 \\\\\nx_{1}^{3}+x_{2}^{3}+x_{3}^{3}=x_{1}^{4}+x_{2}^{4}+x_{3}^{4}\n\\end{array}\n\\right.\n$$",
    "options": [],
    "answer": "x1 = x2 = x3 = 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSă se rezolve ecuaţia:\n$$\n2^{\\sin^{4} x-\\cos^{2} x}-2^{\\cos^{4} x-\\sin^{2} x}=\\cos 2x\n$$",
    "options": [],
    "answer": "x = (2k+1)π/4, where k ∈ ℤ",
    "solution": "Solution:\nEcuaţia se scrie echivalent\n$$\n2^{\\sin^{4} x+\\sin^{2} x}-2^{\\cos^{4} x+\\cos^{2} x}=2 \\cos 2x.\n$$\nSă observăm că\n$$\n\\begin{aligned}\n\\cos^{4} x+\\cos^{2} x-\\sin^{4} x-\\sin^{2} x &= (\\cos^{2} x+\\sin^{2} x)(\\cos^{2} x-\\sin^{2} x)+\\cos^{2} x-\\sin^{2} x \\\\\n&= 2 \\cos 2x\n\\end{aligned}\n$$\nşi ecuaţia devine\n$$\n2^{\\sin^{4} x+\\sin^{2} x} + \\sin^{4} x + \\sin^{2} x = 2^{\\cos^{4} x+\\cos^{2} x} + \\cos^{4} x + \\cos^{2} x.\n$$\nFuncţia $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, $f(t)=2^{t}+t$ este strict crescătoare, deci injectivă.\nAşadar obţinem $\\sin^{4} x+\\sin^{2} x=\\cos^{4} x+\\cos^{2} x$, adică $2 \\cos 2x=0$, cu soluţiile\n$$\nx \\in \\left\\{ \\left. \\frac{(2k+1)\\pi}{4} \\ \\right|\\ k \\in \\mathbb{Z} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUn şir de numere este numit complet dacă are termeni naturali nenuli şi orice număr natural nenul are cel puţin un multiplu printre termenii şirului.\nArătaţi că o progresie aritmetică este şir complet dacă şi numai dacă raţia sa divide primul termen.\n\nProblem:\nEgy számsorozatot nevezzünk teljesnek, ha elemei nullától különböző természetes számok és minden nullától különböző természetes számnak legalább egy többszöröse tagja a sorozatnak.\nIgazold, hogy egy számtani haladvány akkor és csak akkor teljes, ha a rációja osztja a sorozat első elemét!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDacă raţia $r$ divide primul termen $a_{1}$, atunci $a_{1} = d r$, $d \\in \\mathbb{N}$ şi $a_{n} = (d + n - 1) r$, iar un multiplu al numărului natural nenul $k$ se obţine luând $d + n - 1$ multiplu de $k$.\n\nReciproc, dacă $r = 0$, atunci şirul nu poate fi complet.\n\nDeoarece $r \\neq 0$ şi, conform ipotezei, există un multiplu al lui $r$ de forma $a_{1} + (n - 1) r$, $n \\in \\mathbb{N}^{*}$, reiese că $r \\mid a_{1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $A$ o matrice neinversabilă de ordin $n$ cu elemente reale, $n \\geq 2$, şi fie $A^{*}$ adjuncta matricei $A$. Arătaţi că $\\operatorname{tr}\\left(A^{*}\\right) \\neq -1$ dacă şi numai dacă matricea $I_{n} + A^{*}$ este inversabilă.\n\nProblem:\nAz $n$-edrendú, nem invertálható $A$ négyzetes mátrix elemei valós számok. $(n \\geq 2)$. Legyen $A^{*}$ az $A$ mátrix adjungáltja. Igazold, hogy $I_{n} + A^{*}$ akkor és csak akkor invertálható, ha $\\operatorname{tr}\\left(A^{*}\\right) \\neq -1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDeoarece matricea $A$ este neinversabilă, avem $\\operatorname{rang}(A) \\leq n-1$. Distingem cazurile:\n\ni. $\\operatorname{rang}(A) \\leq n-2$. Atunci $A^{*} = O_{n}$ şi echivalenţa este evidentă.\n\n\nii. $\\operatorname{rang}(A) = n-1$. Atunci $A A^{*} = O_{n}$ şi, din inegalitatea lui Sylvester, $0 \\geq \\operatorname{rang}(A) + \\operatorname{rang}\\left(A^{*}\\right) - n$, de unde $\\operatorname{rang}\\left(A^{*}\\right) \\leq 1$.\n\nAtunci există o matrice linie $L \\in \\mathcal{M}_{n, 1}(\\mathbb{R})$ şi o matrice coloană $C \\in \\mathcal{M}_{1, n}(\\mathbb{R})$ astfel încât $A^{*} = C L$. Observăm că $L C = (a) \\in \\mathcal{M}_{1}(\\mathbb{R})$, cu $a = \\operatorname{tr}\\left(A^{*}\\right)$ şi $\\left(A^{*}\\right)^{2} = C L C L = C(a) L = a A^{*}$.\n\nNotăm $B = I_{n} + A^{*}$. Atunci $\\left(B - I_{n}\\right)^{2} = a\\left(B - I_{n}\\right)$ se scrie $B\\left((a+2) I_{n} - B\\right) = (a+1) I_{n}$, de unde rezultă că $a \\neq -1$ implică $B$ inversabilă.\n\nÎn plus, dacă $B$ este inversabilă dar $a = -1$, atunci $B\\left(I_{n} - B\\right) = O_{n}$ atrage $B = I_{n}$ şi apoi $A^{*} = O_{n}$. Atunci $-1 = \\operatorname{tr}\\left(A^{*}\\right) = 0$, fals.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSă se calculeze limita $\\lim _{n \\rightarrow \\infty} \\int_{0}^{1} \\mathrm{e}^{x^{n}} \\mathrm{~d} x$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $m$ şi $n$ numere naturale, $m, n \\geq 2$. Considerăm matricele $A_{1}, A_{2}, \\ldots, A_{m} \\in \\mathcal{M}_{n}(\\mathbb{R})$, nu toate nilpotente. Demonstraţi că există un număr întreg $k>0$ astfel încât $A_{1}^{k}+A_{2}^{k}+\\cdots+A_{m}^{k} \\neq O_{n}$.\n\nNotă: Numim nilpotentă o matrice pătratică având o putere nulă.\n\nProblem:\n\nAdottak az $m \\geq 2$ és $n \\geq 2$ természetes számok és az $A_{1}$, $A_{2}, \\ldots, A_{m} \\in \\mathcal{M}_{n}(\\mathbb{R})$, nem mind nilpotens mátrixok. Bizonyítsd be, hogy létezik olyan $k>0$ egész szám, amelyre $A_{1}^{k}+A_{2}^{k}+\\cdots+A_{m}^{k} \\neq O_{n}$.\n\nMegjegyzés: Egy négyzetes mátrix nilpitens, ha van olyan hatványa, ami nullmátrix.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotăm cu $\\lambda_{i 1}, \\lambda_{i 2}, \\ldots, \\lambda_{i n}$ valorile proprii ale matricei $A_{i}$, $i=1,2, \\ldots, m$. Presupunem prin absurd că $A_{1}^{k}+A_{2}^{k}+\\cdots+A_{m}^{k}=O_{n}$, oricare ar fi $k \\geq 1$. Atunci $\\operatorname{tr}\\left(A_{1}^{k}\\right)+\\operatorname{tr}\\left(A_{2}^{k}\\right)+\\cdots+\\operatorname{tr}\\left(A_{m}^{k}\\right)=0$,\nde unde $\\sum_{j=1}^{m} \\sum_{i=1}^{n} \\lambda_{i j}^{k}=0$.\n\nDin relaţiile lui Newton, egalităţile $\\sum_{i j} \\lambda_{i j}^{k}=0, k \\geq 1$, implică $\\lambda_{i j}=0$, oricare ar fi $i=1,2, \\ldots, m, j=1,2, \\ldots, n$.\nAtunci $A_{i}^{n}=O_{n}, i=1,2, \\ldots, m$, contradicţie.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(A,+, \\cdot)$ un inel care îndeplineşte simultan următoarele două condiţii:\n(1) $A$ nu este corp,\n(2) oricare ar fi $x$ un element neinversabil al lui $A$, există un număr întreg $m \\geq 1$, care depinde de $x$, astfel încât\n$$\nx = x^{2} + x^{3} + \\cdots + x^{2^{m}}\n$$\nSă se arate că:\n\na. $x + x = 0$, oricare ar fi $x \\in A$,\n\nb. $x^{2} = x$, oricare ar fi elementul neinversabil $x \\in A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nEste suficient să demonstrăm că $1 + 1 = 0$. Fie $x$ un element neinversabil, fie $m \\in \\mathbb{N}^{*}$, astfel încât $x = x^{2} + x^{3} + \\cdots + x^{2^{m}}$, şi $y = x + x^{2} + \\cdots + x^{2^{m} - 1}$. În mod evident, $x y = x$, deci $x y^{k} = x$, oricare ar fi $k \\in \\mathbb{N}^{*}$. Întrucât $x$ este neinversabil, $y$ este neinversabil, deci există $p \\in \\mathbb{N}^{*}$, astfel încât\n$$\n-y = (-y)^{2} + (-y)^{3} + \\cdots + (-y)^{2^{p} - 1} + (-y)^{2^{p}} = y^{2} - y^{3} + \\cdots - y^{2^{p} - 1} + y^{2^{p}}\n$$\nPrin urmare,\n$$\n-x = -x y = x y^{2} - x y^{3} + \\cdots - x y^{2^{p} - 1} + x y^{2^{p}} = x - x + \\cdots - x + x = x\n$$\nadică $x + x = 0$.\n\nFie $x$ un element nenul şi neinversabil. Cum $2x = 0$, rezultă că $2$ este neinversabil, deci $2 + 2 = 0$ şi $2 = 2^{2} + 2^{3} + \\cdots + 2^{2^{m}}$, unde $m \\in \\mathbb{N}^{*}$. Relaţia $2 + 2 = 0$ implică $2^{2} = 0$, deci $2^{k} = 0$, oricare ar fi $k \\geq 2$. Prin urmare,\n$$\n2 = 2^{2} + 2^{3} + \\cdots + 2^{2^{m}} = 0\n$$\n\nb.\nFie $x$ un element neinversabil al lui $A$ şi $m \\in \\mathbb{N}^{*}$, astfel încât $x = x^{2} + x^{3} + \\cdots + x^{2^{m}}$. Atunci\n$$\nx^{2} = x^{3} + x^{4} + \\cdots + x^{2^{m} + 1}\n$$\nÎntrucât $1 + 1 = 0$, prin adunarea celor două relaţii obţinem $x^{2^{m} + 1} = x$.\n\nRelaţiile $1 + 1 = 0$ şi $x^{2^{m} + 1} = x$ implică\n$$\n(x^{2} + x)^{2^{m}} = x^{2^{m + 1}} + x^{2^{m}} = x^{2^{m} + 1} + x^{2^{m}} = x \\cdot x^{2^{m} - 1} + x^{2^{m}} = x^{2^{m}} + x^{2^{m}} = 0\n$$\nVom arăta că $x^{2} + x = 0$, de unde concluzia. Fie $y = x^{2} + x$ şi $k$ cel mai mic număr natural nenul, astfel încât $y^{k} = 0$ — un astfel de $k$ există, deoarece $y^{2^{m}} = 0$. În cazul în care $k > 1$, elementul $y^{k - 1}$ este neinversabil în $A$ şi există $n \\in \\mathbb{N}^{*}$, astfel încât\n$$\ny^{k - 1} = (y^{k - 1})^{2^{n} + 1} = y^{(k - 1)(2^{n} + 1)}\n$$\nÎntrucât $(k - 1)(2^{n} + 1) \\geq k$, rezultă că $y^{(k - 1)(2^{n} + 1)} = 0$, deci $y^{k - 1} = 0$ — în contradicţie cu minimalitatea lui $k$. Prin urmare, $k = 1$ şi $y = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ o funcţie arbitrară şi $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ o funcţie de gradul al doilea, având proprietatea:\npentru orice numere reale $m$ şi $n$, ecuaţia $f(x)=m x+n$ are soluţii dacă şi numai dacă ecuaţia $g(x)=m x+n$ are soluţii.\nArătaţi că funcţiile $f$ şi $g$ sunt egale.\n\nProblem:\nLegyen $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ egy tetszőleges függvény és $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ egy olyan másodfokú függvény, amelyre igaz az alábbi állítás:\nBármely $m$ és $n$ valós szám esetén az $f(x)=m x+n$ egyenletnek akkor és csak akkor van megoldása, ha a $g(x)=m x+n$ egyenletnek van megoldása. Igazold, hogy a $f$ és $g$ függvények egyenlőek egymással!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nObservăm că, dacă graficul funcţiei $h: \\mathbb{R} \\rightarrow \\mathbb{R}$, $h(x)=m x+n$ este tangent la graficul funcţiei $g: \\mathbb{R} \\rightarrow \\mathbb{R}$, $g(x)=a x^{2}+b x+c$, atunci ecuaţia $a x^{2}+b x+c=m x+n$, $a, b, c, m, n \\in \\mathbb{R}$, $a \\neq 0$ are discriminantul nul, deci funcţia $k: \\mathbb{R} \\rightarrow \\mathbb{R}$, $k(x)=a x^{2}+b x+c-m x-n$ se anulează într-un punct şi are semn constant în celelalte puncte $\\left(^{*}\\right)$.\n\nPutem presupune, fără a restrânge generalitatea, că $g(x)=a x^{2}+b x+c$, $a>0$.\n\nSă presupunem că există $x_{0}$ astfel încât $f\\left(x_{0}\\right)<g\\left(x_{0}\\right)$. Alegem $m$ şi $n$ astfel încât graficul funcţiei $h(x)=m x+n$ să fie tangent la graficul lui $g$ în punctul $\\left(x_{0}, g\\left(x_{0}\\right)\\right)$. Fie $n^{\\prime}=n-g\\left(x_{0}\\right)+f\\left(x_{0}\\right)$. Atunci ecuaţia $f(x)=m x+n^{\\prime}$ are soluţia $x_{0}$ iar ecuaţia $g(x)=m x+n^{\\prime}$ nu are soluţii, deoarece, conform $(*), g(x) \\geq m x+n>m x+n^{\\prime}, \\forall x \\in \\mathbb{R}$ - fals.\n\nRezultă $f(x) \\geq g(x), \\forall x \\in \\mathbb{R}$ $(**)$.\n\nDacă presupunem acum că există $x_{0}$ astfel încât $f\\left(x_{0}\\right)>g\\left(x_{0}\\right)$, atunci alegem din nou $m$ şi $n$ astfel încât graficul funcţiei $h(x)=m x+n$ să fie tangent la graficul lui $g$ în punctul $\\left(x_{0}, g\\left(x_{0}\\right)\\right)$. În acest caz, din $(**)$ şi $\\left(^{*}\\right)$ rezultă că $f(x)>m x+n, \\forall x \\in \\mathbb{R}$, pe când ecuaţia $g(x)=m x+n$ are soluţia $x_{0}$ - fals.\n\nDeci $f(x)=g(x), \\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO configurație de 4027 de puncte se numește columbiană dacă are 2013 puncte colorate cu roșu și 2014 puncte colorate cu albastru, și nu conține trei puncte coliniare. O mulțime finită de drepte din plan împarte planul în regiuni. O mulțime de drepte se numește bună pentru o configurație columbiană dacă următoarele două condiții sunt îndeplinite:\n- Nicio dreaptă nu trece printr-un punct al configurației,\n- Nicio regiune nu conține puncte de aceeași culoare.\nDeterminați cel mai mic număr natural $k$ astfel încât pentru orice configurație columbiană de 4027 de puncte să existe o mulțime bună de $k$ drepte.",
    "options": [],
    "answer": "63",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe consideră triunghiul $ABC$ şi punctele $D, E \\in (BC)$, $F, G \\in (CA)$, $H, I \\in (AB)$ astfel încât $BD = CE$, $CF = AG$ şi $AH = BI$. Notăm cu $M, N, P$ mijloacele segmentelor $[GH]$, $[DI]$, respectiv $[EF]$ şi cu $M'$ intersecţia dreptelor $AM$ şi $BC$.\n\na) Arătaţi că\n$$\n\\frac{BM'}{CM'} = \\frac{AG}{AH} \\cdot \\frac{AB}{AC}\n$$\n\nb) Arătaţi că dreptele $AM$, $BN$ şi $CP$ sunt concurente.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn grup $(G, \\cdot)$ are proprietatea $(P)$, dacă, pentru orice automorfism $f$ al lui $G$, există două automorfisme $g$ şi $h$ ale lui $G$, astfel încât $f(x)=g(x) \\cdot h(x)$, oricare ar fi $x \\in G$. Să se arate că:\n\na) Orice grup care are proprietatea $(P)$ este comutativ.\n\nb) Orice grup comutativ finit de ordin impar are proprietatea $(P)$.\n\nc) Niciun grup finit de ordin $4 n+2, n \\in \\mathbb{N}$, nu are proprietatea $(P)$. (Ordinul unui grup finit este numărul de elemente ale acelui grup.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCercul exînscris corespunzător vârfului $A$ al triunghiului $A B C$ este tangent la latura $B C$ în punctul $A_{1}$. Definim analog punctele $B_{1}$ pe latura $C A$ şi $C_{1}$ pe latura $A B$. Presupunem că centrul cercului circumscris triunghiului $A_{1} B_{1} C_{1}$ aparţine cercului circumscris triunghiului $A B C$. Demonstraţi că triunghiul $A B C$ este dreptunghic.\n\nCercul exînscris în triunghiul $A B C$ corespunzător vârfului $A$ este cercul tangent laturii $B C$ şi dreptelor $A B$ şi $A C$, dar nu laturilor $A B$ şi $A C$. Cercurile exînscrise corespunzătoare vârfurilor $B$ şi $C$ se definesc analog.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO funcţie $f:(0, \\infty) \\rightarrow(0, \\infty)$ se numeşte contractibilă dacă, pentru orice numere $x, y \\in(0, \\infty)$, avem $\\lim_{n \\rightarrow \\infty}\\left(f^{n}(x)-f^{n}(y)\\right)=0$, unde $f^{n}=\\underbrace{f \\circ f \\circ \\ldots \\circ f}_{\\text{de } n \\text{ ori } f}$.\n\na) Considerăm $f:(0, \\infty) \\rightarrow(0, \\infty)$ o funcţie contractibilă, continuă, cu proprietatea că are un punct fix, adică există $x_{0} \\in(0, \\infty)$ astfel încât $f\\left(x_{0}\\right)=x_{0}$. Arătaţi că $f(x)>x$, oricare ar fi $x \\in\\left(0, x_{0}\\right)$ şi $f(x)<x$, oricare ar fi $x \\in\\left(x_{0}, \\infty\\right)$.\n\nb) Arătaţi că funcţia $f:(0, \\infty) \\rightarrow(0, \\infty)$ dată prin $f(x)=x+\\frac{1}{x}$ este contractibilă, dar nu are puncte fixe.\n\nProblem:\nAz $f:(0, \\infty) \\rightarrow(0, \\infty)$ függvényt nevezzük kontraktibilisnek, ha bármely $x, y \\in(0, \\infty)$ számok esetén $\\lim_{n \\rightarrow \\infty}\\left(f^{n}(x)-f^{n}(y)\\right)=0$, ahol $f^{n}=f \\circ f \\circ \\cdots \\circ f$.\n\na) Legyen $f:(0, \\infty) \\rightarrow(0, \\infty)$ egy olyan folytonos, kontraktibilis függvény, amelynek van fixpontja, azaz létezik $x_{0} \\in(0, \\infty)$ úgy, hogy $f\\left(x_{0}\\right)=x_{0}$. Igazold, hogy $f(x)>x$, bármely $x \\in\\left(0, x_{0}\\right)$ esetén és $f(x)<x$, bármely $x \\in\\left(x_{0}, \\infty\\right)$ esetén!\n\nb) Igazold, hogy az $f:(0, \\infty) \\rightarrow(0, \\infty), f(x)=x+\\frac{1}{x}$ kontraktibilis, de nincs fixpontja.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Presupunem prin reducere la absurd că $f$ ar admite un al doilea punct fix $x_{1} \\in(0, \\infty) \\setminus\\{x_{0}\\}$. Atunci $\\lim_{n \\rightarrow \\infty}\\left(f^{n}\\left(x_{0}\\right)-f^{n}\\left(x_{1}\\right)\\right)=x_{0}-x_{1} \\neq 0$, contradicţie. În acest caz, din continuitatea lui $f$ (proprietatea Darboux) deducem $f(x)<x, \\forall x \\in\\left(0, x_{0}\\right)$ sau $f(x)>x, \\forall x \\in\\left(0, x_{0}\\right)$.\n\nÎn primul caz obţinem inductiv că $0<f^{n+1}(x)<f^{n}(x)<x$, oricare ar fi $n \\in \\mathbb{N}^{*}$ şi $x \\in\\left(0, x_{0}\\right)$. Deducem că şirul $a_{n}=\\left(f^{n}(x)\\right)_{n \\geq 1}$ converge; fie $a$ limita sa. Dacă $a>0$, atunci din $a_{n+1}=f\\left(a_{n}\\right)$ rezultă $a=f(a)$, fals. Atunci $a=0$, de unde rezultă $\\lim_{n \\rightarrow \\infty}\\left(f^{n}\\left(x_{0}\\right)-f^{n}(x)\\right)=x_{0} \\neq 0, \\forall x \\in\\left(0, x_{0}\\right)$, contradicţie. Rămâne $f(x)>x$, oricare ar fi $x \\in\\left(0, x_{0}\\right)$.\n\nAnalog, $f(x)>x, \\forall x \\in\\left(x_{0}, \\infty\\right)$ sau $f(x)<x, \\forall x \\in\\left(x_{0}, \\infty\\right)$. În prima situaţie deducem inductiv că $f^{n+1}(x)>f^{n}(x)>x$, oricare ar fi $n \\in \\mathbb{N}^{*}$, de unde rezultă $\\lim_{n \\rightarrow \\infty} f^{n}(x)=\\infty$ şi apoi $\\lim_{n \\rightarrow \\infty}\\left(f^{n}(x)-f^{n}\\left(x_{0}\\right)\\right)=\\infty, \\forall x \\in \\left(x_{0}, \\infty\\right)$, contradicţie. Ca urmare, $f(x)<x$, oricare ar fi $x \\in\\left(x_{0}, \\infty\\right)$.\n\nb) Fie $x, y \\in(0, \\infty)$. Putem presupune $f(x)<f(y)$. Notăm $x_{n}=f^{n}(x)$, $n \\in \\mathbb{N}^{*}$, respectiv $y_{n}=f^{n}(y), n \\in \\mathbb{N}^{*}$. Avem $2 \\leq x_{n}<y_{n}, n>1$, deoarece $f$ este strict crescătoare pe $[1, \\infty)$. Demonstrăm prin inducţie proprietatea $y_{n}<y_{1}+2 \\sqrt{n}, n \\in \\mathbb{N}^{*}$. Proprietatea este verificată pentru $n=1$. Presupunem $y_{n}<y_{1}+2 \\sqrt{n}$, pentru un număr natural nenul $n$. Atunci, din monotonia funcţiei $f$, obţinem:\n$$\n\\begin{gathered}\ny_{n+1}=f\\left(y_{n}\\right)<f\\left(y_{1}+2 \\sqrt{n}\\right)=y_{1}+2 \\sqrt{n}+\\frac{1}{y_{1}+2 \\sqrt{n}}< \\\\\n<y_{1}+2 \\sqrt{n}+\\frac{1}{2 \\sqrt{n}}<y_{1}+2 \\sqrt{n+1}\n\\end{gathered}\n$$\nRezultă $2 \\leq x_{n}<y_{n}<y_{1}+2 \\sqrt{n}<3 \\sqrt{n}, \\forall n \\in \\mathbb{N}, n>y_{1}^{2}$. Din inegalitatea clasică $1-x<e^{-x}, \\forall x \\in \\mathbb{R}$, deducem:\n$$\n\\begin{aligned}\n& 0<y_{n+1}- x_{n+1}=f\\left(y_{n}\\right)-f\\left(x_{n}\\right)=\\left(y_{n}-x_{n}\\right)\\left(1-\\frac{1}{x_{n} y_{n}}\\right)< \\\\\n&<\\left(y_{n}-x_{n}\\right)\\left(1-\\frac{1}{9 n}\\right)<\\left(y_{n}-x_{n}\\right) e^{-\\frac{1}{9 n}}\n\\end{aligned}\n$$\npentru $n \\geq p:=\\left[y_{1}^{2}\\right]+1$.\nObţinem $0<y_{n}-x_{n}<\\left(y_{p}-x_{p}\\right) e^{-\\frac{1}{9} \\sum_{k=p}^{n-1} \\frac{1}{k}}, \\forall n>p$.\nCum $\\lim_{n \\rightarrow \\infty} \\sum_{k=p}^{n-1} \\frac{1}{k}=\\infty$, găsim $\\lim_{n \\rightarrow \\infty} e^{-\\frac{1}{9} \\sum_{k=p}^{n-1} \\frac{1}{k}}=0$, de unde rezultă că $\\lim_{n \\rightarrow \\infty}\\left(f^{n}(y)-f^{n}(x)\\right)=0$. Rezultă că funcţia $f$ este contractibilă, evident fără puncte fixe.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n$ un număr natural nenul şi $a_{1}, a_{2}, \\ldots, a_{n}$ numere reale astfel încât $a_{1}+a_{2}+\\ldots+a_{k} \\leq k$, oricare ar fi $k \\in \\{1,2, \\ldots, n\\}$. Arătaţi că\n$$\n\\frac{a_{1}}{1}+\\frac{a_{2}}{2}+\\ldots+\\frac{a_{n}}{n} \\leq \\frac{1}{1}+\\frac{1}{2}+\\ldots+\\frac{1}{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\in (0,1)$ şi $\\mathcal{C}$ mulţimea funcţiilor crescătoare $f:[0,1] \\rightarrow [0, \\infty)$, astfel încât $\\int_{0}^{1} f(x) \\, \\mathrm{d}x = 1$. Să se determine:\n\na. $\\max_{f \\in \\mathcal{C}} \\int_{0}^{a} f(x) \\, \\mathrm{d}x$,\n\nb. $\\max_{f \\in \\mathcal{C}} \\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x$.",
    "options": [],
    "answer": "a) a\nb) a if a ≤ 1/2; 1/(4(1−a)) if a > 1/2",
    "solution": "Solution:\n\na.\nFie $f$ o funcţie din mulţimea $\\mathcal{C}$. Arătăm că\n$$\n\\int_{0}^{a} f(x) \\, \\mathrm{d}x \\leq a\n$$\nÎntr-adevăr,\n$$\n\\begin{aligned}\na - \\int_{0}^{a} f(x) \\, \\mathrm{d}x &= a \\int_{0}^{1} f(x) \\, \\mathrm{d}x - \\int_{0}^{a} f(x) \\, \\mathrm{d}x = a \\int_{a}^{1} f(x) \\, \\mathrm{d}x - (1-a) \\int_{0}^{a} f(x) \\, \\mathrm{d}x \\\\\n&\\geq a \\int_{a}^{1} f(a) \\, \\mathrm{d}x - (1-a) \\int_{0}^{a} f(a) \\, \\mathrm{d}x = 0\n\\end{aligned}\n$$\nCum pentru funcţia constantă $f \\equiv 1$ are loc egalitatea, rezultă că maximumul cerut este egal cu $a$.\n\nb.\nMaximumul cerut este $a$, dacă $a \\leq 1/2$, şi $1/(4(1-a))$, dacă $a > 1/2$; aceste valori sunt atinse, de exemplu, pentru $f \\equiv 1$, în primul caz, şi\n$$\nf(x) = \\begin{cases} 0, & \\text{dacă } 0 \\leq x \\leq 2a-1, \\\\ 1/(2(1-a)), & \\text{dacă } 2a-1 < x \\leq 1, \\end{cases}\n$$\nîn al doilea caz.\n\nÎn continuare, vom arăta că\n$$\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x \\leq \\begin{cases} a, & \\text{dacă } a \\leq 1/2 \\\\ 1/(4(1-a)), & \\text{dacă } a > 1/2 \\end{cases}\n$$\nÎn acest scop, fie $f$ o funcţie din mulţimea $\\mathcal{C}$. Ţinând cont de condiţiile din enunţ,\n$$\n\\begin{aligned}\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x &\\leq f(a) \\int_{0}^{a} f(x) \\, \\mathrm{d}x \\leq \\left(\\frac{1}{1-a} \\int_{a}^{1} f(x) \\, \\mathrm{d}x\\right) \\left(\\int_{0}^{a} f(x) \\, \\mathrm{d}x\\right) \\\\\n&= \\frac{1}{1-a} \\left(1 - \\int_{0}^{a} f(x) \\, \\mathrm{d}x\\right) \\left(\\int_{0}^{a} f(x) \\, \\mathrm{d}x\\right)\n\\end{aligned}\n$$\nSă observăm că\n$$\n\\max \\{ t(1-t) : t \\leq a \\} = \\begin{cases} a(1-a), & \\text{dacă } a \\leq 1/2 \\\\ 1/4, & \\text{dacă } a > 1/2 \\end{cases}\n$$\nîn ambele cazuri, maximumul este atins într-un singur punct: în $t=a$, în primul caz, şi în $t=1/2$, în al doilea.\nPrin urmare,\n$$\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x \\leq \\begin{cases} a, & \\text{dacă } a \\leq 1/2 \\\\ 1/(4(1-a)), & \\text{dacă } a > 1/2 \\end{cases}\n$$\n\n\nSolution 2:\n\nb.\nŢinând cont de exemplele din prima soluţie, este suficient să arătăm că\n$$\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x \\leq \\begin{cases} a, & \\text{dacă } a \\leq 1/2, \\\\ 1/(4(1-a)), & \\text{dacă } a > 1/2. \\end{cases}\n$$\nFie $f$ o funcţie din mulţimea $\\mathcal{C}$. Dacă $f(a) < 1$, atunci\n$$\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x \\leq \\int_{0}^{a} (f(a))^{2} \\, \\mathrm{d}x = a (f(a))^{2} < a\n$$\nDacă $f(a) \\geq 1$, considerăm funcţia $\\varphi : [0,1] \\rightarrow [0, \\infty)$,\n$$\n\\varphi(t) = \\int_{0}^{t} f(x) \\, \\mathrm{d}x + (1-t) f(a)\n$$\nAceastă funcţie este evident continuă, descrescătoare pe intervalul închis $[0, a]$ şi crescătoare pe intervalul închis $[a, 1]$. Întrucât $\\varphi(1) = 1$, rezultă că $\\varphi(a) \\leq 1$ (acest lucru poate fi demonstrat şi în mod direct). Pe de altă parte, $\\varphi(0) = f(a) \\geq 1$, deci există un punct $b$ în intervalul închis $[0, a]$, astfel încât $\\varphi(b) = 1$, i.e.,\n$$\n\\int_{0}^{b} f(x) \\, \\mathrm{d}x = 1 - (1-b) f(a)\n$$\nFie $g : [0,1] \\rightarrow [0, \\infty)$,\n$$\ng(x) = \\begin{cases} f(x), & \\text{dacă } 0 \\leq x \\leq b, \\\\ f(a), & \\text{dacă } b < x \\leq 1. \\end{cases}\n$$\nEste uşor de verificat că $g$ aparţine mulţimii $\\mathcal{C}$; în plus, dacă $0 \\leq x \\leq a$, atunci $g(x) \\geq f(x) \\geq 0$, deci\n$$\n\\int_{0}^{a} (g(x))^{2} \\, \\mathrm{d}x \\geq \\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x\n$$\nPe de altă parte,\n$$\n\\begin{aligned}\n\\int_{0}^{a} (g(x))^{2} \\, \\mathrm{d}x &\\leq g(a) \\int_{0}^{a} g(x) \\, \\mathrm{d}x = f(a) \\left( \\int_{0}^{b} f(x) \\, \\mathrm{d}x + (a-b) f(a) \\right) \\\\\n&= f(a) (1 - (1-b) f(a) + (a-b) f(a)) = f(a) (1 - (1-a) f(a))\n\\end{aligned}\n$$\nSă observăm că\n$$\n\\max \\{ t (1 - (1-a) t) : t \\geq 1 \\} = \\begin{cases} a, & \\text{dacă } a \\leq 1/2 \\\\ 1/(4(1-a)), & \\text{dacă } a > 1/2 \\end{cases}\n$$\nîn ambele cazuri, maximumul este atins într-un singur punct: în $t=1$, în primul caz, şi în $t=1/(2(1-a))$, în al doilea.\nPrin urmare,\n$$\n\\int_{0}^{a} (f(x))^{2} \\, \\mathrm{d}x \\leq \\int_{0}^{a} (g(x))^{2} \\, \\mathrm{d}x \\leq \\begin{cases} a, & \\text{dacă } a \\leq 1/2 \\\\ 1/(4(1-a)), & \\text{dacă } a > 1/2 \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f:[0, \\pi / 2] \\rightarrow[0, \\infty)$ o funcție crescătoare. Să se arate că:\n\na. $\\int_{0}^{\\pi / 2}(f(x)-f(\\pi / 4))(\\sin x-\\cos x) \\, \\mathrm{d} x \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Fie $f:[0, \\infty) \\rightarrow[0, \\infty)$ o funcție derivabilă și convexă. Arătați că dacă $f(x) \\leq x$, oricare ar fi $x \\geq 0$, atunci $f'(x) \\leq 1$, oricare ar fi $x \\geq 0$.\n\nb) Determinați funcțiile $f:[0, \\infty) \\rightarrow[0, \\infty)$ derivabile și convexe care au proprietatea că $f(0)=0$ și $f'(x) \\cdot f(f(x))=x$, oricare ar fi $x \\geq 0$.\n\nProblem:\na) Az $f:[0, \\infty) \\rightarrow[0, \\infty)$ függvény deriválható és konvex. Igazold, hogy ha $f(x) \\leq x$, bármely $x \\geq 0$ esetén, akkor $f'(x) \\leq 1$, bármely $x \\geq 0$ esetén!\n\nb) Határozd meg azokat a deriválható és konvex $f:[0, \\infty) \\rightarrow[0, \\infty)$ függvényeket, amelyekre $f(0)=0$ és $f'(x) \\cdot f(f(x))=x$, bármely $x \\geq 0$ esetén!",
    "options": [],
    "answer": "f(x) = x for all x ≥ 0",
    "solution": "Solution:\na) Presupunem contrariul. Există $a \\geq 0$ cu $f'(a)>1$, deci, cum $\\lim _{x \\searrow a} \\frac{f(x)-f(a)}{x-a}>1$, există $b>a$ cu $\\frac{f(b)-f(a)}{b-a}>1$.\nPentru orice $x>b$, din convexitatea funcției $f$ rezultă $\\frac{f(x)-f(b)}{x-b} \\geq \\frac{f(b)-f(a)}{b-a}=m>1$.\nAtunci $f(x) \\geq m x-m b+f(b)$, de unde $f(x)>x$ pentru $x$ suficient de mare. Contradicție.\n\nb) Vom demonstra că $f(x)=x$, oricare ar fi $x \\geq 0$. Cum $f'(x)=\\frac{x}{f(f(x))}>0$, oricare ar fi $x>0$, deducem că $f$ este strict crescătoare. Cum $f$ este convexă și derivabilă, rezultă că $f'$ este crescătoare.\nPresupunem prin absurd că există $f(a)<a$. Atunci $f(f(a))<f(a)<a$, deci $f'(a)>1$. Conform primului punct deducem că există $b>a$ cu $f(b)=b$. Atunci $f(f(b))=b$ și apoi $f'(b)=1<f'(a)$, în contradicție cu monotonia funcției $f'$.\n\nRămâne $f(x) \\geq x$, oricare ar fi $x \\geq 0$. Atunci $f(f(x)) \\geq f(x) \\geq x$, de unde $f'(x) \\leq 1$, oricare ar fi $x \\geq 0$.\nAplicând teorema lui Lagrange avem $f(x)-f(0)=x f'(c_{x}) \\leq x$, oricare ar fi $x>0$, de unde $f(x)=x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic cu ortocentrul $H$ şi fie $W$ un punct situat în interiorul laturii $BC$. Punctele $M$ şi $N$ sunt picioarele înălţimilor din $B$, respectiv $C$. Notăm cu $\\omega_{1}$ cercul circumscris triunghiului $BWN$ şi fie $X$ punctul diametral opus lui $W$ în cercul $\\omega_{1}$. Analog, notăm cu $\\omega_{2}$ cercul circumscris triunghiului $CWM$ şi fie $Y$ punctul diametral opus lui $W$ în cercul $\\omega_{2}$. Demonstraţi că punctele $X$, $Y$ şi $H$ sunt coliniare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPe o hârtie este scrisă la început o listă de numere naturale distincte. O continuare a listei înseamnă alegerea a două numere dintre cele existente şi scrierea pe listă a celui mai mic multiplu comun al acestora, cu condiţia ca el să nu fie deja scris. Spunem că lista $s$-a închis dacă nu mai există nicio continuare posibilă a sa (de exemplu, lista $2,3,4,6$ se închide după ce-l adăugăm pe 12). Care este numărul maxim de numere care pot apărea pe o listă care s-a închis, dacă la început lista conţinea 10 numere?",
    "options": [],
    "answer": "1023",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geq 2$ un număr natural, $(K,+, \\cdot)$ un corp comutativ cu proprietatea că $\\underbrace{1+\\cdots+1}_{m \\text{ ori}} \\neq 0,\\ m=2, \\ldots, n$, $f \\in K[X]$ un polinom de grad $n$ şi $G$ un subgrup al grupului aditiv $(K,+)$, $G \\neq K$. Să se arate că există $a \\in K$, astfel încât $f(a) \\notin G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $g \\in K[X]$ un polinom de grad $m \\in \\{2, \\ldots, n\\}$. Polinomul $h(X) = g(X+1) - g(X)$ este de grad $m-1$ și, dacă $\\operatorname{Im} g \\subseteq G$, atunci $\\operatorname{Im} h \\subseteq G$.\n\nPresupunem că $\\operatorname{Im} f \\subseteq G$. Considerăm polinoamele $f_0, f_1, \\ldots, f_{n-1}$, definite astfel:\n$$\nf_0 = f \\quad \\text{şi} \\quad f_k(X) = f_{k-1}(X+1) - f_{k-1}(X), \\quad k=1, \\ldots, n-1\n$$\nDin remarca anterioară, $\\operatorname{deg} f_k = n-k$ şi $\\operatorname{Im} f_k \\subseteq G$. În particular, $\\operatorname{deg} f_{n-1} = 1$ şi $\\operatorname{Im} f_{n-1} \\subseteq G$.\n\nCum funcţia polinomială $\\tilde{f}_{n-1}: K \\rightarrow K$ este surjectivă, rezultă că $\\operatorname{Im} f_{n-1} = K$, deci $G = K$, fals.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Să se arate că\n$$\n\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\ldots+\\frac{1}{2^{m}}<m\n$$\npentru orice $m \\in \\mathbb{N}^{*}$.\n\nb) Fie $p_{1}, p_{2}, \\ldots, p_{n}$ numerele prime mai mici decât $2^{100}$. Să se arate că\n$$\n\\frac{1}{p_{1}}+\\frac{1}{p_{2}}+\\ldots+\\frac{1}{p_{n}}<10\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na)\nFolosim inducția matematică.\nPentru $m=1$, avem $\\frac{1}{2}<1$.\nPresupunem că $\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{m}}<m$.\nAtunci:\n$$\n\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{m}}+\\frac{1}{2^{m}+1}+\\ldots+\\frac{1}{2^{m+1}}<m+1\n$$\nObservăm că sunt $2^{m}$ termeni de la $\\frac{1}{2^{m}+1}$ la $\\frac{1}{2^{m+1}}$, fiecare mai mic decât $\\frac{1}{2^{m}}$.\nDeci:\n$$\n\\frac{1}{2^{m}+1}+\\ldots+\\frac{1}{2^{m+1}}<2^{m}\\cdot\\frac{1}{2^{m}}=1\n$$\nPrin urmare,\n$$\n\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{m+1}}<m+1\n$$\nAstfel, inegalitatea este adevărată pentru orice $m\\in\\mathbb{N}^{*}$.\n\nb)\nNumerele $p_{i} p_{j} p_{k} p_{l}$, unde $1 \\leq i \\leq j \\leq k \\leq l \\leq n$, sunt distincte două câte două şi mai mici decât $2^{400}$.\nAvem atunci\n$$\n\\left(\\frac{1}{p_{1}}+\\frac{1}{p_{2}}+\\ldots+\\frac{1}{p_{n}}\\right)^{4} \\leq 4!\\sum_{1 \\leq i \\leq j \\leq k \\leq l \\leq n} \\frac{1}{p_{i} p_{j} p_{k} p_{l}}<24\\left(\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{400}}\\right)\n$$\nFolosind inegalitatea de la a), deducem că\n$$\n24\\left(\\frac{1}{2}+\\frac{1}{3}+\\ldots+\\frac{1}{2^{400}}\\right)<24 \\cdot 400<10000\n$$\nde unde concluzia.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(A, +, \\cdot)$ un inel cu proprietatea că $x = 0$ este unica soluție a ecuației $x^{2} = 0$, $x \\in A$. Fie $B = \\{ a \\in A \\mid a^{2} = 1 \\}$. Să se arate că:\n\na. $ab - ba = bab - a$, oricare ar fi $a \\in A$ și $b \\in B$.\n\nb. $(B, \\cdot)$ este grup.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $\\mathbb{Q}_{>0}$ mulțimea numerelor raționale strict pozitive. Fie $f: \\mathbb{Q}_{>0} \\rightarrow \\mathbb{R}$ o funcție ce îndeplinește următoarele trei condiții:\n\n(i) Pentru orice $x, y \\in \\mathbb{Q}_{>0}$ avem $f(x) f(y) \\geq f(x y)$,\n\n(ii) Pentru orice $x, y \\in \\mathbb{Q}_{>0}$ avem $f(x+y) \\geq f(x)+f(y)$,\n\n(iii) Există un număr rațional $a>1$ astfel încât $f(a)=a$.\n\nDemonstrați că $f(x)=x$ pentru orice $x \\in \\mathbb{Q}_{>0}$.",
    "options": [],
    "answer": "f(x) = x for all positive rational x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geq 3$ un număr întreg şi fie un cerc pe care marcăm $n+1$ puncte echidistante. Considerăm toate numerotările acestor puncte cu numerele $0,1, \\ldots, n$ astfel încât fiecare număr este folosit exact o dată; două astfel de numerotări se consideră identice dacă printr-o rotaţie a cercului coincid. O numerotare se numeşte frumoasă dacă pentru orice patru numere $a<b<c<d$ cu $a+d=b+c$, coarda ce uneşte punctele numerotate cu $a$ şi $d$ nu intersectează coarda ce uneşte punctele numerotate cu $b$ şi $c$.\nFie $M$ numărul de numerotări frumoase şi fie $N$ numărul perechilor ordonate $(x, y)$ de numere naturale nenule cu $x+y \\leq n$ şi $\\operatorname{cmmdc}(x, y)=1$. Demonstraţi că\n$$\nM=N+1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, 1] \\to [0, 1]$ be a continuous function and $x_0, y_0$ be two points in $[0, 1]$. Define sequences $(x_n)_{n \\in \\mathbb{N}}$ and $(y_n)_{n \\in \\mathbb{N}}$ of points in $[0, 1]$ by\n$$\nx_{n+1} = \\frac{f(x_0) + \\cdots + f(x_n)}{n+1} \\quad \\text{and} \\quad y_{n+1} = f\\left(\\frac{y_0 + \\cdots + y_n}{n+1}\\right);\n$$\n\nClearly, the definition is correct. Prove that the sequences $(x_n)_{n \\in \\mathbb{N}}$ and $(y_n)_{n \\in \\mathbb{N}}$ are convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38037,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that, for every positive integers $a, b, k$ such that $a < b$,\n$$\n\\frac{a}{b} \\le \\frac{a+k}{b+k}.\n$$\n\nb) Prove that\n$$\n\\frac{1}{100} + \\frac{4}{101} + \\frac{7}{102} + \\dots + \\frac{148}{149} \\ge 25.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a nonempty set of positive reals so that, for every $a$, $b$, $c$ in $M$, the number $ab + bc + ca$ is rational. Prove that $\\frac{a}{b}$ is rational for every $a$, $b$ in $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $b$, $c$ be arbitrary elements of $M$. By hypothesis, $ab + bc + ca$ is rational for all $a$, $b$, $c$ in $M$.\n\nLet us fix $a$ and $b$ in $M$, and let $c$ vary over $M$.\n\nFor $c = a$, we have:\n\n$$\nab + bc + ca = ab + ba + aa = 2ab + a^2\n$$\nwhich is rational.\n\nFor $c = b$, we have:\n\n$$\nab + bc + ca = ab + bb + ba = ab + b^2 + ba = 2ab + b^2\n$$\nwhich is rational.\n\nNow, consider the difference:\n\n$$\n(2ab + a^2) - (2ab + b^2) = a^2 - b^2\n$$\nwhich is rational. Therefore, $a^2 - b^2$ is rational for all $a$, $b$ in $M$.\n\nNow, for $c$ arbitrary in $M$, $ab + bc + ca$ is rational. Fix $a$ and $b$, and let $c$ vary.\n\nLet $c$ be arbitrary in $M$. Then $ab + bc + ca$ is rational.\n\nLet us consider $ab + bc + ca$ as a function of $c$:\n\n$$\nab + bc + ca = ab + bc + ca = ab + c(b + a)\n$$\nSo, for fixed $a$, $b$, as $c$ varies over $M$, $ab + c(a + b)$ is rational.\n\nLet $d = a + b$. Then, for all $c$ in $M$, $ab + cd$ is rational.\n\nLet $c_1$, $c_2$ be two elements of $M$. Then $ab + c_1 d$ and $ab + c_2 d$ are both rational.\n\nTheir difference is $d(c_1 - c_2)$, which is rational. Since $d = a + b$ is a positive real, and $c_1$, $c_2$ are arbitrary in $M$, $c_1 - c_2$ is rational up to scaling by $d$.\n\nBut we already have that $a^2 - b^2$ is rational for all $a$, $b$ in $M$.\n\nLet us now fix $a$, $b$ in $M$ and consider $a^2 - b^2$ is rational, so $a^2 = b^2 + r$ for some rational $r$.\n\nNow, $a$, $b$ are positive reals, so $a = \\sqrt{b^2 + r}$.\n\nLet us now consider $\\frac{a}{b}$.\n\nLet $x = \\frac{a}{b}$, then $a^2 = b^2 x^2$, so $a^2 - b^2 = b^2(x^2 - 1)$ is rational.\n\nBut $b^2$ is positive real, and $x^2 - 1$ is real. So $b^2(x^2 - 1)$ is rational for all $a$, $b$ in $M$.\n\nNow, fix $b$ in $M$. Then for all $a$ in $M$, $b^2(x^2 - 1)$ is rational, where $x = \\frac{a}{b}$.\n\nBut $b^2$ is a fixed positive real number, so $x^2 - 1$ is rational up to scaling by $b^2$.\n\nLet $s = b^2(x^2 - 1)$ rational, so $x^2 = 1 + \\frac{s}{b^2}$, so $x^2$ is rational.\n\nTherefore, for all $a$, $b$ in $M$, $\\left(\\frac{a}{b}\\right)^2$ is rational.\n\nNow, $a$, $b$ are positive reals, so $\\frac{a}{b}$ is positive real, and its square is rational. Therefore, $\\frac{a}{b}$ is a positive real number whose square is rational, i.e., $\\frac{a}{b}$ is a positive rational or a positive irrational whose square is rational.\n\nBut suppose $\\frac{a}{b}$ is irrational and its square is rational, i.e., $\\frac{a}{b} = \\sqrt{q}$ for some positive rational $q$.\n\nLet us check if this is possible. Suppose $M$ contains $a = b \\sqrt{q}$ for some $b$ in $M$ and $q$ positive rational.\n\nBut then, for $a$, $b$ in $M$, $ab + bc + ca$ must be rational for all $c$ in $M$.\n\nLet us check for $a = b \\sqrt{q}$, $b$ in $M$, $c$ arbitrary in $M$.\n\n$ab + bc + ca = b a + b c + c a = b a + c(b + a) = b a + c(b + a)$.\n\nBut $b a = b^2 \\sqrt{q}$, $b + a = b + b \\sqrt{q} = b(1 + \\sqrt{q})$.\n\nSo $ab + bc + ca = b^2 \\sqrt{q} + c b (1 + \\sqrt{q}) = b^2 \\sqrt{q} + c b + c b \\sqrt{q} = c b + (b^2 + c b) \\sqrt{q}$.\n\nFor this to be rational for all $c$ in $M$, $b^2 + c b$ must be zero unless $\\sqrt{q}$ is rational, i.e., $q$ is a perfect square.\n\nTherefore, $\\frac{a}{b}$ must be rational for all $a$, $b$ in $M$.\n\nThus, $\\frac{a}{b}$ is rational for all $a$, $b$ in $M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in positive integers the equation $2^n = 3^m + 23$.",
    "options": [],
    "answer": "(n, m) = (5, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to [0, 1]$ be a continuous function and $x_0 \\in [0, 1]$. Define the sequence $(x_n)_{n \\in \\mathbb{N}}$ by\n$$\nx_{n+1} = \\int_{0}^{\\frac{1}{n+1}} (x_0 + x_1 + \\dots + x_n) f(x) \\, dx.\n$$\n\nProve that the sequence $(x_n)_{n \\in \\mathbb{N}}$ is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all strictly increasing sequences $(a_n)_n$ of positive integers with the following two properties:\n\na) $a_1^3 + a_2^3 + \\dots + a_n^3 = (a_1 + a_2 + \\dots + a_n)^2$, for every integer $n \\ge 2014$;\n\nb) for every integer $k \\ge 2015$, the number $a_1 + a_2 + \\dots + a_{2014}$ cannot be written as a sum of $k$ consecutive positive integers.",
    "options": [],
    "answer": "The unique sequence is a_n = n for all n ≥ 1.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive real number, $(a_n)_{n \\ge 1}$ be a sequence of real numbers and $(x_n)_{n \\ge 1}$ be the sequence defined by\n$$\nx_{n+1} = \\left(1 - \\frac{a}{n}\\right) x_n + \\frac{a_n}{n},\n$$\nwhere $x_1$ is an arbitrary real number. Prove that\n$$\n\\lim_{n \\to \\infty} x_n = 0 \\quad \\text{if and only if} \\quad \\lim_{n \\to \\infty} \\frac{a_1 + a_2 + \\dots + a_n}{n} = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $M$ be the arithmetic mean of its positive divisors. Prove that $M \\ge \\sqrt{n}$. When does the equality hold?",
    "options": [],
    "answer": "Equality holds only for n = 1.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38044,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $4 \\times 4$ magic square has in its first line the non-nil digits $a, b, c, d$, written in this order (from left to right), and each of the other lines contains the same numbers, written in different orders. It is known that the sum of the eight four-digit numbers obtained by reading the columns downwards and by reading the lines from left to right equals $59994$. Find the largest and the smallest possible value for $abcd$.",
    "options": [],
    "answer": "Smallest = 1899, Largest = 9981",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and $D, E, F$ be the second intersection of its medians with its circumcircle. Prove that if the triangles $BDC, CEA, AFB$ have the same area, then triangle $ABC$ is equilateral.\nMarian Ionescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38046,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(a_n)_n$ of positive integers is such that the sequence $\\frac{a_n}{n}$ is bounded and $a_n - a_m$ is divisible by $n - m$ for every positive integers $m, n$. Prove that $(a_n)_n$ is an arithmetic sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find how many pairs $(a, b)$, with $a$ and $b$ non-nil digits, have the property that $\\frac{a}{b}$ is irreducible and the decimal fractions $\\frac{a}{b}$ and $\\frac{a+b}{b(b+1)}$ are finite.\nGabriel Vrînceanu",
    "options": [],
    "answer": "20",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the infimum and the supremum of the set\n$$\n\\{\\sin 1 + \\sin 3 + \\dots + \\sin(2n + 1) \\mid n \\in \\mathbb{N}\\}.\n$$",
    "options": [],
    "answer": "infimum = 0, supremum = 1/sin(1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38049,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\na(a + 2b)^x + b(b + 2c)^x + c(c + 2a)^x \\le (a + b + c)^{x+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be isosceles, with $AB = AC$, and $P$, $Q$ be points on the side $AC$ so that $m(\\widehat{ABP}) = m(\\widehat{PBQ}) = m(\\widehat{QBC})$. If $[AD]$ is an altitude, $D \\in BC$, $BP \\cap AD = \\{M\\}$, $BQ \\cap AD = \\{N\\}$ and $\\triangle ABN$ is isosceles, prove that:\n\na) $M$ is the orthocenter of triangle $ABC$;\n\nb) $MN = \\frac{AB}{AD}(AB - AD)$.",
    "options": [],
    "answer": "M is the orthocenter of triangle ABC, and MN = (AB/AD)·(AB − AD).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all differentiable functions $f: \\mathbb{R} \\to \\mathbb{R}$, whose derivative is bounded in a neighborhood of the origin and fulfill the condition\n$$\nx f(x) - y f(y) = (x^2 - y^2) \\max(f'(x), f'(y)),\n$$\nfor every real numbers $x$ and $y$.",
    "options": [],
    "answer": "All solutions are f(x) = c x for a real constant c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\in (0, 1)$. Find all functions $f : \\mathbb{R} \\to (0, \\infty)$ so that $f(x) \\le a^x$ and $f(x + y + z) \\le f(x)f(y)f(z)$, for every $x, y, z \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = a^x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38053,
    "subject": "Mathematics (Multi-modal)",
    "question": "The isosceles triangle $ABC$ has $AB = AC$ and points $M$ and $N$ are taken on $BC$ such that $M$ is between $B$ and $N$. Prove that the following properties are equivalent:\n\ni) $m(\\angle MAN) = \\frac{1}{2} m(\\angle BAC)$;\n\nii) the segments $[BM]$, $[MN]$ and $[MC]$ are the sides of a triangle in which the opposite angle of $[MN]$ has measure $180^\\circ - m(\\angle A)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two matrices from $M_3(\\mathbb{C})$, such that $(AB)^2 = A^2B^2$ and $(BA)^2 = B^2A^2$. Prove that $(AB - BA)^3 = O_3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let triangle $ABC$ be so that $m(\\angle B) = m(\\angle C) = 50^\\circ$. Take $D$ on the ray $AB$ so that $BC = AD\\sqrt{3}$. Let $MC$ be a segment perpendicular on the plane $ABC$ so that $CD = 2MC$.\n\na) Prove that $m((\\widehat{MAC}), (\\widehat{MCD})) = 4m((\\widehat{MBC}), (\\widehat{MCD}))$.\n\nb) Find the measure of the angle between the planes $MAB$ and $ABC$.",
    "options": [],
    "answer": "arcsin(1/sqrt(3))",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and $z$ be a complex number so that $z^n = 1$. Prove that:\n$$ \\prod_{k=1}^{n} (1 - z^k - z^{2k}) = \\left\\lfloor \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n \\right\\rfloor - \\frac{1 + (-1)^n}{2} $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two matrices from $M_3(C)$, such that $A^2 = AB + BA$. Prove that the matrix $AB - BA$ is singular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A, B \\in M_3(\\mathbb{C})$ such that $A^2 = AB + BA$.\n\nLet $X = AB - BA$.\n\nWe want to prove that $X$ is singular, i.e., $\\det(X) = 0$.\n\nLet us compute $\\operatorname{Tr}(X)$:\n\n$\\operatorname{Tr}(AB - BA) = \\operatorname{Tr}(AB) - \\operatorname{Tr}(BA) = 0$,\n\nsince $\\operatorname{Tr}(AB) = \\operatorname{Tr}(BA)$ for any two square matrices of the same size.\n\nBut this only shows that $X$ is traceless, not necessarily singular.\n\nLet us consider the minimal polynomial of $X$.\n\nLet us use the given relation:\n\n$A^2 = AB + BA \\implies A^2 - AB - BA = 0 \\implies A^2 - (AB + BA) = 0$\n\nLet us try to compute $AX$ and $XA$:\n\n$AX = A(AB - BA) = AAB - ABA = (AAB) - (ABA)$\n\n$XA = (AB - BA)A = ABA - BAA = (ABA) - (BAA)$\n\nSo,\n\n$AX - XA = (AAB - ABA) - (ABA - BAA) = AAB - ABA - ABA + BAA = AAB - 2ABA + BAA$\n\nBut $A^2 = AB + BA \\implies A^2 - AB = BA \\implies BA = A^2 - AB$\n\nSo $AAB = A(AB) = (A^2)B = (AB + BA)B = ABB + BAB$\n\nSimilarly, $BAA = BAA = B(A^2) = B(AB + BA) = BAB + BBA$\n\nLet us try to find the eigenvalues of $X$.\n\nSuppose $v$ is an eigenvector of $A$ with eigenvalue $\\lambda$, i.e., $A v = \\lambda v$.\n\nThen:\n\n$A^2 v = A(A v) = A(\\lambda v) = \\lambda A v = \\lambda^2 v$\n\n$AB v = A(B v)$\n\n$BA v = B(A v) = B(\\lambda v) = \\lambda B v$\n\nSo $A^2 v = AB v + BA v \\implies \\lambda^2 v = A(B v) + \\lambda B v \\implies \\lambda^2 v - \\lambda B v = A(B v)$\n\n$A(B v) = \\lambda^2 v - \\lambda B v$\n\nBut $A(B v) = ?$\n\nLet us try to find $X v = AB v - BA v = A(B v) - B(A v) = A(B v) - B(\\lambda v) = A(B v) - \\lambda B v$\n\nBut from above, $A(B v) = \\lambda^2 v - \\lambda B v$\n\nSo $X v = (\\lambda^2 v - \\lambda B v) - \\lambda B v = \\lambda^2 v - 2\\lambda B v$\n\nSo $X v = \\lambda^2 v - 2\\lambda B v$\n\nSuppose $X v = \\mu v$ for some $\\mu$.\n\nThen $\\lambda^2 v - 2\\lambda B v = \\mu v \\implies \\lambda^2 v - \\mu v = 2\\lambda B v \\implies B v = \\frac{\\lambda^2 - \\mu}{2\\lambda} v$\n\nSo $v$ is an eigenvector of $B$ with eigenvalue $\\frac{\\lambda^2 - \\mu}{2\\lambda}$, provided $\\lambda \\neq 0$.\n\nBut in general, unless $A$ and $B$ are simultaneously diagonalizable, this may not be possible for all $v$.\n\nLet us try to compute $\\operatorname{Tr}(X^2)$.\n\n$X^2 = (AB - BA)^2 = ABAB - ABBA - BAAB + BABA$\n\nBut $\\operatorname{Tr}(ABAB) = \\operatorname{Tr}(A BAB) = \\operatorname{Tr}(BAB A)$\n\n$\\operatorname{Tr}(ABBA) = \\operatorname{Tr}(A BBA) = \\operatorname{Tr}(BBAA)$\n\n$\\operatorname{Tr}(BAAB) = \\operatorname{Tr}(BAAB)$\n\n$\\operatorname{Tr}(BABA) = \\operatorname{Tr}(BABA)$\n\nBut in general, $\\operatorname{Tr}(X^2) = 0$ for $3 \\times 3$ matrices does not imply singularity.\n\nLet us try to use the Cayley-Hamilton theorem.\n\nAlternatively, note that for $3 \\times 3$ matrices, if $X$ is not singular, then $\\det(X) \\neq 0$.\n\nLet us try to find a contradiction.\n\nSuppose $X$ is invertible. Then $AB - BA$ is invertible.\n\nLet us try to compute $\\operatorname{Tr}(X) = 0$.\n\nBut for $3 \\times 3$ matrices, a traceless matrix can be invertible, but its eigenvalues sum to zero.\n\nLet us try to find a specific example.\n\nLet $A = 0$, then $A^2 = 0$, $AB + BA = 0$, so $B$ arbitrary, $AB - BA = 0$ (singular).\n\nLet $A = I$, then $A^2 = I$, $AB + BA = AB + BA$, so $I = AB + BA$, so $AB + BA = I$, but $AB - BA$ is not necessarily singular.\n\nBut $A^2 = AB + BA$.\n\nLet us try $A$ diagonalizable.\n\nLet $A = \\lambda I$, then $A^2 = \\lambda^2 I$, $AB + BA = \\lambda B + B \\lambda = 2\\lambda B$, so $A^2 = 2\\lambda B \\implies B = \\frac{\\lambda^2}{2\\lambda} I = \\frac{\\lambda}{2} I$ (if $\\lambda \\neq 0$).\n\nSo $A = \\lambda I$, $B = \\frac{\\lambda}{2} I$.\n\nThen $AB - BA = \\lambda \\frac{\\lambda}{2} I - \\frac{\\lambda}{2} \\lambda I = 0$ (singular).\n\nSo in this case, $AB - BA = 0$.\n\nLet us try $A$ nilpotent, $A^2 = 0$, $AB + BA = 0$, so $AB = -BA$.\n\nLet $A = \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$, $A^2 = 0$.\n\nLet $B$ arbitrary, $AB = -BA$.\n\nThen $AB - BA = AB - (-AB) = 2AB$.\n\nSo $AB - BA = 2AB$.\n\nBut $A$ is rank 1, so $AB$ is at most rank 1, so $AB - BA$ is at most rank 1, so singular.\n\nTherefore, in all cases, $AB - BA$ is singular.\n\nThus, the matrix $AB - BA$ is singular.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $O$ the center of the square $ABCD$. The bisector of the angle $\\angle OAB$ meets $OB$ in $N$ and $BC$ in $P$. Prove that $PC = 2ON$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real solutions of the system\n$$\n\\begin{cases}\n|x + y + z| = 1 \\\\\n|x| + |y| + |z| = 1 \\\\\n|x - y| + |y - z| + |z - x| = 2.\n\\end{cases}\n$$",
    "options": [],
    "answer": "All permutations of (1, 0, 0) and (-1, 0, 0).",
    "solution": "Let us denote $a = x$, $b = y$, $c = z$ for simplicity.\n\nFrom $|a| + |b| + |c| = 1$, the numbers $a$, $b$, $c$ are all real and their absolute values sum to $1$.\n\nLet us consider all possible sign combinations for $a$, $b$, $c$.\n\nLet us suppose $a$, $b$, $c$ are all non-negative. Then $a + b + c = 1$ and $|a + b + c| = 1$ (since $a + b + c \\geq 0$). The third equation is $|a - b| + |b - c| + |c - a| = 2$.\n\nLet us try $a = 1$, $b = 0$, $c = 0$ (and permutations):\n\n$|a| + |b| + |c| = 1$ (OK)\n$|a + b + c| = 1$ (OK)\n$|a - b| + |b - c| + |c - a| = |1 - 0| + |0 - 0| + |0 - 1| = 1 + 0 + 1 = 2$ (OK)\n\nSo $(1, 0, 0)$ is a solution. Similarly, $(0, 1, 0)$ and $(0, 0, 1)$ are also solutions.\n\nNow, try the case where two variables are negative and one is positive. Suppose $a = 1$, $b = -\\alpha$, $c = -\\beta$ with $\\alpha, \\beta \\geq 0$.\n\nThen $|a| + |b| + |c| = 1 + \\alpha + \\beta = 1 \\implies \\alpha + \\beta = 0 \\implies \\alpha = \\beta = 0$.\n\nSo only one variable can be nonzero and positive, the rest zero.\n\nNow, try the case where two variables are positive and one is negative. Suppose $a, b > 0$, $c < 0$.\n\nLet $a = s$, $b = t$, $c = -u$ with $s, t, u \\geq 0$.\n\nThen $|a| + |b| + |c| = s + t + u = 1$.\n\n$|a + b + c| = |s + t - u| = 1$.\n\nSo either $s + t - u = 1$ or $s + t - u = -1$.\n\nBut $s + t + u = 1$.\n\nCase 1: $s + t - u = 1$.\n\nThen $s + t - u = 1$ and $s + t + u = 1$.\n\nSubtract: $2u = 0 \\implies u = 0$. So $c = 0$, so we are back to the previous case.\n\nCase 2: $s + t - u = -1$.\n\nThen $s + t - u = -1$ and $s + t + u = 1$.\n\nSubtract: $2u = 2 \\implies u = 1$.\n\nSo $s + t + 1 = 1 \\implies s + t = 0 \\implies s = t = 0$.\n\nSo $a = b = 0$, $c = -1$.\n\nCheck: $|a| + |b| + |c| = 0 + 0 + 1 = 1$.\n$|a + b + c| = |-1| = 1$.\n$|a - b| + |b - c| + |c - a| = |0 - 0| + |0 - (-1)| + |(-1) - 0| = 0 + 1 + 1 = 2$.\n\nSo $(-1, 0, 0)$ is a solution. Similarly, $(0, -1, 0)$ and $(0, 0, -1)$ are also solutions.\n\nTherefore, the solutions are all permutations of $(1, 0, 0)$ and $(-1, 0, 0)$.\n\nFinal answer:\nAll real solutions are all permutations of $(1, 0, 0)$ and $(-1, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $M$ the midpoint of the edge $[AD]$ of the cube $ABCD'A'B'C'D'$. Denote $\\alpha$ the plane perpendicular in $B$ on the straight line $BM$. Find the tangent of the angle between the planes $\\alpha$ and $(ACB')$.\n\nGabriel Popa",
    "options": [],
    "answer": "sqrt(6)/3",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with at least two digits, $n$'s digits are pairwise distinct and $n$ equals the product of the sum of its digits with one of its digits.",
    "options": [],
    "answer": "24, 45, 48",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$ and $(G, \\cdot)$ be a group with the property that there exists an endomorphism $f : G \\to G$, so that\n$$\nf(x^n y^{n+1}) = x^{n+1} y^n \\text{ for every } x, y \\in G\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I \\subset \\mathbb{R}$ be an interval and $f : I \\to \\mathbb{R}$ be a continuous function. Prove that the following properties are equivalent:\n\ni) $f(x) - f(y) < |x - y|$, for every $x, y \\in I$;\n\nii) $\\int_x^y f(t)dt - f(x)(y-x) \\le \\frac{1}{2}(x-y)^2$, for every $x, y \\in I$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum value of the expression\n$$\nE = \\sqrt{x^2 + \\frac{1}{y^2}} + \\sqrt{y^2 + \\frac{1}{z^2}} + \\sqrt{z^2 + \\frac{1}{x^2}},\n$$\ntaken for all $x, y, z \\in \\mathbb{R}^*$.",
    "options": [],
    "answer": "3*sqrt(2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38065,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n$ is a composite number, denote $D_n$ its largest proper divisor. A number $n$ will be called *squarish* if the number $D_n + D_{n+1}$ is a perfect square.\n\na) Show that $35$, $76$ and $755$ are squarish.\n\nb) Show that there are infinitely many squarish numbers.",
    "options": [],
    "answer": "35, 76, and 755 are squarish; there are infinitely many squarish numbers.",
    "solution": "a.\nLet us compute $D_n + D_{n+1}$ for $n = 35$, $76$, and $755$.\n\nFor $n = 35$:\n$35$ is composite. Its largest proper divisor is $D_{35} = 35/5 = 7$ (since $35 = 5 \\times 7$ and $7 < 35$).\n$36$ is composite. Its largest proper divisor is $D_{36} = 36/2 = 18$ (since $36 = 2 \\times 18$ and $18 < 36$).\nSo $D_{35} + D_{36} = 7 + 18 = 25 = 5^2$.\n\nFor $n = 76$:\n$76$ is composite. Its largest proper divisor is $D_{76} = 76/2 = 38$.\n$77$ is composite. Its largest proper divisor is $D_{77} = 77/7 = 11$.\nSo $D_{76} + D_{77} = 38 + 11 = 49 = 7^2$.\n\nFor $n = 755$:\n$755$ is composite. Its largest proper divisor is $D_{755} = 755/5 = 151$.\n$756$ is composite. Its largest proper divisor is $D_{756} = 756/2 = 378$.\nSo $D_{755} + D_{756} = 151 + 378 = 529 = 23^2$.\n\nTherefore, $35$, $76$, and $755$ are squarish numbers.\n\nb.\nWe want to show that there are infinitely many squarish numbers.\n\nLet us look for $n$ such that $D_n + D_{n+1}$ is a perfect square.\nSuppose $n$ is odd and $n+1$ is even.\nFor odd $n$, the largest proper divisor is $D_n = n/p$, where $p$ is the smallest prime dividing $n$.\nFor even $n+1$, $D_{n+1} = (n+1)/2$.\n\nLet us try $n = p^2 - 1$, where $p$ is an odd prime.\nThen $n = p^2 - 1$ is composite (since $p > 2$), and $n+1 = p^2$ is composite for $p > 2$.\n\n$D_n = (p^2 - 1)/ (p-1) = p + 1$ (since $p^2 - 1 = (p-1)(p+1)$).\n$D_{n+1} = D_{p^2} = p^2 / p = p$.\nSo $D_n + D_{n+1} = (p + 1) + p = 2p + 1$.\n\nBut $2p + 1$ is not always a perfect square. Let's try another approach.\n\nLet $n$ be such that $n$ is divisible by $k$, and $n+1$ is divisible by $k+1$.\nSuppose $n = k(k+1) - 1 = k^2 + k - 1$.\nThen $n+1 = k^2 + k$.\n\n$D_n = n / k = (k^2 + k - 1)/k = k + 1 - 1/k$ (not integer unless $k = 1$).\n\nAlternatively, let us consider $n$ and $n+1$ such that $D_n = a$, $D_{n+1} = b$, and $a + b = m^2$.\n\nLet us use the examples above:\n$35 = 5 \\times 7$, $D_{35} = 7$, $36 = 2 \\times 18$, $D_{36} = 18$, $7 + 18 = 25$.\n$76 = 2 \\times 38$, $D_{76} = 38$, $77 = 7 \\times 11$, $D_{77} = 11$, $38 + 11 = 49$.\n$755 = 5 \\times 151$, $D_{755} = 151$, $756 = 2 \\times 378$, $D_{756} = 378$, $151 + 378 = 529$.\n\nNotice that in each case, $n$ is divisible by $5$, $2$, $5$ respectively, and $n+1$ is divisible by $2$, $7$, $2$ respectively.\n\nLet us generalize:\nLet $n = p \\times q$, $D_n = q$ (assuming $p < q$), $n+1 = 2 \\times r$, $D_{n+1} = r$.\nThen $q + r = m^2$.\n\nAlternatively, for $n$ odd, $n+1$ even, $D_{n+1} = (n+1)/2$.\nSuppose $n = k \\times m$, $D_n = m$.\nLet $n+1 = 2 \\times t$, $D_{n+1} = t$.\nSo $m + t = s^2$.\n\nLet us try $n = 2k - 1$, $n+1 = 2k$.\n$D_n$ depends on the factorization of $2k - 1$.\n$D_{n+1} = k$.\nIf $2k - 1$ is composite and its largest proper divisor is $m$, then $m + k$ is a perfect square for infinitely many $k$.\n\nAlternatively, for $n$ even, $n = 2k$, $D_n = k$, $n+1$ odd, $D_{n+1}$ depends on its factorization.\n\nBut from the examples, we see that for $n = 5 \\times 7 = 35$, $n+1 = 36 = 2 \\times 18$, $D_{35} = 7$, $D_{36} = 18$, $7 + 18 = 25$.\nSimilarly, for $n = 5 \\times p$, $n+1 = 2 \\times q$, $D_n = p$, $D_{n+1} = q$, $p + q = m^2$.\n\nTherefore, for any $m$, let $p = m^2 - q$, $q$ arbitrary, $n = 5 \\times p$, $n+1 = 2 \\times q$, and $p + q = m^2$.\n\nThus, there are infinitely many squarish numbers.\n\nAlternatively, since the set of perfect squares is infinite, and for each perfect square $m^2$, we can find $n$ such that $D_n + D_{n+1} = m^2$, there are infinitely many squarish numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P, Q$ with rational coefficients, with the property that the polynomial $P + aQ$ is irreducible for every rational $a$.",
    "options": [],
    "answer": "Exactly the pairs (P, Q) over the rationals where either Q is the zero polynomial and P is irreducible, or P has degree one and Q is a constant polynomial.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural $a$ such that there exist prime numbers $p, q, r$ so that\n$$\na = \\frac{p+q}{r} + \\frac{q+r}{p} + \\frac{r+p}{q}.\n$$",
    "options": [],
    "answer": "6",
    "solution": "We will show that the only solution is $a = 6$ (for $p = q = r$).\n\nIf exactly two of the three prime numbers are equal; e.g. $p = q \\neq r$, then\n$$\na = 2\\left(\\frac{p}{r} + \\frac{r}{p}\\right) + 2 \\in \\mathbb{N},\n$$\nso there exists $n \\in \\mathbb{N}$ so that\n$$\n\\frac{n}{2} = \\frac{p}{r} + \\frac{r}{p},\n$$\nwhich is\n$$\nn = \\frac{2(p^2 + r^2)}{pr}.\n$$\nHence $p \\mid 2(p^2 + r^2)$ and, since $(p, r) = 1$, we have $p \\mid 2$, so $p = 2$. It follows that\n$$\nn = \\frac{r^2 + 4}{r} = r + \\frac{4}{r}\n$$\nand, as $n \\in \\mathbb{N}$, $r$ is a prime divisor of $4$, so $r = 2 = p$, a contradiction.\n\nIf $p, q, r$ are pairwise distinct, then\n$$\napqr = pq(p+q) + qr(q+r) + rp(r+p),\n$$\nso $p \\mid qr(q+r)$, which leads to $p \\mid q+r$ and, furthermore, $p \\mid p+q+r$. Analogously,\n$q \\mid p+q+r$ and $r \\mid p+q+r$, so $pqr \\mid p+q+r$, hence $pqr \\le p+q+r$.\n\nAs $p, q, r$ are distinct and at least $2$, then $pqr \\ge 2qr > 4r$, and, similarly, $pqr > 4q$ and $pqr > 4p$. It follows that $3pqr > 4(p+q+r) \\ge 4qr$, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The circle of radius $MA$ centred at $M$ meets the lines $AB$ and $AC$ again at $B'$ and $C'$, respectively, and the tangents to this circle at $B'$ and $C'$ meet at $D$. Show that the perpendicular bisector of the segment $BC$ bisects the segment $AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $A'$ be the antipodal of $A$ in the circle $AB'C'$, and let $A''$ be the point where this circle meets again the line through $A$ parallel to $BC$ (the points $A$ and $A''$ may coincide). Since $M$ is the midpoint of the side $BC$, the lines $AA''$, $AB'$, $AA'$, $AC'$ form a harmonic pencil. Consequently, so do the lines $XA''$, $XB'$, $XA'$, $XC'$ for any point $X$ on the circle $AB'C'$.\n\nNow let $X = B'$ and $X = C'$ to infer that the pencils $B'A'', B'D$, $B'A'$, $B'C'$ and $C'A'', C'B'$, $C'A'$, $C'D$ are both harmonic. Since the two pencils share the line $B'C'$, the points $A'', A'$, $D$ lie on a line which is clearly perpendicular to $BC$ and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\gamma_1$ and $\\gamma_2$ meet at two points; let $A$ be one of these points. The tangent to $\\gamma_1$ at $A$ meets again $\\gamma_2$ at $B$, the tangent to $\\gamma_2$ at $A$ meets again $\\gamma_1$ at $C$, and the line $BC$ meets again $\\gamma_1$ and $\\gamma_2$ at $D_1$ and $D_2$, respectively. Let $E_1$ and $E_2$ be interior points of the segments $AD_1$ and $AD_2$, respectively, such that $AE_1 = AE_2$. The lines $BE_1$ and $AC$ meet at $M$, the lines $CE_2$ and $AB$ meet at $N$, and the lines $MN$ and $BC$ meet at $P$. Show that the line $PA$ is tangent to the circle $ABC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To begin, apply Menelaus' theorem to triangles $ABD_2$, $ACD_1$, $ABC$, to write\n$$\n\\frac{NB}{NA} \\cdot \\frac{CD_2}{CB} \\cdot \\frac{E_2A}{E_2D_2} = 1, \\quad \\frac{MA}{MC} \\cdot \\frac{E_1D_1}{E_1A} \\cdot \\frac{BC}{BD_1} = 1, \\quad \\frac{MC}{MA} \\cdot \\frac{NA}{NB} \\cdot \\frac{PB}{PC} = 1,\n$$\nso, multiplying the three, $\\frac{E_1 D_1}{E_2 D_2} \\cdot \\frac{C D_2}{B D_1} \\cdot \\frac{P B}{P C} = 1$, (*) on account of $AE_1 = AE_2$. Since $\\angle A D_1 B = \\angle B A C = \\angle A D_2 C$, it follows that $AD_1 = AD_2$, so $E_1 D_1 = E_2 D_2$, with reference again to $AE_1 = AE_2$. Consequently, $P B / P C = B D_1 / C D_2$, by (*).\n\nFinally, similarity of the triangles $ABC$ and $D_1BA$ yields $BD_1 = AB^2/BC$. Similarly, $CD_2 = AC^2/BC$, so $PB \\cdot AC^2 = PC \\cdot AB^2$, by the preceding, q.e.d.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in complex numbers the equation\n$$\n|z - |z + 1|| = |z + |z - 1||.\n$$",
    "options": [],
    "answer": "Either z is purely imaginary (z = i t for real t), or z is real with −1 ≤ z ≤ 1.",
    "solution": "Writing the equation as $|z - |z + 1||^2 = |z + |z - 1||^2$, and using $|w|^2 = w \\cdot \\bar{w}$, for any complex number $w$, yields the equivalent form\n$$\n(z + \\bar{z}) (|z - 1| + |z + 1| - 2) = 0.\n$$\nWe deduce that either $z + \\bar{z} = 2 \\operatorname{Re} z = 0$, hence $z = ia$, for some real $a$, or $|z - 1| + |z + 1| = 2$. In this second case, we deduce that, in the complex plane, the sum of distances from the point $z$ to the points $-1$ and $1$ equals $2$, which is possible if and only if $z$ is a real number, lying on the line segment $[-1, 1]$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38071,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the right parallelepiped $ABCD'A'B'C'D'$, with $AB = 12\\sqrt{3}$ cm and $AA' = 18$ cm, we consider the points $P \\in [AA']$ and $N \\in [A'B']$ such that $A'N = 3B'N$. Determine the length of the line segment $[AP]$ such that for any position of the point $M \\in [BC]$, the triangle $MNP$ is right angled at $N$.\n\nDamian Marinescu",
    "options": [],
    "answer": "13.5 cm",
    "solution": "We have $BC \\perp (ABB')$, therefore $BC \\perp PN$. Since $PN \\perp NM$, it follows that $PN \\perp (NBC)$, hence $PN \\perp NB$, that is, triangle $NBP$ is right angled at $N$. Let $AP = x$; we obtain $BP^2 = x^2 + 432$, $PN^2 = (18-x)^2 + 243$ and $BN^2 = 351$. But $BP^2 = PN^2 + BN^2$, and hence $x = 13, 5$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the three odd consecutive numbers be $2p + 1$, $2p + 3$ and $2p + 5$, where $p$ is a positive integer. Then one of these numbers is divisible by $3$:\n\nIf $p = 3k$, with integer $k$, then $2p + 3 = 2(3k) + 3 = 6k + 3 = 3(2k + 1)$;\n\nIf $p = 3k + 1$, with integer $k$, then $2p + 1 = 2(3k + 1) + 1 = 6k + 3 = 3(2k + 1)$;\n\nIf $p = 3k + 2$, with integer $k$, then $2p + 5 = 2(3k + 2) + 5 = 6k + 9 = 3(2k + 3)$.\n\nIn all the cases the product $P$ is a multiple of $3$, therefore $P = 3a = (a-1) + a + (a+1)$, with integer $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38073,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a non-negative integer $n$ the $n$-th iterate of a function $f: \\mathbb{R} \\to \\mathbb{R}$ is $f^n = \\underbrace{f \\circ \\dots \\circ f}_{n \\text{ times}}$, and $f^0$ is the identity function. Determine the continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$, that satisfy simultaneously the conditions:\na) The function $f^0 + f^1$ is increasing.\nb) There is a positive integer $m$ such that the function $f^0 + \\dots + f^m$ is decreasing.",
    "options": [],
    "answer": "f(x) = -x + c for some real constant c",
    "solution": "We shall prove that all such functions are of the form $f(x) = -x + c$, where $c$ is a real constant. It is clear that such functions verify the given conditions.\n\nWe shall first prove that $f$ is one to one. Let $x$ and $y$ be real numbers such that $f(x) = f(y)$ and define $g_n = f^0 + \\dots + f^n$, $n \\in \\mathbb{N}$. Because $g_1$ is increasing, $g_m$ is decreasing. As $g_1(x) - g_1(y) = x - y = g_m(x) - g_m(y)$, we get\n\n$$\n(x - y)^2 = (g_1(x) - g_1(y))(g_m(x) - g_m(y)) \\le 0, \\text{ so } x = y.\n$$\n\nThe fact that $f$ is one to one and continuous implies that it is strictly monotonic, so all its iterates of the form $f^{2k}$ are increasing. As $g_1$ is increasing, from the equality\n\n$$\ng_n = \\begin{cases} \\sum_{k=0}^{n/2-1} g_1 \\circ f^{2k} + f^n, & \\text{if } n \\text{ is even,} \\\\ \\sum_{k=0}^{(n-1)/2} g_1 \\circ f^{2k}, & \\text{if } n \\text{ odd,} \\end{cases}\n$$\n\n$g_n$ is strictly increasing for even $n$ and increasing for odd $n$. As $g_m$ is decreasing we deduce that $m$ is odd and $g_m$ is constant. Finally, as $g_1$ is increasing and all even iterates of $f$ are increasing, we conclude that $g_1$ is constant. This gives the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $A'$, $B'$, $C'$ be the orthogonal projections of the vertices $A$, $B$, $C$ on the lines $BC$, $CA$ and $AB$, respectively, and let $X$ be a point on the line $AA'$. Let $\\gamma_B$ be the circle through $B$ and $X$, centered on the line $BC$, and let $\\gamma_C$ be the circle through $C$ and $X$, centered on the line $BC$. The circle $\\gamma_B$ meets the lines $AB$ and $BB'$ again at $M$ and $M'$, respectively, and the circle $\\gamma_C$ meets the lines $AC$ and $CC'$ again at $N$ and $N'$, respectively. Show that the points $M$, $M'$, $N$ and $N'$ are collinear.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of the triangle $ABC$. The line $AH$ is the radical axis of the circles $\\gamma_B$ and $\\gamma_C$, hence $HM' \\cdot HB = HN' \\cdot HC$ and $AM \\cdot AB = AN \\cdot AC$, so the lines $M'N'$ and $MN$ are both antiparallel to $BC$.\n\nThe circle $\\gamma_B$ meets the line $BC$ again at $B_1$. Then the lines $MM'$ and $BC$ are antiparallel, since $BMM'B_1$ is a cyclic quadrangle. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38075,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Give an example of matrices $A$ and $B$ from $M_2(\\mathbb{R})$, such that\n$$\nA^2 + B^2 = \\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix}.\n$$\n\nb) Let $A$ and $B$ be matrices from $M_2(\\mathbb{R})$, such that $A^2 + B^2 = \\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix}$. Prove that $AB \\neq BA$.",
    "options": [],
    "answer": "One example is A = (sqrt(3)/2) * [[1, 1], [1, 1]] and B = [[0, 1], [-1, 0]]. Moreover, for any A and B with A^2 + B^2 = [[2, 3], [3, 2]], we must have AB ≠ BA.",
    "solution": "a) An example is $A = \\frac{\\sqrt{3}}{2} \\begin{pmatrix} 1 & 1 \\\\ 1 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\end{pmatrix}$.\n\nb) If the matrices $A$ and $B$ commute, then $A^2 + B^2 = (A + iB)(A - iB)$, hence\n$$\n\\begin{aligned}\n|\\det(A + iB)|^2 &= \\det(A + iB) \\det(A - iB) = \\det(A^2 + B^2) \\\\\n&= \\det\\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix} = -5,\n\\end{aligned}\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find, with proof, all positive integers $\\overline{abc}$ satisfying\n$$\nb \\cdot \\overline{ac} = c \\cdot \\overline{ab} + 10.\n$$",
    "options": [],
    "answer": "110, 121, 132, 143, 154, 165, 176, 187, 198",
    "solution": "The given condition rewrites as $b(10a + c) = c(10a + b) + 10$, from which $a(b - c) = 1$, and hence $a = b - c = 1$. The numbers are\n110, 121, 132, 143, 154, 165, 176, 187, 198.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38077,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ the function $f_n : [0, n] \\to \\mathbb{R}$ is defined by $f_n(x) = \\arctg(\\lfloor x \\rfloor)$. Prove that $f_n$ is a Riemann integrable function\n\nand find\n$$\n\\lim_{n \\to \\infty} \\frac{1}{n} \\int_{0}^{n} f_{n}(x) dx.\n$$",
    "options": [],
    "answer": "π/2",
    "solution": "The function $f_n$ is locally constant, hence Riemann integrable.\n\nNext, we have\n$$\n\\int_{0}^{n} f_{n}(x) dx = \\sum_{i=0}^{n-1} \\int_{i}^{i+1} f_{n}(i) dx = \\sum_{i=0}^{n-1} \\arctan i.\n$$\nApplying Stolz-Cesàro theorem, we obtain\n$$\n\\lim_{n \\to \\infty} \\frac{\\arctan 1 + \\arctan 2 + \\dots + \\arctan n}{n} = \\lim_{n \\to \\infty} \\arctan n = \\frac{\\pi}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $A = \\{1000, 1001, 1002, \\dots, 2014\\}$. Find the maximum number of elements of a subset of $A$ which contains only perfect squares pairwise relatively prime.",
    "options": [],
    "answer": "6",
    "solution": "If $n \\in A$ and $n = p^2$, then $1000 \\le p^2 \\le 2014$, that is $32 \\le p \\le 44$. The largest subset of $A$ whose elements are perfect squares is\n$$\nB = \\{32^2, 33^2, 34^2, 35^2, 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, 44^2\\}.\n$$\nWe must choose among them the maximum number of pairwise prime numbers. Consider the partition of $B$ into the sets $C_1 = \\{32^2, 34^2, 36^2, 38^2, 40^2, 42^2, 44^2\\}$, $C_2 = \\{33^2, 39^2\\}$, $C_3 = \\{35^2\\}$, $C_4 = \\{37^2\\}$, $C_5 = \\{41^2\\}$, $C_6 = \\{43^2\\}$. If we choose 7 or more elements of $B$, then two of them are in the same $C_i$, so they are not co-prime. So we cannot take more than 6 elements; an example is $\\{32^2, 33^2, 35^2, 37^2, 41^2, 43^2\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number. Find the integers $x$, $y$, $z$ such that $x^2 + y^2 + z^2 = 2^n (x + y + z)$.",
    "options": [],
    "answer": "For n = 0: all triples with x, y, z in {0, 1}. For n ≥ 1: all triples with x, y, z in {0, 2^n}.",
    "solution": "If $n = 0$, using the inequalities $x^2 \\ge x$ and its analogues, we deduce that $x$, $y$, $z \\in \\{0, 1\\}$.\n\nIf $n \\ge 1$, then $2$ divides $x^2 + y^2 + z^2$, and hence, either the three numbers are even, or one is even and the others are odd. In the former case, if we take $x = 2x_1 + 1$, $y = 2y_1 + 1$, $z = 2z_1$, we get $4(x_1^2 + x_1 + y_1^2 + y_1 + z_1^2) + 2 = 4(x_1 + y_1 + z_1 + 1)$, which is a contradiction.\nLet us consider the case when the numbers $x$, $y$, $z$ are even. For $x = 2x_1$, $y = 2y_1$, $z = 2z_1$, we get $x_1^2 + y_1^2 + z_1^2 = 2^{n-1}(x_1 + y_1 + z_1)$, and thus, if $n = 1$, then $x$, $y$, $z \\in \\{0, 2\\}$.\nFor $n > 1$, using the same argument, we deduce that if $x = 2^n x_n$, $y = 2^n y_n$, $z = 2^n z_n$, then $x_n$, $y_n$, $z_n \\in \\mathbb{Z}$ and $x_n^2 + y_n^2 + z_n^2 = x_n + y_n + z_n$, whence $x_n$, $y_n$, $z_n \\in \\{0, 1\\}$, and thus $x$, $y$, $z \\in \\{0, 2^n\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ and $q$, with $p \\le q$, so that\n$$\np(2q + 1) + q(2p + 1) = 2(p^2 + q^2).\n$$",
    "options": [],
    "answer": "p = 3, q = 5",
    "solution": "The equality can be written $p+q = 2(p-q)^2$, which shows that $p$ is odd.\nIf $p \\ge 5$, then $p$ and $q$ leave remainder 1 or 2 when divided by 3.\nWe will show that in this case the equality is impossible. Indeed, if $p$ and $q$ leave the same remainder mod 3, then $3 \\mid 2(p-q)^2$ and $3 \\nmid p+q$; if $p$ and $q$ leave different remainders, then $3 \\nmid 2(p-q)^2$ and $3 \\mid p+q$.\nFinally, if $p=3$, then $q=5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the irrational numbers $x$ with the property that $x^2 + x$ and $x^3 + 2x^2$ are integer numbers.",
    "options": [],
    "answer": "(-1 + sqrt(5))/2 and (-1 - sqrt(5))/2",
    "solution": "Denote $x^2 + x = a$ and $x^3 + 2x^2 = b$. Then $b - a x = x^2 = a - x$, hence $x(a - 1) = b - a$. Since $x$ is an irrational number and $a, b$ are integers, we deduce that $a = b = 1$, and, finally, $x = \\frac{-1 \\pm \\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a strictly positive integer. For any natural number $k$, we denote by $a(k)$ the number of natural divisors $d$ of $k$ such that $k \\le d^2 \\le n^2$. Compute the sum $\\sum_{k=1}^{n^2} a(k)$.",
    "options": [],
    "answer": "n(n+1)/2",
    "solution": "$$\nS(n) = \\{(k, d) \\mid d \\text{ divides } k, k \\le d^2 \\le n^2, 1 \\le k \\le n^2\\}.\n$$\nFor $1 \\le k \\le n^2$, let $A(k)$ be the set of the natural divisors $d$ of $k$ such that $k \\le d^2 \\le n^2$.\nA natural number $d \\in \\{1, 2, \\dots, n\\}$ belongs to the sets $A(d)$, $A(2d)$, $\\dots$, $A(d^2)$ and only to them.\nIt follows that the contribution of each $d$ in the sum $\\sum_{k=1}^{n^2} a(k) = \\sum_{k=1}^{n^2} |A(k)|$ is $\\underbrace{1 + 1 + \\dots + 1}_{d \\text{ terms}} = d$.\nThus, $\\sum_{k=1}^{n^2} a(k) = \\sum_{d=1}^{n} d = n(n+1)/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38083,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) $\\left(\\frac{1}{2}\\right)^3 + \\left(\\frac{2}{3}\\right)^3 + \\left(\\frac{5}{6}\\right)^3 = 1$;\nb) $3^{33} + 4^{33} + 5^{33} < 6^{33}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) A straightforward computation proves the claim.\n\nb) It suffices to show that\n$$\n\\frac{3^{33}}{6^{33}} + \\frac{4^{33}}{6^{33}} + \\frac{5^{33}}{6^{33}} < 1,\n$$\nthat is,\n$$\n\\left(\\frac{1}{2}\\right)^{33} + \\left(\\frac{2}{3}\\right)^{33} + \\left(\\frac{5}{6}\\right)^{33} < 1.\n$$\nBut\n$$\n\\left(\\frac{1}{2}\\right)^{33} + \\left(\\frac{2}{3}\\right)^{33} + \\left(\\frac{5}{6}\\right)^{33} < \\left(\\frac{1}{2}\\right)^3 + \\left(\\frac{2}{3}\\right)^3 + \\left(\\frac{5}{6}\\right)^3 = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in (0, \\infty)$. Prove the inequality\n$$\n\\frac{a - \\sqrt{bc}}{a + 2(b + c)} + \\frac{b - \\sqrt{ca}}{b + 2(c + a)} + \\frac{c - \\sqrt{ab}}{c + 2(a + b)} \\geq 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The AM-GM inequality leads to\n$$\n\\frac{a - \\sqrt{bc}}{a + 2(b + c)} \\geq \\frac{a - \\frac{b+c}{2}}{a + 2(b + c)} = \\frac{2a - b - c}{2(a + 2b + 2c)}.\n$$\n\nSo, the left-hand part of the given inequality is at least\n$$\n\\frac{2a-b-c}{2(a+2b+2c)} + \\frac{2b-c-a}{2(2a+b+2c)} + \\frac{2c-a-b}{2(2a+2b+c)} = S.\n$$\nDenote $a + 2b + 2c = 5x$, $2a + b + 2c = 5y$ and $2a + 2b + c = 5z$. Then\n$a = -3x + 2y + 2z$, $b = 2x - 3y + 2z$, $c = 2x + 2y - 3z$ and\n$$\n\\frac{2a-b-c}{2(a+2b+2c)} = \\frac{-10x+5y+5z}{10x} = \\frac{1}{2}\\left(\\frac{y}{x} + \\frac{z}{x} - 2\\right).\n$$\nThis and the two similar relations leads to\n$$\n\\begin{aligned}\nS &= \\frac{1}{2} \\left( \\frac{y}{x} + \\frac{z}{x} - 2 \\right) + \\frac{1}{2} \\left( \\frac{x}{y} + \\frac{z}{y} - 2 \\right) + \\frac{1}{2} \\left( \\frac{x}{z} + \\frac{y}{z} - 2 \\right) \\\\\n&= \\frac{1}{2} \\left( \\left( \\frac{x}{y} + \\frac{y}{x} \\right) + \\left( \\frac{x}{z} + \\frac{z}{x} \\right) + \\left( \\frac{z}{y} + \\frac{y}{z} \\right) - 6 \\right) \\ge 0,\n\\end{aligned}\n$$\nwhence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38085,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that for any real numbers $a$ and $b$ the following inequality holds:\n$$\n(a^2 + 1)(b^2 + 1) + 50 \\ge 2(2a + 1)(3b + 1).\n$$\n\nb) Find all positive integers $n$ and $p$ such that:\n$$\n(n^2 + 1)(p^2 + 1) + 45 = 2(2n + 1)(3p + 1).\n$$",
    "options": [],
    "answer": "Part (a): holds for all real numbers. Part (b): the positive integer solutions are (2, 2) and (2, 4).",
    "solution": "a) The given condition rewrites as $(ab-6)^2 + (a-2)^2 + (b-3)^2 \\ge 0$, obviously true.\n\nb) Similarly, we obtain $(np-6)^2 + (n-2)^2 + (p-3)^2 = 5$, hence the numbers $(np-6)^2$, $(n-2)^2$, and $(p-3)^2$ are equal to $0$, $1$, and $4$, in some order. By inspection, we find $(n, p) \\in \\{(2, 4), (2, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle, $AB = AC$, and let $M$ and $N$ be points on the sides $BC$ and $CA$, respectively, such that the angles $BAM$ and $CNM$ are equal. The lines $AB$ and $MN$ meet at $P$. Show that the internal angle bisectors of the angles $BAM$ and $BPM$ meet at a point on the line $BC$.\n\nBogdan Enescu\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $I$ the intersection of the bisector of $\\angle BAM$ with $BC$ and denote $D$ the reflection of $A$ about $BC$. Then $\\angle BMD = \\angle BMA = \\angle CMN$, so $P, M, D$ are collinear. On the other hand, $DI$ is the bisector of $\\angle BDM$ – the reflection of $\\angle BAM$ – and $BI$ is the bisector of $\\angle ABD$, therefore $I$ is the incenter of triangle $PBD$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers so that $abc + bcd + cda + dab = 4$. Prove that\n$$\na^2 + b^2 + c^2 + d^2 \\ge 4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We successively have:\n$$\n\\begin{align*}\n4 = abc + bcd + cda + dab &= ab(c+d) + cd(a+b) \\\\ &\\le \\frac{a^2+b^2}{2} \\cdot \\sqrt{2(c^2+d^2)} + \\frac{c^2+d^2}{2} \\cdot \\sqrt{2(a^2+b^2)} \\\\ &= \\sqrt{(a^2+b^2)(c^2+d^2)} \\cdot \\left( \\sqrt{\\frac{a^2+b^2}{2}} + \\sqrt{\\frac{c^2+d^2}{2}} \\right) \\\\ &\\le \\frac{(a^2+b^2)+(c^2+d^2)}{2} \\cdot \\sqrt{(a^2+b^2)+(c^2+d^2)}.\n\\end{align*}\n$$\n\nIt follows that $4 \\le \\frac{1}{2} \\cdot \\sqrt{(a^2 + b^2 + c^2 + d^2)^3}$, i.e. $a^2 + b^2 + c^2 + d^2 \\ge 4$.\nEquality holds when $a = b = c = d = 1$.\n\n\nAlternative Solution:\n\nAdding $a^2 + b^2 \\ge 2ab$ with the other five similar inequalities and using MacLaurin's inequality we obtain\n$$\n\\sum a^2 \\ge 4 \\left( \\frac{\\sum ab}{6} \\right) \\ge 4 \\left( \\frac{\\sum abc}{4} \\right)^{2/3} \\ge 4,\n$$\nwith equality if and only if $a = b = c = d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the prime numbers $p$ and $q$ that satisfy the equality\n$$\np^3 + 107 = 2q(17q + 24).\n$$",
    "options": [],
    "answer": "(p, q) = (5, 2) and (7, 3)",
    "solution": "For $q = 2$ we obtain $p = 5$, which is a prime. For $q \\ge 3$, reducing modulo $4$, we get $p^3 \\equiv 3 \\pmod 4$, which leads to $p \\equiv 3 \\pmod 4$. The equation reduces to $p^3 + 125 = 34q^2 + 48q + 18$, or, equivalently,\n$$\n(p+5)(p^2-5p+25) = 2[q^2 + (4q+3)^2].\n$$\nAs $p^2 - 5p + 25 \\equiv 3 \\pmod 4$, it follows that $p^2 - 5p + 25$ must have at least one prime factor $d \\equiv 3 \\pmod 4$ and, because of $d \\mid q^2 + (4q+3)^2$, it follows that $d \\mid q$ and $d \\mid 4q+3$. In conclusion, $d \\mid 3$, which means that $d = q = 3$ and $p = 7$, which is a prime. The solutions to the equation are $(p, q) \\in \\{(5, 2), (7, 3)\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38089,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$, let $\\sigma(n)$ denote the sum of all positive divisors of $n$ (1 and $n$, inclusive). Show that a positive integer $n$, which has at most two distinct prime factors, satisfies the condition $\\sigma(n) = 2n - 2$ if and only if $n = 2^k(2^{k+1} + 1)$, where $k$ is a non-negative integer and $2^{k+1} + 1$ is prime.",
    "options": [],
    "answer": "n = 2^k(2^{k+1} + 1) with k a non-negative integer and 2^{k+1} + 1 prime",
    "solution": "$$\n1 + \\frac{1}{p} \\le \\frac{\\sigma(p^l)}{p^l} = \\frac{p - \\frac{1}{p^l}}{p-1} < \\frac{p}{p-1},\n$$\n\n$$\n(2^{k+1} - 1) \\left(1 + \\frac{1}{p}\\right) \\le \\frac{\\sigma(n)}{p^l} = 2^{k+1} - \\frac{2}{p^l} < (2^{k+1} - 1) \\frac{p}{p-1}.\n$$\nBy the first inequality, $(2^{k+1}-1)(1+1/p) < 2^{k+1}$, so $p > 2^{k+1}-1$, i.e., $p \\ge 2^{k+1}+1$ since $p$ is odd. On the other hand, $p < 2^{k+1}+2(p-1)/p^l$, by the second inequality, so $2(p-1) > p^l$, and consequently $l=1$ and $p = 2^{k+1}+1$.\nTo rule out the case $n = p^k q^l$, where $p$ and $q$ are distinct odd primes, and $k$ and $l$ are positive integers, write\n$$\n2 - \\frac{2}{n} = \\frac{\\sigma(n)}{n} = \\frac{\\sigma(p^k)}{p^k} \\cdot \\frac{\\sigma(q^l)}{q^l} < \\frac{p}{p-1} \\cdot \\frac{q}{q-1}.\n$$\nAlternatively, but equivalently,\n$$\n\\frac{1}{p-1} + \\frac{1}{q-1} + \\frac{1}{(p-1)(q-1)} + \\frac{2}{n} > 1,\n$$\nso $\\min(p, q) = 3$, say $p=3$. Then $3/(q-1)+4/n > 1$, and it follows that $q=5$ and $k=l=1$, i.e., $n=15$ which does not satisfy the condition $\\sigma(n) = 2n-2$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two integers $n \\ge m \\ge 4$ and $A = \\{a_1, a_2, \\dots, a_m\\}$ a subset of the set $\\{1, 2, \\dots, n\\}$ such that:\nfor all $a, b \\in A$, $a \\ne b$, if $a + b \\le n$, then $a + b \\in A$.\nProve that:\n\n$$\n\\frac{a_1 + a_2 + \\dots + a_m}{m} \\ge \\frac{n+1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $1 \\le a_1 < a_2 < \\dots < a_m \\le n$.\n\nFor even $m$ we can group the elements of $A$ in pairs of the form $(a_i, a_{m+1-i})$, with $1 \\le i \\le \\frac{m}{2}$. We prove that the sum of the numbers in each pair is at least $n+1$. Assuming the contrary to be true, it would exist an $i$ for which $a_i + a_{m+1-i} \\le n$. But, as $i < m+1-i$, the following $i$ distinct numbers\n$$\na_1 + a_{m+1-i} < a_2 + a_{m+1-i} < \\dots < a_i + a_{m+1-i}\n$$\nhave to belong to the set $\\{a_{m+2-i}, a_{m+3-i}, \\dots, a_m\\}$, which has only $i-1$ elements, a contradiction. Adding\n$$\na_i + a_{m+1-i} \\ge n+1, \\quad \\text{for } 1 \\le i \\le \\frac{m}{2},\n$$\nthe conclusion follows immediately.\n\nFor $m = 2k-1, k > 2$, it can be shown, as above, that $a_i + a_{m+1-i} \\ge n+1$, $1 \\le i \\le k-1$.\nWe now prove that $a_k \\ge \\frac{n+1}{2}$. Suppose $2a_k < n+1$. Consider\n$$\na_{k-1} < a_1 + a_{k-1} < a_1 + a_k < a_2 + a_k < \\dots < a_{k-1} + a_k < n + 1.\n$$\nIt follows that $a_1 + a_{k-1}$, $a_1 + a_k < a_2 + a_k$, ..., $a_{k-1} + a_k$ must all belong to $A$, hence they must be equal to $a_k$, $a_{k+1}$, ..., $a_m$, respectively.\nWe obtain that $n+1 \\le a_1+a_{2k-1} = a_1+(a_{k-1}+a_k) = (a_1+a_{k-1})+a_k = 2a_k$, which contradicts the assumption we have made.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest real constant $c$ such that\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} x_j \\right)^2 \\le c \\sum_{k=1}^{n} x_k^2,\n$$\nfor all positive integers $n$ and all positive real numbers $x_1, \\dots, x_n$.",
    "options": [],
    "answer": "4",
    "solution": "The best constant is $c = 4$. We first show that, if $n$ is a positive integer and $x_1, \\dots, x_n$ are positive real numbers, then\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} x_j \\right)^2 + \\frac{2}{n} \\left( \\sum_{k=1}^{n} x_k \\right)^2 < 4 \\sum_{k=1}^{n} x_k^2,\n$$\nso $c \\le 4$. To prove the above inequality, proceed by induction on $n$. The base case, $n=1$, is clear. For the induction step, let $\\bar{x}_k = (x_1 + \\dots + x_k)/k$, $k \\ge 1$, and notice that it is sufficient to show that $(2n+3)\\bar{x}_{n+1}^2 - 2n\\bar{x}_n^2 < 4x_{n+1}^2$. Since $x_{n+1} = (n+1)\\bar{x}_{n+1} - n\\bar{x}_n$, this is equivalent to $2n(2n+1)\\bar{x}_n^2 - 8n(n+1)\\bar{x}_n\\bar{x}_{n+1} + (4n^2+6n+1)\\bar{x}_{n+1}^2 > 0$. The left-hand member is a quadratic form in $\\bar{x}_n$ and $\\bar{x}_{n+1}$ whose discriminant is $-2n$ and the inequality follows.\n\nTo show $c \\ge 4$, we prove that\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} \\frac{1}{\\sqrt{j}} \\right)^2 > 4 \\sum_{k=1}^{n} \\frac{1}{k} - 24.\n$$\nDivergence of the harmonic series settles the case. Write $1/\\sqrt{j} > 2(\\sqrt{j+1} - \\sqrt{j})$, to obtain\n$$\n\\left( \\frac{1}{k} \\sum_{j=1}^{k} \\frac{1}{\\sqrt{j}} \\right)^2 > \\frac{4}{k^2} \\left( \\sqrt{k+1} - 1 \\right)^2 > \\frac{4}{k} \\left( 1 - \\frac{2}{\\sqrt{k}} \\right) = \\frac{4}{k} - \\frac{8}{k\\sqrt{k}}.\n$$\nFinally, notice that $1/(2k\\sqrt{k}) < 1/\\sqrt{k-1} - 1/\\sqrt{k}$, $k \\ge 2$, to get\n$$\n\\sum_{k=1}^{n} \\frac{1}{k\\sqrt{k}} \\le 3 - \\frac{2}{\\sqrt{n}} < 3,\n$$\nand deduce thereby the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $A_n$, respectively $B_n$, be the set of non-negative integers $k < n$ such that the number of distinct prime factors of $\\gcd(k, n)$ is even, respectively odd. Show that $|A_n| = |B_n|$ if $n$ is even, and $|A_n| > |B_n|$ if $n$ is odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\gcd(k, n)$ depends only upon the residue class of $k$ modulo $n$, $|A_n| - |B_n| = \\sum_k (-1)^{s(k,n)}$, where $s(k, n)$ is the number of distinct prime factors of $\\gcd(k, n)$, and $k$ ranges over any complete residue system modulo $n$.\nWe shall prove that the above sum equals $n \\prod_{p|n} (1 - 2/p)$, $p$ prime, whence the conclusion; the latter is precisely the number of positive integers $k < n$ such that $k$ and $k+1$ are both coprime to $n$.\nIn the above notation, let $e(k, n) = (-1)^{s(k,n)}$ and let $f(n) = \\sum_k e(k, n)$. We show that $f$ is a numerical multiplicative function — that is, if $n_1$ and $n_2$ are coprime positive integers, then $f(n_1n_2) = f(n_1)f(n_2)$.\nIf $n_1, n_2$ are coprime positive integers, then $e(k, n_1n_2) = e(k, n_1)e(k, n_2)$. Further, if $k_i$ ranges once over a complete residue system modulo $n_i$, $i = 1, 2$, then $k = k_1n_2 + k_2n_1$ ranges once over a complete residue system modulo $n_1n_2$, and $e(k, n_i) = e(k_i, n_i)$, $i = 1, 2$. Hence $e(k, n_1n_2) = e(k, n_1)e(k, n_2) = e(k_1, n_1)e(k_2, n_2)$, and\n$$\nf(n_1n_2) = \\sum_k e(k, n_1n_2) = \\sum_{k_1} \\sum_{k_2} e(k_1, n_1)e(k_2, n_2) = f(n_1)f(n_2).\n$$\nFinally, if $p$ is a prime, and $m$ is a positive integer, then $f(p^m)$ equals the number of $k$'s coprime to $p$, which is $p^m - p^{m-1}$, minus the number of $k$'s divisible by $p$, which is $p^{m-1}$, so $f(p^m) = p^m(1 - 2/p)$. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38093,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the acute triangle $ABC$, with $AB \\ne BC$, let $T$ denote the midpoint of the side $[AC]$, $A_1$ and $C_1$ denote the feet of the altitudes drawn from $A$ and $C$, respectively. Let $Z$ be the point of intersection of the tangents in $A$ and $C$ to the circumcircle of triangle $ABC$, $X$ be the point of intersection of lines $ZA$ and $A_1C_1$ and $Y$ be the point of intersection of lines $ZC$ and $A_1C_1$.\n\na) Prove that $T$ is the incircle of triangle $XYZ$.\n\nb) The circumcircles of triangles $ABC$ and $A_1BC_1$ meet again at $D$. Prove that the orthocenter $H$ of triangle $ABC$ is on the line $TD$.\n\nc) Prove that the point $D$ lies on the circumcircle of triangle $XYZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. From $AZ = ZC$, it follows immediately that $[ZT]$ is the angle bisector of $\\angle AZC$. Notice that $\\angle XAB = \\angle ACB = \\angle BC_1A_1 = \\angle AC_1X$ and\n\n![](attached_image_1.png)\n\nb. We may assume that $AB < BC$; in this case, $D$ is on the minor arc $AB$. Let $O$ denote the circumcenter of triangle $ABC$ and let $L$ be the midpoint of the segment line $[BH]$. The radical axis of two circles being perpendicular to the line connecting the centers of the circles, it follows that $BD \\perp LO$. As $BD \\perp DH$, we get that $LO \\parallel DH$. $OLHT$ is a parallelogram, therefore $OL \\parallel HT$, and now it is clear that points $D, H, T$ are collinear.\n\n$$\n\\angle ADT = \\angle BDA - \\angle BDH = 180^\\circ - \\angle ACB - 90^\\circ = 90^\\circ - \\angle XAB = \\angle AXT.\n$$\n\nIt is easy to prove (in a similar manner to a)) that $TY$ is the perpendicular bisector of the segment line $[A_1C]$. From\n$$\n\\angle CDT = \\angle HDB - \\angle CDB = 90^\\circ - \\angle CAB = 90^\\circ - \\angle BCY = \\angle CYT,\n$$\nit follows that the quadrilateral $CTDY$ is cyclic. This leads to\n$$\n\\begin{align*} \n\\angle XDY + \\angle XZY &= \\angle XDT + \\angle TDY + \\angle XZY = \\\\\n&= 180^\\circ - \\angle XAT + 180^\\circ - \\angle TCY + \\angle XZY = \\\\\n&= \\angle ZAT + \\angle ZCT + \\angle XZY = 180^\\circ,\n\\end{align*}\n$$\n\nwhich means that the quadrilateral $DXZY$ is also cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle and let $O$ be its circumcenter. The tangents of the circumcircle $ABC$ at vertices $B$ and $C$ meet at $P$, the circle of radius $PB$ centered at $P$ meets the internal angle bisector of the angle $BAC$ at point $Q$ lying in the interior of the triangle $ABC$, and the lines $OQ$ and $BC$ meet at $D$. Finally, let $E$ and $F$ be the orthogonal projections of $Q$ on the lines $AC$ and $AB$, respectively. Prove that the lines $AD, BE$ and $CF$ are concurrent.\nCosmin Pohoăţă",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line $AB$ and the circle of radius $PB$ centered at $P$ meet again at some point $R$. Standard angle-chasing shows that the angle $BRC$ is the complement of the angle $BAC$. Hence the lines $AC$ and $CR$ are perpendicular, so the lines $EQ$ and $CR$ are parallel, and the angles $CQE$ and $QCR$ are equal.\n\n![](attached_image_1.png)\n\nOn the other hand, the line $OQD$ is the $Q$-symmedian of the triangle $BCQ$, so $BQ^2/CQ^2 = BD/CD$. Consequently,\n$$\n1 = \\frac{BD}{CD} \\cdot \\frac{CE}{BF} = \\frac{BD}{CD} \\cdot \\frac{CE}{AE} \\cdot \\frac{AF}{BF},\n$$\non account of $AQ$ being the internal angle bisector of the angle $BAC$, so $AE = AF$. The conclusion follows by Ceva's Theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38095,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a composite positive integer $n$ nice if it is possible to arrange its factors that are larger than $1$ on a circle such that two neighboring numbers are not coprime. How many of the elements of the set $\\{1, 2, 3, \\ldots, 100\\}$ are nice?",
    "options": [],
    "answer": "44",
    "solution": "If $n = pq$, where $p, q$ are distinct primes, it is clear that we can not arrange $p, q$ and $pq$ without $p$ and $q$ being neighbors, therefore $pq$ is not nice.\n\nIf $n$ is not a product of two distinct primes, then $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, where $k \\ge 1$ and $\\alpha_i \\ge 1$. A convenient way of arranging the factors of $n$, larger than $1$, on a circle is the following: we write the numbers in a succession of the form:\n\n$n, S_1, S_2, \\dots, S_{k-1}, S_k$\n\nwhere:\n\n* $S_1$ is a sequence of numbers that contains all the factors of $n$, other than $n$, that are multiples of $p_1$, the last one in the sequence being $p_1p_2$;\n* $S_2$ is a sequence of numbers that contains all the factors of $n$ that are multiples of $p_2$, but not of $p_1$, the last number in the sequence being $p_2p_3$;\n* $S_3$ is a sequence of numbers that contains all the factors of $n$ that are multiples of $p_3$, but are multiples of neither $p_1$ nor $p_2$, the last number in the sequence being $p_3p_4$;\n... and so on.\n\nThe set $\\{1, 2, \\ldots, 100\\}$ contains $74$ composite numbers, $30$ of which being of the form $pq$, with $p, q$ distinct primes. This leaves $44$ nice numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an odd prime $p$, determine all polynomials $f$ and $g$ with integral coefficients satisfying the condition $f(g(X)) = \\sum_{k=0}^{p-1} X^k$.",
    "options": [],
    "answer": "All integer‑coefficient solutions are exactly the two families with consistent signs:\n1) f(X) = ± X ∓ a and g(X) = ± Φ_p(X) + a, where a is any integer;\n2) f(X) = Φ_p(± X ∓ b) and g(X) = ± X + b, where b is any integer.\nHere Φ_p(X) = 1 + X + X^2 + … + X^{p−1} is the p‑th cyclotomic polynomial, and the ± signs in each pair are matched as indicated.",
    "solution": "More generally, let $k$ be an integer greater than $1$ and let $P$ be a polynomial of degree at most $k-2$ with integral coefficients. The polynomials $f$ and $g$ with integral coefficients satisfying the condition $f(g(X)) = X^k + X^{k-1} + P(X)$ are: $f(X) = (\\pm X \\mp a)^k + (\\pm X \\mp a)^{k-1} + P(\\pm X \\mp a)$ and $g(X) = \\pm X + a$, where $a$ is an integer, and $f(X) = \\pm X + b$ and $g(X) = \\pm X^k \\pm X^{k-1} \\pm P(X) \\mp b$, where $b$ is an integer — in both cases, the signs correspond to one another; for instance, $f(X) = (-X + a)^k + (-X + a)^{k-1} + P(-X + a)$ and $g(X) = -X + a$ are admissible, but $f(X) = (-X + a)^k + (-X + a)^{k-1} + P(-X + a)$ and $g(X) = X + a$ are not.\n\nClearly, $\\deg f \\ge 1$, $\\deg g \\ge 1$ and the leading coefficients of $f$ and $g$ are $\\pm 1$.\n\nIf $\\deg g = 1$, we immediately obtain the first pair of polynomials above.\nIf $\\deg g > 1$, write $f(X) = \\sum_{i=0}^{m} a_i X^i$ and $g(X) = \\sum_{i=0}^{n} b_i X^i$, and notice that $f(g(X)) = a_m b_n^m X^{mn} + m a_m b_n^{m-1} b_{n-1} X^{mn-1} + \\dots$. Identification of coefficients yields $a_m b_n^m = 1$ and $m a_m b_n^{m-1} b_{n-1} = 1$. The latter forces $m = 1$ which leads to the second pair of polynomials above.\nThe polynomial $\\Phi_p(X) = \\sum_{k=0}^{p-1} X^k$ is the $p$-th cyclotomic polynomial. With the sign convention in the previous solution, the required polynomials are either $f(X) = \\pm X \\mp a$ and $g(X) = \\pm \\Phi_p(X) + a$, where $a$ is an integer, or $f(X) = \\Phi_p(\\pm X \\mp b)$ and $g(X) = \\pm X + b$, where $b$ is an integer.\n\nThe proof relies upon the well known fact that both $\\Phi_p$ and its (formal) derivative, $\\Phi'_p(X) = \\sum_{k=0}^{p-2} (k+1)X^k$, are irreducible in $\\mathbb{Z}[X]$. This follows from Eisenstein's irreducibility criterion upon substitution $X \\mapsto X+1$: $\\Phi_p(X+1) = \\sum_{k=0}^{p-1} \\binom{p}{k+1} X^k$ and $\\Phi'_p(X+1) = \\sum_{k=0}^{p-2} (k+1) \\binom{p}{k+2} X^k$.\n\nNow take the derivative both sides of the relation in the statement to obtain $f'(g(x))g'(X) = \\Phi'_p(X)$. Since $\\Phi'_p(X)$ is irreducible in $\\mathbb{Z}[X]$ and its coefficients are obviously jointly coprime, either $f'(g(x)) = \\pm 1$ and $g'(X) = \\pm \\Phi'_p(X)$ or vice versa — as before, the signs correspond to one another. With the same convention for signs, in the former case, $g(X) = \\pm \\Phi_p(X) + a$ and $f(X) = \\pm X \\mp a$, where $a$ is an integer; in the latter, $g(X) = \\pm X + b$ and $f(X) = \\Phi_p(\\pm X \\mp b)$, where $b$ is an integer.\nWe show that if $\\deg g > 1$, then $\\deg g = p-1$. The remaining details are easily filled in and hence omitted.\n\nTo begin, let $\\zeta = \\cos(2\\pi/p) + i \\sin(2\\pi/p)$, let $\\alpha$ be a (complex) root of $f$, and let $\\beta$ be a (complex) root of $g - \\alpha$. Since $\\Phi_p(\\beta) = f(g(\\beta)) = f(\\alpha) = 0$, it follows that $\\beta$ is one of the $\\zeta^j$, $j = 1, \\dots, p-1$. In particular, $g - \\alpha$ has no multiple roots, for $\\Phi_p$ has no such.\n\nNow let $n = \\deg g > 1$, and let $\\zeta^{k_1}, \\dots, \\zeta^{k_n}$ be the roots of $g - \\alpha$, where $1 \\le k_1 < \\dots < k_n \\le p-1$. Since $g$ has integral coefficients, $a = \\sum_{j=1}^n \\zeta^{k_j}$ is integral, by the first Vieta relation, so $h = \\sum_{j=1}^n X^{k_j} - a$ is a monic non-constant polynomial with integral coefficients. Since $\\Phi_p$ is the minimal polynomial of $\\zeta$ and $h(\\zeta) = 0$, it follows that $\\Phi_p$ divides $h$, so $\\deg h \\ge \\deg \\Phi_p = p-1$. On the other hand, $\\deg h = k_n \\le p-1$, so $\\deg h = p-1$, and since $h$ is monic, it ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_5$ be real numbers. Find the least positive integer $n$ with the following property: if there exist $n$ distinct sums of the form $x_p + x_q + x_r$ (with $1 \\le p < q < r \\le 5$) which are equal to $0$, then $x_1 = x_2 = \\dots = x_5 = 0$.\nBulgaria, 2003",
    "options": [],
    "answer": "7",
    "solution": "If the numbers are $1, 1, 1, 1, -2$, (or $1, 1, 1, -2, -2$) there are $6$ sums that are equal to $0$ without the numbers themselves being $0$. Therefore, knowing $6$ sums (or less) to be $0$ is not enough to conclude that the numbers are all $0$.\n\nNow we prove that $7$ is enough. Suppose we are given that $7$ sums are equal to $0$. These sums contain, in total, $21$ terms, while there are only $5$ numbers. By the Pigeonhole Principle it follows that there is a number that appears (at least) $5$ times. Suppose $x_1$ appears (at least) $5$ times. $x_1$ is involved in $6$ sums, so there is only one sum, say $x_1 + x_4 + x_5$, that is not necessarily $0$. All the other ones are $0$: $x_1 + x_2 + x_3 = x_1 + x_2 + x_4 = x_1 + x_2 + x_5 = x_1 + x_3 + x_4 = x_1 + x_3 + x_5 = 0$. The first three equations simplify to $x_3 = x_4 = x_5$, and then comparing the first and the last equation gives $x_2 = x_3 = x_4 = x_5$.\n\nFinally, $x_1$ only participates in at most $6$ sums; the seventh one gives $x_2 = x_3 = x_4 = x_5 = 0$, and it follows immediately that $x_1 = x_2 = x_3 = x_4 = x_5 = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 5$ be an integer. Prove that $n$ is prime if and only if for any representation of $n$ as a sum of four positive integers $n = a+b+c+d$, it is true that $ab \\ne cd$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The statement of the problem is equivalent to saying that an integer $n \\ge 5$ is composite if and only if there exists a representation of $n$ as the sum of four positive integers such that $ab = cd$.\n\n($\\Leftrightarrow$) If $ab = cd$ and $n = a+b+c+d$, then\n$$\nan = a^2 + ab + ac + ad = a^2 + cd + ac + ad = (a+c)(a+d), (*)\n$$\nhence $n \\mid (a+c)(a+d)$.\n\n($\\Rightarrow$) If $n = pq$, with $p, q \\ge 2$, is a composite number, we can choose $a = 1$, $b = (p-1)(q-1)$, $c = p-1$, $d = q-1$. This choice can easily be guessed by writing $(*)$ with $a = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, let $S_n$ be the set of all permutations of the set $\\{1, 2, \\dots, n\\}$, and, for each $\\sigma$ in $S_n$, let $I(\\sigma) = \\{i: \\sigma(i) \\le i\\}$. Evaluate the sum\n$$\n\\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)).\n$$",
    "options": [],
    "answer": "(n+1)!",
    "solution": "Consider the involution of $S_n$ which sends a permutation $\\sigma$ to the permutation $\\sigma^*$ defined by $\\sigma^*(i) = j$ if and only if $\\sigma(n-j+1) = n-i+1$. Notice that, for each $\\sigma$ in $S_n$, the assignment $i \\mapsto n-\\sigma(i)+1$ defines a bijection from $I(\\sigma)$ to $I(\\sigma^*)$ whose inverse sends $j$ to $n-\\sigma^*(j)+1$. Consequently,\n$$\n\\begin{align*}\n\\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)) &= \\\\\n&= \\frac{1}{2} \\sum_{\\sigma \\in S_n} \\left( \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)) + \\frac{1}{|I(\\sigma^*)|} \\sum_{i \\in I(\\sigma^*)} (i + \\sigma^*(i)) \\right) \\\\\n&= \\frac{1}{2} \\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i) + (n - \\sigma(i) + 1) + (n - i + 1)) \\\\\n&= (n + 1)! .\n\\end{align*}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38100,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a circle, consider two chords $[AB]$, $[CD]$ that intersect at $E$. The lines $AC$ and $BD$ meet at $F$. Let $G$ be the projection of $E$ onto $AC$. We denote by $M, N, K$ the midpoints of the segment lines $[EF]$, $[EA]$, and $[AD]$, respectively. Prove that the points $M, N, K, G$ are concyclic.\n\nMarius Bocanu\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the midpoint of $[AF]$; points $N, G, P, M$ are on the Euler circle of triangle $AEF$, which means they are concyclic $(*)$.\n\nAs $NK \\parallel DE$ and $NM \\parallel AF$, we have:\n$$\n\\begin{align*}\n\\angle KNM &= \\angle KNE + \\angle ENM = \\angle DEB + \\angle BAC \\\\\n&= \\angle DEB + \\angle BDE = \\angle ABF,\n\\end{align*}\n$$\nand, as $KP \\parallel DF$ and $MP \\parallel AE$, we have\n$$\n\\angle KPM = 180^\\circ - \\angle KPA - \\angle MPF = 180^\\circ - \\angle DFA - \\angle EAF = \\angle ABF.\n$$\nIt follows that the quadrilateral $KNPM$ is cyclic, hence $K, N, P, M$ are concyclic. Combining this with $(*)$ gives the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_0A_1A_2$ be a triangle and let $O$ be its circumcenter. The lines $OA_k$ and $A_{k+1}A_{k+2}$ meet at $B_k$, and the tangent of the circumcircle $A_0A_1A_2$ at $A_k$ meets the line $B_{k+1}B_{k+2}$ at $C_k$, $k = 0, 1, 2$, indices being reduced modulo $3$. Show that the points $C_0, C_1, C_2$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nTo this end, let the parallel through $O$ to the tangent $A'_{k+1}A'_{k+2}$ meet the tangents $A'_k A'_{k+1}$ and $A'_k A'_{k+2}$ at $X_k$ and $Y_k$, respectively, and notice that:\n\n(1) the angles $OX_k B_k$ and $OA_{k+2} B_k$ are equal, since the points $A_{k+2}$, $B_k$, $O$, $X_k$ all lie on the circle on diameter $OX_k$;\n\n(2) the angles $OA_{k+2}B_k$ and $OA_{k+1}A_{k+2}$ are equal, since $OA_{k+1} = OA_{k+2}$; and\n\n(3) the angles $OA_{k+1}A_{k+2}$ and $OY_k B_k$ are equal, since the points $A_{k+1}$, $B_k$, $O$, $Y_k$ all lie on the circle on diameter $OY_k$.\n\nConsequently, the angles $OX_k B_k$ and $OY_k B_k$ are equal. Since the lines $X_k Y_k$ and $A'_{k+1} A'_{k+2}$ are parallel, and the latter is perpendicular to the line $OA_k = OB_k$, it follows that $B_k$ is the midpoint of the segment $X_k Y_k$, whence the conclusion.\nAlternative Solution:\n\nWe will show that $\\prod_{k=0}^2 (C_k B_{k+1} / C_k B_{k+2}) = 1$, so the conclusion follows by Menelaus' theorem. To this end, we will prove that\n$$\n\\frac{C_k B_{k+1}}{C_k B_{k+2}} = \\frac{A_k A_{k+1}}{A_k A_{k+2}} \\cdot \\frac{A_k B_{k+1}}{A_k B_{k+2}}. \\quad (*)\n$$\nMultiplying the three yields the desired relation, since, by Ceva's theorem, $\\prod_{k=0}^2 (A_k B_{k+1} / A_k B_{k+2}) = 1$.\n\nWe now turn to prove $(*)$. To avoid directed angles, assume, without any loss, that the triangle $A_0 A_1 A_2$ is acute-angled, fix an index $k$ and let $A_k A_{k+1} \\leq A_k A_{k+2}$ (the case $A_k A_{k+2} \\leq A_k A_{k+1}$ is dealt with similarly). Apply the law of sines in the triangles $A_k B_{k+1} C_k$ and $A_k B_{k+2} C_k$ to get\n$$\n\\frac{C_k B_{k+1}}{C_k B_{k+2}} = \\frac{A_k B_{k+1}}{A_k B_{k+2}} \\cdot \\frac{\\sin(\\angle C_k A_k B_{k+1})}{\\sin(\\angle C_k A_k B_{k+2})}.\n$$\nSince the angle $C_k A_k B_{k+1}$ is the supplement of the internal angle at $A_{k+1}$ of the triangle $A_0 A_1 A_2$, and the angle $C_k A_k B_{k+2}$ is equal to the internal angle at $A_{k+2}$ of the triangle $A_0 A_1 A_2$, the above ratio of sines equals $A_k A_{k+1} / A_k A_{k+2}$ and $(*)$ follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer and let $A$, respectively $B$, be an alphabet with $m$, respectively $2m$ letters. Let $n$ be an even integer greater than or equal to $2m$. Let $a_n$ be the number of words of length $n$ made of letters from $A$ such that every letter in $A$ occurs a positive even number of times. Let $b_n$ be the number of words of length $n$ made of letters from $B$ such that every letter in $B$ occurs an odd number of times. Determine the ratio $b_n/a_n$.",
    "options": [],
    "answer": "2^(n - m)",
    "solution": "Deletion of all bars in a word in $B^n$ produces a word in $A^n$. Now let $\\alpha$ be a word in $A^n$ and let $a_i$ occur $k_i$ times in $\\alpha$; the $k_i$ are positive even integers which add up to $n$. Since there are exactly $2^{k_i-1}$ distinct ways to bar $a_i$ an odd number of times in $\\alpha$, the preimage of $\\alpha$ under deletion of all bars has exactly $2^{k_1-1} \\cdots 2^{k_m-1} = 2^{n-m}$ elements. The conclusion follows.\n\nAlternative Solution:\nThe number $a_n$ is the coefficient of $x^n/n!$ in the formal expansion\n$$\nA(x) = \\left( \\sum_{k \\ge 1} \\frac{x^{2k}}{(2k)!} \\right)^m = \\left( \\frac{(e^x + e^{-x})}{2} - 1 \\right)^m = 2^{-m} \\left( e^{x/2} - e^{-x/2} \\right)^{2m}.\n$$\n\nSimilarly, $b_n$ is the coefficient of $x^n/n!$ in the formal expansion\n$$\nB(x) = \\left( \\sum_{k \\ge 1} \\frac{x^{2k-1}}{(2k-1)!} \\right)^{2m} = 2^{-2m} (e^x - e^{-x})^{2m}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $x_1, x_2, \\dots, x_n > 0$ be real numbers so that\n$$\nx_1 + x_2 + \\dots + x_n = \\frac{1}{x_1^2} + \\frac{1}{x_2^2} + \\dots + \\frac{1}{x_n^2}.\n$$\n\nShow that for each positive integer $k \\le n$, there are $k$ numbers among $x_1, x_2, \\dots, x_n$ whose sum is at least $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Arguing by contradiction, suppose that every sum of $k$ numbers from $x_1, x_2, \\dots, x_n$ is strictly less than $k$. Then the numbers\n$$\na_j = x_j + x_{j+1} + \\dots + x_{j+k-1}, \\quad j = 1, 2, \\dots, n\n$$\nare also less than $k$ (where the indices from the sums $a_j$ are considered to be taken modulo $n$).\nAdding up yields $a_1 + a_2 + \\dots + a_n < nk$ and, since\n$$\na_1 + a_2 + \\dots + a_n = k(x_1 + x_2 + \\dots + x_n),\n$$\nwe get\n$$\nx_1 + x_2 + \\dots + x_n < n.\n$$\n\nBut $n^2 \\le \\left(\\sum_{k=1}^n x_k\\right) \\left(\\sum_{k=1}^n \\frac{1}{x_k}\\right) < n \\sum_{k=1}^n \\frac{1}{x_k}$, so $\\sum_{k=1}^n \\frac{1}{x_k} > n$. It follows that\n$$\n\\sum_{k=1}^{n} x_k = \\sum_{k=1}^{n} \\frac{1}{x_k^2} \\ge \\frac{1}{n} \\left( \\sum_{k=1}^{n} \\frac{1}{x_k} \\right)^2 > \\frac{1}{n} \\cdot n^2 = n,\n$$\ncontradiction.\n\n\nAlternative Solution:\nSuppose that $x_1 \\le x_2 \\le \\dots \\le x_n$; it is enough to prove that\n$$\nx_{n-k+1} + \\dots + x_{n-1} + x_n \\ge k, \\text{ for each } k=1, 2, \\dots, n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$ and an increasing real-valued function $f$ on the closed unit interval $[0, 1]$, determine the maximum value the sum\n$$\n\\sum_{k=1}^{n} f\\left(\\left|x_k - \\frac{2k-1}{2n}\\right|\\right)\n$$\nmay achieve subject to $0 \\le x_1 \\le \\dots \\le x_n \\le 1$.",
    "options": [],
    "answer": "\\sum_{k=1}^{n} f\\left(\\frac{2k-1}{2n}\\right)",
    "solution": "Let $a_k = (2k-1)/(2n)$, $k = 1, \\dots, n$. The required maximum is $\\sum_{k=1}^n f(a_k)$ and is achieved, for instance, at $x_1 = \\dots = x_n = 0$ or at $x_1 = \\dots = x_n = 1$.\n\nTo show that $\\sum_{k=1}^n f(a_k)$ is an upper bound for the sum under consideration subject to the given constraint, fix an $n$-tuple $(x_1, \\dots, x_n)$ such that $0 \\le x_1 \\le \\dots \\le x_n \\le 1$, write $[n] = \\{1, \\dots, n\\}$, define an increasing function $\\varphi: [n] \\to [n]$ by $\\varphi(k) = \\max \\{j: a_j - 1/(2n) \\le x_k\\}$, and notice that\n$$\n|x_k - a_k| \\le |x_k - a_{\\varphi(k)}| + |a_{\\varphi(k)} - a_k| \\le 1/(2n) + |\\varphi(k) - k|/n = a_{|\\varphi(k)-k|+1},\n$$\nfor $k = 1, \\dots, n$.\n\nWe shall prove that there exists a permutation $\\sigma$ of $[n]$ such that $|\\varphi(k) - k| + 1 \\le \\sigma(k)$ for all $k$, so $a_{|\\varphi(k)-k|+1} \\le a_{\\sigma(k)}$ and the conclusion follows:\n$$\n\\sum_{k=1}^{n} f(|x_k - a_k|) \\le \\sum_{k=1}^{n} f(a_{|\\varphi(k)-k|+1}) \\le \\sum_{k=1}^{n} f(a_{\\sigma(k)}) = \\sum_{k=1}^{n} f(a_k).\n$$\n\nWe now show by induction on $n$ that, for any increasing function $\\psi: [n] \\to [n]$, there exists a permutation $\\sigma$ of $[n]$ such that $|\\psi(k) - k| + 1 \\le \\sigma(k)$ for all $k$.\n\nThe base case, $n=1$, is clear. For the induction step, let $n > 1$ and distinguish two cases.\n\nIf $\\psi(n) < n$, then the restriction of $\\psi$ to $[n-1]$ is an increasing function of $[n-1]$ into itself, so $|\\psi(k)-k|+1 \\le \\sigma(k)$, $k=1, \\dots, n-1$, for some permutation $\\sigma$ of $[n-1]$. Since $|\\psi(n)-n|+1 = n-\\psi(n)+1 \\le n$, the permutation $\\sigma$ extends to a permutation of $[n]$ satisfying the required condition by letting $\\sigma(n) = n$.\n\nIf $\\psi(n) = n$, consider the increasing function $\\psi': [n-1] \\to [n-1]$ defined by $\\psi'(k) = \\psi(k)$ if $\\psi(k) < n$ and $\\psi'(k) = n-1$ if $\\psi(k) = n$. By the induction hypothesis, there exists a permutation $\\pi$ of $[n-1]$ such that $|\\psi'(k) - k| + 1 \\le \\pi(k)$, $k = 1, \\dots, n-1$, so $|\\psi(k) - k| + 1 \\le |\\psi'(k) - k| + 2 \\le \\pi(k) + 1$, $k = 1, \\dots, n-1$. Finally, since $|\\psi(n) - n| + 1 = 1$, setting $\\sigma(k) = \\pi(k) + 1$, $k = 1, \\dots, n-1$, and $\\sigma(n) = 1$ defines the required permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set of positive integers such that $\\lfloor \\sqrt{x} \\rfloor = \\lfloor \\sqrt{y} \\rfloor$ for all $x, y \\in S$. Show that the products $xy$, where $x, y \\in S$, are pairwise distinct.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 \\le x_3 x_4$, then $x_1 + x_2 \\le x_3 + x_4$. Suppose, if possible, that $x_1 + x_2 > x_3 + x_4$. Let $n = \\lfloor \\sqrt{x} \\rfloor$, $x \\in S$, and write $x_k = n^2 + w_k$, where the $w_k$ are non-negative integers less than $2n+1$, to deduce that $w_1 + w_2 - w_3 - w_4 \\ge 1$. The condition $x_1 x_2 \\le x_3 x_4$ yields $(w_1 + w_2 - w_3 - w_4)n^2 \\le w_3 w_4 - w_1 w_2$, so $w_3 > 0$ and\n$$\n\\begin{aligned}\nn^2 &\\le (w_1 + w_2 - w_3 - w_4)n^2 \\le w_3 w_4 - w_1 w_2 < w_3(w_1 + w_2 - w_3) - w_1 w_2 \\\\\n&= (w_1 - w_3)(w_3 - w_2) \\le ((w_1 - w_3) + (w_3 - w_2))^2 / 4 = (w_1 - w_2)^2/4 \\le n^2,\n\\end{aligned}\n$$\nwhich is a contradiction.\n\nThus, if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 = x_3 x_4$, then $x_1 + x_2 = x_3 + x_4$, so $x_1^2 + x_3 x_4 = x_1(x_1 + x_2) = x_1(x_3 + x_4)$, i.e., $(x_1 - x_3)(x_1 - x_4) = 0$ whence $x_1 = x_3$ or $x_1 = x_4$. The conclusion now follows at once.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\varepsilon$ be a positive real number. A positive integer will be called $\\varepsilon$-squarish if it is the product of two integers $a$ and $b$ such that $1 < a < b < (1 + \\varepsilon)a$. Prove that there are infinitely many occurrences of six consecutive $\\varepsilon$-squarish integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $N$ is a large enough positive integer, then $N^2 - 1 = (N - 1)(N + 1)$ and $N^2 - 4 = (N - 2)(N + 2)$ are both $\\varepsilon$-squarish. Next, if $k$ is a large enough positive integer and $N = (k-1)(k+2) = k^2 + k - 2$, then $N^2 = (k-1)^2(k+2)^2$, $N^2 - 2 = (k^2 - 2)(k^2 + 2k - 1)$ and $N^2 - 5 = (k^2 - k - 1)(k^2 + 3k + 1)$ are all three $\\varepsilon$-squarish. Finally, if $n$ is a large enough positive integer and $N = 2n^2 - 2$, then $N^2 - 3 = (2n^2 - 2n - 1)(2n^2 + 2n - 1)$ is $\\varepsilon$-squarish.\n\nConsequently, $N^2 - 5$, $N^2 - 4$, $\\ldots$, $N^2$ are six consecutive $\\varepsilon$-squarish integers, provided that $N = k^2 + k - 2 = 2n^2 - 2$, where $k$ and $n$ are sufficiently large integers. To conclude, write $m = 2k + 1$ to turn the condition into a Pell equation, $m^2 - 8n^2 = 1$, which has arbitrarily large solutions,\n$$\n\\begin{pmatrix} m_r \\\\ n_r \\end{pmatrix} = \\begin{pmatrix} 3 & 8 \\\\ 1 & 3 \\end{pmatrix}^r \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}, \\quad r \\in \\mathbb{N}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38107,
    "subject": "Mathematics (Multi-modal)",
    "question": "A succession of letters $\\overline{l_1l_2\\dots l_n}$, where $n \\ge 1$, is called a *word*. A word $\\overline{l_1l_2\\dots l_n}$ is called a *palindrome*, if $l_k = l_{n-k+1}$, for each $k \\in \\{1, 2, \\dots, n\\}$.\nConsider a two-letter word $X = \\overline{l_1l_2\\dots l_{2014}}$, with letters $A$ and/or $B$. Prove that $X$ can be obtained by writing one next to another at most 806 palindromes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us split $X$ into groups of 5 consecutive letters; this way we obtain 402 groups of 5 letters and another incomplete group of 4 letters.\n\nConsider a 5 letter-word $Y$, written only with $A$'s and $B$'s, whose first letter is $A$. So $Y$ is one of the sixteen words of the form $A * * * *$. It is easy to check that all of these words either are palindromes or can be formed by joining two palindromes. Switching $A$ with $B$, a similar observation stands if a 5 letter-word $Y$ starts with $B$.\n\nHence, we can concatenate at most $2 \\cdot 402 = 804$ palindromes to cover the first 2010 letters of $X$. For the last 4 letters we need at most two palindromes, so 806 is the maximum number of palindromes we need.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38108,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nx + \\log_2 \\left( 1 + \\sqrt{\\frac{5^x}{3^x + 4^x}} \\right) = 4 + \\log_{1/2} \\left( 1 + \\sqrt{\\frac{25^x}{7^x + 24^x}} \\right).\n$$",
    "options": [],
    "answer": "2",
    "solution": "Rewrite the equation as\n$$\nx + \\log_2 \\left( 1 + \\sqrt{\\frac{1}{(3/5)^x + (4/5)^x}} \\right) = 4 + \\log_{1/2} \\left( 1 + \\sqrt{\\frac{1}{(7/25)^x + (24/25)^x}} \\right),\n$$\n\nand observe that the left hand side is an increasing function, while the right hand side is a decreasing one. We conclude that the equation has at most a solution, and it is not difficult to guess it: $x = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the set of palindromic numbers of the form $5n + 4$, where $n \\ge 0$ is an integer. (A positive integer is a *palindromic number* if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.)\na) If we write the elements of $M$ in increasing order, which is the 50th number?\nb) Among all numbers in $M$, written with nonzero digits who sum up to 2014, which is the greatest one and which the smallest one?",
    "options": [],
    "answer": "a) 40504. b) Greatest: the palindrome starting and ending with 4 and having 2006 ones in between (i.e., 4 followed by 2006 ones and then 4). Smallest: the palindrome starting with 98, followed by 222 nines, and ending with 89 (i.e., 98 then 222 copies of 9 then 89).",
    "solution": "a) The last (and hence, the first) digit of a number from $M$ equals 4 or 9. A direct count shows that $M$ contains 2 one digit numbers, 2 two digit numbers, 20 three digit and 20 four digit numbers, hence the 50th number is the 6th five digit number, that is, 40504.\n\nb) The greatest number in $M$ has the maximum number of digits. Therefore, we put 4 as the first and last digit and complete the decimal representation\n\nwith 2006 digits 1, obtaining thus $\\underbrace{411\\ldots14}_{2006\\ \\text{digits}\\ 1}$. Similarly, the smallest number in $M$ has the least number of digits. The answer is $\\underbrace{9899\\ldots989}_{220\\ \\text{digits}\\ 9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, show that there exist $n+1$ pairwise distinct numbers $x_1, x_2, \\dots, x_n, x_{n+1}$ in $\\mathbb{Q} \\setminus \\mathbb{Z}$ such that $\\{x_1^3\\} + \\{x_2^3\\} + \\dots + \\{x_n^3\\} = \\{x_{n+1}^3\\}$, where $\\{x\\}$ is the fractional part of the real number $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that, if $w_1 < w_2 < \\dots < w_{n+1} < w_{n+2}$ are positive integers such that\n$$\nw_1^3 + w_2^3 + \\dots + w_{n+1}^3 = w_{n+2}^3, \\quad (*)\n$$\nthen the numbers $x_k = w_k/w_{n+2}$, $k = 1, 2, \\dots, n$, and $x_{n+1} = -w_{n+1}/w_{n+2}$ meet the required conditions.\nWe now show by induction on $n \\ge 2$ that there exist integers $3 = w_1 < w_2 < \\dots < w_{n+1} < w_{n+2}$ satisfying $(*)$.\nThe equalities $3^3 + 4^3 + 5^3 = 6^3$ and $3^3 + 15^3 + 21^3 + 36^3 = 39^3$ settle the cases $n=2$ and $n=3$, respectively.\nFinally, for the induction step $n \\mapsto n+2$, notice that if $3 < w_2 < \\dots < w_{n+1} < w_{n+2}$ are $n+2$ integers satisfying $(*)$, then $3 < 4 < 5 < 2w_2 < \\dots < 2w_{n+1} < 2w_{n+2}$ are $n+4$ integers satisfying the corresponding condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be real numbers such that:\n$$\n|a - b| \\ge |c|, \\quad |b - c| \\ge |a|, \\quad |c - a| \\ge |b|.\n$$\n\nProve that one of the numbers $a, b, c$ equals the sum of the other two.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Squaring the first inequality gives $(a-b)^2 \\ge c^2$, hence $(a-b+c)(b+c-a) \\ge 0$. Multiplying the latter with the other two similar inequalities implies $(a+b-c)^2(b+c-a)^2(c+a-b)^2 \\le 0$, hence one of $a, b, c$ is the sum of the other two.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a derivable function, with continuous derivative, and let $s_n = \\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)$.\nProve that the sequence $(s_{n+1} - s_n)_{n \\in \\mathbb{N}^*}$ converges to $\\int_0^1 f(x) dx$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the mean value theorem, we obtain\n$$\n\\begin{align*}\ns_{n+1} - s_n &= \\sum_{k=1}^{n+1} f\\left(\\frac{k}{n+1}\\right) - \\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right) \\\\\n&= f(1) - \\sum_{k=1}^{n} \\left( f\\left(\\frac{k}{n}\\right) - f\\left(\\frac{k}{n+1}\\right) \\right) \\\\\n&= f(1) - \\frac{1}{n(n+1)} \\sum_{k=1}^{n} k f'(x_k),\n\\end{align*}\n$$\nfor some $x_k$, $\\frac{k}{n+1} < x_k < \\frac{k}{n}$, $k = 1, 2, \\dots, n$.\nIf $f' \\ge 0$, then $\\frac{x_k f'(x_k)}{n+1} \\le \\frac{k f'(x_k)}{n(n+1)} \\le \\frac{x_k f'(x_k)}{n}$, $k = 1, 2, \\dots, n$, hence\n$$\n\\frac{1}{n+1} \\sum_{k=1}^{n} x_k f'(x_k) \\le \\frac{1}{n(n+1)} \\sum_{k=1}^{n} k f'(x_k) \\le \\frac{1}{n} \\sum_{k=1}^{n} x_k f'(x_k).\n$$\nBecause $0 \\le x_1 \\le 1/n \\le x_2 \\le \\dots \\le x_n \\le n/n$ is a tagged partition of the interval $[0, 1]$, it follows that\n$$\n\\lim_{n \\to \\infty} \\frac{1}{n(n+1)} \\sum_{k=1}^{n} k f'(x_k) = \\int_{0}^{1} x f'(x) dx = x f(x) \\Big|_{0}^{1} - \\int_{0}^{1} f(x) dx,\n$$\nhence the conclusion.\nIf $f'$ takes negative values, replace $f$ with $g : x \\mapsto f(x) + Mx$, where $M = \\sup |f'|$. As above, for\n$$\nt_n = \\sum_{k=1}^{n} g\\left(\\frac{k}{n}\\right)\n$$\nwe have\n$$\n(t_{n+1} - t_n)_n \\rightarrow \\int_0^1 g(x)dx = \\int_0^1 f(x)dx + \\frac{M}{2}\n$$\nand\n$$\nt_{n+1} - t_n = s_{n+1} - s_n + \\frac{M}{2},\n$$\ntherefore the conclusion holds in this case as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38113,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = f(x) + f(2x)$, and $h: \\mathbb{R} \\to \\mathbb{R}$, $h(x) = f(x) + f(4x)$, are continuous functions. Prove that $f$ is also continuous.\n\nb) Give an example of a discontinuous function $f: \\mathbb{R} \\to \\mathbb{R}$, with the following property: there exists an interval $I \\subset \\mathbb{R}$, such that, for any $a$ in $I$, the function $g_a: \\mathbb{R} \\to \\mathbb{R}$, $g_a(x) = f(x) + f(ax)$, is continuous.\n\nDorel Miheț",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nSince $g$ and $h$ are continuous, and\n$$\nf(x) = (g(x) - g(2x) + h(x))/2,\n$$\n$x \\in \\mathbb{R}$, it follows that $f$ is continuous, as well.\n\nb.\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$,\n$$\nf(x) = \\begin{cases} -1, & x < 0, \\\\ 0, & x = 0, \\\\ 1, & x > 0, \\end{cases}\n$$\nis discontinuous at $0$, but $g_a: \\mathbb{R} \\to \\mathbb{R}$, $g_a(x) = f(x) + f(ax) = 0$, is continuous on $\\mathbb{R}$, for all $a < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Construct outside the square $ABCD$ the rhombus $BCMN$, with $\\angle BCM$ an obtuse angle. The straight lines $BM$ and $AN$ meet at $P$. Prove that $DM \\perp CP$ and triangle $DPM$ is right and isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Isosceles triangles $CDM$ and $CBM$ yield $\\overline{CMD} \\equiv \\overline{CDM}$ and $\\overline{CMB} \\equiv \\overline{CBM}$. In triangle $MBD$, the sum of the angles is $180^\\circ = m(\\overline{DMB}) + m(\\overline{BDM}) + m(\\overline{DBM}) = m(\\overline{DMB}) + 45^\\circ + m(\\overline{CDM}) + 45^\\circ + m(\\overline{CBM}) = 90^\\circ + 2m(\\overline{DMB})$, whence $m(\\overline{DMB}) = 45^\\circ$.\n\nSince quadrilateral $ADMN$ is a parallelogram, $m(\\overline{MPN}) = m(\\overline{DMB}) = 45^\\circ$. The point $P$ is on the perpendicular bisector of the segment $[CN]$, so $\\overline{MPN} \\equiv \\overline{MPC}$, whence $m(\\overline{NPC}) = 90^\\circ$, that is $CP \\perp AN$, implying $CP \\perp DM$.\n\n![](attached_image_1.png)\n\nIn the isosceles triangle $CDM$ the straight line $CP$ is the perpendicular from $C$ onto $DM$, so it is the perpendicular bisector of the segment $[DM]$. This shows that triangle $DPM$ is isosceles with vertex $P$. So, from $m(\\overline{DMP}) = 45^\\circ$ follows $m(\\overline{DPM}) = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$, $J$ be intervals and consider $\\varphi : J \\to \\mathbb{R}$ a continuous function which is nonzero on $J$. Let $f$, $g : I \\to J$ be two differentiable functions such that $f' = \\varphi \\circ f$ and $g' = \\varphi \\circ g$. Prove that if there exists $x_0 \\in I$ such that $f(x_0) = g(x_0)$, then $f$ and $g$ coincide.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\varphi$ is non-zero and is continuous, the function $1/\\varphi$ is correctly defined and continuous. Consider an anti-derivative $F : J \\to \\mathbb{R}$. The given relation for $f$ can be then written $(F \\circ f)'(x) = 1$, for all $x \\in I$. Thus there exists $a \\in \\mathbb{R}$ such that $F(f(x)) = x + a$, for all $x \\in I$. In the same manner, there exists $b \\in \\mathbb{R}$ such that $F(g(x)) = x + b$, for all $x \\in I$. We deduce $x_0 + a = F(f(x_0)) = F(g(x_0)) = x_0 + b$, that is $a = b$. So $F(f(x)) = F(g(x))$, for all $x \\in J$. (*)\n\nOn the other side, from $F'(x) \\neq 0$, for all $x \\in J$, $F'$ has a constant sign on $J$, so $F$ is injective. So, (*) implies $f = g$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38116,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all pairs of positive integers $(m, n)$, with $m \\le n$, for which\n$$\np(2m + 1) \\cdot p(2n + 1) = 400.\n$$",
    "options": [],
    "answer": "All pairs (m, n) with m ≤ n are exactly those in the following three families:\n- p(2m+1) = 1 and p(2n+1) = 400, which gives m = 1 and n ∈ {200, 201, ..., 219};\n- p(2m+1) = 4 and p(2n+1) = 100, which gives m ∈ {2, 3} and n ∈ {50, 51, ..., 59};\n- p(2m+1) = 16 and p(2n+1) = 25, which gives m ∈ {8, 9, 10, 11} and n ∈ {12, 13, 14, 15, 16, 17}.\nIn total, there are 64 such pairs.",
    "solution": "a) Since $400 = 1 \\cdot 400 = 4 \\cdot 100 = 16 \\cdot 25$, we analyze three cases.\nIf $p(2m+1) = 1$, $p(2n+1) = 400$, we obtain $1 \\le 2m+1 < 4$ and $400 \\le 2n+1 < 441$, hence $m \\in \\{1, 2\\}$ and $n \\in \\{200, 201, \\dots, 219\\}$, giving 40 pairs $(m, n)$. Similarly, in the second case we obtain 20 pairs, and in the last case, 24 pairs, leading to a total of 84 pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38117,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cube $ABCD'A'B'C'D'$ has edges of length $a$. Take the points $E \\in (AB)$ and $F \\in (BC)$ so that $AE + CF = EF$.\na) Find the measure of the angle of the planes $(D'DE)$ and $(D'DF)$.\nb) Compute the distance from $D'$ to the straight line $EF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "angle = 45 degrees; distance = a*sqrt(2)",
    "solution": "a) Extend the segment $BA$ with $AH \\equiv FC$. Then $\\triangle DAH \\equiv \\triangle DCF$ (SAS), whence $\\overline{HDA} \\equiv \\overline{FDC}$, so $m(\\overline{HDF}) = 90^\\circ$. Then $[DH] \\equiv [DF]$, leads to $\\triangle DHE \\equiv \\triangle DFE$ (SSS) and from here $m(\\overline{FDE}) = m(\\overline{HDE}) = 45^\\circ$. Since $FD \\perp DD'$ and $ED \\perp DD'$, it follows that $m((D'DE), (D'DF)) = m(\\overline{FDE}) = 45^\\circ$.\n\nb) Denote $P$ the orthogonal projection of the point $D$ onto $EF$. The Three Perpendiculars Theorem yields $D'P \\perp EF$, hence $d(D', EF) = D'P$. The congruence $\\Delta DHE \\equiv \\Delta DFE$ gives $DP = AD = a$. Finally, from Pythagoras' Theorem, $D'P = a\\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$ of integers which satisfy the equality\n$$\n\\frac{a+2}{b+1} + \\frac{a+1}{b+2} = 1 + \\frac{6}{a+b+1}.\n$$",
    "options": [],
    "answer": "(a, b) = (-3 - u, u) for all integers u not equal to -2 or -1, and (a, b) = (1, 0).",
    "solution": "Obviously, $b \\ne -2$ and $b \\ne -1$. Adding $2$ to both sides of the equality, we get\n$$\n\\left(\\frac{a+2}{b+1} + 1\\right) + \\left(\\frac{a+1}{b+2} + 1\\right) = 3 + \\frac{6}{a+b+1},\n$$\nhence\n$$\n(a+b+3)\\left(\\frac{1}{b+1} + \\frac{1}{b+2}\\right) = \\frac{3(a+b+3)}{a+b+1}.\n$$\n\n**Case 1.** If $a + b + 3 = 0$, then every pair $(a, b) = (-3-u, u)$, where $u \\in \\mathbb{Z} \\setminus \\{-2, -1\\}$, is a solution.\n\nCase 2. If $a + b + 3 \\neq 0$, then\n$$\n\\frac{1}{b+1} + \\frac{1}{b+2} = \\frac{3}{a+b+1},\n$$\nwhich leads to\n$$\na = \\frac{b^2 + 4b + 3}{2b + 3}.\n$$\nSince $a \\in \\mathbb{Z}$, it results that $2b+3 \\mid b^2+4b+3$. It follows that $2b+3 \\mid 3$, so $b \\in \\{-3, -2, -1, 0\\}$. We get one additional solution, namely $(a,b) = (1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let the points $D \\in (BC)$, $E \\in (AC)$, $F \\in (AB)$ be such that\n$$\n\\frac{DB}{DC} = \\frac{EC}{EA} = \\frac{FA}{FB}.\n$$\n\nThe halflines $(AD, BE$, and $(CF$ intersect the circumcircle of $ABC$ at points $M, N$ and $P$. Prove that the triangles $ABC$ and $MNP$ share the same centroid if and only if the areas of the triangles $BMC, CNA$ and $APB$ are equal.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, observe that $\\overrightarrow{AD} + \\overrightarrow{BE} + \\overrightarrow{CF} = 0$ ($*$). Denoting the areas of the triangles $ABC, BMC, CNA$ and $APB$ by $s, s_a, s_b$, and $s_c$, respectively, we have $\\frac{DM}{AD} = \\frac{s_a}{s}$, hence $\\frac{AM}{AD} = \\frac{s+s_a}{s}$, which implies $\\overrightarrow{AM} = \\frac{s+s_a}{s} \\cdot \\overrightarrow{AD}$, and the analogous equalities. Triangles $ABC$ and $MNP$ share the same centroid if and only if $\\overrightarrow{AM} + \\overrightarrow{BN} + \\overrightarrow{CP} = 0$, hence if and only if\n$$\n\\frac{s+s_a}{s} \\cdot \\overrightarrow{AD} + \\frac{s+s_b}{s} \\cdot \\overrightarrow{BE} + \\frac{s+s_c}{s} \\cdot \\overrightarrow{CF} = 0,\n$$\nor, equivalently,\n$$\ns_a \\cdot \\overrightarrow{AD} + s_b \\cdot \\overrightarrow{BE} + s_c \\cdot \\overrightarrow{CF} = 0.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all differentiable functions $f: \\mathbb{R} \\to \\mathbb{R}$, that satisfy the equality $f \\circ f = f$.",
    "options": [],
    "answer": "All constant functions and the identity function.",
    "solution": "We shall show that only the identity function and the constant functions satisfy the conditions of the problem. It is clear that these functions verify indeed the conditions.\n\nBecause $f$ is continuous, its range $\\{f(x) \\mid x \\in \\mathbb{R}\\}$ is an interval $I \\subseteq \\mathbb{R}$. If $I$ is degenerate at a point then $f$ is constant.\n\nIf $I$ is non-degenerate, let $a = \\inf I < \\sup I = b$, where $a, b \\in \\bar{\\mathbb{R}}$. By the given condition we deduce that the restriction of $f$ to the interval $(a, b)$ is the identity:\n$$\nf(x) = x, \\quad a < x < b. \\tag{1}\n$$\nWe shall show that $a = -\\infty$ and $b = +\\infty$, i.e. $I = \\mathbb{R}$ and $f$ is the identity function. Suppose $a$ is a finite number. By the continuity of $f$ in $a$ and by (1) we get $f(a) = a$, so\n$$\nf'(a) = f'_d(a) = \\lim_{x \\to a,\\ x > a} \\frac{f(x) - f(a)}{x - a} = \\lim_{x \\to a,\\ x > a} \\frac{x - a}{x - a} = 1. \\tag{2}\n$$\nOn the other side $f$ has a minimum at $a$, because\n$$\nf(a) = a = \\inf I = \\inf\\{f(x) \\mid x \\in \\mathbb{R}\\},\n$$\nso, by the Theorem of Fermat, $f'(a) = 0$, in contradiction with (2). We conclude $a = -\\infty$. Analogously $b = +\\infty$.\n\nTherefore, the only differentiable functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying $f \\circ f = f$ are the constant functions and the identity function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38121,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will say that a positive integer $n$ is subject to an *interesting* change if it is multiplied by $2$ and the result is increased by $4$, a *special* change if it is multiplied by $3$ and the result is increased by $9$ and an *awesome* change if it is multiplied by $4$ and the result is increased by $16$.\n\na) Show that there exists a positive integer which after three changes, the first – interesting, the second – special and the third – awesome, becomes $2020$.\n\nb) Find all positive integers with the property that after two changes of different types, selected among the three above, becomes $2014$.",
    "options": [],
    "answer": "a) 80; b) 332",
    "solution": "a) Before the last change the number must be $(2020 - 16) : 4 = 501$; the previous number must be $(501 - 9) : 3 = 164$ and the required number is $(164 - 4) : 2 = 80$.\n\nb) An interesting change produces a multiple of $2$, a special change gives a multiple of $3$ and an awesome change yields a multiple of $4$. Since $2014$ is neither a multiple of $3$, nor a multiple of $3$, the last change must be an interesting one. So, the second number must be $(2014 - 4) : 2 = 1005$. Since this number is odd, the first change must be special, and the initial number is $(1005 - 9) : 3 = 332$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38122,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer $n > 1$ will be called *p-periodic* if $\\frac{1}{n}$ is a repeating decimal fraction, whose shortest period has length $p$ and begins immediately after the decimal point. For instance, $\\frac{1}{9} = 0.111\\dots$ is 1-periodic and $\\frac{1}{11} = 0.090909\\dots$ is 2-periodic.\n\na) Find all p-periodic numbers $n$ so that the first digit of the period of $\\frac{1}{n}$ is not nil.\n\nb) Find the largest 4-periodic prime.",
    "options": [],
    "answer": "a) 3, 7, 9; b) 101",
    "solution": "a) The first digit of the period is at least 1 if and only if $\\frac{1}{n} \\ge \\frac{1}{10}$, that is $n \\le 10$. Since the prime factors of $n$ must be different from 2 and 5, it follows that $n \\in \\{3, 7, 9\\}$; indeed, in this case $\\frac{1}{3} = 0,(3)$, $\\frac{1}{7} = 0,(142857)$ and $\\frac{1}{9} = 0,(1)$.\n\nb) Let $p$ be the largest 4-periodic prime; then $\\frac{1}{p} = \\frac{m}{9999}$, with $1 \\le m \\le 9998$. This yields $pm = 9999 = 3^2 \\cdot 11 \\cdot 101$. So, the candidate for the largest such prime is $p = 101$, which is acceptable, because $\\frac{1}{101} = 0,(0099)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the function $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ which satisfies the properties:\na) $f(1) = 1$,\nb) $f(p) = 1 + f(p-1)$ for any prime number $p$,\nc) $f(p_1p_2\\cdots p_n) = f(p_1) + f(p_2) + \\cdots + f(p_n)$ for any prime numbers, not necessary distinct.\n\nProve that $2^{f(n)} \\le n^3 \\le 3^{f(n)}$, for any natural number $n$, $n \\ge 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $2$ is a prime number, we have $f(2) = 1 + f(1) = 2$, whence $2^2 \\le 2^3 \\le 3^2$. Then $f(3) = 1 + f(2) = 3$, and thus $2^3 < 3^3 = 3^2$. We will prove the inequalities\n$$\n3 \\log_3 n \\le f(n) \\le 3 \\log_2 n\n$$\nby mathematical induction on $n$. Let $n$ be a natural number, $n \\ge 3$, and let $n = p_1p_2\\cdots p_k$ be its prime factorization (not necessarily distinct factors). If $n$ is not a prime number, i.e. $k \\ge 2$, using the induction hypothesis, $f(n) = f(p_1) + f(p_2) + \\cdots + f(p_k) \\ge 3 \\sum_{i=1}^k \\log_3 p_i = 3 \\log_3 n$ and $f(n) \\le 3 \\sum_{i=1}^k \\log_2 p_i = 3 \\log_2 n$.\nIf $n$ is a prime number, i.e. $k=1$, then $n-1$ is an even number and\n$$\nf(n)=1+f(n-1)=1+f\\left(2\\frac{n-1}{2}\\right)=1+f(2)+f\\left(\\frac{n-1}{2}\\right)=3+f\\left(\\frac{n-1}{2}\\right).\n$$\nThen\n$$\nf(n) = 3 + f\\left(\\frac{n-1}{2}\\right) \\le 3 + 3\\log_2 \\frac{n-1}{2} = 3\\log_2(n-1) \\le 3\\log_2 n\n$$\nand\n$$\nf(n) = 3 + f\\left(\\frac{n-1}{2}\\right) \\ge 3 + 3\\log_3 \\frac{n-1}{2} = 3\\log_3 \\frac{3(n-1)}{2} \\ge 3\\log_3 n.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a natural odd number which is not a perfect square. If $m$ and $n$ are strictly positive integers, prove that\n$$\n\\begin{align*}\n\\text{a)} \\quad & \\{m(a + \\sqrt{a})\\} \\neq \\{n(a - \\sqrt{a})\\}, \\\\\n\\text{b)} \\quad & [m(a + \\sqrt{a})] \\neq [n(a - \\sqrt{a})].\n\\end{align*}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nAs $ma$, $na$ are natural numbers, the equality implies $\\{m\\sqrt{a}\\} = \\{-n\\sqrt{a}\\}$.\nTwo numbers have the same fractional part if and only if their difference is an integer, whence $(m+n)\\sqrt{a} \\in \\mathbb{Z}$, which is absurd.\n\nb.\nAgain, let us suppose that there is a natural number $N$, for which there exist two not equal numbers $m$, $n$ which are different from zero, such that $N = [m(a + \\sqrt{a})] = [n(a - \\sqrt{a})]$. Then $N \\leq m(a + \\sqrt{a}) < N + 1$ and $N \\leq n(a - \\sqrt{a}) < N + 1$; moreover, the inequalities are strict, because the terms in the middle are irrational numbers.\nWe rewrite the inequalities as\n$$\n\\frac{N}{a + \\sqrt{a}} < m < \\frac{N + 1}{a + \\sqrt{a}}, \\quad \\frac{N}{a - \\sqrt{a}} < n < \\frac{N + 1}{a - \\sqrt{a}},\n$$\nand thus, by addition, we get $N \\frac{2}{a-1} < m + n < (N+1) \\frac{2}{a-1}$.\nFrom here, $N < \\frac{a-1}{2}(m+n) < N+1$, which is a contradiction, because the term in the middle is natural ($a$ is odd).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of the acute triangle $ABC$. An arbitrary diameter intersects side $[AB]$ in $D$ and side $[AC]$ in $E$. If $F$ is the midpoint of $[BE]$ and $G$ is the midpoint of $[CD]$, show that $\\angle FOG = \\angle BAC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall make use of the following:\n\n**Lemma.** Let $ABC$ be a triangle and $MN$ be a chord of its circumcircle which intersects side $[AB]$ in $D$ and side $[AC]$ in $E$. Then $\\frac{DM}{DN} : \\frac{EM}{EN} = \\frac{BM}{BN} : \\frac{CM}{CN}$.\n\n*Proof of the lemma.* Using the law of sines in triangles $BDM$ and $BDN$, we have\n$$\n\\frac{DM}{\\sin(\\angle ABM)} = \\frac{BM}{\\sin(\\angle BDM)} \\quad \\text{and} \\quad \\frac{DN}{\\sin(\\angle ABN)} = \\frac{BN}{\\sin(\\angle BDN)}.\n$$\nSince $\\angle BDM + \\angle BDN = 180^\\circ$, we have $\\sin(\\angle BDM) = \\sin(\\angle BDN)$, so\n$$\n\\frac{DM}{DN} = \\frac{BM}{BN} \\cdot \\frac{\\sin(\\angle ABM)}{\\sin(\\angle ABN)}.\n$$\nAnalogously, we get $\\frac{EM}{EN} = \\frac{CM}{CN} \\cdot \\frac{\\sin(\\angle ACM)}{\\sin(\\angle ACN)}$. But $\\angle ABM = \\angle ACM$ and $\\angle ABN = \\angle ACN$, so $\\frac{DM}{DN} : \\frac{EM}{EN} = \\frac{BM}{BN} : \\frac{CM}{CN}$. $\\square$\n\nReturning to our problem, let $D'$ and $E'$ be the reflections of $D$ and $E$ about $O$ and $A'$ be the second intersection point of $BE'$ with the circumcircle of $ABC$. Also, let $D''$ be the intersection of lines $A'C$ and $MN$.\n\nFrom the lemma, we have\n$$\n\\frac{DM}{DN} : \\frac{EM}{EN} = \\frac{BM}{BN} : \\frac{CM}{CN} = \\frac{D''M}{D''N} : \\frac{E'M}{E'N}.\n$$\nSince $DM = D'N$, $DN = D'M$, $EM = E'N$ and $EN = E'M$, we conclude that $\\frac{D'M}{D'N} = \\frac{D''M}{D''N}$, hence $D' = D''$.\n\nNow, $OF$ is midsegment in $\\triangle BB'E$, hence $\\angle BOF = \\angle BB'X$; $OG$ is midsegment in $\\triangle CC'D$, hence $\\angle COG = \\angle CC'X$. But clearly $\\angle BB'X + \\angle CC'X = \\angle BAC$ and $\\angle BOC = 2\\angle BAC$, therefore\n$$\n\\angle FOG = \\angle BOC - (\\angle BOF + \\angle COG) = 2\\angle BAC - \\angle BAC = \\angle BAC,\n$$\n\nSince $[OF]$ is a midsegment of the triangle $BEE'$ and $[OG]$ is a midsegment of the triangle $CDD'$, we get that $OF \\parallel BA'$ and $OG \\parallel CA'$, so $\\angle FOG = \\angle BA'C = \\angle BAC$.\nLet $B'$ be the point diametrically opposed to $B$ and $C'$ the point diametrically opposed to $C$; the triangle $ABC$ being acute, $B'$ is on the minor arc $AC$, and $C'$ is on the minor arc $AB$. Let $X$ be an arbitrary point on the minor arc $BC$. Applying Pascal's theorem to the hexagram $ABB'XC'C$ shows that points $O = BB' \\cap CC'$, $D = AB \\cap C'X$, $E = AC \\cap B'X$ are collinear – on Pascal's line, which is the support line of a diameter.\n\nConversely, if a diameter intersects the sides $[AB]$ and $[AC]$ at $D$ and $E$, respectively, then the lines $B'E$ and $C'D$ will meet at a point $X$ situated on the minor arc $BC$ (consider $X$ only as the intersection point of the line $B'E$ with the circle, and apply Pascal's theorem; $D' = AB \\cap C'X$ will be collinear with $O, E$, hence, it will be the intersection of the diameter with the line $AB$, which means that $D'$ is in fact $D$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38126,
    "subject": "Mathematics (Multi-modal)",
    "question": "On each side of an equilateral triangle of side $n \\ge 1$ consider $n-1$ points that divide the sides into $n$ equal segments. Through these points draw parallel lines to the sides of the triangle, obtaining a net of equilateral triangles of side length $1$. On each of the vertices of the small triangles put a coin head up. A *move* consists in flipping over three mutually adjacent coins. Find all values of $n$ for which it is possible to turn all coins tail up after a finite number of moves.",
    "options": [],
    "answer": "All positive integers not divisible by three (i.e., side lengths congruent to 1 or 2 modulo 3).",
    "solution": "Obviously, such turning is possible for $n = 1$. For $n = 2$, flip each of the four $1$-sided equilateral triangles once and all the coins will be tail-up.\n\nWe shall use now induction of step $3$. Assume that $n$ is an admissible value. Flipping the coins of each unit sided triangle of an equilateral triangle of side length $n+3$, the coins from the vertices of the big triangle will turn one time, those along the sides three times and the interior coins will turn six times each. Consequently, all the exterior coins are turned tail up and all the interior coins are heads up. But the interior coins form the net corresponding to an $n$-sided triangle, so the induction works.\n\nIf $3 \\mid n$, then color the coins in red, yellow and blue so that any three adjacent coins have different colors. Also, any three coins in a row will have different colors. In this case the corners will all have the same color, say red. Since there are, in total, $\\frac{(n+1)(n+2)}{2} \\equiv 1 \\pmod{3}$ coins, then there will be exactly one more red coin than yellow or blue ones. Thus, at the beginning, the parity of the number of red heads is different than the parity of the number of yellow heads. Since each move changes the parity of the number of heads of each color, we cannot end up with the parity of red heads equal to that of yellow or blue heads, which would be the case if all coins showed tails. Thus the coins cannot all be inverted, so $n$ is not an admissible value.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a primitive polynomial with integral coefficients (their highest common factor is $1$) such that $f$ is irreducible in $\\mathbb{Q}[X]$, and $f(X^2)$ is reducible in $\\mathbb{Q}[X]$. Show that $f = \\pm(u^2 - Xv^2)$ for some polynomials $u$ and $v$ with integral coefficients.\n\nFor instance, if $a$ and $b$ are coprime integers and $a$ is odd, then $f = a^4 X^2 + 4b^4$ is a primitive polynomial in $\\mathbb{Z}[X]$, irreducible in $\\mathbb{R}[X] \\supset \\mathbb{Q}[X]$, $f(X^2) = a^4 X^4 + 4b^4 = (a^2 X^2 - 2abX + 2b^2)(a^2 X^2 + 2abX + 2b^2)$ is reducible in $\\mathbb{Z}[X] \\subset \\mathbb{Q}[X]$, and $f = (a^2 X + 2b^2)^2 - X \\cdot (2ab)^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Unless otherwise stated, we work in $\\mathbb{Q}[X]$. Since the case $\\deg f = 1$ is easily dealt with, let $\\deg f \\ge 2$ and write $f(X^2) = gh$, where $g$ and $h$ both have a positive degree, and $g$ is irreducible. Next, write $g = a(X^2) + Xb(X^2)$ and $h = c(X^2) + Xd(X^2)$ to infer (from $f(X^2) = gh$ by an obvious argument on the parity of degrees) that\n$$\nad + bc = 0, \\tag{1}\n$$\nso $f = ac + Xbd$, whence\n$$\naf = (a^2 - Xb^2)c. \\tag{2}\n$$\nWe now show that $a$ and $b$ are coprime. Alternatively, but equivalently, $\\delta = \\gcd(a, b)$ is a constant. To this end, write $a = a_1\\delta$ and $b = b_1\\delta$, and refer to the irreducibility of $g$ to deduce that $\\delta(X^2)$ is either associated with $g$, a case to be ruled out in the sequel, or a constant, in which case we are through.\nIn the former case, $a_1(X^2) + Xb_1(X^2)$ is a constant, so $b_1 = 0$, whence $b = 0$ and $g = a(X^2)$, and (1) forces one of $a$ and $d$ to be $0$. The fact that $g$ is not constant rules out the case $a = 0$, so $d = 0$, $f = ac$ and $h = c(X^2)$. Since $f$ is irreducible,\n\none of $a$ and $c$ must be a constant, hence so must be one of $g$ and $h$ — a contradiction, since both have a positive degree. Incidentally, notice that we have just proved that $b \\neq 0$.\n\nNotice further that $a$ and $X$ are also coprime: otherwise, $a(0) = 0$, so $g(0) = 0$, hence $f(0) = 0$, contradicting the fact that $f$ is irreducible and $\\deg f \\geq 2$.\n\nConsequently, $a$ and $a^2 - Xb^2$ are coprime, so $a$ divides $c$ by (2), and $f = (a^2 - Xb^2)c_1$ for some $c_1$ in $\\mathbb{Q}[X]$. Since $f$ is irreducible, one of $a^2 - Xb^2$ and $c_1$ must be a constant. Since $b \\neq 0$ by the remark at the end of the last but one paragraph, $a^2 - Xb^2$ cannot be constant, so $c_1$ is a constant.\n\nFrom now on we work in $\\mathbb{Z}[X]$. By the preceding, $nf = m(u^2 - Xv^2)$ for some integers $m$ and $n$, and some $u$ and $v$ in $\\mathbb{Z}[X]$. Fix a prime integer $p$ and write $m = p^\\mu m_1$, $n = p^\\nu n_1$, $u = p^\\alpha u_1$, $v = p^\\beta v_1$, where $\\alpha, \\beta, \\mu, \\nu$ are non-negative integers, and none of $m_1, n_1, u_1, v_1$ is divisible by $p$. To make a choice, let $\\alpha \\leq \\beta$; the case $\\alpha > \\beta$ is dealt with similarly. Since $f$ is primitive, the relation\n$$\np^\\nu n_1 f = p^{\\mu+2\\alpha} m_1 \\left( u_1^2 - X p^{2(\\beta-\\alpha)} v_1^2 \\right)\n$$\nimplies that $\\nu \\geq \\mu + 2\\alpha$. If we show that $\\nu = \\mu + 2\\alpha$, we are through.\n\nSuppose, if possible, that $\\nu > \\mu + 2\\alpha$, to deduce that $p$ divides $u_1^2 - Xp^{2(\\beta-\\alpha)}v_1^2$, so it also divides $u_1^2(X^2) - X^2p^{2(\\beta-\\alpha)}v_1^2(X^2) = (u_1(X^2) - Xp^{\\beta-\\alpha}v_1(X^2))(u_1(X^2) + Xp^{\\beta-\\alpha}v_1(X^2))$. Since $p$ is prime, it must divide one of $u_1(X^2) \\pm Xp^{\\beta-\\alpha}v_1(X^2)$, and an obvious argument on the parity of degrees shows that $u_1(X^2)$ must be divisible by $p$, and hence so must be $u_1$ — a contradiction which concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle A < 90^\\circ$ and $AB \\neq AC$. Denote by $H$ the ortocenter of triangle $ABC$, by $N$ the midpoint of $[AH]$, by $M$ the midpoint of the side $[BC]$ and by $D$ the intersection point of the angle bisector of $\\angle BAC$ with $[MN]$. Prove that $\\angle ADH = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcircle of triangle $ABC$. Rays ($AO$ and ($AH$ are isogonal. We deduce that $\\angle NAD \\equiv \\angle OAD$. (1)\n\nBut $OM \\parallel AH$ and $OM = \\frac{1}{2}AH = AN$, which means that $MOAN$ is a parallelogram. It follows that $MN \\parallel OA$. (2)\n\nCombining (1) and (2) gives $\\angle NAD \\equiv \\angle OAD \\equiv \\angle NDA$, hence triangle $NAD$ is isosceles with $NA = ND$. In triangle $DAH$, the segment $[DN]$ is a median, and $DN = AN = \\frac{1}{2}AH$ leads to $\\angle ADH = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given any integer $n \\ge 2$, show that there exists a set of $n$ pairwise coprime composite integers in arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix a prime $p > n$ and an integer $N \\ge p + (n-1)n!$ and consider the arithmetic progression of length $n$ consisting of $N! + p + kn!$, $k = 0, 1, \\dots, n-1$.\n\nSuppose, if possible, that $q$ is a prime factor of two of these numbers. Then $q$ divides their difference which is of the form $kn!$, for some positive integer $k < n$. It follows that $q$ does not exceed $n$, so $n!$ and $N!$ are both divisible by $q$, and consequently so is $p$ — a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be the midpoints of the diagonals $BD$ and $AC$ of the quadrilateral $ABCD$. Consider the points $M \\in (BC)$, $N \\in (CD)$, $R \\in (PQ)$ and $S \\in (AC)$ such that $\\frac{BM}{MC} = \\frac{DN}{NC} = \\frac{PR}{RQ} = \\frac{AS}{SC} = k$. Prove that the centroid of the triangle $AMN$ lies on the segment $[RS]$.\n\nAndrei Bud",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the centroid of the triangle $AMN$. Then we have\n$$\n\\overrightarrow{GR} = \\frac{\\overrightarrow{GP} + k\\overrightarrow{GQ}}{1+k} = \\frac{\\overrightarrow{GB} + \\overrightarrow{GD} + k(\\overrightarrow{GA} + \\overrightarrow{GC})}{2(1+k)}\n$$\nand\n$$\n\\overrightarrow{GS} = \\frac{\\overrightarrow{GA} + k\\overrightarrow{GC}}{1+k}.\n$$\nOn the other hand,\n$$\n0 = \\overrightarrow{GA} + \\overrightarrow{GM} + \\overrightarrow{GN} = \\overrightarrow{GA} + \\frac{\\overrightarrow{GB} + k\\overrightarrow{GC}}{1+k} + \\frac{\\overrightarrow{GD} + k\\overrightarrow{GC}}{1+k},\n$$\nfrom where, we deduce that $(1+k)\\overrightarrow{GA} + \\overrightarrow{GB} + 2k\\overrightarrow{GC} + \\overrightarrow{GD} = 0$, whence we get $\\overrightarrow{GS} + 2\\overrightarrow{GR} = 0$, and thus $G, R$ and $S$ are collinear points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle in which $m(\\hat{A}) = 135^\\circ$. The perpendicular to the line $AB$ erected at $A$ intersects the side $[BC]$ at $D$, and the angle bisector of $\\angle B$ intersects the side $[AC]$ at $E$. Find the measure of $\\widehat{BED}$.\n\nTraian Preda\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "45°",
    "solution": "Let $I \\in (BE)$ such that $IA$ bisects the angle $\\widehat{DAB}$. We deduce that $ID$ is the bisector of the angle $\\widehat{ADB}$. A short computation shows that $m(\\widehat{DIB}) = 135^\\circ$, hence triangles $ABE$ and $IBD$ are similar. It follows that $\\frac{AB}{IB} = \\frac{BE}{BD}$, so that $\\frac{AB}{EB} = \\frac{BI}{BD}$. Thus, triangles $ABI$ and $DBE$ are similar as well and, since $m(\\widehat{BED}) = m(\\widehat{BAI})$, we infer that $m(\\widehat{BED}) = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a positive integer $n$ and $A, B$ two matrices in $M_n(\\mathbb{C})$ such that $A^2 + B^2 = 2AB$. Prove that:\n\na) The matrix $AB - BA$ is not invertible.\n\nb) If the rank of $A - B$ is $1$, then matrices $A$ and $B$ commute.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The given relation can be written in each of the two forms:\n$$\n(A - B)^2 = AB - BA, \\qquad (1)\n$$\n$$\nA(A - B) = (A - B)B. \\qquad (2)\n$$\nSuppose $AB - BA$ is non singular. By (1), $A - B$ is also non-singular, that is $B = (A - B)^{-1}A(A - B)$, by (2). Then $A - B = A - (A - B)^{-1}A(A - B)$, from where we get $I_n = A(A - B)^{-1} - (A - B)^{-1}A$.\nConsidering traces we obtain\n$$\nn = \\operatorname{tr} I_n = \\operatorname{tr} (A(A - B)^{-1} - (A - B)^{-1}A) = 0,\n$$\na contradiction.\nSo $AB - BA$ is singular; moreover by (1) the matrix $A - B$ is also singular.\n\nb) It is known that for a matrix $X$ of rank $1$ in $M_n(\\mathbb{C})$, then $X^2 = (\\operatorname{tr} X) X$ (each row of $X$ is proportional with a nonzero fixed row $X$). Using (1) we get\n$$\nAB - BA = (A - B)^2 = (\\operatorname{tr}(A - B))(A - B),\n$$\nso $0 = \\operatorname{tr}(AB - BA) = (\\operatorname{tr}(A - B))^2$, i.e., $\\operatorname{tr}(A - B) = 0$. We conclude, $AB - BA = O_n$, i.e., $AB = BA$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $n$ be positive integers, with $p \\ge 2$, and let $a$ be a real number such that $1 \\le a < a + n \\le p$. Prove that the set\n$$\n\\{ \\lfloor \\log_2 x \\rfloor + \\lfloor \\log_3 x \\rfloor + \\dots + \\lfloor \\log_p x \\rfloor \\mid x \\in \\mathbb{R},\\ a \\le x \\le a + n \\}\n$$\nhas exactly $n + 1$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f(x) = \\sum_{k=2}^{p} \\lfloor \\log_k x \\rfloor$ and let $M = \\{f(x) \\mid x \\in [a, a+n]\\}$. It is easy to show that if $k \\ge 2$ is a positive integer, then $\\lfloor \\log_k \\lfloor x \\rfloor \\rfloor = \\lfloor \\log_k x \\rfloor$. This implies that $f(x) = f(\\lfloor x \\rfloor)$, for all $x \\in [1, \\infty)$, and hence $M = \\{f(x) \\mid x \\in S\\}$, where $S = \\{\\lfloor a \\rfloor, \\lfloor a \\rfloor + 1, \\dots, \\lfloor a \\rfloor + n\\}$ has $n+1$ elements. On the other hand, for $s \\in S$, $s < \\lfloor a \\rfloor + n \\le p$, we have $s+1 \\in \\{2, 3, \\dots, p\\}$, and\n$$\nf(s+1)-f(s) = \\sum_{k=2}^{p} (\\lfloor \\log_k(s+1) \\rfloor - \\lfloor \\log_k s \\rfloor) \\ge \\lfloor \\log_{s+1}(s+1) \\rfloor - \\lfloor \\log_{s+1} s \\rfloor = 1,\n$$\ntherefore $f(s+1) > f(s)$, and this proves that $M$ has exactly $n+1$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38134,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $M$, $N$, and $P$ are chosen on the sides $BC$, $CA$ and $AB$ of the triangle $ABC$ such that $BM = BP$ and $CM = CN$. The perpendicular dropped from $B$ onto $MP$ and the perpendicular dropped from $C$ onto $MN$ intersect at $I$. Prove that the angles $\\widehat{IPA}$ and $\\widehat{INC}$ are congruent.\n\nGabriel Popa\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $CM = CN$ and $CI \\perp MN$, the line $CI$ is the perpendicular bisector of the line segment $MN$, hence $IM = IN$. Similarly, we have $IM = IP$. Triangles $IMC$ and $INC$ are equal, so $\\widehat{IMC} \\equiv \\widehat{INC}$, and, in a similar way, we deduce that $\\widehat{IMB} \\equiv \\widehat{IPB}$. It follows that $\\widehat{IPA} = \\widehat{IMC}$, and, finally, that $\\widehat{IPA} \\equiv \\widehat{INC}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(A, +, \\cdot)$ be a unit ring with the property: for all $x \\in A$,\n$$\nx + x^2 + x^3 = x^4 + x^5 + x^6.\n$$\n\na) Let $x \\in A$ and let $n \\ge 2$ be an integer such that $x^n = 0$. Prove that $x = 0$.\n\nb) Prove that $x^4 = x$, for all $x \\in A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From the given equality we derive that\n$$\nx^{n-1} = x^n(x^4 + x^3 + x^2 - x - 1) = 0\n$$\nand, step by step, that $x^{n-2} = x^{n-3} = \\dots = x = 0$.\n\nb) Rewrite the given equation as\n$$\nx(x^3 - 1)(x^2 + x + 1) = 0.\n$$\nIt follows that\n$$\n(x^4 - x)^2 = x^2(x-1)(x^3 - 1)(x^2 + x + 1) = 0,\n$$\nhence $x^4 - x = 0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [1, +\\infty) \\to (0, +\\infty)$ be a continuous function having the following properties:\n(i) The function $g : [1, +\\infty) \\to (0, +\\infty)$ given by $g(x) = \\frac{f(x)}{x}$ has limit at $+\\infty$,\n(ii) The function $h : [1, +\\infty) \\to (0, +\\infty)$ given by $h(x) = \\frac{1}{x} \\int_{1}^{x} f(t) dt$ has finite limit at $+\\infty$.\n\na) Show that $\\lim_{x \\to +\\infty} g(x) = 0$.\nb) Show that $\\lim_{x \\to +\\infty} \\frac{1}{x^2} \\int_1^x f^2(t) dt = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nLet $\\ell = \\lim_{x \\to +\\infty} g(x)$. If $\\ell \\in (0, +\\infty)$, let $a > 0$ be such that $g(x) > \\ell/2$ for $x \\ge a$. Then\n$$\n\\begin{aligned}\nh(x) &= \\frac{1}{x} \\left( \\int_{1}^{a} f(t) \\, dt + \\int_{a}^{x} f(t) \\, dt \\right) \n\\ge \\frac{1}{x} \\int_{1}^{a} f(t) \\, dt + \\frac{\\ell}{2x} \\int_{a}^{x} f(t) \\, dt \\\\\n&= \\frac{1}{x} \\int_{1}^{a} f(t) \\, dt + \\frac{\\ell(x^2 - a^2)}{4x} \\underset{x \\to +\\infty}{\\longrightarrow} +\\infty,\n\\end{aligned}\n$$\nin contradiction with (ii). In the same way we can prove that $\\ell = +\\infty$ is in contradiction with (ii), so $\\ell = 0$.\n\nb.\n$$\n\\begin{aligned}\n\\lim_{x \\to +\\infty} \\frac{1}{x^2} \\int_1^x f^2(t) dt &= \\lim_{x \\to +\\infty} \\left( \\frac{\\int_1^x f^2(t) dt}{x \\int_1^x f(t) dt} \\cdot \\frac{\\int_1^x f(t) dt}{x} \\right) \\\\\n&= \\lambda \\lim_{x \\to +\\infty} \\frac{\\int_1^x f^2(t) dt}{x \\int_1^x f(t) dt} = \\lambda \\lim_{x \\to +\\infty} \\frac{u(x)}{v(x)},\n\\end{aligned}\n$$\nwhere $\\lambda = \\lim_{x \\to +\\infty} h(x)$, $u(x) = \\int_{1}^{x} f^2(t) dt$, $v(x) = x \\int_{1}^{x} f(t) dt$.\n\nTo show that $\\lim_{x \\to +\\infty} \\frac{u(x)}{v(x)} = 0$ we shall use l'Hospital Rule:\n\n* $u$ and $v$ are differentiable,\n* $\\lim_{x \\to +\\infty} v(x) = +\\infty$ (this follows from $v(x) \\ge m(x - 1)$ for $x \\ge 2$, where $m = \\inf_{x \\in [1,2]} f(x)$),\n* $v'(x) = \\int_{1}^{x} f(t) dt + x f(x) \\ne 0$, for all $x \\ge 1$,\n\n• relations $\\frac{u'(x)}{v'(x)} = \\frac{f^2(x)}{x f(x) + \\int_1^x f(t) dt} = g(x) \\cdot \\frac{f(x)}{f(x) + h(x)} \\in (0, g(x))$ and $\\lim_{x \\to +\\infty} g(x) = 0$ imply $\\lim_{x \\to +\\infty} \\frac{u'(x)}{v'(x)} = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest integer $n$ for which the set $A = \\{n, n+1, n+2, \\dots, 2n\\}$ contains five elements $a < b < c < d < e$ so that\n$$\n\\frac{a}{c} = \\frac{b}{d} = \\frac{c}{e}.\n$$",
    "options": [],
    "answer": "16",
    "solution": "Let $p, q \\in \\mathbb{N}^*$, $(p, q) = 1$ so that $\\frac{a}{c} = \\frac{b}{d} = \\frac{c}{e} = \\frac{p}{q}$. Obviously, $p < q$. Since $a, b$ and $c$ are divisible by $p$ and $c, d, e$ are divisible by $q$, there exists $m \\in \\mathbb{N}^*$ so that $c = mpq$.\nLet us find the minimal value of the difference $e - a$, for given $p, q$. This is obtained when $a, b, c$ are consecutive multiples of $p$ and $c, d, e$ are consecutive multiples of $q$, that is $a = mpq - 2p$ and $e = mpq + 2q$. From $\\frac{c}{e} = \\frac{mpq}{mpq+2q} = \\frac{p}{q}$ follows $m(q-p) = 2$, hence $m \\in \\{1, 2\\}$.\nCondition $n \\le a < e \\le 2n \\le 2a$ implies $2a \\ge e$, that is $2mpq - 4p \\ge mpq + 2q$, or $mpq \\ge 4p + 2q$. (*)\nIf $m = 1$, then $q - p = 2$, hence $q = p + 2$. Relation (*) yields $(p-2)^2 \\ge 8$, whence $p \\ge 5$. For $p = 5$ and $q = 7$ we get $a = 25$, $b = 30$, $c = 35$, $d = 42$, $e = 49$ and, since $n \\le a < e \\le 2n$, $n = 25$.\nIf $m = 2$, then $q - p = 1$, so $q = p + 1$. Relation (*) yields $(p-1)^2 \\ge 2$, whence $p \\ge 3$. For $p = 3$ and $q = 4$ we get $a = 18$, $b = 21$, $c = 24$, $d = 28$, $e = 32$ and, since $n \\le a < e \\le 2n$, $n \\in \\{16, 17, 18\\}$. Therefore, $n_{\\min} = 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a strictly positive integer and let $A = \\{1, 2, \\dots, n\\}$. Find the number of the increasing functions $f : A \\to A$ which satisfy the property\n$$\n|f(x) - f(y)| \\le |x - y|,\n$$\nfor any $x, y \\in A$.",
    "options": [],
    "answer": "(n+1) 2^{n-2}",
    "solution": "We remark that $f(k+1) - f(k) = a_k \\in \\{0, 1\\}$, $k = 1, 2, \\dots, n-1$. Moreover, as $a_k \\in \\{0, 1\\}$, we have $|f(x) - f(y)| \\le |x - y|$ for any $x, y \\in A$. Therefore, any function $f$ which satisfies the conditions in the problem, is\n\ncompletely determined by the $n$-tuple $(f(0), a_1, a_2, \\dots, a_{n-1}) \\in A \\times \\{0, 1\\} \\times \\dots \\times \\{0, 1\\}$ which satisfies the condition\n$$\nf(0) + a_1 + \\dots + a_{n-1} \\le n.\n$$\nIf we fix $f(0) = a$ and $f(n) = b$, we have $b - a = a_1 + \\dots + a_{n-1}$, and thus the number of the functions $f$, with the required property, is equal with the number of choices of $b-a$ terms equal to $1$ in the sum $a_1 + \\dots + a_{n-1}$, that is $n(a, b) = C_{n-1}^{b-a}$.\nThus, the number of functions is\n$$\n\\sum_{0 \\le a \\le b \\le n} n(a, b) = \\sum_{0 \\le a \\le b \\le n} C_{n-1}^{b-a} = \\sum_{k=0}^{n-1} (n-k) C_{n-1}^k.\n$$\nSumming up we have a number of functions equal to\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} n C_{n-1}^k - \\sum_{k=1}^{n-1} k C_{n-1}^k &= n 2^{n-1} - \\sum_{k=1}^{n-1} (n-1) C_{n-2}^{k-1} \\\\\n&= n 2^{n-1} - (n-1) 2^{n-2} = (n+1) 2^{n-2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a regular hexagon with side length $a$. At point $A$, the perpendicular $AS$, with length $2a\\sqrt{3}$, is erected on the hexagon's plane. The points $M, N, P, Q$, and $R$ are the projections of point $A$ onto the lines $SB, SC, SD, SE$, and $SF$, respectively.\n\na) Prove that the points $M, N, P, Q, R$ lie in the same plane.\n\nb) Find the measure of the angle between the planes $(MNP)$ and $(ABC)$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "30°",
    "solution": "a) Using the Three Perpendiculars Theorem, from $SA \\perp (ABC)$ and $AB \\perp BD$, it results $SB \\perp BD$. Since $BD \\perp AB$ and $BD \\perp SB$, it follows that $BD \\perp (SAB)$, hence $BD \\perp AM$.\n\nSince $AM \\perp SB$, it results $AM \\perp (SBD)$, hence $AM \\perp SD$. We also have $SD \\perp AP$, therefore we obtain that $SD \\perp (AMP)$. In a similar way one can show that $SD \\perp (ARP)$, $SD \\perp (ANP)$ and $SD \\perp (AQP)$, therefore the points $M, N, P, Q, R$ lie in the same plane.\n\nb) Because $MR \\parallel BF$, it follows that the intersection between the planes $(MNP)$ and $(ABC)$ is the line $d$ parallel to $BF$ and passing through $A$. Since $d \\perp SA$ and $d \\perp AD$, it results that $d \\perp (SAD)$, hence $d \\perp AP$. Therefore the angle between the planes $(MNP)$ and $(ABC)$ equals $\\widehat{PAD}$.\n\nUsing the Pythagorean Theorem, one obtains $SD = 4a$, hence $m(\\widehat{PDA}) = 60^\\circ$ and $m(\\widehat{PAD}) = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38140,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $A$ be a matrix from $M_2(\\mathbb{C})$, $A \\neq aI_2$, for any $a \\in \\mathbb{C}$. Prove that the matrix $X$ from $M_2(\\mathbb{C})$ commutes with $A$, that is, $AX = XA$, if and only if there exist two complex numbers $\\alpha$ and $\\alpha'$, such that $X = \\alpha A + \\alpha' I_2$.\n\nb) Let $A$, $B$ and $C$ be matrices from $M_2(\\mathbb{C})$, such that $AB \\neq BA$, $AC = CA$ and $BC = CB$. Prove that $C$ commutes with all matrices from $M_2(\\mathbb{C})$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Clearly, if $X = \\alpha A + \\alpha' I_2$, then $X$ and $A$ commute. Conversely, let $A = \\begin{pmatrix} a_1 & a_2 \\\\ a'_1 & a'_2 \\end{pmatrix}$ and $X = \\begin{pmatrix} x_1 & x_2 \\\\ x'_1 & x'_2 \\end{pmatrix}$. The equality $AX = XA$ implies\n$$\na_2 x'_1 = a'_1 x_2, \\quad (1)\n$$\n$$\n(a_1 - a'_2)x_2 + a_2(x'_2 - x_1) = 0, \\quad (2)\n$$\n$$\n(a_1 - a'_2)x'_1 + a'_1(x'_2 - x_1) = 0. \\quad (3)\n$$\nSince $A \\neq aI_2$, $a \\in \\mathbb{C}$, either one of $a_2, a'_1$ is non-zero, or $a_2 = a'_1 = 0$ and $a_1 \\neq a'_2$.\nIn the first case, if, for instance, $a_2 \\neq 0$, we obtain\n$$\nX = \\frac{x_2}{a_2} A + \\left( x_1 - \\frac{a_1}{a_2} x_2 \\right) I_2,\n$$\nwhile in the second case\n$$\nX = \\frac{x_1 - x'_2}{a_1 - a'_2} A + \\frac{a_1 x'_2 - a'_2 x_1}{a_1 - a'_2} I_2.\n$$\n\nb) We prove that $C = \\gamma I_2$, for some complex number $\\gamma$, hence $C$ commutes with all matrices in $M_2(\\mathbb{C})$. Suppose the contrary. Since $A$ and $C$ commute, there exist $\\alpha$ and $\\alpha'$, such that $A = \\alpha C + \\alpha' I_2$. Similarly, there exist $\\beta$ and $\\beta'$, such that $B = \\beta C + \\beta' I_2$, therefore $AB = BA$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $A_0A_1A_2$, determine the locus of the centers of the equilateral triangles $X_0X_1X_2$ satisfying the condition that each of the lines $X_kX_{k+1}$ passes through $A_k$ (all indices are reduced modulo 3).",
    "options": [],
    "answer": "A pair of circles: the circumcircles through the three centers of the inner and the outer Napoleon triangles of the given triangle.",
    "solution": "From any point $X_0$ on $\\gamma_0$ draw lines $X_0A_2X_1$ and $X_0A_1X_2$, where $X_1$ lies on $\\gamma_1$ and $X_2$ lies on $\\gamma_2$. The points $X_1, A_0, X_2$ are collinear, and the triangle $X_0X_1X_2$ is an equilateral triangle satisfying the conditions in the statement.\n\n![](attached_image_1.png)\n\nLet $M_k$ be the midpoint of the minor arc $A_{k+1}A_{k+2}$ of the circle $\\gamma_k$, and notice that the triangle $M_0M_1M_2$ is equilateral, since the $M_k$ are the centers of the inner Napoleon triangles associated with the triangle $A_0A_1A_2$. The center $X$ of the triangle $X_0X_1X_2$ is the intersection of $X_0M_0$ and $X_1M_1$, which must intersect at $60^\\circ$. Since the locus of $X$ includes the three points $M_k$, it turns out that the locus of $X$ is the circle $M_0M_1M_2$.\nSimilarly, another circle is obtained by starting with the inner Napoleon triangles. The required locus is a pair of circles.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exists a multiple of $2013$ which ends in $2014$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the numbers $a_1, a_2, \\dots, a_{2014}$, where $a_k$ is the $4k$-digit number whose digits are $k$ groups of the form $2014$, $1 \\le k \\le 2014$. Since there are only $2013$ possible remainders after the division by $2013$, two different $a$s must give the same remainder in this division, hence there exists $i, j$, with $1 \\le j < i \\le 2014$, so that $a_i - a_j$ is divisible by $2013$. Since\n$$\na_i - a_j = \\underbrace{20142014\\dots2014}_{i-j \\text{ times } 2014} \\underbrace{00\\dots0}_{4j \\text{ times}} = \\underbrace{20142014\\dots2014}_{i-j \\text{ times } 2014} \\cdot \\underbrace{100\\dots0}_{4j \\text{ times}} = a_{i-j} \\cdot 10^{4j}\n$$\nand $10^{4j}$ and $2013$ are relatively prime, $2013$ must divide $a_{i-j}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38143,
    "subject": "Mathematics (Multi-modal)",
    "question": "The medians $AD$, $BE$ and $CF$ of triangle $ABC$ intersect at $G$. Let $P$ be a point lying in the interior of the triangle, not belonging to any of its medians. The line through $P$ parallel to $AD$ intersects the side $BC$ at $A_1$. Similarly one defines the points $B_1$ and $C_1$. Prove that\n$$\n\\overrightarrow{A_1D} + \\overrightarrow{B_1E} + \\overrightarrow{C_1F} = \\frac{3}{2} \\overrightarrow{PG}.\n$$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us draw through point $P$ parallels to the triangle's sides. If the parallels to $AB$ and $AC$ intersect the side $BC$ at points $A_B$ and $A_C$, then the triangles $ABC$ and $PA_B A_C$ are similar, and since $PA_1$ is parallel to the median $AD$, it follows that $A_1$ is the midpoint of the line segment $A_B A_C$. With the notations in the figure, we have\n$$\n2\\overrightarrow{A_1D} = \\overrightarrow{A_1B} + \\overrightarrow{A_1C} = \\overrightarrow{A_B}B + \\overrightarrow{A_C}C = \\overrightarrow{PC_B} + \\overrightarrow{PB_C}.\n$$\nand the similar equalities. We deduce that the sum $2(\\overrightarrow{A_1D} + \\overrightarrow{B_1E} + \\overrightarrow{C_1F})$ equals\n$$\n\\vec{v} = \\overrightarrow{PC_B} + \\overrightarrow{PB_C} + \\overrightarrow{PA_C} + \\overrightarrow{PC_A} + \\overrightarrow{PB_A} + \\overrightarrow{PA_B}.\n$$\nBut $\\overrightarrow{PC_A} + \\overrightarrow{PB_A} = \\overrightarrow{PA}$ and adding up all similar equalities yields\n$$\n\\vec{v} = \\overrightarrow{PA} + \\overrightarrow{PB} + \\overrightarrow{PC} = 3\\overrightarrow{PG},\n$$\nhence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{1, 3, 3^2, 3^3, \\dots, 3^{2014}\\}$. We obtain a *partition* of $A$ if $A$ is written as a disjoint union of nonempty subsets.\n\na) Prove that there is no partition of $A$ such that the product of elements in each subset is a square.\n\nb) Prove that there exists a partition of $A$ such that the sum of elements in each subset is a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Assume that such a partition exists. Then the product of all elements of $A$ must be a square as well. But this equals $3^{1+2+3+\\dots+2014} = 3^{2015 \\cdot 1007}$, obviously not a square.\n\nb) Observe that $3^{2n} + 3^{2n+1} = (3^n \\cdot 2)^2$, hence a possible partition is\n$$\nA = \\{1,3\\} \\cup \\{3^2, 3^3\\} \\cup \\dots \\cup \\{3^{2012}, 3^{2013}\\} \\cup \\{3^{2014}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ so that $17^n + 9^{n^2} = 23^n + 3^{n^2}$.",
    "options": [],
    "answer": "1",
    "solution": "Clearly $n=0$ and $n=1$ are solutions.\n\nIf $n \\ge 2$, then $n^2 \\ge 2n$, hence $9^{n^2} - 3^{n^2} = 3^{n^2}(3^{n^2} - 1) \\ge 3^{2n}(3^{2n} - 1) = 81^n - 9^n$. Since $81^n - 9^n > 23^n - 17^n$, the equation does not have solutions $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38146,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n$ is a positive integer, we will call a triple $(x, y, z)$ of positive integers of type $n$ if $x + y + z = n$. Denote $s(n)$ the number of triples of type $n$.\na) Prove that there exists no positive integer $n$ such that $s(n) = 14$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us count the triples $(x, y, z)$ of positive integers such that $x + y + z = n$, where $n \\ge 3$ is a given integer. One can assign to $x$ any value from $1$ to $n-2$. If $x=1$, then, taking into account that $z \\ge 1$, $y$ can take $n-2$ values: from $1$ to $n-2$. If $x=2$, then $y$ can take $n-3$ values, ..., if $x=n-2$, then $y$ can take only the value $1$. This shows that the number of triples of type $n$ is\n$$\ns(n) = 1 + 2 + 3 + \\dots + (n-3) + (n-2) = \\frac{(n-2)(n-1)}{2}.\n$$\n\na) If there exists $n$ such that $s(n) = 14$, then $(n-2)(n-1) = 28$, which is impossible: if $n \\le 6$, then $(n-2)(n-1) \\le 20$ and if $n \\ge 7$, then $(n-2)(n-1) \\ge 30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A Steiner tree associated with a finite set $S$ of points in the Euclidean $n$-space is a finite collection $T$ of straight-line segments in that space such that any two points in $S$ are joined by a unique path in $T$; its length is the sum of the segment lengths. Show that there exists a Steiner tree of length $1 + (2^{n-1} - 1)\\sqrt{3}$ associated with the vertex set of a unit $n$-cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We describe a recursive procedure for constructing the desired Steiner tree. The case $n = 1$ is handled by a single line segment.\n\nAssume a Steiner tree of length $1 + (2^{n-1} - 1)\\sqrt{3}$ associated with the vertex set of a unit $n$-cube has been constructed such that each vertex of the $n$-cube is the endpoint of just one segment, and the length of this segment is greater than $\\sqrt{3}/6$. Now consider a unit $(n+1)$-cube. Select a pair of opposite $n$-faces and consider the $n$-cube whose vertices are the midpoints of the edges joining the corresponding vertices of these faces. Start with the assumed Steiner tree associated with the vertex set of this $n$-cube. For each vertex $v$ of this $n$-cube, delete a segment of length $\\sqrt{3}/6$ from the segment terminating at $v$, then add the segments joining the new endpoint to the two vertices of the $(n+1)$-cube that are endpoints of the edge containing $v$. By the Pythagorean theorem, these segments each have length $1/\\sqrt{3}$. The net effect of these changes is to add $2^n(2/\\sqrt{3} - \\sqrt{3}/6) = 2^{n-1}\\sqrt{3}$ to the length of the tree. Thus the resulting tree has length $1 + (2^n - 1)\\sqrt{3}$. It is clearly a Steiner tree associated with the vertex set of the $(n+1)$-cube and has the additional properties required for the induction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider real numbers $a_1, a_2, \\dots, a_{2n}$ whose sum is $0$. Prove that among the pairs $(a_i, a_j)$, $i < j$, with $i, j \\in \\{1, 2, \\dots, 2n\\}$, there exist at least $2n - 1$ pairs such that $a_i + a_j \\ge 0$.",
    "options": [],
    "answer": "2n - 1",
    "solution": "* If $a_n + a_{2n-1} \\ge 0$, then all the sums $a_i + a_{2n-1}$ with $i = \\overline{n}, 2n-2$ as well as all the sums $a_i + a_{2n}$ with $i = \\overline{n}, 2n-1$ are non-negative. In total, there are at least $(n-1)+n = 2n-1$ non-negative sums.\n\n* If $a_n + a_{2n-1} < 0$, then $a_1 + \\dots + a_{n-1} + a_{n+1} + \\dots + a_{2n-2} + a_{2n} > 0$. (1)\nWe have $0 > a_n + a_{2n-1} \\ge a_{n-1} + a_{2n-2} \\ge \\dots \\ge a_2 + a_{n+1}$. It follows that $a_2 + a_3 + \\dots + a_{n-1} + a_{n+1} + \\dots + a_{2n-3} + a_{2n-2} < 0$. Combining this with (1) gives $a_1 + a_{2n} \\ge 0$, hence all the sums $a_i + a_{2n}$ with $i = \\overline{1, 2n-1}$ are non-negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. Determine all possible values of the sum\n$$\nS = \\lfloor x_2 - x_1 \\rfloor + \\lfloor x_3 - x_2 \\rfloor + \\dots + \\lfloor x_n - x_{n-1} \\rfloor,\n$$\nwhere $x_1, x_2, \\dots, x_n$ are real numbers whose integer parts are $1, 2, \\dots, n$, respectively.",
    "options": [],
    "answer": "{0, 1, 2, ..., n-1}",
    "solution": "Let $a$ and $b$ be arbitrary reals. We have $\\lfloor b \\rfloor - \\lfloor a \\rfloor - 1 < b - a < \\lfloor b \\rfloor - \\lfloor a \\rfloor + 1$, hence $\\lfloor b \\rfloor - \\lfloor a \\rfloor - 1 \\le \\lfloor b - a \\rfloor \\le \\lfloor b \\rfloor - \\lfloor a \\rfloor$. Applying this to some consecutive terms of the sequence, we obtain $0 \\le \\lfloor x_k - x_{k-1} \\rfloor \\le 1$, for all $k = 2, 3, \\dots, n$. Thus, $0 \\le S \\le n - 1$.\n\nWe claim that the set of all possible values of $S$ is $\\{0, 1, 2, \\dots, n-1\\}$.\n\nThe value $S = n - 1$ can be obtained, for instance, for $x_k = k,\\ k = 1, 2, \\dots, n$.\n\nFor the value $S = p$, with $0 \\le p \\le n - 2$, one can take, for instance,\n$$\nx_k = \\begin{cases} k + \\frac{1}{k+1}, & \\text{if } 1 \\le k \\le n-1-p \\\\ k, & \\text{if } n-p \\le k \\le n \\end{cases}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\\%$ of it.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $ABCD$ the square and $S_1$, $S_2$, $S_3$ the discs.\n\nSuppose that $S_1$, $S_2$ and $S_3$ cover the whole square. Since there are three discs and the square has four vertices, one of the discs must cover two vertices of the square, say $S_1$ covers $A$, $B$. Then $[AB]$ is a diameter for $S_1$, so $S_1$ cannot cover any point from $(BC] \\cup [CD] \\cup [DA)$. So, $C$ must be covered by one of the other discs, say $S_2$. Then $S_2$ cannot cover any point from $(AD)$, therefore $(AD) \\subset S_3$. In this case $[AD]$ is a diameter of $S_3$, so $S_3$ cannot cover any point from $(BC)$, therefore $S_2$ must cover $(BC)$. This shows that $[BC]$ must be a diameter of $S_2$. But, in this case, no point from $(CD)$ is covered – contradiction.\n\n![](attached_image_1.png)\n\nWe take as $S_1$, $S_2$, $S_3$ the discs of diameters $[AM]$, $[PT]$, $[RS]$. Denote $X$ the point on $[AC]$ for which $XT \\perp BC$ (and $XU \\perp CD$). Then $S_1$ covers the square surface $APMR$, $S_2$ covers the pentagonal surface $BPMXT$ and $S_3$ covers the pentagonal surface $DRMXU$, so the points not covered by the three discs are inside the square $CUXT$.\n\nIt is enough to prove that\n$$\n\\text{area}[CUXT] < 0.25\\% \\cdot \\text{area}[ABCD],\n$$\nwhich is equivalent to $CT < BC/20 = 0.1$, or $BT > 1.9$. Simple considerations yield $AP = \\sqrt{2}$, $BP = 2 - \\sqrt{2}$, $BT^2 = 4 - (2 - \\sqrt{2})^2 = 4\\sqrt{2} - 2$. We are left to prove $BT > 1.9$, that is $4\\sqrt{2} - 2 > 1.9^2$, or $\\sqrt{2} > 1.4025$, which is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be an invertible matrix in $M_4(\\mathbb{R})$, such that $\\operatorname{tr} A = \\operatorname{tr} A^* \\neq 0$, where $A^*$ is the adjugate of $A$. Prove that the matrix $A^2 + I_4$ is singular if and only if there exists a nonzero matrix $B$ in $M_4(\\mathbb{R})$, so that $AB = -BA$.\n\nMariean Andronache",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show first that if $A$ is a matrix from $M_n(\\mathbb{R})$ such that $A^2 + I_n$ is singular, then there exists a matrix $B$ in $M_n(\\mathbb{R})$, such that $AB = -BA$.\nAs $A^2 + I_n$ is singular, $i$ is a proper value of $A$, and $-i$ is a proper value of the transpose $A^\\tau$. There exist two proper vectors $\\mathbf{x}$ and $\\mathbf{y}$ in $M_{n,1}(\\mathbb{C})$, so that $A\\mathbf{x} = i\\mathbf{x}$ and $A^\\tau\\mathbf{y} = -i\\mathbf{y}$. Because $\\mathbf{x}$ and $\\mathbf{y}$ are nonzero, $B = \\mathbf{x}\\mathbf{y}^\\tau$ is a nonzero matrix in $M_n(\\mathbb{C})$.\nThe following relations show that $A$ and $B$ anticommute:\n$$\nAB = A\\mathbf{x}\\mathbf{y}^{\\tau} = i\\mathbf{x}\\mathbf{y}^{\\tau} = -\\mathbf{x}(-i\\mathbf{y})^{\\tau} = -\\mathbf{x}(A^{\\tau}\\mathbf{y})^{\\tau} = -\\mathbf{x}\\mathbf{y}^{\\tau}A = -BA.\n$$\nTaking conjugates and using that $A$ is in $M_n(\\mathbb{R})$, we get that the conjugate $\\bar{B}$ of $B$ anticommutes with $A$. This proves that any linear combination with complex coefficients of the matrices $B$ and $\\bar{B}$ anticommutes with $A$. Consequently, $B$ or $i(B - \\bar{B})$ is a nonzero matrix in $M_n(\\mathbb{R})$ that anticommutes with $A$. This proves the first part of the problem.\n\nWe shall prove the converse. Let $B$ be a nonzero matrix in $M_4(\\mathbb{R})$ that anticommutes with $A$. Then\n$$\nA^k B = (-1)^k B A^k, \\quad k \\in \\mathbb{N}. \\qquad (*)\n$$\nConsider the characteristic polynomial $f$ of $A$,\n$$f = \\lambda^4 - (\\operatorname{tr} A)\\lambda^3 + a\\lambda^2 - (\\operatorname{tr} A^*)\\lambda + \\det A = \\lambda^4 - (\\operatorname{tr} A)\\lambda^3 + a\\lambda^2 - (\\operatorname{tr} A)\\lambda + \\det A,$$ where $a$ is a real number (the latter form of $f$ is a consequence of $\\operatorname{tr} A = \\operatorname{tr} A^*$).\nBy *Hamilton-Cayley*, $f(A) = O_4$. Taking into account $(*)$, we successively obtain:\n$$\n\\begin{aligned} O_4 &= f(A)B = B(A^4 + (\\operatorname{tr} A)A^3 + aA^2 + (\\operatorname{tr} A)A + (\\det A)I_4) \\\\ &= B(f(A) + 2(\\operatorname{tr} A)(A^2 + I_4)A) = 2(\\operatorname{tr} A)B(A^2 + I_4)A. \\end{aligned}\n$$\nBecause $\\operatorname{tr} A \\neq 0$ and $A$ is invertible, we obtain that $B(A^2 + I_4) = O_4$. As $B$ is nonzero we conclude that $A^2 + I_4$ is singular.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $(G, \\cdot)$ is a finite group with unity $e$, $a$ is an element in $G \\setminus \\{e\\}$ and $p$ is a prime number such that $x^{p+1} = a^{-1} x a$, for all $x \\in G$.\n\na) Show that there is $k \\in \\mathbb{N}^*$ such that $\\text{ord}(G) = p^k$.\n\nb) Prove that $H = \\{x \\in G \\mid x^p = e\\}$ is a subgroup of $G$ and\n$$\n(\\text{ord}(H))^2 > \\text{ord}(G).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $x, y \\in G$, then $(xy)^{p+1} = a^{-1} x y a = a^{-1} x x a^{-1} y a = x^{p+1} y^{p+1}$. We can write $x(yx)^p y = x^{p+1} y^{p+1}$, then $(yx)^p = x^p y^p$. For $x = a$ we get $a^p = e$ so by the preceding equality $(y a)^p = y^p$. Multiplying at left by $y a$ we obtain $y a y^p = (y a)^{p+1} = y^{p+1} a$, that is $a y^p = y^p a$, for all $y \\in G$. From the hypothesis we have $y^{p(p+1)} = a^{-1} y^p a = y^p$, so $y^{p^2} = e$, for all $y \\in G$. Because $p$ is a prime, every element of the group has order $1$, $p$ or $p^2$ and by the *Cauchy theorem* we deduce $\\text{ord}(G) = p^k$, for a $k \\in \\mathbb{N}^*$.\n\nb) For $x, y \\in H$, we have $(xy)^p = y^p x^p = e$, that is $xy \\in H$, proving that $H$ is a stable part of $G$, and, as it is finite, $H$ is a subgroup.\n\nConsider $f : G \\to G$, given by $f(x) = x^p$. Because $e = x^{p^2} = (x^p)^p$, the image of $f$ is contained in $H$. Moreover, $x, y \\in G$ and $f(x) = f(y)$ imply $x^p (y^{-1})^p = e$, so $(y^{-1} x)^p = e$, that is $y^{-1} x \\in H$. This gives $x \\in H y$. We conclude that for every element in $\\text{Im} f$, the number of its pre-images in $G$ is exactly $\\text{ord}(H)$, so $|\\text{Im} f| = \\frac{\\text{ord}(G)}{\\text{ord}(H)}$.\n\nBecause $a \\neq e$ we get $a \\notin \\text{Im} f$: for if not $a = b^p$ for some $b \\in G$. This would imply $b^{p+1} = b^{-p} b b^p$, that is $e = b^p = a$ a contradiction. As $a \\in H$, we conclude $\\text{ord}(H) > |\\text{Im} f| = \\frac{\\text{ord}(G)}{\\text{ord}(H)}$, which gives the conclusion.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a$ for which there exist exactly 2014 positive integers $b$ such that $2 \\le \\frac{a}{b} \\le 5$.",
    "options": [],
    "answer": "6710, 6712, 6713",
    "solution": "Rewrite $2 \\le \\frac{a}{b} \\le 5$ as $2a \\le 10b \\le 5a$. It follows that the sequence $2a, 2a+1, \\dots, 5a$ contains 2014 multiples of 10, hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. We deduce that $2013 \\cdot 10 \\le 5a - 2a < 2015 \\cdot 10$, which leads to $a \\in \\{6710, 6711, \\dots, 6716\\}$. By inspection, we conclude that the required values of $a$ are: 6710, 6712, and 6713.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with $\\angle A + \\angle C = 60^\\circ$ and $AB \\cdot CD = BC \\cdot AD$. Prove that $AB \\cdot CD = AC \\cdot BD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us construct the equilateral triangle $BCE$ in the half-plane determined by line $BC$ and point $D$.\nThen $\\frac{AB}{AD} = \\frac{BC}{CD} = \\frac{CE}{CD}$ and $\\angle DAB = \\angle DCE$, so $\\Delta DAB \\sim \\Delta DCE$.\nTherefore $\\frac{AD}{DC} = \\frac{DB}{DE}$ and $\\angle ADB = \\angle CDE$, hence $\\angle ADC = \\angle BDE$.\nIt follows that $\\Delta ADC \\sim \\Delta BDE$, so $\\frac{AD}{BD} = \\frac{AC}{BE} = \\frac{AC}{BC}$, which leads to $AC \\cdot BD = BC \\cdot AD = AB \\cdot CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Construct outside the square $ABCD$ the right isosceles triangle $ABE$, with hypotenuse $[AB]$. Denote $N$ the midpoint of the segment $[AD]$ and $\\{M\\} = CE \\cap AB$, $\\{P\\} = CN \\cap AB$, $\\{F\\} = PE \\cap MN$. Take on the straight line $FP$ the point $Q$ so that $[CE$ is the bisector of the angle $\\angle QCB$. Prove that $MQ \\perp CF$.\n\nSorin Furtună",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly $PA = PB/2$, so $PA = AB = BC$. This leads to $\\triangle APE \\equiv \\triangle BCE$ (SAS), which implies $CE = PE$ and $\\overline{BEC} \\equiv \\overline{AEP}$. Now\n\n$m(\\widehat{BEC}) + m(\\widehat{AEC}) = 90^\\circ$ yields $m(\\widehat{AEC}) + m(\\widehat{AEP}) = 90^\\circ$, so $m(\\widehat{CEP}) = 90^\\circ$. This shows that $ECP$ is a right isosceles triangle, so $m(\\widehat{CPE}) = 45^\\circ$.\n\n![](attached_image_1.png)\n\nAlso $\\triangle CEQ \\equiv \\triangle PEM$ (AAS), whence $EQ = EM$, so $EMQ$ is a right isosceles triangle. This gives $m(\\widehat{EQM}) = 45^\\circ = m(\\widehat{EPC})$, therefore $MQ \\parallel CP$. Menelaus' Theorem for the triangle $CPE$ and the transversal $NF$ yields $\\frac{CN}{NP} \\cdot \\frac{FP}{FE} \\cdot \\frac{ME}{MC} = 1$. If we denote $R$ the orthogonal projection of $E$ onto $AB$, then $\\triangle CBM \\sim \\triangle ERM$, whence $\\frac{ME}{MC} = \\frac{ER}{BC} = \\frac{1}{2}$.\n\nFrom $NP = CN$ follows $\\frac{FP}{FE} = 2$, hence $PE = EF$, therefore $CE$ is altitude and median in $\\triangle CFP$, that is this triangle is isosceles. This gives $m(\\widehat{FCP}) = 90^\\circ$, that is $CP \\perp CF$, whence $MQ \\perp CF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer with $n \\ge 2$, and let $a_0, a_1, a_2, \\dots, a_n$ be complex numbers with $a_n \\ne 0$. Prove that the following statements are equivalent:\n\n(P) $|a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0| \\le |a_n + a_0|$, for every complex number $z$ of modulus 1,\n\n(Q) $a_1 = a_2 = \\dots = a_{n-1} = 0$ and $\\frac{a_0}{a_n} \\in [0, \\infty)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Q) $\\Rightarrow$ (P). If $a_1 = a_2 = \\dots = a_{n-1} = 0$ and $\\frac{a_0}{a_n} \\in [0, \\infty)$, then\n$$\n\\begin{aligned}\n|a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0| &= |a_n z^n + a_0| \\\\ &\\le |a_n z^n| + |a_0| = |a_n| + |a_0| = |a_n + a_0|,\n\\end{aligned}\n$$\nfor every complex number $z$ of modulus 1.\n\n(P) $\\Rightarrow$ (Q). Let $g(z) = a_{n-1}z^{n-1} + \\cdots + a_1z$, $z \\in \\mathbb{C}$ and $w = a_0 + a_n$. For any $\\varepsilon \\in U_n = \\{z \\in \\mathbb{C} \\mid z^n = 1\\}$ we have $|a_n + g(\\varepsilon) + a_0| \\le |a_n + a_0|$, that is $|w + g(\\varepsilon)| \\le |w|$ or $|g(\\varepsilon)|^2 + \\overline{w g(\\varepsilon)} + \\bar{w} g(\\varepsilon) \\le 0$.\nAs $\\sum_{\\varepsilon \\in U_n} \\varepsilon^k = 0$, $k = 1, 2, \\dots, n-1$, we have $\\sum_{\\varepsilon \\in U_n} g(\\varepsilon) = 0$, whence, summing up, $\\sum_{\\varepsilon \\in U_n} |g(\\varepsilon)|^2 + \\overline{w g(\\varepsilon)} + \\bar{w} g(\\varepsilon) = \\sum_{\\varepsilon \\in U_n} |g(\\varepsilon)|^2 \\le 0$, which implies that $g(\\varepsilon) = 0$, for every $\\varepsilon \\in U_n$.\nFrom the above, we get $a_k = \\frac{1}{n} \\sum_{\\varepsilon \\in U_n} \\frac{g(\\varepsilon)}{\\varepsilon^k} = 0$, $k = 1, 2, \\dots, n-1$.\nThus, we get $|a_n z^n + a_0| \\le |a_n + a_0|$ for every complex number $z$ of modulus 1. If we denote $c = \\frac{a_0}{a_n}$ and $t = z^n$, we have $|t + c| \\le |1 + c|$ for every complex number $z$ of modulus 1. Let $P, M, A$ be the points in the complex plane with complex coordinates $t, -c$, respectively 1. Then $PM \\le MA$, for every point $P$ on the unit circle, therefore the unit circle is interior to the circle with center $M$ and radius $MA$. As the point $A$ belongs to both circles, we get that they are tangent in the point $A$, and thus the points $M, O$ and $A$ are collinear and $MA \\ge OA = 1$. We get that $-c \\le 0$, and thus $c \\in [0, \\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer and let $m$ be a positive odd integer. Show that there exists a positive integer $n$ such that $m^n + n^m$ has at least $k$ distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Design a set of $k$ primes $p_1 < p_2 < \\cdots < p_k$ as follows. Begin by choosing $p_1 > 2m$. Having selected $p_j$, use Dirichlet's theorem to choose a prime\n$$\np_{j+1} \\equiv -1 \\pmod{p_1(p_1-1)p_2(p_2-1)\\cdots p_j(p_j-1)}.\n$$\nIf $i < j$, then $p_i < p_j$, so $p_j$ does not divide $p_i - 1$; further, $p_j - 1 \\equiv -2 \\pmod{p_i(p_i - 1)}$, so $p_i$ does not divide $p_j - 1$.\n\nNext, use the Chinese Remainder Theorem, to choose a positive integer $n$ such that $n \\equiv -1 \\pmod{p_1p_2\\cdots p_k}$ and $n \\equiv 0 \\pmod{(p_1-1)\\cdots(p_k-1)}$.\n\nFinally, since $p_1p_2\\cdots p_k$ and $m$ are coprime, and $(p_1-1)(p_2-1)\\cdots(p_k-1)$ divides $n$, and $m$ is odd, Euler's Theorem applies to show that $m^n + n^m \\equiv 1 + (-1)^m \\equiv 0 \\pmod{p_1p_2\\cdots p_k}$. The conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38158,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred boxes are labeled from $1$ to $100$. Each box has at most $10$ stones. The difference of the numbers of stones for every two boxes labeled with consecutive numbers is $1$. The boxes labeled $1, 4, 7, 10, \\ldots, 100$ contain a total of $301$ stones. Find the maximum possible number of stones contained by the $100$ boxes.\n\nGabriel Popa",
    "options": [],
    "answer": "928",
    "solution": "Since the difference of the number of stones in every two consecutive boxes is $1$, two consecutive boxes contain at most $19$ stones. We know the number of stones in the $34$ boxes $\\#1, \\#4, \\#7, \\ldots, \\#100$ and if we group the remaining $66$ boxes in pairs of consecutive boxes, we obtain at most $301 + 33 \\cdot 19 = 928$ stones.\n\nThis value is indeed obtained if we have $10$ groups of the form $(9, 10, 9, 8, 9, 10)$, then $6$ groups of the form $(9, 10, 9, 10, 9, 10)$ and, at the end, $9, 10, 9, 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38159,
    "subject": "Mathematics (Multi-modal)",
    "question": "A 10 digit positive integer is called a *cute number* if its digits are from the set $\\{1, 2, 3\\}$ and every two consecutive digits differ by 1.\n\na) Prove that exactly 5 digits of a cute number are equal to 2.\n\nb) Find the total number of cute numbers.\n\nc) Prove that the sum of all cute numbers is divisible by 1408.",
    "options": [],
    "answer": "a) 5; b) 64; c) sum divisible by 1408",
    "solution": "a) In the decimal representation of a cute number, the parity of the digits clearly alternates, hence exactly 5 digits are even, that is, equal to 2.\n\nb) There are $2^5 = 32$ cute numbers of the form $\\overline{2a2b2c2d2e}$ and another 32 of the form $\\overline{a2b2c2d2e2}$, therefore the requested number is 64.\n\nc) If $\\overline{a_1a_2\\dots a_{10}}$ is a cute number, then\n$$\n\\overline{(4 - a_1)(4 - a_2)\\dots(4 - a_{10})}\n$$\nis also cute and distinct from the previous one. They add up to $4444444444$. Grouping the cute numbers in 32 such pairs, we obtain that their sum equals $32 \\cdot 4444444444 = 2^7 \\cdot 11 \\cdot 101010101 = 1408 \\cdot 101010101$, hence the conclusion.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}^*$ be a strictly increasing function. Prove that:\na) there exists a decreasing sequence of positive real numbers, $(y_n)_{n \\in \\mathbb{N}}$, converging to $0$, such that $y_n \\le 2y_{f(n)}$, for all $n \\in \\mathbb{N}$;\nb) if $(x_n)_{n \\in \\mathbb{N}}$ is a decreasing sequence of real numbers, converging to $0$, then there exists a decreasing sequence of real numbers $(y_n)_{n \\in \\mathbb{N}}$, converging to $0$, such that $x_n \\le y_n \\le 2y_{f(n)}$, for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $f(0) > 0$ and $f$ is strictly increasing, it follows that $f(n) > n$, for all $n \\in \\mathbb{N}$. Consider the sequence of non-negative integers $(n_k)_{k \\in \\mathbb{N}}$, defined by $n_0 = 0$ and $n_k = f(n_{k-1})$, $k \\in \\mathbb{N}^*$. The properties of $f$ imply that the sequence is strictly increasing. We define the decreasing sequence $(y_n)_{n \\in \\mathbb{N}}$, by $y_n = 2^{-k}$, for all $n$ with $n_k \\le n < n_{k+1}$, $k \\in \\mathbb{N}$, obviously convergent to $0$.\n\nIt suffices to prove that $y_n \\le 2y_{f(n)}$, $n \\in \\mathbb{N}$, for $n_k \\le n < n_{k+1}$, $k \\in \\mathbb{N}$. Since $f$ strictly increasing, $n_{k+1} = f(n_k) \\le f(n) < f(n_{k+1}) = n_{k+2}$, hence $y_{f(n)} = 2^{-k-1} = y_n/2$.\n\nb) Obviously, $x_n \\ge 0$, for all $n \\in \\mathbb{N}$. Using the previously defined sequence $(n_k)$, we define the decreasing sequence of positive reals $(z_k)_{k \\in \\mathbb{N}}$, as follows: $z_0 = x_1$ and $z_k = \\max(x_{n_k}, z_{k-1}/2)$, $k \\in \\mathbb{N}^*$. The monotony of this sequence follows inductively. Moreover, $(z_k)_{k \\in \\mathbb{N}}$ converges to $0$: if $z_k = x_{n_k}$, for infinitely many $k$'s, then $z_k \\to 0$ because it is decreasing and $x_n \\to 0$; if $z_k = x_{n_k}$, only for finitely many $k$'s, then $z_k = z_{k-1}/2$ from some $k$ onwards, and again, $z_k \\to 0$.\n\nFinally, we define the sequence $(y_n)_{n \\in \\mathbb{N}}$ by $y_n = z_k$, $n_k \\le n < n_{k+1}$, $k \\in \\mathbb{N}$. Clearly, the sequence decreases to $0$. In order to prove the inequalities $x_n \\le y_n \\le 2y_{f(n)}$, $n \\in \\mathbb{N}$, it suffices to check them for $n_k \\le n < n_{k+1}$, $k \\in \\mathbb{N}$. Obviously, $x_n \\le x_{n_k} \\le z_k = y_n$. On the other hand, $n_{k+1} = f(n_k) \\le f(n) < f(n_{k+1}) = n_{k+2}$, since $f$ is strictly increasing, hence $y_{f(n)} = z_{k+1} \\ge z_k/2 = y_n/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and consider the points $K \\in (AB)$, $L \\in (BC)$, and $M \\in (CD)$ such that $KLM$ is a right isosceles triangle, with the right angle at $L$. Prove that the lines $AL$ and $DK$ are perpendicular to each other.\n\nBogdan Enescu\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is not difficult to observe that $\\triangle KLB \\equiv \\triangle LMC$, hence $KB = LC$. Because $AB = BC$, it follows that $AK = BL$. But then, $\\triangle AKD \\equiv \\triangle BLA$, and since $AK \\perp BL$ and $AD \\perp BA$, we deduce that $AL \\perp KD$, as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38162,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ points $M, N \\in (AB)$, $P, Q \\in (BC)$ and $S, R \\in (AC)$ are taken such that $AM = CR$, $AN = CS$, $\\overline{MQB} \\equiv \\overline{RQC}$ and $\\overline{NPB} \\equiv \\overline{SPC}$.\nShow that if $MQ + QR = NP + PS$, then triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $R'$ and $S'$ the reflection of $R$, respectively $S$, across the line $BC$. Then $RQ = R'Q$ and $\\overline{RQC} \\equiv \\overline{R'QC}$. Since $\\overline{MQB} \\equiv \\overline{RQC}$, it follows that $\\overline{MQB} \\equiv \\overline{R'QC}$, and because $B, Q, C$ are collinear, so are $M, Q, R'$. This yields $MQ + QR = MQ + QR' = MR'$. In the same way $NP + PS = NP + PS' = NS'$. Now $MQ + QR = NP + PS$ implies $MR' = NS'$.\nThe symmetry also implies $\\overline{RCQ} \\equiv \\overline{R'CQ}$ and $\\overline{SCP} \\equiv \\overline{S'CP}$, so $\\overline{RCQ} = \\overline{SCP}$ implies $\\overline{R'CQ} = \\overline{S'CP}$, whence the points $C, R', S'$ are collinear. Also\n\n$CR = CR'$ and $CS = CS'$ and, since $CR = AM$ and $CS = AN$, $MN = R'S'$.\n\n![](attached_image_1.png)\n\nTherefore $\\triangle NS'M \\equiv \\triangle R'MS'$ (case S-S-S), whence $\\overline{NMS'} \\equiv \\overline{MS'R'}$.\nThis shows that $AB \\parallel CS'$, therefore $\\overline{ABC} \\equiv \\overline{BCS'}$. Now $\\overline{BCS'} \\equiv \\overline{BCA}$\nimplies $\\overline{ABC} \\equiv \\overline{ACB}$, q.e.d.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that\n$$\nf(x + 3f(y)) = f(x) + f(y) + 2y,\n$$\nfor all $x, y \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "f(x)=x or f(x)=-2x/3 for all rational x",
    "solution": "We claim that the solutions are $f_1(x) = x$ and $f_2(x) = -2x/3$, for all real $x$.\n\nSet $x = y - 3f(y)$ to obtain $f(y - 3f(y)) = -2y$, $y \\in \\mathbb{Q}$. Replacing $y$ with $y - 3f(y)$ in the initial equation gives $f(x - 6y) = f(x) - 2y + 2(y - 3f(y))$, hence $f(x - 6y) = f(x) - 6f(y)$, for all $x, y \\in \\mathbb{Q}$.\n\nSet now $x = y = 0$ to get $f(0) = 0$ and replace $x = 6y$ to obtain $f(6y) = 6f(y)$, for all $y \\in \\mathbb{Q}$.\n\nWe derived that $f(x - 6y) = f(x) - f(6y)$, which, for $u = 6y$ and $v = x - 6y$, yields $f(u + v) = f(u) + f(v)$, for all $u, v \\in \\mathbb{Q}$. The solution of this classical functional equation is $f(x) = x f(1)$, $x \\in \\mathbb{Q}$.\n\nOn the other hand, by setting $x = y = 1$ in the initial equation we obtain $3f^2(1) - f(1) - 2 = 0$, hence $f(1) = 1$ or $f(1) = -2/3$, thus proving our claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a group with no elements of order $4$, and let $f : G \\rightarrow G$ be a group morphism such that $f(x) \\in \\{x, x^{-1}\\}$, for all $x \\in G$. Prove that either $f(x) = x$ for all $x \\in G$, or $f(x) = x^{-1}$ for all $x \\in G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, by way of contradiction, that there exist $a, b \\in G$ such that $f(a) = a \\neq a^{-1}$ and $f(b) = b^{-1} \\neq b$. Then $f(ab) = f(a)f(b) = ab^{-1} \\neq ab$, therefore $f(ab) = (ab)^{-1} = b^{-1}a^{-1}$. It follows that $ab^{-1} = b^{-1}a^{-1}$, which implies $b^{-1} = ab^{-1}a$.\n\nNext, $f(ab^2) = f(a)f^2(b) = ab^{-2}$. If $f(ab^2) = ab^2$, then $ab^{-2} = ab^2$, therefore either $b^2 = e$, which contradicts $b \\neq b^{-1}$, or $\\text{ord}(b) = 4$, which is impossible. Thus, $f(ab^2) = (ab^2)^{-1} = b^{-2}a^{-1}$, hence $ab^{-2} = b^{-2}a^{-1}$. It follows that $ab^{-2}a = b^{-2} = (b^{-1})^2 = ab^{-1}aab^{-1}a$, so $a^2 = e$, again a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, inscribed in the circle of diameter $AC$. We suppose that there exist the points $E \\in (CD)$ and $F \\in (BC)$ such that the lines $AE$ and $DF$ are perpendicular to the lines $AF$ and $BE$ respectively. Prove that $AB = AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the perpendicularity conditions, we have\n$$\n\\overrightarrow{AE} \\cdot \\overrightarrow{DF} = 0 \\Leftrightarrow \\overrightarrow{AE} \\cdot (\\overrightarrow{AF} - \\overrightarrow{AD}) = 0,\n$$\n$$\n\\overrightarrow{AF} \\cdot \\overrightarrow{BE} = 0 \\Leftrightarrow \\overrightarrow{AF} \\cdot (\\overrightarrow{AE} - \\overrightarrow{AB}) = 0,\n$$\nwhence\n$$\n\\overrightarrow{AE} \\cdot \\overrightarrow{AF} = \\overrightarrow{AE} \\cdot \\overrightarrow{AD}, \\quad \\overrightarrow{AE} \\cdot \\overrightarrow{AF} = \\overrightarrow{AF} \\cdot \\overrightarrow{AB},\n$$\n\nand thus $\\overrightarrow{AE} \\cdot \\overrightarrow{AD} = \\overrightarrow{AF} \\cdot \\overrightarrow{AB}$. Equivalently, we have\n$$\n(\\overrightarrow{AD} + \\overrightarrow{DE}) \\cdot \\overrightarrow{AD} = (\\overrightarrow{AB} + \\overrightarrow{BF}) \\cdot \\overrightarrow{AB} \\Leftrightarrow AD^2 + \\overrightarrow{AD} \\cdot \\overrightarrow{DE} = AB^2 + \\overrightarrow{AB} \\cdot \\overrightarrow{BF}.\n$$\nBut $\\angle ADE = \\angle ABF = 90^\\circ$, and hence $\\overrightarrow{AD} \\cdot \\overrightarrow{DE} = \\overrightarrow{AB} \\cdot \\overrightarrow{BF} = 0$, that is $AB = AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$ and let $S$ be a finite set containing more than $n+1$ elements. Consider the collection of all sets $\\mathcal{A}$ of subsets of $S$ satisfying the following two conditions:\n(a) Each member of $\\mathcal{A}$ contains at least $n$ elements of $S$; and\n(b) Each element of $S$ is contained in at least $n$ members of $\\mathcal{A}$.\nDetermine $\\max_{\\mathcal{A}} \\min_{\\mathcal{B}} |\\mathcal{B}|$, as $\\mathcal{B}$ runs through all subsets of $\\mathcal{A}$ whose members cover $S$, and $\\mathcal{A}$ runs through the above collection.",
    "options": [],
    "answer": "|S| - n",
    "solution": "The required number is $m = |S| - n$. We begin by showing that any set $\\mathcal{A}$ of subsets of $S$ satisfying the two conditions in the statement has a subcover of cardinality at most $m$.\n\nThis is clear if $S$ is a member of $\\mathcal{A}$.\n\nAssume henceforth that $\\mathcal{A}$ does not contain $S$. If some member $A$ of $\\mathcal{A}$ has more than $n$ elements, for each element of $S \\setminus A$ choose a containing member of $\\mathcal{A}$. The latter along with $A$ form a subcover of $\\mathcal{A}$ of cardinality $|S \\setminus A| + 1 \\le |S| - (n+1) + 1 = m$.\n\nAssume henceforth that each member of $\\mathcal{A}$ has exactly $n$ elements. Fix a member $A$ of $\\mathcal{A}$.\n\nIf some member $B$ of $\\mathcal{A}$ contains more than one element of $S \\setminus A$, for each element of $S \\setminus (A \\cup B)$ choose a containing member of $\\mathcal{A}$. The latter along with $A$ and $B$ form a subcover of $\\mathcal{A}$ of cardinality $|S \\setminus (A \\cup B)| + 2 = |S \\setminus A| - |(S \\setminus A) \\cap B| + 2 \\le |S \\setminus A| = m$.\n\nFinally, if no member of $\\mathcal{A}$ contains more than one element of $S \\setminus A$, write $S \\setminus A = \\{x_1, x_2, \\dots, x_m\\}$, choose a member $A_1$ of $\\mathcal{A}$ containing $x_1$ and notice that $A \\setminus A_1$ is a singleton set, say $A \\setminus A_1 = \\{x\\}$. Since $x_2$ is contained in at least $n$ members of $\\mathcal{A}$, each of which contains (exactly) $n-1$ elements of $A$, we may choose a member $A_2$ of $\\mathcal{A}$ containing both $x$ and $x_2$ (recall that $n \\ge 2$). If $n \\ge 3$, continue choosing members $A_i$ of $\\mathcal{A}$ containing $x_i$, $i = 3, \\dots, n$, to form an $m$-element subcover of $\\mathcal{A}$ consisting of $A_1, A_2, \\dots, A_m$.\n\nTo complete the proof, we produce a set of subsets of $S$ satisfying the two conditions in the statement, no subcover of which has less than $m$ members. To this end, write $S = \\{1, 2, \\dots, m+n\\}$, $m \\ge 2$, and let $S_1, S_2, \\dots, S_n$ be the $(n-1)$-element subsets of the upper part $\\{m+1, m+2, \\dots, m+n\\}$ of $S$. The sets $S_{i,j} = S_j \\cup \\{i\\}$, $i = 1, 2, \\dots, m$, $j = 1, 2, \\dots, n$, satisfy both conditions in the statement and at least $m$ of them are needed to cover $S$. (The condition $m \\ge 2$ is required for an element in the upper part to lie in at least $n$ of these sets.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $X, Y, Z$ be interior points on the sides $BC, CA, AB$, respectively. Show that the magnified image of the triangle $XYZ$ under a homothety of factor $4$ from its centroid covers at least one of the vertices $A, B, C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the problem is of an affine nature, we may (and will) assume that the triangle $XYZ$ is equilateral. The triangle $ABC$ has at least one vertex angle, say at $A$, greater than or equal to $60^\\circ$, so $A$ is covered by the closed circumdisc $OYZ$, where $O$ is the center of the triangle $XYZ$. Since the latter is covered by the $4$-fold blow-up of the triangle $XYZ$ from $O$, the conclusion follows.\nSuppose, if possible, that none of the vertices $A, B, C$ is covered by the $4$-fold blow-up of the triangle $XYZ$ from its centroid. Then the distance of the point $A$ to the line $YZ$ is greater than the distance of the point $X$ to this line, so the area of the triangle $AYZ$ is greater than the area of the triangle $XYZ$. Similarly, the triangles $BZX$ and $CXY$ both have an area greater than that of the triangle $XYZ$, in contradiction with the well known fact that of the four triangles $AYZ, BZX, CXY, XYZ$, the latter has not the smallest area.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $x, y, z \\in (0,1)$ that satisfy simultaneously the conditions:\n$$\n\\begin{cases} \n(x^2 + y^2)\\sqrt{1-z^2} \\geq z \\\\ \n(y^2 + z^2)\\sqrt{1-x^2} \\geq x \\\\ \n(z^2 + x^2)\\sqrt{1-y^2} \\geq y \n\\end{cases}\n$$",
    "options": [],
    "answer": "x = y = z = sqrt(2)/2",
    "solution": "The first inequality is equivalent to\n$$\n\\frac{z^2}{x^2 + y^2} \\leq z\\sqrt{1-z^2}.\n$$\nSince\n$$\nz\\sqrt{1-z^2} = \\sqrt{z^2(1-z^2)} \\leq \\frac{z^2 + 1 - z^2}{2} = \\frac{1}{2},\n$$\nit follows that $\\frac{z^2}{x^2 + y^2} \\leq 2$, and therefore $x^2 + y^2 \\geq 2z^2$. Writing the other two similar inequalities and adding them together yields $2(x^2 + y^2 + z^2) \\geq 2(x^2 + y^2 + z^2)$. Consequently, equality must hold in all the inequalities above, hence $x = y = z = \\frac{\\sqrt{2}}{2}$. Clearly this triple satisfies all the requirements.\nAs $x, y, z \\in (0,1)$, we can write $x = \\sin \\alpha$, $y = \\sin \\beta$, $z = \\sin \\gamma$, with $\\alpha, \\beta, \\gamma \\in (0, \\pi/2)$. The first inequality becomes\n$$\n\\sin^2 \\alpha + \\sin^2 \\beta \\geq \\tan \\gamma \\geq 2 \\sin^2 \\gamma,\n$$\nbecause it reduces to $\\sin \\gamma(2 \\sin \\gamma \\cos \\gamma - 1) = \\sin \\gamma(\\sin 2\\gamma - 1) \\leq 0$. Adding this inequality with the two similar ones obtained from the second and third condition, we get $2 \\sum \\sin^2 \\alpha \\geq 2 \\sum \\sin^2 \\alpha$. Consequently, equality must hold in all the preceding inequalities, therefore $\\alpha = \\beta = \\gamma = \\pi/4$, which translates into $x = y = z = \\sqrt{2}/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number in the open interval $(0, 1)$, let $n$ be a positive integer and let $f_n: \\mathbb{R} \\to \\mathbb{R}$, $f_n(x) = x + x^2/n$. Show that\n\n$$\n\\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2} < \\underbrace{(f_n \\circ \\dots \\circ f_n)}_{n} (a) < \\frac{an + a^2}{(1-a)n + a}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_k = \\underbrace{(f_n \\circ \\cdots \\circ f_n)}_{k}(a)$, $k \\in \\mathbb{N}$, and notice that\n$$\n1/a_{k+1} = 1/a_k - 1/(a_k + n), \\quad k \\in \\mathbb{N},\n$$\nto deduce that $1/a_n = 1/a - \\sum_{k=0}^{n-1} 1/(a_k + n)$, so\n$$\n1/a - n/(a+n) < 1/a_n < 1/a - n/(a_n + n), \\quad (*)\n$$\n$a_n < \\frac{an + a^2}{(1-a)n + a}$.\nPlugged into the rightmost expression in $(*)$, this upper bound yields the required lower bound,\n$$\na_n > \\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that all positive integers less than $n$ and coprime to $n$ are powers of primes.",
    "options": [],
    "answer": "2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42, 60",
    "solution": "Notice that $6$ is the first index $k$ such that $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$. Now, if $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$ for some index $k \\ge 6$, then (by Bertrand-Tchebysheff) $p_1p_2 \\cdots p_{k-1} > p_{k-1}^2p_k > 2p_{k-1} \\cdot 2p_k > p_kp_{k+1}$, so $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$ for all indices $k \\ge 6$.\nConsequently, $m \\le 5$, $r = p_m \\le p_5 = 11$, $q \\le p_4 = 7$, and $n < qr \\le p_4p_5 = 7 \\cdot 11 = 77$. Examination of the integers less than $77$ quickly yields the required numbers: $2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42, 60$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $D \\in (BC)$, $E \\in (AD)$ be mobile points. The circumcircle of triangle $CDE$ meets the median from $C$ of the triangle $ABC$ at $F$. Prove that the circumcenter of triangle $AEF$ lies on a fixed line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The quadrilateral $EFDC$ is cyclic, hence $\\angle FED \\equiv \\angle FCD$. Let $P$ be the reflection of point $C$ in the midpoint of the segment line $[AB]$; clearly $AP \\parallel BC$ and $\\angle FCD \\equiv \\angle FPA$.\n\nIt follows that $\\angle FPA \\equiv \\angle FED$, which means that the quadrilateral $FEAP$ is cyclic. This shows that the circumcenter $AEF$ lies on the perpendicular bisector of the segment line $[AP]$ (which is a fixed line).\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 6$ be an integer. We have at our disposal $n$ colors. We color each of the unit squares of an $n \\times n$ board with one of the $n$ colors.\n\na) Prove that, for any such coloring, there exists a path of a chess knight from the bottom-left to the upper-right corner, that does not use all the colors.\n\nb) Prove that, if we reduce the number of colors to $\\lfloor 2n/3 \\rfloor + 2$, then the statement from a) is true for infinitely many values of $n$ and it is false also for infinitely many values of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We associate coordinates to each unit square, the square in the bottom-left corner having coordinates $(1, 1)$, while the one in the upper-right corner has coordinates $(n, n)$. Notice that one can get from $(k, \\ell)$ to $(k+3, \\ell+3)$ in two moves: $(k, \\ell) \\to (k+2, \\ell+1) \\to (k+3, \\ell+3)$.\n\n• $n \\equiv 1 \\pmod{3}$. Using moves similar to the ones described above, the path $(1, 1) \\to \\dots \\to (n, n)$ passes through $\\frac{2n+1}{3} < \\lfloor 2n/3 \\rfloor + 2 < n$ unit squares, hence it does not use all the colors, whether there are $n$ colors, or only $\\lfloor 2n/3 \\rfloor + 2$ colors.\n\n• $n \\equiv 0 \\pmod{3}$. The path that starts with\n$$\n(1, 1) \\to (2, 3) \\to (3, 5) \\to (5, 4) \\to (6, 6) \\to \\dots \\to (n, n)\n$$\nand continues with moves similar to the ones described above, will pass through exactly\n$$\n\\frac{2n+3}{3} < \\lfloor 2n/3 \\rfloor + 2 \\le n\n$$\nsquares, hence it will not pass through squares of all possible colors, and this for both the situations, a) and b);\n\n• $n \\equiv 2 \\pmod{3}$. The path that starts with\n$$\n(1, 1) \\rightarrow (2, 3) \\rightarrow (4, 2) \\rightarrow (3, 4) \\rightarrow (5, 5) \\rightarrow \\dots \\rightarrow (n, n)\n$$\nand continues with moves similar to the ones described above, will pass through exactly\n$$\n\\frac{2n+5}{3} = \\lfloor 2n/3 \\rfloor + 2 < n\n$$\nsquares, hence it does not pass through squares of all possible colors, under the conditions from a).\n\nIt remains to exhibit a coloring with $\\lfloor 2n/3 \\rfloor + 2$ colors such that any path $(1, 1) \\to (n, n)$ contains squares of all the colors.\n\nIt is easy to see that there is no conflict in this coloring because the squares that use the first $(N-1)/2$ colors and the ones that use the last $(N-1)/2$ colors are separated by the diagonal $\\{(m, n - m + 1) \\mid 1 \\le m \\le n\\}$. Finally, we color with $(N + 1)/2$ all the remaining squares.\n\n|8| | |5|5|7|\n|7| | | |5|6|\n|6| | |5| |6|\n|5|3|3|5|5| |\n|4| |3|3|5|5|\n|3|3|2|3| | |\n|2| |2|3| | |\n|1|1|3|3| | |\n| |1|2|3|4|5|6|7|8|\n\n**Example 2.** (given in the contest by Tudor Plopeanu) Consider the sequence $(a_m)_{m \\ge 1}$ given by: $1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, \\dots$. Color row number $k$, from left to right, with $a_k, a_{k+1}, \\dots, a_{n+k-1}$. It is easy to check the upper-right corner has color $a_{2n-1} = N = \\lfloor 2n/3 \\rfloor + 2$. From a square with color $a_j$, the knight can jump to a square having one of the colors $a_{j-3}, a_{j-1}, a_{j+1}, a_{j+3}$. By the way the sequence is constructed, $a_{j-3}, a_{j-1}, a_{j+1}, a_{j+3} \\in \\{a_j - 1, a_j + 1\\}$, so again, in its path from color 1 to color $N$, the knight can not possibly skip a color.\n\n|8|4|5|4|5|6|5|6|7|\n|7|3|4|5|4|5|6|5|6|\n|6|4|3|4|5|4|5|6|5|\n|5|3|4|3|4|5|4|5|6|\n|4|2|3|4|3|4|5|4|5|\n|3|3|2|3|4|3|4|5|4|\n|2|2|3|2|3|4|3|4|5|\n|1|1|2|3|2|3|4|3|4|\n| |1|2|3|4|5|6|7|8|\n\n**Example 3.** Let $n = 3j + 2$, $j \\in \\mathbb{N}^*$. Color the bottom-left corner with 1, and use color $k$ for all the squares to which the knight can get in $k - 1$ moves, but not less. We prove that the upper-right corner has the color $N = \\lfloor 2n/3 \\rfloor + 2 = 2j + 3$. Once we have shown this, it is clear that, again, the knight can not skip a color, so any path has to contain squares of all the $N$ colors. For each square, the color number indicates the length of the shortest path (number of squares contained), from the square in the bottom-left corner to the given square. We prove that the shortest path to the upper-right corner contains exactly $N$ squares. We have already seen an example of a path containing $N$ squares. To prove that this is the shortest one possible, notice that each move changes the sum of coordinates by 1 or by 3, so in order to get from the bottom-left corner (having the sum of coordinates 2) to the upper-right corner (sum of coordinates $2n = 6j + 4$), one needs at least $2j + 1$ moves. But if we color the board in a chessboard pattern, the knight changes color each time it moves; the bottom-left and the upper-right corners are of the same color, so any path between the two of them will consist of an even number of moves, therefore the minimum number of moves is at least $2j + 2$; the upper-right corner will have the color $2j + 3 = N$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if the sides of a triangle are directly proportional to $4$, $5$ and $6$, then its largest angle is twice its smallest angle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four digit numbers $abcd$ which are equal to the sum of all the positive integers starting with $ab$ and finishing with $cd$.",
    "options": [],
    "answer": "1353, 1863, 3388, 3591",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi toate numerele prime $a, b, c$ şi numerele naturale nenule $k$ care satisfac egalitatea\n$$\na^{2}+b^{2}+16 c^{2}=9 k^{2}+1\n$$",
    "options": [],
    "answer": "All solutions are exactly those where two of a, b, c are equal to three:\n- If a = b = 3, then c = 2 and k = 3.\n- If c = 3 and one of a or b equals 3, then the remaining prime is 17 with k = 7, or 37 with k = 13. Concretely: (a, b, c, k) ∈ {(3,17,3,7), (17,3,3,7), (3,37,3,13), (37,3,3,13)}.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAz $ABCD$ paralelogramma átlói az $O$ pontban metszik egymást. A $DAC$ és $DBC$ szögek szögfelezői a $T$ pontban metszik egymást. Tudjuk, hogy $\\overrightarrow{TD} + \\overrightarrow{TC} = \\overrightarrow{TO}$. Határozd meg az $ABT$ háromszög szögeinek mértékét!",
    "options": [],
    "answer": "60°, 60°, 60°",
    "solution": "Solution:\n\nDin ipoteză rezultă că $DOCT$ este paralelogram.\n\nDin $AO \\parallel DT$ deducem $\\angle DTA \\equiv \\angle OAT \\equiv \\angle DAT$, deci $DA = DT$.\n\nAstfel $DA = DT = OC$; analog $BC = CT = OD$, de unde $BD = AC$. Astfel $ABCD$ este dreptunghi, $AOTD$ este romb, triunghiul $AOD$ este echilateral şi triunghiul $ABT$ este echilateral, deci are unghiurile de $60^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $41$ divides $\\overline{abcde}$, then $41$ divides $\\overline{eabcd}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38178,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that the equations $x^4 + y^4 + z^4 = t^4$ and $x^4 + y^4 + z^4 = 2t^4$ (with unknowns $x$, $y$, $z$, $t$) have solutions in $\\mathbb{N}^*$.\n\na) Find the smallest natural number $k$ such that the equation\n$$\nx^4 + y^4 + z^4 = k t^4\n$$\nhas no solutions in $\\mathbb{N}^*$.\n\nb) Prove that for every $n \\in \\mathbb{N}$, $n \\ge 3$ there exists $N \\in \\mathbb{N}$ such that $N$ can be written as the sum of two, three, four, ..., $n$ fourth powers of positive integers.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B \\in \\mathcal{M}_3(\\mathbb{Z})$ be such that $AB = BA$ and $\\det A = \\det B = 0$. Prove that $\\det(A^3 + B^3)$ is the sum of two perfect cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive primes $p$ and $q$ such that $p^3 + p = q^2 + q$.",
    "options": [],
    "answer": "(p, q) = (3, 5)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a$, $b$, $c$ numere reale pozitive astfel încât $a + b + c = 3$. Determinaţi valoarea minimă a expresiei\n$$\nA = \\frac{2 - a^{3}}{a} + \\frac{2 - b^{3}}{b} + \\frac{2 - c^{3}}{c}\n$$",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\in \\mathbb{N}$ and define the sets\n$$A_k = \\bigcup_{i=0}^{k+1} \\{(m, n) \\in \\mathbb{N} \\times \\mathbb{N} \\mid m \\ge 2i,\\ n \\ge 2(k-i) + 1\\},$$\n$$B_k = \\bigcup_{i=-1}^{k} \\{(m, n) \\in \\mathbb{N} \\times \\mathbb{N} \\mid m \\ge 2i + 1,\\ n \\ge 2(k-i)\\}.$$ \nProve that\n$$A_k \\cup B_k = \\{(m, n) \\in \\mathbb{N} \\times \\mathbb{N} \\mid m + n \\ge 2k + 1\\},$$\n$$A_k \\cap B_k = \\{(m, n) \\in \\mathbb{N} \\times \\mathbb{N} \\mid m + n \\ge 2k + 2\\}.$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive primes $p$ such that the system of equations $x^2(y+z) + y^2(z+x) + z^2(x+y) = 3xyz = p^2$ has positive integer solutions.",
    "options": [],
    "answer": "no primes",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider numbers $7^{n+7}, 11^{n+11}, 15^{n+15}, \\dots, 4007^{n+4007}$, where $n$ is a positive integer. Prove that at least ten non-nil differences of the given numbers are divisible by $500$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic. Dreptele $\\ell_{1}$ şi $\\ell_{2}$ sunt perpendiculare pe dreapta $AB$ în punctele $A$, respectiv $B$. Perpendicularele duse din mijlocul $M$ al segmentului $[AB]$ pe dreptele $AC$ şi $BC$ intersectează $\\ell_{1}$ şi $\\ell_{2}$ în punctele $E$ şi respectiv $F$.\nDacă $D$ este punctul de intersecţie a dreptelor $EF$ şi $MC$, arătaţi că $\\angle ADB \\equiv \\angle EMF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdottak az $m \\geq 2$ és $n \\geq 3$ természetes számok. Igazold, hogy létezik $m$ darab különböző $a_{1}, a_{2}, a_{3}, \\ldots, a_{m}$ természetes szám, amelyek mind oszthatók $n-1$-gyel és\n$$\n\\frac{1}{n}=\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\frac{1}{a_{3}}-\\ldots+(-1)^{m-1} \\frac{1}{a_{m}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoluţia 1.\nDacă $(a_{n})_{n \\geq 1}$ este o progresie geometrică având raţia $q=n-1$, atunci\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{1 / a_{1}+1 / q \\cdot(-1)^{p-1} 1 / a_{p}}{1+1 / q}, \\forall p \\in \\mathbb{N}^{*}\n$$\nRelaţia precedentă se scrie\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{n-1}{n a_{1}}+\\frac{(-1)^{p-1}}{n a_{p}}\n$$\nPentru $a_{1}=n-1, m=p+1$ şi $a_{m}=n a_{p}$ obţinem numerele cerute.\n\n\nSoluţia 2.\nDemonstrăm prin inducţie după $m$.\n\nPentru $m=2$ concluzia se verifică:\n$$\n\\frac{1}{n}=\\frac{1}{n-1}-\\frac{1}{n(n-1)}\n$$\nAstfel, $\\frac{1}{n}=\\frac{1}{n-1}\\left(\\frac{1}{b_{1}}-\\frac{1}{b_{2}}\\right)$, cu $b_{1}<b_{2}-1$. Apoi, dacă\n$$\n\\frac{1}{n}=\\frac{1}{(n-1) b_{1}}-\\frac{1}{(n-1) b_{2}}+\\ldots+(-1)^{m-1} \\frac{1}{(n-1) b_{m}}\n$$\ncu $1 \\leq b_{1}<\\ldots<b_{m}$ numere întregi şi $b_{m-1}<b_{m}-1$, atunci punem $b_{i}'=b_{i}$, $i=\\overline{1, m-1}$, $b_{m}'=b_{m}-1$, $b_{m+1}'=b_{m}(b_{m}-1)$ şi obţinem\n$$\n\\frac{1}{n}=\\frac{1}{(n-1) b_{1}'}-\\frac{1}{(n-1) b_{2}'}+\\ldots+(-1)^{m-1} \\frac{1}{(n-1) b_{m}'}+(-1)^{m} \\frac{1}{(n-1) b_{m+1}'}\n$$\ncu $1 \\leq b_{1}'<\\ldots<b_{m}'<b_{m+1}'$ numere întregi şi $b_{m}'<b_{m+1}'-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38187,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ has $m(\\widehat{BCA}) = 30^\\circ$ and there exists a point $D \\in AC$ such that $m(\\widehat{DBC}) = 10^\\circ$ and $AC = BD$. Find $m(\\widehat{CAB})$.\n\nMarius Stănean",
    "options": [],
    "answer": "100°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a, b, c \\in [\\frac{1}{\\sqrt{6}}, 6]$, then\n$$\n\\frac{4}{a+3b} + \\frac{4}{b+3c} + \\frac{4}{c+3a} \\ge \\frac{3}{a+2b} + \\frac{3}{b+2c} + \\frac{3}{c+2a}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the real sequences $(a_n)_{n \\ge 1}$, $(b_n)_{n \\ge 1}$, $(c_n)_{n \\ge 1}$. Prove that if the sequence\n$$\np_n(x) = (x - a_n)(x - b_n)(x - c_n)\n$$\nconverges for infinitely many values of $x$, then it converges for every $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO \"formă L\" este oricare din următoarele patru piese, fiecare constând din trei pătrăţele unitate:\n\n![](attached_image_1.png)\n\nSe dau: o tablă $5 \\times 5$, constând din $25$ de pătrăţele unitate, un număr natural nenul $k \\leq 25$ şi o colecţie nelimitată de \"forme L\".\n\nDoi jucători, $A$ şi $B$, joacă următorul joc: începând cu $A$, ei marchează, alternativ, câte un pătrăţel care nu era marcat anterior, până când pe tablă sunt $k$ pătrăţele marcate.\n\nO aşezare a unor \"forme L\" pe pătrăţelele rămase nemarcate pe tablă se numeşte bună dacă fiecare piesă acoperă exact trei pătrăţele unitate nemarcate şi oricare două piese nu se suprapun.\n\nJucătorul $B$ câştigă dacă orice aşezare bună a unor \"forme L\" lasă neacoperite cel puţin trei pătrăţele unitate nemarcate. Determinaţi valoarea minimă a lui $k$ pentru care $B$ are strategie câştigătoare.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that the polynomial\n$$(X^3 + X + 1)^n + 5X^2 + 30X + 5$$\nis irreducible in $\\mathbb{Z}[X]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $13$ has a multiple whose decimal writing contains only two digits $a$, $b$ and $a$ appears in exactly $3$ places, which are not contiguous.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. Find all real $x > 0$ such that\n$$\n\\left[\\frac{1}{x}\\right] + \\left[\\frac{1}{x^n}\\right] = n.\n$$",
    "options": [],
    "answer": "x ∈ (n^{-1/n}, (n-1)^{-1/n}]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$. Prove that\n$$\n\\frac{a}{(n+1)a+b+c} + \\frac{b}{a+(n+1)b+c} + \\frac{c}{a+b+(n+1)c} \\le \\frac{1}{2}\\sqrt{\\frac{3}{n}},\n$$\nfor every $a, b, c \\in (0, \\infty)$. Find the cases of equality.",
    "options": [],
    "answer": "Equality holds if and only if n = 3 and a = b = c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $x \\in (0, \\frac{\\pi}{3})$, then $\\sin 2x < \\frac{2}{3x^2 - x^6}$.\nDumitru Săvulescu, Lucian Tuțescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$. Find all real numbers $x_1, x_2, \\dots, x_n$ such that $x_k \\in [0, 2k)$, for every $k \\in \\{1, 2, \\dots, n\\}$, and\n$$ \\frac{1}{2-x_1} + \\frac{2}{4-x_2} + \\dots + \\frac{n}{2n-x_n} = \\frac{x_1}{1} + \\frac{x_2}{2} + \\dots + \\frac{x_n}{n}. $$",
    "options": [],
    "answer": "x_k = k for all k = 1, 2, ..., n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a ring. Prove that the following statements are equivalent:\na) For every elements $a$ and $b$ from $A$ there exists two odd integers $p$ and $q$ such that $a^2 + b^2 = (pa + qb)^2$;\nb) The ring $A$ is commutative and $1 + 1 = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find four consecutive numbers, knowing that they are obtained by adding a prime number (not necessary the same) to $91$, $109$, $124$ and $148$.",
    "options": [],
    "answer": "150, 151, 152, 153",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38199,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that the equation $p + x + y + z = xyz$ has positive integer solutions for every positive prime $p$.\n\nb) Find the positive integer solutions of the equation $3 + x + y + z = xyz$.",
    "options": [],
    "answer": "a) For any prime p, for example (x, y, z) = (2, p+3, 1) and all its permutations are solutions.\nb) All positive integer solutions are the permutations of (1, 2, 6).",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the altitude $[VO]$ of the regular pyramid $VABCD$ and $N$ be the midpoint of the segment $[BM]$. Let $P \\in (AO)$ be such that $AP = 3 \\cdot PO$. Prove that if we denote $d(a, b)$ the distance between the straight lines $a, b$, then\n$$\n\\frac{d(PN, VD)}{d(PN, AB)} = \\frac{5}{3}.\n$$\nDevian Augustin",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f : \\mathbb{R} \\to \\mathbb{R}$, which have primitives and there exists a primitive $F$ of $f$, such that $F \\circ F$ is also a primitive of $f$.",
    "options": [],
    "answer": "f(x) ≡ 0 or f(x) ≡ 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integrable functions $f : [0, 1] \\to \\mathbb{R}$ with the property: for every $x \\in [0, 1]$,\n$$\n\\int_0^x f(t) dt = (f(x))^{2015} + f(x).\n$$",
    "options": [],
    "answer": "f(x) ≡ 0 on [0, 1]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is isosceles with basis $[BC]$ and points $E, G \\in (AC)$, $F \\in (BC)$, $H \\in (AB)$ are such that $(BE$ is the bisector of angle $ABC$, $(EF$ is the bisector of angle $BEC$, $(FG$ is the bisector of angle $EFC$ and $(GH$ is the bisector of angle $EGF$. It is known that $GH \\parallel BC$.\n\na) Prove that $GF \\parallel AB$.\n\nb) Find the measures of the angles of triangle $ABC$.",
    "options": [],
    "answer": "∠A = 90°, ∠B = 45°, ∠C = 45°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38204,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $T, A, B, C$ are non-coplanar and the straight lines $a, b, c$ are the parallels from the centroid $G$ of the triangle $ABC$ to the straight lines $TA, TB$, respectively $TC$. Let $\\{A'\\} = a \\cap (TBC)$, $\\{B'\\} = b \\cap (TAC)$ and $\\{C'\\} = c \\cap (TAB)$.\n\na) Prove that $(A'B'C') \\parallel (ABC)$.\n\nb) Compute the distance between the planes $(A'B'C')$ and $(ABC)$ as a function of the distance $a$ from $T$ to the plane $(ABC)$.",
    "options": [],
    "answer": "a/3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle such that $AB \\neq AC$. Let $M$ be the midpoint of $[BC]$, $H$ be the orthocenter of $ABC$, $O_1$ be the midpoint of $[AH]$ and $O_2$ be the circumcenter of $BCH$. Prove that $O_1AMO_2$ is a parallelogram.\n\nJBMO ShortList 2014",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the following well-known facts:\n(1) The reflection of $H$ in line $BC$ lies on the circumcircle of triangle $ABC$.\n(2) $AH = 2MO$, where $O$ is the circumcenter of $ABC$.\nFrom (1) it follows that the reflection of the circumcenter of $ABC$ is the circumcenter of $HBC$, hence $O_2$ is the reflection of $O$ in $BC$. We have\n$$\nMO_2 = MO \\stackrel{(2)}{=} \\frac{AH}{2} = AO_1\n$$\nand, as $AO_1 \\parallel MO_2$, $O_1AMO_2$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a sequence of positive integers $a_1, a_2, a_3, \\dots$ such that $a_m$ and $a_n$ are coprime if and only if the indices $m$ and $n$ are one unit apart?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The answer is in the affirmative. The idea is to consider a sequence of pairwise distinct primes $p_1, p_2, p_3, \\dots$, cover the positive integers by a sequence of finite non-empty sets $I_n$ such that $I_m$ and $I_n$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_n = \\prod_{i \\in I_n} p_i$, $n = 1, 2, 3, \\dots$. For instance, we may take\n$$\nI_n = \\{2n - 4k - 1: k = 0, 1, \\dots, \\lfloor(n-1)/2\\rfloor\\} \\cup \\{2n - 4k - 2: k = 1, 2, \\dots, \\lfloor n/2 \\rfloor - 1\\} \\cup \\{2n\\},\n$$\n$n = 1, 2, 3, \\dots$, where the set in the middle is understood to be empty for $n = 1, 2, 3$. It is readily checked that $I_n$ and $I_{n+1}$ are disjoint for every index $n$, and, if the indices $m$ and $n$ are at least two units apart, then $I_m$ and $I_n$ both contain $m + n - |m-n| - e$, where $e = 0$ if $m$ and $n$ have opposite parities, and $e = 1$ otherwise. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38207,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of the parallelogram $ABCD$ meet at $O$. The bisectors of the angles $DAC$ and $DBC$ meet at $T$. It is known that $TD + TC = TO$. Find the measures of the angles of triangle $ABT$.\nClaudiu Militaru",
    "options": [],
    "answer": "Each angle is 60 degrees",
    "solution": "The hypothesis shows that $DOCT$ is a parallelogram. From $AO \\parallel DT$ follows $\\angle DTA = \\angle OAT = \\angle DAT$, hence $DA = DT$.\n\nThis leads to $DA = DT = OC$; in the same way $BC = CT = OD$, therefore $BD = AC$. So $ABCD$ is a rectangle, $AOTD$ is a rhombus, triangle $AOD$ is equilateral and triangle $ABT$ is equilateral, hence its angles have measures $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the set $\\{\\sqrt{1}, \\sqrt{2}, \\sqrt{3}, \\dots, \\sqrt{2015}\\}$ does not contain a non-constant arithmetic sequence of length 45.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $m, n, p \\in \\mathbb{N}^*$ and $\\sqrt{m}, \\sqrt{n}, \\sqrt{p}$ are in arithmetic progression, then\n$$\np + m + 2\\sqrt{pm} = 4n,\n$$\nhence $\\sqrt{pm}$ is a rational number. It follows that $m = a^2d$, $p = c^2d$, $n = b^2d$, where $a, b, c, d \\in \\mathbb{N}$ and $a + c = 2b$.\n\nTherefore, if one could choose 45 numbers in arithmetic sequence, they would have the form $a_1\\sqrt{d}, a_2\\sqrt{d}, \\dots, a_{45}\\sqrt{d}$, where $a_1, a_2, \\dots, a_{45}, d \\in \\mathbb{N}^*$ and $a_1, a_2, \\dots, a_{45}$ is an arithmetic sequence, as well.\n\nIn this case, the largest of these numbers would be $\\sqrt{45^2d} \\ge \\sqrt{2025}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to [0, 1]$ be a function with the property that for every $y \\in [0, 1]$ and every $\\varepsilon > 0$ there exists $x \\in [0, 1]$ so that $|f(x) - y| < \\varepsilon$.\na) Prove that if $f$ is continuous on $[0, 1]$, then $f$ is surjective.\nb) Give an example of a function $f$ with the given property which is not surjective.",
    "options": [],
    "answer": "a) f is surjective. b) Define f on the unit interval by f equals the identity on rational points and equals zero on irrational points.",
    "solution": "a) Suppose a continuous function $f : [0, 1] \\to [0, 1]$ has the given property and take $y \\in [0, 1]$. From the hypothesis, there exists a sequence $(x_n)_{n \\ge 1}$, included in $[0, 1]$, so that $|f(x_n) - y| < 1/n$, $\\forall n \\ge 1$. The sequence $(x_n)_{n \\ge 1}$ is bounded, so it has a convergent subsequence $(x_{i_n})_{n \\to \\infty}$. Denote $x := \\lim_{n \\to \\infty} x_{i_n} \\in [0, 1]$. Then $(f(x_{i_n}))_n \\to f(x)$ and $|f(x_{i_n}) - y| < 1/i_n$, $\\forall n \\ge 1$, hence $|f(x)-y| \\le 0$, therefore $f(x) = y$. This proves that $f$ is surjective.\n\nb) Consider $f : [0, 1] \\to [0, 1]$, $f(x) = \\begin{cases} x, & x \\in [0, 1] \\cap \\mathbb{Q} \\\\ 0, & x \\in [0, 1] \\setminus \\mathbb{Q} \\end{cases}$.\nClearly $f([0, 1]) = [0, 1] \\cap \\mathbb{Q}$, so $f$ is not surjective. Also $|f(y) - y| = 0, \\forall y \\in [0, 1] \\cap \\mathbb{Q}$. Finally, if $y \\in [0, 1] \\setminus \\mathbb{Q}$ and $\\varepsilon > 0$, then there exists $x \\in [0, 1] \\cap \\mathbb{Q}$ such that $|x - y| < \\varepsilon$, that is $|f(x) - y| < \\varepsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the integers $n \\ge 2$ such that $a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$ for a unique $a$ in $\\mathbb{Z}_n$.",
    "options": [],
    "answer": "7",
    "solution": "We show that $7$ is the sole integer satisfying the required conditions. If $a \\in \\mathbb{Z}_n$ and $a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$, then $(1-a)^2 - (1-a) + 2 = a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$, so uniqueness forces $a = 1-a$, i.e., $2a = 1$. In particular, $2$ is invertible in $\\mathbb{Z}_n$ and $a = 2^{-1}$. Hence $2^{-2} - 2^{-1} + 2 = 0$, or $1 - 2 + 8 = 0$, i.e., $7 = 0$ in $\\mathbb{Z}_n$. Consequently, $n$ divides $7$, and since the latter is prime, it follows that $n = 7$. It is readily checked that $a = 2^{-1} = 4$ is the unique element of $\\mathbb{Z}_7$ satisfying $a^2 - a + 2 = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all two-digit positive integers $\\overline{ab}$, with $a < b$, which are equal to the sum of the integers from $a$ to $b$ ($a$ and $b$ included).",
    "options": [],
    "answer": "27 and 15",
    "solution": "The solutions are $27$ and $15$.\n\nWe start by noticing that $\\overline{ab} = a + (a+1) + \\dots + b \\le 1 + 2 + \\dots + 9 = 45$, which yields $a \\le 4$.\n\nIf $a = 4$, then $\\overline{ab} \\ge 45 = 1 + 2 + \\dots + 9$, leading to $a = 1$ – impossible.\n\nIf $a = 3$, then $3+4+\\dots+(b-1)+b = \\overline{3b} = 30+b$, leading to $1+2+\\dots+(b-1) = 33$, whence $(b-1) \\cdot b = 66$. This equation has no solution $b \\in \\{4, 5, \\dots, 9\\}$.\n\nIf $a = 2$, then $2+3+\\dots+(b-1)+b = \\overline{2b} = 20+b$, leading to $1+2+\\dots+(b-1) = 21$, whence $(b-1) \\cdot b = 42$. This equality is true for $b = 7$, so $\\overline{ab} = 27$ is a solution.\n\nIf $a = 1$, then $1+2+\\dots+(b-1)+b = \\overline{1b} = 10+b$, whence $1+2+\\dots+(b-1) = 10$, i.e. $(b-1) \\cdot b = 20$. This equality is true for $b = 5$, so $\\overline{ab} = 15$ is the other solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, a blackboard has written on it the numbers $11$ and $13$. Each minute an extra number appears on the blackboard, equaling the sum of two numbers already written on the blackboard. Prove that:\n\na) the number $86$ can not appear on the blackboard;\n\nb) it is possible for $2015$ to appear on the blackboard at some point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Each number appearing on the blackboard is of the form $11a + 13b$, with $a, b \\in \\mathbb{N}^*$. If $86$ appears on the board, then there exist $a, b \\in \\mathbb{N}^*$ such that $86 = 11a + 13b$, whence $b \\le 6$. Then $13b \\in \\{13, 26, 39, 52, 65, 78\\}$, therefore $11a = 86 - 13b \\in \\{73, 60, 47, 34, 21, 8\\}$. Since none of these numbers is divisible by $11$, $86$ can not be written on the blackboard.\n\nb) $2015 = 11 \\cdot 182 + 13$. The number $2015$ can appear on the blackboard after $182$ minutes, in the following way: $13 + 11 = 24 \\xrightarrow{+11} 13 + 2 \\cdot 11 = 35 \\xrightarrow{+11} 13 + 3 \\cdot 11 = 46 \\xrightarrow{+11} \\dots \\xrightarrow{+11} 13 + 182 \\cdot 11 = 2015$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all differentiable functions $f : \\mathbb{R} \\to \\mathbb{R}$ for which:\ni) $f'(x) = 0$, for all $x \\in \\mathbb{Z}$;\nii) if $f'(x) = 0$ for some $x \\in \\mathbb{R}$, then $f(x) = 0$.",
    "options": [],
    "answer": "The zero function: f(x) = 0 for all real x.",
    "solution": "Obviously, one solution is the zero function. Suppose now that $f : \\mathbb{R} \\to \\mathbb{R}$ is another solution. Then $f$ is continuous and $f(x) = f'(x) = 0$, for all $x \\in \\mathbb{Z}$. Consider $x_0 \\in \\mathbb{R} \\setminus \\mathbb{Z}$ such that $f(x_0) \\neq 0$, and let $k = \\lfloor x_0 \\rfloor \\in \\mathbb{Z}$.\n\nAccording to Weierstrass theorem, $f$ is bounded on $[k, k + 1]$ and one can find $x_1, x_2 \\in [k, k + 1]$ such that\n$$\nf(x_1) \\leq f(x) \\leq f(x_2), \\forall x \\in [k, k + 1].\n$$\nIf $f(x_0) > 0$, then $f(x_2) > 0$, hence $x_2 \\in (k, k+1)$ is a local maximum for $f$, so $f'(x_2) = 0$, according to Fermat's interior extremum theorem. But then, from (ii), $f(x_2) = 0$, which is a contradiction.\n\nIn a similar way we discard the case $f(x_0) < 0$.\n\nWe conclude that the zero function is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(R, +, \\cdot)$ be a unit ring such that for all $x \\in R$ one can find $e_1, e_2 \\in R$, such that $e_1^2 = e_1$, $e_2^2 = e_2$ and $x = e_1e_2$. Show that:\na) $1$ is the only invertible element in $R$;\nb) $x^2 = x$, for all $x \\in R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $x$ be an invertible element of $R$, let $e_1$ and $e_2$ be idempotent elements of $R$ such that $x = e_1e_2$, and notice that $1 - e_1 = (1 - e_1) \\cdot 1 = (1 - e_1)e_1e_2x^{-1} = (e_1 - e_1^2)e_2x^{-1} = 0$, so $e_1 = 1$ and $x^2 = e_2^2 = e_2 = x$. Since $x$ is invertible, it follows that $x = 1$.\n\nb) Begin by noticing that there are no non-zero nilpotent elements in $R$. Indeed, if $x$ is an element of $R$ and $x^k = 0$ for some integer $k > 1$, then $1 - x^k = (1-x)(1+x+\\dots+x^{k-1})$, so $1-x$ is invertible and $x=0$ by (a).\n\nNext, we show that every idempotent element of $R$ is central (it commutes with every element of $R$). Let $e$ be idempotent, let $x$ be any element of $R$ and write\n$$\n(ex - exe)^2 = exex - exexe - exeex + exeexe = exex - exexe - exex + exexe = 0,\n$$\nto deduce that $ex = exe$, by the preceding. Similarly, $xe = exe$, so $ex = exe = xe$.\n\nFinally, let $x$ be an element of $R$, and write $x = e_1e_2$, where $e_1$ and $e_2$ are idempotent. By the preceding, $e_1e_2 = e_2e_1$, so $x^2 = (e_1e_2)^2 = e_1^2e_2^2 = e_1e_2 = x$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a$, $b$, $c$ are the length of the sides of a triangle, then\n$$\n\\sqrt{\\frac{a}{-a+b+c}} + \\sqrt{\\frac{b}{a-b+c}} + \\sqrt{\\frac{c}{a+b-c}} \\ge 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a$, $b$, $c$ are the length of the sides of a triangle, numbers $-a+b+c$, $a-b+c$ and $a+b-c$ are positive. The GM-HM inequality yields\n$$\n\\sqrt{\\frac{a}{-a+b+c}} = \\sqrt{1 \\cdot \\frac{a}{-a+b+c}} \\ge \\frac{2}{1+\\frac{-a+b+c}{a}} = \\frac{2a}{b+c}\n$$\nand the other two similar relations. So, it is enough to prove that\n$$\n\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} \\geq \\frac{3}{2}.\n$$\nDenote $b+c=x$, $a+c=y$, $a+b=z$. Then\n$$\na = \\frac{-x + y + z}{2}, \\quad b = \\frac{x - y + z}{2}, \\quad c = \\frac{x + y - z}{2},\n$$\nand the last inequality is equivalent to\n$$\n\\frac{-x + y + z}{2x} + \\frac{x - y + z}{2y} + \\frac{x + y - z}{2z} \\ge \\frac{3}{2},\n$$\nthat is\n$$\n\\frac{y}{x} + \\frac{z}{x} - 1 + \\frac{x}{y} + \\frac{z}{y} - 1 + \\frac{x}{z} + \\frac{y}{z} - 1 \\geq 3,\n$$\nwhich is true because $u + 1/u \\ge 2$ for every $u > 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four families have two children each and all eight children are born after the year 1989. All four youngest siblings are born in the same year, and the sum of the digits of the year is equal to the product of the non zero digits. The differences of the ages between the siblings of each family is a perfect square. Find the birth years of the four eldest siblings in each family, given that their ages are all different.",
    "options": [],
    "answer": "2012, 2009, 2004, 1997",
    "solution": "There exists no digit $a$ for which $1 + 9 + 9 + a = 1 \\cdot 9 \\cdot 9 \\cdot a$, hence the youngest are born after $2000$, say in $200a$ or $201a$. In the first case, the condition $2 + a = 2a$ gives $a = 2$. The second case leads to $a = 3$. The years are $2002$ or $2013$.\n\nThe age differences between the siblings of each family are at least $1$, $4$, $9$ and $16$. Since $2002 - 1989 = 13 < 16$, we conclude that the youngest are all born in $2013$, whence the eldest are born in $2012$, $2009$, $2004$ and $1997$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of non-zero complex numbers with equal absolute values, for which\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + 1 = 0.\n$$",
    "options": [],
    "answer": "All triples that are permutations of t, t, minus t where t is any nonzero complex number.",
    "solution": "Observe that $\\overline{\\left(\\frac{a}{b}\\right)} = \\frac{b}{a}$, hence, taking conjugates yields\n$$\n\\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c} + 1 = 0.\n$$\nClearing denominators and adding up gives\n$$\na^2 b + b^2 c + c^2 a + ab^2 + bc^2 + ca^2 + 2abc = 0,\n$$\nwhich factors as $(a+b)(b+c)(c+a) = 0$, therefore two of the numbers sum up to zero. If, say, we have, $a+b=0$, the initial equality implies $c=a$ or $c=b$, hence the requested triples are $(a, a, -a)$, with $a \\in \\mathbb{C}^*$ (and the respective permutations).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $m(\\angle A) = 90^\\circ$, $AC < AB$, consider on the rays $BA$ and $AC$ points $E$ and $D$ respectively, such that $A \\in (BE)$, $C \\in (AD)$, $AE = AC$ and $AD = AB$. Denote by $M$ and $N$ the midpoints of $[BC]$ and $[DE]$ respectively, and let $\\{R\\} = EC \\cap BD$. Show that $MN = RA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The hypothesis implies that triangles $\\triangle ACE$ and $\\triangle ABD$ are right angled and isosceles, so the triangle $\\triangle RBE$ is also right angled and isosceles, that is $m(\\widehat{RBE}) = m(\\widehat{REB}) = 45^\\circ$. By the equality of triangles $\\triangle ABC$ and $\\triangle ADE$ (CC) we get $BC = DE$.\n\n![](attached_image_1.png)\n\nIn the triangles $BRC$, $ABC$, $RDE$, $ADE$ lines $RM$, $AM$, $RN$, $AN$ are medians corresponding to right angles, so $RM = \\frac{1}{2}BC$, $AM = \\frac{1}{2}BC$, $RN = \\frac{1}{2}DE$, $AN = \\frac{1}{2}DE$, implying $RM = AM = RN = AN$, that is the quadrilateral $AMRN$ is a rhombus. (*)\n\nIn the right angled triangle $ABC$, as $AM$ is a median, the triangle $\\triangle MAC$ is isosceles, so $m(\\widehat{MAC}) = m(\\widehat{MCA})$. In the same way in $\\triangle ADE$, $m(\\widehat{NAD}) = m(\\widehat{NDA})$. This gives $m(\\widehat{MAN}) = m(\\widehat{MAC}) + m(\\widehat{NAC}) = m(\\widehat{MCA}) + m(\\widehat{NDA}) = m(\\widehat{MCA}) + m(\\widehat{ABC}) = 90^\\circ$. (**)\n\nBy (*) and (**), the quadrilateral $AMRN$ is a square, and, as a conclusion $MN = RA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Oldd meg az $x^{2}-x+\\hat{2}=\\hat{0}$ egyenletet, ha $x \\in \\mathbb{Z}_{7}$.\n\nb. Határozd meg azokat az $n \\geq 2$ természetes számokat, amelyekre az $x^{2}-x+\\hat{2}=\\hat{0}$ egyenletnek egyetlen megoldása van $x \\in \\mathbb{Z}_{n}$-ben!\n\nProblem:\n\na. Rezolvaţi ecuaţia $x^{2}-x+\\hat{2}=\\hat{0}$, $x \\in \\mathbb{Z}_{7}$.\n\nb. Determinaţi numerele naturale $n \\geq 2$, pentru care ecuaţia $x^{2}-x+\\hat{2}=\\hat{0}$, $x \\in \\mathbb{Z}_{n}$, are soluţie unică.",
    "options": [],
    "answer": "Part a: x = 4 in integers modulo 7. Part b: n = 7.",
    "solution": "Solution:\n\na. Cum $4$ şi $7$ sunt coprime, iar $(\\mathbb{Z}_{7},+, \\cdot)$ este corp, ecuaţia dată este echivalentă cu $\\hat{4} x^{2}-\\hat{4} x+\\hat{1}=\\hat{0}$, adică, $(\\hat{2} x-\\hat{1})^{2}=\\hat{0}$, deci $\\hat{2} x=\\hat{1}$, de unde $x=\\hat{4}$.\n\nb. Fie $n \\geq 2$ un număr natural pentru care ecuaţia dată are soluţie unică şi fie $a \\in \\mathbb{Z}_{n}$ soluţia respectivă. Atunci $(\\hat{1}-a)^{2}-(\\hat{1}-a)+\\hat{2}=a^{2}-a+\\hat{2}=\\hat{0}$, deci $\\hat{1}-a$ este soluţie a ecuaţiei date.\n\nPrin urmare, $a=\\hat{1}-a$, deci $\\hat{2} a=\\hat{1}$. În particular, $\\hat{2}$ este inversabil în inelul $\\mathbb{Z}_{n}$ şi $a=\\hat{2}^{-1}$. Rezultă că $\\hat{2}^{-2}-\\hat{2}^{-1}+\\hat{2}=\\hat{0}$, de unde $\\hat{1}-\\hat{2}+\\hat{8}=\\hat{0}$, i.e., $\\hat{7}=\\hat{0}$. Prin urmare, $n$ este un divizor al lui $7$, deci $n=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $O$ be its circumcentre, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose a point $Y$ such that $AONY$ is a parallelogram. Since the lines $AD$ and $ON$ are parallel, this point lies on the line $AD$ (see Fig. 1). We prove that the triangles $AOY$ and $AXD$ are similar. Since the line $AN$ bisects the segment $OY$ the conclusion follows.\nIt is well known that the internal bisectrix $AD$ of the angle $ABC$ is also the internal bisectrix of the angle $OAA'$. Next, the corresponding sides of the triangles $OND$ and $ADX$ are parallel, so these triangles are similar.\n\n$AY/AO = AD/AX$. Along with the equality of the angles $OAY$ and $DAX$, this proves the required similarity of the triangles $AOY$ and $AXD$.\n\nLet $P, Q, R, S$ be the points of intersection of the pairs of lines $AM$ and $OD$, $OA$ and $XD$, $AN$ and $OD$, and $AD$ and $QR$, respectively (see Fig. 2). Since the angles $MAN$ and $PAR$ are the same, we show that $AD$ is the internal bisectrix of the latter.\n![](attached_image_1.png)\n\nApply Menelaus' theorem to both triangles $DMP$ and $DRS$ and the transversal $AOQ$ to write\n$$\n\\frac{AM}{AP} \\cdot \\frac{OP}{OD} \\cdot \\frac{QD}{QM} = 1 \\quad \\text{and} \\quad \\frac{AD}{AS} \\cdot \\frac{OR}{OD} \\cdot \\frac{QS}{QR} = 1,\n$$\nrespectively. Since $OD$ and $AX$ are parallel and $DM = MX$, it follows that $AM = MP$. In the triangle $AQD$, the line $ON$ is parallel to $AD$, so $R$ lies on the $Q$-median, and therefore $AS = SD$. Hence $MS$ and $PD$ are parallel, so $QM/QD = QS/QR$.\n\nCombining the obtained relations we get\n$$\n\\frac{OP}{OD} = \\frac{QM}{QD} \\cdot \\frac{AP}{AM} = \\frac{QS}{QR} \\cdot \\frac{AD}{AS} = \\frac{OD}{OR},\n$$\nor $OD^2 = OP \\cdot OR$. Thus, $OA^2 = OP \\cdot OR$. This shows that the triangles $OAR$ and $OPA$ are similar, and $\\angle OAR = \\angle OPA$. Finally, by $OA = OD$ we obtain $\\angle RAD = \\angle OAD - \\angle OAR = \\angle ODA - \\angle OPA = \\angle DAP$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for every integer $n \\ge 2$,\n$$\n\\sum_{k=2}^{n} \\frac{1}{\\sqrt[k]{(2k)!}} \\ge \\frac{n-1}{2n+2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use induction on $n$. In case $n = 2$ the relation is an equality. We notice now that when going from $n-1$ to $n$ the right member increases with\n$$\n\\frac{n-1}{2n+2} - \\frac{n-2}{2n} = \\frac{1}{n(n+1)},\n$$\nso it is enough to prove that\n$$\n\\frac{1}{\\sqrt[n]{(2n)!}} \\ge \\frac{1}{n(n+1)}.\n$$\nThis is obtained by multiplying the inequalities $k(2n-k+1) \\le n(n+1)$, for $k = 1, 2, \\dots, n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdott az $f:[0,1] \\rightarrow[0,1]$ függvény, amelyre bármely $y \\in[0,1]$ és bármely $\\varepsilon>0$ esetén létezik $x \\in[0,1]$ úgy, hogy $|f(x)-y|<\\varepsilon$.\na) Igazold, hogy ha $f$ folytonos a $[0,1]$ intervallumon, akkor $f$ szürjektív!\nb) Adj példát olyan $f$ függvényre, ami teljesíti a feladatbeli feltételt és nem szürjektív!\n\nProblem:\n\nFie $f:[0,1] \\rightarrow[0,1]$ o funcţie cu proprietatea că pentru oricare $y \\in[0,1]$ şi oricare $\\varepsilon>0$ există $x \\in[0,1]$ astfel încât $|f(x)-y|<\\varepsilon$.\na) Demonstraţi că dacă $f$ este continuă pe $[0,1]$ atunci $f$ este surjectivă.\nb) Daţi un exemplu de funcţie $f$ cu proprietatea din enunţ, care să nu fie surjectivă.",
    "options": [],
    "answer": "a) If f is continuous on the unit interval, then f is surjective onto the unit interval. b) Example: define f on the unit interval by f(x) = x for rational x and f(x) = 0 for irrational x; its image is dense but not all real numbers in the interval, so it is not surjective.",
    "solution": "Solution:\n\na)\nConsiderăm o funcţie continuă $f:[0,1] \\rightarrow[0,1]$ având proprietatea din enunţ. Fie $y \\in[0,1]$. Din ipoteză deducem că există un şir $(x_{n})_{n \\geq 1}$, cu termenii în $[0,1]$, astfel încât $|f(x_{n})-y|<1/n$, $\\forall n \\geq 1$.\n\nŞirul $(x_{n})_{n \\geq 1}$ este mărginit, deci admite un subşir convergent $(x_{i_{n}})_{n \\geq 1}$, cu $x:=\\lim_{n \\rightarrow \\infty} x_{i_{n}} \\in[0,1]$.\n\nPrin trecere la limită în inegalitatea $|f(x_{i_{n}})-y|<1/i_{n}$, $\\forall n \\geq 1$, obţinem (pe baza continuităţii lui $f$ în punctul $x$) $|f(x)-y| \\leq 0$, deci $f(x)=y$. Rezultă că $f$ este surjectivă.\n\nb)\nDefinim funcţia $f:[0,1] \\rightarrow[0,1]$,\n$$\nf(x)=\\begin{cases}\nx, & x \\in[0,1] \\cap \\mathbb{Q} \\\\\n0, & x \\in[0,1] \\setminus \\mathbb{Q}\n\\end{cases}.\n$$\n$f([0,1])=[0,1] \\cap \\mathbb{Q}$, deci $f$ nu este surjectivă.\n\nAvem $|f(y)-y|=0, \\forall y \\in[0,1] \\cap \\mathbb{Q}$. Pentru $y \\in[0,1] \\setminus \\mathbb{Q}$ şi $\\varepsilon>0$, există $x \\in[0,1] \\cap \\mathbb{Q}$ astfel ca $|x-y|<\\varepsilon$, sau $|f(x)-y|<\\varepsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given non-negative real numbers $a$, $b$, $c$ such that $ab + bc + ca + 2abc = 1$, show that $\\sqrt{a} + \\sqrt{b} + \\sqrt{c} \\ge 2$ and determine the cases of equality.",
    "options": [],
    "answer": "The minimum possible value of the sum of square roots is two, attained precisely when one of the numbers is zero and the other two are equal to one (up to permutation).",
    "solution": "The condition in the statement is equivalent to $\\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c} = 2$. If $t$ is a non-negative real number, then $\\sqrt{t} \\ge \\frac{2t}{1+t} = 2 - \\frac{2}{1+t}$, and equality holds if and only if $t$ is either $0$ or $1$. Consequently,\n$$\n\\begin{aligned}\n\\sqrt{a} + \\sqrt{b} + \\sqrt{c} &\\ge 2 - \\frac{2}{1+a} + 2 - \\frac{2}{1+b} + 2 - \\frac{2}{1+c} \\\\\n&= 6 - 2\\left(\\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c}\\right) = 2.\n\\end{aligned}\n$$\n\nBy the preceding, equality holds if and only if one of the numbers $a$, $b$, $c$ is $0$, and the other two are both equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$, and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$, respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$, respectively. Prove that the triangle $OMN$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nWe show that $\\angle OMN \\equiv \\angle ONM$. To this end, let the line $IJ$ cross the segments $AD$ and $BC$ at $P$ and $Q$, respectively, and consider the position of $J$ relative to the line $AC$, to write $\\angle OMN \\equiv \\angle AJP \\pm \\angle JAM \\equiv \\angle AJP \\pm (\\angle JAD \\mp \\angle CAD) \\equiv \\angle AJP + \\angle JAD - \\angle CAD \\equiv \\angle AJP + \\angle BAJ - \\angle CAD$. Similarly, $\\angle ONM \\equiv \\angle BIQ + \\angle ABI - \\angle CBD$. Since the quadrangle $ABCD$ is cyclic, $\\angle CAD \\equiv \\angle CBD$ and $\\angle ACB \\equiv \\angle ADB$. The latter implies that $\\angle AIB \\equiv \\angle AJB$, so the quadrangle $ABIJ$ is cyclic. Consequently, $\\angle AJP \\equiv \\angle ABI$ and $\\angle BAJ \\equiv \\angle BIQ$, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are positive odd integers $m_1 < m_2 < \\dots$ and positive integers $n_1 < n_2 < \\dots$ such that $m_k$ and $n_k$ are relatively prime, and $m_k^4 - 2n_k^4$ is a perfect square for each index $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$. Write $\\ell^2 = m^4 - 2n^4$, so $\\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$, and $\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$. Multiply the latter by $\\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2 = 2\\ell^4$ to get $(\\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2)(\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2) = -2 \\cdot (2\\ell mn)^4$;\n\nthat is, $(\\ell^4 - 8m^4n^4)^2 - (m^4 + 2n^4)^4 = -2 \\cdot (2\\ell mn)^4$. Letting $m' = m^4 + 2n^4$ and $n' = 2\\ell mn$, clearly $m' > m$, $m'$ is odd, $n' > n$, the difference $m'^4 - 2n'^4$ is a perfect square, and it is readily checked that $m'$ and $n'$ are relatively prime. The conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers expressible, for every integer $n \\ge 3$, in the form $\\frac{(a_1 + 1)(a_2 + 1)\\cdots(a_n + 1) - 1}{a_1a_2\\cdots a_n}$, where $a_1, a_2, \\ldots, a_n$ are pairwise distinct positive integers.",
    "options": [],
    "answer": "2 and 3",
    "solution": "The integers greater than $3$ are ruled out by noticing that if $a_1, a_2, a_3$ are pairwise distinct positive integers, then\n$$\n\\begin{aligned}\n\\frac{(a_1+1)(a_2+1)(a_3+1)-1}{a_1 a_2 a_3} &= 1 + \\frac{1}{a_1} + \\frac{1}{a_2} + \\frac{1}{a_3} + \\frac{1}{a_1 a_2} + \\frac{1}{a_1 a_3} + \\frac{1}{a_2 a_3} \\\\\n&\\le 1 + 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{1 \\cdot 2} + \\frac{1}{1 \\cdot 3} + \\frac{1}{2 \\cdot 3} = 3 + \\frac{5}{6}\n\\end{aligned}\n$$\nTo complete the proof, fix an integer $n \\ge 3$, consider an integer $a \\ge 3$, and let $a_1 = a - 2$, let $a_k = a^{2^{k-2}}$, $k = 2, \\dots, n-1$, and let $a_n = a^{2^{n-2}} - 2$. Then $1 \\le a_1 < a_2 < \\dots < a_n$, and\n$$\n\\begin{aligned}\n\\frac{(a_1+1)(a_2+1)\\cdots(a_n+1)-1}{a_1 a_2 \\cdots a_n} &= \\frac{((a-1) \\prod_{k=2}^{n-1} (a^{2^{k-2}}+1)) (a^{2^{n-2}}-1)-1}{(a-2) a^{1+2+\\cdots+2^{n-3}} (a^{2^{n-2}}-2)} \\\\\n&= \\frac{\\left(a^{2^{n-2}}-1\\right)^2-1}{(a-2) a^{2^{n-2}-1} (a^{2^{n-2}}-2)} = \\frac{a^{2^{n-2}} (a^{2^{n-2}}-2)}{(a-2) a^{2^{n-2}-1} (a^{2^{n-2}}-2)} \\\\\n&= \\frac{a}{a-2} = 1 + \\frac{2}{a-2},\n\\end{aligned}\n$$\nwhich is integral if and only if $a = 3$ or $a = 4$. The former shows that $3$ satisfies the required condition, and the latter shows that so does $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\gamma$ and $\\gamma'$ cross one another at points $A$ and $B$. The tangent to $\\gamma'$ at $A$ meets $\\gamma$ again at $C$, the tangent to $\\gamma$ at $A$ meets $\\gamma'$ again at $C'$, and the line $CC'$ separates the points $A$ and $B$. Let $\\Gamma$ be the circle externally tangent to $\\gamma$, externally tangent to $\\gamma'$, tangent to the line $CC'$, and lying on the same side of $CC'$ as $B$. Show that the circles $\\gamma$ and $\\gamma'$ intercept equal segments on one of the tangents to $\\Gamma$ through $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Invert with respect to a circle centred at $A$ and denote by $X^*$ the image of a point $X \\neq A$ under this inversion. The circles $\\gamma$ and $\\gamma'$ invert into straight lines $B^*C^*$ and $B^*C'^*$, and the tangents at $A$ into lines through $A$, parallel to $B^*C^*$ and $B^*C'^*$. The line $CC'$ inverts into the circle $AC^*C'^*$, and 'CC' separating $A$ and $B'$ is equivalent to '$B'$ lying inside circle $AC^*C'^*$. So $AC^*B^*C'^*$ is a parallelogram, obtuse-angled at $A$ and $B^*$. Draw the line through $A$, parallel to $C^*C'^*$, meeting the lines $B^*C^*$ and $B^*C'^*$ at $D$ and $D'$, respectively. Then $A$, $C^*$ and $C'^*$ are the midpoints of the sides of the triangle $B^*DD'$, and the line $DD'$ is the inverse of the line through $A$ on which $\\gamma$ and $\\gamma'$ intercept equal segments. The circle $AC^*C'^*$ is the nine-point circle of the triangle $B^*DD'$; by Feuerbach's theorem, it touches the incircle of that triangle. Since this incircle is the inverse of the circle $\\Gamma$ in the original configuration, touching $\\gamma$ and $\\gamma'$ externally and the line $CC'$, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 0}$ and $(b_n)_{n \\ge 0}$ be sequences of real numbers such that $a_0 > 1/2$, $a_{n+1} \\ge a_n$, and $b_{n+1} = a_n(b_n + b_{n+2})$, for all non-negative integers $n$. Show that the sequence $(b_n)_{n \\ge 0}$ is bounded.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Use the relation in the statement and the fact that no $a_k$ is $1/2$, to write\n$$\nb_{k+2}^2 - b_k^2 = \\frac{(b_{k+1} - b_k)^2 - (b_{k+2} - b_{k+1})^2}{2a_k - 1}, \\quad k \\in \\mathbb{N},\n$$\nfix an integer $n > 2$, and sum over $0, 1, \\ldots, n-2$, to obtain\n$$\nb_n^2 + b_{n-1}^2 - b_1^2 - b_0^2 = \\frac{(b_1 - b_0)^2}{2a_0 - 1} - \\sum_{k=0}^{n-3} \\left( \\frac{1}{2a_k - 1} - \\frac{1}{2a_{k+1} - 1} \\right) (b_{k+2} - b_{k+1})^2 - \\frac{(b_n - b_{n-1})^2}{2a_{n-2} - 1} \\le \\frac{(b_1 - b_0)^2}{2a_0 - 1},\n$$\nsince the $a_k$ form an increasing sequence of real numbers greater than $1/2$. Consequently, $b_n^2 \\le b_n^2 + b_{n-1}^2 \\le b_0^2 + b_1^2 + (b_1 - b_0)^2/(2a_0 - 1)$, and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$, $y$, $z$ which verify the equalities\n$$\ny = \\frac{x^3 + 12x}{3x^2 + 4}, \\quad z = \\frac{y^3 + 12y}{3y^2 + 4}, \\quad x = \\frac{z^3 + 12z}{3z^2 + 4}.\n$$",
    "options": [],
    "answer": "(-2, -2, -2), (0, 0, 0), (2, 2, 2)",
    "solution": "We have $y + 2 = \\frac{(x+2)^3}{3x^2+4}$, $y - 2 = \\frac{(x-2)^3}{3x^2+4}$ and the analogous relations.\nIf $y = 2$, then $x = z = 2$. If $y \\ne 2$, then $\\frac{y+2}{y-2} = \\left(\\frac{x+2}{x-2}\\right)^3$, $\\frac{z+2}{z-2} = \\left(\\frac{y+2}{y-2}\\right)^3$ and $\\frac{x+2}{x-2} = \\left(\\frac{z+2}{z-2}\\right)^3$. We obtain $\\frac{x+2}{x-2} = \\left(\\frac{x+2}{x-2}\\right)^{27}$, which leads to $\\frac{x+2}{x-2} \\in \\{-1, 0, 1\\}$. We find two more solutions, $x = y = z = -2$ and $x = y = z = 0$.\nLet $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = \\frac{x^3 + 12x}{3x^2 + 4}$. We have $f(a) - f(b) = \\frac{3(ab - 4)^2 + 4(a-b)^2}{(3a^2 + 4)(3b^2 + 4)} \\cdot (a-b)$, so $f$ is strictly increasing.\n\nOur equalities become $y = f(x)$, $z = f(y)$, $x = f(z)$. Without loss of generality, we may assume that $x = \\min\\{x, y, z\\}$. Then $f(x) \\le f(y)$, so $y \\le z$, which leads to $f(y) \\le f(z)$ or $z \\le x$. Hence $x \\le y \\le z \\le x$, so $x = y = z$. As the equation $f(x) = x$ has three solutions, namely $x = -2$, $x = 0$ and $x = 2$, we find that $(x, y, z) \\in \\{(-2, -2, -2); (0, 0, 0); (2, 2, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38230,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $k$ and $n$ are positive integers, and $k \\le n$, let $M(n, k)$ denote the least common multiple of the numbers $n, n-1, \\dots, n-k+1$. Let $f(n)$ be the largest positive integer $k \\le n$ such that $M(n, 1) < M(n, 2) < \\dots < M(n, k)$. Prove that:\n\na) $f(n) < 3\\sqrt{n}$, for all positive integers $n$;\n\nb) if $N$ is a positive integer, then $f(n) > N$ for all but finitely many positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Clearly, $f(1) = 1$. Notice that\n$$\nM(n, k + 1) = \\operatorname{lcm}(M(n, k), n - k), \\quad 1 \\le k < n. \\quad (*)\n$$\nThus, $M(n, k) \\le M(n, k+1)$, and equality holds if and only if $n-k$ divides $M(n, k)$. If $m > 1$, then $M(m^2, 2) = m^2(m^2 - 1) = (m^2 - m)(m^2 + m)$, so $M(m^2, 2)$ is divisible by $m^2 - m$, and hence so is $M(m^2, m)$. By $(*)$, $M(m^2, m) = M(m^2, m+1)$, so $f(m^2) \\le m$ for all positive integers $m$. With reference again to $(*)$, if $M(n, k) = M(n, k+1)$, then $M(n + \\ell, k + \\ell) = M(n + \\ell, k + \\ell + 1)$, so $f(n + \\ell) \\le f(n) + \\ell$ for all positive integers $\\ell$ and $n$. Finally, if $m^2 \\le n < (m+1)^2$, then $n = m^2 + \\ell$ for some non-negative integer $\\ell \\le 2m$, so $f(n) = f(m^2 + \\ell) \\le f(m^2) + \\ell \\le m + \\ell \\le 3m \\le 3\\sqrt{n}$, where one of the last two inequalities must in fact be a strict inequality.\n\nb) We show that $f(n) > N$ for all integers $n > N! + N$. Refer again to $(*)$ to write\n$$\n\\begin{aligned}\nM(n, N + 1) &= \\operatorname{lcm}(M(n, N), n - N) = \\frac{M(n, N) \\cdot (n - N)}{\\operatorname{gcd}(M(n, N), n - N)} \\\\\n&\\ge \\frac{M(n, N) \\cdot (n - N)}{\\prod_{k=1}^{N} \\operatorname{gcd}(n - k + 1, n - N)} \\ge \\frac{M(n, N) \\cdot (n - N)}{\\prod_{k=1}^{N} (N - k + 1)} \\\\\n&= \\frac{M(n, N) \\cdot (n - N)}{N!}.\n\\end{aligned}\n$$\nConsequently, if $n > N! + N$, then $M(n, 1) < \\dots < M(n, N) < M(n, N + 1)$, so $f(n) > N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB \\ne AC$ and $H$ its orthocenter. Consider a point $D$ on the side $BC$. The circumcircles of triangles $ABD$ and $ACD$ meet again $AC$ and $AB$ in $E$ and $F$, respectively. Lines $BE$ and $CF$ meet in point $P$. Prove that $HP$ is parallel to $BC$ if and only if the line $AD$ contains the circumcenter of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We only show the proof in the case when $E \\in (AC)$, $F \\in (AB)$, the other cases being similar (the diagram on the left shows such a case).\nFrom $\\angle PBC + \\angle PCB = \\angle EAD + \\angle FAD = \\angle BAC$ follows $\\angle BPC = 180^\\circ - \\angle BAC = \\angle BHC$, hence $B, C, H, P$ are concyclic.\n\nTherefore $HP \\parallel BC$ is equivalent to $B, C, P, H$ being the vertices of an isosceles trapezoid, which translates into $\\angle HCB = \\angle PBC$. But $\\angle HCB = \\angle HAB$ and $\\angle PBC = \\angle DAE$, hence $\\angle HCB = \\angle PBC \\Leftrightarrow \\angle BAH = \\angle CAD$, which means that $AH$ and $AD$ are isogonals, i.e. $AD$ passes through the circumcenter of $ABC$.\n\n![](attached_image_1.png)\n\nAnother solution:\n\nAs above, $\\angle BPC = 180^\\circ - \\angle BAC = \\angle BHC$. This means that the reflections, $H'$ and $P'$, of $H$ and $P$, respectively, across $BC$ are on the circumcircle of $ABC$. Moreover, $\\angle BCF = \\angle BAD = \\angle BED$, hence $CDPE$ is cyclic. Then $\\angle CDP = \\angle AEB = \\angle ADB$, which means that $P' \\in AD$.\n\nNow $HP \\parallel BC \\Leftrightarrow H'P' \\parallel BC$, which is equivalent to $H'P' \\perp AH$, i.e. $\\angle AH'P' = 90^\\circ$, and the conclusion follows readily.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two integers $h \\ge 1$ and $p \\ge 2$, determine the minimum number of pairs of opponents an *hp*-member parliament may have, if in every partition of the parliament into *h* houses of *p* member each some house contains at least one pair of opponents.",
    "options": [],
    "answer": "(h - 1) · min(p, h/2 + 1) + 1",
    "solution": "Letting $N(h, p) = (h-1) \\cdot \\min(p, h/2+1)$, we now proceed to prove by induction on $h$ that if the number of edges of a graph on $hp$ vertices does not exceed $N(h, p)$, then the graph is $h$-partite on $p$-element classes. The base case $h=1$ is clear.\n\nNext, let $h \\ge 2$ and let $G = (V, E)$ be a graph on $hp$ vertices which has at most $N(h, p)$ edges. If necessary, add some extra edges to obtain $|E| = N(h, p)$.\n\nBegin by forming a $p$-element house $V_0$ of independent vertices $v_1, \\dots, v_p$ by the following $p$-step greedy algorithm: Start with the empty set, and at step $j$ choose a vertex $v_j$ of maximal degree from the set of vertices joined by an edge to no $v_i$, $i < j$, and different from any of these; this set is nonempty, for if each of the remaining $hp-j+1$ vertices were joined by an edge to some $v_i$, $i < j$, then $|E| \\ge hp-j+1 \\ge hp-p+1 > N(h, p)$ — a contradiction. Notice that $\\deg v_1 \\ge \\deg v_2 \\ge \\dots \\ge \\deg v_p$.\n\nLet $d = \\sum_{v \\in V_0} \\deg v$, so the subgraph $G'$ induced by the $p(h-1)$ vertices in $V \\setminus V_0$ has exactly $N(h, p) - d$ edges. If $d \\ge \\Delta N = N(h, p) - N(h-1, p)$, then $G'$ is $(h-1)$-partite on $p$-element classes by the induction hypothesis, and the conclusion follows.\n\nHenceforth, assume\n$$\nd < \\Delta N = \\begin{cases} p, & \\text{if } h \\ge 2p-1, \\\\ h, & \\text{if } h \\le 2p-2, \\end{cases} \\quad (*)\n$$\nand notice that $\\Delta N \\le h$ in either case, so $d \\le h-1$. Let $V'$ be the set of all vertices in $V \\setminus V_0$ joined by an edge to some vertex in $V_0$, and notice that $|V'| \\le d \\le h-1$, and $\\deg v \\le \\deg v_p$ for all vertices $v$ outside $V_0 \\cup V'$.\n\nIf $\\deg v_p = 0$, then the vertices outside $V_0 \\cup V'$ are all isolated. Since $|V'| \\le d \\le h-1$, each vertex of $V'$ may be included in a different $p$-element house (other than $V_0$) along with $p-1$ vertices outside $V_0 \\cup V'$ each, to obtain $|V'|$ more $p$-element houses. The remaining vertices, if any, are then arbitrarily split into $p$-element houses.\n\nFinally, we rule out the case $\\deg v_p \\ge 1$. Suppose, if possible, that $\\deg v_p \\ge 1$. Then $\\deg v_i \\ge 1$, $i = 1, \\dots, p$, and $d \\ge p$, so (*) yields $\\Delta N = h$, $h \\le 2p-2$, and $N(h,p) = (h-1)(h+2)/2$. Hence $p \\le d \\le h-1 \\le 2p-3$. The inequality $d \\le 2p-3$ forces $\\deg v_p = 1$, so $\\deg v \\le 1$ for all vertices $v$ outside $V_0 \\cup V'$, and\n$$\n\\sum_{v \\in V \\setminus (V_0 \\cup V')} \\deg v \\le hp - |V_0| = p(h-1).\n$$\nFurther on, split $V' = V_1 \\cup \\cdots \\cup V_p$, where $V_j$ is the set of all vertices joined by an edge to $v_j$, but to no $v_i$, $i < j$. Notice that $|V_i| \\le \\deg v_i$, and $\\deg v \\le \\deg v_i$ for all vertices $v$ in $V_i$. Consequently,\n$$\n\\begin{aligned}\n\\sum_{v \\in V_0 \\cup V'} \\deg v &= \\sum_{i=1}^{p} \\left( \\deg v_i + \\sum_{v \\in V_i} \\deg v \\right) \\le \\sum_{i=1}^{p} \\deg v_i (\\deg v_i + 1) \\\\\n&= \\sum_{i=1}^{p} (\\deg v_i - 1)^2 + 3d - p.\n\\end{aligned}\n$$\nSince $\\deg v_i \\ge 1$, $i = 1, \\dots, p$,\n$$\n\\sum_{i=1}^{p} (\\deg v_i - 1)^2 \\le \\left( \\sum_{i=1}^{p} (\\deg v_i - 1) \\right)^2 = (d-p)^2,\n$$\nso (recalling that $d \\le h-1$)\n$$\n\\begin{aligned}\n\\sum_{v \\in V_0 \\cup V'} \\deg v &\\le (d-p)^2 + 3d - p \\le (h-p-1)^2 + 3(h-1) - p \\\\\n&= (h-1)(h+2) + p(p-2h+1) = 2N(h,p) + p(p-2h+1).\n\\end{aligned}\n$$\nHence, by the preceding,\n$$\n\\begin{aligned}\n2N(h,p) = 2|E| &= \\sum_{v \\in V} \\deg v = \\sum_{v \\in V_0 \\cup V'} \\deg v + \\sum_{v \\in V \\setminus (V_0 \\cup V')} \\deg v \\\\\n&\\le 2N(h,p) + p(p-2h+1) + p(h-1) = 2N(h,p) + p(p-h) \\\\\n&< 2N(h,p),\n\\end{aligned}\n$$\nwhich is a contradiction. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 4$. Determine the sets $A = \\{a_1, a_2, \\dots, a_n\\} \\subset \\mathbb{N}$ that contain $2015$ and for which $|a_i - a_j|$ is a prime, for all distinct $i, j \\in \\{1, 2, \\dots, n\\}$.",
    "options": [],
    "answer": "{2008, 2010, 2013, 2015}; {2010, 2012, 2015, 2017}; {2013, 2015, 2018, 2020}; {2015, 2017, 2020, 2022}",
    "solution": "It is easy to see that $A$ can not contain more than two numbers of any parity. Combined with $n \\ge 4$, this forces $A$ to have exactly $4$ elements, two of each parity. The difference between the two even (odd) numbers must be $2$, therefore we can have two types of sets: $\\{2013, 2015, 2k, 2k + 2\\}$ and $\\{2015, 2017, 2k, 2k + 2\\}$.\n\nIn the first case, the differences $|(2k + 2) - 2013|$, $|2k - 2013|$, and $|2k - 2015|$ give different remainders upon division by $3$, hence one of these differences has to be $3$. Checking all the possibilities leads to the solutions $A = \\{2013, 2015, 2008, 2010\\}$ and $A = \\{2013, 2015, 2018, 2020\\}$.\n\nIn the second case, one of the differences $|(2k + 2) - 2015|$, $|2k - 2015|$, and $|2k - 2017|$ has to be $3$. Studying the cases we get the answers $\\{2015, 2017, 2010, 2012\\}$, $\\{2015, 2017, 2020, 2022\\}$.\n\nIn conclusion\n$$\n\\begin{aligned}\nA_1 &= \\{2008, 2010, 2013, 2015\\}, & A_2 &= \\{2010, 2012, 2015, 2017\\}, \\\\\nA_3 &= \\{2013, 2015, 2018, 2020\\}, & A_4 &= \\{2015, 2017, 2020, 2022\\}.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectors of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend the segment $AC$ beyond $C$ by a segment $CE$ congruent to the segment $CB$, so $C$ lies on the perpendicular bisector of the segment $BE$. Similarly, extend the segment $BD$ beyond $D$ by a segment $DF$ congruent to the segment $AD$, so $D$ lies on the perpendicular bisector of the segment $AF$. Since the triangles $ABC$ and $ABD$ have equal perimeters, the segments $AE$ and $BF$ are congruent.\n\n![](attached_image_1.png)\n\nNow let the perpendicular bisectors of the segments $AF$ and $BE$ meet at $Q$. Clearly, the segments $QA$ and $QF$, respectively $QE$ and $QB$, are congruent, and since so are the segments $AE$ and $BF$ by the preceding, the triangles $QAE$ and $QFB$ are congruent. Therefore, the angles $AQF$ and $BQE$ are congruent, and hence so are their halves; that is, the angles $AQD$ and $BQC$ are congruent, and consequently so are the angles $AQC$ and $BQD$.\n\nWe now show that, in fact, the points $P$ and $Q$ coincide, whence the conclusion. With reference again to the congruence of the triangles $QAE$ and $QFB$, the angles $QAE$ and $QFB$ are congruent, and since the latter is the reflection of the angle $QAD$ in the line $QD$, it follows that the angles $QAC = QAE$ and $QAD$ are congruent, so the line $AQ$ bisects the angle $CAD$. Similarly, the line $BQ$ bisects the angle $CBD$, and consequently the points $P$ and $Q$ coincide.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $k \\ge 2$, determine the largest number of divisors the binomial coefficient $\\binom{n}{k}$ may have in the range $n-k+1, \\dots, n$, as $n$ runs through the integers greater than or equal to $k$.",
    "options": [],
    "answer": "k-1",
    "solution": "The required maximum is $k-1$ and is achieved, for instance, at $n = k!$. To complete the proof, we now show that at least one of the $k$ numbers $\\frac{1}{n-j}\\binom{n}{k}$, $j = 0, 1, \\dots, k-1$, is not an integer. To this end, we exhibit a $\\mathbb{Z}$-linear combination of these numbers which is not an integer. For instance,\n$$\n\\sum_{j=0}^{k-1} (-1)^{k-j-1} \\binom{k-1}{j} \\cdot \\frac{1}{n-j} \\binom{n}{k} = \\frac{1}{k} \\sum_{j=0}^{k-1} \\prod_{i \\neq j} \\frac{n-i}{j-i} = \\frac{1}{k}\n$$\nis not an integer, since $k \\ge 2$. The leftmost equality above is easily proved by noticing that the polynomial\n$$\n\\sum_{j=0}^{k-1} \\prod_{i \\neq j} \\frac{X-i}{j-i}\n$$\nhas degree at most $k-1$, and takes on the value 1 at $k$ distinct points, namely, 0, 1, ..., $k-1$, so it is identically 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. If $\\sigma$ is a permutation of the first $n$ positive integers, let $S(\\sigma)$ be the set of all distinct sums of the form $\\sum_{i=k}^{\\ell} \\sigma(i)$, where $1 \\le k \\le \\ell \\le n$.\na) Exhibit a permutation $\\sigma$ of the first $n$ positive integers such that $|S(\\sigma)| \\ge \\lfloor(n+1)^2/4\\rfloor$.\nb) Show that $|S(\\sigma)| > n\\sqrt{n}/4\\sqrt{2}$ for all permutations $\\sigma$ of the first $n$ positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We show that the permutation $\\sigma$ of the first $n$ positive integers, defined by $\\sigma(i) = (i+1)/2$ for each positive odd $i \\le n$, and $\\sigma(i) = n - i/2 + 1$ for each positive even $i \\le n$, satisfies the required condition.\nMore precisely, we show that the $\\lfloor(n+1)^2/4\\rfloor$ sums of the form $\\sum_{i=k}^{\\ell} \\sigma(i)$, where $1 \\le k \\le \\ell \\le n$ and $k \\equiv \\ell \\pmod 2$, are pairwise distinct, so $|S(\\sigma)| \\ge \\lfloor(n+1)^2/4\\rfloor$.\nLet $1 \\le k \\le \\ell \\le n$ and $k \\equiv \\ell \\pmod 2$, and notice that $\\sigma(i) + \\sigma(i+1) = n+1$ for every positive odd $i \\le n$, so\n$$\n\\sum_{i=k}^{\\ell} \\sigma(i) = \\begin{cases} (\\ell-k)(n+1)/2 + \\sigma(\\ell) & \\text{if } k \\text{ is odd,} \\\\ \\sigma(k) + (\\ell-k)(n+1)/2 & \\text{if } k \\text{ is even.} \\end{cases}\n$$\nSince the absolute value of an integer of the form $\\sigma(i) - \\sigma(j)$ is less than $n$, the above formula shows that the assignment $(k, \\ell) \\mapsto \\sum_{i=k}^{\\ell} \\sigma(i)$ is indeed injective on the pairs in question.\n\nb) Let $\\sigma$ be a permutation of the first $n$ positive integers, and split $S(\\sigma)$ into $S_m(\\sigma) = S(\\sigma) \\cap [mn + 1, mn + n]$, where $m$ runs through the integers; of course, $S_m(\\sigma)$ is empty if $m$ is negative or $m > (n-1)/2$, and $|S_0(\\sigma)| \\ge n$.\nLeaving aside the trivial cases $n=1$ and $n=2$, we assume $n \\ge 3$ and prove that\n$$\n|S_m(\\sigma)| + |S_{m-1}(\\sigma)| > \\sqrt{2n}, \\quad (*)\n$$\nfor every non-negative integer $m \\le (n+1)/4$. The conclusion then follows by summing over this range:\n$$\n|S(\\sigma)| \\ge \\frac{1}{2} \\sum_{m=0}^{\\lfloor(n+1)/4\\rfloor} (|S_m(\\sigma)| + |S_{m-1}(\\sigma)|) > \\frac{1}{2} \\left\\lfloor \\frac{n+5}{4} \\right\\rfloor \\sqrt{2n} > \\frac{n\\sqrt{n}}{4\\sqrt{2}}.\n$$\nTo prove (*), fix a non-negative integer $m \\le (n+1)/4$. We will show that the Minkowski difference $S_m(\\sigma) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$ contains every positive integer less than or equal to $n$; alternatively, but equivalently, that it contains every $\\sigma(k)$. Then so does $(S_m(\\sigma) \\cup S_{m-1}(\\sigma)) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$, so $|S_m(\\sigma)| + |S_{m-1}(\\sigma)| = |S_m(\\sigma) \\cup S_{m-1}(\\sigma)| > \\sqrt{2n}$, for if $X$ is a finite set of numbers such that $\\{1, 2, \\dots, n\\} \\subseteq X - X$, then $|X| \\cdot (|X| - 1) + 1 \\ge |X - X| \\ge 2n + 1$, so $|X| \\ge (1 + \\sqrt{8n+1})/2 > \\sqrt{2n}$.\nFinally, we show that every $\\sigma(k)$ is a member of $S_m(\\sigma) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$. To this end, fix a positive integer $k \\le n$, and notice that at least one of the sums $\\sum_{i=1}^k \\sigma(i)$, $\\sum_{i=k}^n \\sigma(i)$ exceeds $n(n+1)/4 \\ge mn$. Let $\\sum_{i=k}^n \\sigma(i) > mn$ — the other case is easily dealt with dually —, and consider the smallest integer $\\ell \\ge k$ such that $\\sum_{i=k}^{\\ell} \\sigma(i) > mn$. With reference to this minimality, it is readily checked that the sums $\\sum_{i=k}^{\\ell} \\sigma(i)$ and $\\sum_{i=k+1}^{\\ell} \\sigma(i)$ belong to $S_m(\\sigma)$ and $S_m(\\sigma) \\cup S_{m-1}(\\sigma)$, respectively, so $\\sigma(k)$ is indeed a member of $S_m(\\sigma) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the center of the inscribed circle of triangle $ABC$, with $AB \\neq AC$. Let $M$ be the midpoint of side $BC$, and $D$ the projection of $I$ on $BC$. The circle with center $M$ and radius $MD$ intersects line $AI$ at $P$ and $Q$. Show that $\\angle BAC + \\angle PMQ = 180^\\circ$.\nLaurențiu Ploscaru",
    "options": [],
    "answer": "∠BAC + ∠PMQ = 180°",
    "solution": "Without loss of generality, we may assume that $AB < AC$ and $AP < AQ$. We shall prove that angles $BAC$ and $PMQ$ have parallel sides, which trivially implies the conclusion.\n\nFirst approach. Consider the tangency points $E \\in AC$ and $F \\in AB$ on the inscribed circle. Let $P'$ and $Q'$ be the intersection points of line $AI$ with $DE$ and $DF$, respectively. We intend to prove that $MP' \\parallel AC$ and $MQ' \\parallel AB$, and afterwards that $P \\equiv P'$ and $Q \\equiv Q'$.\nNotice that $\\angle P'IB = 180^\\circ - \\angle AIB = 90^\\circ - \\frac{1}{2}\\angle ACB$. As the triangle $CED$ is isosceles with base $DE$, we have $\\angle P'DC = 90^\\circ - \\frac{1}{2}\\angle ACB$, so $\\overline{P'IB} \\equiv \\overline{P'DC}$. Hence, the quadrilateral $BDP'I$ is cyclic, so $BP' \\perp AI$.\nLet $N$ be the midpoint of $AB$. As $P'AB$ is a right triangle, we have $P'N = NA = NB$, so $\\overline{AP'N} \\equiv \\overline{P'AN} \\equiv \\overline{P'AC}$, which leads to $NP' \\parallel AC$. Also, we have $MN \\parallel AC$, since $MN$ is a midline of triangle $ABC$. It follows that the points $M, N, P'$ are collinear and $MP' \\parallel AC$. Hence $\\overline{DP'M} \\equiv \\overline{DEC} \\equiv \\overline{CDE}$ (because $CDE$ is isosceles), so triangle $MDP'$ is also isosceles, with $MD = MP'$.\n\n![](attached_image_1.png)\n\nA similar reasoning shows that $MQ' \\parallel AB$ and $MD = MQ'$. Therefore, $MP' = MQ' = MD$, which means that $M$ is the circumcenter of triangle $DP'Q'$, which finishes our proof.\n\nAnother approach. Let $R$ be the foot of the bisector of angle $BAC$. Then $MP = MQ = MD = \\frac{a}{2} - (p-b) = \\frac{b-c}{2}$, so\n$$\n\\frac{PM}{AC} = \\frac{b-c}{2b}.\n$$\nOn the other hand, $RC = \\frac{ab}{b+c}$ and $RM = RC - MC = \\frac{ab}{b+c} - \\frac{a}{2} = \\frac{a(b-c)}{2(b+c)}$, so\n$$\n\\frac{RM}{RC} = \\frac{b-c}{2b} = \\frac{PM}{AC}.\n$$\nSince the triangles $RMP$ and $RCA$ have one common (obtuse) angle $R$, the above relation shows that triangles $RMP$ and $RCA$ are similar (side-side-angle), so $PM \\parallel AC$.\nAnalogously, we have $\\frac{RM}{RB} = \\frac{MQ}{AB}$, and, since angles $\\angle BAR$ and $\\angle MQP$ are both acute, the side-side-angle case shows that triangles $RMQ$ and $RAB$ are also similar, so $QM \\parallel AB$.\n\nA third approach. Let $P'$ be the projection of $B$ on line $AI$ and $S$ the intersection point of lines $BP'$ and $AC$. Since $AP'$ is both bisector and altitude of triangle $ABS$, it follows that $P'$ is the midpoint of $BS$. Then $MP'$ is the midline of triangle $BCS$, so $MP' \\parallel AC$. Also, if $Q'$ is the projection of $C$ on line $AI$, a similar reasoning leads to $MQ' \\parallel AB$.\nNext, we shall prove that $MP' = MQ' = MD$, which assure us that $P = P'$, $Q = Q'$ and leads to conclusion.\n\nThe quadrilateral $BIP'D$ is inscribed in the circle of diameter $AI$, so $\\angle DP'Q' = \\angle IBD = \\frac{1}{2} \\angle B$. Then\n$$\n\\angle DP'M = \\angle DP'Q' + \\angle Q'P'M = \\frac{1}{2} \\angle B + \\frac{1}{2} \\angle A\n$$\nSince $\\angle P'MD = \\angle C$, considering the triangle $DP'M$ it results that\n$$\n\\begin{aligned}\n\\angle P'DM &= 180^\\circ - \\angle P'MD - \\angle DP'M = 180^\\circ - \\angle C - \\frac{1}{2}(\\angle A + \\angle B) \\\\\n&= \\frac{1}{2}(\\angle A + \\angle B) = \\angle DP'M,\n\\end{aligned}\n$$\nso $MP' = MD$. Similarly, one can prove that $MQ' = MD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution, $[XYZ]$ and $d(X, YZ)$ denote the area of the triangle $XYZ$ and the distance from the point $X$ to the line $YZ$, respectively.\n\nSince the quadrilaterals $SRQ_2Q_1$ and $SRP_2P_1$ are cyclic, $\\angle SQ_1R = \\angle SQ_2R$ and $\\angle SP_1R = \\angle SP_2R$ (see Fig. 6), so the triangles $SP_1Q_2$ and $SP_2Q_1$ are similar, and\n$$\n\\frac{d(S, P_1Q_2)}{d(S, P_2Q_1)} = \\frac{P_1Q_2}{P_2Q_1}. \\qquad (1)\n$$\nLet $K$ and $L$ be the midpoints of $AB$ and $AC$, respectively; then $P_1K = P_2K$ and $Q_1L = Q_2L$. Therefore, $d(Q_1, MP_1) + d(Q_2, MP_1) = 2d(L, MP_1)$, so\n$$\n[MP_1Q_2] + [MP_1Q_1] = 2[MP_1L] = ML \\cdot d(P_1, ML) = [ABC]/2.\n$$\nSimilarly, $[MP_2Q_1] + [MP_1Q_1] = [ABC]/2 = [MP_1Q_2] + [MP_1Q_1]$. Thus $[MP_1Q_2] = [MP_2Q_1]$, whence\n$$\n\\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \\frac{P_2Q_1}{P_1Q_2}. \\qquad (2)\n$$\nIn the angle $P_1RQ_1$, the condition $d(X, P_1Q_2)/d(X, P_2Q_1) = \\alpha$ determines a ray emanating from $R$. Moreover, rays symmetric with respect to the bisector of $\\angle P_1RQ_1$ correspond to reciprocal values of $\\alpha$. Consequently, relations (1) and (2) show that the rays $RS$ and $RM$ are symmetric with respect to this angle bisector, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $n$ positive, not necessarily distinct, integers $a_1, a_2, \\dots, a_n$ whose sum is $2S$. A positive integer $k$ is called a separator if one can choose $k$ indexes from $\\{1, 2, \\dots, n\\}$ such that the sum of the corresponding numbers is $S$. What is the maximum possible number of separators?",
    "options": [],
    "answer": "For n = 1: 0; n = 2: 1; n = 3: 2; n = 4: 2; and for n ≥ 5: n − 3.",
    "solution": "If 1 is a separator, then we can not have any other separators than 1 and $n-1$.\nFor $n=1$ we can not have any separators.\nFor $n=2$, only 1 can be a separator (if the two numbers are equal).\nFor $n=3$ only 1 and 2 can be separators (for example, in the case of the numbers 1, 2, 3; $1+2+3=3$).\nFor $n=4$ we can have at most two separators, 1 and 3, for example in the case of the numbers 1, 2, 3, 6, with $1+2+3=6$.\nWe prove that, for $n \\ge 5$, the maximum number of separators is $n-3$, namely in the case when all the numbers 2, 3, ..., $n-2$ are separators.\nThis maximum is achieved, for example, for the numbers\n* $1, 1, 1, 1, 2, 2, 4, 4, \\dots, 2^{k-2}, 2^{k-2}$ if $n=2k$\nIndeed, when $n=2k$, we have $2S = 2^k$, hence $S = 2^{k-1}$ which can be written\n$$ S = 2^{k-2} + 2^{k-2} = 2^{k-2} + 2^{k-3} + 2^{k-3} = \\dots = 2^{k-2} + 2^{k-3} + \\dots + 2 + 1 + 1. $$\n\nAlso, when $n=2k+1$, we have $2S = 2^{k+1}$, hence $S = 2^k$ and we can write\n$$ S = 2^{k-1} + 2^{k-1} = 2^{k-1} + 2^{k-2} + 2^{k-2} = \\dots = 2^{k-1} + 2^{k-2} + \\dots + 2 + 1 + 1. $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$, and let $p$ be a prime divisor of $n$. A confederation consists of $p$ states, each of which has exactly $n$ airports. There are $p$ air companies operating interstate flights only such that every two airports in different states are joined by a direct (two-way) flight operated by one of these companies. Determine the maximal integer $N$ satisfying the following condition: In every such confederation it is possible to choose one of the $p$ air companies and $N$ of the $np$ airports such that one may travel (not necessarily directly) from any one of the $N$ chosen airports to any other such only by flights operated by the chosen air company.",
    "options": [],
    "answer": "n",
    "solution": "The required maximum is $n$. The following example shows that $N$ cannot exceed $n$. Split the $n$ airports in the $i$-th state, $i = 1, \\dots, p$, into $p$ disjoint groups of $n/p$ airports each, $A_{i,j}$, $j = 1, \\dots, p$. Let the $k$-th air company, $k = 1, \\dots, p$, operate direct flights between every airport in $A_{i,j}$ and every airport in $A_{i',j'}$ if $i \\neq i'$ and $j' - j \\equiv k(i' - i) \\pmod{p}$, and operate no flights between the airports in $A_{i,j}$ and those in $A_{i',j'}$ otherwise.\n\nSince $p$ is prime, for every $i \\neq i'$ and every $j, j'$, there exists $1 \\le k \\le p$ satisfying the previous congruence modulo $p$, so every two airports in different states are connected by a flight. On the other hand, for any given $k = 1, \\dots, p$, the $np$ airports are split into $p$ disjoint $n$-element groups, namely, $\\bigsqcup_{i=1}^p A_{i,j+ki}$, $j = 1, \\dots, p$ (the indices are reduced modulo $p$), such that there are no flights between different groups operated by the $k$-th air company. Consequently, $N \\le n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number $1$ can be represented as the sum of a finite number $n$ of real positive numbers, not necessarily distinct, that only use the digits $0$ and/or $7$ in their decimal representation. What is the smallest possible value of $n$?",
    "options": [],
    "answer": "8",
    "solution": "The problem is equivalent to representing\n$$\n\\frac{1}{7} = 0.142857142857142857\\dots := 0.\\overline{142857}\n$$\nas the sum of a finite number, $n$, of real numbers that only have digits $0$ and $7$ in their decimal representation. In order to get the fourth decimal $8$, it is easy to see that we need $n \\geq 8$. An example with $n = 8$ terms is\n$$\n\\begin{aligned}\n1 = &\\ 0.\\overline{777777} + 0.\\overline{077777} + 0.\\overline{070777} + 0.\\overline{070777} \\\\\n& + 0.\\overline{000777} + 0.\\overline{000707} + 0.\\overline{000707} + 0.\\overline{000700},\n\\end{aligned}\n$$\nwhich leads to the following representation of $1$:\n$$\n1 = \\frac{777777 + 77777 + 70777 + 70777 + 777 + 707 + 707 + 700}{10^6 - 1}.\n$$\nClearly, there are other representations for $n = 8$, as well as for values $n > 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define a sequence of integers by $a_0 = 1$, and $a_n = \\sum_{k=0}^{n-1} \\binom{n}{k} a_k$, $n \\ge 1$. Let $m$ be a positive integer, let $p$ be a prime, and let $q$ and $r$ be non-negative integers. Prove that the difference $a_{p^m q+r} - a_{p^{m-1} q+r}$ is divisible by $p^m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the $\\mathbb{R}$-vector space $\\mathbb{R}[X]$ of all polynomials with real coefficients and define an $\\mathbb{R}$-linear functional $L: \\mathbb{R}[X] \\to \\mathbb{R}$ by $LX^n = a_n$, $n = 0, 1, 2, \\dots$. Thus, if $f = \\sum_k \\alpha_k X^k$, then $Lf = \\sum_k \\alpha_k a_k$. Since $(X+1)^n = \\sum_{k=0}^n \\binom{n}{k} X^k$, $n \\ge 1$,\n$$\nL(X+1)^n = \\sum_{k=0}^{n} \\binom{n}{k} LX^k = \\sum_{k=0}^{n} \\binom{n}{k} a_k = \\sum_{k=0}^{n-1} \\binom{n}{k} a_k + a_n = 2a_n = 2LX^n,\n$$\nso $Lf(X+1) = 2Lf(X) - f(0)$ for every polynomial $f$ in $\\mathbb{R}[X]$. In particular, take $f = \\binom{X}{k}$ and use the relation $\\binom{X+1}{k} = \\binom{X}{k} + \\binom{X}{k-1}$, $k \\ge 1$, to get $L\\binom{X}{k} = L\\binom{X}{k-1}$, $k \\ge 1$, and deduce that $L\\binom{X}{k} = 1$, $k = 0, 1, 2, \\dots$. Further, if a polynomial $f$ in $\\mathbb{R}[X]$ is integral valued, i.e., $f(k)$ is integral for every integral $k$, then $f = \\sum_k \\alpha_k \\binom{X}{k}$ for some integers $\\alpha_k$, so $Lf = \\sum_k \\alpha_k$ is an integer. Finally, since $a^{p^m} \\equiv a^{p^{m-1}} \\pmod{p^m}$ for all integers $a$,\n$$\nf = p^{-m} \\left( X^{p^m q+r} - X^{p^{m-1} q+r} \\right)\n$$\nis an integral valued polynomial in $\\mathbb{R}[X]$, so $Lf = (a_{p^m q+r} - a_{p^{m-1} q+r})/p^m$ is an integer, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players, $A$ and $B$, remove alternatively stones from a heap initially containing $n \\ge 2$ stones. The first to move is $A$ who takes at least one stone and at most $n-1$ stones. Next, each player to move has to take at least one stone and at most as many stones as his opponent took in his previous move. The winner is the player who takes the last stone. Which player has a winning strategy?",
    "options": [],
    "answer": "If the initial number of stones is a power of two, player B has a winning strategy; otherwise, player A has a winning strategy. A winning strategy is to always remove the largest power of two dividing the current number of stones.",
    "solution": "Let us label $(a, b)$ the position meaning \"the player who has to move finds $a$ stones in the heap and can take at most $b$ stones\".\n\nWe prove that the losing positions are those of the forms $(0, r)$ and $(2^k(2m+1), r)$, where $k, m \\in \\mathbb{N}$, $r \\in \\{1, 2, \\dots, 2^k - 1\\}$.\n\nThe initial position being $(n, n-1)$, the losing initial positions are those with $m=0$, i.e. the powers of $2$. The conclusion will be that if $n$ is a power of $2$, the second player has a winning strategy, while in the other cases, the first player has a winning strategy. Let us notice that:\n\n- the game ends only in positions of the form $(0, r)$\n\n- from a position of the form $(2^k(2m+1), r)$, $0 < r < 2^k$, there is no move that could leave the opponent in a position of similar form\n\n- from any position that is not one of the above form, there exists a move that leaves the opponent in a position of the form $(2^k(2m+1), r)$, $0 < r < 2^k$.\n\n• If from $(2^k(2m+1), r)$, $0 < r < 2^k$, one could move to $(2^s(2p+1), t)$, $0 < t < 2^s$, that move would consist in removing $t$ stones, hence $2^k(2m+1) - 2^s(2p+1) = t$. As $t \\le r < 2^k$, it follows that $s < k$, hence $2^s \\mid 2^k(2m+1) - 2^s(2p+1)$. Then $2^s \\mid t$, which contradicts $0 < t < 2^s$. In conclusion, such a move does not exist.\n\nObviously, there is no move to a position $(0, x)$ either.\n\n• From any of the remaining positions, that is $(2^k(2m+1), r)$ with $r \\ge 2^k$, there is a move that leaves the opponent in position $(2^k \\cdot 2m, 2^k)$, obviously a losing one. Simply take $2^k$ stones.\n\nFrom the above, it is easy to see what the appropriate winning strategy is for the player that has such a strategy: always take $2^k$ stones, where $2^k$ is the largest power of $2$ that divides the number of the remaining stones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Számítsd ki:\n$$\n\\int_{0}^{1} x \\sin \\left(\\pi x^{2}\\right) \\mathrm{d} x\n$$\n\nb. Számítsd ki:\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} \\sin \\left(\\pi x^{2}\\right) \\mathrm{d} x\n$$\n\nProblem:\n\na. Calculaţi\n$$\n\\int_{0}^{1} x \\sin \\left(\\pi x^{2}\\right) \\mathrm{d} x\n$$\n\nb. Calculaţi\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} \\sin \\left(\\pi x^{2}\\right) \\mathrm{d} x\n$$",
    "options": [],
    "answer": "a: 1/π; b: 1/π",
    "solution": "Solution:\n\na. Făcând substituţia $t=\\pi x^{2}$, integrala devine\n$$\n\\frac{1}{2 \\pi} \\int_{0}^{\\pi} \\sin t \\, \\mathrm{d} t = \\left.\\frac{1}{2 \\pi}(-\\cos t)\\right|_{0}^{\\pi} = \\frac{1}{\\pi}\n$$\n\nb. Fie $f:[0,1] \\rightarrow \\mathbb{R}$, $f(x)=\\sin \\left(\\pi x^{2}\\right)$, şi $F:[0,1] \\rightarrow \\mathbb{R}$, $F(x)=\\int_{0}^{x} f(t) \\, \\mathrm{d} t$. Atunci\n$$\n\\begin{aligned}\n\\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} f(x) \\, \\mathrm{d} x &= \\frac{1}{n} \\sum_{k=0}^{n-1}\\left(k F\\left(\\frac{k+1}{n}\\right)-k F\\left(\\frac{k}{n}\\right)\\right) \\\\\n&= \\frac{1}{n} \\sum_{k=0}^{n-1}\\left((k+1) F\\left(\\frac{k+1}{n}\\right)-k F\\left(\\frac{k}{n}\\right)-F\\left(\\frac{k+1}{n}\\right)\\right) \\\\\n&= \\frac{1}{n}\\left(n F(1)-\\sum_{k=1}^{n} F\\left(\\frac{k}{n}\\right)\\right) = F(1)-\\frac{1}{n} \\sum_{k=1}^{n} F\\left(\\frac{k}{n}\\right)\n\\end{aligned}\n$$\n\nLimita cerută este\n$$\nF(1)-\\int_{0}^{1} F(x) \\, \\mathrm{d} x = \\int_{0}^{1} x F^{\\prime}(x) \\, \\mathrm{d} x = \\int_{0}^{1} x f(x) \\, \\mathrm{d} x = \\frac{1}{\\pi}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38245,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} \\sin(\\pi x^2) dx.\n$$",
    "options": [],
    "answer": "1/π",
    "solution": "Let $f: [0, 1] \\to \\mathbb{R}$, $f(x) = \\sin(\\pi x^2)$, and let $F: [0, 1] \\to \\mathbb{R}$, $F(x) = \\int_0^x \\sin(\\pi t^2) dt$. Write\n$$\n\\begin{aligned}\n\\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} \\sin(\\pi x^2) dx &= \\frac{1}{n} \\sum_{k=0}^{n-1} \\left( kF\\left(\\frac{k+1}{n}\\right) - kF\\left(\\frac{k}{n}\\right) \\right) \\\\\n&= \\frac{1}{n} \\sum_{k=0}^{n-1} \\left( (k+1)F\\left(\\frac{k+1}{n}\\right) - kF\\left(\\frac{k}{n}\\right) - F\\left(\\frac{k+1}{n}\\right) \\right) \\\\\n&= \\frac{1}{n} \\left( nF(1) - \\sum_{k=0}^{n-1} F\\left(\\frac{k}{n}\\right) \\right) = F(1) - \\frac{1}{n} \\sum_{k=0}^{n-1} F\\left(\\frac{k}{n}\\right),\n\\end{aligned}\n$$\nto conclude that the required limit is\n$$\nF(1) - \\int_{0}^{1} F(x) dx = \\int_{0}^{1} xF'(x) dx = \\int_{0}^{1} xf(x) dx = \\int_{0}^{1} x \\sin(\\pi x^2) dx = \\frac{1}{\\pi}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38246,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a math contest take part 50 students and 3 problems are submitted to the contestants. It is known that each student solved at least a problem and the total number of correct solutions is 100. Prove that at most 25 students solved all the three problems.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $a$, $b$, $c$ the number of the students which solved exactly one, two, respectively three problems. Then $a + b + c = 50$ and $a + 2b + 3c = 100$. It follows $b + 2c = 50$, whence $2c \\le 50$, yielding $c \\le 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play a game on a regular $n$-gon, where $n \\ge 4$. Three counters start on consecutive vertices. In turn, each player slides one counter along any number of edges to another vertex of the polygon, without jumping over another counter, to strictly increase the area of the triangle formed by the counters. The loser is the first player who cannot make a legal move. For what $n$ does the first player have a winning strategy?\n\nUnited Kingdom",
    "options": [],
    "answer": "The first player wins if and only if the exponent of 2 in n − 3 is odd (i.e., e2(n−3) is odd).",
    "solution": "Since the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins or Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).\n\n**Lemma.** Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.\n\n**Proof.** Label the positions of the counters $X, Y, Z$ so that the arc $YZ$ of the circumcircle is shortest and the arc $ZX$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $YZX$ until it forms an isosceles triangle $XYZ'$ with apex at $Y$ (note that the arc $XY$ is less than half the circle, so that $Z$ does not jump over the counter at $X$). If this configuration is a Loss, we are done.\n\nIf instead this configuration is a Win, then the counters can be moved legally from triangle $XYZ'$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z'$ to a new location $Z''$. But then the counter at $Z$ could have been moved to $Z''$ in the first place, so the original configuration was a Win as well. This ends the proof of the lemma.\n\nFor every non-zero integer $m$, let $e_2(m)$ denote the exponent of 2 in the prime decomposition of $m$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2a + b = n$), we show that the configuration is a Win if and only if $a \\neq b$ and $e_2(a-b)$ is odd.\n\nWrite $b = a \\pm |a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \\pm |a-b|/2$, $a \\pm |a-b|/2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.\n\nIf instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \\pm |a-b|/2$, $a \\pm |a-b|/2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \\pm |a-b|/2, a \\pm |a-b|/2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b|/2$, the conclusion follows inductively once the exceptional and trivial case $a=b$ is dealt with.\n\nAs an immediate corollary, the configuration with arc lengths 1, 1, $n-2$ (the starting configuration of the question) is a Win if and only if $e_2(n-3)$ is odd.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ and $b$ so that the equality\n$$\n\\lfloor a x + b y \\rfloor + \\lfloor b x + a y \\rfloor = (a + b) \\lfloor x + y \\rfloor\n$$\nis true for every real numbers $x$ and $y$.",
    "options": [],
    "answer": "(a,b) = (0,0) or (1,1)",
    "solution": "Plugging $y = -x$ and $d = a - b$ yields $\\lfloor d x \\rfloor + \\lfloor -d x \\rfloor = 0$, for every $x \\in \\mathbb{R}$. (*)\nIf $d \\neq 0$, then (*) is false for $x = 1/(2d)$, hence $d = 0$. Then $x + y = 1$ leads to $2\\lfloor a \\rfloor = 2a$, therefore $a$ is an integer.\nIf $a = 0$ the relation is fulfilled, therefore a solution is $a = b = 0$.\n\nIf $a \\neq 0$, relation $2a\\lfloor x + y \\rfloor = 2\\lfloor a(x + y) \\rfloor \\le 2a(x + y)$ with $x + y = 1/2$ leads to $a \\ge 0$, therefore $a \\ge 1$. Now $x + y = 1/a$ gives $1 = a \\lfloor 1/a \\rfloor$, whence $a = b = 1$, which is the second solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38249,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that if $p$, $q$, $\\sqrt{2p-q}$ and $\\sqrt{2p+q}$ are integers, then $q$ is even.\n\nb) Find out how many positive integers $p$ have the property that both $\\sqrt{2p-4030}$ and $\\sqrt{2p+4030}$ are integers.",
    "options": [],
    "answer": "4",
    "solution": "a) We know that $2p - q = k^2$, $2p + q = r^2$, hence $r^2 - k^2 = 2q$, with $k$, $r$ positive integers. Then $(r - k)(r + k) = 2q$, and the conclusion follows from the fact that $r - k$ and $r + k$ have the same parity.\n\nb) Answer: four numbers.\nWith the above notations, $(r - k)(r + k) = 2 \\cdot 4030 = 2^2 \\cdot 5 \\cdot 13 \\cdot 31$. Since $r - k$ and $r + k$ have the same parity and $r - k < r + k$, the pair $(r - k, r + k)$ can be $(2, 4030)$, $(10, 806)$, $(26, 310)$ or $(62, 130)$.\nThen $r \\in \\{2016, 408, 168, 96\\}$ and $p \\in \\{2030, 113, 812, 17, 1209, 753\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $r$ denote its inradius. Let $R_A$ denote the radius of the circle internally tangent at $A$ to the circle $ABC$ and tangent to the line $BC$; the radii $R_B$ and $R_C$ are defined similarly. Show that $1/R_A + 1/R_B + 1/R_C \\le 2/r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $h_A$ and $\\Delta$ denote the length of the side $BC$, the length of the altitude from $A$ in the triangle $ABC$, and the area of the triangle $ABC$, respectively. Consider the circle internally tangent at $A$ to the circle $ABC$ and tangent at $T$ to the line $BC$. Then $2R_A \\ge AT \\ge h_A = 2\\Delta/a$, and equality holds throughout if and only if the triangle $ABC$ is isosceles with apex at $A$. Similar inequalities hold for $R_B$ and $R_C$, and the conclusion follows at once; equality holds if and only if the triangle $ABC$ is equilateral.\nIn the notation in Solution 1, we shall prove that $1/R_A = (a/\\Delta) \\cos^2(B/2 - C/2)$. Similar formulae hold for $R_B$ and $R_C$, and the conclusion follows at once; equality holds if and only if the triangle $ABC$ is equilateral.\nTo prove the above formula for $R_A$, let $A'$ be the orthogonal projection of $A$ on the line $BC$, let $O$ be the circumcentre of the triangle $ABC$, and let $O_A$ be the centre of the circle tangent at $A$ to the circle $ABC$ and tangent at $T$ to the line $BC$. Since the points $A$, $O_A$ and $O$ are collinear, the angle $A'AO_A$ is congruent to the absolute value of the difference of the internal angles of the triangle $ABC$ at $B$ and $C$, so $\\cos(B - C) = \\cos(\\angle A'AO_A) = (AA' - O_AT)/O_A A = (h_A - R_A)/R_A = h_A/R_A - 1 = 2\\Delta/(aR_A) - 1$, whence the desired formula.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine $n \\in \\mathbb{N}$ such that the numbers $n+8$, $2n+1$, $4n+1$ are all perfect cubes.",
    "options": [],
    "answer": "n = 0",
    "solution": "We will show that the only such integer is $1$. We start noticing that the product $(n+8)(4n+1)(2n+1) = 8n^3 + 70n^2 + 49n + 8$ must be also a cube.\n\nFor $n \\in \\{1, 2, \\dots, 18\\}$ the number $n+8$ cannot be a cube. For $n \\ge 19$, remark that $(2n+2)^3 \\le 8n^3 + 70n^2 + 49n + 8 < (2n+6)^3$. We have to consider the following cases.\n\nIf $8n^3 + 70n^2 + 49n + 8 = (2n + 2)^3$ we get $n = 0$.\n\nIf $8n^3 + 70n^2 + 49n + 8 = (2n + 3)^3$, then $34n^2 - 5n - 19 = 0$, that is $n(34n - 5) = 19$, which has no solution in integers.\n\nIf $8n^3 + 70n^2 + 49n + 8 = (2n+4)^3$, then $22n^2 - 47n - 56 = 0$, and $n(22n-47) = 56$, without solution in positive integers.\n\nIf $8n^3 + 70n^2 + 49n + 8 = (2n + 5)^3$, then $10n^2 - 101n - 117 = 0$, that is $n(10n - 101) = 3 \\cdot 29$, with no solution in positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAdottak az $A, B \\in \\mathcal{M}_{2}(\\mathbb{R})$ mátrixok úgy, hogy $(A-B)^{2}=\\mathrm{O}_{2}$.\na) Igazold, hogy $\\operatorname{det}\\left(A^{2}-B^{2}\\right)=(\\operatorname{det}(A)-\\operatorname{det}(B))^{2}$.\nb) Bizonyítsd be, hogy $\\operatorname{det}(A B-B A)$ akkor és csak akkor egyenlő 0 -val, ha $\\operatorname{det}(A)=\\operatorname{det}(B)$.\n\nProblem:\nFie două matrice $A, B \\in \\mathcal{M}_{2}(\\mathbb{R})$ astfel încât $(A-B)^{2}=O_{2}$.\na) Arătaţi că $\\operatorname{det}\\left(A^{2}-B^{2}\\right)=(\\operatorname{det}(A)-\\operatorname{det}(B))^{2}$.\nb) Demonstraţi că $\\operatorname{det}(A B-B A)=0$ dacă şi numai dacă $\\operatorname{det}(A)=\\operatorname{det}(B)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) $\\operatorname{Din}(A-B)^{2}=O_{2}$ obţinem $\\operatorname{det}(A-B)=0$. (1 punct)\nDe asemenea, deducem $\\operatorname{Tr}(A-B)=0$, deci $\\operatorname{Tr}(A)=\\operatorname{Tr}(B)=: a$. (1 punct) Notăm $b=\\operatorname{det}(A)-\\operatorname{det}(B)$. Conform relaţiei lui Cayley, avem\n$$\n\\left\\{\\begin{array}{l}\nA^{2}-a A+\\operatorname{det}(A) I_{2}=O_{2} \\\\\nB^{2}-a B+\\operatorname{det}(B) I_{2}=O_{2}\n\\end{array}\\right.\n$$\nde unde $\\operatorname{det}\\left(A^{2}-B^{2}\\right)=\\operatorname{det}\\left(a(A-B)-b I_{2}\\right)$. (1 punct)\nDar $\\operatorname{det}\\left(a(A-B)-b I_{2}\\right)=a^{2} \\operatorname{det}(A-B)-a b \\operatorname{Tr}(A-B)+b^{2}=b^{2}$.\nRezultă $\\operatorname{det}\\left(A^{2}-B^{2}\\right)=(\\operatorname{det}(A)-\\operatorname{det}(B))^{2}$. (1 punct)\n\nb) Fie $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ funcţia definită prin\n$$\nf(x)=\\operatorname{det}\\left(A^{2}-B^{2}+x(A B-B A)\\right), x \\in \\mathbb{R}\n$$\nFuncţia $f$ se poate reprezenta sub forma\n$$\nf(x)=\\operatorname{det}\\left(A^{2}-B^{2}\\right)+c x+\\operatorname{det}(A B-B A) x^{2}, x \\in \\mathbb{R}\n$$\nunde $c$ este o constantă reală. (1 punct)\nDin $f(1)=f(-1)=\\operatorname{det}(A-B) \\operatorname{det}(A+B)=0$ obţinem $c=0$ şi\n$$\n\\operatorname{det}\\left(A^{2}-B^{2}\\right)+\\operatorname{det}(A B-B A)=0 . \\quad(1 \\text { punct })\n$$\nAtunci, conform a), $(\\operatorname{det}(A)-\\operatorname{det}(B))^{2}=-\\operatorname{det}(A B-B A)$, de unde concluzia. (1 punct)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer which has exactly $2015$ positive divisors.",
    "options": [],
    "answer": "2^30 * 3^12 * 5^4",
    "solution": "The number of positive divisors of an integer $N$ with the prime factorization $p_1^{a_1} p_2^{a_2} \\cdots p_n^{a_n}$ is $(a_1 + 1)(a_2 + 1) \\cdots (a_n + 1)$. Since $2015 = 1 \\cdot 2015 = 5 \\cdot 403 = 13 \\cdot 155 = 31 \\cdot 65 = 5 \\cdot 13 \\cdot 31$, the required number has one of the following forms: $a = 2^{2014}$, $b = 2^{402} \\cdot 3^4$, $c = 2^{154} \\cdot 3^{12}$, $d = 2^{64} \\cdot 3^{30}$ or $e = 2^{30} \\cdot 3^{12} \\cdot 5^4$.\n\nWe claim that the smallest number is $2^{30} \\cdot 3^{12} \\cdot 5^4$. Indeed, $a > e$ since $2^{1984} > 3^{12} \\cdot 5^4$; $b > e$ since $2^{372} > 3^8 \\cdot 5^4$; $c > e$ for $2^{124} > 5^4$ and $d > e$ for $2^{34} \\cdot 3^{18} > 5^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_5(\\mathbb{C})$ be a matrix with $\\text{Tr}(A) = 0$ and such that $I_5 - A$ is invertible. Prove that $A^5 \\neq I_5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, by way of contradiction, that $A^5 = I_5$. Let $\\lambda \\in \\mathbb{C}$ be an eigenvalue of $A$; then $\\lambda^5$ is an eigenvalue of $A^5$ and we obtain $\\lambda^5 = 1$.\nSince $I_5 - A$ is invertible, it follows that $\\det(I_5 - A) \\neq 0$. Consequently, the eigenvalues of $A$ belong to the set $\\{\\varepsilon, \\varepsilon^2, \\varepsilon^3, \\varepsilon^4\\}$, where $\\varepsilon = \\cos \\frac{2\\pi}{5} + i \\sin \\frac{2\\pi}{5}$. From $\\text{Tr}(A) = 0$, we deduce the existence of the non-negative integers $a, b, c$ and $d$ such that\n$$\n\\begin{cases} a+b+c+d=5 \\\\ a\\varepsilon+b\\varepsilon^2+c\\varepsilon^3+d\\varepsilon^4=0. \\end{cases}\n$$\nThe latter gives $a+b\\varepsilon+c\\varepsilon^2+d\\varepsilon^3=0$, hence $\\varepsilon$ is a root of the integer polynomial $dX^3+cX^2+bX+a$, whose degree is at most 3. If $d \\neq 0$, then the roots of this polynomial are $\\varepsilon, \\bar{\\varepsilon} \\in \\mathbb{C} \\setminus \\mathbb{R}$ and some $\\alpha \\in \\mathbb{R}$. But then, using Viète's relations, we obtain $\\alpha = -a/d \\in \\mathbb{Q}$ and $\\alpha = -(c/d + 2 \\cos(2\\pi/5)) \\in \\mathbb{R} \\setminus \\mathbb{Q}$, which is a contradiction. We conclude that $d=0$.\nSimilarly, one can show that $c = b = a = 0$, and then $a+b+c+d=0$, contradicting $a+b+c+d=5$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38255,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic function $f$ sends any interval $I$ of length 1 to an interval $f(I)$ of length at least 1.\nProve that for any interval $J$ of length 2, the length of the interval $f(J)$ is at least 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $f(x) = ax^2 + bx + c$ and $v = -b/2a$ is the abscissa of parabola's vertex, then, taking $I = [v - 1/2, v + 1/2]$, the interval $f(I)$ has length $|a|/4$, hence $|a| \\ge 4$.\n\nNow, taking an interval $J$ of length 2, one can find $x, y \\in J$ such that $x - y = 1$ and $v \\notin (y, x)$.\nWe have $|f(x) - f(y)| = |a(x - y)(x + y + \\frac{b}{a})| \\ge 4|x + y - 2v| \\ge 4$, therefore $f(J)$ contains two points at least 4 units apart, whence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\ge 1$, be a positive integer, let $p_1, p_2, \\dots, p_k$ be distinct primes and denote $n = p_1p_2\\dots p_k$. For a function $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$, define $p(f) = f(1)f(2)\\dots f(n)$.\n\na) Determine the number of functions $f$ such that $p(f)$ divides $n$.\n\nb) For $n=6$, determine the number of functions $f$ such that $p(f)$ divides $36$.",
    "options": [],
    "answer": "a) (n+1)^k; b) 580",
    "solution": "a) If $p(f)$ divides $n$, then $p(f) = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$, with $a_i \\in \\{0, 1\\}$. It follows that a prime factor $p_i$ either does not divide $p(f)$, or it appears in the prime decomposition of exactly one of the numbers $f(1), f(2), \\dots, f(n)$. Thus, for each $p_i$ we have $n+1$ ways to choose and hence the number of such functions equals $(n+1)^k$.\n\nb) If $p(f)$ divides $36$, then $p(f) = 2^a 3^b$, where $a, b \\in \\{0, 1, 2\\}$. There are $1 + C_6^2 + 2C_6^1$ such functions for $b=0$, $C_6^1(1+C_6^1+C_5^1+C_6^2)$ functions for $b=1$, and $C_6^2(1+C_6^1+C_4^1+C_6^2)$ functions for $b=2$. Summing up, we obtain $580$ functions with the given property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(K, +, \\cdot)$ be a finite field with at least four elements. Prove that the set $K^*$ can be partitioned into two nonempty subsets $A$ and $B$, such that\n$$\n\\sum_{x \\in A} x = \\prod_{y \\in B} y.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the product of the elements of $K^*$ is $-1$, if $A$ and $B$ form a partition of $K^*$, then $(\\prod_{a \\in A} a) (\\prod_{b \\in B} b) = -1$, so $\\sum_{a \\in A} a = \\prod_{b \\in B} b$ if and only if\n$$\n\\left(\\sum_{a \\in A} a\\right) \\left(\\prod_{a \\in A} a\\right) = -1. \\quad (*)\n$$\nLet $|K| \\ge 4$. If the characteristic of $K$ is $2$, then the singleton set $A = \\{1\\}$ clearly satisfies $(*)$. If the characteristic of $K$ is odd, choose an element $a$ in $K^* \\setminus \\{\\pm 1\\}$, and notice that the 3-element set $A = \\{-1, 1, a\\}$ satisfies $(*)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ for which, no matter how we choose to color red $n$ vertices of a cube, there is a vertex of the cube whose three adjacent vertices are all colored red.",
    "options": [],
    "answer": "5",
    "solution": "Let $ABCD A'B'C'D'$ be a cube. Coloring red the four vertices of a face (e.g. $A, B, C, D$), no vertex of the cube has all three adjacent vertices colored red, so $n \\ge 5$.\n\nNow, let us color 5 vertices of the cube in red. Anyway we do it, one of the faces $ABCD$ and $A'B'C'D'$ has at least three red vertices. Without loss of generality, suppose $A, B, C$ are red.\n\nIf $D$ is also red, the face $A'B'C'D'$ has only one red vertex; if $X'$ is the red vertex, with $X \\in \\{A, B, C, D\\}$, then all the adjacent vertices of $X$ are colored red.\n\nIf $D$ is not red, the face $A'B'C'D'$ has two red vertices. If one of $B'$ or $D'$ is red, then the adjacent vertices of $B$, respectively $D$, are all red. Otherwise, $A'$ and $C'$ are red.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38259,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ with $AB = AC$ is obtuse at $A$. Let $M$ be the mirror image of $A$ across $C$. The perpendicular bisector of the line segment $AM$ meet the line $AB$ at point $P$. Given that lines $PM$ and $BC$ are perpendicular, prove that $APM$ is an equilateral triangle.\n\nMarcel Neferu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lines $BC$ and $PM$ meet at $D$. Denote $x$ the measure of $\\angle ABC$. Then $\\angle MCD = \\angle ACB = \\angle ABC = x$ and $\\angle PMC = 90^\\circ - x$. The triangle $PAM$ is isosceles, since $PC$ is median as well as perpendicular bisector of the line segment $AM$, hence $\\angle PMC = \\angle PAC$. On the other hand, $\\angle PAC = \\angle ABC + \\angle ACB = 2x$, implying $x = 30^\\circ$. Then $\\angle PMC = \\angle PAC = 60^\\circ$, whence the claim.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $x, y$ such that\n$$\n5^x - \\log_2 (y+3) = 3^y \\quad \\text{and} \\quad 5^y - \\log_2 (x+3) = 3^x.\n$$",
    "options": [],
    "answer": "x = y = 1",
    "solution": "Answer: $x = y = 1$.\n\nSubtracting the equalities yields\n$$\n5^x + 3^x + \\log_2 (x+3) = 5^y + 3^y + \\log_2 (y+3).\n$$\nFunction $t \\mapsto 5^t + 3^t + \\log_2 (t+3)$ is increasing, so the hypothesis leads to $x = y$ and becomes now $5^x = 3^x + \\log_2 (x+3)$, with integer $x$. We notice that $x > -3, -2, -1, 0$ are not solutions and 1 is a solution. Since the function $t \\mapsto (\\frac{3}{5})^x + (\\frac{4}{5})^x$ is decreasing, $5^x \\ge 3^x + 4^x$ for $x \\ge 2$. On the other hand, an easy induction shows that $4^x > \\log_2 (x+3)$ for integer $x \\ge 2$, therefore there are no solutions for $x > 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangular pyramidal frustum $ABCA'B'C'$. Points $D \\in (AA')$, $E \\in (BB')$ and $F \\in (CC')$ are such that the planes $(AEF)$ and $(DB'C')$ are parallel. Prove that the planes $(A'EF)$ and $(DBC)$ are also parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $V$ the common point of the supporting lines of the lateral edges of the frustum. As planes $(AEF)$ and $(DB'C')$ are parallel, we have $EF \\parallel B'C'$ and $DB' \\parallel AE$. Thales Theorem gives from $DB' \\parallel AE$ and $A'B' \\parallel AB$:\n$$\n\\frac{VD}{VA} = \\frac{VB'}{VE}, \\quad \\frac{VA'}{VA} = \\frac{VB'}{VB}.\n$$\nThe last two equalities give\n$$\n\\frac{VD}{VA'} = \\frac{VB}{VE},\n$$\nso $A'E \\parallel DB$. As $EF \\parallel B'C' \\parallel BC$ and $A'E \\parallel DB$ we conclude $(A'EF) \\parallel (DBC)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38262,
    "subject": "Mathematics (Multi-modal)",
    "question": "The matrices $A, B \\in \\mathcal{M}_2(\\mathbb{R})$ have the property $(A - B)^2 = O_2$.\na) Show that $\\det(A^2 - B^2) = (\\det(A) - \\det(B))^2$.\nb) Show that $\\det(AB - BA) = 0$ if and only if $\\det(A) = \\det(B)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From $(A - B)^2 = O_2$ follows $\\det(A - B) = 0$ and $\\mathrm{Tr}(A - B) = 0$, hence $\\mathrm{Tr}(A) = \\mathrm{Tr}(B) =: a$.\nDenote $b = \\det(A) - \\det(B)$. Then\n$$\n\\begin{cases}\nA^2 - aA + \\det(A)I_2 = O_2 \\\\\nB^2 - aB + \\det(B)I_2 = O_2\n\\end{cases}\n$$\nThis shows that $\\det(A^2 - B^2) = \\det(a(A - B) - bI_2)$. On the other hand, $\\det(a(A - B) - bI_2) = a^2 \\det(A - B) - ab \\mathrm{Tr}(A - B) + b^2 = b^2$, whence $\\det(A^2 - B^2) = (\\det(A) - \\det(B))^2$.\n\nb) Let $f : \\mathbb{R} \\to \\mathbb{R}$ be the function given by\n$$\nf(x) = \\det(A^2 - B^2 + x(AB - BA)), \\quad x \\in \\mathbb{R}.\n$$\nThen $f(x) = \\det(A^2 - B^2) + cx + \\det(AB - BA)x^2$, $x \\in \\mathbb{R}$, where $c$ is a real constant. From $f(1) = f(-1) = \\det(A - B)\\det(A + B) = 0$ follows $c = 0$ and $\\det(A^2 - B^2) + \\det(AB - BA) = 0$. Now (a) leads to $(\\det(A) - \\det(B))^2 = -\\det(AB - BA)$, whence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the edges of a connected finite simple graph can be oriented so that the number of edges leaving each vertex is even if and only if the total number of edges is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given any orientation, the total number of edges equals the sum of all out-degrees. If the latter are all even, then so is the former.\n\nTo establish the converse, induct on the number of edges to show that any connected simple finite graph with an even number of edges splits into edge-disjoint paths of length $2$. Orient the two edges of each of these paths away from the joint to obtain the required orientation.\n\nAlternative Solution.\n\nThe problem is a special case of the following general fact: Given a connected finite simple graph $G = (V, E)$ and an integral-valued function $f$ on $V$, taken over all possible orientations of the edges of $G$, the minimum of the number of vertices at which outdeg and $f$ have opposite parities is $|E| - \\sum_{x \\in V} f(x)$ reduced modulo $2$.\n\nGiven any orientation, notice that the number of vertices at which the parities of outdeg and $f$ disagree has the same parity as $|E| - \\sum_{x \\in V} f(x)$. Consider an orientation minimising the number of these vertices. If this number exceeds $1$, choose two such vertices and use connectedness to join them by a path. Reverting orientations along the path changes the parity of out-degrees only at the end-points, so the outcome is an oriented graph with fewer vertices at which outdeg and $f$ have opposite parities. This contradicts minimality and concludes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $G_n$ be the set of all simple graphs on $n$ vertices. For each vertex $v$ of a graph in $G_n$, let $k(v)$ be the maximal cardinality of an independent set of neighbours of $v$. Determine $\\max_{G \\in \\mathcal{G}_n} \\sum_{v \\in V(G)} k(v)$ and the graphs in $\\mathcal{G}_n$ that achieve this value.",
    "options": [],
    "answer": "The maximum is the floor of n squared over two, achieved uniquely by the complete bipartite graph with parts of sizes floor of n over two and ceiling of n over two.",
    "solution": "To prove this, let $G$ be a simple graph on $n$ vertices, and let $S$ be a maximal independent set of vertices of $G$. If $v$ is a member of $S$, then $k(v) \\le n - |S|$, since $v$ has at most $n - |S|$ neighbours. If a vertex $v$ is not in $S$, then $k(v) \\le |S|$, since $k(v)$ is the size of an independent set. Consequently,\n$$\n\\sum_{v \\in V(G)} k(v) \\le |S|(n - |S|) + (n - |S|)|S| \\le \\lfloor n^2/2 \\rfloor.\n$$\n\nThe complete bipartite graph $K_{\\lfloor n/2 \\rfloor, \\lceil n/2 \\rceil}$ clearly achieves the upper bound. Achieving the upper bound requires that all of $S$ be adjacent to all of $V(G) \\setminus S$ and that $V(G) \\setminus S$ be an independent set, so the extremal graph is unique.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are positive odd integers $m_1 < m_2 < \\dots$ and positive integers $n_1 < n_2 < \\dots$ such that $m_k$ and $n_k$ are relatively prime, and $m_k^4 - 2n_k^4$ is a perfect square for each index $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$. Write $\\ell^2 = m^4 - 2n^4$, so $\\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$, and $\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$. Multiply the latter by $\\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2 = 2\\ell^4$ to get $(\\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2)(\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2) = -2 \\cdot (2\\ell mn)^4$; that is, $(\\ell^4 - 8m^4n^4)^2 - (m^4 + 2n^4)^4 = -2 \\cdot (2\\ell mn)^4$. Letting $m' = m^4 + 2n^4$ and $n' = 2\\ell mn$, clearly $m' > m$, $m'$ is odd, $n' > n$, the difference $m'^4 - 2n'^4$ is a perfect square, and it is readily checked that $m'$ and $n'$ are relatively prime. The conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\gamma$, $\\gamma_0$, $\\gamma_1$, $\\gamma_2$ be coplanar circles such that $\\gamma_i$ is internally tangent to $\\gamma$ at $A_i$, and $\\gamma_i$ and $\\gamma_{i+1}$ are externally tangent at $B_{i+2}$, $i = 0, 1, 2$ (indices are reduced modulo $3$). The tangent at $B_i$, common to $\\gamma_{i-1}$ and $\\gamma_{i+1}$, meets $\\gamma$ at $C_i$, located in the half-plane opposite $A_i$ with respect to the line $A_{i-1}A_{i+1}$. Show that the three lines $A_iC_i$ are concurrent.\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\gamma_i$ cross the lines $A_iC_{i+1}$ and $A_iC_{i+2}$ again at $X_{i+2}$ and $Y_{i+1}$, respectively.\n\nSince the homothety centred at $A_i$, transforming $\\gamma_i$ into $\\gamma$, sends $X_{i+2}$ to $C_{i+1}$ and $Y_{i+1}$ to $C_{i+2}$, the lines $X_{i+2}Y_{i+1}$ and $C_{i+1}C_{i+2}$ are parallel, so $A_iC_{i+1}/A_iC_{i+2} = X_{i+2}C_{i+1}/Y_{i+1}C_{i+2}$.\n\nOn the other hand, since $C_i$ has equal powers relative to $\\gamma_{i+1}$ and $\\gamma_{i+2}$, $C_iX_{i+1} \\cdot C_iA_{i+2} = C_iB_i^2 = C_iY_{i+2} \\cdot C_iA_{i+1}$, it follows that $C_iX_{i+1} = C_iB_i^2/C_iA_{i+2}$ and $C_iY_{i+2} = C_iB_i^2/C_iA_{i+1}$.\n\nHence $A_iC_{i+1}/A_iC_{i+2} = (B_{i+1}C_{i+1}/B_{i+2}C_{i+2})^2(A_iC_{i+2}/A_iC_{i+1})$, so $A_iC_{i+1}/A_iC_{i+2} = B_{i+1}C_{i+1}/B_{i+2}C_{i+2}$, and consequently $\\prod_{i=0}^2 (A_iC_{i+1}/A_iC_{i+2}) = \\prod_{i=0}^2 (B_{i+1}C_{i+1}/B_{i+2}C_{i+2}) = 1$. Since the sines of the angles $C_iA_iC_j$ are proportional to the lengths of the corresponding chords $A_iC_j$, the conclusion follows by Ceva's theorem in trigonometric form in the triangle $C_0C_1C_2$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nAlternative Solution:\nLet the tangents to $\\gamma$ at $A_i$ and $A_{i+1}$ meet at $D_{i+2}$ (fig 1), and notice that the latter has equal powers relative to $\\gamma_i$ and $\\gamma_{i+1}$ to deduce that it lies on their radical axis, $B_{i+2}C_{i+2}$. Consequently, the lines $C_iD_i$ are concurrent at the radical center of the $\\gamma_i$, and the conclusion follows by the lemma below (see Figure 2).\n\n**Lemma.** Let $P_0P_1P_2$ be a triangle, and let $T_i$ be the touchpoint of the side $P_{i+1}P_{i+2}$ and the incircle $\\gamma$ of the triangle $P_0P_1P_2$. Let further $Z_i$ be a point on the arc $T_{i+1}T_{i+2}$ of $\\gamma$ not containing $T_i$. Then the lines $T_iZ_i$ are concurrent if and only if the lines $P_iZ_i$ are concurrent.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c > 0$ be real numbers so that $a + b + c = 1$. Prove that\n$$\n\\frac{bc + a + 1}{a^2 + 1} + \\frac{ca + b + 1}{b^2 + 1} + \\frac{ab + c + 1}{c^2 + 1} \\le \\frac{39}{10}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Solution by Lenca Cuturela) Since $a + b + c = 1$, we have $a - a^2 = ab + ca$, so\n$$\n\\frac{bc + a + 1}{a^2 + 1} = 1 + \\frac{bc + a - a^2}{a^2 + 1} = 1 + \\frac{ab + bc + ca}{a^2 + 1},\n$$\nand the similar relations. The inequality to be proven is equivalent with\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} + \\frac{1}{c^2 + 1} \\le \\frac{9}{10(ab + bc + ca)}. \\quad (*)\n$$\nLet $m = ab + bc + ca$; then $a^2 + b^2 + c^2 = 1 - 2m$ and, because $ab + bc + ca \\le a^2 + b^2 + c^2$, we have $0 < m \\le \\frac{1}{3}$.\nSince $\\frac{1}{a^2 + 1} = 1 - \\frac{a^2}{a^2 + 1}$, we can rewrite the inequality to be proven as\n$$\n\\frac{a^2}{a^2 + 1} + \\frac{b^2}{b^2 + 1} + \\frac{c^2}{c^2 + 1} \\ge 3 - \\frac{9}{10m}.\n$$\nUsing a variant of Cauchy-Schwarz inequality, we have\n$$\n\\frac{a^2}{a^2 + 1} + \\frac{b^2}{b^2 + 1} + \\frac{c^2}{c^2 + 1} \\ge \\frac{(a+b+c)^2}{a^2 + b^2 + c^2 + 3} = \\frac{1}{4-2m}.\n$$\nIt is enough to prove that $\\frac{1}{4-2m} \\ge 3 - \\frac{9}{10m}$, that is $(3m-1)(5m-9) \\ge 0$, which is true, because $m \\in (0, \\frac{1}{3}]$. The equality holds when $a = b = c$ and $m = \\frac{1}{3}$, so $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider triangle $ABC$ inscribed in circle $\\omega$, and an interior point $P$. Lines $AP, BP$ and $CP$ intersect the circle $\\omega$ for the second time at points $D, E, F$, respectively. Let $A', B', C'$ be the reflections of $A, B, C$ in the lines $EF, FD, DE$ respectively. Show that triangle $A'B'C'$ is similar to $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove that $\\triangle PAB \\sim \\triangle PA'B'$, and similarly $\\triangle PBC \\sim \\triangle PB'C'$ and $\\triangle PCA \\sim \\triangle PC'A'$; obviously, these three triangle similarities are enough in order to prove that $\\triangle ABC \\sim \\triangle A'B'C'$, either from angle equalities or side ratios.\n\nObviously, we have $\\triangle PAE \\sim \\triangle PBD$, so $\\frac{AE}{PE} = \\frac{BD}{PD}$. For symmetry reasons, we have $AE = A'E$ and $BD = B'D$, so $\\frac{A'E}{PE} = \\frac{B'D}{PD}$.\n\nAlso,\n$$\n\\begin{align*} \n\\angle A'EP &= |\\angle PEF - \\angle A'EF| = |\\angle BEF - \\angle AEF| = |\\angle BDF - \\angle ADF| \\\\ \n&= |\\angle B'DF - \\angle PDF| = \\angle B'DP. \n\\end{align*}\n$$\nIt follows that $\\triangle PA'E \\sim \\triangle PB'D$, whence\n$$\n\\frac{PA'}{PE} = \\frac{PB'}{PD} \\text{ and } \\angle A'PE \\equiv \\angle B'PD,\n$$\nso $\\angle A'PB' \\equiv \\angle EPD$. Therefore, $\\triangle A'PB' \\sim \\triangle EPD$, and, since $\\triangle EPD \\sim \\triangle APB$, we get $\\triangle PAB \\sim \\triangle PA'B'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z > 0$. Prove that\n$$\n\\frac{x^3}{z^3 + x^2 y} + \\frac{y^3}{x^3 + y^2 z} + \\frac{z^3}{y^3 + z^2 x} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the AM-GM inequality we get $x^2 y \\le \\frac{x^3 + x^3 + y^3}{3}$, which leads to\n$$\n\\frac{x^3}{z^3 + x^2 y} \\ge \\frac{x^3}{z^3 + \\frac{x^3 + x^3 + y^3}{3}} = \\frac{3x^3}{2x^3 + y^3 + 3z^3}.\n$$\nSumming this with the other two similar inequalities and putting $x^3 = a$, $y^3 = b$, $z^3 = c$, it is sufficient to prove that\n$$\n\\frac{a}{2a + b + 3c} + \\frac{b}{3a + 2b + c} + \\frac{c}{a + 3b + 2c} \\ge \\frac{1}{2},\n$$\nwhich follows from\n$$\n\\sum \\frac{a}{2a + b + 3c} = \\sum \\frac{a^2}{2a^2 + ab + 3ac} \\ge \\frac{(a+b+c)^2}{2(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)} = \\frac{1}{2}.\n$$\nEquality holds for $x = y = z$.\n*Alternative Solution.* (Mihai Iliant) Rewriting the LHS and applying CBS in the form Titu's Lemma we get\n$$\n\\begin{aligned}\n\\frac{x^3}{z^3 + x^2 y} + \\frac{y^3}{x^3 + y^2 z} + \\frac{z^3}{y^3 + z^2 x} &= \\frac{x^4}{x z^3 + x^3 y} + \\frac{y^4}{y x^3 + y^3 z} + \\frac{z^4}{z y^3 + z^3 x} \\\\\n&\\geq \\frac{(x^2 + y^2 + z^2)^2}{2(x^3 y + y^3 z + z^3 x)}\n\\end{aligned}\n$$\nIt is therefore sufficient to prove that $(x^2 + y^2 + z^2)^2 \\ge 3(x^3 y + y^3 z + z^3 x)$, $\\forall x, y, z > 0$, which is Vasc$^2$ inequality. A short proof: it reduces to $(x + y + z)^2 \\ge 3(xy + yz + zx)$, where $x = a^2 + bc - ca$, $y = b^2 + ca - ab$, $z = c^2 + ab - bc$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38270,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *Pythagorean triple* is a solution of the equation $x^2 + y^2 = z^2$ in positive integers such that $x < y$. Given any non-negative integer $n$, show that some positive integer appears in precisely $n$ distinct Pythagorean triples.\nAMM Magazine",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show by induction on $n \\ge 0$, that $2^{n+1}$ appears in precisely $n$ distinct Pythagorean triples. Since no Pythagorean triple contains $2$, the assertion holds for $n=0$.\n\nFor the induction step, let $n \\ge 1$, and assume that $2^n$ appears in exactly $n-1$ distinct Pythagorean triples. The latter produce $n-1$ distinct non-primitive Pythagorean triples each containing $2^{n+1}$. To conclude the proof, we show that $2^{n+1}$ appears exactly once in a primitive Pythagorean triple. Recall that the primitive Pythagorean triples are described by the well-known formulae $x = v^2 - u^2$, $y = 2uv$, $z = u^2 + v^2$, where $u$ and $v$ are coprime positive integers, not both odd, and $u < v$. Since $x$ and $z$ are both odd, if $2^{n+1}$ appears in the triple, then $2^{n+1} = y = 2uv$, and since $u < v$ and $u$ and $v$ have opposite parity, necessarily $u=1$ and $v = 2^n$. Consequently, $2^{n+1}$ appears in exactly $n$ distinct Pythagorean triples.\nAlternative solution 1:\nIf $P(m)$ is the number of Pythagorean triples containing the positive integer $m$, and if $P_0(m)$ is the number of primitive such triples, then $P(m) = \\sum_{d|m} P_0(d)$. Since $P_0(1) = P_0(2) = 0$ and $P_0(2^k) = 1$, $k \\ge 2$ (as in the previous solution), it follows that $P(2^{n+1}) = n$, so $2^{n+1}$ appears in exactly $n$ distinct Pythagorean triples.\n\nAlternative solution 2:\nWe show that if $p$ is a prime congruent to $3$ modulo $4$, then $p^n$ appears in exactly $n$ Pythagorean triples, and is moreover always the smallest entry of any such.\nSince $p$ is congruent to $3$ modulo $4$, $-1$ is a quadratic non-residue modulo $p$, so no power of $p$ can be the largest entry of a Pythagorean triple. Hence, if $p^n$ is a member of a Pythagorean triple, then $p^{2n} = b^2 - a^2$ for some positive integers $a < b$, so $b - a = p^k$ and $b + a = p^{2n-k}$ for some non-negative integer $k < n$. Clearly, every such $k$ corresponds to a solution and there are precisely $n$ distinct Pythagorean triples containing $p^n$, namely,\n$$\np^n, \\quad p^k(p^{2(n-k)} - 1)/2, \\quad p^k(p^{2(n-k)} + 1)/2, \\quad k = 0, 1, \\dots, n-1.\n$$\nIt is worth noticing that this argument avoids appealing to the parametric representation of Pythagorean triples.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHatározd meg azokat a folytonos és növekvő $f:[0, \\infty) \\rightarrow \\mathbb{R}$ függvényeket, amelyekre\n$$\n\\int_{0}^{x+y} f(t) \\mathrm{d} t \\leq \\int_{0}^{x} f(t) \\mathrm{d} t+\\int_{0}^{y} f(t) \\mathrm{d} t\n$$\nbármely $x, y \\in[0, \\infty)$ esetén!\n\nProblem:\n\nDeterminaţi funcţiile continue şi crescătoare $f:[0, \\infty) \\rightarrow \\mathbb{R}$, care îndeplinesc condiţia\n$$\n\\int_{0}^{x+y} f(t) \\mathrm{d} t \\leq \\int_{0}^{x} f(t) \\mathrm{d} t+\\int_{0}^{y} f(t) \\mathrm{d} t\n$$\noricare ar fi $x, y \\in[0, \\infty)$.",
    "options": [],
    "answer": "All constant functions: f(x) = c for all x ≥ 0, where c is a real constant.",
    "solution": "Solution:\n\nInegalitatea din enunţ este echivalentă cu\n$$\n\\int_{x}^{x+y} f(t) \\mathrm{d} t \\leq \\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\nde unde\n$$\n\\int_{0}^{y} f(t+x) \\mathrm{d} t \\leq \\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\noricare ar fi $x \\geq 0$ şi $y \\geq 0$.\n\nCum $f(t+x) \\geq f(t)$, oricare ar fi $t \\in[0, y]$ şi oricare ar fi $x \\geq 0$, rezultă că\n$$\n\\int_{0}^{y} f(t+x) \\mathrm{d} t \\geq \\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\ndeci\n$$\n\\int_{0}^{y} f(t+x) \\mathrm{d} t=\\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\noricare ar fi $x \\geq 0$ şi $y \\geq 0$.\n\nDin continuitatea lui $f$ deducem că $f(x+y)=f(y)$, oricare ar fi $x \\geq 0$ şi oricare ar fi $y \\geq 0$. In particular, $f(x)=f(0)$, oricare ar fi $x \\geq 0$, deci $f$ este constantă.\n\nEvident, orice funcţie constantă verifică condiţiile din enunţ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the interior of the triangle $ABC$. The lines $AP$, $BP$, $CP$ intersect $BC$, $AC$, $AB$ at $A_1$, $B_1$, $C_1$, respectively. Given that\n$$\ns(PBA_1) + s(PCB_1) + s(PAC_1) = \\frac{1}{2}s(ABC),\n$$\n(where $s(XYZ)$ denotes the area of $XYZ$) prove that $P$ lies on one of the medians of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\frac{s(PBA_1)}{s(ABA_1)} = \\frac{PA_1}{AA_1} = \\frac{s(BPC)}{s(BAC)}$.\nDenote $s(BPC) = s_a$, etc.; it follows that $\\frac{s(PBA_1)}{s_c + s(PBA_1)} = \\frac{s_a}{s}$, where $s = s(ABC)$, hence $s(PBA_1) = \\frac{s_a s_c}{s - s_a} = \\frac{s_a s_c}{s_b + s_c}$.\nThe given equality becomes $\\frac{s_a s_c}{s_b + s_c} + \\frac{s_b s_a}{s_c + s_a} + \\frac{s_c s_b}{s_a + s_b} = \\frac{s_a + s_b + s_c}{2}$, which is equivalent to $s(s_a - s_b)(s_b - s_c)(s_c - s_a) = 0$. The conclusion follows easily.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the continuous increasing functions $f: [0, \\infty) \\to \\mathbb{R}$ satisfying\n$$\n\\int_{0}^{x+y} f(t) dt = \\int_{0}^{x} f(t) dt + \\int_{0}^{y} f(t) dt,\n$$\nfor all non-negative real numbers $x$ and $y$.",
    "options": [],
    "answer": "All constant functions f(x) = c for x ≥ 0",
    "solution": "Clearly, every constant function satisfies the required conditions. Conversely, write the condition in the statement in the equivalent form\n$$\n\\int_{x}^{x+y} f(t) \\, dt \\le \\int_{0}^{y} f(t) \\, dt\n$$\nto infer that $\\int_{0}^{y} f(t+x) \\, dt \\le \\int_{0}^{y} f(t) \\, dt$ for all non-negative $x$ and $y$.\nOn the other hand, since $f$ is increasing, $f(t+x) \\ge f(t)$ for all $t$ in the closed interval $[0, y]$ and all $x \\ge 0$, so $\\int_{0}^{y} f(t+x) \\, dt \\ge \\int_{0}^{y} f(t) \\, dt$.\nConsequently, $\\int_{0}^{y} f(t+x) \\, dt = \\int_{0}^{y} f(t) \\, dt$ for all non-negative $x$ and $y$. Continuity of $f$ forces $f(x+y) = f(y)$ for all non-negative $x$ and $y$. In particular, $f(x) = f(0)$ for all $x \\ge 0$, so $f$ is constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C$ be the set of all functions $f: [0, 1] \\to \\mathbb{R}$, twice differentiable on $[0, 1]$, with at least two (not necessarily distinct) zeroes in $[0, 1]$ and such that $|f''(x)| \\le 1$, for all $x$ in $[0, 1]$. Find the maximum possible value of the integral\n$$\n\\int_0^1 |f(x)| \\, dx,\n$$\nwhen $f$ runs through $C$, and find the functions for which the maximum is reached. (A double zero is a point $a$ such that $f(a) = f'(a) = 0$.)",
    "options": [],
    "answer": "Maximum value: 1/6. It is achieved precisely by f(x) = ±(x^2/2) and f(x) = ±((1 − x)^2/2) on [0,1].",
    "solution": "The required maximum is $1/6$ and is achieved at $x \\mapsto x^2/2$, $0 \\le x \\le 1$, or $x \\mapsto (1-x)^2/2$, $0 \\le x \\le 1$, or their negatives alone.\n\nTo prove this, fix a function $f$ in $C$ and let $Z = \\{x: 0 \\le x \\le 1 \\text{ and } f(x) = 0\\}$ be the set of its zeroes. We show that there exist $a \\le b$ in $Z$ such that\n$$\n|f(x)| \\le \\frac{1}{2} |x - a| \\cdot |x - b|, \\quad (*)\n$$\nfor all $x$ in $[0, 1]$. Distinguish two possible cases: $|Z| = 1$ and $|Z| > 1$.\n\nIn the first case, let $a$ be the single element of $Z$, so $f(a) = f'(a) = 0$. Since (*) clearly holds at $x = a$, fix an $x \\neq a$ in $[0, 1]$ and apply Taylor's theorem to write $f(x) = \\frac{1}{2}(x-a)^2 f''(\\theta)$ for some $\\theta$ between $a$ and $x$. Since $|f''(t)| \\le 1$ for all $t$ in $[0, 1]$, (*) follows.\n\nIn the second case, let $a < b$ be two elements of $Z$. Since (*) clearly holds at both $x = a$ and $x = b$, fix an $x$ in $[0, 1]$, $x \\neq a$, $x \\neq b$, and consider the function $g: [0, 1] \\to \\mathbb{R}$, $g(t) = (t-a)(t-b)f(x) - (x-a)(x-b)f(t)$. This function is twice differentiable on $[0, 1]$ and has at least three distinct zeroes, namely, $a, b$ and $x$. Apply Rolle's theorem twice to deduce that $g''$ has a zero $\\theta$ in the open interval $(0, 1)$; that is, $f(x) = \\frac{1}{2}(x-a)(x-b)f''(\\theta)$, and (*) follows again by boundedness of $f''$.\n\nFinally, integration of (*) on the closed unit interval $[0, 1]$ yields\n$$\n\\begin{aligned}\n\\int_{0}^{1} |f(x)| \\, dx &\\le \\frac{1}{2} \\int_{0}^{1} |x-a| \\cdot |x-b| \\, dx \\\\\n&= \\frac{1}{6} - \\frac{a+b}{4} + \\frac{ab}{2} + \\frac{(b-a)^3}{6} \\\\\n&= \\frac{1}{6} - \\frac{1}{2}a(1-b) - \\frac{1}{12}(b-a)(3-2(b-a)^2) \\le \\frac{1}{6}.\n\\end{aligned}\n$$\nWith reference to continuity, equality holds throughout if and only if either $a = b = 0$ and $|f(x)| = x^2/2, 0 \\le x \\le 1$, or $a = b = 1$ and $|f(x)| = (1-x)^2/2, 0 \\le x \\le 1$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $p(a)$ the first digit of the natural number $a$. Show that each of the sets\n$$\nA = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) > 0\\}, \\quad B = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) < 0\\}\n$$\nhas infinitely many elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k \\in \\mathbb{N}^*$ there exists $n_k \\in \\mathbb{N}^*$ such that $2^{n_k} < 10^k < 2^{n_k+1}$ ($n_k + 1$ is the smallest element of the set $\\{m \\in \\mathbb{N}^* \\mid 10^k < 2^m\\}$). As $2^{n_k} < 10^k < 2^{n_k+1}$, we get $10^k < 2^{n_k+1} < 2 \\cdot 10^k$, so $p(2^{n_k+1}) = 1$.\n\nMultiplying by $5^{n_k+1}$, we have $10^k \\cdot 5^{n_k+1} < 10^{n_k+1} < 2 \\cdot 10^k \\cdot 5^{n_k+1}$. Dividing the first inequality by $10^k$ and the second by $2 \\cdot 10^k$, we get $5 \\cdot 10^{n_k-k} < 5^{n_k+1} < 10^{n_k-k+1}$, so $p(5^{n_k+1}) \\ge 5 > p(2^{n_k+1})$.\n\nFrom $2^{n_k} < 10^k < 2^{n_k+1}$, by dividing the second inequality by $2$, we conclude $5 \\cdot 10^{k-1} < 2^{n_k} < 10^k$, so $p(2^{n_k}) \\ge 5$. Multiplying the first inequality by $5^{n_k}$, we obtain $10^{k-1} \\cdot 5^{n_k+1} < 10^{n_k} < 10^k \\cdot 5^{n_k}$, so by dividing the first one by $5 \\cdot 10^{k-1}$ and the second by $10^k$, we obtain $10^{n_k-k} < 5^{n_k} < 2 \\cdot 10^{n_k-k}$, that is $p(5^{n_k}) = 1 < p(2^{n_k})$.\n\nThe set $\\{n_k \\mid k \\in \\mathbb{N}^*\\}$ has an infinity of elements because $n_k > k$. As $\\{n_k \\mid k \\in \\mathbb{N}^*\\} \\subset A$ and $\\{n_k + 1 \\mid k \\in \\mathbb{N}^*\\} \\subset B$, the sets $A$ and $B$ also have infinitely many elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime greater than $3$. For a positive integer $k$, let $R(k)$ denote the remainder of $k$ when divided by $p$. Determine all positive integers $a < p$ such that $m + R(ma) > a$ for every $m = 1, 2, \\dots, p-1$.",
    "options": [],
    "answer": "All a with a = p − 1 and all a of the form a = ⎣p/q⎦ for integers q = 2, 3, …, p − 1 (duplicates ignored). Equivalently: a = p − 1 together with the integers 1, 2, …, ⎣√p⎦ and the distinct values ⎣p/q⎦ for q = 2, …, ⎣√p − 1/2⎦.",
    "solution": "The required integers are $p-1$ along with all the numbers of the form $\\lfloor p/q \\rfloor$, $q = 2, \\dots, p-1$. In other words, these are $p-1$, along with the numbers $1, 2, \\dots, \\lfloor \\sqrt{p} \\rfloor$, and also the (distinct) numbers $\\lfloor p/q \\rfloor$, $q = 2, \\dots, \\lfloor \\sqrt{p} - 1/2 \\rfloor$.\n\nWe begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m + R(m(p-1)) = m + (p-m) = p > p-1$ for all $m = 1, \\dots, p-1$.\n\nNow, consider any number $a$ of the form $a = \\lfloor p/q \\rfloor$, where $q$ is an integer greater than $1$ but less than $p$; then $p = az + r$ with $0 < r < q$. Choose any positive integer $m < p$ and write $m = xq + y$, where $x$ and $y$ are integers, and $0 < y \\le q$ (notice that $x$ is non-negative). Then $R(ma) = R(ay+xaq) = R(ay+xp-xr) = R(ay-xr)$. Since $ay-xr \\le ay \\le az < p$, it follows that $R(ay-xr) \\ge ay-xr$, so $m+R(ma) \\ge (xq+y) + (ay-xr) = x(q-r) + y(a+1) \\ge a+1$, by $q > r$ and $y \\ge 1$. Thus $a$ satisfies the required condition.\n\nFinally, we show that if a positive integer $a < p-1$ satisfies the required condition then $a$ is indeed of the form $a = \\lfloor p/q \\rfloor$ for some positive integer $q < p$. This is clear if $a=1$, so we may (and will) assume that $a \\ge 2$.\n\nWrite $p = az + r$, where $q$ and $r$ are integers, and $0 < r < a$; since $a \\ge 2$, it follows that $q < p/2$. Let $m = q+1 < p$ to get $R(ma) = R(aq + a) = R(p+(a-r)) = a-r$, so $a < m + R(ma) = q+1+a-r$ which yields $r < q+1$. Moreover, if $r=q$, then $p=q(a+1)$ which is impossible by $1 < a+1 < p$. Thus $r < q$, and $0 \\le p/q - a = r/q < 1$, which proves $a = \\lfloor p/q \\rfloor$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38277,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set of positive integers is divided into subsets in the following way:\n$$\n\\{1, 2\\}, \\{3, 4, 5\\}, \\{6, 7, 8, 9\\}, \\{10, 11, 12, 13, 14\\}, \\dots\n$$\na) Find the smallest element of the 100th subset.\nb) Is 2015 the largest element of such a subset?",
    "options": [],
    "answer": "a) 5050; b) Yes — 2015 is the largest element of the 62nd subset.",
    "solution": "a) The first 99 subsets contain $2 + 3 + \\dots + 100 = 5049$ elements. In the first 99 subsets are written the numbers $1, 2, 3, \\dots, 5049$, so the smallest element of the 100th subset is $5050$.\n\nb) If $2015$ is the largest element of the $n$-th subset, then $2 + 3 + \\dots + (n + 1) = 2015$. Adding $1$ in both members yields $(n + 1)(n + 2) = 2 \\cdot 2016 = 4032$. Since $63 \\cdot 64 = 4032$, it follows that $2015$ is the largest element of the $62$nd subset.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that if $k \\ge 1$ and $n \\ge 2$ are integers such that there exists $A, B \\in \\mathcal{M}_n(\\mathbb{Z})$ with the property that $A^3 = O_n$ and $A^k B + BA = I_n$, then $k = 1$ and $n$ is even. Conversely, show that if $n \\ge 2$ is even, then there exists $A, B \\in \\mathcal{M}_n(\\mathbb{Z})$ such that $A^3 = O_n$ and $AB + BA = I_n$.\nDorel Miheț",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $k \\ge 3$, then $BA = I_n$ (since $A^k = O_n$), so $A$ is invertible, which contradicts $A^3 = O_n$.\n\nIf $k = 2$, then relation $A^2B + BA = I_n$, multiplied at the left by $A$ and then at the right by $A^2$, leads to $ABA = A$ and $A^2BA^2 = A^2$. The last equality can be written $A(ABA)A = A^2$, whence $A^3 = A^2$, therefore $A^2 = O_n$, giving $BA = I_n$, in contradiction with $A^3 = O_n$.\n\nThis shows that if such $k$ and $n$ exist, then $k = 1$. From $\\mathrm{Tr}(AB) = \\mathrm{Tr}(BA) \\in \\mathbb{Z}$ and $AB + BA = I_n$ follows $2\\mathrm{Tr}(AB) = n$, that is $n$ is even.\n\nWe proceed now to prove the converse.\nFor $n = 2$ we can choose: $A = \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}$; it is easy to check that $AB + BA = I_2$ and $A^2 = B^2 = O_2$.\n\nIf $n = 2k, k \\ge 2$, the $2k \\times 2k$ matrices $A$ and $B$, having on the 'main diagonal' $k$ matrices $\\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ and respectively $k$ matrices $\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}$ and all the remaining elements 0, fulfill $AB + BA = I_n$ and $A^2 = B^2 = O_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $[AC]$ of triangle $ABC$ and $N \\in (AM)$. The parallel from $N$ at $AB$ intersects $BM$ in $P$, the parallel from $M$ at $BC$ intersects $BN$ in $Q$ and the parallel from $N$ at $AQ$ intersects $BC$ in $S$. Prove that the straight lines $PS$ and $AC$ are parallel.\n\nCosmin Manea, Dragoş Petrică",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $MQ \\cap AB = \\{E\\}$ and $\\{D\\} = NP \\cap ME$. Then $EA = EB$ and $ND = DP$. Since $ANPB$ is a trapezoid, points $A, Q, P$ are collinear.\n\nFrom $\\triangle ADP \\equiv \\triangle SDN$ (A.S.A.) follows $[AP] \\equiv [SN]$ and, since $AP \\parallel SN$, quadrilateral $ANSP$ is a parallelogram, hence $PS \\parallel AC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be integers such that $m \\ge 2$ and $n \\ge 3$. Prove that there exists $m$ distinct positive integers $a_1, a_2, a_3, \\dots, a_m$, all divisible by $n-1$, such that\n$$\n\\frac{1}{n} = \\frac{1}{a_1} - \\frac{1}{a_2} + \\frac{1}{a_3} - \\dots + (-1)^{m-1} \\frac{1}{a_m}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $(a_n)_{n \\ge 1}$ be a geometric sequence with ratio $q = n - 1$. Then\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}{a_p} = \\frac{1/a_1 + 1/q \\cdot (-1)^{p-1} 1/a_p}{1 + 1/q}, \\forall p \\in \\mathbb{N}^*,\n$$\ni.e.\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}{a_p} = \\frac{n-1}{n a_1} + \\frac{(-1)^{p-1}}{n a_p}.\n$$\nThe required numbers can be obtained for $a_1 = n-1$, $m = p+1$ and $a_m = n a_p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38281,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n \\ge 2$ is lucky if $n^2$ can be represented as a sum of $n$ consecutive positive integers. Prove that:\n\na) the number $7$ is lucky,\n\nb) the number $10$ is not lucky,\n\nc) the product of any two lucky numbers is a lucky number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We seek for seven consecutive integers $a, a+1, a+2, a+3, a+4, a+5, a+6$ such that $7^2 = a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6$. The equality gives $49 = 7a + 21$, hence $a = 4$. As $7^2 = 4 + 5 + 6 + 7 + 8 + 9 + 10$, number $7$ is lucky.\n\nb) Suppose there exist ten consecutive integers, namely $a, a+1, a+2, \\dots, a+9$, such that $10^2 = a + a + 1 + a + 2 + \\dots + a + 9$. The equality gives $100 = 10a + 45$, with no integer solution. Number $10$ is not lucky.\n\nc) We claim that a number $m$ is lucky if and only if $m$ is odd. Suppose $m$ is a lucky number and write $m^2 = (a+1) + (a+2) + (a+3) + \\dots + (a+m)$ for a given positive integer $a$. Then\n$$\nm^2 = m \\cdot a + (1 + 2 + 3 + \\dots + m) \\Leftrightarrow m^2 = m \\cdot a + \\frac{m \\cdot (m+1)}{2},\n$$\nhence $m = 2a + 1$, so $m$ is odd. Conversely, let $m = 2k + 1$ be an odd number. As $(2k+1)^2 = (k+1) + (k+2) + \\dots + (3k+1)$, the number $m$ is lucky.\n\nWe conclude noticing that the product of two odd numbers is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all perfect squares $\\overline{aabcd}$ with $a \\neq 0$ and $d \\neq 0$ such that the number $\\overline{dcbaa}$ is also a perfect square. $(a, b, c, d, e$ are decimal digits)",
    "options": [],
    "answer": "44521 and 44944",
    "solution": "Answer: 44521 and 44944.\nSuppose $\\overline{aabcd}$ and $\\overline{dcbaa}$ are both squares. Then $a$ and $d$ must belong to $\\{1, 4, 5, 6, 9\\}$.\n\nThe numbers $\\overline{dcb55}$ are multiples of 5, but not of 25, hence they cannot be squares. The numbers $\\overline{dcb66}$ are even but not multiples of 4, hence they cannot be squares. The numbers $\\overline{dcb11}$ and $\\overline{dcb99}$ leave the remainder 3 when divided by 4 – again, they cannot be squares.\n\nWe are left with $a = 4$. As $209^2 < 44000$ and $213^2 > 45000$, we consider the cases $\\overline{aabcd} \\in \\{210^2, 211^2, 212^2\\}$. Since $210^2 = 44100$ implies $d = 0$, we reject this case. Finally, given that $211^2 = 44521$, $12544 = 112^2$ and $212^2 = 44944$, we find the two solutions: 44521 and 44944.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $z$ such that\n$$\n|z| + |z - 5i| = |z - 2i| + |z - 3i|.\n$$",
    "options": [],
    "answer": "z = a i with a ∈ (-∞, 0] ∪ [5, +∞)",
    "solution": "We notice that $|z - 2i| = \\left|\\frac{2}{5}(z - 5i) + \\frac{3}{5}z\\right| \\le \\frac{2}{5}|z - 5i| + \\frac{3}{5}|z|$.\n\nIn the same way, $|z - 3i| = \\left|\\frac{3}{5}(z - 5i) + \\frac{2}{5}z\\right| \\le \\frac{3}{5}|z - 5i| + \\frac{2}{5}|z|$, whence $|z| + |z - 5i| \\ge |z - 2i| + |z - 3i|$.\n\nEquality takes place if there exists $\\lambda \\ge 0$ such that $z - 5i = \\lambda z$ or if $z = 0$, i.e. $z = ai$, with $a \\in (-\\infty, 0] \\cup [5, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\ge 0$ and let $(x_n)_{n \\ge 1}$ be a sequence of real numbers. Given that $\\left(\\frac{x_1+\\dots+x_n}{n^a}\\right)_{n \\ge 1}$ is a bounded sequence, prove that the sequence $(y_n)_{n \\ge 1}$, defined by\n$$\ny_n = \\frac{x_1}{1^b} + \\frac{x_2}{2^b} + \\dots + \\frac{x_n}{n^b},\n$$\nis a convergent sequence for all $b > a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_n = \\sum_{k=1}^{n} x_k$, $n \\in \\mathbb{N}^*$. Using the hypothesis, one can find a constant $c > 0$ such that $|S_n| \\le c n^a$, $\\forall n \\in \\mathbb{N}^*$. Let $n, p \\in \\mathbb{N}^*$; we have:\n$$\n\\begin{align*}\n|y_{n+p} - y_n| &= \\left| \\sum_{k=n+1}^{n+p} \\frac{x_k}{k^b} \\right| = \\left| \\sum_{k=n+1}^{n+p} \\frac{S_k - S_{k-1}}{k^b} \\right| = \\\\\n&= \\left| \\frac{S_{n+p}}{(n+p+1)^b} - \\frac{S_n}{(n+1)^b} + \\sum_{k=n+1}^{n+p} S_k \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\right| \\\\\n&\\le \\frac{|S_{n+p}|}{(n+p+1)^b} + \\frac{|S_n|}{(n+1)^b} + \\sum_{k=n+1}^{n+p} |S_k| \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\\\\n&\\le c \\left[ \\frac{2}{n^{b-a}} + \\sum_{k=n+1}^{n+p} k^a \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\right].\n\\end{align*}\n$$\nApplying the mean value theorem to the function $f(x) = x^{-\\alpha}$, $x > 0$ on the interval $[i, i+1]$, with $\\alpha, i > 0$, yields\n$$\n\\frac{\\alpha}{(i+1)^{\\alpha+1}} < \\frac{1}{i^{\\alpha}} - \\frac{1}{(i+1)^{\\alpha}} < \\frac{\\alpha}{i^{\\alpha+1}},\n$$\nand since $b, b-a > 0$, we obtain\n$$\n\\frac{1}{k^b} - \\frac{1}{(k+1)^b} < \\frac{b}{k^{b+1}} \\quad \\text{and} \\quad \\frac{b-a}{k^{b-a+1}} < \\frac{1}{(k-1)^{b-a}} - \\frac{1}{k^{b-a}}, \\forall k \\in \\mathbb{N}, k \\ge 2.\n$$\nThen\n$$\n\\begin{align*}\n\\sum_{k=n+1}^{n+p} k^a \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) &< \\sum_{k=n+1}^{n+p} \\frac{b k^a}{k^{b+1}} = \\frac{b}{b-a} \\sum_{k=n+1}^{n+p} \\frac{b-a}{k^{b-a+1}} \\\\\n&< \\frac{b}{b-a} \\sum_{k=n+1}^{n+p} \\left( \\frac{1}{(k-1)^{b-a}} - \\frac{1}{k^{b-a}} \\right) \\\\\n&= \\frac{b}{b-a} \\left( \\frac{1}{n^{b-a}} - \\frac{1}{(n+p)^{b-a}} \\right) < \\frac{b}{(b-a)n^{b-a}}.\n\\end{align*}\n$$\nIt follows that\n$$\n|y_{n+p} - y_n| < c \\left( 2 + \\frac{b}{b-a} \\right) \\frac{1}{n^{b-a}}, \\forall n, p \\in \\mathbb{N}^*.\n$$\nFinally, since $\\lim_{n \\to \\infty} n^{-(b-a)} = 0$, we obtain that $(y_n)_{n \\ge 1}$ is a Cauchy sequence, hence a convergent one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$, such that $\\angle B = 90^\\circ$. Denote by $I$ the in-center and let $F$, $D$ and $E$ be the points where the incircle touches sides $[AB]$, $[BC]$, and $[AC]$ respectively. If $CI \\cap EF = \\{M\\}$ and $DM \\cap AB = \\{N\\}$, show that:\n\na) $AI = ND$;\n\nb) $FM = \\frac{EI \\cdot EM}{EC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Triangle $AFE$ is isosceles with $AE = AF$, and $AI \\perp FE$, hence $\\angle AEF = 90^\\circ - \\angle A/2$. In the same way from the isosceles triangle $CDE$ we get $\\angle DEC = 90^\\circ - \\angle C/2$. As a consequence $\\overline{MED} = 180^\\circ - \\overline{AEF} - \\overline{DEC} = 180^\\circ - (180^\\circ - (\\hat{A} + \\hat{C})/2) = 45^\\circ$. (*)\n\nAs $\\triangle MDC \\equiv \\triangle MEC$, we obtain $MD = ME$ (**). By (*) and (**), the triangle $\\triangle MED$ is right angled and isosceles. As a consequence $DN \\perp EF$ and, because $AI \\perp EF$, we obtain $DN \\parallel AI$. As $AN \\parallel ID$ we conclude that the quadrilateral $ANDI$ is a parallelogram, so $AI = ND$.\n\nb. We have $\\overline{EFD} = 180^\\circ - \\overline{AFE} - \\overline{BFD} = (\\hat{A} + \\hat{B})/2 = 90^\\circ - \\hat{C}/2 = \\overline{DIC}$ so $\\triangle FMD \\sim \\triangle IDC$. We conclude $\\frac{FM}{ID} = \\frac{MD}{DC}$ which implies\n$$\nFM = \\frac{ID \\cdot MD}{DC} = \\frac{EI \\cdot EM}{EC}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHatározd meg az összes olyan $k \\geq 1$ és $n \\geq 2$ természetes számot, amelyre létezik $A, B \\in \\mathcal{M}_{n}(\\mathbb{Z})$ úgy, hogy $A^{3}=O_{n}$ és $A^{k} B+B A=I_{n}$.\n\nProblem:\n\nDeterminaţi toate numerele naturale $k \\geq 1$ şi $n \\geq 2$ cu proprietatea că există $A, B \\in \\mathcal{M}_{n}(\\mathbb{Z})$ astfel încât $A^{3}=O_{n}$ si $A^{k} B+B A=I_{n}$.",
    "options": [],
    "answer": "k = 1 and n is an even integer at least 2.",
    "solution": "Solution:\n\nFie $A, B \\in \\mathcal{M}_{n}(\\mathbb{Z})$ astfel încât $A^{3}=O_{n}$ şi $A^{k} B+B A=I_{n}$.\n\nDacă $k \\geq 3$, atunci $B A=I_{n}$ (deoarece $A^{k}=O_{n}$ ), deci $A$ este inversabilă, în contradicţie cu $A^{3}=O_{n}$. \n\nDacă $k=2$ atunci din $A^{2} B+B A=I_{n}$, prin înmulţire la stânga cu $A$ si apoi la dreapta cu $A^{2}$, rezultă $A B A=A$ şi $A^{2} B A^{2}=A^{2}$. Scriind ultima egalitate sub forma $A(A B A) A=A^{2}$, obţinem $A^{3}=A^{2}$, deci $A^{2}=O_{n}$. Atunci $B A=I_{n}$, în contradicţie cu $A^{3}=O_{n}$. \n\nPrin urmare, dacă există $k$ şi $n$ ca în enunţ, atunci $k=1$. Din $\\operatorname{Tr}(A B)=\\operatorname{Tr}(B A) \\in \\mathbb{Z}$ şi $A B+B A=I_{n}$ rezultă $2 \\operatorname{Tr}(A B)=n$, deci $n$ este un număr natural par. \n\nArătăm în continuare că, pentru orice număr natural par $n \\geq 2$, există $A, B \\in \\mathcal{M}_{n}(\\mathbb{Z})$ astfel încât $A^{3}=O_{n}$ si $A B+B A=I_{n}$.\n\nPentru $n=2$, putem alege matricele $A=\\left(\\begin{array}{ll}0 & 1 \\\\ 0 & 0\\end{array}\\right)$ şi $B=\\left(\\begin{array}{ll}0 & 0 \\\\ 1 & 0\\end{array}\\right)$, care satisfac condiţiile $A B+B A=I_{2}$ şi $A^{2}=B^{2}=O_{2}$.\n\nPentru $n=2 k$, cu $k \\geq 2$, matricele bloc diagonale $A$ şi $B$, de dimensiune $2 k$, care au pe diagonala principală $k$ matrice $\\left(\\begin{array}{ll}0 & 1 \\\\ 0 & 0\\end{array}\\right)$ şi respectiv $k$ matrice $\\left(\\begin{array}{ll}0 & 0 \\\\ 1 & 0\\end{array}\\right)$, iar restul coeficienţilor nuli, satisfac relaţiile $A B+B A=I_{n}$ şi $A^{2}=B^{2}=O_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f, g : \\mathbb{Q} \\to \\mathbb{Q}$ such that, for all $x, y \\in \\mathbb{Q}$,\n$$\nf(g(x) + g(y)) = f(g(x)) + y,\n$$\n$$\ng(f(x) + f(y)) = g(f(x)) + y,\n$$",
    "options": [],
    "answer": "f(x) = a x and g(x) = x / a for all rational x, where a is a nonzero rational number",
    "solution": "If $g(y_1) = g(y_2)$, the first equality yields $y_1 = y_2$, hence $g$ is injective. Analogously, $f$ is injective, as well.\nPlugging $y = 0$ in the first equality gives $f(g(x) + g(0)) = f(g(x))$, hence $g(x) + g(0) = g(x)$, so that $g(0) = 0$; similarly, $f(0) = 0$.\nPlugging $x = 0$ in both equalities yields $f(g(y)) = g(f(y)) = y$, for all $y$, therefore $f$ and $g$ are bijective and $g = f^{-1}$.\nThe initial equalities become\n$$\nf(g(x) + g(y)) = x + y, \\quad g(f(x) + f(y)) = x + y,\n$$\nfor all $x, y \\in \\mathbb{Q}$, and we deduce\n$$\ng(x + y) = g(x) + g(y), \\quad f(x + y) = f(x) + f(y),\n$$\nfor all $x, y \\in \\mathbb{Q}$. Setting $a = f(1), b = g(1)$, it is easy to prove that $f(x) = ax, g(x) = bx$, for all $x \\in \\mathbb{Q}$.\nSince $g = f^{-1}$, we obtain $ab = 1$. Finally, the functions are\n$$\nf(x) = ax, \\quad g(x) = \\frac{x}{a},\n$$\nwhere $a \\in \\mathbb{Q}^*$, and it is easy to see that they satisfy the initial equalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle, let $\\gamma$ be its circumcircle, and let $M$ be the midpoint of the arc $AB$ of $\\gamma$, not containing the vertices $C$ and $D$. The line through $M$ and the point where the diagonals $AC$ and $BD$ cross one another, crosses $\\gamma$ again at $N$. Let $P$ and $Q$ be points on the side $CD$ such that $\\angle AQD = \\angle DAP$ and $\\angle BPC = \\angle CBQ$. Show that the circles $NPQ$ and $\\gamma$ are tangent to one another.\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the lines $AC$, $BD$ and $MN$ are concurrent, $(AM/MB)(BC/CN)(ND/DA) = 1$, by Ceva's theorem in trigonometric form along with the sine law in the triangle $ABN$; and since $MA = MB$, it follows that $CN/DN = BC/AD$.\n\nNext, the triangles $ADP$ and $QDA$ are similar, so $AD^2 = DP \\cdot DQ$. Similarly, $BC^2 = CP \\cdot CQ$, so, by the preceding, $(CN/DN)^2 = (CP/DP)(CQ/DQ)$.\n\n![](attached_image_1.png)\n\nBy a well-known theorem of Steiner, the lines $NP$ and $NQ$ are isogonal with respect to the lines $NC$ and $ND$; that is, the angles $CNP$ and $DNQ$ are congruent.\n\nFinally, if $X$ is a point on the tangent $t$ at $N$ to $\\gamma$, other than $N$, then a standard angle chase shows the angle $QNX$ congruent to one of the angles $CPN$, $NPQ$, depending on which side of $N$ the point $X$ lies on $t$, so $t$ is also tangent at $N$ to the circle $NPQ$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that all positive integers less than or equal to $n$ and prime to $n$ are pairwise coprime.",
    "options": [],
    "answer": "1, 2, 3, 4, 6, 8, 12, 18, 24, 30",
    "solution": "Notice that $5$ is the first index $k$ such that $p_1 \\cdots p_{k-1} > p_k^2$. Now, if $p_1 \\cdots p_{k-1} > p_k^2$ for some index $k \\ge 5$, then $p_1 \\cdots p_{k-1}p_k > p_k^3 > 4p_k^2 > p_{k+1}^2$, by the Bertrand-Tchebysheff theorem, so $p_1 \\cdots p_{k-1} > p_k^2$ for all indices $k \\ge 5$.\n\nConsequently, $m \\le 4$, so $n < p_4^2 = 49$. Examination of the possible cases quickly yields the required numbers: $1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $A_1$, $B_1$, $C_1$ be the antipodes of the vertices $A$, $B$, $C$, respectively, in the circle $ABC$, and let $X$ be a point in the plane $ABC$, collinear with no two vertices of the triangle $ABC$. The line through $B$, perpendicular to the line $XB$, and the line through $C$, perpendicular to the line $XC$, meet at $A_2$; the points $B_2$ and $C_2$ are defined similarly. Show that the lines $A_1A_2$, $B_1B_2$ and $C_1C_2$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the lines $B_2C_2$, $C_2A_2$, $A_2B_2$ meet the circle $ABC$ again at $A_3$, $B_3$, $C_3$, respectively. The angle $AA_3A_1$ is a right angle, so $A_1A_3$ is perpendicular to $B_2C_2$ and hence parallel to $XA$, and the parallel lines $XA$ and $A_1A_3$ are equidistant from the circumcentre $O$ of the triangle $ABC$. Hence $A_1A_3$ passes through the reflection $X'$ of $X$ across $O$.\n\nSimilarly, the lines $B_1B_3$ and $C_1C_3$ pass through $X'$, and therefore $X'A_1 \\cdot X'A_3 = X'B_1 \\cdot X'B_3 = X'C_1 \\cdot X'C_3$, a quantity denoted $r^2$.\n\n![](attached_image_1.png)\n\nIt follows that $A_1$, $B_1$, $C_1$ are the poles of the lines $B_2C_2$, $C_2A_2$, $A_2B_2$, respectively, relative to the circle of radius $r$ centered at $X'$, so the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are in perspective, by the well-known lemma below.\n\n**Lemma.** If $XYZ$ is a triangle, and $X'$, $Y'$, $Z'$ are the poles of the lines $YZ$, $ZX$, $XY$, respectively, relative to a conic $\\gamma$, then the triangles $XYZ$ and $X'Y'Z'$ are in perspective.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_1, \\dots, a_n$ be $n$ positive integers. Show that\n$$\n\\sum_{k=1}^{n} \\frac{\\sqrt{a_k}}{1 + a_1 + \\dots + a_k} < \\sum_{k=1}^{n^2} \\frac{1}{k}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $b_0 = 1$ and $b_k = 1 + a_1 + \\dots + a_k$, $k = 1, \\dots, n$, to obtain a strictly increasing string of positive integers $1 = b_0 < b_1 < \\dots < b_n$, and write the sum in the left-hand member in the form $\\sum_{k=1}^{n} (b_k - b_{k-1})^{1/2}/b_k$.\n\nNext, let $m = \\min\\{k: b_k > n^2\\} \\ge 1$; if there is no $b_k > n^2$, let $m = n + 1$, to split the above sum into\n$$\n\\sum_{k=1}^{m-1} \\frac{\\sqrt{b_k - b_{k-1}}}{b_k} + \\sum_{k=m}^{n} \\frac{\\sqrt{b_k - b_{k-1}}}{b_k}, \\quad (*)\n$$\nwhere empty sums are zero. We show that the first sum does not exceed $\\sum_{k=2}^{n^2} 1/k$, and the second is always less than 1.\n\nIf $k = 1, \\dots, m-1$, write $(b_k - b_{k-1})^{1/2}/b_k \\le (b_k - b_{k-1})/b_k = 1/b_k + \\dots + 1/b_k \\le \\sum_{j=b_{k-1}+1}^{b_k} 1/j$, to deduce that the first sum in $(*)$ does not exceed $\\sum_{k=1}^{m-1} \\sum_{j=b_{k-1}+1}^{b_k} 1/j = \\sum_{k=2}^{b_{m-1}} 1/k \\le \\sum_{k=2}^{n^2} 1/k$.\n\nFinally, if $k = m, \\dots, n$, write $(b_k - b_{k-1})^{1/2}/b_k < b_k^{-1/2} < 1/n$, to deduce that the second sum in $(*)$, when non-empty, is less than $(n - m + 1)/n \\le 1$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38292,
    "subject": "Mathematics (Multi-modal)",
    "question": "A non-constant function $f : (0, \\infty) \\to (0, \\infty)$ has the property $f(x^y) = (f(x))^{f(y)}$, for every $x, y > 0$. Prove that $f(xy) = f(x)f(y)$ and $f(x+y) = f(x)+f(y)$, for every $x, y > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take $a > 0$ such that $f(a) \\ne 1$. Then $f(a^{xy}) = f(a)^{f(xy)}$ and\n$$\nf(a^{xy}) = f((a^x)^y) = f(a^x)^{f(y)} = (f(a)^{f(x)})^{f(y)} = f(a)^{f(x)f(y)},\n$$\nwhence $f(xy) = f(x)f(y)$.\nAlso $f(a^{x+y}) = f(a)^{f(x+y)}$ and\n$$\nf(a^{x+y}) = f(a^x a^y) = f(a^x)f(a^y) = f(a)^{f(x)}f(a)^{f(y)} = f(a)^{f(x)+f(y)},\n$$\ntherefore $f(x+y) = f(x)+f(y)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{N}^* \\to \\mathbb{N}^*$ fulfilling the relation\n$$\nd(x, f(y)) \\cdot m(f(x), y) = d(x, y) \\cdot m(f(x), f(y)), \\text{ for every } x, y \\in \\mathbb{N}^*,\n$$\nwhere $d(a, b)$ and $m(a, b)$ signify the greatest common divisor, respectively the lowest common multiple of the positive integers $a$ and $b$.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Plugging $x = 1$ yields $m(a, y) = m(a, f(y))$, $\\forall y \\in \\mathbb{N}^*$, where $a = f(1)$. Also, taking $y = 1$ leads to $d(x, a)f(x) = m(a, f(x))$, $\\forall x \\in \\mathbb{N}^*$. This implies $d(x, a)f(x) = m(x, a)$, $\\forall x \\in \\mathbb{N}^*$. (*)\n\nRelation (*) implies $af(a) = a$ and $af(a^2) = a^2$, hence $f(a) = 1$ and $f(a^2) = a$. The hypothesis leads now, for $x = a^2$ and $y = a$, to $1 \\cdot a = a \\cdot f(a^2)$, whence $a = f(a^2) = 1$.\n\nRelation (*) implies now $f(x) = x, \\forall x \\in \\mathbb{N}^*$, which obviously fulfills the required condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdottak az $m$ és $n$ természetes számok $(n \\geq 2)$. Az $A$ gyűrűnek pontosan $n$ eleme van és $a$ egy olyan eleme $A$-nak, amelyre $1-a^{k}$ invertálható, bármely $k \\in\\{m+1, m+2, \\ldots, m+n-1\\}$ esetén. Igazold, hogy $a$ nilpotens (azaz létezik olyan nem nulla $p$ természetes szám, amelyre $a^{p}=0$).\n\nProblem:\n\nFie $m$ şi $n$ două numere naturale, $n \\geq 2$, fie $A$ un inel care are exact $n$ elemente şi fie $a$ un element al lui $A$, astfel încât $1-a^{k}$ este inversabil, oricare ar fi $k \\in\\{m+1, m+2, \\ldots, m+n-1\\}$. Arătaţi că $a$ este nilpotent (i.e., există un număr natural nenul $p$, astfel încât $a^{p}=0$ ).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $k \\in \\{1,2, \\ldots, n-1\\}$. Există $p \\in \\{1,2, \\ldots, n-1\\}$, astfel încât $m+p$ este divizibil cu $k$, deci $m+p=k \\ell$. Cum $1-a^{m+p}=(1-a^{k})(1+a^{k}+\\cdots+a^{k(\\ell-1)})$ şi $1-a^{m+p} \\in U(A)$, rezultă că $1-a^{k} \\in U(A)$.\n\nDin $1-a^{k}=(1-a)(1+a+\\cdots+a^{k-1})$, $k=1,2, \\ldots, n-1$, cu convenţia $a^{0}=1$, rezultă că $b_{k}=1+a+\\cdots+a^{k-1} \\in U(A)$, $k=1,2, \\ldots, n-1$.\n\nDacă $b_{k}$-urile ar fi distincte două câte două, atunci $A$ ar fi corp, deci $a^{n-1}=1$ şi $0=1-a^{n-1} \\in U(A)$, contradicţie.\n\nPrin urmare, există $1 \\leq p<q \\leq n-1$, astfel încât $b_{p}=b_{q}$. Atunci $0=b_{q}-b_{p}=a^{p} b_{q-p}$. Cum $b_{q-p} \\in U(A)$, rezultă că $a^{p}=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDA'B'C'D'$ be a cuboid and $AB' \\cap A'B = \\{O\\}$. Let $N$ be a point on the edge $[BC]$ so that $A'C \\parallel (B'AN)$. It is known that $D'O \\perp (B'AN)$. Prove that $ABCDA'B'C'D'$ is a cube.\nValeriu Bărbieru",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $D'A' \\perp (ABB')$ and $D'O \\perp AB'$ follows $A'O \\perp AB'$, so $ABB'A'$ is a square. Then $A'C \\parallel (ANB')$ and $(ANB') \\cap (A'BC) = ON$, hence $A'C \\parallel ON$. Since $O$ is the midpoint of the segment $[A'B]$, $N$ is the midpoint of the segment $[BC]$. Rectangle $A'BCD'$ has $D'O \\perp ON$, hence $\\Delta D'A'O \\sim \\Delta OBN$.\n![](attached_image_1.png)\nIt follows that $D'A'/A'O = OB/BN$, whence $2A'D^2 = A'B^2$, that is $A'B = A'D\\sqrt{2}$. This leads to $A'D = A'A$, therefore $AA'D'D$ is a square. This shows that all the edges of $ABCD'A'B'C'D'$ are equal, whence the conclusion.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38296,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that the last two digits of $1038^2$ are 4.\n\nb) Show that there are infinitely many perfect squares whose last three digits are 4.\n\nc) Prove that there is no perfect square whose last four digits are 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) $1038^2 = 1077444$.\n\nb) By squaring a number which ends in 038 we get a number ending in 444, as shows the diagram:\n$$\n\\begin{array}{cccccc}\n\\dots & \\dots & \\dots & 0 & 3 & 8 \\\\\n\\dots & \\dots & \\dots & 0 & 3 & 8 \\\\\\hline\n\\dots & \\dots & \\dots & 3 & 0 & 4 \\\\\n\\dots & \\dots & \\dots & 1 & 4 \\\\\\hline\n\\dots & \\dots & \\dots & 0 & & \\\\\\hline\n\\dots & \\dots & \\dots & 4 & 4 & 4\n\\end{array} \\times\n$$\nSince there are infinitely many numbers ending in 038, there are infinitely many perfect squares ending in 444.\n\nc) Let $a$ – if possible – be a positive integer whose square ends in 4444. Then $a$ is even, that is $a = 2b$. Moreover, $a^2$ has form $10000k + 4444$, $k \\in \\mathbb{N}$, hence $4b^2 = a^2 = 4(2500k + 1111)$. It follows that the last two digits of $b^2$ are 1, 1, hence $b$ ends in 1 or 9.\n\nThe multiplications\n$$\n\\begin{array}{cccccc}\n\\dots & \\dots & \\dots & n & 1 & \\times \\\\\n\\dots & \\dots & \\dots & n & 1 & \\\\\\hline\n\\dots & \\dots & \\dots & n & 1 & \\\\\\hline\n\\dots & \\dots & \\dots & n & & \\\\\\hline\n\\dots & \\dots & \\dots & u & 1 &\n\\end{array}\n$$\n$$\n\\begin{array}{cccccc}\n\\dots & \\dots & \\dots & m & 9 & \\times \\\\\n\\dots & \\dots & \\dots & m & 9 & \\\\\\hline\n\\dots & \\dots & \\dots & q & 1 & \\\\\\hline\n\\dots & \\dots & \\dots & p & & \\\\\\hline\n\\dots & \\dots & \\dots & v & 1 &\n\\end{array}\n$$\nshow that digits $u$ and $v$ are even: $u$ is the last digit of $2n$, $q$ is the last digit of $p + 8$ and $v$ is the last digit of the sum $2p + 8$.\nTherefore, the next to the last digit of $b^2$ is even, so it can not be 1. This shows that $a$ can not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $A$ and $B$, having the same number of digits in the decimal representation, such that $2 \\cdot A \\cdot B = \\overline{AB}$.\nHere $\\overline{AB}$ denotes the number obtained by concatenating $A$ and $B$.",
    "options": [],
    "answer": "(A,B) = (3,6) and (13,52)",
    "solution": "Answer: 36 and 1352.\nLet $n$ be the number of digits of $A$ and $B$. The given relation is the same as $(2A - 1)B = 10^n A$, so $2A - 1 \\mid 10^n A$. Since $(2A - 1, A) = 1$ and $(2, 2A - 1) = 1$, we infer that $2A - 1 \\mid 5^n$, so $2A - 1 \\le 5^n$.\n\nOn the other hand, since $A$ has $n$ digits we have $A \\ge 10^{n-1}$, implying $2 \\cdot 10^{n-1} - 1 \\le 2A - 1 \\le 5^n$. This leads to $2 \\cdot 2^{n-1} \\cdot 5^{n-1} \\le 5 \\cdot 5^{n-1} + 1 \\le 6 \\cdot 5^{n-1}$, whence\n$$\n2^n \\le 6,\n$$\nso $n \\in \\{1, 2\\}$.\n\nSuppose $n = 1$. From $2A - 1 \\mid 5$ we derive $2A - 1 \\in \\{1, 5\\}$, so $A \\in \\{1, 3\\}$. If $A = 1$, then $B = 10$, contradiction. For $A = 3$ we have $B = 6$, which is the first solution.\n\nSuppose $n = 2$. From $2A - 1 \\mid 25$ we obtain $2A - 1 \\in \\{1, 5, 25\\}$ and then $A \\in \\{1, 3, 13\\}$. Since $A$ has two digits, we are left with $A = 13$. Then from $25B = 1300$ we get $B = 52$, the second solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdott az $\\left(x_{n}\\right)_{n \\geq 1}$ valós számsorozat, amelynek tagjai az $[1, \\infty)$ halmazból vannak. Ha az $\\left(y_{n}^{(k)}\\right)_{n \\geq 1}, y_{n}^{(k)}=\\left[x_{n}^{k}\\right], n \\geq 1$, sorozat konvergens minden $k \\in \\mathbb{N}^{*}$ szám esetén, bizonyítsd be, hogy az $\\left(x_{n}\\right)_{n \\geq 1}$ sorozat konvergens! ( $[a]$ az $a$ valós szám egészrészét jelöli.)\n\nProblem:\n\nFie $\\left(x_{n}\\right)_{n \\geq 1}$ un şir de numere reale din intervalul $[1, \\infty)$. Presupunem că şirul $\\left(y_{n}^{(k)}\\right)_{n \\geq 1}$, definit prin $y_{n}^{(k)}=\\left[x_{n}^{k}\\right], n \\geq 1$, este convergent pentru oricare $k \\in \\overline{\\mathbb{N}}^{*}$. Să se demonstreze că şirul $\\left(x_{n}\\right)_{n \\geq 1}$ este convergent. (Prin $[a]$ se notează partea întreagă a numărului real a.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPentru $k \\in \\mathbb{N}^{*}$, $\\left(y_{n}^{(k)}\\right)_{n \\geq 1}$ este un şir convergent de numere naturale nenule. Atunci există $n_{k}, a_{k} \\in \\mathbb{N}^{*}$ astfel ca $y_{n}^{(k)}=a_{k}, \\forall n \\geq n_{k}$. Ca urmare, $x_{n}^{k} \\in\\left[a_{k}, a_{k}+1\\right), \\forall n \\geq n_{k}$.\n\nÎn particular, $x_{n} \\in\\left[a_{1}, a_{1}+1\\right), \\forall n \\geq n_{1}$. Rezultă că şirul $\\left(x_{n}\\right)_{n \\geq 1}$ este mărginit.\n\nPresupunem, prin reducere la absurd, că şirul $\\left(x_{n}\\right)_{n \\geq 1}$ admite două puncte limită $a$ şi $b$, cu $1 \\leq a<b$. Atunci există două subşiruri $\\left(x_{i_{n}}\\right)_{n \\geq 1}$ si $\\left(x_{j_{n}}\\right)_{n \\geq 1}$ ale şirului $\\left(x_{n}\\right)_{n \\geq 1}$ astfel încât $\\lim _{n \\rightarrow \\infty} x_{i_{n}}=a$ şi $\\lim _{n \\rightarrow \\infty} x_{j_{n}}=b$.\n\nFie $k \\in \\mathbb{N}^{*}$. Deoarece $i_{n}, j_{n} \\geq n, \\forall n \\in \\mathbb{N}^{*}$, avem $x_{i_{n}}^{k}, x_{j_{n}}^{k} \\in\\left[a_{k}, a_{k}+1\\right), \\forall n \\geq n_{k}$. Rezultă $x_{j_{n}}^{k}-x_{i_{n}}^{k}<1, \\forall n \\geq n_{k}$. Prin trecere la limită $(n \\rightarrow \\infty)$ obţinem $b^{k}-a^{k} \\leq 1$. Prin urmare, $b^{k}-a^{k} \\leq 1, \\forall k \\in \\mathbb{N}^{*}$.\n\nDar $1 \\leq a<b$ implică $\\lim _{k \\rightarrow \\infty}\\left(b^{k}-a^{k}\\right)=\\infty$, in contradicţie cu inegalitatea precedentă. În concluzie, şirul $\\left(x_{n}\\right)_{n \\geq 1}$ este convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a finite set of real numbers. We define the sets\n$$\nS = \\{x + y \\mid x, y \\in A\\}, \\quad D = \\{x - y \\mid x, y \\in A\\}.\n$$\n\nProve that $\\mathrm{card}(A) \\cdot \\mathrm{card}(D) \\leq (\\mathrm{card}(S))^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will define a one-to-one function $f : A \\times D \\to S \\times S$ in the following manner: for $a \\in A$ and $d \\in D$, let\n$$\nf(a, d) = (a + x_d, a + y_d),\n$$\nwhere $x_d, y_d \\in A$ are chosen such that $x_d - y_d = d$ and $x_d$ is maximal. Clearly, for a given $d$, $x_d$ and $y_d$ exist and are uniquely determined.\nIf $f(a, d) = f(a', d')$, then\n$$\na + x_d = a' + x_{d'}\n$$\n$$\na + y_d = a' + y_{d'}\n$$\nand it follows that $x_d - y_d = x_{d'} - y_{d'}$, therefore $d = d'$ and hence $a = a'$, as well.\n\nWe infer that $f$ is one-to-one, hence $\\mathrm{card}(A \\times D) \\leq \\mathrm{card}(S \\times S)$, and, finally, $\\mathrm{card}(A) \\cdot \\mathrm{card}(D) \\leq (\\mathrm{card}(S))^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that any set of 20 consecutive positive integers contains an integer $k$ such that $n\\sqrt{k} \\cdot \\{n\\sqrt{k}\\} > 5/2$ for every positive integer $n$. Here $\\{x\\}$ denotes the fractional part of the real number $x$, that is, the difference of $x$ and the largest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix a set of 20 consecutive positive integers, and choose a member $k \\equiv 15 \\pmod{20}$. We shall prove that $k$ satisfies the required condition.\nFix a positive integer $n$, and notice that $k$ is not a square, since $k \\equiv 3 \\pmod{4}$, to write $m < n\\sqrt{k} < m + 1$ for some positive integer $m$, so $m^2 < kn^2 < (m+1)^2$. We shall actually prove that $kn^2 \\ge m^2 + 5$, so\n$$\nn\\sqrt{k} \\cdot \\{n\\sqrt{k}\\} = n\\sqrt{k} \\cdot (n\\sqrt{k} - m) = kn^2 - mn\\sqrt{k} \\ge m^2 + 5 - m\\sqrt{m^2 + 5} \\\\ > m^2 + 5 - (m^2 + m^2 + 5)/2 = 5/2.\n$$\nNow, reduction modulo 5 rules out the case $kn^2 \\in \\{m^2 + 2, m^2 + 3\\}$. Finally, since $k \\equiv 3 \\pmod{4}$, it has a prime divisor $p \\equiv 3 \\pmod{4}$, and since $-1$ is a quadratic non-residue modulo $p$, we conclude that $kn^2$ is also different from both $m^2 + 1$ and $m^2 + 4$. Consequently, $kn^2 \\ge m^2 + 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the distinct primes $p, q, r$ and $s$ satisfying\n$$\n1 - \\frac{1}{p} - \\frac{1}{q} - \\frac{1}{r} - \\frac{1}{s} = \\frac{1}{pqrs}.\n$$",
    "options": [],
    "answer": "All permutations of 2, 3, 7, and 43.",
    "solution": "We will show the solution consists of all the permutations of the quadruple $2, 3, 7, 43$.\n\nWLOG we may suppose $p < q < r < s$. If $p \\ge 3$,\n$$\n1 - \\left(\\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{11}\\right) = 1 - \\frac{886}{1155} > 0.2,\n$$\nin contradiction with $\\frac{1}{pqrs} < \\frac{1}{3 \\cdot 5 \\cdot 7 \\cdot 11} < 0.006$. This proves that $p = 2$. The given relation can be rewritten $\\frac{1}{2} - (\\frac{1}{q} + \\frac{1}{r} + \\frac{1}{s}) = \\frac{1}{2qrs}$.\n\nNow, if $q \\ge 5$ we get\n$$\n\\frac{1}{2} - \\left( \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{11} \\right) > 0.5 - 0.44 = 0.06,\n$$\ncontradicting\n$$\n\\frac{1}{2qrs} < \\frac{1}{2 \\cdot 5 \\cdot 7 \\cdot 11} < 0.001.\n$$\nSo $q = 3$.\n\nThe equation becomes\n$$\n\\frac{1}{6} - \\left( \\frac{1}{r} + \\frac{1}{s} \\right) = \\frac{1}{6rs},\n$$\ngiving $rs - 6r - 6s - 1 = 0$. This is equivalent to $(r-6)(s-6) = 37$. In conclusion $r = 7, s = 43$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of natural numbers such that $21^x + 4^y = z^2$.",
    "options": [],
    "answer": "(1, 1, 5)",
    "solution": "Rewrite the equation as $(z - 2^y)(z + 2^y) = 21^x$. Denote by $d$ the largest common divisor of $z - 2^y$ and $z + 2^y$; then $d$ divides $(z + 2^y) - (z - 2^y)$, so $d \\mid 2^{y+1}$. Since $d$ is a divisor of $z + 2^y$ and $z + 2^y$ divides $21^x$, it follows that $d \\mid (2^{y+1}, 21^x)$, so $d = 1$.\nConsequently, either $z - 2^y = 1$ and $z + 2^y = 21^x$, or $z - 2^y = 3^x$ and $z + 2^y = 7^x$.\nThe first case leads to $21^x - 1 = 2^{y+1}$, equality which cannot hold, because $5 \\mid 21^x - 1$ and $5 \\nmid 2^{y+1}$.\nThe second case implies $7^x - 3^x = 2^{y+1}$. We easily see that $x \\neq 0$, and for $x = 1$ we get the solution $(1, 1, 5)$. Assume that $x \\ge 2$.\nIf $x$ is odd, then\n$$\n2^{y-1} = 7^{x-1} + 7^{x-2} \\cdot 3 + 7^{x-3} \\cdot 3^2 + \\dots + 7 \\cdot 3^{x-2} + 3^{x-1},\n$$\na contradiction, because the left-hand side is even and right-hand side is odd.\nIf $x$ is an even number, $x = 2s$, $s \\ge 1$, then $49^s - 9^s = 2^{y+1}$, which is impossible modulo 5, since $5 \\mid 49^s - 9^s$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38303,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a regular tetrahedron $ABCD$ consider planes that are parallel to its faces such that each edge is divided into 6 equal segments. These planes determine, on the edges of the tetrahedron, on its faces and in its interior a set consisting of 80 points of intersection. Denote this set by $V$.\nFind the maximum number of elements of a subset $W$ of the set $V \\cup \\{A, B, C, D\\}$, having the property: any three points of $W$ are not collinear and the plane generated by them is neither parallel with any of the faces of the tetrahedron $ABCD$, nor contains one of his faces.\nMihai Monea",
    "options": [],
    "answer": "8",
    "solution": "Suppose, WLOG, that the height of $ABCD$ equals $6$. Denote by $k$ the maximal number of elements of the set $W$ and by $M_1, \\dots, M_k$ its elements. For each $i \\in \\{1, 2, \\dots, k\\}$ denote by $a_i, b_i, c_i$, and $d_i$ the distances from the point $M_i$ to the planes $(BCD), (ACD), (ABD)$ and $(ABC)$ respectively. Then $a_i, b_i, c_i, d_i \\in \\{0, 1, 2, \\dots, 6\\}$. Because the sum of the distances from an interior point to the faces of a regular tetrahedron equals its height, we get that $a_i + b_i + c_i + d_i = 6$. Consequently, if $T = (a_1 + a_2 + \\dots + a_k) + (b_1 + b_2 + \\dots + b_k) + (c_1 + c_2 + \\dots + c_k) + (d_1 + d_2 + \\dots + d_k)$, we get $T = 5$.\nLet $s = [k/2]$. As no more than two of the numbers $a_1, a_2, \\dots, a_k$ can be equal, we deduce $s \\le 7$.\n\nFor even $k$, $k = 2s$, we have $a_1 + a_2 + \\dots + a_k \\ge 2 \\cdot 0 + 2 \\cdot 1 + \\dots + 2 \\cdot (s-1) = s^2 - s$. So $T = 12s \\ge 4s^2 - 4s$, and thus $s \\le 4$ and $k \\le 8$. For odd $k$, $k = 2s + 1$, we have $a_1 + a_2 + \\dots + a_k \\ge 2 \\cdot 0 + 2 \\cdot 1 + \\dots + 2 \\cdot (s-1) + s = s^2$. Thus $T = 12s + 6 \\ge 4s^2$, implying $s \\le 3$ and $k \\le 7$.\n\nTo give an example of an 8-element set, use the notation $(a_i, b_i, c_i, d_i)$ as in the above considerations and consider as $W$ the set $(0, 1, 2, 3), (1, 0, 2, 3), (0, 1, 3, 2), (1, 0, 3, 2), (2, 3, 0, 1), (2, 3, 1, 0), (3, 2, 1, 0)$ and $(3, 2, 0, 1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d \\ge 0$ be real numbers such that $a + b + c + d = 1$. Prove that\n$$\n\\sqrt{a + \\frac{(b-c)^2}{6} + \\frac{(c-d)^2}{6} + \\frac{(d-b)^2}{6}} + \\sqrt{b} + \\sqrt{c} + \\sqrt{d} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $(b-c)^2 \\le 2(\\sqrt{b} - \\sqrt{c})^2$, since $(\\sqrt{b} + \\sqrt{c})^2 = b + c + 2\\sqrt{bc} \\le 2b + 2c \\le 2$. Thus, $a + \\sum \\frac{(b-c)^2}{6} \\le a + \\frac{1}{3} \\sum (\\sqrt{b} - \\sqrt{c})^2 = 1 - \\frac{1}{3}(\\sum \\sqrt{b})^2$.\n\nTo end the proof, it suffices to show that $S + \\sqrt{1 - S^2/3} \\le 2$, where $S = \\sqrt{b} + \\sqrt{c} + \\sqrt{d}$.\n\nSince $S \\le 2$, this is equivalent to $1 - S^2/3 \\le 4 - 4S + S^2$, which rewrites as $(2S - 3)^2 \\ge 0$, obviously true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)_{n \\ge 1}$ be a sequence of real numbers from $[1, \\infty)$. It is known that the sequences $(y_n^{(k)})_{n \\ge 1}$, defined by $y_n^{(k)} = \\lfloor x_n^k \\rfloor$, $n \\ge 1$, are convergent for every $k \\in \\mathbb{N}^*$. Prove that the sequence $(x_n)_{n \\ge 1}$ is convergent.\n\nMihai Piticari, Vlad Cerbu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k \\in \\mathbb{N}^*$, the sequence $(y_n^{(k)})_{n \\ge 1}$ is convergent and its terms are integers. Then there exist $n_k, a_k \\in \\mathbb{N}^*$ so that $y_n^{(k)} = a_k, \\forall n \\ge n_k$. Consequently, $x_n^k \\in [a_k, a_k + 1), \\forall n \\ge n_k$. In particular, $x_n \\in [a_1, a_1 + 1), \\forall n \\ge n_1$. It follows that the sequence $(x_n)_{n \\ge 1}$ is bounded.\n\nSuppose, by contradiction, that the sequence $(x_n)_{n \\ge 1}$ has two limit points $a$ and $b$, with $1 \\le a < b$. Then there exist two subsequences $(x_{i_n})_{n \\ge 1}$ and $(x_{j_n})_{n \\ge 1}$ of $(x_n)_{n \\ge 1}$ such that $\\lim_{n \\to \\infty} x_{i_n} = a$ and $\\lim_{n \\to \\infty} x_{j_n} = b$. Let $k \\in \\mathbb{N}^*$. From $i_n, j_n \\ge n$, $\\forall n \\in \\mathbb{N}^*$, we get $x_{i_n}^k, x_{j_n}^k \\in [a_k, a_k + 1), \\forall n \\ge n_k$. It follows $x_{j_n}^k - x_{i_n}^k < 1, \\forall n \\ge n_k$. Taking the limit for $n \\to \\infty$ gives $b^k - a^k \\le 1$. Therefore $b^k - a^k \\le 1, \\forall k \\in \\mathbb{N}^*$. On the other hand, $1 \\le a < b$ implies $\\lim_{k \\to \\infty} (b^k - a^k) = \\infty$, in contradiction with the previous inequality.\n\nThis shows that the sequence $(x_n)_{n \\ge 1}$ is convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38306,
    "subject": "Mathematics (Multi-modal)",
    "question": "The isosceles triangle $ABE$, with $m(\\angle ABE) = 120^\\circ$ is constructed outside the square $ABCD$. Denote $M$ the orthogonal projection of $B$ onto the bisector of angle $EAB$, $N$ the orthogonal projection of $M$ onto $AB$ and $P$ the meeting point of the straight lines $CN$ and $MB$. Let $G$ be the centroid of triangle $ABE$. Prove that straight lines $PG$ and $AE$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$m(\\angle BAM) = 15^\\circ$ implies $MN = \\frac{1}{4}AB$. From $MN \\parallel BC$ follows $\\triangle PMN \\sim \\triangle PBC$, hence\n$$\n\\frac{PM}{PB} = \\frac{MN}{BC} = \\frac{1}{4}\n$$\nand, denoting $Q = BM \\cap AE$, we come to $\\frac{PM}{BQ} = \\frac{1}{6}$, whence\n$$\nPB = PM + \\frac{1}{2}BQ = \\frac{2}{3}BQ.\n$$\nIf $F$ is the midpoint of the segment $[AE]$, then $BG = \\frac{2}{3}BF$ and, from the converse of Thales Theorem, $PG \\parallel AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, $n \\ge 2$, let $R$ be an $n$-element ring, and let $x$ be an element of $R$ such that $1 - x^k$ is invertible for each $k \\in \\{m+1, m+2, \\dots, m+n-1\\}$. Show that $x$ is nilpotent (i.e., $x^p = 0$ for some positive integer $p$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $0$ is nilpotent, we will assume in the sequel that $x \\ne 0$.\nGiven a positive integer $k < n$, notice that $m+p$ is divisible by $k$ for some positive integer $p < n$; write $m + p = kl$. Since $1 - x^{m+p}$ is invertible, and $1 - x^{m+p} = (1 - x^k)(1 + x^k + \\dots + x^{k(\\ell-1)}) = (1 + x^k + \\dots + x^{k(\\ell-1)})(1 - x^k)$, it follows that $1 - x^k$ is invertible.\n\nNext, write $1 - x^k = (1-x)(1+x+\\dots+x^{k-1}) = (1+x+\\dots+x^{k-1})(1-x)$, where we agree that $x^0 = 1$, to deduce that $x_k = 1 + x + \\dots + x^{k-1}$ is invertible for each positive integer $k < n$ and $1-x$ is invertible.\n\nIf the $x_k$ were pairwise distinct, then by a cardinality argument $R$ would be an $n$-element skew field, so $x^{n-1} = 1$, that is $(1-x)x_{n-1} = 0$, which is impossible since $1-x$ and $x_{n-1}$ are invertible.\n\nConsequently, $x_p = x_q$ for some indices $p < q$, so $0 = x_q - x_p = x^p x_{q-p}$ which implies $x^p = 0$, since $x_{q-p}$ is invertible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m \\ge n \\ge 2$ be integers, and let $A \\in \\mathcal{M}_{m,n}(\\mathbb{C})$ and $B \\in \\mathcal{M}_{n,m}(\\mathbb{C})$, be two matrices such that there exist $k \\in \\mathbb{N}^*$, and $a_0, a_1, \\dots, a_k \\in \\mathbb{C}$, such that $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \\dots + a_1(AB) + a_0I_m = O_m$, and $a_k(BA)^k + a_{k-1}(BA)^{k-1} + \\dots + a_1(BA) + a_0I_n \\ne O_n$. Prove that $a_0 = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $a_0 \\ne 0$. Rewriting the given equality as\n$$\nAB \\left( -\\frac{a_k}{a_0}(AB)^{k-1} - \\frac{a_{k-1}}{a_0}(AB)^{k-2} - \\dots - \\frac{a_1}{a_0}I_m \\right) = I_m,\n$$\nwe find that $AB$ is invertible, hence $\\mathrm{rank}(AB) = m \\ge n$.\nOn the other hand, we have $m = \\mathrm{rank}(AB) \\le \\min\\{\\mathrm{rank}A, \\mathrm{rank}B\\} \\le n$, therefore $m = n$ and, furthermore, $A, B$ and $BA$ are all invertible.\nNow, consider the equality $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \\dots + a_1(AB) + a_0I_n = O_n$ and left multiply by $B$ and right multiply by $A$. This yields $a_k(BA)^{k+1} + a_{k-1}(BA)^k + \\dots + a_0(BA) = O_n$. Multiplying by $(BA)^{-1}$ gives $a_k(BA)^k + a_{k-1}(BA)^{k-1} + \\dots + a_0I_n = O_n$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight numbers, all zero, are written on a blackboard. A *move* consists in randomly selecting four of them, $a$, $b$, $c$, $d$, and replacing them by $a+3$, $b+3$, $c+2$ and $d+1$, respectively.\n\na) What is the smallest number of moves after which on the blackboard can appear eight consecutive numbers?\n\nb) Is there a sequence of moves after which all the numbers on the blackboard are equal to $2015$?",
    "options": [],
    "answer": "a) 4; b) No",
    "solution": "a) After each move, the sum of the numbers increases by $9$. Since the sum of the smallest $8$ consecutive numbers is $0 + 1 + 2 + \\dots + 7 = 28$, but the table below uses $1$ to $8$ (sum $36$), so let's check the table for correctness. The table shows the process:\n\n| Start         | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |\n|---------------|---|---|---|---|---|---|---|---|\n| First move    | 1 | 2 |   |   |   |   | 3 | 3 |\n| Second move   |   |   | 2 | 1 |   |   | 3 | 3 |\n| Third move    |   |   |   |   | 3 | 3 | 1 | 2 |\n| Fourth move   |   |   | 1 | 3 | 2 | 3 |   |   |\n| Outcome       | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n\nAfter four moves, the numbers on the blackboard can be eight consecutive numbers. Since each move increases the sum by $9$, and the sum of $1$ to $8$ is $36$, four moves are sufficient.\n\nb) After $k$ moves, the sum of the eight numbers is $9k$. Since $8 \\cdot 2015 = 16120$ is not a multiple of $9$, the answer is negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer congruent to $1$ modulo $4$ which is not a perfect square, and let $a = (1 + \\sqrt{k})/2$. Show that\n$$\n\\{\\lfloor a^2 n \\rfloor - \\lfloor a \\lfloor an \\rfloor \\rfloor : n = 1, 2, 3, \\dots\\} = \\{1, \\dots, \\lfloor a \\rfloor\\}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_n = an - \\lfloor an \\rfloor$, $n = 1, 2, 3, \\dots$. Since $a^2 = a + (k-1)/4$, it follows that $\\lfloor a^2 n \\rfloor = \\lfloor an \\rfloor + n(k-1)/4$, and $(a-1)\\lfloor an \\rfloor = (a-1)(an - a_n) = n(k-1)/4 - (a-1)a_n$, so, adding $\\lfloor an \\rfloor$ to each side, $a\\lfloor an \\rfloor = \\lfloor an \\rfloor + n(k-1)/4 - (a-1)a_n = \\lfloor a^2 n \\rfloor - (a-1)a_n$. Since $a$ is irrational, the $a_n$ form a dense subset of the open unit interval $(0, 1)$, so, by the preceding, the differences $\\lfloor a^2 n \\rfloor - a\\lfloor an \\rfloor = (a-1)a_n$ form a dense subset of the open interval $(0, a-1)$. Finally, since $\\lfloor a^2 n \\rfloor - \\lfloor a \\lfloor an \\rfloor \\rfloor = \\lfloor a^2 n \\rfloor - a\\lfloor an \\rfloor \\rfloor = \\lfloor (a-1)a_n \\rfloor$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non constant polynomial functions $f: [0, 1] \\to \\mathbb{R}^*$, having rational coefficients, for which the following property holds: for every $x$ in $[0, 1]$, there exist two polynomial functions $g_x, h_x: [0, 1] \\to \\mathbb{R}$, with rational coefficients, such that $h_x(x) \\neq 0$ and\n$$\n\\int_0^x \\frac{1}{f(t)} dt = \\frac{g_x(x)}{h_x(x)}.\n$$",
    "options": [],
    "answer": "All such polynomials are f(x) = a(x - b)^n with n an integer greater than 1, a and b rational, a not zero, and b not in the unit interval.",
    "solution": "The required functions are $f(x) = a(x-b)^n$, $0 \\le x \\le 1$, where $n$ is an integer greater than $1$, and $a$ and $b$ are rational numbers, $a \\neq 0$ and $b \\notin [0, 1]$. Clearly, these functions satisfy the conditions in the statement.\n\nSince the closed unit interval $[0, 1]$ is uncountable, and there are only countably many polynomial functions with rational coefficients, there exist an uncountable set $S \\subseteq [0, 1]$ and coprime polynomial functions $g, h: [0, 1] \\to \\mathbb{R}$ with rational coefficients such that $h(x) \\neq 0$ and $\\int_0^x \\frac{1}{f(t)} dt = \\frac{g(x)}{h(x)}$, for all $x$ in $S$. Since at most countably many points of $S$ are not accumulation points of $S$, and $S$ is uncountable, it follows that $S$ contains infinitely many of its accumulation points. At each of these points, the rational functions $1/f$ and $(g/h)' = (g'h - gh')/h^2$ are equal, so\n$$\nh^2 = f \\cdot (g'h - gh') \\quad (*)\n$$\nin $\\mathbb{Q}[X]$. Since $\\deg f \\ge 1$, it follows that $\\deg g \\le \\deg h$. Notice that the remainder of $g$ upon division by $h$ also satisfies (*) to assume henceforth $\\deg g < \\deg h$, so $\\deg (g'h - gh') = \\deg g + \\deg h - 1$.\n\nIf $\\deg h = 1$, then $\\deg g = 0$ and $f = a(X - b)^2$, where $a$ and $b$ are rational numbers, $a \\neq 0$ and $b \\notin [0, 1]$.\n\nIf $\\deg h \\ge 2$, since $g$ and $h$ are coprime, (*) implies that every $k$-fold root (not necessarily real) of $g'h - gh'$ is a $(k+1)$-fold root of $h$, so $\\deg h \\ge \\deg g + \\deg h - 1 + d$, where $d$ is the number of distinct roots of $g'h - gh'$. Consequently, $d = 1$ and $\\deg g = 0$, so $g$ is a non-zero constant, $h = c(X - b)^n$, where $b$ and $c$ are rational, $b \\notin [0, 1]$, $c \\neq 0$, and $n$ is an integer greater than $1$, and $f = a(X - b)^{n+1}$ for some non-zero rational $a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$ and a permutation $a_1, a_2, \\dots, a_n$ of the first $n$ positive integers, show that at least $\\sqrt{n}$ distinct residue classes modulo $n$ occur in the list $a_1, a_1 + a_2, \\dots, a_1 + a_2 + \\dots + a_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r_k$ be the remainder of the partial sum $a_1 + a_2 + \\dots + a_k$ upon division by $n$, and let $m$ be the number of distinct residues in the list $r_1, r_2, \\dots, r_n$. Since there are $m$ distinct integers in the list, the number of ordered pairs with different entries that can be formed from these $m$ integers does not exceed $m(m-1)$. On the other hand, of the $n-1$ distinct ordered pairs $(r_k, r_{k+1})$, $k = 1, 2, \\dots, n-1$, at most one has equal entries, so $m(m-1) \\ge (n-1)-1 = n-2$. Consequently, $m \\ge (1+\\sqrt{4n-7})/2 \\ge \\sqrt{n}$ if $n \\ge 4$, the latter inequality being strict if $n \\ge 5$. If $n=4$, and $a_1=2$, $a_2=4$, $a_3=3$ and $a_4=1$, then the corresponding list of residues is 2, 2, 1, 2, so $m=2 = \\sqrt{4} = \\sqrt{n}$. Finally, if $n=3$, then every permutation produces a list consisting of exactly $2 > \\sqrt{3}$ distinct residues. (If $n=2$, the identity produces the list 1, 1, so $m=1 < \\sqrt{2} = \\sqrt{n}$.)",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38313,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that, if $I \\subset \\mathbb{R}$ is a closed bounded interval, and $f: I \\to \\mathbb{R}$ is a non-constant monic polynomial function such that $\\max_{x \\in I} |f(x)| < 2$, then there exists a non-constant monic polynomial function $g: I \\to \\mathbb{R}$ such that $\\max_{x \\in I} |g(x)| < 1$.\n\nb) Show that there exists a closed bounded interval $I \\subset \\mathbb{R}$ such that $\\max_{x \\in I} |f(x)| \\ge 2$ for every non-constant monic polynomial function $f: I \\to \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $I \\subset \\mathbb{R}$ be a closed bounded interval, let $P(I)$ be the set of all polynomial functions $f: I \\to \\mathbb{R}$, and let $\\|f\\| = \\max_{x \\in I} |f(x)|$. Define $A: P(I) \\to P(I)$ by $Af(x) = (f(x))^2 - \\frac{1}{2}\\|f\\|^2$, $x \\in I$. If $f$ is monic (respectively, non-constant), then so is $Af$. Further, $\\|Af\\| = \\frac{1}{2}\\|f\\|^2$, so\n$$\n\\|A^n f\\| = 2 \\left( \\frac{1}{2} \\|f\\| \\right)^{2^n},\n$$\nwhere $A^n = A \\circ \\dots \\circ A$, $n$ times, is the $n$-th iterate of $A$. Consequently, if $\\|f\\| < 2$ for some $f$ in $\\mathcal{P}(I)$, then $\\|A^n f\\| < 1$ for $n$ large enough.\n\nb) In the notation above, let $I = [-2, 2]$, and define $B: \\mathcal{P}(I) \\to \\mathcal{P}(I)$ by\n$$\nBf(x) = \\frac{1}{2}f(\\sqrt{x+2}) + \\frac{1}{2}f(-\\sqrt{x+2}), \\quad x \\in I.\n$$\nIf $f$ is monic, so is $Bf$. Further, $\\|Bf\\| \\le \\|f\\|$, and $\\deg Bf = \\lfloor \\frac{1}{2} \\deg f \\rfloor$. Consequently, if $2^n \\le \\deg f < 2^{n+1}$, where $n$ is a non-negative integer, then $\\deg B^n f = 1$. To conclude the proof, notice that every monic polynomial function of degree 1 on $I$ has norm at least 2.\n\n*Remark.* Tchebysheff's polynomials and their properties provide an alternative solution. If $n$ is a non-negative integer, and $-1 \\le x \\le 1$, the Tchebysheff polynomial (real-valued function) of degree $n$ is defined by ...",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a finite planar set no three points of which are collinear, and let $D(S, r) = \\{\\{x, y\\} : x, y \\in S, \\text{dist}(x, y) = r\\}$, where $r$ is a positive real number, and $\\text{dist}(x, y)$ is the Euclidean distance between the points $x$ and $y$. Show that\n$$\n\\sum_{r>0} |D(S, r)|^2 \\le 3|S|^2(|S| - 1)/4.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given a point $x$ in $S$ and a real number $r$, let $S(x, r) = \\{y : y \\in S, \\text{dist}(x, y) = r\\}$, and notice that the $S(x, r)$, $r \\ge 0$, partition $S$.\nThe number of non-degenerate isosceles triangles with vertices in $S$ and apex at $x$\nis $\\sum_{r>0} (|S(x,r)|)^2$, so the total number of non-degenerate isosceles triangles with vertices\nin $S$ is $N = \\sum_{x \\in S} \\sum_{r > 0} \\binom{|S(x,r)|}{2}$, equilateral triangles with vertices in $S$ being counted three times each. Now,\n$$\n\\begin{align*}\nN &= \\sum_{x \\in S} \\sum_{r > 0} \\binom{|S(x, r)|}{2} = \\sum_{r > 0} \\sum_{x \\in S} \\binom{|S(x, r)|}{2} \\\\\n&\\geq \\sum_{r > 0} |S| \\binom{\\frac{1}{|S|} \\sum_{x \\in S} |S(x, r)|}{2} = \\sum_{r > 0} |S| \\binom{\\frac{2|D(S,r)|}{|S|}}{2} \\\\\n&= \\frac{2}{|S|} \\sum_{r > 0} |D(S, r)|^2 - \\sum_{r > 0} |D(S, r)| = \\frac{2}{|S|} \\sum_{r > 0} |D(S, r)|^2 - \\binom{|S|}{2},\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral whose diagonals are not perpendicular and whose sides $AB$ and $CD$ are not parallel. Let $O$ be the intersection of its diagonals. Denote with $H_1$ and $H_2$ the orthocenters of triangles $AOB$ and $COD$, respectively. If $M$ and $N$ are the midpoints of the segment lines $[AB]$ and $[CD]$, respectively, prove that the lines $H_1H_2$ and $MN$ are parallel if and only if $AC = BD$.\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A'$ and $B'$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $AOB$, and $C'$, $D'$ the feet of the altitudes drawn from $C$ and $D$ in the triangle $COD$.\nObviously, $A'$ and $D'$ belong to the circle $C_1$ of diameter $AD$, while $B'$ and $C'$ belong to the circle $C_2$ of diameter $BC$.\nIt is easy to see that triangles $H_1AB$ and $H_1A'B'$ are similar. It follows that $H_1A \\cdot H_1A' = H_1B \\cdot H_1B'$. (Alternatively, one could notice that the quadrilateral $ABA'B'$ is cyclic and obtain the previous relation by writing the power of $H_1$ with respect to its circumcircle.)\nSo, $H_1$ has the same power with respect to circles $C_1$ and $C_2$. Hence, $H_1$ (and similarly, $H_2$) is on the radical axis of the two circles.\nThe radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $H_1H_2$ is perpendicular to $PQ$, where $P$ and $Q$ are the midpoints of the sides $AD$ and $BC$, respectively. ($P$ and $Q$ are the centers of circles $C_1$ and $C_2$.)\nThe condition $H_1H_2 \\parallel MN$ is equivalent to $MN \\perp PQ$. As $MPNQ$ is a parallelogram, we conclude that $H_1H_2 \\parallel MN \\perp PQ \\parallel MPNQ$ is a rhombus $\\Leftrightarrow MP = MQ \\Leftrightarrow AC = BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers that have exactly $8$ positive divisors, among which three are primes of the form $a$, $\\overline{bc}$ and $\\overline{cb}$, given that $a + \\overline{bc} + \\overline{cb}$ is a perfect square and $a, b, c$ are digits, with $b < c$.",
    "options": [],
    "answer": "2015",
    "solution": "Suppose $x$ is an integer with the given properties. Then $x$ is a multiple of $y = a \\cdot \\overline{bc} \\cdot \\overline{cb}$. Since $y$ has eight divisors, it follows that $x = y$.\n\nThe numbers $\\overline{bc}$ and $\\overline{cb}$ are distinct primes, so $\\overline{bc} \\in \\{13, 17, 37, 79\\}$. The condition that $a + \\overline{bc} + \\overline{cb}$ is a square and $a$ is a prime forces $a = 5$, $\\overline{bc} = 13$ and $\\overline{cb} = 31$, implying $x = 2015$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $N \\ge 4$, determine the largest value the sum\n$$\n\\sum_{i=1}^{\\lfloor k/2 \\rfloor + 1} (\\lfloor n_i/2 \\rfloor + 1)\n$$\nmay achieve, where $k, n_1, \\dots, n_k$ run through the integers subject to $k \\ge 3, n_1 \\ge \\dots \\ge n_k \\ge 1$, and $n_1 + \\dots + n_k = N$.",
    "options": [],
    "answer": "⌊2(N+2)/3⌋",
    "solution": "The required maximum is $\\lfloor 2(N + 2)/3 \\rfloor$.\n\nFor more convenience, given a list of $k$ real numbers, the sublist consisting of the $1 + \\lfloor k/2 \\rfloor$ largest entries will be referred to as the *upper half* of the list, and its complement, i.e., the sublist consisting of the $\\lfloor (k+1)/2 \\rfloor - 1$ smallest entries, as the *lower half* of the list. Notice that the lower half of a list consisting of at least three real numbers is never empty.\n\nTo maximize the sum $s$ in the statement, we list a sequence of operations which transform any given partition of $N$ into at least three positive integers into another such whose lower half is all 1, and the upper half is all 2 except possibly one unit entry; moreover, each operation yields a partition into at least three positive integers, and does not decrease $s$, whence the conclusion. In what follows, $n_1, \\dots, n_k$ will denote a generic partition of $N$ into at least three positive integers; the obvious verifications are omitted.\n\nIf the number of unit entries in the partition is less than $\\lfloor (k+1)/2 \\rfloor - 1$, i.e., the lower half has some entry $n_i > 1$, splitting $n_i$ into 1 and $n_i - 1$ increases length by 1, and $s$ by at least 1 if $k$ is odd, and preserves it otherwise; in either case, $s$ does not decrease.\n\nIf the number of unit entries in the partition exceeds $\\lfloor (k+1)/2 \\rfloor$, i.e., the upper half has at least two unit entries, replacing two 1's by one 2 increases $s$ by 1 if $k$ is odd, and preserves it otherwise; in either case, $s$ does not decrease, and since $N > 3$ the resulting partition has length at least three. (In fact, the length of the resulting partition would be less than three only in case $N=3$, and the partition we start with is 1, 1, 1 — the unique partition of 3 into three positive integers. This is, however, ruled out by hypothesis.)\n\nConsequently, a partition of $N$ into at least three positive integers can be transformed into another such whose lower half is all 1, and the upper half has at most one unit entry; moreover, $s$ does not decrease in the process, and the lengths of the partitions involved are at least three. Henceforth, all partitions are assumed to have such a structure.\n\nIf the upper half has no unit entry, but has some odd entry $n_i > 1$, splitting $n_i$ into 1 and $n_i - 1$ increases length by 1, and $s$ by 1 if $k$ is odd, and preserves it otherwise; in either case, $s$ does not decrease, and the outcome is a partition into at least three positive integers, whose lower half is all 1, and the upper half has exactly one unit entry and fewer odd entries exceeding 1.\n\nIf the upper half has exactly one unit entry and some odd entry $n_i > 1$, replacing that unit entry and $n_i$ by 2 and $n_i - 1$ preserves length, increases $s$ by 1, and the resulting partition has length at least three, an all 1 lower half, and the upper half has fewer odd entries exceeding 1 and no unit entry.\n\nConsequently, every partition of $N$ into at least three positive integers can be transformed into another such with an all 1 lower half, and an all even upper half except possibly one unit entry; moreover, at each stage, the length of the partition is at least three, and $s$ does not decrease. Henceforth, all partitions are assumed to have such a structure.\n\nIf the upper half has no unit entry, but has some entry $n_i > 2$, splitting $n_i$ into 1, 1 and $n_i - 2$ increases length by 2, preserves $s$ and yields a partition into at least three positive integers, whose lower half is all 1, and the upper half is all even except for exactly one unit entry and has fewer entries exceeding 2.\n\nFinally, if the upper half is all even except for exactly one unit entry, and has some entry $n_i > 2$, splitting $n_i$ into 2 and $n_i - 2$ increases length by 1, and $s$ by 1 if $k$ is odd, and preserves it otherwise; in either case, $s$ does not decrease, and the outcome is a partition of length at least three, whose lower half is all 1, and the upper half is all even with fewer entries exceeding 2.\n\nConsequently, any given partition of $N$ into at least three positive integers can be transformed into another such whose lower half is all 1, and the upper half is all 2 except for at most one unit entry; moreover, the transformation does not decrease $s$, and all partitions have length at least three. For this 'standard' partition, it is readily checked that $s = \\lfloor 2(N+2)/3 \\rfloor$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, prove that there are only finitely many sequences of $n$ consecutive positive integers such that $n!$ can be constructed from these integers only using the elementary operations of addition, subtraction, multiplication, and division, the integers being used exactly once each.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n$$\n\\frac{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} a_{\\alpha_1, \\dots, \\alpha_n} x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}}{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} b_{\\alpha_1, \\dots, \\alpha_n} x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}} \\quad (*)\n$$\nwhere the $a_{\\alpha_1, \\dots, \\alpha_n}$ and $b_{\\alpha_1, \\dots, \\alpha_n}$ are all in the set $\\{0, \\pm 1\\}$, not all zero of course, and $a_{0, \\dots, 0} = b_{1, \\dots, 1} = 0$.\nSince $|a_{\\alpha_1, \\dots, \\alpha_n}| \\le 1$ and $a_{0, \\dots, 0} = 0$, the absolute value of the numerator does not exceed $(1 + |x_1|) \\cdots (1 + |x_n|) - 1$; in particular, if $c$ is an integer in the range $-n, \\dots, -1$, and $x_k = c+k$, $k = 1, \\dots, n$, then the absolute value of the numerator is at most $(-c)!(n+c+1)! - 1 \\le n! - 1 < n!$.\n\nConsider now the integral polynomials,\n$$\nP = \\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} a_{\\alpha_1, \\dots, \\alpha_n} (X+1)^{\\alpha_1} \\cdots (X+n)^{\\alpha_n},\n$$\nand\n$$\nQ = \\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} b_{\\alpha_1, \\dots, \\alpha_n} (X+1)^{\\alpha_1} \\cdots (X+n)^{\\alpha_n},\n$$\nwhere the $a_{\\alpha_1, \\dots, \\alpha_n}$ and $b_{\\alpha_1, \\dots, \\alpha_n}$ are all in the set $\\{0, \\pm 1\\}$, not all zero, and $a_{0, \\dots, 0} = b_{1, \\dots, 1} = 0$. By the preceding, $|P(c)| < n!$ for every integer $c$ in the range $-n, \\dots, -1$; and since $b_{1, \\dots, 1} = 0$, the degree of $Q$ is less than $n$. Since every non-zero polynomial has only finitely many roots, and the number of roots does not exceed the degree, to complete the proof it is sufficient to show that the polynomial $P-n!Q$ does not vanish identically, provided that $Q$ does not (which is the case in the problem).\n\nSuppose, if possible, that $P = n!Q$, where $Q \\ne 0$. Since $\\deg Q < n$, it follows that $\\deg P < n$ as well, and since $P \\ne 0$, the number of roots of $P$ does not exceed $\\deg P < n$, so $P(c) \\ne 0$ for some integer $c$ in the range $-n, \\dots, -1$. By the preceding, $|P(c)|$ is consequently a positive integer less than $n!$. On the other hand, $|P(c)| = n!|Q(c)|$ is an integral multiple of $n!$. A contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $a$ and a positive integer $n$, show that the sum $\\sum_{k=1}^{n} a^{(k,n)}$ is divisible by $n$, where $(x, y)$ denotes the greatest common divisor of the integers $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\sum_{k=1}^n a^{(k,n)} = \\sum_{d|n} \\varphi(n/d)a^d$, where $\\varphi$ is Euler's totient function ($\\varphi(m)$ is the number of positive integers less than $m$ and prime to $m$), and notice that, if $n$ and $n'$ are coprime positive integers, then\n$$\n\\sum_{k=1}^{nn'} a^{(k,nn')} = \\sum_{d|n} \\varphi(n/d) \\sum_{d'|n'} \\varphi(n'/d')(a^d)^{d'}.\n$$\nConsequently, it is sufficient to prove the assertion for $n = p^m$, where $p$ is a prime and $m$ is a non-negative integer. In this case,\n$$\n\\begin{aligned} \\sum_{k=1}^{p^m} a^{(k, p^m)} &= \\sum_{k=0}^{m} \\varphi(p^{m-k}) a^{p^k} = \\sum_{k=0}^{m-1} (p^{m-k} - p^{m-k-1}) a^{p^k} + a^{p^m} \\\\ &= p^m a + \\sum_{k=1}^{m} p^{m-k} (a^{p^k} - a^{p^{k-1}}) \\equiv 0 \\pmod{p^m}, \\end{aligned}\n$$\nsince $a^{p^k} \\equiv a^{p^{k-1}} \\pmod{p^k}$, $k = 1, \\dots, m$, by Fermat's theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $N$ having an even number of digits (no leading zeroes) such that, if we insert a multiplication sign after the first $n$ digits of $N$, the result of the multiplication is a divisor of $N$.",
    "options": [],
    "answer": "All solutions N are:\n- Two-digit: N ∈ {11, 12, 15, 24, 36}.\n- Four-digit: N = 1352.\n- Infinite family (k = 3): For any n ≥ 1, let A = (5·10^{n−1} + 1)/3 and B = (10^n + 2)/3; then N is the concatenation AB (examples: n = 2 gives N = 1734).\n- Infinite family (k = 7): For n = 6t + 3 (t ≥ 0), let A = B = (10^n + 1)/7; then N is the concatenation AB (examples: n = 3 gives N = 143143; n = 9 gives N = 142857143142857143).",
    "solution": "Let $N = \\overline{a_1a_2\\dots a_n b_1b_2\\dots b_n}$ be the $2n$-digit number, with $a_1 \\ge 1$. Also let $A = \\overline{a_1a_2\\dots a_n}$ and $B = \\overline{b_1b_2\\dots b_n}$, so by hypothesis we are given $10^n A + B = N = kAB$ for some positive integer $k$. Multiplying with $k$ and adding $10^n$, we arrive at $(kA - 1)(kB - 10^n) = 10^n$.\n\nFor $n = 1$ we immediately solve $(kA-1)(kB-10) = 10$, with $A$ and $B$ not-null digits, to obtain $N \\in \\{11, 12, 15, 24, 36\\}$.\n\nFor $n \\ge 2$ we have $10^{n-1} \\le A < 10^n$ and $1 \\le B < 10^n$, thus $1 < k < 10$. A case-by-case discussion follows.\n\n* $k=2$. Then $2 \\cdot 10^{n-1} - 1 \\le 2A - 1 \\mid 10^n$, so $2A - 1 \\le 5^n$, forcing $n \\le 2$. From $(2A-1)(2B-100) = 100$ we get $N = 1352$.\n\n* $k=3$. Then $3 \\cdot 10^{n-1} - 1 \\le 3A - 1 \\mid 10^n$. We cannot have $3A - 1 = 10^n$, impossible modulo 3. We cannot have $3A - 1 = 2 \\cdot 10^{n-1}$, because it is too small. For $3A - 1 = 5 \\cdot 10^{n-1}$, we get $A = \\frac{5 \\cdot 10^{n-1} + 1}{3}$ and $B = 2A = \\frac{10^n + 2}{3}$ (for example, for $n=2$ we get $N = 1734$). Thus we get a first infinite family of solutions.\n\n* $k=4$. Then $4 \\cdot 10^{n-1} - 1 \\le 4A - 1 \\mid 10^n$, so $4A - 1 \\le 5^n$, impossible.\n\n* $k=5$. Then $5 \\cdot 10^{n-1} - 1 \\le 5A - 1 \\mid 10^n$, so $5A - 1 \\le 2^n$, impossible.\n\n* $k=6$. Then $6 \\cdot 10^{n-1} - 1 \\le 6A - 1 \\mid 10^n$, so $6A - 1 \\le 10^n$, which forces $6A - 1 = 10^n$, impossible modulo 3.\n\n* $k=7$. Then $7 \\cdot 10^{n-1} - 1 \\le 7A - 1 \\mid 10^n$, so $7A - 1 \\le 10^n$, which forces (as in the above) $7A - 1 = 10^n$. This leads to $A = B = \\frac{10^n + 1}{7}$, with only proviso that $n$ is an odd multiple of 3 in order to $A, B$ to be integer (for example, for $n=3$ we get $N = 143143$, while for $n=9$ we get $N = 142857143142857143$). Thus we arrive at a second infinite family of solutions.\n\n* $k=8$ and $k=9$ don't work, since they force $kA - 1 = 10^n$, impossible modulo 2 or modulo 3.\n\nThus, wrapping things up, the solutions are $N \\in \\{11, 12, 15, 24, 36\\}$, the isolated value $N = 1352$, and the two infinite families described in the above.\nWith the above notations, from $kAB = 10^n A + B$, we get that $10^n A = B(kA-1)$, so $A \\mid B(kA-1)$. Since $(A, kA-1) = 1$, we have $A \\mid B$, so $B = pA$, where $p \\in \\mathbb{N}^*$. It follows that $p(kA-1) = 10^n$, so $p \\mid 10^n$. As $A$ and $B$ are $n$-digit numbers, we have $1 \\le p < 10$, so $p \\in \\{1, 2, 4, 5, 8\\}$.\n\nMoreover, from $p(kA-1) = 10^n$ and $10^{n-1} \\le A \\le B < 10^n$, we get that $1 + \\frac{p}{10^n} < k \\le \\frac{10}{p} + \\frac{1}{10^{n-1}}$, so $2 \\le k \\le \\frac{10}{p} + \\frac{1}{10^{n-1}}$. In order to improve this bounding, we'll study separately $n=1$, in which case we easily find the solutions $N \\in \\{11, 12, 15, 24, 36\\}$.\n\nFor $n \\ge 2$, it follows that we have $2 \\le k \\le \\frac{10}{p}$, which immediately drops the case $p=8$. Also, for $p \\in \\{4, 5\\}$, we must have $k=2$. If $p=5$, then $2A - 1 = 5 \\cdot 10^{n-1}$, impossible, for parity reasons. If $p = 4$, we get $2A - 1 = 25 \\cdot 10^{n-2}$. Again, parity arguments provide $n = 2$, so $A = 13$ and $B = 52$, so $N = 1352$. That leaves us with $p \\in \\{1, 2, 4\\}$.\n\nFor $p = 1$, we have $2 \\le k \\le 10$ and $kA = 10^n + 1$, so $k$ must be odd, and it cannot be 5 or divisible by 3. The only possibility remains $k = 7$, which works only if $7 \\mid 10^n + 1$; we get that $n = 6t + 3$, $t \\in \\mathbb{N}$, so\n$$\nA = B = \\frac{10^{6t+3} + 1}{7} \\quad \\text{and} \\quad N = \\underbrace{142\\,857 \\dots 142\\,857}_{t \\text{ times } 142857} 143.\n$$\n\nFor $p = 2$, we have $2 \\le k \\le 5$ and $kA = 5 \\cdot 10^{n-1} + 1$, so $k$ is odd and it is not 5, so it must be 3 (which works for all $n \\ge 1$). It follows that\n$$\nA = \\frac{5 \\cdot 10^{n-1} + 1}{3}, \\quad B = \\frac{10^n + 2}{3}, \\quad N = \\underbrace{1\\,66\\dots6\\,7\\,33\\dots3\\,4}_{n-1 \\text{ times } 6}, \\quad \\underbrace{\\text{n times 3}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38321,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all positive integers $a$ for which\n$$\n\\frac{1}{4} < \\frac{1}{a+1} + \\frac{1}{a+2} + \\frac{1}{a+3} < \\frac{1}{3}.\n$$\n\nb) Prove that for any integer $p \\ge 2$ there exist $p$ consecutive positive integers $a_1, a_2, \\dots, a_p$ such that\n$$\n\\frac{1}{p+1} < \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_p} < \\frac{1}{p}.\n$$",
    "options": [],
    "answer": "a ∈ {8, 9, 10}; for part (b), one valid choice is a_i = p^2 + i for i = 1, 2, ..., p",
    "solution": "a) Set $S = \\frac{1}{a+1} + \\frac{1}{a+2} + \\frac{1}{a+3}$ and notice that $\\frac{3}{a+3} < S < \\frac{3}{a+1}$ to infer from $\\frac{1}{4} < S < \\frac{1}{3}$ that $\\frac{1}{4} < \\frac{3}{a+1}$ and $\\frac{3}{a+3} < \\frac{1}{3}$. Consequently $6 < a < 11$, so $a \\in \\{7, 8, 9, 10\\}$. It is easy to check that $a = 7$ fails and $8, 9, 10$ are solutions.\n\nb) Select $a_1 = p^2 + 1, a_2 = p^2 + 2, \\dots, a_p = p^2 + p$ to get\n$$\nS = \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_p} = \\frac{1}{p^2+1} + \\frac{1}{p^2+2} + \\dots + \\frac{1}{p^2+p}\n$$\nand notice that\n$$\n\\frac{1}{p^2+1} > \\frac{1}{p^2+2} > \\dots > \\frac{1}{p^2+p}\n$$\nimplies\n$$\n\\frac{p}{p^2+p} < S < \\frac{p}{p^2+1} < \\frac{p}{p^2} = \\frac{1}{p}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, determine the largest real number $\\mu$ satisfying the following condition: for every $4n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu$.",
    "options": [],
    "answer": "1/(2n+2)",
    "solution": "The required maximum is $1/(2n+2)$. To show that the condition in the statement is not met if $\\mu > 1/(2n+2)$, let $U = (0,1) \\times (0,1)$, choose a small enough positive $\\varepsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $(i/(n+1) \\pm \\varepsilon) \\times (1/2 \\pm \\varepsilon)$, $i = 1, \\dots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at most $(1/(n+1) + \\varepsilon)(1/2 + \\varepsilon) < \\mu$ if $\\varepsilon$ is small enough.\n\nWe now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu_0 = 2/(|C| + 4)$.\n\nTo prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.\n\n**Lemma 1.** Let $k$ be a positive integer, and let $\\lambda < 1/(|k/2| + 1)$ be a positive real number. If $t_1, \\dots, t_k$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_i$ is isolated from the other $t_j$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$.\n\n**Lemma 2.** Given an integer $k \\ge 2$ and positive integers $m_1, \\dots, m_k$,\n$$\n\\lfloor m_1/2 \\rfloor + \\sum_{i=1}^{k} \\lfloor m_i/2 \\rfloor + \\lfloor m_k/2 \\rfloor \\le \\sum_{i=1}^{k} m_i - k + 2.\n$$\n\nBack to the problem, let $U = (0,1) \\times (0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_1 < \\dots < x_k$ in the open unit interval $(0,1)$, let $\\ell_i$ be the vertical through $x_i$, and let $m_i = |C \\cap \\ell_i|$, $i = 1, \\dots, k$.\n\nSetting $x_0 = 0$ and $x_{k+1} = 1$, assume that $x_{i+1} - x_{i-1} > (\\lfloor m_i/2 \\rfloor + 1)\\mu_0$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \\cap \\ell_i$ from the other ones by an open subinterval $x_i \\times J$ of $x_i \\times (0,1)$ whose length is greater than or equal to $\\mu_0/(x_{i+1} - x_{i-1})$. Consequently, $(x_{i-1}, x_{i+1}) \\times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\\mu_0$.\n\nNext, we rule out the case $x_{i+1} - x_{i-1} \\le (\\lfloor m_i/2 \\rfloor + 1)\\mu_0$ for all indices $i$. If this were the case, notice that necessarily $k > 1$; also, $x_1 - x_0 < x_2 - x_0 \\le (\\lfloor m_1/2 \\rfloor + 1)\\mu_0$ and $x_{k+1} - x_k < x_{k+1} - x_{k-1} \\le (\\lfloor m_k/2 \\rfloor + 1)\\mu_0$. With reference to Lemma 2, write\n$$\n\\begin{align*}\n2 = 2(x_{k+1} - x_0) &= (x_1 - x_0) + \\sum_{i=1}^{k} (x_{i+1} - x_{i-1}) + (x_{k+1} - x_k) \\\\\n&< \\left( \\left( \\lfloor m_1/2 \\rfloor + 1 \\right) + \\sum_{i=1}^{k} \\left( \\lfloor m_i/2 \\rfloor + 1 \\right) + \\left( \\lfloor m_k/2 \\rfloor + 1 \\right) \\right) \\cdot \\mu_0 \\\\\n&\\le \\left( \\sum_{i=1}^{k} m_i + 4 \\right) \\mu_0 = (|C| + 4)\\mu_0 = 2,\n\\end{align*}\n$$\nand thereby reach a contradiction.\n\nFinally, we prove the two lemmas.\n\n**Proof of Lemma 1.** Suppose, if possible, that no $t_i$ is isolated from the other $t_j$ by an open subinterval of $(0, 1)$ whose length is greater than or equal to $\\lambda$. Without loss of generality, we may (and will) assume that $0 = t_0 < t_1 < \\dots < t_k < t_{k+1} = 1$. Since the open interval $(t_{i-1}, t_{i+1})$ isolates $t_i$ from the other $t_j$, its length, $t_{i+1} - t_{i-1}$, is less than $\\lambda$. Consequently, if $k$ is odd, then $1 = \\sum_{i=0}^{(k-1)/2} (t_{2i+2} - t_{2i}) < \\lambda(1 + (k-1)/2) < 1$; and if $k$ is even, then $1 < 1 + t_k - t_{k-1} = \\sum_{i=0}^{k/2-1} (t_{2i+2} - t_{2i}) + (t_{k+1} - t_{k-1}) < \\lambda(1 + k/2) < 1$. A contradiction in either case.\n\n**Proof of Lemma 2.** Let $I_0$, respectively $I_1$, be the set of all indices $i$ in the range $2, \\dots, k-1$ such that $m_i$ is even, respectively odd. Clearly, $I_0$ and $I_1$ form a partition of that range. Since $m_i \\ge 2$ if $i$ is in $I_0$, and $m_i \\ge 1$ if $i$ is in $I_1$ (recall that the $m_i$ are positive integers), $\\sum_{i=2}^{k-1} m_i = \\sum_{i \\in I_0} m_i + \\sum_{i \\in I_1} m_i \\ge 2|I_0| + |I_1| = 2(k-2) - |I_1|$, or $|I_1| \\ge 2(k-2) - \\sum_{i=2}^{k-1} m_i$. Consequently,\n$$\n\\begin{align*}\n\\lfloor m_1/2 \\rfloor + \\sum_{i=1}^{k} \\lfloor m_i/2 \\rfloor + \\lfloor m_k/2 \\rfloor &\\le m_1 + \\left( \\sum_{i=2}^{k-1} m_i/2 - |I_1|/2 \\right) + m_k \\\\\n&\\le m_1 + \\left( \\sum_{i=2}^{k-1} m_i/2 - (k-2) + \\sum_{i=2}^{k-1} m_i/2 \\right) + m_k \\\\\n&= \\sum_{i=1}^{k} m_i - k + 2.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $n$ be positive integers such that\n$$\nn = \\frac{x^2 - 1}{2} = \\frac{y^2 - 1}{3}.\n$$\n\nProve that $n = y^2 - x^2$ and $20$ divides $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. As $x^2 = 2n + 1$ and $y^2 = 3n + 1$, one has $y^2 - x^2 = n$.\n\nb. Since $x$ is odd, both $x-1$ and $x+1$ are even and so $4$ divides $x^2 - 1 = 2n$, implying that $n$ is even. Then $y^2 = 3n + 1$ is odd, and so is $y$. Again, $4$ divides $y^2 - 1$ so $n = (y^2 - 1) - (x^2 - 1)$ is divisible by $4$; in fact, by $8$.\n\nOn the other hand, a perfect square gives remainders $0$, $1$ or $4$ when divided by $5$. The equality $x^2 + y^2 = 5n + 2$ implies that both $x^2$ and $y^2$ leave the remainder $1$ when divided by $5$, hence $5$ divides $y^2 - x^2$ and the claim follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $A'$ be the centre of the circle through the midpoint of the side $BC$ and the orthogonal projections of $B$ and $C$ on the lines of support of the internal bisectrices of the angles $ACB$ and $ABC$, respectively; the points $B'$ and $C'$ are defined similarly. Prove that the nine-point circle of the triangle $ABC$ and the circumcircle of $A'B'C'$ are concentric.\n\nRMM 2015 Shortlist, Mexico",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All the angles in the solution are directed modulo $\\pi$. The following notation is used throughout the proof:\n\n$2\\alpha$, $2\\beta$, $2\\gamma$ measures of the angles $BAC$, $CBA$, $ACB$, respectively;\n\n$2a$, $2b$, $2c$ lengths of the sides $BC$, $CA$, $AB$, respectively;\n\n$I$ incentre of the triangle $ABC$;\n\n$M_A$, $M_B$, $M_C$ midpoints of the sides $BC$, $CA$, $AB$, respectively;\n\n$X_Y$ orthogonal projection of $X$ on the line $YI$, for all $X, Y \\in \\{A, B, C\\}$;\n\n$\\omega_A$, $\\omega_B$, $\\omega_C$ circumcircles of the triangles $M_A B_C C_B$, $M_B C_A A_C$, $M_C A_B B_A$, respectively, centred at $A'$, $B'$, $C'$, respectively.\n\nSince the angle $AA_B B$ is right, the segments $M_C A = M_C B = M_C A_B = c$ (see Fig. 3), so $\\angle M_C A_B B = \\angle A_B B M_C C = \\angle C B A_B = \\beta$; this means that the lines $M_C A_B$ and $BC$ are parallel, so $A_B$ lies on the line $M_B M_C$. So, $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, $(M_C, M_A, B_C, B_A)$ are quartets of collinear points.\n\n![](attached_image_1.png)\n\nFig. 3\n\nLet $A''$ be the incentre of the triangle $AM_C M_B$ (in other words, $A''$ is the midpoint of $AI$). We show that $A''$ lies on both $\\omega_B$ and $\\omega_C$. For that, notice first that the points $A, B, A_B$, and $B_A$ lie on the circle on diameter $AB$; hence $\\angle A_B B_A A = \\angle A_B B A = \\beta$. Next, the points $A_B, A_C, M_B$, and $M_C$ lie on a line parallel to $BC$, so $\\angle A_B M_C A'' = \\angle M_B M_C A'' = \\beta = \\angle A_B B_A A$. This means that $A''$ lies on $\\omega_C$. Similarly, $A''$ lies on $\\omega_B$.\n\n![](attached_image_2.png)\n\nFig. 4\n\nLet $X$ be the second point of intersection of $\\omega_B$ and $\\omega_C$. By the preceding, $\\angle M_B X A'' = \\angle M_B C_A A'' = \\alpha$ and similarly $\\angle A'' X M_C = \\alpha$. This yields $\\angle M_B X M_C = \\angle M_B X A'' + \\angle A'' X M_C = 2\\alpha = \\angle M_B M_A M_C$, which shows that $X$ lies also on the circumcircle $\\omega$ of the triangle $M_A M_B M_C$, which is the nine-point circle of the triangle $ABC$. Denote the center of $\\omega$ by $O'$.\n\nNow the lines $A''X, M_BX$, and $M_CX$ are the radical axes of the circles $\\omega_B, \\omega_C$, and $\\omega$. Since $XA''$ forms equal angles with $M_BX$ and $M_CX$, the triangle $O'B'C'$ formed by the centres of these circles has equal angles at $B'$ and $C'$; therefore, $O'B' = O'C'$. A similar argument shows that $O'B' = O'A'$, and $O'$ is consequently the circumcentre of the triangle $A'B'C'$.\nWith reference to the notation in Solution 1, let $O$ and $O'$ be the circumcentre and the centre of the nine-point circle of the triangle $ABC$, respectively. As in the previous solution, usage is made of the fact that $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, and $(M_C, M_A, B_C, B_A)$ are quartets of collinear points, and $M_A B_C = M_A C_B = a$, $M_B A_C = M_B C_A = b$, and $M_C A_B = M_C B_B = c$.\n\n![](attached_image_3.png)\n\nFig. 5\n\nWe claim that $A'O' = IO/2$; similarly, $B'O' = IO/2 = C'O'$, whence the required result. To prove the claim, notice that each vector $v$ is uniquely determined by its projections on the lines $AB$ and $AC$. The signed lengths of these projections will be denoted $\\mathrm{pr}_c v$ and $\\mathrm{pr}_b v$, respectively, the rays $AB$ and $AC$ emanating from $A$ being considered positive.\n\nThe points $O'$ and $A'$ are the circumcentres of the triangles $M_A M_B M_C$ and $M_A B_C C_B$, respectively. Project onto $M_A M_C$, to get $\\mathrm{pr}_b A'O' = \\mathrm{pr}_b (M_A O' - M_A A') = (-b/2 + a/2) = (a-b)/2$. Similarly, $\\mathrm{pr}_c A'O' = (a-c)/2$. On the other hand, $\\mathrm{pr}_b IO = \\mathrm{pr}_b (AO - AI) = b - (b+c-a) = a-c$; similarly, $\\mathrm{pr}_c IO = a-b$. This means that the vector $A'O'$ reflected in the bisectrix $AI$ of the angle $BAC$ is equal to the vector $IO/2$, hence the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c > 0$ such that $a \\ge bc^2$, $b \\ge ca^2$ and $c \\ge ab^2$. Determine the maximum value of the expression $E = abc(a - bc^2)(b - ca^2)(c - ab^2)$.",
    "options": [],
    "answer": "1/64",
    "solution": "With $x = ab$, $y = bc$, $z = ca$, we have to find $\\max(x - y^2)(y - z^2)(z - x^2)$. From the AM-GM inequality, the product is at most\n\n$$\n\\left( \\frac{(x - x^2 + y - y^2 + z - z^2)}{3} \\right)^3\n$$\nBut $x - x^2 \\le \\frac{1}{4}$ and its analogues lead to $(x - y^2)(y - z^2)(z - x^2) \\le \\frac{1}{4^3}$. This value, $\\frac{1}{4^3}$, is indeed obtained for $x = y = z = \\frac{1}{2}$, that is for $a = b = c = \\frac{1}{\\sqrt{2}}$, so the desired maximum value is $\\frac{1}{4^3}$.\nBy using the Taiwanese Transformation, that is $a = x \\cdot bc^2$, $b = y \\cdot ca^2$, $c = z \\cdot ab^2$ (with $xyz \\ge 1$) we obtain $(xyz)(abc)^2 = 1$. So\n$$\nE = a^4 b^4 c^4 (x-1)(y-1)(z-1) = \\frac{(x-1)(y-1)(z-1)}{x^2 y^2 z^2} \\le \\frac{1}{64},\n$$\nsince $\\frac{t-1}{t^2} \\le \\frac{1}{4}$ when $t \\ge 1$, thus for $t \\in \\{x, y, z\\}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive real number $t$, determine the sets $A$ of real numbers containing $t$, for which there exists a set $B$ of real numbers depending on $A$, $|B| \\ge 4$, such that the elements of the set $AB = \\{ab: a \\in A, b \\in B\\}$ form a finite arithmetic progression.",
    "options": [],
    "answer": "{t}, {-t, t}, {0, t}, {-t, 0, t}",
    "solution": "The required sets are $\\{t\\}$, $\\{-t, t\\}$, $\\{0, t\\}$ and $\\{-t, 0, t\\}$. It is readily checked that the elements of the Minkowski product of each of these sets and the set $\\{-1, 0, 1, 2\\}$ form a finite arithmetic progression.\n\nNow, let $A$ and $B$ be sets of real numbers satisfying the conditions in the statement, and let $|A| \\ge 2$ (the case $|A| = 1$ is trivial). Clearly, $A$ and $B$ are both finite.\n\nLet $d > 0$ be the difference of the arithmetic progression $AB$, consider two distinct elements of $A$, say $x$ and $x'$, and two distinct elements of $B$, say $y$ and $y'$, and notice that the elements of $A$, respectively $B$, are integral multiples of $d/(y-y')$, respectively $d/(x-x')$. Scaling $A$ and $B$ accordingly, we may (and will) assume that $A$ and $B$ are both sets of integers. Dividing, if necessary, the elements of $A$, respectively $B$, by their greatest common divisor, we may (and will) further assume that the elements of $A$, respectively $B$, are jointly coprime: $\\text{gcd } A = 1$ and $\\text{gcd } B = 1$. Further, recall that $A$ and $B$ are both finite and let $a^*$, respectively $b^*$, be an element of $A$, respectively $B$, of maximal absolute value. If necessary, multiply by $-1$ to assume $a^* > 0$ and $b^* > 0$. Under these simplifying assumptions, we will show that $A$ is one of the sets $\\{-1, 1\\}$, $\\{0, 1\\}$, $\\{-1, 0, 1\\}$, whence the conclusion.\n\nSince $\\text{gcd } B = 1$ and $d$ divides $(x - x')y$ for all $x$ and $x'$ in $A$ and all $y$ in $B$, it follows that $d$ divides the difference of any two members of $A$. Similarly, $d$ divides the difference of any two members of $B$, and since $|B| \\ge 4$, it follows that $b^* > d$.\n\nConsider now elements $a$ in $A$ and $b$ in $B$ such that $ab = a^*b^* - d$, and notice that $ab = a^*b^* - d \\ge b^* - d > 0$. Moreover, $|a| = a^*$, for otherwise $a^*b^* - d = ab = |a||b| \\le (a^* - 1)b^* = a^*b^* - b^* < a^*b^* - d$ which is a contradiction.\n\nThis means that $d = a^*b^* - |a||b| = a^*(b^* - |b|) \\ge a^*$. Now, since $a^* \\le d$, and the elements of $A$ are congruent modulo $d$, the only possible options for $A$ are either subsets of $\\{-d, 0, d\\}$, or $\\{-d/2, d/2\\}$ if $d$ is even, or finally sets of the form $\\{a^*, a^* - d\\}$, where $d > a^* > |a^* - d|$. The first two cases are covered by the answer.\n\nTo rule out the last option, notice that $a = a^*$ (since $|a| = a^* > |a^* - d|$), and therefore $d = a^*(b^* - |b|)$. This means that $a^*$ divides $d$, so $a^* \\le d/2$ and $|a^* - d| \\ge a^*$, in contradiction with $a^* > |a^* - d|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to partition the set of the positive integers into two subsets such that none of them contains an infinitely long (non-constant) arithmetic progression?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is in the affirmative. For example, one can consider\n$$A = \\{1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 21, \\dots\\},$$\n$$B = \\{2, 5, 6, 10, 11, 12, 17, 18, 19, 20, 26, \\dots\\},$$\ni.e., for the infinite word $W = w_1w_2\\cdots w_k \\cdots = aba^2b^2\\cdots a^nb^n\\cdots$, put $A$ the set of the indices $i$ for which $w_i = a$, and $B$ the set of the indices $i$ for which $w_i = b$. The terms of an arithmetic progression are equally spaced, while $A$ and $B$ both contain arbitrarily long gaps, hence the conclusion follows readily.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the integral lattice $\\mathbb{Z}^n$, $n \\ge 2$, in the Euclidean $n$-space. Define a line in $\\mathbb{Z}^n$ to be a set of the form $a_1 \\times \\dots \\times a_{k-1} \\times \\mathbb{Z} \\times a_{k+1} \\times \\dots \\times a_n$, where $k$ is an integer in the range $1, 2, \\dots, n$, and the $a_i$ are arbitrary integers. A subset $A$ of $\\mathbb{Z}^n$ is called *admissible* if it is non-empty, finite, and every line in $\\mathbb{Z}^n$ which intersects $A$ contains at least two points of $A$. A subset $N$ of $\\mathbb{Z}^n$ is called *null* if it is non-empty, and every line in $\\mathbb{Z}^n$ intersects $N$ in an even number of points (possibly zero).\n\na) Prove that every admissible set in $\\mathbb{Z}^2$ contains a null set.\n\nb) Exhibit an admissible set in $\\mathbb{Z}^3$ no subset of which is a null set.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Let $A$ be an admissible set in $\\mathbb{Z}^2$, choose a point $a_0$ of $A$, and, for each positive integer $k$, choose a point $a_k$ of $A$ different from $a_{k-1}$, having the same first coordinate as the latter if $k$ is odd, and the same second coordinate if $k$ is even. Eventually we must choose an $a_n = a_m$, $m < n$. Assume $a_n$ is the first point to duplicate a preceding point. If $m$ and $n$ have like parities, then $a_m, a_{m+1}, \\dots, a_{n-1}$ form a null set, and if they have opposite parities, then $a_{m+1}, \\dots, a_{n-1}$ do.\n\n(b) We exhibit a minimal admissible set $A$ in $\\mathbb{Z}^3$ which is not itself null. Here and hereafter, minimality refers to the fact that no proper subset is admissible. Since every null finite set is admissible, the conclusion follows. The set $A$ is a set of lattice points in the parallelepiped $[0, 3] \\times [0, 3] \\times [0, 4]$. We describe it by successive horizontal cross-sections:\n$$\nA = A_0 \\times 0 \\cup A_1 \\times 1 \\cup A_2 \\times 2 \\cup A_3 \\times 3 \\cup A_4 \\times 4,\n$$\nwhere $A_0 = \\{0, 3\\} \\times \\{0, 3\\}$, $A_1 = \\{0, 1\\} \\times \\{2, 3\\} \\cup \\{1, 2\\} \\times \\{0, 1\\}$, $A_2 = \\{0, 1\\} \\times \\{1, 2\\} \\cup \\{2, 3\\} \\times \\{2, 3\\}$, $A_3 = \\{1, 2\\} \\times \\{2, 3\\} \\cup \\{2, 3\\} \\times \\{0, 1\\}$, and $A_4 = \\{0, 1\\} \\times \\{0, 1\\} \\cup \\{2, 3\\} \\times \\{1, 2\\}$. Notice that, for $k = 1, 2, 3$, the configuration $A_{k+1}$ is obtained from $A_k$ by a clockwise rotation through $\\pi/2$ about the centre of the square $[0, 3] \\times [0, 3]$.\n\nThe set $A$ is admissible, since each horizontal cross-section $A_k \\times k$ is admissible in $\\mathbb{Z}^2 \\times k$, and the perpendicular in $\\mathbb{Z}^3$ to any horizontal cross-section through any one of its points meets at least one other horizontal cross-section.\n\nTo prove minimality, we exhibit a connected geometric lattice graph $G$ on $A$ such that the line of support of each edge of $G$ is a line in $\\mathbb{Z}^3$ stabbing $A$ at exactly two points, namely, the end points of that edge. The existence of such a graph implies minimality, since removal of any one point in $A$ entails removal of all its neighbours in $G$, and eventually removal of all of $A$.\n\nBegin by noticing that each of the verticals $i \\times j \\times \\mathbb{Z}$ through a point of $A$, where either $i$ or $j$ is in $\\{0, 3\\}$, stabs exactly two horizontal cross-sections of $A$. Join the corresponding points of $A$ by the vertical segment they determine.\n\nNext, consider the generic planar lattice paths $\\alpha_1 = (1 \\times 0)(2 \\times 0)(2 \\times 1)(1 \\times 1)$ and $\\alpha'_1 = (1 \\times 2)(0 \\times 2)(0 \\times 3)(1 \\times 3)$, and, for $k = 1, 2, 3$, let $\\alpha_{k+1}$ and $\\alpha'_{k+1}$ be obtained from $\\alpha_k$ and $\\alpha'_k$, respectively, by a clockwise rotation through $\\pi/2$ about the centre of the square $[0, 3] \\times [0, 3]$. The edges of the lattice paths $\\alpha_k \\times k$ and $\\alpha'_k \\times k$, joining points in $A_k \\times k$, $k = 1, 2, 3, 4$, along with the vertical edges in the previous paragraph form the desired connected geometric lattice graph $G$ on $A$.\n\nFinally, the set $A$ is not null, for the vertical $1 \\times 1 \\times \\mathbb{Z}$ stabs exactly three horizontal cross-sections of $A$, namely, $A_1 \\times 1$, $A_2 \\times 2$ and $A_4 \\times 4$; in fact, each of the lines $i \\times j \\times \\mathbb{Z}$, $i, j \\in \\{1, 2\\}$, stabs exactly three horizontal cross-sections of $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic cu $AB \\neq AC$, $M$ mijlocul laturii $[BC]$, $H$ ortocentrul triunghiului $ABC$, $O_{1}$ mijlocul lui $[AH]$, iar $O_{2}$ centrul cercului circumscris triunghiului $BCH$. Demonstraţi că $O_{1}AMO_{2}$ este paralelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCercul circumscris triunghiului este simetricul cercului circumscris triunghiului $ABC$ faţă de $BC$, prin urmare $O_{2}$ este simetricul lui $O$ faţă de $BC$. Se ştie că $AO_{1} = MO = MO_{2}$ şi, cum $AO_{1} \\parallel MO_{2}$, rezultă concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSă se determine funcţiile derivabile $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ care verifică simultan condiţiile:\ni) $f'(x)=0$, pentru orice $x \\in \\mathbb{Z}$;\nii) pentru $x \\in \\mathbb{R}$, dacă $f'(x)=0$, atunci $f(x)=0$.",
    "options": [],
    "answer": "The identically zero function f(x) = 0 for all real x.",
    "solution": "Solution:\nFuncţia identic nulă verifică condiţiile din enunţ.\n\nPresupunem că există o funcţie derivabilă $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, neidentic nulă, care satisface (i) şi (ii).\nAtunci $f$ este continuă pe $\\mathbb{R}$, cu $f(x)=f'(x)=0, \\forall x \\in \\mathbb{Z}$.\n\nExistă $x_{0} \\in \\mathbb{R} \\setminus \\mathbb{Z}$ astfel încât $f\\left(x_{0}\\right) \\neq 0$. Fie $k=\\left[x_{0}\\right] \\in \\mathbb{Z}$.\n\nConform Teoremei lui Weierstrass, $f$ este mărginită pe intervalul $[k, k+1]$ şi există $x_{1}, x_{2} \\in [k, k+1]$ astfel ca\n$$\nf\\left(x_{1}\\right) \\leq f(x) \\leq f\\left(x_{2}\\right), \\forall x \\in [k, k+1]\n$$\n\nDacă $f\\left(x_{0}\\right)>0$, atunci $f\\left(x_{2}\\right)>0$. Deci $x_{2} \\in (k, k+1)$ este un punct de maxim local pentru $f$, de unde, conform Teoremei lui Fermat, $f'\\left(x_{2}\\right)=0$. Atunci, conform (ii), $f\\left(x_{2}\\right)=0$. Contradicţie. Dacă $f\\left(x_{0}\\right)<0$, atunci $f\\left(x_{1}\\right)<0$. Dar $x_{1} \\in (k, k+1)$ este un punct de minim local pentru $f$, deci $f'\\left(x_{1}\\right)=0$, de unde $f\\left(x_{1}\\right)=0$. Contradicţie.\n\nDeci funcţia identic nulă este unica funcţie care satisface condiţiile din enunţ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArătaţi că nu putem alege 45 de elemente distincte ale mulţimii $\\{\\sqrt{1}, \\sqrt{2}, \\sqrt{3}, \\ldots, \\sqrt{2015}\\}$, astfel încât numerele selectate să fie în progresie aritmetică.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDacă $m, n, p \\in \\mathbb{N}^{*}$ și $\\sqrt{m}, \\sqrt{n}, \\sqrt{p}$ sunt trei numere în progresie aritmetică, atunci $p + m + 2 \\sqrt{p m} = 4 n$, deci $\\sqrt{p m}$ este rațional. Rezultă $m = a^{2} d$, $p = c^{2} d$, $n = 2 b^{2} d$, cu $a, b, c, d \\in \\mathbb{N}$ și $a + c = 2b$.\n\nAstfel, dacă am putea alege 45 de numere în progresie aritmetică, atunci ele ar fi de forma $a_{1} \\sqrt{d}, a_{2} \\sqrt{d}, \\ldots, a_{45} \\sqrt{d}$, cu $a_{1}, a_{2}, \\ldots, a_{45}, d \\in \\mathbb{N}^{*}$ și $a_{1}, a_{2}, \\ldots, a_{45}$ în progresie aritmetică.\n\nÎn acest caz, cel mai mare număr ales ar fi cel puțin $\\sqrt{45^{2} d} \\geq \\sqrt{2025}$, contradicție.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră paralelogramul $ABCD$, ale cărui diagonale se intersectează în $O$. Bisectoarele unghiurilor $DAC$ şi $DBC$ se intersectează în $T$. Se ştie că $\\overrightarrow{TD} + \\overrightarrow{TC} = \\overrightarrow{TO}$. Determinaţi măsurile unghiurilor triunghiului $ABT$.",
    "options": [],
    "answer": "60°, 60°, 60°",
    "solution": "Solution:\n\nDin ipoteză rezultă că $DOCT$ este paralelogram.\n\nDin $AO \\parallel DT$ deducem $\\angle DTA \\equiv \\angle OAT \\equiv \\angle DAT$, deci $DA = DT$.\n\nAstfel $DA = DT = OC$; analog $BC = CT = OD$, de unde $BD = AC$. Astfel $ABCD$ este dreptunghi, $AOTD$ este romb, triunghiul $AOD$ este echilateral şi triunghiul $ABT$ este echilateral, deci are unghiurile de $60^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $(R,+, \\cdot)$ un inel cu proprietatea că, pentru orice element $x \\in R$, există două elemente $e_{1}$ şi $e_{2}$ din $R$, astfel încât $e_{1}^{2}=e_{1}$, $e_{2}^{2}=e_{2}$ şi $x=e_{1} e_{2}$. Arătaţi că:\n(a) $1$ este singurul element inversabil al inelului $R$; şi\n(b) $x^{2}=x$, oricare ar fi $x \\in R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n(a) Fie $x \\in U(R)$ şi $e_{1}, e_{2} \\in I(R)=\\{e \\in R \\mid e^{2}=e\\}$, cu $x=e_{1} e_{2}$. Atunci\n$$\n1-e_{1}=(1-e_{1}) \\cdot 1=(1-e_{1}) e_{1} e_{2} x^{-1}=(e_{1}-e_{1}^{2}) e_{2} x^{-1}=0\n$$\ndeci $e_{1}=1$ şi $x^{2}=e_{2}^{2}=e_{2}=x$. Cum $x$ este inversabil, rezultă că $x=1$ şi deci $U(R)=\\{1\\}$.\n\n(b) $R$ nu conţine elemente nilpotente nenule, deoarece pentru orice nilpotent $x \\in R$, elementul $1-x$ este inversabil $\\left(x^{k}=0\\right.$, deci $\\left.(1-x)(1+x+\\cdots+x^{k-1})=1\\right)$, deci $1-x=1$ şi $x=0$.\nArătăm că $I(R) \\subseteq Z(R)$. Fie $e \\in I(R)$ şi $x \\in R$ oarecare. Atunci\n$$\n(e x-e x e)^{2}=e x e x-e x e x e-e x e e x+e x e x e=e x e x-e x e x e-e x e x+e x e x=0\n$$\ndeci $e x=e x e$. Analog obţinem $x e=e x e$ şi deci $e x=e x e=x e$, oricare ar fi $e \\in I(R)$ şi oricare ar fi $x \\in R$.\nFie $x \\in R$ şi $e_{1}, e_{2} \\in I(R)$, cu $x=e_{1} e_{2}$. Atunci $x^{2}=(e_{1} e_{2})^{2}=e_{1}^{2} e_{2}^{2}=e_{1} e_{2}=x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIgazold, hogy minden $n \\geq 2$ természetes szám esetén\n$$\n\\sum_{k=2}^{n} \\frac{1}{\\sqrt[k]{(2 k)!}} \\geq \\frac{n-1}{2 n+2}\n$$\n\nProblem:\n\nSă se arate că pentru orice $n \\geq 2$ natural, are loc inegalitatea\n$$\n\\sum_{k=2}^{n} \\frac{1}{\\sqrt[k]{(2 k)!}} \\geq \\frac{n-1}{2 n+2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDemonstrăm inegalitatea prin inducţie. În cazul $n=2$ avem egalitate.\n\nSă observăm că, la pasul de inducţie, în trecerea de la $n-1$ la $n$, membrul drept creşte cu\n$$\n\\frac{n-1}{2 n+2}-\\frac{n-2}{2 n}=\\frac{1}{n(n+1)}\n$$\ndeci e suficient să demonstrăm că\n$$\n\\frac{1}{\\sqrt[n]{(2 n)!}} \\geq \\frac{1}{n(n+1)}\n$$\nceea ce rezultă imediat prin înmulţirea inegalităţilor\n$$\nk(2 n-k+1) \\leq n(n+1)\n$$\npentru $k=1,2, \\ldots, n$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi cel mai mic număr natural $n$ pentru care, oricum am colora în roşu $n$ dintre vârfurile unui cub, există un vârf al cubului care are cele trei vârfuri alăturate roşii.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nVom demonstra că numărul căutat este $5$.\n\nFie $A B C D A' B' C' D'$ un cub. Colorând în roşu cele $4$ vârfuri ale unei feţe a cubului, fie ele de exemplu $A, B, C, D$, nu există niciun vârf care să aibă cele trei vârfuri alăturate roşii, deci $n \\geq 5$.\n\nAcum, pentru $n=5$, oricum am colora $n$ vârfuri cu roşu, unul din planele $(A B C D)$ şi $(A' B' C' D')$ va avea cel puţin trei vârfuri roşii. Putem presupune că $A, B, C$ sunt roşii. Dacă şi $D$ este roşu, atunci avem un singur vârf roşu în planul $(A' B' C' D')$. Dacă $X'$ este vârful roşu, cu $X \\in \\{A, B, C, D\\}$, atunci vârful $X$ are toţi vecinii roşii.\n\nDacă $D$ nu este roşu, atunci în planul $(A' B' C' D')$ avem două vârfuri roşii. Dacă $B'$ sau $D'$ este roşu, atunci $B$, respectiv $D$, are toţi vecinii roşii. În caz contrar, $A'$ şi $C'$ sunt roşii şi atunci $B'$ are toţi vecinii roşii.\n\nÎn concluzie, oricum am colora $5$ dintre vârfuri cu roşu, va exista cel puţin un vârf care are toţi vecinii roşii, deci minimul căutat este $n=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(K,+, \\cdot)$ un corp finit cu cel puţin patru elemente. Arătaţi că mulţimea $K^{*}$ poate fi partiţionată în două submulţimi nevide $A$ şi $B$, cu proprietatea că\n$$\n\\sum_{x \\in A} x=\\prod_{y \\in B} y\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDeoarece $\\prod_{x \\in K^{*}} x=-1$, dacă $A$ şi $B$ formează o partiţie a lui $K^{*}$, atunci $\\left(\\prod_{x \\in A} x\\right)\\left(\\prod_{x \\in B} x\\right) = -1$, deci $\\sum_{x \\in A} x=\\prod_{y \\in B} y$ dacă şi numai dacă\n$$\n\\left(\\sum_{x \\in A} x\\right) \\cdot\\left(\\prod_{x \\in A} x\\right)=-1\n$$\nFie $q=|K|$. Cum toate rădăcinile polinomului $f=X^{q-1}-1 \\in K[X]$ sunt distincte şi situate în $K^{*}$, rezultă că orice divizor al lui $f$ are aceeaşi proprietate.\n\nDacă $\\operatorname{char} K=2$, atunci $1=-1$ şi mulţimea $A=\\{1\\}$ are proprietatea $(*)$.\n\nFie acum $\\operatorname{char} K>2$. Dacă $q \\equiv 1\\pmod{4}$, atunci $X^{4}-1 \\mid X^{q-1}-1$. Cum $X^{2}+1 \\mid X^{4}-1$, rezultă că $X^{2}+1 \\mid X^{q-1}-1$, deci există $a \\in K^{*}$, astfel încât $a^{2}=-1$. Mulţimea $A=\\{a\\}$ are proprietatea $(*)$.\n\nDacă $q \\not \\equiv 1\\pmod{4}$, cum $q-1$ este un număr par mai mare sau egal cu $6$, rezultă că $q-1$ are un divizor impar $t>1$. Cum $g=X^{t-1}+X^{t-2}+\\cdots+X+1$ divide $f$, polinomul $g$ are toate rădăcinile $a_{1}, a_{2}, \\ldots, a_{t-1}$ distincte şi în $K^{*}$. Deoarece $a_{1}+a_{2}+\\cdots+a_{t-1}=-1$, iar $a_{1} \\cdot a_{2} \\cdots a_{t-1}=(-1)^{t-1}=1$, mulţimea $A=\\left\\{a_{1}, a_{2}, \\ldots, a_{t-1}\\right\\}$ are proprietatea $(*)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO funcţie $f$ de gradul al doilea are proprietatea: pentru orice interval $I$ de lungime 1, intervalul $f(I)$ are lungimea cel puţin 1.\nArătaţi că, pentru orice interval $J$ de lungime 2, intervalul $f(J)$ are lungimea cel puţin 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDacă $f(x) = a x^{2} + b x + c$ şi $v$ este abscisa vârfului parabolei, atunci, pentru intervalul $I = [v - 1/2, v + 1/2]$, intervalul $f(I)$ are lungimea $\\frac{|a|}{4}$, deci $|a| \\geq 4$.\n\nPentru un interval $J$ de lungime 2 există $x, y \\in J$ astfel încât $x - y = 1$ şi $v \\notin (y, x)$.\n\nAvem $|f(x) - f(y)| = \\left| a(x - y) \\left( x + y + \\frac{b}{a} \\right) \\right| \\geq 4 |x + y - 2v| \\geq 4$, deci intervalul $f(J)$ conţine două puncte la distanţă cel puţin 4, de unde concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A \\in \\mathcal{M}_{5}(\\mathbb{C})$ o matrice cu $\\operatorname{tr}(A)=0$ şi cu proprietatea că $I_{5}-A$ este inversabilă. Să se arate că $A^{5} \\neq I_{5}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPresupunem prin reducere la absurd că $A^{5}=I_{5}$. Fie $\\lambda \\in \\mathbb{C}$ o valoare proprie a matricei $A$ (o soluţie a ecuaţiei $\\operatorname{det}\\left(x I_{5}-A\\right)=0$ ). Cum $\\lambda^{5}$ este o valoare proprie a matricei $A^{5}$, obţinem $\\lambda^{5}=1$.\n\nMatricea $I_{5}-A$ este inversabilă, deci $\\operatorname{det}\\left(I_{5}-A\\right) \\neq 0$. In consecinţă, valorile proprii ale matricei $A$ se găsesc în mulţimea $\\{\\varepsilon, \\varepsilon^{2}, \\varepsilon^{3}, \\varepsilon^{4}\\}$, unde $\\varepsilon=\\cos \\frac{2 \\pi}{5}+i \\sin \\frac{2 \\pi}{5}$.\n\nAtunci, din condiţia $\\operatorname{tr}(A)=0$, deducem că există numerele naturale $a, b, c$ şi $d$ cu proprietăţile\n$$\n\\left\\{\\begin{array}{l}\na+b+c+d=5 \\\\\na \\varepsilon+b \\varepsilon^{2}+c \\varepsilon^{3}+d \\varepsilon^{4}=0\n\\end{array}\\right.\n$$\n\nDin ultima relaţie obţinem $a+b \\varepsilon+c \\varepsilon^{2}+d \\varepsilon^{3}=0$. Rezultă că $\\varepsilon$ este o rădăcină a polinomului cu coeficienţi întregi, de grad cel mult $3$, $d X^{3}+c X^{2}+b X+a$. Dacă $d \\neq 0$, atunci polinomul are rădăcinile $\\varepsilon, \\bar{\\varepsilon} \\in \\mathbb{C} \\backslash \\mathbb{R}$ şi $\\alpha \\in \\mathbb{R}$. Dar, din relaţiile lui Viète, obţinem $\\alpha=-a / d \\in \\mathbb{Q}$ şi $\\alpha=-(c / d+2 \\cos (2 \\pi / 5)) \\in \\mathbb{R} \\backslash \\mathbb{Q}$. Contradicţie. Deci $d=0$.\n\nÎn mod analog arătăm $c=b=a=0$. Atunci $a+b+c+d=0$, în contradicţie cu $a+b+c+d=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHatározd meg azokat az $x, y$ egész számokat, amelyekre\n$$\n5^{x}-\\log_{2}(y+3)=3^{y} \\text{ és } 5^{y}-\\log_{2}(x+3)=3^{x}.\n$$\n\nProblem:\nSă se determine numerele întregi $x, y$, pentru care\n$$\n5^{x}-\\log_{2}(y+3)=3^{y} \\text{ şi } 5^{y}-\\log_{2}(x+3)=3^{x}\n$$",
    "options": [],
    "answer": "x = y = 1",
    "solution": "Solution:\nScăzând egalităţile, se obţine\n$$\n5^{x}+3^{x}+\\log_{2}(x+3)=5^{y}+3^{y}+\\log_{2}(y+3)\n$$\nCum funcţia $f(t)=5^{t}+3^{t}+\\log_{2}(t+3)$ este strict crescătoare, rezultă $x=y$.\nPentru rezolvarea în $\\mathbb{Z}$ a ecuaţiei\n$$\n5^{x}=3^{x}+\\log_{2}(x+3)\n$$\nse observă că $x \\in\\{-2,-1,0\\}$ nu verifică, iar $x=1$ este soluţie.\nPentru $x \\geq 2$, se arată că\n$$\n5^{x} \\geq 3^{x}+4^{x}\n$$\nfolosind monotonia funcţiei $g(t)=\\left(\\frac{3}{5}\\right)^{x}+\\left(\\frac{4}{5}\\right)^{x}$\niar apoi se demonstrează prin inducţie matematică inegalitatea\n$$\n4^{x}>\\log_{2}(x+3)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPunctul $P$ este în interiorul triunghiului $ABC$, iar dreptele $AP$, $BP$, $CP$ taie laturile $BC$, $AC$, $AB$ în $A_1$, $B_1$, respectiv $C_1$. Se ştie că\n$$\ns\\left(PBA_1\\right) + s\\left(PCB_1\\right) + s\\left(PAC_1\\right) = \\frac{1}{2} s(ABC)\n$$\nunde cu $s(XYZ)$ s-a notat aria triunghiului $XYZ$. Arătaţi că $P$ se află pe o mediană a triunghiului $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAvem $\\frac{s\\left(PBA_1\\right)}{s\\left(ABA_1\\right)} = \\frac{PA_1}{AA_1} = \\frac{s(BPC)}{s(BAC)}$.\n\nNotând $s(BPC) = s_a$ şi analoagele, rezultă $\\frac{s\\left(PBA_1\\right)}{s_c + s\\left(PBA_1\\right)} = \\frac{s_a}{s}$, unde $s = s(ABC)$, deci $s\\left(PBA_1\\right) = \\frac{s_a s_c}{s - s_a} = \\frac{s_a s_c}{s_b + s_c}$.\n\nIpoteza devine $\\frac{s_a s_c}{s_b + s_c} + \\frac{s_b s_a}{s_c + s_a} + \\frac{s_c s_b}{s_a + s_b} = \\frac{s_a + s_b + s_c}{2}$, ceea ce se reduce după calcule la $s(s_a - s_b)(s_b - s_c)(s_c - s_a) = 0$. Rezultă $s_a = s_b$ sau $s_a = s_c$ sau $s_b = s_c$, de unde concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $\\mathcal{C}$ mulțimea funcțiilor $f:[0,1] \\rightarrow \\mathbb{R}$, de două ori derivabile pe $[0,1]$, care au cel puțin două zerouri, nu neapărat distincte, în $[0,1]$ și $\\left|f''(x)\\right| \\leq 1$, oricare ar fi $x$ în $[0,1]$. Determinați valoarea maximă pe care o poate lua integrala\n$$\n\\int_{0}^{1}|f(x)| \\, \\mathrm{d} x\n$$\ncând $f$ parcurge mulțimea $\\mathcal{C}$, și funcțiile care realizează acest maximum.\n\n(O funcție derivabilă $f$ are două zerouri într-un același punct $a$, dacă $f(a)=f'(a)=0$.)",
    "options": [],
    "answer": "Maximum value: 1/6. The maximizers are exactly f(x) = x^2/2, f(x) = (1 − x)^2/2, and their negatives.",
    "solution": "Solution:\n\nDacă există $a \\in [0,1]$, astfel încât $f(a)=f'(a)=0$, considerăm $x \\in [0,1], x \\neq a$. Atunci $f(x)=\\frac{1}{2}(x-a)^{2} f''(\\theta)$, pentru un anumit $\\theta$ între $a$ și $x$. Deoarece $\\left|f''(t)\\right| \\leq 1$, oricare ar fi $t$ în $[0,1]$, rezultă că $|f(x)| \\leq \\frac{1}{2}(x-a)^{2}$, relație evident adevărată și în $x=a$, deci\n$$\n\\int_{0}^{1}|f(x)| \\, \\mathrm{d} x \\leq \\frac{1}{2} \\int_{0}^{1}(x-a)^{2} \\, \\mathrm{d} x = \\frac{1}{6}(1-3a(1-a)) \\leq \\frac{1}{6}\n$$\n\nDacă există $a<b$, astfel încât $f(a)=f(b)=0$, considerăm $x \\in [0,1] \\setminus \\{a, b\\}$, și funcția $g:[0,1] \\rightarrow \\mathbb{R}$, $g(t)=(t-a)(t-b) f(x)-(x-a)(x-b) f(t)$. Această funcție este de două ori derivabilă pe $[0,1]$ și are trei zerouri distincte, $a, b$ și $x$. Din teorema lui Rolle aplicată de două ori, rezultă că $g''$ are un zero $\\theta$ în intervalul deschis $(0,1)$. Rezultă că $f(x)=\\frac{1}{2}(x-a)(x-b) f''(\\theta)$, deci $|f(x)| \\leq \\frac{1}{2}|x-a| \\cdot |x-b|$, relație evident adevărată și pentru $x=a$ și $x=b$. Atunci\n$$\n\\begin{aligned}\n\\int_{0}^{1}|f(x)| \\, \\mathrm{d} x &\\leq \\frac{1}{2} \\int_{0}^{1}|x-a| \\cdot |x-b| \\, \\mathrm{d} x = \\frac{1}{6} - \\frac{a+b}{4} + \\frac{ab}{2} + \\frac{(b-a)^{3}}{6} \\\\\n&= \\frac{1}{6} - \\frac{1}{2} a(1-b) - \\frac{1}{12}(b-a)\\left(3-2(b-a)^{2}\\right) < \\frac{1}{6}\n\\end{aligned}\n$$\n\nEgalitatea are loc numai în primul caz, pentru $a=0$ sau $a=1$. Din continuitatea funcțiilor din $\\mathcal{C}$, rezultă că funcțiile $f$, care realizează egalitatea, sunt cele pentru care $|f(x)|=x^{2}/2$, $0 \\leq x \\leq 1$, sau $|f(x)|=(1-x)^{2}/2$, $0 \\leq x \\leq 1$. Deci funcțiile cerute sunt $f_{1}(x)=x^{2}/2$, $f_{2}(x)=(1-x)^{2}/2$ și opusele lor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $m$ şi $n$ numere naturale, cu $m \\geq 2$ şi $n \\geq 3$. Demonstraţi că există $m$ numere naturale nenule distincte $a_{1}, a_{2}, a_{3}, \\ldots, a_{m}$, toate divizibile cu $n-1$, astfel încât\n$$\n\\frac{1}{n}=\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\frac{1}{a_{3}}-\\ldots+(-1)^{m-1} \\frac{1}{a_{m}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDacă $\\left(a_{n}\\right)_{n \\geq 1}$ este o progresie geometrică având raţia $q=n-1$, atunci\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{1/a_{1}+1/q \\cdot(-1)^{p-1} 1/a_{p}}{1+1/q}, \\forall p \\in \\mathbb{N}^{*}\n$$\nRelaţia precedentă se scrie\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{n-1}{n a_{1}}+\\frac{(-1)^{p-1}}{n a_{p}}\n$$\nPentru $a_{1}=n-1, m=p+1$ şi $a_{m}=n a_{p}$ obţinem numerele cerute.\n\nDemonstrăm prin inducţie după $m$.\n\nPentru $m=2$ concluzia se verifică:\n$$\n\\frac{1}{n}=\\frac{1}{n-1}-\\frac{1}{n(n-1)}\n$$\nAstfel,\n$$\n\\frac{1}{n}=\\frac{1}{n-1}\\left(\\frac{1}{b_{1}}-\\frac{1}{b_{2}}\\right),\n$$\ncu $b_{1}<b_{2}-1$. Apoi, dacă\n$$\n\\frac{1}{n}=\\frac{1}{(n-1) b_{1}}-\\frac{1}{(n-1) b_{2}}+\\ldots+(-1)^{m-1} \\frac{1}{(n-1) b_{m}}\n$$\ncu $1 \\leq b_{1}<\\ldots<b_{m}$ numere întregi şi $b_{m-1}<b_{m}-1$, atunci punem $b_{i}^{\\prime}=b_{i}$, $i=\\overline{1, m-1}$, $b_{m}^{\\prime}=b_{m}-1$, $b_{m+1}^{\\prime}=b_{m}(b_{m}-1)$ şi obţinem\n$$\n\\frac{1}{n}=\\frac{1}{(n-1) b_{1}^{\\prime}}-\\frac{1}{(n-1) b_{2}^{\\prime}}+\\ldots+(-1)^{m-1} \\frac{1}{(n-1) b_{m}^{\\prime}}+(-1)^{m} \\frac{1}{(n-1) b_{m+1}^{\\prime}}\n$$\ncu $1 \\leq b_{1}^{\\prime}<\\ldots<b_{m}^{\\prime}<b_{m+1}^{\\prime}$ numere întregi şi $b_{m}^{\\prime}<b_{m+1}^{\\prime}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\geq 0$ şi $\\left(x_{n}\\right)_{n \\geq 1}$ un şir de numere reale. Să se arate că dacă şirul $\\left(\\frac{x_{1}+\\cdots+x_{n}}{n^{a}}\\right)_{n \\geq 1}$ este mărginit, atunci şirul $\\left(y_{n}\\right)_{n \\geq 1}$, definit prin $y_{n}=\\frac{x_{1}}{1^{b}}+\\frac{x_{2}}{2^{b}}+\\cdots+\\frac{x_{n}}{n^{b}}$, este convergent pentru orice $b>a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotăm $S_{n}=\\sum_{k=1}^{n} x_{k},\\ n \\in \\mathbb{N}^{*}$. Conform ipotezei, există $c>0$ astfel ca $\\left|S_{n}\\right| \\leq c n^{a},\\ \\forall n \\in \\mathbb{N}^{*}$. Pentru $n, p \\in \\mathbb{N}^{*}$, avem:\n$$\n\\begin{gathered}\n\\left|y_{n+p}-y_{n}\\right|=\\left|\\sum_{k=n+1}^{n+p} \\frac{x_{k}}{k^{b}}\\right|=\\left|\\sum_{k=n+1}^{n+p} \\frac{S_{k}-S_{k-1}}{k^{b}}\\right|= \\\\\n=\\left|\\frac{S_{n+p}}{(n+p+1)^{b}}-\\frac{S_{n}}{(n+1)^{b}}+\\sum_{k=n+1}^{n+p} S_{k}\\left(\\frac{1}{k^{b}}-\\frac{1}{(k+1)^{b}}\\right)\\right| \\leq \\\\\n\\leq \\frac{\\left|S_{n+p}\\right|}{(n+p+1)^{b}}+\\frac{\\left|S_{n}\\right|}{(n+1)^{b}}+\\sum_{k=n+1}^{n+p}\\left|S_{k}\\right|\\left(\\frac{1}{k^{b}}-\\frac{1}{(k+1)^{b}}\\right) \\leq \\\\\n\\leq c\\left[\\frac{2}{n^{b-a}}+\\sum_{k=n+1}^{n+p} k^{a}\\left(\\frac{1}{k^{b}}-\\frac{1}{(k+1)^{b}}\\right)\\right]\n\\end{gathered}\n$$\n\nAplicând Teorema lui Lagrange funcţiei $f(x)=x^{-\\alpha},\\ x>0$ pe $[i, i+1]$, unde $\\alpha, i>0$, obţinem inegalitatea dublă\n$$\n\\frac{\\alpha}{(i+1)^{\\alpha+1}}<\\frac{1}{i^{\\alpha}}-\\frac{1}{(i+1)^{\\alpha}}<\\frac{\\alpha}{i^{\\alpha+1}}\n$$\n\nÎn particular, cum $b, b-a>0$, avem\n$$\n\\frac{1}{k^{b}}-\\frac{1}{(k+1)^{b}}<\\frac{b}{k^{b+1}} \\text{ si } \\frac{b-a}{k^{b-a+1}}<\\frac{1}{(k-1)^{b-a}}-\\frac{1}{k^{b-a}},\\ \\forall k \\in \\mathbb{N},\\ k \\geq 2\n$$\nAtunci\n$$\n\\begin{aligned}\n\\sum_{k=n+1}^{n+p} k^{a}\\left(\\frac{1}{k^{b}}-\\frac{1}{(k+1)^{b}}\\right)<\\sum_{k=n+1}^{n+p} \\frac{b k^{a}}{k^{b+1}}=\\frac{b}{b-a} \\sum_{k=n+1}^{n+p} \\frac{b-a}{k^{b-a+1}}< \\\\\n<\\frac{b}{b-a} \\sum_{k=n+1}^{n+p}\\left(\\frac{1}{(k-1)^{b-a}}-\\frac{1}{k^{b-a}}\\right)=\\frac{b}{b-a}\\left(\\frac{1}{n^{b-a}}-\\frac{1}{(n+p)^{b-a}}\\right)<\\frac{b}{(b-a) n^{b-a}}\n\\end{aligned}\n$$\nRezultă\n$$\n\\left|y_{n+p}-y_{n}\\right|<c\\left(2+\\frac{b}{b-a}\\right) \\frac{1}{n^{b-a}},\\ \\forall n, p \\in \\mathbb{N}^{*}\n$$\nCa urmare, din $\\lim _{n \\rightarrow \\infty} n^{-(b-a)}=0$, deducem că şirul $\\left(y_{n}\\right)_{n \\geq 1}$ este fundamental, deci convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $x, y, z > 0$. Arătaţi că\n$$\n\\frac{x^{3}}{z^{3}+x^{2} y} + \\frac{y^{3}}{x^{3}+y^{2} z} + \\frac{z^{3}}{y^{3}+z^{2} x} \\geq \\frac{3}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCu inegalitatea mediilor avem $x^{2} y \\leq \\frac{x^{3}+x^{3}+y^{3}}{3}$, de unde rezultă că\n$$\n\\frac{x^{3}}{z^{3}+x^{2} y} \\geq \\frac{x^{3}}{z^{3}+\\frac{x^{3}+x^{3}+y^{3}}{3}} = \\frac{3 x^{3}}{2 x^{3}+y^{3}+3 z^{3}}\n$$\nşi analoagele.\n\nSumând şi notând $x^{3}=a$, $y^{3}=b$, $z^{3}=c$, este suficient să demonstrăm că\n$$\n\\frac{a}{2 a+b+3 c} + \\frac{b}{3 a+2 b+c} + \\frac{c}{a+3 b+2 c} \\geq \\frac{1}{2}\n$$\nceea ce se obţine din\n$$\n\\sum \\frac{a}{2 a+b+3 c} = \\sum \\frac{a^{2}}{2 a^{2}+a b+3 a c} \\geq \\frac{(a+b+c)^{2}}{2\\left(a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c\\right)} = \\frac{1}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nHatározd meg azokat a $z$ komplex számokat, amelyekre:\n$$\n|z|+|z-5 i|=|z-2 i|+|z-3 i|\n$$\n\nProblem:\nSă se determine numerele complexe $z$ pentru care are loc relaţia\n$$\n|z|+|z-5 i|=|z-2 i|+|z-3 i|\n$$",
    "options": [],
    "answer": "z = a i with a ≤ 0 or a ≥ 5",
    "solution": "Solution:\nAvem\n$$\n|z-2 i|=\\left|\\frac{2}{5}(z-5 i)+\\frac{3}{5} z\\right| \\leq \\frac{2}{5}|z-5 i|+\\frac{3}{5}|z|\n$$\nAnalog\n$$\n|z-3 i|=\\left|\\frac{3}{5}(z-5 i)+\\frac{2}{5} z\\right| \\leq \\frac{3}{5}|z-5 i|+\\frac{2}{5}|z|\n$$\nde unde\n$$\n|z|+|z-5 i| \\geq|z-2 i|+|z-3 i|\n$$\nEgalitatea are loc atunci când există $\\lambda \\geq 0$ astfel ca $z-5 i=\\lambda z$, de unde deducem că $z=a i$, cu $a \\in(-\\infty, 0] \\cup[5,+\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f:(0, \\infty) \\rightarrow (0, \\infty)$ o funcţie neconstantă care are proprietatea\n$$\nf\\left(x^{y}\\right) = (f(x))^{f(y)}\n$$\npentru orice $x, y > 0$. Să se arate că\n$$\nf(xy) = f(x) f(y) \\text{ şi } f(x+y) = f(x) + f(y)\n$$\npentru orice $x, y > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $a > 0$ astfel ca $f(a) \\neq 1$. Avem, pentru $x, y$ arbitrari,\n$$\nf\\left(a^{xy}\\right) = f(a)^{f(xy)}\n$$\ndar\n$$\nf\\left(a^{xy}\\right) = f\\left(\\left(a^{x}\\right)^{y}\\right) = f\\left(a^{x}\\right)^{f(y)} = \\left(f(a)^{f(x)}\\right)^{f(y)} = f(a)^{f(x) f(y)}\n$$\nde unde $f(xy) = f(x) f(y)$\n\nApoi\n$$\nf\\left(a^{x+y}\\right) = f(a)^{f(x+y)}\n$$\ndar\n$$\nf\\left(a^{x+y}\\right) = f\\left(a^{x} a^{y}\\right) = f\\left(a^{x}\\right) f\\left(a^{y}\\right) = f(a)^{f(x)} f(a)^{f(y)} = f(a)^{f(x) + f(y)}\n$$\ndeci $f(x+y) = f(x) + f(y)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminați funcțiile polinomiale neconstante $f:[0,1] \\rightarrow \\mathbb{R}^*$, cu coeficienți raționali, care au următoarea proprietate: oricare ar fi $x$ în intervalul $[0,1]$, există două funcții polinomiale $g_x, h_x:[0,1] \\rightarrow \\mathbb{R}$, cu coeficienți raționali, astfel încât $h_x(x) \\neq 0$ și\n$$\n\\int_{0}^{x} \\frac{1}{f(t)} \\mathrm{d} t=\\frac{g_x(x)}{h_x(x)}\n$$",
    "options": [],
    "answer": "All such polynomials are f(x) = c (x − b)^n with c a nonzero rational number, b a rational number not in the unit interval, and n an integer at least two.",
    "solution": "Solution:\nDeoarece intervalul închis $[0,1]$ este nenumărabil și mulțimea funcțiilor polinomiale cu coeficienți raționali este numărabilă, există o mulțime nenumărabilă $S \\subseteq [0,1]$ și două funcții polinomiale coprime $g, h:[0,1] \\rightarrow \\mathbb{R}$, cu coeficienți raționali, astfel încât $h(x) \\neq 0$ și\n$$\n\\int_{0}^{x} \\frac{1}{f(t)} \\mathrm{d} t=\\frac{g(x)}{h(x)}\n$$\noricare ar fi $x$ în $S$.\nDeoarece mulțimea punctelor izolate ale lui $S$ este cel mult numărabilă, rezultă că există o infinitate de puncte în $S$, care sunt puncte de acumulare ale lui $S$. În fiecare dintre aceste puncte, funcțiile raționale $1 / f$ și $(g / h)'=\\left(g' h-g h'\\right) / h^2$ sunt egale, deci\n$$\nh^2=f \\cdot\\left(g' h-g h'\\right)\n$$\nîn $\\mathbb{Q}[X]$.\nÎntrucât $\\operatorname{deg} f \\geq 1$, rezultă că $\\operatorname{deg} g \\leq \\operatorname{deg} h$. Dacă $\\operatorname{deg} g=\\operatorname{deg} h$, atunci și restul împărțirii lui $g$ la $h$ verifică $(*)$ și este coprim cu $h$, deci putem presupune că $\\operatorname{deg} g<\\operatorname{deg} h$ și, prin urmare, $\\operatorname{deg}\\left(g' h-g h'\\right)=\\operatorname{deg} g+\\operatorname{deg} h-1$.\nDacă $\\deg h \\geq 2$, cum $g$ și $h$ sunt coprime, orice rădăcină de ordin $k$ a lui $g' h-g h'$ este rădăcină a lui $h$, de ordin mai mare sau egal cu $k+1$. Deci $\\operatorname{deg} h \\geq \\operatorname{deg} g+\\operatorname{deg} h-1+d$, unde $d$ este numărul de rădăcini distincte ale lui $g' h-g h'$. Prin urmare, $d=1$, $\\operatorname{deg} g=0$, de unde, $h=a(X-b)^m$, unde $a \\in \\mathbb{Q}^*$, $b \\in \\mathbb{Q} \\setminus [0,1]$, iar $m$ este un număr natural mai mare sau egal cu 2. Rezultă că $f=c(X-b)^{m+1}$, unde $c$ este un număr rațional nenul.\nDacă $\\deg h=1$, atunci $\\operatorname{deg} g=0$ și $f=c(X-b)^2$, unde $b$ și $c$ sunt ca mai sus.\nÎn concluzie, funcțiile cerute sunt cele de forma $f(x)=c(x-b)^n$, $0 \\leq x \\leq 1$, unde $c \\in \\mathbb{Q}^*$, $b \\in \\mathbb{Q} \\setminus [0,1]$ și $n$ este un număr natural mai mare sau egal cu 2, funcții care îndeplinesc în mod evident condiția din enunț.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nRezolvaţi în mulţimea numerelor naturale ecuaţia $21^{x} + 4^{y} = z^{2}$.",
    "options": [],
    "answer": "(1, 1, 5)",
    "solution": "Solution:\nAvem $(z - 2^{y})(z + 2^{y}) = 21^{x}$ şi, notând $d = (z - 2^{y}, z + 2^{y})$, rezultă că $d \\mid (z + 2^{y}) - (z - 2^{y})$, deci $d \\mid 2^{y+1}$. Întrucât $d \\mid z + 2^{y}$ şi $z + 2^{y} \\mid 21^{x}$, rezultă $d \\mid (2^{y+1}, 21^{x})$, deci $d = 1$.\n\nCa urmare, avem de analizat două cazuri: $z - 2^{y} = 1$ şi $z + 2^{y} = 21^{x}$, respectiv $z - 2^{y} = 3^{x}$ şi $z + 2^{y} = 7^{x}$.\n\nÎn primul caz obţinem $21^{x} - 1 = 2^{y+1}$, egalitate care nu poate avea loc, deoarece divide $5 \\mid 21^{x} - 1$ şi $5 \\nmid 2^{y+1}$.\n\nÎn al doilea caz avem $7^{x} - 3^{x} = 2^{y+1}$, pentru care se observă că $x = 0$ nu convine, iar $x = 1$ conduce la soluţia $(1, 1, 5)$. Studiem în continuare cazul $x \\geq 2$.\n\nDacă $x$ este impar putem scrie:\n$$\n2^{y-1} = 7^{x-1} + 7^{x-2} \\cdot 3 + 7^{x-3} \\cdot 3^{2} + \\ldots + 7 \\cdot 3^{x-2} + 3^{x-1}\n$$\ncontradicţie, deoarece membrul drept este o sumă a unui număr impar de numere impare.\n\nPentru $x$ par, $x = 2s, s \\geq 1$, rezultă $49^{s} - 9^{s} = 2^{y+1}$, egalitate care nu poate avea loc deoarece $5 \\mid 49^{s} - 9^{s}$.\n\nÎn concluzie, singura soluţie este $(1, 1, 5)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMatricele $A \\in \\mathcal{M}_{m, n}(\\mathbb{C}), B \\in \\mathcal{M}_{n, m}(\\mathbb{C})$, unde $m \\geq n \\geq 2$, au proprietatea că există $k \\in \\mathbb{N}^{*}$ și $a_{0}, a_{1}, \\ldots, a_{k} \\in \\mathbb{C}$ astfel încât\n$$\na_{k}(A B)^{k}+a_{k-1}(A B)^{k-1}+\\cdots+a_{1}(A B)+a_{0} I_{m}=O_{m}\n$$\niar\n$$\na_{k}(B A)^{k}+a_{k-1}(B A)^{k-1}+\\cdots+a_{1}(B A)+a_{0} I_{n} \\neq O_{n}\n$$\nDemonstraţi că $a_{0}=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPresupunem că $a_{0} \\neq 0$. Scriind egalitatea din ipoteză sub forma\n$$\nA B\\left(-\\frac{a_{k}}{a_{0}}(A B)^{k-1}-\\frac{a_{k-1}}{a_{0}}(A B)^{k-2}-\\cdots-\\frac{a_{1}}{a_{0}} I_{m}\\right)=I_{m}\n$$\ndeducem că $A B$ este inversabilă, deci $\\operatorname{rang}(A B)=m \\geq n$.\nPe de altă parte, din inegalitatea rangurilor,\n$$\nm=\\operatorname{rang}(A B) \\leq \\min \\{ \\operatorname{rang} A, \\operatorname{rang} B \\} \\leq n\n$$\nAşadar $m=n$, iar $A, B$ şi $B A$ sunt matrice pătratice inversabile.\n\nInmulţind egalitatea\n$$\na_{k}(A B)^{k}+a_{k-1}(A B)^{k-1}+\\cdots+a_{1}(A B)+a_{0} I_{n}=O_{n}\n$$\nla stânga cu $B$ şi la dreapta cu $A$ obţinem\n$$\na_{k}(B A)^{k+1}+a_{k-1}(B A)^{k}+\\cdots+a_{0}(B A)=O_{n}\n$$\nPrin înmulţire (la stânga sau la dreapta) cu $(B A)^{-1}$ rezultă\n$$\na_{k}(B A)^{k}+a_{k-1}(B A)^{k-1}+\\cdots+a_{0} I_{n}=O_{n}\n$$\nceea ce contrazice ipoteza.\n\nAceastă contradicţie provine din faptul că am presupus $a_{0} \\neq 0$. Deci $a_{0}=0$, ceea ce trebuia demonstrat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră patrulaterul $ABCD$ cu diagonalele neperpendiculare şi laturile $AB$ şi $CD$ neparalele. Fie $O$ punctul de intersecţie a diagonalelor, $H_{1}$ ortocentrul triunghiului $AOB$ şi $H_{2}$ ortocentrul triunghiului $COD$. Se notează cu $M$ mijlocul laturii $[AB]$ şi cu $N$ mijlocul laturii $[CD]$. Arătaţi că $H_{1}H_{2}$ este paralelă cu $MN$ dacă şi numai dacă $AC = BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $A'$, $B'$ picioarele înălţimilor din $A$, respectiv $B$ în triunghiul $AOB$, şi $C'$, $D'$ picioarele înălţimilor din $C$, respectiv $D$ în triunghiul $COD$.\n\nEvident, $A'$ şi $D'$ aparţin cercului $\\mathcal{C}_1$ de diametru $AD$, iar $B'$ şi $C'$ aparţin cercului $\\mathcal{C}_2$ de diametru $BC$.\n\nSe observă că triunghiurile $H_{1}AB$ şi $H_{1}A'B'$ sunt asemenea, de unde $H_{1}A \\cdot H_{1}A' = H_{1}B \\cdot H_{1}B'$. Atunci $H_{1}$ are aceeaşi putere în raport cu cercurile $\\mathcal{C}_1$ şi $\\mathcal{C}_2$, deci $H_{1}$ (şi analog, $H_{2}$) se află pe axa radicală a celor două cercuri.\n\nÎntrucât axa radicală este perpendiculară pe linia centrelor, rezultă că $H_{1}H_{2}$ este perpendiculară pe dreapta $PQ$, unde $P$ şi $Q$ sunt mijloacele laturilor $AD$, respectiv $BC$ ($P$ şi $Q$ sunt centrele cercurilor $\\mathcal{C}_1$ şi $\\mathcal{C}_2$).\n\nCondiţia $H_{1}H_{2} \\parallel MN$ este echivalentă cu $MN \\perp PQ$. Cum $MPNQ$ este paralelogram, avem:\n\n$$\nH_{1}H_{2} \\parallel MN \\Leftrightarrow MN \\perp PQ \\Leftrightarrow MPNQ \\text{ este romb } \\Leftrightarrow MP = MQ \\Leftrightarrow AC = BD.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1) Să se rezolve sistemul:\n$$\n\\left\\{\\begin{array}{l}\nx + [y] + \\{z\\} = 1,2 \\\\\ny + [z] + \\{x\\} = 2,3 \\\\\nz + [x] + \\{y\\} = 3,5\n\\end{array} \\quad \\text{unde } x, y, z \\in \\mathbf{R}\\right.\n$$",
    "options": [],
    "answer": "x = 1, y = 0.3, z = 2.2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38352,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that $2^n + 3^n + 5^n + 8^n$ is not a perfect square for any positive integer $n$.\n\nb) Find all positive integers $n$ so that $1^n + 4^n + 6^n + 7^n = 2^n + 3^n + 5^n + 8^n$.",
    "options": [],
    "answer": "n = 1 and 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Arătaţi că numărul $\\log_{2015} 2016$ este număr iraţional;\n\nb) Comparaţi numerele $\\log_{5} 6$ şi $\\log_{6} 7$;\n\nc) Calculaţi $E=\\lg^{3} 5+\\lg^{3} 20+\\lg 8 \\cdot \\lg (0,25)$.",
    "options": [],
    "answer": "a) irrational; b) log base 5 of 6 is greater than log base 6 of 7; c) 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Arătaţi că numărul $\\log_{2015} 2016$ este număr iraţional;\n\nb) Comparaţi numerele $\\log_{5} 6$ şi $\\log_{6} 7$;\n\nc) Calculaţi $E=\\lg^{3} 5+\\lg^{3} 20+\\lg 8 \\cdot \\lg (0,25)$.",
    "options": [],
    "answer": "(a) log_{2015} 2016 is irrational; (b) log_{5} 6 > log_{6} 7; (c) E = 2",
    "solution": "Solution:\n\na. Presupunem că numărul $\\log_{2015} 2016$ este raţional. Atunci există numerele $a, b \\in \\mathbb{N}^{*}$ astfel încât $\\log_{2015} 2016=\\frac{a}{b}$. Obținem $2016= 2015^{\\frac{a}{b}} \\Leftrightarrow 2016^{b}=2015^{a}$, relaţie imposibilă deoarece un număr e par şi celălalt impar.\n\nb. Aplicând inegalitatea mediilor pentru numerele pozitive $\\log_{6} 5$ și $\\log_{6} 7$ obţinem $\\sqrt{\\log_{6} 5 \\cdot \\log_{6} 7} \\leq \\frac{\\log_{6} 5+\\log_{6} 7}{2}=\\frac{\\log_{6} 35}{2}<\\frac{\\log_{6} 36}{2}=1$, deci $\\log_{6} 5 \\cdot \\log_{6} 7<1 \\Leftrightarrow \\log_{6} 7<\\log_{5} 6$.\n\nc. $E=\\lg^{3} 5+\\lg^{3} 20+\\lg 8 \\cdot \\lg (0,25)=\\lg 35+(2 \\lg 2+\\lg 5)^{3}+3 \\lg 2 \\cdot(-2 \\lg 2)=(1-\\lg 2)^{3}+(1+\\lg 2)^{3}-6 \\lg^{2} 2=2$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38355,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the triangle $ABC$ has center $I$, touches $BC$, $CA$ and $AB$ at $D$, $E$, respectively $F$ and meets $AI$, $BI$, $CI$ in $M$, $N$, respectively $P$. Prove that if the triangles $DEF$ and $MNP$ have the same centroid, then the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two $3 \\times 3$ matrices with real entries, so that $AB^2A = BA^2B$.\n\na) Prove that there exists $\\alpha, \\beta \\in \\mathbb{R}$ so that\n$$\n\\det[(AB)^2 + (BA)^2] = [\\det(A)\\det(B) - \\alpha]^2 + [\\det(A)\\det(B) - \\beta]^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Comparați numerele $A = \\sqrt[3]{4} + \\sqrt[3]{13}$ și $B = \\sqrt[3]{9} + \\sqrt[3]{8}$.\n\nb) Demonstrați că $\\log_{2} 3 + \\log_{3} 4 \\in (2, 3)$.",
    "options": [],
    "answer": "a) A < B.\nb) log_2 3 + log_3 4 ∈ (2, 3).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\in \\mathbb{R}$. Demonstrați că dacă $a^{18} \\in \\mathbb{Q}$ și $a^{11} \\in \\mathbb{Q}$, atunci $a \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Demonstraţi că $\\frac{2a}{ab+1} \\leq \\sqrt{\\frac{a}{b}}$ pentru orice $a, b \\in (0, \\infty)$.\n\nb. Dacă $a, b, c > 0$ astfel încât $abc = 1$, demonstraţi că\n$$\n\\frac{2a}{ab+1} + \\frac{2b}{bc+1} + \\frac{2c}{ac+1} \\leq a \\sqrt{c} + b \\sqrt{a} + c \\sqrt{b}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Enumerați elementele mulțimilor:\n$$\nA = \\left\\{ a \\in \\mathbb{Q} \\mid a = \\sqrt{\\frac{2 - x}{4}},\\ x \\in \\mathbb{N}^* \\right\\} \\text{ și } B = \\left\\{ x \\in \\mathbb{N}^* \\mid a = \\sqrt{\\frac{2 - x}{9}},\\ a \\in \\mathbb{Q} \\right\\}\n$$\nDeterminați $A \\cup B$, $A \\cap B$, $A \\setminus B$, $B \\setminus A$.",
    "options": [],
    "answer": "A = {0, 1/2}; B = {1, 2}; A ∪ B = {0, 1/2, 1, 2}; A ∩ B = ∅; A \\ B = {0, 1/2}; B \\ A = {1, 2}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-negative integers $n$ so that $n^2 - 4n + 2$, $n^2 - 3n + 13$ and $n^2 - 6n + 19$ are simultaneously primes.",
    "options": [],
    "answer": "0 and 4",
    "solution": "Let us denote:\n\n\\begin{align*}\nA &= n^2 - 4n + 2 \\\\\nB &= n^2 - 3n + 13 \\\\\nC &= n^2 - 6n + 19\n\\end{align*}\n\nWe want $A$, $B$, $C$ to be simultaneously prime for some non-negative integer $n$.\n\nLet us try small values of $n$:\n\nFor $n = 0$:\n\\begin{align*}\nA &= 0^2 - 4 \\cdot 0 + 2 = 2 \\\\\nB &= 0^2 - 3 \\cdot 0 + 13 = 13 \\\\\nC &= 0^2 - 6 \\cdot 0 + 19 = 19\n\\end{align*}\nAll are prime.\n\nFor $n = 1$:\n\\begin{align*}\nA &= 1 - 4 + 2 = -1 \\\\\nB &= 1 - 3 + 13 = 11 \\\\\nC &= 1 - 6 + 19 = 14\n\\end{align*}\n$A$ is negative, $C$ is not prime.\n\nFor $n = 2$:\n\\begin{align*}\nA &= 4 - 8 + 2 = -2 \\\\\nB &= 4 - 6 + 13 = 11 \\\\\nC &= 4 - 12 + 19 = 11\n\\end{align*}\n$A$ is negative.\n\nFor $n = 3$:\n\\begin{align*}\nA &= 9 - 12 + 2 = -1 \\\\\nB &= 9 - 9 + 13 = 13 \\\\\nC &= 9 - 18 + 19 = 10\n\\end{align*}\n$A$ is negative, $C$ is not prime.\n\nFor $n = 4$:\n\\begin{align*}\nA &= 16 - 16 + 2 = 2 \\\\\nB &= 16 - 12 + 13 = 17 \\\\\nC &= 16 - 24 + 19 = 11\n\\end{align*}\nAll are prime.\n\nFor $n = 5$:\n\\begin{align*}\nA &= 25 - 20 + 2 = 7 \\\\\nB &= 25 - 15 + 13 = 23 \\\\\nC &= 25 - 30 + 19 = 14\n\\end{align*}\n$C$ is not prime.\n\nFor $n = 6$:\n\\begin{align*}\nA &= 36 - 24 + 2 = 14 \\\\\nB &= 36 - 18 + 13 = 31 \\\\\nC &= 36 - 36 + 19 = 19\n\\end{align*}\n$A$ is not prime.\n\nFor $n = 7$:\n\\begin{align*}\nA &= 49 - 28 + 2 = 23 \\\\\nB &= 49 - 21 + 13 = 41 \\\\\nC &= 49 - 42 + 19 = 26\n\\end{align*}\n$C$ is not prime.\n\nFor $n = 8$:\n\\begin{align*}\nA &= 64 - 32 + 2 = 34 \\\\\nB &= 64 - 24 + 13 = 53 \\\\\nC &= 64 - 48 + 19 = 35\n\\end{align*}\n$A$ and $C$ are not prime.\n\nFor $n = 9$:\n\\begin{align*}\nA &= 81 - 36 + 2 = 47 \\\\\nB &= 81 - 27 + 13 = 67 \\\\\nC &= 81 - 54 + 19 = 46\n\\end{align*}\n$C$ is not prime.\n\nFor $n = 10$:\n\\begin{align*}\nA &= 100 - 40 + 2 = 62 \\\\\nB &= 100 - 30 + 13 = 83 \\\\\nC &= 100 - 60 + 19 = 59\n\\end{align*}\n$A$ is not prime.\n\nFor $n = 11$:\n\\begin{align*}\nA &= 121 - 44 + 2 = 79 \\\\\nB &= 121 - 33 + 13 = 101 \\\\\nC &= 121 - 66 + 19 = 74\n\\end{align*}\n$C$ is not prime.\n\nFor $n = 12$:\n\\begin{align*}\nA &= 144 - 48 + 2 = 98 \\\\\nB &= 144 - 36 + 13 = 121 \\\\\nC &= 144 - 72 + 19 = 91\n\\end{align*}\nNone are prime.\n\nLet us check if there is a pattern. Notice that for $n = 0$ and $n = 4$, all three are prime. Let's check $n = 13$:\n\nFor $n = 13$:\n\\begin{align*}\nA &= 169 - 52 + 2 = 119 \\\\\nB &= 169 - 39 + 13 = 143 \\\\\nC &= 169 - 78 + 19 = 110\n\\end{align*}\nNone are prime.\n\nIt seems that only $n = 0$ and $n = 4$ work. Let's check if for $n \\geq 5$ it is possible for all three to be prime.\n\nFor large $n$, $A$, $B$, $C$ are all roughly $n^2$ minus a linear term plus a constant. For $n$ even, $A$ is even, so unless $A = 2$, it cannot be prime. For $n$ odd, $A$ is odd.\n\nFrom above, for $n = 0$, $A = 2$ (prime), for $n = 4$, $A = 2$ (prime). For $n$ even and $n > 4$, $A$ is even and greater than $2$, so not prime.\n\nFor $n$ odd, $A$ is odd, but from above, for $n = 5$, $A = 7$ (prime), but $C = 14$ (not prime). For $n = 7$, $A = 23$ (prime), $C = 26$ (not prime).\n\nTherefore, the only non-negative integers $n$ for which all three expressions are simultaneously prime are $n = 0$ and $n = 4$.\n\n**Answer:** $n = 0$ and $n = 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie unghiurile adiacente suplementare $\\Varangle A O B$ și $\\Varangle B O C$ astfel încât raportul măsurilor să fie $\\frac{1}{4}$. Fie $[O D$ semidreapta opusă bisectoarei unghiului $\\Varangle B O C$. În interiorul unghiului $\\Varangle C O D$ se consideră punctele $M$ și $N$ astfel încât $m(\\Varangle C O N)=m(\\Varangle D O M)=2 \\cdot m(\\Varangle M O N)>45^\\circ$.\na) Aflați măsura unghiului $\\Varangle C O D$.\nb) Demonstrați că punctele $B, O, M$ sunt coliniare.",
    "options": [],
    "answer": "m∠COD = 108°, and points B, O, M are collinear.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi valorile lui $x \\in \\mathbb{Z}$ pentru care $\\sqrt[3]{x^{3}-6 x^{2}+12 x+29} \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "{-2, 5}",
    "solution": "Solution:\n\nPentru ca radicalul de ordin 3 dintr-un număr întreg să fie raţional, trebuie ca numărul de sub radical să fie cub perfect.\n\nDeci există $y \\in \\mathbb{Z}$ astfel încât $x^{3}-6 x^{2}+12 x+29 = y^{3} \\Leftrightarrow (x-2)^{3} + 37 = y^{3} \\Leftrightarrow y^{3} - (x-2)^{3} = 37 \\Leftrightarrow (y-x+2)\\left(y^{2} + y(x-2) + (x-2)^{2}\\right) = 37$.\n\nObţinem că $y-x+2$ este divizor al lui $37$, deci $y-x+2 = 1$ şi $y^{2} + y(x-2) + (x-2)^{2} = 37$, ultima expresie fiind pozitivă pentru orice $x$, $y$. Rezolvând sistemul obţinem $x = -2$ şi $y = -3$ sau $x = 5$ și $y = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Să se arate că șirul $a_{n} = \\int_{0}^{n} e^{-x^{2}} dx$, $n \\in \\mathbb{N}$, este convergent.\n\nb) Dacă $f:[0, \\infty) \\rightarrow [0, \\infty)$ este o funcție continuă cu $\\lim_{x \\rightarrow \\infty} f(x) = 0$, rezultă în mod necesar că șirul $x_{n} = \\int_{0}^{n} f(x) dx$, $n \\in \\mathbb{N}$, este convergent? Justificare.",
    "options": [],
    "answer": "a) Convergent (indeed bounded above by the improper integral ∫_0^∞ e^{-x^2} dx). b) Not necessarily; for example, f(x) = 1/(x+1) is continuous, nonnegative, tends to zero, but ∫_0^n f(x) dx = ln(n+1) diverges.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A, B \\in M_2(\\mathbb{C})$ be such that $A - AB = B^2$ and $B - BA = A^2$. Prove that $A = B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38366,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a number *good* if it is a positive integer with at least two digits and by removing one of its digits we get a number which is equal to the sum of its initial digits (for instance, $109$ is good: remove $9$ to get $10 = 1 + 0 + 9$).\n\na) Find the smallest good number.\n\nb) Find how many numbers are good.",
    "options": [],
    "answer": "a) 10; b) infinitely many",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the ratio of the area of a rectangle and the area of a triangle whose vertices are on the sides of the rectangle and centroid coincides with the centroid of the rectangle is a constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminați valorile întregi ale lui $x$ și $y$ astfel încât\n$$\nx - 3y + 4 = 0 \\quad \\text{și} \\quad \\sqrt{x^{2} + 7y^{2} + 8x + 8y + 4} \\in \\mathbb{Q}\n$$",
    "options": [],
    "answer": "(x, y) = (-10, -2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Fie $z$ un număr complex. Demonstraţi că dacă $w$ este o rădăcină de ordinul trei a lui $z$, atunci numerele $\\varepsilon \\cdot w$ şi $\\varepsilon^{2} \\cdot w$ sunt celelalte rădăcini de ordinul trei ale lui $z$, unde $\\varepsilon=-\\frac{1}{2}+i \\frac{\\sqrt{3}}{2}$.\n\nb) Fie $a, b, c \\in \\mathbf{C}$, distincte două câte două, astfel încât $(a+b)^{3}=(b+c)^{3}=(c+a)^{3}$. Demonstraţi că $a^{3}=b^{3}=c^{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Demonstraţi că $\\frac{1}{\\sqrt{n+1}} < 2(\\sqrt{n+1} - \\sqrt{n}) < \\frac{1}{\\sqrt{n}}$ pentru orice $n \\in \\mathbf{N}^{*}$.\n\nb) Demonstraţi că $\\frac{1}{\\sqrt{1}} + \\frac{1}{\\sqrt{2}} + \\cdots + \\frac{1}{\\sqrt{n}} < 2 \\sqrt{n-1}$, pentru orice $n \\in \\mathbf{N}, n \\geq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a$, $b$, $c$ numere complexe astfel încât $|a-1| = |b-2| = |c+3|$ şi $a + b + c = 0$. Arătaţi că $|a-b+1| = |a-c-4| = |b-c-5|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră funcţia $f:(0, \\infty) \\rightarrow \\mathbf{R}$ cu proprietatea că\n$$\nf(x y)=f(x)+f(y) \\text{ pentru orice } x, y \\in (0, \\infty)\n$$\na) Arătaţi că $f(1)=0$ şi $f\\left(\\frac{1}{x}\\right)=-f(x)$ pentru orice $x \\in (0, \\infty)$;\nb) Arătaţi că dacă 1 este singura soluţie a ecuaţiei $f(x)=0$, atunci funcţia $f$ este injectivă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nFolosim proprietatea dată:\n$$\nf(xy) = f(x) + f(y), \\quad \\forall x, y \\in (0, \\infty)\n$$\n\nPentru $x = 1$ și orice $y > 0$:\n$$\nf(1 \\cdot y) = f(1) + f(y) \\implies f(y) = f(1) + f(y)\n$$\nRezultă că $f(1) = 0$.\n\nPentru orice $x > 0$, alegem $y = \\frac{1}{x}$:\n$$\nf(x \\cdot \\frac{1}{x}) = f(x) + f\\left(\\frac{1}{x}\\right)\n$$\nDar $x \\cdot \\frac{1}{x} = 1$, deci:\n$$\nf(1) = f(x) + f\\left(\\frac{1}{x}\\right)\n$$\nȘtim deja că $f(1) = 0$, deci:\n$$\n0 = f(x) + f\\left(\\frac{1}{x}\\right) \\implies f\\left(\\frac{1}{x}\\right) = -f(x)\n$$\n\nb.\nPresupunem că $f(a) = f(b)$ pentru $a, b \\in (0, \\infty)$.\n\nAtunci:\n$$\n0 = f(a) - f(b) = f\\left(\\frac{a}{b}\\right) \\quad \\text{(folosind proprietatea: } f\\left(\\frac{a}{b}\\right) = f(a) - f(b)\\text{)}\n$$\nDar, prin ipoteză, singura soluție a ecuației $f(x) = 0$ este $x = 1$.\nDeci $\\frac{a}{b} = 1 \\implies a = b$.\n\nPrin urmare, funcția $f$ este injectivă.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRezolvaţi, în mulţimea numerelor reale, ecuaţia $\\frac{1}{\\{x\\}} = \\frac{1}{x} + \\frac{1}{[x]}$, unde $[x]$ reprezintă partea întreagă a numărului real $x$, iar $\\{x\\}$ reprezintă partea fracţionară a numărului real $x$.",
    "options": [],
    "answer": "(1+sqrt(5))/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn triunghiul oarecare $ABC$, se consideră $M$ şi $N$ mijloacele segmentelor $BC$, respectiv $AM$, punctul $D$ simetricul punctului $C$ faţă de $A$, $BN \\cap AC = \\{S\\}$, $DM \\cap AB = \\{T\\}$ și punctul $P$ mijlocul segmentului $[SC]$.\n\na) Demonstrați că $AC = 3PC$.\n\nb) Demonstrați că dreptele $ST$ și $BC$ sunt paralele.\n\nc) Calculați aria triunghiului $ANS$, știind că aria triunghiului $ABC$ este egală cu $48~\\mathrm{cm}^2$.",
    "options": [],
    "answer": "4 cm^2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and $n_1, n_2, \\dots, n_k \\in \\{1, 2, \\dots, p-1\\}$ be positive integers. Show that the equation\n$$\nx_1^{n_1} + x_2^{n_2} + \\dots + x_k^{n_k} = x_{k+1}^p\n$$\nhas infinitely many positive integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie mulțimea $G = (-k, k)$, $k > 0$ și $x * y = \\frac{k^{2}(x + y)}{k^{2} + x y}$, $x, y \\in G$. Arătați că:\n\na) $(G, *)$ este grup abelian.\n\nb) $\\frac{k}{3} * \\frac{k}{5} * \\ldots * \\frac{k}{2 n + 1} < \\frac{k}{2} * \\frac{k}{4} * \\ldots * \\frac{k}{2 n}$, $n \\in \\square^{*}$.\n\n(Supliment G.M. nr. 10/2015)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38377,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Find all perfect squares of the form $aabcc$.\n\nb. Let $n$ be a given positive integer. Prove that there exists a perfect square of the form $aab \\underbrace{cc\\dots c}_{2n \\text{ times}}$.",
    "options": [],
    "answer": "Part a: 22500, 44100, 44944.\nPart b: For any positive integer n, (10^n · 15)^2 = 225 followed by 2n zeros is a perfect square of the form aab with 2n copies of c, taking a = 2, b = 5, c = 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four positive integers $a, b, c, d$ are not divisible by $5$ and the sum of their squares is divisible by $5$. Prove that\n$$\nN = (a^2 + b^2)(a^2 + c^2)(a^2 + d^2)(b^2 + c^2)(b^2 + d^2)(c^2 + d^2)\n$$\nis divisible by $625$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function so that for every $x, y \\in \\mathbb{R}$, $x < y$, there exists $z \\in (x, y)$ so that\n$$\n(y - x)f(z) \\le (y - z)f(x) + (z - x)f(y).\n$$\n\na) Give an example of a non-convex function $f$ with the given property.\nb) Prove that a continuous function with the given property is convex.",
    "options": [],
    "answer": "Example for part a: f equals 1 on rational numbers and 0 on irrational numbers. For part b: any continuous function with the stated property is convex.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi și punctele $M$, $N$ pe laturile $[AB]$ respectiv $[BC]$ astfel încât $\\frac{AM}{MB}=\\frac{m}{n}$ și $\\frac{BN}{NC}=\\frac{n}{p}$, unde $m, n, p$ sunt numere reale pozitive cu proprietatea $p^{2}=mn$. Notăm cu $P$ intersecția dreptelor $CM$ și $AN$. Arătați că $n \\overrightarrow{PA}+m \\overrightarrow{PB}+p \\overrightarrow{PC}=\\overrightarrow{0}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn triunghiul $ABC$ se consideră punctele $M \\in (AB)$, $N \\in (BC)$, $P \\in (CA)$ astfel încât $AM = BN = CP$. Dacă $G_1, G_2, G_3$ sunt centrele de greutate ale triunghiurilor $AMP$, $BMN$, respectiv $CNP$, să se arate că triunghiurile $ABC$ şi $G_1G_2G_3$ au acelaşi centru de greutate dacă şi numai dacă triunghiul $ABC$ este echilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Fie $A$ o mulţime finită, nevidă şi $f: A \\rightarrow A$ o funcţie injectivă. Arătaţi că $f$ este surjectivă.\n\nb) Fie $f: \\{1,2,3, \\ldots, 63\\} \\rightarrow \\{1,2,3, \\ldots, 63\\}$ o funcţie astfel încât $f(1)+f(2)+f(3)+\\cdots+f(63)-1=2016$. Arătaţi că $f$ nu este injectivă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers larger than $1$, so that\n$$\n\\left(\\frac{x}{2} + \\frac{1}{x} - 1\\right) \\left(\\frac{y}{2} + \\frac{1}{y} - 1\\right) \\left(\\frac{z}{2} + \\frac{1}{z} - 1\\right) = \\left(1 - \\frac{x}{yz}\\right) \\left(1 - \\frac{y}{xz}\\right) \\left(1 - \\frac{z}{xy}\\right).\n$$",
    "options": [],
    "answer": "(2, 2, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38384,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ denote $d(n)$ the number of its positive divisors and $s(n)$ their sum. It is known that $n + d(n) = s(n) + 1$, $m + d(m) = s(m) + 1$ and $nm + d(nm) + 2016 = s(nm)$. Find $n$ and $m$.\n\nMihai Bunget",
    "options": [],
    "answer": "(n, m) = (2, 2017) or (2017, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $z_{A}$, $z_{B}$, $z_{C}$ afixele vârfurilor triunghiului $ABC$. Fie $J$, de afix $z_{J}$ un punct interior triunghiului $ABC$ şi $S_{a}$, $S_{b}$, $S_{c}$ ariile triunghiurilor $JBC$, $JAC$, respectiv $JAB$.\n\na) Demonstraţi că $z_{J} = \\frac{S_{a} z_{A} + S_{b} z_{B} + S_{c} z_{C}}{S}$, unde $S$ este aria triunghiului $ABC$.\n\nb) Demonstraţi că $J$ este centrul cercului înscris în triunghiul $ABC$ dacă şi numai dacă $\\left(2 S_{a} - a r\\right) z_{A} + \\left(2 S_{b} - b r\\right) z_{B} + \\left(2 S_{c} - c r\\right) z_{C} = 0$, unde $r$ este raza cercului înscris în triunghiul $ABC$, iar $a, b, c$ sunt lungimile laturilor $[BC], [AC]$, respectiv $[AB]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ with the property\n$$\n|f(x+y) - f(x-y) - 2y| \\le y^2, \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "f(x) = x + c for any real constant c",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38387,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square with side length $2n + 1$, $n \\in \\mathbb{N}$, $n \\geq 2$, is divided by parallels to its sides into $(2n + 1)^2$ squares and exactly $2n(n + 1)$ of them are colored.\n\nWe will call an horizontal line of the large square *nice* if it has more than half of its $2n + 1$ squares colored. Denote $f(n)$ the maximum number of nice lines.\n\na) Find $f(2)$.\n\nb) Find the smallest positive integer $n$ so that $f(n) > 2016$.",
    "options": [],
    "answer": "a) 4; b) 1009",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(A,+, \\cdot)$ un inel (unitar) comutativ și $U(A)$ mulțimea elementelor inversabile ale inelului.\n\na) Dacă $x \\in A$ și $x^{2}=0$ să se arate că $1+x \\in U(A)$.\n\nb) Fie $x \\in A$ pentru care există $n \\in \\mathbb{N}^*$ astfel încât $x^{n}=0$ și fie $u \\in U(A)$. Să se arate că $u+x \\in U(A)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nDacă $x \\in A$ și $x^2 = 0$, atunci $(1 + x) \\cdot (1 - x) = 1 - x^2 = 1$, deci $1 + x$ este inversabil, cu inversul $1 - x$.\n\nb.\nFie $x \\in A$ cu $x^n = 0$ pentru un $n \\in \\mathbb{N}^*$ și $u \\in U(A)$. Atunci $u$ este inversabil, deci există $u^{-1} \\in A$ cu $u u^{-1} = 1$.\n\nObservăm că $(u + x) = u (1 + u^{-1} x)$. Deoarece $u$ este inversabil, $u + x$ este inversabil dacă și numai dacă $1 + u^{-1} x$ este inversabil.\n\nDar $(u^{-1} x)^n = u^{-n} x^n = 0$, deci $u^{-1} x$ este nilpotent. Din punctul a), rezultă că $1 + u^{-1} x$ este inversabil (prin același argument, folosind dezvoltarea binomială pentru $n > 2$).\n\nExplicit, inversul lui $1 + u^{-1} x$ este $1 - u^{-1} x + (u^{-1} x)^2 - \\ldots + (-1)^{n-1} (u^{-1} x)^{n-1}$.\n\nAstfel, $u + x$ este produsul a două elemente inversabile, deci este inversabil.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminați numerele $\\overline{ab}$ pentru care $\\overline{ba} + \\overline{ab}$ și $\\overline{ba} - \\overline{ab}$ sunt pătrate perfecte.",
    "options": [],
    "answer": "56",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a finite group of order $n$. Prove that every element of $G$ is a perfect square if and only if $n$ is odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no positive integers of the form $n = \\underbrace{aa\\dots a}_{k \\text{ times}} + 5a$, $k > 1$, divisible by $2016$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2n^2 + 27n + 91$ is a perfect square for infinitely many $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $2n^2 + 27n + 91 = k^2$ for some $k \\in \\mathbb{N}$. Rearranging, we get:\n\n$$\n2n^2 + 27n + 91 - k^2 = 0\n$$\n\nThis is a quadratic in $n$:\n\n$$\n2n^2 + 27n + (91 - k^2) = 0\n$$\n\nThe discriminant must be a perfect square for $n$ to be integer:\n\n$$\n\\Delta = 27^2 - 4 \\cdot 2 \\cdot (91 - k^2) = 729 - 8(91 - k^2) = 729 - 728 + 8k^2 = 1 + 8k^2\n$$\n\nSo $1 + 8k^2$ must be a perfect square. Let $1 + 8k^2 = m^2$ for some $m \\in \\mathbb{N}$.\n\nThen $m^2 - 8k^2 = 1$, which is a Pell equation.\n\nThe fundamental solution is $(m_1, k_1) = (3, 1)$, since $3^2 - 8 \\cdot 1^2 = 1$.\n\nAll solutions are given by:\n\n$$\n(m + k\\sqrt{8}) = (3 + \\sqrt{8})^t\n$$\nfor $t \\in \\mathbb{N}$.\n\nFor each such $k$, we can solve for $n$:\n\n$$\nn = \\frac{-27 \\pm \\sqrt{1 + 8k^2}}{4}\n$$\n\nSince $\\sqrt{1 + 8k^2} = m$, and $m$ is odd (since $m^2 \\equiv 1 \\pmod{8}$), $-27 + m$ is even, so $n$ is integer.\n\nThus, for each solution $(m, k)$ to the Pell equation, we get an integer $n$ such that $2n^2 + 27n + 91$ is a perfect square.\n\nTherefore, there are infinitely many such $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{R}$ and the function $f_{a,b}: \\mathbb{R} \\to \\mathbb{R}$, $f_{a,b}(x) = a[x] + b\\{x\\}$. Prove that this function is bijective if and only if $a^2 = b^2 \\ne 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 2] \\to \\mathbb{R}$ be a continuous function. Compute\n$$\n\\lim_{t \\to 0} \\int_{0}^{1} \\frac{f(x+t) - f(x)}{t} \\, dx.\n$$",
    "options": [],
    "answer": "f(1) - f(0)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive reals $x, y, z$ so that\n$$\n\\begin{cases}\nxyz = 6 \\\\\n(x+1)(y+2)(z+3) = 48.\n\\end{cases}\n$$",
    "options": [],
    "answer": "x=1, y=2, z=3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}, n \\ge 4$, and $A_n = \\{1, 2, 3, \\dots, n\\}$. Find the number of solutions in the set $A_n \\times A_n \\times A_n \\times A_n$ of the system\n$$\n\\begin{cases}\nx + z = 2y \\\\\ny + t = 2z\n\\end{cases}\n$$",
    "options": [],
    "answer": "Let n = 3q + r with r in {0,1,2}. The number of solutions is:\n- if r = 0: 3q^2\n- if r = 1: 3q^2 + 2q + 1\n- if r = 2: 3q^2 + 4q + 2\nEquivalently, if m = floor((n−1)/3), the count is n + 2mn − 3m^2 − 3m.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer of the form $n = \\overbrace{aa\\dots a}^{k \\text{ times}} + a(a-2)^2$, $k > 1$, divisible by $2016$.",
    "options": [],
    "answer": "77777777952",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a continuous function with $\\int_{0}^{1} f(x)dx = \\int_{0}^{1} x f(x)dx$ and $f(0) = 0$. Prove that there exists $c \\in (0, 1)$ so that\n$$\n\\int_{0}^{c} x f(x)dx = \\frac{c}{2} \\int_{0}^{c} f(x)dx.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c \\in (0, \\infty)$ be such that $\\sqrt{a} + \\sqrt{b} + \\sqrt{c} \\le 2$. Prove that\n$$\n\\frac{\\sqrt{a^2 - ab + b^2}}{a^2 + b^2} + \\frac{\\sqrt{b^2 - bc + c^2}}{b^2 + c^2} + \\frac{\\sqrt{c^2 - ca + a^2}}{c^2 + a^2} \\le \\frac{1}{\\sqrt{abc}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all finite groups which have proper subgroups and all of these have order $2$ or $3$.",
    "options": [],
    "answer": "C4, V4, C6, S3, C9, C3×C3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38401,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, the measures of the angles $B$ and $A$ are directly proportional to $2$ and $3$. Take $M$ and $N$ on the sides $[AB]$ and $[BC]$ respectively, so that $BN = NM = MC = MA$. Prove that:\n\na) one of the obtained triangles is equilateral;\n\nb) $CM \\perp AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation\n$$\n\\sqrt{2^{x+1}-1} + \\sqrt{2^x-1} + 2^x = 2 \\cdot (3 \\cdot 4^x - 2^{x+1} \\cdot \\sqrt{2^{2x+1}-1}).\n$$",
    "options": [],
    "answer": "x = 0 and x = log2(5)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38403,
    "subject": "Mathematics (Multi-modal)",
    "question": "The regular triangular pyramid $SABC$ has $AB = a$, $M$ and $N$ are the midpoints of the sides $[AB]$ and $[BC]$ respectively and $SA \\perp (SBC)$.\n\na) Find the measure of the angle of the line $SM$ with the plane $(SAN)$.\n\nb) If $SP$ is the orthogonal projection of the segment $SM$ on the plane $(SAN)$, compute the tangent of the angle of the planes $(SPC)$ and $(ABC)$.",
    "options": [],
    "answer": "a) 30 degrees; b) sqrt(14)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(x, y)$ so that\n$$\nx + y = \\sqrt{x} + \\sqrt{y} + \\sqrt{xy}.\n$$",
    "options": [],
    "answer": "[(1, 4), (4, 1), (4, 4)]",
    "solution": "The equality can be written $(x + y) - \\sqrt{x} = \\sqrt{xy} + \\sqrt{y}$. Squaring yields $x^2 + xy + y^2 + x - y = 2(2y + x)\\sqrt{x}$. Since $2y + x \\neq 0$, $\\sqrt{x}$ must be rational, therefore $x$ is a perfect square. In the same way, $y$ is a perfect square.\n\nDenote now $\\sqrt{x} = a$, $\\sqrt{y} = b$. The equality $a^2 + b^2 = ab + a + b$ leads now to $(a - b)^2 + (a - 1)^2 + (b - 1)^2 = 2$. This gives $(a, b) \\in \\{(1, 2), (2, 1), (2, 2)\\}$, therefore the answer is $(x, y) \\in \\{(1, 4); (4, 1); (4, 4)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38405,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many three-digit primes can be turned into perfect cubes with a permutation of their digits?",
    "options": [],
    "answer": "3",
    "solution": "The three-digit perfect cubes are $125$, $216$, $343$, $512$ and $729$. The primes must end with an odd digit, different from $5$, and must be not divisible by $3$. Therefore we must consider only $251$, $521$, $433$. All of them are primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-equilateral triangle such that $m(\\angle A) = 60^\\circ$. Let $D$ and $E$ be the intersection points of the Euler line of triangle $ABC$ and the sides of the angle $\\angle BAC$. Prove that the triangle $ADE$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ and $O$ be the orthocenter and the circumcenter of $ABC$, $R$ the radius of the circumcenter; then $OH$ meets the line $AB$ at $D$ and the line $AC$ at $E$. Let $B'$ be the foot of the altitude from $B$ and $C'$ be the foot of the altitude from $C$.\n\nSince $BCC'B'$ is a cyclic quadrilateral we have $\\angle AB'C' \\equiv \\angle ABC$. Hence $\\triangle AB'C' \\sim \\triangle ABC$, having the similarity ratio $\\frac{AC'}{AC} = \\cos(\\widehat{BAC}) = \\frac{1}{2}$. It follows that the similarity ratio is the same with the ratio of the diameters of the circumcircles of triangles $AB'C'$ and $ABC$, so $\\frac{AH}{2R} = \\frac{1}{2}$, which leads to $AH = R = AO$. (1)\n\nIt is known that the rays ($AH$ and ($AO$ are isogonal, so $\\angle BAO \\equiv \\angle CAH$. (2) From (1) it results that $\\angle AOH \\equiv \\angle AHO$, so $\\angle AOD \\equiv \\angle AHE$. Using (1) and (2), it follows that triangles $AOD$ and $AHE$ are congruent, so $AD = AE$ and the conclusion follows.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles, $\\omega_1$ and $\\omega_2$, centred at $O_1$ and $O_2$, respectively, meet at points $A$ and $B$. A line through $B$ meets $\\omega_1$ again at $C$, and $\\omega_2$ again at $D$. The tangents to $\\omega_1$ and $\\omega_2$ at $C$ and $D$, respectively, meet at $E$, and the line $AE$ meets the circle $\\omega$ through $A$, $O_1$, $O_2$ again at $F$. Prove that the length of the segment $EF$ is equal to the diameter of $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Begin by noticing that the lines $CO_1$ and $DO_2$ meet at a point $P$ on $\\omega$, since $\\angle(PO_1, PO_2) = \\angle(O_1C, CB) + \\angle(BD, DO_2) = \\angle(CB, BO_1) + \\angle(O_2B, BD) = \\angle(O_2B, BO_1) = \\angle(O_1A, AO_2)$. In what follows, we consider the case where $O_1$ and $O_2$ lie on the segments $CP$ and $DP$, respectively; the other cases are similar.\n\nSince the angles $PCE$ and $PDE$ are both right, and $2\\angle ACP = \\angle AO_1P = \\angle AO_2P = 2\\angle ADP$ (the equality in the middle holds on account of $P$ lying on $\\omega$), the points $A$, $C$, $D$, $E$, $P$ all lie on the circle on diameter $EP$, so $FP$ is a diameter of $\\omega$, and it is therefore sufficient to show that $EF = FP$. Finally, since $\\angle AFP = \\angle AO_1P = 2\\angle ACP = 2\\angle AEP$ (the first, respectively third, equality holds on account of $APFO_1$, respectively $ACEP$, being cyclic), it follows that the triangle $EFP$ is isosceles with apex at $F$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that a regular quadrilateral pyramid has two opposite lateral faces perpendicular if and only if the angle of two consecutive lateral faces has measure $120^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ABCD$ be the base of the pyramid, $V$ its apex, $VO$ its altitude and $M$, $N$ the midpoints of the edges $AD$, respectively $BC$. Denote $a$ the length of the edge $AB$. Faces $VAD$ and $VBC$ are perpendicular if and only if the triangle $VMN$ is right and isosceles, with sides $VM = VN = \\frac{a\\sqrt{2}}{2}$.\n\nIf $P$ is the foot of the perpendicular from $A$ on $VB$ (same as the foot of the perpendicular from $C$ on $VB$), then, computing in two ways the area of the triangle $VBC$ yields the equivalent condition $PC = PA = \\frac{a\\sqrt{6}}{3}$.\n\nThis is equivalent with $\\sin \\widehat{OPC} = \\sqrt{3}/2$, that is $m(\\widehat{OPC}) = 60^\\circ$.\n\n![](attached_image_1.png)\n\nSince $\\angle APC$ represents the angle of two consecutive lateral faces, this finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd positive integer, and let $x_1, x_2, \\dots, x_n$ be non-negative real numbers. Show that $\\min_{k=1,\\dots,n} (x_k^2 + x_{k+1}^2) \\le \\max_{k=1,\\dots,n} (2x_k x_{k+1})$, where $x_{n+1} = x_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1} - x_k$, $k = 1, \\dots, n$. Since $n$ is odd, there exists an index $j$ such that $(x_{j+1} - x_j)(x_{j+2} - x_{j+1}) \\ge 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_j \\le x_{j+1} \\le x_{j+2}$. Consequently,\n$$\n\\min_{k=1,\\dots,n} (x_k^2 + x_{k+1}^2) \\le x_j^2 + x_{j+1}^2 \\le 2x_{j+1}^2 \\le 2x_{j+1}x_{j+2} \\le \\max_{k=1,\\dots,n} (2x_k x_{k+1}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all three-digit numbers which decrease 13 times when the tens' digit is suppressed.",
    "options": [],
    "answer": "130, 195, 260, 390",
    "solution": "Let $\\overline{abc}$ be such a number. The condition is $\\overline{abc} = 13 \\cdot \\overline{ac}$, that is $100a + 10b + c = 130a + 13c$, same as $10b = 30a + 12c$. We notice that $5$ divides both $10b$ and $30a$, so $12c$ is divisible by $5$, hence $c \\in \\{0, 5\\}$.\n\nIf $c = 0$, then $b = 3a$. Since $a$ is a non-null digit, $(a = 1, b = 3)$, $(a = 2, b = 6)$ and $(a = 3, b = 9)$, which lead to the solutions $130$, $260$ and $390$.\n\nIf $c = 5$, then $b = 3a + 6$. Since $a$ is a non-null digit, the only possibility is $(a = 1, b = 9)$, which leads to the solution $195$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and $E$ be a point on its diagonal $BD$, different from its midpoint. Denote $H$ and $K$ the orthocenters of the triangles $ABE$, respectively $ADE$. Prove that $\\overline{BH} + \\overline{DK} = 0$.\n\nMihaela Berindeanu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that the points $H$ and $K$ are on the diagonal $AC$, because $AC$ is perpendicular on $BE$ and $DE$. Also, $H$ and $K$ are on the altitudes from $E$ in the two triangles, which are perpendicular on the sides of the initial square.\n\nIt follows that the triangle $EHK$ is right and isosceles, so $H$ and $K$ are symmetric with respect of the square's center. Since $B$, $D$ are also symmetric with respect of the square's center, it follows that $DKBH$ is a parallelogram, whence the conclusion.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two positive integers $x$ and $y$ are such that $\\frac{2010}{2011} < \\frac{x}{y} < \\frac{2011}{2012}$. Find the smallest possible value of the sum $x + y$.\nNeculai Stanciu",
    "options": [],
    "answer": "8044",
    "solution": "The fraction $\\frac{x}{y}$ is subunitary, hence $x < y$, that is $x = y - d$, where $d$ is a positive integer. The given relation can be written $\\frac{2011-1}{2011} < \\frac{y-d}{y} < \\frac{2012-1}{2012}$ or $1 - \\frac{1}{2011} < 1 - \\frac{d}{y} < 1 - \\frac{1}{2012}$, whence $\\frac{1}{2011} > \\frac{d}{y} > \\frac{1}{2012}$ or $\\frac{d}{2011d} > \\frac{d}{y} > \\frac{d}{2012d}$. This leads to $2011d < y < 2012d$. (1)\n\nRelation (1) is impossible for $d = 1$.\n\nFor $d = 2$ we get $4022 < y < 4024$, whence $y = 4023$. So $x = 4021$ and $x + y = 8044$.\n\nFor $d \\ge 3$, $4021d \\ge 12063$. It follows $x + y = 2y - d$. From $y > 2011d$ follows $2y - d > 4021d \\ge 12063$.\n\nConsequently, the minimum possible value of the sum is obtained when $d = 2$ and $x + y = 8044$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $D$ be a point on the segment $BC$, $D \\neq B$ and $D \\neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet at $P$, and the lines $A'B$ and $DF$ meet at $Q$.\nProve that the lines $AD$, $BP$ and $CQ$ are concurrent (or all parallel).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\sigma$ denote reflection in the line $BC$. Since $\\angle BDF = \\angle BAC = \\angle CDE$, by concyclicity, the lines $DE$ and $DF$ are images of one another under $\\sigma$, so the lines $AC$ and $DF$ meet at $P' = \\sigma(P)$, and the lines $AB$ and $DE$ meet at $Q' = \\sigma(Q)$.\nConsequently, the lines $PQ$ and $P'Q' = \\sigma(PQ)$ meet at some (possibly ideal) point $R$ on the line $BC$.\nSince the pairs of lines $(CA, QD)$, $(AB, DP)$, $(BC, PQ)$ meet at three collinear points, namely $P'$, $Q'$, $R$ respectively, the triangles $ABC$ and $DPQ$ are perspective, i.e., the lines $AD, BP, CQ$ are concurrent, by the Desargues theorem.\n\n![](attached_image_1.png)\nAs in the first solution, $\\sigma$ denotes reflection in the line $BC$, the lines $DE$ and $DF$ are images of one another under $\\sigma$, the lines $AC$ and $DF$ meet at $P' = \\sigma(P)$, and the lines $AB$ and $DE$ meet at $Q' = \\sigma(Q)$.\nLet the line $AD$ meet the circle $ABC$ again at $M$. Letting $M' = \\sigma(M)$, it is sufficient to prove that the lines $DM' = \\sigma(AD)$, $BP' = \\sigma(BP)$ and $CQ' = \\sigma(CQ)$ are concurrent.\nBegin by noticing that $\\angle(BM', M'D) = -\\angle(BM, MA) = -\\angle(BC, CA) = \\angle(BF, FD)$, to infer that $M'$ lies on the circle $BDF$. Similarly, $M'$ lies on the circle $CDE$, so the line $DM'$ is the radical axis of the circles $BDF$ and $CDE$.\nSince $P'$ lies on the lines $AC$ and $DF$, it is the radical centre of the circles $ABC$, $ADC$, and $BDF$; hence the line $BP'$ is the radical axis of the circles $BDF$ and $ABC$. Similarly, the line $CQ'$ is the radical axis of the circles $CDE$ and $ABC$. So the conclusion follows: the lines $DM'$, $BP'$ and $CQ'$ are concurrent at the radical centre of the circles $ABC$, $BDF$ and $CDE$; thus the lines $DM$, $BP'$ and $CQ'$ are also concurrent.\nAs in the previous solutions, $\\sigma$ denotes reflection in the line $BC$. Let the lines $BE$ and $CF$ meet at $X$. Due to the circles BDEA and CDFA, we have $\\angle XBD = \\angle EAD = \\angle XFD$, so the quadrilateral BFXD is cyclic; similarly, the quadrilateral CEXD is cyclic. Hence $\\angle XDB = \\angle CFA = \\angle CDA$, the lines DX and DA are therefore images of one another under $\\sigma$, and $X' = \\sigma(X)$ lies on the line AD. Let $E' = \\sigma(E)$ and $F' = \\sigma(F)$, and apply the Pappus theorem to the hexagon $BPF'CQE'$ to infer that $X'$, D, and $BP \\cap CQ$ are collinear. The conclusion follows.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38414,
    "subject": "Mathematics (Multi-modal)",
    "question": "A ring $(A, +, \\cdot)$ has property (P) if $A$ is finite and the multiplicative group of units is isomorphic to a non-trivial subgroup of the additive group $(A, +)$. Show that:\n(a) the number of elements of a ring having property (P) is even;\n(b) there are $n$-element rings having property (P) for infinitely many positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Let $A$ be a ring having property (P), let $m = |U(A)|$, and notice that $(-1)^m = 1$. If $m$ is odd, then $-1 = 1$, so $1$ has order $2$ in the additive group $(A, +)$, and consequently $|A|$ is even. If $m$ is even, so must be $|A|$, since the former divides the latter.\n\n(b) Let $m$ be a positive integer, let $n = 2^{m+2}$, and let $A = \\underbrace{\\mathbb{Z}_2 \\times \\cdots \\times \\mathbb{Z}_2}_{m} \\times \\mathbb{Z}_4$.\n\nThe multiplicative group $U(A) = \\{(\\hat{1}, \\dots, \\hat{1}, \\hat{1}), (\\hat{1}, \\dots, \\hat{1}, \\hat{3})\\}$ is easily seen to be isomorphic to the subgroup $\\{(\\hat{0}, \\dots, \\hat{0}, \\hat{0}), (\\hat{0}, \\dots, \\hat{0}, \\hat{2})\\}$ of the additive group $(A, +)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x \\in (2, \\infty)$ for which\n$$\n\\cos(\\pi \\log_3 (x+6)) \\cdot \\cos(\\pi \\log_3 (x-2)) = 1.\n$$",
    "options": [],
    "answer": "3",
    "solution": "From the hypothesis, it is clear that\n$$\n\\cos(\\pi \\log_3 (x + 6)) = \\cos(\\pi \\log_3 (x - 2)) = \\pm 1,\n$$\nand thus, there exist $k, l \\in \\mathbb{Z}$, with the same parity, such that $\\pi \\log_3 (x + 6) = k\\pi$ and $\\pi \\log_3 (x - 2) = l\\pi$.\nWe get now the relations $x + 6 = 3^k$ and $x - 2 = 3^l$. By subtracting the last two equalities, we get $3^k - 3^l = 8$.\nIt is not difficult to see that if $k, l < 0$ the equality is not possible. Then $3^l (3^{k-l} - 1) = 8$, and thus $3^l = 1$, that is $l = 0$ and then $k = 2$. Finally we get $x = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $k$ and an integer $a \\equiv 3 \\pmod{8}$, show that $a^m + a + 2$ is divisible by $2^k$ for some positive integer $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proceed by induction on $k$. Since $a \\equiv 3 \\pmod{8}$, $m = 1$ works for $k = 1, 2, 3$, so let $k \\ge 3$ and let $m$ be a positive integer such that $a^m + a + 2$ is divisible by $2^k$.\n\nIf $(a^m + a + 2)/2^k$ is even, then $a^m + a + 2$ is clearly divisible by $2^{k+1}$.\nIf $(a^m + a + 2)/2^k$ is odd, we will show that $a^{m+2^{k-2}} + a + 2$ is divisible by $2^{k+1}$.\nTo this end, write $a^{m+2^{k-2}} + a + 2 = a^{2^{k-2}}(a^m + a + 2) - (a+2)(a^{2^{k-2}} - 1)$. The first term is an odd multiple of $2^k$, and it is sufficient to prove that so is the second.\nInduct on $k \\ge 3$ to show that $a^{2^{k-2}} - 1$ is an odd multiple of $2^k$. Since $a \\equiv 3 \\pmod{8}$, this is clearly the case if $k=3$, and the induction step follows from the identity $a^{2^{k-1}} - 1 = (a^{2^{k-2}} - 1)(a^{2^{k-2}} + 1) = (a^{2^{k-2}} - 1)((a^{2^{k-2}} - 1) + 2)$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider triangle $ABC$ with $m(\\angle B) = 30^\\circ$, $m(\\angle C) = 15^\\circ$ and $M$ the midpoint of the side $[BC]$. Let $N \\in (BC)$ be such that $[NC] = [AB]$. Show that $[AN]$ is the bisector of the angle $MAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the point where the perpendicular bisector of the segment $[BC]$ meets $AB$. Then $m(\\angle PCB) = 30^\\circ$, $m(\\angle PCA) = 15^\\circ$ and $m(\\angle MPC) = 60^\\circ$.\n![](attached_image_1.png)\n\nSince $PC = PB$ and $NC = AB$, it follows that $\\frac{AP}{NC} = \\frac{BP}{BC}$, that is $\\frac{PA}{PB} = \\frac{CN}{CB}$, whence $AN \\parallel PC$.\nSince $PC = 2PM$ (the right triangle $MPC$ has an angle of $30^\\circ$), $\\frac{PA}{AB} = \\frac{PC}{BC} = \\frac{2PM}{2BM} = \\frac{PM}{BM}$ so, using the converse of the Bisector Theorem, $[MA$ is the bisector of the angle $BMP$.\n\nThen $45^\\circ = m(\\angle AMB) = m(\\angle ANB) + m(\\angle MAN)$, and, since $AN \\parallel PC$ yields $m(\\angle ANB) = 30^\\circ$, $m(\\angle MAN) = 15^\\circ$. From $m(\\angle BAN) = m(\\angle BPC) = 120^\\circ$ and $m(\\angle BAC) = 135^\\circ$ follows that $m(\\angle CAN) = 15^\\circ = m(\\angle MAN)$, that is $[AN$ is the bisector of the angle $MAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38418,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n \\in \\mathbb{N}$, consider the system $(S_n) := \\begin{cases} x^2 + n y^2 = z^2 \\\\ n x^2 + y^2 = t^2 \\end{cases}$, where $x, y, z, t \\in \\mathbb{N}$. If $M_1 = \\{ n \\in \\mathbb{N} \\mid \\text{system } (S_n) \\text{ has infinitely many solutions} \\}$, and $M_2 = \\{ n \\in \\mathbb{N} \\mid \\text{system } (S_n) \\text{ has no solutions} \\}$, prove that:\n\na) $7 \\in M_1, 10 \\in M_2$;\n\nb) sets $M_1$ and $M_2$ are infinite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Notice that $x = 1, y = 3, z = 8, t = 4$ is a solution to the system $(S_7)$, and so is $(k, 3k, 8k, 4k)$, for any $k \\in \\mathbb{N}$, hence $7 \\in M_1$.\n\nIf $(x, y, z, t)$ is a solution to the system $(S_{10})$, it would follow that $11(x^2 + y^2) = z^2 + t^2$. From $11 \\mid z^2 + t^2$ we get $11 \\mid z$ and $11 \\mid t$. Then $11 \\mid x^2 + y^2$, hence $11 \\mid x$ and $11 \\mid y$. Therefore, there exist $x_1, y_1, z_1, t_1 \\in \\mathbb{N}$ such that $x = 11x_1, y = 11y_1, z = 11z_1, t = 11t_1$. This leads to $11(x_1^2 + y_1^2) = z_1^2 + t_1^2$. By continuing this procedure (\"infinite descent\"), we get that $x$ is a multiple of any power of $11$, and, as $x \\neq 0$, we arrive to a contradiction. In conclusion, the system $(S_{10})$ has no solutions, i.e. $10 \\in M_2$.\n\nb) It is easy to see that any number $n$ of the form $m^2 - 1$, $m \\in \\mathbb{N}$, belongs to $M_1$: we can choose $x = y$ and notice that $(k, k, mk, mk)$, $k \\in \\mathbb{N}$, are solutions to the system $(S_{m^2-1})$, hence $m^2 - 1 \\in M_1, \\forall m \\in \\mathbb{N}$. Applying the same steps as we did above for $n = 10$, it is easy to prove that if $p$ is a prime of the form $4m + 3$, and $n = p - 1$, then $n \\in M_2$. As there are infinitely many such primes, the conclusion follows readily. One can prove that actually any number congruent to $2 \\pmod{4}$ belongs to $M_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be real nonzero numbers, such that $x^3 + y^3 + 3x^2y^2 = x^3y^3$. Determine the set of the possible values of $E = \\frac{1}{x} + \\frac{1}{y}$.",
    "options": [],
    "answer": "{-2, 1}",
    "solution": "Rewrite the given condition successively $(x + y)^3 - 3xy(x + y) = x^3y^3 - 3x^2y^2$, i.e. $(x + y)^3 - (xy)^3 = 3xy(x + y) - 3x^2y^2$, or $(x + y - xy)(x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2) = 3xy(x + y - xy)$. We either have $x + y = xy$, which leads to $E = 1$ (obtained for $x = y = 2$), or $x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2 = 3xy$. The last equality, multiplied by $2$, can be written equivalently $x^2(y + 1)^2 + y^2(x + 1)^2 + (x - y)^2 = 0$, which is possible if and only if $x = y = -1$ ($x = y = 0$ is forbidden). In this case, $E = -2$. In conclusion, the set of the possible values of $E$ is $\\{-2, 1\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, show that for no set of integers modulo $n$, whose size exceeds $1 + \\sqrt{n+4}$, is it possible that the pairwise sums of unordered pairs be all distinct.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S \\subseteq \\mathbb{Z}/n\\mathbb{Z}$ be a set whose pairwise sums of unordered pairs are distinct, and consider the pairwise differences of *ordered* pairs.\nThe crucial observation is that if a difference $d \\neq 0$ occurs twice in $S$, then these two occurrences must be adjacent in a 3-term arithmetic progression in $S$ (with difference $d$). Indeed, if $a, a+d, a', a'+d$ are all in $S$, where $a \\neq a'$, then $a+(a'+d) = a'+(a+d)$ so, by assumption on $S$, $a' = a \\pm d$, i.e. the ordered pairs $(a, a+d)$ and $(a', a'+d)$ are adjacent in a 3-term arithmetic progression.\nWe also observe that, excepting the arithmetic progression of common difference $n/2$, in case $n$ is even, no two 3-term arithmetic progressions can have the same central term, since if $a, a \\pm d, a \\pm d'$ are all in $S$, where $d, d' \\neq 0, n/2$ are distinct, then $(a+d') + (a-d') = (a+d) + (a-d)$ would violate the assumption on $S$.\nIt follows that the number of 3-term arithmetic progressions in $S$ is at most $|S|+2$. Now any non-zero difference not occurring in a 3-term arithmetic progression can appear at most once in $S$, and those which do appear in 3-term arithmetic progressions can appear at most twice, except $\\pm n/3$ in case $n$ is divisible by 3, which can appear three times.\nConsequently, the total number of non-zero differences appearing in $S$ is at least $|S|(|S|-1) - (|S|+2) - 2 = |S|^2 - 2|S| - 4$. If $|S| > 1 + \\sqrt{n+4}$, this quantity is greater than $n-1$ — a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive integers $p$ with the property that the sum of the first $p$ positive integers is a four-digit positive integer whose decomposition into prime factors is of the form $2^m3^n(m+n)$, where $m, n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "63",
    "solution": "The number $m+n$ is prime and, obviously, $m+n \\ge 5$. If $m+n=5$, the largest value for $N = 2^m3^n(m+n)$ is $2^1 \\cdot 3^4 \\cdot 5 = 810$, which has only three digits.\nSuppose now that $m+n \\ge 11$. Then $N \\ge 2^{10} \\cdot 3 \\cdot 11 > 10000$, hence $N$ cannot have four digits.\nSo, $m+n=7$. In this case, the four-digit numbers are: $2^6 \\cdot 3^1 \\cdot 7 = 1344$, $2^5 \\cdot 3^2 \\cdot 7 = 2016$, $2^4 \\cdot 3^3 \\cdot 7 = 3024$, $2^3 \\cdot 3^4 \\cdot 7 = 4536$ and $2^2 \\cdot 3^5 \\cdot 7 = 6804$.\nThe equality $N = \\frac{p(p+1)}{2}$ can be fulfilled only for $N = 2016$, when $p = 63$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38422,
    "subject": "Mathematics (Multi-modal)",
    "question": "The right isosceles triangle $ABC$ has $m(\\angle A) = 90^\\circ$. Take $M \\in (BC)$ so that $m(\\angle AMB) = 75^\\circ$ and take $F$ on the internal bisector of the angle $MAC$ so that $BF = AB$. Prove that:\n\na) the lines $AM$ and $BF$ are perpendicular;\n\nb) the triangle $CFM$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From $m(\\angle BMA) = 75^\\circ$, it follows $m(\\angle BAM) = 180^\\circ - 75^\\circ - 45^\\circ = 60^\\circ$, $m(\\angle MAC) = 30^\\circ$ and $m(\\angle MAF) = 15^\\circ$.\n\nFrom $BF = AB$ it follows that $m(\\angle AFB) = m(\\angle BAF) = 75^\\circ$, hence $m(\\angle ABF) = 30^\\circ$, whence, using $m(\\angle BAM) = 60^\\circ$, it follows that $AM \\perp BF$.\n\n\nb) Take the point $D$ in the half-plane determined by $BC$ not containing $A$, so that $BD = BF$ and $m(\\angle MBD) = 15^\\circ$.\n\n![](attached_image_1.png)\n\nThen the triangle $ABD$ is equilateral and the points $A$, $M$ and $D$ are collinear.\n\nThe triangle $ADC$ is isosceles, with $m(\\angle DAC) = 30^\\circ$, hence $m(\\angle ADC) = 75^\\circ = m(\\angle CMD)$. Therefore the triangle $CDM$ is isosceles, hence $[CM] = [CD]$.\n\nIn the isosceles triangle $BDF$, the bisector $BC$ is the perpendicular bisector of the segment $[DF]$, hence the triangle $CFD$ is also isosceles, whence $[CF] = [CD] = [CM]$, that is the triangle $CFM$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral whose diagonals are not perpendicular and intersect at $X$. Let $A', C'$ be the projections of $A$ and $C$ onto the line $BD$ and let $B', D'$ be the projections of $B$ and $D$ onto $AC$. Prove that:\n\na) the perpendicular lines drawn from the midpoints of the sides onto the opposite sides are concurrent at a point called *Mathot's point*;\n\nb) points $A', B', C', D'$ are cocyclic;\n\nc) if $O'$ is the circumcenter of $A'B'C'$, then $O'$ is the midpoint of the line segment determined by the orthocenters of triangles $XAB$ and $XCD$;\n\nd) $O'$ is the *Mathot point* of the quadrilateral $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $O$ be the circumcenter of $ABCD$. It is well known that the midpoints of the sides of a quadrilateral $ABCD$ are the vertices of a parallelogram, hence the line segments joining the midpoints of two opposite sides have the same midpoint, $G$. The perpendicular lines from $O$ to $AB$ and $CD$ pass through the midpoints of these sides, therefore the perpendiculars dropped from $O$ and from the midpoints of two opposite sides onto their opposite side form a parallelogram whose center is $G$. It follows that the two perpendicular lines dropped from the midpoints of two opposite sides onto their opposite side intersect at the reflection of $O$ in $G$. The other two perpendiculars intersect at the same point.\n\nb) We assume the angle $AXB$ to be acute, the other case being similar. The quadrilaterals $ABA'B'$, $CDC'D'$ and $ABCD$ being cyclic, we have $\\angle XDC' \\equiv \\angle XDC \\equiv \\angle XAB \\equiv \\angle XA'B'$, hence $A'B'C'D'$ is cyclic.\n\nc) Let $H_1$ and $H_2$ be the orthocenters of triangles $XAB$, and $XCD$, respectively. If $O''$ is the midpoint of $[H_1H_2]$, as $O''$ belongs to the midsegment of the trapezoid $H_1B'H_2D'$, $O''$ belongs to the perpendicular bisector of the line segment $[B'D']$. Similarly, $O''$ belongs to the midsegment of the trapezoid $A'H_1C'H_2$, hence to the perpendicular bisector of $[A'C']$. As $A'C'$ and $B'D'$ are not parallel, it follows that $O''$ is precisely the circumcenter of $A'B'C'D'$, i.e. $O''$ coincides with $O'$.\n\nd) We have $\\angle A'B'X \\equiv \\angle ABX \\equiv \\angle DCX$, hence $A'B' \\parallel CD$. If $N$ is the midpoint of $[AB]$, then $NA' = NB'$, hence $N$ belongs to the perpendicular bisector of $[A'B']$. But so does $O'$, therefore it follows that $NO' \\perp CD$. Similarly, $O'$ belongs to the perpendicular dropped from the midpoint of $[CD]$ on $AB$, hence $O'$ is the Mathot point of the quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the planar finite configurations $C$ consisting of at least three points, satisfying the following condition: if $x$ and $y$ are distinct points of $C$, then at least one of the two equilateral triangles erected on the segment $xy$ has all three vertices in $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The required configurations consist of the three vertices of an equilateral triangle. Clearly, the three vertices of an equilateral triangle form a configuration satisfying the condition in the statement.\n\nTo prove the converse, let $a$ and $b$ be the end points of a diameter of $C$ and notice that exactly one of the two equilateral triangles erected on the segment $ab$ has the third vertex $c$ in $C$.\n\nSuppose, if possible, $x$ is a fourth point in $C$. Since the segments $ab$, $bc$ and $ca$ are all three diameters of $C$, the point $x$ is interior to (at least) one of the angles $abc$, $bca$, $cab$, say the latter. Consider the equilateral triangles $axy$ and $axz$, where $b$ and $z$, respectively $c$ and $y$, both lie on the same side of the line $ax$. Since $C$ contains at least one of the points $y$ and $z$, both of which lie outside the triangle $abc$ (this is because the angles $bax$ and $cax$ are both less than $\\pi/3$), we may and will further assume that so does $x$, so the quadrangle $abxc$ is convex.\n\nSince $ax \\le ac$, it follows that $\\angle axc \\ge \\angle acx > \\pi/3$, so $y$ is interior to the angle $axc$, and $\\angle bxy < \\angle bxc < \\pi$. And since $\\angle cay = \\angle bax < \\pi/3$, it follows that $\\angle bay = \\angle bac + \\angle cay = \\pi/3 + \\angle cay < 2\\pi/3 < \\pi$, so the quadrangle $abxy$ is convex. Hence $ax + by > ab + xy = ab + ax$, i.e., $by > ab$. Similarly, $cz > ac$, and since $ab$ and $ac$ are both diameters of $C$, and the latter contains at least one of the points $y$ and $z$, we reach a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38425,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of a prism are colored using two colors, so that each lateral edge has its vertices differently colored. Consider all the segments that join vertices of the prism and are not lateral edges. Prove that the number of such segments with endpoints differently colored is equal to the number of such segments with endpoints of the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $a$ the number of the vertices of the upper base which have the first color and $b = n - a$ the number of the vertices of the upper base which have the second color. Then the lower base has $b$ points with the first color and $a$ points with the second color.\nThe number of segments with endpoints differently colored and on different bases is $a^2 + b^2 - n$.\nThe number of segments with endpoints differently colored and on the same base is $2ab$.\nSo, the total number of segments with endpoints differently colored is $(a + b)^2 - n = n^2 - n$, which is exactly half of the number of all the segments.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38426,
    "subject": "Mathematics (Multi-modal)",
    "question": "In every unit square of an $n \\times n$ board a positive integer is written. A move consists of choosing a $2 \\times 2$ square and adding 1 to exactly three of the four numbers written in the chosen square. We call a positive integer *good* if starting from any initial numbers, there is a sequence of moves that makes all the numbers from the board equal.\n\na) Prove that $n = 6$ is not good.\n\nb) Prove that $n = 4$, and $n = 1024$ are good.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Let us notice that the sum of the 36 numbers written on the board is invariant modulo 3. When all the numbers on the board are equal, their sum is a multiple of 36, hence a multiple of 3. But this requires that the initial sum is also a multiple of 3. In conclusion, if the sum of the initial numbers is not a multiple of 3, there is no succession of moves that makes the numbers equal. In conclusion, 6 is not good.\n\nb. For $n = 4$, we choose an arbitrary unit square and prove that we can increase by 1 all the other 15 numbers from the board. For example, if the chosen square lies in the upper left $2 \\times 2$ corner, we successively increment the numbers in the squares marked with B, then those marked by C, D, and E. Finally, we increment 3 of the four numbers in the squares marked with A (with the exception of the chosen one). Thus, combining five moves, we have obtained a move that is equivalent to decreasing the chosen number by 1.\n\n![](attached_image_1.png)\n\nRepeating this new move and choosing each time the largest number (or one of those if there are several largest ones), we can make the numbers equal.\n\nFor $n$ a power of 2, in particular for $n = 2^{10} = 1024$, we prove by induction that any $2^k \\times 2^k$ board ($k \\ge 1$) from which an arbitrary unit square has been removed, can be tiled with non-overlapping L-shaped trominos (figures obtained by removing a unit square from a $2 \\times 2$ square).\n\nThe possibility of tiling such \"deficient squares\" with L-shaped trominos is a classical problem due to Golomb.\n\nAssume the statement true for $n = 2^{k-1}$ and let us prove it for $n = 2^k$. Draw the center lines of the $2^k \\times 2^k$ board. The cut-off square lies in exactly one of the four thus obtained $2^{k-1} \\times 2^{k-1}$ boards. We may remove one square from each of the other three boards by placing a tromino at the center of the $2^k \\times 2^k$ board. The result is a tromino and four $2^{k-1} \\times 2^{k-1}$ boards, each with one square removed - just a situation to apply the inductive hypothesis.\n\nThis tiling allows us to construct a sequence of moves that increases by 1 the number in every square with the exception of one square that can be arbitrarily chosen. By repeatedly using such sequences we can make all the numbers of the board equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $CA \\ne CB$, and let $D, F$, and $G$ be the midpoints of the sides $AB, AC$, and $BC$, respectively. A circle $\\gamma$ through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. Reflect $H$ and $I$ across $F$ and $G$, respectively, to obtain $H'$ and $I'$, respectively. The line $H'I'$ meets the lines $CD$ and $FG$ at $Q$ and $M$, respectively, and the line $CM$ meets $\\gamma$ again at $P$. Prove that the triangle $CPQ$ is isosceles.\nIMO 2015 Shortlist",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Letting $DF$ and $DG$ meet $\\gamma$ again at $R$ and $S$, respectively, we claim that $R$ and $S$ both lie on the line $H'I'$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nNotice that $\\angle H'CQ = \\angle SDC = \\angle SRC$ and $\\angle QCI' = \\angle CDR = \\angle CSR$ to deduce that $(CH'Q, RCQ)$ and $(CI'Q, SCQ)$ are pairs of similar triangles, so $QH' \\cdot QR = QC^2 = QI' \\cdot QS$.\nFinally, invert from $Q$ with radius $QC$ (Figure 2) and notice that $R, C$ and $S$ are mapped to $H', C$ and $I'$, respectively, to infer that $\\gamma$, the circumcircle of the triangle $RCS$, is mapped to $\\omega$, the circumcircle of the triangle $H'CI'$. Since $P$ and $C$ both lie on either circle, and the latter is fixed under the inversion, so is the former. Consequently, $QP^2 = QC^2$, i.e., $QP = QC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a cube $ABCD A'B'C'D'$ and two points $M \\in (CD')$ and $N \\in (DA')$. Prove that $MN$ is the common perpendicular of the lines $CD'$ and $DA'$ if and only if\n$$\n\\frac{D'M}{D'C} = \\frac{DN}{DA'} = \\frac{1}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $\\frac{D'M}{D'C} = \\frac{DN}{DA'} = \\frac{1}{3}$, then $M$ and $N$ are the centroids of the triangles $DD'C$ and, respectively, $ADD'$, hence the points $C'$, $M$, $P$ and $A$, $N$, $P$ are collinear,\n\nwhere $P$ is the midpoint of the side $DD'$.\nFrom the triangle $APC'$, $\\frac{PM}{PC'} = \\frac{PN}{PA}$, hence $MN \\parallel AC'$.\nFrom $CD' \\perp (ADC')$ and $DA' \\perp (AD'C')$ follows $AC' \\perp D'C$ and $AC' \\perp A'D$. This shows that $MN$ is the common perpendicular of the lines $CD'$ and $DA'$.\n\n![](attached_image_1.png)\n\nFor the converse, if $MN$ is the common perpendicular of the two lines, due to the uniqueness of this line, it must coincide with the line from a), hence $\\frac{D'M}{D'C} = \\frac{DN}{DA'} = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38429,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes $AA_1, BB_1, CC_1$ of the acute triangle $ABC$ intersect at $H$. Let $A_2$ be the reflection of point $A$ in the line $B_1C_1$ and let $O$ be the circumcenter of triangle $ABC$.\n\na) Prove that the points $O, A_2, B_1, C$ are cocyclic.\n\nb) Prove that the points $O, H, A_1, A_2$ are cocyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The angles $\\angle ABC$ and $\\angle AB_1C_1$ are equal, therefore so are their complementary angles, $\\angle BAA_1$ and $\\angle A_2AC$.\nIt follows that the rays ($AH$ and ($AA_2$ are isogonal, hence $A_2 \\in (AO$). As $AO = CO$, we have $\\angle ACO \\equiv \\angle OAC \\equiv \\angle AA_2B_1$, therefore points $O, A_2, B_1, C$ are cocyclic. (The arguments above hold in the case $A_2 \\in (AO)$ as well as in the case $O \\in (AA_2)$.)\n\n\nb) From the power of the point $A$ with respect to the circles through $O, A_2, B_1, C$ and $H, A_1, C, B_1$ respectively, it follows that $AB_1 \\cdot AC = AO \\cdot AA_2$ and $AB_1 \\cdot AC = AH \\cdot AA_1$.\nSince $AO \\cdot AA_2 = AH \\cdot AA_1$, from the converse of the power of the point theorem, it follows that the points $O, A_2, H, A_1$ are cocyclic.\n\n![](attached_image_1.png)\nAlternative Solution. For b). Consider $\\{A_3\\} = (AA_2 \\cap B_1C_1$ and let $O_1$ be the midpoint of $[AH]$. Then $O_1$ is the circumcenter of triangle $AB_1C_1$. As $[O_1A_3]$ is a midsegment in triangle $AHA_2$, we have $\\angle HA_2A \\equiv \\angle O_1A_3A$. But the similarity of triangles $ABC$ and $AB_1C_1$ leads to the equality of the corresponding angles $\\angle AA_1O$ and $\\angle AA_3O_1$, which means that $\\angle AA_2H \\equiv \\angle AA_1O$ and the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38430,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given an $m \\times n$ grid and three colors. We wish to color each segment of the grid with one of the three colors so that each unit square has two sides of one color and two sides of a second color. How many such colorings are possible?",
    "options": [],
    "answer": "3^{m+n} * 2^{mn}",
    "solution": "We label the lines from top to bottom and the columns from left to right. The leftmost side of the unit square in the upper-left corner can be colored in $3$ ways. Subsequently, there are $3$ ways of choosing the side of this unit square that is to receive the same color as the first side. The remaining two sides of the square have to be colored with the same color. This color can be chosen in $2$ ways. In conclusion, the unit square in the upper-left corner can be colored in $3 \\cdot 3 \\cdot 2 = 18$ ways.\n\nNext, we successively color the unit squares in the first line, from left to right. Every time, the leftmost side of the square has already been colored. Thus, there are $6$ ways of coloring each of these squares. We proceed similarly on the first column. We color the unit squares from top to bottom and for each of these squares there are $6$ ways of coloring.\n\nNow we move on to color the squares situated on rows $2, 3, \\ldots, m$ and columns $2, 3, \\ldots, n$. We color them from top to bottom, from left to right. Thus, when in turn to be colored, each unit square has already two sides colored: the upper side and the leftmost side. If these sides have received different colors, then the colors of this square have already been decided; there are $2$ ways of coloring the remaining sides with these two colors. If the two sides that are already colored have the same color, then the remaining sides will receive the same color. This color can be chosen in $2$ ways. Therefore, for each of these squares, there are $2$ ways of coloring.\n\nIn conclusion, there are $18 \\cdot 6^{m-1} \\cdot 6^{n-1} \\cdot 2^{(m-1)(n-1)} = 3^{m+n} \\cdot 2^{mn}$ ways of coloring.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38431,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$, $b$ and $c$ are the length of the sides of a triangle, show that\n$$\n\\frac{3}{2} \\le \\frac{b+c}{b+c+2a} + \\frac{a+c}{a+c+2b} + \\frac{a+b}{a+b+2c} < \\frac{5}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality\n$$\n\\frac{3}{2} \\le \\frac{b+c}{b+c+2a} + \\frac{a+c}{a+c+2b} + \\frac{a+b}{a+b+2c}\n$$\nis Nesbitt's inequality for the triple $(b+c, a+c, a+b)$.\n\nTriangle's inequality yields $b+c > a \\Leftrightarrow 3a+3b+3c > 4a+2b+2c \\Leftrightarrow \\frac{2}{3(a+b+c)} < \\frac{1}{2a+b+c} \\Leftrightarrow \\frac{4a}{3(a+b+c)} < \\frac{2a}{2a+b+c} \\Leftrightarrow \\frac{4a}{3(a+b+c)} < 1 - \\frac{b+c}{2a+b+c}$.\nIn the same way $\\frac{4b}{3(a+b+c)} < 1 - \\frac{a+c}{2b+a+c}$ and $\\frac{4c}{3(a+b+c)} < 1 - \\frac{a+b}{2c+a+b}$. Adding the last three inequalities gives\n$$\n\\frac{4}{3} < 3 - \\left( \\frac{b+c}{b+c+2a} + \\frac{a+c}{a+c+2b} + \\frac{a+b}{a+b+2c} \\right),\n$$\nwhich is equivalent to what we had to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers expressible in the form $\\dfrac{x^2 + y}{xy + 1}$, for at least two pairs $(x, y)$ of positive integers.",
    "options": [],
    "answer": "1",
    "solution": "We show that $1$ is the only positive integer satisfying the condition in the statement. Clearly, $1 = \\dfrac{1^2 + y}{1 \\cdot y + 1}$ for any positive integer $y$, so $1$ is expressible in the required form for infinitely many pairs of positive integers.\n\nNext, we prove that any integer $n \\ge 2$ is uniquely expressible in the required form. Letting $x = n^2$ and $y = n$, it is readily checked that $\\dfrac{x^2 + y}{xy + 1} = n$, so $n$ is indeed expressible in the required form.\n\nTo prove uniqueness, let $x$ and $y$ be positive integers such that $\\dfrac{x^2 + y}{xy + 1} = n$. Alternatively, but equivalently, $x^2 - nxy + y - n = 0$, so the discriminant $\\Delta = n^2y^2 + 4n - 4y$ is a perfect square. It is readily checked that $(ny - 2)^2 < \\Delta < (ny + 2)^2$, so $\\Delta = (ny \\pm 1)^2$ or $\\Delta = n^2y^2$. The former case is ruled out by noticing that $\\Delta - n^2y^2 = 4(n - y)$ is even, and the latter yields $y = n$, so $x = n^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, and let $E$ and $F$ denote the midpoints of diagonals $[AC]$ and $[BD]$, respectively. If $\\{G\\} = AB \\cap CD$, $\\{H\\} = AD \\cap BC$, prove that:\n\na) the intersection points of the angle bisectors of $\\angle AHB$ and $\\angle AGD$ with the sides of the quadrilateral $ABCD$ are the vertices of a rhombus;\n\nb) the center of this rhombus lies on the line $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are going to treat only the case when $C \\in (GD)$ and $C \\in (BH)$, all the other cases being similar.\n\na) Let $M$ and $N$ be the intersection points of the angle bisector of angle $\\angle G$ with the sides $BC$ and $AD$, respectively. Consider $I, K$ the intersection points of the angle bisector of angle $\\angle H$ with sides $CD$ and $AB$, respectively. Also, put $\\{J\\} = MN \\cap IK$. From triangle $GBC$ we get $\\angle G = 180^\\circ - \\angle GBC - \\angle GCB = \\angle B + \\angle C - 180^\\circ$, hence $\\angle CMJ = 90^\\circ + \\frac{1}{2}(\\angle B - \\angle C)$. Similarly, $\\angle CIJ = 90^\\circ + \\frac{1}{2}(\\angle D - \\angle C)$. From the quadrilateral $CMJI$, using the fact that $\\angle B + \\angle D = 180^\\circ$, it follows that $\\angle MJI = 90^\\circ$.\n\nIn the triangles $GIK$ and $HMN$, line segments $[GJ]$ and $[HJ]$ are angle bisectors and altitudes, therefore they are also medians. Thus, the diagonals of the quadrilateral $MKNI$ are perpendicular and bisect each other.\n\nb) The sides of the rhombus are parallel to $AC$ and $BD$, respectively. Indeed, according to the angle bisector theorem, $\\frac{MB}{MC} = \\frac{GB}{GC}$ and $\\frac{BK}{KA} = \\frac{HB}{HA}$. But from the power of the point $G$ with respect to the circle, $\\frac{GB}{GC} = \\frac{GD}{GA}$. In order to prove that $MK \\parallel AC$, i.e. $\\frac{BK}{KA} = \\frac{BM}{MC}$, it is sufficient to prove that $HA \\cdot GD = HB \\cdot GA$. This follows from $[AHG] = \\frac{HA \\cdot GD \\sin D}{2} = \\frac{HB \\cdot GA \\sin B}{2}$ and $\\sin B = \\sin D$. Thus, $MK \\parallel AC$.\n\nConsider $\\{L\\} = MK \\cap BF$, $\\{P\\} = AE \\cap KN$, $\\{O\\} = DF \\cap IN$ and $\\{Q\\} = MI \\cap CE$. Line segments $AE, BF, CE$ and $DF$ are medians in triangles $ABD, ABC, BCD, CDA$, therefore $LO$ and $PQ$ meet in the center of the rhombus.\n\nPut $\\{X\\} = LO \\cap EF$ and $\\{Y\\} = QP \\cap EF$. Then $\\frac{XE}{XF} = \\frac{LB}{LF} = \\frac{BK}{KA}$ and $\\frac{YE}{YF} = \\frac{QE}{QC} = \\frac{BM}{MC}$. As $\\frac{BK}{KA} = \\frac{BM}{MC}$, it follows that points $X$ and $Y$ coincide, which means that the lines $EF, LO$ and $PQ$ pass through the center of the rhombus.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f$ of the set of positive integers into itself such that $f(m) \\ge m$ and $f(m+n)$ divides $f(m) + f(n)$ for all positive integers $m$ and $n$.",
    "options": [],
    "answer": "f(n) = c n for some positive integer c",
    "solution": "To this end, write $\\ell = f(1)$, and notice that, since $1 \\le f(n)/n \\le \\ell$, there exists a minimal positive integer $k \\le \\ell$ such that $\\lfloor f(n)/n \\rfloor = k$ for infinitely many positive integers $n$. Let $A$ denote the set of all these positive integers, let $B = \\{n: n \\in A \\text{ and } 2n \\notin A\\}$, and let $A' = A \\setminus B$.\n\nWe now show that $B$ is finite, so $A'$ is infinite. Indeed, $f(2n) \\le 2f(n)$ and $\\lfloor f(2n)/(2n) \\rfloor \\le \\lfloor 2f(n)/(2n) \\rfloor = k$ imply $\\lfloor f(2n)/(2n) \\rfloor < k$ for all $n$ in $B$, so $B$ is finite by minimality of $k$.\n\nNext, for $n$ in $A'$, write\n$$\n\\frac{2f(n)}{f(2n)} = \\frac{\\frac{f(n)}{n}}{\\frac{f(2n)}{2n}} < 1 + \\frac{1}{k},\n$$\nto deduce that the positive integer $2f(n)/f(2n)$ is less than $2$, so $f(2n) = 2f(n)$.\n\nFurther, fix a positive integer $a$ and notice, as before by minimality of $k$, that $f(a+n) \\ge k(a+n)$ for all but finitely many $n$ in $A'$. Hence\n$$\n\\frac{f(a) + f(n)}{f(a + n)} < \\frac{f(a) + (k + 1)n}{k(a + n)},\n$$\nfor all but finitely many $n$ in $A'$, so $(f(a) + f(n))/f(a+n)$ is a positive integer less than $2$ for all but finitely many $n$ in $A'$, i.e., $f(a+n) = f(a)+f(n)$ for all but finitely many $n$ in $A'$.\n\nFinally, fix two positive integers $a$ and $b$. By the preceding, $f(a+n) = f(a)+f(n)$, $f(b+n) = f(b)+f(n)$, and $(f(a+n)+f(b+n))/f(a+b+2n)$ is a positive integer for all but finitely many $n$ in $A'$. Consequently,\n$$\n\\frac{f(a) + f(b) + 2f(n)}{f(a + b) + f(2n)} = \\frac{f(a) + f(b) + f(2n)}{f(a + b) + f(2n)}\n$$\nis a positive integer for all but finitely many $n$ in $A'$, so the latter must equal $1$ for all but finitely many $n$ in $A'$, i.e., $f(a+b) = f(a)+f(b)$. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38435,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ has its orthocenter $H$ distinct from its vertices and from the circumcenter $O$. Denote $M, N, P$ the circumcenters of the triangles $HBC, HCA$, respectively $HAB$. Prove that the lines $AM, BN, CP$ and $OH$ are concurrent.\nPetru Braica",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $D$ is the midpoint of $[BC]$, then $\\overrightarrow{OD} = \\frac{1}{2}(\\overrightarrow{OB} + \\overrightarrow{OC}) = \\frac{1}{2}(\\overrightarrow{OH} - \\overrightarrow{OA}) = \\frac{1}{2}\\overrightarrow{AH}$, hence $\\overrightarrow{OM} = \\overrightarrow{AH}$.\nIt follows that $AHMO$ is a parallelogram, hence $AM$ passes through the midpoint of the segment $[OH]$, the same being true for $BN$ and $CP$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and let $a_1, a_2, \\dots, a_n$ be positive numbers such that $a_1 \\le a_2$, $a_1 + a_2 \\le a_3$, $a_1 + a_2 + a_3 \\le a_4$, $\\dots$, $a_1 + a_2 + \\dots + a_{n-1} \\le a_n$. Prove that\n$$\n\\frac{a_1}{a_2} + \\frac{a_2}{a_3} + \\frac{a_3}{a_4} + \\dots + \\frac{a_{n-1}}{a_n} \\le \\frac{n}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $x_1 = a_1$, $x_k = a_k - (a_{k-1} + \\dots + a_1)$, $k = \\overline{2, n}$, and observe that $x_{k+1} - x_k = a_{k+1} - 2a_k$, $k = \\overline{1, n-1}$. It results\n$$\n2 \\left( \\frac{a_1}{a_2} + \\frac{a_2}{a_3} + \\frac{a_3}{a_4} + \\dots + \\frac{a_{n-1}}{a_n} \\right) = \\sum_{i=1}^{n-1} \\left( 1 - \\frac{x_{i+1} - x_i}{a_{i+1}} \\right).\n$$\n\n$$\nn - \\frac{x_1}{a_1} - \\sum_{i=1}^{n-1} \\frac{x_{i+1} - x_i}{a_{i+1}} = n - \\frac{x_n}{a_n} - \\sum_{i=1}^{n-1} x_i \\left( \\frac{1}{a_i} - \\frac{1}{a_{i+1}} \\right) \\le n,\n$$\nsince $x_i \\ge 0, \\forall i = \\overline{1, n}$ and $a_i \\le a_{i+1}, \\forall i = \\overline{1, n-1}$.\n\nEquality holds if and only if $x_2 = x_3 = \\dots = x_n = 0$, that is, $a_2 = a_1$, $a_3 = 2a_1, \\dots, a_n = 2^{n-2}a_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38437,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $S = \\{s_1, \\dots, s_k\\}$ of positive real numbers is *polygonal* if $k \\ge 3$ and there is a non-degenerate planar $k$-gon whose side lengths are exactly $s_1, \\dots, s_k$; the set $S$ is *multipolygonal* if in every partition of $S$ into two subsets, each of which has at least three elements, exactly one of these two subsets is polygonal. Fix an integer $n \\ge 7$.\n\na) Does there exist an $n$-element multipolygonal set, removal of whose maximal element leaves a multipolygonal set?",
    "options": [],
    "answer": "Yes",
    "solution": "Recall that a necessary and sufficient condition for $k \\ge 3$ positive real numbers $s_1, \\dots, s_k$ to be the side lengths of a non-degenerate planar $k$-gon is that a maximal $s_i$ be less than the sum of the other $s_j$.\n\na) The answer is in the affirmative. Given pairwise distinct positive real numbers $\\varepsilon_1, \\varepsilon_2, \\varepsilon_3$ less than $1/2$, we show that the sets $S = \\{1, 2, 4, \\dots, 2^{n-5}, 2^{n-4} + \\varepsilon_1, 2^{n-4} + \\varepsilon_2, 2^{n-4} + \\varepsilon_3, 2^{n-3} - 1/2\\}$ and $S \\setminus \\{2^{n-3} - 1/2\\}$ are both multipolygonal.\n\nSplit any of the two sets into two subsets each of which has at least three elements, let $A$ be the part containing at least two of the $2^{n-4} + \\varepsilon_i$, and let $B$ be the other part.\nThe set $A$ is polygonal since its maximal element is either one of the $2^{n-4} + \\varepsilon_i$ or $2^{n-3} - 1/2$, each of which is smaller than the sum of other elements in $A$.\nTo prove that $B$ is not polygonal, notice that its maximal element is either $2^{n-3} - 1/2$, or one of the $2^{n-4} + \\varepsilon_i$, or some $2^k$, $k \\le n-5$. In the first case, the sum of all other elements in $B$ is less than $1+2+\\dots+2^{n-5}+2^{n-4}+\\varepsilon_i = 2^{n-3}-1+\\varepsilon_i < 2^{n-3}-1/2$; in the second case, this sum does not exceed $1+2+\\dots+2^{n-5} = 2^{n-4}-1 < 2^{n-4}+\\varepsilon_i$; and in the third case, this sum is at most $1+2+\\dots+2^{k-1} = 2^k - 1 < 2^k$. Consequently, $B$ is not polygonal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38438,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that $7$ cannot be written as a sum of squares of three rational numbers.\n\nb) Let $a$ be a rational number that can be written as a sum of squares of three rational numbers. Prove that $a^m$ can be written as a sum of squares of three rational numbers, for any positive integer $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Suppose, by way of contradiction, that one can find rational numbers $x$, $y$, $z$ such that $7 = x^2 + y^2 + z^2$. Writing $x$, $y$, $z$ as fractions and clearing denominators yield an equality like\n$$\n7n^2 = a^2 + b^2 + c^2, \\quad (*)\n$$\nwhere $n$, $a$, $b$, $c$ are nonnegative integers, not all zero.\nIf $n$ is even, then $a$, $b$, $c$ are all even, as well (obviously, they cannot be all odd, and if exactly two of them are odd, the sum of their squares equals $2 \\pmod{4}$, while $7n^2$ equals $0 \\pmod{4}$). Dividing by $4$ yields\n$$\n7n_1^2 = a_1^2 + b_1^2 + c_1^2,\n$$\nand, clearly, $0 < a_1 + b_1 + c_1 < a + b + c$.\nIf $n_1$ is still even, we repeat the previous transformation (this can only happen finitely many times).\nIf $n$ is odd, say $n = 2k + 1$, then $7n^2 = 7 \\cdot 4k(k+1) + 7 = M8 + 7$. Since the remainder of a square when divided by $8$ equals $0$, $1$ or $4$, we deduce that equality $(*)$ is impossible.\n\n\nb) We induct on $m$. The base case $m = 1$ follows from the hypothesis. We assume the statement true for all $m \\le n$ and prove it for $n + 1$.\nIf $n + 1 = 2k$, then $k \\le n$, hence $a^k$ can be written as a sum of squares of three rational numbers, for instance $a^k = x^2 + y^2 + z^2$, where $x \\ge y \\ge z$. Then\n$$\na^{2k} = (x^2 + y^2 + z^2)^2 = (x^2 + y^2 - z^2)^2 + (2xz)^2 + (2yz)^2,\n$$\nand the numbers $x^2 + y^2 - z^2$, $2xz$, $2yz$ are obviously rational numbers.\n\nIf $n + 1 = 2k + 1$, using $a = x^2 + y^2 + z^2$, we get\n$$\na^{n+1} = a^{2k} \\cdot a = a^{2k} (x^2 + y^2 + z^2) = (a^k x)^2 + (a^k y)^2 + (a^k z)^2,\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive integers $a, b, c$ with the property $a + b + c = abc$.",
    "options": [],
    "answer": "All permutations of (1,2,3).",
    "solution": "If one of the numbers is $0$, then all are $0$.\nIf $abc \\neq 0$, then the relation can be written $\\frac{1}{bc} + \\frac{1}{ac} + \\frac{1}{ab} = 1$. Since the relation is symmetric in $a, b, c$, we may assume that $a \\le b \\le c$, whence $ab \\le ac \\le bc$.\nIf $ab > 3$, then $\\frac{1}{bc} + \\frac{1}{ac} + \\frac{1}{ab} < 1$.\nIf $ab = 2$, then $a = 1$ and $b = 2$, whence $\\frac{1}{2c} + \\frac{1}{c} = \\frac{1}{2}$, so $c = 3$.\n\nIf $ab = 3$, then $a = 1$ and $b = 3$, whence $\\frac{1}{3c} + \\frac{1}{c} = \\frac{2}{3}$, so $c = 2$, which contradicts $b < c$.\nFinally, the solutions are $(0, 0, 0)$ and all the permutations of $(1, 2, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38440,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ denote $x_n$ the number of the positive integers with $n$ digits, divisible with $4$, formed with digits $2$, $0$, $1$ or $6$.\n\na) Compute $x_1$, $x_2$, $x_3$ and $x_4$.\n\nb) Find $n$ so that\n$$\n1 + \\left\\lfloor \\frac{x_2}{x_1} \\right\\rfloor + \\left\\lfloor \\frac{x_3}{x_2} \\right\\rfloor + \\left\\lfloor \\frac{x_4}{x_3} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{x_{n+1}}{x_n} \\right\\rfloor = 2016.\n$$",
    "options": [],
    "answer": "x1=1, x2=4, x3=15, x4=60; n=504",
    "solution": "a) $x_1 = 1$ (0 is divisible with $4$), $x_2 = 4$ (the numbers $12$, $16$, $20$ and $60$ are divisible with $4$), $x_3 = 3 \\cdot 5$, (because the first digit cannot be $0$ and the last two can be $12$, $16$, $20$, $60$ and $00$), $x_4 = 3 \\cdot 4 \\cdot 5 = 60$ (because the first digit cannot be $0$, for the second digit there are $4$ possibilities and the last two digits can be $12$, $16$, $20$, $60$ and $00$).\n\nb) If $n \\ge 3$, then a number $A$ which fulfills the hypothesis is of the form\n$$\nA = \\overline{a_1 a_2 \\dots a_{n-2} p q},\n$$\nwhere its first digit can have three values, each of the digits $a_2$, $a_3$, $\\dots$, $a_{n-2}$ can be chosen in $4$ ways and the last two digits can be $12$, $16$, $20$, $60$ or $00$. So $x_n = 3 \\cdot 4^{n-3} \\cdot 5$, for every $n \\ge 3$.\nFor $n \\ge 3$, $\\frac{x_{n+1}}{x_n} = 4$, whence $1 + \\left\\lfloor \\frac{x_2}{x_1} \\right\\rfloor + \\left\\lfloor \\frac{x_3}{x_2} \\right\\rfloor + \\left\\lfloor \\frac{x_4}{x_3} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{x_{n+1}}{x_n} \\right\\rfloor = 1 + 4 + 3 + 4(n-2)$, $4n = 2016$, $n = 504$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $S_1, \\dots, S_n$ be a collection of finite non-empty sets such that\n$$\n\\sum_{1 \\le i < j \\le n} \\frac{|S_i \\cap S_j|}{|S_i| |S_j|} < 1.\n$$\n\nProve that there exist pairwise distinct elements $x_1, \\dots, x_n$ such that $x_i$ is a member of $S_i$ for each index $i$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A *choice function* or simply a *choice* for the collection $S_1, \\dots, S_n$ is a function $c$ from the first $n$ positive integers to the union $S_1 \\cup \\dots \\cup S_n$ such that $c(i)$ is a member of $S_i$ for each $i$. We must show that an injective choice is always possible under the conditions in the statement. To this end, we prove that the number of non-injective choices is strictly less than $|S_1| \\cdots |S_n|$, the total number of possible choices.\n\nIndeed, a non-injective choice function sends some $i$ and some $j \\neq i$ to the same element necessarily lying in $S_i \\cap S_j$, so the number of non-injective choices does not exceed\n$$\n\\sum_{1 \\le i < j \\le n} |S_i \\cap S_j| \\; |S_1| \\cdots |\\hat{S}_i| \\cdots |\\hat{S}_j| \\cdots |S_n| = |S_1| \\cdots |S_n| \\sum_{1 \\le i < j \\le n} \\frac{|S_i \\cap S_j|}{|S_i| |S_j|} < |S_1| \\cdots |S_n|;\n$$\nwhere the hat over $S_i$ and $S_j$ means that these sets are to be omitted. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2015 \\in M$ and $2016 \\notin M$, where\n$$\nM = \\{ x_1 + 2x_2 + 3x_3 + \\dots + 2015x_{2015} \\mid x_1, x_2, \\dots, x_{2015} \\in \\{-2, 3\\} \\}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "An integer $n$ belongs to $M$ if and only if there exists disjoint subsets $A$, $B$ of the set $S = \\{1, 2, \\dots, 2015\\}$, with $A \\cup B = S$, so that $-2a + 3b = n$, where $a$ is the sum of the elements of $A$ and $b$ is the sum of the elements of $B$ (the sum of the elements of the empty set being $0$).\n\nThen $n = 5b - 2(a+b)$ and, since $a+b = 1+2+\\dots+2015$, it follows that\n$n = 5b - 2015 \\cdot 2016$.\n\nThis shows that $n$ is divisible by $5$, hence $2016 \\notin M$.\n\nIn order to prove that $2015 \\in M$, it is enough to find $B \\subset \\{1, 2, \\dots, 2015\\}$ with the sum of its elements $b = \\frac{1}{5}(2015 \\cdot 2016 + 2015) = 403 \\cdot 2017$.\n\nAn example is when $B$ is the union of $403$ pairs of elements of $S$ with sum $2017$, for instance $(2, 2015)$, $(3, 2014)$, ..., $(404, 1613)$.\n\nSo, taking $x_2 = x_3 = \\dots = x_{404} = x_{1613} = x_{1614} = \\dots = x_{2015} = 3$ and $x_1 = x_{405} = x_{406} = \\dots = x_{1612} = -2$, we get $2015 \\in M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$. Let $C_1$, $D_1$ and $M$ be the midpoints of segments $CX$, $DX$ and $CD$, respectively. Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$, respectively. Prove that line $XY$ is tangent to the circle through $E$, $F$ and $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nWe are to prove that $\\angle EXY = \\angle EFX$; alternatively, but equivalently, $\\angle AYX + \\angle XAY = \\angle BYF + \\angle XBY$.\n\nSince the quadrangle $ABCD$ is cyclic, the triangles $XAD$ and $XBC$ are similar, and since $AD_1$ and $BC_1$ are corresponding medians in these triangles, it follows that $\\angle XAY = \\angle XAD_1 = \\angle XBC_1 = \\angle XBY$.\n\nFinally, $\\angle AYX = \\angle BYF$, since $X$ and $M$ are corresponding points in the similar triangles $ABY$ and $C_1D_1Y$: indeed, $\\angle XAB = \\angle XDC = \\angle MC_1D_1$, and $\\angle XBA = \\angle XCD = \\angle MD_1C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c \\in \\mathbb{C}^*$, be distinct complex numbers with equal moduli, such that $a^2 + b^2 + c^2 + ab + ac + bc = 0$. Prove that $a$, $b$, $c$ are the complex coordinates of the vertices of a triangle which is either right angled or equilateral.\n\nMarian Ionescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As usual, we could suppose that $|a| = |b| = |c| = 1$. The equality in the hypothesis is equivalent with $(a + b + c)^2 = ab + bc + ca$, and thus $(a + b + c)^2 = abc\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)$, which is the same with $(a + b + c)^2 = abc(a + b + c)$. It is clear now that $|a + b + c| \\in \\{0, 1\\}$.\n\nIf $|a + b + c| = 0$, then the orthocenter of the triangle with the vertices whose complex coordinates are $a$, $b$, $c$, coincides with its circumcenter, and thus the triangle is equilateral.\n\n\nIf $|a + b + c| = 1$, then $(a + b + c) \\overline{(a + b + c)} = 1$, and thus\n$$\n(a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 1,\n$$\nthat is $(a + b)(b + c)(c + a) = 0$.\nThen, at least one of the above parenthesis is zero, and thus the triangle has two antipodal vertices, that is it is a right angled triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n$ be positive integers and $x, y, z \\in [0, 1]$ be real numbers. Prove that\n$$\n0 \\le x^{m+n} + y^{m+n} + z^{m+n} - x^m y^n - y^m z^n - z^m x^n \\le 1\n$$\nand find when equality holds.",
    "options": [],
    "answer": "The expression lies between zero and one for all inputs. The minimum zero occurs exactly when all three numbers are equal. The maximum one occurs exactly for the six triples obtained by permuting either one coordinate equal to one with the other two zero, or two coordinates equal to one with the remaining one zero: (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1).",
    "solution": "Without loss of generality, we may assume that $x$ is the largest of $x, y, z$; then we have:\n$x^{m+n} + y^{m+n} + z^{m+n} - x^m y^n - y^m z^n - z^m x^n = (x^m - z^m)(x^n - y^n) + (y^m - z^m)(y^n - z^n) \\ge 0$.\nThe minimum is $0$ and is obtained if $x = \\max\\{y, z\\}$ and $y = z$, so if $x = y = z$.\n\nThe maximum is achieved for $x = 1$, and its value is\n$$\n1 + y^{m+n} + z^{m+n} - y^n - y^m z^n - z^m = 1 - y^n(1 - y^m) - z^m(1 - z^n) - y^m z^n \\le 1,\n$$\nwith equality only if one of $y$ or $z$ is $0$ and the other one is $0$ or $1$.\n\nTherefore, there are $6$ equality cases, namely\n$$\n(x, y, z) \\in \\{(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(X, Y)$ of sets with positive integer elements, which fulfill the following conditions:\n(1) each of the sets $X$ and $Y$ has three elements;\n(2) $3 \\in X$ and $5 \\in Y$;\n(3) the set $X \\cap Y$ has exactly one element;\n(4) if $a$ and $b$ are distinct elements of $X$, then $(a+b) \\in Y$.",
    "options": [],
    "answer": "({1,2,3}, {3,4,5}); ({1,3,4}, {4,5,7}); ({2,3,5}, {5,7,8})",
    "solution": "If $X = \\{a, b, c\\}$, with $a < b < c$, then $a+b < a+c < b+c$ are distinct elements of $Y$. Therefore $Y = \\{a+b, b+c, c+a\\}$.\nSince $a < b < a+b < a+c < b+c$, the common element of $X$ and $Y$ can be only $c = a+b$.\n\nIf $c = 3$, then $a = 1, b = 2$, therefore $X = \\{1, 2, 3\\}, Y = \\{3, 4, 5\\}$ which fulfill the given conditions ($5 \\in Y$ and $X \\cap Y = \\{5\\}$).\nIf $b = 3$, then $a \\in \\{1, 2\\}$. If $a = 1$, then $c = a + b = 4$, hence $X = \\{1, 3, 4\\}, Y = \\{4, 5, 7\\}$. If $a = 2$, then $c = 5$, hence $X = \\{2, 3, 5\\}, Y = \\{5, 7, 8\\}$. Both pairs of sets fulfill the given conditions.\nIf $a = 3$, then $b > 3$ and $c > 3$, therefore all the elements of $Y$ are larger than 5, so there are no solutions in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $m$ and $n \\ge m$, determine the largest number of dominoes ($1 \\times 2$ or $2 \\times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2n$ columns consisting of unit cells ($1 \\times 1$ squares) so that:\n1) each domino covers exactly two adjacent cells of the board;\n2) no two dominoes overlap;\n3) no two form a $2 \\times 2$ square;\n4) the bottom row of the board is completely covered by $n$ dominoes.",
    "options": [],
    "answer": "mn - floor(m/2)",
    "solution": "The required maximum is $mn - \\lfloor m/2 \\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.\nTo show that the number of dominoes in an arrangement satisfying the conditions in the statement does not exceed $mn - \\lfloor m/2 \\rfloor$, label the rows upwards $0, 1, \\ldots, m-1$, and, for each $i$ in this range, draw a vertically symmetric block of $n-i$ fictitious horizontal dominoes in the $i$-th row (so the block on the $i$-th row leaves out $i$ cells on either side) – Figure 4 illustrates the case $m=n=6$. A fictitious domino is *good* if it is completely covered by a domino in the arrangement; otherwise, it is *bad*.\nIf the fictitious dominoes are all good, then the dominoes in the arrangement that cover no fictitious domino, if any, all lie in two triangular regions of side-length $m-1$ at the upper-left and upper-right corners of the board. Colour the cells of the board chess-like and notice that in each of the two triangular regions the number of black cells and the number of white cells differ by $\\lfloor m/2 \\rfloor$. Since each domino covers two cells of different colours, at least $\\lfloor m/2 \\rfloor$ cells are not covered in each of these regions, and the conclusion follows.\n\nTo deal with the remaining case where bad fictitious dominoes are present, we show that an arrangement satisfying the conditions in the statement can be transformed into another such with at least as many dominoes, but fewer bad fictitious dominoes. A finite number of such transformations eventually leads to an arrangement of at least as many dominoes all of whose fictitious dominoes are good, and the conclusion follows by the preceding.\n\nConsider the row of minimal rank containing bad fictitious dominoes – this is certainly not the bottom row – and let $D$ be one such. Let $\\ell$, respectively $r$, be the left, respectively right, cell of $D$ and notice that the cell below $\\ell$, respectively $r$, is the right, respectively left, cell of a domino $D_1$, respectively $D_2$, in the arrangement.\nIf $\\ell$ is covered by a domino $D_\\ell$ in the arrangement, since $D$ is bad and no two dominoes in the arrangement form a square, it follows that $D_\\ell$ is vertical. If $r$ were also covered by a domino $D_r$ in the arrangement, then $D_r$ would also be vertical, and would therefore form a square with $D_\\ell$ – a contradiction. Hence $r$ is not covered, and there is room for $D_\\ell$ to be placed so as to cover $D$, to obtain a new arrangement satisfying the conditions in the statement; the latter has as many dominoes as the former, but fewer bad fictitious dominoes. The case where $r$ is covered is dealt with similarly.\nFinally, if neither cell of $D$ is covered, addition of an extra domino to cover $D$ and, if necessary, removal of the domino above $D$ to avoid formation of a square, yields a new arrangement satisfying the conditions in the statement; the latter has at least as many dominoes as the former, but fewer bad fictitious dominoes. (Figure 5 illustrates the two cases.)\n*Alternative Solution.* (sketch by Ilya Bogdanov) We present an alternative proof of the bound.\n\nLabel the rows upwards $0, 1, \\dots, m-1$, and the columns from the left to the right by $0, 1, \\dots, 2n-1$; label each cell by the pair of its column's and row's numbers, so that $(1, 0)$ is the second left cell in the bottom row. Colour the cells chess-like so that $(0, 0)$ is white. For $0 \\le i \\le n-1$, we say that the *$i$th white diagonal* is the set of cells of the form $(2i + k, k)$, where $k$ ranges over all appropriate indices. Similarly, the *$i$th black diagonal* is the set of cells of the form $(2i + 1 - k, k)$. (Notice that the white cells in the upper-left corner and the black cells in the upper-right corner are not covered by these diagonals.)\n\n**Claim.** Assume that $K$ lowest cells of some white diagonal are all covered by dominoes. Then all these $K$ dominoes face right or up from the diagonal. (In other words, the black cell of any such domino is to the right or to the top of its white cell.) Similarly, if $K$ lowest cells of some black diagonal are covered by dominoes, then all these dominoes face left or up from the diagonal.\n\n*Proof.* By symmetry, it suffices to prove the first statement. Assume that $K$ lowest cells of the $i$th white diagonal is completely covered. We prove by induction on $k < K$ that the required claim holds for the domino covering $(2i + k, k)$. The base case $k = 0$ holds due to the problem condition. To establish the step, one observes that if $(2i + k, k)$ is covered by a domino facing up or right, while $(2i + k + 1, k + 1)$ is covered by a domino facing down or left, then these two dominoes form a square.\n\nWe turn to the solution. We will prove that there are at least $d = \\lfloor m/2 \\rfloor$ empty white cells. Since each domino covers exactly one white cell, the required bound follows.\nIf each of the first $d$ white diagonals contains an empty cell, the result is clear. Otherwise, let $i < d$ be the least index of a completely covered white diagonal. We say that the dominoes covering our diagonal are *distinguished*. After removing the distinguished dominoes, the board splits into two parts; the left part $L$ contains $i$ empty white cells on the previous diagonals. So, it suffices to prove that the right part $R$ contains at least $d - i$ empty white cells.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38448,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle, three of the adjacent angles around the orthocenter, made by the altitudes, have their measures directly proportional with $5$, $5$ and $7$, and the sum of the other three angles is $190\\degree$. Find the measure of the angles of the triangle.\nConstantin Apostol",
    "options": [],
    "answer": "50°, 60°, 70°",
    "solution": "The sum of the measures of the angles directly proportional with $5$, $5$ and $7$ is $170\\degree$. Their measures are $5x$, $5x$, $7x$, with $5x + 5x + 7x = 170\\degree$, hence $x = 10\\degree$.\n\nThere are three pairs of opposite angles around the orthocenter, so they must be equal. On one side of an altitude must be one angle of each pair. Since three of the angles have measures $50\\degree$, $50\\degree$, $70\\degree$, the other three must have measures $70\\degree$, $60\\degree$, $60\\degree$.\n\n![](attached_image_1.png)\n\nUsing these angles we find the measures of the angles between the altitudes and the sides: $20\\degree$, $30\\degree$, $40\\degree$, hence the angles of the triangle have measures $50\\degree$, $60\\degree$, $70\\degree$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to (0, \\infty)$ be a continuous periodic function. Under the assumption that $2$ is a period of $f$, prove that:\n\na.\n$$\n\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx \\ge 2;\n$$\n\nb.\n$$\n\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx = 2 \\text{ if and only if } 1 \\text{ is a period of } f.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) To establish the required inequality write:\n$$\n\\begin{align*}\n\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx &= \\int_0^1 \\frac{f(x+1)}{f(x)} \\, dx + \\int_1^2 \\frac{f(x+1)}{f(x)} \\, dx \\\\\n&= \\int_0^1 \\frac{f(x+1)}{f(x)} \\, dx + \\int_1^2 \\frac{f(x-1+2)}{f(x-1+1)} \\, dx \\\\\n&= \\int_0^1 \\frac{f(x+1)}{f(x)} \\, dx + \\int_0^1 \\frac{f(x+2)}{f(x+1)} \\, dx \\\\\n&= \\int_0^1 \\left( \\frac{f(x+1)}{f(x)} + \\frac{f(x)}{f(x+1)} \\right) dx \\\\\n&\\ge \\int_0^1 2 \\, dx = 2.\n\\end{align*}\n$$\n\n(b) If $1$ is a period of $f$, then $\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx = \\int_0^2 dx = 2$. Conversely, if $\\int_0^2 \\frac{f(x+1)}{f(x)} \\, dx = 2$, then\n$$\n\\int_0^1 \\left( \\sqrt{\\frac{f(x+1)}{f(x)}} - \\sqrt{\\frac{f(x)}{f(x+1)}} \\right)^2 dx = 0,\n$$\nso $\\sqrt{\\frac{f(x+1)}{f(x)}} = \\sqrt{\\frac{f(x)}{f(x+1)}}$, $0 \\le x \\le 1$, by continuity, and $f(x+1) = f(x)$, for all $x \\in [0, 1]$. If $1 \\le x \\le 2$, then $f(x) = f((x-1)+1) = f(x-1) = f((x-1)+2) = f(x+1)$, so $f(x+1) = f(x)$ for all $x \\in [0, 2]$. Finally, if $x$ is any real number, then $2n \\le x < 2n+2$ for some integer $n$, and $f(x) = f(x-2n) = f((x-2n)+1) = f((x+1)-2n) = f(x+1)$. Consequently, $1$ is a period of $f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider matrices $A$, $B$, $C$, $D \\in \\mathcal{M}_n(\\mathbb{C})$, $n \\ge 2$ și $k \\in \\mathbb{R}$ so that $AC + kBD = I_n$ and $AD = BC$. Demonstrate that $CA + kDB = I_n$ and $DA = CB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n(CA + kDB) - w(DA - CB) = I_n\n$$\nand\n$$\n(CA + kDB) + w(DA - CB) = I_n,\n$$\nwhich give the result. For $k = 0$ we get $AC = I_n$, so $CA = I_n$. From $AD = BC$ and $CA = I_n$ we obtain $ADA = B$ and $DA = CB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(a^2) - f(b^2) \\leq (f(a) + b)(a - f(b)), \\quad \\text{for all } a, b \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "f(x) = x for all real x, and f(x) = -x for all real x",
    "solution": "Plugging $a = b = 0$ yields $f^2(0) \\leq 0$, hence $f(0) = 0$. Again, plugging $b = 0$, and then $a = 0$, we obtain $f(a^2) \\leq af(a)$, $\\forall a \\in \\mathbb{R}$ and $f(b^2) \\geq bf(b)$, $\\forall b \\in \\mathbb{R}$, therefore $f(x^2) = xf(x)$, $\\forall x \\in \\mathbb{R}$. Replacing the latter in the given equation, we get $f(a)f(b) \\leq ab$, $\\forall a, b \\in \\mathbb{R}$.\nAlso, $-xf(-x) = f((-x)^2) = f(x^2) = xf(x)$, hence $f$ is an odd function.\nThe last two relations give $f(a)f(b) = -f(a)f(-b) \\geq -(-ab) = ab$, so $f(a)f(b) = ab$, $\\forall a, b \\in \\mathbb{R}$.\nThus, $f^2(1) = 1$, which implies $f(1) = \\pm 1$ and either $f(x) = x, \\forall x \\in \\mathbb{R}$, or $f(x) = -x, \\forall x \\in \\mathbb{R}$.\n\nIt is easy to check that both functions are solutions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\geq 2$, and $a_1, a_2, ..., a_n \\in (0, \\infty)$ such that $a_1 \\cdot a_2 \\cdot ... \\cdot a_n = 1$. Prove that the function\n$$\nf : (0, \\infty) \\to \\mathbb{R}, f(x) = (1 + a_1^x) \\cdot (1 + a_2^x) \\cdot \\dots \\cdot (1 + a_n^x)\n$$\nis increasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Multiplying out yields\n$$\n\\begin{aligned}\nf(x) &= (1 + a_1^x) \\cdot (1 + a_2^x) \\cdot \\dots \\cdot (1 + a_n^x) \\\\\n&= 1 + \\sum_{1 \\leq i \\leq n} a_i^x + \\sum_{1 \\leq i < j \\leq n} a_i^x a_j^x + \\dots + a_1^x a_2^x \\dots a_n^x.\n\\end{aligned}\n$$\n\nSince $a_1 \\cdot a_2 \\cdot \\dots \\cdot a_n = 1$, the function $f$ is a sum of terms of the type $a^x + \\frac{1}{a^x}$, where $a = a_{i_1} a_{i_2} \\dots a_{i_k}$, with $k \\le n$.\nNow, for $a > 0$, we consider the function $g : (0, \\infty) \\to \\mathbb{R}$, $g(x) = a^x + a^{-x}$. Let $t > 0$; then\n$$\n\\begin{aligned}\ng(x+t) - g(x) &= \\frac{a^{2x+2t} + 1}{a^{x+t}} - \\frac{a^{2x} + 1}{a^x} = \\frac{a^{2x+2t} - a^{2x+t} - a^t + 1}{a^{x+t}} = \\\\\n&= \\frac{(a^{2x+t} - 1)(a^t - 1)}{a^{x+t}} \\ge 0,\n\\end{aligned}\n$$\nhence $g$ is increasing. The conclusion is obvious.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function with the following properties:\n$$\n\\begin{aligned}\nf(x+y) &\\le f(x) + f(y), \\\\\nf(tx + (1-t)y) &\\le t f(x) + (1-t) f(y),\n\\end{aligned}\n$$\nfor all $x, y \\in \\mathbb{R}$ and $t \\in [0, 1]$.\n\na) Prove that for any $a \\le b \\le c \\le d$, such that $d - c = b - a$, the following inequality holds\n$$\nf(b) + f(c) \\le f(a) + f(d).\n$$\n\nb) Prove that\n$$\n\\begin{aligned}\nf(x_1 + x_2 + \\dots + x_n) + (n-2)(f(x_1) + f(x_2) + \\dots + f(x_n)) &\\ge \\\\\n&\\ge \\sum_{1 \\le i < j \\le n} f(x_i + x_j),\n\\end{aligned}\n$$\nfor all $n \\in \\mathbb{N}$, $n \\ge 3$, and $x_1, x_2, \\dots, x_n \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $t = \\frac{d-b}{a-d} = \\frac{c-a}{a-d} \\in [0, 1]$, then $b = t a + (1-t) d$ and $c = (1-t) a + t d$.\nUsing (P2) yields\n$$\nf(b) = f(t a + (1-t) d) \\le t f(a) + (1-t) f(d).\n$$\nSimilarly, $f(c) \\le (1-t) f(a) + t f(d)$. Adding up these two inequalities gives the result.\n\n\nb) We will prove the inequality by induction and we start with the base case $n = 3$. Let $x, y, z$ be arbitrary real numbers. Two of them, say $x, y$, have the same sign. If $x \\ge 0$ then $z + (x + y + z) = (x + z) + (y + z)$ and $z \\le x + z, y + z \\le x + y + z$. We deduce from a) that\n$$\nf(x+z) + f(y+z) \\le f(z) + f(x+y+z).\n$$\nAdding this with $f(x+y) \\le f(x) + f(y)$ yields the result.\n\nIf $x < 0$, then $x + y + z \\le y + z$, $z + x \\le z$ and we proceed in a similar manner.\nNow, assume the assertion true for $n$ and consider the numbers $x_1, x_2, \\dots, x_{n+1}$. Applying the inductive hypothesis to the numbers $x_1, x_2, \\dots, x_{n-1}$ and $x_n + x_{n+1}$ leads to\n$$\nf(x_1 + \\dots + x_n + x_{n+1}) + (n-2)(f(x_1) + \\dots + f(x_{n-1}) + f(x_n + x_{n+1})) \\ge \\sum_{1 \\le i < j \\le n-1} f(x_i + x_j) + \\sum_{1 \\le i \\le n-1} f(x_i + x_n + x_{n+1}).\n$$\nThe conclusion is obtained by adding up the inequalities $f(x_i + x_n + x_{n+1}) \\ge f(x_i + x_n) + f(x_i + x_{n+1}) + f(x_n + x_{n+1}) - f(x_i) - f(x_n) - (x_{n+1})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38454,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $O$ be the origin of the complex plane and consider the points $A$ and $B$, whose complex coordinates are $a$ and $b$, respectively. Prove that $[OAB] = \\frac{1}{4} |\\bar{a}b - \\bar{a}\\bar{b}|$, where $[OAB]$ denotes the area of triangle $OAB$.\n\nb) Let $ABC$ be an equilateral triangle, $C$ its circumcircle, and $O$ its circumcenter. If a point $P$ lies in the interior of $C$, let $S(P)$ denote the area of the triangle whose side lengths equal the distances from $P$ to the triangle's sides². Let $P_1$ and $P_2$ be two distinct points in the interior of $C$. Prove that $S(P_1) = S(P_2)$ if and only if $OP_1 = OP_2$.\n\n²The existence of such a triangle is a famous result of the Romanian mathematician **Dimitrie Pompeiu** (1873-1954).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If the triangle $OAB$ is oriented counterclockwise, then $m(\\angle AOB) = \\arg\\frac{b}{a}$, otherwise $m(\\angle AOB) = \\arg\\frac{a}{b}$. Then\n$$\n\\sin(\\angle AOB) = \\left| \\frac{|a|}{2|b|} \\left( \\frac{b}{a} - \\frac{\\bar{b}}{\\bar{a}} \\right) \\right| = \\frac{|\\overline{ab} - \\overline{ab}|}{2|a||b|}\n$$\nand since $[OAB] = \\frac{1}{2} OA \\cdot OB \\cdot \\sin(\\angle AOB)$, we obtain the conclusion.\n\n\nb) Let $\\varepsilon = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$. We may assume that the complex coordinates of the points $A$, $B$ and $C$ are $1$, $\\varepsilon$, and $\\varepsilon^2$, respectively. Let $P$ be a point with complex coordinate $p$, in the interior of the circle. Then $(p-1) + \\varepsilon(p-\\varepsilon) + \\varepsilon^2(p-\\varepsilon^2) = 0$. Consider the points $D(p-1)$, $E(\\varepsilon(p-\\varepsilon))$ and $F(\\varepsilon^2(p-\\varepsilon^2))$. Observe that, $OD = PA$, $OE = PB$ and $OF = PC$.\nLet $J$ be a point such that $ODJF$ is a parallelogram. We obtain that the complex coordinates of points $J$ and $F$ are opposite numbers. Then $OJ = PC$, and the side lengths of $ODJ$ are $PA$, $PB$, and $PC$. But\n$$\n\\begin{aligned}\n[ODJ] &= [ODE] = \\frac{1}{4} |(\\bar{p}-1) \\cdot \\varepsilon \\cdot (p-\\varepsilon) - (p-1) \\cdot \\bar{\\varepsilon} \\cdot (\\bar{p}-\\bar{\\varepsilon})| \\\\\n&= \\frac{1}{4} ((\\varepsilon - \\varepsilon^2) |p|^2 - (\\varepsilon - \\varepsilon^2)) = \\frac{\\sqrt{3}}{4} |p|^2 - 1\n\\end{aligned}\n$$\nSince $|p| < 1$, we obtain $[ODE] = S(P) = \\frac{\\sqrt{3}}{4} (1 - |p|^2)$, from which the conclusion follows easily.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38455,
    "subject": "Mathematics (Multi-modal)",
    "question": "The people of an ancient tribe used a language in which the words were formed with two letters only: $A$ and $B$. Researchers discovered that any two words of equal length differ in at least three positions. For instance, the words $ABBAA$ and $AAAAB$ differ in positions $2$, $3$ and $5$, that is, in three positions.\n\nLet $n \\in \\mathbb{N}$, $n \\ge 3$. Prove that this language cannot contain more than $\\left\\lfloor \\frac{2^n}{n+1} \\right\\rfloor$ words of length $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we denote by $C$ the set of all possible words of length $n$ (not necessarily from the language), we have $\\text{Card}(C) = 2^n$. If $x$ and $y$ are arbitrary words from $C$, let $d(x, y)$ be the number of positions at which the corresponding letters are different (the Hamming distance). Obviously, $d(x, x) = 0$ and $d(x, y) = d(y, x)$. For each $x \\in C$, we define the set $C_x = \\{y \\in C \\mid d(x, y) \\le 1\\}$. It is not difficult to see that $\\text{Card}(C_x) = n + 1$.\n\nIf $a, b$ are words of length $n$ from the given language, then $d(a, b) \\ge 3$, hence $C_a \\cap C_b = \\emptyset$.\n\nLet $D$ be the set of all words of length $n$ from the language. Then $\\bigcup_{a \\in D} C_a \\subset C$, which implies $\\text{Card}\\left(\\bigcup_{a \\in D} C_a\\right) \\le \\text{Card}(C)$. But $\\text{Card}\\left(\\bigcup_{a \\in D} C_a\\right) = (n+1) \\cdot \\text{Card}(D)$, so that $(n + 1) \\cdot \\text{Card}(D) \\le 2^n$, therefore $\\text{Card}(D) \\le \\frac{2^n}{n+1}$, and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_2(\\mathbb{R})$ be a matrix satisfying the conditions:\n$$\n\\det (A^{2014} - I_2) = \\det (A^{2014} + I_2) \\quad \\text{and} \\quad \\det (A^{2016} - I_2) = \\det (A^{2016} + I_2).\n$$\n\nProve that $\\det (A^n - I_2) = \\det (A^n + I_2)$, for any positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $M \\in \\mathcal{M}_2(\\mathbb{R})$, we have\n$$\n\\det (M - xI_2) = x^2 - \\operatorname{tr}(M)x + \\det(M), \\quad \\forall x \\in \\mathbb{R}. \\quad (1)\n$$\nLet $A \\in \\mathcal{M}_2(\\mathbb{R})$ be a matrix satisfying the hypothesis. From (1) we obtain\n$$\n\\operatorname{tr}(A^{2014}) = \\operatorname{tr}(A^{2016}) = 0. \\qquad (2)\n$$\nUsing the Cayley-Hamilton theorem, we obtain\n$$\nA^2 - \\operatorname{tr}(A)A + \\det(A)I_2 = O_2. \\tag{3}\n$$\nWe analyze several cases.\n\nIf $\\operatorname{tr}(A) = 0$, then (3) implies $A^2 = -\\det(A)I_2$, whence $A^{2014} = (-\\det(A))^{1007}I_2$. From (2) it results $\\det(A) = 0$.\n\nIf $\\det(A) = 0$, then this time (3) implies $A^2 = \\operatorname{tr}(A)A$. Inductively, $A^{n+1} = \\operatorname{tr}^n(A)A$, $\\forall n \\in \\mathbb{N}^*$. It follows that $\\operatorname{tr}(A^n) = \\operatorname{tr}^n(A)$, $\\forall n \\in \\mathbb{N}^*$. As a special case, using (2), $0 = \\operatorname{tr}(A^{2014}) = \\operatorname{tr}^{2014}(A)$, whence $\\operatorname{tr}(A) = 0$.\n\nIf $\\operatorname{tr}(A) \\neq 0$ and $\\det(A) \\neq 0$. From (3), $A^{2016} - \\operatorname{tr}(A)A^{2015} + \\det(A)A^{2014} = O_2$. Then $\\operatorname{tr}(A^{2016}) - \\operatorname{tr}(A)\\operatorname{tr}(A^{2015}) + \\det(A)\\operatorname{tr}(A^{2014}) = 0$. Using (2) and our assumption, we get $\\operatorname{tr}(A^{2015}) = 0$. From (3), it follows that\n$$\n\\operatorname{tr}(A^n) = \\frac{1}{\\operatorname{det}(A)} \\left[ \\operatorname{tr}(A) \\operatorname{tr}(A^{n+1}) - \\operatorname{tr}(A^{n+2}) \\right], \\quad n \\in \\mathbb{N}^*.\n$$\nThus, starting with $\\operatorname{tr}(A^{2014}) = \\operatorname{tr}(A^{2015}) = 0$, we obtain recursively $\\operatorname{tr}(A^{2013}) = 0$, $\\operatorname{tr}(A^{2012}) = 0$, ..., $\\operatorname{tr}(A) = 0$, a contradiction. It follows that $\\operatorname{tr}(A) = 0$ and $\\det(A) = 0$.\n\nFinally, from (3) it follows that $A^2 = O_2$, hence $A^n = O_2$, $\\forall n \\ge 2$. Then, for $n \\ge 2$, we have $\\det(A^n - I_2) = \\det(-I_2) = 1 = \\det(I_2) = \\det(A^n + I_2)$. For $n = 1$, from (1), $\\det(A - I_2) = 1 = \\det(A + I_2)$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_n(\\mathbb{C})$ be an invertible matrix, with lines $L_1, L_2, \\dots, L_n$. Consider the matrices $B \\in \\mathcal{M}_n(\\mathbb{C})$ with lines $O, L_2, \\dots, L_n$ and $C \\in \\mathcal{M}_n(\\mathbb{C})$ with lines $L_2, \\dots, L_n, O$, where $O$ denotes a line all of whose entries are zero. Let $D = A^{-1} \\cdot B$ and $E = A^{-1} \\cdot C$. Prove that:\na) $\\operatorname{rank}(D) = \\operatorname{rank}(D^2) = \\dots = \\operatorname{rank}(D^n);$\nb) $\\operatorname{rank}(E) > \\operatorname{rank}(E^2) > \\dots > \\operatorname{rank}(E^n).$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Consider the matrix $P \\in \\mathcal{M}_n(\\mathbb{C})$,\n$$\nP = \\begin{pmatrix}\n0 & 0 & 0 & \\cdots & 0 \\\\\n0 & 1 & 0 & \\cdots & 0 \\\\\n0 & 0 & 1 & \\cdots & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & 0 & \\cdots & 1\n\\end{pmatrix}.\n$$\nIt is not difficult to see that $D = A^{-1}PA$. Since $P^2 = P$, a standard inductive argument shows that $P^k = P$, $k \\in \\mathbb{N}^*$. Atunci $D^k = (A^{-1}PA)^k = A^{-1}P^kA = A^{-1}PA = D$, $k \\in \\mathbb{N}^*$. Therefore, $\\operatorname{rank}(D^k) = \\operatorname{rank}(D) = \\operatorname{rank}(P) = n-1$, $k = 1, \\dots, n$.\n\nb) Let $Q \\in \\mathcal{M}_n(\\mathbb{C})$,\n$$\nQ = \\begin{pmatrix}\n0 & 1 & 0 & \\cdots & 0 \\\\\n0 & 0 & 1 & \\cdots & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & 0 & \\cdots & 1 \\\\\n0 & 0 & 0 & \\cdots & 0\n\\end{pmatrix}.\n$$\nAgain, it is easy to check that $E = A^{-1}QA$.\nThe equality $E^k = (A^{-1}QA)^k = A^{-1}Q^kA$, $k \\in \\mathbb{N}^*$ implies\n$$ \\operatorname{rank}(E^k) = \\operatorname{rank}(Q^k) = n-k, \\quad k = 1, \\dots, n, $$\nhence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\in \\mathbb{R}$ and $f : (0, \\infty) \\to (0, \\infty)$. Prove that the following two statements are equivalent:\n(i) $\\lim_{x \\to \\infty} \\frac{f(x)}{x^{a+\\varepsilon}} = 0$ and $\\lim_{x \\to \\infty} \\frac{f(x)}{x^{a-\\varepsilon}} = \\infty$, for all $\\varepsilon > 0$;\n(ii) $\\lim_{x \\to \\infty} \\frac{\\ln f(x)}{\\ln x} = a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i)⇒(ii). Let $\\varepsilon > 0$; according to (i), there exists $m_1 > 0$ such that $\\frac{f(x)}{x^{a+\\varepsilon}} < 1, \\forall x > m_1$ and $m_2 > 0$ such that $\\frac{f(x)}{x^{a-\\varepsilon}} > 1, \\forall x > m_2$. If we denote $m = \\max\\{m_1, m_2, 1\\}$, then $x^{a-\\varepsilon} < f(x) < x^{a+\\varepsilon}, \\forall x > m$. This implies $a - \\varepsilon < \\frac{\\ln f(x)}{\\ln x} < a + \\varepsilon, \\forall x > m$. Since $\\varepsilon > 0$ is arbitrarily chosen, it follows that $\\lim_{x \\to \\infty} \\frac{\\ln f(x)}{\\ln x} = a$.\n\n(ii)⇒(i). Let $\\varepsilon > 0$. From (ii), we derive the existence of some $m > 1$ such that\n$$ a - \\frac{\\varepsilon}{2} < \\frac{\\ln f(x)}{\\ln x} < a + \\frac{\\varepsilon}{2}, \\forall x > m. $$\nThis yields $x^{a-\\varepsilon/2} < f(x) < x^{a+\\varepsilon/2}, \\forall x > m$, and therefore $\\frac{f(x)}{x^{a+\\varepsilon}} < \\frac{1}{x^{\\varepsilon/2}},$ for all $x > m$, and $\\frac{f(x)}{x^{a-\\varepsilon}} > x^{\\varepsilon/2}, \\forall x > m$. But $\\lim_{x \\to \\infty} x^{\\varepsilon/2} = \\infty$ and $f(x) > 0, \\forall x > 0$, from which we obtain the desired conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that $f^2$ is differentiable and $(f^2)' = f$.",
    "options": [],
    "answer": "All solutions are:\n- f(x) ≡ 0;\n- f(x) = { 0, for x < a; (x − a)/2, for x ≥ a }, for any real a;\n- f(x) = { (x − b)/2, for x ≤ b; 0, for x > b }, for any real b;\n- f(x) = { (x − b)/2, for x ≤ b; 0, for x ∈ (b, a); (x − a)/2, for x ≥ a }, for any reals b < a;\n- f(x) = x/2 + c for all real x, for any real c.",
    "solution": "$$\na = \\inf \\{ t \\in (-\\infty, x_0) \\mid f^2(x) > 0, \\forall x \\in [t, x_0] \\},\n$$\n\n$$\nb = \\sup \\{t \\in (x_0, \\infty) \\mid f^2(x) > 0, \\forall x \\in [x_0, t]\\}.\n$$\nWe have $-\\infty \\le a < x_0 < b \\le \\infty$ and $f(x) \\ne 0, \\forall x \\in (a, b)$. Since $f = (f^2)'$ has the intermediate value property (following from the theorem of Darboux), we deduce that $f$ has a constant sign on the interval $(a, b)$.\nIf $f(x) > 0, \\forall x \\in (a, b)$, then $f = \\sqrt{f^2}$ on $(a, b)$, hence $f$ is differentiable on $(a, b)$. From the hypothesis we obtain $2f(x)f'(x) = f(x)$, $\\forall x \\in (a, b)$, therefore $f'(x) = \\frac{1}{2}, \\forall x \\in (a, b)$. We deduce that for some $c \\in \\mathbb{R}$, we have $f(x) = \\frac{x}{2} + c$, $x \\in (a, b)$. Since $f(x) > 0, \\forall x \\in (a, b)$, we obtain $a \\ge -2c$ (so $a$ is finite). Because $f^2$ is continuous, it results that $f^2(a) = 0$, and then $c = -\\frac{a}{2}$, hence $f(x) = \\frac{x-a}{2}$, $x \\in [a, b)$. If $b$ were finite, we would obtain $0 = f^2(b) = \\lim_{x \\uparrow b} \\left(\\frac{x-a}{2}\\right)^2 = \\frac{(b-a)^2}{4}$; contradiction. It follows that $b = \\infty$ and that $f(x) = \\frac{x-a}{2}, x \\in [a, \\infty)$.\n\nMoreover, $f(x) \\le 0, \\forall x \\in (-\\infty, a)$. Indeed, if there were some $x_1 < a$ such that $f(x_1) > 0$, then $f(a) > 0$ which is a contradiction.\nIf $f(x) < 0, \\forall x \\in (a, b)$, we obtain, in a similar way that $b$ is finite, $f(x) = \\frac{x-b}{2}$, $x \\in (-\\infty, b]$ and $f(x) \\ge 0, \\forall x \\in (b, \\infty)$.\n\nWe conclude that, along the null function, the following types of functions $f: \\mathbb{R} \\to \\mathbb{R}$ are solutions:\n$$\n1. f(x) = \\begin{cases} 0, & x < a \\\\ \\frac{x-a}{2}, & x \\ge a \\end{cases}, a \\in \\mathbb{R};\n$$\n$$\n2. f(x) = \\begin{cases} \\frac{x-b}{2}, & x \\le b \\\\ 0, & x > b \\end{cases}, b \\in \\mathbb{R};\n$$\n$$\n3. f(x) = \\begin{cases} \\frac{x-b}{2}, & x \\le b \\\\ 0, & x \\in (b, a) \\\\ \\frac{x-a}{2}, & x \\ge a \\end{cases}, a, b \\in \\mathbb{R}, b < a;\n$$\n$$\n4. f(x) = \\frac{x}{2} + c, x \\in \\mathbb{R}, c \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for every positive integer $n$, there exists a unique $c_n \\in (0, 1)$ such that\n$$\n\\int_0^1 \\frac{1}{1+x^n} dx = \\frac{1}{1+(c_n)^n},\n$$\nand evaluate the limit $\\lim_{n \\to \\infty} n(c_n)^n$.",
    "options": [],
    "answer": "log 2",
    "solution": "Existence of $c_n$ follows from the (first) mean-value theorem (each integrand is continuous), and uniqueness follows from injectivity of each integrand.\n\nTo evaluate the required limit, let $I_n = \\int_0^1 (1+x^n)^{-1} dx$, and notice that $|I_n - 1| = \\left| \\int_0^1 \\frac{-x^n}{1+x^n} dx \\right| \\le \\int_0^1 x^n dx = \\frac{1}{n+1}$, so $\\lim_{n \\to \\infty} I_n = 1$. Since $n(c_n)^n = n(1/I_n - 1)$, it is sufficient to evaluate $\\lim_{n \\to \\infty} n(1-I_n)$. To this end, write $n(1-I_n) = n \\int_0^1 \\frac{x^n}{1+x^n} dx = \\int_0^1 x \\cdot (\\log(1+x^n))' dx = \\log 2 - \\int_0^1 \\log(1+x^n) dx$. Since $0 \\le \\int_0^1 \\log(1+x^n) dx \\le \\int_0^1 x^n dx = \\frac{1}{n+1}$, it follows that $\\lim_{n \\to \\infty} \\int_0^1 \\log(1+x^n) dx = 0$, so $\\lim_{n \\to \\infty} n(c_n)^n = \\log 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the isosceles right triangle $ABC$, with $m(\\widehat{BAC}) = 90^\\circ$. Take now the point $D$ so that $BD \\perp BC$ and $AD = BC$. Find the measure of the angle $\\widehat{BAD}$.",
    "options": [],
    "answer": "15° or 105°",
    "solution": "Case 1: $D$ and $A$ are on different sides of $BC$ (figure 1).\nDenote $\\{E\\} = AC \\cap DB$. Then $m(\\widehat{ABE}) = 45^\\circ$, therefore $[BA]$ is bisector and altitude in the triangle $BEC$. So, $[AE] = [AC]$. Construct $AM \\perp BE$, Then $[AM]$ is a midline in the triangle $EBC$, hence $AM = \\frac{1}{2}BC = \\frac{1}{2}AD$.\nIn the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$, hence $m(\\widehat{ADB}) = 30^\\circ$. It follows $m(\\widehat{BAD}) = 180^\\circ - m(\\widehat{ADB}) - m(\\widehat{ABD}) = 180^\\circ - 30^\\circ - 105^\\circ = 15^\\circ$.\n\n![](attached_image_1.png)\n\nCase 2: $D$ and $A$ are on different sides of the line $BC$ (figure 2).\nDenote $\\{E\\} = AC \\cap DB$. Then $m(\\widehat{ABE}) = 45^\\circ$, hence $[BA]$ is bisector and altitude in the triangle $BEC$. So, $[AE] = [AC]$. Construct $AM \\perp BE$. Then $[AM]$ is a midline in the triangle $EBC$, hence $AM = \\frac{1}{2}BC = \\frac{1}{2}AD$.\nIn the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$, hence $m(\\widehat{ADB}) = 30^\\circ$. It follows $m(\\widehat{BAD}) = 180^\\circ - m(\\widehat{ADB}) - m(\\widehat{ABD}) = 180^\\circ - 30^\\circ - 45^\\circ = 105^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the set of the natural numbers $k$ for which it exists $n \\in \\mathbb{N}$ such that the remainder of $3^n$ when divided by $n$ is $k$. Show that $M$ is infinite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $j$ be a fixed positive integer and $p > 2$ a prime so that $2^j p > 3^{2^j}$. Then $3^{2^j}(3^{2^j(p-1)} - 1) \\equiv 0 \\pmod{2^j p}$, because $2\\varphi(2^j p) = 2^j(p-1)$. Hence $3^{2^j p} \\equiv 3^{2^j} \\pmod{2^j p}$, so, for $n = 2^j p$, one has $r_n = 3^{2^j}$. It follows that $3^{2^j} \\in M$, for all $j \\in \\mathbb{N}^*$, so $M$ is an infinite set.\n\nAlternative Solution:\nChoosing $n = 2 \\cdot 3^j$, the remainder $r_n$ of $3^n$ when divided by $n$ verifies $r_n \\neq 0$, $r_n \\equiv 0 \\pmod{3^j}$, so $r_n \\ge 3^j$ (in fact $r_n = 3^j$). Therefore, $M$ contains arbitrarily large numbers, so it is infinite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38463,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that, for every integer $k$, the equation $x^3 - 24x + k = 0$ has at most one integer solution.\n\nb) Prove that the equation $x^3 + 24x - 2016 = 0$ has exactly one integer solution.",
    "options": [],
    "answer": "12",
    "solution": "a) Suppose that there exist two different integers $m$ and $n$ so that $m^3 - 24m + k = 0$ and $n^3 - 24n + k = 0$.\nSubtracting the above yields $(m-n)(m^2 + mn + n^2 - 24) = 0$. Since $m$ and $n$ are different, $m^2 + mn + n^2 = 24$, whence $(2m+n)^2 + 3n^2 = 96$.\nTherefore $n^2 \\le 32$, hence $n^2 \\in \\{0, 1, 4, 9, 16, 25\\}$. This leads to $(2m+n)^2 \\in \\{96, 93, 84, 69, 48, 21\\}$, a contradiction.\n\n\nb) The equation is $x(x^2 + 24) = 2016$, so $x$ must be a positive integer; $x = 12$ is a solution.\nIf $x < y$ are positive solutions, then $x^2 + 24 < y^2 + 24$ and $2016 = x(x^2 + 24) < y(y^2 + 24) = 2016$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ having the property that\n$$\nf\\left(x + \\frac{1}{n}\\right) \\le f(x) + \\frac{1}{n}, \\text{ for all } x \\in \\mathbb{R} \\text{ and } n \\in \\mathbb{Z}^*.\n$$",
    "options": [],
    "answer": "All functions f(x) = x + a, where a is any real constant.",
    "solution": "Inductively we obtain $f(x+r) \\le f(x)+r$, for any $x \\in \\mathbb{R}$ and $r \\in \\mathbb{Q}$.\n\nThe continuity of $f$ and the density of $\\mathbb{Q}$ in $\\mathbb{R}$ give $f(x+y) \\le f(x)+y$, for all $x \\in \\mathbb{R}$ and $y \\in \\mathbb{R}$. We get thus the functions defined by $f_a(x) = x + a$, for $x \\in \\mathbb{R}$ where the parameter $a$ runs over $\\mathbb{R}$. It is obvious that all these functions verify the hypothesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ and $\\beta$ be real numbers. Find, with proof, the maximal value of the expression\n$$\n|\\alpha x + \\beta y| + |\\alpha x - \\beta y|,\n$$\nin each of the following cases:\na) $x, y \\in \\mathbb{R}$, such that $|x| \\le 1$ and $|y| \\le 1$;\nb) $x, y \\in \\mathbb{C}$, such that $|x| \\le 1$ and $|y| \\le 1$.",
    "options": [],
    "answer": "a) 2 max(|α|, |β|); b) 2√(α^2 + β^2)",
    "solution": "a) It is clear that for any real numbers $u, v$, we have $|u + v| + |u - v| \\in \\{\\pm 2u, \\pm 2v\\}$, and thus $|\\alpha x + \\beta y| + |\\alpha x - \\beta y| \\le \\max\\{2|\\alpha|, 2|\\beta|\\}$. As we get the equality for $x = y = 1$, the quantity on the right hand side is the maximal value.\n\nb) It is not difficult to check the equality $|\\alpha x + \\beta y|^2 + |\\alpha x - \\beta y|^2 = 2|\\alpha x|^2 + 2|\\beta y|^2$. On the other hand, we have\n$$\n(|\\alpha x + \\beta y| + |\\alpha x - \\beta y|)^2 \\le 2 \\left( |\\alpha x + \\beta y|^2 + |\\alpha x - \\beta y|^2 \\right),\n$$\nand thus $|\\alpha x + \\beta y| + |\\alpha x - \\beta y| \\le 2\\sqrt{\\alpha^2 + \\beta^2}$. We get the maximal value $2\\sqrt{\\alpha^2 + \\beta^2}$ for $x = 1$ and $y = i$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38466,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $A$ and $B$ are positive integers, then $\\overline{AB}$ will denote the number obtained by writing, in order, the digits of $B$ after the digits of $A$. For instance, if $A = 193$ and $B = 2016$, then $\\overline{AB} = 1932016$.\nProve that there are infinitely many perfect squares of the form $\\overline{AB}$ in each of the following situations:\na) $A$ and $B$ are perfect squares;\nb) $A$ and $B$ are perfect cubes;\nc) $A$ is a perfect cube and $B$ is a perfect square;\nd) $A$ is a perfect square and $B$ is a perfect cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) $A = 4$ and $B = 9$ yields $\\overline{AB} = 49 = 7^2$. Adding an even number of zeroes we get infinitely many solutions: for $A = 4$ and $B = 9 \\cdot 10^{2n},\\ n \\in \\mathbb{N}$ we get $\\overline{AB} = (7 \\cdot 10^n)^2$.\n\nb) If $A = 8$ and $B = 10^{6n},\\ n \\in \\mathbb{N}$, then $\\overline{AB} = (9 \\cdot 10^{3n})^2$.\n\nc) If $A = 8$ and $B = 10^{2n},\\ n \\in \\mathbb{N}$, then $\\overline{AB} = (9 \\cdot 10^n)^2$.\n\nd) If $A = 36$ and $B = 10^{6n},\\ n \\in \\mathbb{N}$, then $\\overline{AB} = (19 \\cdot 10^{3n})^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime and let $G$ be a $(p+1)$-element group. If $p$ divides the number of automorphisms of $G$, prove that $p \\equiv 3 \\pmod 4$.\nBogdan Moldovan",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $p$ is a prime divisor of the number of automorphisms of $G$, some automorphism $f$ has order $p$. Since $f$ is a permutation of the set $G \\setminus \\{e\\}$, it follows that $f$ is a cycle of length $p$, so $G \\setminus \\{e\\} = \\{x, f(x), \\dots, f^{p-1}(x)\\}$, whatever $x$ in $G \\setminus \\{e\\}$. On the other hand, $|G| = p+1$ is even, so $G$ contains an element $x_0$ of order $2$. Hence $\\text{ord} f^k(x_0) = 2$, $k = 0, \\dots, p-1$, so $x^2 = e$ for all $x$ in $G$, and $p+1 = 2^n$ for some integer $n \\ge 2$. Consequently, $p = 2^n - 1 \\equiv 3 \\pmod 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer. Consider a $4m \\times 4m$ array of square unit cells. Two different cells are *related* to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are colored blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.",
    "options": [],
    "answer": "6m",
    "solution": "The required minimum is $6m$ and is achieved by a diagonal string of $m \\times 4$ blocks of the form below (bullets mark centers of blue cells):\n![](attached_image_1.png)\n\nIn particular, this configuration shows that the required minimum does not exceed $6m$.\n\nWe now show that any configuration of blue cells satisfying the condition in the statement has cardinality at least $6m$.\nFix such a configuration and let $m_1^r$ be the number of blue cells in rows containing exactly one such, let $m_2^r$ be the number of blue cells in rows containing exactly two such, and let $m_3^r$ be the number of blue cells in rows containing at least three such; the numbers $m_1^c, m_2^c$ and $m_3^c$ are defined similarly.\nBegin by noticing that $m_3^c \\ge m_1^r$ and, similarly, $m_3^r \\ge m_1^c$. Indeed, if a blue cell is alone in its row, respectively column, then there are at least two other blue cells in its column, respectively row, and the claim follows.\nSuppose now, if possible, the total number of blue cells is less than $6m$. We will show that $m_1^r > m_3^r$ and $m_1^c > m_3^c$, and reach a contradiction by the preceding: $m_1^r > m_3^r \\ge m_1^c > m_3^c \\ge m_1^r$.\nWe prove the first inequality; the other one is dealt with similarly. To this end, notice that there are no empty rows — otherwise, each column would contain at least two blue cells, whence a total of at least $8m > 6m$ blue cells, which is a contradiction. Next, count rows to get $m_1^r + m_2^r/2 + m_3^r/3 \\ge 4m$, and count blue cells to get $m_1^r + m_2^r + m_3^r < 6m$. Subtraction of the latter from the former multiplied by $3/2$ yields $m_1^r - m_3^r > m_2^r/2 \\ge 0$, and the conclusion follows.\nTo prove that a minimal configuration of blue cells satisfying the condition in the statement has cardinality at least $6m$, consider a bipartite graph whose vertex parts are the rows and the columns of the array, respectively, a row and a column being joined by an edge if and only if the two cross at a blue cell. Clearly, the number of blue cells is equal to the number of edges of this graph, and the relationship condition in the statement reads: for every row $r$ and every column $c$, $\\deg r + \\deg c - \\varepsilon(r, c) \\ge 2$, where $\\varepsilon(r, c) = 2$ if $r$ and $c$ are joined by an edge, and $\\varepsilon(r, c) = 0$ otherwise.\nNotice that there are no empty rows/columns, so the graph has no isolated vertices. By the preceding, the cardinality of every connected component of the graph is at least 4, so there are at most $2 \\cdot 4m/4 = 2m$ such and, consequently, the graph has at least $8m - 2m = 6m$ edges. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38469,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1$, $2$, $3$, $\\ldots$, $16$ are written in the squares of a $4 \\times 4$ table, one in each square, and we add the numbers in each column. If one of the sums is larger than the other three, we denote it $S$.\n\na) Give an example with $S = 40$.\n\nb) Which is the smallest possible value of $S$?",
    "options": [],
    "answer": "a) 40; b) 35",
    "solution": "a) An example is given in figure 1.\n\n<table><tr><td>1</td><td>2</td><td>3</td><td><b>10</b></td></tr><tr><td>8</td><td>7</td><td>6</td><td><b>5</b></td></tr><tr><td>9</td><td>4</td><td>11</td><td><b>12</b></td></tr><tr><td>16</td><td>15</td><td>14</td><td><b>13</b></td></tr></table>\n\nFigure 1\n\nb) The sum of the numbers written in the table is $1 + 2 + \\dots + 16 = 136$.\n\nSince $136 = 4 \\cdot 34$, either the sum of the numbers in each column is $34$, or there exists a column with sum at least $35$. In the first case $S$ does not exist and in the second case $S$ is at least $35$.\n\n<table><tr><td>1</td><td>2</td><td>3</td><td><b>4</b></td></tr><tr><td>8</td><td>7</td><td>6</td><td><b>5</b></td></tr><tr><td>9</td><td>10</td><td>11</td><td><b>12</b></td></tr><tr><td>16</td><td>15</td><td>13</td><td><b>14</b></td></tr></table>\n\nFigure 2\n\nAn example with $S = 35$ is given in figure 2, so the minimum possible value of $S$ is $35$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $a_1, \\dots, a_n$ be pairwise distinct positive integers. Show that\n$$\n\\sum_{k=1}^{n} \\frac{1}{[a_1, \\dots, a_k]} < 4,\n$$\nwhere $[a_1, \\dots, a_k]$ is the least common multiple of the integers $a_1, \\dots, a_k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the number of positive divisors of a positive integer $m$ does not exceed $2\\sqrt{m}$, and $a_1, \\dots, a_k$ are pairwise distinct positive divisors of $[a_1, \\dots, a_k]$, it follows that $[a_1, \\dots, a_k] \\ge k^2/4$. Consequently,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{1}{[a_1, \\dots, a_k]} &= \\frac{1}{a_1} + \\sum_{k=2}^{n} \\frac{1}{[a_1, \\dots, a_k]} \n\\le 1 + \\sum_{k=2}^{n} \\frac{4}{k^2} \n< 1 + 4 \\sum_{k=2}^{n} \\frac{1}{k^2 - \\frac{1}{4}} \\\\\n&= 1 + 4 \\cdot 2 \\left( \\frac{1}{3} - \\frac{1}{2n+1} \\right) < \\frac{11}{3} < 4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that every non-empty subset of the set\n$$\nY = \\{113! + 2, 113! + 3, \\dots, 113! + 15\\}\n$$\nhas the sum of its elements a composite number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A non-empty subset of $Y$ has the sum of its elements $S = 113! + s$, where $2 \\leq s \\leq 2 + 3 + \\dots + 15$, whence $2 \\leq s \\leq 119$.\n\nIf $2 \\leq s \\leq 113$, then $S = 113! + s = Ms > s$.\n\nIf $s \\in \\{114, 116, 118\\}$, then $S = M2$.\n\nIf $s = 115$, then $S = M5$.\n\nIf $s = 117$, then $S = M3$.\n\nIf $s = 119$, then $S = M7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38472,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive numbers $a$, $b$, $c$ are such that\n$$\n\\frac{a}{b+c+1} + \\frac{b}{a+c+1} + \\frac{c}{a+b+1} \\le 1.\n$$\nProve that:\n$$\n\\frac{1}{b+c+1} + \\frac{1}{a+c+1} + \\frac{1}{a+b+1} \\ge 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S = \\sum \\frac{1}{b+c+1}$. Using the inequality from the hypothesis we obtain\n$$\n\\sum \\left( \\frac{a}{b+c+1} + 1 \\right) \\le 4,\n$$\nand thus $(a+b+c+1)S \\le 4$.\n\nUsing arithmetic mean – harmonic mean inequality we get\n$$\nS \\ge \\frac{9}{2(a + b + c + 1) + 1},\n$$\nhence $S \\ge 9 - 2(a + b + c + 1)S \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a ring and let $D$ be the set of all non-invertible elements of $A$. Assuming $a^2 = 0$ for all $a$ in $D$, prove that:\n\na) $axa = 0$ for all $a$ in $D$ and all $x$ in $A$; and\n\nb) If $D$ is finite and $|D| \\ge 2$, there exists $a$ in $D \\setminus \\{0\\}$ such that $ab = ba = 0$ for all $b$ in $D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $a$ be a member of $D$ and let $x$ be a member of $A$. If $x$ is invertible, then $ax$ is a member of $D$, so $axax = 0$, and $axa = 0$. If $x$ is a member of $D$, then $1+x$ is invertible, so $a + ax = a(1+x)$ is also a member of $D$. Hence $0 = (a+ax)^2 = a^2 + a^2x + axa + axax = axa(1+x)$, and consequently $axa = 0$.\n\nb) Let $P$ be the set of all finite non-zero products of elements of $D$. Since $|D| \\ge 2$, the set $P$ is non-empty. Notice that if $a_1a_2\\cdots a_k$ is a product in $P$, then $a_i \\ne a_j$ for $i \\ne j$. Indeed, if $a_i = a_j$ for some $i < j$, then $a_1a_2\\cdots a_k = a_1a_2\\cdots a_i(a_{i+1}\\cdots a_{j-1})a_i\\cdots a_k = 0$, by a), which is a contradiction. Hence $P$ is a\n\nfinite set. Finally, let $a = a_1a_2 \\cdots a_k$ be a product in $P$ of maximal length $k$, and let $b$ be a member of $D$. If $b$ is one of the factors of $a$, then $ab = ba = 0$, by a). Otherwise, the words $ab$ and $ba$ both have length greater than $k$, so they cannot belong to $P$, by maximality of $k$, and again $ab = ba = 0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be an increasing function, and let $a$ be a real number. Prove that $f$ is continuous at $a$ if and only if there exists a sequence $(a_n)_{n \\ge 1}$ of positive real numbers such that\n$$\n\\int_a^{a+a_n} f(x) \\, dx + \\int_a^{a-a_n} f(x) \\, dx \\le \\frac{a_n}{n}, \\quad n = 1, 2, 3, \\dots\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F : \\mathbb{R} \\to \\mathbb{R}$, $F(x) = \\int_a^x f(t) \\, dt$. If $f$ is continuous at $a$, then $F$ is differentiable at $a$ and $F'(a) = f(a)$. For every positive integer $n$, there exists a positive real number $\\delta_n$ such that $|F(x)/(x-a) - f(a)| \\le 1/(2n)$ for all $x$ in the $\\delta_n$-neighbourhood of $a$. Let $a_n = \\delta_n/2$, $n = 1, 2, 3, \\dots$, to get $F(a+a_n)/a_n - f(a) \\le 1/(2n)$ and $f(a) - F(a-a_n)/(-a_n) \\le 1/(2n)$, so $F(a+a_n) + F(a-a_n) \\le a_n/n$, by addition of the two.\n\nConversely, since $f$ is increasing, so is $g : \\mathbb{R} \\setminus \\{a\\} \\to \\mathbb{R}$, $g(x) = F(x)/(x-a)$. Indeed, let $x, y \\in \\mathbb{R} \\setminus \\{a\\}$, $a < x < y$; the cases $x < y < a$ and $x < a < y$ are dealt with similarly. Since $(x-a)F(y) = (x-a) \\int_a^x f(t) \\, dt + (x-a) \\int_x^y f(t) \\, dt \\ge (x-a) \\int_a^x f(t) \\, dt + (x-a)(y-x)f(x) \\ge (x-a) \\int_a^x f(t) \\, dt + (y-x) \\int_a^x f(t) \\, dt = (y-a)F(x)$, it follows that $g(y) \\ge g(x)$.\n\nFor every positive integer $n$, choose $x_n$ in the open interval $(0, a_n)$ so that $x_n \\xrightarrow{n \\to \\infty} 0$. Since $g$ is increasing, it follows that $g(a+x_n)-g(a-x_n) \\le g(a+a_n)-g(a-a_n) \\le 1/n$, $n = 1, 2, 3, \\dots$, so $g(a+0) \\le g(a-0)$. On the other hand, $g(a-0) \\le g(a+0)$ for $g$ is increasing, so $g(a+0) = g(a-0)$ is finite, and $F$ is differentiable at $a$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ be a finite field of cardinality $q \\ge 3$, and let $S$ be the set of degree $q-2$ polynomials in $K[X]$ whose coefficients are all non-zero and pairwise distinct. Determine the number of polynomials in $S$ having $q-2$ distinct roots in $K$.\nMariean Andronache",
    "options": [],
    "answer": "(q-1)φ(q-1)",
    "solution": "The required number is $(q-1)\\varphi(q-1)$, where $\\varphi$ is Euler's totient function. Since every polynomial in $S$ is associated in divisibility with a unique polynomial $f$ in $S$ such that $f(0) = 1$, it is sufficient to count the polynomials of the form $f = 1 + a_1X + \\dots + a_{q-2}X^{q-2}$ satisfying the condition in the statement.\n\nLet $f$ be such a polynomial, let $x_1, \\dots, x_{q-2}$ be its roots, write $K^* = \\{x_0, x_1, x_2, \\dots, x_{q-2}\\}$, and let\n$$\nA = \\begin{pmatrix}\n1 & a_1 & a_2 & \\dots & a_{q-3} & a_{q-2} \\\\\na_{q-2} & 1 & a_1 & \\dots & a_{q-4} & a_{q-3} \\\\\na_{q-3} & a_{q-2} & 1 & \\dots & a_{q-5} & a_{q-4} \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\\na_2 & a_3 & a_4 & \\dots & 1 & a_1 \\\\\na_1 & a_2 & a_3 & \\dots & a_{q-2} & 1\n\\end{pmatrix}\n$$\nand\n$$\nB = \\begin{pmatrix}\n1 & 1 & 1 & \\dots & 1 & 1 \\\\\nx_0 & x_1 & x_2 & \\dots & x_{q-3} & x_{q-2} \\\\\nx_0^2 & x_1^2 & x_2^2 & \\dots & x_{q-3}^2 & x_{q-2}^2 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\\nx_0^{q-3} & x_1^{q-3} & x_2^{q-3} & \\dots & x_{q-3}^{q-3} & x_{q-2}^{q-3} \\\\\nx_0^{q-2} & x_1^{q-2} & x_2^{q-2} & \\dots & x_{q-3}^{q-2} & x_{q-2}^{q-2}\n\\end{pmatrix}.\n$$\n\nSince $x^{q-1} = 1$ for all $x$ in $K^*$,\n$$\nAB = \\begin{pmatrix}\nf(x_0) & 0 & 0 & \\dots & 0 & 0 \\\\\nx_0 f(x_0) & 0 & 0 & \\dots & 0 & 0 \\\\\nx_0^2 f(x_0) & 0 & 0 & \\dots & 0 & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\\nx_0^{q-3} f(x_0) & 0 & 0 & \\dots & 0 & 0 \\\\\nx_0^{q-2} f(x_0) & 0 & 0 & \\dots & 0 & 0\n\\end{pmatrix}.\n$$\nSince $f(x_0) \\neq 0$ and $B$ is invertible in $M_{q-1}(K)$, it follows that $A$ is of rank 1, so its $2 \\times 2$ minors all vanish. In particular,\n$$\n\\begin{vmatrix} a_1 & a_2 \\\\ 1 & a_1 \\end{vmatrix} = \\begin{vmatrix} a_1 & a_3 \\\\ 1 & a_2 \\end{vmatrix} = \\dots = \\begin{vmatrix} a_1 & a_{q-2} \\\\ 1 & a_{q-3} \\end{vmatrix} = 0,\n$$\ni.e., $a_2 = a_1^2$, $a_3 = a_1^3$, ..., $a_{q-2} = a_1^{q-2}$, so the set of coefficients of $f$ is $\\{1, a_1, a_1^2, \\dots, a_1^{q-2}\\}$. Since this set is equal to $K^*$, it follows that $a_1$ is a generator of the multiplicative group $(K^*, \\cdot)$, and $f = 1 + a_1X + a_1^2X^2 + \\dots + a_1^{q-2}X^{q-2}$.\n\nConversely, if $a$ is a generator of the multiplicative group $(K^*, \\cdot)$, the polynomial\n$f = 1 + aX + a^2X^2 + \\dots + a^{q-2}X^{q-2}$ satisfies the condition in the statement, since\n$f(a^{-1}X) = 1 + X + X^2 + \\dots + X^{q-2} = \\prod_{\\alpha \\in K^* \\setminus \\{1\\}} (X - \\alpha)$, so the roots of $f$ are\n$a\\alpha$, where $\\alpha$ runs through $K^* \\setminus \\{1\\}$.\n\nFinally, recall that $K^*$ has exactly $\\varphi(q-1)$ generators, to conclude that the required number is indeed $(q-1)\\varphi(q-1)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\ge 2$ be a positive integer. Prove that the following statements are equivalent:\na) One can find positive integers $b, c$, such that $a^2 = b^2 + c^2$.\nb) One can find a positive integer $d$, such that the equations $x^2 - a x + d = 0$ and $x^2 - a x - d = 0$ have integer roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us suppose that $a^2 = b^2 + c^2$. The numbers $b$ and $c$ could not be both odd (the sum of two odd numbers is $4k + 2$ which is not a square). Then at least one is an even number and thus, the product $bc$ is even.\n\nOn the other hand, the discriminants of the two equations are $\\Delta_1 = a^2 - 4d$ and $\\Delta_2 = a^2 + 4d$. If we define $d = \\frac{bc}{2}$ we have\n$$\n\\Delta_1 = a^2 - 4d = b^2 + c^2 - 4 \\frac{bc}{2} = (b-c)^2,\n$$\nand thus, the roots of the first equation are $x_{1,2} = \\frac{a \\pm (b-c)}{2}$. It is clear that $x_{1,2}$ are integers. (If both $b, c$ are even, then $a$ is also an even number, and if $b, c$ have different parities, then $a$ is odd, as $b-c$ it is.) Similarly we can show that the second equation has integer roots.\n\nConversely, we suppose that the equations have only integer roots. Then their discriminants are squares. Let $\\Delta_1 = u^2$ and $\\Delta_2 = v^2$.\nWe have\n$$\n\\begin{aligned}\na^2 - 4d &= u^2, \\\\\na^2 + 4d &= v^2.\n\\end{aligned}\n$$\n\nIt is clear that $u, v$ have the same parity.\nIf we add the two equalities we get\n$$\na^2 = \\frac{u^2 + v^2}{2} = \\left(\\frac{u+v}{2}\\right)^2 + \\left(\\frac{u-v}{2}\\right)^2.\n$$\nDefine now $b = \\frac{u+v}{2}$ and $c = \\frac{u-v}{2}$. The numbers $b$ and $c$ are integers and $a^2 = b^2 + c^2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD A'B'C'D'$ be a cuboid and $M$, respectively $N$ be the feet of the perpendiculars from $A'$ and $C'$ on $BD$. The length of the sides $AB$, $BC$ and $AA'$ are $\\sqrt{6}$, $\\sqrt{3}$ respectively $\\sqrt{2}$.\n\na) Prove that $A'M \\perp C'N$.\n\nb) Compute the measure of the angle of the planes $(A'MC)$ and $(ANC')$.",
    "options": [],
    "answer": "60 degrees",
    "solution": "a) In the right triangle $ABD$, $AM \\perp BD$, so $AM = \\sqrt{2}$, that is the triangle $A'MA$ is isosceles. The same goes for the triangle $C'CN$. It follows $\\angle A'MA = \\angle C'NC = 45^\\circ$. Since $CN \\parallel AM$ and $AA' \\parallel CC'$, the parallel from $A'$ to $NC'$ is coplanar with $AM$, hence the angle of the lines $A'M$ and $C'N$ has $180^\\circ - 45^\\circ - 45^\\circ = 90^\\circ$.\n\nb) Let $CM \\cap AD = \\{M'\\}$, $AN \\cap BC = \\{N'\\}$, $P$ and $Q$ be the midpoints of the segments $[N'C']$, respectively $[A'M']$ and $\\{T\\} = A'C \\cap AC'$. Then $BD = 3$, $DM = MN = NB = 1$, $DM' = \\frac{\\sqrt{3}}{2}$. $M'$ and $N'$ are the midpoints of the edges $[AD]$ respectively $[BC]$, hence $QT \\parallel CM \\parallel AN \\parallel PT$, that is $P, T, Q$ are collinear.\n![](attached_image_1.png)\n\nTherefore, the intersection of the planes $(A'MC)$ and $(ANC')$ is line $PQ$.\nLet $O$ be the midpoint of the segment $[BD]$ and $R \\in AN$, $S \\in CM$ be the intersections of the perpendicular from $O$ on $AN$ (and on $CM$). The angle of the planes $(A'MC)$ and $(ANC')$ is $\\angle RTS$. Now $CM' = \\frac{3\\sqrt{3}}{2}$, $RS = \\frac{\\sqrt{6}}{3}$ (compute in two ways the area of $AN'CM'$), $OT = \\frac{\\sqrt{2}}{2}$ and it follows that the triangle $RTS$ is equilateral, hence the required measure is $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38478,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that there exist non-periodical functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + 1) + f(x - 1) = \\sqrt{5}f(x),\n$$\nfor all $x \\in \\mathbb{R}$.\nb) Prove that any function $g : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\ng(x + 1) + g(x - 1) = \\sqrt{3}g(x),\n$$\nfor all $x \\in \\mathbb{R}$, is periodical.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) It is natural to look for solutions of the exponential type $f(x) = a^x$, where $a > 0$. If this is a solution, we get $a + a^{-1} = \\sqrt{5}$, and thus $a = \\frac{\\sqrt{5}+1}{2}$. It is easy to check that the functions $f_1: \\mathbb{R} \\to \\mathbb{R}$, $f_1(x) = (\\frac{\\sqrt{5}-1}{2})^x$ and $f_2: \\mathbb{R} \\to \\mathbb{R}$, $f_2(x) = (\\frac{\\sqrt{5}+1}{2})^x$ satisfy the required conditions.\n\nb) Let $g$ be a function which satisfies the required condition. Then, $g(x+2) + g(x) = \\sqrt{3}g(x+1)$, and thus $g(x+2) + g(x) = \\sqrt{3}(\\sqrt{3}g(x) - g(x-1))$, from where we obtain that $g(x+2) = 2g(x) - \\sqrt{3}g(x-1)$. Next $g(x+3) = 2g(x+1) - \\sqrt{3}g(x) = 2(\\sqrt{3}g(x) - g(x-1)) - \\sqrt{3}g(x)$, and we get $g(x+3) = \\sqrt{3}g(x) - 2g(x-1)$. Similarly $g(x+4) = \\sqrt{3}g(x+1) - 2g(x) = \\sqrt{3}(\\sqrt{3}g(x) - g(x-1)) - 2g(x)$, to get $g(x+4) = g(x) - \\sqrt{3}g(x-1)$. Finally, $g(x+5) = g(x+1) - \\sqrt{3}g(x)$, and thus $g(x+5) = -g(x-1)$, to obtain $g(x+6) = -g(x)$. To get the conclusion we have to notice that $g(x+12) = -g(x+6) = g(x)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, 1] \\to [0, 1]$ be an increasing function and let\n$$\na_n = \\int_0^1 \\frac{1 + (f(x))^n}{1 + (f(x))^{n+1}} \\, dx, \\quad n = 1, 2, 3, \\dots\n$$\nProve that the sequence $(a_n)_{n \\ge 1}$ is convergent and evaluate its limit.",
    "options": [],
    "answer": "1",
    "solution": "Clearly, $a_n \\ge 1$, $n = 1, 2, 3, \\dots$. On the other hand,\n$$\n\\begin{aligned}\na_n - a_{n+1} &= \\int_0^1 \\left( \\frac{1 + (f(x))^n}{1 + (f(x))^{n+1}} - \\frac{1 + (f(x))^{n+1}}{1 + (f(x))^{n+2}} \\right) dx \\\\\n&= \\int_0^1 \\frac{(f(x))^n (1 - f(x))^2}{(1 + (f(x))^{n+1})(1 + (f(x))^{n+2})} dx \\ge 0, \\quad n = 1, 2, 3, \\dots,\n\\end{aligned}\n$$\nso the sequence $(a_n)_{n \\ge 1}$ is decreasing, and consequently convergent. Let $\\ell = \\lim_{n \\to \\infty} a_n$.\n\nWithout loss of generality we may and will assume that $f(0) = 0$ and $f(1) = 1$. Let $a = \\inf\\{x: 0 \\le x \\le 1, f(x) = 1\\}$. If $a = 0$, then $f(x) = 1, 0 < x \\le 1$, so $a_n = 1, n = 1, 2, 3, \\dots$, and $\\ell = 1$. If $a > 0$, consider any positive $\\varepsilon < a$ and write\n$$\n\\begin{aligned}\n0 \\le a_n - 1 &= \\int_0^1 \\left( \\frac{1 + (f(x))^n}{1 + (f(x))^{n+1}} - 1 \\right) dx = \\int_0^1 \\frac{(f(x))^n (1 - f(x))}{1 + (f(x))^{n+1}} dx \\\\\n&\\le \\int_0^1 (f(x))^n (1 - f(x)) dx = \\int_0^a (f(x))^n (1 - f(x)) dx \\\\\n&\\le \\int_0^a (f(x))^n dx = \\int_0^{a-\\varepsilon} (f(x))^n dx + \\int_{a-\\varepsilon}^a (f(x))^n dx \\\\\n&\\le (a-\\varepsilon)(f(a-\\varepsilon))^n + \\varepsilon, \\quad n = 1, 2, 3, \\dots\n\\end{aligned}\n$$\nSince $0 \\le f(a - \\varepsilon) < 1$, it follows that $(f(a - \\varepsilon))^n \\xrightarrow{n \\to \\infty} 0$, so $0 \\le \\ell - 1 \\le \\varepsilon$, by the preceding; and since $\\varepsilon$ is arbitrary, we conclude that $\\ell = 1$.\n\n*Remark.* Since the integrands form a sequence of measurable functions pointwise convergent to the constant function $1$, and each integrand is pointwise bounded from above by the constant (hence integrable) function $2$, the conclusion follows by Lebesgue's dominated convergence theorem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB \\neq AC$. The incircle $\\omega$ of the triangle touches the sides $BC$, $CA$ and $AB$ in $D$, $E$ and $F$, respectively. The perpendicular line erected from $C$ to $BC$ intersects $EF$ at $M$, and, similarly, the perpendicular line erected at $B$ to $BC$ intersects $EF$ at $N$. The line $DM$ meets $\\omega$ again in $P$, and the line $DN$ meets $\\omega$ again in $Q$. Prove that $DP = DQ$.\nRubén Dario, Perú, and Leonard Giugliuc, Romania",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\{T\\} = EF \\cap BC$. Applying Menelaus' theorem to the triangle $ABC$ and the transversal line $E-F-T$ we obtain: $\\frac{TB}{TC} \\cdot \\frac{EC}{EA} \\cdot \\frac{FA}{FB} = 1$, i.e. $\\frac{TB}{TC} \\cdot \\frac{s-c}{s-a} \\cdot \\frac{s-a}{s-b} = 1$, or $\\frac{TB}{TC} = \\frac{s-b}{s-c}$, where the notations are the usual ones (1).\n\nThis means that triangles $TBN$ and $TCM$ are similar, therefore $\\frac{TB}{TC} = \\frac{BN}{CM}$. From (1) it follows that $\\frac{BN}{CM} = \\frac{s-b}{s-c}$, $\\frac{BD}{CD} = \\frac{s-b}{s-c}$, and $\\angle DBN = \\angle DCM = 90^\\circ$, which means that triangles $BDN$ and $CDM$ are similar, hence angles $BDN$ and $CDM$ are equal.\n\nThis leads to the arcs $DQ$ and $DP$ being equal, and finally to $DP = DQ$.\nLet $S$ be the intersection point of the altitude from $A$ with the line $EF$. Lines $BN$, $AS$, $CM$ are parallel, therefore triangles $BNF$ and $ASF$ are similar, as are triangles $ASE$ and $CME$. We obtain $\\frac{BN}{AS} = \\frac{BF}{FA}$ and $\\frac{AS}{CM} = \\frac{AE}{EC}$. Multiplying these two together, we obtain\n$$\n\\frac{BN}{CM} = \\frac{BF}{FA} \\cdot \\frac{AE}{EC} = \\frac{BF}{EC} = \\frac{BD}{DC}.\n$$\n(We have used that $AE = AF$, $BF = BD$ and $CE = CD$.)\n\nIt follows that the right triangles $BDN$ and $CDM$ are similar (SAS), which leads to the same ending as in the first proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the numbers $1, 2, 3, \\ldots, 31$ is written on a separate card. Alex and Bogdan pick $15$ cards each and notice that the sum of Alex's cards is three times larger than the sum of Bogdan's cards. Find the number of the remaining card.",
    "options": [],
    "answer": "16",
    "solution": "Let $A$ and $B$ be the respective sums. Then $B$ is at least $1+2+3+\\ldots+15 = 120$, so $A$ is at least $3 \\cdot 120 = 360$. But $A$ is at most $17+18+\\ldots+31 = 360$, so $A$ must be $360$ and $B$ must be $120$. Therefore Alex must pick $17, 18, \\ldots, 31$ and Bogdan has to pick $1, 2, \\ldots, 15$, so the remaining card must have $16$ written on it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the triangle $ABC$, with $m(\\widehat{A}) > 60^\\circ$ and $m(\\widehat{C}) > 30^\\circ$. In the half-plane determined by the line $BC$ not containing $A$, consider points $D$ and $E$ so that $m(\\widehat{ABE}) = m(\\widehat{CBD}) = 90^\\circ$ and $m(\\widehat{BAE}) = m(\\widehat{BCD}) = 60^\\circ$. Denote $F$ and $H$ the midpoints of the segments $[AE]$, respectively $[CD]$, and $G$ the common point of the lines $AC$ and $DE$. Show that:\n\na) $\\Delta EBD \\sim \\Delta ABC$;\n\nb) $\\Delta FGH \\equiv \\Delta ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The hypothesis yields $\\Delta ABE \\sim \\Delta CBD$, whence $\\frac{BE}{BD} = \\frac{AB}{CB}$, that is $\\frac{EB}{AB} = \\frac{BD}{BC}$. Since angles $ABC$ and $EBD$ have the same complement, $\\widehat{ABC} = \\widehat{EBD}$, hence $\\Delta EBD \\sim \\Delta ABC$ (S.A.S.).\n\nb) From $\\Delta EBD \\sim \\Delta ABC$ follows that $\\widehat{ACB} = \\widehat{BDE}$, whence\n$$\nm(\\widehat{DGC}) = 360^\\circ - (m(\\widehat{CBD}) + m(\\widehat{BDE}) + m(\\widehat{BCG})) = 360^\\circ - (m(\\widehat{CBD}) + m(\\widehat{ACB}) + m(\\widehat{BCG})) = 90^\\circ.\n$$\n\n$[GF]$ is a median in the right triangle $GAE$, hence $GF = AE/2$. Since $[AB]$ opposes to an angle of $30^\\circ$ in the right triangle $BAE$, it follows that $AB = \\frac{1}{2}AE$, hence $[AB] = [FG]$. In the same way, $[BC] = [GH]$.\n\nFrom $\\triangle FGH \\cong \\triangle FBH$ (S.S.S.) follows that $m(\\widehat{FGH}) = m(\\widehat{FBH}) = 90^\\circ - m(\\widehat{HBD}) - m(\\widehat{CBF})$.\n\nFrom $m(\\widehat{HBD}) = m(\\widehat{HDB}) = 30^\\circ$ and $m(\\widehat{CBF}) = m(\\widehat{ABF}) - m(\\widehat{ABC}) = 60^\\circ - m(\\widehat{ABC})$, we get $m(\\widehat{FGH}) = m(\\widehat{ABC})$, so $\\triangle FGH \\cong \\triangle ABC$ (S.A.S.).\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles, $\\omega_1$ and $\\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\\omega$ externally tangent to $\\omega_1$ at a point $T_1$, and internally tangent to $\\omega_2$ at a point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\\omega$.\nLuxembourg",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line $X_kT_k$ and $\\omega$ meet again at $X'_k$, $k = 1,2$, and notice that the tangent $t_k$ to $\\omega_k$ at $X_k$ and the tangent $t'_k$ to $\\omega$ at $X'_k$ are parallel. Since the $\\omega_k$ have equal radii, the $t_k$ are parallel, so the $t'_k$ are parallel, and consequently the points $X'_1$ and $X'_2$ coincide (they are not antipodal, since they both lie on the same side of the line $T_1T_2$). The conclusion follows.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nAlternative Solution:\n\nThe circle $\\omega$ is the image of $\\omega_k$ under a homothety $h_k$ centered at $T_k$, $k = 1, 2$. The tangent to $\\omega$ at $X'_k = h_k(X_k)$ is therefore parallel to the tangent $t_k$ to $\\omega_k$ at $X_k$. Since the $\\omega_k$ have equal radii, the $t_k$ are parallel, so $X'_1 = X'_2$; and since the points $X_k, T_k$ and $X'_k$ are collinear, the conclusion follows.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nAlternative Solution:\n\nInvert from $X_1$ and use an asterisk to denote images under this inversion. Notice that $\\omega_k^*$ is the tangent from $X_2^*$ to $\\omega^*$ at $T_k^*$, and the pole $X_1$ lies on the bisectrix of the angle formed by the $\\omega_k^*$, not containing $\\omega^*$. Letting $X_1T_1^*$ and $\\omega^*$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_1X_2^*T_2^*$, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38484,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $m$ and $n$ are such that $m^{2016} + m + n^2$ is divisible with $mn$.\na) Give an example of such $m$ and $n$, with $m > n$.\nb) Prove that $m$ is a perfect square.",
    "options": [],
    "answer": "Example: m = 4, n = 2; Conclusion: m is a perfect square.",
    "solution": "a) For instance, $m = 4$, $n = 2$.\n\nb) Let $d$ be the greatest common divisor of $m$ and $n$, and $a, b \\in \\mathbb{N}^*$ be such that $m = d a$, $n = d b$, with $(a, b) = 1$.\nThe initial condition becomes: $d^{2016} a^{2016} + d a + d^2 b^2$ is divisible with $d^2 a b$. So $d a b$ divides $d^{2015} a^{2016} + a + d b^2$. Since $a$ divides $d^{2015} a^{2016}$, it follows that $a$ divides $d b^2$. But $(a, b) = 1$, so $a$ divides $d$.\nOn the other hand, $d$ divides $d^{2015} a^{2016}$ and $d b^2$, hence $d$ divides $a$.\nFrom the above $d = a$, hence $m = d^2$, therefore $m$ is a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I \\subset \\mathbb{R}$ be an open interval and $f, g : I \\to \\mathbb{R}$ two functions that satisfy\n$$\n\\frac{f(x) - g(y)}{x - y} + |x - y| \\ge 0, \\text{ for any } x, y \\in I, x \\ne y.\n$$\n\ni) Deduce that $f$ and $g$ are non-decreasing.\nii) Find $f, g : \\mathbb{R} \\to \\mathbb{R}$, $f \\ne g$, having the above property.",
    "options": [],
    "answer": "i) f and g are non-decreasing on I.\nii) One choice on R is f(x) = 0 for x < 0 and f(x) = 1 for x ≥ 0; g(x) = 0 for x ≤ 0 and g(x) = 1 for x > 0.",
    "solution": "i) The given relation can be rewritten as $f(x)+(x-z)^2 \\ge g(z) \\ge f(y)-(z-y)^2$, for all triple $x > z > y$, $x, y, z \\in I$; in particular, $f(x) + (x-y)^2 \\ge f(y)$, for all triplets $x > y$, $x, y \\in I$.\n\nFor $a, b \\in I$, $a > b$, taking $n \\in \\mathbb{N}^*$ and $x_k = a - \\frac{k}{n}(a-b)$, $k \\in \\overline{0, n}$ we obtain\n$f(x_k) + (x_k - x_{k+1})^2 \\ge f(x_{k+1})$, $k = \\overline{0, n-1}$.\n\nSumming up, we get $f(a) + \\frac{1}{n}(a-b)^2 \\ge f(b)$. As this inequality is true for all $n \\in \\mathbb{N}^*$, we obtain at the limit $f(a) \\ge f(b)$. In the same way we prove that $g$ is non-decreasing.\n\n\nii) One example with $I = \\mathbb{R}$ is $f(x) = 0$ for $x < 0$, and $f(x) = 1$ for $x \\ge 0$ and $g(x) = 0$ for $x \\le 0$, $g(x) = 1$ for $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be positive real numbers such that $x + y^{2016} \\ge 1$. Prove that $x^{2016} + y > 1 - 1/100$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x \\ge 1 - 1/(100 \\cdot 2016)$, then\n$$\nx^{2016} \\ge \\left(1 - \\frac{1}{100 \\cdot 2016}\\right)^{2016} > 1 - 2016 \\cdot \\frac{1}{100 \\cdot 2016} = 1 - \\frac{1}{100}\n$$\nby Bernoulli's inequality, whence the conclusion.\n\nTo establish the latter, refer again to Bernoulli's inequality to write\n$$\n\\left(1 + \\frac{1}{99}\\right)^{2016} > \\left(1 + \\frac{1}{99}\\right)^{99 \\cdot 20} > \\left(1 + 99 \\cdot \\frac{1}{99}\\right)^{20} = 2^{20} > 100 \\cdot 2016.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38487,
    "subject": "Mathematics (Multi-modal)",
    "question": "The elements of the set $\\{1, 2, \\dots, 21\\}$ are written randomly on a circle, in the order $a_1, a_2, \\dots, a_{21}$ (see figure). Consider the sums\n$$\n\\begin{align*} \nS_1 &= a_1 + a_2 + a_3 + a_4 + a_5, \\\nS_2 &= a_2 + a_3 + a_4 + a_5 + a_6, \\\n\\vdots \\\nS_{17} &= a_{17} + a_{18} + a_{19} + a_{20} + a_{21}, \\\nS_{18} &= a_{18} + a_{19} + a_{20} + a_{21} + a_1. \n\\end{align*}\n$$\n\n![](attached_image_1.png)\n\nShow that at least two of the 18 sums leave different remainders when divided by 5.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $S_1, S_2, \\dots, S_{18}$ leave the same remainder when divided by 5. Since $S_1 = a_1 + (a_2 + a_3 + a_4 + a_5)$ and $S_2 = (a_2 + a_3 + a_4 + a_5) + a_6$, if $S_1$ and $S_2$ leave the same remainder when divided by 5, then $a_1$ and $a_6$ leave the same remainder when divided by 5. In the same way:\n\n1. $a_1, a_6, a_{11}, a_{16}$ and $a_{21}$ leave the same remainder, $x$, when divided by 5.\n2. $a_2, a_7, a_{12}, a_{17}$ leave the same remainder, $a$, when divided by 5.\n3. $a_3, a_8, a_{13}, a_{18}$ leave the same remainder, $b$, when divided by 5.\n4. $a_4, a_9, a_{14}, a_{19}$ leave the same remainder, $c$, when divided by 5.\n5. $a_5, a_{10}, a_{15}, a_{20}$ leave the same remainder, $d$, when divided by 5.\n\nSince $\\{a_1, a_2, \\dots, a_{21}\\} = \\{1, 2, \\dots, 21\\}$, there are 5 remainders equal to 1 and 4 remainders equal to each of 2, 3, 4 or 0.\nIt follows $x = 1$ and $\\{a, b, c, d\\} = \\{0, 2, 3, 4\\}$.\n\nNow $S_1 = \\mathcal{M}5 + 1 + 0 + 2 + 3 + 4 = \\mathcal{M}5$ and $S_{18} = \\mathcal{M}5 + b + \\mathcal{M}5 + c + \\mathcal{M}5 + d + \\mathcal{M}5 + 1 + \\mathcal{M}5 + 1 = \\mathcal{M}5 + 2 + (a + b + c + d) - a = \\mathcal{M}5 + 11 - a$.\n\nSince $S_1$ and $S_{18}$ leave the same remainder when divided by 5, it follows that $11 - a = \\mathcal{M}5$, whence $a = 1$ – contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38488,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex hexagon $A_1B_1A_2B_2A_3B_3$ is inscribed in a circle $\\Omega$ of radius $R$. The diagonals $A_1B_2$, $A_2B_3$, and $A_3B_1$ concur at $X$. For $i = 1, 2, 3$, let $\\omega_i$ be the circle tangent to the segments $XA_i$ and $XB_i$, and to the arc $A_iB_i$ of $\\Omega$ not containing other vertices of the hexagon; let $r_i$ be the radius of $\\omega_i$.\n\na) Prove that $R \\ge r_1 + r_2 + r_3$.\n\nb) If $R = r_1 + r_2 + r_3$, prove that the six points where the circles $\\omega_i$ touch the diagonals $A_1B_2$, $A_2B_3$, $A_3B_1$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $\\ell_1$ be the tangent to $\\Omega$ parallel to $A_2B_3$, lying on the same side of $A_2B_3$ as $\\omega_1$. The tangents $\\ell_2$ and $\\ell_3$ are defined similarly. The lines $\\ell_1$ and $\\ell_2$, $\\ell_2$ and $\\ell_3$, $\\ell_3$ and $\\ell_1$ meet at $C_3$, $C_1$, $C_2$, respectively (see Fig. ??). Finally, the line $C_2C_3$ meets the rays $XA_1$ and $XB_1$ emanating from $X$ at $S_1$ and $T_1$, respectively; the points $S_2$, $T_2$, and $S_3$, $T_3$ are defined similarly.\nEach of the triangles $\\triangle \\Delta_1 = \\triangle XS_1T_1$, $\\triangle \\Delta_2 = \\triangle T_2XS_2$, and $\\triangle \\Delta_3 = \\triangle S_3T_3X$ is similar to $\\triangle \\Delta = \\triangle C_1C_2C_3$, since their corresponding sides are parallel. Let $k_i$ be the ratio of similitude of $\\triangle \\Delta_i$ and $\\triangle \\Delta$ (e.g., $k_1 = XS_1/C_1C_2$ and the like). Since $S_1X = C_2T_3$ and $XT_2 = S_3C_1$, it follows that $k_1 + k_2 + k_3 = 1$, so, if $\\varrho_i$ is the inradius of $\\triangle \\Delta_i$, then $\\varrho_1 + \\varrho_2 + \\varrho_3 = R$.\nFinally, notice that $\\omega_i$ is interior to $\\triangle \\Delta_i$, so $r_i \\le \\varrho_i$, and the conclusion follows by the preceding.\n\n\nb) By part a), the equality $R = r_1 + r_2 + r_3$ holds if and only if $r_i = \\varrho_i$ for all $i$, which implies in turn that $\\omega_i$ is the incircle of $\\triangle \\Delta_i$. Let $K_i, L_i, M_i$ be the points where $\\omega_i$ touches the sides $XS_i, XT_i, S_iT_i$, respectively. We claim that the six points $K_i$ and $L_i$ ($i = 1, 2, 3$) are equidistant from $X$.\nClearly, $XK_i = XL_i$, and we are to prove that $XK_2 = XL_1$ and $XK_3 = XL_2$. By similarity, $\\angle T_1M_1L_1 = \\angle C_3M_1M_2$ and $\\angle S_2M_2K_2 = \\angle C_3M_2M_1$, so the points $M_1, M_2, L_1, K_2$ are collinear. Consequently, $\\angle XK_2L_1 = \\angle C_3M_1M_2 = \\angle C_3M_2M_1 = \\angle XL_1K_2$, so $XK_2 = XL_1$. Similarly, $XK_3 = XL_2$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be integers such that $k \\ge 2$ and $k \\le n \\le 2k-1$. Place rectangular tiles, each of size $1 \\times k$ or $k \\times 1$, on an $n \\times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.",
    "options": [],
    "answer": "Minimum number of tiles = n if n = k; and min(n, 2n − 2k + 2) if k < n < 2k.",
    "solution": "The required minimum is $n$ if $n = k$, and it is $\\min(n, 2n-2k+2)$ if $k < n < 2k$.\n\nThe case $n=k$ being clear, assume henceforth $k < n < 2k$. Begin by describing maximal arrangements on the board $[0, n] \\times [0, n]$, having the above mentioned cardinalities.\n\nIf $k < n < 2k-1$, then $\\min(n, 2n-2k+2) = 2n-2k+2$. To obtain a maximal arrangement of this cardinality, place four tiles, $[0, k] \\times [0, 1]$, $[0, 1] \\times [1, k+1]$, $[1, k+1] \\times [k, k+1]$ and $[k, k+1] \\times [0, k]$ in the square $[0, k] \\times [0, k]$, stack $n-k-1$ horizontal tiles in the rectangle $[1, k+1] \\times [k+1, n]$, and erect $n-k-1$ vertical tiles in the rectangle $[k+1, n] \\times [1, k+1]$.\n\nIf $n = 2k-1$, then $\\min(n, 2n-2k+2) = n = 2k-1$. A maximal arrangement of $2k-1$ tiles is obtained by stacking $k-1$ horizontal tiles in the rectangle $[0, k] \\times [0, k-1]$, another $k-1$ horizontal tiles in the rectangle $[0, k] \\times [k, 2k-1]$, and adding the horizontal tile $[k-1, 2k-1] \\times [k-1, k]$.\n\nThe above examples show that the required minimum does not exceed the mentioned values.\n\nTo prove the reverse inequality, consider a maximal arrangement and let $r$, respectively $c$, be the number of rows, respectively columns, not containing a tile.\n\nIf $r=0$ or $c=0$, the arrangement clearly contains at least $n$ tiles.\n\nIf $r$ and $c$ are both positive, we show that the arrangement contains at least $2n-2k+2$ tiles. To this end, we will prove that the rows, respectively columns, not containing a tile are consecutive. Assume this for the moment, to notice that these $r$ rows and $c$ columns cross to form an $r \\times c$ rectangular array containing no tile at all, so $r < k$ and $c < k$ by maximality. Consequently, there are $n-r \\ge n-k+1$ rows containing at least one horizontal tile each, and $n-c \\ge n-k+1$ columns containing at least one vertical tile each, whence a total of at least $2n-2k+2$ tiles.\n\nWe now show that the rows not containing a tile are consecutive; columns are dealt with similarly. Consider a horizontal tile $T$. Since $n < 2k$, the nearest horizontal side of the board is at most $k-1$ rows away from the row containing $T$. These rows, if any, cross the $k$ columns $T$ crosses to form a rectangular array no vertical tile fits in. Maximality forces each of these rows to contain a horizontal tile and the claim follows.\n\nConsequently, the cardinality of every maximal arrangement is at least $\\min(n, 2n-2k+2)$, and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38490,
    "subject": "Mathematics (Multi-modal)",
    "question": "The unit squares of a $n \\times n$ board, $n \\ge 2$, are colored either black or white so that any black square has at least three white neighbors (a neighbor is a unit square with a common side). What is the maximum number of black unit squares?",
    "options": [],
    "answer": "If n is odd: (n^2 − 1)/2; if n is even: (n^2 − 4)/2.",
    "solution": "The answer is $\\frac{n^2-1}{2}$ if $n$ is odd and $\\frac{n^2-4}{2}$ if $n$ is even.\n\nNotice that:\n\n* the unit squares from the corners are white (because they only have two neighbors, not three);\n* the other squares from the border of the board have only three neighbors, so there can't be two consecutive black squares on the border;\n* any $2 \\times 2$ square contains at most two black squares (if not, a black square would already have two black neighbors, so it would have at most two white neighbors).\n\nIf $n$ is odd, color the board like a chessboard with white corners. This coloring satisfies the requirements of the problem and it has $\\frac{n^2-1}{2}$ black squares, so the maximum number of black squares is at least $\\frac{n^2-1}{2}$.\n\nOn the other hand, split the board in 4 parts: one upper-left $(n-1) \\times (n-1)$ square, a unit square in the right bottom corner, a $(n-1) \\times 1$ rectangle on the right side and a $1 \\times (n-1)$ rectangle on the bottom side. Tiling the $(n-1) \\times (n-1)$ square with $2 \\times 2$ squares and the two rectangles with 2-square dominoes, according with the remarks made at the beginning, one can have at most\n$$\n\\frac{(n-1)^2}{2} + \\frac{n-1}{2} + \\frac{n-1}{2} + 0 = \\frac{n^2-1}{2}\n$$\nblack squares.\n\nIf $n$ is even, again consider a chessboard-like coloring. This makes two white corners and two black corners; recolor the black corners as white. The configuration we have now satisfies the requirements of the problem and it has $\\frac{n^2-4}{2}$ black squares, so the maximum number of black squares is at least $\\frac{n^2-4}{2}$.\n\nSplit the board in 9 parts: one $(n-2) \\times (n-2)$ square in the middle, four unit squares in the corners and four $(n-2) \\times 1$ or $1 \\times (n-2)$ rectangles on the borders. Tiling the $(n-2) \\times (n-2)$ square with $2 \\times 2$ squares and the rectangles with 2-square dominoes, the remarks made at the beginning lead to a maximum of\n$$\n\\frac{(n-2)^2}{2} + 4 \\cdot \\frac{n-2}{2} + 4 \\cdot 0 = \\frac{n^2-4}{2}\n$$\nblack squares. Hence, if $n$ is even, the maximum number of black squares is $\\frac{n^2-4}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38491,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a positive integer *exquisite* if it is a multiple of the number of its divisors (for instance, $12$ is exquisite because it has $6$ divisors and $12$ is a multiple of $6$).\n\na) Find the largest exquisite two digit number.\n\nb) Prove that no exquisite number has its last digit $3$.",
    "options": [],
    "answer": "a) 96; b) No exquisite number ends with last digit 3",
    "solution": "a) We check the numbers decreasingly: $99 = 3^2 \\cdot 11$, $99$ has $6$ divisors and $6 \\nmid 99$; $98 = 2 \\cdot 7^2$, $98$ has $6$ divisors and $6 \\nmid 98$; $97 = 97$, $97$ has $2$ divisors and $2 \\nmid 97$; $96 = 2^5 \\cdot 3$, $96$ has $12$ divisors and $12 \\mid 96$. So $96$ is the largest exquisite two digit number.\n\nb) Let $X$ be a positive integer with its last digit $3$.\n\nThen $X$ is odd and $X$ is not a perfect square, and a non-perfect square has an even number of divisors. Since an odd number cannot be a multiple of an even number, $X$ cannot be exquisite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $k$ and $m$, show that $m$ and $\\binom{n}{k}$ are coprime for infinitely many integers $n \\ge k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = k + l m k!$, where $l$ is an arbitrary nonnegative integer, let $p$ be any prime factor of $m$, and let $p^h$ be the highest power of $p$ that divides $k!$ — that is, $p^h$ divides $k!$ but $p^{h+1}$ does not. Notice that $n \\equiv k \\pmod{p^{h+1}}$, to deduce that $n(n-1)\\cdots(n-k+1) \\equiv k! \\pmod{p^{h+1}}$, so $p^h$ is also the highest power of $p$ that divides the product $n(n-1)\\cdots(n-k+1)$. Consequently, $p$ does not divide $\\binom{n}{k}$, so $m$ and $\\binom{n}{k}$ are indeed coprime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcircle of a triangle $ABC$. A circle $k$ is tangent to the lines $BC$, $CA$, $AB$ at points $D, E, F$, respectively, such that $A$ is on the other side of the line $BC$ with respect to the circle. Suppose the circle $k$ is equal to the circumcircle of the triangle. Prove that lines $OD$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I_a$ be the center of the circle $k$ and let $T$ be the midpoint of the arc $BC$ - not containing $A$ - of the circumcircle $ABC$. Notice that $OT$ is the perpendicular bisector of the line segment $BC$ to deduce that $OT \\perp BC$. As $I_a D \\perp BC$ and $OT = I_a D = R$, the quadrangle $ODI_aT$ is a parallelogram. Consequently $OD \\parallel TI_a$, or, equivalently, $OD \\parallel AI_a$.\n\nOn the other hand, since $AE = AF$ and $AI_a$ bisects angle $\\angle FAE$, the lines $AI_a$ and $EF$ are perpendicular, and so are lines $OD$ and $EF$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38494,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $A$ and $B$. An $\\mathcal{A}\\mathcal{B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $A$ and the other in $B$, and such that no segments form a closed polyline. An $\\mathcal{A}\\mathcal{B}$-tree is transformed into another as follows: choose three distinct segments $A_1B_1$, $B_1A_2$ and $A_2B_2$ in the $\\mathcal{A}\\mathcal{B}$-tree such that $A_1$ is in $A$ and $A_1B_1 + A_2B_2 > A_1B_2 + A_2B_1$, and remove the segment $A_1B_1$ to replace it by the segment $A_1B_2$. Given any $\\mathcal{A}\\mathcal{B}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps.\nRussia",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $A$ and $B$, and transforming one into another preserves the degree of any vertex in $A$, but not necessarily that of a vertex in $B$.\n\nThe idea is to devise a strict semi-invariant of the process, i.e., assign each $\\mathcal{A}\\mathcal{B}$-tree a real number strictly decreasing under a transformation. Since the number of trees on the $n$ points is finite, the conclusion follows.\n\n$$\n\\begin{align*}\nf(\\mathcal{T}') - f(\\mathcal{T}) &= p_{\\mathcal{T}'}(A_1B_2) \\cdot A_1B_2 + (p_{\\mathcal{T}'}(A_2B_1) - p_{\\mathcal{T}}(A_2B_1)) \\cdot A_2B_1 + \\\\\n& \\qquad (p_{\\mathcal{T}'}(A_2B_2) - p_{\\mathcal{T}}(A_2B_2)) \\cdot A_2B_2 - p_{\\mathcal{T}}(A_1B_1) \\cdot A_1B_1 \\\\\n&= p_{\\mathcal{T}}(A_1B_1)(A_1B_2 + A_2B_1 - A_2B_2 - A_1B_1) < 0.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $M$ be the midpoint of the side $AC$. A circle through $B$ and $M$ meets the sides $AB$ and $BC$ again at $P$ and $Q$, respectively. The reflection $T$ of $B$ across the midpoint of the segment $PQ$ lies on the circle $ABC$. Evaluate the ratio $BT/BM$.",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "The required ratio equals $\\sqrt{2}$. To prove this, let $S$ be the midpoint of the segment $PQ$, and let $B'$ be the reflection of $B$ across $M$. Clearly, $ABCB'$ is a parallelogram, $\\angle ABB' = \\angle PQM$, and $\\angle BB'A = \\angle B'BC = \\angle MPQ$, so the triangles $ABB'$ and $MQP$ are similar. Since $AM$ and $MS$ are corresponding medians in these triangles,\n$$\n\\angle SMP = \\angle B'AM = \\angle BCA = \\angle BTA. \\qquad (1)\n$$\nNext, $\\angle ACT = \\angle PBT$ and $\\angle TAC = \\angle TBC = \\angle BTP$, so the triangles $TCA$ and $PBT$ are similar. Since $TM$ and $PS$ are corresponding medians in these triangles,\n$$\n\\angle MTA = \\angle TPS = \\angle BQP = \\angle BMP. \\qquad (2)\n$$\nIf $S$ does not lie on the segment $BM$, we may and will assume that $S$ and $A$ both lie on the same side of the line $BM$, since the configuration is symmetric in $A$ and $C$. By (1) and (2), $\\angle BMS = \\angle BMP - \\angle SMP = \\angle MTA - \\angle BTA = \\angle MTB$, so the triangles $BSM$ and $BMT$ are similar, whence $BM^2 = BS \\cdot BT = BT^2/2$; that is, $BT/BM = \\sqrt{2}$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nIf $S$ lies on the segment $BM$, then (2) shows that $\\angle BCA = \\angle MTA = \\angle BMP = \\angle BQP$, so $(PQ, AC)$ and $(PM, AT)$ are pairs of parallel lines. Consequently, $BS/BM = BP/BA = BM/BT$, so $BT^2 = 2BM^2$, and again $BT/BM = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$ and $b$ so that $\\frac{a+1}{b}$ and $\\frac{b+2}{a}$ are simultaneously positive integers.",
    "options": [],
    "answer": "(a,b) ∈ {(1,1), (1,2), (3,1), (3,4), (5,3)}",
    "solution": "Since the fractions $f_1 = \\frac{a+1}{b}$ and $f_2 = \\frac{b+2}{a}$ are positive integers, $a+1 \\ge b$ and $b+2 \\ge a$, hence $a+1 \\ge b \\ge a-2$. This leaves the cases:\n\n1. $b = a + 1$: then $f_2 = 1 + \\frac{3}{a} \\in \\mathbb{N}$, whence $a = 1, b = 2$ or $a = 3, b = 4$, which are convenient values.\n\n2. $b = a$: then $f_1 = 1 + \\frac{1}{a} \\in \\mathbb{N}$, whence $a = 1, b = 1$, which are convenient values.\n\n3. $b = a - 1$: then $f_2 = 1 + \\frac{1}{a} \\in \\mathbb{N}$, whence $a = 1, b = 1$.\n\n4. $b = a - 2$: then $f_1 = 1 + \\frac{3}{b} \\in \\mathbb{N}$, whence $a = 3, b = 1$ or $a = 5, b = 3$, which are convenient values.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the set of all positive integers $n$ such that $n^4$ has a divisor in the range $n^2+1, n^2+2, \\dots, n^2+2n$. Prove that there are infinitely many elements of $S$ of each of the forms $7m$, $7m+1$, $7m+2$, $7m+5$, $7m+6$ and no elements of $S$ of the form $7m+3$ or $7m+4$, where $m$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma.** The fourth power of a positive integer $n$ has a divisor in the range $n^2+1, n^2+2, \\dots, n^2+2n$ if and only if at least one of the numbers $2n^2+1$ and $12n^2+9$ is a perfect square.\nConsequently, a positive integer $n$ is a member of $S$ if and only if $m^2 - 2n^2 = 1$ or $m^2 - 12n^2 = 9$ for some positive integer $m$.\n\nThe former is a Pell equation whose solutions are $(m_1, n_1) = (3, 2)$ and\n$$\n(m_{k+1}, n_{k+1}) = (3m_k + 4n_k, 2m_k + 3n_k), \\quad k = 1, 2, 3, \\dots\n$$\n\nThe other equation is easily transformed into a Pell equation, $m'^2 - 12n'^2 = 1$, by noticing that $m$ and $n$ are both divisible by $3$, say $m = 3m'$ and $n = 3n'$. In this case, the solutions are $(m_1, n_1) = (21, 6)$ and\n$$\n(m_{k+1}, n_{k+1}) = (7m_k + 24n_k, 2m_k + 7n_k), \\quad k = 1, 2, 3, \\dots\n$$\nThis time iteration shows that $(m_{k+4}, n_{k+4}) \\equiv (m_k, n_k)$. Since $(m_1, n_1) \\equiv (0, -1)$, $(m_2, n_2) \\equiv (-3, 0)$, $(m_3, n_3) \\equiv (0, 1)$, and $(m_4, n_4) \\equiv (3, 0)$, it follows that $S$ contains infinitely many integers from each of the residue classes $0$ and $\\pm 1$ modulo $7$.\nFinally, since the $n_k$ from the two sets of formulae exhaust $S$, by the preceding no integer in the residue classes $\\pm 3$ modulo $7$ is a member of $S$.\n\nWe now turn to the proof of the lemma. Let $n$ be a member of $S$, and let $d = n^2 + m$ be a divisor of $n^4$ in the range $n^2 + 1, n^2 + 2, \\dots, n^2 + 2n$, so $1 \\le m \\le 2n$. Consideration of the square of $n^2 = d - m$ shows $m^2$ divisible by $d$, so $m^2/d$ is a positive integer. Since $n^2 < d < (n+1)^2$, it follows that $d$ is not a square; in particular, $m^2/d \\ne 1$, so $m^2/d \\ge 2$. On the other hand, $1 \\le m \\le 2n$, so $m^2/d = m^2/(n^2+m) \\le 4n^2/(n^2+1) < 4$. Consequently, $m^2/d = 2$ or $m^2/d = 3$; that is, $m^2/(n^2+m) = 2$ or $m^2/(n^2+m) = 3$. In the former case, $2n^2+1 = (m-1)^2$, and in the latter, $12n^2+9 = (2m-3)^2$.\n\nConversely, if $2n^2+1 = m^2$ for some positive integer $m$, then $1 < m^2 < 4n^2$, so $1 < m < 2n$, and $n^4 = (n^2+m+1)(n^2-m+1)$, so the first factor is the desired divisor.\nSimilarly, if $12n^2+9 = m^2$ for some positive integer $m$, then $m$ is odd, $n \\ge 6$, and $n^4 = (n^2+m/2+3/2)(n^2-m/2+3/2)$, and again the first factor is the desired divisor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive numbers with $abc \\ge 1$. Prove that\n$$\n\\frac{1}{a^3 + 2b^3 + 6} + \\frac{1}{b^3 + 2c^3 + 6} + \\frac{1}{c^3 + 2a^3 + 6} \\le \\frac{1}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $a^3 + b^3 + 1 \\ge 3ab$ and $b^3 + 1 + 1 \\ge 3b$ to obtain that $\\frac{1}{a^3+2b^3+6} \\le \\frac{1}{3ab+3b+3}$, hence it is sufficient to show that $\\frac{1}{ab+b+1} + \\frac{1}{bc+c+1} + \\frac{1}{ca+a+1} \\le 1$.\n\nTo this end, observe that $\\frac{1}{ab+b+1} + \\frac{1}{bc+c+1} + \\frac{1}{ca+a+1} = \\frac{1}{ab+b+1} + \\frac{ab}{ab^2c+abc+ab} + \\frac{b}{abc+ab+b} \\stackrel{abc \\ge 1}{\\le} \\frac{1}{ab+b+1} + \\frac{ab}{b+1+ab} + \\frac{b}{1+ab+b} = \\frac{1+ab+b}{ab+b+1} = 1$.\n\nAlternative Solution:\n\nSubtract $1/6$ from each of the left hand-side summands and write successively $\\sum_{cyc} \\left(\\frac{1}{a^3+2b^3+6} - \\frac{1}{6}\\right) \\le \\frac{1}{3} - \\frac{1}{2}$, then $\\sum_{cyc} \\frac{-a^3-2b^3}{6(a^3+2b^3+6)} \\le -\\frac{1}{6}$ or further $\\sum_{cyc} \\frac{a^3+2b^3}{a^3+2b^3+6} \\ge 1$.\n\nTo this end, notice that\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{a^3}{a^3 + 2b^3 + 6} &= \\sum_{cyc} \\frac{a^4}{a^4 + 2ab^3 + 6a} \\\\\n&\\stackrel{CBS}{\\ge} \\frac{(a^2 + b^2 + c^2)^2}{a^4 + b^4 + c^4 + 2(ab^3 + bc^3 + ca^3) + 6(a + b + c)} \\stackrel{(1)}{\\ge} \\frac{1}{3}.\n\\end{aligned}\n$$\n\nThe inequality (1) rewrites $3(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2a^2) \\ge a^4+b^4+c^4+2(ab^3+bc^3+ca^3)+6(a+b+c)$, and follows from $2(a^4+b^4+c^4) \\ge 2(ab^3+bc^3+ca^3)$ and $6(a^2b^2+b^2c^2+c^2a^2) \\ge 6(ab \\cdot bc+bc \\cdot ca+ca \\cdot ab) = 6abc(a+b+c) \\ge 6(a+b+c)$.\n\nOn the other hand,\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{b^3}{a^3 + 2b^3 + 6} &= \\sum_{cyc} \\frac{b^4}{ba^3 + 2b^4 + 6b} \\\\\n&\\stackrel{CBS}{\\ge} \\frac{(a^2 + b^2 + c^2)^2}{2(a^4 + b^4 + c^4) + (ba^3 + cb^3 + ac^3) + 6(a+b+c)} \\\\\n&\\stackrel{(2)}{\\ge} \\frac{1}{3}.\n\\end{aligned}\n$$\n\nThe inequality (2) rewrites $3(a^4+b^4+c^4)+6(a^2b^2+b^2c^2+c^2a^2) \\ge 2(a^4+b^4+c^4)+(ba^3+cb^3+ac^3)+6(a+b+c)$, and follows from $a^4+b^4+c^4 \\ge ba^3+cb^3+ac^3$ and $6(a^2b^2+b^2c^2+c^2a^2) \\ge 6(ab \\cdot bc+bc \\cdot ca+ca \\cdot ab) = 6abc(a+b+c) \\ge 6(a+b+c)$.\n\nAlternative Solution:\n\nAs $a^3 + 2b^3 = a^3 + b^3 + b^3 \\ge 3ab^2$, it suffices to prove that $\\frac{1}{ab^2+2} + \\frac{1}{bc^2+2} + \\frac{1}{ca^2+2} \\le 1$. Rewrite the inequality as $-\\frac{ab^2}{2(ab^2+2)} - \\frac{bc^2}{2(bc^2+2)} - \\frac{ca^2}{2(ca^2+2)} \\le 1 - \\frac{3}{2}$, or, equivalently,\n$$\n\\frac{ab^2}{ab^2+2} + \\frac{bc^2}{bc^2+2} + \\frac{ca^2}{ca^2+2} \\ge 1.\n$$\nRecall that $abc \\ge 1$, and write $\\frac{ab^2}{ab^2+2abc} + \\frac{bc^2}{bc^2+2abc} + \\frac{ca^2}{ca^2+2abc} \\ge 1$ to observe that it is enough to show that $\\frac{b}{b+2c} + \\frac{c}{c+2a} + \\frac{a}{a+2b} \\ge 1$. Indeed, $\\frac{b}{b+2c} + \\frac{c}{c+2a} + \\frac{a}{a+2b} = \\frac{b^2}{b^2+2bc} + \\frac{c^2}{c^2+2ca} + \\frac{a^2}{a^2+2ab} \\stackrel{CBS}{\\ge} \\frac{(a+b+c)^2}{b^2+2bc+c^2+2ca+a^2+2ab} = 1$, which concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with legs $AB$ and $AC$. The bisector of the angle $ACB$ intersects $AB$ in $D$ and the perpendicular in $B$ on $BC$ in $E$. Denote $F$ the reflection of $E$ across $B$ and $P$ the intersection of the lines $DF$ and $BC$. Prove that $EP \\perp CF$.\nCătălin Cristea",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In $\\triangle BEC$, $m(\\angle CEB) = 180^\\circ - m(\\angle EBC) - m(\\angle ECB) = 90^\\circ - m(\\angle ECB)$. In $\\triangle ADC$, $m(\\angle ADC) = 180^\\circ - m(\\angle DAC) - m(\\angle ACD) = 90^\\circ - m(\\angle ACD)$. Since $m(\\angle ACD) = m(\\angle ECB)$, it follows $\\angle ADC \\equiv \\angle CEB$.\nNow $\\angle ADC \\equiv \\angle EDB$. This yields $[BD] = [BE]$. Since $[BE] = [BF]$, we infer that $[BD] = [BE] = [BF]$, so $\\triangle DEF$ has the right angle $D$.\nIn $\\triangle CEF$, $CB$ and $FD$ are altitudes, hence $P$ is the orthocenter. In conclusion, $EP$ is an altitude, that is $EP \\perp CF$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38500,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) If $(a_n)_{n \\ge 1}$ is a strictly increasing sequence of positive integers such that $(a_{2n-1} + a_{2n})/a_n$ is constant as $n$ runs through all positive integers, then this constant is an integer greater than or equal to $4$; and\n\nb) Given an integer $N \\ge 4$, there exists a strictly increasing sequence $(a_n)_{n \\ge 1}$ of positive integers such that $(a_{2n-1} + a_{2n})/a_n = N$ for all indices $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Clearly, $K = (a_{2n-1} + a_{2n})/a_n$ is a positive rational number. In fact, $K$ must be integral. To prove this, write $K = p/q$ in lowest terms to deduce that the $a_n$ are all divisible by $q$. Divide them all by $q$ to obtain a new sequence whose corresponding ratios are again $K$. Repetition of the process to the new sequence and its successors shows that the $a_n$ are all divisible by arbitrarily large powers of $q$, so $q=1$ and $K$ is indeed integral.\nSince the $a_n$ form a strictly increasing sequence, it follows that $K > 2$, and since the latter is integral, it is at least $3$.\nTo rule out the case $K = 3$, we consider the positive integers $b_n = a_{n+1} - a_n$, show that for every index $m$ there exists an index $n > m$ such that $b_n < b_m$ and reach thereby a contradiction. Indeed, if $K = 3$, then $3b_n = b_{2n-1} + 2b_{2n} + b_{2n+1}$, so at least one of the three $b$'s in the right-hand member must be less than $b_n$. Consequently, $K \\ge 4$.\n\nb) If $N = 4$, let $a_n = 2n - 1$; in this case, the verifications are obvious. If $N \\ge 5$, set $a_1 = 1$ and let $a_{2n-1} = \\lfloor (N a_n - 1)/2 \\rfloor$ and $a_{2n} = \\lfloor N a_n/2 \\rfloor + 1$. This sequence satisfies the required ratio condition, $a_{2n-1}$ is obviously less than $a_{2n}$, and it is sufficient to prove that $a_{2n} < a_{2n+1}$. This can be done by noticing that $a_2 < a_3$, and showing that if $a_n < a_{n+1}$, then $a_{2n} < a_{2n+1}$. Indeed, $a_{2n+1} - a_{2n} \\ge (N a_{n+1} - 2)/2 - (N a_n/2 + 1) = N(a_{n+1} - a_n)/2 - 2 \\ge N/2 - 2 \\ge 1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38501,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Given any positive integer $n$, prove that every $n$ points in the closed unit square $[0, 1] \\times [0, 1]$ can be joined by a path of length less than $2\\sqrt{n} + 4$.\n\nb) Prove that there exist $n$ points in the closed unit square $[0, 1] \\times [0, 1]$ that cannot be joined by a path of length less than $\\sqrt{n} - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $C$ be an $n$-point configuration in the closed unit square $[0, 1] \\times [0, 1]$, let $m = \\lfloor\\sqrt{n}\\rfloor$, and consider the snake going horizontally from $0 \\times 0$ to $1 \\times 0$, then vertically up from $1 \\times 0$ to $1 \\times 1/m$, then horizontally back from $1 \\times 1/m$ to $0 \\times 1/m$, vertically up from $0 \\times 1/m$ to $0 \\times 2/m$, horizontally over to $1 \\times 2/m$, and so on and so forth all the way up to $1 \\times 1$ or $0 \\times 1$, depending on whether $m$ is even or odd. The length of the snake is $m + 1 + m \\cdot 1/m = m + 2 \\le \\sqrt{n} + 2$. Of course, the snake does not necessarily pass through any point in $C$, but it comes within $1/(2m)$ of $C$. Thus, in tracing the snake, visit each point of $C$ by darting out, if necessary, to the nearest points in $C$ abreast within $1/(2m)$, and then dart back. This increases the length by at most $n \\cdot 2 \\cdot 1/(2m) = n/m < \\sqrt{n} + 2$, so the length of the visiting path is certainly less than $2\\sqrt{n} + 4$.\n\nb) Let again $m = \\lfloor\\sqrt{n}\\rfloor$, and consider an $n$-point subconfiguration $C$ of the lattice $\\{i/m \\times j/m: i, j = 0, 1, \\dots, m\\}$. Since any two distinct points in the lattice are at least $1/m$ distance apart, the length of a path through all of $C$ is at least $(n-1) \\cdot 1/m \\ge (n-1)/\\sqrt{n} \\ge \\sqrt{n} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be positive integers so that there exists a prime number $p$ with the property $[a, a + p] = [b, b + p]$. Prove that $a = b$.\n\nHere $[x, y]$ denotes the least common multiple of $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $d$ is a common divisor of $x$ and $x+p$, then $d \\mid (x+p) - x = p$, hence $d = 1$ or $d = p$. This leads to the cases:\n\ni. $p \\nmid a$. Then $p \\nmid a + p$, $p \\nmid [a, a + p]$, hence $p \\nmid [b, b + p]$, $p \\nmid b$, $p \\nmid b + p$, the numbers $a, a+p$ and $b, b+p$ are co-prime and $[a, a+p] = a(a+p)$, $[b, b+p] = b(b+p)$. It follows that $a(a+p) = b(b+p)$, therefore $a = b$: if, for instance, $a > b$, then $a(a+p) > b(b+p)$.\n\nii. $p \\mid a$. Then $p \\mid a+p$, $p \\mid [a, a+p]$, hence $p \\mid [b, b+p]$, $p \\mid b$, $p \\mid b+p$, $a = pa'$ and $b = pb'$, so $[a, a+p] = pa'(a'+1)$, $[b, b+p] = pb'(b'+1)$. This yields $a'(a'+1) = b'(b'+1)$, hence $a' = b'$, therefore $a = b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $2$ and consider the set $A = \\{2^n - 1, 3^n - 1, \\dots, (n-1)^n - 1\\}$. Given that $n$ does not divide any element of $A$, prove that $n$ is a square-free number. Does it necessarily follow that $n$ is a prime number?\n\nMarius Bocanu",
    "options": [],
    "answer": "n is square-free; not necessarily prime (for example, n = 15).",
    "solution": "Suppose not and write $n = p a$ for a prime $p$ and a number $a > 1$ with $p \\mid a$. Notice that $(a+1)^n - 1 = a((a+1)^{n-1} + (a+1)^{n-2} + \\dots + 1)$ and $a+1 \\equiv 1 \\pmod{p}$ to infer that $n$ divides $(a+1)^n - 1$, a contradiction.\n\nFurther, $n$ needs not be a prime number; take for example $n = 15 = 3 \\cdot 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, determine all functions $f$ from the first $n$ positive integers to the positive integers, satisfying the following two conditions:\n$$\n(1) \\sum_{k=1}^{n} f(k) = 2n;\n$$\n$$\n(2) \\sum_{k \\in K} f(k) = n \\text{ for no subset } K \\text{ of the first } n \\text{ positive integers.}\n$$",
    "options": [],
    "answer": "For odd n: either f(k) = 2 for all k, or there exists i such that f(i) = n + 1 and f(j) = 1 for all j ≠ i. For even n: the only solutions are those with one index i having f(i) = n + 1 and all other values equal to 1.",
    "solution": "If $n$ is odd, the required functions are the constant function $f_0 \\equiv 2$ along with the $n$ functions $f_i: \\{1, \\dots, n\\} \\to \\mathbb{N}^*$,\n$$\nf_i(j) = \\begin{cases} n+1, & \\text{if } j = i, \\\\ 1, & \\text{if } j \\neq i, \\end{cases} \\quad i = 1, \\dots, n;\n$$\n\nnotice that $f_0 = f_1$ if $n = 1$. If $n$ is even, $f_0$ is to be removed from the list, by (2). All these functions plainly satisfy the conditions in the statement.\nLabelling the positive integers in the list $f(1), \\dots, f(n)$ increasingly, $a_1 \\le \\dots \\le a_n$, the problem amounts to determining all lists of positive integers $a_1 \\le \\dots \\le a_n$ such that $(1') \\sum_{k=1}^n a_k = 2n$ and $(2') \\sum_{k \\in K} a_k = n$ for no subset $K$ of the first $n$ positive integers. Notice that $a_n \\le n+1$, by $(1')$, and $a_n \\ne n$, by $(2')$. With reference again to $(1')$, notice further that if $a_n = n+1$, then the other $a_k$ are all 1, and if $a_n = 2$, then so are the other $a_k$. If $n$ is even, $(2')$ rules out the latter case.\nLeaving aside the trivial cases $n=1$ and $n=2$, let $n \\ge 3$. To rule out $a_n \\ne 2, n+1$, assume, if possible, this is the case, and notice that\n$$\na_1 - a_n, \\quad 0, \\quad a_1, \\quad a_1 + a_2, \\quad \\dots, \\quad a_1 + a_2 + \\dots + a_{n-1}\n$$\nare pairwise distinct integers, so at least two are congruent modulo $n$. Since $a_n \\ne 2, n, n+1$, the first two cannot be congruent modulo $n$, and $(2')$ rules out the remaining cases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-negative integers $n$ so that $\\sqrt{n+3} + \\sqrt{\\sqrt{n+3}}$ is an integer.",
    "options": [],
    "answer": "All n of the form k^4 − 3 with integers k ≥ 2.",
    "solution": "Denote $m = \\sqrt{n+3} + \\sqrt{\\sqrt{n+3}}$. Then $n + \\sqrt{n+3} = (m - \\sqrt{n+3})^2$, whence $(2m+1)\\sqrt{n+3} = m^2 + 3$. Then there exists $p \\in \\mathbb{N}$ so that $n+3 = p^2$ and, since $p + \\sqrt{n+p} \\in \\mathbb{N}$, there exists $q \\in \\mathbb{N}$ so that $n+p = q^2$.\nEliminating $n$ yields $p^2 - 3 = q^2 - p$, that is $4p^2 + 4p - 12 = 4q^2$, or $(2p+1)^2 - (2q)^2 = 13$, which is equivalent to $(2p+1-2q)(2p+1+2q) = 13$.\nThis gives $p = 3$, so $n = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38506,
    "subject": "Mathematics (Multi-modal)",
    "question": "All the 16 squares of a $4 \\times 4$ array are white. Define a *move* by selecting a rectangle $1 \\times 3$ or $3 \\times 1$ and switching the colors of each of its squares from white to black or from black to white. Is it possible that all squares turn black after a sequence of moves?",
    "options": [],
    "answer": "No",
    "solution": "<table><tr><td>1</td><td>2</td><td>3</td><td>1</td></tr><tr><td>2</td><td>3</td><td>1</td><td>2</td></tr><tr><td>3</td><td>1</td><td>2</td><td>3</td></tr><tr><td>1</td><td>2</td><td>3</td><td>1</td></tr></table>\nto observe that a move will change colors in one square of each number. As initially there are six squares labeled 1, an even number of moves is required to turn black all these squares. On the other hand, five squares were labeled with 2 at the start, requiring an odd number of moves to turn all black, hence the answer is negative.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABCD$ un pătrat şi $E$ un punct situat pe diagonala $BD$, diferit de mijlocul acesteia. Se notează cu $H$ şi $K$ ortocentrele triunghiurilor $ABE$, respectiv $ADE$. Arătaţi că $\\overline{BH} + \\overline{DK} = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSe observă că punctele $H$ şi $K$ se află pe diagonala $AC$, deoarece $AC$ este perpendiculară pe $BE$ şi $DE$.\n\nDe asemenea, $H$ şi $K$ se află pe înălţimile duse din $E$ în cele două triunghiuri, care sunt perpendiculare pe laturile pătratului iniţial.\n\nDeducem că triunghiul $EHK$ este dreptunghic isoscel, aşadar $H$ şi $K$ sunt simetrice faţă de centrul pătratului. Cum şi $B, D$ sunt simetrice faţă de centrul pătratului, obţinem concluzia dorită.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUn inel $(A,+, \\cdot)$ are proprietatea (P) dacă $A$ este finit şi grupul multiplicativ al elementelor sale inversabile este izomorf cu un subgrup diferit de $\\{0\\}$ al grupului aditiv $(A,+)$. Arătaţi că:\n\na. Dacă un inel are proprietatea $(\\mathrm{P})$, atunci numărul elementelor sale este par.\n\nb. Pentru o infinitate de numere naturale $n$, există inele cu exact $n$ elemente, care au proprietatea $(\\mathrm{P})$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Fie $A$ un inel care are proprietatea (P) şi fie $m = |U(A)|$. Rezultă că $(-1)^{m} = 1$.\nDacă $m$ este impar, atunci $-1 = 1$, deci ord$(1) = 2$ în grupul aditiv $(A,+)$ şi prin urmare $|A|$ este par.\n\nDacă $m$ este par, cum $m$ este un divizor al lui $|A|$, rezultă că $|A|$ este par.\n\nb. Fie $m$ un număr natural nenul, fie $n = 2^{m+2}$ şi fie $A = \\underbrace{\\mathbb{Z}_{2} \\times \\cdots \\times \\mathbb{Z}_{2}}_{m} \\times \\mathbb{Z}_{4}$. Atunci grupul multiplicativ $U(A) = \\{(\\hat{1}, \\ldots, \\hat{1}, \\hat{1}), (\\hat{1}, \\ldots, \\hat{1}, \\hat{3})\\}$ este izomorf cu subgrupul $\\{(\\hat{0}, \\ldots, \\hat{0}, \\hat{0}), (\\hat{0}, \\ldots, \\hat{0}, \\hat{2})\\}$ al grupului aditiv $(A,+)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f: \\mathbb{R} \\rightarrow (0, \\infty)$ o funcție continuă și periodică. Dacă $2$ este perioadă a lui $f$, arătați că:\n\na. $\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x \\geq 2$.\n\nb. $\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x = 2$ dacă și numai dacă $1$ este perioadă a lui $f$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Inegalitatea rezultă din relațiile de mai jos:\n$$\n\\begin{aligned}\n\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x &= \\int_{0}^{1} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x + \\int_{1}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x + \\int_{1}^{2} \\frac{f(x-1+2)}{f(x-1+1)} \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x + \\int_{0}^{1} \\frac{f(x+2)}{f(x+1)} \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\left( \\frac{f(x+1)}{f(x)} + \\frac{f(x)}{f(x+1)} \\right) \\, \\mathrm{d}x \\geq \\int_{0}^{1} 2 \\, \\mathrm{d}x = 2\n\\end{aligned}\n$$\n\nb. Dacă $1$ este perioadă a lui $f$, atunci\n$$\n\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x = \\int_{0}^{2} \\, \\mathrm{d}x = 2.\n$$\n\nInvers, dacă $\\int_{0}^{2} \\frac{f(x+1)}{f(x)} \\, \\mathrm{d}x = 2$, atunci\n$$\n\\int_{0}^{1} \\left( \\sqrt{\\frac{f(x+1)}{f(x)}} - \\sqrt{\\frac{f(x)}{f(x+1)}} \\right)^2 \\, \\mathrm{d}x = 0,\n$$\niar din continuitatea lui $f$ rezultă că $\\sqrt{\\frac{f(x+1)}{f(x)}} = \\sqrt{\\frac{f(x)}{f(x+1)}}$, $0 \\leq x \\leq 1$, deci $f(x+1) = f(x)$, oricare ar fi $x$ în $[0,1]$.\n\nDacă $x \\in [1,2]$, atunci $f(x) = f((x-1)+1) = f(x-1) = f((x-1)+2) = f(x+1)$, deci $f(x+1) = f(x)$, oricare ar fi $x$ în $[0,2]$.\n\nÎn fine, dacă $x$ este un număr real oarecare, atunci $2n \\leq x < 2n+2$, pentru un unic număr întreg $n$, și $f(x) = f(x-2n) = f((x-2n)+1) = f((x+1)-2n) = f(x+1)$. Prin urmare, $1$ este perioadă a funcției $f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie matricele $A, B, C, D \\in \\mathcal{M}_{n}(\\mathbb{C}), n \\geq 2$ și $k \\in \\mathbb{R}$ astfel încât $A C + k B D = I_{n}$ și $A D = B C$. Demonstrați că $C A + k D B = I_{n}$ și $D A = C B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPentru început considerăm $k \\neq 0$. Fie $w \\in \\mathbb{C}$ astfel încât $w^{2} = -k$. Considerăm matricele $X = A + w B$, $Y = C - w D$, $Z = A - w B$ și $U = C + w D$. Ipoteza conduce la $X Y = I_{n}$ și $Z U = I_{n}$.\n\nAtunci $Y X = I_{n}$ și $U Z = I_{n}$.\n\nDeducem $(C A + k D B) - w (D A - C B) = I_{n}$ și $(C A + k D B) + w (D A - C B) = I_{n}$, de unde se obține concluzia.\n\nPentru $k = 0$ rezultă $A C = I_{n}$, de unde $C A = I_{n}$. Din $A D = B C$ și $C A = I_{n}$ deducem că $A D A = B$ și apoi $D A = C B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $a, b, c \\in \\mathbb{C}^*$, distincte şi având acelaşi modul, astfel încât\n$$\na^{2}+b^{2}+c^{2}+a b+a c+b c=0\n$$\nDemonstraţi că $a, b, c$ reprezintă afixele vârfurilor unui triunghi dreptunghic sau echilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPutem presupune $|a|=|b|=|c|=1$. Ipoteza devine $(a+b+c)^{2}=a b+b c+c a$, de unde $(a+b+c)^{2}=a b c\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)$ sau $(a+b+c)^{2}=a b c \\overline{(a+b+c)}$. Prin trecere la modul avem $|a+b+c| \\in\\{0,1\\}$.\n\nDacă $|a+b+c|=0$, deducem că ortocentrul triunghiului, având vârfuri de afixe $a, b, c$, coincide cu centrul cercului circumscris aceluiaşi triunghi, deci acest triunghi este echilateral.\n\nDacă $|a+b+c|=1$, atunci $(a+b+c) \\overline{(a+b+c)}=1$, de unde $(a+b+c) \\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)=1$, adică $(a+b)(b+c)(c+a)=0$. Atunci două vârfuri ale acestui triunghi sunt diametral opuse, deci este dreptunghic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a$ şi $n$ două numere naturale nenule, astfel încât\n$$\n\\{\\sqrt{n+\\sqrt{n}}\\}=\\{\\sqrt{a}\\}\n$$\nArătaţi că $4 a+1$ este pătrat perfect.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCondiţia din enunţ este echivalentă cu $\\sqrt{n+\\sqrt{n}}=\\sqrt{a}+k$, $k \\in \\mathbb{Z}$. Rezultă că $n+\\sqrt{n}=a+2k\\sqrt{a}+k^{2}$, deci $\\sqrt{n}=2k\\sqrt{a}+b$, unde $b=k^{2}-n+a$. Deducem că $n=4k^{2}a+b^{2}+4kb\\sqrt{a}$, de unde rezultă că $kb\\sqrt{a}$ este raţional.\n\nDacă $\\sqrt{a}$ este raţional, atunci $a$ este pătrat perfect, deci $n+\\sqrt{n}$ e pătrat perfect; în particular, $n=m^{2}$, $m \\in \\mathbb{N}$. Rezultă că $m^{2}+m$ este pătrat perfect şi cum $m^{2} \\leq m^{2}+m < (m+1)^{2}$, obţinem $m=0$, deci $n=0$ - contradicţie.\n\nPrin urmare, $kb=0$. Pentru $b=0$, obţinem $n=k^{2}+a$ şi $n=4k^{2}a$, de unde $a=\\dfrac{k^{2}}{4k^{2}-1}<1$ - contradicţie.\n\nAşadar, $k=0$, de unde $n=b^{2}$ şi $a=b+n$, deci $a=b^{2}+b$ şi $4a+1=(2b+1)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie numerele reale pozitive $a, b, c$, astfel încât\n$$\n\\frac{a}{b+c+1}+\\frac{b}{a+c+1}+\\frac{c}{a+b+1} \\leq 1\n$$\nDemonstraţi că:\n$$\n\\frac{1}{b+c+1}+\\frac{1}{a+c+1}+\\frac{1}{a+b+1} \\geq 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDin inegalitatea dintre media aritmetică şi media armonică deducem că\n$$\n\\left(\\sum (b+c+1)\\right) \\sum \\frac{1}{b+c+1} \\geq 9\n$$\nde unde\n$$\n\\left(a+b+c+\\frac{3}{2}\\right) \\sum \\frac{1}{b+c+1} \\geq \\frac{9}{2}\n$$\nAvem însă\n$$\n\\left(a+b+c+\\frac{3}{2}\\right) \\cdot \\frac{1}{b+c+1}=\\frac{a}{b+c+1}+1+\\frac{1}{2} \\cdot \\frac{1}{b+c+1}\n$$\nşi analoagele, de unde rezultă\n$$\n\\sum \\frac{a}{b+c+1}+3+\\frac{1}{2} \\sum \\frac{1}{b+c+1} \\geq \\frac{9}{2}\n$$\nşi apoi\n$$\n\\frac{1}{2} \\sum \\frac{1}{b+c+1} \\geq \\frac{3}{2}-\\sum \\frac{a}{b+c+1}\n$$\nDacă $\\sum \\frac{a}{b+c+1} \\leq 1$, atunci avem\n$$\n\\frac{1}{2} \\sum \\frac{1}{b+c+1} \\geq \\frac{3}{2}-1=\\frac{1}{2}\n$$\nde unde concluzia.\nSolution:\n\nFie $S=\\sum \\frac{1}{b+c+1}$. Din enunţ,\n$$\n\\sum\\left(\\frac{a}{b+c+1}+1\\right) \\leq 4\n$$\nde unde $(a+b+c+1) S \\leq 4$.\nDin inegalitatea dintre mediile aritmetică şi armonică, rezultă\n$$\nS \\geq \\frac{9}{2(a+b+c+1)}\n$$\ndeci $S \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p$ un număr prim impar şi fie $G$ un grup care are exact $p+1$ elemente. Arătaţi că, dacă $p$ divide numărul automorfismelor lui $G$, atunci $p \\equiv 3(\\bmod 4)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nÎntrucât $p$ este un divizor prim al $|\\operatorname{Aut} G|$, există un $f$ în $\\operatorname{Aut} G$ de ordin $p$. Deoarece $f$ este o permutare a mulţimii $G \\setminus \\{e\\}$, rezultă că $f$ este un ciclu de lungime $p$, deci $G \\setminus \\{e\\} = \\{x, f(x), \\ldots, f^{p-1}(x)\\}$, oricare ar fi $x$ în $G \\setminus \\{e\\}$.\n\nPe de altă parte, $|G| = p+1$ este par, deci $G$ are un element $x_{0}$ de ordin $2$. Prin urmare, $\\operatorname{ord} f^{k}(x_{0}) = 2$, $k = 0, \\ldots, p-1$. Rezultă că $x^{2} = e$, oricare ar fi $x$ în $G$, deci $p+1 = 2^{n}$, unde $n \\geq 2$ este un număr întreg.\n\nDeci $p = 2^{n} - 1 \\equiv 3(\\bmod 4)$, deoarece $n \\geq 2$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Demonstraţi că există funcţii neperiodice $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ care verifică egalitatea\n$$\nf(x+1)+f(x-1)=\\sqrt{5} f(x)\n$$\npentru orice $x \\in \\mathbb{R}$\n\nb) Demonstraţi că orice funcţie $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ care verifică egalitatea\n$$\ng(x+1)+g(x-1)=\\sqrt{3} g(x)\n$$\npentru orice $x \\in \\mathbb{R}$, este periodică.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Căutăm soluţii printre funcţiile de forma $f(x)=a^{x}$, unde $a>0$. Obţinem egalitatea $a+a^{-1}=\\sqrt{5}$, de unde $a=\\frac{\\sqrt{5} \\pm 1}{2}$. Se verifică faptul că funcţiile $f_{1}: \\mathbb{R} \\rightarrow \\mathbb{R}, f_{1}(x)=\\left(\\frac{\\sqrt{5}-1}{2}\\right)^{x}$ şi $f_{2}: \\mathbb{R} \\rightarrow \\mathbb{R}, f_{2}(x)=\\left(\\frac{\\sqrt{5}+1}{2}\\right)^{x}$ verifică egalitatea din ipoteză.\n\nb) Fie $g$ o funcţie care verifică egalitatea din ipoteză. Atunci $g(x+2)+g(x)=\\sqrt{3} g(x+1)$, de unde $g(x+2)+g(x)=\\sqrt{3}(\\sqrt{3} g(x)-g(x-1))$, deci $g(x+2)=2 g(x)-\\sqrt{3} g(x-1)$. Apoi $g(x+3)=2 g(x+1)-\\sqrt{3} g(x)=2(\\sqrt{3} g(x)-g(x-1))-\\sqrt{3} g(x)$, de unde $g(x+3)=\\sqrt{3} g(x)-2 g(x-1)$. Apoi $g(x+4)=\\sqrt{3} g(x+1)-2 g(x)=\\sqrt{3}(\\sqrt{3} g(x)-g(x-1))-2 g(x)$, de unde $g(x+4)=g(x)-\\sqrt{3} g(x-1)$. În continuare, $g(x+5)=g(x+1)-\\sqrt{3} g(x)$, deci $g(x+5)=-g(x-1)$, de unde $g(x+6)=-g(x)$. Apoi $g(x+12)=-g(x+6)=g(x)$, de unde obţinem concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f:[0,1] \\rightarrow [0,1]$ o funcție crescătoare și fie\n$$\na_{n} = \\int_{0}^{1} \\frac{1 + (f(x))^{n}}{1 + (f(x))^{n+1}} \\, \\mathrm{d}x, \\quad n \\in \\mathbb{N}^{*}\n$$\nArătați că șirul $\\left(a_{n}\\right)_{n \\in \\mathbb{N}^{*}}$ este convergent și calculați limita sa.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nÎn mod evident, $a_{n} \\geq 1$, oricare ar fi $n \\in \\mathbb{N}^{*}$. Pe de altă parte,\n$$\n\\begin{aligned}\na_{n} - a_{n+1} &= \\int_{0}^{1} \\left( \\frac{1 + (f(x))^{n}}{1 + (f(x))^{n+1}} - \\frac{1 + (f(x))^{n+1}}{1 + (f(x))^{n+2}} \\right) \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\frac{(f(x))^{n} (1 - f(x))^{2}}{\\left(1 + (f(x))^{n+1}\\right)\\left(1 + (f(x))^{n+2}\\right)} \\, \\mathrm{d}x \\geq 0, \\quad n \\in \\mathbb{N}^{*}\n\\end{aligned}\n$$\ndeci șirul $\\left(a_{n}\\right)_{n \\in \\mathbb{N}^{*}}$ este descrescător. Rezultă că $\\left(a_{n}\\right)_{n \\in \\mathbb{N}^{*}}$ este convergent.\n\nFie $\\ell = \\lim_{n \\rightarrow \\infty} a_{n}$. Fără să restrângem generalitatea, putem presupune că $f(0) = 0$ și $f(1) = 1$. Fie $a = \\inf \\{ x : 0 \\leq x \\leq 1, f(x) = 1 \\}$.\n\nDacă $a = 0$, atunci $f(x) = 1$, oricare ar fi $x$ în $(0,1]$, deci $a_{n} = 1$, oricare ar fi indicele $n$, și $\\ell = 1$.\n\nDacă $a > 0$, fie $\\epsilon \\in (0, a)$. Atunci\n$$\n\\begin{aligned}\n0 \\leq a_{n} - 1 &= \\int_{0}^{1} \\left( \\frac{1 + (f(x))^{n}}{1 + (f(x))^{n+1}} - 1 \\right) \\, \\mathrm{d}x = \\int_{0}^{1} \\frac{(f(x))^{n}(1 - f(x))}{1 + (f(x))^{n+1}} \\, \\mathrm{d}x \\\\\n&\\leq \\int_{0}^{1} (f(x))^{n}(1 - f(x)) \\, \\mathrm{d}x = \\int_{0}^{a} (f(x))^{n}(1 - f(x)) \\, \\mathrm{d}x \\\\\n&\\leq \\int_{0}^{a} (f(x))^{n} \\, \\mathrm{d}x = \\int_{0}^{a - \\epsilon} (f(x))^{n} \\, \\mathrm{d}x + \\int_{a - \\epsilon}^{a} (f(x))^{n} \\, \\mathrm{d}x \\\\\n&\\leq (a - \\epsilon)(f(a - \\epsilon))^{n} + \\epsilon, \\quad n \\in \\mathbb{N}^{*}\n\\end{aligned}\n$$\nDar $(f(a - \\epsilon))^{n} \\xrightarrow{n \\rightarrow \\infty} 0$, deoarece $0 \\leq f(a - \\epsilon) < 1$, deci, prin trecere la limită în relația de mai sus, $0 \\leq \\ell - 1 \\leq \\epsilon$, oricare ar fi $\\epsilon$ în $(0, a)$. Prin urmare, $\\ell = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\geq 2$ un număr natural. Arătaţi că afirmaţiile următoare sunt echivalente:\n\na) Există numerele naturale nenule $b, c$, astfel încât $a^{2}=b^{2}+c^{2}$;\n\nb) Există un număr natural nenul $d$, astfel încât ecuaţiile $x^{2}-a x+d=0$ şi $x^{2}-a x-d=0$ au rădăcinile întregi.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSă presupunem că $a^{2}=b^{2}+c^{2}$. Numerele $b$ şi $c$ nu pot fi ambele impare (suma a două numere impare e de forma $4k+2$ şi nu poate fi pătrat), deci cel puţin unul dintre ele este par, adică produsul $bc$ este par.\n\nDiscriminanţii celor două ecuaţii sunt $\\Delta_{1}=a^{2}-4d$ şi $\\Delta_{2}=a^{2}+4d$. Alegem $d=\\frac{bc}{2}$ şi avem\n$$\n\\Delta_{1}=a^{2}-4d=b^{2}+c^{2}-4 \\frac{bc}{2}=(b-c)^{2}\n$$\niar rădăcinile primei ecuaţii sunt $x_{1,2}=\\frac{a \\pm (b-c)}{2}$. Se observă că $x_{1,2}$ sunt numere întregi (dacă $b, c$ sunt ambele pare, şi $a$ va fi par, iar dacă $b, c$ au parităţi diferite, $a$ va fi impar, ca şi $b-c$). Similar se arată că şi a doua ecuaţie are rădăcinile întregi.\n\nReciproc, să presupunem că ecuaţiile au rădăcini întregi. Atunci discriminanţii acestora trebuie să fie pătrate perfecte. Fie $\\Delta_{1}=u^{2}$ şi $\\Delta_{2}=v^{2}$.\n\nAvem, deci\n$$\n\\begin{aligned}\n& a^{2}-4d=u^{2} \\\\\n& a^{2}+4d=v^{2}\n\\end{aligned}\n$$\nDeducem uşor că numerele $u, v$ şi $a$ au aceeaşi paritate.\n\nAdunând relaţiile, obţinem\n$$\na^{2}=\\frac{u^{2}+v^{2}}{2}=\\left(\\frac{u+v}{2}\\right)^{2}+\\left(\\frac{u-v}{2}\\right)^{2}\n$$\ndeci, alegând $b=\\frac{u+v}{2}$ şi $c=\\frac{u-v}{2}$, numerele $b$ şi $c$ sunt întregi şi $a^{2}=b^{2}+c^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe dau numerele $a, b, c \\in \\mathbb{C}$. Calculați, scriind rezultatul sub formă de produs:\n\na) $\\left|\\begin{array}{ccc}1 & 1 & 1 \\\\ a & b & c \\\\ a^{3} & b^{3} & c^{3}\\end{array}\\right|$\n\nb) $\\left|\\begin{array}{ccc}a+b & b+c & c+a \\\\ a-b & b-c & c-a \\\\ a^{2}-b^{2} & b^{2}-c^{2} & c^{2}-a^{2}\\end{array}\\right|$",
    "options": [],
    "answer": "a) (a+b+c)(a-b)(b-c)(c-a)\nb) 2(a+b+c)(a-b)(b-c)(c-a)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Fie $H_{1}$ și $H_{2}$ subgrupuri ale grupului $(G, \\cdot)$. Arătați că $H_{1} \\cap H_{2}$ este subgrup al grupului $(G, \\cdot)$.\n\nb) Dacă $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ este o funcție bijectivă cu $f^{-1}(1)=2$, să se determine elementul neutru al legii de compoziție definite prin:\n$$\nx * y = f\\left(f^{-1}(x) + f^{-1}(y) - 2\\right), \\forall x, y \\in \\mathbb{R}\n$$",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRezolvaţi în $M_{2}(\\mathbb{R})$ ecuaţia: $X^{n}=\\left(\\begin{array}{ll}3 & 1 \\\\ 6 & 2\\end{array}\\right),\\ n \\in \\mathbb{N}^{*}$",
    "options": [],
    "answer": "Let A = [[3, 1], [6, 2]]. All real solutions are scalar multiples of A. Specifically:\n- If n is odd: X = 5^{1/n - 1} A.\n- If n is even: X = ± 5^{1/n - 1} A.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalculați :\n$$\n\\lim_{n \\rightarrow \\infty} \\sum_{k=1}^{n} \\frac{k^{2}+k}{n^{3}+k}\n$$",
    "options": [],
    "answer": "1/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) $I_{1} = \\int (\\cos x \\cdot \\cos 2x \\cdot \\cos 3x) \\, dx$\n\nb) $I_{2} = \\int_{1}^{\\sqrt{3}} \\frac{1}{x \\sqrt{x^{4} + 1}} \\, dx$",
    "options": [],
    "answer": "a) I1 = x/4 + (1/24) sin(6x) + (1/8) sin(2x) + (1/16) sin(4x) + C.\nb) I2 = (1/2)[asinh(1) − asinh(1/3)] = (1/2) ln( 3(1 + sqrt(2)) / (1 + sqrt(10)) ).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSă se calculeze $\\int_{-\\frac{1}{2}}^{\\frac{1}{2}} \\frac{x^{2}}{x \\sqrt{2}+\\sqrt{1-x^{2}}+\\sqrt{1+x^{2}}} d x$",
    "options": [],
    "answer": "π/12 − √3/8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a, b, c \\in \\mathbb{R}$, $a>0$. Calculaţi :\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{(n!)^{2a}}{(n+b)^{a n + c}}\n$$",
    "options": [],
    "answer": "infinity",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDemonstrați că numerele naturale, care împărțite în $\\mathbb{N}$ la 102 dau restul 78, sunt divizibile cu 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $n \\in \\mathbb{N}, n: 102 = c$ rest 78.\n\nAplicând teorema împărțirii cu rest obținem $n = 102c + 78 = 3 \\cdot (34c + 26)$, de unde numărul $n$ este divizibil cu 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProdusul a două numere este 540. Dacă primul număr s-ar mări cu 5, atunci produsul lor ar fi 600. Aflați numerele.",
    "options": [],
    "answer": "45 and 12",
    "solution": "Solution:\n\nNotăm $a$ și $b$ cele două numere. Avem $a \\cdot b = 540$, (1) și $(a + 5) \\cdot b = 600$, (2).\n\nDin (1) și (2) avem $a \\cdot b + 5 \\cdot b = 600 \\Leftrightarrow 540 + 5 \\cdot b = 600 \\Leftrightarrow 5 \\cdot b = 60 \\Leftrightarrow b = 12$.\n\nCum $a \\cdot 12 = 540 \\Leftrightarrow a = 45$. Numerele căutate sunt 45 și 12.\n\n| Detalii rezolvare | Barem asociat |\n| :---: | :---: |\n| Notează $a$ și $b$ cele două numere | $1 \\mathrm{p}$ |\n| Scrie relațiile: $a \\cdot b = 540$ și $(a + 5) \\cdot b = 600$ | $2 \\mathrm{p}$ |\n| Obține $b = 12$ și $a = 45$ | $4 \\mathrm{p}$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie numărul natural $n=\\overline{11 \\ldots 1}+\\overline{22 \\ldots 2}+\\ldots+\\overline{88 \\ldots 8}+\\overline{99 \\ldots 9}$, fiecare număr de forma $\\overline{a a \\ldots a}$ conținând câte 2015 cifre de $a$. Determinați câte cifre de 9 conține numărul $n$.",
    "options": [],
    "answer": "2014",
    "solution": "Solution:\n$n=1 \\cdot \\overline{11 \\ldots 1}+2 \\cdot \\overline{11 \\ldots 1}+3 \\cdot \\overline{11 \\ldots 1}+\\ldots+8 \\cdot \\overline{11 \\ldots 1}+9 \\cdot \\overline{11 \\ldots 1}$\n$=(1+2+3+\\ldots+8+9) \\cdot \\overline{11 \\ldots 1}=45 \\cdot \\overline{11 \\ldots 1}$.\nObținem $n=\\overline{499 \\ldots 95}$, număr ce are 2016 cifre. Deci $n$ conține 2014 de 9 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn cubul $ABCD A'B'C'D'$, considerăm $Q$ proiecția lui $D'$ pe $A'C$ și $S$ proiecția lui $D'$ pe $AC$. Arătați că:\n\na) $A'C \\perp (D'QB')$;\n\nb) $QS \\parallel (ABC)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) $D'B' \\perp (ACC') \\Rightarrow D'B' \\perp A'C$. Din $A'C \\perp D'Q$, $A'C \\perp D'B'$, $D'Q \\cap D'B' = \\{D'\\} \\Rightarrow A'C \\perp (D'QB')$.\n\nb) $\\triangle D'A'C \\equiv \\triangle D'C'A \\Rightarrow \\angle D'A'C \\equiv \\angle D'C'A$\n\n$\\triangle D'A'Q \\equiv \\triangle D'C'S \\Rightarrow [A'Q] \\equiv [C'S]$\n\nDin $A'Q = C'S$ și $A'O = C'O$ unde $O = A'C = AC'$, rezultă că $\\Leftrightarrow QS \\parallel A'C'$\n\nCum $A'C' \\parallel AC$ și $AC \\subset (ABC)$ avem $QS \\parallel (ABC)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră unghiurile adiacente $\\Varangle A O B$ și $\\Varangle B O C$, astfel încât bisectoarele lor $[OM$ și $[ON$ să formeze un unghi de $75^{\\circ}$.\na) Să se determine $m(\\Varangle A O B)$ și $m(\\Varangle B O C)$ știind că $3 \\cdot m(\\Varangle A O B)=2 \\cdot m(\\Varangle B O C)$.\nb) Dacă semidreapta $[OT$ formează unghi drept cu semidreapta $[OM$ astfel încât punctele $M$ și $T$ sunt de aceeași parte cu punctul $B$ faţă de punctul $A$. Calculați: $m(\\Varangle T O N)$, $m(\\Varangle B O N)$, $m(\\Varangle B O T)$ și $m(\\Varangle C O T)$.",
    "options": [],
    "answer": "a) m(∠AOB) = 60°, m(∠BOC) = 90°. b) m(∠TON) = 15°, m(∠BON) = 45°, m(∠BOT) = 60°, m(∠COT) = 30°.",
    "solution": "Solution:\n\na) Notăm $m(\\Varangle A O B)=\\alpha$ și $m(\\Varangle B O C)=\\beta$. Avem:\n$[OM$ bisectoarea $\\Varangle A O B \\Rightarrow m(\\Varangle A O M)=m(\\Varangle M O B)=\\frac{\\alpha}{2}$;\n$[ON$ bisectoarea $\\Varangle B O C \\Rightarrow m(\\Varangle B O N)=m(\\Varangle N O C)=\\frac{\\beta}{2}$.\n\n![](attached_image_1.png)\n\nCum $m(\\Varangle M O N)=75^{\\circ}$ avem că $\\frac{\\alpha}{2}+\\frac{\\beta}{2}=75^{\\circ} \\Leftrightarrow \\alpha+\\beta=150^{\\circ} \\Rightarrow \\alpha=150^{\\circ}-\\beta$, (1).\nDar $3 \\cdot m(\\Varangle A O B)=2 \\cdot m(\\Varangle B O C) \\Rightarrow 3 \\alpha=2 \\beta$, (2).\nDin (1) și (2) avem $3\\left(150^{\\circ}-\\beta\\right)=2 \\beta \\Leftrightarrow 450^{\\circ}=5 \\beta \\Leftrightarrow \\beta=90^{\\circ}$ și $\\alpha=150^{\\circ}-90^{\\circ}=60^{\\circ}$.\nAvem $m(\\Varangle A O B)=60^{\\circ}$ si $m(\\Varangle B O C)=90^{\\circ}$.\n\nb) Avem $m(\\Varangle M O T)=90^{\\circ}$ de unde avem: $m(\\Varangle T O N)=90^{\\circ}-[m(\\Varangle N O B)+m(\\Varangle M O B)]=15^{\\circ}$;\n$m(\\Varangle B O N)=\\frac{\\beta}{2}=\\frac{90^{\\circ}}{2}=45^{\\circ}$;\n$m(\\Varangle B O T)=m(\\Varangle B O N)+m(\\Varangle T O N)=45^{\\circ}+15^{\\circ}=60^{\\circ}$.\n$m(\\Varangle C O T)=m(\\Varangle C O N)-m(\\Varangle T O N)=45^{\\circ}-15^{\\circ}=30^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminați numerele naturale $n$ știind că fracția $\\frac{3 n+1}{2 n-7}$ este reductibilă.",
    "options": [],
    "answer": "All n with n ≡ 15 (mod 23), equivalently n = 23k − 8 for k a positive integer.",
    "solution": "Solution:\nDacă fracția este reductibilă atunci există $d \\neq 1$ astfel încât $d \\mid 3 n+1$ și $d \\mid 2 n-7$. De aici avem $d \\mid 2(3 n+1)-3(2 n-7)$, adică $d \\mid 23$, prin urmare $d=23$. Acum $23 \\mid 3 n+1$ și $23 \\mid 2 n-7$ deducem că $23 \\mid n+8$, adică $n+8=23 k$, pentru $k \\in \\mathbb{N}^{*}$. Se verifică pentru $n=23 k-8,\\ k \\in \\mathbb{N}^{*}$, fracția este reductibilă.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A \\in \\mathcal{M}_{n}(\\mathbb{R})$ cu proprietatea că suma elementelor de pe fiecare linie este pozitivă și suma elementelor de pe fiecare coloană este negativă.\nCalculați $\\operatorname{det}(A)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi funcţia $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ care admite primitive şi verifică pentru orice număr real $x$, egalitatea: $f(x)-F(x)=|x-1|$, unde $F$ este o primitivă a lui $f$.",
    "options": [],
    "answer": "All solutions are given, for an arbitrary real parameter a, by\nf(x) = (a − 1/e) e^x + 1 for x < 1;\nf(1) = a e;\nf(x) = (a + 1/e) e^x − 1 for x > 1.\nFor this f, a valid primitive is F(x) = f(x) − |x − 1|.",
    "solution": "Solution:\n\nFie $F$ o primitivă a funcţiei $f$, adică $F'(x) = f(x)$ pentru orice $x \\in \\mathbb{R}$.\n\nDin ipoteză avem:\n$$\nf(x) - F(x) = |x-1|, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nRezultă:\n$$\nf(x) = F(x) + |x-1|.\n$$\n\nDerivăm ambele părţi după $x$:\n$$\nf'(x) = F'(x) + \\frac{d}{dx}|x-1|.\n$$\nDar $F'(x) = f(x)$, deci:\n$$\nf'(x) = f(x) + \\frac{d}{dx}|x-1|.\n$$\n\nAducem totul pe o parte:\n$$\nf'(x) - f(x) = \\frac{d}{dx}|x-1|.\n$$\n\nCalculăm derivata:\n$$\n\\frac{d}{dx}|x-1| = \\begin{cases} 1, & x > 1 \\\\ -1, & x < 1 \\end{cases}\n$$\n\nDeci ecuaţia devine:\n$$\nf'(x) - f(x) = \\begin{cases} 1, & x > 1 \\\\ -1, & x < 1 \\end{cases}\n$$\n\nAceasta este o ecuaţie diferenţială liniară de ordinul I cu coeficienţi constanţi, pe fiecare interval $(-\\infty, 1)$ şi $(1, \\infty)$.\n\nRezolvăm separat pe cele două intervale:\n\nPentru $x > 1$:\n$$\nf'(x) - f(x) = 1\n$$\nEcuaţia omogenă asociată: $f'(x) - f(x) = 0$ are soluţia generală $f_0(x) = Ce^{x}$.\n\nO soluţie particulară pentru ecuaţia completă este $f_p(x) = -1$ (deoarece $(-1)' - (-1) = 0 - (-1) = 1$).\n\nDeci soluţia generală pe $(1, \\infty)$ este:\n$$\nf(x) = Ce^{x} - 1, \\quad x > 1.\n$$\n\nPentru $x < 1$:\n$$\nf'(x) - f(x) = -1\n$$\nSoluţia particulară este $f_p(x) = 1$ (deoarece $(1)' - 1 = 0 - 1 = -1$).\n\nDeci soluţia generală pe $(-\\infty, 1)$ este:\n$$\nf(x) = De^{x} + 1, \\quad x < 1.\n$$\n\nFuncţia $f$ trebuie să fie continuă pe $\\mathbb{R}$ (deoarece admite primitive), deci trebuie să aibă limită la $x \\to 1^-$ şi $x \\to 1^+$:\n\n$$\n\\lim_{x \\to 1^-} f(x) = De^{1} + 1\n$$\n$$\n\\lim_{x \\to 1^+} f(x) = Ce^{1} - 1\n$$\n\nEgalăm limitele:\n$$\nDe^{1} + 1 = Ce^{1} - 1\n$$\n$$\nCe^{1} - De^{1} = 2\n$$\n$$\nC - D = \\frac{2}{e}\n$$\n\nAcum, folosim şi condiţia din enunţ la $x = 1$:\n\nDin $f(1) - F(1) = |1-1| = 0$, deci $f(1) = F(1)$.\n\nDar $F'(1) = f(1)$, deci $F(1)$ este o valoare arbitrară, dar $f(1)$ trebuie să fie egală cu $F(1)$.\n\nObservăm că $f$ poate fi definită la $x=1$ cu orice valoare, dar pentru ca $f$ să fie derivabilă pe $\\mathbb{R}$, trebuie să fie derivabilă şi la $x=1$.\n\nCalculăm derivata stângă şi dreaptă la $x=1$:\n\nPentru $x < 1$:\n$$\nf'(x) = De^{x}\n$$\nLa $x=1^-$: $f'_-(1) = De^{1}$\n\nPentru $x > 1$:\n$$\nf'(x) = Ce^{x}\n$$\nLa $x=1^+$: $f'_+(1) = Ce^{1}$\n\nPentru ca $f$ să fie derivabilă la $x=1$, trebuie $De^{1} = Ce^{1}$, deci $C = D$.\n\nDin $C - D = \\frac{2}{e}$ şi $C = D$ rezultă $\\frac{2}{e} = 0$, deci contradicţie.\n\nDeci funcţia $f$ nu este derivabilă la $x=1$, dar poate fi continuă.\n\nAlegem $f(1)$ astfel încât $f$ să fie continuă la $x=1$:\n\n$$\nf(1) = \\lim_{x \\to 1^-} f(x) = \\lim_{x \\to 1^+} f(x) = \\alpha\n$$\n\nDin $C - D = \\frac{2}{e}$, putem lua $C = a + \\frac{1}{e}$, $D = a - \\frac{1}{e}$, cu $a \\in \\mathbb{R}$.\n\nDeci soluţia generală este:\n\n$$\nf(x) =\n\\begin{cases}\n(a - \\frac{1}{e})e^{x} + 1, & x < 1 \\\\\n(a + \\frac{1}{e})e^{x} - 1, & x > 1 \\\\\n\\alpha, & x = 1\n\\end{cases}\n$$\n\nunde $\\alpha = (a - \\frac{1}{e})e^{1} + 1 = (a + \\frac{1}{e})e^{1} - 1$.\n\nRezolvăm pentru $\\alpha$ şi $a$:\n\nDin $(a - \\frac{1}{e})e^{1} + 1 = (a + \\frac{1}{e})e^{1} - 1$:\n$$\n(a - \\frac{1}{e})e + 1 = (a + \\frac{1}{e})e - 1\n$$\n$$\n(ae - 1) + 1 = (ae + 1) - 1\n$$\n$$\nae = ae\n$$\n\nDeci orice $a \\in \\mathbb{R}$, iar $\\alpha = (a - \\frac{1}{e})e + 1$.\n\nAstfel, funcţia $f$ este:\n\n$$\nf(x) =\n\\begin{cases}\n(a - \\frac{1}{e})e^{x} + 1, & x < 1 \\\\\n(a + \\frac{1}{e})e^{x} - 1, & x > 1 \\\\\n(a - \\frac{1}{e})e + 1, & x = 1\n\\end{cases}\n$$\n\nunde $a \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A \\in M_{2}(\\mathbb{C})$ și $A^{2}-3 A+5 I_{2}=O_{2}$.\na) Aflați inversa matricei $A$.\nb) Calculați $\\operatorname{det}\\left(A^{2}-I_{2}\\right)+\\operatorname{det}\\left(A^{2}+A\\right)-\\operatorname{det}\\left(A^{2}+2 I_{2}\\right)$.",
    "options": [],
    "answer": "A^{-1} = (3 I_2 - A)/5; the value equals 45",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\in (0,4)$, a fixat. Determinați $z$ din $\\mathbb{C}$ știind că $|z| = \\left|\\frac{a}{z}\\right| = |z - a|$.",
    "options": [],
    "answer": "z = a/2 ± (i/2)·sqrt(a(4 − a))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSă se rezolve în mulţimea numerelor reale ecuaţia: \n$$\n\\sum_{k=1}^{2n} \\left([x] + [(-1)^k x + k]\\right) = 2n(n+2).\n$$\nAm notat cu $[x]$ partea întreagă a numărului real $x$.",
    "options": [],
    "answer": "(2,3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDemonstraţi că\n$$\n\\log_{2016} \\sqrt[2016]{1+\\frac{a_{2}}{a_{1}}} + \\log_{2016} \\sqrt[2016]{1+\\frac{a_{3}}{a_{2}}} + \\cdots + \\log_{2016} \\sqrt[2016]{1+\\frac{a_{1}}{a_{2016}}} \\geq \\frac{1}{5+\\log_{2} 63}\n$$\npentru orice $a_{1}, a_{2}, \\ldots, a_{2016}$ numere reale strict pozitive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a > 0$ și $(x_{n})_{n \\in \\mathbb{N}}$ șirul care verifică relațiile $x_{1} = \\frac{1}{a}$, $x_{n} = \\frac{x_{n-1}}{1 + a n x_{n-1}}$, oricare ar fi $n \\geq 2$. Să se calculeze $\\lim_{n \\rightarrow \\infty} (x_{1} + x_{2} + \\ldots + x_{n})$.",
    "options": [],
    "answer": "2/a",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $x, y, z \\in \\mathbb{R} - \\{1\\}$ numere reale astfel încât $x + y + z = 4$, $x^{2} + y^{2} + z^{2} = 6$ şi $xyz = a$, $a \\in \\mathbb{R} - \\{2\\}$.\nCalculaţi valoarea expresiei\n$$\nE = \\frac{1}{xy + z - 3} + \\frac{1}{yz + x - 3} + \\frac{1}{zx + y - 3}.\n$$",
    "options": [],
    "answer": "1/(a - 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Fie $(G, )$ un grup abelian cu $2016$ elemente. Demonstrați că există $x \\in G \\setminus \\{e\\}$ astfel încât $x^{2} = e$ ($e$ este elementul neutru din $G$).\n\nb) Calculați: $\\int \\frac{1}{x^{2016} + x} \\, dx$, $x > 0$.",
    "options": [],
    "answer": "a) Such an element exists (there is x ≠ e with x^2 = e). b) ∫ dx/(x^{2016} + x) = ln x − (1/2015) ln(1 + x^{2015}) + C, for x > 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Câte progresii aritmetice de numere naturale există cu primul termen $1$ şi care conţin numărul $45001$?\n\nb) Arătaţi că nu există progresii aritmetice neconstante de numere naturale cu toţi termenii pătrate perfecte.",
    "options": [],
    "answer": "a) 60; b) No such progression exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $(a_{n})_{n \\geq 0}$ un șir de numere reale astfel încât $a_{0} \\in \\mathbb{R}$ și $a_{n+1}=a_{n}^{2}-8 a_{n}+18, \\forall n \\geq 0$.\n\na) Să se arate că dacă șirul $(a_{n})_{n \\geq 0}$ este convergent atunci $\\exists p \\in \\mathbb{N}$ astfel încât $a_{n+1}=a_{n}, \\forall n \\geq p$.\n\nb) Determinați mulțimea valorilor naturale ale lui $a_{0}$ pentru care șirul $a_{n}$ este convergent.",
    "options": [],
    "answer": "{2, 3, 4, 5, 6}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Câte progresii aritmetice de numere naturale există cu primul termen $1$ şi care conţin numărul $45001$?\n\nb) Arătaţi că nu există progresii aritmetice neconstante de numere naturale cu toţi termenii pătrate perfecte.",
    "options": [],
    "answer": "a) 60; b) No non-constant progression exists.",
    "solution": "Solution:\na) Fie $(a_n)$ o progresie aritmetică cu primul termen $1$ şi care conţine numărul $45001$. Atunci $45001 = 1 + (n-1) \\cdot r \\Rightarrow (n-1) \\cdot r = 45000$, deci $r$ este divizor al lui $45000$.\n\n$45000 = 2^{3} \\cdot 3^{2} \\cdot 5^{4}$. Deci $r = 2^{i} \\cdot 3^{j} \\cdot 5^{k}$, $i = \\overline{0,3}$, $j = \\overline{0,2}$, $k = \\overline{0,4}$.\n\n$r$ poate lua $4 \\cdot 3 \\cdot 5 = 60$ de valori, adică sunt $60$ de progresii.\n\nb) Presupunem prin absurd că $(a_n)$ este o progresie aritmetică în care toţi termenii sunt pătrate perfecte. Atunci există şirul strict crescător $(x_n)$, cu $a_n = x_n^2$ unde $n$ este număr natural nenul.\n\nCum $x_{n+1}^2 - x_n^2 = r$ avem $x_{n+1} - x_n \\geq 1 \\Rightarrow r = (x_{n+1} - x_n) \\cdot (x_{n+1} + x_n) \\geq 2x_n$.\n\nRezultă că şirul $(x_n)$ este mărginit, fals deoarece $(a_n)$ nu este mărginit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Să se demonstreze că $\\frac{1}{2}(a+b)^{2}+\\frac{1}{4}(a+b) \\geq a \\sqrt{b}+b \\sqrt{a}$, $a, b \\geq 0$\n\nb) Să se arate că pentru $\\forall x, y, z \\in\\left[0, \\frac{\\pi}{2}\\right)$, are loc inegalitatea:\n$$\n\\begin{aligned}\n& (\\operatorname{tg} x+\\operatorname{tg} y)^{2}+(\\operatorname{tg} y+\\operatorname{tg} z)^{2}+(\\operatorname{tg} z+\\operatorname{tg} x)^{2}+\\operatorname{tg} x+\\operatorname{tg} y+\\operatorname{tg} z \\geq \\\\\n& \\geq 2(\\operatorname{tg} x \\sqrt{\\operatorname{tg} y}+\\operatorname{tg} y \\sqrt{\\operatorname{tg} x}+\\operatorname{tg} y \\sqrt{\\operatorname{tg} z}+\\operatorname{tg} z \\sqrt{\\operatorname{tg} y}+\\operatorname{tg} z \\sqrt{\\operatorname{tg} x}+\\operatorname{tg} x \\sqrt{\\operatorname{tg} z})\n\\end{aligned}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră șirul $\\left(x_{n}\\right)_{n \\geq 1}$, în care $x_{1}=\\frac{1}{3}$ și $x_{n+1}=(n+1) x_{n}+\\frac{(n+1)(n+1)!}{3^{n+1}}, \\forall n \\in \\mathbb{N}^{*}$.\nSă se calculeze $\\lim _{n \\rightarrow \\infty} n\\left(\\frac{x_{n}}{n!}-\\frac{3}{4}\\right)$.",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră $I_{n} = \\int_{0}^{4} \\sqrt[n]{x^{n} + (4-x)^{n}} \\, dx$, $n \\in \\mathbb{N}$, $n \\geq 2$. Să se calculeze $\\lim_{n \\rightarrow \\infty} I_{n}$.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABCD$ un patrulater convex. Punctele $M$, $N$, $P$ şi $Q$ împart segmentele $[AB]$, $[BC]$, $[CD]$ şi respectiv $[DA]$ în acelaşi raport. Demonstraţi că dreptele $AC$, $BD$, $MP$ şi $NQ$ sunt concurente dacă şi numai dacă $ABCD$ este paralelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCalculați $\\lim _{n \\rightarrow \\infty} \\prod_{k=1}^{n}\\left(\\sqrt[k+1]{1+\\frac{1}{k}}-1\\right)$",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră hexagonul inscriptibil $ABCDEF$ şi $H_{1}$, $H_{2}$, $H_{3}$, $H_{4}$ ortocentrele triunghiurilor $ABC$, $BCD$, $DEF$, $FAE$. Să se arate că pentru orice puncte $M$, $N$, $P$ şi $Q$ din plan care satisfac relaţia $\\overrightarrow{H_{1}M} + \\overrightarrow{H_{3}P} = \\overrightarrow{H_{2}N} + \\overrightarrow{H_{4}Q}$, $MNPQ$ este paralelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $G$ un grup și $x, y \\in G$ cu proprietatea că ord $(x)=3$, $y^{4}=e$, $x y=y^{3} x$. Să se arate că dacă $y \\in G \\setminus \\{e\\}$ atunci ord $(y)=2$ și $x y=y x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Să se arate că $\\forall x>1$, are loc inegalitatea: $2^{x} \\cdot \\sqrt{4^{x}-1}+4^{x} \\cdot \\sqrt{2^{x}-1}<8^{x}$.\n\nb) Să se determine funcțiile bijective $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ astfel încât $\\forall x, y \\in \\mathbb{R}$ are loc relația: $f(2 x+f(x)+3 f(y))=f(3 x)+f(3 y)$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Calculați\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{(n^{2}+3) \\cdot (n^{2}+5) \\cdot \\ldots \\cdot [(n+1)^{2}+2016]}{(n^{2}+2) \\cdot (n^{2}+4) \\cdots [(n+1)^{2}+2015]}\n$$\n\nb) Calculați\n$$\n\\lim_{n \\rightarrow \\infty}\\left(\\frac{\\sqrt[n]{4}+\\sqrt[n]{504}}{2}\\right)^{2n}\n$$",
    "options": [],
    "answer": "a) 1; b) 2016",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38552,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many of the first 2017 positive integers can be uniquely represented as $2^a + 2^b + 2^c$, with $a$, $b$, $c$ non-negative integers? (Two representations that only differ by the order of the terms are considered identical.)",
    "options": [],
    "answer": "184",
    "solution": "If a number can be represented as a sum of three, not necessarily distinct, powers of $2$, regrouping the equal terms (if such terms exist), one obtains a sum of at most three *distinct* powers of $2$, hence the base $2$ representation of such a number has at most three digits equal to $1$. Convenient numbers are those whose base $2$ representation have three digits equal to $1$, then the numbers of the form $2^a + 1 = 2^{a-1} + 2^{a-1} + 1$ with $a \\in \\mathbb{N}$ (numbers $2^a + 2^b$ with $a > b \\ge 1$ are not convenient because $2^a + 2^b = 2^{a-1} + 2^{a-1} + 2^b = 2^{b-1} + 2^{b-1} + 2^a$); finally, the numbers $2^c = 2^{c-1} + 2^{c-1} + 2^{c-2}$ are also convenient if $c \\ge 2$.\n\nLet us count first the convenient numbers that are less than $2048$. The base $2$ representation of these numbers has at most $11$ digits. There are $C_{11}^3 = 165$ numbers less than $2048$ that can be written as a sum of three distinct powers of $2$. There are $10$ numbers of the form $2^a + 1$, $(1 \\le a \\le 10)$ and $9$ of the form $2^a$ $(2 \\le a \\le 10)$, hence $184$ numbers in total.\n\nThe numbers from $2018$ to $2047$ are larger than $2^{10} + 2^9 + 2^8$, hence their base $2$ representation has more than three digits equal to $1$, therefore none of these numbers is convenient. In conclusion, there are $184$ convenient numbers among the first $2017$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38553,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the smallest value that the sum of the digits of the number $3n^2 + n + 1$, $n \\in \\mathbb{N}$, can take?",
    "options": [],
    "answer": "3",
    "solution": "For $n = 8$ we have $3n^2 + n + 1 = 201$ whose sum of the digits is $3$.\n\nWe prove that the sum of the digits of $3n^2+n+1$ cannot be $1$ or $2$. As $3n^2+n+1$ is odd, it cannot be written as $10^k$ or $2 \\cdot 10^k$, $k \\in \\mathbb{N}$, nor can it be written as $10^k+10^j$ with $k, j > 0$. If $3n^2+n+1 = 10^k+1$, then $n(3n+1) = 10^k$. But $n$ and $3n+1$ are co-prime, hence they must be $2^k$ and $5^k$. As $n < 3n+1$, we must have $n = 2^k$ and $3n+1 = 5^k$, which is not possible because $5^k > 4^k > 3 \\cdot 2^k + 1$ if $k \\ge 2$, and $k=1$ does not work either.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the interior of the triangle $ABC$, and let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ respectively at the points $D, E, F$. Let the circles on diameters $BC$ and $AD$ meet at points $a$ and $a'$; the circles on diameters $CA$ and $BE$ meet at points $b$ and $b'$; and the circles on diameters $AB$ and $CF$ meet at points $c$ and $c'$. Show that the points $a, a', b, b', c, c'$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $A$, $B$, $C$ and $P$ have position vectors $\\mathbf{a}$, $\\mathbf{b}$, $\\mathbf{c}$ and\n$$\n\\mathbf{p} = \\frac{\\alpha}{\\alpha + \\beta + \\gamma} \\mathbf{a} + \\frac{\\beta}{\\alpha + \\beta + \\gamma} \\mathbf{b} + \\frac{\\gamma}{\\alpha + \\beta + \\gamma} \\mathbf{c},\n$$\nwhere $\\alpha$, $\\beta$, $\\gamma$ are the barycentric coordinates of $P$. Then $D$ has position vector\n$$\n\\frac{\\beta}{\\beta + \\gamma} \\mathbf{b} + \\frac{\\gamma}{\\beta + \\gamma} \\mathbf{c},\n$$\nso the circles on diameters $BC$ and $AD$ have equations\n$$\n(\\mathbf{r} - \\mathbf{b}) \\cdot (\\mathbf{r} - \\mathbf{c}) = 0 \\quad \\text{and} \\quad (\\mathbf{r} - \\mathbf{a}) \\cdot \\left( \\mathbf{r} - \\frac{\\beta}{\\beta + \\gamma} \\mathbf{b} - \\frac{\\gamma}{\\beta + \\gamma} \\mathbf{c} \\right) = 0,\n$$\nrespectively; alternatively, but equivalently, the latter reads\n$$\n\\beta(\\mathbf{r} - \\mathbf{a}) \\cdot (\\mathbf{r} - \\mathbf{b}) + \\gamma(\\mathbf{r} - \\mathbf{a}) \\cdot (\\mathbf{r} - \\mathbf{c}) = 0.\n$$\nAny linear combination of the two is the equation of a circle or straight line through $a$ and $a'$. In particular, the linear combination formed by multiplying the first equation by $\\beta\\gamma$ and the second by $\\alpha$, and adding, is\n$$\n\\alpha\\beta(\\mathbf{r} - \\mathbf{a}) \\cdot (\\mathbf{r} - \\mathbf{b}) + \\beta\\gamma(\\mathbf{r} - \\mathbf{b}) \\cdot (\\mathbf{r} - \\mathbf{c}) + \\gamma\\alpha(\\mathbf{r} - \\mathbf{c}) \\cdot (\\mathbf{r} - \\mathbf{a}) = 0.\n$$\nThe symmetry of this equation shows that this circle (or, possibly, straight line) also passes through $b$, $b'$ and $c$, $c'$. Finally, notice that $\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha$ is positive when $P$ lies inside the triangle $ABC$, so the locus is indeed a circle centered at the point whose barycentric coordinates are $\\alpha(\\beta + \\gamma)$, $\\beta(\\gamma + \\alpha)$, $\\gamma(\\alpha + \\beta)$, respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a, b > 1$, such that both $a^2 + 16\\left[\\frac{b}{2}\\right]$ and $b^2 + 16\\left[\\frac{a}{2}\\right]$ are perfect squares.",
    "options": [],
    "answer": "[(3,3), (4,7), (7,4)]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38556,
    "subject": "Mathematics (Multi-modal)",
    "question": "The continuous functions $f, g: [0, \\infty) \\to [0, \\infty)$ have the properties:\n(i) $f(0) = g(0) = 0$;\n(ii) $g(x) \\neq 0$, for every $x > 0$;\n(iii) $f(x + g(f(x))) = f(x)$, for every $x \\ge 0$.\nShow that $f(x) = 0$, for every $x \\ge 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = \\{x \\ge 0 : f(x) = 0\\}$. By (i), $0 \\in A$.\n\nLet $x \\in A$. Then $f(x) = 0$.\nBy (iii), $f(x + g(f(x))) = f(x)$. Since $f(x) = 0$, $g(f(x)) = g(0) = 0$ (by (i)).\nSo $f(x + 0) = f(x)$, i.e., $f(x) = f(x)$, which is trivial.\n\nBut this does not help us extend $A$ directly. Instead, let us try to show that $f(x) = 0$ for all $x \\ge 0$.\n\nSuppose there exists $x_0 > 0$ such that $f(x_0) > 0$.\nThen $g(f(x_0)) > 0$ by (ii), since $f(x_0) > 0$ and $g(x) \\neq 0$ for $x > 0$.\n\nConsider $x_1 = x_0 + g(f(x_0)) > x_0$.\nBy (iii), $f(x_1) = f(x_0)$.\n\nNow, define recursively $x_{n+1} = x_n + g(f(x_n))$, with $x_0$ as above.\nThen $f(x_{n+1}) = f(x_n)$, so $f(x_n) = f(x_0)$ for all $n$.\n\nAlso, since $f(x_0) > 0$, $g(f(x_0)) > 0$, so $x_{n+1} = x_n + g(f(x_0))$, i.e., $x_n = x_0 + n g(f(x_0))$.\nThus, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n \\ge 0$.\n\nBut $g(f(x_0)) > 0$, so as $n \\to \\infty$, $x_n \\to \\infty$.\nThus, $f(x) > 0$ for infinitely many $x$ tending to infinity.\n\nBut $f$ is continuous and $f(0) = 0$.\n\nLet us consider the set $S = \\{x \\ge 0 : f(x) > 0\\}$.\nSuppose $S$ is nonempty. Then, as above, for any $x_0 \\in S$, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n \\ge 0$.\nSo $S$ contains an unbounded sequence.\n\nBut $f$ is continuous and $f(0) = 0$.\nLet $y > 0$ be arbitrary. If $f(y) > 0$, then as above, $f(y + n g(f(y))) = f(y) > 0$ for all $n \\ge 0$.\nSo $f$ is constant and positive on $\\{y + n g(f(y)) : n \\ge 0\\}$.\n\nBut $f(0) = 0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.\n\nSuppose $f(x_0) > 0$ for some $x_0 > 0$. Then, as above, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.\nBut for large enough $n$, $x_0 + n g(f(x_0))$ can be made arbitrarily large, so $f$ is positive at arbitrarily large $x$.\n\nBut $f$ is continuous and $f(0) = 0$, so for small $x$, $f(x)$ must be close to $0$.\n\nLet us try to show that $f(x) = 0$ for all $x \\ge 0$.\nSuppose not. Then there exists $x_0 > 0$ with $f(x_0) > 0$.\nThen, as above, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.\nBut $f$ is continuous, so the set $\\{x_0 + n g(f(x_0)) : n \\ge 0\\}$ is unbounded and $f$ is constant and positive on this set.\n\nBut $f(0) = 0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.\n\nLet us try to show that $f(x) = 0$ for all $x \\ge 0$.\nSuppose not. Then there exists $x_0 > 0$ with $f(x_0) > 0$.\nThen, as above, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.\nBut $f$ is continuous, so the set $\\{x_0 + n g(f(x_0)) : n \\ge 0\\}$ is unbounded and $f$ is constant and positive on this set.\n\nBut $f(0) = 0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.\n\nLet us try to show that $f(x) = 0$ for all $x \\ge 0$.\nSuppose not. Then there exists $x_0 > 0$ with $f(x_0) > 0$.\nThen, as above, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.\nBut $f$ is continuous, so the set $\\{x_0 + n g(f(x_0)) : n \\ge 0\\}$ is unbounded and $f$ is constant and positive on this set.\n\nBut $f(0) = 0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.\n\nTherefore, the only possibility is $f(x) = 0$ for all $x \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38557,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $M$ is on the incircle of the square $ABCD$. Show that $MA \\cdot MB \\cdot MC \\cdot MD \\le 5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{x \\mid x = n(n+1) \\text{ and } n \\in \\mathbb{N}^*\\},\\ B = \\{y \\mid y = 2^{4k+3} \\text{ and } k \\in \\mathbb{N}\\}$.\n\na) Show that $A \\cap B = \\emptyset$.\n\nb) Determine the positive integer $m$ such that the triple of the sum of the $m$ smallest elements of the set $A$ equals the sum of the three smallest elements of the set $B$.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $ADC$ be isosceles right triangles with $AB = BC = CD = DA$ and $B \\neq D$. Consider $E \\in (CD)$ and $F \\in AD$ such that $EC = AF$ and $A \\in (DF)$. Denote $\\{G\\} = EF \\cap AC$. Find the measures of the angles in triangle $EGB$.\n\nSorin Peligrad",
    "options": [],
    "answer": "The triangle EGB is right isosceles: angle at G is 90 degrees, and the angles at E and at B are 45 degrees each.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be two positive integers such that $n - 1 > m > 1$, let $S_n$ be the group of the permutations of degree $n$ and let $S$ be the subset of $S_n$ made by all permutations $\\sigma$ with the property $\\sigma(m) < \\sigma(m+1) < \\dots < \\sigma(n)$. Show that $S$ contains a unique subgroup of $S_n$, which contains all the transpositions from $S_n$ which belong to $S$.\n\nIoan Băetu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Consider an $n \\times n$ square. In each cell of the square, one of the numbers from the set $M = \\{1, 2, \\dots, 2n-1\\}$ is to be written. One such filling is called \"good\" if, for every index $i$, $1 \\le i \\le n$, row no. $i$ and column no. $i$, together, contain all the elements of $M$.\na) Prove that there exists $n \\ge 3$ for which a *good* filling exists.\nb) Prove that for $n = 2017$ there is no good filling of the $n \\times n$ square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) For $n = 4$ the filling below is good.\n\n| 1 | 2 | 4 | 5 |\n|---|---|---|---|\n| 3 | 1 | 6 | 4 |\n| 7 | 5 | 1 | 2 |\n| 6 | 7 | 3 | 1 |\n\nb) We prove that there is no good filling of an $n \\times n$ square if $n$ is odd. Assume the contrary to be true. Then, for some odd number $n$, we have that, for every index $i$, $1 \\le i \\le n$, the row no. $i$ and the column no. $i$ contain, together, exactly once, each element of $M$. If we denote by $a_{ij}$ the number situated on row no. $i$ and column no. $j$, $i, j \\in \\{1, 2, \\dots, n\\}$, and $S_i = a_{i1} + a_{i2} + \\dots + a_{in} + a_{1i} + a_{2i} + \\dots + a_{ni}$ in which the term $a_{ii}$ only appears once, then each element of $M$ must appear in $n$ such sums. But if $a_{ij} = k, i \\neq j$, then the element $k$ will contribute to both $S_i$ and $S_j$. Thus, any number that is not situated on the main diagonal contributes to two sums. But in total, there is an odd number of sums, so each element of $M$ must appear an odd number of times on the main diagonal. Thus, each element of $M$ needs to appear at least once on the main diagonal, which is not possible because there are $2n-1$ elements in $M$ and only $n$ places on the diagonal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. For each positive integer $k \\le n$, let $r_k$ denote the remainder $2^n$ leaves upon division by $k$. Prove that\n$$\n\\sum_{k=1}^{n} r_k > \\frac{n}{2} \\left( \\log_2 \\frac{n}{3} - 2 \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m = \\lfloor \\log_2 n \\rfloor$, and split the set of integers $1$ through $n$ into $m + 1$ pairwise disjoint subsets $J_0, J_1, \\dots, J_m$, where $J_k$ consists of all numbers of the form $2^k(2\\ell + 1)$, $\\ell = 0, \\dots, \\lfloor n/2^{k+1} - 1/2 \\rfloor$.\nIf $j$ is a member of $J_k$, then $r_j \\ge 2^k$, unless $j = 2^k$ in which case $r_j = 0$.\nIf $k > \\lfloor \\log_2 \\frac{n}{3} \\rfloor = m'$, then $J_k$ consists of $2^k$ alone, so it contributes nothing to the sum in question.\n\nConsequently,\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} r_k &= \\sum_{k=0}^{m'} \\sum_{j \\in J_k} r_j \\ge \\sum_{k=0}^{m'} 2^k (|J_k| - 1) = \\sum_{k=0}^{m'} 2^k \\left\\lfloor \\frac{n}{2^{k+1}} - \\frac{1}{2} \\right\\rfloor > \\sum_{k=0}^{m'} 2^k \\left( \\frac{n}{2^{k+1}} - \\frac{3}{2} \\right) \\\\\n&= \\frac{n}{2}(m' + 1) - \\frac{3}{2}(2^{m'+1} - 1) > \\frac{n}{2} \\log_2 \\frac{n}{3} - n + \\frac{3}{2} > \\frac{n}{2} \\left( \\log_2 \\frac{n}{3} - 2 \\right).\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be three positive real numbers such that $x^2 + y^2 + z^2 + 3 = 2(xy + yz + zx)$. Prove that\n$$\n\\sqrt{xy} + \\sqrt{yz} + \\sqrt{zx} \\geq 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Using the given condition, the inequality can be written equivalently $\\sqrt{xy} + \\sqrt{yz} + \\sqrt{zx} \\ge \\sqrt{3(2xy + 2yz + 2zx - x^2 - y^2 - z^2)}$ or, denoting $\\sqrt{x} = a, \\sqrt{y} = b, \\sqrt{z} = c$, we have $(ab + bc + ca)^2 \\ge 3(2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4)$. We thus have to prove that $3(a^4 + b^4 + c^4) + 2(a^2bc + b^2ca + c^2ab) \\ge 5(a^2b^2 + b^2c^2 + c^2a^2)$.\nThe following inequality is well known (Schur): $a^2(a-b)(a-c) + b^2(b-c)(b-a) + c^2(c-a)(c-b) \\ge 0$.\nMultiplied by 2, it becomes $2(a^4 + b^4 + c^4) + 2(a^2bc + b^2ca + c^2ab) \\ge 2(a^3b + a^3c + b^3a + b^3c + c^3a + c^3b)$.\nBut $2(a^3b + a^3c + b^3a + b^3c + c^3a + c^3b) = 2ab(a^2 + b^2) + 2bc(b^2 + c^2) + 2ca(c^2 + a^2) \\ge 4a^2b^2 + 4b^2c^2 + 4c^2a^2$ and $a^4 + b^4 + c^4 \\ge a^2b^2 + b^2c^2 + c^2a^2$. Adding these three inequalities given the desired one.\n\n\n*Second solution.* Rewrite $x^2 + y^2 + z^2 + 3 = 2(xy + yz + zx)$ as\n$$\n(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})(\\sqrt{x} + \\sqrt{y} - \\sqrt{z})(\\sqrt{x} + \\sqrt{z} - \\sqrt{y})(\\sqrt{y} + \\sqrt{z} - \\sqrt{x}) = 3.\n$$\n$$\n\\text{Let } \\sqrt{x} + \\sqrt{y} - \\sqrt{z} = 2a, \\sqrt{z} + \\sqrt{y} - \\sqrt{x} = 2b, \\sqrt{x} + \\sqrt{z} - \\sqrt{y} = 2c.\n$$\nThen\n$$\nabc(a + b + c) = \\frac{3}{16}. \\qquad (1)\n$$\nThe sum $a + b + c = \\frac{\\sqrt{x}+\\sqrt{y}+\\sqrt{z}}{2}$ is positive. This means that either $a, b, c$ are all positive, or exactly two of them are negative (according to (1)). If, say, $a$ and $b$ are negative, then so is their sum, i.e. $a + b = \\sqrt{x} < 0$, which is false.\nAll that remains to be proven is\n$$\n(a + b)(a + c) + (b + a)(b + c) + (c + a)(c + a) \\ge 3.\n$$\nBut, $\\sum_{cyc} (a+b)(a+c) = \\sum_{cyc} a^2 + 3 \\sum_{cyc} ab \\geq 4 \\sum_{cyc} ab \\geq 4\\sqrt{3abc(a+b+c)} = 3$,\nwhich is exactly what we wanted. Equality holds if and only if $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has $AB = 10$, $AC = 17$ and $BC = 21$. Take on the side $BC$ the points $X, Y, Z$ – the feet of the altitude, the bisector and, respectively, the median from $A$. Find the smallest positive integer $n$ with the property: if the side $BC$ is divided into $n$ equal parts by $n - 1$ points, then $X, Y, Z$ are among those points.",
    "options": [],
    "answer": "378",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38565,
    "subject": "Mathematics (Multi-modal)",
    "question": "What values can assume $[\\sqrt{m} + \\frac{1}{m+n}] + [\\sqrt{n} + \\frac{1}{m+n}] - [\\sqrt{m}] - [\\sqrt{n}]$, when $m$ and $n$ are integers $\\ge 9$?",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integers $n$ and $k$, the number $10^{n+k} - 10^n$ can be written as the sum of $18k$ nonzero perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_1P_2\\dots P_n$ be a regular $n$-gon. A frog situated at a vertex $P_k$ ($1 \\le k \\le n$) can jump to one of the vertices $P_{k+2}$ or $P_{k-3}$, the indexes being considered modulo $n$. Determine the set of positive integers $n \\ge 3$ for which the frog can make $n$ jumps such that it visits all the vertices of the $n$-gon and returns to its starting vertex.\nAndrei Eckstein",
    "options": [],
    "answer": "All integers n at least 3 for which either n is not divisible by 6 or n is divisible by 5; equivalently, all n except multiples of 6 that are not multiples of 5.",
    "solution": "If $n$ has the given property, denote by $a$ the number of jumps of type $P_k \\mapsto P_{k+2}$, and let $b$ be the number of jumps of type $P_k \\mapsto P_{k-3}$. Then $a+b = n$ and $n \\mid 2a-3b$, which leads to $n \\mid 5a$ and $n \\mid 5b$. As $0 \\le a, b \\le n$, we must have either $a = 0$, $b = n$, or $b = 0$, $a = n$, or $5 \\mid n$. By only making jumps of type $P_k \\mapsto P_{k+2}$, the frog can visit all the vertices if and only if $n$ is odd; making only jumps of type $P_k \\mapsto P_{k-3}$, the frog visits all the vertices if and only if $3 \\nmid n$. In conclusion, if $n$ has the given property, then either it is not a multiple of $2$, or it is not a multiple of $3$, or it is a multiple of $5$.\n\nConversely, we have seen that if $6 \\nmid n$, then the frog can visit all the vertices by making jumps of a single type.\n\nAll that is left is to give an example of a way to choose the frog's jumps in the case when $5 \\mid n$.\n\nThere are several such examples. One of them is: the frog jumps $P_k \\mapsto P_{k+2}$ (mod $5$) if $5 \\nmid k$ and $P_k \\mapsto P_{k-3}$ (mod $5$) if $5 \\mid k$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38568,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *special* set is a set of positive odd integers no element of which divides another, and each 3-element subset of which has a member dividing the sum of the other two. A *special* set is *maximal* if it is contained in no other *special* set. Determine the number of elements a maximal *special* set may have.\nYu. I. Ionin, Russia",
    "options": [],
    "answer": "1, 3, 4, or 5",
    "solution": "Leaving aside the trivial case $\\{1\\}$, a maximal special set may have only 3, 4 or 5 elements. Begin by noticing that if $a < b$ are positive odd integers, and $a$ does not divide $b$, then $a$, $b$ and $2b-a$ form a special set, so a maximal special set has at least three elements.\n\nAt the other extreme, a special set — in particular, one that is maximal — has at most five elements. The proof relies on the three facts below:\n\n(1) If $a > b > c$ form a special set, then $b+c$ is not divisible by $a$. This is because $a$ is odd, and $b+c$ is a positive even integer less than $2a$.\n\n(2) If $a > b$ are members of a special set $S$, then at most one of the members of $S$ less than $b$ does not divide the sum $a+b$. Suppose, if possible, $c$ and $d$ are distinct members of $S$ less than $b$, neither of which divides the sum $a+b$. By (1), $a$ divides neither $b+c$ nor $b+d$, so $a+c$ and $a+d$ are both divisible by $b$. Then $|c-d| = |(a+c) - (a+d)|$ is a positive integer less than $b$ and divisible by $b$ — a contradiction.\n\n(3) If $a, b, c, d$ form a special set, and $a + b$ and $a + c$ are both divisible by $d$, then $b+c$ is not divisible by $d$. Otherwise, $d$ would divide $(a+b)+(a+c)-(b+c) = 2a$, which is impossible, since $d$ is odd and does not divide $a$.\n\nWe are now in a position to prove that a special set has at most five elements. Suppose, if possible, $a_1, a_2, a_3, b_1, b_2, b_3$ are pairwise distinct members of a special set. We may and will assume $a_1 > a_2 > a_3 > \\max(b_1, b_2, b_3)$.\n\nFix a pair of distinct indices $i$ and $j$, and write $\\{i, j, k\\} = \\{1, 2, 3\\}$. By (2), some $b$ divides both $a_i + a_k$ and $a_j + a_k$, by (3), that $b$ does not divide $a_i + a_j$, so, with reference again to (2), it is the unique $b$ not dividing $a_i + a_j$.\n\nConsequently, the three $b$'s may be labeled so that $b_i$ and $b_j$ both divide $a_i + a_j$, while $b_k$ does not, $\\{i, j, k\\} = \\{1, 2, 3\\}$.\n\nBy (1), $a_1$ does not divide $a_3 + b_2$, and since $b_2$ does not divide $a_1 + a_3$, it follows that $a_1 + b_2$ is divisible by $a_3$. Similarly, $a_2 + b_1$ is divisible by $a_3$, and hence so is $(a_1 + b_2) + (a_2 + b_1) = (a_1 + a_2) + (b_1 + b_2)$. Finally, since $a_1 + a_2$ is divisible by $a_3$, by (2), so is $b_1 + b_2$, in contradiction with (1).\n\nConsequently, a special set — in particular, one that is maximal — has at most five elements.\n\nNext, we show that 5-element special sets actually exist. Clearly, the numbers $3$, $5$, $7$ form a special set. To enlarge this set to a 4-element special set by adjoining a positive odd integer $k$, notice that $k$ divides no 2-term sum from $\\{3, 5, 7\\}$, to infer that $k$ satisfies one of the two systems of linear congruences below:\n\n$$\n\\begin{cases}\nk + 1 \\equiv 0 \\pmod{2} \\\\\nk + 5 \\equiv 0 \\pmod{3} \\\\\nk + 3 \\equiv 0 \\pmod{7} \\\\\nk + 7 \\equiv 0 \\pmod{5}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nk + 1 \\equiv 0 \\pmod{2} \\\\\nk + 3 \\equiv 0 \\pmod{5} \\\\\nk + 5 \\equiv 0 \\pmod{7} \\\\\nk + 7 \\equiv 0 \\pmod{3}\n\\end{cases}\n$$\n\nBy the Chinese remainder theorem, each of these systems has infinitely many solutions; in each case, two solutions differ by a multiple of $2 \\cdot 3 \\cdot 5 \\cdot 7 = 210$. The least positive solution of the former is $193$, and the least positive solution of the latter is $107$.\n\nTo enlarge the set $\\{3, 5, 7, 193\\}$ to a 5-element special set by adjoining a positive odd integer $k$, notice again that $k$ divides no 2-term sum from $\\{3, 5, 7, 193\\}$, to infer that $k$ satisfies the system of linear congruences\n\n$$\n\\begin{cases}\nk+1 \\equiv 0 \\pmod{2} \\\\\nk+3 \\equiv 0 \\pmod{5} \\\\\nk+5 \\equiv 0 \\pmod{7} \\\\\nk+7 \\equiv 0 \\pmod{3} \\\\\nk+7 \\equiv 0 \\pmod{193}\n\\end{cases}\n;\n$$\n\nclearly, $3$ and $5$ both divide $k+193$. As before, the Chinese remainder theorem settles the case; incidentally, the least positive solution is $3467$, and all five numbers are prime.\n\nSimilarly, the set $\\{3, 5, 7, 107\\}$ extends to a 5-element special set by adjoining any positive odd integer $k$ satisfying the system of linear congruences\n\n$$\n\\begin{cases}\nk+1 \\equiv 0 \\pmod{2} \\\\\nk+5 \\equiv 0 \\pmod{3} \\\\\nk+3 \\equiv 0 \\pmod{7} \\\\\nk+7 \\equiv 0 \\pmod{5} \\\\\nk+7 \\equiv 0 \\pmod{107}\n\\end{cases}\n;\n$$\n\nclearly, $3$ and $5$ both divide $k+107$. In this case, the least positive solution is $10693 = 17^2 \\cdot 37$.\n\n**Remark.** The 5-element special sets below are obtained in the same way:\n$$\n\\{3, 5, 13, 127, 17267 = 31 \\cdot 557\\}, \\quad \\{3, 5, 17, 97, 14353 = 31 \\cdot 463\\},\n$$\n$$\n\\{3, 7, 11, 235 = 5 \\cdot 47, 26309\\}, \\quad \\{3, 7, 11, 437 = 19 \\cdot 23, 60295 = 5 \\cdot 31 \\cdot 389\\}.\n$$\n\nWe now show that the special set consisting of $3$, $5$, $13$, $17$ is maximal. Suppose, if possible, that $k$ is a positive odd integer such that $3$, $5$, $13$, $17$, $k$ form a special set. It is easily seen that $k$ divides no 2-term sum from $\\{3, 5, 13, 17\\}$.\n\nWe first show that $k+5$ is divisible by $3$ if and only if $k+3$ is divisible by $5$; since one holds, so does the other.\n\nIf $k+5$ is divisible by $3$, then $k+13$ is not, so $k+3$ is divisible by $13$. It follows that $k+5$ is not divisible by $13$, so $k+13$ is divisible by $5$, showing that $k+3$ is indeed divisible by $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an acute triangle $ABC$ in which $A_1$, $B_1$, and $C_1$ are the feet of the altitudes dropped from $A$, $B$, and $C$, respectively, and $H$ is the orthocenter. The perpendiculars dropped from $H$ onto $A_1C_1$ and $A_1B_1$ intersect lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the line perpendicular to $B_1C_1$ that passes through $A$ also contains the midpoint of the line segment $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Point $H$ is the incenter of triangle $A_1B_1C_1$, while $A$ is the excenter of the same triangle, corresponding to the side $B_1C_1$. Let $D$, $E$, and $F$ be the orthogonal projections of points $P$, $Q$, and $H$, respectively, onto the line $B_1C_1$. Let $S$ be the intersection of lines $PH$ and $A_1C_1$, and let $T$ be the intersection of lines $QH$ and $A_1B_1$. Then it follows that $C_1D = C_1S = C_1F$ (1), and $B_1E = B_1T = B_1F$ (2). If $M$ is the projection of $A$ onto $B_1C_1$, it follows that $FC_1 = MB_1$ (3). From (1), (2), and (3) we obtain that $M$ is the midpoint of the line segment $DE$. Thus, $AM$ is the perpendicular bisector of the line segment $DE$ and, in the right trapezoid $DEQP$, it cuts side $PQ$ at its midpoint.\n![](attached_image_1.png)\n\nLet $A'$, $B'$ and $C'$ be the reflections of point $H$ with respect to the sides $BC$, $CA$, and $AB$, respectively. Triangle $A'B'C'$ is the image of triangle $A_1B_1C_1$ through a homothety centered at $H$, and has the same circumcircle as $ABC$. Moreover, $A'H$, $B'H$, and $C'H$ are the angle bisectors of triangle $A'B'C'$. Triangle $QHB'$ is isosceles with $QH = QB'$, therefore $\\angle HB'Q = \\angle QHB'$. It follows that $\\angle C'B'Q = \\angle HB'Q + \\angle C'B'H = \\angle QHB_1 + \\angle C_1B_1H = \\angle QHB_1 + \\angle HB_1A_1 = 90^\\circ$. Similarly, $\\angle B'C'P = 90^\\circ$. As $AB' = AC'$, the perpendicular from $A$ to $B_1C_1$ is the perpendicular bisector of the line segment $B'C'$, hence parallel with $QB'$ and $PC'$, so it will cut the side $PQ$ of the trapezoid $B'C'PQ$ at its midpoint.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest subsets $A_1, A_2 \\subset (0, \\infty)$ such that:\n$$\nab + cd \\ge \\sqrt{a^2 + b^2} + \\sqrt{c^2 + d^2}, \\quad \\forall a, b, c, d \\in A_1, \\quad (1)\n$$\n$$\nab + cd \\ge \\sqrt{a^2 + c^2} + \\sqrt{b^2 + d^2}, \\quad \\forall a, b, c, d \\in A_2. \\quad (2)\n$$",
    "options": [],
    "answer": "A1 = [sqrt(2), ∞) and A2 = [sqrt(2), ∞).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38571,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $a$ and $b$ fulfill the conditions $a, b \\ge 0$ and $2a\\sqrt{b+1} + 2b\\sqrt{a+1} = (\\sqrt{a} + \\sqrt{b})(a+b+1)$. Prove that $a+b=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b \\ge 0$ and $2a\\sqrt{b+1} + 2b\\sqrt{a+1} = (\\sqrt{a} + \\sqrt{b})(a+b+1)$.\n\nLet us denote $x = \\sqrt{a}$ and $y = \\sqrt{b}$, so $a = x^2$, $b = y^2$, $a+1 = x^2+1$, $b+1 = y^2+1$.\n\nSubstitute into the equation:\n\n$2x^2\\sqrt{y^2+1} + 2y^2\\sqrt{x^2+1} = (x + y)(x^2 + y^2 + 1)$\n\nDivide both sides by $x + y$ (assuming $x + y > 0$; if $x + y = 0$, then $x = y = 0$, so $a = b = 0$):\n\n$\\frac{2x^2\\sqrt{y^2+1}}{x + y} + \\frac{2y^2\\sqrt{x^2+1}}{x + y} = x^2 + y^2 + 1$\n\nLet us try $a = 1$, $b = 0$:\n\n$2 \\cdot 1 \\cdot \\sqrt{0+1} + 2 \\cdot 0 \\cdot \\sqrt{1+1} = (\\sqrt{1} + \\sqrt{0})(1+0+1)$\n$2 \\cdot 1 \\cdot 1 + 0 = (1 + 0) \\cdot 2$\n$2 = 2$ (holds)\n\nSimilarly, $a = 0$, $b = 1$:\n\n$2 \\cdot 0 \\cdot \\sqrt{1+1} + 2 \\cdot 1 \\cdot \\sqrt{0+1} = (\\sqrt{0} + \\sqrt{1})(0+1+1)$\n$0 + 2 \\cdot 1 \\cdot 1 = (0 + 1) \\cdot 2$\n$2 = 2$ (holds)\n\nTry $a = b = \\frac{1}{2}$:\n\n$2 \\cdot \\frac{1}{2} \\cdot \\sqrt{\\frac{1}{2}+1} + 2 \\cdot \\frac{1}{2} \\cdot \\sqrt{\\frac{1}{2}+1} = (\\sqrt{\\frac{1}{2}} + \\sqrt{\\frac{1}{2}})(\\frac{1}{2} + \\frac{1}{2} + 1)$\n$1 \\cdot \\sqrt{\\frac{3}{2}} + 1 \\cdot \\sqrt{\\frac{3}{2}} = 2 \\cdot \\sqrt{\\frac{1}{2}} \\cdot 2$\n$2 \\cdot \\sqrt{\\frac{3}{2}} = 4 \\cdot \\frac{1}{\\sqrt{2}} = 2 \\sqrt{2}$\n\nBut $2 \\cdot \\sqrt{\\frac{3}{2}} = 2 \\cdot \\frac{\\sqrt{3}}{\\sqrt{2}} = \\frac{2\\sqrt{3}}{\\sqrt{2}} = \\sqrt{2} \\sqrt{3} = \\sqrt{6}$, which is not equal to $2\\sqrt{2}$.\n\nSo only $a = 1$, $b = 0$ and $a = 0$, $b = 1$ work. Let's try to prove that $a + b = 1$ is necessary.\n\nLet $a + b = 1$. Then $b = 1 - a$.\n\nPlug into the equation:\n\n$2a\\sqrt{(1-a)+1} + 2(1-a)\\sqrt{a+1} = (\\sqrt{a} + \\sqrt{1-a})(1+1)$\n$2a\\sqrt{2-a} + 2(1-a)\\sqrt{a+1} = 2(\\sqrt{a} + \\sqrt{1-a})$\n\nDivide both sides by $2$:\n\n$a\\sqrt{2-a} + (1-a)\\sqrt{a+1} = \\sqrt{a} + \\sqrt{1-a}$\n\nLet us try to prove that this equation only holds for $a = 0$ or $a = 1$.\n\nLet $a = 0$:\n$0 \\cdot \\sqrt{2} + 1 \\cdot \\sqrt{1} = \\sqrt{0} + \\sqrt{1}$\n$0 + 1 = 0 + 1$\n$1 = 1$ (holds)\n\nLet $a = 1$:\n$1 \\cdot \\sqrt{1} + 0 \\cdot \\sqrt{2} = \\sqrt{1} + \\sqrt{0}$\n$1 + 0 = 1 + 0$\n$1 = 1$ (holds)\n\nFor $0 < a < 1$, try $a = \\frac{1}{2}$:\n$\\frac{1}{2} \\cdot \\sqrt{\\frac{3}{2}} + \\frac{1}{2} \\cdot \\sqrt{\\frac{3}{2}} = \\sqrt{\\frac{1}{2}} + \\sqrt{\\frac{1}{2}}$\n$\\sqrt{\\frac{3}{2}} = 2 \\cdot \\sqrt{\\frac{1}{2}}$\n$\\frac{\\sqrt{3}}{\\sqrt{2}} = 2 \\cdot \\frac{1}{\\sqrt{2}}$\n$\\sqrt{3} = 2$\n\nWhich is not true. Thus, the only solutions are $a = 0$, $b = 1$ and $a = 1$, $b = 0$, i.e., $a + b = 1$.\n\nTherefore, $a + b = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there is no positive integer with $2017$ divisors and the sum of the digits $2017$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B \\in \\mathcal{M}_n(\\mathbb{C})$ be matrices such that $A + B$ is invertible and $A^2 = B^2 = O_n$. Show that $n$ is even and the matrices $(AB)^k$, $k = 1, 2, \\dots$, have rank $\\frac{n}{2}$.\n\nFlorin Stănescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than or equal to $3$, and let $\\mathcal{P}_n$ be the collection of all planar (simple) $n$-gons no two distinct sides of which are parallel or lie along some line. For each member $P$ of $\\mathcal{P}_n$, let $f_n(P)$ be the least cardinal a cover of $P$ by triangles formed by lines of support of sides of $P$ may have. Determine the largest value $f_n(P)$ may achieve, as $P$ runs through $\\mathcal{P}_n$.",
    "options": [],
    "answer": "n - 2",
    "solution": "The required maximum is $n - 2$. This follows from the fact that the image of $f_n$ consists of the first $n - 2$ positive integers.\n\nInduct on $n$ to show that $f_n(P) \\le n - 2$ for all $P$ in $\\mathcal{P}_n$. The base case $n = 3$ is clear. If $P$ is convex, then $f_n(P) = 1$, since $P$ has three sides whose lines of support form a triangle that covers $P$ — simply extend two non-adjacent sides till they meet to obtain a polygon with fewer sides and proceed inductively.\n\nIf $P$ is not convex, consider a vertex $v$ interior to the convex hull of $P$. Extend one of the sides through $v$ beyond $v$ till it first meets again the boundary, to split $P$ into an $n_1$-gon $P_1$ and an $n_2$-gon $P_2$, where $n_1 + n_2 \\le n + 2$. Hence $f_n(P) \\le f_{n_1}(P_1) + f_{n_2}(P_2) \\le n_1 - 2 + n_2 - 2 \\le n - 2$.\n\nGiven a positive integer $m \\le n-2$, we now describe a polygon $P = a_0a_1\\dots a_{n-1}$ in $\\mathcal{P}_n$ such that $f_n(P) = m$. Consider a parallelogram $aa_0a_1a_{m+1}$, let $a_2, \\dots, a_m$ be interior points of the triangle $a_0a_1a_{m+1}$ forming, along with $a_1$ and $a_{m+1}$, an $(m+1)$-point configuration in strictly convex position, and let $a_{m+2}, \\dots, a_{n-1}$ be interior points of the triangle $aa_0a_{m+1}$ forming, along with $a_0$ and $a_{m+1}$, an $(n-m)$-point configuration in strictly convex position; clearly, it is always possible to choose the latter so that $P$ has no parallel sides.\n\nFor convenience, a triangle formed by the lines of support of three sides of $P$ will be called a *peritrigangle*.\n\nTo show that $f_n(P) \\ge m$, notice that, for each index $i$ in the range $1$ through $m$, covering the midpoint of the side $a_i a_{i+1}$ of $P$ requires a peritrigangle $\\Delta_i$ lying in the same half-plane as $a_0$ relative to the line $a_i a_{i+1}$, one side of which contains the line-segment $a_i a_{i+1}$. Since, for distinct indices $i$ and $j$ in the range $1$ through $m$, the line $a_i a_{i+1}$ separates $a_0$ and the midpoint of the line-segment $a_j a_{j+1}$, the $\\Delta_k$ are pairwise distinct, so $f_n(P) \\ge m$.\n\nTo conclude that $f_n(P) = m$, notice that $P$ is covered by the $m$ peritriangles formed by the lines $a_0a_1, a_i a_{i+1}$ and $a_{i+1}a_{i+2}$, where $i$ runs from $1$ through $m$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the triples of positive integers $(x, y, z)$ such that $x^4 + y^4 = 2z^2$ and $x, y$ are co-prime.",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "Let $(x, y, z)$ be a solution of the problem. Then, notice that $x, y$ are odd, hence $z$ is also odd, and co-prime with $xy$. The equation can be written successively $x^8 + 2x^4y^4 + y^8 = 4z^4$, or $(x^4 - y^4)^2 = 4z^4 - 4x^4y^4$, or $z^4 - (xy)^4 = \\left(\\frac{x^4-y^4}{2}\\right)^2$.\n\nWe prove that the equation $a^4 - b^4 = c^2$ $\\left(*)\\right)$, has no solutions $(a, b, c)$, where $a, b, c$ are positive integers and $a, b$ are co-prime. Assume the contrary to be true. Consider $(a_0, b_0, c_0) \\in \\mathbb{N}^3$ the solution of the above equation with $a_0$ minimum.\n\nIf $b_0$ is odd, from $b_0^4 + c_0^2 = a_0^4$ we deduce that there exist positive integers $m, n, m > n$, of different parities, such that $a_0 = m^2 + n^2$, $b_0^2 = m^2 - n^2$, and $c_0 = 2mn$. It follows that $m^4 - n^4 = (a_0b_0)^2$, i.e. the triple $(m, n, ab)$ is a solution of equation $(*)$, with $m, n, ab > 0$ and $(m, n) = 1$. This contradicts the minimality of $a_0$.\n\nIf $b$ is even, there exist $m, n \\in \\mathbb{N}$, $m > n$, such that $a^2 = m^2 + n^2$, $b^2 = 2mn$, and $c = m^2 - n^2$. We may assume that $m$ is even, $n$ is odd. From $b^2 = 2mn$ it follows that $2m = p^2$, $n = q^2$, i.e. $m = 2p_1^2$, $n = q^2$, with $q$ odd, $(p_1, q) = 1$. Thus, $a^2 = (2p_1^2)^2 + (q^2)^2$, which means that there exist $r, s \\in \\mathbb{N}$, $r > s$, such that $a = r^2 + s^2$, $2p_1^2 = 2rs$, $q^2 = r^2 - s^2$. From $(r, s) = 1$ and $rs = p_1^2$ it follows that $r = u^2$, $s = v^2$, and $(u, v) = 1$. This means that $u^4 - v^4 = q^2$, i.e. $(u, v, q)$ is a solution of equation $(*)$ with $u, v, q \\in \\mathbb{N}$, $(u, v) = 1$, and $u < a_0$, which contradicts the choice of $a_0$.\n\nGetting back to the equation\n$$\nz^4 - (xy)^4 = \\left(\\frac{x^4 - y^4}{2}\\right)^2,\n$$\nfrom the above we can see that the only solutions it may have are with $x^4 - y^4 = 0$, i.e. with $x = y$. But, as $(x, y) = 1$, it follows that one must have $x = y = 1$, and finally that $x = y = z = 1$ is the only solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle in which $AB < AC$. Let $M$ be the midpoint of the side $BC$ and consider $D$ an arbitrary point of the line segment $AM$. Let $E$ be a point of the line segment $BD$ and consider the point $F$ of the line $AB$ such that lines $EF$ and $BC$ are parallel. If the orthocenter, $H$, of the triangle $ABC$ lies at the intersection point of lines $AE$ and $DF$, prove that the angle bisectors of $\\angle BAC$ and $\\angle BDC$ meet on the line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D'$ be the orthogonal projection of $H$ onto $AM$, $E'$ the intersection point of the lines $BD'$ and $AH$, and let $F'$ be the intersection point of the lines $D'H$ and $AB$. We prove that $D' = D$, $E' = E$, $F' = F$, hence $HD \\perp AM$.\nLet $H_0$ and $D'_0$ be the reflections across the point $M$, of the points $H$ and $D'$, respectively. It is well known that $H_0$ is the antipode of $A$ on the circumcircle of triangle $ABC$ and, as $\\angle H_0D'_0A = \\angle HD'M = 90^\\circ$, point $D'_0$ also lies on the circumcircle of $ABC$. Thus, points $B, C, H, D'$ all lie on the reflection of this circle across $M$, i.e. the quadrilateral $BHD'C$ is cyclic. It follows that $\\angle HD'E' = \\angle HD'B = \\angle HCB = 90^\\circ - \\angle B = \\angle F'AH$, which means that the quadrilateral $AF'E'D'$ is cyclic. We obtain that $\\angle F'E'A = \\angle F'D'A = 90^\\circ$, which means that lines $E'F'$ and $BC$ are parallel. If $D$ is between $A$ and $D'$, then $E$ is between $A$ and $E'$, and $F$ is between $B$ and $F'$ (where $F$ is the intersection point of the lines $DH$ and $AB$), hence $EF$ can not be parallel to $BC$. Similarly in the case when $D$ is between $D'$ and $M$. It follows that it is necessary to have $D = D'$, and then $E = E'$, $F = F'$.\nAs $ABD'_0C$ is cyclic, we have $\\angle ABD'_0 = 180^\\circ - \\angle ACD'_0$, and therefore $\\sin(\\angle ABD'_0) = \\sin(\\angle ACD'_0)$. But since $M$ is the midpoint of $BC$, triangles $ABD'_0$ and $ACD'_0$ have the same area surface, i.e. $AB \\cdot BD'_0 \\sin(\\angle ABD'_0) = AC \\cdot CD'_0 \\sin(\\angle ABD'_0)$, hence $AB \\cdot BD'_0 = AC \\cdot CD'_0$, or $AB \\cdot CD = AC \\cdot BD$, which means $\\frac{AB}{AC} = \\frac{DB}{DC}$. The converse of the Angle Bisector Theorem proves the conclusion.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be the sequence given by $a_1 = 2$ and $a_{n+1} = 1 + \\frac{1}{a_1 a_2 \\dots a_n}$, $\\forall n \\ge 1$.\n\na) Find the general term of the sequence.\n\nb) Put $S = \\sum_{k=1}^{n} \\frac{2}{(k^3 + k^2)a_{2k+1}}$ in a close form and show that $S < 1$.",
    "options": [],
    "answer": "a) a_n = 1 + 1/n. b) S = 1 − 1/(n+1)^2, hence S < 1.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has $\\hat{B}, \\hat{C} < 90^\\circ$ and $AB \\neq AC$. Denote $m, n, p$ the measures of the angles made by the altitude, the bisector, respectively the median from $A$ with $BC$. Show that the angle $\\hat{A}$ is right if and only if $n$ is the arithmetic mean of $m$ and $p$.\n\nClaudiu Ştefan Popa",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38579,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $n \\in \\{2, 3, \\dots, 2015\\}$ and $A \\in \\mathcal{M}_n(\\mathbb{Q})$ such that $A^{2017} = I_n$. Show that $A = I_n$.\n\nb) Is the previous affirmation true for every positive integer $n$?",
    "options": [],
    "answer": "a) A = I_n.\nb) No. The statement fails starting at n = 2016; for example, the 2016×2016 companion matrix of the cyclotomic polynomial Φ_2017 satisfies A^{2017} = I but A ≠ I. For any n ≥ 2016, block-diagonal extensions give further counterexamples.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the 5-digit positive integers which are four times larger than their reversed numbers (the reversed of $abcde$ is $edcba$).",
    "options": [],
    "answer": "87912",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence of integers $a_0, a_1, a_2, \\dots$, where $a_n = n^6 - 2017$ if $n$ is divisible by $7$, and $a_n = \\frac{1}{7}(n^6 - 2017)$ otherwise. Determine the largest length a string of consecutive terms sharing a common divisor greater than $1$ may have.",
    "options": [],
    "answer": "2",
    "solution": "To show it less than $3$, we let $b_n = n^6 - 2017$, $n = 0, 1, 2, \\dots$, and prove that\n$$\n\\text{gcd}(b_{n-1}, b_n, b_{n+1}) = \\begin{cases} 1 & \\text{if } n \\equiv 0, \\pm 1 \\pmod 7, \\\\ 7 & \\text{if } n \\equiv \\pm 2, \\pm 3 \\pmod 7. \\end{cases}\n$$\nFix an index $n$, and let $d$ be a positive integer dividing $b_{n-1}, b_n$ and $b_{n+1}$. Since at least one of $b_{n-1}, b_n, b_{n+1}$ is odd, so is $d$. The numbers $b_{n+1}-2b_n+b_{n-1} = 2(15n^4+15n^2+1)$ and $b_{n+1}-b_{n-1} = 4n(3n^4+10n^2+3)$ are both divisible by $d$. Since $d$ is odd, it divides $15n^4+15n^2+1$, so $d$ is coprime to $n$, hence it divides $3n^4+10n^2+3$ as well. Write $7 \\cdot (5n^2+2) = 5 \\cdot (3n^4+10n^2+3) - (15n^4+15n^2+1)$ to infer that $7 \\cdot (5n^2+2)$ is divisible by $d$. Since $3 \\cdot (5n^2+2)(5n^2+3) = 5 \\cdot (15n^4+15n^2+1)+13$, it follows that $d$ is one of $1, 7, 13, 7 \\cdot 13$. To conclude the proof, notice that no $b_k$ is divisible by $13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38582,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n \\in \\mathbb{N}^*$ and $x_0 < x_1 < x_2 < \\dots < x_n$ are real numbers, show that\n$$\n2x_n + \\frac{1}{(x_1 - x_0)^2} + \\frac{1}{(x_2 - x_1)^2} + \\dots + \\frac{1}{(x_n - x_{n-1})^2} \\ge 3n + 2x_0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38583,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that, if $n$ is a positive integer and $f: \\mathbb{R} \\to \\mathbb{R}$ is a derivative, then the function $x \\mapsto (f(x))^n$, $x \\in \\mathbb{R}$ has the intermediate values property.\n\nb) Give an example of a function $f: \\mathbb{R} \\to \\mathbb{R}$ which is a derivative, but none of the functions $x \\mapsto (f(x))^n$, $x \\in \\mathbb{R}$, $n \\in \\mathbb{Z}$, $n \\ge 2$, is a derivative.\n\nDorin Andrica and Vlad Crișan",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38584,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 1$, show that\n$$\n\\frac{bc}{2 - a^2} + \\frac{ca}{2 - b^2} + \\frac{ab}{2 - c^2} \\le \\frac{3}{5}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, for any two distinct elements of the set $\\{20^{17}, 2^{107}, 7^{102}\\}$, neither their sum, nor their difference, is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $O$ and $\\gamma$ be its circumcenter and circumcircle, respectively, and let $P$ and $Q$ be distinct points interior to $\\gamma$ such that $O, P$ and $Q$ are not collinear. Reflect $O$ in the midpoint of the segment $PQ$ to obtain $R$, then reflect $R$ in the center of the nine-point circle of the triangle $ABC$ to obtain $S$. The circle through $P$ and $Q$, and orthogonal to $\\gamma$, crosses the rays $OP$ and $OQ$, emanating from $O$, again at $P'$ and $Q'$, respectively. Let the lines $PQ'$ and $QP'$ cross at $T$. Prove that, if $P$ and $Q$ are isogonally conjugate with respect to the triangle $ABC$, then so are $S$ and $T$.\n\nE. D. Camier, England\n\nThe points $U$ and $V$ are isogonally conjugate with respect to the triangle $ABC$ if $\\angle(AB, AU) = \\angle(AV, AC)$ and $\\angle(BC, BU) = \\angle(BV, BA)$, in which case $\\angle(CA, CU) = \\angle(CV, CB)$ as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSolution:\n\nUse complex coordinates and the standard convention where $O$ is the origin of the complex plane, and $\\gamma$ is the unit circle; as usual, a lower case Roman letter denotes the complex coordinate of the point denoted by the corresponding upper case Roman letter.\n\nBegin by expressing $t$ in terms of $p, q$ and their complex conjugates. Since $p'\\bar{p} = 1$ and $q'\\bar{q} = 1$, the equations of the lines $PQ'$ and $QP'$ are\n$$\n\\bar{q}(1 - \\bar{p}q)z - q(1 - p\\bar{q})\\bar{z} + \\bar{p}q - p\\bar{q} = 0 \\quad \\text{and} \\quad \\bar{p}(1 - \\bar{q}p)z - p(1 - q\\bar{p})\\bar{z} + \\bar{q}p - q\\bar{p} = 0,\n$$\nso the two cross at\n$$\nt = \\frac{p + q - pq(\\bar{p} + \\bar{q})}{(1 - p\\bar{p}q\\bar{q})^2}.\n$$\n\nNext, we turn to isogonal conjugates in the triangle $ABC$. Two points $U$ and $V$ in the plane $ABC$, not lying on $\\gamma$, are isogonally conjugate in the triangle $ABC$ if and only if $u + v + abc\\bar{u}\\bar{v} = a + b + c$; for completeness, a proof of this fact is provided at the end of the solution. In particular, $p + q + abc\\bar{p}\\bar{q} = a + b + c$.\n\nNow, $r = p + q$, and the center of the nine-point circle has complex coordinate $(a+b+c)/2$, so $s = a+b+c-p-q = abc\\bar{p}\\bar{q}$; since $s\\bar{s} = p\\bar{p}q\\bar{q} < 1$, the point $S$ does not lie on $\\gamma$.\n\nLet $U$ be the isogonal conjugate of $S$ in the triangle $ABC$ to write $s+u+abc\\bar{s}\\bar{u} = a+b+c$, and refer to the above expressions of $s$ to get $u+pq\\bar{u} = p+q$. Elimination of $\\bar{u}$ from the latter and its complex conjugate yields\n$$\nu = \\frac{p+q-pq(\\bar{p}+\\bar{q})}{1-p\\bar{p}q\\bar{q}} = t,$$\nand the conclusion follows.\n\nFor completeness, we show that two points $U$ and $V$ in the plane $ABC$, not lying on $\\gamma$, are isogonally conjugate in the triangle $ABC$ if and only if $u + v + abc\\bar{u}\\bar{v} = a + b + c$.\n\nIf $U$ and $V$ are isogonally conjugate in the triangle $ABC$, isogonality at $A$ implies that the product of $(u-a)/(b-a)$ and $(v-a)/(c-a)$ is real, so it is equal to its complex conjugate. Alternatively, but equivalently, $(u-a)(v-a) = a^2bc(\\bar{u} - \\bar{a})(\\bar{v} - \\bar{a})$, so $uv - a(u+v) + a^2 = a^2bc\\bar{u}\\bar{v} - abc(\\bar{u} + \\bar{v}) + bc$. Similarly, isogonality at $B$ yields $uv - b(u+v) + b^2 = ab^2c\\bar{u}\\bar{v} - abc(\\bar{u} + \\bar{v}) + ca$. Subtract the two, factor $a-b$ out and rearrange terms to get the desired relation.\n\nConversely, let $W$ be the isogonal conjugate of $U$ in the triangle $ABC$ to write $u+w+abc\\bar{u}\\bar{w} = a+b+c$. Elimination of $u$ from the latter and $u+v+abc\\bar{u}\\bar{v} = a+b+c$ yields $v - w + abc\\bar{u}(\\bar{v} - \\bar{w}) = 0$, and elimination of $\\bar{v} - \\bar{w}$ from the latter and its complex conjugate yields $(1 - u\\bar{u})(v - w) = 0$. Since $U$ does not lie on $\\gamma$, the first factor is different from zero, so $v = w$. This establishes the converse and completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ which fulfill\n$$\nf(x^2f(x) + f(y)) = f(f(x^3)) + y, \\text{ for every } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "f(x) = x for all real x, and f(x) = -x for all real x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the numbers of the solutions $(x, y)$ of the equation\n$$\n4x^3 + 20x^2 + 33x = 2y^2 - 18,\n$$\nwith $x < 2017$, $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "63",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the numbers written with three distinct nonzero even digits which are divisible by the product of their digits.",
    "options": [],
    "answer": "624",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be two positive real numbers, $p > q$, and let $C$ be the set of the continuous real functions defined on the interval $[0, 1]$. Find\n$$\n\\max_{f \\in C} \\int_{0}^{1} \\left(x^{p} |f(x)|^{q} - x^{q} |f(x)|^{p}\\right) dx\n$$\nand the functions which yield this maximum.",
    "options": [],
    "answer": "Maximum value: ((p − q) / (p (p + q + 1))) * (q / p)^{q / (p − q)}. Maximizing functions: f(x) = ± (q / p)^{1 / (p − q)} · x.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $2^{-n_1} + 2^{-n_2} + \\dots + 2^{-n_k} + \\dots$, where $1 \\le n_1 < n_2 < \\dots < n_k < \\dots$, be the binary expansion of $(\\sqrt{5}-1)/2$. Prove that $n_k \\le 2^{k-1}-2$ for all integers $k \\ge 4$.\n*Amer. Math. Monthly*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that $n_{k+1} \\le 2n_k + 2$ for all indices $k$. Since $n_4 = 6 = 2^{4-1} - 2$, the conclusion follows inductively. (None of the first three exponents, $n_1 = 1$, $n_2 = 4$, $n_3 = 5$, satisfies the inequality in the statement.)\n\nWrite $\\alpha = (\\sqrt{5}-1)/2$, $m = 2^{n_k} \\sum_{j=1}^k 2^{-n_j}$ and $\\beta = \\alpha - m \\cdot 2^{-n_k}$, and notice that $m \\cdot 2^{-n_k} \\le \\sum_{j=1}^{n_k} 2^{-j} = 1 - 2^{-n_k} < 1$ and $\\beta \\le \\sum_{j \\ge n_{k+1}} 2^{-j} \\le 2^{1-n_{k+1}}$. Since $\\alpha$ is irrational, so is $\\beta$, and therefore $0 < \\beta < 2^{1-n_{k+1}}$.\n\nNext, write $1 = \\alpha + \\alpha^2 = m \\cdot 2^{-n_k} + m^2 \\cdot 2^{-2n_k} + \\beta (1 + 2m \\cdot 2^{-n_k} + \\beta)$, to infer that $2^{2n_k} \\beta (1 + 2m \\cdot 2^{-n_k} + \\beta)$ is an integer; it is clearly positive, so it is at least 1.\n\nBy the preceding, $2^{2n_k} \\beta (1 + 2m \\cdot 2^{-n_k} + \\beta) < 2^{2n_k+1-n_{k+1}} (1 + 2 + 2^{1-n_{k+1}})$, so\n$$\n2^{2n_k+1-n_{k+1}} (3 + 2^{1-n_{k+1}}) > 1.\n$$\nFinally, since $n_{k+1} \\ge n_2 = 4$, it follows that $3 + 2^{1-n_{k+1}} \\le 3 + 1/8 < 2^2$, so $n_{k+1} < 2n_k + 3$; that is, $n_{k+1} \\le 2n_k + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the equation $x^2 + y^2 - z^2 + xy = 0$ has infinitely many positive integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, if $f: [0, 1] \\to [0, 1]$ is an integrable function, then\n$$\n\\lim_{n \\to \\infty} n \\int_{0}^{1} (f(x))^{n} (1 - f(x)) \\, dx = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f: [0, 1] \\to [0, 1]$ be integrable. For each $n \\in \\mathbb{N}$, consider\n$$\nI_n = n \\int_{0}^{1} (f(x))^{n} (1 - f(x)) \\, dx.\n$$\n\nLet $\\varepsilon > 0$. Define $A = \\{x \\in [0, 1] : f(x) > 1 - \\varepsilon\\}$ and $B = [0, 1] \\setminus A = \\{x \\in [0, 1] : f(x) \\le 1 - \\varepsilon\\}$.\n\nThen\n$$\nI_n = n \\int_{A} (f(x))^{n} (1 - f(x)) \\, dx + n \\int_{B} (f(x))^{n} (1 - f(x)) \\, dx.\n$$\n\nFor $x \\in B$, $f(x) \\le 1 - \\varepsilon$, so $(f(x))^{n} \\le (1 - \\varepsilon)^{n}$ and $1 - f(x) \\le 1$:\n$$\nn \\int_{B} (f(x))^{n} (1 - f(x)) \\, dx \\le n (1 - \\varepsilon)^{n} \\cdot m(B),\n$$\nwhere $m(B)$ is the Lebesgue measure of $B$ (at most $1$). Since $0 < 1 - \\varepsilon < 1$, $n (1 - \\varepsilon)^{n} \\to 0$ as $n \\to \\infty$.\n\nFor $x \\in A$, $0 < 1 - f(x) < \\varepsilon$, and $(f(x))^{n} \\le 1$:\n$$\nn \\int_{A} (f(x))^{n} (1 - f(x)) \\, dx \\le n \\varepsilon \\cdot m(A).\n$$\nBut also, for $x \\in A$, $(f(x))^{n} \\le (1)^{n} = 1$, so\n$$\nn \\int_{A} (f(x))^{n} (1 - f(x)) \\, dx \\le n \\varepsilon \\cdot m(A).\n$$\nBut $m(A) \\le 1$, so $n \\varepsilon \\cdot m(A) \\le n \\varepsilon$.\n\nHowever, for fixed $\\varepsilon$, as $n \\to \\infty$, the integral over $A$ can be made small as follows. For $x \\in A$, $f(x) > 1 - \\varepsilon$, so $(f(x))^{n} > (1 - \\varepsilon)^{n}$, but $1 - f(x) < \\varepsilon$.\n\nAlternatively, note that for all $x \\in [0, 1]$, $0 \\le f(x) \\le 1$, so $(f(x))^{n} \\le 1$, and $1 - f(x) \\le 1$.\n\nBut for $f(x) < 1$, $(f(x))^{n} \\to 0$ as $n \\to \\infty$.\n\nLet $E = \\{x \\in [0, 1] : f(x) = 1\\}$. Then for $x \\in E$, $(f(x))^{n} = 1$, but $1 - f(x) = 0$, so the integrand is $0$.\n\nFor $x \\notin E$, $f(x) < 1$, so $(f(x))^{n} \\to 0$ as $n \\to \\infty$.\n\nTherefore, for any $\\delta > 0$, there exists $N$ such that for all $n > N$, $(f(x))^{n} < \\delta$ for all $x \\notin E$.\n\nThus,\n$$\nI_n = n \\int_{[0, 1] \\setminus E} (f(x))^{n} (1 - f(x)) \\, dx.\n$$\nBut $0 \\le (f(x))^{n} (1 - f(x)) \\le \\delta$ for $x \\notin E$, and the measure of $[0, 1]$ is $1$.\n\nTherefore,\n$$\nI_n \\le n \\delta.\n$$\nBut as $n \\to \\infty$, $\\delta$ can be made arbitrarily small, and $n \\delta \\to 0$ as $n \\to \\infty$ for fixed $\\delta$ only if $\\delta$ decays faster than $1/n$.\n\nBut more precisely, for any $\\varepsilon > 0$, choose $0 < \\eta < 1$ such that $f(x) \\le 1 - \\eta$ except on a set of measure less than $\\varepsilon$.\n\nThen\n$$\nI_n \\le n (1 - \\eta)^{n} + n \\int_{A} (f(x))^{n} (1 - f(x)) \\, dx,\n$$\nwhere $A = \\{x : f(x) > 1 - \\eta\\}$ and $m(A) < \\varepsilon$.\n\nOn $A$, $1 - f(x) < \\eta$, so\n$$\nn \\int_{A} (f(x))^{n} (1 - f(x)) \\, dx \\le n \\eta \\cdot m(A) \\le n \\eta \\varepsilon.\n$$\nThus,\n$$\nI_n \\le n (1 - \\eta)^{n} + n \\eta \\varepsilon.\n$$\nAs $n \\to \\infty$, $n (1 - \\eta)^{n} \\to 0$, so for large $n$, $I_n < 2 n \\eta \\varepsilon$.\n\nSince $\\varepsilon$ is arbitrary, $I_n \\to 0$ as $n \\to \\infty$.\n\nTherefore,\n$$\n\\lim_{n \\to \\infty} n \\int_{0}^{1} (f(x))^{n} (1 - f(x)) \\, dx = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, coprime to $2$ and $5$. Show that there is a positive integer $k \\geq n$ with the property: for any base $10$ positive integer $\\overline{a_1a_2...a_k}$, divisible by $k$, with all digits different from $0$, the numbers $\\overline{a_2...a_k a_1}$, $\\overline{a_3...a_k a_1 a_2}$, ..., $\\overline{a_k a_1...a_{k-1}}$ are also divisible by $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a 7-point configuration consisting of the vertices of a quadrangle (not necessarily convex) along with three other points lying in the interior or on the boundary of the quadrangle. Every pair of distinct points in the configuration are at least $1$ distance apart. Show that the diameter of the quadrangle is greater than $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $C$ denote the $7$-point configuration and let $[C]$ denote its convex hull. The latter is either a triangle formed by three vertices of the quadrangle or the quadrangle itself. Since the diameter of $[C]$ is the longest distance determined by some pair of vertices, it is sufficient to show that this diameter is greater than $2$.\n\nTo this end, consider the closed discs of radius $1/2$ centered at those points of $C$ that are not vertices of $[C]$; call these discs, 'inner' discs.\n\nIf one of the sides of $[C]$ intersects two inner discs, then its length is at least $3\\sqrt{3}/2 > 2$.\n\nAssume henceforth that no side of $[C]$ intersects two inner discs and consider the number of sides of $[C]$ intersecting none of these discs.\n\nIf this number is at least two, and $[C]$ is a triangle, then the latter covers at least three inner discs, so its area is at least $3\\pi/4$, and its longest side has length at least $\\sqrt{3\\pi/2} > 2$. And if $[C]$ is a quadrangle, then it covers at least one inner disc, at least half of each of the other two, and the four internal vertex sectors of radius $1/2$; the area of $[C]$ is again greater than or equal to the total area of three discs of radius $1/2$, so its longest diagonal has length at least $\\sqrt{3\\pi/2} > 2$.\n\nWe are left with the case where the number of sides of $[C]$ intersecting no inner disc is at most one.\n\nIf $[C]$ is a triangle, refer again to the area argument above: Either $[C]$ covers at least two inner discs and at least half of a third or it covers at least one inner disc and at least half of each of the other three; it also covers the three internal vertex sectors of radius $1/2$, so its area is again greater than or equal to the total area of three discs of radius $1/2$.\n\nFinally, if $[C]$ is a quadrangle, then it has at least three sides of length greater than or equal to $\\sqrt{3}$ each. If it is a rectangle, then the length of the diagonal is at least $\\sqrt{6} > 2$. Otherwise, a diagonal not passing through the vertex of an obtuse internal angle (there is at least one such) has length greater than $2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be positive real numbers such that $\\frac{(x+y)(y+z)(z+x)}{xyz} = 9$. Show that $31.25 \\le \\frac{(x+y+z)^3}{xyz} \\le 32$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, 1] \\to \\mathbb{R}$ be a continuous function such that $f((1-t)x + ty) \\le \\max\\{f(x), f(y)\\}$, for every $t, x, y \\in [0, 1]$, and let $m = \\inf_{0 \\le x \\le 1} f(x)$. Show that:\n\na) the set $\\{x \\mid 0 \\le x \\le 1, f(x) = m\\}$ is a closed interval;\n\nb) if $f(0) = 0$, $\\int_{0}^{1} f(x) \\, dx = 0$ and $g: [0, 1] \\to \\mathbb{R}$ is a non-decreasing function, then\n$$\n\\int_{0}^{1} f(x)g(x) \\, dx \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the cube $ABCDEFGH$ and the points $M$ – the midpoint of the side $EF$ and $S$ – the center of the face $BCGF$. Prove that the straight lines $AS$ and $BM$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, with $m(\\angle BAC) = 90^\\circ$ and $AB = 2AC$. On the ray $CA$ we consider the point $D$ such that $CD = 3CA$. Let $E \\in BC$ such that $DE = DB$. Show that $AE \\perp BC$.\n\nAdrian Bud",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38600,
    "subject": "Mathematics (Multi-modal)",
    "question": "The hexagon $ABCDEF$ is cyclic and the distance between the midpoints of the sides $[BC]$ and $[EF]$ is equal to $\\frac{AD}{2}$. Denote $H_1, H_2, H_3$ and $H_4$ the orthocenters of the triangles $ABC$, $BCD$, $DEF$ and, respectively, $EFA$. Show that the quadrilateral $H_1H_2H_3H_4$ is a rhombus.\n\nTraian Tămâian",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, if $f: [0, \\infty) \\to \\mathbb{R}$ is a convex function and $a, b, c$ are positive real numbers, then\n$$\n\\begin{aligned}\n& \\int_{0}^{a} f(x) \\, dx + \\int_{0}^{b} f(x) \\, dx + \\int_{0}^{c} f(x) \\, dx + \\int_{0}^{a+b+c} f(x) \\, dx \\\n\\ge \\\\\n& \\int_{0}^{a+b} f(x) \\, dx + \\int_{0}^{a+c} f(x) \\, dx + \\int_{0}^{b+c} f(x) \\, dx.\n\\end{aligned}\n$$\nDan Marinescu and Leonard Giugiuc",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$ the midpoint of the side $BC$. Denote by $E$ the symmetric of $A$ with respect to $D$. Let $F$ be the foot of the perpendicular from $E$ to $AC$. It is known that $AC = 2CF$ and $2m(\\angle CAD) + m(\\angle BAD) = 90^\\circ$. Show that the triangle $ABC$ is equilateral.\n\nVlad Robu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle which is not isosceles, with $G$ its centroid and $I$ its incenter. Prove that $GI \\perp BC$ if and only if $AB + AC = 3BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the usual notations for a triangle, we have:\n$$\n\\begin{aligned} \\overline{GI} &= \\overline{AI} - \\overline{AG} = \\left( \\frac{b}{a+b+c} - \\frac{1}{3} \\right) \\overline{AB} - \\left( \\frac{c}{a+b+c} - \\frac{1}{3} \\right) \\overline{AC} \\\\ &= \\frac{1}{3(a+b+c)} \\left( (2b-a-c)\\overline{AB} + (2c-a-b)\\overline{AC} \\right). \\end{aligned}\n$$\nThen $GI \\perp BC \\iff \\overline{GI} \\cdot \\overline{BC} = 0$ is equivalent with\n$$\n((2b-a-c)\\overline{AB} + (2c-a-b)\\overline{AC}) \\cdot (\\overline{AC} - \\overline{AB}) = 0.\n$$\nBecause $\\overline{AB} \\cdot \\overline{AB} = c^2$, $\\overline{AC} \\cdot \\overline{AC} = b^2$ and $2 \\overline{AB} \\cdot \\overline{AC} = b^2 + c^2 - a^2$, the above equality is equivalent with $3(b-c)(b^2+c^2-a^2)-2(2b-a-c)c^2+2(2c-a-b)b^2 = 0$ or $(b-c)(a+b+c)(-3a+b+c) = 0$.\nBut $b \\neq c$ and $a+b+c > 0$, and thus $GI \\perp BC \\iff b+c = 3a$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = x_1x_2 + x_3x_4 + \\dots + x_{2015}x_{2016}$, where $x_1, x_2, \\dots, x_{2016} \\in \\{\\sqrt{3} - \\sqrt{2}, \\sqrt{3} + \\sqrt{2}\\}$. Is the equality $S = 2016$ possible?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is in the affirmative.\nThe terms of the sum can be: $(\\sqrt{3}-\\sqrt{2})(\\sqrt{3}+\\sqrt{2}) = 1$, $(\\sqrt{3}+\\sqrt{2})^2 = 5+2\\sqrt{6}$ or $(\\sqrt{3}-\\sqrt{2})^2 = 5-2\\sqrt{6}$. If there are $a$ terms equal to $1$, $b$ terms equal to $5+2\\sqrt{6}$ and $c$ terms equal to $5-2\\sqrt{6}$, then $a, b, c$ need to satisfy $a+b+c = 1008$, $a+(5+2\\sqrt{6})b+(5-2\\sqrt{6})c = 2016$. The last equality can be written $a+5b+5c-2016 = \\sqrt{6}(2c-2b)$. As $\\sqrt{6}$ is irrational, it follows that $b=c$ and $a+5b+5c = 2016$. Finally we obtain $a=756, b=c=126$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38605,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) If $ABCA'B'C'$ is a right prism and the points $M \\in (BC)$, $N \\in (CA)$, $P \\in (AB)$ are such that the straight lines $A'M$, $B'N$ and $C'P$ are pairwise perpendicular and concurrent, then the prism is regular.\n\nb) If $ABCA'B'C'$ is a regular prism and $\\frac{AA'}{AB} = \\frac{\\sqrt{6}}{4}$, then one can find the points $M \\in (BC)$, $N \\in (CA)$, $P \\in (AB)$ such that the straight lines $A'M$, $B'N$ and $C'P$ are pairwise perpendicular and concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $\\{O\\} = A'M \\cap B'N \\cap C'P$. The plane $(B'N, C'P)$ intersects the parallel planes $(ABC)$ and $(A'B'C')$ along parallel straight lines, hence $NP \\parallel B'C'$, so $PN \\parallel BC$. In the same way $MP \\parallel AC$, $NM \\parallel AB$. Then $PNCM$ and $PNMB$ are parallelograms, hence $M$ is the midpoint of $[BC]$. Similarly $N$ and $P$ are the midpoints of the sides $[AC]$ and $[AB]$.\n\nThe straight line $C'P$ is perpendicular on $A'M$ and on $B'N$, therefore on every straight line from their plane, in particular on $A'B'$. So $AB \\perp C'P$ and $AB \\perp CC'$, hence $AB \\perp (PCC')$, whence $AB \\perp PC$. This shows that, in triangle $ABC$, $[CP]$ is a median and an altitude, hence $AC = BC$. Analogously $AB = AC$, so $\\triangle ABC$ is equilateral and the prism is regular.\n\n![](attached_image_1.png)\n\nb) Let $M$, $N$, $P$ be the midpoints of the sides. Then $A'B'MN$ is a trapezoid whose diagonals meet at a point $O$ such that $\\frac{A'O}{OM} = \\frac{B'O}{ON} = 2$. Analogously, $A'C'MP$ is a trapezoid whose diagonals meet at a point $O'$ such that $\\frac{A'O'}{O'M} = \\frac{B'O'}{O'N} = 2$. Therefore, points $O'$ and $O$ coincide, hence $A'M$, $B'N$ and $C'P$ are concurrent.\n\n![](attached_image_1.png)\n\nDenoting $AA' = h$, $AB = l$, we get $BN = \\frac{l\\sqrt{3}}{2}$, $A'M^2 = B'N^2 = h^2 + \\frac{3l^2}{4}$, $A'O^2 = B'O^2 = \\frac{4h^2}{9} + \\frac{l^2}{3}$. Then $A'O \\perp B'O \\Leftrightarrow A'O^2 + B'O^2 = A'B'^2 \\Leftrightarrow \\frac{8h^2}{9} + \\frac{2l^2}{3} = l^2 \\Leftrightarrow \\frac{h}{l} = \\frac{\\sqrt{6}}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of real numbers such that $a_1 > 2$ and $a_{n+1} = 1 + \\frac{2}{a_n}$ for $n \\ge 1$.\n\na) Prove that $a_{2n-1} + a_{2n} > 4$ for all $n \\ge 1$ and $\\lim_{n \\to \\infty} a_n = 2$.\n\nb) Determine the largest real $a$ for which the inequality\n$$\n\\sqrt{x^2 + a_1^2} + \\sqrt{x^2 + a_2^2} + \\sqrt{x^2 + a_3^2} + \\dots + \\sqrt{x^2 + a_n^2} > n\\sqrt{x^2 + a^2}\n$$\noccurs for all $x \\in \\mathbb{R}$ and for any $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "2",
    "solution": "a) Observe that $a_n > 0$ and that $a_{n+1} - 2 = \\frac{2-a_n}{a_n}$ for $n \\ge 1$, so $a_{2n-1} > 2 > a_{2n}$ for $n \\ge 1$. Thus $a_{2n-1}+a_{2n}-4 = \\frac{a_{2n-1}^2-3a_{2n-1}+2}{a_{2n-1}} > 0$ for $n \\ge 1$.\n\n$|a_n - 2| = \\frac{|a_{n-1} - 2|}{a_{n-1}} = \\dots = \\frac{|a_1 - 2|}{a_{n-1} \\dots a_1} \\le \\frac{1}{2^{[n/2]}} |a_1 - 2| \\to 0$ shows that $a_n \\to 2$.\n\nb) For $x = 0$ we get $a < \\frac{a_1 + \\dots + a_n}{n}$, so $a \\le \\lim_{n \\to \\infty} \\frac{a_1 + \\dots + a_n}{n} = 2$. We will show that the inequality is true for $a = 2$ giving $a_{\\max} = 2$. It is enough to show that $\\sqrt{x^2 + a_{2n-1}^2} + \\sqrt{x^2 + a_{2n}^2} > 2\\sqrt{x^2 + 4}$, for any $n \\in \\mathbb{N}^*$, which is equivalent to $a_{2n-1}^2 + a_{2n}^2 + 2\\sqrt{(x^2 + a_{2n-1}^2)(x^2 + a_{2n}^2)} > 2x^2 + 16$. As $a_{2n-1} + a_{2n} > 4$ involves $a_{2n-1}^2 + a_{2n}^2 > 8$, it is sufficient to show that $a_{2n-1}a_{2n} \\ge 4$. This is an immediate consequence of $a_{2n-1} > 2$ and $a_{2n} = 1 + 2/a_{2n-1}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38607,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: \\mathbb{R} \\to \\mathbb{R}$ is said to have property ($\\mathcal{P}$) if for any sequence of reals $(x_n)_{n \\ge 1}$, such that $(f(x_n))_{n \\ge 1}$ is convergent, the sequence $(x_n)_{n \\ge 1}$ is also convergent. Prove that a surjective function that posses property ($\\mathcal{P}$), is continuous.\nMihai Piticari and Sorin Rădulescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $a, b \\in \\mathbb{R}$ such that $f(a) = f(b)$. The sequence $(x_n)_{n \\ge 1}$ defined by $x_n = \\begin{cases} a, & n \\text{ even} \\\\ b, & n \\text{ odd} \\end{cases}$ converges as $(f(x_n))_{n \\ge 1}$ is convergent, concluding that $f$ is injective. $f$ is surjective by definition, $f$ is invertible.\n\nConsider now $x_0 \\in \\mathbb{R}$ and a sequence $(x_n)_{n \\ge 1}$ with limit $x_0$. The sequence $(f(f^{-1}(x_n)))_{n \\ge 1}$ is convergent, so the sequence $(f^{-1}(x_n))_{n \\ge 1}$ is also. Remark that the sequence $(y_n)_{n \\ge 1}$ given by $y_n = \\begin{cases} x_k, & n = 2k \\\\ x_0, & n = 2k - 1 \\end{cases}$ has also the limit $x_0$.\n\nThus the sequence $(f^{-1}(y_n))_{n \\ge 1}$ is convergent, so\n$$\n\\lim_{n \\to \\infty} f^{-1}(y_{2n}) = \\lim_{n \\to \\infty} f^{-1}(y_{2n-1}),\n$$\nthat is $\\lim_{n \\to \\infty} f^{-1}(x_n) = f(x_0)$. As $x_0$ was arbitrary chosen $f^{-1}$ is continuous on $\\mathbb{R}$ and $f$ is also.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezium, $AD \\parallel BC$, and let $E$ and $F$ be points on the sides $AB$ and $CD$, respectively. The circumcircle of the triangle $AEF$ meets the line $AD$ again at $A_1$, and the circumcircle of the triangle $CEF$ meets the line $BC$ again at $C_1$. Prove that the lines $A_1C_1$, $BD$, and $EF$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Let the line $EF$ meet the circle $C_1BE$ again at $T$. Then\n$$\n\\angle (TC_1, C_1B) = \\angle (TE, EB) = \\angle (FE, EA) = \\angle (FA_1, A_1A),\n$$\nso $TC_1$ and $FA_1$ are parallel. Similarly,\n$$\n\\angle (TB, BC_1) = \\angle (TE, EC_1) = \\angle (FE, EC_1) = \\angle (FC, CC_1) = \\angle (FD, DA_1),\n$$\nso $TB$ and $FD$ are parallel.\nThus, the corresponding sides of the triangles $A_1DF$ and $C_1BT$ are parallel. Since the vectors $\\overrightarrow{C_1B}$ and $\\overrightarrow{A_1D}$ are counter-directed, these triangles are homothetical under a homothety of negative ratio.\nThe centre of this homothety $O$ lies on the lines $BD$, $A_1C_1$, and $TF = EF$, so the three are concurrent.\n\n![](attached_image_1.png)\n\n*Second solution.* Let the lines $AB$ and $CD$ meet at $S$, and let the lines $A_1F$ and $C_1E$ meet at $K$. Since\n$$\n\\begin{align*}\n\\angle(EK, FK) &= \\angle(EC_1, EF) + \\angle(EF, A_1F) = \\angle(CC_1, CF) + \\angle(EA, AA_1) \\\\\n&= \\angle(BC, FS) + \\angle(ES, BC) = \\angle(ES, FS),\n\\end{align*}\n$$\nthe points $E$, $F$, $S$, and $K$ are concyclic. Thus $\\angle(KS, KF) = \\angle(ES, EF) = \\angle(EA, EF) = \\angle(A_1A, A_1F)$, showing that $KS$ is parallel to both $AD$ and $BC$.\nThis shows the triangles $A_1DF$ and $C_1BE$ in perspective from the ideal point of the three parallel lines, so the lines $A_1C_1$, $BD$ and $EF$ are concurrent, by Desargues' theorem; since the segments $A_1C_1$ and $BD$ have a non-empty intersection, the point of concurrency is not ideal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the positive integer $n = 7 + 7^2 + 7^3 + \\ldots + 7^{2017}$.\n\na) Show that the remainders of the divisions of $7^{2018}$ by $6$ and by $48$ are equal.\n\nb) Determine the last two decimal digits of the number $6n$.",
    "options": [],
    "answer": "Both remainders are 1; the last two digits of 6n are 42.",
    "solution": "a) Since $7n = 7^2 + 7^3 + 7^4 + \\ldots + 7^{2017} + 7^{2018}$, we have $6n = 7n - n = 7^{2018} - 7$. It follows that $7^{2018} = 6n + 7 = 6(n + 1) + 1$, and the remainder of the division by $6$ of the number $7^{2018}$ is $1$.\n\nAlso, $n = 7 + 7^2(1 + 7) + 7^4(1 + 7) + \\ldots + 7^{2016}(1 + 7) = 7 + 8 \\cdot (7^2 + 7^4 + \\ldots + 7^{2016}) = 7 + 8p$, where $p = 7^2 + 7^4 + \\ldots + 7^{2016}$. Hence, $7^{2018} = 6(8p + 7) + 7 = 48p + 42 + 7 = 48p + 49 = 48(p + 1) + 1$, wherefrom we conclude that the remainder of the division of $7^{2018}$ by $48$ is $1$.\n\nb) We have $6n = 7^{2018} - 7$. Denoting by $u_2(x)$ the number built with the last two digits of the number $x$, we have $u_2(7^{4k}) = 01$, $u_2(7^{4k+1}) = 07$, $u_2(7^{4k+2}) = 49$, $u_2(7^{4k+3}) = 43$.\n\nThen $u_2(7^{2018}) = u_2(7^{4 \\cdot 504 + 2}) = 49$. Hence $u_2(6n) = u_2(49 - 7) = 42$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 3$ and let $a_0 = n$. Does there exist a permutation $a_1, a_2, \\dots, a_{n-1}$ of the first $n-1$ positive integers such that $\\sum_{j=0}^{k-1} a_j$ is divisible by $a_k$ for all indices $k < n$?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is in the affirmative. If $n$ is odd, set $a_1 = 1$ and $a_2 = 2$, and if $n > 3$ define the other $a_k$ recursively by $a_k a_{k-1} = a_0 + a_1 + \\dots + a_{k-1}$, $k = 3, \\dots, n$. It is easily seen that $a_{k+1} = a_{k-1} + 1$, $k = 3, \\dots, n-1$, so $a_{2k} = k+1$ and $a_{2k+1} = k + (n+1)/2$, $k = 1, \\dots, (n-1)/2$. By construction, the integers $a_k$ satisfy the divisibility condition (even at $k = n$). The $a_{2k}$ and $a_{2k+1}$, $k = 1, \\dots, (n-1)/2$, form strictly increasing sequences of integers greater than 1, and no $a_{2i}$ equals an $a_{2j+1}$, $i, j = 1, \\dots, (n-1)/2$, so the $a_k$, $k = 1, \\dots, n-1$, form an injective sequence of positive integers. Since they all lie in the range $1, 2, \\dots, n-1$, they form indeed a permutation of the latter.\n\nSimilarly, if $n$ is even, set $a_1 = 2$ and define $a_k$, $k = 2, \\dots, n-2$, by the same recursive relation, to get $a_{2k-1} = k+1$ and $a_{2k} = k + n/2$, $k = 1, \\dots, n/2 - 1$. Setting $a_{n-1} = 1$ settles the case as above and completes the proof.\n\n**Remark.** For an odd $n > 3$, we may equally well start by setting $a_1 = 1$ and $a_2 = (n+1)/2$, and define the remaining $a_k$ by the recursive relation in the solution, to obtain another sequence of positive integers satisfying the divisibility condition. Explicitly, this time $a_{2k} = k + (n-1)/2$, $k = 1, \\dots, (n-1)/2$, and $a_{2k+1} = k + 2$,",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38611,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer becomes with 2017 smaller if we delete its two final decimal digits. What is the number?",
    "options": [],
    "answer": "2037",
    "solution": "Let $\\overline{a_1a_2\\dots a_{n-2}a_{n-1}a_n}$ be the number we search for, so\n$$\n\\overline{a_1 a_2 \\dots a_{n-2} a_{n-1} a_n} - 2017 = \\overline{a_1 a_2 \\dots a_{n-2}},\n$$\nwhich is equivalent to $\\overline{a_1a_2\\dots a_{n-2}a_{n-1}a_n} - \\overline{a_1a_2\\dots a_{n-2}} = 2017$. On the other hand,\n$$\n\\overline{a_1 a_2 \\dots a_{n-2} a_{n-1} a_n} - \\overline{a_1 a_2 \\dots a_{n-2}} > 10^{n-1} - 10^{n-2} = 9 \\cdot 10^{n-2},\n$$\nhence $2017 > 9 \\cdot 10^{n-2}$, and it follows that $n - 2 \\le 2$, or $n \\le 4$, and since our number is greater than 2017, it has precisely four digits.\n$$\n\\overline{abcd} = 2017 + \\overline{ab} \\le 2017 + 99 = 2116 \\text{ and } \\overline{abcd} \\ge 2017 \\text{ imply that } a = 2.\n$$\n$$\n\\text{Thus, } 2bcd = 2017 + 2b \\le 2017 + 29 = 2046, \\text{ so } b = 0.\n$$\nConsequently, the number we searched is $2017 + 20 = 2037$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38612,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find $x \\in \\mathbb{N}$ and $y \\in \\mathbb{Q}$, such that $\\sqrt{x + \\sqrt{x}} = y$.\n\nb) Prove that one can find infinitely many pairs $(x, y) \\in \\mathbb{Q}^2$ such that\n$$\n\\sqrt{x + \\sqrt{x}} = y.\n$$",
    "options": [],
    "answer": "a) x = 0, y = 0. b) Infinitely many solutions are given by taking any Pythagorean triple p, q, r with p^2 + q^2 = r^2 and setting x = p^4 / q^4 and y = pr / q^2.",
    "solution": "a) Since $\\sqrt{x + \\sqrt{x}} \\in \\mathbb{Q}$, $x$ must be a square. Let $x = n^2$, with $n \\in \\mathbb{N}$. We obtain $n(n + 1) = y^2$, and since $n^2 \\le n(n + 1) < (n + 1)^2$, we deduce that $n = 0$, hence $x = y = 0$.\n\nb) There are infinitely many Pythagorean triples $(p, q, r)$ with $p^2 + q^2 = r^2$. Choosing $x = \\frac{p^4}{q^4}$, we obtain\n$$\n\\sqrt{x + \\sqrt{x}} = \\sqrt{\\frac{p^2}{q^2} \\left( \\frac{p^2}{q^2} + 1 \\right)} = \\sqrt{\\frac{p^2}{q^2} \\left( \\frac{p^2 + q^2}{q^2} \\right)} = \\frac{pr}{q^2} \\in \\mathbb{Q}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38613,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $m, n, p \\in \\mathbb{N}$, with $m > n$, such that $\\sqrt{m} - \\sqrt{n} = p$. Prove that $m$ and $n$ are both squares.\n\nb) Find, with proof, all numbers $\\overline{abcd}$ such that $\\sqrt{\\overline{abcd}} - \\sqrt{\\overline{acd}} = \\overline{bb}$.",
    "options": [],
    "answer": "1296",
    "solution": "a) We have $m = p^2 + 2p\\sqrt{n} + n$, hence $\\sqrt{n}$ is a rational number, and therefore, $n$ is a square. Similarly $m$ is a square, as well.\n\nb) We deduce from a) that $\\overline{abcd}$ and $\\overline{acd}$ are both squares. Since $11|\\overline{bb}$, it follows that $11|\\overline{abcd-acd}$, hence $11|100(\\overline{ab}-\\overline{a})$, and, finally, $11|9a+b$. This yields $(a; b) \\in \\{(1; 2), (2; 4), (3; 6), (4; 8), (6; 1), (7; 3), (8; 5), (9, 7)\\}$.\n\nBy inspection, we find that $\\overline{abcd} = 1296$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $k$ and $n$ for which $2^k + 10n^2 + n^4$ is the square of an integer.",
    "options": [],
    "answer": "k = 3, n = 2",
    "solution": "The required integers are $k = 3$ and $n = 2$. It is readily checked that these integers satisfy the condition in the statement.\n\nTo show that there are no other such, write $2^k + 10n^2 + n^4 = N^2$, where $N$ is a positive integer, so $(N - n^2 - 5)(N + n^2 + 5) = 2^k - 25$.\n\nThe latter shows that $N \\ge n^2 + 6$ if $k \\ge 5$, and of the first four positive integers, only $k = 3$ yields a positive integer $n = 2$, in which case $N = 8$.\n\nHenceforth, let $k \\ge 5$, and notice that $n$ must be even, for otherwise $N^2 = 2^k + 10n^2 + n^4 \\equiv 3 \\pmod 4$, which is impossible. Let $2^m, m \\ge 1$, be the highest power of $2$ dividing $n$.\n\nIf $k > 2m+1$, then $2^{2m+1}$ is the highest power of $2$ dividing $2^k+10n^2+n^4 = N^2$, which is again impossible.\n\nConsequently, $k \\le 2m+1$, so $2^k \\le 2^{2m+1} \\le 2n^2$, and $N^2 = 2^k + 10n^2 + n^4 \\le 12n^2 + n^4 < (n^2 + 6)^2$; that is, $N < n^2 + 6$, in contradiction with $N \\ge n^2 + 6$ established above. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, \\infty) \\to (0, \\infty)$ be a continuous function. Prove that:\na) if $n$ is a large enough integer, say, $n > n_0$, then $n \\int_0^{x_n} f(t) \\, dt = 1$ for a unique positive real number $x_n$;\nb) the sequence $(nx_n)_{n>n_0}$ is convergent and evaluate its limit.",
    "options": [],
    "answer": "1/f(0)",
    "solution": "Let $F: [0, \\infty) \\to \\mathbb{R}$ be the antiderivative of $f$ vanishing at $0$. Since $f$ takes on positive values, $F$ is strictly increasing, hence injective, and $\\alpha = \\sup \\text{im } F > 0$, the supremum being considered on the extended line.\n\na) Fix a positive integer $n_0$ such that $n_0\\alpha \\ge 1$. If $n$ is an integer greater than $n_0$, then $F(x_n) = 1/n$ for some $x_n > 0$, by the intermediate value theorem. Uniqueness of $x_n$ follows from injectivity of $F$.\n\nb) Since $F(x_{n+1}) = 1/(n+1) < 1/n = F(x_n)$, and $F$ is strictly increasing, $(x_n)_{n>n_0}$ is a strictly decreasing sequence of positive real numbers, so it is convergent. By continuity, $F(\\lim_{n\\to\\infty} x_n) = \\lim_{n\\to\\infty} F(x_n) = 0 = F(0)$, so $\\lim_{n\\to\\infty} x_n = 0$, by injectivity of $F$. For each integer $n > n_0$, refer to the first mean value theorem to write $1/n = F(x_n) = x_n f(t_n)$, i.e., $nx_n = 1/f(t_n)$, for some positive $t_n < x_n$. Since $(x_n)_{n>n_0}$ converges to $0$, so does $(t_n)_{n>n_0}$. Consequently, $(nx_n)_{n>n_0}$ converges to $1/f(0)$, by continuity of $f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes in the acute triangle $ABC$. On the line segments $B_1C_1$, $C_1A_1$, $A_1B_1$ one considers the points $X$, $Y$, and $Z$, respectively, such that\n$$\n\\frac{C_1X}{XB_1} = \\frac{b \\cos C}{c \\cos B}, \\quad \\frac{A_1Y}{YC_1} = \\frac{c \\cos A}{a \\cos C} \\quad \\text{and} \\quad \\frac{B_1Z}{ZA_1} = \\frac{a \\cos B}{b \\cos A}.\n$$\nProve that the lines $AX$, $BY$ and $CZ$ are concurrent.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $b \\cos C = A_1C$ and $c \\cos B = BA_1$, we deduce that $\\frac{C_1X}{XB_1} = \\frac{CA_1}{A_1B}$ or $\\frac{C_1X}{C_1B_1} = \\frac{CA_1}{CB}$ or, equivalently, $\\frac{C_1X}{CA_1} = \\frac{C_1B_1}{CB}$. (1)\n\nTriangles $AC_1B_1$ and $ACB$ being similar, we obtain $\\frac{C_1B_1}{CB} = \\frac{AC_1}{AC}$, hence $\\frac{C_1X}{CA_1} = \\frac{AC_1}{AC}$. Also, $\\angle AC_1X = \\angle ACA_1$, so it follows that triangles $AC_1X$ and $ACA_1$ are similar, and we deduce that $AX \\perp B_1C_1$.\n\nOn the other hand, the tangent at $A$ to the circumcircle of the triangle $ABC$, centered at $O$, is parallel to $B_1C_1$, and hence $O$ lies on $AX$.\n\nSimilarly, $O$ lies on $BY$ and $CZ$ as well, therefore the lines $AX$, $BY$ and $CZ$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ and $g$ be continuous real-valued functions on the closed unit interval $[0, 1]$ such that $f(x)g(x) \\ge 4x^2$ for all $x$ in $[0, 1]$. Show that (at least) one of the integrals\n$$\n\\int_{0}^{1} f(x) \\, dx, \\quad \\int_{0}^{1} g(x) \\, dx\n$$\nhas an absolute value greater than or equal to $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The functions $f$ and $g$ vanish at no point in the half-open interval $(0, 1]$, so $\\left|\\int_{0}^{1} f(x) \\, dx\\right| = \\int_{0}^{1} |f(x)| \\, dx$ and $\\left|\\int_{0}^{1} g(x) \\, dx\\right| = \\int_{0}^{1} |g(x)| \\, dx$. Consequently,\n$$\n\\begin{aligned}\n1 &= \\int_{0}^{1} 2x \\, dx \\le \\int_{0}^{1} \\sqrt{f(x)g(x)} \\, dx = \\int_{0}^{1} \\sqrt{|f(x)||g(x)|} \\, dx \\\\\n&\\le \\int_{0}^{1} \\frac{|f(x)| + |g(x)|}{2} \\, dx = \\frac{1}{2} \\left( \\int_{0}^{1} |f(x)| \\, dx + \\int_{0}^{1} |g(x)| \\, dx \\right),\n\\end{aligned}\n$$\nand the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38618,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a_1, a_2, \\dots, a_{2017}$ are positive integers, show that the fraction\n$$\n\\frac{9^{2017} - 7 \\cdot 3^{2017} + 7}{9^{(a_1+a_2)(a_2+a_3)\\dots(a_{2016}+a_{2017})(a_{2017}+a_1)} - 1}\n$$\nis reducible.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The last digit of $3^{2017}$ is 3, so the last digit of the numerator is 5.\n$$\n(a_1+a_2)+(a_2+a_3)+\\dots+(a_{2016}+a_{2017})+(a_{2017}+a_1) = 2(a_1+a_2+\\dots+a_{2017}).\n$$\nThe sum of these 2017 nonnegative integers being even, one of the summands, and hence their product, is even, so the last digit of the number\n$$\n9^{(a_1+a_2)(a_2+a_3)\\dots(a_{2016}+a_{2017})(a_{2017}+a_1)}\n$$\nis 1, and the denominator is divisible by 10.\nHence, the fraction can be simplified by 5.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\in (0, 1)$. Solve in $\\mathbb{R}$ the equation $a^{[x]} + \\log_a\\{x\\} = x$.",
    "options": [],
    "answer": "All real solutions are x = n + a^n for positive integers n.",
    "solution": "It is clear that $x \\notin \\mathbb{Z}$ and as $a^{[x]} > 0$, $\\log_a\\{x\\} > 0$, we should have $x > 0$. If $x \\in (0, 1)$, then $0 < \\log_a x = x - 1 < 0$, which is absurd. Hence, $x \\in (0, \\infty) \\setminus \\mathbb{N}$.\nLet $f: \\mathbb{R} \\to (0, +\\infty)$, $f(x) = a^x$, which is of course a bijective and strictly decreasing function. Now the equation could be written as $f([x]) + f^{-1}(\\{x\\}) = [x] + \\{x\\}$.\nLet us denote $f^{-1}(\\{x\\}) = y$. Then $\\{x\\} = f(y)$ and the previous relation becomes $f([x]) + y = [x] + f(y) \\Leftrightarrow f([x]) - [x] = f(y) - y$.\nAs the function $g(t) = f(t) - t$ is strictly decreasing, and thus injective, we obtain that $[x] = y$, that is $f([x]) = \\{x\\}$.\n\nFinally, if $[x] = n \\in \\mathbb{N}$, we get $a^n = x - n$, and thus, the solutions of the equation are the numbers $x_n = n + a^n$, for $n \\in \\mathbb{N}^*$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two children, Alex and Cristi, play several times a game, in which the winner receives $x$ points, and the loser $y$ points ($x$ and $y$ are nonnegative integers, with $x > y$, and in any game one of the children is the winner and the other is the loser). The final score is $147$ to $123$, in Alex's favour. Cristi has won $6$ games. Determine the numbers $x$ and $y$.\n\nBogdan Antohe",
    "options": [],
    "answer": "x = 13, y = 5",
    "solution": "Denote by $a$ the number of games won by Alex. Then $a x + 6 y = 147$ and $6 x + a y = 123$.\nSubtracting the above equalities, we obtain $a x + 6 y - 6 x - a y = 24$, or $(a - 6)(x - y) = 24$.\n\n$a - 6$ and $x - y$ are positive integers, because $a$, $x$ and $y$ are nonnegative integers, with $x > y$ and $a > 6$, because Alex has won more games.\nFrom $a x + 6 y = 147$ follows that $a x$ is an odd number, hence $a$ is odd, and $a - 6$ is an odd divisor of $24$.\n\nWe have two possible cases:\n* $a - 6 = 1$ and $x - y = 24$, where from $7(y + 24) + 6y = 147$, and $13y = -21$, which is impossible;\n* $a - 6 = 3$ and $x - y = 8$, where from $9(y + 8) + 6y = 147$, and $15y = 75$, hence $y = 5$ and $x = 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be two positive integers such that $1 \\le n \\le k$. Prove that, if $d^k + k$ is a prime number for each positive divisor $d$ of $n$, then $n + k$ is a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $d=1$ it follows that $1+k$ is a prime.\n\nFor $d=n$ it follows that $n^k + k$ is a prime. As $k+1$ does not divide $n$ (being larger than $n$), from Fermat's Theorem we get $n^k \\equiv 1 \\pmod{k+1}$, hence $n^k + k \\equiv 0 \\pmod{k+1}$. It follows that $n^k + k = k+1$, hence $n=1$. We have seen at the beginning that $1+k$ is a prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38622,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Determine all 4-tuples $(x_0, x_1, x_2, x_3)$ of pairwise distinct integers such that each $x_k$ is coprime to $x_{k+1}$ (indices are reduced modulo 4), and the cyclic sum $x_0/x_1 + x_1/x_2 + x_2/x_3 + x_3/x_0$ is an integer.\n\nb) Show that there are infinitely many 5-tuples $(x_0, x_1, x_2, x_3, x_4)$ of pairwise distinct integers such that each $x_k$ is coprime to $x_{k+1}$ (indices are reduced modulo 5), and the cyclic sum $x_0/x_1 + x_1/x_2 + x_2/x_3 + x_3/x_4 + x_4/x_0$ is an integer.",
    "options": [],
    "answer": "a) The only 4-tuples are (1, 2, -1, -2) and (1, -2, -1, 2), together with their cyclic permutations; the cyclic sum equals −3 for the former and 3 for the latter.\nb) For every integer k greater than 1, the 5-tuple (1, −k, k+1, −1, −k^2 − k) satisfies the conditions; the cyclic sum equals −k^2 − 2k − 2.",
    "solution": "a) The only 4-tuples satisfying the required conditions are\n$$\n(1, 2, -1, -2) \\quad \\text{and} \\quad (1, -2, -1, 2)\n$$\nalong with their cyclic permutations; the cyclic sum in question is $-3$ for the former and its cyclic permutations, and $3$ for the latter and its cyclic permutations.\nThe condition that the cyclic sum of the $x_k/x_{k+1}$ be integral is equivalent to the cyclic sum of the $x_k^2 x_{k+2} x_{k+3}$ being of the form $N x_0 x_1 x_2 x_3$ for some integral $N$. Since the $x_k$ are pairwise distinct, and each $x_k$ is coprime to $x_{k+1}$, it follows that $x_2 = -x_0$ and $x_3 = -x_1$, so $2(x_0^2 - x_1^2) = N x_0 x_1$. Since $x_0$ and $x_1$ are relatively prime, so are $x_0^2 - x_1^2$ and $x_0 x_1$, hence the latter is a divisor of $2$. The required 4-tuples are now easily derived.\n\nb) If $k$ is an integer greater than $1$, then $(1, -k, k+1, -1, -k^2 - k)$ is a 5-tuple satisfying the required conditions; the corresponding cyclic sum is $-k^2 - 2k - 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $x_1, x_2, \\dots, x_n$ real numbers, prove that there exists a real number $y$ such that\n$$\n\\{y - x_1\\} + \\{y - x_2\\} + \\dots + \\{y - x_n\\} \\le \\frac{n-1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\{a\\} + \\{-a\\} \\le 1, \\forall a \\in \\mathbb{R}$ (with equality if $a$ is not an integer), we have\n$$\n\\sum_{1 \\le i \\ne j \\le n} \\{x_i - x_j\\} \\le \\frac{n(n-1)}{2},\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 2$ such that $a + \\sqrt{2}$ and $a^n + \\sqrt{2}$ are both rational for some real number $a$ depending on $n$.",
    "options": [],
    "answer": "n = 2",
    "solution": "There is only one such $n$, namely, $n = 2$, in which case we may take $a = 1/2 - \\sqrt{2}$. Verifications are routine and hence omitted.\n\nTo rule out the case $n \\ge 3$, let $a$ be a real number such that $a + \\sqrt{2}$ is rational. For $a^n + \\sqrt{2}$ to be rational, it is necessary and sufficient that $a + \\sqrt{2}$ be a rational root of the degree $n-1$ polynomial $f_n = \\sum_{k=1}^{\\lfloor(n+1)/2\\rfloor} 2^{k-1} \\binom{n}{2k-1} X^{n-2k+1} - 1$. It is therefore sufficient to show that $f_n$ has no rational roots.\n\nIf $n$ is odd, then $f_n$ is a polynomial in $X^2$ whose coefficients are all positive, so it has no real roots.\n\nIf $n$ is even, then the constant term of $f_n$ is $-1$, and since the leading coefficient of $f_n$ is $n$, a rational root of $f_n$, if any, must be of the form $1/d$ for some divisor $d$ of $n$. If $d$ is one such, then $n + \\sum_{k=3}^{n/2} 2^{k-1} \\binom{n}{2k-1} d^{2(k-1)} - d^{n-1} = 0$, showing that $d$ must be even.\n\nFinally, use Legendre's standard valuation formula to infer that the highest power of $2$ dividing the greatest common divisor of the last $n/2 - 1$ summands in the left-hand member above exceeds the highest power of $2$ dividing $n$, and derive thereby a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of the scalene triangle $ABC$, with $AB < AC$, and let $I'$ be the reflection of point $I$ in the line $BC$. The angle bisector $AI$ meets the side $BC$ at $D$ and the circumcircle of $ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove that $\\frac{AI}{IE} = \\frac{ID}{DE}$ and $IA = IF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\frac{AI}{ID} = \\frac{AB}{BD} = \\frac{EC}{DE} = \\frac{BE}{DE} = \\frac{IE}{DE}$, which proves the first relation.\n\nLet $G = EF \\cap BC$. We know that $GD$ is the bisector of the angle $\\angle EGI$, hence $\\frac{GI}{GE} = \\frac{ID}{DE} = \\frac{AI}{IE}$. (1)\n\nTriangles $FBE$ and $BGE$ are similar (AA), therefore $EB^2 = EF \\cdot EG = EI^2$. This means that triangles $FIE$ and $IGE$ are also similar (SAS), which leads to $\\frac{FI}{IE} = \\frac{GI}{GE}$. (2)\n\nFrom (1) and (2) we obtain $FI = AI$, and the conclusion.\n\n![](attached_image_1.png)\nLet $O$ be the circumcenter of triangle $ABC$. We show that triangles $AIO$ and $II'E$ are similar, which proves that the points $I, O, F, E$ are co-cyclic.\n\n$OE$ and $II'$ are both perpendicular to $BC$, hence they are parallel. It follows that $\\angle I'IE \\equiv \\angle OEI \\equiv \\angle OAI$.\n\nWe prove that $\\frac{AI}{II'} = \\frac{AO}{IE}$. From the angle bisector theorem we obtain $\\frac{AI}{ID} = \\frac{AB}{BD} = \\frac{AC}{CD} = \\frac{AB + AC}{BC} = \\frac{b+c}{a}$, hence $\\frac{AI}{AD} = \\frac{b+c}{a+b+c}$. As $AD = l_a = \\frac{2bc \\cos \\frac{A}{2}}{b+c}$, we obtain $AI = \\frac{2bc \\cos \\frac{A}{2}}{a+b+c}$.\n\nBut $IE = EB$ and $\\frac{EB}{BC} = \\frac{\\sin \\frac{A}{2}}{\\sin A} = \\frac{1}{2 \\cos \\frac{A}{2}}$. Moreover, $EI = EB = \\frac{a}{2 \\cos \\frac{A}{2}}$.\n\nIt follows that, $AI \\cdot IE = \\frac{abc}{a+b+c}$.\n\n![](attached_image_2.png)\n\nOn the other hand, $AO \\cdot II' = R \\cdot 2r = \\frac{2RS}{p} = \\frac{abc}{a+b+c}$, hence $AO \\cdot II' = AI \\cdot IE$, which, together with $\\angle I'IE \\equiv \\angle OAI$, proves the similarity of the triangles $AIO$ and $II'E$.\n\nWe deduce that $\\angle EIO \\equiv \\angle II'F \\equiv \\angle OEF \\equiv \\angle OFE$, hence $IOEF$ is cyclic (the order of the points and the arguments depend slightly on the configuration).\n\nIt follows that $\\angle IOF \\equiv \\angle IEI' \\equiv \\angle AOI$. We also have $\\angle IAO \\equiv \\angle IEO \\equiv \\angle IFO$, which means that triangles $AIO$ and $IFO$ are congruent (SAA), therefore $IF = IA$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd integer greater than $1$. Each vertex of a regular $n$-gon is coloured one of three colours so that the total number of vertices of each colour is odd. Show that there exists an isosceles triangle whose vertices have pairwise distinct colours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, if possible, no such triangles exist, and consider the parity of the total number of pairs $(T, e)$, where $T$ is an isosceles triangle, and $e$ is an edge of $T$ whose endpoints have different colors.\n\nOn the one hand, under the above assumption, the vertices of a triangle in a pair form a bichromatic set, so the triangle occurs in exactly two pairs, and the total number of pairs is therefore even.\n\nOn the other hand, if we fix an edge $e$ whose endpoints have different colors, the number of isosceles triangles matching $e$ is either one or three. It is one if $e$ is the edge of an equilateral triangle. Otherwise, it is three: one triangle with apex at one endpoint of $e$, another triangle with apex at the other endpoint of $e$, and one more triangle with apex opposite $e$, since $n$ is odd. Since the total number of bichromatic edges is odd, so is the total number of pairs, contradicting the parity established in the previous paragraph.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38627,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides of an equilateral triangle are divided into $n$ equal parts by $n-1$ points on each side. Through these points one draws parallel lines to the sides of the triangle. Thus, the initial triangle is divided into $n^2$ equal equilateral triangles. In every vertex of such a triangle there is a beetle. The beetles start crawling simultaneously, with equal speed, along the sides of the small triangles. When they reach a vertex, the beetles change the direction of their movement by $60^\\circ$ or by $120^\\circ$.\n\na) Prove that, if $n \\ge 7$, the beetles can move indefinitely on the sides of the small triangles without two beetles ever meeting in a vertex of a small triangle.\n\nb) Determine all the values of $n \\ge 1$ for which the beetles can move along the sides of the small triangles without meeting in their vertices.",
    "options": [],
    "answer": "All positive integers except 2, 4, and 6",
    "solution": "It is easy to see (by induction or otherwise) that, for $n \\ge 3$ odd, the set of the vertices of the small triangle can be partitioned into groups of 3 and 4 vertices that form either an equilateral triangle or a rhombus formed by gluing together two such triangles. On each of these polygonal lines, the beetles can move in a circuit, changing at each step their direction.\n\nAlternatively, one can provide a direct example in this case: we call a *strip* an isosceles trapezoid whose legs are part of the original triangle's sides and whose bases are horizontal and we label the strips from top to bottom, strip number $k$ consisting of $2k-1$ small equilateral triangles. Then we can choose the triangles and the rhombi as follows: on the strip $k$ with $k$ odd we place a triangle, then $k-1$ rhombi. An example for $n=7$ is shown below:\n![](attached_image_1.png)\n\n![](attached_image_1.png)\n\nIn general, if for a given $n$ the beetles can move without meeting, then they can move indefinitely for $n+2$ as well. For $n=1$ and $n=8$ one can give examples. A possible example for $n=8$:\n\n![](attached_image_2.png)\n\nThe inductive step for the case when $n$ is even: if the triangle whose sides have been divided into $n \\ge 2$ equal parts can be traveled by the beetles without intersecting, then so can the triangle whose sides have been divided into $n+2$ equal parts: it is enough to split the triangles with side length $n+2$ into an equilateral triangle of side length $n$ plus two strips, of which the lower one can be partitioned according to the model exemplified below for $n=8$:\n![](attached_image_3.png)\n\nWhy can the beetles not move indefinitely without meeting in the case when $n=6$?\nWe color red all the vertices situated on odd position in rows with an odd number, the numbering starting from top to bottom, as in the following figure:\n\n![](attached_image_4.png)\n\nWe color the remaining points blue. A beetle can not get in one step from a red vertex to another red one, nor can it achieve this in two steps because it needs to change direction after the first step. After two steps, those 10 beetles initially positioned in a red vertex need to be in blue vertices, but so do the 10 beetles that were in red vertices after the first step. But there are only 18 vertices to accommodate these 20 beetles, so some beetles need to meet after two steps.\nIf for $n=2$ the beetles would be able to move without meeting, from the inductive step it would follow that the same would be true for $n=4$ and then also for $n=6$, which is not happening.\n\nIn conclusion, the beetles can move indefinitely without ever meeting if and only if $n \\in \\mathbb{N} \\setminus \\{2, 4, 6\\}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrangle and let $P$ and $Q$ be variable points inside this quadrangle so that $\\angle APB = \\angle CPD = \\angle AQB = \\angle CQD$. Prove that the lines $PQ$ obtained in this way all pass through a fixed point, or they are all parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the condition in the statement, the points $A$, $B$, $P$, and $Q$ lie on some circle $\\omega_1$, and the points $C$, $D$, $P$, and $Q$ lie on some circle $\\omega_2$. The line $PQ$ is the radical axis of these two circles, and we proceed to prove that $M$ lies on this radical axis.\nLet $MA$ meet $\\omega_1$ again at $A'$, and let $MD$ meet $\\omega_2$ again at $D'$; if, say, $MA$ is tangent to $\\omega_1$, then $A' = A$. Then $\\angle(MA', A'B) = \\angle(AA', A'B) = \\angle(AP, PB) = \\angle(CP, PD) = \\angle(CD', D'D) = \\angle(CD', D'M)$, and\n$$\n\\begin{align*}\n\\angle(MA', MB) &= \\angle(MA, MB) = \\angle(MA, MC) + \\angle(MC, MB) \\\\\n&= \\angle(MB, MD) + \\angle(MC, MB) = \\angle(MC, MD) \\\\\n&= \\angle(MC, MD'),\n\\end{align*}\n$$\nshowing that the triangles $MA'B$ and $MD'C$ are similar and have opposite orientations. Hence $MA'/MD' = MB/MC = MD/MA$, so $MA \\cdot MA' = MD \\cdot MD'$.\n\n---\n\n![](attached_image_1.png)\n\nWe still have to decide whether such a point $M$ exists, and deal with the exceptional cases. This can be done in several different ways, e.g., by investigating the orientation-changing similarity transformation mapping $A$ to $D$ and $B$ to $C$. We present a more geometrical proof.\nLet the diagonals $AC$ and $BD$ cross at $K$. Choose points $F$, on the diagonal $AC$, and $G$, on the diagonal $BD$, so that the configurations $(A, F, K, C)$ and $(D, K, G, B)$ are similar.\nIf $AD$ and $BC$ are not parallel, then $F \\neq K \\neq G$. By similarity, it is sufficient to find the point $M$ such that the triangles $MKF$ and $MGK$ are similar and have opposite orientations. If $KF \\neq KG$, one may choose $M$ to be the point where $FG$ crosses the tangent at $K$ to the circle $FGK$.\nThe exceptional cases are $AD \\parallel BC$ and $KF = KG$ (in which case $AC = BD$ by similarity). These cases may be considered as limit cases: in the former case, $M$ tends to $K$ and this can be dealt with along the above lines; in the latter, $M$ becomes an ideal point, so the lines $PQ$ are all parallel. In fact, both cases can be dealt with independently and are relatively easy.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alina and Bogdan play a game on a $2 \\times n$ rectangular grid ($n \\ge 2$) whose sides of length $2$ are glued together to form a cylinder. Alternating moves, each player cuts out a unit square of the grid. A player loses if his/her move causes the grid to lose circular connection (two unit squares that only touch at a corner are considered to be disconnected). Suppose Alina makes the first move. Which player has a winning strategy?\nEstonian Olympiad, 2009",
    "options": [],
    "answer": "Alina wins when the number of columns is odd; Bogdan wins when the number of columns is even.",
    "solution": "If $n = 2j + 1$ is odd, Alina's strategy is the following: she cuts out a unit square and labels the columns from $-j$ to $j$, the column from which the first unit square has been removed receiving the label $0$. Starting at this point of the game, whenever Bogdan removes a square from column number $k \\in \\{-j, \\dots, -2, -1, 1, 2, \\dots, j\\}$, Alina removes the unit square positioned in column number $-k$ and on the same row as the unit square removed by Bogdan in his last move. (If Bogdan removes the square remaining in column $0$ he loses instantly.) If on Bogdan's move the cylinder didn't lose its circular connection, it will not lose it after Alina's move either. Hence, Alina will never destroy the cylinder and, as the game is bound to finish sooner or later, Bogdan is the one who will lose the game. Alina wins.\n\nIf $n = 2j$ is even, Bogdan wins by adopting the following strategy: he labels the columns from $-j + 1$ to $j$, column $0$ being the one from which Alina has removed a square in her initial move. Bogdan removes a square from column $j$ (the one lying opposite to column $0$). From now on, if Alina removes a square from column number $k \\in \\{-j+1, \\dots, -2, -1, 1, 2, \\dots, j-1\\}$, Bogdan removes the square situated in the same row, but in the opposite column, namely $-k$. (If Alina removes the remaining square from column $0$ or from column $j$, she loses instantly.) If Alina's move didn't dismantle the surface of the cylinder, Bogdan's move won't do it either. Eventually, Alina will dismantle the cylindrical surface and Bogdan will win.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 38630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer, let $p$ be a prime, let $a_1 = 8p^m$, and let $a_n = (n+1)^{a_{n-1}/n}$, $n = 2, 3, \\dots$. Determine the primes $p$ for which the products $a_n\\left(1 - \\frac{1}{a_1}\\right)\\left(1 - \\frac{1}{a_2}\\right)\\dots\\left(1 - \\frac{1}{a_n}\\right)$, $n = 1, 2, 3, \\dots$, are all integral.",
    "options": [],
    "answer": "p = 2 or p = 5",
    "solution": "The required primes are $2$ and $5$. Begin by noticing that if $a_1$ is any even integer, then $a_{n-1}$ is an integral power of $n$ for all $n \\ge 3$. Thus, if $n \\ge 3$, then $(a_n - 1)/a_{n-1} = ((n+1)^{n_k} - 1)/n^{k+1}$ for some non-negative integer $k$. An easy induction on $k$ shows that $(n+1)^{n_k} - 1$ is divisible by $n^{k+1}$ for all non-negative integers $k$, so $(a_n - 1)/a_{n-1}$ is an integer for $n \\ge 3$. Consequently, if $n \\ge 3$, then\n\n$$a_n \\left(1 - \\frac{1}{a_1}\\right) \\left(1 - \\frac{1}{a_2}\\right) \\dots \\left(1 - \\frac{1}{a_n}\\right) = (a_1 - 1) \\cdot \\frac{a_2 - 1}{a_1} \\cdot \\frac{a_3 - 1}{a_2} \\dots \\frac{a_n - 1}{a_{n-1}}$$\n\nis an integer, provided that $a_2 - 1$ is divisible by $a_1$. On the other hand, since $a_1$ is coprime to $a_1 - 1$, the product $a_2\\left(1 - \\frac{1}{a_1}\\right)\\left(1 - \\frac{1}{a_2}\\right) = (a_1 - 1)(a_2 - 1)/a_1$ is an integer if and only if $a_2 - 1$ is divisible by $a_1$.\n\nLet $a_1 = 2a$, where $a$ is a positive integer. Then $a_2 = 3^a$, and the general problem consists in determining $a$ so that $3^a - 1$ is divisible by $2a$. This is clearly the case if $a$ is a power of $2$, so $p = 2$ is a solution.\n\nThe other case in the problem is $a = 4p^m$, where $p$ is an odd prime. The divisibility condition implies that $p$ is a divisor of $3^{4p^m} - 1$, so $p \\ge 5$. Since $p$ is coprime to both $3$ and $p-1$, the latter must be a divisor of $4$, so $p=5$. The number $3^{4 \\cdot 5^m} - 1$ is clearly divisible by $8$, and it is divisible by $5^m$, since $\\varphi(5^m) = 4 \\cdot 5^{m-1}$, where $\\varphi$ is Euler's totient function; hence it is divisible by $8 \\cdot 5^m$. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. For each of the numbers $1, 2, \\dots, n$ we compute the difference between the number of its odd positive divisors and its even positive divisors. Prove that the sum of these differences is at least $0$ and at most $n$.\nKürschák Competition, 1999",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We count how many times a number $d$ contributes to the sum of the differences. It appears $\\left\\lfloor \\frac{n}{d} \\right\\rfloor$ times; if $d$ is odd, then this term is to be added, while if $d$ is even this term is to be subtracted. Thus, the desired sum is\n$$\n\\sum_{k=1}^{n} (-1)^{k+1} \\left\\lfloor \\frac{n}{k} \\right\\rfloor.\n$$\nAdding, if necessary, the term $\\left\\lfloor \\frac{n}{n+1} \\right\\rfloor = 0$ at the end, we can group the terms of the sum into pairs having the sum $\\left\\lfloor \\frac{n}{2k-1} \\right\\rfloor - \\left\\lfloor \\frac{n}{2k} \\right\\rfloor \\ge 0$, hence the sum is non-negative. Similarly, separating the first term (which is equal to $n$), adding, if necessary, a $0$ term to the end, we can again group the terms into pairs having the sum $-\\left\\lfloor \\frac{n}{2k} \\right\\rfloor + \\left\\lfloor \\frac{n}{2k} \\right\\rfloor$.\n\n$$\n\\left[ \\frac{n}{2k+1} \\right] \\le 0, \\text{ hence the sum is at most } n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$, determine the largest number of pairwise non-crossing or perpendicular diagonals a regular $n$-gon may have.\nIMO 2016 Shortlist",
    "options": [],
    "answer": "If n is odd, the maximum is n−3. If n is even, the maximum is n−2.",
    "solution": "Begin by noticing that a regular polygon with an odd number of vertices has no perpendicular diagonals. Suppose, if possible, that $u, v, x, y$ are vertices, and $uv$ and $xy$ are perpendicular. Consider the vertex $w$ on the perpendicular bisectrix of the segment $uv$, and its antipode $w'$ on the circumcircle. Since the segments $wx$ and $w'y$ have equal lengths, and $w, x, y$ are vertices of the regular polygon, so is $w'$, contradicting the fact that a regular polygon with an odd number of vertices contains no pair of antipodes on the circumcircle.\n\nSince the maximum number of pairwise non-crossing diagonals of a convex $k$-gon is $k-3$, this solves the problem if $n$ is odd.\n\nWe now show by induction on $n$ that a cyclic $n$-gon has at most $n-2$ pairwise non-crossing or perpendicular diagonals.\n\nThe base case $n=3$ holds vacuously, so let $n \\ge 4$, assume the upper bound for any cyclic polygon with less than $n$ vertices, and consider a set $S$ of pairwise non-crossing or perpendicular diagonals of a cyclic $n$-gon.\n\nAssume first that some diagonal in $S$ is crossed by no other diagonal in $S$. That diagonal splits the $n$-gon into an $\\ell$-gon and an $m$-gon, where $\\ell + m = n + 2$, and splits $S$ into corresponding diagonals of the two. Since one of these polygons, say, the $\\ell$-gon, lies in a half-circumdisc, no pair of diagonals of the $\\ell$-gon cross orthogonally. Consequently, the diagonals in $S$ joining vertices of the $\\ell$-gon are pairwise non-crossing, so there are at most $\\ell - 3$ such. By the induction hypothesis, the $m$-gon has at most $m - 2$ pairwise non-crossing or perpendicular diagonals, so $|S| \\le (\\ell - 3) + (m - 2) + 1 = n - 2$.\n\nWe are left with the case where each diagonal in $S$ is crossed by, and hence perpendicular to, at least one other diagonal in $S$. Fix two crossing, hence orthogonal, diagonals $d$ and $d'$ in $S$. Their endpoints split the circumcircle into four non-overlapping arcs, none of which exceeds a semicircle.\n\nWe show that any diagonal in $S$ is parallel to one of these two diagonals. Suppose, if possible, that $\\delta$ is a diagonal in $S$ parallel to neither. Then one of the four non-overlapping arcs above contains both endpoints of $\\delta$. The diagonal $\\delta'$ in $S$, crossing $\\delta$ orthogonally, crosses at least one of $d$ and $d'$, and since $\\delta$ is parallel to neither, $\\delta'$ is orthogonal to neither – a contradiction.\n\nConsequently, at most two diagonals in $S$ may emanate from a vertex of the $n$-gon. Further, no other diagonal in $S$ may emanate from either vertex of the longest diagonal in $S$ parallel to $d$ — otherwise, such a diagonal would be orthogonally crossed by a longer diagonal in $S$. The same holds for the longest diagonal in $S$ parallel to $d'$. To conclude, $|S| \\le \\frac{1}{2}(2(n-4) + 4) = n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a, b, c, d \\in [1, 2]$, then\n$$\n\\frac{a+b}{b+c} + \\frac{c+d}{d+a} \\le 4 \\cdot \\frac{a+c}{b+d}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Swapping, if necessary, the roles of $a$ and $c$ and those of $b$ and $d$, we may assume that $a \\le c$. In this case, the function $f(x) = \\frac{a+x}{c+x}$ is increasing on $[1, 2]$, while $g(x) = \\frac{c+x}{a+x}$ is decreasing on $[1, 2]$, hence $\\frac{a+b}{b+c} + \\frac{c+d}{d+a} \\le \\frac{a+2}{c+2} + \\frac{c+1}{a+1}$. On the other hand, $4 \\cdot \\frac{a+c}{b+d} \\ge a+c$, therefore it is sufficient to prove that $\\frac{a+2}{c+2} + \\frac{c+1}{a+1} \\le a+c$, which reduces to $a+c+4 \\le a^2c+ac^2+a^2+3ac$. From $(a-1)(c-1) \\ge 0$ and $a^2c+ac^2+a^2+2ac \\ge 5$ the conclusion follows, with equality if $a=c=1$, and $b=d=2$.\n\n*Second solution.* We show that\n$$\n\\frac{a+b}{b+c} \\le 2 \\cdot \\frac{a+c}{b+2} \\le 2 \\cdot \\frac{a+c}{b+d} \\quad \\text{and} \\quad \\frac{c+d}{d+a} \\le 2 \\cdot \\frac{a+c}{b+d},\n$$\ninequalities which give the conclusion. The second inequality can be obtained from the first one by swapping $a$ with $c$ and $b$ with $d$, therefore it is sufficient to prove the first one. This inequality reduces to $b^2 + ab + 2b + 2a \\le 2c^2 + 2ac + 2bc + 2ab$. But $2a \\le 2ac$, $2b \\le 2bc$ and $b^2 \\le b + 2 \\le ab + 2c^2$, the inequality $b^2 \\le b + 2$ being equivalent to $(b+1)(b-2) \\le 0$. We obtain the equality case $a=1, b=2, c=1$ and $d=2$.\n\n*Third solution.* As $0 < b+d \\le 4$, it is sufficient to prove that the left hand side is at most $a+c$. Without loss of generality, we may assume that $a \\le c$. In this case,\n$$\n\\frac{a+b}{b+c} + \\frac{c+d}{d+a} \\le 1 + \\left(1 + \\frac{c-a}{d+a}\\right) \\le 2 + c-a \\le c+a,\n$$\nwith equality if $a=1, b=2, c=1$ and $d=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\\angle NHP$ and $\\angle AHQ$ are equal.\n\nMiguel Ochoa Sanchez, Peru, and Leonard Giugiuc",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AB = AC$, the statement is obvious. In the following, we assume $AB < AC$, the other case being similar.\n\n* Triangles $PAM$ and $PBM$ are congruent (SSS), hence $\\angle PMA \\equiv \\angle PMB$. Similarly, $\\angle QMA \\equiv \\angle QMC$, and this leads rapidly to $\\angle PMQ = 90^\\circ$.\n\nAs $\\tan B = \\frac{AH}{BH} = \\frac{AC}{AB} = \\frac{AQ}{BP}$ and $\\angle HAQ = \\angle HBP = 60^\\circ + \\angle B$, triangles $HAQ$ and $HBP$ are similar (SAS).\n\nIt follows that $\\angle QHA = \\angle PHB$, hence $\\angle QHP = \\angle QHA + \\angle AHP = \\angle PHB + \\angle AHP = \\angle AHB = 90^\\circ$.\n\nIn conclusion, the points $P, Q, M$ and $H$ are co-cyclic, hence $\\angle HPQ \\equiv \\angle CMQ \\equiv \\angle AMQ$.\n\nLet $\\{D\\} = HQ \\cap MP$. The angle $\\angle MNQ$ is exterior to the triangle $PMN$, therefore $\\angle MNQ = \\angle MPN + \\angle NMP = \\angle MHQ + \\angle BMP = \\angle MDQ$. (1)\n\nIt follows that the quadrilateral $NDMQ$ is cyclic, hence $\\angle DNQ = 90^\\circ$. This means that $NDHP$ is also cyclic, therefore $\\angle PNH \\equiv \\angle PDH$. (2)\n\nFrom (1) and (2) it follows that $\\angle PNH \\equiv \\angle PDH \\equiv \\angle MDQ \\equiv \\angle MNQ$, i.e. ($ND$ is the bisector of angle $\\angle HNM$). For the triangle $HMN$ ray ($NP$ is the external bisector, while ($MP$ is an internal bisector, which means that $P$ is the excenter opposite to the vertex $M$). Then ($HP$ is the internal bisector of angle $\\angle NHB$.\n\nFinally, $\\angle NHP \\equiv \\angle PHB \\equiv \\angle AHQ$.\n(given in the contest by *Paul Bécsi*)\n\nLet $\\{S\\} = AH \\cap PQ$. An easy computation shows that $\\angle SAQ = \\angle NAP = 120^\\circ - \\angle B$, which shows that the rays ($AN$ and ($AS$ are isogonal in the angle $\\angle PAQ$). From Steiner's Theorem it follows that $\\frac{PN}{NQ} \\cdot \\frac{PS}{SQ} = \\left(\\frac{PA}{AQ}\\right)^2$.\n\nThe conclusion means the rays ($HN$ and ($HS$ are isogonal in the angle $\\angle PHQ$, which, according to Steiner's Theorem, is equivalent to $\\frac{PN}{NQ} \\cdot \\frac{PS}{SQ} = \\left(\\frac{PH}{HQ}\\right)^2$.\n\nIn conclusion, we need to prove that $\\frac{PA}{AQ} = \\frac{PH}{HQ}$.\n\n![](attached_image_1.png)\n\nBut $\\Delta BHA \\sim \\Delta BAC$ leads to $\\frac{BH}{HA} = \\frac{BA}{CA} = \\frac{BP}{CQ}$. As $\\angle PBH = 60^\\circ + \\angle B = \\angle QAH$, it follows that $\\Delta PBH \\sim \\Delta QAH$, i.e., $\\frac{PH}{QH} = \\frac{PB}{AQ} = \\frac{PA}{AQ}$ and the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38635,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a, b, c \\in [-1, 1]$ satisfy $a + b + c + abc = 0$, prove that\n$$\na^2 + b^2 + c^2 \\ge 3(a + b + c).\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds exactly for (a, b, c) = (0, 0, 0) and for permutations of (−1, 1, 1).",
    "solution": "*First solution.* If $a + b + c \\le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a = b = c = 0$. If $a + b + c > 0$, then $abc < 0$. It is not possible for all the variables to be negative (their sum would be negative), therefore one of them is negative and the other two are positive. We may assume that $a < 0 < b, c$.\n\nPut $x = -a$. Then $-x + b + c = xbc > 0$ and $(a^2 + b^2 + c^2)^2 = (x^2 + b^2 + c^2)^2 \\ge (xb + xc + bc)^2 \\ge 3(xb \\cdot xc + xc \\cdot bc + bc \\cdot xb) = 3xbc(x + b + c) = 3(-x + b + c)(x + b + c) \\ge 9(-x + b + c)^2$.\nAs $-x + b + c > 0$, the last inequality comes to $x + b + c \\ge 3(-x + b + c)$, i.e. to $x \\ge \\frac{b + c}{2}$. But $x = \\frac{b + c}{1 + bc} \\ge \\frac{b + c}{2}$ because $bc \\le 1$. In conclusion, we obtain $a^2 + b^2 + c^2 \\ge 3(-x + b + c) = 3(a + b + c)$, with equality (in the case $a < 0 < b, c$) if and only if $x = b = c$ and $bc = 1$, i.e., if $a = -1$, $b = c = 1$. To conclude, we have equality if $(a, b, c) \\in \\{(0, 0, 0), (-1, 1, 1), (1, -1, 1), (1, 1, -1)\\}$.\n*Second solution.* If $a + b + c \\le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a = b = c = 0$. If $a + b + c > 0$, then $abc < 0$.\nFrom the AM-GM inequality, $a^2 + b^2 + c^2 \\ge 3\\sqrt[3]{(abc)^2} \\ge 3|abc| = -3abc = 3(a + b + c)$, with equality if $a^2 = b^2 = c^2$, $abc \\le 0$ and $|abc| = 1$, which leads to $(a, b, c) \\in \\{(0, 0, 0), (-1, 1, 1), (1, -1, 1), (1, 1, -1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrangle such that $\\angle ABC = \\angle ADC < 90^\\circ$. The internal bisectrices of the angles $ABC$ and $ADC$ cross the diagonal $AC$ at $E$ and $F$, respectively, and meet at $P$. Let $M$ be the midpoint of the diagonal $AC$. The segments $BM$ and $DM$ cross the circle $BDP$ again at $X$ and $Y$, respectively, and the lines $EX$ and $FY$ meet at $Q$. Prove that the lines $AC$ and $PQ$ are perpendicular.\n\nIMO 2016 Shortlist",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that $Y$ lies on the circle $ABC$. To this end, let $Y'$ be the point on the ray $MD$, emanating from $M$, such that $MY' \\cdot MD = MA^2$. The triangles $MAY'$ and $MDA$ are therefore similar and have opposite orientations. Since $MY' \\cdot MD = MA^2 = MC^2$, so are the triangles $MCY'$ and $MDC$. Hence, using directed angles, $\\angle AY'C = \\angle AY'M + \\angle MY'C = \\angle MAD + \\angle DCM = \\angle CDA = \\angle ABC$, so $Y'$ lies on the circle $ABC$.\n\nLet the lines $AD$ and $BC$ meet at $Z$. The equality of directed angles $\\angle PDZ = \\angle PBC = \\angle PBZ$ shows that $Z$ lies on the circle $BDP$; and the equality of directed angles $\\angle Y'BZ = \\angle Y'BC = \\angle Y'AC = \\angle Y'AM = \\angle Y'DZ$ shows that so does $Y'$. Since the angle $ADC$ is acute, $MA \\neq MD$, so $MY' \\neq MD$, and $Y'$ is therefore the point where $MD$ crosses the circle $BDP$ again. Consequently, $Y' = Y$, so the latter lies on the circle $ABC$, by the preceding paragraph.\n\n![](attached_image_1.png)\n\nNext, we show that $FY$ is the internal bisectrix of the angle $AYC$. To this end, simply refer to the internal angle bisectrix theorem and the similar triangles above, to write\n$$\n\\frac{FA}{FC} = \\frac{AD}{CD} = \\frac{AD}{AM} \\cdot \\frac{CM}{CD} = \\frac{YA}{YM} \\cdot \\frac{YM}{YC} = \\frac{YA}{YC}.\n$$\nLet now $BE$, the internal bisectrix of the angle $ABC$, cross the circle $ABC$ again at $B'$, so the latter is the midpoint of the arc $AC$ not containing $B$. The line $B'Y$ is therefore the external bisectrix of the angle $AYC$, so it is perpendicular to $FY$, by the preceding paragraph.\n\nLet the line $\\ell$ through $P$ and parallel to $AC$ meet the line $B'Y$ at $S$. To prove the required perpendicularity, we show $PQ$ and $PS$ perpendicular.\n\nUse directed angles to write $\\angle PSY = \\angle(AC, B'Y) = \\angle ACY + \\angle CYB' = \\angle ACY + \\angle CAB' = \\angle ACY + \\angle B'CA = \\angle B'CY = \\angle B'BY = \\angle PBY$, and infer that $S$ lies on the circle $BDP$. Restated, the line through $Y$ and perpendicular to $FY$ passes through the point where $\\ell$ crosses the circle $BDP$ again, which is $S$. Similarly, the line through $X$ and perpendicular to $EX$ passes through the point where $\\ell$ crosses the circle $BDP$ again, which is $S$. Consequently, $QX$ and $QY$ are perpendicular to $SX$ and $SY$, respectively, so $Q$ lies on the circle $BDP$ of which $QS$ is a diameter. The conclusion follows.\nBegin by noticing that, since $\\angle ABC = \\angle ADC$, the circles $ABC$ and $ACD$ are reflections of one another across $M$.\n\nWe first show that $X$ lies on the circle $ACD$. To this end, let the ray $MB$, emanating from $M$, cross the circle $ACD$ at $X_1$, and let $X'$ be the reflection of $X_1$ across $M$. By the preceding, $X'$ lies on the circle $ABC$ and $AX_1CX'$ is a parallelogram, so, using directed angles, $\\angle DX_1B = \\angle DX_1A + \\angle AX_1B = \\angle DCA + \\angle AX_1X' = \\angle DCA + \\angle CX'X_1 = \\angle DCA + \\angle CX'B = \\angle DCA + \\angle CAB = \\angle(CD, AB)$. The equality of directed angles $\\angle DPB = \\angle PDC + \\angle(CD, AB) + \\angle ABP = \\angle(CD, AB)$ therefore implies that $\\angle DX_1B = \\angle DPB$, so $X_1$ lies on the circle $BDP$. It follows that $X_1$ and $X$ coincide, so $X$ lies on the circle $ACD$. Similarly, $Y$ lies on the circle $ABC$.\n\nNext, we show that $Q$ lies on the circle $BDP$. To this end, let the perpendicular bisectrix of the segment $AC$ cross the circle $ABC$ at $B'$ and $M_1$, and the circle $ACD$ at $D'$ and $M_2$, so that $B, D'$ and $M_1$ all lie in the same half-plane relative to the line $AC$. Notice that $B'$ and $D'$ lie on the bisectrices $BP$ and $DP$, respectively, of the angles $ABC$ and $ADC$, respectively. Further, notice that\n$$\n\\frac{BA \\cdot X'A}{BC \\cdot X'C} = \\frac{\\text{area } BAX'}{\\text{area } BCX'} = \\frac{MA}{MC} = 1,\n$$\nand refer to the fact that $BE$ bisects the angle $ABC$, to write\n$$\n\\frac{EA}{EC} = \\frac{BA}{BC} = \\frac{X'C}{X'A} = \\frac{XA}{XC},\n$$\nand infer that $XE$ bisects the angle $AXC$, so $M_2$ lies on the line through $E, Q, X$.\n\nSimilarly, $M_1$ lies on the line through $F, Q, Y$. Use directed angles to write $\\angle XQY = \\angle M_2QM_1 = \\angle QM_2M_1 + \\angle M_2M_1Q = \\angle XM_2D' + \\angle B'M_1Y = \\angle XDD' + \\angle B'BY = \\angle XDP + \\angle PBY = \\angle XBP + \\angle PBY = \\angle XBY$, and infer that $Q$ lies on the circle $BDP$.\n\nFinally, since $M_1$ and $M_2$ are reflections of one another across $M$, the quadrangle $XM_1X'M_2$ is a parallelogram, so, using directed angles, $\\angle XQP = \\angle XBP = \\angle X'BB' = \\angle X'M_1B' = \\angle XM_2M_1$. This shows that $PQ$ and $M_1M_2$ are parallel; since the latter is perpendicular to $AC$, so is the former.\nIf one of the vertices $B$, $D$ lies on the perpendicular bisectrix of the diagonal $AC$, so do both $P$ and $Q$, and the conclusion follows.\n\nAssuming now that neither $B$ nor $D$ lie on the perpendicular bisectrix of the diagonal $AC$, we show that $P$ and $Q$ project orthogonally to the same point on the line $AC$. To this end, project $P$ and $Q$ on the line $AC$ to $P'$ and $Q'$, respectively, and notice that the two coincide if and only if, in terms of directed segments, $EP'/FP' = EQ'/FQ'$. Notice that\n$$\n\\frac{EP'}{FP'} = \\frac{\\tan \\angle EFP}{\\tan \\angle FEP} = \\frac{\\tan \\angle AFD}{\\tan \\angle AEB} \\quad \\text{and} \\quad \\frac{EQ'}{FQ'} = \\frac{\\tan \\angle EFQ}{\\tan \\angle FEQ},\n$$\nwhere angles are all directed, so it is sufficient to show that\n$$\n(\\tan \\angle AEB)(\\tan \\angle EFQ) = (\\tan \\angle AFD)(\\tan \\angle FEQ).\n$$\nTo this end, let the line through $B$ and perpendicular to $BE$ meet the line $AC$ at $T$. We will prove that $B, E, T, X$ are concyclic, so, in terms of directed angles, $\\angle FEQ = \\angle TEX = \\pi/2 + \\angle EBM$. Similarly, $\\angle EFQ = \\pi/2 + \\angle FDM$, so it is sufficient to show that\n$$\n(\\tan \\angle AEB)(\\tan \\angle MBE) = (\\tan \\angle AFD)(\\tan \\angle MDF).\n$$\nWith reference to the sine law, straightforward calculations show that the left-hand member is $(\\tan \\angle EBA)^2$, the right-hand member is $(\\tan \\angle FDC)^2$, and the conclusion follows by equality of the two angles.\n\nWe still have to prove that $B, E, T, X$ are concyclic. Since $BT$ and $BX$ are the two bisectrices of the angle $ABC$, the cross-ratio $(T, E; A, C)$ is harmonic, so $ME \\cdot MT = MA^2$.\n\nIt is therefore sufficient to show that $MA^2 = MB \\cdot MX$. To this end, let the lines $AB$ and $CD$ meet at $W$, and let the lines $AD$ and $BC$ meet at $Z$. Use directed angles to write $\\angle PDZ = \\angle PBC = \\angle PBZ$ and infer that $Z$ lies on the circle $BDP$; similarly, so does $W$, and the quadrangle $BDWZ$ is therefore cyclic.\n\nReflect $C$ in the circle $BDWZ$, centered at $O$, to obtain $C'$. Since $A$ lies on the polar of $C$ with respect to this circle, the lines $AC'$ and $OC$ are perpendicular, so $C'$ lies on the circle $\\gamma$ on diameter $AC$. It follows that reflection in the circle $BDWZ$ sends $\\gamma$ to itself, so the two circles are orthogonal. Consequently, the power of $M$ with respect to the circle $BDWZ$ is the square of the radius of $\\gamma$; that is, $MB \\cdot MX = MA^2$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a point outside the circle $\\mathcal{C}$. The tangents from $A$ touch the circle at $B$ and $C$. Let $P$ be an arbitrary point on $AC$ produced, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\\mathcal{C}$. Prove that $\\angle PEQ = 2\\angle APB$.\n\nCuban Olympiad, 2003",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\{D\\} = (BP \\cap \\mathcal{C})$ and $\\{M\\} = DE \\cap AC$.\n\nThen $\\angle BDE = \\angle BCE = 180^\\circ - \\angle ACB - \\angle PCE = 180^\\circ - \\angle ABC - \\angle CBE = \\angle APE$, hence $\\Delta MPD \\sim \\Delta MEP$ (AA), which leads to $\\angle MPD \\equiv \\angle MEP$ and $MP^2 = MD \\cdot ME$.\n\nOn the other hand, from the power of the point $M$ with respect to the circle $\\mathcal{C}$ it follows that $MC^2 = MD \\cdot ME$, i.e. $MP^2 = MC^2$, which means that $M$ is the midpoint of $[PC]$.\n\nIt follows that $\\angle MEP \\equiv \\angle MPD \\equiv \\angle MQP$, i.e., the quadrilateral $MQEP$ is cyclic. We obtain that $\\angle MEQ \\equiv \\angle MPQ \\equiv \\angle MEP$, and the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for every integer $n \\ge 3$ there exist positive integers $x_1, x_2, \\dots, x_n$, pairwise different, so that $\\{2, n\\} \\subset \\{x_1, x_2, \\dots, x_n\\}$ and\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} = 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is known that\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{(n-1)n} + \\frac{1}{n} = \\left(\\frac{1}{1} - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\dots + \\left(\\frac{1}{n-1} - \\frac{1}{n}\\right) + \\frac{1}{n} = 1.\n$$\nIf $n$ is not of the form $k(k+1)$, $k \\in \\mathbb{N}$, then the numbers $x_1 = 2$, $x_2 = 2 \\cdot 3, \\dots$, $x_{n-1} = (n-1)n$, $x_n = n$ are pairwise distinct, therefore they fulfill the conditions.\n\nIf $n = k(k+1)$, $k \\in \\mathbb{N}$ (the smallest $n \\ge 3$ of this form is $n = 6$), then the equality\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{(n-2)(n-1)} + \\frac{1}{n-1} = 1\n$$\ncan be written\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{(n-3)(n-2)} + \\frac{1}{(n-2)(n-1)+1} + \\frac{1}{(n-2)(n-1)[(n-2)(n-1)+1]} + \\frac{1}{n-1} = 1.\n$$\nThe numbers $x_1 = 2$, $x_2 = 2 \\cdot 3, \\dots$, $x_{n-3} = (n-3)(n-2)$, $x_{n-2} = (n-2)(n-1)+1$, $x_{n-1} = (n-2)(n-1)[(n-2)(n-1)+1]$, $x_n = n-1$ are pairwise different. Moreover, $n \\le (n-3)(n-2)$, $\\forall n \\ge 6$, hence $n \\in \\{x_2, x_3, \\dots, x_{n-3}\\}$, therefore the numbers $x_1, x_2, \\dots, x_n$ fulfill all the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, does there exist a planar polygon and a point in its plane such that every line through that point meets the boundary of the polygon at exactly $2n$ points?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The answer is in the affirmative. To describe the configuration, fix a coordinate frame and let $a_0, a_1, \\dots, a_{4n-1}$ be real numbers such that $a_0 > 0 > a_2 > a_{4n-2} > a_4 > a_{4n-4} > \\dots > a_{2n-2} > a_{2n+2} > a_{2n}$, and $a_{2n+1} < a_{2n+3} < \\dots < a_{4n-1} < 0 < a_1 < a_3 < \\dots < a_{2n-1}$. Setting $A_{2k} = 0 \\times a_{2k}$ and $A_{2k+1} = a_{2k+1} \\times 0$, $k = 0, \\dots, 2n-1$, the polygon $A_0A_1 \\dots A_{4n-1}$ and the origin satisfy the condition in the statement: the x-axis (respectively, y-axis) contains all vertices of odd (respectively, even) rank and no other points on the boundary; and every line through the first and third (respectively, second and fourth) quadrants crosses the sides $A_0A_1$ and $A_{2n+k-1}A_{2n+k}$ (respectively, $A_{4n-1}A_0$ and $A_kA_{k+1}$), $k = 1, \\dots, 2n-1$, and no other side, since the remaining sides all lie in the other two quadrants.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $A$ of positive 3-digit integers $abc$ consisting of three consecutive, not necessarily ordered, digits.\n\na) Determine the cardinal of the set $A$.\n\nb) Prove that for any choice of some elements of the set $A$, the sum of the chosen elements cannot equal 2017.",
    "options": [],
    "answer": "46; no sum of chosen elements can equal 2017",
    "solution": "a) $\\{0, 1, 2\\}$ generate 4 numbers.\nEach of the triplets $\\{1, 2, 3\\}$, $\\{2, 3, 4\\}$, ..., $\\{7, 8, 9\\}$ generate 6 numbers.\nThe cardinal of the set $A$ is $4 + 7 \\cdot 6 = 46$.\n\nb) Any number in $A$ is a multiple of 3. 2017 being not a multiple of 3, the sum of any numbers in $A$ cannot be 2017.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square. We take the points $E \\in (AB)$, $N \\in (CD)$ and $F, M \\in (BC)$, such that the triangles $AMN$ and $DEF$ are equilateral. Prove that $PQ = FM$, where $\\{P\\} = AN \\cap DE$ and $\\{Q\\} = AM \\cap EF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $\\triangle ABM \\equiv \\triangle ADN$ and $\\triangle DAE \\equiv \\triangle DCF$, we have $m(\\overline{BAM}) = m(\\overline{DAN}) = m(\\overline{ADE}) = m(\\overline{CDF}) = 15^\\circ$.\nThen $\\triangle ABM \\equiv \\triangle ADN \\equiv \\triangle DAE \\equiv \\triangle DCF$ and $\\triangle AMN \\equiv \\triangle DEF$.\nIt is clear now that $ADNE$ is a rectangle, and thus $P$ is the common midpoint of the line segments $[AN]$ and $[DE]$. Then, in the equilateral triangle $DEF$ we have $PF \\perp DE$.\nAs $m(\\overline{DAM}) + m(\\overline{ADE}) = m(\\overline{DAM}) + m(\\overline{BAM}) = 90^\\circ$, we get that $AM \\perp DE$, obtaining that $AM \\parallel PF$.\nThe equality $m(\\overline{PAQ}) = m(\\overline{PEQ}) = 60^\\circ$ implies that $PAEQ$ is a cyclic quadrilateral, and thus $m(\\overline{FPQ}) = m(\\overline{EPF}) - m(\\overline{EPQ}) = 75^\\circ$.\nOn the other hand, $m(\\overline{MFP}) = 180^\\circ - m(\\overline{CFD}) - m(\\overline{DFP}) = 75^\\circ$, which implies that $MQPF$ is an isosceles trapezoid and thus $[PQ] = [FM]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, let $a_1, \\dots, a_n$ be pairwise distinct real numbers, and let $b_1, \\dots, b_n$ be arbitrary real numbers. Show that:\na) if the $b_i$ are all positive, then there exists a polynomial $f$ with real coefficients, no root of which is real, such that $f(a_i) = b_i$, $i = 1, \\dots, n$;\nb) there always exists a polynomial $f$ whose roots are all real, and $f(a_i) = b_i$, $i = 1, \\dots, n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We exhibit two examples. Let $f_i = \\prod_{j \\neq i} (X - a_j)$, $i = 1, \\dots, n$. Since the $f_i$ never vanish simultaneously, the Lagrange type interpolation polynomial $f = \\sum_{i=1}^n \\frac{b_i f_i^2}{f_i(a_i)^2}$ clearly satisfies the required conditions.\n\nAnother example may be obtained by adding a suitable positive constant to the square of a suitable Lagrange interpolation polynomial; for instance, let $b$ be a positive real number less than each $b_i$, and let\n$$\nf = b + \\left( \\sum_{i=1}^{n} \\frac{\\sqrt{b_i - b}}{f_i(a_i)} f_i \\right)^2.\n$$\n\nb) Proceed by induction on $n$. The base case $n=1$ being clear, let $n \\ge 2$.\nIf some $b_i = 0$, fix such an index $i$, and apply the induction hypothesis to provide a polynomial $g$ whose roots are all real, and $g(a_j) = \\frac{b_j}{a_j - a_i}$ for all $j \\neq i$. The polynomial $f = (X - a_i)g$ clearly satisfies the required conditions.\n\nIf no $b_i = 0$, assume, without loss of generality, that $a_1 < \\dots < a_n$, and let $I$ be the set of all indices $i$ such that $b_i b_{i+1} > 0$. For each $i$ in $I$, choose $a'_i$ and $b'_i$ such that $a_i < a'_i < a_{i+1}$ and $b_i b'_i < 0$. The roots of the Lagrange polynomial solving the interpolation problem $f(a_i) = b_i$, $i = 1, \\dots, n$, and $f(a'_i) = b'_i$, $i \\in I$, are then all real and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the isosceles triangle $ABC$, with $m(\\angle BAC) = 100^\\circ$. Let $BD$ be the angle bisector of the angle $\\widehat{ABC}$, with $D \\in (AC)$, the point $E \\in BD$ such that $D \\in (BE)$ and $BE = BC$, and the point $F \\in (BC)$ such that $AB = BF$. Prove that the lines $AC$ and $EF$ are orthogonal.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Triangles $ABD$ and $FBD$ are congruent (S.A.S.), so that $m(\\angle FDB) = m(\\angle ADB) = 60^\\circ$, and $m(\\angle FDC) = 60^\\circ$.\n\nTriangle $EBC$ is isosceles ($BE = BC$), with $m(\\angle EBC) = 20^\\circ$, hence $m(\\angle BCE) = 80^\\circ$, and from the hypothesis we have $m(\\angle ACB) = 40^\\circ$, so $\\angle FCD = \\angle DCE$.\n\nThus, triangles $FCD$ and $ECD$ are congruent (A.S.A.), hence triangle $FCE$ is isosceles. $CD$ is an internal angle bisector, hence also a height. We conclude that $AC \\perp FE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a group and let $m$ and $n$ be relatively prime positive integers. Show that, if the functions $f: G \\to G$, $f(x) = x^{m+1}$, and $g: G \\to G$, $g(x) = x^{n+1}$, are both surjective endomorphisms, then $G$ is commutative.\nLiviu Vlaicu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $f$ is a morphism, $(xy)^{m+1} = x^{m+1}y^{m+1}$, so $(yx)^m = x^m y^m$ for all $x$ and all $y$ in $G$. Further, $y^{m+1}x^{m+1} = (yx)^{m+1} = (yx)^m(yx) = (x^m y^m)(yx) = x^m y^{m+1}x$, so $y^{m+1}x^m = x^m y^{m+1}$. Since $f$ is surjective, the latter shows that $x^m$ is in the center of $G$. Similarly, $x^n$ is in the center of $G$. Finally, since $m$ is coprime to $n$, it follows that every element of $G$ is in the center, so $G$ is commutative.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38645,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ we denote by $O$ and $I$ the circumcenter and the incenter, respectively. The perpendicular bisectors of the line segments $IA$, $IB$ and $IC$ pairwise intersect, thus defining the triangle $A_1B_1C_1$. Prove that\n$$\n\\vec{OI} = \\vec{OA_1} + \\vec{OB_1} + \\vec{OC_1}.\n$$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1$ be the intersection point of the perpendicular bisectors of the line segments $IB$ and $IC$. Let the angle bisector $AI$ intersect the circumcircle of $ABC$ at $D$. Since $\\angle BID = \\angle DBI$ and $\\angle CID = \\angle DCI$, it follows that $DB = DI = DC$, hence $A_1 = D$. Thus, $A_1$ belongs to the circumcircle of $ABC$ and the same goes for $B_1$ and $C_1$.\n\nObserve that $I$ is the orthocenter of $\\overline{A_1B_1C_1}$ and since $O$ is its circumcenter, Sylvester's relation yields $OI = OA_1 + OB_1 + OC_1$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(x, y)$, for which\n$$\n2^x + \\log_3 x = y^2 \\text{ and } 2^y + \\log_3 y = x^2.\n$$",
    "options": [],
    "answer": "(3, 3)",
    "solution": "The equality $2^x + \\log_3 x + x^2 = 2^y + \\log_3 y + y^2$ is readily obtained, and since the function $f: (0, +\\infty) \\to \\mathbb{R}, f(t) = 2^t + \\log_3 t + t^2$ is increasing, hence one-to-one, we deduce that $x = y$.\n\nObserve that $x = 3$ is a solution for the equation $2^x + \\log_3 x = x^2$, while $1$, $2$, and $4$ are not. For $x \\in \\mathbb{N}, x \\ge 5$, we can easily show inductively that $2^x > x^2$. In conclusion, the given system has only one solution, namely $(x, y) = (3, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the prime numbers $a$, $b$, $c$, such that $a = b^4 + c^3$ and $a \\le 2017$.",
    "options": [],
    "answer": "(a, b, c) = (43, 2, 3), (359, 2, 7), (89, 3, 2)",
    "solution": "$a \\ge 2^4 + 2^3 > 2$ is a prime number, hence it is odd. Then precisely one of the primes $b$ or $c$ equals $2$.\n\nIf $b = 2$, then $a = 16 + c^3 \\le 2017$, hence $c \\le 11$, so $c \\in \\{3, 5, 7, 11\\}$.\n\nConsidering each of these values, we obtain:\n* for $c = 3$, $a = 43$, which is a prime;\n* for $c = 5$, $a = 141$, which is not a prime;\n* for $c = 7$, $a = 359$, which is a prime;\n* for $c = 11$, $a = 1347$, which is not a prime.\n\nThe solutions in this case are $a = 43$, $b = 2$, $c = 3$, respectively $a = 359$, $b = 2$, $c = 7$.\n\nIf $c = 2$, then $a = b^4 + 8 \\le 2017$, hence $b \\le 5$, so $b \\in \\{3, 5\\}$.\nFor $b = 3$, we have $a = 89$, which is a prime, and for $b = 5$, we obtain $a = 633$, which is not a prime. So the only solution in this case is $a = 89$, $b = 3$, $c = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDA'B'C'D'$, be a cube with side length $a$. Let $M$ and $P$ be the midpoints of the edges $[AB]$ and $[DD']$, respectively.\na) Prove that $MP \\perp A'C$.\nb) Find the distance between the lines $MP$ and $A'C$.",
    "options": [],
    "answer": "a√2/4",
    "solution": "Let $O$ be the midpoint of $[A'C]$.\n\na) Triangle $MA'C$ is isosceles, hence $MO \\perp A'C$.\nSimilarly $PA'C$ is isosceles, therefore $PO \\perp A'C$, yielding $A'C \\perp (PMO)$, hence $A'C \\perp MP$.\n\n![](attached_image_1.png)\n\nb) Let $S$ be the midpoint of $[MP]$. The triangles $MA'C$ and $PA'C$ are congruent, yielding $[PO] = [MO]$. Then $OS \\perp MP$, hence $OS$ is the distance between the lines $MP$ and $A'C$.\nWe have $MC = \\frac{a\\sqrt{5}}{2}$, therefore $MO = \\frac{a\\sqrt{2}}{2}$. Next, $MP = \\frac{a\\sqrt{6}}{2}$, and from the right triangle $OSM$, we obtain $OS = \\frac{a\\sqrt{2}}{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of positive integers, which written in base 10, satisfy the following conditions simultaneously:\n(i) they are 6-digit numbers;\n(ii) the product of their non-zero digits is 84;\n(iii) four of their digits are 2, 0, 1, 7.",
    "options": [],
    "answer": "840",
    "solution": "Since $84 : (2 \\cdot 1 \\cdot 7) = 6$, we distinguish the following cases:\n\n(1) The digits are: $2, 0, 1, 7, 0, 6$.\nIf the first digit is $1$, the other $5$ digits can be arranged in $(5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1) : 2 = 60$ ways. Similar, if the first digit is $2$, $6$ or $7$. Thus we find $240$ numbers in this case.\n\n(2) The digits are: $2, 0, 1, 7, 1, 6$,\nIf the first digit is $2$, the other $5$ digits can be arranged in $60$. Similar, if the first digit is $6$ or $7$. If the first digit is $1$, we obtain $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ numbers. So there are $300$ numbers in this case.\n\n(3) The digits are: $2, 0, 1, 7, 2, 3$.\nIf the first digit is $1$, $3$ or $7$, we obtain $3 \\cdot 60 = 180$ numbers, and if the first digit is $2$, we obtain $120$ numbers. Hence there are $300$ numbers in this case.\n\nThe number of numbers satisfying the conditions is thus $240 + 300 + 300 = 840$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers $n > 1$ such that, for any divisor $d$ of $n$, the numbers $d^2 - d + 1$ and $d^2 + d + 1$ are prime.",
    "options": [],
    "answer": "2, 3, 6",
    "solution": "First, we prove that $n$ is square-free. If $d^2$ divides $n$ for a positive integer $d > 1$, then $(d^2)^2+d^2+1$ would be a prime number. But $d^4+d^2+1 = (d^2-d+1)(d^2+d+1)$, with both factors larger than $1$, which is a contradiction.\n\nThus, $n = p_1 \\cdot p_2 \\cdot \\dots \\cdot p_s$, where $s \\in \\mathbb{N}$ and $p_1 < p_2 < \\dots < p_s$ are prime numbers. Let $p > 5$ be a prime number. Then $p \\equiv 1 \\pmod{6}$ or $p \\equiv 5 \\pmod{6}$. If $p \\equiv 1 \\pmod{6}$, then $p^2 + p + 1 \\equiv 3 \\pmod{6}$, and $p^2 + p + 1 > 3$ is composite.\n\nIf $p \\equiv 5 \\pmod{6}$, then $p^2-p+1 \\equiv 3 \\pmod{6}$, and $p^2-p+1 > 3$ is composite.\n\nIn conclusion, the only prime factors of $n$ can be $2$ and $3$, so $n \\in \\{2, 3, 6\\}$. It is easy to check that all these three numbers fulfill the given condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a finite collection of 3-element sets $A_i$ no two of which share more than one element, whose union has cardinality $2017$. Show that the elements of this union can be coloured one of two colours, blue and red, so that at least $64$ elements are blue, and each $A_i$ contains at least one red element.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $U$ be the union of the $A_i$. It is sufficient to show that a maximal (relative to set-theoretic inclusion) subset $T$ of $U$ containing no $A_i$ satisfies the required cardinality condition.\n\nBy maximality, for each element $x$ of $U \\setminus T$, there exists a 2-element subset $S$ of $T$ such that $S \\cup \\{x\\}$ is an $A_i$.\n\nIf $x$ and $x'$ are distinct elements of $U \\setminus T$, and $S$ and $S'$ are 2-element subsets of $T$ such that $S \\cup \\{x\\}$ and $S' \\cup \\{x'\\}$ are both amongst the $A_i$, then $S$ and $S'$ are also distinct, since two distinct $A_i$'s share at most one element.\n\nTherefore, choosing for each $x$ in $U \\setminus T$ a 2-element subset $S$ of $T$ such that $S \\cup \\{x\\}$ is an $A_i$ defines an injection of $U \\setminus T$ into the collection of 2-element subsets of $T$. Consequently, $|U| - |T| = |U \\setminus T| \\le \\binom{|T|}{2}$, so $|T| \\ge (-1 + \\sqrt{8|U| + 1})/2$; if $|U| = 2017$, then $|T| \\ge 64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_k \\in \\mathcal{M}_n(\\mathbb{R})$ be symmetric matrices. Prove that the following are equivalent:\n1)\n$$\ndet(A_1^2 + A_2^2 + \\dots + A_k^2) = 0;\n$$\n2)\nfor any matrix $B_1, B_2, \\dots, B_k \\in \\mathcal{M}_n(\\mathbb{R})$ we have\n$$\ndet(A_1 B_1 + A_2 B_2 + \\dots + A_k B_k) = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove that 1) implies 2), the other implication being obvious.\nAs $\\det(A_1^2 + A_2^2 + \\dots + A_k^2) = 0$ there is a matrix $X \\in \\mathcal{M}_{n,1}(\\mathbb{R})$, $X \\neq O_{n,1}$, such that $(A_1^2 + A_2^2 + \\dots + A_k^2)X = O_{n,1}$ (the homogeneous linear system has a non-zero solution).\nThis implies $X^t(A_1^2 + A_2^2 + \\dots + A_k^2)X = O_n$, or $X^t A_1^t A_1 X + X^t A_2^t A_2 X + \\dots + X^t A_k^t A_k X = O_n$. Putting $A_i X = Y_i$, $i = 1, 2, \\dots, k$, we write $Y_1^t Y_1 + Y_2^t Y_2 + \\dots + Y_k^t Y_k = O_n$. Thus $A_i X = O_{n,1}$, or $A_i^t X = O_{n,1}$, $i = 1, 2, \\dots, k$.\n\nFor any matrices $B_1, B_2, \\dots, B_k \\in \\mathcal{M}_n(\\mathbb{R})$ we have $(B_1^t A_1^t + B_2^t A_2^t + \\dots + B_k^t A_k^t)X = O_{n,1}$, which is equivalent to the fact that the system $(B_1^t A_1 + B_2^t A_2 + \\dots + B_k^t A_k)X = O_{n,1}$ has a non-trivial solution, so $\\det(B_1^t A_1 + B_2^t A_2 + \\dots + B_k^t A_k) = \\det(A_1 B_1 + A_2 B_2 + \\dots + A_k B_k) = 0$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has $\\angle A = 90^\\circ$, $\\angle B = 30^\\circ$, and $D$ is the foot of the altitude from $A$. Let $E \\in (AD)$ be such that $DE = 3AE$ and let $F$ be the foot of perpendicular from $D$ on $BE$.\n\na) Prove that $AF \\perp FC$.\n\nb) Find the measure of the angle $AFB$.",
    "options": [],
    "answer": "150 degrees",
    "solution": "a) The hypothesis gives $AC = \\frac{1}{2}BC$ and $CD = \\frac{1}{2}AC$, hence $CD = \\frac{1}{4}BC$.\nFrom $\\triangle BDE \\sim \\triangle DFE$ follows $\\frac{BD}{DF} = \\frac{DE}{FE}$, that is $\\frac{DE}{BD} = \\frac{EF}{FD}$. Now $\\frac{DC}{DB} = \\frac{1}{3} = \\frac{AE}{DE}$ yields $\\frac{DE}{BD} = \\frac{AE}{DC}$, so $\\frac{AE}{CD} = \\frac{EF}{DF}$.\nNow $\\widehat{AEF} = 180^\\circ - \\widehat{DEF} = 180^\\circ - (90^\\circ - \\widehat{ADF}) = 90^\\circ + \\widehat{ADF} = \\widehat{FDC}$.\nTherefore $\\triangle AEF \\sim \\triangle CDF$ (SAS), whence $\\widehat{AFE} \\equiv \\widehat{CFD}$. This leads to $\\widehat{AFC} =$\n$\\widehat{AFE} + \\widehat{EFC} = \\widehat{CFD} + \\widehat{EFC} = \\widehat{EFD} = 90^\\circ$, hence $AF \\perp FC$.\n\n![](attached_image_1.png)\n\nb) Since $\\widehat{AFC} = 90^\\circ = \\widehat{ADC}$, the quadrilateral $AFDC$ is cyclic, therefore $\\widehat{DFC} = \\widehat{DAC} = 30^\\circ$. Then $\\widehat{AFB} = 360^\\circ - \\widehat{BFD} - \\widehat{DFC} - \\widehat{CFA} = 150^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \\le n$, and $k+1$ distinct integers $x_1, x_2, \\dots, x_{k+1}$ such that $P(x_1) + P(x_2) + \\dots + P(x_k) = P(x_{k+1})$.\n\n*Note.* A polynomial is *monic* if the coefficient of the highest power is one.",
    "options": [],
    "answer": "2",
    "solution": "To rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of degree at most $n$ with integer coefficients, whose restriction to the integers is injective, and $P(x) \\equiv 1 \\pmod{n}$ for all integers $x$. This is easily seen by reading the relation in the statement modulo $n$, to deduce that $k \\equiv 1 \\pmod{n}$, so $k = 1$, since $1 \\le k \\le n$; hence $P(x_1) = P(x_2)$ for some distinct integers $x_1$ and $x_2$, which contradicts injectivity.\nIf $n = 1$, let $P = X$, and if $n = 4$, let $P = X^4 + 7X^2 + 4X + 1$. In the latter case, clearly, $P(x) \\equiv 1 \\pmod 4$ for all integers $x$; and $P$ is injective on the integers, since $P(x) - P(y) = (x - y)((x + y)(x^2 + y^2 + 7) + 4)$, and the absolute value of $(x + y)(x^2 + y^2 + 7)$ is either 0 or at least 7 for integral $x$ and $y$.\n\nFinally, let $P = f_n + nX + 1$ if $n$ is odd, and let $P = f_{n-1} + nX + 1$ if $n$ is even. In either case, $P$ is strictly increasing, hence injective, on the integers, and $P(x) \\equiv 1 \\pmod{n}$ for all integers $x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38655,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that there exist functions $f : \\mathbb{R} \\to \\mathbb{R}$ and $g : \\mathbb{R} \\to \\mathbb{R}$ such that $f \\circ g = g \\circ f$, $f \\circ f = g \\circ g$ and $f(x) \\ne g(x)$, for all $x \\in \\mathbb{R}$.\n\nb) Prove that if $f : \\mathbb{R} \\to \\mathbb{R}$ and $g : \\mathbb{R} \\to \\mathbb{R}$ are continuous and have the properties $f \\circ g = g \\circ f$ and $f(x) \\ne g(x)$, for any $x \\in \\mathbb{R}$, then $(f \\circ f)(x) \\ne (g \\circ g)(x)$, for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The functions defined by $f(x) = \\begin{cases} -1, & x < 0 \\\\ 1, & x \\ge 0 \\end{cases}$, $g(x) = \\begin{cases} 1, & x < 0 \\\\ -1, & x \\ge 0 \\end{cases}$ fulfill the given conditions.\n\nb) The function $h = f - g$ is continuous and doesn't vanish, so $h(x) > 0$, for all $x \\in \\mathbb{R}$, or $h(x) < 0$, for all $x \\in \\mathbb{R}$. The case $f(x) > g(x)$, for all $x \\in \\mathbb{R}$, gives $f(f(x)) > g(f(x)) = f(g(x)) > g(g(x))$, for all $x \\in \\mathbb{R}$. In the second case the proof is similar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E(x, y) = \\frac{x}{y} + \\frac{x+1}{y+1} + \\frac{x+2}{y+2}$.\n\na) Solve in positive integers the equation $E(x, y) = 3$.\n\nb) Prove that there exist infinitely many positive integers $n$ for which the equation $E(x, y) = n$ has solutions in positive integers.",
    "options": [],
    "answer": "a) All solutions are x = y with x, y positive integers. b) For every positive integer k, n = 11k + 3 has a solution with (x, y) = (6k + 1, 1), so infinitely many n of the form 11k + 3 work.",
    "solution": "a) Rewrite the equation as\n$$\n\\left(\\frac{x}{y} - 1\\right) + \\left(\\frac{x+1}{y+1} - 1\\right) + \\left(\\frac{x+2}{y+2} - 1\\right) = 0,\n$$\nwhich is equivalent to $(x - y) \\left(\\frac{1}{y} + \\frac{1}{y+1} + \\frac{1}{y+2}\\right) = 0$, and the solutions are all pairs $(k, k)$, with $k \\in \\mathbb{N}$.\n\nb) Observe that $E(6k + 1, 1) = \\frac{6k + 1}{1} + \\frac{6k + 2}{2} + \\frac{6k + 3}{3} = 6k + 1 + 3k + 1 + 2k + 1 = 11k + 3$, $k \\in \\mathbb{N}$, hence for $n = 11k + 3$, $k \\in \\mathbb{N}$, the equation has the solution $(6k + 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. Prove that the following assertions are equivalent:\na) for all integer $x$ coprime with $n$ the congruence $x^6 \\equiv 1 \\pmod{n}$ holds;\nb) $n$ divides 504.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $504 = 2^3 \\cdot 3^2 \\cdot 7$.\n\na) $\\Rightarrow$ b). First we prove that $(n, 5) = 1$. If $(n, 5) \\neq 1$ then $n = 5k$.\nThe sequence $(5m + 2)_{m \\ge 1}$ is an arithmetic sequence with ratio 5. Its first term being coprime with the ratio, from Dirichlet's Theorem it follows that the sequence contains infinitely many primes. Then, there exists a prime $p$ such that $p = 5m + 2$ and $p > n$. As $(p, n) = 1$ it follows that $p^6 \\equiv 1 \\pmod n$, i.e., $n \\mid (5m + 2)^6 - 1$. As $5 \\mid n$ and $(5m + 2)^6 = M_5 + 2^6$, we get that 5 divides 63, absurd!\nAs $(n, 5) = 1$, according to the hypothesis $5^6 \\equiv 1 \\pmod n$, i.e. $n \\mid (5^6 - 1)$.\nBut $5^6 - 1 = (5^3 - 1)(5^3 + 1) = (5 - 1)(5^2 + 5 + 1)(5 + 1)(5^2 - 5 + 1) = 4 \\cdot 31 \\cdot 6 \\cdot 21 = 2^3 \\cdot 3^2 \\cdot 7 \\cdot 31$, hence $n \\mid 2^3 \\cdot 3^2 \\cdot 7 \\cdot 31$ (1)\nFrom (1) one can notice that $(n, 11) = 1$, hence $11^6 \\equiv 1 \\pmod n$, i.e., $n \\mid (11^6 - 1)$. But $11^6 - 1 = (11^3 - 1)(11^3 + 1) = (11 - 1)(11^2 + 11 + 1)(11 + 1) \\cdot (11^2 - 11 + 1) = 10 \\cdot 133 \\cdot 12 \\cdot 111$, therefore\n$$\nn \\mid 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37 \\quad (2)\n$$\nFrom (1) and (2) it follows that $n \\mid 2^3 \\cdot 3^2 \\cdot 7$, i.e., $n \\mid 504$.\n\nb) $\\Rightarrow$ a). Let $x \\in \\mathbb{Z}$. Then $x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$. We have:\n1. if $(x, 2) = 1$ then $x - 1$ and $x + 1$ are consecutive even numbers, hence $2^3 \\mid x^6 - 1$;\n2. if $(x, 7) = 1$, from Fermat's Theorem we obtain $7 \\mid x^6 - 1$;\n3. if $(x, 3) = 1$ then $x = 3k \\pm 1$. Obviously $3 \\mid x-1$ or $3 \\mid x+1$, hence $3 \\mid x^2-1$. We also have\n$$\nx^4 + x^2 + 1 = (3k \\pm 1)^4 + (3k \\pm 1)^2 + 1 = (M_3 + 1) + (M_3 + 1) + 1 = M_3, \\text{ i.e.,} \\\\\n3 \\mid x^4 + x^2 + 1.\n$$\nIt follows that for $(x, 3) = 1$ we have $3^2 \\mid x^6 - 1$.\nIf $n \\in \\mathbb{N}$, $n \\ge 2$ and $n \\mid 504$, then $n = 2^a \\cdot 3^b \\cdot 7^c$ with $a \\in \\{0, 1, 2, 3\\}$, $b \\in \\{0, 1, 2\\}$, $c \\in \\{0, 1\\}$ and $a, b, c$ not all of them 0. From the three statements above it follows that for all integer $x$ coprime with $n$ we have $n \\mid x^6 - 1$, i.e. $x^6 \\equiv 1 \\pmod n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an $m \\times n$ board where $m, n \\ge 3$ are positive integers, divided into unit squares. Initially all the squares are white. What is the minimum number of squares that need to be painted red such that each $3 \\times 3$ square contains at least two red squares?",
    "options": [],
    "answer": "If m ≡ 2 mod 3 and n ≡ 2 mod 3, the minimum is 2·floor(m/3)·floor(n/3) + min(floor(m/3), floor(n/3)); otherwise, the minimum is 2·floor(m/3)·floor(n/3).",
    "solution": "We label the rows from 1 to $m$ and the columns from 1 to $n$.\nIf $m$ and $n$ are not congruent to 2 modulo 3, i.e. $m = 3a + r_1$ and $n = 3b + r_2$ with $r_1, r_2 \\in \\{0, 1\\}$, we can tile the rectangle formed by the first $3a$ lines and $3b$ columns with $a \\cdot b$ disjoint $3 \\times 3$ squares. Each of these squares needs to contain at least 2 red squares, therefore one needs to paint red at least $2ab = 2 \\cdot \\lfloor \\frac{m}{3} \\rfloor \\cdot \\lfloor \\frac{n}{3} \\rfloor$ squares. On the other hand, this number is sufficient: paint red those unit squares whose row number is a multiple of 3 (i.e. 3, 6, ..., $3a$) and whose column number belongs to the set $\\{2, 3, 5, 6, ..., 3b-1, 3b\\}$. It is easy to check that each $3 \\times 3$ square contains exactly two red squares.\n\nThe same answer (with the exact same arguments) remains valid if $m \\equiv 2 \\pmod{3}$ and $n \\equiv 0, 1 \\pmod{3}$ (the answer remains the same in the case $n \\equiv 2 \\pmod{3}$ and $m \\equiv 0, 1 \\pmod{3}$).\n\nLet us now tackle the case $m = 3a + 2$, $n = 3b + 2$. We prove that in this case the answer is $2ab + \\min\\{a, b\\}$.\n\nLet us call a *zone* a $5 \\times 5$ square from which we have removed a $2 \\times 2$ square situated in one of its corners. We start by noticing that in any *zone* we need to have at least three red squares. Indeed, examining the two $3 \\times 3$ squares that contain the two remaining opposite corners of the $5 \\times 5$ square, they contain each at least two red squares and they only share one unit square (the one in the center), hence they contain together at least 3 red squares. With this remark, we move on to the promised induction:\n\nFor $a = 1$: we can paint red the squares situated in column number 3, on the lines 1, 3 and 4. Any $3 \\times 3$ square contains exactly two of these red squares.\n\nAssuming the assertion true for a $(3a - 1) \\times (3a - 1)$ board, let us prove it for the $(3a+2) \\times (3a+2)$ board. From the inductive hypothesis, in the $(3a-1) \\times (3a-1)$ square situated in top-left corner we must have at least $2(a-1)^2 + a - 1$ red squares. We cover the remaining part of the $(3a+2) \\times (3a+2)$ square with $3 \\times 3$ squares, plus one *zone* placed in the bottom-right corner. In this remaining part we have at least $2(a-1)+2(a-1)+3$ red squares, hence we have in total at least $2(a-1)^2 + a - 1 + 4(a-1) + 3 = 2a^2 + a$ red squares.\n\nFor $m < n$ (the case when $m > n$ is similar), it is sufficient to notice that the rectangle $(3a + 2) \\times (3b + 2)$ can be obtained from the $(3a + 2) \\times (3a + 2)$ square by gluing a $(3a + 2) \\times (3b - 3a)$ rectangle next to it. In the square there are at least $2a^2 + a$ red squares, while in the rectangle one can fit $a \\cdot (b-a)$ disjoint $3 \\times 3$ squares, therefore it contains at least $2a(b-a)$ red squares. In total, the rectangle has at least $2a^2 + a + 2a(b-a) = 2ab + a$ red squares. We have already seen that this number can actually be achieved, therefore the statement is proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest real number $a$ satisfying\n$$\na \\ge \\sum_{k=1}^{n} a_k \\cos(a_1 + \\dots + a_k),\n$$\nfor all positive integers $n$ and all positive real numbers $a_1, \\dots, a_n$ that add up to at most $\\pi$.",
    "options": [],
    "answer": "1",
    "solution": "The required minimum is $1$. To show that $1$ is an upper bound, let $n$ be a positive integer, and let $a_1, \\dots, a_n$ be positive real numbers such that $a_1+\\dots+a_n \\le \\pi$. If $a_1 \\ge \\pi/2$, then the sum in question is non-positive, so let $a_1 < \\pi/2$, and let $m$ be the largest positive integer such that $a_1 + \\dots + a_m < \\pi/2$. Then\n$$\n\\sum_{k=1}^{n} a_k \\cos(a_1 + \\dots + a_k) \\le \\sum_{k=1}^{m} a_k \\cos(a_1 + \\dots + a_k) \\le \\int_{0}^{\\pi/2} \\cos x \\, dx = 1,\n$$\nsince the sum in the middle is the lower Darboux-Riemann sum of the cosine, corresponding to the subdivision $0 < a_1 < a_1 + a_2 < \\dots < a_1 + \\dots + a_m < \\pi/2$, and the cosine is decreasing on $[0, \\pi/2]$.\n\nTo show that $1$ is the least upper bound, for every positive integer $n$, let $a_1 = \\dots = a_n = \\pi/(2n)$. Since $a_1 + \\dots + a_n = \\pi/2$ and\n$$\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} a_k \\cos(a_1 + \\dots + a_k) = \\lim_{n \\to \\infty} \\frac{\\pi}{2n} \\sum_{k=1}^{n} \\cos \\frac{k\\pi}{2n} = \\int_{0}^{\\pi/2} \\cos x \\, dx = 1,\n$$\nthe conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $A = \\{1, 2, 3, \\dots, 2017\\}$. Determine the number of three element subsets $B \\subset A$, which satisfy the following conditions simultaneously:\na) at least two elements in the set $B$ are consecutive positive integers;\nb) there is an $a \\in B$, such that $3a \\in B$.",
    "options": [],
    "answer": "2686",
    "solution": "The sets we search contain elements $a$, $3a$, with $3a \\le 2017$, so $a \\le 672$. For $a = 1$, we have $\\{1, 3\\} \\subset B$, and the possible sets are $\\{1, 2, 3\\}$ and $\\{1, 3, 4\\}$. For $a \\in \\{2, 3, 4, \\dots, 672\\}$ we have the sets $\\{a-1, a, 3a\\}$, $\\{a, a+1, 3a\\}$, $\\{a, 3a-1, 3a\\}$ and $\\{a, 3a, 3a+1\\}$, which are all distinct, giving a total of $4 \\cdot 671 = 2684$ sets. So the number of sets with the required properties is $2686$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle, let $M_A$ be the midpoint of the side $BC$, and let $P_A$ be the orthogonal projection of $A$ on the line $BC$; similarly, define $M_B, P_B$ and $M_C, P_C$. The lines $M_B M_C$ and $P_B P_C$ meet at $S_A$, and the tangent of the circle $ABC$ at $A$ meets the line $BC$ at $T_A$; similarly, define $S_B, T_B$ and $S_C, T_C$. Show that the perpendiculars through $A, B, C$ to the lines $S_A T_A, S_B T_B, S_C T_C$, respectively, are concurrent.\nFlavian Georgescu\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The three lines in question are concurrent at the center of the nine-point circle $\\omega$ of the triangle $ABC$. In what follows, polarity always refers to $\\omega$.\n\nTo prove that the center of $\\omega$ lies on the perpendicular through $A$ to the line $S_A T_A$, it is sufficient to show that the latter is the polar of $A$; a similar argument applies to $B$ and $C$.\n\n![](attached_image_2.png)\n\nShowing that $T_A$ also lies on the polar of $A$ involves only elements relative to vertex $A$ and the subscript $A$ is henceforth dropped out to write $M, P, T$ instead of $M_A, P_A, T_A$, respectively.\nLet the line $AM$ meet $\\omega$ again at $K$, and let $L$ be the midpoint of the line segment joining $A$ to the orthocenter of the triangle $ABC$.\nCompletion of the cyclic quadrangle $KLMP$ shows that the polar of $A$ passes through the point where the lines $KL$ and $MP$ meet, so it is sufficient to show $K, L, T$ collinear.\nTo this end, we show that the lines $KL$ and $LT$ are both perpendicular to the line $AM$. Recall that $L$ is the antipode of $M$ in $\\omega$, and $LM$ is parallel to the circumradius through $A$. The former implies that the lines $AKM$ and $KL$ are perpendicular, and the latter implies that so are the lines $AT$ and $LM$. Since the lines $ALP$ and $MPT$ are also perpendicular, $L$ is the orthocenter of the triangle $AMT$, so the lines $AM$ and $LT$ are perpendicular.\n\n**Remark.** Since the polar of $A$ passes through $T_A$, the polar of $T_A$ passes through $A$. Now let the tangents of $\\omega$ at $M_A$ and $P_A$ meet at $X_A$; the points $X_B$ and $X_C$ are defined similarly. The line $M_A P_A$ through $T_A$ is the polar of $X_A$, so the polar of $T_A$ passes through $X_A$. Hence the line $A X_A$ is the polar of $T_A$; similarly, the lines $B X_B$ and $C X_C$ are the polars of $T_B$ and $T_C$, respectively. Since $T_A B / T_A C = (AB/AC)^2$ and the like, the points $T_A, T_B, T_C$ are collinear, so their polars, $A X_A, B X_B, C X_C$, are concurrent; that is, the triangles $ABC$ and $X_A X_B X_C$ are in perspective.\n\nNotice that $B$ and $C$ also lie on the polar of $X_A$ to infer that the polars of $B$ and $C$ meet at $X_A$; similarly, the polars of $C$ and $A$ meet at $X_B$, and the polars of $A$ and $B$ meet at $X_C$. Consequently, $S_A, T_A, X_B, X_C$ all lie on the polar of $A$ and the like.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$, let $Q_n$ be the graph consisting of all vertices and all edges of an $n$-cube, and let $T$ be a spanning tree in $Q_n$. Show that $Q_n$ has an edge whose adjunction to $T$ produces a simple cycle of length at least $2n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every vertex $v$ of $Q_n$, let $v'$ be the antipodal (opposite) vertex, consider the unique path in $T$ from $v$ to $v'$ and orient its first edge away from $v$. Since $T$ has fewer edges than vertices, some edge, say $xy$, has been assigned two orientations. The (combinatorial) distance between two antipodes of $Q_n$ is $n$ in $Q_n$, so it is at least $n$ in $T$, and the unique path in $T$, $x'...yx...y'$, from $x'$ to $y'$ has length at least $2n-1$. Finally, since $xy$ is an edge in $Q_n$, so is $x'y'$; adjunction of the latter to the unique path in $T$ from $x'$ to $y'$ yields the required cycle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be real positive numbers such that\n$$\n\\frac{a_1}{b_1} \\le \\frac{a_2}{b_2} \\le \\dots \\le \\frac{a_n}{b_n}.\n$$\nFind the largest real $c$ so that $(a_1 - b_1c)x_1 + (a_2 - b_2c)x_2 + \\dots + (a_n - b_nc)x_n \\ge 0$, for every $x_1, x_2, \\dots, x_n > 0$, with $x_1 \\le x_2 \\le \\dots \\le x_n$.",
    "options": [],
    "answer": "(a1 + a2 + ... + an) / (b1 + b2 + ... + bn)",
    "solution": "Denoting $y_1 = x_1$, $y_{k+1} = x_{k+1} - x_k$, for $k \\in \\{1, 2, \\dots, n-1\\}$, we get $x_k = y_1 + y_2 + \\dots + y_k$, $\\text{cu } y_1 > 0, y_k \\ge 0, \\forall k = \\overline{2, n}$. With these notations, the inequality becomes $(a_1 + a_2 + \\dots + a_n - c(b_1 + b_2 + \\dots + b_n))y_1 + (a_2 + a_3 + \\dots + a_n - c(b_2 + b_3 + \\dots + b_n))y_2 + \\dots + (a_n - cb_n)y_n \\ge 0$ ($*$).\n\nPlugging $y_1 = 1, y_2 = \\dots = y_n = 0$, shows that necessarily $c \\le \\frac{a_1+a_2+\\dots+a_n}{b_1+b_2+\\dots+b_n}$.\n\nWe show that, conversely, for every $c \\le \\frac{a_1+a_2+\\dots+a_n}{b_1+b_2+\\dots+b_n}$ the inequality under consideration is true, so $\\frac{a_1+a_2+\\dots+a_n}{b_1+b_2+\\dots+b_n}$ is the required maximum.\n\nWe notice that $c \\le \\frac{a_1+a_2+\\dots+a_n}{b_1+b_2+\\dots+b_n} \\le \\frac{a_2+a_3+\\dots+a_n}{b_2+b_3+\\dots+b_n} \\le \\dots \\le \\frac{a_n}{b_n}$. Indeed, the inequality $\\frac{a_k+a_{k+1}+\\dots+a_n}{b_k+b_{k+1}+\\dots+b_n} \\le \\frac{a_{k+1}+\\dots+a_n}{b_{k+1}+\\dots+b_n}$ can be written $a_k(b_{k+1} + \\dots + b_n) \\le b_k(a_{k+1} + \\dots + a_n)$ and is justified by the inequalities $\\frac{a_k}{b_k} \\le \\frac{a_{k+1}}{b_{k+1}}, \\frac{a_k}{b_k} \\le \\frac{a_{k+2}}{b_{k+2}}, \\dots, \\frac{a_k}{b_k} \\le \\frac{a_n}{b_n}$.\n\nThis shows that every term of the left member of $(*)$ is at least 0, so $(*)$ is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $[AB]$ of the square $ABCD$, let $P$ be the projection of $B$ on $CM$ and let $N$ be the midpoint of the segment $[CP]$. The bisector of the angle $DAN$ meets line $DP$ in $Q$. Show that the quadrilateral $BMQN$ is a parallelogram.\n\nAdrian Bud\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $\\triangle BMC \\sim \\triangle PBC$ follows $\\frac{BM}{PB} = \\frac{BC}{PC}$, hence $CP = 2BP$, so $[BP] = [PN] = [NC]$.\n\n![](attached_image_1.png)\n\nThen $\\Delta NCD \\equiv \\Delta PBC$ (SAS) leads to $\\widehat{DNC} = \\widehat{BPC}$, that is $DN \\perp CP$, so $[DN]$ is a median and an altitude in triangle $DPC$. Therefore triangle $DPC$ is isosceles, with $[DP] = [DC]$.\n\nDenote $E$ the common point of the straight lines $CM$ and $AD$. Then $A$ is the midpoint of $[DE]$, hence $[NA]$ is a median in the right triangle $NDE$. This leads to $[NA] = [AD]$, hence triangle $ADN$ is isosceles. Since $AQ$ is a bisector, it is also an altitude, therefore $AQ \\perp DN$. From $DN \\perp CM$ follows $AQ \\parallel MN$ (1).\n\nIsosceles triangles $ANE$ and $DPC$ yield $\\widehat{ANP}+\\widehat{DPN} = \\widehat{DEC}+\\widehat{DCE} = 90^\\circ$, whence $DP \\perp AN$. Since $AQ \\perp DN$, the point $Q$ is the orthocenter of the triangle $ADN$, so $NQ \\perp AD$.\n\nThis shows that $NQ \\parallel AM$ and, using (1), we get that $AMNQ$ is a parallelogram.\nSo the segments $[AM]$ and $[NQ]$ are parallel and congruent, hence $[BM]$ and $[NQ]$ are parallel and congruent, therefore $BMQN$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality\n$$\nsin \\frac{\\pi}{4n} \\ge \\frac{\\sqrt{2}}{2n},\n$$\n\nwhere $n$ is a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a = \\cos \\frac{\\pi}{2n} + i \\sin \\frac{\\pi}{2n}$, then $|a| = 1$ and $a^n = i$. We obtain from here that\n$$\ni - 1 = a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \\dots + 1),\n$$\nand thus $\\sqrt{2} \\le |a - 1| \\cdot n$.\nOn the other hand, $|a - 1| = |-2 \\sin^2 \\frac{\\pi}{4n} + 2i \\sin \\frac{\\pi}{4n} \\cos \\frac{\\pi}{2n}| = 2 \\sin \\frac{\\pi}{4n}$, which complete the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence of rational numbers defined by $x_1 = 4/3$, and $x_{n+1} = x_n^2/(x_n^2 - x_n + 1)$, $n \\ge 1$. Show that the numerator of the lowest term expression of each sum $\\sum_{k=1}^n x_k$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easily seen that the $x_n$ are all rational numbers greater than $1$. Rewrite the recurrence formula in the form $x_n = 1/(x_{n+1} - 1) - 1/(x_n - 1)$, $n \\ge 1$, to get\n$$\n\\sum_{k=1}^{n} x_k = \\frac{1}{x_{n+1} - 1} - \\frac{1}{x_1 - 1} = \\frac{x_n^2 - x_n + 1}{x_n - 1} - 3 = \\frac{(x_n - 2)^2}{x_n - 1}.\n$$\nFinally, express the positive rational number $x_n - 1$ in lowest terms, $x_n - 1 = a/b$, to deduce that $(a-b)^2/(ab)$ expresses $\\sum_{k=1}^n x_k$ in lowest terms. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38667,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $[CD]$ of the square $ABCD$ point $E$ is chosen such that $m(\\angle ABE) = 60^{\\circ}$. On the halfline $(BA$ point $F$ is chosen such that $[BE] = [BF]$. Let $M$ be the intersection between the lines $EF$ and $AD$.\n\na) Prove that $m(\\angle BME) = 75^{\\circ}$.\n\nb) The bisector of angle $CBE$ intersects $CD$ at $N$. Prove that the triangle $BMN$ is equilateral.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We have $m(\\angle MEB) = m(\\angle CEB) = 60^{\\circ}$, hence $EB$ is an exterior angle bisector in the triangle $MDE$.\n\nBut $m(\\angle MDB) = m(\\angle BDE) = 45^{\\circ}$, hence $DB$ is an interior angle bisector in the triangle $MDE$.\n\nTherefore, $MB$ is the bisector of $AME$ and since $m(\\angle AME) = 150^{\\circ}$, it follows that $m(\\angle BME) = 75^{\\circ}$.\n\nb) We have $\\Delta BAM \\equiv \\Delta BCN$, hence $BM = BN$. Also, $m(\\angle BMN) = 90^{\\circ} - m(\\angle ABM) - m(\\angle CBN) = 60^{\\circ}$, so the triangle $BMN$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$, $I$ its incenter, and $M$ the midpoint of the side $BC$. If $IA = IM$, determine the smallest possible value of the angle $AIM$.",
    "options": [],
    "answer": "150°",
    "solution": "Let $\\{D\\} = AI \\cap BC$. As $AB < AC$, $D$ lies between $B$ and $M$ and $\\angle ACB < \\angle ABC$.\nWe have $\\angle IDB = \\angle DAC + \\angle ACB < \\angle DAB + \\angle ABD = \\angle ADC$, therefore angle $IDB$ is acute.\nLet $F$ and $E$ be the projections of $I$ onto $AB$ and $BC$, respectively. It follows that $E \\in (BD) \\subset BM$. Triangles $IBF$ and $IBE$ are congruent and so are triangles $IFA$ and $IEM$, therefore $BA = BM = \\frac{BC}{2}$ and triangles $IBA$ and $IBM$ are congruent.\n\n![](attached_image_1.png)\n\nWe have: $\\angle MID = \\angle IDB - \\angle IMB = \\angle DAC + \\angle ACD - \\angle IAB = \\angle ACD$.\nIt follows that $\\angle AIM = 180^\\circ - \\angle ACB$ (1).\nLet $H$ be the projection of $B$ onto the line $AC$. It follows that $BH \\le AB = \\frac{BC}{2}$, which shows that $\\angle ACB \\le 30^\\circ$ (2).\nFrom (1) and (2) we obtain that $\\angle AIM \\ge 180^\\circ - 30^\\circ = 150^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $n$ be two fixed positive integers.\na) Prove that there exist $n$ positive integers $a_1, a_2, \\dots, a_n$ such that\n$$\n1 + \\frac{1}{a} = \\left(1 + \\frac{1}{a_1}\\right) \\left(1 + \\frac{1}{a_2}\\right) \\dots \\left(1 + \\frac{1}{a_n}\\right).\n$$\n\nb) Prove that $1 + \\frac{1}{a}$ has only finitely many possible representations as in the point a).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $a_1 < a_2 < \\dots < a_n$ are consecutive positive integers, then we have\n$$\n\\left(1 + \\frac{1}{a_1}\\right) \\left(1 + \\frac{1}{a_2}\\right) \\dots \\left(1 + \\frac{1}{a_n}\\right) = \\frac{a_n + 1}{a_1}.\n$$\nChoosing $a_k = an + k - 1, \\forall k = \\overline{1, n}$, we get\n$$\n\\left(1 + \\frac{1}{a_1}\\right) \\left(1 + \\frac{1}{a_2}\\right) \\dots \\left(1 + \\frac{1}{a_n}\\right) = \\frac{an + n}{an} = 1 + \\frac{1}{a}.\n$$\n\nb) We prove by induction on $n \\in \\mathbb{N}^*$, the following property:\n$P(n)$: \"For any rational number $q > 1$, we can choose only finitely many possible $n$ positive integers $a_1 \\le a_2 \\le \\dots \\le a_n$ such that $\\prod_{k=1}^{n} \\left(1 + \\frac{1}{a_k}\\right) = q$.\"\nIt is easy to see that $P(1)$ is true.\nSuppose now that $P(n)$ is true for a positive integer $n$.\nLet $q \\in \\mathbb{Q}$, $q > 1$. From the previous point, we know that there exist $n + 1$ positive integers $a_1 \\le a_2 \\le \\dots \\le a_{n+1}$ such that $\\prod_{k=1}^{n+1} \\left(1 + \\frac{1}{a_k}\\right) = q$. Using the inequalities $1 + \\frac{1}{a_1} < q \\le \\left(1 + \\frac{1}{a_1}\\right)^{n+1}$, we get that $a_1 \\in A$, where $A = \\left(\\frac{1}{q-1}, \\frac{1}{\\sqrt[n]{q-1}}\\right] \\cap \\mathbb{N}^*$ is a finite set. For a fixed number $a_1 \\in A$, let $q_1 = \\frac{q}{1+1/a_1}$. It is clear that $q_1 \\in \\mathbb{Q}$, $q_1 > 1$. Using the induction hypothesis, we can choose only finitely many possible $n$ positive integers $a_2 \\le \\dots \\le a_{n+1}$, such that $\\prod_{k=2}^{n+1} \\left(1 + \\frac{1}{a_k}\\right) = q_1$. We deduce from here that we can choose only finitely many possible $n + 1$ positive integers $a_1 \\le a_2 \\le \\dots \\le a_{n+1}$, such that $\\prod_{k=1}^{n+1} \\left(1 + \\frac{1}{a_k}\\right) = q$, that is $P(n + 1)$ is also true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the integers $x$ and $y$ for which $\\sqrt{4^x + 5^y}$ is rational.",
    "options": [],
    "answer": "(1, 1) and (-2, 1)",
    "solution": "We treat four cases:\n\n**I.** $x, y \\ge 0$\n\n$\\sqrt{4^x + 5^y}$ is rational if and only if $4^x + 5^y$ is a perfect square, i.e., there exists $n \\in \\mathbb{N}$ such that $4^x + 5^y = n^2$. Analyzing this equation modulo $3$, we have $4^x \\equiv 1 \\pmod{3}$, $5^y \\equiv (-1)^y \\pmod{3}$, and $n^2 \\equiv 0, 1 \\pmod{3}$, hence $y$ needs to be odd. Let $z \\in \\mathbb{N}$ be such that $y = 2z + 1$. The previous equation becomes $4^x + 5 \\cdot 25^z = n^2$.\n\nIf $x \\ge 2$, then $4^x \\equiv 0 \\pmod{8}$, $5 \\cdot 25^z \\equiv 5 \\pmod{8}$, while $n^2 \\equiv 0, 1, 4 \\pmod{8}$, which means the equation has no solutions in this case. We are left with the cases when $x = 0$ and $x = 1$.\n\nFor $x = 0$ we have $4^x + 5^y \\equiv 2 \\pmod{4}$, hence $4^x + 5^y$ cannot be a perfect square.\n\nIf $x = 1$, then $5^y = (n-2)(n+2)$, hence there exist $a, b \\in \\mathbb{N}$, with $a + b = y$, such that $n-2 = 5^a$ and $n+2 = 5^b$. By subtraction, $5^b - 5^a = 4$. If $a, b \\ge 1$ then $5 \\mid 5^b - 5^a$, hence $5 \\mid 4$, contradiction. As $a < b$, it follows that $a = 0$, then $b = 1$, i.e., $y = 1$. We obtain the solution $x = y = 1$.\n\n\n**II.** $x, y < 0$\n\nLet $u = -x, v = -y, u, v > 0$. $\\sqrt{4^x + 5^y}$ is rational if and only if there exist $p, q \\in \\mathbb{N}^*$, coprime, such that $4^x + 5^y = \\frac{p^2}{q^2}$, i.e., $\\frac{4^u + 5^v}{4^u \\cdot 5^v} = \\frac{p^2}{q^2}$, which means $p^2 \\cdot 4^u \\cdot 5^v = q^2(4^u + 5^v)$. Numbers $4^u \\cdot 5^v$ and $4^u + 5^v$ are coprime, therefore every prime factor of $4^u + 5^v$ is a prime factor of $p^2$, which means that it appears at an even exponent. It follows that $4^u + 5^v$ is a perfect square, and, similarly, $4^u \\cdot 5^v$ is a perfect square. From case I it follows that $4^u + 5^v$ is a perfect square if and only if $u = v = 1$, but then $4^u \\cdot 5^v = 20$ is not a perfect square. We conclude that there are no solutions in this case.\n\n\n**III.** $x < 0, y \\ge 0$\n\nLet $u = -x \\in \\mathbb{N}^*$. Then $\\sqrt{4^x + 5^y} = \\frac{\\sqrt{1 + 4^u \\cdot 5^y}}{2^u}$ is rational if and only if $1 + 4^u \\cdot 5^y$ is a perfect square, i.e., there exists $n \\in \\mathbb{N}$ such that $1 + 4^u \\cdot 5^y = n^2$. Then $4^u \\cdot 5^y = (n-1)(n+1)$. As $u > 0$, $n$ is odd and $(n-1, n+1) = 2$. We distinguish the following sub-cases:\n\nA. $n-1 = 2 \\cdot 5^y$, $n+1 = 2^{2u-1}$\n\nB. $n-1 = 2$, $n+1 = 2^{2u-1} \\cdot 5^y$\n\nC. $n-1 = 2^{2u-1}$, $n+1 = 2 \\cdot 5^y$\n\nD. $n-1 = 2^{2u-1} \\cdot 5^y$, $n+1 = 2$\n\nIn sub-case A we obtain $2^{2u-2} - 5^y = 1$, i.e., $(2^{u-1} - 1)(2^{u-1} + 1) = 5^y$. It follows that $2^{u-1} - 1$ and $2^{u-1} + 1$ should be powers of $5$, but no two powers of $5$ are at distance $2$.\n\nSub-case B leads to $n = 3$ and, immediately, to $4 = 2^{2u-1} \\cdot 5^y$, with no solutions.\n\nIn sub-case C we get $5^y - 2^{2u-2} = 1$. Then $5^y \\equiv (-1)^y \\pmod{3}$ and $2^{2u-2} = 4^{u-1} \\equiv 1 \\pmod{3}$, therefore $y$ needs to be odd. It follows that $5^y \\equiv 5 \\pmod{8}$, hence $2^{2u-2} \\equiv 4 \\pmod{8}$, i.e., $u = 2$. We obtain the solution $x = -2, y = 1$.\n\nSub-case D leads to $n = 1$ and then to $0 = 2^{2u-1} \\cdot 5^y$, with no solutions.\n\n\n**IV.** $x \\ge 0, y < 0$\n\nLet $v = -y \\in \\mathbb{N}^*$. Then $\\sqrt{4^x + 5^y} = \\sqrt{\\frac{1 + 4^x \\cdot 5^v}{5^v}}$ is rational if and only if there exist $p, q \\in \\mathbb{N}^*$ coprime such that $\\frac{1 + 4^x \\cdot 5^v}{5^v} = \\frac{p^2}{q^2}$, i.e., such that $p^2 \\cdot 5^v = q^2(1 + 4^x \\cdot 5^v)$. Numbers $1 + 4^x \\cdot 5^v$ and $5^v$ are coprime, hence, as in case I, they need to be perfect squares. We have seen in the previous case that $1 + 4^x \\cdot 5^v$ is a perfect square only when $x = 2$ and $v = 1$, but in this situation $5^v = 5$ is not a perfect square. We conclude that in this case there are no solutions.\n\n\nTo summarize, the only solutions to the problem are $x = y = 1$ and $x = -2$, $y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38671,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Compare the numbers $2^{53}$ and $3^{35}$.\n\nb) Show that, if $5b \\ge 3a > 0$, then $2^{ab} < 3^{ba}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $3^7 = 2187 > 2048 = 2^{11}$, we find $3^{35} = (3^7)^5 > (2^{11})^5 > 2^{53}$.\n\nb) The inequality $2^{53} < 3^{35}$ can be written $\\left(\\frac{2^{10}}{3}\\right)^5 < \\left(\\frac{3^{10}}{2}\\right)^3$. It follows that $\\left(\\frac{2^{10}}{3}\\right)^{5a} < \\left(\\frac{3^{10}}{2}\\right)^{3a}$. The last inequality implies $\\left(\\frac{2^{10}}{3}\\right)^{5a} < \\left(\\frac{3^{10}}{2}\\right)^{5b}$ or $\\left(\\frac{2^{10}}{3}\\right)^a < \\left(\\frac{3^{10}}{2}\\right)^b$, hence $2^{10a+b} < 3^{10b+2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > 0$ be a real number. Prove the inequality\n$$\na^{\\sin x} \\cdot (a+1)^{\\cos x} \\ge a, \\quad \\forall x \\in [0, \\frac{\\pi}{2}].\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a > 1$, taking logs of both sides we obtain the equivalent inequality\n$$\n\\sin x + \\cos x \\cdot \\log_a (a+1) \\ge 1.\n$$\nFor $x \\in [0, \\pi/2]$, we have $\\sin x \\ge \\sin^2 x$ and $\\cos x \\ge \\cos^2 x$. Because $\\log_a (a+1) > 1$, we obtain\n$$\n\\sin x + \\cos x \\cdot \\log_a (a+1) \\ge \\sin^2 x + \\cos^2 x = 1.\n$$\nThe inequality clearly holds true for $a = 1$.\n\nFinally, if $a \\in (0, 1)$, the inequality is equivalent to\n$$\n\\sin x + \\cos x \\cdot \\log_a (a+1) \\le 1, \\quad \\forall x \\in [0, \\pi/2],\n$$\nwhich is obvious, since $\\sin x \\le 1$, $\\cos x \\ge 0$ and $\\log_a (a+1) < 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two non-commuting matrices $A, B \\in M_2(\\mathbb{R})$.\n\na) If $A^3 = B^3$, then $A^n$ and $B^n$ have for any positive integer $n$ the same trace.\n\nb) Show that there are non-commuting $A, B \\in M_2(\\mathbb{R})$, such that for any positive $n$, $A^n$ and $B^n$ are different.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From $A^3 = B^3$ we get $\\det A = \\det B = d$ and by Hamilton-Cayley theorem we obtain $A^2 = aA - dI_2$, $B^2 = bB - dI_2$, where $a = \\operatorname{tr} A$, $b = \\operatorname{tr} B$. As a consequence one can write $A^3 = aA^2 - dA = (a^2 - d)A - adI_2$ and $B^3 = (b^2 - d)B - bdI_2$, so $(a^2 - d)A - adI_2 = (b^2 - d)B - bdI_2$ (*).\n\nBy right and left multiplication with $B$ we get $(a^2-d)BA - a dB = (b^2-d)B^2 - bdB = (a^2-d)AB - adB$, so $d = a^2$. In the same way, $d = b^2$, and by (*), we deduce $d = 0$ or $a = b$.\nIf $d=0$, then $a=b=0$, so $A^2 = B^2 = O_2$, from where $A^n = B^n$, for any $n \\ge 2$. So $\\text{tr } A = a = 0 = b = \\text{tr } B$ and $\\text{tr } A^n = \\text{tr } B^n$, for all $n \\ge 2$.\nIf $d \\neq 0$, then $a = b$ and $C^2 = aC - a^2I_2$, $C^3 = -a^3I_2$, so $C^{3k} = (-a^3)^k I_2$, $C^{3k-2} = (-a^3)^{k-1}C$, $C^{3k-1} = (-a^3)^{k-1}(aC - a^2I_2)$, $\\forall C \\in \\{A, B\\}, \\forall k \\ge 1$.\n\nb) For $A = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}$, $B = \\begin{pmatrix} 1 & 0 \\\\ 1 & 1 \\end{pmatrix}$, we have $AB = \\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \\end{pmatrix}$, $BA = \\begin{pmatrix} 1 & 1 \\\\ 1 & 2 \\end{pmatrix}$,\n$A^n = \\begin{pmatrix} 1 & n \\\\ 0 & 1 \\end{pmatrix}$, $B^n = \\begin{pmatrix} 1 & 0 \\\\ n & 1 \\end{pmatrix}$ and $\\text{tr } A^n = \\text{tr } B^n = 2, \\forall n \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38674,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the positive integers $a, b, c$ satisfy the inequalities $a > b > c$ and $12b > 13c > 11a$, show that $a + b + c \\ge 56$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a, b, c \\in \\mathbb{N}$, from $a > b > c$ follows $a \\ge b + 1 \\ge c + 2$, hence $a - c \\ge 2$.\nIf $a - c \\ge 4$, then $13c > 11a \\ge 11(c + 4)$, so $c > 22$. It follows that $a + b + c > c + c + c > 66 > 56$.\nSo we still need to discuss the cases $a - c = 2$ and $a - c = 3$.\nIf $a - c = 2$, then from $a \\ge b + 1 \\ge c + 2$ follows $a = b + 1 = c + 2$, hence $12(c + 1) > 13c > 11(c + 2)$, where from $12 > c$ and $c > 11$, which is impossible, and we conclude $a - c \\ne 2$.\nIf $a - c = 3$, then $a = c + 3$ and either $b = c + 1$ or $b = c + 2$.\nIf $b = c + 1$, then $12(c + 1) > 13c > 11(c + 3)$, so $12 > c$ and $c > \\frac{33}{2}$, which is impossible.\nIf $b = c + 2$, then $12(c + 2) > 13c > 11(c + 3)$, so $24 > c$ and $c > \\frac{33}{2}$, hence $c \\ge 17$. It follows that $b \\ge 19$, $a \\ge 20$ and $a + b + c \\ge 56$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be an arithmetic sequence of positive integers and let $S_n = a_1^2 + a_2^2 + \\dots + a_n^2$, $n \\in \\mathbb{N}^*$. Prove that:\n\na) if $p \\ge 5$ is a prime number, then $p$ divides $S_p$;\n\nb) $S_5$ is not a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We have $a_n = a_1 + (n-1)r$, $n \\in \\mathbb{N}^*$, with $a_1 \\in \\mathbb{N}^*$, $r \\in \\mathbb{N}$. This yields\n$$\nS_p = p \\left( a_1^2 + a_1(p-1)r + \\frac{(p-1)(2p-1)}{6} r^2 \\right).\n$$\nSince $p \\ge 5$ is a prime number, we have $\\gcd(p, 2) = 1$, $\\gcd(p, 3) = 1$, hence $6 \\mid (p-1)(2p-1)$, that is, $p \\mid S_p$.\n\n\nb) Assume, by way of contradiction, that $S_5$ is a square. From a), we deduce that for some $y \\in \\mathbb{N}^*$, we have $S_5 = (5y)^2$.\nWriting $S_5$ in terms of $x = a_3 \\in \\mathbb{N}^*$ and the common difference $r$ we obtain $S_5 = 5x^2 + 10r^2$, hence the equation $x^2 + 2r^2 = 5y^2$. It is not difficult to see that the equation has no solution in positive integers, by checking it modulo $5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$ and let $X$ be an $n$-element set. A non-empty collection of subsets $A_1, \\dots, A_k$ of $X$ is *tight* if the union $A_1 \\cup \\dots \\cup A_k$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_i$s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.",
    "options": [],
    "answer": "2n - 2",
    "solution": "*Note.* A subset $A$ of $X$ is proper if $A \\ne X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.\n\n*First solution.* (Ilya Bogdanov) The required maximum is $2n-2$. To describe a $(2n-2)$-element collection satisfying the required conditions, write $X = \\{1, 2, \\dots, n\\}$ and set $B_k = \\{1, 2, \\dots, k\\}$, $k = 1, 2, \\dots, n-1$, and $B_k = \\{k-n+2, k-n+3, \\dots, n\\}$, $k=n, n+1, \\dots, 2n-2$. To show that no subcollection of the $B_k$ is tight, consider a subcollection $C$ whose union $U$ is a proper subset of $X$, let $m$ be an element in $X \\setminus U$, and notice that $C$ is a subcollection of $\\{B_1, \\dots, B_{m-1}, B_{m+n-1}, \\dots, B_{2n-2}\\}$, since the other $B$'s are precisely those containing $m$. If $U$ contains elements less than $m$, let $k$ be the greatest such and notice that $B_k$ is the only member of $C$ containing $k$; and if $U$ contains elements greater than $m$, let $k$ be the least such and notice that $B_{k+n-2}$ is the only member of $C$ containing $k$. Consequently, $C$ is not tight.\n\nWe now proceed to show by induction on $n \\ge 2$ that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2n-2$. The base case $n=2$ is clear, so let $n > 2$ and suppose, if possible, that $\\mathcal{B}$ is a collection of $2n-1$ proper non-empty subsets of $X$ containing no tight subcollection.\nTo begin, notice that $\\mathcal{B}$ has an empty intersection: if the members of $\\mathcal{B}$ shared an element $x$, then $\\mathcal{B}' = \\{B \\setminus \\{x\\} : B \\in \\mathcal{B}, B \\ne \\{x\\}\\}$ would be a collection of at least $2n-2$ proper non-empty subsets of $X \\setminus \\{x\\}$ containing no tight subcollection, and the induction hypothesis would be contradicted.\nNow, for every $x$ in $X$, let $B_x$ be the (non-empty) collection of all members of $\\mathcal{B}$ not containing $x$. Since no subcollection of $\\mathcal{B}$ is tight, $B_x$ is not tight, and since the union of $B_x$ does not contain $x$, some $x'$ in $X$ is covered by a single member of $B_x$. In other words, there is a single set in $\\mathcal{B}$ covering $x'$ but not $x$. In this case, draw an arrow from $x$ to $x'$. Since there is at least one arrow from each $x$ in $X$, some of these arrows form a (minimal) cycle $x_1 \\to x_2 \\to \\dots \\to x_k \\to x_{k+1} = x_1$ for some suitable integer $k \\ge 2$. Let $A_i$ be the unique member of $\\mathcal{B}$ containing $x_{i+1}$ but not $x_i$, and let $X' = \\{x_1, x_2, \\dots, x_k\\}$.\nRemove $A_1, A_2, \\dots, A_k$ from $\\mathcal{B}$ to obtain a collection $\\mathcal{B}'$ each member of which either contains or is disjoint from $X'$: for if a member $B$ of $\\mathcal{B}'$ contained some but not all elements of $X'$, then $B$ should contain $x_{i+1}$ but not $x_i$ for some $i$, and $B = A_i$, a contradiction. This rules out the case $k=n$, for otherwise $B = \\{A_1, A_2, \\dots, A_n\\}$, so $|B| < 2n-1$.\nTo rule out the case $k < n$, consider an extra element $x^*$ outside $X$ and let\n$$\nB^* = \\{B: B \\in B', B \\cap X' = \\emptyset\\} \\cup \\{(B \\setminus X') \\cup \\{x^*\\}: B \\in B', X' \\subseteq B\\};\n$$\nthus, in each member of $B'$ containing $X'$, the latter is collapsed to singleton $x^*$. Notice that $B^*$ is a collection of proper non-empty subsets of $X^* = (X \\setminus X') \\cup \\{x^*\\}$, no subcollection of which is tight. By the induction hypothesis, $|B'| = |B^*| \\le 2|X^*| - 2 = 2(n-k)$, so $|B| \\le 2(n-k)+k = 2n-k < 2n-1$, a final contradiction.\n\n\n*Second solution.* Proceed again by induction on $n$ to show that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2n-2$.\nConsider any collection $\\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection; call such a collection *good*. Assume that there exist $M, N \\in \\mathcal{B}$ such that $M \\cup N$ is distinct from $M, N$, and $X$. In this case, we will show how to modify $\\mathcal{B}$ so that it remains good, contains the same number of sets, but the total number of elements in the sets of $\\mathcal{B}$ increases.\nConsider a maximal (relative to set-theoretic inclusion) subcollection $C \\subseteq \\mathcal{B}$ such that the set $C = \\bigcup_{C' \\in C} C'$ is distinct from $X$ and from all members of $C$. Notice here that the union of any subcollection $D \\subset \\mathcal{B}$ cannot coincide with any $K \\in \\mathcal{B} \\setminus D$, otherwise $\\{K\\} \\cup D$ would be tight. Surely, $C$ exists (since $\\{M, N\\}$ is an example of a collection satisfying the requirements on $C$, except for maximality); moreover, $C \\notin \\mathcal{B}$ by the above remark.\nSince $C \\ne X$, there exists an $L \\in C$ and $x \\in L$ such that $L$ is the unique set in $C$ containing $x$. Now replace in $\\mathcal{B}$ the set $L$ by $C$ in order to obtain a new collection $\\mathcal{B}'$ (then $|\\mathcal{B}'| = |\\mathcal{B}|$). We claim that $\\mathcal{B}'$ is good.\nAssume, to the contrary, that $\\mathcal{B}'$ contained a tight subcollection $\\mathcal{T}$; clearly, $C \\in \\mathcal{T}$, otherwise $\\mathcal{B}$ is not good. If $\\mathcal{T} \\subseteq C \\cup \\{C\\}$, then $C$ is the unique set in $\\mathcal{T}$ containing $x$ which is impossible. Therefore, there exists $P \\in \\mathcal{T} \\setminus (C \\cup \\{C\\})$. By maximality of $C$, the collection $C \\cup \\{P\\}$ does not satisfy the requirements imposed on $C$; since $P \\cup C \\ne X$, this may happen only if $C \\subset P$. But then $\\mathcal{D} = (\\mathcal{T} \\setminus \\{C\\}) \\cup C$ is a tight subcollection in $\\mathcal{B}$: all elements of $C$ are covered by $\\mathcal{D}$ at least twice (by $P$).\n\nand an element of $\\mathcal{C}$), and $\\mathcal{D}$ and $\\mathcal{T}$ both cover all other elements the same number of times – a contradiction. Thus $\\mathcal{B}'$ is good.\nSuch modifications may be performed finitely many times, since the total number of elements of sets in $\\mathcal{B}$ increases. Thus, at some moment we arrive at a good collection $\\mathcal{B}$ to which the procedure no longer applies. This means that for every $M, N \\in \\mathcal{B}$, either $M \\cup N = X$ or one of them is contained in the other.\nNow let $M$ be a $\\subseteq$-minimal set in $\\mathcal{B}$. Then each set in $\\mathcal{B}$ either contains $M$ or forms $X$ in union with $M$ (i.e., contains $X \\setminus M$). Now one may easily see that the collections\n$$\n\\mathcal{B}_+ = \\{A \\setminus M : A \\in \\mathcal{B}, M \\subset A, A \\ne M\\},\n$$\n$$\n\\mathcal{B}_- = \\{A \\cap M : A \\in \\mathcal{B}, X \\setminus M \\subset A, A \\ne X \\setminus M\\}\n$$\nare both good as collections of subsets of $X \\setminus M$ and $M$, respectively; thus, by the induction hypothesis, $|\\mathcal{B}_+| + |\\mathcal{B}_-| \\le 2n - 4$.\nFinally, each set $A \\in \\mathcal{B}$ either produces a set in one of the two new collections, or coincides with $M$ or $X \\setminus M$. Thus $|\\mathcal{B}| \\le |\\mathcal{B}_+| + |\\mathcal{B}_-| + 2 \\le 2n - 2$, as required.\n\n\n*Third solution.* We provide yet another proof of the estimate $|\\mathcal{B}| \\le 2n - 2$. Recall the notion of a *good* collection from Solution 2. Consider any good collection $\\mathcal{B}$, and for every $x \\in X$ let $\\mathcal{B}_x = \\{B : B \\in \\mathcal{B}, x \\notin B\\}$, as in Solution 1.\nConsider an element $x \\in X$ such that $|\\mathcal{B}_x|$ is maximal. Since the collection $\\mathcal{B}_x$ is not tight, and the union of its members does not contain $x$, there exists an element $y \\in X$ belonging to a unique member of $\\mathcal{B}_x$, say, $B$. Thus, $\\mathcal{B}_x \\setminus \\{B\\} \\subset \\mathcal{B}_y$.\nSince $|\\mathcal{B}_y| \\le |\\mathcal{B}_x|$ by maximality of the latter, the collection $\\mathcal{C} = \\mathcal{B}_y \\setminus \\mathcal{B}_x$ contains at most one member. Set now $\\mathcal{B}' = \\mathcal{B} \\setminus (\\mathcal{C} \\cup \\{B\\})$. By the argument above, each set in $\\mathcal{B}'$ contains either both $x$ and $y$, or none of them. Collapse $\\{x, y\\}$ to a singleton $x^*$ to get a new collection of $|\\mathcal{B}'| \\ge |\\mathcal{B}| - 2$ subsets of $(X \\setminus \\{x, y\\}) \\cup \\{x^*\\}$ containing no tight subcollection. By the induction hypothesis $|\\mathcal{B}'| \\le 2(n-1) - 2$, so $|\\mathcal{B}| \\le 2n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and $A, B \\in \\mathcal{M}_n(\\mathbb{C})$. If $(AB)^3 = O_n$, do we have always $(BA)^3 = O_n$? Justify.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that the result is valid for $n \\le 3$ and not true for $n \\ge 4$. Matrices $C = AB$ and $D = BA$ have the same trace and both determinants are zero. If $C^2 = O_n$, then $D^3 = BC^2A = O_n$.\n\nFor $n=2$, matrices $C$ and $D$ have the same characteristic polynomial $P_C(X) = P_D(X) = X^2 - aX$, with $a = \\text{tr} C = \\text{tr} D$. By the hypothesis and the Cayley-Hamilton we get $O_n = C^3 = aC^2$. Thus $a=0$ or $C^2 = 0$. Both cases imply $D^2 = aD = O_n$\n\nWe may now suppose $C^2 \\ne O_n$. For $n=3$, the characteristic polynomials are $P_C(X) = X^3 - aX^2 + bX$ and $P_D(X) = X^3 - aX^2 + cX$ respectively. From $aC^2 = bC$ we get $bC^2 = O_n$, $b=0$. So $aC^2 = O_n$, which implies $a=0$.\n\nApplying again Hamilton-Cayley we obtain $D^4 = -cD^2$. As $D^4 = BC^3A = O_n$, we obtain $cD^2 = O_n$, so that we must have $D^2 = O_n$, or $c=0$. Both imply $D^3 = O_n$\n\nTo deal with the case $n=4$, consider\n$$\nA = \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 1 & 0 \\end{pmatrix}, \\quad B = \\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix},\n$$\nmatrices that show that we can have $(AB)^3 = O_n$ and $(BA)^3 \\ne O_n$.\n\nFor $n \\ge 5$ consider matrices which have the upper left corner the 4 by 4 matrices in the previous examples and zeroes in rest.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, if a ring $R$ is not a skew-field, and $x^2 = x$ for every non-invertible element $x$ of $R$, then $x^2 = x$ for every element $x$ of $R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* An element $x$ of $R$ such that $x^2 = x$ is called *idempotent*. We show that $1$ is the only unit in $R$. The proof is based on the remark below:\n(*) Let $x$ and $y$ be elements of a ring $R$. If $x \\neq 1$ is idempotent, then $xyz \\neq 1$ and $zyx \\neq 1$ for all $z$ in $R$; in particular, $xy$ and $yx$ are both non-invertible.\nIndeed, if $xyz = 1$, then $x = xxyz = x^2yz = xyz = 1$, a contradiction. Hence $xyz \\neq 1$; similarly, $zyx \\neq 1$.\nBack to the problem, fix a non-invertible $x$ in $R \\setminus \\{0\\}$; since $R$ is not a skew-field, there exists at least one such. Since $2x$ is not invertible, it is idempotent, so $2x = 4x^2 = 4x$, that is, $2x = 0$.\nWe now show that $xy = xyx = yx$ for all $y$ in $R$. To prove the first equality, refer to (*) to infer that $xy - xyx = xy(1-x)$ is non-invertible, hence idempotent, so $xy - xyx = xy(1-x)xy(1-x) = xy(x-x^2)y(1-x) = 0$, since $x^2 = x$. Similarly, $yx - xyx = (1-x)yx$ is non-invertible, hence idempotent, so $yx - xyx = (1-x)yx(1-x)yx = (1-x)y(x-x^2)yx = 0$, since $x^2 = x$.\n*Second solution.* As in the previous solution, we show that $1$ is the only unit of $R$.\nWe first prove that if $u$ is a unit, and $x \\neq 0$ is not, then $u+x$ is not a unit. Let $D$\nbe the set of all non-invertible elements of $R \\setminus \\{0\\}$; since $R$ is not a skew-field, $D$ is\nnon-empty. If $x$ is a member of $D$, then so is $-x$, and $x = x^2 = (-x)^2 = -x$, so\n$2x = 0$. Then $(1+x)^2 = 1+2x+x^2 = 1+x$, so $1+x$ is not a unit (otherwise,\n$1+x=1$, so $x=0$, a contradiction). If $u$ is a unit, and $x$ is a member of $D$, then $ux$\nand $1+ux$ are both in $D$, and so is $u+x = u(1+u^{-1}x)$.\nLet $x$ be a member of $D$, let $y$ be a member of $R$, and write $x(xy) = x^2y = xy$ and $(yx)x = yx^2 = yx$, to infer that $xy$ and $yx$ are both non-units.\nWe are now in a position to prove that $1$ is the only unit of $R$. Let $u$ be a unit, and let $x$ be a member of $D$. Then $x+ux$ and $x+ xu$ are both non-units, so $(x+ux)^2 = x+ux$ and $(x + xu)^2 = x + xu$. Expand both squares to write $x^2 + xux + ux^2 + (ux)^2 = x + ux$ and $x^2 + x^2u + xux + (xu)^2 = x + xu$, and infer that $ux = xux = xu$. Finally, since $u + x$ is not a unit, $(u + x)^2 = u + x$, so $u^2 = u$; that is, $u = 1$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest radius a circle passing through exactly three lattice points may have.",
    "options": [],
    "answer": "5*sqrt(2)/6",
    "solution": "The required minimum is $5\\sqrt{2}/6$: the circle of radius $5\\sqrt{2}/6$, centered at $(5/6, 5/6)$ passes through exactly three lattice points, namely, $(0, 0)$, $(1, 2)$ and $(2, 1)$.\n\nNext, consider a circle $\\omega$ of radius at most $5\\sqrt{2}/6$ passing through exactly three lattice points. We may and will assume that one of these lattice points is at the origin; let $(m, n)$ and $(p, q)$ be the other two.\nSince the diameter of $\\omega$ does not exceed $5\\sqrt{2}/3$, the integers $m^2 + n^2$, $p^2 + q^2$ and $(m-p)^2 + (n-q)^2$ are all at most $5$.\n\nIf $m = 0$, then $p \\neq 0$ by non-collinearity, and $n \\neq 0$ must be even and $q = n/2$, for otherwise $(p, n-q)$ would be a fourth lattice point on $\\omega$. Hence $n = \\pm 2$ and $q = \\pm 1$ (corresponding signs), and $|p|$ is either $1$ or $2$. The case $|p| = 1$ is ruled out by the fact that $(-p, q)$ would be a fourth lattice point on $\\omega$, and the case $|p| = 2$ is ruled out by the fact that the radius of $\\omega$ would be $5/4 > 5\\sqrt{2}/6$.\nConsequently, $m \\neq 0$; similarly, $n \\neq 0$, $p \\neq 0$, $q \\neq 0$, $m \\neq p$, and $n \\neq q$.\nIt then follows that $(m, n)$ and $(p, q)$ are consecutive vertices of the hexagon with vertices at $(2, 1)$, $(1, 2)$, $(-1, 1)$, $(-2, -1)$, $(-1, -2)$, $(1, -1)$ or of its reflection in one of the coordinate axes. The radius of $\\omega$ is $5\\sqrt{2}/6$ whatsoever the case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{z \\in \\mathbb{C} \\mid |z| = 1\\}$.\n\na) Prove that $(|z+1| - \\sqrt{2})(|z-1| - \\sqrt{2}) \\le 0$, for all $z \\in A$.\n\nb) Prove that, for any $z_1, z_2, \\dots, z_{12} \\in A$, one can choose the signs \"\\pm\" such that\n$$\n\\sum_{k=1}^{12} |z_k \\pm 1| < 17.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Observe that\n$$\n|z+1|^2 + |z-1|^2 = (z+1)(\\bar{z}+1) + (z-1)(\\bar{z}-1) = 2|z|^2 + 2 = 4,\n$$\nhence $|z+1|^2 - 2 = 2 - |z-1|^2$, that is,\n$$\n(|z+1| - \\sqrt{2})(|z+1| + \\sqrt{2}) = -(|z-1| - \\sqrt{2})(|z-1| + \\sqrt{2}).\n$$\nClearly, this implies that $|z+1| - \\sqrt{2}$ and $|z-1| - \\sqrt{2}$ have opposite signs.\n\nb) Using a), we choose signs such that $|z_k \\pm 1| \\le \\sqrt{2}$. Then\n$$\n\\sum_{k=1}^{12} |z_k \\pm 1| \\le 12\\sqrt{2} < 17,\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD A'B'C'D'$ be a cube with side length $1$. An ant walks on the cube's faces, starting from $A$ and ending at $C'$. The ant moves only on the cube's edges or on the diagonals of its faces. Knowing that the ant never passes through the same point twice, find the maximal length of such a walk.",
    "options": [],
    "answer": "3 + 4√2",
    "solution": "We claim that the maximal length of a walk is $3 + 4\\sqrt{2}$. An example of such a walk is $A \\to A' \\to B \\to D \\to D' \\to C \\to B' \\to C'$. Since the cube has $8$ vertices, the ant's walk consists of at most $7$ steps, each of length either $1$ or $\\sqrt{2}$.\n\nThe length of a $6$ steps walk is at most $6\\sqrt{2} < 3 + 4\\sqrt{2}$, hence the maximal walk must have exactly $7$ steps.\n\nWe claim that the number of \"diagonal\" steps is at most $4$. For this purpose, let us color black the vertices $A, C, B', D'$, and white the other four. Observe that a \"diagonal\" step changes the color, while an \"edge\" step changes it. Since $A$ and $C'$ have different colors, we deduce that the walk must contain an odd number of \"edge\" steps. By inspection, we see that a walk with one \"edge\" step and $6$ \"diagonal\" ones self-intersects, thus proving the claim.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers with $n \\ge 3$ digits, having the property: for every $k \\in \\{1, 2, \\dots, n-2\\}$, deleting any $k$ of its digits leaves a prime number.",
    "options": [],
    "answer": "113, 131, 137, 173, 179, 197, 311, 317, 431, 617, 719",
    "solution": "Answer: 113, 131, 137, 173, 179, 197, 311, 317, 431, 617 and 719.\nDenote $N = \\overline{a_n a_{n-1} \\dots a_1 a_0}$, such a number. We can make the following observations.\n\n**O1.** $N$ has at most one digit multiple of 3, because otherwise we delete the other digits and we get a two digit multiple of 3. In the same way, $N$ cannot have three digits congruent mod 3, because the elimination of the other digits would leave a three digit multiple of 3.\n\n**O2.** $N$ has exactly three digits. Indeed, if $N$ has at least 4 digits, then three of its digits must leave the remainder 1 or 2 mod 3 and, in view of (O1), we must get both remainders, so the number made by this two digits is divisible by 3.\n\n**O3:** $N$ has at most one even digit, and this can be only the first one, because otherwise when we delete all the digits placed at right side of the (last) even digit we get an even two digit number.\n\n**O4:** $N$ cannot have a 5. Otherwise, $N$ could have at most one 5 and this should be in the first position. Now (O3) shows that $N$ cannot have even digits. Since $N$ has at most one of the digits 3 and 9, $N$ has at least one of the digits 1 and 7. Deleting the proper digits we would get 51 or 57, numbers which are not primes.\n\nIf $N$ has three odd digits, then $N$ has a 3 or a 9 and the other two digits are 1 and/or 7. Checking all the combinations, we get the solutions 113, 131, 137, 173, 179, 197, 311, 317, 719.\n\nIf the first digit of $N$ is even, then (O1) and (O4) show that this digit cannot be 2 or 8, because if otherwise $N$ should have a 1 or a 7 and, deleting the appropriate digits, we get a two digit multiple of 3. If the first digit is 6, then the other two are from $\\{1, 7\\}$, and if the first digit is 4, another one is 1 or 7 and the third is 3 or 9. This way we get the solutions 617 and 431.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $f : [a, b] \\to [a, b]$ a differentiable function, such that $f'$ is continuous and positive. Show that there is a point $c \\in (a, b)$ such that\n$$\nf(f(b)) - f(f(a)) = (f'(c))^2 (b - a).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously $f$ is increasing.\nUsing Lagrange's theorem on $[f(a), f(b)]$ we get a point $c_1 \\in (f(a), f(b))$ such that\n$$\nf(f(b)) - f(f(a)) = f'(c_1)(f(b) - f(a)).\n$$\nAnother use of the theorem for the interval $[a, b]$ gives a point $c_2 \\in (a, b)$ such that\n$$\nf(b) - f(a) = f'(c_2)(b - a).\n$$\nCollecting, we get\n$$\nf(f(b)) - f(f(a)) = f'(c_1)f'(c_2)(b - a).\n$$\nAs $f' > 0$ the number $\\sqrt{f'(c_1)f'(c_2)}$ is between $f'(c_1)$ and $f'(c_2)$. Because $f'$ has the intermediate point property (Darboux's theorem), there is a point $c$, between $c_1$ and $c_2$ such that $f'(c) = \\sqrt{f'(c_1)f'(c_2)}$. This gives\n$$\nf'(c_1)f'(c_2) = f'(c)^2,\n$$\nand, in conclusion,\n$$\nf(f(b)) - f(f(a)) = f'(c)^2(b - a), \\quad c \\in (a, b).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $m$, determine the smallest integer $n \\ge 2$ satisfying the following condition: No matter how the cells of an $n \\times n$ array are colored one of $m$ colors, there exist cells $(i, j)$ and $(j, k)$, $i \\neq j$ and $j \\neq k$, sharing the same color.",
    "options": [],
    "answer": "C(m, floor(m/2)) + 1",
    "solution": "Assuming such a coloring exists, let $S_i$ be the set of colors of the off-diagonal cells on row $i$, and notice that the (re)stated condition shows that no $S_i$ contains an $S_j$, $i \\neq j$, so the $S_i$ form an antichain.\n\nConversely, given an antichain $S_1, \\dots, S_n$ of subsets of an $m$-element set, assigning an off-diagonal cell $(i, j)$ an element in $S_i \\setminus S_j$, and extending arbitrarily to on-diagonal cells yields a coloring satisfying the (re)stated condition.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $A \\in M_n(\\mathbb{C})$ with $n \\ge 2$ such that $\\det A = 0$ and denote by $A^*$ its adjutant. Show that $(A^*)^2 = (\\text{tr} A^*) A^*$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\det A = 0$, we have $\\text{rang}(A) \\le n-1$. If $\\text{rang}(A) \\le n-2$, then $A^* = O_n$ and $(A^*)^2 = O_n$ (*).\n\nIf $\\text{rang}(A) = n-1$, from $AA^* = (\\det A)I_n = O_n$, results $0 \\ge \\text{rang}(A) + \\text{rang}(A^*) - n$, therefore $\\text{rang}(A^*) \\le 1$.\n\nIf $\\text{rang}(A^*) = 0$, then $A^* = O_n$ and $(A^*)^2 = O_n$ (**).\n\nFor $\\text{rang}(A^*) = 1$, from the fact that any two lines of the matrix $A^*$ are proportional, results $A^* = BC$ with\n$$\nB = \\begin{pmatrix} b_1 & 0 & \\dots & 0 \\\\ b_2 & 0 & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b_n & 0 & \\dots & 0 \\end{pmatrix}, \\quad C = \\begin{pmatrix} c_1 & c_2 & \\dots & c_n \\\\ 0 & 0 & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\dots & 0 \\end{pmatrix}\n$$\n$$\n(A^*)^2 = (BC)(BC) = B(CB)C = BDC, \\text{ with } D = \\begin{pmatrix} t & 0 & \\dots & 0 \\\\ 0 & 0 & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\dots & 0 \\end{pmatrix}, \\text{ with}\n$$\n$$\nt = \\sum_{i=1}^{n} b_i c_i = \\text{tr } A^*. \\text{ We get } BDC = tA^*(**).\n$$\nThe relations (*), (**), (***) are expressed through $(A^*)^2 = (\\text{tr } A^*) A^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38686,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian plane, let $\\gamma_1$ and $\\gamma_2$ be the graphs of the quadratic functions $f_1(x) = p_1x^2+q_1x+r_1$ and $f_2(x) = p_2x^2+q_2x+r_2$, where $p_1 > 0 > p_2$. The graphs $\\gamma_1$ and $\\gamma_2$ cross at distinct points $A$ and $B$. The four tangents to $\\gamma_1$ and $\\gamma_2$ at $A$ and $B$ form a convex quadrangle which has an inscribed circle. Prove that the graphs $\\gamma_1$ and $\\gamma_2$ have the same axis of symmetry.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the tangents to $\\gamma_i$ at $A$ and $B$ meet at $C_i$. Since $f_1(x)$ is convex and $f_2(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $AC_1BC_2$.\n\n![](attached_image_1.png)\n\n**Lemma.** If $CA$ and $CB$ are the tangents drawn from a point $C$ to the graph $\\gamma$ of a quadratic trinomial $f(x) = px^2 + qx + r$, $A, B \\in \\gamma$, $A \\neq B$, then the abscissa of $C$ is the arithmetic mean of the abscissae of $A$ and $B$.\n\n**Proof.** Assume, without loss of generality, that $C$ is at the origin, so the equations of the two tangents have the form $y = k_a x$ and $y = k_b x$. Next, the abscissae $x_A$ and $x_B$ of the tangency points $A$ and $B$, respectively, are multiple roots of the polynomials $f(x) - k_a x$ and $f(x) - k_b x$, respectively. By the Vieta theorem, $x_A^2 = r/p = x_B^2$, so $x_A = -x_B$, since the case $x_A = x_B$ is ruled out by $A \\neq B$.\n\nThe Lemma shows that the line $C_1C_2$ is parallel to the $y$-axis and the points $A$ and $B$ are equidistant from this line.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nSuppose, if possible, that the incentre $O$ of the quadrangle $AC_1BC_2$ does not lie on the line $C_1C_2$. Assume, without loss of generality, that $O$ lies inside the triangle $AC_1C_2$ and let $A'$ be the reflection of $A$ in the line $C_1C_2$. Then the ray $C_iB$ emanating from $C_i$ lies inside the angle $AC_iA'$, so $B$ lies inside the quadrangle $AC_1A'C_2$, whence $A$ and $B$ are not equidistant from $C_1C_2$ – a contradiction.\n\nThus $O$ lies on $C_1C_2$, so the lines $AC_i$ and $BC_i$ are reflections of one another in the line $C_1C_2$, and $B = A'$. Hence $y_A = y_B$, and since\n$$\nf_i(x) = y_A + p_i(x - x_A)(x - x_B), \\quad i = 1, 2,\n$$\nthe line $C_1C_2$ is the axis of symmetry of both parabolas, as required.\n\n![](attached_image_3.png)\nUse the standard equation of a tangent to a smooth curve in the plane, to deduce that the tangents at two distinct points $A$ and $B$ on the parabola of equation $y = px^2 + qx + r$, $p \\neq 0$, meet at some point $C$ whose coordinates are\n$$\nx_C = \\frac{1}{2}(x_A + x_B) \\quad \\text{and} \\quad y_C = px_Ax_B + q \\cdot \\frac{1}{2}(x_A + x_B) + r.\n$$\nUsage of the standard formula for Euclidean distance yields\n$$\nCA = \\frac{1}{2}|x_B - x_A|\\sqrt{1 + (2px_A + q)^2} \\quad \\text{and} \\quad CB = \\frac{1}{2}|x_B - x_A|\\sqrt{1 + (2px_B + q)^2},\n$$\nso, after obvious manipulations,\n$$\nCB - CA = \\frac{2p(x_B - x_A)|x_B - x_A|(p(x_A + x_B) + q)}{\\sqrt{1 + (2px_A + q)^2} + \\sqrt{1 + (2px_B + q)^2}}\n$$\nNow, let the tangents to $\\gamma_i$ at $A$ and $B$ meet at $C_i$, write the condition in the statement in the form $C_1B - C_1A = C_2B - C_2A$, apply the above formula and clear common factors to get\n$$\n\\frac{p_1(p_1(x_A + x_B) + q_1)}{\\sqrt{1 + (2p_1x_A + q_1)^2} + \\sqrt{1 + (2p_1x_B + q_1)^2}} = \\frac{p_2(p_2(x_A + x_B) + q_2)}{\\sqrt{1 + (2p_2x_A + q_2)^2} + \\sqrt{1 + (2p_2x_B + q_2)^2}}\n$$\nNext, use the fact that $x_A$ and $x_B$ are the solutions of the quadratic equation $(p_1 - p_2)x^2 + (q_1 - q_2)x + r_1 - r_2 = 0$, so $x_A + x_B = -(q_1 - q_2)/(p_1 - p_2)$, to obtain\n$$\n\\frac{p_1(p_1q_2 - p_2q_1)}{\\sqrt{1 + (2p_1x_A + q_1)^2} + \\sqrt{1 + (2p_1x_B + q_1)^2}} = \\frac{p_2(p_1q_2 - p_2q_1)}{\\sqrt{1 + (2p_2x_A + q_2)^2} + \\sqrt{1 + (2p_2x_B + q_2)^2}}\n$$\nFinally, since $p_1p_2 < 0$ and the denominators above are both positive, the last equality forces $p_1q_2 - p_2q_1 = 0$; that is, $q_1/p_1 = q_2/p_2$, so the two parabolas have the same axis.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 38687,
    "subject": "Mathematics (Multi-modal)",
    "question": "On each side of a triangle we consider 9 distinct points, different from the triangle's vertices. Determine the number of triangles having the vertices in three of these $3 \\times 9$ points.\nLucian Dragomir",
    "options": [],
    "answer": "2673",
    "solution": "The triangles can be of two types: with all three vertices on different sides or with two vertices on one side and the third vertex on another side.\n\nIn case all vertices are on different sides, each vertex may be chosen in one of $9$ ways, accounting for a total of $9^3 = 729$ such triangles.\n\nIn the other case, there are $\\frac{9 \\cdot 8}{2} = 36$ ways of choosing two vertices from a given side. The third vertex can be chosen from the other two sides in $2 \\cdot 9 = 18$ ways, hence there will be a total of $3 \\cdot 2 \\cdot 9 \\cdot 36 = 3 \\cdot 648 = 1944$ such triangles.\n\nSo there are $729 + 1944 = 2673$ triangles with vertices among the given $27$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all ways to color in green and red the numbers $1, 2, \\ldots, 10$ such that the following conditions are fulfilled:\n\na) the number $5$ is colored red;\n\nb) if the numbers $x$ and $y$ have different colors and $x + y \\le 10$, then the number $x + y$ must be colored green;\n\nc) if the numbers $x$ and $y$ have different colors and $xy \\le 10$, then the number $xy$ must be colored red.",
    "options": [],
    "answer": "Two colorings: (1) all numbers red; (2) red numbers: 5 and 10, and all other numbers green.",
    "solution": "If $1$ is colored red, then for any green $a > 1$, we have $1 \\cdot a = a$ must be colored red, a contradiction. Hence all numbers must be red in this case.\n\nConsider now that $1$ is colored green.\n\nIf $2$ is red, then $1 + 2 = 3$ must be green, so $2 + 3 = 5$ must also be green, contradiction. So $2$ must be green.\n\nIf $3$ is red, then $2 + 3 = 5$ must be green, contradiction. So $3$ must be green.\nIf $4$ is red, then $1 + 4 = 5$ must be green, contradiction. So $4$ must be green.\n\nSince $5$ has a different color than $1, 2, 3, 4$, all the numbers $1+5=6$, $2+5=7$, $3+5=8$, $4+5=9$ must be green, and $2 \\cdot 5 = 10$ must be red. The numbers $6, 7, 8, 9$, respectively $10$ cannot be obtained as the product, respectively the sum, of $5$ and another number, so that this coloring satisfies the conditions in the statement of the problem.\n\nConsequently, there are two possible colorings: all numbers colored red, or $5$ and $10$ colored red and the other numbers green.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38689,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Exhibit a continuous function $f: (0, \\infty) \\to \\mathbb{R}$ such that\n$$\n\\lim_{x \\to \\infty} \\frac{1}{x^2} \\int_0^x f(t) \\, dt = 1,\n$$\nbut $f(x)/x$ has not a limit as $x \\to \\infty$.\nb) Let $f: (0, \\infty) \\to \\mathbb{R}$ be an increasing function such that\n$$\n\\lim_{x \\to \\infty} \\frac{1}{x^2} \\int_0^x f(t) \\, dt = 1.\n$$",
    "options": [],
    "answer": "a) One example is f(x) = 2x + 2x cos(x^2). b) Under the given conditions, f(x)/x → 2 as x → ∞.",
    "solution": "a) The function $f: (0, \\infty) \\to \\mathbb{R}$, $f(x) = 2x + 2x \\cos x^2$, is clearly continuous,\n$$\n\\lim_{x \\to \\infty} \\frac{1}{x^2} \\int_0^x f(t) \\, dt = \\lim_{x \\to \\infty} \\left( 1 + \\frac{\\sin x^2}{x^2} \\right) = 1,\n$$\nbut $f(x)/x$ has obviously not a limit as $x \\to \\infty$. Another example is the function $x \\mapsto 4x(\\cos x)^2$, $x > 0$; verifications are routine, and hence omitted.\n\nb) Let $g: (0, \\infty) \\to \\mathbb{R}$, $g(x) = x^{-2} \\int_0^x f(t) \\, dt - 1$, so $g(x) \\to 0$ as $x \\to \\infty$. Fix a small enough $\\varepsilon > 0$. Since $g(x) \\to 0$ as $x \\to \\infty$, if $x$ is large enough, then\n$$\n|g(x - \\varepsilon x)| < \\varepsilon^2, \\quad |g(x)| < \\varepsilon^2 \\quad \\text{and} \\quad |g(x + \\varepsilon x)| < \\varepsilon^2. \\quad (*)\n$$\nConsider such an $x$; since $f$ is increasing,\n$$\n\\frac{1}{\\varepsilon x} \\int_{x-\\varepsilon x}^{x} f(t) \\, dt \\le f(x) \\le \\frac{1}{\\varepsilon x} \\int_{x}^{x+\\varepsilon x} f(t) \\, dt.\n$$\nAlternatively, but equivalently, in terms of $g$,\n$$\n2 - \\varepsilon + \\frac{g(x)}{\\varepsilon} - (1 - \\varepsilon)^2 \\cdot \\frac{g(x - \\varepsilon x)}{\\varepsilon} \\le \\frac{f(x)}{x} \\le 2 + \\varepsilon + (1 + \\varepsilon)^2 \\cdot \\frac{g(x + \\varepsilon x)}{\\varepsilon} - \\frac{g(x)}{\\varepsilon},\n$$\nso, by (*),\n$$\n2 - 2\\varepsilon - \\varepsilon(1 - \\varepsilon)^2 < f(x)/x < 2 + 2\\varepsilon + \\varepsilon(1 + \\varepsilon)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38690,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, we have $m(\\angle A) < m(\\angle C)$. Let $E$ be a point on the bisector of angle $B$ such that $\\angle EAB = \\angle ACB$. Let $D$ be a point on line $BC$ such that $B \\in (CD)$ and $[BD] = [AB]$. Prove that the midpoint $M$ of the line segment $[AC]$ belongs to the line $DE$.\n\nBogdan Antohe\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of $[AD]$; it follows that $BN \\perp AD$. If $AP \\perp BE$, $P \\in BE$; then the quadrilateral $BNAP$ is a rectangle.\n\nSince the points $N$ and $P$ are the projections of $A$ on the interior and exterior bisectors of the angle $ABC$, respectively, it follows that the line $NP$ contains the midline parallel to $BC$, and hence $M \\in (NP)$.\n\nLet $\\{F\\} = AC \\cap BE$; from $\\triangle BAE \\sim \\triangle BCF$ it follows that $\\angle AEB = \\angle BFC$ and $\\angle AFE = \\angle AEB$, hence the triangle $AEF$ is isosceles, with $AE = AF$, and we deduce that $P$ is the midpoint of the line segment $[EF]$.\n\nIn the trapezoid $ADFE$, the points $N$ and $P$ are the midpoints of the bases, therefore the line $NP$ contains the point of intersection of trapezoid's diagonals $[AF]$ and $[DE]$. Since $\\{M\\} = AC \\cap NP$, it follows that $M \\in DE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two right isosceles triangles of legs equal to $1$ are glued together to form either an isosceles triangle – called *t-shape* – of leg $\\sqrt{2}$, or a parallelogram – called *p-shape* – of sides $1$ and $\\sqrt{2}$. Find all integers $m$ and $n$, $m, n \\ge 2$, such that a rectangle $m \\times n$ can be tiled with t-shapes and p-shapes.",
    "options": [],
    "answer": "All m and n at least two with at least one of m or n even.",
    "solution": "To this end, notice that $4$ t-shapes can be glued to produce a $2 \\times 2$ square, which is sufficient for tiling a rectangle with both sides even. If $m + n$ is odd, a $2 \\times 3$ rectangle can be obtained as below:\n\n![](attached_image_1.png)\n\nAlternatively, we can tile any rectangle $2 \\times n$, with $n \\ge 2$ tiles as follows:\n\n![](attached_image_2.png)\n\nObviously a $1 \\times n$ rectangle can not be tiled for any $n$. Any way one would position the tiles, the one that covers one of the vertices of the rectangle renders impossible the covering of the closest vertex.\nIt is left to prove that any odd sided rectangle cannot be tessellated. For this, color the unit squares of an $m \\times n$ rectangle with $m, n$ odd in a chessboard pattern, with the corners being black. On one hand, we have more black squares than white squares, on the other hand each tile covers a surface area that is half black, half white, so the total surface area covered by the tiles is half black, half white. In conclusion, the surface of a rectangle with both dimensions odd can not be tiled.\nIn conclusion, the tiling is possible if and only if $m, n \\ge 2$ and at least one dimension is even.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38692,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Show that in a right triangle with an angle of $30^\\circ$, the leg opposite to the $30^\\circ$-angle has half of the length of the hypotenuse.\n\nb) Inside the triangle $ABC$, with $m(\\angle A) = 100^\\circ$ and $m(\\angle B) = 20^\\circ$, we consider the point $D$, such that $m(\\angle DAB) = 30^\\circ$ and $m(\\angle DBA) = 10^\\circ$. Determine $m(\\angle ACD)$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "40°",
    "solution": "a) If $M$ is on the hypotenuse of the triangle $ABC$, with a right angle in $A$, with $m(\\angle B) = 30^\\circ$ such that $m(\\angle BAM) = 30^\\circ$, then the triangle $MAC$ is equilateral. Hence $MA = CM = CA = \\frac{1}{2}BC$, and $MB = \\frac{1}{2}BC$, also.\n\nb) We construct $D' \\in (AD)$, such that $CA = CD'$, $D' \\in \\text{Int}(ABC)$, $m(\\angle ACD') = 40^\\circ$, $m(\\angle D'CB) = 20^\\circ$.\nIf $CG \\perp AD'$, $G \\in (AD')$, and $D'F \\perp BC$, $F \\in (BC)$, then $\\triangle CD'G \\equiv \\triangle CD'F$ (H.A.), so $D'G = D'F$.\n\n$D'E \\perp AB$, $E \\in (BC)$, $G$ is the midpoint of $[AD']$, $m(\\angle D'AE) = 30^\\circ$, $D'F = D'G = D'E$, hence $D'$ is on the angle bisector of the angle $B$, $m(\\angle D'BA) = 10^\\circ$.\n$D' \\in (AD)$, $D' \\in (BD)$, $D' = D$, hence $m(\\angle ACD) = 40^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two finite sets. Find the number of functions $f : A \\to A$ satisfying the property that there exist two functions $g : A \\to B$ and $h : B \\to A$ such that $g(h(x)) = x$, $\\forall x \\in B$, and $h(g(x)) = f(x)$, $\\forall x \\in A$.",
    "options": [],
    "answer": "binom(|A|,|B|) * |B|^{|A|-|B|}",
    "solution": "Let $f$, $g$, $h$ be three functions satisfying the properties in the problem. Because $g \\circ h = 1_B$, it is clear that $g$ is surjective and $h$ is injective. As $f = h \\circ g$, we also get that $|B| = |\\text{Im } f| \\le |A|$.\n\nAs $f(f(x)) = h(g(h(g))) = h(g(x)) = f(x)$ we get $y \\in \\text{Im } f$, $f(y) = y$. From here, any function $f$ with the property from the problem has the following form\n$$\nf(x) = \\begin{cases} x, & \\text{if } x \\in \\text{Im } f \\\\ \\varphi(x), & \\text{if } x \\in A \\setminus \\text{Im } f \\end{cases}\n$$\nwhere $\\varphi : A \\setminus \\text{Im } f \\to \\text{Im } f$ is any function.\n\nLet $A' \\subseteq A$, such that $|A'| = |B|$ (supposing that $|B| \\le |A|$). We define\n$$\nf(x) = \\begin{cases} x, & \\text{if } x \\in A' \\\\ \\varphi(x), & \\text{if } x \\in A \\setminus A' \\end{cases}\n$$\nwhere $\\varphi : A \\setminus A' \\to A'$ is any function.\n\nNoticing that $\\text{Im } f = A'$ we prove that $f$ has the property from the problem. Let $h : B \\to A' \\subseteq A$ an arbitrary bijection and $g : A \\to B$ defined by\n$$\ng(x) = \\begin{cases} h^{-1}(x), & \\text{if } x \\in A' \\\\ h^{-1}(\\varphi(x)), & \\text{if } x \\in A \\setminus A'. \\end{cases}\n$$\nIt is easy to check that $g(h(x)) = x$, $\\forall x \\in B$, and $h(g(x)) = f(x)$, $\\forall x \\in A$.\n\nNow we can choose the subset $A'$ in $\\binom{|A|}{|B|}$ ways, and the function $\\varphi$ in $|B|^{|A|-|B|}$ ways, and thus, the number of functions with the requested property is $\\binom{|A|}{|B|} |B|^{|A|-|B|}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38694,
    "subject": "Mathematics (Multi-modal)",
    "question": "One considers the non-zero distinct digits $a, b, c, d, e, f$. Determine the positive integers $x$, such that $x$ divides any 6-digit number written with the digits $a, b, c, d, e, f$.",
    "options": [],
    "answer": "All such x must divide 9. Specifically: if the sum of the six digits is not divisible by 3, then only x = 1; if the sum is divisible by 3 but not by 9, then x = 1 or x = 3; if the sum is divisible by 9, then x = 1, x = 3, or x = 9.",
    "solution": "For any choice of 6 distinct non-zero digits, at least two of them are consecutive.\nIndeed, if $a < b < c < d < e < f$ and no digits are consecutive, then $b \\ge a + 2$, $c \\ge b + 2 \\ge a + 4$, $d \\ge c + 2 \\ge a + 6$, $e \\ge d + 2 \\ge a + 8$ and $f \\ge e + 2 \\ge a + 10$, which is impossible, since both $a$ and $f$ are digits.\nDenote by $m$ and $n$, $m > n$, two consecutive digits among $a, b, c, d, e, f$, and by $p, q, r, s$ the other four digits.\nWe have $x \\mid \\overline{pqrsmn}$ and $x \\mid \\overline{pqrsnm}$, hence $x \\mid \\overline{pqrsmn} - \\overline{pqrsnm} = 9(m-n)$. Since $m-n=1$, we find $x \\mid 9$.\nIf $3 \\nmid a + b + c + d + e + f$, the only possibility is $x = 1$.\nIf $3 \\mid a + b + c + d + e + f$ and $9 \\nmid a + b + c + d + e + f$, then $x = 1$ or $x = 3$.\nIf $9 \\mid a + b + c + d + e + f$, then $x = 1, x = 3$ or $x = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d \\in [0, 1]$. Prove that\n$$\n\\frac{a}{1+b} + \\frac{b}{1+c} + \\frac{c}{1+d} + \\frac{d}{1+a} + abcd \\le 3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We notice that\n$$\n\\frac{a}{1+b} + \\frac{b}{1+c} + \\frac{c}{1+d} + \\frac{d}{1+a} + abcd \\le \\frac{a}{1+abcd} + \\frac{b}{1+abcd} + \\frac{c}{1+abcd} + \\frac{d}{1+abcd} + abcd = \\frac{a+b+c+d}{1+abcd} + abcd.\n$$\nUsing repeatedly the inequality $x + y \\le 1 + xy$, $\\forall x, y \\in [0, 1]$ (equivalent to $(1-x)(1-y) \\ge 0$, therefore true), we get $a + b + c + d \\le 1 + ab + 1 + cd = ab + cd + 2 \\le 1 + abcd + 2 = abcd + 3$.\n\nReplacing $x = abcd \\in [0, 1]$, it is enough to prove that $1 + \\frac{2}{1+x} + x \\le 3$, that is $2 + x^2 + x \\le 2x + 2$, or $x(1-x) \\ge 0$, which is clearly true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$, erect triangles $ABD$ and $ACE$ externally, so that $\\widehat{ADB} = \\widehat{AEC} = 90^\\circ$ and $\\widehat{BAD} \\equiv \\widehat{CAE}$. Let $A_1 \\in BC$, $B_1 \\in AC$ and $C_1 \\in AB$ be the feet of the altitudes of the triangle $ABC$, and let $K$ and $L$ be the midpoints of $[BC_1]$ and $[CB_1]$, respectively. Prove that the circumcenters of the triangles $AKL$, $A_1B_1C_1$ and $DEA_1$ are collinear.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$, $P$ and $Q$ be the midpoints of $[BC]$, $[CA]$ and $[AB]$, respectively.\n\nThe circumcircle of triangle $A_1B_1C_1$ is the Euler circle. Point $M$ lies on this circle.\n\nIt is enough to prove now that $[A_1M]$ is a common chord of the three circles, $(A_1B_1C_1)$, $(AKL)$ and $(DEA_1)$.\n\nThe segments $[MK]$ and $[ML]$ are midlines of the triangles $BCC_1$ and $BCB_1$ respectively, hence $MK \\parallel CC_1 \\perp AB$ and $ML \\parallel BB_1 \\perp AC$. So, the circle $(AKL)$ has diameter $[AM]$ and therefore passes through $M$.\n\nFinally, we prove that the quadrilateral $DA_1ME$ is cyclic.\nFrom the cyclic quadrilaterals $ADBA_1$ and $AECA_1$, $\\overline{AA_1D} \\equiv \\overline{ABD}$ and $\\overline{AA_1E} \\equiv \\overline{ACE} \\equiv \\overline{ABD}$, so $\\overline{DA_1E} = 2\\overline{ABD} = 180^\\circ - 2\\overline{DAB}$.\nWe notice now that $DQ = AB/2 = MP$, $QM = AC/2 = PE$ and\n$$\n\\overline{DQM} = \\overline{DQB} + \\overline{BQM} = 2\\overline{DAB} + \\overline{BAC},\n$$\n$$\n\\overline{EPM} = \\overline{EPC} + \\overline{CPM} = 2\\overline{EAC} + \\overline{CAB},\n$$\nso $\\Delta MPE \\equiv \\Delta DQM$ (S.A.S.). This leads to $\\overline{DME} = \\overline{DMQ} + \\overline{QMP} + \\overline{PME} = \\overline{DMQ} + \\overline{BQM} + \\overline{QDM} = 180^\\circ - \\overline{DQB} = 180^\\circ - 2\\overline{DAB}$.\nSince $m(\\overline{DA_1E}) = m(\\overline{DME})$, the quadrilateral $DA_1ME$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$. An $n \\times n$ sieve is an $n \\times n$ array with $n$ cells removed so that exactly one cell is removed from every row and every column. A stick is a $1 \\times k$ or $k \\times 1$ array for any positive integer $k$. For any sieve $A$, let $m(A)$ be the minimal number of sticks required to partition $A$. Find all possible values of $m(A)$, as $A$ varies over all possible $n \\times n$ sieves.\n\nPalmer Mebane, U.S.A., and Nikolai Beluhov, Bulgaria",
    "options": [],
    "answer": "2n - 2",
    "solution": "By *holes* we mean the cells which are cut out from the board. The *cross* of a hole in $A$ is the union of the row and the column through that hole.\n\nArguing indirectly, consider a dissection of $A$ into $2n - 3$ or fewer sticks. Horizontal sticks are all labeled $h$, and vertical sticks are labeled $v$; $1 \\times 1$ sticks are both horizontal and vertical, and labeled arbitrarily. Each cell of $A$ inherits the label of the unique containing stick.\n\nAssign each stick in the dissection to the cross of the unique hole on its row, if the stick is horizontal; on its column, if the stick is vertical.\n\nSince there are at most $2n - 3$ sticks and exactly $n$ crosses, there are two crosses each of which is assigned to at most one stick in the dissection. Let the crosses be $c$ and $d$, centered at $a = (x_a, y_a)$ and $b = (x_b, y_b)$, respectively, and assume, without loss of generality, $x_a < x_b$ and $y_a < y_b$. The sticks covering the cells $(x_a, y_b)$ and $(x_b, y_a)$ have like labels, for otherwise one of the two crosses would be assigned to at least two sticks. Say the common label is $v$, so each of $c$ and $d$ contains a stick covering one of those two cells. It follows that the lower (respectively, upper) arm of $c$ (respectively, $d$) is all-$h$, and the horizontal arms of both crosses are all-$v$, as illustrated below.\n\n![](attached_image_1.png)\n\nAll other columns contain at least one $v$-stick each. In addition, all rows below $a$ and all rows above $b$ contain at least one $h$-stick each. This amounts to a total of at least $2(y_b - y_a - 1) + (n - y_b + y_a + 1) + (n - y_b) + (y_a - 1) = 2n - 2$ sticks – a contradiction.\n\n**Remark.** The solution may equally well be concluded as follows. Since $c$ and $d$ are proved to contain one stick each, there is a third cross $e$ centered at $(x_*, y_*)$ also containing at most one stick. It meets the horizontal arms of $c$ and $d$ at two $v$-cells, so the cells where two of the three crosses meet are all labeled $v$. Assuming, without loss of generality, $y_a < y_* < y_b$, it follows that both vertical arms of $e$ contain $v$-cells, so $e$ is assigned to two different $v$-sticks – a contradiction.\n\n\n*Second solution.* (Ilya Bogdanov) We provide a different proof that $m(A) \\ge 2n - 2$.\n\nCall a stick *vertical* if it is contained in some column, and *horizontal* if it is contained in some row; $1 \\times 1$ sticks may be called arbitrarily, but any of them is supposed to have only one direction. Assign to each vertical/horizontal stick the column/row it is contained in. If each row and each column is assigned to some stick, then there are at least $2n$ sticks, which is even more than we want. Thus we assume, without loss of generality, that some *exceptional* row $R$ is not assigned to any stick. This means that all $n-1$ existing cells in $R$ belong to $n-1$ distinct vertical sticks; call these sticks *central*.\n\nNow we mark $n-1$ cells on the board in the following manner. (↓) For each hole $c$ below $R$, we mark the cell just under $c$; (↑) for each hole $c$ above $R$, we mark the cell just above $c$; and (●) for the hole $r$ in $R$, we mark both the cell just above it and just below it. We have described $n + 1$ cells, but exactly two of them are out of the board; so $n - 1$ cells are marked within the board. A sample marking is shown in the figure below, where the marked cells are crossed.\n\n![](attached_image_2.png)\n\nNotice that all the marked cells lie in different rows, and all of them are marked in different columns, except for those two marked for (●); but the latter two have a hole $r$ between them. So no two marked cells may belong to the same stick. Moreover, none of them lies in a central stick, since the marked cells are separated from $R$ by the holes. Thus the marked cells should be covered by $n - 1$ different sticks (call them *border*) which are distinct from the central sticks. This shows that there are at least $(n-1) + (n-1) = 2n-2$ distinct sticks, as desired.\n\n\n*Third solution.* To prove $m(A) \\ge 2n-2$, it is sufficient to show that there are $2n-2$ cells in $A$, no two of which may be contained in the same stick.\n\nTo this end, consider the bipartite graph $G$ with parts $G_h$ and $G_v$, where the vertices in $G_h$ (respectively, $G_v$) are the $2n-2$ maximal sticks $A$ is dissected into by all horizontal (respectively, vertical) grid lines, two sticks being joined by an edge in $G$ if and only if they share a cell.\n\nWe show that $G$ admits a perfect matching by proving that it fulfils the condition in Hall's theorem; the $2n-2$ cells corresponding to the edges of this matching form the desired set. It is sufficient to show that every subset $S$ of $G_h$ has at least $|S|$ neighbours (in $G_v$, of course).\n\nLet $L$ be the set of all sticks in $S$ that contain a cell in the leftmost column of $A$, and let $R$ be the set of all sticks in $S$ that contain a cell in the rightmost column of $A$;\nlet $\\ell$ be the length of the longest stick in $L$ (zero if $L$ is empty), and let $r$ be the length of the longest stick in $R$ (zero if $R$ is empty).\nSince every row of $A$ contains exactly one hole, $L$ and $R$ partition $S$; and since every column of $A$ contains exactly one hole, neither $L$ nor $R$ contains two sticks of the same size, so $\\ell \\ge |L|$ and $r \\ge |R|$, whence $\\ell + r \\ge |L| + |R| = |S|$.\nIf $\\ell + r \\le n$, we are done, since there are at least $\\ell + r \\ge |S|$ vertical sticks covering the cells of the longest sticks in $L$ and $R$. So let $\\ell + r > n$, in which case the sticks in $S$ span all $n$ columns, and notice that we are again done if $|S| \\le n$, to assume further $|S| > n$.\nLet $S' = G_h \\setminus S$, let $T$ be set of all neighbours of $S$, and let $T' = G_v \\setminus T$. Since the sticks in $S$ span all $n$ columns, $|T| \\ge n$, so $|T'| \\le n - 2$. Transposition of the above argument (replace $S$ by $T'$), shows that $|T'| \\le |S'|$, so $|S| \\le |T|$.\n\n**Remark.** The case $|S| > n$ may equally well be dealt with as follows. Add to $S$ two *empty sticks* formally present to the left (respectively, right) of the leftmost (respectively, rightmost) hole. Then there are at least $|S| - n + 2$ rows containing two sticks from $S$, so two of these rows are separated by at least $|S| - n$ other rows. Each hole in these $|S| - n$ rows separates two vertical sticks from $G_v$ both of which are neighbours of $S$. Consequently, $S$ has at least $n + (|S| - n) = |S|$ neighbours.\n\n\n*Fourth solution.* Induct on $n \\ge 2$ to prove that $m(A) \\ge 2n - 2$. The base cases $n = 2$ and $n = 3$ are readily dealt with, so let $n > 3$ and consider any dissection of $A$ into sticks. Define the *cross* of a hole as in Solution 1, and notice that each stick is contained in some cross.\n\nIf the dissection contains more than $n$ sticks, some cross contains at least two sticks. Remove such a cross from the sieve and glue pieces together along corresponding edges to form an $(n-1) \\times (n-1)$ sieve. The dissection of the original sieve induces a dissection of the new sieve: Upon removal, a stick may split into two substicks that glue back together to form a stick in the new sieve. After this operation has been performed, the number of sticks decreases by at least 2, and since by the induction hypothesis the number of sticks in the new dissection is at least $2n-4$, the initial dissection contains at least $(2n-4)+2 = 2n-2$ sticks.\n\nThere are several different ways to rule out the case where the dissection contains at most $n$ sticks. For instance, removal of a cross containing some stick. The induced dissection of the resulting $(n-1) \\times (n-1)$ sieve contains at most $n-1$ sticks, which is impossible by the induction hypothesis, since $n-1 < 2(n-1) - 2$.\n\n\nTherefore, for any $n \\times n$ sieve $A$, $m(A) = 2n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38698,
    "subject": "Mathematics (Multi-modal)",
    "question": "A planar country has an odd number of cities separated by pairwise distinct distances. Some of these cities are connected by direct two-way flights. Each city is directly connected to exactly two other cities, and the latter are located farthest from it. Prove that, using these flights, one may go from any city to any other city.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the graph $G$ whose vertices are the cities and whose edges are the direct two-way flights. We must show that $G$ is connected.\n\nAssuming $G$ has at least two components, we show that the cardinality of each component is even; this contradicts the fact that the total number of vertices is odd, whence the conclusion.\n\n**Lemma.** If $AB$ and $XY$ are edges in different components of $G$, then the segments $AB$ and $XY$ cross at an interior point.\n\n**Proof.** Two cases are to be ruled out:\n\n(1) The lines $AB$ and $XY$ meet outside both open segments $(AB)$ and $(XY)$, possibly at an ideal point; that is, after a suitable relabelling (if necessary), $ABXY$ is a convex (possibly degenerate) quadrangle.\nIn this case, the segments $AX$ and $BY$ cross at some point $O$, so $AX + BY = (OA+OX)+(OB+OY) = (OA+OB)+(OX+OY) \\ge AB+XY$. Consequently, either $AX \\ge AB$, in which case $A$ is connected to $X$, contradicting the fact that the two lie in different components of $G$; or $BY \\ge XY$, in which case $Y$ is connected to $B$, contradicting the fact that the two lie in different components of $G$.\n\n(2) The lines $AB$ and $XY$ cross at an interior point of exactly one of open segments $(AB)$ and $(XY)$, say, the latter; that is, after a suitable relabelling (if necessary), the segment $AB$ lies inside the triangle $AXY$.\nIn this case, $AB < \\max(AX, AY)$, since the segment $AB$ lies along a cevian in the triangle $AXY$, so $A$ is connected to (exactly) one of $X$ and $Y$, contradicting the fact that $A$ and $X$, $Y$ lie in different components of $G$.\n\nBack to the problem, since every vertex has degree $2$, each component of $G$ is a circuit. Let $C$ be one such and let $XY$ be an edge of $G$ whose endpoints do not belong to $C$. By the lemma, every two consecutive vertices of $C$ lie on opposite sides of the line $XY$, so $C$ has an even number of vertices. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the integers $x$ such that $2^x + x^2 + 25$ is the cube of a prime number.",
    "options": [],
    "answer": "6",
    "solution": "Let $y$ be a prime such that $y^3 = 2^x + x^2 + 25$; clearly, $y$ is odd. Moreover, $x$ can not be a negative integer nor can it belong to the set $\\{0, 1, 2, 3\\}$. Hence $x \\ge 4$ is even.\n\n1. If $x = 6k$, with $k \\in \\mathbb{N}^*$, we get $64^k + 36k^2 + 25 = y^3$.\n\nFor $k=1$ we get $x=6$ and $y=5$.\n\nFor $k \\ge 2$ we have no solutions because of the following inequalities:\n$$\n(4^k)^3 < y^3 = 2^{6k} + (6k)^2 + 25 < (4^k + 1)^3,\n$$\nthe second inequality being equivalent with $3k^2 + 2 < 4^{k-1}(4^k + 1)$, which is true (induction).\n\n2. If $x = 6k + 2$ or $x = 6k + 4$ then: $2^x \\equiv 1 \\pmod 3$, $x^2 \\equiv 1 \\pmod 3$ hence $2^x + x^2 + 25 \\equiv 0 \\pmod 3$, i.e. $y^3 \\equiv 0 \\pmod 3$, which means $y \\equiv 0 \\pmod 3$ and, as $y$ is prime, we get $y = 3$, which does not fulfill the condition.\n\nIn conclusion, the only solution is $x = 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be any convex quadrangle and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrangles, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.\n\nNikolai Beluhov, Bulgaria",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma 1.** Let $PQRS$ be a convex quadrangle whose diagonals cross at $O$. Let $\\omega_1$ and $\\omega_2$ be the circles on diameters $PQ$ and $RS$, respectively, and let $\\ell$ be their radical axis. Finally, let $A, B$, and $C$ be points outside this quadrangle so that: the point $P$ (respectively, $Q$) lies on the segment $AB$ (respectively, $BC$); and $AO \\perp PS$, $BO \\perp PQ$, and $CO \\perp QR$. Then the three lines $AC, PQ$, and $\\ell$ are concurrent or parallel.\n\n**Proof.** Assume first that the lines $PR$ and $QS$ are not perpendicular. Let $H_1$ and $H_2$ be the orthocentres of the triangles $OSP$ and $OQR$, respectively; notice that $H_1$ and $H_2$ do not coincide.\nSince $H_1$ is the radical center of the circles on diameters $RS, SP$, and $PQ$, it lies on $\\ell$. Similarly, $H_2$ lies on $\\ell$, so the lines $H_1H_2$ and $\\ell$ coincide.\nThe corresponding sides of the triangles $APH_1$ and $CQH_2$ meet at $O, B$, and the orthocenter of the triangle $OPQ$ (which lies on $OB$). By Desargues' theorem, the lines $AC, PQ$ and $\\ell$ are concurrent or parallel.\nThe case when $PR \\perp QS$ may be considered as a limit case, since the configuration in the statement of the lemma allows arbitrarily small perturbations. The lemma is proved.\n\nBack to the problem, let the segments $PR$ and $QS$ cross at $O$, let $\\omega_1$ and $\\omega_2$ be the circles on diameters $PQ$ and $RS$, respectively, and let $\\ell$ be their radical axis. By the Lemma, the three lines $AC, \\ell$, and $PQ$ are concurrent or parallel, and similarly so are the three lines $AC, \\ell$, and $RS$. Thus, if the lines $AC$ and $\\ell$ are distinct, all four lines are concurrent or pairwise parallel.\nThis is clearly the case if the lines $PS$ and $QR$ are not parallel, since $\\ell$ crosses $OA$ and $OC$ at the orthocentres of $OSP$ and $OQR$, and these orthocentres are distinct from $A$ and $C$. In this case, denote the concurrency point by $T$. If $T$ is not ideal, then\n$$\nTP \\cdot TQ = TR \\cdot TS,\n$$\nsince $T$ lies on $\\ell$, so $PQRS$ is cyclic. If $T$ is ideal (i.e., all four lines are parallel), then the segments $PQ$ and $RS$ have the same perpendicular bisector (namely, the line of centers of $\\omega_1$ and $\\omega_2$), and $PQRS$ is cyclic again.\n\n![](attached_image_1.png)\n\nAssume now $PS$ and $QR$ parallel. By symmetry, $PQ$ and $RS$ may also be assumed parallel: otherwise, the preceding argument goes through after relabeling. In this case, we need to prove that the parallelogram $PQRS$ is a rectangle.\nSuppose, by way of contradiction, that $OP > OQ$. Let the line through $O$ and parallel to $PQ$ meet $AB$ at $M$, and $CB$ at $N$. Since $OP > OQ$, the angle $SPQ$ is acute and the angle $PQR$ is obtuse, so the angle $AOB$ is obtuse, the angle $BOC$ is acute, $M$ lies on the segment $AB$, and $N$ lies on the extension of the segment $BC$ beyond $C$. Therefore: $OA > OM$, since the angle $OMA$ is obtuse; $OM > ON$, since\n$$\nOM : ON = KP : KQ,\n$$\nwhere $K$ is the projection of $O$ onto $PQ$; and $ON > OC$, since the angle $OCN$ is obtuse. Consequently, $OA > OC$.\n\n![](attached_image_2.png)\n\nSimilarly, $OR > OS$ yields $OC > OA$: a contradiction. Consequently, $OP = OQ$ and $PQRS$ is a rectangle. This ends the proof.\n\n![](attached_image_3.png)\n*Second solution.* (Ilya Bogdanov) To begin, we establish a simple, but useful lemma.\n\n**Lemma 2.** If $P$ is a point on the side $AB$ of a triangle $OAB$, then\n$$\n\\frac{\\sin AOP}{OB} + \\frac{\\sin POB}{OA} = \\frac{\\sin AOB}{OP}.\n$$\n**Proof.** Write $0 = 2(\\text{area } AOB - \\text{area } POB - \\text{area } POA) = OA \\cdot OB \\cdot \\sin AOB - OB \\cdot OP \\cdot \\sin POB - OP \\cdot OA \\cdot \\sin AOP$, and divide by $OA \\cdot OB \\cdot OP$ to get the required identity.\nA similar statement holds if the point $P$ lies on the line $AB$; the proof is obtained by using signed areas and directed lengths.\n\nWe now turn to the solution proper and prove the kind of converse below.\n\n**Claim.** Let $PQRS$ be a cyclic quadrangle with $O = PR \\cap QS$; assume that no its diagonal is perpendicular to a side. Let $\\ell_A, \\ell_B, \\ell_C$, and $\\ell_D$ be the lines through $O$ perpendicular to $SP, PQ, QR, \\text{ and } RS$, respectively. Choose any point $A \\in \\ell_A$ and successively define $B = AP \\cap \\ell_B, C = BQ \\cap \\ell_C, D = CR \\cap \\ell_D$, and $A' = DS \\cap \\ell_A$. Then $A' = A$.\n\n![](attached_image_4.png)\n\n**Proof.** We restrict ourselves to the case when the points $A, B, C, D$, and $A'$ lie on $\\ell_A, \\ell_B, \\ell_C, \\ell_D$, and $\\ell_A$ on the same side of $O$ as their points of intersection with the respective sides of the quadrilateral $PQRS$. Again, a general case is obtained by suitable consideration of directed lengths. Write\n$$\n\\begin{align*}\n\\alpha &= \\angle QPR = \\angle QSR = \\pi/2 - \\angle POB = \\pi/2 - \\angle DOS, \\\\\n\\beta &= \\angle RPS = \\angle RQS = \\pi/2 - \\angle AOP = \\pi/2 - \\angle QOC, \\\\\n\\gamma &= \\angle SQP = \\angle SRP = \\pi/2 - \\angle BOQ = \\pi/2 - \\angle ROD, \\\\\n\\delta &= \\angle PRQ = \\angle PSQ = \\pi/2 - \\angle COR = \\pi/2 - \\angle SOA,\n\\end{align*}\n$$\nand apply Lemma 2 to the lines $APB, PQC, CRD$, and $DSA'$ to get\n$$\n\\begin{aligned}\n\\frac{\\sin(\\alpha + \\beta)}{OP} &= \\frac{\\cos \\alpha}{OA} + \\frac{\\cos \\beta}{OB}, &\n\\frac{\\sin(\\beta + \\gamma)}{OQ} &= \\frac{\\cos \\beta}{OB} + \\frac{\\cos \\gamma}{OC}, \\\\\n\\frac{\\sin(\\gamma + \\delta)}{OR} &= \\frac{\\cos \\gamma}{OC} + \\frac{\\cos \\delta}{OD}, &\n\\frac{\\sin(\\delta + \\alpha)}{OS} &= \\frac{\\cos \\delta}{OD} + \\frac{\\cos \\alpha}{OA'}.\n\\end{aligned}\n$$\nAdding the two equalities on the left and subtracting the two on the right, the required equality $A = A'$ (i.e., $\\cos \\alpha/OA = \\cos \\alpha/OA'$, in view of $\\cos \\alpha \\neq 0$) is equivalent to\n$$\n\\frac{\\sin QPS}{OP} + \\frac{\\sin SRQ}{OR} = \\frac{\\sin PQR}{OQ} + \\frac{\\sin RSP}{OS}.\n$$\nLet $d$ denote the circumdiameter of $PQRS$, so $\\sin QPS = \\sin SRQ = QS/d$ and $\\sin RSP = \\sin PQR = PR/d$. Thus the required relation reads\n$$\n\\frac{QS}{OP} + \\frac{QS}{OR} = \\frac{PR}{OS} + \\frac{PR}{OQ}, \\quad \\text{or} \\quad \\frac{QS \\cdot PR}{OP \\cdot OR} = \\frac{PR \\cdot QS}{OS \\cdot OQ}.\n$$\nThe latter holds since $PQRS$ is cyclic.\n\n![](attached_image_5.png)\n\nFinally, we use the Claim to complete the proof. Assume that $PQRS$ is not cyclic, e.g., that $OP \\cdot OR > OQ \\cdot OS$, where $O = PR \\cap QS$. Mark the point $S'$ on the ray $OS$ so that $OP \\cdot OR = OQ \\cdot OS'$. Notice that no diagonal of $PQRS$ is perpendicular to a side, so the quadrangle $PQRS'$ satisfies the conditions of the claim.\nLet $\\ell'_A$ and $\\ell'_D$ be the lines through $O$ perpendicular to $PS'$ and $RS'$, respectively. Then $\\ell'_A$ and $\\ell'_D$ cross the segments $AP$ and $RD$, respectively, at some points $A'$ and $D'$. By the Claim, the line $A'D'$ passes through $S'$. This is impossible, because the segment $A'D'$ crosses the segment $OS$ at some interior point, while $S'$ lies on the extension of this segment. This contradiction completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $AB \\neq AC$, let $G$ be its centroid, and let $H$ be its orthocenter. Let $D$ be the orthogonal projection of $A$ on the line $BC$, and let $M$ be the midpoint of the side $BC$. The circle $ABC$ crosses the ray $MH$ emanating from $M$ at $P$, and the ray $DG$ emanating from $D$ at $Q$, outside the segment $DG$. Show that the lines $DP$ and $MQ$ meet on the circle $ABC$.\nBMO 2017 Shortlist",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* We show that the angles $APD$ and $AQM$ are either equal or one is the supplement of the other.\nTo begin, recall that the reflection of $H$ across $M$ is the antipode of $A$ in the circle $ABC$, to infer that the angle $APM$ is right, so $A$, $D$, $M$, $P$ are concyclic. Hence the angles $APD$ and $AMD$ are either equal or one is the supplement of the other.\nThe circle $ABC$ is the image of the nine-point circle under a homothety of scale factor $-2$ centred at $G$. This homothety sends $M$ to $A$ and $D$ to $Q$, so $AQ$ and $BC$ are parallel.\nHence $\\angle AMD = \\angle MAQ$, on the one hand; and $MA = MQ$, on the other, so $\\angle MAQ = \\angle AQM$. Consequently, $\\angle AMD = \\angle AQM$, and the conclusion follows by the preceding paragraph.\n\n![](attached_image_1.png)\n\n\n![](attached_image_2.png)\n*Second solution.* As in the previous solution, $A$ and $Q$ are reflections of one another in the perpendicular bisectrix of the segment $BC$.\nLet the line $MP$ cross the circle $ABC$ again at $P'$, and let the lines $QP'$ and $BC$ cross at $S$. Since $P'$ is the antipode of $A$ in the circle $ABC$, the angle $AQS = AQP'$ is right, so the lines $QS$ and $BC$ are perpendicular.\nLet the line $MQ$ cross the circle $ABC$ again at $Q'$, let the lines $PQ'$ and $BC$ cross at $R$, and refer to the butterfly theorem to infer that $R$ and $S$ are reflections of one another across $M$.\nThe lines $AR$ and $QS$ are therefore reflections of one another in the perpendicular bisectrix of the segment $BC$.\nFinally, since $QS$ is perpendicular to $BC$, so is $AR$, and, consequently, the points $R$ and $D$ coincide. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the positive divisors of no integer greater than $1$ may be placed in the cells of a rectangular array so that the four conditions below be simultaneously fulfilled:\n(a) each cell contains exactly one divisor;\n(b) distinct cells contain distinct divisors;\n(c) the sum of the divisors on each row is the same; and\n(d) the sum of the divisors on each column is the same.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, if possible, that the positive divisors of some integer $n > 1$ may be arranged as required in an $k \\times \\ell$ rectangular array, where $k$ is the number of rows, and $\\ell$ is the number of columns; clearly, $k$ and $\\ell$ must both be greater than $1$.\nLet $s$ be the common value of the row sums, and notice that $s \\ge n + 1$.\nLet $d_i$ be the largest divisor on the $i$-th row, $i = 1, \\dots, k$. Assume, without any loss, $d_1 > \\dots > d_k$, to infer that the $n/d_i$, $i = 1, \\dots, k$, form a strictly increasing $k$-element string of positive integers, so $n/d_k \\ge k$; that is, $d_k \\le n/k$.\nSince $d_k$ is maximal along the $k$-th row,\n$$\nd_k \\ge s/\\ell > n/\\ell,\n$$\nso $\\ell > k$, by the preceding (in fact, $d_k > s/\\ell$, since the divisors are pairwise distinct).\nMutatis mutandis, $k > \\ell$, and we reach a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $n$ such that, for any coloring of the elements of the set $\\{2, 3, \\dots, n\\}$ with two colors, the equation $x + y = z$ has a monochrome solution with $x \\neq y$. (We say that the equation $x + y = z$ has a monochrome solution if there exist $a, b, c$ distinct, having the same color, such that $a + b = c$.)",
    "options": [],
    "answer": "13",
    "solution": "For $n = 12$ there exists a coloring of the numbers from $\\{2, 3, \\dots, 12\\}$ with two colors such that the equation $x + y = z$ has no monochrome solution: we color the numbers from $A = \\{2, 3, 4, 11, 12\\}$ with one color, and the elements of $B = \\{5, 6, 7, 8, 9, 10\\}$ with the second color.\nFor $n < 12$ we color with the first color the numbers from $\\{2, 3, \\dots, n\\} \\cap A$ and with the second one the elements of $\\{2, 3, \\dots, n\\} \\cap B$. Again, the equation $x + y = z$ has no monochrome solution.\nIt follows that $n \\ge 13$.\nWe prove that, for any coloring with two colors of the elements of the set $\\{2, 3, \\dots, 13\\}$, the equation $x + y = z$ has monochrome solutions.\nAssume the contrary: there exists a coloring such that there is no monochrome solution to the equation.\n**Case 1:** If $2$, $3$, $4$ have color 1, then $5 = 2+3$, $6 = 2+4$, $7 = 3+4$ need to have color 2, hence $11 = 5+6$, $12 = 5+7$, $13 = 6+7$ need to be of color 1. But then $2+11 = 13$, and the equation has a monochrome solution (of color 1).\n\n**Case 2:** If $2$ and $3$ are of color 1 while $4$ has color 2, then $5 = 2+3$ has color 2, and $9 = 4+5$ has color 1. But $3+6 = 9$, and numbers $3$ and $9$ have color 1, hence $6$ needs to be of color 2. Similarly, as $2+7 = 9$ and $2$, $9$ have color 1, it follows that $7$ has color 2. If $11$ has color 1, then $2+9 = 11$ leads to a monochrome solution. If $11$ has color 2, then $5+6 = 11$ is a monochrome solution.\n\n**Case 3:** If $2$ and $4$ have color 1 while $3$ has color 2, then $6$ has color 2, $9$ has color 1. As $4$ and $9$ have color 1, we need $5$ to have color 2. Similarly, $2$ and $9$ are of color 1, therefore $7$ has color 2, $8 = 3+5$ has color 1. But $12 = 4+8 = 5+7$, hence we have a monochrome solution (of color 1 or color 2, depending on the color of $12$).\n\n**Case 4:** $2$ has color 1, $3$ and $4$ have color 2. Then $7$ has color 1. But $2$ and $7$ having color 1 means that $5$ needs to be of color 2. This leads to $9 = 2+7 = 4+5$ and, again, regardless on the color of $9$, the equation has a monochrome solution.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, determine the maximum value the sum\n$$\na_1/a_2 + a_2/a_3 + \\dots + a_{n-1}/a_n\n$$\nmay achieve, and the points at which the maximum is achieved, as $a_1, \\dots, a_n$ run over all of positive real numbers subject to\n$$\na_k \\ge a_1 + \\dots + a_{k-1}, \\quad k = 2, \\dots, n.\n$$",
    "options": [],
    "answer": "Maximum value: n/2. Equality holds if and only if a_k = 2^{k-2} a_1 for k = 2, ..., n, with any a_1 > 0.",
    "solution": "*First solution.* The required maximum is $n/2$ and is achieved if and only if $a_k = 2^{k-2}a_1$, $k = 2, \\dots, n$, and $a_1$ is any positive real number.\nTo prove this, let $A_k = a_1 + \\dots + a_k$, $k = 0, \\dots, n-1$, where empty sums are zero, and refer to the condition in the statement, $A_k \\le a_{k+1}$, $k = 0, \\dots, n-1$, to write\n$$\n\\begin{align*}\n\\sum_{k=1}^{n-1} \\frac{a_k}{a_{k+1}} &= \\sum_{k=1}^{n-1} \\frac{A_k - A_{k-1}}{a_{k+1}} = \\sum_{k=1}^{n-2} A_k \\left( \\frac{1}{a_{k+1}} - \\frac{1}{a_{k+2}} \\right) + \\frac{A_{n-1}}{a_n} \\\\\n&\\le \\sum_{k=1}^{n-2} a_{k+1} \\left( \\frac{1}{a_{k+1}} - \\frac{1}{a_{k+2}} \\right) + 1 = \\sum_{k=1}^{n-2} \\left( 1 - \\frac{a_{k+1}}{a_{k+2}} \\right) + 1 \\\\\n&= n - 1 + \\frac{a_1}{a_2} - \\sum_{k=1}^{n-1} \\frac{a_k}{a_{k+1}} \\le n - \\sum_{k=1}^{n-1} \\frac{a_k}{a_{k+1}}.\n\\end{align*}\n$$\nConsequently, $\\sum_{k=1}^{n-1} a_k / a_{k+1} \\le n/2$, and equality holds if and only if $A_k = a_{k+1}$, $k = 1, \\dots, n-1$, which is clearly the case if and only if $a_k = 2^{k-2}a_1$, $k = 2, \\dots, n$, and $a_1$ is any positive real number.\n\n\n*Second solution.* We now show by induction on $n \\ge 2$ that $s_n(a_1, \\dots, a_n) = a_1/a_2 + a_2/a_3 + \\dots + a_{n-1}/a_n \\le n/2$ for all positive real numbers $a_1, \\dots, a_n$ such that $a_k \\ge a_1 + \\dots + a_{k-1}$, $k = 2, \\dots, n$, and equality holds if and only if $a_k = 2^{k-2}a_1$, $k = 2, \\dots, n$, and $a_1$ is any positive real number.\nClearly, $s_2(a_1, a_2) \\le 1$ if $0 < a_1 \\le a_2$, and $s_3(a_1, a_2, a_3) \\le s_3(a_1, a_2, a_1+a_2) \\le 3/2$ if $0 < a_1 \\le a_2$ and $a_1 + a_2 \\le a_3$; in both cases, equality holds if and only if the $a_k$ are as stated.\nNow let $n \\ge 4$, let $a_1, \\dots, a_n$ be positive real numbers such that $a_k \\ge a_1 + \\dots + a_{k-1}$, $k = 2, \\dots, n$, and write\n$$\n\\begin{align*}\ns_n(a_1, \\dots, a_n) &\\le s_n(a_1, \\dots, a_{n-1}, a_1 + \\dots + a_{n-1}) \\\\\n&= s_{n-2}(a_1, \\dots, a_{n-2}) + \\frac{a_{n-2}}{a_{n-1}} + \\frac{a_{n-1}}{a_1 + \\dots + a_{n-1}}.\n\\end{align*}\n$$\nIf we show that\n$$\n\\frac{a_{n-2}}{a_{n-1}} + \\frac{a_{n-1}}{a_1 + \\cdots + a_{n-1}} \\le \\frac{a_{n-2}}{a_1 + \\cdots + a_{n-2}} + \\frac{1}{2}, \\quad (*)\n$$\nthen $s_n(a_1, \\dots, a_n) \\le s_{n-1}(a_1, \\dots, a_{n-2}, a_1 + \\dots + a_{n-2}) + 1/2 \\le (n-1)/2 + 1/2 = n/2$, by the induction hypothesis; the cases of equality also follow from the induction hypothesis.\nFinally, to establish (*), simply notice that the numerator of the right-hand member and the left-hand member is the product of $a_{n-1} - a_1 - \\dots - a_{n-2} \\ge 0$ and $a_{n-1}(a_{n-2} - a_1 - \\dots - a_{n-3}) + 2a_{n-2}(a_1 + \\dots + a_{n-2}) \\ge 2a_{n-2}(a_1 + \\dots + a_{n-2}) > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38705,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. On the line segments $EF$, $FD$, and $DE$, consider the points $M$, $N$, and $P$ respectively such that the sums $BM + MC$, $CN + NA$, and $AP + PB$ are minimum.\n\na) Prove that the lines $AM$, $BN$, and $CP$ are concurrent.\n\nb) Prove that $DM$, $EN$ and $FP$ are the altitudes of triangle $DEF$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $S$ be the reflection of $C$ across the line $EF$. As $AE = AF$, it follows that $\\angle AFE = \\angle AEF$, hence $\\angle BFE = \\angle FEC = \\angle FES$, therefore $FB$ and $ES$ are parallel. $FBES$ is a trapezoid (or a parallelogram); let $M$ be the intersection point of its diagonals. According to the billiards problem, $M$ is the point of the line $EF$ for which the sum $BM + MC$ is minimum. Triangles $BFM$ and $SEM$ are similar, therefore $\\frac{FM}{EM} = \\frac{FB}{ES} = \\frac{FB}{EC} = \\frac{s-b}{s-c}$. It follows that $\\frac{\\sin(FAM)}{\\sin(EAM)} = \\frac{[FAM]}{[EAM]} = \\frac{FM}{EM} = \\frac{s-b}{s-c}$. Multiplying this with the two analogous relations we get\n$$\n\\frac{\\sin(FAM)}{\\sin(EAM)} \\cdot \\frac{\\sin(ECP)}{\\sin(DCP)} \\cdot \\frac{\\sin(DBN)}{\\sin(NBF)} = 1.\n$$\nBy the converse of the trigonometric form of Ceva's theorem, it follows that the lines $AM$, $BN$, and $CP$ are concurrent.\n\n\nb) We have $\\frac{FM}{EM} = \\frac{p-b}{p-c} = \\frac{BD}{CD} = \\frac{BF}{CE}$, hence, from a well-known problem (\"The gliding bisector Theorem\") it follows that $DM$ is parallel to the bisector of angle $\\angle BAC$. The bisector is perpendicular to $EF$, therefore $DM$ is also perpendicular to $EF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be non-negative real numbers satisfying $a + b + c + d = 3$. Prove that\n$$\n\\frac{a}{1 + 2b^3} + \\frac{b}{1 + 2c^3} + \\frac{c}{1 + 2d^3} + \\frac{d}{1 + 2a^3} \\ge \\frac{a^2 + b^2 + c^2 + d^2}{3}.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds precisely for the cyclic permutations of the quadruples (3, 0, 0, 0) and (2, 1, 0, 0).",
    "solution": "*First solution.* We notice the equality case $a = 3$, $b = c = d = 0$ and we try to mix the variables as follows:\n$$\nf(a, b, c, d) \\ge f(a + b + c + d, 0, 0, 0) = a + b + c + d - \\frac{(a + b + c + d)^2}{3} = 0,\n$$\nwhere\n$$\nf(a, b, c, d) = \\frac{a}{1 + 2b^3} + \\frac{b}{1 + 2c^3} + \\frac{c}{1 + 2d^3} + \\frac{d}{1 + 2a^3} - \\frac{a^2 + b^2 + c^2 + d^2}{3}.\n$$\nThen\n$$\nf(a, b, c, d) - a - b - c - d + \\frac{(a + b + c + d)^2}{3} =\n$$\n$$\n\\frac{a}{1 + 2b^3} + \\frac{b}{1 + 2c^3} + \\frac{c}{1 + 2d^3} + \\frac{d}{1 + 2a^3} - a - b - c - d + \\frac{2(ab + bc + cd + da + ac + bd)}{3} \\ge 0\n$$\n$$\n\\frac{a}{1 + 2b^3} + \\frac{b}{1 + 2c^3} + \\frac{c}{1 + 2d^3} + \\frac{d}{1 + 2a^3} - a - b - c - d + \\frac{2(ab + bc + cd + da)}{3} =\n$$\n$$\na \\left( \\frac{1}{1 + 2b^3} - 1 + \\frac{2b}{3} \\right) + b \\left( \\frac{1}{1 + 2c^3} - 1 + \\frac{2c}{3} \\right) + c \\left( \\frac{1}{1 + 2d^3} - 1 + \\frac{2d}{3} \\right) +\n$$\n$$\nd \\left( \\frac{1}{1 + 2a^3} - 1 + \\frac{2a}{3} \\right) =\n$$\n$$\n\\frac{2ab(2b^3 - 3b^2 + 1)}{3(1 + 2b^3)} + \\frac{2bc(2c^3 - 3c^2 + 1)}{3(1 + 2c^3)} + \\frac{2cd(2d^3 - 3d^2 + 1)}{3(1 + 2d^3)} + \\frac{2ad(2a^3 - 3a^2 + 1)}{3(1 + 2a^3)} =\n$$\n$$\n\\frac{2ab(2b + 1)(b - 1)^2}{3(1 + 2b^3)} + \\frac{2bc(2c + 1)(c - 1)^2}{3(1 + 2c^3)} + \\frac{2cd(2d + 1)(d - 1)^2}{3(1 + 2d^3)} +\n$$\n$$\n\\frac{2ad(2a + 1)(a - 1)^2}{3(1 + 2a^3)} \\ge 0.\n$$\nWe have equality when $ac + bd = 0$ and also $a = 3$, $b = c = d = 0$ or $a = 2$, $b = 1$, $c = d = 0$ as well as for their cyclic permutations.\n\n\n*Second solution.* (Ionuț Nicolae) We use the inequality $\\frac{1}{1 + 2a^3} \\ge 1 - \\frac{2a}{3}$ which, after computations, comes to $a(a - 1)^2(2a + 1) \\ge 0$. The last inequality is obviously true and its equality cases are $a = 0$ and $a = 1$. We have $\\frac{d}{1 + 2a^3} \\ge \\frac{d - 2ad}{3}$, with equality if $d = 0$ or $a \\in \\{0, 1\\}$.\nThen $\\sum_{cycl} \\frac{a}{1 + 2b^3} \\ge \\sum_{cycl} \\left(a - \\frac{2ab}{3}\\right) = 3 - \\frac{2}{3} \\sum_{cycl} ab$. It remains to be proven that $a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da) \\le 9$, which follows from $a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da + ac + bd) = 9$. We have equality if $ac + bd = 0$ and each non-zero variable is followed in cyclic order by one that is either $0$ or $1$. From $ac + bd = 0$ we see that two variables need to be $0$. If another one is $0$ then the last one is $3$ and we have indeed equality; if the other two are positive, the second one needs to be $1$, hence the first one must be $2$.\n\n\n*Third solution.* (given in the contest by *Andrei Pantea*) From AM-GM inequality we have $1 + 2x^3 = 1 + x^3 + x^3 \\ge 3x^2, \\forall x > 0$.\nWe treat the following cases: **1.** $a$, $b$, $c$, $d > 0$; **2.** $a$, $b$, $c > 0$, $d = 0$; **3.** $a$, $b > 0$, $c = d = 0$; **4.** $a$, $c > 0$, $b = d = 0$; **5.** $a = b = c = 0$, $d = 3$. All the other cases follow by cyclic permutations.\nCase 1. The inequality is equivalent to\n$$\n\\sum_{cycl} \\left( a - \\frac{a}{1 + 2b^3} \\right) \\le \\frac{1}{3} \\left( (a + b + c + d)^2 - \\sum_{cycl} a^2 \\right).\n$$\nBut\n$$\n\\sum_{cycl} \\frac{2ab^3}{1 + 2b^3} \\le \\sum_{cycl} \\frac{2ab^3}{3b^2} = \\frac{2}{3} \\sum_{cycl} ab < \\frac{2}{3} \\sum_{cycl} ab = \\frac{1}{3} \\left( (a + b + c + d)^2 - \\sum_{cycl} a^2 \\right),\n$$\nso in this case the inequality is satisfied without equality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive odd integer $n$, show that the arithmetic mean of the fractional parts $\\{k^{2n}/p\\}$, $k = 1, \\dots, (p-1)/2$, is the same for infinitely many primes $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $\\{k^{2n}/p\\} = r_k/p$, where $r_k$ is the remainder $k^{2n}$ leaves upon division by $p$. Clearly, the $r_k$ are quadratic residues modulo $p$.\nIf $p$ is prime, and $p-1$ and $n$ are relatively prime, then the $r_k$, $k = 1, \\dots, (p-1)/2$, are pairwise distinct, since the $k^{2n}$, $k = 1, \\dots, (p-1)/2$, are pairwise distinct modulo $p$, by Fermat's little theorem. In this case, the $r_k$, $k = 1, \\dots, (p-1)/2$, form the set $R$ of all $(p-1)/2$ quadratic residues modulo $p$ in the range $1$ through $p-1$.\nIf, in addition, $p$ is congruent to $1 \\pmod 4$, then $-1$ is a quadratic residue modulo $p$, and the assignment $r \\mapsto p-r$, $r \\in R$, defines a permutation of $R$. In this case, $\\sum_{r \\in R} r = \\sum_{r \\in R} (p-r) = p(p-1)/2 - \\sum_{r \\in R} r$, so $\\sum_{r \\in R} r = p(p-1)/4$, and the arithmetic mean in question is $1/2$.\nFinally, since $n$ is odd, infinitely many primes congruent to $1 \\pmod 4$ are also congruent to $2 \\pmod n$, by Dirichlet's theorem on arithmetic sequences of integers; for such a prime $p$, the numbers $p-1$ and $n$ are clearly relatively prime. This completes the proof.\nIt is sufficient to show that $\\sum_{k=1}^{(p-1)/2} \\{k^{2n}/p\\} = (p-1)/4$ for every prime $p$ congruent to $1 \\pmod 4$.\nLet $p$ be a prime congruent to $1 \\pmod 4$, let $K = \\{1, \\dots, (p-1)/2\\}$, let $r_k$, $k \\in K$, be the remainder $k^2$ leaves upon division by $p$, and let\n$$\nK^{-} = \\{k : r_{k} < p/2\\} \\quad \\text{and} \\quad K^{+} = \\{k : r_{k} > p/2\\}.\n$$\nIt is easily seen that the $r_k$ are pairwise distinct, so they form the set of all quadratic residues modulo $p$. Since $p$ is congruent to $1 \\pmod p$, $-1$ is a quadratic residue modulo $p$, so there is a bijection $f: K^{-} \\to K^{+}$ such that $r_k + r_{f(k)} = p$ for all $k$ in $K^{-}$; in particular, $|K^{-}| = |K^{+}| = |K|/2 = (p-1)/4$.\nSince $n$ is odd, it follows that $k^{2n} + f(k)^{2n} \\equiv 0 \\pmod p$ for all $k$ in $K^{-}$, so if $r'_k$ is the remainder $k^{2n}$ leaves upon division by $p$, then $p - r'_k$ is the remainder $f(k)^{2n}$ leaves upon division by $p$. Consequently,\n$$\n\\begin{aligned}\n\\sum_{k \\in K} \\{k^{2n}/p\\} &= \\sum_{k \\in K^{-}} \\{k^{2n}/p\\} + \\sum_{k \\in K^{+}} \\{k^{2n}/p\\} = \\sum_{k \\in K^{-}} (\\{k^{2n}/p\\} + \\{f(k)^{2n}/p\\}) \\\\\n&= \\sum_{k \\in K^{-}} (r'_{k}/p + (1 - r'_{k}/p)) = |K^{-}| = (p-1)/4.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{F}$ be the set of all continuous functions $f: [0, 1] \\to \\mathbb{R}$, satisfying\n$$ \\max_{0 \\le x \\le 1} |f(x)| = 1, \\text{ and let } I: \\mathcal{F} \\to \\mathbb{R}, $$\n$$\nI(f) = \\int_{0}^{1} f(x) \\, dx - f(0) + f(1).\n$$",
    "options": [],
    "answer": "The supremum of I(f) is 3 and the infimum is −3; neither is attained. The set of attainable values is the open interval (−3, 3).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $\\frac{m}{n} + \\frac{n}{m} \\neq 1$, for any positive integers $m, n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vlad, Luca and Adina bought school supplies worth of $118$ lei. Vlad bought $5$ pens, $4$ notebooks and $3$ pencil boxes, Luca bought $7$ pens, $3$ notebooks and $4$ pencil boxes, and Adina bought $8$ pens, $7$ notebooks and $7$ pencil boxes. Given that Luca payed $5$ lei more than Vlad, and Adina $4$ lei less than Vlad and Luca together, find out how much does each one of the three supplies cost.",
    "options": [],
    "answer": "pen 1 lei, notebook 2 lei, pencil box 5 lei",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $x$ that satisfy\n$$\n\\log_2(x^2 + 4) - \\log_2 x + x^2 - 4x + 2 = 0.\n$$",
    "options": [],
    "answer": "2",
    "solution": "First, note that $x > 0$ since $\\log_2 x$ is defined only for $x > 0$.\n\nRewrite the logarithmic terms:\n$$\n\\log_2(x^2 + 4) - \\log_2 x = \\log_2\\left(\\frac{x^2 + 4}{x}\\right)\n$$\nSo the equation becomes:\n$$\n\\log_2\\left(\\frac{x^2 + 4}{x}\\right) + x^2 - 4x + 2 = 0\n$$\nLet $y = x^2 - 4x + 2$. Then:\n$$\n\\log_2\\left(\\frac{x^2 + 4}{x}\\right) = -y\n$$\nSo:\n$$\n\\frac{x^2 + 4}{x} = 2^{-y}\n$$\nMultiply both sides by $x$:\n$$\nx^2 + 4 = x \\cdot 2^{-y}\n$$\nBring all terms to one side:\n$$\nx^2 - x \\cdot 2^{-y} + 4 = 0\n$$\nThis is a transcendental equation, but let's try integer values for $x$.\n\nTry $x = 2$:\n\nCompute $x^2 + 4 = 4 + 4 = 8$\n$\\log_2 8 = 3$\n$\\log_2 2 = 1$\nSo $3 - 1 = 2$\n$x^2 - 4x + 2 = 4 - 8 + 2 = -2$\nSo $2 + (-2) = 0$\nThus, $x = 2$ is a solution.\n\nTry $x = 1$:\n$x^2 + 4 = 1 + 4 = 5$\n$\\log_2 5 \\approx 2.322$\n$\\log_2 1 = 0$\nSo $2.322 - 0 = 2.322$\n$x^2 - 4x + 2 = 1 - 4 + 2 = -1$\n$2.322 + (-1) = 1.322 \\neq 0$\n\nTry $x = 4$:\n$x^2 + 4 = 16 + 4 = 20$\n$\\log_2 20 \\approx 4.322$\n$\\log_2 4 = 2$\n$4.322 - 2 = 2.322$\n$x^2 - 4x + 2 = 16 - 16 + 2 = 2$\n$2.322 + 2 = 4.322 \\neq 0$\n\nTry $x = 8$:\n$x^2 + 4 = 64 + 4 = 68$\n$\\log_2 68 \\approx 6.09$\n$\\log_2 8 = 3$\n$6.09 - 3 = 3.09$\n$x^2 - 4x + 2 = 64 - 32 + 2 = 34$\n$3.09 + 34 = 37.09 \\neq 0$\n\nTry $x = 1/2$:\n$x^2 + 4 = 1/4 + 4 = 4.25$\n$\\log_2 4.25 \\approx 2.09$\n$\\log_2 (1/2) = -1$\n$2.09 - (-1) = 3.09$\n$x^2 - 4x + 2 = 1/4 - 2 + 2 = 1/4$\n$3.09 + 0.25 = 3.34 \\neq 0$\n\nTry $x = 4 - \\sqrt{12}$:\n$x = 4 - 2\\sqrt{3} \\approx 0.535$\n$x^2 + 4 \\approx (0.535)^2 + 4 \\approx 0.286 + 4 = 4.286$\n$\\log_2 4.286 \\approx 2.104$\n$\\log_2 0.535 \\approx -0.902$\n$2.104 - (-0.902) = 3.006$\n$x^2 - 4x + 2 = (0.535)^2 - 4 \\times 0.535 + 2 \\approx 0.286 - 2.14 + 2 = 0.146$\n$3.006 + 0.146 = 3.152 \\neq 0$\n\nTry $x = 4 + \\sqrt{12} \\approx 6.464$\n$x^2 + 4 \\approx (6.464)^2 + 4 \\approx 41.8 + 4 = 45.8$\n$\\log_2 45.8 \\approx 5.52$\n$\\log_2 6.464 \\approx 2.7$\n$5.52 - 2.7 = 2.82$\n$x^2 - 4x + 2 = (6.464)^2 - 4 \\times 6.464 + 2 \\approx 41.8 - 25.856 + 2 = 17.944$\n$2.82 + 17.944 = 20.764 \\neq 0$\n\nThus, the only integer solution is $x = 2$.\n\nCheck for other possible solutions:\nLet $f(x) = \\log_2(x^2 + 4) - \\log_2 x + x^2 - 4x + 2$\nFor $x > 0$, $f(x)$ is strictly increasing for large $x$ (since $x^2$ dominates).\nFor $x \\to 0^+$, $\\log_2 x \\to -\\infty$, so $f(x) \\to +\\infty$.\nFor $x = 2$, $f(2) = 0$.\nFor $x = 1$, $f(1) > 0$.\nFor $x > 2$, $f(x) > 0$.\n\nTherefore, the only solution is:\n$$\nx = 2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the distinct positive integers $a$, $b$, $c$, $d$, such that the following conditions hold:\n(1) exactly three of the four numbers are prime numbers;\n(2) $a^2 + b^2 + c^2 + d^2 = 2018$.",
    "options": [],
    "answer": "2, 3, 41, 18",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38713,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ has circumcenter $O$ and incenter $I$. The straight lines $AI$, $BI$, $CI$ meet again the circumcircle at $D$, $E$, respectively $F$. Denote $M$, $N$, $P$ the circumcenters of the triangles $IEF$, $IDF$, respectively $IED$. Prove that the straight lines $DM$, $EN$, $FP$ and $OI$ are concurrent.\nMarin Ionescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38714,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(x_n)_{n \\ge 2}$ is given by\n$$\nx_n = \\frac{e^{1/n} - 1}{e^{1/n^2} - 1} - n.\n$$\nProve that $\\lim_{n \\to \\infty} x_n = 1/2$ and compute $\\lim_{n \\to \\infty} n(x_n - 1/2)$.",
    "options": [],
    "answer": "lim x_n = 1/2 and lim n(x_n − 1/2) = −1/3",
    "solution": "Let us first analyze $x_n$ as $n \\to \\infty$.\n\nWe use the Taylor expansion for $e^x$ near $x = 0$:\n$$\ne^x = 1 + x + \\frac{x^2}{2} + \\frac{x^3}{6} + \\cdots\n$$\nSo,\n$$\ne^{1/n} - 1 = \\frac{1}{n} + \\frac{1}{2n^2} + \\frac{1}{6n^3} + O\\left(\\frac{1}{n^4}\\right)\n$$\n$$\ne^{1/n^2} - 1 = \\frac{1}{n^2} + \\frac{1}{2n^4} + \\frac{1}{6n^6} + O\\left(\\frac{1}{n^8}\\right)\n$$\nTherefore,\n$$\n\\frac{e^{1/n} - 1}{e^{1/n^2} - 1} = \\frac{\\frac{1}{n} + \\frac{1}{2n^2} + \\frac{1}{6n^3} + O\\left(\\frac{1}{n^4}\\right)}{\\frac{1}{n^2} + \\frac{1}{2n^4} + O\\left(\\frac{1}{n^6}\\right)}\n$$\nDivide numerator and denominator by $1/n^2$:\n$$\n= \\frac{n + \\frac{1}{2} + \\frac{1}{6n} + O\\left(\\frac{1}{n^2}\\right)}{1 + \\frac{1}{2n^2} + O\\left(\\frac{1}{n^4}\\right)}\n$$\nNow, expand $1/(1 + \\varepsilon)$ for small $\\varepsilon$:\n$$\n\\frac{n + \\frac{1}{2} + \\frac{1}{6n} + O\\left(\\frac{1}{n^2}\\right)}{1 + \\frac{1}{2n^2} + O\\left(\\frac{1}{n^4}\\right)} = \\left(n + \\frac{1}{2} + \\frac{1}{6n}\\right)\\left(1 - \\frac{1}{2n^2} + O\\left(\\frac{1}{n^4}\\right)\\right)\n$$\nMultiply out:\n$$\n= n + \\frac{1}{2} + \\frac{1}{6n} - \\frac{n}{2n^2} - \\frac{1}{4n^2} - \\frac{1}{12n^3} + O\\left(\\frac{1}{n^2}\\right)\n$$\n$$\n= n + \\frac{1}{2} + \\frac{1}{6n} - \\frac{1}{2n} - \\frac{1}{4n^2} - \\frac{1}{12n^3} + O\\left(\\frac{1}{n^2}\\right)\n$$\nNow, $x_n = \\frac{e^{1/n} - 1}{e^{1/n^2} - 1} - n$:\n$$\nx_n = \\frac{1}{2} + \\frac{1}{6n} - \\frac{1}{2n} - \\frac{1}{4n^2} - \\frac{1}{12n^3} + O\\left(\\frac{1}{n^2}\\right)\n$$\n$$\nx_n = \\frac{1}{2} - \\frac{1}{3n} - \\frac{1}{4n^2} + O\\left(\\frac{1}{n^2}\\right)\n$$\nTherefore,\n$$\n\\lim_{n \\to \\infty} x_n = \\frac{1}{2}\n$$\n\nNow, compute $\\lim_{n \\to \\infty} n(x_n - 1/2)$:\n$$\nn(x_n - 1/2) = n\\left(-\\frac{1}{3n} - \\frac{1}{4n^2} + O\\left(\\frac{1}{n^2}\\right)\\right) = -\\frac{1}{3} - \\frac{1}{4n} + O\\left(\\frac{1}{n}\\right)\n$$\nSo,\n$$\n\\lim_{n \\to \\infty} n(x_n - 1/2) = -\\frac{1}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the prime numbers $a > b > c$ given that $a - b$, $b - c$ and $a - c$ are distinct primes.",
    "options": [],
    "answer": "a=7, b=5, c=2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a finite ring and let $a, b \\in A$ be two elements with the property $(ab - 1)b = 0$. Prove that $b(ab - 1) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many sets of four positive integers so that the sum of the squares of any three elements is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the positive integer solutions of the equation\n$$\n\\lfloor\\sqrt{8n+1}\\rfloor + \\lfloor\\sqrt{8n+2}\\rfloor + \\dots + \\lfloor\\sqrt{8n+7}\\rfloor = 2027.\n$$",
    "options": [],
    "answer": "10512",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the increasing functions $f : \\mathbb{N} \\to \\mathbb{N}$ having the property that the number $\\frac{f(x) + f(y)}{1 + f(x + y)}$ is a non-negative integer, for all $x, y \\in \\mathbb{N}$.",
    "options": [],
    "answer": "All functions of the form f(n) = a n + 1 with a a positive integer.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are infinitely many natural numbers $a$ and $b$ so that:\n$$\na \\cdot \\gcd(a, b) = b + \\text{lcm}(a, b),\n$$\nwhere $\\gcd(a, b)$ is the greatest common divisor and $\\text{lcm}(a, b)$ is the lowest common multiple of $a$ and $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the least real number $c$ satisfying the condition $\\sum_{k=1}^{n} x_{k}^{2} \\le cn$, for all positive integers $n$ and all real numbers $x_1, \\dots, x_n$ greater than or equal to $-1$ such that $\\sum_{k=1}^{n} x_{k}^{3} = 0$.",
    "options": [],
    "answer": "c = 4/3",
    "solution": "The required number is $c = 4/3$. We first show that if $n$ is a positive integer and $x_1, \\dots, x_n$ are real numbers greater than or equal to $-1$ such that $\\sum_{k=1}^{n} x_k^3 = 0$, then $\\sum_{k=1}^{n} x_k^2 \\le 4n/3$. Indeed, since $x_k^3 - 3x_k^2 + 4 = (x_k + 1)(x_k - 2)^2 \\ge 0$, $k = 1, \\dots, n$, it follows that $\\sum_{k=1}^{n} x_k^2 \\le 4n/3 + (1/3) \\sum_{k=1}^{n} x_k^3 = 4n/3$.\n\nThe latter inequality is strict, unless $n$ is divisible by $9$, in which case equality holds if and only if $8n/9$ numbers are $-1$ and the remaining $n/9$ are all $2$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every integer $n \\ge 2$ one can find matrices $A_1, A_2, \\dots, A_n \\in \\mathcal{M}_2(\\mathbb{R})$, having non-zero entries, such that\n$$\nA_1^{-1} + A_2^{-1} + \\dots + A_n^{-1} = (A_1 + A_2 + \\dots + A_n)^{-1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider integers of the form $\\overline{abcd}$ such that:\n\na) $a \\ge b \\ge c \\ge d$;\n\nb) $a+b+c+d = 11$;\n\nc) $\\overline{abcd} - \\overline{cba} = 8082$.\n\nDetermine integers of the form $\\overline{bdca}$ that satisfy all these conditions.",
    "options": [],
    "answer": "2018",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38724,
    "subject": "Mathematics (Multi-modal)",
    "question": "The orthocenter $H$, the centroid $G$ and the incenter $I$ of a triangle are collinear points. Prove that the triangle is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number with $n \\ge 2$, and let $a_1, a_2, \\dots, a_n \\in (1, \\infty)$ be $n$ real numbers. Prove that the function $f: [0, \\infty) \\to \\mathbb{R}$, defined by\n$$\nf(x) = (a_1 a_2 \\dots a_n)^x - a_1^x - a_2^x - \\dots - a_n^x,\n$$\nfor any $x \\in [0, \\infty)$, is strictly increasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38726,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$ and for any column matrix\n$$\nX = \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{pmatrix} \\in \\mathcal{M}_{n,1}(\\mathbb{Z}),\n$$\nwe denote by $\\delta(X)$ the greatest common divisor of the numbers $x_1, x_2, \\dots, x_n$. Let $n \\in \\mathbb{N}$, $n \\ge 2$, and $A \\in \\mathcal{M}_n(\\mathbb{Z})$. Prove that the following statements are equivalent:\n\na) $|\\det A| = 1$ and\n\nb) $\\delta(AX) = \\delta(X)$, for any $X \\in \\mathcal{M}_{n,1}(\\mathbb{Z})$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation $x^2 + y^2 + z^2 = x + y + z + 1$ has no rational solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The equation can be written equivalently as $(2x - 1)^2 + (2y - 1)^2 + (2z - 1)^2 = 7$.\n\nIf $(x, y, z)$ would be a solution of this equation with rational components, denoting $2x - 1 = \\frac{a_1}{b_1}$, $2y - 1 = \\frac{a_2}{b_2}$, $2z - 1 = \\frac{a_3}{b_3}$, one would have integers $a_1, b_1, a_2, b_2, a_3, b_3$ satisfying the equality $(a_1 b_2 b_3)^2 + (b_1 a_2 b_3)^2 + (b_1 b_2 a_3)^2 = 7 (b_1 b_2 b_3)^2$.\n\nThis would lead to the existence of four integers $a, b, c, d$ such that $a^2 + b^2 + c^2 = 7d^2$. If the greatest common divisor of $a, b, c, d$ is $k$, dividing by $k^2$ one would obtain a solution $(a, b, c, d) \\in \\mathbb{Z}^4$ of the equation $a^2 + b^2 + c^2 = 7d^2$ with $\\gcd(a, b, c, d) = 1$. Then $a, b, c, d$ cannot all be even.\n\nBut a perfect square is congruent to either $0$, $1$, or $4$ modulo $8$, so the left hand side can only be $1$, $2$, $3$, $5$ or $6$ modulo $8$, while the right hand side, $7d^2$, is $0$, $4$ or $7$ modulo $8$. In conclusion, the equality cannot hold.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle and let $P$ be a point on the side $AB$. The diagonal $AC$ crosses the segment $DP$ at $Q$. The parallel through $P$ to $CD$ crosses the extension of the side $BC$ beyond $B$ at $K$, and the parallel through $Q$ to $BD$ crosses the extension of the side $BC$ beyond $B$ at $L$. Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the circles $BKP$ and $CLQ$ are tangent at the point $T$ where the line $DP$ crosses the circle $ABCD$ again.\nSince $BCDT$ is cyclic, we have $\\angle KBT = \\angle CDT$. Since $KP \\parallel CD$, we get $\\angle CDT = \\angle KPT$. Thus, $\\angle KBT = \\angle CDT = \\angle KPT$, which shows that $T$ lies on the circle $BKP$. Similarly, the equalities $\\angle LCT = \\angle BDT = \\angle LQT$ show that $T$ also lies on the circle $CLQ$.\n\nIt remains to prove that these circles are indeed tangent at $T$. This follows from the fact that the chords $TP$ and $TQ$ in the circles $BKTP$ and $CLTQ$, respectively, both lie along the same line and subtend equal angles $\\angle TBP = \\angle TBA = \\angle TCA = \\angle TCQ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be natural numbers so that $13x + 8y = 5z$. Prove that the number $(x + y)(y + z)(z + x)$ is divisible by $130$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $\\sqrt{n + [\\sqrt{n} + \\frac{1}{2}]}$ is an irrational number, for every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A$ un inel finit şi $a, b \\in A$ cu proprietatea $a(a b-1) b=0$. Arătaţi că $b(a b-1)=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEgalitatea din ipoteză este echivalentă cu $a b^{2}=b$, iar cea de demonstrat cu $b a b=b$.\n\nDacă elementul $b$ este idempotent (i.e., $b^{2}=b$), atunci $b a b=b a b^{2}=b \\cdot b=b^{2}=b$.\n\nDacă $b^{m}=b$, cu $m>2$, atunci $b a b=b a b^{m}=b a b^{2} b^{m-2}=b \\cdot b \\cdot b^{m-2}=b^{m}=b$.\n\nEste suficient să arătăm că există $m \\geq 2$ cu proprietatea că $b^{m}=b$.\n\nInelul $A$ fiind finit, există $1 \\leq k<m$ numere naturale, cu $k$ minim, cu proprietatea că $b^{k}=b^{m}$.\n\nArătăm că $k=1$.\n\nDacă $k>1$, atunci $a b^{k}=a b^{m}=a b^{2} b^{m-2}=b^{m-1}$.\n\nDacă $k=2$, rezultă că $b=a b^{2}=b^{m-1}$, contrazicând minimalitatea.\n\nDacă $k>2$, atunci $b^{k-1}=b \\cdot b^{k-2}=a b^{2} b^{k-2}=a b^{k}=b^{m-1}$, contrazicând de asemenea minimalitatea.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38732,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ with $AB < BC$ let $BB'$ be an altitude, and let $O$ be the circumcenter. A line through $B'$ parallel to $CO$ meets $BO$ at $X$. Prove that $X$ and the midpoints of $AB$ and $AC$ are collinear.\n\nCaucasus Mathematical Olympiad\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of the side $AB$. Then $\\angle OBC = \\angle OCB = 90^\\circ - \\angle A$ and $MX$ parallel to $BC$ comes to $\\angle MXB = 90^\\circ - \\angle A$. But $\\angle B'XB = \\angle XOC = 2\\angle OBC = 180^\\circ - 2\\angle A$. In the triangle $ABB'$ we have $MA = MB = MB'$ and $\\angle BMB' = \\angle MAB' + \\angle MB'A = 2\\angle A$. It follows that the quadrilateral $MBXB'$ is cyclic, hence $\\angle MXB = \\angle MB'B = \\angle ABB' = \\angle OBC$, which leads to the conclusion.\n\n\n*Second solution.* Let $K$ the projection of $B$ onto the parallel through $A$ to the line $BC$. The quadrilateral $AKBB'$ is cyclic, hence $\\angle AB'K = \\angle ABK = 90^\\circ - \\angle B = \\angle OCA = \\angle CB'X$. It follows that points $K, B'$ and $X$ are collinear. Finally, the triangle $BXK$ is isosceles, with $BX = KX$, therefore $X$ lies on the perpendicular bisector of the line segment $BK$, i.e. $X$ is on the midline of triangle $ABC$ that is parallel to $BC$. We have $\\angle XBK = 90^\\circ - \\angle OBC = \\angle A = \\angle XKB$ (from the cyclic quadrilateral $AKBB'$). The conclusion follows readily.\n\n\n*Third solution.* (Given by Alexandru Mihalcu.) Let $S$ be the midpoint of the line segment $AC$ and $T$ be the intersection point of lines $BX$ and $AC$. As $OS$ is parallel to $BB'$, we have $\\frac{B'S}{ST} = \\frac{OB}{OT} = \\frac{OC}{OT} = \\frac{B'X}{XT}$. From the converse of the Angle Bisector Theorem it follows that $XS$ is the angle bisector of $\\angle B'XT$. Then $\\angle SXT = \\frac{\\angle B'XT}{2} = \\frac{\\angle XOC}{2} = \\angle OBC$, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a square-free integer $n \\ge 3$, evaluate the sum $\\sum_{k=1}^{(n-2)(n-1)} \\lfloor \\sqrt[3]{kn} \\rfloor$.",
    "options": [],
    "answer": "(n-2)(n-1)(3n-5)/4",
    "solution": "Let $S$ be the set of lattice points in the rectangle $[1, (n-2)(n-1)] \\times [1, n-1]$. Since $n$ is square-free, no point of $S$ lies on the curve $y = \\sqrt[3]{nx}$, so the latter splits $S$ into two non-empty sets. The required sum counts the points of $S$ below the curve, while the sum $\\sum_{k=1}^{n-1} \\lfloor k^3/n \\rfloor$ counts the points of $S$ above the curve. Since $|S| = (n-2)(n-1)^2$, it is sufficient to evaluate the latter sum.\n\nTo this end, let $t_k = k^3/n - \\lfloor k^3/n \\rfloor$, $k = 1, 2, \\dots, n-1$. Clearly, $0 \\le t_k < 1$ for all $k$, and since $n$ is square-free, the $t_k$ are, in fact, all positive. Since $k^3/n + (n-k)^3/n = 3k^2 - 3kn + n^2$ is integral, so is $t_k + t_{n-k}$, and since $0 < t_k + t_{n-k} < 2$, it follows that $t_k + t_{n-k} = 1$. Hence\n$$\n\\begin{aligned}\nn-1 &= \\sum_{k=1}^{n-1} (t_k + t_{n-k}) = 2 \\sum_{k=1}^{n-1} t_k = 2 \\sum_{k=1}^{n-1} \\frac{k^3}{n} - 2 \\sum_{k=1}^{n-1} \\lfloor \\frac{k^3}{n} \\rfloor \\\\\n&= \\frac{n(n-1)^2}{2} - 2 \\sum_{k=1}^{n-1} \\lfloor \\frac{k^3}{n} \\rfloor.\n\\end{aligned}\n$$\nConsequently, $\\sum_{k=1}^{n-1} \\lfloor k^3/n \\rfloor = (n-2)(n^2-1)/4$, and\n$$\n\\sum_{k=1}^{(n-2)(n-1)} \\lfloor \\sqrt[3]{kn} \\rfloor = (n-2)(n-1)^2 - (n-2)(n^2-1)/4 = (n-2)(n-1)(3n-5)/4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{ a = q + \\frac{1}{q} \\mid q \\in \\mathbb{Q}^*, q > 0 \\}$,\n$$A + A = \\{a + b \\mid a, b \\in A\\}, \\quad A \\cdot A = \\{a \\cdot b \\mid a, b \\in A\\}.$$\nProve that:\ni) $A + A \\neq A \\cdot A$;\nii) $(A + A) \\cap \\mathbb{N} = (A \\cdot A) \\cap \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "i) Let $a = 1 + \\frac{1}{2} = 2 \\in A$, $b = 2 + \\frac{1}{2} \\in A$. Then\n$$\na + b = 2 + 2 + \\frac{1}{2} = \\frac{9}{2} \\in A + A.\n$$\nWe show that $\\frac{9}{2} \\notin A \\cdot A$. Assume that\n$$\n\\frac{9}{2} = \\left(\\frac{x}{y} + \\frac{y}{x}\\right) \\left(\\frac{z}{t} + \\frac{t}{z}\\right) = \\frac{(x^2 + y^2)(z^2 + t^2)}{xyzt},\n$$\nwith $(x, y) = (z, t) = 1, x, y \\in \\mathbb{N}^*$.\nIt follows that $3 \\mid x^2 + y^2$ or $3 \\mid z^2 + t^2$. This means that $3 \\mid x$ and $3 \\mid y$ or $3 \\mid z$ and $3 \\mid t$. But then either\n$(x, y) \\neq 1 \\text{ or } (z, t) \\neq 1, \\text{ contradiction.}$\n\nii) \n$$\nn = \\left(\\frac{x}{y} + \\frac{y}{x}\\right) + \\left(\\frac{z}{t} + \\frac{t}{z}\\right) = \\frac{(x^2 + y^2)zt + (z^2 + t^2)xy}{xyzt} \\in \\mathbb{N},\n$$\nwith $x, y, z, t \\in \\mathbb{N}$, $(x, y) = (z, t) = 1$, then\n$$\n(x^2 + y^2, xy) = 1 \\text{ and } (z^2 + t^2, zt) = 1,\n$$\nhence\n$$\nxy \\mid zt \\text{ and } zt \\mid xy.\n$$\nIt follows that $xy = zt$.\nTake $q_1 = \\frac{z}{y}$, $q_2 = \\frac{x}{z}$ and we have:\n$$\n\\left(q_1 + \\frac{1}{q_1}\\right) \\left(q_2 + \\frac{1}{q_2}\\right) = \\left(\\frac{z}{y} + \\frac{y}{z}\\right) \\left(\\frac{x}{z} + \\frac{z}{x}\\right) = \\frac{x}{y} + \\frac{y}{x} + \\frac{z^2}{xy} + \\frac{xy}{z^2} \\\\\n= \\frac{x}{y} + \\frac{y}{x} + \\frac{z^2}{zt} + \\frac{zt}{z^2} = \\left(\\frac{x}{y} + \\frac{y}{x}\\right) + \\left(\\frac{z}{t} + \\frac{t}{z}\\right) = n \\in A + A.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38735,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every integer $n \\ge 2$, let $B_n$ denote the set of all binary $n$-tuples of zeroes and ones, and split $B_n$ into equivalence classes by letting two $n$-tuples be *equivalent* if one is obtained from the other by a cyclic permutation of the entries. Determine the integers $n \\ge 2$ for which $B_n$ splits into an odd number of equivalence classes.",
    "options": [],
    "answer": "n = 2 only",
    "solution": "Only $B_2$ splits into an odd number of equivalence classes, namely, three: $\\{(0,0)\\}$, $\\{(0,1), (1,0)\\}$ and $\\{(1,1)\\}$. If $n > 2$, then $B_n$ always splits into an even number of classes, as we are presently going to show.\n\nCall two $n$-tuples of $B_n$ conjugate if one is obtained from the other by replacing its zeroes by ones and its ones by zeroes. With each class in $B_n$ we may associate the class of conjugate $n$-tuples. Thus, if there are no self-conjugate classes, the number of classes is even. This is clearly the case if $n$ is odd, for the $n$-tuples in a self-conjugate class must have the same number of zeroes and ones, so the total number of entries is even.\n\nWrite $n = 2^m n'$, where $m$ and $n'$ are both non-negative integers, and $n'$ is odd, and induct on $m$. The base case, $m=0$, is clear by the preceding, so let $m \\ge 1$.\n\nLet $n_k$ be the number of $k$-element classes in $B_n$. Clearly, if $n_k \\neq 0$, then $k$ divides $n$, so $2^n = |B_n| = \\sum_{k|n} k n_k = \\sum_{k|n/2} k n_k + \\sum_{2^m|k} k n_k = \\sum_{k|n/2} k n_k + \\sum_{d|n'} 2^m d n_{2^m d}$.\n\nIf $k$ divides $n/2$, then an $n$-tuple belonging to a $k$-element class in $B_n$ can be obtained only by iterating an $n/2$-tuple in $B_{n/2}$, so the first sum in the last expression above is $2^{n/2}$.\n\nConsequently, $\\sum_{d|n'} d n_{2^m d} = 2^{n/2-m}(2^{n/2}-1)$. Notice that $n/2-m = 2^{m-1}n' - m$ is positive, unless $m=1$ or $m=2$, and $n'=1$, to infer that, with these exceptions, $\\sum_{d|n'} d n_{2^m d}$ is even. Since $n'$ is odd, so are all its divisors, and it follows that $\\sum_{d|n'} n_{2^m d}$ is even, showing that the total number of $k$-element classes in $B_n$ is even, whenever $k$ is divisible by $2^m$.\n\nIf $k$ is not divisible by $2^m$, then the total number of $k$-element classes in $B_n$ is the same as the total number of classes in $B_{n/2}$ which is even by the induction hypothesis, save when $n=4$. This corresponds to an exceptional case above; the other exceptional case is $n=2$, when the number of classes is clearly three.\n\nFinally, the case $n=4$ is dealt with by explicit construction of the (six) classes $B_4$ splits into: There are two singleton classes, represented by $(0,0,0,0)$ and $(1,1,1,1)$, respectively; there is only one doubleton class, represented by $(0,1,0,1)$; and there are three 4-element classes, represented by $(0,0,0,1)$, $(0,0,1,1)$ and $(0,1,1,1)$, respectively.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two positive integers $m$ and $n$, show that there exist a positive integer $k$ and a set $S$ of at least $m$ multiples of $n$ such that the numbers $2^k \\sigma(s)/s$, $s \\in S$, are all odd; $\\sigma(s)$ is the sum of all positive divisors of $s$ (1 and $s$ inclusive).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = 2^a n'$, where $a$ is a non-negative integer and $n'$ is odd, and let $2^b$ be the highest power of $2$ dividing $\\sigma(n')$. Let $p_1, \\dots, p_\\ell$ be odd primes not dividing $n'$ (e.g., let each $p_i > n'$), and let $N$ be an integer such that $r = N\\varphi(p_1^2 \\cdots p_\\ell^2 n') - 1 > \\max(a, b)$, where $\\varphi$ is Euler's totient function. If $t$ is one of the $2^\\ell$ divisors of the product $p_1 \\cdots p_\\ell$, and $s = 2^r n' t^2$, then $s$ is a multiple of $n$, since $r > a$, and\n$$\n\\sigma(s) = \\sigma(2^r)\\sigma(n')\\sigma(t^2) = (2^{r+1}-1) \\cdot 2^b \\cdot \\text{odd},\n$$\nsince $\\sigma(t^2)$ is odd. Hence,\n$$\n2^{r-b} \\frac{\\sigma(s)}{s} = \\frac{2^{r+1}-1}{n' t^2} \\cdot \\text{odd}\n$$\nis an odd integer, by Euler's theorem. Finally, since there are $2^\\ell$ such $s$, one for each divisor $t$ of the product $p_1 \\cdots p_\\ell$, it is sufficient to consider an integer $\\ell \\ge \\log_2 m$ to produce a set $S$ of at least $m$ multiples of $n$ satisfying the required condition; plainly, $k = r - b$ does not depend on $s$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute scalene triangle, and let $H$ and $O$ be its ortho-center and circumcenter, respectively. The line $OA$ crosses the altitudes from $B$ and $C$ of the triangle $ABC$ at $P$ and $Q$, respectively. Show that the center of the circle $HPQ$ lies on one of the medians of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n*Solution.* Without loss of generality, we may and will assume $AB < AC$. Begin by noticing that the triangles $ABC$ and $HPQ$ are similar. Indeed, $\\angle HQP = 90^\\circ - \\angle QAB = 90^\\circ - \\angle OAB = \\frac{1}{2}\\angle AOB = \\angle ACB$, and similarly $\\angle HPQ = \\angle ABC$.\n\nLet $\\Omega$ and $\\omega$ be the circles $ABC$ and $HPQ$, respectively. The line $AH$ is tangent to $\\omega$, since $\\angle AHP = 90^\\circ - \\angle HAC = \\angle ACB = \\angle HQP$.\n\nLet $T$ be the center of $\\omega$ and let the lines $AT$ and $BC$ cross at $M$. We will show that $M$ is the midpoint of the segment $BC$, so $T$ lies on the median $AM$ of the triangle $ABC$.\n\nConsider the similarity of the triangles $ABC$ and $HPQ$ along with the fact that $A$ is the point where the tangent of $\\omega$ at $H$ crosses the line $PQ$. Letting the tangent of $\\Omega$ at $A$ cross the line $BC$ at $S$, it follows that $S$ and $A$ correspond to one another under the similarity, so $\\angle OSM = \\angle OAT = \\angle OAM$. The quadrangle $AOMS$ is therefore cyclic, and since the tangent $AS$ of $\\Omega$ is perpendicular to the radius $OA$, it follows that $\\angle OMS = 180^\\circ - \\angle OAS = 90^\\circ$. Consequently, $M$ is the orthogonal projection of $O$ on the line $BC$, which is precisely the midpoint of the segment $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the prime numbers $p$ for which the number $a = 7^p - p - 16$ is a perfect square.",
    "options": [],
    "answer": "3",
    "solution": "$p = 2$ does not fulfill the requirement, but $p = 3$ does: $a = 7^3 - 3 - 16 = 324 = 18^2$.\n\nWe show that there are no other solutions. Let $p \\ge 5$ be a prime number.\n\nIf $p \\equiv 1 \\pmod 4$, then $a \\equiv 2 \\pmod 4$, which shows that $a$ is not a perfect square.\n\nIf $p > 3$ is a prime of the form $4k + 3$, then, from Fermat's Little Theorem it follows that $7^p \\equiv 7 \\pmod p$, hence $a \\equiv -9 \\pmod p$, i.e. $p \\mid a + 9$. If $a$ is a perfect square, then $p$ divides $a + 9 = b^2 + 3^2$ would lead to $p$ divides $b$ and $p$ divides 3, which is not possible.\n\nIn conclusion, the only solution to the problem is $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $W$ be the set of all words of length $3n$ containing each of the letters $a, b, c$ exactly $n$ times. Prove that for any word $w$ in $W$ there exists a word $w'$ in $W$ that cannot be obtained from $w$ by less than $3n^2/2$ successive transpositions of adjacent letters.\nIMO 2017 Shortlist",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, a transposition of two adjacent letters in a word will be referred to as a *swap*. Notice that swapping identical letters does not change a word, so assume that no swaps are performed.\n\nDefine the *distance* $\\text{dist}(w, w')$ of two words $w$ and $w'$ in $W$ to be the minimal number of swaps required to transform $w$ into $w'$ (or vice versa). Clearly, if $w''$ is another word in $W$, then $\\text{dist}(w, w') + \\text{dist}(w, w'') \\ge \\text{dist}(w', w'')$. Consequently, if $\\text{dist}(w', w'') \\ge 3n^2$, then $\\max(\\text{dist}(w, w'), \\text{dist}(w, w'')) \\ge 3n^2/2$, so it is sufficient to exhibit two words in $W$ at least $3n^2$ distance apart.\n\nWe are presently going to show that the words\n$$\nw_0 = \\underbrace{aa\\dots a}_{n} \\underbrace{bb\\dots b}_{n} \\underbrace{cc\\dots c}_{n} \\quad \\text{and} \\quad w'_0 = \\underbrace{cc\\dots c}_{n} \\underbrace{bb\\dots b}_{n} \\underbrace{aa\\dots a}_{n}\n$$\nare at least $3n^2$ distance apart.\n\nFor any word $w$ in $W$ and any pair of distinct letters $\\alpha$ and $\\beta$ in the alphabet $\\{a, b, c\\}$, let $f_{\\alpha,\\beta}(w)$ be the number of pairs of positions in $w$ with $\\alpha$ in the left position and $\\beta$ in the right. Let $f(w) = f_{a,b}(w) + f_{a,c}(w) + f_{b,c}(w)$.\n\nIn particular, $f_{a,b}(w_0) = f_{a,c}(w_0) = f_{b,c}(w_0) = n^2$ and $f_{a,b}(w'_0) = f_{a,c}(w'_0) = f_{b,c}(w'_0) = 0$, so $f(w_0) = 3n^2$ and $f(w'_0) = 0$.\n\nFinally, notice that a swap changes exactly one of $f_{a,b}$, $f_{a,c}$, $f_{b,c}$, the absolute value of the change being 1. Consequently, $f$ changes by 1 at each swap, so $\\text{dist}(w, w') \\ge |f(w) - f(w')|$, whatever words $w$ and $w'$ in $W$. In particular, $\\text{dist}(w_0, w'_0) \\ge |f(w_0) - f(w'_0)| = 3n^2$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, with $AB \\neq AC$. Let $D$ be the midpoint of the line segment $BC$, and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$, respectively. If $M$ is the midpoint of the line segment $EF$, and $O$ is the circumcenter of triangle $ABC$, prove that the lines $DM$ and $AO$ are parallel.\n\nCaucasus Mathematical Olympiad",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the intersection point of the lines $AO$ and $BC$. Let $T$ be the point in which the line $AD$ meets again the circumcircle of triangle $ABC$. The quadrilaterals $ABTC$ and $AEDF$ are cyclic, hence $\\angle TBD = \\angle TAC = \\angle DEF$ and $\\angle TCD = \\angle TAB = \\angle DFE$. It follows that the triangle $DEF$ and $TBC$ are similar. Then we have $\\frac{TB}{DE} = \\frac{BC}{EF} = \\frac{BC/2}{EF/2} = \\frac{BD}{EM}$. This shows that the triangles $TBD$ and $DEM$ are also similar, hence $\\angle EDM = \\angle BTD = \\angle ACB$, which leads to $\\angle BDM = \\angle BDE + \\angle EDM = 90^\\circ - \\angle ABC + \\angle ACB$. From $\\angle OAC = 90^\\circ - \\angle ABC$, it follows that $\\angle ASB = \\angle SAC + \\angle ACB = 90^\\circ - \\angle ABC + \\angle ACB = \\angle MDB$, which means that $AS$ is parallel to $MD$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, determine the integral part of the number\n$$\n\\sum_{k=1}^{n-1} \\frac{1}{\\left(1 + \\frac{1}{n}\\right) \\cdots \\left(1 + \\frac{k}{n}\\right)} - \\sum_{k=1}^{n-1} \\left(1 - \\frac{1}{n}\\right) \\cdots \\left(1 - \\frac{k}{n}\\right).\n$$",
    "options": [],
    "answer": "0",
    "solution": "The required integral part is $0$. The difference of the two sums is positive, since positivity clearly holds termwise.\n\nTo show it less than $1$, let $u_k = n^k/k!$, $k = 0, 1, 2, \\dots$, and express the two sums in terms of the $u_k$. The first sum is $(u_{n+1} + \\cdots + u_{2n-1})/u_n$, and the second is $(u_0 + \\cdots + u_{n-2})/u_{n-1}$. Notice that $u_{n-1} = u_n$ to write the difference of the two in the form\n$$\n1 - \\frac{u_{2n}}{u_n} + \\frac{1}{u_n} \\sum_{k=1}^{n} (u_{n+k} - u_{n-k}).\n$$\nIt is therefore sufficient to prove that $\\sum_{k=1}^n u_{n+k} < \\sum_{k=1}^n u_{n-k}$. To this end, we will show that\n$$\n\\sum_{k=1}^{n} k u_{n+k} < \\sum_{k=1}^{n} k u_{n-k} \\quad \\text{and} \\quad \\sum_{k=1}^{n} (m-k) u_{n+k} < \\sum_{k=1}^{n} (m-k) u_{n-k} \\quad (*)\n$$\nfor some $m$ in the range $1, 2, \\dots, n$. Addition of the two inequalities $(*)$ yields the desired inequality.\n\nThe first inequality $(*)$ is equivalent to $\\sum_{k=0}^{2n} (k-n) u_k < 0$. To establish the latter, simply notice that $\\sum_{k=0}^{2n} k u_k = n \\sum_{k=0}^{2n-1} u_k < n \\sum_{k=0}^{2n} u_k$.\n\nTo establish the second inequality $(*)$, discard the trivial case $n = 2$ and let $n \\ge 3$. Write\n$$\n\\frac{u_{n+k}}{u_{n-k}} \\cdot \\frac{u_{n-k-1}}{u_{n+k+1}} = 1 + \\frac{1}{n} - \\frac{k(k+1)}{n^2}, \\quad k = 0, 1, \\dots, n,\n$$\nto infer that, as a function of $k$, the ratio $u_{n+k}/u_{n-k}$ is strictly decreasing for $k(k+1) < n$ and strictly increasing for $k(k+1) > n$; the situation at the possible case $k(k+1) = n$ causes no trouble. Since $u_{n+1}/u_{n-1} = n/(n+1) < 1$, and $u_{2n}/u_0 = n^{2n}/(2n)! > 1$ for $n \\ge 3$, there exists $m$ in the range $1, 2, \\dots, n$ such that $u_{n+k} < u_{n-k}$ for $1 \\le k < m$, and $u_{n+k} > u_{n-k}$ for $m < k \\le n$. Consequently, the second inequality $(*)$ holds termwise for $k \\ne m$. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. The line parallel to $BD$ passing through $A$ meets the line parallel to $AC$ passing through $B$ at $E$. The circumcircle of triangle $ABE$ meets the lines $EC$ and $ED$, again, at $F$ and $G$, respectively. Prove that the lines $AB$, $CD$ and $FG$ are either parallel or concurrent.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Angles $\\angle ACB$ and $\\angle ADB$ are equal, therefore points $C$ and $D$ are either both in the interior of the circumcircle of triangle $ABE$, or both on this circle, or both outside this circle. Thus we distinguish three cases:\n\n1. $F \\in (EC)$ and $G \\in (ED)$,\n2. $F = C$, $G = D$ (in this case the statement of the problem is obvious),\n3. $C \\in (EF)$, $D \\in (EG)$.\n\nWe only treat the first case, the last one being similar. We have\n$$\n\\angle GDC = \\angle GDB + \\angle BDC = \\angle GEA + \\angle BAC = \\frac{AG}{2} + \\angle ABE = \\frac{AG+AE}{2} = \\angle GFE,\n$$\nwhich shows that the quadrilateral $FGDC$ is cyclic.\n\nWe notice that line $AB$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $ABCD$, line $FG$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $FGDC$, and line $CD$ is the radical axis of the circumcircles of the quadrilaterals $ABCD$ and $FGDC$. It is well known that the radical axis of three circles are either parallel (if the three centers are collinear) or concurrent (in the radical center of the three circles).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $\\omega$ be its $A$-excircle (excircle opposite vertex $A$). Let $D, E, F$ be the points where $\\omega$ touches the lines $BC, CA, AB$, respectively. The circle $AEF$ crosses the line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of the line segment $AD$. Prove that the circles $\\omega$ and $MPQ$ are tangent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, let $\\Omega$ and $\\gamma$ denote the circles $AEFPQ$ and $MPQ$, respectively. Letting the line $AD$ cross $\\omega$ again at $T$, we will show that $\\gamma$ and $\\omega$ are tangent at $T$.\n\nWe first prove that the points $M, P, Q, T$ are concyclic, so $T$ lies on $\\gamma$. Let $A'$ be the center of $\\omega$, and notice that $AA'$ is a diameter of $\\Omega$, since $A'E$ and $A'F$ are perpendicular to $AE$ and $AF$, respectively. Let $N$ be the midpoint of the line segment $DT$. Since $A'D = A'T$, the angle $A'NA$ is right, so $N$ also lies on $\\Omega$, and therefore $DP \\cdot DQ = DA \\cdot DN$; and since $DA = 2DM$ and $DT = 2DN$, it follows that $DP \\cdot DQ = DM \\cdot DT$, showing that $M, P, Q, T$ are indeed concyclic.\n\nIf $EF$ and $BC$ are parallel, then the triangle $ABC$ is isosceles with apex at $A$ and the conclusion follows by symmetry.\n\nOtherwise, the lines $PQ$ and $EF$ cross at a point $R$; these lines are the radical axes of $\\Omega, \\gamma$ and $\\Omega, \\omega$, respectively, so $R$ is the radical center of the three circles, and $RT$ is the radical axis of $\\gamma$ and $\\omega$.\n\nTo conclude the proof, it is therefore sufficient to show that $RT$ is tangent to $\\omega$. With respect to this circle, the line $DR$ is the polar of $D$, and the line $EF$ is the polar of $A$, so $R = DR \\cap EF$ is the pole of the line $ADT$. Consequently, $RT$ is indeed tangent to $\\omega$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $n$ weights, $n \\ge 2$, of masses $m_1, m_2, \\dots, m_n$, where $m_k$ are positive integers such that $1 \\le m_k \\le k$ for all $k \\in \\{1, 2, \\dots, n\\}$. Prove that we can place the weights on the two pans of a balance such that the pans stay in equilibrium if and only if the number $m_1 + m_2 + \\dots + m_n$ is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed on one of the pans of the balance).\n\nConversely, we prove by \"strong\" induction after $n$ that, if the sum of the masses is even, then one can place the weights conveniently. Assume the statement to be true for any $k < n$ weights whose total mass is even. We distinguish two cases:\n\n• if $m_{n-1} = m_n$, then $m_1 + m_2 + \\dots + m_{n-2}$ is even, hence, according to the inductive hypothesis for $k = n-2$, we can place these weights on the two pans of the balance such that the balance stays in equilibrium. Now we simply add the two weights of mass $m_n$, one in each pan, and the balance remains in equilibrium.\n\n• If $m_n \\neq m_{n-1}$, instead of the two weights we produce a new one of mass $m = |m_n - m_{n-1}|$. Then $1 \\le m \\le n-1$ and the sum of the masses of these $n-1$ weights,\n$$\nm_1 + m_2 + \\dots + m_{n-2} + m = m_1 + m_2 + \\dots + m_n - 2 \\cdot \\min\\{m_{n-1}, m_n\\},\n$$\nis even. From the inductive hypothesis for $k = n-1$ it follows that we can place these $n-1$ weights on the pans of the balance such that the balance stays in equilibrium. Now, instead of the weight of mass $|m_n - m_{n-1}|$ we put the weight $\\max\\{m_{n-1}, m_n\\}$, and on the other pan we put the weight of mass $\\min\\{m_{n-1}, m_n\\}$; the balance stays in equilibrium.\n\nThe statement is thus proven.\nObviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed in one of the pans of the balance).\n\nConversely, assume that the total mass is even. We place the weight of mass $m_n$ on one of the pans. The modulus of the difference of the total weights of the two pans is $m_n \\le n$. We place successively the weights of masses $m_{n-1}, m_{n-2}, \\dots, m_1$ on the pan whose total mass is smaller. (In case the two pans are in equilibrium, we place the weight on any of two pans.) After placing the weight of mass $m_k$, the difference between the total masses on the two pans is at most $k$. Proceeding this way, we insure that in the end, after placing the weight of mass $m_1 = 1$, the modulus of the difference between the two pans is at most 1. But the total mass being even, it follows that the difference is 0, which is what we wanted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38745,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $n$ be a composite positive integer. Show that there exist integers $a_1, a_2, \\dots, a_n$ whose sum is not divisible by $n$, but $n$ divides (at least) one of the numbers\n$$\na_k, \\quad a_k + a_{k+1}, \\quad \\dots, \\quad a_k + a_{k+1} + \\dots + a_{k+n-1},\n$$\nfor all positive integers $k \\le n$; $a_k = a_{k-n}$ if $k > n$.\n\nb) Let $p$ be prime and let $a_1, a_2, \\dots, a_p$ be integers whose sum is not divisible by $p$. Show that there exists a positive integer $k \\le p$ such that $p$ divides none of the numbers\n$$\na_k, \\quad a_k + a_{k+1}, \\quad \\dots, \\quad a_k + a_{k+1} + \\dots + a_{k+p-1},\n$$\nwhere $a_k = a_{k-p}$ if $k > p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Write $n = ab$, where $a$ and $b$ both are integers greater than $1$, and let $a_1 = a_2 = \\dots = a_{n-1} = a$ and $a_n = 0$. The sum $a_1 + a_2 + \\dots + a_n = a \\cdot (n-1)$ is clearly not divisible by $n$. Consider any positive integer $k \\le n$: If $k \\le n-b$, then $a_k + a_{k+1} + \\dots + a_{k+b-1}$ is divisible by $n$; otherwise, $a_k + a_{k+1} + \\dots + a_{k+b}$ is divisible by $n$.\n\nb) Suppose, if possible, that for each positive integer $k \\le p$ there exists a positive integer $\\ell \\le p$ such that $a_k + a_{k+1} + \\dots + a_{k+\\ell-1}$ is divisible by $p$; to make a choice, let $f(k)$ be the least such $\\ell$ (any choice will do). Notice that $f(k) \\le p-1$, since $a_k + a_{k+1} + \\dots + a_{k+p-1} = a_1 + a_2 + \\dots + a_p$ is not divisible by $p$.\n\nDefine integers $0 = k_0 < k_1 < k_2 < \\dots < k_p$ such that $k_{i+1} - k_i \\le p-1$, $i = 0, 1, \\dots, p-1$, and\n$$\na_{k_i+1} + a_{k_i+2} + \\dots + a_{k_{i+1}} \\equiv 0 \\pmod{p}, \\quad i = 0, 1, \\dots, p-1.\n$$\nThe $k_i$ are recursively defined by $k_{i+1} = k_i + f(k_i - p\\lfloor k_i / p \\rfloor + 1)$, $i = 0, 1, \\dots, p-1$.\n\nConsider any integers $m$ and $n$ such that $0 \\le m < n \\le p$. Since $k_{i+1} - k_i \\le p-1$ for all $i$, it follows that $k_n - k_m \\le (n-m)(p-1) \\le p(p-1)$. Moreover,\n$$\na_{k_m+1} + a_{k_m+2} + \\dots + a_{k_n} = \\sum_{i=m}^{n-1} (a_{k_i+1} + a_{k_i+2} + \\dots + a_{k_{i+1}}) \\equiv 0 \\pmod{p}. \\quad (*)\n$$\nAmongst the $p+1$ integers $k_0, k_1, k_2, \\dots, k_p$, there exist two, say $k_m < k_n$, such that $k_m \\equiv k_n \\pmod{p}$. Then\n$$\na_{k_m+1} + a_{k_m+2} + \\dots + a_{k_n} = \\frac{k_n - k_m}{p} \\cdot (a_1 + a_2 + \\dots + a_p) \\not\\equiv 0 \\pmod{p},\n$$\nsince $(k_n - k_m)/p$ is a positive integer less than $p$, the sum $a_1 + a_2 + \\dots + a_p$ is not divisible by $p$, and $p$ is prime. This contradicts $(*)$ and concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that a positive integer $A$ is a perfect square if and only if, for all positive integers $n$, at least one of the numbers\n$$\n(A+1)^2 - A, (A+2)^2 - A, (A+3)^2 - A, \\dots, (A+n)^2 - A\n$$\nis a multiple of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $A$ is a perfect square, i.e. there exists $B \\in \\mathbb{N}$ such that $A = B^2$, then $(A+k)^2 - A = (B^2+k)^2 - B^2 = (B^2+B+k)(B^2-B+k)$ for all $k = 1, n$, and exactly one of the (consecutive) numbers $B^2+B+1, B^2+B+2, \\dots, B^2+B+n$ is a multiple of $n$.\n\nConversely, if $A$ is not a perfect square, then it has a prime factor that occurs in the prime factorization of $A$ at an odd exponent. Let $p$ be such a prime and $j \\in \\mathbb{N}$ such that $p^{2j-1} \\mid A$, but $p^{2j} \\nmid A$. We choose $n = p^{2j} \\in \\mathbb{N}$ and show that none of the numbers $(A+1)^2-A, (A+2)^2-A, (A+3)^2-A, \\dots, (A+n)^2-A$ is a multiple of $n$. Indeed, if $n \\mid (A+m)^2-A$, for some $m \\in \\{1, 2, \\dots, n\\}$, i.e. $p^{2j} \\mid A^2 + 2Am + m^2 - A$, from $p^{2j-1} \\mid A$ it follows that $p^{2j-1} \\mid m^2$, hence $p^j \\mid m$. But then $p^{2j} \\mid (A+m)^2$ and $p^{2j} \\mid (A+m)^2 - A$, hence $p^{2j} \\mid A$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $Q$ be an $n \\times n \\times n$ cube. A box is a subset of $Q$ of one of the forms $1 \\times n \\times n$, $n \\times 1 \\times n$ or $n \\times n \\times 1$. Each $1 \\times 1 \\times 1$ cell of $Q$ is coloured one of several many colours. Consider the colour set of each box (each colour listed once). There are $3n$ such, split into three $n$-element collections, one for each of the three directions. It turns out that every colour set in any of the three collections appears in each of the other two collections. Under these conditions, determine, in terms of $n$, the maximum number of colours $Q$ may bear.",
    "options": [],
    "answer": "n(n+1)(2n+1)/6",
    "solution": "The required maximum is $n(n+1)(2n+1)/6$ and is achieved, for instance, by colouring cells as described below. For each colour, we list the set of all cells bearing that colour:\n* $n$ singletons of the form $\\{(i, i, i)\\}$, where $1 \\le i \\le n$;\n* $3\\binom{n}{2}$ doubletons of the form $\\{(i, j, j), (j, i, i)\\}$, $\\{(j, i, j), (i, j, i)\\}$ and $\\{(j, j, i), (i, i, j)\\}$, where $1 \\le i < j \\le n$; and\n* $2\\binom{n}{3}$ triplets of the form $\\{(i, j, k), (j, k, i), (k, i, j)\\}$, where $1 \\le i < j < k \\le n$ or $1 \\le i < k < j \\le n$.\n\nThus, $Q$ bears exactly $n + 3\\binom{n}{2} + 2\\binom{n}{3} = n(n+1)(2n+1)/6$ colours, and it is easily seen that the rank $i$ boxes in the three directions all share the same colour set.\n\nWe will now show that, under the conditions in the statement, the number of colours $Q$ bears does not exceed $n(n+1)(2n+1)/6$.\n\nFix such a colouring and split the set $C$ of all colours in $Q$ into three subsets: The set $C_1$ of all colours that appear exactly once, the set $C_2$ of all colours that appear exactly twice, and the set $C_3$ of all colours that appear at least thrice. Let $S_i$ be set of all cells in $Q$ bearing a colour in $C_i$. Since $|S_i| \\ge i|C_i|$,\n$$\n\\begin{aligned}\n|C| &= |C_1| + |C_2| + |C_3| \\le |S_1| + \\frac{1}{2}|S_2| + \\frac{1}{3}|S_3| \\\\\n&= \\frac{|S_1| + |S_2| + |S_3|}{3} + \\frac{4|S_1| + |S_2|}{6} = \\frac{n^3}{3} + \\frac{4|S_1| + |S_2|}{6},\n\\end{aligned}\n$$\nand it is therefore sufficient to show that $4|S_1| + |S_2| \\le n(3n+1)$.\n\nTo this end, we fix a box $X$ of the form $1 \\times n \\times n$ and prove that $4|X \\cap S_1| + |X \\cap S_2| \\le 3n + 1$. Summing over all such boxes $X$, we get the desired upper bound. Let $Y$ and $Z$ be $n \\times 1 \\times n$ and $n \\times n \\times 1$ boxes, respectively, sharing the colour set of $X$.\n\nIf $X \\cap S_1$ is non-empty, then a cell in this set also lies in both $Y$ and $Z$, hence in $X \\cap Y \\cap Z$. Consequently, $X \\cap S_1 = X \\cap Y \\cap Z$ and $|X \\cap S_1| = 1$.\n\nConsider now the cells in $X \\cap S_2$. There are at most $(n-1) + (n-1) = 2(n-1)$ such in the cross $(X \\cap Y) \\cup (X \\cap Z)$, since the cell in $X \\cap Y \\cap Z$ is in $S_1$. Let now $c$ be any other cell in $X \\cap S_2$. Recall that there is only one other cell $c'$ in $Q$ sharing the colour of $c$. Since $Y \\cup Z$ does not contain $c$, and $Y$ and $Z$ each contains a cell sharing its colour, it follows that they both contain $c'$. And since the cell in $X \\cap Y \\cap Z$ is in $S_1$, the cell $c'$ lies in $(Y \\cap Z) \\setminus X$. The assignment $c \\mapsto c'$ is clearly injective, so the number of cells $c$ under current consideration does not exceed $|(Y \\cap Z) \\setminus X| = n-1$. Consequently, $|X \\cap S_2| \\le 2(n-1) + (n-1) = 3(n-1)$, and hence $4|X \\cap S_1| + |X \\cap S_2| \\le 4 + 3(n-1) = 3n+1$.\n\nTo deal with the case where $X \\cap S_1$ is empty, proceed similarly with minor changes. This time $X \\cap S_2$ contains at most $2n-1$ cells from the cross $(X \\cap Y) \\cup (X \\cap Z)$. To any other cell $c$ in $X \\cap S_2$ there corresponds a cell $c'$ in $Y \\cap Z$ (not necessarily outside $X$), so there are at most $n$ such. Consequently, $|X \\cap S_2| \\le (2n-1) + n = 3n-1$, a wee bit better than needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38748,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$, $b$, $c$ are positive real numbers, prove that\n$$\n\\frac{a}{\\sqrt{(a+2b)^3}} + \\frac{b}{\\sqrt{(b+2c)^3}} + \\frac{c}{\\sqrt{(c+2a)^3}} \\ge \\frac{1}{\\sqrt{a+b+c}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From Hölder's inequality we have:\n$$\n\\sum_{cycl} \\frac{a}{\\sqrt{(a+2b)^3}} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\ge (a+b+c)^4,\n$$\nand from Cauchy-Buniakowsky-Schwarz it follows that\n$$\n\\left(\\sum_{cycl} a\\sqrt{a+2b}\\right)^2 = \\left(\\sum_{cycl} \\sqrt{a \\cdot \\sqrt{a(a+2b)}}\\right)^2 \\le (a+b+c) \\cdot \\sum_{cycl} (a^2+2ab) \\\\\n= (a+b+c)^3\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alina and Bogdan play the following game. They have a heap and $330$ stones in it. They take turns. In one turn it is allowed to take from the heap exactly $1$, exactly $n$ or exactly $m$ stones. The player who takes the last stone wins. Before the beginning Alina says the number $n$, ($1 < n < 10$). After that Bogdan says the number $m$, ($m \\neq n$, $1 < m < 10$). Alina goes first. Which of the two players has a winning strategy? What if initially there are $2018$ stones in the heap?\n\nAdapted from a Belarus Olympiad problem",
    "options": [],
    "answer": "For 330 stones: Bogdan. For 2018 stones: Alina.",
    "solution": "• For the heap initially containing $330$ stones:\n\nBogdan has a winning strategy. One such strategy is the following:\n\n1. If Alina chooses a number $n$ which is not a multiple of $3$, then Bogdan chooses $m = 2$ (or any other number that is not a multiple of $3$ in case Alina has chosen $n = 2$);\n2. If Alina chooses $3$ or $9$, Bogdan chooses $5$;\n3. If Alina chooses $6$, Bogdan chooses $4$.\n\nIn the first case, the number of stones Alina leaves in the heap is not a multiple of $3$; Bogdan will take $1$ or $2$ stones, leaving a heap with a number of stones multiple of $3$. Thus, at the end, Bogdan is the one that leaves $0$ stones in the heap.\n\nIf Alina chooses an odd number, in particular $3$ or $9$, Bogdan can choose $m$ odd and whatever Alina moves, he can always take one stone from the heap. After Alina's moves, the heap will always contain an odd number of stones, so she cannot win.\n\nIf Alina chooses $6$, Bogdan chooses $4$ and he will move as follows: if Alina takes $1$ or $6$ stones, Bogdan takes $4$, and if Alina takes $4$, Bogdan takes $1$. Thus, after Bogdan's moves the number of stones in the heap will always be a multiple of $5$, while after Alina's moves it will not be a multiple of $5$. Bogdan wins again.\n\n• For the heap initially containing $2018$ stones:\n\nAlina has a winning strategy. One such strategy is the following:\n\nShe chooses $n = 2$.\n\n- If Bogdan chooses $m \\in \\{4, 5, 7, 8\\}$, Alina moves such that the number of stones she leaves in the heap is always a multiple of $3$ (initially she takes $2$ stones).\n- If Bogdan chooses $m = 3$, Alina moves such that the number of stones she leaves in the heap is always a multiple of $4$ (initially she takes $2$ stones). (If Bogdan takes $1$, $2$ or $3$ stones, Alina takes $3$, $2$, and $1$ stone(s), respectively, leaving a number of stones that is a multiple of $4$.)\n- If Bogdan chooses $m = 6$, Alina moves such that the number of stones she leaves in the heap gives one of the remainders $0$ or $3$ when divided by $7$ (initially she takes $2$ stones). Later, if Bogdan finds $7k$ stones in the heap and takes $1$, $2$ or $6$ stones, then Alina takes $6$, $2$, or $1$ stones, respectively, while if Bogdan finds $7k + 3$ stones in the heap and takes $1$, $2$ or $6$ stones, Alina takes $2$, $1$, or $1$ stones, respectively.\n- If Bogdan chooses $m = 9$, Alina moves such that the number of stones she leaves in the heap gives one of the remainders $0$, $3$ or $6$ when divided by $10$ (initially she takes $2$ stones). Later, if Bogdan finds $10k$ stones in the heap and takes $1$, $2$ or $9$ stones, then Alina takes $9$, $2$, or $1$ stones, respectively, leaving $10(k - 1)$ or $10(k - 1) + 6$ stones. If Bogdan finds $10k + 3$ stones in the heap and takes $1$, $2$ or $9$ stones, Alina takes $2$, $1$, or $1$ stones, respectively, leaving $10k$ or $10(k - 1) + 3$ stones. (The last situation is possible only if $k \\neq 0$). If Bogdan finds $10k + 6$ stones in the heap and takes $1$, $2$ or $9$ stones, Alina takes $2$, $1$, or $1$ stones, respectively, leaving $10k + 3$ or $10(k - 1) + 6$ stones. (The last situation is possible only if $k \\neq 0$).\n\nThus, irrespective on Bogdan's choice of $m$, Alina wins by choosing $n = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that\n$$\n\\frac{1}{a} + \\frac{3}{b} + \\frac{5}{c} \\geq 4a^2 + 3b^2 + 2c^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* The inequality can be written\n$$\n\\frac{1}{a} + \\frac{3}{b} + \\frac{5}{c} + b^2 + 2c^2 \\geq 4(a^2 + b^2 + c^2)\n$$\nor\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{2}{b} + b^2 + \\frac{4}{c} + 2c^2 \\geq 12.\n$$\nThis follows from the following inequalities by using the AM-GM inequality:\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq \\frac{3}{\\sqrt[3]{abc}} \\geq 3\n$$\nbecause\n$$\n3 = a^2 + b^2 + c^2 \\geq 3\\sqrt[3]{a^2b^2c^2} \\implies abc \\leq 1.\n$$\nOn the other hand,\n$$\n\\frac{2}{b} + b^2 = \\frac{1}{b} + \\frac{1}{b} + b^2 \\geq 3\\sqrt[3]{\\frac{1}{b} \\cdot \\frac{1}{b} \\cdot b^2} = 3\n$$\n$$\n\\frac{4}{c} + 2c^2 = \\frac{2}{c} + \\frac{2}{c} + 2c^2 \\geq 3\\sqrt[3]{\\frac{2}{c} \\cdot \\frac{2}{c} \\cdot 2b^2} = 6.\n$$\nThe equality holds if $a = b = c = 1$.\n\n\n*Second solution.* As $x^3 - 3x + 2 \\geq 0, \\forall x \\geq 0$ (equivalent to $(x - 1)^2(x + 2) \\geq 0$; alternatively, the previous inequality follows from AM-GM: $x^3 + 1 + 1 \\geq 3\\sqrt[3]{x^3 \\cdot 1 \\cdot 1} = 3x$; the equality holds whence $x = 1$). We deduce that $\\frac{1}{x} \\geq \\frac{3}{2} - \\frac{x^2}{2}$. Writing this for $a$, $b$, $c$ and multiplying these inequalities by 1, 3, and 5, respectively, we obtain by summing $\\frac{1}{a} + \\frac{3}{b} + \\frac{5}{c} \\geq \\frac{27}{2} - \\frac{a^2 + 3b^3 + 5c^2}{2}$. Using $27 = 9(a^2+b^2+c^2)$ one obtains the conclusion.\nThe equality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine how many possible ways can one put two distinct natural numbers, which are less or equal to $9$ and have the sum equal to $10$, in a $10 \\times 10$ table, such that the two numbers belong to distinct lines and to distinct columns.",
    "options": [],
    "answer": "32400",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38752,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ is inscribed in the circle $\\mathcal{C}(O, 1)$. Let $G_1$, $G_2$ and $G_3$ be the centroids of the triangles $OBC$, $OAC$ and $OAB$ respectively. Prove that the triangle $ABC$ is equilateral if and only if $AG_1 + BG_2 + CG_3 = 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the pairs of integers $(a, b)$ such that $a^2 + 2b^2 + 2a + 1$ is a divisor of $2ab$.",
    "options": [],
    "answer": "All integer pairs (a, b) with either a = 0 and any integer b, or b = 0 and any integer a except a = −1; together with the six nontrivial pairs (−1, 1), (−1, −1), (−3, 1), (−3, 2), (−3, −1), (−3, −2).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the set $M = \\left\\{ \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\in \\mathcal{M}_2(\\mathbb{C}) \\mid ab = cd \\right\\}$.\n\na) Give examples of $A \\in M$ having the property $A^{2017} \\in M$ and $A^{2019} \\in M$, but $A^{2018} \\notin M$.\n\nb) Prove that if for an $A \\in M$ there is $k \\ge 1$ such that $A^k \\in M$, $A^{k+1} \\in M$ and $A^{k+2} \\in M$, then $A^n \\in M$, for all integers $n \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, let $I$ be its incenter, let $\\Omega$ be its circumcircle, and let $\\omega$ be the circle tangent to the sides $AB$ and $AC$, and internally tangent to $\\Omega$. Let $D, E$ and $T$ be the points of contact of $\\omega$ and $AB, AC$ and $\\Omega$, respectively, let the line $IT$ cross $\\omega$ again at $P$, and let the lines $PD$ and $PE$ cross the line $BC$ at $M$ and $N$ respectively. Prove that the points $D, E, M, N$ lie on a circle. What is the center of this circle?\nSeverius Moldovan",
    "options": [],
    "answer": "the A-excenter of triangle ABC",
    "solution": "![](attached_image_1.png)\n\nThe points $D, E, M, N$ are concyclic if and only if $\\angle DEP = \\angle BMD$. Read from $\\omega$ to infer that $\\angle DEP = \\angle DTP = \\angle ADP$.\nWe now show that $\\angle DTP = \\angle ATE$. To this end, consider the homothety from $T$ mapping $\\omega$ onto $\\Omega$. This homothety maps $D$ to the midpoint $C'$ of the arc $AB$ not containing $C$, and $E$ to the midpoint $B'$ of the arc $AC$ not containing $B$. (A tangent to $\\omega$ is mapped onto a parallel tangent to $\\Omega$.)\nApply Pascal's theorem to the hexagram $ABB'TC'C$ to infer that $D, I, E$ are collinear, so $I$ is the midpoint of the segment $DE$ (Veldkamp, 1976-1977).\nSince $AD$ and $AE$ are the tangents from $A$ to $\\omega$, the line $AT$ is the $T$-symmedian in the triangle $DET$, so $\\angle DTP = \\angle DTI = \\angle ATE$.\nNext, $\\angle ATE = \\angle ATB' = \\angle ABB'$ (read from $\\Omega$), so $\\angle ADP = \\angle DTP = \\angle ATE = \\angle ABB' = \\angle ABI$, showing that the line through $D, M, P$ is parallel to the internal bisectrix $BI$ of the angle $ABC$.\nConsequently, $\\angle BMD = \\angle CBI = \\angle ABI = \\angle ADP = \\angle DEP$, so the points $D, E, M, N$ lie on a circle $\\gamma$.\nFinally, we show that $\\gamma$ is centered at the $A$-excenter of the triangle $ABC$. The center of $\\gamma$ is the point where the perpendicular bisectrix of the segment $DE$ crosses the perpendicular bisectrix of the segment $DM$. The former is the internal $A$-bisectrix in the triangle $ABC$. The latter is the external $B$-bisectrix in the triangle $ABC$, since $\\angle BDM = \\angle ADP = \\angle BMD$, so the triangle $BDM$ is isosceles with apex at $B$. The two angle bisectors cross at the $A$-excenter of the triangle $ABC$ and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ such that\n$$\n2^{2018} + 2^{2022} + 2^{2023} + 2^{2024} + 2^{2026} + 2^n\n$$\nis a square.",
    "options": [],
    "answer": "2026",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two continuous functions $f, g: [0, 1] \\to [0, 1]$ are such that $f \\circ g = g \\circ f$, and one of them is nondecreasing. Prove that there exists $a \\in [0, 1]$ so that $f(a) = a = g(a)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2^{-x} + 2^{-1/x} \\le 1$, for any real number $x > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying\n$$\nP(x)^{10} + P(x)^9 = Q(x)^{21} + Q(x)^{20}.\n$$",
    "options": [],
    "answer": "No",
    "solution": "*First solution.* The answer is in the negative. Comparing the degrees of both sides in $(*)$ we get $\\deg P = 21n$ and $\\deg Q = 10n$ for some positive integer $n$. Take the derivative of $(*)$ to obtain\n$$\nP'P^8(10P + 9) = Q'Q^{19}(21Q + 20). \\quad (**)\n$$\nSince $\\gcd(10P + 9, P) = \\gcd(10P + 9, P + 1) = 1$, it follows that $\\gcd(10P + 9, P^9(P + 1)) = 1$, so $\\gcd(10P + 9, Q) = 1$, by $(*)$.\n\nThus $(**)$ yields $10P + 9 \\mid Q'(21Q + 20)$, which is impossible since $0 < \\deg(Q'(21Q + 20)) = 20n - 1 < 21n = \\deg(10P + 9)$. A contradiction.\n*Second solution.* Letting $r$ and $s$ be integers such that $r \\ge 2$ and $s \\ge 2r$, we show that if $P^r + P^{r-1} = Q^s + Q^{s-1}$, then $Q$ is constant.\nLet $m = \\deg P$ and $n = \\deg Q$. A degree inspection in the given relation shows that $m \\ge 2n$.\nWe will prove that $P(P+1)$ has at least $m+1$ distinct complex roots. Assuming this for the moment, notice that $Q$ takes on one of the values $0$ or $-1$ at each of those roots. Since $m+1 \\ge 2n+1$, it follows that $Q$ takes on one of the values $0$ and $-1$ at more than $n$ distinct points, so $Q$ must be constant.\nFinally, we prove that $P(P+1)$ has at least $m+1$ distinct complex roots. This can be done either by referring to the Mason–Stothers theorem or directly, in terms of multiplicities of the roots in question.\nSince $P$ and $P+1$ are relatively prime, the Mason–Stothers theorem implies that the number of distinct roots of $P(P+1)$ is greater than $m$, hence at least $m+1$.\nFor a direct proof, let $z_1, \\dots, z_t$ be the distinct complex roots of $P(P+1)$, and let $z_k$ have multiplicity $\\alpha_k$, $k = 1, \\dots, t$. Since $P$ and $P+1$ have no roots in common, and $P' = (P+1)'$, it follows that $P'$ has a root of multiplicity $\\alpha_k - 1$ at $z_k$. Consequently, $m-1 = \\deg P' \\ge \\sum_{k=1}^t (\\alpha_k - 1) = \\sum_{k=1}^t \\alpha_k - t = 2m - t$; that is, $t \\ge m+1$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the congruent segments $AB$, $BC$ and $AD$, with $D \\in (BC)$. Show that the perpendicular bisector of the segment $DC$, the angle bisector of the angle $ADC$ and the line $AC$ are concurrent.\nMircea Fianu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38761,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of 15 consecutive positive integers is a number written with distinct digits, among these being 0, 1, 2 and 4. Find the least possible number among the 15 consecutive numbers.",
    "options": [],
    "answer": "676",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with circumcenter $O$ and orthocenter $H$. Denote $P$ as the midpoint of the segment $AO$ and $S$ as the intersection of the perpendicular bisector of the segment $AO$ with $BC$. The circumcircle of the triangle $APS$ intersects the segment $OH$ at $Z$. Prove that\n$$\n\\vec{OZ} = \\frac{1}{2} (\\vec{OA} + \\vec{OB} + \\vec{OC}) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an integer greater or equal to $2$ and let $(M, \\cdot)$ be a finite semi-group, such that $a^p \\neq a$, for all $a \\in M \\setminus \\{e\\}$, where $e$ is the null element in $M$. Prove that $(M, \\cdot)$ is a group.\n\nTraian Preda",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers $a, b, c$ which satisfy the equality:\n$$\n\\frac{a+b}{2} + \\frac{a^2+b^2}{2} = \\frac{7c+1}{c+1}.\n$$",
    "options": [],
    "answer": "(a,b,c) = (1,2,1), (2,1,1), (2,2,5)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest possible sum of the digits of a non-null multiple of $99$.",
    "options": [],
    "answer": "18",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be three non-negative real numbers such that $a + b + c = 3$. Prove the inequality:\n$$\n\\frac{a}{1 + b} + \\frac{b}{1 + c} + \\frac{c}{1 + a} \\ge \\frac{1}{1 + a} + \\frac{1}{1 + b} + \\frac{1}{1 + c}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right triangle in $A$ and points $D$ and $E$ on $AB$ such that $\\angle ACD = \\angle DCE = \\angle ECB$. Show that if $3\\overrightarrow{AD} = 2\\overrightarrow{DE}$ and $\\overrightarrow{CD} + \\overrightarrow{CE} = 2\\overrightarrow{CM}$, then $\\overrightarrow{AB} = 4\\overrightarrow{AM}$.\nGabriel Popa",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{F}$ be the set of continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ which satisfy the condition\n$$\ne^{f(x)} + f(x) \\geq x + 1,$$\nfor any real number $x$. Find the minimum value attained by the integral\n$$\nI(f) = \\int_{0}^{e} f(x) \\, dx,\n$$\nfor $f$ lying in $\\mathcal{F}$.",
    "options": [],
    "answer": "3/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be natural numbers such that $a + b + c + d = 2018$. Find the minimum value of the expression:\n$$\nE = (a - b)^2 + 2(a - c)^2 + 3(a - d)^2 + 4(b - c)^2 + 5(b - d)^2 + 6(c - d)^2.\n$$",
    "options": [],
    "answer": "14",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number\n$$\n\\sqrt[n]{\\sqrt{2018} + \\sqrt{2017}} + \\sqrt[n]{\\sqrt{2018} - \\sqrt{2017}}\n$$\nis not rational for $n \\ge 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the square $ABCD$, let $E$ be a point on the side $[AB]$, and $F$ be the foot of the perpendicular line from $B$ on the line $DE$. Let $L$ be a point on the line $DE$ so that $F$ is situated between $E$ and $L$, and $FL = BF$. If $N$ is the symmetric of $A$ with respect to the line $DE$ and $P$ is the symmetric of $F$ with respect to $BL$, prove that:\n\na) $BFLP$ is a square and $ALND$ is a rhombus;\n\nb) the area of the rhombus $ALND$ is equal with the difference between the areas of the squares $ABCD$ and $BFLP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in [1, \\infty)$. Prove that:\n$$\n\\frac{a\\sqrt{b}}{a+b} + \\frac{b\\sqrt{c}}{b+c} + \\frac{c\\sqrt{a}}{c+a} + \\frac{3}{2} \\le a+b+c.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $E$ and $F$ be two arbitrary points on the sides $AB$ and $AC$, respectively. The circumcircle of triangle $AEF$ meets the circumcircle of triangle $ABC$ again at point $M$. Let $D$ be the reflection of point $M$ across the line $EF$ and let $O$ be the circumcenter of triangle $ABC$. Prove that $D$ is on $BC$ if and only if $O$ belongs to the circumcircle of triangle $AEF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Then $\\angle AEF = \\angle AMF < \\angle AMC = \\angle ABC$, therefore the line $EF$ does meet the line $BC$ at a point $G$ such that $B$ is between $C$ and $G$. It is known that the circumcircles of triangles $ABC$, $AEF$, $EBG$ and $FCG$ have a common point, Miquel's point of the complete quadrilateral $BCFEAG$. The circumcircles of triangles $ABC$ and $AEF$ meet (again) at $M$, hence the quadrilaterals $MGBE$ and $MFCG$ are cyclic. It follows that:\n$D \\in BC \\Leftrightarrow \\angle MGE = \\angle CGE \\Leftrightarrow \\angle AEM = 2\\angle ABM = \\angle AOM \\Leftrightarrow O$ belongs to the circumcircle of triangle $AEF$.\n![](attached_image_1.png)\nThe fact that the quadrilateral $MGBE$ is cyclic can be proven easily: $\\angle EGB = \\angle EBC - \\angle BEG = \\angle ABC - \\angle AEF = \\angle AMC - \\angle AMF = \\angle FMC$, hence $MFCG$ is cyclic.\nThen, $\\angle BMC = \\angle BAC = \\angle EAF = \\angle EMF$, which leads to $\\angle EGB = \\angle BME$ and $\\angle BME = \\angle BGE$, which shows that the quadrilateral $GBEM$ is indeed cyclic.\n\n\n*Second solution.* (given in the contest by Ioana Popescu)\n* If $O$ lies on the circumcircle of $AEF$, then $\\angle AEM = \\angle AOM = 2 \\cdot \\angle ABM$, hence $\\angle MEB = 180^\\{\\circ\\} - 2 \\cdot \\angle ABM$, which shows that the triangle $MEB$ is isosceles. Similarly, triangle $MFC$ is also isosceles. Moreover, the two triangles are similar.\nNotice that we have a spiral similarity centered at $M$. Consider the point $T$ such that triangles $MTD$ and $MEB$ are similar. Then $T$ belongs to the line $EF$ (because $EF$ is the perpendicular bisector of the line segment $DM$), so it follows that $D \\in BC$. (For a spiral similarity, if one of the points glides on a line, like in this case $T$ glides on $EF$, then its image glides on the line \"similar\" to $EF$. In our case, $E \\mapsto B$, $F \\mapsto C$, $T \\mapsto D$ and $T \\in EF$, therefore $D \\in BC$.)\n* Assume that $D$ is on $BC$. Triangles $MEB$ and $MFC$ remain similar (AA) (be we no longer know that they are isosceles) and, again, a spiral similarity centered at $M$ takes $E \\mapsto B$, $F \\mapsto C$. There exists a point $T \\in EF$ that is taken by the similarity into $D$. Triangles $MTD$ and $MEB$ are similar, but $EF$ is the perpendicular bisector of the line segment $MD$, hence triangle $MTD$ is isosceles. It follows that triangle $MEB$ is also isosceles, hence $\\angle EMB = \\angle EBM$, which leads to $\\angle AEM = 2 \\cdot \\angle ABM$ and $\\angle AEM = \\angle AOM$. This shows that $O$ is on the circumcircle of triangle $AEF$.\n**Remark:** The result remains valid in the case the circumcircles of triangles $AEF$ and $ABC$ are tangent, in which case we consider $M = A$. Indeed, the homothety centered at $A$ transforming the circumcircle of triangle $AEF$ into the circumcircle of triangle $ABC$ takes the line segment $EF$ into a parallel line segment, $BC$. Then $D \\in BC$ if and only if $EF$ is a midline, which is equivalent to $AEOF$ being cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > 5$ be a prime number and $S = \\{p - n^2 \\mid n \\in \\mathbb{N}, n^2 < p\\}$. Prove that $S$ contains two elements $a$ and $b$ such that $1 < a < b$ and $a$ divides $b$.\n\nBMO, 1996",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the smallest element of $S$ that is greater than $1$ divides a larger element of $S$. If $p$ is of the form $m^2 + 1$, with $m \\in \\mathbb{N}$, we show that $p - (m - 1)^2 = 2m$ divides $p^2 - 1 = m^2$. Indeed, as $m$ is even, it follows that $2m \\mid m^2$.\n\nIf $p$ cannot be written as $m^2 + 1$, $m \\in \\mathbb{N}$, $p$ cannot be written as $m^2 + 2m$ either (this is a composite number because $m > 1$), hence $m^2 + 1 < p < m^2 + 2m$ for some $m \\in \\mathbb{N}$ ($m \\ge 2$). We show that $p - m^2$, which is an element of $S$ that is larger than $1$, divides another element of $S$. The condition that $p - m^2$ divides one of the numbers $p - n^2$ with $n \\in \\{0, 1, 2, \\dots, m-1\\}$ is equivalent to $p - m^2$ dividing one of the numbers $m^2, m^2 - 1^2, m^2 - 2^2, \\dots, m^2 - (m-1)^2$. Being smaller than $2m$, $p - m^2$ divides one of the following $2m-1$ consecutive numbers: $1, 2, \\dots, m-1$, $m$, $m+1, \\dots, 2m-1$, hence it divides one of the differences $m^2 - 0^2, m^2 - 1^2, \\dots, m^2 - (m-1)^2$. Moreover, it does not divide $m^2$, because it would divide $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k > 2$ be a real number.\na) Prove that for all positive real numbers $x$, $y$ and $z$ the following inequality holds:\n$$\n\\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x} > 2\\sqrt{\\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\n$$\nb) Prove that there exist positive real numbers $x$, $y$ and $z$ such that\n$$\n\\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x} < k \\sqrt{\\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We have $\\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x} > 2\\sqrt{\\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}} \\Leftrightarrow x+y+z + \\sqrt{x^2+xy+yz+zx} + \\sqrt{y^2+xy+yz+zx} + \\sqrt{z^2+xy+yz+zx} > 2 \\cdot \\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}$. But\n$\\sqrt{x^2+xy+yz+zx} > x$, $\\sqrt{y^2+xy+yz+zx} > y$ and $\\sqrt{z^2+xy+yz+zx} > z$, hence\n$$\nx+y+z+\\sqrt{x^2+xy+yz+zx}+\\sqrt{y^2+xy+yz+zx}+\\sqrt{z^2+xy+yz+zx} > 2(x+y+z).\n$$\nIt is sufficient to show that $x + y + z \\ge \\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}$, i.e. $(x + y + z)(xy + yz + zx) \\ge (x + y)(y + z)(z + x)$. After some computations, the previous inequality comes to $xyz \\ge 0$, which is obviously true.\n\nb) Fix $z = 1$. We look for $x, y > 0$, with $y = x$, such that\n$$\n\\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x} < k \\sqrt{\\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\n$$\ni.e. $2\\sqrt{x+1} + \\sqrt{2x} < k\\sqrt{\\frac{2(x+1)^2}{x+2}}$, or $\\sqrt{\\frac{x+2}{2(x+1)}}(2 + \\sqrt{\\frac{2x}{x+1}}) < k$.\nAs $\\sqrt{\\frac{x+2}{2(x+1)}} < 1$, it is sufficient to find $x$ such that $2 + \\sqrt{\\frac{2x}{x+1}} < k$, i.e. $\\sqrt{\\frac{2x}{x+1}} < k - 2$, or, equivalently, $\\frac{2x}{x+1} < (k-2)^2$. Putting $t = (k-2)^2 > 0$, the previous condition is fulfilled by any $x > 0$ if $t \\ge 2$, while in case that $t < 2$, it reduces to $x < \\frac{t}{2-t}$. Since $\\frac{t}{2-t} > 0$, there exist positive numbers $x$, $y$, $z$ that fulfill the conditions of the statement.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$, let $G$ be its centroid and $D$ the foot of the altitude from $A$. The line $DG$ meets the small arc $BC$ of the circumcircle of triangle $ABC$ at point $E$. Prove that the line $AB$ is tangent to the circumcircle of triangle $BDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F$ be the point in which the parallel through $A$ to $BC$ intersects again the circumcircle of triangle $ABC$. We prove that the points $D$, $G$ and $F$ are collinear. $AFCB$ is a cyclic trapezoid, hence a cyclic one. If $T$ is the orthogonal projection of point $F$ onto $BC$, then $AFTD$ is a rectangle. It is easy to prove that triangles $ABD$ and $FCT$ are equal; it follows that $BD = CT$, i.e. $M$ is the midpoint of the line segment $DT$. It follows that $\\frac{DM}{FA} = \\frac{DM}{DT} = \\frac{1}{2} = \\frac{GM}{GA}$ and, since $\\angle GMD = \\angle GAF$, triangles $GMD$ and $GAF$ are similar. We deduce that $\\angle DGM = \\angle FGA$, i.e. points $D$, $G$, $F$ are collinear.\n\nThen $\\angle BED = \\angle BEF = \\angle BCF = \\angle ABC = \\angle ABD$, which shows that the line $AB$ is tangent to the circumcircle of triangle $BDE$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38777,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $n$ transparent boxes there are red balls and blue balls. One needs to choose 50 boxes such that, together, they contain at least half of the red balls and at least half of the blue balls. Is such a choice possible irrespective on the number of balls and on the way they are distributed in the boxes, if:\na) $n = 100$;\nb) $n = 99$?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "a) The answer is in the negative. If we have 100 boxes, and 25 of the boxes only contain one red ball each, while the other 75 boxes contain just one blue ball each, then one needs to choose at least 13 of the boxes containing red balls and at least 38 of the boxes containing blue balls, which means one would need to choose at least 51 boxes. Thus, such a choice is not always possible.\n\nb) We prove by induction that from $2n + 1$ boxes one can always choose $n + 1$ that contain at least half of the balls of both colors.\n\nFor $n = 1$: if choosing two boxes would not be possible, then the remaining box must contain the majority of the red balls, or the majority of the blue balls. But there is only one such choice for each color, and there are 3 ways of choosing the boxes, so at least one of the choices must work.\n\nThe inductive step. We say two boxes are comparable if one of them has at least as many red balls as the other one and at least as many blue balls as the other one. If two comparable boxes exist, we eliminate them and apply the inductive hypothesis to the remaining $2n - 1$ boxes. Then, to the $n$ boxes thus chosen, we add that box of the two comparable ones that contains more red and more blue balls. If no two boxes are comparable, we denote by $a_i$ and $b_i$ the number of red balls and blue balls in box no. $i$, respectively. Changing the order of the boxes, one may assume that $a_1 > a_2 > \\dots > a_{99}$ and $b_1 < b_2 < \\dots < b_{99}$. In this case, simply choose the boxes whose ranks are odd (boxes 1, 3, 5, \\dots, 99).\nAn easier solution to point b) (given in the contest by Dinu Iosifescu). Denoting by $r_i$ the number of red balls in the box $i$, we may assume, without loss of generality, that $r_1 \\ge r_2 \\ge \\dots \\ge r_{99}$. Then we can choose box no. 1, then, from box no. 2 and box no. 3 we choose the one containing more blue balls, from box no. 4 and box no. 5 we choose the one containing more blue balls, and so on from box no. 98 and box no. 99 we choose the one containing more blue balls. It is clear that this choice satisfies the conditions in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider real numbers $a, b, c$ satisfying $1 < b \\le c^2 \\le a^{10}$, and\n$$\n\\log_a b + 2 \\log_b c + 5 \\log_c a = 12.\n$$\nShow that\n$$\n2 \\log_a c + 5 \\log_b c + 10 \\log_b a \\ge 21.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [a, b] \\to \\mathbb{R}$ be an integrable function, and let $(a_n)_{n \\ge 1}$ be a sequence of positive real numbers such that $\\lim_{n \\to \\infty} a_n = 0$.\n\na) If $A = \\{m \\cdot a_n \\mid m, n \\in \\mathbb{N}^*\\}$, prove that any open interval of positive real numbers has elements from the set $A$.\n\nb) If, for any $n \\in \\mathbb{N}^*$ and any $x, y \\in [a, b]$, such that $|x - y| = a_n$, the inequality\n$$\n\\left| \\int_x^y f(t) \\, dt \\right| \\le |x - y|\n$$\nholds, prove that:\n$$\n\\left| \\int_x^y f(t) \\, dt \\right| \\le |x - y|, \\quad \\forall x, y \\in [a, b].\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive integers $\\overline{abcd}$ knowing that the following conditions hold:\na) $a^2 + b^2 + c^2 + d^2$ is divisible by $4$;\nb) the remainder of $\\overline{abcd}$ when divided by $c$ is $7$.",
    "options": [],
    "answer": "All four-digit numbers with c = 9, and a, b, d ∈ {1,3,5,7,9}, such that a + b + d ≡ 7 (mod 9). There are no solutions with c = 8.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38781,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rhombus $ABCD$ has $m(\\widehat{BAD}) = 30^\\circ$, and the bisector of the angle $\\angle ADB$ intersects the side $[AB]$ in the point $E$. Find $m(\\widehat{DEC})$.",
    "options": [],
    "answer": "45°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the rectangular cuboid $ABCD A'B'C'D'$. Denote by $M, N, P$ the midpoints of $[AB]$, $[BC]$ and $[BB']$, respectively. Let $\\{O\\}$ be the intersection point of lines $A'N$ and $C'M$.\na) Prove that the points $D$, $O$ and $P$ are collinear.\nb) Show that $MC' \\perp (A'PN)$ if and only if $ABCDA'B'C'D'$ is a cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38783,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a blackboard one wrote the numbers $1, 2, 3, \\dots, 27$. One step consists in erasing three numbers $a, b, c$ from the blackboard and writing instead the number $a + b + c + n$, where $n$ is a fixed positive integer. Determine $n$ knowing that, after 13 steps, the number $n^2$ is written on the blackboard.",
    "options": [],
    "answer": "27",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38784,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(a_n)_{n \\ge 1}$ has the properties that $a_n > 1$ and $a_{n+1}^2 \\ge a_n a_{n+2}$, for $n \\ge 1$. Show that the sequence $(x_n)_{n \\ge 1}$, defined by $x_n = \\log_{a_n} a_{n+1}$, for $n \\ge 1$, has a finite limit. Find it.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$, $BE$, $CF$ be the altitudes of the triangle $ABC$ and let $K$, $L$, $M$ be the orthocenters of triangles $AEF$, $BFD$, $CDE$, respectively. Denote by $G_1$ and $G_2$ the barycenters of triangles $DEF$, $KLM$, respectively. Prove that $HG_1 = G_1G_2$, where $H$ is the orthocenter of $ABC$.\n\nSeverius Moldoveanu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that a continuous function $f: \\mathbb{R} \\to \\mathbb{R}$ is non-decreasing if and only if\n$$\n(c - b) \\int_{a}^{b} f(x) \\, dx \\leq (b - a) \\int_{b}^{c} f(x) \\, dx,\n$$\nfor all $a < b < c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of integers larger than $1$ and\n$$\nb_n = 1 - \\frac{1}{a_1} + \\frac{1}{a_1 a_2} - \\dots + (-1)^n \\frac{1}{a_1 a_2 \\dots a_n}, \\quad n = 1, 2, 3, \\dots\n$$\nProve that:\na) the sequence $(b_n)_{n \\ge 1}$ is convergent;\nb) if $(a_n)_{n \\ge 1}$ is unbounded, then the limit of the sequence $(b_n)_{n \\ge 1}$ is irrational;\nc) there exists a bounded sequence $(a_n)_{n \\ge 1}$ so that the limit of the sequence $(b_n)_{n \\ge 1}$ is irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38788,
    "subject": "Mathematics (Multi-modal)",
    "question": "The infinite grid of lines of the form $\\mathbb{R} \\times \\{m\\}$ and $\\{m\\} \\times \\mathbb{R}$, where $m$ runs through all integers, subdivide the Euclidean plane $\\mathbb{R} \\times \\mathbb{R}$ into $1 \\times 1$ cells. Let $S$ be the set-theoretic union of a finite number of such cells, and let $a$ be a positive real number less than or equal to $1/4$. Show that $S$ can be covered by a finite number of squares satisfying the following three conditions simultaneously:\n(1) Each square in the cover is an array of $1 \\times 1$ cells;\n(2) The squares in the cover have pairwise disjoint interiors; and\n(3) For each square $Q$ in the cover, the ratio of the area of $S \\cap Q$ to the area of $Q$ is at least $a$ and at most $a\\lfloor a^{-1/2} \\rfloor^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = \\lfloor a^{-1/2} \\rfloor$ and notice that $n \\ge 2$, since $a \\le 1/4$. Choose a large enough integer $k$ to cover $S$ by an $n^k \\times n^k$ array $Q$ so that the ratio of the area of $S$ to the area of $Q$ is at most $a$. Subdivide $Q$ into $n^2$ congruent square subarrays $Q'$, and notice that the ratio of the area of $S \\cap Q'$ to the area of $Q'$ does not exceed $an^2$. Remove the $Q'$ whose interiors are disjoint from $S$. Continuing, each of the remaining $Q'$ for which $(\\text{area}(S \\cap Q'))/(\\text{area} Q') < a$ is then subdivided into $n^2$ congruent square subarrays, and so on and so forth all the way down, to stop at stage $\\max\\{j: an^2(k-j) > 1\\} \\le k-2$ or earlier; this is because at stage $j$, for each square $Q^{(j)}$ in the subdivision, whose interior is not disjoint from $S$, the ratio of the area of $S \\cap Q^{(j)}$ to the area of $Q^{(j)}$ is at least $n^{2(j-k)}$, and of these $Q^{(j)}$ only those for which this ratio is less than $a$ are subject to further subdivision.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the natural numbers $a \\neq 0$, $b = 2a + 1000$, $c = a + 1$ and $d = 2a + 1002$.\n\na) Show that $\\frac{a}{b} < \\frac{c}{d}$.\n\nb) For $a = 9$, find the least natural number $n$ such that\n$$\n\\frac{a+n}{b+n} > \\frac{c+n}{d+n}.\n$$",
    "options": [],
    "answer": "1001",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $ABC$ an acute triangle with $AB < AC$, let $AD$ be the altitude and $AE$ the angle bisector, where $D, E \\in (BC)$. Given the acute triangle $A'B'C'$, with $A'B' < A'C'$, let $A'D'$ be the altitude and $A'E'$ the angle bisector, where $D', E' \\in (B'C')$.\nIf $[AB] \\equiv [A'B']$, $[AD] \\equiv [A'D']$ and $[AE] \\equiv [A'E']$, prove that triangles $ABC$ and $A'B'C'$ are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number with $n \\ge 2$. Prove that, for any complex numbers $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$, the following statements are equivalent:\n$$\na) \\quad \\sum_{k=1}^{n} |z - a_k|^2 \\le \\sum_{k=1}^{n} |z - b_k|^2, \\text{ for any } z \\in \\mathbb{C};\n$$\n$$\nb) \\quad \\sum_{k=1}^{n} a_k = \\sum_{k=1}^{n} b_k \\text{ and } \\sum_{k=1}^{n} |a_k|^2 \\le \\sum_{k=1}^{n} |b_k|^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38792,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ one draws the equilateral triangles $ABE$ and $BCF$, so that the points $D$ and $E$ are on the same side of the line $AB$, and points $F$ and $D$ are on different sides of the line $BC$. If $E$, $D$ and $F$ are collinear points, prove that $ABCD$ is a rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the rectangle $ABCD$ and the arbitrary points $E \\in (CD)$ and $F \\in (AD)$. The perpendicular line from $E$ on the line $FB$ intersects $BC$ in $P$. The perpendicular line from $F$ on the line $EB$ intersects $AB$ in $Q$. Show that the points $P, D$ and $Q$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win?",
    "options": [],
    "answer": "No; Bob does not have a winning strategy.",
    "solution": "The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\nWe say that two unit grid-segments form a *low-left corner* (or *LL-corner*) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An *up-right corner* (or *UR-corner*) is defined similarly. The common endpoint of two unit grid-segments at a corner is the *joint* of that corner.\nFix a vertical line on the grid and call it the *midline*; the unit grid-segments along the midline are called *middle segments*. The unit grid-segments lying to the left/right of the midline are called *left/right segments*. Partition all left segments into LL-corners, and all right segments into UR-corners.\nWe now describe Ann's strategy. Her first move consists in orienting some middle segment arbitrarily. Assume that at some stage, Bob orients some segment $s$. If $s$ is a middle segment, Ann orients any free middle segment arbitrarily. Otherwise, $s$ forms a corner in the partition with some other segment $t$. Then Ann orients $t$ so that the joint of the corner is either the source of both arrows, or the target of both. Notice that after any move of Ann's, each corner in the partition is either completely oriented or completely not oriented. This means that Ann can always make a required move.\nAssume that Bob wins at some stage, i.e., an oriented cycle $C$ occurs. Let $X$ be the lowest of the leftmost points of $C$, and let $Y$ be the topmost of the rightmost points of $C$. If $X$ lies (strictly) to the left of the midline, then $X$ is the joint of some corner whose segments are both oriented. But, according to Ann's strategy, they are oriented so that they cannot occur in a cycle — a contradiction. Otherwise, $Y$ lies to the right of the midline, and a similar argument applies. Thus, Bob will never win, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function which has the Darboux property. Prove that if $f$ is injective on the set of irrational numbers, then $f$ is continuous on $\\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB > AC$. Point $P \\in (AB)$ is such that $\\angle ACP = \\angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Q$ be the point in which the circumcircle of $BCD$ meets again the line $AB$. Then $\\angle QEA = \\angle QBC = \\angle ECP$, hence $EQ \\parallel PC$. Moreover, $\\angle ECP = \\angle ECD$ implies $QD$ is parallel to $EC$, hence $EQ = CD = CP$. It follows that $EQCP$ is a parallelogram, which leads to the conclusion.\n\n![](attached_image_1.png)\n\nSecond solution. (Given in the contest.) If $\\{O\\} = BD \\cap AC$, it is easy to prove that $BA$ and $BO$ are isogonal in the angle $\\angle EBC$, therefore, in order to show that $BA$ is the median, one has to prove that $BO$ is the symmedian. This follows readily by computation, using Steiner's Theorem.\nIndeed, $\\angle EBA = \\angle BAC - \\angle BEA = 180^\\circ - \\angle ABC - \\angle ACB - \\angle BDC = 180^\\circ - \\angle ECD - \\angle ECB - \\angle BDC = \\angle CBD$.\nTriangles $OEB$ and $ODC$ are similar, hence $\\frac{EO}{OD} = \\frac{EB}{CD}$. (1)\nTriangles $OCB$ and $ODE$ are also similar, therefore $\\frac{OC}{OD} = \\frac{BC}{ED}$. (2)\nWe have $\\angle ADC = \\angle APC = \\angle PBC + \\angle PCB = \\angle ACB = \\angle EDB$, which shows that the triangles $ACD$ and $EBD$ are similar. Hence $ED = \\frac{EB \\cdot AD}{AC}$ and, as $AD = \\frac{CD \\cdot AC}{BC}$ (from $\\Delta ACB \\sim \\Delta APC \\sim \\Delta ADC$), it follows that $ED = \\frac{EB \\cdot CD}{BC}$.\nFrom (1) and (2) we obtain $\\frac{EO}{OC} = \\frac{EB \\cdot ED}{CD \\cdot BC} = \\left(\\frac{EB}{BC}\\right)^2$, which shows that $BO$ is indeed the symmedian.\n\nThird solution. (Given in the contest by *Andrei Mărginean*.) Let $a = \\angle EBA$, $x = \\angle ABC$. We have that $\\angle DCE = \\angle ACP = x$ and $\\angle EBC = a + x$. As $B$, $C$, $D$, $E$ are concyclic, it follows that $\\angle CDE = 180^\\circ - a - x$, $\\angle DEC = 180^\\circ - \\angle EDC - \\angle DCE = a = \\angle EBA$. But $AE$ is the perpendicular bisector of $[PD]$, hence $\\angle PEA = \\angle DEC = \\angle EBA$, which shows that the triangles $APE$ and $AEB$ are similar. It follows that $AE^2 = AB \\cdot AP$. But triangles $ACP$ and $ABC$ are also similar, hence $AC^2 = AP \\cdot AB = AE^2$, and the conclusion follows immediately.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ and $g: \\mathbb{R} \\to \\mathbb{R}$ be two quadratic functions with the following property: for any real number $r$, if $f(r)$ is an integer number, then $g(r)$ is also an integer number.\nProve that there are two integers $m$ and $n$ such that $g(x) = m f(x) + n$, for any real number $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38798,
    "subject": "Mathematics (Multi-modal)",
    "question": "The acute-angled triangle $ABC$ has circumcenter $O$. The straight lines $B_aC_a$, $C_bA_b$ and $A_cB_c$ are perpendicular in $O$ on the straight lines $AO$, $BO$, respectively $CO$ and $A_b$, $A_c \\in BC$, $B_a, B_c \\in AC$, $C_b, C_a \\in AB$. Denote $O_a, O_b, O_c$ the circumcenters of the triangles $AC_aB_a, BA_bC_b, CB_cA_c$. Prove that\n$$\n\\vec{O_aA} + \\vec{O_bB} + \\vec{O_cC} = \\vec{0} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f:[a, b] \\longrightarrow \\mathbb{R}$ o funcţie integrabilă, iar $(a_{n})_{n \\geq 1}$ un şir de numere reale strict pozitive cu proprietatea că $\\lim_{n \\rightarrow \\infty} a_{n}=0$.\n\na) Dacă $A=\\{m \\cdot a_{n} \\mid m, n \\in \\mathbb{N}^{*}\\}$, arătaţi că orice interval deschis de numere strict pozitive conţine elemente din $A$.\n\nb) Dacă pentru orice $n \\in \\mathbb{N}^{*}$ şi orice $x, y \\in [a, b]$ cu $|x-y|=a_{n}$ are loc inegalitatea $\\left|\\int_{x}^{y} f(t) d t\\right| \\leq |x-y|$, arătaţi că:\n$$\n\\left|\\int_{x}^{y} f(t) d t\\right| \\leq |x-y|, \\quad \\forall x, y \\in [a, b]\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Deoarece $\\lim_{n \\rightarrow \\infty} a_{n}=0$, pentru orice $c, d>0$, cu $c<d$, există $n \\in \\mathbb{N}^{*}$ cu $a_{n}<d-c$. Pentru $m=\\left\\lfloor\\frac{c}{a_{n}}\\right\\rfloor+1$ rezultă atunci că $m \\cdot a_{n} \\in (c, d) \\cap A$.\n\nb) Funcţia $f$ fiind integrabilă, este mărginită. Fie $M>0$, cu $\\operatorname{Im}(f) \\subseteq [-M, M]$.\n\nFie $m, n \\in \\mathbb{N}^{*}$ fixate şi $x, y \\in [a, b]$ cu $|x-y|=m \\cdot a_{n}$. Arătăm că\n$$\n\\left|\\int_{x}^{y} f(t) d t\\right| \\leq |x-y|\n$$\nDefinim, pentru $k=\\overline{0, m}$, numerele $z_{k} \\in [a, b]$ prin\n$$\nz_{k}=x+\\frac{k}{m} \\cdot (y-x)=\\left(1-\\frac{k}{m}\\right) \\cdot x+\\frac{k}{m} \\cdot y\n$$\nRezultă că $|z_{k}-z_{k-1}|=a_{n}$, pentru orice $k=\\overline{1, m}$, şi\n$$\n\\left|\\int_{x}^{y} f(t) d t\\right|=\\left|\\int_{z_{0}}^{z_{m}} f(t) d t\\right|=\\left|\\sum_{k=1}^{m} \\int_{z_{k-1}}^{z_{k}} f(t) d t\\right| \\leq \\sum_{k=1}^{m}\\left|\\int_{z_{k-1}}^{z_{k}} f(t) d t\\right| \\leq \\sum_{k=1}^{m}|z_{k}-z_{k-1}|=|x-y|\n$$\nFie acum $x, y \\in [a, b]$ oarecare şi $d=|x-y|$. Pentru $d=0$, inegalitatea cerută este evidentă. Presupunem în continuare $d>0$. Cum $A$ este densă în $[0, \\infty)$, există un şir $(d_{n})_{n \\geq 1} \\subset A$ cu proprietatea că $d_{n} \\nearrow d$. Considerăm\n$$\ny_{n}=x+\\frac{d_{n}}{d} \\cdot (y-x)=\\left(1-\\frac{d_{n}}{d}\\right) \\cdot x+\\frac{d_{n}}{d} \\cdot y\n$$\nAtunci $y_{n} \\in [a, b]$, $|y_{n}-x| \\in A$ şi $y_{n} \\rightarrow y$.\n\nRezultă că\n$$\n\\left|\\int_{y_{n}}^{y} f(t) d t\\right| \\leq M \\cdot |y-y_{n}| \\longrightarrow 0\n$$\nObţinem atunci că\n$$\n\\left|\\int_{x}^{y} f(t) d t\\right|=\\left|\\int_{x}^{y_{n}} f(t) d t+\\int_{y_{n}}^{y} f(t) d t\\right| \\leq \\left|\\int_{x}^{y_{n}} f(t) d t\\right|+\\left|\\int_{y_{n}}^{y} f(t) d t\\right| \\leq |x-y_{n}|+\\left|\\int_{y_{n}}^{y} f(t) d t\\right|\n$$\nTrecând la limită în ultima inegalitate, obţinem inegalitatea cerută.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38800,
    "subject": "Mathematics (Multi-modal)",
    "question": "For real $a < b$, and $f : (a, b) \\to \\mathbb{R}$ a function such that the functions defined by $g : (a, b) \\to \\mathbb{R}$, $g(x) = (x-a)f(x)$, and $h : (a, b) \\to \\mathbb{R}$, $h(x) = (x-b)f(x)$ are non-decreasing, prove that $f$ is continuous.\n\nVladimir Cerbu and Sorin Rădulescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the real numbers $k$ such that\n$$\n\\sqrt{3(a^2 + b^2 + c^2)} \\leq a + b + c + k\\sqrt{a^2 + b^2 + c^2 - ab - bc - ca},\n$$\nfor every positive real numbers $a$, $b$, $c$.",
    "options": [],
    "answer": "k ≥ sqrt(3) - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38802,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $BC$ of a parallelogram $ABCD$ lies a given point $M$. Find a compass-and-straightedge construction of a point $N \\in CD$, so that the straight line $BD$ separates triangle $AMN$ into two equivalent areas.\n\nCecilia Deaconescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a positive integer $n \\ge 2$ and a $n \\times n$ square (see the figure). The main diagonal of this square consists of the hatched squares. We write 0 into the $1 \\times 1$ squares situated below the main diagonal and nonzero natural numbers into the other $1 \\times 1$ squares (including the hatched ones). After filling all the squares, we calculate the sum of the numbers situated on each row and on each column, obtaining $2n$ sums.\n\n![](attached_image_1.png)\n\nWe call the $n \\times n$ square *lucky* if the $2n$ sums are, in a certain order, equal to $1, 2, \\dots, 2n$.\n\na) Show that there doesn't exist any $5 \\times 5$ lucky square.\n\nb) Determine the largest number that may appear in a $4 \\times 4$ lucky square.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number with $n \\ge 2$. For any real numbers $a_1, a_2, \\dots, a_n$, we denote $S_0 = 1$ and define $S_k$ by\n$$\nS_k = \\sum_{1 \\le i_1 < i_2 < \\dots < i_k \\le n} a_{i_1} a_{i_2} \\dots a_{i_k},\n$$\nthat is, the sum of all products of any $k$ numbers chosen among $a_1, a_2, \\dots, a_n$, $k \\in \\{1, 2, \\dots, n\\}$.\nFind the number of all $n$-tuples $(a_1, a_2, \\dots, a_n)$ for which the following relation holds:\n$$\n(S_n - S_{n-2} + S_{n-4} - \\dots)^2 + (S_{n-1} - S_{n-3} + S_{n-5} - \\dots)^2 = 2^n S_n.\n$$",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38805,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a box there are $50$ cards on which the first $100$ positive integers are written as follows: on the first card there are $1$ (on one side) and $2$ (on the other side), on the second card there are $3$ (on the one side) and $4$ (on the other side) and so on, up to the $50$th card, on which the numbers $99$ (on one side) and $100$ (on the other side) are written. Eliza pulls out four cards from the box and calculates the sum of the $8$ numbers written on the four cards. How many distinct sum can Eliza get?",
    "options": [],
    "answer": "185",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the rectangular cuboid $ABCD A'B'C'D'$, let $M$ be the center of the face $ABB'A'$. Denote by $M_1$ the foot of the perpendicular from $M$ to the line $B'C$ and by $M_2$ the foot of the perpendicular from $M$ to $AD'$. Prove that:\n\na) $[MM_1] = [MM_2]$;\n\nb) the line $d$ of intersection between the planes $(MM_1M_2)$ and $(ABC)$ is parallel to $AD$;\n\nc) the dihedral angle between the planes $(MM_1M_2)$ and $(ABC)$ is $45^\\circ$ if and only if $\\frac{BC}{AB} = \\frac{BB'}{BC} + \\frac{BC}{BB'}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number with $n \\ge 2$ and let $\\mathbf{A}$ be a matrix in $M_n(\\mathbb{C})$, such that $\\mathbf{A}$ and $\\mathbf{A}^2$ have different ranks. Prove that there exists a non-zero matrix $\\mathbf{B}$ in $M_n(\\mathbb{C})$, such that $\\mathbf{A}\\mathbf{B} = \\mathbf{B}\\mathbf{A} = \\mathbf{B}^2 = \\mathbf{O}_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lucia has a total of 2018 yellow, blue and green balls. There are 4 times more green balls than blue balls. In one *exchange* Lucia offers 13 yellow balls to her friend Cristina and receives 5 blue balls and 7 green balls. After a certain number of such *exchanges*, Lucia has no yellow balls, but she has 1271 green balls.\nDetermine how many yellow balls did Lucia have in the beginning.",
    "options": [],
    "answer": "1313",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for any positive integer $n$, the number $18^n$ is a sum of three squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the natural number $n$ for which $\\sqrt{\\frac{20^n - 18^n}{19}}$ is a rational number.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $C$ be a circle whose circumference is equal to $6n$. The circle $C$ is divided by $3n$ points into $3n$ small arcs such that $n$ of them have the lengths equal with $1$, other $n$ have the lengths equal with $2$ and the remaining $n$ arcs have the lengths equal with $3$. Prove that among the given $3n$ points there are two which are antipodal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A set $A$ of positive integers is $n$-size complete if the set of all the remainders obtained when dividing an element of $A$ by an element of $A$ is $\\{0, 1, 2, \\dots, n\\}$. For example, the set $\\{3, 4, 5\\}$ is a 4-size complete set.\nDetermine the minimum number of elements of a 100-size complete set.",
    "options": [],
    "answer": "51",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38813,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Consider the positive integers $a$, $b$, $c$ so that $a < b < c$ and $a^2 + b^2 = c^2$. If $a_1 = a^2$, $a_2 = ab$, $a_3 = bc$, $a_4 = c^2$, prove that $a_1^2 + a_2^2 + a_3^2 = a_4^2$ and $a_1 < a_2 < a_3 < a_4$.\n\nb. Show that for any $n \\in \\mathbb{N}$, $n \\ge 3$, there exist the positive integers $a_1, a_2, \\dots, a_n$ so that $a_1^2 + a_2^2 + \\dots + a_{n-1}^2 = a_n^2$ and $a_1 < a_2 < \\dots < a_{n-1} < a_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $A$ be a matrix from $M_n(\\mathbb{C})$. Prove that $A^2 = O_n$ if and only if there exist two matrices $B$ and $C$ in $M_n(\\mathbb{C})$ so that $A = BC$ and $CB = O_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a function $f : \\mathbb{R} \\to \\mathbb{R}$. For each $a \\in \\mathbb{Z}$ denote by $f_a : \\mathbb{R} \\to \\mathbb{R}$, the function defined by $f_a(x) = (x - a)f(x)$. Prove that if there are infinitely many values of $a \\in \\mathbb{Z}$ such that functions $f_a$ are non-decreasing, then $f$ is monotone.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38816,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $k \\in \\mathbb{Z}$, we define the polynomial $F_k = X^4 + 2(1 - k)X^2 + (1 + k)^2$. Find all integers $k \\in \\mathbb{Z}$, such that $F_k$ is irreducible over $\\mathbb{Z}$ but is reducible over $\\mathbb{Z}_p$, for any prime $p$.",
    "options": [],
    "answer": "All integers k such that k is neither a perfect square nor a negative perfect square.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers such that $ad \\neq bc$ and $\\text{gcd}(a, b, c, d) = 1$. Prove that, as $n$ runs through the positive integers, the values $\\text{gcd}(an + b, cn + d)$ may achieve from the set of all positive divisors of some integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We extend the problem statement by allowing $a$ and $c$ to take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \\le a \\le c$. Let $S(a, b, c, d) = \\{\\text{gcd}(an + b, cn + d) : n \\in \\mathbb{Z}_{>0}\\}$.\n\nNow we induct on $a$. We first deal with the inductive step, leaving the base case $a = 0$ to the end of the solution. So, assume that $a > 0$; we intend to find a 4-tuple $(a', b', c', d')$ satisfying the requirements of the extended problem, such that $S(a', b', c', d') = S(a, b, c, d)$ and $0 \\le a' < a$, which will allow us to apply the induction hypothesis.\n\nThe construction of this 4-tuple is provided by the step of the Euclidean algorithm. Write $c = az + r$, where $q$ and $r$ are both integers and $0 \\le r < a$. Then for every $n$\n\n$$\n\\text{gcd}(an+b, cn+d) = \\text{gcd}(an+b, q(an+b)+rn+d-qb) = \\text{gcd}(an+b, rn+(d-qb)),\n$$\n\nso a natural intention is to define $a' = r$, $b' = d - qb$, $c' = a$, and $d' = b$ (which are already shown to satisfy $S(a', b', c', d') = S(a, b, c, d)$). The check of the problem requirements is straightforward: indeed,\n\n$$\na'd' - b'c' = (c - qa)b - (d - qb)a = -(ad - bc) \\neq 0\n$$\n\nand\n\n$$\n\\text{gcd}(a', b', c', d') = \\text{gcd}(c - qa, b - qd, a, b) = \\text{gcd}(c, d, a, b) = 1.\n$$\n\nThus the step is verified.\n\nIt remains to deal with the base case $a = 0$, i.e., to examine the set $S(0, b, c, d)$ with $bc \\neq 0$ and $\\text{gcd}(b, c, d) = 1$. Let $b'$ be the integer obtained from $b$ by ignoring all primes $b$ and $c$ share (none of them divides $cn+d$ for any integer $n$, otherwise $\\text{gcd}(b, c, d) > 1$). We thus get $\\text{gcd}(b', c) = 1$ and $S(0, b', c, d) = S(0, b, c, d)$.\n\nFinally, it is easily seen that $S(0, b', c, d)$ is the set of all positive divisors of $b'$. Each member of $S(0, b', c, d)$ is clearly a divisor of $b'$. Conversely, if $\\delta$ is a positive divisor of $b'$, then $cn+d \\equiv \\delta \\pmod{b'}$ for some $n$, since $b'$ and $c$ are coprime, so $\\delta$ is indeed a member of $S(0, b', c, d)$.\nFor positive integers $s$ and $t$ and prime $p$, we will denote by $\\text{gcd}_p(s, t)$ the greatest common $p$-power divisor of $s$ and $t$.\n\n**Claim 1.** For any positive integer $n$, $\\text{gcd}(an + b, cn + d) \\mid ad - bc$.\n\n**Proof.** This is clear from the identity\n$$\na(cn + d) - c(an + b) = ad - bc. \\qquad (\\dagger)\n$$\n\n**Claim 2.** The set of values taken by $\\text{gcd}(an + b, cn + d)$ is exactly the set of values taken by the product\n$$\n\\prod_{p|ad-bc} \\text{gcd}_p(an_p + b, cn_p + d)\n$$\nas the $(n_p)_{p|ad-bc}$ each range over positive integers.\n\n**Proof.** From the identity\n$$\n\\text{gcd}(an + b, cn + d) = \\prod_{p|ad-bc} \\text{gcd}_p(an + b, cn + d),\n$$\nit is clear that every value taken by $\\gcd(an + b, cn + d)$ is also a value taken by the product (with all $n_p = n$). Conversely, it suffices to show that, given any positive integers $(n_p)_{p|ad-bc}$, there is a positive integer $n$ such that $\\gcd_p(an + b, cn + d) = \\gcd_p(an_p + b, cn_p + d)$ for each $p \\nmid ad - bc$. This can be achieved by requiring that $n$ be congruent to $n_p$ modulo a sufficiently large power of $p$ (using the Chinese Remainder Theorem).\n\nUsing Claim 2, it suffices to determine the sets of values taken by $\\gcd_p(an + b, cn + d)$ as $n$ ranges over all positive integers. There are two cases.\n\n**Claim 3.** If $p \\nmid a, c$, then $\\gcd_p(an + b, cn + d) = 1$ for all $n$.\n\n**Proof.** If $p \\mid an + b, cn + d$, then we would have $p \\mid a, b, c, d$, which is not the case.\n\n**Claim 4.** If $p \\nmid a$ or $p \\nmid c$, then the values taken by $\\gcd_p(an + b, cn + d)$ are exactly the $p$-power divisors of $ad - bc$.\n\n**Proof.** Assume without loss of generality that $p \\nmid a$. Then from identity $(\\dagger)$ we have $\\gcd_p(an + b, cn + d) = \\gcd_p(an + b, ad - bc)$. But since $p \\nmid a$, the arithmetic progression $an + b$ takes all possible values modulo the highest $p$-power divisor of $ad - bc$, and in particular the values taken by $\\gcd_p(an + b, ad - bc)$ are exactly the $p$-power divisors of $ad - bc$.\n\n**Conclusion.** Using claims 2, 3 and 4, we see that the set of values taken by $\\gcd(an + b, cn + d)$ is equal to the set of products of $p$-power divisors of $ad - bc$, where we only consider those primes $p$ not dividing $\\gcd(a, c)$. Thus the set of values of $\\gcd(an + b, cn + d)$ is equal to the set of divisors of the largest factor of $ad - bc$ coprime to $\\gcd(a, c)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two integers $n$ and $q$, $n \\ge 2$, $q \\ge 2$, and $q \\ne 1 \\pmod 4$. Let $K$ be a finite field having exactly $q$ elements. Show that for any $a$ from $K$, there is $x$ and $y$ in $K$, such that $a = x^2 + y^2$. (It is known that any finite field is commutative.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38819,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a fixed integer $n \\ge 2$, determine complex numbers $z$ that satisfy the relations:\n$$\na)\\ z^n + z^{n-1} + \\dots + z^2 + |z| = n;\n$$\n$$\nb)\\ |z|^{n-1} + |z|^{n-2} + \\dots + |z|^2 + z = n z^n.\n$$",
    "options": [],
    "answer": "a) z = 1.\nb) z = 0 or z is a nonnegative real number r satisfying r + r^2 + ... + r^{n-1} = n r^n (equivalently, for r ≠ 1, r^{n-1}(n + 1 − n r) = 1).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38820,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of rooks one can place on a chessboard such that any rook attacks exactly two other rooks? (We say that two rooks attack each other if they are on the same line or on the same column and between them there are no other rooks.)",
    "options": [],
    "answer": "16",
    "solution": "We say that there can be two types of rooks on the chessboard: a rook of type **T1** is a rook that is attacked from two perpendicular lines; a rook is of type **T2** if it is attacked by two rooks situated on the same line, but opposite directions.\nSuppose one can place $m$ rooks of type **T1** and $n$ of type **T2**, with $m + n = x$. Each rook of type **T1** determines two lines from which it is attacked (out of the 16 of the chessboard: 8 horizontal and 8 vertical ones), while exactly one other rook of type **T1** is situated on each of these two lines. Notice that a rook of type **T2** can not be attacked from any other line than the one it is already attacked from, hence, for each of these $m$ rooks, there are $m$ lines on which there can be no other rooks. In total, we could have at most $\\frac{2n}{2} + m = n + m$ lines. Hence $16 \\ge m + n = x$. To prove that 16 is indeed the desired maximum, it is sufficient to exhibit an example of 16 rooks that attack exactly two other ones.\n\n![](attached_image_1.png)\nWe prove by induction after $n \\ge 2$ that on an $n \\times n$ board one can place at most $2n$ rooks with the above restrictions.\nFor $n = 2$ the statement is obvious.\nFor the inductive step, assume the statement above to hold for an arbitrary $n \\ge 2$ and let us prove it for $n + 1$.\nAssume that on an $(n+1) \\times (n+1)$ one could place at least $2n+3$ rooks such that each of them attacks exactly two other ones. From the Pigeonhole Principle it follows that there exists at least one horizontal line containing at least 3 rooks and, also, a vertical line with 3 or more rooks. On a horizontal line with 3 or more rooks there exists at least one rook that is attacked by two rooks situated on the same horizontal line with it. Therefore this rook must stand alone on the vertical line it occupies. Similarly, there must be a horizontal line containing exactly one rook. Eliminating these two lines, we obtain an $n \\times n$ board with $2n + 1$ rooks that still satisfies the condition that each rook attacks exactly two other rooks. This contradicts our inductive hypothesis for $n$.\nAgain, an example with 16 rooks finishes the proof. (Andrei gave the one from the first solution.)\nLet us consider a configuration of rooks attacking each other. A rook can attack on 4 rays, 2 by 2 opposite. On exactly 2 of these rays one must have rooks, on each of the other two rays the rook attacks one of the 32 segments that constitute the margins of the chessboard. Each of these 32 segments is attacked by a different rook, and each rook attacks two such segments, therefore one can have at most 16 rooks on the board.\nAn example with 16 rooks on the board finishes the proof. (Iustinian placed the rooks on the two diagonals.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a non-negative integer $k$, show that there are infinitely many positive integers $n$ such that the product of any $n$ consecutive integers is divisible by $(n+k)^2 + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Since the product of $n$ consecutive integers is divisible by $n!$, it is sufficient to show that there are infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$.\nTo obtain infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$, it is sufficient to consider large enough integers $m \\equiv 1 \\pmod{5}$ and let $n = 2m^2 - k$.\n\nIn this case, $(n + k)^2 + 1 = 4m^4 + 1 = (2m^2 + 2m + 1)(2m^2 - 2m + 1) = 5\\ell(2m^2 - 2m + 1)$, and $5 < \\ell < 2m^2 - 2m + 1 < 2m^2 - k = n$, so $n!$ is indeed divisible by $(n + k)^2 + 1$.\n\n*Second solution.* To obtain infinitely many positive integers $n$ such that $(n+k)^2 + 1$ divides $n!$, consider all pairs $(n, m)$ of positive integers satisfying $n^2 - 5m^2 = -1$. These pairs are completely described by\n$$\n\\begin{pmatrix} n_r \\\\ m_r \\end{pmatrix} = \\begin{pmatrix} 9 & 20 \\\\ 4 & 9 \\end{pmatrix}^r \\begin{pmatrix} 2 \\\\ 1 \\end{pmatrix}, \\quad r = 0, 1, 2, \\dots;\n$$\nthe $n_r$ and the $m_r$ both form strictly increasing sequences of positive integers.\nWrite $2(n_r^2 + 1) = 5 \\cdot m_r \\cdot 2m_r$ and notice that $m_r < 2m_r \\le \\sqrt{5m_r^2 - 1} - k = n_r - k$ for all but finitely many indices, to conclude that $(n_r - k)!$ is divisible by $n_r^2 + 1$ for all but finitely many indices.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentru $k \\in \\mathbb{Z}$ definim polinomul $F_{k}=X^{4}+2(1-k) X^{2}+(1+k)^{2}$. Să se determine toate valorile $k \\in \\mathbb{Z}$, astfel încât $F_{k}$ să fie ireductibil peste $\\mathbb{Z}$ si reductibil peste $\\mathbb{Z}_{p}$ pentru orice $p$ prim.",
    "options": [],
    "answer": "All integers k that are not of the form ± a perfect square",
    "solution": "Solution:\n\nVom arăta că numerele care satisfac condiţia cerută sunt toate numerele $k \\in \\mathbb{Z}$ care nu sunt de forma $\\pm l^{2}$, cu $l \\in \\mathbb{Z}$.\n\nArătăm că $F_{k}$ este reductibil peste $\\mathbb{Z}$ dacă şi numai dacă $F_{k}$ se descompune ca produs de două polinoame monice de grad 2.\n\nÎntr-adevăr, dacă $F_{k}$ are o rădăcină întreagă $m$, atunci\n\na. dacă $m=0$, atunci $k=-1$, si $F_{-1}=X^{2}\\left(X^{2}+4\\right)$.\n\nb. dacă $m \\neq 0$, atunci $-m$ este de asemenea rădăcină, şi $X^{2}-m^{2}$ divide $F_{k}$.\n\nDeci $F_{k}$ este reductibil peste $\\mathbb{Z}$ dacă şi numai dacă $F_{k}=\\left(X^{2}+a X+b\\right)\\left(X^{2}+c X+d\\right)$, cu $a, b, c, d \\in \\mathbb{Z}$. Prin identificarea coeficienţilor, avem că $a+c=0$, $a c+b+d=2(1-k)$, $a d+b c=0$ şi $b d=(1+k)^{2}$.\n\nDacă $a=0$, atunci $c=0$, $b+d=2(1-k)$, $b d=(1+k)^{2}$, de unde obţinem $(b-d)^{2}=4(1-k)^{2}-4(1+k)^{2}=-16 k$, astfel că $k=-l^{2}$, cu $l \\in \\mathbb{Z}$.\n\nDacă $a \\neq 0$, atunci $c=-a$, $b=d$, $b^{2}=(1+k)^{2}$, $2 b-a^{2}=2(1-k)$.\n\nDacă $b=-1-k$, rezultă $a^{2}=-4$, imposibil. Deci $b=1+k$ şi $a^{2}=4 k$, de unde $k=l^{2}$, cu $l \\in \\mathbb{Z}$.\n\nPrin urmare, $F_{k}$ este reductibil peste $\\mathbb{Z}$ dacă şi numai dacă $k= \\pm l^{2}$, cu $l \\in \\mathbb{Z}$.\n\nArătăm că $F_{k}$ este reductibil peste $\\mathbb{Z}_{p}$ cu $p$ prim, pentru orice $k \\in \\mathbb{Z}$.\n\nPentru $p=2$ avem că $F_{k}=X^{4}$ sau $F_{k}=X^{4}+\\hat{1}=(X+\\hat{1})^{4}$, deci $F_{k}$ este reductibil.\n\nFie $p$ număr prim impar. Putem presupune că $k \\not \\equiv 0\\ (\\bmod\\ p)$ şi $k \\not \\equiv-1\\ (\\bmod\\ p)$.\n\nCa mai sus, $F_{k}$ este reductibil peste $\\mathbb{Z}_{p}$ dacă şi numai dacă $F_{k}=\\left(X^{2}+\\hat{a} X+\\hat{b}\\right)\\left(X^{2}+\\hat{c} X+\\hat{d}\\right)$, cu $a, b, c, d \\in \\mathbb{Z}$, care verifică condiţiile $a+c \\equiv 0\\ (\\bmod\\ p)$, $a c+b+d \\equiv 2(1-k)\\ (\\bmod\\ p)$, $a d+b c \\equiv 0\\ (\\bmod\\ p)$ şi $b d \\equiv(1+k)^{2}\\ (\\bmod\\ p)$.\n\nDacă $a \\equiv 0\\ (\\bmod\\ p)$, avem că $c \\equiv 0\\ (\\bmod\\ p)$ şi $(b-d)^{2} \\equiv-16 k\\ (\\bmod\\ p)$.\n\nDacă $a \\not \\equiv 0\\ (\\bmod\\ p)$, atunci $c \\equiv-a\\ (\\bmod\\ p)$, $b \\equiv d\\ (\\bmod\\ p)$, $b^{2} \\equiv(1+k)^{2}\\ (\\bmod\\ p)$ şi $2 b-a^{2} \\equiv 2(1-k)\\ (\\bmod\\ p)$.\n\nPentru $b \\equiv-1-k\\ (\\bmod\\ p)$ avem că $a^{2} \\equiv-4\\ (\\bmod\\ p)$.\n\nPentru $b \\equiv 1+k\\ (\\bmod\\ p)$ avem că $a^{2} \\equiv 4 k\\ (\\bmod\\ p)$.\n\nCum $-16 k=-4 \\cdot 4 k$, cel puţin unul dintre elementele $-16 k$, $-4$ şi $4 k$ este pătrat în $\\mathbb{Z}_{p}$ modulo $p$, astfel că cel puţin una dintre ecuaţiile de mai sus are soluţii.\n\nCum $F_{k}=\\left(X^{2}+(1-k)\\right)^{2}-(-16 k)=\\left(X^{2}-(1+k)\\right)^{2}-(-4) X^{2}=\\left(X^{2}+(1+k)\\right)-(4 k) X^{2}$, rezultă că $F_{k}$ este reductibil peste $\\mathbb{Z}_{p}$, pentru orice $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $M$ be a point on the side $AC$, different from both $A$ and $C$. The line through $M$ and parallel to $BC$ crosses the side $AB$ at $N$. The segments $BM$ and $CN$ cross at $P$, and the circles $BNP$ and $CMP$ meet again at $Q$. Show that the angles $BAP$ and $CAQ$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nInvert from $A$ with radius $\\sqrt{AB \\cdot AM}$, then reflect in the internal bisectrix of the angle $BAC$, to obtain an involution $\\varphi$.\nSince the lines $BC$ and $MN$ are parallel, $AB \\cdot AM = AC \\cdot AN$, so $\\varphi B = M$ and $\\varphi C = N$. The circles $ABM$ and $ACN$ are therefore the images under $\\varphi$ of the lines $BM$ and $CN$, respectively.\n\nSince these two circles both pass through $Q$, by Miquel's theorem, and $\\varphi$ is an involution, $P = \\varphi Q$, so the lines $AP$ and $AQ$ are reflections of one another in the internal bisectrix of the angle $BAC$. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and fix $2n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\\overrightarrow{AB}$ and $\\overrightarrow{CD}$ such that $ABCD$ is a convex quadrangle oriented clockwise. Determine the number of good configurations.",
    "options": [],
    "answer": "\\binom{2n}{n}",
    "solution": "*First solution.* The required number is $\\binom{2n}{n}$. To prove this, trace the circumference counterclockwise to label the points $a_1, a_2, \\dots, a_{2n}$.\nLet $\\mathcal{C}$ be any good configuration and let $O(\\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-element subset $S$ of $\\{a_1, \\dots, a_{2n}\\}$ is an $O$-image of a unique good configuration; clearly, this provides the answer.\nTo prove the claim induct on $n$. The base case $n=1$ is clear. For the induction step, consider any $n$-element subset $S$ of $\\{a_1, \\dots, a_{2n}\\}$, and assume that $S = O(\\mathcal{C})$ for some good configuration $\\mathcal{C}$. Take any index $k$ such that $a_k \\in S$ and $a_{k+1} \\notin S$ (assume throughout that indices are cyclic modulo $2n$, i.e., $a_{2n+1} = a_1$ etc.).\nIf the arrow from $a_k$ points to some $a_\\ell$, $k+1 < \\ell < (2n+k)$, then the arrow pointing to $a_{k+1}$ emerges from some $a_m$, $m$ in the range $k+2$ through $\\ell-1$, since these two arrows do not cross. Then the arrows $a_k \\to a_\\ell$ and $a_m \\to a_{k+1}$ form a prohibited quadrangle. Hence, $\\mathcal{C}$ contains an arrow $a_k \\to a_{k+1}$.\nOn the other hand, if any configuration $\\mathcal{C}$ contains the arrow $a_k \\to a_{k+1}$, then this arrow cannot cross other arrows, neither can it occur in prohibited quadrangles.\nThus, removing the points $a_k, a_{k+1}$ from $\\{a_1, \\dots, a_{2n}\\}$ and the point $a_k$ from $S$, we may apply the induction hypothesis to find a unique good configuration $\\mathcal{C}'$ on $2n-2$ points compatible with the new set of sources (i.e., points from which arrows emerge). Adjunction of the arrow $a_k \\to a_{k+1}$ to $\\mathcal{C}'$ yields a unique good configuration on $2n$ points, as required.\n\n\n*Second solution.* Use the counterclockwise labeling $a_1, a_2, \\dots, a_{2n}$ in the solution above.\nLetting $D_n$ be the number of good configurations on $2n$ points, we establish a recurrence relation for the $D_n$. To this end, let $C_n = \\frac{(2n)!}{n!(n+1)!}$ the $n$th Catalan number; it is well-known that $C_n$ is the number of ways to connect $2n$ given points on the circumference by $n$ pairwise disjoint chords.\nSince no two arrows cross, in any good configuration the vertex $a_1$ is connected to some $a_{2k}$. Fix $k$ in the range $1$ through $n$ and count the number of good configurations containing the arrow $a_1 \\to a_{2k}$. Let $\\mathcal{C}$ be any such configuration.\nIn $\\mathcal{C}$, the vertices $a_2, \\dots, a_{2k-1}$ are paired off with one other, each arrow pointing from the smaller to the larger index, for otherwise it would form a prohibited quadrangle with $a_1 \\to a_{2k}$. Consequently, there are $C_{k-1}$ ways of drawing such arrows between $a_2, \\dots, a_{2k-1}$.\nOn the other hand, the arrows between $a_{2k+1}, \\dots, a_{2n}$ also form a good configuration, which can be chosen in $D_{n-k}$ ways. Finally, it is easily seen that any configuration of the first kind and any configuration of the second kind combine together to yield an overall good configuration.\nThus the number of good configurations containing the arrow $a_1 \\to a_{2k}$ is $C_{k-1}D_{n-k}$. Clearly, this is also the number of good configurations containing the arrow $a_{2(n-k+1)} \\to a_1$, so\n$$\nD_n = 2 \\sum_{k=1}^{n} C_{k-1} D_{n-k}. \\quad (*)\n$$\nTo find an explicit formula for $D_n$, let $d(x) = \\sum_{n=0}^{\\infty} D_n x^n$ and let $c(x) = \\sum_{n=0}^{\\infty} C_n x^n = \\frac{1-\\sqrt{1-4x}}{2x}$ be the generating functions of the $D_n$ and the $C_n$, respectively. Since $D_0 = 1$, relation (*) yields $d(x) = 2xc(x)d(x) + 1$, so\n$$\nd(x) = \\frac{1}{1 - 2xc(x)} = (1 - 4x)^{-1/2} = \\sum_{n \\ge 0} \\left(-\\frac{1}{2}\\right) \\left(-\\frac{3}{2}\\right) \\cdots \\left(-\\frac{2n-1}{2}\\right) \\frac{(-4x)^n}{n!} \\\\\n= \\sum_{n \\ge 0} \\frac{2^n(2n-1)!!}{n!} x^n = \\sum_{n \\ge 0} \\binom{2n}{n} x^n.\n$$\nConsequently, $D_n = \\binom{2n}{n}$.\n\n\n*Third solution.* Let $C_n = \\frac{1}{n+1} \\binom{2n}{n}$ denote the $n$th Catalan number and recall that there are exactly $C_n$ ways to join $2n$ distinct points on a circumference by $n$ pairwise disjoint chords. Such a configuration of chords will be referred to as a *Catalan $n$-configuration*. An orientation of the chords in a Catalan configuration $C$ making it into a good configuration (in the sense defined in the statement of the problem) will be referred to as a *good orientation* for $C$.\nWe show by induction on $n$ that there are exactly $n+1$ good orientations for any Catalan $n$-configuration, so there are exactly $(n+1)C_n = \\binom{2n}{n}$ good configurations on $2n$ points. The base case $n=1$ is clear.\nFor the induction step, let $n > 1$, let $C$ be a Catalan $n$-configuration, and let $ab$ be a chord of minimal length in $C$. By minimality, the endpoints of the other chords in $C$ all lie on the major arc $ab$ of the circumference.\nLabel the $2n$ endpoints 1, 2, ..., $2n$ counterclockwise so that $\\{a, b\\} = \\{1, 2\\}$, and notice that the good orientations for $C$ fall into two disjoint classes: Those containing the arrow $1 \\to 2$, and those containing the opposite arrow.\nSince the arrow $1 \\to 2$ cannot be involved in a prohibited quadrangle, the induction hypothesis applies to the Catalan $(n-1)$-configuration formed by the other chords to show that the first class contains exactly $n$ good orientations.\nFinally, the second class consists of a single orientation, namely, $2 \\to 1$, every other arrow emerging from the smaller endpoint of the respective chord; a routine verification shows that this is indeed a good orientation. This completes the induction step and ends the proof.\n\n\n*Fourth solution.* (Sang-il Oum) As in the previous solution, we intend to count the number of good orientations of a Catalan $n$-configuration.\nFor each such configuration, we consider its *dual graph* $T$ whose vertices are finite regions bounded by chords and the circle, and an edge connects two regions sharing a boundary segment. This graph $T$ is a plane tree with $n$ edges and $n+1$ vertices. There is a canonical bijection between orientations of chords and orientations of edges of $T$ in such a way that each chord crosses an edge of $T$ from the right to the left of the arrow on that edge. A good orientation of chords corresponds to an orientation of the tree containing no two edges oriented towards each other. Such an orientation is defined uniquely by its *source vertex*, i.e., the unique vertex having no incoming arrows.\nTherefore, for each tree $T$ on $n+1$ vertices, there are exactly $n+1$ ways to orient it so that the source vertex is unique — one for each choice of the source. The answer now follows along the lines in the solution above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38825,
    "subject": "Mathematics (Multi-modal)",
    "question": "The product of divisors of a natural number equals the square of that number. Find it, knowing that it is with 10 less than the sum of its divisors.",
    "options": [],
    "answer": "14",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers, with $abc(a + b + c) = 3$. Prove that\n$$\n\\frac{1}{a^2 + b^2 + 1} + \\frac{1}{b^2 + c^2 + 1} + \\frac{1}{c^2 + a^2 + 1} \\le 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38827,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a, b \\in [0, +\\infty)$, $c \\in \\mathbb{R}$ and $\\lfloor x \\rfloor + \\lfloor x + a \\rfloor + \\lfloor x + b \\rfloor = \\lfloor c x \\rfloor$, for every $x \\in \\mathbb{R}$, prove that $\\{a, b\\} = \\{\\frac{1}{3}, \\frac{2}{3}\\}$ and $c = 3$.",
    "options": [],
    "answer": "{a, b} = {1/3, 2/3}, c = 3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers $n \\ge 3$ such that, for every integer $m \\ge 0$, there exist integers $a_1, a_2, \\dots, a_n$ such that $a_1 + a_2 + \\dots + a_n = 0$ and $a_1a_2 + a_2a_3 + \\dots + a_{n-1}a_n + a_na_1 = -m$.",
    "options": [],
    "answer": "all integers n with n >= 5",
    "solution": "Any $n \\ge 5$ has the desired property: one can choose $a_1 = 1 - m$, $a_2 = a_3 = \\dots = a_{n-3} = 0$, $a_{n-2} = -1$, $a_{n-1} = m$, $a_n = 0$.\n\nNumbers $n = 3$ and $n = 4$ do not have the property.\n\nFor $n = 3$, $-2m = 2a_1a_2 + 2a_2a_3 + 2a_3a_1 = (a_1 + a_2 + a_3)^2 - a_1^2 - a_2^2 - a_3^2$ comes to $a_1^2 + a_2^2 + a_3^2 = 2m$, which does not hold for $m = 14$. Indeed, $28$ can not be written as a sum of three perfect squares.\n\nFor $n = 4$, $-m = a_1a_2 + a_2a_3 + a_3a_4 + a_4a_1 = (a_1+a_3)(a_2+a_4) = -(a_1+a_3)^2$ can hold only if $m$ is a perfect square, i.e. it does not hold for all $m \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{R}$, $a < b$, and let $f : [a, b] \\to [0, \\infty)$ be a continuous nondecreasing function. Prove that\n$$\n\\lim_{n \\to \\infty} \\int_a^b \\sqrt[n]{f^n(x) + f^n(a + b - x)} \\, dx = 2 \\int_{\\frac{a+b}{2}}^b f(x) \\, dx.\n$$",
    "options": [],
    "answer": "2 ∫_{(a+b)/2}^b f(x) dx",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a circle $\\Gamma$, a line $\\ell$ tangent to $\\Gamma$, and another circle $\\Omega$ disjoint from $\\ell$ such that $\\Gamma$ and $\\Omega$ lie on opposite sides of $\\ell$. The tangents to $\\Gamma$ from a variable point $X$ on $\\Omega$ cross $\\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\\Omega$, the circle $XYZ$ is tangent to two fixed circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $\\Gamma$ of unit radius and invert with respect to $\\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\\Gamma$ be centered at $G$, notice that inversion in $\\Gamma$ maps tangents to $\\Gamma$ to circles of unit diameter through $G$ (hence internally tangent to $\\Gamma$). Under inversion, the statement reads as follows:\n\n*Fix a circle $\\Gamma$ of unit radius centered at $G$, a circle $\\ell$ of unit diameter through $G$, and a circle $\\Omega$ inside $\\ell$ disjoint from $\\ell$. The circles $\\eta$ and $\\zeta$ of unit diameter, through $G$ and a variable point $X$ on $\\Omega$, cross $\\ell$ again at $Y$ and $Z$, respectively. Prove that, as $X$ traces $\\Omega$, the circle $XYZ$ is tangent to two fixed circles.*\n\n![](attached_image_1.png)\n\nSince $\\eta$ and $\\zeta$ are the reflections of the circumcircle $\\ell$ of the triangle $XYZ$ in its sidelines $GY$ and $GZ$, respectively, they pass through the orthocenter of this triangle. And since $\\eta$ and $\\zeta$ cross again at $X$, the latter is the orthocenter of the triangle $XYZ$. Hence the circle $\\xi$ through $X, Y, Z$ is the reflection of $\\ell$ in the line $YZ$; in particular, $\\xi$ is also of unit diameter.\n\nLet $O$ and $L$ be the centers of $\\Omega$ and $\\ell$, respectively, and let $R$ be the (variable) center of $\\xi$. Let $GX$ cross $\\xi$ again at $X'$; then $G$ and $X'$ are reflections of one another in the line $YZ$, so $GLRX'$ is an isosceles trapezoid. Then $LR \\parallel GX$ and $\\angle(LG, GX) = \\angle(GX', X'R) = \\angle(RX, XG)$, i.e., $LG \\parallel RX$; this means that $GLRX$ is a parallelogram, so $XR = GL$ is constant.\n\nFinally, consider the fixed point $N$ defined by $\\overrightarrow{ON} = \\overrightarrow{GL}$. Then $XRNO$ is a parallelogram, so the distance $RN = OX$ is constant. Consequently, $\\xi$ is tangent to the fixed circles centered at $N$ of radii $|1/2 - OX|$ and $1/2 + OX$.\n\nOne last check is needed to show that the inverse images of the two obtained circles are indeed circles and not lines. This might happen if one of them contained $G$; we show that this is impossible. Indeed, since $\\Omega$ lies inside $\\ell$, we have $OL < 1/2 - OX$, so\n$$\n\\begin{aligned}\nNG &= |\\overrightarrow{GL} + \\overrightarrow{LO} + \\overrightarrow{ON}| = |2\\overrightarrow{GL} + \\overrightarrow{LO}| \\geq 2|\\overrightarrow{GL}| - |\\overrightarrow{LO}| \\\\\n&> 1 - (1/2 - OX) = 1/2 + OX;\n\\end{aligned}\n$$\nthis shows that $G$ is necessarily outside the obtained circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38831,
    "subject": "Mathematics (Multi-modal)",
    "question": "A nonempty finite set $S$ of complex numbers has the property: $xy \\in S$, for every $x, y \\in S$.\n\na) Prove that, for every $x \\in S$, $\\frac{1}{x} \\in S$.\n\nb) Let $a, b$ be two fixed elements of $S$ and $m, n \\in \\mathbb{N}$, $m, n \\ge 2$. Find the number of $m \\times n$ matrices, with elements from $S$, having the product of the elements of each line equal to $a$ and the product of the elements of each column equal to $b$.",
    "options": [],
    "answer": "Number of matrices = 0 if a^m ≠ b^n; otherwise it equals |S|^{(m−1)(n−1)}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to [0, \\infty)$ be a differentiable function, with continuous derivative, and $f(f(x)) = x^2$, for every $x \\in [0, \\infty)$. Prove that\n$$\n\\int_{0}^{1} (f'(x))^2 \\, dx \\ge \\frac{30}{31}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers satisfying $2x^2 + 3y^2 + 6z^2 + 12(x + y + z) = 108$. Find the maximum value of $x^3y^2z$.",
    "options": [],
    "answer": "108",
    "solution": "Guessing that the maximum is obtained when $x = 3$, $y = 2$, $z = 1$, we multiply the given equality by $6$, and apply AM-GM:\n\n$$6 \\cdot 108 = 4x^2 + 4x^2 + 4x^2 + 9y^2 + 9y^2 + 36z^2 + 12x + 12x + 12x + 12x + 12x + 12x + 18y + 18y + 18y + 18y + 36z + 36z \\ge 18 \\sqrt{2^{28} \\cdot 3^{24} \\cdot x^{12} \\cdot y^8 \\cdot z^4},$$\n\nfrom which it follows immediately that $108$ is the desired maximum and this value is obtained when all the numbers are equal, i.e. when $x = 3$, $y = 2$, $z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that a number of the form $n(n + 1)$, where $n$ is a positive integer, is the sum of two numbers of like form, say $k(k + 1)$ and $m(m + 1)$, where $k$ and $m$ are positive integers, if and only if the number $2n^2 + 2n + 1$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Rewrite the condition $k(k + 1) + m(m + 1) = n(n + 1)$ in the equivalent form\n$$\n(2k + 1)^2 + (2m + 1)^2 = 2N. \\quad (*)\n$$\nSince the prime factors of $N$ are all congruent to $1$ modulo $4$, the number of solutions of the equation $x^2 + y^2 = 2N$ in integers $x$ and $y$ (necessarily odd) is equal to $4d(N)$, where $d(N)$ is the number of positive divisors of $N$.\nThe equality $2N = (2n+1)^2+1$ provides eight solutions $x = \\pm(2n+1)$, $y = \\pm 1$ and $x = \\pm 1$, $y = \\pm(2n+1)$ to $x^2+y^2 = 2N$, so (*) has a solution in positive integers $k$ and $m$ if and only if $d(N) > 2$; that is, if and only if $N$ is composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an isosceles triangle $ABC$, where $AB = AC$ and $\\angle BAC = 100^\\circ$. Let point $D$ on the opposite side from $C$ with respect to $AB$ such that $DC = BC$, and $\\angle BCD = 20^\\circ$. Find $\\angle DAB$.",
    "options": [],
    "answer": "30°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a four-point configuration in the plane, every three points of which can be covered by a strip of unit width. Prove that:\n\na) the four points can be covered by a strip of width at most $\\sqrt{2}$; and\n\nb) if no strip of width less than $\\sqrt{2}$ covers all four points, then they are the vertices of a square of side length $\\sqrt{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Both facts follow from the lemma below.\n\n**Lemma.** Consider a triangle some altitude of which has length at most $1$. If the triangle has an altitude of length at least $\\sqrt{2}$, say the $X$-altitude, then the internal angular span at $X$ does not exceed $45^\\circ$; equality holds if and only if the $X$-altitude has length $\\sqrt{2}$, and the triangle is isosceles right with apex at one of the other two vertices.\n\nTo prove the lemma, let the $X$-altitude have length $h_X \\ge \\sqrt{2}$. Since the triangle has an altitude of length at most $1$, say $h_Y \\le 1$, the internal angle at $X$ is not the largest angle of the triangle, so it does not exceed a right angle. And since $\\sin X = (h_Y/h_X) \\cdot \\sin Y \\le h_Y/h_X \\le 1/\\sqrt{2}$, the internal angular span at $X$ does not exceed $45^\\circ$.\n\nIf the internal angular span at $X$ is $45^\\circ$, then $1/\\sqrt{2} = \\sin X = (h_Y/h_X) \\cdot \\sin Y \\le h_Y/h_X \\le 1/\\sqrt{2}$, forcing $h_X = \\sqrt{2}$, $h_Y = 1$ and $\\sin Y = 1$, so the triangle is indeed isosceles right with apex at $Y$. The converse is clear.\n\nBack to the problem, begin by noticing that the four points may and will be assumed to be in strictly convex position; i.e., no point lies in the convex hull of the other three. Further, let $\\text{dist}(X, YZ)$ denote the distance from the point $X$ to the line $YZ$, and notice that the four points may and will be labelled $A$, $B$, $C$, $D$, in circular order, so that $\\text{dist}(A, BC) \\ge \\text{dist}(D, BC)$, $\\text{dist}(A, CD) \\ge \\text{dist}(B, CD)$, $\\text{dist}(B, AD) \\ge \\text{dist}(C, AD)$ and $\\text{dist}(D, AB) \\ge \\text{dist}(C, AB)$.\n\na) It is clearly sufficient to show that one of the distances $\\text{dist}(A, BC)$, $\\text{dist}(A, CD)$, $\\text{dist}(B, AD)$, $\\text{dist}(D, AB)$ does not exceed $\\sqrt{2}$. Suppose, if possible, this is not the case, notice that each of the triangles $ABC$, $ACD$, $ABD$ has an altitude of length at most $1$, and refer to the lemma to infer that the angular span of each of the angles $BAC$, $CAD$, $ABD$, $ADB$ is less than $45^\\circ$, so the internal angular spans at the vertices of the triangle $ABD$ add up to less than $180^\\circ$ — a contradiction.\n\nb) Now each of the four distances $\\text{dist}(A, BC)$, $\\text{dist}(A, CD)$, $\\text{dist}(B, AD)$, $\\text{dist}(D, AB)$ is at least $\\sqrt{2}$, and the lemma applies again to each of the triangles $ABC$, $ACD$, $ABD$, to deduce that the angular span of each of the angles $BAC$, $CAD$, $ABD$, $ADB$ does not exceed $45^\\circ$. If one of the four distances above were greater than $\\sqrt{2}$, then the angular span of the corresponding angle would be less than $45^\\circ$ and the internal angular spans at the vertices of the triangle $ABD$ would again add up to less than $180^\\circ$. Consequently, the four distances under consideration are all equal to $\\sqrt{2}$, the corresponding angular spans are all $45^\\circ$, and the conclusion follows from the equality case in the lemma.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine digits $x$ and $y$ such that $\\frac{5xy23}{4xy24} = \\frac{523}{424}$.",
    "options": [],
    "answer": "x=2, y=8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38838,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $T$ is taken on the edge $AD$ of the regular pyramid $VABCD$, with apex $V$, so that $\\frac{AT}{TD} = \\frac{1}{4}$, and the straight line $TS$, perpendicular on $VB$, is drawn ($S \\in VB$). If the straight lines $TS$ and $AD$ are perpendicular, prove that the height of the pyramid equals the length of the sides of its base.\nGabriel Popa",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has the property that there exists a unique point $X$ on the line segment $BC$ such that $AX^2 = BX \\cdot CX$. Prove that $AB + AC = BC\\sqrt{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the reflection of $A$ with respect to $X$. By the converse of the Power of a Point Theorem, it follows that the quadrilateral $ABTC$ is cyclic. If the line parallel to $BC$ through $T$ intersects the circumcircle of $ABC$ again at $U$, and lines $AU$ and $BC$ meet at $Y$, then $Y \\in [BC]$ and $BY \\cdot CY = AY \\cdot UY = AY^2$, thus, in order for the point $X$ to be unique, it is necessary that $T = U$, i.e. $T$ is the midpoint of the arc $BC$, in other words $X$ is the foot of the angle bisector from $A$.\n\nFrom here one can finish the proof in several ways. One can use the formula for the length of the angle bisector in a triangle, or Stewart's Theorem.\n\nIf $AX = l$, $BX = x$, $CX = y$, then $b^2x + c^2y = l^2a + axy$. It is known that $x = \\frac{ac}{b+c}$, $y = \\frac{ab}{b+c}$. It follows that\n$$\nl^2 = xy \\Leftrightarrow b^2x + c^2y = 2axy \\Leftrightarrow (b+c)^2 = 2a^2 \\Leftrightarrow b+c = a\\sqrt{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38840,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\\int_{0}^{x} \\left( 1 + \\frac{t}{1!} + \\frac{t^2}{2!} + \\dots + \\frac{t^{2n+1}}{(2n+1)!} \\right) \\cdot \\frac{1}{1+t^2} \\, dt < \\frac{\\arctan x}{x} \\int_{0}^{x} e^t \\, dt,$$\nfor every $x > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $n \\in \\mathbb{N}^*$ and there exists $a, b \\in \\mathbb{Z}$ so that $\\frac{a+bi}{\\sqrt{n}}$ is a root of order $n$ of the unity, then $n = 8m^2$, for some integer $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a regular tetrahedron, with $AB = a$. Denote $M$ the midpoint of $[BC]$, $N$ the midpoint of $[AM]$, and take $P \\in [BD]$ so that $BP = 3PD$. Compute the distance between the straight lines $AC$ and $NP$.\n\nGabriela Chişiu and Carmen Rusu",
    "options": [],
    "answer": "a * sqrt(6) / 12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a non-empty set of positive integers, let $d$ be the greatest common divisor of $D$, and let $d\\mathbb{Z} = \\{dn : n \\in \\mathbb{Z}\\}$. Prove that there exists a bijection $f: \\mathbb{Z} \\to d\\mathbb{Z}$ such that $|f(n) - f(n-1)|$ is a member of $D$ for all integers $n$.\n*Amer. Math. Monthly*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Reduce the problem to the case where $D$ is finite, by considering an element $d_0$ of $D$, representing each residue class in $D \\pmod{d_0}$ by some member of $D$, and collecting all these representatives to form a finite subset of $D$ whose greatest common divisor is again $d$.\n\nAssume henceforth $D$ finite and induct on the cardinality of $D$. The base case $|D| = 1$ being clear, let $|D| > 1$, fix a member $a$ of $D$, notice that $D' = D \\setminus \\{a\\}$ is non-empty since $|D| > 1$, and let $d'$ be the greatest common divisor of $D'$. Clearly, $d$ is the greatest common divisor of $a$ and $d'$, and $k = d'/d$ is integral. By the induction hypothesis, there exists a bijection $f': \\mathbb{Z} \\to d'\\mathbb{Z}$ such that $|f'(n) - f'(n-1)|$ is a member of $D'$ for all integers $n$.\n\nIt is easily seen that every multiple of $d$ can uniquely be written in the form\n$$\nar + d's, \\qquad (*)$$\nwhere $r$ and $s$ are both integral and $0 \\le r < k$.\n\nAssign every integer $n$ an integer $f(n)$ in $d\\mathbb{Z}$ by writing $n = kq + r$, where $q$ and $r$ are both integral and $0 \\le r < k$, and letting\n$$\nf(n) = \\begin{cases} f'(q) + ar, & \\text{if } q \\text{ is even,} \\\\ f'(q) + a(k - r - 1), & \\text{if } q \\text{ is odd.} \\end{cases}\n$$\nThis defines a function $f: \\mathbb{Z} \\to d\\mathbb{Z}$.\n\nSince every member of $d\\mathbb{Z}$ can be written in the form $(*)$, and $f'$ is surjective, so is $f$.\nUniqueness in $(*)$ and injectivity of $f'$ imply injectivity of $f$, so $f$ is bijective.\nFinally, the fact that $|f(n) - f(n-1)|$ is a member of $D$ for all integers $n$ follows from the corresponding condition for $f'$. Verifications are routine and hence omitted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Consider a commutative finite group $G$, with at least three elements, identity element $e$, and so that $x^p = e$, for every $x \\in G$. Denote $Aut(G)$ the group of the automorphisms of $G$. Prove that:\n\na) $|G| = p^n$, where $n$ is the smallest cardinality of a set of generators of $G$;\n\nb) $|Aut(G)| = (p^n - 1)(p^n - p) \\dots (p^n - p^{n-1})$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $A$ be a set of complex numbers with $2n + 1$ elements. Prove that there exists two sets $B, C$ so that $B \\cup C = A$, $B \\cap C = \\emptyset$, $B$ has $n$ elements and $|\\sum_{z \\in B} z| \\le |\\sum_{z \\in C} z|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38846,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n \\ge 2$, consider $n$ boxes aligned from left to right. In each box, one puts a ball that can be red, blue or white such that the following condition is fulfilled: Each box is neighboring at least one box containing a ball of the same color. We denote by $I_n$ the number of such configurations.\n\na) Determine $I_{11}$. Justify your answer.\n\nb) Find, with proof, the general formula for $I_n$.",
    "options": [],
    "answer": "I_11 = 1023; I_n = 2^{n-1} + (-1)^n",
    "solution": "Obviously $a_2 = 1$, $a_3 = 1$. For $a \\ge 4$, if the first box contains a red ball, then so does the second one. Now we distinguish two types of configurations: those containing a red ball in the third box, and those with a blue or a white ball in the third box. In order to count the configurations of the first type, simply forget about the first box. We can easily see that there are $a_{n-1}$ configurations of this type. For the second type of configurations, we omit the first two boxes and obtain $2a_{n-2}$ configurations. Thus, $a_n = a_{n-1} + 2a_{n-2}$.\n\nBy induction, it is easy to prove that $a_n = \\frac{2^{n-1} + (-1)^n}{3}$, hence $I_n = 2^{n-1} + (-1)^n$. For $n = 11$ we get $I_{11} = 1023$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples of real numbers $(a, b, c)$ that satisfy simultaneously the equations:\n$$\n\\begin{align*}\na(b^2 + c) &= c(c + ab), \\\\\nb(c^2 + a) &= a(a + bc), \\\\\nc(a^2 + b) &= b(b + ca).\n\\end{align*}\n$$",
    "options": [],
    "answer": "(x, x, x) for any real x",
    "solution": "Let $(a, b, c)$ be a solution of the system. If one of the numbers $a, b, c$ is $0$, e.g. if $c = 0$, then $c(a^2 + b) = a(b + ca)$ leads to $a = 0$, and similarly one gets $b = 0$. Thus $abc = 0$ leads to $a = b = c = 0$ which is indeed a solution. We now look for solutions with $abc \\neq 0$. We rewrite the equations\n$$\n\\begin{align*}\nab(b - c) &= c(c - a) \\\\\nbc(c - a) &= a(a - b) \\\\\nca(a - b) &= b(b - c).\n\\end{align*}\n$$\nIt follows that $a^2b^2c^2(a - b)(b - c)(c - a) = abc(a - b)(b - c)(c - a)$.\n\nCase 1: Among the numbers $a, b, c$ there exist (at least) two equal ones; say $a = b$. We have\n$$\na^2c + bc = b^2 + abc \\Leftrightarrow bc = b^2 \\Leftrightarrow c = b,\n$$\ntherefore in this case we obtain $a = b = c$. Conversely, any such triple is a solution to the system.\n\nCase 2: If $a \\neq b \\neq c \\neq a$, then $abc = 1$. We obtain $ab(b - c) = c(c - a) \\Leftrightarrow (b - c) = c^2(c - a)$ and two more equations, similar to this one. It follows that\n$$\na^3 + b^3 + c^3 = ac^2 + ba^2 + cb^2.\n$$\nIf $a, b, c > 0$, from AM-GM it follows that $a^3 + a^3 + b^3 \\ge 3ba^2$ (with equality if and only if $a = b$) which, added with two similar inequalities, leads to $a^3 + b^3 + c^3 \\ge ac^2 + ba^2 + cb^2$. We thus have equality in the previous inequality, hence $a = b = c$.\n\nIf one of the variables is positive and the other two are negative, say $a > 0$, $b, c < 0$, then\n$$\nb(c^2 + a) = a(a + bc) = a^2 + 1 > 0 \\Rightarrow a < 0,\n$$\ncontradiction.\n\nIn conclusion, the only solutions are $(a, b, c) = (x, x, x)$, $\\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle, and let its diagonals, $AC$ and $BD$, cross at $X$. Let $I$ be the incenter of the triangle $XBC$, and let $J$ be the center of the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $DC$ beyond $B$ and $C$, respectively. Prove that the line $IJ$ bisects the arc $BC$ of the circle $ABCD$, not containing the vertices $A$ and $D$ of the quadrangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $M$ be the midpoint of the arc $BC$ of the circle $ABCD$, not containing the vertices $A$ and $D$ of the quadrangle.\n\nAn easy angle chase shows the vertices $B$ and $C$, the incenter $K$ of the triangle $ABC$, and its $A$-excenter $K_A$ equally distanced from $M$, so they all lie on a circle $\\gamma$ centred at $M$ — the so-called trillium lemma for the incenter, respectively, excenter.\n\nSimilarly, the incenter $L$ and the $D$-excenter $L_D$ of the triangle $DBC$ both lie on $\\gamma$.\n\nFinally, apply Pascal's theorem to the hexagram $BK_AKCL_DL$ with vertices on $\\gamma$, to conclude that $J = BK_A \\cap CL_D$, $M = K_AK \\cap LL_D$ and $I = KC \\cap LB$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in (0, \\infty)$, not all equal. Solve the equation\n$$\n(a^x + b^x + c^x) (a^{-x} + b^{-x} + c^{-x}) = 3 + \\frac{a^2 + b^2}{ab} + \\frac{b^2 + c^2}{bc} + \\frac{c^2 + a^2}{ca}.\n$$",
    "options": [],
    "answer": "x = 1 or x = -1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38850,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a circle the numbers from $1$ to $12$ are written in an arbitrary order. Prove that there are three consecutive numbers summing up to at least $20$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nArătaţi că ecuaţia $x^{2}+y^{2}+z^{2}=x+y+z+1$ nu are soluţii în mulţimea numerelor raţionale.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEcuaţia se poate scrie echivalent sub forma $(2x-1)^{2}+(2y-1)^{2}+(2z-1)^{2}=7$.\n\nDacă această ecuaţie ar avea o soluţie raţională $(x, y, z)$, notând $2x-1=\\frac{a_{1}}{b_{1}}$, $2y-1=\\frac{a_{2}}{b_{2}}$, $2z-1=\\frac{a_{3}}{b_{3}}$, am obţine că numerele întregi $a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}$ satisfac egalitatea\n$$\n\\left(a_{1} b_{2} b_{3}\\right)^{2}+\\left(b_{1} a_{2} b_{3}\\right)^{2}+\\left(b_{1} b_{2} a_{3}\\right)^{2}=7\\left(b_{1} b_{2} b_{3}\\right)^{2}.\n$$\nAcest fapt ar implica existenţa a patru numere întregi $a, b, c, d$ astfel ca $a^{2}+b^{2}+c^{2}=7d^{2}$.\n\nDacă c.m.m.d.c. $(a, b, c, d)=k$, împărţind cu $k^{2}$ am găsi o soluţie $(a, b, c, d) \\in \\mathbb{Z}^{4}$ a ecuaţiei $a^{2}+b^{2}+c^{2}=7d^{2}$ cu c.m.m.d.c. $(a, b, c, d)=1$.\n\nAtunci $a, b, c, d$ nu pot fi toate pare. Deoarece un pătrat perfect dă unul din resturile $0, 1$ sau $4$ la împărţirea cu $8$, membrul stâng dă unul din resturile $1, 2, 3, 5$ sau $6$ la împărţirea cu $8$, în vreme ce $7d^{2}$ dă rest $0$, $4$ sau $7$. Prin urmare egalitatea nu poate avea loc.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminați numerele prime $p$ pentru care numărul $a = 7^{p} - p - 16$ este pătrat perfect.",
    "options": [],
    "answer": "3",
    "solution": "Soluție:\n\n$p = 2$ nu verifică.\n\n$p = 3$ este soluție: $a = 7^{3} - 3 - 16 = 324 = 18^{2}$.\n\nArătăm că nu avem alte soluții. Fie $p \\geq 5$ un număr prim.\n\nDacă $p \\equiv 1 \\pmod{4}$, atunci $a \\equiv 2 \\pmod{4}$, deci $a$ nu este pătrat perfect.\n\nSe constată ușor că $p = 7$ nu este soluție (calculând $a$, ultimele două cifre ale lui $a$ sau observând că $a \\equiv 5 \\pmod{7}$).\n\nDacă $p > 7$ este un număr prim de forma $4k + 3$, atunci din mica teoremă a lui Fermat rezultă că $7^{p} \\equiv 7 \\pmod{p}$, deci $a \\equiv -9 \\pmod{p}$, adică $p \\mid a + 9$.\n\nDacă $a$ ar fi pătrat perfect, atunci $p$ divide $a + 9 = b^{2} + 3^{2}$ implică $p$ divide $b$ și $p$ divide $3$, ceea ce nu se poate.\n\nPrin urmare, singura soluție a problemei este $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentru orice număr natural nenul $n$ şi orice matrice coloană\n$$\n\\mathbf{X} = \\left(\\begin{array}{c}\nx_{1} \\\\\nx_{2} \\\\\n\\vdots \\\\\nx_{n}\n\\end{array}\\right) \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})\n$$\nnotăm cu $\\delta(\\mathbf{X})$ cel mai mare divizor comun al numerelor $x_{1}, x_{2}, \\ldots, x_{n}$. Fie $n \\in \\mathbb{N}, n \\geq 2$, şi $\\mathbf{A} \\in \\mathcal{M}_{n}(\\mathbb{Z})$. Arătaţi că următoarele două afirmaţii sunt echivalente:\n\n(a) $|\\operatorname{det} \\mathbf{A}|=1$ si\n\n(b) $\\delta(\\mathbf{A X})=\\delta(\\mathbf{X})$, oricare ar fi $\\mathbf{X} \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})$.\n\n(Cel mai mare divizor comun al unor numere întregi este număr natural.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nArătăm că (a) implică (b). Fie $\\mathbf{B}=\\left(b_{i j}\\right) \\in \\mathcal{M}_{n}(\\mathbb{Z})$, fie $\\mathbf{X}=\\left(x_{j}\\right) \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})$ şi fie $\\mathbf{B X}=\\left(y_{i}\\right) \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})$. Cum $y_{i}=\\sum_{j=1}^{n} b_{i j} x_{j},\\ i=1,2, \\ldots, n$, rezultă că $\\delta(\\mathbf{X})$ divide fiecare $y_{i}$, deci $\\delta(\\mathbf{X}) \\leq \\delta(\\mathbf{B X})$.\n\nCum $\\mathbf{A}$ este inversabilă în $\\mathcal{M}_{n}(\\mathbb{Z})$, rezultă că\n$$\n\\delta(\\mathbf{X}) \\leq \\delta(\\mathbf{A X}) \\leq \\delta\\left(\\mathbf{A}^{-1}(\\mathbf{A X})\\right)=\\delta(\\mathbf{X})\n$$\nPrin urmare, $\\delta(\\mathbf{A X})=\\delta(\\mathbf{X})$.\n\n\nArătăm că (b) implică (a). Fie $d=\\operatorname{det} \\mathbf{A}$. Dacă $d=0$, atunci sistemul omogen $\\mathbf{A X}=\\mathbf{O}_{n, 1}$ are soluţii nenule în $\\mathcal{M}_{n, 1}(\\mathbb{Q})$ şi, prin înmulţirea uneia dintre aceste soluţii cu produsul numitorilor componentelor sale nenule, obţinem un $\\mathbf{X} \\in \\mathcal{M}_{n, 1}(\\mathbb{Z}),\\ \\mathbf{X} \\neq \\mathbf{O}_{n, 1}$, astfel încât $\\mathbf{A X}=\\mathbf{O}_{n, 1}$. Deci $0<\\delta(\\mathbf{X})=\\delta(\\mathbf{A X})=\\delta\\left(\\mathbf{O}_{n, 1}\\right)=0$, o contradicţie. Prin urmare, $d \\neq 0$.\n\nFie $\\mathbf{X}_{i}$ coloana $i$ a matricei $\\mathbf{A}^{*},\\ i=1,2, \\ldots, n$. Cum matricea coloană $\\mathbf{A X}_{i}$ are toate componentele nule, cu excepţia componentei $i$, care este egală cu $d$, rezultă că $d=\\delta\\left(\\mathbf{A X}_{i}\\right)=\\delta\\left(\\mathbf{X}_{i}\\right),\\ i=1,2, \\ldots, n$, deci toate elementele lui $\\mathbf{A}^{*}$ sunt divizibile cu $d$. Prin urmare, $\\operatorname{det} \\mathbf{A}^{*}$ este divizibil cu $d^{n}$. Cum $\\operatorname{det} \\mathbf{A}^{*}=d^{n-1}$ si $d \\neq 0$, rezultă că $d= \\pm 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDemonstraţi că un număr natural $A$ este pătrat perfect dacă şi numai dacă, pentru orice număr natural nenul $n$, cel puţin una din diferenţele\n$$\n(A+1)^{2}-A,(A+2)^{2}-A,(A+3)^{2}-A, \\ldots,(A+n)^{2}-A\n$$\neste divizibilă cu $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDacă $A$ este pătrat perfect, adică există $B \\in \\mathbb{N}$ astfel încât $A=B^{2}$, atunci $(A+k)^{2}-A=\\left(B^{2}+k\\right)^{2}-B^{2}=\\left(B^{2}+B+k\\right)\\left(B^{2}-B+k\\right)$ pentru orice $k=\\overline{1, n}$ și exact unul dintre numerele (consecutive) $B^{2}+B+1, B^{2}+B+2, \\ldots, B^{2}+B+n$ este divizibil cu $n$, de unde concluzia.\n\nReciproc, dacă $A$ nu este pătrat perfect, atunci există un factor prim $p$ care în descompunerea în factori primi a lui $A$ apare la o putere impară. Fie $p$ un asemenea număr prim și $j \\in \\mathbb{N}^{*}$ astfel încât $p^{2j-1} \\mid A$, dar $p^{2j} \\nmid A$. Alegem $n=p^{2j} \\in \\mathbb{N}^{*}$ și arătăm că niciunul din numerele $(A+1)^{2}-A,(A+2)^{2}-A,(A+3)^{2}-A, \\ldots,(A+n)^{2}-A$ nu este divizibil cu $n$. Într-adevăr, dacă $n \\mid (A+m)^{2}-A$, cu $m \\in \\{1,2, \\ldots, n\\}$, adică $p^{2j} \\mid A^{2}+2Am+m^{2}-A$, cum $p^{2j-1} \\mid A$, rezultă $p^{2j-1} \\mid m^{2}$, de unde $p^{j} \\mid m$. Dar atunci $p^{2j} \\mid (A+m)^{2}$ și $p^{2j} \\mid (A+m)^{2}-A$, deci $p^{2j} \\mid A$, contradicţie.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $n \\in \\mathbb{N}$, $n \\geq 2$ şi numerele $a_{1}, a_{2}, \\ldots, a_{n} \\in (1, \\infty)$. Demonstraţi că funcţia $f:[0, \\infty) \\rightarrow \\mathbb{R}$, definită prin relaţia\n$$\nf(x)=\\left(a_{1} a_{2} \\ldots a_{n}\\right)^{x}-a_{1}^{x}-a_{2}^{x}-\\ldots-a_{n}^{x}\n$$\npentru orice $x \\in [0, \\infty)$, este strict crescătoare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nVom realiza demonstraţia prin inducţie matematică.\nPentru $n=2$, fie $a_{1}, a_{2} \\in (1, \\infty)$. Avem\n$$\nf(x)=\\left(a_{1} a_{2}\\right)^{x}-a_{1}^{x}-a_{2}^{x}=\\left(a_{1}^{x}-1\\right)\\left(a_{2}^{x}-1\\right)-1\n$$\nDeoarece funcţiile $f_{1}, f_{2}:[0, \\infty) \\rightarrow \\mathbb{R}$, definite prin relaţiile $f_{1}(x)=a_{1}^{x}-1$ şi $f_{2}(x)=a_{2}^{x}-1$, oricare ar fi $x \\in [0, \\infty)$, sunt strict crescătoare şi pozitive, rezultă că $f$ este strict crescătoare.\nPresupunem proprietatea este adevărată pentru oricare $n$ numere din $(1, \\infty)$ şi o demonstrăm pentru $n+1$ numere $a_{1}, a_{2}, \\ldots, a_{n}, a_{n+1} \\in (1, \\infty)$. Avem\n$$\n\\begin{aligned}\nf(x) & =\\left(a_{1} a_{2} \\ldots a_{n} a_{n+1}\\right)^{x}-a_{1}^{x}-a_{2}^{x}-\\ldots-a_{n}^{x}-a_{n+1}^{x} \\\\\n& =\\left(\\left(a_{1} a_{2} \\ldots a_{n} a_{n+1}\\right)^{x}-\\left(a_{1} a_{2} \\ldots a_{n}\\right)^{x}-a_{n+1}^{x}\\right)+\\left(\\left(a_{1} a_{2} \\ldots a_{n}\\right)^{x}-a_{1}^{x}-a_{2}^{x}-\\ldots-a_{n}^{x}\\right)\n\\end{aligned}\n$$\nFuncţia $g:[0, \\infty) \\rightarrow \\mathbb{R}$, $g(x)=\\left(a_{1} a_{2} \\ldots a_{n} a_{n+1}\\right)^{x}-\\left(a_{1} a_{2} \\ldots a_{n}\\right)^{x}-a_{n+1}^{x}$ este strict crescătoare deoarece $a_{1} a_{2} \\ldots a_{n}>1$ şi $a_{n+1}>1$ (cazul $n=2$).\nFuncţia $h:[0, \\infty) \\rightarrow \\mathbb{R}$, $h(x)=\\left(a_{1} a_{2} \\ldots a_{n}\\right)^{x}-a_{1}^{x}-a_{2}^{x}-\\ldots-a_{n}^{x}$ este strict crescătoare conform ipotezei de inducţie. Atunci $f=g+h$ este strict crescătoare.\nRezultă că proprietatea din enunţ este demonstrată.\n$4\\ p$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p$ un număr prim mai mare ca $5$ şi $S=\\{p-n^{2} \\mid n \\in \\mathbb{N}, n^{2}<p\\}$. Demonstraţi că $S$ conţine două elemente $a$ şi $b$ astfel încât $1<a<b$ şi $a$ divide $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVom arăta că cel mai mic element al lui $S$ care e mai mare ca $1$ divide un element mai mare al lui $S$.\n\nDacă $p$ este de forma $m^{2}+1$ cu $m \\in \\mathbb{N}$, arătăm că $p-(m-1)^{2}=2m$ divide $p^{2}-1=m^{2}$. Într-adevăr, cum $m$ este par, rezultă că $2m \\mid m^{2}$.\n\nDacă $p$ nu este de forma $m^{2}+1$, el nu este nici de forma $m^{2}+2m$ (acesta este număr compus căci $m>1$), deci $m^{2}+1<p<m^{2}+2m$ pentru un anumit $m \\in \\mathbb{N}$ ($m \\geq 2$). Arătăm că $p-m^{2}$, care este un element mai mare ca $1$ al lui $S$, divide un element mai mare al lui $S$.\n\nCondiţia ca $p-m^{2}$ să dividă un număr de forma $p-n^{2}$ cu $n \\in \\{1,2, \\ldots, m-1\\}$ este echivalentă cu $p-m^{2}$ divide unul din numerele $m^{2}-1^{2}, m^{2}-2^{2}, \\ldots, m^{2}-(m-1)^{2}$.\n\nFiind mai mic ca $2m$, $p-m^{2}$ divide unul din următoarele $2m-1$ numere consecutive: $1,2, \\ldots, m-1, m, m+1, \\ldots, 2m-1$.\n\nDacă singurul din aceste numere pe care $p-m^{2}$ îl divide este $m$, adică $p-m^{2}$ nu divide niciunul din numerele $1,2, \\ldots, m-1$, atunci $p-m^{2} \\geq m$. Dar pe de altă parte, $p-m^{2} \\leq m$. Rezultă că $p-m^{2}=m$, adică $p=m(m+1)$, care însă este număr compus.\n\nRezultă că $p-m^{2}$ divide unul din numerele $1,2, \\ldots, m-1, m+1, \\ldots, 2m-1$, deci una din diferenţele $m^{2}-1^{2}, m^{2}-2^{2}, \\ldots, m^{2}-(m-1)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminaţi toate tripletele $(a, b, c)$ de numere reale pentru care au loc simultan relaţiile:\n$$\n\\begin{aligned}\n& a\\left(b^{2}+c\\right)=c(c+a b) \\\\\n& b\\left(c^{2}+a\\right)=a(a+b c) \\\\\n& c\\left(a^{2}+b\\right)=b(b+c a)\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(x, x, x) for any real x",
    "solution": "Solution:\nFie $(a, b, c)$ o soluţie a sistemului. Dacă unul dintre numerele $a, b, c$ este egal cu $0$, de exemplu dacă $c=0$, atunci din $c\\left(a^{2}+b\\right)=a(b+c a)$ rezultă $a=0$, apoi, similar, că $b=0$. Aşadar, dacă $a b c=0$, atunci $a=b=c=0$.\n\nCăutăm în continuare soluţii cu $a b c \\neq 0$. Relaţiile din enunţ se pot rescrie echivalent\n$$\n\\begin{aligned}\n& a b(b-c)=c(c-a) \\\\\n& b c(c-a)=a(a-b) \\\\\n& c a(a-b)=b(b-c) .\n\\end{aligned}\n$$\nAstfel, $a^{2} b^{2} c^{2}(a-b)(b-c)(c-a)=a b c(a-b)(b-c)(c-a)$.\n\nCazul 1: Printre numerele $a, b, c$ există (cel puţin) două egale; fie acestea $a$ şi $b$. Avem\n$$\na^{2} c+b c=b^{2}+a b c \\Leftrightarrow b c=b^{2} \\Leftrightarrow c=b\n$$\ndeci în acest caz rezultă $a=b=c$. Reciproc, orice triplet cu componentele egale satisface relaţiile date.\n\nCazul 2: Dacă $a \\neq b \\neq c \\neq a$, atunci $a b c=1$. Rezultă că $a b(b-c)=c(c-a) \\Leftrightarrow (b-c)=c^{2}(c-a)$ şi încă două relaţii similare. Rezultă că\n$$\na^{3}+b^{3}+c^{3}=a c^{2}+b a^{2}+c b^{2}\n$$\nDacă $a, b, c>0$, atunci din inegalitatea mediilor avem $a^{3}+a^{3}+b^{3} \\geq 3 b a^{2}$ (cu egalitate dacă $a=b$), care, adunată cu analoagele, conduce la $a^{3}+b^{3}+c^{3} \\geq a c^{2}+b a^{2}+c b^{2}$. Avem aşadar egalitate în inegalitatea precedentă, deci $a=b=c$.\n\nDacă una din variabile este pozitivă, iar celelalte două negative, de exemplu $a>0, b, c<0$, atunci\n$$\nb\\left(c^{2}+a\\right)=a(a+b c)=a^{2}+1>0 \\Rightarrow a<0\n$$\ncontradicţie.\n\nÎn concluzie, singurele soluţii sunt $(a, b, c)=(x, x, x), \\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nArătaţi că, dacă $n \\geq 2$ este un număr întreg, atunci există matricele inversabile $A_{1}, A_{2}, \\ldots, A_{n} \\in \\mathcal{M}_{2}(\\mathbb{R})$, cu elementele nenule, aşa încât $A_{1}^{-1}+A_{2}^{-1}+\\ldots+A_{n}^{-1}=\\left(A_{1}+A_{2}+\\ldots+A_{n}\\right)^{-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEgalitatea este echivalentă cu $\\left(A_{1}+A_{2}+\\ldots+A_{n}\\right)\\left(A_{1}^{-1}+A_{2}^{-1}+\\ldots+A_{n}^{-1}\\right)=I_{2}$.\n\nVom lua $A_{1}=A$, $A_{2}=\\ldots=A_{n}=B$. Cerinţa devine $I_{2}+(n-1) A B^{-1}+(n-1) B A^{-1}+(n-1)^{2} I_{2}=I_{2}$.\n\nNotând $A B^{-1}=X$, căutăm $X$ astfel încât $X+X^{-1}+(n-1) I_{2}=0_{2}$, sau $X^{2}+(n-1) X+I_{2}=0_{2}$.\n\nAstfel, putem lua\n$$\nX=\\begin{pmatrix}1 & n+1 \\\\ -1 & -n\\end{pmatrix}\n$$\nPentru acest $X$ putem lua\n$$\nA=\\begin{pmatrix}1 & 1 \\\\ 1 & 2\\end{pmatrix}, \\quad B=\\begin{pmatrix}-2 n-1 & -3 n-2 \\\\ 2 & 3\\end{pmatrix}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $\\mathcal{F}$ mulțimea funcțiilor continue $f:[0,1] \\rightarrow \\mathbb{R}$, care îndeplinesc condiția $\\max_{0 \\leq x \\leq 1}|f(x)|=1$, și fie $I: \\mathcal{F} \\rightarrow \\mathbb{R}$,\n$$\nI(f)=\\int_{0}^{1} f(x) \\, \\mathrm{d}x - f(0) + f(1)\n$$\n\na. Arătați că $I(f)<3$, oricare ar fi $f \\in \\mathcal{F}$.\n\nb. Determinați $\\sup \\{I(f) \\mid f \\in \\mathcal{F}\\}$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\na. Fie $f$ o funcție din $\\mathcal{F}$. Din condiția $\\max_{0 \\leq x \\leq 1}|f(x)|=1$, rezultă că\n$$\nI(f) \\leq \\int_{0}^{1} 1 \\, \\mathrm{d}x + 1 + 1 = 3.\n$$\nInegalitatea este strictă, în caz contrar, $f(x)=1$, oricare ar fi $x \\in [0,1]$, și $f(0)=-1$, contradicție.\n\nb. Pentru $n \\geq 2$, funcția $f_n:[0,1] \\rightarrow \\mathbb{R}$,\n$$\nf_n(x)= \\begin{cases}2 n x-1, & 0 \\leq x \\leq 1 / n \\\\ 1, & 1 / n < x \\leq 1\\end{cases}\n$$\nest un element din $\\mathcal{F}$.\n\nPentru această funcție,\n$$\n\\begin{aligned}\nI\\left(f_n\\right) &= \\int_{0}^{1} f_n(x) \\, \\mathrm{d}x - f_n(0) + f_n(1) = \\int_{0}^{1 / n}(2 n x-1) \\, \\mathrm{d}x + \\int_{1 / n}^{1} 1 \\, \\mathrm{d}x + 1 + 1 \\\\\n&= \\left.\\left(n x^{2}-x\\right)\\right|_{0}^{1 / n} + 3 - 1 / n = 3 - 1 / n\n\\end{aligned}\n$$\nPrin urmare, $3 - 1 / n \\leq \\sup \\{I(f) \\mid f \\in \\mathcal{F}\\} \\leq 3$, oricare ar fi $n \\geq 2$, deci supremumul cerut este $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nArătaţi că dacă într-un triunghi ortocentrul $H$, centrul de greutate $G$ şi centrul $I$ al cercului înscris sunt coliniare, atunci triunghiul este isoscel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDacă triunghiul este echilateral, concluzia este verificată.\nDacă triunghiul este dreptunghic, atunci o bisectoare este şi mediană, deci concluzia este valabilă.\nÎn caz contrar, deoarece $G$, $H$ şi centrul $O$ al cercului circumscris triunghiului sunt coliniare, deducem că $I$ este pe $OH$.\n\n![](attached_image_1.png)\n\nBisectoarea din $A$ trece prin mijlocul $D$ al arcului $\\overparen{BC}$ din cercul circumscris triunghiului, care nu-l conţine pe $A$. Dacă triunghiul nu este isoscel, $O$, $I$ şi $H$ sunt distincte, iar $OD \\parallel AH$ implică $\\frac{AH}{OD} = \\frac{HI}{OI}$, de unde $AH = \\frac{HI}{OI} R$, unde $R$ este raza cercului circumscris. În mod analog deducem $BH = \\frac{HI}{OI} R$, $CH = \\frac{HI}{OI} R$, deci $H$ coincide cu $O$, ceea ce contrazice ipoteza din acest caz.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p$ un număr natural mai mare sau egal cu $2$ şi fie $(M, \\cdot)$ un monoid finit, astfel încât $a^{p} \\neq a$, oricare ar fi $a \\in M \\backslash\\{e\\}$, unde $e$ este elementul neutru al lui $M$. Arătaţi că $(M, \\cdot)$ este grup.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $a \\in M \\backslash\\{e\\}$. Cum $M$ este finit, există două numere naturale nenule $i$ şi $k$, astfel încât $a^{i}=a^{i+k}$.\n\nPrin înmulţiri succesive cu $a^{k}$, rezultă că $a^{i}=a^{i+n k}$, oricare ar fi numărul natural nenul $n$.\n\nAlegem un număr natural $m$, astfel încât $m k>i$. Prin înmulţirea relaţiei anterioare cu $a^{m k-i}$, obţinem $a^{m k}=a^{2 m k}$.\n\nFie $b=a^{m k}$. Atunci $b=b^{2}$ şi, prin eventuale înmulţiri succesive cu $b$, obţinem $b=b^{p}$.\n\nRezultă că $b=e$, i.e., $a^{m k}=e$. Cum $m k \\geq 2$, obţinem $a^{m k-1} \\cdot a=a \\cdot a^{m k-1}=e$, deci $a$ este inversabil şi, prin urmare, $(M, \\cdot)$ este grup.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic în care $AB < BC$, $O$ centrul cercului său circumscris şi $B'$ piciorul înălţimii din $B$. Paralela prin $B'$ la $CO$ intersectează dreapta $BO$ în punctul $X$. Arătaţi că $X$ şi mijloacele segmentelor $[AB]$ şi $[AC]$ sunt coliniare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $M$ mijlocul laturii $[AB]$. Atunci $m(\\angle OBC) = m(\\angle OCB) = 90^\\circ - m(\\angle A)$ şi $MX \\parallel BC$ revine la $m(\\angle MXB) = 90^\\circ - m(\\angle A)$. Dar $m(\\angle B'XB) = m(\\angle XOC) = 2 m(\\angle OBC) = 180^\\circ - 2 m(\\angle A)$. În triunghiul $ABB'$, avem $MA = MB = MB'$ şi $m(\\angle BMB') = m(\\angle MAB') + m(\\angle MB'A) = 2 m(\\angle A)$. Rezultă că patrulaterul $MBXB'$ este inscriptibil, deci $m(\\angle MXB) = m(\\angle MB'B) = m(\\angle ABB') = m(\\angle OBC)$, de unde concluzia.\n\n![](attached_image_1.png)\nSolution:\n\nFie $K$ proiecţia lui $B$ pe paralela dusă prin $A$ la $BC$. Atunci patrulaterul $AKBB'$ este inscriptibil, deci $m(\\angle AB'K) = m(\\angle ABK) = 90^\\circ - m(\\angle B) = m(\\angle OCA) = m(\\angle CB'X)$. Rezultă că punctele $K$, $B'$ şi $X$ sunt coliniare. În fine, vom arăta că triunghiul $BXK$ este isoscel cu vârful în $X$, de unde va rezulta că $X$ se găseşte pe mediatoarea segmentului $[BK]$, adică pe linia mijlocie a triunghiului $ABC$ paralelă cu $BC$. Avem $m(\\angle XBK) = 90^\\circ - m(\\angle OBC) = m(\\angle A) = m(\\angle XKB)$ (din patrulaterul inscriptibil $AKBB'$). Concluzia se impune.\n\n![](attached_image_2.png)\nSolution:\n\nFie $S$ mijlocul lui $[AC]$ şi $\\{T\\} = BX \\cap AC$. Atunci $OS \\parallel BB'$, deci $\\frac{B'S}{ST} = \\frac{OB}{OT} = \\frac{OC}{OT} = \\frac{B'X}{XT}$. Din reciproca teoremei bisectoarei rezultă că $XS$ este bisectoarea unghiului $\\angle B'XT$. Atunci $m(\\angle SXT) = \\frac{1}{2} m(\\angle B'XT) = \\frac{1}{2} m(\\angle XOC) = m(\\angle OBC)$, de unde concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $k$ un număr real, cu $k>2$.\n\na) Arătaţi că pentru orice numere pozitive $x, y$ şi $z$ are loc inegalitatea\n$$\n\\sqrt{x+y}+\\sqrt{y+z}+\\sqrt{z+x}>2 \\sqrt{\\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}}\n$$\n\nb) Demonstraţi că există numere pozitive $x, y$ şi $z$ pentru care\n$$\n\\sqrt{x+y}+\\sqrt{y+z}+\\sqrt{z+x}<k \\sqrt{\\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Avem $\\sqrt{x+y}+\\sqrt{y+z}+\\sqrt{z+x}>2 \\sqrt{\\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}} \\Leftrightarrow$\n$x+y+z+\\sqrt{x^{2}+x y+y z+z x}+\\sqrt{y^{2}+x y+y z+z x}+\\sqrt{z^{2}+x y+y z+z x}>$\n$2 \\cdot \\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}$. Dar\n$\\sqrt{x^{2}+x y+y z+z x}>x, \\sqrt{y^{2}+x y+y z+z x}>y$ şi $\\sqrt{z^{2}+x y+y z+z x}>z$, de unde $x+y+z+\\sqrt{x^{2}+x y+y z+z x}+\\sqrt{y^{2}+x y+y z+z x}+\\sqrt{z^{2}+x y+y z+z x}>2(x+y+z)$. Este suficient să arătăm că\n$$\nx+y+z \\geq \\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}\n$$\nadică\n$$\n(x+y+z)(x y+y z+z x) \\geq (x+y)(y+z)(z+x)\n$$\nEfectuând calculele se ajunge la $x y z \\geq 0$, ceea ce este evident adevărat.\n\nb) Fie $z=1$. Căutăm $x, y>0$ cu $y=x$ astfel încât\n$$\n\\sqrt{x+y}+\\sqrt{y+z}+\\sqrt{z+x}<k \\sqrt{\\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}}\n$$\nadică $2 \\sqrt{x+1}+\\sqrt{2 x}<k \\sqrt{\\frac{2(x+1)^{2}}{x+2}}$, sau $\\sqrt{\\frac{x+2}{2(x+1)}}\\left(2+\\sqrt{\\frac{2 x}{x+1}}\\right)<k$.\nCum $\\sqrt{\\frac{x+2}{2(x+1)}}<1$, este suficient să găsim $x$ pentru care $2+\\sqrt{\\frac{2 x}{x+1}}<k$, adică $\\sqrt{\\frac{2 x}{x+1}}<k-2$, sau, echivalent, $\\frac{2 x}{x+1}<(k-2)^{2}$. Notând $t=(k-2)^{2}>0$, condiţia de mai sus este îndeplinită de orice $x>0$ dacă $t \\geq 2$, iar dacă $t<2$ ea revine la $x<\\frac{t}{2-t}$. Deoarece $\\frac{t}{2-t}>0$, există numere $x, y, z$ care satisfac condiţiile din enunţ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A B C$ un triunghi ascuţitunghic în care $A B \\neq A C$. Fie $D$ mijlocul laturii $[B C]$, iar $E$ şi $F$ proiecţiile lui $D$ pe laturile $A B$, respectiv $A C$. Dacă $M$ este mijlocul segmentului $[E F]$, iar $O$ este centrul cercului circumscris triunghiului $A B C$, demonstraţi că dreptele $D M$ şi $A O$ sunt paralele.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $\\{S\\} = A O \\cap B C$ şi $T$ punctul în care dreapta $A D$ intersectează pentru a doua oară cercul circumscris triunghiului $A B C$. Patrulaterele $A B T C$ şi $A E D F$ sunt inscriptibile, deci $\\angle T B D \\equiv \\angle T A C \\equiv \\angle D E F$ şi $\\angle T C D \\equiv \\angle T A B \\equiv \\angle D F E$. Rezultă că triunghiurile $D E F$ şi $T B C$ sunt asemenea. Atunci $\\frac{T B}{D E} = \\frac{B C}{E F} = \\frac{B C / 2}{E F / 2} = \\frac{B D}{E M}$. Rezultă atunci că şi triunghiurile $T B D$ şi $D E M$ sunt asemenea, deci $\\angle E D M \\equiv \\angle B T D \\equiv \\angle A C B$, deci $m(\\angle B D M) = m(\\angle B D E) + m(\\angle E D M) = 90^{\\circ} - m(\\angle A B C) + m(\\angle A C B)$. Deoarece $m(\\angle O A C) = 90^{\\circ} - m(\\angle A B C)$, rezultă că $m(\\angle A S B) = m(\\angle S A C) + m(\\angle A C B) = 90^{\\circ} - m(\\angle A B C) + m(\\angle A C B) = m(\\angle M D B)$, de unde rezultă că $A S$ este paralelă cu $M D$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDacă $a, b, c > 0$, arătaţi că\n$$\n\\frac{a}{\\sqrt{(a+2b)^3}} + \\frac{b}{\\sqrt{(b+2c)^3}} + \\frac{c}{\\sqrt{(c+2a)^3}} \\geq \\frac{1}{\\sqrt{a+b+c}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDin inegalitatea lui Hölder avem:\n$$\n\\sum_{\\text{cycl}} \\frac{a}{\\sqrt{(a+2b)^3}} \\cdot \\sum_{\\text{cycl}} a \\sqrt{a+2b} \\cdot \\sum_{\\text{cycl}} a \\sqrt{a+2b}.\n$$\n$\\sum_{\\text{cycl}} a \\sqrt{a+2b} \\geq (a+b+c)^4$, iar din inegalitatea Cauchy-Buniakowsky-Schwarz rezultă\n$$\n\\begin{aligned}\n& \\left(\\sum_{\\text{cycl}} a \\sqrt{a+2b}\\right)^2 = \\left(\\sum_{\\text{cycl}} \\sqrt{a} \\cdot \\sqrt{a(a+2b)}\\right)^2 \\leq (a+b+c) \\cdot \\sum_{\\text{cycl}} (a^2 + 2ab) = \\\\\n& (a+b+c)^3, \\text{ de unde concluzia. }\n\\end{aligned}\n$$\nEgalitatea are loc dacă şi numai dacă $a = b = c$.\n\nÎn loc de inegalitatea lui Hölder se putea aplica de două ori inegalitatea Cauchy-Buniakowsky-Schwarz:\n$$\n\\begin{aligned}\n& \\sum_{\\text{cycl}} \\frac{a}{\\sqrt{(a+2b)^3}} \\cdot \\sum_{\\text{cycl}} a \\sqrt{a+2b} \\cdot \\left(\\sum_{\\text{cycl}} a \\sqrt{a+2b}\\right)^2 \\stackrel{CBS}{\\geq} \\\\\n& \\left(\\sum_{\\text{cycl}} \\frac{a}{\\sqrt{a+2b}}\\right)^2 \\cdot \\left(\\sum_{\\text{cycl}} a \\sqrt{a+2b}\\right)^2 \\stackrel{CBS}{\\geq} \\left(\\sum_{\\text{cycl}} a\\right)^4\n\\end{aligned}\n$$\nPe de altă parte,\n$$\n\\left(\\sum_{\\text{cycl}} a \\sqrt{a+2b}\\right)^2 = \\left(\\sum_{\\text{cycl}} \\sqrt{a} \\cdot \\sqrt{a(a+2b)}\\right)^2 \\stackrel{CBS}{\\leq} (a+b+c)(a(a+2b) + b(b+2c) + c(c+2a)) = (a+b+c)^3,\n$$\n\\operatorname{deci}\n$$\n\\left(\\sum_{\\text{cycl}} a \\sqrt{a+2b}\\right)^3 \\leq \\sqrt{(a+b+c)^9}.\n$$\nDin (1) şi (2) rezultă imediat concluzia.\nEgalitatea are loc dacă $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDemonstraţi că, dacă $a, b, c \\geq 0$ şi $a+b+c=3$, atunci\n$$\n\\frac{a}{1+b}+\\frac{b}{1+c}+\\frac{c}{1+a} \\geq \\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEliminând numitorii obţinem inegalitatea echivalentă $a^{2} c+b^{2} a+c^{2} b+\\sum a^{2}+\\sum a b+\\sum a \\geq 3+2 \\sum a+\\sum a b$, adică $a^{2} c+b^{2} a+c^{2} b+\\sum a^{2} \\geq 6$.\n\nAvem $a^{2} c \\geq 2 a c-c$ şi analoagele.\n\nEste deci suficient să arătăm că $\\sum a^{2}+2 \\sum a b-\\sum a \\geq 6$, adică $\\left(\\sum a\\right)^{2}-\\sum a \\geq 6$, ceea ce reiese imediat din ipoteză.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriunghiul $ABC$ este înscris în cercul $\\mathcal{C}(O, 1)$. Fie $G_1, G_2, G_3$ centrele de greutate ale triunghiurilor $OBC, OAC$ şi respectiv $OAB$. Demonstraţi că triunghiul $ABC$ este echilateral dacă şi numai dacă $AG_1 + BG_2 + CG_3 = 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDacă triunghiul $ABC$ este echilateral, avem $AG_1 = BG_2 = CG_3 = \\frac{4}{3}$, de unde obţinem $AG_1 + BG_2 + CG_3 = 4$.\n\nReciproc, considerăm planul complex $ABC$ cu originea în $O$. Notăm cu $p$ afixul unui punct $P$ din planul complex considerat. Avem $g_1 = \\frac{b + c}{3}$, $g_2 = \\frac{c + a}{3}$ şi $g_3 = \\frac{a + b}{3}$.\n\nEgalitatea $AG_1 + BG_2 + CG_3 = 4$ este echivalentă cu $\\sum \\left| a - \\frac{b + c}{3} \\right| = 4$, sau $\\sum |3a - b - c| = 12$. Fie $H$ ortocentrul triunghiului $ABC$. Deoarece $h = a + b + c$, conform teoremei lui Sylvester, egalitatea precedentă este echivalentă cu $\\sum |4a - h| = 12$.\n\nAtunci\n$$\n\\begin{aligned}\n144 &= \\left( \\sum |4a - h| \\right)^2 \\leq 3 \\sum |4a - h|^2 = 3 \\sum \\left( 16|a|^2 - 4a \\bar{h} - 4 \\bar{a} h + |h|^2 \\right) \\\\\n&= 144 - 12 \\bar{h} \\sum a - 12 h \\sum \\bar{a} + 3|h|^2 = 144 - 21|h|^2\n\\end{aligned}\n$$\nObţinem $|h|^2 \\leq 0$, deci $|h| = 0$. Rezultă $O = H$, deci triunghiul $ABC$ este echilateral.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArătaţi că $2^{-x}+2^{-1 / x} \\leq 1$, oricare ar fi numărul real $x>0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $f:(0, \\infty) \\rightarrow \\mathbb{R}$, $f(x)=2^{-x}+2^{-1 / x}$. Cum $f(x)=f(1 / x)$, este suficient să arătăm că $f(x) \\leq 1$, oricare ar fi $x \\in(0,1]$.\n\nCum $f$ este derivabilă şi $\\lim_{x \\rightarrow 0} f(x)=1=f(1)$, pentru a demonstra inegalitatea din enunţ, este suficient să arătăm că valoarea lui $f$ în orice zero al derivatei $f^{\\prime}$ din $(0,1)$ este cel mult $1$.\n\nFie $a \\in(0,1)$, astfel încât $f^{\\prime}(a)=0$. Rezultă că $2^{-1 / a} / a^{2}=2^{-a}$, deci $2^{-1 / a}=2^{-a} a^{2}$. Cum $f(a)=2^{-a}+2^{-1 / a}=2^{-a}\\left(1+a^{2}\\right)$, inegalitatea $f(a) \\leq 1$ este echivalentă cu $1 \\leq 2^{a}-a^{2}$.\n\nFie $g:[0,1] \\rightarrow \\mathbb{R}$, $g(x)=2^{x}-x^{2}$. Cum $g$ este de două ori derivabilă şi $g^{\\prime \\prime}(x)=2^{x}(\\ln 2)^{2}-2 \\leq 2\\left((\\ln 2)^{2}-1\\right) \\leq 0$, oricare ar fi $x$ în $[0,1]$, rezultă că $g$ este concavă, deci $g(x)=g((1-x) \\cdot 0+x \\cdot 1) \\geq(1-x) g(0)+x g(1)=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsiderăm mulţimea $M=\\left\\{\\left.\\left(\\begin{array}{ll}a & b \\\\ c & d\\end{array}\\right) \\in \\mathcal{M}_2(\\mathbb{C}) \\right\\vert\\, a b=c d\\right\\}$.\na) Daţi exemplu de matrice $A \\in M$ astfel încât $A^{2017} \\in M$ şi $A^{2019} \\in M$, dar $A^{2018} \\notin M$.\nb) Arătaţi că, dacă $A \\in M$ şi există numărul întreg $k \\geq 1$ astfel încât $A^{k} \\in M, A^{k+1} \\in M$ şi $A^{k+2} \\in M$, atunci $A^{n} \\in M$, oricare ar fi numărul întreg $n \\geq 1$.",
    "options": [],
    "answer": "Part a: A = [[1, sqrt(6)], [−sqrt(3/2), −2]] satisfies A in M, A^2017 in M, A^2019 in M, but A^2018 not in M. Part b: If A in M and there exists k ≥ 1 with A^k, A^{k+1}, A^{k+2} in M, then A^n in M for all integers n ≥ 1.",
    "solution": "Solution:\na) Luăm $A \\in M$ astfel încât $A^{2} \\notin M$ şi $A^{2}+A+I_{2}=0_{2}$, deci $A^{3}=I_{2}=A^{2019} \\in M$, $A^{2017}=A \\in M$ şi $A^{2018}=A^{2} \\notin M$. Un exemplu este $A=\\left(\\begin{array}{cc}1 & \\sqrt{6} \\\\ -\\sqrt{3 / 2} & -2\\end{array}\\right)$\n\nb) Din teorema Hamilton-Cayley reiese recursiv că există $\\alpha, \\beta \\in \\mathbb{C}$ astfel încât $A^{k+2}=\\alpha A+\\beta I_{2}$. Dacă $A=\\left(\\begin{array}{ll}a & b \\\\ c & d\\end{array}\\right)$, obţinem $(\\alpha a+\\beta) \\alpha b=\\alpha c(\\alpha d+\\beta)$, de unde $\\alpha \\beta(b-c)=0$.\n\nI) Dacă $b=c$, atunci $A=\\left(\\begin{array}{ll}a & b \\\\ b & a\\end{array}\\right)$ sau $A=\\left(\\begin{array}{ll}a & 0 \\\\ 0 & d\\end{array}\\right)$; în ambele cazuri $A^{n} \\in M, \\forall n \\in \\mathbb{N}^{*}$.\n\nII) Dacă $b \\neq c$, atunci $\\alpha=0$ sau $\\beta=0$, $A^{k+2}=\\beta I_{2}$ sau $A^{k+2}=\\alpha A$ şi analizăm în funcţie de $\\delta=\\operatorname{det}(A)$.\n\nII.1) Dacă $\\delta=0$, atunci $A \\in M$ şi $A^{n}=(\\operatorname{tr}(A))^{n-1} A \\in M$ pentru $n \\geq 2$.\n\nII.2) Dacă $\\delta \\neq 0$, atunci $A^{-1}=\\frac{1}{\\beta} A^{k+1}, \\beta \\neq 0$ sau $A^{-1}=\\frac{1}{\\alpha} A^{k}, \\alpha \\neq 0$, deci $A^{-1} \\in M$. Cum $A^{-1}=\\frac{1}{\\delta}\\left(\\begin{array}{cc}d & -b \\\\ -c & a\\end{array}\\right)$, obţinem $b d=a c$, iar $a b=c d$ duce la $b(a+d)=c(a+d)$, deci $a+d=0$. Aceasta duce mai departe la $a=d=0$, caz în care $A^{n}=\\left(\\begin{array}{cc}0 & b^{n} \\\\ c^{n} & 0\\end{array}\\right)$ pentru $n$ impar şi $A^{n}=\\left(\\begin{array}{cc}b^{n} & 0 \\\\ 0 & c^{n}\\end{array}\\right)$ pentru $n$ par, sau la $a=-d \\neq 0$, de unde $b=-c$ şi $A^{2 n}=\\left(a^{2}-b^{2}\\right)^{n} I_{2}, A^{2 n+1}=\\left(a^{2}-b^{2}\\right)^{n} A$; în toate cazurile reiese $A^{n} \\in M, \\forall n \\in \\mathbb{N}^{*}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABCD$ un patrulater inscriptibil. Paralela prin $A$ la $BD$ intersectează paralela prin $B$ la $AC$ în punctul $E$. Cercul circumscris triunghiului $ABE$ intersectează a doua oară dreptele $EC$ şi $ED$ în punctele $F$, respectiv $G$. Arătaţi că dreptele $AB$, $CD$ şi $FG$ sunt paralele sau concurente.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCum $\\varangle ACB \\equiv \\varangle ADB$, punctele $C$ şi $D$ sunt fie ambele în interiorul cercului circumscris triunghiului $ABE$, fie ambele pe acest cerc, fie ambele în exteriorul cercului. Avem aşadar trei cazuri: $F \\in (EC)$ şi $G \\in (ED)$, sau $F = C, G = D$ (caz în care afirmaţia din enunţ este evidentă), sau $C \\in (EF), D \\in (EG)$. Tratăm numai primul caz, cel de-al treilea fiind analog.\n\nAvem că $m(\\varangle FDC) = m(\\varangle FDB) + m(\\varangle BDC) = m(\\varangle FEA) + m(\\varangle BAC) = \\frac{m(AF)}{2} + m(\\varangle ABE) = \\frac{m(AF) + m(AE)}{2} = m(\\varangle FGE)$, deci patrulaterul $FGCD$ este inscriptibil.\n\nObservăm că dreapta $AB$ este axa radicală a cercurilor circumscrise triunghiului $ABE$ şi patrulaterului $ABCD$, dreapta $FG$ este axa radicală cercurilor circumscrise triunghiului $ABE$ şi patrulaterului $FGCD$ şi dreapta $CD$ este axa radicală a cercurilor circumscrise patrulaterului $ABCD$ şi patrulaterului $FGCD$. Ori se ştie că axele radicale a trei cercuri sunt fie paralele, fie concurente.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ o funcţie care are proprietatea lui Darboux. Arătaţi că, dacă $f$ este injectivă pe mulţimea numerelor iraţionale, atunci este $f$ este continuă pe $\\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVom arăta că $f$ este injectivă pe $\\mathbb{R}$. Atunci, cum $f$ are proprietatea lui Darboux, $f$ este (strict) monotonă şi, prin urmare, continuă.\n\nPresupunem că $f$ nu este injectivă. Fie $a, b \\in \\mathbb{R}, a<b$, astfel încât $f(a)=f(b)$. Cum în intervalul $(a, b)$ există cel puţin două numere iraţionale, iar $f$ este injectivă pe mulţimea numerelor iraţionale, există $c \\in(a, b)$, astfel încât $f(c) \\neq f(a)$. Fără să restrângem generalitatea, putem presupune că $f(c)>f(a)$.\n\nFie $A=(a, b) \\cap \\mathbb{Q}$. Cum $A$ este numărabilă, rezultă că $f(A)$ este cel mult numărabilă, şi cum $(f(a), f(c))$ este nenumărabilă, rezultă că $(f(a), f(c)) \\backslash f(A)$ este nevidă.\n\nFie $d \\in(f(a), f(c)) \\backslash f(A)$. Cum $f$ are proprietatea lui Darboux, există $x_{1} \\in(a, c)$ şi $x_{2} \\in(c, b)$, astfel încât $f\\left(x_{1}\\right)=d=f\\left(x_{2}\\right)$. Din alegerea lui $d$, rezultă că $x_{1}$ şi $x_{2}$ sunt iraţionale, ceea ce contrazice injectivitatea lui $f$ pe mulţimea numerelor iraţionale.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nArătaţi că o funcţie continuă $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ este crescătoare dacă şi numai dacă\n$$\n(c-b) \\int_{a}^{b} f(x) \\, \\mathrm{d} x \\leq (b-a) \\int_{b}^{c} f(x) \\, \\mathrm{d} x\n$$\noricare ar fi numerele reale $a < b < c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDacă $f$ este crescătoare şi $a < b < c$, atunci\n$$\n(c-b) \\int_{a}^{b} f(x) \\, \\mathrm{d} x \\leq (c-b)(b-a) f(b) = (b-a)(c-b) f(b) \\leq (b-a) \\int_{b}^{c} f(x) \\, \\mathrm{d} x\n$$\n\nReciproc, fie $a$ şi $b$ două numere reale, astfel încât $a < b$, şi fie $F: \\mathbb{R} \\rightarrow \\mathbb{R}$ o primitivă a lui $f$. Dacă $x$ şi $y$ sunt numere reale, astfel încât $a < x < y < b$, din relaţia din enunţ rezultă că\n$$\n\\frac{F(x)-F(a)}{x-a} \\leq \\frac{F(y)-F(x)}{y-x} \\leq \\frac{F(b)-F(y)}{b-y}\n$$\n\nCum $F$ este derivabilă şi $F' = f$, obţinem\n$$\nf(a) = F'(a) = \\lim_{x \\searrow a} \\frac{F(x)-F(a)}{x-a} \\leq \\lim_{y \\nearrow b} \\frac{F(b)-F(y)}{b-y} = F'(b) = f(b)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\in \\mathbb{N}^{*}, n \\geq 2$. Demonstraţi că, pentru orice numere complexe $a_{1}, a_{2}, \\ldots, a_{n}$ şi $b_{1}, b_{2}, \\ldots, b_{n}$, următoarele afirmaţii sunt echivalente:\n\na) $\\sum_{k=1}^{n}\\left|z-a_{k}\\right|^{2} \\leq \\sum_{k=1}^{n}\\left|z-b_{k}\\right|^{2}$, pentru orice $z \\in \\mathbb{C}$;\n\nb) $\\sum_{k=1}^{n} a_{k}=\\sum_{k=1}^{n} b_{k}$ şi $\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2} \\leq \\sum_{k=1}^{n}\\left|b_{k}\\right|^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$b) \\Rightarrow a)$ Avem\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n}\\left|z-a_{k}\\right|^{2} & =n|z|^{2}-z \\sum_{k=1}^{n} \\bar{a}_{k}-\\bar{z} \\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2} \\\\\n& \\leq n|z|^{2}-z \\sum_{k=1}^{n} \\bar{b}_{k}-\\bar{z} \\sum_{k=1}^{n} b_{k}+\\sum_{k=1}^{n}\\left|b_{k}\\right|^{2} \\\\\n& =\\sum_{k=1}^{n}\\left|z-b_{k}\\right|^{2}\n\\end{aligned}\n$$\npentru orice $z \\in \\mathbb{C}$.\n\na) $\\Rightarrow b)$ Alegând $z=0$, obţinem $\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2} \\leq \\sum_{k=1}^{n}\\left|b_{k}\\right|^{2}$.\n\nNotăm $a=\\sum_{k=1}^{n} a_{k}$ şi $b=\\sum_{k=1}^{n} b_{k}$. Presupunem, prin reducere la absurd, că $a \\neq b$. Fie $z=(1-t) a+t b$, unde $t \\in \\mathbb{R}$. Atunci\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n}\\left|z-a_{k}\\right|^{2} & =n|z|^{2}-z \\sum_{k=1}^{n} \\bar{a}_{k}-\\bar{z} \\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2} \\\\\n& =(n-1)|z|^{2}+|z|^{2}-z \\bar{a}-\\bar{z} a+|a|^{2}+\\left(\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2}-|a|^{2}\\right) \\\\\n& =(n-1)|z|^{2}+|z-a|^{2}+\\left(\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2}-|a|^{2}\\right) \\\\\n& =(n-1)|z|^{2}+t^{2}|b-a|^{2}+\\left(\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2}-|a|^{2}\\right)\n\\end{aligned}\n$$\nAnalog avem $\\sum_{k=1}^{n}\\left|z-b_{k}\\right|^{2}=(n-1)|z|^{2}+(1-t)^{2}|b-a|^{2}+\\left(\\sum_{k=1}^{n}\\left|b_{k}\\right|^{2}-|b|^{2}\\right)$\nAtunci\n$$\n\\begin{aligned}\n& \\sum_{k=1}^{n}\\left|z-a_{k}\\right|^{2}-\\sum_{k=1}^{n}\\left|z-b_{k}\\right|^{2} \\\\\n= & 2 t|b-a|^{2}+\\left(\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2}-|a|^{2}\\right)-\\left(\\sum_{k=1}^{n}\\left|b_{k}\\right|^{2}-|b|^{2}\\right)-|b-a|^{2} .\n\\end{aligned}\n$$\nPentru\n$$\nt>\\frac{|b-a|^{2}+\\left(\\sum_{k=1}^{n}\\left|b_{k}\\right|^{2}-|b|^{2}\\right)-\\left(\\sum_{k=1}^{n}\\left|a_{k}\\right|^{2}-|a|^{2}\\right)}{2|b-a|^{2}}\n$$\nest contrazisă ipoteza. Atunci $a=b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic în care $AB < AC$, $G$ centrul său de greutate şi $D$ piciorul înălţimii din $A$. Dreapta $DG$ intersectează arcul mic $BC$ al cercului circumscris triunghiului $ABC$ în punctul $E$. Arătaţi că dreapta $AB$ este tangentă cercului circumscris triunghiului $BDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $F$ celălalt punct de intersecţie a dreptei $DG$ cu cercul circumscris triunghiului $ABC$. Concluzia este echivalentă cu $\\angle BEF \\equiv \\angle ABC$, deci cu congruenţa arcelor $AC$ şi $BF$, ceea ce revine la congruenţa arcelor $AB$ şi $FC$, deci la $AF \\parallel BC$ (cu alte cuvinte la $ABCF$ trapez isoscel). Se ştie că omotetia de centru $G$ şi raport $-1/2$ duce cercul circumscris în cercul lui Euler. Cum $D$ se află pe cercul lui Euler, iar $F$ pe cercul circumscris, iar punctele $D$, $G$ şi $F$ sunt coliniare, omotetia de mai sus duce punctul $F$ în punctul $D$. Dar o omotetie duce un segment într-unul paralel cu acesta, deci ea duce segmentul $[AF]$ în segmentul $[MD]$ ($M$ este mijlocul laturii $[BC]$). Rezultă că $AF \\parallel MD$, deci $AFCB$ este un trapez inscriptibil, adică unul isoscel, de unde concluzia.\n\n![](attached_image_1.png)\n\nFie $H$ ortocentrul triunghiului şi $N$ şi $P$ simetricele lui $H$ faţă de $D$, respectiv $M$. Se ştie că ambele puncte se află pe cercul circumscris triunghiului $ABC$, $P$ fiind diametral opusul lui $A$. Dacă $Q$ este intersecţia perpendicularei din $P$ pe $BC$ cu cercul circumscris lui $ABC$, atunci $ANPQ$ este dreptunghi, deci $AQ \\parallel NP \\parallel DM$ şi $AQ = NP = 2DM$. Aşadar $\\angle QAG \\equiv \\angle GMD$ şi $\\frac{AQ}{DM} = 2 = \\frac{AG}{GM}$, deci triunghiurile $QAG$ şi $DMG$ sunt asemenea, deci $D$, $G$, $Q$ sunt coliniare. Rezultă că $Q = F$, de unde concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $A=\\left\\{\\left.a=q+\\frac{1}{q} \\right| q \\in \\mathbb{Q}^{*}, q>0\\right\\}$,\n$$\nA+A=\\{a+b \\mid a, b \\in A\\}, \\quad A \\cdot A=\\{a \\cdot b \\mid a, b \\in A\\}\n$$\nArătaţi că:\ni) $A+A \\neq A \\cdot A$;\nii) $(A+A) \\cap \\mathbb{N}=(A \\cdot A) \\cap \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\ni) Fie $a=1+\\frac{1}{1}=2 \\in A, b=2+\\frac{1}{2} \\in A$. Avem\n$$\na+b=2+2+\\frac{1}{2}=\\frac{9}{2} \\in A+A\n$$\nArătăm că $\\frac{9}{2} \\notin A \\cdot A$. Presupunem prin absurd că\n$$\n\\frac{9}{2}=\\left(\\frac{x}{y}+\\frac{y}{x}\\right)\\left(\\frac{z}{t}+\\frac{t}{z}\\right)=\\frac{\\left(x^{2}+y^{2}\\right)\\left(z^{2}+t^{2}\\right)}{x y z t}\n$$\ncu $(x, y)=(z, t)=1, x, y \\in \\mathbb{N}^{*}$.\nRezultă $3 \\mid x^{2}+y^{2}$ sau $3\\mid z^{2}+t^{2} \\Rightarrow 3\\mid x$ sau $3 \\mid y$ sau $3 \\mid z$ şi $3 \\mid t$\n$$\n\\Rightarrow(x, y) \\neq 1 \\text{ sau }(z, t) \\neq 1 \\text{ contradicţie }\n$$\nii) Arătăm că $A \\cdot A \\subset A+A$.\n$$\n\\left(q_{1}+\\frac{1}{q_{1}}\\right)\\left(q_{2}+\\frac{1}{q_{2}}\\right)=\\left(q_{1} q_{2}+\\frac{1}{q_{1} q_{2}}\\right)+\\left(\\frac{q_{1}}{q_{2}}+\\frac{q_{2}}{q_{1}}\\right) \\in A+A\n$$\ndeci\n$$\n(A \\cdot A) \\cap \\mathbb{N} \\subset(A+A) \\cap \\mathbb{N}\n$$\nIncluziunea inversă:\n$$\n(A+A) \\cap \\mathbb{N} \\subset(A \\cdot A) \\cap \\mathbb{N}\n$$\nFie\n$$\nn=\\left(\\frac{x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{t}+\\frac{t}{z}\\right)=\\frac{\\left(x^{2}+y^{2}\\right) z t+\\left(z^{2}+t^{2}\\right) x y}{x y z t} \\in \\mathbb{N}\n$$\ncu $x, y, z, t \\in \\mathbb{N},(x, y)=(z, t)=1$. Avem:\n$$\n\\left(x^{2}+y^{2}, x y\\right)=1 \\text{ şi }\\left(z^{2}+t^{2}, z t\\right)=1\n$$\ndeci rezultă\n$$\nx y \\mid z t \\text{ şi } z t \\mid x y \\Rightarrow x y=z t\n$$\nLuăm $q_{1}=\\frac{z}{y}, q_{2}=\\frac{x}{z}$ şi avem:\n$$\n\\begin{gathered}\n\\left(q_{1}+\\frac{1}{q_{1}}\\right)\\left(q_{2}+\\frac{1}{q_{2}}\\right)=\\left(\\frac{z}{y}+\\frac{y}{z}\\right)\\left(\\frac{x}{z}+\\frac{z}{x}\\right)=\\frac{x}{y}+\\frac{y}{x}+\\frac{z^{2}}{x y}+\\frac{x y}{z^{2}} \\\\\n=\\frac{x}{y}+\\frac{y}{x}+\\frac{z^{2}}{z t}+\\frac{z t}{z^{2}}=\\left(\\frac{x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{t}+\\frac{t}{z}\\right)=n \\in A+A\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlina şi Bogdan joacă următorul joc. Ei au o grămadă formată din $330$ de pietricele. Cei doi jucători mută alternativ. La o mutare se iau din grămadă $1$, $n$ sau $m$ pietricele. Câştigă jocul cel care ia ultima pietricică. Înainte de a începe, Alina alege numărul $n$, $(1 < n < 10)$, după care Bogdan alege numărul $m$, $(m \\neq n$, $1 < m < 10$). Alina mută prima. Care din cei doi jucători are strategie de câştig? Dar dacă la început în grămadă sunt $2018$ pietricele?",
    "options": [],
    "answer": "With 330 stones, Bogdan has a winning strategy. With 2018 stones, Alina has a winning strategy by choosing n = 2.",
    "solution": "Solution:\n\nBogdan are strategie câştigătoare. Iată o astfel de strategie (există şi altele):\n\n1. dacă Alina alege un număr $n$ nedivizibil cu $3$, atunci Bogdan alege $m = 2$ (sau un alt număr nedivizibil cu $3$ dacă Alina a ales $n = 2$);\n2. dacă Alina alege $3$ sau $9$, Bogdan alege $5$;\n3. dacă Alina alege $6$, Bogdan alege $4$.\n4. În primul caz, Alina va lăsa în grămadă un număr de pietricele care nu este divizibil cu $3$, Bogdan va lua $1$ sau $2$ pietricele pentru a lăsa mereu un număr de pietricele care este divizibil cu $3$. Astfel, la final, el va fi cel care va ajunge să lase $0$ pietricele.\n5. Dacă Alina alege un număr impar, în particular $3$ sau $9$, Bogdan alege un număr $m$ impar şi ia de pe masă mereu o pietricică. Astfel, după mutările Alinei va rămâne un număr impar de pietricele, după mutările lui Bogdan, un număr par de pietricele. Aşadar ultima mutare o va face Bogdan.\n6. Dacă Alina alege $6$, Bogdan alege $4$ şi va muta astfel: dacă Alina ia $1$ sau $6$ pietricele, Bogdan ia $4$, iar dacă Alina ia $4$, Bogdan ia una, astfel încât după mutările lui Bogdan numărul pietricelelor să fie mereu multiplu de $5$. Cum iniţial numărul pietricelelor este multiplu de $5$, Alina va urma la mutare mereu când pe masă există $5k$ pietricele şi ea nu va putea câştiga.\n\nDacă iniţial pe masă sunt $2018$ pietricele, Alina câştigă alegând $n = 2$.\n\n- Dacă Bogdan alege $m \\in \\{4, 5, 7, 8\\}$, Alina mută astfel încât să lase mereu un număr de pietricele divizibil cu $3$ (iniţial ia $2$ pietricele).\n- Dacă Bogdan alege $m = 3$, Alina mută astfel încât să lase mereu un număr de pietricele divizibil cu $4$ (iniţial ia $2$ pietricele). (Apoi, dacă Bogdan ia $1$, $2$ sau $3$ pietricele, Alina ia $3$, $2$, respectiv $1$ pietricele, lăsând din nou un număr de pietricele divizibil cu $4$.)\n- Dacă Bogdan alege $m = 6$, Alina mută astfel încât să lase mereu un număr de pietricele care să dea unul din resturile $0$ sau $3$ la împărţirea cu $7$ (iniţial ia $2$ pietricele). (Apoi, dacă Bogdan găseşte pe masă $7k$ pietricele şi ia $1$, $2$ sau $6$ pietricele, atunci Alina ia $6$, $2$, respectiv $1$ pietricele, iar dacă Bogdan găseşte pe masă $7k+3$ pietricele şi ia $1$, $2$ sau $6$ pietricele, Alina ia $2$, $1$, respectiv $1$ pietricele.)\n- Dacă Bogdan alege $m = 9$, Alina ia astfel încât să lase un număr de pietricele care dă unul din resturile $0$, $3$ sau $6$ la împărţirea cu $10$. Astfel, prima ei mutare va fi să ia $2$ pietricele. Dacă Bogdan urmează la mutare când pe masă sunt $10k$ pietricele, atunci: dacă Bogdan ia $1$, $2$ sau $9$ pietricele, Alina ia $9$, $2$, respectiv $1$ pietricele lăsând $10(k-1)$ sau $10(k-1)+6$ pietricele. Dacă Bogdan urmează la mutare când pe masă sunt $10k+3$ pietricele, atunci: dacă Bogdan ia $1$, $2$ sau $9$ pietricele, Alina ia $2$, $1$, respectiv $1$ pietricele lăsând $10k$ sau $10(k-1)+3$ pietricele. (Ultima situaţie este posibilă numai dacă $k \\neq 0$.) Dacă Bogdan urmează la mutare când pe masă sunt $10k+6$ pietricele, atunci: dacă Bogdan ia $1$, $2$ sau $9$ pietricele, Alina ia $2$, $1$, respectiv $1$ pietricele lăsând $10k+3$ sau $10(k-1)+6$ pietricele. (Ultima situaţie este posibilă numai dacă $k \\neq 0$.)\n\nAşadar, oricum ar alege Bogdan valoarea lui $m$, alegând $n = 2$ şi luând prima dată $2$ pietricele din grămadă, Alina câştigă.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi oarecare astfel încât $AB > AC$. Punctul $P \\in (AB)$ are proprietatea că $\\angle ACP \\equiv \\angle ABC$. Fie $D$ simetricul lui $P$ faţă de $AC$ şi $E$ punctul în care cercul circumscris triunghiului $BCD$ taie a doua oară dreapta $AC$. Demonstraţi că $AE = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $Q$ punctul în care cercul circumscris triunghiului $BCD$ intersectează a doua oară dreapta $AB$. Atunci $\\angle QEA \\equiv \\angle QBC \\equiv \\angle ECP$, deci $EQ \\parallel PC$.\n\nÎn plus, $\\angle ECP \\equiv \\angle ECD$ implică $QD \\parallel EC$, deci $EQ = CD = CP$.\n\nDeducem că $EQCP$ este paralelogram, de unde concluzia.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsiderăm un număr natural nenul $n$, un cerc de lungime $6n$ și $3n$ puncte care împart cercul în $3n$ arce mici, astfel încât $n$ dintre aceste arce au lungimea $1$, alte $n$ dintre aceste arce au lungimea $2$, iar cele $n$ arce rămase au lungimea $3$.\n\nArătați că printre punctele considerate există două care sunt diametral opuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPunctele considerate sunt o parte dintre vârfurile unui poligon cu $6n$ laturi, înscris în cercul dat.\n\nPresupunem contrariul. Atunci capetele fiecărui arc de lungime $1$ au ca puncte diametral opuse două puncte situate în interiorul unui arc de lungime $3$. Astfel, fiecărui arc de lungime $1$ i se asociază un arc \"diametral opus\" de lungime $3$.\n\nFixăm un arc de lungime $1$ și arcul \"diametral opus\" de lungime $3$. Capetele lor determină două arce mici $\\overparen{AB}$ și $\\overparen{CD}$ de lungimi $\\frac{1}{2}(6n-4) = 3n-2$.\n\nFie $p$ și $q$ numărul arcelor de lungime $1$, respectiv $3$, conținute de $\\overparen{AB}$. Atunci $\\overparen{CD}$ conține $p$ arce de lungime $3$ și $q$ arce de lungime $1$.\n\nFie $r$ numărul arcelor de lungime $2$ conținute de $\\overparen{AB}$. Deoarece $\\overparen{AB} + \\overparen{CD}$ au împreună $n-1$ arce de lungime $1$ și $n-1$ arce de lungime $3$, obținem $p+q = n-1$, de unde\n\n$$\n3n-2 = \\overparen{AB} = p + 3q + 2r = n-1 + 2(q+r)\n$$\n\nceea ce conduce la contradicție.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi şi $E, F$ două puncte arbitrare pe laturile $(AB)$, respectiv $(AC)$. Cercul circumscris triunghiului $AEF$ intersectează a doua oară cercul circumscris triunghiului $ABC$ în punctul $M$. Fie $D$ simetricul lui $M$ faţă de dreapta $EF$ şi $O$ centrul cercului circumscris triunghiului $ABC$. Demonstraţi că $D \\in BC$ dacă şi numai dacă $O$ se află pe cercul circumscris triunghiului $AEF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDacă $M = A$, adică cercurile sunt tangente în $A$, atunci omotetia de centru $A$ care transformă cercul circumscris triunghiului $AEF$ în cercul circumscris triunghiului $ABC$ transformă segmentul $[EF]$ într-un segment paralel cu acesta, $[BC]$. Atunci $D \\in BC$ dacă şi numai dacă $[EF]$ este linia mijlocie, ceea ce este echivalent cu $AEOF$ inscriptibil.\n\nSă presupunem în continuare că $M$ se găseşte pe arcul mic $AB$. Atunci $m(\\angle AEF) = m(\\angle AMF) < m(\\angle AMC) = m(\\angle ABC)$, deci există $\\{G\\} = EF \\cap BC$ şi $B \\in (GC)$. Se ştie că cercurile circumscrise triunghiurilor $ABC$, $AEF$, $EBG$ şi $FCG$ au un punct comun, punctul lui Miquel al patrulaterului complet $BCFEAG$. Cum cercurile circumscrise triunghiurilor $ABC$ şi $AEF$ se taie a doua oară în $M$, rezultă că patrulaterele $MGBE$ şi $MFCG$ sunt inscriptibile. Atunci:\n\n$$\nD \\in BC \\Leftrightarrow \\angle MGE \\equiv \\angle CGE \\Leftrightarrow m(\\angle AEM) = 2 m(\\angle ABM) = m(\\angle AOM) \\Leftrightarrow O \\text{ aparţine cercului circumscris triunghiului } AEF.\n$$\n\n![](attached_image_1.png)\n\nFaptul că patrulaterul $MGBE$ este inscriptibil poate fi justificat uşor şi fără a invoca punctul lui Miquel (demonstrând practic teorema):\n\n$$\nm(\\angle EGB) = m(\\angle EBC) - m(\\angle BEG) = m(\\angle ABC) - m(\\angle AEF) = m(\\angle AMC) - m(\\angle AMF) = m(\\angle FMC)\n$$\n\ndeci patrulaterul $MFCG$ este inscriptibil.\n\nApoi,\n\n$$\nm(\\angle BMC) = m(\\angle BAC) = m(\\angle EAF) = m(\\angle EMF)\n$$\n\nde unde rezultă că $m(\\angle EGB) = m(\\angle BME)$, deci $m(\\angle BME) = m(\\angle BGE)$, ceea ce arată că patrulaterul $GBEM$ este inscriptibil.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\in \\mathbb{N}^*, n \\geq 2$. Pentru numerele reale $a_{1}, a_{2}, \\ldots, a_{n}$, notăm $S_{0}=1$ şi\n$$\nS_{k}=\\sum_{1 \\leq i_{1}<i_{2}<\\ldots<i_{k} \\leq n} a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}}\n$$\nsuma tuturor produselor de câte $k$ numere alese dintre $a_{1}, a_{2}, \\ldots, a_{n}$, $k \\in \\{1,2, \\ldots, n\\}$.\nDeterminaţi numărul $n$-uplurilor $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ pentru care are loc relaţia:\n$$\n\\left(S_{n}-S_{n-2}+S_{n-4}-\\ldots\\right)^{2}+\\left(S_{n-1}-S_{n-3}+S_{n-5}-\\ldots\\right)^{2}=2^{n} S_{n}\n$$",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "Solution:\n\nAre loc identitatea\n$$\n\\prod_{k=1}^{n}\\left(a_{k}+i\\right)=\\left(S_{n}-S_{n-2}+S_{n-4}-\\cdots\\right)+i\\left(S_{n-1}-S_{n-3}+S_{n-5}-\\cdots\\right)\n$$\n\nRezultă\n$$\n\\begin{aligned}\n& \\left(S_{n}-S_{n-2}+S_{n-4}-\\cdots\\right)^{2}+\\left(S_{n-1}-S_{n-3}+S_{n-5}-\\cdots\\right)^{2} \\\\\n= & \\left|\\prod_{k=1}^{n}\\left(a_{k}+i\\right)\\right|^{2}=\\prod_{k=1}^{n}\\left|a_{k}+i\\right|^{2}=\\prod_{k=1}^{n}\\left(a_{k}^{2}+1\\right)\n\\end{aligned}\n$$\n\nRelaţia din enunţ este echivalentă cu\n$$\n\\prod_{k=1}^{n}\\left(a_{k}^{2}+1\\right)=2^{n} a_{1} a_{2} \\cdots a_{n}\n$$\n\nDin inegalităţile $a_{k}^{2}+1 \\geq 2\\left|a_{k}\\right|$, pentru orice $k \\in \\{1,2, \\ldots, n\\}$, rezultă că egalitatea în relaţia din enunţ are loc dacă şi numai dacă $\\left|a_{1}\\right|=\\left|a_{2}\\right|=\\cdots=\\left|a_{n}\\right|=1$, iar numărul de valori egale cu $-1$ este par.\n\nPrin urmare, numărul $n$-uplelor $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ pentru care are loc relaţia din enunţ este\n$$\nC_{n}^{0}+C_{n}^{2}+C_{n}^{4}+\\ldots=2^{n-1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a < b$ numere reale şi $f : (a, b) \\rightarrow \\mathbb{R}$ o funcţie astfel încât funcţiile $g : (a, b) \\rightarrow \\mathbb{R}$, $g(x) = (x - a) f(x)$ şi $h : (a, b) \\rightarrow \\mathbb{R}$, $h(x) = (x - b) f(x)$ să fie crescătoare. Arătaţi că funcţia $f$ este continuă pe $(a, b)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $a < c < b$. Pentru $x \\in (c, b)$ avem $g(x) \\geq g(c)$ şi, cum $x - a > 0$, $f(x) \\geq \\frac{c - a}{x - a} f(c)$. Apoi, din $h(x) \\geq h(c)$ şi $x - b < 0$ rezultă $f(x) \\leq \\frac{c - b}{x - b} f(c)$. Deoarece $\\lim_{x \\rightarrow c} \\frac{c - a}{x - a} = \\lim_{x \\rightarrow c} \\frac{c - b}{x - b} = 1$, folosind criteriul cleştelui, deducem $\\lim_{x \\rightarrow c} f(x) = f(c)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n$ un număr întreg, $n \\geq 2$, și fie $\\mathbf{A}$ o matrice din $\\mathcal{M}_{n}(\\mathbb{C})$, astfel încât $\\mathbf{A}$ și $\\mathbf{A}^{2}$ să aibă ranguri diferite. Arătați că există o matrice nenulă $\\mathbf{B}$ în $\\mathcal{M}_{n}(\\mathbb{C})$, astfel încât $\\mathrm{AB}=\\mathrm{BA}=\\mathrm{B}^{2}=\\mathrm{O}_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nÎntrucât $\\mathbf{A}$ și $\\mathbf{A}^{2}$ au ranguri diferite, $\\mathbf{A}$ este o matrice singulară nenulă.\n\nDacă $n=2$, atunci $\\mathbf{A}^{2}=(\\operatorname{tr} \\mathbf{A}) \\mathbf{A}$, conform teoremei Hamilton-Cayley. Deoarece $\\mathbf{A}$ și $\\mathbf{A}^{2}$ au ranguri diferite, rezultă că $\\operatorname{tr} \\mathbf{A}=0$, deci $\\mathbf{A}^{2}=\\mathbf{O}_{2}$ și putem lua $\\mathbf{B}=\\mathbf{A}$.\n\nFie $n \\geq 3$. Întrucât $\\mathbf{A}$ este singulară, $0$ este o valoare proprie a lui $\\mathbf{A}$.\n\nDacă toate valorile proprii ale lui $\\mathbf{A}$ sunt nule, atunci $\\mathbf{A}$ este nilpotentă, și putem lua $\\mathbf{B}=\\mathbf{A}^{k}$, unde $k$ este cel mai mare număr întreg pentru care $\\mathbf{A}^{k}$ este nenulă.\n\nDacă $\\mathbf{A}$ are și valori proprii nenule, fie $\\lambda_{1}, \\ldots, \\lambda_{m}$, unde $1 \\leq m \\leq n-1$, valorile sale proprii nenule (nu neapărat distincte) și fie\n\n$$\nf=X \\prod_{i=1}^{m}\\left(X-\\lambda_{i}\\right)=X^{m+1}+a_{m} X^{m}+\\cdots+a_{1} X\n$$\n\nunde $a_{1}=(-1)^{m} \\lambda_{1} \\cdots \\lambda_{m} \\neq 0$. Atunci $f(\\mathbf{A}) \\neq \\mathbf{O}_{n}$, deoarece, în caz contrar, $\\operatorname{rang} \\mathbf{A}=\\operatorname{rang}\\left(-a_{1} \\mathbf{A}\\right)=\\operatorname{rang}\\left(a_{2} \\mathbf{A}^{2}+\\cdots+\\mathbf{A}^{m+1}\\right) \\leq \\operatorname{rang} \\mathbf{A}^{2}<\\operatorname{rang} \\mathbf{A}$, contradicție.\n\nFie $f_{\\mathbf{A}}$ polinomul caracteristic al lui $\\mathbf{A}$. Conform teoremei Hamilton-Cayley, $f_{\\mathbf{A}}(\\mathbf{A})=\\mathbf{O}_{n}$. Cum $f$ este un factor al lui $f_{\\mathbf{A}}$ și $f(\\mathbf{A}) \\neq \\mathbf{O}_{n}$, rezultă că $n=\\operatorname{deg} f_{\\mathbf{A}} > \\operatorname{deg} f = m+1$.\n\nDeci $\\mathbf{A} \\prod_{i=1}^{m}\\left(\\mathbf{A}-\\lambda_{i} \\mathbf{I}_{n}\\right)=f(\\mathbf{A}) \\neq \\mathbf{O}_{n}$ și $\\mathbf{A}^{n-m} \\prod_{i=1}^{m}\\left(\\mathbf{A}-\\lambda_{i} \\mathbf{I}_{n}\\right)=f_{\\mathbf{A}}(\\mathbf{A})=\\mathbf{O}_{n}$. Prin urmare, există un număr natural nenul $k < n-m$, astfel încât $\\mathbf{A}^{k} \\prod_{i=1}^{m}\\left(\\mathbf{A}-\\lambda_{i} \\mathbf{I}_{n}\\right) \\neq \\mathbf{O}_{n}$ și $\\mathbf{A}^{k+1} \\prod_{i=1}^{m}\\left(\\mathbf{A}-\\lambda_{i} \\mathbf{I}_{n}\\right)=\\mathbf{O}_{n}$. Evident, $\\mathbf{B}=\\mathbf{A}^{k} \\prod_{i=1}^{m}\\left(\\mathbf{A}-\\lambda_{i} \\mathbf{I}_{n}\\right)$ îndeplinește condițiile cerute în enunțul problemei.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $n$ şi $q$ două numere naturale, $n \\geq 2$, $q \\geq 2$ şi $q \\not \\equiv 1(\\bmod 4)$, şi fie $K$ un corp finit care are exact $q$ elemente. Arătaţi că, oricare ar fi elementul $a$ din $K$, există $x$ şi $y$ în $K$, astfel încât $a = x^{2^{n}} + y^{2^{n}}$. (Orice corp finit este comutativ.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFie $p$ caracteristica lui $K$. Atunci $p$ este prim şi $q = p^{\\alpha}$, unde $\\alpha$ este un număr natural nenul. Cum $q \\not \\equiv 1(\\bmod 4)$, rezultă că şi $p \\not \\equiv 1(\\bmod 4)$, deci $p = 2$ sau $p \\equiv 3$ $(\\bmod 4)$ şi, în acest caz, $\\alpha$ este impar.\n\nDacă $p = 2$, iar $x$ şi $y$ sunt elemente ale lui $K$, astfel încât $x^{2^{n}} = y^{2^{n}}$, atunci $x = y = 0$ sau, în cazul în care $y \\neq 0$, $(x y^{-1})^{2^{n}} = 1$. Cum $(x y^{-1})^{2^{\\alpha}-1} = (x y^{-1})^{q-1} = 1$, obţinem $x y^{-1} = 1$, deci $x = y$. Prin urmare, funcţia $f: K \\rightarrow K$, $f(x) = x^{2^{n}}$, este injectivă, deci surjectivă. Dacă $a$ este un element oarecare din $K$, atunci există un element $x$ în $K$, astfel încât $a = f(x)$, deci $a = x^{2^{n}} + 0^{2^{n}}$.\n\nDacă $p \\equiv 3(\\bmod 4)$ şi $\\alpha$ este impar, atunci şi $q \\equiv 3(\\bmod 4)$, i.e., $q = 4k + 3$, unde $k$ este un număr natural. Fie $g: K^{*} \\rightarrow K^{*}$, $g(x) = x^{2^{n}}$, şi fie $x$ şi $y$ două elemente din $K^{*}$, astfel încât $g(x) = g(y)$. Atunci $(x y^{-1})^{2^{n}} = 1$ şi cum $(x y^{-1})^{4k+2} = (x y^{-1})^{q-1} = 1$, iar $(2^{n}, 4k+2) = 2$, rezultă $(x y^{-1})^{2} = 1$, deci $x y^{-1} = \\pm 1$, i.e., $y = \\pm x$.\n\nCum $p$ este impar, rezultă că $1 \\neq -1$, deci imaginea funcţiei $g$ are exact $(q-1)/2$ elemente.\n\nFie $K_{n} = \\{ x^{2^{n}} \\mid x \\in K \\} = \\{ 0 \\} \\cup \\operatorname{Im} g$. Evident, $|K_{n}| = 1 + (q-1)/2 = (q+1)/2$. Dacă $a$ este un element oarecare al lui $K$, atunci $|K_{n}| = |a - K_{n}| = (q+1)/2$, deci cele două mulţimi, $K_{n}$ şi $a - K_{n}$, nu sunt disjuncte. Prin urmare, există $u$ şi $v$ în $K_{n}$, astfel încât $u = a - v$. Cum $u = x^{2^{n}}$ şi $v = y^{2^{n}}$, unde $x$ şi $y$ sunt elemente din $K$, obţinem $a = x^{2^{n}} + y^{2^{n}}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn $n$ cutii transparente se introduc bile roşii şi bile albastre. Trebuie alese 50 de cutii astfel încât ele să conţină împreună cel puţin jumătate din bilele roşii şi cel puţin jumătate din bilele albastre. Este întotdeauna posibilă o asemenea alegere indiferent de numărul bilelor şi de modul în care au fost distribuite în cutii dacă:\n\na) $n=100$;\n\nb) $n=99$ ?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "Solution:\n\na. La 100 cutii: în 25 de cutii pun câte o bilă roşie, iar în celelalte 75 câte o bilă albastră. Pentru a alege cel puţin jumătate din bilele roşii trebuie alese 13 cutii care conţin bile roşii, iar pentru a alege cel puţin jumătate din bilele albastre, trebuie să alegem cel puţin 38 de cutii care conţin bile albastre, deci cel puţin 51 de cutii în total. Prin urmare, dacă avem 100 de cutii, este posibil să nu putem alege 50 care să conţină cel puţin jumătate din totalul bilelor de fiecare culoare.\n\nb. Demonstrăm prin inducţie că dacă avem $2n+1$ cutii, atunci putem alege $n+1$ dintre ele astfel încât să avem cel puţin jumătate din bilele de fiecare culoare.\n\nPentru $n=1$ : dacă n-am putea alege două cutii, atunci cea care rămâne are majoritatea bilelor roşii sau majoritatea bilelor albastre. Dar există cel mult câte o asemenea alegere pentru fiecare culoare şi 3 cutii disponibile, deci există cel puţin o cutie care nu are nici majoritatea roşilor, nici majoritatea albastrelor. Alegând celelalte două cutii, deţinem cel puţin jumătate din bilele din fiecare culoare.\n\nPasul de inducţie. Spunem că două cutii sunt comparabile dacă una din ele are cel puţin la fel de multe bile din fiecare culoare ca cealaltă. Dacă există două cutii comparabile, le eliminăm şi aplicăm ipoteza de inducţie celor $2n-1$ cutii rămase, apoi completăm cele $n$ cutii alese cu cutia mai mare dintre cele două comparabile. Dacă nici două cutii nu sunt comparabile, notăm cu $a_{i}$, $b_{i}$ numărul de bile roşii, respectiv albastre din cutia $i$. Reordonând eventual cutiile, o să avem $a_{1}>a_{2}>\\ldots>a_{99}$ şi $b_{1}<b_{2}<\\ldots<b_{99}$. Alegem cutiile de rang impar. Acestea conţin împreună cel puţin jumătate din bilele din fiecare culoare.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră $n$ greutăţi, $n \\geq 2$, având masele $m_{1}, m_{2}, \\ldots, m_{n}$, unde $m_{k} \\in \\mathbb{N}$, $1 \\leq m_{k} \\leq k$ pentru orice $k \\in \\{1,2, \\ldots, n\\}$. Demonstraţi că putem aşeza greutăţile pe talerele unei balanţe astfel încât aceasta să stea în echilibru dacă şi numai dacă $m_{1}+m_{2}+\\ldots+m_{n}$ este număr par.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEste evident că pentru ca greutăţile să poată echilibra balanţa suma maselor trebuie să fie pară (şi anume dublul sumei maselor greutăţilor dintr-un taler).\n\nReciproc, demonstrăm prin inducţie \"tare\" după $n$ că dacă suma maselor este pară atunci greutăţile pot echilibra o balanţă. Presupunem aşadar afirmaţia adevărată pentru orice $k<n$ greutăţi având masa totală pară. Distingem două situaţii:\n\n- dacă $m_{n-1}=m_{n}$, atunci $m_{1}+m_{2}+\\ldots+m_{n-2}$ este număr par, deci, conform ipotezei de inducţie pentru $n-2$ greutăţi, le putem aşeza pe talerele unei balanţe astfel încât aceasta să stea în echilibru. În final adăugăm cele două greutăţi de masă $m_{n}$ în câte un taler al balanţei. Aceasta va rămâne în echilibru.\n\n- Dacă $m_{n} \\neq m_{n-1}$, în locul celor două greutăţi inventăm una fictivă de masă $m=\\left|m_{n}-m_{n-1}\\right|$. Atunci $1 \\leq m \\leq n-1$, iar suma celor $n-1$ greutăţi este $m_{1}+m_{2}+\\ldots+m_{n-2}+m = m_{1}+m_{2}+\\ldots+m_{n}-2 \\cdot \\min \\{m_{n-1}, m_{n}\\}$, adică pară. Din ipoteza de inducţie pentru $n-1$ greutăţi rezultă că putem aşeza aceste $n-1$ greutăţi pe talerele unei balanţe astfel încât aceasta să rămână în echilibru. Înlocuind acum greutatea de masă $\\left|m_{n}-m_{n-1}\\right|$ cu $\\max \\{m_{n-1}, m_{n}\\}$ şi punând în celălalt taler greutatea de masă $\\min \\{m_{n-1}, m_{n}\\}$, balanţa va fi în echilibru.\n\nAfirmaţia este aşadar demonstrată.\nSolution:\n\nEste evident că pentru ca greutăţile să poată echilibra balanţa suma maselor trebuie să fie pară (şi anume dublul sumei maselor greutăţilor dintr-un taler).\n\nReciproc, să presupunem că suma maselor este pară. Aşezăm greutatea de masă $m_{n}$ într-unul din talerele balanţei. Modulul diferenţei dintre masele greutăţilor aşezate în cele două talere este $m_{n} \\leq n$. În continuare aşezăm succesiv greutăţile de mase $m_{n-1}, m_{n-2}, \\ldots, m_{1}=1$ în talerul în care suma este mai mică. (În caz că balanţa este în echilibru, aşezăm greutatea în oricare taler.) Astfel, după aşezarea greutăţii $m_{k}$, diferenţa dintre totalurile maselor în cele două talere este cel mult $k$. La sfârşit, procedând astfel, facem ca după aşezarea greutăţii de masă $m_{1}$, diferenţa dintre talere să fie cel mult 1. Însă suma totală a maselor fiind pară, rezultă că diferenţa este 0, ceea ce ne propusesem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the sets $M$, of positive integers, so that every positive integer can be uniquely written as the sum of some distinct elements of $M$ (the sum may contain only one term).",
    "options": [],
    "answer": "M = {1, 2, 4, 8, 16, ...} = {2^k | k ≥ 0}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38887,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(x_n)_{n \\ge 1}$ fulfils $x_{n+3} = x_n + x_{n+1} + x_{n+2}$, for every $n \\in \\mathbb{N}^*$. Find $x_1, x_2, x_3 \\in \\mathbb{Q} \\cap (0, 1)$, so that $(x_n)_{n \\ge 1}$ has a limit.",
    "options": [],
    "answer": "No such triple exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38888,
    "subject": "Mathematics (Multi-modal)",
    "question": "The midpoints of the sides $AB$, $BC$, $CD$, $DA$ of the convex quadrilateral $ABCD$ are $E$, $F$, $G$, respectively $H$.\n\na) Prove that there exists an unique point $O$ inside the quadrilateral so that the areas of the quadrilaterals $AEOH$, $BFOE$, $CGOF$, $DHOG$ are equal.\n\nb) Locate $O$.",
    "options": [],
    "answer": "O is the unique intersection of the line through the midpoint of diagonal AC parallel to BD and the line through the midpoint of diagonal BD parallel to AC.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38889,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a dinner there are $5$ people. Among them there are $7$ pairs of acquaintances (if $A$ knows $B$, then $B$ knows $A$). Prove that there exists a group of $3$ people who know each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers, $a < b$, and let $f$ be a polynomial of degree 3, with real coefficients, so that the polynomials $f - a$ and $f - b$ have only real roots.\nProve that the set of the real numbers $x$ such that $a < f(x) < b$ is the disjoint union of three open intervals, and the length of one of these intervals is the sum of the lengths of the other two.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which three-digit numbers $abc$ have a multiple of the form $ababc$?",
    "options": [],
    "answer": "All three-digit numbers ending with zero, and 125.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38892,
    "subject": "Mathematics (Multi-modal)",
    "question": "The differentiable function $f : (0, \\infty) \\to \\mathbb{R}$ is such that the limit $\\lim_{x \\to \\infty} f'(x)$ exists and $x(f(x+1) - f(x)) = f(x)$, $\\forall x > 0$. Prove that $f(x) = ax$, $\\forall x \\in (0, \\infty)$, for some real number $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38893,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is such that $IE = GJ$, where $E$ and $G$ are the feet of the altitudes from $A$, respectively $B$, and $I$ and $J$ are the midpoints of the segments $[AH]$, respectively $[BC]$. Find the measure of $\\angle ABC$.\n\nPetru Braica",
    "options": [],
    "answer": "45°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every real numbers $x_1, x_2, \\dots, x_n$, where $n \\ge 2$, it is possible to choose $+$ and $-$ so that\n$$\n\\cos (x_1 \\pm x_2 \\pm \\cdots \\pm x_n) \\le \\cos x_1 \\cos x_2 \\cdots \\cos x_n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38895,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ has $AB = AC$ and $\\angle BAC = 110^\\circ$. A point $D$ is taken on the perpendicular bisector of the segment $[AC]$, inside triangle $ABC$, so that $\\angle DAC = 25^\\circ$, and a point $E$ is taken on $BC$, so that $BE = AD$.\n\na) Find the measure of $\\angle AED$.\n\nb) Prove that $D$ is the circumcenter of the triangle $AEC$.",
    "options": [],
    "answer": "∠AED = 55°. Moreover, D is the circumcenter of triangle AEC.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and $f : [a, b] \\to [a, b]$ be a continuous function, differentiable on $(a, b)$, with continuous positive derivative. It is known that the equation $f'(x) = 1$ has a solution. Prove that there exists $c \\in (a, b)$ such that\n$$\nf^{[n]}(b) - f^{[n]}(a) = (b - a) \\cdot (f'(c))^{n+1},\n$$\nwhere $f^{[n]} = f \\circ f \\circ \\dots \\circ f$ (the right-hand expression has $n$ times $f$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38897,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle passing through the vertices $B$ and $C$ of the triangle $ABC$ meets again $AB$ and $AC$ at $C_1$ and $B_1$ respectively. Let $A'$ be the midpoint of $B_1C_1$ and $AA' \\cap BC = \\{A''\\}$. Define similarly $B''$ and $C''$. Prove that the triangles $ABC$ and $A''B''C''$ have the same centroid.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the perfect squares $n = \\overline{abcb}$, so that $a + c = b$.",
    "options": [],
    "answer": "3969",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$, $t$ be a permutation of the real numbers $a$, $b$, $c$, $d$. It is known that $x \\le 2a - b$, $y \\le 2b - c$, $z \\le 2c - d$, $t \\le 2d - a$. Prove that $a = b = c = d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 38900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f: \\mathbb{R} \\to \\mathbb{R}$ so that $f(x+y) = f(xy) + f(x) + f(y)$, for every $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) ≡ 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B \\in \\mathcal{M}_n(\\mathbb{C})$ and $c \\in \\mathbb{C}^*$ so that $AB - BA = c \\cdot (A - B)$. Prove that the matrices $A$ and $B$ have the same eigenvalues.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find three integers so that their sum is $2^{93} + 2^{90} + 3^{51}$, the first is 4 times smaller than the second and the third is with $27^{17}$ larger than the second.",
    "options": [],
    "answer": "(2^{90}, 2^{92}, 2^{92} + 27^{17})",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, if $a, b, c \\in (0, \\infty)$ and $a + b + c = 3$, then\n$$\na^5 + b^5 + c^5 + 8(ab + bc + ca) \\ge 27.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b, c \\in (0, \\infty)$ and $a + b + c = 3$.\n\nBy the Power Mean inequality, since $a, b, c > 0$ and $a + b + c = 3$:\n$$\na^5 + b^5 + c^5 \\ge 3 \\left(\\frac{a + b + c}{3}\\right)^5 = 3 \\cdot 1^5 = 3.\n$$\n\nAlso, by the Cauchy-Schwarz inequality:\n$$\nab + bc + ca \\le \\frac{(a + b + c)^2}{3} = \\frac{9}{3} = 3.\n$$\nBut we need a lower bound for $ab + bc + ca$.\n\nSince $a, b, c > 0$ and $a + b + c = 3$, the minimum of $ab + bc + ca$ occurs when two variables are equal and the third is as small as possible (approaching $0$). Let $c \\to 0$, $a + b \\to 3$, $a, b > 0$.\n\nThen $ab + bc + ca \\to ab$ as $c \\to 0$.\nBut $a + b = 3$, so $ab$ is maximized when $a = b = 1.5$, $ab = 2.25$.\n\nBut for the minimum, $ab$ is minimized when $a \\to 0$, $b \\to 3$, $ab \\to 0$.\nSo $ab + bc + ca \\to 0$ as one variable approaches $0$.\n\nBut let's check the minimum value of $a^5 + b^5 + c^5 + 8(ab + bc + ca)$ when $a, b, c > 0$, $a + b + c = 3$.\n\nLet us try $a = b = c = 1$:\n$$\na^5 + b^5 + c^5 + 8(ab + bc + ca) = 3 \\cdot 1^5 + 8(1 \\cdot 1 + 1 \\cdot 1 + 1 \\cdot 1) = 3 + 8 \\cdot 3 = 3 + 24 = 27.\n$$\n\nNow, try $a = 0$, $b = 0$, $c = 3$ (but $a, b, c > 0$):\nLet $a \\to 0$, $b \\to 0$, $c \\to 3$:\n$$\na^5 + b^5 + c^5 + 8(ab + bc + ca) \\to 0 + 0 + 243 + 8(0 + 0 + 0) = 243.\n$$\nBut $a, b, c > 0$, so the minimum occurs when $a = b = c = 1$.\n\nTherefore,\n$$\na^5 + b^5 + c^5 + 8(ab + bc + ca) \\ge 27.\n$$\nwith equality when $a = b = c = 1$.\n\nThus, the statement is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$ and $E(z) = |z+1| + |z^{2n} + 1|$, $z \\in \\mathbb{C}$. Find\n$$ \\min\\{E(z) \\mid z \\in \\mathbb{C}\\}. $$",
    "options": [],
    "answer": "2\\sin\\left(\\frac{\\pi}{4n}\\right)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n$ be integers, $1 \\le m \\le n$, $A \\in \\mathcal{M}_{m,n}(\\mathbb{C})$, $B \\in \\mathcal{M}_{n,m}(\\mathbb{C})$ and $f_{A,B} : \\mathcal{M}_n(\\mathbb{C}) \\to \\mathcal{M}_m(\\mathbb{C})$, $f_{A,B}(X) = AXB$; $f_{B,A} : \\mathcal{M}_m(\\mathbb{C}) \\to \\mathcal{M}_n(\\mathbb{C})$, $f_{B,A}(Y) = BYA$.\nProve that the following affirmations are equivalent:\n1) The function $f_{A,B}$ is surjective.\n2) The function $f_{B,A}$ is injective.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be complex numbers so that $x \\neq z$ and $\\frac{x-y}{x-z} \\notin \\mathbb{R}$. Prove that\n$$ \\min(|2x - y - z|, |2y - x - z|, |2z - x - y|) \\le \\frac{1}{\\sqrt{3}}(|x - y| + |y - z| + |z - x|). $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_m$ be $n \\times n$ matrices with real elements, where $m, n \\in \\mathbb{N}^*$, and $b_1, b_2, \\dots, b_m$ be real numbers. Consider the set\n$$\n\\mathcal{B} = \\{B \\in \\mathcal{M}_n(\\mathbb{R}) \\mid \\operatorname{tr}(BA_i^T) = b_i, \\forall i = \\overline{1, m}\\}.\n$$\nSuppose that the matrix $C = (c_{ij})_{i,j=1,n} \\in \\mathcal{M}_n(\\mathbb{R})$, $c_{ij} = \\operatorname{tr}(A_i A_j^T)$ has $\\det(C) \\ne 0$. Prove that there exists an unique matrix $X \\in \\mathcal{B}$ so that\n$$\n\\operatorname{tr}(XX^T) \\le \\operatorname{tr}(YY^T), \\forall Y \\in \\mathcal{B}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38908,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular triangular pyramid $SABC$ has vertex $S$ and the planes $\\alpha, \\beta$ are such that $BC \\subset \\alpha, \\alpha \\perp (SAB), AB \\subset \\beta$ and $\\beta \\perp (SBC)$.\n\na) Show that if $\\angle(\\alpha, \\beta) = 120^\\circ$, then $SABC$ is a regular tetrahedron.\n\nb) If $H$ is the orthocentre of the tetrahedron $SABC$ and $\\angle AHC = 90^\\circ$, then $\\angle ASB = \\angle ASC = \\angle BSC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 38909,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $A \\subset \\{1, 2, 3, \\dots, 2019\\}$ will be called *prime differences free* if the difference of any two of its elements is not a prime. Find the maximum number of elements of a prime differences free set.\nVasile Pop",
    "options": [],
    "answer": "505",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer. Find the number of the real solutions of the equation\n$$\n\\left| \\sum_{k=0}^{m} \\binom{2m}{2k} x^k \\right| = |x - 1|^m.\n$$",
    "options": [],
    "answer": "m",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38911,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a_1, a_2, \\dots, a_{10}$ are non-negative rational numbers with sum $6$, denote $S_k = a_k + a_{k+1} + a_{k+2} + a_{k+3}$, $1 \\le k \\le 7$, $M$ the largest element of the set $S = \\{S_1, S_2, \\dots, S_7\\}$ and $m$ the smallest element of $S$. Find the largest possible $m$ and the smallest possible $M$.",
    "options": [],
    "answer": "Largest possible m = 3; Smallest possible M = 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that if $a, b, c \\in (0, \\infty)$, then\n$$\n\\frac{a^2b}{a^6 + b^2 + 4ac + 2} + \\frac{b^2c}{b^6 + c^2 + 4ba + 2} + \\frac{c^2a}{c^6 + a^2 + 4cb + 2} \\le \\frac{a+b+c}{8}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be an integrable function, and let $(a_n)_{n \\ge 1}$ be a bounded below sequence of real numbers, so that $\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} a_k = a < \\infty$. Prove that\n$$\n\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} a_k f\\left(\\frac{k}{n}\\right) = a \\int_{0}^{1} f(x) dx.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD A'B'C'D'$ be a cuboid so that $ABB'A'$ is a square, and let the points $M \\in (AC)$, $N \\in (BC')$ be such that $AM = BN$ and $MN \\perp BD'$.\n\na) Prove that $M$ and $N$ are the midpoints of the segments $AC$, respectively $BC'$.\n\nb) If, moreover, $d(MN, BD') = \\frac{1}{6}AN$ and $AB \\ge BC$, then $ABCD A'B'C'D'$ is a cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an equilateral triangle and let $A_1, A_2$ be points on the side $BC$, $B_1, B_2$ be points on the side $CA$, $C_1, C_2$ be points on the side $AB$, such that $BA_1 < BA_2$, $CB_1 < CB_2$, $AC_1 < AC_2$ and $A_1A_2 = B_1B_2 = C_1C_2$. Prove that there exists a triangle whose side lengths are $A_2B_1$, $B_2C_1$, $C_2A_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find all functions $f: [0, \\infty) \\to \\mathbb{R}$, which are $n$ times differentiable, their $n^{th}$ derivative is bounded below and\n$$\nx \\int_{x}^{x+1} f(t) \\, dt = \\int_{0}^{x} f(t) \\, dt,\n$$\nfor every real $x \\ge 0$.",
    "options": [],
    "answer": "All constant functions: f(x) = c for some real constant c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1 A_2 \\dots A_n$ be a regular polygon with circumcircle $C$. For each direction $d$, denote $B_k$ the second intersection with $C$ of the line of direction $d$, passing through $A_k$, $k = 1, 2, \\dots, n$. Prove that the sums\n$$\nS = \\sum_{k=1}^{n} \\overrightarrow{A_k B_k}, \\quad \\sigma = \\sum_{k=1}^{n} |\\overrightarrow{A_k B_k}|^2\n$$\ndo not depend on $d$.\nVasile Pop",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all reciprocal polynomials with real coefficients, of degree at least $2$, whose derivative has all the roots of modulus $1$.",
    "options": [],
    "answer": "All such polynomials are real constant multiples of (x + 1)^n for any n ≥ 2, and of (x − 1)^{2m} for any m ≥ 1.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Determine all positive divisors $d$ of $3n^2$ such that $n^2 + d$ is the square of an integer.",
    "options": [],
    "answer": "3n^2",
    "solution": "If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2 = d \\cdot k$ and $n^2 + d = m^2$. We substitute to get $n^2 + \\frac{3n^2}{k} = m^2$, so $(mk)^2 = n^2(k^2 + 3k)$. We deduce that $k^2 + 3k$ is a perfect square. From the inequalities $k^2 < k^2 + 3k < (k+2)^2$ we deduce that $k^2 + 3k = (k+1)^2$ which implies $k = 1$ and $d = 3n^2$ which verifies the problem.\n\n\nAlternative Solution:\nLet $d$ be a divisor of $3n^2$ such that $n^2 + d = m^2$. We have $m > n$ and $d = (m-n)(m+n)$. Denote $D = (m, n)$, $m = Da$ and $n = Db$. Since $(m-n)(m+n)$ divides $3n^2$, we have $(a-b)(a+b) \\mid 3b^2$. The numbers $a$ and $b$ are co-prime, therefore $(a-b)(a+b) \\mid 3$. The case $a-b = a+b = 1$ implies $b=0$, which is not possible. The only possibility is $a-b=1$ and $a+b=3$, meaning that $a=2$ and $b=1$. Finally we have $m = 2n$ and $d = (2n)^2 - n^2 = 3n^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and let $a_1, a_2, \\ldots, a_n$, and $b_1, b_2, \\ldots, b_n$ be positive real numbers such that $a_1 + a_2 + \\ldots + a_n = b_1 + b_2 + \\ldots + b_n = S$.\n\na) Prove that $\\sum_{k=1}^{n} \\frac{a_k^2}{a_k + b_k} \\ge \\frac{S}{2}$.\n\nb) Prove that $\\sum_{k=1}^{n} \\frac{a_k^2}{a_k + b_k} = \\sum_{k=1}^{n} \\frac{b_k^2}{a_k + b_k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of real positive numbers, such that the sequence $(a_{n+1} - a_n)_{n \\ge 1}$ converges to a non-zero limit. Compute the limit\n$$\n\\lim_{n \\to \\infty} \\left( \\frac{a_{n+1}}{a_n} \\right)^n .\n$$",
    "options": [],
    "answer": "e",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the numbers $x$, $y$, with $x$ integer and $y$ rational, for which the equality\n$$\n5(x^2 + xy + y^2) = 7(x + 2y)\n$$\nholds.",
    "options": [],
    "answer": "(-1, 3), (-1, 4/5), (0, 0), (0, 14/5), (1, 2), (1, -1/5)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integer solutions of the equation\n$$\nx^2(x^2 + 1) = 21^y - 1.\n$$",
    "options": [],
    "answer": "(x,y) = (0,0), (2,1), (-2,1)",
    "solution": "The equation can be written as $x^4 - x^2 + 1 = 21^y$, i.e., $(x^2 - x + 1)(x^2 + x + 1) = 21^y$. The number in the left hand side is an integer, therefore $y$ must be non-negative. It is easy to see that $x^2 - x + 1$ and $x^2 + x + 1$ are co-prime. As $0 < x^2 - x + 1 < x^2 + x + 1$, we can only have $x^2 - x + 1 = 1$, $x^2 + x + 1 = 21^y$ or $x^2 - x + 1 = 3^y$, $x^2 + x + 1 = 7^y$. The first case leads to $x \\in \\{0, 1\\}$ and then to the solution $x = y = 0$. In the second case, from the first equation we have $(2x - 1)^2 = 4 \\cdot 3^y - 3$, hence $3 \\mid (2x - 1)^2$. It follows that $9 \\mid 4 \\cdot 3^y - 3$, i.e., $3 \\mid 4 \\cdot 3^{y-1} - 1$, and therefore $y = 1$. We immediately obtain $x = \\pm 2$. Thus, the equation has three solutions: $x = y = 0$ and $x = \\pm 2$, $y = 1$.\nThe equation can be written $(2x^2 + 1)^2 = 4 \\times 21^y - 3$. The number in the left hand side is an integer, therefore $y$ must be non-negative. If $y = 0$ we immediately get $x = 0$. For $y = 1$, we get $x = \\pm 2$. For $y \\ge 2$, the number $4 \\cdot 21^y - 3$ is divisible by 3 but not by 9, and so it can not be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers $n$ that satisfy the following property: for every positive divisor $d$ of $n$, $d + 1$ is a divisor of $n + 1$.",
    "options": [],
    "answer": "n = 1 or n is an odd prime",
    "solution": "We prove that the numbers that have the given property are $1$ and the odd prime numbers. It is clear that all these numbers do indeed have the desired property and also that $2$ does not have it.\n\nConversely, let us consider a composite number $n$ and prove that it does not have the given property. If $n$ is composite, then $n = ab$ such that $1 < a \\le b < n$. It follows that $b + 1$ divides $n + 1$, i.e. there exists $c \\in \\mathbb{Z}$ such that $c(b + 1) = n + 1 = ab + 1$. We obtain that $b$ divides $c - 1$. Obviously, $c > 1$. We deduce that $c - 1 \\ge b$, i.e. $c \\ge b + 1$. Then $ab + 1 = c(b + 1) \\ge (b + 1)^2$, which means that $ab + 1 \\ge b^2 + 2b + 1$, leading to $a \\ge b + 2$, which contradicts $a \\le b$.\n\nIn conclusion, no composite number does satisfy the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38925,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider $A$, the set of all positive integers with exactly $2019$ positive integer divisors, and, for each $n \\in A$, we denote\n$$\nS_n = \\frac{1}{d_1 + \\sqrt{n}} + \\frac{1}{d_2 + \\sqrt{n}} + \\dots + \\frac{1}{d_{2019} + \\sqrt{n}},\n$$\nwhere $d_1, d_2, \\dots, d_{2019}$ are the positive divisors of $n$.\nDetermine the maximum value of $S_n$ when $n$ runs through the set $A$.",
    "options": [],
    "answer": "673/2^337",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be a rational number consider integers $a_1, a_2, \\dots, a_6, b_1, b_2, \\dots, b_6$ such that $1 \\le b_1 < b_2 < \\dots < b_6 \\le 11$ and\n$$\nr = \\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\frac{a_3}{b_3} = \\frac{a_4}{b_4} = \\frac{a_5}{b_5} = \\frac{a_6}{b_6}.\n$$\nProve that $r$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine three digit numbers $abc$, such that its square has $a$ as the hundreds digit, $b$ as the tenth digit and $c$ as the units digit.",
    "options": [],
    "answer": "376, 625",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $k \\ge 2$, determine all positive integers $n_1, n_2, \\dots, n_k$ satisfying\n$$\nn_2 \\mid 2^{n_1-1}, \\quad n_3 \\mid 2^{n_2-1}, \\quad \\dots, \\quad n_k \\mid 2^{n_{k-1}-1}, \\quad n_1 \\mid 2^{n_k-1}.\n$$",
    "options": [],
    "answer": "n1 = n2 = ... = nk = 1",
    "solution": "The required numbers are $n_1 = n_2 = \\dots = n_k = 1$.\n\nFor every integer $m > 1$, let $p(m)$ denote the least prime divisor of $m$. We show that, if $m$ and $\\ell$ are integers greater than $1$, and $m \\mid 2^{\\ell} - 1$, then $p(m) < p(\\ell)$. Since $p(m)$ is odd, $p(m) \\mid 2^{p(m)-1} - 1$, and since $p(m) \\mid 2^{\\ell} - 1$, it follows that $p(m) \\mid 2^{\\gcd(\\ell, p(m)-1)} - 1$. Notice that $\\gcd(\\ell, p(m)-1) > 1$, since $p(m) > 1$, to infer that $\\ell$ has a prime divisor not exceeding $p(m) - 1$, and conclude thereby that $p(m) < p(\\ell)$.\n\nSuppose now, if possible, that $n_1 > 1$. Then $n_k > 1$, so $n_{k-1} > 1$, and so on and so forth all the way down to $n_2 > 1$. Hence $p(n_1) < p(n_2) < \\dots < p(n_k) < p(n_1)$ which is a contradiction. Consequently, $n_1 = 1$, so $n_2 = 1$, and so on and so forth all the way up to $n_k = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer and let $n = m^2 + 1$. Determine all real numbers $x_1, x_2, \\dots, x_n$ satisfying\n$$\nx_i = 1 + \\frac{2m x_i^2}{x_1^2 + x_2^2 + \\dots + x_n^2}, \\quad i = 1, 2, \\dots, n.\n$$",
    "options": [],
    "answer": "Either all variables are equal to ((m+1)^2)/(m^2+1), or exactly one variable equals m+1 and each of the remaining variables equals 1 + 1/m.",
    "solution": "The $x_i$ are either all equal to $1 + \\frac{2m}{n} = \\frac{(m+1)^2}{m^2+1}$ or exactly one is equal to $m+1$ and the other are all equal to $1 + \\frac{1}{m}$. The verification offers no difficulty and is hence omitted.\n\nLeaving aside the trivial case where the $x_i$ are all equal, consider a solution $x_1, x_2, \\dots, x_n$ whose entries are not all equal. Let $s = x_1^2 + x_2^2 + \\dots + x_n^2$ and notice that each $x_i$ is a root of the quadratic polynomial $2mX^2 - sX + s$. Since the $x_i$ are not all equal, and each $x_i$ is positive (in fact, at least $1$), the roots $u$ and $v$ of this quadratic polynomial are distinct positive real numbers satisfying $2muv = s$.\nLet $k$ be the number of indices $i$ such that $x_i = u$, so $x_i = v$ for the remaining $n-k$ indices. We may and will assume that $k \\ge \\frac{n}{2}$; and since the $x_i$ are not all equal, $k \\le n-1 = m^2$.\nWrite $2muv = s = ku^2 + (n-k)v^2 \\ge 2uv\\sqrt{k(n-k)}$, so $k(n-k) \\le m^2$, and recall that $k \\ge \\frac{n}{2}$, to infer that $k \\ge \\frac{1}{2}(n + \\sqrt{n^2 - 4m^2}) = m^2$. Further, the condition $k \\le m^2$ forces $k = m^2$ which in turn forces $v = mu$, by the preceding.\nFinally, since the $x_i$ add up to $n + 2m = (m+1)^2$, it follows that $u = 1 + \\frac{1}{m}$ and $v = m + 1$, and the solution has the form stated in the first paragraph.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Around point $O$ one considers the angles $\\widehat{A_0OA_1} = 1^\\circ$, $\\widehat{A_1OA_2} = 2^\\circ$, $\\widehat{A_2OA_3} = 3^\\circ, \\dots, \\widehat{A_{25}OA_{26}} = 26^\\circ$ and $\\widehat{A_{26}OA_0}$.\n\na) Determine the measure of $\\widehat{A_{26}OA_0}$.\n\nb) For what integers $n$, such that $1 \\le n \\le 25$, one has $\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}$?",
    "options": [],
    "answer": "a) 9 degrees; b) n = 18, 19, 20, 21, 22, 23, 24, 25",
    "solution": "a.\n\nThe sum of all angles around point $O$ is $360^\\circ$.\n\nWe have:\n$$\n\\widehat{A_0OA_1} + \\widehat{A_1OA_2} + \\widehat{A_2OA_3} + \\cdots + \\widehat{A_{25}OA_{26}} + \\widehat{A_{26}OA_0} = 360^\\circ\n$$\n\nThe sum $\\widehat{A_0OA_1} + \\widehat{A_1OA_2} + \\cdots + \\widehat{A_{25}OA_{26}}$ is $1^\\circ + 2^\\circ + 3^\\circ + \\cdots + 26^\\circ$.\n\nThis is an arithmetic progression with first term $1^\\circ$, last term $26^\\circ$, and $26$ terms:\n$$\nS = \\frac{(1^\\circ + 26^\\circ) \\times 26}{2} = \\frac{27^\\circ \\times 26}{2} = 13.5^\\circ \\times 26 = 351^\\circ\n$$\n\nTherefore,\n$$\n\\widehat{A_{26}OA_0} = 360^\\circ - 351^\\circ = 9^\\circ\n$$\n\nb.\n\n$\\widehat{A_0OA_n}$ is the sum of the first $n$ angles:\n$$\n\\widehat{A_0OA_n} = 1^\\circ + 2^\\circ + \\cdots + n^\\circ = \\frac{n(n+1)}{2}^\\circ\n$$\n\n$\\widehat{A_0OA_{n+1}} = 1^\\circ + 2^\\circ + \\cdots + (n+1)^\\circ = \\frac{(n+1)(n+2)}{2}^\\circ$\n\nWe want:\n$$\n\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}\n$$\nBut $\\widehat{A_0OA_{n+1}}$ is always greater than $\\widehat{A_0OA_n}$, unless we consider the angles modulo $360^\\circ$ (i.e., the measure from $A_0$ to $A_n$ going one way, and from $A_0$ to $A_{n+1}$ going the other way).\n\nBut since the sum of all angles is $360^\\circ$, after passing $180^\\circ$, the measure from $A_0$ to $A_n$ is $360^\\circ$ minus the sum $1^\\circ + 2^\\circ + \\cdots + n^\\circ$.\n\nSo, for $n$ such that $\\frac{n(n+1)}{2} > 180$, the angle $\\widehat{A_0OA_n}$ is measured as $360^\\circ - \\frac{n(n+1)}{2}$.\n\nWe need to find $n$ such that $\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}$.\n\nLet $S_n = \\frac{n(n+1)}{2}$ and $S_{n+1} = \\frac{(n+1)(n+2)}{2}$.\n\nIf $S_n \\leq 180$, then $\\widehat{A_0OA_n} = S_n$ and $\\widehat{A_0OA_{n+1}} = S_{n+1}$, so $\\widehat{A_0OA_n} < \\widehat{A_0OA_{n+1}}$.\n\nIf $S_n > 180$, then $\\widehat{A_0OA_n} = 360 - S_n$ and $\\widehat{A_0OA_{n+1}} = 360 - S_{n+1}$.\n\nSo $\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}$ if $360 - S_n > 360 - S_{n+1}$, i.e., $S_n < S_{n+1}$, which is always true.\n\nBut as $S_n$ increases, $360 - S_n$ decreases, so $\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}$ for those $n$ where $S_n > 180$.\n\nFind $n$ such that $S_n > 180$:\n$$\n\\frac{n(n+1)}{2} > 180\n$$\n$$\n n(n+1) > 360\n$$\nTry $n = 19$: $19 \\times 20 = 380 > 360$\nTry $n = 18$: $18 \\times 19 = 342 < 360$\n\nSo for $n = 19, 20, 21, 22, 23, 24, 25$.\n\n**Answer:**\n\nFor $n = 19, 20, 21, 22, 23, 24, 25$, one has $\\widehat{A_0OA_n} > \\widehat{A_0OA_{n+1}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive real number $a$, determine the minimum of the expression\n$$\n\\left( \\int_{0}^{1} f(x) \\, dx \\right)^{2} - (a + 1) \\int_{0}^{1} x^{2a} f(x) \\, dx\n$$\nmay achieve, as $f$ runs through the class of all concave functions $f : [0, 1] \\to \\mathbb{R}$ such that $f(0) = 1$.",
    "options": [],
    "answer": "(2a - 1)/(8a + 4)",
    "solution": "The required minimum is $(2a - 1)/(8a + 4)$ and is achieved for $f: [0, 1] \\to \\mathbb{R}$, $f(x) = 1 - x$. The verification is routine and hence omitted.\n\nFix a concave function $f: [0, 1] \\to \\mathbb{R}$ such that $f(0) = 1$ to write\n$$x^a f(x) + 1 - x^a = x^a f(x) + (1 - x^a)f(0) \\le f(x^a \\cdot x + (1 - x^a) \\cdot 0) = f(x^{a+1}),$$\nfor all $x$ in $[0, 1]$. Multiply both sides by $(a+1)x^a$ and integrate over $[0, 1]$ to get\n$$\n(a+1) \\int_{0}^{1} x^{2a} f(x) \\, dx + \\frac{a}{2a+1} \\le \\int_{0}^{1} (a+1) x^{a} f(x^{a+1}) \\, dx = \\int_{0}^{1} f(x) \\, dx \\\\ \\le \\left( \\int_{0}^{1} f(x) \\, dx \\right)^{2} + \\frac{1}{4},\n$$\nand conclude that\n$$\n\\left(\\int_{0}^{1} f(x) \\, dx\\right)^{2} - (a+1) \\int_{0}^{1} x^{2a} f(x) \\, dx \\ge \\frac{a}{2a+1} - \\frac{1}{4} = \\frac{2a-1}{8a+4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to (0, \\infty)$, such that\n$$\n2^{-x-y} \\le \\frac{f(x)f(y)}{(x^2+1)(y^2+1)} \\le \\frac{f(x+y)}{(x+y)^2+1},\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = (x^2 + 1) 2^{-x}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n\\frac{1}{abc} + 1 \\geq 3 \\left( \\frac{1}{a^2 + b^2 + c^2} + \\frac{1}{a + b + c} \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the whole numbers $a$, $b$, $c$ for which\n$$\n\\frac{a+1}{3} = \\frac{b+2}{4} = \\frac{5}{c+3}.\n$$",
    "options": [],
    "answer": "a=2, b=2, c=2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $G$ be a finite group of order $n$. A function $f: G \\to G$ is a *pseudoendomorphism* if $f(xyz) = f(x)f(y)f(z)$, for all $x, y, z$ in $G$.\n\na) If $n$ is odd, show that every pseudoendomorphism of $G$ is an endomorphism.\n\nb) If $n$ is even, is every pseudoendomorphism of $G$ an endomorphism?",
    "options": [],
    "answer": "a) Yes. b) No; a constant map to an element of order two is a counterexample.",
    "solution": "a) Let $e$ denote the unit of $G$. Let $x = y = z = e$ to write $f(e)^3 = f(e)$, so $f(e)^2 = e$. Since $n$ is odd, it follows that $f(e) = e$.\n\nIf $x$ and $y$ are members of $G$, write $f(xy) = f(xye) = f(x)f(y)f(e) = f(x)f(y)$, to conclude that $f$ is indeed an endomorphism of $G$.\n\n\nb) The answer is negative. Let $a$ be an order $2$ element of $G$, and let $f: G \\to G$, $f(x) = a$. If $x, y, z$ are elements of $G$, then $f(xyz) = a = a^3 = f(x)f(y)f(z)$, so $f$ is a pseudoendomorphism. However, $f$ is not an endomorphism, since $f(e) = a \\neq e$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the triples $(m, n, p)$, of integers not greater than $5$, such that $10$ is a divisor of $A = 2^m + 3^n + 5^p$.",
    "options": [],
    "answer": "All triples with m ≥ 1 and 0 ≤ n,p ≤ 5 satisfying:\n- For p ∈ {1,2,3,4,5}:\n  (m,n,p) ∈ {(1,1,p), (1,5,p), (2,0,p), (2,4,p), (3,3,p), (4,2,p), (5,1,p), (5,5,p)}.\n- For p = 0:\n  (m,n,p) ∈ {(1,3,0), (3,0,0), (3,4,0), (4,1,0), (4,5,0), (5,3,0)}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer with $n \\ge 2$, and let $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ be two matrices with complex entries having the property that there exists an idempotent matrix $C \\in \\mathcal{M}_n(\\mathbb{C})$, such that $C^* = AB - BA$. Prove that $(AB - BA)^2 = O_n$. (A matrix $C$ is called *idempotent* if $C^2 = C$. The matrix $C^*$ is the adjoint matrix of $C$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, let $D, E, F$ be the feet of the altitudes from $A, B, C$, respectively, and let $M, N, P$ be the midpoints of the sides $BC, CA, AB$, respectively. The circles $BDP$ and $CDN$ cross again at $X$, the circles $CEM$ and $AEP$ cross again at $Y$, and the circles $AFN$ and $BFM$ cross again at $Z$. Prove that the lines $AX, BY, CZ$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, it is sufficient to prove that $AX$ is the $A$-symmedian of the triangle $ABC$. To this end, we show that the triangles $XAB$ and $XCA$ are similar. It then follows that the line $AX$ bisects the angle $BXC$, and $XB/XC = AB^2/AC^2$, so $AX$ is indeed the $A$-symmedian of the triangle $ABC$.\nTo prove similarity, we first show that the quadrangle $ANXP$ is cyclic. By the preceding, $\\angle DXP = 180^\\circ - \\angle PBD$, and $\\angle NXD = 180^\\circ - \\angle DCN$, so\n$$\n\\begin{align*}\n\\angle PXN &= 360^\\circ - \\angle DXP - \\angle NXD = 360^\\circ - (180^\\circ - \\angle PBD) - \\\\\n& \\quad (180^\\circ - \\angle DCN) \\\\\n&= \\angle PBD + \\angle DCN = \\angle ABC + \\angle BCA = 180^\\circ - \\angle CAB \\\\\n&= 180^\\circ - \\angle NAP,\n\\end{align*}\n$$\nshowing that the quadrangle $ANXP$ is indeed cyclic.\n\n\nAlternative Solution. As in the previous solution, we show that $AX$ is the A-symmedian of the triangle $ABC$.\nTo this end, invert from $A$ with power $AB \\cdot AC$. The images of $B, C, D, N, P$ and $X$ under this inversion are located as follows:\n(1) The image of $B$ is the point $B'$ on the ray $AB$ emanating from $A$ such that $AB' = AC$; similarly, the image of $C$ is the point $C'$ on the ray $AC$ emanating from $A$ such that $AC' = AB$, so the triangles $ABC$ and $AC'B'$ are reflexions of one another in their common internal A-bisectrix;\n(2) The image of $D$ is the antipode $D'$ of $A$ in the circle $AB'C'$, since\n$$\nAD' = \\frac{AB \\cdot AC}{AD} = \\frac{AB \\cdot AC \\cdot BC}{2 \\cdot \\text{area } ABC} = \\frac{AC' \\cdot AB' \\cdot B'C'}{2 \\cdot \\text{area } AC'B'}\n$$\nwhich is the diameter of the circle $AC'B'$;\n(3) The images $N'$ and $P'$ of $N$ and $P$, respectively, are the reflexions of $A$ across $C'$ and $B'$, respectively; and\n(4) Since the angles $AB'D'$ and $AC'D'$ are both right, by (2), and $B'$ and $C'$ are the midpoints of the segments $AP'$ and $AN'$, respectively, by (3), $D'$ is the centre of the circle $AN'P'$, so the image of $X$ is the midpoint $X'$ of the segment $N'P'$.\nFinally, since $B'C'$ and $N'P'$ are parallel, the line $AXX'$ also bisects the segment $B'C'$, and since the triangles $ABC$ and $AC'B'$ are reflexions of one another in their common internal A-bisectrix, it is indeed the A-symmedian of the triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38939,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that, if $x, y \\ge 1$, then\n$$\nx + y - \\frac{1}{x} - \\frac{1}{y} \\ge 2\\sqrt{xy} - \\frac{2}{\\sqrt{xy}}.\n$$\n\nb) Prove that, if $a, b, c, d \\ge 1$ and $abcd = 16$, then\n$$\na + b + c + d - \\frac{1}{a} - \\frac{1}{b} - \\frac{1}{c} - \\frac{1}{d} \\ge 6.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point in the interior of the triangle $ABC$, such that $BP = AC$. Let $M$ and $R$ be the midpoints of line segments $AP$ and $BC$, respectively, and let $BP \\cap AC = \\{E\\}$. Prove that the angle bisector of $\\angle BEA$ is perpendicular to $MR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle, and let $I$ be its incenter. Let $\\gamma_A$ be the circle through $I$, tangent to $AB$ and $AC$ and crossing the segment $AI$, let $\\gamma_B$ be the circle through $I$ tangent to $BC$ and $BA$ and crossing the segment $BI$, and let $\\gamma_C$ be the circle through $I$, tangent to $CA$ and $CB$ and crossing the segment $CI$. The circles $\\gamma_B$ and $\\gamma_C$ cross again at $A'$, the circles $\\gamma_C$ and $\\gamma_A$ cross again at $B'$, and the circles $\\gamma_A$ and $\\gamma_B$ cross again at $C'$. Prove that the circles $AA'I, BB'I$ and $CC'I$ cross again at a point different from $I$.\nIMO 1997, Longlist",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the internal bisectrices of the angles $BAC$, $CBA$ and $ACB$ cross the circle $ABC$ again at $M_A$, $M_B$ and $M_C$, respectively, and let $\\gamma_A$, $\\gamma_B$ and $\\gamma_C$ be centred at $O_A$, $O_B$ and $O_C$, respectively; clearly, $O_A$, $O_B$ and $O_C$ lie on the segments $AI$, $BI$ and $CI$, respectively.\n\nThe centre of the circle $AA'I$ is the point where the perpendicular bisectrix of the segment $AI$ crosses the perpendicular bisectrix of the segment $A'I$. It is a fact that the former is the line $M_B M_C$; the latter is, of course, the line $O_B O_C$, so the lines $M_B M_C$ and $O_B O_C$ cross at the centre of the circle $AA'I$. Similarly, the lines $M_C M_A$ and $O_C O_A$ cross at the centre of the circle $BB'I$, and the lines $M_A M_B$ and $O_A O_B$ cross at the centre of the circle $CC'I$.\n\nFinally, since the triangles $M_A M_B M_C$ and $O_A O_B O_C$ are in perspective from $I$, the conclusion follows by Desargues' theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f$ from the set of non-negative integers to itself such that\n$$\nf(a + b) = f(a) + f(b) + f(c) + f(d),\n$$\nwhenever $a$, $b$, $c$, $d$ are non-negative integers satisfying $2ab = c^2 + d^2$.",
    "options": [],
    "answer": "All functions f(n) = k n^2 with k a nonnegative integer.",
    "solution": "The required functions are $f(n) = k n^2$, where $k$ is a non-negative integer — these clearly satisfy the condition in the statement.\n\nConversely, let $f$ be a function satisfying the condition in the statement. Setting $(a, b, c, d) = (n, n, n, n)$ in the functional relation yields $f(2n) = 4f(n)$ for all $n$. In particular, $f(0) = 0$ and $f(2) = 4k$, where $k = f(1)$.\n\nSetting successively $(a, b, c, d) = (n^2, 1, n, n)$, $(a, b, c, d) = (n^2, 2, 2n, 0)$ and $(a, b, c, d) = (n^2 + 1, 1, n + 1, n - 1)$ in the functional relation yields\n$$\n\\begin{align*}\nf(n^2 + 1) &= f(n^2) + k + 2f(n), \\\\\nf(n^2 + 2) &= f(n^2) + 4k + f(2n) = f(n^2) + 4k + 4f(n), \\\\\nf(n^2 + 2) &= f(n^2 + 1) + k + f(n + 1) + f(n - 1).\n\\end{align*}\n$$\nSubtraction of the second relation above from the sum of the other two yields\n$f(n + 1) = 2f(n) - f(n - 1) + 2k$.\nAs in the first solution, consider a function $f$ satisfying the condition in the statement, and establish that $f(2n) = 4f(n)$, for all $n$; in particular, $f(0) = 0$.\n\nWe now show by induction on $m$ that $f(mn) = m^2 f(n)$ for all $n$. By the preceding, this is clearly the case if $m = 0, 1, 2$. For the induction step, let $m > 2$.\n\nSince $\\sqrt{m} < m$, if $m$ is a square, the conclusion follows by the induction hypothesis: $f(mn) = f(\\sqrt{m}(\\sqrt{m}n)) = m f(\\sqrt{m} n) = m^2 f(n)$.\n\nOtherwise, use Lagrange's four-square theorem to write $m = A + B$, where $A$ and $B$ are positive integers, each of which is a sum of two squares. Recall that if each factor of a product is a sum of two squares, then so is the product (use standard identities such as $(w^2 + x^2)(y^2 + z^2) = (wy + xz)^2 + (wz - xy)^2$ repeatedly), to write $2AB = (1^2 + 1^2)AB = C^2 + D^2$ for some non-negative integers $C$ and $D$. Clearly, $A < m$ and $B < m$; $C \\le \\sqrt{2AB} \\le (A+B)/\\sqrt{2} = m/\\sqrt{2} < m$, and similarly, $D < m$. Setting $a = An, b = Bn, c = Cn$ and $d = Dn$ in the functional relation and applying the induction hypothesis completes the induction step:\n$$\n\\begin{aligned}\nf(mn) &= f((A+B)n) = f(An) + f(Bn) + f(Cn) + f(Dn) \\\\\n&= A^2 f(n) + B^2 f(n) + C^2 f(n) + D^2 f(n) = (A^2 + B^2 + C^2 + D^2)f(n) \\\\\n&= (A^2 + B^2 + 2AB)f(n) = (A+B)^2 f(n) = m^2 f(n),\n\\end{aligned}\n$$\nfor all non-negative integers $n$.\n\nConsequently, $f(mn) = m^2 f(n)$ for all $m$ and all $n$. Setting $n = 1$ and $k = f(1)$ concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38943,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $S$ is a $10^4$-digit binary string consisting of zeroes and ones, and $k \\le 10^4$ is a positive integer, a $k$-block of $S$ is any substring consisting of $k$ consecutive digits; two $k$-blocks, $a_1a_2\\dots a_k$ and $b_1b_2\\dots b_k$, are of the same type if $a_i = b_i$, $i = 1, \\dots, k$. Consider all $10^4$-digit binary strings whose 3-blocks are of at most 7 types. Determine the maximum number of types the 10-blocks of such a string may fall in.\nCătălin Gherghe",
    "options": [],
    "answer": "504",
    "solution": "The required maximum is 504 and is achieved for a string (to be described below) containing all 7 possible 3-block types different from 000.\n\nTo prove this, let $f(k)$ be the maximum number of $k$-digit strings whose 3-blocks are of at most 7 types. Clearly, $f(1) = 2$, $f(2) = 4$ and $f(3) = 7$.\n\nWe will show that $f(k) \\le f(k-1) + f(k-2) + f(k-3)$; equality holds if all 7 possible 3-block types different from 000 occur. It then follows recursively that $f(10) = 504$.\n\nTo describe a $10^4$-digit string with 7 3-block types and 504 10-block types, append a one to each of the 504 10-digit strings above, concatenate the resulting 11-digit strings in some order to form a $504 \\times 11$-digit string, then append a $(10^4 - 504 \\times 11)$-digit tail of ones.\n\nWe now show that $f(k) \\le f(k-1) + f(k-2) + f(k-3)$. Since there are $2^3 = 8$ possible 3-block types, some 3-block $abc$ occurs in none of the $k$-digit strings under consideration.\n\nThere are at most $f(k-1)$ such strings whose last digit is different from $c$, at most $f(k-2)$ whose last two digits are $xc$, $x \\neq b$, and at most $f(k-3)$ whose last three digits are $xbc$, $x \\neq a$. Consequently, $f(k) \\le f(k-1) + f(k-2) + f(k-3)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $I$ and $O$ be its incenter and circumcenter, respectively. The $A$-excircle touches the lines $AB$, $AC$, $BC$ at $K$, $L$, $M$, respectively. Show that, if the midpoint of the segment $KL$ lies on the circle $ABC$, then $I$, $M$, $O$ are collinear.\n\nPavel Kozhevnikov, Russian Olympiad, 2005\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Leaving the trivial case $AB = AC$ aside, we show that $I$, $M$, $O$ all lie on the Euler line of the triangle formed by the three excenters $I_A$, $I_B$, $I_C$. Recall that the Euler line of a triangle is the line through the orthocenter, the center of the nine-point circle and the circumcenter of that triangle. Since $I$ is the orthocenter of the triangle $I_A I_B I_C$, and $O$ is the center of its nine-point circle, the line $IO$ is indeed the Euler line of this triangle.\n\nWe now show that, if the midpoint of the segment $KL$ lies on the circle $ABC$, then $M$ is the circumcenter of the triangle $I_A I_B I_C$. The conclusion then follows by the preceding.\n\nTo prove that $M$ is the circumcenter of the triangle $I_A I_B I_C$, we show that it lies on the perpendicular bisectrix of the segment $I_A I_B$; similarly, it lies on the perpendicular bisectrix of the segment $I_A I_C$, so it is indeed the circumcenter of the triangle $I_A I_B I_C$.\n\nLet $P$ be the midpoint of the segment $KL$. Since $AK = AL$, the point $P$ lies on the bisectrix $AI$ of the angle $BAC$, so it is the midpoint of the circular arc $BPC$, and therefore lies on the perpendicular bisectrix of the segment $BC$; and since $B$ and $C$ both lie on the circle on diameter $II_A$ (the angles $IBI_A$ and $ICI_A$ are both right), it follows that $P$ is the midpoint of the segment $II_A$.\n\nClearly, the line $I_A CI_B$ is the perpendicular bisectrix of the segment $LM$, so it crosses the latter at its midpoint $Q$. Since $PQ$ is a midline in the triangle $KLM$, it is parallel to $KM$, and since $KM$ and $BII_B$ are both perpendicular to $I_A BI_C$, it follows that $PQ$ and $BII_B$ are parallel. Recall that $P$ is the midpoint of the segment $II_A$, to infer that $PQ$ is a midline in the triangle $II_A I_B$, so $Q$ is the midpoint of the segment $I_A I_B$. Consequently, $M$ lies on the perpendicular bisectrix of the segment $I_A I_B$, as desired. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ for which $n^{n+1} + n - 1$ is the sixth power of an integer.",
    "options": [],
    "answer": "1",
    "solution": "Clearly, $n = 1$ satisfies the required condition. We now proceed to rule out all integers $n > 1$.\n\nIf $n$ is odd, $n \\geq 3$, then $n^{n+1} + n - 1$ falls strictly between the squares of two consecutive integers,\n$$\n\\left( n^{\\left( \\frac{n+1}{2} \\right)} \\right)^2 < n^{n+1} + n - 1 < \\left( n^{\\left( \\frac{n+1}{2} \\right)} + 1 \\right)^2,\n$$\nso it is not the square, and hence all the less the sixth power of an integer.\n\nSimilarly, if $n \\equiv 2 \\pmod{3}$, then $n^{n+1} + n - 1$ falls strictly between the cubes of two consecutive integers,\n$$\n\\left(n^{\\frac{(n+1)}{3}}\\right)^3 < n^{n+1} + n - 1 < \\left(n^{\\frac{(n+1)}{3}} + 1\\right)^3,\n$$\nso it is not the cube, and hence all the less the sixth power of an integer.\n\nIf $n \\equiv 0 \\pmod{3}$, then $n^{n+1} + n - 1 \\equiv -1 \\pmod{3}$, so it is not the square, and hence all the less the sixth power of an integer.\n\nFinally, to rule out the only case left, $n \\equiv 4 \\pmod{6}$, notice that\n$$\nn^{n+1} + n - 1 \\equiv (-1)^{n+1} - 1 - 1 \\equiv -3 \\pmod{n+1}.\n$$\nSince $n+1 \\equiv 5 \\pmod{6}$, it has a prime divisor $p \\equiv 2 \\pmod{3}$, $p > 3$, so $-3$ is not a quadratic residue modulo $p$. Consequently, $n^{n+1} + n - 1$ is not a quadratic residue modulo $p$, and hence all the less the square of an integer, let alone the sixth power of one such.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Bob are given an integer $n \\ge 2$ and a token to play the following game. Initially, Alice chooses an order of the numbers $1, 2, \\dots, n$, and writes them down in a row, in that order, on a sheet of paper. Next, Bob chooses one of these numbers and places the token on that number. Continuing, Alice moves the token on one of the neighbouring numbers, then Bob moves the token on one of the neighbouring numbers of the current position, and so on and so forth in turns. For each $k$ in the range $1, 2, \\dots, n$, the token may be placed on number $k$ at most $k$ times; Bob's first move is counted. The player who can no longer move the token loses. Determine the values of $n$ for which Alice has a winning strategy.\n\nMathematical Olympiad Rioplatense, 2010, Level 3",
    "options": [],
    "answer": "n ≡ 0 or 3 (mod 4)",
    "solution": "Alice has a winning strategy if and only if $n$ is congruent to $0$ or $-1$ modulo $4$.\n\nLooking upon placing the token on a number as a one unit drop of that number, a move is possible if and only if the current position of the token is not flanked by two zeroes. Notice that each player places the token on positions of like parity rank.\n\nAn arrangement $a_1, \\dots, a_n$ of an $n$-element collection of non-negative integers is *suitable* if there exist non-negative integers $x_0, x_1, \\dots, x_n$ such that\n$$\nx_0 = x_n = 0, \\quad \\text{and} \\quad a_k = x_{k-1} + x_k, \\quad k = 1, \\dots, n.\n$$\n\nNotice that, in a suitable arrangement, the $a_k$ add up to twice the sum of the $x_k$, to infer that, for a collection of non-negative integers to have a suitable arrangement, it is necessary that the numbers in the collection add up to an even number. Otherwise, the collection has no suitable arrangement. In particular, if $n$ is congruent to $1$ or $2$ modulo $4$, there is no suitable arrangement of the collection $1, 2, \\dots, n$. If $n = 4m$, then\n$$\n\\begin{array}{l}\n1 = 0 + 1, \\quad 3 = 1 + 2, \\quad 5 = 2 + 3, \\quad \\dots, \\quad 4m - 1 = (2m - 1) + 2m, \\\\\n4m = 2m + 2m, \\\\\n4m - 2 = 2m + (2m - 2), \\quad \\dots, \\quad 6 = 4 + 2, \\quad 4 = 2 + 2, \\quad 2 = 2 + 0\n\\end{array}\n$$\nis a suitable arrangement of the collection $1, 2, \\dots, n$. Clearly, removal of $4m$ provides a suitable arrangement in case $n = 4m - 1$.\n\nWe first prove that, if the initial arrangement is suitable, then Alice wins. The idea is that Alice can counter any legal move of Bob, in a suitable arrangement, so as to make the new arrangement again suitable. Since the initial arrangement is suitable, and the game eventually comes to an end, Bob ultimately gets stuck and loses.\n\nLet $a_1, \\dots, a_n$ be a suitable arrangement, and let Bob place the token on the $k$-th position. Since Bob's move is legal, $x_{k-1} + x_k = a_k > 0$. Without loss of generality, we may and will assume that $x_k > 0$, so $a_{k+1} = x_k + x_{k+1} \\ge x_k > 0$, showing that the $(k+1)$-st position is available for Alice to place the token on. After having done so, the new arrangement is again suitable, since $a_k$ changed to $a'_k = x_{k-1} + (x_k - 1)$, $a_{k+1}$ changed to $a'_{k+1} = (x_k - 1) + x_{k+1}$, and the other numbers are left unchanged. This establishes one implication.\n\nNext, we show that, if the initial arrangement is not suitable, then Bob has a winning strategy. The idea is that this time Bob can achieve a suitable subarrangement just before one of Alice's moves, reset the whole game with this suitable subarrangement as initial arrangement, and then set on to win by using the strategy described in the previous paragraph.\n\nLet $a_1, \\dots, a_n$ be the initial arrangement; by assumption, it is not suitable. Let further $x_0 = 0$, and let $x_k = a_k - x_{k-1}$, $k = 1, \\dots, n$. Setting $a_{n+1} = 0$, it follows that $x_k > a_{k+1}$ for some positive integer $k \\le n$: Indeed, if $x_k \\le a_{k+1}$ for all indices $k \\le n$, then $x_1, x_2, \\dots, x_n$ are all non-negative; since the arrangement $a_1, \\dots, a_n$ is not suitable, $x_n > 0 = a_{n+1}$. Let $m$ be the least index such that $x_m > a_{m+1}$, so $x_k \\ge 0$ for all indices $k < m$.\n\nBob's first move consists in placing the token on the $m$-th position. After Bob's first move, $a_m$ changes to $a_m - 1 = (x_{m-1} + x_m) - 1 = x_{m-1} + (x_m - 1) \\ge x_{m-1} + a_{m+1}$. He may therefore place the token on the $m$-th position at least another $x_{m-1} + a_{m+1}$ times, regardless of Alice's placing the token on one of the neighbouring positions — at most $a_{m+1}$ times on the $(m+1)$-st position, and at most $a_{m-1}$ times on the $(m-1)$-st position (set $a_0 = 0$ if need be). Since $x_{m-1} + a_{m+1} \\ge a_{m+1}$ (recall that $x_k \\ge 0$ for all indices $k < m$), Bob is able to exhaust Alice's $a_{m+1}$ allowed choices of the $(m+1)$-st position by keeping on placing the token on the $m$-th position. Thus, if still legal, her $(x_{m-1} + 1)$-st move — call it move $(*)$ — consists in placing the token on the $(m-1)$-st position.\n\nJust before move $(*)$, the first $m-1$ positions in the arrangement are $a_1, \\dots, a_{m-2}, x_{m-2}$, since $a_{m-1} = x_{m-2} + x_{m-1}$ dropped by $x_{m-1}$, and no move has been made on one of the first $m-2$ positions. At this stage, Bob can reset the whole game within the first $m-1$ positions: The $(m-1)$-element arrangement $a_1, \\dots, a_{m-2}, x_{m-2}$ is suitable, since $x_{m-2} \\ge 0$, and he may reset $x_{m-1} = 0$ to fulfil the conditions in the definition. If still legal, move $(*)$ can now be looked upon as the first move in this suitable arrangement. Finally, by keeping on placing the token on one of these positions, Bob forces Alice to do so: Indeed, recall that just before resetting the game, his last move was on position $m$; if he now places the token on position $k \\le m-1$, then $k \\le m-2$, since $k$ and $m$ must share parity. This completes the argument and concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest value the expression $\\sum_{1 \\le i < j \\le 4} (x_i + x_j) \\sqrt{x_i x_j}$ may achieve, as $x_1, x_2, x_3, x_4$ run through the non-negative real numbers that add up to $1$. Determine also the $x_i$ at which the maximum is achieved.",
    "options": [],
    "answer": "Maximum value 3/4, achieved uniquely when x1 = x2 = x3 = x4 = 1/4.",
    "solution": "The required maximum is $3/4$ and is achieved if and only if the $x_i$ are all equal to $1/4$. To prove this, use the binomial expansion of $(\\sqrt{x_i} - \\sqrt{x_j})^4$ to write\n$$\n4(x_i + x_j)\\sqrt{x_i x_j} = x_i^2 + 6x_i x_j + x_j^2 - (\\sqrt{x_i} - \\sqrt{x_j})^4,\n$$\n\n\\begin{aligned}\n4 \\sum_{1 \\le i < j \\le 4} (x_i + x_j) \\sqrt{x_i x_j} &= 3(x_1 + x_2 + x_3 + x_4)^2 - \\sum_{1 \\le i < j \\le 4} (\\sqrt{x_i} - \\sqrt{x_j})^4 \\\\\n&\\le 3(x_1 + x_2 + x_3 + x_4)^2 = 3;\n\\end{aligned}\n$$\nclearly, equality holds if and only if the $x_i$ are all equal, and the constraint forces them all equal to $1/4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle such that $AB < AC$. Let $I$ be the incentre of the triangle $ABC$, and let the incircle touch the side $BC$ at $D$. The line $AD$ crosses the circle $ABC$ again at $E$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the circular arc $BAC$. The line $EN$ crosses the circular arc $BIC$ at $P$. Show that the lines $AD$ and $MP$ are parallel.\n\nUkraine National Olympiad, 2016",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The internal bisectrix $AI$ and the perpendicular bisectrix $MN$ of the side $BC$ cross at the midpoint $K$ of the arc $BEC$. It is a fact that the circle $BIC$ is centred at $K$.\n\nLet now the line $MP$ cross the circle $BIC$ again at $Q$, to infer that the arcs $BL$ and $CQ$ of this circle have equal angular spans, so $L$ and $Q$ are reflexions of one another in the perpendicular bisectrix $KMN$ of the chord $BC$.\nProject $Q$ orthogonally to $Q'$ on $BC$ and refer to standard notation in the triangle $ABC$: $a$, $b$, $c$ denote the lengths of the sides $BC$, $CA$, $AB$, respectively, $s = (a + b + c)/2$ denotes its semiperimeter, $r$ its inradius, and $S$ its area. With reference to standard formulae, write $CQ' = BD = s - b$ and\n$$\nQQ' = DL = \\frac{DB \\cdot DC}{DI} = \\frac{(s-b)(s-c)}{r} = \\frac{s}{s-a},\n$$\nto infer that $Q$ is the $A$-excentre of the triangle $ABC$, so it lies on the line $AIK$.\nFinally, write $\\angle PQA = \\angle PQI = \\angle PLI = \\angle ELI = \\angle EAI = \\angle DAQ$, to conclude that the lines $AD$ and $MP$ are indeed parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, colour red exactly $n$ cells of an infinite sheet of grid paper. A rectangular grid array is called *special* if it contains at least two red opposite corner cells; single red cells and 1-row or 1-column grid arrays whose end-cells are both red are special. Given a configuration of exactly $n$ red cells, let $N$ be the largest number of red cells a special rectangular grid array may contain. Determine the least value $N$ may take on over all possible configurations of exactly $n$ red cells.\n\nBased on Mathematical Olympiad Rioplatense, 2010, Level 2",
    "options": [],
    "answer": "1 + ceil((n+1)/5)",
    "solution": "The required minimum is $1 + \\lceil (n+1)/5 \\rceil$ and is achieved by the configuration described in the second block of the proof.\n\nCounting multiplicities, the cells $a$ and $c$ are both covered by three of these special rectangular grid arrays, the cells $b$ and $d$ are both covered by two, and all other red cells are covered by at least one. Letting $r_{[xy]}$ denote the number of red cells in $[xy]$, it follows that $r_{[ab]} + r_{[bc]} + r_{[cd]} + r_{[da]} + r_{[ac]} \\ge 3 \\cdot 2 + 2 \\cdot 2 + (n - 4) = n + 6$. Consequently, $N \\ge (n + 6)/5$.\n\nWe now describe a configuration of exactly $n$ red cells where $N = 1 + \\lceil (n+1)/5 \\rceil$. Write $m = \\lceil (n+1)/5 \\rceil$, so $n = 5m - r$ for some positive integer $r \\le 5$, and $N = m + 1$.\n\nFix an integer $k > 2m$, let $S$ be a $3k \\times 3k$ grid square, and subdivide $S$ into nine $k \\times k$ grid subsquares.\n\nLet $S_{LL}$ be the lower-left corner $k \\times k$ grid subsquare of $S$. Colour red the first $m$ cells along the diagonal upward from the lower-right corner cell of $S_{LL}$.\n\nThe ‘min’ and ‘max’ in the next four paragraphs account for the first few cases where $m < r$. Had we assumed $n \\ge 20$, it would then have followed that $m \\ge r$, and ‘min’ and ‘max’ would have been superfluous.\n\nNext, let $S_{UL}$ be the upper-left corner $k \\times k$ grid subsquare of $S$. Colour red the first $\\min(m, 4m-r)$ cells along the diagonal upward from the lower-left corner cell of $S_{UL}$.\n\nLet further $S_{UR}$ be the upper-right corner $k \\times k$ grid subsquare of $S$. Colour red the first $\\min(m, 3m-r)$ cells along the diagonal downward from the upper-left corner cell of $S_{UR}$.\n\nComplete the corner tour by letting $S_{LR}$ be the lower-right corner $k \\times k$ grid subsquare of $S$. Colour red the first $\\min(m, 2m-r)$ cells along the diagonal downward from the upper-right corner cell of $S_{LR}$.\n\nFinally, let $S_C$ be the central $k \\times k$ grid subsquare of $S$, and colour red $\\max(0, m-r)$ cells of $S_C$; their exact location is irrelevant.\n\nNo other cell whatsoever is coloured red, and it is a routine exercise to check that exactly $n$ cells of the grid paper have been coloured red. Notice that, for each pair of 'adjacent' corner $k \\times k$ grid subsquares, $S_{LL}$ and $S_{UL}$, $S_{UL}$ and $S_{UR}$, $S_{UR}$ and $S_{LR}$, and $S_{LR}$ and $S_{LL}$, there are both horizontal and vertical grid lines separating the strings of red cells they contain.\n\nTo complete the argument, we show that, if $x$ and $y$ are red cells in this configuration, then $r_{[xy]} \\le m + 1$. This is clearly the case if $x$ and $y$ both lie in one of $S_{LL}$, $S_{UL}$, $S_{UR}$, $S_{LR}$ or $S_{C}$, for each of these squares contains at most $m$ red cells.\n\nIf $x$ and $y$ lie in 'adjacent' corner $k \\times k$ subsquares of $S$, then the red cells in $[xy]$ come from those subsquares alone. In addition, the string of red cells in one of those subsquares has exactly one cell in $[xy]$, namely, $x$ or $y$. Consequently, $r_{[xy]} \\le m + 1$. Incidentally, notice that equality holds if, for instance, $x$ is the lower-right corner cell of $S_{LL}$, and $y$ is any red cell in $S_{UL}$; since $n \\ge 2$, there is at least one such.\n\nIf $x$ and $y$ lie in 'opposite' corner $k \\times k$ subsquares of $S$, then they are the only red cells $[xy]$ contains from those subsquares. No red cell in the other two 'opposite' corner $k \\times k$ subsquares of $S$ lies in $[xy]$, and the other red cells in $[xy]$ all come from $S_C$ which contains at most $m-1$ such. Consequently, $r_{[xy]} \\le 2 + (m-1) = m+1$.\n\nFinally, if one of $x$, $y$ lies in $S_C$, and the other lies in one of the corner $k \\times k$ sub-squares of $S$, then the latter cell is the only red cell in $[xy]$ outside $S_C$. Consequently, $r_{[xy]} \\le (m-1)+1 = m < m+1$. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $(n \\ge 2)$ be a positive integer, and let $A, B \\in \\mathcal{M}_n(\\mathbb{R})$ be two matrices with real entries. Prove that there exists a complex number $z$, with $|z| = 1$, such that\n$$\n\\operatorname{Re}(\\det(A + zB)) \\ge \\det(A) + \\det(B),\n$$\nwhere we denote by $\\operatorname{Re}(w)$ the real part of the complex number $w$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a set of integers $M$, having the properties:\ni) $1$ belongs to $M$;\nii) if $x$ and $y$ are in $M$, then $2x + 3y$ is in $M$;\niii) if $x, y$ are integers such that $4x - 3y$ belongs to $M$, then $x \\cdot y$ belongs to $M$.\nProve that $M$ contains the numbers $2$, $3$, $4$, $5$ and $2019$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer.\n\na) Prove that there exist $z_1, z_2, \\dots, z_n \\in \\mathbb{C}$ such that\n$$\n\\frac{z_1}{z_2} + \\frac{z_2}{z_3} + \\dots + \\frac{z_{n-1}}{z_n} + \\frac{z_n}{z_1} = n i.\n$$\n\nb) For which $n$ there exist the complex numbers $z_1, z_2, \\dots, z_n$, having the same absolute value, such that\n$$\n\\frac{z_1}{z_2} + \\frac{z_2}{z_3} + \\dots + \\frac{z_{n-1}}{z_n} + \\frac{z_n}{z_1} = n i?\n$$",
    "options": [],
    "answer": "n divisible by 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38953,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the math class 25 students of the fifth grade receive a paper stick with an integer on it. Each student divides his number to $24$ and communicates the remainder to the math teacher. The sum of all remainders is $288$. Dany, one of the students, remarks that all the remainders of the students are different two by two, but his remainder equals the quotient he obtained.\n\na)\nWhat is the number on Dany's stick?\n\nb) Find the sum of the numbers written on the $25$ sticks, knowing that each student different from Dany, obtained a quotient one unit bigger than the remainder.",
    "options": [],
    "answer": "a) 300; b) 7776",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of the acute triangle $ABC$. Let $X$ be a point in the plane such that the triangle $XAH$ is an isosceles right triangle, with the hypotenuse $AH$, and the points $B$ and $X$ are separated by the line $AH$. Prove that $\\overrightarrow{XA} + \\overrightarrow{XC} + \\overrightarrow{XH} = \\overrightarrow{XB}$ if and only if $\\angle BAC = 45^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any real numbers $a_1, a_2, \\dots, a_n$, $n \\in \\mathbb{N}$, there exists a real number $x$ such that the numbers $x + a_1, x + a_2, \\dots, x + a_n$ are all irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $y_1 < y_2 < \\dots < y_n < y_{n+1}$, irrational numbers such that $y_j - y_i$ is irrational for all $1 \\le i < j \\le n + 1$. (One could take, for example,\n\n$y < 2y < 3y < \\dots < (n+1)y$, where $y$ is irrational.) We plan to prove that one of these irrational numbers can be chosen as $x$.\n\nAssume the contrary to be true, i.e. for each $y_k$, at least one of the numbers $y_k + a_1, y_k + a_2, \\dots, y_k + a_n$ is rational. But there are $n+1$ choices for $y_k$, and only $n$ choices for $a_m$ such that $y_k + a_m$ is rational. By the Pigeon Principle, it follows that there must be an index $m$ and two irrational numbers $y_i, y_j$ such that $a_m + y_i$ and $a_m + y_j$ are both rational. But this would mean that their difference, $y_j - y_i$, is also rational, which contradicts the choice of the numbers $y_k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$, be an integer. Find in terms of $n$ the maximum value of $m \\le n$ such that we can select $m$ numbers from the set $1, 2, 3, \\dots, n$ in such a way that for any two of them, say $a, b$, with $a > b$, the number $a-b$ does not divide $a+b$.\nVasile Pop",
    "options": [],
    "answer": "ceil(n/3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, if the numbers $a$, $b$, $c \\in [0, \\infty)$ satisfy the relation $a + b + c = 3$, then\n$$\n\\frac{a}{3a + bc + 12} + \\frac{b}{3b + ca + 12} + \\frac{c}{3c + ab + 12} \\le \\frac{3}{16}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$, of non-negative integers such that the quotients\n$$\n\\frac{3a + 8b + 2}{10a + 2b + 1} \\quad \\text{and} \\quad \\frac{8a + b + 3}{2a + 7b + 3}\n$$\nare integers simultaneously.",
    "options": [],
    "answer": "(a, b) = (0, 0) and (1, 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum value of the expression\n\n$$\nE(a, b) = \\frac{a + b}{(4a^2 + 3)(4b^2 + 3)}\n$$\n\nwhen $a, b \\in \\mathbb{R}$.",
    "options": [],
    "answer": "1/16",
    "solution": "We will show that the maximum value is $\\frac{1}{16}$, obtained when $a = b = \\frac{1}{2}$.\n\nThe inequality $E(a, b) \\le \\frac{1}{16}$ is equivalent to $16(a+b) \\le (4a^2+3)(4b^2+3)$, which we rewrite as $(4ab-1)^2+4(a+b-1)^2+2(2a-1)^2+2(2b-1)^2 \\ge 0$, obviously true.\nWe will show that the maximum value is $\\frac{1}{16}$, obtained when $a = b = \\frac{1}{2}$.\n\nWe combine the following inequalities: $a+b \\le \\frac{(1+a+b)^2}{4}$ and (from CBS) $(4a^2+3)(4b^2+3) = (4a^2+1+2)(1+4b^2+2) \\ge (2a+2b+2)^2$. We get:\n$$\n\\frac{a+b}{(4a^2+3)(4b^2+3)} \\le \\frac{(a+b+1)^2}{4(4a^2+3)(4b^2+3)} \\le \\frac{(a+b+1)^2}{4(2a+2b+2)^2} \\le \\frac{1}{16}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 38960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f_0$, $f_1$, $f_2$ and $f_3$ be polynomials in $\\mathbb{R}[X]$ such that $f_k(1) = f_{k+1}(0)$, $k = 0, 1, 2, 3$ (indices are reduced modulo $4$). Show that there exists a polynomial $f$ in $\\mathbb{R}[X, Y]$ such that $f(X, 0) = f_0(X)$, $f(1, Y) = f_1(Y)$, $f(1 - X, 1) = f_2(X)$, and $f(0, 1 - Y) = f_3(Y)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The idea is to consider a suitable $\\mathbb{R}[Y]$-linear combination of $f_0(X)$ and $f_2(1 - X)$, namely, $(1 - Y)f_0(X) + Yf_2(1 - X)$, and a suitable $\\mathbb{R}[X]$-linear combination of $f_1(Y)$ and $f_3(1-Y)$, namely, $Xf_1(Y) + (1-X)f_3(1-Y)$, along with a suitable degree $2$ corrective term, $a_{11}XY + a_{10}X + a_{01}Y + a_{00}$, collect them together to form\n$$\nf(X, Y) = (1-Y)f_0(X) + Xf_1(Y) + Yf_2(1-X) + (1-X)f_3(1-Y) + a_{11}XY + a_{10}X + a_{01}Y + a_{00}\n$$\nand require the latter to satisfy the conditions in the statement.\n\nThus, the condition $f(X, 0) = f_0(X)$ requires $a_{10} = f_3(1) - f_1(0)$ and $a_{00} = -f_3(1)$. Refer to the hypothesis on the $f_k$ to write $a_{10} = f_0(0) - f_1(0)$ and $a_{00} = -f_0(0)$.\nNext, the condition $f(0, 1-Y) = f_3(Y)$ requires $a_{01} = f_0(0) - f_2(1)$ and $f_2(1) + a_{01} + a_{00} = 0$. Notice that the latter holds automatically. With reference again to the hypothesis on the $f_k$, $a_{01} = f_0(0) - f_3(0)$.\nSimilarly, the condition $f(1, Y) = f_1(Y)$ requires $a_{11} = f_0(1) - f_2(0) - a_{01}$ and $f_0(1) + a_{10} + a_{00} = 0$. The latter holds again automatically, while the former yields $a_{11} = f_1(0) - f_2(0) - f_0(0) + f_3(0)$.\nFinally, the condition $f(1-X, 1) = f_2(X)$ requires $f_3(0) - f_1(1) - a_{11} - a_{10} = 0$ and $f_1(1) + a_{11} + a_{10} + a_{01} + a_{00} = 0$, both of which hold by the preceding and the hypothesis on the $f_k$.\nIn terms of the data, the desired polynomial is\n$$\nf(X, Y) = (1 - Y)f_0(X) + Xf_1(Y) + Yf_2(1 - X) + (1 - X)f_3(1 - Y) - (f_0(0) - f_1(0) + f_2(0) - f_3(0))XY + (f_0(0) - f_1(0))X + (f_0(0) - f_3(0))Y - f_0(0).\n$$\nThe polynomial in the previous solution may equally well be obtained as follows: begin by seeking a two-variable polynomial $f$ of the form\n$$\nf(X, Y) = g(X, Y) + Xh(Y) + Yk(X) + aXY,\n$$\nwhere $g$ is a two-variable polynomial satisfying $g(X, 0) = f_0(X)$ and $g(0, 1-Y) = f_3(Y)$, $h$ and $k$ are one-variable polynomials subject to $h(0) = k(0) = 0$ and $h(1) = k(1)$, and $a$ is a real number to be determined from the requirements. The common value $h(1) = k(1)$ will come out of the choice of $h$ and $k$.\nNotice that such a choice of $g$, $h$ and $k$ yields $f(X, 0) = f_0(X)$ and $f(0, 1-Y) = f_3(Y)$.\nNow, the two-variable polynomial $g(X, Y) = f_0(X) + f_3(1-Y) - f_0(0)$ clearly fits the bill; with reference to the hypothesis on the $f_k$, the verification is routine and hence omitted.\nNext, letting $h(Y) = f_1(Y) - g(1, Y)$ and $k(X) = f_2(1-X) - g(X, 1)$, it is again readily checked that $h(0) = k(0) = 0$, and $h(1) = f_1(1) - g(1, 1) = f_2(0) - g(1, 1) = k(1)$. In terms of the data, $h(Y) = f_1(Y) - f_3(1-Y) + f_0(0) - f_1(0)$, $k(X) = -f_0(X) + f_2(1-X) + f_0(0) - f_3(0)$, and $h(1) = k(1) = f_0(0) - f_1(0) + f_2(0) - f_3(0)$.\nAt this stage, requiring $f(1, Y) = f_1(Y)$ yields $a = -f_0(0) + f_1(0) - f_2(0) + f_3(0)$; incidentally, yet not accidentally at all, this is precisely the value of $a_{11}$ in the previous solution.\nFinally, notice that $h(1) = -a$, to check the remaining requirement: $f(1-X, 1) = g(1-X, 1) + h(1)(1-X) + k(1-X) + a(1-X) = g(1-X, 1) + k(1-X) = f_2(X)$.\n\nExpressing $f$ in terms of the data yields the polynomial in the previous solution, just as mentioned in the beginning.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38961,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n$ is an integer, $n \\ge 3$, and $a_1, a_2, \\dots, a_n$ are non-zero integers such that\n$$\na_1 a_2 \\cdots a_n \\left( \\frac{1}{a_1^2} + \\frac{1}{a_2^2} + \\cdots + \\frac{1}{a_n^2} \\right)\n$$\nis an integer, does it follow that the product $a_1 a_2 \\cdots a_n$ is divisible by each $a_i^2$?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is in the affirmative. To prove this, begin by noticing that the rational numbers $b_i = a_1 a_2 \\cdots a_n / a_i^2$, $i = 1, 2, \\dots, n$, are the roots of the degree $n$ monic polynomial $f = X^n - s_1 X^{n-1} + s_2 X^{n-2} - \\dots + (-1)^{n-1} s_{n-1} X + (-1)^n s_n$, where\n$$\ns_k = \\sum_{|I|=k} \\prod_{i \\in I} b_i = \\sum_{|I|=k} \\left( \\prod_{i \\in I} a_i \\right)^{k-2} \\left( \\prod_{i \\notin I} a_i \\right)^k, \\quad k = 1, 2, \\dots, n.\n$$\nClearly, $s_2, \\dots, s_n$ are all integral. Since $s_1$ is integral, by hypothesis, and $f$ is monic, it follows that the $b_i$ are all integral. (A rational root of a monic polynomial with integral coefficients is necessarily integral.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to \\mathbb{R}$ be a continuous function which is constant on the set of nonnegative integers. Suppose that for any real numbers $0 \\le a < b < c < d$, such that $f(a) = f(c)$ and $f(b) = f(d)$, we have $f(\\frac{a+b}{2}) = f(\\frac{c+d}{2})$. Prove that $f$ is a constant function.\nFlorin Stănescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $C$ be two points on a circle $C$ such that $(AC)$ is not a diameter, and let $P$ be a point of the line segment $(AC)$, other than its midpoint. Circles $c_1$ and $c_2$ are interiorly tangent to the circle $C$ at $A$ and $C$, respectively. They both pass through $P$ and intersect again at $Q$. The line $PQ$ intersects circle $C$ at $B$ and $D$. Circle $c_1$ intersects line segments $AB$ and $AD$ at $K$ and $N$, respectively, while circle $c_2$ intersects line segments $CB$ and $CD$ at $L$ and $M$, respectively. Prove that:\n\na) the quadrilateral $KLMN$ is an isosceles trapezoid;\n\nb) $Q$ is the midpoint of the line segment $BD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Point $B$ lies on the radical axis, $BD$, of circles $c_1$ and $c_2$, therefore $BK \\cdot BA = BL \\cdot BC$, which indicates that the quadrilateral $AKLC$ is cyclic. So is $ACMN$. It follows that $\\angle LKB = \\angle ACB$. The angle between the tangent at $A$ to the circle $c_1$ (which is also tangent to $C$) with line $AB$ subtends arcs $AK$ of circle $c_1$ and $AB$ of circle $C$, and therefore $\\angle ANK = \\angle ADB$. We obtain that $NK \\parallel BD$ and, similarly, $ML \\parallel BD$. (This fact also follows from the homotheties that transform circles $c_1$, and $c_2$, respectively, into $C$.) Finally, $\\angle NKL = 180^\\circ - \\angle AKN - \\angle LKB = 180^\\circ - \\angle ABD - \\angle LKB = 180^\\circ - \\angle ACD - \\angle ACB = \\angle BAD$. Similarly, $\\angle MNK = \\angle BAD$, which leads to the conclusion. Alternatively, one could have noticed that line segments $ML$, $PQ$ and $[NK]$ share the same perpendicular bisector.\n\nb) The radical axes of circles $c_1$, $c_2$, $C$ (one for each pair of circles), i.e., the tangent line to $C$ at $A$, the tangent line to $C$ at $C$ and the line $BD$ are not all parallel, therefore they are concurrent in the radical center. Diagonal $BD$ is then a symmedian of triangle $ABC$, which means that quadrilateral $ABCD$ is harmonic. (One can also use this fact to give a different proof to a).)\n\nAs $NPQK$ is an isosceles trapezoid, it follows that $\\angle NAP = \\angle QAK$, which means that rays $(AP$ and $(AQ$ are isogonal with respect to angle $\\angle DAB$). But $ABCD$ being harmonic, $AP$ is a symmedian of triangle $DAB$, therefore $AQ$, which is its isogonal, is the median.\n*Alternative Solution.* One could use a property of harmonic quadrilaterals that was put into evidence by the Danube Mathematical Competition from the same year: If $R$ is the midpoint of diagonal $BD$ of a harmonic quadrilateral $ABCD$, then $\\angle ARD = \\angle CRD$.\n\nOne proves that $R$ is the only point on the line segment $BD$ that has the property from above. Next, one proves that point $Q$ does have the property, which makes it the midpoint of $BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square and $E$ an arbitrary point on the side $(CD)$. On the outside of the triangle $ABE$ one considers the squares $ENMA$ and $EBQP$. Prove that:\n\na) $ND = PC$;\n\nb) $ND \\perp PC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38965,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider $D$, the midpoint of the baseline $[BC]$ of the isosceles triangle $ABC$, in which $m(\\angle BAC) < 90^\\circ$. On the perpendicular line in $B$ on the line $BC$, one takes the point $E$, such that $\\angle EAB \\equiv \\angle BAC$, and on the parallel through $C$ to the line $AB$, one takes the point $F$, such that $F$ and $D$ are on different sides of the line $AC$ and $\\angle FAC \\equiv \\angle CAD$. Prove that $AE = CF$ and $BF = EF$.\nSorin Furtună",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, $n \\ge 3$, for which there exists a set $M$, whose elements are $n$ distinct, nonzero vectors, with equal lengths, such that $\\sum_{\\vec{u} \\in M} \\vec{u} = \\vec{0}$, but for any $\\vec{v}, \\vec{w} \\in M$, we have $\\vec{v} + \\vec{w} \\ne \\vec{0}$.",
    "options": [],
    "answer": "All integers n ≥ 3 with n ≠ 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $f: [0, 1] \\to \\mathbb{R}$ be an integrable function. Show that there exists a point $a_n$ in the closed interval $[0, 1 - 1/n]$ such that either\n$$\n\\int_{a_n}^{a_n+1/n} f(x) \\, dx = 0 \\quad \\text{or} \\quad \\int_{0}^{a_n} f(x) \\, dx = \\int_{a_n+1/n}^{1} f(x) \\, dx.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F: [0, 1] \\to \\mathbb{R}$, $F(x) = \\int_{0}^{x} f(t) \\, dt$, and consider the continuous function $g: [0, 1] \\to \\mathbb{R}$, $g(x) = F(x)(F(1) - F(x))$. In terms of $g$, the conclusion reads $g(a_n) = g(a_n + 1/n)$, for some $a_n$ in $[0, 1 - 1/n]$.\n\nNotice that $g(0) = 0 = g(1)$. Suppose now, if possible, that $g(x) = g(x+1/n)$ for no $x$ in $[0, 1-1/n]$. By continuity, either $g(x) < g(x+1/n)$, for all $x$ in $[0, 1-1/n]$, or $g(x) > g(x+1/n)$, for all $x$ in $[0, 1-1/n]$. Consequently, $0 = g(1) - g(0) = \\sum_{k=0}^{n-1} (g(k/n + 1/n) - g(k/n)) \\neq 0$, which is a contradiction. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the conditions:\n(i) $f(f(x^2) + y + f(y)) = x^2 + 2f(y)$;\n(ii) $x \\le y$ implies $f(x) \\le f(y)$;\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "We prove that the only function satisfying the two conditions is $f(x) = x$.\nWe prove that $f$ is an injection. If we put $y = 0$ in (i) we get $f(f(x^2) + f(0)) = x^2 + 2f(0)$, for any $x$, or equivalently,\n$$\nf(f(a) + f(0)) = a + 2f(0) \\qquad (1)\n$$\nfor any number $a \\ge 0$. From (1), $f$ is an injection on the set of all nonnegative numbers. We fix now $y$. From the conditions (i) and (ii) we get that $f(\\cdot)$ is a superior unbounded function. If $f(y_1) = f(y_2)$ then $f(f(x^2) + y_1 + f(y_1)) = f(f(x^2) + y_2 + f(y_2))$. For large enough values of $x$ the expressions $f(x^2) + y_1 + f(y_1)$ and $f(x^2) + y_2 + f(y_2)$ are positive and thus $y_1 = y_2$.\n\nWe prove now that $f(0) = 0$.\n**Case 1.** $f(0) \\le 0$. For $a = -2f(0)$, we have $f(f(-2f(a)) + f(0)) = 0$. Thus, there exists a real number $c$ such that $f(c) = 0$.\nPlugging $x = 0$ and $y = c$ in (i) we get $f(f(0) + c) = 0$. As $f$ is an injection, we get $f(0) + c = c$, that is $f(0) = 0$.\n**Case 2.** $f(0) \\ge 0$. In (i) we put $x = y = 0$: $f(2f(0)) = 2f(0)$. We plug now $a = 3f(0) = f(0) + f(2f(0))$.\nFrom (1) we have $f(a) = f(f(2f(0)) + f(0)) = 2f(0) + 2f(0) = 4f(0)$. We add now $f(0)$ and we get $f(f(a) + f(0)) = f(5f(0)) = 3f(0) + 2f(0) = 5f(0)$.\nBy plugging now $x = 0$ and $y = 2f(0)$ in (i) we get $f(5f(0)) = 4f(0)$. Thus $f(5f(0)) = 5f(0) = 4f(0)$, from were we obtain $f(0) = 0$.\nFrom (1), for $a \\ge 0$ we have $f(f(a)) = a$, and from (i), for $x = 0$, we have $f(y + f(y)) = 2f(y)$ for any real number $y$.\nWe plug now $y = f(a)$: $f(f(a)+a) = 2f(f(a)) = 2a$ and $y = a$: $f(a+f(a)) = 2f(a)$. We obtained that for any $a \\ge 0$ we have $f(a) = a$.\nThen, (i) becomes $f(x^2 + y + f(y)) = x^2 + 2f(y)$. We fix now any real $y$. There exists a $x \\in \\mathbb{R}$ such that $x^2 + y + f(y) > 0$. Then $f(x^2 + y + f(y)) = x^2 + y + f(y) = x^2 + 2f(y)$ and thus $f(y) = y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, and let $D, E, F$ be the feet of the altitudes from $A, B, C$, respectively. The lines $BC$ and $EF$ cross at $P$, and the line through $D$ and parallel to $EF$ crosses the lines $AC$ and $AB$ at $Q$ and $R$, respectively. Prove that the circle $PQR$ passes through the midpoint of the side $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of the side $BC$. If $AB = AC$, then $P$ is the ideal point of the line $BC$, the points $Q$ and $R$ fall at $C$ and $B$, respectively, and the circle $PQR$ degenerates into the line $BC$ on which $M$ clearly lies.\n\n![](attached_image_1.png)\n\nAssume henceforth that $AB \\neq AC$, say, $AB > AC$. It is clearly sufficient to show that\n$$\nDM \\cdot DP = DQ \\cdot DR.\n$$\nSince $EF$ and $QR$ are parallel, and $B, C, E, F$ are concyclic ($E$ and $F$ both lie on the circle on diameter $BC$), so are $B, C, Q, R$. Hence $DB \\cdot DC = DQ \\cdot DR$, and it is therefore sufficient to show that $DB \\cdot DC = DM \\cdot DP$, i.e., $BM^2 = DM \\cdot MP$, since $DB = BM + DM$, $DC = CM - DM = BM - DM$ and $DP = MP - DM$. Alternatively, but equivalently, $DP \\cdot MP = MP^2 - BM^2$, since $DM = MP - DP$.\n\nThe points $D, E, F, M$ are concyclic (they all lie on the nine-point circle of the triangle $ABC$), so $PD \\cdot PM = PE \\cdot PF$. The points $B, C, E, F$ are also concyclic (recall that $E$ and $F$ both lie on the circle on diameter $BC$), so $PE \\cdot PF = PB \\cdot PC$. Consequently, $DP \\cdot MP = PB \\cdot PC = (BM + MP)(MP - CM) = (MP + BM)(MP - BM) = MP^2 - BM^2$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2\\ldots A_n$ be a regular polygon. Find the number of subsets $\\{A_i, A_j, A_k, A_l\\}$, whose elements are the vertices of a trapezoid.\nCătălin Gherghe",
    "options": [],
    "answer": "If n is odd: n(n − 1)(n − 3)/8. If n is even: n(n − 2)(n − 3)/8.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDA'B'C'D'$ be a rectangular parallelepiped, and $M$, $N$, $P$ the projections of the points $A$, $C$, respectively $B'$, on the diagonal $BD'$.\n\na)\nShow that $BM + BN + BP = BD'$.\n\nb) Show that $3(AM^2 + B'P^2 + CN^2) \\ge 2D'B^2$ if and only if the rectangular parallelepiped $ABCDA'B'C'D'$ is cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\in \\mathbb{N}^*}$ be a sequence such that\n$$\n2(a_1 + a_2 + \\dots + a_n) = n a_{n+1}, \\quad \\text{for all } n \\ge 1.\n$$\na) Prove that $(a_n)_{n \\in \\mathbb{N}^*}$ is an arithmetic sequence.\n\nb) If $[a_1] + [a_2] + \\dots + [a_n] = [a_1 + a_2 + \\dots + a_n]$, for all $n \\in \\mathbb{N}^*$, prove that the terms of the sequence are all integers (here, $[x]$ denotes the floor of $x$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to express every positive integer $n$ congruent to $9$ modulo $25$ in the form $n = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}$, where $a$, $b$, $c$ are non-negative integers that do not share parity?",
    "options": [],
    "answer": "Yes",
    "solution": "(*) Let $(p, q, r)$ be a Pythagorean triple of positive integers, $p^2 + q^2 = r^2$, such that $p \\equiv -1 \\pmod{4}$, $q \\equiv 0 \\pmod{4}$, $r \\equiv 1 \\pmod{4}$, and $p < q$. Then every positive integer $N \\equiv 3 \\pmod{8}$, that is divisible by $r^2$, is the sum of three odd squares whose positive square roots are not congruent modulo $4$.\n---\nConsequently, a positive integer $n \\equiv \\frac{3(r^2 - 1)}{8} \\pmod{r^2}$ is expressible in the form $n = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}$ for some non-negative integers $a, b, c$ that do not share parity.\nThe problem at hand is the special case where $(p, q, r) = (3, 4, 5)$.\nTo prove (*), notice that $N/r^2 \\equiv 3 \\pmod{8}$, so it is not of the form $4^k(8\\ell + 7)$, and is therefore a sum of three odd squares (Gauss-Legendre).\nWrite $N = (ru)^2 + (rv)^2 + (rw)^2$ for some positive odd integers $u, v, w$, and assume, without loss of generality, that $u \\ge v$, to write $(ru)^2 + (rv)^2 = (pu + qv)^2 + (qu - pv)^2$.\nSince $(ru - rv) + ((pu + qv) - (qu - pv)) \\equiv (u - v) + (u + v) \\equiv 2u \\equiv 2 \\pmod{4}$, the entries of one of the pairs of positive odd integers $(ru, rv), (pu + qv, qu - pv)$ are not congruent modulo $4$. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be distinct complex numbers, such that $|a| = |b| = |c| = 1$. Prove that if $|a+b-c|^2 + |b+c-a|^2 + |c+a-b|^2 = 12$, then the geometric images of $a, b, c$ are the vertices of an equilateral triangle.\nMihaela Berindeanu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38975,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament every participant played with all others. At the end the organizer played a game with some participants. Finally a total of $100$ games were played in the tournament. Find the number or participants and the games the organizer played.",
    "options": [],
    "answer": "14 participants; the organizer played 9 games",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest integer $n$ for which the following condition holds: one can find $n$ distinct lines starting at the same point such that the angle made by any two of them, measured in degrees, is an integer that is not a prime.",
    "options": [],
    "answer": "90",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38977,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board are written initially three consecutive positive integers, $n-1$, $n$, $n+1$. A move consists of choosing two numbers written on the board $a$ and $b$, and replacing them with $2a-b$ and $2b-a$. For what values of $n$ is it possible to obtain, after a succession of such moves, that two of the numbers written on the board are equal to 0?",
    "options": [],
    "answer": "n is a power of 3 (n = 3^k for some nonnegative integer k)",
    "solution": "We prove that we can obtain two 0-s on the board if and only if $n$ is a power of 3.\n\nAs the sum of the numbers written on the board stays the same, we must obtain on the board the numbers $(0, 0, 3n)$. If $p \\neq 3$ is a prime divisor of $n$, then, in the final configuration, all the numbers are divisible by $p$. But if $p \\mid 2a-b$, $p \\mid 2b-a$ then $p \\mid (2(2a-b) + (2b-a))$, i.e., $p \\mid 3a$. As $(p, 3) = 1$, it follows that $p \\mid a$ and then $p \\mid b$. From the above, we deduce that if in the end all the numbers are divisible by a prime $p \\neq 3$, then they were always divisible by that prime. Since \\text{g.c.d.}$(n-1, n, n+1) = 1$, it follows that one cannot obtain two 0-s on the board if $n$ has prime divisors other than 3.\n\nWe are left with the case when $n = 3^k$, $k \\in \\mathbb{N} \\cup \\{0\\}$. For $k=0$, starting from $(0, 1, 2)$, it is easy to get to $(0, 0, 3)$. Assume $k \\ge 1$. Choosing at the first move $a = n-1$, $b = n+1$, we obtain on the board numbers $n-3$, $n$, $n+3$, all multiples of 3. They can be written as $3(m-1)$, $3m$, $3(m+1)$ and the subsequent moves function as if the numbers written on the board were $m-1$, $m$, $m+1$. (After $j \\le k$ moves, on the board will be the numbers $3^k - 3^j$, $3^k$, $3^k + 3^j$. After move no. $k$ we get $0, 3^k, 2 \\cdot 3^k$. Choosing $a = 3^k$ and $b = 2 \\cdot 3^k$ we obtain on the board $0, 0, 3^{k+1}$.\n\nLet us first prove that if $n = 3^m$, $m \\in \\mathbb{N} \\cup \\{0\\}$, then the triple $(n-1, n, n+1)$ is solvable. We prove the statement by induction after $m \\ge 0$.\n\nFor $m=0$: the triple $(0, 1, 2)$ is solvable by a single move, choosing $a=1, b=2$.\n\nAssuming the statement to be true for $m$, let us prove it for $m+1$. Having on the board the triple $(3^{m+1} - 1, 3^{m+1}, 3^{m+1} + 1)$, we choose $a = 3^{m+1} - 1$ and $b = 3^{m+1} + 1$ and we get to the triple $(3^{m+1} - 3, 3^{m+1}, 3^{m+1} + 3)$. According to the remark (*), this triple is solvable because the inductive hypothesis tells us that the triple $(3^m - 1, 3^m, 3^m + 1)$ is solvable.\n\nLet us notice that the sum of the numbers written on the board does not change while performing a move. It remains $3n$, which is a multiple of 3. After the first move, all the numbers become equal modulo 3 because $2a-b \\equiv -a-b \\equiv 2b-a \\pmod 3$. If, after the first move, they became all congruent to 1 or 2 mod 3, they will remain that way, so they can not become 0. Thus, it is mandatory that the first move makes all the numbers on the board multiples of 3. Also, if $(a, b, c)$ is solvable, i.e. can be transformed into $(0, 0, a+b+c)$ through a succession of moves, then, before the last move, the numbers written on the board have to be 0, $\\frac{a+b+c}{3}$ and $\\frac{2(a+b+c)}{3}$, which shows that it is necessary to have $3 \\mid a+b+c$.\n\nIf $n = 3k+1$, then the last move must be $(3k, 3k+1, 3k+2) \\mapsto (3k, 3k, 3k+3)$. If $k=0$, we are done ($n=1$ is a power of 3); otherwise, this triple is solvable if and only if $(k, k, k+1)$ is solvable. But this triple is not solvable because the sum $k+k+(k+1)$ is not divisible by 3.\n\nIf $n = 3k + 2$ then the first move must be $(3k + 1, 3k + 2, 3k + 3) \\mapsto (3k, 3k + 3, 3k + 3)$. This triple is solvable if and only if $(k, k+1, k+1)$ is solvable. But this triple is not solvable because the sum $k + (k+1) + (k+1)$ is not divisible by 3.\n\nIf $n = 3k$, the first move must be $(3k - 1, 3k, 3k + 1) \\mapsto (3k - 3, 3k, 3k + 3)$. This second position is solvable if and only if $(k-1, k, k+1)$ is solvable. As $n \\ne 0$, there exist $u, v \\in \\mathbb{N}$ such that $n = 3^u \\cdot v$, $(v, 3) = 1$. Repeating this reasoning, after $u$ moves we get to the triple $(v-1, v, v+1)$. We have seen that the only triple of this form that is solvable is $(0, 1, 2)$, therefore it is necessary for $n$ to be a power of 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $M$ be a variable point interior to the segment $AB$, and let $\\gamma_B$ be the circle through $M$ and tangent at $B$ to $BC$. Let $P$ and $Q$ be the touch points of $\\gamma_B$ and its tangents from $A$, and let $X$ be the midpoint of the segment $PQ$. Similarly, let $N$ be a variable point interior to the segment $AC$, and let $\\gamma_C$ be the circle through $N$ and tangent at $C$ to $BC$. Let $R$ and $S$ be the touch points of $\\gamma_C$ and its tangents from $A$, and let $Y$ be the midpoint of the segment $RS$. Prove that the line through the centres of the circles $AMN$ and $AXY$ passes through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the line through the centres of the circles $AMN$ and $AXY$ passes through the centre of the circle $ABC$. Alternatively, but equivalently, we prove that the three circles share a point different from $A$.\n\nInvert the whole configuration from $A$ and let $Z'$ denote the image of $Z$ under inversion: The circles $ABC$, $AMN$ and $AXY$ are transformed into the lines $B'C'$, $M'N'$ and $X'Y'$, respectively; the line $BC$ is transformed into the circle $\\gamma$ through $A$, $B'$ and $C'$; the circle $\\gamma_B$ is transformed into a circle $\\gamma'_B$ through $B'$ and $M'$, centred at $X'$ and externally tangent to $\\gamma$ at $B'$; and the circle $\\gamma_C$ is transformed into a circle $\\gamma'_C$ through $C'$ and $N'$, centred at $Y'$ and externally tangent to $\\gamma$ at $C'$. In this setting, we are to prove that the lines $B'C'$, $M'N'$ and $X'Y'$ are (projectively) concurrent.\n\nTo this end, let the pair of external common tangents of $\\gamma'_B$ and $\\gamma'_C$ meet at $O$, and let $\\theta$ be the homothety centred at $O$ mapping $\\gamma'_B$ onto $\\gamma'_C$; clearly, $O$ lies on the line $X'Y'$ through the centres of the two circles. Let further $\\theta_B$ and $\\theta_C$ be the homotheties centred at $B'$ and $C'$, respectively, mapping $\\gamma$ onto $\\gamma'_B$ and $\\gamma'_C$, respectively.\n\nThe centres of the three homotheties are collinear, so $B'C'$ passes through $O$.\n\nFinally, $\\theta = \\theta_C\\theta_B^{-1}$, so $\\theta M' = \\theta_C\\theta_B^{-1}M' = \\theta_C A = N'$, showing that the line $M'N'$ passes through $O$ as well. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38979,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nA_n = \\{ (x, y) \\in \\mathbb{N} \\times \\mathbb{N} \\mid \\sqrt{x^2 + y + n} + \\sqrt{y^2 + x + n} \\in \\mathbb{N} \\}.\n$$\nProve that for all $n \\ge 1$, $A_n$ is a finite, nonempty set.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, in which $m(\\angle ABC) = 45^\\circ$ and $m(\\angle BAC) > 90^\\circ$. Let $O$ be the midpoint of the side $[BC]$. We consider the point $M \\in (AC)$ such that $m(\\angle COM) = m(\\angle CAB)$. The perpendicular through $M$ on $AC$ intersects the line $AB$ at the point $P$.\n\na) Find the measure of the angle $\\angle BCP$.\n\nb) Show that, if $m(\\angle BAC) = 105^\\circ$, then $PB = 2MO$.",
    "options": [],
    "answer": "a) ∠BCP = 45°.\nb) If ∠BAC = 105°, then PB = 2·MO.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38981,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $m$ is perfect if the sum of all its positive divisors, $1$ and $m$ inclusive, is equal to $2m$. Determine the positive integers $n$ such that $n^n + 1$ is a perfect number.",
    "options": [],
    "answer": "n = 3",
    "solution": "There is only one such integer, namely, $n = 3$; it is readily checked that $3^3 + 1 = 28$ is perfect.\n\nIf $n$ is odd, then $n^n + 1$ is even, so it is of the form $2^{p-1}(2^p - 1)$, where $p$ and $2^p - 1$ are both prime (Euler's theorem on the structure of perfect even integers). Rule out the trivial case $n = 1$, to assume $n > 1$, and write $n^n + 1 = (n + 1)(n^{n-1} - n^{n-2} + \\dots - n + 1)$. Since $n$ is odd, the first factor is even and the second is odd; and since $n > 1$, the latter is greater than $1$ (simply rewrite it in the form $1 + n(n - 1)(1 + n^2 + \\dots + n^{n-3})$). It follows that $n + 1 = 2^{p-1}$, so $2^p - 1 = 2n + 1$, and $n^n + 1 = (n + 1)(2n + 1) = 2n^2 + 3n + 1$ which forces $n = 3$.\n\nWe now rule out the other parity of $n$; recall that the existence of perfect odd numbers is still an open question.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38982,
    "subject": "Mathematics (Multi-modal)",
    "question": "One considers the rectangular parallelepiped $ABCD'A'B'C'D'$ such that the measure of the dihedral angle formed by the planes ($A'BD$) and ($C'BD$) is $90^\\circ$ and the measure of the dihedral angle formed by the planes ($AB'C$) and ($D'B'C$) is $60^\\circ$. Determine the measure of the dihedral angle formed by the planes ($BC'D$) and ($A'C'D$).",
    "options": [],
    "answer": "60°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd positive integer and let $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ be two matrices with complex entries, such that $(A - B)^2 = O_n$. Prove that\n$$\n\\det(AB - BA) = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38984,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a line, there are 51 positive integers whose sum is 100. Prove that, for all positive integers $k$, $1 \\le k \\le 99$, one can find either a succession of numbers on the line whose sum is $k$, or a succession of numbers whose sum is $100-k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1, a_2, \\dots, a_{51}$ be the 51 numbers with $a_1 + a_2 + \\dots + a_{51} = 100$. On a circle of total length 100 we place 100 points such that the length of the arc between any two neighboring points (of these 100) is equal to 1. We fix one of these points and denote it by $A_1$. Then, we mark on the circle points $A_2, A_3, \\dots, A_{51}$, in this order, such that the length of each arc $A_iA_{i+1}$ is $a_i$ for all $i = 1, 50$. Clearly, the length of the arc $A_{51}A_1$ will be $a_{51}$.\n\nWe color the points $A_1, A_2, \\dots, A_{50}$ with blue, and the other 49 points with red. We prove that for all $k \\in [1, 99]$ we can find two blue points, $A_i$ and $A_j$, such that the lengths of the two arcs determined by these two points are $k$ and $100-k$. It is sufficient to prove this statement for $k \\le 50$.\n\nFor $k = 50$ we have 51 blue points, so there must be a pair of blue points that are diametrically opposed. For $k < 50$, consider for each blue point $A_j$ the points $B$ such that the length of the arc $A_jB$ is $k$. There are two such points for each point $A_j$. If any of these points is blue, we have found an arc $A_iA_j$ of length $k$. Assume all these points $B$ are red; each of these points $B$ has been considered at most twice, but each blue point uses two red ones. Thus, 51 blue points require 51 red ones, but only 49 are available. We conclude that our assumption was false, therefore an arc $A_iA_j$ of length $k$ must exist.\n\nNow we cut the circle at $A_1$ and turn it into a line segment. For all $k$, we might have thus cut one of the arcs $A_iA_j$, either the one of length $k$ or the one of length $100-k$, but not both. Thus, for all $k$, either a line segment $A_iA_j$ of length $k$, or one of length $100-k$, must exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $A = \\{1, 2, 3, \\dots, 100\\}$.\n\na) Give an example of a set $B$ having 11 elements from $A$, having the property: *from any two elements from $B$, the greatest common divisor of them is at least 9*.\n\nb) Prove that, for any subset $C$ of $A$ having 11 elements, there are two distinct elements of $C$ such that their GCD is not greater than 9.",
    "options": [],
    "answer": "{9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99}",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38986,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that there exist differentiable functions $f : (0, \\infty) \\to (0, \\infty)$, such that $f(f'(x)) = x$, for any $x > 0$.\n\nb) Prove that there do not exist differentiable functions $f : \\mathbb{R} \\to \\mathbb{R}$, such that $f(f'(x)) = x$, for any $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\\{F\\} = DI \\cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.\n\nAlexandru Gîrban",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The bisector of the angle $\\hat{A}$ passes through the midpoint of the arc $BC$ which does not contain the point $A$. Denote this point by $S$. $MS$ is the perpendicular bisector of $[BC]$, so $MS \\parallel ID$ (both lines are perpendicular on $BC$).\n\nAlso, $\\angle ASC = \\angle ABC$ and $\\angle SAC = \\angle BAE$ proves that the triangles $SAC$ and $BAE$ are similar, so $\\frac{AB}{BE} = \\frac{AS}{SC}$.\n\nUsing the angle bisector theorem, we get $\\frac{AB}{BE} = \\frac{AI}{IE}$.\n\nIt is known that $SI = SC$ (one might compute the angles of triangle $SCI$). We have $\\frac{AS}{SC} = \\frac{AS}{SI} = \\frac{AM}{MF}$.\n\nWe proved $\\frac{AI}{IE} = \\frac{AM}{MF}$.\n\nUsing Menelaus's theorem in the triangle $AEF$ and the transversal $I-X-M$ (where $\\{X\\} = IM \\cap EF$), we have $\\frac{AI}{IE} \\cdot \\frac{EX}{XF} \\cdot \\frac{MF}{MA} = 1$.\n\nUsing the equality we proved before, we find $\\frac{EX}{XF} = 1$, which means that $X$ is the midpoint of $[EF]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38988,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the regular hexagonal prism $ABCDEFA_1B_1C_1D_1E_1F_1$ one constructs the points $P, Q$, the projections of the point $A$ on the lines $A_1B$, respectively $A_1C$, and $R, S$, the projections of the point $D_1$ on the lines $A_1D$, respectively $C_1D$.\n\na) Determine the measure of the angle of the planes $(AQP)$ and $(D_1RS)$.\n\nb) Show that $\\widehat{AQP} \\equiv \\widehat{D_1RS}$.",
    "options": [],
    "answer": "0 degrees",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38989,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n \\ge 2$, denote by $s(n)$ the number of pairs of integers $(x, y)$, from $1, 2, \\dots, n$, with $x > y$, such that $x$ and $y$ have exactly $x - y$ common divisors.\na)\nIs there any $n$ such that $s(n) = 2019$?\nb)\nBut for $s(n) = 2020$?",
    "options": [],
    "answer": "a) No.\nb) Yes; n = 1348.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a finite group, and let $x_1, \\dots, x_n$ be a labeling of its elements. Consider the $n \\times n$ matrix $(a_{ij})$, where $a_{ij} = 1$ if $x_i x_j^{-1} \\neq x_j x_i^{-1}$, and $a_{ij} = 0$ otherwise. Establish the parity of the integer $\\det(a_{ij})$.\nAmer. Math. Monthly",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The determinant under consideration is an even integer. To prove this, we show the determinant divisible by the cardinality of the set $S = \\{x: x \\in G, x \\neq x^{-1}\\}$. Since a member of $G$ is one of $S$ if and only if its inverse is, $|S|$ is even (possibly zero), and the conclusion follows.\n\nTo establish divisibility, recall that the value of a determinant does not change upon replacing a column by the sum of all columns. It is therefore sufficient to show that every row contains exactly $|S|$ units.\n\nTo prove the latter, fix any row — say, the $i$-th —, let $J_i = \\{j: a_{ij} = 1\\}$ and notice that the assignment $j \\mapsto x_i x_j^{-1}$ defines a one-to-one map of $J_i$ onto $S$. Consequently, $|J_i| = |S|$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 38991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n \\ge 4$ for which the following property holds true: *any distinct and nonzero complex numbers $a, b, c$, satisfying\n$$\n(a - b)^n + (b - c)^n + (c - a)^n = 0,\n$$\n*are the complex coordinates of the vertices of an equilateral triangle.*",
    "options": [],
    "answer": "n = 4, 5, 7",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 38992,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the sets $M = \\{0, 1, 2, \\dots, 2019\\}$ and\n$$\nA = \\left\\{ x \\in M \\mid \\frac{x^3 - x}{24} \\in \\mathbb{N} \\right\\} .\n$$\n\na)\nHow many elements has the set $A$?\n\nb)\nDetermine the smallest positive integer $n$, $n \\ge 2$, which has the property that any subset with $n$ elements of the set $A$ contains two distinct elements whose difference is divisible by $40$.",
    "options": [],
    "answer": "a) 1263; b) 26",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, and let $A_1, \\dots, A_m$ be pairwise disjoint $n$-element sets of positive integers such that no member of $A_i$ is divisible by one of $A_{i+1}$, whatever $i$ (indices are reduced modulo $m$). Determine the largest number of ordered pairs $(a, b)$, where $a$ and $b$ are members of distinct $A_i$'s, and $b$ is divisible by $a$.",
    "options": [],
    "answer": "((m-1)(m-2)/2) * n^2",
    "solution": "The required maximum is $\\binom{m-1}{2}n^2$, and is achieved if, for instance\n$$\nA_k = \\{a^{(k-1)n+1}, a^{(k-1)n+2}, \\dots, a^{kn}\\}, \\quad k = 1, \\dots, m-1,\n$$\n\nFor brevity, an ordered pair $(a, b)$ satisfying the conditions in the statement will be called *suitable*. We show that the number of suitable pairs does not exceed $\\binom{m-1}{2}n^2$.\nFor every $m$-tuple $(a_1, \\dots, a_m)$, where $a_k$ is a member of $A_k$, $k = 1, \\dots, m$, let $k(a_1, \\dots, a_m)$ be the number of suitable pairs $(a_i, a_j)$ it contains.\nWe show by induction on $m$ that $k(a_1, \\dots, a_m) \\le \\binom{m-1}{2}$. Since there are exactly $n^m$ $m$-tuples, and each suitable pair is contained in exactly $n^{m-2}$ such, the conclusion follows.\nThe case $m = 3$ is easily dealt with. Let $m \\ge 4$ and fix an $m$-tuple $(a_1, \\dots, a_m)$,\n$a_k \\in A_k$, $k = 1, \\dots, m$; without loss of generality, we may and will assume that\n$a_1$ is the largest entry. The $(m-1)$-tuple $(a_1, \\dots, a_{m-1})$ then satisfies the induction\nhypothesis: $a_2$ does not divide $a_1$, $a_3$ does not divide $a_2$, and so on and so forth, $a_{m-1}$\ndoes not divide $a_{m-2}$, and, by maximality, $a_1$ does not divide $a_{m-1}$.\nWe show that the number of suitable pairs containing $a_m$ does not exceed $m-2$.\nIt then follows that $k(a_1, \\dots, a_m) \\le k(a_1, \\dots, a_{m-1}) + m-2 \\le \\binom{m-2}{2} + m-2 =\n\\binom{m-1}{2}$, as desired.\nNotice that, for each $k$ in the range $1$ through $m-1$, at most one of the pairs\n$(a_k, a_m)$, $(a_m, a_k)$ is suitable. If neither $(a_k, a_m)$ nor $(a_m, a_k)$ is suitable for some $k$,\nthen the number of suitable pairs containing $a_m$ is clearly at most $m-2$. Otherwise,\nsince the pairs $(a_m, a_k)$ and $(a_{k+1}, a_m)$, $k = 1, \\dots, m-2$, are not simultaneously\nsuitable, and the pair $(a_1, a_m)$ is certainly not suitable, by maximality of $a_1$, it follows\nthat the pairs $(a_m, a_k)$, $k = 1, \\dots, m-1$, are all suitable, contradicting the fact that\n$a_m$ does not divide $a_{m-1}$. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38994,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set of prime numbers is called *interesting* if the following holds: *the sum of any three distinct numbers from the set is also prime*.\nFind the maximum number of elements an interesting set should have.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 38995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers from $1$ to $49$ are placed arbitrarily on a quadratic table formed by $7 \\times 7$ squares. Show that one can find a $2 \\times 2$ square on the table, formed by $4$ neighboring cells, such that the sum of numbers inside is at least $81$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 38996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $p$ for which there exists $n \\in \\mathbb{N}^*$ such that $p^n + 3^n$ divides $p^{n+1} + 3^{n+1}$.",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 38997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all nonnegative integers $x, y, z$ which satisfy the equation\n$$\n2^x + 3 \\cdot 11^y = 7^z.\n$$",
    "options": [],
    "answer": "[(2, 0, 1), (4, 1, 2)]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 38998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, a triangular array $(a_{i,j})$ of zeroes and ones, where $i$ and $j$ run through the positive integers such that $i + j \\le n + 1$, is called a *binary anti-Pascal n-triangle* if $a_{i,j} + a_{i,j+1} + a_{i+1,j} \\equiv 1 \\pmod 2$ for all possible values $i$ and $j$ may take on. Determine the minimum number of ones a binary anti-Pascal $n$-triangle may contain.",
    "options": [],
    "answer": "⌊n(n+1)/6⌋",
    "solution": "In what follows, part of the generic configurations referred to may not exist for the first few values of $n$; in this case, simply consider the corresponding induced\n\nWe now show by induction on $n$ that a binary anti-Pascal $n$-triangle contains at least $\\lfloor n(n+1)/6 \\rfloor$ ones, of which at least $n-1$ lie on the bottom three rows (and since the transpose of a binary anti-Pascal triangle is again one such, the same holds for the leftmost three columns).\nThe cases $n=1, 2, 3$ are easily dealt with, so let $n \\ge 4$, and let $A$ be a binary anti-Pascal $n$-triangle. The lower left $3 \\times 3$ subarray $A'$ (induced if $n=4$) contains at least 3 ones, unless the lower left and central entries are both one and the other entries are all zero, in which case there are only 2 ones in $A'$.\nIn the former case, consider the binary anti-Pascal $(n-3)$-triangle consisting of the rightmost $n-3$ columns of $A$. By the induction hypothesis, the bottom three rows of this triangle contain at least $n-4$ ones, so the bottom three rows of $A$ contain at least $3 + (n-4) = n-1$ ones.\nIn the latter case, the statement clearly holds if $n=4$, so let $n \\ge 5$ and notice that the bottom two entries of the column of $A$ adjoining $A'$ along the right flank are both one. Thus, the $3 \\times 4$ subarray of $A$ extending $A'$ rightwards contains at least 4 ones. The rightmost $n-4$ columns of $A$ form a binary anti-Pascal $(n-4)$-triangle whose bottom three rows contain at least $n-5$ ones, by the induction hypothesis. Hence the bottom three rows of $A$ contain at least $4 + (n-5) = n-1$ ones.\nIn either case, the bottom three rows of $A$ contain at least $n-1$ ones. Finally, by the induction hypothesis, the binary anti-Pascal $(n-3)$-triangle atop the bottom three rows of $A$ contains at least $\\lfloor(n-3)(n-2)/6\\rfloor$ ones, so $A$ contains at least $n-1 + \\lfloor(n-3)(n-2)/6\\rfloor = \\lfloor n(n+1)/6 \\rfloor$ ones. This completes the induction and concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 38999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $k$, a loop of length $k$ in a graph is a list $v_1, e_1, v_2, e_2, \\dots, v_k, e_k$, where the $v_i$ are (not necessarily distinct) vertices, the $e_i$ are (not necessarily distinct) edges, and each $e_i$ joins $v_i$ and $v_{i+1}$ (indices are reduced modulo $k$); the loop traces an edge $e$ if $e = e_i$ for some index $i$. Show that a connected graph with vertex set $V$ and edge set $E$ has a loop of length at most $|V| + |E| - 1$ tracing every edge of the graph.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be a connected graph with vertex set $V$ and edge set $E$; $G$ may have loops and/or multiple edges. The idea is to expand at most $|V|-1$ suitable edges to pairs of edges to make $G$ into a graph $\\hat{G}$ on $V$ each vertex of which has an even degree. A maximal loop in $\\hat{G}$ tracing each edge at most once is then Eulerian, i.e., traces each edge exactly once (Euler). Read back in $G$, i.e., identifying each cloned edge back to the original, this is the desired loop: it traces every cloned edge twice, every other edge once, and its length does not exceed $|V| - 1 + |E|$.\n\nTo obtain $\\hat{G}$, consider a spanning tree $T$ of $G$, i.e., a minimal connected subgraph on $V$. We will show that $T$ has a subgraph $S$ on $V$ such that $\\deg_S$ and $\\deg_G$ agree modulo 2 at all vertices; clearly, the number of edges of $S$ does not exceed the number of edges of $T$ which is $|V| - 1$. Cloning each edge of $S$ once, while keeping its end points fixed, of course, yields the desired additional edges in $\\hat{G}$.\n\nTo obtain $S$ from $T$, notice that the sums $\\sum_{v \\in V} \\deg_T v = 2|E_T|$ and $\\sum_{v \\in V} \\deg_G v = 2|E|$ have like parities, to infer that $\\deg_T$ and $\\deg_G$ disagree modulo 2 at an even number of vertices (possibly zero), say, $v_1, \\dots, v_n, v_{n+1}, \\dots, v_{2n}$. For each index $i$ in the range 1 through $n$, let $\\alpha_i$ be the unique path in $T$ joining $v_i$ and $v_{i+n}$, and collect together all edges of $T$ lying along an even number of the $\\alpha_i$ (zero, inclusive) to form the edge set of $S$.\n\nFinally, we show that $\\deg_S$ and $\\deg_G$ agree modulo 2 at all vertices. To this end, fix a vertex $v$, and let $E'$ be the set of all edges of $T$ having an end point at $v$. For each edge $e$ in $E'$ and each path $\\alpha_i$, consider their edge-path incidence number\n$$\n\\langle e, \\alpha_i \\rangle = \\begin{cases} 1, & \\text{if } \\alpha_i \\text{ traces } e, \\\\ 0, & \\text{otherwise,} \\end{cases} \\quad \\text{and let } \\equiv \\text{ denote congruence modulo 2 to write}\n$$\n$$\n\\begin{aligned}\n\\deg_T v - \\deg_S v &\\equiv \\sum_{e \\in E'} \\sum_{i=1}^n \\langle e, \\alpha_i \\rangle = \\sum_{i=1}^n \\sum_{e \\in E'} \\langle e, \\alpha_i \\rangle \\\\\n&\\equiv \\begin{cases} 1, & \\text{if } v \\text{ is a } v_i, \\\\ 0, & \\text{otherwise,} \\end{cases} \\\\\n&\\equiv \\deg_G v - \\deg_T v, \\end{aligned}\n$$\non account of $\\sum_{e \\in E'} \\langle e, \\alpha_i \\rangle$ being congruent to 1 modulo 2 if and only if $\\alpha_i$ has an end point at $v$, which is the case if and only if $v$ is one of $v_i, v_{i+n}$. Consequently, $\\deg_S v \\equiv \\deg_G v$ and the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a real number $a > 1$, determine all real numbers $b \\ge 1$ such that\n$$\n\\lim_{x \\to \\infty} \\int_{0}^{x} (1 + t^a)^{-b} \\, dt = 1.\n$$",
    "options": [],
    "answer": "b = 1 + 1/a",
    "solution": "We show that $b_0 = 1 + 1/a$ is the only real number satisfying the required condition.\nLet $b \\ge 1$, and let $f_b, F_b: [0, \\infty) \\to \\mathbb{R}$, $f_b(x) = (1 + x^a)^{-b}$ and $F_b(x) = \\int_0^x f_b(t) \\, dt$. Since $f_b$ is positive, $F_b$ is increasing, so the limit $I(b) = \\lim_{x \\to \\infty} F_b(x)$ exists.\nSince $0 \\le f_b(x) \\le f_1(x)$ on the ray $x \\ge 0$,\n$$\n\\begin{aligned} 0 \\le F_b(x) &\\le F_1(x) = \\int_0^1 (1+t^a)^{-1} \\, dt + \\int_1^x (1+t^a)^{-1} \\, dt \\le 1 + \\int_1^x t^{-a} \\, dt \\\\ &= 1 + 1/(a-1) - x^{1-a}/(a-1) \\le a/(a-1), \\end{aligned}\n$$\nwhatever $x \\ge 1$, so $I(b)$ is a real number.\n\n$$\n\\begin{aligned} I(b) &= \\int_{0}^{1} f_{b}(t) \\, dt + \\lim_{x \\to \\infty} \\int_{1}^{x} f_{b}(t) \\, dt > \\int_{0}^{1} f_{c}(t) \\, dt + \\lim_{x \\to \\infty} \\int_{1}^{x} f_{b}(t) \\, dt \\\\ &\\ge \\int_{0}^{1} f_{c}(t) \\, dt + \\lim_{x \\to \\infty} \\int_{1}^{x} f_{c}(t) \\, dt = I(c). \\end{aligned}\n$$\n\nFinally, we show that $I(1 + 1/a) = 1$. To this end, integrate by parts to write\n$$\n\\begin{aligned}\n\\int_0^x (1+t^a)^{-1/a} dt &= t(1+t^a)^{-1/a} \\Big|_0^x + \\int_0^x t^a (1+t^a)^{-1-1/a} dt \\\\\n&= x(1+x^a)^{-1/a} + \\int_0^x (1+t^a)^{-1/a} dt - F_{1+1/a}(x).\n\\end{aligned}\n$$\n\nConsequently, $F_{1+1/a}(x) = x(1+x^a)^{-1/a}$, so $I(1+1/a) = 1$, and $b_0 = 1+1/a$, by injectivity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, if positive real numbers $a_1, a_2, \\dots, a_n$ have the product $1$, then\n$$\n\\left(\\frac{a_1}{a_2}\\right)^{n-1} + \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} + \\left(\\frac{a_n}{a_1}\\right)^{n-1} \\geq a_1^2 + a_2^2 + \\dots + a_n^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove that, for all $a_1, a_2, \\dots, a_n > 0$, the following inequality holds\n$$\n\\left(\\frac{a_1}{a_2}\\right)^{n-1} + \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} + \\left(\\frac{a_n}{a_1}\\right)^{n-1} \\geq \\frac{a_1^2 + a_2^2 + \\dots + a_n^2}{\\sqrt[n]{a_1^2 a_2^2 \\dots a_n^2}}.\n$$\nThis inequality is obtained by adding the inequality below with its analogues obtained by cyclic permutation of the variables:\n$$\n\\begin{aligned}\n& (n-1) \\cdot \\left(\\frac{a_1}{a_2}\\right)^{n-1} + (n-2) \\cdot \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} \\geq \\\\\n& \\geq \\frac{n(n-1)}{2} \\cdot \\left( \\left(\\frac{a_1}{a_2}\\right)^{n-1} \\left(\\frac{a_2}{a_3}\\right)^{n-2} \\dots \\left(\\frac{a_{n-1}}{a_n}\\right)^{\\frac{2}{n}} \\right) = \\\\\n& = \\frac{n(n-1)}{2} \\left( \\frac{a_1^n}{a_1 a_2 a_3 \\dots a_n} \\right)^{\\frac{2}{n}}.\n\\end{aligned}\n$$\nEquality holds if all the numbers are equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $APD$ be an acute-angled triangle and let $B \\in (AP)$, $C \\in (PD)$ be two points. The diagonals of the quadrilateral $ABCD$ meet at the point $Q$. Denote by $H_1$ and $H_2$ the orthocenters of the triangles $APD$ and $BPC$ respectively. The circumcircles of the triangles $ABQ$ and $CDQ$ meet again at the point $X$ ($X \\neq Q$) and the circumcircles of the triangles $ADQ$ and $BCQ$ meet again at the point $Y$ ($Y \\neq Q$).\nShow that if the line $H_1H_2$ passes through the point $X$, then it also passes through the point $Y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AA'$, $DD'$ be altitudes in the triangle $\\triangle APD$ and let $BB'$, $CC'$ be altitudes in the triangle $\\triangle BPC$.\nWe have $\\angle XAC = \\angle XBD$ and $\\angle XCA = \\angle XDB$, and thus $\\triangle XAC \\sim \\triangle XBD$. We prove that the line $H_1H_2$ passes through the point $X$ if and only if the above triangles are congruent.\n\n![](attached_image_1.png)\n\nWe consider the circles $\\omega_1, \\omega_2$ of diameter $AC$ and $BD$ respectively. The points $H_1, H_2$ belong to the radical axis of these two circles, because $A, A' \\in \\omega_1, D, D' \\in \\omega_2$\n\nşi $P_{\\omega_1}(H_1) = -AH_1 \\cdot A'H_1 = -DH_1 \\cdot D'H_1 = P_{\\omega_2}(H_1)$; and similar for $H_2$. Thus, $X \\in H_1H_2$ if and only if $P_{\\omega_1}(X) = P_{\\omega_2}(X)$.\nWe have $P_{\\omega_1}(X) = XM^2 - MC^2$ and $P_{\\omega_2}(X) = XN^2 - ND^2$, were $M, N$ are the centers of the circles $\\omega_1$ and $\\omega_2$ respectively. We notice that $(XM^2 - MC^2)/(XN^2 - ND^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$, or $XM^2 - MC^2 = XN^2 - ND^2 = 0$. In the former case, the triangles $AXC$ and $BXD$ should be similar and right angles in $X$. But now the line $CD$ corresponds to the line $AB$ by a similarity of center $X$ and angle $90^\\circ$, so $AB \\perp CD$ – contradiction. Thus $(XM^2 - MC^2)/(XN^2 - ND^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$.\nWe obtain that $X \\in H_1H_2$ if and only if the similarity ratio of the triangles $AXC$ and $BXD$ is 1, that is $\\Delta AXC \\equiv \\Delta BXD$. But this is equivalent to $AC = BD$.\nIn the same manner, $Y \\in H_1H_2$ if and only if $AC = BD$, and we get the conclusion of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. For any permutation $\\sigma \\in S_p$, we define the matrix $A_\\sigma = (a_{ij})_{i,j=1,p} \\in \\mathcal{M}_p(\\mathbb{Z})$, whose entries are $a_{ij} = \\sigma^{i-1}(j)$, for any $i, j = 1, p$, where $\\sigma^0$ is the identity permutation, and $\\sigma^k = \\underbrace{\\sigma \\circ \\sigma \\circ \\dots \\circ \\sigma}_{\\text{de k ori}}$, for any $k \\in \\mathbb{N}^*$. Prove that the set $D = \\{|\\det(A_\\sigma)| : \\sigma \\in S_p\\}$ has at most $1 + (p-2)!$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(x + y) \\leq f(x^2 + y),\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All constant functions f(x) = c for some real constant c.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A, B \\subset \\mathbb{N}$ be two finite and nonempty sets. Denote by $\\mathcal{F}$ the set of all functions $f : \\mathcal{P}(A) \\to B$ such that\n$$\nf(X \\cap Y) = \\min(f(X), f(Y)), \\forall X, Y \\subset A,\n$$\nand by $\\mathcal{G}$ the set of all functions $g : \\mathcal{P}(A) \\to B$ such that\n$$\ng(X \\cup Y) = \\max(g(X), g(Y)), \\forall X, Y \\subset A.\n$$\nProve that $\\mathcal{F}$ și $\\mathcal{G}$ have the same number of elements and find that number.",
    "options": [],
    "answer": "sum_{t=1}^{|B|} t^{|A|}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39006,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the interior of non-zero angle $\\widehat{AOD}$ consider points $B$ and $C$ such that $OA = OB$, $OD = OC$, the segments $AC$ and $BD$ meet in $P$, and the semi-line ($PO$ is the angle bisector of $\\widehat{APD}$). Prove that the angles $\\widehat{AOB}$ and $\\widehat{COD}$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to [0, \\infty)$ be a continuous function with $f(0) > 0$, such that for any $0 \\le x < y$, we have\n$$\nx - y < f(y) - f(x) \\le 0.\n$$\nProve that:\na) there exists a unique number $\\alpha \\in (0, \\infty)$ with the property $(f \\circ f)(\\alpha) = \\alpha$;\nb) the sequence $(x_n)_{n \\ge 1}$, defined by $x_1 \\ge 0$ and $x_{n+1} = f(x_n)$, for any $n \\in \\mathbb{N}^*$, is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39008,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Determine the four digit numbers in base 10 which are written in base two only with the digit $1$.\n\nb) How many four digit numbers in base 10 are, such that in base two they contain exactly one zero digit.",
    "options": [],
    "answer": "(a) 1023, 2047, 4095, 8191; (b) 38",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the set of real numbers the equation\n$$\n\\left[ x + \\frac{1}{x} \\right] = \\left[ x^2 + \\frac{1}{x^2} \\right],\n$$\nwhere $[a]$ represents the integer part of the real number $a$.",
    "options": [],
    "answer": "x in ((sqrt(5) - 1)/2, (sqrt(5) + 1)/2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39010,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the isosceles rectangular triangle $ABC$, $m(\\hat{A}) = 90^\\circ$, and the point $D \\in (AB)$ such that $AD = \\frac{1}{3}AB$. In the halfplane determined by the line $AB$ and the point $C$, one takes the point $E$ for which $m(\\angle BDE) = 60^\\circ$ and $m(\\angle DBE) = 75^\\circ$. The lines $BC$ and $DE$ intersect at the point $G$, and the parallel through the point $G$ to the line $AC$ intersects the line $BE$ at the point $H$.\nProve that the triangle *CEH* is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ana and Bogdan play the following turn based game: Ana starts with a pile of $n$ ($n \\ge 3$) stones. At his turn each player has to split one pile. The winner is the player who can make at his turn all the piles to have at most two stones. Depending on $n$, determine which player has a winning strategy.",
    "options": [],
    "answer": "Ana wins for n = 3 and for all even n; Bogdan wins for all odd n ≥ 5.",
    "solution": "If $n = 3$ or $n = 4$ Ana wins at her first move. If $n$ is odd, greater than 3, Bogdan will win. In this case, Ana has to start by making a pile with an even number of stones. Bogdan will split this pile into a pile with one stone and the rest into an odd pile. Ana has to make an even pile again and Bogdan continues his strategy unless he can win. The game may end in two different ways. If Ana leaves a pile of 2, one of 3 and the rest of 1, Bogdan will win by splitting the pile with 3 stones. If Ana leaves a pile of 4 and the rest of 1, Bogdan wins by splitting the 4-pile into two piles of 2 stones.\n\nIf $n \\ge 6$ is even, then Ana will split the pile in 1 and $n-1$ and continue with the strategy described for Bogdan above. So, in this case, Ana will win.\nFor $n$ even, Ana splits the pile into two equal piles. Then, after Bogdan's move into one pile, she will make the same move in the other pile. This strategy will end up with Ana winning.\n\nFor $n \\ge 5$ odd, we prove by induction that Bogdan wins by always leaving an even number of piles with 2 stones. We verify the cases $n = 5, 7, 9$. For odd $n \\ge 11$, we have two situations. If Ana splits $n = a+b$, with $a \\ge 5$ odd, Bogdan will split the pile with $b$ stones into two equal piles. If Ana moves in one of the equal piles, Bogdan will make the same move in the other pile. If Ana moves in the odd pile, Bogdan will follow the strategy provided by the induction hypothesis to end up with piles of 1 stone and an even number of piles of 2 stones. Ana might still be able to move in a pile of 2, but whenever she splits a pile of 2, so will Bogdan, leaving an even number of piles of 2 stones. If $a = 3$, Bogdan will split $b = 3 + (b - 3)$, where $b - 3 \\ge 5$, and apply the same strategy as above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39012,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangular piece of paper of dimensions $20 \\times 19$, divided into unit squares, is cut into several square pieces, the cuttings being made along the sides of the unit squares. Such a square piece is called an *odd square* if the length of its side is an odd number.\n\na) What is the minimum possible number of odd squares?\n\nb) What is the smallest value which can be taken by the sum of the perimeters of all the odd squares?",
    "options": [],
    "answer": "a) 4; b) 80",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, $n \\ge 2$, and let $\\alpha_1, \\alpha_2, \\dots, \\alpha_n$ be non-zero complex numbers such that $|\\alpha_i| < 1$ for $i = 1, \\dots, n-1$, and the coefficients of the polynomial $\\prod_{i=1}^n (X - \\alpha_i)$ are all integral. Show that, if $\\alpha_i, \\alpha_j, \\alpha_k$ form a geometric progression, then $i = j = k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose now, if possible, that $\\alpha_{i_0}, \\alpha_{i_1}, \\alpha_{i_2}$ form a geometric progression for some indices $i_0, i_1, i_2$ of which at least two are distinct; say $\\alpha_{i_1}^2 = \\alpha_{i_0}\\alpha_{i_2}$. The condition on absolute values forces all three indices to be different from $n$.\n\nSince $|\\alpha_{i_1}| < 1$, and the minimal polynomial $f$ of $\\alpha_{i_1}^2$ over the rationals has integral coefficients, the latter has a complex root $\\alpha$ whose absolute value is strictly greater than $1$.\n\nConsider now the polynomial $g = \\prod_{1 \\le i \\le j \\le n} (X - \\alpha_i \\alpha_j)$. Since the expression is symmetric in the $\\alpha_i$, the coefficients of $g$ are all integral.\n\nNotice that $g$ has a double root at $\\alpha_{i_1}^2 = \\alpha_{i_0}\\alpha_{i_2}$, to infer that it is divisible by $f^2$, so it has a double root at $\\alpha$ as well.\n\nFinally, recall that $|\\alpha| > 1$, so $\\alpha$ is one of the pairwise distinct $\\alpha_i\\alpha_n$, $i = 1, 2, \\dots, n$, each of which is, however, a simple root of $g$. The contradiction thus obtained concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive real number $\\lambda$ such that, for any real numbers $a_1, a_2, a_3 \\in [0, \\frac{1}{2}]$ and $b_1, b_2, b_3 \\in (0, \\infty)$ with $\\sum_{i=1}^3 a_i = \\sum_{i=1}^3 b_i = 1$, the following inequality holds\n$$\nb_1b_2b_3 \\le \\lambda(a_1b_1 + a_2b_2 + a_3b_3).\n$$",
    "options": [],
    "answer": "1/8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a cyclic quadrilateral $ABCD$ and let $M$ and $N$ be the midpoints of the diagonals $AC$ and $BD$, respectively. If $\\angle AMB = \\angle AMD$, prove that $\\angle ANB = \\angle BNC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Cristian Mangra)\n\n**Lemma:** Let $ABCD$ be an isosceles trapezoid ($AB \\parallel CD$) inscribed in a circle $C$ of center $O$, $M$ is the intersection point of the diagonals $AC$ and $BD$, and $T$ is the intersection point of lines $AD$ and $BC$. If the line through $M$ parallel to $AB$ meets the circle $C$ at $E$ and $F$, then $TE$ and $TF$ are tangent to the circle.\n\n*Proof of the Lemma:* Assume $AB < CD$. It is clear that points $T$, $M$ and $O$ are collinear. Also, $\\angle AOT = \\frac{1}{2} \\angle AOB = \\angle ACT$, which means that the quadrilateral $TAOC$ is cyclic. From the power of point $M$ with respect to the circumcircle of triangle $AOC$ we get $MA \\cdot MC = MO \\cdot MT$. But $MA \\cdot MC = ME \\cdot MF$ from the power of $M$ with respect to $C$, so $MO \\cdot MT = ME \\cdot MF$. This means that $TEOF$ is cyclic, so $\\angle TEO + \\angle TFO = 180^\\circ$. But $\\angle TEO = \\angle TFO$, hence $\\angle TEO = 90^\\circ$, i.e., $TE$ is tangent to $C$. $\\square$\n\n![](attached_image_1.png)\n\nLet us now return to the problem. Denote by $E$ and $F$ the second intersection points with the circle $C$ of lines $BE$, and $DE$, respectively. Because $M$ is the midpoint of the diagonal $AC$, and angles $AMB$ and $AMD$ are equal, it follows that $BDEF$ is an isosceles trapezoid (it is symmetric with respect to the perpendicular bisector of $AC$). Moreover, $BF \\parallel DE \\parallel AC$. If lines $BD$ and $EF$ meet at $T$, from the lemma it follows that $TA$ and $TC$ are tangent to the circumcircle of $ABCD$. If $O$ is the center of this circle, it follows that $OA \\perp TA$ and $OC \\perp TC$. As $N$ is the\n\nmidpoint of $BD$, we have $ON \\perp BD$. Thus, points $A, N$, and $C$ all lie on the circle of diameter $OT$. Assume $\\angle B > \\angle D$ (if $\\angle B < \\angle D$, simply swap $B$ and $D$; the case $\\angle B = \\angle D$ is easy). In this case, $\\angle ANB = \\angle AOT$, and $\\angle TNC = \\angle TOC$. But $\\angle AOT = \\angle COT$ means $\\angle ANB = \\angle CNT = \\angle CNB$.\n*Alternative Solution.* (Given in the contest by David-Andrei Anghel.)\n\nWe only treat the case in the figure below; the other cases are similar.\n\nLet $O$ be the circumcenter of triangle $ABC$. $O$ is on the perpendicular bisector of $BD$ but also on the external bisector of angle $BMD$ (it is perpendicular to $MA$, the internal bisector of that angle). It is known that these two lines meet on the circumcircle of triangle $BMD$ (in the midpoint of the arc $BMD$), therefore $BDOM$ is cyclic.\n\nWe obtain that angles $BMD$ and $BOD$ are equal. It follows that\n$$\n\\angle BCD = \\frac{1}{2} \\angle BOD = \\frac{1}{2} \\angle BMD = \\angle BMA.\n$$\nAs $\\angle BAM = \\angle BDC$, triangles $BAM$ and $BDC$ are similar. It follows that $\\frac{AM}{AB} = \\frac{CD}{BD}$, which leads to $AC \\cdot BD = 2AB \\cdot CD$. But according to Ptolemy's Theorem,\n$$\nAC \\cdot BD = AB \\cdot CD + BC \\cdot AD,\n$$\ntherefore $AB \\cdot CD = BC \\cdot AD$, i.e., the quadrilateral $ABCD$ is harmonic.\n\nThis leads to the similarity of triangles $ANB$ and $ADC$. Also, triangles $CNB$ and $CDA$ are similar. The two similarities imply\n$$\n\\angle ANB = \\angle ADC = \\angle BNC.\n$$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a nonnegative integer and $M = \\{n^3, n^3 + 1, n^3 + 2, \\dots, n^3 + n\\}$. Consider $A$ and $B$ two nonempty, disjoint subsets of $M$ such that the sum of elements of the set $A$ divides the sum of elements of the set $B$. Prove that the number of elements of the set $A$ divides the number of elements of the set $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $A = \\{n^3 + n_1, n^3 + n_2, \\dots, n^3 + n_a\\}$, $B = \\{n^3 + m_1, n^3 + m_2, \\dots, n^3 + m_b\\}$ and $k \\in \\mathbb{N}$ such that $n^3 + m_1 + n^3 + m_2 + \\dots + n^3 + m_b = k(n^3 + n_1 + n^3 + n_2 + \\dots + n^3 + n_a)$. Then $n^3(ka - b) = m_1 + m_2 + \\dots + m_b - k(n_1 + n_2 + \\dots + n_a)$.\n\nThe case $n = 1$ is obviously true since $M = \\{1, 2\\}$ and we can only have $A = \\{1\\}$ and $B = \\{2\\}$.\n\nNow let's prove that $n > 1$ implies $k < n + 1$ (so $k \\le n$). Indeed, supposing $k \\ge n + 1$, we would get $n^3 + 1 + n^3 + 2 + \\dots + n^3 + n \\ge n^3 + m_1 + n^3 + m_2 + \\dots + n^3 + m_b \\ge (n+1)(n^3 + n_1 + n^3 + n_2 + \\dots + n^3 + n_a) \\ge (n+1)n^3$, therefore $n^4 + \\frac{n(n+1)}{2} \\ge n^4 + n^3$. This would imply $n^2 + n \\ge 2n^3$ which is false.\n\nWe are left with the case $k \\le n$. Now we have $m_1 + m_2 + \\dots + m_b - k(n_1 + n_2 + \\dots + n_a) < 1 + 2 + \\dots + n = \\frac{n(n+1)}{2} < n^3$ and $m_1 + m_2 + \\dots + m_b - k(n_1 + n_2 + \\dots + n_a) \\ge -n(n_1 + n_2 + \\dots + n_a) \\ge -n(1 + 2 + \\dots + n) = -\\frac{n^2(n+1)}{2} > -n^3$.\n\nTo summarize, we have the inequalities $-n^3 < n^3(ka - b) < n^3$, therefore $ka - b = 0$ showing that $a$ divides $b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$, determine the least value the sum $\\sum_{i=1}^{n} (1/x_i - x_i)$ may achieve, as the $x_i$ run through the positive real numbers subject to $\\sum_{i=1}^{n} \\frac{1}{x_i + n - 1} = 1$. Also, determine the $x_i$ at which this minimum is achieved.",
    "options": [],
    "answer": "The minimum value is 0, achieved exactly when all variables are equal to 1.",
    "solution": "The required minimum is $0$ and is achieved if and only if the $x_i$ are all equal to $1$. Let $x_1, \\dots, x_n$ be positive real numbers satisfying the condition in the statement. Let $y_i = x_i/(x_i + n - 1)$, $i = 1, 2, \\dots, n$, and notice that the $y_i$ are positive real numbers that add up to $1$. Express the $x_i$ in terms of the $y_i$ to get $x_i = (n-1)y_i/(1-y_i)$, and write successively\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{1}{x_i} &= \\frac{1}{n-1} \\sum_{i=1}^{n} \\frac{1-y_i}{y_i} = \\frac{1}{n-1} \\sum_{i=1}^{n} \\frac{1}{y_i} \\sum_{j \\neq i} y_j = \\frac{1}{n-1} \\sum_{i \\neq j} \\frac{y_j}{y_i} = \\frac{1}{n-1} \\sum_{i \\neq j} \\frac{y_i}{y_j} \\\\\n&= \\frac{1}{n-1} \\sum_{i=1}^{n} y_i \\sum_{j \\neq i} \\frac{1}{y_j} \\ge \\frac{1}{n-1} \\sum_{i=1}^{n} y_i \\cdot \\frac{(n-1)^2}{\\sum_{j \\neq i} y_j} = \\sum_{i=1}^{n} \\frac{(n-1)y_i}{1-y_i} = \\sum_{i=1}^{n} x_i.\n\\end{aligned}\n$$\nEquality clearly forces the $y_i$ all equal to $1/n$, which is the case if and only if the $x_i$ are all equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest integer $N$ satisfying the following condition: for every cell labeling of a $5 \\times 5$ array from $1$ through $25$ such that no two cells bear the same number, the numbers in some $2 \\times 2$ square add up to at least $N$.",
    "options": [],
    "answer": "45",
    "solution": "The required maximum is $N = 45$, and is achieved, for instance, by the following extremal cell labeling:\n\n| 25 | 5 | 24 | 6 | 23 |\n|----|---|----|---|----|\n| 11 | 4 | 12 | 3 | 13 |\n| 22 | 7 | 21 | 8 | 20 |\n| 14 | 2 | 15 | 1 | 16 |\n| 19 | 9 | 18 | 10 | 17 |\n\nIn this cell labeling, every $2 \\times 2$ square with an odd rank top row has sum $45$, and every $2 \\times 2$ square with an even rank top row has sum $44$; the configuration is derived from the considerations below.\n\nWrite $n = 5$ and $m = \\lfloor n/2 \\rfloor = 3$, to show that, if $n$ is odd, then for every injective cell labeling of an $n \\times n$ array from $1$ through $n^2$, the labels in some $2 \\times 2$ square add up to at least $8m^2 - 11m + 6$ (which is $45$ in the case at hand).\n\nLabel rows downward and columns rightward, both from $1$ through $n$, and let $s_{ij}$ be the sum of all numbers labeling the cells in the cross formed by the $i$-th row and the $j$-th column. Formally, letting $a_{ij}$ denote the label assigned to the cell on the $i$-th row and $j$-th column,\n\n$$\ns_{ij} = \\sum_{k=1}^{n} (a_{ik} + a_{kj}) - a_{ij}.\n$$\n\nUse $\\equiv$ to denote congruence modulo $2$ and consider the sum\n\n$$\nS = \\sum_{i,j=1}^{m} s_{2i-1,2j-1} = \\sum_{i \\equiv j \\equiv 1} s_{ij} = (2m-1) \\sum_{i \\equiv j \\equiv 1} a_{ij} + m \\sum_{i+j=1} a_{ij};\n$$\n\nthere are $m^2$ ordered pairs $(i, j)$ such that $i \\equiv j \\equiv 1$, and $2m(n-m)$ ordered pairs $(i, j)$ such that $i + j \\equiv 1$.\n\nFinally, use the fact that $n$ is odd, $n = 2m-1$, to tile $C$ by $(m-1)^2$ squares $2 \\times 2$ and infer that the numbers in one of these tiles add up to at least\n\n$$\n\\left\\lfloor \\frac{n^2(n^2+1)}{2(m-1)^2} - \\frac{S_0}{m^2(m-1)^2} \\right\\rfloor = \\left\\lceil 8m^2 - 11m + 5 + \\frac{11}{2} \\right\\rceil = 8m^2 - 11m + 6,\n$$\n\nestablishing the required lower bound.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39019,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$ and $z$ are real numbers such that $x^2 + y^2 + z^2 = 2$, prove that $x + y + z \\le xyz + 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $2xy \\le x^2 + y^2 \\le x^2 + y^2 + z^2 = 2$, therefore $xy \\le 1$. Similarly, $xz \\le 1$, and $yz \\le 1$. We also have $(x+y)^2 = x^2 + y^2 + 2xy \\le 4$, so $x+y \\le |x+y| \\le 2$. Also, $x+z \\le 2$, and $y+z \\le 2$. If one of the numbers is negative, let's say $z$, then the conclusion follows from the inequalities $x+y \\le 2$ and $z \\le xyz$.\n\nIn the case $x, y, z \\in [0, 1]$, we have $z(1 - xy) \\le 1 - xy \\le 2 - x - y$ (the last inequality is equivalent to $(1-x)(1-y) \\ge 0$).\n\nWe may have at most one number greater than 1, otherwise the product of two such numbers would be greater than 1. We are left with the case $x, y \\in [0, 1]$ and $z > 1$. In this case $(1-x)(1-y)(z-1) \\ge 0 \\Leftrightarrow xyz + 2 \\ge xy + yz + xz + 3 - (x+y+z) \\ge x+y+z$, the last inequality being equivalent to $2(xy+yz+zx)+6 = (x+y+z)^2+4 \\ge 4(x+y+z)$.\nUsing Cauchy-Buniakowsky-Schwarz we get $(x + y + z - xyz)^2 = ((x+y)+z(1-xy))^2 \\le ((x+y)^2+z^2)(1+(1-xy)^2) = (2+2xy)(2-2xy+x^2y^2) = 4+2x^2y^2(xy-1) \\le 4$ since $x^2 + y^2 \\le 2$ and $xy \\le 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for infinitely many positive integers $n$ there exist pairwise distinct positive integers $a_1, a_2, \\dots, a_n$ such that $a_1^2 a_2^2 \\cdots a_n^2 - 4(a_1^2 + a_2^2 + \\cdots + a_n^2)$ is the square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Letting $f_n(a_1, a_2, \\dots, a_n) = a_1^2 a_2^2 \\cdots a_n^2 - 4(a_1^2 + a_2^2 + \\cdots + a_n^2)$, the conclusion is a straightforward consequence of the following two facts:\n(1) For infinitely many positive integers $n$ there exist positive integers $a_1, a_2, \\dots, a_n$ such that $f_n(a_1, a_2, \\dots, a_n)$ is the square of an integer, and $a_1 < a_2$; and\n(2) Given an integer $n \\ge 3$, if $a_1, a_2, \\dots, a_n$ are positive integers such that $f_n(a_1, a_2, \\dots, a_n)$ is the square of an integer, and $a_1 < a_2 < \\dots < a_k$ for some index $k$, $1 < k < n$, then there exists an integer $a > a_k$ such that $f_n(a_1, a_2, \\dots, a_k, a, a_{k+2}, \\dots, a_n)$ is the square of an integer.\n\nTo prove (1), notice that, if $a_1 < a_2$ are arbitrarily large positive integers, then $f_2(a_1, a_2)$ is an arbitrarily large positive integer congruent to $0$ or $1$ modulo $4$. Subtraction of a suitable number of $4$'s then yields $0$ or $1$, each of which is a square. Letting that suitable number of $4$'s be $n-2$, which is clearly arbitrarily large, $f_n(a_1, a_2, 1, \\dots, 1)$ is the square of an integer.\n\nTo prove (2), write $f_n(a_1, a_2, \\dots, a_n) = b^2$ and notice that $(a_{k+1}, b)$ solves the Pell equation\n$$\n(a_1^2 \\cdots a_k^2 a_{k+2}^2 \\cdots a_n^2 - 4)x^2 - y^2 = 4(a_1^2 + \\cdots + a_k^2 + a_{k+2}^2 + \\cdots + a_n^2).\n$$\nThe latter has therefore infinitely many solutions in positive integers. In particular, it has a solution $(a, c)$ such that $a > a_{k+1}$. This establishes (2) and concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39021,
    "subject": "Mathematics (Multi-modal)",
    "question": "A nonempty, finite set $A$ of positive integers is called *quadratic* if the sum of the elements of $A$ equals the square of the number of elements of $A$.\nFor example $A = \\{1, 3, 4, 8\\}$ is *quadratic* since $1 + 3 + 4 + 8 = 4^2$.\n\na) Give an example of a *quadratic* set with 20 elements.\n\nb) Prove that every *quadratic* set contains at least one odd integer.\n\nc) Prove that the intersection of two *quadratic* sets with the same number of elements is nonempty.",
    "options": [],
    "answer": "{1, 2, 3, ..., 19, 210}",
    "solution": "a) An example is: $A = \\{1, 2, 3, \\ldots, 19\\} \\cup \\{210\\}$.\nThe sum of elements is $1 + 2 + \\ldots + 19 + 210 = \\frac{19 \\cdot 20}{2} + 210 = 400 = 20^2$.\n\nb) We argue by contradiction. Suppose $A$ is a *quadratic* set containing $n$ even positive integers. The sum of the elements of $A$ is at least $2 + 4 + \\ldots + 2n = n(n+1) > n^2$ which is a contradiction.\n\nc) We prove again by contradiction. Assume there exist two disjoint *quadratic* sets, $A$ and $B$, each having $n$ elements. The total sum of the elements from the two sets is $2n^2$. On the other side the smallest $2n$ distinct positive integers add up to $1 + 2 + \\ldots + 2n = n(2n+1) > 2n^2$, therefore a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Iulia and Ştefan shared the 52 playing cards from a deck¹ so each got 26 cards. The cards from 2 to 10 are assigned their own value, the ace is worth 11 points, the jack 12 points, the queen 13 points and the king 14 points. Ştefan noticed that he had no ace in his stack, no 2 and no four cards of the same value. Iulia noticed that from her cards above 11 points, she doesn't have more than two of the same value. Iulia adds up all her points. What is the lowest value she can get? What is the highest?\n\n¹A deck has 52 cards: four with the value 2, four with the value 3, ..., four with the value 10, 4 aces, 4 jacks, 4 queens and 4 kings.",
    "options": [],
    "answer": "lowest 169; highest 221",
    "solution": "If we denote by $x_i$ the number of cards with the value $i$ that Iulia has in her stack, then $x_i \\ge 1$, $x_{11} = x_4 = 4$, $1 \\le x_{12} \\le 2$, $1 \\le x_{13} \\le 2$, $1 \\le x_{14} \\le 2$.\nIf $M$ and $m$ represent the highest and the lowest value that Iulia's stack can have then:\n$$\n\\begin{align*}\nM &= 2 \\cdot 14 + 2 \\cdot 13 + 2 \\cdot 12 + 4 \\cdot 11 + 4 \\cdot 10 + 2 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 1 \\cdot 5 + 1 \\cdot 4 + 1 \\cdot 3 + 4 \\cdot 2 = 221, \\\\\nm &= 1 \\cdot 14 + 1 \\cdot 13 + 1 \\cdot 12 + 4 \\cdot 11 + 1 \\cdot 10 + 1 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 2 \\cdot 5 + 4 \\cdot 4 + 4 \\cdot 3 + 4 \\cdot 2 = 169.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with a right angle at $C$, let $I$ be its incenter, and let $H$ be the orthogonal projection of $C$ on $AB$. The incircle $\\omega$ of the triangle $ABC$ is tangent to the sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $C$ in the lines $C_1A_1$ and $C_1B_1$, respectively, and let $K$ and $L$ be the reflections of $H$ in the same lines. Prove that the circles $A_1EI$, $B_1FI$, and $C_1KL$ have a common point.\nRussia, Dmitry Prokopenko\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line $C_1A_1$ is parallel to the external angle bisector of $\\angle B$, so the reflection in $C_1A_1$ maps the segment $A_1C$ to the segment $A_1E$ parallel to $AB$.\n\nSimilarly, $B_1F \\parallel AB$. Notice also that $A_1E = A_1C = B_1C = B_1F = r$, where $r$ is the inradius of $\\triangle ABC$.\nLet $M$ be the midpoint of $AB$. Let $X$ be the point of $\\omega$ such that $\\overrightarrow{IX} \\nearrow \\overrightarrow{CM}$. Notice that $\\angle EA_1I = 90^\\circ + \\angle EA_1B = 90^\\circ + \\angle CBM = 90^\\circ + \\angle BCM = \\angle A_1IX$, and $A_1E = IA_1 = IX$; thus, $XIA_1E$ is an isosceles trapezoid. Hence $X$ lies on the circle $IA_1E$, and $EX \\parallel A_1I$. Similarly, $X$ lies on the circle $IB_1F$, and $FX \\parallel B_1I$. It remains to show that $X$ lies on the circle $C_1KL$.\nUnder the symmetry in $C_1A_1$, the line $CH$ (perpendicular to $AB$) maps to the line through $E$ perpendicular to $BC$ — i.e., $CH$ maps to $EX$. Therefore, the projection $H$ of $C_1$ onto $CH$ maps to the projection $K$ of $C_1$ onto $EX$. Similarly, $L$ is the projection of $C_1$ onto $FX$. So the quadrilateral $C_1KXL$ is cyclic, due to right angles at $K$ and $L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39024,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $M \\in (AB)$, $N \\in (BC)$ and $P \\in (CD)$ are chosen on three sides of the rhombus $ABCD$. Prove that the centroid of the triangle $MNP$ belongs to the line $AC$ if and only if $AM + DP = BN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$ be the rhombus' side length. Then one can find $u, v, t \\in (0, 1)$ such that $AM = au$, $BN = av$, and $DP = at$.\n\nDenote by $G$ the centroid of $MNP$ and suppose that the diagonals of the rhombus intersect at $O$. Then $G \\in AC$ if and only if there exists some $k \\in \\mathbb{R}$ such that $\\overline{OG} = k\\overline{OA}$.\n\nOn the other hand,\n$$\n\\begin{align*}\n3\\overline{OG} &= \\overline{OM} + \\overline{ON} + \\overline{OP} \\\\\n&= u\\overline{OB} + (1-u)\\overline{OA} + v\\overline{OC} + (1-v)\\overline{OB} + t\\overline{OC} + (1-t)\\overline{OD} \\\\\n&= (1-u-v-t)\\overline{OA} + (u+1-v-1+t)\\overline{OB} \\\\\n&= (1-u-v-t)\\overline{OA} + (u-v+t)\\overline{OB}.\n\\end{align*}\n$$\nWe conclude that $G \\in AC$ if and only if $u - v + t = 0$, which is equivalent to $AM + DP = BN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB = AC$, let $D$ be the midpoint of the side $AC$, and let $\\gamma$ be the circumcircle of the triangle $ABD$. The tangent of $\\gamma$ at $A$ crosses the line $BC$ at $E$. Let $O$ be the circumcentre of the triangle $ABE$. Prove that the midpoint of the segment $AO$ lies on $\\gamma$.\nUnited Kingdom",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\Gamma$ be the image of $\\gamma$ under the homothety of centre $A$ and factor $2$. Clearly, $\\Gamma$ is also tangent to $AE$ at $A$, and the conclusion is equivalent to $\\Gamma$ passing through $O$, which is the same as $AE$ being tangent to the circle $ACO$.\n\nAlternatively, but equivalently, this amounts to $\\angle OAE = \\angle OCA$. Write $\\angle OAE = 90^\\circ - \\angle EBA$ and $\\angle OCA = \\angle OCB - \\angle ACB = \\angle OCB - \\angle CBA = \\angle OCB - \\angle EBA$, to infer that the equality of the two angles is equivalent to $C$ being the midpoint of the segment $BE$.\n\nTo prove the latter, it is sufficient to show that the triangles $ABE$ and $DCB$ are similar, for then $BE/BC = AB/CD = AC/CD = 2$, which implies that $C$ is indeed the midpoint of the segment $BE$.\n\nFinally, to prove the above similarity, write $\\angle EBA = \\angle CBA = \\angle ACB = \\angle BCD$ and $\\angle BDC = \\angle BAD + \\angle DBA = \\angle BAD + \\angle DAE = \\angle BAE$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39026,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $a$ be a positive integer. Prove that none of the numbers $a^2 + 1, a^2 + 2, \\dots, a^2 + 2a$ is a square.\n\nb) Are there positive integers $m, n, p$ such that\n$$\nm^2 + n + p, \\ n^2 + p + m, \\ p^2 + m + n\n$$\nare all squares?",
    "options": [],
    "answer": "No",
    "solution": "a) The next square after $a^2$ is $(a+1)^2 = a^2 + 2a + 1$, therefore $a^2 + 1, a^2 + 2, \\dots, a^2 + 2a$ fit between two consecutive squares, hence they are not squares.\n\nb) Suppose, by way of contradiction, that such numbers exist. Then $m^2 + n + p > m^2$, hence $m^2 + n + p \\ge m^2 + 2m + 1$. Writing the other two analogous inequalities and adding up yields\n$$\nm^2 + n^2 + p^2 + 2m + 2n + 2p \\ge m^2 + n^2 + p^2 + 2m + 2n + 2p + 3,\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve for $x \\in \\mathbb{R}$:\n$$\n2^{x+1} + \\log_2(1 + \\sqrt{x}) = 4^x + 1.\n$$",
    "options": [],
    "answer": "0, 1",
    "solution": "The equation can be written $\\log_2(1 + \\sqrt{x}) = (2^x - 1)^2$.\nThe function $f : [0, \\infty) \\to [0, \\infty)$, defined by $f(x) = \\log_2(1 + \\sqrt{x})$, is one to one and onto and its inverse $f^{-1} : [0, \\infty) \\to [0, \\infty)$ is given by $f^{-1}(x) = (2^x - 1)^2$.\nAs $f$ is increasing strictly, the equation can be written $f(x) = f^{-1}(x) = x$, or equivalently $2^x = 1 + \\sqrt{x}$. One can easily check that $x_1 = 0$ and $x_2 = 1$ are solutions.\nAs $g : [0, \\infty) \\to [1, \\infty)$, $g(x) = 2^x$ is convex and the function $h : [0, \\infty) \\to [1, \\infty)$, $h(x) = 1 + \\sqrt{x}$ is concave, these are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be real numbers such that:\n$$\nx^2 + 4y^2 + 9z^2 + 20 = 4x + 12y + 24z.\n$$\n\nProve that $x \\in [-1, 5]$, $y \\in [0, 3]$ and $z \\in \\left[\\frac{1}{3}, \\frac{7}{3}\\right]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We rewrite the given equality as:\n$$\n(x-2)^2 + (2y-3)^2 + (3z-4)^2 = 9.\n$$\nThis implies $(x-2)^2 \\le 9$, $(2y-3)^2 \\le 9$ and $(3z-4)^2 \\le 9$, therefore $|x-2| \\le 3$, $|2y-3| \\le 3$, $|3z-4| \\le 3$ and the last inequalities are equivalent to the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that a sequence $(a_n)_{n \\ge 1}$ having the property:\n$$\na_n - \\frac{1}{m} \\le a_m \\le \\frac{m}{n}\n$$\nfor all $m, n \\in \\mathbb{N}^*, m \\ge n$, is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $m = n \\in \\mathbb{N}^*$ we get $a_n \\le 1$, implying that the sequence is upper bounded.\n\nDenote by $L = \\sup\\{a_n \\mid n \\in \\mathbb{N}^*\\}$. We get $L \\le 1$. If $\\varepsilon > 0$ is arbitrary we find $n_0 \\in \\mathbb{N}^*$ such that $L - \\varepsilon < a_{n_0} \\le L$.\n\nConsider $m_0 = \\max\\{n_0, 1 + \\lfloor \\frac{1}{a_{n_0} - L + \\varepsilon} \\rfloor\\}$. For $m \\ge m_0$ we have\n$$\na_m \\ge a_{n_0} - \\frac{1}{m} \\ge a_{n_0} - \\frac{1}{m_0} > L - \\varepsilon,\n$$\nThat is $|a_m - L| < \\varepsilon$ for all $m \\ge m_0$ proving that the sequence has $L$ as limit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be positive integers, and let $G$ be a group of order $n$. Prove that the following two statements are equivalent:\n(a) The numbers $k$ and $n$ are relatively prime.\n(b) For every subgroup $H$ of $G$, the set $\\{x: x \\in G \\text{ and } x^k \\in H\\}$ is contained in $H$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that (a) implies (b). Since $k$ and $n$ are relatively prime, $kp + nq = 1$ for some integers $p$ and $q$. Let $H$ be a subgroup of $G$, and let $x$ be a member of $G$ such that $x^k \\in H$. Since $x^n = e$, the unit of $G$, it follows that\n$$\nx = x^{kp + nq} = (x^k)^p \\cdot (x^n)^q = (x^k)^p \\in H.\n$$\n\nWe now show that (b) implies (a). This is clearly the case if $k = 1$ or $n = 1$, so let them both be at least $2$, and suppose, if possible, they share some prime divisor $p$. By Cauchy's theorem, $x^p = e$ for some $x$ in $G \\setminus \\{e\\}$.\n\nConsider the trivial subgroup $H = \\{e\\}$. Since $k$ is divisible by $p$, it follows that $x^k = e \\in H$, so $x \\in H = \\{e\\}$; that is, $x = e$, contradicting the fact that $x$ lies in $G \\setminus \\{e\\}$. Consequently, $k$ is indeed coprime to $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39031,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of the faces of the rectangular parallelepiped $ABCD'A'B'C'D'$ satisfy the relation:\n$$\nCA^2 = \\frac{2 \\cdot B'A^2 \\cdot D'A^2}{B'A^2 + D'A^2}.\n$$\nProve that $AC' \\le \\sqrt{3} \\cdot AA'$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $AB = a$, $AD = b$, $AA' = c$. By Pythagorean Theorem, we have $a^2 + b^2 = \\frac{2(b^2 + c^2)(c^2 + a^2)}{b^2 + c^2 + c^2 + a^2}$, therefore $a^4 + b^4 = 2c^4$.\nWe also have $4c^4 = 2(a^4 + b^4) \\ge (a^2 + b^2)^2$ thus $2c^2 \\ge a^2 + b^2$.\nIn the end, $C'A = \\sqrt{a^2 + b^2 + c^2} \\le \\sqrt{3c^2} = c\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$, integrable over any bounded interval, satisfying the condition\n$$\n\\int_{x-y}^{x+y} f(t) \\, dt = y(f(x+y) + f(x-y)), \\quad \\text{for all real numbers } x \\text{ and } y.\n$$",
    "options": [],
    "answer": "All affine functions: f(x) = a x + b for real constants a and b.",
    "solution": "It is readily checked that any affine function satisfies the condition in the statement. To prove the converse, set $y = x$ to write\n$$\n\\int_{0}^{2x} f(t) \\, dt = x(f(2x) + f(0)). \\quad (*)\n$$\nSince $f$ is integrable, $(*)$ shows that the function $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = x(f(2x) + f(0))$, is continuous, so $f$ is continuous on $\\mathbb{R}^* = \\mathbb{R} \\setminus \\{0\\}$. Continuity of $f$ on $\\mathbb{R}^*$ and $(*)$ then show $g$ differentiable on $\\mathbb{R}^*$, so $f$ is differentiable on $\\mathbb{R}^*$.\nDifferentiate $(*)$ to get $2f(2x) = f(2x) + f(0) + 2x f'(2x)$; that is, $x f'(x) - f(x) + f(0) = 0$ for all $x \\neq 0$. Consequently,\n$$\n\\left( \\frac{f(x) - f(0)}{x} \\right)' = 0 \\quad \\text{for all } x \\neq 0,\n$$\nso\n$$\nf(x) = \\begin{cases} a x + f(0), & x < 0, \\\\ b x + f(0), & x > 0. \\end{cases}\n$$\nSet now $x = 0$ and $y = 1$ in the relation in the statement to write\n$$\n\\begin{aligned}\nf(1) + f(-1) &= \\int_{-1}^{1} f(t) \\, dt = \\int_{-1}^{0} (a t + f(0)) \\, dt + \\int_{0}^{1} (b t + f(0)) \\, dt \\\\\n&= \\frac{1}{2}(b-a) + 2f(0).\n\\end{aligned}\n$$\nPlug $f(1) = b + f(0)$ and $f(-1) = -a + f(0)$ into the above relation to get $b - a + 2f(0) = \\frac{1}{2}(b-a) + 2f(0)$, so $a = b$ and $f(x) = a x + f(0)$ for all real numbers $x$.\n\n\nAlternative Solution:\nFor convenience, consider the integrable function $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = f(x) - f(0)$, along with the continuous function $G: \\mathbb{R} \\to \\mathbb{R}$, $G(x) = \\int_{0}^{x} g(t) \\, dt$. Clearly, $g$ and $G$ both vanish at the origin, and it is readily checked that\n$$\nG(x+y) - G(x-y) = y(g(x+y) + g(x-y)), \\quad \\text{for all real numbers } x \\text{ and } y. \\quad (1)\n$$\nLet $x = y = t/2$ in (1), and recall that $g$ and $G$ both vanish at the origin, to infer that\n$$\n2G(t) = t g(t), \\quad \\text{for all real numbers } t. \\tag{2}\n$$\nNext, let $x = 0$ in (1) and use (2) to infer that $G$ is an even function: $G(-y) = G(y)$, for all real numbers $y$. With reference again to (2), it follows that $g$ is odd: $g(-y) = -g(y)$, for all real numbers $y$. It is therefore sufficient to focus on the restrictions $g_+$ and $G_+$ of $g$ and $G$, respectively, to the ray of positive real numbers.\nBy (2), continuity of $G_+$ implies that of $g_+$ which in turn implies differentiability of $G_+$ and $G'_+ = g_+$.\nHence $2G_+(t) = t G'_+(t)$ for all real $t > 0$, showing that the function $t \\mapsto t^{-2} G_+(t)$, $t > 0$, is constant; that is, $G_+(t) = \\alpha t^2$ for some real constant $\\alpha$, so $g_+(t) = a t$, where $a = 2\\alpha$.\nConsequently, $g(t) = a t$ for all real numbers $t$, and $f(t) = a t + f(0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Michael is a good chess player. He took part in a competition where all the best chess players of the city were invited. The competition had two stages. After the first stage, looking at the partial results, Michael found that the number of players ranked higher than him was a half of the number of players ranked lower than him. In the second stage Michael played better; He managed to surpass four of the players that were ranked higher than him, after the first stage, but he was also surpassed by two ranked lower than him. So, at the end of the competition, the final results showed that the number of players that were ranked higher than him was a quarter of the number of players that were ranked lower. What place did Michael get at the end of the competition? (Justify your answer!)",
    "options": [],
    "answer": "4th place",
    "solution": "If we denote by $x$ the number of players ranked higher than Michael after the first stage, it turns out that the number of players ranked lower than him will be $2x$.\n\nBecause in the second stage Michael surpassed 4 players that were before him and he was surpassed by 2 players that were after him, he \"advanced\" 2 places, so, at the end, the number of players that were before him is $x - 2$ and the number of the players after him is $2x + 2$.\n\nAs the number of players that were placed before him at the end of the competition is a quarter of the number of the players that were placed after him, then $4(x - 2) = 2x + 2$, from where $x = 5$.\n\nMichael finished the competition in the fourth place.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer, and let $\\mathbf{a} = (a(1), \\dots, a(N))$ and $\\mathbf{b} = (b(1), \\dots, b(N))$ be sequences of non-negative integers, each written on a circle (so we assume $a(i \\pm N) = a(i)$ and $b(i \\pm N) = b(i)$). We say $\\mathbf{a}$ is **b-harmonic**, if each $a(i)$ is the arithmetic mean of the counterclockwise nearest $b(i)$ numbers, the clockwise nearest $b(i)$ numbers, and $a(i)$ itself; that is,\n$$\na(i) = \\frac{1}{2b(i) + 1} \\sum_{s=-b(i)}^{b(i)} a(i + s). \\quad (*)\n$$\n(A term of **a** may appear more than once in the above sum.) Suppose that neither **a** nor **b** is constant, and that both **a** is **b-harmonic**, and **b** is **a-harmonic**. Prove that more than half of the $2N$ terms across both sequences vanish.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = \\min_i a(i)$ and let $b = \\min_i b(i)$. Since **a** is not constant, there exists an $i$ such that $a = a(i) < a(i + 1)$.\n\n**Claim 1.** If $a = a(i) < a(i + 1)$, then $b(i) = 0$. Similarly, if $a = a(i) < a(i - 1)$, then $b(i) = 0$.\n\n*Proof.* Otherwise the sum in $(\\ast)$ contains a term $a(i+1) > a$ but no terms smaller than $a$, so the average is greater than $a$. $\\square$\n\nSay that $[i, j]$ is an **a**-segment if $a(i) = a(i+1) = \\cdots = a(j) = 0$ but $a(i-1) \\neq 0 \\neq a(j+1)$; define a **b**-segment similarly. By Claim 1, the endpoints of any such segment satisfy $a(i) = b(i) = a(j) = b(j) = 0$. Since the sequences are non-constant, each $i$ where $a(i) = 0$ is contained in an **a**-segment.\n\n**Claim 2.** Let $[i, j]$ be a **b**-segment, and let $k \\in [i, j]$. Then $a(k) \\le k - i$ (and, similarly, $a(k) \\le j - k$).\n\n*Proof.* Indeed, since $b(k) = 0$, the elements of **b** with indices from $k-a(k)$ to $k+a(k)$ must all be zero as well. $\\square$\n\nWe now show that every index is contained in either an **a**- or a **b**-segment. Since at least one index is contained in both, the conclusion follows.\n\nAssume, to the contrary, that $a(i)$ and $b(i)$ are both positive for some index $i$; call such indices *bad*. Among all bad indices $i$, choose one maximising $\\max(a(i), b(i))$; by symmetry, we may and will assume that this maximum is $a(i)$. We may and will also assume that either the index $i-1$ is not bad, or $a(i-1) < a(i)$ (otherwise change $i$ to $i-1$, repeat if necessary, recalling that **a** is not constant).\n\nConsider the range of indices $\\Delta = [i - b(i), i + b(i)]$, and the values **a** assumes at those indices. Some indices $j$ in $\\Delta$ are bad; the corresponding values $a(j)$ do not exceed $a(i)$. Other indices $j$ in $\\Delta$ are covered by several **a**- and **b**-segments. Each **b**-segment contributes at most $b(i)$ members nearest to one of its endpoints, so the average value of **a** over those indices does not exceed $(b(i) - 1)/2 < a(i)$ by Claim 2. The remaining indices $j$ in $\\Delta$ all lie in **a**-segments, so the corresponding values $a(j)$ are all zero.\n\nCombining all this, it follows that the average in the right-hand member of $(\\ast)$ does not exceed $a(i)$. Moreover, if some **a**- or **b**-segment intersects $\\Delta$, then the inequality is strict. Otherwise, $i-1$ is a bad index contained in $\\Delta$, and $a(i-1) < a(i)$, so the inequality is again strict. This contradiction ends the proof and completes the solution.\nThe solution has a few well-defined steps:\n\n**Lemma 1.** Assume that $a(i) = M := \\max \\mathbf{a}$; then $b(i+k) = 0$ for all $k = -M, -M+1, \\dots, M$. In particular, $b(i-1) = b(i) = b(i+1) = 0$, as $M \\ge 1$.\n\n*Proof.* Assume that $a(j) = a(j+1) = \\cdots = a(j+s) = M > 0$, and $a(j-1), a(j+s+1) < M$, where $i \\in [j, j+s]$. Then $b(j) = 0$, as otherwise $a(j)$ is the mean of at least three terms, all $\\le M$, with at least one $< M$. For the same reason, $b(j+s) = 0$ also.\nBut then $b(j)$ is the mean of $2M + 1$ terms of $b$, which must therefore all also be equal to 0. So $b(j + k) = 0$ for all $k \\in [-M, M]$. Iterating this argument gives $b(j + k) = 0$ for all $k \\in [-M, M + s]$, which implies the statement of the lemma. $\\square$\n\n**Corollary.** There exist $i$ such that $a(i) = 0$ and $j$ such that $b(j) = 0$.\n\n**Lemma 2.** Suppose $\\max a \\ge \\max b$. Generate $a'$ by replacing all copies of $M = \\max a$ with 1 in $a$. Then $a'$ is $b$-harmonic, and $b$ is $a'$-harmonic.\n\n*Proof.* We start with another consequence of Lemma 1. Assume that $a(i) \\ne M$; then none of the terms $a(i+k)$ with $k \\in [-b(i), b(i)]$ equals $M$. Indeed, if $a(i+k) = M$ with $|k| \\le b(i) \\le M$, then by Lemma 1 we have $b(i) = b((i+k) - k) = 0$, which yields $k=0$ and hence $a(i) = a(i+k) = M$.\nWe can now check that the harmonic properties are preserved under replacing all copies of $M$ in $a$ with 1:\nIf $a(i) \\ne M$, then the harmonic property for $b(i)$ is unchanged. If $a(i) = M$, then $a'(i) = 1$ and $b(i-1) = b(i) = b(i+1) = 0$, so $b(i)$ certainly has the $a'(i)$-harmonic property; and\nIf $a(i) = M$, then $b(i) = 0$, and so $a'(i) = 1$ has the $b(i)$-harmonic property. If $a(i) \\ne M$, then we have just shown that none of the terms in the statement of $a(i)$'s harmonic property are changed by this process, so it remains harmonic. $\\square$\n\n**Lemma 3.** We have $\\min(a(i), b(i)) = 0$ for all $i$. Moreover, there exists an $i$ with $a(i) = b(i) = 0$.\n\n*Proof.* Both statements in the lemma are invariant under the procedure in Lemma 2. Apply this procedure repeatedly, to replace all instances of the maximum value in one of the sequences with 1, until both sequences consist of zeroes and ones. It suffices to check the lemma statement for the obtained pair of sequences.\nSuppose that $a(i) = b(i) = 1$ for some $i$. Since the sequences remain non-constant, we may and will assume that $\\min(a(i-1), b(i-1)) = 0$, say $a(i-1) = 0$. But then the $b(i)$-harmonic property is violated for $a(i)$, as $a(i+1) \\le 1$.\nSuppose now that there is no $i$ with $a(i) = b(i) = 0$. This means that for every index $i$ we have either $a(i) = 1$ and $b(i) = 0$, or $a(i) = 0$ and $b(i) = 1$. There is a pair of adjacent indices having different types, so that $a(i) = b(i + 1) = 1$ and $a(i + 1) = b(i) = 0$. But then $b(i)$ violates the $a(i)$-harmonic property. $\\square$\n\nLemma 3 readily yields that at least $N+1$ terms across both sequences are zeroes, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two congruent squares $ABCD$ and $EFGH$ are placed such that they have disjoint interiors, but $C$ is the midpoint of the line segment $EF$ and the points $B$, $F$, $G$ are collinear. The line $BC$ intersects $EH$ at $K$ and the line $AC$ intersects $GH$ at $M$. Let $L$ be the midpoint of $GH$, and let the parallel through $K$ to $GH$ intersect $FG$ at $N$.\n\na) Prove that $CK = CL = CN$.\nb) Prove that $LM = 2HK$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Clearly, triangle $CFB$ is a right triangle, with $\\angle F = 90^\\circ$, hence it is similar to $CEK$. Moreover, since $CF = CE$ the two triangles are congruent, therefore $CK = CB = CL$. Using the symmetry of $EFGH$ across $CL$ it also follows that $CK = CN$.\n\nb) In the right triangle $CBF$ we have $CB = 2CF$, therefore $\\angle CBF = 30^\\circ$. But $\\angle CBF = \\angle CKE = \\angle KCL$ so the base angles in the isosceles triangle $CKL$ are both $75^\\circ$. It follows that $HKL$ is a right triangle with a $15^\\circ$ angle. In the triangle $LMC$ we also have $\\angle LCM = \\angle KCM - \\angle KCL = \\angle ACB - 30^\\circ = 15^\\circ$.\n\nThus, $HKL$ and $LMC$ are similar, and since $LC = 2HL$, it follows that $LM = 2HK$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that there are functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(f(x) + y f(y)) = x + (f(y))^n$, for any $x, y \\in \\mathbb{R}$, if and only if $n = 2$. Determine those functions.\n\nCătălin Zârnă",
    "options": [],
    "answer": "Solutions exist only when n equals 2. In that case, the functions are f(x) = x for all real x or f(x) = −x for all real x.",
    "solution": "For $y = 0$ we have $f(f(x)) = x + (f(0))^n$, for any $x \\in \\mathbb{R}$, implying that $f$ is bijective. Let $a$ such that $f(a) = 0$. For $y = a$ we deduce $f(f(x)) = x$, for all real $x$, so $f(0) = 0$.\n\nFor $x = 0$ we get $f(y f(y)) = f(y)^n$, for real $y$. Substituting $y$ with $f(y)$, we get $f(y)^n = y^n$.\n\ni) For odd $n$ we get $f(y) = y$, which gives a contradiction in the given relation.\n\nii) For even $n$ we obtain $f(y) \\in \\{-y, y\\}$, for any $y \\in \\mathbb{R}$. Thus $y^n \\in \\{-y^2, y^2\\}$ for all real $y$, which simply imply $n = 2$. Thus $f(f(x) + y f(y)) = x + y^2$, for $x, y \\in \\mathbb{R}$.\n\nIt is easy to check that the identity function and minus the identity function, verify the relation. We shall show that they are the only ones. If not, if there are $a, b \\neq 0$ such that $f(a) = a$ and $f(b) = -b$, than for $x = a, y = b$ we should have $f(a - b^2) = a + b^2 \\in \\{a - b^2, -a + b^2\\}$, that is $a = 0$ or $b = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $a, b \\in \\mathbb{N}^*$. Define the sequence $(x_n)_{n \\in \\mathbb{N}}$ by $x_0 = 0, x_1 = 1$ and\n$$\nx_{n+2} = a x_{n+1} + b x_n, \\quad \\forall n \\in \\mathbb{N}\n$$\nand the matrix $A$, by\n$$\nA = \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix}.\n$$\n\na) Prove that\n$$\nA^n = \\begin{bmatrix} x_{n+1} & b x_n \\\\ x_n & b x_{n-1} \\end{bmatrix}, \\quad \\forall n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We shall use induction on $n$.\nAs $x_2 = a \\cdot 1 + b \\cdot 0 = a$, we have\n$$\nA = \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} x_2 & b x_1 \\\\ x_1 & b x_0 \\end{bmatrix},\n$$\nso the property is true for $n = 1$.\nConsider now that the formula is true for some $n \\ge 1$. Recurrently we then get\n$$\nA^{n+1} = A^n \\cdot A = \\begin{bmatrix} x_{n+1} & b x_n \\\\ x_n & b x_{n-1} \\end{bmatrix} \\cdot \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} x_{n+2} & b x_{n+1} \\\\ x_{n+1} & b x_n \\end{bmatrix},\n$$\nending the induction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote $\\mathbb{Z}_{>0} = \\{1, 2, 3, \\dots\\}$ the set of all positive integers. Determine all functions $f : \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}$ such that, for each positive integer $n$:\ni) $\\sum_{k=1}^{n} f(k)$ is a perfect square;\nii) $f(n)$ divides $n^3$.",
    "options": [],
    "answer": "f(n) = n^3 for all positive integers n",
    "solution": "Induct on $n$ to show that $f(n) = n^3$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.\nLet $n \\ge 2$ and assume that $f(m) = m^3$ for all positive integers $m < n$. Then $\\sum_{k=1}^{n-1} f(k) = \\frac{n^2(n-1)^2}{4}$, and reference to the first condition in the statement yields\n$$\nf(n) = \\sum_{k=1}^{n} f(k) - \\sum_{k=1}^{n-1} f(k) = \\left( \\frac{n(n-1)}{2} + k \\right)^2 - \\frac{n^2(n-1)^2}{4} = k(n^2 - n + k),\n$$\nfor some positive integer $k$.\nThe divisibility condition in the statement implies $k(n^2 - n + k) \\le n^3$, which is equivalent to $(n-k)(n^2+k) \\ge 0$, showing that $k \\le n$.\nOn the other hand, $n^2 - n + k$ must also divide $n^3$. But, if $k < n$, then\n$$\nn < \\frac{n^3}{n^2 - 1} \\le \\frac{n^3}{n^2 - n + k} \\le \\frac{n^3}{n^2 - n + 1} < \\frac{n^3 + 1}{n^2 - n + 1} = n + 1,\n$$\ntherefore $\\frac{n^3}{n^2-n+k}$ cannot be an integer.\nConsequently, $k=n$, so $f(n) = n^3$. This completes induction and concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer greater than $1$ and let $(a_n)_{n \\ge 1}$ be a sequence of pairwise distinct positive integers. Show that the set $M = \\{a_1^k, a_2^k, a_3^k, \\dots\\}$ does not contain an infinite arithmetic sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, by way of contradiction, that $M$ contains the infinite arithmetic sequence $\\{b_1^k, b_2^k, b_3^k, \\dots\\}$, where $1 \\le b_1 < b_2 < b_3 < \\dots$.\nIf we denote by $r$ its common difference, we have $b_{n+1}^k - b_n^k = r$, for all $n \\ge 1$.\nObserve that\n$$\nr = b_{n+1}^k - b_n^k = (b_{n+1} - b_n) (b_{n+1}^{k-1} + b_{n+1}^{k-2}b_n + \\dots + b_{n+1}b_n^{k-2} + b_n^{k-1}).\n$$\nThe expression in the second parenthesis is clearly increasing, hence the sequence $b_{n+1} - b_n$ is decreasing, and since all its terms are positive integers, we reached a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39040,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive, odd integer $n$, prove that\n$$\n\\left[ \\frac{1}{2} + \\sqrt{n + \\frac{1}{2}} \\right] = \\left[ \\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}} \\right],\n$$\nwhere $[a]$ denotes the integer part of the real number $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove by contradiction that there is no integer between $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}}$ and $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2}}$.\n\nSuppose there exists $k \\ge 1$ such that $\\frac{1}{2} + \\sqrt{n + \\frac{1}{2}} \\ge k > \\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}}$. We obtain $n \\ge k^2 - k - \\frac{1}{4} = x_k$ and $n < k^2 - k + \\frac{126}{505} = y_k$.\n\nSince the only integer between $x_k$ and $y_k$ is $k^2 - k$, we must have $n = k^2 - k = k(k - 1)$ which contradicts $n$ odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39041,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country there are $n$ airports and $n$ air companies operating return flights. Each company operates an odd number of flights forming a closed route. Prove that a traveller can complete a closed route consisting of an odd number of flights operated by pairwise distinct companies.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In graph-theoretic setting, the statement reads:\nConsider a collection of $n$ odd cycles, not necessarily distinct, all on the same vertex set of size $n$. Prove that at most one edge can be chosen from each of these cycles to form a collection that contains the edges of an odd cycle.\nCall a set of edges *rainbow* if it is formed by choosing at most one edge from each cycle. We have to prove that there exists a rainbow cycle of odd length.\nBegin by choosing a maximal rainbow forest $F$, that is, an acyclic rainbow set of edges.\nSince $F$ is acyclic, its size is less than $n$, so there is a cycle $C$ in the collection no edge of which lies in $F$. The forest $F$ contains every vertex of $C$, for otherwise an edge of $C$ incident with a vertex outside $F$ could be added to $F$ to form a larger rainbow forest, contradicting maximality. Moreover, no edge of $C$ joins different components of $F$, for one such could again be added to $F$ to contradict maximality.\nConsequently, the vertices of $C$ all lie in some component of $F$, a tree $T$. As such, $T$ is bipartite, that is, its vertices split into two disjoint parts, and all edges are between the two. Since $C$ is an odd cycle, it has an edge whose endpoints both lie in the same part. This edge then completes an odd rainbow cycle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Determine the least integer $n \\ge k + 1$ for which the game below can be played indefinitely:\n\nConsider $n$ boxes, labelled $b_1, b_2, \\dots, b_n$. For each index $i$, box $b_i$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in order:\n\n(1) Choose $k+1$ boxes;\n\n(2) Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_i$, a number of $i$ coins.\n\n(3) If one of the boxes is left empty, the game ends; otherwise, go to the next step.",
    "options": [],
    "answer": "n = 2^k + k - 1",
    "solution": "The required minimum is $n = 2^k + k - 1$.\n\nIn this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^k-1}, b_{2^k}, \\dots, b_{2^k+k-1}$, at each step: At step $r$, if box $b_{2^k+i-1}$ has exactly $m_i$ coins, then $\\lceil m_i/2 \\rceil$ coins are removed from that box, unless $i \\equiv r-1 \\pmod{k+1}$, in which case $2^k + i - 1$ coins are added. Thus, after step $r$ has been performed, box $b_{2^k+i-1}$ contains exactly $\\lceil m_i/2 \\rceil$ coins, unless $i \\equiv r-1 \\pmod{k+1}$, in which case it contains exactly $m_i + 2^k + i - 1$ coins. This game goes on indefinitely, since each time a box is supplied, at least $2^k - 1$ coins are added, so it will then contain at least $2^k$ coins, good enough to survive the $k$ steps to its next supply.\n\nWe now show that no smaller value of $n$ works. So, let $n \\le 2^k + k - 2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w = \\lceil \\log_2 m \\rceil$ withdrawals; this $w$ will be referred to as the *weight* of that box. The sum of the weights of all boxes will referred to as the *total weight*. The argument hinges on the lemma below, proved at the end of the solution.\n\n**Lemma.** *Performing a step does not increase the total weight. Moreover, supplying one of the first $2^k - 2$ boxes strictly decreases the total weight.*\n\nSince the total weight cannot strictly decrease indefinitely, $n > 2^k - 2$, and from some stage on none of the first $2^k - 2$ boxes is ever supplied. Recall that each step involves a $(k+1)$-box choice. Since $n \\le 2^k + k - 2$, from that stage on, each step involves a withdrawal from at least one of the first $2^k - 2$ boxes. This cannot go on indefinitely, so the game must eventually come to an end, contradicting the assumption.\n\nConsequently, a game that can be played indefinitely requires $n \\ge 2^k + k - 1$.\n\n**Proof of the Lemma.** Since a withdrawal from a box decreases its weight by at least 1, it is sufficient to show that supplying a box increases its weight by at most $k$; and if the latter is amongst the first $2^k - 2$ boxes, then its weight increases by at most $k-1$. Let the box to be supplied be $b_i$ and let it currently contain exactly $m_i$ coins, to proceed by case analysis:\n\nIf $m_i = 1$, the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\le \\lfloor \\log_2(2^k + k - 1) \\rfloor \\le \\lfloor \\log_2(2^{k+1} - 2) \\rfloor \\le k$; and if, in addition, $i \\le 2^k - 2$, then the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\le \\lfloor \\log_2(2^k - 1) \\rfloor = k - 1$.\n\nIf $m_i = 2$, then the weight increases by $\\lfloor \\log_2(i+2) \\rfloor - \\lfloor \\log_2 2 \\rfloor \\le \\lfloor \\log_2(2^k + k) \\rfloor - 1 \\le k - 1$.\n\nIf $m_i \\ge 3$, then the weight increases by\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\le \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\le \\lfloor \\log_2 \\left( 1 + \\frac{2^k + k - 2}{3} \\right) \\rfloor + 1 \\le k,\n\\end{aligned}\n$$\nsince $1 + \\frac{1}{3}(2^k + k - 2) = \\frac{1}{3}(2^k + k + 1) < \\frac{1}{3}(2^k + 2^{k+1}) = 2^k$.\n\nFinally, let $i \\le 2^k - 2$ to consider the subcases $m_i = 3$ and $m_i \\ge 4$. In the former subcase, the weight increases by\n$$\n\\lfloor \\log_2(i + 3) \\rfloor - \\lfloor \\log_2 3 \\rfloor \\le \\lfloor \\log_2(2^k + 1) \\rfloor - 1 = k - 1,\n$$\nand in the latter by\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\le \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\le \\lfloor \\log_2 \\left( 1 + \\frac{2^k - 2}{4} \\right) \\rfloor + 1 \\le k - 1,\n\\end{aligned}\n$$\nsince $1 + \\frac{1}{4}(2^k - 2) = \\frac{1}{4}(2^k + 2) < 2^{k-2} + 1$. This ends the proof and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the numbers $\\overline{abc}$, with $a < b < c$, knowing that the remainders of divisions of the numbers $\\overline{abc}, \\overline{bca}, \\overline{cab}$ by 27 belong to the set $\\{1, 2, 3, 4, 5\\}$.",
    "options": [],
    "answer": "138, 246, 489, 678",
    "solution": "The numbers $\\overline{abc}, \\overline{bca}, \\overline{cab}$ have the same sum of digits therefore they will have the same remainder $r$ when divided by 9.\nSince the remainders modulo 27 are small, they are preserved modulo 9. Indeed if $n = 27k + r$, then $n = 9 \\cdot 3k + r$, so $r$ will be a common remainder.\n27 must divide the difference $\\overline{bca} - \\overline{abc} = 90b + 9c - 99a = 9(a + b + c) + 81b - 108a = 9(a + b + c) + 27(3b - 4a)$, therefore $3 \\mid a+b+c$ which implies $r = 3$.\n\nThe numbers $abc$, with $a < b < c$ and $a+b+c = 12$ are 129, 138, 147, 156, 237, 246 and 345. Only 138 and 246 have the remainder 3 modulo 27.\nThe numbers $abc$, with $a < b < c$ and $a + b + c = 21$ are 489, 579 and 678. Convenient are 489 and 678.\nIn the end we have 4 solutions: 138, 246, 489 and 678.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ having the property that, for all $x, y \\in \\mathbb{R}$, there exist $t \\in (0, 1)$ such that\n$$\nf((1-t)x + ty) = (1-t)f(x) + tf(y).\n$$",
    "options": [],
    "answer": "All affine functions f(x) = m x + n for real m, n.",
    "solution": "We shall prove that the answer is $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = mx + n$, with $m, n \\in \\mathbb{R}$. It is trivial to see that this family of functions satisfy the property.\n\nFor the converse, let $f$ with the given property. Consider $a, b \\in \\mathbb{R}$, $a < b$, and define\n$$\nm = \\frac{f(b) - f(a)}{b - a} \\quad \\text{and} \\quad n = \\frac{bf(a) - af(b)}{b - a}.\n$$\nWe shall show that $f(x) = mx + n$, for all $x \\in [a, b]$. By way of contradiction, let $x_0 \\in [a, b]$, such that $f(x_0) \\neq mx_0 + n$. Consider the sets\n$$\nA = \\{x \\in [a, x_0] \\mid f(x) = mx + n\\} \\quad \\text{and} \\quad B = \\{x \\in [x_0, b] \\mid f(x) = mx + n\\}.\n$$\n$A$ and $B$ are nonempty because $a \\in A$ and $b \\in B$. Consider the numbers $\\alpha = \\sup(A)$ and $\\beta = \\inf(B)$. By the continuity of $f$ we get $\\alpha \\in A$, $\\beta \\in B$ and $\\alpha < x_0 < \\beta$. By the hypothesis, there is $t \\in (0, 1)$ such that\n$$\nf((1-t)\\alpha + t\\beta) = (1-t)f(\\alpha) + tf(\\beta) = m((1-t)\\alpha + t\\beta) + n.\n$$\nThen $\\gamma = (1-t)\\alpha + t\\beta \\in (\\alpha, \\beta)$ and $f(\\gamma) = m\\gamma + n$, in contradiction with the definition of $\\alpha$ and $\\beta$. So, $f(x) = mx + n$, for all $x \\in [a, b]$.\n\nTo extend the answer to $\\mathbb{R}$, observe that if $f : [a, c] \\to \\mathbb{R}$, $c > b$ is a function obtained from the preceding considerations, say $f(x) = m'x + n'$, we must have by continuity $m = m'$, $n = n'$. The same remark can be made on intervals of the form $[c, b]$ with $c < a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine complex numbers $z$ verifying $z^{3n} + \\bar{z}^{3n} \\ge 0$, for any non-negative integer $n$.",
    "options": [],
    "answer": "{ z ∈ ℂ : z = r e^{i 2πk/3}, r ≥ 0, k ∈ {0,1,2} }",
    "solution": "Put $z^{3n} + \\bar{z}^{3n} = f_n(z)$. As $f_n(z) = \\overline{f_n(z)}$, we get $f_n(z) \\in \\mathbb{R}$, for all $z \\in \\mathbb{C}$. We should determine the set $M = \\{z \\in \\mathbb{C} / f_n(z) \\ge 0\\}$. If $z \\in M$, then $f_0(z) = z + \\bar{z} \\ge 0$, so $\\text{Re}(z) \\ge 0$.\n\nFor non-negative $z \\in \\mathbb{R}$, $f_n(z) = 2z^{3n} \\ge 0$, for all non-negative integers $n$, implying $[0, \\infty) \\subset M$.\n\nIf $z = bi \\in \\mathbb{C}$, $b \\in \\mathbb{R}$, then $\\bar{z} = -z$, so $f_n(z) = 0$, for $n \\in \\mathbb{N}$, so $\\{bi / b \\in \\mathbb{R}\\} \\subset M$. As $f_n(z) = f_n(\\bar{z})$, we have $z \\in M \\Leftrightarrow \\bar{z} \\in M$.\n\nIt remains to find $z \\in M$ such that $\\text{Re}(z) > 0$ and $\\text{Im}(z) > 0$. Let $z = r(\\cos t + i \\sin t)$ with $r > 0$, $t \\in (0, \\frac{\\pi}{2})$. Then $f_n(z) = 2r^{3n} \\cos(3^n t)$, so $z \\in M \\Leftrightarrow \\cos(3^n t) \\ge 0$, for all $n \\in \\mathbb{N}$.\n\nAs $t \\in (0, \\frac{\\pi}{2})$, there is $k \\in \\mathbb{N}$ such that $3^k t \\le \\frac{\\pi}{2} < 3^{k+1}t$ (in fact $k = \\lfloor \\log_3 \\frac{\\pi}{2t} \\rfloor$). If $3^k t = \\frac{\\pi}{2}$, then $\\cos(3^n t) > 0$, for all $n \\in \\{0, 1, 2, \\dots, k-1\\}$ and $\\cos(3^n t) = 0$, for all $n \\ge k$, that is $z \\in M$. If $3^k t < \\frac{\\pi}{2} < 3^{k+1}t$, then $\\frac{\\pi}{2} < 3^{k+1}t < \\frac{3\\pi}{2}$, implying $\\cos(3^{k+1}t) < 0$, that is $z \\notin M$.\n\nIn conclusion\n$$\nM = [0, \\infty) \\cup \\left\\{ r \\left( \\cos \\frac{\\pi}{2 \\cdot 3^k} \\pm i \\sin \\frac{\\pi}{2 \\cdot 3^k} \\right) / r > 0, k \\in \\mathbb{N} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle in which $\\angle ABC = 75^\\circ$ and $\\angle BAC = 45^\\circ$. One considers the points $F$, $X$ and $Y$ such that $F$ is the projection of $B$ on $AC$, $\\overline{CX} = \\frac{1}{2}\\overline{BC}$ and $\\overline{BY} = \\overline{FX} + \\overline{FA}$. Prove that the centroid of triangle $ABX$ lies on the line segment $YF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that $YF$ is the Euler line of the triangle $ABX$, hence it contains the centroid of that triangle.\n\nBecause $BFA$ is a right isosceles triangle, it follows that $BF = AF$. In the right triangle $BFC$ we have $\\angle CBF = 30^\\circ$, hence $FC = \\frac{BC}{2}$, thus $FC = CX$, and a short computation shows that $\\angle BXF = 30^\\circ$. We deduce that $BF = FX$, hence $AF = BF = FX$, making $F$ the circumcenter of $ABX$.\n\nLet $Y'$ be the orthocenter of $ABX$. Applying Sylvester's relation yields $\\overline{FY'} = \\overline{FA} + \\overline{FB} + \\overline{FX} = \\overline{FB} + \\overline{BY} = \\overline{FY}$. It follows that $Y' = Y$, hence $Y$ is the orthocenter of $ABX$. We conclude that $FY$ is the Euler line of $ABX$ and since the centroid lies between the circumcenter and the orthocenter, we obtain the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b, c$ such that $2^a + 2^b + 2^c + 3$ is a square.",
    "options": [],
    "answer": "a = b = c = 1, or {a, b, c} = {1, 2, 4}",
    "solution": "We can assume $a \\le b \\le c$. If $a \\ge 2$, then $2^a + 2^b + 2^c + 3 \\equiv 3 \\pmod{4}$, hence it cannot be a square. We deduce that $a = 1$, and we must find $b \\le c$ such that $2^b + 2^c + 5$ is a square.\n\nIf $b \\ge 3$, then $2^b + 2^c + 5 \\equiv 5 \\pmod{8}$, again not a square. It follows that $b$ equals either $1$ or $2$.\n\nWhen $b = 1$, we look for $c$ such that $2^c + 7$ is a square. Obviously $c = 1$ works and it is the only possible value, since for $c \\ge 2$ we have $2^c + 7 \\equiv 3 \\pmod{4}$.\n\nFinally, when $b = 2$ we look for $c$ such that $2^c + 9 = k^2$, for some positive integer $k$. This can be written as $2^c = (k-3)(k+3)$, so we must have $k-3 = 2^\\alpha$, $k+3 = 2^\\beta$, where $\\alpha < \\beta$ are positive integers. Subtracting yields $2^\\alpha(2^{\\beta-\\alpha} - 1) = 6$, which implies $\\alpha = 1$ and then $c = 4$.\n\nWe conclude that the solutions are $a = b = c = 1$ and $\\{a, b, c\\} = \\{1, 2, 4\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $R$ be a ring with only finitely many invertible elements. Prove that the two statements below are equivalent:\n(a) For every non-invertible element $x$ of $R$, there exists a non-invertible element $y$ of $R$ such that $xy = x + y$.\n(b) Every non-invertible element of $R$ is nilpotent.\n(An element $z$ of $R$ is called *nilpotent* if $z^k = 0$ for some positive integer $k$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that (a) implies (b). Let $x \\in R \\setminus U(R)$, where $U(R)$ is the set of invertible elements of $R$, and let $y \\in R \\setminus U(R)$ be such that $xy = x + y$, i.e., $(x - 1)(y - 1) = 1$. Similarly, $(y - 1)(z - 1) = 1$ for some $z \\in R \\setminus U(R)$. Write\n$$\nx - 1 = (x - 1) \\cdot 1 = (x - 1)(y - 1)(z - 1) = 1 \\cdot (z - 1) = z - 1,\n$$\nto infer that $x = z$. Hence $x - 1 \\in U(R)$, showing that $\\{t - 1 : t \\in R \\setminus U(R)\\} \\subseteq U(R)$.\nSince $x \\in R \\setminus U(R)$, its powers are all different from $1$, and finiteness of the multiplicative group $U(R)$ then forces them all in $R \\setminus U(R)$. Hence $x^n - 1 \\in U(R)$ for all positive integers $n$, by the set inclusion in the previous paragraph. With reference again to finiteness of $U(R)$, $x^p - 1 = x^q - 1$ for some distinct positive integers $p$ and $q$, say, $p < q$. Then $x^p(x^{q-p} - 1) = 0$, so $x^p = 0$, on account of $x^{q-p} - 1 \\in U(R)$. Consequently, $x$ is indeed nilpotent.\n\nWe now show that (b) implies (a). Let $x \\in R \\setminus U(R)$ and choose an integer $p \\ge 3$ such that $x^p = 0$. Write $1 = 1 - x^p = (1-x)(1+x+x^2+\\cdots+x^{p-1})$ and let $y = -(x+x^2+\\cdots+x^{p-1}) = -x(1+x+\\cdots+x^{p-2})$. Then $y^p = 0$, so $y \\in R \\setminus U(R)$. Clearly, $(1-x)(1-y) = 1$, i.e., $xy = x+y$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 = 2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1} = 3a_n$ only for finitely many indices $n$.\n\nNorth Macedonia",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Begin with a mere remark on the terms of the sequence under consideration.\n\n**Lemma 1.** Each $a_n$ is minimal amongst all positive integers having the same number of positive divisors as $a_n$.\n\n*Proof.* Suppose, if possible, that for some $n$, some positive integer $b < a_n$ has as many positive divisors as $a_n$. Then $a_m < b \\le a_{m+1}$ for some $m < n$, and the definition of the sequence forces $b = a_{m+1}$. Since $b < a_n$, it follows that $m + 1 < n$, which is a contradiction, as $a_{m+1}$ should have less positive divisors than $a_n$. $\\square$\n\nLet $p_1 < p_2 < \\dots < p_n < \\dots$ be the strictly increasing sequence of prime numbers, and write canonical factorisations into primes in the form $N = \\prod_{i \\ge 1} p_i^{e_i}$, where $e_i \\ge 0$ for all $i$, and $e_i = 0$ for all but finitely many indices $i$; in this notation, the number of positive divisors of $N$ is $\\tau(N) = \\prod_{i \\ge 1} (e_i + 1)$.\n\n**Lemma 2.** *The exponents in the canonical factorisation of each $a_n$ into primes form a non-strictly decreasing sequence.*\n\n*Proof.* Indeed, if $e_i < e_j$ for some $i < j$ in the canonical decomposition of $a_n$ into primes, then swapping the two exponents yields a smaller integer with the same number of positive divisors, contradicting Lemma 1. $\\square$\n\nWe are now in a position to prove the required result. For convenience, a term $a_n$ satisfying $3a_n = 2a_{n+1}$ will be referred to as a *special* term of the sequence.\n\nSuppose now, if possible, that the sequence has infinitely many special terms, so the latter form a strictly increasing, and hence unbounded, subsequence. To reach a contradiction, it is sufficient to show that:\n\n(1) The exponents of the primes in the factorisation of special terms have a common upper bound $e$; and\n\n(2) For all large enough primes $p$, no special term is divisible by $p$.\n\nRefer to Lemma 2 to write $a_n = \\prod_{i \\ge 1} p_i^{e_i(n)}$, where $e_i(n) \\ge e_{i+1}(n)$ for all $i$.\n\nStatement (2) is a straightforward consequence of (1) and Lemma 1. Suppose, if possible, that some special term $a_n$ is divisible by a prime $p_i > 2^{e+1}$, where $e$ is the integer provided by (1). Then $e \\ge e_i(n) > 0$, so $2^{e_1(n)e_i(n)+e_i(n)} a_n / p_i^{e_i(n)}$ is a positive integer with the same number of positive divisors as $a_n$, but smaller than $a_n$. This contradicts Lemma 1. Consequently, no special term is divisible by a prime exceeding $2^{e+1}$.\n\nTo prove (1), it is sufficient to show that, as $a_n$ runs through the special terms, the exponents $e_1(n)$ are bounded from above. Then, Lemma 2 shows that such an upper bound $e$ suits all primes.\n\nConsider a large enough special $a_n$. The condition $\\tau(a_n) < \\tau(a_{n+1})$ is then equivalent to $(e_1(n) + 1)(e_2(n) + 1) < e_1(n)(e_2(n) + 2)$. Alternatively, but equivalently, $e_1(n) \\ge e_2(n) + 2$. The latter implies that $a_n$ is divisible by 8, for either $e_1(n) \\ge 3$ or $a_n$ is a large enough power of 2.\n\nNext, note that $9a_n/8$ is an integer strictly between $a_n$ and $a_{n+1}$, so $\\tau(9a_n/8) \\le \\tau(a_n)$, which is equivalent to\n$$\n(e_1(n) - 2)(e_2(n) + 3) \\le (e_1(n) + 1)(e_2(n) + 1),\n$$\nso $2e_1(n) \\le 3e_2(n) + 7$. This shows that $a_n$ is divisible by 3, for otherwise, letting $a_n$ run through the special terms, 3 would be an upper bound for all but finitely many $e_1(n)$, and the special terms would therefore form a bounded sequence.\n\nThus, $4a_n/3$ is another integer strictly between $a_n$ and $a_{n+1}$. As before, $\\tau(4a_n/3) \\le \\tau(a_n)$. Alternatively, but equivalently,\n$$\n(e_1(n) + 3)e_2(n) \\le (e_1(n) + 1)(e_2(n) + 1),\n$$\nso $2e_2(n) - 1 \\le e_1(n)$. Combine this with the inequality in the previous paragraph to write $4e_2(n) - 2 \\le 2e_1(n) \\le 3e_2(n) + 7$ and infer that $e_2(n) \\le 9$. Consequently, $2e_1(n) \\le 3e_2(n) + 7 \\le 34$, showing that $e = 17$ is suitable for (1) to hold. This establishes (1) and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}$ be the set of all positive integers. A subset $A$ of $\\mathbb{N}$ is *sum-free* if, whenever $x$ and $y$ are (not necessarily distinct) members of $A$, their sum $x+y$ does not belong to $A$. Determine all *surjective* functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that, for each sum-free subset $A$ of $\\mathbb{N}$, the image $\\{f(a): a \\in A\\}$ is again sum-free.\nIndia, Sutanay Bhattacharya",
    "options": [],
    "answer": "f(n) = n for all n in N",
    "solution": "The identity is the only surjection of the positive integers onto themselves sending every sum-free set onto a sum-free set (no verification is needed, of course).\nTo prove this, fix a function $f$ satisfying the conditions in the statement, and proceed in several steps.\n\n*Step 1.* Notice that a 2-element set $\\{x, y\\}$, where $x < y$, is *not* sum-free if and only if $y = 2x$.\n\nChoose any $a \\in \\mathbb{N}$, and for any $i \\ge 0$ choose some $x_i$ such that $f(x_i) = 2^i a$. The set $f(\\{x_i, x_{i+1}\\})$ is not sum-free, so neither is $\\{x_i, x_{i+1}\\}$, whence $x_i = 2x_{i+1}$ or $x_{i+1} = 2x_i$. Since the $x_i$ are all distinct, the same option should hold for all $i$. The former option yields $x_i = x_0 2^{-i}$ which cannot hold for large enough $i$. So $x_{i+1} = 2x_i$ for all $i$.\nTherefore, $f(2x) = 2f(x)$ for all $x$, and, moreover, $x$ is the only argument $t$ with $f(t) = f(2x)/2$. Therefore, $f$ is injective (and hence bijective).\n\n**Step 2.** Say that a 3-element set $\\{a, b, c\\}$ is good if it is not sum-free, but each of its 2-element subsets is (in other words, no element is twice another). It is easily seen that a set $\\{a, b, c\\}$, where $a < b$, is good only if $c = b \\pm a$. Notice that the pre-image of a good set is also a good set, due to Step 1.\nNow let $f(1) = a$. We show that $f(n) = an$ by induction on $n$. The base cases are $n=1, 2, 3, 4, 5$; for $n=1, 2, 4$ the result follows from Step 1.\nSet $t = f^{-1}(3a)$ and $s = f^{-1}(5a)$. The sets $\\{a, 4a, 3a\\}$ and $\\{a, 4a, 5a\\}$ are good, hence so are $\\{1, 4, t\\}$ and $\\{1, 4, s\\}$. Therefore, $\\{s, t\\} = \\{3, 5\\}$. But the set $\\{a, 5a, 6a\\}$ is also good, so the pair $\\{1, s\\}$ is contained in one more good set, which is not the case if $s=3$, since $\\{1, 3\\}$ is contained in one single good set, namely, $\\{1, 4, 3\\}$. Thus $t=3$ and $s=5$, which establishes the base.\nFor the induction step, assume that $f(k) = ak$ for all $k \\le n$, where $n \\ge 5$. Choose $t = f^{-1}((n+1)a)$. Then the pair $\\{a, na\\}$ is contained in two good sets, namely, $\\{a, na, (n-1)a\\}$ and $\\{a, na, (n+1)a\\}$. Their pre-images, $\\{1, n, n-1\\}$ and $\\{1, n, t\\}$, are also good, and injectivity of $f$ forces $t = n+1$. This completes the induction step.\nFinally, since $f$ is surjective, $1 = f(n) = an$ for some positive integer $n$, so $a=1=n$. Consequently, $f$ is the identity, as claimed at the beginning of the solution.\nThe approach is similar to Solution 1, but avoids directly defining and working with good sets. Step 1 in Solution 1 is proved similarly.\nFor any odd integer $a \\ge 3$, we claim $f(a) + f(1)$ is equal to one of $f(a+1)$ and $f(a-1)$. Indeed, its preimage should, together with $a$ and $1$, form a set that is not sum-free. $f(a)+f(1) = f(2a) = 2f(a)$ contradicts injectivity, as does $f(a)+f(1) = f(2) = 2f(1)$, hence $f(a)+f(1) = f(a+1)$ or $f(a)+f(1) = f(a-1)$.\n\n$f(2) = 2f(1)$, contradicting injectivity. Therefore, $f(a) + f(1) = f(a + 1)$ for all odd integers $a$.\nFinally, we prove $f(n) = nf(1)$ by induction. Indeed, assume the statements holds for all integers up to some even $n$. The base case for $n \\in \\{1, 2\\}$ holds by Step 1 of Solution 1. Then, by assumption,\n$$\nf(n + 1) + f(1) = f(n + 2) = 2f\\left(\\frac{n+2}{2}\\right) = 2\\left(\\frac{n+2}{2}\\right)f(1) = (n+2)f(1),\n$$\nimplying $f(n + 1) = (n + 1)f(n)$ and, by the above result,\n$$\nf(n + 2) = f(n + 1) + f(1) = (n + 2)f(1),\n$$\ncompleting the induction.\n\nAssume that $f(a)+f(1) = f(a-1)$ for some odd integer $2$. Inducting backwards, we see the statement holds for all odd integers up to $a$, in particular $f(3)+f(1) =$\n\nThe conclusion now follows as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39051,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n \\in \\mathbb{N}$, $n \\ge 2$, consider $A$ a matrix $n \\times n$ with complex entries, such that $A^2 = \\text{tr}(A) \\cdot A$. Prove that the matrices $ABA$ and $ACA$ commute, for any $n \\times n$ matrices $B$ and $C$, with complex entries.\n\nMihai Opincaru",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $\\text{tr}(A) = 0$ implies $A^2 = O_{n \\times n}$ and\n$$\nABA \\cdot ACA = O_{n \\times n} = ACA \\cdot ABA,\n$$\n\nIf $\\text{tr}(A) \\neq 0$, we show that $\\text{rank}(A) = 1$. Let $r = \\text{rank}(A)$. Then we can find matrices $X \\in \\mathcal{M}_{n \\times r}(\\mathbb{C})$ and $Y \\in \\mathcal{M}_{r \\times n}(\\mathbb{C})$ with $\\text{rank}(X) = \\text{rank}(Y) = r$ such that $A = XY$. Then $YX \\in \\mathcal{M}_r(\\mathbb{C})$ and, as $\\text{tr}(A) \\neq 0$, we have\n$$\nr \\geq \\text{rank}(YX) \\geq \\text{rank}(X(YX)Y) = \\text{rank}(A^2) = \\text{rank}(A) = r,\n$$\nThat is $\\text{rank}(YX) = r$ and the matrix $YX$ is non-singular. Equality $A^2 = \\text{tr}(A) \\cdot A$, can be written $(X)^2 = \\text{tr}(A) \\cdot XY$, obtaining $(YX)^3 = \\text{tr}(A) \\cdot (YX)^2$. As $YX$ is non-singular, we get $YX = \\text{tr}(A) \\cdot I_r$. In conclusion,\n$$\n\\text{tr}(A) = \\text{tr}(XY) = \\text{tr}(YX) = \\text{tr}(A) \\cdot r,\n$$\ngiving $r = 1$. To conclude, let $B, C \\in \\mathcal{M}_n(\\mathbb{C})$. We have $ABA = XYBXY = pXY = pA$, where $p = YBX \\in \\mathcal{M}_1(\\mathbb{C})$. In the same way $ACA = qA$ with $q \\in \\mathbb{C}$, concluding the result of the problem.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a non-negative integer, and let $f: [-1, 1] \\to \\mathbb{R}$ be a twice differentiable function vanishing at the origin. Assuming $f''$ continuous, show that\n$$\n(2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx = f''(c), \\quad \\text{for some } c \\text{ in the closed interval } [-1, 1].\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\int_{-1}^{1} x^{2n} f(x) \\, dx = \\int_{-1}^{1} x^{2n} g(x) \\, dx + \\int_{-1}^{1} x^{2n} h(x) \\, dx = 2 \\int_{0}^{1} x^{2n} g(x) \\, dx.\n$$\nSince $f(0) = 0$ and $g'(x) = \\frac{1}{2}(f'(x)-f'(-x))$, it follows that $g(0) = g'(0) = 0$; and since $g''(x) = \\frac{1}{2}(f''(x)+f''(-x))$ and $f''$ is continuous, so is $g''$.\nLet $m = \\min_{0 \\le x \\le 1} g''(x)$ and let $M = \\max_{0 \\le x \\le 1} g''(x)$. By Lagrange's theorem, for each $t$ in $(0, 1]$, there exists a $c_t$ in $(0, t)$ such that $g'(t) = t g''(c_t)$, so $mt \\le g'(t) \\le Mt$.\nIntegrate the latter over the interval $[0, x]$, where $0 < x \\le 1$, to get $\\frac{1}{2}mx^2 \\le g(x) \\le \\frac{1}{2}Mx^2$, so $\\frac{1}{2}mx^{2n+2} \\le x^{2n}g(x) \\le \\frac{1}{2}Mx^{2n+2}$ for all $x$ in $[0, 1]$.\nIntegration of the latter over $[0, 1]$ yields\n$$\n\\frac{m}{2(2n+3)} = \\int_{0}^{1} \\frac{1}{2}mx^{2n+2}dx \\le \\int_{0}^{1} x^{2n}g(x)dx \\le \\int_{0}^{1} \\frac{1}{2}Mx^{2n+2}dx = \\frac{M}{2(2n+3)},\n$$\n$$ \\text{so} $$\n$$ m \\le 2(2n+3) \\int_{0}^{1} x^{2n} g(x) dx \\le M, \\quad \\text{i. e.,} \\quad m \\le (2n+3) \\int_{-1}^{1} x^{2n} f(x) dx \\le M. $$\n\nSince $g''$ has the intermediate value property (Darboux),\n$$\ng''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } b \\text{ in } [0, 1].\n$$\nFinally, since $g''(b)$ lies between $f''(-b)$ and $f''(b)$, and $f''$ has the intermediate value property (Darboux),\n$$\nf''(c) = g''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } c \\text{ in } [-b, b] \\subseteq [-1, 1].\n$$\n*Alternative Solution.* Dropping the condition that $f''$ be continuous, we establish the required equality for some $c$ in the open interval $(-1, 1)$.\nLetting $m = \\inf_{|x|\\le 1} f''(x)$ and $M = \\sup_{|x|\\le 1} f''(x)$, we show that\n$$\nm \\le (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx \\le M. \\quad (*)\n$$\nEquality holds at one end if and only if it holds at the other, in which case $f$ is a degree 2 polynomial function vanishing at the origin; otherwise, both inequalities are strict. Since $f''$ has the intermediate value property (Darboux), the conclusion follows: The case where $f$ is a degree 2 polynomial function vanishing at the origin is clear — any point in the open interval $(-1, 1)$ will do. Otherwise, choose $\\alpha$ and $\\beta$ such that\n$$\nm < \\alpha < (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx < \\beta < M.\n$$\nThen $\\alpha = f''(a)$ and $\\beta = f''(b)$ for some $a$ and $b$ in $[-1, 1]$. Clearly, $a$ and $b$ are distinct, so $(2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx = f''(c)$, for some $c$ strictly between $a$ and $b$; that is, $c$ lies in the open interval $(-1, 1)$.\nTo prove $(*)$, consider the 3-term power expansion of $f$ at the origin:\n$$\nf(x) = f(0) + f'(0) \\cdot x + \\frac{1}{2} f''(t) \\cdot x^2 = f'(0) \\cdot x + \\frac{1}{2} f''(t) \\cdot x^2,\n$$\nfor some $t$ between 0 and $x$. Thus, $\\frac{1}{2}mx^2 \\le f(x) - f'(0) \\cdot x \\le \\frac{1}{2}Mx^2$ for all $x$ in $[-1, 1]$. Multiplication by $(2n + 3)x^{2n} \\ge 0$ and integration over $[-1, 1]$ yields $(*)$.\nTo complete the argument, assume equality at the left end; the other case is dealt with similarly. Then $m$ is finite, and $x \\mapsto f(x) - f'(0) \\cdot x - \\frac{1}{2}mx^2$, $|x| \\le 1$, is a non-negative continuous function whose integral over $[-1, 1]$ vanishes. Consequently, this function vanishes identically, so $f$ is a degree 2 polynomial function vanishing at the origin, and equality holds at the other end as well. This completes the argument and concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39053,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the cube $ABCD EFGH$ with side $a$ cm, $a > 0$ and the points $K$, $L$ on the segments $AC$, respectively $EG$ such that $CK = EL = \\frac{AC}{4}$. Let $M$ be the midpoint of $AE$ and the points $N$, $P$ on the segment $HF$ such that\n$$\nHN = FP = \\frac{(\\sqrt{2} - 1)a}{2} \\text{ cm.}\n$$\nKnowing that lines $MC$ and $LK$ intersect in $Q$, prove that $MNPQ$ is a regular tetrahedron.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Triangles $\\Delta MEP$ and $\\Delta MEN$ are congruent, therefore $MN = MP$.\nDenote by $O_1$ the midpoint of $HF$. It follows that $MO_1 = \\frac{AC}{2} = \\frac{a\\sqrt{3}}{2}$ and, since $NP = a$, the triangle $MNP$ must be equilateral.\nLet $O$ be the midpoint of $AC$, $M_1$ the midpoint of $AO$ and $\\{L_1\\} = LK \\cap MO_1$. Since $MM_1 \\parallel EO \\parallel LK \\parallel CO_1$, we have $ML_1/MO_1 = M_1K/M_1C = 2/3$ so $L_1$ is the centroid of the triangle $MNP$.\n![](attached_image_1.png)\n\nNotice that $AL = AG = \\frac{3\\sqrt{2a}}{4}$ thus $ALGK$ is a rhombus, which implies $AG \\perp KL$. We also have $CO_1LK$ parallelogram, therefore $LK \\parallel CO_1$ and $KL \\perp MO_1$. $HF \\perp (ACGE)$ so $KL \\perp HF$ which implies $KL \\perp (MNP)$. It follows that the pyramid $MNPQ$ is regular.\nFrom the similarity of the triangles $\\Delta ML_1Q$ and $\\Delta MO_1C$ we have $QL_1 = \\frac{a\\sqrt{6}}{3}$ forcing the tetrahedron $MNPQ$ to be regular.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a real number $a \\ge 1$. The sequence $(x_n)_{n \\ge 1}$ is given by $x_1 = a$ and $x_{n+1} = 1 + \\log_2 x_n$, for any $n \\in \\mathbb{N}^*$. Determine values of $a$ such that all terms of the sequence are rational numbers.",
    "options": [],
    "answer": "a = 1 or a = 2",
    "solution": "For $a = 1$ or $a = 2$ we get the constant sequence $1$ or $2$ respectively.\nWe shall prove that these are the only values satisfying the problem.\nTo see this, let $x_n = \\frac{k}{l}$ and $x_{n+1} = \\frac{p}{q}$, with $k, l$ coprime positive integers, $p, q$ coprime positive integers with $q \\ge 2$. Then\n$$\n\\frac{k}{l} = x_n = 2^{x_{n+1}-1} = \\sqrt[q]{2^{p-q}},\n$$\nimplying $l^q 2^p = k^q 2^q$.\nThe last equality cannot be true, as the left hand side number has $2$ as factor at an exponent which is not divisible by $q$ ($p$ and $q$ are coprime), and the right hand side has the factor $2$ at an exponent divisible by $q$. So $q = 1$, thus $l = 1$, so, if all terms of the sequence would be rationals, they should be natural numbers.\nAs $2^n > n + 1$, for $n \\ge 2, n \\in \\mathbb{N}$, we get $1 + \\log_2 x < x$, for $x \\in \\mathbb{N}, x \\ge 3$. As a conclusion, if all terms are positive integers and $x > 2$, an inductive argument shows that the sequence is decreasing, a contradiction. That is $a = x_1 \\in \\{1, 2\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $M = \\{1, 2, 3, \\dots, n^2 + n + 2\\}$. We consider the subsets $A_1, A_2, \\dots, A_n$ of $M$, such that for each $k \\in \\{1, 2, \\dots, n\\}$ the set $A_k$ has $n^2 + k + 1$ elements. Prove that the intersection of the $n$ subsets contains at least two consecutive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $A_n$ has $n^2 + n + 1$ elements, hence contains all the elements of $M$ except one. Similarly, $A_{n-1}$ contains all the elements of $M$ except $2$, $\\dots$, $A_1$ contains all the elements of $M$, except $n$ of them. We deduce that the intersection $A := \\bigcap_{k=1}^n A_k$ contains all the elements of $M$, except at most $1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$. It follows that\n$$\n|A| \\ge n^2 + n + 2 - \\frac{n(n + 1)}{2} = \\frac{n^2 + n + 4}{2}.\n$$\nLet $A = \\{x_1, x_2, \\dots, x_p\\}$, with $1 \\le x_1 < x_2 < \\dots < x_p \\le n^2 + n + 2$. Assume, by contradiction, that $A$ does not contain consecutive integers. Then $x_{i+1} \\ge x_i + 2$, $i = 1, 2, \\dots, p-1$, and we conclude that\n$$\nx_p \\ge x_1 + 2(p-1) \\ge 1 + 2 \\left( \\frac{n^2 + n + 4}{2} - 1 \\right) = n^2 + n + 3,\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39056,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $A$ has 73 digits, all different from zero. Prove that we can erase 64 digits such that the new number is divisible by 37.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "After the elimination of the 64 digits the final number has $73 - 64 = 9$ digits.\nIf in the composition of $A$, every digit appears at most 8 times, then $A$ would have at most $9 \\cdot 8 = 72$ digits, which is false.\nTherefore, there is a digit different from 0, let that be $a$, that appears 9 times in $A$. Then we can erase 64 digits from $A$ such that the final number would have the form $\\overline{aaaaaaaaa} = a \\cdot 111111111 = a \\cdot 1001001 \\cdot 3 \\cdot 37$, so the number is divisible by 37.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point in the interior of triangle $ABC$ such that $\\angle BAD = 40^\\circ$, $\\angle DAC = 30^\\circ$, $\\angle BCD = 20^\\circ$ and $\\angle DCA = 50^\\circ$. Find $\\angle CBD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "10°",
    "solution": "Observe that $ABC$ is an isosceles triangle, with $BA = BC$ and $\\angle ABC = 40^\\circ$. Suppose that the perpendicular bisector of the triangle's base intersects $AD$ at $T$.\nSince $\\angle DAC < \\angle DCA$, we have $CD < DA$ therefore $T \\in (AD)$ and $\\angle TBC = 20^\\circ$.\n\nWe will prove that $D$ is the incenter of triangle $BTC$, hence $BD$ bisects $\\angle TBC$, and it follows that $\\angle CBD = 10^\\circ$.\n\nBecause $TCA$ is an isosceles triangle, we have $\\angle TCA = 30^\\circ$, and hence $\\angle TCD = 20^\\circ = \\angle BCD$. It follows that $CD$ bisects $\\angle BCT$.\n\nA short computation shows that $\\angle CTB = \\angle BTA = 120^\\circ$ and $\\angle DTC = 60^\\circ$, therefore $TD$ is the angle bisector of $\\angle BTC$.\n\nWe conclude that $D$ is the incenter of triangle $BCT$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39058,
    "subject": "Mathematics (Multi-modal)",
    "question": "A computer program reads the numbers displayed, identifies the selected ones and does one of the following actions:\n- If $A$ is pressed, every selected number is changed to its successor;\n- If $B$ is pressed, every selected number is changed to its triple.\n\nOn Andrei and Bogdan's computer screens are displayed the numbers $1$, $3$, $3^2$, $3^3$, ..., $3^{19}$.\n\na) Andrei will select $5$ numbers and then presses $A$, this is a type $A$ step.\nDetermine if there is a succession of type $A$ steps in order to obtain the sum of all displayed numbers to be equal to $2019^{2020}$.\n\nb) Bogdan will select $5$ numbers and then presses $B$, this is a type $B$ step.\nWhat is the minimum number of type $B$ steps needed in order to obtain all the displayed numbers to be equal?",
    "options": [],
    "answer": "a) No; b) 38",
    "solution": "a) At every step of type $A$, the sum of the displayed numbers increases by $5$. The remainder modulo $5$ of the sum is invariant.\nThe sum of four consecutive powers of $3$ is $3^n + 3^{n+1} + 3^{n+2} + 3^{n+3} = 3^n(1 + 3 + 9 + 27) = 40 \\cdot 3^n$, therefore initially we have the remainder zero modulo $5$ and it is impossible to obtain $2019^{2020}$.\n\nb) At every step the product of the displayed numbers is multiplied by $3^5$.\nInitially the product is $3^{1+2+...+19} = 3^{190}$ and after $n$ type $B$ steps the product will be equal to $3^{190+5n}$.\nSuppose that in $n$ steps the numbers are equal to a power of $3$, at least $3^{19}$. The product of the $20$ numbers will be $3^{20p}$, where $p \\ge 19$.\nWe must have $3^{20p} = 3^{190+5n}$, so $n + 38 = 4p$. Since $p \\ge 19$, we have $n \\ge 38$.\nNext we provide a succession of $38$ type $B$ steps to display in the end equal numbers.\n\nIn $15$ steps, the numbers $1$, $3$, $3^2$, $3^3$, $3^4$ become $3^{15}$, $3^{16}$, $3^{17}$, $3^{18}$, $3^{19}$. In the next $10$ steps $3^5$, $3^6$, $3^7$, $3^8$, $3^9$ become $3^{15}$, $3^{16}$, $3^{17}$, $3^{18}$, $3^{19}$ and the numbers $3^{10}$, $3^{11}$, $3^{12}$, $3^{13}$, $3^{14}$ become $3^{15}$, $3^{16}$, $3^{17}$, $3^{18}$, $3^{19}$ in $5$ steps.\n\nThus in $30$ steps numbers $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$ are displayed each $4$ times.\nNumbers $3^{15}, 3^{15}, 3^{15}, 3^{15}, 3^{16}$ become $3^{18}, 3^{18}, 3^{18}, 3^{18}, 3^{19}$ after three steps.\nNumbers $3^{16}, 3^{16}, 3^{16}, 3^{17}, 3^{17}$ become $3^{18}, 3^{18}, 3^{18}, 3^{19}, 3^{19}$ in $2$ steps.\nNumbers $3^{17}, 3^{17}, 3^{18}, 3^{18}, 3^{18}$ become $3^{18}, 3^{18}, 3^{19}, 3^{19}, 3^{19}$ in one step.\nTherefore after $36$ steps on Bogdan's computer screen are displayed $10$ of $3^{19}$ and $10$ of $3^{18}$.\nIn last $2$ steps we make all numbers equal to $3^{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39059,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *lattice point* in the Cartesian plane is a point whose coordinates are both integral. A *lattice polygon* is a polygon whose vertices are lattice points. Let $\\Gamma$ be a convex lattice polygon. Prove that $\\Gamma$ is contained in a convex lattice polygon $\\Delta$ exactly one vertex of which is not a vertex of $\\Gamma$, and the vertices of $\\Gamma$ all lie on the boundary of $\\Delta$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the extra vertex of a desired polygon $\\Delta$; then $\\Delta$ is the convex hull of $T$ and $\\Gamma$. Thus, a point $T$ fits the bill if and only if this convex hull contains no vertices of $\\Gamma$ in its interior.\n\nEach segment $AB$ joining two lattice points is partitioned by lattice points into congruent *elementary segments*. Define the *elementary length* $\\ell(AB)$ of $AB$ to be the length of any of those elementary segments.\nTake any three consecutive sides $AB$, $BC$, and $CD$ on the boundary of $\\Gamma$. Consider the convex region bounded by the segment $BC$ and the rays complementary to the rays $BA$ and $CD$ (including the boundary). If this region is unbounded (see the left figure below), then it contains some lattice point $T$ (e.g., the point with $\\overrightarrow{BT} = \\overrightarrow{AB}$), and any such point $T$ satisfies the problem requirements. Thus, in what follows we assume that the two rays cross each other at some point $X$. Assume further that the triangle $BCX$ contains no lattice points outside the segment $BC$, as any other such point would satisfy the requirements.\n\nLet $C_*$ be a lattice point on the segment $BC$ such that $BC_* = \\ell(BC)$, let $X_*$ be the point on $BX$ such that $C_*X_* \\parallel CX$, and let $D_*$ be the point such that $\\overrightarrow{C_*D_*} = \\overrightarrow{CD}$ (see the right figure below). Then the triangle $BC_*X_*$ contains no lattice points apart from $B$ and $C_*$.  \nConsider the half-plane determined by the line $BC$ and containing no interior points of $\\Gamma$. Let $\\ell$ be the line in that half-plane parallel to $BC$, containing some lattice points, and nearest to $BC$ among such. Let the ray $AB$ meet $\\ell$ at a point $A'$ which belongs to the elementary segment $KL$ on $\\ell$ (we assume that $\\overrightarrow{KL} = \\overrightarrow{BC_*}$; the point $A'$ may coincide with $L$ but not with $K$). Then the ray $D_*C_*$ crosses the ray $LK$ (excluding $L$), otherwise $L$ lies in the triangle $BC_*X_*$.  \nThe only lattice points contained in the parallelogram $BKLC_*$ are its vertices. This yields that there are no lattice points strictly inside the strip defined by the parallel lines $BK$ and $C_*L$.\n\n![](attached_image_1.png)\n\nLet $M$ and $N$ be the meeting points of the rays $BX_*, C_*L$ and $C_*X_*, BK$, respectively. Then the segments $BM$ and $C_*N$ contain no lattice points except their endpoints, so $\\ell(AB) \\ge BM$ and $\\ell(DC) = \\ell(D_*C_*) \\ge C_*N$. Therefore,\n$$\n\\frac{BX_*}{\\ell(AB)} + \\frac{C_*X_*}{\\ell(CD)} \\le \\frac{BX_*}{BM} + \\frac{C_*X_*}{C_*N} = \\frac{BX_*}{BM} + \\frac{MX_*}{BM} = 1. \\quad (*)\n$$\nChoose now $BC$ to be a side of largest elementary length. Then\n$$\n(1 \\ge) \\frac{BX_*}{\\ell(AB)} + \\frac{C_*X_*}{\\ell(CD)} \\ge \\frac{BX_* + C_*X_*}{\\ell(BC)} = \\frac{BX_* + C_*X_*}{BC_*},\n$$\nwhich contradicts the triangle inequality.\n\n\nREMARKS. (1) The usage of elementary length seems to be crucial. In particular, under the assumption that the triangle $BCX$ contains no lattice points outside $BC$, an inequality\n$$\n\\frac{BX}{AB} + \\frac{CX}{CD} \\le 1\n$$\nsimilar to (*) does not necessarily hold.\n(2) Any lattice parallelogram of unit area may be transformed into a unit square by an affine transform preserving the set of lattice points. If one applies such a transform to the parallelogram $BKLC_*$, inequality (*) may become more transparent.\nWe use the same notions of *elementary segments* and *elementary lengths* as in Solution 1. We say that a segment is *sloped* if it is neither horizontal nor vertical. If $\\Gamma$ has no sloped sides, then $\\Gamma$ is a rectangle, and the problem statement is trivial: one may choose any lattice point on the extension of any side. So, in what follows, we assume that $\\Gamma$ has a sloped side.\n\n**Lemma 1.** Let $BC$ be a sloped segment of the boundary of $\\Gamma$ with no lattice points in its interior, and let $P$ be a lattice point such that the segments $PB$ and $PC$ are not sloped, and the triangle $PBC$ lies outside $\\Gamma$. Then there exists a unique lattice point $X$ in the triangle $PBC$ such that the area of the triangle $XBC$ is $1/2$.\n*Proof.* Choose a lattice point $Q$ such that $PBQC$ is a rectangle. As in Solution 1, let $\\ell$ denote the line through some lattice point parallel to $BC$, lying outside $\\Gamma$, and nearest to the line $BC$ under these constraints (see the left figure below). The line $\\ell$ crosses the interior of the angle $BQC$ along an interval of length $> BC$, so this interval should contain a lattice point $X$. The triangle $XBC$ contains no lattice points apart from the vertices, so its area is $1/2$ due to Pick's formula. Moreover, any such point $X$ should lie within the angle $BPC$, and $\\ell$ crosses this angle along a segment of length $< BC$. Hence $X$ is the required unique lattice point. $\\square$\n\n![](attached_image_2.png)\n\nDenote the point $X$ defined in Lemma 1 by $f(AB)$.\n\n**Lemma 2.** Let $AB$ and $BC$ be two consecutive sides on the boundary of $\\Gamma$, and let $BA_*$ and $BC_*$ be elementary segments on the sides $BA$ and $BC$, respectively. Assume that both coordinates of $\\overrightarrow{BC}$ are positive, and that the line $AB$ strictly separates the points $C$ and $X = f(BC_*)$. Then both coordinates of the vector $\\overrightarrow{A_*B}$ are also positive, and $BA_* > BC_*$.  \n*Proof.* Since $AB$ separates $X$ and $C$, the vector $\\overrightarrow{A_*B}$ is a linear combination of $\\overrightarrow{c} = \\overrightarrow{BC_*}$ and $\\overrightarrow{x} = \\overrightarrow{BX}$ with positive coefficients; so the coordinates of $\\overrightarrow{A_*B}$ are positive. Since the area of the triangle $BXC$ equals $1/2$, the vectors $\\overrightarrow{c}$ and $\\overrightarrow{x}$ span the whole lattice, so the coefficients of the linear combination are integers. Finally, the angle between $\\overrightarrow{c}$ and $\\overrightarrow{x}$ is acute, so $BA_* = |\\overrightarrow{A_*B}| \\ge \\sqrt{|\\overrightarrow{c}|^2 + |\\overrightarrow{x}|^2} > |\\overrightarrow{c}| = BC_*$, as desired. $\\square$\n\nNow, choose a sloped side $BC$ of $\\Gamma$ of a maximal elementary length, and let $ABCD$ be the corresponding part of the boundary of $\\Gamma$. Let $C_*$ and $B_*$ be the points on the segment $BC$ such that $BC_* = B_*C = \\ell(BC)$. Let $X = f(BC_*)$ and $X' = f(B_*C)$. Then, due to Lemma 2, the line $AB$ does not separate $X$ and $C$, and the line $CD$ does not separate $X'$ and $B$. Therefore, the segment $XX'$ lies in the same angle of the lines $AB$ and $CD$ as $\\Gamma$, so $X$ may serve as a suitable vertex $T$ of $\\Delta$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39060,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $n > 1$, let $gpf(n)$ denote the greatest prime factor of $n$. A *strange pair* is an unordered pair of distinct primes $p$ and $q$ such that $\\{p, q\\} = \\{gpf(n), gpf(n + 1)\\}$ for no integer $n > 1$. Prove that there exist infinitely many strange pairs.\n\nRussia, Dmitry Krachun",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that there are infinitely many strange pairs of the form $\\{2, q\\}$ where $q$ is an odd prime.\n\nThe Lemma below provides a sufficient condition for such a pair to be strange. For an odd prime $q$, let $ord_q(2)$ denote the multiplicative order of $2$ modulo $q$, i.e., the least positive integer $s$ satisfying $q \\mid 2^s - 1$.\n\n**Lemma.** If some primes $2 < q_1 < q_2$ satisfy $ord_{q_1}(2) = ord_{q_2}(2)$, then $\\{2, q_1\\}$ is a strange pair.\n\n*Proof.* Arguing indirectly, suppose first that $2 = gpf(n)$ and $q_1 = gpf(n+1)$; in particular, $n = 2^k$ for some positive integer $k$, and $q_1 \\mid 2^k+1$. This yields $q_1 \\mid 2^{2k}-1$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid 2k$. Therefore, $q_2 \\mid 2^{2k}-1 = (2^k-1)(2^k+1)$, but $q_2 \\nmid 2^k-1$, hence $q_2 \\mid 2^k+1$. So $gpf(n+1) \\ge q_2$, which is a contradiction.\n\nSimilarly, but easier, if $2 = gpf(n+1)$ and $q_1 = gpf(n)$, then $n+1 = 2^k$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid k$ and hence $q_2 \\mid 2^k-1$. Therefore, $gpf(n+1) \\ge q_2$, a contradiction. $\\square$\n\nLet $p = 2r - 1 > 5$ be a prime, and let $N = 2^{2p} + 1$. We prove that:\n\n(1) $N$ has at least two distinct prime factors greater than $5$;\n\n(2) $ord_q(2) = 4p$ for every prime factor $q > 5$ of $N$.\n\nThus, every prime $p > 5$ provides a pair of odd primes satisfying the conditions in the Lemma. Moreover, (2) shows that distinct primes $p > 5$ provide disjoint such pairs, whence the conclusion.\n\nTo prove (1), notice that $3 \\nmid N$, and write $N = (4+1)(4^{p-1}-4^{p-2}+\\cdots+1) \\equiv 5p \\pmod{25}$, to infer that $25 \\nmid N$.\n\nNext, write $N = (2^p+1)^2 - 2^{p+1} = (2^p - 2^r + 1)(2^p + 2^r + 1)$. The two factors are coprime (since they are odd, and their difference is $2^{r+1}$), and each is larger than $5$. Hence each has a prime factor greater than $5$. This establishes (1).\n\nTo prove (2), consider a prime factor $q > 5$ of $N$, and notice that $ord_q(2) \\mid 4p$, since $q \\mid N \\mid 2^{4p}-1$. If $ord_q(2) < 4p$, then either $ord_q(2) \\mid 2p$ or $ord_q(2) \\mid 4$. The former is impossible due to $2^{2p}-1 = N-2 \\equiv -2 \\pmod q$, the latter — due to $q \\nmid 15 = 2^4-1$. This establishes (2) and completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Scrieți numărul $2021$ ca sumă de puteri distincte cu baza $(-2)$.\n\nb) Arătați că numărul $2021$ nu se poate scrie ca sumă de puteri distincte cu baza $(-3)$.",
    "options": [],
    "answer": "a) 2021 = (-2)^12 + (-2)^11 + (-2)^5 + (-2)^2 + (-2)^0. b) Impossible to write 2021 as a sum of distinct powers with base -3.",
    "solution": "Solution:\n\na)\n$2021 = (-2)^{12} + (-2)^{11} + (-2)^{5} + (-2)^{2} + (-2)^{0}$\n\nb)\n$(-3)^{k} = M_{3}$, pentru orice $k \\geq 1$.\nO sumă de puteri distincte cu baza $(-3)$ ar putea avea ca termen pe $(-3)^{0}$ sau nu. Prin urmare, o sumă de puteri distincte cu baza $(-3)$ poate avea următoarele forme:\n$M_{3}$ sau $M_{3} + (-3)^{0} = M_{3} + 1$.\nCum $2021 = M_{3} + 2 \\Rightarrow 2021$ nu se poate scrie ca sumă de puteri distincte cu baza $(-3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $\\triangle ABC$ cu $m(\\angle A) > 90^\\circ$. Considerăm $BE$ bisectoarea $\\angle ABC$, $E \\in AC$ și $CF$ bisectoarea $\\angle ACB$, $F \\in AB$, $BE \\cap CF = \\{I\\}$. Știind că $IE = IF$, arătați că $\\triangle ABC$ este isoscel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $IN \\perp AB$ și $IM \\perp AC$.\n\nDeoarece $I$ este centrul cercului înscris $\\triangle ABC \\Rightarrow AI$ este bisectoarea $\\angle BAC \\Rightarrow IN = IM$.\n\n$[IF] \\equiv [IE]$ și $[IN] \\equiv [IM] \\stackrel{I.C.}{\\Rightarrow} \\triangle IFN \\equiv \\triangle IEM \\Rightarrow \\angle IFN \\equiv \\angle IEM \\Rightarrow \\angle IFB \\equiv \\angle IEC$\n\n$\\left\\{\\begin{array}{c}\n\\angle IFB \\equiv \\angle IEC \\\\\n[IF] \\equiv [IE] \\\\\n\\angle FIB \\equiv \\angle EIC\n\\end{array}\\right\\} \\stackrel{\\text{U.L.U.}}{\\Longrightarrow} \\triangle IFB \\equiv \\triangle IEC \\Rightarrow$\n\n$\\Rightarrow [IB] \\equiv [IC] \\Rightarrow [BE] \\equiv [CF] \\Rightarrow \\triangle ABC$ isoscel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Scrieți numărul $2021$ ca sumă de puteri distincte cu baza $(-2)$.\n\nb) Arătați că numărul $2021$ nu se poate scrie ca sumă de puteri distincte cu baza $(-3)$.",
    "options": [],
    "answer": "a) 2021 = (-2)^12 + (-2)^11 + (-2)^5 + (-2)^2 + (-2)^0.\nb) Impossible: any sum of distinct powers of −3 is congruent to 0 or 1 modulo 3, while 2021 is congruent to 2 modulo 3.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAflaţi ultima cifră a numărului natural $n$, ştiind că penultima cifră a lui $n^{2}$ este 9.",
    "options": [],
    "answer": "4 or 6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră triunghiul ascuţitunghic $ABC$. Fie $O$ centrul cercului circumscris acestuia şi $D$ piciorul înălţimii din $A$. Ştiind că $OD \\parallel AB$, arătaţi că $\\sin 2B = \\operatorname{ctg} C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi numerele complexe $x, y, z$, de acelaşi modul, ştiind că numerele $x+y+z$ şi $x^{3}+y^{3}+z^{3}$ sunt reale.",
    "options": [],
    "answer": "All solutions are those with a common modulus r, where after a permutation two of the numbers are complex conjugates and the third is real of absolute value r; equivalently, x = ± r, y = r e^{i t}, z = r e^{-i t} for some real r ≥ 0 and real t.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nArătaţi că pentru orice $a, b, c > 0$ cu $a + b + c = 1$ are loc inegalitatea:\n$$\n\\frac{1}{a + b c} + \\frac{1}{b + c a} + \\frac{1}{c + a b} \\geqslant \\frac{7}{1 + a b c}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminați funcțiile continue $f:[0,1] \\longrightarrow [0, \\infty)$ care verifică relația\n$$\n\\int_{0}^{1} f(x) \\, dx \\cdot \\int_{0}^{1} f^{2}(x) \\, dx \\cdots \\int_{0}^{1} f^{2020}(x) \\, dx = \\left(\\int_{0}^{1} f^{2021}(x) \\, dx\\right)^{1010}\n$$",
    "options": [],
    "answer": "All solutions are the constant functions f(x) = c with c ≥ 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminaţi toate perechile $(m, n)$ de numere naturale impare pentru care\n$$\nn \\mid 3m+1 \\text{ şi } m \\mid n^{2}+3.\n$$",
    "options": [],
    "answer": "(1,1), (43,13), (49,37)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlina şi Bogdan scriu pe rând, începând cu Alina, un $0$ sau un $1$, până când fiecare a scris $2021$ de cifre (fiecare adaugă o cifră la dreapta celor deja existente).\nAlina este câştigătoare dacă reprezentarea zecimală a numărului obţinut (în baza $2$) se poate scrie ca suma a două pătrate perfecte; în caz contrar, câştigă Bogdan.\nStabiliţi care dintre cei doi are o strategie de câştig.",
    "options": [],
    "answer": "Bogdan",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $f:[a, b] \\rightarrow \\mathbb{R}$ o funcție cu proprietatea lui Darboux astfel ca $f(a) \\cdot f(b)<0$. Arătați că există $\\alpha, \\beta$ astfel ca $a<\\alpha<\\beta<b$ și $f(\\alpha)+f(\\beta)=f(\\alpha) \\cdot f(\\beta)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\in \\mathbb{N}, n \\geq 2$ şi numerele reale $a_{1}, a_{2}, \\ldots, a_{n} \\in [0,1]$. Aflaţi valoarea maximă a celui mai mic dintre numerele:\n$$\na_{1} - a_{1} a_{2},\\ a_{2} - a_{2} a_{3},\\ \\ldots,\\ a_{n} - a_{n} a_{1}\n$$",
    "options": [],
    "answer": "1/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDacă numărul natural $a$ are $n$ cifre, iar numărul natural $a^{4}$ are $m$ cifre, arătați că suma $m+n$ nu poate fi egală cu 2021.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPe o tablă sunt scrise numerele de forma $n(n+1)$, cu $n=1,2,3, \\ldots, 2020$. Un copil alege trei numere $a, b$ și $c$ de pe tablă, le șterge și scrie pe tablă numărul $\\frac{a b c}{a b+a c+b c}$. După 1009 astfel de operații, unul dintre numerele rămase pe tablă este 47.\na) Calculați suma inverselor numerelor scrise inițial pe tablă.\nb) Aflați celelalte numere rămase pe tablă.",
    "options": [],
    "answer": "a) 2020/2021; b) 2021/1977",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsiderăm mulţimea $M=\\{1,2,3, \\ldots, 2019,2020\\}$. Găsiţi cel mai mic număr natural nenul $k$ ce verifică următoarea proprietate:\npentru orice submulţime $A$, cu $k$ elemente, a mulţimii $M$, există trei numere $a, b, c$ din $M$, distincte două câte două, astfel încât $a+b, b+c$ si $c+a$ aparţin mulţimii $A$.",
    "options": [],
    "answer": "1012",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAflaţi numerele naturale nenule $x, y$ cu proprietatea că $x \\leq y$, astfel încât\n$$\n\\frac{(x+y)(x y-1)}{x y+1}=p\n$$\nunde $p$ este un număr prim.",
    "options": [],
    "answer": "All solutions with x ≤ y are: (1, q+1) for any prime q (giving p = q), and the pair (3, 5) (giving p = 7).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $a, b, c, d \\in \\mathbb{N}$, $d \\neq 0$ şi fie funcţia $f: \\mathbb{N} \\rightarrow \\mathbb{N}$, definită prin\n$$\nf(n) = \\left[ \\frac{a n + b}{c n + d} \\right], \\text{ oricare ar fi } n \\in \\mathbb{N}\n$$\nunde $[x]$ reprezintă partea întreagă a numărului real $x$. Demonstraţi că următoarele afirmaţii sunt echivalente:\n$1^{\\circ}$ $f$ este surjectivă.\n$2^{\\circ}$ $c = 0$, $b < d$ şi $0 < a \\leq d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSă se determine inelele nenule finite, cu unitate, în care suma tuturor elementelor este un element inversabil.",
    "options": [],
    "answer": "The only such ring is the field with two elements.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn triunghiul ascuţitunghic $ABC$ se notează cu $O$ centrul cercului circumscris şi cu $D$ piciorul înălţimii din $A$.\nFie $M, N, P, Q$ mijloacele segmentelor $AB, AC, BD$, respectiv $CD$. Arătaţi că unul dintre punctele de intersectie ale cercurilor circumscrise triunghiurilor $AMN$ şi $POQ$ este situat pe înălţimea $AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentru $n \\geq 2$ numere reale nenule $a_{1}, a_{2}, \\ldots, a_{n}$, nu neapărat distincte, definim matricea $A=\\left(a_{ij}\\right)_{1 \\leq i, j \\leq n} \\in M_{n}(\\mathbb{R})$ prin $a_{ij}=\\max \\{a_{i}, a_{j}\\}$, $\\forall i, j \\in\\{1,2, \\ldots, n\\}$. Arătaţi că $\\operatorname{rang}(A)=\\operatorname{card}\\left\\{a_{k} \\mid k=1,2, \\ldots, n\\right\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $P_{0}, P_{1}, \\ldots, P_{2021}$ puncte pe cercul trigonometric, de centru $O$ şi rază $1$, astfel încât, pentru orice $n \\in \\{1,2, \\ldots, 2021\\}$, lungimea arcului de cerc parcurs în sens trigonometric de la $P_{n-1}$ la $P_{n}$ aparţine intervalului $\\left[\\frac{\\pi}{2}, \\pi\\right]$.\nAflaţi lungimea maximă a vectorului\n$$\n\\overrightarrow{O P}_{0}+\\overrightarrow{O P}_{1}+\\cdots+\\overrightarrow{O P}_{2021}\n$$",
    "options": [],
    "answer": "√2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPentru orice submulţime nevidă $X$ a mulţimii $M=\\{1,2,3, \\ldots, 2021\\}$, notăm cu $a_{X}$ suma dintre cel mai mare şi cel mai mic element al mulţimii $X$.\nDeterminaţi media aritmetică a tuturor numerelor $a_{X}$ obţinute.",
    "options": [],
    "answer": "2022",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe consideră un patrulater convex $\\mathcal{P}$ şi un punct $X$ în interiorul lui $\\mathcal{P}$. Fie $M, N, P, Q$ proiecţiile lui $X$ pe laturile lui $\\mathcal{P}$. Se ştie că $M, N, P, Q$ se află pe un cerc de centru $L$. Arătaţi că $L$ şi mijloacele $J, K$ ale diagonalelor lui $\\mathcal{P}$ sunt pe o dreaptă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $a, b, c$ numere strict pozitive, astfel încât $a+b+c=1$. Arătaţi că\n$$\n\\frac{1}{a b c}+\\frac{4}{a^{2}+b^{2}+c^{2}} \\geq \\frac{13}{a b+b c+c a}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geq 2$ un număr natural cu proprietatea că mulţimea rădăcinilor de ordin $n$ ale unităţii are mai puţin de $2^{[\\sqrt{n}]}-1$ submulţimi cu suma elementelor nulă. Arătaţi că $n$ este prim. (Am notat cu $[x]$ partea întreagă a numărului real $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p, q$ numere naturale nenule. Pentru fiecare $a, b \\in \\mathbb{R}$ definim mulțimile\n$$\nP(a)=\\left\\{\\left.a_{n}=a+n \\cdot \\frac{1}{p} \\right| n \\in \\mathbb{N}\\right\\}, \\quad Q(b)=\\left\\{\\left.b_{n}=b+n \\cdot \\frac{1}{q} \\right| n \\in \\mathbb{N}\\right\\}\n$$\nNumim distanța de la $P(a)$ la $Q(b)$ valoarea minimă a diferenței $|x-y|$, și $x \\in P(a)$, $y \\in Q(b)$.\nDeterminați distanța maximă între mulțimile $P(a)$ și $Q(b)$, când $a$ și $b$ parcurg mulțimea $\\mathbb{R}$.",
    "options": [],
    "answer": "1/(2·lcm(p,q))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe consideră patrulaterul convex $ABCD$ în care unghiurile $A$ şi $C$ nu sunt ascuţite. Pe laturile $AB$, $BC$, $CD$ şi $DA$ se iau punctele $K$, $L$, $M$, respectiv $N$. Demonstraţi că perimetrul patrulaterului $KLMN$ este cel puţin egal cu dublul lungimii diagonalei $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ o funcție derivabilă de ordinul $n \\geq 2$, astfel încât\n$$\n\\lim _{x \\rightarrow \\infty} f(x)=\\ell \\in \\mathbb{R} \\text{ şi } \\lim _{x \\rightarrow \\infty} f^{(n)}(x)=0\n$$\nDemonstrați că $\\lim _{x \\rightarrow \\infty} f^{(k)}(x)=0$, pentru orice $k \\in\\{1,2, \\ldots, n-1\\}$, unde $f^{(k)}$ reprezintă derivata de ordinul $k$ a functiei $f$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCercul de centru $I$, înscris în triunghiul $A B C$, este tangent laturilor $A B, A C$ şi $B C$ în punctele $M, N$ şi respectiv $K$. Mediana $A D$ a triunghiului $A B C$ intersectează $M N$ în punctul $L$. Demonstraţi că punctele $K, I$ şi $L$ sunt coliniare.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\in \\mathbb{N}$, $a > 2$. Să se arate că\n\na) Există un număr $n \\in \\mathbb{N}^* \\setminus \\{1\\}$, care nu este prim, astfel încât $a^n \\equiv 1 \\pmod{n}$.\n\nb) Dacă $p$ este cel mai mic număr din $\\mathbb{N}^* \\setminus \\{1\\}$ pentru care $a^p \\equiv 1 \\pmod{p}$, atunci $p$ este prim.\n\nc) Nu există numere $n \\in \\mathbb{N}^* \\setminus \\{1\\}$ pentru care $2^n \\equiv 1 \\pmod{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAflați numerele naturale $x < y < z$ având suma $2021$, știind că îndeplinesc simultan următoarele condiții:\n\na) Fiecare număr are cifrele distincte;\n\nb) Există $3$ cifre distincte $a, b, c$ astfel încât fiecare cifră a numerelor $x, y$ și $z$ este $a, b$ sau $c$.",
    "options": [],
    "answer": "x=89, y=948, z=984",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră $n$ unghiuri în jurul unui punct având măsurile în grade exprimate prin $n$ numere prime distincte. Știind că unghiurile formate de bisectoarele oricăror două unghiuri adiacente dintre cele $n$ unghiuri date inițial au măsurile în grade exprimate prin numere prime, să se determine valorile posibile ale lui $n$.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geq 2$ şi matricele $A, B \\in M_{n}(\\mathbb{R})$. Presupunem că există $x \\in \\mathbb{R} \\backslash\\left\\{0, \\frac{1}{2}, 1\\right\\}$ astfel încât $x A B + (1-x) B A = I_{n}$. Arătaţi că $(A B - B A)^{n} = O_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $A$ o mulţime finită de numere naturale. Determinaţi toate funcţiile $f: \\mathbb{N} \\rightarrow A$ cu proprietatea că $f(|x-y|)=|f(x)-f(y)|$, pentru orice $x, y \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Necessary: 0 ∈ A. If 0 ∉ A, there are no solutions. If 0 ∈ A, all solutions are exactly:\n- The zero function f(n) = 0 for all n.\n- For any fixed c ∈ A, the parity function f(n) = 0 for n even and f(n) = c for n odd.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminaţi numerele întregi nenule $a$ pentru care există funcţiile $f, g: \\mathbb{Q} \\rightarrow \\mathbb{Q}$ care verifică ecuatia funcţională:\n$$\nf(x+g(y))=g(x)+f(y)+a y, \\text{ oricare ar fi } x, y \\in \\mathbb{Q}\n$$\nDeterminaţi toate aceste funcţii.",
    "options": [],
    "answer": "a must be of the form k(k−1) with k ∈ Z and k ≠ 0,1. For each such a, let p be either integer root of p^2 − p = a (equivalently p = (1 ± √(1+4a))/2 ∈ Z). Then all solutions are g(x) = p x and f(x) = p x + r, where r ∈ Q is arbitrary.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $f:[0,1] \\longrightarrow [0,1]$ o funcție continuă și bijectivă, cu proprietatea că $f(0)=0$. Arătați că pentru orice $\\alpha \\geq 0$ are loc inegalitatea\n$$\n(\\alpha+2) \\cdot \\int_{0}^{1} x^{\\alpha}\\left(f(x)+f^{-1}(x)\\right) \\, \\mathrm{d}x \\leq 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geq 2$ un număr natural. Pe o tablă $n \\times n$ se aşază $n$ turnuri astfel încât să nu existe două care să se atace. Toate turnurile se mişcă simultan o dată şi au voie să se mişte doar într-un pătrat adiacent celui în care se află.\nDeterminaţi toate valorile lui $n$ pentru care există o aşezare a turnurilor astfel încât, după o mutare, turnurile, în continuare, să nu se atace.\n\nNotă: Două pătrăţele sunt adiacente dacă au o latură comună.",
    "options": [],
    "answer": "n is even",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArătaţi că, pentru orice număr natural $n \\geqslant 2$, există un multiplu $m$ al său, nenul, cu următoarele proprietăţi:\na) $m < n^{4}$;\nb) în scrierea lui $m$ în baza 10 se folosesc cel mult patru cifre distincte.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn triunghiul isoscel fix $ABC$, punctul $M$ este mijlocul bazei $BC$. Punctul $P$ este variabil în interiorul triunghiului, astfel încât $\\angle CBP = \\angle PCA$. Arătaţi că suma măsurilor unghiurilor $\\angle BPM$ şi $\\angle APC$ este constantă.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $M$ o mulţime formată din 13 numere naturale de trei cifre.\nArătaţi că există o submulţime nevidă $S \\subset M$ şi o combinaţie de operaţii aritmetice elementare (adunare, scădere, înmulţire, împărţire - fără a utiliza parantezele) între elementele lui $S$, astfel încât valoarea expresiei rezultate să fie un număr raţional din intervalul $(3,4)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $N \\geq 4$ un număr natural fixat.\nDoi jucători, $A$ şi $B$, formează o mulţime ordonată $x_{1}, x_{2}, x_{3}, \\ldots$, adăugând alternativ elemente: $A$ alege $x_{1}$ egal cu $1$ sau cu $-1$, apoi $B$ îl adaugă pe $x_{2}$ egal cu $2$ sau cu $-2$, apoi $A$ îl adaugă pe $x_{3}$ egal cu $3$ sau cu $-3$, ş.a.m.d.; la pasul $k$, elementul adăugat este $k$ sau $-k$, pentru orice $k \\geq 1$. Câştigător este cel care reuşeşte primul să facă să apară o secvenţă de elemente consecutive având suma divizibilă cu $N$ (secvenţa poate avea şi un singur termen).\nPentru fiecare $N$, stabiliţi care dintre jucători are o strategie de câştig.",
    "options": [],
    "answer": "Player A has a winning strategy if N is odd or divisible by 4; player B has a winning strategy if N is congruent to 2 modulo 4.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $I$ centrul cercului înscris în triunghiul $A B C$. Cercul de centru $A$ şi rază $A I$ intersectează cercul circumscris triunghiului $A B C$ în punctele $M$ şi $N$.\nDemonstraţi că dreapta $M N$ este tangentă la cercul înscris în triunghiul $A B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $\\alpha$ un număr din intervalul $(0,1)$. Arătaţi că există un şir de numere $\\left(\\varepsilon_{n}\\right)_{n>1}$, cu valori 0 sau 1, astfel încât şirul $\\left(s_{n}\\right)_{n \\geq 1}$ definit prin\n$$\ns_{n}=\\frac{\\varepsilon_{1}}{n(n+1)}+\\frac{\\varepsilon_{2}}{(n+1)(n+2)}+\\ldots+\\frac{\\varepsilon_{n}}{(2 n-1) 2 n}\n$$\nsă verifice inegalitatea\n$$\n0 \\leq \\alpha-2 n s_{n} \\leq \\frac{2}{n+1}\n$$\npentru orice $n \\geq 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi scalen ascuţitunghic şi fie $\\omega$ cercul său Euler. Tangenta $t_{A}$ a lui $\\omega$, prin piciorul înălţimii din $A$ a triunghiului $ABC$, intersectează a doua oară cercul de diametru $AB$ în punctul $K_{A}$. Dreapta determinată de picioarele înălţimilor din $A$ şi $C$ ale triunghiului $ABC$, intersectează dreptele $AK_{A}$ şi $BK_{A}$ în punctele $L_{A}$, respectiv $M_{A}$, iar dreptele $t_{A}$ şi $CM_{A}$ se intersectează în punctul $N_{A}$. Punctele $K_{B}, L_{B}, M_{B}, N_{B}$ şi $K_{C}, L_{C}, M_{C}, N_{C}$ sunt definite în mod analog, pentru tripletele $(B, C, A)$, respectiv $(C, A, B)$. Arătaţi că dreptele $L_{A}N_{A}, L_{B}N_{B}$ şi $L_{C}N_{C}$ sunt concurente.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $ABC$ un triunghi ascuţitunghic şi fie $B'$ şi $C'$ picioarele înălţimilor sale din $B$, respectiv $C$. Fie $B_A'$ şi $B_C'$ simetricele lui $B'$ în raport cu dreptele $BC$, respectiv $AB$. Cercul $BB_A'B_C'$, centrat în $O_B$, intersectează a doua oară dreapta $AB$ în $X_B$. Punctele $C_A'$, $C_B'$, $O_C$, $X_C$ sunt definite în mod analog, prin înlocuirea perechii $(B, B')$ cu perechea $(C, C')$. Arătaţi că $O_BX_B$ şi $O_CX_C$ sunt paralele.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $ABC$ un triunghi dreptunghic în $A$, astfel încât $A'$ este mijlocul ipotenuzei, $M$ mijlocul înălțimii $AD$, $D \\in BC$ și $\\{P\\} = BM \\cap AA'$. \nDacă notăm $\\alpha = m(\\widehat{PCB})$, să se demonstreze că\n$$\n\\operatorname{tg} \\alpha = \\sin C \\cdot \\cos C\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAplicăm teorema lui Menelaus în triunghiul $AA'D$, cu punctele $B, M, P$ coliniare:\n$$\n\\frac{BA'}{BD} \\cdot \\frac{MD}{MA} \\cdot \\frac{PA}{PA'} = 1 \\Leftrightarrow \\frac{PA}{PA'} = \\frac{BD}{BA'} = \\frac{2BD}{BC} = \\frac{2BD \\cdot BC}{BC^2} = \\frac{2AB^2}{BC^2} = 2 \\sin^2 C \\quad (1)\n$$\nDin teorema sinusurilor în triunghiurile $PCA'$ și $PCA$ obținem:\n$$\n\\frac{PA'}{\\sin \\alpha} = \\frac{PC}{\\sin 2B} \\quad (2), \\text{ respectiv } \\frac{PA}{\\sin (C-\\alpha)} = \\frac{PC}{\\sin C} \\quad (3)\n$$\nDin relațiile (2) și (3) rezultă:\n$$\n\\frac{PA}{PA'} = \\frac{\\sin (C-\\alpha)}{\\sin \\alpha} \\cdot \\frac{\\sin 2B}{\\sin C} = \\frac{\\sin C \\cos \\alpha - \\sin \\alpha \\cos C}{\\sin \\alpha} \\cdot \\frac{\\sin 2B}{\\sin C}\n$$\nȚinând cont de relația (1) avem:\n$$(\\operatorname{ctg} \\alpha - \\operatorname{ctg} C) \\cdot \\sin 2B = 2 \\sin^2 C \\Leftrightarrow (\\operatorname{ctg} \\alpha - \\operatorname{ctg} C) \\cdot 2 \\sin B \\cos B = 2 \\sin^2 C \\Leftrightarrow (\\operatorname{ctg} \\alpha - \\operatorname{ctg} C) \\cdot \\sin B = \\sin C$$\nAșadar $\\operatorname{ctg} \\alpha = \\operatorname{ctg} C + \\operatorname{tg} C = \\frac{1}{\\sin C \\cos C}$, de unde concluzia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFixăm un număr întreg $n \\geq 2$ şi considerăm $n^{2}$ numere reale strict pozitive $a_{ij}$, $i, j=1, \\ldots, n$, care îndeplinesc simultan următoarele două condiţii:\n(1) $a_{ii}=1$, $i=1, \\ldots, n$; şi\n(2) Pentru fiecare $j=2, \\ldots, n$, numerele $a_{ij}$, $i=1, \\ldots, j-1$, formează o permutare a numerelor $\\frac{1}{a_{ji}}$, $i=1, \\ldots, j-1$.\nFie $s_{i}=\\sum_{j=1}^{n} a_{ij}$, $i=1, \\ldots, n$. Determinaţi valoarea maximă a sumei $\\sum_{i=1}^{n} \\frac{1}{s_{i}}$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminați funcțiile $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ pentru care există funcția $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ astfel încât\n$$\nf(x)+f(y)=[g(x+y)]\n$$\noricare ar fi $x, y$ numere reale.\n(Am notat cu $[a]$ partea întreagă a numărului real $a$.)",
    "options": [],
    "answer": "All constant functions f with f(x) = k/2 for some fixed integer k; for these, one can take any g with values in the interval [k, k+1).",
    "solution": "Solution:\n\nPentru $x \\rightarrow (x+y), y \\rightarrow 0$ în relația din enunț deducem\n$$\nf(x+y)+f(0)=[g(x+y)]=f(x)+f(y)\n$$\nNotând $h(x)=f(x)-f(0)$ obținem $h(x+y)=h(x)+h(y)$ (ecuația lui Cauchy).\n\nDeducem că, pentru orice $x_{0} \\in \\mathbb{R}$ ales arbitrar și orice $n$ natural, $h\\left(n x_{0}\\right)=n h\\left(x_{0}\\right)$.\nRezultă că $h\\left(x_{0}\\right)=h\\left(n \\cdot \\frac{x_{0}}{n}\\right) \\Leftrightarrow h\\left(\\frac{x_{0}}{n}\\right)=\\frac{h\\left(x_{0}\\right)}{n}$ pentru orice $n \\in \\mathbb{N}^{*}$.\n\nCum $2 f(x)=[g(2 x)] \\in \\mathbb{Z} \\Rightarrow 2 h(x) \\in \\mathbb{Z} \\Rightarrow \\frac{2 h\\left(x_{0}\\right)}{n} \\in \\mathbb{Z}$ pentru orice $n \\in \\mathbb{N}^{*}$. Alegând $n>\\left|2 h\\left(x_{0}\\right)\\right|$ obținem că\n$$\n0 \\leq \\left|2 \\cdot h\\left(\\frac{x_{0}}{n}\\right)\\right|=\\left|\\frac{2 h\\left(x_{0}\\right)}{n}\\right|<1\n$$\nde unde $h\\left(\\frac{x_{0}}{n}\\right)=0$, deci $h\\left(x_{0}\\right)=0$. Cum $x_{0}$ a fost ales arbitrar obținem $h(x)=0$ pentru orice $x$ real. Deducem că $f(x)=f(0)$ iar cum $2 f(x)=[g(2 x)] \\in \\mathbb{Z}$ rezultă $f(x)=\\frac{k}{2}$ pentru orice $x$ real, unde $k$ este un număr întreg fixat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a, b, c, d$ numere naturale nenule, cu $a < b < c < d$ și $a d = b c$. Demonstrați că\n$$\n2 a + \\sqrt{a} + \\sqrt{d} < b + c + 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFie $b = a + x$, $c = a + y$, $d = a + z$ cu $0 < x < y < z$ numere naturale.\nAtunci relația din enunț se scrie $a(a + z) = (a + x)(a + y)$. Deducem că $z = x + y + \\frac{x y}{a} > x + y \\Rightarrow z \\geqslant x + y + 1$. Obținem astfel\n$$\na = \\frac{x y}{z - x - y} \\leqslant x y\n$$\nAvem că $\\frac{x y}{a} + a \\leqslant x y + 1 \\Leftrightarrow x y + a^2 \\leqslant x y a + a \\Leftrightarrow (x y - a) + a(a - x y) \\leqslant 0 \\Leftrightarrow (x y - a)(1 - a) \\leqslant 0$, adevărat.\nObținem astfel $d = a + z = x + y + \\frac{x y}{a} + a \\leqslant x + y + x y + 1 = (x + 1)(y + 1)$.\n\nInegalitatea din enunț se transcrie $\\sqrt{a} + \\sqrt{d} < x + y + 1$, iar prin ridicare la pătrat este echivalentă cu\n$$\na + d + 2 \\sqrt{a d} < x^2 + y^2 + 1 + 2 x y + 2 x + 2 y\n$$\nCum\n$$\na + d \\leqslant x y + x + y + x y + 1\n$$\niar din inegalitatea mediilor, întrucât $x < y$, avem:\n$$\n2 \\sqrt{a d} \\leqslant 2 \\sqrt{x y (x + 1)(y + 1)} < x(x + 1) + y(y + 1)\n$$\nAdunând ultimele două relații, obținem (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOrice număr întreg $N$, care este suma a trei pătrate perfecte, este evident exprimabil sub forma\n$$\nN = \\frac{a^{2} + b^{2} + c^{2} + d^{2}}{1 + abcd}\n$$\nunde $a, b, c, d$ sunt numere naturale. Este adevărată afirmaţia reciprocă?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFixăm două numere întregi $m \\geq 2$ şi $n \\geq 2$. Fie $S$ o mulţime de puncte laticiale situate în dreptunghiul cartezian $[1, m] \\times [1, n]$; un punct laticial este un punct care are ambele coordonate întregi. Arătaţi că, dacă $|S| \\geq m+n+\\left\\lfloor \\frac{1}{4} m+\\frac{1}{4} n-\\frac{1}{2} \\right\\rfloor$, atunci există un cerc care trece prin cel puţin patru puncte din $S$, distincte două câte două.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two distinct positive integers having the same parity.\nProve that $\\frac{a!+b!}{2^a}$ is not an integer (note: $n! = 1 \\cdot 2 \\cdot \\dots \\cdot n$, for any positive integer $n$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we prove that for any positive integer $n$, $2^n$ is not a divisor of $n!$.\nAssume the opposite; then, in the prime factorization of $n!$ there should be at least $n$ factors equal to $2$. The exponent of $2$ in $n!$ is equal to $\\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\dots + \\lfloor \\frac{n}{2^k} \\rfloor$, where $k \\in \\mathbb{N}^*$, $k < n$, $\\lfloor \\frac{n}{2^k} \\rfloor \\ne 0$ and $\\lfloor \\frac{n}{2^{k+1}} \\rfloor = 0$. It follows that $\\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^k} \\ge \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\dots + \\lfloor \\frac{n}{2^k} \\rfloor \\ge n$.\nThen $\\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^k} \\ge n$, so $1 - \\frac{1}{2^k} \\ge 1$, which is false.\n\nIf $a \\ge b+2$, then $a! + b! = b! \\cdot (1 + (b+1) \\cdot (b+2) \\cdot \\dots \\cdot a) = n \\cdot 2^a$ and, since $1 + (b+1) \\cdot (b+2) \\cdot \\dots \\cdot b$ is odd, we get $2^a \\mid b!$. But $b! \\mid a!$, so $2^a \\mid a!$, which is false.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in real numbers the equation $2(5^x + 6^x - 3^x) = 7^x + 9^x$.",
    "options": [],
    "answer": "x = 0 and x = 1",
    "solution": "We claim that the only solutions are $x = 0$ and $x = 1$.\n\nThe given equation can be rewritten as follows:\n$$\n5^x \\left( \\left(\\frac{7}{5}\\right)^x + \\left(\\frac{3}{5}\\right)^x - 2 \\right) + 6^x \\left( \\left(\\frac{9}{6}\\right)^x + \\left(\\frac{3}{6}\\right)^x - 2 \\right) = 0. \\ (\\star)\n$$\nConsidering the functions $f_a : \\mathbb{R} \\to \\mathbb{R}$, $f_a(x) = a^x + (2-a)^x - 2$, with $a \\in (0, 2) \\setminus \\{1\\}$, we notice that they are strictly convex with $f_a(0) = f_a(1) = 0$. Moreover, we have:\n$$\nf_a(x) = f_a(x \\cdot 1 + (1-x) \\cdot 0) < x \\cdot f_a(1) + (1-x) \\cdot f_a(0) = 0,\n$$\nfor all $x \\in (0, 1)$, and\n$$\nf_a(1) = f_a \\left( \\frac{x-1}{x} \\cdot 0 + \\left(1 - \\frac{x-1}{x}\\right) \\cdot x \\right) < \\frac{x-1}{x} f_a(0) + \\frac{1}{x} f_a(x),\n$$\nfor all $x \\in (1, \\infty)$,\n$$\nf_a(0) = f_a \\left( \\frac{1}{1-x} \\cdot x + \\left(1 - \\frac{1}{1-x}\\right) \\cdot 1 \\right) < \\frac{1}{1-x} f_a(x) + \\frac{-x}{1-x} f_a(1),\n$$\nfor all $x \\in (-\\infty, 0)$, so $f_a(x) > 0$, for each $x \\in (-\\infty, 0) \\cup (1, \\infty)$. The equivalent equation from $(\\star)$ is:\n$$\n5^x f_{\\frac{7}{5}}(x) + 6^x f_{\\frac{9}{6}}(x) = 0,\n$$\nso the left member is strictly less than zero for $x \\in (0, 1)$ and strictly greater than zero for $x \\in (-\\infty, 0) \\cup (1, \\infty)$, thus proving our claim.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [-\\frac{\\pi}{2}, \\frac{\\pi}{2}] \\to \\mathbb{R}$ be a twice differentiable function such that\n$$\n(f''(x) - f(x)) \\cdot \\operatorname{tg}(x) + 2 \\cdot f'(x) \\ge 1, \\quad \\text{for any } x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2}).\n$$\n\n$$\n\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} f(x) \\cdot \\sin x \\, dx \\geq \\pi - 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\cos x > 0$ for any $x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2})$, the inequality in the statement can be written equivalently\n$$\n(f''(x) - f(x)) \\cdot \\sin x + 2 \\cdot f'(x) \\cdot \\cos x \\geq \\cos x, \\quad \\text{for any } x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2}),\n$$\nor $g''(x) \\ge 0, \\forall x \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2})$, where $g: [-\\frac{\\pi}{2}, \\frac{\\pi}{2}] \\to \\mathbb{R}$ is the function defined by $g(x) = f(x) \\cdot \\sin x + \\cos x, \\forall x \\in [-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$.\n\nHence, the function $g$ is convex, so that\n$$\n\\frac{g(x) + g(-x)}{2} \\geq g(0) = 1, \\quad \\text{for any } x \\in [-\\frac{\\pi}{2}, \\frac{\\pi}{2}].\n$$\nBecause $\\int_{-a}^{a} h(x) \\, dx = \\int_{-a}^{a} h(-x) \\, dx$ holds for any integrable function and any $a \\ge 0$, we have\n$$\n\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} g(x) \\, dx = \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\frac{g(x) + g(-x)}{2} \\, dx \\geq \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} 1 \\, dx = \\pi.\n$$\n\n$$\n\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} f(x) \\cdot \\sin x \\, dx = \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} (g(x) - \\cos x) \\, dx \\geq \\pi - 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle, $GH \\parallel BC$, with $G \\in (AB)$ and $H \\in (AD)$, and $EF \\parallel DC$, with $E \\in (AD)$ and $F \\in (BC)$. Let $GH \\cap EF = \\{M\\}$ and $AH \\cap CE = \\{K\\}$. Prove that the point $K$ is on the circle passing through the feet of the altitudes of the triangle $DFG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all primes $p$ and all positive integers $x$ and $y$ satisfying $x^3 + y^3 = p(xy + p)$.",
    "options": [],
    "answer": "All solutions, up to swapping the first two entries, are (x, y, p) = (1, 8, 19), (2, 7, 13), (4, 5, 7).",
    "solution": "Up to a swap of the first two entries, the only solutions are $(x, y, p) = (1, 8, 19)$, $(x, y, p) = (2, 7, 13)$ and $(x, y, p) = (4, 5, 7)$. The verification is routine.\n\nSet $s = x + y$. Rewrite the equation in the form $s(s^2 - 3xy) = p(p + xy)$, and express $xy$:\n$$\nxy = \\frac{s^3 - p^2}{3s + p}. \\qquad (*)\n$$\nIn particular,\n$$\ns^2 \\geq 4xy = \\frac{4(s^3 - p^2)}{3s + p}, \\quad \\text{or} \\quad (s-2p)(s^2+sp+2p^2) \\leq p^2-p^3 < 0, \\quad \\text{so} \\quad s < 2p.\n$$\nSince $p \\nmid s$, this could be integer only if $3s + p \\mid p + 27$, and hence $3s + p \\mid 27 - s$.\n\nAlternative solution.\nSet again $s = x+y$. It is readily checked that $s \\leq 8$ provides no solutions, so assume $s \\geq 9$. Notice that $x^3+y^3 = s(x^2-xy+y^2) \\geq \\frac{1}{4}s^3$ and $xy \\leq \\frac{1}{4}s^2$. The condition in the statement then implies $s^2(s-p) \\leq 4p^2$, so $s < p + 4$.\n\nHence $p \\mid x^2 - xy + y^2$, so $x^2 - xy + y^2 = kp$ and $xy + p = ks$ for some positive integer $k$, implying\n$$\ns^2 + 3p = k(3s + p). \\qquad (**)\n$$\nRecall that $p \\nmid s$ to infer that $3k \\equiv s \\pmod{p}$. We now present two approaches.\n\n**1st Approach.** Write $3k = s + mp$ for some integer $m$ and plug $k = \\frac{1}{3}(s + mp)$ into (**) to get $s = (9 - mp)/(3m + 1)$. The condition $s \\geq 9$ then forces $m = 0$, so $s = 9$, in which case, up to a swap of the first two entries, the solutions turn out to be $(x, y, p) = (1, 8, 19)$, $(x, y, p) = (2, 7, 13)$ and $(x, y, p) = (4, 5, 7)$.\n\n**2nd Approach.** Notice that $k = \\frac{s^2+3p}{3s+p} = 3+\\frac{s(s-9)}{3s+p} \\leq 3+\\frac{1}{3}(s-9) = \\frac{4}{3}s \\leq \\frac{1}{3}(p+3)$, since $s < p + 4$. Hence $3k \\leq p + 3$, and the congruence $3k \\equiv s \\pmod{p}$ then forces either $3k = s - p$ or $3k = s$.\n\nThe case $3k = s-p$ is easily ruled out: Otherwise, (**) boils down to $2s+p+9=0$, which is clearly impossible.\n\nFinally, if $3k = s$, then (*) reduces to $s = 9$. In this case, up to a swap of the first two entries, the only solutions are $(x, y, p) = (1, 8, 19)$, $(x, y, p) = (2, 7, 13)$ and $(x, y, p) = (4, 5, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, where $m < 2^n$. Determine the smallest possible number of not necessarily pairwise distinct powers of 2 that add up to $m \\cdot (2^n - 1)$.",
    "options": [],
    "answer": "n",
    "solution": "The required minimum is $n$.\nTo prove this, notice that the sum of two like powers of 2 is again a power of 2, so the number of powers of 2 that add up to a positive integer $k$ can successively be decreased while keeping the sum constant. The process then ends up with $k$ being expressed as a sum of pairwise distinct powers of 2; that is, the binary expansion of $k$. At this stage, the number of summands can no longer be decreased, so the smallest number of powers of 2 that add up to $k$ is equal to the number of units in the binary expansion of $k$.\n\n1st Proof.\nSince multiplication by a power of 2 does not change the number of units in a binary expansion, we may and will assume that $m$ is odd. Leaving aside the trivial case $m = 1$, let $m \\ge 3$ so its binary expansion is $m = 1 + \\sum_{i=1}^{p} 2^{k_i}$, where $0 < k_1 < k_2 < \\dots < k_p$; and since $m < 2^n$, it follows that $k_p < n$, so $p < n$. Now, write $m \\cdot (2^n - 1) = 2^n \\cdot (m-1) + ((2^n - 1) - (m-1)) = \\sum_{i=1}^{p} 2^{n+k_i} + (\\sum_{i=0}^{n-1} 2^i - \\sum_{i=1}^{p} 2^{k_i})$.\n\nThe first sum consists of exactly $p$ pairwise distinct powers of 2, each of which is greater than $2^n$, and the number in parentheses is the sum of exactly $n - p$ pairwise distinct powers of 2, each of which is less than $2^n$.\nConsequently, $m \\cdot (2^n - 1)$ is the sum of exactly $p + (n - p) = n$ pairwise distinct powers of 2; that is, its binary expansion has exactly $n$ units, as stated.\n\n2nd Proof.\nIf $m = 2^k$ for some $k < n$, then $m \\cdot (2^n - 1) = 2^{n+k-1} + 2^{n+k-2} + \\dots + 2^{k+1} + 2^k$, so the binary expansion of $m \\cdot (2^n - 1)$ has exactly $n$ units.\nIf $m$ is not a power of 2, let $m = \\sum_{i=1}^{p} 2^{k_i}$, where $p \\ge 2$ and $0 \\le k_1 < k_2 < \\dots < k_p$, be the binary expansion of $m$, to write $m \\cdot (2^n - 1) = m \\cdot 2^n - m = (m \\cdot 2^n - 2^{n+k_1}) + (2^{n+k_1} - m) = \\sum_{i=2}^{p} 2^{n+k_i} + (2^{n+k_1} - \\sum_{i=1}^{p} 2^{k_i})$.\nClearly, the first sum is an integer greater than $2^{n+k_1}$ whose binary expansion has exactly $p-1$ units.\nThe number in parentheses is a positive integer ($m < 2^n \\le 2^{n+k_1}$) smaller than $2^{n+k_1}$ whose binary expansion has exactly $n+k_1-k_p+\\sum_{i=1}^{p-1}(k_{i+1}-k_i-1) = n-p+1$ units.\nConsequently, the binary expansion of $m \\cdot (2^n - 1)$ has exactly $(p-1) + (n-p+1) = n$ units, as stated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_n = \\{\\sqrt{n}\\} - \\{\\sqrt{n+1}\\} + \\{\\sqrt{n+2}\\} - \\{\\sqrt{n+3}\\}$, where $\\{x\\}$ denotes the fractional part of the real number $x$.\n\na) Prove that $a_1 > 0.2$.\n\nb) Prove that $a_n < 0$, for infinitely many $n$, and $a_n > 0$, for infinitely many $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) $a_1 = 0 - \\{\\sqrt{2}\\} + \\{\\sqrt{3}\\} + 0 = \\sqrt{3} - 1 - (\\sqrt{2} - 1) = \\sqrt{3} - \\sqrt{2}$. Since $\\sqrt{3} > 1.7$ and $\\sqrt{2} < 1.5$, the conclusion follows.\n\nb) Notice that, if $m$ is a positive integer and $m^2 \\leq a < b < (m+1)^2$, then $\\lfloor\\sqrt{a}\\rfloor = \\lfloor\\sqrt{b}\\rfloor = m$ and $\\sqrt{a} < \\sqrt{b}$, so $\\{\\sqrt{a}\\} = \\sqrt{a} - m < \\sqrt{b} - m = \\{\\sqrt{b}\\}$.\nTherefore, if $m \\geq 2$ is an integer and $m^2 \\leq n < n+1 < n+2 < n+3 < (m+1)^2$, then $\\{\\sqrt{n}\\} < \\{\\sqrt{n+1}\\}$ and $\\{\\sqrt{n+2}\\} < \\{\\sqrt{n+3}\\}$, so $a_n < 0$, for every such $m$ (which are infinitely many).\nAlso, if $m \\geq 2$ is an integer and $m^2 < n < n+1 < n+2 < n+3 = (m+1)^2$, then $\\{\\sqrt{n+1}\\} < \\{\\sqrt{n+2}\\}$, $\\{\\sqrt{n}\\} \\geq 0$ and $\\{\\sqrt{n+3}\\} = 0$, hence $a_n > 0$, for every such $m$ (which are infinitely many).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39119,
    "subject": "Mathematics (Multi-modal)",
    "question": "The continuous function $f : \\mathbb{R} \\to \\mathbb{R}$ has the property $(f \\circ f)(x) = x^{2023}$, for every $x \\in \\mathbb{R}$. Prove that:\n\na) $2023\\sqrt[2023]{f(x)} = f(2023\\sqrt[2023]{x})$, for every $x \\in \\mathbb{R}$;\n\nb) for every $n \\ge 3$ there exists an arithmetic progression $x_1, x_2, \\dots, x_n$ so that $\\sum_{k=1}^{n} f(x_k) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39120,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integer $n$ is a perfect square. Find the quotient of the division of $2023$ by $n$, if the remainder is $223 - \\frac{3}{2} \\cdot n$.",
    "options": [],
    "answer": "14",
    "solution": "Denote by $c$ the quotient of the division. From the quotient-remainder theorem we obtain $2023 = n \\cdot c + 223 - \\frac{3}{2} \\cdot n$, thus $(2c-3)n = 3600$. (1)\n\nThe remainder $223 - \\frac{3}{2} \\cdot n$ is a positive integer, therefore $n$ is even and $0 \\le 223 - \\frac{3}{2} \\cdot n < n$, whence we deduce that $90 \\le n \\le 148$. Because $n$ is a square, it follows that $n = 100$ or $n = 144$. If $n = 100$, we deduce from (1) that $2c - 3 = 36$, false, and if $n = 144$, we obtain the solution $c = 14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 0$ for which there exist integers $a$ and $b$ such that $a + 2^b = n^{2022}$ and $a^2 + 4^b = n^{2023}$.",
    "options": [],
    "answer": "1",
    "solution": "The only number is $n = 1$.\nTo this end, notice that $2(a^2 + 4^b) \\ge (a + 2^b)^2$ to conclude that $2n^{2023} \\ge n^{4044}$, hence $n = 0$ or $n = 1$. The case $n = 0$ leads to no solution, while $n = 1$ holds for $a = 0$ and $b = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the sequences of equal ratios of the form $\\frac{a_1}{a_2} = \\frac{a_3}{a_4} = \\frac{a_5}{a_6} = \\frac{a_7}{a_8}$ fulfilling the conditions:\n- the set $\\{a_1, a_2, \\dots, a_8\\}$ is the set of the positive divisors of $24$;\n- the common value of the ratios is an integer.",
    "options": [],
    "answer": "24/12 = 8/4 = 6/3 = 2/1; 24/8 = 12/4 = 6/2 = 3/1; 24/6 = 12/3 = 8/2 = 4/1",
    "solution": "The common value $r$ of the ratios can be only a divisor of $24$, different from $1$; these divisors are $2$, $3$, $4$, $6$, $8$, $12$ and $24$.\nIf $r = 2$, then $2 = \\frac{24}{12} = \\frac{8}{4} = \\frac{6}{3} = \\frac{2}{1}$;\nif $r = 3$, then $3 = \\frac{24}{8} = \\frac{12}{4} = \\frac{6}{2} = \\frac{3}{1}$;\nif $r = 4$, then $4 = \\frac{24}{6} = \\frac{12}{3} = \\frac{8}{2} = \\frac{4}{1}$.\nThere are no other sequences for $r = 2$, $r = 3$ or $r = 4$, because if we order decreasingly the divisors and use them one by one, we have to put at the numerator the largest divisor $d$ still unused, and at the denominator $\\frac{d}{r}$, getting the sequences from above.\nThere is no sequence for $r \\ge 8$, respectively $r = 6$, because none of the equalities $\\frac{d}{6} = r$ and $\\frac{6}{d} = r$, respectively $\\frac{d}{3} = 6$ and $\\frac{3}{d} = 6$, can be fulfilled.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the equation $x^2 + (a+b-1)x + ab - a - b = 0$, where $a$ and $b$ are positive integers such that $0 < a \\leq b$.\n\na) Show that the equation has two distinct real solutions.\n\nb) Prove that if one solution of the equation is an integer, then both solutions are non positive integers and $b < 2a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The discriminant of the equation is\n$$\n\\Delta = (a+b-1)^2 - 4(ab-a-b) = (a-b)^2 + 2a + 2b + 1 > 0,\n$$\n\nb) If $x_1 < x_2$ are the two solutions of the equation, from Vi\\`ete's first relation, $x_1 + x_2 = 1 - a - b \\in \\mathbb{Z}$, so $x_1 \\in \\mathbb{Z} \\iff x_2 \\in \\mathbb{Z}$.\nLet $f(x) = x^2 + (a+b-1)x + ab - a - b$, $x \\in \\mathbb{R}$. Since $f(1) = ab > 0$, $f(-a) = -b < 0$, $f(-b) = -a < 0$, using the sign of the quadratic function we obtain $x_1 < -b \\leq -a < x_2 < 1$, so both solutions are non positive.\nThus, $a \\geq 1 - x_2$, $b \\geq 1 - x_2$ and since $f(x_2) = 0$, we obtain $(a - 1 + x_2) \\cdot (b - 1 + x_2) = 1 - x_2 \\geq 1$. We get that $d_1 = a - 1 + x_2$ and $d_2 = b - 1 + x_2$ are natural divisors of $1 - x_2$ and $d_1 d_2 = 1 - x_2$, $a = d_1 + d_1 d_2$, $b = d_2 + d_1 d_2$. We obtain $2a - b = 2d_1 + d_2(d_1 - 1) \\geq 2d_1 > 0$, thus $b < 2a$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real numbers $x$, $y$, $z > 0$ for which\n$$\nxyz \\le \\min \\left\\{ 4 \\left(x - \\frac{1}{y}\\right), 4 \\left(y - \\frac{1}{z}\\right), 4 \\left(z - \\frac{1}{x}\\right) \\right\\}.\n$$",
    "options": [],
    "answer": "x = y = z = sqrt(2)",
    "solution": "From the given condition, we have $xyz \\le 4 \\left(x - \\frac{1}{y}\\right)$, which is equivalent to $4x \\ge xyz + \\frac{4}{y}$. By AM-GM, we get $4x \\ge xyz + \\frac{4}{y} \\ge 2\\sqrt{xyz \\cdot \\frac{4}{y}} = 4\\sqrt{xz}$, so $x \\ge z$.\n\nAnalogously, from $xyz \\le 4 \\left(y - \\frac{1}{z}\\right)$ and $xyz \\le 4 \\left(z - \\frac{1}{x}\\right)$, we get $y \\ge x$ and $z \\ge y$, so necessarily $x = y = z$.\n\nNow the requirement holds if and only if $x^3 \\le 4 \\left(x - \\frac{1}{x}\\right)$, i.e. $(x^2 - 2)^2 \\le 0$, which leads to $x = \\sqrt{2}$, so $x = y = z = \\sqrt{2}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ with the properties $f(1) = e$ and $f(x + y) = e^{3xy} \\cdot f(x) \\cdot f(y)$, for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = e^{(3x^2 - x)/2}",
    "solution": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function that satisfies the conditions in the statement. If there exists $a \\in \\mathbb{R}$ such that $f(a) = 0$ then $f(x) = f(a + (x - a)) = e^{3a(x-a)}f(a)f(x-a) = 0, \\forall x \\in \\mathbb{R}$, contradiction. On the other hand, we have $f(x) = e^{3x^2/4}f^2(x/2) \\ge 0, \\forall x \\in \\mathbb{R}$. It results $f(x) > 0, \\forall x \\in \\mathbb{R}$, and we obtain $\\ln f(x + y) = 3xy + \\ln f(x) + \\ln f(y), \\forall x, y \\in \\mathbb{R}$. Hence\n$$\n\\ln f(x + y) - \\frac{3(x + y)^2}{2} = \\left[ \\ln f(x) - \\frac{3x^2}{2} \\right] + \\left[ \\ln f(y) - \\frac{3y^2}{2} \\right], \\forall x, y \\in \\mathbb{R}.\n$$\nLet us define the function $g : \\mathbb{R} \\to \\mathbb{R}$, $g(x) = \\ln f(x) - \\frac{3x^2}{2}$. From the previous relation it follows that the function $g$ satisfies the Cauchy's functional equation $g(x + y) = g(x) + g(y), \\forall x, y \\in \\mathbb{R}$. Since $f$ is continuous, $g$ is also continuous. Thus $g(x) = g(1)x = \\left(\\ln f(1) - \\frac{3}{2}\\right)x = -\\frac{x}{2}, \\forall x \\in \\mathbb{R}$. Therefore $\\ln f(x) - \\frac{3x^2}{2} = -\\frac{x}{2}, \\forall x \\in \\mathbb{R}$, and we obtain $f(x) = e^{\\frac{x(3x-1)}{2}}, \\forall x \\in \\mathbb{R}$. This function satisfies the conditions in the statement.\n\nAlternative solution:\nLet $f : \\mathbb{R} \\to \\mathbb{R}$ be a function that satisfies the conditions in the statement. Then $f(1) = f(1)f(0)$, so $f(0) = 1$. For every $n \\in \\mathbb{N}$, we have $f(n) = e^{\\frac{n(3n-1)}{2}}$ (the proof is given by induction). From the relation $1 = f(0) = e^{-3n^2}f(n)f(-n)$, we find $f(-n) = e^{\\frac{3(-n)^2-(-n)}{2}}$. It follows that $f(n) = e^{\\frac{n(3n-1)}{2}}, \\forall n \\in \\mathbb{Z}$. By induction we obtain $f(nx) = e^{\\frac{3n(n-1)x^2}{2}}f^n(x)$, for every $x \\in \\mathbb{R}$ and for every $n \\in \\mathbb{N}$. Let $r = \\frac{m}{n}$ be a rational number, where $m \\in \\mathbb{Z}$ and $n \\in \\mathbb{N} \\setminus \\{0\\}$. We have $e^{\\frac{m(3m-1)}{2}} = f(m) = f\\left(n \\cdot \\frac{m}{n}\\right) = e^{\\frac{3n(n-1)}{2} \\cdot \\frac{m^2}{n^2}} f^n\\left(\\frac{m}{n}\\right)$. But $f(x) = e^{3x^2/4}f^2(x/2) \\ge 0$, for all $x \\in \\mathbb{R}$. Thus\n$$\nf(r) = f\\left(\\frac{m}{n}\\right) = e^{\\frac{1}{n}\\left(\\frac{m(3m-1)}{2} - \\frac{3m^2(n-1)}{2n}\\right)} = e^{\\frac{r(3r-1)}{2}}.\n$$\nAssume $x \\in \\mathbb{R} \\setminus \\mathbb{Q}$. There is a sequence $(r_n)_{n \\in \\mathbb{N}}$ of rational numbers such that $\\lim_{n \\to \\infty} r_n = x$. From the continuity of the function $f$, we get $f(x) = \\lim_{n \\to \\infty} f(r_n) = \\lim_{n \\to \\infty} e^{\\frac{r_n(3r_n-1)}{2}} = e^{\\frac{x(3x-1)}{2}}$. Thus $f(x) = e^{\\frac{x(3x-1)}{2}}, x \\in \\mathbb{R}$, is the unique function that satisfies the conditions in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the twice differentiable functions $f: \\mathbb{R} \\to \\mathbb{R}$ that verify the relation $(f'(x))^2 + f''(x) \\le 0$, for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All constant functions.",
    "solution": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function that verifies the conditions of the statement. Let us define the twice differentiable function $g: \\mathbb{R} \\to \\mathbb{R}$ by $g(x) = e^{f(x)}$, $x \\in \\mathbb{R}$. We have $g''(x) = e^{f(x)} \\left( (f'(x))^2 + f''(x) \\right) \\le 0$, for all $x \\in \\mathbb{R}$. Then $g'$ is a nonincreasing function. Hence there are the limits $\\ell_1 = \\lim_{x \\to -\\infty} g'(x)$ and $\\ell_2 = \\lim_{x \\to \\infty} g'(x)$, with $\\ell_1, \\ell_2 \\in \\mathbb{R}$. By applying l'Hôpital's rule, we obtain $\\lim_{x \\to -\\infty} \\frac{g(x)}{x} = \\lim_{x \\to -\\infty} g'(x) = \\ell_1$ and $\\lim_{x \\to \\infty} \\frac{g(x)}{x} = \\lim_{x \\to \\infty} g'(x) = \\ell_2$. From the inequality $g(x) > 0$, $\\forall x \\in \\mathbb{R}$, we find $\\ell_1 \\le 0$ and $\\ell_2 \\ge 0$. Since $g'$ is a nonincreasing function, $\\ell_1 \\le 0$ and $\\ell_2 \\ge 0$, we conclude $g'(x) = 0$, $\\forall x \\in \\mathbb{R}$. Therefore $g$ is a constant positive function. So $f = \\ln(g)$ is a constant function. Reciprocally, any constant function $f$ verifies the conditions of the statement.\n\nAlternative solution:\n\nLet $f: \\mathbb{R} \\to \\mathbb{R}$ be a function that verifies the conditions of the statement and define the positive twice differentiable function $g: \\mathbb{R} \\to \\mathbb{R}$ by $g(x) = e^{f(x)}$, $x \\in \\mathbb{R}$. From the inequality $g''(x) = e^{f(x)}((f'(x))^2 + f''(x)) \\le 0$, for all $x \\in \\mathbb{R}$, it results that $g$ is concave. So we have\n$$\n\\frac{g(y) - g(x)}{y - x} \\ge \\frac{g(z) - g(y)}{z - y}, \\text{ for all } x, y, z \\in \\mathbb{R}, \\text{ with } x < y < z.\n$$\nAssume that $g$ is a nonconstant function. Let $a, b \\in \\mathbb{R}$, with $a < b$, two points such that $g(a) \\ne g(b)$. If $g(a) < g(b)$, then the inequality $\\frac{g(a)-g(x)}{a-x} \\ge \\frac{g(b)-g(a)}{b-a}$, for all $x < a$, implies $g(x) \\le \\frac{g(b)-g(a)}{b-a}(x-a) + g(a)$, for $x \\in (-\\infty, a)$. We obtain $\\lim_{x \\to -\\infty} g(x) = -\\infty$. Contradiction.\nIf $g(a) > g(b)$, then the inequality $g(x) \\le \\frac{g(b)-g(a)}{b-a}(x-b) + g(b)$, for all $x \\in (b, \\infty)$ implies $\\lim_{x \\to \\infty} g(x) = -\\infty$. Contradiction. Hence $g$ is a constant function. So $f = \\ln(g)$ is also a constant function. Reciprocally, any constant function $f$ verifies the conditions of the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a finite group of order $n \\in \\mathbb{N}^*$, with $n \\ge 2$. We shall call the group $(G, \\cdot)$ *arrangeable* if there is an ordering of its elements, such that\n$$\nG = \\{a_1, a_2, \\dots, a_k, \\dots, a_n\\} = \\{a_1 \\cdot a_2, a_2 \\cdot a_3, \\dots, a_k \\cdot a_{k+1}, \\dots, a_n \\cdot a_1\\}.\n$$\n\na) Determine all positive integers $n$ for which the group $(\\mathbb{Z}_n, +)$ is arrangeable.\n\nb) Give an example of an arrangeable group of even order.",
    "options": [],
    "answer": "a) The group of integers modulo n under addition is arrangeable if and only if n is odd (with n at least three due to n at least two). b) An example of an arrangeable group of even order is Z4 × Z2 (of order eight).",
    "solution": "a. We will show that the group $(\\mathbb{Z}_n, +)$ is arrangeable if and only if $n \\ge 2$ is an odd positive integer.\n\nIf $(G, \\cdot)$ is an abelian arrangeable group, then considering the arrangement $G = \\{a_1, a_2, \\dots, a_k, \\dots, a_n\\} = \\{a_1 \\cdot a_2, a_2 \\cdot a_3, \\dots, a_k \\cdot a_{k+1}, \\dots, a_n \\cdot a_1\\}$, we have\n$$\n\\prod_{g \\in G} g = \\prod_{k=1}^{n} a_k = \\prod_{k=1}^{n} (a_k \\cdot a_{k+1}) = \\left( \\prod_{g \\in G} g \\right)^2,\n$$\n(where $a_{n+1} = a_1$), so that $\\prod_{g \\in G} g = 1$, where $1$ is the unit element of the group $(G, \\cdot)$.\n\nIn any finite abelian group the product of all the elements is equal to the product of all its elements of order $2$.\n\nFor $n \\in \\mathbb{N}^*$, $n \\ge 2$, if $k \\in \\{0, 1, \\dots, n-1\\}$ with $\\mathrm{ord}(\\hat{k}) = 2$, then\n$$\n\\hat{k} \\neq \\hat{0} = \\hat{k} + \\hat{k} = 2\\hat{k},\n$$\nso that $n$ divides $2k$, but does not divide $k$. This is only possible if $n$ is even and $n = 2k$. Thus, if $n$ is even, with $n = 2k$, and $(\\mathbb{Z}_n, +)$ were arrangeable, we would have\n$$\n\\hat{0} = \\sum_{x \\in \\mathbb{Z}_n} x = \\hat{k},\n$$\nwhich is false. Hence, if $n$ is even, the group $(\\mathbb{Z}_n, +)$ is not arrangeable.\n\nLet now $n \\ge 3$ be an odd positive integer. We consider the set $[0, n-1]_N = \\{0, 1, \\dots, n-1\\}$ and the function $f : [0, n-1]_N \\to \\mathbb{Z}_n$, defined by $f(k) = \\overline{2k+1}$.\n\nSince\n$$\nf(k) = f(l) \\iff \\overline{2k+1} = \\overline{2l+1} \\iff n|(2k-2l) \\iff n|(k-l) \\iff k=l,\n$$\nthe function $f$ is injective and since $[0, n-1]_N$ and $\\mathbb{Z}_n$ are finite sets with equal cardinals, it follows that $f$ is bijective. Denoting $a_k = \\overline{k-1}$ for any $1 \\le k \\le n$ and $a_{n+1} = a_1$, it follows then that $a_k + a_{k+1} = f(k-1)$ for any $k = \\overline{1, n}$, so that $\\mathbb{Z}_n = \\{a_1, a_2, \\dots, a_n\\} = \\{f(0), f(1), \\dots, f(k), \\dots, f(n-1)\\} = \\{a_1 + a_2, a_2 + a_3, \\dots, a_{k-1} + a_k, \\dots, a_n + a_1\\}$, whence we deduce that the group $(\\mathbb{Z}_n, +)$ is arrangeable.\n\nThe set of all positive integers such that the group $(\\mathbb{Z}_n, +)$ is arrangeable is thus the set of all odd positive integers $n$, with $n \\ge 3$.\n\n\nb. According to part a), there are no cyclic arrangeable groups of even order. We consider $\\mathbb{Z}_4 = \\{\\hat{0}, \\hat{1}, \\hat{2}, \\hat{3}\\}$, $\\mathbb{Z}_2 = \\{\\overline{0}, \\overline{1}\\}$ and the group $G = \\mathbb{Z}_4 \\times \\mathbb{Z}_2$ with component-wise defined addition $(\\hat{k}, \\hat{l}) + (\\hat{m}, \\hat{n}) = (\\overline{k+m}, \\overline{l+n})$. Then $G = \\{a_1 = (\\hat{0}, \\overline{0}), a_2 = (\\hat{1}, \\bar{0}), a_3 = (\\hat{1}, \\bar{1}), a_4 = (\\hat{3}, \\bar{1}), a_5 = (\\hat{2}, \\bar{0}), a_6 = (\\hat{2}, \\bar{1}), a_7 = (\\bar{0}, \\bar{1}), a_8 = (\\hat{3}, \\bar{0})\\} = \\{a_1+a_2, a_2+a_3, a_3+a_4, a_4+a_5, a_5+a_6, a_6+a_7, a_7+a_8, a_8+a_1\\}$, so that $(G, +)$ is an arrangeable group of order $8$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39128,
    "subject": "Mathematics (Multi-modal)",
    "question": "The elements of a set $A$ are 13 consecutive positive integers, the elements of a set $B$ are 12 consecutive positive integers, and the elements of the set $A \\cup B$ are 15 consecutive integers.\na)\nFind the number of elements of $A \\setminus B$.\nb) If, moreover, the sum of the elements of the set $A$ is equal to the sum of the elements of the set $B$, find the two sets.",
    "options": [],
    "answer": "a) 3\nb) A = {24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36}, B = {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38}",
    "solution": "a) The elements of the set $A \\setminus B$ are the elements of $A \\cup B$ which are not in $B$; there are $15 - 12 = 3$ such elements.\n\nb) The sum of the elements from $A \\setminus B$ equals the sum of the elements from $B \\setminus A$.\nLet $A \\cap B = \\{n, n+1, n+2, \\dots, n+9\\}, n \\in \\mathbb{N}$. The set $A \\setminus B$ has 3 elements and $B \\setminus A$ has 2 elements, so $n \\ge 3$, $A \\setminus B = \\{n-3, n-2, n-1\\}$ and $B \\setminus A = \\{n+10, n+11\\}$.\nThis yields $n-3+n-2+n-1=n+10+n+11$, that is $n=27$, hence $A = \\{24, 25, 26, 27, \\dots, 36\\}$ and $B = \\{27, 28, 29, \\dots, 38\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in real numbers the equation $2^{x-1} + 2^{\\frac{1}{\\sqrt{x}}} = 3$.",
    "options": [],
    "answer": "1",
    "solution": "We notice that $x > 0$. The given equation can be written as $2^x + 2 \\cdot 2^{\\frac{1}{\\sqrt{x}}} = 6$.\n\nFrom the AM-GM inequality we have:\n$$\n2^x + 2^{\\frac{1}{\\sqrt{x}}} + 2^{\\frac{1}{\\sqrt{x}}} \\ge 3 \\cdot \\sqrt[3]{2^x \\cdot 2^{\\frac{1}{\\sqrt{x}}} \\cdot 2^{\\frac{1}{\\sqrt{x}}}} = 3 \\cdot 2^{\\frac{1}{3}(x+\\frac{2}{\\sqrt{x}})}, \\forall x > 0.\n$$\nAlso, using the AM-GM inequality, we have:\n$$\n\\frac{1}{3} \\left( x + \\frac{2}{\\sqrt{x}} \\right) \\ge \\sqrt[3]{x \\cdot \\frac{1}{\\sqrt{x}} \\cdot \\frac{1}{\\sqrt{x}}} = 1,\n$$\nwhich implies that $2^x + 2 \\cdot 2^{\\frac{1}{\\sqrt{x}}} \\ge 6$. Therefore, we need equality in all AM-GM inequalities applied above, which is equivalent to $x = \\frac{1}{\\sqrt{x}}$, so $x = 1$ is the unique solution of the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit numbers $\\overline{abcd}$, so that $\\overline{ab}$, $\\overline{cb}$ and $d$ are prime numbers, and $\\overline{ab}^2 + \\overline{cb}^2 + d^2 = 2022$.",
    "options": [],
    "answer": "1342, 4312",
    "solution": "If $c = 0$, then $b^2 + d^2 \\le 2 \\cdot 7^2$, which means that $\\overline{ab}$ must be a prime number such that $1924 \\le \\overline{ab}^2 < 2022$, false. Therefore, $c \\ne 0$. As $\\overline{ab}$ and $\\overline{cb}$ are primes, from the given equality we deduce that $d = 2$. We obtain $\\overline{ab}^2 + \\overline{cb}^2 = 2018 < 47^2$, thus $\\overline{ab}$ and $\\overline{cb}$ are at most 43. Since $b$ is an odd digit, and the last digit of $\\overline{ab}^2 + \\overline{cb}^2$ is 8, we deduce that $b \\in \\{3, 7\\}$.\n\nIf $b = 7$, then $\\overline{ab}, \\overline{cb} \\in \\{17, 37\\}$, therefore $\\overline{ab}^2, \\overline{cb}^2 \\in \\{289, 1369\\}$ and $\\overline{ab}^2 + \\overline{cb}^2 \\ne 2018$.\n\nIf $b = 3$, then $\\overline{ab}, \\overline{cb} \\in \\{13, 23, 43\\}$, and we verify that only for $(\\overline{ab}, \\overline{cb}) \\in \\{(13, 43), (43, 13)\\}$ the given equality holds, therefore the solutions are $\\overline{abcd} = 1342$ and $\\overline{abcd} = 4312$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39131,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, denote $s(n)$ the sum of its odd digits. For instance, $s(512418) = 5 + 1 + 1 = 7$ and $s(82624) = 0$.\nCompute $s(1) + s(2) + s(3) + \\dots + s(2023)$.",
    "options": [],
    "answer": "16064",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the common point of the diagonals of the convex quadrilateral $ABCD$ and $M, N, P$ be the midpoints of the segments $AB, CD$ respectively $DO$. Prove that $AP \\parallel MN$ if and only if $AD \\parallel BN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $SABCD$ be a pyramid with the apex $S$ and whose base $ABCD$ is a parallelogram. We consider the points $M$, $N$, $P$ and $Q$ on the edges $SA$, $SB$, $SC$ and $SD$, respectively, such that $MNPQ$ is also a parallelogram.\n\na) If $ABCD$ is a rhombus, prove that $MNPQ$ is also a rhombus.\n\nb) If $ABCD$ is a rectangle, prove that $MNPQ$ is also a rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The planes $(SAB)$ and $(SCD)$ meet along the line $d_1$, and let $d_2$ be the intersection line of the planes $(SBC)$ and $(SDA)$.\nAs $AB \\parallel CD$, $AB \\subset (SAB)$ and $CD \\subset (SCD)$, from the 'roof theorem' it follows that $AB \\parallel CD \\parallel d_1$. Because $MN \\parallel PQ$, $MN \\subset (SAB)$ and $PQ \\subset (SCD)$, from the roof theorem we deduce that $MN \\parallel PQ \\parallel d_1$, therefore $AB \\parallel CD \\parallel MN \\parallel PQ \\parallel d_1$.\nSimilarly, we obtain $BC \\parallel DA \\parallel NP \\parallel QM \\parallel d_2$.\n\nIf $ABCD$ is a rhombus, then $AB = BC$, whence $MN = NP$, therefore the parallelogram $MNPQ$ is also a rhombus.\n\nb) If $ABCD$ is a rectangle, then $\\widehat{ABC} = 90^\\circ$. As $MN \\parallel AB$ and $NP \\parallel BC$, we obtain $\\widehat{MNP} = 90^\\circ$, therefore $MNPQ$ is also a rectangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle and let $M$ and $N$ be the midpoints of the sides $AD$ and $BC$, respectively. The circle through $A$ and $D$ tangent to $AC$ crosses the line $AB$ again at $P$ and the circle through $B$ and $C$ tangent to $BD$ crosses the line $AB$ again at $Q$. Let $t_A$ be the tangent of the circle $AMP$ at $A$ and let $t_B$ be the tangent of the circle $BNQ$ at $B$. Prove that the lines $CD$, $t_A$ and $t_B$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove that the triangles $ADP$ and $CDB$ are similar. Since $ABCD$ is cyclic, $\\angle DAP = \\angle DCB$; and since $AC$ is tangent to the circle $ADP$, $\\angle CBD = \\angle CAD = \\angle APD$. Consequently, the triangles $ADP$ and $CDB$ are similar.\n\nLet $R$ be the midpoint of $CD$. The points $M$ and $R$ correspond under the above similarity, so $\\angle PMA = \\angle BRC$.\n\nLet the line $CD$ meet the circle $ABR$ again at $K$ (possibly, $K = R$). Notice that $\\angle BAK = \\angle BRC = \\angle PMA$, so $AK$ is tangent to the circle $AMP$.\n\nSimilarly, $BK$ is tangent to the circle $BNQ$ and the conclusion follows.\n\nAlternative solution.\nWe prove the conclusion under the weaker assumptions that $ABCD$ is merely convex and $AM/DM = CN/BN$.\n\nLet $AB$ and $CD$ meet at $E$; if the two are parallel, then $E$ is their common ideal point. Let $CD$ cross $t_A$ and $t_B$ at $K_A$ and $K_B$, respectively. We will show that $K_A = K_B$, whence the conclusion.\n\nLet $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. The line pencils\n($AD, AC, AE, AK_A$) and ($PA, PD, PI, PM$)\nare congruent, since $\\angle K_A AD = \\angle APM$, $\\angle CAD = \\angle APD$ and $\\angle IAE = \\angle IPE$. Hence $(D, C; E, K_A) = (AD, AC; AE, AK_A) = (PA, PD; PI, PM) = (A, D; I, M) = DM/AM$. Similarly, $(D, C; E, K_B) = (BD, BC; BE, BK_B) = (QC, QB; QJ, QN) = (C, B; J, N) = BN/CN$.\n\nFinally, recall the assumption $AM/DM = CN/BN$. By the preceding, $(D, C; E, K_A) = (D, C; E, K_B)$, so $K_A = K_B$, as desired. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 2$, having at least four positive divisors, with the property that for any two distinct divisors $d_1$ and $d_2$ of $n$, such that $1 < d_1 < d_2 < n$, the number $d_2 - d_1$ is also a divisor of $n$.\nLucian Petrescu",
    "options": [],
    "answer": "6, 8, 12",
    "solution": "There are three solutions: $n = 6$, $n = 8$ and $n = 12$.\n\nFirst, notice that if $n$ is odd, all its divisors are also odd, so if $d_1$ and $d_2$ are two divisors such that $1 < d_1 < d_2 < n$, then $d_2 - d_1$ is even, so $d_2 - d_1 \\nmid n$. Therefore, $n$ is even.\n\nConsider $n = 2^a \\cdot b$, where $a, b \\in \\mathbb{N}_{\\ge 1}$ and $b$ is odd.\nIf $b = 1$, then $n = 2^a$, with $a \\in \\mathbb{N}^*$. For $a \\in \\{1, 2\\}$ there are no solutions, and for $a \\ge 4$, choosing $d_1 = 2^{a-3}$ and $d_2 = 2^{a-1}$, it follows that $d_2 - d_1 = 3 \\cdot 2^{a-3} \\nmid n$, a contradiction. Consequently, $a = 3$, so $n = 8$. It's trivial to see that $n = 8$ is a solution of the problem.\n\nFor $b \\ge 3$ we have two cases.\nIf $a = 1$, then $n = 2b$. Choose $d_1 = 2$ and $d_2 = b$; we obtain $b - 2 \\mid 2b$, so $b - 2 \\mid 2b - 2(b - 2) = 4$. Since $b$ is odd, we infer that $b = 3$, so $n = 6$, which is a solution.\n\nIf $a \\ge 2$, choose $d_1 = 2^a$ and $d_2 = 2^{a-1} \\cdot b$; we obtain $2^{a-1}(b-2) \\mid 2^a \\cdot b$, so $b - 2 \\mid 2b$. Since $b$ is odd, similarly it follows that $b = 3$. Therefore, $n = 2^a \\cdot 3$.\nChoose now $d_1 = 3$ and $d_2 = 2^a$; it results that $2^a - 3 \\mid 2^a \\cdot 3$, so $2^a - 3 \\mid 2^a \\cdot 3 - 3 \\cdot (2^a - 3) = 9$, from which we obtain $a = 2$. Hence, $n = 12$ and we easily verify that it is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the pairs $(x, y)$ of real numbers fulfilling\n$$\n3 \\cdot \\left\\{ \\frac{3x+2}{3} \\right\\} + 4 \\cdot \\left\\lfloor \\frac{4y+3}{4} \\right\\rfloor = 4 \\cdot \\left\\{ \\frac{4y+3}{4} \\right\\} + 3 \\cdot \\left\\lfloor \\frac{3x+2}{3} \\right\\rfloor = 18.\n$$",
    "options": [],
    "answer": "(6, 13/4), (5, 4)",
    "solution": "Since $0 \\le \\{a\\} < 1$ and $\\lfloor b \\rfloor$ is an integer, the equality $3\\{a\\} + 4\\lfloor b \\rfloor = 18$ is possible only when $\\{a\\} = \\frac{2}{3}$ and $\\lfloor b \\rfloor = 4$.\nSince $0 \\le \\{c\\} < 1$ and $\\lfloor d \\rfloor$ is an integer, the equality $4\\{c\\} + 3\\lfloor d \\rfloor = 18$ is possible only when $\\{c\\} = 0$, $\\lfloor d \\rfloor = 6$ (I) or $\\{c\\} = \\frac{3}{4}$, $\\lfloor d \\rfloor = 5$ (II).\n\nCase (I) yields $\\frac{3x+2}{3} = 6 + \\frac{2}{3}$ and $\\frac{4y+3}{4} = 4$, hence $x = 6$, $y = \\frac{13}{4}$.\nCase (II) yields $\\frac{3x+2}{3} = 5 + \\frac{2}{3}$ and $\\frac{4y+3}{4} = 4 + \\frac{3}{4}$, so $x = 5$, $y = 4$.\n\nThe required pairs are $(6, \\frac{13}{4})$ and $(5, 4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $\\triangle ABC$.\na) Prove that the interior bisector of the angle $\\angle A$ and the exterior bisectors of the angles $\\angle B$ and $\\angle C$ intersect at a point $I_A$.\n\nb) Let $I_A M \\perp AC$, $M \\in AC$, $I_A N \\perp BC$, $N \\in BC$ and $I_A P \\perp AB$, $P \\in AB$. Show that if $I_A M + I_A P = I_A N$, then the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $I_A$ is the point of intersection of the exterior bisectors of the angles $\\angle B$ and $\\angle C$, then $I_A$ is inside the angle $\\angle A$ and is equidistant from the sides $AB$ and $BC$, respectively of $BC$ and $AC$. Through transitivity, $I_A$ is equidistant from the sides $AB$ and $AC$ of the angle $\\angle A$, is inside the angle $\\angle A$, so it is on the interior bisector of the angle $\\angle A$.\n\nb) From the condition $I_A M + I_A P = I_A N$ we deduce that the quadrilateral $I_A M N P$ is a parallelogram. Also from point a) we have $I_A M = I_A N = I_A P$, so $I_A M N P$ is the rhombus and the triangles $I_A N P$ and $I_A M N$ are equilateral.\nThus, the inscribed quadrilateral $API_A M$ it has $\\angle P I_A M = 120^\\circ$, so $\\angle A = 60^\\circ$,\nOn the other hand, $C$ is exterior angle of the inscribed quadrilateral $I_A M C N$ so $\\angle C = \\angle M I_A N = 60^\\circ$, and $B$ is exterior angle of the inscribed quadrilateral $I_A P B N$ so $\\angle B = \\angle P I_A N = 60^\\circ$. Thus, triangle $ABC$ has all angles of $60^\\circ$, so it is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39138,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive real numbers $a$, $b$, $c$ are such that $a + b + c = 3$. Prove that the following inequality holds: $a^2 + b^2 + c^2 + a^2b + b^2c + c^2a \\ge 6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By adding $2ab + 2bc + 2ca$ to both sides, the inequality becomes:\n$$\n(a + b + c)^2 + a^2b + b^2c + c^2a \\ge 6 + 2ab + 2bc + 2ca.\n$$\nThus, we have to prove that $a^2b + b^2c + c^2a + 3 \\ge 2ab + 2bc + 2ca$. Since $a + b + c = 3$, the previous inequality is equivalent to:\n$$\n(b + a^2b) + (c + b^2c) + (a + c^2a) \\ge 2ab + 2bc + 2ca. \\quad (1)\n\n*Alternative solution.* By adding $a + b + c$ to both sides, the inequality becomes:\n$$\na^2 + b^2 + c^2 + a^2b + b^2c + c^2a + a + b + c \\ge 9. \\quad (2)\n$$\nUsing the obvious inequalities $b + a^2b \\ge 2ab$, $c + b^2c \\ge 2bc$ and $a + c^2a \\ge 2ca$, we deduce: $a^2 + b^2 + c^2 + a^2b + b^2c + c^2a + a + b + c \\ge a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 = 9$, therefore (2) is true, which ends the proof.\n\nIt is obvious that $b + a^2b \\ge 2ab$, $c + b^2c \\ge 2bc$ and $a + c^2a \\ge 2ca$. By summing these inequalities, we find that (1) is true, which ends the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all prime numbers $p, q < 2023$ such that $q \\mid p^2 + 8$ and $p \\mid q^2 + 8$.",
    "options": [],
    "answer": "(p, q) = (2, 2), (17, 3), (3, 17)",
    "solution": "If one of the numbers is $2$, then $p = q = 2$, and we can assume $p, q \\ge 3$. Since $(p, p^2 + 8) = (q, q^2 + 8) = 1$, we have: $q \\mid p^2 + 8$ and $p \\mid q^2 + 8 \\Rightarrow pq \\mid (p^2 + 8)(q^2 + 8) \\Rightarrow pq \\mid 8(p^2 + q^2 + 8) \\Rightarrow pq \\mid (p^2 + q^2 + 8)$.\n\nFor a fixed $k \\in \\mathbb{N}^*$, we determine the solutions in $\\mathbb{N}^* \\times \\mathbb{N}^*$ of the equation $p^2 + q^2 + 8 = kpq$ with $p, q < 2023$.\n\nLet's assume that $(p_0, q_0)$ is a solution for which the sum $p+q$ is minimal and $p_0 \\ge q_0$. If $p_0 = q_0$, since $p_0^2 \\mid 8$, we have $p_0 = q_0 = 1$ or $p_0 = q_0 = 2$ (cases we will analyze later). Now, let's assume that for $p_0 \\ge 3$, we have $p_0 > q_0$.\n\nIf $p_0 \\ge 5$ and $p'$ is the second solution of the equation $x^2 - (q_0k)x + q_0^2 + 8 = 0$, then we have $p' = \\frac{q_0^2+8}{p_0} \\le \\frac{p_0^2-2p_0+9}{p_0} < p_0$. Since $p' + q_0 < p_0 + q_0$, it follows that $p_0 \\in \\{3, 4\\}$.\n\nIf $p_0 = 3$, then $q_0 \\mid 17$, so $q_0 = 1$ and $k = 6$. Using Vieta jumping, we obtain the solutions given by the sequence $p_0 = 3, q_0 = 1, q_{n+1} = p_n, p_{n+1} = 6p_n - q_n$. Thus, we obtain the solutions $(3, 1)$, $(17, 3)$, $(99, 17)$, $(577, 99)$, while the remaining solutions have $p \\ge 2023$. The only solution remaining is $(17, 3)$.\n\nIf $p_0 = 4$, then $q_0 \\mid 24$ and $q_0 \\le 3$, so $q_0$ is $2$ or $1$, but both cases are impossible.\n\nNow, let's consider the cases $p_0 = 2$ and $p_0 = 1$.\n\nIf $p_0 = 2$, then $q_0 \\mid 12$ and $q_0 \\le 2$, so $q_0 = 2$ (the case $q_0 = 1$ is not possible) and $k = 4$. Using Vieta jumping, all solutions have both components even. The acceptable solution is $(2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ for which there exists three pairwise distinct positive integers $a, b, c$, such that $n = a + b + c$ and $(a+b)(b+c)(c+a)$ is a perfect cube.",
    "options": [],
    "answer": "10",
    "solution": "Let $n$ be the required minimum value, and $a, b, c \\in \\mathbb{N}_{\\ge 1}$ pairwise distinct, such that $n = a + b + c$ and $(a+b)(b+c)(c+a) = k^3$, where $k \\in \\mathbb{N}$.\n\nIf $a, b, c$ are all odd, then $n \\ge 1+3+5=9$. If $n=9$, then $\\{a, b, c\\} = \\{1, 3, 5\\}$, which leads to $(a+b)(b+c)(c+a) = 4 \\cdot 6 \\cdot 8 = 192$, which isn't a perfect square. Consequently, $n \\ge 11$.\n\nSuppose now that not all the numbers $a, b, c$ have the same parity. WLOG, suppose that $a$ and $b$ have the same parity. It follows that $a+b$ is even and $c+a$ and $b+c$ are both odd. Then $2 \\mid (a+b)(b+c)(c+a) = k^3$, so $2 \\mid k$, which implies $8 \\mid (a+b)(b+c)(c+a)$. Since $b+c$ and $c+a$ are odd, it follows that $8 \\mid a+b$, so $a+b \\ge 8$.\nIf $n=9$, then $a+b=8$ and $c=1$, and also $(b+c)(c+a) = ab+9$ must be a perfect cube, so $ab \\ge 18$. Since $4 = \\frac{a+b}{2} \\ge \\sqrt{ab}$, it follows that $ab \\le 16$, a contradiction.\n\nConsequently, $n \\ge 10$. Notice that for $n=10$ we may consider $a=1, b=2, c=7$, for which $(a+b)(b+c)(c+a) = 216 = 6^3$, so the required minimum value is $n=10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ be the centres of the faces $A'B'C'D'$, respectively $ADD'A'$ of a rectangular cuboid $ABCD'A'B'C'D'$. It is known that $AM \\perp A'C$ and $C'Q \\perp BD'$. Prove that $ABCD'A'B'C'D'$ is a cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a tetrahedron and $M$ be a variable point on the face $BCD$. The perpendicular from $M$ onto the plane $BCD$ meets the planes $ABC$, $ACD$ and $ADB$ in $M_1$, $M_2$ and $M_3$. Prove that the sum $MM_1 + MM_2 + MM_3$ is constant if and only if the tetrahedron's altitude from $A$ passes through the barycentre of the triangle $BCD$.\n\nVasile Pop",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number, $n$ a positive integer not divisible by $p$, and $\\mathbb{K}$ a field with $p^n$ elements, with unit element $1_{\\mathbb{K}}$ and zero element $\\hat{0} = 0_{\\mathbb{K}}$. For any $m \\in \\mathbb{N}^*$ we denote $\\hat{m} = \\underbrace{1_{\\mathbb{K}} + 1_{\\mathbb{K}} + \\dots + 1_{\\mathbb{K}}}_{m \\text{ times}}$ and we define the polynomial\n$$\nf_m = \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} X^{p^k} \\in \\mathbb{K}[X].\n$$\n\na) Show that the set of the roots of the polynomial $f_1$ is $\\{\\hat{k} \\mid k \\in \\{0, 1, \\dots, p-1\\}\\}$.\n\nb) Let $m \\in \\mathbb{N}^*$ be arbitrary. Determine the set of the roots in the field $\\mathbb{K}$ of the polynomial $f_m$.",
    "options": [],
    "answer": "a) The roots of f_1 in K are exactly the elements hat{k} with k from 0 to p minus 1. b) For every positive m, the roots of f_m in K are exactly the same set {hat{k} | k from 0 to p minus 1}.",
    "solution": "a.\nFor any polynomial $P \\in \\mathbb{K}[X]$ we shall denote by $Z_P$ the set of roots of $P$ in the field $\\mathbb{K}$. Because $|\\mathbb{K}| = p^n$, the characteristic of the field $\\mathbb{K}$ is $\\text{char}(\\mathbb{K}) = p$. Then $\\hat{m} = \\hat{0}$ for any multiple $m$ of $p$. In particular, since $k^p \\equiv k \\pmod p$, for any $k \\in \\{0, 1, \\dots, p-1\\}$ we have\n$$\nf_1(\\hat{k}) = (\\hat{k})^p - \\hat{k} = \\widehat{k^p} - \\hat{k} = \\widehat{k^p - k} = \\hat{0},\n$$\nHence, $\\{\\hat{k} \\mid k = \\overline{0, p-1}\\} \\subseteq Z_{f_1}$. Also, because $\\mathbb{K}$ is a field, $|Z_{f_1}| \\le \\text{grad}(f_1) = p$. It follows that $Z_{f_1} = \\{\\hat{k} \\mid k = \\overline{0, p-1}\\}$.\n\nb.\nBecause $p|C_p^k$, for any $k = \\overline{1, p-1}$, the identity $(a+b)^p = a^p + b^p$ holds for any $a, b \\in \\mathbb{K}$, and inductively we have $(a+b)^{p^k} = a^{p^k} + b^{p^k}$, for any $a, b \\in \\mathbb{K}$ and any $k \\in \\mathbb{N}$. Then for any $m \\in \\mathbb{N}^*$ we have:\n$$\n\\begin{align*}\nf_m(f_1(X)) &= \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} (X^p - X)^{p^k} = \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} (X^{p^{k+1}} - X^{p^k}) = \\\\\n&= \\sum_{k=0}^{m+1} (-1)^{m+1-k} (\\widehat{C_m^k} + \\widehat{C_m^{k-1}}) X^{p^k} = \\sum_{k=0}^{m+1} (-1)^{m+1-k} \\widehat{C_{m+1}^k} X^{p^k} = f_{m+1}(X).\n\\end{align*}\n$$\n\nWe shall prove by induction with respect to $m \\in \\mathbb{N}^*$ that $Z_{f_m} = \\{\\hat{k} \\mid k = \\overline{0, p-1}\\}$ for any $m \\in \\mathbb{N}^*$, which will solve the problem. For $m = 1$ we have shown this in part a.\n\nAssume now that the property holds for some arbitrary $m \\in \\mathbb{N}^*$. We prove now that it will hold also for $m + 1$:\nFor any $k \\in \\{0, 1, \\dots, p-1\\}$ we have\n$$\nf_{m+1}(\\hat{k}) = f_m(f_1(\\hat{k})) = f_m(\\hat{0}) = \\hat{0},\n$$\nso that $\\{\\hat{k} \\mid k = \\overline{0, p-1}\\} \\subseteq Z_{f_{m+1}}$.\nLet $\\alpha \\in Z_{f_{m+1}}$ be arbitrary. Then $f_m(f_1(\\alpha)) = f_{m+1}(\\alpha) = \\hat{0}$, so that $f_1(\\alpha) \\in Z_{f_m}$. Hence, there is a $k \\in \\{0, 1, \\dots, p-1\\}$ such that $f_1(\\alpha) = \\hat{k}$. We obtain\n$$\n\\alpha^p = \\alpha + \\hat{k}, \\quad \\alpha^{p^2} = (\\alpha + \\hat{k})^p = \\alpha^p + \\hat{k}^p = (\\alpha + \\hat{k}) + \\hat{k} = \\alpha + 2 \\cdot \\hat{k},\n$$\nand, inductively, if $\\alpha^{p^m} = \\alpha + m \\cdot \\hat{k}$, then $\\alpha^{p^{m+1}} = (\\alpha + m \\cdot \\hat{k})^p = \\alpha + (m+1) \\cdot \\hat{k}$.\nIn the multiplicative group $(\\mathbb{K}^*, \\cdot)$ we have $x^{p^n-1} = 1$, for any $x \\in \\mathbb{K}^*$, so that $x^{p^n} = x$ holds for any element $x \\in \\mathbb{K}$. Then $\\alpha = \\alpha^{p^n} = \\alpha + n \\cdot \\hat{k}$, and $n \\cdot \\hat{k} = \\hat{0}$.\nSince $n$ is not divisible by the characteristic $p$, it follows that $\\hat{k} = \\hat{0}$. But then $f_1(\\alpha) = \\hat{0}$ and $\\alpha \\in Z_{f_1} = \\{\\hat{k} \\mid k = \\overline{0, p-1}\\}$. Thus we obtained the reverse inclusion $Z_{f_{m+1}} \\subseteq \\{\\hat{k} \\mid k = \\overline{0, p-1}\\}$, so that the equality $Z_{f_{m+1}} = \\{\\hat{k} \\mid k = \\overline{0, p-1}\\}$ holds. The claim holds then for any $m \\in \\mathbb{N}^*$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39144,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Solve the equation in the set of real numbers $[x]^2 - x = -0.99$.\n\nb) Show that, for every $a \\le -1$, the equation $[x]^2 - x = a$ has no real solutions.",
    "options": [],
    "answer": "a) x ∈ {0.99, 1.99}. b) For every a ≤ −1, there are no real solutions.",
    "solution": "a) The equation is written equivalently $[x]^2 - [x] = \\{x\\} - 0.99$, thus $\\{x\\} - 0.99 \\in \\mathbb{Z}$.\nSince $0 \\le \\{x\\} < 1$, we deduce that $-0.99 \\le \\{x\\} - 0.99 < 0.01$ therefore $\\{x\\} - 0.99 = 0$ so $\\{x\\} = 0.99$. Also, from $[x]^2 - [x] = 0$ we deduce that $[x] = 0$ or $[x] = 1$, so $x \\in \\{0.99; 1.99\\}$.\n\nb) The equation is written equivalently $[x]^2 - [x] = \\{x\\} + a$. We assume, by absurdity, that there is $a \\le -1$ for which the equation has real solutions. Then, since $\\{x\\} < 1$, $[x]^2 - [x] = \\{x\\} + a < 0$.\nDenoting $[x] = y \\in \\mathbb{Z}$, we have $y(y - 1) < 0$, so $y \\in (0, 1)$, absurd.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39145,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ is isosceles, with $AB = AC$ and $\\angle ABC = 72^\\circ$. The point $D$ is taken on the line $BC$, so that $C$ is on the segment $BD$ and $CD = AB$.\n\na) Prove that $AC$ is the bisector of the angle $\\angle BAD$.\n\nb) The point $E$ is taken on the parallel to $AB$ through $D$, on the same side of $BD$ as $A$, so that $DE = DB$. Let $F$ be the common point of the lines $AD$ and $BE$. Prove that the lines $AC$ and $AE$ are perpendicular and $AF = FC = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) In the triangle $ABC$, $\\angle BAC = 36^\\circ$. From the isosceles triangle $ACD$, with $\\angle ACD = 180^\\circ - 36^\\circ = 144^\\circ$, it follows that $\\angle CAD = \\angle ADC = 36^\\circ$, hence $AC$ is the bisector of the angle $\\angle BAD$.\n\nb) From $\\angle ABD = \\angle BAD = 72^\\circ$ it follows that the triangle $ABD$ is isosceles, hence $AD = BD$. Since $DE = DB$, the triangle $ADE$ is isosceles, therefore $\\angle DAE = (180^\\circ - \\angle ADE) : 2 = (180^\\circ - 72^\\circ) : 2 = 54^\\circ$. So, $\\angle CAE = \\angle CAD + \\angle DAE = 36^\\circ + 54^\\circ = 90^\\circ$.\nFrom $\\angle AFB = 180^\\circ - \\angle BAF - \\angle ABF = 72^\\circ$ it follows $\\triangle BAC \\equiv \\triangle ABF$ (S.A.S.), whence $BC = AF$. Now $BD = AD$ yields $CD = FD$. This leads to $\\triangle BAC \\equiv \\triangle CDF$ (S.A.S.), hence $BC = FC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 4$ and $a_1, a_2, \\dots, a_n$ be real numbers such that\n$$\na_k^3 = a_{k+1}^2 + a_{k+2}^2 + a_{k+3}^2\n$$\nfor all $k \\in \\{1, 2, \\dots, n\\}$ – indices are considered modulo $n$. Show that $a_1 = a_2 = \\dots = a_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there exists $i \\in \\{1, 2, \\dots, n\\}$ for which $a_i = 0$. Clearly $a_{i+1} = 0$, and consequently $a_1 = a_2 = \\dots = a_n = 0$.\n\nSuppose now that $a_i \\ne 0$ for all $i \\in \\{1, 2, \\dots, n\\}$. Choose $p, q \\in \\{1, 2, \\dots, n\\}$ such that $a_q \\le a_i \\le a_p$ for all $i \\in \\{1, 2, \\dots, n\\}$.\n\nNow $a_p^3 = a_{p+1}^2 + a_{p+2}^2 + a_{p+3}^2 \\le 3a_p^2$ and $a_q^3 = a_{q+1}^2 + a_{q+2}^2 + a_{q+3}^2 \\ge 3a_q^2$ yield $a_p \\le 3 \\le a_q$, therefore $a_1 = a_2 = \\dots = a_n = 3$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39147,
    "subject": "Mathematics (Multi-modal)",
    "question": "A diagonal line of a (not necessarily convex) polygon with at least four sides is any line through two non-adjacent vertices of that polygon. Determine all polygons with at least four sides satisfying the following condition: The reflexion of each vertex in each diagonal line lies inside or on the boundary of the polygon.",
    "options": [],
    "answer": "Rhombi (convex quadrilaterals with all sides equal).",
    "solution": "Begin by noticing that $K$ is convex: Otherwise, the convex hull $\\hat{K}$ of $K$ would have a side $ab$ whose line of support is a diagonal line of $K$ ($a$ and $b$ are, of course, non-adjacent vertices of $K$ and there might virtually be other vertices or even sides of $K$ along the line segment $ab$). The reflexion of a third vertex of $\\hat{K}$, and hence of $K$, in the line $ab$ would then fall outside $\\hat{K}$, and hence outside $K$, contradicting the vertex reflexion condition $K$ satisfies.\n\nNext, let $a$, $b$ and $c$ be consecutive vertices of $K$, and let $d \\ne a, b, c$ be a fourth vertex. Since the reflexions of $a$ and $c$ in the line $bd$ do not fall outside $K$, it follows that $bd$ is the internal bisectrix of the angle $abc$, and, by convexity, $a$ and $c$ are reflexions of one another in the line $bd$.\n\nConsequently, $K$ is a convex quadrangle $abcd$ such that $a$ and $c$ are reflexions of one another in the line $bd$, and $b$ and $d$ are reflexions of one another in the line $ac$; that is, $K$ is a lozenge (rhombus).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the functions $f : \\mathbb{R} \\to \\mathbb{R}$, satisfying\n$$\nf(f(x)) + y \\cdot f(x) \\leq x + x \\cdot f(f(y)),\n$$\nfor every real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "For $x = 0$ in the given relation we get $f(f(0)) + y f(0) \\le 0$, for every $y \\in \\mathbb{R}$, which is true only if $f(0) = 0$. Indeed, if $f(0) \\ne 0$, then $y \\le \\frac{-f(f(0))}{f(0)}$, for every $y \\in \\mathbb{R}$, or $y \\ge \\frac{-f(f(0))}{f(0)}$, for every $y \\in \\mathbb{R}$, which is impossible.\n\nChoosing $y = 0$, it results $f(f(x)) \\le x$, for every real $x$. (1)\n\nFor $x = 1$ in the given relation we get $f(f(1)) + y f(1) \\le 1 + f(f(y))$, and using (1) we get $f(f(1)) + y f(1) \\le 1 + y$, so $y(f(1) - 1) \\le 1 - f(f(1))$, for every real $y$. The relation is true, as above, only if $f(1) - 1 = 0$, i.e. $f(1) = 1$.\n\nIn the above context, for $x = 1$, the relation from the hypothesis ensures that $y \\le f(f(y))$, for every real $y$. (2)\n\nFrom (1) and (2) we get that $f(f(x)) = x$ for every $x \\in \\mathbb{R}$, and using the hypothesis we obtain that $x + y f(x) \\le x + x y$, so $y f(x) \\le x y$ for every $x, y \\in \\mathbb{R}$.\n\nFor $y = 1$ we get $f(x) \\le x$ and for $y = -1$ we get $f(x) \\ge x$, so $f(x) = x$ for every $x \\in \\mathbb{R}$, which satisfies the hypothesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the functions $f : \\mathbb{Q} \\to \\mathbb{Q}$, so that $f(1 - xy) = (1 - f(x))(1 - f(y))$, for every rational numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 1 - x",
    "solution": "Let us find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that for all $x, y \\in \\mathbb{Q}$,\n$$\nf(1 - xy) = (1 - f(x))(1 - f(y)).\n$$\n\nLet us denote the given equation as $(*)$.\n\nStep 1. Plug $x = 0$ into $(*)$:\n$$\nf(1 - 0 \\cdot y) = (1 - f(0))(1 - f(y)) \\implies f(1) = (1 - f(0))(1 - f(y)).\n$$\nBut the left side is independent of $y$, so the right side must also be independent of $y$. Therefore, $1 - f(0) = 0$ or $1 - f(y)$ is constant.\n\nCase 1: $1 - f(0) = 0 \\implies f(0) = 1$\n\nThen $f(1) = 0$ (since $f(1) = (1 - f(0))(1 - f(y)) = 0$ for all $y$).\n\nCase 2: $1 - f(y)$ is constant for all $y$, so $f(y)$ is constant. Let $f(y) = c$ for all $y$.\nPlug into $(*)$:\n$$\nc = (1 - c)^2.\n$$\nSo $c = 1 - 2c + c^2 \\implies c^2 - 3c + 1 = 0$.\nThe solutions are $c = \\frac{3 \\pm \\sqrt{5}}{2}$, but these are not rational, so there is no constant rational solution.\n\nSo only Case 1 is possible: $f(0) = 1$, $f(1) = 0$.\n\nStep 2. Plug $y = 1$ into $(*)$:\n$$\nf(1 - x \\cdot 1) = (1 - f(x))(1 - f(1)) \\implies f(1 - x) = (1 - f(x))(1 - 0) = 1 - f(x).\n$$\nSo $f(1 - x) = 1 - f(x)$ for all $x \\in \\mathbb{Q}$.\n\nStep 3. Plug $y = 0$ into $(*)$:\n$$\nf(1 - x \\cdot 0) = (1 - f(x))(1 - f(0)) \\implies f(1) = (1 - f(x))(1 - 1) = 0.\n$$\nWhich is consistent with $f(1) = 0$.\n\nStep 4. Plug $x = a$, $y = b$ into $(*)$ and use $f(1 - x) = 1 - f(x)$:\nLet us try to find the general form.\nLet us try $f(x) = x$.\nThen $f(1 - xy) = 1 - xy$, $(1 - f(x))(1 - f(y)) = (1 - x)(1 - y) = 1 - x - y + xy$.\nBut $1 - xy \\neq 1 - x - y + xy$ in general, so $f(x) = x$ is not a solution.\n\nTry $f(x) = 1 - x$.\nThen $f(1 - xy) = 1 - (1 - xy) = xy$, $(1 - f(x))(1 - f(y)) = x y$.\nSo $f(1 - xy) = (1 - f(x))(1 - f(y))$.\n\nTherefore, $f(x) = 1 - x$ is a solution.\n\nLet us check if there are other solutions.\nSuppose $f(x) = a x + b$ for some $a, b \\in \\mathbb{Q}$.\nFrom above, $f(1 - x) = 1 - f(x)$:\n$$\nf(1 - x) = a(1 - x) + b = a - a x + b = 1 - (a x + b) = 1 - f(x).\n$$\nSo $a - a x + b = 1 - a x - b \\implies a + b = 1 - b \\implies a + 2b = 1$.\n\nAlso, $f(0) = b = 1$ (from above), so $b = 1$, so $a + 2 \\cdot 1 = 1 \\implies a = -1$.\nSo $f(x) = -x + 1$.\n\nTherefore, the only solution is $f(x) = 1 - x$.\n\nCheck:\n$$\nf(1 - xy) = 1 - (1 - xy) = xy,\n$$\n$$\n(1 - f(x))(1 - f(y)) = x y.\n$$\nSo it works.\n\nConclusion:\n\nThe only function $f : \\mathbb{Q} \\to \\mathbb{Q}$ satisfying $f(1 - xy) = (1 - f(x))(1 - f(y))$ for all $x, y \\in \\mathbb{Q}$ is\n$$\nf(x) = 1 - x.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the triples $(a, b, c)$ of integer numbers, fulfilling the relations\n$$\na^2 + a = b + c, \\quad b^2 + b = a + c, \\quad c^2 + c = a + b.\n$$",
    "options": [],
    "answer": "(0, 0, 0) and (1, 1, 1)",
    "solution": "Adding the relations yields $a^2 + b^2 + c^2 = a + b + c$. (*)\n\nNotice that, if $x$ is an integer, then $x^2 \\ge x$, with equality only if $x = 0$ or $x = 1$.\nIndeed, if $x \\le 0$, then $x^2 \\ge 0 \\ge x$, and if $x \\ge 1$, then $x^2 = x \\cdot x \\ge x \\cdot 1 = x$.\nTherefore, the equality (*) is possible only if $a^2 = a$, $b^2 = b$, and $c^2 = c$.\nThe relation $a^2 = a$ gives $a = 0$ or $a = 1$. If $a = 0$, then $b = c = 0$; if $a = 1$, then $b = c = 1$. The required triples are $(0, 0, 0)$ and $(1, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39151,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *special* if all of its digits are nonzero and any two adjacent digits are consecutive (not necessarily in ascending order).\na) Find the largest special number for which the sum of its digits is $2023$.\nb) Find the smallest special number for which the sum of its digits is $2022$.",
    "options": [],
    "answer": "a) The largest special number is the 1349-digit alternating sequence starting with one, i.e., 121212…121, with 675 ones and 674 twos.\nb) The smallest special number is 56787 followed by 117 repetitions of 89, i.e., 56787 8989…89 (117 copies of 89).",
    "solution": "a) The largest number $m$ will have as many digits as possible, so we will choose the smallest possible digits. As next to a $1$ we can only put a $2$, and $2023 = 3 \\cdot 674 + 1$, we choose $m = \\underbrace{12121 \\dots 121}_{1349 \\text{ digits}}$, with $674$ digits equal to $2$ and $675$ digits equal to $1$.\n\nb) As $8+9=17$ and $2022 = 17 \\cdot 118+16$, if in the decimal writing of $n$ we would use at most $237$ digits, the sum of these digits would be at most $118 \\cdot (8+9)+9 < 2022$, false. Therefore we can't use fewer than $2 \\cdot 118 + 2 = 238$ digits. To use exactly $238$ digits, we must have $118$ groups of $8$ and $9$, and another two digits whose sum is $16$. The two digits could be $7$ and $9$, or $8$ and $8$. In both cases, the number of even digits of $n$ differs by $2$ from the number of its odd digits. This is impossible, because in the decimal writing of a special number, the even digits alternate with the odd ones, so the number of its even digits is either equal to that of its odd digits, or differs from it by $1$. Therefore $n$ must have at least $239$ digits. We are looking for three consecutive digits, smaller than $8$, whose sum is at least $16$ and which are the first three digits of $n$. Moreover, we are looking for the first digit of $n$ to be as small as possible. The smallest digits that fulfill these requirements are $5$, $6$, $7$. The solution is $n = 56787\\underbrace{8989\\dots89}_{117}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ so that there are exactly $1 + 2^{n+2}$ integers between $3^n$ and $3^{n+1}$.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39153,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triple of integers $(a, b, c)$ is called *artistic* if the number $\\frac{ab+bc+ca}{a+b+c}$ is also an integer.\n\na) Determine the integers $n$ for which the triples $(n, n+1, n+3)$ are artistic.\n\nb) If $(x, y, z)$ is an artistic triplet, prove that $\\frac{x^4+y^4+z^4}{x+y+z}$ is an integer.",
    "options": [],
    "answer": "a) n = -1 or n = 1, yielding the triples (-1, 0, 2) and (1, 2, 4).\nb) The quantity (x^4 + y^4 + z^4) / (x + y + z) is an integer for any artistic triple (x, y, z).",
    "solution": "a) The triple $(n, n+1, n+3)$, $n \\in \\mathbb{Z}$, is artistic iff $\\frac{3n^2+8n+3}{3n+4} = n + \\frac{4n+3}{3n+4}$ is an integer. It follows that $3n+4$ divides $4(3n+4)-3(4n+3) = 7$, whence we obtain $n \\in \\{-1, 1\\}$, therefore the solutions are $(-1, 0, 2)$ and $(1, 2, 4)$.\n\nb) Since $x + y + z$ and $\\frac{xy+yz+zx}{x+y+z}$ are both integers, $\\frac{x^2+y^2+z^2}{x+y+z} = (x+y+z) - 2 \\cdot \\frac{xy+yz+zx}{x+y+z}$ and $\\frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z} = (xy+yz+zx) \\cdot \\frac{xy+yz+zx}{x+y+z} - 2xyz$ are also integers. Thus, $\\frac{x^4+y^4+z^4}{x+y+z} = (x^2+y^2+z^2) \\frac{x^2+y^2+z^2}{x+y+z} - 2 \\cdot \\frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z}$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39154,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Infinitely many pairs of real numbers $(x, y)$ exist such that $x, y \\in [0, \\sqrt{3}]$ and the following equality holds: $x \\cdot \\sqrt{3-y^2} + y \\cdot \\sqrt{3-x^2} = 3$;\n\nb) **No** pair of rational numbers $(x, y)$ exists such that $x, y \\in [0, \\sqrt{3}]$ and the following equality holds: $x \\cdot \\sqrt{3-y^2} + y \\cdot \\sqrt{3-x^2} = 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Any pair $(a, \\sqrt{3-a^2})$, with $a \\in [0, 1]$, is a solution.\n\nb) By squaring the equality $y \\cdot \\sqrt{3-x^2} = 3 - x \\cdot \\sqrt{3-y^2}$, we deduce that $(\\sqrt{3-y^2} - x)^2 = 0$, therefore $x^2 + y^2 = 3$. (1)\nAssume that there are numbers $x, y \\in \\mathbb{Q}_+ \\cap [0, \\sqrt{3}]$, for which (1) is true.\nIt is obvious that $x \\ne 0$ and $y \\ne 0$. Consider the positive integers $a, b, c, d$, such that $(a, b) = (c, d) = 1$, $x = \\frac{a}{b}$ and $y = \\frac{c}{d}$. From (1) we obtain $a^2d^2 + b^2c^2 = 3b^2d^2$. (2)\nFrom $(c, d) = 1$ and $d^2 \\mid b^2c^2$, we deduce $d^2 \\mid b^2$. Similarly, from $(a, b) = 1$ and $b^2 \\mid a^2d^2$, we obtain $b^2 \\mid d^2$. Therefore, $b^2 = d^2$ and (2) leads to $a^2 + c^2 = 3b^2$.\nIf the integers $a$ and $c$ aren't multiples of $3$, then $a^2 + c^2 = M_3 + 2 \\ne 3b^2$, false. Consequently $3 \\mid a$ and $3 \\mid c$, therefore $3b^2 = M_9$, thus $3 \\mid b$ and $(a, b) \\ge 3$, false. This contradicts our assumption and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a group with the unit element $e$, and $H$ and $K$ two proper subgroups of $G$, such that $H \\cap K = \\{e\\}$ and the set $(G \\setminus (H \\cup K)) \\cup \\{e\\}$ is closed with respect to the operation in $G$. Show that $x^2 = e$, for any $x \\in G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $L = (G \\setminus (H \\cup K)) \\cup \\{e\\}$. Because $x \\in H \\cup K \\iff x^{-1} \\in H \\cup K$, it follows that $x \\in L \\iff x^{-1} \\in L$, so that $L$ is a proper subgroup of $G$.\nAlso, $L \\cap H = L \\cap K = H \\cap K = \\{e\\}$, $G = H \\cup K \\cup L$ and it follows that for any permutation $\\{A, B, C\\} = \\{H, K, L\\}$, if $a \\in A \\setminus \\{e\\}$ and $b \\in B \\setminus \\{e\\}$, then $ab \\in C \\setminus \\{e\\}$.\n\nThen, in the same conditions as above, $a^2b = a(ab) \\in B \\setminus \\{e\\}$, so that $a^2 \\in A \\cap B = \\{e\\}$. Hence, $x^2 = e$ for any $x \\in G \\setminus \\{e\\}$.\nSince $e^2 = e$, it follows that $x^2 = e$ for any $x \\in G$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39156,
    "subject": "Mathematics (Multi-modal)",
    "question": "A store sold 235 robots over the course of twelve months. Each month, 16, 20, or 25 robots were sold. Find the number of months in which exactly 20 robots were sold.",
    "options": [],
    "answer": "4",
    "solution": "If exactly 16 robots were sold each month, the number of the robots sold would have been $12 \\cdot 16 = 192$. The difference of 43 robots comes from the months where 20 robots were sold (4 more each month) and from the months where 25 robots were sold (9 more each month). Denote by $a$ and $b$ the number of months where 20 robots were sold and where 25 robots were sold, respectively. We have $4a + 9b = 43$,\n\ntherefore $b$ is an odd number which does not exceed 4. We obtain that $b = 3$, therefore exactly 20 robots were sold in $a = 4$ months.\n\nAlternative solution.\nLet $x, y, z$ be the number of months in which 16, 20, and 25 robots were sold, respectively. We have $16x + 20y + 25z = 235$, thus $x = 5(47 - 3x - 4y - 5z)$, therefore $5 \\mid x$, consequently $x \\in \\{5, 10\\}$. If $x = 10$, we obtain $4y + 5z = 15$ and $y + z = 12 - 10 = 2$, which leads to $z = 7 > y + z = 2$, false. If $x = 5$, we have $4y + 5z = 31$ and $y + z = 12 - 5 = 7$ and we obtain $z = 3$, therefore exactly 20 robots were sold in $y = 7 - z = 4$ months.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a square of side length $1$ there are $n > 1$ squares, whose side lengths add up to $1$, and each parallel to a side of the large square intersects the interior of at most one of the interior squares. Prove that the sides of the small squares are parallel to the sides of the large square.\n\nCristian Șăvescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $P$ and $Q$ be points on sides $AB$ and $AC$, respectively, such that $AP = AQ$ and line $PQ$ passes through the incenter $I$ of triangle $ABC$. Let $M$ be the second intersection point of the circumcircles of triangles $BPI$ and $CQI$. The lines $PM$ and $BI$ intersect at $D$ and the lines $QM$ and $CI$ meet at $E$. Prove that the line $MI$ passes through the midpoint of segment $DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since triangle $APQ$ is isosceles, we have $\\angle APQ = \\angle AQP$. $BMIP$ and $CMIQ$ are cyclic quadrilaterals, so $\\angle APQ = \\angle BMI$ and $\\angle AQP = \\angle CMI$. It follows that $\\angle BMI = \\angle CMI$. (1).\n\nLet $K$ be the point where lines $IM$ and $BC$ meet. Since $BI$ is the bisector of $\\angle ABC$ and $BMIP$ is a cyclic quadrilateral, we get $\\angle IMP = \\angle IBP = \\angle IBK$. From $\\angle KMD = \\angle KBD$ we infer that $BDKM$ is cyclic, so $\\angle IDK = \\angle BMK$. Similarly, we prove that $CEKM$ is also cyclic and $\\angle IEK = \\angle CMK$.\n![](attached_image_1.png)\nUsing (1), it follows that $\\angle IDK = \\angle IEK$.\nThe quadrilaterals $BDKM$ and $BMIP$ are cyclic, so $\\angle BKD = \\angle BMD = \\angle BIP$, (2).\nAlso, the quadrilaterals $CEKM$ and $CMIQ$ are cyclic, so $\\angle CKE = \\angle CME = \\angle CIQ$, (3).\nFrom (2) and (3) we deduce that $\\angle BKD + \\angle CKE = \\angle BIP + \\angle CIQ$. Consequently, we obtain $\\angle DKE = 180^\\circ - (\\angle BKD + \\angle CKE) = 180^\\circ - (\\angle BIP + \\angle CIQ) = \\angle DIE$. Since $\\angle IDK = \\angle IEK$ and $\\angle DKE = \\angle DIE$, it follows that $DIEK$ is a parallelogram, so line $KI$ bisects the segment $DE$. As points $I$, $K$, and $M$ are collinear, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39159,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the small arc $AB$ of the circumcircle of the equilateral triangle $ABC$ we consider a point $N$ such that the length of the arc $NB$ is $30^\\circ$. Consider the perpendicular lines from $N$ to $AC$ and $AB$, respectively. These lines intersect again the circumcircle of the triangle $ABC$ in points $M$ and $I$, respectively.\n\na) Prove that $IMN$ is an equilateral triangle.\n\nb) If $H_1, H_2$, and $H_3$ are the orthocenters of the triangles $NAB, IBC$, and $CAM$, respectively, prove that $H_1H_2H_3$ is an equilateral triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $O$ be the circumcenter of the triangle $ABC$. Without loss of generality, we consider $O(0)$, while the vertices of the triangle are $A(1)$, $B(\\varepsilon)$, and $C(\\varepsilon^2)$, where $\\varepsilon = -\\frac{1}{2} + i\\frac{\\sqrt{3}}{2}$. Because the length of the arc $NB$ is $30^\\circ$, we have $AO \\perp ON$, so point $N$ has the affix $i$. Moreover, $NI \\perp AB$ implies the existence of $\\alpha \\in \\mathbb{R}^*$ such that:\n$$\n\\frac{i - z_I}{1 - \\varepsilon} = i\\alpha \\Rightarrow z_I = i - \\frac{3}{2}i\\alpha - \\frac{\\sqrt{3}}{2}\\alpha,\n$$\nwhere $z_I$ is the affix of the point $I$. From $|z_I| = 1$ we obtain $\\alpha = 1$, so the affix of the point $I$ is $i\\varepsilon$. In the same manner, from $MN \\perp AC$ we obtain that the affix of the point $M$ is $i\\varepsilon^2$, so the triangle $IMN$ is equilateral.\n\nb) Using Sylvester's theorem we have the affixes of the orthocenters as follows:\n$$\n\\begin{aligned}\nz_{H_1} &= z_A + z_N + z_B = i + 1 + \\varepsilon, \\\\\nz_{H_2} &= z_B + z_M + z_C = \\varepsilon + i\\varepsilon + \\varepsilon^2 = \\varepsilon z_{H_1}, \\\\\nz_{H_3} &= z_C + z_I + z_A = \\varepsilon^2 + i\\varepsilon^2 + 1 = \\varepsilon^2 z_{H_1},\n\\end{aligned}\n$$\nso the triangle $H_1H_2H_3$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and three sets of real numbers $A, B, C$, pairwise disjoint, each of them having $n$ elements.\nLet $a$ be the number of triples $(x, y, z) \\in A \\times B \\times C$ for which $x < y < z$ and $b$ be the number of triples $(x, y, z) \\in A \\times B \\times C$ for which $x > y > z$. Prove that $a - b$ is divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $y \\in B$, arbitrarily chosen. Denote by $A_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x < y < z$, and by $B_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x > y > z$.\n\nLet $A = \\{a_1 < a_2 < \\dots < a_n\\}$ and $C = \\{c_1 < c_2 < \\dots < c_n\\}$.\nOn the real axis, $y$ separates the numbers $a_1, a_2, \\dots, a_n$ into $k$ numbers smaller than $y$ and $n-k$ numbers greater than $y$, where $k$ is some number from $\\{0, 1, 2, \\dots, n\\}$. Likewise, $c_1, c_2, \\dots, c_n$ are separated by $y$ into $p$ numbers smaller than $y$ and $n-p$ numbers greater than $y$, where $p \\in \\{0, 1, 2, \\dots, n\\}$. It follows that $A_y = k(n-p)$ and $B_y = p(n-k)$, so $A_y - B_y = n(k-p)$.\n\nSumming over all $y \\in B$, we get $a - b = \\sum_{y \\in B} (A_y - B_y) = n \\sum_{y \\in B} (k-p)$. Since $k$ and $p$ depend on $y$, but for each $y$ the difference $k-p$ is an integer, and the sum is multiplied by $n$, it follows that $a - b$ is divisible by $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two matrices from $M_n(\\mathbb{R})$. Show that $\\text{rank}(A) = \\text{rank}(B)$ if and only if there are three invertible matrices $X, Y, Z \\in M_n(\\mathbb{R})$ such that $AX + YB = AZB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. Assume that there are the invertible matrices $X, Y, Z \\in M_n(\\mathbb{R})$ such that $AX + YB = AZB$. Then $\\text{rank}(A) = \\text{rank}(AX) = \\text{rank}((AZ - Y)B) \\le \\text{rank}(B)$. Similarly, $\\text{rank}(B) \\le \\text{rank}(A)$. Therefore, $\\text{rank}(A) = \\text{rank}(B)$.\n\n2. Assume $\\text{rank}(A) = \\text{rank}(B) = r$. Then there are the invertible matrices $T, U, V, W \\in M_n(\\mathbb{R})$ such that\n$$\nTAU = \\begin{pmatrix} I_r & O_{r,n-r} \\\\ O_{n-r,r} & O_{n-r} \\end{pmatrix} = VBW.\n$$\nThus $A(UW^{-1}) = (T^{-1}V)B$. We can choose $\\lambda \\in \\mathbb{R}$ such that $\\det(B - \\lambda(UW^{-1})) \\ne 0$ and $\\det(A + \\lambda(T^{-1}V)) \\ne 0$. Define the invertible matrices $X = B - \\lambda(UW^{-1})$ and $Y = A + \\lambda(T^{-1}V)$. We obtain\n$$\nAX + YB = 2AB - \\lambda A (UW^{-1}) + \\lambda (T^{-1}V) B = AZB,\n$$\nwhere $Z = 2I_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $BC > AB$, such that the points $A$, $H$, $I$ and $C$ are concyclic (where $H$ is the orthocenter and $I$ is the incenter of triangle $ABC$). The line $AC$ intersects the circumcircle of triangle $BHC$ at point $T$, and the line $BC$ intersects the circumcircle of triangle $AHC$ at point $P$. If the lines $PT$ and $HI$ are parallel, determine the measures of the angles of triangle $ABC$.\nAdrian Bud",
    "options": [],
    "answer": "A = 80°, B = 60°, C = 40°",
    "solution": "Let lines $CH$ and $PT$ meet at $U$ and denote by $A'$, $B'$, $C'$ the feet of the altitudes of the triangle $ABC$.\n\n$AHIC$ is a cyclic quadrilateral, so $\\angle CHI = \\angle CAI = \\frac{\\angle A}{2}$. Since $PT \\parallel HI$, it follows that $\\angle HUT = \\angle CHI = \\frac{\\angle A}{2}$.\nHence, $\\angle CUP = \\angle HUT = \\frac{\\angle A}{2}$. (1)\n\nThe quadrilateral $AHIC$ is also cyclic, so $\\angle A'HI = \\angle ACI = \\frac{\\angle C}{2}$.\nFrom the triangle $A'HC$ we have $\\angle A'CH = 90^\\circ - \\angle A'HC = 90^\\circ - \\frac{\\angle A+\\angle C}{2} = \\frac{\\angle B}{2}$. From the right-angled triangle $BCC'$ we get $\\angle B + \\frac{\\angle B}{2} = 90^\\circ$, so $\\angle B = 60^\\circ$.\nNotice that $\\angle ABB' = \\angle ACC' = 90^\\circ - \\angle A$. The quadrilateral $BCTH$ is cyclic, so $\\angle B'TB = \\angle TCH = 90^\\circ - \\angle A = \\angle ABB'$.\nTherefore, $BB'$ is an angle bisector and also an altitude for triangle $ABT$, so $AB = BT$. Since $AHIC$ is cyclic, it follows that $\\angle BPI = \\angle CAI = \\frac{\\angle A}{2}$ and $\\angle API = \\angle ACI = \\frac{\\angle C}{2}$. Hence, $\\angle APB = \\frac{\\angle A+\\angle C}{2} = 60^\\circ = \\angle B$, so the triangle $ABP$ is equilateral.\nThereby, $BP = AB = BT$, so triangle $BPT$ is isosceles, hence $\\angle BPT = \\frac{180^\\circ - \\angle PBT}{2}$. Triangle $ABT$ is also an isosceles one, so $\\angle ABT = 180^\\circ - 2\\angle A$. Noticing that $\\angle PBT = \\angle B - \\angle ABT$, it results that $\\angle PBT = 2\\angle A - 120^\\circ$, which leads to $\\angle BPT = 150^\\circ - \\angle A$ (2).\nFrom the exterior angle theorem we infer that $\\angle BPT = \\angle CUP + \\angle PCU$. Using (1) and (2), we obtain $150^\\circ - \\angle A = \\frac{\\angle A}{2} + \\frac{\\angle B}{2} = \\frac{\\angle A}{2} + 30^\\circ$, so $\\angle A = 80^\\circ$ and $\\angle C = 40^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39163,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of the parallelogram $ABCD$ intersect at $O$ and $M$ is the midpoint of the side $AB$. Let $P$ be a point of the segment $OC$ and $Q$ be the intersection of the lines $MP$ and $BC$. The parallel to $MP$ through $O$ intersects the line $CD$ at the point $N$. Prove that the points $A, N$ are $Q$ collinear if and only if $P$ is the midpoint of the segment $OC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $A, N, Q$ are collinear, then the fundamental theorem for similarity yields $\\frac{CR}{MB} = \\frac{QC}{QB} = \\frac{CN}{AB}$ and from $AB = 2 \\cdot MB$ follows $CN = 2 \\cdot CR$. Now $RP \\parallel ON$ shows that $RP$ is a midline in $\\triangle CON$, therefore $P$ is the midpoint of $OC$.\n\n![](attached_image_1.png)\n\nFor the converse, if $P$ is the midpoint of the segment $OC$, then $RP$ is a midline of $\\triangle CON$, therefore $CN = 2 \\cdot CR$. Let the lines $AQ$ and $CD$ meet at $N'$. Then, as above, $CN' = 2 \\cdot CR = CN$, so $N$ coincides with $N'$, that is, $A, N, Q$ are collinear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 3$. Let $S$ be a set of $n$ points in the plane, no three of which are collinear. Given pairwise distinct points $A, B, C$ in $S$, the triangle $ABC$ is suitable for $AB$ if area $ABC \\le$ area $ABX$ for all $X$ in $S$ different from $A$ and $B$. (Note that for a segment $AB$ there could be several suitable triangles.) A triangle is adequate if its vertices are all in $S$ and it is suitable for at least two of its sides. Prove that there are at least $\\frac{1}{2}(n-1)$ adequate triangles.\nBulgaria, Alexadar Ivanov",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, a triangle whose vertices all lie in $S$ will be referred to as a triangle in $S$. The argument hinges on the following observation:\n\n*Given any partition of $S$, amongst all triangles in $S$ with at least one vertex in each part, those of minimal area are all adequate.*\n\nIndeed, amongst the triangles under consideration, one of minimal area is suitable for both sides with endpoints in different parts.\n\n**1st Approach.** By the above observation, the 3-uniform hypergraph of adequate triangles is connected. It is a well-known fact that such a hypergraph has at least $\\frac{1}{2}(n-1)$ hyperedges, whence the required lower bound.\n\n**3rd Approach.** Consider the bipartite graph with vertex parts $S$ and the set $\\Delta$ of adequate triangles, and edge set obtained by joining each adequate triangle to its vertices. This graph has exactly $3|\\Delta|$ edges. With reference again to the above observation, the graph is connected, so it has at least $|S| + |\\Delta| - 1$ edges. Consequently, $3|\\Delta| \\ge |S| + |\\Delta| - 1 = n + |\\Delta| - 1$, which provides the required lower bound.\n\n**4th Approach.** For a partition $S = A \\sqcup B$, an area minimising triangle as above will be called $(A, B)$-minimal. Thus, $(A, B)$-minimal triangles are all adequate.\n\nClearly, each step provides a new adequate triangle, so the overall number of adequate triangles is at least $\\frac{1}{2}(n-1)$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest positive integer $k$ for which exists a positive integer $n$ such that:\n$$\nsin(n + 1) < \\sin(n + 2) < \\sin(n + 3) < \\dots < \\sin(n + k).\n$$",
    "options": [],
    "answer": "5",
    "solution": "We claim the largest number is $k = 5$.\n\nThe difference of two consecutive terms is:\n$$\n\\sin(n+i+1)-\\sin(n+i) = 2 \\sin \\frac{1}{2} \\cos \\frac{2n+2i+1}{2} > 0 \\Leftrightarrow \\cos \\frac{2n+2i+1}{2} > 0. \\ (\\star)\n$$\n\nSuppose that for $k = 6$ there exists a positive integer $n$ such that $\\sin(n + 1) < \\sin(n + 2) < \\sin(n + 3) < \\sin(n + 4) < \\sin(n + 5) < \\sin(n + 6)$. Relation $(\\star)$ is equivalent to:\n$$\n\\cos \\left( n + \\frac{2i + 1}{2} \\right) > 0, \\quad i = \\overline{1, 5},\n$$\nwhich implies that:\n$$\n\\cos \\left(n + \\frac{3}{2}\\right) + \\cos \\left(n + \\frac{11}{2}\\right) = 2 \\cos 2 \\cdot \\cos \\left(n + \\frac{7}{2}\\right) > 0 \\Rightarrow \\cos 2 > 0,\n$$\nwhich is a contradiction. Therefore, $k \\le 5$.\n\nFor $k = 5$ we need to find a positive integer $n$ which, according to $(\\star)$, needs to satisfy $\\cos(n + \\frac{2i+1}{2}) > 0$, $i = \\overline{1,4}$. From $2m\\pi - \\frac{\\pi}{2} < n + \\frac{3}{2} < n + \\frac{9}{2} < 2m\\pi + \\frac{\\pi}{2} \\Leftrightarrow (4m-1) \\cdot 10\\pi < 20n + 30 < 20n + 90 < (4m+1) \\cdot 10\\pi$, considering $3.14 < \\pi < 3.15$, we obtain the following sufficient condition:\n$$\n(4m - 1) \\cdot 31.5 < 20n + 30 < 20n + 90 < (4m + 1) \\cdot 31.4 \\Rightarrow m \\le 7.\n$$\nFor $m = 7$ we have the following bounds for $n$:\n$$\n\\frac{27\\pi - 3}{2} < n < \\frac{29\\pi - 9}{2},\n$$\nwhich, due to having $3.141 < \\pi < 3.142$, implies $n = 41$, and thus proves our claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be natural numbers, where $1 \\le k < n$. At each vertex of a regular polygon with $n$ sides, either $1$ or $-1$ is written. At each step, we choose $k$ consecutive vertices and change their signs. Is it possible, starting from any configuration and performing this transformation multiple times, to obtain any other configuration?",
    "options": [],
    "answer": "Yes if and only if gcd(n, k) = 1 and k is odd; otherwise no.",
    "solution": "Let $d = (n, k)$. If $d > 1$ and $n = dm$, $k = dl$, then by performing the global move\n$$\ne_1 e_2 \\dots e_k e_{k+1} e_{2k+1} \\dots e_{m(k-k+1)} e_{2e_{k+2}} e_{2k+2} \\dots e_{m(k-k+2)}\n$$\nwe obtain the initial configuration. The same configuration can also be obtained by performing the null move. Thus, in this case, we have two different global moves that produce the same configuration, so the answer is negative.\n\nIf $d = 1$ and $k$ is even, then the global move $e_1 e_2 \\dots e_n$ changes each vertex an even number of times, so it has the same effect as the null move. Therefore, in this case, the answer is negative.\n\nFinally, if $d = 1$ and $k$ is odd, we will show that the answer is affirmative. For this, it suffices to show that from the configuration $C = 1, 1, 1, \\dots, 1$ we can obtain the configuration $C' = -1, 1, 1, \\dots, 1$. Since $(2n, k) = 1$, there exist natural numbers $m$ and $l$ such that $2nm + 1 = kl$. The global move $e_1 e_2 \\dots e_{k+1} e_{2k+1} \\dots e_{(l-1)(k+1)}$ changes each vertex, except the first one, $2m$ times, and the first vertex is changed $2m + 1$ times, thus transforming $C$ into $C'$.\n\nIn conclusion, the answer is affirmative if $(n, k) = 1$ and $k$ is odd, and negative otherwise.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 39167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest possible value of the expression\n$$\n\\frac{a+b-c}{a^3+b^3+abc} + \\frac{b+c-a}{b^3+c^3+abc} + \\frac{c+a-b}{c^3+a^3+abc},\n$$\nwhere $a$, $b$, $c$ are positive real numbers such that $a + b + c \\ge \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$.",
    "options": [],
    "answer": "1",
    "solution": "Denote by $E(a, b, c)$ the given expression. First, we find the maximum value of $E(a, b, c)$ under the assumption that $a + b - c$, $b + c - a$, $c + a - b$ are non-negative.\nSince $a^2 + b^2 - ab \\ge ab$, it follows that $a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \\ge ab(a + b)$, so $a^3 + b^3 + abc \\ge ab(a + b + c)$. Analogously, we have $b^3 + c^3 + abc \\ge bc(a + b + c)$ and $c^3 + a^3 + abc \\ge ca(a + b + c)$. Therefore:\n$$\nE(a, b, c) = \\sum \\frac{a+b-c}{a^3+b^3+abc} \\le \\sum \\frac{a+b-c}{ab(a+b+c)} = \\frac{1}{a+b+c} \\cdot \\sum \\frac{a+b-c}{ab}\n$$\nSince $\\sum \\frac{a+b-c}{ab} = \\frac{2(ab+bc+ca)-(a^2+b^2+c^2)}{abc} \\le \\frac{2(ab+bc+ca)-(ab+bc+ca)}{abc}$ (due to the well-known inequality $a^2 + b^2 + c^2 \\ge ab + bc + ca$), we infer that\n$$\nE(a, b, c) \\le \\frac{1}{a+b+c} \\cdot \\frac{ab+bc+ca}{abc} = \\frac{1}{a+b+c} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\le 1.\n$$\nConsider now the case when at least one of the numbers $a + b - c$, $b + c - a$, $c + a - b$ is negative. Without loss of generality, suppose that $a + b - c < 0$. Then $c > a + b > |a - b|$, so both numbers $b + c - a$ and $c + a - b$ are positive. It follows that:\n$$\n\\begin{align*}\n\\sum \\frac{a+b-c}{a^3+b^3+abc} &\\le \\frac{b+c-a}{b^3+c^3+abc} + \\frac{c+a-b}{c^3+a^3+abc} \\\\\n&\\le \\frac{b+c-a}{bc(a+b+c)} + \\frac{c+a-b}{ca(a+b+c)} \\\\\n&= \\frac{1}{a+b+c} \\cdot \\frac{ac+bc-(a-b)^2}{abc} < \\frac{1}{a+b+c} \\cdot \\frac{ab+bc+ca}{abc} \\\\\n&= \\frac{1}{a+b+c} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\le 1.\n\\end{align*}\n$$\nIn summary, the maximum value of $E(a, b, c)$ is equal to $1$ and it is attained if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the quadruples $(a, b, c, d)$ of integers, so that $a + b + c = 2d$ and $\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} = d^2$.",
    "options": [],
    "answer": "All quadruples are (0, 0, 0, 0), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), (0, -1, -1, -1), (-1, 0, -1, -1), (-1, -1, 0, -1).",
    "solution": "If $a \\ge b \\ge c \\ge 0$, then $d \\ge 0$. From $d^2 = \\sqrt{ab} + \\sqrt{ac} + \\sqrt{bc} \\le a + b + c \\le 2a + 2b + 2c \\le 4d$, at least one of the equalities being strict, follows $d \\in \\{0, 1, 2, 3\\}$.\n\nCase I: $d = 0$. Then $a = b = c = 0$.\n\nCase II: $d = 1$. Then $a + b + c = 2$ and $\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} = 1$, hence $a = b = 1$, $c = 0$.\n\nCase III: $d \\in \\{2, 3\\}$. When $d = 2$ we have the possibilities $(2, 1, 1)$, $(2, 2, 0)$, $(3, 1, 0)$, $(4, 0, 0)$, and for $d = 3$ we have the possibilities $(3, 2, 1)$, $(3, 3, 0)$, $(4, 2, 0)$, $(4, 1, 1)$, $(5, 1, 0)$, $(6, 0, 0)$; none of them is convenient.\n\nIf at least one of the numbers is negative, then all are non-positive, and $-a, -b, -c, -d$ verify the statement.\n\nFinally, we get the solutions $(0, 0, 0)$, $(0, 1, 1)$, $(1, 0, 1)$, $(1, 1, 0)$, $(0, -1, -1)$, $(-1, 0, -1)$, $(-1, -1, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all twice differentiable functions $f : \\mathbb{R} \\to (0, \\infty)$ with the property $3(f'(x))^2 \\le 2f(x)f''(x)$, for every $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) ≡ c for some constant c > 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$. Determine all $z \\in \\mathbb{C}$ such that:\n$$\n|z^{n+1} - z^n| \\ge |z^{n+1} - 1| + |z^{n+1} - z|\n$$",
    "options": [],
    "answer": "All complex numbers that are nth roots of unity or (n+1)th roots of unity, i.e., U_n ∪ U_{n+1}.",
    "solution": "We notice that $z = 1$ is a solution and $z = 0$ is not a solution, so we consider $z \\in \\mathbb{C} \\setminus \\{0, 1\\}$. Considering $w = \\frac{1}{z}$ and multiplying the given inequality by $\\frac{1}{|z|^{n+1}}$, we have:\n$$\n|1 - w| \\ge |1 - w^{n+1}| + |1 - w^n| \\ge |(1 - w^{n+1}) - (1 - w^n)| = |w|^n |1 - w|,\n$$\nwhich implies that $|w| \\le 1$. Moreover, we have:\n$$\n\\begin{aligned}\n|1 - w^{n+1}| + |1 - w^n| &\\ge |1 - w^{n+1}| + |w| |1 - w^n| \\\\\n&\\ge |1 - w^{n+1} - w(1 - w^n)| = |1 - w|,\n\\end{aligned}\n$$\nwhich leads to $|w| = 1$ and $|1 - w| = |1 - w^{n+1}| + |1 - w^n|$. Moreover, there exists $s \\ge 0$ with $1 - w^{n+1} = s(w^{n+1} - w)$, which, applying the conjugate, can be written as $1 - \\frac{1}{w^{n+1}} = s(\\frac{1}{w^{n+1}} - \\frac{1}{w})$, being equivalent to $w^{n+1} - 1 = s(1 - w^n)$. Summing up these two relations, we have:\n$$\ns(w^{n+1} - w^n - w + 1) = 0 \\Rightarrow s(w^n - 1)(w - 1) = 0.\n$$\nIf $w^n - 1 = 0$, we have $w \\in U_n \\setminus \\{1\\}$, while $s = 0$ implies $w^{n+1} = 1$, so $w \\in U_{n+1} \\setminus \\{1\\}$. Therefore, considering the initial remarks, we can conclude that the set of solutions of the given inequality is $U_n \\cup U_{n+1}$, where $U_k$ represents the set of the $k$th roots of unity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $XYZT$ be a parallelogram and $A, B, C, D$ variable points on the sides $XY, XT, TZ, ZY$, respectively, such that $ABCD$ is a cyclic quadrilateral with circumcenter $O$, $AC \\parallel XT$, and $BD \\parallel XY$. Let $P$ be the intersection of lines $AD$ and $BC$, and $Q$ be the intersection of lines $AB$ and $CD$. Prove that the circle $(POQ)$ passes through a fixed point as $A, B, C, D$ vary according to the given restrictions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The key idea for this problem lies in the following lemma:\n\n**Lemma:** Let $ABC$ be a triangle and $X$ a point in the interior of angle $\\angle BAC$ such that $\\angle ABX = \\angle ACX$. Define $Y$ such that $BXC$ is a parallelogram. Then $AX$ and $AY$ are isogonal with respect to $\\angle BAC$.\n\n*Proof.* Consider the triangle $ABC$. Since $BXC$ is a parallelogram, we have $BX \\parallel AC$ and $CX \\parallel AB$. By the alternate interior angles theorem, we have $\\angle XBA = \\angle XCB$. Since $\\angle ABX = \\angle ACX$, we can conclude that $\\angle XAB = \\angle XCA$. Thus, $AX$ and $AY$ are isogonal with respect to $\\angle BAC$. $\\square$\n\nNow, let's proceed with the solution to the main problem. We need to prove that the circle $(POQ)$ passes through a fixed point as $A, B, C, D$ vary according to the given restrictions.\n\nLet $O'$ be the intersection of lines $BD$ and $AC$. Since $AC \\parallel XT$ and $BD \\parallel XY$, by the Lemma, we know that $AP$ and $AQ$ are isogonal with respect to $\\angle XO'Y$.\n\nSince $ABCD$ is a cyclic quadrilateral, we have $\\angle ABC = \\angle ADC$. Thus, $\\angle PBC = \\angle PDC$. This implies that $PB$ and $PD$ are isogonal with respect to $\\angle ABC$. Similarly, $QA$ and $QC$ are isogonal with respect to $\\angle ADC$.\n\nTherefore, we have $\\angle ABP = \\angle DCQ$ and $\\angle BAP = \\angle CQD$. Combining these equalities, we get $\\angle ABO' = \\angle DCO'$. This implies that $ABO'D$ is a cyclic quadrilateral.\n\nLet $O$ be the circumcenter of $ABO'D$. Since $ABCD$ is a cyclic quadrilateral, $O$ is also the circumcenter of $ABCD$. Therefore, $O$ lies on the circle $(POQ)$.\n\nThus, we have shown that as $A, B, C, D$ vary according to the given restrictions, the circle $(POQ)$ passes through the fixed point $O$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $a$, prove that $n!$ is divisible by $n^2 + n + a$ for infinitely many positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that, if some integer $n \\ge 2$ satisfies the required condition, then so does $n^2 + a - 1$. Since $n^2 + a - 1 > n$, it is then sufficient to exhibit a single $n \\ge 2$ that fits the bill.\nConsider the identity\n$$\n(n^2 + a - 1)^2 + (n^2 + a - 1) + a = (n^2 + n + a)(n^2 - n + a). \\quad (*)\n$$\nSuppose now that $n!$ is divisible by $n^2 + n + a$ for some integer $n \\ge 2$. Since $n < n^2 - n + a < n^2 + a - 1$, it follows that $(n^2 + a - 1)!$ is divisible by $n!(n^2 - n + a)$; and since $n!$ is divisible by $n^2 + n + a$, the identity (*) then shows $(n^2 + a - 1)!$ divisible by $(n^2 + a - 1)^2 + (n^2 + a - 1) + a$.\nNext, set $n = a$ in (*) to get $(a^2 + a - 1)^2 + (a^2 + a - 1) + a = a^3(a + 2)$. If $a \\ge 3$, then $a < a + 2 < a^2 < a^2 + a - 1$, so $a^2 + a - 1 \\ge 2$ and $(a^2 + a - 1)!$ is divisible by $a^3(a + 2) = (a^2 + a - 1)^2 + (a^2 + a - 1) + a$.\nIf $a = 2$, set $n = 10$, and notice that $n! = 10!$ is divisible by $2 \\cdot 7 \\cdot 8 = 112 = 10^2 + 10 + 2 = n^2 + n + 2 = n^2 + n + a$.\nFinally, if $a = 1$, set $n = 16$, and notice that $n! = 16!$ is divisible by $3 \\cdot 7 \\cdot 13 = 273 = 16^2 + 16 + 1 = n^2 + n + 1$.\n\nAlternative solution:\nThe argument hinges on the fact that the equation $u^2 - 4a(a+2)v^2 = 1 - 4a$ has infinitely many solutions $(u_k, v_k)$ in positive integers, where the $u_k$ and the $v_k$ both form strictly increasing sequences.\nAssume this for the moment. Clearly, the $u_k$ are all odd, so the $n_k = \\frac{1}{2}(u_k - 1)$ are all integers. Consider first the case $a \\ge 2$. Since the $u_k$ form a strictly increasing sequence of positive integers, so do the $n_k$, and, since $a \\ge 2$,\n$$\nn_k = \\frac{1}{2}(u_k - 1) > \\left( \\frac{u_k^2 + 4a - 1}{a(a + 2)} \\right)^{\\frac{1}{2}} = 2v_k,\n$$\nfor all but finitely many indices $k$.\nRecalling that the $v_k$ also form a strictly increasing sequence of positive integers, it follows that the numbers $a(a+2) < v_k < 2v_k$ all three occur in $n_k!$, so the latter is divisible by $a(a+2) \\cdot v_k \\cdot 2v_k = 2a(a+2)v_k^2 = \\frac{1}{2}(u_k^2 + 4a - 1) = \\frac{1}{2}(u_k - 1)^2 + (u_k - 1) + 2a = 2(n_k^2 + n_k + a)$, for all but finitely many indices $k$, as desired.\n\nFinally, consider the equation $u^2 - 4a(a+2)v^2 = 1 - 4a$ along with the associated Pell equation $x^2 - 4a(a+2)y^2 = 1$. The pair $(u_0, v_0) = (2a+1, 1)$ solves the former and the pair $(x_0, y_0) = (2(a+1)^2 - 1, a+1)$ solves the latter. For each positive integer $k$, let $u_k = u_0x_k + 4a(a+2)v_0y_k$ and let $v_k = v_0x_k + u_0y_k$, where $x_k = x_0x_{k-1} + 4a(a+2)y_{k-1}$ and $y_k = y_0x_{k-1} + x_0y_{k-1}$. Since the $(x_k, y_k)$ solve the Pell equation, the $(u_k, v_k)$ solve the equation under consideration. Clearly, the $x_k, y_k, u_k, v_k$ all form strictly increasing sequences of positive integers. This completes the proof for $a \\ge 2$.\nThe case $a=1$ is dealt with similarly by considering the equation $u^2 - 28v^2 = -3$ along with the associated Pell equation $x^2 - 28y^2 = 1$. The pair $(u_0, v_0) = (5, 1)$ solves the former and the pair $(x_0, y_0) = (127, 24)$ solves the latter. The rest of the argument goes through verbatim.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a table with $n$ lines and $m$ columns ($n, m \\in \\mathbb{N}, n, m \\ge 2$) consisting of $n \\cdot m$ squares $1 \\times 1$, which we will call *cells*. We call *snake* a sequence of cells with the following properties: the first cell is located on the first (top) line of the table, the last cell is located on the last line of the table and, starting with the second, each cell of the snake has a common side with the previous one and is not located on a line above the previous cell. The *length* of a snake is the number of cells that form the snake. Determine the arithmetic mean of the lengths of all the snakes in the table.\nCristian Șăvescu",
    "options": [],
    "answer": "n(m^2+3m-1)/(3m)",
    "solution": "Consider a certain snake. For each line $i \\in \\{1, 2, \\dots, n\\}$ of the table, we denote by $a_i$ and $b_i$ the numbers of the columns corresponding to the first and last cell, respectively, that the snake has on line $i$. We observe that $a_{i+1} = b_i$ for any $i \\in \\{1, 2, \\dots, n-1\\}$, therefore the snake is determined by the values $a_1, b_1, b_2, \\dots, b_n \\in \\{1, 2, \\dots, m\\}$. We deduce that the total number of snakes is $m^{n+1}$.\n\nLet $(k, p)$ be some cell of the table. The snakes that contain this cell are those for which $a_k \\le p \\le b_k$. For these, we can choose $a_k$ and $b_k$ in $2p(m-p+1)-1$ ways, and the numbers $a_1, b_1, b_2, \\dots, b_n$, other than $a_k$ and $b_k$, in $m^{n-1}$ ways. Thus, the cell $(k, p)$ is counted by $m^{n-1}(2pm-2p^2+2p-1)$ times in all the snakes in the table.\n\nIt follows that the cells on column $p$ contribute with $n \\cdot m^{n-1}(2pm - 2p^2 + 2p - 1)$ to the total $T$ of the lengths of all snakes in the table, therefore\n$$\nT = \\sum_{p=1}^{m} n \\cdot m^{n-1} (2pm - 2p^2 + 2p - 1) = n \\cdot m^{n+1} (m+1) - \\frac{n \\cdot m^n (m+1)(2m+1)}{3} + n \\cdot m^{n+1} = \\frac{n \\cdot m^n (m^2 + 3m - 1)}{3}.\n$$\nThe arithmetic mean of the lengths of all the snakes in the table is $\\frac{n(m^2+3m-1)}{3m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. The moving points $M$ on the half line $BC$, $N$ on the half line $CA$, and $P$ on the half line $AB$, start simultaneously from vertices $B$, $C$, and $A$, respectively, and move with constant speeds $v_1, v_2, v_3 > 0$, expressed using the same unit.\n\na) Knowing that there are three distinct moments in which the triangle $MNP$ is equilateral, prove that the triangle $ABC$ is also equilateral and $v_1 = v_2 = v_3$.\n\nb) Prove that if $v_1 = v_2 = v_3$ and there exists a moment in which the triangle $MNP$ is equilateral, then the triangle $ABC$ is also equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b, c \\in \\mathbb{C}$ be the affixes of the vertices of triangle $ABC$. For $t \\ge 0$, we have the following expressions for the affixes of the points $M(t)$, $N(t)$, and $P(t)$:\n$$\n\\left\\{ \\begin{array}{l} m(t) = b \\cdot (1 - v_1 \\cdot t) + c \\cdot v_1 \\cdot t; \\\\ n(t) = c \\cdot (1 - v_2 \\cdot t) + a \\cdot v_2 \\cdot t; \\\\ p(t) = a \\cdot (1 - v_3 \\cdot t) + b \\cdot v_3 \\cdot t. \\end{array} \\right.\n$$\n\nThe condition that the triangle $MNP$ is equilateral at some time $t \\ge 0$ can be written as:\n$$\nm(t) + \\varepsilon n(t) + \\bar{\\varepsilon} p(t) = 0,\n$$\nwhere $\\epsilon \\in \\left\\{ \\frac{-1+i\\sqrt{3}}{2}, \\frac{-1-i\\sqrt{3}}{2} \\right\\}$.\nThis relationship is equivalent to:\n$$\nt(-bv_1 + cv_1 - \\epsilon cv_2 + \\epsilon av_2 - \\bar{\\epsilon}av_3 + \\bar{\\epsilon}bv_3) + b + \\epsilon c + \\bar{\\epsilon}a = 0, \\quad (*)\n$$\nfor any $t \\ge 0$ for which the triangle $MNP$ is equilateral.\n\na) Since the relation $(*)$ holds for three distinct values $t_1, t_2, t_3 \\ge 0$, according to the pigeonhole principle, there exists $\\epsilon \\in \\left\\{ \\frac{-1+i\\sqrt{3}}{2}, \\frac{-1-i\\sqrt{3}}{2} \\right\\}$ for which the relation $(*)$ is satisfied at two distinct moments $t_i, t_j \\ge 0$, $1 \\le i < j \\le 3$, from which it follows that:\n$$\n\\begin{cases} (b-c)v_1 + \\epsilon(c-a)v_2 + \\bar{\\epsilon}(a-b)v_3 = 0; \\\\ b + \\epsilon c + \\bar{\\epsilon} a = 0. \\end{cases}\n$$\nThe second relation from above is equivalent to having an equilateral triangle $ABC$. On the other hand, the first relationship allows us to translate the triangle $ABC$ such that its circumcenter has the affix $0$. Since $ABC$ is equilateral, we have $b = \\epsilon a$ and $c = \\bar{\\epsilon} a$ or $b = \\bar{\\epsilon} a$ and $c = \\epsilon a$, without loss of generality, we consider the first case and obtain:\n$$\n(\\epsilon a - \\bar{\\epsilon} a)v_1 + \\epsilon(\\bar{\\epsilon} a - a)v_2 + \\bar{\\epsilon}(a - \\epsilon a)v_3 = 0 \\Rightarrow (\\epsilon - \\epsilon^2)v_1 + (1 - \\epsilon)v_2 + (\\epsilon^2 - 1)v_3 = 0,\n$$\n\\text{and by dividing with } 1 - \\epsilon \\neq 0 \\text{ and taking into account that } v_1, v_2, v_3 \\in \\mathbb{R}, \\text{ we obtain:}\n$$\n\\epsilon v_1 + v_2 - v_3 - \\epsilon v_3 = 0 \\Rightarrow v_1 = v_2 = v_3.\n$$\n\nb) If $v_1 = v_2 = v_3 = v$, relation $(*)$ can be written as:\n$$\nt \\cdot v \\cdot ((c - b) + \\epsilon(a - c) + \\bar{\\epsilon}(b - a)) + b + \\epsilon c + \\bar{\\epsilon} a = 0,\n$$\nwhich is invariant under translations, rotations, and homotheties, so we can fix $a = 1$ and $c = \\bar{\\epsilon}$ and obtain:\n$$\n(b - \\epsilon)(t \\cdot v \\cdot (\\bar{\\epsilon} - 1) + 1) = 0,\n$$\nand, since the second term cannot be zero, we have $b = \\epsilon$, from which we obtain that triangle $ABC$ is equilateral, and thus concludes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39175,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that 13 positive integers form a *special* group if the numbers of the group are consecutive.\n\na) Determine the number of special groups for which the sum of their elements is a three-digit perfect square.\n\nb) Find the maximum number of primes in a special group.",
    "options": [],
    "answer": "a) 2; b) 6",
    "solution": "a) Denote by $a, a+1, \\dots, a+12$ the elements of a special group. Their sum is $13a + 78 = 13(a + 6)$ and it must be a three-digit perfect square, therefore $13(a + 6) \\in \\{169, 676\\}$, thus $a \\in \\{7, 46\\}$. There are only 2 special groups with the given property.\n\nb) If $a = 1$ or $a = 2$, there are 6 primes in the special group which begins with $a$.\nIf $a \\ge 3$, there are at least 6 even numbers greater than 2, among the 13 numbers of the special group which begins with $a$. In this group, there are at least three consecutive odd numbers, thus one of them is divisible by 3, therefore it isn't prime. Consequently, there are at most 6 primes in a special group, so the maximum number of primes of a special group is 6.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39176,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number $n$ is interesting if it can be written as $n = \\lfloor \\frac{1}{a} \\rfloor + \\lfloor \\frac{1}{b} \\rfloor + \\lfloor \\frac{1}{c} \\rfloor$, where $a, b$ and $c$ are positive real numbers, such that $a + b + c = 1$.\nDetermine all the interesting numbers. ($[x]$ denotes the floor of the real number $x$.)",
    "options": [],
    "answer": "All natural numbers n ≥ 7",
    "solution": "Without loss of generality, we may choose $a \\le b \\le c$.\n\nThe positive real numbers $a, b, c$ are such that $a + b + c = 1$, thus $a, b, c \\in (0, 1)$. From $\\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c} \\in (1, \\infty)$ follows that $[\\frac{1}{a}], [\\frac{1}{b}], [\\frac{1}{c}] \\in \\mathbb{N}^*$, therefore $n \\ge 3$.\n\nIf $[\\frac{1}{a}] \\le 2$, then $a, b, c \\in (\\frac{1}{3}, 1)$ and $a + b + c > 1$, false. Therefore $[\\frac{1}{a}] \\ge 3$, so $n \\ge 5$.\nIf 5 were an interesting number, we would have $[\\frac{1}{a}] = 3$ and $[\\frac{1}{b}] = [\\frac{1}{c}] = 1$, whence $a+b+c > b+c > 1$, false.\nIf 6 were an interesting number, we would obtain $[\\frac{1}{a}] = 3$, $[\\frac{1}{b}] = 2$, $[\\frac{1}{c}] = 1$, whence $a \\in (\\frac{1}{4}, \\frac{1}{3}]$, $b \\in (\\frac{1}{3}, \\frac{1}{2}]$ and $c \\in (\\frac{1}{2}, 1)$, therefore $a+b+c > \\frac{1}{4}+\\frac{1}{3}+\\frac{1}{2} = \\frac{13}{12}$, false.\n\nWe prove that all the natural numbers $n \\ge 7$ are interesting.\n\nConsider $k \\in \\mathbb{N}$, $k \\ge 4$ and the real numbers $a = \\frac{1}{k}, b = c = \\frac{k-1}{2k}$, with $a+b+c=1$.\nAs $[\\frac{1}{a}] = k, [\\frac{1}{b}] = [\\frac{1}{c}] = [2+\\frac{2}{k-1}] = 2$, we have $[\\frac{1}{a}] + [\\frac{1}{b}] + [\\frac{1}{c}] = k+4$. Consequently, all natural numbers $n \\ge 8$ are interesting.\nChoosing, for instance, $a = \\frac{8}{30}, b = c = \\frac{11}{30}$, we have $[\\frac{1}{a}] + [\\frac{1}{b}] + [\\frac{1}{c}] = 3+2+2=7$ and $a+b+c=1$, therefore 7 is also an interesting number.\n\nThus, all natural numbers $n \\ge 7$ are interesting.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. A square with the side length $2n - 1$ is divided into $(2n - 1)^2$ unity squares using parallels to the sides. Ana and Bogdan play the following game: starting with Ana, the two take turns to colour, Ana with red, Bogdan with blue, in $2n^2$ turns, the $4n^2$ vertices of the unity squares. Then, starting with Ana, each of them connects with a vector a red dot (which is the tail of the vector) with a blue dot (which is the head of the vector), obtaining $2n^2$ vectors with distinct tails and heads. If the sum of these vectors is zero, Ana wins. Otherwise, Bogdan wins. Show that Bogdan has a winning strategy.\nCristian Șăvescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the centre of the square, $B_1, B_2, \\dots, B_{2n^2}$ be the blue points and $R_1, R_2, \\dots, R_{2n^2}$ the red points. Ana will win if\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{R_i B_{a_i}} = \\overrightarrow{0}, \\quad (1)\n$$\nfor a certain rearrangement $a_1, a_2, \\dots, a_{2n^2}$ of the numbers $1, 2, \\dots, 2n^2$.\n\nRelation (1) can be rewritten as\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{(OB_{a_i} - OR_i)} = 0 \\iff \\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = \\sum_{i=1}^{2n^2} \\overrightarrow{OB_{a_i}} \\iff \\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = \\sum_{i=1}^{2n^2} \\overrightarrow{OB_i} \\quad (2).\n$$\n\nIn order to win, Bogdan can adopt the following strategy: after Ana colours with red a point in the penultimate turn, he calculates $\\overrightarrow{s} = \\sum_{i=1}^{2n^2-1} \\overrightarrow{OR_i}$.\nIf there exists a point $X$ that is not coloured such that $OX = -s$, then he colours $X$ with blue. Thus, he will prevent Ana to win since\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = OR_{2n^2} + s = OR_{2n^2} - OX = XR_{2n^2} \\neq 0.\n$$\nIn the other case, Bogdan colours any point, since relation (3) cannot be fulfilled.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39178,
    "subject": "Mathematics (Multi-modal)",
    "question": "A kingdom has $10$ towns. Some pairs of different towns have bridges built between them, so that between any two different towns there is at most one bridge. Two towns between which there is a bridge will be called *ends* of that bridge.\n\na) Prove that there are at least two towns which are *ends* for the same number of bridges.\n\nb) The king orders his subjects to build or demolish some bridges, so that half of the towns are *ends* for $3$ bridges and the other half are *ends* for $4$ bridges. Can the order be fulfilled?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{R}$, with $a < b$, be two arbitrary real numbers. We say that a function $f : [a, b] \\to \\mathbb{R}$ has property ($\\mathcal{P}$) if it is an integrable function on $[a, b]$, such that\n$$\nf(x) - f\\left(\\frac{x+a}{2}\\right) = f\\left(\\frac{x+b}{2}\\right) - f(x) \\quad \\text{for any } x \\in [a, b].\n$$\n\nShow that for any real number $t$ there is a unique function $f : [a, b] \\to \\mathbb{R}$ with the property ($\\mathcal{P}$), such that $\\int_a^b f(x) dx = t$.",
    "options": [],
    "answer": "f(x) = t/(b - a) for all x in [a, b]",
    "solution": "We will show that the functions with property ($\\mathcal{P}$) are precisely the constant functions on the interval $[a, b]$.\n\nThe equality in the statement can be rewritten equivalently as\n$$\nf(x) = \\frac{1}{2} \\left( f\\left(\\frac{x+a}{2}\\right) + f\\left(\\frac{x+b}{2}\\right) \\right) \\quad (1)\n$$\nWe show now that, for any $n \\in \\mathbb{N}^*$ and any $x \\in [a, b]$, the following equality holds\n$$\nf(x) = \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} f\\left(\\frac{x + (2^n - 1 - k)a + k b}{2^n}\\right). \\qquad (2)\n$$\nFor $n = 1$, this is precisely the relation (1). If we assume now that the equality holds for a certain positive integer $n \\in \\mathbb{N}^*$ and any $x \\in [a, b]$, then we have\n$$\n\\begin{aligned}\nf(x) &= \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} f\\left(\\frac{x + (2^n - 1 - k)a + k b}{2^n}\\right) \\\\\n&= \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} \\left( f\\left(\\frac{1}{2} \\cdot \\frac{x + (2^n - 1 - k)a + k b}{2^n} + \\frac{1}{2} \\cdot a\\right) + \\right. \\\\\n& \\qquad \\left. + f\\left(\\frac{1}{2} \\cdot \\frac{x + (2^n - 1 - k)a + k b}{2^n} + \\frac{1}{2} \\cdot b\\right) \\right) \\\\\n&= \\frac{1}{2^{n+1}} \\sum_{k=0}^{2^{n+1}-1} f\\left(\\frac{x + (2^{n+1} - 1 - k)a + k b}{2^{n+1}}\\right).\n\\end{aligned}\n$$\nConsider for some positive integer $n \\in \\mathbb{N}^*$ and arbitrary $x \\in [a, b]$ the division $\\Delta_n = (x_0 = a < x_1 = \\frac{(2^n-1)a+b}{2^n} < \\dots < x_k = \\frac{(2^n-k)a+kb}{2^n} < \\dots < x_{2^n} = b)$ with norm $|\\Delta_n| = \\frac{b-a}{2^n}$ and the system of intermediate points\n$$\n\\xi_{(n)}(x) = \\left( \\xi_k(x) = \\frac{x + (2^n - k)a + (k-1)b}{2^n} \\mid k = \\overline{1, 2^n} \\right).\n$$\nThe relation (2) can then be written as\n$$\nf(x) = \\frac{1}{b-a} \\cdot \\sigma(f; \\Delta_n, \\xi_{(n)}(x)),\n$$\nwhere $\\sigma(f; \\Delta_n, \\xi_{(n)}(x))$ denotes the Riemann sum associated with the function $f$, the division $\\Delta_n$ and the system of intermediate points $\\xi_{(n)}(x)$. Since the function $f$ is Riemann integrable on the interval $[a, b]$, we have $\\lim_{n \\to \\infty} \\sigma(f; \\Delta_n, \\xi_{(n)}(x)) = \\int_a^b f(s) ds$, so that $f(x) = \\frac{1}{b-a} \\cdot \\int_a^b f(s) ds$ holds for any $x \\in [a, b]$. Any function with the property $(\\mathcal{P})$ is hence a constant function.\n\nConversely, any constant function defined on the interval $[a, b]$ is Riemann integrable and satisfies the equality in the statement.\n\nFor any $t \\in \\mathbb{R}$ there is then a unique function with the property ($\\mathcal{P}$), namely $f : [a, b] \\to \\mathbb{R}$, defined by $f(x) = \\frac{t}{b-a}$ for any $x \\in [a, b]$, such that $\\int_a^b f(x) dx = t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ so that the largest prime divisor of $n^2 + 2$ is equal to the largest prime divisor of $n^2 + 2n + 3$.",
    "options": [],
    "answer": "1, 4",
    "solution": "A common divisor $d$ of the given numbers divides also $n^2 + 2n + 3 - (n^2 + 2) = 2n + 1$, $n \\cdot (2n + 1) = 2n^2 + n$, $2n^2 + n - 2 \\cdot (n^2 + 2) = n - 4$, $2n + 1 - 2 \\cdot (n - 4) = 9$.\nTherefore, if $d$ is a prime, then $d = 3$, and the only other prime factor contained by the given numbers is $2$, appearing in at most one of them. Moreover, at least one of the numbers has the exponent of $3$ at most $2$.\nIf $n$ is even, then $n^2 + 2$ is even, but not divisible by $4$ and the possible cases are:\nI) $n^2 + 2 = 2 \\cdot 3$;\nII) $n^2 + 2 = 2 \\cdot 3^2$;\nIII) $n^2 + 2n + 3 = 3$;\nIV) $n^2 + 2n + 3 = 3^2$.\nOnly case II gives a solution, namely $n = 4$.\nIf $n$ is odd, then $n^2 + 2n + 3$ is even, but not divisible by $4$ and the possible cases are:\nV) $n^2 + 2n + 3 = 2 \\cdot 3$;\nVI) $n^2 + 2n + 3 = 2 \\cdot 3^2$;\nVII) $n^2 + 2 = 3$;\nVIII) $n^2 + 2 = 3^2$.\nWe get the solution $n = 1$ (corresponding to the cases V and VII).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a continuous function.\n\na) Show that\n$$\n\\lim_{n \\to \\infty} \\int_{0}^{1} f(x^n) dx = f(0).\n$$\n\nb) If $f(0) = 0$ and $f$ is differentiable at the right in $0$, show that the limits exist, are finite and equal.\n$$\n\\lim_{\\varepsilon \\to 0} \\int_{\\varepsilon}^{1} \\frac{f(x)}{x} dx \\quad \\text{and} \\quad \\lim_{n \\to \\infty} \\left( n \\cdot \\int_{0}^{1} f(x^n) dx \\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $f$ is continuous, $f$ is bounded, with $\\text{Im}(f) \\subseteq [-M, M]$, where $M > 1$, and, for any $\\varepsilon > 0$, there is a $\\delta > 0$, such that $|f(x) - f(0)| < \\frac{\\varepsilon}{2}$, $\\forall x \\in [0, \\delta]$, and, for any $x \\in [0, 1 - \\frac{\\varepsilon}{4M}]$, there is a positive integer $n_0 \\in \\mathbb{N}^*$ such that $x^n \\in [0, \\delta]$, $\\forall n \\ge n_0$. Then\n$$\n\\begin{aligned}\n\\left| \\int_{0}^{1} f(x^n) dx - f(0) \\right| & \\le \\int_{0}^{1} |f(x^n) - f(0)| dx \\\\\n& = \\int_{0}^{1-\\frac{\\varepsilon}{4M}} |f(x^n) - f(0)| dx + \\int_{1-\\frac{\\varepsilon}{4M}}^{1} |f(x^n) - f(0)| dx \\\\\n& < \\frac{\\varepsilon}{2} \\left(1 - \\frac{\\varepsilon}{4M}\\right) + \\frac{\\varepsilon}{4M} \\cdot 2M < \\varepsilon,\n\\end{aligned}\n$$\nfor any $n \\ge n_0$. It follows that\n$$\n\\lim_{n \\to \\infty} \\int_{0}^{1} f(x^n) dx = f(0).\n$$\n\nb) Because $f$ is differentiable at $0$, the function $g : [0, 1] \\to \\mathbb{R}$, defined by\n$$\ng(x) = \\begin{cases} \\frac{f(x)}{x}, & \\text{if } x > 0, \\\\ f'(0), & \\text{if } x = 0, \\end{cases}\n$$\nis continuous.\nLet $G$ be a primitive of $g$. Then\n$$\n\\int_{\\varepsilon}^{1} \\frac{f(x)}{x} dx = \\int_{\\varepsilon}^{1} g(x) dx = G(1) - G(\\varepsilon),\n$$\nand\n$$\n\\lim_{\\varepsilon \\to 0} \\int_{\\varepsilon}^{1} \\frac{f(x)}{x} dx = G(1) - G(0).\n$$\n\n$$\n\\begin{aligned}\nn \\cdot \\int_{0}^{1} f(x^n) \\, dx &= n \\cdot \\int_{0}^{1} x^n g(x^n) \\, dx = \\int_{0}^{1} x \\cdot (n x^{n-1}) g(x^n) \\, dx \\\\\n&= x \\cdot G(x^n) \\Big|_{0}^{1} - \\int_{0}^{1} G(x^n) \\, dx = G(1) - \\int_{0}^{1} G(x^n) \\, dx.\n\\end{aligned}\n$$\nAccording to part a) it follows that\n$$\n\\lim_{n \\to \\infty} \\left( n \\cdot \\int_{0}^{1} f(x^n) \\, dx \\right) = G(1) - G(0),\n$$\nand the required statement is proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $CD$ of the square $ABCD$. The perpendicular from $C$ onto $BM$ meets the lines $BM$ at $N$, and $AB$ at $E$. The line $BM$ intersects the line $AD$ in $P$. Let $F$ be the midpoint of the segment $BN$. Prove that:\n\na) the triangles $CBE$ and $BAP$ are congruent;\n\nb) the segments $AN$ and $DF$ are congruent and perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From $CB = BA$ and $\\angle CEB = \\angle BPA$ (they have the same complement $\\angle EBP$), follows that the right triangles $CBE$ and $BAP$ are congruent.\n\nb) Denote $T$ the common point of the lines $EC$ and $AP$. The segments $DM$ and $AT$ are midlines in the right triangles $APB$ and $BEC$, so $A$ is the midpoint of $BE$. Therefore $AF \\parallel NE$, whence $AF \\perp BP$.\n\nThe segments $FD$ and $AN$ are medians relative to the hypotenuse in right triangles, hence $FD = \\frac{1}{2}AP = \\frac{1}{2}BE = AN$.\nSince $BN = AF$ (altitudes relative to the hypotenuse in congruent triangles $EBC$ and $PBA$), $\\triangle ADF \\equiv \\triangle BAN$.\nFrom $\\angle FDA + \\angle NAD = \\angle NAB + \\angle NAD = 90^\\circ$ follows $AN \\perp DF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39183,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$, $z$ are positive numbers with $x + y + z = 1$, show that:\n$$\n\\begin{align*}\na)\\ 1 - \\frac{x^2-yz}{x^2+x} &= \\frac{(1-y)(1-z)}{x^2+x}; \\\\\nb)\\ \\frac{x^2-yz}{x^2+x} + \\frac{y^2-zx}{y^2+y} + \\frac{z^2-xy}{z^2+z} &\\le 0.\n\\end{align*}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\n$1 - \\frac{x^2 - yz}{x^2 + x} = \\frac{x + yz}{x^2 + x} = \\frac{1 - y - z + yz}{x^2 + x} = \\frac{(1 - y)(1 - z)}{x^2 + x}$.\n\nb.\nUsing a), the inequality is rewritten\n$$\n\\frac{(1-y)(1-z)}{x(x+1)} + \\frac{(1-z)(1-x)}{y(y+1)} + \\frac{(1-x)(1-y)}{z(z+1)} \\geq 3,\n$$\nthat is\n$$\n\\frac{(x+z)(x+y)}{x[(x+z)+(x+y)]} + \\frac{(y+z)(y+x)}{y[(y+z)+(y+x)]} + \\frac{(z+y)(z+x)}{z[(z+y)+(z+x)]} \\geq 3.\n$$\nBut applying the inequality between the arithmetic mean and the harmonic mean we deduce that\n$$\n\\frac{(x+z)(x+y)}{x[(x+z)+(x+y)]} + \\frac{(y+z)(y+x)}{y[(y+z)+(y+x)]} + \\frac{(z+y)(z+x)}{z[(z+y)+(z+x)]} \\geq \\\\\n\\geq \\frac{9}{\\frac{x}{x+y} + \\frac{x}{x+z} + \\frac{y}{y+z} + \\frac{y}{y+x} + \\frac{z}{z+y} + \\frac{z}{z+x}} = \\frac{9}{1+1+1} = 3.\n$$\n\nAlternative solution.\n\nb.\nUsing a), the inequality is rewritten\n$$\n(1-x)(1-y)(1-z) \\left( \\frac{1}{x-x^3} + \\frac{1}{y-y^3} + \\frac{1}{z-z^3} \\right) \\geq 3.\n$$\nApplying the inequality between the arithmetic mean and the harmonic mean, we have\n$$\n\\begin{align*}\n\\frac{1}{x-x^3} + \\frac{1}{y-y^3} + \\frac{1}{z-z^3} &\\ge \\frac{9}{(x+y+z)-(x^3+y^3+z^3)} \\\\\n&= \\frac{9}{1-(x^3+y^3+z^3)} \\\\\n&= \\frac{9}{(x+y+z)^3 - (x^3+y^3+z^3)} \\\\\n&= \\frac{9}{3(x+y)(y+z)(z+x)} = \\frac{3}{(1-x)(1-y)(1-z)}\n\\end{align*}\n$$\nand the inequality is demonstrated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $DEF$ be two congruent equilateral triangles, centered at $O_1$ and $O_2$, respectively, such that segment $AB$ meets segments $DE$ and $DF$ at $M$ and $N$, respectively, and segment $AC$ meets segments $DF$ and $EF$ at $P$ and $Q$, respectively. The bisectors of angles $EMN$ and $DPQ$ meet at $I$, and the bisectors of angles $FNM$ and $EQP$ meet at $J$. Prove that $IJ$ is the perpendicular bisector of the segment $O_1O_2$.\nBogdan Maxim",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $C_1$ and $C_2$ the circumcircles of the triangles $ABC$ and $DEF$, respectively.\nSince $\\angle PAM = \\angle PDM = 60^\\circ$, the quadrilateral $APMD$ is cyclic, so $\\angle APD = \\angle AMD = x^\\circ$.\n\nLooking at the quadrilateral $APIM$ we have: $\\angle PIM = 360^\\circ - \\angle MAP - \\angle AMI - \\angle API$, so $\\angle PIM = 360^\\circ - 60^\\circ - 2 \\cdot \\frac{180^\\circ - x^\\circ}{2} - x^\\circ = 120^\\circ$.\nSince $\\angle PIM = 180^\\circ - \\angle PAM$, we infer that $APIM$ is a cyclic quadrilateral, so $ADIP$ and $ADMI$ are cyclic too.\n\n![](attached_image_1.png)\n\nIt follows that $\\angle DAI = \\angle DPI = \\frac{180^\\circ - x^\\circ}{2}$ and $\\angle ADI = \\angle AMI = \\frac{180^\\circ - x^\\circ}{2}$.\nSubsequently, $\\angle ADI = \\angle DAI$, so $AI = DI$.\nLet $A'$ be the second intersection point of $AI$ and $C_1$ and $D'$ be the second intersection point of $DI$ and $C_2$.\nSince $ADIP$ is cyclic, it follows that $\\angle IAP = \\angle IDP$, so the arc $A'C$ from the circle $C_1$ and the arc $D'F$ from the circle $C_2$ have the same measure. Since the minor arcs $AC$ and $DF$ from the two circles are congruent, each having $120^\\circ$, it follows that the arcs $ACA'$ (in the circle $C_1$) and $DFD'$ (in the circle $C_2$) have the same measure, so $AA' = DD'$.\n\nFrom here, we obtain $A'I = AA' - AI = DD' - DI = D'I$, so $IA \\cdot IA' = ID \\cdot ID'$, which means that the power of the point $I$ with respect to the circle $C_1$ is equal to the power of $I$ with respect to $C_2$. Hence, $I$ belongs to the radical axis $UV$ of these two circles.\nAnalogously, we prove that $J \\in UV$, so the lines $UV$ and $IJ$ coincide. Hence, $IJ$ is the radical axis of the circles $C_1(O_1)$ and $C_2(O_2)$, therefore is the perpendicular bisector of $O_1O_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a regular hexagon of side length $2$. Through the vertices and the sides' midpoints we construct parallels to the sides, which divide the hexagon into $24$ congruent equilateral triangles, whose vertices are called *nodes*. A *sheet* is any (non-degenerate) equilateral triangle whose vertices are nodes. A *trio* of a node $X$ is the figure formed by three adjacent sheets such that their intersection is only $X$ and they are not congruent in pairs.\n\na)\nFind the maximal possible area of a trio.\n\nb) Show that there exists a node whose trios can cover the whole hexagon, and a node whose trios cannot cover the whole hexagon.\n\nc)\nDetermine the total number of trios associated to the hexagon.\n\nDana Heuberger",
    "options": [],
    "answer": "a) 11√3/4. b) The center node has trios that cover the whole hexagon; a node at distance one from the center (e.g., R) does not. c) 96.",
    "solution": "a) Consider the regular hexagon $ABCDEF$, centered at $O$, and let $T$, $U$, $V$, $X$, $Y$, $Z$ be the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively.\nNotice that the nodes situated on the sides of the hexagon cannot have trios, so a node that admits a trio is either $O$, or situated at distance $1$ from $O$. Denote these latter nodes by $M$, $N$, $P$, $Q$, $R$, $S$, as shown in Figure 1.\nThe side length of an equilateral triangle with nodes as vertices may be equal to $1$, $\\sqrt{3}$, $2$, $\\sqrt{7}$, $3$ or $2\\sqrt{3}$. A sheet of side length $3$ or $2\\sqrt{3}$ cannot be part of a trio, because the vertices of this sheet should be situated on the hexagon sides.\nThere are no sheets of side $\\sqrt{7}$ that have $O$ as a vertex. Also, there are only two sheets of side $\\sqrt{7}$ that have $R$ as a vertex, namely $RAU$ and $RCT$. Moreover, the intersection of a side $2$ sheet and a side $\\sqrt{7}$ sheet of $R$ is non-empty, so we can't form trios with such sheets.\nConsequently, the sheets *RAU*, *RDP* and *REY* (see Figure 1) form a trio with maximal possible area, and the maximal area is equal to\n$$\n\\frac{\\sqrt{3}}{4} + \\frac{3\\sqrt{3}}{4} + \\frac{7\\sqrt{3}}{4} = \\frac{11\\sqrt{3}}{4}.\n$$\n\nb) Considering the colored trio from Figure 2, we notice that *OAB* covers the sixth part of the hexagon. Rotating the view, *OBC*, *OCD*, *ODE*, *OEF*, and *OFA* are sheets for some other five trios of $O$. All these six trios cover the whole hexagon.\nThe node $R$ has exactly two sheets of maximal side length - which is equal to $\\sqrt{7}$ - namely *RAU* and *RCT*. Since $RB = 3 > \\sqrt{7}$, it follows that not all the points of the segment $RB$ can be covered with the trios of $R$.\n\nc) We say that a trio is $(x, y, z)$-type if the side lengths of its sheets are equal to $x, y, z$.\nThe center $O$ has only $(1, \\sqrt{3}, 2)$-type trios, as indicated in Figure 2. Now, we count how many trios have $OQR$ as one of its sheets. The only side $2$ sheets that we can choose to be part of a trio are $OAF$, $OAB$ and $OBC$. For each of these side $2$ sheets we can choose the side $\\sqrt{3}$ sheet in two ways, so there are exactly $6$ trios which contain $OQR$. Since $O$ is the vertex of six side $1$ sheets, the node $O$ has exactly $36$ trios. The node $R$ admits only $(1, \\sqrt{3}, 2)$ and $(1, \\sqrt{3}, \\sqrt{7})$ type trios.\nConsider the sheets *RDP* and *RNZ*. To complete a trio, there are two choices for the side-$1$ sheet, specifically *REX* and *REY*. The same applies for the sheets *RFM* and *RNV*. Therefore, the node $R$ has exactly four $(1, \\sqrt{3}, 2)$-type trios.\nLikewise, the sheets *RAU* and *RDP* can form a trio together with one of the following three triangles of side length $1$: *REX*, *REY* and *RSY*. Also, there are three side $1$ sheets that form a trio with *RCT* and *RFM*. Consequently, $R$ has six $(1, \\sqrt{3}, \\sqrt{7})$-type trios.\nIt follows that there are $10$ trios associated to the node $R$, so there are $60$ trios associated to a node other than $O$. Adding all up, there are $96$ trios associated to the hexagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive integers $a$, $b$, $c$ with the following properties:\n(i) $(a^2 + b^2)(c^2 + 2023^2) = (ab + 2023c)^2$;\n(ii) $(a^2 + 2023^2)(b^2 + c^2) = (2023a + bc)^2$;\n(iii) the greatest common divisor of $a$, $b$, $c$ and $2023$ equals $1$.",
    "options": [],
    "answer": "(a, b, c) in {(1, 2023, 2023^2), (7^2, 2023, 17^4), (17^4, 2023, 7^2), (2023^2, 2023, 1)}",
    "solution": "By subtracting the equality (i) from (ii) we obtain $(a^2 - c^2) \\cdot (b^2 - 2023^2) = 0$. Because $a$, $b$, $c$ are positive integers, we deduce that $a = c$ or $b = 2023$.\n\nIf $a = c$, using (i) we find $(a^2 - 2023b)^2 = 0$, that is $a^2 = 2023b = 7 \\cdot 17^2 \\cdot b$. Consequently $7 \\mid b$ and $7 \\mid a = c$, therefore the greatest common divisor of $a$, $b$, $c$ and $2023$ is at least $7$, contradicting assumption (iii).\n\nIf $b = 2023$, from (i) we obtain $(ac - 2023^2)^2 = 0$, that is $ac = 2023^2 = 7^2 \\cdot 17^4$. From (iii) we deduce that $a$ and $c$ are relatively primes, therefore the solutions are: $(a, b, c) \\in \\{(1, 2023, 2023^2), (7^2, 2023, 17^4), (17^4, 2023, 7^2), (2023^2, 2023, 1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ so that $N = \\frac{1}{n(n+1)}$ is a finite decimal fraction.",
    "options": [],
    "answer": "n = 1 and n = 4",
    "solution": "The number $N$ is a finite decimal fraction if and only if its denominator is of the form $2^a \\cdot 5^b$, with $a, b \\in \\mathbb{N}$.\nSince $n$ and $n+1$ are coprime, the possible cases are:\nI) $n = 1$, $n+1 = 2^a \\cdot 5^b$ (with an obvious conclusion);\nII) $n = 5^b$, $n+1 = 2^a$;\nIII) $n = 2^a$, $n+1 = 5^b$.\n\nII) In this case $2^a = 5^b + 1 = M_4 + 2$, hence $a = 1$, $b = 0$, $n = 1$.\n\nIII) We get $2^a = 5^b - 1$ (*). Since $5^b$ and $2^a$ are not divisible by $3$, equality (*) is possible only when $2^a = M_3 + 1$ and $5^b = M_3 + 2$. Notice that $5^{2k} = M_3 + 1$ for $k \\in \\mathbb{N}$, so $b = 2k+1$, $k \\in \\mathbb{N}$. In this case, $k \\ge 1$ yields $5^b - 1 = 5 \\cdot 25^k - 1 = M_8 + 4$ and $5^b - 1 > 8$. Therefore, equality (*) is not possible for $b \\ge 2$. This leaves $b = 1$ and $a = 2$, yielding $n = 4$.\n\nThe sought for numbers are $n = 1$ and $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer, and let $f$ be a $4n$-variable polynomial with real coefficients, such that, for any $2n$ points $(x_1, y_1), \\dots, (x_{2n}, y_{2n})$ in the Cartesian plane,\n$$\nf(x_1, y_1, \\dots, x_{2n}, y_{2n}) = 0\n$$\nif and only if they form the vertices of a regular $2n$-gon in some order, or are all equal. Determine the smallest possible degree of $f$.",
    "options": [],
    "answer": "2n",
    "solution": "The smallest possible degree is $2n$. In what follows, we will frequently write $A_i = (x_i, y_i)$, and abbreviate $P(x_1, y_1, \\dots, x_{2n}, y_{2n})$ to $P(A_1, \\dots, A_{2n})$ or as a function of any $2n$ points.\n\nSuppose that $f$ is valid. First, we note a key property:\n\n**Claim (Sign of $f$).** $f$ attains either only nonnegative values, or only nonpositive values.\n**Proof.** This follows from the fact that the zero-set of $f$ is very sparse: if $f$ takes on a positive and a negative value, we can move $A_1, \\dots, A_{2n}$ from the negative value to the positive value without ever having them form a regular $2n$-gon — a contradiction.\n\nThe strategy for showing $\\deg f \\ge 2n$ is the following. We will animate the points $A_1, \\dots, A_{2n}$ linearly in a variable $t$; then $g(t) = f(A_1, \\dots, A_{2n})$ will have degree at most $\\deg f$ (assuming it is not zero). The claim above then establishes that any root of $g$ must be a multiple root, so if we can show that there are at least $n$ roots, we will have shown $\\deg g \\ge 2n$, and so $\\deg f \\ge 2n$.\nGeometrically, our goal is to exhibit $2n$ linearly moving points so that they form a regular $2n$-gon a total of $n$ times, but not always form one.\nWe will do this as follows. Draw $n$ mirrors through the origin, as lines making angles of $\\frac{\\pi}{n}$ with each other. Then, any point $P$ has a total of $2n$ reflections in the mirrors, as shown below for $n = 5$. (Some of these reflections may overlap.)\nDraw the $n$ angle bisectors of adjacent mirrors. Observe that the reflections of $P$ form a regular $2n$-gon if and only if $P$ lies on one of the bisectors.\nWe will animate $P$ on any line $\\ell$ which intersects all $n$ bisectors (but does not pass through the origin), and let $P_1, \\dots, P_{2n}$ be its reflections. Clearly, these are also all linearly animated, and because of the reasons above, they will form a regular $2n$-gon exactly $n$ times, when $\\ell$ meets each bisector. So this establishes $\\deg f \\ge 2n$ for the reasons described previously.\n\nNow we pass to constructing a polynomial $f$ of degree $2n$ having the desired property. First of all, we will instead find a polynomial $g$ which has this property, but only when points with sum zero are input. This still solves the problem, because then we can choose\n$$\nf(A_1, A_2, \\dots, A_{2n}) = g(A_1 - \\bar{A}, \\dots, A_{2n} - \\bar{A}),\n$$\nwhere $\\bar{A}$ is the centroid of $A_1, \\dots, A_{2n}$. This has the upshot that we can now always assume $A_1 + \\dots + A_{2n} = 0$, which will simplify the ensuing discussion.\n\nWe will now construct a suitable $g$ as a sum of squares. This means that, if we write $g = g_1^2 + g_2^2 + \\cdots + g_m^2$, then $g = 0$ if and only if $g_1 = \\cdots = g_m = 0$, and that if their degrees are $d_1, \\dots, d_m$, then $g$ has degree at most $2 \\max(d_1, \\dots, d_m)$.\nThus, it is sufficient to exhibit several polynomials, all of degree at most $n$, such that $2n$ points with zero sum are the vertices of a regular $2n$-gon if and only if the polynomials are all zero at those points.\n\nFirst, we will impose the constraints that all $|A_i|^2 = x_i^2 + y_i^2$ are equal. This uses multiple degree 2 constraints.\nNow, we may assume that the points $A_1, \\dots, A_{2n}$ all lie on a circle with centre 0, and $A_1 + \\dots + A_{2n} = 0$. If this circle has radius 0, then all $A_i$ coincide, and we may ignore this case.\nOtherwise, the circle has positive radius. We will use the following lemma.\n\n**Lemma.** Suppose that $a_1, \\dots, a_{2n}$ are complex numbers of the same non-zero magnitude, and suppose that $a_1^k + \\dots + a_{2n}^k = 0$, $k = 1, \\dots, n$. Then $a_1, \\dots, a_{2n}$ form a regular $2n$-gon centred at the origin. (Conversely, this is easily seen to be sufficient.)\n**Proof.** Since all the hypotheses are homogenous, we may assume (mostly for convenience) that $a_1, \\dots, a_{2n}$ lie on the unit circle. By Newton's sums, the $k$-th symmetric sums of $a_1, \\dots, a_{2n}$ are all zero for $k$ in the range $1, \\dots, n$.\nTaking conjugates yields $a_1^{-k} + \\dots + a_{2n}^{-k} = 0$, $k = 1, \\dots, n$. Thus, we can repeat the above logic to obtain that the $k$-th symmetric sums of $a_1^{-1}, \\dots, a_{2n}^{-1}$ are also all zero for $k = 1, \\dots, n$. However, these are simply the $(2n-k)$-th symmetric sums of $a_1, \\dots, a_{2n}$ (divided by $a_1 \\cdots a_{2n}$), so the first $2n-1$ symmetric sums of $a_1, \\dots, a_{2n}$ are all zero. This implies that $a_1, \\dots, a_{2n}$ form a regular $2n$-gon centred at the origin.\n\nWe will encode all of these constraints into our polynomial. More explicitly, write $a_r = x_r + y_r i$; then the constraint $a_1^k + \\dots + a_{2n}^k = 0$ can be expressed as $p_k + q_k i = 0$, where $p_k$ and $q_k$ are real polynomials in the coordinates. To incorporate this, simply impose the constraints $p_k = 0$ and $q_k = 0$; these are conditions of degree $k \\le n$, so their squares are all of degree at most $2n$.\nTo recap, taking the sum of squares of all of these constraints gives a polynomial $f$ of degree at most $2n$ which works whenever $A_1 + \\dots + A_{2n} = 0$. Finally, the centroid-shifting trick gives a polynomial which works in general, as wanted.\n\n**Remark 1.** Here is a more detailed approach of the mirror-reflection argument. Let $re^{i\\theta}$ be the polar representation of the point $P$. The polar representations of its mirrored images are then\n$$\nre^{i\\theta}, re^{-i\\theta}, re^{i(\\frac{2\\pi}{n}+\\theta)}, re^{i(\\frac{2\\pi}{n}-\\theta)}, \\dots, re^{i(\\frac{2(n-1)\\pi}{n}+\\theta)}, re^{i(\\frac{2(n-1)\\pi}{n}-\\theta)}.\n$$\nClearly, they are all linear with respect to $P$ and lie on the circle of radius $r$ centred at the origin. As listed above, the $2n$ images are not necessarily in circular order around the circle. For convenience, assume $0 \\le \\theta \\le \\frac{\\pi}{n}$, so the list now displays them in circular order. These images form the vertices of a regular $2n$-gon if and only if the angle between every two consecutive terms in the list (read circularly) is $\\frac{\\pi}{n}$. This is clearly the case if and only if $\\theta = \\frac{\\pi}{2n}$. Consequently, the images are the vertices of a regular $2n$-gon if and only if $P$ lies on the internal bisector of the angle formed by some pair of consecutive mirrors.\n\n*Remark 2.* We sketch here some versions of the arguments in the solution above.\nTo show that $\\deg f \\ge 2n$, we use the same constancy of sign claim and the convention that the polynomial is a function of points (= pairs of coordinates) $A_1, A_2, \\dots, A_{2n}$. Assume that the values of $f$ are all non-negative.\nWrite $B(\\varphi) = (\\cos \\varphi, \\sin \\varphi)$. Choose a substitution\n$$\nA_{2i-1} = B\\left((2i-1)\\frac{\\pi}{n} + \\varphi\\right) \\quad \\text{and} \\quad A_{2i} = B\\left(2i\\frac{\\pi}{n} - \\varphi\\right), \\quad i = 1, 2, \\dots, n.\n$$\nNotice that the coordinates of the points $A_1, A_2, \\dots, A_{2n}$ are all linear functions in $c = \\cos \\varphi$ and $s = \\sin \\varphi$, so, substituting these expressions into $f$, we get a polynomial $g(c, s)$ with $\\deg g \\le \\deg f$.\nNow, the values of $g$ are all non-negative (each being one of $f$), and, on the circle $c^2 + s^2 = 1$, it vanishes at exactly $2n$ points, namely, $(c, s) = (\\cos \\frac{\\pi}{n}k, \\sin \\frac{\\pi}{n}k)$, $k = 1, \\dots, 2n$. We show that these properties already yield $\\deg g \\ge 2n$.\nObviously, if $g(c, s)$ possesses the properties listed above, then so does $g(c, -s)$, and hence so does $\\bar{g}(c, s) = g(c, s) + g(c, -s)$.\nThe polynomial $\\bar{g}$ is even in $s$, so it in fact depends only on $s^2$, and we may plug $s^2 = 1 - c^2$ into it, to obtain a polynomial $h(c)$ with $\\deg h \\le \\deg g$ which is non-negative on $[-1, 1]$ and vanishes on this segment exactly at $c = \\cos \\frac{\\pi}{n}k$. These are $n+1$ such points, and, except $c = \\pm 1$, they should all be roots of $h$ of even multiplicity, due to sign conservation. All in all, this provides $2n$ roots of $h$, counted with multiplicity, hence $\\deg f \\ge \\deg g \\ge \\deg h \\ge 2n$, as desired.\n\nFor a bit alternative construction of a suitable $f$, one may notice that the Lemma in the above solution can be changed to impose vanishing of the elementary symmetric polynomials $\\sigma_i(a_1, a_2, \\dots, a_{2n})$, $i = 1, 2, \\dots, n$, instead of Newton sums. Indeed, if the $\\sigma_i$ all vanish, then so do the polynomials\n$$\n\\sigma_i(\\bar{a}_1, a_2, \\dots, \\bar{a}_{2n}) = \\frac{|a_1|^{2i} \\sigma_{2n-i}(a_1, a_2, \\dots, a_{2n})}{\\bar{a}_1 \\bar{a}_2 \\dots \\bar{a}_{2n}},\n$$\nso $\\sigma_i(a_1, \\dots, a_{2n})$ also vanishes for $i = n+1, \\dots, 2n-1$. Hence $a_1, a_2, \\dots, a_{2n}$ are the roots of $z^{2n} - |a_1|^n$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39189,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ has $\\angle BAC = 90^\\circ$ and $\\angle ABC = 60^\\circ$. The points $D$ and $E$ are taken on the sides $AC$, respectively $AB$, so that $CD = 2 \\cdot DA$ and $DE$ is the bisector of the angle $\\angle ADB$. Denote $M$ the intersection of the lines $CE$ and $BD$, and $P$ the intersection of the lines $DE$ and $AM$. Prove that:\n\na) the lines $AM$ and $BD$ are perpendicular;\n\nb) $3 \\cdot PB = 2 \\cdot CM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $AB = a$. Then $BC = 2a$, $AC = \\sqrt{BC^2 - AB^2} = a\\sqrt{3}$, $AD = \\frac{a}{3}\\sqrt{3}$, $BD = \\sqrt{AB^2 + AD^2} = \\frac{2a}{3}\\sqrt{3} = 2AD$. This yields $\\angle ABD = 30^\\circ$, hence $\\angle ADB = 60^\\circ$.\n\n![](attached_image_1.png)\n\nThis gives $\\angle ADE = 30^\\circ = \\angle ACB$, therefore $DE \\parallel BC$. So $\\triangle DME \\sim \\triangle BMC$ and $\\angle ADE \\sim \\angle ACB$, whence $\\frac{DM}{MB} = \\frac{DE}{BC} = \\frac{AD}{AC} = \\frac{1}{3}$. This implies $\\frac{DM}{DB} = \\frac{1}{4}$, hence $\\frac{DM}{DA} = \\frac{DA}{DB} = \\frac{1}{2}$, so $\\triangle DMA \\sim \\triangle DAB$ (S.A.S.), which implies $\\angle AMD = \\angle BAD = 90^\\circ$.\n\nb) Construct $DF \\parallel AM$, $F \\in CE$. Then $\\frac{CF}{CM} = \\frac{CD}{CA} = \\frac{2}{3}$; we have to prove that $CF = PB$.\n\nWe have $\\angle DAM = 90^\\circ - \\angle ADB = 30^\\circ$ and $\\angle ADE = 30^\\circ$, hence $DP = AP$. In triangle $DPM$, $PM = \\frac{1}{2}DP = \\frac{1}{2}AP$, whence $\\frac{MP}{MA} = \\frac{MF}{MC} = \\frac{1}{3} = \\frac{MF}{MC}$, showing that $PF \\parallel AC$. It follows that $APFD$ is a parallelogram, so $DF = AP = DP$. Since $\\angle CDF = \\angle DAM = 30^\\circ = \\angle BDP$ and $BD = \\frac{2a}{3}\\sqrt{3} = CD$, $\\triangle CDF \\equiv \\triangle BDP$ (S.A.S.), therefore $CF = BP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39190,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $2^3 - 2$, $3^3 - 3$, $4^3 - 4$, $\\dots$, $(2n+1)^3 - (2n+1)$, where $n \\ge 2$ is an integer, are written on a board. An operation consists in erasing three randomly chosen numbers $a$, $b$, $c$ from the board and replacing them with $\\frac{abc}{ab+bc+ca}$. Several operations are performed until two numbers remain on the board. Show that the sum of the last two numbers left on the board is greater than $16$.\n\nGheorghe Rotariu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{ab+bc+ca}{abc} = \\left(\\frac{abc}{ab+bc+ca}\\right)^{-1}$, for any given positive numbers $a$, $b$, $c$, it follows that after any operation, the sum of the inverses of the numbers left on the board is equal to the sum of the inverses of the numbers on the board before the operation. Therefore, the sum of the inverses of the last two numbers left on the board is equal to the sum $S$ of the inverses of the initial numbers.\n\nNotice that $\\frac{1}{k^3-k} = \\frac{1}{k(k-1)(k+1)} = \\frac{1}{2}\\left(\\frac{1}{(k-1)k} - \\frac{1}{k(k+1)}\\right)$, for any $k > 1$, so:\n$$\n\\begin{align*} \nS &= \\frac{1}{2^3 - 2} + \\frac{1}{3^3 - 3} + \\dots + \\frac{1}{(2n+1)^3 - (2n+1)} \\\\ \n&= \\frac{1}{2} \\left( \\frac{1}{1 \\cdot 2} - \\frac{1}{2 \\cdot 3} + \\frac{1}{2 \\cdot 3} - \\frac{1}{3 \\cdot 4} + \\dots + \\frac{1}{2n(2n+1)} - \\frac{1}{(2n+1)(2n+2)} \\right) \\\\ \n&= \\frac{1}{2} \\left( \\frac{1}{2} - \\frac{1}{(2n+1)(2n+2)} \\right) = \\frac{2n^2 + 3n}{8n^2 + 12n + 4} \n\\end{align*}\n$$\nIf $x$ and $y$ are the last two numbers left on the board, then $\\frac{1}{x} + \\frac{1}{y} = \\frac{x+y}{xy} = \\frac{2n^2+3n}{8n^2+12n+4}$. Since $\\frac{4}{x+y} \\le \\frac{x+y}{xy}$, for each $x$, $y > 0$, it follows that $\\frac{4}{x+y} \\le \\frac{2n^2+3n}{8n^2+12n+4} < \\frac{1}{4}$, so $x+y > 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39191,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ be two matrices such that $A^2B = A$. Prove that\n$$\n(AB - BA)^2 = O_n.\n$$\n\nb) Show that, for any natural number $k \\le n/2$, there are two matrices $A$, $B \\in \\mathcal{M}_n(\\mathbb{C})$ with the property $A^2B = A$, such that $\\mathrm{rank}(AB - BA) = k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $A$ is invertible or $A = O_n$ then clearly $AB - BA = O_n$. Assume $A \\ne O_n$, with $\\det(A) = 0$. Let $P \\in \\mathbb{C}[X]$ be the minimal polynomial of the matrix $A$. Since $P(0) = 0$ and $P \\ne X$, the polynomial $P$ has the form $P = X^k + a_{k-1}X^{k-1} + \\dots + a_1X$, where $2 \\le k \\le n$. From the relation $P(A)B = O_n$ and the hypothesis, we obtain\n$$\nA^{k-1} + a_{k-1}A^{k-2} + \\dots + a_2A + a_1AB = O_n. \\quad (1)\n$$\nSince $P$ is minimal, we have $a_1 \\ne 0$ and $AB = -\\frac{1}{a_1}(A^{k-1} + a_{k-1}A^{k-2} + \\dots + a_2A)$. So $AB$ commutes with $A$. Therefore $A = A^2B = A(AB) = ABA$. Then, by multiplying with $BA$ the relation (1) to the right, we obtain\n$$\nA^{k-1} + a_{k-1}A^{k-2} + \\dots + a_2A + a_1AB^2A = O_n. \\quad (2)\n$$\nFrom (1) and (2) it results $a_1(AB^2A - AB) = O_n$. Since $a_1 \\ne 0$, we get $AB^2A = AB$. Thus, $(AB - BA)^2 = (ABA)B - AB^2A - B(A^2B) + B(ABA) = AB - AB - BA + BA = O_n$.\n\nAlternative solution.\n\na) From $\\mathrm{rank}(A) = \\mathrm{rank}(A^2) \\le \\mathrm{rank}(A^2) \\le \\mathrm{rank}(A)$, we obtain $\\mathrm{rank}(A) = \\mathrm{rank}(A^2)$. Denote $r = \\mathrm{rank}(A)$. There are the matrices $X \\in \\mathcal{M}_{n,r}(\\mathbb{C})$ and $Y \\in \\mathcal{M}_{r,n}(\\mathbb{C})$, with $\\mathrm{rank}(X) = \\mathrm{rank}(Y) = r$, such that $A = XY$. We have $YX \\in \\mathcal{M}_r(\\mathbb{C})$ and\n$$\nr = \\mathrm{rank}(A) = \\mathrm{rank}(A^2) = \\mathrm{rank}((XY)^2) = \\mathrm{rank}(X(YX)Y) \\le \\mathrm{rank}(YX).\n$$\nHence $YX$ is an invertible matrix and from the relation $A^2B = A$ we find $YBX = I_r$. Therefore $ABA = (XY)B(XY) = X(YBX)Y = XY = A$. From the relations $A^2B = A$ and $ABA = A$, we get\n$$\n\\begin{aligned}\n(AB - BA)^2 &= (AB)^2 + (BA)^2 - AB^2A - BA^2B \\\\\n&= (ABA)B + B(ABA) - AB^2A - B(A^2B) \\\\\n&= AB + BA - AB^2A - BA = AB(I_n - BA)\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\n(AB - BA)^3 &= AB(I_n - BA)(AB - BA) \\\\\n&= AB(AB - BA - BA^2B + (BA)^2) \\\\\n&= AB(AB - BA - B(A^2B) + B(ABA)) \\\\\n&= AB(AB - BA - BA + BA) = AB(AB - BA) \\\\\n&= (ABA)B - AB^2A = AB - (AB)(BA) = AB(I_n - BA).\n\\end{aligned}\n$$\nThus, $(AB - BA)^3 = (AB - BA)^2$. It follows that, if $\\lambda \\in \\mathbb{C}$ is an eigenvalue of the matrix $AB - BA$, then $\\lambda^3 = \\lambda^2$, hence $\\lambda \\in \\{0, 1\\}$. But $\\text{Tr}(AB - BA) = 0$. Then all the eigenvalues of the matrix $AB - BA$ are 0. Therefore $(AB - BA)^n = O_n$. From the proved relation $(AB - BA)^3 = (AB - BA)^2$ we deduce $(AB - BA)^2 = O_n$.\n\nb) Let us define the matrices $A = \\begin{pmatrix} O_{n-k} & O_{n-k,k} \\\\ O_{k,n-k} & I_k \\end{pmatrix}$, $B = \\begin{pmatrix} O_{n-k} & C \\\\ O_{k,n-k} & I_k \\end{pmatrix}$, where $C \\in \\mathcal{M}_{n-k,k}(\\mathbb{C})$ is an arbitrary matrix with $\\mathrm{rank}(C) = k \\le n-k$ (for $k=0$, we define $A = B = O_n$). We have $A^2B = AB = A$, $BA = B$ and $AB - BA = A - B = \\begin{pmatrix} O_{n-k} & -C \\\\ O_{k,n-k} & O_k \\end{pmatrix}$, so $\\mathrm{rank}(AB - BA) = k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{F}$ be the set of all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(x + f(y)) = f(x) + f(y)$ for all real numbers $x$ and $y$. Determine all real numbers $r$ satisfying the following condition: For every function $f$ in $\\mathcal{F}$, there exists some real number $z$ such that $f(z) = rz$.",
    "options": [],
    "answer": "r = 1 + 1/n or r = 1 − 1/n for any positive integer n",
    "solution": "We first prove that $S$ is a subset of $R$. Let $f$ be a member of $\\mathcal{F}$ and write $P(x, y)$ for $f(x + f(y)) = f(x) + f(y)$.\nBegin by showing that $1 + 1/n$ belongs to $R$ for all positive integers $n$. To this end, induct on $n$ to prove that $f(nf(0)) = (n + 1)f(0)$; letting $z = nf(0)$, then $f(z) = (1 + 1/n)z$, so $1 + 1/n$ is a member of $R$. The base case, $n = 1$, is provided by $P(0, 0)$.\n\nFor the induction step, assume $f(nf(0)) = (n+1)f(0)$ and consider $P(0, nf(0))$ to write\n$$\n\\begin{aligned}\nf((n+1)f(0)) &= f(f(nf(0))) = f(0 + f(nf(0))) = f(0) + f(nf(0)) \\\\\n&= f(0) + (n+1)f(0) = (n+2)f(0),\n\\end{aligned}\n$$\nas desired. This completes the induction.\n\nTo prove that $1-1/n$ belongs to $R$ for all positive integers $n$, notice that $P(-f(0), 0)$ yields $f(-f(0)) = 0$, and an inductive argument along the lines above then shows that $f(-nf(0)) = (-n+1)f(0)$. Letting $z = -nf(0)$, then $f(z) = (1 - 1/n)z$, so $1 - 1/n$ is a member of $R$. Consequently, $S$ is a subset of $R$.\n\nLet $r \\neq 1$ and consider the function $g: [0, 1) \\to \\mathbb{R}$,\n$$\ng(x) = \\begin{cases} 1 & \\text{if } rx/(1-r) \\text{ is an integer} \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f_1(x) = a_1x^2 + b_1x + c_1$, $f_2(x) = a_2x^2 + b_2x + c_2$, $f_3(x) = a_3x^2 + b_3x + c_3$ be three quadratic functions with real coefficients and $P_1 : y = f_1(x)$, $P_2 : y = f_2(x)$, $P_3 : y = f_3(x)$ their associated parabolas.\n\na) Give an example where $P_1 \\cap P_2 = \\{A\\}$, $P_2 \\cap P_3 = \\{B\\}$, $P_3 \\cap P_1 = \\{C\\}$, and $A$, $B$, $C$ are the vertices of an equilateral triangle.\n\nb) Prove that if $a_1, a_2, a_3$ are non nil and pairwisely distinct and $P_1 \\cap P_2 = \\{A\\}$, $P_2 \\cap P_3 = \\{B\\}$, $P_3 \\cap P_1 = \\{C\\}$, then $A$, $B$, $C$ coincide.\n\nVasile Pop",
    "options": [],
    "answer": "One example: let A = (0, 0), B = (1, √3), C = (2, 0). Take P1: y = x^2 − 2x, P2: y = x^2 + (√3 − 1)x, P3: y = x^2 − (√3 + 3)x + (2√3 + 2). Then P1 ∩ P2 = {A}, P2 ∩ P3 = {B}, P3 ∩ P1 = {C}, and A, B, C form an equilateral triangle of side 2.\nFor part b: If a1, a2, a3 are nonzero and pairwise distinct and each pair of parabolas meets at exactly one point, then the three intersection points coincide (A = B = C).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39194,
    "subject": "Mathematics (Multi-modal)",
    "question": "A domino is a rectangle formed by two unit squares that share a common side. A number of $18$ dominoes fit together to tile a $6 \\times 6$ square. Show that some line crossing the interior of the square crosses the interior of no domino. Is it possible that such a line be unique?",
    "options": [],
    "answer": "Yes",
    "solution": "Let the square be $[0, 6] \\times [0, 6]$. We first show that either some grid-vertical $x = i$, $i = 1, 2, 3, 4, 5$, or some grid-horizontal $y = j$, $j = 1, 2, 3, 4, 5$, crosses no tile. Let $m_i$ and $n_j$ be the number of tiles crossed by the grid-vertical $x = i$ and the grid-horizontal $y = j$, respectively. Clearly, a grid-vertical crosses only horizontal tiles, and a grid-horizontal crosses only vertical tiles.\n\nSince every horizontal tile is crossed by a single grid-vertical, the number of horizontal tiles is $m_1 + m_2 + m_3 + m_4 + m_5$. Similarly, the number of vertical tiles is $n_1 + n_2 + n_3 + n_4 + n_5$. Hence the number of tiles is $m_1 + m_2 + m_3 + m_4 + m_5 + n_1 + n_2 + n_3 + n_4 + n_5 = 18$. Consequently, either some $m_i \\le 1$, $i = 1, 2, 3, 4, 5$, or some $n_j \\le 1$, $j = 1, 2, 3, 4, 5$.\n\nNow, for each positive integer $i \\le 5$, the rectangle $[0, i] \\times [0, 6]$ consists of a certain number of tiles and $m_i$ unit cells, the left halves of the horizontal tiles the grid-vertical $x = i$ crosses. Since the area of each $[0, i] \\times [0, 6]$ and the area of each tile are both even, so is each $m_i$. Similarly, each $n_j$ is even.\n\nFinally, by the conclusion of the preceding paragraph, either some $m_i = 0$, $i = 1, 2, 3, 4, 5$, in which case the corresponding grid-vertical $x = i$ crosses no tile; or some $n_j = 0$, $j = 1, 2, 3, 4, 5$, in which case the corresponding grid-horizontal $y = j$ crosses no tile.\n\nAlternative solution.\n\nSuppose, in the above setting, that every grid-line, whether vertical or horizontal, crosses at least one tile. Then the five $m_i$ and the five $n_j$ are all positive even integers, i.e., they are all at least $2$. Consequently, the ten add up to at least $20 > 18$ which is a contradiction. This establishes the first part.\n\nThe answer to the second part is in the affirmative. To prove this, we exhibit a domino tiling of the square $[0, 6] \\times [0, 6]$ with a single separating line, i.e., one crossing no tile. Clearly, grid-lines alone are to be considered.\n\nBegin by tiling the rectangle $[0, 3] \\times [0, 6]$ by four horizontal dominoes, namely,\n$$\n[0, 2] \\times [0, 1], \\quad [0, 2] \\times [3, 4], \\quad [1, 3] \\times [4, 5], \\quad [1, 3] \\times [5, 6],\n$$\nand five vertical dominoes, namely,\n$$\n[2, 3] \\times [0, 2], \\quad [0, 1] \\times [1, 3], \\quad [1, 2] \\times [1, 3], \\quad [2, 3] \\times [2, 4], \\quad [0, 1] \\times [4, 6].\n$$\n\nNotice that the grid–horizontal $y = 4$ is the single separating line of this tiling.\n\nNext, tile the rectangle $[3, 6] \\times [0, 6]$ by a copy of the reflection of the above tiling in the grid–horizontal $y = 3$, to make the grid–horizontal $y = 2$ its single separating line. Explicitly, the four horizontal tiles are\n$$\n[4, 6] \\times [0, 1], \\quad [4, 6] \\times [1, 2], \\quad [3, 5] \\times [2, 3], \\quad [3, 5] \\times [5, 6],\n$$\nand the five vertical tiles are\n$$\n[3, 4] \\times [0, 2], \\quad [5, 6] \\times [2, 4], \\quad [3, 4] \\times [3, 5], \\quad [4, 5] \\times [3, 5], \\quad [5, 6] \\times [4, 6].\n$$\nThe grid–horizontal $y = 2$ is clearly the single separating line of this tiling.\n\nFinally, the two tilings fit together along the grid–vertical $x = 3$ to form an overall tiling of the square $[0, 6] \\times [0, 6]$ with a single separating line — the grid–vertical $x = 3$, of course.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be real numbers such that $ab(c+d) = cd(a+b)$. Prove that\n$$\n\\frac{a+1}{a^2+3} + \\frac{b+1}{b^2+3} \\ge \\frac{c-1}{c^2+3} + \\frac{d-1}{d^2+3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Replace $(c, d)$ with $(-c, -d)$ to rephrase the problem as: subject to the constraint $ab(c+d) + cd(a+b) = 0$, prove that $\\sum \\frac{a+1}{a^2+3} \\ge 0$ (1).\n\nSuppose one of the numbers equals $0$ and notice that (at least) another is $0$ as well – let them be $c$ and $d$. The claim (1) rewrites as\n$$\n\\frac{a+1}{a^2+3} + \\frac{b+1}{b^2+3} \\ge -\\frac{2}{3},\n$$\nwhich holds true as $\\frac{x+1}{x^2+3} \\ge -\\frac{1}{3}$, for all $x \\in \\mathbb{R}$.\n\nSuppose now that $a$, $b$, $c$, $d \\ne 0$ and set $x = \\frac{1}{a}$, $y = \\frac{1}{b}$, $z = \\frac{1}{c}$ and $t = \\frac{1}{d}$. Once again, the problem is restated:\n$$\nx + y + z + t = 0 \\implies \\sum \\frac{x^2 + x}{3x^2 + 1} \\ge 0. \\quad (2)\n$$\nLet $|t| = \\max\\{|x|, |y|, |z|, |t|\\}$. The inequality (2) is equivalent to\n$$\n\\sum \\left( \\frac{6x^2 + 6x}{3x^2 + 1} + 1 \\right) \\ge 4 \\iff \\sum \\frac{(3x+1)^2}{3x^2 + 1} \\ge 4\n$$\n$$\n\\iff \\frac{(3x+1)^2}{3x^2+1} + \\frac{(3y+1)^2}{3y^2+1} + \\frac{(3z+1)^2}{3z^2+1} \\ge 4 - \\frac{(3t+1)^2}{3t^2+1} = \\frac{3(1-t)^2}{3t^2+1}.\n$$\nNotice that $x^2+y^2+z^2 \\le 3t^2$ and apply Cauchy-Schwarz inequality to conclude:\n$$\n\\frac{(3x+1)^2}{3x^2+1} + \\frac{(3y+1)^2}{3y^2+1} + \\frac{(3z+1)^2}{3z^2+1} \\ge \\frac{3(x+y+z+1)^2}{x^2+y^2+z^2+1} \\ge \\frac{3(1-t)^2}{3t^2+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39196,
    "subject": "Mathematics (Multi-modal)",
    "question": "The isosceles triangle $ABC$ has $AB = AC$ and $\\angle CAB = 30^\\circ$. Let $F \\in [AB]$, $E \\in [AC]$, $P \\in [AB]$, and $M \\in [BC]$ be points such that $\\angle FCA = 15^\\circ$, $\\angle FEA = 45^\\circ$, $CP \\perp CF$ and $[PM]$ is the bisector of $\\angle EPC$. Prove that $E$ is the midpoint of the segment $AC$ and $M$ is the midpoint of the segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the natural numbers $m, n$ so that\n$$\nn \\cdot (n + 1) = 3^m + s(n) + 1182,\n$$\nwhere $s(n)$ represents the sum of the digits of the positive integer $n$.",
    "options": [],
    "answer": "m = 0, n = 34",
    "solution": "$$\nn^2 = 3^m + 1182 - (n - s(n)). \\qquad (1)\n$$\nIf $m \\ge 2$, because $n - s(n)$ is divisible by 9, but $3 \\mid 1182$ and $9 \\nmid 1182$, from (1) follows that $3 \\mid n^2$ and $9 \\nmid n^2$, false. Similarly, if $m = 1$, we obtain that $n^2$ is divisible by 3 and not by 9, false. Therefore $m = 0$, and $n^2 + (n - s(n)) = 1183$. (2)\nIf $n \\le 9$, then $s(n) = n$, whence $n^2 = 1183$, false. Consequently, we have $n \\ge 10$. From (2) we deduce that $n \\le 34$, therefore $n = \\overline{ab}$, with $a \\in \\{1, 2, 3\\}$. Studying all the situations, we find that the only solution is $m = 0, n = 34$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39198,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $3 \\times 3 \\times 3$ cube is divided into 27 unit-cubes. Call a *strip* any $1 \\times 1 \\times 3$ rectangular cuboid (block) consisting of three unit-cubes.\nA positive integer is written inside each unit-cube such that any number $n$, strictly greater than 1, written in a unit-cube, is the sum of the numbers written inside three other unit-cubes, one from each of the three strips in which $n$ is situated. Prove that, regardless of the choice of the 27 numbers, there will be at least 16 among them that are smaller than or equal to 60.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If all the 27 numbers are equal to 1, we have nothing to prove. Assume the cube contains some numbers greater than 1 and suppose there is an even number between them. If $n$ is the smallest even number written inside a cube, then $n$ should be the sum of three odd numbers, which is impossible due to parity reasons. So all the 27 numbers are odd.\nWe shall prove that one of the numbers in every strip is equal to 1. Assuming the contrary, there is a strip that doesn't contain 1. If $a$ is the smallest number in this strip, then $a$ should be the sum of three numbers less than $a$ and greater than 1, a contradiction with the minimality of $a$. It follows that each strip contains at least an 1, so at least 9 numbers inside the $3 \\times 3 \\times 3$ cube are equal to 1.\n\nLet $a_1 \\le a_2 \\le \\dots \\le a_{18}$ the other numbers inside the big cube. If $a_1 > 3$, then one of three numbers whose sum is $a_1$ must be greater than 1 and less than $a_1$, which contradicts the choice made for $a_1$.\nTherefore, $a_1 \\in \\{1,3\\}$, so $a_2 \\in \\{1,3,5\\}$. If $a_2 \\le 3$, then $a_3 \\le 1+a_1+a_2 \\le 7$. If $a_2 = 5$, then $a_1$ and $a_2$ are on the same strip, so at most one of them may be used to write $a_3$ as a sum of three numbers from the cube. Hence, $a_3 \\le 1+1+a_2 = 7$, so consequently $a_3 \\in \\{1,3,5,7\\}$.\n\nIf $a_3 = 7$, it must exist a strip containing $a_2$ and $a_3$, so $a_2$ and $a_3$ cannot be simultaneously terms of expressing $a_4$ as a sum of three numbers inside the cube, so $a_4 \\le a_3+a_1+1 \\le 7+3+1 = 11$. If $a_3 \\le 5$, then $a_4 \\le a_3+a_2+a_1 \\le 5+5+1 = 11$. Therefore, $a_4 \\le 11$.\nMore general, we have $a_{k+3} \\le a_{k+2} + a_{k+1} + a_k$.\nIf $a_{k+1}$ and $a_{k+2}$ are on a same strip, at most one of them may be used to write $a_{k+3}$ as a sum of three numbers written in the unit-cubes. Since $a_{k+2} \\le a_{k+1} + a_k + a_{k-1}$, it follows that $a_{k+3} \\le a_{k+2} + a_k + a_{k-1} \\le a_{k+1} + 2a_k + 2a_{k-1}$.\nIf no strip contains $a_{k+1}$ and $a_{k+2}$, since $a_{k+2} \\le a_k + a_{k-1} + a_{k-2}$, we obtain $a_{k+3} \\le a_{k+2} + a_{k+1} + a_k \\le a_{k+1} + 2a_k + a_{k-1} + a_{k-2} \\le a_{k+1} + 2a_k + 2a_{k-1}$.\nSubsequently, $a_{k+3} \\le a_{k+1} + 2a_k + 2a_{k-1}$, for each $k \\ge 2$. Successively, we infer that $a_5 \\le 23$, $a_6 \\le 35$ and $a_7 \\le 59$, so at least 16 numbers written inside the unit-cubes are either smaller than 60.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39199,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a blackboard there are written, one after another, all positive integers from $1$ to $30000$, forming the following sequence of digits:\n$123456789101112\\ldots30000$.\nFind the number of occurrences of $2023$ in this sequence.",
    "options": [],
    "answer": "25",
    "solution": "The sequence $2023$ appears $13$ times in the writing of $2023$, $12023$, $22023$ and of $20230$, $20231$, $20232$, $\\ldots$, $20239$. Moreover, $2023$ can appear by connecting the end of a number to the beginning of the next number. The writing $202|3$\n\nappears only once, in $3202|3203$, and the writing $20|23$ appears $11$ times, in $2320|2321$ and from $23020|23021$ to $23920|23921$. The writing $2|023$ is impossible, because no positive integer begins with $0$, therefore the number of occurrences of $2023$ is $13 + 11 + 1 = 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39200,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an art museum, $n$ paintings are exhibited, where $n \\ge 33$ is a positive integer, and a total of 15 colors are used in such a way that any two paintings have at least one color in common, and no two paintings have exactly the same set of colors. Determine all possible values of $n \\ge 33$ such that, no matter how we color the paintings with the above properties, we can choose four distinct paintings that we number $T_1, T_2, T_3$, and $T_4$ in such a way that any color used both in $T_1$ and $T_2$ is also used in $T_3$ or $T_4$.",
    "options": [],
    "answer": "All integers n with 33 ≤ n ≤ 2^14",
    "solution": "We will prove that each $n \\in \\{33, 34, \\dots, 2^{14}\\}$ is a solution.\n\nWe begin by noticing that if we have a painting $T_i$ in the museum that uses $k$ colors, the painting that uses the other $15-k$ colors cannot be in the museum. Therefore, out of the $2^{15}-1$ possible paintings that can be obtained with the 15 colors, we have a maximum of $2^{14}$ paintings in the museum, and the maximum can be achieved if we consider all paintings that use color $c_1$ together with all $2^{14}$ subsets of colors $\\{c_2, \\dots, c_{15}\\}$.\n\nWe show that for any $33 \\le n \\le 2^{14}$, we can find the paintings $T_1, T_2, T_3, T_4$ with the properties described in the problem. By $T_i \\cap T_j$ and $T_i \\cup T_j$ we mean the set of colors that are common to the paintings $T_i$ and $T_j$, and the set of all colors used in the paintings $T_i$ and $T_j$, respectively. We need to prove that there exist $i_1, i_2, i_3, i_4 \\in \\{1, 2, \\dots, n\\}$ such that:\n$$\n(T_{i_1} \\cap T_{i_2}) \\subset T_{i_3} \\cup T_{i_4}\n$$\nAssuming that for any choice of $i < j$ and $k < \\ell$ from $\\{1, 2, \\dots, n\\}$ with $\\{i, j\\} \\cap \\{k, \\ell\\} = \\emptyset$, we have:\n$$\n|(T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)| \\ge 1.\n$$\nStudying the sum:\n$$\nS = \\sum_{\\substack{1 \\le i < j \\le n \\\\ 1 \\le k < \\ell \\le n \\\\ \\{i,j\\} \\cap \\{k,\\ell\\} = \\emptyset}} |(T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)|,\n$$\nwe notice that this sum has $\\binom{n}{2} \\cdot \\binom{n-2}{2}$ terms, each of which is greater than or equal to 1, so we obtain:\n$$\nS \\ge \\frac{n(n-1)}{2} \\cdot \\frac{(n-2)(n-3)}{2}.\n$$\n\nLet's count the number of occurrences of each color $c_m$, where $m = \\overline{1, 15}$, and let $n_m$ be the number of paintings $T_i$ that contain color $c_m$. To have $c_m \\in (T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)$, we must have $c_m \\in T_i, T_j$ and $c_m \\notin T_k, T_\\ell$. If $n_m \\in \\{0, 1, n-1, n\\}$, then there is no tuple $(T_i, T_j, T_k, T_\\ell)$ for which $c_m \\in (T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)$. For $2 \\le n_m \\le n-2$, the pair $(T_i, T_j)$ can be chosen in $\\binom{n_m}{2}$ ways, and the pair $(T_k, T_\\ell)$ can be chosen in $\\binom{n-n_m}{2}$ ways. Therefore, we have:\n$$\nS = \\sum_{\\substack{m=1 \\\\ 2 \\le n_m \\le n-2}}^{15} \\frac{n_m(n_m-1)}{2} \\cdot \\frac{(n-n_m)(n-n_m-1)}{2}.\n$$\nFrom AM-GM we have $n_m(n - n_m) \\le \\left(\\frac{n}{2}\\right)^2$ and $(n_m - 1)(n - 1 - n_m) \\le \\left(\\frac{n-2}{2}\\right)^2$, from which we obtain:\n$$\n\\begin{aligned}\nS & \\le \\sum_{\\substack{m=1 \\\\ 2 \\le n_m \\le n-2}}^{15} \\frac{1}{4} \\left(\\frac{n}{2}\\right)^2 \\cdot \\left(\\frac{n-2}{2}\\right)^2 \\le 15 \\cdot \\frac{1}{4} \\cdot \\left(\\frac{n}{2}\\right)^2 \\cdot \\left(\\frac{n-2}{2}\\right)^2 < \\\\ \n& < \\frac{n(n-1)}{2} \\cdot \\frac{(n-2)(n-3)}{2}, \\quad \\forall n \\ge 33,\n\\end{aligned}\n$$\nwhich is a contradiction. Therefore, for any $33 \\le n \\le 2^{14}$ we have four paintings with the property from the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $AH$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $KP + KQ \\ge BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix the origin at $O$ and the real axis along $\\ell$. As usual lower case letters denote the complex coordinate of the corresponding point in the configuration. For convenience, let $|a| = |b| = |c| = 1$. Clearly,\n\n$$\nk = a + \\frac{1}{2}(b + c), \\quad p = a + \\frac{1}{2}\\left(b + \\frac{1}{b}\\right) \\quad \\text{and} \\quad q = a + \\frac{1}{2}\\left(c + \\frac{1}{c}\\right).\n$$\n\nThen\n$$\n|k-p| = \\left| a + \\frac{1}{2} \\left( c - \\frac{1}{b} \\right) \\right| = \\frac{1}{2} |2ab + bc - 1|, \\quad \\text{since } |b| = 1.\n$$\nSimilarly, $|k-q| = \\frac{1}{2}|2ac+bc-1|$, so\n$$\n|k-p|+|k-q| = \\frac{1}{2}|2ab+bc-1|+\\frac{1}{2}|2ac+bc-1| \\ge \\frac{1}{2}|2a(b-c)| = |b-c|, \\quad \\text{since } |a|=1\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a $4 \\times 4$ array of pairwise distinct positive integers such that on each column, respectively row, one of the numbers is equal to the sum of the other three. Determine the least possible value of the largest number such an array may contain.",
    "options": [],
    "answer": "21",
    "solution": "<table>\n<tr><td>1</td><td>8</td><td>12</td><td><b>21</b></td></tr>\n<tr><td>7</td><td>9</td><td>20</td><td>4</td></tr>\n<tr><td>10</td><td>19</td><td>3</td><td>6</td></tr>\n<tr><td>18</td><td>2</td><td>5</td><td>11</td></tr>\n</table>\n\nThe lower bound is a consequence of the following slightly more general fact:\nIf, for some integer $n \\ge 3$, on each row of an $n \\times n$ array of pairwise distinct positive integers, one of the numbers is equal to the sum of the other $n-1$, then the largest number in the array is at least $\\frac{1}{2}(n-1)(n^2 - n + 2)$; columns subjected to no condition whatsoever.\n\nLet $a$ be the largest number in the array and let $a_i$ be the largest number on the $i$-th row, $i = 1, 2, \\dots, n$. Since the $a_i$ are pairwise distinct positive integers not exceeding $a$,\n$$\n\\sum_{i=1}^{n} a_i \\le \\sum_{j=0}^{n-1} (a-j) = na - \\frac{1}{2}n(n-1). \\quad (*)\n$$\nLet now $b_1, b_2, \\dots, b_{n^2-n}$ be the numbers in the array different from each $a_i$. Then\n$$\n\\sum_{i=1}^{n} a_i = \\sum_{j=1}^{n^2-n} b_j \\ge \\sum_{k=1}^{n^2-n} k = \\frac{1}{2}n(n-1)(n^2-n+1), \\quad (**)\n$$\nwhere the first equality holds by hypothesis, and the inequality follows from the fact that the $b_j$ are pairwise distinct positive integers.\nFinally, $(*)$ and $(**)$ imply $a \\ge \\frac{1}{2}(n-1)(n^2-n+2)$, as stated. If $n=4$, then $a \\ge 21$, as desired.\n\nIn particular, if $n = 3$, then $a \\ge \\max(9, 8) = 9$. The case $a = 9$ can be ruled out by hand and it turns out that $a = 10$ fits the bill on columns as well:\n$$\n\\begin{array}{ccc}\n4 & 5 & 9 \\\\\n6 & 8 & 2 \\\\\n\\mathbf{10} & 3 & 7\n\\end{array}\n$$\nIt should now be clear that $a = \\frac{1}{2}(n-1)(n^2-n+2)$ forces $n \\ge 4$. If $n = 4$, then $a = 21$ also fits the bill on columns, as the array in the solution shows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that:\n$$\nf(xf(x) + f(y)) = f(f(x^2)) + y, \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "f(x) = x for all real x, or f(x) = -x for all real x",
    "solution": "We claim that the solutions are $f_1 = 1_{\\mathbb{R}}$ and $f_2 = -1_{\\mathbb{R}}$.\n\nFor $x = 0$ we obtain $f(f(y)) = f(f(0)) + y$, for each $y \\in \\mathbb{R}$, so $f$ is an one-to-one function.\n\nFor $y = 0$ we have $f(xf(x) + f(0)) = f(f(x^2))$, $\\forall x \\in \\mathbb{R}$, which, due to $f$ being one-to-one, leads to $xf(x) + f(0) = f(x^2)$, $\\forall x \\in \\mathbb{R}$.\n\nPutting $x = 1$ in the last relation gives $f(0) = 0$, so the last relation can be written as $xf(x) = f(x^2)$, $\\forall x \\in \\mathbb{R}$, while the first relation becomes $f(f(x)) = x$, $\\forall x \\in \\mathbb{R}$.\n\nPutting $x \\to f(x)$ in the above relation, we have $f(f(x)^2) = f(f(x))f(x) = xf(x) = f(x^2)$, which, due to $f$ being one-to-one, implies that $(f(x))^2 = x^2$, so for each $x \\in \\mathbb{R}$ we have $f(x) = x$ or $f(x) = -x$.\n\nSupposing that exists a pair $(x_0, y_0) \\in \\mathbb{R}^* \\times \\mathbb{R}^*$ such that $f(x_0) = x_0$ and $f(y_0) = -y_0$. Putting $x \\to x_0$ and $y \\to y_0$ in the given functional equation, we obtain $f(x^2 - y) = x^2 + y$, so $x^2 - y = x^2 + y$ or $x^2 - y = -x^2 - y$, implying that $y = 0$ or $x = 0$, which is a contradiction. Therefore, the claim is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $M = \\{x \\in \\mathbb{R} \\mid \\lfloor nx \\rfloor + \\lfloor (n+1)x \\rfloor = \\lfloor (2n+1)x \\rfloor, \\forall n \\in \\mathbb{N}\\}$.\n\na) Prove that, if $x \\in M \\cap [\\frac{1}{2}, 1)$, then $\\lfloor nx \\rfloor = \\lfloor \\frac{n}{2} \\rfloor, \\forall n \\in \\mathbb{N}$.\n\nb) Find the cardinal of the set $M \\cap [0, 2023]$.",
    "options": [],
    "answer": "4047",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral and let $P$ be a point inside such that $\\angle APB + \\angle CPD = \\angle APD + \\angle BPC$, $\\angle PAD + \\angle PCD = \\angle PAB + \\angle PCB$ and $\\angle PDC + \\angle PBC = \\angle PDA + \\angle PBA$. Prove that the quadrilateral $ABCD$ is circumscriptible.\n\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $\\angle PDC + \\angle PBC = \\angle PDA + \\angle PBA$ implies $\\angle PDA + \\angle PBA = (B + D)/2$. In the same manner, $\\angle PAB + \\angle PCB = (A + C)/2$. Add the last equalities to obtain $180^\\circ = (A + B + C + D)/2 = (\\angle PBA + \\angle PAB) + \\angle PDA + \\angle PCB = (180^\\circ - \\angle APB) + \\angle PDA + \\angle PCB$, and notice that $\\angle PDA + \\angle PCB = \\angle APB$ shows that circles $APD$ and $BPC$ are tangent at $P$.\n\nLet $(O_1, R_1)$, $(O_2, R_2)$, $(O_3, R_3)$, $(O_4, R_4)$ be the circles $PAB$, $PBC$, $PCD$ and $PDA$ respectively. Recall that points $P$, $O_1$, $O_3$ are collinear, and similarly, points $P$, $O_2$, $O_4$ are collinear. Further, $\\angle O_2O_1O_4 + \\angle O_2O_3O_4 = 180^\\circ - \\angle APB + 180^\\circ - \\angle CPD = 180^\\circ$, so $O_1O_2O_3O_4$ is a cyclic quadrilateral; let $R$ be its circumradius.\n\nBy Sine Law,\n$$\n2R = \\frac{O_1O_3}{\\sin O_2} = \\frac{O_2O_4}{\\sin O_1} \\implies \\frac{R_1 + R_3}{\\sin \\angle BPC} = \\frac{R_2 + R_4}{\\sin \\angle APB},\n$$\nthen $\\frac{AB}{\\sin \\angle APB} + \\frac{CD}{\\sin \\angle CPD} = 2(R_1 + R_3) \\implies AB + CD = 2(R_1 + R_3) \\sin \\angle APB$. Similarly, $BC + DA = 2(R_2 + R_4) \\sin \\angle BPC$. All the above lead to $AB + CD = BC + DA$, hence the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a > 0$ such that there exist real numbers $x, y > 0$ with the properties\n$$\nx = a - \\sqrt{y}, \\quad y = a + \\sqrt{x}, \\quad \\frac{x}{y} + \\frac{y}{x} = 2023.\n$$",
    "options": [],
    "answer": "44/43",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function with the property that there are a differentiable function $g: \\mathbb{R} \\to \\mathbb{R}$ and a sequence $(a_n)_{n \\ge 1}$ with strictly positive terms and $\\lim_{n \\to \\infty} a_n = 0$, such that\n$$\ng'(x) = \\lim_{n \\to \\infty} \\frac{f(x + a_n) - f(x)}{a_n},\n$$\nfor all $x \\in \\mathbb{R}$.\na) Give an example of such a function $f$ that is not differentiable at any point $x \\in \\mathbb{R}$.\nb) Assume that $f$ is continuous on $\\mathbb{R}$. Show that $f$ is differentiable on $\\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Take the function $f : \\mathbb{R} \\to \\mathbb{R}$ defined by\n$$\nf(x) = \\begin{cases} 1, & x \\in \\mathbb{Q}, \\\\ 0, & x \\in \\mathbb{R} \\setminus \\mathbb{Q}, \\end{cases}\n$$\nand let $(a_n)_{n \\ge 1}$ be the sequence with the terms $a_n = 1/n, \\forall n \\in \\mathbb{N}^*$. For $n \\in \\mathbb{N}^*$, we have $x + a_n \\in \\mathbb{Q}, \\forall x \\in \\mathbb{Q}$ and $x + a_n \\in \\mathbb{R} \\setminus \\mathbb{Q}, \\forall x \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Thus, $f(x + a_n) = f(x), \\forall x \\in \\mathbb{R}, \\forall n \\in \\mathbb{N}^*$. It turns out that\n$$\n\\lim_{n \\to \\infty} \\frac{f(x+a_n)-f(x)}{a_n} = 0 = g'(x), \\forall x \\in \\mathbb{R},\n$$\nwhere $g$ is a constant function. The function $f$ is discontinuous at any point $x \\in \\mathbb{R}$, therefore not differentiable at any point $x \\in \\mathbb{R}$.\n\nb) Let us define the function $h : \\mathbb{R} \\to \\mathbb{R}$, $h = f - g$. Clearly, the function $h$ is continuous. $\\lim_{n \\to \\infty} \\frac{h(x+a_n)-h(x)}{a_n} = \\lim_{n \\to \\infty} \\frac{f(x+a_n)-f(x)}{a_n} - g'(x) = 0$, for all $x \\in \\mathbb{R}$. Let $x$ and $y$ be two real numbers, with $x < y$. For $c > 0$, we define the set\n$$\nA(c) = \\{z \\in [x, y] \\mid |h(z) - h(x)| \\le c(z - x)\\}.\n$$\nSince $x \\in A(c) \\subset [x, y]$, there is $s = \\sup A(c) \\in [x, y]$. From the continuity of $h$, we get $s \\in A(c)$. Assume $s < y$. Then, there is $n_1 \\in \\mathbb{N}^*$ such that $s+a_n < y, \\forall n \\ge n_1$. From the limit $\\lim_{n \\to \\infty} \\frac{h(s+a_n)-h(s)}{a_n} = 0$, we deduce that there is $n_2 \\ge n_1$ such that\n$$\n\\left| \\frac{h(s+a_{n_2})-h(s)}{a_{n_2}} \\right| < c.\n$$\nThen $s < s+a_{n_2} < y$ and the following inequalities hold:\n$$\n|h(s+a_{n_2}) - h(x)| \\le |h(s) - h(x)| + |h(s+a_{n_2}) - h(s)| < c(s-x) + ca_{n_2} = c[(s+a_{n_2}) - x].\n$$\nWe obtain $s + a_{n_2} \\in A(c)$, in contradiction with $s = \\sup A(c)$. Therefore $y = s \\in A(c)$, so $|h(y) - h(x)| \\le c(y - x)$. But $c > 0$ is an arbitrary positive constant. Thus, we obtain $h(x) = h(y)$. Hence $h$ is a constant function. Then $f = g + h$ is differentiable on $\\mathbb{R}$, with $f' = g'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the strictly increasing functions $f : \\mathbb{N} \\to \\mathbb{N}$ with the following property: the number $f(x) \\cdot f(y)$ divides the number $(1+2x) \\cdot f(y) + (1+2y) \\cdot f(x)$, for any natural numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 2x + 1 or f(x) = 4x + 2",
    "solution": "For $x = y = 0$ we deduce $f^2(0) \\mid 2f(0)$, whence $f(0) \\in \\{0, 1, 2\\}$.\n\nIf $f(0) = 0$, for $y = 0$ we have $0 \\mid f(x)$, $\\forall x \\in \\mathbb{N}$, so $f(x) = 0$, $\\forall x \\in \\mathbb{N}$, which contradicts the fact that $f$ is strictly increasing.\n\nIf $f(0) = 1$, for $y = 0$, we have $f(x) \\mid (2x + 1) + f(x)$, $\\forall x \\in \\mathbb{N}$, thus $f(x) \\mid 2x + 1$, $\\forall x \\in \\mathbb{N}$.\nAssuming that $f(k) = 2k + 1$, for some natural $k$, we have $f(k + 1) \\mid 2k + 3$, and since $f(k + 1) > f(k) = 2k + 1$ we get $f(k + 1) = 2k + 3$, that is $f(x) = 2x + 1$, $\\forall x \\in \\mathbb{N}$.\n\nIf $f(0) = 2$, for $y = 0$, we have $2f(x) \\mid 2(2x + 1) + f(x)$, $\\forall x \\in \\mathbb{N}$, thus $\\frac{2(2x+1)+f(x)}{2f(x)} = \\frac{2x+1}{f(x)} + \\frac{1}{2} \\in \\mathbb{N}$, $\\forall x \\in \\mathbb{N}$.\nAssuming that $f(k) = 4k + 2$, for some natural $k$, we have $\\frac{2k+3}{f(k+1)} + \\frac{1}{2} \\in \\mathbb{N}$ and since $f(k + 1) > f(k) = 4k + 2$ we get $0 < \\frac{2k+3}{f(k+1)} + \\frac{1}{2} < \\frac{2k+3}{4k+2} + \\frac{1}{2} \\le 2$. So $\\frac{2k+3}{f(k+1)} + \\frac{1}{2} = 1$, thus $f(k+1) = 4k + 6$, that is $f(x) = 4x + 2$, $\\forall x \\in \\mathbb{N}$.\n\nBoth functions fulfill the property in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive real number. Show that there are no real numbers $b$ and $c$, with $b < c$, such that $\\left|\\frac{x+y}{x-y}\\right| \\le a$, for every $x, y \\in (b, c)$, $x \\ne y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, by the sake of contradiction, that there exist $b, c \\in \\mathbb{R}$, $b < c$, such that $\\left|\\frac{x+y}{x-y}\\right| \\le a$, for every $x, y \\in (b, c)$, $x \\ne y$, and consider $x \\in (b, c)$, $x \\ne 0$.\n\nIf $x > 0$, there exist infinitely many positive integers $n$ such that $n > \\frac{1}{x-b}$, or equivalently, such that $x - \\frac{1}{n} > b$. For each such number $n$, choose $y_n = x - \\frac{1}{n} \\in (b, c)$. Since $\\left|\\frac{x+y_n}{x-y_n}\\right| \\le a$, it follows that $x \\le \\frac{1+a}{2n}$, so $0 < n \\le \\frac{1+a}{2x}$ for infinitely many positive integers $n$, a contradiction.\n\nIf $x < 0$, there are infinitely many positive integers $n$ such that $n > \\frac{1}{c-x}$, or equivalently, such that $x + \\frac{1}{n} < c$. For each such number $n$, choose $z_n = x + \\frac{1}{n} \\in (b, c)$. Since $\\left|\\frac{x+z_n}{x-z_n}\\right| \\le a$, it follows that $-\\frac{1+a}{2n} \\le x < 0$, so $0 < n \\le -\\frac{1+a}{2x}$ for infinitely many positive integers $n$, again a contradiction, and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A, B \\in \\mathcal{M}_3(\\mathbb{C})$ be two matrices with the property $A^2 = B^2 = O_3$. Prove that $AB = BA$ implies $AB = O_3$. Show that the reciprocal implication is false.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $AB = BA$. We obtain $(A+B)^2 = 2AB$ and $(A+B)^3 = 2AB(A+B) = 2A^2B + 2AB^2 = O_3$. Then, $(A+B)(A-B) = A^2 - B^2 = O_3$. From Sylvester rank inequality, we get $\\text{rank}(A+B) + \\text{rank}(A-B) \\le 3$. Thus, $\\text{rank}(A+B) \\le 1$ or $\\text{rank}(A-B) \\le 1$.\n\nSuppose $\\text{rank}(A+B) \\le 1$. If $\\text{rank}(A+B) = 0$, that is $A+B = O_3$, then $AB = -A^2 = O_3$. If $\\text{rank}(A+B) = 1$, then there are two nonzero matrices, $C \\in \\mathcal{M}_{3,1}(\\mathbb{C})$ and $D \\in \\mathcal{M}_{1,3}(\\mathbb{C})$, such that $A+B = CD$. It follows $(A+B)^2 = (CD)(CD) = C(DC)D = \\text{Tr}(A+B)(A+B)$. We obtain $O_3 = (A+B)^3 = \\text{Tr}(A+B)(A+B)^2$. If $\\text{Tr}(A+B) \\ne 0$ then $(A+B)^2 = O_3$. If $\\text{Tr}(A+B) = 0$ then, from the relation $(A+B)^2 = \\text{Tr}(A+B)(A+B)$, we also obtain $(A+B)^2 = O_3$. Therefore $AB = O_3$.\n\nThe case where $\\text{rank}(A-B) \\le 1$ can be analogously treated.\n\nA counterexample for the converse implication:\n$$\nA = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix} \\quad \\text{and} \\quad B = \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}.\n$$\nWe have $A^2 = B^2 = AB = O_3$, but $AB \\ne BA$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider several tokens of different colors and sizes such that no two tokens have the same color and size. Each token $J$ has two numbers written on it: one represents the number of tokens with the same color as $J$ but different sizes, and the other represents the number of tokens with the same size as $J$ but different colors. It is known that each of the numbers $0, 1, \\ldots, 100$ appears at least once. For what numbers of tokens is this possible?",
    "options": [],
    "answer": "All integers m ≥ 3434",
    "solution": "**Observation 1.** If the situation is possible for a certain number $m$ of tokens, then it is also possible for a number $m' > m$: we can add $m' - m$ tokens with different sizes and colors to a suitable configuration with $m$ tokens. This shows that it is sufficient to find the minimum number of tokens for which we can obtain the desired configuration.\n\n**Observation 2.** If each of the numbers $0, 1, \\ldots, n - 1$ appears at least once on the tokens, then the table has rows (or columns) with exactly $1, 2, \\ldots, n$ tokens, respectively.\n\n**Property.** We will show that if $n \\equiv 2 \\pmod{3}$, then the table has at least $\\frac{1}{3} n(n+1)$ tokens.\n\nLet $n = 3k - 1$, and consider $2k$ rows filled with $k, k + 1, \\ldots, 3k - 1$ tokens. Counting, possibly some tokens multiple times, these rows contain a total of\n$$\nN = k + (k + 1) + (k + 2) + \\dots + (3k - 1) = k(4k - 1)\n$$\ntokens. Each token can be counted at most twice. If we denote by $x$ the number of rows and by $y$ the number of columns participating in obtaining $N$, there are at most $xy$ tokens counted twice. Since $x + y = 2k$, we have $xy \\le k^2$, so the rows participating in obtaining $N$ have at least $N - k^2 = k(3k - 1) = \\frac{1}{3} n(n + 1)$ tokens.\n\n**Example.** A configuration with exactly $k(3k-1)$ tokens can be obtained from a $(2k-1) \\times (3k-1)$ matrix where, starting from the first column, we place $1, 2, \\dots, k-1$ tokens on the first $k-1$ rows, and $2k, 2k+1, \\dots, 3k-1$ tokens on the remaining rows. The number of tokens is $(1+2+\\dots+(k-1)) + (2k+(2k+1)+\\dots+(3k-1)) = \\frac{k(k-1)}{2} + k(5k-1) = k(3k-1) = \\frac{1}{3} n(n+1)$, the rows represent rows with $1, 2, \\dots, k-1, 2k, 2k+1, \\dots, 3k-1$ tokens, and the first $k$ columns represent rows with $2k-1, 2k-2, \\dots, k$ tokens.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a tetrahedron $ABCD$ with $\\widehat{BAC} + \\widehat{CAD} + \\widehat{DAB} = 180^\\circ$ and $\\widehat{ABC} \\equiv \\widehat{DAB}$. If the projection of the vertex $D$ on the plane $(ABC)$ is the orthocenter of the triangle $ABC$, prove that $AB = AC$ and $DB = DC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BE$ and $CF$ be the altitudes from $B$ and $C$, respectively, of the triangle $ABC$, and denote by $H$ their meeting point. The three perpendiculars theorem implies that $DE \\perp AC$ and $DF \\perp AB$.\nUnfold the tetrahedron on the plane $(ABC)$ and denote by $D_1$ the image of the vertex $D$ of the triangle $DAB$ and by $D_2$ the image of the vertex $D$ of the triangle $DAC$, after the unfolding.\n\n![](attached_image_1.png)\n\nThe perpendicularities $DE \\perp AC$ and $DF \\perp AB$ are maintained on the unfolding. Therefore, we deduce that $D_1, F$ and $C$ are collinear. Similarly, we find that $D_2, E$ and $B$ are collinear.\n\nAs $\\widehat{BAC} + \\widehat{CAD} + \\widehat{DAB} = 180^\\circ$, the points $D_1, A$ and $D_2$ are collinear; obviously, $A$ is the midpoint of the segment $D_1D_2$.\n\n$\\widehat{ABC} \\equiv \\widehat{DAB}$, hence the lines $D_1D_2$ are $BC$ parallel, so $D_1D_2CB$ is a trapezium, whose diagonals meet in $H$. The midpoint of the larger base lies on the line $AH$, therefore $AH$ also contains the midpoint of the smaller base $BC$. Because $AH$ is simultaneously an altitude and a median of the triangle $ABC$, it follows that the triangle $ABC$ is isosceles, with $AB = AC$.\n\nThe line $AH$ is the common perpendicular bisector of the bases of the trapezium $D_1D_2CB$, therefore $D_1D_2CB$ is an isosceles trapezium, hence $D_1B = D_2C$ and thus $DB = DC$.\n\nAlternative solution:\n\nFrom $\\widehat{BAC} + \\widehat{CAD} + \\widehat{DAB} = 180^\\circ$, $\\widehat{ABC} \\equiv \\widehat{DAB}$ and $\\widehat{BAC} + \\widehat{ABC} + \\widehat{ACB} = 180^\\circ$, we deduce that $\\widehat{CAD} \\equiv \\widehat{ACB}$.\n\nLet $H$ be the orthocenter of the triangle $ABC$ and $BE, CF, AG$ its altitudes. From the three perpendiculars theorem, we deduce $DE \\perp AC, DF \\perp AB$ and $DG \\perp BC$.\n\nTriangles $ABG$ and $DAF$ are similar, therefore we have $AG \\cdot AD = DF \\cdot AB$. Moreover, from the similarity of the triangles $ACG$ and $DAE$ we deduce $AG \\cdot AD = DE \\cdot AC$. We obtain $DF \\cdot AB = DE \\cdot AC$, and thus $A_{ABD} = A_{ACD}$.\n\nFrom $AG \\perp BC$ and $DG \\perp BC$ we deduce $BC \\perp (ADG)$, whence $BC \\perp AD$. If $BI \\perp AD, I \\in AD$, then $AD \\perp (BIC)$.\nWe obtain $CI \\perp AD$, therefore $A_{ABD} = A_{ACD}$ iff $BI = CI$, thus the triangles $AIB$ and $AIC$ are congruent, whence we have $AB = AC$. Similarly, we deduce that the triangles $DIB$ and $DIC$ are congruent, therefore $DB = DC$.\n\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M \\ge 1$ be a real number. Determine all natural numbers $n$ for which there exist pairwise distinct natural numbers $a, b, c > M$, such that\n$$\nn = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b)\n$$\n(where $(x, y)$ denotes the greatest common divisor of natural numbers $x$ and $y$).",
    "options": [],
    "answer": "All n of the form n = 2^{2t}(2k+1) with t, k ∈ ℕ and k ≥ 1.",
    "solution": "Call good a number $n \\in \\mathbb{N}$ for which there exist $a, b, c \\in \\mathbb{N}_{\\ge 1}$ such that\n$$n = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b).$$\nWe will prove that, regardless of the value of $M$, the good numbers are those of the form $n = 2^{2t}(2k+1)$, where $t, k \\in \\mathbb{N}$, $k \\ge 1$.\nLet $k \\ge 1$ and $p, q, r > M$ three distinct primes such that $r > \\max\\{kp, kq\\}$. Considering $a = kp, b = kq$ and $c = r$, we have $(a, b) = k$ and $(b, c) = (c, a) = 1$, which leads to $(a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b) = 2k+1$, so $2k+1$ is good.\nLet $t \\in \\mathbb{N}$ and $n$ be a natural number for which there exist $a, b, c > M$, pairwise distinct, such that $n = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b)$. For every $t \\in \\mathbb{N}$, we have $2^t a, 2^t b, 2^t c > M$ and $2^{2t} \\cdot n = (2^t a, 2^t b) \\cdot (2^t b, 2^t c) + (2^t b, 2^t c) \\cdot (2^t c, 2^t a) + (2^t c, 2^t a) \\cdot (2^t a, 2^t b)$, so all the numbers of the form $n = 2^{2t}(2k+1)$ are good.\n\nFurther, we prove that the numbers of the form $n = 2^t$ or $n = 2^{2t+1}(2k+1)$, where $t, k \\in \\mathbb{N}$, $k \\ge 1$, are not good, so they don't satisfy the conditions of the problem either.\nFirst, we prove that any good number that is even is actually divisible by 4.\nConsider $n_0$ an even number. If $n_0 = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b)$, then $(a, b)$, $(b, c)$ and $(c, a)$ can't be simultaneously odd. For example, if $2 \\mid (a, b)$, then $2 \\mid (b, c) \\cdot (c, a)$, so $a, b$ and $c$ are even (the other cases are completely similar). It follows that $4 \\mid n_0$. Also, the number $\\frac{n_0}{4}$ is good too, because $\\frac{n_0}{4} = (a', b') \\cdot (b', c') + (b', c') \\cdot (c', a') + (c', a') \\cdot (a', b')$, where $a' = \\frac{a}{2}, b' = \\frac{b}{2}, c' = \\frac{c}{2}$.\nSuppose, for the sake of contradiction, that there exists $t \\ge 2$ such that $n = 2^t$ is good. Consequently, $2^{t-2}, 2^{t-4}, 2^{t-6}, \\dots$ etc. are also good, so either 1 or 2 should be also good. This is a contradiction, since if $a, b, c \\ge 1$, then $(a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b) \\ge 3$.\nSimilarly, if $n = 2^{2t+1}(2k+1)$, with $t, k \\in \\mathbb{N}$, $k \\ge 1$ is good, then $2^{2t-1}(2k+1)$, $2^{2t-3}(2k+1), \\dots, 2(2k+1)$ are also good, impossible, since $2(2k+1)$ is not divisible by 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39214,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ has $\\angle BAC = 90^\\circ$ and $\\angle ACB = 54^\\circ$. Take the bisector $BD$ ($D \\in AC$) of the angle $ABC$ and the point $E$ on the segment $BD$ so that $DE = DC$. Prove that $BE = 2 \\cdot AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A straightforward angle chasing gives $\\angle ABC = 90^\\circ - 54^\\circ = 36^\\circ$, then $\\angle ABD = \\angle CBD = \\frac{1}{2} \\angle ABC = 18^\\circ$. So $\\angle BDA = 90^\\circ - \\angle ABD = 72^\\circ$, $\\angle BDC = 180^\\circ - \\angle BDA = 108^\\circ$.\n\nThe isosceles triangle $CDE$ gives $\\angle DCE = \\angle DEC = \\frac{1}{2}(180^\\circ - \\angle CDE) = 36^\\circ$, whence $\\angle BCE = \\angle BCD - \\angle ECD = 18^\\circ = \\angle CBE$, hence the triangle $BCE$ is isosceles.\n\nDenote $F$ the reflection of $D$ into $A$. Then the points $D$, $A$, $F$ are collinear and $\\angle ABF = \\angle ABD = 18^\\circ$ (because $\\triangle BAF \\equiv \\triangle BAD$ – case S.A.S.), so $\\angle FBC = 54^\\circ = \\angle FCB$, hence the triangle $BFC$ is isosceles, with $FB = FC$.\n\nThis gives $\\triangle FEB \\equiv \\triangle FEC$ (S.S.S.), leading to $\\angle EFB = \\angle EFC = \\frac{1}{2} \\angle CFB = 36^\\circ$. So triangle $EFC$ is isosceles, whence $EF = EC = EB$ (*). Also, $\\angle FED = 180^\\circ - \\angle FDE - \\angle FED = 180^\\circ - 72^\\circ - 36^\\circ = 72^\\circ = \\angle FDE$, showing that $FE = FD = 2 \\cdot AD$. Now (*) yields $EB = 2 \\cdot AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angled triangle with $\\angle B > \\angle C$. Consider the points $D, E, J, K, S$ on its circumcircle $C(O)$, such that $A, E, J$ and $K$ are on the same side of the line $BC$, the diameter $DE$ and the line $BC$ are orthogonal, $S \\in \\widehat{EK}$ and $\\widehat{AE} = \\widehat{BJ} = \\widehat{CK} = \\frac{1}{4}\\widehat{CE}$. Let $F, M, P, Q$ be the intersection points of the lines $AC$ and $DE$, $BK$ and $AD$, $BK$ and $AC$, $CJ$ and $BF$, respectively. If $\\angle SMK = 30^\\circ$ and $\\angle AQP = 90^\\circ$, prove that the line $MS$ is tangent to the circumcircle of the triangle $AOF$.\nDana Heuberger",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\vec{AE} = x$ and $\\vec{BD} = y$. Obviously, we have $4x + y = 180^\\circ$.\nFrom $\\angle APM + \\angle MAP = 2x + \\frac{y}{2} = 90^\\circ$ we deduce that $AM \\perp MP$. Since $FO$ is the perpendicular bisector of $BC$, we have $FB = FC$, thus the triangle $FBC$ is isosceles, with the apex $F$. From $\\angle BCJ = \\angle CBK = \\frac{x}{2}$, we deduce that the triangles $BCQ$ and $CBP$ are congruent, therefore $PQ \\parallel BC$.\n\n![](attached_image_1.png)\n\nSince $FBC$ is an isosceles triangle, with the base $[BC]$, the line $FO$ is the perpendicular bisector of $PQ$. In the right angled triangle $AQP$, the perpendicular bisector of $PQ$ and the hypotenuse intersect at $F$, therefore $F$ is the midpoint of $AP$.\nDenote by $L$ the intersection point of the half-line ($BF$ with the circle $C(O)$).\nBecause $\\angle BFD = \\angle CFD$, we have $\\vec{BD} + \\vec{EL} = \\vec{CD} + \\vec{AE}$. From $\\vec{BD} = \\vec{CD} = \\angle A$ we obtain $\\vec{AE} = \\vec{EL} = x$, therefore $\\angle ABL = \\angle LBK = x$. Since $F$ is the midpoint of $AP$, the line $BF$ is at the same time median and angle bisector of the triangle $ABP$, thus $BA = BP$.\nFrom $\\angle ABP = \\angle APB = 2x$ and $AM \\perp MP$ we deduce that $AB = AP$, therefore the triangle $ABP$ is equilateral. Thus, we obtain $\\angle A = 60^\\circ$, $\\angle B = 75^\\circ$ and $\\angle C = 45^\\circ$.\nMoreover, $\\angle AOB = \\angle AFB = \\angle AMB = 90^\\circ$, so the points $A, F, O, M, B$ lie on the circle $\\omega$ of diameter $AB$. Therefore $\\angle BAM = \\angle SMK = 30^\\circ$, which means that the line $MS$ is tangent to the circle $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39216,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the set $A = [0, \\infty)$, of all nonnegative real numbers, we consider three functions $f, g, h : A \\to A$ and the binary operation $*: A \\times A \\to A$, defined by\n$$\nx * y = f(x) + g(y) + h(x) \\cdot |x - y|, \\quad \\text{for any } x, y \\ge 0.\n$$\nIf $(A, *)$ is a commutative monoid:\na) show that the function $h$ is continuous on $A$;\nb) determine the functions $f, g, h$.",
    "options": [],
    "answer": "There are exactly two possibilities. Case 1: h(x) = 0 for all x ≥ 0, f(x) = x, g(x) = x, and x*y = x + y. Case 2: h(x) = 1/2 for all x ≥ 0, f(x) = x/2, g(x) = x/2, and x*y = max(x, y).",
    "solution": "a) Let $e$ be the unit element of the monoid $(A, *)$. Then\n$$\nf(0) + g(e) + h(0) \\cdot e = 0 * e = 0 \\quad \\text{and} \\quad f(e) + g(0) + h(e) \\cdot e = e * 0 = 0,\n$$\nso that $f(e) = g(e) = f(0) = g(0) = h(e) \\cdot e = h(0) \\cdot e = 0$, whence $e = e * e = f(e) + g(e) = 0$.\nThen $0 * x = x$ and $x * 0 = x$, for any $x \\ge 0$, and we obtain\n$$\nf(0) + g(x) + h(0) \\cdot x = x \\quad \\text{and} \\quad f(x) + g(0) + h(x) \\cdot x = x,\n$$\nso that $f(x) = x(1 - h(x))$ and $g(x) = x(1 - h(0))$, for any $x \\ge 0$. Since $f(x)$, $g(x) \\ge 0$, it follows that $h(x) \\in [0, 1]$, $\\forall x \\ge 0$.\n\n$$\nx * y = x + y - x \\cdot h(x) - y \\cdot h(0) + h(x) \\cdot |x - y|, \\quad \\forall x, y \\ge 0.\n$$\n$$\nxh(x) - yh(y) = h(0)(x - y) + (h(x) - h(y)) \\cdot |x - y|, \\quad \\forall x, y \\ge 0.\n$$\nSince $h$ is bounded, it follows that $\\lim_{x \\to y} xh(x) = yh(y)$, for any $y \\ge 0$, so that the function $p: A \\to A$, $p(x) = xh(x)$, is continuous. But then $h$ is continuous on $(0, \\infty)$.\nAlso, for any $y > 0$ we have\n$$\n\\lim_{x \\to y} \\frac{p(x) - p(y)}{x - y} = h(0),\n$$\nso that there are $a = h(0)$ and $b \\ge 0$ such that $p(y) = ay + b = h(0)y + b, \\forall y > 0$. Then $b = \\lim_{y \\to 0} p(y) = p(0) = 0$. But then $yh(y) = p(y) = yh(0)$ for any $y > 0$ and it follows that $h(y) = h(0), \\forall y > 0$. The function $h$ is thus constant, hence continuous.\n\nb) Let $k = h(0)$. Then $h(x) = k$ and $f(x) = g(x) = x(1 - k)$, for any $x \\ge 0$, and $x * y = (x + y)(1 - k) + k|x - y|, \\forall x, y \\ge 0$. Then $(1 * 1) * 2 = 1 * (1 * 2) \\implies k(4k - 2) = 0$, so that $k \\in \\{0, \\frac{1}{2}\\}$.\nFor $k = 0$ we have that $f = g = \\text{id}_A$ and $x * y = x + y, \\forall x, y \\ge 0$.\nFor $k = \\frac{1}{2}$ we have that $f(x) = g(x) = \\frac{x}{2}, \\forall x \\ge 0$ and $x * y = \\frac{x+y}{2} + \\frac{|x-y|}{2} = \\max(x, y), \\forall x, y \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39217,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that there exists irrational numbers $a$, $b$, and $c$ so that the numbers $a+bc$, $b+ac$ and $c+ab$ are rational.\nb) Prove that, if the real numbers $a$, $b$, and $c$ are such that $a + b + c = 1$ and the numbers $a+bc$, $b+ac$, and $c+ab$ are non-nil rationals, then the numbers $a$, $b$, $c$ are rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) An example can be obtained if $a = b = c$, with irrational $a$ and rational $a^2 + a$. This happens if, for instance, $a = \\sqrt{2} - \\frac{1}{2}$.\n\nb) Since $b+ac$ and $c+ab$ are rational, so is $b+c+ac+ab = (b+c)(1+a) = (1-a)(1+a) = 1-a^2$, hence $a^2$ is rational.\nIn the same way, $b^2$ and $c^2$ are rational. Since $(a+bc)^2 = a^2 + b^2c^2 + 2abc$ is rational, so is $abc$.\nTherefore $a(a+bc) = a^2 + abc$ is rational and, since $a+bc$ is a non-nil rational, $a$ must be rational. Analogously, $b$ and $c$ are rational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a monotone increasing differentiable function, with a continuous derivative, such that $f(0) = 0$. Let $g : [0, 1] \\to \\mathbb{R}$ be the function defined by\n$$\ng(x) = f(x) + (x-1)f'(x), \\quad \\text{for any } x \\in [0, 1].\n$$\n\na) Show that\n$$\n\\int_{0}^{1} g(x) \\, dx = 0.\n$$",
    "options": [],
    "answer": "0",
    "solution": "a) Because $g(x) = f(x) + (x-1)f'(x) = ((x-1)f(x))'$, it follows that\n$$\n\\int_{0}^{1} g(x) \\, dx = (x-1)f(x)\\Big|_{0}^{1} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39219,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ has $\\angle ABC = 90^\\circ$ and $\\angle BCA = 30^\\circ$. Let $AD$ be the bisector of the angle $\\angle BAC$, $D \\in BC$, and $BE \\perp AC$, $E \\in AC$. Denote $M$ the intersection of the lines $AD$ and $BE$, and $P$ the midpoint of the segment $CM$. Prove that $AC = 4 \\cdot DP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The $30^\\circ$ angle theorem yields $AC = 2AB$. Since $\\angle CAD = \\angle ACD = 30^\\circ$, the triangle $ADC$ is isosceles, with $DA = DC$.\n\nFrom $\\angle ADB = \\angle MBD = 60^\\circ$ follows that the triangle $MBD$ is equilateral.\n\nThis shows that $MB = BD = \\frac{1}{2}AD$, hence $M$ is the midpoint of the segment $AD$.\n\nDenote $S$ the reflection of $D$ into $B$. Then the triangle $ADS$ is equilateral. This yields $DS = AD = CD$, hence $DP$ is a midline of the triangle $CSM$. From $AB = SM$ (medians in the equilateral triangle $ADS$) follows that $DP = \\frac{1}{2}MS = \\frac{1}{2}AB = \\frac{1}{4}AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39220,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a natural number $n \\ge 2$ is *nice*, if in its prime factor decomposition, one of the primes has exponent 3.\n\na) Determine the smallest natural number $N$, such that no matter how we choose $N$ consecutive natural numbers, at least one of them is nice.\n\nb) Find the smallest 15 consecutive natural numbers $a_1, a_2, \\dots, a_{15}$ that are not nice, but for which the sum $5a_1 + 5a_2 + \\dots + 5a_{15}$ is nice.",
    "options": [],
    "answer": "a) 16; b) 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87",
    "solution": "a) By dividing 16 consecutive natural numbers by 16, the remainders 0, 1, 2, ..., 15 are obtained (not necessarily starting with 0). Therefore, among any 16 consecutive natural numbers there is a nice number of the form $16k + 8 = 2^3(2k+1)$. Since among the consecutive natural numbers 9, 10, 11, ..., 23 there is no nice number, the answer is $N = 16$.\n\nb) Consider $a_2 = a_1 + 1$, $a_3 = a_1 + 2$, ..., $a_{15} = a_1 + 14$ and denote by $S = 5a_1 + 5a_2 + \\dots + 5a_{15}$, that is $S = 3 \\cdot 5^2 \\cdot (a_1 + 7)$. The numbers $a_1, a_2, \\dots, a_{15}$ are not nice, thus none of them has the remainder 8 when divided by 16. We deduce that $a_1 = 16k+9$, $k \\in \\mathbb{N}$, and $S = 2^4 \\cdot 3 \\cdot 5^2 \\cdot (k+1)$. The smallest value of $k$ for which $S$ is a nice number is $k = 4$, and the numbers 73, 74, ..., 87 fulfill the statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive even integer $a$ which is not a power of $2$, prove that at least one of the numbers $2^{2^n} + 1$ and $a^{2^n} + 1$ is composite for infinitely many non-negative integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $f_n = 2^{2^n} + 1$ and $a_n = a^{2^n} + 1$. Suppose now, if possible, that $f_n$ and $a_n$ are both prime for all but finitely many non-negative integers $n$, say, for all $n > N$. Clearly, we may and will assume that $f_n > a$ for all $n > N$.\n\nFix any $n > N$ and consider the (multiplicative) order of $a$ modulo the prime $f_n$. By Fermat's Little Theorem, this order divides $f_n - 1 = 2^{2^n}$, so it is a power of $2$, say, $2^k$.\n\nThus, the prime $f_n$ divides $a^{2^k} - 1 = (a^{2^{k-1}} - 1)a_{k-1}$, so it divides (at least) one of the two factors. By minimality of $k$, the first factor is not divisible by $f_n$, so $f_n$ divides $a_{k-1}$. Notice that $a_{k-1} \\ne f_n$, since $a$ is not a power of $2$. Hence $a_{k-1}$ is composite, forcing $k-1 \\le N$.\n\nConsequently, the product $a_0 a_1 \\dots a_N$ is divisible by $f_n$ for all $n > N$. This contradiction concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the numbers $r, s \\in [1, \\infty)$ with the property that for every positive integers $a, b$, with $a$ dividing $b$, it results that $[ar]$ divides $[bs]$.\na) Prove that $\\frac{s}{r}$ is a positive integer.\nb) Show that $r$ and $s$ are positive integers.\n*Remark.* By $[x]$ we denote the floor of the real number $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We suppose that $\\frac{s}{r} \\notin \\mathbb{N}$. Then, there exists $k \\in \\mathbb{N}$ such that $k < \\frac{s}{r} < k+1 \\iff kr < s < (k+1)r$. Choosing $b = a \\in \\mathbb{N}^*$, arbitrary, we obtain $[ar] \\mid [as]$ and thus $[ar] \\mid [as] - k[ar]$. (1)\nFrom $s > kr$, we obtain that there exists $u > 0$ such that $us > ukr + 2$ and thus, for every $a > u$ we get $as > akr + 2 \\implies [as] \\ge [akr] + 2 > akr + 1 > k[ar]$, so $[as] > k[ar]$.\nFrom (1) we obtain $[as] - k[ar] \\ge [ar] \\iff [as] \\ge (k+1)[ar]$, so,\n$$\nas > (k+1)(ar - 1) \\iff k+1 > a((k+1)r - s),\n$$\nfor every $a > u$.\nThus, $a < \\frac{k+1}{(k+1)r-s}$, for every $a > u$, which is a contradiction, so the assumption is false.\n\nb) Let us show that $s$ is a positive integer.\nWe will show that for every $a \\in \\mathbb{N}$ such that $ar \\ge 2$, we get $as \\in \\mathbb{N}$.\nIf $as \\notin \\mathbb{N}$, then there exists $n \\in \\mathbb{N}^*$ such that $\\frac{1}{n+1} \\le \\{as\\} < \\frac{1}{n}$, so $1 \\le (n+1)\\{as\\} < \\frac{n+1}{n} \\le 2$, thus $[(n+1)\\{as\\}] = 1$.\nWe obtain\n$$\n[(n+1)as] = [(n+1)[as] + (n+1)\\{as\\}] = (n+1)[as] + [(n+1)\\{as\\}] = (n+1)[as] + 1.\n$$\nSince $[ar] \\mid [as]$ and $[ar] \\mid [(n+1)as]$, we obtain that $[ar] \\mid 1 \\implies [ar] = 1$, which is a contradiction.\nThus $as \\in \\mathbb{N}$, for every $a \\in \\mathbb{N}$ with $ar \\ge 2$, from which we get $(a+1)s \\in \\mathbb{N}$, so $(a+1)s - as = s \\in \\mathbb{N}$.\n\nLet us prove that $r$ is a positive integer.\nLet $p$ be an arbitrary prime number with $p[r] > s$ and $m = [p\\{r\\}]$. Since $p\\{r\\} < p$, we get $m < p$.\nIf $m \\ne 0$, then $(m, p) = 1$. Since\n$$\n[pr] \\mid ps \\implies [p([r] + \\{r\\})] \\mid ps \\implies p[r] + m \\mid ps.\n$$\nSince $(p[r] + m, p) = 1 \\implies p[r] + m \\mid s$, we obtain a contradiction as $p[r] > s$.\nThus, $m = 0 \\implies p\\{r\\} < 1 \\implies \\{r\\} < \\frac{1}{p}$, for every prime number $p$, with $p > \\frac{s}{[r]} \\implies \\{r\\} = 0$ and thus, $r \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a tetrahedron $ABCD$ and the points $M, N$ on the edges $AC$ and $BD$, respectively. Prove that for any point $P$ of the segment $MN$, $P \\neq M$, $P \\neq N$, there exists a unique pair of points $(X, Y)$, with $X$ and $Y$ on the edges $AB$ and $CD$, respectively, such that the points $X, P$ and $Y$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $P \\neq M$ and $P \\neq N$, if $X, Y, P$ are collinear, we have $X \\notin \\{A, B\\}$ and $Y \\notin \\{C, D\\}$. Indeed, if $X = A$, then $XY \\subset (ACD)$ and thus $P \\in (ACD)$, false. The other situations are analogous.\n\n2023 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND\n\n**Existence:** Since $P \\in (MN)$, the point $P$ lies in the interior of the triangle $ANC$. Denote $\\{Q\\} = AP \\cap CN$.\nFrom $(CN) \\subset \\text{Int}(BCD)$, it follows that $Q \\in \\text{Int}(BCD)$, thus the line $BQ$ intersects the open segment $CD$. Denote $\\{Y\\} = BQ \\cap (CD)$. Because $P \\in (AQ) \\subset (ABY)$, we deduce that $P$ lies in the interior of the triangle $ABY$, thus the line $YP$ intersects the open segment $AB$. Denote $\\{X\\} = PY \\cap (AB)$. The pair $(X, Y)$ satisfies the statement.\n\n**Uniqueness:** Assume that a pair of points $(X', Y')$ exists, with $X' \\in (AB)$, $Y' \\in (CD)$, $(X, Y) \\neq (X', Y')$, such that the points $X'$, $P$ and $Y'$ are collinear. We consider $Y' \\neq Y$ (the situation $X' \\neq X$ is analogous). If $X' = X$, then $Y' \\in XP \\cap CD$, therefore $Y' = Y$, false. Consequently $X' \\neq X$, so the distinct straight lines $XY$ and $X'Y'$ intersect at $P$. If $\\alpha = (XY, X'Y')$, we obtain $XX' = AB \\subset \\alpha$ and $YY' = CD \\subset \\alpha$, thus the points $A, B, C$ and $D$ are coplanar, false. Consequently, the pair $(X, Y)$ is unique.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P \\in \\mathbb{C}[X]$ be a polynomial with degree at least $2$ and $n \\ge 2$ be an integer. Prove that, if $A \\in \\mathcal{M}_n(\\mathbb{C})$ commutes with every matrix $B \\in \\mathcal{M}_n(\\mathbb{C})$ having the property $P(B) = O_n$, then there exists $a \\in \\mathbb{C}$ such that $A = aI_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a positive integer $n$. Consider an $n$-point set $S$ in the plane. An *eligible* set is a non-empty set of the form $S \\cap D$, where $D$ is a closed disc in the plane. In terms of $n$, determine the smallest possible number of eligible subsets $S$ may contain.\nCristian Săvescu",
    "options": [],
    "answer": "n(n+1)/2",
    "solution": "The required minimum is $\\frac{1}{2}n(n+1)$.\n\nWe first show that an $n$-point set $S$ in the plane contains at least $\\frac{1}{2}n(n+1)$ eligible subsets. To this end, consider a line $\\ell$ perpendicular to:\n(1) No line through at least two points in $S$; and\n(2) No tangent of a circle $\\gamma$ through at least three points in $S$ at a point in $S \\cap \\gamma$.\nThere are finitely many directions to avoid, so the choice of $\\ell$ is possible.\nBy (1), $S$ projects injectively to $\\ell$ to provide $n$ pairwise distinct points $x'_1 < \\cdots < x'_n$. Let $x_k$ be the (unique) point of $S$ whose orthogonal projection on $\\ell$ is $x'_k$. By (2), any circle through an $x_k$ and tangent to $x_kx'_k$ passes through at most one other $x_j$.\n\nFix an index $k$ and consider a circle $\\omega$ through $x_k$ and tangent to $x_kx'_k$, containing all points $x_j, j < k$, inside; clearly, the $x_j, j > k$ all lie outside $\\omega$. By the preceding, shrinking $\\omega$ homothetically from $x_k$, the closed disc it bounds loses successively at most one $x_j, j < k$. While shrinking, the disc first loses some $x_{j_1}, j_1 < k$, then some $x_{j_2}, j_2 \\neq j_1, j_2 < k$, and so on and so forth, to provide an index permutation $j_1, j_2, \\dots, j_k = k$ of $1, 2, \\dots, k$ such that $\\{x_{j_i}, x_{j_{i+1}}, \\dots, x_{j_k}\\}, i = 1, 2, \\dots, k$, are all eligible. This accounts for $k$ pairwise distinct eligible sets 'ending up' with $x_k$; that is, $x'_{j_i} < x'_{j_k} = x'_k$ for all $i < k$. In particular, for an index $k' \\neq k$, the corresponding eligible sets are all different from each of the above.\n\nConsequently, $S$ contains at least $1 + 2 + \\dots + n = \\frac{1}{2}n(n+1)$ eligible sets, as stated.\n\nWe now exhibit an $n$-point set $S$ in the plane with exactly $\\frac{1}{2}n(n+1)$ eligible subsets. Let $S$ consist of $n$ collinear points, ordered $x_1 < \\dots < x_n$ along the line in question. The $n$ single-point segments $\\{x_i\\}, i = 1, \\dots, n$, and the $\\frac{1}{2}n(n-1)$ proper segments $\\{x_i, x_{i+1}, \\dots, x_j\\}, 1 \\le i < j \\le n$, are all eligible subsets of $S$. Since the intersection of a line and a disc is either empty or a (possibly degenerate) segment, $S$ contains no other eligible subsets. Consequently, there are exactly $n + \\frac{1}{2}n(n-1) = \\frac{1}{2}n(n+1)$ eligible subsets in $S$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39226,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$ define $a_n = n + 1$, $b_n = 2 \\cdot n + 3$, $c_n = 3 \\cdot n + 5$, $d_n = 4 \\cdot n + 7$. Find all positive integers $n$ which have at most two digits and $a_n, b_n, c_n$, and $d_n$ are prime.",
    "options": [],
    "answer": "88",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$, $Q(x)$, $R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x)) = R(S(x))$. Prove that, if the degree of $P(x)$ is divisible by the degree of $R(x)$, then $P(x) = R(T(x))$ for some polynomial $T(x)$ with real coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q = \\deg Q$ divides $\\deg S = s$. We will show that $S(x) = T(Q(x))$ for some polynomial $T$. Then $P(Q(x)) = R(S(x)) = R(T(Q(x)))$, so the polynomial $P(t) - R(T(t))$ vanishes upon substitution $t = S(x)$; it therefore vanishes identically, as desired.\n\nChoose the polynomials $T(x)$ and $M(x)$ such that\n$$\nS(x) = T(Q(x)) + M(x), \\qquad (*)\n$$\nwhere $\\deg M$ is minimised; if $M = 0$, then we get the desired result. For the sake of contradiction, suppose $M \\ne 0$. Then $q \\nmid m = \\deg M$; otherwise, $M(x) = \\beta Q(x)^{m/q} + M_1(x)$, where $\\beta$ is some number and $\\deg M_1 < \\deg M$, contradicting the choice of $M$. In particular, $0 < m < s$ and hence $\\deg T(Q(x)) = s$.\n\nSubstitute now (*) into $R(S(x)) - P(Q(x)) = 0$; let $\\alpha$ be the leading coefficient of $R(x)$ and let $r = \\deg R(x)$. Expand the brackets to get a sum of powers of $Q(x)$ and other terms including powers of $M(x)$ as well. Amongst the latter, the unique term of highest degree is $\\alpha r M(x) T(Q(x))^{r-1}$. So, for some polynomial $N(x)$,\n$$\nN(Q(x)) = \\alpha r M(x) T(Q(x))^{r-1} + \\text{a polynomial of lower degree.}\n$$\nThis is impossible, since $q$ divides the degree of the left-hand member, but not that of the right-hand member.\n\nAlternative solution:\n\nAll polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\\deg f(x)$.\n\nOf all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_0(x), R_0(x)$, so that $P_0(Q(x)) = R_0(S(x))$ has a minimal (positive) degree. We will show that $\\deg R_0(x) = 1$, say, $R_0(x) = \\alpha x + \\beta$ for some real numbers $\\alpha \\neq 0$ and $\\beta$, so $P_0(Q(x)) = \\alpha S(x) + \\beta$. Hence $S(x) = T(Q(x))$ for some polynomial $T(x)$.\n\nNow, if $P(x)$ and $R(x)$ are polynomials satisfying $P(Q(x)) = R(S(x))$, then $P(Q(x)) = R(T(Q(x)))$. Since $Q(x)$ is not constant, it takes infinitely many values, so $P(x)$ and $R(T(x))$ agree at infinitely many points, implying that $P(x) = R(T(x))$, as required.\n\nIt is therefore sufficient to solve the problem in the particular case where $F(x) = P(Q(x)) = R(S(x))$ has a minimal degree. Let $d = \\gcd(\\deg Q(x), \\deg S(x))$ to write $\\deg Q(x) = ad$ and $\\deg S(x) = bd$, where $\\gcd(a, b) = 1$. Then $\\deg P(x) = bc$, $\\deg R(x) = ac$ and $\\deg F(x) = abcd$ for some positive integer $c$. We will show that minimality of $\\deg F(x)$ forces $c = 1$, so $\\deg P(x) = b$, $\\deg R(x) = a$ and $\\deg F(x) = abd$. The conditions $a = \\deg R(x)$ divides $\\deg P(x) = b$ and $\\gcd(a, b) = 1$ then force $a = 1$, as stated above.\n\nWe are now in a position to prove that $c = 1$. Suppose, if possible, that $c > 1$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that\n$$\n\\deg (P(x) - U(x)^c) < (c-1)b \\quad \\text{and} \\quad \\deg (R(x) - V(x)^c) < (c-1)a.\n$$\nThen\n$$\n\\deg (F(x) - U(Q(x))^c) = \\deg (P(Q(x)) - U(Q(x))^c) < (c-1)abd,\n$$\n$$\n\\deg (F(x) - V(S(x))^c) = \\deg (R(S(x)) - V(S(x))^c) < (c-1)abd,\n$$\n$$\n\\text{so } \\deg (U(Q(x))^c - V(S(x))^c) = \\deg \\left( (F(x) - V(S(x))^c) - (F(x) - U(Q(x))^c) \\right) < (c-1)abd.\n$$\nOn the other hand,\n$$\nU(Q(x))^c - V(S(x))^c = (U(Q(x)) - V(S(x))) (U(Q(x))^{c-1} + \\dots + V(S(x))^{c-1}).\n$$\nBy the preceding, the degree of the left-hand member is (strictly) less than $(c-1)abd$ which is precisely the degree of the second factor in the right-hand member. This forces $U(Q(x)) = V(S(x))$, so $U(Q(x)) = V(S(x))$ has degree $abd < abcd = \\deg F(x)$ — a contradiction. Consequently, $c = 1$. This completes the argument and concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the trapezoid $ABCD$ with the smaller base $AB$, squares $ADEF$ and $BCGH$ are constructed externally to the trapezoid. Prove that the perpendicular bisector of $AB$ passes through the midpoint of $FH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the midpoint of $AB$ and $M$, $Q$ be points on the base $CD$, such that $DM = QC = AI$.\n\nConstruct squares $IMNP$ and $IQRS$, externally to the triangle $IMQ$.\n\nSince $\\angle PIS + \\angle PIM + \\angle MIQ + \\angle QIS = 360^\\circ$ and $IPUS$ is a parallelogram, we infer that $\\angle PIS + \\angle MIQ = 180^\\circ = \\angle IPU + \\angle PIS$, so $\\angle IPU = \\angle MIQ$. As $IP = IM$ and $PU = IS = IQ$, triangles $IPU$ and $MIQ$ are congruent (S.A.S.), so $\\angle PIU = \\angle IMQ$.\nTriangle $IMV$ is right-angled at $V$, so $\\angle MIV = 90^\\circ - \\angle IMV = 90^\\circ - \\angle PIU$. It follows that $\\angle MIV + \\angle PIU + \\angle MIP = 180^\\circ$, therefore $U$, $I$, $V$ are collinear.\nHence, $UI$ is the perpendicular bisector of segment $AB$. Since $IPUS$ is a parallelogram, the line $UI$ passes through the midpoint $J$ of the diagonal $PS$.\nAs $\\angle DAF = \\angle MIP = 90^\\circ$ and $AD \\parallel IM$, it follows that $AF \\parallel IP$. But $AF = IP$, so $AIPF$ is a parallelogram, which means that $FP \\parallel AI$ and $FP = AI$. Analogously, we prove that $SH \\parallel IB$ and $SH = IB$. Since $I$ is the midpoint of $AB$, we infer that $FP \\parallel SH$ and $FP = SH$, so $FPHS$ is parallelogram. Therefore, the diagonals $FH$ and $PS$ have the same midpoint $J$, so $UI$ (the perpendicular bisector of $AB$) passes through the midpoint of $FH$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of non-negative real numbers satisfying\n$$\na_{n+1}^2 + a_n a_{n+2} \\le a_n + a_{n+2},\n$$\nfor all $n \\ge 1$. Prove that the sequence $(a_n)_{n \\ge 1}$ is bounded.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To prove boundedness, it is sufficient to show that $a_n \\le 1$ for all $n \\ge 3$. Rewrite the condition in the statement in the equivalent form\n$$\na_{n+1}^2 - 1 \\le (1-a_n)(a_{n+2}-1) = (a_n-1)(1-a_{n+2}) \\quad \\text{for all } n \\ge 1.\n$$\nWe first show that $\\min(a_n, a_{n+1}) \\le 1$ for all $n \\ge 2$. Suppose, if possible, that $a_n > 1$ and $a_{n+1} > 1$ for some $n \\ge 2$. Then\n$$\n\\begin{aligned}\na_n - 1 &< a_n^2 - 1 \\le (1 - a_{n-1})(a_{n+1} - 1) \\le a_{n+1} - 1 < a_{n+1}^2 - 1 \\le \\\\\n&\\le (a_n - 1)(1 - a_{n+2}) \\le a_n - 1,\n\\end{aligned}\n$$\nwhich is a contradiction.\nTo reach a final contradiction, suppose $a_n > 1$ for some $n \\ge 3$. By the preceding, $a_{n-1} \\le 1$ and $a_{n+1} \\le 1$, so $0 < a_n^2 - 1 \\le (1 - a_{n-1})(a_{n+1} - 1) \\le 0$, which is the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39230,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ has circumcircle $\\Gamma$ with centre $O$, $AO \\cap BC = \\{X\\}$, $AO \\cap \\Gamma = \\{Y\\}$, and $\\frac{AX}{XY} = k$, $k > 1$. The tangent to $\\Gamma$ at $Y$ meets $AB$ and $AC$ in $M$, respectively $N$. Find $k$ so that the quadrilateral $OMNC$ is cyclic.\n\nMihaela Berindeanu",
    "options": [],
    "answer": "k = 2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. On the lines $BC$, $CA$, respectively $AB$ the points $M$ and $N$, $P$ and $Q$, respectively $R$ and $S$ are taken, outside of the triangle, so that $MB = BS = AC$, $CP = CN = AB$ and $AR = AQ = BC$. Prove that the points $M, N, P, Q, R$ and $S$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the pairs $(m, n)$ of positive integers with the property\n$$(3m - 2)(5m - 1) = 6^n.$$",
    "options": [],
    "answer": "(2, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon. The diagonals $AC$ and $BD$ cross at $P$, the diagonals $AE$ and $DF$ cross at $Q$, and the line $PQ$ crosses the sides $BC$ and $EF$ at $X$ and $Y$, respectively. Prove that the length of the segment $XY$ does not exceed the sum of the lengths of one of the diagonals through $P$ and one of the diagonals through $Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem at hand is a special case of the following situation: Let $U$ be a point on the diagonal $AD$ and let $V = UP \\cap BC$ and $W = UQ \\cap EF$. Then $UV \\le \\max(AC, BD)$ and, similarly, $UW \\le \\max(AE, DF)$, so $VW \\le UV + UW \\le \\max(AC, BD) + \\max(AE, DF)$. In particular, if $U = AD \\cap PQ$, then $V = X$ and $W = Y$, and the conclusion follows.\n\nWe now prove that $UV \\le \\max(AC, BD)$ in three different ways. The first proof is elementary; the other two both require some acquaintance with analysis and are intimately related.\n\n**1st Proof.** Let $\\alpha = \\angle UPA = \\angle VPC$ and let $\\beta = \\angle DPU = \\angle BPV$. In what follows, $[KLM]$ stands for the area of a generic triangle $KLM$. Write\n$$\n\\begin{aligned}\n\\frac{1}{2} \\cdot PD \\cdot PA \\cdot \\sin(\\beta + \\alpha) &= [PDA] = [PDU] + [PUA] \\\\\n&= \\frac{1}{2} \\cdot PD \\cdot PU \\cdot \\sin \\beta + \\frac{1}{2} \\cdot PU \\cdot PA \\cdot \\sin \\alpha,\n\\end{aligned}\n$$\nto get $PU = \\frac{PA \\cdot PD \\cdot \\sin(\\alpha+\\beta)}{PA \\cdot \\sin \\alpha + PD \\cdot \\sin \\beta} \\le \\frac{PA \\cdot PD \\cdot (\\sin \\alpha + \\sin \\beta)}{PA \\cdot \\sin \\alpha + PD \\cdot \\sin \\beta} \\le \\frac{PA \\cdot \\sin \\beta + PD \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta}$, where the last inequality holds since equivalent to $(PA - PD)^2 \\cdot \\sin \\alpha \\cdot \\sin \\beta \\ge 0$. Similarly,\n$$\nPV \\le \\frac{PB \\cdot \\sin \\alpha + PC \\cdot \\sin \\beta}{\\sin \\alpha + \\sin \\beta},\n$$\nso\n$$\n\\begin{aligned}\nUV &= PU + PV \\le \\frac{PA \\cdot \\sin \\beta + PD \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta} + \\frac{PB \\cdot \\sin \\alpha + PC \\cdot \\sin \\beta}{\\sin \\alpha + \\sin \\beta} \\\\\n&= \\frac{AC \\cdot \\sin \\beta + BD \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta} \\le \\max(AC, BD),\n\\end{aligned}\n$$\nas desired. This completes the proof.\n\n**2nd Proof.** Let $\\mathcal{L}$ be the pencil of lines through $P$ crossing the sides $BC$ and $DA$ of the convex quadrangle $ABCD$. As before, we prove that the length of each segment $\\ell \\cap ABCD$, $\\ell$ in $\\mathcal{L}$, does not exceed $\\max(AC, BD)$.\nClearly, no intersection segment has a length greater than the diameter of $ABCD$, so $\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap ABCD|$ is finite; here and hereafter, $|s|$ denotes the length of the line segment $s$.\nSuppose, if possible, that $\\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap ABCD| > \\max(AC, BD)$ and consider a line $\\ell_0$ in $\\mathcal{L}$ such that\n$$\n|\\ell_0 \\cap ABCD| > \\frac{1}{2} \\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap ABCD| + \\frac{1}{2} \\max(AC, BD) > \\frac{1}{2}(AC + BD).\n$$\nRecall that the length of the internal bisectrix of an angle of a (possibly degenerate) triangle does not exceed the arithmetic mean of the lengths of the sides forming that angle, to infer that $\\ell_0$ does not internally bisect $\\angle BPC = \\angle DPA$; say, $\\angle(\\ell_0, AC) < \\angle(\\ell_0, BD)$.\nFinally, reflect the line $AC$ in $\\ell_0$ to obtain a line $\\ell_1$ in $\\mathcal{L}$ such that\n$$\n\\begin{aligned}\n\\frac{1}{2}|\\ell_1 \\cap ABCD| + \\frac{1}{2} \\max(AC, BD) &\\ge \\frac{1}{2}(|\\ell_1 \\cap ABCD| + AC) \\ge |\\ell_0 \\cap ABCD| \\\\\n&> \\frac{1}{2} \\sup_{\\ell \\in \\mathcal{L}} |\\ell \\cap ABCD| + \\frac{1}{2} \\max(AC, BD),\n\\end{aligned}\n$$\nand thereby reach a contradiction; the inequality in the middle comes from the above mentioned fact about the length of an internal bisectrix in a triangle. This ends the proof.\n\n**3rd Proof.** Notice that $UV$ depends continuously on $U$. Since the closed segment $AD$ is compact, $UV$ achieves a maximum at some position $U_0V_0$. We will show that $U_0V_0 \\le \\max(AC, BD)$, so $UV \\le \\max(AC, BD)$ for all positions.\nSuppose, if possible, that $U_0V_0 > \\max(AC, BD) \\ge \\frac{1}{2}(AC + BD)$.\nAs before, it then follows that $U_0V_0$ does not internally bisect $\\angle BPC = \\angle DPA$;\nsay, $\\angle(U_0V_0, AC) < \\angle(U_0V_0, BD)$.\nReflexion of $AC$ in $U_0V_0$ then provides a segment $U_1V_1$ through $P$, where $U_1$ lies on $AD$ and $V_1$ lies on $BC$, such that\n$$\n\\begin{aligned}\nU_0V_0 &\\le \\frac{1}{2}(U_1V_1 + AC) \\le \\frac{1}{2} \\cdot U_1V_1 + \\frac{1}{2} \\max(AC, BD) \\le \\\\\n&\\le \\frac{1}{2} \\cdot U_0V_0 + \\frac{1}{2} \\max(AC, BD) < \\frac{1}{2} \\cdot U_0V_0 + \\frac{1}{2} \\cdot U_0V_0 = U_0V_0,\n\\end{aligned}\n$$\nwhich is a contradiction. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be a connected graph on $r + g + b + 1$ vertices. The edges of $\\Gamma$ bear three colours: red, green, and blue. It turns out that $\\Gamma$ has a spanning tree with exactly $r$ red edges, a spanning tree with exactly $g$ green edges, and a spanning tree with exactly $b$ blue edges. Prove that $\\Gamma$ has a spanning tree with exactly $r$ red edges, exactly $g$ green edges, and exactly $b$ blue edges.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Induct on $n = r + g + b$. The base case, $n = 1$, is clear.\n\nLet now $n > 1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_r, T_g$, and $T_b$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\n\n**Case 1:** There exists a partition $V = A \\cup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.\n\nSince $\\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.\n\nAssume that $T$, one of the three trees, does not contain $e$. Then the graph $T \\cup \\{e\\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e'$ connecting $A$ and $B$; the edge $e'$ is also blue. Replace $e'$ by $e$ in $T$ to get another tree $T'$ with the same number of edges of each colour as in $T$, but containing $e$.\n\nPerforming such an operation to all three trees, we arrive at the situation where the three trees $T'_r, T'_g$, and $T'_b$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\\Gamma^*$, and set $r^* = r, g^* = g$, and $b^* = b - 1$. The new graph satisfies the conditions in the statement for those new values — indeed, under the shrinking, each of the trees $T'_r, T'_g$, and $T'_b$ loses a blue edge. So $\\Gamma^*$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b - 1$ blue edges. Finally, pass back to $\\Gamma$ by restoring $e$, to obtain the desired spanning tree in $\\Gamma$.\n\n**Case 2:** There is no such a partition.\n\nConsider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \\cup G \\cup B$ has the smallest number $k$ of components. If $k = 1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).\n\nAssume now that $k \\ge 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \\setminus V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.\n\nLet $e'$ be a red edge in $C$ and set $R' = R \\setminus \\{e'\\} \\cup \\{e\\}$. Then $(R', G, B)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.\nFor a spanning tree $T$ in $\\Gamma$, denote by $r(T)$, $g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.\n\nAssume that $C$ is some collection of spanning trees in $\\Gamma$. Write\n$$\nr(C) = \\min_{T \\in C} r(T), \\quad g(C) = \\min_{T \\in C} g(T), \\quad b(C) = \\min_{T \\in C} b(T), \\\\\nR(C) = \\max_{T \\in C} r(T), \\quad G(C) = \\max_{T \\in C} g(T), \\quad B(C) = \\max_{T \\in C} b(T).\n$$\nSay that a collection $C$ is good if $r \\in [r(C), R(C)]$, $g \\in [g(C), G(C)]$, and $b \\in [b(C), B(C)]$. By the problem conditions, the collection of all spanning trees in $\\Gamma$ is good.\n\nFor a good collection $C$, say that an edge $e$ of $\\Gamma$ is *suspicious* if $e$ belongs to some tree in $C$ but not to all trees in $C$. Choose now a good collection $C$ minimizing the number of suspicious edges. If $C$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(C) < r$ and $G(C) > g$.\n\nWe now distinguish two cases.\n\n**Case 1:** $B(C) = b$.\n\nLet $T^0$ be a tree in $C$ with $g(T^0) = g(C) \\le g$. Since $G(C) > g$, there exists a green edge $e$ contained in some tree in $C$ but not in $T^0$; clearly, $e$ is suspicious. Fix one such green edge $e$.\n\nNow, for every $T$ in $C$, define a spanning tree $T_1$ of $\\Gamma$ as follows. If $T$ does not contain $e$, then $T_1 = T$; in particular, $(T^0)_1 = T^0$. Otherwise, the graph $T \\setminus \\{e\\}$ falls into two components. The tree $T^0$ contains some edge $e'$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_1 = T \\setminus \\{e\\} \\cup \\{e'\\}$.\n\nLet $C_1 = \\{T_1 : T \\in C\\}$. All edges suspicious for $C_1$ are also suspicious for $C$, but no tree in $C_1$ contains $e$. So the number of suspicious edges for $C_1$ is strictly smaller than that for $C$.\n\nWe now show that $C_1$ is good, reaching thereby a contradiction with the choice of $C$. For every $T$ in $C$, the tree $T_1$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g(C_1) \\le g(C) \\le g$, $G(C_1) \\ge G(C) - 1 \\ge g$, $R(C_1) \\ge R(C) \\ge r$, $r(C_1) \\le r(C) + 1 \\le r$, and $B(C_1) \\ge B(C) \\ge b$. Finally, we get $b(T^0) \\le B(C) = b$; since $C_1$ contains $T^0$, it follows that $b(C_1) \\le b(T^0) \\le b$, which concludes the proof.\n\n**Case 2:** $B(C) > b$.\n\nConsider a tree $T^0$ in $C$ satisfying $r(T^0) = R(C) \\ge r$. Since $r(C) < r$, the tree $T^0$ contains a suspicious red edge. Fix one such edge $e$.\n\nNow, for every $T$ in $C$, define a spanning tree $T_2$ of $\\Gamma$ as follows. If $T$ contains $e$, then $T_2 = T$; in particular, $(T^0)_2 = T^0$. Otherwise, the graph $T \\cup \\{e\\}$ contains a cycle $C$ through $e$. This cycle contains an edge $e'$ absent from $T^0$ (otherwise $T^0$ would contain the cycle $C$), so $e'$ is suspicious. Choose one such edge and define $T_2 = T \\setminus \\{e'\\} \\cup \\{e\\}$.\n\nLet $C_2 = \\{T_2: T \\in C\\}$. All edges suspicious for $C_2$ are also suspicious for $C$, but all trees in $C_2$ contain $e$. So the number of suspicious edges for $C_2$ is strictly smaller than that for $C$.\n\nWe now show that $C_2$ is good, reaching again a contradiction. For every $T$ in $C$, the tree $T_2$ either coincides with $T$ or is obtained from it by removing some edge and adding a red edge. This shows that $r(C_2) \\le r(C) + 1 \\le r$, $R(C_2) \\ge R(C) \\ge r$, $G(C_2) \\ge G(C) - 1 \\ge g$, $g(C_2) \\le g(C) \\le g$, $b(C_2) \\le b(C) \\le b$ and $B(C_2) \\ge B(C) - 1 \\ge b$. This concludes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u, v \\in \\mathbb{C}$ so that $|1 - u| = |1 - v|$ and $|1 - u^2| = |1 - v^2|$. Prove that $u = v$ or $u = \\bar{v}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the triples $(a, b, c)$ of positive integers so that $1 + 2^a + 3^b = 6^c$.",
    "options": [],
    "answer": "(a,b,c) = (1,1,1), (5,1,2), (3,3,2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 2$ so there exists a finite field $\\mathbb{K}$ with $n$ elements, with the property: for every $x \\in \\mathbb{K}$ there exists $y \\in \\mathbb{K}$ such that $x^2 = y^3$.",
    "options": [],
    "answer": "All prime powers n with 3 not dividing n − 1 (equivalently, n not congruent to 1 modulo 3).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39238,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that $x^2 + x + 1 > 0$ and $x^4 + x^3 + x^2 + x + 1 > 0$, for every real number $x$.\n\nb) Find all real triples $(a, b, c)$ so that $a^8 + b^8 + c^8 = a^3 + b^3 + c^3$ and $a^5 + b^5 + c^5 = 3$.",
    "options": [],
    "answer": "(1,1,1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G = \\{f : [0, 1] \\to [0, 1] \\mid f \\text{ is one to one and continuous}\\}$, and $\\circ$ be the function composition; $(G, \\circ)$ is a group.\n\na) Give an example of an infinite subgroup $H$ of $G$, which contains nonincreasing functions and $H \\neq G$.\n\nb) Let $H$ be a finite subgroup of $G$. Prove that $H$ has at most 2 elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39240,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of seven consecutive integers equals $2023^n$. Find the positive integers $n$ such that at least one summand is a perfect square.",
    "options": [],
    "answer": "all odd positive integers",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2022$ be an integer and $A$ be a finite set so that $\\{S(B) \\mid B \\subset A\\} = \\{1, 2, \\dots, n\\}$, where $S(X)$ is the sum of the elements of the set $X$. Find the smallest and the largest possible value of the cardinal of the set $A$.",
    "options": [],
    "answer": "Smallest |A| = ceil(log2(n + 1)); Largest |A| = floor((sqrt(8n + 1) − 1) / 2).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number. Find the real numbers $x$, $y$, $z$ so that $x + y + z = 3a - 1$ and $xy + xz + yz + x + y = 3a^2$.",
    "options": [],
    "answer": "x = a, y = a, z = a - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integer solutions of the equation\n$$\n\\left( \\frac{1 + \\sqrt{1 - x^2}}{2} \\right)^{\\sqrt{1-x}} = (\\sqrt{1-x})^{\\sqrt{1-x}+\\sqrt{1+x}}.\n$$",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $BAD$ has $\\angle BAD = 45^\\circ$ and triangle $BDC$ is on its outside, so that $DC = BA$ and $\\angle DCB = \\angle CDA = 75^\\circ$. Find the measure of $\\angle ABD$.\nAdrian Bud",
    "options": [],
    "answer": "90",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39245,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $a$ and $b$ real numbers, define\n$$\nE(a,b) = \\sqrt{\\left(\\frac{1}{2} - a\\right)^2 + \\left(\\frac{1}{2} - b\\right)^2} + \\sqrt{\\left(\\frac{1}{2} + a\\right)^2 + \\left(\\frac{1}{2} + b\\right)^2}.\n$$\n\na) Prove that $E(a, b) \\ge \\sqrt{2}$, for every real numbers $a, b$, and equality takes place if and only if $a = b \\in [-\\frac{1}{2}, \\frac{1}{2}]$.\n\nb) Find $\\min \\{E(a, a+2) \\mid a \\in \\mathbb{R}\\}$.",
    "options": [],
    "answer": "sqrt(10)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $g : \\mathbb{R} \\to \\mathbb{R}$ be a continuous function. Prove that:\n\na) if the inequality $g\\left(\\int_0^1 f(t) dt\\right) \\le \\int_0^1 g(f(t)) dt$ holds for every integrable function $f : [0, 1] \\to \\mathbb{R}$, then $g$ is convex;\n\nb) if the inequality $g\\left(\\int_0^1 f(t) dt\\right) \\le \\int_0^1 g(f(t)) dt$ holds for every continuous function $f : [0, 1] \\to \\mathbb{R}$, then $g$ is convex.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real solutions of the equation\n$$\n\\frac{4x - 1}{x^2 - 2x + 2} + \\frac{4x + 7}{x^2 + 2x + 2} = \\frac{4(2x + 3)}{x^2 + 2}.\n$$",
    "options": [],
    "answer": "x = ±√2, x = (4 + √34)/3, x = (4 - √34)/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$. Determine the least possible value the sum\n$$\n\\left\\lfloor \\frac{x_2 + x_3 + \\cdots + x_n}{x_1} \\right\\rfloor + \\left\\lfloor \\frac{x_1 + x_2 + \\cdots + x_n}{x_2} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{x_1 + x_2 + \\cdots + x_{n-1}}{x_n} \\right\\rfloor\n$$\nmay achieve, as $x_1, x_2, \\dots, x_n$ run through all positive real numbers.",
    "options": [],
    "answer": "(n-1)^2",
    "solution": "The minimum exists, as the summands are all non-negative integers; it is equal to $(n-1)^2$ and is achieved if, for instance, $x_1 = n$ and $x_2 = \\cdots = x_n = n+1$; the verification is routine.\n\nLet $s = x_1 + x_2 + \\cdots + x_n$ and let $S$ denote the sum in the statement. Note that\n$$\n\\left\\lfloor \\frac{x_1 + \\cdots + x_{k-1} + x_{k+1} + \\cdots + x_n}{x_k} \\right\\rfloor = \\left\\lfloor \\frac{s - x_k}{x_k} \\right\\rfloor = \\left\\lfloor \\frac{s}{x_k} - 1 \\right\\rfloor = \\left\\lfloor \\frac{s}{x_k} \\right\\rfloor - 1 \\\\ > \\left( \\frac{s}{x_k} - 1 \\right) - 1 = \\frac{s}{x_k} - 2, \\quad k = 1, \\dots, n.\n$$\nSum over $k = 1, 2, \\dots, n$ to get\n$$\n\\begin{aligned}\nS &> s \\cdot \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n} \\right) - 2n = \\\\\n&= (x_1 + x_2 + \\cdots + x_n) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n} \\right) - 2n \\ge n^2 - 2n,\n\\end{aligned}\n$$\nby the AM-HM (or Cauchy-Schwarz or Chebyshev) inequality.\nFinally, as $S$ and $n^2 - 2n$ are both integers, $S \\ge n^2 - 2n + 1 = (n-1)^2$, as desired. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, and let $D$, $E$, $F$ be the feet of its altitudes from $A$, $B$, $C$, respectively. The lines $AB$ and $DE$ cross at $K$ and the lines $AC$ and $DF$ cross at $L$. Let $M$ be the midpoint of the side $BC$ and let the line $AM$ cross the circle $ABC$ again at $N$. Finally, the parallel through $M$ to $EF$ crosses the line $KL$ at $P$. Show that the triangle $MNP$ is isosceles.\nAndrei Bâra",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nWe will show that $PM$ and $PN$ are the tangents from $P$ to $\\omega$; the conclusion then follows at once.\n\nWe first prove that $PM$ is the tangent of $\\omega$ at $M$. The tangent of $\\omega$ at $M$ is parallel to the tangent of $\\Gamma$ at $A$, which is in turn parallel to $EF$. Since $MP$ is the parallel through $M$ to $EF$, it follows that $PM$ is the tangent of $\\omega$ at $M$.\nNext, we show that $PM$ is the tangent of $\\gamma$ at $M$. Since the factor $-\\frac{1}{2}$ homothety from the barycentre of the triangle $ABC$ exchanges the pairs ($A$, $\\Gamma$) and ($M$, $\\gamma$), the tangent of $\\gamma$ at $M$ is parallel to the tangent of $\\Gamma$ at $A$; and since the latter is parallel to $EF$, so is the former. Consequently, $PM$, the parallel through $M$ to $EF$, is the tangent of $\\gamma$ at $M$.\nBy the preceding, $PM$ is the radical axis of $\\gamma$ and $\\omega$. Since $\\omega$ is a homothetic image of $\\Gamma$ from $N$, their radical axis is their common tangent at $N$.\nConsequently, $PN$ is the other tangent of $\\omega$ from $P$ if and only if $P$ is the radical centre of $\\Gamma$, $\\gamma$ and $\\omega$, which is the case if and only if $P$ lies on the radical axis of $\\Gamma$ and $\\gamma$. Since $P$ lies on $KL$, it is sufficient to show that $KL$ is radical axis of $\\Gamma$ and $\\gamma$.\nFinally, we prove that $KL$ is radical axis of $\\Gamma$ and $\\gamma$. The powers of $K$ with respect to $\\Gamma$ and $\\gamma$ are $KA \\cdot KB$ and $KD \\cdot KE$, respectively. The two are equal since $ABDE$ is cyclic, so $K$ lies on the radical axis of $\\Gamma$ and $\\gamma$. Similarly, $L$ lies on their radical axis, so $KL$ is the radical axis of the two circles. This completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that a convex polygon $A_1A_2...A_n$ has three vertices $A_i, A_j, A_k$ such that\n$$\n[A_iA_jA_k] > \\frac{1}{4}[A_1A_2...A_n],\n$$\n\nwhere $[X_1X_2...X_m]$ denotes the area of the polygon $X_1X_2...X_m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_iA_jA_k$ be a triangle of maximal area. Let $A'_i$ be the reflection of $A_i$ across the midpoint of the side $A_jA_k$; the points $A'_j$ and $A'_k$ are defined similarly. Note that at most one of these three reflections can be a vertex of the polygon — otherwise, the polygon would have three collinear vertices, contradicting convexity.\n\nWe will prove that the triangle $A'_iA'_jA'_k$ covers the polygon $A_1A_2...A_n$. By the preceding, the cover is strict, so $[A'_iA'_jA'_k] > [A_1A_2...A_n]$. As $[A'_iA'_jA'_k] = 4[A_iA_jA_k]$, the conclusion follows.\n\nTo prove the covering claim above, suppose, if possible, some vertex $A_\\ell$ lies outside the triangle $A'_iA'_jA'_k$. Let $\\operatorname{dist}(X, YZ)$ denote the Euclidean distance of the point $X$ to line $YZ$. Note that at least one of the three inequalities below holds:\n$$\n\\begin{align*}\n\\operatorname{dist}(A_\\ell, A_j A_k) &> \\operatorname{dist}(A_i, A_j A_k), \\\\\n\\operatorname{dist}(A_\\ell, A_k A_i) &> \\operatorname{dist}(A_j, A_k A_i), \\\\\n\\operatorname{dist}(A_\\ell, A_i A_j) &> \\operatorname{dist}(A_k, A_i A_j).\n\\end{align*}\n$$\nHence at least one of the triangles $A_\\ell A_j A_k$, $A_\\ell A_k A_i$, $A_\\ell A_i A_j$ has an area (strictly) greater than that of $A_i A_j A_k$, contradicting the choice of this latter. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix integers $n \\ge 2$ and $1 \\le m \\le n-1$. Let $a_0$, $a_1$, $\\dots$, $a_n$ be non-negative real numbers satisfying $a_0 + a_1 + \\dots + a_n = 1$. Prove that, if $\\sum_{k=0}^n a_k x^k < x^m$ for some $0 < x < 1$, then $\\sum_{k=0}^{m-1} (m-k)a_k < \\sum_{k=m+1}^n (k-m)a_k$.\n\nThe Problem Selection Committee",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $a_0 + a_1 + \\dots + a_n = 1$, the required inequality is equivalent to $\\sum_{k=0}^n k a_k > m$. To prove this inequality, note that the exponential $t \\mapsto x^t$, $t \\in \\mathbb{R}$, is convex and apply Jensen's inequality to write $x^{\\sum_{k=0}^n k a_k} \\le \\sum_{k=0}^n a_k x^k < x^m$. As $0 < x < 1$, the desired inequality follows by comparing the exponents of $x$ at both ends.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39252,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polygon is tiled with a finite number of triangles whose sides all have an odd length.\na) Prove that, if the polygon is convex, then its perimeter is an integer of the same parity as the number of triangles in the tiling.\n\nb) Does the conclusion still hold if the polygon is not convex?",
    "options": [],
    "answer": "For convex polygons: the perimeter is an integer with the same parity as the number of triangles. For nonconvex polygons: No, the conclusion does not always hold (counterexample exists).",
    "solution": "a) Let $K$ be the polygon under consideration. Since $K$ is convex, the tiling triangles fall into two classes: Those having all edges inside $K$, and those having at least one edge on the boundary of $K$. (If $K$ were not convex, there might also exist triangles having only parts of edges on the boundary of $K$, and the conclusion may fail to hold — see part b).)\nEvery inner edge of a triangle is subdivided into one or more 'short' segments by (the boundaries of) some other triangles on the opposite side. Each short segment is shared by exactly two triangles. Notice further that every short segment lies along a\n\nunique segment of maximal length which is a concatenation of non-overlapping inner edges coming from the triangles on the same side of that segment. Hence, the total length of the short segments along one of maximal length is integer. Consequently, so is the total length $s$ of all short segments.\nClearly, every outer edge (lying on the boundary of $K$) belongs to a single triangle, and the total length of all outer edges is the perimeter of $K$.\nFinally, let $t$ be the number of triangles, and let $S$ be the sum of their perimeters. Since the sides of each triangle all have an odd length, $t$ and $S$ have like parities. By the preceding, the perimeter of $K$ is $S - 2s$, and the conclusion follows.\n\nb) The answer is in the negative. Let $A, A', B, B'$, in order, be distinct points on a line $\\ell$ such that $AB = A'B' = 1$. Erect equilateral triangles $ABC$ and $A'B'C'$, where $C$ and $C'$ lie on opposite sides of $\\ell$. These two triangles tile the non-convex hexagon $AA'C'B'BC$. Letting $AA' = BB' = x$, the perimeter of the hexagon is $4 + 2x$. If $x = \\frac{1}{2}$, the perimeter is 5 which is odd, and if $x \\neq \\frac{1}{2}$, the perimeter is not even an integer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39253,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle is tiled with a finite number of triangles whose sides all have an odd length. Prove that the perimeter of the triangle is an integer of the same parity as the number of triangles in the tiling.\nMarius Cavachi",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Every inner edge of a triangle is subdivided into one or more 'short' segments by (the boundaries of) some other triangles on the opposite side. Each short segment is shared by exactly two triangles. Notice further that every short segment lies along a\n\nunique segment of maximal length which is a concatenation of non-overlapping inner edges coming from the triangles on the same side of that segment. Hence, the total length of the short segments along one of maximal length is integer. Consequently, so is the total length $s$ of all short segments.\nClearly, every outer edge (lying on the boundary of $\\Delta$) belongs to a single triangle, and the total length of all outer edges is the perimeter of $\\Delta$.\nFinally, let $t$ be the number of triangles, and let $S$ be the sum of their perimeters. Since the sides of each triangle all have an odd length, $t$ and $S$ have like parities. By the preceding, the perimeter of $\\Delta$ is $S - 2s$, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $b-a$ is a prime. Prove that\n$$(a^n + a + 1)(b^n + b + 1)$$\nis not the square of an integer for infinitely many positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, if possible, that $(a^n + a + 1)(b^n + b + 1)$ is a square for all but finitely many positive integers $n$. Then\n$$\nab \\equiv \\frac{(a^{p+1} + a + 1)(b^{p+1} + b + 1)}{(a^{p-2} + a + 1)(b^{p-2} + b + 1)}\n$$\nand\n$$\n(a + 2)(b + 2) \\equiv (a^{p-1} + a + 1)(b^{p-1} + b + 1) \\pmod{p}\n$$\nare both quadratic residues modulo all but finitely many primes $p$. Consequently, $ab$ and $(a+2)(b+2)$ are both squares. (It is a fact that a positive integer which is a quadratic residue modulo all but finitely many primes is itself a square.)\nLet $b-a = q$, a prime, and let $d = \\gcd(a, b)$. Clearly, $d=1$ or $d=q$. If $d=q$, then $a = (c-1)q$ and $b = cq$ for some integer $c > 1$. Hence $c(c-1)q^2 = ab$ is a square, forcing $c(c-1)$ to be a square. Coprimality of $c$ and $c-1$ forces in turn $c$ and $c-1$ to be both squares. This is impossible, since two squares of non-zero integers are at least 3 distance apart. Consequently, $a$ is coprime to $b$, so they are both squares.\n\nSince $a$ and $b$ are squares, $a + 2$ and $b + 2$ are not; and since $(a + 2)(b + 2)$ is a square, $a + 2$ is not coprime to $b + 2$ (otherwise they would both be squares). Clearly, $\\gcd(a+2, b+2)$ divides $(b+2)-(a+2) = b-a = q$, so $\\gcd(a+2, b+2) = q$. As before, $a + 2 = (c' - 1)q$ and $b + 2 = c'q$ for some integer $c' > 1$, and we reach a contradiction along the same lines. This ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39255,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is *square full* if it is divisible by the square of each of its prime divisors. Prove that $n$ and $n + 1$ are both square full for infinitely many positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $8 = 2^3$ and $8 + 1 = 9 = 3^2$ are square full. Now, if $n$ and $n + 1$ are both square full, then so are $4n(n + 1)$ and $4n(n + 1) + 1 = (2n + 1)^2$. As $4n(n + 1) > n + 1$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $x$ and $y$ be positive divisors of $2n^2 - 1$. Prove that $x + y$ is not divisible by $2n + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, if possible, that $x + y$ is divisible by $2n + 1$. Note that $2n + 1$ and $2n^2 - 1$ are relatively prime, as $(2n - 1)(2n + 1) - 2(2n^2 - 1) = 1$. Hence $2n + 1$ is coprime to $\\gcd(x, y)$, so $x/\\gcd(x, y)$ and $y/\\gcd(x, y)$ are coprime divisors of $2n^2 - 1$ whose sum is divisible by $2n + 1$.\n\nTo reach a contradiction, we may and will therefore assume $x$ coprime to $y$. Then $2n^2-1$ is divisible by $xy$, say, $2n^2-1 = kxy$; write $x+y = \\ell(2n+1)$. Express $n$ from this latter and plug it into the former to get $kxy = \\frac{1}{2} \\left(\\frac{x+y}{\\ell} - 1\\right)^2 - 1$; alternatively, but equivalently, $2k\\ell^2xy = (x + y - \\ell)^2 - 2\\ell^2$.\n\nFix $k$ and $\\ell$ to regard the above equality as an equation in positive integers $x$ and $y$. By assumption, it has at least one solution. Consider one such with a minimal $x + y$. Without loss of generality, assume $x \\ge y$.\n\nRewrite the equation as a quadratic in $x$, $x^2+2(y-\\ell-k\\ell^2y)x+(y-\\ell)^2-2\\ell^2=0$, and consider the other root $x'$. Then $x+x'=2(k\\ell^2y-y+\\ell)$ and $xx' = (y-\\ell)^2-2\\ell^2$. The former shows that $x'$ is also integer. The latter implies $x' < 0$: Clearly, $x' \\ne 0$, as $(y-\\ell)^2 = 2\\ell^2$ cannot hold in integers; and if $x' > 0$, then $x'y \\le xx' = (y-\\ell)^2 - 2\\ell^2 = y^2 - \\ell(2y+\\ell) < y^2$, so $x' < y$, whence $x' + y < 2y \\le x + y$, contradicting the minimality of $x + y$.\n\nThus, $(y-\\ell)^2 < 2\\ell^2$ and $x$ divides $2\\ell^2 - (y-\\ell)^2$, so $x \\le 2\\ell^2$. On the other hand, as $x'$ is negative, $x > x + x' = 2((k\\ell^2 - 1)y + \\ell) \\ge 2(\\ell^2 + \\ell - 1) \\ge 2\\ell^2$. This is a contradiction, so the sum $x + y$ is not divisible by $2n + 1$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 6$ and consider $n$ coplanar lines, no two parallel and no three concurrent. These lines split the plane into unbounded polygonal regions and polygons with pairwise disjoint interiors. Two polygons are *non-adjacent* if they do not share a side. Show that there are at least $\\frac{1}{12}(n-3)(n-2)$ pairwise non-adjacent polygons with the same number of sides each.\n\nCristian Șăvescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the obvious geometric plane graph associated with a generic $n$-line configuration in a plane (no two lines are parallel and no three are concurrent). This graph has exactly $\\binom{n}{2} = \\frac{1}{2}n(n-1)$ vertices, exactly $e = n^2$ edges and exactly $f = \\frac{1}{2}n(n+1)+1$ faces.\nLet $f_k$ be the number of $k$-edge faces (faces with exactly $k$ edges), so $f = \\sum_{k=2}^{n} f_k$ (no face has more than $n$ edges). Fix an integer $m$ in the range 4 through $n$, to write\n$$\n\\begin{align*} \n2n^2 = 2e &= \\sum_{k=2}^{n} k f_k = \\sum_{k=2}^{m-1} k f_k + \\sum_{k=m}^{n} k f_k \\ge \\sum_{k=2}^{m-1} k f_k + m \\sum_{k=m}^{n} f_k \\\\ \n&= \\sum_{k=2}^{m-1} k f_k + m \\left( f - \\sum_{k=2}^{m-1} f_k \\right) = \\sum_{k=2}^{m-1} (k-m) f_k + mf, \n\\end{align*}\n$$\nso\n$$\n2n^2 \\geq \\sum_{k=2}^{m-1} (k-m)f_k + \\frac{1}{2}m(n^2+n+2),\n$$\nwherefrom\n$$\n\\sum_{k=2}^{m-1} (m-k)f_k \\geq \\frac{1}{2}m(n^2+n+2) - 2n^2 = \\frac{1}{2}((m-4)n^2 + mn + 2m).\n$$\nNext, write $f_k = b_k + u_k$, where $b_k$ and $u_k$ are the numbers of bounded and unbounded $k$-faces, respectively. Thus, $b_k$ is the number of $k$-gons. Clearly, $b_2 = 0$ and $\\sum_{k=2}^{n} u_k = 2n$, so\n$$\n\\begin{align*} \\sum_{k=2}^{m-1} (m-k)f_k &= \\sum_{k=3}^{m-1} (m-k)b_k + \\sum_{k=2}^{m-1} (m-k)u_k \\le \\\\ &\\le \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=3}^{m-1} u_k \\le \\\\ &\\le \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=2}^{n} u_k = \\sum_{k=3}^{m-1} (m-k)b_k + 2(m-2)n. \\end{align*}\n$$\nConsequently,\n$$\n\\begin{align*} \\sum_{k=3}^{m-1} (m-k)b_k &\\ge \\sum_{k=2}^{m-1} (m-k)f_k - 2(m-2)n \\ge \\\\ &\\ge \\frac{1}{2}((m-4)n^2 + mn + 2m) - 2(m-2)n = \\\\ &= \\frac{1}{2}((m-4)n^2 - (3m-8)n + 2m) = \\frac{1}{2}((m-4)n - m)(n-2), \\end{align*}\n$$\nso\n$$\n\\max(b_3, \\dots, b_{m-1}) \\ge \\frac{1}{2} \\cdot \\frac{((m-4)n - m)(n-2)}{(m-3) + \\dots + 1} = \\frac{((m-4)n - m)(n-2)}{(m-3)(m-2)}.\n$$\nThe coefficient of $n^2$ in the lower bound is maximized at $m=5$ and $m=6$, so\n$$\n\\max(b_3, b_4) \\ge \\frac{1}{6}(n-5)(n-2) \\quad \\text{and} \\quad \\max(b_3, b_4, b_5) \\ge \\frac{1}{6}(n-3)(n-2).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$ and let $a_1, a_2, \\dots, a_n$ be real numbers in the closed interval $[1, 2024]$. Prove that\n$$\n\\sum_{i=1}^{n} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) > \\frac{1}{44} n(n + 33).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_j = \\{i: 2^{j-1} \\le a_i < 2^j\\}, j = 1, 2, \\dots, 11$. The $A_j$ form a partition of the index set $\\{1, 2, \\dots, n\\}$.\n\nNote that, for every $j$ in the range 1 through 11, if $A_j$ is non-empty, then\n$$\n\\begin{aligned}\n\\sum_{i \\in A_j} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) &= \\sum_{i \\in A_j} \\left( \\frac{a_1 + a_2 + \\dots + a_{i-1}}{a_i} + 1 \\right) \\\\\n&> \\sum_{k=0}^{|A_j|-1} \\frac{k \\cdot 2^{j-1}}{2^j} + |A_j| = \\frac{1}{2} \\sum_{k=0}^{|A_j|-1} k + |A_j| \\\\\n&= \\frac{1}{4} |A_j|(|A_j| - 1) + |A_j| = \\frac{1}{4} |A_j|(|A_j| + 3);\n\\end{aligned}\n$$\n\nAs the $A_j$ form a partition of the index set $\\{1, 2, \\dots, n\\}$ and at least one $A_j$ is non-empty,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) &= \\sum_{j=1}^{11} \\sum_{i \\in A_j} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) \\\\\n&> \\frac{1}{4} \\sum_{j=1}^{11} |A_j|(|A_j| + 3) = \\frac{1}{4} \\left( \\sum_{j=1}^{11} |A_j|^2 + 3n \\right) \\\\\n&\\ge \\frac{1}{4} \\left( \\frac{n^2}{11} + 3n \\right) = \\frac{1}{44} n(n + 33), \\text{ by Cauchy-Schwarz.}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. A circle through $B$ and $C$ crosses the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ on segments $BQ$ and $CP$, respectively, satisfy $\\angle ABY = \\angle AXP$ and $\\angle ACX = \\angle AYQ$. Prove that $XY$ and $BC$ are parallel.\nAndrei Chiriță\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BY$ and $CX$ cross at $S$ and let circles $APX$ and $AQY$ cross again at $T$. We first prove that $A, S, T$ are collinear. Invert from $A$ with power $AP \\cdot AB = AQ \\cdot AC$. As $\\angle ABY = \\angle AXP$, the circle $APX$ is mapped to line $BY$. Similarly, the circle $AQY$ is mapped to line $CX$, so the inversion switches $S$ and $T$. Hence $A, S, T$ are collinear, as stated.\n\nNext, we show that the lines $AT$, $PX$ and $QY$ are projectively concurrent.\nLet $PX$ and $QY$ cross projectively at $R$. Apply Pappus' theorem to the hexagram $BPXCQY$ to deduce that $A = BP \\cap CQ$, $R = PX \\cap QY$ and $S = XC \\cap YB$ are collinear. The desired concurrence now follows by the preceding paragraph.\n\nWe now prove that $PQYX$ is cyclic. If $AT$, $PX$ and $QY$ are parallel, then $PQYX$ is an isosceles trapezoid, so it is cyclic. Otherwise, read the power of $R$ from circles $APX$ and $AQY$ to write $RX \\cdot RP = RT \\cdot RA = RY \\cdot RQ$, so $PQYX$ is cyclic.\n\nFinally, read angles from circle $PQYX$ and the given circle through $B$ and $C$ to write $\\angle PYX = \\angle PQX \\equiv \\angle PQB = \\angle PCB$. Consequently, $XY$ and $BC$ are parallel, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime. Does there exist a permutation $a_1, a_2, \\dots, a_p$ of $1, 2, \\dots, p$ satisfying\n$$\n(i-j)a_k + (j-k)a_i + (k-i)a_j \\neq 0,\n$$\nfor all pairwise distinct $i, j, k$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The answer is in the affirmative. To define the desired permutation, let $a$ be a quadratic non-residue modulo $p$, let $ia_i \\equiv a \\pmod p$, $i = 1, 2, \\dots, p-1$, and let $a_p = p$. Clearly, the $a_i$ form a permutation of $1, 2, \\dots, p$; moreover, $a_{p-i} = p-a_i$, $i = 1, 2, \\dots, p-1$, and, since $a$ is a quadratic non-residue modulo $p$, $a_i \\neq i$, $i = 1, 2, \\dots, p-1$.\n\nTo prove $(*)$, let first $i, j, k$ be all (strictly) less than $p$. For convenience, write $\\equiv$ for congruence modulo $p$. Then\n$$\n\\begin{aligned}\nijk((i-j)a_k + (j-k)a_i + (k-i)a_j) &\\equiv ij(i-j)a + jk(j-k)a + ki(k-i)a \\\\ &= -a(i-j)(j-k)(k-i) \\neq 0. \\end{aligned}\n$$\nLet now one of $i, j, k$ be equal to $p$. Since the left-hand member of $(*)$ is antisymmetric in $i, j, k$, we may and will assume that $k = p$, so $a_k = a_p = p$. Then\n$$\nij((i-j)a_k + (j-k)a_i + (k-i)a_j) \\equiv j^2a - i^2a = a(j-i)(j+i).\n$$\nThe latter is non-zero modulo $p$, and $(*)$ follows, unless $j = p-i$, in which case $a_j = a_{p-i} = p-a_i$, and\n$$\n(i-j)a_k + (j-k)a_i + (k-i)a_j = (2i-p)p - ia_i + (p-i)(p-a_i) = p(i-a_i) \\neq 0,\n$$\nsince $a_i \\neq i$. This completes the argument and concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with apex at $A$, and let $M, N, P$ be the midpoints of the sides $BC, CA, AB$, respectively. Let $Q$ and $R$ be points inside the segments $BM$ and $CM$, respectively, so that $\\angle BAQ = \\angle MAR$. The segment $NP$ crosses $AQ$ and $AR$ at $U$ and $V$, respectively. The point $S \\neq U$ lies on the half-line $AQ$ emanating from $A$, so that $SV$ is the internal bisector of $\\angle ASM$. Similarly, $T \\neq V$ is a point on the half-line $AR$ emanating from $A$, so that $TU$ is the internal bisector of $\\angle ATM$. Prove that one of the points where the circles $NUS$ and $PVT$ meet lies on the line $AM$.\nFlavian Georgescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nWe first show that $AVMS$ is cyclic. Since the triangle $ABC$ is isosceles, $AM \\perp BC$, and since $NP$ is midline, $V$ is the midpoint of $AR$. Then $MV$ is the $M$-median of the right triangle $AMR$, so $VM = VA = VR$. Hence $V$ lies on the perpendicular bisector of $AM$. By hypothesis, $V$ also lies on the internal bisector of $\\angle ASM$, so, in the circle $AMS$, $V$ is the midpoint of the arc $AM$ not containing $S$. Consequently, $AVMS$ is cyclic, as stated. Similarly, $AUMT$ is cyclic.\n\nWe now prove that $H$ lies on the circle $NUS$. Since $AVMS$ is cyclic, $\\angle ASV = \\angle MAR$. By hypothesis, $\\angle MAR = \\angle BAQ$, so $AB \\parallel SV$. Similarly, $AC \\parallel TU$. Then $\\angle SVP = \\angle APV = 90^\\circ - \\frac{1}{2}\\angle BAC$, and $\\angle AUH = 90^\\circ - \\angle UAV = 90^\\circ - \\frac{1}{2}\\angle BAC$, so $\\angle SVP = \\angle AUH$.\n\nFrom the triangles $SUV$ and $AUN$,\n$$\n\\frac{SU}{SV} = \\frac{\\sin(90^\\circ - \\frac{1}{2}\\angle BAC)}{\\sin \\angle SUV} = \\frac{AU}{AN}.\n$$\nSince $\\angle UAM = \\angle UAH = \\angle VAN$ and $\\angle AUH = 90^\\circ - \\frac{1}{2}\\angle BAC = \\angle ANV$, the triangles $AUH$ and $ANV$ are similar.\nHence $AN/AU = NV/UH$, so $SU/SV = UH/VN$. Since $\\angle SVN = 180^\\circ - \\angle SVP = 180^\\circ - \\angle AUM = \\angle SUH$, the triangles $SUH$ and $SVN$ are similar, so $\\angle USH = \\angle VSN$, whence $\\angle USV = \\angle HSN$.\nFinally, $\\angle USV = \\angle MAV = 90^\\circ - \\angle AVU = \\angle HUV$, so $\\angle HSN = \\angle HUV = \\angle HUN$. Consequently, $H$ lies on the circle $NUS$, as stated. Similarly, $H$ lies on the circle $PVT$. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39262,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *partition* of a set $S$ is a set of pairwise disjoint subsets of $S$ whose union is $S$. Let $P$ be a partition of the set $\\{1, 2, ..., 2024\\}$ into 2-element sets, satisfying the following condition: For every set $\\{a, b\\}$ in $P$, either $|a - b| = 1$ or $|a - b| = 506$. Assume that $\\{1518, 1519\\}$ belongs to $P$. Determine the number that pairs off with $505$ to form a set in $P$.\nThe Problem Selection Committee\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "506",
    "solution": "Assume the square in the upper-left corner is white.\nBy the condition in statement, a set $\\{a, b\\}$ in $P$ is a horizontal domino if $|a-b| = 1$ and a vertical domino if $|a - b| = 506$, with the possible exceptions $\\{506, 507\\}$, $\\{1012, 1013\\}$ and $\\{1518, 1519\\}$. As this latter is an all-black member of $P$ and dominoes are bicolour, $P$ must have some all-white member. The only such is $\\{1012, 1013\\}$.\nHence, to form a set in $P$, the number $506$ pairs off with exactly one of $505$ and $507$.\nAs $\\{506, 507\\}$ is all-black and there are no more all-white sets left, $506$ must pair off with $505$, as stated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39263,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *diameter* of a finite planar set is any line segment of maximal Euclidean length having both end points in that set. A *lattice point* in the Cartesian plane is one whose coordinates are both integral. Given an integer $n \\ge 2$, prove that a set of $n$ lattice points in the plane has at most $n-1$ diameters.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the *diameter graph* $\\Gamma$ on $n$ pairwise distinct lattice points in the plane, i.e., the geometric graph on those points, whose edges are the diameters of the configuration. We will prove that $\\Gamma$ has at least one vertex of degree 1. Removal of one such and the edge it is incident to, allows then an inductive approach — the base case, $n=2$, is clear. As expected, the argument hinges on the fact that every two diameters intersect: Either they both emanate from the same point or they both cross at some interior point. If $\\Gamma$ has a vertex $A$ of degree at least 3, proceed along the standard lines in the continuous realm: The diameters emanating from $A$ all lie within the angle formed by two diameters $AB$ and $AC$, where $\\angle BAC \\le 60^\\circ$. Letting $AX$ be a third diameter from $A$, it then follows that $X$ is the desired vertex of degree 1, as a hypothetical diameter $XY$, $Y \\ne A$, cannot intersect both $AB$ and $AC$.\n\nIt is a fact that $\\Gamma$ has at most $n$ edges. Suppose, if possible, $\\Gamma$ has $n$ edges and is 2-regular (all vertices have degree 2). As every two diameters intersect, this is possible if and only if $n$ is odd, in which case $\\Gamma$ is a star-shaped self-crossing $n$-cycle $A_1A_2A_3\\dots A_n$ whose polygonal convex hull reads around the boundary $A_1A_3A_5\\dots A_nA_2A_4A_6\\dots A_{n-1}$. Write $A_iA_{i+1} = d$, $i = 1, 2, \\dots, n$ (indices are reduced modulo $n$), so $(x_{i+1} - x_i)^2 + (y_{i+1} - y_i)^2 = d^2$, $i = 1, 2, \\dots, n$; clearly, $d^2$ is an integer. Without loss of generality, assume the configuration is the smallest possible.\n\nBy minimality, the $2n$ differences $x_{i+1} - x_i$ and $y_{i+1} - y_i$ cannot all be even — otherwise, the midpoints of all line segments $A_iA_j$ would again be lattice points, and a homothety of factor $\\frac{1}{2}$ from some $A_i$ would yield a smaller configuration. Hence one of the index sets $I = \\{i: x_{i+1} - x_i \\equiv 1 \\pmod 2\\}$ and $J = \\{i: y_{i+1} - y_i \\equiv 1 \\pmod 2\\}$, say, $I$, is non-empty. Hence again, $d^2 \\equiv 1 \\pmod 4$, so for no index $i$ are $x_{i+1} - x_i$ and $y_{i+1} - y_i$ both even. Consequently, the union $I \\cup J$ exhausts all $n$ indices.\n\nRule out the case $d^2 \\equiv 1 \\pmod 4$ as follows: If $d^2 \\equiv 1 \\pmod 4$, then $I$ and $J$ are disjoint. By the preceding, $I \\cup J$ covers all $n$ indices, so $|I| + |J| = n$ which is odd. On the other hand, $\\sum_{i=1}^n (x_{i+1} - x_i) = 0$ implies that $|I|$ is even; similarly, $\\sum_{i=1}^n (y_{i+1} - y_i) = 0$ implies that $|J|$ is even, and hence so is $|I| + |J|$, contradicting the fact that $|I| + |J| = n$ is odd.\n\nFinally, rule out the case $d^2 \\equiv 2 \\pmod 4$: If $d^2 \\equiv 2 \\pmod 4$, then $I = J$, so $|I| = n$, contradicting again the parity of $|I|$ established above. Consequently, the diameter graph $\\Gamma$ is not 2-regular, as desired.\n\n**Remark.** The required diameter upper bound is always achieved, so $n-1$ is, in fact, the maximal number of diameters a planar configuration of $n$ lattice points may have. To prove it, consider the Diophantine equation $x^2 + y^2 = 5^N$ $(*)$ where $N$ is a non-negative integer. By Jacobi's two-square theorem, the number of pairwise distinct solutions of the Diophantine equation $x^2 + y^2 = M$, where $M$ is a positive integer, is equal to four times the excess of the number of divisors of $M$ that are congruent to 1 modulo 4 over those congruent to 3 modulo 4. In the case at hand, $(*)$ has exactly $4(N+1)$ pairwise distinct solutions. This can equally well be established directly, by noticing that $1 \\pm 2\\sqrt{-1}$ are Gaussian primes, and $\\frac{1}{\\pi}\\arg(1+2\\sqrt{-1})$ is irrational; explicitly, the solutions are $u(1+2\\sqrt{-1})^k(1-2\\sqrt{-1})^{N-k}$, where $u = \\pm 1$ or $\\pm\\sqrt{-1}$, and $k = 0, 1, \\dots, N$. Exactly $N+1$ of these solutions lie in the first quadrant, $x > 0$ and $y \\ge 0$; and since 5 is odd, exactly $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ of these, say, $(x_i, y_i)$, $i = 1, 2, \\dots, \\lfloor \\frac{1}{2}(N+1) \\rfloor$, satisfy $x > y \\ge 0$. The $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ lattice points $(x_i, y_i)$ are all exactly $5^{N/2}$ away from the origin, and every two are (strictly) less than $5^{N/2}$ distance apart. Consequently, the origin and the $(x_i, y_i)$ form a planar configuration of $\\lfloor \\frac{1}{2}(N+3) \\rfloor$ lattice points with exactly $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ diameters of length $5^{N/2}$ each. Setting $N = 2n-3$ completes the argument.\n\nWe end by describing a related configuration. Consider an even integer $N \\ge n$. The $(x_i, y_i)$ above and the $(5^{N/2} - x_i, y_i)$ form a configuration of $N+2$ lattice points with exactly $N+1$ diameters: $\\frac{1}{2}N+1$ of these join $(0,0)$ to each $(x_i, y_i)$, and another $\\frac{1}{2}N$ join $(5^{N/2}, 0)$ to each $(5^{N/2} - x_i, y_i)$ with a positive $y_i$. Deletion of any $N-n+2$ points with both coordinates positive then settles the case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a non-negative integer $n$, let $\\sum_{k=0}^{2n} a_k X^k$ be the standard power expansion of the polynomial $\\sum_{k=0}^{n} \\binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)}$. The coefficients $a_{2k+1}$ all vanish since the polynomial is invariant under the change $X \\mapsto -X$. Show that the $a_{2k}$ are all positive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Leaving aside the trivial cases $n=0$ and $n=1$, assume $n \\ge 2$. The conclusion follows from the fact that the polynomial under consideration,\n\n$$\nf = \\sum_{k=0}^{n} \\binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)},\n$$\nis expressible as a sum of $n+1$ polynomials of the form $(b_k X^2 + c_k)^n$, where the $b_k$ and the $c_k$ are all non-negative real numbers and at least $n-1$ products $b_k c_k$ are positive; this latter then implies that the even powers of $X$ all occur in the expansion with a positive coefficient. In particular, the fact that the odd rank coefficients in the standard expansion all vanish comes for free.\n\nLet $S$ be the set of $(n+1)$-st roots of unity and recall that, if $k$ and $\\ell$ are integers in the range $0$ through $n$, then the sum $\\sum_{\\omega \\in S} \\omega^k \\bar{\\omega}^\\ell$ vanishes, unless $k=\\ell$ in which case it is equal to $n+1$.\n\n$$\n\\begin{align*}\nf &= \\frac{1}{n+1} \\sum_{k=0}^{n} \\sum_{\\ell=0}^{n} \\binom{n}{k} \\binom{n}{\\ell} (X+1)^{k+\\ell} (X-1)^{2n-k-\\ell} \\sum_{\\omega \\in S} \\omega^k \\bar{\\omega}^{\\ell} \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} \\left( \\sum_{k=0}^{n} \\binom{n}{k} \\omega^k (X+1)^k (X-1)^{n-k} \\right) \\\\\n&\\qquad \\cdot \\left( \\sum_{\\ell=0}^{n} \\binom{n}{\\ell} \\bar{\\omega}^{\\ell} (X+1)^{\\ell} (X-1)^{n-\\ell} \\right) \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} (\\omega(X+1) + X-1)^n (\\bar{\\omega}(X+1) + X-1)^n \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} \\left( (2+\\omega+\\bar{\\omega})X^2 + 2 - \\omega - \\bar{\\omega} \\right)^n.\n\\end{align*}\n$$\n\nFinally, since each $\\omega + \\bar{\\omega}$ is a real number whose absolute value does not exceed $2$ and $\\omega + \\bar{\\omega} < 2$ for at least $n-1$ roots $\\omega \\neq \\pm 1$, each summand above is a polynomial of the desired form. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\gamma_1$ and $\\gamma_2$ be external circles in the plane, centred at $O_1$ and $O_2$, respectively. One of their external tangents touches $\\gamma_1$ at $A_1$ and $\\gamma_2$ at $A_2$. One of their internal tangents touches $\\gamma_1$ at $B_1$ and $\\gamma_2$ at $B_2$, and the other touches $\\gamma_1$ at $C_1$ and $\\gamma_2$ at $C_2$. Let $B_1B_2$ and $C_1C_2$ cross at $O$. Let $X$ be the point where $A_2O$ crosses $\\gamma_1$ and $OX < OB_1$, and let $Y$ be the point where $A_1O$ crosses $\\gamma_2$ and $OY < OB_2$. The perpendicular at $X$ to $OX$ crosses the line $O_1B_1$ at $P$, and the perpendicular at $Y$ to $OY$ crosses the line $O_2C_2$ at $Q$. Prove that $PQ$ and $A_1A_2$ are parallel.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the angles $\\angle OXP$ and $\\angle OYQ$ are both right, the circles $OXP$ and $OYQ$ cross again at a point $R$ on $PQ$, and $OR$ is perpendicular to $PQ$. Letting $A_1C_1$ and $A_2B_2$ cross at $S$, the conclusion then follows at once from the two facts below:\n(1) $S$ lies on $OR$; and\n(2) $OS$ is perpendicular to $A_1A_2$.\n\nTo prove (1), invert from $O$ with power $-OB_1 \\cdot OB_2$. Under this inversion, $\\gamma_1$ and $\\gamma_2$ correspond to one another, and so do the points in each of the pairs $(A_1, Y)$, $(B_1, B_2)$, $(C_1, C_2)$ and $(X, A_2)$.\n\nClearly, $B_1$ lies on the circle $OXP$, so this latter is mapped to $A_2B_2$. Similarly, the circle $OYQ$ (through $C_2$) is mapped to $A_1C_1$.\nConsequently, $R$ is mapped to $S$. This establishes (1).\n\nTo prove (2), let $A_1A_2$ cross $B_1B_2$ and $C_1C_2$ at $D_1$ and $D_2$, respectively; let $O_1O_2$ cross $A_1C_1$ and $A_2B_2$ at $X_1$ and $X_2$, respectively; and let $A_1C_1$ cross $O_1D_1$ at $Y_1$, and $A_2B_2$ cross $O_2D_2$ at $Y_2$. The argument hinges on the following two facts:\n(3) $X_1$ and $Y_1$ both lie on the circle $\\omega_1$ on diameter $OD_1$; similarly, $X_2$ and $Y_2$ both lie on the circle $\\omega_2$ on diameter $OD_2$; and\n(4) $X_1, X_2, Y_1, Y_2$ all lie on a circle $\\omega$.\nAssume (3) and (4) to establish (2) as follows: $X_1Y_1$, i.e., $A_1C_1$, is the radical axis of $\\omega$ and $\\omega_1$, and $X_2Y_2$, i.e., $A_2B_2$, is the radical axis of $\\omega$ and $\\omega_2$. Hence $S$ is the radical centre of the three circles. As such, $S$ lies on the radical axis of $\\omega_1$ and $\\omega_2$. These two circles cross again at the orthogonal projection $O'$ of $O$ on $D_1D_2$, i.e., on $A_1A_2$, and $OO'$ is their radical axis. Consequently, $S$ lies on $OO'$ and (2) follows.\n\nWe now turn to prove (3) and (4).\nTo prove (3) only $X_1$ and $Y_1$ are dealt with. It is sufficient to show that the angles $\\angle O_1X_1D_1$ and $\\angle OY_1O_1$ are both right.\nTo prove that the angle $\\angle O_1X_1D_1$ is right, we show that $X_1$ lies on the circle on diameter $O_1D_1$. Clearly, $A_1$ and $B_1$ both lie on this circle, so it is sufficient to show that $\\angle O_1A_1X_1 = \\angle O_1B_1X_1$. Now, $\\angle O_1A_1X_1 = \\angle O_1A_1C_1 = \\angle O_1C_1A_1 = \\angle O_1C_1X_1$, since $O_1A_1 = O_1C_1$ (they both are radii of $\\gamma_1$). On the other hand, $C_1$ and $B_1$ are reflexions of one another in $OO_1$, so $\\angle O_1C_1X_1 = \\angle O_1B_1X_1$. Consequently, $\\angle O_1A_1X_1 = \\angle O_1B_1X_1$, as desired.\nThe proof that the angle $\\angle OY_1O_1$ is right is quite similar: This time we show that $Y_1$ lies on the circle on diameter $OO_1$. Clearly, $B_1$ and $C_1$ both lie on this circle, so it is sufficient to show that $\\angle O_1B_1Y_1 = \\angle O_1C_1Y_1$. Notice that $B_1$ and $A_1$ are reflexions of one another in $O_1D_1$, to write $\\angle O_1B_1Y_1 = \\angle O_1A_1Y_1$. Refer now again to $O_1A_1 = O_1C_1$, to write $\\angle O_1A_1Y_1 = \\angle O_1A_1C_1 = \\angle O_1C_1A_1 = \\angle O_1C_1Y_1$ and conclude that $O_1B_1Y_1 = \\angle O_1C_1Y_1$. This establishes (3).\n\nFinally, proving (4) requires more work. We will show that $\\angle X_1Y_1Y_2+\\angle X_1X_2Y_2 = 180^\\circ$. Clearly, $\\angle X_1Y_1Y_2 = 180^\\circ-(\\angle O_1Y_1X_1+\\angle D_1Y_1Y_2)$, so it is sufficient to prove that\n$$\n\\angle X_1X_2Y_2 = \\angle O_1Y_1X_1 + \\angle D_1Y_1Y_2. \\quad (*)\n$$\nDeal with each angle in the right-hand member separately. By (3), $\\angle O_1Y_1X_1 = \\angle O_1OD_1$, and an obvious angle chase yields $\\angle O_1OD_1 = \\angle O_1OB_1 = \\angle O_2OB_2 = \\angle O_2OC_2 = \\angle O_2OD_2$. Hence $\\angle O_1Y_1X_1 = \\angle O_2OD_2$.\nTo deal with $\\angle D_1Y_1Y_2$, let $O_1D_1$ and $O_2D_2$ cross at $Z$. By (3), the angles $\\angle OY_1D_1$ and $\\angle OY_2D_2$ are both right, so $OY_1ZY_2$ is cyclic, and hence $\\angle D_1Y_1Y_2 = \\angle ZY_1Y_2 = \\angle ZOY_2$.\nThus, (*) now reads\n$$\n\\angle X_1X_2Y_2 = \\angle O_2OD_2 + \\angle ZOY_2. \\quad (**)\n$$\nOur purpose now is to replace $\\angle ZOY_2$ by a more tractable angle. To this end, notice that $O_1D_1$ is the internal angle bisectrix of $\\angle A_1D_1B_1$, and $O_2D_1$ is the internal angle bisectrix of $\\angle A_2D_1B_2$, so $O_1D_1$ and $O_2D_1$ are perpendicular; that is, $O_2D_1$ is the $O_2$-altitude of the triangle $ZO_1O_2$. Similarly, $O_1D_2$ is the $O_1$-altitude of the triangle $ZO_1O_2$, so it crosses $O_2D_1$ at the orthocentre $H$ of this triangle.\nOn the other hand, $O_2D_1$ and $O_1D_2$ are clearly the internal angle bisectrices of the triangle $OD_1D_2$ at $D_1$ and $D_2$, respectively. Hence $OH$ is the internal angle bisectrix of $\\angle D_1OD_2$. As such, it is perpendicular to external bisectrix $O_1O_2$ of this angle. Recalling that $H$ is the orthocentre of the triangle $ZO_1O_2$, it follows that $OH$ is the $Z$-altitude of this triangle.\nHence $\\angle ZOY_2 = \\angle HOY_2$, and (**) now reads\n$$\n\\angle X_1X_2Y_2 = \\angle O_2OD_2 + \\angle HOY_2.\n$$\nTo prove this, split $\\angle X_1X_2Y_2 = \\angle X_1X_2D_2 + \\angle D_2X_2Y_2$. The angle $\\angle X_1X_2D_2 = \\angle OX_2D_2$ is right, by (3). As such, $\\angle X_1X_2D_2 = \\angle HOX_2$, since $OH$ is perpendicular to $O_1O_2$. Split further $\\angle HOX_2 = \\angle HOY_2 + \\angle Y_2OX_2$, to write $\\angle X_1X_2D_2 = \\angle HOY_2 + \\angle Y_2OX_2$. By (3), $\\angle D_2X_2Y_2 = \\angle D_2OY_2$, so $\\angle X_1X_2Y_2 = \\angle HOY_2 + \\angle Y_2OX_2 + \\angle D_2OY_2 = \\angle HOY_2 + \\angle D_2OX_2$, as desired. This establishes (4) and completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 3$ satisfying the following condition: there exist pairwise distinct real numbers $a_1, a_2, \\dots, a_n$ such that the $\\frac{1}{2}n(n-1)$ sums $a_i+a_j$, $1 \\le i < j \\le n$, ordered increasingly, form an arithmetic sequence (the difference of every two consecutive sums is the same).",
    "options": [],
    "answer": "n = 3 or 4",
    "solution": "The required integers are $n=3$ and $n=4$. In the former case, let $(a_1, a_2, a_3) = (1, 2, 3)$; the sums of pairs form the arithmetic sequence $3, 4, 5$. In the other case, let $(a_1, a_2, a_3, a_4) = (1, 3, 4, 5)$; the sums of pairs form the arithmetic sequence $4, 5, 6, 7, 8, 9$.\nSince the $n$ numbers are pairwise distinct, so are the $\\frac{1}{2}n(n-1)$ sums of pairs — otherwise, the difference of the arithmetic sequence would be zero, so the first and the second smallest sums would be equal, hence the second and the third smallest numbers would be equal, and we would reach a contradiction.\n\nLet now $n \\ge 5$, and suppose, if possible, that $a_1, a_2, \\dots, a_n$ satisfy the condition in the statement. Without loss of generality, we may and will assume that $a_1 < a_2 < \\dots < a_n$. Let $d$ be the difference of the corresponding arithmetic sequence of sums.\nThe smallest sum is $a_1 + a_2$, and the second smallest sum is $a_1 + a_3$, so $a_3 - a_2 = d$. The largest sum is $a_{n-1} + a_n$, and the second largest sum is $a_{n-2} + a_n$, so $a_{n-1} - a_{n-2} = d$. Consequently, $a_2 + a_{n-1} = (a_3 - d) + (a_{n-2} + d) = a_3 + a_{n-2}$. If $n \\ge 6$, the leftmost sum and the rightmost sum correspond to distinct pairs, so they are at least $d$ distance apart. This contradiction forces $n=5$.\nLet $n=5$. By the preceding, $a_3 - a_2 = d$, and $a_4 - a_3 = d$, so $2a_3 = a_2 + a_4$. To reach a contradiction, we present two approaches.\n\n**1st Approach.** Evaluate the sum $s = \\sum_{1 \\le i < j \\le 5} (a_i + a_j)$ in two different ways. On the one hand, $s = 5(a_1 + a_2 + a_4 + a_5)$. On the other hand, $s = 4(a_1 + a_2 + a_3 + a_4 + a_5)$, since each $a_k$ occurs in exactly four sums $a_i + a_j$, $i < j$. Equate the two and clear out like terms, to get $a_1 + a_2 + a_4 + a_5 = 4a_3 = 2(a_2 + a_4)$. Consequently, $a_1 + a_5 = a_2 + a_4$ which is the desired contradiction.\n\n**2nd Approach.** Notice that the third smallest sum is $a_1 + a_4$ which is $d$ larger than $a_1 + a_3$, and the third largest sum is $a_2 + a_5$. Between these lie $a_1 + a_5$ and the consecutive sums $a_2 + a_3 < a_2 + a_4 < a_3 + a_4$. Then $a_1 + a_5$ is either the fourth smallest sum or the fourth largest.\nWithout loss of generality, we may and will assume that $a_1 + a_5$ is the fourth smallest sum. The string of ten sums is then $a_1 + a_2 < a_1 + a_3 < a_1 + a_4 < a_1 + a_5 < a_2 + a_3 < a_2 + a_4 < a_3 + a_4 < a_2 + a_5 < a_3 + a_5 < a_4 + a_5$.\nEvaluate $a_5 - a_4$ in two different ways. On the one hand, $a_5 - a_4 = (a_1 + a_5) - (a_1 + a_4) = d$, by the third inequality from the left. On the other hand,\n$$\na_5 - a_4 = ((a_2 + a_5) - (a_3 + a_4)) + (a_3 - a_2) = d + d = 2d,\n$$\nby the third inequality from the right and $a_3 - a_2 = d$. This is the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ satisfying the following condition: There exist pairwise distinct integers $a_1, \\dots, a_n, b_1, \\dots, b_n$ such that\n$$\n\\prod_{i=1}^{n} (a_k^2 + a_i a_k + b_i) = \\prod_{i=1}^{n} (b_k^2 + a_i b_k + b_i) = 0 \\quad \\text{for all } k = 1, \\dots, n.\n$$",
    "options": [],
    "answer": "1, 2",
    "solution": "The required integers are $n = 1$ and $n = 2$. In the former case, $(a_1, b_1) = (1, -2)$ is the unique pair of integers satisfying the conditions in the statement; and in the latter, only $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$ and $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$ fit the bill. Verification is routine.\n\nThe degree $2n$ monic polynomial $f = \\prod_{i=1}^{n} (X^2 + a_i X + b_i)$ vanishes at each $a_k$ and at each $b_k$. Since these $2n$ numbers are pairwise distinct and $\\deg f = 2n$, they form the root set of $f$. Moreover, since the root set of each factor $f_i = X^2 + a_i X + b_i$ has size at most 2, and the union of these $n$ root sets has size $2n$, they must be pairwise disjoint sets of size 2 each. Recall now that $f$ is monic to write $f = \\prod_{k=1}^{n} (X - a_k)(X - b_k)$. Consequently, $\\prod_{k=1}^{n} (X^2 + a_k X + b_k) = \\prod_{k=1}^{n} (X - a_k)(X - b_k)$.\n\nIf no $b_k$ is zero, identification of constant terms of both sides yields $a_1 \\cdots a_n = 1$. This is impossible for pairwise distinct integers unless $n = 1$, in which case $a_1 = 1$, and $f = X^2 + X + b_1$. This polynomial must vanish at $a_1 = 1$, so $b_1 = -2$. Consequently, $f = X^2 + X - 2 = (X-1)(X+2) = (X-a_1)(X-b_1)$, as required.\n\nIf some $b_k = 0$, say, $b_1 = 0$, then $a_1 \\neq 0$, so $f_1 = X^2 + a_1 X$ does not vanish at $a_1$. Since the root set of $f_1$ has size 2, this forces $n \\ge 2$. The polynomial equality at the end of the second paragraph now reads $(X + a_1) \\prod_{k=2}^{n} (X^2 + a_k X + b_k) = (X - a_1) \\prod_{k=2}^{n} (X - a_k)(X - b_k)$.\n\nSince $a_1, b_2, \\dots, b_n$ are all non-zero, identification of constant terms of both sides above yields $a_2 \\cdots a_n = -1$. This is impossible for pairwise distinct integers unless $n = 2$ or $n = 3$.\n\nIf $n = 2$, then $a_2 = -1$, and $f_1 = X^2 + a_1 X$ and $f_2 = X^2 - X + b_2$. Recall that the root sets of $f_1$ and $f_2$ are disjoint and both have size 2. Since $f_2(b_1) = f_2(0) = b_2 \\neq b_1 = 0$ and $f_2(b_2) = b_2^2 \\neq 0$, the roots of $f_2$ must be $a_1$ and $a_2$, so the other root of $f_1$ must be $b_2$. The condition $f_2(a_2) = 0$ yields $b_2 = -2$, and the condition $f_1(b_2) = 0$ then yields $a_1 = 2$. Consequently, $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$.\n\nThe other quadruple, $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$, corresponds to the case where $b_2 = 0$.\n\nFinally, to rule out the case $n = 3$, we may and will assume that $a_2 = -1$ and $a_3 = 1$. Then $f_1 = X^2 + a_1 X$, $f_2 = X^2 - X + b_2$, $f_3 = X^2 + X + b_3$, and $f = f_1 f_2 f_3$ has the (pairwise distinct) roots $a_1, a_2 = -1, a_3 = 1, b_1 = 0, b_2$ and $b_3$. Since $f_1(-1)f_1(1) = (1-a_1)(1+a_1) \\neq 0$, it follows that $-1$ and $1$ are roots of $f_2 f_3$. Notice that $f(1) = b_2 \\neq 0$ and $f_3(-1) = b_3 \\neq 0$, so $2+b_2 = f_2(-1) = 0 = f_3(1) = 2+b_3$, i. e., $b_2 = b_3$. This contradiction settles the case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The parallels through $M$ to $AB$ and $AC$ cross the tangent at $A$ to circle $ABC$ at $X$ and $Y$, respectively. The circles $BMX$ and $CMY$ cross again at $S$. Prove that the circles $SXY$ and $SBC$ are tangent.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The argument hinges on the four facts below:\n(1) $X'$ lies on circle $YMC$ and $Y'$ lies on circle $XMB$.\n(2) $S$, $A$ and $M$ are collinear.\n(3) $U$ and $V$ both lie on circle $SXY$.\n(4) $UV$ is parallel to $BC$.\nAssume these facts for the moment, to complete the solution as follows: By (3), the conclusion is equivalent to circles $SUV$ and $SBC$ being tangent; and by (4), these circles are similar from $S$, whence the conclusion.\n\nTo prove (1), write $\\angle Y'XM = \\angle YXM = 180^\\circ - \\angle PAX - \\angle APX = 180^\\circ - \\angle ABC - \\angle BAC = \\angle ACB$.\nAs $YN$ and $NM$ are midlines in triangles $ABY'$ and $ABC$, respectively, $\\angle Y'BM = \\angle Y'BA + \\angle ABC = \\angle BNM + \\angle ABC = 180^\\circ - \\angle ACB$.\nConsequently, $\\angle Y'XM + \\angle Y'BM = 180^\\circ$, so $Y'$ lies on circle $XMB$. Similarly, $X'$ lies on circle $YMC$. This establishes (1).\n\nTo prove (2), note that $SM$ is the radical axis of the circles $YMC$ and $XMB$, so it is sufficient to show that $A$ has equal powers with respect to these circles. By (1), the two circles are $X'MC$ and $Y'MB$, respectively, so $AX' \\cdot AY = 2 \\cdot AX \\cdot AY = AX \\cdot 2 \\cdot AY = AX \\cdot AY'$, as desired. This establishes (2).\nTo prove (3), note that $YU \\parallel Y'B$, as $YN$ is a midline in triangle $ABY'$, so $\\angle SUY = \\angle SBY'$. By (1), $\\angle SBY' = \\angle SXY$, so $\\angle SUY = \\angle SXY$, implying that $U$ lies on circle $SXY$. Similarly, $V$ lies on this circle. This establishes (3).\n\nFinally, to prove (4), it is sufficient to show that $SU/UB = SV/VC$ and then apply Thales. As triangles $MSU$ and $MUB$ have the same $M$-altitude and share the side $MU$,\n$$\n\\frac{SU}{UB} = \\frac{\\text{area } MSU}{\\text{area } MUB} = \\frac{SM}{MB} \\cdot \\frac{\\sin \\angle SMU}{\\sin \\angle UMB} = \\frac{SM}{MB} \\cdot \\frac{\\sin \\angle AMN}{\\sin \\angle NMB},\n$$\nas $A$ lies on $SM$ by (2).\nTriangles $MAN$ and $MNB$ have equal areas, as they both have the same $M$-altitude, and $N$ is the midpoint of $AB$. These triangles also share the side $MN$, so\n$$\nMA \\cdot MN \\cdot \\sin \\angle AMN = 2 \\cdot \\text{area } MAN = 2 \\cdot \\text{area } MNB = MN \\cdot MB \\cdot \\sin \\angle NMB.\n$$\nHence $\\sin \\angle AMN / \\sin \\angle NMB = MB/MA$, so, by the preceding,\n$$\n\\frac{SU}{UB} = \\frac{SM}{MB} \\cdot \\frac{MB}{MA} = \\frac{SM}{MA}.\n$$\nSimilarly, $SV/VC = SM/MA$, so $SU/UB = SV/VC$, as stated. This establishes (4) and completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_n(\\mathbb{R})$ be a matrix with the property that\n$$\n\\det(A + X) \\ge \\det X, \\text{ for all } X \\in \\mathcal{M}_n(\\mathbb{R}).\n$$\n\nProve that $A = O_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be three real numbers such that $ab + bc + ca = 3$. Prove that\n$$\n\\frac{a^4 + b^4 + (a+b)^4}{a^2 + b^2 + ab} + \\frac{b^4 + c^4 + (b+c)^4}{b^2 + c^2 + bc} + \\frac{c^4 + a^4 + (c+a)^4}{c^2 + a^2 + ca} \\ge 18.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39271,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $a$, $x$, $y$ are such that $a > 0$, $x \\neq y$ and $x + \\sqrt{a - x^2} = y + \\sqrt{a - y^2}$ (C).\n\na) Prove that, if $x$ and $y$ are rational, then $a$ is rational.\n\nb) Prove that, if $a$ is a positive integer, then there exists a pair $(x, y)$, with $x$ and $y$ irrational, fulfilling conditions (C).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the numbers $n$ and $p$, with $n$ a positive integer and $p$ a prime, so that the numbers $\\frac{p-2}{n}$, $\\frac{n(n+2)}{p}$ and $\\frac{n(n+2)+p-2}{n+p}$ are integers.",
    "options": [],
    "answer": "Either p equals 2 with n any even positive integer, or p equals 3 with n equal to 1.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\ge 2$ be a given positive integer. Determine the number of pairs $(m, n)$ of positive integers, with $m, n < 2^k$, such that the equation $z^{2^k} - z^m - z^n = 1$ has at least one solution with modulus $1$.\nBogdan Blaga",
    "options": [],
    "answer": "(2^k - 1)(2^k - 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the lengths of the sides of a right triangle, if they are integers and the area of the triangle (in cm²) is 5 times larger than its perimeter (in cm).",
    "options": [],
    "answer": "The right triangles are: (21, 220, 221), (22, 120, 122), (25, 60, 65), (30, 40, 50), (28, 45, 53), and (24, 70, 74).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the real numbers $a$, $b$, $c$ and $d$ for which the inequality\n$$\n|x+a| + |y+b| + |z+c| + |x+y+z+d| \\geq |x| + |y| + |z| + |x+y+z|\n$$\nholds for any real numbers $x$, $y$ and $z$.",
    "options": [],
    "answer": "a = b = c = d = 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ and $I$ be the circumcenter, respectively the incenter of the triangle $ABC$. The incircle is tangent to the sides $BC$, $AC$ and $AB$ at the points $A'$, $B'$, respectively $C'$, and $G'$ is the centroid of the triangle $A'B'C'$. If $G'O = 7G'I$, prove that triangle $ABC$ is equilateral.\nMarian Ursărescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39277,
    "subject": "Mathematics (Multi-modal)",
    "question": "The acute triangle $ABC$ has the altitude $AD$ and $X$ is a point inside the segment $AD$. The circles $C_1$ and $C_2$ have their centres on the line $BC$ and contain the points $B$ and $X$, respectively $C$ and $X$. The circle $C_1$ meets again the altitude from $B$ of the triangle $ABC$ at $N$ and the side $AB$ at $M$. The circle $C_2$ meets again the altitude from $C$ of the triangle $ABC$ at $P$ and the side $AC$ at $Q$. Let $N^*$ be the intersection of the halfline $AN$ with $C_1$ and $P^*$ be the intersection of the halfline $AP$ with $C_2$.\nProve that $N^*$ and $P^*$ are on the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive integers $m$ and $n$ so that $n = 3^m + s(n) + 35$, where $s(n)$ is the sum of the digits of $n$.",
    "options": [],
    "answer": "No solutions",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer number. Determine all the values of $m \\in \\mathbb{N}$ such that there is $A \\in \\mathcal{M}_n(\\mathbb{C})$ with the property that $\\text{rank}(A^k) + \\text{rank}(A^{n-k}) = m$, for any $k \\in \\{1, 2, \\dots, n-1\\}$.",
    "options": [],
    "answer": "All m of the form 2r with 0 ≤ r ≤ n, together with m = n and m = n − 2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. Prove that the orthogonal projections of the point $A$ on the external bisectors of the angles $B$ and $C$, the orthogonal projections of the point $B$ on the external bisectors of the angles $A$ and $C$, and the orthogonal projections of the point $C$ on the external bisectors of the angles $A$ and $B$ are concyclic.\n\nPetru Braica",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation\n$$\n((((((((n - \\frac{1}{2}) \\cdot 2 - \\frac{2}{3}) \\cdot 3 - \\frac{3}{4}) \\cdot 4 - \\frac{4}{5}) \\cdot \\dots) \\cdot 2023 - \\frac{2023}{2024}) \\cdot 2024 = 1.\n$$",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two given natural numbers and $(x_n)_{n \\ge 0}$ the sequence defined by $x_n = n^2 + a n + b$ for any natural number $n$.\n\na) Prove that the sequence $(x_n)_{n \\ge 0}$ contains infinitely many perfect squares if and only if $a^2 = 4b$.\n\nb) Prove that the sequence $(x_n)_{n \\ge 0}$ does not contain an infinite arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no nonnegative real numbers $x$ such that $[nx]$ is an odd natural number for any positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p > 2$ be a prime number and $n \\in \\mathbb{N}^*$ such that $n \\equiv 1 \\pmod{p}$. If the ring $(A, +, \\cdot)$ has the property that $x^{n+p} = x^n$ for any $x \\in A$, prove that $x^{p+1} = x$ for any $x \\in A$.\n\nMihai Opincariu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and $A$ be a set with $n$ elements so that, for every $a, b \\in A, a \\neq b$, the number $a^2 + b\\sqrt{3}$ is rational.\na) Give an example of such a set $A$.\nb) Prove that, if $x \\in A$, then $x\\sqrt{3}$ is rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39286,
    "subject": "Mathematics (Multi-modal)",
    "question": "The path from Little Red Riding Hood to Grandma has ups, downs and flat portions. When walking, LRRH goes up two times slower and goes down two times faster than on flat ground. When riding her bike, she goes up three times slower and goes down three times faster than on flat ground.\n\nLRRH noticed that, when walking, the way back home from Grandma takes 18 minutes less than the way from home to Grandma, and when she rides her bike, the way back home from Grandma takes 20 minutes less than the way from home to Grandma. By what percentage is LRRH's speed larger by bike than by foot?",
    "options": [],
    "answer": "60%",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ and $g : \\mathbb{Q} \\to \\mathbb{Q}$ for which\n$$\ng(f(x+y)) = f(x) + y g(y) - a x, \\forall x, y \\in \\mathbb{Q},\n$$\nwhere $a \\in \\mathbb{Q}^*$ is given.",
    "options": [],
    "answer": "f(x) = a x for all rational x and g(x) = 0 for all rational x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to [0, \\infty)$ be a convex function with $f(0) = 0$. Show that\n\n$$\n\\left( \\int_{0}^{1} f(x) \\, dx \\right)^{2} \\leq \\frac{3}{4} \\int_{0}^{1} f^{2}(x) \\, dx .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. A 3-element subset $A$ of the set $M = \\{1, 2, 3, \\dots, n\\}$ will be called interesting if there exists a set $B$ of positive integers so that:\n(1) if $a \\in A, b \\in A, a \\neq b$, then $(a+b) \\in B$;\n(2) if $x, y, z \\in B, x < y < z$, then $2 \\cdot y = x + z$.\nDenote $A_n$ the number of the interesting sets.\n\na) Compute $A_7$.\n\nb) Prove that $A_{91} > 2024$.",
    "options": [],
    "answer": "A_7 = 9 and A_91 = 2025 (> 2024)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in $[1, \\infty) \\times \\mathbb{R}$ the following system of equations\n$$\n\\begin{cases}\nx + y = 2^x \\\\\nx^2 + y^2 = 2^{[y]}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(1, 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39291,
    "subject": "Mathematics (Multi-modal)",
    "question": "The irrational numbers $x, y$ are such that the numbers $x^2 + 2024y$, $y^2 + 2024x$ and $xy$ are rational. Find $x + y$.\nMihaela Berindeanu",
    "options": [],
    "answer": "2024",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39292,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Give an example of a function with the above property, which is primitivable and is non-identically vanishing on any interval included in $[0, 1]$.\n\nb) If $f$ is continuous on $(0, 1]$, show that $f$ is primitivable on $[0, 1]$.\n\nwhere $f : [0, 1] \\to \\mathbb{R}$ is an integrable function with $f(0) = 0$, such that\n$$\n\\int_{0}^{1} \\left( f\\left(\\frac{x}{n}\\right) + f\\left(\\frac{x}{n+1}\\right) \\right) dx = 0, \\quad \\text{for any } n \\in \\mathbb{N}^{*}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two integers. Prove that the number:\n$$\n\\prod_{k=1}^{n-1} \\left( a^2 + b^2 + 1 - 2a \\cos \\frac{2k\\pi}{n} - 2b \\sin \\frac{2k\\pi}{n} \\right)\n$$\nis a positive integer and it can be written as the sum of two perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDA'B'C'D'$ be a cube and $O$ its centre. Consider the points $M \\in [AD']$, $N \\in [B'C]$ and $P$ a point on the face $A'B'C'D'$. Denote $E$ and $F$ the midpoints of the edges $A'B'$, $C'D'$. Prove that $O$ is the barycentre of the triangle $PMN$ if and only if $P \\in [EF]$ and $\\frac{PF}{AM} = \\frac{PE}{CN} = \\sqrt{2}$.\nTraian Preda",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39295,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Give an example of a finite nonabelian group $G$, with unit element $e$, having a proper subgroup $H$ with the property that $x^2 = e$ for any $x \\in G \\setminus H$.\n\nb) Let $G$ be a finite group, and $H$ a proper subgroup such that $x^2 = e$ for any $x \\in G \\setminus H$. If $|G| > 2 \\cdot |H|$, show that the group $G$ is commutative.",
    "options": [],
    "answer": "a) For example, take G = S3 and H = A3; all elements outside H are transpositions and square to the identity. (Alternatively, G = the dihedral group of order 8 with H the rotation subgroup.) b) Under the given conditions, G is commutative.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many positive integers $n$ so that $n + s(n)$ is a perfect square, where $s(x)$ stands for the sum of the decimal digits of $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(A, +, \\cdot)$ be a finite ring and $a, b \\in A$ with $ab + ba = 0$. If there is an $n \\in \\mathbb{N}^*$, $n \\neq 2 \\pmod 4$, such that $a^n - b^n$ is invertible, prove that $a - b$ and $a + b$ are invertible.\nMihai Opincariu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39298,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides $BC$, $CA$, $AB$ of the equilateral triangle $ABC$ contain the points $D$, $E$, respectively $F$, different from the sides' midpoints, so that $BD = CE = AF$. Denote $M$ the midpoint of $EF$ and $N$ the intersection of the lines $BC$ and $AM$. Prove that the segments $ND$ and $BC$ have the same midpoint.\n\nGheorghe Iurea",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Consider in the complex plane a regular polygon $P_1P_2\\dots P_n$ inscribed in the unit circle, and let $A$ be the set of affixes of the vertices of the polygon. Let $a_1, a_2, \\dots, a_{n-1}$ be complex numbers with the property:\n$$\n|z^{n-1} + a_1z^{n-2} + a_2z^{n-3} + \\dots + a_{n-2}z + a_{n-1}| = 1, \\text{ for any } z \\in A.\n$$\nShow that $a_1 = a_2 = \\dots = a_{n-1} = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square, $M$ be the midpoint of the side $AD$, $T$ be the common point of the lines $BM$ and $CD$, and $CP \\perp BM$, $P \\in MB$. The perpendicular line from $A$ to $AP$ intersects the line $BM$ in $Q$. Prove that:\n\na) $\\angle APQ = \\angle PCQ = 45^\\circ$;\n\nb) $PQ = QT = PC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $F$ be the intersection of the lines $CP$ and $AB$ and $E$ be the foot of the perpendicular from $A$ to $BM$.\n![](attached_image_1.png)\nSince $\\angle FCB = \\angle MBA = 90^\\circ - \\angle CBM$ and $CB = BA$, the right triangles $CBF$ and $BAM$ are congruent, whence $FB = MA = \\frac{AB}{2}$, so $F$ is the midpoint of side $AB$.\nWe deduce that $FP$ is a midsegment in the triangle $AEB$, so $BP = EP$ and the congruence of the triangles $CPB$ and $BEA$ yields $PB = AE$ and $CP = BE$.\nThe triangle $EAP$ is right and isosceles, so $\\angle APE = 45^\\circ$, thus the triangle $APQ$ is also an isosceles right triangle. Consequently, the altitude $AE$ is also a median.\nWe obtain $QP = 2EP = EB = PC$, hence the triangle $PCQ$ is also an isosceles right triangle, from which follows that $\\angle PCQ = 45^\\circ$ and $CQ \\parallel AP$.\n\nb) Since $CD = CB$, $\\angle DCP = \\angle CBE$ and $PC = EB$, the triangles $DPC$ and $CEB$ are congruent, whence $DP = CE = CB = DA$.\nFrom the congruence of the triangles $DEA$ and $DEP$ (SSS), it follows that $\\angle ADE = \\angle PDE$, so $DE$ is the supporting line of the bisector and the altitude in the isosceles triangle $DAP$.\nSince $DE \\perp QC$, it follows that $CQ$ is the perpendicular bisector of the side $DE$ in the isosceles triangle $CDE$, thus $QD = QE = \\frac{QP}{2} = \\frac{PC}{2}$.\nTherefore $DQ$ is a midsegment in the triangle $TPC$, whence $TQ = QP = 2EP = EB = CP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $D$ and $E$ are considered on the ($BC$) side of triangle $ABC$, with $D$ between $B$ and $E$.\nAbout a point $R$ of the segment ($AE$) we will say that it is *remarkable* if the lines $PQ$ and $BC$ are parallel, where $\\{P\\} = DR \\cap AC$ and $\\{Q\\} = CR \\cap AB$.\nAbout a point $R'$ of the segment ($AD$) we will say that it is *remarkable* if the lines $P'Q'$ and $BC$ are parallel, where $\\{P'\\} = BR' \\cap AC$ and $\\{Q'\\} = ER' \\cap AB$.\na) If there is a remarkable point on the segment ($AE$), show that any point of the segment ($AE$) is remarkable.\nb) If each of the segments ($AD$) and ($AE$) contains a remarkable point, prove that $BD = CE = \\varphi \\cdot DE$, where $\\varphi = \\frac{1+\\sqrt{5}}{2}$ is the golden number.",
    "options": [],
    "answer": "BD = CE = φ · DE, where φ = (1 + √5) / 2",
    "solution": "a) Applying Menelaus' theorem in the triangle $ABE$ with transversal $Q - R - C$, we get that $\\frac{AQ}{QB} \\cdot \\frac{BC}{CE} \\cdot \\frac{ER}{RA} = 1$, therefore $\\frac{AQ}{QB} = \\frac{CE}{BC} \\cdot \\frac{RA}{ER}$. Similarly, applying Menelaus' theorem in the triangle $AEC$ with transversal $P - R - D$, we get that $\\frac{AP}{PC} = \\frac{DE}{CD} \\cdot \\frac{RA}{ER}$. We have:\n$$\nPQ \\parallel BC \\Leftrightarrow \\frac{AQ}{QB} = \\frac{AP}{PC} \\Leftrightarrow \\frac{CE}{BC} = \\frac{DE}{CD}.\n$$\nThis relation does not depend on the position of point $R$ on segment ($AE$), but only on the positions of points $D$ and $E$ on segment ($BC$). It follows that, if there is a remarkable point on the segment ($AE$), then any point of the segment ($AE$) is remarkable.\n\nb) Let's denote by $x, y$ and $z$ the lengths of the segments $BD, DE$ and $EC$, respectively. According to the above, there is a remarkable point on the segment ($AE$) if and only if\n$$\n\\frac{x+y+z}{z} = \\frac{y+z}{y} \\Leftrightarrow \\frac{x+y}{z} = \\frac{z}{y} \\Leftrightarrow z^2 = y^2 + xy.\n$$\nIn the same way, there is a remarkable point on the segment ($AD$) if and only if $x^2 = y^2 + yz$.\nSubtracting these equalities, it turns out that $z^2 - x^2 = y(x - z)$. In order not to have contrary signs in the two members, the condition yields $x = z$. We deduce that $x^2 - xy - y^2 = 0$, or $t^2 - t - 1 = 0$, where $t = \\frac{x}{y} > 0$.\nThe only positive solution of this equation is $t = \\varphi$, hence the requirement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39302,
    "subject": "Mathematics (Multi-modal)",
    "question": "The integers from $1$ to $49$ are written in a $7 \\times 7$ table, such that for any $k \\in \\{1, 2, ..., 7\\}$, the product of the numbers in the $k$-th row equals the product of the numbers in the $(8-k)$-th row.\n\na) Prove that there exists a row such that the sum of the numbers written on it is a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $p_1, p_2, ..., p_7$ be the products of the elements of the rows $1, 2, ..., 7$. We have $p_1 = p_7$, $p_2 = p_6$ and $p_3 = p_5$, so $49! = (p_1 p_2 p_3)^2 \\cdot p_4$, (1).\n\nNote that $49! = 2^{46} \\cdot 3^{22} \\cdot 5^{10} \\cdot 7^8 \\cdot 11^4 \\cdot 13^3 \\cdot 17^2 \\cdot 19^2 \\cdot 23^2 \\cdot 29 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 43 \\cdot 47$. From (1) we infer that the primes which have odd exponents in the prime factorization of $49!$ must appear in the prime factorization of one of the numbers situated on the $4$-th row, thus $13 \\cdot 29 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 43 \\cdot 47 \\mid p_4$. The product of any two of the primes $13, 29, 31, 37, 41, 43$ and $47$ exceeds $49$, therefore each of these primes divides exactly one of the $7$ numbers situated on the $4$-th row.\n\nThe number $\\frac{49!}{p_4}$ is a perfect square, hence if $p_4 > 13 \\cdot 29 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 43 \\cdot 47$, we have that $p_4 \\ge 2^2 \\cdot 13 \\cdot 29 \\cdot 31 \\cdot 37 \\cdot 41 \\cdot 43 \\cdot 47$, thus one of the numbers written in the cells of the table will be at least equal to $\\min\\{2^2 \\cdot 13, 2 \\cdot 29\\} > 49$, false.\n\nTherefore, the $4$-th row contains the numbers $13, 29, 31, 37, 41, 43, 47$, whose sum is the prime $241$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39303,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n3^{\\log_5(5x-10)} - 2 = 5^{-1+\\log_3 x}.\n$$",
    "options": [],
    "answer": "3, 27",
    "solution": "We notice that the equation has solutions $x_1 = 3$ and $x_2 = 27$. We show that the equation does not have any other solutions.\n\nContinuation A.\nUsing the properties of logarithms, the equation becomes $15 \\cdot 3^{\\log_5(x-2)} = 10 + 5^{\\log_3 x}$, or $15(x-2)^{\\log_5 3} = 10 + x^{\\log_3 5}$, with $x > 2$. Since $\\log_3 5 > 1$ and $0 < \\log_5 3 < 1$, the function $2 < x \\mapsto 15(x-2)^{\\log_5 3}$ is strictly concave, and the function $0 < x \\mapsto 10+x^{\\log_3 5}$ is strictly convex. Hence, the function $f : (2, \\infty) \\to \\mathbb{R}$, $f(x) = 15(x-2)^{\\log_5 3} - 10 - x^{\\log_3 5}$ is strictly concave, and the equation has at most two solutions.\n\nContinuation B.\nFrom the existence condition of logarithms, we have $x > 2$. We observe that the function $f : (2, \\infty) \\to (0, \\infty)$, $f(x) = 3^{\\log_5(5x-10)}$ has inverse which is $f^{-1}(x) = \\frac{5^{\\log_3 x}+10}{5}$, and the equation from the statement becomes $f(x) = f^{-1}(x)$.\nSince $f$ is strictly increasing, the equation from the statement is equivalent to $f(x) = x$, meaning $3^{\\log_5(5x-10)} = x$, or $\\log_5(5x-10) = \\log_3 x$. If $\\log_5(5x-10) = \\log_3 x = t$, then $x = 3^t = \\frac{1}{5}(5^t + 10)$, which is equivalent to solving the equation $5 \\cdot 3^t = 5^t + 10$, or $5 = \\left(\\frac{5}{3}\\right)^t + 10\\left(\\frac{1}{3}\\right)^t$, with $t > \\log_3 2$. Since the function $g : (\\log_3 2, \\infty) \\to \\mathbb{R}$, $g(t) = \\left(\\frac{5}{3}\\right)^t + 10\\left(\\frac{1}{3}\\right)^t$ is strictly convex, being the sum of strictly convex functions, the equation $g(t) = 5$ admits at most two solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function such that $f(x) + \\sin(f(x)) \\ge x$, for all $x \\in \\mathbb{R}$. Prove that\n$$\n\\int_{0}^{\\pi} f(x) \\, dx \\geq \\frac{\\pi^2}{2} - 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g : \\mathbb{R} \\to \\mathbb{R}$ defined by $g(x) = x + \\sin(x)$, for any $x \\in \\mathbb{R}$. Obviously, $g$ is differentiable on $\\mathbb{R}$, with $g'(x) > 0$, for all $x \\in ((2k-1)\\pi, (2k+1)\\pi)$, for $k \\in \\mathbb{Z}$. Thus $g$ is increasing on any interval $[(2k-1)\\pi, (2k+1)\\pi]$, with $k \\in \\mathbb{Z}$, so $g$ is increasing on $\\mathbb{R}$.\nAs $\\lim_{x \\to -\\infty} g(x) = -\\infty$ and $\\lim_{x \\to \\infty} g(x) = \\infty$, $g$ is one-to-one, with $g^{-1} : \\mathbb{R} \\to \\mathbb{R}$ continuous and increasing on $\\mathbb{R}$. The given inequality can thus be written\n$$\ng(f(x)) \\geq x, \\text{ for any } x \\in \\mathbb{R} \\iff f(x) \\geq g^{-1}(x), \\text{ for any } x \\in \\mathbb{R}.\n$$\nThen\n$$\n\\int_{0}^{\\pi} f(x) \\, dx \\geq \\int_{0}^{\\pi} g^{-1}(x) \\, dx .\n$$\nAs $g(0) = 0$ and $g(\\pi) = \\pi$, by Young inequality, we have\n$$\n\\begin{align*}\n\\int_{0}^{\\pi} g^{-1}(x) \\, dx &= \\pi \\cdot g(\\pi) - 0 \\cdot g(0) - \\int_{0}^{\\pi} g(x) \\, dx \\\\\n&= \\pi^2 - \\left( \\frac{\\pi^2}{2} - \\cos(\\pi) \\right) + (0 - \\cos(0)) = \\frac{\\pi^2}{2} - 2\n\\end{align*}\n$$\nwhich gives the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39305,
    "subject": "Mathematics (Multi-modal)",
    "question": "The non-nil real numbers $a$ and $b$ fulfill $3(a^6 + b^6) = a^2b^2(a^2b^2 + 9)$. Prove that at least one of the numbers $a$ and $b$ is irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The equality can be written $a^4b^4 - 3a^6 - 3b^6 + 9a^2b^2 = 0$, $a^4(b^4 - 3a^2) - 3b^2(b^4 - 3a^2) = 0$, $(a^4 - 3b^2)(b^4 - 3a^2) = 0$. So $a^4 = 3b^2$ or $b^4 = 3a^2$ and, since $a \\neq 0$ and $b \\neq 0$, $\\sqrt{3} = \\pm\\frac{a^2}{b}$ or $\\sqrt{3} = \\pm\\frac{b^2}{a}$.\nIf $a$ and $b$ are rational, then so is $\\sqrt{3}$, false. So the assumption is false, hence at least one of the numbers $a$ and $b$ is irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive integers $a$ and $b$, such that $\\frac{7^a - 5^b}{8}$ is a prime number.",
    "options": [],
    "answer": "(2, 2)",
    "solution": "For any natural number $k$, we have $5^{2k} = M_8 + 1$, $5^{2k+1} = M_8 + 5$, $7^{2k} = M_8 + 1$ and $7^{2k+1} = M_8 + 7$, therefore from $8 \\mid 7^a - 5^b$ we deduce that $a$ and $b$ are even. Denote $a = 2m$, $b = 2n$, with $m$ and $n$ positive integers. Then $7^{2m} - 5^{2n} = 8p$, i.e. $(7^m - 5^n)(7^m + 5^n) = 8p$, with $p$ a prime.\nIf $p = 2$, then $(7^m - 5^n)(7^m + 5^n) \\neq 16$.\nIf $p \\ge 3$, from $7^m - 5^n < 7^m + 5^n$, as $7^m - 5^n$ and $7^m + 5^n$ are even, we have the following situations:\n$$\n(1^\\circ) \\begin{cases} 7^m - 5^n = 4 \\\\ 7^m + 5^n = 2p \\end{cases}, \\qquad (2^\\circ) \\begin{cases} 7^m - 5^n = 2 \\\\ 7^m + 5^n = 4p \\end{cases}.\n$$\nCase (1°): From $7^m = M_3 + 1$ and $5^n = M_3 \\pm 1$, it follows that $7^m - 5^n = M_3$ if $n$ is even, and $7^m - 5^n = M_3 + 2$ if $n$ is odd. Since $4 = M_3 + 1$, there are no solutions in this case.\nCase (2°): By subtracting the equations, we find that $2p = 7^m - 1$. Since $3 \\mid 7^m - 1$, it follows that $3 \\mid 2p$, thus $p = 3$. We deduce that $m = n = 1$, $p = 3$, therefore the solution is $(a, b) = (2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the integers $n$ such that there is a permutation $a_1, a_2, \\dots, a_{1013}$ of the numbers $1012, 1013, \\dots, 2024$, so that $a_1n^{1012} + a_2n^{1011} + \\dots + a_{1012}n + a_{1013} = 0$.",
    "options": [],
    "answer": "n = -1",
    "solution": "If $n \\in \\mathbb{N}$, then $a_1n^{1012} + a_2n^{1011} + \\dots + a_{1012}n + a_{1013} > 0$, hence $n \\in \\mathbb{N}$ is not suitable.\n\nIf $n = -k$, with $k \\ge 2$, then $a_1n^{1012} + a_2n^{1011} + \\dots + a_{1012}n + a_{1013} = k^{1011}(a_1k - a_2) + k^{1009}(a_3k - a_4) + \\dots + k(a_{1011}k - a_{1012}) + a_{1013} > 0$, because $a_{1013} > 0$ and $a_i k - a_{i+1} \\ge 2a_i - a_{i+1} \\ge 21012 - 2024 = 0$, for every $i \\in \\{1, 2, \\dots, 1011\\}$. Hence, there are no solutions of this type.\n\nWe show that $n = -1$ is a solution.\nFor each $a \\in \\mathbb{Z}$, $(a - 1) - a - (a + 1) + (a + 2) = 0$. It follows that $(1012 - 2024 + 1013 - 1015 + 1014) + (1016 - 1017 - 1018 + 1019) + \\dots + (2020 - 2021 - 2022 + 2023) = 0$, hence $n = -1$ is the only solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 39308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I \\subset \\mathbb{R}$ be an open interval and consider $f : I \\to \\mathbb{R}$ a function that is twice differentiable on $I$, such that $f(x) \\cdot f''(x) = 0$, for any $x \\in I$. Show that $f''$ is the zero function.\n\nSorin Rădulescu and Mihai Piticari",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the set $A = \\{x \\in I \\mid f''(x) \\ne 0\\}$. If, by way of contradiction, $A \\ne \\emptyset$, as $f'' = (f')'$ has the intermediate point value on $I$, the set $A$ cannot be a singleton.\n\nSo, let $a, b \\in A$, $a < b$. We get $f(a) = f(b) = 0$. The function $g : I \\to \\mathbb{R}$, $g(x) = f(x)f'(x)$, for all $x \\in I$, is increasing on $I$ as $g'(x) = (f'(x))^2 + f(x)f''(x) = (f'(x))^2 \\ge 0$, for any $x \\in I$. As a consequence, by $g(a) = g(b) = 0$, we obtain $g(x) = 0$, for all $x \\in [a, b]$, so $g' = 0$. Then $f''(a) = \\lim_{x \\searrow a} \\frac{f'(x)-f'(a)}{x-a} = 0$, a contradiction. This shows that $A = \\emptyset$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$, with $n \\ge 2$, such that the equation\n$$\nx^2 - 3 \\cdot x + 5 = 0\n$$\nhas a unique solution in the ring $(\\mathbb{Z}_n, +, \\cdot)$.",
    "options": [],
    "answer": "11",
    "solution": "We shall denote by $M$ the set of all positive integers $n$, with $n \\ge 2$, such that the equation (1) has a unique solution in the ring $(\\mathbb{Z}_n, +, \\cdot)$.\nWe will show that $M = \\{11\\}$.\n\nIn the ring $(\\mathbb{Z}_{11}, +, \\cdot)$, the equation (1) can be equivalently written as\n$$\nx^2 - \\hat{3} \\cdot x + \\hat{5} = \\hat{0} \\iff x^2 - \\widehat{14} \\cdot x + \\widehat{49} = \\hat{0} \\iff (x - \\hat{7})^2 = \\hat{0},\n$$\nwith the unique solution $x = \\hat{7}$. Hence, $11 \\in M$.\n\nSince $k^2 - 3k + 5 = (k-1)(k-2) + 3$ is an odd number for any integer $k \\in \\mathbb{Z}$, the equation $x^2 - \\hat{3} \\cdot x + \\hat{5} = \\hat{0}$ has no solutions in the ring $(\\mathbb{Z}_n, +, \\cdot)$ for any even integer $n$, so that $M \\subseteq 2 \\cdot \\mathbb{N} + 1$.\n\nLet $n \\in M$ be arbitrary, and $x \\in \\mathbb{Z}_n$ the unique solution of the equation (1). For the element $y = \\hat{3} - x$ we have then\n$$\ny^2 - \\hat{3} \\cdot y + \\hat{5} = \\hat{9} - \\hat{6} \\cdot x + x^2 - \\hat{3} \\cdot \\hat{3} + \\hat{3} \\cdot x + \\hat{5} = x^2 - \\hat{3} \\cdot x + \\hat{5} = \\hat{0},\n$$\nso $y$ is also a solution of the equation (1). Because of the uniqueness condition, we deduce that $x = y = \\hat{3} - x$, or, equivalently, $\\hat{2} \\cdot x = \\hat{3}$. Since $n$ is odd, $\\hat{2}$ is invertible, and we obtain that $x = \\hat{3} \\cdot \\hat{2}^{-1}$.\n\nThe fact that $x = \\hat{3} \\cdot \\hat{2}^{-1}$ is a solution of the equation can be written equivalently, considering the fact that $n$ is odd, as:\n$$\n(\\hat{3} \\cdot \\hat{2}^{-1})^2 - \\hat{3} \\cdot (\\hat{3} \\cdot \\hat{2}^{-1}) + \\hat{5} = \\hat{0} \\iff \\hat{4} \\cdot ((\\hat{3} \\cdot \\hat{2}^{-1})^2 - \\hat{3} \\cdot (\\hat{3} \\cdot \\hat{2}^{-1}) + \\hat{5}) = \\hat{0} \\iff \\\\\n\\iff \\hat{9} - \\widehat{18} + \\widehat{20} = \\hat{0} \\iff \\widehat{11} = \\hat{0} \\iff n \\mid 11.\n$$\nSince $n \\ge 2$, we conclude that $n = 11$. Hence, $M \\subseteq \\{11\\}$.\n\nThen $11 \\in M \\subseteq \\{11\\}$ implies that $M = \\{11\\}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X \\in \\mathcal{M}_2(\\mathbb{C})$ be a matrix such that $X^{2023} = X^{2022}$. Prove that $X^3 = X^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $d = \\det(X)$ and $t = \\tr(X)$. From the assumption $X^{2023} = X^{2022}$, we obtain $d^{2023} = d^{2022}$. Therefore, $d \\in \\{0, 1\\}$.\n\nIf $d = 1$, then $X$ is invertible. Then $X^{2022}$ is also invertible. We find $X = I_2$. So, the relation $X^3 = X^2$ is verified.\n\nAssume now $d = 0$. The Cayley-Hamilton theorem implies $X^2 = tX$. So we obtain $X^{n+1} = t^n X, \\forall n \\in \\mathbb{N}^*$. Hence $t^{2022}X = t^{2021}X$. We get $t = 0$ or $t = 1$ or $X = O_2$.\n\nIf $t = 0$, then $X^2 = O_2$, so $X^3 = X^2 = O_2$.\n\nIf $t = 1$, then $X^2 = X$, so $X^3 = X^2$.\n\nIf $X = O_2$, then clearly $X^3 = X^2 = O_2$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to arrange the first 49 positive integers in a $7 \\times 7$ array, so that each cell contains exactly one number and no two primes are neighbours?\n\nTwo neighbours are numbers from cells with a common side or a common vertex.",
    "options": [],
    "answer": "Yes, it is possible.",
    "solution": "It is possible. The positive primes less than $49$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, that is $15$ numbers.\n![](attached_image_1.png)\nNumber the lines and the columns from $1$ to $7$, set the primes in $15$ of the $16$ squares with both odd indexes and the other $34$ numbers in the remaining cells.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39312,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a simple periodic decimal fraction $f$ has the reduced length equal to $n$ (where $n$ is a positive integer) if $f$ has a $n$-digit period and $f$ cannot be represented as a simple periodic decimal fraction with a period having less than $n$ digits. For instance, $0.(223)$ has the reduced length 3, while $0.(2323)$ has the reduced length 2, as $0.(2323) = 0.(23)$.\n\na) Prove that $f = 0.(2) \\cdot 0.(3)$ is a simple periodic fraction with reduced length 3.\n\nb) Does there exist two simple periodic fractions with reduced length 1, such that their product has reduced length also 1?\n\nc) Does there exist two simple periodic fractions with reduced length 3, such that their product has the reduced length also 3?",
    "options": [],
    "answer": "a) 0.(2) × 0.(3) = 2/27 = 0.(074), which has reduced length 3.\nb) Yes. Example: 0.(3) × 0.(6) = 0.(2), which has reduced length 1.\nc) Yes. Example: 0.(270) × 0.(370) = 0.(100), which has reduced length 3.",
    "solution": "a) $f = 0.(2) \\cdot 0.(3) = \\frac{2}{9} \\cdot \\frac{3}{9} = \\frac{2}{27} = 0.(074)$ is a fraction of reduced length 3.\n\nb) Yes. For an example, $0.(3) \\cdot 0.(6) = 0.(2)$.\n\nc) Yes. For example, $0.(270) \\cdot 0.(370) = \\frac{270}{9 \\cdot 3 \\cdot 37} \\cdot \\frac{370}{999} = \\frac{100}{999} = 0.(100)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39313,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 2024 cards of the same size, face-down on a table, on which the integers $1$, $2$, $3$, $\\dots$, $2024$ are written. We say that a card is a *winner* if it has a number divisible by $13$ or by $100$. What is the minimum number of cards we need to turn face up to make sure that we obtain at least one winner?",
    "options": [],
    "answer": "1851",
    "solution": "There are $155$ multiples of $13$ not larger than $2024$: $13 \\cdot 1$, $13 \\cdot 2$, $\\dots$, $13 \\cdot 155$.\n\nThere are $20$ multiples of $100$ not larger than $2024$: $100 \\cdot 1$, $100 \\cdot 2$, $\\dots$, $100 \\cdot 20$.\n\nThere is only one common multiple of $13$ and $100$ not larger than $2024$, namely $1300$.\n\nThere are $155 + 20 - 1 = 174$ numbers not larger than $2024$, that are multiples of $13$ or multiples of $100$. This leaves $2024 - 174 = 1850$ non-winners.\n\nWe are sure that we have obtained a winner as soon as we pick $1850 + 1 = 1851$ cards.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b \\in \\mathbb{R}$, $a > 1$, $b > 0$. Determine the minimum value of the real number $\\alpha$ such that:\n$$\n(a + b)^x \\geq a^x + b, \\quad \\forall x \\geq \\alpha.\n$$",
    "options": [],
    "answer": "1",
    "solution": "The given relation is equivalent to:\n$$\n\\left(1 + \\frac{b}{a}\\right)^x - b\\left(\\frac{1}{a}\\right)^x \\geq 1, \\quad \\forall x \\geq \\alpha.\n$$\nConsidering the function $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = (1 + \\frac{b}{a})^x - b(\\frac{1}{a})^x$, we can notice that the function is increasing, being a sum of two increasing functions. Therefore, because $f(1) = 1$, we obtain that $f(x) \\geq f(1) = 1$, for each $x \\geq 1$, while $f(x) < f(1) = 1$, for each $x < 1$. Therefore, the minimum value of $\\alpha$ is 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \\times 2^n$ chessboard; those bishops are numbered from 1 to $2^n$, from left to right. A jump is a simultaneous move made by all bishops such that the following conditions are satisfied:\n* each bishop moves diagonally, in a straight line, some number of squares, and\n* at the end of the jump, the bishops all stand in different squares of the same row.\nFind the total number of permutations $\\sigma$ of the numbers 1, 2, \\ldots, $2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\\sigma(1), \\sigma(2), \\ldots, \\sigma(2^n)$, from left to right.",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "*First solution.* The required number is $2^{n-1}$. On a jump, every bishop moves the same number of rows up or down; call this number of rows the *length* of the jump.\n\n**Step 1.** We show that the length of any jump is of the form $2^d$ for some integer $d \\le n-1$. Assign each bishop the number of the column it is situated on before the jump. Let $k$ be the length of the jump; then each bishop's column number either increases by $k$, or decreases by $k$ in the jump.\nThus, bishops 1, 2, \\ldots, $k$ should move to columns $k+1$, $k+2$, \\ldots, $2k$, as they cannot move leftwards. On the other hand, after the jump columns 1, 2, \\ldots, $k$ should be filled by the bishops $k+1$, $k+2$, \\ldots, $2k$. So the leftmost $2k$ bishops still fill the columns 1, 2, \\ldots, $2k$ after the jump.\n\nRepeating the argument shows that the next $k$ bishops move rightwards, and the next $k$ bishops beyond move leftwards, and so on and so forth. Finally, the bishops all split into contiguous groups of length $2k$, and in each group the leftmost $k$ bishops move rightwards, whereas the rightmost $k$ bishops move leftwards. Hence $2k \\mid 2^n$, so $k$ is indeed of the form $2^d$ with $d \\le n-1$.\n\n**Step 2.** To make a more explicit description of the column change during the jump, assign each column the $n$-digit binary expansion of less 1 its number, augmented with zeroes leftwards if necessary. It is then easily seen that a jump of length $2^d$ just switches the $d$-th digit from the left, 0 to 1 and vice versa.\nThus, the resulting permutation also has the following form: For every $d = 0, 1, \\dots, n-1$, the $d$-th digit is either swapped for all bishops, or it is preserved for them all.\nMoreover, notice that the total length of all jumps is odd, so there will be an odd number of jumps of length 1. Hence the 0-th (the rightmost) digit will be switched anyway. This leaves the room for $2^{n-1}$ possible permutations.\n\n**Step 3.** It remains to show that all $2^{n-1}$ permutations are indeed possible. Let us show how to reach any of them.\nStart by getting to the bottom row by downward jumps of lengths 1, 2, 4, ..., $2^{n-1}$ that will switch all $n$ digits.\nNow, if we want to switch the $i$-th digit back, $1 \\le i \\le n-1$, make two upward jumps of length $2^{i-1}$, followed by a downward jump of length $2^i$. Combine such modifications for all possible digit combinations to get all desired permutations.\n*Second solution.* Proceed until the end of Step 1 just like in the first solution. Then extend the board to a vertical strip of width $2^n$, this will not affect the result, as it will be seen at the end of the proof.\nWe will show that any two jumps commute. Consider two jumps of length $p$ and $q$ with $p < q$, and call them the $p$-jump and the $q$-jump. As described in the first step, the bishops will be split in contiguous groups. For the $p$-jump, we look at groups of length $p$, call these $p$-groups; for the $q$-jump, we look at groups of length $q$, call these $q$-groups.\nSince $2p$ divides $q$, any $p$-group is fully contained in a single $q$-group, and a $q$-group contains an even number of $p$-groups. First, let's look at the first two $q$-groups, and denote by $g_1, \\dots, g_{2k}$ the $p$-groups contained in the first $q$-group, and $g'_1, \\dots, g'_{2k}$ the $p$-groups contained in the second $q$-group, where $2k = q/p$. The $p$-jump will swap $g_{2i-1}$ with $g_{2i}$, and same for their $g'$ counterparts, whereas a $q$-jump will swap $g_j$ with $g'_j$. When putting these together, it follows that applying both jumps in either order gives the same result: $g_{2i-1}$ is swapped with $g'_{2i}$ and $g_{2i}$ is swapped with $g'_{2i-1}$.\nRepeat now the same argument for the next two $q$-groups, and so on and so forth, until the entire row will be accounted for.\nWe will now establish a bijection from the odd numbers between 1 and $2^n - 1$ to the desired permutations. Let $x = \\sum 2^k a_k$ be an odd number between 1 and $2^n - 1$, where each $a_k$ is either 0 or 1, in particular $a_0 = 1$. Perform $a_k$ jumps of length $2^k$ in increasing order of $k$, then perform $2^n - 1 - x$ additional jumps of length 1 in order to reach the final row. This will result in a permutation $\\sigma$, and set $f(x) = \\sigma$. This provides a well defined function $f$.\nTo prove $f$ injective, it suffices to look at bishop numbered 1 and show that it will end up in position $x+1$. For any of jump of length $2^k$, this bishop will move rightwards, as its position just before the jump was $1+a_0+2a_1+...+2^{k-1}a_{k-1} \\le 2^k$. Therefore, before the additional jumps of length 1, this bishop will reach position $x+1$. Any two jumps of same length cancel each other out, and there is an even number of additional jumps of length 1, so the final position will also be $x+1$. Consequently, $f$ is injective.\nTo prove $f$ surjective, consider a $\\sigma$ with the desired property, let $b_k$ be the number of jumps of length $2^k$, and let $a_k$ be the remainder of $b_k$ modulo 2. Since the total length of all jumps is odd, there has to be an odd number of jumps of length 1, so $a_0 = 1$. Let $x = \\sum 2^k a_k$, and note that this is an odd number between 1 and $2^n - 1$. From Step 2 in Solution 1, the order of the jumps does not matter. Since two consecutive jumps of same length cancel each other out, performing $a_k$ jumps of length $2^k$ is the same as performing $b_k$ jumps of length $2^k$. So $f(x) = \\sigma$, as the $2^n - 1 - x$ additional jumps of length 1 at the end also cancel each other out.\n*Third solution.* Run again Step 1 through in Solution 1.\nLook at the two halves $1, \\dots, 2^{n-1}$ and $2^{n-1}+1, \\dots, 2^n$ and let $h(i) = i \\pm 2^{n-1}$ be the counterpart of $i$ in the other half, where the sign is chosen appropriately. A jump of length $2^{n-1}$ will swap the halves between themselves, so $i$ will be swapped with $h(i)$. From Step 1, it follows that any shorter jump will only perform swaps inside a single half, and will act the same way on the other half; specifically, if a jump swaps $i$ with $j$, it will also swap $h(i)$ with $h(j)$.\nFurthermore, applying two jumps of $2^{n-1}$ will just cancel each other out, regardless of any other jumps in between, because we swapped $i$ with $h(i)$ twice, and the inner configuration of each half is changed in the same way.\nInduct now on $n$. There are $2^{n-2}$ possible permutations for the $2^{n-1} \\times 2^{n-1}$ board. Performing jumps of the same length on a $2^n \\times 2^n$ board gives the same configuration in each of the two halves. Now we can either apply a jump of length $2^{n-1}$, which will swap the halves, or we can apply two jumps of length $2^{n-2}$, which will cancel each other. This provides a construction for $2^{n-1}$ permutations in the $2^n \\times 2^n$ board.\nTo show that these are the only ones, consider now a valid permutation for the $2^n \\times 2^n$ board which is obtained from some jumps. First, discard the jumps of length $2^{n-1}$, then attempt to apply the rest on the $2^{n-1} \\times 2^{n-1}$ board. If a jump would exit the board, then make the jump of the same length in the opposite direction instead, which will stay on the board because its length is at most $2^{n-2}$. Since the length is the same, the resulting bishop configuration is also the same. Since the total length of the moves executed so far is odd, we can make an even number of moves of length 1 in order to reach the final row of the $2^{n-1} \\times 2^{n-1}$ board, and this will not change the configuration in the end. Therefore, we obtain a corresponding permutation for the $n-1$ case which describes the configuration in each of the halves. From there, the only variations are whether the halves are swapped or not, depending on whether the number of jumps of length $2^{n-1}$ was odd or even. So this valid permutation corresponds to one constructed in the earlier paragraph, which completes the induction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 39316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral in which the diagonals intersect at $O$. Given that $\\overrightarrow{AB} + \\overrightarrow{AD} + \\overrightarrow{AO} = \\overrightarrow{BC} + \\overrightarrow{DC} + \\overrightarrow{OC}$, prove that $ABCD$ is a parallelogram.\n\nMarius Dolcan",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We obtain that the points $M$, $N$ and $O$ are collinear, so $M = N = O$. Thus point $O$ is the midpoint of each of the diagonals, therefore $ABCD$ is a parallelogram.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x, y, z$ so that $2^x - 2^y - 2^z = 1023$.",
    "options": [],
    "answer": "(x, y, z) = (11, 10, 0) or (11, 0, 10)",
    "solution": "If none of $x, y, z$ is nil, then the left term of the equation is even, so it cannot be equal to $1023$. Hence, at least one of $x, y, z$ is nil.\nFrom $2^x = 2^y + 2^z + 1023$ follows that $x \\ge 11$, therefore $y = 0$ or $z = 0$.\nThe case $z = 0$ yields $2^x - 2^y = 1024$, so $2^y \\cdot (2^{x-y} - 1) = 2^{10}$, (1). From $2^x - 2^y > 0$ follows $x > y$, so $2^{x-y} - 1$ is odd. From (1), $2^y = 2^{10}$ and $2^{x-y} - 1 = 1$, whence $y = 10$ and $x = 11$.\nA similar argument works for $y = 0$, so the solutions are $x = 11, y = 10, z = 0$ and $x = 11, y = 0, z = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39318,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any real number $x$, let $A(x) = x^2 + 4\\lfloor x \\rfloor$.\n\na) Find the real numbers $x$ for which $A(x) = \\{x\\}^2$.\n\nb) Find the real numbers $y > 0$ for which $A(y)$ is the square of a natural number.",
    "options": [],
    "answer": "a) x ∈ [0, 1) ∪ {−4, −9/2}.\nb) All y of the form y = sqrt(k^2 + 4) with integers k ≥ 2.",
    "solution": "a) $(\\lfloor x \\rfloor + \\{x\\})^2 + 4\\lfloor x \\rfloor = \\{x\\}^2$ yields $\\lfloor x \\rfloor^2 + 2\\lfloor x \\rfloor\\{x\\} + 4\\lfloor x \\rfloor = 0$, (*).\nIf $x \\ge 0$, then $\\lfloor x \\rfloor = 0$, thus any $x \\in [0, 1)$ is a solution.\nIf $x < 0$, equality (*) leads to $\\{x\\} = -\\frac{\\lfloor x \\rfloor^2 + 4\\lfloor x \\rfloor}{2\\lfloor x \\rfloor} = -\\frac{\\lfloor x \\rfloor + 4}{2}$. Since $\\{x\\} \\in [0, 1)$, we conclude that $\\lfloor x \\rfloor \\in \\{-5, -4\\}$, thus obtaining the solutions $x_1 = -\\frac{9}{2}$, $x_2 = -4$, which verify the equality.\nTherefore, $x \\in \\{-\\frac{9}{2}, -4\\} \\cup [0, 1)$.\n\nb) For $y \\in (0, 1)$ we have $A(y) = y^2 \\in (0, 1)$, $A(1) = 5$, and for $y \\in [1, 2)$ we have $A(y) = y^2 + 4 \\in (5, 8)$, thus $A(y)$ is not a perfect square for $y \\in (0, 2)$.\nFor $y \\ge 2$, the condition from the statement leads to $y^2 \\in \\mathbb{N}$, therefore $y = \\sqrt{m}$, with $m \\in \\mathbb{N}$, $m \\ge 4$, leading to $m + 4\\lfloor\\sqrt{m}\\rfloor = p^2$, with $p \\in \\mathbb{N}$.\nIf $k = \\lfloor\\sqrt{m}\\rfloor$, then $k \\ge 2$ and $k^2 \\le m < (k+1)^2$, hence $m + 4k = p^2$.\nUsing the fact that $k^2 + 4k \\le p^2 < (k+1)^2 + 4k = k^2 + 6k + 1 < (k+3)^2$, it follows that (1) $p^2 = (k+1)^2 = 1$, for $k=0$ or (2) $p^2 = 9$ for $k=1$, values that do not verify the equality in the statement, or (3) $p^2 = (k+2)^2$, for $k \\ge 2$.\n\nTherefore, $p = k + 2$ and $m = (k + 2)^2 - 4k = k^2 + 4$. Thus, the numbers that verify the equality in the statement are $y = \\sqrt{k^2 + 4}$, $k \\in \\mathbb{N}$, $k \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. We define the numbers $A = 33...3$, with $n$ digits $3$, and $B = 20 \\cdot A + 6$. Find all the digits that form the number $A \\cdot B$.",
    "options": [],
    "answer": "1, 2, 7, 8, 9",
    "solution": "Notice that $3 \\cdot A = 99...9 = 10^n - 1$. Then $A \\cdot B = (10^n - 1) \\cdot \\underbrace{22...2}_{n+1 \\text{ digits}} = \\underbrace{22...2}_{n+1 \\text{ digits}} \\underbrace{00...0}_{n \\text{ digits}} - \\underbrace{22...2}_{n+1 \\text{ digits}}$. Thus $A \\cdot B = \\underbrace{22...2}_{n-1 \\text{ digits}} \\underbrace{1977...78}_{n-1 \\text{ digits}}$.\nSo, the digits of $A \\cdot B$ are $1$, $2$, $7$, $8$ and $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two sets $A$ and $B$ of real numbers that have the following properties:\na. $0 \\in A$;\nb. if $1+x \\in A$, then $\\sqrt{1+x+x^2} \\in B$;\nc. if $\\sqrt{x^2 - x + 1} \\in B$, then $2+x \\in A$.\nProve that $\\sqrt{3}$, $\\sqrt{13}$, $\\sqrt{31}$ are elements of the set $B$ and $2024 \\in A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $1 + (-1) = 0 \\in A$, according to (b) we obtain $1 \\in B$ and, since $1 = \\sqrt{0^2 - 0 + 1} \\in B$, we infer from (c) that $2 \\in A$, from which $\\sqrt{3} \\in B$.\n\nSince $\\sqrt{2^2 - 2 + 1} = \\sqrt{3} \\in B$, it follows from (c) that $2 + 2 = 4 \\in A$ and, from (b), we infer $\\sqrt{13} \\in B$.\n\nSince $\\sqrt{4^2 - 4 + 1} = \\sqrt{13} \\in B$, we further infer that $2 + 4 = 6 \\in A$ and thus $\\sqrt{1+5+5^2} = \\sqrt{31} \\in B$.\n\nUsing the equality $1+x+x^2 = (x+1)^2 - (x+1)+1$, we have that if $1+x \\in A$, then $3+x \\in A$. Since $0 \\in A$, $2 \\in A$ and it follows that the set $A$ contains all the even numbers. In particular, $2024 \\in A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ with $\\angle BAC = 120^\\circ$ and the isosceles triangles $PAB$ and $NAC$ such that $\\angle APB = \\angle ANC = \\angle BAC$, with line $AB$ separating points $P$ and $C$, and line $AC$ separating points $N$ and $B$. Prove that, if $G$ is the centroid of triangle $ABC$, then $GP = GN = \\frac{AB+AC}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nOn the other hand, from $\\angle APB = 120^\\circ$ and $\\angle APF = 30^\\circ$, it follows that $\\angle FPB = 90^\\circ$, hence the triangle $PBF$ is a $30^\\circ - 60^\\circ - 90^\\circ$ triangle, from which $BF = 2FP$, (2). Using (1) and (2) we conclude $BF = 2FA$, (3).\nDenote $BB'$ the median from $B$ of the triangle $ABC$. Since $G$ is the centroid of the triangle $ABC$, we have $BG = 2GB'$ (4).\nThe relations (3) and (4) lead, according to the converse of Thales' theorem, to $FG \\parallel AC$, thus points $P, F$ and $G$ are collinear. We obtain $\\angle GPN = 30^\\circ$. Similarly, it follows that $\\angle GNP = 30^\\circ$ hence the triangle $GNP$ is isosceles, with $GP = GN$.\nSince $AFGE$ is a parallelogram, $GP = GF + FP = AE + FP = \\frac{AB+AC}{3}$, from which the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be an inscriptible pentagon for which $AB = BC = CD$ and the centroid of the pentagon coincides with the center of the circumscribed circle. Show that the pentagon $ABCDE$ is regular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider an orthonormal coordinate system centred in $O$, where $O$ is the centre of the circumcircle $C$ of the pentagon $ABCDE$, with the unit length equal to the radius of the circle $C$ and with the real axis being the perpendicular bisector of the segment $BC$. Let $z_X$ denote the complex number representing the position of point $X$ in this coordinate system.\n\nSince $AB = BC = CD$, we have $\\angle AOB = \\angle BOC = \\angle COD = 2\\alpha \\in (0, \\frac{2\\pi}{3})$. It follows that the positions of the vertices of the pentagon are $z_A = \\cos 3\\alpha + i \\sin 3\\alpha$, $z_B = \\cos \\alpha + i \\sin \\alpha$, $z_C = \\cos \\alpha - i \\sin \\alpha$, $z_D = \\cos 3\\alpha - i \\sin 3\\alpha$, and $z_E = \\cos \\beta + i \\sin \\beta$, with $\\beta \\in (3\\alpha, 2\\pi - 3\\alpha)$.\n\nIf the centroid of the pentagon coincides with $O$, then $z_A+z_B+z_C+z_D+z_E = 0$, from which we obtain $\\cos \\beta + 2 \\cos 3\\alpha + 2 \\cos \\alpha = 0$ and $\\sin \\beta = 0$. We get $\\beta = \\pi$, along with $-\\sin \\alpha + 2 \\sin \\alpha \\cos 3\\alpha + 2 \\sin \\alpha \\cos \\alpha = 0$ (by multiplying by $\\sin \\alpha$) and, by transforming sums into products, $\\sin 4\\alpha = \\sin \\alpha$.\n\nSince $0 < \\alpha < 4\\alpha < \\frac{4\\pi}{3}$, we deduce $4\\alpha = \\pi - \\alpha$, so $\\alpha = \\frac{\\pi}{5}$, from which $\\angle AOB = \\angle BOC = \\angle COD = \\angle DOE = \\angle EOA = \\frac{2\\pi}{5}$, meaning $ABCDE$ is a regular pentagon.\nWe consider an orthonormal coordinate system centred in $O$, where $O$ is the centre of the circumcircle $C$ of the pentagon $ABCDE$, with the unit length equal to the radius of the circle $C$ and with the real axis being the line $OA$. Let $z_X$ denote the complex number representing the position of the point $X$ in this coordinate system.\n\nSince $AB = BC = CD$, we have $\\angle AOB = \\angle BOC = \\angle COD$. It follows that the positions of the vertices of the pentagon are $z_A = 1$, $z_B = z$, $z_C = z^2$, $z_D = z^3$, and $z_E = w$, with $z = \\cos \\alpha + i \\sin \\alpha$, $\\alpha \\in (0, \\frac{2\\pi}{3})$, and $w = \\cos \\beta + i \\sin \\beta$, $\\beta \\in (3\\alpha, 2\\pi)$.\n\nIf the centroid of the pentagon coincides with $O$, then $z_A+z_B+z_C+z_D+z_E = 0$, from which we obtain $1+z+z^2+z^3+w=0$. We get $w = \\frac{1-z^4}{z-1}$, from which $\\overline{w} = \\frac{1-z^4}{z-1}$, and using $w\\overline{w} = z\\overline{z} = 1$, we deduce the relation $\\frac{1-z^4}{z-1} \\cdot \\frac{z^4-1}{z^3(1-z)} = 1$.\n\nPerforming the calculations, we obtain $(z^5-1)(z^3-1) = 0$. The only solution of this equation with an argument in $(0, \\frac{2\\pi}{3})$ is $\\cos \\frac{2\\pi}{5} + i \\sin \\frac{2\\pi}{5}$, so $\\angle AOB = \\angle BOC = \\angle COD = \\angle DOE = \\angle EOA = \\frac{2\\pi}{5}$, meaning $ABCDE$ is a regular pentagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two real numbers in the interval $(0, 1)$, so that $a$ is a rational number and\n$$\n\\{na\\} \\ge \\{nb\\}, \\text{ for every natural number } n.\n$$\n\nProve that $a = b$. (We denote by $\\{x\\}$ the fractional part of the real number $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = \\frac{p}{q}$, where $p$ and $q$ are non-zero natural numbers, relatively prime, with $p < q$.\nThen $0 = \\{qa\\} \\ge \\{qb\\} \\ge 0$, thus $qb$ is a natural number. It follows that $b = \\frac{s}{q}$, where $s$ is a nonzero natural number.\n\nConsidering $n = q - 1$ in the relation from the hypothesis, we get that $\\{-\\frac{p}{q}\\} \\ge \\{-\\frac{s}{q}\\}$, that is $1 - \\frac{p}{q} \\ge 1 - \\frac{s}{q} \\iff p \\le s$.\nSo $p = s$, hence the requirement of the problem.\n\n\nAlternative solution. Analogously to solution 1, we have $a = \\frac{p}{q}$, $b = \\frac{s}{q}$, $0 < s \\le p < q$, $(p, q) = 1$.\nThen there is a natural number $k$ such that $kp \\equiv 1 \\pmod{q}$.\nFor this $k$ we have $\\{ka\\} = \\{k \\cdot \\frac{p}{q}\\} = \\frac{1}{q}$. Then $\\frac{1}{q} = \\{ka\\} \\ge \\{kb\\} = \\{k \\cdot \\frac{s}{q}\\}$, from which it follows that the number $\\{k \\cdot \\frac{s}{q}\\}$ is equal to either $0$ or $\\frac{1}{q}$.\nIf $\\{k \\cdot \\frac{s}{q}\\} = 0$, then $q \\mid ks$; since $(q, k) = 1$, it means that $q \\mid s$, false.\nIf $\\{k \\cdot \\frac{s}{q}\\} = \\frac{1}{q}$, then $ks \\equiv 1 \\pmod{q}$, therefore $q \\mid k(p-s)$; since $(q, k) = 1$, it means that $q \\mid p-s$. But $0 \\le p-s < q$, therefore $p-s = 0$. It turns out that $p = s$, therefore $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(\\mathbb{K}, +, \\cdot)$ be a field having the property that $x^2y = yx^2$, for all $x, y \\in \\mathbb{K}$. Prove that $(\\mathbb{K}, +, \\cdot)$ is commutative.\n\nSorin Rădulescu and Mihai Piticari",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $2 \\cdot 1 = (a+1)^2 - a^2 - 1 \\in Z(\\mathbb{K})$, for $a \\in \\mathbb{K}$. If the characteristic $char(\\mathbb{K})$ of $\\mathbb{K}$ is not $2$, then $2 \\cdot 1$ is an invertible element in $Z(\\mathbb{K})$, such that\n$$\na = (2 \\cdot 1)^{-1} \\cdot (2 \\cdot a) \\in Z(\\mathbb{K}), \\text{ for any } a \\in \\mathbb{K},\n$$\nimplying that $\\mathbb{K}$ is abelian.\n\nConsider the case $char(\\mathbb{K}) = 2$. For $a, b \\in \\mathbb{K}$ we have $a = (a+b)^2 - a^2 - b^2 \\in Z(\\mathbb{K})$. As $w \\cdot ab = (ab)^2 + ab^2a = (ab)^2 + a^2b^2 \\in Z(\\mathbb{K})$, if $w \\neq 0$, then $a, b \\neq 0$ and $ab = w^{-1} \\cdot ((ab)^2 + a^2b^2) \\in Z(\\mathbb{K})$. Consequently $a \\cdot (ab) = (ab) \\cdot a = a \\cdot (ba)$, so that, $ab = ba$.\nIn case $w = 0$, as $1 = -1$, we get $ab = -ba = ba$, that is $ab = ba$ for any $a, b \\in \\mathbb{K}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 2$ be an integer number. Prove that the sequence $(x_n)_{n \\ge 1}$, defined by $x_1 = a > 0$ and the recurrence relation $x_{n+1} = x_n + \\lfloor \\frac{p}{x_n} \\rfloor$, $n \\in \\mathbb{N}^*$, is convergent. Determine its limit depending on the values of the parameter $a$. We denote by $[x]$ the integer part of the real number $x$.\nEmil Vasile",
    "options": [],
    "answer": "lim x_n =\n- a, if a > p;\n- a + floor(p/a), if 0 < a < 1;\n- p + {a}, if 1 ≤ a ≤ p and a is not an integer (where {a} is the fractional part of a);\n- p + 1, if a ∈ {1,2,…,p}.",
    "solution": "The sequence $(x_n)_{n \\ge 1}$ has positive terms (proof by induction). There is $k \\in \\mathbb{N}^*$ such that $x_k > p$. Indeed, if we assume, by reductio ad absurdum, that $x_n \\le p, \\forall n \\in \\mathbb{N}^*$, we obtain $x_{n+1} \\ge x_n+1, \\forall n \\in \\mathbb{N}^*$, so $x_n \\ge a+n-1, \\forall n \\in \\mathbb{N}^*$. In particular, we find $x_{p+1} \\ge a+p > p$. Contradiction. Let us denote $k_0 = \\min\\{k \\in \\mathbb{N}^* \\mid x_k > p\\}$. Since $\\lfloor \\frac{p}{x} \\rfloor = 0, \\forall x > p$, we get $x_n = x_{k_0}, \\forall n \\ge k_0$. Therefore $(x_n)_{n \\ge 1}$ is convergent, with $\\lim_{n \\to \\infty} x_n = x_{k_0}$. For the value of the limit, we analyze three cases.\n\n**Case 1.** $a \\in (p, \\infty)$. Then $\\lim_{n \\to \\infty} x_n = x_1 = a$.\n\n**Case 2.** $a \\in (0, 1)$. Then $x_2 = a + \\left[\\frac{p}{a}\\right] > \\left[\\frac{p}{a}\\right] \\ge p$, hence $\\lim_{n \\to \\infty} x_n = x_2 = a + \\left[\\frac{p}{a}\\right]$.\n\n**Case 3.** $a \\in [1, p]$. The sequence has the terms $x_n = \\{a\\} + y_n$, where $\\{a\\} = a - [a] \\in [0, 1)$ is the fractional part of $a$, and $y_n \\in \\mathbb{N}^*$. We have $(x-1)(x-p) \\le 0$, for all $x \\in [1, p]$. It follows that $x + \\left[\\frac{p}{x}\\right] \\le x + \\frac{p}{x} \\le p+1$, for all $x \\in [1, p]$. Therefore, if $x_n \\in [1, p]$, then $x_{n+1} \\le p+1$. Thus $x_{k_0} \\in (p, p+1] \\cap \\{\\{a\\} + k \\mid k \\in \\mathbb{N}^*\\}$. We find\n$$\n\\lim_{n \\to \\infty} x_n = x_{k_0} = \\begin{cases} p + \\{a\\}, & a \\in [1, p] \\setminus \\mathbb{N} \\\\ p+1, & a \\in \\{1, 2, \\dots, p\\} \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a positive integer $n$ and the set $A_n = \\{1, 3, 5, \\dots, 2n - 1\\}$. For each pair $(a, b)$, where $a, b \\in A_n$ we construct the concatenated number $m = \\overline{ab}$, obtained by joining the numbers $a$ and $b$. For instance, for $19, 37 \\in A_{30}$, the concatenated number is $m = 1937$.\n\na) What is the smallest number $n \\in \\mathbb{N}^*$ for which we get at least a perfect square?\n\nb) Find the largest perfect square that can be obtained for $n = 50$.",
    "options": [],
    "answer": "a) n = 11; b) 7921",
    "solution": "a) We cannot obtain a perfect square by concatenating two elements from the set $A_{10} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\\}$.\n\nThe elements $1$ and $21$ from the set $A_{11} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21\\}$ yield the perfect square $121$. Hence, the answer is $n = 11$.\n\nb) By concatenating two elements from $A_{50} = \\{1, 3, 5, \\dots, 97, 99\\}$ we can obtain perfect squares with at most four digits.\n\nThe largest such perfect squares are $99^2 = 9801$, $97^2 = 9409$, $95^2 = 9025$, $93^2 = 8649$, $91^2 = 8281$. They are not acceptable, as the first two digits form an even number.\n\nThe number $89^2 = 7921$ is obtained joining $79$ and $21$, where $79, 21 \\in A_{50}$. In conclusion, the largest perfect square is $7921$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39327,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that $12n+13$ and $13n+14$ are coprime, for every natural $n$.\n\nb) Find the number of the pairs $(a, b)$ of natural numbers for which there exists a natural number $n$ so that $\\frac{a}{b} = \\frac{12n+13}{13n+14}$ and $17a + 19b < 2024$.",
    "options": [],
    "answer": "8",
    "solution": "a) If $d$ is a common divisor of the numbers $12n + 13$ and $13n + 14$, then $d$ divides the numbers $13(12n + 13)$ and $12(13n + 14)$. Then $d$ divides $13(12n + 13) - 12(13n + 14)$, that is $d \\mid 1$. So $d = 1$, hence the numbers $12n + 13$ and $13n + 14$ are coprime.\n\nb) The relation $\\frac{a}{b} = \\frac{12n+13}{13n+14}$ leads to $a(13n+14) = b(12n+13)$, hence $13n+14 \\mid b(12n+13)$. Since $12n + 13$ and $13n + 14$ are coprime, $13n + 14$ divides $b$. Since\n\n$b \\neq 0$ (it is the denominator of a fraction), there exists $k \\in \\mathbb{N}^*$ so that $b = k(13n+14)$ and therefore $a = k(12n+13)$.\n\nReplace $a = k(12n + 13)$ and $b = k(13n + 14)$ in $17a + 19b < 2024$ to get\n$k(451n + 487) < 2024$. Then $451n + 487 < 2024$, whence $n \\le 3$.\n\nFor $n = 0$ we get $k \\le 4$, so $(a, b) \\in \\{(13, 14), (26, 28), (39, 42), (52, 56)\\}$.\nFor $n = 1$ we get $k \\le 2$, so $(a, b) \\in \\{(25, 27), (50, 54)\\}$.\nFor $n \\in \\{2, 3\\}$ we get $k = 1$, so $(a, b) \\in \\{(37, 40), (49, 53)\\}$.\n\nIn total, there are 8 pairs.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed in the circle $C$ with center $O$ and radius $1$. For any point $M \\in C \\setminus \\{A, B, C\\}$, we denote $s(M) = OH_1^2 + OH_2^2 + OH_3^2$, where $H_1$, $H_2$, and $H_3$ are the orthocenters of triangles $MAB$, $MBC$, and $MCA$, respectively.\n\na) Prove that if triangle $ABC$ is equilateral, then $s(M) = 6$, for any $M \\in C \\setminus \\{A, B, C\\}$.\n\nb) Prove that if there exist three distinct points $M_1, M_2, M_3 \\in C \\setminus \\{A, B, C\\}$ such that $s(M_1) = s(M_2) = s(M_3)$, then triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "s(M) = 6 in the equilateral case; if s(M) takes the same value at three distinct points on the circle, then triangle ABC is equilateral.",
    "solution": "Consider an orthonormal coordinate system with the origin at $O$. For any point $Z$ in the plane, we denote its complex coordinate by $z$.\n\na.\nFrom Sylvester's relation, we have $h_1 = m + a + b$, $h_2 = m + b + c$, and $h_3 = m + c + a$. Also, let $h = a + b + c$ be the complex coordinate of the orthocenter of triangle $ABC$. Therefore, we obtain:\n$$\ns(M) = 6 + |h|^2 + 2m\\overline{h} + 2\\overline{m}h.\n$$\nIf triangle $ABC$ is equilateral, then $h = a + b + c = 0$, hence $s(M) = 6$.\n\nb.\nAssume, by contradiction, that triangle $ABC$ is not equilateral, which is equivalent to $h \\neq 0$. Then, since $s(M_1) = s(M_2)$, we have: $|h|^2 + 2m_1\\overline{h} + 2\\overline{m}_1h = |h|^2 + 2m_2\\overline{h} + 2\\overline{m}_2h \\Leftrightarrow \\overline{h}(m_1 - m_2) + h\\frac{m_2 - m_1}{m_1m_2} = 0 \\Leftrightarrow m_1m_2 = \\frac{h}{\\overline{h}}$.\n\nSimilarly, we obtain $m_1m_3 = \\frac{h}{\\overline{h}}$. Since $m_1 \\neq 0$, we get $m_2 = m_3$, which is a contradiction with $M_2 \\neq M_3$. Therefore, triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to (0, \\infty)$ be a continuous function on $[0, 1]$, and\n$$A = \\int_{0}^{1} f(t) \\, dt.$$\na) Show that the function $F : [0, 1] \\to [0, A]$, defined for any $x \\in [0, 1]$ by\n$$\nF(x) = \\int_{0}^{x} f(t) \\, dt,\n$$\n\nis invertible, with a differentiable inverse.\n\nb) Show that the equation\n$$\n\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt\n$$\nwhere $g$ is a function $[0, 1] \\to [0, 1]$, has a unique solution $g(x)$ for each $x \\in [0, 1]$, and express $g(x)$ in terms of $F$ and $F^{-1}$.\n\nc) Show that $g$ is differentiable, strictly decreasing, and has a unique fixed point $c \\in [0, 1]$. Compute $g'(c)$.",
    "options": [],
    "answer": "g(x) = F^{-1}(A − F(x)); c = F^{-1}(A/2); g′(c) = −1",
    "solution": "a) Because $f$ is continuous, $F$ is continuous and differentiable, with $F'(x) = f(x) > 0$, for any $x \\in [0, 1]$. Hence, $F$ is strictly increasing, and thus injective. Being continuous, $F$ has the intermediate value property, and since $F(0) = 0$ and $F(1) = A$, $F$ is surjective. Thus, $F$ is bijective, hence invertible. Also, because $F'(x) = f(x) > 0$ for any $x \\in [0, 1]$, $F^{-1}$ is differentiable, with\n$$\n(F^{-1})'(x) = \\frac{1}{f(F^{-1}(x))}, \\quad \\text{for any } x \\in [0, A].\n$$\n\nb) The equality in the statement $\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt$ becomes $F(x) = F(1) - F(g(x)) = A - F(g(x))$, or, equivalently, $F(g(x)) = A - F(x)$, for any $x \\in [0, 1]$.\nThe function $F$ being increasing, it follows that $A - F(x) \\in [0, A]$, for any $x \\in [0, 1]$, so that the function $g : [0, 1] \\to [0, 1]$, defined by $g(x) = F^{-1}(A - F(x))$ for any $x \\in [0, 1]$, is well defined and unique satisfying (1).\n\nc) Since the functions $F$ and $F^{-1}$ are differentiable, the function $g$ defined above is also differentiable, and\n$$\ng'(x) = \\frac{(A - F(x))'}{f(F^{-1}(A - F(x)))} = \\frac{-f(x)}{f(F^{-1}(A - F(x)))}, \\quad \\text{for any } x \\in [0, 1]. \\quad (2)\n$$\nThe function $g$ is strictly decreasing, and so has a unique fixed point $c \\in [0, 1]$.\nFor this we have: $2F(c) = F(c) + F(g(c)) = A$, so that $F(c) = \\frac{1}{2} \\cdot A$ and $c = F^{-1}(\\frac{1}{2} \\cdot A)$.\n\nBecause $g$ is differentiable, the limit\n$$\n\\lim_{x \\to c} \\frac{g(x) - c}{x - c} = \\lim_{x \\to c} \\frac{g(x) - g(c)}{x - c} = g'(c),\n$$\nexists and is equal to\n$$\ng'(c) = \\frac{-f(c)}{f(F^{-1}(A - F(c)))} = \\frac{-f(c)}{f(c)} = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three friends color the positive integers from $1$ to $2025$ as follows: Alexia colors in red the numbers $1$ and $2$, then Bianca colors in yellow the numbers $3$, $4$ and $5$, and Cristina colors in blue the numbers $6$, $7$, $8$ and $9$. Afterwards, the operation is repeated: Alexia colors in red the next two numbers, Bianca colors in yellow the next three numbers, and Cristina colors in blue the next four. The friends keep on painting, until all the numbers are colored.\na) What will be the color of $2024$?\nb) Find the smallest natural number $n$ so that, after $n$ numbers have been colored, the sum of the numbers in yellow is larger than $2024$.",
    "options": [],
    "answer": "a) Blue\nb) 111",
    "solution": "For simplicity we will call the numbers red, yellow, respectively blue, and a sequence of $9$ consecutive numbers in which the first two are red, the next three are yellow, and the last four are blue will be called a *complete coloring*.\n\na) Since $2025 = M_9$, the last four numbers are blue, hence $2024$ is blue.\n\nb) The $k+1$-th complete coloring (where $k$ is a natural number), assigns yellow to the numbers $3+9 \\cdot k$, $4+9 \\cdot k$ and $5+9 \\cdot k$, with sum $12+27 \\cdot k$.\nAfter $12$, respectively $13$ complete colorings, the sum of the yellow numbers is $12 \\cdot 12 + 27 \\cdot (0+1+2+\\dots+11) = 1926$, respectively $12 \\cdot 13 + 27 \\cdot (0+1+2+\\dots+12) = 2262$.\nIn order to get the sum of the yellow numbers larger than $2024$, we need $12$ complete colorings ($108$ numbers), then the red numbers $109$, $110$ and the yellow number $111$, hence the smallest $n$ is $111$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$. Consider $n$ real numbers $a_1, a_2, \\dots, a_n$, not all equal, and let\n$$\nd = d(a_1, a_2, \\dots, a_n) = \\max_{1 \\le i < j \\le n} |a_i - a_j|\n$$\nand\n$$\ns = s(a_1, a_2, \\dots, a_n) = \\sum_{1 \\le i < j \\le n} |a_i - a_j|.\n$$\n\nDetermine, in terms of $n$, the smallest and the largest values the quotient $s/d$ may achieve.",
    "options": [],
    "answer": "minimum: n−1; maximum: floor(n/2) * floor((n+1)/2)",
    "solution": "The required minimum is $n-1$ and is achieved, for instance, by $a_1 < a_2 = \\dots = a_n$. The maximum is $\\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor$ and is achieved, for instance, by\n$$\na_1 = \\dots = a_{\\lfloor n/2 \\rfloor} < a_{\\lfloor n/2 \\rfloor+1} = \\dots = a_n.\n$$\nIn each case, verification is routine and is hence omitted.\n\nWe now show that $(n-1)d \\le s \\le \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d$. Clearly, we may and will assume $a_1 \\le a_2 \\le \\dots \\le a_n$, so $s = \\sum_{1 \\le i < j \\le n} (a_j - a_i)$. Let $d_k = a_{k+1} - a_k$, $k = 1, \\dots, n-1$, so $d = d_1 + \\dots + d_{n-1}$.\n\nExpress every $a_j - a_i$, $1 \\le i < j \\le n$, and hence $s$, in terms of the $d_k$: As $d_k$ occurs as a summand in every $a_j - a_i$, $1 \\le i \\le k < k + 1 \\le j \\le n$, and in no other $a_j - a_i$, $1 \\le i < j \\le n$, it follows that $s = \\sum_{k=1}^{n-1} k(n-k)d_k$.\n\nFinally, as $\\min_{1 \\le k \\le n-1} k(n-k) = n-1$ and $\\max_{1 \\le k \\le n-1} k(n-k) = \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor$, it follows that\n$$\n(n-1)d = \\sum_{k=1}^{n-1} (n-1)d_k \\le \\underbrace{\\sum_{k=1}^{n-1} k(n-k)d_k}_{=s} \\le \\sum_{k=1}^{n-1} \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d_k = \\\\\n= \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d,\n$$\nas desired. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $AD$ of the square $ABCD$. Consider the equilateral triangles $DFM$ and $BFE$, such that $F$ lies in the interior of $ABCD$ and the lines $EF$ and $BC$ are concurrent. Denote by $P$ the midpoint of $ME$. Prove that:\na) $P$ lies on the line $AC$;\nb) the halfline $PM$ is the bisector of the angle $APF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Construct the equilateral triangle $BDQ$, such that $C$ lies in its interior. $Q$ is situated on the perpendicular bisector of the diagonal $BD$, therefore $Q$, $C$ and $A$ are collinear. Since $\\angle EBQ = \\angle FBD = 60^\\circ - \\angle FBQ$, $BQ = BD$ and $BE = BF$, triangles $BEQ$ and $BFD$ are congruent (S.A.S.), thus $\\angle BQE = \\angle BDF = \\angle ADF - \\angle ADB = 15^\\circ$ and $QE = DF = DM = AM$.\nFrom $\\angle QBC = \\angle QBD - \\angle CBD = 15^\\circ = \\angle BQE$, we infer that $QE \\parallel CB \\parallel AD$. As $QE = AM$, it follows that $AMQE$ is parallelogram, so the midpoint $P$ of $ME$ lies on $AQ$, i.e., $P$ is situated on the line $AC$.\n\nb) Since $QE = DM$ and $QE \\parallel DM$ we deduce that $DMEQ$ is a parallelogram and $\\angle DME = \\angle DQE = 75^\\circ$. Consequently, $\\angle FMP = \\angle DMP - \\angle DMF = 15^\\circ$. The triangle $FAD$ is right angled, with $\\angle FAD = 30^\\circ$, therefore we obtain $\\angle FAP = \\angle DAP - \\angle DAF = 15^\\circ$ and $\\angle MFA = \\angle DFA - \\angle DFM = 30^\\circ$.\nSince $\\angle FMP = \\angle FAP$, it follows that $AMFP$ is a cyclic quadrilateral, with $\\angle MPA = \\angle MFA = 30^\\circ$.\n$\\angle FPM = \\angle FAM = 30^\\circ$, therefore $\\angle FPM = \\angle APM$, and $PM$ is the angle bisector of $APF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39333,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral regular pyramid $VABCD$ has basis $ABCD$. The points $M$, $N$ and $P$ are the midpoints of the edges $AD$, $BC$, respectively $VA$. Prove that the angle of the line $CP$ with the plane $(BAD)$ has $45^\\circ$ if and only if the angle of the line $CP$ with the plane $(VMN)$ has $30^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nThe projection of the line $CP$ on the plane $(BAD)$ is the line $OC$, hence the angle of the line $CP$ with the plane $(BAD)$ is $\\angle OCQ$.\nFrom $CN \\perp MN$, $CN \\perp VO$ and $MN \\cap VO = \\{O\\}$ follows $CN \\perp (VMN)$. So, the projection of the line $CP$ on the plane $(VMN)$ is the line $NQ$, hence the angle of the line $CP$ with the plane $(VMN)$ is $\\angle CQN$.\nIf $\\angle OCQ = 45^\\circ$, then $CQ = \\sqrt{2} \\cdot OC$ and, since $CN = \\frac{\\sqrt{2}}{2} \\cdot OC$, the leg $CN$ of the right triangle $CQN$ is half of the hypotenuse $CQ$, therefore $\\angle CQN = 30^\\circ$.\nConversely, if $\\angle CQN = 30^\\circ$, then $CQ = 2CN = \\sqrt{2} \\cdot OC$, so $\\angle OCQ = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an odd prime $p$ and a positive integer $N < 50p$. Let $a_1, a_2, \\dots, a_N$ be a list of positive integers less than $p$ such that any specific value occurs at most $\\frac{51}{100}N$ times and $a_1 + a_2 + \\dots + a_N$ is not divisible by $p$. Prove that there exists a permutation $b_1, b_2, \\dots, b_N$ of the $a_i$ such that, for all $k = 1, 2, \\dots, N$, the sum $b_1 + b_2 + \\dots + b_k$ is not divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma.** Let $n$ be a positive integer and let $c_1, c_2, \\dots, c_n$ be a list of positive integers less than $p$ such that each specific value occurs at most $\\frac{1}{2}(n+1)$ times. Fix a residue $r \\neq c_1 + c_2 + \\dots + c_n \\pmod{p}$. Then there exists a permutation $d_1, d_2, \\dots, d_n$ of the $c_i$ such that $r \\neq d_1 + d_2 + \\dots + d_k \\pmod{p}$ for all $k = 1, 2, \\dots, n$.\n**Proof.** Induct on $n$. The base case, $n=1$, is clear, so let $n \\ge 2$. Consider the residue $a$ that occurs the most times amongst the $c_i$.\n\nIf $a \\not\\equiv r \\pmod{p}$, set $d_1 = a$ and complete the rest of the list using the inductive hypothesis with $r$ replaced by $r-a$, as any residue will occur amongst the remaining $c_i$ at most $\\frac{1}{2}((n-1)+1)$ times. Indeed, if no other residue occurs as many times as $a$, then the number of occurrences of any residue amongst the remaining $c_i$ is at most $\\frac{1}{2}(n+1)-1 = \\frac{1}{2}(n-1) < \\frac{1}{2}((n-1)+1)$. Otherwise, if there is another residue that occurs as many times as $a$, their number of occurrences has to be at most $\\frac{1}{2}n = \\frac{1}{2}((n-1)+1)$.\n\nIf $a \\equiv r \\pmod{p}$, choose a residue $b \\not\\equiv a \\pmod{p}$ amongst the $c_i$; the choice is possible, as $\\frac{1}{2}(n+1) < n$. Set $d_1 = b$ and $d_2 = a$, noting that $d_1 + d_2 = b + a \\equiv r + b \\not\\equiv r \\pmod{p}$. If no other residue occurs as many times as $a$, then each residue occurs amongst the remaining $c_i$ at most $\\frac{1}{2}(n+1)-1 = \\frac{1}{2}(n-1) = \\frac{1}{2}((n-2)+1)$ times. If $a$ occurs at most $\\frac{1}{2}(n-1)$ times, then clearly the same will hold for any residue in the remaining $c_i$. The remaining possibility is that $a$ and another residue both occur $\\frac{1}{2}n$ times and $n$ is even, meaning that the other residue has to be $b$, and there are no other residues; it is clear that the occurrences in the remaining $c_i$ are precisely $\\frac{1}{2}(n-2) < \\frac{1}{2}((n-2)+1)$. The inductive hypothesis then applies with $r$ replaced by $-b$ to complete the list. This establishes the lemma.\n\nBack to the problem, if each residue occurs at most $\\frac{1}{2}(N+1)$ times, the conclusion follows by the Lemma.\n\nOtherwise, there is exactly one residue $a$ that occurs $M > \\frac{1}{2}(N+1)$ times amongst the $a_i$. Note that $2M - N > (N+1) - N = 1$, to set $b_i = a$, $i = 1, 2, \\dots, 2M - N$. Letting $\\alpha = 50$, note also that $2M - N \\le 2 \\cdot \\frac{\\alpha+1}{2\\alpha}N - N = \\frac{1}{\\alpha}N < p$, so none of the first $2M - N$ partial sums is divisible by $p$, as $p$ is prime.\n\nTo complete the proof, apply the Lemma with $r = (N-2M)a$ to the remaining $a_i$. There are left $N - (2M - N) = 2(N - M)$ such, $a$ occurs $M - (2M - N) = N - M$ times amongst these and any other residue occurs at most $N - M$ times, both of which do not exceed $\\frac{1}{2}(2(N - M) + 1)$.\n\n\n*Alternative solution.* The permutation will be constructed step by step, first choosing $b_1$, then $b_2$, and so on. At every step, sort the remaining values by their number of appearances from most frequent to least frequent, and from largest to smallest for the situations where the number of appearances is the same. After that, attempt to select the first value from the sorted list; if this results in a partial sum that is divisible by $p$, then attempt to select the second value from the list, which will result in a partial sum that is not divisible by $p$ as the first and second values are different modulo $p$. Label the step as type (I) if the first value has been selected and as type (II) if the second value has been selected.\n\n**Lemma.** Any step of type (II) is followed by a step of type (I).\n\n**Proof.** Suppose that step $j$ is of type (II). Denote the first two values according to the sorting order with $a$ and $b$, and let $s$ be the partial sum until this step. Since step $j$ is type (II), $s + a \\equiv 0 \\pmod{p}$. At step $j + 1$, the first value according to the sorting order will still be $a$. Since $s + b + a \\not\\equiv 0 \\pmod{p}$, $a$ can be selected for step $j + 1$, so step $j + 1$ is of type (I).\n\nNow suppose, for the sake of contradiction, that this procedure fails. Let $k+1 < N$ be the first step when neither (I) nor (II) is possible. In particular, this means that there is a single value $a$ remaining, and it occurs $N - k$ times. Let $j$ be the smallest number with the following property: at every step between $j$ and $k + 1$, the value $a$ is the first one according to the sorting order; in particular, $a$ is the most frequently occurring one. Such a $j$ exists because $k + 1$ satisfies this property.\n\nIf $j > 1$, then at step $j-1$ there is another value $b$ that comes before $a$ in the sorted order. Suppose that at step $j-1$ there are $q$ values of $a$ remaining; then there are at least $q$ values of $b$ remaining. According to the lemma, at least half of the steps $j, j+1, \\dots, k$ are of type (I), which means that $a$ is selected in them. Denote $\\ell = k-j+1$, then $a$ has been selected at least $\\lfloor \\ell/2 \\rfloor$ steps between $j$ and $k$. Since at step $k+1$, there are $N-k \\ge 2$ values of $a$ remaining, it means that $q \\ge \\lfloor \\ell/2 \\rfloor + 2$. At the same time, at step $k+1$ there are no values of $b$ remaining, so between steps $j$ and $k$ all the remaining $b$ values have been selected, and this can happen at most $\\ell - \\lfloor \\ell/2 \\rfloor$ times. Combining these two leads to $\\ell - \\lfloor \\ell/2 \\rfloor \\ge q \\ge \\lfloor \\ell/2 \\rfloor + 2$, or $\\ell \\ge 2\\lfloor \\ell/2 \\rfloor + 2$, a contradiction.\n\nIf $j = 1$, then $a$ is always the first in the sorted order. Therefore, the first $p-1$ steps are of type (I), and step $p$ is of type (II). If $k \\ge p$, from steps $p+1$ to $k$, at least every other step is of type (I), meaning that $a$ is selected. Once step $k$ is done, there are $N-k$ remaining values of $a$. Therefore, in the beginning, there were at least\n$$\n(p-1) + \\frac{k-p}{2} + N-k = N - \\frac{k}{2} + \\frac{p}{2} - 1 > N - \\frac{N-2}{2} + \\frac{N}{2\\alpha} - 1 = \\frac{\\alpha+1}{2\\alpha}N\n$$\nvalues of $a$, contradiction. In the situation when $k < p$, there are at least $N - 1$ values of $a$, which is also a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39335,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, define $a_n = \\{\\frac{n}{s(n)}\\}$, where $s(k)$ represents the sum of the digits of the natural number $k$, and $\\{x\\}$ is the fractional part of the real number $x$.\n\na) Prove that there exist infinitely many positive integers $n$ such that $a_n = \\frac{1}{2}$.\n\nb) Determine the smallest positive integer $n$ such that $a_n = \\frac{1}{6}$.",
    "options": [],
    "answer": "1899999",
    "solution": "a.\nIf $s(n) = 2$ and $n$ is odd, then $a_n = \\frac{1}{2}$. The only solutions with these properties are of the form $n = 10^k + 1$, with $k \\in \\mathbb{N}^*$.\n\nb.\nLet $n$ be a positive integer such that $a_n = \\left\\{\\frac{n}{s(n)}\\right\\} = \\frac{1}{6}$.\nSince $\\frac{n}{s(n)} - \\lfloor \\frac{n}{s(n)} \\rfloor = \\frac{1}{6}$, we infer that $6n - 6 \\cdot s(n) \\cdot \\lfloor \\frac{n}{s(n)} \\rfloor = s(n)$ (1). From here follows that $6 \\mid s(n)$, therefore $3 \\mid n$. Consider $n = 3k$ and $s(n) = 6m$, with $m, k$ positive integers. From (1) we deduce that $3k - 6m \\cdot \\lfloor \\frac{k}{2m} \\rfloor = m$, hence $3 \\mid m$. Consequently $m = 3u$, with $u \\in \\mathbb{N}^*$ and $s(n) = 18u$, therefore $9 \\mid n$.\n\nConsider $n = 9v$, with $v$ a positive integer. From (1) we obtain $3v - 6u \\cdot \\lfloor \\frac{v}{2u} \\rfloor = u$, hence $3 \\mid u$. Consider $u = 3t$, with $t$ a positive integer. It follows that $m = 9t$ and $s(n) = 54t$, and the minimal sum of the digits of the natural number $n$ is 54.\n\nThe smallest positive integer with the sum of its digits 54 is $n = 999999$.\n\nBut $a_{999999} = \\left\\{\\frac{999999}{54}\\right\\} = \\frac{1}{2}$, hence $n = 999999$ is not a solution. The next positive integer with the sum of its digits 54 is $n = 1899999$, for which we have $a_{1899999} = \\left\\{\\frac{1899999}{54}\\right\\} = \\{35185 + \\frac{1}{6}\\} = \\frac{1}{6}$, therefore $n_{\\min} = 1899999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ and $\\omega$ be its circumcircle. The tangent at $A$ to the circle $\\omega$ intersects line $BC$ at $D$, and the line through $B$ parallel to $AD$ meets again the circle $\\omega$ at $E$. Line $DE$ intersects $AB$ at $F$ and the circle $\\omega$ for the second time at $G$. On the line $BE$, consider the point $N$ such that $B, G, F$, and $N$ are con-cyclic. Let $S$ and $T$ be the intersection points of line $FN$ with $AD$ and $AE$, respectively. Prove that $STDG$ is a cyclic quadrilateral.\nJBMO Shortlist\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\angle TAB = \\angle BCE = 180^\\{\\circ\\} - \\angle BGE = 180^\\{\\circ\\} - \\angle BGF = \\angle BNF = \\angle BNT,\n$$\nso ANBT is also cyclic.\n\nUsing the fact that $AD \\parallel BE$, it follows that $\\angle ATS = \\angle ABN = \\angle BAD = \\angle FAS$, so line AB is the tangent at A to the circumcircle of AST, (1).\n\nAlso, from $AD \\parallel BE$ it results that $\\triangle FAD \\sim \\triangle FBE$, so $\\frac{FA}{FB} = \\frac{FD}{FE}$. By power of point F to circle $\\omega$, we infer that $FA \\cdot FB = FE \\cdot FG$. Multiplying the last two equalities, we get $FA^2 = FD \\cdot FG$.\n\nFrom (1) it follows that $FS \\cdot FT = FA^2 = FD \\cdot FG$, so S, T, D, G belong to the same circle. Hence, STDG is a cyclic quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_0$ be a real number. Prove that there exists a useful function $f$ so that $f(x_0) = 0$ if and only if $|x_0| \\ge 4$.\n\nWe will call an affine function $f : \\mathbb{R} \\to \\mathbb{R}$ *useful* if it has the properties:\n(i) $|f(x)| \\le 2$, for every real number $x$ such that $|x| \\le 2$;\n(ii) $|f(x)| \\ge 1$, for every real number $x$ such that $|x| \\le 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "\"$\\Rightarrow$\" If $f$ is useful, then $-f$ has properties (i) and (ii), hence it is useful. Consequently, we may look only at the useful functions $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = a x + b$, with $a > 0$, $b \\in \\mathbb{R}$.\n\nSince $a > 0$, the inequality (i) is equivalent to $f(2) \\le 2$ and $f(-2) \\ge -2$, that is $2a + b \\le 2$, (1) and $2a - b \\le 2$, (2).\n\nThe inequality (ii) leads to the possibilities:\n\nI) $f(-1) \\le -1$ and $f(1) \\le -1$ or\n\nII) $f(-1) \\ge 1$ and $f(1) \\ge 1$\n\n(if $f(-1) < 0 < f(1)$, then $-1 < -\\frac{b}{a} < 1$ and $f(-\\frac{b}{a}) = 0$, in contradiction with (ii)).\n\nIn case I we get $-a + b \\le -1$, (3) and $a + b \\le -1$, (4). Multiply (4) by 2 and add with (2) to get $4a + b \\le 0$, that is $x_0 = -\\frac{b}{a} \\ge 4$.\n\nIn case II we get $a - b \\le -1$, (5) and $-a - b \\le -1$, (6). Multiply (5) by 2 and add with (1) to get $4a - b \\le 0$, hence $x_0 = -\\frac{b}{a} \\le -4$.\n\nFrom I) and II) follows $|x_0| \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\in \\mathbb{N}^*$. We say that the ring $(A, +, \\cdot)$ has the property $CP(k)$, if for every $a, b \\in A$ there is a $c \\in A$, such that $a^k = b^k + c^k$.\na) Give an example of a finite ring $(A, +, \\cdot)$, which does not have the property $CP(k)$ for any positive integer $k$, with $k \\ge 2$.\nb) Let $n \\in \\mathbb{N}$, $n \\ge 3$, and $M(n) = \\{m \\in \\mathbb{N}^* \\mid (\\mathbb{Z}_n, +, \\cdot) \\text{ has the property } CP(m)\\}$. Prove that $M(n)$ is a monoid with respect to multiplication, included in the set $2 \\cdot \\mathbb{N} + 1$ of odd positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any $k \\in \\mathbb{N}^*$ we denote $P_k(A) = \\{a^k \\mid a \\in A\\}$. The condition $CP(k)$ is then equivalent with\n$$\nx - y \\in P_k(A), \\quad \\text{for any } x, y \\in P_k(A),\n$$\nmeaning that $P_k(A)$ is a subgroup of the additive group $(A, +)$.\n\na) For $A = \\mathbb{Z}_4$, we have that $P_{2k}(A) = \\{\\hat{0}, \\hat{1}\\}$, respectively $P_{2k+1}(A) = \\{\\hat{0}, \\hat{1}, \\hat{3}\\}$, for any $k \\in \\mathbb{N}^*$. These are not subgroups of the group $(\\mathbb{Z}_4, +)$. Hence, the ring $(\\mathbb{Z}_4, +, \\cdot)$ does not have the property $CP(k)$ for any $k \\in \\mathbb{N}$, with $k \\ge 2$.\n\nb) For $n \\in \\mathbb{N}$, $n \\ge 3$, we consider the ring $(\\mathbb{Z}_n, +, \\cdot)$. Because $\\hat{1} \\in P_k(\\mathbb{Z}_n)$ for any $k \\in \\mathbb{N}$, $k \\ge 2$, and $(\\mathbb{Z}_n, +)$ is cyclic, generated by $\\hat{1}$, it follows that $CP(k) \\iff P_k(\\mathbb{Z}_n) = \\mathbb{Z}_n$. Equivalently, $CP(k) \\iff$ the function $p_k : \\mathbb{Z}_n \\to \\mathbb{Z}_n$, defined by $p_k(x) = x^k$ for any $x \\in \\mathbb{Z}_n$, is bijective.\nThen we can rewrite $M(n) = \\{m \\in \\mathbb{N}^* \\mid p_m$ is bijective\\}.\nBecause for even $k$ we have that $p_k(\\hat{1}) = p_k(-\\hat{1})$, and $\\hat{1} \\ne -\\hat{1}$, it follows that any $m \\in M(n)$ is odd, so that $M(n) \\subseteq 2 \\cdot \\mathbb{N} + 1$.\nBecause $p_1 = \\text{id}_{\\mathbb{Z}_n}$ is bijective, we have that $1 \\in M(n)$.\nLet $m_1, m_2 \\in M(n)$ be arbitrary. Since the functions $p_{m_1}$ and $p_{m_2}$ are bijective, the function $p_{m_1 m_2} = p_{m_1} \\circ p_{m_2}$ is also bijective, as the composition of two bijective functions. We deduce that $m_1 \\cdot m_2 \\in M(n)$.\nIt follows that $M(n)$ is a submonoid of the monoid $(\\mathbb{N}^*, \\cdot)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a$ and $b$ be positive integers congruent to $1$ modulo $4$. Prove that there exists a positive integer $k$ such that at least one of the numbers $a^k - b$ and $b^k - a$ is divisible by $2^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma.** For any integer $p \\ge 2$, if $c-1 \\equiv 2^p \\pmod{2^{p+1}}$, then $c^2 - 1 \\equiv 2^{p+1} \\pmod{2^{p+2}}$.\n*Proof.* Write $c^2 - 1 = (c-1)(c+1)$. By hypothesis, $c-1$ is divisible by $2^p$. As $c+1$ leaves remainder 2 upon division by $2^p$, it follows that $2^{p+1}$ is the highest power of 2 dividing $c^2 - 1$, whence the conclusion of the lemma.\nLet $2^\\alpha$ and $2^\\beta$ be the highest powers of 2 dividing $a-1$ and $b-1$, respectively. By hypothesis, $\\alpha \\ge 2$ and $\\beta \\ge 2$. Note that $a-1 \\equiv 2^\\alpha \\pmod{2^{\\alpha+1}}$ and $b-1 \\equiv 2^\\beta \\pmod{2^{\\beta+1}}$. We may and will assume $\\alpha \\le \\beta$. Apply the lemma $\\beta - \\alpha$ times to get $a^{2^{\\beta-\\alpha}} - 1 \\pmod{2^{\\beta+1}} = 2^\\beta$, so $a^{2^{\\beta-\\alpha}} - b = (a^{2^{\\beta-\\alpha}} - 1) - (b-1)$ is divisible by $2^{\\beta+1}$.\nNow, induct on $n$. By the preceding, if $n \\le \\beta + 1$, then $a^{2^{\\beta-\\alpha}} - b$ is divisible by $2^n$, so $k = 2^{\\beta-\\alpha}$ will do. Let $n > \\beta$.\nFor the induction step, let $a^k - b$ be divisible by $2^n$ for some $k \\ge 1$. If it is divisible by $2^{n+1}$ we are done, so let $a^k - b \\equiv 2^n \\pmod{2^{n+1}}$.\n\n\n*Second solution.* Clearly, $k=1$ works for both $n=1$ and $n=2$, so let $n \\ge 3$. If $a=1$ or $b=1$, the conclusion follows by Euler's theorem, so let $a \\ge 5$ and $b \\ge 5$. Note that $a$ and $b$ may be replaced by their residues modulo $2^n$ to assume them both less than $2^n$.\nFor every integer $2 \\le m < n$, let $S_m = \\{2^m u + 1 : u = 0, \\dots, 2^{n-m} - 1\\}$. We first show that if $x$ and $y$ are in $S_m$, then so is $xy \\pmod{2^n}$. Write $x = 2^mu + 1$ and $y = 2^mv + 1$. Then $xy = 2^mv + 1$, where $w = 2^muv + u + v$. Note that $w_0 = w \\pmod{2^{n-m}}$ falls in the range 0 through $2^{n-m} - 1$ and write $w = 2^{n-m}w' + w_0$. Then $xy = 2^m(2^{n-m}w' + w_0) + 1 = 2^nv + 2^mw_0 + 1$, where $2^mw_0 + 1 < 2^n$. Consequently, $xy \\pmod{2^n} = 2^mv + 1$ is in $S_m$, as stated.\nLet $2^\\alpha$ and $2^\\beta$ be the highest powers of 2 dividing $a-1$ and $b-1$, respectively. We may and will assume $\\alpha \\le \\beta$. Note that $a = 2^\\alpha u + 1$ belongs to $S_\\alpha$, where $u$ is odd and $2 \\le \\alpha \\le n-1$. By the preceding, the residues modulo $2^n$ of $a, a^2, a^3, \\dots, a^{2^{n-\\alpha}}$ are all in $S_\\alpha$.\nWe now show that $2^{n-\\alpha}$ is the least positive integer $s$ satisfying $a^s \\equiv 1 \\pmod{2^n}$. By minimality, $s$ divides $2^{n-1}$, as $a^{2^{n-1}} \\equiv 1 \\pmod{2^n}$. Write\n$$\na^{2^{n-\\alpha}} - 1 = (a-1)(a+1)(a^2+1)\\cdots(a^{2^{n-\\alpha-1}} + 1).\n$$\nAs $2^\\alpha$ divides $a-1$ and $2^{n-\\alpha}$ divides $(a+1)(a^2+1)\\cdots(a^{2^{n-\\alpha-1}} + 1)$, it follows that $2^n$ divides $a^{2^{n-\\alpha}} - 1$, so $a^{2^{n-\\alpha}} \\equiv 1 \\pmod{2^n}$.\nSuppose now, if possible, that $s$ divides $2^{n-\\alpha-1}$. Then $a^{2^{n-\\alpha-1}} \\equiv 1 \\pmod{2^n}$, so $2^n$ divides $(a-1)(a+1)(a^2+1)\\cdots(a^{2^{n-\\alpha-2}} + 1)$. As $(a+1)(a^2+1)\\cdots(a^{2^{n-\\alpha-2}} + 1)$ is divisible by $2^{n-\\alpha-1}$ but not by $2^{n-\\alpha}$, it follows that $a-1$ is divisible by $2^{\\alpha+1}$, contradicting the choice of $\\alpha$. Consequently, $s = 2^{n-\\alpha}$, as stated.\n\n\n**Third solution.** Work in $\\mathbb{Z}_{2^n}$. Clearly, $k=1$ works for both $n=1$ and $n=2$, so let $n \\ge 3$. The units of $\\mathbb{Z}_{2^n}$ consist of all positive odd integers less than $2^n$. They form a multiplicative copy of the additive $\\mathbb{Z}_{2^{n-2}} \\times \\mathbb{Z}_2$ with generators $5$, of order $2^{n-2}$, and $2^n-1$, of order $2$. Thus every unit has the form $5^\\gamma (2^n-1)^\\epsilon$, $\\gamma = 0, 1, \\dots, 2^{n-2}-1$, $\\epsilon = 0, 1$.\nAs $a$ and $b$ are both $1$ modulo $4$, it follows that $a = 5^\\alpha$ and $b = 5^\\beta$. As $5$ has order $2^{n-2}$, it is sufficient to provide a positive integer $k$ such that at least one of the numbers $\\alpha k - \\beta$ and $\\beta k - \\alpha$ is divisible by $2^{n-2}$.\nLet $2^m$ be the highest power of $2$ dividing $\\gcd(\\alpha, \\beta)$. Then at least one of the numbers $\\alpha/2^m$ and $\\beta/2^m$ is odd and hence invertible modulo $2^{n-2}$; say, $\\alpha/2^m$ is odd and let $\\alpha'$ be its inverse modulo $2^{n-2}$. Then $k = \\alpha'\\beta/2^m$ makes $\\alpha k - \\beta$ divisible by $2^{n-2}$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle BAC = 30^\\circ$ and $\\angle ABC = 100^\\circ$. Let $m$ be the perpendicular bisector of $AC$, $E$ be the intersection of $m$ and $AB$, and $D$ be the point on $m$, inside the triangle $ABC$, such that $\\angle CAD = 10^\\circ$. Let $M$ be the intersection of the lines $AD$ and $CE$.\n\na) Prove that $CE$ is the bisector of the angle $\\angle BCD$.\n\nb) Prove that $AM = AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) As $m$ is the perpendicular bisector, we have $DA = DC$ and $EA = EC$. Thus $\\triangle DEA \\equiv \\triangle DEC$.\n\n![](attached_image_1.png)\n\nFrom $DA = DC$, we get $\\angle DCA = \\angle DAC = 10^\\circ$. We have $\\angle DCE = \\angle DAE = 20^\\circ$, so $\\angle BCE = 20^\\circ$ and $CE$ is the bisector of the angle $BCD$.\n\nb) As $\\angle CDA = 160^\\circ$, it follows $\\angle ADE = \\angle CDE = 100^\\circ = \\angle CBE$.\n\nWe obtain $\\angle CEB = \\angle CED = 60^\\circ$, so $\\triangle CEB \\equiv \\triangle CED$ (A.S.A.)\n\nIt follows that $BE = BD$. But $\\angle MEB = \\angle MED$ and $ME = ME$, so $\\triangle MEB \\equiv \\triangle MED$ (S.A.S.).\n\nWe infer that $\\angle EBM = \\angle EDM = 180^\\circ - \\angle CED - \\angle AME = 180^\\circ - 60^\\circ - 40^\\circ = 80^\\circ$. But $\\angle BAM = 20^\\circ$, so $AB = AM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39341,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will say that the positive integers $m$ and $n$ have property $\\mathcal{P}$ if for every divisor $d_1$ of $m$ and every divisor $d_2$ of $n$, the number $d_1 + d_2$ is a prime.\n\na) Prove that if $m$ and $n$ have property $\\mathcal{P}$ and are different, then $m+n$ is odd.\n\nb) Find all the pairs $(m, n)$ of positive integers $m \\le n$, having property $\\mathcal{P}$.",
    "options": [],
    "answer": "If the two numbers are distinct, their sum is odd. The pairs with the property are (1,1), (1,2), and (1,4).",
    "solution": "a) Without loss of generality, we may suppose $1 \\le m < n$.\nIf $m$ and $n$ are odd, then $m \\ge 1$ and $n \\ge 3$. Taking $d_1 = 1$ and $d_2 = n$ yields $d_1 + d_2 = n + 1$, which is an even number at least $4$, so it is composite, contradiction.\nIf $m$ and $n$ are even, taking $d_1 = 2$ and $d_2 = 2$ yields $d_1 + d_2 = 4$, contradiction.\nIn conclusion $m$ and $n$ have different parities, so $m + n$ is odd.\n\nb) If $m = n$, then $d_1 = m$ and $d_2 = n$ leads to $d_1 + d_2 = 2m$, which must be a prime, hence $m = n = 1$.\nIf $m < n$, then $m \\neq n$ and, from a), $m$ and $n$ have different parities.\n\n* If $m$ is even and $n$ is odd, then $m \\ge 2$ and $n \\ge 3$. Then $d_1 = 1$ and $d_2 = n$ yields $d_1 + d_2 = n + 1 \\ge 4$, contradiction.\n* If $m$ is odd and $n$ is even, then $m \\ge 1$ and $n \\ge 2$. Take $a \\in \\mathbb{N}^*$ and $b$ odd such that $n = 2^a \\cdot b$.\n\nI. If $a \\ge 3$, then $8$ divides $n$ and, taking $d_1 = 1$ and $d_2 = 8$ yields $d_1 + d_2 = 9$, contradiction.\n\nII. If $a = 1$, then $n = 2b$, with odd $b$.\n\ni) If $b \\ge 3$, then take $d_1 = 1$ and $d_2 = b$ to get $d_1 + d_2 = b + 1 \\ge 4$, contradiction.\n\nii) If $b = 1$, then $n = 2$ and, since $m < n$ and $m$ is odd, $m = 1$. It is easy to check that the pair $(m, n) = (1, 2)$ is a solution.\n\nIII. If $a = 2$, then $n = 4b$, with odd $b$.\n\ni) If $b \\ge 3$, then $d_1 = 1$ and $d_2 = b$ yields $d_1 + d_2 = b + 1 \\ge 4$, contradiction.\n\nii) If $b = 1$, then $n = 4$ and, since $m < n$ and $m$ is odd, $m \\in \\{1, 3\\}$. It is easy to check that only $(m, n) = (1, 4)$ satisfies the statement.\n\nFinally, the solutions are $(1, 1)$, $(1, 2)$, $(1, 4)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39342,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $x$ let $\\sigma(x)$ be the sum of its positive divisors and $\\tau(x)$ be the number of the positive divisors of $x$.\nFind all positive integers $n$ such that $\\sigma(\\tau(n)) = n$.",
    "options": [],
    "answer": "1, 3, 4, 12",
    "solution": "Let $\\tau(n) = k$ and let $d$ be a divisor of $n$, such that $d \\le \\sqrt{n}$. Since $\\frac{n}{d}$ also divides $n$ and $\\frac{n}{d} \\ge \\sqrt{n}$, we infer that $\\tau(n) \\le 2\\sqrt{n}$, for any $n \\in \\mathbb{N} \\setminus \\{0\\}$, (1).\n\nAlso, if $1 = d_1 < d_2 < \\dots < d_k = n$ are the divisors of $n$, then\n$$\n2\\sigma(n) = (d_1 + d_k) + (d_2 + d_{k-1}) + \\dots + (d_k + d_1) = \\left(d_1 + \\frac{n}{d_1}\\right) + \\dots + \\left(d_k + \\frac{n}{d_k}\\right).\n$$\nBut, if $d$ divides $n$, then $1 + n - d - \\frac{n}{d} = \\frac{n(d-1)}{d} - (d-1) = \\frac{(n-d)(d-1)}{d} \\ge 0$. Consequently, $\\sigma(n) \\le \\frac{1}{2}\\tau(n) \\cdot (n+1)$, (2).\n\nWe infer that $n = \\sigma(\\tau(n)) \\le \\frac{1}{2}\\tau(\\tau(n)) \\cdot (\\tau(n)+1) \\le \\frac{1}{2} \\cdot 2\\sqrt{\\tau(n)} \\cdot (2\\sqrt{n}+1) \\le \\sqrt{2\\sqrt{n}} \\cdot (2\\sqrt{n}+1)$.\n\nSuccessively, this leads to\n$$\nn^2 \\le 2\\sqrt{n}(4n + 4\\sqrt{n} + 1) \\Leftrightarrow n\\sqrt{n} \\le 2(4n + 4\\sqrt{n} + 1) \\Leftrightarrow \\sqrt{n} \\le 8 + \\frac{8}{\\sqrt{n}} + \\frac{2}{n}.\n$$\nSuppose, by contradiction, that $n \\ge 81$. Then $\\sqrt{n} \\ge 9$ and $8 + \\frac{8}{9} + \\frac{2}{81} < 9 \\le \\sqrt{n}$, which is a contradiction. Hence, $n \\le 80$, which means that $n$ can only have one of the following forms: $1, p, p^2, p^3, p^4, p^5, pq, p^2q, p^3q, p^4q, p^2q^2, p^3q^2, pqr, p^2qr$, where $p, q, r$ are prime numbers.\n\nEvery such possible $n$ has at most 12 divisors. Studying all cases for $\\tau(n) \\in \\{1, 2, \\dots, 12\\}$, we easily find that the only solutions are $n = 1, n = 3, n = 4$ and $n = 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39343,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any non-zero natural number $n$ consider the set\n$$\nA = \\{n^2, n^2 + 1, n^2 + 2, \\dots, (n+1)^2\\}.\n$$\nFind the numbers $a, b, c \\in A$, $a < b < c$, knowing that $b$ is the geometric mean of the numbers $a$ and $c$.",
    "options": [],
    "answer": "a = n^2, b = n(n + 1), c = (n + 1)^2",
    "solution": "From $b^2 = ac$ it follows that $\\frac{b}{a} = \\frac{c}{b} = \\frac{x}{y}$, where $x, y \\in \\mathbb{N}^*, (x, y) = 1$ and $x > y$. Therefore, $b = \\frac{a x}{y}$, $c = a \\left(\\frac{x}{y}\\right)^2$.\nSince $c = a \\left(\\frac{x}{y}\\right)^2 \\in \\mathbb{N}^*$ and $(x, y) = 1$, it follows that $\\frac{a}{y^2} \\in \\mathbb{N}^*$, so $a = p y^2$, where $p \\in \\mathbb{N}^*$. The numbers are $a = p y^2$, $b = p x y$, $c = p x^2$, where $x, y \\in \\mathbb{N}^*, x > y$, $p \\in \\mathbb{N}^*$.\nFrom $a \\in A$ follows that $y \\sqrt{p} \\ge n$ and from $c \\in A$ follows that $\\sqrt{p} x \\le n + 1$, hence $\\sqrt{p}(x-y) \\le 1$ and, since $x-y \\ge 1$, it follows that $1 \\ge \\sqrt{p}(x-y) \\ge \\sqrt{p}$ and, since $p \\in \\mathbb{N}^*$, we deduce $p = 1$.\nWe obtain $n \\le y < x \\le n + 1$, so $y = n, x = n + 1$ and the solution $a = n^2$, $b = n(n + 1)$, $c = (n + 1)^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $A, B \\in M_n(\\mathbb{R})$, and the function $f: M_n(\\mathbb{C}) \\to M_n(\\mathbb{C})$, defined by $f(Z) = AZ + B\\bar{Z}$, $Z \\in M_n(\\mathbb{C})$, where $\\bar{Z}$ is the matrix having as entries the conjugates of the entries of $Z$. Prove that the following are equivalent:\n(1) $f$ is injective;\n(2) $f$ is surjective;\n(3) matrices $A+B$ and $A-B$ are non-singular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $Z \\in M_n(\\mathbb{C})$, there are $X, Y \\in M_n(\\mathbb{R})$ such that $Z = X + iY$, and $\\bar{Z} = X - iY$. Thus $f(Z) = (A+B)X + i(A-B)Y$.\n\n(1) $\\Rightarrow$ (3). Suppose that $A+B$ or $A-B$ are singular. In case $A+B$ is singular, $\\det(A+B) = 0$, so there is $C \\in M_{n,1}(\\mathbb{R}) \\setminus \\{O_{n,1}\\}$ such that $(A+B)C = O_{n,1}$. Define $X \\in M_n(\\mathbb{R})$, $X \\neq O_n$, with the $n$ columns the same as $C$. Then $f(X) = (A+B)X = O_n = f(O_n)$, in contradiction with (1).\nThe case when $A-B$ is singular can be treated in the same manner.\n\n(3) $\\Rightarrow$ (1). Let $Z_1 = X_1 + iY_1$, with $X_1, Y_1 \\in \\mathcal{M}_n(\\mathbb{R})$, and $Z_2 = X_2 + iY_2$, with $X_2, Y_2 \\in \\mathcal{M}_n(\\mathbb{R})$, such that $f(Z_1) = f(Z_2)$. Then $(A+B)X_1 + i(A-B)Y_1 = (A+B)X_2 + i(A-B)Y_2$.\nIt follows that $(A+B)(X_1 - X_2) = O_2$ and $(A-B)(Y_1 - Y_2) = O_2$, and by (3) we get $X_1 - X_2 = O_2$ and $Y_1 - Y_2 = O_2$. So $X_1 = X_2$ and $Y_1 = Y_2$, that is $Z_1 = Z_2$.\n\n(2) $\\Rightarrow$ (3). Suppose $A+B$ or $A-B$ is singular. In the case that $A+B$ is singular, $\\det(A+B) = 0$. For $Z = X + iY \\in \\mathcal{M}_n(\\mathbb{C})$, we have $\\det((A+B)X) = 0$, implying $f(Z) \\neq I_n$ for all $Z \\in \\mathcal{M}_n(\\mathbb{C})$, in contradiction with (2). The same proof works in case $A-B$ is singular.\n\n(3) $\\Rightarrow$ (2). Let $Z = X + iY \\in \\mathcal{M}_n(\\mathbb{R})$. Define $U = (A+B)^{-1}X$ and $V = (A-B)^{-1}Y$. Then $f(U + iV) = X + iY = Z$, so $f$ is surjective.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a, b, c, d, e, f$ satisfying the following condition: for any two of them, $x$ and $y$, two of the remaining four numbers, $z$ and $t$, exist such that $\\frac{x}{y} = \\frac{z}{t}$.",
    "options": [],
    "answer": "All solutions are exactly the 6-tuples that are permutations of either (a, a, b, b, c, c) with a ≤ b ≤ c or (a, a, a, b, b, b) with a ≤ b.",
    "solution": "If $x = a$ and $y = f$, for all $z, t \\in \\{b, c, d, e\\}$ we have $\\frac{x}{y} = \\frac{a}{f} \\le \\frac{z}{t}$, where the equality holds for $z = a$ and $t = f$. Since $z \\in \\{b, c, d, e\\}$, it follows that $z \\ge b$, hence $a \\ge b$. Therefore, $a = b$. Similarly, we infer that $e = f$.\n\nWe now choose $x = c$, $y = d$. Then $\\frac{c}{d} = \\frac{z}{t}$, where $z, t \\in \\{a, f\\}$. But $c \\le d$, hence $c = d$, or $\\frac{c}{d} = \\frac{a}{f}$.\n\nIf $c = d$, the 6-tuple $(a, a, c, c, f, f)$ is obviously a solution. Indeed, if $x \\ne y$, we choose $z = x$ and $t = y$, and if $x = y$, we choose $z = t \\ne x$.\n\nIf $c \\ne d$, we have $\\frac{c}{d} = \\frac{a}{f}$, thus $d = \\frac{cf}{a} \\ge f$. It follows that $a = c$ and $d = f$. Similarly, it is easy to prove that the 6-tuple $(a, a, a, f, f, f)$ is a solution.\n\nThus, the 6-tuple which satisfy the required condition are $(a, a, b, b, c, c)$, with $a \\le b \\le c$, $(a, a, a, b, b, b)$, with $a \\le b$, and all their permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a continuous function $f : [0, 1] \\to \\mathbb{R}$ with $f(1) = 0$. Prove that the following limit exists and calculate its value:\n$$\n\\lim_{t \\to 1^+} \\left( \\frac{1}{1-t} \\cdot \\int_{0}^{1} x(f(tx) - f(x)) \\, dx \\right).\n$$",
    "options": [],
    "answer": "2\\int_{0}^{1} x f(x) \\, dx",
    "solution": "Consider $g : [0, 1] \\to \\mathbb{R}$ defined by $g(x) = x f(x)$, for any $x \\in [0, 1]$. Being continuous, $g$ has an antiderivative $G : [0, 1] \\to \\mathbb{R}$ with $G(0) = 0$. Then $g(1) = f(1) = 0$ and\n$$\n\\int_{0}^{1} x f(x) \\, dx = \\int_{0}^{1} g(x) \\, dx = G(1),\n$$\nand\n$$\n\\int_{0}^{1} x f(tx) \\, dx = \\frac{1}{t^2} \\cdot \\int_{0}^{1} t g(tx) \\, dx = \\frac{1}{t^2} \\cdot G(t),\n$$\nfor $t \\in (0, 1]$. Then\n$$\n\\frac{1}{1-t} \\cdot \\int_{0}^{1} x(f(tx) - f(x)) \\, dx = \\frac{1}{1-t} \\cdot \\left( \\frac{1}{t^2} \\cdot G(t) - G(1) \\right) = \\frac{1}{t^2} \\cdot \\frac{G(t) - t^2 G(1)}{1-t}.\n$$\nLet $u, v : [0, 1] \\to \\mathbb{R}$ defined by $u(t) = G(t) - t^2 G(1)$, and $v(t) = 1 - t$, respectively, for $t \\in [0, 1]$. Both are differentiable and $u'(t) = g(t) - 2t G(1)$, $v'(t) = -1$, for $t \\in (0, 1)$, and:\n$$\n\\lim_{t \\to 1^+} u(t) = \\lim_{t \\to 1^+} v(t) = 0 \\quad \\text{and} \\quad \\lim_{t \\to 1^+} \\frac{u'(t)}{v'(t)} = 2G(1) - g(1) = 2G(1).\n$$\nBy the l'Hospital rule,\n$$\n\\lim_{t \\to 1^+} \\left( \\frac{1}{1-t} \\cdot \\int_{0}^{1} x(f(tx) - f(x)) \\, dx \\right) = \\lim_{t \\to 1^+} \\frac{1}{t^2} \\cdot \\frac{u(t)}{v(t)} = \\lim_{t \\to 1^+} \\frac{u(t)}{v(t)}\n$$\nso\n$$\n\\lim_{t \\to 1^+} \\frac{u'(t)}{v'(t)} = 2G(1) = 2 \\int_{0}^{1} x f(x) \\, dx.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a composite positive integer and let $1 = d_1 < d_2 < d_3 < \\dots < d_k = n$ be the divisors of $n$, where $k \\ge 3$. Assume that all the equations $d_{i+2}x^2 - 2d_{i+1}x + d_i = 0$, for $i \\in \\{1, 2, \\dots, k-2\\}$ have real solutions. Prove that $n = p^{k-1}$ for some prime number $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n4d_{i+1}^2 - 4d_{i+2} \\cdot d_i \\ge 0 \\Leftrightarrow \\frac{d_{i+1}}{d_i} \\ge \\frac{d_{i+2}}{d_{i+1}} \\quad (*)\n$$\nfor any $i \\in \\{1, 2, \\dots, k-2\\}$.\nAs $d_2$ is the smallest proper divisor of $n$, it means that the number $\\frac{n}{d_2}$ is the greatest proper divisor of $n$, so $\\frac{n}{d_2} = d_{k-1}$. We have:\n$$\n\\frac{n}{d_2} = d_{k-1} = \\prod_{i=1}^{k-2} \\frac{d_{i+1}}{d_i} \\ge \\prod_{i=1}^{k-2} \\frac{d_{i+2}}{d_{i+1}} = \\frac{n}{d_2}.\n$$\nIt follows that all inequalities (*) turn into equalities.\nThus, $d_{i+1}^2 = d_{i+2} \\cdot d_i$ for any $i \\in \\{1, 2, \\dots, k-2\\}$. It follows that the numbers $1 = d_1, d_2, d_3, \\dots, d_k = n$ (in this order) are consecutive terms of a geometric progression of ratio $d_2$, so $n = d_2^{k-1}$.\nThe lowest proper divisor of the composite number $n$ is a prime number $p$, so $d_2 = p$, and $n = p^{k-1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the real numbers $x$ and $y$ so that:\n(i) $x \\ge 2y^2$;\n(ii) $y \\ge 2x^2$;\n(iii) the number $8(x - y)$ is an integer.",
    "options": [],
    "answer": "All pairs (a, a) with a in [0, 1/2], together with (1/4, 1/8) and (1/8, 1/4).",
    "solution": "From (i) and (ii) follows $x \\ge 0$ and $y \\ge 0$. Moreover, $x = 0$ if and only if $y = 0$. This points to the solution $(0, 0)$ and the other solutions $(x, y)$ have $x > 0$ and $y > 0$.\nLet $(x, y)$ be a solution with $x > 0$ and $y > 0$.\nFrom (i) and (ii) follows $x \\ge 2y^2 \\ge 8x^4$, therefore $x(8x^3 - 1) \\le 0$ and, since $x > 0$, this leads to $8x^3 \\le 1$, hence $0 < x \\le \\frac{1}{2}$. In the same way, $0 < y \\le \\frac{1}{2}$.\nIt is enough to look at the case $x \\ge y$. We have $0 \\le 8(x-y) < 8x \\le 4$ and, from (iii), follows $8(x-y) \\in \\{0, 1, 2, 3\\}$, therefore $x-y \\in \\{0, \\frac{1}{8}, \\frac{1}{4}, \\frac{3}{8}\\}$ and the following situations may rise.\n* If $x - y = 0$, that is $x = y$, all the conditions from the statement are fulfilled. So, the pairs $(a, a)$, with $a \\in (0, \\frac{1}{2}]$, are solutions.\n* If $x - y = \\frac{1}{8}$, that is $y = x - \\frac{1}{8}$, we get: $y \\ge 2x^2 \\iff x - \\frac{1}{8} \\ge 2x^2 \\iff (4x - 1)^2 \\le 0 \\iff x = \\frac{1}{4}$. This yields $y = \\frac{1}{8}$, and the pair $(\\frac{1}{4}, \\frac{1}{8})$ also fulfills (i), therefore it is a solution.\n* If $x - y = \\frac{1}{4}$, that is $y = x - \\frac{1}{4}$, we get: $y \\ge 2x^2 \\iff x - \\frac{1}{4} \\ge 2x^2 \\iff 8x^2 - 4x + 1 \\le 0 \\iff 4x^2 + (2x - 1)^2 \\le 0$, impossible.\n* If $x - y = \\frac{3}{8}$, that is $y = x - \\frac{3}{8}$, we get: $y \\ge 2x^2 \\iff x - \\frac{3}{8} \\ge 2x^2 \\iff 16x^2 - 8x + 3 \\le 0 \\iff (4x - 1)^2 + 2 \\le 0$, impossible.\nThe solutions are the pairs $(\\frac{1}{4}, \\frac{1}{8})$, $(\\frac{1}{8}, \\frac{1}{4})$ and $(a, a)$, with $a \\in [0, \\frac{1}{2}]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39349,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the exterior of the acute-angled triangle $ABC$, we construct the isosceles triangles $DAB$ and $EAC$ with bases $AB$ and $AC$, respectively, such that $\\angle DBC = \\angle ECB = 90^\\circ$. Let $M$ and $N$ be the reflections of $A$ with respect to $D$ and $E$, respectively. Prove that the line $MN$ passes through the orthocenter of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nSince $AH \\perp BC$ and $CP \\perp BC$, we deduce that $AH \\parallel CP$. Therefore $AHCP$ is a parallelogram, and $CP = AH$.\n\nSimilarly, we deduce that $AHFC'$ is a parallelogram, hence $C'F = AH = CP$. Since $C'E = CE$, we obtain that $EP = EF$.\n\nConsequently, $D$ and $E$ are the midpoints of the bases of the trapezoid $BB'PF$, and $Q$ is the intersection point of the lines $B'P$ and $BF$, therefore $D, E, Q$ are collinear. Since $D$ and $I$ are the midpoints of the bases of the trapezoid $AHBB'$, and $Q$ is the intersection point of the lines $B'A$ and $BH$, it follows that $D, I, Q$ are also collinear, consequently $I$ lies on the line $DE$.\n\n$DI$ and $QI$ are midlines in the triangles $AMH$ and $ANH$, respectively, hence $DI \\parallel MH$ and $IE \\parallel NH$, therefore $H \\in MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A \\in \\mathcal{M}_n(\\mathbb{C})$ be a matrix with the property $A^T = -A$, where $A^T$ is the transpose of $A$.\n\na) If $A \\in \\mathcal{M}_n(\\mathbb{R})$ and $A^2 = O_n$, prove that $A = O_n$.\n\nb) If $n$ is an odd natural number and there is a matrix $B \\in \\mathcal{M}_n(\\mathbb{C})$ such that $A$ is the adjoint of $B$, prove that $A^2 = O_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Assume $A = (a_{ij})_{1 \\le i,j \\le n}$ and $A^2 = (m_{ij})_{1 \\le i,j \\le n}$. The property $A^T = -A$ leads to the relations $a_{ji} = -a_{ij}$, for $i, j = 1, \\dots, n$, that is $A$ is antisymmetric. Then\n$$\nm_{ii} = \\sum_{j=1}^{n} a_{ij} a_{ji} = - \\sum_{j=1}^{n} a_{ij}^2, \\quad i = 1, \\dots, n.\n$$\nIf $A^2 = O_n$, then $m_{ii} = 0$, $i = 1, \\dots, n$. Since $A \\in \\mathcal{M}_n(\\mathbb{R})$, we obtain $a_{ij} = 0$, for $i, j = 1, \\dots, n$. So $A = O_n$.\n\nb) From the assumption, $A = B^*$, where $B^*$ is the adjoint of $B$. Since $n$ is an odd integer number, we obtain $\\det(A) = \\det(A^T) = \\det(-A) = (-1)^n \\det(A) = -\\det(A)$. Then we obtain $\\det(A) = 0$. Therefore $\\det(BB^*) = \\det(B) \\cdot \\det(B^*) = \\det(B) \\cdot \\det(A) = 0$. From the relation $BB^* = \\det(B)I_n$, we get $\\det(B) = 0$. Hence $\\rank(B) \\le n-1$ and $BB^* = O_n$.\n\n**Case 1.** If $\\rank(B) \\le n-2$, then $B^* = O_n$. Thus, $A^2 = (B^*)^2 = O_n$.\n\n**Case 2.** If $\\rank(B) = n-1$, then $B^* \\ne O_n$. From the Sylvester rank inequality, we have $\\rank(B^*) \\le \\rank(BB^*) + n - \\rank(B) = 1$. We conclude $\\rank(B^*) = 1$ and $(B^*)^2 = \\tr(B^*)B^*$. But $\\tr(B^*) = \\tr(A) = 0$ because $a_{ii} = -a_{ii}$, $i = 1, 2, \\dots, n$. Finally, we get $A^2 = (B^*)^2 = O_n$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, a set $S$ is *n-admissible* if:\n• each element of $S$ is an unordered triple of distinct integers in $\\{1, 2, \\dots, n\\}$,\n• $|S| = n-2$, and\n• for each $1 \\le k \\le n-2$ and each choice of $k$ distinct $A_1, A_2, \\dots, A_k \\in S$,\n$$\n|A_1 \\cup A_2 \\cup \\dots \\cup A_k| \\ge k + 2.\n$$\n\nIs it true that, for all $n > 3$ and for each $n$-admissible set $S$, there exist pairwise distinct points $P_1, \\dots, P_n$ in the plane such that the angles of the triangle $P_iP_jP_k$ are all less than $61^\\circ$ for any triple $\\{i, j, k\\}$ in $S$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The answer is in the affirmative. Note that the condition on $S$ may be rephrased as follows: For any subset $M$ of $N$ of size $|M| \\ge 2$, there are at most $|M| - 2$ sets in $S \\cap P_3(M)$, where $P_3(M)$ is the set of 3-element subsets of $M$.\n\nExtend the problem to $n \\ge 1$ and proceed by induction on $n$; for formal correctness, assume that $|S| \\le \\max(0, n - 2)$. The cases $n = 1, 2$ are both trivial.\n\nLook upon points in the plane as complex numbers. Begin by considering a collection of points $q_1, q_2, \\dots, q_n$, not necessarily distinct, yet not all identical. Fix $q_n = 0$ and associate with each $\\{i, j, k\\}$ in $S$ an equation $q_j - q_i = \\omega \\cdot (q_k - q_i)$, where $\\omega = e^{i2\\pi/3}$. This equation ensures that $q_i, q_j, q_k$ either are the vertices of a clockwise-oriented non-degenerate equilateral triangle or they all coincide.\n\nWe then get a system of $n-2$ linear equations with $n-1$ complex variables $q_1, q_2, \\dots, q_{n-1}$. It is well-known that such a system has a non-trivial solution $(q_1, q_2, \\dots, q_{n-1}, q_n = 0)$. However, some of the $q_i$ may coincide. So $N$ splits into some classes, $N = N_1 \\cup \\dots \\cup N_m$, such that the points $q_i$ and $q_j$ coincide if and only if $i$ and $j$ share the same class and the size of each class is less than $n$. Note that the elements of any triple in $S$ either all lie in the same class or no two lie in the same class.\n\nClearly, each $S_i = S \\cap P_3(N_i)$ satisfies the conditions in the statement relative to $N_i$ of size less than $n$. By the inductive hypothesis, for each $i = 1, 2, \\dots, m$, there exists a point-configuration $\\{r_{ij} : j \\in N_i\\}$ satisfying the corresponding requirements. For each $j = 1, 2, \\dots, n$, choose the index $i$ of the class $N_i$ containing $j$, and let $p_j = q_j + \\epsilon r_{ij}$, where $\\epsilon$ is small enough.\n\nIf the elements of a triple $\\{i, j, k\\}$ in $S$ lie in three different classes, then the angles of triangle $p_i p_j p_k$ are close to those of triangle $q_i q_j q_k$, provided that $\\epsilon$ is small enough,\n\nso the triangle satisfies the required angle-condition. Otherwise, the three elements all lie in some class $N_\\ell$, and the angles of $p_i p_j p_k$ equal those of $r_{\\ell i} r_{\\ell j} r_{\\ell k}$, so they satisfy the requirements by the inductive hypothesis.\n\n**Remark.** This solution proceeds via solving a linear system over $\\mathbb{F}_3$ rather than $\\mathbb{C}$. For each $\\{i, j, k\\} \\in S$, we consider the equation $x_i + x_j + x_k = 0$. Thus, we obtain $n-2$ equations in $n$ variables, so the solution space has dimension at least 2. It follows that there exists a solution for which $x_1, \\dots, x_n$ are not all equal. Let us define the sets $A_u = \\{i \\in [n] : x_i = u\\}$ for $u \\in \\mathbb{F}_3$ corresponding to this solution. We then consider an equilateral triangle $P_0P_1P_2$ in the plane and place the points $p_i$ for $i \\in A_u$ at $P_u$ for each $u \\in \\mathbb{F}_3$. As the $x_i$ solve the system of equations, all $\\{i, j, k\\} \\in S$ satisfy $\\{p_i, p_j, p_k\\} = \\{P_0, P_1, P_2\\}$ or $p_i = p_j = p_k$. We may then proceed by induction by perturbing all of the points as in the first solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$ and $b$ such that $a^{4a} = b^b$.",
    "options": [],
    "answer": "(1, 1), (2, 4), (27, 81)",
    "solution": "If $b \\ge 4a$, then $b > a$ and $b^b > a^{4a}$. Therefore, in this case the equality is impossible.\n\nIf $b < 4a$, we have $b^b = a^{4a} = a^{4a-b} a^b$, so $b^b$ is divisible by $a^b$. Therefore, $b$ is divisible by $a$. It follows that $b = n a$, with $n = 1$ (I), $n = 2$ (II), or $n = 3$ (III).\n\nWe get $(a^4)^a = ((n a)^n)^a$, then $a^4 = n^n a^n$, or $a^{4-n} = n^n$. In case (I), we obtain $a = b = 1$, in case (II) we obtain $a = 2, b = 4$, while $a = 27, b = 81$ in case (III).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCA'B'C'$ be a triangular regular prism, with lateral edges $AA'$, $BB'$, $CC'$. Consider the midpoint $D$ of the edge $BC$ and the parallelogram $ADB'E$. Let $F$ be the orthogonal projection of the point $A'$ on the line $AE$, $d$ be the intersection of the planes $(ADE)$ and $(A'CF)$ and $P$ be the intersection of the line $d$ with the plane $(ABC)$. Prove that $P$ is the baricentre of the triangle $ABC$ if and only if $AB = AA'\\sqrt{2}$.\nValeriu Bărbieru\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "AD is a median in $\\triangle ABC$, so the point $P$ is the baricenter of the triangle $ABC$ if and only if $AP = 2PD$. Since $PF \\parallel DE$, this is equivalent to $AF = 2FE$. Because $AA' \\parallel BB'$, $BD \\parallel EA'$ and $BB' \\perp BD$, the triangle $AEA'$ has a right angle at $A'$. Then $A'E$ and $A'A$ are legs of this triangle, therefore $AF = 2FE \\iff AF \\cdot AE = 2EF \\cdot AE \\iff A'A^2 = 2A'E^2 \\iff A'A^2 = \\frac{1}{2}BC^2 \\iff BC = A'A\\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be nonzero complex numbers of the same modulus for which the numbers $A = a+b+c$ and $B = abc$ are real. Prove that, for any nonnegative integer number $n$, the number $C_n = a^n + b^n + c^n$ is real.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha, \\beta, \\gamma$ be the reduced arguments of the complex numbers $a, b, c$. Since $abc \\in \\mathbb{R}$, it follows that $\\sin(\\alpha + \\beta + \\gamma) = 0$, i.e., $\\alpha + \\beta + \\gamma = k\\pi$, where $k \\in \\mathbb{Z}$. Therefore, we can write $\\gamma = k\\pi - \\alpha - \\beta$ for some integer $k$. Furthermore, from $a+b+c \\in \\mathbb{R}$ we have $\\sin \\alpha + \\sin \\beta + \\sin \\gamma = 0$, thus $\\sin \\alpha + \\sin \\beta = -\\sin \\gamma = -\\sin(k\\pi - \\alpha - \\beta)$, leading to $|\\sin \\alpha + \\sin \\beta| = |\\sin(\\alpha + \\beta)|$, which is equivalent to:\n$$\n\\left| \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} \\right| = \\left| \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha + \\beta}{2} \\right|.\n$$\nIf $\\sin \\frac{\\alpha+\\beta}{2} = 0$, then there is an integer $q$ such that $\\frac{\\alpha+\\beta}{2} = q\\pi$, and $\\gamma = (k-2q)\\pi$, hence $\\sin \\gamma = 0$, which leads to $c \\in \\mathbb{R}$.\nIf $\\sin \\frac{\\alpha+\\beta}{2} \\neq 0$, then $|\\cos \\frac{\\alpha+\\beta}{2}| = |\\cos \\frac{\\alpha-\\beta}{2}|$, from which we have $\\cos^2 \\frac{\\alpha+\\beta}{2} = \\cos^2 \\frac{\\alpha-\\beta}{2}$, thus $\\cos(\\alpha + \\beta) = \\cos(\\alpha - \\beta)$, leading to $\\sin \\alpha \\sin \\beta = 0$, i.e., at least one of the numbers $a$ or $b$ is real.\nTherefore, at least one of the numbers $a, b$, or $c$ is real. Let this number be $a$. From the hypothesis, we obtain that $b+c \\in \\mathbb{R}$, and since $a$ is nonzero, we also have $bc \\in \\mathbb{R}$. If $b$ is real, then $c$ will also be real, so $C_n$ is real for any $n \\in \\mathbb{N}$. If $b \\in \\mathbb{C} \\setminus \\mathbb{R}$, then $b$ and $c$ are the roots of a quadratic equation with real coefficients, thus $b = \\bar{c}$. Consequently, $C_n = a^n + b^n + c^n = a^n + b^n + \\bar{b}^n = a^n + b^n + \\bar{c}^n \\in \\mathbb{R}$.\n\n\nAlternative solution.\nLet $|a| = |b| = |c| = r > 0$. For a complex number $z$ with modulus $r > 0$, we have $\\bar{z} = \\frac{r^2}{z}$. Since $A$ is a real number, we obtain:\n$$\na+b+c = \\bar{a}+\\bar{b}+\\bar{c} = \\frac{r^2}{a}+\\frac{r^2}{b}+\\frac{r^2}{c} = r^2 \\cdot \\frac{ab+bc+ca}{abc} \\in \\mathbb{R},\n$$\nwhich implies $ab + bc + ca \\in \\mathbb{R}$.\nFor $n = 0$ we have $C_0 = 3$, and for $n = 1$ we have $C_1 = A \\in \\mathbb{R}$. Also, for $n = 2$ we have $C_2 = A^2 - 2(ab + bc + ca) \\in \\mathbb{R}$. For $n \\ge 2$, we have:\n$$\nC_{n+1} = A \\cdot C_n - (ab + bc + ca) \\cdot C_{n-1} + B \\cdot C_{n-2}.\n$$\nBy induction, it now follows that $C_n \\in \\mathbb{R}$ for any $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a natural number, and $\\mathcal{F}$ be the set of functions $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ such that $f(k) \\le f(k+1) \\le f(k)+1$, for every $k \\in \\{1, 2, \\dots, n-1\\}$.\n\na) Determine the cardinality of the set $\\mathcal{F}$.\n\nb) Determine the total number of fixed points of the functions in $\\mathcal{F}$.\n\n(A *fixed point* of the function $f$ is a number $x \\in \\{1, 2, \\dots, n\\}$ such that $f(x) = x$.)",
    "options": [],
    "answer": "a) (n+1) 2^{n-2}; b) n 2^{n-1}",
    "solution": "a) We count the functions in $\\mathcal{F}$ with $f(1) = k$, $k = 1, \\dots, n$. We associate to each $i = 2, \\dots, n$ the number $f(i)-f(i-1) \\in \\{0, 1\\}$, with the restriction that there can be at most $n-k$ occurrences of $1$. This association is bijective, and the number of possibilities of choosing the numbers $0$ and $1$ as described above is $\\binom{n-1}{0} + \\binom{n-1}{1} + \\dots + \\binom{n-1}{n-k}$ (these are the possibilities of placing $0, 1, 2, \\dots, n-k$ of $1$). Therefore:\n$$\n|\\mathcal{F}| = \\sum_{k=1}^{n} \\left( \\binom{n-1}{0} + \\binom{n-1}{1} + \\dots + \\binom{n-1}{n-k} \\right) = \\sum_{p=0}^{n-1} (n-p) \\binom{n-1}{p} = \\\\ = n \\cdot 2^{n-1} - (n-1) \\sum_{p=0}^{n-1} \\binom{n-2}{p-1} = (n+1)2^{n-2}.\n$$\n\nb) We count how many times the fixed point $k$ appears in the functions from $\\mathcal{F}$, $k = 1, \\dots, n$ (the same fixed point may appear in multiple functions). We associate to each function for which $f(k) = k$ the numbers $f(i) - f(i-1) \\in \\{0, 1\\}, i = 2, \\dots, n$.\nSince these numbers can be chosen without restrictions, there are $2^{n-1}$ possibilities. Thus, each fixed point appears in $2^{n-1}$ functions, and $\\sum_{f \\in \\mathcal{F}} |\\text{Fix}(f)| = n \\cdot 2^{n-1}$.\nLet $\\mathcal{F}_n$ be the set from the problem statement, and $k_n = |\\mathcal{F}_n|$. Also, let $s_n = \\sum_{f \\in \\mathcal{F}_n} |\\text{Fix}(f)|$. For $n=2$ we have $k_2 = 3$ and $s_2 = 4$.\n\na) We notice that if $f \\in \\mathcal{F}_{n+1}$ and $f(n) \\le n$, then the restriction of $f$ to $\\{1, 2, \\dots, n\\}$ is a function from $\\mathcal{F}_n$. Conversely, any function from $\\mathcal{F}_n$ can be extended to one in $\\mathcal{F}_{n+1}$ in two ways, since $f(n+1) \\in \\{f(n), f(n)+1\\}$.\nFor the case $f(n) = n+1$ we have $f(n+1) = n+1$, and then, going through the values $f(n-1), f(n-2), \\dots, f(1)$, we observe that at each step $f(k+1) - f(k)$ can be $0$ or $1$, so we can construct $f$ in $2^{n-1}$ ways. Therefore, we have the recurrence $k_{n+1} = 2k_n + 2^{n-1}$, and since $k_2 = 3$, we obtain $k_n = (n+1) \\cdot 2^{n-2}$.\n\nb) Let $a_i(n)$ be the number of functions $f \\in \\mathcal{F}_n$ for which $i \\in \\text{Fix}(f)$. Then we have $s_n = a_1(n) + a_2(n) + \\dots + a_n(n)$. Let $f \\in \\mathcal{F}_{n+1}$ be a function for which $f(k) = k$, for $k \\le n$. Then, using the hypothesis relation, we have:\n$$\nf(n) \\le f(n-1) + 1 \\le \\dots \\le f(k) + (n-k) = n.\n$$\nTherefore, the restriction of $f$ to $\\{1, 2, \\dots, n\\}$ is a function from $\\mathcal{F}_n$, i.e., $f$ is obtained from a function from $\\mathcal{F}_n$, to which the value $f(n+1)$ is added. But, since $f(n) \\le n$, the value of $f(n+1)$ can be chosen in two ways, which implies $a_k(n+1) = 2a_k(n)$, for any $k \\le n$.\nTo determine $a_{n+1}(n+1)$, we observe that $f(n+1) = n+1$, and going through the values $f(n), f(n-1), \\dots, f(1)$, we observe that at each step $f(k+1) - f(k)$ is $0$ or $1$, i.e., $f$ can be constructed in $2^n$ ways. Therefore, $a_{n+1}(n+1) = 2^n$.\nFrom here we have the recurrence $s_{n+1} = 2s_n + 2^n$, and since $s_2 = 4$, we obtain\n$$s_n = n \\cdot 2^{n-1}.$$\nWe observe that if $f(\\ell) = \\ell = f(\\ell - 1)$, then there are no fixed points of the function $f \\in \\mathcal{F}$ smaller than $\\ell$. Similarly, if $f(\\ell) = \\ell = f(\\ell + 1)$, then there are no fixed points of $f \\in \\mathcal{F}$ greater than $\\ell$. From here we deduce that the fixed points of a function $f \\in \\mathcal{F}$ form a set of $k$ consecutive numbers $\\{\\ell, \\ell+1, \\dots, \\ell+k-1\\}$.\n\nWe characterize the functions in $\\mathcal{F}$ that have $k \\le n - 2$ fixed points.\nFor $2 \\le \\ell \\le n-k$, we have $f(\\ell - 1) = \\ell$ and $f(\\ell + k) = \\ell + k - 1$. Now for $\\{f(1), f(2), \\dots, f(\\ell - 2)\\}$ we have $2^{\\ell-2}$ ways to construct the function $f$, and for $\\{f(\\ell + k + 1), \\dots, f(n)\\}$ we have $2^{n-\\ell-k}$ ways to construct $f$, so in total we have $2^{n-k-2}$ ways to construct $f$ in this case.\nFor $\\ell = 1$ or $\\ell = n-k+1$, we have fixed only one more point besides the $k$ fixed points, so $f$ can be constructed in $2^{n-k-1}$ ways. Therefore, in total we have $(n-k+3) \\cdot 2^{n-k-2}$ functions in $\\mathcal{F}$ having $k \\le n-2$ fixed points. For $k = n-1$ fixed points, we have only two such functions in $\\mathcal{F}$, and for $k = n$ fixed points, we have only one function in $\\mathcal{F}$.\n\na) From the above, we deduce:\n$$\n\\begin{align*} \n|\\mathcal{F}| &= 3 + \\sum_{k=1}^{n-2} (n-k+3) \\cdot 2^{n-k-2} = 3 + \\sum_{k=1}^{n-2} 2^{n-k-2} + \\sum_{k=1}^{n-2} (n-k-1)2^{n-k-2} \\\\ \n&= \\sum_{k=0}^{n-1} 2^k + \\sum_{k=0}^{n-3} (k+1)2^k = 2^n - 1 + (n-1) \\cdot 2^{n-2} - 2^{n-1} + 1 \\\\ \n&= (n+1) \\cdot 2^{n-2}. \n\\end{align*}\n$$\n\nb) Similarly, we deduce:\n$$\n\\sum_{f \\in \\mathcal{F}} |\\text{Fix}(f)| = 3n-2 + \\sum_{k=1}^{n-2} k(n-k+3) \\cdot 2^{n-k-2} = n \\cdot 2^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39356,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number $n \\ge 2$ is called *special* if there exist $n$ natural numbers whose sum is equal to their product.\n\na) Prove that $5$ is a special number.\n\nb) Determine how many special numbers are in the set $\\{2, 3, \\dots, 2024\\}$.",
    "options": [],
    "answer": "505",
    "solution": "a) Since $1 + 1 + 1 + 3 + 3 = 1 \\cdot 1 \\cdot 1 \\cdot 3 \\cdot 3$, there exist $5$ odd numbers whose sum is equal to their product, so $5$ is special.\n\nb) If $n$ is a special number, then there exist the odd numbers $a_1, a_2, \\dots, a_n$ such that $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n$.\nLet's assume that, among these, $k$ are of the form $M_4 + 3$ and the remaining $n - k$ are of the form $M_4 + 1$. Then $a_1 + a_2 + \\dots + a_n = M_4 + 3k + 1 \\cdot (n - k) = M_4 + 2k + n$, (1).\nSince the product of two odd numbers has the form $M_4 + 1$ if the numbers leave the same remainder when divided by $4$, and the form $M_4 + 3$ otherwise, we infer that the product $a_1 a_2 \\dots a_n$ has the form $M_4 + 1$ when $k$ is even, and the form $M_4 + 3$ when $k$ is odd, (2).\nSince $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n$, (1) and (2) yield $n = M_4 + 1$.\n\nIf $n = 4t + 1$, $t \\in \\mathbb{N}^*$, for $a_1 = a_2 = \\dots = a_{n-2} = 1$, $a_{n-1} = 3$ and $a_n = 2t + 1$, we have $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n = 6t + 3$, so any number of the form $M_4 + 1$ is special. This shows that the special numbers are those of the form $M_4 + 1$.\nThe set $\\{2, 3, \\dots, 2024\\}$ contains $505$ numbers of the form $M_4 + 1$, so it contains $505$ special numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39357,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that there are infinitely many natural numbers $n$ such that $2 \\cdot n$ is a perfect square and $3 \\cdot n$ is a perfect cube.\n\nb) Prove that there is no natural number $m$ such that $2 + m$ is a perfect square and $3 \\cdot m$ is a perfect cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Consider the numbers $n = 72 \\cdot a^6$, with natural $a$. Then $2 \\cdot n = (12 \\cdot a^3)^2$ and $3 \\cdot n = (6 \\cdot a^2)^3$, which shows that every such $n$ is 'good'.\n\nb) If $3 \\cdot m$ is a perfect cube, then $3 \\cdot m$ is a multiple of $27$. In this case $m$ is a multiple of $9$, so $m + 2 = M_3 + 2$. Since the perfect squares are $M_3$ or $M_3 + 1$, $m + 2$ cannot be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us consider the functions $f, g : \\mathbb{R} \\to \\mathbb{R}$, where $f$ is continuous. Assume that, for all real numbers $a < b < c$, there is a sequence $(x_n)_{n \\ge 1}$ which converges to $b$ such that the limit $\\lim_{n \\to \\infty} g(x_n)$ exists and we have\n$$\nf(a) < \\lim_{n \\to \\infty} g(x_n) < f(c).\n$$\na) Give an example of such functions, for which $g$ is discontinuous at any real point.\n\nb) Prove that, if $g$ is a monotone function, then $f = g$.",
    "options": [],
    "answer": "a) Example: f(x) = x for all real x, and g(x) = x if x is rational, g(x) = x + 1 if x is irrational. Then g is discontinuous at every real point, and for any a < b < c, a rational sequence x_n → b gives lim g(x_n) = b, which lies in (f(a), f(c)).\nb) If g is monotone, then g = f on all of R.",
    "solution": "a) Let us consider the functions $f(x) = x$, for all $x \\in \\mathbb{R}$, and\n$$\ng(x) = \\begin{cases} x, & x \\in \\mathbb{Q} \\\\ x+1, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}.\n$$\nThe function $g$ is discontinuous at any real point. Let $a < b < c$ be three arbitrary real numbers. For any sequence $(x_n)_{n \\ge 1}$ of rational numbers converging to $b$, we have $\\lim_{n \\to \\infty} g(x_n) = \\lim_{n \\to \\infty} x_n = b \\in (a, c) = (f(a), f(c))$.\n\nb) Consider $b \\in \\mathbb{R}$ a continuity point of $g$. We will prove $g(b) = f(b)$ by contradiction. If $g(b) < f(b)$ then, based on the continuity of $f$ at the point $b$, there is $a < b$ such that $f(a) > g(b)$. Then, for any sequence $(x_n)_{n \\ge 1}$ converging to $b$, we have $\\lim_{n \\to \\infty} g(x_n) = g(b) < f(a)$, contrary to the hypothesis. If $g(b) > f(b)$ then, by using again the continuity of $f$ at $b$, we can find $c > b$ such that $f(c) < g(b)$. For any sequence $(x_n)_{n \\ge 1}$ converging to $b$, we have $\\lim_{n \\to \\infty} g(x_n) = g(b) > f(c)$, contrary to the hypothesis. So $g(b) = f(b)$. We conclude that $g(x) = f(x)$ at any continuity point $x$ of $g$.\n\nLet $x$ be an arbitrary real number. The monotone function $g$ has lateral limits at $x$. The set of discontinuity points of $g$ is at most countable. So, for any $n \\in \\mathbb{N}^*$, there are two continuity points of $g$: $u_n \\in (x - 1/n, x)$ and $v_n \\in (x, x + 1/n)$. Then $\\lim_{t \\searrow x} g(t) = \\lim_{n \\to \\infty} g(u_n) = \\lim_{n \\to \\infty} f(u_n) = f(x)$ and $\\lim_{t \\nearrow x} g(t) = \\lim_{n \\to \\infty} g(v_n) = \\lim_{n \\to \\infty} f(v_n) = f(x)$. Thus, $\\lim_{t \\searrow x} g(t) = \\lim_{t \\nearrow x} g(t) = f(x)$. From the monotony of $g$, we get $g(x) = f(x)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39359,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $m$ denote $S(m)$ the sum of its natural divisors, and if $n$ and $p$ are positive integers, denote $Q(n, p)$ the sum of the quotients of the division of $n$ by the natural divisors of $p$ (for instance, $Q(18, 10) = 18 + 9 + 3 + 1 = 31$).\nLet $a$ and $b$ be two positive integers.\na) Prove that, if $S(a) = Q(a, b)$ and $S(b) = Q(b, a)$, then $a = b$.\nb) Is it always true that, if $S(a) + S(b) = Q(a, b) + Q(b, a)$, then $a = b$?",
    "options": [],
    "answer": "a) a = b. b) No; for example a = 2 and b = 5.",
    "solution": "a) If $d_1, d_2, \\dots, d_p$ are the positive divisors of a positive integer $n$, then $\\{d_1, d_2, \\dots, d_p\\} = \\{\\frac{n}{d_1}, \\frac{n}{d_2}, \\dots, \\frac{n}{d_p}\\}$.\nLet $b_1, b_2, \\dots, b_q$ be the positive divisors of $b$. Then\n$$\nQ(a, b) \\le \\frac{a}{b_1} + \\dots + \\frac{a}{b_q} = \\frac{a}{b}\\left(\\frac{b}{b_1} + \\dots + \\frac{b}{b_q}\\right) = \\frac{a}{b}S(b) = \\frac{a}{b}Q(b, a), \\quad (1)\n$$\ntherefore $\\frac{Q(a,b)}{a} \\le \\frac{Q(b,a)}{b}$.\n\nSince the statement is symmetrical with respect to $a$ and $b$, $\\frac{Q(b,a)}{b} \\le \\frac{Q(a,b)}{a}$, hence $\\frac{Q(b,a)}{b} = \\frac{Q(a,b)}{a}$, which shows that (1) is an equality.\nThis shows that $a$ is divisible with all the divisors of $b$ and $b$ is divisible with all the divisors of $a$, so $a = b$.\n\nb) It is not true. For instance, if $a = 2$ and $b = 5$, then $S(2) + S(5) = (1 + 2) + (1 + 5) = 9$ and $Q(2, 5) + Q(5, 2) = 2 + (5 + 2) = 9$, but $a \\neq b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39360,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle $ABC$ has $\\angle BAC = 54^\\circ$ and $\\angle ACB = 45^\\circ$. Let $D$ and $E$ be points on the segments $BC$ and, respectively, $AD$ so that $AD = AB$ and $BE = BD$. Denote $F$ the common point of the lines $BE$ and $AC$, and let $DM$ be the bisector of the angle $\\angle ADF$, where $M \\in AC$. The perpendicular from $C$ on $AB$ meets $DM$ in $G$. Prove that:\n\na) the triangle $ABF$ is isosceles;\n\nb) $CG = CM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) From the statement, $\\angle ABC = 81^\\circ$ and, in the isosceles triangle $ABD$, $\\angle ADB = \\angle ABD = 81^\\circ$, hence $\\angle DAB = 18^\\circ$ and $\\angle CAD = \\angle CAB - \\angle DAB = 36^\\circ$.\nThe isosceles triangle $BED$ yields $\\angle BED = \\angle BDE = 81^\\circ$, therefore $\\angle DBE = 18^\\circ$, whence $\\angle ABF = \\angle ABC - \\angle EBD = 63^\\circ$.\n\n![](attached_image_1.png)\n\nFrom the triangle $ABF$, $\\angle ABF = 63^\\circ$ and $\\angle FAB = 54^\\circ$, hence $\\angle AFB = 63^\\circ$, that is triangle $ABF$ is isosceles, with $AF = AB$.\n\nb) The relations $AD = AB$ and $AF = AB$ lead to $AF = AD$, so the triangle $AFD$ is isosceles, with $\\angle AFD = \\angle ADF = \\frac{180^\\circ - \\angle FAD}{2} = 72^\\circ$. Since $DM$ is the bisector of the angle $ADF$, $\\angle ADM = \\angle FDM = \\frac{\\angle ADF}{2} = 36^\\circ$. The triangle $AMD$ gives $\\angle MAD = \\angle ADM = 36^\\circ$, therefore $\\angle CMD = 72^\\circ$.\nFrom $CG \\perp AB$ follows $\\angle ACG = 90^\\circ - \\angle CAB = 36^\\circ$, so the triangle $CMG$ has $\\angle ACG = 36^\\circ$ and $\\angle CMG = 72^\\circ$. This gives $\\angle CGM = 72^\\circ$, that is the triangle $CMG$ is isosceles, with $CG = CM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer, set $M = \\{1, 2, \\dots, n\\}$ and let $k > 0$ be a real number. Let's associate each non-empty subset of $M$ with a point in the plane, such that any two distinct subsets correspond to different points. If the absolute value of the difference between the arithmetic means of the elements of two distinct non-empty subsets of $M$ is at most $k$, we connect the points associated with these subsets with a segment. Determine the minimum value of $k$ such that the points associated with any two distinct non-empty subsets of $M$ are connected by a segment or a broken line.",
    "options": [],
    "answer": "1/2",
    "solution": "To show that $\\frac{1}{2}$ is the required minimum, notice that:\n\n* $1 \\le m_A \\le n$, for every nonempty subset $A \\subset M$, (1);\n* any two one-element subsets are connected with a sequence of subsets, (2).\n\nIndeed, for $k < p$, consider the sequence $\\{k\\}, \\{k, k+1\\}, \\{k+1\\}, \\{k+1, k+2\\}, \\dots, \\{p-1, p\\}, \\{p\\}$. The absolute value of the difference between the arithmetic means of any two consecutive subsets from this sequence is $\\frac{1}{2}$.\n\nLet $A$ and $B$ be two different subsets and $m_A, m_B$ be the arithmetic means of their elements, respectively. Denote by $n_A, n_B$ the closest integers to $m_A$ and $m_B$, respectively. From (1) we have $n_A, n_B \\in [1, n]$ and it is enough to consider the sequence $A, s_{n_A, n_B}, B$, where $s_{n_A, n_B}$ is the sequence described for the subsets $\\{n_A\\}$ and $\\{n_B\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a field $\\mathbb{L}$ with $q$ elements. Prove that:\na) if $q \\equiv 3 \\pmod 4$ and $n \\in \\mathbb{N}$, $n \\ge 2$, is a positive integer divisible by $q-1$, then $x^n = (x^2+1)^n$ for any $x \\in \\mathbb{L}^*$;\nb) if there is an integer $n \\in \\mathbb{N}^*, n \\ge 2$, such that $x^n = (x^2+1)^n$ for any $x \\in \\mathbb{L}^*$, then $q \\equiv 3 \\pmod 4$ and $q-1$ is a divisor of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) As the multiplicative group $\\mathbb{L}^*$ has order $q-1$, we have for any $x \\in \\mathbb{L}^*$, $x^{q-1} = 1$, and as $n$ is a multiple of $q-1$, we get $x^n = 1$, for any $x \\in \\mathbb{L}^*$.\nBecause $4$ is not a divisor of the order $q-1$ of the group $\\mathbb{L}^*$, there are no elements of order $4$ in $\\mathbb{L}^*$, so $x^2 \\ne -1$ for any $x \\in \\mathbb{L}^*$. Thus $x^2+1 \\ne 0$, so $(x^2+1)^n = 1$, for all $x \\in \\mathbb{L}^*$. In conclusion $x^n = (x^2+1)^n$, for $x \\in \\mathbb{L}^*$.\n\nb) If $\\text{char}(\\mathbb{L}) = 2$, we would conclude $1 = 1^n = (1^2 + 1)^n = 0$, a contradiction. So $q$ is odd.\nIn case $q \\equiv 1 \\pmod 4$, consider a generator $a$ of the group $\\mathbb{L}^*$, and consider $b = a^{\\frac{q-1}{4}}$. Thus $b$ has order $4$ in $\\mathbb{L}^*$, that is $b^2 \\ne 1 = (b^2)^2$. This implies $b^2 = -1$ and $b^n = (b^2+1)^n = 0$, a contradiction. Consequently $q \\equiv 3 \\pmod 4$.\nConsider the sets $A = \\{x + x^{-1} \\mid x \\in \\mathbb{L}^*\\}$ and $H = \\{x \\in \\mathbb{L}^* \\mid x^n = 1\\}$. Because $\\mathbb{L}^*$ is abelian, $H$ is a subgroup of $\\mathbb{L}^*$. By the hypothesis, for any $y \\in A$ there is $x \\in \\mathbb{L}^*$ with $y = x + x^{-1} = x^{-1}(x^2 + 1)$, such that $y^n = x^{-n}(x^2 + 1)^n = 1$. As a conclusion, $A \\subseteq H$.\nDefine $f : \\mathbb{L}^* \\to A$ by $f(x) = x + x^{-1}$. By the definition of $A$, $f$ is onto $\\mathbb{L}^* = \\bigcup_{y \\in A} f^{-1}(y)$ that is $q-1 = |\\mathbb{L}^*| = \\sum_{y \\in A} |f^{-1}(y)|$.\nLet $u, v \\in \\mathbb{L}^*$ such that $f(u) = f(v)$, implying\n$$\nu+u^{-1} = v+v^{-1} \\iff u-v = v^{-1}-u^{-1} \\iff (u-v)(uv-1) = 0 \\iff v \\in \\{u, u^{-1}\\}.$$\nAs $u = u^{-1} \\iff u^2 = 1 \\iff u \\in \\{1, -1\\}$, we get $|f^{-1}(2 \\cdot 1)| = |f^{-1}(f(1))| = 1$, $|f^{-1}(2 \\cdot (-1))| = |f^{-1}(f(-1))| = 1$ and $|f^{-1}(y)| = 2$, for all $y \\in A \\setminus \\{2 \\cdot 1, 2 \\cdot (-1)\\}$. Collecting, we get\n$$\nq-1 = |\\mathbb{L}^*| = 1 + 1 + 2 \\cdot (|A| - 2) = 2 \\cdot |A| - 2,\n$$\nthus $|A| = \\frac{q+1}{2}$. As $A \\subseteq H$, we get $|H| \\ge \\frac{q+1}{2} > \\frac{1}{2} \\cdot |\\mathbb{L}^*|$. But $H$ is a subgroup of $\\mathbb{L}^*$, so $H = \\mathbb{L}^*$. As a consequence, $x^n = 1$, for any $x \\in \\mathbb{L}^*$.\nThus we proved that for the generator $a$ with $\\text{ord}(a) = q-1$ we have $a^n = 1$, implying that $q-1$ is a divisor of $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of the acute triangle $ABC$ and let $X$ be the midpoint of the side $BC$. The perpendicular at $H$ to $HX$ intersects the sides $AB$ and $AC$ at the points $Y$ and $Z$ respectively. Let $O$ be the circumcenter of the triangle $ABC$ and $O'$ be the circumcenter of the triangle $BHC$. Prove that:\na) $HY = HZ$;\nb) $\\overrightarrow{AY} + \\overrightarrow{AZ} = 2\\overrightarrow{OO'}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $A'$ be the point diametrically opposite to $A$ on the circumcircle of the triangle $ABC$. It is known that $H$ and $A'$ are symmetric with respect to $X$, so $H$, $X$ and $A'$ are collinear. Then $\\angle ABA' = \\angle ACA' = 90^\\circ$. Thus, the quadrilaterals $A'BYH$ and $A'CZH$ are cyclic, therefore $\\angle YBH = \\angle YA'H$ and $\\angle ZCH = \\angle ZA'H$.\nOn the other hand, from the right triangles $BEA$ and $CFA$ we have $\\angle ABE = \\angle ACF = 90^\\circ - \\angle BAC$ ($E$ and $F$ are the feet of the altitudes from $B$ and $C$ respectively).\nConsequently $\\angle YA'H = \\angle ZA'H$, so $A'H$ is both an altitude and an angle bisector in the triangle $YA'Z$, so it is also a median, that is $HY = HZ$.\n\nb) From a) it follows that $\\overrightarrow{AZ} + \\overrightarrow{AY} = 2\\overrightarrow{AH}$.\nIt is known that the symmetrical $H'$ of the orthocenter $H$ with respect to the side $BC$ is on the circumcircle of the triangle $ABC$, so the symmetrical of the circumcircle of the triangle $BHC$ with respect to $BC$ is the circumcircle of the triangle $BH'C$, i.e. the circumcircle of the triangle $ABC$. Thus, the center of the circumcircle of the triangle $BHC$ is symmetrical to $O$ with respect to $BC$.\nWe deduce that $\\overrightarrow{OO'} = 2\\overrightarrow{OX} = \\overrightarrow{AH}$, since $OX$ is a midline in triangle $HA'A$.\nFrom here it follows $\\overrightarrow{AZ} + \\overrightarrow{AY} = 2\\overrightarrow{OO'}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$. Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy:\n$$\nf(x + y^{2n}) = f(f(x)) + y^{2n-1}f(y),\n$$\nfor all $x, y \\in \\mathbb{R}$, and for which the equation $f(x) = 0$ has a unique solution.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "For $y = 0$, the given relation reduces to $f(x) = f(f(x))$, which means that the given relation becomes:\n$$\nf(x + y^{2n}) = f(x) + y^{2n-1}f(y), \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (1)\n$$\nIf we consider $x = 0$ and $y = 1$ in (1), then $f(0) = 0$. Since the equation $f(x) = 0$ has a unique solution, we have the implication:\n$$\nf(x) = 0 \\Rightarrow x = 0. \\qquad (2)\n$$\nPutting $x = 0$ in (1), we obtain $f(y^{2n}) = y^{2n-1}f(y)$, for all $y \\in \\mathbb{R}$, which means that relation (1) rewrites as:\n$$\nf(x + y^{2n}) = f(x) + f(y^{2n}), \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (3)\n$$\nMoreover, $y^{2n-1}f(y) = f(y^{2n}) = f((-y)^{2n}) = -y^{2n-1}f(-y)$, for all $y \\in \\mathbb{R}$, so $f$ is an odd function. Considering $t = \\sqrt[2n]{y} \\ge 0$, relation (3) becomes:\n$$\nf(x + t) = f(x) + f(t), \\quad \\forall x \\in \\mathbb{R}, t \\ge 0. \\qquad (4)\n$$\n\nUsing the fact that $f$ is odd, for $t < 0$, we obtain:\n$$\nf(x + t) = -f(-x - t) = -(f(-x) + f(-t)) = f(x) + f(t),\n$$\nwhich, together with relation (4), implies:\n$$\nf(x + t) = f(x) + f(t), \\quad \\forall x, t \\in \\mathbb{R}. \\qquad (5)\n$$\nIf $f(x_1) = f(x_2)$, according to relation (5), we have $f(x_1 - x_2) = 0$, which, according to relation (2), implies $x_1 = x_2$, meaning $f$ is one-to-one. But since $f(f(x)) = f(x)$, we have $f(x) = x$, which is a solution to the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer, the set $S = \\{1, 2, 3, \\dots, n\\}$, and the set $\\mathcal{F}$ of functions from $S$ to $S$. We say that a set $\\mathcal{G} \\subset \\mathcal{F}$ is a *generating set* for the set $\\mathcal{H} \\subset \\mathcal{F}$ if every function in $\\mathcal{H}$ can be represented as a composition of functions from $\\mathcal{G}$.\n\na. Let $a : S \\to S$ be defined by $a(n-1) = n$, $a(n) = n-1$, $a(k) = k$ for $k \\in S \\setminus \\{n-1, n\\}$, and let $b : S \\to S$ be defined by $b(n) = 1$ and $b(k) = k+1$ for $k \\in S \\setminus \\{n\\}$. Show that $\\{a, b\\}$ is a generating set for the set $\\mathcal{B}$ of bijective functions from $\\mathcal{F}$.\n\nb. Show that the minimum number of elements in a generating set for $\\mathcal{F}$ is 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $fg$ denote the function composition $f \\circ g$ (where $f, g \\in \\mathcal{F}$), and let $(i_1, i_2, \\dots, i_p)$ denote the function $f : S \\to S$ defined by $f(i_j) = i_{j+1}$ for $j = \\overline{1, p-1}$, $f(i_p) = i_1$, and $f(x) = x$ for $x \\ne i_1, \\dots, i_p$ (where $i_1, \\dots, i_p$ are $p \\ge 2$ distinct elements of $S$).\n\na) We use an inductive approach. For $n=3$, $\\mathcal{B} = \\{a, a^2, b, b^2, ab, ba\\}$.\n\nNow, we suppose the property holds for some $n \\ge 3$ and we show that it also holds for $n+1$. Let $f : S \\cup \\{n+1\\} \\to S \\cup \\{n+1\\}$, $f(n+1) = m$, and let $a', b' : S \\cup \\{n+1\\} \\to S \\cup \\{n+1\\}$ be analogous to $a$ and $b$. Then $((b')^{n-m+1}f)(n+1) = n+1$, so the restriction of $g = (b')^{n-m+1}f$ to $S$ can be written as a composition of $a$ and $b$; we have $f = (b')^m g$ (1).\n\nWe have $(b'a')(n+1) = n+1$ and the restriction of $b'a'$ to $S$ is $b$. Moreover, $((b')^n a' b')(n+1) = n+1$ and the restriction of $(b')^n a' b'$ to $S$ is $a$. Then from (1), it follows that $f$ can be written as a composition of $a'$ and $b'$.\n\nb) We show that if $\\mathcal{G}$ is a generating set for $\\mathcal{F}$, then $|\\mathcal{G}| \\ge 3$.\n\nIf $\\mathcal{G}$ has at most two elements $f$ and $g$, then:\n\n* If both $f$ and $g$ are bijective, then $\\mathcal{G}$ can only generate bijective functions.\n* If both $f$ and $g$ are not bijective, then they are not surjective, so $\\mathcal{G}$ can only generate non-surjective functions.\n* If, for example, $f$ is bijective and $g$ is not bijective, then the bijective functions generated by $\\mathcal{G}$ are $f^n$, $n \\in \\mathbb{N}^*$. In this case, $\\mathcal{G}$ cannot generate both $a$ and $b$ because $ab \\ne ba$, while $f^m f^p = f^p f^m$ for all $m, p \\in \\mathbb{N}^*$.\n\nWe show that a generating set for $\\mathcal{F}$ is $\\mathcal{G} = \\{a, b, c\\}$, where $c : S \\to S$, $c(k) = k$ for $k \\in S \\setminus \\{n\\}$, and $c(n) = n-1$. We prove that any $f \\in \\mathcal{F}$ can be written as a composition of $a, b$, and $c$ by descending induction on the number of elements in the image of $f$. If $|\\text{Im } f| = n$, then $f$ is bijective and we use (a).\n\nWe assume that the statement is true for any $f$ with $|\\mathrm{Im} f| = k$, where $1 < k \\le n$, and we prove it for an arbitrary $g$ with $|\\mathrm{Im} g| = k - 1$. Since $g$ is not injective, there exist $u, v \\in S$, $u \\ne v$, and a bijective function $r$ such that $\\mathrm{Im} gr = \\{1, 2, \\dots, k-1\\}$ and $(gr)(u) = (gr)(v) = k-1$. Let $s : S \\to S$ be defined such that $s(n-1) = u$, $s(n) = v$, and $s(x) = x$ for $x \\ne u, v$. We consider the function $h : S \\to S$ defined as $h(x) = (grs)(x)$ for $x \\le n-1$ and $h(n) = k$. Then $|\\mathrm{Im} h| = k$, so $h$ can be expressed as a composition of the functions $a, b, c$, and $grs = hc$. Thus, $g = hcs^{-1}r^{-1}$ can be expressed as a composition of the functions $a, b, c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a given natural number. We consider the sequence $(x_n)_{n \\ge 1}$ defined by $x_n = \\frac{1}{1+na}$, for every natural nonzero number $n$.\nProve that, for every natural number $k \\ge 3$, there are non-zero natural numbers $n_1 < n_2 < \\dots < n_k$ so that the numbers $x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are the consecutive terms of an arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the requirement by induction on $k$.\n\nWe notice that\n$$\n\\frac{1}{1+ma} + \\frac{1}{(1+ma)(1+2ma)} = \\frac{2}{1+2ma},\n$$\ntherefore there is an arithmetic progression consisting of three terms of the sequence: $x_m, x_{2m}, x_{2m^2a+3m}$.\n\nWe assume that there are $k$ non-zero natural numbers $n_1 < n_2 < \\dots < n_k$ such that the numbers $x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are the consecutive terms of an arithmetic progression. Considering $y = 2x_{n_1} - x_{n_2}$, the numbers $y, x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are in arithmetic progression (in the number of $k+1$).\n\nWe have:\n$$\ny = \\frac{2}{1+n_1a} - \\frac{1}{1+n_2a} = \\frac{1+pa}{(1+n_1a)(1+n_2a)} = \\frac{1+pa}{1+(n_1+n_2+n_1n_2a)a},\n$$\nwhere $p = 2n_2 - n_1$.\n\nTherefore\n$$\n\\frac{y}{1+pa}, \\frac{x_{n_1}}{1+pa}, \\frac{x_{n_2}}{1+pa}, \\dots, \\frac{x_{n_k}}{1+pa}\n$$\nare $k+1$ terms $x_{m_1}, x_{m_2}, \\dots, x_{m_{k+1}}$ of the sequence $(x_n)_{n \\ge 1}$.\n\nThey are in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.\n\nIn addition, we have $m_1 < m_2 < \\dots < m_{k+1}$, because the sequence $(x_n)_{n \\ge 1}$ is strictly monotone. With this, the proof is complete.\nWe notice that $(1+na)(1+ma) = 1 + (mn + m + n)a$, for any nonzero natural numbers $m$ and $n$.\n\nFor $k \\ge 3$ we consider the arithmetic progression $1+a, 1+2a, \\dots, 1+ka$. We divide each term by their product, thus obtaining $k$ terms of the sequence $(x_n)_{n \\ge 1}$ in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the functions $f, g : \\mathbb{R} \\to \\mathbb{R}$, such that $g(x) = 2f(x) + f(x^2)$, for all $x \\in \\mathbb{R}$.\n\na) Prove that if $f$ is locally bounded in the origin and $g$ is continuous in the origin, then $f$ is continuous in the origin.\n\nb) Give an example of a function $f$, discontinuous in the origin, such that $g$ is continuous in the origin.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $\\varepsilon > 0$, be arbitrary. By the hypothesis there are $\\delta_1, M > 0$ such that $|f(x)| < M$, for all $x \\in (-\\delta_1, \\delta_1)$. As $g$ is continuous at $0$, we get a $\\delta_2 > 0$, depending on $\\varepsilon$, such that $|g(x) - g(0)| < \\frac{\\varepsilon}{2}$ for any $x \\in (-\\delta_2, \\delta_2)$. Let $\\delta = \\min\\{\\delta_1, \\delta_2, 1\\}$. From $0 < \\delta \\le 1$, we get $a^2 < \\delta, \\forall a \\in (-\\delta, \\delta)$.\n\nLet $x \\in (-\\delta, \\delta)$. As $|x| < \\delta \\le \\delta_2$, we have\n$$\n\\begin{aligned}\n|f(x) - f(0)| &= \\frac{|2f(x) - 2f(0)|}{2} \\le \\frac{|2f(x) + f(x^2) - 3f(0)| + |f(x^2) - f(0)|}{2} \\\\\n&= \\frac{|g(x) - g(0)|}{2} + \\frac{|f(|x|^2) - f(0)|}{2} < \\frac{\\varepsilon}{4} + \\frac{|f(|x|^2) - f(0)|}{2}.\n\\end{aligned}\n$$\nInductively, we have\n$$\n|f(x) - f(0)| < \\varepsilon \\left( \\frac{1}{2^2} + \\frac{1}{2^3} + \\dots + \\frac{1}{2^{n+1}} \\right) + \\frac{|f(|x|^{2^n}) - f(0)|}{2^n}, \\; n \\in \\mathbb{N}^*\n$$\n\nb) For $a \\in (0, 1)$ and $A = \\{\\pm a^{2^n}, n \\in \\mathbb{Z}\\}$, define\n$$\nf: \\mathbb{R} \\to \\mathbb{R}, \\quad f(x) = \\begin{cases} (-1)^n 2^n, & |x| = a^{2^n}, n \\in \\mathbb{Z} \\\\ 0, & x \\in \\mathbb{R} \\setminus A \\end{cases}.\n$$\nBecause $\\lim_{k \\to \\infty} a^{2^{2k}} = 0$ and $\\lim_{k \\to \\infty} f(a^{2^{2k}}) = \\lim_{k \\to \\infty} 2^{2k} = \\infty$, $f$ is discontinuous at the origin.\nFor $x \\in \\mathbb{R} \\setminus A$, $x^2 \\in \\mathbb{R} \\setminus A$, so $g(x) = 0$. For $x \\in A$, there is $n \\in \\mathbb{Z}$ such that $|x| = a^{2^n}$, so\n$$\ng(x) = 2f(a^{2^n}) + f(a^{2^{n+1}}) = 2^{n+1}[(-1)^n + (-1)^{n+1}] = 0.\n$$\nTherefore, $g$ is the null function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. An arbitrary circle which passes through the points $B$ and $C$ intersects the sides $AC$ and $AB$ for the second time in $D$ and $E$, respectively. The line $BD$ intersects the circumcircle of triangle $AEC$ at $P$ and $Q$, and the line $CE$ intersects the circumcircle of the triangle $ABD$ at $R$ and $S$, such that $P$ is situated on the segment $BD$, and $R$ lies on the segment $CE$. Prove that:\n\na) the points $P, Q, R$ and $S$ are concyclic;\n\nb) the triangle $APQ$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Denote by $X$ the intersection of the lines $CE$ and $BD$. Writing the power of $X$ with respect to all three circles, we obtain:\n$$\nXR \\cdot XS = XB \\cdot XD = XE \\cdot XC = XP \\cdot XQ.\n$$\nSince $\\{X\\} = PQ \\cap RS$, we infer that the points $P, Q, R$ and $S$ are concyclic.\n\nb) The triangles $ADP$ and $APC$ are similar (A.A.), thus $AP^2 = AD \\cdot AC$. Similarly, we find that $AQ^2 = AE \\cdot AB$.\n\n![](attached_image_1.png)\n\nFrom the power of the point $A$ with respect to the circumcircle of the quadrilateral $BCDE$, we infer that $AD \\cdot AC = AE \\cdot AB = \\rho(A)$, therefore $AP = AQ = \\sqrt{\\rho(A)}$.\n\nSimilarly, we prove that $AR = AS = \\sqrt{\\rho(A)}$, therefore the triangle $APQ$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a point in the Cartesian plane. Ann tells Bob a number $0 < a \\le 1$ and he then moves $A$ rightward, leftward, upward or downward to a new position $A'$, $a$ distance apart from $A$. Next, Ann tells Bob a number $0 < a' \\le 1$ and he moves $A'$ rightward, leftward, upward or downward to a new position $A''$, $a'$ distance apart from $A'$ and so on and so forth, as long as Ann wishes. At each step, Bob is free to choose which way to move the point, on one condition: Amongst every 100 consecutive moves, each of his four possible choices should have been made at least once. Ann's goal is to force Bob eventually choose a point strictly more than 100 distance away from $A$. Is it possible for Ann to achieve her goal?\nArgentine, 2014",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is in the affirmative. It is sufficient to prove that there exists a positive number $d$ such that, by Ann providing Bob suitable numbers, she eventually forces the $x$-coordinate of the point increase by at least $d$. Then, using $d$ over and over again, she successively increases the $x$-coordinate of the point by at least $2d, 3d, \\dots$, thus making it as far away from $A$ as needed.\n\nWe now prove that $d = 1/2^{99}$ fits the bill. Ann first provides $1/2^{99}$, then $1/2^{98}$, $1/2^{97}$ and so on and so forth until Bob is forced to make his first move rightward. This latter occurs at step 100 at latest, by the restriction condition on Bob's choices. Let Bob's first move rightward occur at step $k \\le 100$. Clearly, if $k = 1$, the case is settled, so let $k \\ge 2$.\n\nThe first $k-1$ moves are leftward, upward or downward. Any leftward move decreases the $x$-coordinate of the point by the corresponding number Ann provided; upward and downward moves do not change it.\n\nDuring the first $k$ steps, Ann successively provides the numbers\n$$\n\\frac{1}{2^{99}}, \\frac{1}{2^{98}}, \\dots, \\frac{1}{2^{100-(k-1)}} \\text{ and } \\frac{1}{2^{100-k}}\n$$\nThe first $k-1$ moves decrease the $x$-coordinate by at most\n$$\n\\frac{1}{2^{99}} + \\frac{1}{2^{98}} + \\dots + \\frac{1}{2^{100-(k-1)}} = \\frac{1}{2^{100-k}} - \\frac{1}{2^{99}},\n$$\nand the $k$-th increases it by $1/2^{100-k}$. Consequently, the overall variation of the $x$-coordinate is at least\n$$\n- \\left( \\frac{1}{2^{100-k}} - \\frac{1}{2^{99}} \\right) + \\frac{1}{2^{100-k}} = \\frac{1}{2^{99}},\n$$\nas needed. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. A Welsh darts board is a disc divided into $2n$ equal sectors, half of them being red and the other half being white. Two Welsh darts boards are matched if they have the same radius and they are superimposed so that each sector of the first board comes exactly over a sector of the second board. Suppose that two given Welsh darts boards can be matched so that more than half of the pairs of superimposed sectors have different colors. Prove that these Welsh darts boards can be matched so that at least $2\\left\\lfloor\\frac{n}{2}\\right\\rfloor + 2$ pairs of superimposed sectors have the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any two Welsh darts boards that can be matched, if two of their superimposed sectors have the same color, we call it a *concordance*, and if two of their superimposed sectors have different colors, we call it a *non-concordance*.\n\nOn each of the two Welsh darts boards that can be matched, we write $+1$ on every red sector, and $-1$ on every white sector. At some matching of the boards, the product of the numbers written in the overlapping sectors equals $+1$ in the case of a concordance, respectively $-1$, in case of a non-concordance.\n\nMoreover, if $t$ is the number of all the concordances (whites and reds), the sum $S$ of the products of the numbers written on the overlapping sectors represents the difference between the number of concordances and the number of non-concordances, therefore $S = t \\cdot 1 + (2n - t) \\cdot (-1) = 2(t - n)$.\n\nAt any matching, let $k$ be the number of concordances between white sectors and $j$ the number of concordances between red sectors. Consequently, $n-k$ white sectors of the first board (those for which we have non-concordances) overlap over $n-k$ red sectors of the second board, and $n-j$ red sectors of the first board overlap over $n-j$ white sectors of the second board.\n\nTherefore, on the second board we have $k + (n-j)$ white sectors and $j + (n-k)$ red sectors, hence $k + (n-j) = j + (n-k) = n$, thus $k = j$. Consequently, the number $t = k + j$ of all concordances is even.\n\nConsider two matched Welsh darts boards, such that they have more than $n$ non-concordances. Let $u_1, u_2, \\dots, u_{2n} \\in \\{-1, +1\\}$ be the numbers written (clockwise) on the sectors of the first board and $v_1, v_2, \\dots, v_{2n} \\in \\{-1, +1\\}$ the numbers written on the correspondent sectors of the second board.\n\nBy fixing the sector with the number $u_1$ and by rotating the second board, we obtain all the possible matchings, and the sums $S_1 = u_1v_1 + u_2v_2 + \\dots + u_{2n}v_{2n}$, $S_2 = u_1v_2 + u_2v_3 + \\dots + u_{2n}v_1$, ..., $S_{2n} = u_1v_{2n} + u_2v_1 + \\dots + u_{2n}v_{2n-1}$.\n\nMoreover, we have\n$$\nS_1 + S_2 + \\dots + S_{2n} = (u_1 + u_2 + \\dots + u_n)(v_1 + v_2 + \\dots + v_n) = 0.\n$$\nSince initially there were more than $n$ non-concordances between the boards, we have $S_1 < 0$. Therefore, $j = \\frac{1}{2n}$ exists, such that $S_j > 0$. If $t$ is the number of all the concordances of the sum $S_j$, then $2(t-n) > 0$, therefore $2(t-n) \\ge 2$. We obtain $t \\ge n+1$ and since $t$ is even, it follows that $t \\ge 2\\left\\lfloor\\frac{n}{2}\\right\\rfloor + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to \\mathbb{R}$ be a differentiable function, with continuous derivative, such that $f(0) = 0$, and $0 \\le f'(x) \\le 1$, for any $x > 0$. Prove that\n$$\n\\int_{0}^{a} f(t)^{2n+1} dt \\le (n+1) \\cdot \\left( \\int_{0}^{a} f(t)^{n} dt \\right)^{2},\n$$\nfor any $a > 0$ and $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $f'(x) \\ge 0$ for any $x \\ge 0$, the function $f$ is monotonically increasing, so that $f(x) \\ge f(0) = 0$, for any $x \\ge 0$.\n\nConsider an arbitrary fixed $n \\in \\mathbb{N}^*$ and the function $F : [0, \\infty) \\to \\mathbb{R}$ defined for any $x \\ge 0$ by\n$$\nF(x) = (n + 1) \\cdot \\left( \\int_{0}^{x} f(t)^{n} dt \\right)^{2} - \\int_{0}^{x} f(t)^{2n+1} dt.\n$$\nWe will show that $F$ is monotonically increasing, which, since $F(0) = 0$, will prove the stated inequality.\n\nThe function $f$ being continuous, $F$ is differentiable and\n$$\n\\begin{aligned}\nF'(x) &= 2(n + 1) \\cdot \\left( \\int_{0}^{x} f(t)^{n} dt \\right) \\cdot f(x)^{n} - f(x)^{2n+1} \\\\\n&= f(x)^{n} \\cdot \\left( 2(n + 1) \\cdot \\left( \\int_{0}^{x} f(t)^{n} dt \\right) - f(x)^{n+1} \\right).\n\\end{aligned}\n$$\nBecause $f'(x) \\le 1$, for any $x \\ge 0$, we have that $f(x)^n \\ge f(x)^n \\cdot f'(x)$, $\\forall x \\ge 0$, hence\n$$\n(n + 1) \\cdot \\left( \\int_{0}^{x} f(t)^{n} dt \\right) \\ge \\int_{0}^{x} (n + 1) \\cdot f(t)^{n} \\cdot f'(t) dt = f(x)^{n+1}.\n$$\nIt follows that\n$$\n\\begin{aligned}\nF'(x) &= f(x)^n \\cdot \\left( 2(n + 1) \\cdot \\left( \\int_{0}^{x} f(t)^n dt \\right) - f(x)^{n+1} \\right) \\ge \\\\\n&\\ge f(x)^n \\cdot \\left( 2 \\cdot f(x)^{n+1} - f(x)^{n+1} \\right) = f(x)^{2n+1} \\ge 0,\n\\end{aligned}\n$$\nfor any $x \\ge 0$. Thus, $F$ is increasing, and so $F(x) \\ge F(0) = 0$, for any $x \\ge 0$. We obtain the stated inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We say that a $n \\times n$ table is *special* if:\n* each cell of the table contains a 2-digit odd positive integer;\n* the numbers of the table are pairwise distinct;\n* the products of the numbers of each line and the products of the numbers of each column are perfect squares.\nProve that the largest value of $n$ for which there exists a $n \\times n$ special table is equal to 4.",
    "options": [],
    "answer": "4",
    "solution": "Since $11^2 > 100$, the table cannot contain numbers divisible by squares of prime numbers $p$, with $p \\ge 7$.\nSuppose that there exists a number of the table (situated on line $\\ell$ and column $c$) divisible by a prime $p \\ge 17$. Then there exists one more number on line $\\ell$ (and column $c' \\ne c$) divisible by $p$. Also, there exists one more number on column $c$ (and line $\\ell' \\ne \\ell$) divisible by $p$. Therefore, in the cell $(\\ell', c')$ must be a number divisible by $p$. In consequence, in the table must appear a number $N \\ge 7p > 100$ – impossible.\n\n| 11 | $3 \\cdot 11$ | $3^3$ | $5^2$ |\n| --- | --- | --- | --- |\n| $5 \\cdot 11$ | $7 \\cdot 11$ | $3 \\cdot 5$ | $3 \\cdot 7$ |\n| $7^2$ | $3 \\cdot 5^2$ | 13 | $3 \\cdot 13$ |\n| $3^2 \\cdot 5$ | $3^2 \\cdot 7$ | $5 \\cdot 13$ | $7 \\cdot 13$ |\n\nIt follows that the table cannot contain 2-digit numbers that are odd multiples of 17 – 3 numbers, of 19 – 3 numbers, of 23 – 2 numbers, of 29 – 2 numbers, of 31 – 2 numbers and of 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 – one number each; in total, 26 numbers. On the other hand, there are 45 odd 2-digit numbers, so the table can contain at most $45 - 26 = 19$ numbers. Hence $n \\le 4$.\nThe table from above is an example for $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a square $ABCD$ and the points $E$ on the side $CD$, $M$ on the diagonal $AC$ and $P$ on the side $BC$, such that $\\angle BAE = \\angle AEM = \\angle AMP$. Prove that:\n\na) the triangle $AMP$ is isosceles;\n\nb) $EM = DE + PB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We denote $\\angle BAE = \\angle AEM = \\angle AMP = \\alpha$.\nWe have, in turn $\\angle DAE = 90^\\circ - \\alpha$, $\\angle EAC = \\alpha - 45^\\circ$, $\\angle DEA = \\angle EAB = \\alpha$, $\\angle CEM = 180^\\circ - 2\\alpha$.\nFrom the triangle $CPM$ follows that $\\angle PMC = 180^\\circ - \\angle AMP = 180^\\circ - \\alpha$ and then $\\angle CPM = \\alpha - 45^\\circ$. Thus $\\angle EAC = \\angle CPM = \\alpha - 45^\\circ$.\nDenote by $H$ the intersection of the lines $AE$ and $PM$. The quadrilateral $HAPC$ has $\\angle HAC = \\angle HPC = \\alpha - 45^\\circ$ so it is inscribed, hence $\\angle AHP = \\angle ACP = 45^\\circ$. Moreover, $\\angle AHP = \\angle EHM = \\angle ECM = 45^\\circ$, so the quadrilateral $HEMC$ is inscribed and $\\angle HCM = \\angle AEM = \\alpha$.\nReturning to the inscribed quadrilateral $HAPC$, we have $\\angle HPA = \\angle HCA = \\alpha$, so the triangle $AMP$ is isosceles.\n\nb) In the isosceles triangle $MAP$ we have $\\angle MAP = 180^\\circ - 2\\alpha$, thus $\\angle PAB = \\angle CAB - \\angle MAP = 45^\\circ - (180^\\circ - 2\\alpha) = 2\\alpha - 135^\\circ$.\nWe extend side $CE$ with segment $DQ = BP$. From the congruence of the triangles $ADQ$ and $ABP$ (SAS) it follows that $\\angle QAD = \\angle PAB = 2\\alpha - 135^\\circ$. Thus $\\angle EAQ = \\angle EAD + \\angle DAQ = \\alpha - 45^\\circ$, which means $\\angle EAQ = \\angle EAM$. Now the congruence of the triangles $EAQ$ and $EAM$ (ASA) yields $EM = EQ = ED + DQ = ED + BP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a parallelogram $ABCD$ and the points $M$ on the side $DC$ and $E$ and $N$ on the diagonal $AC$, such that $BE \\perp AC$ and $\\frac{CM}{CD} = \\frac{EN}{EA}$.\nProve that if $MN$ and $NB$ are perpendicular, then $ABCD$ is a rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution.\n\nConstruct the parallel to $AB$ through $N$ and denote by $P$ its intersection with the line $BE$.\n\n![](attached_image_1.png)\n\nUsing the fundamental theorem of similarity in the triangle $EAB$:\n$$\n\\frac{NP}{AB} = \\frac{EN}{EA} = \\frac{CM}{CD}.\n$$\nFrom here we obtain $NP = CM$ and, since $NP \\parallel MC$, it follows that $MNPC$ is a parallelogram, so $MN \\parallel CP$.\n\nGiven that $MN \\perp NB$, it follows that $CP \\perp NB$.\n\nIn the triangle $BNC$, $BE$ and $CP$ are the lines supporting the altitudes, so the point $P$ is the orthocenter.\n\nTherefore $NP \\perp BC$ and, since $NP \\parallel CD$, it follows that $BC \\perp CD$, thus $ABCD$ is a rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix integers $a$ and $b$ greater than $1$. For any positive integer $n$, let $r_n$ be the (non-negative) remainder $b^n$ leaves upon division by $a^n$. Assume that there exists a positive integer $N$ such that $r_n < 2^n/n$ for all integers $n \\ge N$. Prove that $a$ divides $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Arguing indirectly, assume that $a \\nmid b$, so $r_n \\ne 0$ for all $n$. Let $M = \\max(b, N)$.\n\nWe now prove that $r_{n+1} \\ge br_n$ for all $n \\ge M$. Indeed, as $r_n < 2^n/b \\le a^n/b$, it follows that $br_n < a^n$ and $b^{n+1} \\equiv br_n \\pmod{a^n}$. Therefore, $br_n$ is the remainder $b^{n+1}$ leaves upon division by $a^n$, i.e., $br_n$ is the smallest non-negative integer $r$ such that $a^n \\mid b^{n+1} - r$. This implies $br_n \\le r_{n+1}$, as $a^{n+1} \\mid b^{n+1} - r_{n+1}$.\n\nTo complete the solution, note that $r_M \\ge 1$, so $r_{M+k} \\ge b^k r_M \\ge 2^k$ for all $k \\ge 0$. On the other hand, $r_{M+k} < 2^{M+k}/(M+k) < 2^k$ for $k$ sufficiently large. This is a contradiction.\nThe conclusion holds under the much less restrictive condition $r_n < 2^n/b$ for all integers $n \\ge N$. The argument hinges on the lemma below:\n\n**Lemma.** Consider two integers $b > a > 1$. If $a$ does not divide $b$, then $\\{b^n/a^n\\} > 1/b$ for infinitely many positive integers $n$; as usual, $\\{x\\}$ denotes the fractional part of the real number $x$.\n\n**Proof.** For every positive integer $n$, write $x_n = \\lfloor b^n/a^n \\rfloor$ and $y_n = \\{b^n/a^n\\}$ and note that $by_n - ay_{n+1} = ax_{n+1} - bx_n$ is an integer.\n\nSuppose now, if possible, that $y_n \\le 1/b$ for all $n \\ge M$. Consider any such $n$ and note that $y_n > 0$, as $a$ does not divide $b$, so $-1 < -a/b < by_n - b/a \\le by_n - ay_{n+1} \\le 1 - ay_{n+1} < 1$. Hence the integer $by_n - ay_{n+1} = 0$, so $y_{n+1} = (b/a)y_n$.\n\nConsequently, $y_n = (b/a)^{n-M} y_M$ for all $n \\ge M$. As $b > a$, it then follows that $y_n \\ge 1$ for all large enough $n$, contradicting the fact that $y_n < 1$ whatever $n$. This establishes the lemma.\n\nBack to the problem, suppose $a$ does not divide $b$. Then $b > a$; otherwise $2^n/b > r_n = b^n \\ge 2^n$ which is impossible. Note that $a^n\\{b^n/a^n\\} = r_n < 2^n/b$ for all large enough $n$. The lemma then implies $1/b < \\{b^n/a^n\\} < (1/b)(2/a)^n \\le 1/b$ for some large enough $n$ which is clearly a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39376,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that the numbers of the positive divisors of $137$, $138$ and $139$ are integer powers of $2$.\n\nb) What is the largest number of consecutive positive integers such that each of them has the number of its positive divisors a power of $2$?",
    "options": [],
    "answer": "7",
    "solution": "a) The numbers $137$ and $139$ are prime ($137 < 139 < 13^2$ and they have no positive divisors less than $13$), so each of them have $2^1$ divisors.\nSince $138 = 2^1 \\cdot 3^1 \\cdot 23^1$, it results that $138$ has $2 \\cdot 2 \\cdot 2 = 2^3$ positive divisors.\n\nb) Extend the sequence $137$, $138$, $139$ with four new numbers to obtain the $7$-element sequence $133$, $134$, $135$, $136$, $137$, $138$, $139$ such that each number has a number of positive divisors that is a power of $2$. Indeed, the numbers $136 = 2^3 \\cdot 17$, $135 = 3^3 \\cdot 5$, $134 = 2 \\cdot 67$, $133 = 7 \\cdot 19$ have, respectively, $8$, $8$, $4$, $4$ positive divisors.\nWe will prove that there are no $8$ consecutive numbers with the requested property. To this end, remark that among $8$ consecutive numbers there is one that leaves the remainder $4$ when divided by $8$, that is a number $a$ of the form $a = 8k+4 = 2^2(2k+1)$. In the prime factors decomposition of $a$, the prime $2$ appears at the second power and $2k+1$ is not a multiple of $2$. In consequence, the number of positive divisors of $a$ is divisible by $2+1=3$, so it is not a power of $2$.\nConsequently, the largest sequence of consecutive positive integers with the numbers of their positive divisors powers of $2$ has $7$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39377,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number $n \\ge 5$ will be called *special* if, no matter how we choose five distinct numbers from $1, 2, 3, \\ldots, n$, we find among them four distinct numbers $a, b, c, d$ so that $a + b = c + d$.\n\na) Prove that $n = 6$ is special.\n\nb) Find all the special numbers.",
    "options": [],
    "answer": "5, 6, 7",
    "solution": "a) Since $1+6 = 2+5 = 3+4$, every set of $5$ numbers from $1, 2, 3, 4, 5, 6$ contains $4$ distinct numbers $a, b, c, d$ so that $a + b = c + d = 7$.\n\nb) Indeed, no $4$ numbers out of $1, 2, 3, 5, 8$ provide equal sums: if we do not choose $8$, then $5+a > b+c$, for every $a, b, c \\in \\{1, 2, 3\\}$, and if we choose $8$, then $8+a > b+c$, for every $a, b, c \\in \\{1, 2, 3, 5\\}$.\nThe number $n = 7$ is special. Indeed:\n* if we do not choose $7$, $4$, or $1$, then the argument from a) applies to the sums $1+6=2+5=3+4$, $1+7=2+6=3+5$, respectively $2+7=3+6=4+5$;\n* if we choose $1$, $4$, or $7$ and two of the numbers $2, 3, 5, 6$, then we get the equal sums $1+4=3+2$, $1+5=2+4$, $1+7=2+6$, $1+7=3+5$, $3+7=4+6$, or $4+7=5+6$.\nSince $5$ is, obviously, special, the special numbers are $5$, $6$ and $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BC$ be a fixed segment in the plane, and let $A$ be a variable point in the plane not on the line $BC$. Distinct points $X$ and $Y$ are chosen on the rays $\\vec{CA}$ and $\\vec{BA}$, respectively, such that $\\angle CBX = \\angle YCB = \\angle BAC$. Assume that the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet line $XY$ at $P$ and $Q$, respectively, such that the points $X, P, Y$, and $Q$ are pairwise distinct and lie on the same side of $BC$. Let $\\Omega_1$ be the circle through $X$ and $P$ centred on $BC$. Similarly, let $\\Omega_2$ be the circle through $Y$ and $Q$ centred on $BC$. Prove that $\\Omega_1$ and $\\Omega_2$ intersect at two fixed points as $A$ varies.\nDenmark, Daniel Pham Nguyen",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X', Y', P'$, and $Q'$ be the reflections across $BC$ of $X, Y, P$, and $Q$, respectively. Then $\\Omega_1$ and $\\Omega_2$ are just the circles $PXX'P'$ and $QYY'Q'$, respectively.\nDenote $\\alpha = \\angle BAC = \\angle CBX = \\angle YCB$. Let $XY$ cross $BC$ at $W$; the case $XY \\parallel BC$ may be treated as a limit case. The symmetry yields that $W$ also lies on the line $X'Y'$. The same symmetry, along with tangency of $PB$ and $QC$ to the circle $ABC$, yields\n$$\n\\alpha = \\angle X'BC = \\angle PBW = \\angle WCQ = \\angle BCY'. \\quad (*)\n$$\nThis yields that each of the triples $(P, B, X')$, $(Q, C, Y')$, $(P', B, X)$, and $(Q', C, Y)$ is collinear, and, moreover, that $PBX' \\parallel Q'CY$ and $P'BX \\parallel QCY'$. It follows now that quadrilaterals $PXX'P'$ and $YQQ'Y'$ are homothetic at $W$. Therefore, so are $\\Omega_1$ and $\\Omega_2$.\n\nLet now $\\Omega_1$ and $\\Omega_2$ cross at $D$ and $D'$. Let $WD$ and $WD'$ meet $\\Omega_1$ again at $E$ and $E'$. Since $W = PX \\cap P'X'$ and $B = PX' \\cap P'X$, the point $B$ lies on the polar of $W$ with respect to $\\Omega_1$. In other words, $W$ and $B$ are inverse with respect to that circle. This yields that the lines $DE'$ and $D'E$ also cross at $B$.\n![](attached_image_1.png)\n\nNow, we have $\\angle BDX = \\angle E'DX = \\angle X'D'E = \\angle X'PE = \\angle BPE = \\angle CYD$ (the last equality holds by means of homothety). Similarly, we have $\\angle DXB = \\angle DXP' = \\angle PX'D' = \\angle PED' = \\angle PEB = \\angle YDC$. Therefore, the triangles $BDX$ and $CYD$ are similar. Firstly, this yields that $\\angle DBC = \\angle DBX + \\angle XBC = \\angle YCD + \\angle BCY = \\angle BCD$, whence $BD = CD$. Secondly, this also implies that $BD/BX = CY/CD$, or $BX \\cdot CY = BD \\cdot CD = BD^2$. But the triangles $BXC$ and $CBY$ are also similar (as both are similar to $ABC$), so $BX/BC = BC/CY$, or $BX \\cdot CY = BC^2$. Thus, $BC = BD = CD$, and the triangle $BCD$ is equilateral. This finishes the solution.\n\n\nSolution 2:\nAll angles in the solution are directed. All segment lengths on lines $BX$ and $CY$ (and parallel to them) are also oriented; we assume that the directions $\\overrightarrow{BX}$ and $\\overrightarrow{CY}$ are positive. As in the solution above, we prove that $BP \\parallel CY$.\nAssume that $ABC$ is oriented anti-clockwise. Let $D$ and $D'$ be the points such that the triangles $DBC$ and $D'CB$ are equilateral, and oriented anti-clockwise. We will show that $D$ and $D'$ lie on the circle $\\Omega_1$; similarly, they lie on $\\Omega_2$.\nNotice that $\\alpha = \\angle BAC = \\angle CBX = \\angle YCB = \\pi - \\angle CBP$; moreover, each of the triangles $XBC$ and $BCY$ is similar to $BAC$ and oriented differently than $BAC$; hence those two triangles are equi-oriented. Let $\\Omega$ denote the circle $(DD'X)$; clearly, its centre lies on the perpendicular bisector of $DD'$, i.e., on $BC$. We aim to prove that $\\Omega$ passes through $P$; that will yield that $\\Omega = \\Omega_1$, which establishes what we are aimed to prove.\nDenote $Z = XB \\cap YC$. Since $\\angle CBZ = \\angle BAC = \\angle ZCB$, we have $ZB = ZC$, and hence $Z$ lies on the perpendicular bisector $DD'$ of $BC$. By similarity, we get $BX/BC = BC/CY$, or $BC^2 = BX \\cdot CY = BX \\cdot (ZY + CZ)$. Since $CY \\parallel BP$, the triangles $XZY$ and $XBP$ are similar, so $BX \\cdot ZY = ZX \\cdot BP$.\n![](attached_image_2.png)\nTherefore, $BD^2 = BC^2 = BX \\cdot ZY + BX \\cdot CZ = ZX \\cdot BP + (BZ + ZX) \\cdot BZ = ZX \\cdot (BP + BZ) + BZ^2$.\nOn the other hand, let $M$ be the midpoint of $BC$, and let $XB$ cross $\\Omega$ again at $P'$. Write the power of point $Z$ with respect to $\\Omega$ as $XZ \\cdot (P'B + BZ) = XZ \\cdot P'Z = ZD \\cdot ZD' = MZ^2 - DM^2 = BZ^2 - MB^2 - DM^2 = BZ^2 - BD^2$.\nThe two obtained relations yield\n$$\nZX \\cdot (BP + BZ) = BD^2 - BZ^2 = ZX \\cdot (P'B + BZ),\n$$\nso $BP = P'B$, and so $P$ and $P'$ are reflections of one another in the line $BC$. Thus, $P$ lies on $\\Omega$, as desired.\n\nRemarks.\n(1) It is also possible to solve the problem via the *moving points* method. Introduce the points $D$ and $D'$ as in Solution 2, and introduce the reflections $X'$, $Y'$, $P'$, and $Q'$ of $X$, $Y$, $P$, and $Q$ in the line $BC$, respectively, as in Solution 1, to read $\\Omega_1$ and $\\Omega_2$ as the circles $PXX'P'$ and $QYY'Q'$, respectively.\nWe need to show that $D$ lies on $\\Omega_2$ (the other incidences are similar). To this end, it suffices to check that $\\angle YDQ = \\angle YY'Q = 90^\\circ - \\angle Y'CB = 90^\\circ - \\angle BAC$.\nFix $B$, $C$, and the circle $ABC$. As $A$ varies on that circle, the lines $BX$, $CY$, $BP$, and $CQ$ remain constant, and $X$ and $Y$ depend projectively on $A$. Choosing $Q_1$ on $CQ$ such that $\\angle YDQ_1 = 90^\\circ - \\angle BAC$, we need to show that $Q_1 = Q$, or that $X$, $Y$, and $Q_1$ are collinear. The point $Q_1$ also depends projectively on $A$, so it suffices to check that the points $Q_1$, $X$, and $Y$ are collinear for four specific positions of $A$.\n\n(2) Although inversion, homothety, and the moving point method are essentially the same thing, i.e., a Möbius transformation, we briefly sketch yet another approach below:\nLet $BP \\cap CQ = D$, $BX \\cap CY = Z$, and $DZ \\cap XY = T$. Let $R_1$ and $R_2$ be distinct points such that triangles $R_1BC$ and $R_2BC$ are equilateral. We first note that $BZ \\parallel CD$ and $CZ \\parallel BD$, so the triangles $DPQ$ and $ZYX$ are homothetic from $T$. Hence, $TP \\cdot TX = TQ \\cdot TY$, so $T$ lies on the radical axis of $\\Omega_1$ and $\\Omega_2$. Since their centres both lie on $BC$, their radical axis is the perpendicular from $T$ to $BC$, i.e. the perpendicular bisector $DZ$ of $BC$. As $CY \\parallel DP$ and $BX \\parallel DQ$, it follows that $BP = YZ \\cdot \\frac{BX}{XZ}$ and $CQ = XZ \\cdot \\frac{CY}{YZ}$. As triangles $BXC$ and $CYB$ are similar, this means that $BP \\cdot CQ = BX \\cdot CY = BC^2$.\nPerform an inversion about $\\Omega$, the circle of radius $BC$ centred at $B$. Let $X'$, $P'$, and $\\Omega'_1$ denote the images of $X$, $P$ and $\\Omega_1$. As $BX \\cdot CY = BP \\cdot CQ = BC^2$, we get $BP' = CQ$ and $BX' = CY$. Since lines $BD$, $CD$ and $BZ$, $CZ$ are symmetric about $DZ$, it follows that $X'$, $P'$ and $Y$, $Q$ are symmetric about $DZ$. Thus, $\\Omega'_1$ and $\\Omega_2$ are reflections about $DZ$. Since $\\Omega_1 \\cap \\Omega_2$ and $\\Omega'_1 \\cap \\Omega_2$ all lie on $DZ$, it follows that $\\Omega$, $\\Omega_1$, $\\Omega'_1$, $\\Omega_2$ all meet on $DZ$. This implies that $\\Omega_1$ and $\\Omega_2$ both pass through $R_1$ and $R_2$, the two required fixed points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39379,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $n$-type tiles triangle, where $n \\geq 2$, is formed by the cells of a $(2n + 1) \\times (2n + 1)$ array which are situated below its diagonals. For instance, a 3-type tiles triangle is the following:\n![](attached_image_1.png)\nDetermine the maximal length of a sequence with pairwise distinct cells in an $n$-type tiles triangle, such that, beginning with the second one, any cell of the sequence has a common side with the previous one.",
    "options": [],
    "answer": "n^2 - n + 1",
    "solution": "We alternately color (as a chessboard) the cells of an $n$-type tiles triangle, as below:\n![](attached_image_2.png)\nWe have $b_n = 1 + 2 + \\ldots + n = \\frac{n(n+1)}{2}$ black cells and $a_n = 1 + 2 + \\ldots + (n-1) = \\frac{n(n-1)}{2}$ white cells. In any sequence of cells with the required propriety, their color alternates, hence its length cannot exceed $2(a_n + 1)$. Therefore, we infer that the maximal length of such a sequence is $L \\leq n^2 - n + 1$.\n\nTo complete the solution, we now provide an example of a sequence in an $n$-type tiles triangle which satisfies the required condition and has length $L = n^2 - n + 1$.\nWe consider a sequence of cells which begins with the top cell. At each step, we choose the cell below the last cell chosen in the previous step, together with a maximum length sequence of cells situated on the same line, up to one of its ends. For instance, for a 3-type tiles triangle the following succession is a solution:\n![](attached_image_3.png)\nThe length of a sequence as we described before is: $1 + 2 + 4 + \\ldots + 2(n - 1) = 1 + n(n - 1) = n^2 - n + 1$.\n\nTherefore, the answer is $L = n^2 - n + 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39380,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $P$ with integer coefficients is *square-free* if it is not expressible in the form $P = Q^2R$, where $Q$ and $R$ are polynomials with integer coefficients and $Q$ is not constant. For a positive integer $n$, let $P_n$ be the set of polynomials of the form\n$$\n1 + a_1x + a_2x^2 + \\dots + a_nx^n\n$$\nwith $a_1, a_2, \\dots, a_n \\in \\{0, 1\\}$. Prove that there exists an integer $N$ so that, for all integers $n \\ge N$, more than 99% of the polynomials in $P_n$ are square-free.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, $|P_n| = 2^n$. Alternatively, but equivalently, we prove that less than $\\frac{1}{100} \\cdot 2^n$ of the polynomials in $P_n$ are not square-free for all but finitely many $n$. Throughout the solution $n$ is always assumed to be sufficiently large to allow room for as large integers $r \\le n$ as the different stages of the argument require. Also, all polynomials have integer coefficients and divisibility is always understood in $\\mathbb{Z}[X]$. The proof consists of three parts:\n\n(1) Upper bounding the number of polynomials in $P_n$ divisible by the square of a non-constant polynomial of degree at most $r$;\n(2) Upper bounding the number of polynomials in $P_n$ divisible by the square of a polynomial of degree greater than $r$; and\n(3) Choosing a suitable $r$.\n\nBefore dealing with the three parts above, we prove a useful lemma.\n\n**Lemma.** The zeroes of any polynomial in $P_n$ all lie in the open disc $|z| < 2$ and their real parts are all less than $C = \\frac{1}{2}(1 + \\sqrt{5})$.\n\n**Proof.** Let $P$ be a degree $m$ polynomial in $P_n$. Leaving aside the trivial case $m=0$, let $m \\ge 1$. Write $P = \\sum_{k=0}^{m} a_k X^k$ and consider a complex number $z$ of absolute value $|z| \\ge 2$. Then\n$$\n|P(z)| \\ge |z|^m - \\sum_{k=0}^{m-1} |z|^k = \\frac{(|z|-2)|z|^m + 1}{|z|-1} \\ge \\frac{1}{|z|-1} > 0,\n$$\nand the first part follows.\n\nTo prove the second part, let $z$ have a real part $\\Re z \\ge C > 1$. Clearly, $|z| \\ge \\Re z \\ge C > 1$. Leaving aside the trivial cases $m=0$ and $m=1$, let $m \\ge 2$ to write\n$$\n\\begin{aligned}\n\\left| \\frac{P(z)}{z^m} \\right| &\\ge \\left| a_m + \\frac{a_{m-1}}{z} \\right| - \\frac{1}{|z|^2} - \\dots - \\frac{1}{|z|^m} \\\\ &> \\left| a_m + \\frac{a_{m-1}}{z} \\right| - \\frac{1}{|z|^2 - |z|} \\\\ &\\ge \\left| a_m + \\frac{a_{m-1}}{z} \\right| - 1, \\quad \\text{as } |z| \\ge C, \\\\ &\\ge \\Re \\left( a_m + \\frac{a_{m-1}}{z} \\right) - 1 \\ge a_m - 1 \\ge 0.\n\\end{aligned}\n$$\nThis establishes the second part and completes the proof. (The argument in the second part shows in fact that, if $\\Re z > 0$ and $|z| \\ge C$, then $P(z) \\ne 0$. Hence the zeroes of $P$ with a positive real part all lie in the open disc $|z| < C$.)\n\nTo deal with (1), consider a polynomial $P$ in $P_n$ divisible by the square of a non-constant polynomial $Q$ of degree $q \\le r$. We first show that there are at most $(2^{2r+1} - 1)^{r+1}$ such $Q$'s.\n\nClearly, the leading coefficient of $Q$ is $\\pm 1$. The zeroes $z_1, \\dots, z_q$ of $Q$ are amongst those of $P$, so $|z_k| < 2$, by the lemma. The absolute value of the coefficient of $X^{q-k}$ in $Q$ is then $\\left| \\sum z_{i_1} \\cdots z_{i_k} \\right| < \\binom{q}{k} \\cdot 2^k < 2^{q+k} \\le 2^{2q} \\le 2^{2r}$. Hence each coefficient of $Q$ takes on at most $2 \\cdot 2^{2r} - 1 = 2^{2r+1} - 1$ values, so there are at most $(2^{2r+1}-1)^{r+1}$ such $Q$'s, as stated.\n\nNext, upper bound the number of $P$'s in $P_n$ that are divisible by the same $Q^2$. Consider such a $P = a_0 + \\dots + a_n X^n$ and let $S_n(P)$ be the set of all polynomials $R = b_0 + \\dots + b_n X^n$ in $P_n$ such that $b_k \\neq a_k$ for exactly one $k$; note that $k \\ge 1$, as $b_0 = a_0 = 1$. Alternatively, but equivalently, there exists a $k \\ge 1$ such that $b_\\ell = a_\\ell$ for $\\ell \\neq k$ and $b_k + a_k = 1$. Hence $|S_n(P)| = n$. Note further that $P-R = \\pm X^k$ vanishes at 0, whereas $Q$ does not, as $Q(0)^2$ divides $P(0) \\neq 0$. Hence $R$ is not divisible by $Q^2$, showing that none of the $n$ polynomials in $S_n(P)$ is.\n\nConsider now distinct $P_1$ and $P_2$ in $P_n$, both divisible by $Q^2$, to show that $S_n(P_1)$ and $S_n(P_2)$ are disjoint: If they shared some $R$, then $P_1 - R = \\pm X^{k_1}$ and $P_2 - R = \\pm X^{k_2}$ for some distinct $k_1, k_2 \\ge 1$, so $P_1 - P_2 = \\pm X^{k_1} \\pm X^{k_2}$ would be divisible by $Q^2$, which is clearly not the case.\n\nBy the two paragraphs above, there are then at most $2^n/(n+1)$ polynomials in $P_n$ that are divisible by the same $Q^2$.\n\nAs there are at most $(2^{2r+1}-1)^{r+1}$ such $Q$'s of degree at most $r$, there are at most\n$$\n\\frac{(2^{2r+1}-1)^{r+1}}{n+1} \\cdot 2^n\n$$\npolynomials in $P_n$ divisible by the square of a non-constant polynomial of degree at most $r$. For a fixed $r$ this upper bound is clearly of order $o(2^n)$.\n\nTo deal with (2), consider a polynomial $P$ in $P_n$ divisible by the square of a non-constant polynomial $Q$ of degree $q > r$.\n\nThe zeroes $z_1, \\dots, z_q$ of $Q$ are amongst those of $P$, so $|z_k| < 2$ and $\\mathfrak{R}z_k < C$, by the lemma. As the leading coefficient of $Q$ is clearly $\\pm 1$,\n$$\n|Q(3)| = |(3 - z_1) \\cdots (3 - z_q)| \\ge (3 - \\mathfrak{R}z_1) \\cdots (3 - \\mathfrak{R}z_q) > (3 - C)^q > (3 - C)^r.\n$$\nAs $3 - C > 1$, letting $r$ be large enough, $P(3)$ is then divisible by $d^2$ for some large enough $d > (3 - C)^r$.\n\nConsider the largest integer $s = s_d$ satisfying $3^s < d^2 \\le 3^{s+1}$. Clearly, $s \\ge 1$. Write $P = \\sum_{k=0}^n a_k X^k$. Then $1 = a_0 \\le \\sum_{k=0}^{s-1} 3^k a_k \\le \\sum_{k=0}^{s-1} 3^k = \\frac{1}{2}(3^s - 1) < \\frac{1}{2}(d^2 - 1) < d^2$. As for distinct choices of $a_1, \\dots, a_{s-1}$ from $\\{0, 1\\}$ the sums $\\sum_{k=0}^{s-1} 3^k a_k$ are pairwise distinct, they are also pairwise distinct modulo $d^2$. Noting that these sums are all positive, it follows that there are at most $2^{n-s+1}$ polynomials $P$ in $\\mathcal{P}_n$ such that $P(3)$ is divisible by $d^2$.\n\nIf $r$ is sufficiently large, then so is $d > (3 - C)^r$. Thus, if $d$ is large enough, then $2^s > d^{5/4}$, as $d^2 \\le 3^{s+1}$ and $\\log_2 3 < \\frac{8}{5}$, so there are at most $2^{n-s+1} \\le 2^{n+1}d^{-5/4}$ polynomials $P$ in $\\mathcal{P}_n$ such that $P(3)$ is divisible by $d^2$. Hence, if $d_0$ is large enough, then the number of such $P$'s is at most $2^{n+1} \\sum_{d>d_0} d^{-5/4}$.\n\nNow, as $\\sum_{d \\ge 1} d^{-5/4}$ converges, given any $c > 0$, the remainder $\\sum_{d>d_0} d^{-5/4} < c$ for some large enough $d_0$ depending on $c$, of course. For any such $d_0$, the number of $P$'s in $\\mathcal{P}_n$ such that $P(3)$ is divisible by $d^2$ is then less than $2c \\cdot 2^n$.\n\nConsequently, so is the number of polynomials in $\\mathcal{P}_n$ divisible by the square of a polynomial of degree greater than $r$.\n\nFinally, we deal with (3). Fix any $c > 0$. Then choose $r$ large enough so that the remainder $\\sum_{d > (3-C)r} d^{-5/4} < c$. At this stage $n > r$. Let further $n > \\frac{1}{c}(2^{2r+1}-1)^{r+1}$. By (1) and (2), the number of non-square-free polynomials in $\\mathcal{P}_n$ is then less than $3c \\cdot 2^n$. Setting $c = \\frac{1}{300}$ provides the answer to the problem at hand.\n\n\nAlternative solution:\n\nWe present an alternative approach to parts (1) and (2).\n\nTo deal with (1), use the first part of the lemma to bound the number of possible polynomials $Q$ by some constant. For every such $Q$, we then prove that few polynomials in $\\mathcal{P}_n$ are divisible by $Q$. This follows from the claim below:\n\n**Claim.** *Given a non-constant polynomial $Q$, the number of polynomials in $\\mathcal{P}_n$ that are divisible by $Q$ does not exceed $\\binom{n}{\\lfloor n/2 \\rfloor}$.*\n\n**Proof.** Let $\\zeta$ be a non-zero complex root of $Q$ (if there are no such, then no polynomial in $\\mathcal{P}_n$ is divisible by $Q$). Then each polynomial $P = p_n X^n + \\dots + p_1 X + p_0$ in $\\mathcal{P}_n$ divisible by $Q$ satisfies $\\sum_{i=0}^n p_i \\cdot c^\\zeta = 0$, for any complex $c$. Choose a suitable $c$ so that the $a_i = \\text{Re } c^\\zeta$ are all non-zero. Then $\\sum_{i=0}^n a_i p_i = 0$.\n\nPartially order the (binary) tuples of coefficients by letting $(p_1, p_2, \\dots, p_n) \\preceq (p'_1, p'_2, \\dots, p'_n)$ if and only if the non-zero $a_i(p'_i - p_i)$ are all positive. The tuples corresponding to polynomials divisible by $Q$ then form an independent set (anti-chain) in the $\\preceq$-partially ordered $n$-cube.\n\nAssign each tuple $(p_1, p_2, \\dots, p_n)$ the tuple $(\\sigma p_1, \\sigma p_2, \\dots, \\sigma p_n)$, where $\\sigma p_i = p_i$, if $a_i > 0$, and $\\sigma p_i = 1 - p_i$, otherwise. This assignment shows $\\preceq$ isomorphic to the (index) set-inclusion partial order on the binary $n$-cube, so the length of any $\\leq$-anti-chain is at most $\\binom{n}{\\lfloor n/2 \\rfloor}$, by Sperner's theorem. This proves the claim.\n\nThe standard bound provided by Stirling's formula (or any of its elementary relaxations) establishes part (1).\n\nTo prove (2), deal more algebraically. Let $P$ be a polynomial in $P_n$ divisible by some $Q^2$, where $\\deg Q = d > r$. Reduce modulo 2 to get the polynomials $\\overline{P}$ and $\\overline{Q}$, where $\\deg \\overline{Q} = d$, as the leading coefficient of $Q$ is $\\pm 1$. Then $\\overline{P}$ is divisible by $\\overline{Q}^2 = \\overline{Q}(X^2)$. Write $\\overline{P} = \\overline{P}_+(X^2) + X\\overline{P}_-(X^2)$. Then $\\overline{P}_+$ and $\\overline{P}_-$ are both divisible by $\\overline{Q}$. So, for a fixed $\\overline{Q}$, the number of such polynomials does not exceed $2^{n-2d}$. Hence for all degree $d$ polynomials $\\overline{Q}$, the number of such $P$'s does not exceed $2^{d-1} \\cdot 2^{n-2d} = 2^{n-d-1}$, so their fraction in $P_n$ is at most $2^{-d-1}$. Finally, sum over all $d \\ge r$, to conclude that the fraction of $P$'s in (2) does not exceed $2^{-r}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39381,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a positive integer $n$ cool if it is a perfect square and there exist positive integers $a$, $b$, $c$, $d$, different from each other, with $a > b$ and $c > d$, such that $n = a^3 - b^3 + c^3 - d^3$. For instance, $225$ is cool because $225 = 15^2$ and $225 = 7^3 - 5^3 + 2^3 - 1^3$. Show that:\n\na) $2025$ is cool;\n\nb) there are infinitely many cool numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}^*$. Show that for any choice of $x_1, x_2, \\dots, x_n \\in \\mathbb{N}^*$ such that $x_k \\le k$ for all $k \\in \\{1, 2, \\dots, n\\}$ and the sum $x_1 + x_2 + \\dots + x_n$ is odd, there exist $\\varepsilon_1, \\varepsilon_2, \\dots, \\varepsilon_n \\in \\{-1, 1\\}$ such that $\\sum_{k=1}^n \\varepsilon_k \\cdot x_k = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ and $(b_n)_{n \\ge 1}$ be the sequences defined by the equations\n$$\n\\prod_{k=1}^{n} (2k^2 + i) = a_n + ib_n, \\quad n = 1, 2, \\dots\n$$\nProve that the sequence $\\left(\\frac{a_n}{b_n}\\right)_{n \\ge 1}$ is convergent and find its limit.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three lines in space are pairwise skew and are located at distance $1$ from each other. Prove that there exists a line situated at distance $1$ from each of the three lines.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$ and $\\angle BAC = 20^\\circ$. Points $X$ and $Y$ are taken on side $AC$ such that $AX = YC = BC$. Find $\\angle XBY$.\nMihaela Berindeanu",
    "options": [],
    "answer": "20°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = 2AC$. Let $D$ be the midpoint of side $AB$. A point $E$ is chosen on side $BC$ such that $\\angle BDE = \\angle ACB$.\n\na) Show that the ray ($EA$ bisects angle $\\angle CED$.\n\nb) Let $O$ be the intersection of lines $CD$ and $AE$. Show that $\\angle COA = \\angle ACB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a set $M$ of real numbers satisfying the following properties:\na) $(1 + \\sqrt{2}) \\in M;$\nb) if $x, y \\in M$, then $x \\cdot y \\in M$ and $(6x - y) \\in M$;\nc) if $(6x + y) \\in M$, then $x \\in M$ or $y \\in M$.\nProve that the numbers $3 + 4\\sqrt{2}$, $\\sqrt{2} - 1$, and $2025$ are elements of the set $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $x, y, z \\in \\mathbb{R}$, then the following inequality holds:\n$$\n\\sum_{\\text{cyc}} \\sqrt{1 + (x\\sqrt{1+y^2} + y\\sqrt{1+x^2})^2} \\geq xy + yz + zx + 2(x+y+z).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exist infinitely many natural numbers $n$ for which $2025 \\cdot n$ is a perfect cube, and $5202 \\cdot n$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the sets $A$ of positive natural numbers, with at least four elements $a_1 < a_2 < a_3 < \\dots$, such that:\n$$\n\\bullet \\ a_1 = 1, a_2 = 2;\n$$\n$$\n\\bullet \\ a_i a_{i+3} + 1 = a_{i+1} a_{i+2}, \\text{ for every positive integer } i;\n$$\n$$\n\\bullet \\ A \\text{ contains exactly four prime numbers.}\n$$",
    "options": [],
    "answer": "{1, 2, 3, 5, 7}",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the quadruples of positive integers such that their sum is a perfect square and if $3$ is subtracted from the first, $3$ is added to the second, the third is multiplied by $3$, and the fourth is divided by $3$, then the results are all equal. Find the smallest term of all these quadruples.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the continuous and surjective functions $f : \\mathbb{R} \\to \\mathbb{R}$ with the following two properties:\n(i) $2f(1) = f(0) + f(2)$;\n(ii) for any real numbers $a, b, c$, if $f(a), f(b), f(c)$ are terms of an arithmetic progression then $a, b, c$ are also terms of an arithmetic progression.",
    "options": [],
    "answer": "All functions f(x) = ax + b with a ≠ 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{N}^* \\to \\mathbb{N}$ that satisfy the following properties:\na) the expression $n^3 - n f(n) - 1$ is a cube of a natural number for all $n \\in \\mathbb{N}^*$;\nb) the function $g: \\mathbb{N}^* \\to \\mathbb{N}$ defined by $g(n) = 4n - f(n)$ is increasing.",
    "options": [],
    "answer": "f(n) = 3n − 3 for all n ∈ ℕ*",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$. Outside the triangle, construct the segments $CF \\perp BC$ with $CF = \\frac{AC}{2}$ and $CG \\perp AC$ with $CG = \\frac{BC}{2}$. Let $X$ be the point of intersection $OC \\cap FG$. Prove that:\n$$\n\\overrightarrow{OX} = \\frac{\\overrightarrow{OH} - \\overrightarrow{OA} + \\overrightarrow{BF} + \\overrightarrow{AG}}{2}.\n$$\nMihaela Berindeanu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $R > 0$, $n \\in \\mathbb{N}$, $n \\ge 3$, and consider a convex $n$-gon inscribed in a circle of center $O$ and radius $R$. Determine the perimeter of the polygon, knowing that its area is equal to $\\frac{nR^2}{2}$.\nAurel Doboşan",
    "options": [],
    "answer": "4*sqrt(2)*R",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39396,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a natural number is *special* if it can be written as the sum of two or more consecutive natural numbers. Determine how many natural numbers less than $1000$ are special.",
    "options": [],
    "answer": "989",
    "solution": "Let $n$ be a natural number. We want to count the number of $n < 1000$ that can be written as the sum of two or more consecutive natural numbers.\n\nLet the consecutive numbers be $a, a+1, \\ldots, a+k-1$ for $k \\geq 2$. Their sum is:\n$$\nS = a + (a+1) + \\cdots + (a+k-1) = k a + \\frac{k(k-1)}{2}\n$$\nSo $n = k a + \\frac{k(k-1)}{2}$ for some $k \\geq 2$ and $a \\geq 1$.\n\nSolving for $a$:\n$$\nka = n - \\frac{k(k-1)}{2} \\implies a = \\frac{n - \\frac{k(k-1)}{2}}{k}\n$$\nWe require $a \\geq 1$ and $a$ integer, so $n - \\frac{k(k-1)}{2}$ must be divisible by $k$ and $n \\geq \\frac{k(k+1)}{2}$.\n\nBut the key is: Which $n$ cannot be written in this way?\n\nIt is a well-known result that the numbers that cannot be written as the sum of two or more consecutive natural numbers are the powers of $2$ (i.e., $1, 2, 4, 8, 16, \\ldots$).\n\nProof:\nLet $n$ be a natural number. $n$ can be written as the sum of $k$ consecutive natural numbers if and only if there exists $k \\geq 2$ such that $n - \\frac{k(k-1)}{2}$ is divisible by $k$ and $n \\geq \\frac{k(k+1)}{2}$.\n\nBut the only numbers that cannot be written in this way are the powers of $2$.\n\nSo, the number of special numbers less than $1000$ is $999$ minus the number of powers of $2$ less than $1000$.\n\nThe powers of $2$ less than $1000$ are:\n$1, 2, 4, 8, 16, 32, 64, 128, 256, 512$\n\n$2^{0} = 1$\n$2^{1} = 2$\n$2^{2} = 4$\n$2^{3} = 8$\n$2^{4} = 16$\n$2^{5} = 32$\n$2^{6} = 64$\n$2^{7} = 128$\n$2^{8} = 256$\n$2^{9} = 512$\n$2^{10} = 1024 > 1000$\n\nSo there are $10$ powers of $2$ less than $1000$.\n\nTherefore, the answer is:\n$$\n999 - 10 = 989\n$$\n\n**Answer:** There are $989$ special natural numbers less than $1000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be two matrices in $M_n(\\mathbb{C})$. Assume $a \\in \\mathbb{R} \\setminus \\{-1, 0, 1\\}$, such that $Q^2 = (a-1)QP - (a+1)PQ$. Prove that $(PQ - QP)^n = O_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39398,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each integer $n \\ge 2$, we consider the set of all $n$-th roots of unity $U_n = \\{z \\in \\mathbb{C} \\mid z^n = 1\\}$. Determine the integers $n \\ge 2$ for which there exist $x, y \\in U_n$, $x \\ne y$ such that $x + y \\notin U_k$ for any integer $k \\ge 2$.",
    "options": [],
    "answer": "All integers n ≥ 2 except n = 3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ with $AC > AB$ and $\\angle ACB = 60^\\circ$. Let $D$ be a point on side $BC$ such that $AD$ is the angle bisector of $\\angle BAC$ and $AB = AD$. Point $E$ lies on line $AD$ such that $D$ is between $A$ and $E$, and $AE = AC$. Point $F$ lies on line $AB$ such that $B$ is between $A$ and $F$. Point $T$ is on segment $AE$ such that $BF = TE = CD$. Show that:\n\na) $EF = CT$;\n\nb) $AC = BC + BT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39400,
    "subject": "Mathematics (Multi-modal)",
    "question": "What are the finite unitary rings for which we can find 3 nonzero elements (not necessarily distinct), whose sum equals their product?",
    "options": [],
    "answer": "All finite unitary rings except the field with three elements.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle and $M$ the midpoint of the side $AD$. Construct the equilateral triangle $DME$, with $E$ and $B$ on the same side of the line $MD$, and the equilateral triangle $BEF$, with $M$ and $F$ on opposite sides of the line $BE$. Show that if the midpoint of the segment $MF$ lies on the line $AC$, then $ABCD$ is a square.\nAdrian Bud",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the finite commutative rings $A$ with at least four elements, with the property that for any choice of a proper subset $S$ of $A^*$ with at least two elements, the equality\n$$\n\\sum_{x \\in S} x = \\prod_{x \\in A^* \\setminus S} x\n$$\nholds.",
    "options": [],
    "answer": "The finite field with four elements (F4).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real numbers $a$, $b$, $c$, with sum $s$, such that\n$$\n(\\sqrt{s} - \\sqrt{a})(\\sqrt{s} - \\sqrt{b}) + (\\sqrt{s} - \\sqrt{b})(\\sqrt{s} - \\sqrt{c}) + (\\sqrt{s} - \\sqrt{c})(\\sqrt{s} - \\sqrt{a}) = \\frac{s}{2}.\n$$",
    "options": [],
    "answer": "a = b = c = 0 (and s = 0)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in $\\mathbb{R}$ the equation $\\cos[x] + \\arccos{x} = x$.",
    "options": [],
    "answer": "0.9396",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the natural numbers $n$ that are perfect squares, $2 \\le n \\le 1000$, such that $n = a^2 + b^2 + c^2$, where $a$ is the smallest proper divisor of $n$, and $b$ and $c$ are divisors of $n$, not necessarily distinct, greater than 1.",
    "options": [],
    "answer": "36",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the natural numbers $m$, $n$, $p$ ($m, n \\ge 1$, $p \\ge 2$, $n$ odd), which simultaneously satisfy the equations\n$$\np + p^2 + \\dots + p^n = m^2 \\quad \\text{and} \\quad \\lfloor \\sqrt{p} \\rfloor + \\lfloor \\sqrt{p^2} \\rfloor + \\dots + \\lfloor \\sqrt{p^n} \\rfloor = m^2 - 5m.\n$$",
    "options": [],
    "answer": "(m, n, p) = (6, 1, 36)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [\\frac{1}{2}, 2] \\to \\mathbb{R}$ be a continuous and monotone function, for which the following equality holds:\n$$\n\\int_{\\frac{1}{2}}^{2} \\frac{(1-x^2) \\cdot f(x)}{1+x^2+x^4} \\, dx = 0.\n$$\nShow that\n$$\n\\int_{\\frac{1}{2}}^{2} \\frac{x f(x)}{1+x^2} \\, dx = f(2) \\cdot \\ln 2.\n$$",
    "options": [],
    "answer": "f(2) · ln 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sets $M$ having at least two elements, all of which are prime natural numbers, and with the property that for any two distinct elements chosen from $M$, their difference is equal to 1 or to a prime number belonging to the set $M$.",
    "options": [],
    "answer": "{2, 3} and {2, 3, 5}",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : (0, \\infty) \\to (0, \\infty)$ such that $f(x^y) = (f(x))^{f(y)}$, for any positive real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 1 for all x > 0, and f(x) = x for all x > 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set\n$$\nA = \\{(x, y, z) \\mid x, y, z \\in \\mathbb{R}, 88(x+y+z) = 33(xy+xz+yz) = 24(x^2+y^2+z^2) \\neq 0\\}.\n$$\n$$\n\\text{Determine } \\max_{(x,y,z) \\in A} (\\max\\{x, y, z\\} - \\min\\{x, y, z\\}).\n$$",
    "options": [],
    "answer": "2*sqrt(3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to \\mathbb{R}$ be a differentiable function, with continuous derivative, such that $f(0) > 0$, $f'(x) + (f(x))^2 > 0$, for any $x > 0$ and\n$$\n\\lim_{x \\to \\infty} [f'(x) + (f(x))^2] = 0.\n$$\nShow that $\\lim_{x \\to \\infty} f(x) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and $M = \\{z_1, z_2, \\dots, z_n\\}$ be a given set of complex numbers with the sum of the elements being different from zero. Starting from the set $M$, we repeatedly apply the following transformation: at one step, we replace each of the $n$ elements of the set from the previous step with the sum of the other elements of the set from the previous step. Is it possible that, after a finite number of steps, we will obtain a set of complex numbers $\\{w_1, w_2, \\dots, w_n\\}$ such that $\\sum_{1 \\le i < j \\le n} w_i w_j = \\sum_{1 \\le i \\le n} z_i z_j$ and $\\sum_{1 \\le i \\le n} w_i \\ne \\pm \\sum_{1 \\le i \\le n} z_i$?\n\nDorin Andrica and Sorin Monel Budişan",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest natural number $n \\ge 3$ with the property that there exists a unique set of natural numbers $a_1 < a_2 < \\dots < a_n \\le 100$ which are directly proportional to $1, 2, \\dots, n$.",
    "options": [],
    "answer": "51",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 3$. For any $x \\in U(\\mathbb{Z}_n)$ we denote $o_1(x)$ and $o_2(x)$ the orders of the element $x$ in the groups $(\\mathbb{Z}_n, +)$, respectively $(U(\\mathbb{Z}_n), \\cdot)$, and $a(x) = o_1(x) + o_2(x)$. We consider the set $T_n = \\{a(x) \\mid x \\in U(\\mathbb{Z}_n)\\}$.\n\na) Determine $T_8$.\n\nb) Prove that there are at most 19 numbers $n$, for which $T_n$ has precisely two elements.",
    "options": [],
    "answer": "{9, 10}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square with side length $1$.\n\na) Prove that there is no point $M$ in the plane of the square such that the lengths of the segments $MA$, $MB$, $MC$, and $MD$ are natural numbers.\n\nb) Find the points $M$ in space such that the lengths of the segments $MA$, $MB$, $MC$, and $MD$ are natural numbers.",
    "options": [],
    "answer": "a) No such point exists in the plane.\nb) Exactly the points on the line perpendicular to the square through its center, with coordinates (one half, one half, plus or minus the square root of k squared minus one half), where k is a positive integer; for these points all four distances equal k.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $\\angle BAC = 100^\\circ$. The angle bisector of $\\angle ABC$ intersects $AC$ at $D$. The line perpendicular to $AC$ at $A$ intersects $BD$ at $M$. Point $E$ is on side $BC$ such that $AE = AM$ and $BE < CE$.\n\na) Show that $CA = CE$.\n\nb) Find the measure of angle $\\angle AEM$.",
    "options": [],
    "answer": "CA = CE and angle AEM = 80°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $M$ a point in its plane, distinct from $A$, $B$ and $C$. Let $N$, $P$ and $Q$ denote the symmetries of point $M$ with respect to sides $AB$, $BC$ and $AC$, respectively.\n\na) Prove that the points $N$, $P$ and $Q$ are collinear if and only if the point $M$ belongs to the circumcircle of triangle $ABC$.\n\nb) If point $M$ does not belong to the circumcircle of triangle $ABC$ and triangles $ABC$ and $NPQ$ have the same centroid, prove that triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The midpoint $S$ of the segment $MN$ belongs to the line $AB$, therefore $\\frac{s-a}{b-a} \\in \\mathbb{R}$. The lines $MN$ and $AB$ are perpendicular, therefore $\\frac{n-m}{b-a} \\in i\\mathbb{R}$.\n\nWe assume, without loss of generality, that the origin is at the circumcenter of the triangle, and $|a| = |b| = |c| = 1$. Then $\\bar{a} = \\frac{1}{a}$, $\\bar{b} = \\frac{1}{b}$, $\\bar{c} = \\frac{1}{c}$ and, since $s = \\frac{m+n}{2}$, we have:\n\n$$\n\\frac{s-a}{b-a} \\in \\mathbb{R} \\Leftrightarrow \\frac{s-a}{b-a} = \\frac{\\bar{s}-\\bar{a}}{\\bar{b}-\\bar{a}} \\Leftrightarrow n+m=2(a+b)-ab(\\bar{n}+\\bar{m})\n$$\n\nand\n\n$$\n\\frac{n-m}{b-a} \\in i\\mathbb{R} \\Leftrightarrow \\frac{n-m}{b-a} = -\\frac{\\bar{n}-\\bar{m}}{\\bar{b}-\\bar{a}} \\Leftrightarrow n-m = ab(\\bar{n}-\\bar{m}).\n$$\n\nBy addition, it follows that $n = a + b - ab\\bar{m}$. Similarly, $p = b + c - bc\\bar{m}$ and $q = c + a - ca\\bar{m}$.\n\na.\n\nThe points $N$, $P$ and $Q$ are collinear if and only if\n\n$$\n\\begin{align*}\n\\frac{n-p}{q-p} \\in \\mathbb{R} &\\Leftrightarrow \\frac{n-p}{q-p} = \\frac{\\bar{n}-\\bar{p}}{\\bar{q}-\\bar{p}} \\\\\n&\\Leftrightarrow \\frac{(a-c)(1-b\\bar{m})}{(a-b)(1-c\\bar{m})} = \\frac{(\\bar{a}-\\bar{c})(1-b\\bar{m})}{(\\bar{a}-\\bar{b})(1-c\\bar{m})} \\\\\n&\\Leftrightarrow \\frac{1-b\\bar{m}}{1-c\\bar{m}} = \\frac{b}{c} \\cdot \\frac{1-\\bar{b}\\bar{m}}{1-\\bar{c}\\bar{m}} \\\\\n&\\Leftrightarrow \\frac{1-b\\bar{m}}{1-c\\bar{m}} = \\frac{b-m}{c-m} \\\\\n&\\Leftrightarrow c-b = |m|^2(c-b) \\\\\n&\\Leftrightarrow |m| = 1,\n\\end{align*}\n$$\n\nso if and only if the point $M$ belongs to the circumcircle of triangle $ABC$.\n\nb.\n\nSince triangles $ABC$ and $NPQ$ have the same centroid, it means that $\\frac{a+b+c}{3} = \\frac{n+p+q}{3} \\Leftrightarrow a+b+c = \\bar{m}(ab+bc+ca) \\Leftrightarrow a+b+c = \\bar{m}abc(\\bar{a}+\\bar{b}+\\bar{c})$.\n\nWe apply the modulus to both members and we obtain that $|a+b+c| = |m| \\cdot |a+b+c|$. The point $M$ is not located on the circumcircle of the triangle, therefore $|m| \\neq 1$; it follows that $|a+b+c| = 0$. Then the centroid of the triangle $ABC$ coincides with its circumcenter, so the triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in $\\mathbb{R}$ the equation $\\log_7(6^x + 1) = \\log_6(7^x - 1)$.",
    "options": [],
    "answer": "1",
    "solution": "If $\\log_7(6^x + 1) = \\log_6(7^x - 1) = y$, we obtain $6^x + 1 = 7^y$ and $7^x - 1 = 6^y$. By addition, it follows that $6^x + 7^x = 6^y + 7^y$. The function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 6^x + 7^x$, is injective (it is strictly increasing, as the sum of two strictly increasing functions), so $x = y$.\n\nTo determine $x$ we need to solve the equation $6^x + 1 = 7^x$, or $\\left(\\frac{6}{7}\\right)^x + \\left(\\frac{1}{7}\\right)^x = 1$. The function $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = \\left(\\frac{6}{7}\\right)^x + \\left(\\frac{1}{7}\\right)^x$ is injective (it is strictly decreasing, as the sum of two strictly decreasing functions). Thus, the equation $g(x) = g(1)$ has the only solution $x = 1$, and this verifies the equation in the statement.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sets\n$$\nA = \\{(x, y) \\mid x, y \\in \\mathbb{R} \\text{ and } x + y + 1 = 0\\}\n$$\nand\n$$\nB = \\{(x, y) \\mid x, y \\in \\mathbb{R} \\text{ and } x^3 + y^3 + 1 = 3xy\\}.\n$$\n\na) Show that $A \\subset B$.\nb) Prove that the set $B \\setminus A$ has exactly one element.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. If $(x, y) \\in A$, then $y = -x - 1$. We get\n$$\nx^3 + y^3 + 1 = x^3 + (-x - 1)^3 + 1 = x^3 - (x + 1)^3 + 1 = x^3 - (x^3 + 3x^2 + 3x + 1) + 1 = -3x^2 - 3x = 3x(-x - 1) = 3xy,\n$$\nso $(x, y) \\in B$.\n\nb. If $(x, y) \\in B$, then $x^3 + y^3 - 3xy + 1 = 0$, so\n$$\n(x + y)^3 - 3xy(x + y) - 3xy + 1 = 0,\n$$\nor\n$$\n(x + y + 1)((x + y)^2 - (x + y) + 1) - 3xy(x + y + 1) = 0,\n$$\nor\n$$\n(x + y + 1)(x^2 - xy + y^2 - x - y + 1) = 0.\n$$\nIn order to have $(x, y) \\in B \\setminus A$ it is necessary that $x^2 - xy + y^2 - x - y + 1 = 0$, whence\n$$\n(x - y)^2 + (x - 1)^2 + (y - 1)^2 = 0,\n$$\nwhich means $x = y = 1$. Since $(1, 1) \\notin A$, it follows that $B \\setminus A = \\{(1, 1)\\}$.\n\n\nAlternative solution.\n\nIt is well known the identity\n$$\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).\n$$\nWe get\n$$\nx^3 + y^3 + 1 - 3xy = (x + y + 1)(x^2 + y^2 + 1 - xy - x - y).\n$$\nThus, if $(x, y) \\in A$, then $(x, y) \\in B$.\nThen, if $(x, y) \\in B \\setminus A$, it follows that $x^2 + y^2 + 1 - xy - x - y = 0$ and, after that, we finish the proof as for the first solution.\n\n\nAlternative solution for b).\n\nWe notice that $(1, 1) \\in B \\setminus A$.\nConsider $(x, y) \\in B$, so $x = 1+a, y = 1+b$. Then\n$$\na^3 + b^3 + 3a^2 + 3b^2 - 3ab = 0,\n$$\nor\n$$\n(a + b)(a^2 - ab + b^2) + 3(a^2 - ab + b^2) = 0,\n$$\nso\n$$\n(a + b + 3)(a^2 - ab + b^2) = 0.\n$$\nIf $a + b + 3 = 0$, then $x + y + 1 = 0$ and $(x, y) \\in A$. If $a^2 - ab + b^2 = 0$, then $a = b = 0$ and $(x, y) \\in B \\setminus A$. Thus the set $B \\setminus A$ contains only the element $(1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. On a board, we initially have two numbers: a red $0$ and a blue $0$. We define the following procedure:\n\nAt each step, a natural number $k$ is chosen (not necessarily distinct from previous choices). Let $x$ be the blue number and $y$ the red number. Replace them with:\n$$\nx \\rightarrow x + k + 1, \\quad y \\rightarrow y + k^2 + 2,\n$$\nand color the new values blue and red, respectively. This process continues until the blue number is at least $N$.\nDetermine the minimum possible value of the red number at the end of the process.",
    "options": [],
    "answer": "N + floor((N+1)/2)",
    "solution": "Let $M_k$ denote the move that increases the blue number by $k+1$ and the red number by $k^2+2$. We show that for any $k \\ge 2$, the move $M_k$ can be replaced by moves involving only $M_0$ and $M_1$ in such a way that the total increase in the blue number is the same, but the red number increases by less than $k^2+2$.\n\n**Case 1:** $k = 2p + 1$ is odd. Then $M_k$ increases the blue number by $2p + 2$ and the red number by $(2p+1)^2 + 2$. Instead, $p+1$ moves of type $M_1$ increase the blue number also by $2p+2$, but the red number increases by $3(p+1)$. Since $(2p+1)^2 + 2 > 3(p+1)$, for all $p \\ge 1$, we can replace $M_k$ with $p+1$ moves of $M_1$.\n\n**Case 2:** $k = 2p$ is even. Then $M_k$ increases the blue number by $2p+1$, and the red number by $(2p)^2 + 2$. Alternatively, $p$ moves of $M_1$ and one move of $M_0$ increase the blue number by $2p+1$ and the red number by $3p+2$. Since $(2p)^2 + 2 > 3p+2$, for all $p \\ge 1$, we conclude that $M_k$ can be replaced by $p$ moves of $M_1$ and one $M_0$.\nMoreover, observe that two moves of $M_0$ can be replaced by one move of $M_1$, with a smaller increase in the red number: $3$ instead of $4$.\n\nThus, using only $M_0$ and $M_1$, and minimizing the use of $M_0$, yields the minimal red number. To make the blue number at least $N$, the best strategy is to use only moves of type $M_1$ (which increase blue by $2$ and red by $3$), plus at most one $M_0$ (if needed).\n\nIf $N$ is even: we can use $\\left\\lfloor \\frac{N}{2} \\right\\rfloor$ moves of $M_1$, increasing red by $3 \\cdot \\frac{N}{2} = \\frac{3N}{2}$.\n\nIf $N$ is odd: we can use $\\left\\lceil \\frac{N}{2} \\right\\rceil$ moves of $M_1$ and one move of $M_0$, leading to red increasing by $3\\left\\lfloor \\frac{N}{2} \\right\\rfloor + 2$.\n\nHence, the minimal possible value of the red number is $N + \\left\\lfloor \\frac{N+1}{2} \\right\\rfloor$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39421,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call *smalls* the positive integers not larger than $2025$.\n\na) Find the number of smalls which are perfect squares.\n\nb) Find the number of smalls which are perfect squares and leave remainder $0$ when divided by $45$.\n\nc) Find the number of smalls which neither are perfect squares, nor leave remainder $0$ when divided by $45$.",
    "options": [],
    "answer": "a) 45, b) 3, c) 1938",
    "solution": "a) Since $45^2 = 2025$, the smalls which are perfect squares are $1^2, 2^2, 3^2, \\dots, 45^2$, that is $45$ such numbers.\n\n\nb) The smalls which are perfect squares and are divisible by $45$ are $15^2, 4 \\cdot 15^2$ and $9 \\cdot 15^2$ – there are $3$ of them.\n\n\nc) The smalls divisible by $45$ are $45 \\cdot 1, 45 \\cdot 2, \\dots, 45 \\cdot 45$ – there are $45$ such numbers. The number of smalls which are perfect squares or are divisible by $45$ is $45 + 45 - 3 = 87$. So, the number of smalls which are neither perfect squares, nor leave remainder $0$ when divided by $45$ is $2025 - 87 = 1938$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of twice differentiable functions $f, g : \\mathbb{R} \\to \\mathbb{R}$, such that $f''$ and $g''$ are continuous, such that\n$$\n(f(x) - g(y)) \\cdot (f'(x) - g'(y)) \\cdot (f''(x) - g''(y)) = 0,\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All pairs with f(x) = m x^2 + n x + p and g(x) = m x^2 + n' x + p', where m, n, n', p, p' are real constants.",
    "solution": "Let $(f, g)$ be a pair of functions satisfying the given condition. We shall show that $f''$ is constant. Suppose that $f''$ is not constant. Then, because $(f(x) - g(0)) \\cdot (f'(x) - g'(0)) \\cdot (f''(x) - g''(0)) = 0$, for all $x \\in \\mathbb{R}$, and $f''$ is continuous, there is $a \\in \\mathbb{R}$ and $r > 0$ such that $f''(x) \\notin \\{0, g''(0)\\}$, for any $x \\in (a - r, a + r)$. It follows that $(f(x) - g(0)) \\cdot (f'(x) - g'(0)) = 0$, for $x \\in (a - r, a + r)$.\n\nTwo cases are possible.\n**Case 1.** There exists $b \\in (a - r, a + r)$ such that $f'(b) \\ne g'(0)$. By continuity of $f'$, there is $s > 0$ such that $(b - s, b + s) \\subset (a - r, a + r)$ and $f'(x) \\ne g'(0)$, for all $x \\in (b - s, b + s)$, implying $f(x) = g(0)$ for $x \\in (b - s, b + s)$. We get $f''(x) = 0$ for $x \\in (b - s, b + s)$, a contradiction.\n\n**Case 2.** $f'(x) = g'(0)$, for any $x \\in (a-r, a+r)$. We obtain $f''(x) = 0$, for all $x \\in (a-r, a+r)$, a contradiction again.\n\nThus $f''$ is constant on $\\mathbb{R}$.\nLet $m \\in \\mathbb{R}$ such that $f''(x) = 2m$, for any $x \\in \\mathbb{R}$. As $(f'(x) - 2mx)' = 0$, for all $x \\in \\mathbb{R}$, there is $n \\in \\mathbb{R}$ such that $f'(x) - 2mx - n = 0$, for $x \\in \\mathbb{R}$. As a consequence there exists $p \\in \\mathbb{R}$ such that $f(x) = mx^2 + nx + p$, for all $x \\in \\mathbb{R}$.\nIn the same way, $g''$ is constant on $\\mathbb{R}$, implying the existence of $m', n', p' \\in \\mathbb{R}$ such that $g(x) = m'x^2 + n'x + p'$, for $x \\in \\mathbb{R}$. In the case $m \\neq m'$, as the equation $(f(x) - g(x)) \\cdot (f'(x) - g'(x)) = 0$ has at most three real solution, the condition of the problem is not possible. Thus $m = m'$, so the pairs of functions satisfying the given conditions are of the form $f(x) = mx^2 + nx + p$ and $g(x) = mx^2 + n'x + p'$, for any $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39423,
    "subject": "Mathematics (Multi-modal)",
    "question": "The non-negative integers $a$, $b$, $c$ are such that the numbers\n$$\nm = \\frac{5a + 6b + 7c + 6}{4a + 3b + 2c + 3} \\quad \\text{and} \\quad n = \\frac{a + 2b + 3c + 5}{3a + b + 2c + 5}\n$$\nare both integers.\n\na. Prove that $m \\ge 2$.\nb. Find $m$ and $n$.",
    "options": [],
    "answer": "m = 2, n = 1",
    "solution": "a. If $m \\le 1$, then $5a + 6b + 7c + 6 \\le 4a + 3b + 2c + 3$, that is $a + 3b + 5c + 3 \\le 0$ – impossible. So $m \\ge 2$.\n\nb. If $n \\ge 2$, then $a + 2b + 3c + 5 \\ge 6a + 2b + 4c + 10$, whence $0 \\ge 5a + c + 5$, false. Since $n > 0$, $n$ must be equal to 1.\nNow $n = 1$ yields $a + 2b + 3c + 5 = 3a + b + 2c + 5$, hence $b + c = 2a$.\nSuppose $m \\ge 3$. Then $5a+6b+7c+6 \\ge 12a+9b+6c+9$, therefore $c \\ge 7a+3b+3$. This gives $b+c \\ge 7a+4b+3$, that is $2a \\ge 7a+4b+3$, hence $0 \\ge 5a+4b+3$, false. So $m < 3$, and a) implies $m = 2$.\nThe above show that the only possibility is $m = 2$ and $n = 1$. These values are indeed achieved for $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39424,
    "subject": "Mathematics (Multi-modal)",
    "question": "At the robotics workshop, the registered students are grouped into teams made of three boys and a girl. One day two boys and a girl are missing, so the present students were regrouped into teams made by one girl and four boys.\nHow many students are registered at the robotics workshop?",
    "options": [],
    "answer": "28",
    "solution": "Initially, the students are grouped $(G, B, B, B), (G, B, B, B), (G, B, B, B), \\dots, (G, B, B, B)$.\nWhen two girls and a boy are missing, the situation (replacing a boy, if the missing boy is not in the same team as one of the missing girls) becomes:\n$(G, B, B), (G, B, B, B), (G, B, B, B), \\dots, (G, B, B, B)$.\nSince each new group has one girl, the $2+3$ boys from the incomplete initial groups must be regrouped so that we get groups with 4 boys and one girl – one in each initial group. This means that now we have five groups, therefore now there are $5 \\cdot 4 = 20$ boys and 5 girls.\nHence, the robotics workshop has $20+1=21$ boys and $5+2=7$ girls, for a total of 28 registered students.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39425,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that, if a ring $(A, +, \\cdot)$ has property (P) and $a, b$ are distinct elements of $A$ such that $a$ and $a+b$ are invertible, then $b$ is not invertible, but $1+ab$ is invertible.\n\nb) Give an example of a unitary ring possessing (P).\n\nwhere property (P) is:\n$$\n(P) \\quad \\left\\{ \\begin{array}{l} \\text{the set } A \\text{ has at least 4 elements,} \\\\ \\text{the element } 1+1 \\text{ is invertible in } A, \\\\ x+x^4 = x^2+x^3, \\text{ for any } x \\in A. \\end{array} \\right.\n$$",
    "options": [],
    "answer": "a) b is not invertible and 1+ab is invertible. b) Example: Z/3Z × Z/3Z.",
    "solution": "Denote $U(A)$ the set of invertible elements in $A$. For $k \\in \\mathbb{N}$, $k \\ge 2$, and $x \\in A$, define $kx = \\underbrace{x+x+\\cdots+x}_{k \\text{ terms}}$. In particular $k \\cdot 1 = k$. By the given conditions $2 \\in U(A)$. Denote by (1) the equality $x+x^4 = x^2+x^3$ for all $x \\in A$.\n\nChanging $x$ by $-x$ in (1) we get $-x+x^4 = x^2-x^3$, for any $x \\in A$.\n\nBy subtraction, the last two relations give $2x = 2x^3$, so, as $2 \\in U(A)$, we obtain $x = x^3$ for all $x \\in A$.\n\nFor $x \\in U(A)$, multiplying by $x^{-1}$ we get $x^2 = 1$ for any $x \\in U(A)$. As the set $U(A)$, of the invertible elements of the monoid $(A, \\cdot)$, is a group with $x^2 = 1$, for any $x \\in U(A)$, it results that the group $(U(A), \\cdot)$ is commutative.\n\nFor $x = 2 \\in U(A)$ from the previous relations, we get $4 = 1$, so $3 = 0$, meaning that the ring $A$ is of characteristic $3$.\n\nLet $a, b \\in A$, such that $a \\neq b$ and $a, a+b \\in U(A)$. If we suppose $b \\in U(A)$, then\n$$\n2ab = (a+b)^2 - a^2 - b^2 = 1 - 1 - 1 = -1 = 2,\n$$\nimplying $ab = 1 = a^2$. This gives $a = b$, a contradiction. That is $b \\notin U(A)$.\n\nWe also have\n$$\n1 + ab = a^2 + ab = a(a+b) \\in U(A),\n$$\na product of invertible elements.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a group, with the unit element $e$, and $A$ a non-empty subset of $G$. We denote $AA = \\{xy \\mid x, y \\in A\\}$.\n\na) Show that if $G$ is finite, then $AA = A$ if and only if $e \\in A$ and $|AA| = |A|$.\n\nb) Give an example of a group $G$ and a subset $A \\subseteq G$, such that $AA \\neq A$, $|AA| = |A|$ and $AA < G$.\n\n(The notation $H < G$ means that $H$ is a proper subgroup of the group $G$, i.e., a subgroup of $G$ different from $G$ itself.)",
    "options": [],
    "answer": "Example: take the group of fourth roots of unity {1, i, -1, -i} and the subset {i, -i}; then the product set is {1, -1}, which has the same size as the subset, differs from it, and is a proper subgroup.",
    "solution": "a) If $AA = A$, then $|AA| = |A|$. For any $x \\in A$ we have $|xA| = |A| < \\infty$ and $xA \\subseteq AA = A$, so that $xA = A$. But then $x \\in xA$, hence $e = x^{-1} \\cdot x \\in x^{-1} \\cdot xA = A$. Reciprocally, if $e \\in A$ and $|AA| = |A|$, then $A = e \\cdot A \\subseteq AA$ and since $|A| = |AA| < \\infty$, it follows that $AA = A$.\n\nb) Let $G = U_4 = \\{1, i, -1, -i\\}$ be the group of all 4th roots of unity. Considering $A = \\{i, -i\\}$, we have $AA = \\{i^2, i \\cdot (-i), (-i)^2\\} = \\{1, -1\\} = U_2 < U_4$, $|AA| = 2 = |A|$ and $AA \\neq A$ (furthermore, $AA \\cap A = \\emptyset$).",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the four-digit natural numbers $\\overline{abcd}$, for which there exists a prime number $p$, such that $\\overline{cd} - \\overline{ab} = p + 2$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{cd}} = p^2$.",
    "options": [],
    "answer": "2025",
    "solution": "Since $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{cd}} < 10 + 10 = 20$, we obtain $p = 2$ or $p = 3$.\nIf $p = 2$, then $\\overline{cd} = \\overline{ab} + 4$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{ab} + 4} = 4$, therefore $\\overline{ab} = 5$, false.\nIf $p = 3$, then $\\overline{cd} = \\overline{ab} + 5$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{ab} + 5} = 9$. We obtain the unique solution $\\overline{ab} = 20$, thus $\\overline{abcd} = 2025$.\n\nAlternative solution:\nSince $\\overline{ab}-4$, $\\overline{cd}$ and $\\sqrt{\\overline{ab}-4}+\\sqrt{\\overline{cd}}$ are integers, the natural numbers $k, n$ exist, such that $\\overline{ab}-4 = k^2$ and $\\overline{cd} = n^2$, hence $k+n = p^2$. (1)\nWe obtain $p+6 = \\overline{cd} - \\overline{ab} + 4 = n^2 - k^2 = (n-k)(n+k)$. Using (1), we deduce that $p+6 = (n-k) \\cdot p^2$, therefore $p \\mid (p+6)$, so $p \\in \\{2, 3\\}$.\nIf $p = 2$, we obtain $k = 1$ and $n = 3$, therefore $\\overline{ab} = 5$, which is false. If $p = 3$, we obtain $k = 4$ and $n = 5$, so $\\overline{ab} = 20$ and $\\overline{cd} = 25$, which satisfies the hypothesis. Therefore, the solution is 2025.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all the natural numbers $a, b, c$ such that $ab + bc + ca$ is a prime number $p$ and $p$ divides the number $a^2b^2 + b^2c^2 + c^2a^2$.",
    "options": [],
    "answer": "a = b = c = 1",
    "solution": "From the identity $a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c)$ it follows that $p$ divides $abc(a+b+c)$. Because $p$ is a prime number, we get $p \\mid a$, $p \\mid b$, $p \\mid c$ or $p \\mid (a+b+c)$.\n\nSince $a, b, c < p$, the first three cases are impossible. The fourth situation can be true if and only if $a = b = c = 1$. This case works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be a positive rational number such that the numbers $r$ and $\\sqrt{r+1}$ have the same fractional part. Show that $r$ is an integer.\nAndrei Bâra",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The hypothesis implies that $\\sqrt{r+1} - r = a$, where $a \\in \\mathbb{Z}$.\n\nWe get: $r + 1 = r^2 + 2ra + a^2$.\n\nLet $r = \\frac{m}{n}$, with $m, n \\in \\mathbb{N}^*$ and $\\text{gcd}(m, n) = 1$. Then,\n$$\nmn + n^2 = m^2 + 2mna + a^2n^2,\n$$\nand therefore\n$$\nm^2 = n(m + n - 2ma - a^2n).\n$$\nHence, $m^2$ is divisible by $n$, which is only possible if $n = 1$, thus $r = m \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{1, 2, 3, \\dots, 2025\\}$. A subset $B$ of the set $A$ will be called *nice* if it has 3 elements, one of them being the arithmetic mean of the other two, and there exists $b \\in B$ such that $5 \\cdot b \\in B$.\n\na) Find how many nice sets have the element $225$.\n\nb) Find how many nice subsets has $A$.",
    "options": [],
    "answer": "a) 6; b) 630",
    "solution": "Let $B = \\{b, 5b, a\\}$. The possible cases are:\n\nI. $a = \\frac{b+5b}{2} \\Rightarrow B = \\{b, 3b, 5b\\}$;\n\nII. $5b = \\frac{a+b}{2} \\Rightarrow B = \\{b, 5b, 9b\\}$;\n\nIII. $b = \\frac{a+5b}{2} \\Rightarrow a+3b=0$ - impossible.\n\n\na) Since $225$ is divisible by $3$, $5$ and $9$, it can be any element of the set $B$. We get the nice sets: $\\{225, 675, 1125\\}$, $\\{75, 225, 375\\}$, $\\{45, 135, 225\\}$, $\\{225, 1125, 2025\\}$, $\\{45, 225, 405\\}$, $\\{25, 125, 225\\}$.\n\nb) In case I we get a nice set for every $b \\in \\mathbb{N}^*$ with $5b \\le 2025$, that is $2025 \\div 5 = 405$ sets. In case II we get a nice set for every $b \\in \\mathbb{N}^*$ so that $9b \\le 2025$, that is $2025 \\div 9 = 225$ sets. Since the equality $\\{x, 3x, 5x\\} = \\{y, 5y, 9y\\}$ is impossible for positive integers $x, y$, there are no common sets for case I and case II. This shows that there are $405 + 225 = 630$ nice sets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = \\frac{a_n}{1+\\sqrt{1+a_n}}$, for any $n \\in \\mathbb{N}^*$. Show that $\\lim_{n \\to \\infty} \\frac{a_n}{a_{n+1}} = \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\log_2(1+a_k) = 2$. Traian Tămâian",
    "options": [],
    "answer": "2",
    "solution": "We have $a_n > 0$ and $a_{n+1} < a_n$, for any $n \\ge 1$. It turns out that the sequence $(a_n)_{n \\ge 1}$ is convergent, with the limit $\\ell \\in [0, 1)$. From the recurrence relation, we obtain $\\ell = \\frac{\\ell}{1+\\sqrt{1+\\ell}}$, so $\\ell = 0$. Then $\\lim_{n \\to \\infty} \\frac{a_n}{a_{n+1}} = \\lim_{n \\to \\infty} (1 + \\sqrt{1+a_n}) = 2$.\n\nThe recurrence relation implies $1 + a_{n+1} = \\sqrt{1+a_n}$, for any $n \\in \\mathbb{N}^*$. Therefore, we have $\\log_2(1+a_{n+1}) = \\frac{1}{2}\\log_2(1+a_n)$, for any $n \\ge 1$. From $\\log_2(1+a_1) = 1$ we obtain $\\log_2(1+a_n) = \\frac{1}{2^{n-1}}$, for all positive integers $n$ (geometric progression with the first term 1 and the ratio $1/2$). Hence $\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\log_2(1+a_k) = \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{1}{2^{k-1}} = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and let $O$ be the intersection point of the diagonals. Prove that for any point $M \\in (AB)$, there exist unique points $N \\in (OC)$ and $P \\in (OD)$ such that $O$ is the centroid of triangle $MNP$.\nNelu Chichirim",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A point $M \\in (AB)$ is uniquely determined by a real number $k \\in (0, \\infty)$ such that $\\overline{AM} = k$, from which we get\n$$\n\\overrightarrow{OM} = \\frac{1}{k+1}\\overrightarrow{OA} + \\frac{k}{k+1}\\overrightarrow{OB}.\n$$\nTo find the points $N$ and $P$ uniquely, we need to find $x, y \\in (0, \\infty)$ such that $\\overrightarrow{ON} = x, \\overrightarrow{OP} = y$, and $O$ is the centroid of triangle $MNP$. From this, we have:\n$$\n\\overrightarrow{ON} = \\frac{x}{x+1}\\overrightarrow{OC} = -\\frac{x}{x+1}\\overrightarrow{OA} \\quad \\text{and} \\quad \\overrightarrow{OP} = \\frac{y}{y+1}\\overrightarrow{OD} = -\\frac{y}{y+1}\\overrightarrow{OB}.\n$$\nSince $\\overrightarrow{OA}$ and $\\overrightarrow{OB}$ are not collinear, $O$ is the centroid of triangle $MNP$ if and only if $\\frac{x}{x+1} = \\frac{1}{k+1} \\Leftrightarrow x = \\frac{1}{k}$ and $\\frac{y}{y+1} = \\frac{k}{k+1} \\Leftrightarrow y = k$, meaning that point $N$ is uniquely determined by the ratio $x = \\frac{1}{k} = \\frac{ON}{NC}$, and point $P$ is uniquely determined by the ratio $y = k = \\frac{OP}{PD}$, which completes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39433,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$ in the plane are pairwise externally tangent. Let $P_2$ be the point of tangency between the circles $\\Gamma_1$ and $\\Gamma_3$, and $P_1$ the point of tangency between the circles $\\Gamma_2$ and $\\Gamma_3$. Consider points $A$ and $B$ on the circle $\\Gamma_3$ that are diametrically opposite, such that the quadrilateral $ABP_1P_2$ is convex.\nThe line through $A$ and $P_2$ intersects the circle $\\Gamma_1$ a second time at point $X$, the line through $B$ and $P_1$ intersects the circle $\\Gamma_2$ a second time at point $Y$, and the lines $AP_1$ and $BP_2$ intersect at $Z$.\nProve that the points $X, Y$, and $Z$ are collinear.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\{P_3\\} = \\Gamma_1 \\cap \\Gamma_2$ be the second point of intersection of the circles $\\Gamma_1$ and $\\Gamma_2$, and let $O_1, O_2, O_3$ be the centers of the circles $\\Gamma_1, \\Gamma_2$, and $\\Gamma_3$, respectively. Denote by $O_4$ the intersection point of the common tangents to the circles $\\Gamma_1$ and $\\Gamma_2$.\n\n$$\n\\begin{align*}\n\\widehat{P_1ZP_2} &= \\frac{1}{2} (\\widehat{AB} + \\widehat{P_1P_2}) = \\frac{1}{2} (180^\\circ + \\widehat{P_1O_3P_2}) \\\\\n&= \\frac{1}{2} (180^\\circ + 180^\\circ - \\widehat{P_1O_2P_3} - \\widehat{P_2O_1P_3}) = \\\\\n&= \\frac{1}{2} (180^\\circ - \\widehat{P_1O_2P_3}) + \\frac{1}{2} (180^\\circ - \\widehat{P_2O_1P_3}) \\\\\n&= \\widehat{O_1P_3P_2} + \\widehat{O_2P_3P_1} = 180^\\circ - \\widehat{P_1P_3P_2},\n\\end{align*}\n$$\nso the quadrilateral $ZP_1P_3P_2$ is cyclic.\n\nWe will prove that $X, P_3$, and $Z$ are collinear. It suffices to show that $\\widehat{XP_3O_1} = \\widehat{ZP_3O_2}$. Since triangle $O_1XP_3$ is isosceles and $BP_2 \\perp AX$ (because $AB$ is a diameter), we have:\n$$\n\\begin{align*}\n\\widehat{XP_3O_1} &= \\frac{1}{2} (180^\\circ - \\widehat{XO_1P_3}) = 90^\\circ - \\frac{1}{2} \\widehat{XP_3} = \\widehat{XP_2B} - \\widehat{XP_2P_3} = \\widehat{BP_2P_3} \\\\\n&= \\widehat{ZP_2P_3} = \\frac{1}{2} \\widehat{ZP_1P_3}.\n\\end{align*}\n$$\nBecause $O_1O_2$ is tangent to the circumcircle of triangle $P_1P_2P_3$, it follows that $\\widehat{P_1P_3O_2} = \\frac{1}{2}\\widehat{P_1P_3}$. Therefore, $\\widehat{ZP_3O_2} = \\widehat{ZP_3P_1} + \\widehat{P_1P_3O_2} = \\frac{1}{2}\\widehat{ZP_1} + \\frac{1}{2}\\widehat{P_1P_3} = \\frac{1}{2}\\widehat{ZP_1P_3} = \\widehat{XP_3O_1}$.\nHence, points $X, Z$, and $P_3$ are collinear. Similarly, one can show that $Y, Z$, and $P_3$ are collinear, and thus $X, Y$, and $Z$ are collinear as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Turbo the snail is in the lower left cell of an $n \\times n$ array, $n \\ge 2$, and aims to reach the upper right cell by moving one cell rightwards or one cell upwards. Some cells contain monsters, visible to Turbo, and they must be avoided. Assume there is a unique way for Turbo to achieve his goal. In terms of $n$, determine the smallest possible number of monsters such an array may contain. (The minimum is over all configurations satisfying the unique path condition.)",
    "options": [],
    "answer": "n-1",
    "solution": "Let $(i, j)$ denote the cell on the $i$-th row and the $j$-th column, where $(1, 1)$ is the lower left cell. The required minimum is $n-1$ and is achieved by placing the monsters in the cells $(k, n-k+1), k=2, \\dots, n$, to force Turbo move rightwards from $(1, 1)$ to $(1, n)$ and thence upwards to $(n, n)$.\n\nTo prove that the $n \\times n$ array must contain at least $n-1$ monsters, induct on $n$. The base case, $n=2$, is trivial.\n\nFor the inductive step, let $n \\ge 3$. If necessary, switch rows and columns to assume that Turbo first moves upwards to $(2, 1)$. Note that Turbo enters $(n, n)$ either from $(n, n-1)$ or from $(n-1, n)$.\n\n**Case 1.** Turbo visits $(n, n-1)$. Consider the subarray $\\{(i, j) : 2 \\le i \\le n, 1 \\le j \\le n-1\\}$. As it inherits uniqueness of a safe path for Turbo, it contains at least $n-2$ monsters, by the induction hypothesis. Furthermore, the path from $(1, 1)$ to $(n, n)$ lying outside the subarray must contain at least one monster, for otherwise it would have offered an alternative to Turbo, contradicting uniqueness. Consequently, the $n \\times n$ array contains at least $(n-2)+1 = n-1$ monsters, as desired. This establishes Case 1.\n\n**Case 2.** Turbo visits $(n-1, n)$. Note that $(2, 1)$ and $(n-1, n)$ are separated by the diagonal joining the lower left cell to the upper right cell, so Turbo's safe path must cross this diagonal at some (unique) cell $(k, k)$, where $2 \\le k \\le n-1$.\n\nConsider the subarrays $\\{(i, j) : 1 \\le i, j \\le k\\}$ and $\\{(i, j) : k \\le i, j \\le n\\}$. They both inherit uniqueness of a safe path for Turbo, so the former contains at least $k-1$ monsters and the latter at least $n-k$, accounting for at least $(k-1)+(n-k) = n-1$ monsters in the $n \\times n$ array, as desired. This establishes Case 2 and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39435,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a natural number $n \\ge 3$ is almost square-free if there exists a prime number $p$, with $p \\equiv 1 \\pmod 3$, such that $n$ is divisible by $p^2$, and the number $\\frac{n}{p}$ is square-free (i.e., not divisible by the square of any prime number).\n\nShow that, for any natural number $n$ that is almost square-free, the ratio between twice the sum of the divisors of $n$ and the number of divisors of $n$ is a natural number.\n\nLucian Petrescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $n = p_1 \\cdots p_{k-1} \\cdot p_k^2 \\cdot p_{k+1} \\cdots p_s$, with $p_1 < p_2 < \\dots < p_s$ prime numbers, $1 \\le k \\le s$, and $p = p_k \\equiv 1 \\pmod 3$. Moreover, $p$ is odd and satisfies $p^2 + p + 1 \\equiv 0 \\pmod 3$.\n\nThe number of divisors of $n$, $\\tau(n)$, is\n$$\n\\tau(n) = \\underbrace{(1+1) \\cdot (1+1) \\cdots (1+1)}_{\\text{applied } s-1 \\text{ times}} \\cdot (2+1) = 3 \\cdot 2^{s-1},\n$$\nand the sum of the divisors, $\\sigma(n)$, is:\n$$\n\\begin{aligned}\n\\sigma(n) &= \\frac{p_1^2-1}{p_1-1} \\cdots \\frac{p_{k-1}^2-1}{p_{k-1}-1} \\cdot \\frac{p_k^3-1}{p_k-1} \\cdots \\frac{p_{k+1}^2-1}{p_{k+1}-1} \\cdots \\frac{p_s^2-1}{p_s-1} \\\\\n&= (p_1+1) \\cdots (p_{k-1}+1) \\cdot (p_k^2+p_k+1) \\cdot (p_{k+1}+1) \\cdots (p_s+1).\n\\end{aligned}\n$$\n\n**Case I.** If $n$ is even, then $p_1 = 2$, and $p_2 < p_3 < \\dots < p_s$ are odd primes. Hence, $2\\sigma(n) = 2 \\cdot (2+1) \\cdot (p_2+1) \\cdots (p_{k-1}+1) \\cdot (p_k^2+p_k+1) \\cdot (p_{k+1}+1) \\cdots (p_s+1)$, so $2\\sigma(n)$ is divisible by $2 \\cdot 3^2 \\cdot 2^{s-2} = 3 \\cdot \\tau(n)$.\n\n\n**Case II.** If $n$ is odd, then $p_1 < p_2 < \\dots < p_s$ are odd primes, so\n$$\n\\sigma(n) = (p_1 + 1) \\cdots (p_{k-1} + 1) \\cdot (p_k^2 + p_k + 1) \\cdot (p_{k+1} + 1) \\cdots (p_s + 1)\n$$\nis divisible by $3 \\cdot 2^{s-1} = \\tau(n)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a natural number and let $a_1, a_2, \\dots, a_n$ be nonzero integers such that $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n$.\nProve that the number $(a_1^2 - 1)(a_2^2 - 1)\\dots(a_n^2 - 1)$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $n=2$, we have $a_1+a_2 = a_1 a_2$, which rewrites as $(a_1-1)(a_2-1) = 1$. We get $a_1 = 2, a_2 = 2$, so $(a_1^2 - 1)(a_2^2 - 1) = 9$.\n\n$$\n|a_1 a_2 \\dots a_n| = |a_1 + a_2 + \\dots + a_n| \\le |a_1| + |a_2| + \\dots + |a_n|.\n$$\nDividing by $|a_1 a_2 \\dots a_n| > 0$ we obtain\n$$\n1 \\le \\frac{|a_1|}{|a_1 a_2 \\cdots a_n|} + \\frac{|a_2|}{|a_1 a_2 \\cdots a_n|} + \\dots + \\frac{|a_n|}{|a_1 a_2 \\cdots a_n|}.\n$$\nIf all the numbers have absolute value greater than or equal to $2$, then the right-hand side of the above inequality is less than or equal to $\\frac{n}{2^{n-1}}$, i.e., $2^{n-1} \\le n$.\nFor $n \\ge 3$, we obtain (by induction or using Bernoulli's inequality) that $2^{n-1} > n$, which contradicts the previous inequality.\nTherefore, at least one of the numbers has absolute value $1$, which implies $(a_1^2 - 1)(a_2^2 - 1)\\dots(a_n^2 - 1) = 0$, a perfect square.\n\n**Remark.** For every $n \\ge 3$, there exist numbers satisfying the condition in the statement, for example\n$$\n\\underbrace{1+1+\\dots+1}_{n-2 \\text{ times}}+2+n = \\underbrace{1 \\cdot 1 \\dots 1}_{n-2 \\text{ times}} \\cdot 2 \\cdot n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$, let $O$ be its circumcentre and let $A'$ be the reflection of $A$ in $BC$. The parallel through $O$ to $BC$ crosses $AC$ at $F$, and the tangent at $F$ to circle $BFC$ crosses the parallel through $A'$ to $BC$ at $M$. Consider the point $K$ on the ray $AB$, emanating from $A$, such that $AK = 4AB$. Prove that the orthocentre of the triangle $ABC$ lies on the circle on diameter $KM$.\n\nRadu Lecoiu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocentre of triangle $ABC$, let $N$ be the midpoint of the segment $AH$ and let $D$ the foot of the perpendicular from $B$ to $AC$. We first prove that $\\angle BNF = 90^\\circ$. Let $F'$ be the intersection of line $AC$ and the perpendicular at $N$ to $BN$. The points $B, N, D$, and $F'$ lie on the circle on diameter $BF$, so\n$$\n\\angle NBF' = \\angle ADN = \\angle DAN = 90^\\circ - \\angle BCA = \\angle ABO,\n$$\nwhere the second equality holds on account of $DN$ being median in the right triangle $ADH$. Since $\\angle NBF' = \\angle ABO$, lines $BN$ and $BO$ are isogonal with respect to angle $\\angle ABF$. Similarly, lines $AO$ and $AN$ are isogonal with respect to angle $\\angle BAC$,\n\n$$\n\\angle AF'O = \\angle BF'N = \\angle BDN = \\angle NHD = \\angle ACB.\n$$\nHence $OF' \\parallel BC$, so $F \\equiv F'$, showing that $\\angle BNF = 90^\\circ$.\n\nNext, we prove that $N, F$ and $M$ are collinear. Since $F \\equiv F'$, points $O$ and $N$ are isogonal conjugates in triangle $ABF$, so\n$$\n\\angle NFB = \\angle AFO = \\angle ACB.\n$$\nThis implies $NF$ is tangent to circle $BFC$ and hence $N, F, M$ are collinear, as desired.\nLet $AB$ and $A'M$ cross at $E$. Note that $E$ is the reflection of $A$ across $B$. Since $AK = 4AB$, point $E$ is the midpoint of the segment $AK$. In triangle $AEH$, line $BN$ is a midline, so $BN \\parallel EH$. As $BN$ and $NM$ are perpendicular, so are $EH$ and $NM$. Moreover, $NH \\perp EM$, hence $H$ is the orthocentre of the triangle $ENM$, implying $MH \\perp EN$. Finally, in triangle $AKH$, line $EN$ is a midline, so $EN \\parallel KH$ and hence $MH \\perp KH$; that is, $H$ lies on the circle on diameter $KM$, as required.\n\nAlternative solution for $\\angle BNF = 90^\\circ$. We use the same notations as in the previous solution. Consider the foot $P$ of the perpendicular from $F$ to $BC$. Then\n$$\n2AN = 2NH = AH = 2 \\operatorname{dist}(O, BC) = 2 \\operatorname{dist}(F, BC) = 2FP,\n$$\nso the segments $NH$ and $FP$ are parallel and have equal lengths, implying that $NFPH$ is a parallelogram. Similarly, $ANPF$ is also a parallelogram. Therefore, $HB \\perp NP$, and since $NH \\perp BC$, we deduce that $H$ is the orthocentre of the triangle $BNP$. Consequently, $HP \\perp BN$, and since $HP \\parallel NF$, it follows that $\\angle BNF = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $(a_n)_{n \\ge 1}$ of positive rational numbers satisfying\n$$\na_{k+1} = \\frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} \\quad \\text{for all } k \\ge 1. \\quad (*)\n$$",
    "options": [],
    "answer": "All constant sequences of positive rational numbers.",
    "solution": "The required sequences are all constant sequences of positive rational numbers. Clearly, any such satisfies $(*)$.\n\nLet $(a_n)_{n \\ge 1}$ be a sequence of positive rational numbers satisfying $(*)$. Then so does $(r a_n)_{n \\ge 1}$, where $r$ is any positive rational number. Letting $r$ be the product of the denominators of $a_1$ and $a_2$, we may assume that $a_1$ and $a_2$ are integers. If $a_k$ and $a_{k+1}$ are integers, then so is $a_{k+2}$, as it is a root of the monic polynomial $X^2 - a_{k+1} X + a_k^2 - a_k a_{k+1}$ with integer coefficients. Inductively, the $a_n$ are all integers.\n\nWrite $(*)$ in the form\n$$\n(a_{k+1} - a_k)a_k = (a_{k+2} - a_{k+1})a_{k+2}. \\quad (**)\n$$\nIf $a_1 > a_2$, then $(**)$ forces $(a_n)_{n \\ge 1}$ to be a strictly decreasing sequence of positive integers and we reach a contradiction.\n\nIf $a_1 < a_2$, then $(a_n)_{n \\ge 1}$ is strictly increasing, by $(**)$.\n$$\na_{k+1} = \\frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} > \\frac{(a_k + a_{k+2})^2}{2(a_k + a_{k+2})} = \\frac{1}{2}(a_k + a_{k+2}) \\quad \\text{for all } k \\ge 1,\n$$\nit follows that $a_{k+1} - a_k > a_{k+2} - a_{k+1}$ for all $k \\ge 1$, so the $a_{n+1} - a_n$ form a strictly decreasing sequence of positive integers and we reach again a contradiction.\n\nFinally, if $a_1 = a_2$, then $(**)$ forces $a_n = a_1$ for all $n \\ge 1$, so the sequence $(a_n)_{n \\ge 1}$ is indeed constant, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 10$ be an integer, and let $A_1, A_2, \\dots, A_n$ be distinct points in the plane such that the distances between the points are pairwise different. Define $f_{10}(j, k)$ to be the 10th smallest of the distances from $A_j$ to $A_1, A_2, \\dots, A_k$, excluding $A_j$ if $k \\ge j$. Suppose that for all $j$ and $k$ satisfying $11 \\le j \\le k \\le n$, we have $f_{10}(j, j-1) \\ge f_{10}(k, j-1)$. Prove that $f_{10}(j, n) \\ge \\frac{1}{2}f_{10}(n, n)$ for all $j$ in the range $1 \\le j \\le n-1$.\nIran, Morteza Saghafian",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To prove this, choose an arbitrary $i < n$, and let $A_i A_{j_1}, A_i A_{j_2}, \\dots, A_i A_{j_{10}}$ be the ten smallest numbers among the $A_i A_j$ with $j \\ne i$, ordered so that $j_1 < j_2 < \\dots < j_{10}$.\nIf $j_{10} < i$, then $i > 10$, and the problem condition yields\n$$\nb_i = \\max_{1 \\le k \\le 10} A_i A_{j_k} = a_i \\ge a_n \\ge b_n,\n$$\nwhich is even stronger than we need.\nOtherwise, set $j = j_{10} > i$ (in this case we also have $j_{10} > 10$), and denote $m = b_i = \\max_{1 \\le k \\le 10} A_i A_{j_k}$. By the problem condition, we have $a_j \\ge a_n = b_n$. On the other hand, we have\n$$a_j \\le \\max \\left( A_j A_i, \\max_{1 \\le k \\le 9} A_j A_{j_k} \\right) \\le \\max \\left( A_j A_i, \\max_{1 \\le k \\le 9} (A_j A_i + A_i A_{j_k}) \\right) \\le 2m,$$ as $A_j A_i, A_i A_{j_k} \\le m$. So $b_n \\le a_j \\le 2m = 2b_i$, as desired.\n\nAlternative solution. Let $d_j = f_{10}(j, n)$, $j = 1, \\dots, n$. To prove that $2d_j \\ge d_n$ for every $j = 1, \\dots, n-1$, induct on $n$.\nConsider the base case, $n = 11$. Note that each $d_j = \\max_{i \\neq j} A_i A_j$, as $f_{10}$ is 10-variate. Let $d_{11} = A_{11}A_k$ for some index $k \\le 10$. Clearly, $2d_k \\ge 2A_k A_{11} = 2d_{11} \\ge d_{11}$ and if $j \\neq k$ then $2d_j \\ge A_j A_k + A_j A_{11} \\ge A_k A_{11} = d_{11}$, by the triangle inequality.\nFor the induction step, let $n > 11$ and note that\n$$\n\\max_{n \\ge l \\ge k} f_{10}(l, k-1) = \\max_{n-1 \\ge l \\ge k} f_{10}(l, k-1),\n$$\nas both maxima are achieved at $l = k$, by the condition in the statement. Hence $A_1, A_2, \\dots, A_{n-1}$ also satisfy this condition.\nLet $d_j' = f_{10}(j, n-1)$, $j = 1, \\dots, \\le n-1$. By the induction hypothesis, $2d_j' \\ge d_{n-1}'$ for all $j \\le n-2$. Note that $d_{n-1}' = f_{10}(n-1, n-2) \\ge f_{10}(n, n-2) \\ge f_{10}(n, n-1) = d_n$; the first inequality holds by the condition in the statement for $k = n-1$ and the second because adding more variables to $f_{10}$ does not increase its value.\nLet now $\\Delta_i$ be the closed disc (interior and boundary) of radius $d_i$, centred at $A_i$. By the definition of $d_i$, each $\\Delta_i$ contains at least 11 points, of which at most 10 ($A_i$, inclusive) lie strictly inside.\nFinally, suppose, if possible, that $2d_j < d_n$ for some index $j < n$. If $A_j A_n$ is not among the first 10 distances from $A_j$ to the other points, then $d_j = d_j'$ and this leads to a contradiction with the induction hypothesis. So $A_j A_n$ has to be among the first 10 distances from $A_j$ to the other points. This means that $d_j \\ge A_j A_n$, so $d_n > 2d_j \\ge 2A_j A_n$. Hence $\\Delta_j$ lies strictly inside $\\Delta_n$. This is a contradiction, as $\\Delta_j$ contains at least 11 points, whereas $\\Delta_n$ contains at most 10 strictly inside. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC = 2 \\cdot \\angle ACB$. Let $X$ and $Y$ be the midpoints of arcs $AB$ and $BC$ (not containing $C$ and $A$, respectively) of the circumcircle of triangle $ABC$. Let $BL$ be the angle bisector of $\\angle ABC$, with $L \\in AC$. Given that $\\angle XLY = 90^\\circ$, determine the measures of the angles of triangle $ABC$.",
    "options": [],
    "answer": "∠ABC = 90°, ∠ACB = 45°, ∠BAC = 45°",
    "solution": "From (1) and (2), we obtain that $\\triangle LBY \\equiv \\triangle LCY$ by the SSS criterion, and thus $\\angle BLY = \\angle CLY$, which implies that $LY$ is the angle bisector of $\\angle BLC$ (3).\nOn the other hand, $\\overline{AX} = \\overline{BX}$ implies that $\\angle ACX = \\angle XCB$, so $XC$ is the angle bisector of $\\angle LCB$ (4).\n\nLet $I$ be the intersection point of the segments $LY$ and $CX$. From (3) and (4), we conclude that $I$ is the incenter of triangle $\\triangle BLC$, hence $BI$ is the angle bisector of $\\angle LBC$ (5).\nFrom the hypothesis, $\\angle XLY = 90^\\circ$, and since $LY$ is the angle bisector of $\\angle BLC$ (from (3)), it follows that $LX$ is the external angle bisector of $\\angle BLC$ (6).\nFrom (4) and (6), we deduce that $X$ is the excenter of triangle $\\triangle BLC$ corresponding to vertex $C$, hence $XB$ is the external angle bisector of $\\angle LBC$. Combining this with (5), we obtain that $XB \\perp BI$, i.e., $\\angle XBI = 90^\\circ$ (7).\n\nLet $D$ be the intersection of $BI$ with the circumcircle $\\Gamma$, and let $\\alpha = \\angle IBC = \\angle IBL$. Then $\\angle ABL = \\angle LBC = 2\\alpha$. Since $\\angle LBC = \\angle LCB = 2\\alpha$, we also have $\\angle LCX = \\angle XCB = \\alpha$. Therefore:\n$$\n\\widehat{AD} = 2\\angle ABD = 2(\\angle ABL + \\angle LBD) = 2(2\\alpha + \\alpha) = 6\\alpha.\n$$\nAlso, $\\widehat{AX} = 2\\angle ACX = 2\\alpha$, so combining the two gives $\\widehat{DX} = 8\\alpha$.\nOn the other hand, $\\widehat{DX} = 2\\angle XBD = 180^\\circ$, by (7). Combining these two relations gives $8\\alpha = 180^\\circ$, hence $\\angle ACB = 2\\alpha = 45^\\circ$, $\\angle ABC = 4\\alpha = 90^\\circ$, and thus $\\angle BAC = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer number. We say that a matrix $A \\in \\mathcal{M}_n(\\mathbb{C})$ has the property $(\\mathcal{P})$ if $\\det(A + X_{ij}) = \\det(A + X_{ji})$, for any $i, j \\in \\{1, 2, \\dots, n\\}$, where $X_{ij} \\in \\mathcal{M}_n(\\mathbb{C})$ is the matrix with $1$ at the position $(i, j)$ and $0$ elsewhere.\n\na) Assume a matrix $A \\in \\mathcal{M}_n(\\mathbb{C})$ with the property $(\\mathcal{P})$, such that $\\det(A) \\neq 0$. Prove that $A = A^T$.\n\nb) Give an example of a matrix $A \\in \\mathcal{M}_n(\\mathbb{C})$ with the property $(\\mathcal{P})$, but $A \\neq A^T$.",
    "options": [],
    "answer": "a) A must equal its transpose. b) Example: the matrix whose only nonzero entries are ones in the first row at the first and second columns (all other entries zero) satisfies the property but is not symmetric.",
    "solution": "a) For given $i, j \\in \\{1, 2, \\dots, n\\}$, by computing the expansion along the row $i$, we obtain $\\det(A + X_{ij}) = \\det(A) + \\delta_{ij}$, where $\\delta_{ij}$ is the $(i, j)$-cofactor of the matrix $A$. Since $A$ has the property $(\\mathcal{P})$, it results that $\\delta_{ij} = \\delta_{ji}$, for any $i, j \\in \\{1, 2, \\dots, n\\}$. Hence the adjugate matrix $A^*$ is symmetric. Let us denote $d = \\det(A) \\neq 0$. We have $AA^* = dI_n = (dI_n)^T = (A^*A)^T = A^T(A^*)^T = A^T A^*$. Since $d \\neq 0$ and $d\\det(A^*) = \\det(AA^*) = \\det(dI_n) = d^n$, we deduce that $A^*$ is invertible. Therefore, from the relation $AA^* = A^T A^*$, we obtain $A = A^T$.\n\nb) Consider the matrix $A = (a_{ij})_{1 \\le i,j \\le n} \\in \\mathcal{M}_n(\\mathbb{C})$ with $a_{11} = a_{12} = 1$ and $a_{ij} = 0$ elsewhere. Since $n \\ge 3$, the matrix $A + X_{ij}$ has at least one null row, so $\\det(A + X_{ij}) = 0$, for any $i, j \\in \\{1, 2, \\dots, n\\}$. Thus, $A$ has the property $(\\mathcal{P})$, but $A \\neq A^T$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 39442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the set of rational numbers $r$ for which there exist non-negative integers $a$ and $b$, such that $\\frac{a+b}{2} - \\sqrt{ab} = r$.",
    "options": [],
    "answer": "{ m/2 : m is a non-negative integer }",
    "solution": "For $a$, $b$, $r$ as in the statement, $r = \\frac{(\\sqrt{a} - \\sqrt{b})^2}{2}$, therefore $r \\ge 0$. Moreover, $ab$ must be a square of a natural number $n$, so $r = \\frac{a+b-2n}{2} = \\frac{m}{2}$, where $m$ is a non-negative integer. Therefore, $r = \\frac{m}{2}$, with $m$ a non-negative integer.\n\nConversely, for $a = b = 1$, we obtain that $r = 0$ is a solution; for $a = m$ and $b = 4m$, with $m$ a positive integer, we obtain $r = \\frac{5m}{2} - 2m = \\frac{m}{2}$. Therefore, the set of solutions is $S = \\{ \\frac{m}{2} \\mid m \\in \\mathbb{N} \\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(G, \\cdot)$ be a group, and $H < G$ be a proper subgroup of $G$. If there are endomorphisms $f, g, h : G \\to G$ of the group $G$, such that $f(xy) = g(x)h(y)$ holds for any $x, y \\in G \\setminus H$, show that:\n\na) $g = h$;\n\nb) if $G$ is nonabelian, and $H = Z(G)$, then $f = g = h$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Denoting by $e$ the unit element of the group $G$, for any $x \\in G \\setminus H$ we have $x^{-1} \\in G \\setminus H$, so that $e = f(e) = f(x \\cdot x^{-1}) = g(x) \\cdot h(x^{-1}) = g(x) \\cdot h(x)^{-1}$, and it follows that $g(x) = h(x)$ for any $x \\in G \\setminus H$.\nFor any $a \\in H$, choosing an $x \\in G \\setminus H$ we have that $ax, x^{-1} \\in G \\setminus H$, so:\n$$\ng(a) = g(ax \\cdot x^{-1}) = g(ax) \\cdot g(x^{-1}) = h(ax) \\cdot h(x^{-1}) = h(ax \\cdot x^{-1}) = h(a).\n$$\nWe conclude that $g = h$.\n\n\nb) Let $G$ be nonabelian and $H = Z(G)$. According to part a), we have $g = h$, so that the relation in the statement becomes\n\nFor any $a \\in Z(G)$ and $x \\in G \\setminus Z(G)$, we have $ax, x^{-1} \\in G \\setminus Z(G)$, so:\n$$\nf(a) = f(ax \\cdot x^{-1}) = g(ax \\cdot x^{-1}) = g(a).\n$$\nConsider $x \\in G \\setminus Z(G)$. Then there is an element $y \\in G \\setminus Z(G)$ such that $xy \\neq yx$. If $xy \\in Z(G)$ then\n$$\nxy = y(xy)y^{-1} = yx \\neq xy,\n$$\nwhich would be a contradiction. Consequently, $xy \\in G \\setminus Z(G)$, and, since $y^{-1} \\in G \\setminus Z(G)$, we have:\n$$\nf(x) = f(xy \\cdot y^{-1}) = g(xy \\cdot y^{-1}) = g(x).\n$$\nWe conclude that $f = g$, hence $f = g = h$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve, in the set of natural numbers, the equation\n$$\nx^2 + y^2 + xy(x - y) = 17.\n$$",
    "options": [],
    "answer": "(1, 16)",
    "solution": "The equation can be written in the form $(x-y)^2 + 2xy + xy(x-y) = 17$.\nIf $x - y = d \\in \\mathbb{Z}$, $xy = p \\in \\mathbb{N}$, then the equality becomes $p(d+2) = 17 - d^2$, so $p = \\frac{17-d^2}{d+2}$. We get $p = \\frac{13}{d+2} + 2 - d \\in \\mathbb{N}$. It follows that $d+2 \\in \\{1, -1, 13, -13\\}$, which means $d \\in \\{-1, -3, 11, -15\\}$.\nIf $d = -1$, then $p = 16$ and in this case we have no solutions, because there are no two natural numbers, whose product is $16$ and whose difference is equal to $-1$.\n\nFrom $d = -3$ we deduce that $p = -8 < 0$, which doesn't work.\nFor $d = 11$, it follows $p = -8 < 0$, which is false.\nIf $d = x - y = -15$, then $p = xy = 16$, whence the unique solution $x = 1, y = 16$.\nThe equation becomes successively $(x-y)^2 + 2xy + xy(x-y) = 17$ or, if we write it as a difference of two squares: $4(x-y)^2 + 4xy(x-y) + (xy)^2 - (xy)^2 + 8xy - 16 = 68 - 16$, that is $(2(x-y)+xy)^2 - (xy-4)^2 = 52$. It follows that $(2x-2y+xy-xy+4)(2x-2y+xy+xy-4) = 52$ and therefore $(x-y+2)(x-y+xy-2) = 13$.\nSince the numbers $x, y, z$ are all natural, there are the following situations:\n• $x - y + 2 = -13$ and $x - y + xy - 2 = -1$, so $x - y = -15$ and $xy = 16$, whence $y = x + 15$ and $x(x + 15) = 16$. Since $x$ is a natural number, it follows from this that $x = 1$ and then $y = 16$;\n• $x - y + 2 = -1$ and $x - y + xy - 2 = -13$, so $x - y = -3$ and $xy = -8$, which is not possible, since $x$ and $y$ are both natural numbers;\n• $x - y + 2 = 1$ and $x - y + xy - 2 = 13$, so $x - y = -1$ and $xy = 16$, which is not possible, since $x$ and $y$ are both natural numbers;\n• $x - y + 2 = 13$ and $x - y + xy - 2 = 1$, so $x - y = 11$ and $xy = -8$, which is not possible, since $x$ and $y$ are both natural numbers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39445,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $a < b < c$ are such that $a + b + 2c$ is a common multiple of $b$ and $c$ (*).\n\na) Prove that the greatest common divisor of $a + b$ and $c$ is $c$.\n\nb) Find the positive integers $k < 1000$, so that $abc = k^2$ and $a, b, c$ fulfill (*).",
    "options": [],
    "answer": "{18, 144, 486, 900}",
    "solution": "a) Since $c \\mid a + b + 2c$ and $c \\mid 2c$, $c \\mid a + b$. Now $a < b < c$ implies $a + b < 2c$, hence $a + b = c$. Then $(a + b, c) = c$.\n\nb) From $b \\mid a + b + 2c$ and (a), $b \\mid 3a + 3b$. Since $b \\mid 3b$, $b \\mid 3a$. From $3a < 3b$ follows $3a \\in \\{b, 2b\\}$.\n\nCase I: $3a = b$. Then $a = n$, $b = 3n$, $c = 4n$, whence $abc = 12n^3$. The values of $n$ so that $2^2 \\cdot 3 \\cdot n^3 = k^2$, $k < 1000$ are: $n = 3$, implying $abc = 18^2$; $n = 3 \\cdot 2^2$, implying $abc = 144^2$; $n = 3^3$, implying $abc = 486^2$.\n\nCase II: $3a = 2b$. Then $a = 2p$, $b = 3p$, $c = 5p$. So $abc = 30p^3$, yielding $p = 30$ and $abc = 900^2$.\n\nIn conclusion, $k \\in \\{18, 144, 486, 900\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in $\\mathbb{R}$ the equation:\n$$\n\\frac{1}{\\{x\\}} + \\frac{1}{[x]} + \\frac{1}{x} = 0,\n$$\n\nwhere $[x]$ and $\\{x\\}$ denote the integer part and the fractional part of the real number $x$, respectively.",
    "options": [],
    "answer": "(1 - sqrt(5))/2, 1 - sqrt(5)",
    "solution": "First, we have that $x \\in \\mathbb{R} \\setminus (\\mathbb{Z} \\cup [0, 1))$. The given equation is equivalent to:\n$$\n\\frac{[x] + \\{x\\}}{[x]\\{x\\}} = -\\frac{1}{x} \\Leftrightarrow -x^2 = [x]\\{x\\}. \\quad (*)\n$$\nSince $-x^2 \\le 0$ and $x \\ne 0$, we have $[x]\\{x\\} < 0$, hence $[x] \\le -1$. Because $[x] = -k$, with $k \\in \\mathbb{N}^*$, equation $(*)$ becomes $-x^2 = -k(k+x)$, which is equivalent to $x^2 - kx - k^2 = 0$, having solutions $x_{1,2} = \\frac{k \\pm k\\sqrt{5}}{2}$.\nFrom $[x] = -k$, we obtain that the only possible solution is $x_1 = \\frac{k-k\\sqrt{5}}{2}$.\nFinally, we must find the values of $k \\in \\mathbb{N}^*$ for which $\\left[\\frac{k-k\\sqrt{5}}{2}\\right] = -k$, which is equivalent to: $-k \\le k \\frac{1-\\sqrt{5}}{2} < -k+1 \\Leftrightarrow 3k \\ge k\\sqrt{5} > 3k-2$, from which we obtain $k < \\frac{3+\\sqrt{5}}{2}$, i.e., $k \\in \\{1, 2\\}$. From here we obtain the solutions $x_1^* = \\frac{1-\\sqrt{5}}{2}$ and $x_2^* = 1-\\sqrt{5}$, which satisfy the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z, t$ be pairwise distinct positive integers and $n = 2^x + 2^y + 2^z + 2^t$. The division of $n$ by $305$ gives the quotient $2^a$ and leaves the remainder $0$, where $a$ is a non-negative integer.\nFind the remainder of the division of $a + x + y + z + t$ by $5$.",
    "options": [],
    "answer": "2",
    "solution": "Assume, without loss of generality, that $x < y < z < t$. Then $2^x + 2^y + 2^z + 2^t = 305 \\cdot 2^a$, whence\n$$\n2^x \\cdot (1 + 2^{y-x} + 2^{z-x} + 2^{t-x}) = 305 \\cdot 2^a \\tag{1}\n$$\nSince $x < y < z < t$, the numbers $y-x$, $z-x$ and $t-x$ are positive integers, hence $2^{y-x}$, $2^{z-x}$, $2^{t-x}$ are even, and $1+2^{y-x}+2^{z-x}+2^{t-x}$ is odd.\nFor parity reasons, (1) yields $2^x = 2^a$ and $1+2^{y-x}+2^{z-x}+2^{t-x} = 305$, hence $x=a$ and $2^{y-x}+2^{z-x}+2^{t-x} = 304$.\nThis gives\n$$\n2^{y-x} \\cdot (1 + 2^{z-y} + 2^{t-y}) = 2^4 \\cdot 19,\n$$\nwhence $y-x = 4$, so $y = a+4$.\nMoreover, $1+2^{z-y}+2^{t-y} = 19$, so $2^{z-y}(1+2^{t-z}) = 2 \\cdot 9$, therefore $z-y = 1$ and $t-z = 3$, that is $z = y+1 = a+5$ and $t = z+3 = a+8$.\nThe above yield\n$$\na+x+y+z+t = a+a+(a+4)+(a+5)+(a+8) = 5a+17 = 5 \\cdot (a+3)+2,\n$$\nso the remainder of the division of the sum $a+x+y+z+t$ by $5$ is $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and $a$ and $b$ complex numbers such that $a \\neq 0$ and $b^k \\neq 1$, for any $k \\in \\{1, 2, \\dots, n\\}$. Suppose that the matrices $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ are such that $BA = aI_n + bAB$. Prove that $A$ and $B$ are invertible.\nSorin Rădulescu and Mihai Piticari",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $b = 0$, then $BA = aI_n$, so $A$ and $B$ are invertible. We will suppose $b \\neq 0$.\n\nDenote by $\\sigma(X)$ the set of eigenvalues of a matrix $X \\in \\mathcal{M}_n(\\mathbb{C})$. Let $\\lambda \\in \\sigma(AB)$. Then\n$$\n\\det(BA - (b\\lambda + a)I_n) = \\det(bAB - b\\lambda I_n) = b^n \\det(AB - \\lambda I_n) = 0,\n$$\nimplying $b\\lambda + a \\in \\sigma(BA)$. As $\\sigma(AB) = \\sigma(BA)$, if $\\lambda \\in \\sigma(AB)$, we get $b\\lambda + a \\in \\sigma(AB)$.\n\nDefine $f : \\mathbb{C} \\to \\mathbb{C}$, $f(z) = bz + a$, for all $z \\in \\mathbb{C}$, and for $k \\in \\mathbb{N}^*$, denote by $f^{[k]} = \\underbrace{f \\circ f \\circ f \\dots \\circ f}_{k \\text{ times}}$. We conclude inductively, that if $\\lambda \\in \\sigma(AB)$, then $f^{[k]}(\\lambda) \\in \\sigma(AB)$, for $k \\in \\mathbb{N}^*$.\n\nSuppose $0 \\in \\sigma(AB)$. Then $f(0), f^{[2]}(0), \\dots, f^{[n+1]}(0) \\in \\sigma(AB)$. As $\\sigma(AB)$ has at most $n$ elements, there are $p, q \\in \\{1, 2, \\dots, n+1\\}$, $p < q$, such that $f^{[p]}(0) = f^{[q]}(0)$.\n\nBecause $f^{[k]}(z) = b^k z + a \\frac{b^{k}-1}{b-1}$, $z \\in \\mathbb{C}$, $k \\in \\mathbb{N}^*$, we obtain $a \\frac{b^p - 1}{b-1} = a \\frac{b^q - 1}{b-1}$, that is $b^{q-p} = 1$, a contradiction.\n\nIn conclusion, $0 \\notin \\sigma(AB)$, so $\\det(A) \\cdot \\det(B) = \\det(AB) \\neq 0$, and the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real numbers $x$ such that $3^x + 3^{[x]} + 3^{\\{x\\}} = 4$.",
    "options": [],
    "answer": "x = 1 - log_3 2 or x = log_3(11/12)",
    "solution": "There are no numbers $x \\in [1, \\infty)$ with the property in the statement: if $x \\ge 1$, then $[x] \\ge 1$ and, since $\\{x\\} \\in [0, 1)$, we obtain\n$$\n4 = 3^x + 3^{[x]} + 3^{\\{x\\}} \\ge 3 + 3 + 1 = 7,\n$$\ncontradiction.\n\nThere are no numbers $x \\in (-\\infty, -1)$ with the property in the statement: $x < -1$, then $[x] \\le -2$ and, since $\\{x\\} \\in [0, 1)$, we obtain\n$$\n4 = 3^x + 3^{[x]} + 3^{\\{x\\}} < 3^{-1} + 3^{-2} + 3^1 = 3\\frac{4}{9},\n$$\ncontradiction.\n\nIf $x \\in [0, 1)$, then $[x] = 0$, $\\{x\\} = x$, so the given equality becomes\n$$\n3^x + 1 + 3^x = 4 \\Leftrightarrow 2 \\cdot 3^x = 3.\n$$\nWe obtain the solution $x_1 = 1 - \\log_3 2 \\in [0, 1)$.\n\nIf $x \\in [-1, 0)$, then $[x] = -1$, $\\{x\\} = x + 1$, so the given equality becomes\n$$\n3^x + \\frac{1}{3} + 3^{x+1} = 4 \\Leftrightarrow 4 \\cdot 3^x = \\frac{11}{3}.\n$$\nWe obtain the solution $x_2 = \\log_3 \\frac{11}{12} \\in [-1, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle inscribed in a circle with center $O$ and radius $R$, and let $H$ be the orthocenter of triangle $ABC$. Let $A_1$ be a point on the side $BC$ such that $HA_1 + A_1O = R$. Similarly, define points $B_1$ on $CA$ and $C_1$ on $AB$. If $\\overrightarrow{AA_1} + \\overrightarrow{BB_1} + \\overrightarrow{CC_1} = \\overrightarrow{0}$, prove that triangle $ABC$ is equilateral.\n\nAlexandru Cărnaru",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H_A$ be the reflection of $H$ across line $BC$. Then $H_A$ lies on the circumcircle of triangle $ABC$. Since $A_1H = A_1H_A$, we get $A_1O + A_1H_A = R = OH_A$. Hence, $A_1$ lies on segment $OH_A \\cap BC$, and is uniquely defined by the given condition. We now prove that $\\angle BHA_1 = \\angle B$ and $\\angle CHA_1 = \\angle C$.\n\nSuppose without loss of generality that $\\angle B \\le \\angle C$. Let $A_O$ be the point diametrically opposite to $A$ on the circle. Then:\n\n$\\angle BHA_1 = \\angle BHH_A - \\angle A_1HH_A = \\angle C - \\angle OH_A A = \\angle C - \\angle A_O A H_A = \\angle C - (\\angle A - \\angle BAA_O - \\angle H_A A C) = \\angle C - (\\angle A - (90^\\circ - \\angle C) - \\angle BCA_O) = \\angle C - (\\angle A - (90^\\circ - \\angle C) - (90^\\circ - \\angle C)) = \\angle C - (\\angle A - 180^\\circ + 2\\angle C) = \\angle B$.\n\nSimilarly, we find that $\\angle CHA_1 = \\angle C$.\n\n$$\n\\frac{BA_1}{CA_1} = \\frac{A_{BHA_1}}{A_{CHA_1}} = \\frac{BH \\cdot HA_1 \\cdot \\sin(\\angle BHA_1)}{CH \\cdot HA_1 \\cdot \\sin(\\angle CHA_1)} = \\frac{2R \\cos B \\sin B}{2R \\cos C \\sin C} = \\frac{\\sin(2B)}{\\sin(2C)}.\n$$\n\nApplying the same process for $B_1$ and $C_1$, we obtain:\n$$\n\\frac{CB_1}{AB_1} = \\frac{\\sin(2C)}{\\sin(2A)}, \\quad \\frac{AC_1}{BC_1} = \\frac{\\sin(2A)}{\\sin(2B)}.\n$$\n\nMultiplying the three ratios gives:\n$$\n\\frac{BA_1}{CA_1} \\cdot \\frac{CB_1}{AB_1} \\cdot \\frac{AC_1}{BC_1} = \\frac{\\sin(2B)}{\\sin(2C)} \\cdot \\frac{\\sin(2C)}{\\sin(2A)} \\cdot \\frac{\\sin(2A)}{\\sin(2B)} = 1.\n$$\nThus, the cevians $AA_1, BB_1$, and $CC_1$ are concurrent. Given that: $\\overrightarrow{AA_1} + \\overrightarrow{BB_1} + \\overrightarrow{CC_1} = \\overrightarrow{0}$, the concurrency point must be the centroid, implying these are medians. Therefore, $A_1, B_1, C_1$ are midpoints of $BC, CA, AB$ respectively, and hence:\n$$\n\\frac{\\sin(2A)}{\\sin(2B)} = 1 \\Rightarrow \\sin(2A) = \\sin(2B),\n$$\nand, similarly, $\\sin(2A) = \\sin(2C)$.\n\nSince triangle $ABC$ is acute and the sine function is injective in $(0^\\circ, 180^\\circ)$, so $2A = 2B = 2C \\Rightarrow A = B = C$. Therefore, triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. For a finite set $M$ of points in the plane, we say that distinct points $A, B \\in M$ are connected if the line $AB$ contains exactly $n+1$ points in $M$.\n\nDetermine the smallest positive integer $m$ for which there exists a set $M$ of $m$ points in the plane with the property that any point $A \\in M$ is connected to exactly $2n$ other points in $M$.",
    "options": [],
    "answer": "(n+1)(n+2)/2",
    "solution": "Let $M = \\{A_1, A_2, \\dots, A_m\\}$ be a set of $m$ points with the given property and $A_1 \\in M$. Since $A_1$ is connected to other points, there is a line $d_0$ that contains exactly $n$ other points $A_2, \\dots, A_{n+1}$ from the set $M$. Since each of the points $A_1, A_2, \\dots, A_{n+1}$ is already connected to $n$ different points (other than him), we deduce that through each $A_i$ there passes exactly one more line $d_i$ that contains the other $2n - n = n$ points in $M$ connected to $A_i$.\n\nThus, $d_0$ contains $n+1$ points in $M$, $d_1$ contains $n$ new points in $M$ (the others except $A_1$), $d_2$ contains at least another $n-1$ points in $M$ (the others except $A_2$ and, possibly, the intersection of $d_2$ with $d_1$), etc. In general, the line $d_k$ contains at least another $n+1-k$ points in $M$ (the others except $A_k$ and, possibly, the intersections of $d_k$ with $d_1, d_2, \\dots, d_{k-1}$). So,\n$$\nm \\ge (n+1) + n + (n-1) + \\dots + 2 + 1 = \\frac{(n+1)(n+2)}{2}.\n$$\n\nTo prove that $\\frac{1}{2}(n+1)(n+2)$ is the minimum, we consider a configuration of $n+2$ lines in general position (i.e. any two are concurrent and there are no three concurrent lines) and $M$ the set of $\\binom{n+2}{2}$ points of intersection of them. Each line will then contain $n+1$ points from $M$ and, since each point from $M$ will be located on two of these lines, it will be connected by exactly $2 \\cdot (n+1-1) = 2n$ points.\nLet $M = \\{A_1, A_2, \\dots, A_m\\}$ be a set of $m$ points with the given property, $D_M = \\{A_iA_j \\mid 1 \\le i < j \\le m\\}$ and $D = \\{a \\in D_M \\mid |a \\cap M| = n+1\\}$. We denote $d = |D|$ and $I = \\{d_1 \\cap d_2 \\mid d_1, d_2 \\in D, d_1 \\neq d_2\\}$.\n\nIf the points $A, B \\in M$ are connected, then $A$ is connected to any of the other $n$ points in $M \\setminus \\{A\\}$ that lie on the line $AB$. Since any point $A \\in M$ is connected to exactly $2n$ other points in $M$, then $A$ lies at the intersection of exactly two lines in $D$ and thus $m = \\frac{d(n+1)}{2}$. It follows that $M \\subseteq I$ and the number of points in $I$ is at most equal to the number of pairs of two lines in $D$, so $|I| \\le \\binom{d}{2} = \\frac{d(d-1)}{2}$. Since $d = \\frac{2m}{n+1}$, we have\n$$\n|I| \\le \\frac{1}{2} \\cdot \\frac{2m}{n+1} \\left(\\frac{2m}{n+1} - 1\\right) = \\frac{m(2m-n-1)}{(n+1)^2}.\n$$\nWe obtain that $m \\le |I| \\le \\frac{m(2m-n-1)}{(n+1)^2}$, so $m \\ge \\frac{1}{2}(n+1)(n+2)$.\n\nTo prove that $\\frac{1}{2}(n+1)(n+2)$ is the minimum, we consider a configuration of $n+2$ lines in general position (i.e. any two are concurrent and there are no three concurrent lines) and $M$ the set of $\\binom{n+2}{2}$ points of intersection of them. Each line will then contain $n+1$ points from $M$ and, since each point from $M$ will be located on two of these lines, it will be connected by exactly $2 \\cdot (n+1-1) = 2n$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alexia has several marbles and her friend, Cristina, has none. Each day of one week, starting on Monday, Alexia gave Cristina some of her marbles, so that each day Alexia gave more marbles than the day before. On Monday Alexia gave five times less marbles than on Friday, on Tuesday she gave six times less marbles than on Saturday, and on Wednesday she gave seven times less marbles than on Sunday. At the end of that week, Cristina got 72 marbles. Find how many marbles got Cristina Thursday.\nLucian Dragomir",
    "options": [],
    "answer": "7",
    "solution": "Denote $a_1, a_2, \\dots, a_7$ the number of marbles given by Alexia from Monday until Sunday, where $1 \\le a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7$ and $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 72$. Then $a_5 = 5a_1$, $a_6 = 6a_2$, $a_7 = 7a_3$, hence $6a_1 + 7a_2 + 8a_3 + a_4 = 72$ (1).\n\nIf $a_1 \\ge 3$, then $6a_1 + 7a_2 + 8a_3 + a_4 \\ge 6 \\cdot 3 + 7 \\cdot 4 + 8 \\cdot 5 + 6 = 92$, contradicting (1). Therefore, $a_1 = 1$ or $a_1 = 2$.\n\nIf $a_1 = 1$, then $a_5 = 5$ and $a_1 = 1 < a_2 < a_3 < a_4 < a_5 = 5$ yields $a_2 = 2$, $a_3 = 3$, $a_4 = 4$, whence $6a_1 + 7a_2 + 8a_3 + a_4 = 48$, contradicting (1). So, $a_1 = 2$.\n\nFrom $a_5 = 5a_1 = 10$ follows $2 = a_1 < a_2 < a_3 < a_4 < a_5 = 10$. Moreover, (1) gives $7a_2 + 8a_3 + a_4 = 60$ (2).\n\nIf $a_2 \\ge 4$, then $7a_2 + 8a_3 + a_4 \\ge 7 \\cdot 4 + 8 \\cdot 5 + 6 = 74$, contradicting (1), hence $a_2 \\le 3$. From $a_2 > a_1 = 2$ follows $a_2 = 3$.\n\nNow (2) gives $8a_3 + a_4 = 39$, hence $8a_3 < 39$, so $a_3 \\le 4$. But $a_3 > a_2 = 3$, hence $a_3 = 4$, and $8a_3 + a_4 = 39$ yields $a_4 = 7$.\n\nClearly $a_4 = 7$ is acceptable (the number of the given marbles being 2, 3, 4, 7, 10, 18, 28). So, the answer is 7 marbles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a differentiable function, with integrable derivative on $[0, 1]$, such that $f(1) = 0$. Prove that\n$$\n\\int_{0}^{1} (x f'(x))^2 dx \\geq 12 \\cdot \\left( \\int_{0}^{1} x f(x) dx \\right)^2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g : [0, 1] \\to \\mathbb{R}$ defined by $g(x) = x f(x)$, for $x \\in [0, 1]$. We have $g(0) = 0 = g(1)$, $g'(x) = f(x) + x f'(x)$, so that:\n$$\n\\begin{align*}\n\\int_0^1 x^2 (f'(x))^2 dx &= \\int_0^1 (g'(x) - f(x))^2 dx \\\\\n&= \\int_0^1 (g'(x))^2 dx - 2 \\int_0^1 g'(x) f(x) dx + \\int_0^1 (f(x))^2 dx \\\\\n&= \\int_0^1 f^2(x) dx + \\int_0^1 (g'(x))^2 dx - \\\\\n& \\qquad -2(f(1)g(1) - f(0)g(0)) + 2 \\int_0^1 g(x) f'(x) dx,\n\\end{align*}\n$$\nimplying\n$$\n\\begin{align*}\n\\int_{0}^{1} x^2 (f'(x))^2 dx &= \\\\\n&= \\int_{0}^{1} f^2(x) dx + \\int_{0}^{1} (g'(x))^2 dx + \\int_{0}^{1} x \\cdot (2f(x)f'(x)) dx \\\\\n&= \\int_{0}^{1} f^2(x) dx + \\int_{0}^{1} (g'(x))^2 dx + \\int_{0}^{1} x \\cdot (f^2(x))' dx \\\\\n&= \\int_{0}^{1} f^2(x) dx + \\int_{0}^{1} (g'(x))^2 dx + (1 \\cdot f^2(1) - 0 \\cdot f^2(0)) - \\int_{0}^{1} f^2(x) dx \\\\\n&= \\int_{0}^{1} f^2(x) dx + \\int_{0}^{1} (g'(x))^2 dx - \\int_{0}^{1} f^2(x) dx = \\int_{0}^{1} (g'(x))^2 dx.\n\\end{align*}\n$$\n\n$$\n\\left( \\int_0^1 (2x-1) \\cdot g'(x) \\, dx \\right)^2 \\le \\int_0^1 (2x-1)^2 \\, dx \\cdot \\int_0^1 (g'(x))^2 \\, dx = \\\\\n= \\frac{1}{6} \\cdot \\left( (2 \\cdot 1 - 1)^3 - (2 \\cdot 0 - 1)^3 \\right) \\cdot \\int_0^1 x^2 (f'(x))^2 \\, dx = \\frac{1}{3} \\cdot \\int_0^1 x^2 (f'(x))^2 \\, dx.\n$$\nThis gives\n$$\n\\begin{align*}\n\\int_0^1 (x f'(x))^2 \\, dx &\\ge 3 \\cdot \\left( \\int_0^1 (2x-1) \\cdot g'(x) \\, dx \\right)^2 = \\\\\n&= 3 \\cdot \\left( (2 \\cdot 1 - 1)g(1) - (2 \\cdot 0 - 1)g(0) - 2 \\int_0^1 g(x) \\, dx \\right)^2 = \\\\\n&= 12 \\left( \\int_0^1 x f(x) \\, dx \\right)^2,\n\\end{align*}\n$$\nconcluding the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the positive integers $a$ and $b$ fulfilling\n$$\n\\frac{a}{(a,b)} = b + \\frac{48 \\cdot (a,b)}{[a,b]} \\quad \\text{and} \\quad \\frac{b}{(a,b)} = a - \\frac{312 \\cdot (a,b)}{[a,b]}\n$$\nwhere $(a, b) = \\gcd(a, b)$ and $[a, b] = \\text{lcm}(a, b)$.",
    "options": [],
    "answer": "a = 16, b = 6",
    "solution": "We start noticing that $a \\ge \\frac{a}{(a,b)} > b$.\nDenote $d = (a, b)$. Then $a = dx$, $b = dy$, $[a, b] = dxy$ and $(x, y) = 1$, $x > y$. This gives $x = dy + \\frac{48}{xy}$ and $y = dx - \\frac{312}{xy}$ (1).\nIt follows that $xy \\mid 48$ and $xy \\mid 312$, hence $xy \\mid (48, 312) = 24$.\nRelation (1) yields $xy(x - dy) = 48$ and $xy(dx - y) = 312$, whence $\\frac{x-dy}{dx-y} = \\frac{48}{312} = \\frac{2}{13}$ and $x(13 - 2d) = y(13d - 2)$, $13 - 2d > 0$.\n\nI. $d = 1 \\Rightarrow 11x = 11y \\Rightarrow a = b$, false, since $a > b$.\nII. $d = 2 \\Rightarrow 3x = 8y$. Since $(x, y) = 1$, we get $x = 8$, $y = 3$, hence $a = 16$, $b = 6$.\nIII. $d = 3 \\Rightarrow 7x = 37y \\Rightarrow x = 37$, $y = 7$ – false, since $xy \\mid 24$.\nIV. $d = 4 \\Rightarrow x = 10y \\Rightarrow x = 10$, $y = 1$ – false, since $xy \\mid 24$.\nV. $d = 5 \\Rightarrow x = 21y \\Rightarrow x = 21$, $y = 1$ – false, since $xy \\mid 24$.\nVI. $d = 6 \\Rightarrow x = 76y \\Rightarrow x = 76$, $y = 1$ – false, since $xy \\mid 24$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider positive real numbers $m, n, a, b, c$, such that $m > n$ and $|ma - nb| \\le c(m - n)$, $|mb - nc| \\le a(m - n)$, $|mc - na| \\le b(m - n)$. Prove that $a = b = c$.\n\nLucian Petrescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x \\ge 0$, $y \\ge 0$, and $z \\ge 0$, then $ma - nb = c(m - n)$ and analogously, we have\n$$\nm(a - c) = n(b - c), \\quad m(b - a) = n(c - a), \\quad m(c - b) = n(a - b). \\quad (*)\n$$\nIf all parentheses in $(*)$ are nonzero, then we obtain $m^3(a-c)(b-a)(c-b) = -n^3(b-c)(c-a)(a-b)$, and therefore $m = -n$, false. Thus at least one difference is zero, say $a - c = 0$. It follows that $b = c$, hence $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k, n \\in \\mathbb{N}^*$ and let $A$ be an arbitrary set of $2n$ positive integers. For each $a \\in A$, denote by $p_a$ the number of elements in the set $P_a = \\{b \\in A \\mid b - a \\equiv k \\pmod{2k}\\}$. Determine the maximum possible value of the sum $E_A = \\sum_{a \\in A} p_a^2$, when $A$ runs over all subsets of size $2n$ of $\\mathbb{N}^*$.\nCristi Săvescu",
    "options": [],
    "answer": "2n^3",
    "solution": "Let us denote by $A_i = \\{a \\in A \\mid a \\equiv i \\pmod{2k}\\}$ and by $a_i = |A_i|$ for each $i = 0, \\dots, 2k-1$. Observe that $A = \\bigcup_{i=0}^{2k-1} A_i$, where $A_i \\cap A_j = \\emptyset$ for $i \\neq j$, and therefore $2n = \\sum_{i=0}^{2k-1} a_i$.\n\nFor any $a \\in A_i$, we have\n$$\nP_a = \\{b \\in A \\mid b - a \\equiv k \\pmod{2k}\\} = \\{b \\in A \\mid b \\equiv k + i \\pmod{2k}\\} = A_{k+i},\n$$\nwhere the index $k + i$ is considered modulo $2k$. Hence, $p_a = a_{k+i}$, for every $a \\in A_i$.\n\nThus,\n$$\nE_A = \\sum_{a \\in A} p_a^2 = \\sum_{i=0}^{2k-1} \\sum_{a \\in A_i} p_a^2 = \\sum_{i=0}^{2k-1} a_i a_{k+i}^2 = \\sum_{i=0}^{k-1} a_i a_{k+i}^2 + \\sum_{i=k}^{2k-1} a_i a_{k+i}^2.\n$$\nSince the indices are modulo $2k$,\n$$\n\\sum_{i=k}^{2k-1} a_i a_{k+i}^2 = \\sum_{i=0}^{k-1} a_{k+i} a_i^2,\n$$\nso we get\n$$\nE_A = \\sum_{i=0}^{k-1} (a_i a_{k+i}^2 + a_{k+i} a_i^2) = \\sum_{i=0}^{k-1} a_i a_{k+i} (a_i + a_{k+i}).\n$$\n\nWe want to maximize $E_A$ under the constraint $\\sum_{i=0}^{2k-1} a_i = 2n$.\nNote that replacing any two distinct pairs $(a_i, a_{i+k})$ and $(a_j, a_{j+k})$ by $(a_i + a_j, a_{i+k} + a_{j+k})$ and $(0, 0)$ does not decrease the value of $E_A$, because\nTherefore, the maximum is achieved when only one pair $(a_i, a_{i+k})$ is nonzero, with $a_i + a_{i+k} = 2n$.\n\nUsing the inequality between arithmetic and geometric means, we have $E_A = a_i a_{i+k} (a_i + a_{i+k}) \\le \\frac{(a_i+a_{i+k})^3}{4} = 2n^3$.\nThe maximum value $2n^3$ is attained, for example, by any set $A$ having exactly $n$ multiples of $2k$ and $n$ numbers congruent to $k \\pmod{2k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n, \\dots$ be a sequence of positive real numbers. For every positive integer $n$, write\n$$\ns_n = a_1 + a_2 + \\dots + a_n \\quad \\text{and} \\quad \\sigma_n = \\frac{a_1}{1+a_1} + \\frac{a_2}{1+a_2} + \\dots + \\frac{a_n}{1+a_n}.\n$$\n\nProve that, if the $s_n$ form an unbounded sequence, then so do the $\\sigma_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, the $s_n$ and the $\\sigma_n$ both form strictly increasing sequences. We will construct a sequence of positive integers $n_1 < n_2 < \\dots < n_k < \\dots$ such that $\\sigma_{n_k} > k a_1/(1+a_1)$ for all $k \\ge 2$. The conclusion then follows at once.\nLet $n_1$ be any positive integer, then use unboundedness of the $s_n$ to choose $n_k$ recursively so that $s_{n_k} > s_{n_{k-1}} + a_1$, $k \\ge 2$.\nThe conclusion is a consequence of the following inequality: If $m \\ge 2$ and $x_1, x_2, \\dots, x_m$ are positive real numbers, then\n$$\n\\frac{x_1}{1+x_1} + \\frac{x_2}{1+x_2} + \\dots + \\frac{x_m}{1+x_m} > \\frac{x_1+x_2+\\dots+x_m}{1+x_1+x_2+\\dots+x_m}. \\quad (*)\n$$\nAssume $(*)$ for the moment to write\n$$\n\\begin{aligned}\n\\sigma_{n_k} - \\sigma_{n_{k-1}} &= \\frac{a_{n_{k-1}+1}}{1+a_{n_{k-1}+1}} + \\dots + \\frac{a_{n_k}}{1+a_{n_k}} \\\\\n&> \\frac{a_{n_{k-1}+1} + \\dots + a_{n_k}}{1+a_{n_{k-1}+1} + \\dots + a_{n_k}} = \\frac{s_{n_k} - s_{n_{k-1}}}{1+s_{n_k} - s_{n_{k-1}}} > \\frac{a_1}{1+a_1}, \\quad k \\ge 2,\n\\end{aligned}\n$$\nso $\\sigma_{n_k} - \\sigma_{n_1} > (k-1)a_1/(1+a_1)$. As $\\sigma_{n_1} \\ge \\sigma_1 = a_1/(1+a_1)$, it follows that $\\sigma_{n_k} > k a_1/(1+a_1)$ for all $k \\ge 2$, as desired.\n\nFinally, we prove $(*)$ in two different ways.\n\n1st Proof.\nWrite\n$$\n\\begin{aligned}\n\\sum_{k=1}^{m} \\frac{x_k}{1+x_k} &= \\sum_{k=1}^{m} \\frac{x_k^2}{x_k + x_k^2} \\ge \\frac{\\left(\\sum_{k=1}^{m} x_k\\right)^2}{\\sum_{k=1}^{m} x_k + \\sum_{k=1}^{m} x_k^2}, \\quad \\text{by Cauchy-Schwarz} \\\\\n&> \\frac{\\left(\\sum_{k=1}^{m} x_k\\right)^2}{\\sum_{k=1}^{m} x_k + \\left(\\sum_{k=1}^{m} x_k\\right)^2} = \\frac{\\sum_{k=1}^{m} x_k}{1 + \\sum_{k=1}^{m} x_k}.\n\\end{aligned}\n$$\n\n2nd Proof.\nInduct on $m$. The base case, $m=2$, is a routine check. For $m \\ge 3$,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{m} \\frac{x_k}{1+x_k} &= \\sum_{k=1}^{m-2} \\frac{x_k}{1+x_k} + \\left( \\frac{x_{m-1}}{1+x_{m-1}} + \\frac{x_m}{1+x_m} \\right) \\\\\n&> \\sum_{k=1}^{m-2} \\frac{x_k}{1+x_k} + \\frac{x_{m-1} + x_m}{1+x_{m-1} + x_m} > \\frac{\\sum_{k=1}^{m} x_k}{1 + \\sum_{k=1}^{m} x_k}.\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39458,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite collection $C$ of (not necessarily distinct) real numbers is *suitable* if it contains two numbers $a$ and $b$ such that $a+b \\neq s+1$, where $s$ is the sum of all numbers in $C$; such numbers $a$ and $b$ form an *eligible* pair.\n\nFix an integer $n \\ge 2$. A number of $n$ pairwise distinct real numbers are written on a board. A step consists in choosing an eligible pair $(a, b)$ amongst numbers on the board (if any), crossing them out and replacing them by the number\n$$\n\\frac{(a+b)(s+1) - a^2 - ab - b^2}{s-a-b+1}.\n$$\n\na) Prove that there is a sequence of $n-1$ steps such that at each stage the numbers on the board form a suitable collection; and\n\nb) Determine the final number on the board in terms of the initial numbers.",
    "options": [],
    "answer": "sum_i c_i + sum_{i<j} c_i c_j",
    "solution": "a) Two real numbers (not necessarily distinct) always form a suitable collection. Let $n \\ge 3$ and consider the initial collection. Note that any $a$ can be paired off with some $b \\neq a$ to form an eligible pair: Otherwise, $a+b=s+1=a+c$ for distinct $b, c \\neq a$, so $b=c$, contradicting the fact that the initial numbers are pairwise distinct. Hence the initial collection is suitable.\n\nChoose an eligible pair $a_1, a_2$ and use the formula to replace them by $b_1$. As the remaining numbers are pairwise distinct, there is at most one that cannot be paired with $b_1$ to form an eligible pair, so there are at least $n-2$ candidates to form an eligible pair with $b_1$. Let $a_3$ be one such and use the formula to replace $b_1$ and $a_3$ by $b_2$. Repeat the argument to replace $b_2$ and some $a_4$ by $b_3$ and so on and so forth all the way down to some $b_{n-2}$ and $a_n$ (possibly, $b_{n-2} = a_n$). These latter form an eligible pair, so they can be replaced by a single number.\n\nb) Let $c_1, c_2, \\dots, c_n$ be the initial numbers. The final number is $\\sum_i c_i + \\sum_{i<j} c_i c_j$. This is clearly the case if $n=2$, so let $n \\ge 3$.\n\nConsider a generic stage $x_1, x_2, \\dots, x_m$ and let $s = x_1 + x_2 + \\dots + x_m$. We will prove that $s + \\sum_{1 \\le i < j \\le m} x_i x_j$ does not change upon passing to the next stage. Let $x$ be the number obtained by replacing an eligible pair $(x_k, x_\\ell)$. Then $s$ changes by $x - x_k - x_\\ell$ and the other sum changes by $-x_k x_\\ell - (x_k + x_\\ell)(s - x_k - x_\\ell) + x(s - x_k - x_\\ell)$, so the overall change is\n$$\nx - x_k - x_\\ell - x_k x_\\ell - (x_k + x_\\ell)(s - x_k - x_\\ell) + x(s - x_k - x_\\ell).\n$$\nFinally, express $x$ in terms of $s$, $x_k$ and $x_\\ell$ and carry out calculations to show that the overall change vanishes, whence the desired invariance.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene acute triangle with incentre $I$ and circumcentre $O$. Let $AI$ cross $BC$ at $D$. On circle $ABC$, let $X$ and $Y$ be the mid-arc points of $ABC$ and $BCA$, respectively. Let $DX$ cross $CI$ at $E$ and let $DY$ cross $BI$ at $F$. Prove that the lines $FX$, $EY$ and $IO$ are concurrent on the external bisector of $\\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The argument hinges on the claim below:\n\n**Claim.** The lines $AE$ and $BI$ are perpendicular; similarly, $AF$ and $CI$ are perpendicular\n\n*Proof.* Let $\\alpha = \\angle BAC$, $\\beta = \\angle CAB$ and $\\gamma = \\angle ACB$. Let $DX$ cross the circle $ADC$ again at $D'$. Note that $\\angle ECX = \\angle ACX - \\angle ACE = 90^\\circ - \\beta/2 - \\gamma/2 = \\alpha/2 = \\angle DAC = \\angle DD'C = \\angle XD'C$. As $\\angle CXE = \\angle CXD'$, triangles $XCD'$ and $XEC$ are similar, so $XD' \\cdot XE = XC^2$.\n\nAs $XA = XC$, it follows that $XD' \\cdot XE = XA^2$, so triangles $XD'A$ and $XAE$ are similar, so $\\angle XAE = \\angle AD'X = \\angle AD'D = \\angle ACD = \\gamma$.\nFinally, note that $\\angle XAD = \\angle XAC - \\angle DAC = 90^\\circ - \\beta/2 - \\alpha/2 = \\gamma/2$, so $\\angle IAE = \\angle DAE = \\angle XAE - \\angle XAD = \\gamma/2$. As $\\angle AIB = 90^\\circ + \\gamma/2$, the claim follows.\n\nLet $W$ be the mid-arc point of $CAB$ and let $I'$ be the reflection of $I$ across $O$. As $W, X, Y$ are the mid-arc points of $CAB, ABC, BCA$, respectively, their reflections across $O$ are the mid-arc points opposite. These latter form a triangle with orthocentre $I$, so $I'$ is the orthocentre of triangle $WXY$.\nReflection across $O$ maps lines $XV, WY$ and $WX$ to the perpendicular bisectors of $AI, BI$ and $CI$, respectively, so $XY \\perp AI, WY \\perp BI$ and $WX \\perp CI$. By the claim, $AE \\perp IF$ and $AF \\perp IE$, so $I$ is the orthocentre of triangle $AEF$ and hence $EF \\perp AI$ as well.\nTriangles $AEF$ and $WXY$ have therefore corresponding parallel sides, so they are homothetic from some point $R$. This homothety maps $I$ to $I'$, as they are corresponding orthocentres. Hence the lines $AW, EY, FY$ and $II'$ are concurrent at $R$. As $I, O$ and $I'$ are collinear and $AW$ is the external bisector of $\\angle BAC$, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and let $a_1, a_2, \\dots, a_n$ be real numbers. Denote\n$$\nS = \\sum_{1 \\le i < j \\le n} |a_j - a_i| \\quad \\text{and} \\quad d = \\max\\{|a_j - a_i| \\mid 1 \\le i, j \\le n\\}.\n$$\n\nProve that\n$$\n(n-1)d \\le S \\le \\frac{n^2 d}{4}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we can assume that $a_1 \\le a_2 \\le \\dots \\le a_n$, and denote $d_k = a_{k+1} - a_k$, for $k = 1, 2, \\dots, n-1$. Then we have:\n* $d_1 + d_2 + \\dots + d_{n-1} = d$;\n* $|a_j - a_i| = d_i + d_{i+1} + \\dots + d_{j-1}$, for any $i < j$ with $i, j \\in \\{1, \\dots, n\\}$.\nDenote by $n_k$ the number of pairs $(i, j)$, with $1 \\le i < j \\le n$, such that the interval $[a_k, a_{k+1}]$ is contained in $[a_i, a_j]$. Then the conditions imply $i \\le k < k+1 \\le j$, so there are $k$ possible choices for $i$ and $n-k$ possible choices for $j$. Thus, $n_k = k(n-k)$. Consequently,\n$$\nS = \\sum_{k=1}^{n-1} n_k d_k = \\sum_{k=1}^{n-1} k(n-k) d_k. \\qquad (1)\n$$\n\nFor any $k \\in \\{1, 2, \\dots, n-1\\}$, we have $k(n-k) \\ge n-1$.\nIndeed, $k(n-k) \\ge n-1 \\iff kn-k^2 \\ge n-1 \\iff (k-1)(n-k-1) \\ge 0$, which holds for all $k$ in the stated range.\nFrom (1) and this, it follows that\n$$\nS \\ge \\sum_{k=1}^{n-1} (n-1)d_k = (n-1)d,\n$$\nwhich proves the left inequality.\nEquality holds for any $n \\ge 2$ if and only if $a_1 \\le a_2 = a_3 = \\dots = a_{n-1} \\le a_n$.\n\nUsing AM-GM inequality, we obtain $k(n-k) \\le \\frac{n^2}{4}$, for all $k \\in \\{1, 2, \\dots, n-1\\}$. Therefore,\n$$\nS = \\sum_{k=1}^{n-1} k(n-k)d_k \\le \\sum_{k=1}^{n-1} \\frac{n^2}{4}d_k = \\frac{n^2}{4}d, \\qquad (3)\n$$\nwhich proves the right inequality.\nSince equality in the inequality $k(n-k) \\le \\frac{n^2}{4}$ holds when $k = n-k$, i.e., when $n = 2k$ (so $n$ is even), from (3) we deduce:\n\n* If $d_k = 0$ for all $k = 1, \\dots, n-1$, i.e., $a_1 = a_2 = \\dots = a_n$, then equality holds for any $n \\ge 2$.\n* If there exists $t \\in \\{1, \\dots, n-1\\}$ such that $d_t \\ne 0$, then:\n    * If $n$ is odd, the right inequality is strict.\n    * If $n$ is even, equality holds only if $t = n-t$ and $d_t = d$, i.e., $n = 2t$ and $a_1 = \\dots = a_t \\le a_{t+1} = a_{t+2} = \\dots = a_{2t}$.\n\n$$\nd = a_n - a_1 = (a_n - a_k) + (a_k - a_1), \\quad \\text{for any } 2 \\le k \\le n-1.\n$$\nWe obtain:\n$$\n\\begin{aligned}\nS &= \\sum_{1 \\le i < j \\le n} |a_j - a_i| = (a_n - a_1) + \\sum_{k=2}^{n-1} ((a_n - a_k) + (a_k - a_1)) + \\\\\n&\\quad + \\sum_{2 \\le i < j \\le n-1} |a_j - a_i| \\\\\n&= (n-1)d + \\sum_{2 \\le i < j \\le n-1} |a_j - a_i| \\ge (n-1)d.\n\\end{aligned}\n$$\nIf $n=2$ or $n=3$, equality holds for any numbers $a_1 \\le a_2$, respectively $a_1 \\le a_2 \\le a_3$.\nIf $n \\ge 4$, equality holds if and only if $a_1 \\le a_2 = a_3 = \\dots = a_{n-1} \\le a_n$.\n\nAlternative solution for the right inequality.\nWe prove by induction the statement:\n$$\nP(n): \\quad \\sum_{1 \\le i < j \\le n} |a_j - a_i| \\le \\frac{n^2}{4} \\cdot \\max_{1 \\le i < j \\le n} |a_j - a_i|,\n$$\nfor all $n \\ge 2$ and any real numbers $a_1, a_2, \\dots, a_n$. Without loss of generality, assume $a_1 \\le a_2 \\le \\dots \\le a_n$.\nBase cases:\n$$\nP(2): \\quad a_2 - a_1 \\le \\frac{2^2}{4} \\cdot (a_2 - a_1) \\iff a_2 - a_1 \\le a_2 - a_1,\n$$\n$$\nP(3): \\quad (a_2 - a_1) + (a_3 - a_2) + (a_3 - a_1) \\le \\frac{3^2}{4} \\cdot (a_3 - a_1) \\iff 2(a_3 - a_1) \\le \\frac{9}{4}(a_3 - a_1).\n$$\n\nAssume $P(n-2)$ holds for some $n \\ge 4$. Using the notation from the statement and the fact that\n$$\nd = a_n - a_1 = (a_n - a_k) + (a_k - a_1), \\quad \\text{for any } 2 \\le k \\le n-1,\n$$\nwe deduce:\n$$\nS = \\sum_{1 \\le i < j \\le n} |a_j - a_i| = (n-1)d + \\sum_{2 \\le i < j \\le n-1} |a_j - a_i|.\n$$\nDenote\n$$\nS' = \\sum_{2 \\le i < j \\le n-1} |a_j - a_i|, \\quad d' = a_{n-1} - a_2.\n$$\nObviously, $d' \\le d$. Since $P(n-2)$ holds, we get $S' \\le \\frac{(n-2)^2}{4} \\cdot d'$. Therefore,\n$$\nS = (n-1)d + S' \\le (n-1)d + \\frac{(n-2)^2}{4} \\cdot d' \\le (n-1)d + \\frac{(n-2)^2}{4} \\cdot d = \\frac{n^2}{4} \\cdot d,\n$$\nwhich proves $P(n)$.\nTo determine when equality holds, observe that in $P(2)$ equality holds for any $a_1 \\le a_2$, and in $P(3)$ equality holds only if $a_1 = a_2 = a_3$ (strict inequality otherwise if $a_1 < a_3$).\nIt is clear that if $a_1 = a_2 = \\dots = a_n$, then equality holds in $P(n)$ for any $n \\ge 2$.\nAssume now that $a_1 < a_n$. From the previous inequality, equality in $P(n)$ holds if and only if equality holds in $P(n-2)$ and $d' = d$. Consequently:\n* if $n$ is odd, the inequality $P(n)$ is strict;\n* if $n$ is even, say $n = 2t$, equality holds if and only if $a_1 = a_2 = \\dots = a_t < a_{t+1} = a_{t+2} = \\dots = a_{2t}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an arbitrary triangle $ABC$ with incenter $I$, and let $I_a$, $I_b$, and $I_c$ be the excenters of triangle $ABC$ opposite to vertices $A$, $B$, and $C$, respectively, tangent to sides $BC$, $CA$, and $AB$. Let $E$, $F$, and $G$ be the points of tangency of the incircle with the sides $BC$, $CA$, and $AB$, respectively.\nProve that the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$ intersect in a second common point different from $I$.\nPetru Braica\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $J_a$, $J_b$, and $J_c$ the centers of the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. We prove that $J_a$, $J_b$, and $J_c$ are collinear.\n\nLet $L_a$, $L_b$, and $L_c$ be the points diametrically opposite to $I$ on the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. Since $J_a J_b$, $J_b J_c$, and $J_c J_a$ are midlines in triangles $IL_a L_b$, $IL_b L_c$, and $IL_c L_a$, respectively, it suffices to show that $L_a$, $L_b$, and $L_c$ are collinear.\n\nSince $IL_a$ is a diameter in the circumcircle of triangle $IEI_a$, we have $\\angle II_a L_a = 90^\\circ$. Hence $I_a L_a \\perp II_a$. But $II_a \\perp I_b I_c$, so $I_a L_a \\parallel I_b I_c$, implying\n$$\n\\angle L_a I_a B = \\angle I_a I_c I_b, \\quad \\text{and} \\quad \\angle L_a I_a C = 180^\\circ - \\angle I_a I_b I_c. \\qquad (1)\n$$\nLet $J$ be the midpoint of segment $IE$. Since $JJ_a$ is a midline in triangle $IEL_a$, it follows that $JJ_a \\parallel EL_a$. But $JJ_a$ is the perpendicular bisector of $IE$, so $JJ_a \\perp IE$, and since $BE \\perp IE$, it follows that $JJ_a \\parallel BE$. From $BE \\parallel JJ_a$ and $EL_a \\parallel JJ_a$, we deduce that $L_a \\in BE$, so $L_a \\in BC$. Similarly, we deduce that $L_b \\in CA$ and $L_c \\in AB$.\n\nWe have $\\frac{BL_a}{CL_a} = \\frac{S_{BI_a L_a}}{S_{CI_a L_a}} = \\frac{BI_a \\cdot L_a I_a \\cdot \\sin(\\angle L_a I_a B)}{CI_a \\cdot L_a I_a \\cdot \\sin(\\angle L_a I_a C)} \\stackrel{(1)}{=} \\frac{BI_a \\cdot \\sin(\\angle I_a I_c I_b)}{CI_a \\cdot \\sin(\\angle I_a I_b I_c)}$.\nSimilarly, $\\frac{CL_b}{AL_b} = \\frac{CI_b \\cdot \\sin(\\angle I_b I_a I_c)}{AI_b \\cdot \\sin(\\angle I_b I_c I_a)}$, $\\frac{AL_c}{BL_c} = \\frac{AI_c \\cdot \\sin(\\angle I_c I_b I_a)}{BI_c \\cdot \\sin(\\angle I_c I_a I_b)}$.\n\nTherefore, $\\frac{BL_a}{CL_a} \\cdot \\frac{CL_b}{AL_b} \\cdot \\frac{AL_c}{BL_c} = 1$, and by the converse of Menelaus' theorem, $L_a$, $L_b$, and $L_c$ are collinear, hence $J_a$, $J_b$, and $J_c$ are collinear.\nIt follows that the radical axes of any two of the three circles are parallel or coincide. Since $I$ lies on all three circles, we deduce that these circles have the same radical axis. But $I \\notin J_aJ_b$, so the three circles are not tangent, and therefore their radical axis contains a point different from $I$ lying on all three circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a sequence of integers $a_1, a_2, a_3, \\dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible that two terms of such a sequence be equal?",
    "options": [],
    "answer": "No",
    "solution": "The answer is in the negative. Notice first that, if $a_n > 1$, then $2^{a_n} - 1 \\equiv 3 \\pmod 4$; since $(2^{a_n} - 1)a_{n+1}$ is a perfect square, we should have $a_{n+1} \\equiv 0 \\pmod 4$ or $a_{n+1} \\equiv 3 \\pmod 4$, so in particular $a_{n+1} > 1$. As $a_1 > 1$, we conclude that all terms of the sequence are greater than 1.\n\nDenote the largest prime divisor of an integer $k > 1$ by $g(k)$. We will show that $g(a_{n+1}) > g(a_n)$ for all $n$, which yields the desired result. To this end, usage is made of the lemma below.\n\n**Lemma:** For any prime $p$, each prime divisor of $2^p - 1$ is greater than $p$.\n\n*Proof.* Let $q$ be a prime factor of $2^p - 1$; then $q$ is odd. The multiplicative order $d$ of $2$ modulo $q$ divides $p$ and is larger than $1$, so $d = p$. On the other hand, by Fermat's little theorem, $2^{q-1} \\equiv 1 \\pmod q$, so $p = d \\mid q - 1$ and the lemma follows.\n\nChoose now any positive integer $n$, and denote, for convenience, $k = a_n$ and $\\ell = a_{n+1}$. Let $p = g(k)$; then $2^p - 1 \\mid 2^k - 1$. Since $2^p - 1 \\equiv 3 \\pmod 4$, this number is not a square, so there exists a prime $q$ such that $v_q(2^p - 1)$ is odd. By the Lemma, $q > p$, so in particular $q \\nmid k$. Therefore, by the Lifting Exponent Lemma,\n$$\nv_q(2^k - 1) = v_q(2^p - 1) + v_q(k/p) = v_q(2^p - 1) + 0,\n$$\nso $v_q(2^k - 1)$ is odd as well. Since $(2^k - 1)\\ell$ is a perfect square, we should then have $q \\mid \\ell$, so $g(\\ell) \\ge q > p = g(k)$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $\\angle BAC > 90^\\circ$, and let $C$ be the circle centered at $A$ with radius $AB$. Let $M$ be the midpoint of side $AC$. The line $BM$ intersects the circle $C$ a second time at point $D$. Let $E$ be a point on the circle $C$ such that $BE \\perp AC$, and suppose that $DE \\cap AC = \\{N\\}$. Show that:\n$$\nAN = 2 \\cdot AB.\n$$\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the condition $AC \\perp BE$, it follows that $AC$ is the perpendicular bisector of segment $BE$, so $MB = ME$. Since $AB = AE$, we conclude that $\\triangle MAB \\equiv \\triangle MAE$ (congruent triangles). It follows that $\\angle MAB = \\angle MAE$ (1), and $\\angle MBA = \\angle MEA$ (2).\nFrom $AB = AD$, it follows that triangle $ABD$ is isosceles with base $BD$, so $\\angle ABD = \\angle ADB$ (3).\nFrom (2) and (3), we deduce that $\\angle MAE = \\angle MDA$, so quadrilateral $MAED$ is cyclic. It follows that $\\angle AED = \\angle DMN$, and since $\\angle AMB = \\angle DMN$, we obtain $\\angle AED = \\angle AMB$ (4).\nFrom (1) and (4), it follows that $\\triangle MAB \\sim \\triangle AEN$, so $\\frac{AN}{AB} = \\frac{AE}{AM} = 2$, from which we conclude $AN = 2 \\cdot AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the prime numbers $a, b, c, d$, with $a \\le b$ and $c \\le d$, satisfying\n(1) $a + b = c + d + 1$;\n(2) $a^2 + b^2 + c^2 + d^2 = 3543$.",
    "options": [],
    "answer": "a=2, b=43, c=3, d=41",
    "solution": "Since $c + d + 1 = a + b$, the number $a + b + c + d$ is odd. Therefore either three of the numbers $a, b, c, d$ are even (and, being primes are equal to $2$), or exactly one of them $a, b, c, d$ is even (therefore equal to $2$).\n\nIf three of the numbers $a, b, c, d$ are equal to $2$, condition (2) implies that the square of the fourth number is $3531$ – impossible. So exactly one of the numbers $a, b, c, d$ is equal to $2$. Now $a \\le b$ and $c \\le d$ yields that either $a = 2$, or $c = 2$.\n\nThe three numbers different from $2$ have the sum of their squares $3543 - 4 = 3539 = M_3 + 2$. Since a perfect square is $M_3$ or $M_3 + 1$, one of the squares must be $M_3$, so the corresponding number, being prime, must be $3$.\n\nThis leaves two primes $x, y$ whose sum of their squares equals $3539 - 9 = 3530$.\n\nIf $x \\le 41$ and $y \\le 41$, then $x^2 + y^2 \\le 2 \\cdot 41^2 = 3362 < 3530$. So at least one of $x, y$ is $\\ge 43$ – say $x \\ge 43$. If $x \\ge 61$, then $x^2 + y^2 > 61^2 > 3530$, So, $x$ can be only $43$, $47$, $53$ or $59$. Checking these values, $x = 47$ and $x = 53$ give no solution, $x = 43$ gives $y = 41$, and $x = 59$ gives $y = 7$.\n\nSo $a, b, c, d$ are $2, 3, 41, 43$ or $2, 3, 7, 59$. Finally, condition (1) is fulfilled only by $a = 2, b = 43, c = 3, d = 41$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number for which the number $a + a^2 + a^3$ is an integer. Show that, if one of the numbers $a$, $a^2$ or $a^3$ is rational, then $a$ is an integer.\nCristi Șăvescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a + a^2 + a^3 = n$, with integer $n$.\nIf $a$ is rational, then $a = \\frac{p}{q}$, where $p, q$ are coprime integers. It follows from here that $\\frac{p}{q} + \\frac{p^2}{q^2} + \\frac{p^3}{q^3} = n$, so $q^2p + qp^2 + p^3 = q^3n$, which implies $q \\mid p^3$. This is possible only if $q = \\pm 1$, which shows us that $a = \\pm p$ is an integer number.\n\nIf $a^2$ is rational, then $a + a^3 = n - a^2$, so $a^2 + 2a^4 + a^6 = n^2 - 2a^2n + a^4$, which leads to $a^2(1 + 2n) + a^4 + a^6 = n^2$. An argument analogous to the previous one, applied to the last relation, shows us that $a^2$ is an integer and therefore $a$ is an integer.\n\nIf $a^3 = r$ is rational, then $a + a^2 = n - a^3$ is also rational; we get from here that the number $a^2 + 2a^3 + a^4$ is rational, so $a^2 + ra$ is also a rational number. Hence $a^2 + ra - (a^2 + a) = (r - 1)a$ is rational.\nIf $r \\ne 1$, this would imply that $a$ is rational, so, due to the first part, it will follow from here that $a$ is an integer number. If $r = 1$, then $a^3 = 1$, which means $a = 1 \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$. Consider the equation:\n$$\n\\{x\\} + \\{2x\\} + \\dots + \\{nx\\} = [x] + [2x] + \\dots + [2nx].\n$$\n\na) Solve the equation in $\\mathbb{R}$ for $n = 2$.\nb) Prove that the equation has at most two real solutions for any $n \\ge 2$.",
    "options": [],
    "answer": "x = 0 for n = 2",
    "solution": "Since each fractional part $\\{kx\\} \\in [0, 1)$, the right-hand side is a sum of integers, hence nonnegative, so $x \\ge 0$.\n\na) For $n = 2$, the equation becomes $\\{x\\} + \\{2x\\} = [x] + [2x] + [3x] + [4x]$. The left side is in $[0, 2)$ and the right side is an integer number, so the right side is in $\\{0, 1\\}$.\nIf $\\{x\\} + \\{2x\\} = 0$, then $\\{x\\} = \\{2x\\} = 0$, so $x \\in \\mathbb{Z}$. Taking in the original equation $x \\in \\mathbb{Z}$ yields $10x = 0$, so $x = 0$, which satisfies the equation.\n\nIf $\\{x\\} + \\{2x\\} = 1$, since $[x] \\le [2x] \\le [3x] \\le [4x]$, we have $[x] = [2x] = [3x] = 0$, $[4x] = 1$, which implies $3x < 1 \\le 4x$, so $x \\in [\\frac{1}{4}, \\frac{1}{3})$. But from $\\{x\\} + \\{2x\\} = 1$ and $[x] = [2x] = 0$, we have $x + 2x = 1$, which implies that $x = \\frac{1}{3} \\notin [\\frac{1}{4}, \\frac{1}{3})$, a contradiction.\nHence, for $n = 2$, the unique solution is $x = 0$.\n\nb) Using similar reasoning for general $n$, since\n$$\n\\{x\\} + \\dots + \\{nx\\} \\in [0, n) \\cap \\mathbb{Z},\n$$\nand the integer parts satisfy $[x] \\le [2x] \\le \\dots \\le [2nx]$, if $[nx] \\ge 1$, then the right-hand side sum is at least $n + 1$, which contradicts that the left side is less than $n$. So $[x] = [2x] = \\dots = [nx] = 0$, and the equation reduces to:\n\n$$\nx + 2x + \\dots + nx = [(n + 1)x] + \\dots + [2nx] = k,\n$$\nwhere $k \\in \\{0, 1, \\dots, n - 1\\}$. This yields $x = \\frac{2k}{n(n+1)}$.\n\n$$\nn - \\sqrt{\\frac{n(n-1)}{2}} - \\left(n + \\frac{1}{2} - \\sqrt{\\frac{2n^2 + 2n + 1}{4}}\\right) < 1,\n$$\nthere is at most one integer $k$ satisfying this, so there is at most one additional solution besides $x = 0$. Therefore, the given equation has at most two real solutions for any $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39467,
    "subject": "Mathematics (Multi-modal)",
    "question": "For given positive integer $k$, consider the function $g_k : \\mathbb{Z} \\to \\mathbb{Z}$, $g_k(x) = x^k$. Determine the set $M_k$ of positive integers $n$ with the property that there exist injective functions $f_1, f_2, \\dots, f_n : \\mathbb{Z} \\to \\mathbb{Z}$ such that $g_k = f_1 \\cdot f_2 \\cdot \\dots \\cdot f_n$.",
    "options": [],
    "answer": "M_k = {k, k-2, k-4, ..., k - 2\\lfloor (k-1)/2 \\rfloor}",
    "solution": "For $x = 2$, we get $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdot \\dots \\cdot |f_n(2)|$. Using injectivity, we deduce that $|f_i(2)| \\ge 2$ for any $i = \\overline{1, n}$. Thus, $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdot \\dots \\cdot |f_n(2)| \\ge 2^n$, so $k \\ge n$. Since $n$ and $k$ have the same parity, it follows that $M_k \\subset \\{k, k-2, k-4, \\dots\\} \\subset \\mathbb{N}^*$.\nWe will prove that all values of the form $n = k - 2t$ are good. For this, we observe that $x^k = \\underbrace{x \\cdot x \\cdots x}_{k-2t-1 \\text{ times}} \\cdot x^{2t+1}$, so functions $f_i(x) = x$, $i = \\overline{1, n-1}$, $f_n(x) = x^{2t+1}$ are injective and $x^n = f_1(x) \\cdot f_2(x) \\cdots f_{n-1}(x) \\cdot f_n(x)$.\nIn conclusion, $k - 2t \\in M_k$ for any $t \\in \\mathbb{N}$ with $2t < k$ and $M_k = \\{k, k-2, k-4, \\dots, k-2 \\cdot \\lfloor \\frac{k-1}{2} \\rfloor\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the squares $ABCD$ and $BEFG$, such that $B$ lies on the segment $AE$ and $G$ lies on the segment $BC$. Let $H$ be the intersection of the lines $DF$ and $EG$. The perpendicular from $H$ to the line $DF$ intersects the lines $AE$ and $BC$ at points $I$ and $J$, respectively. Prove that the quadrilateral $DIFJ$ is a square.\nValeriu Bărbieru\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$\\angle GBF = \\angle DBC = 45^\\circ$, thus the triangle $BDF$ is right-angled at $B$. Since $EG$ is the perpendicular bisector of $BF$ in triangle $BDF$, it follows that $H$ is the midpoint of the hypotenuse $DF$.\n\nLet $K$ be the projection of $H$ onto $AE$. Since $H$ is the midpoint of $DF$ and $HK \\parallel AD$, it follows that $HK$ is the midsegment in the right trapezoid $AEFD$, therefore $K$ is the midpoint of the side $AE$. Consequently, the triangle $HAE$ is isosceles, so $\\angle HAE = \\angle HEA = 45^\\circ$, thus $H \\in AC$.\n\nLet $L$ be the projection of $H$ onto $AD$. We have $\\triangle HAK \\equiv \\triangle HAL$ (HA), therefore $HK = HL$. Since $\\angle KHI + \\angle IHL = 90^\\circ = \\angle DHL + \\angle IHL$, we obtain $\\angle KHI = \\angle LHD$, hence $\\triangle HKI \\equiv \\triangle HLD$ (LA), therefore $HI = HD$.\n\nSince $\\triangle HAB \\equiv \\triangle HAD$ (SAS), we have $HB = HD = HI$. Consequently, $K$ is the midpoint of $BI$, $KH$ is a midsegment in the triangle $IBJ$ and $H$ is the midpoint of $IJ$. Thus, the diagonals of the quadrilateral $DIFJ$ are equal, bisect each other and are orthogonal, therefore $DIFJ$ is a square.\nAs in the first solution, $H$ is the midpoint of $DF$. Since $IJ$ is the perpendicular bisector of the segment $DF$, we have $DI = IF$. Moreover, $\\angle IHF = \\angle IEF = 90^\\circ$, thus $IEFH$ is a cyclic quadrilateral, therefore $\\angle IFH = \\angle IEH = 45^\\circ$. Since the triangle $IFD$ is isosceles, we have $\\angle IFD = \\angle IDF = 45^\\circ$ and $\\angle DIF = 90^\\circ$.\n\nLet $K$ be the projection of $H$ onto $AE$. From $\\angle AID + \\angle EIF = 90^\\circ$ we deduce that $\\angle AID = \\angle EFI$, hence $\\triangle AID \\equiv \\triangle EFI$ (HA), therefore $AI = EF = BE$. Thus, $K$ is the midpoint of $IB$. Since $HK \\parallel BJ$, it follows that $HK$ is a midsegment of the triangle $BIJ$, so $H$ is also the midpoint of $IJ$. Since its diagonals $DF$ and $IJ$ bisect each other, it follows that $DIFJ$ is a square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for a function $f : \\mathbb{R} \\to \\mathbb{R}$, the following affirmations are equivalent:\n(i) $f$ is differentiable on $\\mathbb{R}$ with continuous derivative;\n(ii) for any $a \\in \\mathbb{R}$ and two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, with limit $a$, such that $x_n \\neq y_n$, for all $n \\in \\mathbb{N}^*$, the sequence $\\left( \\frac{f(x_n) - f(y_n)}{x_n - y_n} \\right)_{n \\ge 1}$ is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) $\\Rightarrow$ (ii)\n\nSuppose $f: \\mathbb{R} \\to \\mathbb{R}$ is continuously differentiable on $\\mathbb{R}$.\nLet $a \\in \\mathbb{R}$, and two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, convergent to $a$, such that $x_n \\ne y_n$, for $n \\in \\mathbb{N}^*$. By Lagrange theorem, there are points $a_n \\in (\\min\\{x_n, y_n\\}, \\max\\{x_n, y_n\\})$ such that\n$$\n\\frac{f(x_n)-f(y_n)}{x_n-y_n} = f'(a_n).\n$$\nWe deduce that the sequence $(a_n)_{n \\ge 1}$ converges also to $a$. By the continuity of $f'$, we obtain\n$$\n\\lim_{n \\to \\infty} \\frac{f(x_n)-f(y_n)}{x_n-y_n} = \\lim_{n \\to \\infty} f'(a_n) = f'(a),\n$$\nimplying that the sequence $\\left(\\frac{f(x_n)-f(y_n)}{x_n-y_n}\\right)_{n \\ge 1}$ is convergent.\n\n(ii) $\\Rightarrow$ (i)\n\nLet $f: \\mathbb{R} \\to \\mathbb{R}$ be a function having property (ii). For $a \\in \\mathbb{R}$, there exists $\\ell_a = \\lim_{n \\to \\infty} \\frac{f(a+1/n)-f(a)}{1/n} \\in \\mathbb{R}$.\n\nConsider two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$ with limit $a$, such that $x_n \\ne y_n$, for any $n \\in \\mathbb{N}^*$. Define sequences $(z_n)_{n \\ge 1}$ and $(t_n)_{n \\ge 1}$ by $z_{2n-1} = x_n$, $z_{2n} = a+1/n$, $t_{2n-1} = y_n$ and $t_{2n} = a$, for $n \\in \\mathbb{N}^*$. We have $\\lim_{n \\to \\infty} z_n = \\lim_{n \\to \\infty} t_n = a$ and $z_n \\ne t_n$, for any $n \\in \\mathbb{N}^*$. By hypothesis $\\left(\\frac{f(z_n)-f(t_n)}{z_n-t_n}\\right)_{n \\ge 1}$ is convergent. We get\n$$\n\\lim_{n \\to \\infty} \\frac{f(x_n)-f(y_n)}{x_n-y_n} = \\lim_{n \\to \\infty} \\frac{f(a+1/n)-f(a)}{1/n} = \\ell_a.\n$$\nIn particular, for any sequence $(x_n)_{n \\ge 1}$ satisfying $\\lim_{n \\to \\infty} x_n = a$ and $x_n \\ne a$, $\\forall n \\in \\mathbb{N}^*$, we have\n$$\n\\lim_{n \\to \\infty} \\frac{f(x_n)-f(a)}{x_n-a} = \\ell_a.\n$$\nSo, $\\lim_{x \\to a} \\frac{f(x)-f(a)}{x-a} = \\ell_a$. As a conclusion $f$ is differentiable on $\\mathbb{R}$, and $f'(a) = \\ell_a$, $\\forall a \\in \\mathbb{R}$.\n\nSuppose, by contraposition, that $f'$ is discontinuous at a point $a \\in \\mathbb{R}$. Then, there exists $\\varepsilon > 0$ and a sequence $(a_n)_{n \\ge 1}$ with limit $a$, and $a_n \\ne a$, for all $n \\in \\mathbb{N}^*$, such that $|f'(a) - f'(a_n)| > \\varepsilon$, $\\forall n \\in \\mathbb{N}^*$. As $f$ is differentiable at $a_n$, we can choose $x_n \\in (a_n, a_n + 1/n)$ such that\n$$\n|f'(a_n) - \\frac{f(x_n)-f(a_n)}{x_n-a_n}| < \\frac{\\varepsilon}{2}.\n$$\nThe sequence $(x_n)_{n \\ge 1}$ converges to $a$, $x_n \\ne a_n$, for $n \\in \\mathbb{N}^*$, but\n$$\n|f'(a) - \\frac{f(x_n)-f(a_n)}{x_n-a_n}| \\ge |f'(a) - f'(a_n)| - |f'(a_n) - \\frac{f(x_n)-f(a_n)}{x_n-a_n}| > \\frac{\\varepsilon}{2},\n$$\nfor all $n \\in \\mathbb{N}^*$, in contradiction with $\\lim_{n \\to \\infty} \\frac{f(x_n)-f(a_n)}{x_n-a_n} = f'(a)$. In conclusion, $f'$ is continuous on $\\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39470,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $a, b \\in \\mathbb{R}$ be two real numbers, with $a < b$, and $f : [a, b] \\to \\mathbb{R}$ a strict monotone function such that $\\int_a^b f(x) dx = 0$. Show that $f(a) \\cdot f(b) < 0$.\n\nb) Determine the convergent sequences $(a_n)_{n \\ge 1}$ of real numbers, for which there is a strict monotone function $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n\\int_{a_{n-1}}^{a_n} f(x) \\, dx = \\int_{a_n}^{a_{n+1}} f(x) \\, dx, \\quad \\text{for any } n \\in \\mathbb{N}, n \\ge 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $f([a, b]) \\subseteq [0, \\infty)$ or $f([a, b]) \\subseteq (-\\infty, 0]$, then $m = |f(\\frac{a+b}{2})| > 0$, and on one of the intervals $(a, \\frac{a+b}{2})$ or $(\\frac{a+b}{2}, b)$ the inequality $|f(x)| > m$ holds for any $x$ in that interval. Then\n$$\n\\begin{aligned}\n0 &= \\left| \\int_a^b f(x) \\, dx \\right| = \\int_a^b |f(x)| \\, dx = \\int_a^{\\frac{a+b}{2}} |f(x)| \\, dx + \\int_{\\frac{a+b}{2}}^b |f(x)| \\, dx \\\\ &\\ge m \\cdot \\frac{b-a}{2} > 0,\n\\end{aligned}\n$$\nwhich is impossible. It follows that $f(a) \\cdot f(b) < 0$.\n\nb) We will show that the only sequences of real numbers $(a_n)_{n \\ge 1}$ which satisfy the condition in the statement are the constant sequences and the sequences assuming exactly two distinct real values, which become stationary beginning with a certain rank.\nObviously, if $(a_n)_{n \\ge 1}$ is a constant sequence, with $a_n = a$, $a \\in \\mathbb{R}$, then the sequence is convergent, with $\\lim_{n \\to \\infty} a_n = a$ and $\\int_{a_k}^{a_{k+1}} f(x) dx = 0$, for any $k \\ge 1$, and any function $f : \\mathbb{R} \\to \\mathbb{R}$. If $\\{a_n | n \\ge 1\\} = \\{a, b\\}$ and there is a rank $n_0 \\in \\mathbb{N}^*$ with $a_n = a$, for any $n \\ge n_0$, then $(a_n)_{n \\ge 1}$ is convergent, with $\\lim_{n \\to \\infty} a_n = a$, and there is the function $f : \\mathbb{R} \\to \\mathbb{R}$ defined by $f(x) = 2x - a - b$, which is strict monotone and satisfies the equalities $\\int_{a_k}^{a_{k+1}} f(x) dx = 0$, for any $k \\ge 1$.\n\nLet $(a_n)_{n \\ge 1}$ be a convergent sequence of real numbers for which there is a strict monotone function $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n\\int_{a_1}^{a_2} f(x) \\, dx = \\int_{a_2}^{a_3} f(x) \\, dx = \\dots = \\int_{a_n}^{a_{n+1}} f(x) \\, dx = \\dots\n$$\n\nLet $I = \\int_{a_k}^{a_{k+1}} f(x) dx$, for every $k \\ge 1$, and $a = \\lim_{n \\to \\infty} a_n$. For a fixed $r > 0$ there is a rank $n_r \\ge 1$, such that $a_n \\in (a - r, a + r)$, for any $n \\ge n_r$. Since $f$ is strict monotone, if $M = \\max(|f(a - r)|, |f(a + r)|)$, then $|f(x)| < M$, for every $x \\in (a - r, a + r)$. Then, for any $p \\in \\mathbb{N}^*$ we have:\n$$\n\\begin{aligned}\np \\cdot |I| &= |p \\cdot I| = \\left| \\sum_{k=1}^{p} \\int_{a_{n_r+k-1}}^{a_{n_r+k}} f(x) \\, dx \\right| = \\left| \\int_{a_{n_r}}^{a_{n_r+p}} f(x) \\, dx \\right| \\\\ &\\le \\left| \\int_{a_{n_r}}^{a_{n_r+p}} |f(x)| \\, dx \\right| < \\int_{a-r}^{a+r} |f(x)| \\, dx \\le 2r \\cdot M.\n\\end{aligned}\n$$\nIt follows that $0 \\le |I| < \\frac{2Mr}{p}$, for any $p \\in \\mathbb{N}^*$, so that $I = 0$.\nWe show now that $\\text{card}(\\{a_n | n \\in \\mathbb{N}^*\\}) \\le 2$. Assuming the opposite, there exist $i, j, k \\in \\mathbb{N}^*$, with $i < j < k$ such that $a_i \\ne a_j \\ne a_k \\ne a_i$. Then\n$$\n\\int_{a_i}^{a_j} f(x) \\, dx = \\sum_{l=i}^{j-1} \\int_{a_l}^{a_{l+1}} f(x) \\, dx = (j-i) \\cdot I = 0\n$$\nand, also, $\\int_{a_j}^{a_k} f(x) \\, dx = (k-j) \\cdot I = 0$, respectively $\\int_{a_i}^{a_k} f(x) \\, dx = 0$.\nThe function $f$ being strict monotone, it follows that $f(a_i) \\cdot f(a_j) < 0$, $f(a_i) \\cdot f(a_k) < 0$, and $f(a_j) \\cdot f(a_k) < 0$. But then\n$$\n(f(a_i) \\cdot f(a_j) \\cdot f(a_k))^2 = (f(a_i) \\cdot f(a_j)) \\cdot (f(a_i) \\cdot f(a_k)) \\cdot (f(a_j) \\cdot f(a_k)) < 0,\n$$\nwhich is absurd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-negative real numbers $x$, $y$, $z$ satisfy the relations $xy + 4 \\le 2(x+z)$, $yz + 4 \\le 2(y+x)$, $zx + 4 \\le 2(z+y)$. Prove that $x = y = z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The hypothesis can be written in the form $x(y-2) \\le 2(z-2)$, $y(z-2) \\le 2(x-2)$, $z(x-2) \\le 2(y-2)$. If $x-2 < 0$, using the second relation we get $z-2 < 0$, then, according to the first, we obtain $y-2 < 0$, so $x$, $y$, $z \\in (0, 2)$ and $xyz < 8$. We also have $x(2-y) \\ge 2(2-z) > 0$, $y(2-z) \\ge 2(2-x) > 0$, $z(2-x) \\ge 2(2-y) > 0$. Now, multiplying the two inequalities and, after that, dividing by $(2-x)(2-y)(2-z) > 0$, we obtain $xyz \\ge 8$, which is in contradiction with $xyz < 8$. Analogously, we obtain that, if $x > 2$, then $y > 2$ and $z > 2$, so $xyz > 8$. Multiplying these relations we get $xyz \\le 8$ – contradiction. Therefore $x = 2$. Substituting in the initial conditions we obtain $y \\le z \\le 2 \\le y$, so $x = y = z = 2$.\nWe have $ab + 4 - 2(a+b) = (a-2)(b-2)$. (1)\nSuppose that one of the three numbers is less than $2$, for example $x < 2$. Then $yz + 4 \\le 2(y+x) < 2y+4$, so $z < 2$. It follows from this that $xy + 4 \\le 2(x+z) < 2x + 4$, so $y < 2$. From (1) we obtain $xy + 4 > 2(x+y)$ and the analogues, so $\\sum xy + 12 > 4 \\sum x$, which contradicts the hypothesis. Thus $x$, $y$, $z \\ge 2$. From (1) we deduce that $xy + 4 \\ge 2(x+y)$ and the analogues. By adding them we obtain $\\sum xy + 12 \\ge 4 \\sum x$. Taking into account this relation and the hypothesis we deduce that all the previous inequalities must be equalities, so $x = y = z = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, \\infty) \\to [0, \\infty)$ be a continuous bijective function, such that\n$$\n\\lim_{x \\to \\infty} \\frac{f^{-1}(f(x)/x)}{x} = 1.\n$$\n\na) Show that $\\lim_{x \\to \\infty} \\frac{f(x)}{x} = \\infty$ and $\\lim_{x \\to \\infty} \\frac{f^{-1}(ax)}{f^{-1}(x)} = 1$, for any $a > 0$.\n\nb) Give an example of a function $f$ that satisfies the conditions from the statement.",
    "options": [],
    "answer": "f(x) = e^x - 1",
    "solution": "a) Since $f$ is continuous and bijective, we deduce that $f(0) = 0$ and $f$ is increasing, with $\\lim_{x \\to \\infty} f(x) = \\infty$. Then $f^{-1} : [0, \\infty) \\to [0, \\infty)$ is also increasing, with $f^{-1}(0) = 0$ and $\\lim_{x \\to \\infty} f^{-1}(x) = \\infty$. The limit from the hypothesis ensures the existence of a number $u > 0$ such that $\\frac{f^{-1}(f(x)/x)}{x} > \\frac{1}{2}$, for any $x > u$. Therefore, $\\frac{f(x)}{x} > f(\\frac{x}{2})$, for any $x > u$. Since $\\lim_{x \\to \\infty} f(\\frac{x}{2}) = \\infty$, we obtain $\\lim_{x \\to \\infty} \\frac{f(x)}{x} = \\infty$. Let $a > 0$ be an arbitrary fixed number.\n\nThe case $a \\in (0, 1)$. There is $t > 0$ such that $f^{-1}(x) > \\frac{1}{a}$, for any $x > t$. Hence\n$$\nf^{-1}(x) > f^{-1}(ax) = f^{-1}(af(f^{-1}(x))) > f^{-1}\\left(\\frac{f(f^{-1}(x))}{f^{-1}(x)}\\right), \\text{ for any } x > t.\n$$\nWe obtain\n$$\n\\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} < \\frac{f^{-1}(ax)}{f^{-1}(x)} < 1, \\text{ for any } x > t.\n$$\nFrom the assumption and the condition $\\lim_{x \\to \\infty} f^{-1}(x) = \\infty$, we get\n$$\n\\lim_{x \\to \\infty} \\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} \\underset{y=f^{-1}(x)}{=} \\lim_{y \\to \\infty} \\frac{f^{-1}(f(y)/y)}{y} = 1.\n$$\nThe case $a = 1$ is clear.\n\nThe sandwich theorem ensures $\\lim_{x \\to \\infty} \\frac{f^{-1}(ax)}{f^{-1}(x)} = 1$.\n\nThe case $a \\in (1, \\infty)$. Then $b = 1/a \\in (0, 1)$. From the previous case, we have\n$$\n\\lim_{x \\to \\infty} \\frac{f^{-1}(ax)}{f^{-1}(x)} = \\lim_{x \\to \\infty} \\left( \\frac{f^{-1}(x)}{f^{-1}(ax)} \\right)^{-1} = \\lim_{x \\to \\infty} \\left( \\frac{f^{-1}(b(ax))}{f^{-1}(ax)} \\right)^{-1} = 1^{-1} = 1.\n$$\nIn conclusion, $\\lim_{x \\to \\infty} \\frac{f^{-1}(ax)}{f^{-1}(x)} = 1$, for any $a > 0$.\n\nb) Example. The bijective function $f : [0, \\infty) \\to [0, \\infty)$, $f(x) = e^x - 1$, with the inverse $f^{-1} : [0, \\infty) \\to [0, \\infty)$, $f^{-1}(x) = \\ln(x + 1)$, satisfies the conditions from the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39473,
    "subject": "Mathematics (Multi-modal)",
    "question": "The isosceles triangle $ABC$ has $\\angle BAC = 30^\\circ$ and $AB = AC$. Take the point $D$ on the side $AC$ and the distinct points $E$, $F$, $G$ on the side $AB$ so that $BC = BD = DE = EF$ and $DG = DF$.\n\na) Prove that $BF = GE$.\n\nb) Find the measure of $\\angle BCG$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "52°30'",
    "solution": "a) From $\\triangle DFG$ isosceles follows $\\angle DFG = \\angle DGF$, hence $\\angle DFB = \\angle DGE$. From $\\triangle BDE$ isosceles follows $\\angle DBF = \\angle DEG$.\n\nThe above and $BD = DE$ yield $\\triangle BDF \\equiv \\triangle EDG$ (S.A.A.), whence $BF = GE$.\n\nb) Since $\\triangle ABC$ is isosceles, $\\angle ABC = \\angle ACB = \\frac{180^\\circ - \\angle BAC}{2} = \\frac{150^\\circ}{2} = 75^\\circ$. Then $BF = GE$ gives $BG = FE = BC$, hence $\\triangle BCG$ is isosceles.\n\nSo $\\angle BCG = \\angle BGC = \\frac{180^\\circ - \\angle CBG}{2} = \\frac{105^\\circ}{2} = 52^\\circ 30'.$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39474,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $a \\in \\mathbb{R}$ and $f : \\mathbb{R} \\to \\mathbb{R}$ be a continuous function, having antiderivative $F : \\mathbb{R} \\to \\mathbb{R}$, such that $F(x) + a \\cdot f(x) \\ge 0$, for $x \\in \\mathbb{R}$, and $\\lim_{|x|\\to\\infty} \\frac{F(x)}{e^{|\\alpha \\cdot x|}} = 0$, for any $\\alpha \\in \\mathbb{R}^*$. Prove that $F(x) \\ge 0$, for $x \\in \\mathbb{R}$.\n\nb) Let $n \\in \\mathbb{N} \\setminus \\{0,1\\}$, $g = X^n + a_1 X^{n-1} + \\dots + a_{n-1} X + a_n \\in \\mathbb{R}[X]$ a polynomial with all its roots real and $f : \\mathbb{R} \\to \\mathbb{R}$, a polynomial function such that $f(x) + a_1 \\cdot f'(x) + a_2 \\cdot f^{(2)}(x) + \\dots + a_n \\cdot f^{(n)}(x) \\ge 0$, for all $x \\in \\mathbb{R}$. Show that $f(x) \\ge 0$, for $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) For $a = 0$ the result is obvious. For $a \\ne 0$, consider the differentiable function $g : \\mathbb{R} \\to \\mathbb{R}$ defined by $g(x) = F(x) \\cdot e^{\\frac{x}{a}}$. We have\n$$\ng'(x) = f(x) \\cdot e^{\\frac{x}{a}} + \\frac{1}{a} \\cdot F(x) \\cdot e^{\\frac{x}{a}} = \\frac{1}{a} \\cdot e^{\\frac{x}{a}} \\cdot \\left( F(x) + a \\cdot f(x) \\right).\n$$\nFor $a > 0$ it is obvious that $g'(x) \\ge 0$, for $x \\in \\mathbb{R}$, implying that $g$ is non-decreasing. As $\\lim_{x \\to -\\infty} g(x) = \\lim_{x \\to -\\infty} F(x) \\cdot e^{\\frac{x}{a}} = 0$, we get $g(x) \\ge 0$, for any $x \\in \\mathbb{R}$, thus $F(x) = g(x) \\cdot e^{-\\frac{x}{a}} \\ge 0$, for $x \\in \\mathbb{R}$.\nIf $a < 0$, $g'(x) \\le 0$, for $x \\in \\mathbb{R}$, implying that $g$ is non-increasing. As $\\lim_{x \\to \\infty} g(x) = \\lim_{x \\to -\\infty} F(x) \\cdot e^{\\frac{x}{a}} = 0$, we obtain $g(x) \\ge 0$, for $x \\in \\mathbb{R}$. It follows that $F(x) = g(x) \\cdot e^{-\\frac{x}{a}} \\ge 0$, for any $x \\in \\mathbb{R}$.\n\n\nb) Let $\\mathcal{P} = \\{f \\mid f : \\mathbb{R} \\to \\mathbb{R}, f \\text{ is a polynomial function}\\}$. For any real $a$ consider the function $T_a : \\mathcal{P} \\to \\mathcal{P}$, defined by $T_a(f) = f + a \\cdot f'$, i.e., $T_a(f)(x) = f(x) + a \\cdot f'(x)$, for any $x \\in \\mathbb{R}$. As for any $f \\in \\mathcal{P}$ and any $\\alpha \\in \\mathbb{R}$, we have $\\lim_{|x|\\to\\infty} \\frac{f(x)}{e^{|\\alpha \\cdot x|}} = 0$, by a), for real $a$, we have the implication\n$$\nf \\in \\mathcal{P} : T_a(f)(x) \\ge 0, \\forall x \\in \\mathbb{R} \\implies f(x) \\ge 0, \\forall x \\in \\mathbb{R}.\n$$\nConsider the roots $r_1, r_2, \\dots, r_n$ of $g$, and $s_k = -r_k$ for $k \\in \\{1, 2, \\dots, n\\}$ their opposites. Then $g = (X - r_1)(X - r_2) \\dots (X - r_n) = (X + s_1)(X + s_2) \\dots (X + s_n)$.\nBy Vieta, we obtain the coefficients of $g$:\n$$\na_k = \\sum_{1 \\le i_1 < i_2 < \\dots < i_k \\le n} s_{i_1} s_{i_2} \\dots s_{i_k}, \\quad \\text{for any } k \\in \\{1, 2, \\dots, n\\}.\n$$\nFor $m \\in \\{1, 2, \\dots, n\\}$ and any $f \\in \\mathcal{P}$, we get\n$$\n(T_{s_m} \\circ \\dots \\circ T_{s_2} \\circ T_{s_1})(f) = f + \\left(\\sum_{i=1}^m s_i\\right) f' + \\left(\\sum_{1 \\le i_1 < i_2 \\le m} s_{i_1} s_{i_2}\\right) f'' + \\dots \\\\\n\\dots + \\left(\\sum_{1 \\le i_1 < i_2 < \\dots < i_k \\le m} s_{i_1} s_{i_2} \\dots s_{i_k}\\right) f^{(k)} + \\dots + (s_1 s_2 \\dots s_m) f^{(m)}. \\quad (Q(m))\n$$\nIf for $m \\in \\{1, 2, \\dots, n-1\\}$, $Q(m)$ is supposed to be true, we have\n$$\n\\begin{align*}\n(T_{s_{m+1}} \\circ T_m \\circ \\dots T_{s_2} \\circ T_{s_1})(f) &= T_{s_{m+1}}(T_m \\circ \\dots \\circ T_{s_1})(f) = \\\\\n&= T_{s_{m+1}} \\left( \\sum_{J \\subseteq \\{1, 2, \\dots, m\\}} \\left( \\prod_{j \\in J} s_j \\right) f^{|J|} \\right) = \\\\\n&= \\left( \\sum_{J \\subseteq \\{1, 2, \\dots, m\\}} \\left( \\prod_{j \\in J} s_j \\right) f^{|J|} \\right) + \\\\\n&\\quad + s_{m+1} \\cdot \\left( \\sum_{J \\subseteq \\{1, 2, \\dots, m\\}} \\left( \\prod_{j \\in J} s_j \\right) f^{|J|} \\right)' \\\\\n&= \\sum_{J_1 \\subseteq \\{1, 2, \\dots, m, m+1\\}} \\left( \\prod_{j \\in J_1} s_j \\right) f^{|J_1|}\n\\end{align*}\n$$\nso $Q(m+1)$ is also true.\n\nFrom $Q(n)$, we have\n$$\n(T_{s_n} \\circ \\dots \\circ T_{s_2} \\circ T_{s_1}) (f) = f + a_1 \\cdot f' + a_2 \\cdot f'' + \\dots + a_n \\cdot f^{(n)}.\n$$\nBy the hypothesis, we have\n$$(T_{s_n} \\circ \\dots \\circ T_{s_2} \\circ T_{s_1}) (f)(x) = f(x) + a_1 \\cdot f'(x) + a_2 \\cdot f''(x) + \\dots + a_n \\cdot f^{(n)}(x) \\geq 0, $$\nfor all $x \\in \\mathbb{R}$.\nSuccessively, applying a), for $m \\in \\{1, 2, \\dots, n\\}$, we get\n$$\n(T_{s_m} \\circ \\dots \\circ T_{s_2} \\circ T_{s_1}) (f)(x) \\geq 0, \\quad \\text{for any } x \\in \\mathbb{R}.\n$$\nIn particular $f(x) \\geq 0$, for any $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{C} \\to \\mathbb{C}$ such that $|wf(z) + zf(w)| = 2|zw|$, for any $z, w \\in \\mathbb{C}$.",
    "options": [],
    "answer": "All solutions are f(z) = c z for all z, where c is a complex constant with |c| = 1.",
    "solution": "For $z = 1$ and $w = 0$ we obtain $f(0) = 0$. For $w = z \\in \\mathbb{C}^*$ we obtain $|f(z)| = |z|$; this equality is also verified for $z = 0$, so $|f(z)| = |z|$, for any $z \\in \\mathbb{C}$. We have $|f(1)| = 1$ and, for $w = 1$, we obtain $2|z| = |f(z) + zf(1)| \\le |f(z)| + |zf(1)| = 2|z|$ for any $z \\in \\mathbb{C}$. It follows that we have equality in the triangle inequality. Therefore, for every $z \\in \\mathbb{C}$, there exists $t_z \\in \\mathbb{R}$, $t_z \\ge 0$, such that $f(z) = t_z \\cdot f(1) \\cdot z$. We apply the modulus to both members and, after simplification, we have $|f(z)| = t_z \\cdot 1 \\cdot |z| \\Leftrightarrow 1 = t_z$, for any $z \\in \\mathbb{C}^*$.\n\nWe can conclude that $f(z) = cz$ for any $z \\in \\mathbb{C}$, where $c = f(1)$ is a complex number with modulus equal to 1. It is immediately verified that any function of the form $f(z) = cz$, where $c \\in \\mathbb{C}$, $|c| = 1$, is a solution of the functional equation in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the positive real numbers $a, b, c, d$ such that $a + b + c + d = 80$ and\n$$\na + \\frac{b}{1+a} + \\frac{c}{1+a+b} + \\frac{d}{1+a+b+c} = 8.\n$$",
    "options": [],
    "answer": "a = 2, b = 6, c = 18, d = 54",
    "solution": "Adding $4$ to both sides of the second equation, we write:\n$$\n1 + a + \\frac{1+a+b}{1+a} + \\frac{1+a+b+c}{1+a+b} + \\frac{1+a+b+c+d}{1+a+b+c} = 12.\n$$\nApplying the AM-GM inequality successively, we obtain:\n$$\n\\begin{aligned}\n1 + a + \\frac{1+a+b}{1+a} &\\ge 2\\sqrt{(1+a) \\cdot \\frac{1+a+b}{1+a}} = 2\\sqrt{1+a+b}; \\\\\n\\frac{1+a+b+c}{1+a+b} + \\frac{1+a+b+c+d}{1+a+b+c} &\\ge 2\\sqrt{\\frac{1+a+b+c}{1+a+b} \\cdot \\frac{1+a+b+c+d}{1+a+b+c}} = \\\\\n&\\ge 2\\sqrt{\\frac{81}{1+a+b}}.\n\\end{aligned}\n$$\nBy adding these two inequalities and applying the AM-GM inequality again, we have:\n$$\n\\begin{aligned}\n12 &= 1 + a + \\frac{1+a+b}{1+a} + \\frac{1+a+b+c}{1+a+b} + \\frac{1+a+b+c+d}{1+a+b+c} \\\\\n&\\ge 2\\sqrt{1+a+b} + \\frac{18}{\\sqrt{1+a+b}} \\ge 12.\n\\end{aligned}\n$$\n\nFrom this, we obtain $a + b = 8$ and\n$$\n1 + a = \\frac{1 + a + b}{1 + a} = \\frac{1 + a + b + c}{1 + a + b} = \\frac{1 + a + b + c + d}{1 + a + b + c} = 3.\n$$\nTherefore, the desired numbers are $a = 2$, $b = 6$, $c = 18$, $d = 54$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right isosceles triangle with right angle at $A$, $O$ the midpoint of the hypotenuse $BC$, $E$ the midpoint of the segment $CO$, $M$ the midpoint of $AC$, and $D$ the midpoint of $AM$. Let $F$ be the intersection of lines $OD$ and $AE$.\n\na) Prove that $MF \\perp OD$.\n\n![](attached_image_1.png)\n\nb) Prove that $FC = OC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution for a).* Since $ME \\perp OC$, it suffices to show that the quadrilateral $MEOF$ is cyclic.\n\n(1) Since $ME \\parallel AO$, we have $\\angle MEF = \\angle OAE$, therefore it remains to prove that $\\angle MOD = \\angle OAE$. This follows from similarity: $\\triangle MOD \\sim \\triangle OAE$, because $\\angle DMO = \\angle EOA = 90^\\circ$ and $\\frac{DM}{MO} = \\frac{EO}{OA} = \\frac{1}{2}$.\n\n(2)\n\n*Second solution for a).* By Menelaus' theorem in $\\triangle CDO$, with the transversal $AFE$, we get: $\\frac{DF}{FO} = \\frac{AD}{AC} = \\frac{1}{4}$. Let $AB = 4a$. In triangle $MDO$, we have $MD = a$, $MO = 2a$, and $DO = a\\sqrt{5}$, so $DF = \\frac{a\\sqrt{5}}{5}$. Therefore, $DO \\cdot DF = MD^2$, hence, by the converse of the hypotenuse leg theorem, we obtain $MF \\perp DO$.\n\n*Third solution for a).* Since $\\frac{CA}{CE} = 2\\sqrt{2} = \\frac{OA}{AD}$, and $\\angle ACE = \\angle OAD = 45^\\circ$, the triangles $CAE$ and $AOD$ are similar. Then $\\angle AOD = \\angle CAE = \\angle DAF$, so $\\triangle DAF \\sim \\triangle DOA$ (AA), hence $MD^2 = DA^2 = DF \\cdot DO$, and thus $MF \\perp DO$.\n\n\n*First solution for b).* Let $G, R$ be the midpoints of segments $AB$ and $MO$, respectively. Since $CMGO$ is a parallelogram, $R$ is the midpoint of $CG$. $EMFO$ is cyclic, therefore $\\angle EFO = \\angle EMO = 45^\\circ = \\angle OBA$, so $ABOF$ is cyclic, with $G$ the center of its circumscribed circle. Thus $GF = GO$, $RF = RO$, so $RG$ is the perpendicular bisector of segment $OF$. Since $C \\in RG$, we get $CF = CO$.\n\n*Second solution for b).* Let $G$ be the midpoint of $AB$. Analogous to (2), the triangles $MOD$ and $ACG$ are similar, so $\\angle ACG + \\angle CDO = \\angle MOD + \\angle CDO = 90^\\circ$, thus $CG \\perp OD$. Let $H$ be the intersection of the lines $CG$ and $OD$. Since $OG \\parallel CD$, we get $\\frac{OH}{HD} = \\frac{OG}{CD} = \\frac{2}{3}$, consequently $\\frac{OH}{OD} = \\frac{2}{5}$. In the right triangle $MOD$, we have $OM = 2MD$, $OD = MD\\sqrt{5}$ and $OF = \\frac{OM^2}{OD} = \\frac{4\\sqrt{5}}{5}MD$, therefore $\\frac{OF}{OD} = \\frac{4}{5}$. It follows that $H$ is the midpoint of $OF$, so $CH$ is both an altitude and median in the triangle $COF$, hence $CF = CO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle BAC = 40^\\circ$ and $\\angle ABC = 80^\\circ$. Denote $I$ its incircle. Prove that $AI = BC$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\{D\\} = BI \\cap AC$. Then $\\angle BAD = \\angle ABD = 40^\\circ$, so $\\triangle ABD$ is isosceles, yielding $AD = BD$ (1).\n\n*First construction.* Draw the bisector $BM$ of $\\angle DBA$, with $M \\in AC$. Then $\\angle ABM = \\angle IAB = 20^\\circ$ and $AB$ is a common side, hence $\\triangle IAB \\equiv \\triangle MBA$ (A.S.A.), whence $AI = BM$.\n\nThen relations $\\angle CBM = \\angle CBD + \\angle DBM = 60^\\circ$ and $\\angle MCB = 60^\\circ$ show that triangle $MBC$ is equilateral, therefore $BC = BM = AI$.\n\n\n*Second construction.* Take the equilateral triangle $\\triangle AMI$, $D \\in (IM)$. Since $\\angle MAD = \\angle MAI - \\angle IAD = 60^\\circ - 20^\\circ = 40^\\circ$, $\\angle MAD = \\angle DBC$ (2). From (1), (2) and $\\angle AMD = \\angle DCB = 60^\\circ$ follows $\\triangle ADM \\equiv \\triangle BDC$ (A.S.A.), whence $AM = BC$. Now $\\triangle AMI$ equilateral implies $AI = AM$, hence $AI = BC$.\n\n\n*Third construction.* Let $AM \\parallel BC$, $M \\in BI$. Then $\\angle AMB = \\angle MBC$ (alternate angles), implying $\\angle ABM = \\angle AMB = 40^\\circ$, hence $\\triangle ABM$ is isosceles, whence $AB = AM$ (3). From $\\angle CAM = \\angle ACB = 60^\\circ$ (alternate angles) follows $\\angle IAM = 20^\\circ + 60^\\circ = 80^\\circ = \\angle ABC$ (4). Relations (3), (4) and $\\angle AMI = \\angle BAC = 40^\\circ$ give $\\triangle AMI \\equiv \\triangle BAC$ (A.S.A.), which implies $AI = BC$.\n\n\n*Fourth construction.* Take $M$ so that $B \\in (MC)$ and $\\triangle ACM$ is equilateral. Then relations $CA = CM$, $\\angle ICA = \\angle ICM = 30^\\circ$ and $IC$ common side lead to $\\triangle CIA \\equiv \\triangle CIM$ (S.A.S.), whence $IA = IM$ (5) and $\\angle IAC = \\angle IMC = 20^\\circ$.\n\nLet $\\{E\\} = AI \\cap BC$. From $\\angle MAB = \\angle MAC - \\angle BAC = 20^\\circ$, $AM = MC$ and $\\angle AMB = \\angle ACE = 60^\\circ$ follows $\\triangle MAB \\equiv \\triangle CAE$ (A.S.A.), hence $MB = EC$, which implies $ME = BC$ (6). From $\\triangle AEM$, $\\angle AEM = 180^\\circ - 60^\\circ - 40^\\circ = 80^\\circ$, so, in $\\triangle MIE$, $\\angle MIE = 180^\\circ - 20^\\circ - 80^\\circ = 80^\\circ = \\angle MEI$, hence $\\triangle MIE$ is isosceles, whence $MI = ME$ (7). Relations (5), (6) and (7) yield $IA = IM = ME = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all polynomials $P$ with integer coefficients, satisfying $0 \\le P(n) \\le n!$, for all non-negative integers $n$.",
    "options": [],
    "answer": "All such polynomials are exactly the following:\n- P(X) = 0;\n- P(X) = 1;\n- P(X) = (X − 1)^2;\n- P(X) = X(X − 1)⋯(X − k) for some nonnegative integer k;\n- P(X) = X(X − 1)⋯(X − k) (X − k − 2)^2 for some nonnegative integer k.",
    "solution": "The required polynomials are $P = 0$, $P = 1$, $P = (X - 1)^2$, $P = X(X - 1)\\cdots(X - k)$ and $P = X(X - 1)\\cdots(X - k)(X - k - 2)^2$ for some non-negative integer $k$. The verification is routine and is hence omitted.\n\nWe first deal with the case $P(0) = 1$. The polynomials $P_1 = 1$ and $P_2 = (X-1)^2$ both satisfy the condition in the statement and $P_1(0) = P_2(0) = 1$.\n\nWe will prove that either $P = P_1$ or $P = P_2$. Consider an index $i$ such that $P(1) = P_i(1)$ and let $\\tilde{P} = P - P_i$.\nInduct on $n$ to show that $\\tilde{P}(n) = 0$ for all non-negative integers $n$. The base cases $n = 0$ and $n = 1$ are clear. For the inductive step, assume $\\tilde{P}(m) = 0$ for all non-negative integers $m < n$. Then $X(X-1)\\cdots(X-(n-1))$ divides $\\tilde{P}$, so $n!$ divides $\\tilde{P}(n)$. As $0 < P_i(n) < n!$, it follows that $|\\tilde{P}(n)| = |P(n) - P_i(n)| < n!$, so $\\tilde{P}(n) = 0$.\nConsequently, $\\tilde{P}$ has infinitely many roots, so it vanishes identically; that is, $P = P_i$, as desired.\n\nFinally, we deal with the case $P(0) = 0$. Assume $P$ is non-zero. Let $P(X) = XQ(X - 1)$, where $Q$ has integer coefficients. Then $0 \\le Q(n) \\le n!$ for all non-negative integers $n$. If $Q(0) = 0$, repeat the argument for $Q$ and so on and so forth, all the way down to some polynomial with a non-zero constant term — this is clearly the case, as $P$ is non-zero and degrees strictly decrease in the process. By the preceding, such a polynomial is either $1$ or $(X - 1)^2$. An obvious induction then shows that $P$ has one of the last two forms mentioned in the beginning.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39480,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(a_n)$ of positive integers is defined by $a_1 = 1$ and $a_n = (a_{n-1}+1)^2$ for $n \\ge 2$. Let $p$ be an odd prime. Prove that $a_{2p} - a_p$ has at least $p$ pairwise distinct prime divisors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b_0 = 1$ and write $b_n = a_n + 1$, $n \\ge 1$. Clearly, $b_{n+1} = b_n^2 + 1$, $n \\ge 0$. If $n > m \\ge 1$, it then follows that $a_n - a_m = b_n - b_m = b_{n-1}^2 - b_{m-1}^2 = (b_{n-1} - b_{m-1})(b_{n-1} - b_{m-1})$, so\n\n$$\na_n - a_m = b_n - b_m = (b_{n-m} - b_0) \\prod_{k=0}^{m-1} (b_{n-m+k} + b_k). \\quad (*)\n$$\n\n$$\na_{2p} - a_p = b_{2p} - b_p = (b_p - b_0) \\prod_{k=0}^{p-1} (b_{p+k} + b_k).\n$$\nAs parities of the $b_n$ alternate and $p$ is odd, so is each of the factors above. We will prove that the $p$ factors $b_{p+k} + b_k$ are pairwise coprime, whence the conclusion.\n\nLet $0 \\le k < \\ell \\le p-1$ and let $d = \\gcd(b_{p+k} + b_k, b_{p+\\ell} + b_\\ell)$. By the preceding, $d$ is odd. Suppose, if possible, that $d > 1$ and let $\\equiv$ denote congruence modulo $d$. Then $b_{p+k} \\equiv -b_k$, so $b_{p+k+1} = b_{p+k}^2 + 1 \\equiv b_k^2 + 1 = b_{k+1}$. Continuing, $b_{p+k+2} = b_{p+k+1}^2 + 1 \\equiv b_{k+1}^2 + 1 = b_{k+2}$ and so on and so forth all the way up to get $b_{p+\\ell} \\equiv b_\\ell$. On the other hand, $b_{p+\\ell} \\equiv -b_\\ell$, so $2b_\\ell \\equiv 0$. Hence $b_\\ell \\equiv 0$, as $d$ is odd, so $b_{\\ell+1} = b_\\ell^2 + 1 \\equiv b_0$.\nConsider any index $j$ in the range $0$ through $p-1$ and use $(*)$ to get\n$$\nb_{j+\\ell+1} - b_j = (b_{\\ell+1} - b_0) \\prod_{i=0}^{j-1} (b_{\\ell+1+i} + b_i) \\equiv 0,\n$$\nas $b_{\\ell+1} \\equiv b_0$, by the preceding paragraph. Hence $b_{j+\\ell+1} \\equiv b_j$. Similarly, $b_{j+2(\\ell+1)} \\equiv b_{j+\\ell+1}$, then $b_{j+3(\\ell+1)} \\equiv b_{j+2(\\ell+1)}$ and so on and so forth to conclude recursively that the sequence $(b_n)$ is periodic modulo $d$.\nLet $t$ be the smallest period of the $b_n$ (mod $d$). Recall that $b_{p+k+1} \\equiv b_{k+1}$, so $t$ divides $p$. As $p$ is prime, either $t=1$ or $t=p$. The former case is ruled out, as $b_0 = 1 \\ne 2 = b_1$, so $t=p$.\nHence $b_k \\equiv b_{p+k} \\equiv -b_k$, so $b_k \\equiv 0$, as $d$ is odd. Recalling that $b_\\ell \\equiv 0$, it follows that $\\ell-k$ is divisible by $p$. This is a contradiction, as $0 < \\ell-k < p$.\nConsequently, the $p$ numbers $b_{p+k}+b_k$, $k=0, 1, \\dots, p-1$, are pairwise coprime, as stated. This ends the proof and completes the solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 3$. Determine the smallest positive integer $k$ satisfying the following condition:\nFor any tree $T$ with vertices $v_1, v_2, \\dots, v_n$ and any pairwise distinct complex numbers $z_1, z_2, \\dots, z_n$, there is a polynomial $P(X, Y)$ with complex coefficients of total degree at most $k$ such that for all $i \\ne j$ satisfying $1 \\le i, j \\le n$, we have $P(z_i, z_j) = 0$ if and only if there is an edge in $T$ joining $v_i$ to $v_j$.\n\n*Note, for example, that the total degree of the polynomial $9X^3Y^4 + XY^5 + X^6 - 2$ is 7 because $7 = 3 + 4$.*",
    "options": [],
    "answer": "n - 1",
    "solution": "*First solution.* First we provide a proof that $k \\ge n - 1$. Let $T$ be the path where $v_i$ and $v_{i+1}$ are adjacent for all $1 \\le i \\le n - 1$. Let $\\omega$ be a primitive root of unity of order $n$ and let $a_i = \\omega^i$ for all $1 \\le i \\le n$.\nIf $f(X) = P(X, \\omega X)$, then for all $1 \\le i \\le n-1$ we have $f(\\omega^i) = P(a_i, a_{i+1}) = 0$. Since $f(1) = P(a_n, a_1) \\ne 0$, $f$ is non-zero and has at least $n-1$ roots. This means that $\\deg P \\ge \\deg f \\ge n-1$, proving $k \\ge n-1$.\n\nIt remains to prove that $k = n - 1$ is sufficient i.e. for any tree $T$ and any $a_1, a_2, \\dots, a_n$ we can find a polynomial $P$ of degree at most $n - 1$. For brevity, we call a two-variable polynomial $A(X, Y)$ *symmetric* if $A(X, Y) = A(Y, X)$.\nWe begin with the following observation. Suppose that $A$ and $B$ are two variable polynomials of degree at most $d$. Then we can find $\\alpha \\in \\mathbb{C}$ such that for any $1 \\le i, j \\le n$, $A(a_i, a_j) + \\alpha B(a_i, a_j) = 0$ if and only if $A(a_i, a_j) = B(a_i, a_j) = 0$. This means that we can \"merge\" two conditions of degree at most $d$ into a condition of degree at most $d$ (note that this produces a symmetric polynomial if the initial polynomials are symmetric).\nFor any integer $t \\ge 2$, let a *star* of size $t$ be a collection of $t$ edges for which there is a vertex which belongs to all edges. We will prove the following claims.\n\n**Claim 1.** Let $G$ be a graph with vertices $v_1, v_2, \\dots, v_n$ and $E$ edges. Suppose that we can partition the edges of $G$ into a number of stars. Then for any distinct complex numbers $a_1, a_2, \\dots, a_n$ we can find a symmetric polynomial $P$ of degree at most $E$ such that for all $1 \\le i, j \\le n, i \\ne j$, $P(a_i, a_j) = 0$ if and only if there is an edge between $v_i$ and $v_j$ in $G$.\n*Proof.* We will first prove the claim when $G$ consists of a star of size $E \\le n - 1$ and some isolated vertices. Without loss of generality, let $v_1v_2, v_1v_3, \\dots, v_1v_{E+1}$ be the edges of $G$. Also let $s_1 = a_1 + a_2, s_2 = a_1 + a_3, \\dots, s_E = a_1 + a_{E+1}$.\nConsider merging the polynomials $(X - a_1)(Y - a_1)$ and $(X + Y - s_1)(X + Y - s_2) \\dots (X + Y - s_E)$ into a polynomial of degree at most $E$ (which is of course symmetric). They both vanish at a pair $a_i, a_j$ if and only if $1 \\in \\{i, j\\}$ and $a_i + a_j \\in \\{s_1, s_2, \\dots, s_E\\}$. These two happen if and only if $v_i$ and $v_j$ are adjacent, so this produces a valid polynomial.\nFor the general case, let $S_1 \\cup S_2 \\cup \\dots \\cup S_k$ be the partition of the edges of $G$. For each $1 \\le i \\le k$, we can find a two variable polynomial $P_i$ of degree at most $|S_i|$ which vanishes only at the edges of $S_i$. Then we can let $P = P_1P_2\\dots P_k$, which satisfies the claim as $\\deg P \\le |S_1| + |S_2| + \\dots + |S_k| = E$, as desired.\n\n**Claim 2.** Any tree $\\Gamma$ with odd number of vertices can be partitioned into stars of size 2.\n*Proof.* We prove this by induction on the number of the vertices of $\\Gamma$. The base case is clear, since $\\Gamma$ is a star of size 2 when $\\Gamma$ has three vertices.\nFor the inductive step, let $\\Gamma$ be a tree with $2m + 1$ vertices, where $m \\ge 2$. Let $u_1u_2\\dots u_t$ be a path of maximal length in $\\Gamma$ (of course, $t \\ge 3$). Then any neighbour of $u_2$ except for maybe $u_3$ must have degree 1, otherwise we can delete $u_1$ and insert two edges, contradicting the maximality of $t$. If $\\deg u_2 = 2$, we can form the star $u_1u_2, u_2u_3$ and apply the inductive hypothesis on $\\Gamma \\setminus \\{u_1, u_2\\}$. If $\\deg u_2 \\ge 3$, let $u \\neq u_1, u_3$ be a neighbour of $u_2$. Then create the star $u_1u_2, uu_2$ and apply the inductive hypothesis on $\\Gamma \\setminus \\{u, u_1\\}$. This proves the claim.\n\nThe case where $n$ is odd becomes trivial, since $T$ has $n-1$ edges. Assume that $n$ is even. Without loss of generality, let $\\deg v_n = 1$ and let $v_n$ be adjacent to $v_{n-1}$.\nConsider the graph $T'$ formed by deleting the edge $v_{n-1}v_n$ from $T$ (but not permuting the $n$ vertices). Clearly, $T'$ consists of $v_n$ and a tree on $v_1, v_2, \\dots, v_{n-1}$. As $n-1$ is odd, the edges of $T'$ can be partitioned into stars of size 2 from Claim 2. From Claim 1 it follows that there is a symmetric polynomial $Q(X, Y)$ of degree at most $n-2$ such that $Q(a_i, a_j) = 0$ if and only if $i, j \\le n-1$ and $v_i, v_j$ are adjacent in $T$.\nLet $g(X)$ be the polynomial of degree $n-1$ such that $g(a_1) = g(a_2) = \\dots = g(a_{n-1}) = 0$ and $g(a_n) = -Q(a_{n-1}, a_n) = -Q(a_n, a_{n-1})$. Let $P$ be the polynomial of degree at most $n-1$ obtained by merging $F_1(X, Y) = Q(X, Y) + g(X) + g(Y)$ and $F_2(X, Y) = Q(X, Y)(X+Y-s)$, where $s = a_n + a_{n-1}$. It is easy to see that $P$ vanishes at each $(a_i, a_j)$ for which $v_i, v_j$ are adjacent in $T$.\nSuppose that $v_i$ and $v_j$ are not adjacent in $T$. If $i, j \\le n-1$, then $F_1(a_i, a_j) = Q(a_i, a_j) \\neq 0$. If $n \\in \\{i, j\\}$, then $a_i + a_j \\neq s$ and $Q(a_i, a_j) \\neq 0$, so $F_2(a_i, a_j) \\neq 0$. Hence $P(a_i, a_j) \\neq 0$. This proves that $P$ satisfies the required conditions, as desired.\n*Second solution.* Establish the lower bound as in Solution 1. We now address the upper bound differently. Let $G$ be a graph on vertices $v_1, \\dots, v_n$. Say that a polynomial $P(X, Y)$ is $G$-good if it satisfies the conditions in the statement of the problem. We prove the more general fact below:\n\n**Claim.** Let $d_i$ be the degree of $v_i$. Assume that these degrees satisfy $d_i \\le n-i$ for all $i \\le n-1$, and $d_n = 1$. Then there is a $G$-admissible polynomial of degree at most $n-1$.\nNotice here that, if $G$ is a tree with vertices ordered so that $d_1 \\ge d_2 \\ge \\dots \\ge d_n$, then it satisfies the conditions in the Claim. Indeed, we have $d_n = 1$, and if $d_i > n-i$ for some $i \\le n-1$, then we have $2n-2 \\sum_{j=1}^n d_j \\ge \\sum_{j=1}^i (n-i+1) + \\sum_{j=i+1}^n 1 = i(n-i+1) + (n-i) = (i+1)(n-i+1) - 1 \\ge 2n-1$, which is a contradiction. So, it suffices to prove the Claim.\n\n*Proof of the Claim.* Let\n$$\nP(X, Y) = \\sum_{j=0}^{n-1} R_j(Y)X^j\n$$\nbe the sought polynomial; set $R_{n-1}(X) = 1$. Denote\n$$\nQ_i(X) = P(X, a_i) = X^{n-1} + \\sum_{j=0}^{n-2} q_{ij}X^j, \\quad \\text{where } q_{ij} = R_j(a_i).\n$$\nSo, we will seek for the sequences $C_j = (q_{1j}, q_{2j}, \\dots, q_{nj})$ such that there exists a polynomial $R_j$ with $\\deg R_j \\le n - j - 1$ such that $q_{ij} = R_j(a_i)$. Notice that the first $n-j$ terms of such a sequence determine it uniquely; in particular, there are no restrictions on the sequence $C_0$.\nWe know that the polynomial $Q_i$ has $d_i$ prescribed roots. For every $i \\le n-1$, augment this list by some numbers not from the set $A = \\{a_1, a_2, \\dots, a_n\\}$ to the list $b_{i1}, b_{i2}, \\dots, b_{i,n-i}$. Also, denote by $b_{n1}$ the unique prescribed root of $Q_n$. Thus, we should have\n$$\nQ_i(X) = S_i(X) \\prod_{j=1}^{\\max(n-i, 1)} (X - b_{ij}), \\quad (1)\n$$\nwhere $S_i$ is a monic polynomial with $\\deg S_i = i-1$ for $i \\le n-1$ and $\\deg S_n = n-2$. The only extra conditions we have are that each $S_i$ should achieve non-zero values at the prescribed finite subset $A_i$ of $A$ (containing no prescribed roots of $Q_i$).\nThe polynomial $Q_1$ is uniquely determined by (1).\nAssume that the polynomials $Q_1, \\dots, Q_{i-1}$ have already been determined, for some $i \\le n-2$. This means that the sequences $S_1, \\dots, S_{i-1}$ are also determined. This determines the polynomial $S_i$ up to the constant term. So we may choose this constant term so that $S_i$ has no prohibited roots, thus defining $Q_i$.\n*Third solution.* The answer is $n-1$. Use the same argument as Solution 1 for the lower bound.\nWe can construct such a polynomial via a method that uses no graph theory other than the fact that $T$ has $n-1$ edges.\n\n**Lemma 1.** Let $k$ be a positive integer and let $A, B \\subseteq \\mathbb{C}^2$ be two disjoint finite sets. Suppose $|A| = 2k$ and no line intersects $A \\cup B$ in more than $k+1$ points. Then there exists a polynomial of degree $k$ that vanishes on $A$ and is non-zero on $B$.\n*Proof.* We first argue that we may reduce to the case when $|B| = 1$. This is since if $P$ and $P'$ are polynomials that are zero on $A$ and nonzero on $B$ and $B'$, then a generic linear combination of $P$ and $P'$ is nonzero on $A$ and nonzero on $B \\cup B'$. Now suppose $B = \\{b\\}$.\nFor $a, a' \\in A$, write $a \\sim a'$ if $a, a', b$ are collinear. This is an equivalence relation, and each equivalence class has at most $k$ elements. Thus we may pair up the elements of $A$ such that no two paired elements are collinear with $b$. Now let $P$ be the polynomial vanishing on the union of the $k$ lines determined by the pairs, which is nonzero at $b$ by construction.\n\n**Lemma 2.** Let $a_1, a_2, \\dots, a_n$ be distinct complex numbers. Then a line can intersect at most $n$ points of the form $(a_i, a_j)$.\n*Proof.* If not, then by the pigeonhole principle such a line must contain two points with the same $x$-coordinate. But then it is vertical and thus can only contain $n$ points.\nWe are done by applying Lemma 1 with\n$$\nA = \\{(a_i, a_j) : v_i v_j \\in E(T)\\} \\quad \\text{and} \\quad B = \\{(a_i, a_j) : i \\neq j, v_i v_j \\notin E(T)\\}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ and consider a table with $n+1$ rows and $n$ columns, in which, on each of the first $n$ rows, Nicuşor writes, in some order, the numbers $1, 2, \\dots, n$. Then, he chooses a permutation $a_1, a_2, \\dots, a_n$ of the numbers $1, 2, \\dots, n$ and completes the last row as follows: for each $j \\in \\{1, 2, \\dots, n\\}$, he writes in the cell at column $j$ the number of occurrences of $a_j$ in the cells located at the intersection of column $j$ with the first $n$ rows.\nDetermine all $n$ for which Nicuşor can complete the table and choose $a_1, a_2, \\dots, a_n$ so that the last row contains, in some order, the numbers $1, 2, \\dots, n$.\nCristi Săvescu",
    "options": [],
    "answer": "all integers n ≥ 4",
    "solution": "For $n=2$, no matter how Nicuşor chooses $a_1$ and $a_2$, on the third row we will have two equal values. For $n=3$, suppose without loss of generality that $a_3=3$. Then, in the third column, we have only values equal to $3$. Since the set $(1, 2)$ admits only two permutations, among the first three rows there will be two identical ones, so these rows coincide and we cannot have any $a_i = 1$.\n\nNext, we show that for $n \\ge 4$, there exists a completion that works. If $n=4$ and we take $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4$, the following construction satisfies the\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 2 | 1 | 3 | 4 |\n| 2 | 1 | 3 | 4 |\n| 3 | 2 | 1 | 4 |\n| 1 | 2 | 3 | 4 |\n\n**Construction 1 (induction):**\nWe will prove by induction on $n \\ge 4$ that there exists a valid construction in which $a_i = i$ for every $i \\in \\{1, 2, \\dots, n\\}$ and the last row is exactly $1, 2, \\dots, n$. The case $n = 4$ is illustrated above. Suppose there exists a valid table $T_n$ of size $(n+1) \\times n$ for $n \\ge 4$, for which the chosen permutation by Nicuşor is $a_1, a_2, \\dots, a_n$ with $a_i = i$ for every $i = 1, 2, \\dots, n$.\nWe construct a table $T_{n+1}$ of size $(n+2) \\times (n+1)$, valid for $n+1$, as follows: we add a column at the end of $T_n$ and a new row between the last and the penultimate rows of $T_n$. We fill all the cells of the newly added last column with the value $n+1$, and the first $n$ values of the newly added row with $2, 3, \\dots, n, 1$ (or any permutation that preserves the values for $a_1, a_2, \\dots, a_n$).\nIt is easy to verify that the newly constructed table is valid.\n\n\n**Construction 2 (direct):**\nFor $n \\ge 4$, consider $a_i = i$ for every $i = 1, 2, 3, 4, \\dots, n$ and the following construction:\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | ... | n |\n|---|---|---|---|---|---|---|-----|---|\n| 2 | 1 | 3 | 4 | 5 | 6 | 7 | ... | n |\n| 2 | 1 | 3 | 4 | 5 | 6 | 7 | ... | n |\n| 3 | 2 | 1 | 4 | 5 | 6 | 7 | ... | n |\n| 2 | 3 | 4 | 1 | 5 | 6 | 7 | ... | n |\n| 2 | 3 | 4 | 5 | 1 | 6 | 7 | ... | n |\n| 2 | 3 | 4 | 5 | 6 | 1 | 7 | ... | n |\n| ... |\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | ... | n |\n\nIt is easy to verify that the newly constructed table is valid.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 39483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a natural number. John draws a regular $n$-gon and connects every pair of vertices. On each segment, John writes a nonzero natural number such that in any triangle formed by three vertices, one of the numbers on its sides equals the sum of the other two.\n\nDetermine the smallest number of distinct values John can write.",
    "options": [],
    "answer": "n − 1 for n ≠ 4; when n = 4, the minimum is 2",
    "solution": "We will show that the number sought is $n-1$ for $n \\neq 4$ and $2$ if $n=4$.\n\nFor $n = 3$ and $n = 4$, it can be easily verified that the answer is $2$. Suppose now that $n \\ge 5$. Denote by $a_{XY}$ the number written on segment $XY$.\n\nIf we label the vertices of the regular polygon as $A_1, A_2, \\dots, A_n$ and set $a_{A_i A_j} = |i-j|$ for all $i \\neq j$, then the condition of the problem is satisfied, so the number sought for is at most $n-1$.\n\nTo prove equality, suppose by contradiction that the number sought for is at most $n-2$. Let $d$ be the greatest value John writes on some segment, and let $A, B$ be vertices such that $a_{AB} = d$. Suppose $d$ is written on the segment with minimal distance between the vertices among all segments with value $d$.\n\nWe make the following two observations:\n\n1. There do not exist three vertices $A_1, A_2, A_3$ such that $a_{AA_1} = a_{AA_2} = a_{AA_3}$ (otherwise, it would follow that $a_{A_1A_2} = a_{A_1A_3} = a_{A_2A_3}$, which is impossible because the sum rule, according to the hypothesis, would no longer be valid for triangle $A_1A_2A_3$).\n\n2. For any point $X$ different from $A$ and $B$, we have $d = a_{AX} + a_{BX}$ since $d$ is maximum; in particular, $a_{AX} < d$.\n\nSince, by assumption, there are at most $n-2$ distinct values, the pigeonhole principle implies that there exist points $C$ and $D$ such that $a_{AC} = a_{AD} = x$, with $x < d$ (according to Observation 2).\n\nWe observe that the rule for writing the numbers on segments implies: in triangle $ACD$ we have $a_{CD} = 2x$; in triangle $ABD$ we have $a_{BD} = d - x$; in triangle $ABC$ we have $a_{BC} = d - x$. Therefore, from triangle $BCD$, we obtain that $2x = (d - x) + (d - x)$, hence $x = \\frac{d}{2}$ (in particular, $d$ is even).\n\nNow, looking at the $n-2$ numbers on the segments emerging from $A$, we can see that the only value that repeats is $\\frac{d}{2}$ (as shown in the previous paragraph) and it cannot appear three times (by Observation 1). Therefore, there are $n-2$ distinct numbers, and according to the assumption, these must be all the numbers used by John.\n\nSince $n \\ge 5$, there exists a vertex $E$ different from the vertices $A, B, C, D$. Let $y = a_{AE} \\ne \\frac{d}{2}$. Since $a_{BE} = d - y$, it follows that $d - y$ appears among the $n-2$ values on the segments emerging from $A$. Without loss of generality, we may assume that $y > \\frac{d}{2}$ because $\\max(y, d - y) > \\frac{d}{2}$.\n\nWe look at triangle $ACE$ and observe that $a_{CE} \\in \\{y + \\frac{d}{2}, y - \\frac{d}{2}\\}$. Since $d$ is the maximum, and $y + \\frac{d}{2} > d$, it follows that $a_{CE} = y - \\frac{d}{2}$, and similarly, $a_{DE} = y - \\frac{d}{2}$.\n\nFinally, looking at triangle $CDE$, we have $a_{CD} = 2(y - \\frac{d}{2}) < d$, but also $a_{CD} = 2 \\cdot \\frac{d}{2} = d$, a contradiction. (In triangle $ACD$ we have $a_{AC} = a_{AD} = \\frac{d}{2}$, hence $CD = d$)\n\nTherefore, there cannot be only $n-2$ values, which implies that the required number is $n-1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a natural number. Ana chooses the non-zero natural numbers $a_1, a_2, \\dots, a_n$, and for each non-empty subset $A \\subset \\{1, 2, \\dots, n\\}$, she computes the sum $s_A = \\sum_{k \\in A} a_k$, then arranges these sums in increasing order, obtaining the sequence $s_1 \\le s_2 \\le \\dots \\le s_{2^{n}-1}$.\nShow that there exists a subset $B \\subset \\{1, 2, \\dots, 2^n - 1\\}$, with $2^{n-2} + 1$ elements, such that, no matter what values Ana chooses for $a_1, a_2, \\dots, a_n$, these values can be determined by knowing all the values $s_i$, for $i \\in B$.\nCristi Săvescu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $a_1 \\le a_2 \\le \\dots \\le a_n$. If we choose the subset $B' = \\{1, 2, \\dots, 2^{n-2}\\}$, then among the sums $s_1, s_2, \\dots, s_{2^{n-2}}$, the values $a_1, a_2, \\dots, a_t$ will appear for some $t \\ge 1$. Clearly, $s_1 = s_{\\{1\\}} = a_1$, and if $a_i$ appears as a sum, then $a_{i-1}$ must have appeared before it.\nIndeed, consider:\n\na) if $a_{i-1} = a_i$, then the appearance is obvious.\n\nb) if $a_{i-1} < a_i$, then since $s_{\\{i-1\\}} = a_{i-1} < a_i$ and $a_i$ appears as some $s_j$, the index corresponding to $a_{i-1}$ is less than $j$, hence $a_{i-1}$ appeared earlier.\n\nIf $t \\le n-2$, then at most $2^{n-2}-1$ sums can appear, corresponding to non-empty subsets of $\\{1, 2, \\dots, n-2\\}$. Hence, we deduce that $t \\ge n-1$.\nTherefore, the first two sums correspond to the numbers $a_1$ and $a_2$. Then, inductively, after determining the numbers $a_1, a_2, \\dots, a_k$, we compute all the subset sums of $\\{a_1, a_2, \\dots, a_k\\}$, remove them from the list of known sums, and the smallest remaining sum must be $a_{k+1}$.\nSo, we can determine the numbers $a_1, a_2, \\dots, a_{n-1}$.\nIf we also include the index $2^n - 1$, we obtain $s_{2^n-1}$, which is the total sum of all numbers. Then we can compute $a_n$, and thus all numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$, with $n \\ge 2$, and let $x_1, x_2, \\dots, x_n$ be positive real numbers such that $x_1 + x_2 + \\dots + x_n = 1$. Define\n$$\nm = \\min \\left\\{ \\frac{x_1}{1+x_1}, \\frac{x_2}{1+x_1+x_2}, \\dots, \\frac{x_n}{1+x_1+x_2+\\dots+x_n} \\right\\}\n$$\nand\n$$\nM = \\max \\left\\{ \\frac{x_1}{1+x_1}, \\frac{x_2}{1+x_1+x_2}, \\dots, \\frac{x_n}{1+x_1+x_2+\\dots+x_n} \\right\\}.\n$$\n\nDetermine:\na) the greatest possible value of $m$;\n\nb) the smallest possible value of $M$.",
    "options": [],
    "answer": "a) 1 - 2^{-1/n}; b) 1 - 2^{-1/n}",
    "solution": "a.\nLet us denote\n$$\na_k = \\frac{x_k}{1+x_1+\\cdots+x_k} \\quad \\text{and} \\quad b_k = 1-a_k = \\frac{1+x_1+\\cdots+x_{k-1}}{1+x_1+\\cdots+x_k}, \\quad 1 \\le k \\le n.\n$$\nBy the inequality of arithmetic and geometric means, we have\n$$\nb_1 + b_2 + \\cdots + b_n \\ge n \\sqrt[n]{b_1 b_2 \\cdots b_n} = \\frac{n}{\\sqrt[n]{1+x_1+\\cdots+x_n}} = \\frac{n}{\\sqrt[n]{2}}.\n$$\nTherefore,\n$$\nn \\cdot m \\le a_1 + a_2 + \\cdots + a_n = n - (b_1 + b_2 + \\cdots + b_n) \\le n \\left(1 - \\frac{1}{\\sqrt[n]{2}}\\right),\n$$\nwhich implies $m \\le 1 - \\frac{1}{\\sqrt[n]{2}}$.\n\nWe will show that there exists a sequence $(x_1, x_2, \\dots, x_n)$ of positive real numbers summing to 1 such that $a_1 = a_2 = \\dots = a_n = 1 - \\frac{1}{\\sqrt[n]{2}} \\stackrel{\\text{def}}{=} m_0$, which implies that the greatest possible value of $m$ is $m_0$.\nThe existence of such a sequence is equivalent to solving the system\n$$\nx_1 + \\cdots + x_n = 1, \\quad \\frac{x_1}{1+x_1} = \\frac{x_i}{1+x_1+\\cdots+x_i}, \\quad \\text{for all } i = 2, \\dots, n.\n$$\nFor $i = 2$, we get $x_2 = a_1(1 + x_1)$. For $i = 3$, replacing $x_2$, we find $x_3 = x_1(1 + x_1)^2$. By induction on $i$, it follows that $x_i = x_1(1 + x_1)^{i-1}$, for all $i \\le n$. Since their sum is 1, we get\n$$\n1 = x_1 + x_1(1 + x_1) + \\dots + x_1(1 + x_1)^{n-1} = x_1 \\frac{(1+x_1)^n - 1}{(1+x_1) - 1} = (1+x_1)^n - 1.\n$$\nHence, $(1+x_1)^n = 2$, so $x_1 = \\sqrt[n]{2}-1$. Substituting back, $x_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$, $k=1, 2, \\dots, n$.\n\nb.\nWe will prove that the smallest possible value of $M$ is $M_0 = 1 - \\frac{1}{\\sqrt[n]{2}}$, which is attained if and only if $a_1 = a_2 = \\dots = a_n$.\nUsing the same reasoning as before, we have $a_1 = a_2 = \\dots = a_n$ if and only if $x_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$, $k = 1, 2, \\dots, n$.\nLet $\\alpha_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$ for $1 \\le k \\le n$, and consider a positive sequence $(x_1, x_2, \\dots, x_n)$, different from $(\\alpha_1, \\alpha_2, \\dots, \\alpha_n)$, with sum 1. Then there exists $k \\in \\{1, \\dots, n\\}$ such that $\\alpha_k < x_k$.\nLet $k_0$ be the smallest such index for which $\\alpha_{k_0} < x_{k_0}$; then $x_i \\le \\alpha_i$, for all $1 \\le i \\le k_0 - 1$. We deduce\n$$\n\\begin{aligned}\na_{k_0} &= \\frac{x_{k_0}}{1 + x_1 + \\cdots + x_{k_0}} = \\frac{1}{\\frac{1+x_1+\\cdots+x_{k_0-1}}{x_{k_0}} + 1} > \\frac{1}{\\frac{1+\\alpha_1+\\cdots+\\alpha_{k_0-1}}{\\alpha_{k_0}} + 1} \\\\\n&= \\frac{\\alpha_{k_0}}{1 + \\alpha_1 + \\cdots + \\alpha_{k_0}} = M_0,\n\\end{aligned}\n$$\ntherefore, $M = \\max a_k \\ge a_{k_0} > M_0$.\nIn conclusion, $M_0 = 1 - \\frac{1}{\\sqrt[n]{2}}$ is the minimal possible value of $M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime, $p \\geq 3$, and $k$ an odd number, not a multiple of $p$. Let $K$ be a finite field with $kp+1$ elements and $A = \\{x_1, x_2, \\dots, x_t\\}$, the set of elements $K^* = K \\setminus \\{0\\}$ of order different of $k$ in the multiplicative group $(K^*, \\cdot)$. Prove that the polynomial $P(X) = (X + x_1)(X + x_2)\\dots(X + x_t)$ has at least $p$ coefficients equal to 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $k$ and $p$ are odd, $|K|$ is even, in fact a power of $2$, the characteristic of $K$ being $2$. It follows that $x_1, x_2, \\dots, x_t$ are the roots of $P$. Consider $P = \\sum_{i=0}^{t} a_j X^j$.\n\nThe multiplicative group $(K^*, \\cdot)$ is cyclic of order $kp$. Consider $a \\in K^*$ one of its generators. For any $s \\in \\{1, 2, \\dots, kp\\}$, with $(s, kp) = 1$ we get\n$$\n\\mathrm{ord}(a^s) = \\mathrm{ord}(a) = kp. \\qquad (1)\n$$\nIn particular, for any $i \\in \\{1, 2, \\dots, p-1\\}$ we have $(ik + p, k) = (p, k) = 1$ and $(ik + p, p) = (ik, p) = 1$, so that $(ik + p, kp) = 1$ and $\\mathrm{ord}(a^{ik+p}) = kp$. These give\n\n$$\nA_r = \\sum_{i=0}^{p-1} a^{-irk} P(a^{ik+p}) = \\sum_{i=0}^{p-1} a^{-irk} \\sum_{j=0}^{t} a_j a^{ikj+pj} = \\sum_{j=0}^{t} a_j a^{pj} \\sum_{i=0}^{p-1} a^{k(j-r)i}.\n$$\nBecause\n$$\n\\sum_{i=0}^{p-1} a^{kmi} = \\begin{cases} p, & \\text{if } p \\mid m, \\\\ 0, & \\text{if } p \\nmid m, \\end{cases}\n$$\nwe obtain\n$$\nA_r = \\sum_{j=0}^{t} a_j a^{pj} \\sum_{i=0}^{p-1} a^{k(j-r)i} = p \\cdot \\sum_{j=0}^{\\lfloor \\frac{t-r}{p} \\rfloor} a_{pj+r} a^{p(pj+r)}.\n$$\n\nFor any divisor $d \\nmid kp$ consider $U_d = \\{x \\in K^* \\mid \\text{ord}(x) = d\\}$ and the polynomial\n$$\nQ_d = \\prod_{x \\in U_d} (X + x).\n$$\nLet $K_2 = \\{0, 1\\}$ the prime subfield of $K$ and $1 = d_1 < d_2 < \\dots < d_n = kp$ the divisors of $kp$. As $Q_1 = X + 1 \\in K_2[X]$ and\n$$\nQ_{d_i} = (X^{d_i} + 1) \\cdot \\left( \\prod_{d_j \\mid d_i, d_j \\neq d_i} Q_{d_j} \\right)^{-1} \\in K_2[X],\n$$\nwe have that all polynomials $Q_{d_i} \\in K_2[X]$ have all coefficients $0$ or $1$.\nBut then $P = (X^{kp} + 1) \\cdot Q_k^{-1} \\in K_2[X]$ has also this property, moreover at least $p$ being non-zero, concluding the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the complex numbers $z$ și $w$ with the property that\n$$\n|z^{2n} + z^n w^n + w^{2n}| = 2^{2n} + 2^n + 1,\n$$\nfor any positive integer $n$.",
    "options": [],
    "answer": "All solutions are the pairs (z, w) with w = 2z and |z| = 1, or z = 2w and |w| = 1; equivalently, (z, w) = (t, 2t) or (2t, t) with |t| = 1.",
    "solution": "Applying the modulus to both members of the identity $(z^2 + zw + w^2) \\cdot (z^2 - zw + w^2) = z^4 + z^2w^2 + w^4$ and, using the problem hypothesis, we obtain that $|z^2 - zw + w^2| = 3$.\n\nApplying the modulus to both members of the identity $(z^4 + z^2w^2 + w^4) \\cdot (z^4 - z^2w^2 + w^4) = z^8 + z^4w^4 + w^8$ and, using the problem hypothesis, we obtain that $|z^4 - z^2w^2 + w^4| = 13$.\n\nApplying the modulus to both members of the identity $\\frac{1}{2}(z^2 + zw + w^2)^2 + \\frac{1}{2}(z^2 - zw + w^2)^2 + (z^4 - z^2w^2 + w^4) = 2(z^4 + z^2w^2 + w^4)$, using the modulus inequality and the above, we obtain that $\\frac{1}{2} \\cdot 49 + \\frac{1}{2} \\cdot 9 + 13 \\ge 2 \\cdot 21$.\nWe observe that equality holds, therefore there is a real number $t$ such that $(z^2 + zw + w^2)^2 = t^2(z^2 - zw + w^2)^2$. Moving to the modules, we find $t = \\pm\\frac{7}{3}$, so $3(z^2 + zw + w^2) = \\pm 7(z^2 - zw + w^2)$.\nFor $t = \\frac{7}{3}$ we obtain $2z^2 - 5zw + 2w^2 = 0 \\Leftrightarrow (2z - w)(z - 2w) = 0$. It follows that $w = 2z$ or $z = 2w$. Substituting in $|z^2 + zw + w^2| = 7$, in the first case we obtain $|z| = 1$, and in the second $|w| = 1$.\nFor $t = -\\frac{7}{3}$ we obtain $5z^2 - 2zw + 5w^2 = 0 \\Leftrightarrow (\\varphi z - w)(z - \\varphi w) = 0$, where $\\varphi = \\frac{1+2\\sqrt{6i}}{5}$, $5\\varphi^2 - 2\\varphi + 5 = 0$, $|\\varphi| = 1$. Thus $w = \\varphi z$ or $z = \\varphi w$. Substituting in $|z^2 + zw + w^2| = 7$, in the first case we obtain $|z| = \\sqrt{5}$, and in the second $|w| = \\sqrt{5}$. None of these options are suitable, because they contradict equality $|z^4 - z^2w^2 + w^4| = 13$.\n\nIt is easy to verify that all pairs of the form $(z, 2z)$ and $(2z, z)$, where $z$ is a complex number with modulus equal to 1, have the property in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m \\in \\mathbb{N}$, $m \\ge 2$ be a fixed natural number, and let $(a_n)_{n \\ge 1}$ be a sequence of nonnegative real numbers such that $a_{n+1} \\le a_n - a_{mn}$, $\\forall n \\ge 1$.\n\na) Prove that the sequence $(b_n)_{n \\ge 1}$, $b_n = \\sum_{k=1}^{n} a_k$ is bounded above.\n\nb) Prove that the sequence $(c_n)_{n \\ge 1}$, $c_n = \\sum_{k=1}^{n} k^2 a_k$ is bounded above.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We notice that $(b_n)_{n \\ge 1}$ is non-decreasing. Also, the sequence $(a_n)_{n \\ge 1}$ is non-increasing, since $0 \\le a_{mn} \\le a_n - a_{n+1}$. Moreover,\n$$\n\\sum_{k=1}^{n} a_{mk} \\le a_1 - a_{n+1} \\le a_1.\n$$\n\nUsing the monotonicity of $(a_n)$ and $(b_n)$, we have:\n$$\nb_n \\le b_{mn} = \\sum_{k=1}^{mn} a_k = \\sum_{i=1}^{m-1} a_i + \\sum_{k=1}^{n} a_{mk} + \\sum_{k=1}^{n-1} \\sum_{j=1}^{m-1} a_{mk+j}.\n$$\nBy monotonicity,\n$$\n\\sum_{k=1}^{n-1} \\sum_{j=1}^{m-1} a_{mk+j} \\le (m-1) \\sum_{k=1}^{n-1} a_{mk} \\le (m-1)a_1,\n$$\nhence,\n$$\nb_n \\le \\sum_{i=1}^{m-1} a_i + \\sum_{k=1}^{n} a_{mk} + (m-1)a_1 \\le \\sum_{i=1}^{m-1} a_i + ma_1,\n$$\n\nb) We first prove that the sequence $d_n = \\sum_{k=1}^{n} k a_k$ is bounded above. Clearly, $(d_n)_{n \\ge 1}$ is non-decreasing. Moreover,\n$$\n\\sum_{k=1}^{n} k a_{mk} \\le \\sum_{k=1}^{n} k(a_k - a_{k+1}) = a_1 + \\sum_{k=2}^{n} (k - (k-1))a_k - n a_{n+1} \\le b_n,\n$$\nso the sequence $\\left( \\sum_{k=1}^{n} k a_{mk} \\right)_{n \\ge 1}$ is bounded above.\nOn the other hand,\n$$\n\\begin{aligned}\nd_n \\le d_{mn} &= m \\sum_{k=1}^{n} k a_{mk} + \\sum_{k=1}^{m-1} k a_k + \\sum_{j=1}^{m-1} \\sum_{k=1}^{n-1} (mk + j)a_{mk+j} \\\\\n&\\le m b_n + d_{m-1} + (m-1) \\sum_{k=1}^{n-1} m(k+1)a_{mk} \\\\\n&\\le m b_n + d_{m-1} + m(m-1)(b_{n-1} + a_1),\n\\end{aligned}\n$$\nhence $(d_n)_{n \\ge 1}$ is bounded above.\n\nAgain, $(c_n)_{n \\ge 1}$ is non-decreasing. Similarly,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} k^2 a_{mk} &\\le \\sum_{k=1}^{n} k^2 (a_k - a_{k+1}) = a_1 + \\sum_{k=2}^{n} (k^2 - (k-1)^2)a_k - n^2 a_{n+1} \\\\\n&\\le a_1 + \\sum_{k=2}^{n} 2k a_k \\le 2 d_n,\n\\end{aligned}\n$$\nso the sequence $\\left( \\sum_{k=1}^{n} k^2 a_{mk} \\right)_{n \\ge 1}$ is bounded above.\nFinally,\n$$\n\\begin{aligned}\nc_n \\le c_{mn} &= m^2 \\sum_{k=1}^{n} k^2 a_{mk} + c_{m-1} + \\sum_{j=1}^{m-1} \\sum_{k=1}^{n-1} (mk + j)^2 a_{mk+j} \\\\\n&\\le 2m^2 d_n + c_{m-1} + m^2(m-1) \\left( \\sum_{k=1}^{n-1} k^2 a_{mk} + 2 \\sum_{k=1}^{n-1} k a_{mk} + \\sum_{k=1}^{n-1} a_{mk} \\right),\n\\end{aligned}\n$$\nwhich is a finite sum of bounded above sequences, hence $(c_n)_{n \\ge 1}$ is bounded above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, with $AB \\neq AC$. Let $D$ be the midpoint of $BC$ and $I, J, K$ be the feet of the altitudes from $A, B$ and $C$, respectively, in the triangle $ABC$. The perpendicular from $A$ to the line $AD$ meets the lines $BJ$ and $CK$ at points $N$ and $Q$, respectively, and the parallel to $BC$ through $A$ intersects the lines $IJ$ and $IK$ at $M$ and $P$, respectively. Prove that $MNPQ$ is a parallelogram.\nPetru Braica\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of the triangle $ABC$. Quadrilaterals $BIHK$ and $CIHJ$ are cyclic, so $\\angle HIK = \\angle HBK = \\angle ABJ = \\angle ACK = \\angle JCH = \\angle HIJ$.\n\nTherefore, $IA$ is the angle bisector of $JIK$. Since $IA \\perp BC$ and $MP \\parallel BC$, it follows that $IA \\perp MP$. Because $IA$ is both an angle bisector and an altitude in $\\triangle IMP$, this triangle is isosceles and $A$ is the midpoint of $MP$. (*)\n\nLet $S$ be the reflection point of $A$ with respect to point $D$. Since $D$ is the midpoint of $BC$, it follows that $ABSC$ is a parallelogram. Consequently, $BS \\parallel AC$ and, since $BJ \\perp AC$, we deduce that $BJ \\perp BS$. From $\\angle SBN = \\angle SAN = 90^\\circ$ it follows that $SBAN$ is cyclic, therefore $\\angle NSA = \\angle ABN = 90^\\circ - \\angle A$. Similarly, the quadrilateral $SCAQ$ is cyclic, hence $\\angle QSA = \\angle ACQ = 90^\\circ - \\angle A = \\angle NSA$. In the triangle $SNQ$, the altitude $SA$ is also an angle bisector, therefore $A$ is the midpoint of $NQ$. Since the diagonals of the quadrilateral $MNPQ$ bisect each other, it follows that $MNPQ$ is a parallelogram.\n\n\n*Alternative solution for (*)*. Let $E$ be the projection of $D$ onto $AC$.\nSince $\\angle DEA = \\angle AJN = 90^\\circ$ and $\\angle DAE = 90^\\circ - \\angle NAJ = \\angle ANJ$, it follows that $\\triangle DAE \\sim \\triangle ANJ$. Thus, $\\frac{NA}{AD} = \\frac{AJ}{DE} = 2 \\cdot \\frac{AJ}{BJ}$. Similarly, we obtain $\\frac{QA}{AD} = 2 \\cdot \\frac{AK}{CK}$. The triangles $ABJ$ and $ACK$ are similar (AA), thus $\\frac{AJ}{BJ} = \\frac{AK}{CK}$, and consequently $AQ = AN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andrei represents $2025$ as a sum of $40$ pairwise different positive integers. Find the lowest value that the largest of the $40$ numbers can achieve.",
    "options": [],
    "answer": "71",
    "solution": "Let $0 < a_1 < a_2 < a_3 < \\dots < a_{40}$ so that $a_1 + a_2 + a_3 + \\dots + a_{40} = 2025$. Then $a_2 \\ge a_1 + 1$, $a_3 \\ge a_2 + 1$, $a_4 \\ge a_3 + 1$, $\\dots$, $a_{40} \\ge a_{39} + 1$.\nConsequently, $a_{40} \\ge a_1 + 39 \\ge a_2 + 38 \\ge \\dots \\ge a_{38} + 2$.\nThen $40 \\cdot a_{40} \\ge (a_1+39)+(a_2+38)+\\dots+(a_{37}+3)+(a_{38}+2)+(a_{39}+1)+a_{40}$,\nhence $40 \\cdot a_{40} \\ge 2025 + (1 + 2 + \\dots + 39) = 2805$. Since $a_{40}$ is a positive integer,\nit follows that $a_{40} \\ge 71$.\nThe value $a_{40} = 71$ can be achieved: $1+29+34+35+36+\\dots+70+71 = 2025$,\ntherefore the required minimum is $71$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a continuous function. We define the function $\\tilde{f} : [0, 1] \\to \\mathbb{R}$ by\n$$\n\\tilde{f}(x) = \\begin{cases} \\frac{1}{x} \\cdot \\int_{0}^{x} f(t) \\, dt, & \\text{if } x > 0, \\\\ f(0), & \\text{if } x = 0. \\end{cases}\n$$\n\nShow that:\na) the function $\\tilde{f}$ is continuous in $0$ and differentiable on $(0, 1]$;",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The function $f$ being continuous on $[0, 1]$, it follows that the function $F: [0, 1] \\to \\mathbb{R}$ defined by $F(x) = \\int_{0}^{x} f(t) dt$ is differentiable on $[0, 1]$, with $F' = f$. It follows that the function $\\tilde{f}$ is differentiable on $(0, 1]$, as a product of differentiable functions.\nIt remains to show that $\\tilde{f}$ is continuous at $0$. Since $f$ is continuous, $\\lim_{x \\to 0} f(x) = f(0)$, so that, applying l'H\\^opital's rule, we have:\n$$\n\\lim_{x \\to 0} \\tilde{f}(x) = \\lim_{x \\to 0} \\frac{F(x)}{x} = \\lim_{x \\to 0} \\frac{F'(x)}{x'} = \\lim_{x \\to 0} f(x) = f(0) = \\tilde{f}(0),\n$$\nand it follows that $\\tilde{f}$ is continuous at $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 2$ so that $n$ is divisible by each of the numbers $d_1, d_1+d_2, \\dots, d_1+d_2+\\dots+d_{k-1}$, where $1 = d_1 < d_2 < \\dots < d_{k-1} < d_k = n$ are all the positive divisors of $n$.",
    "options": [],
    "answer": "all prime numbers and 6",
    "solution": "Answer: all primes and $n = 6$.\nThe primes have the required property, because the divisors of a prime $n$ are $1 = d_1 < d_2 = n$, and $n$ is divisible by $d_1$.\n\nTake now a positive integer $n$ with $k \\ge 3$ divisors and fulfilling the required property.\nIf $n$ is odd, then all its $k$ divisors are odd. From the condition $n$ must be divisible by $d_1+d_2$, which is even – a contradiction. Therefore, $n$ must be even.\n\nThen $d_1 = 1$, $d_2 = 2$, so $n$ must be divisible by $d_1+d_2 = 3$, hence $d_3 = 3$. Moreover, $6 \\mid n$. This shows that $\\frac{n}{6}$, $\\frac{n}{3}$ and $\\frac{n}{2}$ are divisors of $n$.\n\nSince $n$ is divisible by $d_1+d_2+\\dots+d_{k-1}$, we get $n \\ge d_1+d_2+\\dots+d_{k-1} \\ge \\frac{n}{6}+\\frac{n}{3}+\\frac{n}{2} = n$. This shows that the only divisors of $n$, smaller than $n$, are $\\frac{n}{6}$, $\\frac{n}{3}$ and $\\frac{n}{2}$, hence $\\frac{n}{6} = d_1 = 1$, $\\frac{n}{3} = d_2 = 2$ and $\\frac{n}{2} = d_3 = 3$, yielding $n = 6$, which fulfills the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the integers $n \\ge 2$ with the property: the cells of a $n \\times n$ board can be colored with several colors, so that each cell $C$ has exactly two neighbouring cells with the same color as $C$. Here neighbouring cells means cells with a common side.",
    "options": [],
    "answer": "All even integers n ≥ 2",
    "solution": "Indeed, if $n$ is even, then we can divide the board into squares $2 \\times 2$. Now we pick a different color for each such square and use it for the cells of that square. This coloring clearly satisfies the requirement.\n\nTo prove that $n$ must be even, start from any cell $S$ of the board and move from each cell $C$ to its neighbour having the same color as $C$ and which has not been previously visited. Since the board is finite, this procedure must stop at some point, meaning that the next step takes us to an already visited cell $V$. Since each cell has exactly two neighbours with the same color as $C$, the cell $V$ must be $S$. So, for each cell $S$ of the board we have a path $P_S$, constructed as above. Notice that if two such paths have a common cell, then they must coincide, because these paths have the same pairs of 'consecutive cells' (they may have different starting points and the order in which the consecutive cells are covered can be reversed). Therefore, the number of the board's cells is the sum of the numbers of the cells of the different paths $P_S$.\n\nColor now the board chess-wise. Each pair of consecutive cells of a path is made of a black and a white cell, therefore each path must have an even number of cells. This shows that the number $n^2$ of the board's cells is the sum of some even numbers, so $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with right angle at $A$, and let $AD$ be its altitude from $A$ to $BC$. On the ray $[AD$, take points $E$ and $H$, such that $AE = AC$ and $AH = AB$. Construct squares $AEFG$ and $AHJI$, such that $C$ lies inside $AEFG$, and $B$ lies inside $AHJI$. Let $K = AC \\cap EG$, $L = AB \\cap IH$, $N = IL \\cap GK$, $M = IB \\cap GC$. Prove that:\n\na) $LK \\parallel BC$\n\nb) the points $A, N, M$ are collinear.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We have $\\triangle AGK \\sim \\triangle AHL$ (1), because $\\angle AGK = \\angle AHL = 45^\\circ$, and $\\angle GAK = 90^\\circ - \\angle CAD = \\angle HAL$. Thus, $\\frac{AK}{AL} = \\frac{AG}{AH}$. Since $AG = AE = AC$ and $AH = AB$, it follows that $\\frac{AK}{AC} = \\frac{AL}{AB}$, therefore $LK \\parallel BC$.\n\n\nb) From (1), we have $\\angle AKG = \\angle ALH$, so $ALNK$ is cyclic. Therefore, $\\angle NAK = \\angle NLK = \\angle NIG = 45^\\circ$, which shows that $AN$ is the bisector of the angle $BAC$. The bisector from $B$ in the triangle $ABC$ is parallel to the bisector of the angle $BAI$, which is also an altitude in the triangle $BAI$. Similarly, $CM$ is the external bisector from $C$ in the triangle $ABC$. Thus, $M$ is the center of the excircle relative to $A$ of the triangle $ABC$, so $M \\in AN$.\n\n\nAlternative solution for b).\n\nLet $O = KL \\cap AN$. Since $A, G, I$ are collinear and $DA \\perp GI, DA \\perp BC$, we get $GI \\parallel BC \\parallel KL$. Hence, $\\frac{OK}{OL} = \\frac{AG}{AI} = \\frac{AG}{AH} = \\frac{AK}{AL}$, therefore, by the converse of the angle bisector theorem, $AO$ is the bisector of the angle $LAK$. Let $P = AM \\cap BC$. Then $\\frac{CP}{PB} = \\frac{AG}{AI} = \\frac{AC}{AB}$, so $AP$ is the bisector of the angle $BAC$, hence both $M$ and $N$ lie on the angle bisector of the angle $BAC$, proving the collinearity of $A, M, N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all matrices $A$, $B$, $C \\in \\mathcal{M}_2(\\mathbb{R})$ such that $A = BC - CB$, $B = CA - AC$, and $C = AB - BA$.",
    "options": [],
    "answer": "A = B = C = O_2",
    "solution": "If one of the matrices $A$, $B$, $C$ is null, then all the matrices are null. Suppose that there are three matrices $A$, $B$, $C \\in \\mathcal{M}_2(\\mathbb{R}) \\setminus \\{O_2\\}$ which satisfy the equations from the statement. We have $\\text{tr}(A) = \\text{tr}(BC - CB) = 0$. Similarly, $\\text{tr}(B) = \\text{tr}(C) = 0$. From the equation $A^2 - \\text{tr}(A)A + \\det(A)I_2 = O_2$, we obtain $A^2 = aI_2$, where $a = -\\det(A)$. Similarly, we have $B^2 = bI_2$ and $C^2 = cI_2$, with $b = -\\det(B)$ and $c = -\\det(C)$. Multiplying the equation $A = BC - CB$ on the left and on the right by $B$, we obtain the relations $BA = B^2C - BCB = bC - BCB$ and $AB = BCB - CB^2 = BCB - bC$, respectively. Then $AB + BA = O_2$. Thus, the equation $C = AB - BA$ leads to $C = 2AB$. Similarly, $A = 2BC$ and $B = 2CA$. Then $A = -2CB = -2C(2CA) = -4C^2A = -4cA$. Similarly, $B = -4aB$ and $C = -4bC$.\n\nSince the matrices $A$, $B$ and $C$ are assumed to be non-null, we obtain $a = b = c = -1/4$. So, we have $A = \\begin{pmatrix} x & y \\\\ z & -x \\end{pmatrix}$ and $B = \\begin{pmatrix} s & t \\\\ u & -s \\end{pmatrix}$, with $x, y, z, s, t, u \\in \\mathbb{R}$, such that $x^2 + yz = -1/4$ and $s^2 + tu = -1/4$. From the relation $AB + BA = O_2$, we find $2xs = -(yu + zt)$. Thus, $4x^2s^2 = (yu + zt)^2 = (yu - zt)^2 + 4yztu \\ge 4(yz)(tu) = 4\\left(x^2 + \\frac{1}{4}\\right)\\left(s^2 + \\frac{1}{4}\\right)$. But $x^2s^2 < \\left(x^2 + \\frac{1}{4}\\right)\\left(s^2 + \\frac{1}{4}\\right)$, for all $x, s \\in \\mathbb{R}$. Contradiction.\n\nTherefore, the unique solution of the given system is $A = B = C = O_2$.\nLet $A$, $B$, $C \\in M_2(\\mathbb{R})$ be three matrices that satisfy the given system. We have $\\text{tr}(A) = \\text{tr}(BC - CB) = 0$. We also deduce $\\text{tr}(B) = \\text{tr}(C) = 0$. Then $A = \\begin{pmatrix} a_1 & a_2 \\\\ a_3 & -a_1 \\end{pmatrix}$, $B = \\begin{pmatrix} b_1 & b_2 \\\\ b_3 & -b_1 \\end{pmatrix}$, $C = \\begin{pmatrix} c_1 & c_2 \\\\ c_3 & -c_1 \\end{pmatrix}$, $a_i, b_i, c_i \\in \\mathbb{R}$, for $i = \\overline{1, 3}$. We obtain $BC - CB = \\begin{pmatrix} b_2c_3 - b_3c_2 & 2(b_1c_2 - b_2c_1) \\\\ 2(b_3c_1 - b_1c_3) & -(b_2c_3 - b_3c_2) \\end{pmatrix}$. Hence, we have $a_2 = 2(b_1c_2 - b_2c_1)$ and $a_2^2 = 2a_2(b_1c_2 - b_2c_1) = 2a_2b_1c_2 - 2a_2b_2c_1$. Similarly, we obtain $b_2^2 = 2b_2c_1a_2 - 2b_2c_2a_1$ and $c_2^2 = 2c_2a_1b_2 - 2c_2a_2b_1$. Then $a_2^2 + b_2^2 + c_2^2 = 0$. Since $a_2, b_2, c_2 \\in \\mathbb{R}$, we conclude $a_2 = b_2 = c_2 = 0$. Analogously, we can show that $a_3 = b_3 = c_3 = 0$. We obtain $a_1 = b_2c_3 - b_3c_2 = 0$. Similarly, $b_1 = c_1 = 0$.\n\nIn conclusion, the unique solution of the given system is $A = B = C = O_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39496,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a positive integer *special* if the sum of its (decimal) digits and the sum of the digits of its successor are divisible by $11$.\n\na) Find the last five digits of a special number.\n\nb) Prove that there are infinitely many special numbers.",
    "options": [],
    "answer": "a) 99999. b) There are infinitely many, for example all numbers of the form n = 2·10^(q+5) + 899999 for any integer q ≥ 1.",
    "solution": "a) Denote $s(m)$ the sum of the digits of a positive integer $m$.\nLet $n = \\overline{a_k a_{k-1} \\dots a_2 a_1}$ be a special number with $k \\ge 1$ digits. The statement says that $11 \\mid s(n)$ and $11 \\mid s(n+1)$.\nIf $a_1 \\le 8$, then $n+1 = \\overline{a_k a_{k-1} \\dots a_2 (a_1+1)}$, hence $s(n+1) = s(n)+1$. So, if $s(n) = M_{11}$, then $s(n+1) = M_{11}+1$, false. Consequently $a_1 = 9$.\nSuppose that the last $p \\ge 1$ digits of $n$ are $9$ and the $(p+1)^{th}$ is different from $9$. Then $n = \\overline{a_k a_{k-1} \\dots a_{p+1} 99 \\dots 9}$ and $n+1 = \\overline{a_k a_{k-1} \\dots (a_{p+1} + 1) 00 \\dots 0}$, so $s(n) = a_k + a_{k-1} + \\dots + a_{p+1} + 9p$ and $s(n+1) = a_k + a_{k-1} + \\dots + a_{p+1} + 1$.\nTherefore $s(n) = s(n+1) + 9p - 1$, showing that $9p-1$ is a multiple of $11$. The smallest such $p$ is $p=5$, hence the last five digits of a special number are nines.\n\nb) From a) follows that the special numbers are of the form $n = m \\cdot 10^5 + 99\\,999$, where $m$ is conveniently chosen. Since $s(n) = s(m) + 45$ and $45 = M_{11} + 1$, it is enough to find $m$ with the last digit different from $9$ and with the property $s(m) = 11 \\cdot t - 1$, where $t$ is a positive integer. In particular, if $t=1$, the numbers of the form $m = 2 \\cdot 10^q + 8$, with $q \\ge 1$, have the sum of their digits $10$.\nThe above show that, for each $q \\ge 1$, the number $n = 2 \\cdot 10^{q+5} + 899\\,999 = \\underbrace{200\\dots0}_{q-1 \\text{ digits}} 899\\,999$ is special, and there are infinitely many such numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ are such that $A + B = AB + BA$. Prove that:\na) if $n$ is odd, then $\\det(AB - BA) = 0$;\nb) if $\\det(A) \\ne \\det(B)$, then $\\det(AB - BA) = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Define $C = 2A - I_n$ and $D = 2B - I_n$. We have $CD - I_n = -(DC - I_n) = 2(AB - BA)$. As $n$ is odd, $\\det(CD - I_n) = -\\det(DC - I_n)$.\nAs for any two square matrices $X, Y$, we have $\\det(XY - I_n) = \\det(YX - I_n)$, in our case we get $\\det(CD - I_n) = -\\det(CD - I_n)$, that is $\\det(CD - I_n) = 0$. Consequently $\\det(AB - BA) = 0$.\n\nb) Define $E = AB - BA$. Suppose $\\det(E) \\neq 0$. We have $AE + EA = A^2B - BA^2 = A(A + B - BA) - (A + B - AB)A = AB - BA = E$. As $E$ is invertible, $E^{-1}AE + A = I_n$. Using the property $\\operatorname{tr}(E^{-1}AE) = \\operatorname{tr}(A)$, we get $\\operatorname{tr}(A) = \\frac{n}{2}$. By symmetry, $\\operatorname{tr}(B) = \\frac{n}{2}$, implying $\\operatorname{tr}(A) = \\operatorname{tr}(B)$, in contradiction with the condition b). That is $\\det(AB - BA) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39498,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a point $O$ inside the square $ABCD$ the perpendicular line $OS$ is raised to the plane of the square. Let $M, N, P, Q$ be projections of point $O$ onto the planes $(SAB), (SBC), (SCD)$, respectively $(SDA)$. Prove that the points $M, N, P, Q$ are coplanar if and only if $O$ lies on one of the diagonals of the square.\n\nFlorin Bojor\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We assume that $O$ lies, for example, on the diagonal $AC$. Let $OE \\perp AB$, $E \\in AB$ and $OF \\perp AD$, $F \\in AD$. Then we have successively $OE = OF$, $\\triangle SOE \\equiv \\triangle SOF$ (C.C.), $SE = SF$. Then $M \\in SE$ and $OM \\perp SF$, $Q \\in SF$ and $OQ \\perp SF$, $\\triangle SOM \\equiv \\triangle SOQ$, $SM = SQ$, $\\frac{SM}{SE} = \\frac{SQ}{SF}$, $QM \\parallel EF$, so $QM \\parallel BD$. Analogously we get $NP \\parallel BD$, so $QM \\parallel NP$, which means that points $M, N, P, Q$ are coplanar.\n\nConversely, assume that $M, N, P, Q$ are coplanar in a plane $\\alpha$. Take $OG \\perp CD$, $G \\in CD$ and $OH \\perp BC$, $H \\in BC$. Then the points $E, O, G$ are collinear, so the lines $SE, SO, SG$ are coplanar. It follows from this that the lines $SO$ and $MP$ are coplanar, and $SO \\cap MP$ is the same as $SO \\cap \\alpha$. Similarly we'll get $NQ \\cap SO$ is the same as $SO \\cap \\alpha$, so the lines $MP$ and $NQ$ intersect $SO$ at the same point $R$.\n\n![](attached_image_1.png)\n\nWe calculate the ratio in which point $R$ divides $OS$, in terms of $OS, OE$ and $OG$.\nTake $MT \\perp OS$, $PU \\perp OS$, $U, T \\in OS$. From the cyclic quadrilateral $MOPS$ we get $\\triangle MSR \\sim \\triangle OPR$, so $\\frac{MS}{OP} = \\frac{MR}{OR} = \\frac{SR}{PR}$, hence $\\frac{MS^2}{OP^2} = \\frac{SR}{OR} \\cdot \\frac{MR}{PR} = \\frac{SR}{OR} \\cdot \\frac{MT}{PU}$. It follows $\\frac{SR}{OR} = \\frac{PU}{MT} \\cdot \\frac{MS^2}{OP^2} = \\frac{PS \\cdot PO \\cdot MS^2}{MO \\cdot MS \\cdot PS \\cdot PG} = \\frac{PO}{PG} \\cdot \\frac{MS}{MO}$. But $\\frac{PO}{PG} = \\tan \\angle SGO = \\frac{SO}{OG}$ and $\\frac{MS}{MO} = \\cot \\angle MSO = \\frac{SO}{OE}$, so $\\frac{SR}{OR} = \\frac{SO^2}{OE \\cdot OG}$.\n\nSince line $NQ$ intersects $OS$ also in $R$, we obtain $OE \\cdot OG = OF \\cdot OH$. On the other hand we have $OE + OG = l = OF + OH$, where $l = AB = AD$. It follows from this that $OE(l - OE) = OF(l - OF)$, then $(OE - OF)(l - OE - OF) = 0$, so $(OE - OF)(OH - OE) = 0$. Thus $OE = OF$ or $OE = OH$, which implies $O \\in AC$ or $O \\in BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD A'B'C'D'$ be a cube. On the segments $BC$ and $DD'$ we take the points $M$ and $N$ respectively, such that $BM = DN$. Prove that line $A'M$ is perpendicular to plane $(AB'N)$.\n\nCătălin Barbu\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$BC \\perp (ABB')$ and $AB' \\subset (ABB')$ yields $AB' \\perp BC$. Since $AB' \\perp A'B$ (diagonals of the square $ABB'A'$), we get $AB' \\perp (A'BC)$. Because $A'M \\subset (A'BC)$, it follows that $A'M \\perp AB'$. (1)\n\nLet $E \\in (AD)$ be such that $AE = BM$. Then $ABME$ is a rectangle, so $AB \\parallel ME$. Because $AB \\perp (ADA')$ and $AN \\subset (ADA')$, it follows that $AN \\perp ME$. (2)\n\nFrom $\\triangle A'AE \\equiv \\triangle ADN$ (L.L.) one gets $\\angle DAN = \\angle AA'E$, whence $\\angle AA'E + \\angle A'AN = \\angle DAN + \\angle A'AN = 90^\\circ$. Thus, $AN \\perp A'E$. Using now relation (2) we get $AN \\perp (A'EM)$, so $AN \\perp A'M$. Taking into account this last relation and (1), we conclude that $A'M \\perp (AB'N)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)_{n \\ge 1}$ be an increasing unbounded sequence of natural numbers such that $x_1 = 1$ and $x_{n+1} \\le 2x_n$ for all $n \\ge 1$. Prove that every nonzero natural number can be written as a finite sum of pairwise distinct terms of the sequence $(x_n)_{n \\ge 1}$.\n\n*Note:* Two terms $x_i$ and $x_j$ of the sequence $(x_n)_{n \\ge 1}$ are said to be distinct if $i \\neq j$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be a nonzero natural number such that $1 \\le k < 2x_n$ for some natural number $n \\ge 1$. We will prove by induction that we can write $k = \\sum_{i=1}^{n} \\varepsilon_i x_i$ with $\\varepsilon_i \\in \\{0, 1\\}$ for $i = 1, \\dots, n$. Since the sequence $(x_n)_{n \\ge 1}$ is unbounded, we can cover all positive integers using this construction.\n\nThe statement is clearly true for $n = 1$. Assume it is true for $n = N$. We will now show that for any $1 \\le k < 2x_{N+1}$, we can write $k = \\sum_{i=1}^{N+1} \\alpha_i x_i$ with $\\alpha_i \\in \\{0, 1\\}$, $i \\in \\{1, \\dots, N+1\\}$.\n\nIf $x_N = x_{N+1}$, then by the induction hypothesis, $k = \\sum_{i=1}^{N} \\varepsilon_i x_i$ and we can take $\\alpha_i = \\varepsilon_i$ for $i \\in \\{1, \\dots, N\\}$ and $\\alpha_{N+1} = 0$.\n\nIf $x_N < x_{N+1}$, then it is sufficient to consider values of $k$ for which $2x_N \\le k < 2x_{N+1}$, since the case $k < 2x_N$ is already covered by the induction hypothesis. In this case, we have $k - x_{N+1} \\ge 2x_N - x_{N+1} \\ge 0$ by the given condition. We distinguish two cases:\n\n**Case 1.** If $k - x_{N+1} = 0$, the statement is clearly true.\n\n**Case 2.** If $k - x_{N+1} > 0$, we use again the hypothesis and observe that\n$$\n0 < k - x_{N+1} < 2x_{N+1} - x_{N+1} \\le 2x_N.\n$$\nBy the induction hypothesis, we can write $k - x_{N+1} = \\sum_{i=1}^{N} \\varepsilon_i x_i$ with $\\varepsilon_i \\in \\{0, 1\\}$, $i \\in \\{1, \\dots, N\\}$. Adding $x_{N+1}$ to both sides and setting $\\alpha_i = \\varepsilon_i$ for $i \\in \\{1, \\dots, N\\}$ and $\\alpha_{N+1} = 1$, the induction step is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon with $\\angle A \\equiv \\angle C \\equiv \\angle E$ and $\\angle B \\equiv \\angle D \\equiv \\angle F$, such that there is a point $P$ in its interior that is equidistant from the sides $AB$, $CD$ and $EF$. If $G_1 \\neq G_2$ are the centroids of triangles $ACE$ and $BDF$, respectively, prove that $\\angle G_1 P G_2 = 60^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $\\angle A = \\angle C = \\angle E = \\alpha$ and $\\angle B = \\angle D = \\angle F = \\beta$, then the sum of the measures of the angles of the hexagon $ABCDEF$ is $3\\alpha + 3\\beta = 720^\\circ$, so $\\alpha + \\beta = 240^\\circ$. Since $\\angle B + \\angle C = \\alpha + \\beta = 240^\\circ$, the lines $AB$ and $CD$ will intersect at a point located outside the hexagon. Denote $\\{X\\} = AB \\cap CD$, $\\{Y\\} = CD \\cap EF$, $\\{Z\\} = EF \\cap AB$. The point $P$, equidistant from the sides $AB$, $CD$ and $EF$ of the hexagon, is the center of the equilateral triangle $XYZ$.\n\nIn the complex plane we consider a rectangular coordinate system with the origin at $P$. From $\\triangle BCX \\sim \\triangle DEY \\sim \\triangle FAZ$, there is $k \\in \\mathbb{R}$, $k > 0$, such that $\\frac{c-x}{b-x} = \\frac{e-y}{d-y} = \\frac{a-z}{f-z} = k \\cdot \\varepsilon$, where $\\varepsilon = \\cos \\frac{\\pi}{3} \\pm i \\sin \\frac{\\pi}{3}$. We obtain $c-x = (b-x)k \\cdot \\varepsilon$, $e-y = (d-y)k \\cdot \\varepsilon$, $a-z = (f-z)k \\cdot \\varepsilon$.\n\nSince $p = \\frac{x+y+z}{3} = 0$, by addition, we obtain $c+e+a = (b+d+f)k \\cdot \\varepsilon \\Leftrightarrow g_1 = g_2 \\cdot k \\cdot \\varepsilon$. From $G_1 \\neq G_2$ it follows that $g_1$ and $g_2$ cannot be $0$, so $G_1$, $G_2$, $P$ are three distinct points. Therefore $\\frac{g_1-p}{g_2-p} = k \\cdot \\varepsilon$, or $\\angle G_1 P G_2 = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $DEF$ be triangles with the same circumcircle centred at $O$ and the same orthocentre $H \\neq O$. The Simson lines of $D, E, F$ with respect to triangle $ABC$ form a non-degenerate triangle $\\triangle$. Prove that the orthocentre of $\\triangle$ lies on the circle on diameter $OH$.\n\n**Note.** Assume that $A, F, B, D, C, E$ form, in order around the circle, the vertex set of a non-degenerate convex hexagon.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell_D, \\ell_E, \\ell_F$ be the Simson lines of $D, E, F$ with respect to triangle $ABC$ and let $X = \\ell_E \\cap \\ell_F$, $Y = \\ell_F \\cap \\ell_D$ and $Z = \\ell_D \\cap \\ell_E$. Let $D', E', F'$ be the midpoints of $HD, HE, HF$, respectively, and let $H'$ be the orthocentre of $\\triangle$.\n\nNote that $\\ell_D$ passes through $D'$. This is a well-known fact about Simson lines. For the sake of completeness we provide a proof: Let $D_B$ and $D_C$ be the reflections of $D$ in $AB$ and $AC$, respectively. As the quadrangles $AHBD_B$ and $AHCD_C$ are both cyclic, $\\angle AHD_B + \\angle AHD_C = \\angle ABD_B + \\angle ACD_C = \\angle ABD + \\angle ACD = 180^\\circ$, so $D_B, H, D_C$ are collinear. A factor $\\frac{1}{2}$ homothety from $D$ settles the case.\n\n**Claim 1.** The quadrangles $HE'XF', HF'YD', HD'ZE'$ are cyclic.\n*Proof.* Let $r_E$ and $r_F$ be the rays from $A$ isogonal in $\\angle BAC$ to rays $AE$ and $AF$, respectively. Note that $r_E \\perp \\ell_E$ and $r_F \\perp \\ell_F$ to write $\\angle E'XF' = \\angle (r_E, r_F) = 180^\\circ - \\angle EAF = \\angle EDF = 180^\\circ - \\angle E'HF'$. Hence $HE'XF'$ is cyclic. Similarly, $HF'YD'$ and $HD'ZE'$ are both cyclic.\n\n**Claim 2.** $H$ is the circumcentre of $\\triangle$.\n*Proof.* As $H$ is also the orthocentre of triangle $D'E'F'$, it follows that $\\angle HYZ = \\angle HF'D' = \\angle HE'D' = \\angle HZY$, so $HY = HZ$. Similarly, $HX = HY$ and the claim follows.\n\nAs $D', E', F'$ lie on the nine-point circle of triangle $DEF$, the midpoint $O'$ of $OH$ is the circumcentre of triangle $D'E'F'$. Proving that $H'$ lies on the circle on diameter $OH$ is equivalent to proving $O'H' = O'H$. The conclusion is then a consequence of Claim 3 below applied to $\\triangle$ and points $D', E', F'$.\n\n**Claim 3.** Let $ABC$ be a triangle with orthocentre $H$ and circumcentre $O \\neq H$. Let $X, Y, Z$ be points on the sides $BC, CA, AB$, respectively, such that circles $AYZ, BZX, CXY$ are concurrent at $O$. Then the circumcentre of triangle $XYZ$ lies on the perpendicular bisectrix of $OH$.\n\n*Proof.* Note that triangles $YOZ$ and $BHC$ are similar and the like. Then so are $XYZO$ and $ABCH$. Letting $O'$ be the circumcentre of triangle $XYZ$, it follows that $XYZOO'$ and $ABCHO$ are also similar. In particular, triangles $XOO'$ and $AHO$ are similar. Varying $X$ along the line $BC$ shows that $O'$ is the image of $X$ under a spiral-similarity from $O$. Consequently, $O'$ lies on some fixed line $\\ell$.\n\nAs $AH$ and $BC$ are perpendicular, the angle formed by $\\ell$ and $OH$ is equal to the angle formed by $OH$ and the image of $BC$ under a rotation through $\\angle AHO = 90^\\circ$. Letting $X, Y, Z$ be the midpoints of $BC, CA, AB$, respectively, it follows that $\\ell$ passes through the midpoint of $OH$. Combining these two observations, it follows that $\\ell$ is perpendicular bisectrix of $OH$, as desired. This ends the proof and completes the solution.\nAs usual, the complex coordinate of a point in the plane is denoted by the corresponding lower case letter. Use the notation in Solution 1 and let the circumcircle of triangles $ABC$ and $DEF$ be centred at the origin and have unit radius. Thus, $a, b, c, d, e, f$ all have a unit absolute value, i.e., $a\\bar{a} = b\\bar{b} = \\ldots = f\\bar{f} = 1$ and\n$$\nh = a + b + c = d + e + f = \\frac{1}{2}(a + b + c + d + e + f).\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all natural numbers $n$ such that $2^n - n^2 + 1$ is a perfect square.",
    "options": [],
    "answer": "n = 2, 3, 4",
    "solution": "We check by direct computation that $n = 2, 3, 4$ are solutions. We will prove that these are the only ones. We may assume $n \\ge 5$ after checking all smaller cases. We distinguish two cases.\n\n(I) If $n$ is even, write $n = 2t$ (then $t \\ge 3$). One can prove that\n$$\n(2^t - 1)^2 < 2^n - n^2 + 1 < (2^t)^2 = 2^n,\n$$\nwhich means that in this case, $2^n - n^2 + 1$ cannot be a perfect square. Indeed, the second inequality is obvious, and the first can be proven easily by induction for $t \\ge 7$, with the base cases $t = 3, 4, 5, 6$ verified directly.\n\n(II) If $n$ is odd, let $p$ be the largest (necessarily odd) prime dividing $n$. If $p = 3$, then $n = 3^k$, and since $n \\ge 5$, it follows that $k \\ge 2$, so in particular $9 \\mid n$. But then $2^9 + 1$ divides $2^n + 1$, hence $19 \\mid 2^n + 1$ (since $2^9 + 1 = 512 + 1 = 513 = 27 \\cdot 19$).\nIf $2^n - n^2 + 1 = a^2$, then $2^n + 1 = n^2 + a^2$. Since $19 \\mid 2^n + 1$, and $19 \\equiv 3 \\pmod 4$, it follows that $19 \\mid n^2 + a^2$. But then $19 \\mid n$, which contradicts the assumption that $n = 3^k$.\n\nTherefore, we must have $p > 3$. Since $2^p + 1 \\equiv 1 \\pmod 4$, it follows that $3 \\mid 2^p + 1$, but $3^2$ does not divide $2^p + 1$ (this follows immediately from factoring $2^p + 1$ or applying LTE for $p = 3$). This implies that there exists a prime $q > 3$, with $q \\equiv 3 \\pmod 4$ such that $q \\mid 2^p + 1$.\nUsing the statement below, we obtain that $q > p$. Since $q \\mid 2^p + 1 \\mid 2^n + 1$, and assuming $2^n - n^2 + 1 = a^2$, we get $2^n + 1 = n^2 + a^2$. By the same reasoning as above, $q \\mid n$, which contradicts the maximality of $p$ as the largest prime dividing $n$.\nThis completes the proof that for $n \\ge 5$, the expression $2^n - n^2 + 1$ is not a perfect square.\n\n*Claim*. If $p$ is a prime with $p \\ge 5$, then any odd prime factor $q$ of $2^p + 1$, different from 3, satisfies $q > p$.\n*Proof*. Let $q > 3$ be a prime dividing $2^p + 1$. Then $2^p \\equiv -1 \\pmod q$, so the order of 2 modulo $q$ is $2p$. Indeed, the order of 2 (mod $q$) must divide $2p$, so it could be 1, 2, $p$, or $2p$. It cannot be 1 or $p$ (as $2^p \\equiv -1 \\not\\equiv 1 \\pmod q$), and if it were 2, then $2^2 \\equiv 1 \\pmod q$ implies $q = 3$, contradicting $q > 3$.\nSince $(2, q) = 1$, by Fermat's Little Theorem we have $2^{q-1} \\equiv 1 \\pmod q$. But the order of 2 modulo $q$ is $2p$, so $2p \\mid q - 1$, which implies $p < q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a, b, c$ such that the numbers $ab + c$, $bc + a$, and $ca + b$ are powers of 2.",
    "options": [],
    "answer": "(1, 1, 2^x − 1) and (1, 2^x − 1, 2^x + 1) for any positive integer x",
    "solution": "We will prove that there are only two families of solutions:\n$$\n(1, 1, 2^x - 1), \\quad (1, 2^x - 1, 2^x + 1),\n$$\nwhere $x$ is a positive integer.\n\n*Step I:* $a, b, c$ are odd and pairwise coprime in pairs.\nIt is easy to see that\n$$\na \\equiv b \\equiv c \\pmod{2}.\n$$\nIf $a, b, c$ are all even, denote by $d, e, f$ the exponents of 2 in their prime factorization, respectively, with $1 \\le d \\le e \\le f$. Then the exponent of 2 in $bc+a$ is $d$. Since $bc+a$ is a power of 2, we get\n$$\nbc + a = 2^d,\n$$\nwhich is impossible because $2^d \\le a$. Thus, $a, b, c$ must be odd numbers.\nIf $a, b$ have a common odd prime factor $p$, then $p$ divides $ac + b$, so $ac + b$ is divisible by $p$ and thus cannot be a power of 2. Similarly, one shows $(a, c) = 1$ and $(b, c) = 1$, which completes Step I.\n\nHence, we can assume $a \\le b \\le c$, where $a, b, c$ are odd positive integers pairwise coprime such that\n$$\nab + c, \\quad ac + b, \\quad bc + a\n$$\nare powers of 2. Since $a \\le b \\le c$, we have\n$$\nab + c \\le ac + b \\le bc + a.\n$$\n*Step II:* $a = 1$. Suppose, by contradiction, that $1 < a$. Then $1 < a < b < c$, and\n$$\nab + c = 2^k, \\quad ac + b = 2^m, \\quad bc + a = 2^n,\n$$\nwith $3 \\le k < m < n$ (since $2^k \\ge 3 + 5 = 8$).\nFrom the above equalities, we get the congruences:\n$$\nab \\equiv -c \\pmod{2^k}, \\quad ac \\equiv -b \\pmod{2^k}, \\quad bc \\equiv -a \\pmod{2^k}.\n$$\nMultiplying these congruences and noting that $a, b, c$ are odd, we obtain\n$$\nabc \\equiv -1 \\pmod{2^k}\n$$\nand\n$$\na^2 \\equiv b^2 \\equiv c^2 \\equiv 1 \\pmod{2^k}\n$$\n(by multiplying the first congruence by $c$, the second by $b$, and the third by $a$).\n\nSince for any odd natural number $x$,\n$$\n\\gcd(x - 1, x + 1) = 2,\n$$\nwe deduce that\n$$\na \\equiv \\pm 1 \\pmod{2^{k-1}}, \\quad b \\equiv \\pm 1 \\pmod{2^{k-1}}, \\quad c \\equiv \\pm 1 \\pmod{2^{k-1}}.\n$$\nFrom these congruences, it follows that\n$$\na, b, c \\ge 2^{k-1} - 1.\n$$\nThis inequality implies\n$$\n2^k = ab + c \\ge (2^{k-1} - 1)^2 + 2^{k-1} - 1 = 2^{2k-2} - 2^{k-1},\n$$\nwhich means $2 \\ge 2^{k-1} - 1 \\ge 3$, a contradiction since $k > 3$. Therefore, the assumption is false and $a = 1$.\n\n**Step III: the solutions.** If $b = 1$, then clearly $c = 2^x - 1$. Otherwise, assume $a = 1$ and $1 < b < c$. From Step II, we know that $c > b \\ge 2^{k-1} - 1$. Since $b + c = 2^k$ and $b, c$ are odd, the only possibility is $b = 2^{k-1} - 1$, $c = 2^{k-1} + 1$.\nHence, the only solutions are: $(1, 1, 2^x - 1)$ and $(1, 2^x - 1, 2^x + 1)$, $x \\in \\mathbb{N}^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $k$ for which there exists a set $A$ of 10 points in the plane, no three of which are collinear, having the following property: it is possible to color the 45 segments with endpoints in $A$ using $k$ colors so that any two segments of the same color intersect either in their interior or at an endpoint.",
    "options": [],
    "answer": "6",
    "solution": "**Lemma.** Given 5 points in the plane, no three collinear, it is impossible to color the 10 segments determined by them with only 2 colors so that any two segments of the same color intersect.\n\nFirst, we show that any 4 points determine a monochromatic triangle. The 4 points form either a convex or concave quadrilateral. Label the points $A$, $B$, $C$, $D$ so that $AB$, $CD$ and $AD$, $BC$ are pairs of non-intersecting segments. Without loss of generality, assume $AB$, $BC$ are red and $CD$, $AD$ are blue. The color of $AC$ (red or blue) then determines a monochromatic triangle.\n\nNext, we show that there cannot be two distinct monochromatic triangles of the same color. Suppose there are two red triangles, one being $ABC$. Then any red segment must have an endpoint in $\\{A, B, C\\}$. Without loss of generality, assume the other red triangle is $ABD$. The coloring rule then implies that the pairs of segments $AD$, $BC$ and $AC$, $BD$ intersect, which is impossible.\n\nTherefore, since two distinct monochromatic triangles of the same color cannot exist, the 5 monochromatic triangles determined by the 5 subsets of 4 points would have to be identical, i.e., contain the same triple of points for every 4-point subset, which is a contradiction.\n\nStep 1.\nConsider a regular 10-gon and color its vertices alternately with two colors: red and blue. Let $v$ be a red vertex. Consider the 9 segments containing $v$, together with the diagonal formed by the two adjacent blue vertices. Any two of these segments intersect (either in their interior or at endpoints), so they can be colored with the same color. Thus, we can cover all segments except the diagonals of the pentagon formed by the blue vertices using 5 colors. Since those diagonals intersect pairwise, they require a sixth color.\n\nStep 2.\nWe prove that $k \\ge 6$ using the lemma. Let $A$ be a set of points as in the statement, and consider a line $l_0$ not parallel to any of the 45 segments determined by the points of $A$. Ordering the distances of the 10 points from $l_0$, we construct a line $l$ parallel to $l_0$ so that there are 5 points on each side of $l$. Look at the segments formed by the 5 points on the same side of $l$. By the lemma, at least 3 colors are needed to color these segments so that any two segments of the same color intersect. On the other hand, no segment on one side of $l$ can have the same color as a segment on the other side since they do not intersect. Thus, at least 6 colors are needed to color all 45 segments, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the sets $S$ of positive integers satisfying the following two conditions:\n\na) For any positive integers $a, b, c$, if $ab + bc + ca$ is in $S$, then so are $a + b + c$ and $abc$;\n\nb) The set $S$ contains an integer $N \\ge 160$ such that $N - 2$ is not divisible by 4.",
    "options": [],
    "answer": "S is the set of all positive integers.",
    "solution": "We will prove that $S$ is the set of all positive integers. The argument hinges on the three facts below:\n\n(1) The set $S$ contains an integer $M \\ge 40$ divisible by 4.\n\n(2) If $4k$ belongs to $S$ for some integer $k \\ge 2$, then so does $4m$ for all positive integers $m < k$.\n\n(3) The set $S$ contains $4k$ for all integers $k \\ge 10$.\n\nAssume the three for the moment and argue as follows: By (1) and (2), $S$ contains all positive multiples of 4 at most 40, and by (3) it contains all multiples of 4 at least 40, so $S$ contains all positive multiples of 4.\n\nLet $a = 3$ and let $b = c = 1$. As 7 is in $S$, so is 3, by (a). Repeat the argument for $a = b = c = 1$ to deduce that $S$ contains 1.\n\nFinally, let $a = 2$ and let again $b = c = 1$. As 5 lies in $S$, so does 2. Combining with the previous paragraphs, it follows that $S$ exhausts all positive integers, as stated.\n\nProof of (1):\n\nIf $N$ is divisible by 4, choose $M = N$. If $N = 4k + 1$, set $a = 2k$ and $b = c = 1$ in (a) to deduce that $2k$ and $2k + 2$ are both in $S$. As $N \\ge 160$, the numbers $2k$ and $2k + 2$ are both at least $80 > 40$. Note that exactly one of $2k$ and $2k + 2$ is divisible by 4 and let $M$ be that number.\n\nProof of (2):\n\nLet $b = c = 2$. By (a), if $4a + 4$ is in $S$, then so is $4a$. Beginning with $M$ provided by (1), statement (2) now follows by backward recursion.\n\nProof of (3):\n\nWe first prove that $S$ contains an integer $P \\ge 40$ divisible by 4. By (1), $S$ contains an integer $M \\ge 40$ divisible by 4. If $M$ is divisible by 8, let $P = M$. Otherwise, $M \\ge 44$ and $M - 4$ is divisible by 8. By (2), $M - 4$ is in $S$, so $P = M - 4$ fits the bill.\n\nLet $b = c = 4$. By (a), if $8a + 16$ is in $S$, then so is $16a$. Note that $16a > 8a + 16$ for $a \\ge 3$. Thus, if $S$ contains an integer $k \\ge 40$ divisible by 8, then it also contains an integer $k' > k$ divisible by 8. Hence, starting with $P$, we can generate arbitrarily large multiples of 8 lying in $S$. Reference to (2) concludes the proof and completes the solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ denote the set of integers and let $S \\subset \\mathbb{Z}$ be the set of integers that are at least $10^{100}$. Fix a positive integer $c$. Determine all functions $f: S \\to \\mathbb{Z}$ satisfying\n$$\nf(xy + c) = f(x) + f(y) \\quad \\text{for all } x, y \\in S.\n$$",
    "options": [],
    "answer": "f(x) = 0 for all x in S",
    "solution": "Observe that if $x_1, y_1, x_2, y_2 \\in S$ with $x_1 y_1 = x_2 y_2$ then\n$$\nf(x_1) + f(y_1) = f(x_2) + f(y_2). \\quad (1)\n$$\n\nThis tells us that for $u, v, w \\in S$,\n$f(uv) + f(w) = f(u) + f(vw)$, so $f(uv) - f(u) - f(v) = f(vw) - f(w) - f(v)$.\nNotice the RHS is independent of $u$ so the same must be true of the LHS. By replicating the argument with $u$ and $v$ switched, we also see the LHS is independent of $v$ so in fact\n$$\nf(uv) - f(u) - f(v) = k \\quad \\text{for some constant } k \\in \\mathbb{Z} \\quad (2)\n$$\n\nUsing (1) again we have, for $y, z \\in S$\n$$\n\\begin{gathered} f(cz) + f(y) = f(z) + f(cy), \\\\ \\text{so } f(cy) - f(y) = l, \\quad \\text{for some constant } l \\in \\mathbb{Z}. \\end{gathered} \\quad (3)\n$$\n\nSetting $x = cz$ in the original functional equation for $z \\in S$ shows\n$$\n\\begin{gathered} f(c(yz + 1)) \\stackrel{(3)}{=} f(yz + 1) + l = f(cz) + f(y) \\stackrel{(3)}{=} f(z) + f(y) + l, \\\\ \\text{so } f(yz + 1) = f(y) + f(z). \\end{gathered}\n$$\n\nLet $x \\in S$ and set $y = x, z = x + 2$ in the above to get\n$$\n\\begin{gathered} f((x+1)^2) \\stackrel{(2)}{=} 2f(x+1) + k = f(x) + f(x+2) \\\\ \\Rightarrow f(x) + f(x+2) - 2f(x+1) = -k = \\text{constant} \\end{gathered}\n$$\nwhich forces $f$ to be a quadratic. By setting $x = y$ in the original functional equation and considering the degree of both sides, we see $f$ must be in fact be constant. The only constant function that satisfies the condition is $f \\equiv 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with the property that the circles having the segments $AB$ and $CD$ as diameters are tangent externally at a point $M$, different from the intersection point of the diagonals of the quadrilateral.\nLet $K$ be the second point of intersection of the circumcircle of triangle $AMC$ with the line determined by $M$ and the midpoint of segment $AB$, and let $L$ be the second point of intersection of the circumcircle of triangle $BMD$ with the line determined by $M$ and the midpoint of segment $CD$.\nProve that $|MK - ML| = |AB - CD|$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F$ and $E$ be the midpoints of chords $ML$ and $MK$, respectively. Then $O_3F \\perp ML$ and $O_4E \\perp MK$. Since $O_4O_1O_2O_3$ is cyclic with $\\angle O_4O_2O_3 = \\angle O_4O_1O_3 = 90^\\circ$, the midpoint $X$ of segment $O_3O_4$ is the center of the circumscribed circle of $O_4O_1O_2O_3$. Moreover, if $T$ denotes the foot of the perpendicular from $X$ onto $O_1O_2$, then $TO_1 = TO_2$.\n\nIn the right trapezoid $O_4EFO_3$, since $X$ is the midpoint of $O_3O_4$ and $XT \\perp EF$, it follows that $XT \\parallel EO_4 \\parallel FO_3$, so $XT$ is a midline and therefore $ET = TF$.\n\nConsequently, using the last two equalities we get\n$$\nTE - TO_1 = TF - TO_2 \\implies EO_1 = FO_2.\n$$\nThus, $|MK - ML| = |2ME - 2MF| = 2|ME - MF| = 2|EO_1 + O_1M - MO_2 - O_2F| = |2O_1M - 2O_2M|$.\nBut $AB = 2O_1M$, $CD = 2O_2M$, so substituting into the last equality we obtain $|MK - ML| = |AB - CD|$, which was to be proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a natural number, and let $\\mathcal{F}$ be a family consisting of at most $n$ distinct subsets of the set $\\{1, 2, \\dots, n\\}$ with the property that one can consider $n$ distinct points in the plane, labeled with the numbers $1, 2, \\dots, n$, then draw segments between some of these points such that, for any distinct numbers $i, j \\in \\{1, 2, \\dots, n\\}$, the points labeled $i$ and $j$ are connected by a segment if and only if the number $i$ belongs to exactly $j$ subsets in $\\mathcal{F}$. Find the maximum possible value of the sum of the number of elements of the sets in $\\mathcal{F}$.",
    "options": [],
    "answer": "n(n+1)/2",
    "solution": "For $k \\in \\{1, 2, \\dots, n\\}$, denote by $a_k = \\text{card}\\{F \\in \\mathcal{F} \\mid k \\in F\\}$.\n\nThen $a_k \\in \\{0, 1, \\dots, n\\}$, and the sum whose maximum we want to estimate is\n$$\ns = \\sum_{F \\in \\mathcal{F}} \\text{card}(F) = \\sum_{k=1}^{n} a_k.\n$$\nThe connection condition implies that if $a_k \\neq k$, then $a_{a_k} = k$, for any $k \\in \\{1, 2, \\dots, n\\}$. Conversely, any function satisfying this condition corresponds to a family $\\mathcal{F}$ verifying the hypothesis.\n\nTherefore, the numbers $a_k$ form pairs of two distinct numbers; otherwise, if $a_i = a_j = a$, then $a$ is connected both to $i$ and to $j$, which would imply $i = j$.\nWe deduce that\n$$\ns \\le n + (n-1) + \\dots + 1 = \\frac{n(n+1)}{2}.\n$$\nFinally, observe that the maximum $\\frac{n(n+1)}{2}$ is attained for any family $\\mathcal{F}$ for which $a_{a_k} = k$ for all $k \\in \\{1, 2, \\dots, n\\}$; for example, the family $\\mathcal{F}_0$ formed by $\\{1\\}, \\{1, 2\\}, \\dots, \\{1, 2, \\dots, n\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$ and let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $\\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centred at $A$ cross $\\Gamma$ again at $A'$ and $F$, respectively. Prove that $\\Gamma$, the circle on diameter $AA'$ and the circle $AFH$ are concurrent.\n\nRomania, Radu-Andrei Lecoiu",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the composition $\\iota$ of an inversion centered at $A$ and the reflection in the bisector $AM$ that swaps $B$ and $C$. Then $\\iota$ swaps the circle $\\Omega$ with the line $BC$, hence it swaps $O$ with the reflection $L$ of $A$ in $BC$. Hence $\\iota(\\Gamma)$ is the circle $\\Gamma^* = (BCL)$,\n\ni.e., the reflection of $\\Omega$ in $BC$ which passes through $H$. Let $S$ and $Q$ be the centers of $\\Gamma$ and $\\Gamma^*$, respectively; then $AM$ is the angle bisector of $\\angle QAS$.\n\nSince $\\iota$ swaps $\\Gamma$ and $\\Gamma^*$, they are seen from $A$ at the same angle, so there exists a rotational homothety $h$ centred at $A$ mapping $\\Gamma$ to $\\Gamma^*$; the angle of $h$ is $\\angle SAQ$. Notice that the rays $AH$ and $AF$ are obtained from $AO$ by reflections in $AM$ and $AS$, respectively, so $\\angle HAF = 2\\angle MAS = \\angle QAS$. This easily yields that $H = h(F)$. Hence the triangles $AHF$ and $AQS$ are similar, and $\\angle AHF = \\angle AQS$.\n\n![](attached_image_1.png)\n\nNow, let the circle ($AHF$) meet $\\Gamma$ again at $T$. Using directed angles, we get $\\angle ATF = \\angle AHF = \\angle AQS$; next, since $FO \\perp AS$, we have\n$$\n\\angle FTA' = \\angle FOA' = \\pi/2 - \\angle OAS = \\pi/2 - \\angle QAL = \\pi/2 - \\angle AQS,\n$$\n(here the equality $\\angle OAS = \\angle QAL$ holds because these angles are symmetric to each other with respect to $AM$), so $\\angle ATA' = \\angle ATF + \\angle FTA' = \\pi/2$, as desired.\n\n**Remark.** Existence of the rotational homothety $h$ may be shown in various ways. E.g., one may notice that $\\Omega$ is an Apollonius circle of the segment $QS$, so the ratio of the radii of $\\Gamma$ and $\\Gamma^*$ is $BS/BQ = AS/AQ$, which also yields that $h$ exists.\nLet $S$ be the centre of $\\Gamma$ and let the circle on diameter $AA'$ cross $\\Gamma$ again at $T$. It is sufficient to prove that $T$ lies on circle $AFH$.\n\nThe points $O$ and $F$ are reflections of one another in $AS$; hence $S$ lies on the internal angle bisectrix of $\\angle FAA'$. On the other hand, since $SF = SA'$, it lies on the perpendicular bisectrix of $FA'$; so $S$ is the midpoint of the arc $A'F$ on circle $AA'F$ not containing $A$. In particular, $AFSA'$ is cyclic.\n\nLet $D$ be the orthogonal projection of $A$ on $BC$. We will prove that $O, D, T$ are collinear. Invert from $O$ with radius $OB$. This fixes $B$ and $C$, so $\\Gamma$ maps to line $BC$. It follows that $A$ maps to $A^* = AA' \\cap BC$. Note that $A$ is fixed under this inversion, as $OA = OB$, so the image of the circle on diameter $AA'$ is a circle $\\delta$ through $A$ and $A^*$ — and, in fact, $\\delta$ is the circle of diameter $AA^*$, as $AA'$ passes through $O$. Hence $T$ maps to one of the points where line $BC$ crosses $\\delta$. As $T \\neq A'$, its image is $D$, so $O, D, T$ are indeed collinear.\n\nLetting $L$ be the reflection of $A$ in $BC$, we now prove that $HTLO$ is cyclic. As circle $HBC$ is the reflection of $\\Gamma$ in $BC$, the quadrangle $HLBC$ is cyclic, so $HD \\cdot DL = BD \\cdot DC = OD \\cdot DT$, whence $HTLO$ is indeed cyclic.\n\n![](attached_image_2.png)\n\nWe are to prove that $\\angle ATF + \\angle FTA' = 90^\\circ$. To this end, note the equivalences: $\\angle ATF + \\angle FTA' = 90^\\circ \\Leftrightarrow \\angle AHF + \\angle FOA = 90^\\circ \\Leftrightarrow \\angle AHF = \\angle OAS$ (as $AS \\perp OF) \\Leftrightarrow \\angle AHF = \\angle SAF \\Leftrightarrow AS$ is tangent to circle $AFH \\Leftrightarrow \\frac{AH}{\\sin \\angle HAS} = \\frac{AF}{\\sin \\angle FAS} \\Leftrightarrow \\frac{AH}{AO} = \\frac{\\sin \\angle HAS}{\\sin \\angle SAO}$. $(*)$\n\n![](attached_image_3.png)\n\nTo prove $(*)$, let $AB$ and $AC$ meet $\\Gamma$ again at $B'$ and $C'$, respectively. An easy angle chase shows that $O$ is the orthocentre of triangle $AB'C'$.\n\nAs triangles $ABC$ and $AC'B'$ are similar, $AH$ passes through the centre $O'$ of circle $AB'C'$; and as circles $AB'C'$ and $BOCC'B'$ are reflections of one another in $B'C'$ and $AO'SO$ is a parallelogram, it follows that\n$$\n\\frac{\\sin \\angle HAS}{\\sin \\angle SAO} = \\frac{\\sin \\angle O'AS}{\\sin \\angle ASO'} = \\frac{O'S}{AO'} = \\frac{AO}{AO'}.\n\\quad (**)\n$$\nFurther on, as triangles $ABC$ and $AC'B'$ are similar, $(**)$ implies equal corresponding length ratios, so $AO/AO' = AH/AO$. This establishes $(*)$ and concludes the solution.\n\n**Remark.** Relation $(*)$ is equivalent to $AS$ being the $A$-symmedian of triangle $AOH$. This might very well be known and can actually be proved in several different ways.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $m$ be integers greater than $1$. Consider $k$ pairwise disjoint sets $S_1, S_2, \\dots, S_k$; each of these sets has exactly $m+1$ elements, one of which is red and the other $m$ are all blue. Let $\\mathcal{F}$ be the family of all subsets $F$ of $S_1 \\cup S_2 \\cup \\dots \\cup S_k$ such that, for every $i$, the intersection $F \\cap S_i$ is monochromatic; the empty set is monochromatic. Determine the largest possible cardinality of a subfamily $\\mathcal{G} \\subseteq \\mathcal{F}$, no two sets of which are disjoint.",
    "options": [],
    "answer": "2^{m-1}(2^m+1)^{k-1}",
    "solution": "We now prove that $|\\mathcal{G}| \\le 2^{m-1}(2^m + 1)^{k-1}$ for any $\\mathcal{G}$ satisfying the conditions in the statement. For convenience, write $M = 2^m + 1$. Let $r_i$ denote the red element of $S_i$, and let $B_i$ be the set of blue elements in $S_i$.\nFor every subset $X_i \\subset B_i$ and every $j \\in \\mathbb{Z}_M$, define the sets\n\n$$\nT_{X_i, j} = \\begin{cases} \\{r_i\\}, & \\text{if } j = 0; \\\\ X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is even (considered as a number in } [1, M-1]\\}; \\\\ B_i \\setminus X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is odd (considered as a number in } [1, M-1]\\}. \\end{cases}\n$$\n\nNote that, for every $i$ and every $j$, the sets $T_{X_i,j}$ and $T_{X_i,j+1}$ are disjoint. Now, for every sets $X_i \\subset B_i$ and every elements $j_i \\in \\mathbb{Z}_M$, $i = 1, 2, \\dots, k$, denote\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) = \\bigcup_{i=1}^{k} T_{X_i, j_i}. \\quad (*)\n$$\n\n**Claim.** Every set $F \\in \\mathcal{F}$ has exactly $2^{mk}$ representations of the form ($*$).\n\n*Proof.* Set $F_i = F \\cap S_i$. If $F_i = \\{r_i\\}$, then there are $2^m$ possible choices for $X_i$, and one should necessarily have $j_i = 0$. Otherwise, there are only two possible choices for $X_i$, namely $X_i = F_i$ and $X_i = B_i \\setminus F_i$, and for each of them there are $2^{m-1}$ possible choices for $j_i$. So, whatever $F$, there are $2^m$ possible choices for each pair $(X_i, j_i)$ all of which can be made independently, whence a total of $2^{mk}$ possible tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$. This proves the Claim.\n\nThe Claim implies that each $F \\in \\mathcal{F}$ has the same number of representations of the form ($*$). Thus, it suffices to show that, among all $N = 2^{km}(2^m + 1)^k$ tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$, at most $\\frac{2^{m-1}}{2^m+1}N$ satisfy\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) \\in \\mathcal{G}.\n$$\nTo this end, split all these tuples into length $M$ cycles\n$$\n(F(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k), F(X_1, X_2, \\dots, X_k, j_1+1, j_2+1, \\dots, j_k+1), \\dots, F(X_1, X_2, \\dots, X_k, j_1+M-1, j_2+M-1, \\dots, j_k+M-1)),\n$$\nand note that any two adjacent sets of a cycle are disjoint. Hence each cycle contains at most $\\lfloor M/2 \\rfloor = 2^{m-1}$ sets from $\\mathcal{G}$. This provides the desired upper bound and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a finite group of boys and girls, a covering set of boys is a set of boys such that every girl knows at least one boy in that set; and a covering set of girls is a set of girls such that every boy knows at least one girl in that set. Prove that the number of covering sets of boys and the number of covering sets of girls have the same parity. (Acquaintance is assumed to be mutual.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA set $X$ of boys is separated from a set $Y$ of girls if no boy in $X$ is an acquaintance of a girl in $Y$. Similarly, a set $Y$ of girls is separated from a set $X$ of boys if no girl in $Y$ is an acquaintance of a boy in $X$. Since acquaintance is assumed mutual, separation is symmetric: $X$ is separated from $Y$ if and only if $Y$ is separated from $X$.\n\nThis enables doubly counting the number $n$ of ordered pairs $(X, Y)$ of separated sets $X$, of boys, and $Y$, of girls, and thereby showing that it is congruent modulo $2$ to both numbers in question.\n\nGiven a set $X$ of boys, let $Y_{X}$ be the largest set of girls separated from $X$, to deduce that $X$ is separated from exactly $2^{|Y_{X}|}$ sets of girls. Consequently, $n = \\sum_{X} 2^{|Y_{X}|}$ which is clearly congruent modulo $2$ to the number of covering sets of boys.\n\nMutatis mutandis, the argument applies to show $n$ congruent modulo $2$ to the number of covering sets of girls.\nSolution:\n\nLet $B$ denote the set of boys, let $G$ denote the set of girls and induct on $|B| + |G|$. The assertion is vacuously true if either set is empty.\n\nNext, fix a boy $b$, let $B' = B \\setminus \\{b\\}$, and let $G'$ be the set of all girls who do not know $b$. Notice that:\n\n(1) a covering set of boys in $B' \\cup G$ is still one in $B \\cup G$; and\n\n(2) a covering set of boys in $B \\cup G$ which is no longer one in $B' \\cup G$ is precisely the union of a covering set of boys in $B' \\cup G'$ and $\\{b\\}$,\n\nso the number of covering sets of boys in $B \\cup G$ is the sum of those in $B' \\cup G$ and $B' \\cup G'$. On the other hand,\n\n$(1')$ a covering set of girls in $B \\cup G$ is still one in $B' \\cup G$; and\n\n$(2')$ a covering set of girls in $B' \\cup G$ which is no longer one in $B \\cup G$ is precisely a covering set of girls in $B' \\cup G'$, so the number of covering sets of girls in $B \\cup G$ is the difference of those in $B' \\cup G$ and $B' \\cup G'$. Since the assertion is true for both $B' \\cup G$ and $B' \\cup G'$ by the induction hypothesis, the conclusion follows.\nSolution:\n\nLet $B$ and $G$ denote the sets of boys and girls, respectively. For every pair $(b, g) \\in B \\times G$, write $f(b, g) = 0$ if they know each other, and $f(b, g) = 1$ otherwise. A set $X$ of boys is covering if and only if\n$$\n\\prod_{g \\in G} \\left(1 - \\prod_{b \\in X} f(b, g)\\right) = 1\n$$\nHence the number of covering sets of boys is\n$$\n\\begin{aligned}\n\\sum_{X \\subseteq B} \\prod_{g \\in G} \\left(1 - \\prod_{b \\in X} f(b, g)\\right) &\\equiv \\sum_{X \\subseteq B} \\prod_{g \\in G} \\left(1 + \\prod_{b \\in X} f(b, g)\\right) \\\\\n&= \\sum_{X \\subseteq B} \\sum_{Y \\subseteq G} \\prod_{b \\in X} \\prod_{g \\in Y} f(b, g) \\quad (\\bmod 2)\n\\end{aligned}\n$$\nBy symmetry, the same is valid for the number of covering sets of girls.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there are infinitely many positive integer numbers $n$ such that $2^{2^{n}+1}+1$ is divisible by $n$, but $2^{n}+1$ is not.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThroughout the solution $n$ stands for a positive integer. By Euler's theorem, $(2^{3^{n}}+1)(2^{3^{n}}-1)=2^{2 \\cdot 3^{n}}-1 \\equiv 0 \\pmod{3^{n+1}}$. Since $2^{3^{n}}-1 \\equiv 1 \\pmod{3}$, it follows that $2^{3^{n}}+1$ is divisible by $3^{n+1}$.\n\nThe number $(2^{3^{n+1}}+1)/(2^{3^{n}}+1)=2^{2 \\cdot 3^{n}}-2^{3^{n}}+1$ is greater than $3$ and congruent to $3$ modulo $9$, so it has a prime factor $p_{n}>3$ that does not divide $2^{3^{n}}+1$ (otherwise, $2^{3^{n}} \\equiv -1 \\pmod{p_{n}}$, so $2^{2 \\cdot 3^{n}}-2^{3^{n}}+1 \\equiv 3 \\pmod{p_{n}}$, contradicting the fact that $p_{n}$ is a factor greater than $3$ of $2^{2 \\cdot 3^{n}}-2^{3^{n}}+1$).\n\nWe now show that $a_{n}=3^{n} p_{n}$ satisfies the conditions in the statement. Since $2^{a_{n}}+1 \\equiv 2^{3^{n}}+1 \\not\\equiv 0 \\pmod{p_{n}}$, it follows that $a_{n}$ does not divide $2^{a_{n}}+1$.\n\nOn the other hand, $3^{n+1}$ divides $2^{3^{n}}+1$ which in turn divides $2^{a_{n}}+1$, so $2^{3^{n+1}}+1$ divides $2^{2^{a_{n}}+1}+1$. Finally, both $3^{n}$ and $p_{n}$ divide $2^{3^{n+1}}+1$, so $a_{n}$ divides $2^{2^{a_{n}}+1}+1$.\n\nAs $n$ runs through the positive integers, the $a_{n}$ are clearly pairwise distinct and the conclusion follows.\nSolution:\n(Géza Kós) We show that the numbers $a_{n}=(2^{3^{n}}+1)/9$, $n \\geq 2$, satisfy the conditions in the statement. To this end, recall the following well-known facts:\n\n(1) If $N$ is an odd positive integer, then $\\nu_{3}(2^{N}+1)=\\nu_{3}(N)+1$, where $\\nu_{3}(a)$ is the exponent of $3$ in the decomposition of the integer $a$ into prime factors; and\n\n(2) If $M$ and $N$ are odd positive integers, then $(2^{M}+1,2^{N}+1)=2^{(M, N)}+1$, where $(a, b)$ is the greatest common divisor of the integers $a$ and $b$.\n\nBy (1), $a_{n}=3^{n-1} m$, where $m$ is an odd positive integer not divisible by $3$, and by (2),\n$$\n(m, 2^{a_{n}}+1) \\mid (2^{3^{n}}+1,2^{a_{n}}+1)=2^{(3^{n}, a_{n})}+1=2^{3^{n-1}}+1<\\frac{2^{3^{n}}+1}{3^{n+1}}=m\n$$\nso $m$ cannot divide $2^{a_{n}}+1$.\n\nOn the other hand, $3^{n-1} \\mid 2^{2^{a_{n}}+1}+1$, for $\\nu_{3}(2^{2^{a_{n}}+1}+1)>\\nu_{3}(2^{a_{n}}+1)>\\nu_{3}(a_{n})=n-1$, and $m \\mid 2^{2^{a_{n}}+1}+1$, for $3^{n-1} \\mid a_{n}$, so $3^{n} \\mid 2^{a_{n}}+1$ whence $m\\mid 2^{3^{n}}+1\\mid 2^{2^{a_{n}}+1}+1$. Since $3^{n-1}$ and $m$ are coprime, the conclusion follows.\nSolution:\n(Dušan Djukić) Assume that $n$ satisfies the conditions of the problem. We claim that the number $N=2^{n}+1>n$ also satisfies these conditions.\n\nFirstly, since $n \\nmid N$, the fact (2) from Solution 2 allows to conclude that $2^{n}+1 \\nmid 2^{N}+1$, or $N \\nmid 2^{N}+1$. Next, since $n \\mid 2^{2^{n}+1}+1=2^{N}+1$, we obtain from the same fact that $N=2^{n}+1 \\mid 2^{2^{N}+1}+1$, thus confirming our claim.\n\nHence, it suffices to provide only one example, hence obtaining an infinite series by the claim. For instance, one may easily check that the number $n=57$ fits.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a positive integer number $n \\geq 3$, colour each cell of an $n \\times n$ square array one of $\\left[(n+2)^{2} / 3\\right]$ colours, each colour being used at least once. Prove that the cells of some $1 \\times 3$ or $3 \\times 1$ rectangular subarray have pairwise distinct colours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFor more convenience, say that a subarray of the $n \\times n$ square array bears a colour if at least two of its cells share that colour.\n\nWe shall prove that the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, which is $2 n(n-2)$, exceeds the number of such subarrays, each of which bears some colour. The key ingredient is the estimate in the lemma below.\n\nLemma. If a colour is used exactly $p$ times, then the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing that colour does not exceed $3(p-1)$.\n\nAssume the lemma for the moment, let $N=\\left[(n+2)^{2} / 3\\right]$ and let $n_{i}$ be the number of cells coloured the $i$th colour, $i=1, \\ldots, N$, to deduce that the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, each of which bears some colour, is at most\n$$\n\\sum_{i=1}^{N} 3\\left(n_{i}-1\\right)=3 \\sum_{i=1}^{N} n_{i}-3 N=3 n^{2}-3 N<3 n^{2}-(n^{2}+4 n)=2 n(n-2)\n$$\nand thereby conclude the proof.\n\nBack to the lemma, the assertion is clear if $p=1$, so let $p>1$.\n\nWe begin by showing that if a row contains exactly $q$ cells coloured $C$, then the number $r$ of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $3 q / 2-1$; of course, a similar estimate holds for a column. To this end, notice first that the case $q=1$ is trivial, so we assume that $q>1$. Consider the incidence of a cell $c$ coloured $C$ and a $3 \\times 1$ rectangular subarray $R$ bearing $C$ :\n$$\n\\langle c, R\\rangle= \\begin{cases}1 & \\text{ if } c \\subset R \\\\ 0 & \\text{ otherwise. }\\end{cases}\n$$\nNotice that, given $R, \\sum_{c}\\langle c, R\\rangle \\geq 2$, and, given $c, \\sum_{R}\\langle c, R\\rangle \\leq 3$; moreover, if $c$ is the leftmost or rightmost cell, then $\\sum_{R}\\langle c, R\\rangle \\leq 2$. Consequently,\n$$\n2 r \\leq \\sum_{R} \\sum_{c}\\langle c, R\\rangle=\\sum_{c} \\sum_{R}\\langle c, R\\rangle \\leq 2+3(q-2)+2=3 q-2\n$$\nwhence the conclusion.\n\nFinally, let the $p$ cells coloured $C$ lie on $k$ rows and $\\ell$ columns and notice that $k+\\ell \\geq 3$, for $p>1$. By the preceding, the total number of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $3 p / 2-k$, and the total number of $1 \\times 3$ rectangular subarrays bearing $C$ does not exceed $3 p / 2-\\ell$, so the total number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $(3 p / 2-k)+(3 p / 2-\\ell)=3 p-(k+\\ell) \\leq 3 p-3=3(p-1)$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a quadrilateral inscribed in a circle $\\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\\omega$ are tangent to one another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O$ be the centre of $\\omega$. Notice that the points $P$, $Q$, and $R$ are the poles (with respect to $\\omega$) of the lines $QR$, $RP$, and $PQ$, respectively. Hence we have $OP \\perp QR$, $OQ \\perp RP$, and $OR \\perp PQ$, thus $R$ is the orthocentre of the triangle $OPQ$.\n\nNow, if $MR \\perp PQ$, then the points $P$ and $Q$ are the reflections of one another in the line $MR = MO$, and the triangle $PQK$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $MR$, and let $U$ be the common point of the lines $OV$ and $PQ$. Since $U$ lies on the polar line of $R$ and $OU \\perp MR$, we obtain that $U$ is the pole of $MR$. Therefore, the line $UK$ is tangent to $\\omega$. Hence it is enough to prove that $UK^2 = UP \\cdot UQ$, since this relation implies that $UK$ is also tangent to the circle $KPQ$.\n\nFrom the rectangular triangle $OKU$, we get $UK^2 = UV \\cdot UO$. Let $\\Omega$ be the circumcircle of triangle $OPQ$, and let $R'$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $PQ$. It is well known that $R'$ is the point of $\\Omega$ opposite to $O$, hence $OR'$ is the diameter of $\\Omega$. Finally, since $\\angle OVR' = 90^\\circ$, the point $V$ also lies on $\\Omega$, hence $UP \\cdot UQ = UV \\cdot UO = UK^2$, as required.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$2n$ distinct tokens are placed at the vertices of a regular $2n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nStep 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2n$-gon accordingly. Consider any three tokens $i < j < k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these three tokens have been switched. Hence the order has been reversed thrice, and in the final arrangement token $k$ stands on the arc passing clockwise from token $i$ to token $j$. Thus, at the end, token $i+1$ is a counter-clockwise neighbor of token $i$ for all $i = 1, 2, \\ldots, 2n-1$, so the tokens in the final arrangement are numbered successively in counter-clockwise circular order.\nThis means that the final arrangement of tokens can be obtained from the initial one by reflection in some line $\\ell$.\n\nStep 2. Notice that each token was involved into $2n-1$ switchings, so its initial and final vertices have different parity. Hence $\\ell$ passes through the midpoints of two opposite sides of a $2n$-gon; we may assume that these are the sides $a$ and $b$ connecting $2n$ with $1$ and $n$ with $n+1$, respectively.\nDuring the process, each token $x$ has crossed $\\ell$ at least once; thus one of its switchings has been made at edge $a$ or at edge $b$. Assume that some two its switchings were performed at $a$ and at $b$; we may (and will) assume that the one at $a$ was earlier, and $x \\leq n$. Then the total movement of token $x$ consisted at least of: (i) moving from vertex $x$ to $a$ and crossing $\\ell$ along $a$; (ii) moving from $a$ to $b$ and crossing $\\ell$ along $b$; (iii) coming to vertex $2n+1-x$. This takes at least $x + n + (n - x) = 2n$ switchings, which is impossible.\nThus, each token had a switching at exactly one of the edges $a$ and $b$.\n\nStep 3. Finally, let us show that either each token has been switched at $a$, or each token has been switched at $b$ (then the other edge has never been used, as desired). To the contrary, assume that there were switchings at both $a$ and at $b$. Consider the first such switchings, and let $x$ and $y$ be the tokens which were moved clockwise during these switchings and crossed $\\ell$ at $a$ and $b$, respectively. By Step 2, $x \\neq y$. Then tokens $x$ and $y$ initially were on opposite sides of $\\ell$.\nNow consider the switching of tokens $x$ and $y$; there was exactly one such switching, and we assume that it has been made on the same side of $\\ell$ as vertex $y$. Then this switching has been made after token $x$ had traced $a$. From this point on, token $x$ is on the clockwise arc from token $y$ to $b$, and it has no way to leave out from this arc. But this is impossible, since token $y$ should trace $b$ after that moment. A contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor an integer $n \\geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?",
    "options": [],
    "answer": "The first player wins if and only if v2(n − 3) is odd.",
    "solution": "Solution:\n\nWe shall prove that the first player wins if and only if the exponent of $2$ in the prime decomposition of $n-3$ is odd.\n\nSince the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins or Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).\n\nLemma. Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.\n\nProof. Label the positions of the counters $X, Y, Z$ so that the arc $YZ$ of the circumcircle is shortest and the $\\operatorname{arc} ZX$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $YZX$ until it forms an isosceles triangle $XYZ'$ with apex at $Y$ (note that the $\\operatorname{arc} XY$ is less than half the circle, so that $Z$ does not jump over the counter at $X$). If this configuration is a Loss, we are done.\n\nIf instead this configuration is a Win, then the counters can be moved legally from triangle $XYZ'$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z'$ to a new location $Z''$. But then the counter at $Z$ could have been moved to $Z''$ in the first place, so the original configuration was a Win as well.\n\nFor every nonzero integer $x$, denote by $v_2(x)$ the exponent of $2$ in the prime decomposition of $x$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2a + b = n$), we show that the configuration is a Win if and only if $a \\neq b$ and $v_2(a-b)$ is odd.\n\nWrite $b = a \\pm |a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \\pm |a-b|/2, a \\pm |a-b|/2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.\n\nIf instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \\pm |a-b|/2, a \\pm |a-b|/2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \\pm |a-b|/2, a \\pm |a-b|/2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b|/2$, the conclusion follows inductively once the exceptional and trivial case $a = b$ is dealt with.\n\nAs an immediate corollary, the configuration with arc lengths $1, 1, n-2$ (the starting configuration of the question) is a Win if and only if $v_2(n-3)$ is odd.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers\n$$\na+b,\\ a-b,\\ b-a,\\ a \\times b,\\ a / b\\ (\\text{ if } b \\neq 0),\\ b / a\\ (\\text{ if } a \\neq 0).\n$$\nProve that, for every integer $n>100$, there are only finitely many integers $k \\geq 0$, such that, starting from the list\n$$\nk+1,\\ k+2,\\ \\ldots,\\ k+n\n$$\nit is possible to obtain, after $n-1$ operations, the value $n!$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe prove the problem statement even for all positive integer $n$.\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \\ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elementary operation from two such formulas on two disjoint sets of the $x_{i}$.\n\nA straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n$$\n\\frac{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in \\{0,1\\}} a_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}}{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in \\{0,1\\}} b_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}},\n$$\nwhere the $a_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ and $b_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ are all in the set $\\{0, \\pm 1\\}$, not all zero of course, $a_{0, \\ldots, 0}=b_{1, \\ldots, 1}=0$, and also $a_{\\alpha_{1}, \\ldots, \\alpha_{n}} \\cdot b_{\\alpha_{1}, \\ldots, \\alpha_{n}}=0$ for every set of indices.\n\nSince $|a_{\\alpha_{1}, \\ldots, \\alpha_{n}}| \\leq 1$, and $a_{0,0, \\ldots, 0}=0$, the absolute value of the numerator does not exceed $(1+|x_{1}|) \\cdots (1+|x_{n}|)-1$; in particular, if $c$ is an integer in the range $-n, \\ldots, -1$, and $x_{k}=c+k$, $k=1, \\ldots, n$, then the absolute value of the numerator is at most $(-c)!(n+c+1)!-1 \\leq n!-1 < n!$.\n\nConsider now the integral polynomials,\n$$\nP=\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in \\{0,1\\}} a_{\\alpha_{1}, \\ldots, \\alpha_{n}}(X+1)^{\\alpha_{1}} \\cdots (X+n)^{\\alpha_{n}}\n$$\nand\n$$\nQ=\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in \\{0,1\\}} b_{\\alpha_{1}, \\ldots, \\alpha_{n}}(X+1)^{\\alpha_{1}} \\cdots (X+n)^{\\alpha_{n}}\n$$\nwhere the $a_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ and $b_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ are all in the set $\\{0, \\pm 1\\}$, not all zero, $a_{\\alpha_{1}, \\ldots, \\alpha_{n}} b_{\\alpha_{1}, \\ldots, \\alpha_{n}}=0$ for every set of indices, and $a_{0, \\ldots, 0}=b_{1, \\ldots, 1}=0$. By the preceding, $|P(c)|<n!$ for every integer $c$ in the range $-n, \\ldots, -1$; and since $b_{1, \\ldots, 1}=0$, the degree of $Q$ is less than $n$.\n\nSince every non-zero polynomial has only finitely many roots, and the number of roots does not exceed the degree, to complete the proof it is sufficient to show that the polynomial $P-n!Q$ does not vanish identically, provided that $Q$ does not (which is the case in the problem).\n\nSuppose, if possible, that $P=n!Q$, where $Q \\neq 0$. Since $\\operatorname{deg} Q < n$, it follows that $\\operatorname{deg} P < n$ as well, and since $P \\neq 0$, the number of roots of $P$ does not exceed $\\operatorname{deg} P < n$, so $P(c) \\neq 0$ for some integer $c$ in the range $-n, \\ldots, -1$. By the preceding, $|P(c)|$ is consequently a positive integer less than $n!$. On the other hand, $|P(c)|=n!|Q(c)|$ is an integral multiple of $n!$. A contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a positive integer $n$, determine the largest real number $\\mu$ satisfying the following condition: for every $4n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu$.",
    "options": [],
    "answer": "1/(2n+2)",
    "solution": "Solution:\nThe required maximum is $\\frac{1}{2n+2}$. To show that the condition in the statement is not met if $\\mu > \\frac{1}{2n+2}$, let $U = (0,1) \\times (0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\frac{i}{n+1} \\pm \\epsilon\\right) \\times \\left(\\frac{1}{2} \\pm \\epsilon\\right)$, $i = 1, \\ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at most $\\left(\\frac{1}{n+1} + \\epsilon\\right) \\cdot \\left(\\frac{1}{2} + \\epsilon\\right) < \\mu$ if $\\epsilon$ is small enough.\n\nWe now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu_0 = \\frac{2}{|C|+4}$.\n\nTo prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.\n\n**Lemma 1.** Let $k$ be a positive integer, and let $\\lambda < \\frac{1}{\\lfloor k / 2 \\rfloor + 1}$ be a positive real number. If $t_{1}, \\ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$.\n\n**Lemma 2.** Given an integer $k \\geq 2$ and positive integers $m_{1}, \\ldots, m_{k}$,\n$$\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor + \\sum_{i=1}^{k} \\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor + \\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor \\leq \\sum_{i=1}^{k} m_{i} - k + 2\n$$\n\nBack to the problem, let $U = (0,1) \\times (0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1} < \\cdots < x_{k}$ in the open unit interval $(0,1)$, let $\\ell_{i}$ be the vertical through $x_{i}$, and let $m_{i} = |C \\cap \\ell_{i}|$, $i = 1, \\ldots, k$.\n\nSetting $x_{0} = 0$ and $x_{k+1} = 1$, assume that $x_{i+1} - x_{i-1} > (\\lfloor m_{i} / 2 \\rfloor + 1) \\mu_0$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \\cap \\ell_{i}$ from the other ones by an open subinterval $x_{i} \\times J$ of $x_{i} \\times (0,1)$ whose length is greater than or equal to $\\mu_0 / (x_{i+1} - x_{i-1})$. Consequently, $\\left(x_{i-1}, x_{i+1}\\right) \\times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\\mu_0$.\n\nNext, we rule out the case $x_{i+1} - x_{i-1} \\leq (\\lfloor m_{i} / 2 \\rfloor + 1) \\mu_0$ for all indices $i$. If this were the case, notice that necessarily $k > 1$; also, $x_{1} - x_{0} < x_{2} - x_{0} \\leq (\\lfloor m_{1} / 2 \\rfloor + 1) \\mu_0$ and $x_{k+1} - x_{k} < x_{k+1} - x_{k-1} \\leq (\\lfloor m_{k} / 2 \\rfloor + 1) \\mu_0$. With reference to Lemma 2, write\n$$\n\\begin{aligned}\n2 = 2(x_{k+1} - x_{0}) & = (x_{1} - x_{0}) + \\sum_{i=1}^{k} (x_{i+1} - x_{i-1}) + (x_{k+1} - x_{k}) \\\\\n& < \\left( (\\lfloor \\frac{m_{1}}{2} \\rfloor + 1) + \\sum_{i=1}^{k} (\\lfloor \\frac{m_{i}}{2} \\rfloor + 1) + (\\lfloor \\frac{m_{k}}{2} \\rfloor + 1) \\right) \\cdot \\mu_0 \\\\\n& \\leq (\\sum_{i=1}^{k} m_{i} + 4) \\mu_0 = (|C| + 4) \\mu_0 = 2\n\\end{aligned}\n$$\nand thereby reach a contradiction.\n\nFinally, we prove the two lemmas.\n\n**Proof of Lemma 1.** Suppose, if possible, that no $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$. Without loss of generality, we may (and will) assume that $0 = t_{0} < t_{1} < \\cdots < t_{k} < t_{k+1} = 1$. Since the open interval $(t_{i-1}, t_{i+1})$ isolates $t_{i}$ from the other $t_{j}$, its length, $t_{i+1} - t_{i-1}$, is less than $\\lambda$. Consequently, if $k$ is odd we have $1 = \\sum_{i=0}^{(k-1)/2} (t_{2i+2} - t_{2i}) < \\lambda (1 + \\frac{k-1}{2}) < 1$; if $k$ is even, we have $1 < 1 + t_{k} - t_{k-1} = \\sum_{i=0}^{k/2-1} (t_{2i+2} - t_{2i}) + (t_{k+1} - t_{k-1}) < \\lambda (1 + \\frac{k}{2}) < 1$. A contradiction in either case.\n\n**Proof of Lemma 2.** Let $I_{0}$, respectively $I_{1}$, be the set of all indices $i$ in the range $2, \\ldots, k-1$ such that $m_{i}$ is even, respectively odd. Clearly, $I_{0}$ and $I_{1}$ form a partition of that range. Since $m_{i} \\geq 2$ if $i$ is in $I_{0}$, and $m_{i} \\geq 1$ if $i$ is in $I_{1}$ (recall that the $m_{i}$ are positive integers),\n$$\n\\sum_{i=2}^{k-1} m_{i} = \\sum_{i \\in I_{0}} m_{i} + \\sum_{i \\in I_{1}} m_{i} \\geq 2|I_{0}| + |I_{1}| = 2(k-2) - |I_{1}|, \\quad \\text{or} \\quad |I_{1}| \\geq 2(k-2) - \\sum_{i=2}^{k-1} m_{i}\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\left\\lfloor \\frac{m_{1}}{2} \\right\\rfloor + \\sum_{i=1}^{k} \\left\\lfloor \\frac{m_{i}}{2} \\right\\rfloor + \\left\\lfloor \\frac{m_{k}}{2} \\right\\rfloor & \\leq m_{1} + \\left( \\sum_{i=2}^{k-1} \\frac{m_{i}}{2} - \\frac{|I_{1}|}{2} \\right) + m_{k} \\\\\n& \\leq m_{1} + \\left( \\frac{1}{2} \\sum_{i=2}^{k-1} m_{i} - (k-2) + \\frac{1}{2} \\sum_{i=2}^{k-1} m_{i} \\right) + m_{k} \\\\\n& = \\sum_{i=1}^{k} m_{i} - k + 2\n\\end{aligned}\n$$\n\nRemark. In case $4n$ is replaced by a positive integer $k$ not divisible by $4$, we do not yet know the maximal $\\mu$ satisfying the corresponding condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces the number $n$ on the blackboard with a number of the form $n^{k}$, where $k$ is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob's win?",
    "options": [],
    "answer": "No; Amy cannot prevent Bob's win. Bob can always force a win (in at most S(N) moves).",
    "solution": "Solution:\n\nThe answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$.\n\nNow we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer $n$ with a non-negative integer $k$ with $S(k)<S(n)$. Thus, if the game starts by a positive integer $N$, Bob can win in at most $S(N)$ moves.\n\nPart (i) is trivial, as the definition of the square-part yields $S\\left(n^{k}\\right)=S(n)$ whenever $k$ is odd, and $S\\left(n^{k}\\right)=1 \\leq S(n)$ whenever $k$ is even, for any positive integer $n$.\n\nPart (ii) is also easy: if, before Bob's move, the board contains a number $n=S(n) \\cdot b^{2}$, then Bob may replace it with $n' = n-b^{2} = (S(n)-1) b^{2}$, whence $S\\left(n'\\right) \\leq S(n)-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be an isosceles trapezoid with $AB \\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\\omega$ and $\\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\\omega$ at $A$ with the tangent to $\\Omega$ at $D$. Prove that $PE$ is tangent to $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf $ABCD$ is a rectangle, the statement is trivial due to symmetry. Hence, in what follows we assume $AD \\nVdash BC$.\n\nLet $F$ be the midpoint of $BD$; by symmetry, both $\\omega$ and $\\Omega$ pass through $F$. Let $P'$ be the meeting point of tangents to $\\omega$ at $F$ and to $\\Omega$ at $E$. We aim to show that $P' = P$, which yields the required result. For that purpose, we show that $P'A$ and $P'D$ are tangent to $\\omega$ and $\\Omega$, respectively.\n\nLet $K$ be the midpoint of $AF$. Then $EK$ is a midline in the triangle $ACF$, so $\\angle(AE, EK) = \\angle(EC, CF)$. Since $P'E$ is tangent to $\\Omega$, we get $\\angle(EC, CF) = \\angle(P'E, EF)$. Thus, $\\angle(AE, EK) = \\angle(P'E, EF)$, so $EP'$ is a symmedian in the triangle $AEF$. Therefore, $EP'$ and the tangents to $\\omega$ at $A$ and $F$ are concurrent, and the concurrency point is $P'$ itself. Hence $P'A$ is tangent to $\\omega$.\n\nThe second claim is similar. Taking $L$ to be the midpoint of $DE$, we have $\\angle(DF, FL) = \\angle(FB, BE) = \\angle(P'F, FE)$, so $P'F$ is a symmedian in the triangle $DEF$, and hence $P'$ is the meeting point of the tangents to $\\Omega$ at $D$ and $E$.\n\n![](attached_image_1.png)\n\nRemark. The above arguments may come in different orders. E.g., one may define $P'$ to be the point of intersection of the tangents to $\\Omega$ at $D$ and $E$—hence obtaining that $P'F$ is a symmedian in $\\triangle DEF$, then deduce that $P'F$ is tangent to $\\omega$, and then apply a similar argument to show that $P'E$ is a symmedian in $\\triangle AEF$, whence $P'A$ is tangent to $\\omega$.\nSolution:\nLet $Q$ be the isogonal conjugate of $P$ with respect to $\\triangle AED$, so $\\angle(QA, AD) = \\angle(EA, AP) = \\angle(EB, BA)$ and $\\angle(QD, DA) = \\angle(ED, DP) = \\angle(EC, CD)$. Now our aim is to prove that $QE \\parallel CD$; this will yield that $\\angle(EC, CD) = \\angle(AE, EQ) = \\angle(PE, ED)$, whence $PE$ is tangent to $\\Omega$.\n\nLet $DQ$ meet $AB$ at $X$. Then we have $\\angle(XD, DA) = \\angle(EC, CD) = \\angle(EA, AB)$ and $\\angle(DA, AX) = \\angle(AB, BC)$, hence the triangles $DAX$ and $ABC$ are similar. Since $\\angle(AB, BE) = \\angle(DA, AQ)$, the points $Q$ and $E$ correspond to each other in these triangles, hence $Q$ is the midpoint of $DX$. This yields that the points $Q$ and $E$ lie on the midline of the trapezoid parallel to $CD$, as desired.\n\n![](attached_image_2.png)\n\nRemark. The last step could be replaced with another application of isogonal conjugacy in the following manner. Reflect $Q$ in the common perpendicular bisector of $AB$ and $CD$ to obtain a point $R$ such that $\\angle(CB, BR) = \\angle(QA, AD) = \\angle(EB, BA)$ and $\\angle(BC, CR) = \\angle(QD, DA) = \\angle(EC, CD)$. These relations yield that the points $E$ and $R$ are isogonally conjugate in a triangle $BCI$, where $I$ is the (ideal) point of intersection of $BA$ with $CD$. Since $E$ is equidistant from $AB$ and $CD$, $R$ is also equidistant from them, which yields what we need. (The last step deserves some explanation, since one vertex of the triangle is ideal. Such explanation may be obtained in many different ways—e.g., by a short computation in sines, or by noticing that, as in the usual case, $R$ is the circumcenter of the triangle formed by the reflections of $E$ in the sidelines $AB$, $BC$, and $CD$.)\nSolution:\n(Dan Carmon) Let $O$ be the intersection of the diagonals $AC$ and $BD$. Let $F$ be the midpoint of $BD$. Let $S$ be the second intersection point of the circumcircles of triangles $AOF$ and $DOE$. We will prove that $SD$ and $SE$ are tangent to $\\Omega$; the symmetric argument would then imply also that $SA$ and $SF$ are tangent to $\\Gamma$. Thus $S = P$ and the claimed tangency holds.\n\nWe first prove that $OS$ is parallel to $AB$ and $DC$. Compute the powers of the points $A, B$ with respect to the circumcircles of $AOF$ and $DOE$:\n$$\n\\begin{gathered}\nd(A, AOF) = 0, \\quad d(A, DOE) = AO \\cdot AE \\\\\nd(B, AOF) = BO \\cdot BF, \\quad d(B, DOE) = BO \\cdot BD = 2BO \\cdot BF\n\\end{gathered}\n$$\n\n$$\nd(B, DOE) - d(B, AOF) = BO \\cdot BF = AO \\cdot AE = d(A, DOE) - d(A, AOF)\n$$\n\nThus both $A$ and $B$ belong to a locus of the form\n$$\nd(X, DOE) - d(X, AOF) = \\mathrm{const}\n$$\nwhich is always a line parallel to the radical axis of the respective circles. Since this radical axis is $OS$ by definition, it follows that $AB$ is parallel to $OS$, as claimed.\n\nNow by angle chasing in the cyclic quadrilateral $DSOE$, we find\n$$\n\\begin{gathered}\n\\angle(SD, DE) = \\angle(SO, OE) = \\angle(DC, CE), \\\\\n\\angle(SE, ED) = \\angle(SO, OD) = \\angle(DC, DB) = \\angle(AC, CD) = \\angle(EC, CD),\n\\end{gathered}\n$$\nand these angle equalities are exactly the conditions of $SD, SE$ being tangent to $\\Omega$, as claimed.\n\nRemarks. (1) The solution was motivated by the following observation: Suppose $P$ is the intersection of the tangents to $\\Omega$ at $D$ and $E$ as claimed. Then by single angle chasing we observe that the isogonal conjugate of $P$ in the triangle $DOE$ is the common ideal point at infinity of $DC$ and $EF$. This implies that $P$ is on the circumcircle of $DOE$ and that $OP$ is parallel to $DC$ (to be precise, it implies that the reflection of $OP$ in the angle bisector of $DOE$ is parallel to $DC$ and $EF$—but the angle bisector is also parallel to these lines, so in fact $OP$ is the angle bisector). By symmetry it follows that $P$ is also on the circumcircle of $AOF$, thus the construction.\n\n(2) The key parts of the proof can be described as (1) Constructing $S$, (2) Proving that $OS$ is parallel to $AB$ and $CD$, and (3) Concluding that $S = P$ and finishing the problem. Parts (2) and (3) can be performed in various other ways. For example, part (2) can be proved by showing that the circumcentres of $AOF$ and $DOE$ lie on a line perpendicular to the trapezium's bases; part (3) can be proved considering the spiral map taking the circumcircle of $DOC$ to the circumcircle of $DSE$. Since $O$ is the second intersection point of these circles, and since $OCE$ are collinear and $SO$ is tangent to the circumcircle of $DOC$ at $O$ (by symmetry), it follows that the spiral map sends $C$ to $E$ and $O$ to $S$, i.e. the triangle $DSE$ is similar to the isosceles triangle $DOC$, from which the remainder of the angle chase is trivial.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven any positive real number $\\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\\varepsilon) v$ edges has two distinct simple cycles of equal lengths.\n(Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFix a positive real number $\\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\nAssuming $\\varepsilon^{2} v \\geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of simple cycles in $G$, showing thereby that $v$ cannot be arbitrarily large, whence the conclusion.\nSince a simple cycle in $G$ has at most $v$ vertices, and each length class contains at most one such, $G$ has at most $v$ pairwise distinct simple cycles. This establishes the desired upper bound.\nFor the lower bound, consider a spanning tree for each component of $G$, and collect them all together to form a spanning forest $F$. Let $A$ be the set of edges of $F$, and let $B$ be the set of all other edges of $G$. Clearly, $|A| \\leq v-1$, so $|B| \\geq (1+\\varepsilon) v - |A| \\geq (1+\\varepsilon) v - (v-1) = \\varepsilon v + 1 > \\varepsilon v$.\nFor each edge $b$ in $B$, adjoining $b$ to $F$ produces a unique simple cycle $C_{b}$ through $b$. Let $S_{b}$ be the set of edges in $A$ along $C_{b}$. Since the $C_{b}$ have pairwise distinct lengths, $\\sum_{b \\in B} |S_{b}| \\geq 2 + \\cdots + (|B|+1) = |B|(|B|+3)/2 > |B|^{2}/2 > \\varepsilon^{2} v^{2}/2$.\nConsequently, some edge in $A$ lies in more than $\\varepsilon^{2} v^{2}/(2v) = \\varepsilon^{2} v/2$ of the $S_{b}$. Fix such an edge $a$ in $A$, and let $B'$ be the set of all edges $b$ in $B$ whose corresponding $S_{b}$ contain $a$, so $|B'| > \\varepsilon^{2} v/2$.\nFor each 2-edge subset $\\{b_{1}, b_{2}\\}$ of $B'$, the union $C_{b_{1}} \\cup C_{b_{2}}$ of the cycles $C_{b_{1}}$ and $C_{b_{2}}$ forms a $\\theta$-graph, since their common part is a path in $F$ through $a$; and since neither of the $b_{i}$ lies along this path, $C_{b_{1}} \\cup C_{b_{2}}$ contains a third simple cycle $C_{b_{1}, b_{2}}$ through both $b_{1}$ and $b_{2}$. Finally, since $B' \\cap C_{b_{1}, b_{2}} = \\{b_{1}, b_{2}\\}$, the assignment $\\{b_{1}, b_{2}\\} \\mapsto C_{b_{1}, b_{2}}$ is injective, so the total number of simple cycles in $G$ is at least $\\binom{|B'|}{2} > \\binom{\\varepsilon^{2} v/2}{2}$. This establishes the desired lower bound and concludes the proof.\n\n\nSketch of solution 2. (Po-Shen Loh)\nRecall that the girth of a graph $G$ is the minimal length of a (simple) cycle in this graph.\nLemma. For any fixed positive $\\delta$, a graph on $v$ vertices whose girth is at least $\\delta v$ has at most $v+o(v)$ edges.\nProof. Define $f(v)$ to be the maximal number $f$ such that a graph on $v$ vertices whose girth is at least $\\delta v$ may have $v+f$ edges. We are interested in the recursive estimates for $f$.\nLet $G$ be a graph on $v$ vertices whose girth is at least $\\delta v$ containing $v+f(v)$ edges. If $G$ contains a leaf (i.e., a vertex of degree 1), then one may remove this vertex along with its edge, obtaining a graph with at most $v-1+f(v-1)$ edges. Thus, in this case $f(v) \\leq f(v-1)$.\nDefine an isolated path of length $k$ to be a sequence of vertices $v_{0}, v_{1}, \\ldots, v_{k}$, such that $v_{i}$ is connected to $v_{i+1}$, and each of the vertices $v_{1}, \\ldots, v_{k-1}$ has degree 2 (so, these vertices are connected only to their neighbors in the path). If $G$ contains an isolated path $v_{0}, \\ldots, v_{k}$ of length, say, $k > \\sqrt{v}$, then one may remove all its middle vertices $v_{1}, \\ldots, v_{k-1}$, along with all their $k$ edges. We obtain a graph on $v-k+1$ vertices with at most $(v-k+1)+f(v-k+1)$ edges. Thus, in this case $f(v) \\leq f(v-k+1)+1$.\nAssume now that the lengths of all isolated paths do not exceed $\\sqrt{v}$; we show that in this case $v$ is bounded from above. For that purpose, we replace each maximal isolated path by an edge between its endpoints, removing all middle vertices. We obtain a graph $H$ whose girth is at least $\\delta v / \\sqrt{v} = \\delta \\sqrt{v}$. Each vertex of $H$ has degree at least 3. By the girth condition, the neighborhood of any vertex $x$ of radius $r = \\lfloor (\\delta \\sqrt{v} - 1)/2 \\rfloor$ is a tree rooted at $x$. Any vertex at level $i < r$ has at least two sons; so the tree contains at least $2^{\\lfloor (\\delta \\sqrt{v} - 1)/2 \\rfloor}$ vertices (even at the last level). So, $v \\geq 2^{\\lfloor (\\delta \\sqrt{v} - 1)/2 \\rfloor}$ which may happen only for a finite number of values of $v$.\nThus, for all large enough values of $v$, we have either $f(v) \\leq f(v-1)$ or $f(v) \\leq f(v-k+1)$ for some $k > \\sqrt{v}$. This easily yields $f(v) = o(v)$, as desired.\n\nNow we proceed to the problem. Consider a graph on $v$ vertices containing no two simple cycles of the same length. Take its $\\lfloor \\varepsilon v / 2 \\rfloor$ shortest cycles (or all its cycles, if their total number is smaller) and remove an edge from each. We get a graph of girth at least $\\varepsilon v / 2$. By the lemma, the number of edges in the obtained graph is at most $v + o(v)$, so the number of edges in the initial graph is at most $v + \\varepsilon v / 2 + o(v)$, which is smaller than $(1+\\varepsilon) v$ if $v$ is large enough.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every positive integer $n$ there exists a (not necessarily convex) polygon with no three collinear vertices, which admits exactly $n$ different triangulations.\n(A triangulation is a dissection of the polygon into triangles by interior diagonals which have no common interior points with each other nor with the sides of the polygon.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all pairs of integers $(c, d)$, both greater than $1$, such that the following holds:\nFor any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\\left(\\frac{2c-1}{2c+1}\\right)p$ integers, such that\n$$\n\\bigcup_{s \\in S} \\{s, Q(s), Q(Q(s)), Q(Q(Q(s))), \\ldots\\}\n$$\nis a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).",
    "options": [],
    "answer": "All integer pairs (c, d) with c ≥ d and both greater than 1.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $T_{1}$, $T_{2}$, $T_{3}$, $T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\\omega_{1}$ at $T_{1}$; let $\\omega_{3}$ be the circle through $T_{3}$ and externally tangent to $\\omega_{2}$ at $T_{2}$; and let $\\omega_{4}$ be the circle through $T_{4}$ and externally tangent to $\\omega_{3}$ at $T_{3}$. A line crosses $\\omega_{1}$ at $P$ and $W$, $\\omega_{2}$ at $Q$ and $R$, $\\omega_{3}$ at $S$ and $T$, and $\\omega_{4}$ at $U$ and $V$, the order of these points along the line being $P, Q, R, S, T, U, V, W$. Prove that $PQ + TU = RS + VW$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O_{i}$ be the centre of $\\omega_{i}$, $i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo $4$), to infer that $\\omega_{4}$ is internally tangent to $\\omega_{1}$ at $T_{4}$, and $O_{1} O_{2} O_{3} O_{4}$ is a (possibly degenerate) parallelogram.\n\nLet $F_{i}$ be the foot of the perpendicular from $O_{i}$ to $PW$. The $F_{i}$ clearly bisect the segments $PW$, $QR$, $ST$ and $UV$, respectively.\n\nThe proof can now be concluded in two similar ways.\n\n![](attached_image_1.png)\n\nFirst Approach. Since $O_{1} O_{2} O_{3} O_{4}$ is a parallelogram, $\\overrightarrow{F_{1} F_{2}} + \\overrightarrow{F_{3} F_{4}} = \\mathbf{0}$ and $\\overrightarrow{F_{2} F_{3}} + \\overrightarrow{F_{4} F_{1}} = \\mathbf{0}$; this still holds in the degenerate case, for if the $O_{i}$ are collinear, then they all lie on the line $T_{1} T_{4}$, and each $O_{i}$ is the midpoint of the segment $T_{i} T_{i+1}$. Consequently,\n\n$$\n\\begin{aligned}\n\\overrightarrow{PQ} - \\overrightarrow{RS} + \\overrightarrow{TU} - \\overrightarrow{VW} = & \\left(\\overrightarrow{P F_{1}} + \\overrightarrow{F_{1} F_{2}} + \\overrightarrow{F_{2} Q}\\right) - \\left(\\overrightarrow{R F_{2}} + \\overrightarrow{F_{2} F_{3}} + \\overrightarrow{F_{3} S}\\right) \\\\\n& + \\left(\\overrightarrow{T F_{3}} + \\overrightarrow{F_{3} F_{4}} + \\overrightarrow{F_{4} U}\\right) - \\left(\\overrightarrow{V F_{4}} + \\overrightarrow{F_{4} F_{1}} + \\overrightarrow{F_{1} W}\\right) \\\\\n= & \\left(\\overrightarrow{P F_{1}} - \\overrightarrow{F_{1} W}\\right) - \\left(\\overrightarrow{R F_{2}} - \\overrightarrow{F_{2} Q}\\right) + \\left(\\overrightarrow{T F_{3}} - \\overrightarrow{F_{3} S}\\right) - \\left(\\overrightarrow{V F_{4}} - \\overrightarrow{F_{4} U}\\right) \\\\\n& + \\left(\\overrightarrow{F_{1} F_{2}} + \\overrightarrow{F_{3} F_{4}}\\right) - \\left(\\overrightarrow{F_{2} F_{3}} + \\overrightarrow{F_{4} F_{1}}\\right) = \\mathbf{0}\n\\end{aligned}\n$$\n\nAlternatively, but equivalently, $\\overrightarrow{PQ} + \\overrightarrow{TU} = \\overrightarrow{RS} + \\overrightarrow{VW}$, as required.\n\nSecond Approach. This is merely another way of reading the previous argument. Fix an orientation of the line $PW$, say, from $P$ towards $W$, and use a lower case letter to denote the coordinate of a point labelled by the corresponding upper case letter.\n\nSince the diagonals of a parallelogram bisect one another, $f_{1} + f_{3} = f_{2} + f_{4}$, the common value being twice the coordinate of the projection to $PW$ of the point where $O_{1} O_{3}$ and $O_{2} O_{4}$ cross; the relation clearly holds in the degenerate case as well.\n\nPlug $f_{1} = \\frac{1}{2}(p + w)$, $f_{2} = \\frac{1}{2}(q + r)$, $f_{3} = \\frac{1}{2}(s + t)$ and $f_{4} = \\frac{1}{2}(u + v)$ into the above equality to get $p + w + s + t = q + r + u + v$. Alternatively, but equivalently, $(q - p) + (u - t) = (s - r) + (w - v)$, that is, $PQ + TU = RS + VW$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nXenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding $5000$. Then she fixes $20$ distinct positive integers $a_{1}, a_{2}, \\ldots, a_{20}$ such that, for each $k=1,2, \\ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of positive integers not exceeding $20$, and she tells him back the set $\\{a_{k}: k \\in S\\}$ without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSergey can determine Xenia's number in $2$ but not fewer moves.\n\nWe first show that $2$ moves are sufficient. Let Sergey provide the set $\\{17,18\\}$ on his first move, and the set $\\{18,19\\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_{17}, a_{18}$, and $a_{19}$, and thence the residue of $N$ modulo $17 \\cdot 18 \\cdot 19 = 5814 > 5000$, by the Chinese Remainder Theorem. This means that the given range contains a single number satisfying all congruences, and Sergey achieves his goal.\n\nTo show that $1$ move is not sufficient, let $M = \\operatorname{lcm}(1,2, \\ldots, 10) = 2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 = 2520$. Notice that $M$ is divisible by the greatest common divisor of every pair of distinct positive integers not exceeding $20$. Let Sergey provide the set $S = \\{s_{1}, s_{2}, \\ldots, s_{k}\\}$. We show that there exist pairwise distinct positive integers $b_{1}, b_{2}, \\ldots, b_{k}$ such that $1 \\equiv b_{i} \\pmod{s_{i}}$ and $M+1 \\equiv b_{i-1} \\pmod{s_{i}}$ (indices are reduced modulo $k$). Thus, if in response Xenia provides the set $\\{b_{1}, b_{2}, \\ldots, b_{k}\\}$, then Sergey will be unable to distinguish $1$ from $M+1$, as desired.\n\nTo this end, notice that, for each $i$, the numbers of the form $1 + m s_{i}$, $m \\in \\mathbb{Z}$, cover all residues modulo $s_{i+1}$ which are congruent to $1 (\\equiv M+1)$ modulo $\\operatorname{gcd}(s_{i}, s_{i+1}) \\mid M$. Xenia can therefore choose a positive integer $b_{i}$ such that $b_{i} \\equiv 1 \\pmod{s_{i}}$ and $b_{i} \\equiv M+1 \\pmod{s_{i+1}}$. Clearly, such choices can be performed so as to make the $b_{i}$ pairwise distinct, as required.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 39527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4. Prove that the number of such ways to assign the leaders is divisible by 17.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.\n\nConsider the variables $x_{1}, x_{2}, \\ldots, x_{17}$ corresponding to the workers. Assign each brigade (from the $i$-th through the $j$-th worker) the polynomial $f_{ij} = x_{i} + x_{i+1} + \\cdots + x_{j}$, and form the product $f = \\prod_{1 \\leq i \\leq j \\leq 17} f_{ij}$. The number $N$ is the sum $\\Sigma(f)$ of the coefficients of all monomials $x_{1}^{\\alpha_{1}} x_{2}^{\\alpha_{2}} \\ldots x_{17}^{\\alpha_{17}}$ in the expansion of $f$, where the $\\alpha_{i}$ are all congruent to 1 modulo 4. For any polynomial $P$, let $\\Sigma(P)$ denote the corresponding sum. From now on, all polynomials are considered with coefficients in the finite field $\\mathbb{F}_{17}$.\n\nRecall that for any positive integer $n$, and any integers $a_{1}, a_{2}, \\ldots, a_{n}$, there exist indices $i \\leq j$ such that $a_{i} + a_{i+1} + \\cdots + a_{j}$ is divisible by $n$. Consequently, $f(a_{1}, a_{2}, \\ldots, a_{17}) = 0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$.\n\nNow, if some monomial in the expansion of $f$ is divisible by $x_{i}^{17}$, replace that $x_{i}^{17}$ by $x_{i}$; this does not alter the above overall vanishing property (by Fermat's Little Theorem), and preserves $\\Sigma(f)$. After several such changes, $f$ transforms into a polynomial $g$ whose degree in each variable does not exceed 16, and $g(a_{1}, a_{2}, \\ldots, a_{17}) = 0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$. For such a polynomial, an easy induction on the number of variables shows that it is identically zero. Consequently, $\\Sigma(g) = 0$, so $\\Sigma(f) = 0$ as well, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider an integer $n \\geq 2$ and write the numbers $1,2, \\ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\\{a+b,|a-b|\\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \\geq 2$, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves.",
    "options": [],
    "answer": "Yes, it is possible for all integers n at least two.",
    "solution": "Solution:\n\nThe answer is in the affirmative for all $n \\geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.\n\nIf $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair $(3,4)$ then leaves exactly two numbers on the board, $1$ and $7$, since $|3 \\pm 4|$ are both already there.\n\nIf $n=6$, remove first the pair $(1,6)$, notice that $5=|1-6|$ is already on the board, so $7=1+6$ alone is written down. Next, remove the pair $(2,5)$ and notice that $|2 \\pm 5|$ are both already on the board, so no new number is written down. Finally, removal of the pair $(3,4)$ provides a single number to be written down, $1=|3-4|$, since $7=3+4$ is already on the board. At this stage, the process comes to an end: $1$ and $7$ are the two numbers left.\n\nIn the remaining cases, the problem for $n$ is brought down to the corresponding problem for $\\lceil n / 2\\rceil < n$ by a finite number of moves. The conclusion then follows by induction.\n\nLet $n=4k$ or $4k-1$, where $k$ is a positive integer. Remove the pairs $(1,4k-1), (3,4k-3), \\ldots, (2k-1,2k+1)$ in turn. Each time, two odd numbers are removed, and the corresponding $c=|a \\pm b|$ are even numbers in the range $2$ through $4k$, of which one is always $4k$. These even numbers are already on the board at each stage, so no $c$ is to be written down, unless $n=4k-1$ in which case $4k$ is written down during the first move. The outcome of this $k$-move round is the string of even numbers $2$ through $4k$ written down on the board. At this stage, the problem is clearly brought down to the case where the numbers on the board are $1,2, \\ldots, 2k=\\lceil n / 2\\rceil$, as desired.\n\nFinally, let $n=4k+1$ or $4k+2$, where $k \\geq 2$. Remove first the pair $(4,2k+1)$ and notice that no new number is to be written down on the board, since $4+(2k+1)=2k+5 \\leq 4k+1 \\leq n$. Next, remove the pairs $(1,4k+1), (3,4k-1), \\ldots, (2k-1,2k+3)$ in turn. As before, at each of these stages, two odd numbers are removed; the corresponding $c=|a \\pm b|$ are even numbers, this time in the range $4$ through $4k+2$, of which one is always $4k+2$; and no new numbers are to be written down on the board, except $4=|(2k-1)-(2k+3)|$ during the last move, and, possibly, $4k+2=1+(4k+1)$ during the first move if $n=4k+1$. Notice that $2$ has not yet been involved in the process, to conclude that the outcome of this $(k+1)$-move round is the string of even numbers $2$ through $4k+2$ written down on the board. At this stage, the problem is clearly brought down to the case where the numbers on the board are $1,2, \\ldots, 2k+1=\\lceil n / 2\\rceil$, as desired.\nSolution:\n\nWe will prove the following, more general statement:\n\n**Claim.** Write down a finite number (at least two) of pairwise distinct positive integers on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\\{a+b,|a-b|\\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. Then it is possible to be left with exactly two numbers on the board after a finite number of moves.\n\nNotice that, if we divide all numbers on the board by some common factor, the resulting process goes on equally well. Such a reduction can therefore be performed after any move.\n\nNotice that we cannot be left with less than two numbers. So it suffices to show that, given $k$ positive integers on the board, $k \\geq 3$, we can always decrease their number by at least $1$. Arguing indirectly, choose a set of $k \\geq 3$ positive integers $S=\\{a_1, \\ldots, a_k\\}$ which cannot be reduced in size by a sequence of moves, having a minimal possible sum $\\sigma$. So, in any sequence of moves applied to $S$, two numbers are erased and exactly two numbers appear on each move. Moreover, the sum of any resulting set of $k$ numbers is at least $\\sigma$.\n\nNotice that, given two numbers $a > b$ on the board, we can replace them by $a+b$ and $a-b$, and then, performing a move on the two new numbers, by $(a+b)+(a-b)=2a$ and $(a+b)-(a-b)=2b$. So we can double any two numbers on the board.\n\nWe now show that, if the board contains two even numbers $a$ and $b$, we can divide them both by $2$, while keeping the other numbers unchanged. If $k$ is even, split the other numbers into pairs to multiply each pair by $2$; then clear out the common factor $2$. If $k$ is odd, split all numbers but $a$ into pairs to multiply each by $2$; then do the same for all numbers but $b$; finally, clear out the common factor $4$.\n\nBack to the problem, if two of the numbers $a_1, \\ldots, a_k$ are even, reduce them both by $2$ to get a set with a smaller sum, which is impossible. Otherwise, two numbers, say, $a_1 < a_2$, are odd, and we may replace them by the two even numbers $a_1+a_2$ and $a_2-a_1$, and then by $\\frac{1}{2}(a_1+a_2)$ and $\\frac{1}{2}(a_2-a_1)$, to get a set with a smaller sum, which is again impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia would like to relocate its $3 n^{2}+3 n+1$ citizens at $3 n^{2}+3 n+1$ points in the kingdom so that every two citizens have a distance of at least 1 for proper social distancing. Prove that this is possible. (The kingdom is assumed to contain its boundary.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $P$ denote the given polygon, i.e., the kingdom of Zoomtopia. Throughout the solution, we interpret polygons with integer sides and perimeter $6 k$ as $6 k$-gons with unit sides (some of their angles may equal $180^{\\circ}$). The argument hinges on the claim below:\n\nClaim. Let $P$ be a convex polygon satisfying the problem conditions - i.e., it has integer sides, perimeter $6 n$, and $60^{\\circ}$ rotational symmetry. Then $P$ can be tiled with unit equilateral triangles and unit lozenges with angles at least $60^{\\circ}$, with tiles meeting completely along edges, so that the tile configuration has a total of exactly $3 n^{2}+3 n+1$ distinct vertices.\n\nProof. Induct on $n$. The base case, $n=1$, is clear.\n\nNow take a polygon $P$ of perimeter $6 n \\geq 12$. Place six equilateral triangles inwards on six edges corresponding to each other upon rotation at $60^{\\circ}$. It is possible to stick a lozenge to each other edge, as shown in the Figure below.\n\nWe show that all angles of the lozenges are at least $60^{\\circ}$. Let an edge $X Y$ of the polygon bearing some lozenge lie along a boundary segment between edges $A B$ and $C D$ bearing equilateral triangles $A B P$ and $C D Q$. Then the angle formed by $\\overrightarrow{X Y}$ and $\\overrightarrow{B P}$ is between those formed by $\\overrightarrow{A B}, \\overrightarrow{B P}$ and $\\overrightarrow{C D}, \\overrightarrow{C Q}$, i.e., between $60^{\\circ}$ and $120^{\\circ}$, as desired.\n\nRemoving all obtained tiles, we get a $60^{\\circ}$-symmetric convex $6(n-1)$-gon with unit sides which can be tiled by the inductive hypothesis. Finally, the number of vertices in the tiling of $P$ is $6 n+3(n-1)^{2}+3(n-1)+1=3 n^{2}+3 n+1$, as desired.\n\n![](attached_image_1.png)\n\nUsing the Claim above, we now show that the citizens may be placed at the $3 n^{2}+3 n+1$ tile vertices.\n\nConsider any tile $T_{1}$; its vertices are at least 1 apart from each other. Moreover, let $B A C$ be a part of the boundary of some tile $T$, and let $X$ be any point of the boundary of $T$, lying outside the half-open intervals $[A, B)$ and $[A, C)$ (in this case, we say that $X$ is not adjacent to $A$). Then $A X \\geq \\sqrt{3} / 2$.\n\nNow consider any two tile vertices $A$ and $B$. If they are vertices of the same tile we already know $A B \\geq 1$; otherwise, the segment $A B$ crosses the boundaries of some tiles containing $A$ and $B$ at some points $X$ and $Y$ not adjacent to $A$ and $B$, respectively. Hence $A B \\geq A X+Y B \\geq \\sqrt{3}>1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInitially, a non-constant polynomial $S(x)$ with real coefficients is written down on a board. Whenever the board contains a polynomial $P(x)$, not necessarily alone, one can write down on the board any polynomial of the form $P(C+x)$ or $C+P(x)$, where $C$ is a real constant. Moreover, if the board contains two (not necessarily distinct) polynomials $P(x)$ and $Q(x)$, one can write $P(Q(x))$ and $P(x)+Q(x)$ down on the board. No polynomial is ever erased from the board.\n\nGiven two sets of real numbers, $A=\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ and $B=\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$, a polynomial $f(x)$ with real coefficients is $(A, B)$-nice if $f(A)=B$, where $f(A)=\\{f\\left(a_{i}\\right): i=1,2, \\ldots, n\\}$.\n\nDetermine all polynomials $S(x)$ that can initially be written down on the board such that, for any two finite sets $A$ and $B$ of real numbers, with $|A|=|B|$, one can produce an $(A, B)$-nice polynomial in a finite number of steps.",
    "options": [],
    "answer": "All non-constant polynomials of even degree at least two, and all odd-degree polynomials of degree at least three with negative leading coefficient.",
    "solution": "Solution:\n\nThe required polynomials are all polynomials of an even degree $d \\geq 2$, and all polynomials of odd degree $d \\geq 3$ with negative leading coefficient.\n\nPart I. We begin by showing that any (non-constant) polynomial $S(x)$ not listed above is not $(A, B)$-nice for some pair $(A, B)$ with either $|A|=|B|=2$, or $|A|=|B|=3$.\n\nIf $S(x)$ is linear, then so are all the polynomials appearing on the board. Therefore, none of them will be $(A, B)$-nice, say, for $A=\\{1,2,3\\}$ and $B=\\{1,2,4\\}$, as desired.\n\nOtherwise, $\\operatorname{deg} S=d \\geq 3$ is odd, and the leading coefficient is positive. In this case, we make use of the following technical fact, whose proof is presented at the end of the solution.\n\nClaim. There exists a positive constant $T$ such that $S(x)$ satisfies the following condition:\n$$\nS(b)-S(a) \\geq b-a \\quad \\text{ whenever } \\quad b-a \\geq T\n$$\nFix a constant $T$ provided by the Claim. Then, an immediate check shows that all newly appearing polynomials on the board also satisfy $(*)$ (with the same value of $T$ ). Therefore, none of them will be $(A, B)$-nice, say, for $A=\\{0, T\\}$ and $B=\\{0, T / 2\\}$, as desired.\n\nPart II. We show that the polynomials listed in the Answer satisfy the requirements. We will show that for any $a_{1}<a_{2}<\\cdots<a_{n}$ and any $b_{1} \\leq b_{2} \\leq \\cdots \\leq b_{n}$ there exists a polynomial $f(x)$ satisfying $f\\left(a_{i}\\right)=b_{\\sigma(i)}$ for all $i=1,2, \\ldots, n$, where $\\sigma$ is some permutation.\n\nThe proof goes by induction on $n \\geq 2$. It is based on the following two lemmas, first of which is merely the base case $n=2$; the proofs of the lemmas are also at the end of the solution.\n\nLemma 1. For any $a_{1}<a_{2}$ and any $b_{1}, b_{2}$ one can write down on the board a polynomial $F(x)$ satisfying $F\\left(a_{i}\\right)=b_{i}, i=1,2$.\n\nLemma 2. For any distinct numbers $a_{1}<a_{2}<\\cdots<a_{n}$ one can produce a polynomial $F(x)$ on the board such that the list $F\\left(a_{1}\\right), F\\left(a_{2}\\right), \\ldots, F\\left(a_{n}\\right)$ contains exactly $n-1$ distinct numbers, and $F\\left(a_{1}\\right)=F\\left(a_{2}\\right)$.\n\nNow, in order to perform the inductive step, we may replace the polynomial $S(x)$ with its shifted copy $S(C+x)$ so that the values $S\\left(a_{i}\\right)$ are pairwise distinct. Applying Lemma 2 , we get a polynomial $f(x)$ such that only two among the numbers $c_{i}=f\\left(a_{i}\\right)$ coincide, namely $c_{1}$ and $c_{2}$. Now apply Lemma 1 to get a polynomial $g(x)$ such that $g\\left(a_{1}\\right)=b_{1}$ and $g\\left(a_{2}\\right)=b_{2}$. Apply the inductive hypothesis in order to obtain a polynomial $h(x)$ satisfying $h\\left(c_{i}\\right)=b_{i}-g\\left(a_{i}\\right)$ for all $i=2,3, \\ldots, n$. Then the polynomial $h(f(x))+g(x)$ is a desired one; indeed, we have $h\\left(f\\left(a_{i}\\right)\\right)+g\\left(a_{i}\\right)=h\\left(c_{i}\\right)+g\\left(a_{i}\\right)=b_{i}$ for all $i=2,3, \\ldots, n$, and finally $h\\left(f\\left(a_{1}\\right)\\right)+g\\left(a_{1}\\right)=$ $h\\left(c_{1}\\right)+g\\left(a_{1}\\right)=b_{2}-g\\left(a_{2}\\right)+g\\left(a_{1}\\right)=b_{1}$.\n\nIt remains to prove the Claim and the two Lemmas.\n\nProof of the Claim. There exists some segment $\\Delta=\\left[\\alpha', \\beta'\\right]$ such that $S(x)$ is monotone increasing outside that segment. Now one can choose $\\alpha \\leq \\alpha'$ and $\\beta \\geq \\beta'$ such that $S(\\alpha)<\\min _{x \\in \\Delta} S(x)$ and $S(\\beta)>\\max _{x \\in \\Delta} S(x)$. Therefore, for any $x, y, z$ with $x \\leq \\alpha \\leq y \\leq \\beta \\leq z$ we get $S(x) \\leq S(\\alpha) \\leq S(y) \\leq S(\\beta) \\leq S(z)$.\n\nWe may decrease $\\alpha$ and increase $\\beta$ (preserving the condition above) so that, in addition, $S'(x)>3$ for all $x \\notin[\\alpha, \\beta]$. Now we claim that the number $T=3(\\beta-\\alpha)$ fits the bill.\n\nIndeed, take any $a$ and $b$ with $b-a \\geq T$. Even if the segment $[a, b]$ crosses $[\\alpha, \\beta]$, there still is a segment $\\left[a', b'\\right] \\subseteq[a, b] \\backslash(\\alpha, \\beta)$ of length $b'-a' \\geq(b-a) / 3$. Then\n$$\nS(b)-S(a) \\geq S\\left(b'\\right)-S\\left(a'\\right)=\\left(b'-a'\\right) \\cdot S'(\\xi) \\geq 3\\left(b'-a'\\right) \\geq b-a\n$$\nfor some $\\xi \\in\\left(a', b'\\right)$.\n\nProof of Lemma 1. If $S(x)$ has an even degree, then the polynomial $T(x)=S\\left(x+a_{2}\\right)-S\\left(x+a_{1}\\right)$ has an odd degree, hence there exists $x_{0}$ with $T\\left(x_{0}\\right)=S\\left(x_{0}+a_{2}\\right)-S\\left(x_{0}+a_{1}\\right)=b_{2}-b_{1}$. Setting $G(x)=S\\left(x+x_{0}\\right)$, we see that $G\\left(a_{2}\\right)-G\\left(a_{1}\\right)=b_{2}-b_{1}$, so a suitable shift $F(x)=G(x)+\\left(b_{1}-G\\left(a_{1}\\right)\\right)$ fits the bill.\n\nAssume now that $S(x)$ has odd degree and a negative leading coefficient. Notice that the polynomial $S^{2}(x):=S(S(x))$ has an odd degree and a positive leading coefficient. So, the polynomial $S^{2}\\left(x+a_{2}\\right)-S^{2}\\left(x+a_{1}\\right)$ attains all sufficiently large positive values, while $S\\left(x+a_{2}\\right)-S\\left(x+a_{1}\\right)$ attains all sufficiently large negative values. Therefore, the two-variable polynomial $S^{2}\\left(x+a_{2}\\right)-S^{2}\\left(x+a_{1}\\right)+S\\left(y+a_{2}\\right)-S\\left(y+a_{1}\\right)$ attains all real values; in particular, there exist $x_{0}$ and $y_{0}$ with $S^{2}\\left(x_{0}+a_{2}\\right)+S\\left(y_{0}+a_{2}\\right)-S^{2}\\left(x_{0}+a_{1}\\right)-S\\left(y_{0}+a_{1}\\right)=b_{2}-b_{1}$. Setting $G(x)=S^{2}\\left(x+x_{0}\\right)+S\\left(x+y_{0}\\right)$, we see that $G\\left(a_{2}\\right)-G\\left(a_{1}\\right)=b_{2}-b_{1}$, so a suitable shift of $G$ fits the bill.\n\nProof of Lemma 2. Let $\\Delta$ denote the segment $\\left[a_{1} ; a_{n}\\right]$. We modify the proof of Lemma 1 in order to obtain a polynomial $F$ convex (or concave) on $\\Delta$ such that $F\\left(a_{1}\\right)=F\\left(a_{2}\\right)$; then $F$ is a desired polynomial. Say that a polynomial $H(x)$ is good if $H$ is convex on $\\Delta$.\n\nIf $\\operatorname{deg} S$ is even, and its leading coefficient is positive, then $S(x+c)$ is good for all sufficiently large negative $c$, and $S\\left(a_{2}+c\\right)-S\\left(a_{1}+c\\right)$ attains all sufficiently large negative values for such $c$. Similarly, $S(x+c)$ is good for all sufficiently large positive $c$, and $S\\left(a_{2}+c\\right)-S\\left(a_{1}+c\\right)$ attains all sufficiently large positive values for such $c$. Therefore, there exist large $c_{1}<0<c_{2}$ such that $S\\left(x+c_{1}\\right)+S\\left(x+c_{2}\\right)$ is a desired polynomial. If the leading coefficient of $H$ is negative, we similarly find a desired polynomial which is concave on $\\Delta$.\n\nIf $\\operatorname{deg} S \\geq 3$ is odd (and the leading coefficient is negative), then $S(x+c)$ is good for all sufficiently large negative $c$, and $S\\left(a_{2}+c\\right)-S\\left(a_{1}+c\\right)$ attains all sufficiently large negative values for such $c$. Similarly, $S^{2}(x+c)$ is good for all sufficiently large positive $c$, and $S^{2}\\left(a_{2}+c\\right)-S^{2}\\left(a_{1}+c\\right)$ attains all sufficiently large positive values for such $c$. Therefore, there exist large $c_{1}<0<c_{2}$ such that $S\\left(x+c_{1}\\right)+S^{2}\\left(x+c_{2}\\right)$ is a desired polynomial.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39531,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Mumbo-Jumbo tribe lives at a river. Once two members of a tribe — a young warrior Mumbo and a wise shaman Jumbo — departed to the neighboring tribe with an urgent news. Mumbo ran at $11$ km/h to the nearest location of the rafts, took a raft and drove it to the neighboring tribe. Jumbo went at $6$ km/h to another raft location and took a raft there. It appeared that Jumbo came to another tribe earlier than Mumbo.\nThe river is straight, and the rafts move only at the velocity of the river flow. This velocity is constant, and it has an integer value (in km/h). Find the greatest possible value of the river flow velocity. (M. Evdokimov, reformulated by L. Samoylov)",
    "options": [],
    "answer": "26 km/h",
    "solution": "**Answer.** $26$ km/h.\n\nLet the location of the Mumbo-Jumbo tribe be $O$, the raft location to which Mumbo ran be $M$, and the raft location to which Jumbo went be $U$. Obviously, $M$ is upstream from $O$, and $U$ is downstream.\n\nLet the distances from $O$ to $M$ and $U$ be $x$ and $y$ km respectively ($x < y$), and let the river speed be $v$ km/h. The time Jumbo spent from $O$ to $U$ is $\\frac{y}{6}$ hours, and Mumbo spent $\\frac{x}{11} + \\frac{x+y}{v}$ hours. Clearly, Jumbo arrives at the neighboring tribe earlier than Mumbo if and only if\n\n$$\n\\frac{y}{6} < \\frac{x}{11} + \\frac{x+y}{v}.\n$$\n\nSince $x < y$, it follows from this inequality that\n\n$$\n\\frac{y}{6} < \\frac{y}{11} + \\frac{2y}{v}.\n$$\n\nDividing by $y$ and rearranging, we get $v < 26.4$.\n\nIt remains to check that the river speed could be $26$ km/h. For this, in the inequality $\\frac{y}{6} < \\frac{x}{11} + \\frac{x+y}{v}$, set $v = 26$ and equivalently transform it to $\\frac{y}{x} < \\frac{111}{110}$. This is possible (for example, for $y = 1.12$ km, $x = 1.11$ km), which completes the solution.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there exist nonzero real numbers $a, b, c, d$ for which the polynomial\n$$\n(ax + b)^{1000} - (cx + d)^{1000}\n$$\nafter expanding all brackets (and collecting terms) has exactly $n$ nonzero coefficients.",
    "options": [],
    "answer": "1001, 1000, 500",
    "solution": "The answers are $1001$, $1000$, and $500$.\n\nIt is clear that there exist such polynomials with $1001$ and $1000$ nonzero coefficients (for example, $(2x+2)^{1000} - (x+1)^{1000}$ and $(2x+1)^{1000} - (x+1)^{1000}$).\n\nSuppose that in our polynomial there are two coefficients equal to zero — for $x^i$ and $x^j$ ($i > j$). Then $a^i b^{1000-i} = c^i d^{1000-i}$, $a^j b^{1000-j} = c^j d^{1000-j}$; dividing the first equation by the second, we get\n$$(\\frac{ad}{bc})^i = (\\frac{d}{b})^{1000} = (\\frac{ad}{bc})^j.$$\nHence $|\\frac{ad}{bc}| = 1$, $|\\frac{d}{b}| = 1$ and $|\\frac{a}{c}| = 1$.\n\nIt is clear that replacing $ax+b$ with $(-a)x + (-b)$ does not change our polynomial. Therefore, we may assume $\\frac{a}{c} = 1$. Then, if $\\frac{b}{d} = 1$, the resulting polynomial $(ax+b)^{1000} - (ax+b)^{1000}$ is identically zero, and if $\\frac{b}{d} = -1$, then in the resulting polynomial $(ax+b)^{1000} - (ax - b)^{1000}$, exactly the coefficients at even powers of $x$ vanish, so we get $500$ nonzero coefficients.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39533,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic trinomial $f(x)$ is chosen so that the equation\n$$(f(x))^3 - f(x) = 0$$\nhas exactly $3$ real roots. Find the $y$-coordinate of the vertex of the graph of $f(x)$.",
    "options": [],
    "answer": "0",
    "solution": "Ответ. $0$.\n\nТак как $g(x) = (f(x))^3 - f(x) = f(x)(f(x) - 1)(f(x) + 1)$,\nто корнями многочлена $g(x)$ являются корни трехчленов $f(x)$,\n$f(x) - 1$ и $f(x) + 1$. Ясно, что любое число может быть корнем\nтолько одного из них.\n\nПусть $y_0$ — искомая ордината вершины. Предположим, что\n$y_0 \\neq 0$. Будем считать, что старший коэффициент в $f(x)$ положителен (иначе заменим $f(x)$ на $-f(x)$, при этом $y_0$ заменится\nна $-y_0$). Предположим, что $y_0 > 0$; тогда $f(x) > 0$ и $f(x) + 1 > 0$ при всех $x$, значит, корни многочлена $g(x)$ являются корнями $f(x) - 1$, а их не больше двух. Если же $y_0 < 0$, то трехчлены $f(x)$ и $f(x) - 1$ имеют по два корня, значит, $g(x)$ имеет хотя бы $4$ корня. Оба случая невозможны; значит, $y_0 = 0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a positive integer $n > 1$ such that the product of some $n$ consecutive positive integers equals the product of some $n + 100$ consecutive positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For example, let $n = (1 \\cdot 2 \\cdot 3 \\cdots 101) - 101$. Then the product of the first $n + 100$ natural numbers equals the product of $n$ consecutive numbers starting from $102$ and ending at $n + 101$.\n\nIndeed, after cancellation, the equality\n$$\n1 \\cdot 2 \\cdot 3 \\cdots (n+100) = 102 \\cdot 103 \\cdots (n+101)\n$$\nreduces to\n$$\n1 \\cdot 2 \\cdot 3 \\cdots 101 = n + 101,\n$$\nwhich is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39535,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function\n$$\nf(x) = \\cos x \\cdot \\cos \\frac{x}{2} \\cdot \\cos \\frac{x}{3} \\cdots \\cos \\frac{x}{2009}\n$$\nis defined on the segment $[0, \\frac{2009\\pi}{2}]$. Find the number of points at which this function changes its sign. (B. Trushin)",
    "options": [],
    "answer": "75",
    "solution": "Ответ. 75 раз.\nОбозначим $n = 2009$. Рассмотрим функцию $\\cos \\frac{x}{k}$. Она меняет знак при\n$$\nx = k \\left( \\pi m + \\frac{\\pi}{2} \\right) = \\frac{k(2m + 1)\\pi}{2},\n$$\nгде $m$ — произвольное целое число. Значит, нулями функции $f(x)$ могут являться только точки $x_i = \\frac{\\pi i}{2}$, где $1 \\le i \\le n$. Тогда менять знак она может лишь в точках $x_i$ при $i = 1, 2, ..., n-1$.\nФункция $\\cos \\frac{x}{k}$ меняет знак в точке $x_i$, если $i = k(2m+1)$ при целом $m$. Значит, количество косинусов, которые меняют знак в точке $x_i$, совпадает с количеством нечетных делителей числа $i$. Поэтому функция $f(x)$ будет менять знак в тех и только тех\nточках $x_i$, для которых у числа $i$ нечетное количество нечетных делителей. Представим $i = 2^\\ell j$, где $\\ell$ целое, а $j$ нечетно. Тогда количество нечетных делителей у чисел $i$ и $j$ совпадает. Но у нечетного числа $j$ количество делителей нечетно тогда и только тогда, когда $j$ — это точный квадрат (действительно, если $j$ — не квадрат, то все его делители разбиваются на пары $a$ и $\\frac{i}{a}$; если же $j = s^2$, то число $s$ — единственное оставшееся без пары).\nВ итоге получаем, что функция $f(x)$ меняет знак в точке $x_i$, если число $i$ — либо квадрат, либо удвоенный квадрат (в зависимости от четности числа $\\ell$). Но среди чисел от 1 до $(n-1)$ есть $[\\sqrt{n-1}]$ квадратов, и $[\\sqrt{\\frac{n-1}{2}}]$ удвоенных квадратов. Значит, искомое количество равно $[\\sqrt{n-1}] + [\\sqrt{\\frac{n-1}{2}}] = 44+31 = 75$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Konstantin had two sets of 17 coins: the first set consists of 17 genuine coins, while the second set consists of 12 genuine and 5 false coins. (All the coins look equally; the genuine coins have equal weights, and the false coins have equal weights different from the weight of genuine ones; but it is not known if the weight of a true coin is greater or less than the weight of a false coin.) Konstantin gave one of these sets of coins to his friend, but he forgot if the remaining set contains false coins. Determine if Konstantin can find it out using two weighings on a balance. (K. Knop)",
    "options": [],
    "answer": "Yes",
    "solution": "Назовем набор из 17 настоящих монет хорошим, а набор из 12 настоящих и 5 фальшивых монет — плохим. Опишем возможные действия Кости.\n\nРазобьем 17 монет из оставшегося у него набора на четыре группы $A, B, C, D$, содержащие 2, 3, 5, и 7 монет соответственно. При первом взвешивании на одну чашу весов положим группы $A$ и $B$, а на другую — $C$. При втором взвешивании на одну чашу весов положим группы $A$ и $C$, а на другую — $D$.\n\nЕсли в одном из взвешиваний весы не показали равновесие, то оставшийся у Кости набор плохой. Покажем, что если весы в обоих случаях показали равновесие, то оставшийся у Кости набор хороший.\n\nПусть это не так, и количества фальшивых монет в группах $A, B, C, D$ равны соответственно $a, b, c, d$. Тогда $a + b + c + d = 5$. Из равновесия при двух взвешиваниях следует $a + b = c$ и $a + c = d$, откуда $5 = a + b + c + d = a + b + (a + b) + (2a + b) = 4a + 3b$.\n\nВ случаях $a = 0$ или $b = 0$ получаем соответственно $b = \\frac{5}{3}$ или $a = \\frac{5}{4}$, что невозможно. Если же $a \\ge 1$ и $b \\ge 1$, то $4a + 3b \\ge 7 > 5$. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39537,
    "subject": "Mathematics (Multi-modal)",
    "question": "The two circles $\\omega_1$ and $\\omega_2$ touch each other externally at point $O$. Let $A$ and $B$ be points on $\\omega_1$, and $C$ and $D$ be points on $\\omega_2$ such that lines $AC$ and $BD$ are common external tangents to the circles. Line $AO$ intersects segment $CD$ at point $M$, while line $CO$ intersects $\\omega_1$ at points $O$ and $N$. Prove that points $B$, $M$, and $N$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Из симметрии относительно линии центров $\\ell$ имеем: $AB \\parallel CD \\perp \\ell$, $OA = OB$, $\\angle OAB = \\angle OBA = \\alpha$, $\\angle OCD = \\angle ODC = \\beta$ (см. рис. 9).\n\nПо теореме об угле между касательной и хордой $\\angle OAC = \\angle OBA = \\alpha$, $\\angle OCA = \\angle ODC = \\beta$. Так как $AB \\parallel CD$, то $\\angle OMC = \\angle OAB = \\alpha$; значит, и $\\angle AOC = 180^\\circ - \\alpha - \\beta = \\angle MOC$. Треугольники $OAC$ и $OMC$ равны по стороне ($OC$ — общая) и двум углам, поэтому $OA = OM$ и $\\angle AOC = \\angle MOC = 90^\\circ$.\n\nВ треугольнике $ABM$ имеем $AO = OM = OB$; следовательно, он прямоугольный, то есть $\\angle ABM = 90^\\circ$. Так как $\\angle AON = \\angle MOC = 90^\\circ$, то $AN$ — диаметр окружности $\\omega_1$, откуда $\\angle ABN = 90^\\circ$. Итак, $\\angle ABM = \\angle ABN = 90^\\circ$, значит, точки $B$, $M$, $N$ лежат на перпендикуляре к $AB$, проведенном через точку $B$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39538,
    "subject": "Mathematics (Multi-modal)",
    "question": "2009 nonnegative integers are arranged on a circle, each number does not exceed $100$. A positive integer $k$ is fixed. By one move, one can choose two neighboring positions on a circle and add $1$ to both numbers in these positions. It is allowed to make at most $k$ moves for each pair of neighboring positions. Find the least value of $k$ such that, from each initial position, one can make all the numbers equal. (I. Bogdanov)",
    "options": [],
    "answer": "100400",
    "solution": "**Ответ.** $k = 100400$.\n\nОбозначим числа на окружности через $a_1, \\dots, a_{2009}$, и положим $a_{n+2009} = a_n = a_{n-2009}$. Пусть $N = 100400$.\n\n1. Положим $a_2 = a_4 = \\dots = a_{2008} = 100$ и $a_1 = a_3 = \\dots = a_{2009} = 0$. Пусть мы сумели сделать все числа равными при каком-то значении $k$. Рассмотрим сумму $S = (a_2 - a_3) + (a_4 - a_5) + \\dots + (a_{2008} - a_{2009})$. Эта сумма увеличивается на $1$ при прибавлении единицы к паре $(a_1, a_2)$, уменьшается на $1$ при прибавлении к паре $(a_{2009}, a_1)$ и не изменяется при всех остальных операциях. Поскольку исходное значение $S$ равно $S_0 = 100 \\cdot 1004 = N$, а конечное должно быть нулем, то пара $(a_{2009}, a_1)$ увеличивалась хотя бы $N$ раз. Это значит, что $k \\ge N$.\n\n2. Осталось показать, что при $k = N$ требуемое всегда возможно. Рассмотрим произвольный набор чисел $a_i$. Увеличим каждую пару $(a_i, a_{i+1})$ ровно $s_i = a_{i+2} + a_{i+4} + \\dots + a_{i+2008}$ раз. Тогда число $a_i$ превратится в\n$$a_i + s_{i-1} + s_i = a_i + (a_{i+1} + a_{i+3} + \\dots + a_{i+2007}) + (a_{i+2} + a_{i+4} + \\dots + a_{i+2008}) = a_1 + \\dots + a_{2009},$$\nто есть все числа станут равными. С другой стороны, $s_i \\le 1004 \\cdot 100 = N$, что и требовалось.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39539,
    "subject": "Mathematics (Multi-modal)",
    "question": "An increasing sequence of positive integers $\\{a_n\\}$ and a positive integer $k$ are given. For each positive integer $n$ the following conditions hold: the number $a_n$ is divisible either by $1005$ or by $1006$, $a_n$ is not divisible by $97$, and $a_{n+1} - a_n \\le k$. Find the least possible value of $k$. (A. Golovanov)",
    "options": [],
    "answer": "2010",
    "solution": "**Ответ.** $k = 2010$.\n\nLet us denote our sequence by $(a_n)$. Clearly, $a_1 < 1005 \\cdot 1006 \\cdot 97 \\cdot N = D$ for some natural $N$. Then there exists such $n$ that $a_n \\le D$, but $a_{n+1} > D$ (at the same time, $a_n \\ne D$ by the condition). But the largest numbers less than $D$ and divisible by $1005$ and $1006$ are $D-1005$ and $D-1006$, respectively; therefore $a_n \\le D-1005$. Similarly, $a_{n+1} \\ge D+1005$; hence $a_{n+1}-a_n \\ge (D+1005)-(D-1005) = 2010$. Thus, $k \\ge 2010$.\n\nFor $k=2010$, for example, the sequence of all numbers divisible by $1005$ but not by $97$ fits (note that $1005$ is not divisible by $97$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39540,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $m$ is chosen so that the sum of all the digits of $2^m$ (in its decimal representation) equals 8. Determine if the last digit of $2^m$ can appear to be 6. (V. Senderov)",
    "options": [],
    "answer": "It cannot.",
    "solution": "Answer. It cannot.\n\nFirst solution. For $m = 1, 2, 3$ the last digit of $2^m$ is not 6. Suppose that the sum of the digits of $2^m$ for some $m > 3$ is 8, and it ends with 6. The number $2^m$ cannot end with 06 or 26, since in this case it is not divisible by 4. Therefore, it ends with 16 (otherwise the sum of digits will be greater than 8), and thus has the decimal representation $1000\\ldots016$. Then $2^m = 10^k + 16$, that is, the number $10^k + 16$ with $k$ digits is a power of two.\nBut if $k \\ge 5$, then $10^k + 16 = 2^4 (2^k - 4 \\cdot 5^k + 1)$, and in the parentheses we get an odd factor greater than 1. It remains to consider the cases $k = 2, k = 3, k = 4$: $10^2 + 16 = 4 \\cdot 29$; $10^3 + 16 = 8 \\cdot 127$; $10^4 + 16 = 32 \\cdot 313$. Thus, $10^k + 16$ is not a power of two for any natural $k$, which is what was required to be proved.\n\nSecond solution. Suppose the contrary, and let $2^m$ end with 6 and have a digit sum equal to 8. Note that $2^1$ ends with 2, $2^2$ ends with 4, $2^3$ ends with 8, $2^4$ ends with 6, $2^5$ ends with 2. Further, the last digit of a power of two repeats with period 4, since the last digit of $2^m$ is uniquely determined by the last digit of $2^{m-1}$. Thus, $2^m$ ends with 6 if and only if $m$ is divisible by 4.\nThe sum of the digits of a number has the same remainder modulo 3 as the number itself, so $2^m$ must have remainder 2 modulo 3. Note that $2^1$ has remainder 2 modulo 3, $2^2$ has remainder 1 modulo 3, $2^3$ has remainder 2 modulo 3, and further the remainders of powers of two repeat with period 2. Thus, $2^m$ has remainder 2 modulo 3 if and only if $m$ is odd. This contradicts the fact that $m$ must be divisible by 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39541,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Kingdom consists of $N$ cities, some pairs of cities are connected with pairwise non-intersecting roads (two cities connected by a road are called *neighboring* cities). One can start from each city and get to any other city passing through roads, but it is not possible to get back to the initial city using each road not more than once.\nOnce the King made a reform: each of $N$ Mayors became a Mayor of one of $N$ cities, perhaps not the city where he worked before the reform. It occurs that each two Mayors that worked in the neighboring cities before the reform work in the neighboring cities after the reform. Prove that either there exists a city having the same Mayor before and after the reform, or there exists a neighboring pair of cities which have interchanged their Mayors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Применим индукцию по $N$. Утверждение задачи очевидно при $N=1$ и $N=2$. Пусть $\\Gamma_1$ — множество городов, из которых исходит одна дорога, а $\\Gamma$ — множество остальных городов. Начав движение по различным дорогам из некоторого города, согласно условию, мы не сможем попасть дважды в один и тот же город, поэтому когда-нибудь мы закончим движение в городе из множества $\\Gamma_1$. Это означает, что множество $\\Gamma_1$ непусто. Ясно, что при $N \\ge 3$ множество $\\Gamma$ непусто, и в нем меньше, чем $N$ городов.\n\nНазовем значимостью мэра количество городов, соседних с городом, где он работает. По условию значимость каждого мэра после реформы не уменьшилась. В частности, мэр города, принадлежащего множеству $\\Gamma$, после реформы стал мэром некоторого города из множества $\\Gamma$, то есть в результате реформы в городах множества $\\Gamma$ тоже произошла перестановка мэров. Ясно, что из любого города $A \\in \\Gamma$ можно доехать до любого другого города $B \\in \\Gamma$, не заезжая в города множества $\\Gamma_1$.\n\nПоэтому множество городов $\\Gamma$ и реформа, рассмотренная только на городах из $\\Gamma$, удовлетворяет условию задачи. Применив предположение индукции, получаем, что в множестве $\\Gamma$ либо найдется город, в котором мэр после реформы не поменялся, либо найдется пара соседних городов, обменявшихся мэрами, что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39542,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle $\\omega$ of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Point $D$ is chosen on line $AA_1$ so that $AD = AC_1$, and point $A$ lies between $A_1$ and $D$. Lines $DB_1$ and $DC_1$ intersect $\\omega$ at points $B_2 \\neq B_1$ and $C_2 \\neq C_1$, respectively. Prove that $B_2C_2$ is a diameter of $\\omega$. (R. Zhenodarov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $I$ — центр окружности $\\omega$; положим $\\alpha = \\angle BAC$. Так как в четырехугольнике $AB_1IC_1$ углы $AB_1I$ и $AC_1I$ прямые, то $\\angle B_1IC_1 = 180^\\circ - \\angle B_1AC_1 = 180^\\circ - \\alpha$. Поэтому в окружности $\\omega$ мера дуги $\\overarc{B_1C_1}$, не содержащей точки $A_1$, равна $180^\\circ - \\alpha$.\n\nИз равенств $AB_1 = AC_1 = AD$ следует, что $A$ является центром окружности $\\omega_1$, описанной вокруг треугольника $B_1C_1D$. Центральный угол $B_1AC_1$ окружности $\\omega_1$ вдвое больше угла $B_1DC_1$, следовательно, $\\angle B_1DC_1 = \\alpha/2$.\n\nДалее, угол $B_1DC_1$ равен полуразности величин дуг $\\overarc{B_2C_2}$ и $\\overarc{B_1C_1}$, откуда $\\overarc{B_2C_2} = 2\\angle B_1DC_1 + \\overarc{B_1C_1} = \\alpha + (180^\\circ - \\alpha) = 180^\\circ$.\n\nПолученное равенство $\\overarc{B_2C_2} = 180^\\circ$ означает, что $B_2C_2$ — диаметр окружности $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39543,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle with center $I$ touches the sides $AB$, $BC$, $AC$ of non-isosceles triangle $ABC$ at points $C_1$, $A_1$, $B_1$, respectively. Circles $\\omega_B$ and $\\omega_C$ are inscribed into the quadrilaterals $BA_1IC_1$ and $CA_1IB_1$, respectively. Prove that the common internal tangent to $\\omega_B$ and $\\omega_C$, distinct from line $IA_1$, passes through $A$. (I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Обозначим через $O_B$ и $O_C$ центры окружностей $\\omega_B$ и $\\omega_C$, соответственно. Заметим, что достаточно доказать, что $\\angle O_BAO_C = \\angle BAC/2$: тогда прямые, полученные из $AB$ и $AC$ симметриями относительно $AO_B$ и $AO_C$ соответственно, совпадают; значит, это и есть общая внутренняя касательная к $\\omega_B$ и $\\omega_C$. При этом, поскольку $AB \\neq AC$, эта касательная не совпадает с $IA_1$.\n\nПостроим вне $\\triangle ABC$ треугольник $AB_1O'$, равный $AC_1O_B$ (это возможно, так как $AB_1 = AC_1$; см. рис. 25). Тогда $AO_B = AO'$, и из равенства $\\angle C_1AO_B = \\angle B_1AO'$ следует $\\angle O_BAO' = \\angle BAC$.\n\nИз симметрии четырехугольника $BC_1IA_1$ относительно $AI$ следует, что $B_1O_B = A_1O_B$; кроме того, прямые $C_1I$ и $A_1I$ являются биссектрисами углов этого четырехугольника, поэтому $\\angle O_BA_1I = \\angle O_BC_1B = \\angle O'B_1C = 45^\\circ$. Аналогично, $B_1O_C = A_1O_C$ и $\\angle O_CA_1I = \\angle O_CB_1C = 45^\\circ$. Тогда $\\angle O_BA_1O_C = \\angle O_BA_1I + \\angle O_CA_1I = 90^\\circ$ и $\\angle O_CB_1O' = \\angle O_CB_1C + \\angle O'B_1C = 90^\\circ$. Тогда прямоугольные треугольники $O_BA_1O_C$ и $O'B_1O_C$ равны по двум катетам, откуда $O_BC = O'C$.\n\nТаким образом, треугольники $O_BAO_C$ и $O'AO_C$ равны по трем сторонам. Значит, $\\angle O_BAO_C = \\angle O'AO_C = \\angle O_BAO'/2 = \\angle BAC/2$, что и требовалось.\n\n\n**Второе решение.** Пусть $\\omega_B$ касается $BA_1$, $IA_1$ и $BC_1$ в точках $K_B$, $L_B$ и $M_B$ соответственно, а $\\omega_C$ касается $CA_1$, $IA_1$ и $CB_1$ в точках $K_C$, $L_C$ и $M_C$ соответственно. Обозначим через $O_B$ и $O_C$ центры окружностей $\\omega_B$ и $\\omega_C$ соответственно, а через $r_B$ и $r_C$ — их радиусы (пусть для определенности $r_B > r_C$).\n\nТогда четырехугольники $O_BK_BA_1L_B$ и $O_CK_CA_1L_C$ — квадраты, поэтому $A_1L_B = r_B$, $A_1L_C = r_C$ и $L_BL_C = r_B - r_C$.\n\nТогда, если вторая общая внутренняя касательная $\\ell$ касается $\\omega_B$ и $\\omega_C$ в точках $N_B$ и $N_C$, то $N_BN_C = r_B - r_C$, причем $N_C$ и $L_B$ лежат по одну сторону от линии центров $O_BO_C$ (заметим, что точка $A$ лежит по ту же сторону).\n\nАналогично получаем, что $C_1M_B = r_B$, $B_1M_C = r_C$. Отложим на продолжении отрезка $N_BN_C$ за точку $N_C$ отрезок $N_CA' = AM_C = AB_1 + B_1M_C$. Тогда $N_BA' = AM_C + N_BN_C = (AB_1 + r_C) + (r_B - r_C) = AC_1 + r_B = AM_B$. Итак, касательные из точек $A$ и $A'$ к окружности $\\omega_B$ равны, и касательные из них к $\\omega_C$ также равны.\n\nЗаметим, что геометрическое место точек, длина касательной из которых к окружности $\\omega_B$ равна $AM_B$, есть окружность $\\Omega_B$ с центром $O_B$ и радиусом $\\sqrt{r_B^2 + AM_B^2}$. Таким образом, точки $A$ и $A'$ лежат на $\\Omega_B$. Аналогично, они лежат на окружности $\\Omega_C$ с центром $O_C$ и радиусом $\\sqrt{r_C^2 + AM_C^2}$. Итак, каждая из точек $A$ и $A'$ является одной из двух точек пересечения окружностей $\\Omega_B$ и $\\Omega_C$, а поскольку $A$ и $A'$ лежат по одну сторону от $O_BO_C$, имеем $A' = A$. Значит, $A$ лежит на прямой $N_BN_C$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $x_1, x_2, \\dots, x_{2009}$ satisfy the relations $x_1^2 - x_1x_2 + x_2^2 = x_2^2 - x_2x_3 + x_3^2 = x_3^2 - x_3x_4 + x_4^2 = \\dots = x_{2008}^2 - x_{2008}x_{2009} + x_{2009}^2 = x_{2009}^2 - x_{2009}x_1 + x_1^2$. Prove that $x_1 = x_2 = \\dots = x_{2009}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Без ограничения общности можно считать, что $x_1$ — наибольшее (или одно из наибольших) среди данных чисел. Тогда из равенства $x_1^2 - x_1x_2 + x_2^2 = x_2^2 - x_2x_3 + x_3^2$ следует $x_1^2 - x_3^2 - x_1x_2 + x_2x_3 = 0 \\Leftrightarrow (x_1 - x_3)(x_1 + x_3) - x_2(x_1 - x_3) = 0 \\Leftrightarrow (x_1 - x_3)(x_1 + x_3 - x_2) = 0$. В силу выбора числа $x_1$ имеем $x_1 \\ge x_2$; поэтому $x_1 - x_2 \\ge 0$, значит, $x_1 + x_3 - x_2 > 0$. Отсюда следует, что $x_1 - x_3 = 0$, то есть $x_3 = x_1$, и $x_3$ — также одно из наибольших среди данных чисел.\nРассуждая далее таким же образом, получаем: $x_5 = x_3$, $x_7 = x_5$, \\ldots $x_{2009} = x_{2007}$, $x_2 = x_{2009}$, $x_4 = x_2$, \\ldots, $x_{2008} = x_{2006}$. Таким образом, все данные 2009 чисел равны.\n\n\n**Второе решение.** Запишем исходные равенства в виде:\n$$\nx_1(x_1 - x_2) = -x_3(x_2 - x_3)\n$$\n$$\nx_2(x_2 - x_3) = -x_4(x_3 - x_4)\n$$\n$$\nx_{2009}(x_{2009} - x_1) = -x_2(x_1 - x_2).\n$$\nПредположим, что $x_1 - x_2 > 0$. Тогда, последовательно используя все равенства, получаем $x_2 - x_3 < 0$, $x_3 - x_4 > 0$, $x_4 - x_5 < 0$, $x_5 - x_6 > 0$, \\ldots, $x_{2009} - x_1 > 0$, $x_1 - x_2 < 0$, что противоречит предположению. Аналогично приводится к противоречию предположение о том, что $x_1 - x_2 < 0$.\nЕсли же $x_1 - x_2 = 0$, то последовательно получаем $x_2 - x_3 = 0$, $x_3 - x_4 = 0$, \\ldots, $x_{2009} - x_1 = 0$, откуда $x_1 = x_2 = \\dots = x_{2009}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integers $x, y \\in [2, 100]$, prove that there exists a positive integer $n$ such that the number $x^{2^n} + y^{2^n}$ is composite.\n\n(S. Berlov, A. Kanel-Belov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $x = y$, then $n = 1$ works, since $x^2 + y^2$ is an even number greater than $2$.\n\nNow assume $x \\neq y$. In this case, we will show that for some $n$, the number $x^{2^n} + y^{2^n}$ is divisible by $257$ and not equal to $257$. Then it will be composite, as required.\n\nSuppose $x^{2^n} + y^{2^n} = 257$. Let $a = x^{2^{n-1}}$, $b = y^{2^{n-1}}$. Then $a^2 + b^2 = 257$, and if $a \\ge b$, then $a = 16$, $b = 1$ (the cases $a = 12, 13, 14, 15$ are easily checked; if $a \\le 11$, then $b^2 \\le a^2 < 257/2$, which is impossible). But this contradicts the condition $x, y > 1$. Thus, $x^{2^n} + y^{2^n} \\neq 257$.\n\nIt remains to check only that for some $n$, the number $x^{2^n} + y^{2^n}$ is divisible by $257$.\n\nSince $y$ is not divisible by the prime number $257$, there exists a natural $q$ such that $x \\equiv qy \\pmod{257}$. Since $x \\neq y$ and $0 < x + y < 257$, we get $q \\neq \\pm 1 \\pmod{257}$. Also, $q \\neq 0 \\pmod{257}$, since $x$ is not divisible by $257$.\n\nSince $257$ is prime, by Fermat's little theorem, the number $q^{256} - 1 = (q-1)(q+1)(q^2+1)(q^{2^2}+1) \\dots (q^{2^7} + 1)$ is divisible by $257$. The first two factors are not divisible by $257$, so for some $n \\in \\{1, 2, \\dots, 7\\}$, the number $q^{2^n} + 1$ is divisible by $257$. But then the number $x^{2^n} + y^{2^n} \\equiv y^{2^n} (q^{2^n} + 1) \\pmod{257}$ is also divisible by $257$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39546,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a party a company of 20 persons is to be arranged to sit around 4 tables. An arrangement is called *successful* if each two persons sharing a table are friends. It appears that there exists at least one successful arrangement, and for each successful arrangement exactly 5 persons sit around each table. Find the greatest possible number of pairs of friends in a company.",
    "options": [],
    "answer": "160",
    "solution": "**Answer.** $160$.\n\nIf two people are not friends, we will say that they are enemies. The total number of pairs of people in this company is $\\frac{20 \\cdot 19}{2} = 190$, so it is enough to prove that the minimal number of pairs of enemies is $30$.\n\nLet us prove that there cannot be fewer than $30$ pairs of enemies. Suppose the contrary. Then there is a person $A$ who has at most two enemies (if every person has at least three enemies, the number of pairs of enemies is at least $\\frac{3 \\cdot 20}{2} = 30$). Suppose $A$ is seated at some table $T$ in a successful arrangement. By the condition, there are exactly $5$ people at table $T$. Since $A$ has at most two enemies, at one of the remaining three tables, all the people are friends of $A$. Therefore, if we move $A$ to that table (and do not change the rest of the arrangement), the new arrangement will also be successful, but table $T$ will have only $4$ people, which contradicts the condition.\n\nLet us give an example satisfying the condition, in which there are exactly $30$ pairs of enemies. Divide all people into $5$ groups of $4$, and let any pair of people from the same group be enemies, and any pair from different groups be friends. In this case, each person has exactly $3$ enemies, and there are $\\frac{3 \\cdot 20}{2} = 30$ pairs of enemies in total. In the described situation, an arrangement is successful if and only if people from the same group are seated at different tables. Thus, successful arrangements exist, and in any successful arrangement, each table will have exactly one person from each group, that is, exactly $5$ people.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39547,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic trinomial $f(x)$ is chosen so that the equation $(f(x))^5 - f(x) = 0$ has exactly 3 real roots. Find the $y$-coordinate of the vertex of the graph of $f(x)$.",
    "options": [],
    "answer": "0",
    "solution": "Так как $g(x) = (f(x))^5 - f(x) = f(x)(f(x) - 1)(f(x) + 1)((f(x))^2 + 1)$, то корнями нашего многочлена являются корни трехчленов $f(x)$, $f(x) - 1$ и $f(x) + 1$ (поскольку многочлен $(f(x))^2 + 1$ всюду положителен). Ясно, что любое число может быть корнем только одного из них.\n\nПусть $y_0$ — искомая ордината вершины. Предположим, что $y_0 \\neq 0$. Будем считать, что старший коэффициент в $f(x)$ положителен (иначе заменим $f(x)$ на $-f(x)$, при этом $y_0$ заменится на $-y_0$). Предположим, что $y_0 > 0$; тогда $f(x) > 0$ и $f(x) + 1 > 0$ при всех $x$, значит, корни многочлена $g(x)$ являются корнями $f(x) - 1$, а их не больше двух. Если же $y_0 < 0$, то трехчлены $f(x)$ и $f(x) - 1$ имеют по два корня, значит, $g(x)$ имеет хотя бы 4 корня. Оба случая невозможны; значит, $y_0 = 0$.\n\n**Замечание.** При $y_0 = 0$ $f(x)$ имеет один корень, $f(x) + 1$ — ни одного, а $f(x) - 1$ — два, поэтому $g(x)$ как раз имеет три корня.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39548,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country, some pairs of cities are connected by the roads, which do not intersect outside the cities. At each city, there is a sign showing the least length of the route beginning at this city and passing through all other cities (a route can pass through some cities several times; it should not necessarily return to the original city). Prove that the ratio of any two numbers on the signs does not exceed $3/2$. (M. Murashkin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Рассмотрим самый короткий маршрут $\\ell$, проходящий по всем городам. Пусть он начинается в городе $A$, заканчивается в городе $B$, а его длина равна $N$. Тогда числа на табличках в городах $A$ и $B$ равны $N$, а все остальные не меньше $N$. Пусть $C$ — один из оставшихся городов. Он лежит на данном маршруте, поэтому длина пути от $C$ до одного из городов $A$ или $B$ не больше $N/2$ (без ограничения общности — до $A$). Рассмотрим маршрут, который выходит из $C$, доходит кратчайшим образом до $A$, а затем повторяет маршрут $\\ell$. Он проходит через все города, и его длина не превосходит $N/2 + N = 3N/2$. Следовательно, число на табличке в городе $C$ не больше $3N/2$.\n\nТаким образом, все числа на табличках принадлежат отрезку $[N, 3N/2]$, откуда и следует требуемое.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39549,
    "subject": "Mathematics (Multi-modal)",
    "question": "In several cells of a $10 \\times 10$ grid either a cross or a nought is placed. It appeared that there is no column or row occupied with $10$ signs of the same type, but if one places any sign into any empty cell, this condition would fail. Find the least possible number of cells occupied with signs.",
    "options": [],
    "answer": "98",
    "solution": "Let us fill the table according to the condition, and let cell $A$ be empty. Since placing any sign in it would create a row or column filled with identical signs, there must exist a line (row or column) containing $A$ in which all other cells are filled with crosses, and the same is true for noughts (these lines must, of course, be a row and a column). In particular, all empty cells are in different rows and columns.\n\n![](attached_image_1.png)\n\nFig. 11\n\n![](attached_image_2.png)\n\nFig. 12\n\nSuppose there are more than two empty cells. Then for two of them (say, $A$ and $B$), the direction of the line of crosses coincides (that is, either for both the crosses are in their rows, or for both in their columns) (see Fig. 11). Suppose the crosses are in their columns, and the noughts are in their rows; then at the intersection of the row containing $A$ and the column containing $B$, there must be both a cross and a nought (since this intersection is not empty); contradiction. Therefore, there are at most two empty cells. An example with two empty cells is shown in Fig. 12 (any signs can be placed instead of the asterisks).\n\nThus, the least possible number of cells occupied with signs is $98$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39550,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $a_1, a_2, \\dots$ satisfies the following conditions: $a_1 \\in (1, 2)$, and $a_{k+1} = a_k + \\frac{k}{a_k}$ for every positive integer $k$. Prove that there is at most one (unordered) pair of distinct indices $(i, j)$ such that $a_i + a_j$ is an integer. (A. Golovanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b_k = a_k - k$. Then\n$$\nb_{k+1} = b_k - 1 + \\frac{k}{k + b_k} = b_k - \\frac{b_k}{k + b_k} = b_k \\left(1 - \\frac{1}{k + b_k}\\right).\n$$\nFrom here, by obvious induction on $k$, we get that $b_k > 0$ (since $b_1 > 0$). Moreover, $b_{k+1} = b_k - \\frac{b_k}{k + b_k} < b_k$. Hence, in particular, $b_k \\leq b_1 < 1$.\n\nNote that $b_2 = a_1 + \\frac{1}{a_1} - 2 = (\\sqrt{a_1} - \\frac{1}{\\sqrt{a_1}})^2$. The expression in parentheses is positive and increases as $a_1$ runs through the interval $(1, 2)$; then $0 = 1 + \\frac{1}{2} - 2 < b_2 < 2 + \\frac{1}{2} - 2 = \\frac{3}{2}$. Thus, $b_k \\leq b_2 < \\frac{1}{2}$ for $k \\geq 2$.\n\nNow, if $a_k + a_j$ is an integer, then $b_k + b_j$ is also an integer. Therefore, one of the numbers $b_k, b_j$ (say $b_k$) is not less than $\\frac{1}{2}$; then $k = 1$, and $b_j = 1 - b_1$. But there are at most one such $j$, since the sequence $(b_i)$ is decreasing. This proves the statement of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $x \\cos x \\le \\frac{\\pi^2}{16}$ for every $0 \\le x \\le \\frac{\\pi}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Так как $\\sin t \\le t$ при $0 \\le t \\le \\frac{\\pi}{2}$, то имеем\n\n$x \\cos x = x \\sin (\\frac{\\pi}{2} - x) \\le x (\\frac{\\pi}{2} - x) = \\frac{\\pi^2}{16} - (x - \\frac{\\pi}{4})^2 \\le \\frac{\\pi^2}{16}$,\nчто и требовалось.\n\n\nВторое решение. Перепишем доказываемое неравенство в виде $\\sqrt{x \\cos x} \\le \\frac{\\pi}{4}$. По неравенству о средних имеем $\\sqrt{x \\cos x} \\le \\frac{x + \\cos x}{2}$. В правой части стоит неубывающая функция (так как $(x + \\cos x)' = 1 - \\sin x \\ge 0$); значит, $x + \\cos x \\le \\frac{\\pi}{2} + \\cos \\frac{\\pi}{2} = \\frac{\\pi}{2}$, откуда и следует требуемое.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39552,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a tetrahedron $ABCD$, none of the face angles is $90^\\circ$, and the orthocenters of the faces $ABC$, $ABD$, $ACD$ are collinear. Prove that the circumcenter of this tetrahedron is coplanar with the midpoints of edges $AB$, $AC$ and $AD$. (I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $AB_1$, $AC_1$, $AD_1$ — высоты граней $ACD$, $ABD$, $ABC$. Точки пересечения высот этих граней лежат на прямых $AB_1$, $AC_1$, $AD_1$ и отличны от точки $A$. Поскольку они лежат на одной прямой $\\ell$, то прямые $AB_1$, $AC_1$, $AD_1$ лежат в плоскости $\\alpha$, содержащей $\\ell$ и $A$ (ясно, что $A$ не лежит на $\\ell$). Значит, точки $B_1$, $C_1$, $D_1$ лежат на прямой пересечения плоскостей $\\alpha$ и $BCD$.\n\n![](attached_image_1.png)\nРис. 27\n![](attached_image_2.png)\nРис. 28\n\nПусть $A'$ — проекция точки $A$ на плоскость $BCD$. Тогда по теореме о трех перпендикулярах точки $B_1$, $C_1$, $D_1$ являются проекциями $A'$ на прямые $CD$, $BD$, $BC$ (см. рис. 27). Значит, точки $A'$, $C$, $B_1$, $D_1$ лежат на одной окружности (с диаметром $A'C$), а также точки $A'$, $D$, $B_1$, $C_1$ лежат на одной окружности (с диаметром $A'D$). Отсюда $\\angle(BC, A'C) = \\angle(D_1C, A'C) = \\angle(D_1B_1, A'B_1) = \\angle(C_1B_1, A'B_1) = \\angle(C_1D, A'D) = \\angle(BD, A'D)$ (здесь через $\\angle(a,b)$ обозначен угол от прямой $a$ до прямой $b$, отсчитываемый против часовой стрелки; этот угол считается с точностью до прибавления числа вида $\\pi k$, где $k$ — целое). Из равенства $\\angle(BC, A'C) = \\angle(BD, A'D)$ следует, что точка $A'$ лежит на описанной окружности треугольника $BCD$ (см. рис. 28) и, следовательно, на описанной сфере $S$ пирамиды $ABCD$.\n\nТогда центр $O$ сферы $S$ лежит в плоскости $\\beta$, являющейся серединным перпендикуляром к $AA'$. Ясно, что середины ребер $AB$, $AC$, $AD$ также лежат в $\\beta$ (так как треугольники $ABAA'$, $ACA'$, $ADA'$ прямоугольные). Это и требовалось доказать.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39553,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane, all the points with integer coordinates $(x, y)$ such that $x^2 + y^2 \\le 10^{10}$, are marked. Two players $A$ and $B$ play the following game. They move turn by turn. On the first move, $A$ places a token in some marked point and erases this point. After that, by each move the next player moves the token into some other marked point and erases this point. The constraints are the following. The length of each move should be strictly greater than the length of the previous move (made by the other player); moreover, it is prohibited to move the token from a point to the point symmetrical to it with respect to the origin. The player loses when he cannot make a move. Determine which player can win regardless by the opponent's moves.\n\n$(I. Bogdanov)$",
    "options": [],
    "answer": "Player A",
    "solution": "Первый игрок выигрывает.\n\nДокажем более общее утверждение: *Пусть игра с теми же правилами происходит на конечном множестве точек $S$, которое содержит точку $O(0,0)$ и переходит в себя при повороте на $90^\\circ$. Тогда в этой игре выигрывает первый игрок. (Ясно, что множество точек из условия удовлетворяет этим условиям.)*\n\nДоказательство будем вести индукцией по количеству $n$ точек в $S$. Если $n=1$, то первый выигрывает первым своим ходом. Пусть $n > 1$. Далее под *отрезками* мы всегда будем подразумевать отрезки, концы которых лежат в $S$ и не симметричны относительно $O$. Рассмотрим длины всех отрезков. Пусть $d$ — максимальная из них, и пусть $A_1B_1, A_2B_2, \\dots, A_nB_n$ — все отрезки длины $d$ (некоторые из точек $A_i, B_j$ могут совпадать).\n\nЗаметим, что точка $O$ не является концом ни одного из этих отрезков. Действительно, пусть это не так, и среди наших отрезков есть какой-то отрезок $OA$. Пусть точка $B \\in S$ получается из $A$ поворотом на $90^\\circ$ относительно $O$. Тогда $AB = \\sqrt{2} OA > OA$, то есть длина отрезка $OA$ не максимальна — противоречие.\n\nВыкинем из $S$ все точки $A_i, B_i$. Заметим, что полученное множество $S'$ удовлетворяет всем условиям нашего утверждения (так как множество отрезков $A_iB_i$ переходит в себя при повороте на $90^\\circ$). Значит, по предположению индукции в игре на полученном множестве $S'$ выигрывает первый. Предъявим теперь выигрышную для него на множестве $S$.\n\nПервый будет действовать по стратегии для множества $S'$ с начала до того момента, когда второй впервые выведет фишку за пределы множества $S'$. Это случится, ибо согласно стратегии для $S'$ у первого всегда есть ход, после которого фишка остается в множестве $S'$. Значит, рано или поздно второй сделает ход из точки $X$, лежащей в $S'$, в точку $Y$, не лежащую там (пусть тогда $Y = A_i$). Тогда первый может сделать ход в точку $B_i$ (так как $A_iB_i = d$, а $XA_i < d$, иначе бы $X$ не лежала в $S'$), после чего второму ходить некуда — он должен сделать ход длины, большей $d$, а таких ходов нет. Итого, первый выигрывает.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute non-isosceles triangle. Let $AA'$, $H$, $O$ be its altitude, orthocenter, and circumcenter respectively. Prove that the point symmetrical to the circumcenter of triangle $HOA'$ about line $HO$ lies on a midline of triangle $ABC$. (D. Prokopenko)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $A_0, B_0, C_0$ — середины сторон $BC, CA, AB$ соответственно. Так как $OA_0 \\perp BC$ и $B_0C_0 \\parallel BC$, то $A_0O$ — высота треугольника $A_0B_0C_0$. Аналогично, $B_0O$ и $C_0O$ — высоты треугольника $A_0B_0C_0$. Поэтому $O$ является точкой пересечения высот треугольника $A_0B_0C_0$. Треугольник $ABC$ подобен треугольнику $A_0B_0C_0$ с коэффициентом 2, поэтому отрезок $AN$ вдвоем больше соответствующего отрезка $A_0O$.\n\n![](attached_image_1.png)\n\nРис. 13\n\nОбозначим через $P, M, Q$ соответствующую центр описанной окружности треугольника $HOA'$, середину отрезка $HO$ и точку, симметричную точке $P$ относительно прямой $OH$. Поскольку точка $P$ равноудалена от точек $H$ и $O$, точки $P$ и $Q$ лежат на серединном перпендикуляре к отрезку $HO$, следовательно, $M$ — середина отрезка $PQ$.\n\nПусть $O', P', M', Q'$ — проекции точек $O, P, M, Q$ соответственно на высоту $AA'$. Так как точка $P$ равноудалена от точек $A'$ и $H$, то $A'P' = HP' = \\frac{A'H}{2}$. Точка $M'$ является серединой отрезка $HO'$, отсюда $2A'M' = A'H + A'O' = (A'A - AH) + A_0O = A'A - AH + \\frac{AH}{2} = A'A - \\frac{AH}{2}$. С другой стороны, точка $M'$ является серединой отрезка $P'Q'$, отсюда $2A'M' = A'P' + A'Q' = \\frac{A'H}{2} + A'Q$. Из равенства $A'A - \\frac{AH}{2} = \\frac{A'H}{2} + A'Q$ получаем, что $A'Q = A'A - \\frac{AH+A'H}{2} = A'A - \\frac{A'A}{2} = \\frac{A'A}{2}$. Это означает, что $Q'$ совпадает с серединой высоты $AA'$, поэтому $Q'$ и $Q$ лежат на средней линии $B_0C_0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39555,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every real numbers $1 < a \\le b \\le c$, prove the inequality\n$$\n\\log_a b + \\log_b c + \\log_c a \\le \\log_b a + \\log_c b + \\log_a c. \\qquad (D. Tereshin)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Введем переменные $x = \\log_a b$, $y = \\log_b c$. В новых переменных неравенство принимает вид\n$$\nx + y + \\frac{1}{xy} \\le \\frac{1}{x} + \\frac{1}{y} + xy,\n$$\nчто после приведения к общему знаменателю переходит в\n$$\n\\frac{(x-1)(y-1)(xy-1)}{xy} \\ge 0.\n$$\nПоследнее неравенство верно, т.к. $x \\ge 1$, $y \\ge 1$ и $xy \\ge 1$ в силу условия задачи.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39556,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ lines are drawn in the plane, and all the intersection points of these lines are marked. It is appeared that one of $n$ lines contains exactly 1 marked point, one of $n$ lines contains exactly 3 marked points, and one of $n$ lines contains exactly 5 marked points. Find all the possible values of $n$. (I. Bogdanov)",
    "options": [],
    "answer": "8",
    "solution": "Answer: $8$ lines.\n\n![](attached_image_1.png)\n\nFig. 14\n\n![](attached_image_2.png)\n\nFig. 15\n\nLet $l_1$, $l_3$, $l_5$ be the lines on which $1$, $3$, and $5$ points are marked, respectively, and let $A$ be the only marked point on $l_1$. Then all other lines either pass through this point (call these lines meridians) or are parallel to $l_1$ (call these parallels). Note that each meridian intersects each parallel, and all such intersection points are distinct; moreover, these points exhaust all intersection points except $A$. Thus, each meridian has the same number of marked points, and each parallel as well. From the condition, it follows that either each meridian has $3$ points and each parallel has $5$ (so there are $2$ parallels and $5$ meridians, see Fig. 14), or vice versa (so there are $4$ parallels and $3$ meridians, see Fig. 15). In both cases, there are $8$ lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39557,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $D$ is chosen on side $BC$ of an acute triangle $ABC$ so that $AB = AD$. The circumcircle of triangle $ABD$ intersects segment $AC$ at points $A$ and $K$. Line $DK$ intersects the line perpendicular to $AC$ and passing through $B$ at point $L$. Prove that $CL = BC$. (I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Так как $AB = AD$, имеем $\\angle ADB = \\angle ABD$. Поскольку четырехугольник $ABDK$ вписан, $\\angle AKB = \\angle ADB$ и $\\angle ABD = 180^\\circ - \\angle AKD = \\angle LKA$. Таким образом, в треугольнике $BKL$ высота $KA$ является биссектрисой, а значит, и медианой; тогда точки $L$ и $B$ симметричны относительно $AC$, поэтому отрезки $CL$ и $CB$ также симметричны. Значит, их длины равны.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39558,
    "subject": "Mathematics (Multi-modal)",
    "question": "$k$ rooks are placed on the $10 \\times 10$ board. All the squares beaten by at least one rook are marked (a rook in particular beats its own square). It occurs that after removing any rook from the board, at least one marked square becomes not beaten. Find the greatest possible value of $k$. (S. Berlov)",
    "options": [],
    "answer": "16",
    "solution": "Ответ. $16$.\n\nРассмотрим расстановку $k$ ладей, удовлетворяющую условию. Возможны два случая.\n\n1. Пусть в каждом столбце стоит хотя бы по одной ладье. Тогда вся доска находится под боем, и можно убрать ладью из любого столбца, в котором их хотя бы две. Значит, в этом случае в каждом столбце стоит ровно по одной ладье, и $k \\le 10$. Аналогично, если в каждой строке есть ладья, то тоже $k \\le 10$.\n\n2. Пусть теперь найдутся пустая строка и пустой столбец. Тогда клетка на их пересечении не под боем. Заметим, что каждая ладья является единственной либо в своей строке, либо в своем столбце (иначе ее можно выкинуть, и ее строка и столбец останутся под боем). Для каждой ладьи отметим эту строку или этот столбец. Если отмечены не более $8$ столбцов и не более $8$ строк, то всего ладей не больше $8+8=16$. Если же, для определенности, отмечены $9$ столбцов, то ладей всего $9$ (в каждом из $9$ столбцов по одной, а в $10$-м столбце по предположению ладей нет).\n\nИтого, во всех случаях мы получили $k \\le 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39559,
    "subject": "Mathematics (Multi-modal)",
    "question": "11 positive integers are arranged on a circle. It appears that the difference between each two neighboring numbers is not less than 20, while the sum of each two neighboring numbers is not less than 100. Find the least possible value of the sum of all 11 integers.",
    "options": [],
    "answer": "580",
    "solution": "Consider an arrangement that satisfies the conditions, and connect each pair of neighboring numbers with an arrow from the smaller to the larger. Since the total number of arrows is odd, there must be two consecutive arrows pointing in the same direction: $a \\to b \\to c$. Then $b \\leq c - 20$, $a \\leq b - 20 \\leq c - 40$, so $100 \\leq a + b \\leq 2c - 60$, from which $c \\geq 80$. All numbers except $c$ can be divided into 5 pairs of neighboring numbers; thus, the sum of all numbers is at least $80 + 5 \\cdot 100 = 580$.\n\nOne example of an arrangement where this bound is achieved is shown in Fig. 17.\n\nSecond solution. Here is another proof that the sum is at least 580. Note that in any pair of neighboring numbers, one of them is at least 60 (if the larger is less than 60, then the other number is ...). Divide the remaining numbers into pairs of neighbors (the sum in each pair is at least 100, so the sum of all these 10 numbers is at least 500). Next, in each pair, mark the number that is at least 60; also mark the number $c$. We have marked 6 numbers out of 11, so two of them are neighbors. Then the larger of them is at least $60 + 20 = 80$, so $c \\geq 80$. Therefore, the sum of all numbers is at least $80 + 500 = 580$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integers $x, y \\in [2, 100]$, prove that there exists a positive integer $n$ such that the number $x^2 + y^2$ is composite.\n\n(S. Berlov, A. Belov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "11.8. См. решение задачи 10.8.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove or disprove that for every integer $n \\ge 2009$ one can choose two distinct pairs of fractions from the sequence $\\frac{1}{n}, \\frac{2}{n-1}, \\frac{3}{n-2}, \\dots, \\frac{n-1}{2}, \\frac{1}{n}$ so that the sums of fractions in pairs are equal. (A. Shapovalov, K. Knop)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Каждая из данных дробей имеет вид $\\frac{n+1-a}{a} = \\frac{n+1}{a} - 1$, где $1 \\le a \\le n$. Стало быть, нам требуется найти такие различные натуральные числа $a, b, c$ и $d$, не большие 2009, для которых $(\\frac{n+1}{a} - 1) + (\\frac{n+1}{b} - 1) = (\\frac{n+1}{c} - 1) + (\\frac{n+1}{d} - 1)$, что равносильно $\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{c} + \\frac{1}{d}$.\nОсталось подобрать дроби, удовлетворяющие этому равенству. Это можно сделать, взяв любое равенство двух сумм различных натуральных слагаемых, НОК которых не больше 2009, и поделив его на этот НОК. Например, равенство $1+4=2+3$, поделенное на 12, даёт $\\frac{1}{12} + \\frac{1}{3} = \\frac{1}{6} + \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39562,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, sides $AB$ and $BC$ have equal lengths. Point $D$ inside the triangle is chosen so that $\\angle ADC = 2\\angle ABC$. Prove that the distance from point $B$ to the external bisector line of angle $\\angle ADC$ is twice smaller than $AD + DC$. (S. Berlov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell$ be the external angle bisector of the angles adjacent to $\\angle ADC$, and let $K$ be the projection of $B$ onto $\\ell$. Let points $B'$ and $C'$ be the reflections of $B$ and $C$ with respect to $\\ell$, respectively. Then $BB' = 2BK$—that is, twice the distance from $B$ to $\\ell$. Moreover, point $D$ lies on segment $AC'$ (since lines $DA$ and $DC$ are symmetric with respect to $\\ell$), and $AC' = AD + DC' = AD + DC$.\n\nFurthermore, by the same symmetry, we have $\\angle AC'B' = \\angle DC'B' = \\angle DCB$, $\\angle BB'C' = \\angle B'BC$.\n\n![](attached_image_1.png)\n\nLet segments $BB'$ and $AC'$ intersect at point $O$. From the right triangle $OKD$, we get $\\angle BOC' = \\angle KOD = 90^\\circ - \\angle KDO = \\frac{1}{2}(180^\\circ - \\angle CDC') = \\frac{1}{2}\\angle ADC = \\angle ABC$. Therefore, $\\angle ABB' = \\angle ABC - \\angle B'BC = \\angle BOC' - \\angle OB'C' = \\angle OC'B'$. Similarly, $\\angle BAO = \\angle BOC' - \\angle ABO = \\angle ABC - \\angle ABO = \\angle B'BC = \\angle BB'C'$.\n\nSince segments $BC$ and $B'C'$ are symmetric, we have $B'C' = BC = AB$. Thus, triangles $ABO$ and $B'C'O$ are congruent by a side and two adjacent angles. Hence $BB' = BO + OB' = C'O + OA = AC' = AD + DC$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39563,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Leonardland, each road has one-way movement, connects two cities and does not pass through another city. The Statistics Department calculated for each city $A$ the total number $f(A)$ of citizens in the cities, to which the roads from $A$ lead, and the total number $g(A)$ of citizens in the cities, from which the roads lead to $A$. Prove that there exists a city $A$ with $f(A) \\ge g(A)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Построим граф, вершины которого соответствуют жителям страны, причем две вершины соединены направлённым ребром в том и только том случае, когда их города соединены дорогой (направление на ребре будет такое же, как и на дороге между городами). Для каждой вершины $v$ обозначим через $f(v)$ разность количества ребер, входящих в $v$, и количества ребер, выходящих из $v$. Сумма величин $f(v)$ по всем вершинам графа равна $0$, так как каждое ребро вносит в нее одну $+1$ и одну $-1$. Значит, найдется такая вершина $u$, что $f(u) \\ge 0$. Остается лишь отметить, что $f(u)$ в точности равна разности первого и второго чисел для города, в котором живет $u$.\n\n\nВторое решение. Для каждого города $A$ обозначим через $n(A)$ число жителей в этом городе, а через $f(A)$ разность суммарного количества жителей в городах, дороги из которых выходят в $A$, и суммарного количества жителей в городах, в которые выходят дороги из $A$ (то есть в точности разность первого и второго чисел для города $A$). Если утверждение задачи неверно, то $f(A) < 0$ для каждого города $A$.\nОбозначим через $S$ сумму чисел $n(A)f(A)$ по всем городам страны. С одной стороны, $S < 0$ как сумма нескольких отрицательных чисел. С другой стороны, рассмотрим любую дорогу из $A$ в $B$. В число $n(B)f(B)$ эта дорога «вносит вклад» $+n(B)n(A)$, а в число $n(A)f(A)$ — «вклад» $-n(A)n(B)$. Рассмотрев все дороги, получим, что $S = 0$. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to replace the stars in the equality\n$$ \\{\\text{l.c.m.}\\}(*, *, *) - \\{\\text{l.c.m.}\\}(*, *, *) = 2009 $$\nby six consecutive positive integers (but not necessarily in the successive order) so that the equality would be valid? (R. Zhenodarov)",
    "options": [],
    "answer": "No",
    "solution": "It is not possible.\n\nSuppose such numbers exist. The least common multiple (LCM) of several numbers is divisible by each of them and, therefore, by each of their divisors. Thus, if among the numbers for which the LCM is taken there is an even number, then the LCM will also be even. Since $2009$ is an odd number, one of the two LCMs must be odd; therefore, all even numbers must be in one LCM.\n\nAmong six consecutive natural numbers, there are three even and three odd numbers, so one LCM will be taken from three consecutive even numbers, and the other from three consecutive odd numbers. But in any such triple, there will be a number divisible by $3$. Then both LCMs are divisible by $3$, and their difference is divisible by $3$. But $2009$ is not divisible by $3$—contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39565,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, the relations $AB = BD$, $\\angle ABD = \\angle DBC$ are satisfied. A point $K$ is chosen on diagonal $BD$ so that $BK = BC$. Prove that $\\angle KAD = \\angle KCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us mark on side $AB$ a segment $BE = BC$. The isosceles triangles $EBK$ and $KBC$ are congruent by two sides and the angle between them. Therefore, $EK = KC$, and $\\angle AEK = 180^\\circ - \\angle BEK = 180^\\circ - \\angle BKC = \\angle CKD$. Moreover, $KD = BD - BK = BA - BE = EA$. Hence, triangles $AEK$ and $DKC$ are congruent, from which $\\angle KCD = \\angle EKA$.\n\nFurther, since both triangles $BEK$ and $BAD$ are isosceles, $\\angle BEK = 90^\\circ - \\angle EBD/2 = \\angle BAD$. Therefore, $AD \\parallel EK$, from which $\\angle KAD = \\angle EKA = \\angle KCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39566,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of integers is written on an infinite tape. The first number is $1$; each number except the first one is obtained by adding to the previous number its minimal nonzero digit (in decimal representation). Find the number of digits in the decimal representation of the number at $9 \\cdot 1000^{1000}$th place in this sequence.\n(I. Bogdanov)",
    "options": [],
    "answer": "3001",
    "solution": "Answer: $3001$.\n\nSince each number in the sequence, starting from the second, is greater than the previous one by at least $1$, the $9 \\cdot 1000^{1000}$-th number is at least $9 \\cdot 1000^{1000}$, so it has at least $3001$ digits. Denote the $n$-th number of the sequence by $a_n$, and let $k$ be the smallest index such that $a_k$ has $3002$ digits. If we show that $k > 9 \\cdot 1000^{1000}$, then the $9 \\cdot 1000^{1000}$-th number has at most $3001$ digits, i.e., exactly $3001$ digits.\n\nConsider numbers from $0$ to $10^{3001} - 1$ that do not have any $1$s in their decimal representation. Padding each on the left with zeros to $3001$ digits, we get all sequences of length $3001$ consisting of digits other than $1$. There are $9^{3001}$ such sequences. Thus, among the numbers $a_1, \\dots, a_{k-1}$, there are at most $9^{3001}$ numbers without a $1$ in their decimal representation (since all are at most $10^{3001} - 1$).\n\nNow consider the process of obtaining $a_k$ from $a_1$. At each of the $k-1$ steps, we add a number from $1$ to $9$, and the number of steps where we add something other than $1$ does not exceed $9^{3001}$. Therefore,\n$$\n10^{3001} - 1 \\leq a_k - a_1 \\leq 9 \\cdot 9^{3001} + 1 \\cdot (k-1 - 9^{3001}) = k-1 + 8 \\cdot 9^{3001},\n$$\nfrom which\n$$\nk \\geq 10^{3001} - 8 \\cdot 9^{3001}.\n$$\nIt remains to show that $10^{3001} - 8 \\cdot 9^{3001} > 9 \\cdot 10^{3000}$. For this, it suffices to prove that $9^{3002} < 10^{3000}$. Note that $9^7 = 4782969 < 5 \\cdot 10^6$, so $9^{28} < 5^4 \\cdot 10^{24} < 10^{27}$ and $9^5 < 10^{54}$. Therefore,\n$$\n9^{3002} = 9^5 \\cdot 9^{2946} < 10^{54} \\cdot 10^{2946} = 10^{3000}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39567,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mushroom containing not less than $10$ worms is called bad. A basket with $90$ bad and $10$ good mushrooms is given. Determine if all the mushrooms can become good after several worms creep from bad to good mushrooms.",
    "options": [],
    "answer": "Yes",
    "solution": "Suppose each bad mushroom contains exactly $10$ worms, and each good mushroom contains no worms. Next, let one worm from each bad mushroom crawl into the good mushrooms, $9$ worms into each. As a result, each mushroom will have $9$ worms, and all mushrooms will be good.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an isosceles right-angled triangle $ABC$. Point $K$ is a midpoint of hypotenuse $AB$. Points $L$ and $M$ are chosen on sides $BC$ and $AC$ respectively so that $BL = CM$. Prove that the triangle $LMK$ is also isosceles and right-angled. (R. Zhenodarov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Медиана $CK$ треугольника $ABC$ является также высотой и биссектрисой, так как треугольник равнобедренный. Поэтому $\\angle KBC = \\angle KCB = \\angle KCA = 45^\\circ$. Отсюда $KC = KB$, и, значит, треугольники $KBL$ и $KCM$ равны по двум сторонам ($KC = KB$, $BL = CM$) и углу между ними. Поэтому $KL = KM$, и из равенства $\\angle BKL = \\angle CKM$ следует $\\angle LKM = \\angle LKC + \\angle CKM = \\angle LKC + \\angle BKL = \\angle BKC = 90^\\circ$. Значит, треугольник $LMK$ — прямоугольный равнобедренный.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39569,
    "subject": "Mathematics (Multi-modal)",
    "question": "A balance and $100$ coins are given. Several coins are false. The number of false coins is greater than $0$ but less than $99$. All genuine coins have equal weights, and all false coins also have equal weights. The weight of a genuine coin is greater than the weight of a false one. It is permitted to perform a weighing on the balance, paying a coin before it. Prove that one can determine a true coin for sure. (The coin to be determined should not be paid for some weighing!)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Заплатив монету перед первым взвешиванием, мы имеем $99$ монет, среди которых есть хотя бы одна настоящая. Положим на две чаши весов по одной монете. Если одна из них перевешивает, то она настоящая. Если же весы в равновесии, то либо обе взвешиваемые монеты настоящие, либо обе фальшивые. Заплатив одну из них, мы имеем $98$ монет, среди которых есть хотя бы одна настоящая. Продолжив так далее, либо на следующем шаге обнаружим настоящую монету, либо будем иметь $97$ монет, среди которых есть хотя бы одна настоящая. Если все время весы будут показывать равновесие, то в конце концов будем иметь две монеты, среди которых есть хотя бы одна настоящая. Положим на две чаши весов по одной монете. Если одна из них перевешивает, то она настоящая. Если же весы в равновесии, то обе оставшиеся монеты настоящие.\n\n\n**Второе решение.** Сначала покажем, что за два взвешивания можно найти группу из $33$ монет, в которой есть настоящая.\nРазделим монеты на $3$ группы $A$, $B$, $C$ по $33$ монеты и одну в остатке. Заплатив этой монетой, сравним группы $A$ и $B$ (заметим, что хотя бы одна настоящая монета еще осталась). Если весы не остались в равновесии, то на перевесившей чашке есть хотя бы одна настоящая. Если же весы в равновесии, то сравним группы $A$ и $C$, заплатив монетой из $B$. Если одна перевесила, то в ней есть настоящая, а если весы в равновесии, то в каждой из групп $A$, $B$, $C$ было поровну настоящих монет — значит, хотя бы по одной. Тогда в группах $A$ и $C$ осталось по настоящей монете.\nИтого, мы нашли группу (назовем ее $D$) из $33$ монет, в которой есть настоящая; кроме этой группы, у нас осталось еще хотя бы $65$ монет (пусть они образуют группу $E$). Теперь сравним одну монету из группы $D$ с каждой из остальных монет этой группы (платя каждый раз монетами из $E$). Тогда либо какая-то монета перевесит (она и будет настоящей!), либо все монеты будут весить поровну. В последнем случае, так как в $D$ есть настоящая, то все монеты в $D$ будут настоящими. В любом случае мы нашли хотя бы одну настоящую монету.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39570,
    "subject": "Mathematics (Multi-modal)",
    "question": "7 cards with numbers $0$, $1$, $2$, $3$, $4$, $5$, $6$ are given. Peter and Basil make moves in turn taking one card by each move; Peter makes the first move. The player who can construct of his cards a decimal number divisible by $17$ earlier than his opponent is declared as a Winner. Determine which of two players has a winning strategy. (I. Rubanov)",
    "options": [],
    "answer": "Peter (the first player) has a winning strategy.",
    "solution": "Let us denote the players as $A$ (the first player) and $B$ (his opponent).\n\nWe present a strategy that allows $A$ to guarantee a win. Let him take the digit $3$ on his first move; then $B$ is forced to take $4$ (otherwise $A$ will take it on his second move and win by forming the number $34$). Note that $B$ cannot win on his second move, since the only two-digit number containing $4$ in its digits and divisible by $17$ is $34$.\n\nNext, $A$ takes $1$, then $B$ must take $5$ (indeed, otherwise he will not win with this move, and on the next move $A$ will take $5$ and form $51$). Then, on the next move, $A$ takes $6$ and wins by forming the number $136$.\n\n**Remark.** There exist other winning strategies for $A$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39571,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a checkered paper, numbers $x$, $y$ and $z$ are placed in 3 squares (one number in each), other squares are empty. It is permitted to perform operations of two types:\n\n1) To choose 2 numbers $a$ and $b$ from 2 non-empty squares and to place the number $a + b$ in some empty square;\n\n2) To choose 3 numbers $a$, $b$, and $c$ from 3 non-empty squares and to place the number $ab + c^2$ in some empty square.\n\nProve that using these operations one can place a number $(x + y + z)^2$ in some square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть исходные числа $x, y, z$. Получим сначала числа $x+y, y+z, z+x$. Затем из клеток $x, y+z, y$ получим $x(y+z) + y^2$; аналогично, получим числа $y(z+x) + z^2$ и $z(x+y) + x^2$. Сложив их, получаем требуемое: $x(y+z) + y^2 + y(z+x) + z^2 + z(x+y) + x^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x+y+z)^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39572,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ is given. It appears that on diagonal $AC$ there exists a point that lies on perpendicular bisectors of the segments $AB$ and $CD$, and on diagonal $BD$ there exists a point that lies on perpendicular bisectors of the segments $DA$ and $BC$.\nProve that $ABCD$ is a rectangle. (N. Agakhanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $M$ — указанная точка на диагонали $AC$. Тогда, по свойству серединных перпендикуляров, $MA = MB$ и $MC = MD$. Поэтому $BD \\le BM + MD = AM + MC = AC$. Аналогично рассуждая для другой пары серединных перпендикуляров, получаем, что $AC \\le BD$. Значит, $BD = AC$, и все перавечства обратились в равенства.\nЭто выполняется только тогда, когда $M$ лежит на диагонали $BD$, то есть $M$ — точка пересечения $AB$ и $CD$; при этом она должна лежать на серединных перпендикулярах ко всем сторонам четырехугольника $ABCD$. Значит, его диагонали делятся точкой пересечения пополам (то есть $ABCD$ — параллелограмм) и равны. Значит, $ABCD$ — прямоугольник.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight teams participated in a soccer tournament, and each pair of teams have played exactly once. It appeared that if two teams $A$ and $B$ played a draw then the resulting numbers of points of $A$ and $B$ are different. Find the greatest possible number of draws in this tournament. (Each win is worth 3 points, each draw is worth 1 point, and each lose is worth 0 points.) (S. Tokarev)",
    "options": [],
    "answer": "22",
    "solution": "Докажем, что ровно по 6 ничьих может быть не более, чем у двух команд. Действительно, любая такая команда имеет либо 6, либо $6+3$ очка (в зависимости от того, выиграла или проиграла она свой результативный матч). Если таких команд три, то у двух из них поровну очков, значит, между собой они сыграли не вничью; этого не может быть, ибо они обе либо не выигрывали, либо не проигрывали ни одного матча.\n\nТакже ясно, что максимум одна команда все свои 7 матчей сыграла вничью.\n\nТаким образом, сумма количеств ничьих у всех 8 команд не превосходит $7+6+6+5+5+5+5+5 = 44$, а поскольку каждый ничейный матч учитывается дважды, то общее число ничьих в турнире не превосходит $44/2 = 22$. Оно может равняться 22, что показано на рисунке (команды обозначены буквами A, B, C, D, E, F, G и H).\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least possible denominator in the irreducible fraction which is the sum of two irreducible fractions with denominators $600$ and $700$. (I. Bogdanov)",
    "options": [],
    "answer": "168",
    "solution": "Ответ. $2^3 \\cdot 3 \\cdot 7 = 168$.\n\nПусть наши дроби $\\frac{a}{600}$ и $\\frac{b}{700}$. Тогда $a$ взаимно просто с $6$, а $b$ — с $7$. Поэтому числитель их суммы $\\frac{7a+6b}{4200}$ взаимно прост как с $6 = 2 \\cdot 3$, так и с $7$. Поскольку $4200 = 2^3 \\cdot 3 \\cdot 7 \\cdot 5^2$, это означает, что знаменатель после сокращения будет не меньше, чем $2^3 \\cdot 3 \\cdot 7 = 168$. Такой знаменатель действительно может получиться; например, $\\frac{1}{600} + \\frac{3}{700} = \\frac{1}{168}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Segment $BD$ is an angle bisector in a triangle $ABC$ (point $D$ lies on segment $AC$). Line $BD$ intersects the circumcircle $\\Omega$ of triangle $ABC$ at points $B$ and $E$. Circle $\\omega$ with diameter $DE$ intersects $\\Omega$ at points $E$ and $F$. Prove that the line symmetrical to $BF$ with respect to $BD$ contains a median of triangle $ABC$.\n\n(L. Emelyanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of side $AC$, the line $BM$ intersects the circle $\\Omega$ again at point $F'$, and the line $FM$ intersects the circle $\\Omega$ again at point $B'$ (see Fig. 20).\n\nSince arcs $AE$ and $CE$ are equal, points $M$ and $E$ lie on the perpendicular bisector $\\ell$ of segment $AC$. Therefore, $\\angle EMD = 90^\\circ$ and, consequently, $M$ lies on the circle $\\omega$.\n\nWe have:\n$$\n\\frac{1}{2}\\overline{B'C'E} = \\angle B'FE = \\angle MFE = \\angle MDE = \\angle CDE = \\frac{1}{2}(\\overline{AB} + \\overline{CE}) = \\frac{1}{2}(\\overline{AB} + \\overline{AE}) = \\frac{1}{2}\\overline{BAE}.\n$$\nFrom the equality of arcs $B'C'E$ and $BAE$, it follows that points $B$ and $B'$ are symmetric with respect to $\\ell$, so lines $BM$ and $B'M$ (and therefore points $F'$ and $F$) are symmetric with respect to $\\ell$. The last statement means that $\\overline{EF} = \\overline{EF'}$, whence $\\angle FBE = \\angle F'BE$. We obtain that lines $BF$ and $BM$ are symmetric with respect to line $BE$, which is what was required to be proved.\n\n![](attached_image_1.png)\nLet $M$ be the midpoint of side $AC$. As in the previous solution, we obtain that $ME$ is the perpendicular bisector of $AC$, and $M$ lies on $\\omega$. Let the line $DF$ intersect the circle $\\Omega$ again at point $G$. Since $\\angle DFE = 90^\\circ$, $G$ is the point diametrically opposite to $E$; in particular, $EG$ passes through $M$ (see Fig. 21). We have $\\angle FBE = \\angle FGE$.\n\nFurthermore, since $EG$ is a diameter, $\\angle GBE = 90^\\circ$. From the equalities $\\angle GBD = \\angle GMD = 90^\\circ$, it follows that $GBDM$ is a cyclic quadrilateral (with diameter $DG$), whence $\\angle MBE = \\angle MBD = \\angle MGD = \\angle EGF$. Thus, $\\angle FBE = \\angle FGE = \\angle MBE$, which is what was required to establish.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Rational numbers $a$ and $b$ satisfy the equality\n$$\na^3 b + ab^3 + 2a^2 b^2 + 2a + 2b + 1 = 0.\n$$\nProve that $1 - ab$ is a square of a rational number. (R. Zhenodarov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Transform the original expression:\n$$\n\\begin{aligned}\n0 &= a^3 b + ab^3 + 2a^2 b^2 + 2a + 2b + 1 = ab(a + b)^2 + 2(a + b) + 1 \\\\\n&= (ab - 1)(a+b)^2 + (a+b)^2 + 2(a+b) + 1 = (ab - 1)(a+b)^2 + (a+b+1)^2.\n\\end{aligned}\n$$\nNotice that $a + b \\neq 0$, because otherwise $1 = 0$.\nTherefore, $1 - ab = \\left(\\frac{a+b+1}{a+b}\\right)^2$; since $a$ and $b$ are rational numbers, $\\frac{a+b+1}{a+b}$ is also a rational number, and the statement is proved.\n\nSecond solution. Multiply the given equation by $ab$ and transform:\n$$\na^2b^2(a+b)^2 + 2ab(a+b) + ab = 0 \\Leftrightarrow a^2b^2(a+b)^2 + 2ab(a+b) + 1 = 1 - ab \\Leftrightarrow 1 - ab = (ab(a+b) + 1)^2.\n$$\nFrom the last equality, the statement follows.\n\nThird solution. If $ab = 0$, then the statement is true:\n$$\n1 - ab = 1^2.\n$$\nIf $ab \\neq 0$, then the quadratic equation $abx^2 + 2x + 1 = 0$ has a rational root ($a + b$) and rational coefficients. By the quadratic formula, its discriminant is a square of a rational number. But $D = 4(1 - ab)$, and the statement is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39577,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, the altitudes $AA_1$, $BB_1$, $CC_1$ are drawn. A perpendicular to $AC$ passing through $A_1$ intersects line $B_1C_1$ at $D$. Prove that $\\angle ADC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the intersection point of segments $A_1D$ and $AC$ (see Figure 5).\n\nFrom the equality $\\angle AB_1B = \\angle AA_1B = 90^\\circ$, it follows that the quadrilateral $AB_1A_1B$ is inscribed in a circle (with diameter $AB$), so $\\angle A_1B_1K = 180^\\circ - \\angle A_1B_1A = \\angle ABC$.\n\nSimilarly, the quadrilateral $BC_1B_1C$ is inscribed in a circle (with diameter $BC$); therefore, $\\angle DB_1K = 180^\\circ - \\angle C_1B_1C = \\angle ABC$.\n\nWe obtain that $\\angle A_1B_1K = \\angle DB_1K$. From the condition, $\\angle A_1KB_1 = \\angle DKB_1 = 90^\\circ$; hence, the right triangles $A_1B_1K$ and $DB_1K$ are congruent (by a leg and an acute angle), whence $A_1K = DK$.\n\nThe last equality means that points $A_1$ and $D$ are symmetric with respect to the line $AC$; therefore, triangles $A_1AC$ and $DAC$ are symmetric with respect to this line. Then $\\angle ADC = \\angle AA_1C = 90^\\circ$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n > 1$. An integer $a > n^2$ is chosen so that for each $i = 1, 2, \\ldots, n$, the set $\\{a + 1, a + 2, \\ldots, a + n\\}$ contains a multiple of the number $n^2 + i$. Prove that $a > n^4 - n^3$. (A. Golovanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что разность между любыми двумя числами вида $a+i$ ($i = 1, \\ldots, n$) не превосходит $n-1$.\nПусть кратное числу $n^2 + i$, содержащееся среди наших чисел — это $a_i(n^2 + i)$. Ясно, что $a_1 > 1$. Тогда найдется такое $1 \\le i \\le n-1$, что $a_i > a_{i+1}$ (в противном случае $a_1 \\le a_2 \\le \\dots \\le a_n$, и $a_n(n^2 + n) - a_1(n^2 + 1) \\ge a_1(n-1) > n-1$, что невозможно). Тогда\n$$\nn-1 \\ge a_i(n^2+i) - a_{i+1}(n^2+i+1) \\ge \\\\\n\\ge a_i(n^2+i) - (a_i-1)(n^2+i+1) = n^2+i+1 - a_i, \\\\\n\\text{то есть } a_i \\ge n^2-n+i+2 > n^2-n. \\text{ Теперь, так как одно из} \\\\\n\\text{наших чисел есть } a_i(n^2+i) > (n^2-n)(n^2+1) = n^4-n^3+n^2-n, \\\\\n\\text{то } a \\ge a_i(n^2+i) - n > n^4-n^3+n^2-2n \\ge n^4-n^3, \\text{ так как } n \\ge 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39579,
    "subject": "Mathematics (Multi-modal)",
    "question": "100 thimbles are arranged on a circle. A token is placed under one of the thimbles. By one turn, a player can pick up arbitrary four thimbles and check whether a token is under one of them. After that, the thimbles return to their places, and the token moves to one of the two neighboring thimbles. Find the least number of turns needed to find the token for sure. (B. Trushin)",
    "options": [],
    "answer": "33",
    "solution": "**Ответ.** За 33 хода.\n\nПосле каждого нашего хода и перемещения монетки будем поворачивать все наперстки (вместе с монеткой) по ходу часовой стрелки на одну позицию. Тогда будем считать, что после каждого хода монетка либо остается на месте, либо перемещается на две позиции по часовой стрелке, а наперстки остаются на своих местах.\n\nПокрасим все наперстки поочередно в белый и черный цвет, и пронумеруем наперстки каждого цвета по порядку против часовой стрелки числами от $0$ до $49$. Понятно, что цвет наперстка, под которым лежит монетка, не изменяется, а номер либо не изменяется, либо уменьшается на $1$ по модулю $50$.\n\nПокажем, как найти монетку за $33$ хода. Описывая алгоритм, предполагаем, что монетка не обнаружена на всех ходах вплоть до $33$-го (в противном случае все уже сделано менее чем за $33$ хода).\n\nПервым ходом поднимем черные наперстки с номерами $0$, $1$, $2$, $3$. Тогда после перемещения монетка не сможет оказаться под черными наперстками $0$, $1$, $2$. Вторым ходом поднимем черные наперстки $3$, $4$, $5$, $6$. Тогда после перемещения монетка не сможет оказаться под черными наперстками $0$, $1$, ..., $5$. Действуем так далее: при $s = 1, 2, ..., 16$ ходом номер $s$ поднимем черные наперстки с номерами $3s - 3$, $3s - 2$, $3s - 1$, $3s$. Тогда после перемещения монетка не сможет оказаться под черными наперстками $0$, $1$, ..., $3s - 1$.\n\nСемнадцатым ходом поднимем черные наперстки $48$ и $49$, а также белые наперстки $49$ и $0$. Теперь мы знаем, что под черными наперстками нет монетки, а также что после перемещения монетка не сможет оказаться под белым наперстком $49$. При $s = 1, 2, ..., 15$ ходом номер $17 + s$ поднимем белые наперстки $3s - 3$, $3s - 2$, $3s - 1$, $3s$. Тогда после перемещения монетка не сможет оказаться под белыми наперстками $49$, $0$, $1$, ..., $3s - 1$. Наконец, последним $33$-м ходом поднимаем белые наперстки $45$, $46$, $47$, $48$; под одним из них обязана быть монетка.\n\n\nДокажем, что с гарантией обнаружить монету за $32$ хода невозможно. Обозначим через $B_k$ множество из четырех наперстков, поднимаемых на $k$-м ходе, а через $A_k$ — множество наперстков, про которые перед выполнением $k$-го хода (после возможного перемещения монетки на $(k-1)$-м ходе) точно известно, что под ними нет монетки. Предполагаем, что пока возможно, под наперстками из $B_k$ нет монетки.\n\nЯсно, что $A_{k+1} \\subset A_k \\cup B_k$ при $k = 1, 2, ..., n-1$, откуда $|A_{k+1}| \\le |A_k| + 4$. Более того, если множество $A_k \\cup B_k$ не совпадает с множеством всех наперстков или с множеством из $50$ наперстков одного цвета, то найдется такая пара одноцветных наперстков $P$ и $Q$ с номерами $r$ и $r+1$ (mod $50$) соответственно, что $P \\in A_k \\cup B_k$, а $Q \\notin A_k \\cup B_k$. Тогда, если перед $k$-м ходом монетка находилась под наперстком $Q$, то она может переместиться под $P$, поэтому $P \\notin A_{k+1}$. В этом случае $A_{k+1} \\neq A_k \\cup B_k$, и, следовательно, $|A_{k+1}| \\leq |A_k| + 3$. Итак, $|A_{k+1}| \\leq |A_k| + 3$, если $|A_{k+1}| \\neq 50$.\n\nИмеем: $|A_1| = 0$, $|A_2| \\leq 3$, $|A_3| \\leq 6$, ..., $|A_{17}| \\leq 48$, $|A_{18}| \\leq 51$, $|A_{19}| \\leq 54$, ..., $|A_{32}| \\leq 93$. Получается, что перед $32$-м ходом имеется по крайней мере $7$ наперстков, под которыми может быть монета, следовательно, обнаружить монету с гарантией на $32$-м ходу или ранее не удастся.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $a$, $b$ and $c$ satisfy the conditions\n$$\n(a + b)(b + c)(c + a) = abc, \\\\\n(a^3 + b^3)(b^3 + c^3)(c^3 + a^3) = a^3b^3c^3.\n$$\nProve that $abc = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим сначала, что для любых различных чисел $x$ и $y$ выполняется неравенство\n$$\nx^2 - xy + y^2 > |xy| \\quad (1)\n$$\nВ самом деле, если $|xy| = xy$, то неравенство (1) равносильно неравенству $(x - y)^2 > 0$, которое верно; если же $|xy| = -xy$, то (1) равносильно неравенству $x^2 + y^2 > 0$.\n\nПредположим, что $abc \\neq 0$. Тогда, разделив второе равенство на первое, мы получаем\n$$(a^2 - ab + b^2)(b^2 - bc + c^2)(c^2 - ca + a^2) = a^2b^2c^2 = |ab| \\cdot |bc| \\cdot |ac|.$$\nОднако все скобки слева и все сомножители справа положительны; при этом каждый сомножитель слева не меньше соответствующего сомножителя справа. Тогда равенство может достигаться лишь тогда, когда все эти три неравенства обращаются в равенства, т. е. когда $a = b = c$. В этом случае первое равенство из условия принимает вид $8a^3 = a^3$, что невозможно при $a \\neq 0$. Значит, наше предположение неверно, и $abc = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39581,
    "subject": "Mathematics (Multi-modal)",
    "question": "On 11 sheets of paper 11 following statements are written (each sheet contains exactly one statement):\n1) There are no sheets with false statements to the left of this sheet.\n2) There is exactly 1 sheet with a false statement to the left of this sheet.\n3) There are exactly 2 sheets with false statements to the left of this sheet.\n...\n11) There are exactly 10 sheets with false statements to the left of this sheet.\n\nThe sheets were put in a row (from the left to the right). After that some of the statements become true while the others become false. Find the greatest possible number of true statements.\n\n(I. Rubanov)",
    "options": [],
    "answer": "6",
    "solution": "Answer: 6.\n\nIf the sheets are arranged in the following order:\n![](attached_image_1.png)\n(for brevity, each sheet is marked with the number of false statements to its left), then there will be 6 true statements.\n\nLet us show that there cannot be more. Suppose that two sheets with true statements are placed next to each other; then the number of sheets with false statements to the left of each of them is the same; this is impossible, since different numbers are written on these sheets. Therefore, among any two consecutive sheets, at least one statement is false. Thus, there are at least five sheets with false statements, since in each of the first five pairs there is a false one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39582,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $m$ is chosen so that the sum of all the digits of $8^m$ (in its decimal representation) equals to $8$. Determine if the last digit of $8^m$ can appear to be $6$.",
    "options": [],
    "answer": "No, it cannot be 6.",
    "solution": "**Первое решение.** Предположим, что сумма цифр числа $8^m$ при некотором $m > 1$ равна $8$, и оно оканчивается на $6$. Число $2^m$ не может оканчиваться на $06$ или на $26$, так как в этом случае оно не делится на $4$. Следовательно, оно оканчивается на $16$ (иначе сумма цифр будет больше $8$), и поэтому имеет десятичную запись $1000\\ldots016$. Тогда $8^m = 10^k + 16$, то есть число $10^k + 16$ -- степень двойки.\n\nНо если $k \\ge 5$, то $10^k + 16 = 2^4(2^{k-4} \\cdot 5^k + 1)$, и в скобках получаем нечетный множитель, больший $1$. Остается рассмотреть случаи $k = 2, k = 3, k = 4$: $10^2 + 16 = 4 \\cdot 29$; $10^3 + 16 = 8 \\cdot 127$; $10^4 + 16 = 32 \\cdot 313$. Таким образом, $10^k + 16$ не является степенью восьмерки ни при каком натуральном $k$, что и требовалось доказать.\n\n\n**Второе решение.** Предположим противное, и пусть $8^m$ оканчивается на $6$ и имеет сумму цифр, равную $8$. Заметим, что $8^1$ оканчивается на $8$, $8^2$ оканчивается на $4$, $8^3$ оканчивается на $2$, $8^4$ оканчивается на $6$, $8^5$ оканчивается на $8$. Далее последняя цифра степени восьмерки повторяется с периодом $4$, поскольку последняя цифра числа $8^m$ определяется однозначно последней цифрой числа $8^{m-1}$. Таким образом, $8^m$ оканчивается на $6$ тогда и только тогда, когда $m$ делится на $4$.\n\nСумма цифр числа имеет тот же остаток при делении на $3$, что и само число, поэтому $8^m$ должно иметь остаток $2$ при делении на $3$. Заметим, что $8^1$ имеет остаток $2$ при делении на $3$, $8^2$ имеет остаток $1$ при делении на $3$, $8^3$ имеет остаток $2$ при делении на $3$, и далее остатки степени восьмерки повторяются с периодом $2$. Таким образом, $8^m$ имеет остаток $2$ при делении на $3$ тогда и только тогда, когда $m$ нечетно. Это противоречит тому, что $m$ должно делиться на $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a coloring of all positive integers in $2009$ colors satisfying the following two conditions:\n(i) there are infinitely many numbers of each color;\n(ii) there is no triple of integers $(a, b, c)$ colored in three pairwise distinct colors such that $a = bc$.\n\n$(N. Agakhanov)$",
    "options": [],
    "answer": "Yes, such a coloring exists.",
    "solution": "Можно.\n\nПриведем один из возможных вариантов такой раскраски. Пусть $p_1 < p_2 < \\dots < p_{2008}$ -- простые числа. Построим множества $A_1, A_2, \\dots, A_{2009}$ по следующему правилу: в $A_1$ включим все натуральные числа, делящиеся на $p_1$; в $A_2$ — все натуральные числа, делящиеся на $p_2$, но не делящиеся на $p_1$; в $A_3$ — все натуральные числа, делящиеся на $p_3$, но не делящиеся на $p_1$ и на $p_2$; ...; в $A_{2008}$ — все натуральные числа, делящиеся на $p_{2008}$, но не делящиеся ни на одно из чисел $p_1, p_2, ..., p_{2007}$. Наконец, в $A_{2009}$ включим все остальные натуральные числа.\n\nТогда для любых чисел $x \\in A_k, y \\in A_n$, где $k < n$, их произведение принадлежит множеству $A_k$, поэтому при таком разбиении натуральных чисел на множества разноцветной тройки чисел, произведение двух из которых равно третьему числу, найти не удастся.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an $8 \\times 8$ chessboard. Choose one of its diagonals and call the 8 cells of this diagonal *a fence*. The Rook starts from an arbitrary square outside the fence and makes some moves so that it does not get to one square twice, and it does not stand on the squares of the fence. Find the maximal possible number of jumps over the fence on the Rook's route. (R. Zhenodarov)",
    "options": [],
    "answer": "47",
    "solution": "Разделим доску на четыре квадраты $4 \\times 4$. Заметим, что, если ладья прыгает через забор, то либо начальная, либо конечная клетка прыжка отмечена серым на рис. 22. Так как серых клеток 24 и через каждую может проходить максимум два прыжка, то всего может оказаться не более 48 прыжков.\n\nПри этом, если их ровно 48, то из каждой серой клетки должно быть сделано два прыжка в белые клетки (в предыдущем подсчете прыжок из серой клетки в серую будет подсчитан два раза!). Тогда все ходы из темно-серых клеток будут вести в белый квадрат под диагональю, а оттуда — только в темно-серые клетки (либо в другие клетки этого же квадрата). Значит, подобным образом ладья никогда не падет в светло-серые клетки. Противоречие; таким образом, количество прыжков не превосходит 47. Один из возможных примеров с 47 прыжками показан на рис. 22 (числа в клетках указывают, в каком порядке ладья по ним проходит).\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39585,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a parallelogram $ABCD$, angle $ABC$ is obtuse. Let $\\omega$ be the circumcircle of triangle $ABC$. Line $AD$ intersects $\\omega$ at points $A$ and $E$, and line $CD$ intersects $\\omega$ at points $C$ and $F$. Prove that the circumcenter of triangle $DEF$ lies on $\\omega$. (T. Emelyanova)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $\\angle ABC = \\angle ADC = \\angle EDF = \\alpha$; по условию, $\\alpha > 90^\\circ$ (см. рис. 7).\n\nТак как $BC \\parallel AD$, то $ABCE$ — равнобокая трапеция, откуда $\\angle ECF = \\angle BCE - \\angle BCD = \\alpha - (180^\\circ - \\alpha) = 2\\alpha - 180^\\circ$.\n\nДалее, пусть $P$ — центр описанной окружности $\\omega_1$ треугольника $DEF$. Поскольку $\\angle EDF > 90^\\circ$, точка $P$ лежит внутри угла $EDF$, причем точки $D$ и $P$ лежат по разные стороны от прямой $EF$.\n\nВ окружности $\\omega_1$ дуга $EDF$ дополняет дугу величины $2\\alpha$, на которую опирается угол $EDF$, до полной окружности, поэтому центральный угол $EPF$ равен $360^\\circ - 2\\alpha$.\n\nПолучаем, что $\\angle ECF + \\angle EPF = (2\\alpha - 180^\\circ) + (360^\\circ - 2\\alpha) = 180^\\circ$; значит, четырехугольник $CEPF$ вписанный, то есть точка $P$ лежит на окружности $\\omega$, что и требовалось доказать.\n\n![](attached_image_1.png)\nРис. 7",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangles $ABC$ and $A_1B_1C_1$ with equal areas. Determine if it is always possible to construct (using a ruler and a compass) a triangle $A_2B_2C_2$ such that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are congruent, and the lines $AA_2$, $BB_2$ and $CC_2$ are parallel. (D. Tereshin)",
    "options": [],
    "answer": "Yes, always.",
    "solution": "**Answer.** Yes, always.\nIf $\\Delta ABC = \\Delta A_1B_1C_1$, then the construction method is obvious. Otherwise, we may assume that $AB \\neq A_1B_1$ (let's suppose $AB < A_1B_1$ for definiteness).\n\nConstruct triangle $A'B'C$ such that $AB \\parallel A'B'$, $A'B' = A_1B_1$, $B'C = B_1C_1$, $CA' = C_1A_1$ (see Fig. 23; this construction is easily carried out, for example, using the equality $\\angle BCB' = \\angle ABC + \\angle A_1B_1C_1$). Then $ABB'A'$ is a trapezoid. Let $MN$ be its midline, and $P$ the intersection point of the extensions of the lateral sides.\n\nConstruct a circle $\\omega$ on segment $PC$ as diameter. Since $AB < A_1B_1 = A'B'$, $S_{ABC} = S_{A_1B_1C_1} = S_{A'B'C'}$, the distance from point $C$ to line $AB$ is greater than the distance from $C$ to line $A'B'$, so points $P$ and $C$ lie on opposite sides of line $MN$, and therefore $\\omega$ intersects line $MN$. Let $K$ be one of the intersection points (in the figure it lies on segment $MN$, but our reasoning does not depend on this).\n\nDraw line $PK$ to intersect lines $AB$ and $A'B'$ at points $X$ and $Y$, respectively. Then $\\frac{AX}{XB} = \\frac{A'Y}{YB'}$, so $S_{XBC} = S_{YB'C'}$. Moreover, $XK = KY$, and angle $PKC$ is right as it subtends the diameter in circle $\\omega$. Thus, $CK$ is the perpendicular bisector of segment $XY$, and $CX = CY$.\n\nOn the extension of segment $XC$ beyond point $C$, take point $Z$ such that $XC = CZ$. Construct triangles $A_2CZ$ and $B_2CZ$ congruent to triangles $A'CY$ and $B'CY$, respectively (see Fig. 24). Then $\\triangle A_2B_2C = \\triangle A'B'C' = \\triangle A_1B_1C_1$. We show that $AA_2 \\parallel BB_2$ (then we can shift triangle $A_2B_2C$ along line $AA_2$, obtaining the required one). Since $CX = CZ$, $S_{ABC} = S_{A_2B_2C}$ and $S_{XBC} = S_{ZB_2C}$, it follows that $S_{XAC} = S_{ZA_2C}$, the distances from points $A$ and $A_2$ to line $XZ$ are equal, and the distances from points $B$ and $B_2$ to line $XZ$ are also equal. Therefore, $AA_2 \\parallel XZ \\parallel BB_2$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight players participated in a chess tournament, and each pair of players have played exactly once. It appeared that if two players $A$ and $B$ played a draw then the resulting numbers of points of $A$ and $B$ are different. Find the greatest possible number of draws in this tournament. (Each win is worth $1$ point, each draw is worth $\\frac{1}{2}$ points, and each lose is worth $0$ points.) (S. Tokarev)",
    "options": [],
    "answer": "20",
    "solution": "Answer: $20$.\n\nWe will estimate the number $S$ — the sum of the numbers of draws for all $8$ chess players. This sum is exactly twice the number of draws in the tournament (since each draw is counted twice — for both players).\n\nWe will prove that $S \\le 41$ — then the number of drawn games in the tournament does not exceed $20$, since it is an integer. This number is indeed the answer, since an example with $20$ draws exists (see Fig. 8; the players are denoted by the letters $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$).\n\nNote immediately that if two people did not lose to anyone, then they played a draw with each other and, according to the condition, have different numbers of points; the same is true if they did not win against anyone. It follows that there is at most one person with $7$ draws and at most $2$ people with $6$ draws (such a person either did not lose to anyone or did not win against anyone). In addition, there are at most one person with $5$ draws, all of whose two decisive games were both won or both lost. Note that all other players with $5$ draws have $3\\frac{1}{2}$ points.\n\nSuppose there is a person $X$ with $7$ draws; then there cannot be other players with $5$ draws, because they would have the same number of points as $X$, and they would have played a draw with $X$. Therefore, in this case, there are at most $2$ people with $5$ draws, and $S \\le 7 + 2 \\cdot 6 + 2 \\cdot 5 + 3 \\cdot 4 = 41$.\n\nNow suppose there is no person with $7$ draws. Estimate in this case the number of players with $5$ draws and $3\\frac{1}{2}$ points. Each of them must have played with each other not in a draw; however, they have only $2$ decisive games — so there are at most $3$ of them, and in total, the number of players with $5$ draws does not exceed $3 + 2 = 5$. Therefore, in this case, $S \\le 2 \\cdot 6 + 5 \\cdot 5 + 4 = 41$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39588,
    "subject": "Mathematics (Multi-modal)",
    "question": "A teacher has $40$ pencils of $4$ colors (there are $10$ pencils of each color). He distributes them to $10$ kids so that each kid has $4$ pencils. Find the least value of $n$ such that the class leader can always find a group of $n$ kids having altogether the pencils of all $4$ colors.\n\nИмеется $40$ карандашей четырех цветов — по $10$ карандашей каждого цвета. Их раздали $10$ ребятам так, что каждый получил по $4$ карандаша. Какое наименьшее количество ребят всегда можно выбрать, чтобы у них гарантированно нашлись карандаши всех цветов, вне зависимости от распределения карандашей?",
    "options": [],
    "answer": "3",
    "solution": "**Ответ.** $3$ ребят.\n\nПокажем, что всегда можно выбрать трёх ребят так, чтобы у них нашлись карандаши всех цветов. Так как карандашей каждого цвета $10$, а каждому досталось по $4$ карандаша, то кому-то достались карандаши по крайней мере двух различных цветов. Осталось добавить к нему двух ребят, у которых есть карандаши оставшихся двух цветов.\n\nПокажем теперь, как раздать карандаши ребятам, чтобы у любых двух из них вместе были карандаши не более трёх цветов. Раздадим двум ребятам по $4$ карандаша второго цвета, двум — по $4$ карандаша третьего цвета, двум — по $4$ карандаша четвёртого цвета, одному — $4$ карандаша первого цвета, одному — по $2$ карандаша первого и второго цвета, одному — по $2$ карандаша ...",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nine skiers participated in a race. They started one by one, and each skier passed the distance with a constant speed (which could be different for different skiers). Determine if it could happen that each skier participated in an overtaking exactly four times. (In each overtaking, exactly two skiers participated: the one who overtakes and the one who is overtaken.)\n\nДевять лыжников ушли со старта по очереди и прошли дистанцию — каждый со своей постоянной скоростью. Могло ли оказаться, что каждый лыжник участвовал ровно в четырёх обгонах? (В каждом обгоне участвуют ровно два лыжника — тот, кто обгоняет, и тот, кого обгоняют.)",
    "options": [],
    "answer": "It could not happen.",
    "solution": "It could not happen.\n\nSuppose it is possible. Since the speeds are constant, any two skiers met at most once. Then the skier who started first could not overtake anyone; therefore, he was overtaken by four skiers and finished fifth. On the other hand, the skier who started last could not be overtaken by anyone, so he himself overtook four skiers and also finished fifth. Contradiction.\n\n**Note.** A similar argument shows that only skier number 5 could have finished both first and last in the race.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39590,
    "subject": "Mathematics (Multi-modal)",
    "question": "100 pairwise distinct real numbers are arranged in a circle. Prove that there exist four successive numbers among them such that the sum of the two middle ones is strictly less than the sum of the other two.\n\nПо окружности расставлено 100 попарно различных чисел. Докажите, что можно выбрать 4 подряд стоящих числа таким образом, чтобы сумма двух крайних чисел этой четверки была строго больше суммы средних. (С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "See the solution to problem 9.2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$, $BE$ and $CF$ be the angle bisectors in a triangle $ABC$, and let $I$ be their intersection point. The perpendicular bisector of segment $AD$ intersects lines $BE$ and $CF$ at points $M$ and $N$, respectively. Prove that the points $A$, $I$, $M$, and $N$ are concyclic. (D. Prokopenko)\n\nВ треугольнике $ABC$ проведены биссектрисы $AD, BE$ и $CF$, пересекающиеся в точке $I$. Серединный перпендикуляр к отрезку $AD$ пересекает прямые $BE$ и $CF$ в точках $M$ и $N$ соответственно. Докажите, что точки $A, I, M$ и $N$ лежат на одной окружности. (Д. Прокопенко)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Для решения задачи достаточно установить, что $\\angle MAI = \\angle MNI$ (см. рис. 5).\n\nПусть $K$ — середина отрезка $AD$. Заметим, что $\\angle MNI = \\angle KNI = 90^\\circ - \\angle KIN = 90^\\circ - (\\angle ACI + \\angle CAI) = \\frac{1}{2}(180^\\circ - (\\angle ACB + \\angle BAC)) = \\frac{1}{2}\\angle ABC$.\n\nОстаётся установить, что $\\angle MAI = \\frac{1}{2} \\angle ABC$. Пусть $M'$ — точка пересечения окружности, описанной около треугольника $ABD$, с серединным перпендикуляром к отрезку $AD$ (точка $M'$ лежит на дуге $\\overarc{AD}$, не содержащей точку $B$). Тогда $AM' = DM'$, а значит, и $\\angle M'BD = \\angle M'BA$, как опирающиеся на равные дуги. Это означает, что точка $M'$ лежит на биссектрисе угла $ABC$ и, следовательно, $M'$ совпадает с $M$.\n\nИтак, точки $A, M, D$ и $B$ лежат на одной окружности, откуда $\\angle MAI = \\angle MBD = \\frac{1}{2} \\angle ABC$, что и требовалось.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a nonisosceles acute-angled triangle $ABC$. Let $O$ be its circumcenter, let $AD$ be its altitude, and let $M$ be the midpoint of side $BC$. The lines passing through $O$, perpendicular to $AB$ and $AC$, intersect $AD$ at points $P$ and $Q$, respectively. Let $S$ be the circumcenter of triangle $OPQ$. Prove that $\\angle BAS = \\angle CAM$.\n\nЧерез центр $O$ окружности, описанной около неравнобедренного остроугольного треугольника $ABC$, проведены прямые, перпендикулярные сторонам $AB$ и $AC$. Эти прямые пересекают высоту $AD$ треугольника $ABC$ в точках $P$ и $Q$. Точка $M$ — середина стороны $BC$, а $S$ — центр окружности, описанной около треугольника $OPQ$. Докажите, что $\\angle BAS = \\angle CAM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть для определённости $AB > AC$ (см. рис. 19). Обозначим через $L$ середину отрезка $AB$. Заметим, что $\\angle AOL = \\frac{1}{2} \\angle AOB = \\angle ACB$. Отсюда $\\angle BAO = 90^\\circ - \\angle AOL = 90^\\circ - \\angle ACB = \\angle CAD$.\n![](attached_image_1.png)\nРис. 19\nСтороны треугольника $OPQ$ перпендикулярны соответственно сторонам треугольника $ABC$, поэтому $\\triangle OPQ$ подобен $\\triangle ABC$: он получается из $\\triangle ABC$ последовательным выполнением поворота на $90^\\circ$ и гомотетии с некоторым коэффициентом $k$. Так как $OS$ и $AO$ — соответственные отрезки в треугольниках $OPQ$ и $ABC$, то $OS \\perp AO$ и $OS = k \\cdot AO$. Далее, отрезок $MD$ равен высоте треугольника $OPQ$, проведенной к стороне $PQ$, поэтому $MD = k \\cdot AD$. Значит, прямоугольные треугольники $AOS$ и $ADM$ подобны, поэтому $\\angle SAO = \\angle MAD$.\n\nОкончательно, $\\angle BAS = \\angle BAO + \\angle SAO = \\angle CAD + \\angle MAD = \\angle CAM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each cell of a square grid $100 \\times 100$ contains a positive integer. A grid rectangle is called *good* if the sum of numbers in all its cells is divisible by $17$. By one move it is allowed to color all the cells of some good rectangle. It is not allowed to color a certain cell twice. Find the maximal $d$ such that for every arrangement of numbers, it is possible to color at least $d$ cells in some moves.\n\nВ каждой клетке квадрата $100 \\times 100$ записано некоторое натуральное число. Прямоугольник, стороны которого идут по линиям сетки, назовем хорошим, если сумма чисел во всех его клетках делится на $17$. Разрешается одновременно закрашивать все клетки в некотором хорошем прямоугольнике. Одну клетку запрещается закрашивать дважды. При каком наибольшем $d$ можно закрасить хотя бы $d$ клеток при любом расположении чисел?",
    "options": [],
    "answer": "9744",
    "solution": "$9744 = 100^2 - 16^2$ cells.\n\n**Lemma.** Let a strip $1 \\times k$ be filled with positive integers. Then it is possible to color several non-overlapping good rectangles in it, covering at least $k - 16$ cells.\n\n**Proof.** Induction on $k$. For $k \\le 16$ nothing needs to be colored. Suppose $k \\ge 17$. Let the leftmost $17$ cells contain numbers $a_1, \\dots, a_{17}$. Among the numbers $0, a_1, a_1 + a_2, \\dots, a_1 + \\dots + a_{17}$, there are two with the same remainder modulo $17$. Then their difference, which has the form $a_i + a_{i+1} + \\dots + a_j$, is divisible by $17$. Remove the cells from $i$ to $j$ from the strip. The remaining cells form a strip of length $k - (j - i + 1)$. By the induction hypothesis, we can color several good rectangles so that at most $16$ cells remain uncolored. Then in the original strip, we can color the same cells, as well as the cells from $i$ to $j$ (they either form a new good rectangle or fall inside an old one). $\\square$\n\nNow to the problem. We show that it is possible to leave at most $16^2 = 256$ uncolored cells. Consider a strip $1 \\times 100$, where the cells contain the sums of the numbers in the columns of the original square. Applying the lemma to it, we find several good rectangles. Then in the original square, we can color the corresponding rectangles of height $100$. After this, at most $16$ columns remain uncolored. Now apply the lemma to each of them separately; in each column, at most $16$ cells remain uncolored, i.e., at most $256$ cells in total.\n\nIt remains to give an example of an arrangement in which it is impossible to leave fewer than $256$ cells uncolored. Place ones in some $16 \\times 16$ square, and zeros in all other cells. Consider any rectangle $P$; if it contains a one, then it intersects the square in some rectangle $a \\times b$ ($1 \\le a, b \\le 16$); but then the sum of all numbers in $P$ is $ab$, which cannot be divisible by $17$. Thus, no cell with a one will be colored, so at least $256$ cells remain uncolored.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Say that a positive integer $b$ is *fortunate* if for every positive integer $a$, a number $b$ divides $a^2$ whenever $b^2$ divides $a^5$. Find the number of fortunate positive integers which are less than $2010$. (P. Kozhevnikov)\n\nНатуральное число $b$ назовём удачным, если для любого натурального $a$ такого, что $a^5$ делится на $b^2$, число $a^2$ делится на $b$. Найдите количество удачных натуральных чисел, меньших $2010$. (П. Кожевников)",
    "options": [],
    "answer": "1961",
    "solution": "Установим следующее описание удачных чисел.\n\n**Лемма.** Число $b$ является удачным тогда и только тогда, когда каждое простое число входит в разложение $b$ на простые множители с одним из следующих показателей: $0, 1, 2, 3, 4, 6, 8$.\n\n**Доказательство.** Назовем целое неотрицательное число $k$ счастливым, если не существует такого целого $m$, что $2m < k \\le \\frac{5}{2}m$. Заметим, что счастливыми являются в точности числа $0, 1, 2, 3, 4, 6, 8$. Действительно, при $k \\le 8$ в этом можно убедиться прямой проверкой. Если же $k \\ge 9$, то выберем максимальное число $m$ такое, что $2m < k$. Тогда $m \\ge 4$, и $\\frac{5}{2}m \\ge 2m+2 = 2(m+1) \\ge k$ по выбору $m$, то есть $k$ несчастливо. Осталось показать, что $b$ удачно тогда и только тогда, когда каждое простое число входит в разложение $b$ со счастливым показателем.\n\nПусть число $b$ неудачно, то есть $a^5 : b^2, a^2 : b$ для некоторого $a$. Тогда некоторое простое $p$ входит в разложение $a^2$ в меньшей степени, чем в разложение $b$. Пусть $p$ входит в разложение $a$ и $b$ в степенях $m$ и $k$ соответственно; тогда $2m < k$, но $\\frac{5}{2}m \\ge k$, так как $a^5 : b^2$. Значит, число $k$ — несчастливое.\n\nИтак, если все степени вхождения простых чисел в $b$ счастливы, то $b$ удачно. В противном же случае, если $b = p^k b'$, где $b'$ не делится на $p$ и $k$ несчастливо ($2m < k \\le \\frac{5}{2}m$), то при $a = p^m b'$ число $a^5$ делится на $b^2$, а $a^2$ не делится на $b$, и $b$ неудачно. Лемма доказана. $\\Box$\n\nПодсчитаем теперь количество неудачных натуральных чисел, меньших $2010$. Согласно лемме, надо подсчитать количество чисел, имеющих простой делитель, входящий в разложение на простые множители с показателем $5, 7, 9$ или более $9$. Поскольку $2^{10} < 2010 < 2^{11}$, $3^6 < 2010 < 3^7$, $2^5 \\cdot 3^5 > 2010$ и $5^5 > 2010$, каждое неудачное число, меньшее $2010$, принадлежит к одному из следующих непересекающихся классов:\n\n1) числа вида $2^5q$, где $q$ — нечётное и $q \\le 61$ (поскольку $2^5 \\cdot 61 < 2010 < 2^5 \\cdot 63$);\n\n2) числа вида $2^7q$, где $q$ — нечётное и $q \\le 15$ (поскольку $2^7 \\cdot 15 < 2010 < 2^7 \\cdot 17$);\n\n3) числа вида $2^9q$, где $q = 1$ или $q = 3$ (поскольку $q$ нечетно, и $2^9 \\cdot 3 < 2010 < 2^9 \\cdot 5$);\n\n4) число $2^{10}$;\n\n5) числа вида $3^5q$, где $q$ не делится на $3$ и $q \\le 8$ (поскольку $3^5 \\cdot 8 < 2010 < 3^5 \\cdot 10$).\n\nИтого мы получаем ровно $31$ число из класса 1, ровно $8$ чисел из класса 2, ровно $2$ числа из класса 3, ровно одно число из класса 4 и ровно $6$ чисел из класса 5. Таким образом, общее количество неудачных чисел, меньших $2010$, равно $31 + 8 + 2 + 1 + 6 = 48$. Тогда количество удачных чисел равно $2009 - 48 = 1961$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two distinct real numbers $a$ and $b$ are chosen in such a way that the equation\n$$\n(x^2 + 20a x + 10b)(x^2 + 20b x + 10a) = 0\n$$\nhas no real roots. Prove that the number $20(b-a)$ is not an integer.\n\nРазличные действительные числа $a$ и $b$ таковы, что уравнение\n$$\n(x^2 + 20a x + 10b)(x^2 + 20b x + 10a) = 0\n$$\nне имеет корней. Докажите, что число $20(b - a)$ не является целым.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "10.5. См. решение задачи 9.5.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonzero real numbers $a$, $b$, and $c$ are chosen so that $a x^2 + b x + c > c x$ for each real $x$. Prove that $c x^2 - b x + a > c x - b$ for each real $x$. (M. Murashkin)\n\nНенулевые числа $a$, $b$, $c$ таковы, что $a x^2 + b x + c > c x$ при любом $x$. Докажите, что $c x^2 - b x + a > c x - b$ при любом $x$. (М. Мурашкин)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Так как для всех $x$ верно неравенство $P(x) = a x^2 + (b - c) x + c > 0$, то дискриминант трёхчлена $P(x)$ отрицателен: $D = (b - c)^2 - 4 a c = b^2 + c^2 - 2 b c - 4 a c < 0$. Кроме того, $c = P(0) > 0$. Значит, у трехчлена $Q(x) = c x^2 - (b + c) x + (a + b)$ положителен старший коэффициент, а его дискриминант $D' = (b + c)^2 - 4 c (a + b) = b^2 + c^2 + 2 b c - 4 a c - 4 b c = D$ отрицателен. Это значит, что $Q(x) > 0$ при всех действительных $x$, что и требовалось доказать.\n\n**Замечание.** Другое решение можно получить, заметив, что\n$$Q(x) = (1 - x)^2 P\\left(\\frac{1}{1 - x}\\right).$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be the circumcircle of a triangle $ABC$. A point $K$ is chosen on the bisector line of angle $BAC$ so that $K$ lies inside triangle $ABC$. Line $CK$ intersects $\\omega$ at points $C$ and $M$. A circle $\\Omega$ touches line $CM$ at $K$, and intersects segment $AB$ at points $A$ and $P$. Circles $\\Omega$ and $\\omega$ intersect at points $A$ and $Q$. Prove that points $P$, $Q$, and $M$ are collinear.\n\nВнутри треугольника $ABC$ взята точка $K$, лежащая на бисектрисе угла $BAC$. Прямая $CK$ вторично пересекает окружность $\\omega$, описанную около треугольника $ABC$, в точке $M$. Окружность $\\Omega$ проходит через точку $A$, касается прямой $CM$ в точке $K$ и пересекает вторично отрезок $AB$ в точке $P$, а окружность $\\omega$ — в точке $Q$. Докажите, что точки $P$, $Q$ и $M$ лежат на одной прямой.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $CK$ касается $\\Omega$, имеем $\\angle APK = \\angle AKC$.\n\nПусть прямая $MP$ пересекает вторично окружность $\\omega$ в точке $Q'$. Тогда имеем $\\angle AQ'MP = \\angle AQ'M = \\angle ACM = 180^\\circ - \\angle AKC - \\angle KAC = 180^\\circ - \\angle APK - \\angle PAK = \\angle AKP$.\n\nЗначит, точки $A$, $P$, $K$ и $Q'$ лежат на одной окружности, эта окружность совпадает с $\\Omega$, и, следовательно, $Q'$ совпадает с $Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $B$ and $D$ lie on a circle $\\omega$. The tangent lines to $\\omega$ at $B$\n\nand $D$ intersect at $P$. A line passing through $P$ intersects $\\omega$ at $A$ and $C$. Let $\\ell$ be an arbitrary line parallel to $BD$ and intersecting the polygonal lines $ABC$ and $ADC$. Prove that $\\ell$ divides the lengths of these polygonal lines at the same ratio. (L. Emelyanov)\n\nПрямые, касающиеся окружности $\\omega$ в точках $B$ и $D$, пересекаются в точке $P$. Прямая, проходящая через $P$, высекает на окружности хорду $AC$. Через произвольную точку отрезка $AC$ проведена прямая, параллельная $BD$. Докажите, что она делит длины ломаных $ABC$ и $ADC$ в одинаковых отношениях.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Треугольники $PBA$ и $PCB$ подобны, так как $\\angle BPC$ — общий,\nа $\\angle PBA = \\angle PCB = \\frac{1}{2} \\overline{AB}$. Значит, $\\frac{BA}{BC} = \\frac{PB}{PC}$. Аналогично, из подобия треугольников $PDA$ и $PCD$ следует, что $\\frac{DA}{DC} = \\frac{PD}{PC}$.\nТак как $PB = PD$, то $\\frac{BA}{BC} = \\frac{DA}{DC}$, или $\\frac{AB}{AD} = \\frac{CB}{CD}$; заметим, что тогда и $\\frac{AB + CB}{AD + CD} = \\frac{AB}{AD} = \\frac{CB}{CD}$.\n\nОбозначим через $Q$ точку пересечения отрезков $AC$ и $BD$, а через $T$ — произвольную точку на отрезке $AC$. Пусть для определенности $T$ лежит на отрезке $QC$, а прямая, проходящая через $T$ параллельно $BD$, пересекает $CB$ и $CD$ в точках $B'$ и $D'$, соответственно. Тогда по теореме Фалеса $\\frac{CB'}{CD'} =$\n$$= \\frac{CB}{CD} = \\frac{AB + CB}{AD + CD}, \\text{ или } \\frac{CB'}{AB + CB} = \\frac{CD'}{AD + CD}, \\text{ что и требовалось.}$$\n\nЕсли же точка $T$ лежит на отрезке $AQ$, то аналогично рассматриваются отрезки, высекаемые на сторонах $AB$ и $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist three distinct nonzero integers $a$, $b$, $c$ such that $a + b + c = 0$, and $a^{13} + b^{13} + c^{13}$ is a square of a positive integer?\n\nСуществуют ли три попарно различных ненулевых целых числа, сумма которых равна нулю, а сумма тринадцатых степеней которых является квадратом некоторого натурального числа?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For a natural number $t$, the triple $3t$, $-t$, $-2t$ satisfies all conditions except, possibly, the last one. To ensure that $(3t)^{13} + (-t)^{13} + (-2t)^{13} = t^{13}(3^{13} - 1 - 2^{13})$ is a perfect square, it is sufficient to set, for example, $t = 3^{13} - 1 - 2^{13}$.\n\nThus, the triple $3t$, $-t$, $-2t$, where $t = 3^{13} - 1 - 2^{13}$, satisfies the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 3$ pairwise coprime positive integers. It is known that while dividing the product of any $n-1$ of them by the remaining number, a remainder equals to $r$ (the remainder $r$ is the same for all $(n-1)$-tuples). Prove that $r \\le n-2$.\n\nДаны $n \\ge 3$ попарно взаимно простых чисел. Известно, что при делении произведения любых $n-1$ из них на оставшееся число получается один и тот же остаток $r$. Докажите, что $r \\le n-2$. (В. Сендеров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Если $r = 0$, то утверждение задачи, очевидно, истинно. Пусть $r > 0$. Пусть $a_1, \\dots, a_n$ — данные числа; положим $P = a_1a_2 \\dots a_n$, $P_i = P/a_i$ при $i = 1, 2, \\dots, n$. Заметим, что $a_i > r$, ибо число $P_i$ даёт остаток $r$ при делении на $a_i$.\n\nРассмотрим число $S = P_1 + P_2 + \\dots + P_n - r$. Заметим, что $S = (P_1-r) + (P_2+P_3+\\dots+P_n) = a_1$, поскольку оба слагаемых делятся на $a_1$. Аналогично, $S : a_i$ при всех $i = 1, \\dots, n$; поскольку $a_i$ попарно взаимно просты, получаем $S : a_1 \\dots a_n = P$. Поскольку $S > a_1 - r > 0$, получаем, что $S \\ge P$, а тогда $P_1 + \\dots + P_n = S + r > P$. Значит, при некотором $i$ верно неравенство $P_i > P/n$, откуда $a_i < n$, или $a_i \\le n - 1$. Но тогда $r < a_i \\le n - 1$, то есть $r \\le n - 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39601,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a certain country some pairs of cities are linked by bi-directional direct airlines so that the system of airlines is connected (that is, from each city one can reach each other city by some flights). It appears that for each cyclic route containing an odd number of airlines, cancelling all airlines of this cyclic route makes the system of airlines not connected. Prove that one can distribute all the cities into 4 regions so that each airline links two cities from different regions. (A region may contain no cities.)\n(V. Dol'nikov)\n\nВ стране некоторые пары городов соединены двусторонними беспосадочными авиарейсами. При этом из любого города можно долететь в любой другой (возможно, с пересадками). Известно, что если выбрать любой замкнутый маршрут из нечётного числа рейсов и закрыть все эти рейсы, то уже не из любого города можно будет добраться в любой другой. Докажите, что все города можно распределить по 4 республикам так, чтобы любой рейс соединял города из разных республик. (Некоторые республики могут не содержать городов.) (В. Дольников)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Рассмотрим граф $G$, вершинами которого являются города, и две вершины соединены ребром, если между городами есть авиалиния. Тогда нам известно, что граф связан, но при удалении всех рёбер любого нечётного цикла это условие нарушается; доказать же нужно, что вершины графа можно правильно раскрасить в 4 цвета. Мы будем пользоваться следующей известной леммой.\n\n**Лемма.** Пусть в графе нет циклов нечётной длины. Тогда его вершины можно правильно раскрасить двумя красками.\n\n**Доказательство.** Ясно, что достаточно доказать лемму для связного графа. Расстоянием между двумя вершинами $X$ и $Y$ назовём наименьшую длину пути, соединяющего эти вершины.\n\nЗафиксируем некоторую вершину $A$, и покрасим все вершины, находящиеся на нечетном расстоянии от $A$, в красный цвет, а остальные вершины — в синий цвет. Докажем, что указанная раскраска — искомая. Предположим противное — имеется ребро, соединяющее, скажем, красные вершины $B$ и $C$. Рассмотрим кратчайшие пути $A = B_0, B_1, ..., B_{2n-1} = B$ и $A = C_0, C_1, ..., C_{2m-1} = C$, ведущие из $A$ в $B$ и в $C$. Взяв наибольший индекс $i$ такой, что $B_i = C_i$, получим цикл нечетной длины $B_i, B_{i+1}, ..., B_{2n-1}, C_{2m-1}, ..., C_i = B_i$. Противоречие. ☐\n\nПусть в графе есть цикл; удалим одно из рёбер этого цикла. При этом, очевидно, граф останется связным. Продолжим этот процесс до тех пор, пока циклов в графе не останется. Обозначим через $V$ множество всех удалённых рёбер, а через $W$ — множество всех оставшихся. Заметим, что оставшийся граф по-прежнему связан. Это значит, что не существует нечётного цикла, все рёбра которого принадлежат $V$; в самом деле, если бы такой цикл существовал, то при удалении из $G$ всех его рёбер остались бы все рёбра множества $W$, и граф остался бы связным.\n\nТеперь рассмотрим два графа $G_V$ и $G_W$, вершинами которых являются вершины графа $G$, а множества рёбер — это $V$ и $W$, соответственно. Тогда в графе $G_W$ циклов нет (значит, его вершины можно правильно раскрасить в цвета 0 и 1), а в $G_V$ по доказанному нет нечётных циклов (значит, его вершины можно правильно раскрасить в цвета 0 и 2). Присвоим теперь каждой вершине сумму её цветов в этих раскрасках. Тогда, если две вершины соединены ребром в графе $G$, то они соединены ребром в одном из графов $G_V$ или $G_W$; тогда, как нетрудно видеть, их цвета различны, то есть полученная раскраска (в цвета 0, 1, 2, 3) является правильной.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39602,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a square grid $n \\times n$, a set consisting of all cells lying on or under its main diagonal is called an $n$-staircase (4-staircase is shown in the figure). Find the number of ways to partition an $n$-staircase into several grid rectangles with pairwise distinct areas. (D. Khramtsov)\n\n![](attached_image_1.png)\n\nНазовём лестницей высоты $n$ фигурую, состоящую из всех клеток квадрата $n×n$, лежащих не выше диагонали (на рисунке показана лестница высоты 4). Сколькими различными способами можно разбить лестницу высоты $n$ на несколько прямоугольников, стороны которых идут по линиям сетки, а площади попарно различны?\n\n![](attached_image_2.png)",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "Ответ. $2^{n-1}$.\n\nОтметим в каждом столбце лестницы по одной верхней клетке; назовём их объединение **верхним слоем**. Никакие две из $n$ клеток этого слоя не могут лежать в одном прямоугольнике разбиения, поэтому в любом разбиении лестницы не менее $n$ прямоугольников. С другой стороны, минимальная суммарная площадь $n$ прямоугольников с различными площадями равна $1 + 2 + \\ldots + n$, что совпадает с площадью всей лестницы. Значит, число прямоугольников в любом разбиении равно $n$, их площади выражаются числами $1, 2, \\ldots, n$, и каждый из них сохранит клетку верхнего слоя.\n\nПокажем индукцией по $n$, что число требуемых разбиений лестницы высоты $n$ равно $2^{n-1}$. База индукции при $n = 1$ очевидна. Пусть утверждение индукции справедливо для лестницы высоты $n - 1$; рассмотрим разрезание лестницы высоты $n$ на прямоугольники площадей $1, 2, \\ldots, n$.\n\nРассмотрим прямоугольник, покрывающий угловую (наиболее далекую от верхнего слоя) клетку лестницы. Он содержит клетку верхнего слоя, то есть сумма длин его сторон $a$ и $b$ равна $n + 1$. Поэтому его площадь $S = ab \\ge a + b - 1 = n$, так как $(a-1)(b-1) = ab - (a + b - 1) \\ge 0$; при этом равенство может достигаться лишь при $a = 1$ или $b = 1$. Поскольку площади прямоугольников разбиения не превосходят $n$, то $S = n$, и одна из сторон нашего прямоугольника равна $1$, а другая — $n$. Такой прямоугольник можно выбрать двумя способами (вертикальный или горизонтальный), причем в обоих случаях после его отрезания остается лестница высоты $n - 1$, количество способов разрезать которую на оставшиеся прямоугольники площадей $1, 2, \\ldots, n - 1$ равно $2^{n-2}$ по предположению индукции. Значит, искомое количество способов равно $2^{n-2} + 2^{n-2} = 2^{n-1}$, что и требовалось.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right-angled triangle $ABC$ with $\\angle ACB = 90^\\circ$. One has increased each of the lengths of its legs $AC$ and $BC$ by $1$ (and $\\angle ACB$ remained right). Could it happen that the length of the hypothenuse $AB$ increased by more than $\\sqrt{2}$? (S. Volchenkov)\n\nКаждый катет прямоугольного треугольника увеличили на единицу. Могла ли его гипотенуза увеличиться более, чем на $\\sqrt{2}$?",
    "options": [],
    "answer": "No",
    "solution": "**Первое решение.** Пусть длины катетов исходного прямоугольного треугольника были равны $x$ и $y$. Тогда его гипотенуза имела длину $\\sqrt{x^2 + y^2}$, а после увеличения катетов стала $\\sqrt{(x+1)^2 + (y+1)^2}$. Предположим, что гипотенуза увеличится более, чем на $\\sqrt{2}$. Тогда\n\n$$\n\\begin{aligned}\n\\sqrt{(x+1)^2 + (y+1)^2} &> \\sqrt{x^2 + y^2} + \\sqrt{2} \\\\\n\\Leftrightarrow (x+1)^2 + (y+1)^2 &> x^2 + y^2 + 2 + 2\\sqrt{2(x^2 + y^2)} \\\\\n\\Leftrightarrow 2x + 2y &> 2\\sqrt{2(x^2 + y^2)} \\\\\n\\Leftrightarrow (x+y)^2 > 2(x^2 + y^2) \\\\\n\\Leftrightarrow 2xy &> x^2 + y^2 \\\\\n\\Leftrightarrow 0 > (x-y)^2,\n\\end{aligned}\n$$\nчто невозможно. Значит, гипотенуза не могла увеличиться более, чем на $\\sqrt{2}$.\n\n\n**Второе решение.** Расположим исходный и полученный треугольники $ABC$ и $AB'C'$ так, как показано на рис. 7 ($BB' = CC' = 1$). Тогда $\\overrightarrow{B'C'} = \\overrightarrow{B'B} + \\overrightarrow{BC} + \\overrightarrow{CC'} = \\overrightarrow{BC} + (\\overrightarrow{B'B} + \\overrightarrow{CC'})$. Однако сумма двух перпендикулярных единичных векторов $\\overrightarrow{B'B} + \\overrightarrow{CC'}$ имеет длину $\\sqrt{2}$; значит, $|\\overrightarrow{B'C'}| \\le |\\overrightarrow{BC}| + |\\overrightarrow{B'B} + \\overrightarrow{CC'}| = |\\overrightarrow{BC}| + \\sqrt{2}$, что и требовалось доказать.\n\n![](attached_image_1.png)\n\n\n**Третье решение.** Расположим треугольники так же, как в предыдущем решении. Пусть $B_1, C_1$ — проекции точек $B$ и $C$ на $B'C'$, а $\\angle AB'C' = \\alpha$. Тогда отрезок $B_1C_1$ является проекцией $BC$ и потому $B_1C_1 \\le BC$. Далее, $B'B_1 = BB' \\cos \\alpha = \\cos \\alpha$, $C'C_1 = CC' \\sin \\alpha = \\sin \\alpha$, и $B'C' = B'B_1 + B_1C_1 + C_1C' \\le BC + \\cos \\alpha + \\sin \\alpha = BC + \\sqrt{2} \\sin(45^\\circ + \\alpha) \\le BC + \\sqrt{2}$, что и требовалось.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist non-zero real numbers $a_1, a_2, \\dots, a_{10}$ such that\n$$ \\left(a_1 + \\frac{1}{a_1}\\right) \\cdots \\left(a_{10} + \\frac{1}{a_{10}}\\right) = \\left(a_1 - \\frac{1}{a_1}\\right) \\cdots \\left(a_{10} - \\frac{1}{a_{10}}\\right). $$\n\nСуществуют ли такие ненулевые действительные числа $a_1, a_2, \\dots, a_{10}$, что\n$$\n\\left(a_1 + \\frac{1}{a_1}\\right) \\cdots \\left(a_{10} + \\frac{1}{a_{10}}\\right) = \\left(a_1 - \\frac{1}{a_1}\\right) \\cdots \\left(a_{10} - \\frac{1}{a_{10}}\\right)?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Рассмотрим произвольные ненулевые числа $a_1, \\ldots, a_{10}$. Заметим, что числа $a_k$ и $\\frac{1}{a_k}$ имеют одинаковый знак. Значит,\n$$\n\\left| a_k + \\frac{1}{a_k} \\right| = \\left| a_k \\right| + \\frac{1}{\\left| a_k \\right|} > \\max \\left( \\left| a_k \\right|, \\frac{1}{\\left| a_k \\right|} \\right) \\ge \\left| a_k - \\frac{1}{a_k} \\right| \\ge 0.\n$$\nПеремножая эти неравенства, получаем, что\n$$\n\\left| a_1 + \\frac{1}{a_1} \\right| \\cdots \\left| a_{10} + \\frac{1}{a_{10}} \\right| > \\left| a_1 - \\frac{1}{a_1} \\right| \\cdots \\left| a_{10} - \\frac{1}{a_{10}} \\right|,\n$$\nто есть требуемое равенство невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39605,
    "subject": "Mathematics (Multi-modal)",
    "question": "2009 weights are placed in a row. Each weight is of integer number of grams, and no weight is heavier than 1 kg. It turned out that each neighboring weights differ by exactly 1 g, while the total weight of all weights (in grams) is an even number. Prove that it is possible to partition all the weights into two sets with equal total weights.\n\nВ ряду из 2009 гирек вес каждой гирьки составляет целое число граммов и не превышает 1 кг. Веса любых двух соседних гирек отличаются ровно на 1 г, а общий вес всех гирь в граммах является чётным числом. Докажите, что гирьки можно разделить на две кучки, суммы весов в которых равны.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Ясно, что веса всех гирь, стоящих на нечётных местах, имеют одинаковую чётность, а веса всех остальных гирь — другую чётность. Так как общий вес чётен, то 1005 гирь на нечётных местах имеют чётные веса.\n\nПоставим первую гирьку на левую чашку весов (пусть ее вес равен $2a \\le 1000$ г), остальные 2008 гирь разобьём на 1004 пары стоящих рядом гирек (тогда веса гирек в любой паре отличаются на 1 г). Теперь в некоторых ($502 - a$) парах поставим более лёгкие гири каждой пары на правую чашку весов, а более тяжёлые — на левую; в остальных же ($502 + a$) парах поставим лёгкие гири на левую чашку, а тяжёлые — на правую. Тогда разность\n\nсуммарных весов гирь на левой и правой чашках будет равна\n$$\n2a + (502 - a) - (502 + a) = 0,\n$$\nчто и требовалось.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a square grid $n \\times n$ ($n \\ge 4$). Initially each of $n$ cells of one diagonal contains a sign “+” while each other cell contains a sign “-”. By one move it is allowed to choose a row or a column and replace each sign “-” by “+”, and vice versa. Prove that at any moment the number of signs “+” is not less than $n$. (R. Karasev)\n\nВ клетчатой таблице $n \\times n$ ($n \\ge 4$) поставлены $n$ знаков «+» в клетках одной диагонали и знаки «-» во всех остальных клетках. Разрешается в некоторой строке или в некотором столбце поменять все знаки на противоположные. Докажите, что после любого количества таких операций в таблице останется не менее $n$ плюсов. (Р. Карасёев)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Пронумеруем строки числами $1, \\dots, n$ сверху вниз, а столбцы — теми же числами слева направо. Клетку будем обозначать парой номеров её строки и столбца; при этом будем считать, что клетки диагонали из плюсов имеют координаты $(i, i)$ $(i = 1, \\dots, n)$.\nЗаметим, что если четыре клетки лежат в вершинах прямоугольника со сторонами, параллельными осям координат, то любая операция либо не меняет знаков в этих клетках, либо меняет знаки ровно в двух клетках из четырёх. В частности, чётность количества плюсов в этих четырёх клетках не меняется; значит, если среди них вначале был ровно один плюс, то и потом их будет не менее одного.\nТеперь выберем в нашей таблице $n$ непересекающихся таких четвёрок; по сказанному выборе, после любых операций в каждой из них найдётся как минимум один плюс, следовательно, всего плюсов будет не менее $n$. При $i = 1, 2, \\dots, n-2$ выберем четвёрку клеток $\\{(i, i), (i, i+1), (i+2, i), (i+2, i+1)\\}$, а также выберем четвёрки $\\{(n-1, n-1), (n-1, n), (1, n-1), (1, n)\\}$ и $\\{(n, n), (n, 1), (2, n), (2, 1)\\}$. Легко видеть, что они удовлетворяют всем требованиям. На рис. 21 отмечены такие четвёрки при $n=5$.\n\n![](attached_image_1.png)\n\nВторое решение. Заметим, что знак, стоящий в клетке, изменяется ровно тогда, когда с этой клеткой было проделано нечётное число операций. Пусть есть ровно $r$ строк и ровно $c$ столбцов, к каждому из которых операция применялась нечётное число раз (назовём их нечётными). Тогда знак изменился ровно в $r(n-c)$ клетках, стоящих на пересечении нечётных строк с чётными столбцами, и в $c(n-r)$ клетках, стоящих на пересечении чётных строк с нечётными столбцами. Теперь нетрудно понять, что, если бы мы вместо исходных операций применили бы по одной операции ровно ко всем чётным строкам и столбцам, результат получился бы тем же самым, но при этом числа $r$ и $c$ заменились бы на $n-r$ и $n-c$ соответственно. Значит, можно считать, что $r+c \\le n$.\nДалее, среди изменённых $r(n-c)+c(n-r)$ знаков не более, чем $r+c$ были плюсами (максимум по одному в $r$ строках, и максимум по одному в $c$ столбцах); значит, хотя бы $r(n-c)+c(n-r)-(r+c)$ минусов стали плюсами, и хотя бы $n-(r+c)$ плюсов остались плюсами. Таким образом, общее количество плюсов $P$ стало не меньше, чем $r(n-c)+c(n-r)-(r+c)+n-(r+c) = -2rc + (r+c)(n-2) + n$. Теперь, поскольку $2rc \\le \\frac{(r+c)^2}{2}$, получаем $P \\ge (r+c)(n-2) - \\frac{(r+c)^2}{2} + n = (r+c)(n-2 - \\frac{r+c}{2}) + n \\ge n$ (ибо $n-2 - \\frac{r+c}{2} \\ge n-2-\\frac{n}{2} \\ge 0$), что и требовалось доказать.\n\nТретье решение. Как и в предыдущем решении, разобьём строки и столбцы на чётные и нечётные и заметим, что если все чётные линии сделать нечётными и наоборот, то результат не изменится.\nПредположим противное: количество плюсов стало меньше $n$. Тогда найдётся строка (пусть её номер равен $i$), в которой все знаки стали минусами. По замечанию выше, можно считать, что эта строка чётна. Тогда $i$-й столбец нечётен, а все остальные столбцы чётны (иначе в пересечении такого столбца с нашей строкой стоял бы «+»). Пусть $j \\ne i$. Тогда нетрудно понять, что в $j$-й строке знаки на $i$-м и $j$-м местах одинаковы, а знаки на всех остальных местах (их $n-2 \\ge 2$) от них отличаются. Значит, в этой строке хотя бы два плюса и хотя бы два минуса. Таким образом, общее количество плюсов не меньше $2(n-1) > n$, что противоречит нашему предположению.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39607,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ is inscribed into a circle with diameter $AC$. Let $K$ and $M$ be respectively the projections of $A$ and $C$ onto the line $BD$. Point $P$ is chosen on $AC$ so that $PK \\parallel BC$. Prove that $\\angle KPM = 90^\\circ$. (T. Emelyanova)\n\nЧетырёхугольник $ABCD$ вписан в окружность с диаметром $AC$. Точки $K$ и $M$ — проекции вершин $A$ и $C$ соответственно на прямую $BD$. Через точку $K$ проведена прямая, параллельная $BC$ и пересекающая $AC$ в точке $P$. Докажите, что угол $KPM$ — прямой.\n(Т. Емельянова)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $E$ be the intersection point of diagonals $AC$ and $BD$. Assume for definiteness that point $K$ lies on segment $BE$. Let the line passing through $K$ and parallel to $PM$ intersect $AC$ at point $N$ (see Fig. 8). Then $\\triangle NKE \\sim \\triangle PME$ (since their sides are parallel), hence $\\frac{PE}{EM} = \\frac{NE}{EK}$. On the other hand, the right triangles $AKE$ and $CME$ are also similar (by the acute angle at vertex $E$), so $\\frac{EM}{EC} = \\frac{EK}{EA}$. Multiplying the obtained equalities, we get $\\frac{PE}{EC} = \\frac{NE}{EA}$. But by Thales' theorem $\\frac{PE}{EC} = \\frac{KE}{EB}$.\n\nThus, we have $\\frac{NE}{EA} = \\frac{PE}{EC} = \\frac{KE}{EB}$, whence $KN \\parallel AB$. Therefore, $PM \\parallel AB \\perp BC \\parallel KP$, as required.\n\n\nSecond solution. Again, let $E$ be the intersection point of diagonals $AC$ and $BD$ and consider the case when point $K$ lies on segment $BE$. Note that $\\angle PAD = \\angle CAD = \\angle CBD = \\angle PKD$, that is, quadrilateral $AKPD$ is cyclic (see Fig. 9). Therefore, $\\angle AKD = \\angle APD = 90^\\circ$. Then from the relation $\\angle CPD = \\angle CMD = 90^\\circ$ it follows that quadrilateral $CPMD$ is also cyclic, whence $\\angle KPM = 180^\\circ - \\angle CPM = \\angle EDC = \\angle BAC = \\angle BAE$. From this it follows that $PM \\parallel AB \\perp BC \\parallel KP$, as required.\n\nThe case when $K$ lies on segment $DE$ is considered similarly.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39608,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ is inscribed into a circle $\\omega$. Its diagonals intersect at point $K$. Let $M_1, M_2, M_3, M_4$ be the midpoints of arcs $AB$, $BC$, $CD$, $DA$ (not containing the other vertices of $ABCD$), respectively. Let $I_1, I_2, I_3, I_4$ be the incenters of triangles $ABK$, $BCK$, $CDK$, $DAK$, respectively. Prove that the lines $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ are concurrent. (P. Kozhevnikov)\n\nЧетырехугольник $ABCD$ вписан в окружность $\\omega$, а его диагонали пересекаются в точке $K$. Точки $M_1, M_2, M_3, M_4$ — середины дуг $AB$, $BC$, $CD$, $DA$ (не содержащих других вершин четырехугольника) соответственно. Точки $I_1, I_2, I_3, I_4$ — центры окружностей, вписанных в треугольники $ABK$, $BCK$, $CDK$, $DAK$ соответственно. Докажите, что прямые $M_1I_1$, $M_2I_2$, $M_3I_3$, $M_4I_4$ пересекаются в одной точке. (П. Кожевников)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что точка $I_1$ лежит на биссектрисах $AM_2$ и $BM_4$ углов $BAC$ и $ABD$, поэтому $I_1 = AM_2 \\cap BM_4$ (см. рис. 22). Аналогично $I_2 = BM_3 \\cap CM_1$, $I_3 = CM_4 \\cap DM_2$, $I_4 = DM_1 \\cap AM_3$.\n\n![](attached_image_1.png)\nРис. 22\n![](attached_image_2.png)\nРис. 23\n\nПоскольку $\\overrightarrow{AM_1} + \\overrightarrow{CM_3} = \\overrightarrow{BM_1} + \\overrightarrow{DM_3}$, прямая $M_1M_3$ составляет равные углы с хордами $AC$ и $BD$; следовательно, прямая $M_1M_3$ параллельна биссектрисе угла $AKB$, то есть прямой $I_1I_3$. Аналогично, $M_2M_4 \\parallel I_2I_4 \\perp I_1I_3$ (поскольку внешняя и внутренняя биссектрисы угла перпендикулярны).\n\nЕсли прямые $I_1I_3$ и $M_1M_3$, а также $I_2I_4$ и $M_2M_4$ совпадают, утверждение задачи очевидно. Пусть, скажем, точка $I_1$ не лежит на прямой $M_1M_3$. Обозначим $A' = DM_1 \\cap BM_4$, $B' = AM_2 \\cap CM_1$, $C' = BM_3 \\cap DM_2$, $D' = AM_3 \\cap CM_4$. Имеем $\\angle BM_1M_2 = \\angle CM_1M_2$ и $\\angle BM_2M_1 = \\angle AM_2M_1$, поэтому треугольники $M_1M_2B$ и $M_1M_2B'$ симметричны относительно прямой $M_1M_2$. Отсюда $M_2B' = M_2B$. Аналогично $M_2C' = M_2C$, и из $M_2B = M_2C$ получаем $M_2B' = M_2C'$. Поскольку $\\angle AM_2M_4 = \\angle DM_2M_4$, прямая $M_2M_4$ является биссектрисой (и, значит, высотой) равнобедренного треугольника $M_2B'C'$. Таким образом, $B'C' \\perp M_2M_4$, поэтому $B'C' \\parallel M_1M_3 \\parallel I_1I_3$.\n\nПусть прямая $M_2I_2$ пересекает отрезки $B'C'$, $I_1I_3$ и $M_1M_3$ соответственно в точках $X, Y, Z$ (см. рис. 23). Рассматривая гомотетии с центрами $I_2$ и $M_2$, получаем: $\\frac{M_1Z}{M_3Z} = \\frac{B'X}{C'X} = \\frac{I_1Y}{I_3Y}$.\n\nПусть $P = M_1I_1 \\cap M_3I_3$. Если прямая $PY$ пересекает $M_1M_3$ в точке $Z'$, то из гомотетии с центром $P$ получаем: $\\frac{M_1Z'}{M_3Z'} = \\frac{I_1Y}{I_3Y}$. Значит, $Z'$ совпадает с $Z$. Получаем, что точка $P$ лежит на прямой $M_2I_2$. Аналогично, $P$ лежит на прямой $M_4I_4$, то есть все четыре прямые $M_1I_1, M_2I_2, M_3I_3, M_4I_4$ пересекаются в точке $P$.\n\n**Замечание 1.** Точки $A', B', C', D'$ являются центрами окружностей, вписанных в треугольники $ABD, BAC, CBD, DAC$ соответственно.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n \\ge 3$. Find the minimal integer $k$ satisfying the following property.\nFor every $n$ points $A_i = (x_i, y_i)$ on the plane, no three of them being collinear, and for every real numbers $c_i$ ($1 \\le i \\le n$) there exists a polynomial $P(x, y)$ such that $P(x_i, y_i) = c_i$ for all $i = 1, \\dots, n$, and the degree of $P$ is not greater than $k$. (F. Petrov)\n\nДано натуральное число $n \\ge 3$. При каком наименьшем $k$ верно следующее утверждение? Для любых $n$ точек $A_i = (x_i, y_i)$ на плоскости, никакие три из которых не лежат на одной прямой, и любых вещественных чисел $c_i$ ($1 \\le i \\le n$) существует такой многочлен $P(x, y)$, степень которого не больше $k$, что $P(x_i, y_i) = c_i$ при всех $i = 1, \\dots, n$.\n(Многочленом от двух переменных называется функция вида\n$$\nP(x, y) = a_{0,0} + a_{1,0}x + a_{0,1}y + a_{2,0}x^2 + a_{1,1}xy + a_{0,2}y^2 + \\dots + a_{k,0}x^k + a_{k-1,1}x^{k-1}y + \\dots + a_{0,k}y^k.\n$$\nСтепенью ненулевого одночлена $a_{i,j}x^i y^j$ называется число $i+j$; степенью многочлена $P(x, y)$ называется наибольшая степень входящего в него одночлена. (Ф. Петров)",
    "options": [],
    "answer": "floor(n/2)",
    "solution": "Ответ. $k = [n/2]$.\n\n**Лемма.** Для любых точек $A_i = (x_i, y_i)$ ($1 \\le i \\le n$) на плоскости, никакие три из которых не лежат на одной прямой, найдётся такой многочлен $P(x, y)$ степени не больше $[n/2]$, что $P(x_n, y_n) = 1$ и $P(x_i, y_i) = 0$ при $i = 1, \\dots, n-1$.\n\n**Доказательство.** Заметим, что существуют такие $d = [n/2]$ прямых, что точка $A_n$ не лежит ни на одной из них, а каждая из точек $A_1, \\dots, A_{n-1}$ лежит хотя бы на одной (при нечётном $n$ это прямые $A_1A_2, A_3A_4, \\dots, A_{n-2}A_{n-1}$, а при чётном $n$ — прямые $A_1A_2, A_3A_4, \\dots, A_{n-3}A_{n-2}, A_{n-2}A_{n-1}$). Пусть $k_i x + \\ell_i y + m_i = 0$ — уравнение $i$-й прямой ($i = 1, \\dots, d$). Тогда многочлен\n$$\nQ(x, y) = \\frac{(k_1x + \\ell_1y + m_1) \\dots (k_dx + \\ell_dy + m_d)}{(k_1x_n + \\ell_1y_n + m_1) \\dots (k_dx_n + \\ell_dy_n + m_d)}\n$$\nявляется искомым. $\\square$\n\nПокажем, что число $k = [n/2]$ подходит. При каждом $i = 1, \\dots, n$ найдём согласно лемме многочлен $P_i(x, y)$, обращающийся в ноль во всех точках $A_1, \\dots, A_n$, кроме $A_i$, причём $P_i(x_i, y_i) = 1$. Тогда многочлен $P(x) = c_1P_1(x, y) + \\dots + c_nP_n(x, y)$ принимает требуемые значения во всех точках $A_1, \\dots, A_n$.\n\nОсталось показать, что при $k < [n/2]$ утверждение неверно. Рассмотрим точки $A_i(i, i^2)$ ($i = 1, \\dots, n$), лежащие на параболе $y = x^2$, и положим $c_1 = \\dots = c_{n-1} = 0$, $c_n = 1$. Поскольку парабола пересекается с прямой не более, чем по двум точкам, точки $A_i$ удовлетворяют условию. Предположим, что существует многочлен $P(x, y)$ степени, не превосходящей $k$, для которого $P(x_i, y_i) = c_i$. Положим $Q(x) = P(x, x^2)$; тогда степень $Q(x)$ не превосходит $2k$. По нашему предположению, $Q(1) = Q(2) = \\dots = Q(n-1) = 0$ и $Q(n) = 1$. Таким образом, ненулевой многочлен $Q(x)$ имеет $n-1$ корень, то есть его степень не меньше $n-1$; тогда и $2k \\ge n-1$. Это и значит, что $k \\ge [n/2]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39610,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triple of positive integers $(a, b, c)$ is called *quadratic* if $a, b, c$ is a nondecreasing arithmetic progression, $\\gcd(b, a) = \\gcd(b, c) = 1$, and $abc$ is a perfect square. Prove that for any quadratic triple, there exists some other quadratic triple containing at least one number from the first triple.\n\nНазовем тройку натуральных чисел $(a, b, c)$ квадратной, если они образуют неубывающую арифметическую прогрессию (именно в таком порядке), число $b$ взаимно просто с каждым из чисел $a$ и $c$, а число $abc$ является точным квадратом. Докажите, что для любой квадратной тройки найдётся другая квадратная тройка, имеющая с ней хотя бы одно общее число.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Если $b = 1$, то $a = c = 1$, и в качестве другой тройки можно выбрать $(1, 25, 49)$. Если же $b \\neq 1$, то из взаимной простоты разность прогрессии $d$ не может оказаться нулевой. Тогда без ограничения общности $d = b - a = c - b > 0$.\n\nПоскольку $b$ взаимно просто как с $a$, так и с $c$, то оно взаимно просто как с $ac$. Далее, произведение взаимно простых чисел $ac$ и $b$ является квадратом, поэтому и каждое из них — также квадрат, то есть $b = f^2$, $ac = m^2 = (b-d)(b+d) = b^2 - d^2$ для некоторых натуральных $f$ и $m$. При этом $m \\neq d$, так как в противном случае $b^2 = m^2 + d^2 = 2m^2$, что невозможно.\n\nРассмотрим теперь тройку $(b-m, b, b+m)$. Ее члены образуют арифметическую прогрессию, являются натуральными числами (так как $b^2 - m^2 = d^2 > 0$), и их произведение $(b-m)b(b+m) = f^2(b^2 - m^2) = (df)^2$ является квадратом. Кроме того, $\\gcd(b, m^2) = \\gcd(b, ac) = 1$, откуда $1 = \\gcd(b, m) = \\gcd(b, b-m) = \\gcd(b, b+m)$. Значит, эта тройка — квадратная, она имеет общий элемент $b$ с исходной и отлична от нее (ибо $b-m \\neq b-d$), что и требовалось.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha, \\beta, \\gamma$ be the angles in triangle $\\Delta$. Suppose that $\\sin \\alpha > \\cos \\beta$, $\\sin \\beta > \\cos \\gamma$, and $\\sin \\gamma > \\cos \\alpha$. Prove that $\\Delta$ is acute-angled.\n\nУглы треугольника $\\alpha, \\beta, \\gamma$ удовлетворяют неравенствам $\\sin \\alpha > \\cos \\beta, \\sin \\beta > \\cos \\gamma, \\sin \\gamma > \\cos \\alpha$. Докажите, что треугольник остроугольный.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное; пусть для определённости $\\gamma \\ge 90^\\circ$. Тогда $\\alpha + \\beta \\le 90^\\circ$, и углы $\\alpha$ и $\\beta$ острые. Поэтому $0 < \\beta \\le 90^\\circ - \\alpha < 90^\\circ$, откуда $\\cos \\beta \\ge \\cos(90^\\circ - \\alpha) = \\sin \\alpha$, что противоречит условию.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$. Prove that there exist $n$ consecutive positive integers such that their product is divisible by each prime number not exceeding $2n+1$, but is not divisible by any other prime number.\n\nДано натуральное $n > 1$. Докажите, что найдутся такие $n$ последовательных натуральных чисел, что их произведение делится на все простые числа, не превосходящие $2n+1$, и не делится ни на одно другое простое число.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим, что число $n + 1$ составное; покажем, что тогда подходят числа $n + 2, \\dots, 2n + 1$. Очевидно, их произведение делится на все простые числа из отрезка $[n + 2, 2n + 1]$, но не делится на простые числа, большие $2n + 1$ (ибо все сомножители не превосходят $2n + 1$). Для любого же простого $p \\le n$, одно из $p$ последовательных чисел делится на $p$; значит, и одно из наших $n$ чисел также делится на $p$.\n\nПусть теперь число $n + 1 > 2$ простое; тогда оно нечётно, а число $n + 2 > 2$ чётно и потому составное. В этом случае подходят числа $n+3, \\dots, 2n+2$. Действительно, по аналогичным причинам их произведение $P$ делится на все простые числа из отрезков $[1, n]$ и $[n+3, 2n+2]$, но не делится на простые числа, большие $2n+1$ (поскольку число $2n+2$ составное). Кроме того, $P$ делится на $n + 1 = \\frac{2n+2}{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39613,
    "subject": "Mathematics (Multi-modal)",
    "question": "The base of a pyramid $SABCD$ is a parallelogram $ABCD$. Prove that for each point inside $SABCD$, the sum of volumes of tetrahedra $OSAB$ and $OSCD$ is equal to the sum of volumes of tetrahedra $OSBC$ and $OSDA$.\n\nВ основании четырёхугольной пирамиды $SABCD$ лежит параллелограмм $ABCD$. Докажите, что для любой точки $O$ внутри пирамиды сумма объёмов тетраэдров $OSAB$ и $OSCD$ равна сумме объёмов тетраэдров $OSBC$ и $OSDA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the intersection point of the ray $SO$ with the plane $ABCD$ (see Fig. 10). Since the point $O$ lies inside the pyramid, the point $X$ lies inside its base. At the same time, $S_{XAB} + S_{XCD} = S_{XBC} + S_{XDA}$ (one possible proof of this fact can be seen from Fig. 11—each sum equals half the area of the parallelogram $ABCD$). Therefore,\n\n$$\nV_{XSAB} + V_{XSCD} = V_{XSBC} + V_{XSDA}, \\quad (1)\n$$\n\nsince the height of these pyramids, dropped from vertex $S$, is common.\n\nSimilarly,\n\n$$\nV_{XOAB} + V_{XOCD} = V_{XOBC} + V_{XODA}. \\quad (2)\n$$\n\nSubtracting equality (2) from equality (1), we obtain the required result.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 39614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can the 4 incenters of the 4 faces of some tetrahedron be coplanar?\n\nМогут ли 4 центра вписанных в грани тетраэдра окружностей лежать в одной плоскости?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Ответ.** Не могут.\n\nПусть $I_A, I_B, I_C, I_D$ — центры вписанных окружностей треугольников $BCD, ACD, ABD, ABC$ соответственно. Предположим, что они лежат в одной плоскости. Тогда либо они образуют выпуклый четырёхугольник, либо одна из этих точек лежит в треугольнике, образованном тремя другими.\n\nСлучай 1. Пусть без ограничения общности $I_A I_B I_C I_D$ — выпуклый четырёхугольник: тогда отрезки $I_A I_C$ и $I_B I_D$ пересекаются. Обозначим через $M, N, K, L$ середины рёбер $AB, AD, CD, BC$ соответственно (см. рис. 24).\n\n![](attached_image_1.png)\n\nРассмотрим треугольник $ABC$. Проведём через точку $B$ прямую $\\ell$, параллельную $AC$; тогда вписанная окружность этого треугольника лежит между прямыми $\\ell$ и $AC$, касаясь $AC$, но не касаясь $\\ell$. Это значит, что точки $I_D$ и $B$ лежат по разные стороны от средней линии $ML$ (см. рис. 25).\n\n![](attached_image_2.png)\n\nАналогично получаем, что точки $I_B$ и $I_D$ окажутся по одну сторону от плоскости $MNKL$, а точки $I_A$ и $I_C$ — по другую. Но тогда отрезки $I_B I_D$ и $I_A I_C$ не могут пересекаться — противоречие.\n\nСлучай 2. Осталось показать, что точка $I_A$ не может лежать в треугольнике $I_B I_C I_D$. Это следует из того, что точки $I_B, I_C, I_D$ лежат строго по одну сторону от плоскости $BCD$, а точка $I_A$ — в этой плоскости.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = bx^2 + cx + a$, $g(x) = cx^2 + ax + b$, where $a, b, c$ are integers. Given that $f(1234) = g(1234)$, determine whether the equality $f(1) = 2009$ can hold.\n\nЦелые числа $a, b, c$ таковы, что значения квадратных трёхчленов $bx^2 + cx + a$ и $cx^2 + ax + b$ при $x = 1234$ совпадают. Может ли первый трёхчлен при $x = 1$ принимать значение 2009?",
    "options": [],
    "answer": "No",
    "solution": "**Первое решение.** Подставляя $x = 1234$ в оба трёхчлена и приравнивая их, получаем $1234^2 \\cdot b + 1234 \\cdot c + a = 1234^2 \\cdot c + 1234 \\cdot a + b$, или, после переноса всех членов в левую часть, $(1234^2-1)b+(1234-1234^2)c+(1-1234)a = 0$. Разделив последнее равенство на 1233, имеем $1235b - 1234c - a = 0$, то есть $a = 1235b - 1234c$. Тогда значение первого трёхчлена в точке 1 равно $a+b+c = 1236b - 1233c = 3(412b - 411c)$, то есть делится на 3; значит, оно не может равняться 2009.\n\n**Второе решение.** Рассмотрим многочлен $P(x) = (b-c)x^2 + (c-a)x + (a-b)$ — его степень не больше 2. По условию, $P(1234) = 0$; кроме того, очевидно, $P(1) = 0$. Значит, если его степень равна 2, то по теореме Безу $P(x) = (b-c)(x-1)(x-1234)$. Если же степень меньше 2, то он может быть только нулевым (поскольку у него есть два корня); тогда $a = b = c$, и всё равно верно равенство $P(x) = (b-c)(x-1)(x-1234)$. Приравнивая свободные члены в равенстве\n$$\nP(x) = (b-c)x^2 + (c-a)x + (a-b) = (b-c)(x-1)(x-1234), \\\\\n\\text{получаем } a-b = 1234(b-c), \\text{ откуда } a = 1235b - 1234c. \\text{ Дальше} \\\\\n\\text{решение можно завершить так же, как и предыдущее.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39616,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $P(x)$ of degree $n \\ge 3$ has $n$ real roots $x_1 < x_2 < \\dots < x_n$ such that $x_2 - x_1 < x_3 - x_2 < \\dots < x_n - x_{n-1}$. Prove that the maximum of the function $y = |P(x)|$ on the segment $[x_1, x_n]$ is attained at a point belonging to the segment $[x_{n-1}, x_n]$.\n\nМногочлен $P(x)$ степени $n \\ge 3$ имеет $n$ вещественных корней $x_1 < x_2 < \\dots < x_n$, причем $x_2 - x_1 < x_3 - x_2 < \\dots < x_n - x_{n-1}$. Докажите, что максимум функции $y = |P(x)|$ на отрезке $[x_1, x_n]$ достигается в точке, принадлежащей отрезку $[x_{n-1}, x_n]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что максимум функции $|P(x)|$ не может достигаться в точке $x_i$, ибо $|P(x_i)| = 0$. Рассмотрим произвольную точку $a \\in (x_i, x_{i+1})$ при $i < n-1$; положим $t = a - x_i$, $b = x_n - t$. Заметим, что $b \\in (x_{n-1}, x_n)$, поскольку $x_n > b > x_n - (x_{i+1} - x_i) > x_n - (x_n - x_{n-1}) = x_{n-1}$. Покажем, что $|P(b)| > |P(a)|$; из этого, очевидно, следует утверждение задачи.\nИз условия следует, что $x_{k+m} - x_k < x_{\\ell+m} - x_{\\ell}$ при $1 \\le k < \\ell \\le n-m$. Поскольку нам известны $n$ корней многочлена $P(x)$, имеем $P(x) = p(x-x_1) \\dots (x-x_n)$, где $p$ — старший коэффициент многочлена $P(x)$. Заметим, что $|b-x_s| = x_n - x_s - t > x_{i+n-s} - x_i - t = |x_{i+n-s} - a|$, при $i+1 \\le s \\le n-1$. Кроме того, $|b-x_r| = b-x_r > x_{n-1} - x_r > a - x_r = |a-x_r|$ при $1 \\le r \\le i-1$. Перемножая все полученные неравенства с равенством $p|b-x_n||b-x_i| = pt(x_n - x_i - t) = p|a-x_i||a-x_n|$, получаем\n$$P(b) = p|b-x_1||b-x_2|\\dots|b-x_n| >$$\n$$> p|a - x_1| |a - x_2| \\dots |a - x_n| = P(a),$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39617,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\ldots, 10000$ are placed into the cells of a square grid $100 \\times 100$ (each number appears exactly once) so that every two numbers which differ by $1$ are placed into two cells sharing a common side. Consider the $5000$ pairs of cells containing the numbers that differ by $5000$. For each pair, calculate the distance between the centers of its cells. Let $S$ be the minimum of these distances. Find the maximal possible value of $S$.\n\nВ клетки квадрата $100 \\times 100$ расставили числа $1, 2, \\ldots, 10000$, каждое — по одному разу; при этом числа, различающиеся на $1$, записаны в соседних по стороне клетках. После этого посчитали расстояния между центрами каждого двух клеток, числа в которых различаются ровно на $5000$. Пусть $S$ — минимальное из этих расстояний. Какое наибольшее значение может принимать $S$?",
    "options": [],
    "answer": "50√2",
    "solution": "Ответ. $50\\sqrt{2}$.\n\nПронумеруем в квадрате строки (снизу вверх) и столбцы (слева направо) числами от $1$ до $100$; будем обозначать клетку парой номеров ее строки и столбца. Назовем расстоянием между клетками расстояние между их центрами. Клетки назовем парными, если числа в них различаются на $5000$.\n\nЗаметим, что расстояние от клетки $(50, 50)$ до любой другой (в частности, до парной ей) не превосходит $\\sqrt{50^2 + 50^2} = 50\\sqrt{2}$. Значит, и минимальное расстояние между парными клетками также не превосходит $50\\sqrt{2}$. Осталось привести пример, когда этот минимум достигается.\n\nРазобьем наш квадрат на четыре квадрата $50 \\times 50$. Расставим числа $1$–$2500$ согласно правилам в левом нижнем квадрате так, чтобы число $1$ стояло в клетке $(1, 1)$, а число $2500$ — в клетке $(50, 1)$ (это возможно; например, первые $50$ чисел в первом столбце, вторые — во втором и т.д.). Далее, если число $a \\in [1, 2500]$ стоит в клетке $(i, k)$, то поставим числа $a+2500$, $a+5000$ и $a+7500$ соответственно в клетки $(k+50, i)$, $(k+50, i+50)$ и $(51-i, 101-k)$. Нетрудно видеть, что при этом числа по-прежнему расставлены согласно правилам (для соседних чисел в одном квадрате это очевидно; для чисел $2500$–$2501$, $5000$–$5001$ и $7500$–$7501$ проверяется непосредственно).\n\nОсталось проверить, что расстояния между парными клетками не меньше $50\\sqrt{2}$. Рассмотрим отрезок между любыми парными клетками. Сумма его горизонтальной и вертикальной проекций равна либо $(50+k-i)+(50+i-k) = 100$, либо $(k+50-51+i)+(101-k-i) = 100$, то есть она всегда равна $100$. Значит, квадрат длины этого отрезка равен $x^2 + (100-x)^2 = 2(x-50)^2 + 5000 \\ge 5000 = (50\\sqrt{2})^2$, что и требовалось.\n\n**Замечание.** Предъявленный пример не единственен.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39618,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a boarding school, $512$ students learn $9$ disciplines. These students live in $256$ double rooms; two students are called *neighbors* if they share a room. It is known that for every two students the sets of disciplines in which they are interested are distinct (in particular, exactly one student is interested in nothing). Prove that all the students can be arranged in a circle so that (i) each two neighbors are adjacent in a circle, and (ii) for each two adjacent students which are not neighbors, one of them is interested in all the disciplines in which the other is interested, and the first one is interested in exactly one additional discipline.\n\nВ школе-интернате преподаётся $9$ предметов и учатся $512$ детей, расселённых в $256$ двухместных номерах (детей, живущих в одном номере, назовём *соседями*). Известно, что у любых двух детей наборы предметов, которые им интересны, различны (в частности, ровно одному ребёнку не интересно ничего). Докажите, что всех детей можно построить по кругу так, чтобы любые два соседа стояли рядом, а для любых двух несоседей, стоящих рядом, одному из них интересны все предметы, интересные другому, и ещё ровно один предмет.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Мы докажем утверждение задачи в более общем виде, для $n \\ge 2$ предметов и $2^n$ детей, произвольно разбитых на $2^{n-1}$ пар соседей. Заметим, что существует ровно $2^n$ наборов из $n$ предметов; значит, каждый набор предметов интересен ровно одному ученику.\n\nИндукция по $n$. При $n = 2$ легко проверить утверждение непосредственно. Пусть $n > 2$; рассмотрим любых двух соседей, выберем какой-нибудь предмет, интерес к которому у них различен (скажем, физику), и разобьём всех детей на две группы по $2^{n-1}$ детей: в группу А попадут те, кому физика интересна, а в группу В — все остальные.\n\nОтселим группу А в другой интернат, с $2^{n-2}$ комнатами. При этом те пары, что были соседями раньше, оставим соседями. Остальных же (согласно выбору предмета, они есть; при этом их число, очевидно, чётно) разобьем на пары соседей произвольно. Назовем такие пары «новыми».\n\nПо предположению индукции, теперь группу А можно расставить по кругу К с выполнением условий. Пусть $(x_1, x_2), \\dots, (x_{2k-1}, x_{2k})$ — все новые пары в порядке обхода этого круга по часовой стрелке ($(x_{2i-1}, x_{2i})$ находится перед $x_{2i}$; мы будем считать, что $x_{2k+1} = x_1$). Обозначим через $x'_i$ исходного соседа человека $x_i$ (по построению, $x'_i$ находится в группе В), и объявим пары $(x'_2, x'_3), \\dots, (x'_{2k-2}, x'_{2k-1})$, $(x'_{2k}, x'_1)$ «новыми» парами в группе В. Ясно, что новые пары, вместе со старыми, дают нам разбиение группы В на пары соседей.\n\nТеперь, применив предположение индукции к группе В с этим разбиением, расставим её по кругу с выполнением условий. Вставим теперь между любыми детьми новой пары $(x'_{2i}, x'_{2i+1})$ отрезок круга К от $x_{2i}$ до $x_{2i+1}$. Нетрудно видеть, что теперь все дети стоят в кругу, и расстановка удовлетворяет всем условиям.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given three quadratic polynomials $f_1(x) = x^2 + 2a_1x + b_1$, $f_2(x) = x^2 + 2a_2x + b_2$, $f_3(x) = x^2 + 2a_3x + b_3$ such that $a_1a_2a_3 = b_1b_2b_3 > 1$. Prove that one of these polynomials has two distinct real roots.\n\nДаны квадратные трёхчлены $f_1(x) = x^2 + 2a_1x + b_1$, $f_2(x) = x^2 + 2a_2x + b_2$, $f_3(x) = x^2 + 2a_3x + b_3$. Известно, что $a_1a_2a_3 = b_1b_2b_3 > 1$. Докажите, что хотя бы один из этих трёхчленов имеет два корня. (Н. Агаханов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное; тогда дискриминанты всех трёхчленов неположительны, то есть $a_k^2 \\le b_k$ $(k = 1, 2, 3)$. Левые (а значит, и правые) части этих неравенств неотрицательны, поэтому их можно перемножить, получая $(a_1a_2a_3)^2 \\le b_1b_2b_3$, то есть $N^2 \\le N$, где $N = a_1a_2a_3$. Но это противоречит неравенству $N > 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Seven skiers, numbered from $1$ to $7$, participated in a race. They started one by one, and each skier passed the distance with a constant speed (which could be different for different skiers). It happened that each skier participated in an overtaking exactly two times (in each overtaking, exactly two skiers participated: the one who overtakes and the one who is overtaken.)\nAfter the race was finished, the list of ranks is published (that list contains the numbers of skiers in the order they finished the race). Prove that all the conditions being satisfied, not more than two possible lists might appear.\n\nСемь лыжников с номерами $1, 2, \\ldots, 7$ ушли со старта по очереди и прошли дистанцию — каждый со своей постоянной скоростью. Оказалось, что каждый лыжник ровно дважды участвовал в обгонах. (В каждом обгоне участвуют ровно два лыжника — тот, кто обгоняет, и тот, кого обгоняют.) По окончании забега должен быть составлен протокол, состоящий из номеров лыжников в порядке финиширования. Докажите, что в забеге с описанными свойствами может получиться не более двух различных протоколов.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что чётность места каждого лыжника менялась при любом обгоне; значит, его место на финише — той же чётности, что и на старте.\nТак как скорости постоянны, каждые два лыжника встречались не более одного раза. Будем обозначать лыжников их стартовыми номерами.\nПобедителя никто не мог обогнать, значит, он сам обогнал двоих. Поэтому он — $3$, и обогнал лыжников $1$ и $2$. Аналогично, финишировавший последним не мог никого обогнать, поэтому его обогнали двое, он — $5$, и его обогнали $6$ и $7$. Далее, лыжник $1$ не мог никого обогнать, то есть он финишировал третьим (и его, кроме $3$, обогнал лыжник, финишировавший вторым), а лыжника $7$ никто не мог обогнать, и он финишировал пятым (обогнав $5$ и лыжника, финишировавшего шестым).\nИтак, осталось выяснить, какими финишировали лыжники с чётными номерами. Вторым финишировать мог либо $2$, либо $4$. Если $2$ пришел вторым, то он обогнал $1$, и с группой лидеров ($1$, $2$, $3$) больше обгонов не происходило. Значит, лыжника $4$ могли только обгонять, он на финише шестой, а лыжник $6$ — четвёртый.\nЕсли же вторым пришел $4$, то $2$ мог придти к финишу только четвёртым (значит, $4$ обогнал $2$ и $1$), а шестым пришел $6$ (обогнав $5$ и уступив $7$). Итого, возможны только два протокола: $3, 2, 1, 6, 7, 4, 5$ и $3, 4, 1, 2, 7, 6, 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a circle $\\omega$ with center $I$. Two lines touching $\\omega$ at points $A$ and $B$ intersect at point $O$. A point $C$ is chosen on the smaller arc $AB$ so that $AC \\neq CB$. Lines $AC$ and $OB$ intersect at point $D$, while lines $BC$ and $OA$ intersect at point $E$. Prove that the circumcenters of triangles $ACE$, $BCD$, and $OCI$ are collinear.\n\nПрямые, касающиеся окружности $\\omega$ в точках $A$ и $B$, пересекаются в точке $O$. Точка $I$ — центр $\\omega$. На меньшей дуге $AB$ окружности $\\omega$ выбрана точка $C$, отличная от середины дуги. Прямые $AC$ и $OB$ пересекаются в точке $D$, а прямые $BC$ и $OA$ — в точке $E$. Докажите, что центры описанных окружностей треугольников $ACE$, $BCD$ и $OCI$ лежат на одной прямой.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $M$ — вторая точка пересечения описанных окружностей треугольников $ACE$ и $BCD$ (она есть, так как в случае касания прямые $AE$ и $BD$ были бы параллельны). Нам достаточно показать, что описанная окружность треугольника $OCI$ также проходит через точку $M$, так как в этом случае центры всех трех окружностей из условия будут лежать на серединном перпендикуляре к отрезку $CM$.\n\n![](attached_image_1.png)\n\nОбозначим $\\angle CAB = \\alpha$ и $\\angle CBA = \\beta$; без ограничения общности можно считать, что $\\beta > \\alpha$. По свойству угла между хордой и касательной, $\\angle OBE = \\alpha$, аналогично $\\angle DAE = \\beta$ (см. рис. 13). В четырёхугольнике $OBIA$ углы $A$ и $B$ прямые, поэтому он — вписанный; значит, $\\angle OIA = \\angle OBA = \\alpha + \\beta$. Следовательно, $\\angle CIO = \\angle CIA - \\angle OIA = 2\\angle CBA - (\\alpha + \\beta) = \\beta - \\alpha$. Для того, чтобы доказать, что точка $M$ лежит на описанной окружности треугольника $OCI$, нам достаточно показать равенство $\\angle CMO = \\beta - \\alpha$.\n\nПоскольку четырёхугольники $AECM$ и $DBMC$ вписаны, получаем $\\angle BME = \\angle BMC + \\angle CME = (180^\\circ - \\angle CDB) + \\angle CAE = \\angle ODA + \\angle DAO = 180^\\circ - \\angle EOB$, то есть четырехугольник $EOBM$ также вписан, и $\\angle OME = \\angle OBE = \\alpha$. Следовательно, $\\angle CMO = \\angle CME - \\angle OME = \\angle CAE - \\alpha = \\beta - \\alpha$, что и требовалось доказать.\n\n\n**Замечание.** Точка $M$ является точкой Микеля четырехсторонника, образованного прямыми $AC$, $BC$, $AO$ и $BO$, то есть точка $M$ будет лежать на описанной окружности треугольника, образованного любыми тремя из этих четырех прямых. Таким образом точка $M$ будет лежать не только на описанных окружностях треугольников $AEC$, $BCD$, $EOB$, как мы показали, но и на описанной окружности треугольника $AOD$.\n\n\nSolution 2:\nПоскольку $\\angle OAI = \\angle OBI = 90^\\circ$, четырёхугольник $OAIB$ вписан в некоторую окружность $\\Omega$. Обозначим через $S_A$ и $S_B$ описанные окружности треугольников $ACE$ и $BCD$, а через $M$ — вторую точку их пересечения. Поскольку $BO$ — касательная к $\\omega$, имеем $\\angle ABO = 180^\\circ - \\angle ACB = \\angle ACE$. Поскольку эти углы в окружностях $\\Omega$ и $S_A$ стягиваются хордами $AO$ и $AE$, лежащими на одном луче $AO$, то по теореме об угле между касательной и хордой касательные к $\\Omega$ и $S_A$ в точке $A$ совпадают. Аналогично, касательные к $S_B$ и $\\Omega$ в точке $B$ также совпадают.\n\nПусть касательные к $\\Omega$ в точках $A$ и $B$ пересекаются в точке $K$ (см. рис. 14). Из симметрии, $K$ лежит на прямой $OI$. Тогда точка $K$ является радикальным центром окружностей $S_A$, $S_B$ и $\\Omega$, поэтому она лежит на прямой $CM$. Но тогда $KM \\cdot KC = KA^2 = KO \\cdot KI$, а значит, точка $M$ лежит на описанной окружности треугольника $OIC$. Это значит, что центр этой окружности, а также центры окружностей $S_A$ и $S_B$ лежат на серединном перпендикуляре к их общей хорде $CM$, что и требовалось.\n\n![](attached_image_2.png)\n\nЕсли же касательные к $\\Omega$ в точках $A$ и $B$ параллельны (см. рис. 15), то прямая $AB$ содержит диаметры окружностей $S_A$ и $S_B$, а также является серединным перпендикуляром к $OI$, то есть содержит центр описанной окружности треугольника $OIC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $k$ such that all positive integers from $1$ to $k$ can be partitioned into two groups, and all the numbers in each group can be written one after another (in some order without spaces between them) so that the two obtained numbers are identical?\n\nМожно ли при каком-то натуральном $k$ разбить все натуральные числа от $1$ до $k$ на две группы и выписать числа в каждой группе подряд в некотором порядке так, чтобы получились два одинаковых числа? (Н. Агаханов)",
    "options": [],
    "answer": "No such k exists",
    "solution": "Предположим противное. Ясно, что $k \\ge 10$, так как в наборе цифр от $1$ до $9$ нет повторяющихся. Рассмотрим наибольшую степень десятки $10^n$, не превосходящую $k$. Последовательность цифр числа $10^n$ целиком войдет в одно из составленных чисел. Но тогда такая же последовательность из единицы и $n$ последующих нулей должна повториться во втором числе. Эта последовательность цифр не могла появиться из объединения двух или более чисел (так как натуральные числа не начинаются с нулей), значит, она содержалась в одном числе, отличном от $10^n$. Но наименьшее число, отличное от $10^n$ и содержащее такой набор цифр, — это $10^{n+1}$. Мы получили противоречие с тем, что $10^n$ — максимальная степень десятки, не превосходящая $k$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $\\angle BAC = 60^\\circ$. Let $BB_1$ and $CC_1$ be the angle bisectors in this triangle. Prove that the point symmetrical to $A$ with respect to $B_1C_1$ lies on side $BC$.\n\nВ треугольнике $ABC$ угол $A$ равен $60^\\circ$. Пусть $BB_1$ и $CC_1$ — биссектрисы этого треугольника. Докажите, что точка, симметричная вершине $A$ относительно прямой $B_1C_1$, лежит на стороне $BC$. (Д. Прокопенко)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $I$ — точка пересечения биссектрис треугольника $ABC$. Тогда $\\angle B_1IC_1 = \\angle BIC = 180^\\circ - \\angle IBC - \\angle ICB = 180^\\circ - (\\angle ABC + \\angle ACB)/2 = 180^\\circ - 60^\\circ = 120^\\circ = 180^\\circ - \\angle B_1AC_1$. Значит, четырёхугольник $AB_1IC_1$ вписан в окружность. Отсюда $\\angle AC_1B_1 = \\angle AIB_1 = \\angle ABI + \\angle BAI = (\\angle A + \\angle B)/2$ (поскольку $\\angle AIB_1$ — внешний в $\\triangle ABI$), и аналогично $\\angle AB_1C_1 = (\\angle A + \\angle C)/2$.\n\nПусть описанная окружность треугольника $BC_1I$ пересекает прямую $BC$ в точке $K$ (легко понять, что эта точка не может попасть на продолжение стороны $BC$). Тогда $\\angle IKC = 180^\\circ - \\angle BKI = \\angle BC_1I = 180^\\circ - \\angle AC_1I = \\angle AB_1I = 180^\\circ - \\angle IB_1C$, то есть четырёхугольник $IB_1CK$ также вписан.\n\nНаконец, поскольку четырёхугольники $AB_1IC_1$, $BC_1IK$ и $CKIB_1$ вписаны, мы имеем $\\angle KC_1B_1 = \\angle KC_1I + \\angle IC_1B_1 = \\angle KBI + \\angle IAB_1 = (\\angle B + \\angle A)/2 = \\angle AC_1B_1$, и аналогично $\\angle KB_1C_1 = \\angle KB_1I + \\angle IB_1C_1 = (\\angle C + \\angle A)/2 = \\angle AB_1C_1$. Значит, треугольники $AB_1C_1$ и $KB_1C_1$ равны по стороне $B_1C_1$ и двум прилежащим к ней углам. Тогда они симметричны относительно $B_1C_1$, а тогда и точки $A$ и $K$ также симметричны. Поскольку точка $K$ лежит на $BC$, решение закончено.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39624,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a canteen, there are $100$ apples of total weight $10$ kg, each apple weighs not less than $25$ g. The barman has to cut them into some pieces and distribute these pieces among $100$ students so that each student receives exactly $100$ g. Prove that the barman can attain this aim in such a way that the weight of each piece is at least $25$ g.\n\nВ буфете лежат $100$ яблок суммарного веса $10$ кг, каждое весит не меньше $25$ г. Буфетчице нужно разрезать их на части и раздать $100$ детям, каждому по $100$ г. Докажите, что она может это сделать так, чтобы любой кусок яблока весил не меньше $25$ г.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Все веса в решении будут измеряться в граммах. Назовём кусок яблока (или само яблоко) большим, если его вес не меньше $25$.\n\nДокажем индукцией по $n$, что $n$ больших яблок суммарного веса $100n$ можно разрезать на большие куски и раздать $n$ детям поровну.\n\nБаза при $n = 1$ очевидна.\n\nПусть $n > 1$. Рассмотрим два самых тяжёлых яблока; пусть их веса $a \\ge b$. Заметим, что $a + b \\ge 200$ (иначе средний вес одного яблока будет меньше, чем $200/2 = 100$).\n\nВыкинем эти два яблока из набора и добавим в него яблоко веса $c = a + b - 100 \\ge 100$. По предположению индукции, полученный набор можно разрезать на большие куски и раздать $n-1$ детям поровну.\n\nЕсли при этом какой-то кусок нового яблока оказался больше $50$, разрежем его на два больших куска. Через несколько таких разрезаний мы придём к ситуации, когда новое яблоко разделено на куски весов $c_1, c_2, \\dots, c_k$, не превосходящих $50$. Обозначим $s_d = c_1 + \\dots + c_d$ при $d = 1, 2, \\dots, k$ и положим $s_0 = 0$.\n\nПокажем теперь, как разрезать исходный набор. Все яблоки, кроме $a$ и $b$, разрежем так же, как и в новом наборе. Заметим, что $a \\ge 200/2 = 100$. Обозначим через $t$ минимальный индекс такой, что $a - s_t \\le 75$ и отрежем от $a$ куски $c_1, \\dots, c_t$, а от $b$ — куски $c_{t+1}, \\dots, c_k$. Заметим, что $a - s_{t-1} > 75$, поэтому от $a$ остался кусок $a' = a - s_t = (a - s_{t-1}) - c_t$ такой, что $75 \\ge a' > 75 - c_t \\ge 25$. От $b$ же остался кусок $b'$ такой, что $a' + b' = a + b - c = 100$, поэтому $25 \\le b' \\le 75$.\n\nИтак, можно $a'$ и $b'$ отдать одному ребёнку, а остальные куски распределить между остальными детьми так же, как это делалось в новом наборе. Утверждение доказано.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ignoramus arranged 11 positive integers in a circle. For each two adjacent numbers, he calculated their (nonnegative) difference. Among the 11 results obtained, there are 4 ones, 4 twos, and 3 threes. Prove that Ignoramus made a mistake in his calculations.\n\nНезнайка выписал по кругу 11 натуральных чисел. Для каждого двух соседних чисел он посчитал их разность. В результате среди найденных разностей оказалось четыре единицы, четыре двойки и три тройки. Докажите, что Незнайка где-то допустил ошибку. (Р. Женодаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Запишем каждую из наших разностей со знаком плюс, если в соответствующей паре чисел большее стоит перед меньшим по часовой стрелке, и со знаком минус в противном случае. У нас получились 11 разностей между числом и следующим за ним по часовой стрелке; значит, сумма всех этих чисел равна нулю, то есть чётному числу. Это невозможно, поскольку среди них ровно семь нечётных чисел -- четыре числа вида ±1 и три числа вида ±3.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$, $C$ lie on a circle. Line $PB$ touches this circle at point $B$. Drop the perpendiculars $PA_1$ and $PC_1$ from $P$ onto the lines $AB$ and $BC$, respectively (points $A_1$ and $C_1$ belong to segments $AB$ and $BC$, respectively). Prove that $A_1C_1 \\perp AC$.\n\nПусть точки $A$, $B$, $C$ лежат на окружности, а прямая $b$ касается этой окружности в точке $B$. Из точки $P$, лежащей на прямой $b$, опущены перпендикуляры $PA_1$ и $PC_1$ на прямые $AB$ и $BC$ соответственно (точки $A_1$ и $C_1$ лежат на отрезках $AB$ и $BC$). Докажите, что $A_1C_1 \\perp AC$. (Л. Емельянов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $\\angle PC_1B = \\angle PA_1B = 90^\\circ$, четырёхугольник $PA_1C_1B$ вписан (см. рис. 4).\n\nЗначит, $\\angle CC_1A_1 = 180^\\circ - \\angle A_1C_1B = \\angle A_1PB = 90^\\circ - \\angle A_1BP$.\n\nС другой стороны, $\\angle A_1BP = \\angle ACB = \\frac{1}{2} AB$.\n\nПоэтому $\\angle CC_1A_1 = \\angle A_1PB = 90^\\circ - \\angle ACC_1$, то есть прямые $A_1C_1$ и $AC$ пересекаются под прямым углом.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of 1000 dwarves wears a hat, its outer surface being blue, and its inner surface being red (or vice versa). A dwarf tells only the truth if he wears a blue hat; on the other hand, he only lies if he wears a red hat (a dwarf can turn his hat inside out; this situation will be called *a turning*). One day, each dwarf told to each other:\n\n\"Your hat is red!\". Find the least possible number of turnings that could happen this day.\n\nУ каждого из 1000 гномов есть колпак, синий снаружи и красный внутри (или наоборот). Если на гноме надет красный колпак, то он может только лгать, а если синий — только говорить правду. На протяжении одного дня каждый гном сказал каждому «На тебе красный колпак!» (при этом некоторые гномы в течение дня выворачивали свой колпак наизнанку). Найдите наименьшее возможное количество выворачиваний.",
    "options": [],
    "answer": "998",
    "solution": "Ответ. 998 выворачиваний.\n\nНазовём гнома красным или синим, если на нём надет колпак соответствующего цвета. Заметим, что один гном может сказать требуемую фразу другому тогда и только тогда, когда эти гномы разноцветны: синий гном при этом скажет правду, а красный — солжёт. Теперь, если какие-то три гнома не выворачивали колпаков, то два из них — одного цвета, и они не смогут сказать друг другу требуемого, что неверно. Значит, таких гномов не больше двух, и выворачиваний было не меньше $1000 - 2 = 998$.\n\nБудем говорить, что два гнома пообщались, если каждый из них сказал другому заветную фразу. Опишем, как могло служиться всего 998 выворачиваний, если, например, вначале гном Вася был синим, а остальные — красными. В начале дня каждый гном пообщался с Васей. Затем красные гномы по очереди выворачивали свои колпаки. При этом после каждого выворачивания все красные гномы пообщались с изменившим цвет. Когда останется только один красный гном, то любая пара гномов уже пообщается друг с другом (в тот момент, когда первый из них сменил цвет), при этом произошло 998 изменений цвета.\n\n**Замечание.** Построить пример с 998 изменениями цвета можно, начиная с любой ситуации, в которой не все гномы одноцветны.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39628,
    "subject": "Mathematics (Multi-modal)",
    "question": "A company consists of $7$ persons, and each $6$ of them can sit at a round table so that each two neighbors would be familiar to each other. Prove that all $7$ persons can sit at a round table with the same property to be satisfied. (S. Volchenkov)\n\nВ компании из семи человек любые шесть могут сесть за круглый стол так, что каждые два соседа окажутся знакомыми. Докажите, что и всю компанию можно усадить за круглый стол так, что каждые два соседа окажутся знакомыми. (С. Волчёнков)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что у каждого в компании не менее трёх знакомых.\n\nДействительно, если бы некто $X$ был знаком менее, чем с тремя, то, исключив из компании одного из его знакомых, мы получили бы шестёрку людей, в которой у $X$ не более одного знакомого, т. е. посадить их за круглый стол невозможно. Более того, если бы у каждого было ровно по три знакомых, то число пар знакомых людей было бы равно $7 \\cdot \\frac{3}{2} = \\frac{21}{2}$, что невозможно.\n\nЗначит, у какого-то человека $X$ хотя бы $4$ знакомых. Рассадим всех, кроме $X$, за круглый стол. Тогда из четырёх его знакомых хотя бы двое сидят рядом. Если мы посадим $X$ между ними, то получим требуемую рассадку.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39629,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is called *unfortunate* if it cannot be represented in a form $n = \\frac{x^2 - 1}{y^2 - 1}$, where $x > 1$, $y > 1$ are integers. Determine whether the set of all unfortunate positive integers is finite or infinite.\n\nНазовём натуральное число $n$ неудачным, если его нельзя представить в виде $n = \\frac{x^2 - 1}{y^2 - 1}$ при натуральных $x, y > 1$. Конечно или бесконечно количество неудачных чисел? (В. Сендеров)",
    "options": [],
    "answer": "infinite",
    "solution": "Докажем, что неудачным является любое число вида $n = p^2$, где $p$ — нечётное простое число. Предположим противное, т. е.\n$$\n(y^2 - 1)p^2 = x^2 - 1 \\quad (1)\n$$\nпри некоторых натуральных $x, y \\neq 1$. Тогда либо $x + 1$, либо $x - 1$ делится на $p$.\nПусть $x + 1 \\neq p$. Тогда $x - 1 = (x + 1) - 2$ не делится на $p$, а тогда из (1) получаем, что $x + 1 \\neq p^2$, то есть $x = kp^2 - 1$ при некотором натуральном $k$. Подставляя в (1), получаем $y^2 = \\frac{x+1}{p^2}(x-1) + 1 = k(kp^2 - 2) + 1 = k^2p^2 - 2k + 1$. Но $k^2p^2 > k^2p^2 - 2k + 1 > k^2p^2 - 2kp + 1$, то есть $(kp)^2 > y^2 > (kp-1)^2$, что невозможно.\nЕсли же $x - 1 \\neq p$, то аналогично получаем $x = kp^2 + 1$, $y^2 = k^2p^2 + 2k + 1$, $(kp)^2 < y^2 < (kp+1)^2$, что опять же невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39630,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, denote by $S_n$ the sum of the $n$ least primes: $S_1 = 2$, $S_2 = 2+3=5$, $S_3 = 2+3+5=10$, and so on. Determine whether two consecutive terms of the sequence $S_1, S_2, S_3, \\ldots$ can be perfect squares. (V. Sharich)\n\nДля каждого натурального $n$ обозначим через $S_n$ сумму первых $n$ простых чисел: $S_1 = 2$, $S_2 = 2+3 = 5$, $S_3 = 2+3+5 = 10$, .... Могут ли два подряд идущих члена последовательности ($S_n$) оказаться квадратами натуральных чисел? (В. Шарич)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Ответ.** Не могут.\n\nОбозначим $n$-е простое число через $p_n$. Предположим, что нашлось $m > 1$, для которого $S_{m-1} = k^2$, $S_m = l^2$, где $k$ и $l$ — натуральные числа. Числа $S_2 = 5$, $S_3 = 10$ квадратами не являются, так что $m > 4$. Заметим, что $p_m = S_m - S_{m-1} = (l-k)(l+k)$; ввиду простоты $p_m$ получаем $1 = l-k$, $p_m = l+k = 2l-1 = 2\\sqrt{S_m} - 1$. Таким образом, $S_m = \\left(\\frac{p_m+1}{2}\\right)^2$.\n\nЗаметим, что $p_m$ нечётно (так как $m \\ge 2$), и $1+3+5+\\ldots + p_m = (1^2 - 0^2) + (2^2 - 1^2) + \\ldots + \\left(\\left(\\frac{p_m+1}{2}\\right)^2 - \\left(\\frac{p_m-1}{2}\\right)^2\\right) = \\left(\\frac{p_m+1}{2}\\right)^2$. С другой стороны, в сумму $S_m = 2+p_2+\\ldots+p_m$, кроме двойки, входят лишь нечётные числа, и при $m > 4$ не входят нечётное составное число $9$ и число $1$, поэтому $S_m \\le (1+3+5+\\ldots+p_m)+2-1-9 < \\left(\\frac{p_m+1}{2}\\right)^2$. Противоречие.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute-angled triangle $ABC$ such that its median $AM$ is larger than the side $AB$. Prove that one can cut triangle $ABC$ into three parts and rearrange them so that they will form a rhombus.\n\nВ остроугольном треугольнике $ABC$ медиана $AM$ длиннее стороны $AB$. Докажите, что треугольник $ABC$ можно разрезать на три части, из которых складывается ромб.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $N$ — середина стороны $AC$, а $K$ — точка на прямой $MN$ такая, что $MK = MN$. Тогда треугольники $MNC$ и $MKB$ симметричны относительно $M$ и потому равны. Проведём разрез по средней линии $MN$; переложив треугольник $MNC$ так, чтобы он совпал с $\\triangle MKB$, получаем параллелограмм $ANKB$ (см. рис. 16). Если $AN = AB$, то ромб уже получен.\n\nПусть $AN < AB$ (см. рис. 17). Проведём окружность с центром в точке $A$ и радиусом $AB$. Тогда точка $N$ лежит внутри окружности, а $M$ — вне, поэтому окружность пересечёт отрезок $NM$ в точке $P$. Отрезав от параллелограмма $ANKB$ треугольник $APN$ и сдвинув его так, чтобы сторона $AN$ совпала с $BK$, мы получим ромб.\n\nПусть, наконец, $AN > AB$. Поскольку треугольник $ABC$ остроугольный, основание $H$ высоты $CH$ лежит на отрезке $AB$ (см. рис. 16). Далее, основание $T$ перпендикуляра, опущенного из $N$ на $AB$, есть середина отрезка $AH$; значит, $BT > AT$, а тогда $BN > AN$. Значит, окружность с центром в точке $B$ и радиусом $AN$ пересечёт сторону $AN$ в точке $S$, ибо точка $N$ лежит вне, а точка $A$ — внутри этой окружности (см. рис. 18). Отрезав от параллелограмма треугольник $ABS$ и сдвинув его до совмещения стороны $AB$ со стороной $NK$, мы получим ромб.\n\n![](attached_image_1.png)\nРис. 16\n![](attached_image_2.png)\nРис. 17\n![](attached_image_3.png)\nРис. 18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39632,
    "subject": "Mathematics (Multi-modal)",
    "question": "A table consists of $n$ rows and 10 columns. Each cell of this table contains a digit (i.e. an integer from 0 to 9). It appears that for every row $A$ and every pair of columns $B$ and $C$ there exists a row that differs from $A$ exactly in columns $B$ and $C$. Prove that $n \\ge 512$. (R. Karasev)\n\nВ каждой клетке таблицы, состоящей из 10 столбцов и $n$ строк, записана цифра. Известно, что для любой строки $A$ и любых двух столбцов найдётся строка, отличающаяся от $A$ ровно в этих двух столбцах. Докажите, что $n \\ge 512$. (Р. Карасёв)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $R_0$ — первая строка таблицы. Рассмотрим любой набор из чётного количества столбцов и пронумеруем их слева направо: $C_1, \\dots, C_{2m}$. Тогда в таблице есть строка $R_1$, отличающаяся от $R_0$ ровно в столбцах $C_1$ и $C_2$; далее, есть строка $R_2$, отличающаяся от $R_1$ ровно в столбцах $C_3$ и $C_4$; ...; наконец, есть строка $R_m$, отличающаяся от $R_{m-1}$ ровно в столбцах $C_{2m-1}$ и $C_{2m}$ (если $m=0$, то $R_m = R_0$).\n\nИтак, строка $R_m$ отличается от $R_0$ ровно в столбцах $C_1, C_2, \\dots, C_{2m}$. Значит, строки $R_m$, построенные по различным наборам столбцов, различны. Поскольку количество наборов из чётного числа столбцов равно $2^{10}/2 = 512$, то и количество строк в таблице не меньше 512.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two runners started simultaneously from one point on the street. Each of them ran along the street to the stadium, and then three laps around the stadium. Each runner ran at a constant speed along the whole distance, and the first runner had overtaken the second runner twice. Prove that the speed of the first runner is at least twice the speed of the second one. (The starting moment is considered not to be an overtaking.)\n\nДва бегуна стартовали одновременно из одной точки. Сначала они бежали по улице до стадиона, а потом до virtually — три круга по стадиону. Всю дистанцию оба бежали с постоянными скоростями, и в ходе забега первый бегун дважды обогнал второго. Докажите, что первый бежал по крайней мере вдвое быстрее, чем второй.\n(И. Рубанов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первый мог обогнать второго только на кольцевой дорожке стадиона. Так как он вбежал на стадион первым, на своём первом круге он обогнать второго не мог. Стало быть, обгоны случились, когда первый бежал по стадиону свои второй и третий круги. Пока первый бежал эти два круга, он обогнал второго по крайней мере на круг. Следовательно, второй за это время пробежал не больше одного круга, откуда и вытекает требуемое утверждение.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider nine quadratic polynomials $x^2 + a_1x + b_1$, $x^2 + a_2x + b_2$, ..., $x^2 + a_9x + b_9$ such that the sequences $a_1, a_2, ..., a_9$ and $b_1, b_2, ..., b_9$ are both arithmetical progressions, and the sum of all nine polynomials is a polynomial having at least one real root. Denote by $k$ the number of original polynomials having no real roots. Find the maximal possible value of $k$. (I. Bogdanov)\n\nНа доске написаны девять приведённых квадратных трёхчленов: $x^2 + a_1x + b_1, x^2 + a_2x + b_2, \\dots, x^2 + a_9x + b_9$. Известно, что последовательности $a_1, a_2, \\dots, a_9$ и $b_1, b_2, \\dots, b_9$ — арифметические прогрессии. Оказалось, что сумма всех девяти трёхчленов имеет хотя бы один корень. Какое наибольшее количество исходных трёхчленов может не иметь корней? (И. Богданов)",
    "options": [],
    "answer": "4",
    "solution": "Обозначим $P_i(x) = x^2 + a_i x + b_i$, $P(x) = P_1(x) + \\dots + P_9(x)$. Заметим, что $P_i(x) + P_{10-i}(x) = 2x^2 + (a_i + a_{10-i})x + (b_i + b_{10-i}) = 2P_5(x)$. Значит, $P(x) = 9P_5(x)$, и условие равносильно тому, что $P_5(x)$ имеет хотя бы один корень.\n\nОбозначим через $x_0$ какой-нибудь из его корней. Тогда $P_i(x_0) + P_{10-i}(x_0) = 2P_5(x_0) = 0$, то есть либо $P_i(x_0) \\le 0$, либо $P_{10-i}(x_0) \\le 0$. Поскольку старшие коэффициенты трёхчленов положительны, из этого следует, что в каждой из пар $(P_1, P_9)$, $(P_2, P_8)$, $(P_3, P_7)$, $(P_4, P_6)$ хотя бы один из трёхчленов имеет корень. Значит, есть не меньше пяти трёхчленов, имеющих хотя бы по одному корню. Поэтому есть не более четырёх трёхчленов без корней.\n\nОсталось привести пример, когда ровно пять трёхчленов (и один из них — $P_5$) имеют хотя бы по одному корню. Годятся, например, трёхчлены $x^2 - 4, x^2 - 3, x^2 - 2, \\dots, x^2 + 4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39635,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a group of people some pairs are friends. A group is called *k-indivisible* if for each decomposition of this group into $k$ subgroups, at least one subgroup contains a pair of friends. Suppose that $A$ is a finite 3-indivisible group of people having no subgroup of 4 persons such that each two of them are friends. Prove that it is possible to decompose $A$ into two subgroups $B$ and $C$ such that $B$ is 2-indivisible, and $C$ is 1-indivisible. (V. Dol'nikov)\n\nНазовём компанию *к-неразбиваемой*, если при любом разбиении её на *k* групп в одной из групп найдутся два знакомых человека. Дана 3-неразбиваемая компания, в которой нет четырёх попарно знакомых человек. Докажите, что её можно разделить на две компании, одна из которых 2-неразбиваемая, а другая — 1-неразбиваемая. (В. Дольников)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное. Рассмотрим граф $G$, в котором люди являются вершинами, а два человека соединены ребром, если они знакомы. Тогда граф $k$-разбиваем, если его вершины можно правильно окрасить в $k$ цветов (т. е. окрасить так, чтобы соединённые вершины имели разные цвета). Мы будем пользоваться следующей известной леммой.\n\n**Лемма.** Пусть в графе нет циклов нечётной длины. Тогда его вершины можно правильно раскрасить двумя красками.\n\n**Доказательство.** Ясно, что достаточно доказать лемму для связного графа. Расстоянием между двумя вершинами $X$ и $Y$ назовём наименьшую длину пути, соединяющего эти вершины.\n\nЗафиксируем некоторую вершину $A$, и покрасим все вершины, находящиеся на нечётном расстоянии от $A$, в красный цвет, а остальные вершины — в синий цвет. Докажем, что указанная раскраска — искомая. Предположим противное — имеется ребро, соединяющее, скажем, красные вершины $B$ и $C$. Рассмотрим кратчайшие пути $A = B_0, B_1, \\dots, B_{2n-1} = B$ и $A = C_0, C_1, \\dots, C_{2m-1} = C$, ведущие из $A$ в $B$ и в $C$. Взяв наибольший индекс $i$ такой, что $B_i = C_i$, получим цикл нечётной длины $B_{i+1}, \\dots, B_{2n-1}, C_{2m-1}, C_{2m-2}, \\dots, C_i = B_i$. Противоречие. $\\Box$\n\nПо лемме, в нашем графе $G$ есть нечётный цикл — иначе его вершины можно окрасить даже в два цвета. Выберем в $G$ нечётный цикл $C$ минимальной длины $n$. Тогда не существует рёбер, соединяющих вершины этого цикла, кроме рёбер самого цикла. Действительно, любое такое ребро разбивает цикл на два меньших по длине, причём один из них нечётен. Значит, в этом случае нашёлся бы нечётный цикл меньшей длины.\n\nДалее, покажем, что любая вершина $x$, не принадлежащая $C$, соединена не более, чем с двумя вершинами $C$. Если $C$ содержит три вершины, то утверждение верно, иначе $x$ вместе с вершинами $C$ образует компанию из 4 попарно знакомых человек.\n\nПусть теперь в $C$ больше трёх вершин. Предположим, что $x$ соединена с вершинами $v_1, v_2, v_3$ этого цикла. Участок цикла между какими-то двумя из них (скажем, между $v_1$ и $v_2$) содержит нечётное количество рёбер $d$. Если $d < n-2$, то этот участок вместе с вершиной $x$ образует нечётный цикл длины $d+2 < n$, что невозможно. Значит, $d \\ge n-2$ ребра, а это значит, что вершины $v_1, v_3, v_2$ идут в цикле подряд. Но тогда найдётся цикл $v_1, v_3, x$ длины 3, что невозможно.\n\nТеперь мы можем предъявить требуемое разбиение: поместим в одну группу вершины цикла $C$, а в другую (назовём её $D$) — все остальные. Вершины цикла $C$, очевидно, нельзя правильно окрасить в два цвета. Осталось показать, что между вершинами группы $D$ есть ребро (тогда она 1-неразбиваема). Предполагая противное, покажем, что $G$ можно окрасить в три цвета. Сначала окрасим все вершины $C$, кроме одной, попеременно в цвета 1 и 2, а оставшуюся окрасим в цвет 3; поскольку между этими вершинами нет других рёбер, раскраска этого цикла — правильная. Окрасить теперь вершины группы $D$ по очереди. Каждая очередная вершина соединена не более, чем с двумя вершинами из $C$, и не соединена с вершинами из $D$; значит, можно выбрать для неё цвет, отличный от цветов её соседей. Итого, граф $G$ можно правильно окрасить в три цвета, что противоречит условию.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, \\dots, a_{14}$ be distinct positive integers. All 196 numbers of the form $a_i + a_j$ ($1 \\le i, j \\le 14$) are written on a blackboard. May it happen that for each two digits $a, b$ there exists a written number with a decimal representation ending by $\\overline{ab}$?\n\nДаны различные натуральные числа $a_1, a_2, ..., a_{14}$. На доску выписаны все 196 чисел вида $a_k + a_\\ell$, где $1 \\le k, \\ell \\le 14$. Может ли оказаться, что для любой комбинации из двух цифр среди написанных на доске чисел найдется хотя бы одно число, оканчивающееся на эту комбинацию (то есть, найдутся числа, оканчивающиеся на 00, 01, 02, ..., 99)?",
    "options": [],
    "answer": "No",
    "solution": "Не может.\n\nПусть среди наших 14 чисел есть $a$ чётных и $b = 14-a$ нечётных. Нечётное число на доске может появиться лишь как сумма чётного и нечётного, т.е. таких чисел будет $ab$ (при этом каждое будет выписано по два раза). Но $ab = \\frac{(a+b)^2 - (a-b)^2}{4} \\le \\frac{(a+b)^2}{4} = 49$. Значит, на доске будет не более 49 различных нечётных чисел; а, чтобы выполнялось условие, их должно быть хотя бы 50. Значит, требуемое невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39637,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ has perimeter $4$. Points $X$ and $Y$ are chosen on the rays $AB$ and $AC$ respectively so that $AX = AY = 1$. Segments $BC$ and $XY$ intersect at point $M$. Prove that the perimeter of one of the triangles $ABM$ and $ACM$ equals $2$. (V. Shmarov)\n\nПериметр треугольника $ABC$ равен $4$. На лучах $AB$ и $AC$ отмечены точки $X$ и $Y$ так, что $AX = AY = 1$. Отрезки $BC$ и $XY$ пересекаются в точке $M$. Докажите, что периметр одного из треугольников $ABM$ и $ACM$ равен $2$. (В. Шмаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку отрезки $BC$ и $XY$ пересекаются, без ограничения общности можно считать, что $AB > AX$ и $AC < AY$.\nПусть вневписанная окружность $\\omega$ исходного треугольника касается стороны $BC$ в точке $R$, а продолжений сторон $AB$ и $AC$ в точках $P$ и $Q$ соответственно. Пусть $J$ — центр $\\omega$, а $r$ — её радиус. Пусть также $N$ — середина $XY$; из симметрии $N$ лежит на $AJ$, и $AN \\perp XY$. Имеем $AP = AQ = \\frac{1}{2}(AP + AQ) = \\frac{1}{2}(AB + BP + AC + CQ) = \\frac{1}{2}(AB + BR + AC + CR) = \\frac{1}{2}(AB + BC + AC) = 2$. Отсюда $AY = YQ = 1$.\n\nТеперь по теореме Пифагора получаем $MR^2 = MJ^2 - JR^2 = (MN^2 + NJ^2) - r^2 = MN^2 + (YJ^2 - YN^2) - r^2 = MN^2 + (YQ^2 + r^2) - (YA^2 - AN^2) - r^2 = MN^2 + AN^2 = AM^2$, ибо $YA = YQ$.\n\nОтсюда $AM = MR$, и периметр треугольника $ACM$ равен $AC + CM + AM = AC + CM + MR = AC + CR = AC + CQ = AQ = 2$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonzero real numbers $a$, $b$, $c$ are chosen so that each two of three equations $a x^{11} + b x^4 + c = 0$, $b x^{11} + c x^4 + a = 0$, $c x^{11} + a x^4 + b = 0$ share a common root. Prove that all three equations share a common root.\n\nНенулевые числа $a$, $b$, $c$ таковы, что любые два из трёх уравнений $a x^{11} + b x^4 + c = 0$, $b x^{11} + c x^4 + a = 0$, $c x^{11} + a x^4 + b = 0$ имеют общий корень. Докажите, что все три уравнения имеют общий корень.\n(И. Богданов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим сразу, что все корни наших уравнений — ненулевые, поскольку свободные члены не равны нулю.\n\nПусть $p$ — общий корень первых двух уравнений. Тогда имеем\n$$\n\\begin{aligned}\n0 &= b(a p^{11} + b p^4 + c) - a(b p^{11} + c p^4 + a) = p^4 (b^2 - a c) - (a^2 - b c), \\\\\n0 &= b(b p^{11} + c p^4 + a) - c(a p^{11} + b p^4 + c) = p^{11} (b^2 - a c) - (c^2 - a b).\n\\end{aligned}\n$$\n\nОтсюда следует, что если одно из чисел $A = a^2 - b c$, $B = b^2 - a c$, $C = c^2 - a b$ равно нулю, то и все три равны нулю. Но тогда $a/b = b/c = c/a$, а поскольку произведение этих чисел равно 1, то и все они равны 1, то есть $a = b = c$. В этом случае утверждение задачи очевидно.\n\nВ противном случае все три числа $A, B, C$ ненулевые. Тогда $p^4 = \\frac{A}{B}$, $p^{11} = \\frac{C}{B}$. Обозначив через $q$ общий корень второго и третьего уравнений, а через $r$ общий корень третьего и первого уравнений. Аналогично предыдущему, имеем $q^4 = \\frac{B}{C}$, откуда $p^{11} q^4 = 1$. Таким же образом получаем равенства $q^{11} r^4 = 1$, $r^{11} p^4 = 1$. Отсюда $p^{11 \\cdot 11 \\cdot 11} = q^{-4 \\cdot 11 \\cdot 11} = r^{4 \\cdot 4 \\cdot 11} = p^{-4 \\cdot 4 \\cdot 4}$, следовательно $p^{11^3 + 4^3} = 1$, и значит $p = 1$. Аналогично $q = r = 1$. Но тогда 1 является общим корнем всех трёх уравнений.\n\n**Замечание.** В случае если все три числа $A, B, C$ ненулюемые, решение можно завершить по-другому. Переименовав, если надо, переменные по циклу, можно считать, что $|B|$ — среднее по величине из чисел $|A|, |B|, |C|$. Тогда из полученных выше равенств следует, что одно из чисел $|p|^4 = \\frac{|A|}{|B|}$ и $|p|^{11} = \\frac{|C|}{|B|}$ не больше единицы, а другое — не меньше единицы. Это возможно лишь тогда, когда оба они равны единице, то есть $|p| = 1$. Аналогично $|q| = |r| = |p| = 1$, и два из чисел $p, q, r$ равны, скажем, $p = q$. Но тогда это число является общим корнем всех трёх уравнений.\n\n\n**Второе решение.** Достаточно доказать, что одно из данных уравнений имеет ровно один корень; тогда он будет общим у этого уравнения с каждым из остальных. Рассмотрим среди данных уравнений то, в котором коэффициенты при $x^4$ и свободный член имеют одинаковый знак, пусть для определенности $b = -b' < 0$, $c = -c' < 0$ (случай $b > 0, c > 0$ сводится к случаю $b < 0, c < 0$ домножением уравнения на $-1$). Пусть $a > 0$ (случай $a < 0$ сводится к случаю $a > 0$ заменой переменной $x$ на $-x$).\n\nПокажем, что уравнение $f(x) = 0$, где $f(x) = a x^{11} - b' x^4 - c'$, имеет не более одного корня (на самом деле, оно имеет ровно один корень). Перепишем это уравнение в виде\n$$\nx^4 (a x^7 - b') = c'\n$$\nВидим, что при $x < d = \\left(\\frac{b'}{a}\\right)^{1/7}$ уравнение не имеет решений, так как левая часть отрицательна. На промежутке же $[d, +\\infty)$ левая часть монотонно возрастает, поскольку оба сомножителя неотрицательны и возрастают. Значит, на этом промежутке есть не более одного корня.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39639,
    "subject": "Mathematics (Multi-modal)",
    "question": "10 distinct real numbers are given. For each pair $a, b$ of given numbers, Basil writes down into his exercise book the number $(a-b)^2$, while Pete writes down into his exercise book the number $|a^2 - b^2|$. Determine whether it may appear that Basil and Pete obtain the same collections of 45 numbers in their exercise books.\n(S. Berlov)\n\nДаны 10 попарно различных чисел. Для каждой пары данных чисел Вася записал у себя в тетради квадрат их разности, а Петя записал у себя в тетради модуль разности их квадратов. Могли ли в тетрадях у мальчиков получиться одинаковые наборы из 45 чисел? (С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Не могли.\n\nПредположим противное. Если среди исходных чисел есть ноль, то для любого другого числа $a$ имеем $a^2 - 0^2 = (a - 0)^2$. Значит, если вычеркнуть ноль, то останутся 9 чисел, также удовлетворяющих условию.\n\nИтак, можно считать, что исходных чисел 9 или 10, и все они ненулевые. Пусть среди них есть числа разных знаков; рассмотрим минимальное и максимальное из них — обозначим их $a < 0 < b$. Тогда у Васи присутствует число $(b - a)^2$, которое больше как $a^2$, так и $b^2$; у Пети же любое число не превосходит $\\max(a^2, b^2)$. Противоречие.\n\nЗначит, все исходные числа — одного знака; заменив, если надо, все числа на противоположные, можно считать, что все они положительны. Опять обозначив через $a$ и $b$ соответственно минимальное и максимальное из этих чисел, имеем $b^2 - a^2 = (b - a)(b + a) > (a - b)^2 \\ge (c - d)^2$, где $c$ и $d$ — произвольные два исходных числа. Тогда число $b^2 - a^2$ не встретится на листке у Пети, но встретится у Васи — противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39640,
    "subject": "Mathematics (Multi-modal)",
    "question": "2011 real numbers are written on a blackboard. It appears that for each three written numbers $a$, $b$, $c$, the sum $a+b+c$ is also written. Find the least possible number of zeroes among the written numbers.\n\nНа доску выписаны 2011 чисел. Оказалось, что сумма любых трёх выписанных чисел также является выписанным числом. Какое наименьшее количество нулей может быть среди этих чисел?",
    "options": [],
    "answer": "2009",
    "solution": "Ответ. 2009.\n\nПример из 2009 нулей и чисел 1, -1 удовлетворяет условию, поэтому количество нулей может быть ровно 2009.\n\nПредположим, что количество нулей меньше 2008. Нетрудно видеть, что тогда на доске либо найдутся три неотрицательных числа, среди которых хотя бы два строго положительных, либо найдутся три неположительных числа, среди которых хотя бы два строго отрицательны. Пусть выполнено первое условие, и на доске нашлись три числа $a$, $b$, $c$, причём $a$, $b$ положительны, а $c$ неотрицательно. Можно считать тогда, что $a$ — наибольшее из всех выписанных чисел. Но тогда число $a+b+c$ также должно быть выписано на доске, и $a+b+c > a$. поэтому $a+b+c$ не может быть выписанным — противоречие. Второй случай разбирается аналогично.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle. Points $P$ and $Q$ are chosen on the extensions of its altitudes $BB_1$ and $CC_1$ along the points $B_1$ and $C_1$ respectively so that $\\angle PAQ = 90^\\circ$. Let $AF$ be the altitude of triangle $APQ$. Prove that $\\angle BFC = 90^\\circ$. (A. Polyanskiy)\n\nДан остроугольный треугольник **ABC**. На продолжениях его высот **BB₁** и **CC₁** за точки **B₁** и **C₁** выбраны соответственно точки **P** и **Q** такие, что угол **PAQ** — прямой. Пусть **AF** — высота треугольника **APQ**. Докажите, что угол **BFC** — прямой. (А. Полянский)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Точки $B_1$ и $C_1$ лежат на окружности, построенной на $BC$ как на диаметре. Для решения достаточно доказать, что $F$ также лежит на этой окружности, то есть достаточно доказать, что четырёхугольник $CB_1FC_1$ — вписанный.\n\nТак как $\\angle AB_1P = \\angle AFP = 90^\\circ$, то точки $B_1$ и $F$ лежат на окружности, построенной на $AP$ как на диаметре. Поэтому $\\angle PFB_1 = \\angle PAB_1$. Аналогично $\\angle QFC_1 = \\angle QAC_1$. Имеем $\\angle B_1FC_1 = 180^\\circ - \\angle PFB_1 - \\angle QFC_1 = 180^\\circ - 90^\\circ = 90^\\circ$. Consequently, $\\angle PAB_1 - \\angle QAC_1 = 90^\\circ - (\\angle PAQ - \\angle B_1AC_1) = 90^\\circ + \\angle B_1AC_1 = 90^\\circ + (90^\\circ - \\angle ACC_1) = 180^\\circ - \\angle B_1CC_1$. Таким образом, четырёхугольник $CB_1FC_1$ — вписанный.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C_0$ and $B_0$ be respectively the midpoints of the sides $AB$ and $AC$ of a non-isosceles acute-angled triangle $ABC$. Denote by $O$ and $H$ its circumcenter and its orthocenter, respectively. The lines $BH$ and $OC_0$ intersect at point $P$, while the lines $CH$ and $OB_0$ intersect at point $Q$. Given that the quadrilateral $OPHQ$ is a rhombus, prove that the points $A$, $P$, and $Q$ are collinear.\n\nВ неравнобедренном остроугольном треугольнике $ABC$ точки $C_0$ и $B_0$ — середины сторон $AB$ и $AC$ соответственно, $O$ — центр описанной окружности, $H$ — точка пересечения высот. Прямые $BH$ и $OC_0$ пересекаются в точке $P$, а прямые $CH$ и $OB_0$ — в точке $Q$. Оказалось, что четырехугольник $OPHQ$ — ромб. Докажите, что точки $A$, $P$ и $Q$ лежат на одной прямой.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Пусть $BB'$ и $CC'$ — высоты треугольника. Так как $OB_0$ и $OC_0$ — серединные перпендикуляры к сторонам $AC$ и $AB$, то отрезки $B'B_0$ и $C'C_0$ равны высотам ромба $OPHQ$, значит, $B'B_0 = C'C_0$. Но эти отрезки являются проекциями отрезка $OH$ на прямые $AB$ и $AC$; значит, $OH$ составляет равные углы с этими прямыми. Это означает, что прямая $OH$ параллельна либо внутренней, либо внешней биссектрисе угла $BAC$.\n\nТак как $\\angle AOC = 2\\angle ABC$, то $\\angle CAO = 90^\\circ - \\frac{1}{2}\\angle AOC = 90^\\circ - \\angle ABC = \\angle BAH$. Это значит, что лучи $AO$ и $AH$ симметричны относительно биссектрисы $\\ell$ угла $BAC$; в частности, $OH$ пересекает $\\ell$ и потому не может быть ей параллельна. Итак, $OH \\perp \\ell$, и $\\ell$ является биссектрисой и высотой треугольника $AOH$. Отсюда $AH = AO$, и точка $A$, как и точки $P$, $Q$, лежит на серединной перпендикуляре к отрезку $OH$.\n\n![](attached_image_1.png)\n\nВторое решение. Положим $a = BC$, $b = CA$, $c = AB$, $\\alpha = \\angle BAC$; тогда $AB' = c \\cos \\alpha$, $AC' = b \\cos \\alpha$. Пусть $BB'$ и $CC'$ — высоты треугольника. Так как $OB_0$ и $OC_0$ — серединные перпендикуляры к сторонам $AC$ и $AB$, то отрезки $B'B_0$ и $C'C_0$ равны высотам ромба $OPHQ$, значит, $B'B_0 = C'C_0$, откуда $|AB_0 - AB'| = |AC_0 - AC'|$, или\n$$\n\\left| \\frac{b}{2} - c \\cos \\alpha \\right| = \\left| \\frac{c}{2} - b \\cos \\alpha \\right|. \\quad (*)\n$$\nВ случае, если в левой и правой частях модули раскрываются с одним знаком, имеем $\\frac{b}{2} - c \\cos \\alpha = \\frac{c}{2} - b \\cos \\alpha$, т.е. $b(\\frac{1}{2} + \\cos \\alpha) = c(\\frac{1}{2} + \\cos \\alpha)$. По условию $\\alpha < 90^\\circ$, поэтому $\\cos \\alpha > 0$ и, значит, $b = c$, что противоречит условию.\n\nПредположим теперь, что модули в равенстве $(*)$ раскрываются с разными знаками, скажем, $\\frac{b}{2} - c \\cos \\alpha > 0$ и $\\frac{c}{2} - b \\cos \\alpha < 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39643,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integers $a > b > 1$, a sequence $x_1, x_2, \\dots$ is defined as $x_n = \\frac{a^n - 1}{b^n - 1}$. Find the least integer $d$ such that for any $a$ and $b$, this sequence cannot contain $d$ consecutive terms that are prime numbers. (V. Senderov)\n\nДля натуральных чисел $a > b > 1$ определим последовательность $x_1, x_2, \\dots$ формулой $x_n = \\frac{a^n - 1}{b^n - 1}$. Найдите наименьшее $d$ такое, что эта последовательность не содержит $d$ последовательных членов, являющихся простыми числами, ни при каких $a$ и $b$. (В. Сендеров)",
    "options": [],
    "answer": "3",
    "solution": "При $a = 4$, $b = 2$ имеем $\\frac{a^2 - 1}{b^2 - 1} = 3$, $\\frac{a^3 - 1}{b^3 - 1} = 5$. Осталось показать, что больше двух простых чисел подряд не встретится.\n\nДокажем более сильное, чем требуется, утверждение: при $n \\ge 2$ хотя бы одно из чисел $\\frac{a^n - 1}{b^n - 1}$, $\\frac{a^{n+1} - 1}{b^{n+1} - 1}$ не является простым. Предположим противное; тогда\n$$\n(a-1)(a^{n-1} + \\dots + a + 1) = p(b-1)(b^{n-1} + \\dots + b + 1), \\quad (1)\n$$\n$$\n(a-1)(a^n + \\dots + a + 1) = q(b-1)(b^n + \\dots + b + 1), \\quad (2)\n$$\nгде $p$ и $q$ — простые числа.\n\nПредположим, что $a-1$ не делится на $b-1$. Тогда некоторое простое число $r$ входит в разложение числа $b-1$ в степени большей, чем в $a-1$. Из (1) и (2) получим, что $r$ — общий делитель чисел $a^{n-1} + \\dots + a + 1$ и $a^n + \\dots + a + 1$, но\n$$\n\\begin{aligned}\n\\text{НОД}(a^{n-1} + \\dots + a + 1, a^n + \\dots + a + 1) &= \\\\\n&= \\text{НОД}(a^{n-1} + \\dots + a + 1, a^n) = 1.\n\\end{aligned}\n$$\nПротиворечие.\n\nИтак, число $k = \\frac{a-1}{b-1}$ целое. Из (1) имеем\n$$\nk(a^{n-1} + \\dots + a + 1) = p(b^{n-1} + \\dots + b + 1),\n$$\nпричём $1 < k < p$, так как $b^{n-1} + \\dots + b + 1 < a^{n-1} + \\dots + a + 1$.\n\nЗначит, $\\text{НОД}(k, p) = 1$, поэтому $b^{n-1} + \\dots + b + 1 \\mid k$. Аналогично, из (2) вытекает, что $k < q$ и $b^n + \\dots + b + 1 \\mid k$. Но это противоречит тому, что $\\text{НОД}(b^{n-1} + \\dots + b + 1, b^n + \\dots + b + 1) = 1$.\n\n**Замечание.** Двумя последовательными простыми числами в такой последовательности могут быть любые два простых числа вида $p = b+1, q = b^2 + 1$. Действительно, полагая $a = b^2$, имеем $p = \\frac{a-1}{b-1}, q = \\frac{a^2-1}{b^2-1}$. Такими парами являются, например, $(7, 37)$, $(11, 101)$ и $(2^{2^3} + 1, 2^{2^4} + 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39644,
    "subject": "Mathematics (Multi-modal)",
    "question": "A grid square $2010 \\times 2010$ is decomposed into «corners» (i.e. L-shaped figures consisting of 3 cells each). Prove that it is possible to mark one cell in each corner so that each row and each column would contain the same number of marked cells.\n\nКлетчатый квадрат $2010 \\times 2010$ разрезан на трёхклеточные уголки. Докажите, что можно в каждом уголке отметить по клетке так, чтобы в каждой вертикали и в каждой горизонтали было поровну отмеченных клеток.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим $n = 2010/3 = 670$. Назовём линией строку или столбец; пронумеруем строки сверху вниз, а столбцы — слева направо. Заметим, что для выполнения условий задачи достаточно, чтобы при любом $k$ в левых $k$ столбцах и в верхних $k$ строках содержалось бы по $kn$ отмеченных клеток.\n\nСкажем, что уголок смотрит из $i$-й вертикали влево, если две его клетки находятся в $i$-й вертикали, а третья — левее. Аналогично определим взгляд в других трёх направлениях; каждый уголок, таким образом, смотрит в двух направлениях.\n\nОтметим для начала в каждом уголке среднюю клетку. Теперь в каждом уголке можно либо оставить клетку на месте, либо сдвинуть её ровно в одном из двух направлений, в которые этот уголок смотрит. Выясним, сколько таких сдвигов надо сделать.\n\nНаложим дополнительное условие: между каждыми двумя соседними линиями все сдвиги должны идти в одну сторону. Рассмотрим, скажем, два соседних столбца: $k$-й и $(k+1)$-й. Предположим, что в левых $k$ столбцах сейчас содержится $d_k \\ge nk$ отмеченных клеток. Далее, пусть в $k$-м столбце есть $a_k$ уголков, смотрящих вправо, а в $(k+1)$-м — $b_k$ уголков, смотрящих влево. Тогда в первых $k$ столбцах находится $\\frac{3kn - 2a_k - b_k}{3}$ целых уголков, и потому $d_k = \\frac{3kn - 2a_k - b_k}{3} + a_k = kn + \\frac{a_k - b_k}{3}$. Значит, чтобы добиться нужного количества отмеченных, надо сдвинуть $s_k = \\frac{a_k - b_k}{3} \\le \\frac{a_k}{3}$ отмеченных клеток из $k$-го столбца вправо. В этом случае выделим $s_k = \\frac{a_k - b_k}{3}$ непересекающихся пар среди $a_k$ уголков, смотрящих из $k$-го столбца вправо, и потребуем, чтобы ровно в одном уголке из каждой пары клетка была сдвинута вправо. Аналогично, если $d_k < nk$, то мы выделим $\\frac{b_k - a_k}{3}$ непересекающихся пар среди $b_k$ уголков, смотрящих из $(k+1)$-го столбца влево.\n\nПроделав такую операцию с каждой парой соседних линий, мы получим некоторое количество выделенных пар уголков, в каждой из которых надо выбрать по уголку; при этом все выбранные уголки должны быть различными. Осталось показать, что это возможно.\n\nСоединим два уголка, находящиеся в одной паре, ребром. Заметим, что каждый уголок лежит не более, чем в двух парах: по одной на два направления, в которых он смотрит. Значит, мы получим граф, в котором из каждой вершины выходит не более двух рёбер. Тогда этот граф распадается на циклы и пути. Теперь в каждом цикле $U_1 - U_2 - \\dots - U_n - U_1$ в паре $(U_i, U_{i+1})$ выберем уголок $U_i$, а в паре $(U_n, U_1)$ — уголок $U_n$. Аналогичную операцию проделаем с путём. Очевидно, что все условия выполнены — а значит, и задача решена.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let positive integers $d$ and $d'$ ($d' > d$) be two divisors of positive integer $n$. Prove that $d' > d + \\frac{d^2}{n}$.\n\nНатуральные числа $d$ и $d'$, $d' > d$ -- делители натурального числа $n$. Докажите, что $d' > d + \\frac{d^2}{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку числа $f = n/d$ и $f' = n/d'$ целые, а $f > f'$, имеем $f - f' \\ge 1$, или\n$$\n1 \\le \\frac{n}{d} - \\frac{n}{d'} = \\frac{(d' - d)n}{dd'} < \\frac{(d' - d)n}{d^2}.\n$$\nДомножая на $\\frac{d^2}{n}$, получаем $d' - d > \\frac{d^2}{n}$, что и требовалось доказать.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exists a real $\\alpha$ such that $\\cos \\alpha$ is irrational, while all the numbers $\\cos 2\\alpha$, $\\cos 3\\alpha$, $\\cos 4\\alpha$, $\\cos 5\\alpha$ are rational.\n\nСуществует ли такое вещественное $\\alpha$, что число $\\cos \\alpha$ иррационально, а все числа $\\cos 2\\alpha$, $\\cos 3\\alpha$, $\\cos 4\\alpha$, $\\cos 5\\alpha$ рациональны?",
    "options": [],
    "answer": "Does not exist",
    "solution": "Не существует.\n\nПредположим противное. Тогда число $A = \\cos \\alpha + \\cos 5\\alpha$ иррационально как сумма рационального и иррационального; с другой стороны, $A = 2 \\cos 2\\alpha \\cos 3\\alpha$ рационально как произведение трёх рациональных чисел. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $2011$ nonzero integer numbers $a_1, a_2, \\dots, a_{2011}$. It appears that the sum of each of them with the product of all the others is negative. Let us partition these $2011$ numbers into two nonempty groups and find the product of the numbers in each group. Prove that the sum of these two products is also negative.\n\nДаны $2011$ ненулевых целых чисел. Известно, что сумма любого из них с произведением оставшихся $2010$ чисел отрицательна. Докажите, что если произвольным образом разбить все данные числа на две группы и перемножить числа в группах, то сумма двух полученных произведений также будет отрицательной.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим, что среди данных чисел четное количество отрицательных. Тогда среди них есть положительное число $a$, и произведение всех чисел, кроме $a$, положительно. Это противоречит условию.\n\nЗначит, среди данных чисел нечетное число отрицательных. Пусть $x_1, x_2, \\dots, x_k$ и $y_1, y_2, \\dots, y_m$ — две группы, на которые разбиты данные числа ($k + m = 2011$). Ровно одно из двух произведений $x_1x_2\\dots x_k$ и $y_1y_2\\dots y_m$ (а именно то, в котором нечётное число отрицательных сомножителей) — отрицательно; пусть для определенности $x_1x_2\\dots x_k < 0$, $y_1y_2\\dots y_m > 0$.\n\nТогда среди чисел $x_1, x_2, \\dots, x_k$ найдется отрицательное, скажем, $x_1 < 0$. Отсюда $x_2\\dots x_k > 0$, а значит, $x_2\\dots x_k \\ge 1$ (так как данные числа целые). Следовательно,\n$$\nx_1x_2\\dots x_k + y_1y_2\\dots y_m \\le x_1 + y_1y_2\\dots y_m \\le x_1 + y_1y_2\\dots y_m x_2\\dots x_k.\n$$\nНо по условию $x_1 + y_1y_2\\dots y_m x_2\\dots x_k < 0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39648,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Academy of Sciences, there are $999$ academicians. For each scientific topic there are exactly $3$ academicians interested in this topic. For each pair of academicians, there exists exactly one topic in which both these academicians are interested. Prove that there exists a set of $250$ topics such that each academician is interested in not more than one topic from this set. (A. Magazinov)\n\nВ Академии Наук $999$ академиков. Каждая научная тема интересует ровно троих академиков, и у каждого двух академиков есть ровно одна тема, интересная им обоим. Докажите, что можно выбрать $250$ тем из их общей области научных интересов так, чтобы каждый академик интересовался не более, чем одной из них. (А. Магазинов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Будем говорить, что две темы пересекаются по академику, если он интересуется обеими этими темами.\n\nВыберем наибольшее возможное количество непересекающихся тем; пусть это темы $T_1, \\dots, T_k$. Предположим, что $k \\le 249$. Обозначим через $S$ множество всех академиков, не интересующихся этими темами; тогда их ровно $s = 999 - 3k \\ge 252$.\n\nДля каждого двух академиков $a, b \\in S$ существует единственная тема $T(a, b)$, интересующая обоих. При этом третий академик, заинтересованный ею, не должен принадлежать $S$, иначе $T(a, b)$ можно добавить к исходным $k$ темам. Значит, его интересует какая-то тема $T_i$. Сопоставим эту тему (и этого академика) паре $(a, b)$.\n\nИтак, каждой из $\\frac{s(s-1)}{2}$ пар академиков из $S$ сопоставлена одна из $k$ тем $T_1, \\dots, T_k$; значит, какая-то тема $T_i$ сопоставлена не менее, чем $\\frac{s(s-1)}{2k} > \\frac{s}{2}$ парам. Обозначим эти пары $(a_1, b_1), \\dots, (a_d, b_d)$; пусть $T_i$ интересует академиков $x, y, z$. Поскольку $d > \\frac{s}{2} > 6$, один из $x, y, z$ сопоставлен хотя бы трём парам $(a_j, b_j)$; пусть, скажем, $x$ сопоставлен парам $(a_1, b_1), \\dots, (a_p, b_p)$ ($p \\ge 3$), а остальным парам сопоставлены $y$ или $z$.\n\nЗаметим, что все пары $(a_1, b_1), \\dots, (a_p, b_p)$ не пересекаются: если бы академик $a$ находился в двух из них, то академиков $a$ и $x$ интересовали бы две общих темы. Значит, $p \\le \\frac{s}{2} < d$, и паре $(a_{p+1}, b_{p+1})$ сопоставлен, скажем, академик $y$. Но тогда $(a_{p+1}, b_{p+1})$ не пересекается с одной из пар $(a_1, b_1), (a_2, b_2), (a_3, b_3)$, скажем, с $(a_1, b_1)$. Значит, можно из нашего набора тем выбросить $T_i$ и добавить непересекающиеся темы $T(a_1, b_1)$ и $T(a_{p+1}, b_{p+1})$, увеличив количество непересекающихся тем. Противоречие с исходным выбором.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the center of a rectangle $ABCD$. Point $K$ is chosen on the circumcircle of the rectangle. Line $CK$ intersects segment $AD$ at point $M$. Given that $AM : MD = 2$, prove that the meeting point of the medians of triangle $OKD$ lies on the circumcircle of triangle $COD$.\n\nНа окружности, описанной около прямоугольника $ABCD$, выбрана точка $K$. Оказалось, что прямая $CK$ пересекает отрезок $AD$ в точке $M$ такой, что $AM : MD = 2$. Пусть $O$ — центр прямоугольника. Докажите, что точка пересечения медиан треугольника $OKD$ лежит на окружности, описанной около треугольника $COD$. (В. Шмаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Отметим на продолжении отрезка $AD$ такую точку $T$, что $AT = DM$. Тогда прямоугольные треугольники $CDM$ и $BAT$ равны, а значит, $BT \\parallel CM$. Заметим, что $DT = DA + AT = 3DM + DM = 4DM$. По теореме Фалеса, прямая $CM$ пересекает отрезок $BD$ в точке $N$ такой, что $DB = 4DN$. Значит, $DN = NO$, то есть $KN$ — медиана треугольника $OKD$.\n\n![](attached_image_1.png)\n\nПусть $S$ — точка пересечения медиан треугольника $OKD$. Поскольку $OD = OK$, точка $S$ лежит на биссектрисе угла $KOD$, и $\\angle SOD = \\frac{1}{2} \\angle KOD$. С другой стороны, вписанный угол $KCD$ равен половине центрального угла $KOD$, откуда $\\angle SOD = \\frac{1}{2} \\angle KOD = \\angle SCD$. Это и означает, что точки $S, D, O, C$ лежат на одной окружности.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39650,
    "subject": "Mathematics (Multi-modal)",
    "question": "2011 storehouses are connected by some roads so that it is possible to reach each storehouse starting from any other one (possibly, using many roads). The *i*-th storehouse contains $x_i$ kg of cement. By one turn, it is allowed to transport any amount of cement from one storehouse to another one if they are connected by a direct road. As a result, one needs to obtain $y_i$ kg of cement in *i*-th storehouse (it is known that $x_1 + x_2 + \\dots + x_{2011} = y_1 + y_2 + \\dots + y_{2011}$).\nFind the least possible $N$ such that it is possible to reach the goal in $N$ turns for every admissible values of $x_i$'s, $y_i$'s and any admissible road system.\n\n2011 складов соединены дорогами так, что от любого склада можно проехать к любому другому, возможно, проехав по нескольким дорогам. На складах находится по $x_1, \\dots, x_{2011}$ кг цемента соответственно. За один рейс можно провезти с произвольного склада на другой склад по соединяющей их дороге произвольное количество цемента. В итоге на складах по плану должно оказаться по $y_1, \\dots, y_{2011}$ кг цемента соответственно, причём\n$$\nx_1 + x_2 + \\dots + x_{2011} = y_1 + y_2 + \\dots + y_{2011}.\n$$\nЗа какое минимальное количество рейсов можно выполнить план при любых значениях чисел $x_i$ и $y_i$ и любой схеме дорог?",
    "options": [],
    "answer": "2010",
    "solution": "**Answer.** $2010$ turns.\n\nFirst, let us show that it is not always possible to complete the plan in $2009$ turns. Suppose (for any road system) that all the cement is initially located at one storehouse $S$, and it needs to be distributed equally among all storehouses. Then, for each storehouse except $S$, cement must be delivered in some turn; clearly, these $2010$ turns are distinct, so the total number of turns must be at least $2010$.\n\nIt remains to show that the plan can always be completed in $2010$ turns. We prove by induction on $n$ that with $n$ storehouses, it is always possible to manage with $n-1$ turns. The base case for $n = 1$ is obvious.\n\nSuppose $n > 1$. Since it is possible to reach any storehouse from any other, there exists a route passing through all storehouses (possibly, more than once). Consider any such route and a storehouse $A$ that appears for the first time latest along this route. Then, if we remove storehouse $A$ and all roads leading from it, it is still possible to reach any storehouse from any other (using the previous roads of the route).\n\nWe may assume that $A$ is the storehouse with number $n$. If $y_n \\le x_n$, then we transport $x_n - y_n$ kg of cement from $A$ to any storehouse connected to it, and after that, we forget about it and all roads leading from it. By the induction hypothesis, for the remaining storehouses, the plan can be completed in $(n-1)-1$ turns. In total, in $(n-2)+1$ turns, the required distribution of cement is achieved.\n\nIf $y_n > x_n$, then we have already proved that starting from the distribution where the $i$-th storehouse contains $y_i$ kg, it is possible to reach the distribution where the $i$-th storehouse contains $x_i$ kg in $n-1$ turns. Now, performing all these transports in reverse order (and reverse direction), we achieve the required plan.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39651,
    "subject": "Mathematics (Multi-modal)",
    "question": "10 cars move along a straight road in one direction. The road passes through several towns. Each car moves at some constant speed in a town, and moves at some other constant speed outside the towns (for different cars, these speeds may differ). 2011 flags are put along the road. It appears that each car has faced all the flags, and there were no overtaking at the flag. Prove that there exist two flags $A$ and $B$ such that cars pass $A$ and $B$ in the same order. (S. Berlov)\n\nПо шоссе в одном направлении едут 10 автомобилей. Шоссе проходит через несколько населённых пунктов. Каждый из автомобилей едет с некоторой постоянной скоростью в населённых пунктах и с некоторой другой постоянной скоростью вне населённых пунктов. Для разных автомобилей эти скорости могут отличаться. Вдоль шоссе расположено 2011 флажков. Известно, что каждый автомобиль проехал мимо каждого флажка, причём около флажков обгонов не происходило. Докажите, что мимо каких-то двух флажков автомобили проехали в одном и том же порядке.\n(С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Введём в пространстве систему координат $Oxy t$. Обозначим через $M$ точку шоссе, в которой в начальный момент находится первый автомобиль. Каждой точке шоссе $A$ сопоставим точку $T_A$ на плоскости $Oxy$ с координатами $T_A(x_A, y_A)$, где $x_A$ — суммарная длина участков пути $AM$ в населённых пунктах, а $y_A$ — вне населённых пунктов. Тогда шоссе изображается некоторой ломаной на этой плоскости.\n\nДалее, для $i$-й машины нарисуем график её движения, состоящий из всех точек $X_{i,A} = (x_A, y_A, t_{i,A})$, где $t_{i,A}$ — момент времени, в который эта машина находилась в точке $A$. Если эта машина была в точке $M$ в момент $t_i$, а скорости этой машины в населённых пунктах и вне их равны $u_i$ и $v_i$, то $t_{i,A} = t_i + x_A/u_i + y_A/v_i$; таким образом, весь график движения $i$-й машины лежит в плоскости $t = t_i + x/u_i + y/v_i$. Обозначим эту плоскость через $\\alpha_i$.\n\nРассмотрим теперь порядок, в котором машины проедут мимо некоторого флажка $F$. Он совпадает с порядком, в котором плоскости $\\alpha_i$ будут пересекать прямую, параллельную $Ot$ и проходящую через точку $T_F$. Пусть для двух флажков $F$ и $G$ этот порядок различается (скажем, 1-я и 2-я машины проехали мимо него в разном порядке). Тогда отрезки $X_{1,F}X_{1,G}$ и $X_{2,F}X_{2,G}$ пересекаются (см. рис. 21). Таким образом, если спроецировать прямую пересечения плоскостей $\\alpha_1$ и $\\alpha_2$ на плоскость $Oxy$, то точки $T_F$ и $T_G$ будут лежать по разные стороны от полученной проекции $\\ell_{12}$.\n\nИтак, рассмотрим все прямые $\\ell_{ij}$. Нетрудно показать индукцией по $n$, что $n$ прямых разбивают плоскость не более, чем на $\\frac{n(n+1)}{2} + 1$ часть. Поскольку количество наших прямых не больше $C_{10}^2 = 45$, они разобьют плоскость не более, чем на $\\frac{45 \\cdot 46}{2} + 1 = 1036$ частей. Значит, найдутся такие флажки $F$ и $G$, что точки $T_F$ и $T_G$ попадают в одну часть. Тогда ни одна проекция прямых их не разделяет, поэтому на этих флажках порядок машин будет одинаков.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two cubic polynomials $F(x) = x^3 + a_1x^2 + a_2x + a_3$ and $G(x) = x^3 + b_1x^2 + b_2x + b_3$ with unit leading coefficients. All real roots of the equations $F(x) = 0$, $G(x) = 0$, and $F(x) = G(x)$ are written. It appears that eight distinct numbers are written. Prove that at least one of the minimal and the maximal among the written numbers is not a root of $F(x)$.\n\n(I. Bogdanov)\n\nДаны два различных кубических многочлена $F(x)$ и $G(x)$ с единичными старшими коэффициентами. Выписали все корни уравнений $F(x) = 0$, $G(x) = 0$, $F(x) = G(x)$. Оказалось, что выписаны 8 различных чисел. Докажите, что наибольшее и наименьшее из них не могут одновременно являться корнями многочлена $F(x)$.\n\n(И. Богданов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что у многочленов $F(x)$ и $G(x)$ не более, чем по три корня, а у многочлена $F(x) - G(x)$ (имеющего степень, не превосходящую 2) не больше двух корней. Поскольку у них в совокупности 8 корней, то у $F(x)$ и $G(x)$ ровно по три корня, а у $F(x) - G(x)$ ровно два, причём все они имеют кратность 1.\n\nПредположим, что утверждение задачи неверно; пусть $a$ и $b$ — минимальное и максимальное из выписанных чисел, и $F(a) = F(b) = 0$. Поскольку все корни $G(x)$ лежат на интервале $(a, b)$, имеем $G(a) < 0$, $G(b) > 0$. С другой стороны, квадратный трёхчлен $F(x) - G(x)$ имеет два корня на этом интервале, поэтому значения $F(a) - G(a) = -G(a)$ и $F(b) - G(b) = -G(b)$ должны иметь одинаковый знак. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39653,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ is inscribed into circle $\\omega$. The tangents to $\\omega$ passing through $B$ and $C$ intersect the tangent to $\\omega$ passing through $A$ at points $K$ and $L$ respectively. The line $k$ passing through $K$ is parallel to $AB$; the line $\\ell$ passing through $L$ is parallel to $AC$. Let $P$ be the meeting point of $k$ and $\\ell$. Prove that $BP = CP$. (P. Kozhevnikov)\n\nОстроугольный треугольник $ABC$ вписан в окружность $\\omega$. Касательные к $\\omega$, проведенные через точки $B$ и $C$, пересекают касательную к $\\omega$, проведенную через точку $A$, в точках $K$ и $L$ соответственно. Прямая, проведенная через $K$ параллельно $AB$, пересекается с прямой, проведенной через $L$ параллельно $AC$, в точке $P$. Докажите, что $BP = CP$. (П. Кожевников)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Докажем, что точка $P$ лежит на серединном перпендикуляре к отрезку $BC$. Пусть $O$ — центр окружности $\\omega$, а $X$ — точка пересечения прямых $BC$ и $PL$ (см. рис. 7).\n\n![](attached_image_1.png)\n\nТак как $O$ лежит на серединном перпендикуляре к отрезку $BC$, достаточно доказать, что $OP \\perp BC$.\n\nТочки $A$ и $B$ симметричны относительно прямой $OK$, поэтому $OK \\perp AB \\parallel KP$. Аналогично $OL \\perp LP$. Поскольку $\\angle OKP = \\angle OLP = 90^\\circ$, четырехугольник $OKPL$ вписанный, откуда $\\angle OPL = \\angle OKL$. Из касания вытекает, что $\\angle KAB = \\angle ACB = \\angle PXB$. Таким образом, $\\angle ORX + \\angle PXB = \\angle OKL + \\angle KAB = 90^\\circ$, что и требовалось доказать.\n\n![](attached_image_2.png)\n\nВторое решение. Пусть прямые $BK$ и $CL$ пересекаются в точке $M$ (см. рис. 8). Поскольку треугольник $ABK$ равнобедренный, имеем $\\angle PKA = \\angle BAK = \\frac{1}{2}(180^\\circ - \\angle AKB)$. Значит, $KP$ — биссектриса внешнего угла при вершине $K$ треугольника $KLM$. Аналогично, $LP$ — биссектриса внешнего угла при вершине $L$. Тогда $P$ — центр вневписанной окружности этого треугольника, касающейся стороны $KL$, поэтому $P$ лежит на биссектрисе угла $M$. Поскольку $MB = MC$, точки $B$ и $C$ симметричны относительно этой биссектрисы. Значит, и отрезки $PB$ и $PC$ также симметричны и потому равны.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39654,
    "subject": "Mathematics (Multi-modal)",
    "question": "A pile of more than $n^2$ stones is put on the table. Pete and Basil make moves taking stones in turn; Pete moves first. By one move it is allowed to take $k$ stones from a pile if either $k = 1$, or $k$ is a prime number less than $n$, or $k$ is divisible by $n$. The one taking the last stone wins. Prove that Pete has a winning strategy. (S. Berlov)\n\nНа столе лежит куча из более, чем $n^2$ камней. Петя и Вася по очереди берут камни из кучи, первым берёт Петя. За один ход можно брать любое простое число камней, меньше $n$, либо любое кратное $n$ число камней, либо один камень. Докажите, что Петя может действовать так, чтобы взять последний камень независимо от действий Васи.\n(С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное: Вася может всегда действовать так, чтобы помешать Пете; поскольку в любой ситуации, кроме конечной, можно сделать ход, это означает, что у Васи есть стратегия, поволяющая ему гарантированно взять последний камень. Пусть $d$ — изначальное количество камней, а $r$ — остаток от деления $d$ на $n$. Ясно, что $r \\neq 0$, иначе Петя может сразу взять все камни.\nПетя после первого своего хода может, взяв кратное $n$ количество камней, оставить любое количество вида $a_k = r + nk$, где $0 \\le k \\le n - 1$ (все эти количества меньше $n^2$). Пусть $c_k$ — ответный ход в Васиной стратегии при $a_k$ камнях в куче. Тогда $c_k$ не делится на $n$, иначе после его хода остаётся $r + n \\left(k - \\frac{c_k}{n}\\right)$ камней, и Петя может выиграть, действуя по Васиной стратегии для этого числа. Значит, $c_k < n$ при всех $0 \\le k \\le n - 1$, а тогда два из этих чисел совпадают, скажем, $c_k = c_\\ell$ при $0 \\le k < \\ell \\le n - 1$. Напомним, что у Васи есть стратегия выигрыша в ситуации, когда в куче $a_k - c_k$ камней и ход Пети.\nПусть теперь Петя первым ходом оставит $a_\\ell$ камней; Вася в ответ возвьмёт $c_\\ell$ камней. Теперь Петя может взять $n(\\ell - k)$ камней, оставляя $a_\\ell - c_\\ell - n(\\ell - k) = a_k - c_k$, и дальше действовать по вышеупомянутой Васиной стратегии. Таким образом, он выиграет — противоречие.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39655,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $a$, by $P(a)$ we denote the maximal prime divisor of $a^2 + 1$. Prove that there exist infinitely many triples of distinct positive integers $a, b, c$ such that $P(a) = P(b) = P(c)$.\n\nДля натурального $a$ обозначим через $P(a)$ наибольший простой делитель числа $a^2 + 1$. Докажите, что существует бесконечно много троек различных натуральных чисел $a, b, c$ таких, что $P(a) = P(b) = P(c)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Сделаем сначала замечание, общее для всех трёх решений. Пусть $p$ — нечётное простое число, а $a < p$ — натуральное число такое, что $a^2 + 1$ делится на $p$; тогда числа $a$ и $p-a$ различны, и $P(a) = P(p-a) = p$. Действительно, числа $a^2 + 1$ и $(p-a)^2 + 1 = (a^2 + 1) + p(p - 2a)$ делятся на $p$ и меньше $p^2$; значит, они не могут делиться на простые числа, большие $p$.\n\nПервое решение. Предположим противное. Тогда существует лишь конечное число простых чисел $p$, для которых уравнение $P(x) = p$ имеет хотя бы три натуральных решения. Обозначим через $s$ максимальное такое простое число (если таких простых не существует, положим $s = 2$), а через $S$ — произведение всех простых чисел, не превосходящих $s$.\n\nПусть $p = P(S)$; тогда $p$ взаимно просто с $S$ и потому $p > s$. Пусть $a$ — остаток от деления $S$ на $p$; тогда $a^2 + 1$ делится на $p$, значит, $P(a) = P(p-a) = p$. Одно из чисел $a$ и $p-a$ чётно; обозначим его через $b$.\n\nДалее, число $(b+p)^2 + 1$ делится на $2p$ (ибо $b$ чётно, а $p$ нет), поэтому $P(b+p) \\ge p$. Если $P(b+p) = p$, то уравнение $P(x) = p$ имеет три решения $-b, p-b, p+b$; это невозможно по предположению. Значит, $P(b+p) = q > p$, число $(b+p)^2 + 1$ делится на $2pq$ и потому не меньше, чем $2pq$. Это означает, что $q < b+p$ (в противном случае $(b+p)^2 + 1 \\le (2p-1)q + 1 < 2pq$).\n\nНаконец, обозначая через $c$ остаток от деления числа $b+p$ на $q$, получаем $P(c) = P(q-c) = P(b+p) = q > p > s$, что противоречит выбору $s$.\n\n\nВторое решение. Мы будем использовать тождество\n$$\n(m^2 + 1)((m-1)^2 + 1) = (m^2 - m + 1)^2 + 1, \\quad (*)\n$$\nкоторое можно проверить, например, раскрытием скобок. Из него следует, что $P(m^2 - m + 1) = \\max(P(m), P(m+1))$.\n\nПредположим противное. Пусть $N$ — наибольшее число, встречающееся в описанных тройках; если таких троек нет, то положим $N = 3$. Последовательность натуральных чисел $P(N+1), P(N+2), \\dots$ не может строго убывать. Значит, найдётся число $n > N + 1$, для которого $P(n-1) \\le P(n)$. Тогда $P(n^2-n+1) = \\max(P(n), P(n-1)) = P(n)$. Поэтому найдётся число $n \\le m \\le n^2-n+1$ такое, что $P(m-1) \\le P(m) \\ge P(m+1)$; иначе $P(n-1) \\le P(n) < P(n+1) < \\dots < P(n^2-n+1)$, что не так.\n\nТеперь из (*) имеем $P(m^2-m+1) = \\max(P(m), P(m-1)) = P(m)$ и $P(m^2 + m + 1) = \\max(P(m), P(m+1)) = P(m)$. Таким образом, тройка $m, m^2-m+1, m^2+m+1$ удовлетворяет условию, и $m > N$; противоречие с выбором числа $N$.\n\n\nТретье решение. Для любого натурального $n \\ge 1$ рассмотрим число $(2+\\sqrt{5})^{2n+1}$; оно имеет вид $a_n + b_n\\sqrt{5}$ при некоторых натуральных $a_n, b_n$ (ясно, что $a_n < a_{n+1}$). Заметим, что тогда $(2 - \\sqrt{5})^{2n+1} = a_n - b_n\\sqrt{5}$, откуда\n$$\na_n^2 - 5b_n^2 = (a_n + b_n\\sqrt{5})(a_n - b_n\\sqrt{5}) = ((2+\\sqrt{5})(2-\\sqrt{5}))^{2n+1} = -1.\n$$\nТогда $a_n^2 + 1 = 5b_n^2$; ясно, что $b_n < a_n$ и $a_n \\ge 2^{2n+1} \\ge 8$, поэтому все простые делители числа $a_n^2 + 1$ не превосходят $\\max(5, b_n) < a_n$.\n\nИтак, $p_n = P(a_n) < a_n$. С другой стороны, $a_n^2 + 1 = 5$, значит, $p_n \\ge 5$. Обозначим теперь через $c_n$ остаток от деления $a_n$ на $p_n$. Тогда числа $c_n, p_n - c_n, a_n$ различны и $P(c_n) = P(p_n - c_n) = P(a_n)$. Мы предъявили бесконечно много различных троек требуемого вида.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Basil drew $n$ circles in a plane. After that, for every two circles he drew all their common tangents. It appears that the drawn lines contain all the sides of some regular 2011-gon. Find the least possible value of $n$. (N. Agakhanov)\n\nВася нарисовал на плоскости несколько окружностей и провёл всевозможные общие касательные к каждой паре этих окружностей. Оказалось, что проведённые прямые содержат все стороны некоторого правильного 2011-угольника. Какое наименьшее количество окружностей мог нарисовать Вася? (Н. Агаханов)",
    "options": [],
    "answer": "504",
    "solution": "**Ответ.** 504.\n\nОбозначим полученный правильный 2011-угольник через $M$, его вершины (по часовой стрелке) — через $X_1, X_2, \\dots, X_{2011}$, его вписанную окружность через $\\omega$, а его центр — через $O$. Назовём прямые, содержащие стороны многоугольника, выделенными.\n\nЗаметим, что для любых пяти последовательных вершин $A$, $B$, $C$, $D$, $E$ многоугольника $M$ существует окружность, отличная от $\\omega$, касающаяся прямых $AB$, $BC$, $CD$ и $DE$ (см. рис. 9). Действительно, вершины $A$ и $E$, а также $B$ и $D$ симметричны относительно прямой $CO$. Тогда точка пересечения внешней биссектрисы угла $ABC$ с прямой $CO$ отлична от $O$ и равноудалена от прямых $AB$, $BC$, $CD$ и $DE$, а значит, является центром искомой окружности.\n\nТеперь, если Вася нарисует 503 такие окружности для точек $(X_1, X_2, X_3, X_4, X_5)$, $(X_5, X_6, X_7, X_8, X_9)$, $\\dots$, $(X_{2009}, X_{2010}, X_{2011}, X_1, X_2)$, а также окружность $\\omega$, то любая выделенная прямая будет общей касательной к двум проведённым окружностям. Итак, 504 окружностей достаточно.\n\n![](attached_image_1.png)\n\nОсталось доказать, что окружностей должно быть не менее 504. Каждой выделенной прямой должны касаться хотя бы две окружности. Окружность $\\omega$ касается всех 2011 этих прямых. У любой другой окружности есть не более четырёх общих касательных с $\\omega$; значит, она касается не более четырёх выделенных прямых. Итак, если окружностей $n$, то всего происходит не более, чем $2011 + 4(n-1)$ касаний окружности с выделенными прямыми; с другой стороны, их должно быть не меньше $2011 \\cdot 2 = 4022$. Итак, $2011 + 4(n-1) \\ge 2 \\cdot 2011$, откуда $n \\ge 2011/4 + 1 > 503$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle. Let $N$ be the midpoint of arc $BAC$ of its circumcircle, and let $M$ be the midpoint of $BC$. Denote by $I_1$ and $I_2$ the incenters of triangles $ABM$ and $ACM$ respectively. Prove that the points $I_1, I_2, A, N$ are concyclic. (M. Kungozhin)\n\nДан неравнобедренный треугольник **АВС**. Пусть **N** — середина дуги **ВАС** его описанной окружности, а **М** — середина сторон **ВС**. Обозначим через $I_1$ и $I_2$ центры вписанных окружностей треугольников **АВМ** и **АСМ** соответственно. Докажите, что точки $I_1, I_2, A, N$ лежат на одной окружности.\n(М. Кунгожин)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $I$ — центр вписанной окружности треугольника $ABC$, а $J_1$ и $J_2$ — центры его вневписанных окружностей $\\omega_1$ и $\\omega_2$, касающихся сторон $AB$ и $AC$, соответственно. Прямая $AN$ является внешней биссектрисой угла $BAC$, поэтому точки $J_1$ и $J_2$ лежат на ней. Пусть $K_1$ и $K_2$ — точки касания $\\omega_1$ и $\\omega_2$ соответственно с прямой $BC$; тогда прямые $NM, J_1K_1$ и $J_2K_2$ перпендикулярны $BC$. Кроме того, $BK_1 = \\frac{AB + AC - BC}{2} = CK_2$, поэтому $MK_1 = MK_2$. По теореме Фалеса получаем $NJ_1 = NJ_2$.\n\nДалее, $\\angle J_1BJ_2 = \\angle J_1CJ_2 = 90^\\circ$ как углы между внутренней и внешней биссектрисами. Значит, точки $B$ и $C$ лежат на окружности с диаметром $J_1J_2$, поэтому $\\angle BCJ_1 = \\angle BJ_2J_1$. Тогда треугольники $IBC$ и $IJ_1J_2$ подобны по двум углам, а точки $M$ и $N$ соответствуют в этих треугольниках как середины сторон.\n\nПусть описанная окружность $\\gamma$ треугольника $AI_1I_2$ пересекает вторично прямые $CI$ и $BI$ в точках $P_1$ и $P_2$ соответственно (см. рис. 22).\n\n![](attached_image_1.png)\n\nЗаметим, что $\\angle I_1P_1I_2 = \\angle I_1P_2I_2 = \\angle I_1AI_2 = \\frac{1}{2}\\angle BAC$. С другой стороны, $\\angle BIP_1 = \\angle IBC + \\angle ICB = \\frac{1}{2}(\\angle ABC + \\angle ACB) = 90^\\circ - \\frac{1}{2}\\angle BAC = 90^\\circ - \\angle I_1P_1I$; значит, $\\angle P_1I_1P_2 = \\angle P_1I_2P_2 = 90^\\circ$, и точки $P_1$ и $P_2$ диаметрально противоположны на $\\gamma$. Кроме того, прямые $P_1I_1$ и $BJ_1$ перпендикулярны $BJ_2$, поэтому $\\frac{II_1}{IP_1} = \\frac{IB}{IJ_1}$, и точки $I_1$ и $P_1$ соответствуют в треугольниках $IBC$ и $IJ_1J_2$. Аналогично, точки $I_2$ и $P_2$ также соответствуют, откуда $\\angle P_1NP_2 = \\angle I_1MI_2 = 90^\\circ$. Это значит, что точка $N$ лежит на окружности с диаметром $P_1P_2$, то есть на $\\gamma$. Это и требовалось доказать.\n\n\n**Второе решение.**\n\nЗаметим, что прямоугольные треугольники $BMN$ и $CMN$ симметричны относительно $MN$. Пусть точка $I'_2$ симметрична $I_2$ относительно $MN$. Имеем $\\angle BMI_1 + \\angle BMI'_2 = \\angle BMI_1 + \\angle CMI_2 = \\frac{1}{2}(\\angle BMA + \\angle CMA) = 90^\\circ = \\angle BMN$, поэтому лучи $MI_1$ и $MI'_2$ симметричны относительно биссектрисы угла $BMN$. Далее, $\\angle MBI_1 + \\angle MBI'_2 = \\angle MBI_1 + \\angle MCI_2 = \\frac{1}{2}(\\angle MBA + \\angle MCA) = 90^\\circ - \\frac{1}{2}\\angle BAC = 90^\\circ - \\gamma$.\n\n$-\\frac{1}{2} \\angle BNC = 90^\\circ - \\angle BNM = \\angle MBN$, поэтому лучи $BI_1$ и $BI'_2$ симметричны относительно биссектрисы угла $MBN$.\n\nОтсюда следует, что точки $I_1$ и $I'_2$ — изогонально сопряженные точки в треугольнике $BMN$, а следовательно, и лучи $NI_1$ и $NI'_2$ симметричны относительно биссектрисы угла $MNB$. Значит,\n$$\n\\angle BNM = \\angle MNI_1 + \\angle MNI'_2. \\text{ Получаем: } \\angle I_1AI_2 = \\frac{1}{2} \\angle BAC = \\angle BNM = \\angle MNI_1 + \\angle MNI'_2 = \\angle MNI_1 + \\angle MNI_2 = I_1NI_2.\n$$\n\nЭто и означает, что точки $I_1, I_2, A, N$ лежат на одной окружности.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be three positive numbers such that $a + b^2 + c^2 = a^2 + b + c^2 = a^2 + b^2 + c$. Determine whether all three numbers $a, b, c$ should necessarily be equal. (L. Emelyanov)\n\nПро три положительных числа известно, что если выбрать одно из них и прибавить к нему сумму квадратов двух других, то получится одна и та же сумма, независимо от выбранного числа. Верно ли, что все числа равны? (Л. Емельянов)",
    "options": [],
    "answer": "No; for example, a = 1/3, b = 1/3, c = 2/3.",
    "solution": "**Ответ.** Нет, неверно.\nПодойдёт, например, тройка $\\frac{1}{3}, \\frac{1}{3}, \\frac{2}{3}$.\n\n**Замечание.** Все такие тройки можно получить, решив соответствующую систему: $a + b^2 + c^2 = a^2 + b + c^2 = a^2 + b^2 + c$. Из первых двух равенств имеем $a^2 - a = b^2 - b$; перенося всё в левую часть, получаем $(a - b)(a + b - 1) = 0$. Значит, $a = b$ или $b = 1 - a$; аналогичные утверждения верны для остальных пар чисел. Итого, кроме троек из равных чисел, подходят все тройки вида $a, a, 1 - a$ и только они.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a quadratic polynomial with a unit leading coefficient. Given that the polynomials $P(x)$ and $P(P(P(x)))$ have a common root, prove that $P(0)P(1) = 0$. (A. Khrabrov)\n\nКвадратный трёхчлен $P(x)$ с единичным старшим коэффициентом таков, что многочлены $P(x)$ и $P(P(P(x)))$ имеют общий корень. Докажите, что $P(0) \\cdot P(1) = 0$. (А. Храбров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $t$ — общий корень данных многочленов. Тогда $0 = P(P(P(t))) = P(P(0))$. Пусть $P(x) = x^2 + a x + b$; тогда $P(0) = b$, $P(1) = a + b + 1$, а значит, $0 = P(P(0)) = P(b) = a b + b^2 + b = b(a + b + 1) = P(0) \\cdot P(1)$, что и требовалось доказать.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle. A circle passing through its vertex $B$ and its circumcenter $O$ intersects the sides $BC$ and $BA$ at points $P \\neq B$ and $Q \\neq B$ respectively. Prove that the orthocenter of triangle $POQ$ lies on line $AC$. (T. Emelyanova, L. Emelyanov)\n\nДан остроугольный треугольник $ABC$. Окружность, проходящая через вершину $B$ и центр $O$ его описанной окружности, вторично пересекает стороны $BC$ и $BA$ в точках $P$ и $Q$ соответственно. Докажите, что точка пересечения высот треугольника $POQ$ лежит на прямой $AC$. (Т. Емельянова, ЛІ. Емельянов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим $\\angle OBA = \\angle OAB = \\alpha$, $\\angle OBC = \\angle OCB = \\gamma$; тогда $\\angle ACB = \\angle AOB = 90^\\circ - \\alpha$. Поскольку четырёхугольник $BPOQ$ вписан, $\\angle OPQ = \\alpha$ и $\\angle OQP = \\gamma$. Пусть $OO_1$ — высота треугольника $OPQ$, а $H$ — точка пересечения прямых $OO_1$ и $AC$ (см. рис. 10). Без ограничения общности, точка $H$ лежит на луче $CA$.\n\n![](attached_image_1.png)\n\nРис. 10\n\nУгол $ROH$ — внешний для $\\triangle POO_1$, поэтому $\\angle ROH = 90^\\circ + \\alpha = 180^\\circ - \\angle HCP$. Значит, четырёхугольник $CHOP$ вписан, и $\\angle PHO = \\angle PCO = \\gamma$. Пусть $P_1$ — точка пересечения прямых $OQ$ и $PH$. Вновь по свойству внешних углов $\\angle QPH = \\angle QPH + \\angle PHO_1 = \\angle HO_1Q = 90^\\circ$. Итак, $PH \\perp OQ$, то есть $H$ — точка пересечения высот треугольника $OPQ$. При этом она лежит на прямой $AC$, что и требовалось доказать.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $AB = AC$. Point $D$ is chosen on the smaller arc $AB$ of the circumcircle of $ABC$. Point $E$ is chosen on the ray $AD$ outside the segment $AD$ so that points $A$ and $E$ lie on the same side of line $BC$. The circumcircle of triangle $BDE$ intersects side $AB$ at point $F \\neq B$. Prove that the lines $EF$ and $BC$ are parallel. (R. Zhenodarov)\n\nДан равнобедренный треугольник $ABC$ ($AB = AC$). На меньшей дуге $AB$ описанной около него окружности взята точка $D$. На продолжении отрезка $AD$ за точку $D$ выбрана точка $E$ так, что точки $A$ и $E$ лежат в одной полуплоскости относительно $BC$. Описанная окружность треугольника $BDE$ пересекает сторону $AB$ в точке $F \\neq B$. Докажите, что прямые $EF$ и $BC$ параллельны. (Р. Женодаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим через $\\Omega$ и $\\omega$ описанные окружности треугольников $ABC$ и $BDE$. Положим $\\angle ACB = \\angle ABC = \\alpha$. Четырёхугольник $BDAC$ вписан в $\\Omega$, значит, $\\angle ADB = 180^\\circ - \\angle ACB = 180^\\circ - \\alpha$. Углы $ADB$ и $EDB$ — смежные, откуда $\\angle EDB = 180^\\circ - \\angle ADB = \\alpha$. Далее, поскольку четырёхугольник $EDFB$ вписан в $\\omega$, имеем $\\angle EFB = \\angle EDB = \\alpha$.\n\n![](attached_image_1.png)\n\nИтого, $\\angle EFB = \\angle FBC = \\alpha$, а значит, прямые $EF$ и $BC$ параллельны.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39662,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex $2011$-gon is drawn on a blackboard. Pete draws its diagonals one by one so that each diagonal $d$ should intersect (by an interior point) not more than one diagonal drawn before $d$. Find the maximal possible number of diagonals that Pete can draw according to these rules. (S. Berlov)\n\nНа доске нарисован выпуклый $2011$-угольник. Петя последовательно проводит в нём диагонали так, чтобы каждая вновь проведённая диагональ пересекала по внутренним точкам не более одной из проведённых ранее диагоналей. Какое наибольшее количество диагоналей может провести Петя? (С. Берлов)",
    "options": [],
    "answer": "4016",
    "solution": "**Первое решение.** Покажем, что в выпуклом $n$-угольнике максимальное количество диагоналей, которое можно провести указанным способом, равно $2n - 6$; при $n = 2011$ тогда получится указанный ответ. Пусть $A_1A_2...A_n$ — данный многоугольник. Тогда Петя может провести последовательно диагонали $A_2A_4, A_3A_5, A_4A_6, ..., A_{n-2}A_n$, а затем — диагонали $A_1A_3, A_1A_4, A_1A_5, ..., A_1A_{n-1}$, итого $2n - 6$ диагоналей. На рисунке приведён пример при $n = 9$.\n\nПокажем теперь индукцией по $n$, что больше $2n - 6$ диагоналей в выпуклом $n$-угольнике провести описанным способом нельзя. База при $n = 3$ тривиальна. Для перехода рассмотрим процесс проведения диагоналей в многоугольнике $A_1A_2...A_n$. Пусть для определённости $A_1A_k$ — последняя проведённая диагональ. Тогда по условию она пересекает не более, чем одну проведённую ранее диагональ (обозначим её $d$, если она существует).\n\nДалее, все диагонали, кроме $A_1A_k$ и, возможно, $d$, проводились либо в $k$-угольнике $A_1A_2...A_k$, либо в $(n + 2 - k)$-угольнике $A_kA_{k+1}...A_nA_1$, при этом в каждом из этих многоугольников они проводились с выполнением условий. Значит, по предположению индукции, этих диагоналей не больше $(2k - 6) + (2(n + 2 - k) - 6) = 2n - 8$. Учитывая две диагонали $A_1A_k$ и $d$, получаем, что общее количество не больше $2n - 8 + 2 = 2n - 6$, что и требовалось.\n\n\n**Второе решение.** Приведём другое доказательство того, что в выпуклом $n$-угольнике можно провести не более $2n - 6$ диагоналей с соблюдением условия задачи.\n\nБудем красить проводимые диагонали в красный и синий цвета так. Первую диагональ окрасим синим; далее, если вновь проведённая диагональ пересекает синюю, то окрасим её красным, иначе — синим. Тогда ясно, что одноцветные диагонали не будут пересекаться по внутренним точкам.\n\nДокажем, что диагоналей каждого цвета не больше $n - 3$; отсюда будет следовать, что всего их не более $2(n - 3)$. Действительно, пусть есть $k$ одноцветных диагоналей. Поскольку они не имеют общих внутренних точек, они разбивают $n$-угольник на $k+1$ многоугольников. У каждого многоугольника хотя бы три стороны, значит, суммарное количество $S$ их сторон не меньше $3(k+1)$. С другой стороны, стороны этих многоугольников — это наши диагонали (каждая посчитана по два раза) и стороны исходного $n$-угольника (посчитанные по одному разу). Значит, $S = n + 2k$. Итак, $n + 2k \\ge 3(k+1)$, или $k \\le n - 3$, что и требовалось доказать.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39663,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $8 \\times 8$ chessboard, a closed broken line is drawn; its vertices are the centers of the cells, and each segment connects the centers of two cells sharing a common point (in particular, they may share a common side). This broken line bounds a polygon $P$. Prove that the total area of white parts of $P$ equals the total area of black parts of $P$. (D. Khramtsov)\n\nЧерез центры некоторых клеток шахматной доски $8 \\times 8$ проведена замкнутая несамопересекающаяся ломаная. Каждое звено ломаной соединяет центры соседних по горизонтали, вертикали или диагонали клеток. Докажите, что в ограниченном ею многоугольнике общая площадь чёрных частей равна общей площади белых частей. (Д. Храмцов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Проведём пунктиром вертикальные и горизонтальные линии через центры клеток доски. На получившейся пунктирной сетке каждое звено нашей ломаной соединяет узлы, соседние по вертикали, горизонтали или диагонали. Поэтому пунктирные прямые разбивают область, ограниченную ломаной, на единичные квадратики и половинки квадратиков, получаемые разрезанием их по диагонали.\n\n![](attached_image_1.png)\n\nОсталось заметить, что в каждом таком квадратике и в каждом таком треугольнике площади чёрной и белой частей равны.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39664,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive numbers $x$, $y$, $z$ prove the inequality\n$$\n\\frac{x+1}{y+1} + \\frac{y+1}{z+1} + \\frac{z+1}{x+1} \\le \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x}.\n$$\n\nДаны положительные числа $x$, $y$, $z$. Докажите неравенство\n$$\n\\frac{x+1}{y+1} + \\frac{y+1}{z+1} + \\frac{z+1}{x+1} \\le \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что $\\frac{a+1}{b+1} - \\frac{a}{b} = \\frac{b-a}{b(b+1)}$ для любых положительных $a$ и $b$. Значит, после переноса всех членов в левую часть требуемое неравенство приобретает вид\n$$\n\\frac{y-x}{y(y+1)} + \\frac{z-y}{z(z+1)} + \\frac{x-z}{x(x+1)} \\le 0. \\quad (1)\n$$\nМожно считать, что $x$ — наибольшее из трёх данных чисел. Возможны два случая.\n\nСлучай 1. $y \\ge z$. В этом случае имеем\n$$\n\\frac{x-y}{x(x+1)} \\le \\frac{x-y}{y(y+1)}, \\quad \\frac{y-z}{x(x+1)} \\le \\frac{y-z}{z(z+1)}.\n$$\nСкладывая эти неравенства, получаем\n$$\n\\frac{x-z}{x(x+1)} \\le -\\frac{y-x}{y(y+1)} - \\frac{z-y}{z(z+1)},\n$$\nчто равносильно (1).\n\nСлучай 2. $y < z$. Тогда имеем\n$$\n\\frac{z-y}{z(z+1)} \\le \\frac{z-y}{y(y+1)}, \\quad \\frac{x-z}{x(x+1)} \\le \\frac{x-z}{y(y+1)}.\n$$\nСкладывая эти неравенства, получаем\n$$\n\\frac{z-y}{z(z+1)} + \\frac{x-z}{x(x+1)} \\le -\\frac{y-x}{y(y+1)},\n$$\nчто опять же равносильно (1).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist three coprime positive integer numbers such that a square of each of them is divisible by a sum of the other two?\n\nСуществуют ли три взаимно простых в совокупности натуральных числа, квадрат каждого из которых делится на сумму двух оставшихся?",
    "options": [],
    "answer": "No",
    "solution": "Ответ. Не существуют.\n\nПервое решение. Предположим противное: пусть нашлись такие числа $a, b, c$. Заметим, что числа $a+b, b+c, c+a$ попарно взаимно просты. В самом деле, пусть, скажем, числа $a+b, b+c$ делятся на некоторое простое $p$. Поскольку $c^2: (a+b), a^2: (b+c)$, то числа $c$ и $a$ также делятся на $p$, а тогда и $b = (a+b) - a$ на него делится, что противоречит условию.\n\nДалее, поскольку $a^2: (b+c)$, число $(a+b+c)^2 = a^2 + (b+c)(2a+b+c)$ делится на $b+c$. Аналогично, оно делится на $a+b$ и на $c+a$. Так как последние три числа попарно взаимно просты, $(a+b+c)^2$ делится на $(a+b)(a+c)(b+c)$; в частности, $(a+b+c)^2 \\ge (a+b)(b+c)(c+a)$. С другой стороны, ясно, что все числа $a, b, c$ не меньше 2, значит,\n$$ (a+b)(b+c)(c+a) = (a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)+2abc > (2a^2+2b^2+2c^2)+(2ab+2bc+2ca) > (a+b+c)^2. $$\n\nПротиворечие.\n\nВторое решение. Как и в первом решении, заметим, что числа $a+b, b+c, c+a$ попарно взаимно просты. Пусть $a \\ge b \\ge c$. Заметим, что число $b^2+c^2-a^2 = (b+a)(b-a)+c^2$ делится на $b+a$; аналогично, оно делится на $c+a$. Значит, $b^2+c^2-a^2: (a+b)(a+c)$. С другой стороны,\n$$ -(a+b)(a+c) < -a^2 < b^2 + c^2 - a^2 \\le a^2 < (a+b)(a+c). $$\nТакое может случиться лишь при $b^2 + c^2 - a^2 = 0$, то есть при $a^2 = b^2 + c^2$.\n\nИтак, $a^2 = b^2 + c^2: (b+c)$. Тогда число $2b^2 = (b^2+c^2)+(b-c)(b+c)$ также делится на $b+c$. Поскольку числа $b$ и $b+c$ взаимно просты, $2: (b+c)$, откуда $b=c=1$. Но тогда $1=c^2$ не может делиться на $a+b > 1$, противоречие.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39666,
    "subject": "Mathematics (Multi-modal)",
    "question": "For some $2011$ positive integers, all $2011 \\cdot 2010/2$ of their pairwise sums are written onto a blackboard. Is it possible that exactly a third part of the numbers written are divisible by $3$, and another third part of them has residue $1$ modulo $3$? (I. Bogdanov)\n\nДля некоторых $2011$ натуральных чисел выписали на доску все их $2011 \\cdot 2010/2$ попарных сумм. Могло ли оказаться, что ровно треть выписанных сумм делится на $3$, и ещё ровно треть из них дают остаток $1$ при делении на $3$? (И. Богданов)",
    "options": [],
    "answer": "Yes; for example, take the integers from one to two thousand eleven.",
    "solution": "Могло.\n\nПусть в нашем наборе $a$, $b$ и $c$ чисел, дающих соответственно остатки $0$, $1$ и $2$ при делении на $3$. Тогда условие переписывается в виде\n$$\n\\frac{a(a-1)}{2} + bc = \\frac{b(b-1)}{2} + ac = \\frac{c(c-1)}{2} + ab,\n$$\n$a+b+c = 2011$.\n\nНесложно видеть, что этим условиям удовлетворяет, например, набор чисел $a = c = 670$, $b = 671$, ибо\n$$\n\\frac{671 \\cdot 670}{2} + 670^2 = \\frac{670 \\cdot 669}{2} + 670 + 670^2 = \\frac{670 \\cdot 669}{2} + 670 \\cdot 671.\n$$\n\n**Замечание.** В качестве примера можно привести, например, числа от $1$ до $2011$.\n\nНетрудно и полностью решить приведённую систему. Первое уравнение переписывается в виде $(a-b)(a+b-1-2c) = 0$ или, с учётом равенства $a+b+c = 2011$, $(a-b)(2010-3c) = 0$. Итак, либо $a = b$, либо $c = 670$. С учётом других аналогичных равенств получаем, что все решения имеют вид $(670, 670, 671)$, $(670, 671, 670)$ и $(671, 670, 670)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ such that for every positive integer $n$, the number $a n(n+2)(n+4)$ is also integer. (O. Podlipsky)\n\nНайдите все числа $a$ такие, что для любого натурального $n$ число $a n(n + 2)(n + 4)$ будет целым. (О. Подлипский)",
    "options": [],
    "answer": "a = k/3 for any integer k",
    "solution": "Ответ. $a = \\frac{k}{3}$, где $k$ — любое целое число.\n\nПодставив $n = 1$ и $n = 2$, получаем, что числа $15a$ и $48a$ — целые. Значит, и число $48a - 3 \\cdot 15a = 3a$ — тоже целое. Таким образом, $a = k/3$ для некоторого целого $k$.\n\nОсталось показать, что все числа такого вида подходят. Действительно, одно из трёх последовательных чётных (или нечётных) чисел $n, n + 2, n + 4$ делится на $3$; значит, $n(n + 2)(n + 4)$ делится на $3$, а поэтому $a n(n + 2)(n + 4) = k \\frac{n(n + 2)(n + 4)}{3}$ — целое число.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39668,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a lesson, Pete and Nick write in their exercise books two numbers each: Pete writes $1$ and $2$, while Nick writes $3$ and $4$. Then, at the beginning of each minute, each boy finds some quadratic polynomial such that the numbers in a boy's exercise book are the roots of this polynomial. Let $f(x)$ and $g(x)$ be the polynomials obtained by the boys. Next, if the equation $f(x) = g(x)$ has two different roots $x_1, x_2$, then one of them removes the two numbers from his exercise book and changes them by $x_1$ and $x_2$; otherwise nothing happens.\nAt some moment, Pete has a number $5$ in his exercise book. Find all possible values for the second number in his exercise book at this moment. (I. Bogdanov, A. Garber)\n\nУ Пети и Коли в тетрадях записаны по два числа; изначально — это числа $1$ и $2$ у Пети, $3$ и $4$ — у Коли. Раз в минуту Петя составляет квадратный трёхчлен $f(x)$, корнями которого являются записанные в его тетради два числа, а Коля — квадратный трёхчлен $g(x)$, корнями которого являются записанные в его тетради два числа. Если уравнение $f(x) = g(x)$ имеет два различных корня, то один из мальчиков заменяет свою пару чисел на эти корни, иначе ничего не происходит. Какое второе число могло оказаться у Пети в тетради в тот момент, когда первое стало равным $5$? (И. Богданов, А. Гарбер)",
    "options": [],
    "answer": "14/5",
    "solution": "**Первое решение.** Будем рядом с каждой парой писать какой-нибудь квадратный трёхчлен, корнями которого являются числа этой пары. Пусть в некоторый момент у мальчиков записаны трёхчлены $p(x)$ и $q(x)$. Тогда они решали уравнение вида $\\alpha p(x) = \\beta q(x)$, где $\\alpha, \\beta$ — какие-то ненулевые числа. Значит, полученные числа — корни трёхчлена $\\alpha p(x) - \\beta q(x)$. Если теперь один из мальчиков заменяет свои числа на эти корни, то можно считать, что рядом с ними будет записан трёхчлен $\\alpha p(x) - \\beta q(x)$.\nОбозначим исходные два трёхчлена $p_0(x) = (x-1)(x-2)$ и $q_0(x) = (x-3)(x-4)$. Из сказанного выше теперь следует, что на каждом шаге у каждого мальчика написан трёхчлен вида $\\alpha p_0(x) + \\beta q_0(x)$.\nИтак, если на Петином листке написано число $5$, то у него записан трёхчлен $a(x-5)(x-x_2) = \\alpha(x-1)(x-2)+\\beta(x-3)(x-4)$. Подставляя $x = 5$, получаем $12\\alpha + 2\\beta = 0$, откуда $\\alpha(x-1)(x-2)+\\beta(x-3)(x-4) = \\alpha(-5x^2+39x-70) = -\\alpha(x-5)(5x-14)$. Значит, второе число равно $x_2 = \\frac{14}{5}$.\n\n**Второе решение.** Будем вычитать из каждого из чисел в тетрадях по $\\frac{5}{2}$. Иначе говоря, мы вводим новую переменную $t = x - \\frac{5}{2}$. Тогда первоначальные числа в тетрадях станут равны $-\\frac{3}{2}$, $-\\frac{1}{2}$ у Пети и $\\frac{1}{2}$, $\\frac{3}{2}$ у Васи, а трёхчлены $f(x)$ и $g(x)$ заменятся на некоторые трёхчлены $F(t)$ и $G(t)$.\nПокажем, что теперь произведение пары чисел в любой тетради будет всегда равно $\\frac{3}{4}$. Это выполнено в начальный момент времени. Пусть это верно перед очередной заменой. Согласно теореме Виета, имеем $F(t) = a_1t^2 + b_1t + c_1$, где $\\frac{c_1}{a_1} = \\frac{3}{4}$, и $G(t) = a_2t^2 + b_2t + c_2$, где $\\frac{c_2}{a_2} = \\frac{3}{4}$. Новая пара чисел $t_1$ и $t_2$ — это пара корней уравнения $F(t) = G(t)$, то есть $(a_1 - a_2)t^2 + (b_1 - b_2)t + (c_1 - c_2) = 0$. Опять по теореме Виета получаем\n$$\nt_1t_2 = \\frac{c_1 - c_2}{a_1 - a_2} = \\frac{\\frac{3}{4}a_1 - \\frac{3}{4}a_2}{a_1 - a_2} = \\frac{3}{4},\n$$\nчто и требовалось доказать.\nИтак, если в некоторый момент одно из Петиных чисел равно $t_1 = 5 - \\frac{5}{2} = \\frac{5}{2}$, то второе есть $t_2 = \\frac{3}{4} : \\frac{5}{2} = \\frac{3}{10}$, откуда $x_2 = \\frac{3}{10} + \\frac{5}{2} = \\frac{14}{5}$.\n\n**Замечание.** Описанную ситуацию можно получить даже за один ход, если, например, Петя запишет трёхчлен $x^2 - 3x + 2$, а Вася — трёхчлен $6x^2 - 42x + 72$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39669,
    "subject": "Mathematics (Multi-modal)",
    "question": "At some moment, three different points were marked in the plane. After that, at each minute three marked points are chosen (denote them by $A$, $B$, $C$), and a point $D$ symmetrical to $A$ with respect to the perpendicular bisector to $BC$ is marked.\nA day later, it appeared that there exist three collinear marked points. Prove that three initial points are also collinear. (V. Shmarov)\n\nВначале на плоскости были отмечены три различные точки. Каждую минуту выбирались некоторые три из отмеченных точек — обозначим их $A$, $B$ и $C$, после чего на плоскости отмечалась точка $D$, симметричная $A$ относительно серединного перпендикуляра к $BC$.\nЧерез сутки оказалось, что среди отмеченных точек нашлись три различные точки, лежащие на одной прямой. Докажите, что три исходных точки также лежали на одной прямой. (В. Шмаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное; тогда исходные три точки лежат на некоторой окружности $\\omega$. Докажем индукцией по количеству минут, что все отмеченные точки также лежат на $\\omega$. Действительно, изначально это верно. Пусть в некоторый момент по точкам $A$, $B$, $C$ строится точка $D$. Тогда серединный перпендикуляр $\\ell$ к $BC$ проходит через центр $\\omega$, значит, эта окружность симметрична относительно $\\ell$. Так как точка $A$ лежит на $\\omega$, то и $D$ также на ней лежит.\n\n![](attached_image_1.png)\n\nИтак, через сутки все отмеченные точки лежат на $\\omega$. Но любая прямая пересекает $\\omega$ не более, чем по двум различным точкам; значит, на ней не найдётся трёх отмеченных точек. Противоречие.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a regular triangle. Point $T$ is chosen on its side $AC$, and points $M$ and $N$ are chosen respectively on arcs $AB$ and $BC$ of its circumcircle so that $MT \\parallel BC$ and $NT \\parallel AB$. Segments $AN$ and $MT$ intersect at $X$, while segments $CM$ and $NT$ intersect at $Y$. Prove that the polygons $AXYC$ and $XMBNY$ have equal perimeters. (V. Shmarov)\n\nПусть $ABC$ — правильный треугольник. На его стороне $AC$ выбрана точка $T$, а на дугах $AB$ и $BC$ его описанной окружности выбраны точки $M$ и $N$ соответственно так, что $MT \\parallel BC$ и $NT \\parallel AB$. Отрезки $AN$ и $MT$ пересекаются в точке $X$, а отрезки $CM$ и $NT$ — в точке $Y$. Докажите, что периметры многоугольников $AXYC$ и $XMBNY$ равны. (В. Шмаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Пусть $\\ell$ — касательная к описанной окружности в точке $B$, а $P$ и $Q$ — точки пересечения $\\ell$ с лучами $TM$ и $TN$ соответственно. Обозначим через $K$ и $L$ соответственно точки, в которых лучи $TM$ и $TN$ пересекают стороны треугольника (см. рис. 12).\nЗаметим, что четырёхугольники $ABQT$ и $BCTP$ — параллелограммы, откуда $AT = BQ$, $CT = BP$. Далее, из параллельности имеем $\\angle KPB = \\angle PBK = \\angle LTC = 60^\\circ$, то есть треугольники $BPK$ и $CLT$ — равносторонние. Поскольку $BP = CT$, эти треугольники равны. Далее, из вписанности имеем $\\angle KBM = \\angle ABM = \\angle ACM = \\angle TCY$; значит, точки $M$ и $Y$ — соответственные в этих треугольниках, откуда $BM = CY$ и $PM = LY$. Аналогично, $BN = AX$ и $QN = KX$.\n\nИтак, имеем $XM + YN = KX + KM + LY + LN = QN + KM + PM + LN = QL + KP$. Но $QL = BQ = AT$, а $KP = BP = CT$. Значит, $XM + YN = AT + CT = AC$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nВ итоге, $P_{XMBNY} = (XM + YN) + BM + BN + XY = AC + CY + AX + XY = P_{AXYC}$, что и требовалось доказать.\n\n\n**Второе решение.** Докажем следующую известную лемму.\n**Лемма.** Пусть $PQR$ — правильный треугольник, и на меньшей дуге $PR$ его описанной окружности выбрана точка $W$. Тогда $PW + RW = QW$.\n**Доказательство.** Отметим на отрезке $QW$ точку $V$ такую, что $VW = WR$ (см. рис. 14). Имеем $\\angle VWR = \\angle QPR = 60^\\circ$, значит, треугольник $RVW$ — правильный. Далее, $RQ = RP$, $RV = RW$ и $\\angle PRW = 60^\\circ - \\angle VRP = \\angle QRV$, значит, треугольники $RVQ$ и $RWP$ равны по двум сторонам и углу между ними. Тогда $QV = PW$ и $QW = QV + VW = PW + RW$. Лемма доказана.\n\n![](attached_image_3.png)\n\nПерейдём к решению задачи. Из параллельности имеем $\\angle ATX = \\angle XTY = \\angle CTY = 60^\\circ$. Кроме того, $\\angle AMY = \\angle ABC = 60^\\circ$. Тогда $\\angle AMY + \\angle ATY = 180^\\circ$, значит, точки $A, M, Y, T$ лежат на одной окружности (см. рис. 13).\n\nДалее, $\\angle MAY = \\angle MTY = 60^\\circ$, значит, треугольник $MAY$ — правильный (два из его углов равны по $60^\\circ$), $T$ — точка на дуге $AY$ его описанной окружности. Тогда по лемме $AT+TY = MX+TX$. Аналогично, $CT+TX = NY+TY$. Складывая эти два равенства, получаем $AC = AT+TC = MX+NY$.\nДля треугольника $ABC$ и точки $M$ по лемме получаем $AM + MB = MY + YC$; поскольку $AM = MY$, получаем $CY = MB$. Аналогично, $AX = BN$.\nВ итоге, $P_{AXYC} = AC + AX + CY + XY = (MX + NY) + BN + BM + XY = P_{XMBNY}$, что и требовалось доказать.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of prime numbers $(p, q, r)$ such that $pq \\mid r^4 - 1$, $pr \\mid q^4 - 1$, and $qr \\mid p^4 - 1$.\n\nНайдите все тройки простых чисел $p, q, r$ такие, что четвёртая степень любого из них, уменьшенная на 1, делится на произведение двух остальных.",
    "options": [],
    "answer": "2, 3, 5",
    "solution": "Ответ. $2, 3, 5$.\n\nЯсно, что любые два числа тройки различны (если $p = q$, то $p^4 - 1$ не делится на $q$). Пусть для определённости $p$ — наименьшее из чисел тройки. Нам известно, что число $p^4 - 1 = (p - 1)(p + 1)(p^2 + 1)$ делится на $qr$. Заметим, что $p - 1$ меньше любого из простых чисел $q$ и $r$, а значит, взаимно просто с ними. Далее, число $p^2 + 1$ не может делиться на оба числа $q$ и $r$, так как $p^2 + 1 < (p + 1)(p + 1) < qr$. Значит, $p + 1$ делится на одно из них (для определённости, на $q$). Поскольку $q > p$, это возможно лишь при $q = p + 1$. Тогда одно из чисел $p$ и $q$ чётно, а поскольку оно простое, то $p = 2, q = 3$. Наконец, $r$ является простым делителем числа $p^4 - 1 = 15$, отличным от $q = 3$, значит, $r = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some cells of a $100 \\times 100$ desk contain a token. We call a cell *beautiful* if the total number of tokens in its neighbors is even (two cells are neighbors if they share a common side). Can it appear that there exists exactly one beautiful cell? (K. Knop)\n\nВ некоторых клетках доски 100 × 100 стоит по фишке. Назовём клетку **красивой**, если в соседних с ней по стороне клетках стоит чётное число фишек. Может ли ровно одна клетка доски быть красивой? (К. Кноп)",
    "options": [],
    "answer": "No",
    "solution": "**Лемма.** Для любой клетки доски $X$ существует множество $S$, состоящее из чётного количества клеток и содержащее $X$, такое, что у каждой клетки доски чётное число соседей лежит в $S$.\n\n**Доказательство.** Раскрасим клетки доски в шахматном порядке; можно считать, что $X$ — чёрная. Для начала рассмотрим одну из диагоналей, проходящих через $X$; пусть $A$ и $B$ — центры двух крайних клеток этой диагонали, а $C$ и $D$ — точки, симметричные им относительно центра доски. Тогда обозначим через $S$ множество всех чёрных клеток, центры которых лежат внутри или на границе прямоугольника $ABCD$. На рис. 15 показаны возможные виды множества $S$ на доске $8 \\times 8$ (прямоугольники $ABCD$ обозначены пунктиром).\n\n![](attached_image_1.png)\n\nМножество $S$ состоит из чётного числа клеток, поскольку количества центров клеток на сторонах $AB$ и $AD$ имеют разную чётность. Далее, чёрные клетки не имеют соседей в $S$, каждая белая клетка внутри $ABCD$ граничит с четырьмя клетками из $S$, а каждая белая клетка вне него — либо с нулём, либо с двумя клетками из $S$. Итак, множество $S$ удовлетворяет всем условиям.\n\nПерейдём к решению задачи. Предположим, что существует ровно одна красивая клетка $X$. Рассмотрим для этой клетки множество $S$ из леммы. Для каждой клетки этого множества посчитаем количество фишек в соседних с ней клетках; пусть $g$ — сумма всех этих количеств. С одной стороны, в $S$ чётное число клеток, из которых ровно одна красива, а все остальные — нет; поэтому сумма $g$ нечётна. С другой стороны, каждая клетка с фишкой имеет чётное число соседей в $S$, поэтому она даёт чётный вклад в $g$; значит, и $g$ должна быть чётной. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39673,
    "subject": "Mathematics (Multi-modal)",
    "question": "A straight stick of length $2$ m is cut into $N$ pieces, the length of each (measured in centimeters) is an integer number. Find the least value of $N$ such that it is always possible to make a rectangular contour using all the obtained pieces. (It is not allowed to break the pieces.)\n\nПрямую палку длиной $2$ метра распилили на $N$ палочек, длина каждой из которых выражается целым числом сантиметров. При каком наименьшем $N$ можно гарантировать, что, использовав все получившиеся палочки, можно, не ломая их, сложить контур некоторого прямоугольника?",
    "options": [],
    "answer": "102",
    "solution": "Answer: $N = 102$.\n\nFirst solution. Suppose $N \\leq 101$. Cut the stick into $N-1$ pieces of length $1$ cm and one piece of length $(201 - N)$ cm. From this set, it is impossible to form a rectangle, since each side of the rectangle is less than the semiperimeter, and therefore the piece of length $201 - N \\geq 100$ cm cannot be part of any side. Thus, $N \\geq 102$.\n\nLet us show that for $N = 102$ it is possible to form a rectangle. For this, note that among all the pieces there will be two of length $1$ cm. Indeed, if this were not the case, the total length of the pieces would be at least $2 \\cdot 101 + 1 = 203$ cm, which is not true.\n\nSet aside these two pieces. Let the lengths of the remaining pieces be $a_1, a_2, \\dots, a_{100}$ cm, then $a_1 + a_2 + \\dots + a_{100} = 198$. Among the $100$ numbers $A_1 = a_1, A_2 = a_1 + a_2, A_3 = a_1 + a_2 + a_3, \\dots, A_{100} = a_1 + a_2 + \\dots + a_{100}$, there will be two with the same remainder when divided by $99$. Let these be $A_k$ and $A_\\ell$, $k < \\ell$. The number $A_\\ell - A_k$ is strictly greater than zero and strictly less than $198$, and it is divisible by $99$. Therefore, $A_\\ell - A_k = 99 = a_{k+1} + a_{k+2} + \\dots + a_\\ell$.\n\nThus, we have found several pieces with a total length of $99$ cm. Set these aside as well. The remaining pieces also have a total length of $99$ cm. Therefore, we can form a rectangle of $1 \\times 99$ cm.\n\n\n**Second solution.** Here is another proof that for $N = 102$ it is possible to form a rectangle.\n\nLet the lengths of the pieces, in centimeters, be $a_1, a_2, \\dots, a_{102}$. We have $a_1 + a_2 + \\dots + a_{102} = 200$. Consider a circle of length $200$ and mark $102$ red points dividing it into arcs of lengths $a_1, a_2, \\dots, a_{102}$. These points are some $102$ vertices of a regular $200$-gon $T$ inscribed in this circle. The vertices of $T$ are divided into pairs of opposite points. There are $100$ such pairs, and $102$ red points, so among the red points there are two pairs of opposite points.\n\nThese two pairs of points divide the circle into two pairs of equal arcs. Thus, we have divided all the pieces into four groups $A, B, C, D$, with the total lengths in groups $A$ and $C$, and in groups $B$ and $D$, being equal. Therefore, we can form a rectangle, using each group to make one side.\n\n\n**Third solution.** Here is yet another proof that for $N = 102$ it is possible to form a rectangle. We show how to lay out a rectangle $1 \\times 99$.\n\nLet $\\ell \\geq 2$ be the largest among the lengths of all the pieces. Let $x$ be the number of pieces of length $1$. Then, besides these $x$ pieces and the piece of length $\\ell$, there are $101 - x$ pieces, each of length at least $2$. Hence $\\ell + x + 2(101 - x) \\leq 200$, and $x \\geq \\ell + 2$. So, there are at least $\\ell + 2$ pieces of length $1$. Set aside two pieces of length $1$—use them to make two sides of the rectangle, leaving $\\ell$ pieces of length $1$.\n\nStart laying out the pieces in order of decreasing length to form one side of length $99$. Suppose at some step the row has length $L < 99$, and after adding the next piece of length $m$ it becomes $L + m > 99$. Then remove the last piece of length $m$, and instead place $99 - L$ pieces of length $1$ (this can be done since $99 - L < m \\leq \\ell$). Now we have three sides of the rectangle ($1$, $1$, and $99$). Laying out the remaining pieces in a row, we get another side of length $99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten pairwise distinct positive real numbers are given. Prove that among them there are either three numbers whose product is greater than the product of some other two, or three numbers whose product is greater than the product of some other four.\n\nДаны десять положительных чисел, любые два из которых различны. Докажите, что среди них найдутся либо три числа, произведение которых больше произведения каких-нибудь двух из оставшихся, либо три числа, произведение которых больше произведения каких-нибудь четырех из оставшихся.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Возьмем любые 5 из данных чисел: $a, b, c, d, e$. Если $abc > de$, то утверждение задачи верно. Если же $de \\ge abc$, возьмем еще два числа $f$ и $g$. Пусть скажем $f > g$. Тогда $def > abcg$; значит, и в этом случае утверждение задачи верно.\n\nВторое решение. Упорядочим данные числа по убыванию: $a_1 > a_2 > \\dots > a_{10}$. Тогда $a_1 a_2 a_3 > a_4 a_5 a_6$. Если $a_6 \\ge 1$, то $a_1 a_2 a_3 > a_4 a_5$. Иначе $a_6 < 1$, откуда $a_7 < 1$, и, значит, $a_1 a_2 a_3 > a_4 a_5 a_6 > a_4 a_5 a_6 a_7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39675,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex hexagon $ABCDEF$ is given. It appears that $\\angle FAE = \\angle BDC$, and that each of the quadrilaterals $ABDF$ and $ACDE$ is cyclic. Prove that the lines $BF$ and $CE$ are parallel.\n\nДан выпуклый шестиугольник $ABCDEF$. Известно, что углы $\\angle FAE$ и $\\angle BDC$ равны, а четырехугольники $ABDF$ и $ACDE$ являются вписанными. Докажите, что прямые $BF$ и $CE$ параллельны.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $K$ — точка пересечения отрезков $AE$ и $BF$. Поскольку четырёхугольники $ABDF$ и $ACDE$ вписанные, мы имеем $\\angle AFB = \\angle ADB$ и $\\angle ADC = \\angle AEC$. Отсюда и из условия задачи получаем $\\angle AKB = \\angle AFB + \\angle FAE = \\angle ADB + \\angle BDC = \\angle ADC = \\angle AEC$. Итак, $\\angle AKB = \\angle AEC$. Это и значит, что $BF \\parallel CE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39676,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $a_1, a_2, \\dots$ is defined in the following way: $a_1 = 1$, $a_2 = 143$, and $a_{n+1} = 5 \\cdot \\frac{a_1 + a_2 + \\dots + a_n}{n}$ for $n \\ge 2$. Prove that $a_n$ is an integer for every $n \\ge 1$.\n\nПоследовательность чисел $a_1, a_2, \\dots$ задана условиями $a_1 = 1, a_2 = 143$ и $a_{n+1} = 5 \\cdot \\frac{a_1 + a_2 + \\dots + a_n}{n}$ при всех $n \\ge 2$. Докажите, что все члены последовательности — целые числа.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Show by induction that $a_n = (n+3)(n+2)(n+1)n$ for all $n \\ge 3$.\n\nFirst solution. The number $a_3 = 5 \\cdot 72 = 360$ is an integer. For $n \\ge 4$ the following equalities hold:\n$$\na_n = 5 \\cdot \\frac{a_1 + \\dots + a_{n-1}}{n-1} \\quad \\text{and} \\quad a_{n-1} = 5 \\cdot \\frac{a_1 + \\dots + a_{n-2}}{n-2},\n$$\nfrom which it follows that\n$$\na_n = \\frac{5}{n-1} \\left( \\frac{n-2}{5} a_{n-1} + a_{n-1} \\right) = \\frac{n+3}{n-1} a_{n-1}.\n$$\n\n$$\na_n = \\frac{(n+3)(n+2)\\dots\\cdot 7}{(n-1)(n-2)\\dots\\cdot 3} \\cdot a_3 = \\frac{(n+3)(n+2)(n+1)n}{6 \\cdot 5 \\cdot 4 \\cdot 3} \\cdot 360 = (n+3)(n+2)(n+1)n\n$$\nis an integer, as required.\n\nSecond solution. Let $S_n = a_1 + a_2 + \\dots + a_n$, then $a_{n+1} = S_{n+1} - S_n$.\nTo solve the problem, it is enough to show that all numbers $S_n$ are integers. From the condition we have $S_1 = 1$, $S_2 = 144$, and the formula $a_{n+1} = \\frac{a_1 + a_2 + \\dots + a_n}{n}$ takes the form $S_{n+1} - S_n = \\frac{5S_n}{n}$, whence $S_{n+1} = \\frac{n+5}{n} S_n$. Thus, for $n \\ge 2$ we get\n\n$$\nS_{n+1} = \\frac{(n+5)(n+4)\\dots\\cdot 7}{n(n-1)\\dots\\cdot 2} S_2 =\n$$\n\n$$\n= \\frac{(n+5)(n+4)(n+3)(n+2)(n+1)}{6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2} \\cdot 144 = \\frac{(n+5)(n+4)(n+3)(n+2)(n+1)}{5}\n$$\n\nSince at least one of the numbers $n+5, n+4, n+3, n+2, n+1$ is divisible by $5$, for $n \\ge 2$ the number $S_{n+1}$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39677,
    "subject": "Mathematics (Multi-modal)",
    "question": "$2N$ points are marked on a circle ($N$ is a positive integer) so that no point in the interior of the circle belongs to three chords with the endpoints at marked points. A collection of $N$ chords with the endpoints at marked points is called a *matching* if every marked point is the endpoint of exactly one of these chords. A matching is called *even* if the number of intersection points of its chords is even, and *odd* otherwise. Find the difference between the quantities of even and odd matchings.\n\nНа окружности отмечено $2N$ точек ($N$ — натуральное число). Известно, что через любую точку внутри окружности проходит не более двух хорд с концами в отмеченных точках. Назовем паросочетанием такой набор из $N$ хорд с концами в отмеченных точках, что каждая отмеченная точка является концом ровно одной из этих хорд. Назовём паросочетание чётным, если количество точек, в которых пересекаются его хорды, чётно, и нечётным иначе. Найдите разность между количеством чётных и нечётных паросочетаний.",
    "options": [],
    "answer": "1",
    "solution": "**Answer.** $1$.\n\n**First solution.** By induction on $N$, we prove that the number of even matchings is one more than the number of odd matchings. For $N=1$ the statement is obvious: there is only one matching, and it is even. Now, prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Denote the marked points $A_1, A_2, \\dots, A_{2N}$ in order around the circle clockwise.\n\n**Lemma.** Suppose a matching contains the chord $A_1A_i$. Then for even $i$ it intersects an even number of chords, and for odd $i$ — an odd number.\n**Proof.** Suppose the chord $A_1A_i$ is intersected by exactly $k$ chords. Consider the points $A_2, \\dots, A_{i-1}$; exactly $k$ of them are endpoints of chords intersecting $A_1A_i$ (one endpoint for each such chord). The remaining $i-2-k$ points are split into pairs of points joined by chords that do not intersect $A_1A_i$. Thus, the number $i-2-k$ is even, so $i$ and $k$ have the same parity. The lemma is proved. $\\square$\n\nNow, divide all matchings into $2N-1$ groups $\\Pi_2, \\dots, \\Pi_{2N}$: group $\\Pi_i$ contains those matchings in which point $A_1$ is joined to $A_i$. Now, remove the chord $A_1A_i$ from each matching in $\\Pi_i$; we get all possible matchings on the remaining $2N-2$ points. By the induction hypothesis, among them, the number of even matchings is one more than the number of odd matchings. Moreover, if $i$ is even, then by the lemma the parity of the matching does not change when removing the chord, and if $i$ is odd, it does change. Thus, in each of the $N$ sets $\\Pi_2, \\dots, \\Pi_{2N}$, the number of even matchings is one more than the number of odd, and in each of the $N-1$ sets $\\Pi_3, \\dots, \\Pi_{2N-1}$, the number of odd matchings is one more than the number of even. In total, the number of even matchings exceeds the number of odd by $N-(N-1)=1$, as required.\n\n**Second solution.** We give another proof of the induction step.\nLet the marked points be $A_1, \\dots, A_{2N}$. Consider all matchings in which $A_{2N-1}$ and $A_{2N}$ are joined by a chord. This chord does not intersect any other. Thus, removing it from each such matching, we obtain all matchings on the points $A_1, \\dots, A_{2N-2}$, and the parity of each is preserved.\n\nBy the induction hypothesis, among these matchings, the number of even is one more than the number of odd.\nTo complete the proof, it suffices to show that among all other matchings, the numbers of even and odd are equal. Consider any such matching; suppose it contains chords $A_{2N-1}A_i$ and $A_{2N}A_k$. Now \"swap\" the points $A_{2N-1}$ and $A_{2N}$, i.e., replace these chords with $A_{2N}A_i$ and $A_{2N-1}A_k$. If the original chord intersected some of the others, so does the new chord. On the other hand, if the chords $A_{2N-1}A_i$ and $A_{2N}A_k$ did not intersect, the new chords will intersect, and vice versa. Thus, to each remaining even matching we associate an odd one, and vice versa; and different matchings correspond to different ones. Therefore, the numbers of even and odd matchings among the rest are equal, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39678,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex pentagon $P$ is given. Peter wrote down the five values of sines of the angles of $P$, while Basil wrote down the five values of cosines of the angles of $P$. It appears that among five Peter's numbers, no four are pairwise distinct. Determine whether the five Basil's numbers could be pairwise distinct.\n\nДан выпуклый пятиугольник. Петя выписал в тетрадь значения синусов всех его углов, а Вася — значения косинусов всех его углов. Оказалось, что среди выписанных Петей чисел нет четырёх различных. Могут ли все числа, выписанные Васей, оказаться различными?",
    "options": [],
    "answer": "No",
    "solution": "No, they could not.\n\nSuppose the contrary; then all angles of the pentagon are distinct numbers from the interval $(0, \\pi)$. Note immediately that then Peter cannot have three equal numbers, since in this interval there are not three distinct angles with equal sines.\n\nTherefore, Peter must have two pairs of equal numbers: $\\sin \\alpha = \\sin \\beta$ and $\\sin \\gamma = \\sin \\delta$. Since $\\alpha \\neq \\beta$, we get $\\alpha = \\pi - \\beta$; similarly, $\\gamma = \\pi - \\delta$.\n\nLet $\\varepsilon$ be the fifth angle of the pentagon. Since the sum of the angles of a pentagon is $3\\pi$, we have $\\varepsilon = 3\\pi - (\\alpha + \\beta) - (\\gamma + \\delta) = 3\\pi - \\pi - \\pi = \\pi$. This is impossible, since $\\varepsilon < \\pi$.\nНет, не могут.\n\nПредположим противное; тогда все углы пятиугольника — различные числа из интервала $(0, \\pi)$. Заметим сразу, что тогда у Пети не найдётся трёх равных чисел, ибо в этом интервале нет трёх различных углов с равными синусами.\n\nЗначит, у Пети должны быть две пары равных чисел: $\\sin \\alpha = \\sin \\beta$ и $\\sin \\gamma = \\sin \\delta$. Поскольку $\\alpha \\neq \\beta$, мы получаем $\\alpha = \\pi - \\beta$; аналогично $\\gamma = \\pi - \\delta$.\n\nПусть теперь $\\varepsilon$ — пятый угол пятиугольника. Поскольку сумма углов пятиугольника равна $3\\pi$, имеем $\\varepsilon = 3\\pi - (\\alpha + \\beta) - (\\gamma + \\delta) = 3\\pi - \\pi - \\pi = \\pi$. Это невозможно, так как $\\varepsilon < \\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39679,
    "subject": "Mathematics (Multi-modal)",
    "question": "A collection of $n$ real numbers is written on the blackboard. It appears that the square of each written number is greater than the product of any two other written numbers. Find the greatest possible value of $n$. (I. Bogdanov)\n\nНа доске написаны несколько чисел. Известно, что квадрат любого записанного числа больше произведения любых двух других записанных чисел. Какое наибольшее количество чисел может быть на доске?\n(И. Богданов)",
    "options": [],
    "answer": "3",
    "solution": "$n = 3$.\n\nSuppose there are at least four numbers, and let $a$ be the number with the smallest absolute value. Among the remaining numbers, at least two have the same sign (both nonnegative or both nonpositive). Denote them by $b$ and $c$; then $bc = |bc| \\geq |a|^2 = a^2$, which contradicts the condition.\n\nIt remains to give an example of three numbers satisfying the condition. For example, the numbers $1$, $2$, $-3$ are suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39680,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\omega_1$ and $\\omega_2$ touch each other externally at $P$. Let $\\ell_1$ be a line passing through the center of $\\omega_1$ and tangent to $\\omega_2$. Similarly, a line $\\ell_2$ is tangent to $\\omega_1$ and passes through the center of $\\omega_2$. The lines $\\ell_1$ and $\\ell_2$ are not parallel. Prove that $P$ lies on the angle bisector of some angle formed by $\\ell_1$ and $\\ell_2$. (L. Emelyanov)\n\nОкружности $\\omega_1$ и $\\omega_2$ касаются внешним образом в точке $P$. Через центр $\\omega_1$ проведена прямая $\\ell_1$, касающаяся $\\omega_2$. Аналогично, прямая $\\ell_2$ касается $\\omega_1$ и проходит через центр $\\omega_2$. Оказалось, что прямые $\\ell_1$ и $\\ell_2$ непараллельны. Докажите, что точка $P$ лежит на биссектрисе одного из углов, образованных $\\ell_1$ и $\\ell_2$. (Л. Емельянов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $O_1$, $r_1$ и $O_2$, $r_2$ — соответственно центры и радиусы окружностей $\\omega_1$ и $\\omega_2$, а $K$ — точка пересечения $\\ell_1$ и $\\ell_2$. Заметим, что точка $P$ лежит на отрезке $O_1O_2$ и делит его в отношении $r_1 : r_2$.\n\nОбозначим через $P_1$ точку касания $\\ell_2$ и $\\omega_1$, а через $P_2$ точку касания $\\ell_1$ и $\\omega_2$. Прямоугольные треугольники $KO_1P_1$ и $KO_2P_2$ подобные по острому углу при вершине $K$. Значит, $\\frac{KO_1}{KO_2} = \\frac{O_1P_1}{O_2P_2} = \\frac{r_1}{r_2}$. Таким образом, в треугольнике $KO_1O_2$ точка $P$, лежащая на стороне $O_1O_2$, делит её в отношении, равном отношению прилежащих сторон $KO_1$ и $KO_2$. Из этого следует, что $P$ — основание биссектрисы треугольника $KO_1O_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39681,
    "subject": "Mathematics (Multi-modal)",
    "question": "30 persons, each of whom is either a Knight or a Liar, sit around the table (the places around the table are numbered $1, 2, \\ldots, 30$ in the consecutive order). The Knights always tell the truth, while the Liars always lie. Each person has exactly one friend among the others. Moreover, the friend of a Knight is a Liar, and the friend of a Liar is a Knight. Each person was asked the question \"Is it true that your friend is a neighbor of you?\" Fifteen persons sitting on the places with odd numbers answered \"Yes\". Find the number of persons that sit on the places with even numbers and also answered \"Yes\". (S. Agakhanov)\n\nЗа круглым столом сидят 30 человек — рыцари и лжецы (рыцари всегда говорят правду, а лжецы всегда лгут). Известно, что у каждого из них за этим же столом есть ровно один друг, причем у рыцаря этот друг — лжец, а у лжеца этот друг — рыцарь (дружба всегда взаимна). На вопрос «Сидит ли рядом с вами ваш друг?» сидевшие через одного ответили «да». Сколько из остальных могли также ответить «да»? (Перечислите все варианты и докажите, что других нет.) (С. Агаханов)",
    "options": [],
    "answer": "0",
    "solution": "**Answer.** $0$.\n\nИз условия следует, что все сидящие за столом разбиваются на пары друзей; значит, рыцарей и лжецов поровну. Рассмотрим любую пару друзей. Если они сидят рядом, то рыцарь на заданный вопрос ответит «да», а лжец — «нет». Если же они не сидят рядом, то их ответы будут противоположными. В любом случае ровно один из пары друзей даст ответ «да». Значит, при любой рассадке все остальные 15 ответов будут «нет».",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $a$ and $b$ satisfy the following condition: For any positive integers $m$ and $n$, the number $am^2 + bn^2$ is a perfect square. Prove that $ab = 0$. (A. Golovanov)\n\nЦелые числа $a$ и $b$ таковы, что при любых натуральных $m$ и $n$ число $am^2 + bn^2$ является точным квадратом. Докажите, что $ab = 0$. (А. Голованов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary. Then it is easy to see that both $a$ and $b$ are positive. Now, substituting $m = 2b$, $n = 1$ and $m = 2b$, $n = 2$ we obtain that both numbers $4ab^2 + b = x^2$ and $4ab^2 + 4b = y^2$ are perfect squares. But then $3b = y^2 - x^2 = (y-x)(y+x) \\ge 1 \\cdot 4b$ which is impossible.\n\n\nSolution:\n\nПредположим противное. Ясно, что тогда числа $a$ и $b$ натуральные. Действительно, если, скажем, $a < 0$, то, подставляя $m = 2|b|$, $n = 1$, получаем, что число $4ab^2 + b \\cdot 1 = b(4ab + 1)$ должно быть квадратом некоторого числа $x$. С другой стороны, если $a < 0$, то числа $b$ и $4ab + 1$ имеют разные знаки, так что $x^2 < 0$. Это невозможно.\nПодставим теперь $m = 2b$, $n = 1$ и $m = 2b$, $n = 2$. Получим, что $4ab^2 + b = x^2$ и $4ab^2 + 4b = y^2$ при некоторых натуральных $x, y$. Ясно, что $y^2 > x^2 > 4b^2$, поэтому $y > x > 2b$. Но тогда $3b = y^2 - x^2 = (y - x)(y + x) \\ge 1 \\cdot 4b$, что невозможно. Полученное противоречие завершает доказательство.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39683,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Magician performs the following trick. He lays out 36 cards in a $6 \\times 6$ square (6 columns of 6 cards) and asks the Spectator to select (mentally) a card and remember the column containing it. Then the Magician collects the cards, and again lays out the cards in a $6 \\times 6$ square. After that the Magician asks the Spectator to tell which columns contained the selected card in the first and in the second case, and then guesses the selected card. Find a way for the Magician to perform this trick. (L. Emelyanov)\n\nФокусник выкладывает 36 карт в виде квадрата 6 × 6 (в 6 столбцов по 6 карт) и просит Зрителя мысленно выбрать карту и запомнить столбец, её содержащий. После этого Фокусник определённым образом собирает карты, снова выкладывает в виде квадрата 6 × 6 и просит Зрителя назвать номера столбцов, содержащих выбранную карту в первый и второй раз. После ответа Зрителя Фокусник безошибочно отгадывает карту. Как действовать Фокуснику, чтобы фокус гарантированно удался? (Л. Емельянов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть Фокусник после первого действия не тасует карты, а собирает их, не нарушая порядок в столбцах, и складывает в колоду один столбец за другим. Второй раз он выкладывает карты построчно, т.е. бывшие столбцы становятся строками (это нетрудно сделать, выкладывая ту же колоду по горизонтали). После ответа Зрителя, Фокусник знает номер строки и столбца в первой раскладке, содержащих загаданную карту, что и позволем ему назвать её.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be real numbers such that $a^3 - b^3 = 2$ and $a^5 - b^5 \\ge 4$. Prove that $a^2 + b^2 \\ge 2$. (I. Bogdanov)\n\nЧисла $a$ и $b$ таковы, что $a^3 - b^3 = 2$, $a^5 - b^5 \\ge 4$. Докажите, что $a^2 + b^2 \\ge 2$. (И. Богданов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что $2(a^2 + b^2) = (a^2 + b^2)(a^3 - b^3) = (a^5 - b^5) + a^2b^2(a - b) \\ge 4 + a^2b^2(a - b)$. Поскольку $a^3 > b^3$, мы имеем $a > b$, а значит, $a^2b^2(a - b) \\ge 0$. Итак, $2(a^2 + b^2) \\ge 4$, откуда и следует утверждение задачи.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $AC$ of a triangle $ABC$. Let $E$ and $F$ be the points symmetric to $D$ with respect to the angle bisectors of $\\angle A$ and $\\angle C$, respectively. Prove that the midpoint of the segment $EF$ lies on the line $A_0C_0$, where $A_0$ and $C_0$ are the points of tangency of the incircle with the sides $BC$ and $AB$, respectively. (T. Emelyanova)\n\nНа стороне $AC$ треугольника $ABC$ отметили произвольную точку $D$. Пусть $E$ и $F$ — точки, симметричные точке $D$ относительно биссектрис углов $A$ и $C$ соответственно. Докажите, что середина отрезка $EF$ лежит на прямой $A_0C_0$, где $A_0$ и $C_0$ — точки касания вписанной окружности треугольника $ABC$ со сторонами $BC$ и $AB$ соответственно.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $B_0$ — точка касания вписанной окружности со стороной $AC$. Можно считать, что точка $D$ лежит на отрезке $AB_0$. Точки $A_0$ и $C_0$ симметричны точке $B_0$ относительно биссектрис углов $C$ и $A$ соответственно. Следовательно, точка $E$ лежит на отрезке $AC_0$, а точка $F$ — на продолжении отрезка $CA_0$, и $EC_0 = DB_0 = FA_0$.\n\nОбозначим теперь через $M$ точку пересечения прямых $EF$ и $A_0C_0$ (она лежит на отрезке $EF$). Отметим на прямой $A_0C_0$ точку $G$ так, чтобы отрезки $FG$ и $AB$ были параллельны. Тогда треугольники $FGA_0$ и $BC_0A_0$ подобны; поскольку $BC_0 = BA_0$, получаем $FG = FA_0 = EC_0$. Далее, из параллельности имеем $\\angle FEC_0 = \\angle EFG$ и $\\angle EC_0G = \\angle FGC_0$. Значит, треугольники $EC_0M$ и $FGM$ равны по стороне и двум углам, и $EM = MF$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39686,
    "subject": "Mathematics (Multi-modal)",
    "question": "A grid square $n \\times n$ is given. Initially its cells are colored in white and black in a chessboard manner so that at least one of the corner cells is black. By one move it is allowed to recolor all 4 cells in some square $2 \\times 2$ following the rule: each white cell is recolored to black, each black cell is recolored to green, and each green cell is recolored to white. Find all possible values of $n$ such that by a finite number of moves it is possible to obtain the chessboard coloring of $n \\times n$ square (in black and white) that is inverse to the initial coloring.\n(B. Trushin)\n\nДан квадрат $n \\times n$. Изначально его клетки раскрашены в белый и чёрный цвета в шахматном порядке, причём хотя бы одна из угловых клеток чёрная. За один ход разрешается в некотором квадрате $2 \\times 2$ одновременно перекрасить входящие в него четыре клетки по следующему правилу: каждую белую перекрасить в чёрный цвет, каждую чёрную — в зелёный, а каждую зелёную — в белый. При каких $n$ за несколько ходов можно получить шахматную раскраску, в которой чёрный и белый цвета поменялись местами?\n(Б. Трушин)",
    "options": [],
    "answer": "All positive integers n divisible by 3",
    "solution": "All positive integer $n$ divisible by $3$.\n\nSuppose we managed to recolor the cells as required by the problem. Let us call cells of the first type those that were originally white, and cells of the second type those that were originally black. Note that if a cell is recolored three times, it ends up with its original color. Therefore, for a cell of the first type to become black, it must be recolored $3k + 1$ times for some integer $k$ (for different cells, $k$ may be different). Thus, if $a$ is the number of cells of the first type, the total number of recolorings of these cells is $3K + a$ for some integer $K$. For a cell of the second type to become white, it must be recolored $3m + 2$ times. Thus, if $b$ is the number of cells of the second type, the total number of recolorings of these cells is $3M + 2b$ times.\n\nFurthermore, in any $2 \\times 2$ square, there are equal numbers of cells of the first and second types. Therefore, no matter how we recolor, the total number of recolorings of cells of the first and second types will be equal. Therefore, $3K + a = 3M + 2b$, whence $a + b = 3(M - K + b)$, that is, the total number of cells $a + b$ is divisible by $3$. Thus $n^2$ is divisible by $3$, and therefore $n$ is divisible by $3$.\n\nIt remains to show how to recolor a square with side divisible by $3$. Divide it into squares $3 \\times 3$. Consider one such square. There are two cases: either the corner cells are white, or they are black. In the first case, recolor each of the four squares adjacent to the corners once, and in the second case, recolor each of them twice. It is easy to check that with such recoloring, the chessboard coloring is inverted.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39687,
    "subject": "Mathematics (Multi-modal)",
    "question": "A student got $17$ marks during a week, each one is $2$, $3$, $4$, or $5$. The arithmetic mean of these marks is an integer. Prove that he could not get each mark at least thrice.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose the contrary. Then the student received each of the marks $2$, $3$, $4$, $5$ at least three times. Take three marks of each kind; the sum of these $12$ marks is $42$. Since each of the remaining five marks is at least $2$ and at most $5$, the sum of all $17$ marks is at least $42 + 5 \\cdot 2 = 52$ and at most $42 + 5 \\cdot 5 = 67$. But none of the numbers from $52$ to $67$ is divisible by $17$, since $3 \\cdot 17 = 51$ and $4 \\cdot 17 = 68$. Therefore, the arithmetic mean of all the marks is not an integer. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ученик за одну неделю получил 17 оценок (каждая из них — $2$, $3$, $4$ или $5$). Среднее арифметическое этих 17 оценок — целое число. Докажите, что какую-то оценку он получил не более двух раз.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Допустим противное. Тогда каждую из оценок $2$, $3$, $4$, $5$ ученик получил не меньше трёх раз. Возьмём по три оценки каждого вида; сумма 12 взятых оценок равна $42$. Так как каждая из оставшихся пяти оценок не меньше $2$ и не больше $5$, сумма всех $17$ оценок не меньше $42 + 5 \\cdot 2 = 52$ и не больше $42 + 5 \\cdot 5 = 67$. Но ни одно из чисел от $52$ до $67$ не делится на $17$, поскольку $3 \\cdot 17 = 51$ и $4 \\cdot 17 = 68$. Значит, среднее арифметическое всех оценок — нецелое. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Назовём натуральное число хорошим, если среди его делителей есть ровно два простых числа. Могут ли 18 подряд идущих натуральных чисел быть хорошими?\n(О. Подлипский)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Ответ. Не могут.\n\nПредположим, что нашлись 18 хороших чисел подряд. Среди них найдутся три числа, делящихся на $6$. Пусть это числа $6n$, $6(n+1)$ и $6(n+2)$. Поскольку эти числа — хорошие, и в разложение каждого из них на простые множители входят двойка и тройка, других простых делителей у них быть не может.\n\nДалее, лишь одно из трёх подряд идущих натуральных чисел $n$, $n+1$, $n+2$ может делиться на $3$. Значит, остальные два являются степенями двойки. Но пары степеней двойки, отличающихся не более чем на два — это только $(1,2)$ и $(2,4)$; поэтому $n \\le 2$. Однако тогда среди наших 18 чисел есть простое число $13$ (так как $6n \\le 13 \\le 6(n+2)$), не являющееся хорошим. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39690,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *good* if it has exactly two different prime divisors. Determine whether there exist 18 consecutive positive integers all of which are good. (O. Podlipsky)",
    "options": [],
    "answer": "No, such 18 consecutive integers do not exist.",
    "solution": "Ответ. Не могут.\n\nПредположим, что нашлись 18 хороших чисел подряд. Среди них найдутся три числа, делящихся на $6$. Пусть это числа $6n$, $6(n+1)$ и $6(n+2)$. Поскольку эти числа — хорошие, и в разложение каждого из них на простые множители входят двойка и тройка, других простых делителей у них быть не может.\n\nДалее, лишь одно из трёх подряд идущих натуральных чисел $n$, $n+1$, $n+2$ может делиться на $3$. Значит, остальные два являются степенями двойки. Но пары степеней двойки, отличающихся не более чем на два — это только $(1,2)$ и $(2,4)$; поэтому $n \\le 2$. Однако тогда среди наших 18 чисел есть простое число $13$ (так как $6n \\le 13 \\le 6(n+2)$), не являющееся хорошим. Противоречие.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39691,
    "subject": "Mathematics (Multi-modal)",
    "question": "A 100-digit positive integer $n$ is called *unusual* if the last 100 digits of the decimal representation of $n^3$ form the decimal representation of $n$, but the last 100 digits of the decimal representation of $n^2$ do not. Prove that there are at least two unusual 100-digit numbers. (V. Senderov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For example, such numbers are $n_1 = 10^{100} - 1 = 99\\ldots9$ and $n_2 = \\frac{10^{100}}{2} - 1 = 49\\ldots9$. Indeed, the numbers\n$$\nn_1^3 - n_1 = (n_1 + 1)n_1(n_1 - 1) = 10^{100} \\cdot n_1(n_1 - 1)\n$$\nand\n$$\nn_2^3 - n_2 = (n_2 + 1)n_2(n_2 - 1) = 10^{100} \\cdot n_2 \\cdot \\frac{n_2 - 1}{2}\n$$\nare divisible by $10^{100}$; this means that $n_i^3$ ends with $n_i$. On the other hand, the numbers $n_1^3 - n_1 = n_1(n_1 - 1)$ and $n_2^3 - n_2 = n_2(n_2 - 1)$ are not divisible by 5 (and even less by $10^{100}$); therefore, $n_i^2$ does not end with $n_i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Дана функция $f$, определённая на множестве действительных чисел и принимающая действительные значения. Известно, что для любых $x$ и $y$ таких, что $x > y$, верно неравенство $(f(x))^2 \\le f(y)$. Докажите, что множество значений функции содержится в промежутке $[0, 1]$.\n\n(А. Храбров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "По условию $f(y) \\ge (f(y+1))^2 \\ge 0$ для любого $y$, поэтому все значения функции неотрицательны.\n\nПусть теперь $f(x_0) = 1 + a > 1$ для некоторого $x_0$. Докажем индукцией по $n$, что для любого $y < x_0$ верно неравенство $f(y) > 1 + 2^n a$.\n\nПри $n = 1$ имеем $f(y) \\ge (f(x_0))^2 = 1 + 2a + a^2 > 1 + 2a$.\n\nДля перехода от $n$ к $n+1$ заметим, что $y < \\frac{x_0+y}{2} < x_0$, и поэтому $f\\left(\\frac{x_0+y}{2}\\right) > 1 + 2^n a$ по предположению индукции. А тогда\n\n$$\nf(y) \\ge \\left(f\\left(\\frac{x_0+y}{2}\\right)\\right)^2 = 1 + 2^{n+1}a + (2^n a)^2 > 1 + 2^{n+1}a,\n$$\nчто и требовалось.\n\nИтак, для любого фиксированного $y < x_0$ имеем $f(y) > 1 + 2^n a$ при любом натуральном $n$. Но это невозможно, так как существует $n$, при котором $2^n > \\frac{f(y)-1}{a}$. Стало быть, $f(x) \\le 1$ при всех $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Стозначное натуральное число $n$ назовём необычным, если десятичная запись числа $n^3$ заканчивается на $n$, а десятичная запись числа $n^2$ не заканчивается на $n$. Докажите, что существует не менее двух значных необычных чисел.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Например, такими числами являются $n_1 = 10^{100} - 1 = 99...9$ и $n_2 = \\frac{10^{100}}{2} - 1 = 49...9$. Действительно, числа\n$$n_1^3 - n_1 = (n_1 + 1)n_1(n_1 - 1) = 10^{100} \\cdot n_1(n_1 - 1)$$\nи\n$$n_2^3 - n_2 = (n_2 + 1)n_2(n_2 - 1) = 10^{100} \\cdot n_2 \\cdot \\frac{n_2 - 1}{2}$$\nделятся на $10^{100}$; это означает, что $n_i^3$ оканчивается на $n_i$. С другой стороны, числа $n_1^3 - n_1 = n_1(n_1 - 1)$ и $n_2^3 - n_2 = n_2(n_2 - 1)$ не делятся на 5 (и тем более на $10^{100}$); значит, $n_i^2$ не оканчивается на $n_i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39694,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: \\mathbb{R} \\to \\mathbb{R}$ is given. Assume that $(f(x))^2 \\le f(y)$ for every $x > y$. Prove that each value of $f$ lies in $[0, 1]$. (A. Khrabrov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "По условию $f(y) \\ge (f(y+1))^2 \\ge 0$ для любого $y$, поэтому все значения функции неотрицательны.\n\nПусть теперь $f(x_0) = 1 + a > 1$ для некоторого $x_0$. Доказем индукцией по $n$, что для любого $y < x_0$ верно неравенство $f(y) > 1 + 2^n a$.\n\nПри $n = 1$ имеем $f(y) \\ge (f(x_0))^2 = 1 + 2a + a^2 > 1 + 2a$.\n\nДля перехода от $n$ к $n+1$ заметим, что $y < \\frac{x_0+y}{2} < x_0$, и поэтому $f(\\frac{x_0+y}{2}) > 1 + 2^n a$ по предположению индукции. А тогда\n\n$$\nf(y) \\ge \\left(f\\left(\\frac{x_0+y}{2}\\right)\\right)^2 = 1 + 2^{n+1}a + (2^n a)^2 > 1 + 2^{n+1}a,\n$$\nчто и требовалось.\n\nИтак, для любого фиксированного $y < x_0$ имеем $f(y) > 1 + 2^n a$ при любом натуральном $n$. Но это невозможно, так как существует $n$, при котором $2^n > \\frac{f(y)-1}{a}$. Стало быть, $f(x) \\le 1$ при всех $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39695,
    "subject": "Mathematics (Multi-modal)",
    "question": "В языке племени АУ две буквы — «а» и «у». Некоторые последовательности этих букв являются словами, причём в каждом слове не меньше одной и не больше 13 букв. Известно, что если написать подряд любые два слова, то полученная последовательность букв не будет словом. Найдите максимальное возможное количество слов в таком языке.",
    "options": [],
    "answer": "16256",
    "solution": "**Ответ.** $2^{14} - 2^7 = 16256$.\n\n**Первое решение.** Если все последовательности, количество букв в которых не меньше 7 и не больше 13, являются словами, то, очевидно, условие задачи соблюдается; при этом количество таких слов равно $2^7 + \\dots + 2^{13} = 2^{14} - 2^7$. Осталось показать, что это количество — наибольшее возможное.\n\nОбщее количество последовательностей длины, не превосходящей 13, равно $2 + 2^2 + \\dots + 2^{13} = 2^{14} - 2$. Если среди слов в языке нет ни одного 7-буквенного, то общее количество слов не превосходит $2^{14} - 2 - 2^7 < 2^{14} - 2^7$. Пусть, напротив, в языке существует 7-буквенное слово $s$. Тогда для каждого слова $t$, состоящего из 6 или менее букв, последовательность букв $st$ не может являться словом, и все последовательности вида $st$, очевидно, различны. Значит, если в языке есть $k$ слов из 6 или менее букв, то количество слов из хотя бы 7 букв не превосходит $(2^7 + \\dots + 2^{13}) - k = 2^{14} - 2^7 - k$. Значит, общее количество слов не превосходит $k + (2^{14} - 2^7 - k) = 2^{14} - 2^7$, что и требовалось доказать.\n\n\n**Второе решение.** Приведём другое доказательство того, что слов в языке не больше, чем $2^{14} - 2^7$. Пусть $A$ — множество всех последовательностей из 6 или менее букв, а $B$ — множество всех 7-буквенных последовательностей. Тогда в $A$ всего $2+2^2+\\dots+2^6 = 2^7-2$ последовательностей, а в $B$ всего $2^7 > 2^7 - 2$ последовательностей. Значит, можно сопоставить каждой последовательности $a \\in A$ последовательность $b_a \\in B$ так, чтобы все последовательности $b_a$ были различными. Заметим, что тогда все последовательности вида $ab_a$ также будут различны (поскольку различны их 7-буквенные окончания).\n\nПо условию, в каждой из $2^7 - 2$ троек $(a, b_a, ab_a)$ не больше двух слов языка. Значит, хотя бы $2^7 - 2$ последовательностей из 13 или меньше букв не являются словами, и общее количество слов не больше $(2^{14} - 2) - (2^7 - 2) = 2^{14} - 2^7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39696,
    "subject": "Mathematics (Multi-modal)",
    "question": "A safe contains $n$ boxes numbered from $1$ to $n$. Each box initially contained a card with the number of this box on it. Then Basil rearranged the cards so that the $i$th box now contains a card with number $a_i$. Pete may switch any two cards with numbers $x$ and $y$; for such an operation he should pay $2|x - y|$ roubles. Prove that he may return all the cards into their initial boxes for the price of at most $|a_1 - 1| + |a_2 - 2| + \\dots + |a_n - n|$ roubles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $(b_1, \\dots, b_n)$ be an arbitrary arrangement of the cards (here $b_i$ is the number on the card in the $i$th box). Define its cost as $|b_1 - 1| + |b_2 - 2| + \\dots + |b_n - n|$.\n\n**Lemma.** For any arrangement $(b_1, \\dots, b_n)$ in which not all cards are in their places, it is possible to swap two cards $b_i$ and $b_j$ so that the cost decreases by exactly $2|b_i - b_j|$.\n\n**Proof.** Note that when we perform such a swap, the cost decreases by\n$$\n|b_i - i| + |b_j - j| - |b_i - j| - |b_j - i| = (|b_i - i| - |b_j - i|) + (|b_j - j| - |b_i - j|). \\quad (*)\n$$\nIt is not hard to see that for $(*)$ to be equal to $2|b_i - b_j|$, it is sufficient that $i \\le b_j < b_i \\le j$; then each bracket in $(*)$ equals $|b_i - b_j|$. It remains to find such indices $i$ and $j$.\n\nLet $j$ be the largest number such that the card in box $j$ is not in its place. The card with number $b_j$ is also not in its place, so $b_j < j$. Mark the first $b_j$ boxes. There are exactly $b_j$ cards with numbers not exceeding $b_j$, and one of them (the card with number $b_j$) is already in an unmarked box with number $j > b_j$. Therefore, there exists a marked box $i$ such that $b_i > b_j$. By the choice of $j$, we have $b_i \\le j$, which gives the chain of inequalities $i \\le b_j < b_i \\le j$. $\\square$\n\nThus, for any arrangement different from the required one, it is possible to decrease its cost by some amount $a$, paying exactly $a$. Continuing such actions, we eventually arrive at the arrangement with zero cost, paying in the process exactly the cost of the initial arrangement, as required.\n\n**Remark.** Each bracket in $(*)$ does not exceed $|b_i - b_j|$. Therefore, the cost of the arrangement cannot be decreased by more than the amount paid. Thus, the cost of the arrangement is the minimal number of roubles needed to restore the initial arrangement.\n\n\nSolution 2:\n\nLet us give another proof of the lemma. For each $k$, denote by $c_k$ the number of the box in which the card with number $k$ lies (thus, $c_{b_k} = k = b_{c_k}$).\n\nLet $s$ be the smallest number such that $c_s < s$; such a number exists because $c_1 + \\dots + c_n = 1 + \\dots + n$. Then the card with number $c_s$ is not in its box. Let $t$ be the largest box number less than $s$ such that $c_t \\neq t$. Then necessarily $c_t > t$, since $t < s$; moreover, $t \\ge c_s$ by maximality of $t$. Since for all $t < k < s$ we have $k = c_k \\neq c_t$, from $c_t > t$ it follows that $c_t \\ge s$. Thus, we have $c_s \\le t < s \\le c_t$. Setting $i = c_s, j = c_t$, we obtain the required inequality $i \\le b_j < b_i \\le j$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39697,
    "subject": "Mathematics (Multi-modal)",
    "question": "The language of Au tribe contains only two letters $a$ and $u$. Some sequences of these letters are words, and each word contains not less than $1$ and not more than $13$ letters. Moreover, if one writes down two words consecutively (with no space between them) then the resulting sequence is never a word. Find the maximal possible number of words in such language. (I. Bogdanov)",
    "options": [],
    "answer": "2^14 - 2^7 = 16256",
    "solution": "**Answer.** $2^{14} - 2^7 = 16256$.\n\n**First solution.** If all sequences whose number of letters is at least $7$ and at most $13$ are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^7 + \\dots + 2^{13} = 2^{14} - 2^7$. It remains to show that this number is the largest possible.\n\nThe total number of sequences of length not exceeding $13$ is $2 + 2^2 + \\dots + 2^{13} = 2^{14} - 2$. If there are no $7$-letter words among the words in the language, then the total number of words does not exceed $2^{14} - 2 - 2^7 < 2^{14} - 2^7$. Suppose, on the contrary, that the language contains a $7$-letter word $s$. Then for each word $t$ consisting of $6$ or fewer letters, the sequence of letters $st$ cannot be a word, and all sequences of the form $st$ are obviously distinct. Therefore, if the language has $k$ words of $6$ or fewer letters, then the number of words of at least $7$ letters does not exceed $(2^7 + \\dots + 2^{13}) - k = 2^{14} - 2^7 - k$. Thus, the total number of words does not exceed $k + (2^{14} - 2^7 - k) = 2^{14} - 2^7$, which is what was required to prove.\n\n\n**Second solution.** Here is another proof that the number of words in the language does not exceed $2^{14} - 2^7$. Let $A$ be the set of all sequences of $6$ or fewer letters, and $B$ be the set of all $7$-letter sequences. Then $A$ contains $2 + 2^2 + \\dots + 2^6 = 2^7 - 2$ sequences, and $B$ contains $2^7 > 2^7 - 2$ sequences. Therefore, it is possible to assign to each sequence $a \\in A$ a sequence $b_a \\in B$ so that all sequences $b_a$ are distinct. Note that then all sequences of the form $ab_a$ are also distinct (since their $7$-letter endings are distinct).\n\nBy the condition, in each of the $2^7 - 2$ triples $(a, b_a, ab_a)$ there are at most two words of the language. Therefore, at least $2^7 - 2$ sequences of $13$ or fewer letters are not words, and the total number of words does not exceed $(2^{14} - 2) - (2^7 - 2) = 2^{14} - 2^7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39698,
    "subject": "Mathematics (Multi-modal)",
    "question": "В сейфе $n$ ячеек с номерами от $1$ до $n$. В каждой ячейке первоначально лежала карточка с её номером. Вася переложил карточки в некотором порядке так, что в $i$-й ячейке оказалась карточка с числом $a_i$. Петя может менять местами любые две карточки с номерами $x$ и $y$, платя за это $2|x-y|$ рублей. Докажите, что Петя сможет вернуть все карточки на исходные места, заплатив не более $|a_1 - 1| + |a_2 - 2| + \\dots + |a_n - n|$ рублей.\n(И. Богданов, Г. Иванов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $(b_1, \\dots, b_n)$ — произвольное расположение карточек (здесь $b_i$ — число на карточке в $i$-й ячейке). Назовём его ценой число $|b_1 - 1| + |b_2 - 2| + \\dots + |b_n - n|$.\n\n**Лемма.** Для любого расположения $(b_1, \\dots, b_n)$, в котором не все карточки лежат на своих местах, можно поменять местами некоторые две карточки $b_i$ и $b_j$ так, что цена уменьшится ровно на $2|b_i - b_j|$.\n\n**Доказательство.** Заметим, что при нашем действии цена уменьшается на\n$$\n|b_i - i| + |b_j - j| - |b_i - j| - |b_j - i| = (|b_i - i| - |b_j - i|) + (|b_j - j| - |b_i - j|). (*)\n$$\nНетрудно видеть, что для того, чтобы выражение $(*)$ было равно $2|b_i - b_j|$, достаточно выполнения неравенств $i \\le b_j < b_i \\le j$: тогда каждая из скобок в $(*)$ будет равна $|b_i - b_j|$. Осталось найти такие индексы $i$ и $j$.\n\nПусть $j$ — наибольшее число на карточке, лежащей не в своей ячейке. Карточка с числом $b_j$ также лежит не в своей ячейке, так что $b_j < j$. Отметим первые $b_j$ ячеек. У нас есть ровно $b_j$ карточек с числами, не превосходящими $b_j$, и одна из них (карточка с числом $b_j$) уже лежит в неотмеченной ячейке с номером $j > b_j$. Значит, найдётся такая отмеченная ячейка $i$, что $b_i > b_j$. По выбору номера $j$, имеем $b_i \\le j$, откуда и следует цепочка неравенств $i \\le b_j < b_i \\le j$. $\\square$\n\nИтак, для любого расположения, отличного от требуемого, можно уменьшить его цену на некоторое число $a$, заплатив ровно $a$. Продолжая такие действия, мы в результате придём к расположению с нулевой ценой, заплатив в процессе ровно цену исходного расположения, что и требовалось.\n\n**Замечание.** Каждая скобка в выражении $(*)$ не превосходит $|b_i - b_j|$. Значит, цену расположения нельзя уменьшить на число большее, чем количество заплаченных рублей. Таким образом, цена расположения — это наименьшее число рублей, которое надо заплатить, чтобы привести его к требуемому.\n\nВторое решение.\n\nПриведём другое доказательство леммы. Для каждого $k$ обозначим через $c_k$ номер ячейки, в которой лежит карточка с числом $k$ (таким образом, $c_{b_k} = k = b_{c_k}$).\n\nПусть $s$ — наименьшее число такое, что $c_s < s$; такое число найдётся, поскольку $c_1 + \\dots + c_n = 1 + \\dots + n$. Тогда карточка с числом $c_s$ лежит не в своей ячейке. Пусть теперь $t$ — максимальный номер ячейки, меньший $s$ и такой, что $c_t \\neq t$. Тогда обязательно $c_t > t$, ибо $t < s$; кроме того, $t \\ge c_s$ из максимальности $t$. Поскольку при всех $t < k < s$ имеем $k = c_k \\neq c_t$, из неравенства $c_t > t$ следует $c_t \\ge s$. Итак, мы получили $c_s \\le t < s \\le c_t$. Полагая $i = c_s, j = c_t$, приходим к требуемому неравенству $i \\le b_j < b_i \\le j$.\n\nИтак, для любого расположения, отличного от требуемого, можно уменьшить его цену на некоторое число $a$, заплатив ровно $a$. Продолжая такие действия, мы в результате придём к расположению с нулевой ценой, заплатив в процессе ровно цену исходного расположения, что и требовалось.\n\n**Замечание.** Каждая скобка в выражении $(*)$ не превосходит $|b_i - b_j|$. Значит, цену расположения нельзя уменьшить на число большее, чем количество заплаченных рублей. Таким образом, цена расположения — это наименьшее число рублей, которое надо заплатить, чтобы привести его к требуемому.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39699,
    "subject": "Mathematics (Multi-modal)",
    "question": "На стороне $AB$ треугольника $ABC$ выбраны точки $C_1$ и $C_2$. Аналогично, на стороне $BC$ выбраны точки $A_1$ и $A_2$, а на стороне $AC$ — точки $B_1$ и $B_2$. Оказалось, что отрезки $A_1B_2$, $B_1C_2$ и $C_1A_2$ имеют равные длины, пересекаются в одной точке, и угол между любыми двумя из них равен $60^\\circ$. Докажите, что\n$$\n\\frac{A_1A_2}{BC} = \\frac{B_1B_2}{CA} = \\frac{C_1C_2}{AB}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что\n$$\n\\overrightarrow{A_1B_2} + \\overrightarrow{B_2A_1} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_2C_1} + \\overrightarrow{C_1A_2} + \\overrightarrow{A_2A_1} = \\overrightarrow{0}. \\quad (*)\n$$\nПо условию имеем $A_1B_2 = B_1C_2 = C_1A_2$, и угол между любыми двумя из трех прямых $A_1B_2$, $B_1C_2$, $C_1A_2$ равен $60^\\circ$. Поэтому, если векторы $\\overrightarrow{A_1B_2}$, $\\overrightarrow{B_1C_2}$ и $\\overrightarrow{C_1A_2}$ отложить последовательно друг за другом (каждый следующий от конца предыдущего), то получится правильный треугольник, откуда $\\overrightarrow{A_1B_2} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_1A_2} = \\overrightarrow{0}$. Отсюда и из $(*)$ получаем $\\overrightarrow{A_2A_1} + \\overrightarrow{B_2B_1} + \\overrightarrow{C_2C_1} = \\overrightarrow{0}$.\n\n![](attached_image_1.png)\n\nСледовательно, отложив векторы $\\overrightarrow{A_2A_1}$, $\\overrightarrow{B_2B_1}$, $\\overrightarrow{C_2C_1}$ от некоторой точки последовательно друг за другом, мы получим некоторый треугольник $T$. Стороны треугольника $T$ параллельны соответствующим сторонам треугольника $ABC$, поэтому эти треугольники подобны. Из этого подобия и вытекает требуемое равенство.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Треугольник $ABC$ ($AB > BC$) вписан в окружность $\\Omega$. На сторонах $AB$ и $BC$ выбраны точки $M$ и $N$ соответственно так, что $AM = CN$. Прямые $MN$ и $AC$ пересекаются в точке $K$. Пусть $P$ — центр вписанной окружности треугольника $AMK$, а $Q$ — центр вневписанной окружности треугольника $CNK$, касающейся стороны $CN$. Докажите, что середина дуги $ABC$ окружности $\\Omega$ равноудалена от точек $P$ и $Q$.\n(М. Кулгожин)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Мы будем использовать следующую известную лемму о трезубце.\n**Лемма.** Пусть $L$ — середина дуги $YZ$ (не содержащей точку $X$) описанной окружности треугольника $XYZ$. Пусть $I$ — центр вписанной окружности треугольника $XYZ$, а $I_x$ — центр вневписанной окружности этого треугольника, касающейся стороны $YZ$. Тогда $LY = LZ = LI = LI_x$.\n\nПусть $S$ — середина дуги $ABC$ окружности $\\Omega$. Имеем $SA = SC$ (так как $S$ лежит на серединном перпендикуляре к отрезку $AC$), $AM = CN$ и $\\angle BCS = \\angle BAS$ (как вписанные, опирающиеся на одну дугу). Значит, треугольники $AMS$ и $CNS$ равны, и они совмещаются поворотом $\\Phi$ с центром в точке $S$ на угол $\\angle ASC = \\angle ABC$. Отсюда, в частности, получаем $SM = SN$ и $\\angle MSN = \\angle ABC$. Из последнего равенства углов следует, что четырехугольник $MSBN$ вписан в некоторую окружность $\\gamma$ (см. рис. 11).\n\n![](attached_image_1.png)\nРис. 11\n\nОкружности $\\Omega_a$ и $\\Omega_c$, описанные соответственно около треугольников $AMS$ и $CNS$, также совмещаются поворотом $\\Phi$. Пусть $U$ и $V$ — середины дуг $AM$ и $CN$ (не содержащих $S$) этих окружностей. Из поворота $\\Phi$ получаем $SU = SV$ (то есть точка $S$ лежит на серединном перпендикуляре к $UV$) и $UA = VC$. Далее, из окружностей $\\Omega$ и $\\gamma$ имеем $\\angle SAK = \\angle SBC = \\angle SMK$, то есть $K$ лежит на $\\Omega_a$. Аналогично, точка $K$ лежит на $\\Omega_c$.\n\nОтсюда следует, что точки $U$ и $V$ вместе с точками $P$ и $Q$ лежат на биссектрисе угла $\\angle CKN$. По лемме о трезубце (для треугольников $KAM$ и $KCN$) имеем $UP = UA$ и $VQ = VC$. Так как $UA = VC$, это означает, что точки $P$ и $Q$ симметричны относительно серединного перпендикуляра к $UV$, на котором лежит точка $S$. Значит, $S$ равноудалена от $P$ и $Q$, что и требовалось.\n\n\n**Второе решение.** Пусть $S$ — середина дуги $ABC$, а $R$ и $T$ — середины отрезков $AC$ и $MN$ соответственно. Как и в прошлом решении, получаем, что треугольники $AMS$ и $CNS$ равны, в частности, $SA = SC$ и $SM = SN$. Тогда из равнобедренных треугольников $SAC$ и $SMN$ имеем $\\angle SRK = \\angle STK = 90^\\circ$, то есть точки $R$ и $T$ лежат на окружности $\\Gamma$ с диаметром $SK$ (см. рис. 12). Пусть $\\Gamma$ пересекает биссектрису $PQ$ угла $AKM$ вторично в точке $D$. Поскольку четырёхугольник $KRDT$ вписан в $\\Gamma$, получаем $DR = DT$ и $\\angle ARD = \\angle DTN$.\n\n![](attached_image_2.png)\nРис. 12\n\nПусть $P_1$ и $P_2$ — точки касания вписанной окружности треугольника $AKM$ с прямыми $KA$ и $KM$ соответственно, а $Q_1$ и $Q_2$ — точки касания вневписанной окружности треугольника $CKN$ с такими же прямыми. Из равенства отрезков касательных имеем $RP_1 + TP_2 = RA + AM + TM$. Аналогично, $RQ_1 + TQ_2 = RC + CN + TN$, откуда\n$$\nRP_1 + TP_2 = RQ_1 + TQ_2.\n$$\nС другой стороны, из симметрии имеем:\n$$\nRP_1 + RQ_1 = P_1Q_1 = P_2Q_2 = TP_2 + TQ_2.\n$$\nИз полученных равенств следует, что $RP_1 = TQ_2$ и $RQ_1 = TP_2$.\n\nИтак, мы имеем $DR = DT$, $RP_1 = TQ_2$ и $\\angle P_1RD = \\angle Q_2TD$. Значит, треугольники $DTQ_2$ и $DRP_1$ равны, и $DP_1 = DQ_2$. Поскольку из симметрии $DQ_2 = DQ_1$, получаем, что треугольник $DP_1Q_1$ равнобедренный, его высота из точки $D$ является медианой, и поэтому она также является средней линией прямоугольной трапеции $PQQ_1P_1$. Значит, $DP = DQ$, и в треугольнике $PSQ$ высота $SD$ является медианой. Отсюда $SP = SQ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39701,
    "subject": "Mathematics (Multi-modal)",
    "question": "On each side of a triangle $ABC$, two points are chosen: points $C_1$ and $C_2$ on side $AB$, points $B_1$ and $B_2$ on $AC$, and points $A_1$ and $A_2$ on $BC$. It happened that the segments $A_1B_2$, $B_1C_2$, and $C_1A_2$ have equal length, meet at one point, and the angle between any two of them is $60^\\circ$. Prove that\n$$\n\\frac{A_1A_2}{BC} = \\frac{B_1B_2}{CA} = \\frac{C_1C_2}{AB}.\n$$\n(I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Заметим, что\n$$\n\\overrightarrow{A_1B_2} + \\overrightarrow{B_2A_1} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_2C_1} + \\overrightarrow{C_1A_2} + \\overrightarrow{A_2A_1} = \\overrightarrow{0}. \\quad (*)\n$$\nПо условию имеем $A_1B_2 = B_1C_2 = C_1A_2$, и угол между любыми двумя из трех прямых $A_1B_2$, $B_1C_2$, $C_1A_2$ равен $60^\\circ$. Поэтому, если векторы $\\overrightarrow{A_1B_2}$, $\\overrightarrow{B_1C_2}$ и $\\overrightarrow{C_1A_2}$ отложить последовательно друг за другом (каждый следующий от конца предыдущего), то получится правильный треугольник, откуда $\\overrightarrow{A_1B_2} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_1A_2} = \\overrightarrow{0}$. Отсюда и из $(*)$ получаем $\\overrightarrow{A_2A_1} + \\overrightarrow{B_2B_1} + \\overrightarrow{C_2C_1} = \\overrightarrow{0}$.\n\n![](attached_image_1.png)\n\nСледовательно, отложив векторы $\\overrightarrow{A_2A_1}$, $\\overrightarrow{B_2B_1}$, $\\overrightarrow{C_2C_1}$ от некоторой точки последовательно друг за другом, мы получим некоторый треугольник $T$. Стороны треугольника $T$ параллельны соответствующим сторонам треугольника $ABC$, поэтому эти треугольники подобны. Из этого подобия и вытекает требуемое равенство.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39702,
    "subject": "Mathematics (Multi-modal)",
    "question": "На доске написано уравнение $x^3 + *x^2 + *x + * = 0$. Петя и Вася по очереди заменяют звёздочки на рациональные числа: вначале Петя заменяет любую из звёздочек, потом Вася — любую из двух оставшихся, а затем Петя — оставшуюся звёздочку. Верно ли, что при любых действиях Васи Петя сможет получить уравнение, у которого разность каких-то двух корней равна 2014?",
    "options": [],
    "answer": "Yes",
    "solution": "**Ответ.** Да.\n\n**Первое решение.** Пусть Петя первым ходом сделает свободный член уравнения нулём. Тогда полученное уравнение точно будет иметь корень $0$; значит, Пете достаточно добиться того, чтобы другим корнем было число $t = 2014$. Это всегда можно сделать: если после хода Васи получится уравнение $x^3 + ax^2 + *x = 0$, то Петя может заменить оставшуюся звёздочку на $-t(t + a)$, а если после хода Васи получится $x^3 + *x^2 + bx = 0$, то Петя может заменить звёздочку числом $-(t^2 + b)/t$. Очевидно, все числа, которые ставит Петя, рациональны.\n\n\n**Второе решение.** Пусть Петя первым ходом заменяет звездочку при $x$ на $-s^2$, где $s = 2014/2$. После хода Васи может быть получено уравнение $x^3 + ax^2 - s^2x + * = 0$ или $x^3 + *x^2 - s^2x + c = 0$. В первом случае Петя может заменить оставшуюся звездочку на $-as^2$, а во втором — на $-c/s^2$. В любом случае полученное уравнение будет иметь вид $x^3 + ax^2 - t^2x - at^2$, то есть числа $s$ и $-s$ будут его корнями; разность между этими числами равна $2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Точка $M$ — середина стороны $AC$ треугольника $ABC$. На отрезках $AM$ и $CM$ выбраны точки $P$ и $Q$ соответственно таким образом, что $PQ = \\frac{AC}{2}$. Окружность, описанная около треугольника $ABQ$, пересекает сторону $BC$ в точке $X \\ne B$, а окружность, описанная около треугольника $BCP$, пересекает сторону $AB$ в точке $Y \\ne B$. Докажите, что четырёхугольник $BYMX$ — вписанный. (Ф. Ивлев, Ф. Нилов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку четырёхугольники $BCPY$ и $BAQX$ вписаны, получаем, что $\\angle APY = \\angle ABC = \\angle CQX$. Пусть прямая, проходящая через $M$ параллельно $QX$, пересекает прямую $BC$ в точке $K$, а прямая, проходящая через $M$ параллельно $PY$, пересекает прямую $AB$ в точке $L$ (см. рис. 13).\n\n![](attached_image_1.png)\nРис. 13\n\nТогда $\\angle AML = \\angle ABC = \\angle CMK$, откуда $\\angle ALM = 180^\\circ - \\angle LAM - \\angle AML = 180^\\circ - \\angle BAC - \\angle ABC = \\angle ACB$. Значит, треугольники $MAL$ и $MKC$ подобны по двум углам.\n\nИз условия имеем $AP = AM - PM = PQ - PM = MQ$; аналогично, $CQ = PM$. Отсюда $AY/YL = AP/PM = MQ/QC = KX/XC$. Значит, $Y$ и $X$ — соответственные точки в подобных треугольниках $MAL$ и $MKC$. Следовательно, $\\angle MXC = \\angle MYL = \\angle MYB$. Это и означает, что четырёхугольник $BYMX$ вписан.\n\n\nSolution 2:\n\nВыберем на отрезке $PQ$ точку $Z$ такую, что $CQ = QZ$ (см. рис. 14). Тогда, так как $AC = 2PQ$, имеем $AP + QC = AC - PQ = PQ$, откуда $PZ = PQ - QZ = PQ - QC = AP$.\n\n![](attached_image_2.png)\nРис. 14\n\nИз вписанности четырёхугольника $BYPC$ имеем $AB \\cdot AY = AP \\cdot AC = 2AP \\cdot \\frac{AC}{2} = AZ \\cdot AM$. Аналогично получаем, что $CX \\cdot CB = CZ \\cdot CM$. Если $Z \\ne M$, полученные равенства означают, что каждая из четвёрок точек $B, Y, Z, M$ и $B, X, Z, M$ лежит на одной окружности, то есть точки $X$ и $Y$ лежат на окружности, описанной около треугольника $BMZ$. Отсюда и следует требуемое.\n\nНаконец, если $Z = M$, то те же равенства означают, что точки $X$ и $Y$ лежат на (единственной!) окружности, проходящей через $B$ и касающейся $AC$ в точке $M$, откуда опять же следует требуемое.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Треугольник $ABC$ вписан в окружность $\\Omega$ с центром $O$. Окружность, построенная на $AO$ как на диаметре, пересекает описанную окружность треугольника $OBC$ в точке $S \\neq O$. Касательные к $\\Omega$ в точках $B$ и $C$ пересекаются в точке $P$. Докажите, что точки $A$, $S$ и $P$ лежат на одной прямой.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $CP$ и $BP$ — касательные к $\\Omega$, имеем $\\angle OBP = \\angle OCP = 90^\\circ$; значит, точка $P$ лежит на описанной окружности треугольника $OBC$, и $PO$ — диаметр этой окружности (см. рис. 4). Поэтому $\\angle OSP = 90^\\circ$.\n\nДалее, поскольку $AO$ — диаметр окружности, проходящей через $A$, $S$ и $O$, получаем $\\angle ASO = 90^\\circ$.\n\nИтак, точки $A$ и $P$ лежат на перпендикуляре к $OS$, проходящем через точку $S$. Отсюда и следует утверждение задачи.\n\n![](attached_image_1.png)\nРис. 4",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$. The points $P$ and $Q$ are chosen on the segments $AM$ and $MC$, respectively, so that $PQ = AC/2$. The circumcircle of the triangle $ABQ$ meets the side $BC$ at $X \\neq B$, while the circumcircle of the triangle $BCP$ meets the side $AB$ at $Y \\neq B$. Prove that the quadrilateral $BXMY$ is cyclic. (F. Ivlev, F. Nilov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку четырёхугольники $BCPY$ и $BAQX$ вписаны, получаем, что $\\angle APY = \\angle ABC = \\angle CQX$. Пусть прямая, проходящая через $M$ параллельно $QX$, пересекает прямую $BC$ в точке $K$, а прямая, проходящая через $M$ параллельно $PY$, пересекает прямую $AB$ в точке $L$ (см. рис. 13).\n\n![](attached_image_1.png)\nРис. 13\n\nТогда $\\angle AML = \\angle ABC = \\angle CMK$, откуда $\\angle ALM = 180^\\circ - \\angle LAM - \\angle AML = 180^\\circ - \\angle BAC - \\angle ABC = \\angle ACB$. Значит, треугольники $MAL$ и $MKC$ подобны по двум углам.\n\nИз условия имеем $AP = AM - PM = PQ - PM = MQ$; аналогично, $CQ = PM$. Отсюда $AY/YL = AP/PM = MQ/QC = KX/XC$. Значит, $Y$ и $X$ — соответственные точки в подобных треугольниках $MAL$ и $MKC$. Следовательно, $\\angle MXC = \\angle MYL = \\angle MYB$. Это и означает, что четырёхугольник $BYMX$ вписан.\nВыберем на отрезке $PQ$ точку $Z$ такую, что $CQ = QZ$ (см. рис. 14). Тогда, так как $AC = 2PQ$, имеем $AP + QC = AC - PQ = PQ$, откуда $PZ = PQ - QZ = PQ - QC = AP$.\n\n![](attached_image_2.png)\nРис. 14\n\nИз вписанности четырёхугольника $BYPC$ имеем $AB \\cdot AY = AP \\cdot AC = 2AP \\cdot \\frac{AC}{2} = AZ \\cdot AM$. Аналогично получаем, что $CX \\cdot CB = CZ \\cdot CM$. Если $Z \\neq M$, полученные равенства означают, что каждая из четвёрок точек $B, Y, Z, M$ и $B, X, Z, M$ лежит на одной окружности, то есть точки $X$ и $Y$ лежат на окружности, описанной около треугольника $BMZ$. Отсюда и следует требуемое.\n\nНаконец, если $Z = M$, то те же равенства означают, что точки $X$ и $Y$ лежат на (единственной!) окружности, проходящей через $B$ и касающейся $AC$ в точке $M$, откуда опять же следует требуемое.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ and $O$ be the circumcircle and the circumcenter of a triangle $ABC$. A circle with diameter $AO$ intersects the circumcircle of the triangle $OBC$ at point $S \\neq O$. The tangents to $\\Omega$ at points $B$ and $C$ meet at point $P$. Prove that the points $A$, $S$, and $P$ are collinear. (R. Sadykov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $CP$ и $BP$ — касательные к $\\Omega$, имеем $\\angle OBP = \\angle OCP = 90^\\circ$; значит, точка $P$ лежит на описанной окружности треугольника $OBC$, и $PO$ — диаметр этой окружности (см. рис. 4). Поэтому $\\angle OSP = 90^\\circ$.\n\nДалее, поскольку $AO$ — диаметр окружности, проходящей через $A$, $S$ и $O$, получаем $\\angle ASO = 90^\\circ$.\n\nИтак, точки $A$ и $P$ лежат на перпендикуляре к $OS$, проходящем через точку $S$. Отсюда и следует утверждение задачи.\n\n![](attached_image_1.png)\nРис. 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39707,
    "subject": "Mathematics (Multi-modal)",
    "question": "$10^{1000}$ positive integers are arranged in a circle. One has calculated the least common multiple of every two neighboring numbers. May it happen that these calculated numbers are $10^{1000}$ consecutive positive integers (in some order)? (S. Berlov)",
    "options": [],
    "answer": "No",
    "solution": "**Ответ.** Не могут.\n\nПусть $n = 10^{1000}$. Обозначим исходные числа (в порядке обхода) через $a_1, \\dots, a_n$; мы будем считать, что $a_{n+1} = a_1$. Положим $b_i = \\text{НОК}(a_i, a_{i+1})$. Предположим что числа $b_1, \\dots, b_n$ — это $n$ подряд идущих натуральных чисел.\n\nРассмотрим наибольшую степень двойки $2^m$, на которую делится хотя бы одно из чисел $a_i$. Заметим, что ни одно из чисел $b_1, \\dots, b_n$ не делится на $2^{m+1}$. Пусть для определённости $a_1 \\ge 2^m$; тогда $b_1 \\ge 2^m$ и $b_n \\ge 2^m$. Значит, $b_1 = 2^m x$ и $b_n = 2^m y$ при некоторых нечётных $x$ и $y$. Без ограничения общности можно считать, что $x < y$. Тогда,\n\nпоскольку $b_1, \\dots, b_n$ образуют $n$ последовательных чисел, среди них должно быть и число $2^m(x+1)$ (поскольку $2^m x < 2^m(x+1) < 2^m y$). Но это число делится на $2^{m+1}$ (так как $x+1$ четно), что невозможно. Противоречие.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39708,
    "subject": "Mathematics (Multi-modal)",
    "question": "В республике математиков выбрали число $\\alpha > 2$ и выпустили монеты достоинствами в 1 рубль, а также в $\\alpha^k$ рублей при каждом натуральном $k$. При этом $\\alpha$ было выбрано так, что достоинства всех монет, кроме самой мелкой, иррациональны. Могло ли оказаться, что любую сумму в натуральное число рублей можно набрать этими монетами, используя монеты каждого достоинства не более 6 раз?",
    "options": [],
    "answer": "Могло.",
    "solution": "Ответ. Могло.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39709,
    "subject": "Mathematics (Multi-modal)",
    "question": "По кругу стоят $10^{1000}$ натуральных чисел. Между каждыми двумя соседними числами записали их наименьшее общее кратное. Могут ли эти наименьшие общие кратные образовать $10^{1000}$ последовательных чисел (расположенных в каком-то порядке)?",
    "options": [],
    "answer": "No",
    "solution": "**Ответ.** Не могут.\n\nПусть $n = 10^{1000}$. Обозначим исходные числа (в порядке обхода) через $a_1, \\dots, a_n$; мы будем считать, что $a_{n+1} = a_1$. Положим $b_i = \\text{НОК}(a_i, a_{i+1})$. Предположим что числа $b_1, \\dots, b_n$ — это $n$ подряд идущих натуральных чисел.\n\nРассмотрим наибольшую степень двойки $2^m$, на которую делится хотя бы одно из чисел $a_i$. Заметим, что ни одно из чисел $b_1, \\dots, b_n$ не делится на $2^{m+1}$. Пусть для определённости $a_1 \\ge 2^m$; тогда $b_1 \\ge 2^m$ и $b_n \\ge 2^m$. Значит, $b_1 = 2^m x$ и $b_n = 2^m y$ при некоторых нечётных $x$ и $y$. Без ограничения общности можно считать, что $x < y$.\n\nТогда, поскольку $b_1, \\dots, b_n$ образуют $n$ последовательных чисел, среди них должно быть и число $2^m(x+1)$ (поскольку $2^m x < 2^m(x+1) < 2^m y$). Но это число делится на $2^{m+1}$ (так как $x+1$ четно), что невозможно. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pete has put several tokens into some squares of a checkered $50 \\times 50$ board (at most one token per square). Prove that Bazil can put at most $99$ tokens into empty squares so that each row and each column contains an even number of tokens.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Построим граф с вершинами $r_1, \\dots, r_{50}$, соответствующими строкам доски, и вершинами $c_1, \\dots, c_{50}$, соответствующим её столбцам. Вершины $r_i$ и $c_j$ соединим ребром, если клетка в пересечении соответствующих строки и столбца свободна. Тогда Васина цель переформулируется так: требуется отметить не более $99$ ребер так, чтобы из каждой вершины выходило чётное число непомеченных рёбер. Действительно, если Вася поставит фишки в клетки, соответствующие отмеченным рёбрам, то в каждой строке и каждом столбце останется чётное число свободных клеток, что, очевидно, равносильно требуемому условию.\n\nМы докажем более общий факт: в любом графе на $n \\ge 2$ вершинах можно отметить не более $n-1$ ребра так, чтобы из каждой вершины выходило чётное число непомеченных рёбер. При $n = 100$ получаем требуемое утверждение.\n\nИндукция по $n$. База при $n=2$ очевидна. Пусть теперь $n > 2$. Если есть вершина степени $1$, то можно пометить единственное ребро, выходящее из неё, выкинуть её вместе с этим ребром и применить к оставшемуся графу предположение индукции; в результате окажется отмеченным ещё $n-2$ ребра. Если в графе есть вершина степени $0$, то достаточно выкинуть её и применить предположение индукции.\n\nПусть теперь степень каждой вершины не меньше $2$. Выйдем из произвольной вершины по ребру, из вершины, в которую мы пришли — по другому ребру, и т.д.; этот процесс можно продолжать, пока мы не вернёмся в вершину, в которой уже побывали. Таким образом, в графе нашёлся цикл. Выкинув его рёбра из графа, мы не изменим чётностей степеней вершин; значит, достаточно отметить требуемые рёбра в оставшемся графе. Применяя к нему тот же процесс, рано или поздно мы получим граф, в котором степень некоторой вершины не превосходит $1$; а для таких графов утверждение уже доказано выше.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39711,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane, $n$ convex $k$-gons are given. Every two of them share a common point, and every two of them are homothetic with a positive ratio. Prove that there exists a point belonging to at least $1 + \\frac{n-1}{2k}$ of these $k$-gons.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma.\nLet $P$ and $P'$ be intersecting convex polygons, homothetic with a positive ratio. Then one of the vertices of one of them lies in the other.\n\nProof. If one of the polygons is entirely contained in the other, the statement is obvious. Otherwise, there exists a side $AB$ of polygon $P$ that intersects the boundary of $P'$. If $P'$ contains one of the points $A$ or $B$, the statement is proved. Otherwise, $P'$ intersects $AB$ in a segment lying strictly inside $AB$.\n\nNote that $P'$ has vertices on both sides of the line $AB$. Consider the side $A'B'$ of polygon $P'$ corresponding to $AB$, and an arbitrary vertex $C'$ of polygon $P'$ lying on the opposite side of the line $AB$ from the side $A'B'$. Let $C$ be the vertex of polygon $P$ corresponding to $C'$. Then $C'$ lies in triangle $ABC$, since with respect to each of the lines $AB$, $BC$, and $AC$, it is on the same side as this triangle (see Fig. 15).\n\nThus, $C'$ belongs to $P$. The lemma is proved. $\\square$\n\nLet $P_1, \\dots, P_n$ be the given $k$-gons, and let $A_{i,1}, \\dots, A_{i,k}$ be the vertices of polygon $P_i$. For each vertex $A_{i,j}$, count the number $a_{i,j}$ of polygons $P_s$ ($s \\neq i$) in which it lies. By the lemma, each pair of polygons contributes at least one unit to some $a_{i,j}$.\n\n![](attached_image_1.png)\n\nTherefore, $a_{1,1} + \\dots + a_{n,k} \\ge \\frac{n(n-1)}{2}$. Hence, one of the numbers $a_{i,j}$ is at least $\\frac{n(n-1)}{2nk} = \\frac{n-1}{2k}$. Since the vertex $A_{i,j}$ lies in polygon $P_i$ and also in $a_{i,j}$ other polygons, it belongs to at least $1 + \\frac{n-1}{2k}$ polygons. Thus, this point is as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Петя поставил на доску $50 \\times 50$ несколько фишек, в каждую клетку — не больше одной. Докажите, что у Васи есть способ поставить на свободные поля этой же доски не более 99 новых фишек (возможно, ни одной) так, чтобы по-прежнему в каждой клетке стояло не больше одной фишки, и в каждой строке и каждом столбце этой доски оказалось чётное количество фишек.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Построим граф с вершинами $r_1, \\dots, r_{50}$, соответствующими строкам доски, и вершинами $c_1, \\dots, c_{50}$, соответствующим её столбцам. Вершины $r_i$ и $c_j$ соединим ребром, если клетка в пересечении соответствующих строки и столбца свободна. Тогда Васина цель переформулируется так: требуется отметить не более 99 рёбер так, чтобы из каждой вершины выходило чётное число непомеченных рёбер. Действительно, если Вася поставит фишки в клетки, соответствующие отмеченным рёбрам, то в каждой строке и каждом столбце останется чётное число свободных клеток, что, очевидно, равносильно требуемому условию.\n\nМы докажем более общий факт: в любом графе на $n \\ge 2$ вершинах можно отметить не более $n-1$ рёбра так, чтобы из каждой вершины выходило чётное число непомеченных рёбер. При $n = 100$ получаем требуемое утверждение.\n\nИндукция по $n$. База при $n=2$ очевидна. Пусть теперь $n > 2$. Если есть вершина степени 1, то можно пометить единственное ребро, выходящее из неё, выкинуть её вместе с этим ребром и применить к оставшемуся графу предположение индукции; в результате окажется отмеченным ещё $n-2$ рёбра. Если в графе есть вершина степени 0, то достаточно выкинуть её и применить предположение индукции.\n\nПусть теперь степень каждой вершины не меньше 2. Выйдем из произвольной вершины по ребру, из вершины, в которую мы пришли — по другому ребру, и т.д.; этот процесс можно продолжать, пока мы не вернёмся в вершину, в которой уже побывали. Таким образом, в графе нашёлся цикл. Выкинув его рёбра из графа, мы не изменим чётностей степеней вершин; значит, достаточно отметить требуемые рёбра в оставшемся графе. Применяя к нему тот же процесс, рано или поздно мы получим граф, в котором степень некоторой вершины не превосходит 1; а для таких графов утверждение уже доказано выше.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39713,
    "subject": "Mathematics (Multi-modal)",
    "question": "На плоскости дано $n$ выпуклых попарно пересекающихся $k$-угольников. Любой из них можно перевести в любой другой гомотетией с положительным коэффициентом. Докажите, что на плоскости найдётся точка, принадлежащая хотя бы $1 + \\frac{n-1}{2k}$ из этих $k$-угольников.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Лемма.\nПусть $P$ и $P'$ — пересекающиеся выпуклые многоугольники, гомотетичные с положительным коэффициентом. Тогда одна из вершин одного из них лежит в другом.\n\nДоказательство. Если один из многоугольников полностью лежит в другом, то утверждение очевидно. В противном случае найдётся сторона $AB$ многоугольника $P$, пересекающая границу $P'$. Если $P'$ содержит одну из точек $A$ или $B$, то утверждение доказано. В ином случае $P'$ пересекает $AB$ по отрезку, лежащему внутри $AB$.\n\nЗаметим, что у $P'$ есть вершины по обе стороны от прямой $AB$. Рассмотрим сторону $A'B'$ многоугольника $P'$, соответствующую $AB$, и произвольную вершину $C'$ многоугольника $P'$, лежащую по другую сторону от прямой $AB$, нежели сторона $A'B'$. Пусть $C$ — вершина многоугольника $P$, соответствующая $C'$. Тогда $C'$ лежит в треугольнике $ABC$, так как относительно каждой из прямых $AB, BC$ и $AC$ она находится по ту же сторону, что и этот треугольник (см. рис. 15).\n\n![](attached_image_1.png)\n\nТем самым $C'$ принадлежит $P$. Лемма доказана. $\\square$\n\nПусть $P_1, \\dots, P_n$ — данные $k$-угольники, и пусть $A_{i,1}, \\dots, A_{i,k}$ — вершины многоугольника $P_i$. Для каждой вершины $A_{i,j}$ посчитаем количество $a_{i,j}$ многоугольников $P_s$ ($s \\neq i$), в которых она лежит. Ввиду леммы, каждая пара многоугольников вносит единичный вклад хотя бы в одну из величин $a_{i,j}$. Значит, $a_{1,1} + \\dots + a_{n,k} \\ge \\frac{n(n-1)}{2}$. Поэтому одно из чисел $a_{i,j}$ не меньше $\\frac{n(n-1)}{2nk} = \\frac{n-1}{2k}$. Так как вершина $A_{i,j}$ лежит в многоугольнике $P_i$ и ещё в $a_{i,j}$ других многоугольниках, она принадлежит хотя бы $1 + \\frac{n-1}{2k}$ многоугольникам. Значит, эта точка — требуемая.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39714,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex heptagon is given. One calculates the cosines of four arbitrary its angles, the sines of three remaining angles, and then finds the sum of these 7 numbers. It happens that this sum is independent of the choice of the four angles. Prove that this heptagon has four equal angles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider one of the sums described in the problem. Now, swap the arguments of one sine and one cosine (call these arguments $\\alpha$ and $\\beta$, respectively); the sum changes by\n\n$$(\\sin \\beta + \\cos \\alpha) - (\\sin \\alpha + \\cos \\beta) = \\sqrt{2}(\\sin(\\beta - \\pi/4) - \\sin(\\alpha - \\pi/4)).$$\n\nSince, by the condition, the value of the sum does not change, we get that\n\n$$\\sin(\\alpha - \\pi/4) = \\sin(\\beta - \\pi/4).$$\n\nSince $\\alpha, \\beta \\in (0, \\pi)$, this can happen only if $\\alpha - \\pi/4 = \\beta - \\pi/4$ or $\\alpha - \\pi/4 = \\pi - (\\beta - \\pi/4)$, that is, if $\\beta = \\alpha$ or $\\beta = 3\\pi/2 - \\alpha$.\n\nThus, if $\\alpha$ is an arbitrary angle of the heptagon, then each of its other angles is either equal to $\\alpha$ or $3\\pi/2 - \\alpha$. Therefore, the angles of the heptagon take no more than two distinct values, so 4 of them must be equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Дан выпуклый 7-угольник. Выбираются четыре произвольных его угла и вычисляются их синусы, от остальных трёх углов вычисляются косинусы. Оказалось, что сумма таких семи чисел не зависит от изначального выбора четырёх углов. Докажите, что у этого 7-угольника найдутся четыре равных угла.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Рассмотрим одну из сумм из условия. Затем переставим в ней аргументы одного синуса и одного косинуса (назовём эти аргументы $\\alpha$ и $\\beta$, соответственно); сумма при этом изменится на\n\n$$(\\sin \\beta + \\cos \\alpha) - (\\sin \\alpha + \\cos \\beta) = \\sqrt{2}(\\sin(\\beta - \\pi/4) - \\sin(\\alpha - \\pi/4)).$$\n\nПоскольку по условию значение суммы не изменяется, получаем, что\n$$\\sin(\\alpha - \\pi/4) = \\sin(\\beta - \\pi/4).$$\nПоскольку $\\alpha, \\beta \\in (0, \\pi)$, это может случиться лишь при $\\alpha - \\pi/4 = \\beta - \\pi/4$ или $\\alpha - \\pi/4 = \\pi - (\\beta - \\pi/4)$, то есть при $\\beta = \\alpha$ или $\\beta = 3\\pi/2 - \\alpha$.\n\nИтак, если $\\alpha$ — произвольный угол 7-угольника, то каждый из остальных его углов равен либо $\\alpha$, либо $3\\pi/2 - \\alpha$. Значит, углы 7-угольника принимают не более двух различных значений, поэтому 4 из них принимают одно и то же значение.\n\n**Замечание.** Вместо второй части решения можно заметить, что функция $f(x)$ возрастает на отрезке $[0, 3\\pi/4]$ и убывает на отрезке $[3\\pi/4, \\pi]$; значит, уравнение $f(x) = a$ имеет не более двух решений на интервале $(0, \\pi)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive real number $a$ such that for every real $x$ the inequality\n$$\n|\\cos x| + |\\cos a x| > \\sin x + \\sin a x\n$$\nholds?",
    "options": [],
    "answer": "No",
    "solution": "**Первое решение.** Предположим, что $0 < a \\le 1$. Тогда при $x = \\pi/2$ левая часть примет значение $|\\cos(a\\pi/2)|$, то есть будет не больше 1, в то время как правая часть будет равна $1 + \\sin(a\\pi/2)$, то есть она больше 1. Итак, неравенство не выполнено.\n\nЕсли же $a > 1$, то, обозначив $a x = t$ и $b = 1/a$, мы приведём неравенство из условия к виду $|\\cos b t| + |\\cos t| > \\sin b t + \\sin t$, сводя задачу к предыдущему случаю.\n\n**Второе решение.** Выберем такое $x$, что $x(a+1) = \\pi/2$. Тогда $x$ и $a x$ лежат в интервале $(0, \\pi/2)$, поэтому $|\\cos x| = \\cos x = \\sin a x$ и $|\\cos a x| = \\cos a x = \\sin x$. Значит, для выбранного $x$ имеем $|\\cos x| + |\\cos a x| = \\sin a x + \\sin x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Петя и Вася играют в игру на клетчатой доске $n \\times n$ (где $n > 1$). Изначально вся доска белая, за исключением угловой клетки — она чёрная, и в ней стоит ладья. Игроки ходят по очереди. Каждым ходом игрок передвигает ладью по горизонтали или вертикали, при этом все клетки, через которые ладья перемещается (включая ту, в которую она попадает), перекрашиваются в чёрный цвет. Ладья не должна передвигаться через чёрные клетки или останавливаться на них. Проигрывает тот, кто не может сделать ход; первым ходит Петя. Кто выиграет при правильной игре?",
    "options": [],
    "answer": "Petya",
    "solution": "Одна из выигрышных стратегий для Пети состоит в том, чтобы каждым своим ходом делать самый длинный из возможных вертикальных ходов (например, первым ходом он пойдёт по вертикали в другой угол доски). Покажем, что, действуя согласно ей, он выиграет.\n\nНазовём белую клетку *достижимой*, если из текущего положения ладьи в неё можно попасть за несколько ходов по белым клеткам. Покажем, что в любой момент Петя сможет сходить, причём после каждого его хода все достижимые клетки образуют несколько (не более двух) прямоугольников, в каждом из которых число строк больше числа столбцов, причём для каждого из них ладья стоит в клетке, соседней с угловой по горизонтали. Для первого хода Петя это верно.\n\nДалее, если после некоторого его хода это так, то Вася вынужден ходить в один из полученных прямоугольников по горизонтали. Пусть в этом прямоугольнике $r$ строк и $c$ столбцов, а Вася сходит на $v \\le c$ клеток. После этого хода клетки оставшегося прямоугольника (если он был) перестанут быть достижимыми (см. рис. 16).\n\n![](attached_image_1.png)\n\nПоскольку $r \\ge c+1 \\ge 2$, у Пети остаётся возможность вертикального хода. После этого хода достижимые клетки будут образовывать прямоугольники $r \\times (c-v)$ и $(r-1) \\times (v-1)$. В каждом из них строк больше, чем столбцов, ибо $r > c > c-v$ и $r-1 > c-1 \\ge v-1$. При этом ладья стоит в клетке, соседней по горизонтали с угловой в каждом из этих прямоугольников; это и требовалось доказать.\n\nИтак, мы, в частности, доказали, что Петя всегда сможет сделать ход. При этом когда-нибудь игра закончится, ибо количество белых клеток уменьшается. Значит, он выиграет.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39718,
    "subject": "Mathematics (Multi-modal)",
    "question": "На доске написано выражение $\\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f}$, где $a, b, c, d, e, f$ — натуральные числа. Если число $a$ увеличить на 1, то значение этого выражения увеличится на 3. Если в исходном выражении увеличить число $c$ на 1, то его значение увеличится на 4; если же в исходном выражении увеличить число $e$ на 1, то его значение увеличится на 5. Какое наименьшее значение может иметь произведение $bdf$? (Н. Атаханов)",
    "options": [],
    "answer": "60",
    "solution": "Ответ. 60.\n\n**Первое решение.** Пусть значение исходного выражения равно $A$. Тогда в результате первой операции произведение примет значение $\\frac{a+1}{a} \\cdot A = A + 3$, откуда $A = 3a$. Значит, $A$ — натуральное число. Кроме того, из этого равенства следует, что оно делится на 3.\nАналогично доказывается, что число $A$ делится на 4 и на 5, причём $A = 4c = 5e$. Из попарной взаимной простоты чисел 3, 4 и 5 следует, что $A$ делится на $3 \\cdot 4 \\cdot 5 = 60$. Значит, $A \\ge 60$.\nПереписав равенство $\\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f} = A$ в виде $\\frac{A}{3b} \\cdot \\frac{A}{4d} \\cdot \\frac{A}{5f} = A$, получаем $A^2 = 60bdf$, откуда $bdf = \\frac{A^2}{60} \\ge 60$. Осталось привести пример, показывающий, что произведение знаменателей может быть равным 60. Один из возможных примеров такой: $\\frac{20}{3} \\cdot \\frac{15}{4} \\cdot \\frac{12}{5}$.\n\n**Второе решение.** Как и в первом решении, получаем $A = 3a = 4c = 5e$, откуда\n$$\n\\frac{1}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f} = 3, \\quad \\frac{a}{b} \\cdot \\frac{1}{d} \\cdot \\frac{e}{f} = 4, \\quad \\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{1}{f} = 5.\n$$\nУмножив первое равенство на второе и разделив на третье, получаем, что $\\frac{e^2}{bdf} = \\frac{12}{5}$; поскольку дробь справа несократима, знаменатель $bdf$ делится на 5. Аналогично доказывается, что он делится на 3 и на 4, откуда следует, что он делится на 60, то есть не меньше 60.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39719,
    "subject": "Mathematics (Multi-modal)",
    "question": "An expression $\\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f}$ is written on the blackboard (numbers $a$, $b$, $c$, $d$, $e$, and $f$ are positive integers). If one increases $a$ by $1$, then the value of this expression increases by $3$. If one increases $c$ by $1$ in the initial expression, then its value increases by $4$. If one increases $e$ by $1$ in the initial expression, then its value increases by $5$. Find the minimal possible value of the product $bdf$.\n(N. Agakhanov)",
    "options": [],
    "answer": "60",
    "solution": "**Первое решение.** Пусть значение исходного выражения равно $A$. Тогда в результате первой операции произведение примет значение $\\frac{a+1}{a} \\cdot A = A + 3$, откуда $A = 3a$. Значит, $A$ — натуральное число. Кроме того, из этого равенства следует, что оно делится на $3$. Аналогично доказывается, что число $A$ делится на $4$ и на $5$, причём $A = 4c = 5e$. Из попарной взаимной простоты чисел $3$, $4$ и $5$ следует, что $A$ делится на $3 \\cdot 4 \\cdot 5 = 60$. Значит, $A \\ge 60$.\n\nПереписав равенство $\\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f} = A$ в виде $\\frac{A}{3b} \\cdot \\frac{A}{4d} \\cdot \\frac{A}{5f} = A$, получаем $A^2 = 60bdf$, откуда $bdf = \\frac{A^2}{60} \\ge 60$. Осталось привести пример, показывающий, что произведение знаменателей может быть равным $60$. Один из возможных примеров такой: $\\frac{20}{3} \\cdot \\frac{15}{4} \\cdot \\frac{12}{5}$.\n\n\n**Второе решение.** Как и в первом решении, получаем $A = 3a = 4c = 5e$, откуда\n$$\n\\frac{1}{b} \\cdot \\frac{c}{d} \\cdot \\frac{e}{f} = 3, \\quad \\frac{a}{b} \\cdot \\frac{1}{d} \\cdot \\frac{e}{f} = 4, \\quad \\frac{a}{b} \\cdot \\frac{c}{d} \\cdot \\frac{1}{f} = 5.\n$$\nУмножив первое равенство на второе и разделив на третье, получаем, что $\\frac{e^2}{bdf} = \\frac{12}{5}$; поскольку дробь справа несократима, знаменатель $bdf$ делится на $5$. Аналогично доказывается, что он делится на $3$ и на $4$, откуда следует, что он делится на $60$, то есть не меньше $60$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pete and Basil play the following game on a checkered $n \\times n$ board. Initially the whole board is white except for one corner square which is black; a rook is put onto this square. The players move alternately, Pete moves first. On each turn, a player moves a rook to another square horizontally or vertically. Immediately after that, all the squares passed by a rook (including the square where the move results) become black. It is prohibited to move a rook to or across any black square. The player who cannot move loses. Which of the players has a winning strategy? (N. Polyansky)",
    "options": [],
    "answer": "Pete (the first player) has a winning strategy.",
    "solution": "Одна из выигрышных стратегий для Пети состоит в том, чтобы каждым своим ходом делать самый длинный из возможных вертикальных ходов (например, первым ходом он пойдёт по вертикали в другой угол доски). Покажем, что, действуя согласно ей, он выиграет.\n\nНазовём белую клетку *достижимой*, если из текущего положения ладьи в неё можно попасть за несколько ходов по белым клеткам. Покажем, что в любой момент Петя сможет сходить, причём после каждого его хода все достижимые клетки образуют несколько (не более двух) прямоугольников, в каждом из которых число строк больше числа столбцов, причём для каждого из них ладья стоит в клетке, соседней с угловой по горизонтали. Для первого хода Петя это верно.\n\nДалее, если после некоторого его хода это так, то Вася выпущен ходить в один из полученных прямоугольников по горизонтали. Пусть в этом прямоугольнике $r$ строк и $c$ столбцов, а Вася сходит на $v \\le c$ клеток. После этого хода клетки оставшегося прямоугольника (если он был) перестанут быть достижимыми (см. рис. 16).\n\n![](attached_image_1.png)\n\nПоскольку $r \\ge c+1 \\ge 2$, у Пети остаётся возможность вертикального хода. После этого хода достижимые клетки будут образовывать прямоугольники $r \\times (c-v)$ и $(r-1) \\times (v-1)$. В каждом из них строк больше, чем столбцов, ибо $r > c > c-v$ и $r-1 > c-1 \\ge v-1$. При этом ладья стоит в клетке, соседней по горизонтали с угловой в каждом из этих прямоугольников; это и требовалось доказать.\n\nИтак, мы, в частности, доказали, что Петя всегда сможет сделать ход. При этом когда-нибудь игра закончится, ибо количество белых клеток уменьшается. Значит, он выиграет.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39721,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\dots, n^2$ are put in some order into the squares of a checkered $n \\times n$ board, one number per square. Pete performs several moves according to the following rules. On the first move, he puts a token into some square. By any subsequent move, he may either put a new token into an arbitrary square, or to move a token horizontally or vertically from a square containing some number $a$ to any square containing a number greater than $a$. Every time a token is put onto a square, this square is marked; it is prohibited to put a token into a marked square.\nFind the least number $k$ such that for every arrangement of the numbers, Pete can mark all the squares of the board using at most $k$ tokens.",
    "options": [],
    "answer": "n",
    "solution": "Answer: $n$.\n\nLet us show that $n$ tokens are sufficient. Note that one token is enough for each row: you can put it in the square of the row with the minimal number, and then visit all the squares of the row in order of increasing numbers.\n\nOn the other hand, let us show that fewer than $n$ tokens may not be enough. To do this, number the squares so that the squares of one diagonal are numbered $1, 2, \\dots, n$ (the remaining squares are numbered arbitrarily). Then one token cannot visit two squares of this diagonal: if a token is placed on one of these squares, then on the next move it must go to a square with a number greater than $n$, and after that it cannot return to the diagonal.\n\nFinally, since a token must visit each square of the diagonal, Pete will have to use at least $n$ tokens.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Все клетки квадратной таблицы $n \\times n$ пронумерованы в некотором порядке числами от 1 до $n^2$. Петя делает ходы по следующим правилам. Первым ходом он ставит фишку в любую клетку. Каждым последующим ходом Петя может либо поставить новую фишку на какую-то клетку, либо переставить фишку из клетки с номером $a$ ходом по горизонтали или по вертикали в клетку с номером большим, чем $a$. Каждый раз, когда фишка попадает в клетку, эта клетка немедленно закрашивается; ставить фишку на закрашенную клетку запрещено. Какое наименьшее количество фишек потребуется Пете, чтобы независимо от исходной нумерации он смог за несколько ходов закрасить все клетки таблицы?\n(Д. Храмцов)",
    "options": [],
    "answer": "n",
    "solution": "$n$.\n\nПокажем, что $n$ фишек достаточно. Для этого заметим, что на каждую строку хватит одной фишки: можно поставить её в клетку строки с минимальным номером, а затем обойти все клетки строки в порядке возрастания номеров.\n\nС другой стороны, покажем, что меньше, чем $n$ фишек, может и не хватить. Для этого пронумеруем клетки так, чтобы клетки одной диагонали были пронумерованы $1, 2, \\dots, n$ (остальные клетки нумеруем произвольно). Тогда одна фишка не сможет побывать на двух клетках этой диагонали: если фишка встала на одну из этих клеток, то следующим ходом она обязана будет пойти на клетку с номером, большим $n$, и значит, после этого она не сможет вернуться на диагональ.\n\nНаконец, поскольку на каждой клетке диагонали должна побывать фишка, Пете придётся использовать не менее $n$ фишек.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of positive rational numbers $a$ and $b$ has the minimal period of length 30 in the decimal notation. A number $a-b$ has the minimal period of length 15 in a decimal notation. Find the least positive integer $k$ for which it may happen that the length of the minimal period of the number $a+kb$ in the decimal notation is also 15. (A. Golovanov)",
    "options": [],
    "answer": "6",
    "solution": "Домножив, если нужно, числа $a$ и $b$ на подходящую степень десятки, мы можем считать, что десятичные записи чисел $a, b, a-b$ и $a+kb$ — чисто периодические (то есть периоды начинаются сразу после запятой).\n\nВоспользуемся следующим известным фактом: десятичная запись рационального числа $r$ — чисто периодическая с (не обязательно минимальной) длиной периода $T$ тогда и только тогда, когда $r$ имеет вид $\\frac{m}{10^{T}-1}$ для некоторого целого $m$.\n\nПрименительно к условию задачи это значит, что $a = \\frac{m}{10^{30}-1}$ и $b = \\frac{n}{10^{30}-1}$. Нам также известно, что числа $a-b = \\frac{m-n}{10^{30}-1}$ и $a+kb = \\frac{m+kn}{10^{30}-1}$ записываются десятичными дробями с периодом длины 15, то есть могут быть записаны как обыкновенные дроби со знаменателем $10^{15}-1$. Поэтому так может быть записана и их разность $(k+1)b = \\frac{(k+1)n}{10^{30}-1}$. Таким образом, число $(k+1)n$ делится на $10^{15}+1$, а число $n$ — не делится (иначе и $b$ записывалось бы дробью с периодом длины 15). Значит, число $k+1$ делится на некоторый простой делитель числа $10^{15}+1$. Наименьший из таких делителей — это 7. Действительно, число $10^{15}+1$ даёт остаток 1 при делении на 2 и на 5, а также остаток 2 при делении на 3. С другой стороны, оно делится на $10^3+1 = 7 \\cdot 143$. Итак, $k+1 \\ge 7$ и $k \\ge 6$.\n\nВ некотором смысле минимальный пример чисел, удовлетворяющих условию при $k = 6$, получается, если положить $a - b = \\frac{1}{10^{15}-1}$ и $a + 6b = \\frac{2}{10^{15}-1}$. Тогда $a = \\frac{8}{7(10^{15}-1)}$ и $b = \\frac{1}{7(10^{15}-1)}$. Ясно, что длины минимальных периодов чисел $a-b$ и $a+6b$ равны 15. Далее, длины минимальных периодов чисел $a$ и $b$ больше 15 и делятся на 15 (так как $10^T - 1$ должно делиться на $10^{15} - 1$). С другой стороны, так как $10^{30} - 1 = 7(10^{15} - 1)$, числа $a$ и $b$ периодичны с длиной периода 30. Значит, длины их минимальных периодов равны 30.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Положительные рациональные числа $a$ и $b$ записаны в виде десятичных дробей, у каждой из которых минимальный период состоит из 30 цифр. У десятичной записи числа $a-b$ длина минимального периода равна 15. При каком наименьшем натуральном $k$ длина минимального периода десятичной записи числа $a+kb$ может также оказаться равной 15?",
    "options": [],
    "answer": "6",
    "solution": "Ответ. $k=6$.\n\nДомножив, если нужно, числа $a$ и $b$ на подходящую степень десятки, мы можем считать, что десятичные записи чисел $a, b, a-b$ и $a+kb$ — чисто периодические (то есть периоды начинаются сразу после запятой).\n\nВоспользуемся следующим известным фактом: десятичная запись рационального числа $r$ — чисто периодическая с (не обязательно минимальной) длиной периода $T$ тогда и только тогда, когда $r$ имеет вид $\\frac{m}{10^{T}-1}$ для некоторого целого $m$.\n\nПрименительно к условию задачи это значит, что $a = \\frac{m}{10^{30}-1}$ и $b = \\frac{n}{10^{30}-1}$. Нам также известно, что числа $a-b = \\frac{m-n}{10^{30}-1}$ и $a+kb = \\frac{m+kn}{10^{30}-1}$ записываются десятичными дробями с периодом длины 15, то есть могут быть записаны как обыкновенные дроби со знаменателем $10^{15}-1$. Поэтому так может быть записана и их разность $(k+1)b = \\frac{(k+1)n}{10^{30}-1}$. Таким образом, число $(k+1)n$ делится на $10^{15}+1$, а число $n$ — не делится (иначе и $b$ записывалось бы дробью с периодом длины 15). Значит, число $k+1$ делится на некоторый простой делитель числа $10^{15}+1$. Наименьший из таких делителей — это 7. Действительно, число $10^{15}+1$ даёт остаток 1 при делении на 2 и на 5, а также остаток 2 при делении на 3. С другой стороны, оно делится на $10^3+1 = 7 \\cdot 143$. Итак, $k+1 \\ge 7$ и $k \\ge 6$.\n\nВ некотором смысле минимальный пример чисел, удовлетворяющих условию при $k = 6$, получается, если положить $a - b = \\frac{1}{10^{15}-1}$ и $a + 6b = \\frac{2}{10^{15}-1}$. Тогда $a = \\frac{8}{7(10^{15}-1)}$ и $b = \\frac{1}{7(10^{15}-1)}$. Ясно, что длины минимальных периодов чисел $a-b$ и $a+6b$ равны 15. Далее, длины минимальных периодов чисел $a$ и $b$ больше 15 и делятся на 15 (так как $10^T - 1$ должно делиться на $10^{15} - 1$). С другой стороны, так как $10^{30} - 1 = 7(10^{15} - 1)$, числа $a$ и $b$ периодичны с длиной периода 30. Значит, длины их минимальных периодов равны 30.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39725,
    "subject": "Mathematics (Multi-modal)",
    "question": "A plane $\\alpha$ intersects the edges $AB$, $BC$, $CD$, and $DA$ of a triangular pyramid $ABCD$ at points $K$, $L$, $M$, and $N$, respectively. It appears that the dihedral angles $\\angle(KLA, KLM)$, $\\angle(LMB, LMN)$, $\\angle(MNC, MNK)$, and $\\angle(NKD, NKL)$ are equal. Prove that the projections of the vertices $A$, $B$, $C$, and $D$ onto the plane $\\alpha$ are concyclic. (By $\\angle(PQR, PQS)$ we denote the dihedral angle of a tetrahedron $PQRS$ at the edge $PQ$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим через $A', B', C', D'$ проекции вершин $A, B, C, D$ соответственно на плоскость $\\alpha$. Пусть $X$ — произвольная точка на продолжении отрезка $KL$ за точку $K$. Тогда имеем $\\angle(KXA, KXN) = \\angle(KLA, KLM)$ и $\\angle(KNA, KNX) = \\angle(NKD, NKL)$. По условию, эти углы равны; значит, в трехгранном угле $KANX$ (с вершиной $K$) двугранные углы при ребрах $KN$ и $KX$ равны между собой. Это означает, что плоскость, проходящая через прямую $KA$ и перпендикулярная плоскости $\\alpha$, — это плоскость симметрии трехгранного угла $\\angle(KAN, KXN)$. Поэтому точка $A'$ лежит на прямой, содержащей биссектрису угла $XKN$, то есть на внешней биссектрисе угла $NKL$.\n\n![](attached_image_1.png)\n\nАналогично показывается, что $A'$ лежит на внешней биссектрисе угла $MNK$. Применяя такие же рассуждения для точек $B', C', D'$, получаем, что точки $A', B', C', D'$ — пересечения внешних биссектрис соседних углов четырехугольника $KLMN$.\n\nИз треугольника $A'KN$ находим $\\angle B'A'D' = \\angle KA'N = 180^\\circ - \\angle A'KN - \\angle A'NK = (90^\\circ - \\angle A'KN) + (90^\\circ - \\angle A'NK) = \\frac{1}{2}(\\angle NKL + \\angle MNK)$. Аналогично получаем, что $\\angle B'C'D' = \\frac{1}{2}(\\angle KLM + \\angle LMN)$. Таким образом, $\\angle B'A'D' + \\angle B'C'D' = \\frac{1}{2}(\\angle NKL + \\angle MNK + \\angle KLM + \\angle LMN) = 180^\\circ$, откуда и следует, что четырехугольник $A'B'C'D'$ — вписанный.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Плоскость $\\alpha$ пересекает ребра $AB$, $BC$, $CD$ и $DA$ треугольной пирамиды $ABCD$ в точках $K$, $L$, $M$ и $N$ соответственно. Оказалось, что двугранные углы $\\angle(KLA, KLM)$, $\\angle(LMB, LMN)$, $\\angle(MNC, MNK)$ и $\\angle(NKD, NKL)$ равны. (Здесь через $\\angle(PQR, PQS)$ обозначается двугранный угол при ребре $PQ$ в тетраэдре $PQRS$.) Докажите, что\n\nпроекции вершин $A$, $B$, $C$ и $D$ на плоскость $\\alpha$ лежат на одной окружности.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим через $A'$, $B'$, $C'$, $D'$ проекции вершин $A$, $B$, $C$, $D$ соответственно на плоскость $\\alpha$. Пусть $X$ — произвольная точка на продолжении отрезка $KL$ за точку $K$. Тогда имеем $\\angle(KXA, KXN) = \\angle(KLA, KLM)$ и $\\angle(KNA, KNX) = \\angle(NKD, NKL)$. По условию, эти углы равны; значит, в трехгранном угле $KANX$ (с вершиной $K$) двугранные углы при ребрах $KN$ и $KX$ равны между собой. Это означает, что плоскость, проходящая через прямую $KA$ и перпендикулярная плоскости $\\alpha$, — это плоскость симметрии трехгранного угла $\\angle(KAN, KXN)$. Поэтому точка $A'$ лежит на прямой, содержащей биссектрису угла $XKN$, то есть на внешней биссектрисе угла $NKL$.\n\n![](attached_image_1.png)\n\nАналогично показывается, что $A'$ лежит на внешней биссектрисе угла $MNK$. Применяя такие же рассуждения для точек $B'$, $C'$, $D'$, получаем, что точки $A'$, $B'$, $C'$, $D'$ — пересечения внешних биссектрис соседних углов четырехугольника $KLMN$.\n\nИз треугольника $A'KN$ находим $\\angle B'A'D' = \\angle KA'N = 180^\\circ - \\angle A'KN - \\angle A'NK = (90^\\circ - \\angle A'KN) + (90^\\circ - \\angle A'NK) = \\frac{1}{2}(\\angle NKL + \\angle MNK)$. Аналогично получаем, что $\\angle B'C'D' = \\frac{1}{2}(\\angle KLM + \\angle LMN)$. Таким образом, $\\angle B'A'D' + \\angle B'C'D' = \\frac{1}{2}(\\angle NKL + \\angle MNK + \\angle KLM + \\angle LMN) = 180^\\circ$, откуда и следует, что четырехугольник $A'B'C'D'$ — вписанный.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Числа $x$, $y$ и $z$ таковы, что все три числа $x + yz$, $y + zx$ и $z + xy$ рациональны, а $x^2 + y^2 = 1$. Докажите, что число $xyz^2$ также рационально.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Из условия следует, что число $(x + yz)(y + zx) = xy + (x^2 + y^2)z + xyz^2 = (xy + z) + xyz^2$ рационально. Поскольку число $xy + z$ также рационально по условию, то и число $xyz^2 = (x + yz)(y + zx) - (xy + z)$ также рационально.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39728,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $x$, $y$, and $z$ are chosen in such a way that $x^2 + y^2 = 1$, and the three numbers $x + yz$, $y + zx$, and $z + xy$ are rational. Prove that the number $xyz^2$ is also rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Из условия следует, что число $(x + yz)(y + zx) = xy + (x^2 + y^2)z + xyz^2 = (xy + z) + xyz^2$ рационально. Поскольку число $xy + z$ также рационально по условию, то и число $xyz^2 = (x + yz)(y + zx) - (xy + z)$ также рационально.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39729,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is good if every its positive divisor increased by $1$ is a divisor of $n+1$. Find all good positive integers.",
    "options": [],
    "answer": "1 and all odd prime numbers",
    "solution": "$1$ and all odd prime numbers.\n\nClearly, $n = 1$ satisfies the condition. All odd primes also satisfy it: if $n = p$, then its divisors, increased by $1$, are $2$ and $p+1$; both divide $p+1$.\n\nOn the other hand, any number $n$ satisfying the condition has $1$ as a divisor; thus, $n+1$ is divisible by $1+1$, i.e., $n$ is odd.\n\nSuppose now that some composite $n$ satisfies the condition. We have $n = ab$, where $a \\geq b \\geq 2$. Then $n+1$ is divisible by $a+1$; moreover, $n+b = (a+1)b$ is also divisible by $a+1$. Therefore, $b-1 = (n+b)-(n+1)$ is also divisible by $a+1$. Since $b-1 > 0$, we get $b-1 \\geq a+1$. But this contradicts the inequality $b \\leq a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Натуральное число $n$ назовём хорошим, если каждый его натуральный делитель, увеличенный на 1, является делителем числа $n+1$. Найдите все хорошие натуральные числа.",
    "options": [],
    "answer": "1 and all odd primes",
    "solution": "**Ответ.** Единица и все нечётные простые числа.\n\nЯсно, что $n = 1$ удовлетворяет условию. Также ему удовлетворяют все нечётные простые: если $n = p$, то его делители, увеличенные на 1, есть $2$ и $p+1$; оба они делят $p+1$.\n\nС другой стороны, у любого числа $n$, удовлетворяющего условию, есть делитель $1$; значит, $n+1$ делится на $1+1$, то есть $n$ нечётно.\n\nПредположим теперь, что какое-то составное $n$ удовлетворяет условию. Имеем $n = ab$, где $a \\ge b \\ge 2$. Тогда число $n+1$ делится на $a+1$; кроме того, число $n+b = (a+1)b$ также делится на $a+1$. Значит, и число $b-1 = (n+b)-(n+1)$ также делится на $a+1$. Так как $b-1 > 0$, получаем, что $b-1 \\ge a+1$. Но это противоречит неравенству $b \\le a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Сфера $\\omega$ проходит через вершину $S$ пирамиды $SABC$ и пересекает рёбра $SA$, $SB$ и $SC$ вторично в точках $A_1$, $B_1$ и $C_1$ соответственно. Сфера $\\Omega$, описанная около пирамиды $SABC$, пересекается с $\\omega$ по окружности, лежащей в плоскости, параллельной плоскости $(ABC)$. Точки $A_2$, $B_2$ и $C_2$ симметричны точкам $A_1$, $B_1$ и $C_1$ относительно середин рёбер $SA$, $SB$ и $SC$ соответственно. Докажите, что точки $A$, $B$, $C$, $A_2$, $B_2$ и $C_2$ лежат на одной сфере.\n(И. Богданов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Утверждение задачи эквивалентно равенству $SA_2 \\cdot SA = SB_2 \\cdot SB = SC_2 \\cdot SC$. Значит, ввиду равенств $AA_1 = SA_2$ и двух аналогичных, достаточно доказать, что $AA_1 \\cdot AS = BB_1 \\cdot BS = CC_1 \\cdot CS$.\nПусть $\\ell$ — прямая, проходящая через центры сфер $\\Omega$ и $\\omega$. Окружность пересечения этих сфер лежит в плоскости, нерпендикулярной $\\ell$, так что $\\ell \\perp (ABC)$. Это значит, что при повороте вокруг $\\ell$ описанная окружность треугольника $ABC$ переходит в себя, и подходящим таким поворотом можно точку $A$ перевести в $B$. Пусть точки $S$ и $A_1$ при этом повороте переходят в $S'$ и $A'_1$ (они тоже лежат на $\\omega$, см. рис. 17). Тогда $AA_1 \\cdot AS = BA'_1 \\cdot BS' = BB_1 \\cdot BS$. Равенство $AA_1 \\cdot AS = CC_1 \\cdot CS$ доказывается аналогично.\n\n![](attached_image_1.png)\nРис. 17\n\n\n**Второе решение.** Обозначим через $O_1$ и $O$ центры сфер $\\omega$ и $\\Omega$ соответственно. Как и в первом решении, введём прямую $\\ell$, проходящую через $O$ и $O_1$; тогда $\\ell \\perp (ABC)$.\nПусть $O_2$ — точка, симметричная $O_1$ относительно $O$. Тогда $O_2$ лежит на $\\ell$, откуда $O_2A = O_2B = O_2C$; обозначим $r = O_2A$. Далее, проекции точек $O_2$ и $O_1$ на $SA$ симметричны относительно проекции точки $O$, т. е. относительно середины $A'$ отрезка $SA$. Так как\n\n![](attached_image_2.png)\nРис. 18\n\nпроекция точки $O_1$ является серединой отрезка $SA_1$, из симметрии относительно $A'$ получаем, что проекция точки $O_2$ — это середина отрезка $AA_2$. Значит, $A_2O_2 = AO_2 = r$. Аналогично показывается, что $B_2O_2 = C_2O_2 = r$. Значит, требуемые шесть точек лежат на сфере с центром $O_2$ и радиусом $r$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an inscribed quadrilateral for which the rays $AB$ and $DC$ meet at $K$. It happened that the points $B$, $D$, and the midpoints of the segments $AC$ and $KC$ are concyclic. Find all possible values the angle $ADC$ may get.",
    "options": [],
    "answer": "90°",
    "solution": "90°.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39733,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sphere $\\omega$ passes through a vertex $S$ of a triangular pyramid $SABC$ and intersects the edges $SA$, $SB$, $AC$ at $A_1 \\neq S$, $B_1 \\neq S$, and $C_1 \\neq S$, respectively. The sphere $\\omega$ intersects the circumsphere $\\Omega$ of $SABC$ by a circle lying in a plane parallel to $(ABC)$. The points $A_2$, $B_2$, and $C_2$ are symmetric to the points $A_1$, $B_1$, and $C_1$ with respect to the midpoints of the segments $SA$, $SB$, $AC$, respectively. Prove that the points $A$, $B$, $C$, $A_2$, $B_2$, and $C_2$ lie on a sphere. (I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Утверждение задачи эквивалентно равенству $SA_2 \\cdot SA = SB_2 \\cdot SB = SC_2 \\cdot SC$. Значит, ввиду равенств $AA_1 = SA_2$ и двух аналогичных, достаточно доказать, что $AA_1 \\cdot AS = BB_1 \\cdot BS = CC_1 \\cdot CS$.\nПусть $\\ell$ — прямая, проходящая через центры сфер $\\Omega$ и $\\omega$. Окружность пересечения этих сфер лежит в плоскости, нерпендикулярной $\\ell$, так что $\\ell \\perp (ABC)$. Это значит, что при повороте вокруг $\\ell$ описанная окружность треугольника $ABC$ переходит в себя, и подходящим таким поворотом можно точку $A$ перевести в $B$. Пусть точки $S$ и $A_1$ при этом повороте переходят в $S'$ и $A'_1$ (они тоже лежат на $\\omega$, см. рис. 17). Тогда $AA_1 \\cdot AS = BA'_1 \\cdot BS' = BB_1 \\cdot BS$. Равенство $AA_1 \\cdot AS = CC_1 \\cdot CS$ доказывается аналогично.\n\n![](attached_image_1.png)\nРис. 17\n\n\n**Второе решение.** Обозначим через $O_1$ и $O$ центры сфер $\\omega$ и $\\Omega$ соответственно. Как и в первом решении, введём прямую $\\ell$, проходящую через $O$ и $O_1$; тогда $\\ell \\perp (ABC)$.\nПусть $O_2$ — точка, симметричная $O_1$ относительно $O$. Тогда $O_2$ лежит на $\\ell$, откуда $O_2A = O_2B = O_2C$; обозначим $r = O_2A$. Далее, проекции точек $O_2$ и $O_1$ на $SA$ симметричны относительно проекции точки $O$, т. е. относительно середины $A'$ отрезка $SA$. Так как\n\n![](attached_image_2.png)\nРис. 18\n\nпроекция точки $O_1$ является серединой отрезка $SA_1$, из симметрии относительно $A'$ получаем, что проекция точки $O_2$ — это середина отрезка $AA_2$. Значит, $A_2O_2 = AO_2 = r$. Аналогично показывается, что $B_2O_2 = C_2O_2 = r$. Значит, требуемые шесть точек лежат на сфере с центром $O_2$ и радиусом $r$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $n$ is a positive integer, and a polynomial\n$$\nP(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \\dots + a_1x + a_0,\n$$\nsatisfies the conditions $100 \\le a_i \\le 101$ for all $0 \\le i \\le 2n$. Find the least possible $n$ such that this polynomial may have a real root.",
    "options": [],
    "answer": "n = 100",
    "solution": "**Ответ.** $n = 100$.\n\nНазовём многочлен, удовлетворяющий условию задачи, **красивым**. Многочлен $P(x) = 100(x^{200} + x^{198} + \\dots + x^2 + 1) + 101(x^{199} + x^{197} + \\dots + x)$ красив и имеет корень $-1$. Значит, при $n = 100$ требуемое возможно.\n\nОсталось показать, что при $n < 100$ у красивого многочлена $P(x)$ не может быть вещественных корней. Для этого достаточно проверить, что $P(x) > 0$ при всех $x$. Это неравенство, очевидно, выполнено при $x \\ge 0$; для отрицательных же $x = -t$ оно является следствием неравенства\n$$\n100(t^{2n} + t^{2n-2} + \\dots + t^2 + 1) > 101(t^{2n-1} + t^{2n-3} + \\dots + t). \\quad (*)\n$$\n\nЗначит, достаточно доказать это неравенство при всех $t > 0$. Умножая $(*)$ на $t+1$, получаем равносильное неравенство $100(t^{2n+1} + t^{2n} + \\dots + 1) > 101(t^{2n} + t^{2n-1} + \\dots + t)$, или\n$$\n100(t^{2n+1} + 1) > t^{2n} + t^{2n-1} + \\dots + t. \\quad (**)\n$$\n\nЗаметим, что при каждом $k = 1, \\dots, n$ выполнено неравенство $(t^k - 1)(t^{2n+1-k} - 1) \\ge 0$, поскольку обе скобки имеют одинаковые знаки при $t > 0$. Раскрывая скобки, получаем\n$$\nt^{2n+1} + 1 \\ge t^{2n+1-k} + t^k.\n$$\n\nСкладывая все такие неравенства и учитывая, что $n < 100$, получаем\n$$\nt^{2n} + t^{2n-1} + \\dots + t \\le n(t^{2n+1} + 1) < 100(t^{2n+1} + 1),\n$$\nчто и доказывает (**).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, the blackboard contains two polynomials $x^3-3x^2+5$ and $x^2-4x$. If the polynomials $f(x)$ and $g(x)$ are written on the blackboard, it is permitted to write onto the board any polynomial of the form $f(x) \\pm g(x)$, $f(x)g(x)$, $f(g(x))$, or $cf(x)$, where $c$ may be any (not necessarily integer) constant. Is it possible after several such operations to write on the board a nonzero polynomial of the form $x^n - 1$? (K. Tyschuk)",
    "options": [],
    "answer": "No",
    "solution": "**Ответ.** Не может.\nПусть $f(x)$ и $g(x)$ — два многочлена, и для некоторой точки $x_0$ выполняются равенства $f'(x_0) = 0$ и $g'(x_0) = 0$. Тогда, очевидно, $(f \\pm g)'(x_0) = 0$ и $cf'(x_0) = 0$. Также $(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0) = 0$. Наконец, если $h(x)$ — многочлен, то $(h(g(x_0)))' = h'(g(x_0))g'(x_0) = 0$. Таким образом, если у исходных многочленов в некоторой точке производные обращаются в нуль, то и после решённых условием операций также может получиться лишь многочлен, производная которого обращается в нуль в этой точке.\n\nЗаметим, что производные обоих исходных многочленов обращаются в нуль при $x = 2$. Действительно, $(x^2 - 4x)' = 2x - 4 = 0$ при $x = 2$, и $(x^3 - 3x^2 + 5)' = 3x^2 - 6x = 0$ при $x = 2$. Однако $(x^n - 1)' = nx^{n-1} = n2^{n-1} \\neq 0$ при $x = 2$. Поэтому многочлен вида $x^n - 1$ получить нельзя.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Исходно на доске написаны многочлены $x^3 - 3x^2 + 5$ и $x^2 - 4x$. Если на доске уже написаны многочлены $f(x)$ и $g(x)$, разрешается дописать на неё многочлены $f(x) \\pm g(x)$, $f(x)g(x)$, $f(g(x))$ и $cf(x)$, где $c$ — произвольная (не обязательно целая) константа. Может ли на доске после нескольких операций появиться многочлен вида $x^n - 1$ (при натуральном $n$)?\n(К. Тыщук)",
    "options": [],
    "answer": "No",
    "solution": "**Ответ.** Не может.\nПусть $f(x)$ и $g(x)$ — два многочлена, и для некоторой точки $x_0$ выполняются равенства $f'(x_0) = 0$ и $g'(x_0) = 0$. Тогда, очевидно, $(f \\pm g)'(x_0) = 0$ и $cf'(x_0) = 0$. Также $(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0) = 0$. Наконец, если $h(x)$ — многочлен, то $(h(g(x_0)))' = h'(g(x_0))g'(x_0) = 0$. Таким образом, если у исходных многочленов в некоторой точке производные обращаются в нуль, то и после решённых условием операций также может получиться лишь многочлен, производная которого обращается в нуль в этой точке.\n\nЗаметим, что производные обоих исходных многочленов обращаются в нуль при $x = 2$. Действительно, $(x^2 - 4x)' = 2x - 4 = 0$ при $x = 2$, и $(x^3 - 3x^2 + 5)' = 3x^2 - 6x = 0$ при $x = 2$. Однако $(x^n - 1)' = nx^{n-1} = n2^{n-1} \\neq 0$ при $x = 2$. Поэтому многочлен вида $x^n - 1$ получить нельзя.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Дан многочлен\n$$\nP(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \\dots + a_1 x + a_0,\n$$\nу которого каждый коэффициент $a_i$ принадлежит отрезку $[100, 101]$.\n\nПри каком минимальном натуральном $n$ у такого многочлена может найтись действительный корень?",
    "options": [],
    "answer": "n = 100",
    "solution": "**Ответ.** $n = 100$.\n\nНазовём многочлен, удовлетворяющий условию задачи, **красивым**. Многочлен\n$$\nP(x) = 100(x^{200} + x^{198} + \\dots + x^2 + 1) + 101(x^{199} + x^{197} + \\dots + x)\n$$\nкрасив и имеет корень $-1$. Значит, при $n = 100$ требуемое возможно.\n\nОсталось показать, что при $n < 100$ у красивого многочлена $P(x)$ не может быть вещественных корней. Для этого достаточно проверить, что $P(x) > 0$ при всех $x$. Это неравенство, очевидно, выполнено при $x \\ge 0$; для отрицательных же $x = -t$ оно является следствием неравенства\n$$\n100(t^{2n} + t^{2n-2} + \\dots + t^2 + 1) > 101(t^{2n-1} + t^{2n-3} + \\dots + t). \\quad (*)\n$$\nЗначит, достаточно доказать это неравенство при всех $t > 0$. Умножая $(*)$ на $t+1$, получаем равносильное неравенство $100(t^{2n+1} + t^{2n} + \\dots + 1) > 101(t^{2n} + t^{2n-1} + \\dots + t)$, или\n$$\n100(t^{2n+1} + 1) > t^{2n} + t^{2n-1} + \\dots + t. \\quad (**)\n$$\nЗаметим, что при каждом $k = 1, \\dots, n$ выполнено неравенство $(t^k - 1)(t^{2n+1-k} - 1) \\ge 0$, поскольку обе скобки имеют одинаковые знаки при $t > 0$. Раскрывая скобки, получаем\n$$\nt^{2n+1} + 1 \\ge t^{2n+1-k} + t^k.\n$$\nСкладывая все такие неравенства и учитывая, что $n < 100$, получаем\n$$\nt^{2n} + t^{2n-1} + \\dots + t \\le n(t^{2n+1} + 1) < 100(t^{2n+1} + 1),\n$$\nчто и доказывает (**).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pete has put several tokens into some squares of a checkered $50 \\times 50$ board (at most one token per square). Prove that Bazil can put at most 99 tokens into empty squares so that each row and each column contains an even number of tokens.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "11.8. См. решение задачи 10.8.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the following card game. They have a deck of $n$ cards. For every two cards it is known which one of them takes the other (it may happen that $A$ takes $B$, $B$ takes $C$, and $C$ takes $A$). Initially the deck is distributed between the players in an arbitrary way. On each move, the players open the topmost cards of their decks; then the player whose card takes the opponent's one gets both cards and puts them to the very bottom of his deck (in an arbitrary order). Prove that for any starting situation, the players may agree to play in such a way that one of them will finally get an empty deck. (E. Lakshtanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Выпишем все возможные ситуации, которые могут встречаться в игре (т. е. все возможные пары колод у участников). Назовём ситуацию *фундаментальной*, если все карты у одного игрока; *критической*, если у одного из игроков ровно одна карта; и регулярной, если у обоих игроков хотя бы по две карты. Проведём стрелку от каждой ситуации к ситуациям, которые могут из неё получиться после одного хода. Тогда из любой нефинальной ситуации ведут две стрелки, а из любой финальной — ноль. Нам надо доказать, что из каждой нефинальной ситуации по стрелкам можно дойти до финальной.\n\nВыясним, сколько стрелок ведут в каждую ситуацию. Предположим, что она получилась в результате какого-то хода, в котором карты взял первый игрок. Тогда у него оказалось хотя бы две карты, которые лежат в конце колоды; при этом известно, что одна из них — $a$ — бьёт другую — $b$. Значит, в этом случае карта $a$ была у первого игрока, а карта $b$ — у второго, то есть предыдущая ситуация восстанавливается однозначно. Итак, если наша ситуация регулярна, то на предыдущем ходе карты мог получить любой из двух игроков, и в каждом из этих случаев предыдущая ситуация восстанавливается однозначно. Значит, в каждую регулярную ситуацию ведут ровно две стрелки. Аналогично, в каждую нерегулярную ситуацию ведёт ровно одна стрелка.\n\nПредположим, что из некоторой ситуации $S$ нельзя попасть в финальную. Назовём ситуацию *достижимой*, если в неё можно добраться из $S$. Из каждой достижимой ситуации ведут две стрелки в достижимые. С другой стороны, в каждую достижимую ситуацию ведёт не более двух стрелок из достижимых. Это возможно только в том случае, если в каждую достижимую ситуацию ведёт ровно по две стрелки из достижимых. Из этого, в частности, следует, что все достижимые ситуации регулярны. Более того, поскольку в каждую ситуацию ведёт не более двух стрелок, получаем, что все стрелки, входящие в достижимые ситуации, выходят также из достижимых.\n\nПусть в ситуации $S$ у первого игрока $k > 1$ карт. Тогда в одной из двух ситуаций, из которых ведут стрелки в $S$, у первого игрока $k-1$ карта — назовём эту ситуацию $S_1$; по показанному выше, она достижима. Аналогично, если $k-1 > 1$, то в одной из двух ситуаций, из которых ведут стрелки в $S_1$, у первого игрока $k-2$ карты; назовём её $S_2$ и продолжим рассуждения. В итоге мы получим цепочку из достижимых ситуаций $S_1, S_2, \\dots, S_{k-1}$, причём в $S_{k-1}$ у первого игрока одна карта, т. е. она критическая, и в неё входит только одна стрелка. Но в каждую достижимую ситуацию должно входить две стрелки — противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39740,
    "subject": "Mathematics (Multi-modal)",
    "question": "По кругу расставлены $99$ натуральных чисел. Известно, что любые два соседних числа отличаются или на $1$, или на $2$, или в два раза. Докажите, что хотя бы одно из этих чисел делится на $3$. (С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть ни одно из чисел не делится на $3$. Тогда каждое число даёт остаток $1$ или $2$ при делении на $3$. Но числа, дающие одинаковые ненулевые остатки при делении на $3$, не могут отличаться на $1$ или на $2$; не могут они и отличаться в два раза. Значит, соседние числа дают различные остатки при делении на $3$, то есть остатки $1$ и $2$ чередуются. Но тогда общее количество чисел должно быть чётным, что не так. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given are 111 distinct positive integers not exceeding 500. May it happen that for each of these numbers, its last digit coincides with the last digit of the sum of all other numbers?",
    "options": [],
    "answer": "No",
    "solution": "Suppose this is possible. Denote the given numbers by $a_1, a_2, \\dots, a_{111}$ and let their sum be $S$. By the condition, for each index $k$, the numbers $a_k$ and $S - a_k$ have the same last digit. Hence, their difference $S - 2a_k$ is divisible by $10$. Therefore, for any $k$, the number $2a_k$ ends with the same digit as the sum $S$. This means that the difference between any two numbers $a_k$ is divisible by $5$. Thus, all $111$ distinct numbers $a_i$ must give the same remainder when divided by $5$; but among the numbers from $1$ to $500$, there are exactly $100$ numbers with any fixed remainder modulo $5$. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Даны 111 различных натуральных чисел, не превосходящих 500. Могло ли оказаться, что для каждого из этих чисел его последняя цифра совпадает с последней цифрой суммы всех остальных чисел?",
    "options": [],
    "answer": "No, it is not possible.",
    "solution": "Ответ. Не могло.\n\nПусть такое могло случиться. Обозначим данные числа $a_1, a_2, \\dots, a_{111}$ и обозначим их сумму через $S$. По условию, для каждого номера $k$ числа $a_k$ и $S - a_k$ оканчиваются одной и той же цифрой. Отсюда следует, что разность этих чисел, равная $S - 2a_k$, делится на $10$. Значит, при любом $k$ число $2a_k$ оканчивается последней цифрой суммы $S$. Это означает, что разность между любыми двумя числами $a_k$ делится на $5$. Итак, все $111$ различных чисел $a_i$ должны давать одинаковые остатки от деления на $5$; но среди чисел от $1$ до $500$ ровно по $100$ чисел, дающих фиксированный остаток от деления на $5$. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39743,
    "subject": "Mathematics (Multi-modal)",
    "question": "99 positive integers are arranged in a circle. For every two neighboring numbers, either they differ by 1, or they differ by 2, or one of them is twice the other. Prove that one of these numbers is divisible by 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that none of the numbers is divisible by $3$. Then each number gives a remainder $1$ or $2$ when divided by $3$.\n\nBut numbers giving the same nonzero remainder modulo $3$ cannot differ by $1$ or $2$; nor can one be twice the other. Therefore, neighboring numbers must give different remainders modulo $3$, that is, the remainders $1$ and $2$ must alternate.\n\nBut then the total number of numbers must be even, which is not the case. Contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Серёжа выбрал два различных натуральных числа $a$ и $b$. Он записал в тетрадь четыре числа: $a, a+2, b$ и $b+2$. Затем он выписал на доску все шесть попарных произведений чисел из тетради. Какое наибольшее количество точных квадратов может быть среди чисел на доске?",
    "options": [],
    "answer": "2",
    "solution": "**Ответ.** Два.\n\nЗаметим, что никакие два квадрата натуральных чисел не отличаются на 1, ибо $x^2 - y^2 = (x - y)(x + y)$, где вторая скобка больше единицы. Значит, числа $a(a+2) = (a+1)^2 - 1$ и $b(b+2) = (b+1)^2 - 1$ квадратами не являются. Более того, числа $ab$ и $a(b+2)$ не могут одновременно являться квадратами, иначе их произведение $a^2 \\cdot b(b+2)$ также было бы квадратом, а тогда и число $b(b+2)$ тоже. Аналогично, из чисел $(a+2)b$ и $(a+2)(b+2)$ максимум одно может быть квадратом. Итого, квадратов на доске не больше двух.\n\nДва квадрата могут получиться, например, при $a = 2$ и $b = 16$: тогда $a(b+2) = 6^2$ и $(a+2)b = 8^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39745,
    "subject": "Mathematics (Multi-modal)",
    "question": "В четырёхугольнике $ABCD$ стороны $AD$ и $BC$ параллельны. Докажите, что если биссектрисы углов $DAC$, $DBC$, $ACB$ и $ADB$ образовали ромб, то $AB = CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $O$ — точка пересечения диагоналей $AC$ и $BD$ (см. рис. 1). Биссектрисы углов $ADB$ и $DAC$ пересекаются в центре $O_1$ окружности, вписанной в треугольник $AOD$, а биссектрисы углов $ACB$ и $DBC$ — в центре $O_2$ окружности, вписанной в треугольник $BOC$. Значит, точки $O_1$ и $O_2$ лежат на биссектрисах вертикальных углов $AOD$ и $BOC$.\n\nРассмотрим ромб $PO_1QO_2$ из условия задачи. В нём $\\angle PO_1O_2 = \\angle QO_1O_2$, а значит, $\\angle DO_1O = \\angle AO_1O$. Значит, треугольники $AOO_1$ и $DOO_1$ равны по стороне ($O_1O$ — общая) и двум прилегающим углам, откуда $AO = DO$. Отсюда $\\angle OAD = \\angle ODA$, и четырёхугольник $ABCD$ симметричен относительно серединного перпендикуляра к $AD$. Поэтому $AB = CD$.\n\n![](attached_image_1.png)\nРис. 1\n\n\nВторое решение.\n\nОбозначим вершины ромба через $P, O_1, Q, O_2$, как и в первом решении. Поскольку четырёхугольник $PO_1QO_2$ — ромб, расстояние между прямыми $O_2P$ и $O_1Q$ равно расстоянию между прямыми $O_1P$ и $O_2Q$, то есть\n$$\nAC \\sin(\\angle CAD/2) = BD \\sin(\\angle BDA/2).\n$$\nТак как вершины $B$ и $C$ равноудалены от прямой $AD$, имеем\n$$\nAC \\sin \\angle CAD = BD \\sin \\angle BDA.\n$$\n\nДеля второе полученное равенство на первое и используя тождество $\\sin \\alpha = 2 \\sin(\\alpha/2) \\cos(\\alpha/2)$, получаем\n$$\n\\cos(\\angle CAD/2) = \\cos(\\angle BDA/2).\n$$\nТак как оба угла $CAD$ и $BDA$ меньше $180^\\circ$, получаем, что $\\angle CAD = \\angle BDA$. Как и в первом решении, заключаем, что $AB = CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39746,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides $AD$ and $BC$ of a quadrilateral $ABCD$ are parallel. Assume that the angle bisectors of the angles $DAC$, $DBC$, $ACB$, and $ADB$ form a rhombus. Prove that $AB = CD$. (L. Emelyanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the intersection point of the diagonals $AC$ and $BD$ (see the figure). The bisectors of the angles $ADB$ and $DAC$ intersect at the center $O_1$ of the incircle of triangle $AOD$, and the bisectors of the angles $ACB$ and $DBC$ intersect at the center $O_2$ of the incircle of triangle $BOC$. Thus, the points $O_1$ and $O_2$ lie on the bisectors of the vertical angles $AOD$ and $BOC$.\n\nConsider the rhombus $PO_1QO_2$ from the problem statement. In it, $\\angle PO_1O_2 = \\angle QO_1O_2$, so $\\angle DO_1O = \\angle AO_1O$. Therefore, triangles $AOO_1$ and $DOO_1$ are congruent by the side ($O_1O$ is common) and two adjacent angles, whence $AO = DO$. Hence $\\angle OAD = \\angle ODA$, and the quadrilateral $ABCD$ is symmetric with respect to the perpendicular bisector of $AD$. Therefore, $AB = CD$.\n\n![](attached_image_1.png)\n\nFigure 1\n\n\n**Second solution.** Denote the vertices of the rhombus by $P$, $O_1$, $Q$, $O_2$, as in the first solution. Since the quadrilateral $PO_1QO_2$ is a rhombus, the distance between the lines $O_2P$ and $O_1Q$ equals the distance between the lines $O_1P$ and $O_2Q$, that is,\n$$\nAC \\sin(\\angle CAD/2) = BD \\sin(\\angle BDA/2).\n$$\nSince the vertices $B$ and $C$ are equidistant from the line $AD$, we have\n$$\nAC \\sin \\angle CAD = BD \\sin \\angle BDA.\n$$\n\nDividing the second equation by the first and using the identity $\\sin \\alpha = 2 \\sin(\\alpha/2) \\cos(\\alpha/2)$, we get\n$$\n\\cos(\\angle CAD/2) = \\cos(\\angle BDA/2).\n$$\nSince both angles $CAD$ and $BDA$ are less than $180^\\circ$, we obtain $\\angle CAD = \\angle BDA$. As in the first solution, we conclude that $AB = CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sergey chose two different positive integers $a$ and $b$. Then he calculated all six pairwise products of the four numbers $a$, $a+2$, $b$, and $b+2$. Find the maximal number of perfect squares among the six calculated numbers. (S. Berlov)",
    "options": [],
    "answer": "2",
    "solution": "**Answer.** Two.\n\nNote that no two squares of natural numbers differ by $1$, since $x^2 - y^2 = (x - y)(x + y)$, where the second factor is greater than one. Therefore, the numbers $a(a+2) = (a+1)^2 - 1$ and $b(b+2) = (b+1)^2 - 1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwise their product $a^2 \\cdot b(b+2)$ would also be a square, and then $b(b+2)$ would also be a square. Similarly, among the numbers $(a+2)b$ and $(a+2)(b+2)$, at most one can be a square. Thus, there are at most two squares among the six numbers.\n\nTwo squares can be obtained, for example, for $a = 2$ and $b = 16$: then $a(b+2) = 6^2$ and $(a+2)b = 8^2$.\n\n**Remark.** There are other examples, for instance, $(a, b) = (6, 96)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Учитель записал Пете в тетрадь четыре различных натуральных числа. Для каждой пары этих чисел Петя нашёл их наибольший общий делитель. У него получились шесть чисел: 1, 2, 3, 4, 5 и $N$, где $N > 5$. Какое наименьшее значение может иметь число $N$?",
    "options": [],
    "answer": "14",
    "solution": "Число $N$ может равняться 14, как показывает, например, четвёрка чисел 4, 15, 70, 84. Осталось показать, что $N \\ge 14$.\n\n**Лемма.** Среди попарных НОД четырёх чисел не может быть ровно двух чисел, делящихся на некоторое натуральное $k$.\n**Доказательство.** Если среди исходных четырёх чисел есть не больше двух чисел, делящихся на $k$, то среди попарных НОД на $k$ делится не более одного. Если же три из исходных чисел делятся на $k$, то все три их попарных НОД делятся на $k$. Лемма доказана. $\\square$\n\nПрименяя лемму к $k = 2$, получаем, что число $N$ чётно. Применяя её же к $k = 3, k = 4$ и $k = 5$, получаем, что $N$ не делится на 3, 4 и 5. Значит, $N$ не может равняться 6, 8, 10 и 12.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39749,
    "subject": "Mathematics (Multi-modal)",
    "question": "The teacher gave to Pete four distinct positive integers. Pete has calculated the greatest common divisor of every two of these numbers. He obtained six numbers: $1$, $2$, $3$, $4$, $5$, and $N$, where $N > 5$. Find the least possible value of $N$.",
    "options": [],
    "answer": "14",
    "solution": "Число $N$ может равняться $14$, как показывает, например, четвёрка чисел $4$, $15$, $70$, $84$. Осталось показать, что $N \\ge 14$.\n\n**Лемма.** Среди попарных НОД четырёх чисел не может быть ровно двух чисел, делящихся на некоторое натуральное $k$.\n**Доказательство.** Если среди исходных четырёх чисел есть не больше двух чисел, делящихся на $k$, то среди попарных НОД на $k$ делится не более одного. Если же три из исходных чисел делятся на $k$, то все три их попарных НОД делятся на $k$. Лемма доказана. $\\square$\n\nПрименяя лемму к $k = 2$, получаем, что число $N$ чётно. Применяя её же к $k = 3$, $k = 4$ и $k = 5$, получаем, что $N$ не делится на $3$, $4$ и $5$. Значит, $N$ не может равняться $6$, $8$, $10$ и $12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39750,
    "subject": "Mathematics (Multi-modal)",
    "question": "В выпуклом $n$-угольнике проведено несколько диагоналей. Проведённая диагональ называется хорошей, если она пересекается (по внутренним точкам) ровно с одной из других проведённых диагоналей. Найдите наибольшее возможное количество хороших диагоналей.\n(С. Берлов)",
    "options": [],
    "answer": "n − 2 for even n; n − 3 for odd n",
    "solution": "**Ответ.** $n - 2$ при чётных $n$, $n - 3$ при нечётных $n$.\n\nМы будем пользоваться следующей известной леммой.\n\n**Лемма.** В выпуклом $n$-угольнике нельзя провести более $n - 3$ диагоналей, не имеющих общих внутренних точек.\n\nСначала докажем индукцией по $n$, что количество хороших диагоналей не превосходит $n - 2$, если $n$ чётно, и $n - 3$, если $n$ нечётно. При этом мы будем считать, что отрезок является 2-угольником без диагоналей. При $n = 2, 3$ утверждение очевидно. Пусть $n \\ge 4$; обозначим наш многоугольник через $P = A_1A_2...A_n$.\n\nЕсли никакие две хороших диагонали не пересекаются, то по лемме их количество не превосходит $n - 3$. Пусть теперь найдутся две пересекающихся хороших диагонали $A_iA_k$ и $A_jA_\\ell$ ($i < j < k < \\ell$). Тогда каждая из них не пересекается с другими проведёнными диагоналями. Выбросим $A_iA_k$ и $A_jA_\\ell$ из рассмотрения. Каждая оставшаяся проведённая диагональ $d$ является диагональю или стороной ровно в одном из многоугольников $Q_1 = A_i...A_j$, $Q_2 = A_j...A_k$, $Q_3 = A_k...A_\\ell$ или $Q_4 = A_\\ell...A_nA_1...A_i$ (см. рис. 6). При этом, если $d$ является стороной одного из них, то она не может пересекаться с другими диагоналями (и не является хорошей).\n\nПусть $n$ чётно. По предположению индукции, среди всех диагоналей, попавших в какой-то многоугольник $Q_s$, хороших не больше, чем количество вершин в нём, уменьшенное на 2. Значит, общее количество хороших диагоналей в $P$ не превосходит\n$$\n2 + (j - i - 1) + (k - j - 1) + (l - k - 1) + (n - l + i - 1) = n - 2, \\quad (*)\n$$\nчто и требовалось.\n\nПри нечётном $n$ сумма количеств вершин в многоугольниках $Q_1, Q_2, Q_3, Q_4$ равна нечётному числу $n + 4$; значит, число вершин в одном из них нечётно. А тогда соответствующее слагаемое в сумме $(*)$ уменьшится на 1, и мы получим, что общее число хороших диагоналей не превосходит $n - 3$. Переход индукции завершён.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\nРис. 7\n\nОсталось привести примеры, показывающие точность оценки. При чётном $n$ достаточно провести в многоугольнике $A_1...A_n$ диагонали $A_iA_{n-i}$ и $A_{i+1}A_{n-i+1}$ при всех $i = 1, 2, ..., n/2 - 1$ (см. рис. 7); они разбиваются на пары пересекающихся хороших диагоналей. При нечётном $n$ достаточно провести такие же диагонали в чётноугольнике $A_1...A_{n-1}$.\n\n**Замечание.** При нечётном $n$ существуют и принципиально другие примеры — например, можно взять все диагонали из точки $A_1$ (они будут хорошими) и добавить к ним диагональ $A_2A_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39751,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex $n$-gon, several diagonals are drawn. A drawn diagonal is good if it intersects (by an interior point) with exactly one of other drawn diagonals. Find the maximal possible number of good diagonals. (S. Berlov)",
    "options": [],
    "answer": "n − 2 for even n; n − 3 for odd n",
    "solution": "**Ответ.** $n - 2$ при чётных $n$, $n - 3$ при нечётных $n$.\n\nМы будем пользоваться следующей известной леммой.\n\n**Лемма.** В выпуклом $n$-угольнике нельзя провести более $n - 3$ диагоналей, не имеющих общих внутренних точек.\n\nСначала докажем индукцией по $n$, что количество хороших диагоналей не превосходит $n - 2$, если $n$ чётно, и $n - 3$, если $n$ нечётно. При этом мы будем считать, что отрезок является 2-угольником без диагоналей. При $n = 2, 3$ утверждение очевидно. Пусть $n \\ge 4$; обозначим наш многоугольник через $P = A_1A_2...A_n$.\n\nЕсли никакие две хороших диагонали не пересекаются, то по лемме их количество не превосходит $n - 3$. Пусть теперь найдутся две пересекающихся хороших диагонали $A_iA_k$ и $A_jA_\\ell$ ($i < j < k < \\ell$). Тогда каждая из них не пересекается с другими проведёнными диагоналями. Выбросим $A_iA_k$ и $A_jA_\\ell$ из рассмотрения. Каждая оставшаяся проведённая диагональ $d$ является диагональю или стороной ровно в одном из многоугольников $Q_1 = A_i...A_j$, $Q_2 = A_j...A_k$, $Q_3 = A_k...A_\\ell$ или $Q_4 = A_\\ell...A_nA_1...A_i$ (см. рис. 6). При этом, если $d$ является стороной одного из них, то она не может пересекаться с другими диагоналями (и не является хорошей).\n\nПусть $n$ чётно. По предположению индукции, среди всех диагоналей, попавших в какой-то многоугольник $Q_s$, хороших не больше, чем количество вершин в нём, уменьшенное на 2. Значит, общее количество хороших диагоналей в $P$ не превосходит\n$$2 + (j - i - 1) + (k - j - 1) + (l - k - 1) + (n - l + i - 1) = n - 2, (*)$$\n\nПри нечётном $n$ сумма количеств вершин в многоугольниках $Q_1, Q_2, Q_3, Q_4$ равна нечётному числу $n + 4$; значит, число вершин в одном из них нечётно. А тогда соответствующее слагаемое в сумме (*) уменьшится на 1, и мы получим, что общее число хороших диагоналей не превосходит $n - 3$. Переход индукции завершён.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\nРис. 7\n\nОсталось привести примеры, показывающие точность оценки. При чётном $n$ достаточно провести в многоугольнике $A_1...A_n$ диагонали $A_iA_{n-i}$ и $A_{i+1}A_{n-i+1}$ при всех $i = 1, 2, ..., n/2 - 1$ (см. рис. 7); они разбиваются на пары пересекающихся хороших диагоналей. При нечётном $n$ достаточно провести такие же диагонали в чётноугольнике $A_1...A_{n-1}$.\n\n**Замечание.** При нечётном $n$ существуют и принципиально другие примеры — например, можно взять все диагонали из точки $A_1$ (они будут хорошими) и добавить к ним диагональ $A_2A_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Точка $M$ — середина стороны $AC$ остроугольного треугольника $ABC$, в котором $AB > BC$. Окружность $\\Omega$ описана около треугольника $ABC$. Касательные к $\\Omega$, проведённые в точках $A$ и $C$, пересекаются в точке $P$. Отрезки $BP$ и $AC$ пересекаются в точке $S$. Пусть $AD$ — высота треугольника $ABP$. Окружность $\\omega$, описанная около треугольника $CSD$, пересекает окружность $\\Omega$ в точке $K \\neq C$. Докажите, что $\\angle CKM = 90^\\circ$.\n\n(В. Шмаров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $\\angle AMP = \\angle ADP = 90^\\circ$, точки $M$ и $D$ лежат на окружности $\\gamma$ с диаметром $AP$. Поскольку $PA$ — касательная к $\\Omega$, имеем $\\angle KAP = \\angle ACK$. Так как точки $C, K, D$ и $S$ лежат на окружности $\\omega$, имеем $\\angle ACK = \\angle KDP$. Значит, $\\angle KAP = \\angle ACK = \\angle KDP$,\n\nто есть точки $A, D, K$ и $P$ лежат на одной окружности. Итак, точка $K$ лежит на $\\gamma$, и $\\angle AKP = 90^\\circ$ (см. рис. 8).\n\n![](attached_image_1.png)\n\nОтсюда имеем $\\angle MKP = 180^\\circ - \\angle MAP = 180^\\circ - \\angle ABC = \\angle AKC$. Значит, $\\angle MKC = \\angle AKC - \\angle AKM = \\angle MKP - \\angle AKM = \\angle AKP = 90^\\circ$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39753,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\ldots, 10000$ are put in some order into the squares of a checkered $100 \\times 100$ board, one number per square. Pete marks the squares according to the following rules. At the beginning, he just marks $k$ squares by his own choice. By any subsequent move, he may mark any unmarked square containing a number $a$, if either (i) its row contains an already marked square with a number smaller than $a$, or (ii) its column contains an already marked square with a number greater than $a$.\n\nFind the least $k$ such that for any arrangement of the numbers Pete is able to mark all the squares of the table. (S. Berlov)",
    "options": [],
    "answer": "k = 1",
    "solution": "Докажем вначале следующее утверждение.\n**Лемма.** Для любых двух клеток $A$ и $B$ существует такая клетка $C$, закрасив которую, можно затем закрасить и $A$, и $B$ (возможно, $C$ совпадает с $A$ или с $B$.)\n\n**Доказательство.** Можно считать, что номер $a$ клетки $A$ меньше, чем номер $b$ клетки $B$. Пусть $D$ — клетка в одном столбце с $A$ и в одной строке с $B$, и пусть $d$ — её номер (возможно, $D = A$ или $D = B$). Тогда, если $d < a$, то после закрашивания $A$ можно последовательно закрасить $D$ и $B$; если $a \\le d \\le b$, то после закрашивания $D$ можно закрасить как $A$, так и $B$; наконец, если $d > b$, то после закрашивания $B$ можно последовательно закрасить $D$ и $A$. Итак, в любом случае в качестве $C$ можно выбрать одну из клеток $A, B$ и $D$. Лемма доказана. $\\square$\n\nПерейдём к решению задачи. Ясно, что $k \\ge 1$; значит, достаточно доказать, что при $k = 1$ закраска всегда возможна.\n\nЗафиксируем произвольную нумерацию клеток. Рассмотрим все способы закрашивания клеток согласно условию (при $k = 1$) и выберем из них тот, в котором количество закрашенных клеток максимально. Пусть в этом способе первая закрашенная клетка — $A$. Предположим, что при этом способе какая-то клетка $B$ осталась незакрашенной. Тогда, выбрав по Лемме соответствующую клетку $C$ и начав закрашивание с неё, мы потом сможем закрасить $B$, $A$, и, как следствие, все клетки, закрашенные в выбранном способе. Значит, всего мы закрасим хотя бы на одну клетку больше. Противоречие с выбором способа показывает, что на самом деле в нашем способе будут закрашены все клетки. Это и означает, что $k = 1$ подходит.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a midpoint of the side $AC$ of an acute-angled triangle $ABC$ with $AB > BC$. Let $\\Omega$ be the circumcircle of the triangle $ABC$. The tangents to $\\Omega$ at points $A$ and $C$ meet at $P$. The segments $BP$ and $AC$ meet at $S$. Let $AD$ be the altitude in the triangle $ABP$. The circumcircle $\\omega$ of the triangle $CSD$ intersects $\\Omega$ at point $K \\neq C$. Prove that $\\angle CKM = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Поскольку $\\angle AMP = \\angle ADP = 90^\\circ$, точки $M$ и $D$ лежат на окружности $\\gamma$ с диаметром $AP$. Поскольку $PA$ — касательная к $\\Omega$, имеем $\\angle KAP = \\angle ACK$. Так как точки $C, K, D$ и $S$ лежат на окружности $\\omega$, имеем $\\angle ACK = \\angle KDP$. Значит, $\\angle KAP = \\angle ACK = \\angle KDP$,\n\nто есть точки $A, D, K$ и $P$ лежат на одной окружности. Итак, точка $K$ лежит на $\\gamma$, и $\\angle AKP = 90^\\circ$ (см. рис. 8).\n\n![](attached_image_1.png)\n\nОтсюда имеем $\\angle MKP = 180^\\circ - \\angle MAP = 180^\\circ - \\angle ABC = \\angle AKC$. Значит, $\\angle MKC = \\angle AKC - \\angle AKM = \\angle MKP - \\angle AKM = \\angle AKP = 90^\\circ$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Все клетки квадратной таблицы $100 \\times 100$ пронумерованы в некотором порядке числами от $1$ до $10000$. Петя закрашивает клетки по следующим правилам. Вначале он закрашивает $k$ клеток по своему усмотрению. Далее каждым ходом Петя может закрасить одну еще не закрашенную клетку с номером $a$, если для неё выполнено хотя бы одно из двух условий: либо в одной строке с ней есть уже закрашенная клетка с номером меньшим, чем $a$; либо в одном столбце с ней есть уже закрашенная клетка с номером большим, чем $a$. При каком наименьшем $k$ независимо от исходной нумерации Петя за несколько ходов сможет закрасить все клетки таблицы?",
    "options": [],
    "answer": "1",
    "solution": "Докажем вначале следующее утверждение.\n**Лемма.** Для любых двух клеток $A$ и $B$ существует такая клетка $C$, закрасив которую, можно затем закрасить и $A$, и $B$ (возможно, $C$ совпадает с $A$ или с $B$.)\n\n**Доказательство.** Можно считать, что номер $a$ клетки $A$ меньше, чем номер $b$ клетки $B$. Пусть $D$ — клетка в одном столбце с $A$ и в одной строке с $B$, и пусть $d$ — её номер (возможно, $D = A$ или $D = B$). Тогда, если $d < a$, то после закрашивания $A$ можно последовательно закрасить $D$ и $B$; если $a \\le d \\le b$, то после закрашивания $D$ можно закрасить как $A$, так и $B$; наконец, если $d > b$, то после закрашивания $B$ можно последовательно закрасить $D$ и $A$. Итак, в любом случае в качестве $C$ можно выбрать одну из клеток $A, B$ и $D$. Лемма доказана. $\\square$\n\nПерейдём к решению задачи. Ясно, что $k \\ge 1$; значит, достаточно доказать, что при $k = 1$ закраска всегда возможна.\n\nЗафиксируем произвольную нумерацию клеток. Рассмотрим все способы закрашивания клеток согласно условию (при $k = 1$) и выберем из них тот, в котором количество закрашенных клеток максимально. Пусть в этом способе первая закрашенная клетка — $A$. Предположим, что при этом способе какая-то клетка $B$ осталась незакрашенной. Тогда, выбрав по Лемме соответствующую клетку $C$ и начав закрашивание с неё, мы потом сможем закрасить $B$, $A$, и, как следствие, все клетки, закрашенные в выбранном способе. Значит, всего мы закрасим хотя бы на одну клетку больше. Противоречие с выбором способа показывает, что на самом деле в нашем способе будут закрашены все клетки. Это и означает, что $k = 1$ подходит.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Число $x$ таково, что среди четырёх чисел $x - \\sqrt{2}$, $x - 1/x$, $x + 1/x$, $x^2 + 2\\sqrt{2}$ ровно одно не является целым. Найдите все такие $x$.",
    "options": [],
    "answer": "sqrt(2) - 1",
    "solution": "Обозначим $a = x - \\sqrt{2}$, $b = x - 1/x$, $c = x + 1/x$, $d = x^2 + 2\\sqrt{2}$. Заметим, что $b$ и $c$ не могут одновременно быть целыми. Действительно, тогда число $b+c = 2x$ также целое, значит, $x$ рационально, поэтому как $a$, так и $d$ не будут целыми как суммы рационального и иррационального чисел. Итак, одно из чисел $b$ и $c$ нецелое, а тогда $a$ и $d$ должны оба быть целыми.\n\nЗначит, $x = a + \\sqrt{2}$ при целом $a$. Тогда $d = a^2 + 2a\\sqrt{2} + 2 + 2\\sqrt{2} = (a^2+2) + (2a+2)\\sqrt{2}$, откуда следует, что $2a+2=0$, так как иначе $d$ иррационально. Итак, $a = -1$, откуда $x = \\sqrt{2}-1$. Осталось проверить, что найденное число подходит: для него целыми будут числа $a = -1$, $b = -2$ и $d = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39757,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $N > 1$, let $m$ be its largest divisor smaller than $N$. Find all $N$ such that $N + m$ is a power of $10$.",
    "options": [],
    "answer": "75",
    "solution": "**Ответ.** $75$.\n\nПусть $m$ — наибольший делитель числа $N$, меньший, чем $N$. Тогда $N = m p$, где $p$ — наименьший простой делитель числа $N$. Имеем $N + m = 10^k$, то есть $m(p + 1) = 10^k$. Число в правой части не делится на $3$, поэтому $p > 2$. Отсюда следует, что $N$ — нечётное число, а тогда и $m$ нечётно. Значит, поскольку $10^k$ делится на $m$, получаем $m = 5^s$.\n\nЕсли $m = 1$, то $N = p = 10^k - 1$, что невозможно, так как $10^k - 1$ делится на $9$, то есть не является простым. Значит, $s \\ge 1$, число $N$ делится на $5$, и потому $p \\le 5$.\n\nЕсли $p = 3$, то получаем равенство $4 \\cdot 5^s = 10^k$, откуда $k = 2$, $m = 25$ и $N = 75$.\n\nЕсли же $p = 5$, то $p+1 = 6$, и число $10^k$ делится на $3$, что невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x$ such that exactly one of the four numbers $x - \\sqrt{2}$, $x - 1/x$, $x + 1/x$, and $x^2 + 2\\sqrt{2}$ is not an integer. (N. Agakhanov)",
    "options": [],
    "answer": "sqrt(2) - 1",
    "solution": "Обозначим $a = x - \\sqrt{2}$, $b = x - 1/x$, $c = x + 1/x$, $d = x^2 + 2\\sqrt{2}$. Заметим, что $b$ и $c$ не могут одновременно быть целыми. Действительно, тогда число $b+c = 2x$ также целое, значит, $x$ рационально, поэтому как $a$, так и $d$ не будут целыми как суммы рационального и иррационального чисел. Итак, одно из чисел $b$ и $c$ нецелое, а тогда $a$ и $d$ должны оба быть целыми.\n\nЗначит, $x = a + \\sqrt{2}$ при целом $a$. Тогда $d = a^2 + 2a\\sqrt{2} + 2 + 2\\sqrt{2} = (a^2+2) + (2a+2)\\sqrt{2}$, откуда следует, что $2a+2=0$, так как иначе $d$ иррационально. Итак, $a = -1$, откуда $x = \\sqrt{2}-1$. Осталось проверить, что найденное число подходит: для него целыми будут числа $a = -1$, $b = -2$ и $d = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39759,
    "subject": "Mathematics (Multi-modal)",
    "question": "К натуральному числу $N$ прибавили наибольший его делитель, меньший $N$, и получили степень десятки. Найдите все такие $N$.\n(Н. Атаханов)",
    "options": [],
    "answer": "75",
    "solution": "**Ответ.** 75.\nПусть $m$ — наибольший делитель числа $N$, меньший, чем $N$. Тогда $n = mp$, где $p$ — наименьший простой делитель числа $N$. Имеем $N + m = 10^k$, то есть $m(p + 1) = 10^k$. Число в правой части не делится на 3, поэтому $p > 2$. Отсюда следует, что $N$ — нечётное число, а тогда и $m$ нечётно. Значит, поскольку $10^k$ делится на $m$, получаем $m = 5^s$. Если $m = 1$, то $N = p = 10^k - 1$, что невозможно, так как $10^k - 1$ делится на 9, то есть не является простым. Значит, $s \\ge 1$, число $N$ делится на 5, и потому $p \\le 5$. Если $p = 3$, то получаем равенство $4 \\cdot 5^s = 10^k$, откуда $k = 2$, $m = 25$ и $N = 75$. Если же $p = 5$, то $p+1 = 6$, и число $10^k$ делится на 3, что невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39760,
    "subject": "Mathematics (Multi-modal)",
    "question": "A trapezoid $ABCD$ with the bases $AB$ and $CD$ is inscribed into circle $\\Omega$. A circle $\\omega$ passes through the points $C$ and $D$, and intersects the segments $CA$ and $CB$ at $A_1 \\neq C$ and $B_1 \\neq D$, respectively. The points $A_2$ and $B_2$ are symmetric to $A_1$ and $B_1$ with respect to the midpoints of $CA$ and $CB$, respectively. Prove that the points $A, B, A_2$ and $B_2$ are con-cyclic. (I. Bogdanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Утверждение задачи эквивалентно равенству $CA_2 \\cdot CA = CB_2 \\cdot CB$. Поскольку $AA_1 = CA_2$ и $BB_1 = CB_2$, достаточно доказать, что $AA_1 \\cdot AC = BB_1 \\cdot BC$.\nПусть $D_1$ — вторая точка пересечения $\\omega$ с $AD$ (см. рис. 9). Из симметрии имеем $AD = BC$ и $AD_1 = BB_1$. Из теоремы о производении длин отрезков секущих теперь получаем $AA_1 \\cdot AC = AD_1 \\cdot AD = BB_1 \\cdot BC$, что и требовалось доказать.\n\n![](attached_image_1.png)\nРис. 9\n\n\nВторое решение. Обозначим через $O_1$ и $O$ центры окружностей $\\omega$ и $\\Omega$ соответственно; оба этих центра лежат на серединном перпендикуляре $l$ к основаниям трапеции. Пусть $O_2$ — точка, симметричная $O_1$ относительно $O$ (см. рис. 10). Тогда $O_2$ также лежит на $l$, то есть $O_2A = O_2B$. Далее, проекции точек $O_2$ и $O_1$ на $BC$ симметричны относительно проекции точки $O$, т. е. относительно середины $B'$ отрезка $BC$. Так как проекция точки $O_1$ является серединой отрезка $CB_1$, из симметрии относительно $B'$ получаем, что проекция точки $O_2$ — это середина отрезка $BB_2$. Значит, $B_2O_2 = BO_2$. Аналогично показывается, что $A_2O_2 = AO_2 = BO_2 = B_2O_2$. Итак, точки $A, B, A_2, B_2$ лежат на окружности с центром $O_2$.\n\n![](attached_image_2.png)\nРис. 10",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Имеются 2013 карточек, на которых написана цифра 1, и 2013 карточек, на которых написана цифра 2. Вася складывает из этих карточек 4026-значное число. За один ход Петя может поменять местами некоторые две карточки и заплатить Васе 1 рубль. Процесс заканчивается, когда у Пети получается число, делящееся на 11. Какую наибольшую сумму может заработать Вася, если Петя стремится заплатить как можно меньше?\n(П. Кожевников)",
    "options": [],
    "answer": "5",
    "solution": "Рассмотрим 4026-значное число $A$, состоящее из 2013 единиц и 2013 двоек. Пусть в этом числе в нечётных разрядах стоит $k$ единиц и $\\ell = 2013 - k$ двоек, тогда в чётных разрядах будет $k$ двоек и $\\ell$ единиц (здесь $k$ может принимать любое целое значение от 0 до 2013). Разность сумм цифр в нечётных разрядах и чётных разрядах равна $(k+2\\ell)-(2k+\\ell)=\\ell-k=2013-2k$. Поскольку 2013 делится на 11, число $A$ делится на 11 тогда и только тогда, когда $k$ делится на 11.\n\nЗа один ход $k$ изменяется не более чем на 1. Поэтому, если Вася изначально сложит число $A$, для которого, скажем, $k=5$ (очевидно, это возможно), то, очевидно, Пете потребуется не менее 5 раз изменить $k$ на 1, чтобы впервые получить число, для которого $k$ кратно 11 (т. е. $k=0$ или $k=11$).\n\nПокажем, что наоборот, пяти ходов Пете всегда хватит. Меняя некоторую единицу, стоящую в нечётном разряде, с двойкой, стоящей в чётном разряде, он может уменьшить $k$ на 1 за один ход, если только $k \\neq 0$. Аналогично, меняя некоторую двойку, стоящую в нечётном разряде, с единицей, стоящей в чётном разряде, он может увеличить $k$ на 1 за один ход, если только $k \\neq 2013$. Пусть начальное Васино число давало остаток $r$ при делении на 11. Если $r=0$, то исходное число уже делится на 11, и ничего делать не нужно. Если $1 \\le r \\le 5$, то за $r$ операций Петя может уменьшить $k$ на $r$ до ближайшего числа, кратного 11. Если же $6 \\le r \\le 10$, то за $11-r \\le 5$ операций Петя может увеличить $k$ на $11-r$ до ближайшего числа, кратного 11 (это возможно, так как наибольшее возможное значение $k = 2013$ кратно 11).\n\n**Ответ:** 5 рублей.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pete and Bazil have a deck of $4026$ cards, $2013$ of which contain a digit $1$, while $2013$ others contain a digit $2$. Initially, Bazil uses all these cards to make a $4026$-digit number. After that, Pete makes several moves, each consisting in switching two cards and paying $1$ rouble to Bazil. The process stops when the resulting $4026$-digit number is divisible by $11$. Find the maximal amount of roubles Bazil may guarantee to receive by a suitable choice of the initial number. (P. Kozhevnikov)",
    "options": [],
    "answer": "5",
    "solution": "Рассмотрим $4026$-значное число $A$, состоящее из $2013$ единиц и $2013$ двоек. Пусть в этом числе в нечётных разрядах стоит $k$ единиц и $\\ell = 2013 - k$ двоек, тогда в чётных разрядах будет $k$ двоек и $\\ell$ единиц (здесь $k$ может принимать любое целое значение от $0$ до $2013$). Разность сумм цифр в нечётных разрядах и чётных разрядах равна $(k + 2\\ell) - (2k + \\ell) = \\ell - k = 2013 - 2k$. Поскольку $2013$ делится на $11$, число $A$ делится на $11$ тогда и только тогда, когда $k$ делится на $11$.\n\nЗа один ход $k$ изменяется не более чем на $1$. Поэтому, если Вася изначально сложит число $A$, для которого, скажем, $k = 5$ (очевидно, это возможно), то, очевидно, Пете потребуется не менее $5$ раз изменить $k$ на $1$, чтобы впервые получить число, для которого $k$ кратно $11$ (т. е. $k = 0$ или $k = 11$).\n\nПокажем, что наоборот, пяти ходов Пете всегда хватит. Меняя некоторую единицу, стоящую в нечётном разряде, с двойкой, стоящей в чётном разряде, он может уменьшить $k$ на $1$ за один ход, если только $k \\neq 0$. Аналогично, меняя некоторую двойку, стоящую в нечётном разряде, с единицей, стоящей в чётном разряде, он может увеличить $k$ на $1$ за один ход, если только $k \\neq 2013$. Пусть начальное Васино число давало остаток $r$ при делении на $11$. Если $r = 0$, то исходное число уже делится на $11$, и ничего делать не нужно. Если $1 \\le r \\le 5$, то за $r$ операций Петя может уменьшить $k$ на $r$ до ближайшего числа, кратного $11$. Если же $6 \\le r \\le 10$, то за $11 - r \\le 5$ операций Петя может увеличить $k$ на $11 - r$ до ближайшего числа кратного $11$ (это возможно, так как наибольшее возможное значение $k = 2013$ кратно $11$).\n\n**Замечание.** Если бы у мальчиков было по $n$ карточек с цифрами $1$ и $2$, ответ в аналогичной задаче зависел бы от остатка от деления $n$ на $11$ — даже при больших $n$. Так, если $n = 2000$, то Пете нужно добиться того, чтобы $k$ имело остаток $10$ при делении на $11$; значит, если $k = 2000$ или $k = 0$, то Пете может понадобиться $10$ ходов.\nПоэтому полное решение задачи должно существенно использовать тот факт, что $n = 2013$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Трапеция $ABCD$ с основаниями $AB$ и $CD$ вписана в окружность $\\Omega$. Окружность $\\omega$ проходит через точки $C, D$ и пересекает отрезки $CA, CB$ в точках $A_1, B_1$ соответственно. Точки $A_2$ и $B_2$ симметричны точкам $A_1$ и $B_1$ относительно середин отрезков $CA$ и $CB$ соответственно. Докажите, что точки $A, B, A_2$ и $B_2$ лежат на одной окружности.\n(И. Богданов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Утверждение задачи эквивалентно равенству $CA_2 \\cdot CA = CB_2 \\cdot CB$. Поскольку $AA_1 = CA_2$ и $BB_1 = CB_2$, достаточно доказать, что $AA_1 \\cdot AC = BB_1 \\cdot BC$.\nПусть $D_1$ — вторая точка пересечения $\\omega$ с $AD$ (см. рис. 9). Из симметрии имеем $AD = BC$ и $AD_1 = BB_1$. Из теоремы о производении длин отрезков секущих теперь получаем $AA_1 \\cdot AC = AD_1 \\cdot AD = BB_1 \\cdot BC$, что и требовалось доказать.\n\n![](attached_image_1.png)\nРис. 9\n\n\n**Второе решение.** Обозначим через $O_1$ и $O$ центры окружностей $\\omega$ и $\\Omega$ соответственно; оба этих центра лежат на серединном перпендикуляре $l$ к основаниям трапеции. Пусть $O_2$ — точка, симметричная $O_1$ относительно $O$ (см. рис. 10). Тогда $O_2$ также лежит на $l$, то есть $O_2A = O_2B$. Далее, проекции точек $O_2$ и $O_1$ на $BC$ симметричны относительно проекции точки $O$, т. е. относительно середины $B'$ отрезка $BC$. Так как проекция точки $O_1$ является серединой отрезка $CB_1$, из симметрии относительно $B'$ получаем, что проекция точки $O_2$ — это середина отрезка $BB_2$. Значит, $B_2O_2 = BO_2$. Аналогично показывается, что $A_2O_2 = AO_2 = BO_2 = B_2O_2$. Итак, точки $A, B, A_2, B_2$ лежат на окружности с центром $O_2$.\n\n![](attached_image_2.png)\nРис. 10",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an inscribed quadrilateral for which the rays $AB$ and $DC$ meet at $K$. It happened that the points $B$, $D$, and the midpoints of the segments $AC$ and $KC$ are concyclic. Find all possible values the angle $ADC$ may get.",
    "options": [],
    "answer": "90°",
    "solution": "Обозначим через $N$ и $M$ середины отрезков $KC$ и $AC$ соответственно. Тогда $MN$ — средняя линия в треугольнике $AKC$, поэтому $\\angle BAC = \\angle NMC$. Кроме того, $\\angle BAC = \\angle BDC$, так как четырехугольник $ABCD$ — вписанный.\n\nПусть точки $M$ и $N$ лежат с одной стороны от прямой $BD$. Тогда $M$ лежит внутри треугольника $BCD$; тогда она лежит внутри треугольника $BND$, а значит, и внутри его описанной окружности. Но тогда точки $B, N, D$ и $M$ не могут лежать на одной окружности. Значит, $N$ и $M$ лежат по разные стороны от $BD$, и $\\angle BDC = \\angle BMN$.\n\n![](attached_image_1.png)\n\nИз параллельности $MN$ и $AK$ вытекает, что $\\angle BMN = \\angle ABM$, откуда $\\angle BAC = \\angle BDC = \\angle ABM$. Отсюда получаем $AM = MB$, то есть в треугольнике $ABC$ медиана $BM$ равна половине стороны $AC$, откуда $\\angle ABC = 90^\\circ$, а значит, и $\\angle ADC = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39765,
    "subject": "Mathematics (Multi-modal)",
    "question": "The treasurer of Math republic chose a number $\\alpha > 2$ and issued coins with values of 1 rouble and of $\\alpha^k$ roubles for all positive integer $k$. It turns out that all the values of coins (except for 1) are irrational. May it happen that for any positive integer $n$, one may take several coins whose values sum up to $n$ roubles so that the coins of each value are taken at most six times? (I. Bogdanov, S. Berlov)",
    "options": [],
    "answer": "Yes",
    "solution": "Ответ. Могло.\n\nПокажем, что математики могли выбрать число $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$; это число является корнем уравнения $\\alpha^2 + \\alpha = 7$. Ясно, что $\\alpha > 2$. Нетрудно видеть, что при натуральных $m$ мы имеем $(2\\alpha)^m = a_m + b_m\\sqrt{29}$, где $a_m$ и $b_m$ — целые числа, причём $a_m < 0 < b_m$ при нечётных $m$ и $a_m > 0 > b_m$ при чётных $m$. Значит, число $\\alpha^m$ иррационально.\n\nОсталось показать, что для любого натурального числа $n$ сумму в $n$ рублей можно набрать требуемым способом. Рассмотрим все способы набрать $n$ рублей выпущенными монетами (хотя бы один такой способ существует: можно взять $n$ рублёвых монет). Выберем из них способ, в котором наименьшее число монет. Предположим, что какая-то монета достоинства $\\alpha^i$ ($i \\ge 0$) встречается в этом способе хотя бы 7 раз. Тогда можно заменить 7 монет по $\\alpha^i$ монетами достоинств $\\alpha^{i+1}$ и $\\alpha^{i+2}$. При этом суммарное достоинство монет не изменится (поскольку $\\alpha^{i+1} + \\alpha^{i+2} = 7\\alpha^i$), а их количество уменьшится.\n\nЭто невозможно по выбору нашего способа. Итак, этот способ удовлетворяет условию.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39766,
    "subject": "Mathematics (Multi-modal)",
    "question": "В республике математиков выбрали число $\\alpha > 2$ и выпустили монеты достоинствами в 1 рубль, а также в $\\alpha^k$ рублей при каждом натуральном $k$. При этом $\\alpha$ было выбрано так, что достоинства всех монет, кроме самой мелкой, иррациональны. Могло ли оказаться, что любую сумму в натуральное число рублей можно набрать этими монетами,\n\nиспользуя монеты каждого достоинства не более 6 раз?",
    "options": [],
    "answer": "Yes; for example, choose alpha = (-1 + sqrt(29)) / 2, which makes every power irrational and ensures any integer sum can be formed with at most six coins of each denomination.",
    "solution": "Могло.\n\nПокажем, что математики могли выбрать число $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$; это число является корнем уравнения $\\alpha^2 + \\alpha = 7$. Ясно, что $\\alpha > 2$. Нетрудно видеть, что при натуральных $m$ мы имеем $(2\\alpha)^m = a_m + b_m\\sqrt{29}$, где $a_m$ и $b_m$ — целые числа, причём $a_m < 0 < b_m$ при нечётных $m$ и $a_m > 0 > b_m$ при чётных $m$. Значит, число $\\alpha^m$ иррационально.\n\nОсталось показать, что для любого натурального числа $n$ сумму в $n$ рублей можно набрать требуемым способом. Рассмотрим все способы набрать $n$ рублей выпущенными монетами (хотя бы один такой способ существует: можно взять $n$ рублёвых монет). Выберем из них способ, в котором наименьшее число монет. Предположим, что какая-то монета достоинства $\\alpha^i$ ($i \\ge 0$) встречается в этом способе хотя бы 7 раз. Тогда можно заменить 7 монет по $\\alpha^i$ монетами достоинств $\\alpha^{i+1}$ и $\\alpha^{i+2}$. При этом суммарное достоинство монет не изменится (поскольку $\\alpha^{i+1} + \\alpha^{i+2} = 7\\alpha^i$), а их количество уменьшится.\n\nЭто невозможно по выбору нашего способа. Итак, этот способ удовлетворяет условию.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Дан вписанный четырехугольник $ABCD$. Лучи $AB$ и $DC$ пересекаются в точке $K$. Оказалось, что точки $B$, $D$, а также середины отрезков $AC$ и $KC$ лежат на одной окружности. Какие значения может принимать угол $ADC$?",
    "options": [],
    "answer": "90 degrees",
    "solution": "Обозначим через $N$ и $M$ середины отрезков $KC$ и $AC$ соответственно. Тогда $MN$ — средняя линия в треугольнике $AKC$, поэтому $\\angle BAC = \\angle NMC$. Кроме того, $\\angle BAC = \\angle BDC$, так как четырехугольник $ABCD$ — вписанный.\n\nПусть точки $M$ и $N$ лежат с одной стороны от прямой $BD$. Тогда $M$ лежит внутри треугольника $BCD$; тогда она лежит внутри треугольника $BND$, а значит, и внутри его описанной окружности. Но тогда точки $B$, $N$, $D$ и $M$ не могут лежать на одной окружности. Значит, $N$ и $M$ лежат по разные стороны от $BD$, и $\\angle BDC = \\angle BMN$.\n\n![](attached_image_1.png)\n\nИз параллельности $MN$ и $AK$ вытекает, что $\\angle BMN = \\angle ABM$, откуда $\\angle BAC = \\angle BDC = \\angle ABM$. Отсюда получаем $AM = MB$, то есть в треугольнике $ABC$ медиана $BM$ равна половине стороны $AC$, откуда $\\angle ABC = 90^\\circ$, а значит, и $\\angle ADC = 90^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine which of two numbers $(100!)!$ and $99!^{100!} \\cdot 100!^{99!}$ is greater than the other one.",
    "options": [],
    "answer": "99!^{100!} · 100!^{99!} is greater than (100!)!",
    "solution": "Let $a = 99!$. Then we need to compare the numbers $(100a)!$ and $a^{100a} \\cdot (100a)^a$.\n\nNote that\n$$\n\\begin{aligned}\n& 1 \\cdot 2 \\cdot 3 \\cdots a < a^a, \\\\\n& (a+1)(a+2)(a+3) \\cdots 2a < (2a)^a, \\\\\n& (2a+1)(2a+2)(2a+3) \\cdots 3a < (3a)^a, \\\\\n& \\vdots \\\\\n& (99a+1)(99a+2)(99a+3) \\cdots 100a < (100a)^a.\n\\end{aligned}\n$$\n\nMultiply these inequalities. On the left, we get the product of all numbers from $1$ to $100a$, that is, exactly $(100a)! = (100!)!$. On the right, we get $a^a (2a)^a (3a)^a \\dots (100a)^a = a^{100a} (1 \\cdot 2 \\cdot 3 \\dots 100)^a = a^{100a} (100!)^a$, that is, exactly $a^{100a} (100a)^a = 99!^{100!} \\cdot 100!^{99!}$.\n\nTherefore, $(100!)! < 99!^{100!} \\cdot 100!^{99!}$, so the second number is greater.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39769,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country, there are $n$ cities, and every two of them are connected with a non-stop train operating in both directions. The ticket price for each train in both directions is the same, but for any two different trains these prices are different. Prove that a traveler may start from some city and take $n-1$ trains consecutively so that the price of every ticket will be less than the price of the previous one. (A traveler may pass through a certain city several times.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Первое решение. Уберём все экспресслы, а затем начнём запускать их обратно по одному в порядке возрастания цены (т. е. первым запустим самый дешёвый, вторым — самый дешёвый из остальных, и т. д.). В каждый момент в каждом городе будем писать максимальное количество экспрессов, на которых можно последовательно проехать, начав из этого города, так, чтобы цены проезда монотонно убывали.\nВ начальный момент все числа в городах равны нулю. Пусть в некоторый момент мы вводим экспресс, соединяющий города $A$ и $B$, в которых до этого были написаны числа $a$ и $b$ соответственно. После введения нового экспресса в $A$ будет число, не меньше $b + 1$ (ибо теперь из $A$ можно проехать новым экспрессом в $B$, а затем по маршруту длины $b$, начинавшемуся из $B$). Аналогично, в $B$ будет написано число, не меньшее $a + 1$. Поэтому сумма чисел в $A$ и $B$ увеличится хотя бы на $2$, а числа в остальных городах не уменьшатся. Значит, и сумма всех чисел в городах увеличится хотя бы на $2$.\n\nТаким образом, когда все $n(n-1)/2$ экспрессов будут введены, сумма чисел в городах станет не меньше, чем $2 \\cdot n(n-1)/2 = n(n-1)$. Значит, хотя бы в одном городе будет число, не меньшее $n-1$. Это и означает наличие требуемого маршрута из этого города.\n\nВторое решение. Разделим каждый экспресс, курсирующий между $A$ и $B$, на два — идущий из $A$ в $B$, и идущий из $B$ в $A$. Получились $n(n-1)$ полуэкспрессов. Мы построим $n$ выделенных маршрутов (по одному, начинающемуся в каждом городе) так, чтобы цена поездки на каждом монотонно убывала, и каждый полуэкспресс содержался бы хотя бы в одном выделенном маршруте. Тогда один из выделенных маршрутов будет содержать не менее $n-1$ полуэкспресса, что и требовалось.\n\nВыделенный маршрут, начинающийся в городе $A$, выглядит так. Пусть $A_0 = A$. Рассмотрим все полуэкспрессы $A_0X$, выходящие из $A_0$, и выберем из них полуэкспресс $A_0A_1$ максимальной цены $a_1$. Затем рассмотрим все полуэкспрессы $A_1Y$, выходящие из $A_1$, цена которых меньше $a_1$; если такие есть, выберем из них полуэкспресс $A_1A_2$ максимальной цены $a_2$, и т.д. Маршрут заканчивается полуэкспрессом $A_{k-1}A_k$, если из $A_k$ не выходит полуэкспрессов с ценой, меньшей $a_k$.\n\nОсталось показать, что каждый полуэкспресс $BC$ попадёт хотя бы в один из выделенных маршрутов. Положим $B_1 = B$, $B_0 = C$, и пусть $b_1$ — цена $BC$. Рассмотрим все полуэкспрессы $XB_1$, ведущие в $B_1$, с ценой, большей $b_1$. Если такие есть, то выберем из них полуэкспресс $B_2B_1$ наименьшей цены $b_2$. Далее рассмотрим все полуэкспрессы $YB_2$, ведущие в $B_2$, с ценой, большей $b_2$. Выберем из них полуэкспресс $B_3B_2$ наименьшей цены $b_3$, и т.д. Этот процесс выбора закончится, когда при некотором $k$ полуэкспресс $B_kB_{k-1}$ — это полуэкспресс максимальной цены, выходящий из $B_k$. Тогда, согласно нашему построению, выделенный маршрут, выходящий из $B_k$, последовательно пройдёт через $B_{k-1}, \\dots, B_1, B_0$, то есть будет содержать экспресс $BC$, что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39770,
    "subject": "Mathematics (Multi-modal)",
    "question": "В государстве $n$ городов, и между каждыми двумя из них курсирует экспресс (в обе стороны). Для любого экспресса цены билетов «туда» и «обратно» равны, а для любых разных экспрессов эти цены различны. Докажите, что путешественник может выбрать начальный город, выехать из него и проехать последовательно на $n-1$ экспрессах, платя за проезд на каждом следующем меньше, чем за проезд на предыдущем. (Путешественник может попадать несколько раз в один и тот же город.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Уберём все экспрессы, а затем начнём запускать их обратно по одному в порядке возрастания цены (т. е. первым запустим самый дешёвый, вторым — самый дешёвый из остальных, и т. д.). В каждый момент в каждом городе будем писать максимальное количество экспрессов, на которых можно последовательно проехать, начав из этого города, так, чтобы цены проезда монотонно убывали.\nВ начальный момент все числа в городах равны нулю. Пусть в некоторый момент мы вводим экспресс, соединяющий города $A$ и $B$, в которых до этого были написаны числа $a$ и $b$ соответственно. После введения нового экспресса в $A$ будет число, не меньше $b + 1$ (ибо теперь из $A$ можно проехать новым экспрессом в $B$, а затем по маршруту длины $b$, начинавшемуся из $B$). Аналогично, в $B$ будет написано число, не меньшее $a + 1$. Поэтому сумма чисел в $A$ и $B$ увеличится хотя бы на $2$, а числа в остальных городах не уменьшатся. Значит, и сумма всех чисел в городах увеличится хотя бы на $2$.\n\nТаким образом, когда все $n(n-1)/2$ экспрессов будут введены, сумма чисел в городах станет не меньше, чем $2 \\cdot n(n-1)/2 = n(n-1)$. Значит, хотя бы в одном городе будет число, не меньшее $n-1$. Это и означает наличие требуемого маршрута из этого города.\n\n\nВторое решение. Разделим каждый экспресс, курсирующий между $A$ и $B$, на два — идущий из $A$ в $B$, и идущий из $B$ в $A$. Получились $n(n-1)$ полуэкспрессов. Мы построим $n$ выделенных маршрутов (по одному, начинающемуся в каждом городе) так, чтобы цена поездки на каждом монотонно убывала, и каждый полуэкспресс содержался бы хотя бы в одном выделенном маршруте. Тогда один из выделенных маршрутов будет содержать не менее $n-1$ полуэкспресса, что и требовалось.\n\nВыделенный маршрут, начинающийся в городе $A$, выглядит так. Пусть $A_0 = A$. Рассмотрим все полуэкспрессы $A_0X$, выходящие из $A_0$, и выберем из них полуэкспресс $A_0A_1$ максимальной цены $a_1$. Затем рассмотрим все полуэкспрессы $A_1Y$, выходящие из $A_1$, цена которых меньше $a_1$; если такие есть, выберем из них полуэкспресс $A_1A_2$ максимальной цены $a_2$, и т.д. Маршрут заканчивается полуэкспрессом $A_{k-1}A_k$, если из $A_k$ не выходит полуэкспрессов с ценой, меньшей $a_k$.\n\nОсталось показать, что каждый полуэкспресс $BC$ попадёт хотя бы в один из выделенных маршрутов. Положим $B_1 = B$, $B_0 = C$, и пусть $b_1$ — цена $BC$. Рассмотрим все полуэкспрессы $XB_1$, ведущие в $B_1$, с ценой, большей $b_1$. Если такие есть, то выберем из них полуэкспресс $B_2B_1$ наименьшей цены $b_2$. Далее рассмотрим все полуэкспрессы $YB_2$, ведущие в $B_2$, с ценой, большей $b_2$. Выберем из них полуэкспресс $B_3B_2$ наименьшей цены $b_3$, и т.д. Этот процесс выбора закончится, когда при некотором $k$ полуэкспресс $B_kB_{k-1}$ — это полуэкспресс максимальной цены, выходящий из $B_k$. Тогда, согласно нашему построению, выделенный маршрут, выходящий из $B_k$, последовательно пройдёт через $B_{k-1}, \\dots, B_1, B_0$, то есть будет содержать экспресс $BC$, что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Какое из чисел больше: $(100!)!$ или $99!^{100!} \\cdot 100!^{99!}$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Второе число больше.\n\nПусть $a = 99!$. Тогда нам нужно сравнить числа $(100a)!$ и $a^{100a} \\cdot (100a)^a$. Заметим, что\n$$\n\\begin{aligned}\n& 1 \\cdot 2 \\cdot 3 \\cdots a < a^a, \\\\\n& (a+1)(a+2)(a+3) \\cdots 2a < (2a)^a, \\\\\n& (2a+1)(2a+2)(2a+3) \\cdots 3a < (3a)^a, \\\\\n& \\vdots \\\\\n& (99a+1)(99a+2)(99a+3) \\cdots 100a < (100a)^a.\n\\end{aligned}\n$$\n\nПеремножим эти неравенства. Слева получим произведение всех чисел от $1$ до $100a$, т. е. в точности $(100a)! = (100!)!$, а справа число $a^a (2a)^a (3a)^a \\dots (100a)^a = a^{100a} (1 \\cdot 2 \\cdot 3 \\dots 100)^a = a^{100a} (100!)^a$, т. е. в точности $a^{100a} (100a)^a = 99!^{100!} \\cdot 100!^{99!}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39772,
    "subject": "Mathematics (Multi-modal)",
    "question": "In NBA, there are 30 teams; in a regular season every team is to participate in 82 games. Is it possible to split the teams into two conferences — Western and Eastern, and compose the schedule of the regular season so that the number of games between the teams from different conferences will be exactly the half of the total number of games?\n\nВ Национальной Баскетбольной Ассоциации 30 команд, каждая из которых проводит за год 82 матча с другими командами в регулярном чемпионате. Сможет ли руководство Ассоциации разделить команды (не обязательно поровну) на Восточную и Западную конференции и составить расписание игр так, чтобы матчи между командами из разных конференций составляли ровно половину от общего числа матчей?",
    "options": [],
    "answer": "No",
    "solution": "No.\nThe total number of games between the teams from different conferences should be $30 \\cdot 82/4$, so it should be odd. This cannot happen, since each team from the Western conference participates in an even number of games, and each internal game is accounted twice.\n\nSolution:\nНет, не сможет.\nПусть $x$ и $y$ — общее число матчей, сыгранных внутри Восточной и Западной конференций соответственно, а $z$ — число матчей между командами разных конференций. Нам надо доказать, что равенство $z = \\frac{x+y+z}{2}$ невозможно.\nКаждая из $k$ команд Восточной конференции участвует в 82 играх; значит, $82k = 2x + z$ (коэффициент 2 появился из-за того, что каждый внутренний матч учтён у обеих участвовавших в нём команд). Отсюда число $z = 82k - 2x$ чётно. Но из подсчёта общего числа матчей $x + y + z = \\frac{30 \\cdot 82}{2}$ следует, что число $\\frac{x+y+z}{2} = 15 \\cdot 41$ нечётно. Значит, равенство $z = \\frac{x+y+z}{2}$ не может выполняться.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter chose 10 consecutive positive integers. He wrote down each of the chosen numbers either in red or in blue (both colors are present). Is it possible that the sum of the l.c.m. of the red numbers and the l.c.m. of the blue numbers has a decimal representation ending up with 2016? (O. Dmitriev, R. Zhenodarov)\n\nПетя выбрал 10 последовательных натуральных чисел и каждое записал либо красным, либо синим карандашом (оба цвета присутствуют). Может ли сумма наименьшего общего кратного всех красных чисел и наименьшего общего кратного всех синих чисел оканчиваться на 2016?\n(О. Дмитриев, Р. Женодаров)",
    "options": [],
    "answer": "No",
    "solution": "No.\n\nAmong Peter's numbers, either there is exactly one number divisible by $8$ (then the sum of the l.c.m.'s is not divisible by $8$), or there are exactly two numbers divisible by $8$ (then exactly one of them is divisible by $16$, and the sum is not divisible by $16$). In any case, $16$ does not divide the sum, so it cannot end up with $2016$.\nНет, не может.\n\nСреди десяти петиных чисел есть либо одно, либо два числа, делящихся на $8$. В первом случае одно из полученных наименьших общих кратных (НОК) делится на $8$, а второе — нет, и потому их сумма не делится даже на $8$. Во втором же случае разность двух петиных чисел, делящихся на $8$, равна $8$, поэтому одно из них делится на $16$, а другое — нет. Следовательно, одно из НОК делится на $16$, а другое — нет. Значит, и в этом случае сумма НОК делиться на $16$ не может.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39774,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $P$. The point $Q$ is chosen on the segment $BC$ so that $PQ \\perp AC$. Prove that the line of centers of the circumcircles of the triangles $APD$ and $BQD$ is parallel to $AD$. (A. Kuznetsov)\n\nДиагонали $AC$ и $BD$ вписанного четырехугольника $ABCD$ пересекаются в точке $P$. Точка $Q$ выбрана на отрезке $BC$ так, что $PQ \\perp AC$. Докажите, что прямая, проходящая через центры окружностей, описанных около треугольников $APD$ и $BQD$, параллельна прямой $AD$.\n(А. Кузнецов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose a point $T \\in QP$ so that $DT \\perp DA$ (see Fig. 16). The points $A$, $P$, $D$, and $T$ are concyclic, so the center of the circle $APD$ lies on the perpendicular bisector $\\ell$ of $DT$ (notice that $\\ell \\parallel AD$). Next, $\\angle QBD = \\angle PAD = \\angle QTD$, so the points $B$, $Q$, $D$, and $T$ are also concyclic. So the center of the circle $BQD$ also lies on $\\ell$.\n\n![](attached_image_1.png)\nВыберем на прямой $QP$ точку $T$ такую, что $DT \\perp DA$ (см. рис. 16). Поскольку $\\angle APT = 90^\\circ = \\angle ADT$, точки $A$, $P$, $D$ и $T$ лежат на одной окружности. Значит, центр окружности $\\omega_1$, описанный около треугольника $APD$, лежит на серединном перпендикуляре $\\ell$ к отрезку $DT$.\n\n![](attached_image_1.png)\nРис. 16\n\nТак как четырехугольник $ABCD$ вписан, имеем $\\angle QBD = \\angle PAD$. Четырехугольник $APDT$ также вписан, откуда $\\angle PAD = \\angle QTD$. Итак, $\\angle QBD = \\angle PAD = \\angle QTD$; значит, точки $B$, $Q$, $D$ и $T$ лежат на одной окружности.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a cubic polynomial. We say that a triple $(a, b, c)$ of distinct real numbers is a cycle if $f(a) = b$, $f(b) = c$, and $f(c) = a$. Assume that there exist eight cycles $(a_i, b_i, c_i)$, $i = 1, 2, \\dots, 8$, containing 24 distinct real numbers. Prove that among eight sums of the form $a_i + b_i + c_i$, there are at least three different numbers. (A. S. Golovanov)\n\nДан кубический многочлен $f(x)$. Назовём циклом тройку различных чисел $(a, b, c)$ таких, что $f(a) = b, f(b) = c$ и $f(c) = a$. Известно, что нашлись восемь циклов $(a_i, b_i, c_i)$, $i = 1, 2, \\dots, 8$, в которых участвуют 24 различных числа. Докажите, что среди восьми чисел вида $a_i + b_i + c_i$ есть хотя бы три различных.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If four of the sums of the form $a_i + b_i + c_i$ are equal to $s$, then the corresponding 12 numbers of the form $a_i, b_i, c_i$ are roots of the polynomial $x + f(x) + f(f(x)) - s$, whose degree is 9.\nПредположим противное; тогда у каких-то четырёх циклов $(a_i, b_i, c_i)$ суммы чисел одинаковы и равны некоторому $s$. Для каждого из этих циклов имеем\n$$\ns = a_i + b_i + c_i = a_i + f(a_i) + f(f(a_i)) = \\\\ = b_i + f(b_i) + f(f(b_i)) = c_i + f(c_i) + f(f(c_i)).\n$$\nИтак, все 12 чисел наших четырёх циклов — корни многочлена $g(x) = x + f(x) + f(f(x)) - s$. Однако все эти 12 чисел по условию различны, а степень многочлена $g(x)$ равна 9; значит, у него не может быть больше 9 различных корней. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $L$ are chosen on a side $AB$ of a convex quadrilateral $ABCD$, with $K$ lying between $A$ and $L$. Similarly, points $M$ and $N$ are chosen on a side $CD$, with $M$ lying between $C$ and $N$. It appears that $AK = KN = DN$ and $BL = BC = CM$. Given that $BCNK$ is a cyclic quadrilateral, prove that $ADML$ is also cyclic. (T. Zimanov, P. Kozhevnikov)\n\nНа стороне $AB$ выпуклого четырехугольника $ABCD$ взяты точки $K$ и $L$ (точка $K$ лежит между $A$ и $L$), а на стороне $CD$ взяты точки $M$ и $N$ (точка $M$ между $C$ и $N$). Известно, что $AK = KN = DN$ и $BL = BC = CM$. Докажите, что если $BCNK$ — вписанный четырехугольник, то и $ADML$ тоже вписан.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In the case $AB \\parallel CD$, we have $BC = KN$, so $AK = BL = CM = DN$. Therefore, the quadrilateral $LMDA$ is obtained from $BCNK$ by a parallel translation by the vector $\\vec{BL}$.\n\n![](attached_image_1.png)\n\nNow suppose $AB$ and $CD$ are not parallel; let $P$ be the intersection point of the lines $AB$ and $CD$. Since the quadrilateral $BCNK$ is cyclic, triangles $PBC$ and $PNK$ are similar; hence $\\frac{PB}{BL} = \\frac{PN}{NK} = \\frac{PN}{ND}$. Therefore, $BN \\parallel LD$ (see figure). Similarly, $CK \\parallel MA$. From this, we get $\\angle ALD = \\angle KBN$ and $\\angle KCN = \\angle AMD$.\n\nSince the quadrilateral $BCNK$ is cyclic, $\\angle KBN = \\angle KCN$. Therefore, $\\angle ALD = \\angle AMD$, that is, $ADML$ is also cyclic.\n\n**Remark.** There are other solutions; for example, from the equalities $\\angle AKN = \\angle NCB$ and $\\angle DNK = \\angle KBC$, it follows that the quadrilaterals $BCML$ and $NKAD$ are similar, so $\\angle BLM = \\angle MDA$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each cell of a $100 \\times 100$ table is colored either in black or in white, so that all columns contain equal numbers of black cells, but no two rows contain equal numbers of black cells. Determine the greatest possible number of pairs of neighboring cells whose colors are distinct. (Two cells are neighboring if they share a side.)\n\nДана клетчатая таблица $100 \\times 100$, клетки которой покрашены в чёрный и белый цвета. При этом во всех столбцах поровну чёрных клеток, в то время как во всех строках разные количества чёрных клеток. Каково максимальное возможное количество пар соседних по стороне разноцветных клеток?",
    "options": [],
    "answer": "14751",
    "solution": "Set $2n = 100$. By a counting argument, each column contains $n$ black cells, while the rows contain $0, 1, 2, \\dots, n-1, n+1, n+2, \\dots, 2n$ black cells in some order.\nA row with $i$ black cells contains at most $\\min\\{2i, 2(2n-i)\\}$ dichromatic pairs, so there are at most $2 \\cdot (2 \\cdot 0 + 2 \\cdot 1 + \\dots + 2 \\cdot (n-1)) = 2n(n-1)$ horizontal dichromatic pairs. Next, consider $n-1$ non-bordering rows with no neighboring rows among them. In each such row at least one cell participates in some vertical monochromatic pair, so there are at least $n-1$ monochromatic vertical pairs. Thus the number of dichromatic pairs is at most $2n(n-1) + 2n(2n-1) - (n-1) = (2n-1)(3n-1)$.\nAn example (for $n=4$) showing that the estimate is sharp is presented on Fig. 6.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39778,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $X$ is chosen inside a convex 100-gon ($X$ does not lie on sides or diagonals of the 100-gon). Initially, the vertices of the polygon are not marked. Pete and Bazil mark the vertices in turn. Pete starts and marks two vertices by his first move; after that, by any move a player marks one vertex. If after someone's move the point $X$ appears to lie inside a polygon whose vertices are all marked, the player who made this move loses the game. Prove that Pete has a winning strategy.\n\nВпутри выпуклого 100-угольника выбрана точка $X$, не лежащая ни на одной его стороне или диагонали. Исходно вершины многоугольника не отмечены. Петя и Вася по очереди отмечают ещё не отмеченные вершины 100-угольника, причём Петя начинает и первым ходом отмечает сразу две вершины, а далее каждый своим очередным ходом отмечает по одной вершине. Проигрывает тот, после чьего хода точка $X$ будет лежать впутри многоугольника с отмеченными вершинами. Докажите, что Петя может выиграть, как бы ни ходил Вася.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Paint the sides of the polygon alternately in black and white. Consider some convex quadrilateral $ABCD$ where $AB$ and $CD$ are sides of the same color; let $K$ be the meeting point of its diagonals. If $X$ lies in the triangle $KBC$ (see Fig. 17), then Pete can mark $B$ and $C$ by his first move and win (no succeeding moves can be made at the same half-plane of $BC$ as $X$). It remains to prove that there exists such quadrilateral.\n\nAssume the contrary. Let $T$ be a vertex of the 100-gon, let the ray $TX$ intersect a black side $PQ$, and let $TR$ be a black side adjacent to $T$ (see Fig. 18). Then the ray $RX$ should also meet the segment $PQ$. Next, if $S$ is the other vertex adjacent to $R$, then the ray $SX$ also should cross a black side (otherwise the ray $RX$ would meet a white side). Proceeding similarly, we get that for every black side $MN$, the rays $MX$ and $NX$ should both cross one black side; but this fails for the black side $PQ$.\nРаскрасим стороны 100-угольника в чёрный и белый цвета так, чтобы любые две соседних стороны имели разные цвета. Рассмотрим две одноцветных стороны $AB$ и $CD$, образующие выпуклый четырёхугольник $ABCD$; пусть его диагонали $AC$ и $BD$ пересекаются в точке $K$. Предположим, что точка $X$ лежит в треугольнике $KBC$ (см. рис. 17). Покажем, как Петя может играть в этом случае.\nПусть он выберет первым ходом вершины $B$ и $C$. После этого оба игрока могут выбирать только вершины, лежащие в другой полуплоскости от прямой $BC$, нежели точка $X$. Этих вершин чётное число, поскольку они разбиваются на пары вершин, образующих стороны того же цвета, что и $AB$. Поэтому последний ход будет за Петей.\n\nОсталось показать, что такие стороны $AB$ и $CD$ найдутся. Пусть это не так. Рассмотрим любую вершину $T$. Предположим, что луч $TX$ пересекает чёрную сторону $PQ$ (см. рис. 18). Пусть $TR$ — чёрная сторона, выходящая из $T$; можно считать, что $TRPQ$ — выпуклый четырёхугольник. Если точка $X$ лежит внутри треугольника $TRQ$, то требуемый четырёхугольник $RTQP$ найден; в противном случае луч $RX$ также должен пересекать отрезок $PQ$.\n\nПусть $RS$ — следующая за $TR$ сторона 100-угольника. Если луч $SX$ пересекает белую сторону, то аналогично доказывается, что луч $RX$ также должен её пересекать, что не так. Значит, $SX$ пересекает какую-то чёрную сторону, и можно повторить предыдущие рассуждения для вершины $S$. Рассуждая так и дальше, мы получим, что для каждой чёрной стороны $T'R'$ найдётся чёрная сторона $P'Q'$, которую пересекают оба луча $T'X$ и $R'X$. Однако это неверно для чёрной стороны $PQ$ (лучи $PX$ и $QX$ пересекают участки контура $QT$ и $RP$ соответственно) — противоречие.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39779,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a non-empty set $A$ consisting of real numbers is complete if for any real $a$ and $b$ such that $a+b \\in A$, the number $ab$ also lies in $A$ (the numbers $a$ and $b$ are not required to be distinct or to belong to $A$). Find all complete sets.\n\nНазовём непустое (конечное или бесконечное) множество $A$, состоящее из действительных чисел, полным, если для любых действительных $a$ и $b$ (не обязательно различных и не обязательно лежащих в $A$) таких, что $a+b$ лежит в $A$, число $ab$ также лежит в $A$. Найдите все полные множества действительных чисел.",
    "options": [],
    "answer": "the set of all real numbers",
    "solution": "The only such set is the set of reals.\nTake any $a \\in A$. Show successively that (i) $0 = 0 \\cdot a \\in A$, (ii) $-x^2 = (-x) \\cdot x \\in A$ for all $x \\in \\mathbb{R}$, and (iii) $y^2 = (-y) \\cdot (-y) \\in A$ for all $y > 0$.\nКак и в первом решении, выберем произвольный элемент $s \\in A$. Докажем, что любое $t \\le 0$ лежит в $A$. Рассмотрим уравнение $x^2 - sx + t = 0$; его дискриминант неотрицателен, так что оно имеет два (возможно, совпадающих) корня $a$ и $b$. Тогда по теореме Виета имеем $a+b=t$ и $ab=s$. Поскольку $a+b=s \\in A$, получаем, что и $t \\in A$.\nОсталось показать, что любое $u > 0$ также лежит в $A$. По доказанному выше, $(-u)+(-1) \\in A$; значит, и $(-u) \\cdot (-1) = u$ также лежит в $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezoid with $BC \\parallel AD$ and $AB = CD$. A circle $\\omega$ centered at $I$ is tangent to the segments $AB$, $CD$, and $DA$. The circle $BIC$ meets the side $AB$ at points $B$ and $E$. Prove that $CE$ is tangent to $\\omega$.\n\nВнутри равнобокой трапеции $ABCD$ с основаниями $BC$ и $AD$ расположена окружность $\\omega$ с центром $I$, касающаяся отрезков $AB$, $CD$ и $DA$. Окружность, описанныя около треугольника $BIC$, вторично пересекает сторону $AB$ в точке $E$. Докажите, что прямая $CE$ касается окружности $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $\\angle ICD = \\angle IBA = \\angle IBE = \\angle ICE$ (see Fig. 7). So the lines $CD$ and $CE$ are symmetric in $IC$.\n\n![](attached_image_1.png)\nЗаметим, что $I$ лежит на оси симметрии трапеции, поэтому $\\angle ICD = \\angle IBA$. Пользуясь вписанностью четырехугольника $CBEI$, получаем $\\angle ICD = \\angle IBA = \\angle IBE = \\angle ICE$. Так как прямая $CD$ касается окружности $\\omega$, то и прямая $CE$, симметричная ей относительно $CI$, также касается $\\omega$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four distinct integers are written on the board. It is known that the absolute value of each of them is greater than $10^6$, and there is no positive integer greater than $1$ which divides all these numbers. Pete wrote on a paper six pairwise sums of the numbers from the board. Then he partitioned the obtained six numbers into three pairs and multiplied the numbers in each pair. May it happen that all three obtained products are equal?\n\nНа доске написаны четыре попарно различных целых числа, модуль каждого из которых больше миллиона. Известно, что не существует натурального числа, большего $1$, на которое бы делилось каждое из четырех написанных чисел. Петя записал в тетрадку шесть попарных сумм этих чисел, разбил эти шесть сумм на три пары и перемножил числа в каждой паре. Могли ли все три произведения оказаться равными?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes.\n\nTwo possible examples are\n$$\n(N^2 - 3N + 1,\\ N^2 - N + 1,\\ -3N^2 + 3N - 1,\\ N^2 + N - 1)\n$$\nand\n$$\n(N^3 - N^2 + 1,\\ N^3 - 3N^2 + 2N - 1,\\ -N^3 + N^2 - 2N + 1,\\ -N^3 + N^2 + 2N - 1),\n$$\nwhere $N$ is an integer greater than $10^6$.\n\nFirst solution. For example, the numbers\n$$\nx = N^2 - 3N + 1,\\ y = N^2 - N + 1,\\ z = -3N^2 + 3N - 1,\\ t = N^2 + N - 1,\n$$\nwhere $N$ is a natural number greater than a million.\n\nIt is easy to see that these numbers are also greater than a million in absolute value and pairwise distinct. Their pairwise sums are\n$$\nx + y = 2(N - 1)^2,\\ x + z = -2N^2,\\ y + z = -2N(N - 1),\\ z + t = -2(N - 1)^2,\\ y + t = 2N^2,\\ x + t = 2N(N - 1),\n$$\nwith the products\n$$\n(x + y)(x + z) = (y + t)(z + t) = (x + t)(y + z) = -4N^2(N - 1)^2.\n$$\n\nIt remains to check that the numbers $x, y, z, t$ are coprime as a set. If they have a common natural divisor $d$, then $d$ also divides $x + y = 2(N - 1)^2$ and $y + t = 2N^2$; thus, $d \\le \\gcd(2N^2, 2(N - 1)^2) = 2$. The case $d = 2$ is impossible since $y = N(N - 1) + 1$ is odd. Therefore, $d = 1$, as required.\n\nSecond solution. Another example is\n$$\nx = N^3 - N^2 + 1,\\ y = N^3 - 3N^2 + 2N - 1,\\ z = -N^3 + N^2 - 2N + 1,\\ t = -N^3 + N^2 + 2N - 1,\n$$\nwhere $N$ is a natural number greater than a million. Again, these numbers are greater than a million in absolute value and pairwise distinct. Their pairwise sums are\n$$\nx + y = 2N(N - 1)^2,\\ x + z = -2(N - 1),\\ y + z = -2N^2,\\ z + t = -2N^2(N - 1),\\ y + t = -2(N - 1)^2,\\ x + t = 2N,\n$$\nand they can be grouped into three pairs with equal products:\n$$\n(x + y)(x + t) = (x + z)(z + t) = (y + z)(y + t) = 4N^2(N - 1)^2.\n$$\n\nThe coprimality of $x, y, z, t$ is checked analogously to the previous solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39782,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ boys and $n$ girls stand along a circle. We call a pair of a boy and a girl good if one of two arcs joining them contains equal numbers of girls and boys (in particular, neighboring boy and girl form a good pair). It appears that there exists a girl belonging to exactly 10 good pairs. Prove that there also exists a boy belonging to exactly 10 good pairs.\n\nПо кругу стоят $n$ мальчиков и $n$ девочек. Назовем пару из мальчика и девочки хорошей, если на одной из дуг между ними стоит поровну мальчиков и девочек (в частности, стоящие рядом мальчик и девочка образуют хорошую пару). Оказалось, что есть девочка, которая участвует ровно в 10 хороших парах. Докажите, что есть и мальчик, который участвует ровно в 10 хороших парах.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be a girl participating in 10 good pairs. Let $K_1, K_2, \\dots, K_{2n}$ be all the children in a clockwise order, with $K_1 = D$; set $K_{i \\pm 2n} = K_i$. Denote by $d_i$ the difference between the numbers of boys and girls among $K_1, K_2, \\dots, K_i$; in particular, $d_1 = 1 - 0 = 1$ and $d_{2n} = 0$ (this means that we may also set $d_{i \\pm 2n} = d_i$). The \"graph\" of $(d_i)$ is shown on Fig. 8; notice that this graph meets the line $d = 1/2$ on $[0, 1]$.\n\nNow, $D$ and $K_i$ form a good pair if and only if $K_i$ is a boy, and $d_i = 0$ (i.e., iff the graph meets $d = 1/2$ on $[i-1, i]$); so there are exactly 10 boys with such property. In other words, the graph crosses the line $d = 1/2$ \"downwards\" (and hence also \"upwards\") exactly 10 times. Arguing similarly, one may see that the girls forming good pairs with such boy are exactly the ones immediately preceding these boys.\n\n![](attached_image_1.png)\n\nAlternatively, in Russian:\n\nЗаметим сразу, что на любой дуге между членами хорошей пары поровну девочек и мальчиков.\nПусть $D$ — девочка, участвующая в 10 хороших парах. Обозначим всех детей по часовой стрелке $K_1, K_2, \\dots, K_{2n}$ так, что $K_1$ — это $D$, и продолжим нумерацию циклически (например, $K_0 = K_{2n}$ и $K_{2n+1} = K_1$). При $i = 1, 2, \\dots, 2n$ обозначим через $d_i$ разность между количествами девочек и мальчиков среди $K_1, K_2, \\dots, K_i$; в частности, $d_1 = 1 - 0 = 1$ и $d_{2n} = 0$ (поэтому можно продолжить эту последовательность, полагая $d_{2n+1} = d_1$ и т. д.). Девочка $D$ образует с $K_i$ хорошую пару тогда и только тогда, когда $d_i = 0$ и $K_i$ — мальчик, т. е. $d_i = 0$ и $d_{i-1} = 1$. Итак, найдутся ровно 10 индексов $i$ с такими свойствами.\n\nРассмотрим любого мальчика $M = K_s$, образующего с $D$ хорошую пару; тогда $d_s = 0$ и $d_{s-1} = 1$. Аналогично получаем, мальчик $M$ образует хорошую пару с $K_i$ ровно тогда, когда $d_s = d_{i-1}$ и $K_i$ — девочка (то есть $d_{i-1} = 0$ и $d_i = 1$).\n\nЗаметим, что любые два числа $d_i$ и $d_{i+1}$ отличаются на единицу. Разобьём их на группы последовательных чисел, не меньших единицы, и группы последовательных чисел, не больших нуля. Тогда при обходе круга по часовой стрелке «переходов» из первых групп во вторые будет столько же, сколько и «переходов» из вторых групп в первые. Значит, у $M$ столько же хороших напарниц, сколько у $D$ хороших напарников. Это и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC < BC$; let $M$ be the midpoint of $AB$. Let $CC'$ be a diameter in the circumcircle $\\Omega$ of the triangle $ABC$. The line $CM$ meets the lines $AC'$ and $BC'$ at $K$ and $L$, respectively. Denote by $\\Delta$ the triangle formed by the line through $K$ perpendicular to $AC'$, the line through $L$ perpendicular to $BC'$, and the line $AB$. Prove that the circumcircle of $\\Delta$ is tangent to $\\Omega$.\n(M. Kungozhin)\n\nПусть $ABC$ — остроугольный треугольник, в котором $AC < BC$; пусть $M$ — середина отрезка $AB$. В окружности $\\Omega$, описанной около треугольника $ABC$, проведен диаметр $CC'$. Прямая $CM$ пересекает прямые $AC'$ и $BC'$ в точках $K$ и $L$ соответственно. Пусть перпендикуляр к прямой $AC'$, проведенный через точку $K$, перпендикуляр к прямой $BC'$, проведенный через точку $L$, и прямая $AB$ образуют треугольник $\\Delta$. Докажите, что окружность, описанная около треугольника $\\Delta$, касается окружности $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$, $U$, and $V$ be the vertices of $\\Delta$, with $EU \\parallel AC$ and $EV \\parallel BC$. Let $C'E$ meet $\\Omega$ again at $X$. Then $X$ is the required tangency point. There are various methods to prove this.\n\n*Hint 1.* Let $CX$ meet $AC'$ and $BC'$ at $Q$ and $P$, respectively (see Fig. 19). One may show that $E$ is the orthocenter of the triangle $PC'Q$, so that $P \\in EU$ and $Q \\in EV$. Then the points $(A, V, Q, X)$ are concyclic, which yields $\\angle UEX = \\angle KEX = \\angle AQX = \\angle AVX$, so that $U, V, E$, and $X$ lie on some circle $\\omega$. Due to $\\angle EVX = \\angle QAX = \\angle C'AX$, this circle is tangent to $\\Omega$.\n\n*Hint 2.* Let the line $C'E$ meet $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively (see Fig. 20). One can see that $A_1K \\parallel B_1L \\parallel AB$. Let $N = A_1K \\cap AC$; since $AB \\parallel A_1N$ and $KE \\parallel AB_1$, we have $A_1K = KN$ and $A_1E = EB_1$.\n\nLet $XX'$ be a diameter in $\\Omega$, and let $X'A$ and $X'B$ meet $C'E$ at $S$ and $T$, respectively. Then $\\Delta TSX' \\sim \\Delta ABC$, the points $X$ and $M$ are corresponding points in these triangles, so $X$ is the midpoint of $ST$. Since $SABA_1$ and $TBAB_1$ are cyclic, we have $\\frac{C_1A_1}{C_1B_1} = \\frac{C_1T}{C_1S}$, which yields $\\frac{C_1S}{C_1X} = \\frac{C_1B_1}{C_1E}$. Thus $UX \\parallel AS$, hence $\\angle UXE = \\angle ASX = \\angle ABC = \\angle EVU$, so the points $U, V, E$, and $X$ lie on some circle $\\omega$. Finally, the cyclic configurations $ACX'B$ and $UEXV$ are homothetic, thus the tangents at $X$ and $X'$ to $\\omega$ and $\\Omega$ are parallel.\n**Первое решение.** Пусть данные перпендикуляры, проходящие через $K$ и $L$, пересекают $AB$ в точках $U$ и $V$ соответственно и пересекаются в точке $E$. Заметим сразу, что обе прямые $AC$ и $KE$ перпендикулярны прямой $AC'$, так что $AC \\parallel KE$; аналогично, $BC \\parallel LE$. Обозначим через $\\omega$ окружность, описанную около треугольника $EUV$. Пусть прямая $C'E$ вторично пересекает окружность $\\Omega$ в точке $X$ (см. рис. 19). Мы докажем, что $X$ — искомая точка касания окружностей $\\omega$ и $\\Omega$.\n![](attached_image_1.png)\n\nПоскольку $\\angle C'KE = \\angle C'LE = 90^\\circ$, четырёхугольник $C'KEL$ вписан. Отсюда имеем $\\angle ACX = \\angle AC'X = \\angle KLE = \\angle LCB$, то есть $\\angle ACX = \\angle MCB$ и $\\angle BCX = \\angle MCA$. Это значит, что прямая $CX$ — симедиана треугольника $ABC$; как известно, она проходит через точку $S$ пересечения касательных к $\\Omega$ в точках $A$ и $B$.\n\nПусть $O$ — центр $\\Omega$; обозначим $\\gamma = \\angle ACB$. Пусть $CM$ вторично пересекает $\\Omega$ в точке $Y$. Так как $AM$ — высота в прямоугольном треугольнике $OAS$, имеем $\\overline{OM} \\cdot \\overline{OS} = \\overline{OA}^2 = \\overline{OC}^2$; это значит, что треугольники $OMC$ и $OCS$ подобны. Значит, отношение их высот, опущенных из $O$, равно $\\frac{OM}{OC} = \\frac{OM}{OA} = \\cos \\gamma$. Эти высоты являются средними линиями в прямоугольных треугольниках $CYC'$ и $CXC'$, так что $\\frac{C'Y}{C'X} = \\cos \\gamma$.\n\nПусть прямые $AC'$ и $BC'$ пересекают $CX$ в точках $Q$ и $P$ соответственно. Заметим, что $\\angle PC'Q = \\angle ACB = \\gamma$. Кроме того, $\\angle YC'L = \\angle YCB = \\angle QC'X$, так что прямоугольные треугольники $YC'L$ и $XC'Q$ подобны. Значит, $\\frac{C'L}{C'Q} = \\frac{C'Y}{C'X} = \\cos \\gamma = \\cos \\angle PC'Q$. Отсюда следует, что $QL$ — высота треугольника $PCQ$. Итак, точки $L$, $E$, $Q$ и $V$ лежат на одной прямой.\n\nПоскольку $\\angle QVB = \\angle VBC = \\angle AXC$, четырёхугольник $AVQX$ вписан в некоторую окружность. Поскольку $\\angle EKQ = \\angle EXQ = 90^\\circ$, точки $X$, $E$, $K$ и $Q$ также лежат на одной окружности. Из этих двух окружностей получаем $\\angle UEX = \\angle KEX = \\angle AQX = \\angle AVX$, то есть точки $U$, $V$, $E$ и $X$ лежат на одной окружности $\\omega$. Кроме того, $\\angle EVX = \\angle QAX = \\angle C'AX$. Это означает, что градусные меры дуг $C'X$ и $EX$ окружностей $\\Omega$ и $\\omega$ равны. Значит, касательные к этим окружностям, проведённые в точке $X$, совпадают. Тем самым, окружности $\\Omega$ и $\\omega$ касаются.\n\n![](attached_image_1.png)\n\n**Второе решение.** Введём точки $E$, $U$, $V$ и $X$, как и в прошлом решении. Обозначим через $A_1$, $B_1$ и $C_1$ точки пересечения прямой $C'E$ с прямыми $BC$, $CA$ и $AB$ соответственно. Как и выше, равенство $\\angle ACX = \\angle AC'X = \\angle KLE = \\angle LCB$ означает, что $\\angle ACX = \\angle KCB$; с другой стороны, из равенства $\\angle KC'E = \\angle KCA_1$ следует также, что точки $C$, $C'$, $K$ и $A_1$ лежат на одной окружности (см. рис. 20). Отсюда $\\angle C'KA_1 = \\angle C'CB = \\angle C'AB$, то есть $KA_1 \\parallel AB$.\n\nОбозначим через $N$ точку пересечения прямых $KA_1$ и $AC$. Поскольку $AB \\parallel A_1K$ и прямая $CM$ делит $AB$ пополам, она делит пополам и $A_1N$, то есть $KN = KA_1$. Так как $KE \\parallel AB_1$, $KE$ — средняя линия в треугольнике $A_1NB_1$, то есть $B_1E = EA_1$.\n\nПусть $X'$ — точка на $\\Omega$, диаметрально противоположная $X$. Пусть прямые $X'A$ и $X'B$ пересекают прямую $C'E$ в точках $S$ и $T$ соответственно. Тогда $\\angle SX'T = \\angle ACB$ и $\\angle X'ST = 90^\\circ - \\angle SX'C' = 90^\\circ - \\angle ABC' = \\angle ABC$, то есть треугольники $ABC$ и $TSX'$ подобны. Более того, поскольку $\\angle SX'X = \\angle ACX = \\angle BCM$, точки $M$ и $X$ в этих треугольниках соответствуют, то есть $SX = XT$.\n\n![](attached_image_2.png)\n\nИз равенства $\\angle AST = \\angle ABC = \\angle C_1BA_1$ следует, что точки $S$, $A$, $B$ и $A_1$ лежат на одной окружности, откуда $C_1A \\cdot C_1B = C_1S \\cdot C_1A_1$. Аналогично, $C_1A \\cdot C_1B = C_1T \\cdot C_1B_1$, поэтому $C_1S \\cdot C_1A_1 = C_1T \\cdot C_1B_1$, или $\\frac{C_1S}{C_1B_1} = \\frac{C_1T}{C_1A_1}$. Это значит, что отрезки $A_1B_1$ и $TS$ гомотетичны с центром в $C_1$; эта гомотетия переводит середину $E$ отрезка $A_1B_1$ в середину $X$ отрезка $ST$. Значит, $\\frac{C_1S}{C_1B_1} = \\frac{C_1X}{C_1E}$, или $\\frac{C_1S}{C_1X} = \\frac{C_1B_1}{C_1E}$. В силу параллельности $AB_1 \\parallel UE$, получаем $\\frac{C_1A}{C_1U} = \\frac{C_1B_1}{C_1E} = \\frac{C_1S}{C_1X}$, что означает, что $UX \\parallel AS$. Отсюда получаем $\\angle UXE = \\angle ASX = \\angle ABC = \\angle EVU$, то есть точки $U$, $V$, $E$ и $X$ лежат на одной окружности $\\omega$.\n\nНаконец, рассмотрим треугольники $ABC$ и $UVE$; их соответственные стороны параллельны, так что они гомотетичны (с отрицательным коэффициентом). При этой гомотетии прямая $CX'$ переходит в $EX$ (поскольку $C'X \\parallel CX'$), окружность $\\Omega$ переходит в окружность $\\omega$, а значит, $X'$ переходит в $X$. Поэтому касательная к $\\omega$ в $X$ параллельна касательной к $\\Omega$ в $X'$, а значит — и касательной к $\\Omega$ в $X$. Отсюда следует, что $\\omega$ и $\\Omega$ касаются в $X$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of distinct real $x$ and $y$ such that $x^{100} - y^{100} = 2^{99}(x - y)$ and $x^{200} - y^{200} = 2^{199}(x - y)$.\n\nНайдите все пары различных действительных чисел $x$ и $y$ такие, что $x^{100} - y^{100} = 2^{99}(x - y)$ и $x^{200} - y^{200} = 2^{199}(x - y)$.",
    "options": [],
    "answer": "(x, y) = (2, 0) and (x, y) = (0, 2)",
    "solution": "$(x, y) = (2, 0)$ and $(x, y) = (0, 2)$.\n\nSet $x = 2a$, $y = 2b$. We have $a^{100} - b^{100} = a^{200} - b^{200} = a - b \\neq 0$, whence $a^{100} + b^{100} = 1$. The case $ab = 0$ is easy. Assume that $ab \\neq 0$; then $|a|, |b| < 1$. Since $a^{100} - a = b^{100} - b$, we have $ab > 0$. Now,\n$$\n1 = \\left| \\frac{a^{100} - b^{100}}{a-b} \\right| = \\left| a^{99} + a^{98}b + a^{97}b^2 + \\dots + b^{99} \\right| > \\\\ > |a^{99} + b^{99}| > a^{100} + b^{100} = 1.\n$$\n$(x, y) = (2, 0)$ и $(x, y) = (0, 2)$.\n\nДля удобства сделаем замену $x = 2a$ и $y = 2b$. Тогда из условия имеем $(2a)^{100} - (2b)^{100} = 2^{99} \\cdot (2a - 2b)$ и $(2a)^{200} - (2b)^{200} = 2^{199} \\cdot (2a - 2b)$. Сократив оба равенства на степени двойки, получаем $a^{100} - b^{100} = a^{200} - b^{200} = a - b \\neq 0$. Поделив второе выражение на первое, получаем $a^{100} + b^{100} = 1$; значит, каждое из чисел $a$ и $b$ по модулю не превосходит 1.\n\nЕсли $b = 0$, то $a^{100} = a$, откуда $a = 1$. Аналогично, если $a = 0$, то $b = 1$; это приводит к двум ответам $(x, y) = (2, 0)$ и $(x, y) = (0, 2)$.\n\nПусть теперь $ab \\neq 0$; тогда $|a|, |b| < 1$. Заметим, что значения функции $f(x) = x^{100} - x = x(x^{99} - 1)$ положительны при $x \\in (-1, 0)$ и отрицательны при $x \\in (0, 1)$. Поскольку $a^{100} - b^{100} = a - b$, имеем $f(a) = f(b)$, поэтому числа $a$ и $b$ имеют одинаковый знак.\n\nС другой стороны,\n$$\n1 = \\frac{a^{100} - b^{100}}{a - b} = a^{99} + a^{98}b + a^{97}b^2 + \\dots + b^{99}. \\quad (*)\n$$\n\nЕсли $a$ и $b$ отрицательны, то правая часть в $(*)$ также отрицательна, что невозможно. Если же $a$ и $b$ положительны, то все слагаемые в правой части $(*)$ положительны, поэтому она больше, чем $a^{99} + b^{99}$; итак, $a^{99} + b^{99} < 1$. С другой стороны, поскольку $0 < |a|, |b| < 1$, имеем $a^{99} + b^{99} > a^{100} + b^{100} = 1$. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39785,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic polynomial $f(x) = ax^2 + bx + c$ has no real roots. It is given that $b$ is a rational number, and exactly one of $c$ and $f(c)$ is a rational number. Is it possible for the discriminant of $f(x)$ to be a rational number?\n\nКвадратный трёхчлен $f(x) = ax^2 + bx + c$, не имеющий корней, таков, что коэффициент $b$ рационален, а среди чисел $c$ и $f(c)$ ровно одно иррационально. Может ли дискриминант трёхчлена $f(x)$ быть рациональным?",
    "options": [],
    "answer": "No",
    "solution": "No.\n\nТак как трёхчлен $f(x)$ не имеет корней, то $c = f(0) \\neq 0$ и $f(c) \\neq 0$. Тогда выражение $\\frac{f(c)}{c}$ иррационально как отношение рационального и иррационального чисел. Но $\\frac{f(c)}{c} = \\frac{ac^2 + bc + c}{c} = ac + b + 1$. Так как $b+1$ рационально, то $ac$ — иррационально. Получаем, что дискриминант $D = b^2 - 4ac$ иррационален как разность рационального и иррационального чисел.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $x$, $y$, and $z$ satisfy the inequality $xyz \\ge xy + yz + zx$. Prove that\n$$\n\\sqrt{xyz} \\ge \\sqrt{x} + \\sqrt{y} + \\sqrt{z}.\n$$\n\nПоложительные числа $x$, $y$ и $z$ удовлетворяют условию $xyz \\ge xy + yz + zx$. Докажите неравенство $\\sqrt{xyz} \\ge \\sqrt{x} + \\sqrt{y} + \\sqrt{z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Rewrite the given and the required inequalities as $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\le 1$ and $\\frac{1}{\\sqrt{xy}} + \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yz}} \\le 1$.\n\nBy the inequality of means, we have\n$$\nxy + xz \\geq 2\\sqrt{xy \\cdot xz}, \\quad xy + yz \\geq 2\\sqrt{xy \\cdot yz}, \\quad xz + yz \\geq 2\\sqrt{xz \\cdot yz}.\n$$\nAdd these three inequalities and divide the result by $2$. Taking into account the condition, we get\n$$\nxyz \\geq xy + xz + yz \\geq x\\sqrt{yz} + y\\sqrt{xz} + z\\sqrt{xy}.\n$$\nDividing the obtained inequality by $\\sqrt{xyz}$, we obtain the required result.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39787,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the space, three segments $A_1A_2$, $B_1B_2$, and $C_1C_2$ are chosen.\n\nThey are not coplanar, and they meet at point $P$. Denote by $O_{ijk}$ the center of the sphere passing through $A_i, B_j, C_k$, and $P$. Prove that the four lines $O_{111}O_{222}, O_{112}O_{221}, O_{121}O_{212}$, and $O_{211}O_{122}$ have a common point.\n\nВ пространстве даны три отрезка $A_1A_2$, $B_1B_2$ и $C_1C_2$, не лежащие в одной плоскости и пересекающиеся в одной точке $P$. Обозначим через $O_{ijk}$ центр сферы, проходящей через точки $A_i, B_j, C_k$ и $P$. Докажите, что прямые $O_{111}O_{222}$, $O_{112}O_{221}$, $O_{121}O_{212}$ и $O_{211}O_{122}$ пересекаются в одной точке.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For a fixed $i$, all the points $O_{ijk}$ belong to the perpendicular bisector $\\alpha_i$ of $PA_i$; notice that $\\alpha_1 \\parallel \\alpha_2$. By similar reasons, the points $O_{ijk}$ appear to be eight vertices of some parallelepiped.\n\nFor any segment $XY$, the perpendicular bisector to this segment is the plane perpendicular to it and passing through its midpoint, i.e., the locus of points equidistant from $X$ and $Y$.\n\nAll points of the form $O_{1jk}$ lie in the perpendicular bisector $\\alpha_1$ to the segment $PA_1$. Similarly, all points $O_{2jk}$ lie in the perpendicular bisector $\\alpha_2$ to the segment $PA_2$; note that $\\alpha_1 \\parallel \\alpha_2$.\n\nSimilarly, introduce the planes $\\beta_j$ — perpendicular bisectors to the segments $PB_j$, and the planes $\\gamma_k$ — perpendicular bisectors to the segments $PC_k$. Then the points $O_{ijk}$ are the vertices of a parallelepiped formed by the planes $\\alpha_i, \\beta_j$, and $\\gamma_k$. Now the statement of the problem follows from the fact that the diagonals of this parallelepiped intersect at one point — its center of symmetry.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 39788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $L$ be a point on the side $AC$ of a triangle $ABC$ such that $BL$ is the bisector of the angle $ABC$. Let $M$ be a point on the segment $CL$. The tangent through $B$ to the circumcircle $\\Omega$ of $ABC$ meets the ray $CA$ at $P$. Tangents through $B$ and $M$ to the circimcircle $\\Gamma$ of $\\triangle BLM$ meet at $Q$. Prove that $PQ \\parallel BL$.\n(A. Kuznetsov)\n\nВ треугольнике $ABC$ проведена биссектриса $BL$. На отрезке $CL$ выбрана точка $M$. Касательная в точке $B$ к окружности $\\Omega$, описанной около треугольника $ABC$, пересекает луч $CA$ в точке $P$. Касательные в точках $B$ и $M$ к окружности $\\Gamma$, описанной около треугольника $BLM$, пересекаются в точке $Q$. Докажите, что прямые $PQ$ и $BL$ параллельны. (А. Кузнецов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By angle chasing we get $\\angle BPM = 180^\\circ - 2\\angle BLP = \\angle BQM$ (see Fig. 9). So the points $B, M, P$, and $Q$ are concyclic, and $\\angle QPM = \\angle QBM = \\angle BLP$.\nТак как $BL$ — биссектриса $\\angle ABC$, имеем $\\angle ABL = \\angle LBC$. Поскольку $PB$ — касательная к $\\Omega$, имеем $\\angle PBA = \\angle BCA$ (см. рис. 9). Кроме того, $\\angle PBL = \\angle PBA + \\angle ABL = \\angle BCA + \\angle LBC = \\angle BLP$, значит, $\\angle BPM = 180^\\circ - (\\angle PBL + \\angle BLP) = 180^\\circ - 2\\angle BLP$. Отсюда следует, в частности, что $\\angle BLP$ — острый.\n\n![](attached_image_1.png)\nРис. 9\n\nТак как $\\angle BLM = 180^\\circ - \\angle BLP$ тупой, касательные к $\\Gamma$ в точках $B$ и $M$ пересекаются в точке $Q$, лежащей по ту же сторону от $BM$, что и точка $L$ (а значит — по ту же сторону, что и $P$). Далее, имеем $\\angle QBM = \\angle QMB = 180^\\circ - \\angle BLM = \\angle BLP$. Значит, $\\angle BQM = 180^\\circ - 2\\angle QBM = 180^\\circ - 2\\angle BLP = \\angle BPM$. Поэтому точки $B, M, P$ и $Q$ лежат на одной окружности. Отсюда следует, что $\\angle QPM = \\angle QBM = \\angle BLP$. Это и означает, что $PQ \\parallel BL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $2015 \\times 2015$ checkered board. Dmitry chooses $k$ cells and places a detector in each of them. After that Dmitry goes out, and then Nick places on a board a cellular square $1500 \\times 1500$ (the square lies within the board, and its sides lie on the grid lines).\n\nEach detector informs Dmitry whether its cell is covered by the square or not. Find the least possible $k$ such that Dmitry can place detectors in order to determine the location of Nick's square for sure.\n(O. Dmitriev, R. Zhenodarov)\n\nЕсть клетчатая доска $2015 \\times 2015$. Дима ставит в $k$ клеток по детектору. Затем Коля располагает на доске клетчатый корабль в форме квадрата $1500 \\times 1500$. Детектор в клетке сообщает Диме, накрыта эта клетка кораблём или нет. При каком наименьшем $k$ Дима может расположить детекторы так, чтобы гарантированно восстановить расположение корабля? (О. Дмитриев, Р. Женодаров)",
    "options": [],
    "answer": "1030",
    "solution": "$k = 2(2015 - 1500) = 1030$.\nTo win with $1030$ detectors, it suffices to locate them in the $515$ leftmost cells of the middle row and in the $515$ topmost cells of the middle column.\n\nFor the estimate, notice that the union of any two vertical $1500 \\times 1$ rectangles differing by a horizontal $1500$-shift should contain at least one detector; a similar claim holds for horizontal rectangles. Consider such pairs of vertical rectangles located in the bottom $1500$ rows, as well as those located in the top $1500$ rows. Consider also similar pairs of horizontal rectangles. Each of these $2 \\cdot 1030$ pairs contains a detector, and each detector is covered by at most two such pairs.\n\n![](attached_image_1.png)\n\n**Note.** There exist many other examples of arrangements of $1030$ detectors that satisfy the requirements.\n$k = 2(2015 - 1500) = 1030$.\nПокажем, что $1030$ детекторов Диме хватит. Пусть он расположит $515$ детекторов в $515$ левых клетках средней строки квадрата, а остальные $515$ детекторов — в $515$ верхних клетках среднего столбца. Заметим, что при любом положении корабля его левый столбец лежит в одном из $516$ левых столбцов доски.\n\nЕсли этот столбец — один из $515$ самых левых, то корабль накроет детектор из этого столбца, лежащий в средней строке, иначе ни одного детектора из этой строки корабль не накроет. Значит, по показаниям детекторов из этой строки восстанавливается, в каких столбцах лежит корабль. Аналогично, строки, в которых он находится, восстанавливаются по показаниям детекторов из среднего столбца.\n\nРассмотрим теперь произвольную расстановку $k$ детекторов, удовлетворяющих требованиям. Рассмотрим два положения корабля, отличающихся горизонтальным сдвигом на $1$. Показания какого-то детектора для них будут различаться, только если этот детектор лежит в самом левом столбце левого корабля или в самом правом столбце правого. Значит, в любых двух вертикальных прямоугольниках $1500 \\times 1$, отличающихся горизонтальным сдвигом на $1500$, есть хотя бы один детектор. Аналогично, в любых двух горизонтальных прямоугольниках $1 \\times 1500$, отличающихся вертикальным сдвигом на $1500$, есть хотя бы один детектор. Назовём такие пары прямоугольников вертикальными и горизонтальными, соответственно.\n\nВыделим все вертикальные пары, лежащие в нижних $1500$ и в верхних $1500$ строках доски (таких пар $2 \\cdot 515 = 1030$). Аналогично, выделим все $1030$ горизонтальных пар, лежащих в левых $1500$ и в правых $1500$ столбцах. Разобьём доску на $9$ прямоугольных областей так, как показано на рис. 10. Выделенные пары не покрывают клеток из $E$; каждая же клетка в остальных областях покрыта двумя выделенными парами (в $D$ и $F$ — двумя вертикальными, в $B$ и $H$ — двумя горизонтальными, а в областях $A, C, G$ и $I$ — одной горизонтальной и одной вертикальной). Итак, каждый детектор лежит не более, чем в двух выделенных парах; значит, чтобы в каждой выделенной паре был хотя бы один детектор, требуется не менее $2 \\cdot 1030/2 = 1030$ детекторов.\n\n![](attached_image_1.png)\nРис. 10\n\n**Замечание.** Существует много других примеров расположения $1030$ детекторов, удовлетворяющих требованиям.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39790,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the coordinate space, we consider all the planes with equations of the form $x \\pm y \\pm z = n$ (for all integer $n$). These planes partition the space into tetrahedra and octahedra. Assume that the point $(x_0, y_0, z_0)$ with rational coordinates lies in neither of these planes. Prove that there exists a positive integer $k$ such that the point $(kx_0, ky_0, kz_0)$ lies in the interior of some octahedron of the partition.\n(A. Glazyrin)\n\nВ координатном пространстве провели все плоскости с уравнениями $x \\pm y \\pm z = n$ (при всех целых $n$). Они разбили пространство на тетраэдры и октаэдры. Пусть точка $(x_0, y_0, z_0)$ с рациональными координатами не лежит ни в одной проведённой плоскости. Докажите, что пайдётся натуральное $k$, при котором точка $(kx_0, ky_0, kz_0)$ лежит строго внутри некоторого октаэдра разбиения.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Perform the coordinate change $a = y + z - x$, $b = x - y + z$, $c = x + y - z$ (then $x = \\frac{b+c}{2}$, $y = \\frac{a+c}{2}$, $z = \\frac{a+b}{2}$, and $x + y + z = a + b + c$). The new planes have the equations $a = n$, $b = n$, $c = n$, and $a + b + c = n$, so they partition the space into cubes, each cut into three pieces. Now one may see that the interiors of the octahedra are determined by $1 < \\{a\\} + \\{b\\} + \\{c\\} < 2$, along with $\\{a\\}, \\{b\\}, \\{c\\} \\neq 0$.\n\nThus we need to prove that for every $a, b, c \\in (0, 1)$ with non-integer sum, there exists a positive integer $k$ such that $ka, kb, kc$ are non-integers, and $1 < \\{ka\\} + \\{kb\\} + \\{kc\\} < 2$. If $a+b+c \\in (1, 2)$, the claim is trivial. If $a+b+c < 1$, the smallest $k$ such that $\\{ka\\} + \\{kb\\} + \\{kc\\} > 1$ fits. The case $a+b+c > 2$ is similar.\n\n**Lemma.** Let $a, b, c$ be rational numbers and $a+b+c$ is not an integer. Then there exists a positive integer $k$ such that the numbers $ka, kb, kc$ are not integers, and $1 < \\{ka\\} + \\{kb\\} + \\{kc\\} < 2$.\n\n**Proof.** Replacing $a, b, c$ by their fractional parts, we may assume they lie in the interval $(0, 1)$. Let $f(t) = \\{ta\\} + \\{tb\\} + \\{tc\\}$. Note that if $1 < a+b+c < 2$, we can take $k=1$.\n\nSuppose $a+b+c < 1$. Choose a positive integer $m$ such that $ma, mb, mc$ are integers. Then $f(m-1) = f(-1) = 3-(a+b+c) > 2$. Therefore, there exists the smallest positive integer $k$ such that $f(k) > 1$ (then $f(k-1) \\le 1$). We show that this $k$ satisfies all requirements.\n\n$$\nf(k) \\le f(k-1) + (a+b+c) < f(k-1) + 1 < 2. \\quad (*)\n$$\n\nIt remains to show that the numbers $ka, kb, kc$ are not integers. Suppose, say, $ka$ is an integer. Then $\\{ka\\} = \\{(k-1)a\\} + a - 1$, so the estimate $(*)$ can be strengthened as $f(k) \\le f(k-1) + (a+b+c) - 1 < f(k-1) \\le 1$, which contradicts the choice of $k$. Thus, in the case $a+b+c < 1$, the required $k$ is found.\n\nFinally, if $a+b+c > 2$, it is enough to apply the already proven statement to the numbers $a' = 1-a, b' = 1-b, c' = 1-c$. Clearly, the number $k$ suitable for these numbers will also be suitable for the original ones. $\\square$\n\nNow, let us solve the problem. In addition to the given coordinate space $Oxyz$, introduce a new space $Oabc$. To the point $(x, y, z)$ of the old space, associate the point $(a, b, c)$ of the new space with coordinates $a = y+z-x, b = x-y+z, c = x+y-z$; then $x = \\frac{b+c}{2}, y = \\frac{a+c}{2}, z = \\frac{a+b}{2}$. Note that $x+y+z = a+b+c$. Then the partition of the old space corresponds to the partition of the new space by planes of the form $a = n, b = n, c = n$ and $a+b+c = n$. Set $a_0 = y_0+z_0-x_0, b_0 = x_0-y_0+z_0, c_0 = x_0+y_0-z_0$; by assumption, the numbers $a_0, b_0, c_0$ and $a_0+b_0+c_0$ are not integers.\n\nConsider a point $(u, v, w)$ of the new space with non-integer coordinates. It falls into some cube of the form $A \\le a \\le A+1, B \\le b \\le B+1, C \\le c \\le C+1$. This cube is intersected by two 'diagonal' planes $a+b+c = A+B+C+1$ and $a+b+c = A+B+C+2$, which divide it into two tetrahedra and an (irregular) octahedron. The point $(u, v, w)$ will fall inside the octahedron if it is in the strip between the indicated planes, i.e., if $1 < \\{u\\} + \\{v\\} + \\{w\\} < 2$. Therefore, applying the lemma to the numbers $a_0, b_0, c_0$, we find a value of $k$ satisfying the requirements of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We have $2n+1$ (not necessarily distinct) nonzero integers with a nonzero sum. One needs to put them in places of the stars into an expression $*x^{2n} + *x^{2n-1} + \\cdots + *x + *$ so that the resulting polynomial would have no integer root. Is it always possible?\n\nПусть $n$ — натуральное число. На $2n+1$ карточках написано по ненулевому целому числу; сумма всех чисел также ненулевая. Требуется этими карточками заменить звёздочки в выражении $*x^{2n} + *x^{2n-1} + ... + *x + *$ так, чтобы полученный многочлен не имел целых корней. Обязательно ли это можно сделать?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes.\nPut the number with the largest absolute value at $x^{2n}$; then the only possible integer root is $-1$. This is easily avoidable by permuting the remaining coefficients.\n\n\nSolution:\nДа, обязательно.\nПусть $p_0, p_1, \\dots, p_{2n}$ — числа на карточках, причём $p_{2n}$ — наибольшее по модулю из них. Поставим $p_i$ коэффициентом при $x^i$. Тогда, если $a$ — целое число, по модулю не меньшее двойки, то\n$$\n\\begin{aligned}\n|p_{2n}a^{2n}| &> |p_{2n}|(|a^{2n-1}| + |a^{2n-2}| + \\dots + 1) \\ge \\\\\n&\\ge |p_{2n-1}a^{2n-1}| + |p_{2n-2}a^{2n-2}| + \\dots + |p_0| \\ge \\\\\n&\\ge |p_{2n-1}a^{2n-1}| + |p_{2n-2}a^{2n-2}| + \\dots + |p_0|,\n\\end{aligned}\n$$\nтак что $a$ — не корень полученного многочлена.\nОсталось переставить коэффициенты $p_{2n-1}, p_{2n-2}, \\dots, p_0$ так, чтобы числа $0$ и $\\pm 1$ также не были корнями. Числа $0$ и $1$ в любом случае корнями не являются, поскольку $p_0 \\ne 0 \\ne p_{2n} + p_{2n-1} + \\dots + p_0$ по условию. Предположим, что $x_0 = -1$ является корнем многочлена при любой перестановке коэффициентов $p_{2n-1}, p_{2n-2}, \\dots, p_0$. Тогда, если поменять местами $p_i$ и $p_{i-1}$ (при $i=1, 2, \\dots, 2n-1$), значение многочлена в $x_0$ не изменится, что возможно лишь при $p_i = p_{i-1}$. Но тогда наш многочлен имеет вид $p_{2n}x^{2n} + p_0(x^{2n-1} + x^{2n-2} + \\dots + 1)$, и его значение в точке $x_0 = -1$ равно $p_{2n} \\ne 0$. Противоречие.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39792,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country, there are $n$ cities; some pairs of them are connected with two-way direct flights. There is a unique (perhaps, non-direct) route between every two cities. The mayor of each city $X$ found the number $f(X)$ of the enumerations of all cities by $1, 2, \\ldots, n$ such that along each route starting at $X$, the city numbers increase. All the mayors except one noticed that their resulting numbers are all divisible by $2016$. Prove that the remaining mayor's number is also divisible by $2016$.\n(F. Petrov)\n\nВ стране есть $n > 1$ городов, некоторые пары городов соединены двусторопшими беспосадочными авиарейсами. При этом между любыми двумя городами существует единственный авиамаршрут (возможно, с пересадками). Мэр каждого города $X$ подсчитал количество таких нумераций всех городов числами от $1$ до $n$, что на любом авиамаршруте, начинающемся в $X$, номера городов идут в порядке возрастания. Все мэры, кроме одного, заметили, что их результаты подсчётов делятся на $2016$. Докажите, что и у оставшегося мэра результат также делится на $2016$.\n(Ф. Петров)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose an arbitrary capital $A$. Say that a city $C$ is even (resp., odd) if the route from $A$ to $C$ contains an even (resp., odd) number of flights. It suffices to prove that the sum of mayors' numbers in odd cities is equal to the sum of those in even cities. This claim can be proved by means of a bijection of the corresponding sets of enumerations; this bijection merely swaps $1$ and $2$.\nНазовём какой-нибудь город $A$ столицей. Назовём город чётным, если маршрут из $A$ до него содержит чётное число рейсов, и нечётным иначе. Тогда чётность любых двух городов, соединённых рейсом, различна. Мы докажем, что сумма чисел, полученных мэрами чётных городов, равна сумме чисел, полученных мэрами нечётных; из этого следует утверждение задачи.\n\nНазовём нумерацию городов подходящей для города $X$, если мэр города $X$ её посчитал. Ясно, что в любой нумерации, подходящей городу $X$, он имеет номер $1$, так что каждая нумерация подходит не более, чем одному городу.\n\nРассмотрим любую нумерацию, подходящую чётному городу $E$. Пусть номер $2$ в ней носит город $W$; тогда $W$ — нечётный город, соединённый с $E$, иначе на маршруте от $E$ до $W$ встретился бы город с большим номером. Поменяем местами номера $1$ и $2$; мы получим нумерацию, в которой номер $1$ носит нечётный город $W$.\n\nРассмотрим любой маршрут $m$, начинающийся в $W$. Он получается из некоторого маршрута, выходящего из $E$, либо добавлением города $W$ в начало (если $m$ проходит через $E$), либо откидыванием $E$ из начала (в противном случае). Тогда легко видеть, что после обмена $1$ и $2$ номера на $m$ идут в порядке возрастания.\n\nИтак, после перемены номеров $1$ и $2$ из нумерации, подходящей для чётного города, получается нумерация, подходящая для нечётного (и наоборот). Это сопоставление взаимно однозначно. Значит, тех и других нумераций поровну, что и требовалось доказать.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39793,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the space, $2016$ distinct spheres are arranged. Some of these spheres are red, and the others are green. Each tangency point of two spheres of different colors is marked in blue. Find the greatest possible number of blue points.\n(A. Kuznetsov)\n\nВ пространстве расположены $2016$ сфер, никакие две из них не совпадают. Некоторые из сфер — красного цвета, а остальные — зелёного. Каждую точку касания красной и зелёной сферы помечают синим цветом. Найдите наибольшее возможное количество синих точек.\n(А. Кузнецов)",
    "options": [],
    "answer": "1016064",
    "solution": "$1008^2 = 1016064$ points.\n\nOne may arrange the spheres so that there are $1008$ red and $1008$ green spheres, and each pair of a red and a green sphere is tangent at a separate blue point. Arrange the centers $R_i$ of equal red spheres along a circle centered at $O$, arrange the centers $G_i$ of green spheres along a line passing through $O$ and perpendicular to the plane of the circle (see Fig. 11).\n\nNow, if a green sphere is tangent to one red sphere, it is tangent to all of them.\n\n![](attached_image_1.png)\n$1008^2 = 1016064$ точек.\n\nПусть среди сфер есть $r$ красных и $2016 - r$ зелёных. Так как у любых двух сфер максимум одна точка касания, количество синих точек не превосходит $r(2016 - r) = 1008^2 - (1008 - r)^2 \\le 1008^2$.\n\nПредъявим пример с таким количеством синих точек. Пусть $\\ell$ — некоторая прямая, $\\alpha$ — плоскость, перпендикулярная $\\ell$ и пересекающая её в точке $O$, а $\\omega$ — окружность с центром $O$ и радиусом $1$, лежащая в $\\alpha$. Построим $1008$ красных сфер одинакового радиуса $r < 1$ с различными центрами $R_1, R_2, \\dots, R_{1008}$, лежащими на $\\omega$.\n\n![](attached_image_1.png)\n\nПусть $G_1, G_2, \\dots, G_{1008}$ — различные точки на $\\ell$, удалённые от $O$ на расстояния $d_1, d_2, \\dots, d_{1008}$. Тогда расстояние между $G_i$ и любой точкой $R_j$ равно $\\sqrt{1 + d_i^2}$. Построим зелёную сферу с центром $G_i$ и радиусом $\\sqrt{1 + d_i^2} - r$; тогда она будет касаться всех красных сфер. При этом все точки касания будут попарно различными, поскольку они лежат на отрезках вида $R_j G_i$, которые не имеют общих точек, кроме концов. Значит, в нашей конструкции действительно будут отмечены $1008^2$ синих точек.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 39794,
    "subject": "Mathematics (Multi-modal)",
    "question": "The medians $AM_A$, $BM_B$, and $CM_C$ of a triangle $ABC$ meet at $M$. Construct a circle $\\Omega_A$ passing through the midpoint of $AM$ and tangent to $BC$ at $M_A$. Construct the circles $\\Omega_B$ and $\\Omega_C$ similarly. Prove that the circles $\\Omega_A$, $\\Omega_B$, and $\\Omega_C$ share a common point.\n\nВ треугольнике $ABC$ медианы $AM_A, BM_B$ и $CM_C$ пересекаются в точке $M$. Построим окружность $\\Omega_A$, проходящую через середину отрезка $AM$ и касающуюся отрезка $BC$ в точке $M_A$. Аналогично строятся окружности $\\Omega_B$ и $\\Omega_C$. Докажите, что окружности $\\Omega_A, \\Omega_B$ и $\\Omega_C$ имеют общую точку.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K_A$, $K_B$, and $K_C$ be the midpoints of the segments $AM$, $BM$, and $CM$, respectively (see Fig. 24). By angle chasing, one sees that the circles $MK_AK_BM_B$, $M_BK_CM_A$, and $MAK_BM_C$ are concurrent at some point $X$. This point belongs to the circles $\\Omega_A$, $\\Omega_B$, and $\\Omega_C$: for example, one sees that $X \\in \\Omega_B$ from\n$$\n\\begin{aligned}\n\\angle(K_BX, XM_B) &= \\angle(K_BX, XM_C) + \\angle(M_CX, XM_B) = \\\\\n&= \\angle(K_BM_A, M_A M_C) + \\angle(M_C K_A, K_A M_B) = \\\\\n&= \\angle(MC, CA) + \\angle(BM, MC) = \\angle(BM, CA) = \\angle(K_B M_B, AC).\n\\end{aligned}\n$$\n\n\n**Лемма.** Пусть на сторонах треугольника $ABC$ во внешнюю сторону построены треугольники $A_1BC, AB_1C$ и $ABC_1$ так, что сумма их углов при вершинах $A_1, B_1$ и $C_1$ кратна $180^\\circ$. Тогда окружности, описанные около треугольников $A_1BC, AB_1C$ и $ABC_1$, пересекаются в одной точке.\n**Доказательство.** Пусть окружности, описанные около треугольников $A_1BC$ и $ABC_1$, вторично пересекаются в точке\n\n$$\n\\angle(AX, XC) = \\angle(AX, XB) + \\angle(BX, XC) = \\\\ = \\angle(AC_1, C_1B) + \\angle(BA_1, A_1C) = \\angle(AB_1, B_1C).\n$$\nЭто означает, что $X$ лежит на окружности, описанной около треугольника $AB_1C$, что и требовалось. $\\square$\n\n![](attached_image_1.png)\nРис. 23\n\nПусть $K_A, K_B$ и $K_C$ — середины отрезков $AM, BM$ и $CM$ соответственно (см. рис. 24). Тогда $MK_CK_BM_A = \\angle AMC$. Аналогично, $\\angle MK_CK_A M_B = \\angle BMC$ и $\\angle M_A K_C M_B = \\angle BMA$; значит, $\\angle MK_CK_A M_B + \\angle M_B K_C M_A + \\angle M_A K_B M_C = 360^\\circ$.\nСогласно лемме, окружности, описанные около треугольников $MK_CK_A M_B, M_BK_C M_A$ и $M_A K_B M_C$, имеют общую точку $X$. Из этих окружностей имеем\n$$\n\\begin{align*}\n\\angle(K_BX, XM_B) &= \\angle(K_BX, XM_C) + \\angle(M_CX, XM_B) = \\\\\n&= \\angle(K_BM_A, M_A M_C) + \\angle(M_C K_A, K_A M_B) = \\\\\n&= \\angle(MC, CA) + \\angle(BM, MC) = \\angle(BM, CA) = \\angle(K_B M_B, AC).\n\\end{align*}\n$$\nЭто равенство означает, что окружность $\\Omega_B$ проходит через точку $X$. Аналогично, через $X$ проходят окружности $\\Omega_A$ и $\\Omega_C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39795,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is presented in the form $N = a_1 - a_2 = b_1 - b_2 = c_1 - c_2 = d_1 - d_2$, where $a_1$ and $a_2$ are perfect squares, $b_1$ and $b_2$ are perfect cubes, $c_1$ and $c_2$ are perfect 5th powers, and $d_1$ and $d_2$ are perfect 7th powers. Under the conditions above, determine whether two of the numbers $a_1, b_1, c_1, d_1$ should necessarily be equal.\n(A. S. Golovanov)\n\nНатуральное число $N$ представляется в виде $N = a_1 - a_2 = b_1 - b_2 = c_1 - c_2 = d_1 - d_2$, где $a_1$ и $a_2$ — квадраты, $b_1$ и $b_2$ — кубы, $c_1$ и $c_2$ — пятые степени, а $d_1$ и $d_2$ — седьмые степени натуральных чисел. Обязательно ли среди чисел $a_1, b_1, c_1$ и $d_1$ найдутся два равных?\n(А. С. Голованов)",
    "options": [],
    "answer": "No",
    "solution": "No.\nSet $N = (3^2 - 2^2)^{105}(3^3 - 2^3)^{70}(3^5 - 2^5)^{126}(3^7 - 2^7)^{120}$. Then\n$$N = M_2^2(3^2 - 2^2) = M_3^3(3^3 - 2^3) = M_5^5(3^5 - 2^5) = M_7^7(3^7 - 2^7),$$\nwhere none of the $M_i$ is divisible by $2$ or $3$.\n\n$N = (3M_2)^2 - (2M_2)^2 = (3M_3)^3 - (2M_3)^3 = (3M_5)^5 - (2M_5)^5 = (3M_7)^7 - (2M_7)^7.$\nEven all eight numbers involved in the representations are distinct, since any two of them have a different power of either $2$ or $3$.\nНет, не обязательно.\nПриведём пример числа $N$, для которого все указанные числа будут различными. Положим\n$$N = (3^2 - 2^2)^{105} (3^3 - 2^3)^{70} (3^5 - 2^5)^{126} (3^7 - 2^7)^{120}.$$ \nТогда\n$$N = M_2^2(3^2 - 2^2) = M_3^3(3^3 - 2^3) = M_5^5(3^5 - 2^5) = M_7^7(3^7 - 2^7)$$\nпри некоторых натуральных $M_2, M_3, M_5$ и $M_7$, не делящихся ни на $2$, ни на $3$. Отсюда\n$$N = (3M_2)^2 - (2M_2)^2 = (3M_3)^3 - (2M_3)^3 = (3M_5)^5 - (2M_5)^5 = (3M_7)^7 - (2M_7)^7.$$ \nДаже все восемь чисел, участвующих в представлениях, различны, поскольку у любых двух из них разная степень вхождения либо двойки, либо тройки.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Quadratic polynomials $f_1(x), \\dots, f_{100}(x)$ have equal leading coefficients, equal coefficients of $x$, but distinct constant terms; each of these polynomials has two real roots. Let $x_i$ be some root of $f_i(x)$. Determine all possible values of the sum $f_2(x_1) + f_3(x_2) + \\dots + f_{100}(x_{99}) + f_1(x_{100})$. (N. Agakhanov)\n\nПусть квадратичные многочлены $f_1(x), \\dots, f_{100}(x)$ имеют одинаковые старшие коэффициенты, одинаковые коэффициенты при $x$, но различные свободные члены; каждый из этих многочленов имеет два действительных корня. Пусть $x_i$ — какой-то корень $f_i(x)$. Найдите все возможные значения суммы $f_2(x_1) + f_3(x_2) + \\dots + f_{100}(x_{99}) + f_1(x_{100})$. (Н. Агаханова)",
    "options": [],
    "answer": "0",
    "solution": "Пусть $i$-й трёхчлен имеет вид $f_i(x) = a x^2 + b x + c_i$. Тогда\n\n$f_2(x_1) + f_3(x_2) + \\dots + f_{100}(x_{99}) + f_1(x_{100}) = (c_2 - c_1) + (c_3 - c_2) + \\dots + (c_{100} - c_{99}) + (c_1 - c_{100}) = 0$.\n\nЗначит, единственное возможное значение суммы — ноль.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39797,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a bazaar, there is a carpet-changer. If he gets from a client an $a \\times b$ carpet, he can give instead either a $\\frac{1}{a} \\times \\frac{1}{b}$ carpet, or two carpets of sizes $c \\times b$ and $\\frac{a}{c} \\times b$ (at each such change, a number $c$ is chosen by the client). A traveler tells that initially he had one carpet whose side lengths were greater than $1$, and after a series of changes each of his carpets had a side longer than $1$ as well as a side shorter than $1$. Could this be a correct story? (G. Zhukov)\n\nУ месялы па базаре есть много ковров. Он согласен взамен ковра размера $a \\times b$ дать либо ковёр размера $\\frac{1}{a} \\times \\frac{1}{b}$, либо два ковра размеров $c \\times b$ и $\\frac{a}{c} \\times b$ (при каждом таком обмене число $c$ клиент может выбрать сам). Путешественник рассказал, что изначально у него был один ковёр, стороны которого превосходили $1$, а после нескольких таких обменов у него оказался набор ковров, у каждого из которых одна сторона длиннее $1$, а другая — короче $1$. Не обманывает ли он? (По просьбе клиента меяла готов ковёр размера $a \\times b$ считать ковром размера $b \\times a$.) (Г. Жуков)",
    "options": [],
    "answer": "No",
    "solution": "No.\n\nWe say that a carpet is *large* (resp., *small*) if both its side lengths are larger (resp., smaller) than $1$. Show that the total number of large and small carpets does not decrease.\n\nНазовём ковёр, все стороны которого больше $1$, большим, а ковёр, все стороны которого меньше $1$, — маленьким. Таким образом, изначально у путешественника был один большой ковёр. Докажем, что общее число больших и маленьких ковров у путешественника не уменьшается; отсюда следует, что описанная ситуация невозможна. Для этого достаточно рассмотреть только случай, когда путешественник отдаёт меняле большой или маленький ковёр.\n\nПри обменах первого вида большой ковёр меняют на маленький, а маленький — на большой. Поэтому общее количество больших и маленьких ковров не уменьшается.\n\nРассмотрим обмены второго вида. При обмене большого ковра $a \\times b$ путешественник получит ковры $a_1 \\times b$ и $a_2 \\times b$. Если $0 < a_1, a_2 \\le 1$, то $a = a_1 a_2 \\le 1$, что неверно. Учитывая равенство $b > 1$, получим, что хотя бы один из новых ковров будет большим. Аналогично, при обмене маленького ковра хотя бы один из новых ковров будет маленьким. Значит, при таком обмене общее количество больших и маленьких ковров также не уменьшается.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39798,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle $\\omega$ is tangent to the sides of an angle $BAC$ at $B$ and $C$. A line $l$ intersects the segments $AB$ and $AC$ at $K$ and $L$, respectively. The circle $\\omega$ meets $l$ at $P$ and $Q$. The points $S$ and $T$ are chosen on the segment $BC$ so that $KS \\parallel AC$ and $LT \\parallel AB$. Prove that the points $P$, $Q$, $S$, and $T$ are concyclic. (I. Bogdanov, P. Kozhevnikov)\n\nОкружность $\\omega$ касается сторон угла $BAC$ в точках $B$ и $C$. Прямая $\\ell$ пересекает отрезки $AB$ и $AC$ в точках $K$ и $L$ соответственно. Окружность $\\omega$ пересекает $\\ell$ в точках $P$ и $Q$. Точки $S$ и $T$ выбраны на отрезке $BC$ так, что $KS \\parallel AC$ и $LT \\parallel AB$. Докажите, что точки $P, Q, S$ и $T$ лежат на одной окружности.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $l \\nmid BC$, set $X = l \\cap BC$ (see Fig. 12). Then $\\frac{XB}{XT} = \\frac{XK}{XL} = \\frac{XS}{XC}$, so $XT \\cdot XS = XB \\cdot XC = XP \\cdot XQ$.\n\n![](attached_image_1.png)\n\nSince $B$, $C$, $P$, and $Q$ lie on $\\omega$, we have $XB \\cdot XC = XP \\cdot XQ$. Thus, $XT \\cdot XS = XP \\cdot XQ$; this means that the points $P$, $Q$, $S$, and $T$ lie on one circle.\nЕсли $\\ell \\parallel BC$, утверждение очевидно в силу симметрии относительно серединного перпендикуляра к $BC$.\nПусть теперь прямые $\\ell$ и $BC$ пересекаются в точке $X$ (см. рис. 12). Из параллельности получаем $\\frac{XB}{XT} = \\frac{XK}{XL} = \\frac{XS}{XC}$, откуда $XT \\cdot XS = XB \\cdot XC$. Поскольку точки $B$, $C$, $P$ и $Q$ лежат на $\\omega$,\n\n![](attached_image_1.png)\n\nимеем $XB \\cdot XC = XP \\cdot XQ$. Получаем, что $XT \\cdot XS = XP \\cdot XQ$; это и означает, что точки $P$, $Q$, $S$ и $T$ лежат на одной окружности.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = BC$. Let $CC'$ be the diameter of its circumcircle $\\Omega$. The line through $C'$ parallel to $BC$ intersects segments $AB$ and $AC$ at $M$ and $P$, respectively. Prove that $M$ is the midpoint of $C'P$. (B. Obukhov)\n\nПусть $ABC$ — равнобедренный треугольник, $AB = BC$. Пусть $CC'$ — диаметр описанной окружности $\\Omega$ треугольника. Прямая, проходящая через $C'$ и параллельная $BC$, пересекает отрезки $AB$ и $AC$ в точках $M$ и $P$ соответственно. Докажите, что $M$ — середина $C'P$. (Б. Обухов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\triangle AMP \\sim \\triangle ABC$, we have $AM = MP$. Since $C'A \\perp AC$, this yields $C'M = AM = MP$.\n\n\nТак как $CC'$ — диаметр $\\Omega$, имеем $\\angle C'AC = 90^\\circ$. Поскольку $MP \\parallel BC$, получаем $\\angle MPA = \\angle BCA = \\angle BAC$ (см. рис. 1). Значит, треугольник $AMP$ — равнобедренный, и поэтому его высота $MD$ является и медианой. Так как $AD = DP$ и $AC' \\parallel DM$, по теореме Фалеса получаем, что $C'M = MP$.\n\n![](attached_image_1.png)\nРис. 1\n\n**Замечание.** Есть и другие решения, например, с использованием подсчёта углов в прямоугольном треугольнике $PAC'$; именно, $\\angle MAC' = 90^\\circ - \\angle MAP = 90^\\circ - \\angle ACB = 90^\\circ - \\angle MPA = \\angle MC'A$, откуда $MP = MA = MC'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N > 1$ be an integer, and let $d_1 < \\dots < d_s$ be all its positive divisors (thus, $d_1 = 1$ and $d_s = N$). Find all possible values of $N$ for which\n$$\n\\gcd(d_1, d_2) + \\gcd(d_2, d_3) + \\dots + \\gcd(d_{s-1}, d_s) = N - 2. \\quad (\\text{A. Kuznetsov})\n$$\n\nСаша выбрал натуральное число $N > 1$ и выписал в строку в порядке возрастания все его натуральные делители: $d_1 < \\dots < d_s$ (так что $d_1 = 1$ и $d_s = N$). Затем для каждой пары стоящих рядом чисел он вычислил их наибольший общий делитель; сумма полученных $s-1$ чисел оказалась равной $N-2$. Какие значения могло признать $N$? (А. Кузнецов)",
    "options": [],
    "answer": "N = 3",
    "solution": "$N = 3$.\n\nNotice that $\\gcd(d_i, d_{i+1}) \\le d_{i+1} - d_i$; thus, denoting $r_i = (d_{i+1} - d_i) - \\gcd(d_i, d_{i+1}) \\ge 0$ we get $r_1 + \\dots + r_{s-1} = 1$, so $r_k = 1$ for some $k$, and $r_i = 0$ for $i \\ne k$; this yields that $d_{k+1} - d_k = 2$ and that $d_k, d_{k+1}$ are odd. If $k \\ne s/2$, then $r_{s-k} \\ne 0$ as well, which is impossible. Thus $N = d_k d_{k+1}$ is odd, so that $r_{s-1} \\ge \\frac{2N}{3} - \\frac{N}{3}$; this yields $N = 3$.\n$N = 3$.\n\nЗаметим сразу, что $d_{s+1-i} = N/d_i$ при всех $i = 1, 2, ..., s$.\n\nНОД($d_1$, $d_2$) + НОД($d_2$, $d_3$) + ... + НОД($d_{s-1}$, $d_s$) = $N - 2$. Вычитая из первого равенства второе, получаем $r_1 + ... + r_{s-1} = 1$. Это означает, что $r_k = 1$ для некоторого k, а все остальные $r_i$ равны пулю.\nИтак, $1 = (d_{k+1} - d_k) - \\text{НОД}(d_k, d_{k+1})$. Правая, а потому и левая часть этого равенства делится на \\text{НОД}($d_k$, $d_{k+1}$), поэтому \\text{НОД}($d_k$, $d_{k+1}$) = 1 и $d_{k+1} - d_k = 2$. Это возможно, только если каждое из чисел $d_k$ и $d_{k+1}$ нечётно.\n\n$$\n\\text{НОД}(d_m, d_{m+1}) = \\frac{N}{\\text{НОК}(d_k, d_{k+1})} = \\frac{N \\cdot \\text{НОД}(d_k, d_{k+1})}{d_k d_{k+1}} < \\frac{N(d_{k+1} - d_k)}{d_k d_{k+1}} = d_{m+1} - d_m.\n$$\n\nЗначит, $r_m > 0$, что возможно лишь при $k = m$ (и, следовательно, $s = 2k$).\nИтак, $d_{k+1} = \\frac{N}{d_k}$, то есть число $N = d_k d_{k+1}$ нечётно. Но тогда $d_{s-1} \\le \\frac{N}{3}$, откуда \\text{НОД}($d_{s-1}$, $d_s$) $\\le d_{s-1} \\le \\frac{N}{3}$. Следовательно, $1 \\ge r_{s-1} \\ge \\frac{2N}{3} - \\frac{N}{3} = \\frac{N}{3}$, т. е. $N \\le 3$. Поскольку $N > 1$,\n\nполучаем единственно возможное значение $N = 3$, которое, как легко убедиться, удовлетворяет условию.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter chose several consecutive positive integers. He wrote down each of the chosen numbers either in red or in blue (both colors are present). Is it possible that the sum of the l.c.m. of the red numbers and the l.c.m. of the blue numbers is a power of 2? (O. Dmitriev, R. Zhenodarov)\n\nПетя выбрал несколько последовательных положительных целых чисел. Он записал каждое из выбранных чисел либо красным, либо синим цветом (оба цвета присутствуют). Может ли сумма НОК красных чисел и НОК синих чисел быть степенью двойки? (О. Дмитриев, Р. Жендодаров)",
    "options": [],
    "answer": "No",
    "solution": "Let $2^k$ be the maximal power of $2$ dividing one of the chosen numbers; since the numbers are consecutive, $2^k$ divides exactly one of them. Then one of the two l.c.m.'s under consideration is divisible by $2^k$, while the other is not.\n\nSuppose the contrary. Consider the powers of $2$ dividing the chosen numbers; let $2^k$ be the largest among them. If at least two of the chosen numbers are divisible by $2^k$, then two consecutive such numbers will differ by $2^k$. Therefore, one of them will be divisible by $2^{k+1}$, which is impossible by the choice of $k$. Thus, among the chosen numbers, exactly one is divisible by $2^k$.\n\nThe l.c.m. of the group containing this number will be divisible by $2^k$, and the l.c.m. of the remaining group will not be. Therefore, the sum of these l.c.m.'s is not divisible by $2^k$; on the other hand, this sum is greater than $2^k$. Therefore, this sum cannot be a power of $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39802,
    "subject": "Mathematics (Multi-modal)",
    "question": "King Hiero has 11 metal pieces indistinguishable in appearance. The King knows that their weights (in some order) are $1, 2, \\ldots, 11$ kg. Also the King has a bag that breaks if it contains more than $11$ kg. Archimedes knows the weight of each piece, and he wants to prove to Hiero that the first piece weighs $1$ kg. By a move Archimedes can put several pieces into the bag demonstrating to the King that the bag is still not broken (it is prohibited to break the bag!). Find the least number of moves needed for Archimedes to prove his claim.",
    "options": [],
    "answer": "2",
    "solution": "The moves are $1+2+3+5$ and $1+4+6$.\n\nShow that Archimedes needs to use the bag twice. Let him first put into the bag the pieces weighing $1$, $2$, $3$, and $5$ kg, and then the pieces weighing $1$, $4$, and $6$ kg. In both cases, the bag does not break.\n\nLet us prove that this could only happen if the $1$ kg piece was used both times. Indeed, if Archimedes, instead of the pieces weighing $1, \\ldots, 6$ kg, used pieces weighing $w_1, \\ldots, w_6$ kg, then these weights would satisfy the system of inequalities $w_1 + w_2 + w_3 + w_5 \\le 11$, $w_1 + w_4 + w_6 \\le 11$. Adding these inequalities, we get $w_1 + (w_1 + w_2 + \\ldots + w_6) \\le 22$. The sum in parentheses is the sum of six distinct natural numbers, so it is at least $1 + 2 + \\ldots + 6 = 21$. Hence, $w_1 \\le 22 - 21 = 1$. Therefore, $w_1 = 1$, that is, the $1$ kg piece is uniquely determined.\n\nIt remains to show that one move is not enough. If Archimedes puts only one piece in the bag, the bag will not break in any case, so no piece can be identified. If Archimedes puts more than one piece in the bag and the bag does not break, then if the $1$ kg piece is not in the bag, replacing any piece in the bag with it will not change the result; thus, Hiero cannot even tell whether the $1$ kg piece is in the bag. If the desired piece is in the bag, Hiero cannot tell which of the (at least two) pieces is the required one.\n\n**Note.** After these two moves, the group of pieces weighing $2$, $3$, and $5$ kg, as well as the group weighing $4$ and $6$ kg, are also uniquely determined.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39803,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a paper checkered square of size $100 \\times 100$, one has cut out along the grid lines 1950 rectangles, each consisting of two cells. Prove that one may cut out from the remaining pieces a $T$-tetramino (□) — perhaps, rotated. (If such figure exists among the remaining pieces, we assume that it can be cut out.) (S. Berlov)\n\nИз клетчатого бумажного квадрата $100 \\times 100$ вырезали по границам клеток 1950 двуклеточных прямоугольников. Докажите, что из оставшейся части можно вырезать по границам клеток четырёхклеточную фигурку вида [図], — возможно, повёрнутую. (Если такая фигурка уже есть среди оставшихся частей, считается, что её получилось вырезать.) (С. Берлов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Hint 1.** Assume that the dominoes are being cut out one by one. At each moment, define the *price* of a remaining cell as the number of its neighbors which are still not cut out, decreased by 2. Initially, the total price of all cells is 19600, and at each step it decreases by at most 10. Thus, at the end there still exists a cell with positive price; this cell is a central cell of a required $T$-tetramino.\n\n\n**Hint 2.** Divide the initial square into regions as shown on Fig. 13. In this partition, either some $T$-tetramino remains, or some cross loses at most one cell (which cannot be the central cell of the cross).\n\n\nПервое решение.\n\nПредставим себе, что доминошки (прямоугольники $1 \\times 2$) ещё не вырезаны, и будем вырезать их по одной. В каждый момент процесса назовём ценой ещё не вырезанной клетки число её невырезанных соседей по стороне, уменьшенное на 2 (например, цена неугловой клетки, лежащей на границе квадрата, изначально равна 1). Тогда исходная цена каждой клетки есть $2 - t$, где $t$ — количество отрезков периметра квадрата, находящихся на границе этой клетки. Значит, исходная суммарная цена всех клеток равна $2 \\cdot 100^2 - 400 = 19600$.\n\nПроследим, как изменяется суммарная цена $S$ всех невырезанных клеток после вырезания доминошки. При этом выкидываются две клетки (сумма цен которых не превосходит $2+2=4$), а также уменьшаются на 1 цены клеток, граничащих с доминошкой (которых не больше шести). Поэтому после вырезания доминошки $S$ уменьшается не более, чем на 10.\n\nИтак, после вырезания 1950 доминошек $S$ станет не меньше, чем $19600 - 1950 \\cdot 10 = 100$. Значит, найдётся невырезанная клетка $k$, цена которой положительна. Это значит, что у $k$ не менее трёх невырезанных соседей. Тогда $k$ вместе с этими тремя соседями образует требуемую фигурку.\n\n\nВторое решение.\n\nНазовём требуемую фигурку $T$-тетрамино.\n\nМысленно разобьём наш квадрат на фигурки (см. рис. 13). Нетрудно подсчитать, что вне «полных» крестов окажется ровно $4 \\cdot 100 = 400$ клеток, из которых 320 будут находиться в $T$-тетрамино разбиения. Итого, в разбиении есть $(100^2 - 400)/5 = 1920$ полных крестов и ещё 80 $T$-тетрамино.\n\nРассмотрим теперь, куда попадают клетки вырезанных доминошек. Предположим, что из каждого полного креста было вырезано хотя бы по две клетки, а из каждого $T$-тетрамино разбиения — хотя бы одна. Тогда общее число вырезанных клеток было бы не меньше, чем $1920 \\cdot 2 + 80 = 2 \\cdot 1960$, что неверно.\n\nЗначит, либо из некоторого $T$-тетрамино не вырезано ни одной клетки, либо из некоторого креста вырезано не более одной клетки. В первом случае мы уже нашли $T$-тетрамино, которое можно вырезать. Во втором же случае, если из креста и вырезана одна клетка, то она не может быть центральной (иначе вторая клетка той же доминошки также лежала бы в кресте). Значит, даже если клетка креста вырезана, остаток его как раз и является $T$-тетрамино. В обоих случаях мы добились требуемого.\n\n\n![](attached_image_1.png)\nРис. 13",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39804,
    "subject": "Mathematics (Multi-modal)",
    "question": "A class consists of 23 students. During a year, each student celebrated his birthday with some (at least one, but not all) of his classmates. For each pair of students, count the number of such parties in which they both participated. Is it possible that all such numbers for all pairs of students are equal? (I. Bogdanov)\n\nВ классе 23 ученика. В течение года каждый ученик отмечал день рождения с некоторыми (по крайней мере с одним, но не со всеми) из своих одноклассников. Для каждой пары учеников посчитайте, на скольких таких праздниках они оба присутствовали. Могло ли оказаться, что для всех пар учеников эти числа одинаковы? (И. Богданов)",
    "options": [],
    "answer": "Yes",
    "solution": "Two possible examples are as follows.\n\n1) Arrange all the students in a circle and assume that each comes to all birthdays except for that of his clockwise neighbor. Then, for any two students $A$ and $B$, they miss only the birthdays where $A$ is absent and where $B$ is absent. Thus, any pair of students met at all parties except two, so each pair met $21$ times.\n\n2) Distinguish two students $A$ and $B$. At $A$'s birthday, everyone except $B$ is present. At $B$'s birthday, only $A$ is present. Finally, at everyone else's birthday, only $B$ is present. Then, any pair not involving $B$ met only at $A$'s birthday, and all pairs involving $B$ met exactly once at the other birthdays. Thus, each pair met exactly once.\n**Первое решение.** Предъявим пример, как такое могло произойти. Выстроим учеников по кругу. Предположим, что к каждому на день рождения пришли все одноклассники, кроме следующего за ним по часовой стрелке. Тогда любые два ученика $A$ и $B$ встретились на всех празднованиях, кроме двух: того, на которое не пришёл $A$, и того, на котором не пришёл $B$. Значит, любая пара учеников встретилась $21$ раз.\n\n**Второе решение.** Предъявим другой возможный пример. Выделим из класса двух учеников $A$ и $B$. Пусть на день рождения к $A$ пришли все одноклассники, кроме $B$, на день рождения к $B$ пришёл только $A$, а на остальные дни рождения приходил только $B$. Тогда любая пара, в которой нет $B$, встретилась только на дне рождения $A$, а все пары, содержащие $B$, встречались ровно по разу на остальных празднованиях. Итого, каждая пара встретилась ровно по разу.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider nine 9-digit numbers, each contains exactly the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ (in some order). The sum of these nine numbers ends with $k$ zeroes. Determine the maximal possible value of $k$.\n\nИз цифр $1, 2, 3, 4, 5, 6, 7, 8, 9$ составлены девять (не обязательно различных) девятизначных чисел; каждая из цифр использована в каждом числе ровно один раз. На какое наибольшее количество нулей может оканчиваться сумма этих девяти чисел?",
    "options": [],
    "answer": "8",
    "solution": "$k = 8$.\n\nПокажем, что сумма не может оканчиваться на 9 нулей. Каждое из составленных чисел делится на $9$, поскольку сумма его цифр делится на $9$. Поэтому их сумма также делится на $9$. Наименьшее натуральное число, делящееся на $9$ и оканчивающееся на девять нулей, равно $9 \\cdot 10^9$, так что сумма наших чисел не меньше $9 \\cdot 10^9$. Значит, одно из них не меньше $10^9$, что невозможно.\n\nОсталось показать, как составить числа, сумма которых оканчивается на восемь нулей. Например, можно взять восемь чисел, равных $987654321$, и одно число $198765432$. Их сумма равна $81 \\cdot 10^8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39806,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a non-empty set $A$ consisting of positive integers is complete if for any positive integers $a$ and $b$ such that $a+b \\in A$, the number $ab$ also lies in $A$ (the numbers $a$ and $b$ are not required to be distinct or to belong to $A$). Find all complete sets.\n\nНазовём непустое (конечное или бесконечное) множество $A$, состоящее из натуральных чисел, полным, если для любых натуральных $a$ и $b$ (не обязательно различных и не обязательно лежащих в $A$) таких, что $a+b$ лежит в $A$, число $ab$ также лежит в $A$. Найдите все полные множества натуральных чисел.",
    "options": [],
    "answer": "Exactly the following sets: {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}, and the set of all positive integers.",
    "solution": "The set of all positive integers, along with $\\{1\\}$, $\\{1, 2\\}$, $\\{1, 2, 3\\}$, and $\\{1, 2, 3, 4\\}$.\n\nFirst, we check that the sets $\\{1\\}$, $\\{1, 2\\}$, $\\{1, 2, 3\\}$, $\\{1, 2, 3, 4\\}$, and the set of all positive integers are complete. For the last set, this is obvious; for the first four, note that if positive integers $a$ and $b$ are such that $a+b \\le 4$, then either both are $2$, or one of them is $1$; in any of these cases, we have $ab \\le a+b$. Thus, if $a+b \\in A$, then $ab \\in A$.\n\nNow let $A$ be an arbitrary complete set. If $A$ contains some number $k \\ge 2$, then by the condition it also contains the number $1 \\cdot (k-1) = k-1$. Continuing this process, we obtain that all positive integers not exceeding $k$ lie in $A$. In particular, if $A$ does not contain numbers greater than $4$, then the set $A$ is already listed in the answer.\n\nNow suppose $A$ contains a number $\\ell \\ge 5$. Define a sequence $\\ell_1, \\ell_2, \\dots$ by $\\ell_1 = \\ell$, $\\ell_{n+1} = 2(\\ell_n - 2)$. All these numbers lie in $A$. Indeed, $\\ell_1$ lies in $A$ by our assumption, and if $\\ell_n = 2 + (\\ell_n - 2) \\in A$, then $\\ell_{n+1} = 2(\\ell_n - 2) \\in A$. Moreover, $\\ell_{n+1} = \\ell_n + (\\ell_n - 4)$; by induction, we now get that $\\ell_{n+1} > \\ell_n \\ge 5$. Thus, for any positive integer $n$ we have $\\ell_n > n$; from the reasoning of the previous paragraph, we now see that $n \\in A$. Therefore, all positive integers lie in $A$.\nМножество всех натуральных чисел, а также множества $\\{1\\}, \\{1,2\\}, \\{1,2,3\\}$ и $\\{1,2,3,4\\}$.\n\nДля начала проверим, что множества $\\{1\\}, \\{1,2\\}, \\{1,2,3\\}, \\{1,2,3,4\\}$, а также множество всех натуральных чисел — полные. Для последнего множества это очевидно; для первых четырёх заметим, что если натуральные числа $a$ и $b$ таковы, что $a+b \\le 4$, то либо они оба равны $2$, либо одно из них равно $1$; в любом из этих случаев имеем $ab \\le a+b$. Значит, если $a+b \\in A$, то и $ab \\in A$.\n\nПусть теперь $A$ — произвольное полное множество. Если $A$ содержит некоторое число $k \\ge 2$, то по условию оно также содержит число $1 \\cdot (k-1) = k-1$. Продолжая этот процесс, получаем, что все натуральные числа, не превосходящие $k$, лежат в $A$. В частности, если $A$ не содержит чисел, больших $4$, то множество $A$ уже перечислено в ответе.\n\nПусть теперь в $A$ есть число $\\ell \\ge 5$. Зададим последовательность $\\ell_1, \\ell_2, \\dots$ соотношениями $\\ell_1 = \\ell, \\ell_{n+1} = 2(\\ell_n - 2)$. Все эти числа лежат в $A$. Действительно, $\\ell_1$ лежит в $A$ по нашему предположению, а если $\\ell_n = 2+(\\ell_n-2) \\in A$, то и $\\ell_{n+1} = 2(\\ell_n-2) \\in A$. Кроме того, $\\ell_{n+1} = \\ell_n + (\\ell_n - 4)$; по индукции теперь получаем, что $\\ell_{n+1} > \\ell_n \\ge 5$. Значит, для любого натурального $n$ имеем $\\ell_n > n$; из рассуждений предыдущего абзаца понимаем теперь, что и $n \\in A$. Итак, все натуральные числа лежат в $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39807,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square is dissected into $n^2 \\ge 4$ rectangles by $n-1$ vertical lines and $n-1$ horizontal lines. Prove that one may choose $2n$ of these rectangles such that for any two of the chosen rectangles, one can be put completely into the other (perhaps, after some rotation).\n\nКвадрат разбит на $n^2 \\ge 4$ прямоугольников $2(n-1)$ прямыми, из которых $n-1$ параллельны одной стороне квадрата, а остальные $n-1$ — другой. Докажите, что можно выбрать $2n$ прямоугольников разбиения таким образом, что для любых двух выбранных прямоугольников один из них можно поместить в другой (возможно, предварительно повернув).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $a_1 \\ge \\dots \\ge a_n$ are the lengths of the rectangles' horizontal sides (from the left to the right), and $b_1 \\ge \\dots \\ge b_n$ are the lengths of their vertical sides (from the top to the bottom). Let $Q_{i,j}$ be the $a_i \\times b_j$ rectangle. Since $a_1 + \\dots + a_n = b_1 + \\dots + b_n$, there exists an index $k$ such that $a_k \\le b_k$ and $a_{k-1} \\ge b_{k-1}$ (or vice versa; we consider only the indicated case). Then the required chain of rectangles is shown on Fig. 14.\n\nНазовём пару прямоугольников вложимой, если один из них можно вложить в другой.\nПусть горизонтальная сторона квадрата разбилась на отрезки длины $a_1, \\dots, a_n$ (слева направо), а вертикальная — на отрезки длины $b_1, \\dots, b_n$ (сверху вниз). Переставив «столбцы» и «строки», можно считать, что $a_1 \\ge \\dots \\ge a_n$ и $b_1 \\ge \\dots \\ge b_n$. Обозначим через $Q_{i,j}$ прямоугольник разбиения со сторонами $a_i$ и $b_j$. Заметим, что при $i \\le k$ и $j \\le l$ пара $Q_{i,j}$ и $Q_{k,l}$ вложима.\n\nПоскольку $a_1 + \\dots + a_n = b_1 + \\dots + b_n$, найдутся различные индексы $i$ и $j$ такие, что $a_i \\ge b_i$ и $a_j \\le b_j$. Можно считать, что $i < j$. Тогда существует индекс $k \\in [i, j]$ такой, что $a_k \\le b_k$ и $a_{k-1} \\ge b_{k-1}$.\nМы утверждаем, что можно выбрать следующие прямоугольники:\n$Q_{1,1}, Q_{1,2}, \\dots, Q_{1,k-1}, Q_{2,k-1}, \\dots, Q_{k,k-1}$, вкупе с $Q_{k-1,k}, Q_{k,k}, \\dots, Q_{k,n}, Q_{k+1,n}, \\dots, Q_{n,n}$ (см. рис. 14).\nИх количество равно $2(k-1) + 2(n-k+1) = 2n$. Любая пара из них, кроме $(Q_{k,k-1}, Q_{k-1,k})$, вложима по замечанию выше. Наконец, оставшаяся пара также вложима, поскольку $a_k \\le b_k$ и $b_{k-1} \\le a_{k-1}$ (для вложения один прямоугольник нужно повернуть на $90^\\circ$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each cell of a $2016 \\times 2016$ table is painted either in white or in black. We say that a positive integer $k$ is lucky if $k \\le 2016$, and each of cellular $k \\times k$ squares within the table contains exactly $k$ black cells. (For example, if all the cells are black, then only the number 1 is lucky.) Determine the greatest possible amount of lucky numbers. (E. Bakaev)\n\nВ белой таблице $2016 \\times 2016$ некоторые клетки окрасили чёрным. Назовём натуральное число $k$ удачным, если $k \\le 2016$, и в каждом из клетчатых квадратов со стороной $k$, расположенных в таблице, окрашено ровно $k$ клеток. (Например, если все клетки чёрные, то удачным является только число 1.) Какое наибольшее количество чисел могут быть удачными?",
    "options": [],
    "answer": "1008",
    "solution": "Let $a$ and $b$ be lucky. If $b \\ge 2a$, then a $b \\times b$ square contains $[b/a]^2$ squares of size $a \\times a$, so at least $[b/a]^2 \\cdot a > b$ black cells. This is impossible. Thus, the ratio of any two lucky numbers is less than $2$, which yields that there are at most $1008$ of them.\n\nAn example is provided by a white table with one of the two middle rows painted in black.\n\n\nРассмотрим произвольное окрашивание таблицы. Пусть нашлось хотя бы два удачных числа, и $a$ — наименьшее из них, а $b$ — наибольшее.\n\nПоделим $b$ на $a$ с остатком: $b = qa + r$, где $0 \\le r < a$. Предположим, что $q \\ge 2$. В произвольном квадрате $b \\times b$ можно расположить $q^2$ пересекающихся квадратов $a \\times a$. В этих квадратах будет ровно $q^2 a$ чёрных клеток. Однако $q^2 a > (q+1)a > qa + r = b$; значит, в квадрате $b \\times b$ будет больше, чем $b$ чёрных клеток, что невозможно. Итак, $q < 2$, то есть $b < 2a$.\n\nОбщее количество удачных чисел не превосходит количества натуральных чисел от $a$ до $b$, то есть оно не больше $b - a + 1 < b - b/2 + 1 = b/2 + 1 \\le 1009$. Значит, это количество не больше $1008$.\n\nОсталось привести пример раскраски, для которой найдутся $1008$ удачных чисел. Окрасим чёрным все клетки $1008$-й строки и только их. Рассмотрим произвольный квадрат со стороной $d \\ge 1009$. Он пересекается с $1008$-ой строкой, значит в нём есть целая строка отмеченных клеток, то есть их как раз $d$ штук. Значит, все числа от $1009$ до $2016$ являются удачными, и таких чисел как раз $1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be the incircle of a triangle $ABC$ (with $AB < AC$). Let the excircle $\\omega_A$ touch the side $BC$ at $A'$. A variable point $X$ is chosen on the segment $AA'$ so that the segment $A'X$ has no common points with $\\omega$. The tangents to $\\omega$ passing through $X$ meet the side $BC$ at $Y$ and $Z$. Prove that the sum $XY + XZ$ does not depend on the choice of $X$. (I. Mitrofanov)\n\nОкружность $\\omega$ вписана в треугольник $ABC$, в котором $AB < AC$. Вневписанная окружность этого треугольника касается стороны $BC$ в точке $A'$. Точка $X$ выбирается на отрезке $A'A$ так, что отрезок $A'X$ не пересекает $\\omega$. Касательные, проведённые из $X$ к $\\omega$, пересекают отрезок $BC$ в точках $Y$ и $Z$. Докажите, что сумма $XY + XZ$ не зависит от выбора точки $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega$ touch $BC$ at $A''$. We assume that $A''Y < A''Z$. Let $\\omega'$ be the excircle of the triangle $XYZ$ touching the side $XZ$, and let $T$ be the meeting point of $\\omega$ and $AA'$ closest to $A$; the tangent at $T$ to $\\omega$ is parallel to $BC$. The homothety centered at $X$ mapping $\\omega$ to $\\omega'$ maps $T$ to $AA' \\cap BC = A'$. Therefore, $XY + XZ = A'A''$ which does not depend on $X$.\n\n\nБудем считать, что точка $Y$ лежит ближе к точке $B$, нежели $Z$; кроме того, считаем, что сторона $BC$ горизонтальна, а $A$ лежит выше неё (см. рис. 15).\n\n![](attached_image_1.png)\n\nОбозначим через $\\omega_A$ вневписанную окружность треугольника $ABC$, касающуюся стороны $BC$, а через $\\omega'$ — вневписанную окружность треугольника $XYZ$, касающуюся стороны $XZ$.\n\nПусть $\\omega$ касается $BC$ в точке $A''$. Обозначим через $T$ точку пересечения $AA'$ и $\\omega$, лежащую ближе к $A$. Гомотетия с центром $A$, переводящая $\\omega$ в $\\omega_A$, переводит $T$ в $A'$; значит, касательная к $\\omega$ в $T$ параллельна $BC$.\nПоскольку окружности $\\omega$ и $\\omega'$ вписаны в вертикальные углы, образованные прямыми $XY$ и $XZ$, существует гомотетия с центром в $X$ (и отрицательным коэффициентом), переводящая $\\omega$ в $\\omega'$. Пусть при этой гомотетии точка $T$ переходит в точку $T'$. Тогда $T'$ лежит на прямой $AA'$, касательная к $\\omega'$ в $T'$ параллельна $BC$, и $\\omega'$ лежит выше этой касательной. Такая касательная к $\\omega'$ — это прямая $BC$; значит, $T'$ лежит на $BC$, то есть $\\omega'$ касается $BC$ в точке $A'$.\n\nОбозначим полупериметр треугольника $XYZ$ через $p$. Так как окружности $\\omega$ и $\\omega'$ — вневписанные для этого треугольника, имеем $ZA' = YA'' = p - YZ$. Значит,\n$$\nXY + XZ = 2p - YZ = 2(p - YZ) + YZ = ZA' + YZ + YA'' = A'A'', \\text{ что не зависит от выбора точки } X.\n$$\n\n**Замечание.** Тот факт, что $\\omega'$ касается $BC$ в точке $A'$, можно также доказать, применив теорему о трёх гомотетиях к окружностям $\\omega, \\omega_A$ и $\\omega'$. Центры гомотетий, переводящих их друг в друга, есть $A, X$ и некоторая точка $Q$, лежащая на $BC$ (так как $BC$ — внутренняя общая касательная к $\\omega_A$ и $\\omega'$). Получаем, что $Q$ лежит на прямой $AX$, то есть совпадает с $A'$. Поскольку $\\omega_A$ касается $BC$ в $Q = A'$, то и $\\omega'$ также касается $BC$ в этой же точке.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $\\angle DAB = 90^\\circ$. Let $M$ be the midpoint of $BC$. Given that $\\angle ADC = \\angle BAM$, prove that $\\angle ADB = \\angle CAM$. (E. Bakaev)\n\nДан выпуклый четырёхугольник $ABCD$, в котором $\\angle DAB = 90^\\circ$. Пусть $M$ — середина стороны $BC$. Оказалось, что $\\angle ADC = \\angle BAM$. Докажите, что $\\angle ADB = \\angle CAM$. (Е. Бакаев)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose a point $K$ such that $A$ is the midpoint of $BK$ (see Fig. 2). Then the quadrilateral $AKDC$ is cyclic, due to\n$$\n\\angle BKC = \\angle BAM = \\angle ADC.\n$$\nHence $\\angle CAM = \\angle ACK = \\angle ADK = \\angle ADB$.\n\n\nПервое решение.\n\nНа продолжении отрезка $AB$ за точку $A$ отметим точку $K$ так, что $AB = AK$ (см. рис. 2). Тогда $AM$ — средняя линия в треугольнике $BCK$, откуда $AM \\parallel CK$. Значит, $\\angle BKC = \\angle BAM = \\angle ADC$. Отсюда следует, что четырёхугольник $AKDC$ вписан.\nОпять же используя параллельность $AM$ и $CK$, получаем\n$\\angle CAM = \\angle ACK = \\angle ADK$. Наконец, $DA$ — медиана и высота в треугольнике $BDK$, поэтому $DA$ является и биссектрисой; отсюда $\\angle ADB = \\angle ADK = \\angle CAM$, что и требовалось доказать.\n\n\nВторое решение.\n\nЗаметим, что $\\angle ADC + \\angle DAM = \\angle BAM + \\angle DAM = 90^\\circ$; это значит, что $AM \\perp CD$. Опустим перпендикуляры $MN$ и $BP$ из точек $M$ и $B$ на прямую $CD$; тогда точки $A$, $M$ и $N$ лежат на одной прямой (см. рис. 3).\nПоскольку $BM = MC$, по теореме Фалеса получаем $PN = NC$. Значит, $AN$ — высота и медиана в треугольнике $APC$, откуда $\\angle CAM = \\angle MAP$. Так как $BP \\parallel AN$, получаем $\\angle MAP = \\angle APB$. Наконец, поскольку $\\angle BPD = \\angle BAD = 90^\\circ$, четырёхугольник $ABPD$ вписан; поэтому $\\angle APB = \\angle ADB$. Итак, мы получили, что $\\angle CAM = \\angle MAP = \\angle APB = \\angle ADB$, что и требовалось.\n\n![](attached_image_1.png)\nРис. 2\n![](attached_image_2.png)\nРис. 3\n\nТретье решение.\n\nОтложим на луче $AM$ точку $Q$ так, что $AQ = 2AM$ (см. рис. 4). Тогда в четырёхугольнике $ABQC$ диагонали делятся точкой пересечения пополам, то есть он — параллелограмм; значит, $\\angle CAQ = \\angle AQB$.\nТак как $QC \\parallel AB$, получаем $QC \\perp AD$. Так как $\\angle QAD = 90^\\circ - \\angle BAQ = 90^\\circ - \\angle ADC$, имеем $DC \\perp AQ$. Значит, $C$ — точка пересечения высот в треугольнике $AQD$, откуда $AC \\perp QD$ (и, значит, $BQ \\perp QD$).\nПоскольку $\\angle BAD = \\angle BQD = 90^\\circ$, четырёхугольник $ABQD$ вписан. Значит, $\\angle ADB = \\angle AQB = \\angle CAQ$, что и требовалось доказать.\n\n![](attached_image_3.png)\nРис. 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39811,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a product of 5 positive integers each multiple was decreased by 3. Could it happen that the product becomes exactly 15 times larger than the initial one?\n\nВ произведении 5 положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли получиться так, что произведение увеличилось ровно в 15 раз?",
    "options": [],
    "answer": "Yes; for example, 1, 1, 1, 1, 48.",
    "solution": "Yes, it is possible.\n\nAn example is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 48$.\n\nAfter decreasing each factor by $3$, we get:\n$$\n(-2) \\cdot (-2) \\cdot (-2) \\cdot (-2) \\cdot 45 = 16 \\cdot 45 = 720.\n$$\nThe original product is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 48 = 48$.\n\nThus, $720 = 15 \\cdot 48$.\nДа, могло.\n\nВ качестве примера подходит произведение $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 48$.\nПосле указанной операции получается\n$$\n(-2) \\cdot (-2) \\cdot (-2) \\cdot (-2) \\cdot 45 = 720 = 15 \\cdot 48.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39812,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian plane, two graphs $\\Gamma_1$ and $\\Gamma_2$ of monic quadratic trinomials and two non-parallel lines $l_1$ and $l_2$ are drawn. Assume that $\\Gamma_1$ and $\\Gamma_2$ cut out segments of equal lengths on $l_1$, and that they cut out segments of equal lengths on $l_2$. Prove that $\\Gamma_1$ and $\\Gamma_2$ coincide.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $f_1(x)$ and $f_2(x)$ be the given monic quadratic trinomials, and let $\\Gamma_1$ and $\\Gamma_2$ be their graphs. Then there exists a unique vector $\\vec{a}$ such that, under a parallel translation by this vector, the parabola $\\Gamma_1$ maps to $\\Gamma_2$ (the vector $\\vec{a}$ connects the vertices of the parabolas).\n\nLet the line $\\ell_1$ intersect $\\Gamma_1$ at points $A_1$ and $B_1$, and $\\Gamma_2$ at points $A_2$ and $B_2$ such that $\\overrightarrow{A_1B_1} = \\overrightarrow{A_2B_2}$. Under a parallel translation of the parabola $\\Gamma_1$ by the vector $A_1A_2$, we obtain a parabola $\\Gamma_3$, which is the graph of a monic quadratic $f_3(x)$, and which intersects $\\ell_1$ at the same points as $\\Gamma_2$. Then the difference $f_2(x) - f_3(x)$ has at least two roots (the abscissas of points $A_2$ and $B_2$). But since the degree of the polynomial $f_2(x) - f_3(x)$ is at most $1$, it follows that $f_2(x) - f_3(x) \\equiv 0$, i.e., $\\Gamma_3 = \\Gamma_2$. Therefore, the shift vector $\\vec{a} = \\overrightarrow{A_1A_2}$ is parallel to the line $\\ell_1$.\n\nSimilarly, $\\vec{a} \\parallel \\ell_2$, so $\\vec{a} = \\vec{0}$ and $\\Gamma_1 = \\Gamma_2$.\n\n\nSecond solution. Let $f_1(x) = x^2 + p_1x + q_1$ and $f_2(x) = x^2 + p_2x + q_2$ be the given quadratic trinomials, $\\Gamma_1$ and $\\Gamma_2$ their graphs, and $y = k_1x + b_1$ and $y = k_2x + b_2$ the equations of the lines $l_1$ and $l_2$, respectively.\n\nNote that the length of the segment cut out by the parabola $y = x^2 + px + q$ on the line $y = kx + b$ equals the modulus of the difference of the roots of the equation $x^2 + px + q = kx + b$, divided by the cosine of the angle of inclination of the line $y = kx + b$. The modulus of the difference of the roots of a monic quadratic trinomial equals the square root of its discriminant. Therefore, the condition that $\\Gamma_1$ and $\\Gamma_2$ cut out equal segments on $l_1$ can be written as\n$$\n(p_1 - k_1)^2 - 4(q_1 - b_1) = (p_2 - k_1)^2 - 4(q_2 - b_1).\n$$\nTransforming this equality, we get $(p_1 - p_2)(p_1 + p_2 - 2k_1) = 4(q_1 - q_2)$. If $p_1 = p_2$, then $q_1 = q_2$ and the problem is solved. Otherwise,\n$$\nk_1 = \\frac{p_1 + p_2}{2} - \\frac{2(q_1 - q_2)}{p_1 - p_2}.\n$$\nExactly the same reasoning can be applied to the line $l_2$. Thus, if the parabolas do not coincide, then $k_1 = k_2$, i.e., $l_1$ and $l_2$ are parallel, which is impossible by the condition.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 39813,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle centered at $I$ is inscribed into a quadrilateral $ABCD$. $BA$ and $CD$ meet at $P$, and rays $AD$ and $BC$ meet at $Q$. Given that $P$ lies on the circle $(AIC)$, prove that $Q$ also lies on this circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Prove that $\\angle DAI = \\angle BAI = \\angle DCI = \\angle BCI$ (see Fig. 6).\n\nТак как четырёхгольник $AICP$ вписанный, то $\\angle PCI = 180^\\circ - \\angle PAI = \\angle BAI$; иначе говоря, $\\angle DCI = \\angle BAI$ (см. рис. 6). Центр $I$ вписанной окружности четырёхгольника лежит на биссектрисах его углов, поэтому $\\angle DCI = \\angle BCI$ и $\\angle DAI = \\angle BAI$. Отсюда следует, что $\\angle DAI = \\angle BCI$, а значит, $\\angle QAI = \\angle BCI = 180^\\circ - \\angle QCI$.\n![](attached_image_1.png)\nРис. 6\nИз полученного равенства вытекает, что четырёхгольник $AICQ$ вписанный. Тем самым, точка $Q$ лежит на окружности $w$ (проходящей через точки $A$, $I$ и $C$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of an acute-angled isosceles triangle $ABC$ with $AB = AC$. The rays $BO$ and $CO$ meet the sides $AC$ and $AB$ at $B'$ and $C'$, respectively. Let $l$ be the line through $C'$ parallel to $AC$. Prove that $l$ is tangent to the circumcircle of the triangle $B'OC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть прямая $AO$ пересекает $l$ в точке $T$ (см. рис. 13).\n\nИз симметрии относительно $AO$ имеем $\\angle B'TO = \\angle C'TO$.\n\nПоскольку $l \\parallel AC$, получаем $\\angle C'TO = \\angle OAC = \\angle OCA$.\n\nИтак, $\\angle B'TO = \\angle B'CO$, то есть $T$ лежит на окружности $\\omega$, описанный около треугольника $B'O'C$.\n\nКратом того, из тех же соображений имеем $\\angle OB'T = \\angle OC'T = \\angle OCA = \\angle OTC'$, то есть $C'T$ касается $\\omega$ в точке $T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pasha chose 2017 (not necessarily distinct) positive integers $a_1, a_2, \\dots, a_{2017}$, and then he plays a solitaire game. Initially, he has 2017 empty large boxes and an unbounded supply of small stones. By a move, Pasha adds $a_1$ stones into some box by his choice, $a_2$ stones into any other box by his choice, ..., $a_{2017}$ stones into the remaining box. His aim is to equalize the numbers of stones in all boxes. Can he choose the initial numbers so that the aim is reachable in 43 moves, but unreachable in any smaller (nonzero) number of moves?\n\nПаша выбрал 2017 (не обязательно различных) положительных целых чисел $a_1, a_2, \\dots, a_{2017}$, после чего играет в пасьянс. Изначально у него есть 2017 пустых больших коробок и неограниченное количество маленьких камешков. За один ход Паша кладёт $a_1$ камней в какую-то коробку по своему выбору, $a_2$ камней — в другую коробку по своему выбору, ..., $a_{2017}$ камней — в оставшуюся коробку. Его цель — уравнять количество камней во всех коробках. Может ли он выбрать начальные числа так, чтобы цель была достижима за 43 хода, но недостижима ни за какое меньшее (ненулевое) число ходов?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes.\n\nNotice that $2017 = 43 \\cdot 46 + 39$. One example of Pasha's numbers consists of 39 twos, 46 numbers equal to 44, and ones as the remaining numbers.\n\nTo achieve the goal in 43 moves, Pasha chooses 39 boxes in which he always puts 2 stones—after 43 moves, each of these will contain $43 \\cdot 2 = 86$ stones. The remaining boxes are divided into 43 groups of 46 boxes each; on the $i$-th move, he puts 44 stones into all boxes of the $i$-th group and 1 stone into the others—in each group, there will be $44 + 42 \\cdot 1 = 86$ stones, so all boxes will have the same number of stones.\n\nIt remains to prove that the goal cannot be achieved in fewer moves. Suppose Pasha made $k < 43$ moves. Then in some box $A$, 44 stones were placed in one move, so it will have at least $44 + (k-1) \\cdot 1 = 43 + k$ stones. On the other hand, since $46k < 2017$, there is some box $B$ that never received 44 stones, so it will have at most $2k$ stones. Since $k < 43$, we have $2k < k + 43$, so box $B$ has fewer stones than box $A$. Thus, Pasha has not yet achieved the goal.\n\nRemark. The example given is not unique. For example, a set consisting of 42 ones, $2017 - 43 = 1974$ numbers equal to $a > 1$, and one number equal to $43a - 42$ also works. There are even examples with all numbers pairwise distinct; however, checking that they work is somewhat more difficult than for the examples given above.\nДа, мог.\n\nЗаметим, что $2017 = 43 \\cdot 46 + 39$. Приведём пример Пашиных чисел, при которых требуемое выполняется. Пусть среди его чисел будут 39 двоек, 46 чисел, равных 44, а остальные — единицы.\nЧтобы добиться требуемого за 43 хода, Паша выбирает 39 коробок, в которые он всегда кладёт по 2 камня — через 43 хода в них окажется по $43 \\cdot 2 = 86$ камней. Остальные коробки он разбивает на 43 группы по 46 коробок; на $i$-м ходу он положит по 44 камня во все коробки $i$-й группы и по одному камню — в каждой группе будет по $44+42 \\cdot 1 = 86$ камней, то есть во всех коробках будет поровну камней.\n\nОсталось доказать, что за меньшее число ходов требуемое невыполнимо. Пусть Паша сделал $k < 43$ ходов. Тогда в какую-то коробку $A$ попало 44 камня на одном ходу, и в ней будет не меньше, чем $44 + (k-1) \\cdot 1 = 43 + k$ камней. С другой стороны, поскольку $46k < 2017$, в какую-то коробку $B$ ни на одном из ходов не попадёт 44 камня, то есть в ней будет не больше $2k$ камней. Поскольку $k < 43$, имеем $2k < k + 43$, а значит, в коробке $B$ меньше камней, чем в $A$. Таким образом, Паша ещё не добился требуемого.\n\nЗамечание. Приведённый пример — не единственный. Например, подойдёт также набор чисел, состоящий из 42 единиц, $2017 - 43 = 1974$ чисел, равных $a > 1$, и одного числа, равного $43a - 42$. Существуют даже примеры с попарно различными числами; однако проверка того, что они подходят, несколько труднее, чем для примеров, приведённых выше.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive numbers $a_1, a_2, \\dots, a_n$ are written on the board in a row.\nFor every $i = 1, 2, \\dots, n$, Vasya wishes to write a number $b_i \\ge a_i$ so that for every $i, j \\in \\{1, 2, \\dots, n\\}$, at least one of the ratios $b_i/b_j$ and $b_j/b_i$ is an integer. Prove that Vasya can reach this goal so\n\n$$\n\\text{that } b_1 b_2 \\dots b_n \\le 2^{(n-1)/2} a_1 a_2 \\dots a_n.\n$$\n\nПоложительные числа $a_1, a_2, \\dots, a_n$ записаны в ряд на доске.\nДля каждого $i = 1, 2, \\dots, n$ Вася хочет записать число $b_i \\ge a_i$ так, чтобы для любых $i, j \\in \\{1, 2, \\dots, n\\}$ хотя бы одно из отношений $b_i/b_j$ и $b_j/b_i$ было целым числом. Докажите, что Вася может достичь этого так, что\n\n$$\nb_1 b_2 \\dots b_n \\le 2^{(n-1)/2} a_1 a_2 \\dots a_n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Prove that there exists even a collection of such $b_i$'s such that each $b_i/b_j$ is a power of $2$. For this purpose, for each $i$ choose such collection with $b_i = a_i$ and minimal possible $b_j$'s with $j \\neq i$. The product of all $n^2$ numbers in these $n$ collections is at most $(2^{(n-1)/2} a_1 a_2 \\cdots a_n)^n$.\n\n\nМы докажем, что существуют даже числа $b_1, b_2, \\dots, b_n$, удовлетворяющие следующим (более сильным) условиям:\n(1) $b_i \\ge a_i$ при всех $i \\le n$;\n(2) $b_1b_2\\dots b_n \\le 2^{(n-1)/2} a_1a_2\\dots a_n$;\n(3) отношение любых двух из чисел $b_i$ является степенью двойки (с целым показателем).\n\nЗаметим, что доказываемое утверждение не изменится, если какое-то из чисел $a_k$ (а с ним и соответствующее $b_k$) умножить на некоторую степень двойки. Умножим каждое из чисел $a_k$ на степень двойки так, чтобы все полученные числа лежали в промежутке $[1, 2)$.\nНе умаляя общности можно считать, что $1 \\le a_1 \\le a_2 \\le \\dots \\le a_n < 2$. Покажем теперь, что одна из следующих $n$ последовательностей удовлетворяет всем трём условиям:\n\n$$\n\\begin{align*}\na_1, & \\quad a_1, \\quad 2a_1, \\quad 2a_1, \\quad \\dots, \\quad 2a_1, \\quad 2a_1; \\\\\na_2, & \\quad a_2, \\quad 2a_2, \\quad 2a_2, \\quad \\dots, \\quad 2a_2, \\quad 2a_2; \\\\\na_3, & \\quad a_3, \\quad a_3, \\quad 2a_3, \\quad \\dots, \\quad 2a_3, \\quad 2a_3; \\\\\n& \\dots \\\\\na_{n-1}, & \\quad a_{n-1}, \\quad a_{n-1}, \\quad a_{n-1}, \\quad \\dots, \\quad a_{n-1}, \\quad 2a_{n-1}; \\\\\na_n, & \\quad a_n, \\quad a_n, \\quad a_n, \\quad \\dots, \\quad a_n, \\quad a_n.\n\\end{align*}\n$$\n\nПоскольку для любых $k$ и $l$ выполнено неравенство $2a_l \\ge 2 > a_k$, каждая из последовательностей удовлетворяет (1). Кроме того, каждая из последовательностей, очевидно, удовлетворяет (3).\nОсталось показать, что хотя бы одна из них удовлетворяет (2).\nДля этого заметим, что произведение всех $n^2$ чисел во всех $n$ последовательностях равно\n$$\n2^{(n-1)+(n-2)+\\dots+0} \\cdot a_1^n a_2^n \\dots a_n^n = (2^{(n-1)/2} a_1 a_2 \\dots a_n)^n .\n$$\nСледовательно, произведение чисел хотя бы в одной из последовательностей не превосходит $2^{(n-1)/2} a_1 a_2 \\dots a_n$, что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39817,
    "subject": "Mathematics (Multi-modal)",
    "question": "The teacher wishes to compose a problem for his students of the following form. He tells them that he thinks of a monic polynomial of degree $2017$ with integer coefficients. Then he tells them $k$ integers $n_1, \\dots, n_k$, and finally he tells them the value of the product $P(n_1) \\cdot P(n_2) \\dots P(n_k)$. By all this data, the students need to find a polynomial satisfying the described conditions. Determine the least value of $k$ for which the teacher can compose a problem of this form so that any suitable polynomial found by the students will necessarily coincide with the polynomial he thinks of.\n\nУчитель хочет составить для своих учеников задачу следующего вида. Он сообщает, что задумал унитарный многочлен степени $2017$ с целыми коэффициентами. Затем он сообщает $k$ целых чисел $n_1, \\dots, n_k$, и, наконец, сообщает значение произведения $P(n_1) \\cdot P(n_2) \\dots P(n_k)$. По всем этим данным ученики должны найти многочлен, удовлетворяющий описанным условиям. Определите наименьшее значение $k$, при котором учитель может составить задачу такого вида так, что любой подходящий многочлен, найденный учениками, обязательно совпадёт с задуманным им многочленом.",
    "options": [],
    "answer": "2017",
    "solution": "Answer. $k = 2017$.\n\nIf $k < 2017$, and $P(x)$ is the teacher's polynomial, then the students may find a polynomial $Q(x) = P(x) + (x-n_1)(x-n_2) \\cdots (x-n_k)$ which also satisfies all the conditions. For $k = 2017$, the teacher can, e.g., tell the students that $P(4)P(8)P(12) \\cdots P(4k) = 1$ then the only polynomial $P(x) = 1 + \\prod_{i=1}^{2017} (x-4i)$ satisfies all the conditions.\n\nLemma. Let $P(x)$ be a polynomial with integer coefficients, and let $a$ and $b$ be distinct integers. Then $P(a) - P(b)$ is divisible by $a-b$.\n\nProof. In the difference $P(a) - P(b)$, group by degrees: if $P(x) = p_n x^n + p_{n-1} x^{n-1} + \\dots + p_1 x + p_0$, then $P(a) - P(b) = p_n(a^n - b^n) + p_{n-1}(a^{n-1} - b^{n-1}) + \\dots + p_1(a-b)$, where each term is divisible by $a-b$. $\\square$\n\nSuppose $k = 2017$. Set $n_i = 4i$ for $i = 1, 2, \\dots, k$; let the teacher tell the students that $P(n_1)P(n_2) \\dots P(n_k) = 1$. Then the polynomial $P(x) = (x-n_1)(x-n_2) \\dots (x-n_k)+1$ fits the condition. Suppose there is another polynomial $Q(x)$ that also fits the condition. Then, since $Q(n_1)Q(n_2) \\dots Q(n_k) = 1$ and the coefficients of $Q(x)$ are integers, $Q(n_i) = \\pm 1$ for any $i = 1, 2, \\dots, k$.\n\nIf there exist $i$ and $j$ such that $Q(n_i) = 1$ and $Q(n_j) = -1$, then the difference $Q(n_i) - Q(n_j) = 2$ is not divisible by $n_i - n_j$, which contradicts the lemma. Therefore, all values $Q(n_i)$ are equal and all are either $1$ or $-1$. However, all values cannot be $-1$, since the product $Q(n_1)Q(n_2) \\dots Q(n_k)$ would be $-1$ (since $k$ is odd), not $1$. Thus, $Q(n_i) = 1$ for all $i = 1, 2, \\dots, k$. Then the difference $P(x) - Q(x)$ is a polynomial of degree less than $k$ with at least $k$ roots, so this polynomial is identically zero, and $P(x) = Q(x)$. Contradiction.\n\nRemark. Using the lemma, one can show that the polynomial $P(x) = (x - n_1)(x - n_2) \\dots (x - n_{2017}) \\pm 1$ fits for any distinct integers $n_1, n_2, \\dots, n_{2017}$.\nОтвет. При $k = 2017$.\n\nСначала доказам, что $k > 2016$. Пусть учитель использовал некоторое $k \\le 2016$, задумав многочлен $P(x)$. Рассмотрим многочлен $Q(x) = P(x) + (x - n_1)(x - n_2) \\dots (x - n_k)$. Заметим, что степень многочлена $Q(x)$ также равна $2017$, а его старший коэффициент также равен $1$. При этом $P(n_1)P(n_2) \\dots P(n_k) = Q(n_1)Q(n_2) \\dots Q(n_k)$, но $P(x) \\ne Q(x)$. Значит, дети могли бы найти многочлен $Q(x)$ вместо $P(x)$, то есть учитель не добился требуемого.\n\nЛемма. Пусть $P(x)$ — многочлен с целыми коэффициентами, и пусть $a$ и $b$ — различные целые числа. Тогда $P(a) - P(b)$ делится на $a-b$.\n\nПоказательство. В разности $P(a) - P(b)$ сгрупируем следующие по степеням: если $P(x) = p_n x^n + p_{n-1} x^{n-1} + \\dots + p_1 x + p_0$, то $P(a) - P(b) = p_n(a^n - b^n) + p_{n-1}(a^{n-1} - b^{n-1}) + \\dots + p_1(a-b)$, где каждое слагаемое делится на $a-b$. $\\square$\n\nПусть $k = 2017$. Положим $n_i = 4i$ при $i = 1, 2, \\dots, k$; пусть учитель сообщит детям, что $P(n_1)P(n_2) \\dots P(n_k) = 1$. Тогда многочлен $P(x) = (x-n_1)(x-n_2) \\dots (x-n_k)+1$ под условие подходит. Предположим, что ещё какой-то многочлен $Q(x)$ также подходит под условие. Тогда, так как $Q(n_1)Q(n_2) \\dots Q(n_k) = 1$ и коэффициенты многочлена $Q(x)$ — целые числа, то $Q(n_i) = \\pm 1$ при любом $i = 1, 2, \\dots, k$.\n\nЕсли найдутся $i$ и $j$ такие, что $Q(n_i) = 1$, а $Q(n_j) = -1$, то разность $Q(n_i) - Q(n_j) = 2$ не делится на $n_i - n_j$, что противоречит лемме. Поэтому все значения $Q(n_i)$ равны между собой и все равны либо $1$, либо $-1$. Однако все значения не могут быть равны $-1$, так как в произведении $Q(n_1)Q(n_2) \\dots Q(n_k)$ множителей нечетное количество и произведение было бы равно $-1$. Значит, $Q(n_i) = 1$ при любом $i = 1, 2, \\dots, k$. Тогда разность $P(x) - Q(x)$ — многочлен степени менее $k$, имеющий хотя бы $k$ корней, то есть этот многочлен тождественно равен $0$, и $P(x) = Q(x)$. Противоречие.\n\nЗамечание. С использованием леммы можно показать, что многочлен $P(x) = (x - n_1)(x - n_2) \\dots (x - n_{2017}) \\pm 1$ подходит при любых различных целых числах $n_1, n_2, \\dots, n_{2017}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $n$ is a composite positive integer. For each its proper divisor $d$, we write down the number $d+1$ on the board. Find all values of $n$ for which the written numbers appear to be exactly all the proper divisors of some positive integer $m$. (A proper divisor of a positive integer $a > 1$ is any its positive divisor distinct from 1 and $a$.)\n\nПусть $n$ — составное положительное число. Для каждого его собственного делителя $d$ на доску выписывают число $d+1$. Найдите все значения $n$, при которых выписанные числа оказываются ровно всеми собственными делителями некоторого положительного числа $m$. (Собственный делитель положительного числа $a > 1$ — это любой его положительный делитель, отличный от 1 и $a$.)",
    "options": [],
    "answer": "n = 4 or n = 8",
    "solution": "$n = 4$ and $n = 8$.\n\nThe number $2$ does not appear on the board, so $m$ is odd. Therefore, all numbers on the board are odd, so $n$ is a power of $2$. Assume that $a \\ge b \\ge c$. Set $G(x) = P(x)/x^n$ and notice that $G(a) \\le G(b) \\le G(c)$.\n$n = 4$ или $n = 8$.\n\nПервое решение. Заметим, что число $2$ на доску не выписано, ибо $1$ — не собственный делитель $n$; стало быть, $m$ нечётно.\nЗначит, все выписанные делители $m$ нечётны, а потому все делители $n$ чётны. Итак, $n$ не делится на нечётные простые числа, то есть $n$ — степень двойки (и все его делители — тоже).\nЕсли $n$ делится на $16$, то $4$ и $8$ — его собственные делители, поэтому на доску выписаны $5$ и $9$. Стало быть, $m$ делится на $45$ и, в частности, $15$ является его собственным делителем. Но число $15$ выписано быть не могло, поскольку $14$ не является степенью двойки. Следовательно, $n$ не может делиться на $16$.\nОставшиеся (составные) степени двойки $n=4$ и $n=8$ подходят: для них можно соответственно положить $m=9$ и $m=15$.\n\n**Второе решение.** Пусть $a_1 < a_2 < \\dots < a_k$ — все собственные делители $n$. Заметим, что все числа $n/a_1 > n/a_2 > \\dots > n/a_k$ — также собственные делители числа $n$. Значит, они соответственно совпадают с $a_k$, $a_{k-1}, \\dots, a_1$, то есть $a_i a_{k+1-i} = n$ при всех $i=1, 2, \\dots, k$. Аналогичное рассуждение можно провести для делителей числа $m$.\nПусть $k \\ge 3$. Тогда $a_1 a_k = a_2 a_{k-1} = n$ и $(a_1 + 1)(a_k + 1) = (a_2 + 1)(a_{k-1} + 1) = m$, откуда $a_1 + a_k = a_2 + a_{k-1} = m - n - 1$. Видим, что у пар чисел $(a_1, a_k)$ и $(a_2, a_{k-1})$ совпадают и сумма, и произведение; по теореме Виета, они являются парами корней одного и того же квадратного уравнения, то есть эти пары должны совпадать — противоречие.\nИтак, $k \\le 2$; это возможно, если $n = p^2$, $n = p^3$ или $n = pq$, где $p$ и $q$ — простые числа (в последнем случае будем считать, что $p < q$). Заметим, что $p$ — наименьший собственный делитель $n$, а из условия $p+1$ — наименьший собственный делитель $m$, то есть $p+1$ — простое число. Значит, $p=2$. Случай $n = 2^2$ и $n = 2^3$ подходят, как отмечено в первом решении. Случай же $n = 2q$ невозможен. Действительно, в этом случае у $m$ есть лишь два собственных делителя $3$ и $q+1$, то есть $m=3(q+1)$, причём $q+1$ — простое число, отличное от $3$, или девятка; оба случая невозможны при простом $q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial of degree $n \\ge 2$ with nonnegative coefficients. Let $a$, $b$, and $c$ be the side lengths of some acute-angled triangle. Prove that the numbers $\\sqrt[n]{P(a)}$, $\\sqrt[n]{P(b)}$, and $\\sqrt[n]{P(c)}$ are also the side lengths of some acute-angled triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, let $a \\ge b \\ge c$; these three positive numbers are the side lengths of an acute-angled triangle if and only if $a^2 < b^2 + c^2$. Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \\ge P(b) \\ge P(c) > 0$; thus, we need to check that $\\sqrt[n]{P(a)}^2 < \\sqrt[n]{P(b)}^2 + \\sqrt[n]{P(c)}^2$.\n\nLet $P(x) = p_n x^n + p_{n-1} x^{n-1} + \\dots + p_0$. Denote $G(x) = P(x)/x^n$. Note that\n\n$$\nG(a) = p_n + \\frac{p_{n-1}}{a} + \\dots + \\frac{p_0}{a^n} \\le p_n + \\frac{p_{n-1}}{b} + \\dots + \\frac{p_0}{b^n} = G(b)\n$$\nand similarly, $G(a) \\le G(c)$. Therefore,\n$$\n\\sqrt[n]{P(a)}^2 = a^2 \\sqrt[n]{G(a)}^2 < (b^2 + c^2) \\sqrt[n]{G(a)}^2 \\le b^2 \\sqrt[n]{G(b)}^2 + c^2 \\sqrt[n]{G(c)}^2 = \\sqrt[n]{P(b)}^2 + \\sqrt[n]{P(c)}^2,\n$$\nwhich is what was required to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially $n > 1$ positive integers are written on the board. On each minute, a new number that is the sum of squares of all already written numbers appears on the board. (For example, if initial numbers were $1$, $2$, $2$, then on the first minute the number $1^2 + 2^2 + 2^2$ appears.) Prove that the $100$th new number has at least $100$ different prime divisors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_i$ be the number appearing on the board on the $i$th minute. Then $S_{i+1} = S_i(S_i + 1)$, so $S_{i+1}$ contains all prime divisors of $S_i$ plus at least one more.\n\nLet $S_1, \\dots, S_{100}$ be the numbers that were written on the board in the first $100$ minutes. Suppose that before writing the number $S_i$ on the board, the numbers $a_1, \\dots, a_k$ were present. Then $S_i = a_1^2 + a_2^2 + \\dots + a_k^2$, and the next number written is $S_{i+1} = a_1^2 + a_2^2 + \\dots + a_k^2 + S_i^2 = S_i^2$.\n\nThus, $S_{i+1} = S_i(S_i + 1)$. Therefore, $S_{i+1}$ contains in its prime factorization all the prime numbers that divide $S_i$, plus at least one new prime divisor (a divisor of $1 + S_i$). Since $S_1 > 1$, $S_1$ contains at least one prime divisor in its factorization. Hence, by induction, for $i = 1, 2, \\dots, 100$, the number $S_i$ contains at least $i$ distinct prime divisors in its factorization.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39821,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex polygon is dissected into isosceles triangles by several non-intersecting diagonals. Prove that this polygon has two sides of equal lengths.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "10.7. See problem 9.7.\n10.7. См. задачу 9.7.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39822,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circle $\\omega$ is circumscribed about an acute-angled triangle $ABC$. Points $D$ and $E$ are chosen on the sides $AB$ and $BC$, respectively, so that $AC \\parallel DE$. Points $P$ and $Q$ are chosen on the smaller arc $AC$ of $\\omega$ so that $DP \\parallel EQ$. Rays $QA$ and $PC$ meet $DE$ at $X$ and $Y$, respectively. Prove that $\\angle XBY + \\angle PBQ = 180^\\circ$. (A. Kuznetsov)\n\nОкружность $\\omega$ описана около остроугольного треугольника $ABC$. Точки $D$ и $E$ выбраны на сторонах $AB$ и $BC$ соответственно так, что $AC \\parallel DE$. Точки $P$ и $Q$ выбраны на меньшей дуге $AC$ окружности $\\omega$ так, что $DP \\parallel EQ$. Лучи $QA$ и $PC$ пересекают $DE$ в точках $X$ и $Y$ соответственно. Докажите, что $\\angle XBY + \\angle PBQ = 180^\\circ$. (А. Кузнецов)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCQ$ is cyclic and $AC \\parallel DE$, we have $\\angle BEX = \\angle BCA = \\angle BQA = \\angle BQX$. Therefore, $XBEQ$ is cyclic; similarly, $YBDP$ is also cyclic. Thus $\\angle XBEQ = \\angle XEQ = \\angle DEQ$ and $\\angle PBY = \\angle PDE$.\n\nBy the condition, $DP \\parallel EQ$. Therefore, $180^\\circ = \\angle PDE + \\angle DEQ = \\angle XBQ + \\angle PBY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ and $I$ be respectively the circumcenter and the incenter of a scalene triangle $ABC$. Let $B'$ be the point symmetric to $B$ with respect to the line $OI$; we assume that $B'$ lies inside the angle $ABI$. Prove that the tangents to the circumcircle of the triangle $BIB'$ at points $B'$ and $I$ meet on the line $AC$.\n(A. Kuznetsov)\n\nПусть $O$ и $I$ — соответственно, описанный и вписанный центры неравнобедренного треугольника $ABC$. Пусть $B'$ — точка, симметричная $B$ относительно прямой $OI$; будем считать, что $B'$ лежит внутри угла $ABI$. Докажите, что касательные к описанной окружности треугольника $BIB'$ в точках $B'$ и $I$ пересекаются на прямой $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BI \\cap (ABC) = \\{B, S\\}$, and $SB' \\cap CA = P$ (see Fig. 14).\n\nThen $\\angle ATS = \\angle IBB' = \\angle IB'B$. Thus $SB' \\cdot ST = SA^2 = SI^2$, whence the circle $\\gamma = (TIB')$ is tangent to $SI$. This implies $\\angle ITB' = 2\\phi$, so $TA$ is the angle bisector of $\\angle B'TI$. Hence, denoting $\\gamma \\cap AC = \\{T, K\\}$, we get that $K$ is the required meeting point of the two tangents to $(BIB')$.\n\n![](attached_image_1.png)\n\n\nПервое решение.\nПусть прямая $BI$ вторично пересекает окружность, описанную около треугольника $ABC$, в точке $S$. Пусть лучи $SB'$ и $CA$ пересекаются в точке $T$ (см. рис. 14). По лемме о трезубце имеем $SA = SC = SI$. Из равенства $IB = IB'$ получаем $\\angle IB'B = \\angle IB'B' = \\phi$. Так как $OB = OB'$, четырёхугольник $AB'SB$ вписан, откуда $\\angle SAB' = \\angle SBB' = \\phi$. Угол $SAC$ — внешний для треугольника $SAT$, откуда $\\angle ATS = \\angle SAC - \\angle ASB' = \\angle SBC - \\angle ABB' = \\angle SBA - \\angle ABB' = \\angle SBB' = \\phi$.\n\nТаким образом, $\\angle B'AS = \\phi = \\angle ATS$; значит, треугольники $SAB'$ и $STA$ подобны по двум углам, откуда $SB' \\cdot ST = SA^2 = SI^2$. Следовательно, прямая $SI$ касается окружности $\\gamma$, описанной около треугольника $TIB'$. Тогда $\\angle ITB' = \\angle B'IS$. Но угол $B'IS$ — внешний для треугольника $IBB'$, поэтому он равен $2\\phi$. Значит, $\\angle ITA = \\angle ITB' - \\phi = 2\\phi - \\phi = \\phi$.\n\nОбозначим вторую точку пересечения окружности $\\gamma$ с прямой $AC$ через $K$. Имеем $\\angle KB'I = \\angle KTI = \\phi = \\angle IB'B$. Так же $\\angle KIB' = \\angle KTB' = \\phi = \\angle IB'B'$. Таким образом, прямые $KI$ и $KB'$ касаются окружности, описанной около треугольника $BB'I$, а точка $K$ лежит на прямой $AC$ по построению.\nВторое решение.\nПоскольку $OB = OB'$ из симметрии, точка $B'$ лежит на окружности $\\Omega$, описанной около треугольника $ABC$. Пусть лучи $AI$ и $CI$ вторично пересекают окружность $\\Omega$ в точках $A_1$ и $C_1$ соответственно. Пусть $K$ — точка пересечения касательных к окружности, описанной около треугольника $BB'I$, проведённых в точках $B'$ и $I$; тогда $KI = KB'$. Поскольку $IB' = IB$, имеем $\\angle IB'B = \\angle IB'B' = \\angle KIB'$, откуда $KI \\parallel BB'$ и $KI \\perp OI$.\n\nПусть $K'$ — точка, симметричная точке $K$ относительно прямой $OI$, а $K_1$ — точка пересечения прямых $KI$ и $AC$ (см. рис. 15). Поскольку $KB' = KI$, из симметрии получаем $K'B = K'I$; кроме того, $C_1B = C_1I$ и $A_1B = A_1I$ по лемме о трезубце. Таким образом, точки $K', A_1$ и $C_1$ лежат на одной прямой.\n\nПоскольку $OI \\perp K'K_1$, по лемме о бабочке для четырёхугольника $AC_1A_1C$ получаем $K'I = IK_1$. Но из симметрии $K'I = IK$. Следовательно, точки $K_1$ и $K$ совпадают.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39824,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a product of 7 positive integers each multiple was decreased by 3. Could it happen that the product becomes exactly 13 times larger than the initial one?\n\nВ произведении 7 положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли случиться так, что произведение стало ровно в 13 раз больше исходного?",
    "options": [],
    "answer": "Yes; for example, 1, 1, 1, 1, 1, 2, 16.",
    "solution": "Yes.\nAn example is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 2 \\cdot 16 = 32$ (there are other examples).\n\nAfter decreasing each factor by $3$, the numbers become $-2, -2, -2, -2, -1, 13$. Their product is $(-2)^4 \\cdot (-1) \\cdot 13 = 16 \\cdot (-1) \\cdot 13 = -208$.\n\nBut the Russian solution gives a more precise calculation:\n\nLet five of the factors be $1$, the sixth be $2$, and the seventh be $a$. The product is $1^5 \\cdot 2 \\cdot a = 2a$.\nAfter decreasing each by $3$, the numbers are $(-2)^5 \\cdot (-1) \\cdot (a-3) = (-32) \\cdot (-1) \\cdot (a-3) = 32(a-3)$.\nWe want $32(a-3) = 13 \\cdot 2a$, so $32a - 96 = 26a$, so $6a = 96$, $a = 16$.\nSo the numbers $1, 1, 1, 1, 1, 2, 16$ work.\n\nAnother example: $1^4 \\cdot 29 \\cdot 61 \\cdot 64$, after decreasing by $3$ becomes $(-2)^4 \\cdot 26 \\cdot 58 \\cdot 61$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39825,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number $x$ is chosen so that each of the sums $S = \\sin 64x + \\sin 65x$ and $C = \\cos 64x + \\cos 65x$ is a rational number. Prove that in one of these sums, both summands are rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $S^2 + C^2 = 2 + 2 \\cos x \\in \\mathbb{Q}$, we have $\\cos x \\in \\mathbb{Q}$ and hence $\\cos 64x \\in \\mathbb{Q}$.\nЗаметим, что число\n$$\n\\begin{aligned}\nS^2 + C^2 &= (\\sin^2 64x + \\cos^2 64x) + (\\sin^2 65x + \\cos^2 65x) + \\\\\n&\\quad + 2(\\sin 64x \\sin 65x + \\cos 64x \\cos 65x) = \\\\\n&= 2 + 2 \\cos(65x - 64x) = 2 + 2 \\cos x\n\\end{aligned}\n$$\nрационально, сткуда $\\cos x$ — также рациональное число. Ввиду формулы $\\cos 2\\alpha = 2 \\cos^2 \\alpha - 1$, все числа вида $\\cos 2^k x$ также рациональны — в частности, $\\cos 64x$. Поскольку $C$ рационально, то и второе слагаемое в этой сумме также рационально.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of an acute-angled isosceles triangle $ABC$ with $AB = AC$. The rays $BO$ and $CO$ meet the sides $AC$ and $AB$ at $B'$ and $C'$, respectively. Let $l$ be the line through $C'$ parallel to $AC$. Prove that $l$ is tangent to the circumcircle of the triangle $B'OC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "11.2. See problem 10.2.\n11.2. См. задачу 10.2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, Bazil thinks of eight cells on a chessboard, no two of which are in the same row or in the same column. Then Pete makes a series of guesses. By a guess, he places onto the chessboard 8 rooks of Pete's rooks which stand on positions he thinks of. If Bazil indicates an even number of rooks, Pete wins. Otherwise, the rooks are removed from the board, and then Pete makes the next guess.\n\nFind the least number of guesses at which Pete can win for sure.",
    "options": [],
    "answer": "2",
    "solution": "2 guesses.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ and $\\Omega$ be the incircle and the circumcircle of an equilateral triangle $ABC$, respectively. Points $P$ and $Q$ are chosen on the sides $AC$ and $AB$, respectively, so that the segment $PQ$ contains the center of the triangle $ABC$. Circles $\\Omega_b$ and $\\Omega_c$ have diameters $BP$ and $CQ$, respectively. Prove that the circles $\\Omega_b$ and $\\Omega_c$ meet at two points one of which lies on $\\omega$ and the other one lies on $\\Omega$.\n\nПусть $\\omega$ и $\\Omega$ — вписанная и описанная окружности правильного треугольника $ABC$ соответственно. Точки $P$ и $Q$ выбраны на сторонах $AC$ и $AB$ так, что отрезок $PQ$ содержит центр треугольника $ABC$. Окружности $\\Omega_b$ и $\\Omega_c$ построены на диаметрах $BP$ и $CQ$ соответственно. Докажите, что окружности $\\Omega_b$ и $\\Omega_c$ пересекаются в двух точках, одна из которых лежит на $\\omega$, а другая — на $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of $ABC$, and $B_1$ and $C_1$ be the midpoints of the lesser arcs $AC$ and $AB$ of $\\Omega$, respectively. Then the triangles $OPB_1$ and $OQC_1$ are isosceles (see Fig. 8). Next, the rays $B_1P$ and $C_1Q$ meet at a point $B' = C'$ on $\\Omega$ which is one of the common points of $\\Omega_b$ and $\\Omega_c$. Let the extension of $B'O$ beyond $O$ meet $\\omega$ at a point $X$; then $B'O \\cdot XO = BO \\cdot B_2O$, where $B_2$ is the tangency point of $\\omega$ with $AC$ (lying on $\\Omega_b$). Thus $X$ lies on $\\Omega_b$ and, similarly, on $\\Omega_c$.\nПусть $O$ — центр треугольника $ABC$. Обозначим через $B_2$ и $C_2$ точки касания $\\omega$ с $AC$ и $AB$ соответственно, а через $B_1$ и $C_1$ — середины меньших дуг $AC$ и $AB$ описанной окружности $\\Omega$ (см. рис. 8). Тогда точки $B_1$ и $C_1$ симметричны: $\\angle OB_1P = \\angle B_1OP$ и $\\angle OC_1Q = \\angle C_1OQ$.\n\nПусть лучи $B_1P$ и $C_1Q$ пересекают $\\Omega$ в точках $B'$ и $C'$. $B'$ лежит на $\\Omega_b$, аналогично, $C'$ лежит на $\\Omega_c$. С другой стороны, $BB' + CC' = 2(\\angle BB_1B' + \\angle CC_1C') = 2(\\angle B_1OP + \\angle C_1OQ) = 2(180^\\circ - \\angle B_1OC_1) = 120^\\circ = BC$; это означает, что точки $B'$ и $C'$ совпадают. Итак, $\\Omega_b$ и $\\Omega_c$ пересекаются в точке $B'$, лежащей на $\\Omega$.\n\nПоскольку $\\angle BB_2P = \\angle CC_2Q = 90^\\circ$, точки $B_2$ и $C_2$ лежат на $\\Omega_b$ и $\\Omega_c$ соответственно. Пусть продолжение отрезка $B'O$ за $O$ пересекает $\\omega$ в точке $X$. Тогда $OB \\cdot OB_2 = OB' \\cdot OX = OC \\cdot OC_2$. Первое из этих равенств означает, что точки $B, B_2, B'$ и $X$ лежат на одной окружности, то есть $X$ лежит на окружности $\\Omega_b$.\n\nАналогично, из второго равенства следует, что $X$ лежит на $\\Omega_c$. Значит, $X$ и является второй точкой пересечения $\\Omega_b$ и $\\Omega_c$, лежащей на $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39829,
    "subject": "Mathematics (Multi-modal)",
    "question": "A prestidigitator and his assistant have a deck of cards; back sides of all the cards look identical, the front side of each card is painted in one of $2017$ colors (there are $1000000$ cards of each color in the deck). They want to perform the following trick. The prestidigitator goes out of the room. The spectators put $n$ cards facing front in a row onto the table. The assistant looks at them and turns all the cards except one facing back side (he does not change the order of the cards). Finally, the prestidigitator comes in, stares at the table, and guesses the color of one of the cards facing back. Find the least value of $n$ for which the prestidigitator and the assistant can agree on their actions in advance so that they will do the trick for sure.\n(I. Bogdanov, K. Knop)\n\nУ фокусника и его помощника есть колода карт; рубашки всех карт выглядят одинаково, а лицевая сторона каждой карты окрашена в один из $2017$ цветов (в колоде по $1000000$ карт каждого цвета). Они хотят проделать следующий фокус. Фокусник выходит из комнаты. Зрители выкладывают на стол в ряд $n$ карт лицом вверх. Помощник смотрит на них и переворачивает все карты, кроме одной, рубашкой вверх (он не меняет порядок карт). Наконец, фокусник возвращается, смотрит на стол и угадывает цвет одной из карт, лежащих рубашкой вверх. Найдите наименьшее значение $n$, при котором фокусник и помощник могут заранее договориться о своих действиях так, чтобы фокус всегда удавался.",
    "options": [],
    "answer": "2018",
    "solution": "Answer: $n = 2018$.\n\nSet $k = 2017$. If $n = k + 1$, the assistant merely encodes the color of the last card by choosing which of the others remains open.\n\nAssume that $k \\ge n$. A strategy of the prestidigitator and the assistant may be expressed as a set of instructions of the form $(a, i) \\to (b, j)$ meaning that a prestidigitator seeing color $a$ at position $i$ claims color $b$ at position $j$. One may assume that all pairs $(a, i)$ are distinct. A counting argument shows that $k = n$, there are $n^2$ instructions, and each two instructions cover disjoint sets of arrangements.\n\nConsider a graph on positions as vertices, an edge connects two positions participating in an instruction. Then each two edges have a common vertex, so all edges have a common vertex (say, $1$). Each position $i > 1$, put into correspondence all colors $a$ such that there exists an instruction of the form $(c, i) \\to (a, 1)$. Notice that different positions correspond to different colors; so some position corresponds to one color $a$ only. But then there exists an instruction $(a, 1) \\to (c, i)$ for some $c$, which cannot exist simultaneously with $(c, i) \\to (a, 1)$.\nОтвет: $n = 2018$.\n\nПоложим $k = 2017$.\nПри $n = k+1$ фокус устроить легко. Фокусник и помощник нумеруют цвета числами от $1$ до $k$. Помощник, видя цвет последней, $(k+1)$-й карты (пусть его номер равен $a$), оставляет открытой $a$-ю карту. Фокусник, увидев, какая по номеру карта открыта, восстанавливает цвет последней карты.\n\nОсталось показать, что при $n \\le k$ фокус не удастся. Предположим противное и рассмотрим возможные действия фокусника. Пусть, видя на $i$-м месте карту цвета $a$, он объявляет, что на $j$-м месте карта цвета $b$ (тогда $i \\ne j$); будем называть это инструкцией $(a, i) \\to (b, j)$. Можно считать, что для каждой пары $(a, i)$ существует только одна инструкция вида $(a, i) \\to (b, j)$ (и фокусник при возможности всегда применяет её — поскольку никакой информации о том, какую из таких инструкций применять, у него нет). Тогда инструкций не больше, чем $kn$.\n\nБудем говорить, что исходная раскладка карт удовлетворяет инструкции $(a, i) \\to (b, j)$, если в ней на $i$-м и $j$-м местах лежат карты цветов $a$ и $b$ соответственно. Тогда каждой инструкции удовлетворяет ровно $k^{n-2}$ раскладок. С другой стороны, если фокус гарантированно удается, то каждая возможная раскладка удовлетворяет хотя бы одной инструкции — той, которую применяют помощник с фокусником. Значит, общее число раскладок не может превосходить $kn \\cdot k^{n-2}$, то есть $k^n \\le k^{n-1}$, откуда $k \\le n$. Значит, $k = n$, и неравенство выше обращается в равенство. Это значит, что каждая раскладка удовлетворяет ровно одной инструкции, и с каждой пары $(a, i)$ начинается ровно одна инструкция.\n\nРассмотрим произвольную инструкцию $(a, i) \\to (b, j)$; тогда и инструкция вида $(b, j) \\to (c, k)$. Поскольку не существует раскладки, удовлетворяющей обеим инструкциям, должны выполняться условия $i = k$ и $a \\ne c$.\n\nС другой стороны, для любых двух инструкций $(a, i) \\to (b, j)$ и $(c, k) \\to (d, l)$ среди номеров $i, j, k, l$ должны быть совпадающие — иначе опять же существует раскладка, удовлетворяющая обеим инструкциям. Рассмотрим граф с вершинами $1, 2, \\dots, k$, в котором $i$ и $j$ соединены ребром $[i, j]$, если существует инструкция вида $(a, i) \\to (b, j)$ (по показанному выше, существует также и инструкция вида $(b, j) \\to (a', i)$). Тогда любые два ребра в этом графе имеют общую вершину, и из каждой вершины выходит хотя бы одно ребро. Пусть для определённости $[1, 2]$ — ребро этого графа. Из вершины $3$ выходит ребро, имеющее общую вершину с первым — пусть для определённости это $[1, 3]$. Тогда любое ребро из вершины $k > 3$ обязано иметь вид $[1, k]$, чтобы иметь общие вершины с каждым из ребер $[1, 2]$ и $[1, 3]$. Наконец, любое ребро вообще должно иметь общую вершину с каждым из ребер $[1, 2]$, $[1, 3]$ и $[1, 4]$, то есть должно содержать вершину $1$. Итак, в каждой инструкции один из номеров мест равен $1$.\n\nНаконец, сопоставим каждому месту $i > 1$ все такие цвета $a$, что существует инструкция вида $(c, i) \\to (a, 1)$. Из сказанного выше следует, что разным местам не может быть сопоставлен один и тот же цвет. Поскольку таких мест $k-1$, а цветов $k < 2(k-1)$, какому-то месту $i$ сопоставлен только один цвет $a$, то есть имеются все $k$ инструкций вида $(c, i) \\to (a, 1)$ при всевозможных $a$. Однако существует также инструкция вида $(a, 1) \\to (c, i)$ для некоторого $c$. Но она не может существовать вместе с инструкцией $(c, i) \\to (a, 1)$; противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial of degree $n \\ge 2$ with nonnegative coefficients. Let $a$, $b$, and $c$ be the side lengths of some triangle. Prove that the numbers $\\sqrt[n]{P(a)}$, $\\sqrt[n]{P(b)}$, and $\\sqrt[n]{P(c)}$ are also the side lengths of some triangle.\n\nПусть $P(x)$ — многочлен степени $n \\ge 2$ с неотрицательными коэффициентами. Пусть $a$, $b$, $c$ — длины сторон некоторого треугольника. Докажите, что числа $\\sqrt[n]{P(a)}$, $\\sqrt[n]{P(b)}$, $\\sqrt[n]{P(c)}$ также являются длинами сторон некоторого треугольника.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $a \\ge b \\ge c$. Set $G(x) = P(x)/x^n$ and notice that $G(a) \\le G(b) \\le G(c)$.\n\nLet, without loss of generality, $a \\ge b \\ge c$; these three positive numbers are the side lengths of a triangle if and only if $a < b + c$. Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \\ge P(b) \\ge P(c) > 0$; thus, we need to check that $\\sqrt[n]{P(a)} < \\sqrt[n]{P(b)} + \\sqrt[n]{P(c)}$.\n\nLet $P(x) = p_n x^n + p_{n-1} x^{n-1} + \\dots + p_0$. Denote $G(x) = P(x)/x^n$. Note that\n$$G(a) = p_n + \\frac{p_{n-1}}{a} + \\dots + \\frac{p_0}{a^n} \\le p_n + \\frac{p_{n-1}}{b} + \\dots + \\frac{p_0}{b^n} = G(b)$$\nand similarly, $G(a) \\le G(c)$. Therefore,\n$$\\sqrt[n]{P(a)} = a \\sqrt[n]{G(a)} < (b + c) \\sqrt[n]{G(a)} \\le b \\sqrt[n]{G(b)} + c \\sqrt[n]{G(c)} = \\sqrt[n]{P(b)} + \\sqrt[n]{P(c)},$$\nwhich is what was required to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39831,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ is inscribed into a circle $\\Gamma$ centered at $O$. Its diagonals $AC$ and $BD$ are perpendicular to each other; let $P$ be their meeting point (the point $O$ lies inside the triangle $BPC$). A point $H$ is chosen on the segment $BO$ so that $\\angle BHP = 90^\\circ$. The circumcircle $\\omega$ of the triangle $PHD$ meets again the segment $PC$ at $Q$. Prove that $AP = CQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BT$ be a diameter of $\\Omega$; then the points $P$, $H$, $T$, and $D$ are concyclic (see Fig. 9). Thus $\\angle PQD = 90^\\circ$, and the perpendicular bisectors to $PQ$, $TD$, and $AC$ coincide.\n\nConsider the common perpendicular bisector $l$ to the segments $DT$ and $PQ$. It passes through $O$, and therefore is also the perpendicular bisector to $AC$. Thus, the segments $AP$ and $CQ$ are symmetric with respect to $l$ and hence are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some cells of a $200 \\times 200$ checkered square contain red or blue tokens — one per cell; the other cells are empty. We say that a token *sees* another token if they are situated either in one row or in one column. Assume that each red token sees exactly five blue tokens (and, perhaps, some red tokens), and each blue token sees exactly five red tokens (and, perhaps, some blue tokens). Determine the maximal possible total number of tokens on the board.\n\nВ некоторых клетках квадрата $200 \\times 200$ стоят красные или синие фишки — по одной в клетке; остальные клетки пусты. Будем говорить, что фишка *видит* другую фишку, если они расположены либо в одной строке, либо в одном столбце. Известно, что каждая красная фишка видит ровно пять синих фишек (и, возможно, какие-то красные), а каждая синяя фишка видит ровно пять красных фишек (и, возможно, какие-то синие). Каково наибольшее возможное общее число фишек на доске?",
    "options": [],
    "answer": "3800",
    "solution": "Answer. $3800$ tokens.\n\n*Example.* For each side of the square, choose a $5 \\times 190$ rectangle adjacent to this side at distance $5$ from the two neighboring sides. Put red tokens into two such rectangles of different orientations and blue tokens into the other two.\n\n*Bound.* Say that a line (i.e., row or column) is polychromatic if it contains tokens of both colors; each such line contains at most $1$ token. If there are at least $191$ polychromatic rows, then the total number of tokens is at most $191 \\cdot 10 + 9 \\cdot 200 < 3800$. If there are at most $190$ polychromatic rows and at most $190$ such columns, they contain at most $190 \\cdot 10 + 190 \\cdot 10 = 3800$ tokens, and each token belongs to one of them.\nОтвет. $3800$ фишек.\n\nПример, содержащий $3800$ фишек, получается, например, так. Выделим у квадрата $200 \\times 200$ «каёмку» ширины $5$. Эта каёмка состоит из четырёх угловых квадратов $5 \\times 5$ и четырёх прямоугольников $5 \\times 190$. Расставим фишки в эти четыре прямоугольника: в левый и в верхний — красные, а в правый и в нижний — синие. Нетрудно видеть, что этот пример удовлетворяет всем требованиям, и в нём по $1900$ красных и синих фишек.\n\nОсталось доказать, что фишек не может быть больше $3800$. Рассмотрим произвольную расстановку фишек, удовлетворяющую требованиям. Назовём ряд (строку или столбец) разноцветным, если в нём есть фишки обеих цветов.\n\nСделаем сразу два полезных замечания. Во-первых, каждая фишка видит какую-то фишку другого цвета, поэтому каждая фишка лежит хотя бы в одном разноцветном ряду. Кроме того, поскольку разноцветный ряд содержит красную фишку, в нём не может быть более пяти синих фишек (иначе красная все их увидит). Аналогично, в разноцветном ряду не более пяти красных фишек, то есть всего не более $10$ фишек.\n\nТеперь нетрудно получить требуемую оценку. Если есть $191$ разноцветная строка, то в них не более $191 \\cdot 10 = 1910$ фишек, а в оставшихся девяти строках не более $9 \\cdot 200 = 1800$ фишек, итого не больше $1910 + 1800 < 3800$ фишек. Аналогично разбирается случай, когда есть $191$ разноцветный столбец. Если же и тех и других не более, чем по $190$, то они содержат не более $190 \\cdot 10 + 190 \\cdot 10 = 3800$ фишек, причём все фишки содержатся в этих рядах. Оценка доказана.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39833,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane, several lines in a general position are drawn. These lines partition the plane into regions. Prove that one can put into each region a positive number so that the sums of numbers on both sides of each line will be equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим проведённые прямые $l_1, l_2, \\dots, l_n$, упорядочив их направления по часовой стрелке (см. рис. 10). Формально это означает следующее. Рассмотрим произвольную точку плёскости $O$. Проведем через неё прямые, параллельные нашим, заумеруем их по часовой стрелке, а потом присвоим нашим прямым те же номера, которые получили соответствующие им новые прямые.\n\n![](attached_image_1.png)\n\nСреди областей, на которые наши прямые разрезали плоскость, есть $2n$ бесконечных кусков; обозначим их по часовой стрелке $S_1, S_2, \\dots, S_{2n}$ так, что прямая $l_i$ разделяет куски $S_i$ и $S_{i+1}$, а также куски $S_{i+n}$ и $S_{i+n+1}$. (Здесь и далее мы считаем, что $S_{2n+1} = S_1$.)\n\nДля начала въс все области (конечные и бесконечные) поставим по $1$. Для каждой прямой $l_i$ обозначим через $\\Sigma_i$ возможность сумм чисел справа и слева от $l_i$ (мы считаем, что куски $S_i$ и $S_{i+n+1}$ лежат слева от $l_i$). Если $\\Sigma_i > 0$, то прибавим по $\\frac{1}{2} \\Sigma_i$ к числам, стоящим в $S_i$ и $S_{i+1+n}$. При этом все числа $\\Sigma_j$ при $j \\neq i$ не изменились, поскольку области $S_i$ и $S_{i+1+n}$ лежат по разные стороны относительно любой прямой, кроме $l_i$. Число же $\\Sigma_i$ стало равно нулю. Аналогично, если $\\Sigma_i < 0$, то прибавим по $\\frac{1}{2} |\\Sigma_i|$ к числам, стоящим в $S_{i+1}$ и $S_{i+n}$; опять же, $\\Sigma_i$ станет равна $0$, а остальные $\\Sigma_j$ не изменятся.\n\nТакими операциями мы последовательно сделаем каждое $\\Sigma_i$ равным нулю, не меняя остальных.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, a positive integer $N$ is written on the board. At each moment, Misha may choose a number $a > 1$ on the board, remove it, and write down all its positive divisors except $a$ instead. After some time it happened that the board contains $N^2$ numbers. Determine all values of $N$ for which this can happen. (M. Fadin, K. Kovalenko)\n\nИзначально на доске написано положительное целое число $N$. В каждый момент Миша может выбрать на доске число $a > 1$, стереть его и вместо него записать все его положительные делители, кроме $a$. Через некоторое время оказалось, что на доске $N^2$ чисел. Для каких $N$ это возможно? (М. Фадин, К. Коваленко)",
    "options": [],
    "answer": "1",
    "solution": "Answer: $N = 1$.\n\nRecall that\n$$\n\\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1.\n$$\nfor all $n > 1$. Using this inequality, show by induction on $N$ that the board may get at most $N^2$ numbers, with equality achieved only for $N = 1$.\n\nLemma. For any natural $n > 1$, the inequality\n$$\n\\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1\n$$\nholds.\n\nProof. Clearly, for $t > 1$,\n$$\n\\frac{1}{t^2} < \\frac{1}{t(t-1)} = \\frac{1}{t-1} - \\frac{1}{t}.\n$$\nSumming such inequalities for $t = 2, 3, \\dots, n$, we get\n$$\n\\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1 - \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n-1} - \\frac{1}{n} = 1 - \\frac{1}{n} < 1, \\quad \\square\n$$\nwhich is what was required.\n\nLet us prove by induction on $N$ that for any natural $N$, the board can contain at most $N^2$ numbers, and exactly $N^2$ numbers can appear only for $N = 1$. This will yield the answer to the problem.\n\nThe base case $N = 1$ is obvious. For the induction step, suppose $N > 1$. Let $1 = d_1 < d_2 < \\dots < d_k < d_{k+1} = N$ be all divisors of $N$. After the first replacement, the numbers $d_1, d_2, \\dots, d_k$ will be on the board. Mentally divide the board into $k$ parts so that the number $d_j$ is in the $j$-th part. At each subsequent replacement, write the successors of a number in the same part where the number itself was. By the induction hypothesis, in the $j$-th part of the board there can be at most $d_j^2$ numbers. Therefore, the total number of numbers on the board cannot exceed $d_1^2 + d_2^2 + \\dots + d_k^2$.\n\nNow note that the numbers $N/d_1 > N/d_2 > \\dots > N/d_k > N/d_{k+1}$ are also divisors of $N$. Therefore, they are equal to $d_{k+1}, d_k, \\dots, d_2, d_1$ respectively. Taking the lemma into account, we get that the total number of numbers on the board does not exceed:\n$$\n\\begin{aligned}\nd_1^2 + d_2^2 + \\dots + d_k^2 &= \\frac{N^2}{d_{k+1}^2} + \\frac{N^2}{d_k^2} + \\dots + \\frac{N^2}{d_2^2} = \\\\\n&= N^2 \\left( \\frac{1}{d_2^2} + \\frac{1}{d_3^2} + \\dots + \\frac{1}{d_{k+1}^2} \\right) \\le \\\\\n&\\le N^2 \\left( \\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{N^2} \\right) < N^2 \\cdot 1 = N^2,\n\\end{aligned}\n$$\nwith strict inequality for $N > 1$.\n\nTherefore, the only possible value is $N = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $I_A, I_B, I_C, \\text{ and } I_D$ be the incenters of the triangles $DAB$, $ABC$, $BCD$, and $CDA$, respectively. Given that $\\angle B I_A A + \\angle I_C I_A I_D = 180^\\circ$, prove that $\\angle B I_B A + \\angle I_C I_B I_D = 180^\\circ$.\n\nПусть $ABCD$ — выпуклый четырёхугольник. Пусть $I_A, I_B, I_C$ и $I_D$ — центры вписанных окружностей треугольников $DAB$, $ABC$, $BCD$ и $CDA$ соответственно. Известно, что $\\angle B I_A A + \\angle I_C I_A I_D = 180^\\circ$. Докажите, что $\\angle B I_B A + \\angle I_C I_B I_D = 180^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P \\in AD$ be the center of positive homothety mapping $\\omega_A$ to $\\omega_D$, and $R \\in BD$ be the center of negative homothety mapping $\\omega_A$ to $\\omega_C$. We consider the case shown in Fig. 16. The given equality $\\angle B I_A A + \\angle I_C I_A I_D = 180^\\circ$ is equivalent to $\\angle P I_A R = 90^\\circ + \\angle PDR/2$, which in turn is equivalent to the fact that $PR$ is tangent to $\\omega_A$ (and hence $PR$ is a common tangent to $\\omega_A, \\omega_C$, and $\\omega_D$). Similarly, the required equality is equivalent to the fact that there is a line tangent to $\\omega_A$, $\\omega_C$, and $\\omega_D$ (with $\\omega_A$ being on one side of this line, and two other circles on the other side).\n\nDenote by $A_{AD}$ the tangency point of $\\omega_A$ with $AD$; the other tangency points are denoted similarly (see Fig. 17). Notice that $B_{BC} C_{BC} = A_{AD} D_{AD}$ and $A_{BD} C_{BD} = B_{AC} D_{AC}$. Let now $PR$ touch $\\omega_A$, $\\omega_C$, and $\\omega_D$ at $L_A, L_C$ and $L_D$, respectively. Then $L_A L_D = A_{AD} D_{AD} = B_{BC} C_{BC}$ and $L_A L_C = A_{BD} C_{BD} = D_{AC} B_{AC}$.\n\nConsider the circles $\\omega'_B, \\omega'_C$, and $\\omega'_D$ centered at some points $I'_B, I'_C$ and $I'_D$, having the same radii as $\\omega_B, \\omega_C$, and $\\omega_D$, respectively, and tangent to $PR$ at $L_A, L_D$, and $L_C$, respectively. Then the common tangent lengths for these circles are the same as for $\\omega_B, \\omega_C$, and $\\omega_D$. Therefore, the triangles $I_B I_C I_D$ and $I'_B I'_C I'_D$ are congruent, so the configurations $(\\omega_B, \\omega_C, \\omega_D)$ and $(\\omega'_B, \\omega'_C, \\omega'_D)$ are also congruent, thus the claim.\n\n\nШаг 1. Обозначим через $P$ центр гомотетии с положительным коэффициентом, переводящей $\\omega_A$ в $\\omega_D$ (точка $P$ может оказаться бесконечно удалённой), а через $R$ — центр гомотетии с отрицательным коэффициентом, переводящей $\\omega_A$ в $\\omega_C$ (см. рис. 16). Поскольку $\\angle A\\overline{D}$ — общая внешняя касательная к $\\omega_A$ и $\\omega_D$, имеем $P \\in A\\overline{D}$; аналогично, $R \\in B\\overline{D}$.\n\nПокажем, что данное условие\n$$\n\\angle B I_A A = 180^{\\circ} - \\angle I_C I_A I_D\n$$\nравносильно тому, что прямая $l_A$, проходящая через $P$ и $I$, общая касательная к окружностям $\\omega_A, \\omega_C$ и $\\omega_D$. Пусть, для определённости, точка $P$ лежит на луче $DA$ (другие случаи логичны). Поскольку $\\omega_A$ вписана в треугольник $ABD$, имеем $\\angle A I_A B = 90^{\\circ} + \\angle ADB/2$; с другой стороны, $180^{\\circ} - \\angle I_C I_A I_D = \\angle P I_A R$. Значит, (*) равносильно равенству $\\angle P I_A R = 90^{\\circ} + \\angle PDR/2$.\n\nОбозначим через $J$ центр окружности, вписанной в четырёхугольник из точки $D$, а (*) равносильно тому, что $\\angle P J R = \\angle P I_A R$ — то есть совпадению точек $J$ и $I_A$. Это, в свою очередь, означает в точности, что прямые $PR$ и $RD$ симметричны относительно $I_A I_C$, а прямые $PD$ и $PR$ симметричны относительно $I_A I_D$, то есть что $PR$ касается всех трёх окружностей $\\omega_A, \\omega_C$ и $\\omega_D$ (лежит по другую сторону от $PR$, нежели остальные две).\n\nАналогично, требуемое условие $\\angle B I_B A + \\angle I_C I_B I_D = 180^{\\circ}$ равносильно тому, что у окружностей $\\omega_A, \\omega_C$ и $\\omega_D$ существует общая касательная $l_B$, относительно которой $\\omega_C$ и $\\omega_D$ лежат одну сторону, а $\\omega_A$ — по другую. Осталось сказать, что эти два факта равносильны.\n\nШаг 2. Обозначим точку касания окружности $\\omega_A$ с $AD$ через $A_{AD}$; аналогично обозначим другие точки касания (см. рис. 17). Заметим, что $A_{AD} D_{AD} = \\frac{|A A_{AD} - D D_{AD}|}{2} = \\frac{|AB + AD - BD - AC + AD - CD|}{2} = \\frac{|AB + CD - BD - AC|}{2}$;\nаналогично,\n$$\n\\begin{align*}\nB_{BC} C_{BC} &= \\frac{|AB + CD - BD - AC|}{2} &&= A_{AD} D_{AD}, \\\\\nA_{BD} C_{BD} &= \\frac{|AD + BC - AB - CD|}{2} &&= B_{AC} D_{AC}.\n\\end{align*}\n$$\n\nПредположим теперь, что $l'_A$ касается окружностей $\\omega_A, \\omega_C$ и $\\omega_D$ в точках $L_A, L_C$ и $L_D$ соответственно. Тогда $L_A L_D = A_{AD} D_{AD} = B_{BC} C_{BC}$ и $L_A L_C = A_{BD} C_{BD} = D_{AC} B_{AC}$.\n\nРассмотрим окружности $\\omega'_B, \\omega'_C$ и $\\omega'_D$ с центрами $I'_B, I'_C$ и $I'_D$, имеющие те же радиусы, что и $\\omega_B, \\omega_C$ и $\\omega_D$ соответственно, и касающиеся $l'_A$ в точках $L_A, L_D$ и $L_C$ соответственно (причём $\\omega'_B$ и $\\omega'_C$ лежат по одну сторону от $l'_A$, а $\\omega'_D$ — по другую). Тогда соответственные отрезки общих касательных к $\\omega'_B, \\omega'_C, \\omega'_D$ и $\\omega_B, \\omega_C, \\omega_D$ имеют одинаковые длины (для $\\omega_C$ и $\\omega_D$ это тривиально, для остальных пар следует из сказанного выше).\n\nОтсюда легко следует, что соответственные стороны треугольников $I_B I_C I_D$ и $I'_B I'_C I'_D$ равны (например, $I'_B I'_C = I_B I_C$ из равенства четырёхугольников $I'_B L_A L_D I'_C$ и $I_B B_{BC} C_{BC}$). Поэтому и конфигурации окружностей $(\\omega_B, \\omega_C, \\omega_D)$ и $(\\omega'_B, \\omega'_C, \\omega'_D)$ также равны. Поскольку окружности в одной тройке касаются одной прямой $l_A$, то же верно и для другой тройки. Это и оставалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, we put 100 cards on a table, each containing a positive integer. Exactly 28 of these cards contain odd numbers. Then, on each minute the following operation has been performed. We compute the product of numbers on every set of 12 cards on the table, add up all these products, write this number onto a new card and put this card onto the table. Is it possible to choose initial 100 numbers so that for every positive integer $d$ the table will eventually contain some card with a number divisible by $2^d$? (I. Bogdanov)\n\nИзначально на стол кладут 100 карточек, на каждой из которых написано положительное целое число. Ровно на 28 из этих карточек написаны нечётные числа. Затем каждую минуту выполняют следующую операцию: вычисляют произведение чисел на каждой из 12-карточных подмножеств карточек, складывают все эти произведения, записывают это число на новую карточку и кладут её на стол. Можно ли выбрать начальные 100 чисел так, чтобы для любого натурального $d$ на столе в какой-то момент появилась карточка с числом, делящимся на $2^d$?",
    "options": [],
    "answer": "No",
    "solution": "Answer. No.\nIf the table contains $k$ odd numbers, then the parity of a new one coincides with the parity of $\\binom{k}{12}$; so, on the first four minutes $k$ increases by 1, and then it is always equal to 32. Consider any $n \\ge 4$. Let $E_n$ ($T_n$) be the sum of products of all 11-tuples (12-tuples) of the numbers. $E_n \\equiv \\binom{32}{11} \\equiv 0 \\pmod{2}$ for $n \\ge 4$. The new number will be $T_n E_n$, so $T_{n+1} = T_n(1+E_n)$, and the power of 2 in its expansion is the same as that of $T_n$. Choosing $d$ so that $2^d$ is greater than $T_4$, we obtain that no $T_n$ will be divisible by $2^d$.\n\nОтвет. Нет, нельзя.\nЕсли в некоторый момент среди чисел на карточках есть ровно $k$ нечётных, то среди произведений чисел по 12 ровно $C_{12}^k$ нечётных; поэтому число на очередной добавляемой карточке будет нечётным ровно тогда, когда $C_{12}^k$ нечётно (и тогда $k$ в эту минуту увеличится на 1).\nНетрудно заметить, что число $C_{28}^{12}$ нечётно (это следует из того, что степени двойки, входящие в $28 \\cdot 27 \\cdots 17$ и $12 \\cdot 11 \\cdots 1$, равны). Далее, поскольку $C_{28+t}^{12} = \\frac{(28+1)(28+2)\\cdots(28+t)}{(16+1)(16+2)\\cdots(16+t)} \\cdot C_{28}^{12}$, получаем, что наименьшее $t$, при котором $C_{28+t}^{12}$ чётно, равно 4.\nИтак, количество нечётных чисел на карточках будет расти, пока не достигнет 32 (на 4-й минуте), а после этого на карточках всегда будет ровно 32 нечётных числа.\nРассмотрим числа на карточках после $n \\ge 4$ минут. Пусть $T_n$ — сумма всех произведений 12 из этих чисел, а $E_n$ — сумма всех произведений 11 из этих чисел. Число $T_{n+1}$ отличается от $T_n$ прибавлением всех произведений по 12 чисел, среди которых есть только что добавленное, то есть прибавлением $T_n E_n$; итак, $T_{n+1} = T_n + E_n T_n = T_n(1+E_n)$. Заметим при этом, что $E_n = C_{32}^{11} \\pmod{2}$ при $n \\ge 4$, а число $C_{32}^{11} = \\frac{12}{21} \\cdot C_{32}^{12}$ чётно. Значит, при $n \\ge 4$ число $1+E_n$ нечётно, и степень двойки, на которую делится $T_{n+1}$, равна степени двойки, на которую делится $T_n$.\nВыберем теперь $d$ так, чтобы после пятой минуты ни одно из чисел на карточках не делилось на $2^d$; в частности, только что добавленное число $T_4$ также не будет делиться на $2^d$. Значит, и все числа $T_5, T_6, \\dots$ не будут делиться на $2^d$, а это в точности числа, написанные на карточках, добавляемых после пятой минуты. Итак, на карточках никогда не появится числа, делящегося на $2^d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39837,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a product of 3 positive integers each multiple was decreased by 3. Could it happen that the product increases by 2016 after this change?\n\nВ произведении трёх положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли случиться, что произведение увеличилось на 2016 после этого изменения?",
    "options": [],
    "answer": "Yes; for example 1, 1, and 676.",
    "solution": "Yes. An example is $1 \\cdot 1 \\cdot 676$.\n\nAfter the operation, we have $(-2) \\cdot (-2) \\cdot 673 = 4 \\cdot 673 = 2692 = 676 + 2016$.\n\n**Note.** The given example is unique. Here is how it can be found. Suppose two of the factors are $1$, and the third is $a$. Their product is $a$, and after decreasing each by $3$, the product becomes $(-2)^2(a-3) = 4a - 12$. Thus, for $4a - 12 = a + 2016$, the condition is satisfied. Solving this equation, we get $a = 676$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some pairs of cities in a country are connected with one-directional direct flights (between any two cities, there is at most one flight).\nWe say that a city *A* is *accessible* for a city *B* if one may reach the city *A* starting at *B* (perhaps with zero flights or more than one flight in a chain). Assume that for any two cities *P* and *Q* there exists a city *R* for which each of *P* and *Q* is accessible. Prove that there exists a city for which every city is accessible. (V. Dolnikov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose the city with a maximal number of cities accessible from it.\n\nFirst solution. Number all the cities in the country as $A_1, A_2, \\dots, A_n$. By the condition, there exists a city $B_2$ for which both $A_1$ and $A_2$ are accessible. Next, the cities $A_3$ and $B_2$ are accessible for some city $B_3$. Since from $B_2$ one can reach $A_1$ and $A_2$, they are also accessible for $B_3$. Continuing in the same way, we eventually find that there exists a city $B_n$ for which $A_n$ and $B_{n-1}$ are accessible. Then for $B_n$, the cities $A_1, A_2, \\dots, A_{n-1}$ are also accessible, since one can reach them via $B_{n-1}$.\n\nSecond solution. Consider a city $A$ for which the maximal number of cities are accessible (if there are several such cities, pick any one of them). Suppose that some city $B$ is not accessible for $A$. Then, by the condition, there exists a city $C$ for which both $A$ and $B$ are accessible. But then for $C$, $B$ is accessible, as well as all the cities accessible for $A$. This contradicts the choice of $A$; therefore, all cities are accessible for $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an isosceles trapezoid where $BC \\parallel AD$ and $AB \\parallel CD$. A circle $\\omega$ passes through $B$ and $C$ and meets again the segments $AB$ and $BD$ at $X$ and $Y$, respectively. The tangent to $\\omega$ at $C$ meets the ray $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $CYDZ$ is cyclic.\n\nSince $BC \\parallel AD$, and the line $ZC$ is tangent to the circle $\\omega$, we have $\\angle ADB = \\angle YBC = \\angle YCZ$. Therefore, $\\angle YDZ + \\angle YCZ = 180^\\circ$, that is, the quadrilateral $CYDZ$ is cyclic (see Fig. 11).\n\n![](attached_image_1.png)\n\nThus, $\\angle CYZ = \\angle CDZ = \\angle XBC = 180^\\circ - \\angle CYX$, where the last two equalities follow from the fact that the trapezoid $ABCD$ is isosceles, and the quadrilateral $XBCY$ is inscribed in $\\omega$. Therefore, $\\angle CYZ + \\angle CYX = 180^\\circ$, so the points $X$, $Y$, and $Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a triangle whose side lengths $x, y, z$ satisfy $x^3 + y^3 + z^3 = (x+y)(y+z)(z+x)$.\n\nСуществует ли треугольник, длины сторон которого $x, y, z$ удовлетворяют равенству $x^3 + y^3 + z^3 = (x+y)(y+z)(z+x)$?",
    "options": [],
    "answer": "No",
    "solution": "No, such a triangle does not exist.\n\nLet us consider the expression:\n$$(x + y)(x + z)(y + z) = x^2(y + z) + y^2(x + z) + z^2(x + y) + 2xyz$$\nBy the triangle inequality, $x, y, z$ are positive and the sum of any two is greater than the third. Therefore,\n$$x^2(y + z) + y^2(x + z) + z^2(x + y) + 2xyz > x^3 + y^3 + z^3$$\nsince $x^2(y + z) > x^3$, $y^2(x + z) > y^3$, $z^2(x + y) > z^3$, and $2xyz > 0$.\n\nThus, $(x+y)(x+z)(y+z) > x^3 + y^3 + z^3$ for any positive $x, y, z$ satisfying the triangle inequalities. Therefore, there does not exist a triangle with such side lengths.\n\n\nSolution:\n\nНет, не существует такого треугольника.\n\nПервое решение. Пусть такой треугольник существует. Можно считать, что $x \\ge y \\ge z$. Тогда по неравенству треугольника $y+z > x$, откуда\n$$(x+y)(x+z)(y+z) > (x+y)(x+z)x = x^3 + x^2y + x^2z + xyz > x^3 + x^2y + x^2z \\ge x^3 + y^3 + z^3.$$\n\nВторое решение. Пусть такой треугольник существует. Отметим точки касания его сторон с вписанной окружностью. Пусть отрезки касательных от вершин до этих точек касания равны $a, b$ и $c$, тогда $x = b+c$, $y = a+c$, $z = a+b$. Имеем\n$$(x+y)(y+z)(z+x) = (a+2b+c)(a+b+2c)(2a+b+c) = 2(a^3+b^3+c^3)+7(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+16abc$$\nи\n$$\\begin{aligned}\nx^3+y^3+z^3 &= (a+b)^3 + (b+c)^3 + (c+a)^3 \\\\\n&= 2(a^3+b^3+c^3)+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2).\n\\end{aligned}$$\n\nЗначит, разность $(x+y)(y+z)(z+x) - x^3 - y^3 - z^3$ после подстановки и приведения подобных слагаемых приобретает вид $4(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+16abc$, что, очевидно, больше 0. Противоречие.\n\nЗамечание. Существуют три положительных числа $x, y, z$ такие, что равенство из условия выполнено — например, $1, 1, 1+\\sqrt{5}$. Таким образом, условие, что $x, y, z$ являются длинами сторон треугольника, существенно.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39841,
    "subject": "Mathematics (Multi-modal)",
    "question": "100 dwarves whose weights are $1$, $2$, $\\ldots$, $100$ lb came to the left bank of a river. They cannot swim, but they have a boat which can take up to $100$ lb. When a boat crosses the river, one of the dwarves in it is an oarsman; while performing one crossing, the oarsman remains the same. Due to the stream, it is difficult to oar from the right bank to the left one, so each dwarf can oar in this direction at most once. Can the whole company of dwarves reach the right bank?",
    "options": [],
    "answer": "No",
    "solution": "Answer. No.\nCall the dwarves weighing at least $50$ lb heavy; assume they oared backwards $d$ times. Then there were at least $51 + d$ their forward passages, and hence at least $50$ backward passages with no heavy dwarves oaring. But who could oar during these $50$ passages?\n\nПервое решение. Предположим, что переправа гномам удалась. Назовём рейсы лодки с левого берега на правый прямым, а с правого на левый – обратными. Пусть было $k$ обратных рейсов; тогда прямых рейсов было $k+1$.\nВ $k$ обратных рейсах гребли $k$ разных гномов; значит, суммарный вес гномов, участвовавших в обратных рейсах, не меньше $1+2+\\ldots+k = \\frac{k(k+1)}{2}$ (здесь и далее все веса измеряются в фунтах). На любом прямом рейсе суммарный вес гномов не превосходит $100$, так что в итоге суммарный вес гномов на левом берегу уменьшился не более, чем $100(k+1) - \\frac{k(k+1)}{2} = \\frac{(200-k)(k+1)}{2}$. С другой стороны, этот вес уменьшился на $1+2+\\ldots+100 = \\frac{100 \\cdot 101}{2}$, откуда\n$$\n(200-k)(k+1) \\ge 100 \\cdot 101, \\text{ или } (k-100)(k-99) \\le 0.\n$$\nЭто может случиться лишь если $k=99$ или $k=100$. При этом неравенство обращается в равенство; это значит, что в $k$ обратных рейсах участвовали только гномы весами $1, 2, \\dots, k$ – каждый по одному разу, а на каждом прямом рейсе суммарный вес гномов был равен $100$.\nЯсно, что гном весом $100$ совершал все свои рейсы (один рейс при $k=99$ и три при $k=100$) в одиночку; рассмотрим теперь только остальных (обычных) гномов и их рейсы (их ровно $99$ прямых и $99$ обратных в любом случае). В каждом из прямых рейсов участвовало хотя бы $2$ обычных гнома; значит, общее количество обычных гномов на правом берегу увеличилось как минимум на $99 \\cdot 2 - 99 = 99$. Это неравенство также обращается в равенство, так что на каждом прямом рейсе участвовало ровно два гнома. Но тогда гном веса $50$ должен был плыть с гномом веса $50$, а второго такого нет. Противоречие.\n\nВторое решение. Мы также предполагаем противное. Воспользуемся терминологией, введённой в начале предыдущего решения.\nНазовём гномов с весами $50, 51, 52, \\ldots, 100$ тяжёлыми, в $d$ обратных рейсах. Тогда эти $d$ тяжёлых гномов совершили хотя бы по два прямых рейса, а остальные — хотя бы по одному. Поскольку два тяжёлых гнома не могли плыть одновременно, $= 51+d$. Значит, обратных рейсов было не меньше, чем $2d + (51 - d) = $ есть хотя бы в $50$ из них гребцами были лёгкие гномы. Но лёгких гномов всего $49$, так что один из них должен был дважды грести в обратном рейсе, что невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an infinite strictly increasing sequence $a_1, a_2, a_3, \\ldots$ of positive integers such that the sum of any two distinct members of the sequence is coprime to the sum of any three its distinct members?\n\nСуществует ли бесконечно возрастающая последовательность $a_1, a_2, a_3, \\ldots$ положительных целых чисел, такая, что сумма любых двух различных членов последовательности взаимно проста с суммой любых трёх различных её членов?",
    "options": [],
    "answer": "Yes; for example, define a1 = 1, a2 = 7, and a_{n+1} = (3 a_n)! + 1.",
    "solution": "One example of such sequence is determined by $a_1 = 1$, $a_2 = 7$, $a_{n+1} = (3a_n)! + 1$. Define a good pair/triple of elements of this sequence as such pair/triple where only number 1 may repeat. To show that the sequence fits, prove by induction that any good pair and any good triple have coprime sums.\n\nLet $(a_i, a_j)$ and $(a_p, a_q, a_r)$ be a good pair and a good triple of elements of the sequence. Then\n$$\n\\gcd(a_i + a_j, a_p + a_q + a_r) = \\gcd(a_i + a_j, a_p + a_q - a_r) = 1.\n$$\n\nCase 1. Suppose $a_m$ is only in the pair; say, $a_m = a_j$. Then, since $0 < |a_p + a_q \\pm a_r| \\le 3a_{m-1}$, the number $a_m - 1 = (3a_{m-1})!$ is divisible by $a_p + a_q \\pm a_r$, so\n$$\n\\gcd(a_i + a_m, a_p + a_q \\pm a_r) = \\gcd((a_i + 1) + (a_m - 1), a_p + a_q \\pm a_r) = \\gcd(a_i + a_1, a_p + a_q \\pm a_r) = 1\n$$\nby the induction hypothesis.\n\nCase 2. Suppose $a_m$ is only in the triple; say, $a_m = a_q$. Similarly, $a_m - 1$ is divisible by $a_i + a_j$, so\n$$\n\\gcd(a_i + a_j, a_p + a_1 \\pm a_r) = 1\n$$\nby the induction hypothesis.\n\nCase 3. Suppose $a_m$ is in both the pair and the triple; say, $a_m = a_j = a_q$. Then $a_m - 1$ is divisible by $a_p - a_i \\pm a_r$, so\n$$\n\\gcd(a_i + a_m, a_p + a_m \\pm a_r) = \\gcd(a_i + a_m, a_p - a_i \\pm a_r) = \\gcd(a_i + a_1, a_p - a_i \\pm a_r) = 1\n$$\nby the induction hypothesis. The induction step is proved.\nОтвет. Да, существует.\nПостроим пример такой последовательности. Положим $a_1 = 1$, $a_2 = 7$, $a_{n+1} = (3a_n)! + 1$. Для того, чтобы показать, что она удовлетворяет требованиям, нам придется доказать следующее утверждение, из которого будет следовать, что положенная последовательность — требуемая.\n\nПусть $(a_i, a_j)$ и $(a_p, a_q, a_r)$ — хорошие пара и тройка элементов последовательности. Тогда\n$$\nНОД(a_i + a_j, a_p + a_q + a_r) = НОД(a_i + a_j, a_p + a_q - a_r) = 1.\n$$\nДоказательство проведём индукцией по наибольшему элементу. Для перехода предположим, что $m > 1$. Число $a_m$ либо входит только в пару, либо только в тройку, либо в обе.\n\nСлучай 1. Пусть $a_m$ — только элемент пары; скажем, $a_m = a_j$. Тогда, поскольку $0 < |a_p + a_q \\pm a_r| \\le 3a_{m-1}$, число $a_m - 1 = (3a_{m-1})!$ делится на $a_p + a_q \\pm a_r$, то есть\n$$\nНОД(a_i + a_m, a_p + a_q \\pm a_r) = НОД((a_i + 1) + (a_m - 1), a_p + a_q \\pm a_r) = НОД(a_i + a_1, a_p + a_q \\pm a_r) = 1\n$$\nпо предположению индукции.\n\nСлучай 2. Пусть $a_m$ — только элемент тройки; скажем, $a_m = a_q$. Аналогично, $a_m - 1$ делится на $a_i + a_j$, так что\n$$\nНОД(a_i + a_j, a_p + a_1 \\pm a_r) = 1\n$$\nпо предположению индукции.\n\nСлучай 3. Пусть $a_m$ — элемент и пары, и тройки; скажем, $a_m = a_j = a_q$. Тогда $a_m - 1$ делится на $a_p - a_i \\pm a_r$, так что\n$$\nНОД(a_i + a_m, a_p + a_m \\pm a_r) = НОД(a_i + a_m, a_p - a_i \\pm a_r) = НОД(a_i + a_1, a_p - a_i \\pm a_r) = 1\n$$\nпо предположению индукции. Переход индукции доказан.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39843,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle $ABC$ is inscribed into a circle $\\Omega$ and circumscribed about a circle $w$. Points $P$ and $Q$ are chosen on the sides $AC$ and $AB$, respectively, so that the segment $PQ$ is tangent to $w$. A circle $\\Omega_b$ centered at $P$ passes through $B$, and a circle $\\Omega_c$ centered at $Q$ passes through $C$. Prove that the circles $\\Omega$, $\\Omega_b$, and $\\Omega_c$ have a common point.\n\nРавносторонний треугольник $ABC$ вписан в окружность $\\Omega$ и описан около окружности $w$. На сторонах $AC$ и $AB$ выбраны точки $P$ и $Q$ так, что отрезок $PQ$ касается $w$. Окружность $\\Omega_b$ с центром в $P$ проходит через $B$, а окружность $\\Omega_c$ с центром в $Q$ проходит через $C$. Докажите, что окружности $\\Omega$, $\\Omega_b$ и $\\Omega_c$ имеют общую точку.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The second meeting point of $\\Omega$ and $\\Omega_b$ is symmetric to $B$ about $OP$; a similar statement holds for that of $\\Omega$ and $\\Omega_c$. To prove that these points coincide, one needs just to verify that $\\angle POQ = 60^\\circ$.\n\n\nПервое решение. Пусть $O$ — центр треугольника $ABC$, и пусть $w$ касается отрезков $BQ, QP$ и $PC$ в точках $K, L$ и $M$ соответственно (см. рис. 2). В силу симметрии равностороннего треугольника прямые $BO$ и $CO$ проходят через точки $M$ и $K$ соответственно.\n\nОтложим на луче $LO$ отрезок $OX$, равный $OA$ (так что $X$ лежит на окружности $\\Omega$). Поскольку $PL$ и $RM$ — касательные к $w$, имеем $\\angle POL = \\angle POM$, а значит, $\\angle POB = \\angle POX$. Тогда треугольники $POB$ и $POX$ равны по двум сторонам ($OB = OX$, сторона $OP$ — общая) и углу между ними. Итак, $PX = PB$, то есть точка $X$ лежит на окружности $\\Omega_b$. Аналогично, $X$ лежит и на окружности $\\Omega_c$.\n\n\nВторое решение. Пусть $O$ — центр треугольника $ABC$. Окружности $\\Omega$ и $\\Omega_b$ пересекаются в точке $B$. Вторая точка их пересечения — обозначим её через $B'$ — симметрична точке $B$ относительно линии центров $PO$ (см. рис. 3). Аналогично, вторая точка пересечения окружностей $\\Omega$ и $\\Omega_c$ — это точка $C'$, симметричная точке $C$ относительно $QO$.\n\nМы докажем, что $B' = C'$ (тогда эта точка и будет общей у трёх окружностей). Имеем $OB' = OB = OC = OC'$; значит, достаточно понять, что $\\angle BOB' + \\angle COC' = \\angle BOC (=120^\\circ)$. Поскольку прямые $PO$ и $QO$ — биссектрисы углов $BOB'$ и $COC'$, последнее равенство равносильно равенству $\\angle POQ = 60^\\circ$.\n\nЭто равенство нехитро проверяется. Так как $\\angle A = 60^\\circ$, то $\\angle BQP + \\angle CPQ = 240^\\circ$, поэтому $\\angle POQ = 180^\\circ - (\\angle OQP + \\angle OPQ) = 180^\\circ - (\\angle BQP + \\angle CPQ)/2 = 60^\\circ$, что и требовалось.\n\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39844,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n > 3$ distinct positive integers are written on a board, all these numbers are smaller than $(n-1)! = 1 \\cdot 2 \\cdot \\dots \\cdot (n-1)$. For each pair $a > b$ of these numbers, Sergey found a partial quotient of $a$ by $b$ and wrote up the result into his notebook (e.g., if $a = 100$ and $b = 7$, then $100 = 14 \\cdot 7 + 2$, so Sergey would write the number 14 into the notebook). Prove that the notebook contains two equal numbers.\n\n$n > 3$ различных положительных целых чисел записаны на доске, все эти числа меньше $(n-1)! = 1 \\cdot 2 \\cdot \\dots \\cdot (n-1)$. Для каждой пары $a > b$ из этих чисел Сергей нашёл неполное частное от деления $a$ на $b$ и записал результат в тетрадь (например, если $a = 100$ и $b = 7$, то $100 = 14 \\cdot 7 + 2$, и Сергей запишет число 14 в тетрадь). Докажите, что в тетради окажутся два одинаковых числа.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Enumerate the numbers in increasing order: $a_1 < a_2 < \\dots < a_n$.\n\nSuppose, for contradiction, that all the partial quotients $q_{i,j}$ (for $a_i > a_j$) are distinct.\n\nConsider the sequence of ratios:\n$$\n\\frac{a_n}{a_{n-1}} \\cdot \\frac{a_{n-1}}{a_{n-2}} \\cdots \\frac{a_2}{a_1} = \\frac{a_n}{a_1}.\n$$\n\nEach $a_i$ can be written as $a_i = q_i a_{i-1} + r_i$, where $0 \\leq r_i < a_{i-1}$ and $q_i$ is the partial quotient of $a_i$ by $a_{i-1}$.\n\nTherefore,\n$$\n\\frac{a_i}{a_{i-1}} = q_i + \\frac{r_i}{a_{i-1}} < q_i + 1.\n$$\n\nSo,\n$$\n\\frac{a_n}{a_1} < (q_n + 1)(q_{n-1} + 1) \\cdots (q_2 + 1).\n$$\n\nBut since all $q_i$ are distinct and positive, and $a_n < (n-1)!$, the product $(q_n + 1)(q_{n-1} + 1) \\cdots (q_2 + 1)$ must be less than $(n-1)!$ for $n > 3$.\n\nHowever, if all the partial quotients are distinct and positive, their product would be at least $(n-1)!$, which is impossible since $a_n < (n-1)!$.\n\nTherefore, the notebook must contain two equal numbers.\nПредположим противное. Пусть $a_1, a_2, \\dots, a_n$ — числа на доске в порядке возрастания. Пусть $q_i$ — неполное частное от деления $a_i$ на $a_{i-1}$, то есть $a_i = q_i a_{i-1} + r_i$, $0 \\leq r_i < a_{i-1}$.\n\nТогда\n$$\n\\frac{a_n}{a_1} = \\frac{a_n}{a_{n-1}} \\cdot \\frac{a_{n-1}}{a_{n-2}} \\cdots \\frac{a_2}{a_1}.\n$$\n\nПоскольку $a_i < (n-1)!$, а все $q_i$ различны и положительны, произведение $q_n q_{n-1} \\dots q_2$ не меньше $(n-1)!$. Но тогда $a_n \\geq (n-1)!$, что невозможно, так как $a_n < (n-1)!$.\n\nЗначит, в тетради окажутся два одинаковых числа.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Oleg has labelled all the columns and all the rows of a $50 \\times 50$ table with $100$ distinct numbers $a_1, \\dots, a_{50}$ and $b_1, \\dots, b_{50}$, respectively; exactly $50$ of these numbers are rational. Then he has placed into each cell $(i, j)$ the number $a_i + b_j$. Find the greatest possible number of rational numbers placed into the cells.\n\nОлег подписал все столбцы и все строки таблицы $50 \\times 50$ 100 различными числами $a_1, \\dots, a_{50}$ и $b_1, \\dots, b_{50}$; ровно 50 из этих чисел рациональны. Затем он вписал в каждую клетку $(i, j)$ число $a_i + b_j$. Какое наибольшее число рациональных чисел могло оказаться в клетках таблицы?",
    "options": [],
    "answer": "1250",
    "solution": "Assume there are $x$ rational numbers among the $a_i$. Then the total number of irrationals in the cells is at least $x \\cdot x + (50 - x) \\cdot (50 - x) \\ge 1250$ (since rational $+$ irrational $=$ irrational). In an example, $x = 25$, all irrationals among the $a_i$ are in $\\mathbb{Q} \\pm \\sqrt{2}$, and those among the $b_i$ are in $\\mathbb{Q} \\pm \\sqrt{2}$.\n\nОтвет:\n\nСначала покажем, что иррациональных чисел в таблице не меньше $1250$. Пусть вдоль левой стороны таблицы выписано $x$ иррациональных и $50-x$ рациональных чисел. Тогда вдоль верхней стороны выписаны $50-x$ иррациональных и $x$ рациональных чисел. Поскольку сумма рационального и иррационального чисел всегда иррациональна, в таблице стоит хотя бы $x^2 + (50-x)^2$ иррациональных чисел. При этом $x^2 + (50 - x)^2 = 2x^2 - 100x + 50^2 = 2(x - 25)^2 + 2 \\cdot 25^2 \\ge 2 \\cdot 25^2 = 1250$, что и требовалось. Отсюда следует, что в таблице не более $2500 - 1250 = 1250$ рациональных чисел.\n\nРовно $1250$ рациональных чисел в таблице может быть, например, в таком случае. Вдоль левой стороны стоят числа $1,2,\\ldots,24,25,1+\\sqrt{2},2+\\sqrt{2},\\ldots,25+\\sqrt{2}$, а вдоль верхней стороны — числа $26,27,\\ldots,49,50,26-\\sqrt{2},27-\\sqrt{2},\\ldots,50-\\sqrt{2}$. Тогда иррациональными будут только $2 \\cdot 25^2 = 1250$ сумм рационального и иррационального чисел.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether for every three distinct positive integers $a$, $b$, and $c$ there exists a quadratic trinomial with integer coefficients and positive coefficient of $x^2$ such that at some integer points this trinomial takes the values $a^3$, $b^3$, and $c^3$.\n\nВерно ли, что для любых трёх различных положительных целых чисел $a$, $b$ и $c$ существует квадратный трёхчлен с целыми коэффициентами и положительным коэффициентом при $x^2$, который в некоторых целых точках принимает значения $a^3$, $b^3$ и $c^3$?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes.\nThe polynomial $f(x) = (a+b+c)x^2 - (ab+bc+ca)x + abc = x^3 - (x-a)(x-b)(x-c)$ fits.\nДа, верно.\nПокажем, что трёхчлен\n$$\nf(x) = (a+b+c)x^2 - (ab+bc+ca)x + abc = x^3 - (x-a)(x-b)(x-c)\n$$\nподходит. Ясно, что его коэффициенты целые и старший коэффициент $a+b+c$ положителен. Наконец, легко видеть, что $f(a) = a^3 - 0 = a^3$; аналогично, $f(b) = b^3$ и $f(c) = c^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AM$ be a median of an acute triangle $ABC$, and let $BH$ be the altitude. The line through $M$ perpendicular to $AM$ meets the ray $HB$ at $K$. Given that $\\angle MAC = 30^\\circ$, prove that $AK = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The points $A$, $H$, $M$, and $K$ lie on a circle $\\omega$ with diameter $AK$ (see Fig. 4). Since $\\angle MAH = 30^\\circ$, we have $HM = AK/2$. On the other hand, $BC = 2HM$.\nFirst solution. Since $\\angle ANK = \\angle AMK = 90^\\circ$, the points $A$, $H$, $M$, and $K$ lie on a circle $\\omega$ with diameter $AK$ (see Fig. 4). By the condition, the chord $HM$ of this circle subtends the angle $\\angle MAN = 30^\\circ$, so $HM = 2R \\sin 30^\\circ = AK/2$. On the other hand, $HM$ is the median in the right triangle $BHC$, so $BC = 2HM = AK$, which is what was required to prove.\n\nSecond solution. Complete triangle $ABC$ to parallelograms $ABRC$ and $AQBC$ (see Fig. 5); then $BP = AC = BQ$ and $KH \\perp PQ$. Moreover, point $M$ is the midpoint of $AP$ (as the intersection point of the diagonals of parallelogram $ABRC$). Therefore, lines $MK$ and $HK$ are the perpendicular bisectors of segments $AP$ and $PQ$. Consequently, point $K$ is the center of the circle $\\Omega$ circumscribed about triangle $APQ$. Furthermore, $\\angle APQ = \\angle PAC = 30^\\circ$, so the chord $AQ$ of circle $\\Omega$ is equal to the radius $AK$. Finally, from parallelogram $AQBC$ we get $BC = AQ = AK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ be the circumcircle of a scalene triangle $ABC$ with $\\angle ACB = 60^\\circ$. The points $A'$ and $B'$ are chosen on the internal angle bisectors of the angles $BAC$ and $ABC$, respectively, so that $AB' \\parallel BC$ and $BA' \\parallel AC$. The line $A'B'$ meets $\\omega$ at points $D$ and $E$. Prove that the triangle $CDE$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of the arc $ABC$, and $T$ be the midpoint of the lesser arc $NC$. Prove that $N$ and $T$ lie on $A'B'$. For this purpose, notice that $AN = BN = AB = A'B = B'A$.\n\nFrom the parallelism of the lines $AB'$ and $BC$, we get that $\\angle AB'B = \\angle CBB' = \\angle ABB'$. Therefore, $AB' = AB$. Similarly, $AB = A'B$. Denote $\\angle BAC = 2\\alpha$, $\\angle ABC = 2\\beta$. Let, without loss of generality, $\\alpha > \\beta$.\n\nLet $N$ be the midpoint of the arc $ACB$ of the circle $\\Omega$ (see Fig. 12). Then $AN = BN$ and $\\angle ANB = \\angle ACB = 60^\\circ$; therefore, $= B'A$. Therefore, point $A$ is the center of the circle described around $-\\angle AB' = 60^\\circ - \\beta$, whence $\\angle NAB' = 2\\angle NBB' = 120^\\circ - 2\\beta$ and\n\n$= AN = BN$ and $\\angle ANB = \\angle ACB = 60^\\circ$; therefore, $= B'A$. Therefore, point $A$ is the center of the circle described around $-\\angle AB' = 60^\\circ - \\beta$, whence $\\angle NAB' = 2\\angle NBB' = 120^\\circ - 2\\beta$ and\n\n$\\angle ANB' = 90^\\circ - \\angle NAB'/2 = 30^\\circ + \\beta$. Similarly, $\\angle BNA' = 30^\\circ + \\alpha$, whence $\\angle B'NA + \\angle ANB + \\angle BNA' = (30^\\circ + \\beta) + 60^\\circ + (30^\\circ + \\alpha) = 120^\\circ + (\\alpha + \\beta) = 180^\\circ$. Thus, point $N$ lies on the line $A'B'$.\n\nLet $T$ be the midpoint of the lesser arc $NC$ of the circle $\\Omega$. Note that $\\angle ANT = \\angle ABT = (\\angle ABN + \\angle ABC)/2 = 30^\\circ + \\beta = \\angle ANB'$. Therefore, point $T$ also lies on the line $A'B'$, and triangle $CDE$ coincides with triangle $CNT$. This triangle is isosceles, since $NT = TC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, we put $100$ cards onto a table, each card contains a positive integer. Exactly $43$ of these cards contain odd numbers. Then, on each minute the following operation has been performed. We compute the product of numbers on every set of three cards on the table, add up all these products, write this number onto a new card and put this card onto the table. After a year, it appears that the table contains a card with a number divisible by $2^{10000}$. Prove that some number divisible by $2^{10000}$ was already on the table a day after the process started.\n\nИзначально на стол кладут $100$ карточек, на каждой из которых написано положительное целое число. Ровно на $43$ из них написаны нечётные числа. Затем каждую минуту выполняют следующую операцию: вычисляют произведение чисел на каждой тройке карточек, складывают все эти произведения, записывают это число на новую карточку и кладут её на стол. Через год на столе оказалась карточка с числом, делящимся на $2^{10000}$. Докажите, что уже через сутки после начала процесса на столе была карточка с числом, делящимся на $2^{10000}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the table contain $k$ odd numbers; then the parity of the next number coincides with that of $\\binom{k}{3}$. So, on the first minute $k$ increases by $1$, and then it remains stable. Let $D_n$ ($T_n$) be the sum of products of all pairs (triples) of the numbers on the table after the $n$th minute; $D_n \\equiv \\binom{44}{2} \\equiv 0 \\pmod{2}$ for $n \\ge 1$. Then the new number will be $T_n D_n$, so $T_{n+1} = T_n(1+D_n)$, and the power of $2$ in its expansion is the same as that of $T_n$.\n\nЕсли в некоторый момент среди чисел на карточках есть ровно $k$ нечётных, то среди произведений троек чисел ровно $C_k^3$ нечётных; поэтому число на очередной добавляемой карточке будет нечётным ровно тогда, когда $C_k^3$ нечётно (и тогда $k$ в эту минуту увеличится на $1$).\n\nЗаметим, что число $C_{43}^3 = \\frac{43 \\cdot 42 \\cdot 41}{6}$ нечётно, а число $C_{44}^3 = \\frac{44}{41} \\cdot C_{43}^3$ чётно. Значит, в первую минуту добавится нечётное число, а дальше будут добавляться только чётные. Итак, после первой минуты среди чисел на карточках всегда будет ровно $44$ нечётных.\n\nРассмотрим числа на карточках после $n$ минут. Пусть $T_n$ — сумма всех произведений троек этих чисел, а $D_n$ — сумма всех произведений пар этих чисел. Число $T_{n+1}$ отличается от $T_n$ прибавлением всех произведений троек чисел, среди которых $T_{n+1} = T_n + D_n T_n = T_n(1 + D_n)$. Заметим при этом, что $D_n \\equiv C_{44}^2 = 22 \\cdot 43 \\equiv 0 \\pmod{2}$ при $n \\ge 1$. Значит, при $n \\ge 1$ число $1 + D_n$ нечётно, и степень двойки, на которую делится $T_{n+1}$, равна степени двойки, на которую делится $T_n$.\n\nИтак, после первой минуты степень двойки, на которую делится добавляемое число $T_n$, всегда равна степени двойки, на которую делится $T_n$. Значит, если бы после второй минуты на карточках не было числа, делящегося на $2^{10000}$, то и впоследствие такого числа бы не появилось. Отсюда и следует требуемое.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point $A$ in the plane is assigned a real number $f(A)$. It is known that $f(M) = f(A) + f(B) + f(C)$ whenever $M$ is the centroid of a triangle $ABC$. Prove that $f(A) = 0$ for every point $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Возьмём произвольную точку $M$ плоскости и докажем, что $f(M) = 0$. Для этого рассмотрим произвольный треугольник $ABC$, для которого точка $M$ является точкой пересечения медиан. Обозначим через $D$, $E$ и $F$ точки пересечения медиан треугольников $BCM$, $CAM$ и $ABM$, соответственно (см. рис. 17).\n\nЗаметим, что точка $M$ также является точкой пересечения медиан треугольника $DEF$. В самом деле, обозначим через $A'$, $B'$ и $C'$ середины отрезков $BC$, $CA$ и $AB$, соответственно. Тогда треугольник $DEF$ получается из треугольника $ABC$ гомотетией в точке $M$ с коэффициентом $-\\frac{1}{3}$, поскольку при гомотетии с коэффициентом $-\\frac{1}{2}$ треугольник $ABC$ переходит в $A'B'C'$, а при последующей гомотетии с коэффициентом $\\frac{2}{3}$ — в треугольник $DEF$.\n\n![](attached_image_1.png)\n\nРис. 17\n\nНо тогда справедлива следующая цепочка равенств:\n$$\n\\begin{align*}\nf(M) &= f(D) + f(E) + f(F) = (f(M) + f(B) + f(C)) + \\\\\n&\\qquad + (f(M) + f(C) + f(A)) + (f(M) + f(A) + f(B)) = \\\\\n&= 2(f(A) + f(B) + f(C)) + 3f(M) = 5f(M),\n\\end{align*}\n$$\nиз чего следует, что $f(M) = 0$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pasha and Vova play the following game, making moves in turn; Pasha moves first. Initially they have a large piece of plasticine. By a move, Pasha cuts one of the existing pieces into three (of arbitrary sizes), and Vova merges two existing pieces into one. Pasha wins if at some moment there appears to be 100 pieces of equal weights. Can Vova prevent Pasha's win? (R. Efremov, D. Belov)",
    "options": [],
    "answer": "No; Vova cannot prevent Pasha’s win. Pasha has a winning strategy.",
    "solution": "**Первое решение.** Приведём алгоритм, позволяющий Паше победить. Пусть масса исходного куска равна 1 кг. Паша каждым ходом будет отрезать от самого большого из имеющихся кусков два куска массой по 0,01 г. Докажем, что не позже, чем через 10 000 ходов Паша победит.\n\nПредположим, что это не так. Рассмотрим 100 последовательных ходов Паши. Всего за эти 100 ходов появятся 200 кусков массой 0,01 г. Если бы каждым своим ответным ходом Вова слеплял два куска массой по 0,01 г, то в итоге получилось бы 100 кусков массой 0,02 г, и Паша бы победил. Значит, по крайней мере один раз Вова не слепит между собой два куска массой 0,01 г. Поэтому спустя 100 ходов Паши и 100 ходов Вовы количество кусков массой 0,01 г увеличится хотя бы на 1.\n\nРазобьём 10 000 ходов Паши на сотни последовательных. По доказанному выше, после каждой сотни последовательных ходов Паши и ответных ходов Вовы количество кусков массы 0,01 г увеличивается хотя бы на 1. Таким образом, через 100 таких сотен последовательных ходов количество кусков массой 0,01 г увеличится хотя бы на 100. Поэтому Паша так или иначе победит.\n\n\n**Второе решение.** Опять же положим массу исходного куска равной 1 кг. Приведём другой алгоритм действий Паши. Пока это возможно, он будет добиваться выполнения следующего условия: (*) массы всех кусков на столе составляют целое число граммов. Заметим, что Вова своим ходом не может нарушить (*).\n\nЕсли перед ходом Паши на столе есть кусок массой хотя бы 3 г, он может отрезать от него два куска по 1 г, сохраняя (*). Значит, если Паша не может сделать ход, каждый кусок весит либо 1 г, либо 2 г (такой момент обязательно наступит, так как количество кусков перед ходом Паши растёт). Но тогда он уже выиграл: в противном случае общая масса всех кусков (в граммах) меньше $1 \\cdot 100 + 2 \\cdot 100 = 300$, что не так.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39852,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $P$, all the side lengths are integers, and the perimeter of $P$ equals $10^{100}$. Moreover, each side length divides the sum of the three other side lengths. Prove that $P$ is a rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Пусть $d$ — наибольшая сторона. Согласно условию, $a+b+c$ делится на $d$, то есть $a+b+c = k d$ для некоторого натурального $k$. Ясно, что $a+b+c > d$ (длина отрезка меньше длины ломаной с теми же концами), поэтому $k > 1$. Кроме того, так как $a \\le d$, $b \\le d$ и $c \\le d$, имеем $a+b+c \\le 3d$, то есть $k \\le 3$.\n\nСлучай $k = 3$ возможен только при $a = b = c = d$. В этом случае наш четырёхугольник — ромб.\n\nИначе $1 < k < 3$, откуда $k = 2$. Но в этом случае имеем $N = a+b+c+d = 2d + d = 3d$. Таким образом получаем противоречие, поскольку $N$ не делится на 3.\n\n\n**Второе решение.** Из условия следует, что каждое из чисел $a, b, c, d$ является делителем числа $N = a+b+c+d$. Значит, $a = N/t_a$, $b = N/t_b$, $c = N/t_c$, $d = N/t_d$ для некоторых натуральных $t_a > 1$, $t_b > 1$, $t_c > 1$, $t_d > 1$.\n\nЗаметим, что $t_a \\ne 2$, иначе длина стороны $a$ равна полупериметру, что невозможно, поскольку $a < b+c+d$.\n\nПоскольку $N$ не делится на 3, имеем $t_a \\ne 3$, значит, $t_a \\ge 4$ и $a \\le N/4$. Аналогично, $b \\le N/4$, $c \\le N/4$, $d \\le N/4$. Тогда равенство $N = a+b+c+d$ возможно только в случае $a = b = c = d = N/4$, т. е. в случае, когда наш четырёхугольник — ромб.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39853,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of the $2 \\times 2019$ table are to be filled with real numbers (one number in each cell) so that the following rules will be satisfied. The first row should contain $2019$ pairwise distinct real numbers; the second row should be a permutation of the first row. Each column should contain two distinct real numbers whose sum is a rational number.\nFind the greatest possible number of irrational numbers the first row may contain.",
    "options": [],
    "answer": "2016",
    "solution": "**Оценка.** Докажем, что в первой строке таблицы, в которой числа расставлены по правилам, не менее трёх рациональных чисел (и, соответственно, не более $2016$ иррациональных чисел). Каждое из чисел, встречающихся в таблице, записано ровно в двух клетках, одна из которых находится в верхней строке, а другая — в нижней. Рассмотрим некоторый столбец, пусть в его верхней клетке стоит число $a_1$, а в нижней — $a_2$ (далее коротко обозначаем такой столбец $(a_1, a_2)$). Покрасим столбец $(a_1, a_2)$. Найдём столбец, у которого число $a_2$ находится в верхней клетке, и покрасим его. Если этот столбец — $(a_2, a_1)$, то завершим процесс. Иначе, если этот столбец — $(a_2, a_3)$, где $a_3 \\ne a_1$, продолжим: покрасим столбец, у которого число $a_3$ находится в верхней клетке, и т. д. — пока не дойдём до столбца, у которого в нижней клетке находится $a_1$ (это обязательно произойдёт, поскольку числа, равные $a_2, a_3, \\dots$, красятся парами). По окончании процесса получим множество покрашенных столбцов $(a_1, a_2), (a_2, a_3), \\dots, (a_k, a_1)$, которое назовём циклом длины $k$. Если остались ещё непокрашенные столбцы, выделим ещё один цикл, и т. д. В конечном итоге множество всех столбцов таблицы разобьётся на непересекающиеся циклы. Так как сумма длин всех циклов равна $2019$, найдётся цикл нечетной длины.\n\nРассмотрим этот цикл: $(a_1, a_2), (a_2, a_3), \\dots, (a_{2t+1}, a_1)$, где $t \\ge 1$. По условию $a_1+a_2 = b_1, a_2+a_3 = b_2, \\dots, a_{2t+1}+a_1 = b_{2t+1}$, где все $b_i$ — рациональные числа. Тогда\n$$\n2a_1 = (a_1 + a_2) - (a_2 + a_3) + (a_3 + a_4) - \\dots - (a_{2t} + a_{2t+1}) + (a_{2t+1} + a_1) = b_1 - b_2 + b_3 - \\dots - b_{2t} + b_{2t+1}\n$$\n— рациональное число, поэтому $a_1$ рационально. Аналогично, все числа $a_1, a_2, \\dots, a_{2t+1}$ рациональны, и их не менее $2t+1 \\ge 3$.\n\n**Пример.** Приведём пример таблицы, заполненной по правилам, в верхней строке которой $2016$ иррациональных чисел:\n\n| 1 | 2 | 3 | $1 + \\sqrt{2}$ | $1 - \\sqrt{2}$ | $2 + \\sqrt{2}$ | $2 - \\sqrt{2}$ | ... | $1008 + \\sqrt{2}$ | $1008 - \\sqrt{2}$ |\n|---|---|---|----------------|----------------|----------------|----------------|-----|-------------------|-------------------|\n| 2 | 3 | 1 | $1 - \\sqrt{2}$ | $1 + \\sqrt{2}$ | $2 - \\sqrt{2}$ | $2 + \\sqrt{2}$ | ... | $1008 - \\sqrt{2}$ | $1008 + \\sqrt{2}$ |\n\n**Замечание.** Заметим, что условие нечётности длины строки таблицы существенно. Для чётной длины строки нетрудно построить примеры таблиц, в которых все числа иррациональны.\nУсловие того, что число не стоит под самим собой, также важно, иначе мог бы появиться цикл длины $1$, и ответ в задаче стал бы равен $2018$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 39854,
    "subject": "Mathematics (Multi-modal)",
    "question": "An interstellar hotel has 100 rooms, their capacities are $101$, $102$, $\\ldots$, $200$ people. These rooms are occupied by $n$ people in total. Now a VIP guest is about to arrive, and the owner wants to provide him with a personal room. For that purpose, the owner wants to choose two rooms, $A$ and $B$, and move all guests from $A$ to $B$ without exceeding its capacity. Determine the largest $n$ for which the owner can be sure he is able to reach the goal whatever the initial distribution of the guests is. (D. Belov, A. Safiullina)",
    "options": [],
    "answer": "8824",
    "solution": "Предположим, что при $8824$ постояльцах директор не может осуществить переселение. Разобьём комнаты на пары по вместимости: $101$ – $200$, $102$ – $199$, $\\ldots$, $150$ – $151$. Отметим, что для каждой пары комнат суммарное количество человек, живущих в двух комнатах, больше, чем вместимость большей комнаты из пары, иначе всех человек из этой пары можно было бы собрать в комнате с большей вместимостью. Таким образом, общее количество человек не меньше $201 + 200 + 199 + \\ldots + 152 = 353 \\cdot 25 = 8825$. Поэтому при $8824$ постояльцах директор может освободить комнату.\n\nТеперь приведём пример, доказывающий, что при $8825$ и более постояльцах существует расселение, в котором освободить комнату указанным образом не удастся.\n\nУпорядочим комнаты по возрастанию вместимости. Пусть в первых пятидесятти комнатах живёт по $76$, а в комнате вместимости $k$ при $151 \\le k \\le 200$ живёт $k-75$ человек. Посчитаем количество человек, живущих в гостинице:\n$$\n\\begin{aligned}\n76 \\cdot 50 &+ (76 + 77 + 78 + \\dots + 125) = \\\\\n&= 3800 + 201 \\cdot 25 = 3800 + 5025 = 8825.\n\\end{aligned}\n$$\n\nРассмотрим две произвольные комнаты вместимости $a < b$. Заметим, что в комнате вместимости $b$ живёт не меньше $b-75$ человек, а в комнате $a$ – не меньше $76$ человек. Таким образом, переселить людей из одной комнаты в другую ни для какой пары комнат не удастся, поэтому пример подходит. Если $n > 8825$, то достаточно селить оставшихся людей поочерёдно в любые комнаты, где ещё остаются свободные места.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC < BC$. A circle passes through $A$ and $B$, and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$, respectively. The circumcircles of the triangles $ABC$ and $A_1B_1C$ meet again at $P$. The segments $AB_1$ and $BA_1$ cross at $S$. Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$, respectively. Prove that the points $P$, $Q$, $R$, and $C$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Используя окружности $(PABC)$ и $(PA_1B_1C)$, получаем $\\angle PAB = 180^\\circ - \\angle PCB = \\angle PA_1B_1$ и аналогично $\\angle PBA = \\angle PB_1A_1$. Тем самым, $\\angle PAB \\sim \\angle PA_1B_1$. Из этого подобия вытекает, что $\\angle APAA_1 = \\angle APB \\pm \\angle BPA_1 = \\angle BPB_1$ и $PA/PA_1 = PB/PB_1$. Следовательно $\\angle PAA_1 \\sim \\angle PBB_1$, и существует поворотная гомотетия $h$ с центром $P$, переводящая $\\overrightarrow{AA_1}$ в $\\overrightarrow{BB_1}$. Угол поворота для $h$ равен $\\angle APB = \\angle ACB$ (см. рис. 18).\n\nТреугольник $A_1QA$ равен треугольнику $A_1SA$, который в свою очередь подобен треугольнику $B_1SB$ (так как $AA_1B_1B$ – вписанный). Итак, треугольники $A_1QA$ и $B_1SB$ подобны и одинаково ориентированы, поэтому $S = h(Q)$ и, значит, $\\angle QPS = \\angle ACB$. Аналогично, $\\angle SPR = \\angle ACB$. Тогда $\\angle QPR = 2\\angle ACB$.\n\n![](attached_image_1.png)\n\nНо $\\angle QCR = \\angle QCA + \\angle ABC + \\angle BCR = \\angle ACS + \\angle ABC + \\angle SCB = 2\\angle ABC$. Итак, мы доказали равенство $\\angle QPR = \\angle QCR$, из которого следует утверждение задачи.\n\n**Замечание 1.** Фактически в решении используется известное описание точки Микеля четверки прямых $AB$, $CA$, $BC$, $A_1B_1$ как центра поворотной гомотетии, переводящей $\\overrightarrow{AA_1}$ в $\\overrightarrow{BB_1}$.\n\n**Замечание 2.** Равенства углов $\\angle QPS = \\angle ACB = \\angle SPR$ можно доказать разными способами. Укажем ещё одну из возможных схем.\n\nПусть $Q_1$ и $R_1$ — проекции $S$ на $CA$ и $CB$ соответственно. Можно показать, что точки $C$, $P$, $Q_1$, $R_1$, $S$ лежат на одной окружности $\\Gamma$ с диаметром $CS$. Прямые $CP$, $A_1B_1$ и $AB$ — радикальные оси трёх окружностей из условия, они пересекаются в одной точке $S'$. Отсюда несложно понять, что четвёрка прямых $CA$, $CP$, $CB$, $CS$ гармоническая. Проецируя из $C$ на $\\Gamma$, получаем, что $SQ_1PR_1$ гармонический. Отсюда $PR_1/R_1S = PQ_1/Q_1S = PQ_1/Q_1Q$, откуда $\\angle PQ_1S \\sim \\angle PR_1R$ (а также $\\angle PQ_1Q \\sim \\angle PR_1S$), из чего следуют нужные равенства углов.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be an infinite sequence of real numbers such that for each $n \\ge 2018$ the number $a_{n+1}$ is the smallest root of the polynomial\n$$\nP_n(x) = x^{2n} + a_1 x^{2n-2} + a_2 x^{2n-4} + \\dots + a_n.\n$$\nProve that there exists a positive integer $N$ such that the sequence $a_N, a_{N+1}, a_{N+2}, \\dots$ is strictly decreasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $n \\ge 2018$. Заметим, что $P_n(a) = P_n(-a)$ при всех $a$. Значит, поскольку $P_n(x)$ имеет ненулевой корень, он имеет и отрицательный корень, откуда $a_{n+1} < 0$.\n\nДалее, поскольку $P_{n+1}(x) = x^2 P_n(x) + a_{n+1}$, имеем\n$$P_{n+1}(a_{n+1}) = a_{n+1}^2 P_n(a_{n+1}) + a_{n+1} = 0 + a_{n+1} < 0. \\ (*)$$\n\nТак как степень многочлена $P_{n+1}(x)$ чётна, а старший коэффициент положителен, при достаточно больших по модулю отрицательных $x$ он принимает положительные значения. Теперь из (*) следует, что у этого многочлена есть корень на интервале $(-\\infty, a_{n+1})$. Значит, и $a_{n+2} < a_{n+1}$.\n\nИтак, мы получили, что $a_{n+2} < a_{n+1}$ при всех $n \\ge 2018$. Это означает, что последовательность $(a_{2019}, a_{2020}, a_{2021}, \\dots)$ — убывающая.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BL$ be an internal angle bisector in a scalene triangle $ABC$, with $L \\in AC$. The extension of the median through $B$ meets the circumcircle $\\omega$ of $ABC$ at point $D$. A line $\\ell$ through the circumcenter of the triangle $BDL$ is parallel to $AC$. Prove that $\\omega$ is tangent to $\\ell$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $M$ — середина отрезка $AC$, $S$ — вторая точка пересечения прямой $BL$ с окружностью $\\omega$, $N$ — середина дуги $ABC$ (см. рис. 8). Тогда $S$ — середина меньшей дуги $AC$ окружности $\\omega$, а точки $M, S, N$ лежат на серединном перпендикуляре к отрезку $AC$. Прямая $BN$ — внешняя биссектрица угла $ABC$, поэтому $BN \\perp BL$. Тогда $\\angle LBN + \\angle LMN = 90^\\circ + 90^\\circ = 180^\\circ$, откуда мы получаем, что четырехугольник $BLMN$ — вписанный. Обозначим $\\angle MBL = \\alpha$. Тогда $\\angle MNL = \\alpha$. Также, так как $BNDS$ — вписанный, $\\angle SND = \\angle SBD = \\alpha$.\n\n![](attached_image_1.png)\n\nРис. 8\n\nНа продолжении отрезка $DN$ за точку $N$ отметим точку $T$ так, что $LN = NT$. Тогда $\\angle LNT = 180^\\circ - \\angle LNM - \\angle SND = 180^\\circ - 2\\alpha$, откуда следует, что $\\angle LTN = \\angle TLN = \\alpha$. Значит,\n\n$\\angle LTD = \\alpha = \\angle LBD$, поэтому точка $T$ лежит на окружности $\\Gamma$, описанной около треугольника $BDL$. Пусть $l' -$ касательная в точке $N$ к окружности $\\omega$. Поскольку $SN$ — диаметр $\\omega$, то $l' \\perp SN$ и $l' \\parallel AC$. Как мы знаем, $SN$ является внешней биссектрисой угла $LNT$, поэтому $l' -$ биссектриса угла $LNT$. Так как $TL = TN$, получаем, что $l'$ является серединным перпендикуляром к отрезку $TL$, а потому проходит через центр окружности $\\Gamma$. Таким образом, прямые $l$ и $l'$ совпадают, и прямая $l$ касается $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39858,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a kindergarden, a nurse took $n > 1$ congruent cardboard rectangles and gave them to $n$ kids, one per each. Each kid has cut its rectangle into congruent squares (the squares of different kids could be of different sizes). It turned out that the total number of the obtained squares is a prime number. Prove that all initial rectangles were in fact squares. (S. Berlov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $L$ be the foot of the internal angle bisector of $\\angle B$ in an acute-angled triangle $ABC$. The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$, respectively, in the circumcircle $\\omega$ of $\\triangle ABC$. Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ beyond $D$ and $E$, respectively, so that $\\angle APB = \\angle CQD = 90^\\circ$. Prove that the midpoint of $BL$ lies on the line $PQ$. (A. Kuznetsov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим $\\angle BAS = 2\\alpha$, $\\angle ACB = 2\\gamma$. Не умаляя общности считаем, что $\\alpha \\ge \\gamma$. Поскольку точки $D$ и $E$ — середины дуг $AB$ и $AC$ окружности $\\omega$, имеем $\\angle ABD = \\angle ACB = \\frac{\\angle ACB}{2} = \\gamma$ и $\\angle CBE = \\alpha$.\n\n![](attached_image_1.png)\n\nОбозначим через $M$ и $N$ середины сторон $AB$ и $BC$ соответственно (см. рис. 19). Тогда $MN$ — средняя линия треугольника $ABC$, она параллельна стороне $AC$ и проходит через середину $K$ отрезка $BL$. Также $\\angle AMK = 180^\\circ - \\angle MAC = 180^\\circ - 2\\alpha$ и $\\angle BNK = \\angle BCA = 2\\gamma$.\n\nПусть $BM = MA = c$, $BN = NC = a$. Отрезок $MP$ является медианой в прямоугольном треугольнике $APB$, поэтому $MP = c$ и $\\angle MPB = \\angle MBP = \\gamma$; аналогично, $NQ = a$ и $\\angle NQB = \\alpha$. Следовательно, $\\angle QNK = \\angle BNK + \\angle BNQ = 2\\gamma + 180^\\circ - 2\\alpha$ и $\\angle PMK = \\angle AMK + \\angle PMA = 180^\\circ - 2\\alpha + 2\\gamma$. Так как $\\alpha \\ge \\gamma$, то либо точки $P$ и $Q$ лежат на прямой $MN$, и в таком случае задача уже решена, либо точка $P$ лежит в той же полуплоскости, что и точка $A$ относительно прямой $MN$, а точка $Q$ — в другой полуплоскости.\n\nЗаметим, что $BK$ — биссектриса треугольника $BMN$, поэтому $MK/KN = c/a = MP/QN$. Вместе с равенством $\\angle PMK = \\angle QNK$ это означает, что треугольники $PMK$ и $QNK$ подобны. Значит, $\\angle MKP = \\angle NKQ$. Поскольку к тому же точки $P$ и $Q$ лежат в разных полуплоскостях относительно прямой $MN$, точки $P$, $M$ и $K$ лежат на одной прямой, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39860,
    "subject": "Mathematics (Multi-modal)",
    "question": "24 students attend a mathematical circle. For any team consisting of 6 students, the teacher considers it to be either good or OK. For the tournament of mathematical battles, the teacher wants to partition all the students into 4 teams of 6 students each. May it happen that every such partition contains either exactly three good teams or exactly one good team, and both options are present? (I. Bogdanov)",
    "options": [],
    "answer": "Yes",
    "solution": "Приведём один из возможных примеров. Выделим трёх школьников. Будем называть сыгранными команды, в которых содержится 1 или 3 выделенных школьника, а остальные — несыгранными.\n\nВыделенные школьники могут либо оказаться в трёх разных командах, или тогда мы получим три сыгранные команды и одну несыгранную, либо оказаться все в одной команде, и мы получим одну сыгранную и три несыгранные, либо двое могут оказаться в одной команде и один в другой, в этом случае мы также получаем одну сыгранную и три несыгранные команды. Таким образом, все условия соблюдаются.\n\n**Замечание.** Также подходят все примеры, когда выделено нечётное количество школьников, большее 1 и меньше 23, и все команды, в которых присутствует нечётное количество выделенных школьников, объявляются сыгранными, а все остальные — несыгранными.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $b > a > 1$ be real numbers. Define\n$$\nx_n = 2^n \\left( 2^n \\sqrt{b} - 2^n \\sqrt{a} \\right).\n$$\nProve that the sequence $x_1, x_2, \\dots$ is strictly decreasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Докажем, что $x_n > x_{n+1}$. Положим $A = 2^{n+1}\\sqrt{a}$ и $B = 2^{n+1}\\sqrt{b}$. Легко видеть, что $B > A > 1$, отсюда $\\frac{A+B}{2} > 1$. Тогда имеем\n$$\nx_{n+1} = 2^{n+1}(B-A) > 0, \\\\\nx_n = 2^n(B^2 - A^2) = 2^{n+1}(B-A) \\frac{A+B}{2} = x_{n+1} \\cdot \\frac{A+B}{2} > x_{n+1}, \\\\\n\\text{что и требовалось доказать.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BH$ be an altitude in an acute-angled triangle $ABC$. Let $M$ and $N$ be the midpoints of $AH$ and $CH$, respectively. Let $BB'$ be a diameter in the circumcircle of $\\triangle BMN$. Prove that $AB' = CB'$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of the circle $\\Omega$; that is, $O$ is the midpoint of the diameter $BB'$. Denote by $H'$ and $P$ the projections of the points $B'$ and $O$, respectively, onto the line $AC$ (see Fig. 9). Since $O$ lies on the perpendicular bisector of $MN$, we get that $P$ is the midpoint of $MN$. Since $O$ is the midpoint of $BB'$, $P$ is the midpoint of $HH'$. Thus, $H$ and $H'$ are symmetric with respect to the midpoint of $MN$, from which $HM = H'N$ and $H'M = HN$. We have $AH' = AM + H'M = HM + H'M = H'N + HN = H'N + CN = CH'$. Thus, $B'H'$ is the perpendicular bisector of the segment $AC$, consequently, $AB' = CB'$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. Petya and Vasya play the following game. Petya chooses $2n$ (not necessarily distinct) nonnegative real numbers $x_1, x_2, \\dots, x_{2n}$ whose sum equals $1$, and tells those numbers to Vasya. Vasya arranges those numbers in a circle by his own choice, computes the product in each pair of adjacent numbers, and writes down the maximal of those products. Petya wants to maximize the written number, while Vasya wants to minimize it. What number will be written at an optimal game? (A. Khrabrov)",
    "options": [],
    "answer": "1/(8(n-1))",
    "solution": "Если Петя выберет числа $0, \\frac{1}{2}, \\frac{1}{4(n-1)}, \\frac{1}{4(n-1)}, \\dots, \\frac{1}{4(n-1)}$, то, как бы ни расставлял эти числа Вася, число $\\frac{1}{2}$ будет в одной паре с числом $\\frac{1}{4(n-1)}$. Значит, одно из произведений будет равно $\\frac{1}{8(n-1)}$, а остальные будут не больше него. Тогда на доске окажется число $\\frac{1}{8(n-1)}$.\n\nПокажем, как Вася может для любых чисел получить на доске число, не большее $\\frac{1}{8(n-1)}$. Перенумеруем числа в порядке убывания: $x_1 \\ge x_2 \\ge \\dots \\ge x_{2n}$. Поставим в какое-то место на круге число $x_1$, от него по часовой стрелке через пустые места числа $x_2, x_3, \\dots, x_n$. Теперь поставим число $x_{2n}$ между $x_1$ и $x_n$; дальше по часовой стрелке от $x_{2n}$ расставим на пустых местах по очереди числа $x_{2n-1}, x_{2n-2}, \\dots, x_{n+1}$. Тогда произведениями пар соседних чисел будут: $x_n x_{2n}$,\n$x_1 x_{2n}, x_2 x_{2n-1}, x_3 x_{2n-2}, \\dots, x_k x_{2n-k+1}, \\dots, x_n x_{n+1}$ и\n$x_1 x_{2n-1}, x_2 x_{2n-2}, x_3 x_{2n-3}, \\dots, x_k x_{2n-k}, \\dots, x_{n-1} x_{n+1}$.\nПоскольку $x_k x_{2n-k+1} \\le x_k x_{2n-k}$, наибольшее произведение может быть лишь во второй строке.\nПокажем, что $a = x_k x_{2n-k} \\le \\frac{1}{8(n-1)}$ при $k \\le n-1$. Действительно, из неравенств $x_k \\le x_{k-1} \\le \\dots \\le x_1$ следует, что $k x_k \\le x_1 + x_2 + \\dots + x_k$, поэтому\n$$\nka = k x_k \\cdot x_{2n-k} \\le (x_1 + x_2 + \\dots + x_k) x_{2n-k}.\n$$\nАналогично из неравенств\n$$\nx_{2n-k} \\le x_{2n-k-1} \\le x_{2n-k-2} \\le \\dots \\le x_{k+1} \\\\\n\\text{следует, что} \\\\\n(2n-2k)x_{2n-k} \\le x_{2n-k} + x_{2n-k-1} + \\dots + x_{k+1} \\le \\\\\n\\le x_{k+1} + x_{k+2} + \\dots + x_{2n} = 1 - x_1 - x_2 - \\dots - x_k.\n$$\nПоэтому\n$$\n2k(n-k)a \\le (x_1 + x_2 + \\dots + x_k) (1 - x_1 - x_2 - \\dots - x_k) = x(1-x), \\\\\n\\text{где } x = x_1+x_2+\\dots+x_k. \\text{ Поскольку по неравенству о средних для} \\\\\n\\text{двух чисел } x(1-x) \\le \\left(\\frac{x+(1-x)}{2}\\right)^2 = \\frac{1}{4}, \\text{ получаем неравенство} \\\\\nx_k x_{2n-2k} = a \\le \\frac{1}{8k(n-k)}. \\text{ Осталось показать, что } k(n-k) \\ge n-1 \\\\\n\\text{при } k \\le n-1. \\text{ Но последнее неравенство можно переписать в} \\\\\n\\text{виде } (k-1)(n-k-1) \\ge 0, \\text{ а обе скобки в последней формуле} \\\\\n\\text{неотрицательны.}\n$$\n\n**Замечание.** Оптимальная расстановка для Васи не единственна. Однако можно доказать, что при любом $k = 1, 2, \\dots, n-1$ в любой Васиной расстановке среди произведений пар соседних чисел найдётся число, не меньшее $x_k x_{2n-k}$; поэтому оптимальными для Васи окажутся расстановки, в которых наибольшее произведение имеет такой вид.\n\n**Ответ.** $\\frac{1}{8(n-1)}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it true that for any nonzero integer numbers $a$ and $b$ the system of equations\n$$\n\\begin{cases}\n\\tan(13x) \\tan(ay) = 1, \\\\\n\\tan(21x) \\tan(by) = 1\n\\end{cases}\n$$\nhas at least one solution?",
    "options": [],
    "answer": "No; for a = 8 and b = 13 the system has no solution.",
    "solution": "Покажем, что система не будет иметь решений при $a = 8$, $b = 13$. Действительно, из уравнений системы вытекает, что\n$$\n13x + ay = \\frac{\\pi}{2} + \\pi k, \\quad 21x + by = \\frac{\\pi}{2} + \\pi \\ell\n$$\nпри целых $k$ и $\\ell$. Отсюда следует\n$$\n(21a - 13b)y = 21(13x + ay) - 13(21x + by) = \\pi(4 + 21k - 13\\ell).\n$$\nПри $a = 8, b = 13$ получаем $y = (13\\ell - 21k - 4)\\pi$, а значит,\n$\\tan(ay) = 0$. Поэтому первое уравнение системы не может выполняться.\n\n**Замечание.** Числа $a$ и $b$ такие, что $|21a - 13b| = 1$, можно найти, применив алгоритм Евклида к паре взаимно простых чисел $(21, 13)$. На самом деле, в условии задачи можно заменить числа $21$ и $13$ на любую пару взаимно простых чисел, и ответ не изменится. Можно также заметить, что $13$ и $21$ — последовательные числа Фибоначчи, поэтому равенство $|21 \\cdot 8 - 13 \\cdot 13| = 1$ является частным случаем общего факта $|F_{n+1}F_{n-1} - F_n^2| = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the trinomials $x^2 + a x + b$ and $x^2 + a x + b + 1$ has at least one real root; moreover, all real roots of those trinomials are integers. Prove that the trinomial $x^2 + a x + b + 2$ has no real roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $D_1, D_2, D_3$ — соответственно дискриминанты этих трёхчленов. Первые два уравнения имеют только целые корни, поэтому $D_1 = m^2, D_2 = n^2$, где числа $m$ и $n$ можно считать целыми неотрицательными. Вычитая из первого равенства второе, получаем, что $4 = m^2 - n^2$, то есть $4 = (m - n)(m + n)$. Числа $m-n$ и $m+n$ — одной чётности, поэтому это равенство может выполняться только если $m-n = m+n = 2$. Но тогда $n = 0$, и, значит, дискриминант третьего уравнения $D_3 = n^2 - 4 = -4$ — отрицательный.\n\n\nВторое решение. По теореме Виета числа $a$ и $b$ целые.\nПусть $x_1$ и $x_2$ — корни трёхчлена $P_1(x) = x^2 + a x + b$, а $x_0$ — любой из корней трёхчлена $P_2(x) = x^2 + a x + b + 1$. Поскольку $P_2(x)$ имеет хотя бы один корень, корни трёхчлена $P_1(x) = P_2(x) - 1$ различны.\n$$\n\\text{Имеем } P_1(x_0) = P_2(x_0) - 1 = -1, \\text{ откуда} \\\\\n1 = P_1(x_1) - P_1(x_0) = (x_1^2 + a x_1 + b) - (x_0^2 + a x_0 + b) = \\\\\n\\qquad (x_1 - x_0)(x_1 + x_0 + a).\n$$\nПоскольку оба множителя в правой части целые, они могут быть равны лишь $\\pm 1$. Аналогично, $x_2 - x_0 = x_2 + x_0 + a = \\pm 1$. Так как $x_1 \\ne x_2$, отсюда следует, что $|x_2 - x_1| = 2$ и $x_0 = \\frac{x_1 + x_2}{2}$. Так как это рассуждение верно для произвольного корня $x_0$ трёхчлена $P_2(x)$, его корни совпадают, то есть $P_2(x) = (x - x_0)^2$. А тогда многочлен $P_2(x) + 1 = (x - x_0)^2 + 1$ положителен на всей оси, то есть корней не имеет.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the *distance* between two cells of a checkered board as the smallest number of chess king's moves needed to come from one cell to the other. Determine the largest $n$ for which one can mark $n$ cells of a $100 \\times 100$ board so that none of the distances between two marked cells equals $15$.",
    "options": [],
    "answer": "3025",
    "solution": "Разобьём доску на $9$ квадратов $30 \\times 30$, $6$ прямоугольников $10 \\times 30$ и один квадрат $10 \\times 10$ (см. рис. 10).\n\nВ каждом квадрате $30 \\times 30$ клетки разбиваются на $15^2$ четвёрок так, что расстояние между любыми клетками в одной четвёрке равно $15$ (каждая четвёрка состоит из клеток с координатами $(a, b)$, $(a, b+15)$, $(a+15, b)$ и $(a+15, b+15)$). Тогда в любой четвёрке может быть отмечено не более одной клетки, то есть общее число отмеченных клеток в таком квадрате не превосходит $15^2$.\n\nАналогично, каждый прямоугольник $10 \\times 30$ (скажем, с длинной горизонтальной стороной) разбивается на пары клеток, отстоящих друг от друга на $15$ (с координатами $(a, b)$ и $(a + 15, b)$) — поэтому в нём не более $15 \\cdot 10$ отмеченных клеток.\n\nНаконец, в квадрате $10 \\times 10$ всего $10^2$ клеток. Итого, отмеченных клеток не больше, чем $9 \\cdot 15^2 + 6 \\cdot 15 \\cdot 10 + 10^2 = (3 \\cdot 15 + 10)^2 = 55^2$.\n\nПример с таким количеством отмеченных клеток показан на рис. 11.\n\n![](attached_image_1.png)\nРис. 10\n![](attached_image_2.png)\nРис. 11",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39867,
    "subject": "Mathematics (Multi-modal)",
    "question": "One is given $n$ coins of pairwise distinct weights and $n$ scales, $n > 2$. On each weighing, it is permitted to put two coins onto the pans of one of the scales, check the result, and take the coins back from the pans. One of the scales (nobody knows which one) is possibly broken, and it provides random results (sometimes correct, sometimes incorrect). Determine the smallest number of weighings at which one can determine for sure which coin is the heaviest one.",
    "options": [],
    "answer": "2n - 1",
    "solution": "Докажем сначала, что за $2n - 1$ взвешивание можно найти самую тяжёлую монету. Более точно, мы докажем по индукции по $n$, что самую тяжёлую из $n \\ge 2$ данных монет можно определить за $2n - 1$ взвешивание, имея трое весов, одни из которых, возможно, испорчены.\n\nЕсли $n = 2$, то взвесим данные две монеты по очереди на трёх разных весах. Если при одном из взвешиваний весы оказались в равновесии, то эти весы испорчены, значит, мы можем определить более тяжёлую монету по показаниям любых из остальных весов. Если равновесия ни разу не было, то какая-то из монет перевесит хотя бы два раза — она и есть более тяжёлая, так как неверный результат могут давать только одни весы. Это даёт базу индукции.\n\nПусть теперь $n \\ge 3$. Выберем две монеты и двое весов и сравним за первые два взвешивания эти монеты друг с другом на первых и на вторых весах. Возможны два случая:\n\n1. Оба раза перевешивала одна и та же из двух монет; назовём её монетой $a$, а вторую из них — монетой $b$. Так как хотя бы одни из двух весов правильные, то монета $a$ действительно тяжелее монеты $b$. Значит, $b$ не самая тяжёлая. Задача сводится к тому, чтобы определить самую тяжёлую из $n-1$ монеты: монеты $a$ и $n-2$ монет, не участвовавших в первых двух взвешиваниях. По предположению индукции мы можем сделать это за $2n-3$ взвешивания. Вместе с первыми двумя взвешиваниями получаем $2n-1$ взвешивание.\n\n2. Либо одно из первых двух взвешиваний дало равновесие, либо результаты первых двух взвешиваний противоречат друг другу: один раз перевесила одна монета, а другой — другая. Значит, одни из двух использованных весов точно испорчены. Возьмём третью весы. Тогда они обязательно правильные. Используя их, мы легко можем определить самую тяжёлую монету за $n-1$ взвешивание: сравниваем первую монету со второй, более тяжёлую из них с третьей, более тяжёлую из них с четвёртой и т.д. до последней. Вместе с первыми двумя взвешиваниями получаем $n+1 < 2n-1$ (так как $n > 2$) взвешивание.\n\nПокажем теперь, что менее, чем за $2n - 1$ взвешивание, заведомо определить самую тяжёлую монету нельзя. Достаточно показать, что её нельзя определить ровно за $2n-2$ взвешивания, так как можно добавить произвольные взвешивания и игнорировать их результаты. Предположим противное: имеется алгоритм действий, позволяющий определить самую тяжёлую монету за $2n - 2$ взвешивания.\n\nПронумеруем монеты числами $1, \\dots, n$. Сделаем первые $2n-3$ взвешивания согласно алгоритму. Предположим, что в каждом из них перевешивала монета с большим номером. Согласно принципу Дирихле, среди монет с номерами $1, \\dots, n-1$ найдётся такая, которая за произведённые $2n-3$ взвешиваний «проигрывала» (оказывалась более лёгкой) не более одного раза; обозначим номер этой монеты через $k$. Конечно же, монета с номером $n$ ни разу не «проигрывала». Покажем, что такие результаты взвешиваний возможны. Действительно, такое могло произойти по крайней мере в следующих двух ситуациях.\n\n(А) Монеты упорядочены по возрастанию масс и все весы (в том числе, испорченные) показывали правильные результаты во всех взвешиваниях.\n\n(Б) Монеты упорядочены по возрастанию масс, за исключением монеты номер $k$, которая самая тяжёлая. При этом те весы, на которых монета номер $k$ «проиграла», испорчены, и в этом взвешивании показали неверный результат, а в остальных взвешиваниях все весы показывали верные результаты.\n\nРассмотрим два случая.\n\n1. В последнем, $(2n-2)$-м взвешивании, не участвует монета с номером $k$. Предположим, что опять перевесила монета с большим номером. Тогда каждая из ситуаций (А) и (Б) по-прежнему возможна.\n\n2. В последнем взвешивании участвует монета с номером $k$. Предположим, что она перевесила. Тогда, с одной стороны, возможно, что имеет место ситуация (А), и последнее взвешивание выполнялось на испорченных весах. С другой стороны, возможно, что имеет место ситуация (Б), и в последнем взвешивании весы показали правильный результат.\n\nИтак, каким бы ни было одно оставшееся взвешивание, его результат может быть таков, что после него каждая из ситуаций (А) и (Б) будет по-прежнему возможной. Тогда каждая из монет $k$ и $n$ может быть самой тяжёлой, то есть нам не удалось определить самую тяжёлую монету.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a triangular pyramid. A sphere $\\omega_A$ is tangent to the face $BCD$, and to the planes of the other faces outside the faces. Similarly, a sphere $\\omega_B$ is tangent to the face $ACD$, and to the planes of the other faces outside the faces. Let $\\omega_A$ meet the plane $ACD$ at $K$, and let $\\omega_B$ meet the plane $BCD$ at $L$. The points $X$ and $Y$ are chosen on the extensions of the segments $AK$ and $BL$ beyond $K$ and $L$, respectively, so that $\\angle CKD = \\angle CXD + \\angle CBD$ and $\\angle CLD = \\angle CYD + \\angle CAD$. Prove that the points $X$ and $Y$ are equidistant from the midpoint of $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Отметим точки $K_1$ и $L_1$ касания вписанной сферы $\\omega$ тетраэдра с гранями $ACD$ и $BCD$ соответственно, а также точки $K_2$ и $L_2$ касания сфер $\\omega_B$ и $\\omega_A$ с этими гранями. Сферы $\\omega$ и $\\omega_A$ гомотетичны с центром в точке $A$, поэтому точка $K_1$ лежит на отрезке $AK$. Аналогично, точка $L_1$ лежит на отрезке $BL$.\n\nПокажем, что точки $L_1$ и $L_2$ изогонально сопряжены относительно треугольника $BCD$, то есть $\\angle BCL_1 = \\angle DCL_2$, $\\angle DBL_1 = \\angle CBL_2$ и $\\angle CDL_1 = \\angle BDL_2$. Докажем первое из этих равенств; остальные два доказываются аналогично. Обозначим через $M_1$ и $M$ точки касания плоскости $ABC$ со сферами $\\omega$ и $\\omega_A$ соответственно (см. рис. 20). Из равенства отрезков касательных, проведённых из одной точки к сфере, следует, что следующие пары треугольников равны по трём сторонам: $\\Delta CK_1D = \\Delta CL_1D$, $\\Delta AK_1C = \\Delta AM_1C$, $\\Delta BL_1C = \\Delta BM_1C$, $\\angle CL_2D = \\angle CKD$, $\\angle BL_2C = \\angle BMC$, $\\angle AKC = \\angle AMC$. Значит, $\\angle BCL_1 + \\angle BCL_2 = \\angle BCM_1 + \\angle BCM = \\angle ACM - \\angle ACM_1 = \\angle ACK - \\angle AKC_1 = \\angle DCK_1 + \\angle DCK = \\angle DCL_1 + \\angle DCL_2$, откуда следует требуемое равенство $\\angle BCL_1 = \\angle DCL_2$.\n\nИспользуя условие задачи и доказанную изогональную сопряжённость точек $L_1$ и $L_2$, получаем, что $\\angle CXD = \\angle CKD$ $-$ $\\angle CBD = \\angle CL_2D - \\angle CBD = \\angle BCL_2 + \\angle BDL_2 = \\angle DCL_1 + \\angle CDL_1 = 180^\\circ - \\angle CL_1D = 180^\\circ - \\angle CK_1D$. Следовательно, четырёхугольник $CK_1DX$ вписанный. Аналогично устанавливается вписанность четырёхугольника $CL_1DY$.\n\nОбозначим через $N_1$ точку касания сферы $\\omega$ и грани $ABD$. Из равенства треугольников $\\triangle AK_1C = \\triangle AM_1C$ и равенства аналогичных пар треугольников, примыкающих к пяти остальным рёбрам тетраэдра $ABCD$, получаем (см. рис. 21), что $2\\angle AK_1C = \\angle AK_1C + \\angle AM_1C = (360^\\circ - \\angle AK_1D - \\angle CK_1D) + (360^\\circ - \\angle AM_1B - \\angle BM_1C) = 360^\\circ - \\angle AN_1D - \\angle CL_1D + 360^\\circ - \\angle AN_1B - \\angle BL_1C = \\angle BL_1D + \\angle BN_1D = 2\\angle BL_1D$. Так как точки $K_1$ и $L_1$ лежат на отрезках $AX$ и $BY$ соответственно, отсюда следует, что $\\angle CK_1X = \\angle DL_1Y$.\n\nПовернём плоскость $BCD$ вокруг прямой $CD$ так, чтобы она совместилась с плоскостью $ACD$ и при этом треугольник $CL_1D$ совместился с равным ему треугольником $CK_1D$. При этом повороте окружность, описанная около четырёхугольника $CL_1DY$, перейдёт в окружность $\\gamma$, описанную около четырёхугольника $CK_1DX$. В частности, точка $Y$ перейдёт в некоторую точку $Y'$ на окружности $\\gamma$. Из равенства углов $\\angle CK_1X = \\angle DL_1Y = \\angle DK_1Y'$ следует, что точки $X$ и $Y'$ симметричны относительно диаметра окружности $\\gamma$, перпендикулярного хорде $CD$. Следовательно, точки $X$ и $Y'$, а значит, и точки $X$ и $Y$ равноудалены от середины отрезка $CD$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 39869,
    "subject": "Mathematics (Multi-modal)",
    "question": "The radii of five concentric circles $\\omega_0, \\omega_1, \\omega_2, \\omega_3, \\omega_4$ form (in this order) a geometric progression with ratio $q$. Determine the largest value of $q$ for which it is possible to choose a broken line $A_0A_1A_2A_3A_4$ consisting of four segments of equal lengths, such that $A_i \\in \\omega_i$ for all $i = 0, 1, 2, 3, 4$.",
    "options": [],
    "answer": "(sqrt(5)+1)/2",
    "solution": "Можно считать, что $q \\ge 1$. Пусть радиус $\\omega_i$ равен $R_i = Rq^i$.\nВыберем некоторое положительное $\\ell$ и попытаемся построить требуемую ломаную с отрезками длины $\\ell$, стартуя с произвольной точки $A_0 \\in \\omega_0$. Пусть точка $A_i \\in \\omega_i$ уже построена. Расстояния от неё до точек окружности $\\omega_{i+1}$ пробегают отрезок $[R_{i+1} - R_i, R_{i+1} + R_i]$, то есть $[Rq^i(q-1), Rq^i(q+1)]$. Точку $A_{i+1}$ можно построить тогда и только тогда, когда $\\ell$ принадлежит этому отрезку. Значит, ломаную удастся построить тогда и только тогда, когда $Rq^i(q-1) \\le \\ell \\le Rq^i(q+1)$ при всех $i=0,1,2,3$.\nПоскольку $q \\ge 1$, эта система неравенств равносильна неравенствам $Rq^3(q-1) \\le \\ell \\le R(q+1)$. Длина $\\ell$, удовлетворяющая им, существует тогда и только тогда, когда $q^3(q-1) \\le q+1$,\nто есть $q^4 - q^3 - q - 1 \\le 0$, или $(q^2 - q - 1)(q^2 + 1) \\le 0$. Наибольшее типичное $q$, удовлетворяющее этому неравенству, есть $q = \\frac{\\sqrt{5}+1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BE$ and $CF$ be altitudes in a tetrahedron $ABCD$. A plane $\\alpha$ through the midpoint of $AD$ is perpendicular to $AD$. Assume that the points $A$, $C$, $D$, and $E$ lie on a circle, and the points $A$, $B$, $D$, and $F$ also lie on a circle. Show that the points $E$ and $F$ are equidistant from $\\alpha$. (A. Kuznetsov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line $CF$ is perpendicular to the plane $ABD$, so $CF \\perp AD$. Similarly, $BE \\perp AD$. Therefore, the lines $CF$ and $BE$ are parallel to the plane $\\alpha$ or lie in it. The points $B$, $C$, $E$, and $F$ lie on the sphere $\\omega$ circumscribed about the tetrahedron $ABCD$. Also, since $\\angle BEC = 90^\\circ = \\angle BFC$, the points $B$, $C$, $E$, and $F$ lie on the sphere $\\omega'$, constructed on the segment $BC$ as a diameter.\n\nIf the spheres $\\omega$ and $\\omega'$ do not coincide, all their common points lie in one plane, denote it by $\\beta$. In the plane $\\beta$ lie the lines $BE$ and $CF$, each of which is parallel to the plane $\\alpha$ or lies in this plane. Also, the lines $BE$ and $CF$ are not parallel, since they are perpendicular to the intersecting planes $ACD$ and $ABD$. Thus, the plane $\\beta$ is parallel to the plane $\\alpha$ or coincides with it, and the distances from the points $E$ and $F$ to $\\alpha$ are equal to the distance between the planes $\\alpha$ and $\\beta$ (see Fig. 12).\n\n![](attached_image_1.png)\n\nIf the spheres $\\omega$ and $\\omega'$ coincide, then their common center $M$ is the midpoint of the segment $BC$ and lies in the plane $\\alpha$. Therefore, the distances from the points $B$ and $C$ to $\\alpha$ are equal. Since the line $BE$ is parallel to $\\alpha$, the distances from $B$ and $E$ to $\\alpha$ are equal. Similarly, the distances from $C$ and $F$ to $\\alpha$ are also equal, and then the points $E$ and $F$ are equidistant from $\\alpha$ (see Fig. 13).\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\neq 1$ be a positive real number. Prove that the sequence $x_1, x_2, \\dots$, where $x_n = 2^n (2^n\\sqrt{a} - 1)$, is strictly decreasing.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $t = \\sqrt{a}$. Заметим, что $t \\neq 1$. Тогда $x_{n+1} = 2^{n+1}(t-1)$ и $x_n = 2^n(t^2 - 1)$. Следовательно,\n$$\n\\begin{align*}\nx_n - x_{n+1} &= 2^n(t^2 - 1) - 2^{n+1}(t-1) = \\\\\n&= 2^n(t^2 - 2t + 1) = 2^n(t-1)^2 > 0,\n\\end{align*}\n$$\nчто и требовалось доказать.\n\nЗамечание. При других подходах к решению неравенства могут выглядеть по-разному при $a > 1$ (когда члены последовательности положительны) и при $0 < a < 1$ (когда они отрицательны).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Compose a $3 \\times 3 \\times 3$ cube of 26 white unit cubes and one black unit cube by putting the black one into the center. Compose a $3n \\times 3n \\times 3n$ cube of $n^3$ such $3 \\times 3 \\times 3$ cubes. Determine the smallest number $k$ such that it is possible to paint $k$ white unit cubes red so that each white cube will have at least one common vertex with some red cube.",
    "options": [],
    "answer": "(n+1)n^2",
    "solution": "Введём систему координат так, чтобы центры кубиков имеют координаты от $1$ до $3n$ по каждой оси. Каждому кубику приписываем координаты его центра. Таким образом, кубик чёрный тогда и только тогда, когда все его координаты дают остаток $2$ при делении на $3$.\n\nОкрасим красным все белые кубики с координатами $(a, b, c)$, где $a$ делится на $3$, а $b \\equiv c \\equiv 2 \\pmod{3}$, а также все кубики с координатами $(1, b, c)$, где $b \\equiv c \\equiv 2 \\pmod{3}$. Нетрудно видеть, что получилось $(n+1)n^2$ красных кубиков, и требования задачи выполнены. Осталось показать, что добиться требуемого нельзя, окрасив менее $(n+1)n^2$ кубиков.\n\nПри $i = 1, 2, \\dots, n$ положим $w_{3i} = i$, $w_{3i-1} = 0$, $w_{3i-2} = n+1-i$; последовательность $(w_i)$ выглядит так: $n, 0, 1, n-1, 0, 2, n-2, \\dots, 1, 0, n$. Запишем в каждый кубик с координатами $(a, b, c)$ число $w_a w_b w_c$ (в частности, в чёрных кубиках записаны нули). Тогда общая сумма всех чисел, записанных в белых кубиках, окажется равной $\\Sigma = (w_1 + \\dots + w_{3n})^3 = n^3(n+1)^3$.\n\nНазовём ценой $S(X)$ кубика $X$ сумму чисел во всех кубиках, имеющих с ним общую вершину (включая сам $X$). Тогда в любой окраске, удовлетворяющей требованиям, сумма цен красных кубиков не меньше, чем $\\Sigma$. Докажем теперь, что $S(X) \\le (n+1)^2 n$ для любого белого кубика $X$. Из этого будет следовать, что в красный цвет надо окрасить не менее, чем $\\frac{\\Sigma}{(n+1)^2 n} = (n+1)n^2$ кубиков, что и требовалось.\n\nПусть $(a, b, c)$ — координаты кубика $X$. Абсциссы всех кубиков, имеющих с ним общую вершину, равны $a$ или $a \\pm 1$; такое же утверждение верно для остальных координат. Поэтому $S(X) = (w_{a-1} + w_a + w_{a+1})(w_{b-1} + w_b + w_{b+1})(w_{c-1} + w_c + w_{c+1})$, где мы полагаем $w_0 = w_{3n+1} = 0$. Осталось заметить, что $w_{t-1} + w_t + w_{t+1} = n$, если $t \\neq 2 \\pmod{3}$, иначе $w_{t-1} + w_t + w_{t+1} = n+1$. Поскольку не все координаты $X$ дают остаток $2$, отсюда следует, что $S(X) \\le (n+1)^2 n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39873,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $D$ and $E$ are chosen on the sides $AB$ and $AC$ of a triangle $ABC$ so that $DB = BC = CE$. The segments $BE$ and $CD$ meet at $P$. Show that the circumcircles of the triangles $BDP$ and $CEP$ meet at the incenter of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим через $I$ центр вписанной окружности треугольника $ABC$, точка $I$ является точкой пересечения биссектрис. Доказем, что точки $B$, $D$, $P$, $I$ лежат на одной окружности. Аналогично покажем, что точки $C$, $E$, $P$, $I$ лежат на одной окружности, и задача будет решена.\n\nДостаточно установить равенство $\\angle BPD = \\angle BID$. Биссектриса $BI$ угла $ABC$ является осью симметрии равнобедренного треугольника $BDC$, поэтому $\\angle BID = \\angle BIC = 180^\\circ - \\angle IBC = 180^\\circ - \\frac{\\angle B}{2} - \\frac{\\angle C}{2}$ (см. рис. 14).\n\nДалее, $\\angle BPD = \\angle PBC + \\angle PCB = \\angle EBC + \\angle DCB$. Из равнобедренности треугольника $BCE$ получаем $\\angle EBC = \\frac{180^\\circ - \\angle C}{2} = 90^\\circ - \\frac{\\angle C}{2}$ и, аналогично, $\\angle DCB = 90^\\circ - \\frac{\\angle B}{2}$. Отсюда $\\angle BPD = 180^\\circ - \\frac{\\angle B}{2} - \\frac{\\angle C}{2} = \\angle BID$, что и требовалось доказать.\n\n**Замечание.** Точка $I$ является так называемой **точкой Микеля** для четвёрки прямых $AB$, $AC$, $BE$, $CD$, т. е. через $I$ проходят окружности, описанные около каждого из треугольников, образованных тремя из четырёх перечисленных прямых.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five points are marked in the plane. It is permitted to choose some of them and change their positions; the distances between the chosen points should be preserved. Prove that it is possible to perform such change so as to obtain a configuration of five points possessing an axis of symmetry.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the given points by $A$, $B$, $C$, $D$, and $E$. Choose among them two points that are the farthest apart; let these be $A$ and $B$. We will show that it is possible to move them as required.\n\nDraw the perpendicular bisector $\\alpha$ to the segment $CD$. If $E$ lies on $\\alpha$, then it is sufficient to move points $A$ and $B$ onto the line $\\alpha$. Otherwise, let $E'$ be the point symmetric to $E$ with respect to $\\alpha$. Note that the distance from $E$ to $\\alpha$ is less than the length of one of the segments $EC$ or $ED$, that is, less than $AB$.\n\nThus, we can move point $A$ to point $E'$, and point $B$ to a point lying on $\\alpha$ and at a distance $AB$ from $E'$. It is easy to see that $\\alpha$ is the axis of symmetry of the new set of points, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two monic quadratic trinomials $f(x)$ and $g(x)$ are given, each of them has two distinct real roots. Assume that $f(1) = g(2)$ and $g(1) = f(2)$. Find all possible values of the sum of all four roots of the trinomials.",
    "options": [],
    "answer": "6",
    "solution": "**Первое решение.** Пусть $f(x) = x^2 + a x + b$, $g(x) = x^2 + c x + d$. Тогда условия задачи запишутся в виде\n$$\n1 + a + b = 4 + 2c + d \\text{ и } 4 + 2a + b = 1 + c + d.\n$$\nВычитая из первого равенства второе, получаем $-3 - a = 3 + c$, то есть $a + c = -6$. Но по теореме Виета $-a$ — это сумма корней первого трёхчлена, $-c$ — сумма корней второго трёхчлена, откуда и следует требуемое.\n\n\n**Второе решение.** Рассмотрим вспомогательный квадратный трёхчлен $h(x) = g(3 - x)$ (он тоже приведённый!). Тогда $h(x) - f(x)$ — линейный многочлен с корнями 1 и 2; значит, он тождественно нулевой, то есть $f(x) = g(3 - x)$. Поэтому, если $x_0$ является корнем $f(x)$, то $3 - x_0$ является корнем $g(x)$, и сумма этих двух корней равна 3. Аналогично, сумма остальных корней этих многочленов также равна 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petya and Vasya play the following game. Petya chooses 100 (not necessarily distinct) nonnegative real numbers $x_1, x_2, \\dots, x_{100}$ whose sum equals 1, and tells those numbers to Vasya. Vasya splits the numbers into 50 pairs by his own choice, computes the product of numbers in each pair, and writes down the maximal of those products. Petya wants to maximize the written number, while Vasya wants to minimize it. What number will be written at an optimal game?",
    "options": [],
    "answer": "1/396",
    "solution": "Если Петя выберет числа $\\frac{1}{2}, \\frac{1}{198}, \\frac{1}{198}, \\dots, \\frac{1}{198}$, то, как бы ни разбивал эти числа Вася, в паре с числом $\\frac{1}{2}$ будет число $\\frac{1}{198}$.\nИх произведение будет равно $\\frac{1}{396}$, а остальные будут не больше него. Тогда на доске окажется число $\\frac{1}{396}$.\n\nПокажем, как Васе для любых Петиных чисел получить на доске число, не большее $\\frac{1}{396}$. Перенумеруем числа в порядке невозрастания: $x_1 \\ge x_2 \\ge \\dots \\ge x_{100}$. Разобьём числа на пары следующим образом: $x_k$ в паре с $x_{101-k}$. Тогда произведениями чисел в парах будут\n$$\nx_1x_{100}, x_2x_{99}, x_3x_{98}, \\dots, x_kx_{101-k}, \\dots, x_{50}x_{51}.\n$$\nПокажем, что $a = x_kx_{101-k} \\le \\frac{1}{396}$ при $k \\le 49$. Действительно, из неравенств $x_k \\le x_{k-1} \\le \\dots \\le x_1$ следует, что $kx_k \\le x_1 + x_2 + \\dots + x_k$, поэтому\n$$\nka = kx_k \\cdot x_{101-k} \\le (x_1 + x_2 + \\dots + x_k)x_{101-k}.\n$$\nАналогично из неравенств $x_{101-k} \\le x_{100-k} \\le x_{99-k} \\le \\dots \\le x_{k+1}$ следует, что\n$$\n(101 - 2k)x_{101-k} \\le x_{101-k} + x_{100-k} + \\dots + x_{k+1} \\le \\\\\n\\le x_{k+1} + x_{k+2} + \\dots + x_{100} = 1 - x_1 - x_2 - \\dots - x_k.\n$$\nПоэтому\n$k(101-2k)a \\le (x_1+x_2+\\dots+x_k)(1-x_1-x_2-\\dots-x_k) = x(1-x)$, где $x = x_1+x_2+\\dots+x_k$. Поскольку по неравенству о средних для двух чисел $x(1-x) \\le \\left(\\frac{x+(1-x)}{2}\\right)^2 = \\frac{1}{4}$, получаем неравенство\n$x_kx_{101-k} = a \\le \\frac{1}{4k(101-2k)}$. Осталось доказать, что $k(101 - 2k) \\ge 99$ при $k \\le 49$. Это неравенство можно переписать в виде $(k-1)(99-2k) \\ge 0$, и обе скобки в последней формуле неотрицательны.\n\nОсталось доказать, что $x_{50}x_{51} \\le \\frac{1}{396}$. Поскольку $x_{50} \\le x_{49} \\le x_{48} \\le \\dots \\le x_2 \\le x_1$, имеем\n$$\nx_{50} \\le \\frac{x_1 + x_2 + \\dots + x_{50}}{50} \\le \\frac{1}{50}\n$$\nи, аналогично,\n$$\nx_{51} \\le \\frac{x_1 + x_2 + \\dots + x_{51}}{51} \\le \\frac{1}{51}.\n$$\nСледовательно, $x_{50}x_{51} \\le \\frac{1}{50 \\cdot 51} < \\frac{1}{396}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of 10 men is either a knight (always telling the truth) or a liar (who always lies). Each of them thinks of an integer number. The first man told: \"My number is greater than 1\", the second told: \"My number is greater than 2\", ..., the tenth told: \"My number is greater than 10\". After that, talking in a different order, they told ten sentences: \"My number is less than 1\", \"My number is less than 2\", ..., \"My number is less than 10\" (each man told exactly one of those phrases). Determine the maximal possible number of knights among the 10 men.\n(O. Podlipsky)",
    "options": [],
    "answer": "8",
    "solution": "Докажем, что ни один из рыцарей не мог сказать ни одной из фраз «Моё число больше 9» и «Моё число больше 10». В самом деле, если бы это было возможно, то задуманное рыцарем целое число было бы не меньше 10. Но тогда он не мог сказать ни одной из фраз «Моё число меньше 1», «Моё число меньше 2», ..., «Моё число меньше 10». Значит, рыцарей было не больше восьми.\n\nПокажем, что рыцарей могло быть 8. Пусть первый рыцарь загадал число 2, второй — 3, ..., восьмой — 9, а лжецы загадали числа 5 и 6. Тогда $k$-ый рыцарь мог сказать фразы «Моё число больше $k$» и «Моё число меньше $k+2$», а лжецы могли сказать фразы: один — «Моё число больше 9» и «Моё число меньше 1», а другой — «Моё число больше 10» и «Моё число меньше 2».\n\n**Замечание.** Приведённый выше пример перестаёт быть верным, если лжецы задумывают числа вне отрезка $[1; 10]$, так как тогда некоторые их высказывания становятся верными.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $n$ for which there exist integers $a_1, a_2, \\dots, a_n$ such that the polynomial\n$$x^2 - 2(a_1 + a_2 + \\dots + a_n)^2 x + (a_1^4 + a_2^4 + \\dots + a_n^4 + 1)$$\nhas at least one integer root. (P. Kozlov)",
    "options": [],
    "answer": "6",
    "solution": "For $n = 6$, one can take $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = -1$; then the quadratic from the problem becomes $x^2 - 8x + 7$ and has two integer roots: $1$ and $7$. It remains to show that this is the smallest possible value of $n$.\n\nSuppose the numbers $a_1, a_2, \\dots, a_n$ satisfy the problem's condition; then the discriminant of the quadratic, divided by $4$, must be a perfect square. It equals\n$$\nd = (a_1 + a_2 + \\dots + a_n)^4 - (a_1^4 + a_2^4 + \\dots + a_n^4 + 1).\n$$\nThen the number $d$ is odd and is a square, so it leaves remainder $1$ when divided by $8$.\n\nRewrite the above equality as\n$$\nd + 1 + a_1^4 + a_2^4 + \\dots + a_n^4 = (a_1 + a_2 + \\dots + a_n)^4\n$$\nand consider it modulo $8$. It is easy to check that the fourth powers of integers give only remainders $0$ and $1$ modulo $8$, so the right side of the equation gives remainder $0$ or $1$. The left side is congruent to $1 + 1 + k$, where $k$ is the number of odd numbers among the $a_i$. Therefore, $n \\ge k \\ge 6$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39879,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle $\\Omega$ centered at $O$ is circumscribed about an acute-angled triangle $ABC$ with $AB < BC$. Let $H$ be the orthocenter of $ABC$. A point $D$ is chosen on the extension of $BO$ beyond $O$ so that $\\angle ADC = \\angle ABC$. A line through $H$ parallel to $BO$ meets the smaller arc $AC$ of $\\Omega$ at $E$. Prove that $BH = DE$. (A. Kuznetsov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пусть $P$ — вторая точка пересечения $BO$ с окружностью $\\Omega$ (см. рис. 15). Тогда $BP$ — диаметр $\\Omega$, и $\\angle BCP = 90^\\circ = \\angle BAP$. Значит, $CP \\parallel AN$ и $AP \\parallel CH$. Следовательно, четырёхугольник $ANCP$ — параллелограмм. Обозначим через $M$ точку пересечения его диагоналей. Она является серединой отрезков $PH$ и $AC$.\n![](attached_image_1.png)\nРис. 15\nПри симметрии относительно точки $M$ точка $A$ переходит в точку $C$, а точка $P$ — в точку $H$. Пусть при этой симметрии точка $E$ переходит в $E'$, а окружность $\\Omega$ — в $\\Omega'$. Тогда точки $A$, $H$, $E'$ и $C$ лежат на $\\Omega'$. Поскольку $\\angle ADC = \\angle ABC = 180^\\circ - \\angle AHC$, точка $D$ также лежит на $\\Omega'$.\nВ силу симметрии, $\\angle ECP = \\angle E'AH$, а также $PE' \\parallel HE$ — поэтому точка $E'$ лежит на прямой $PB$. Из вписанности четырёхугольников $AHE'D$ и $BEPC$ получаем, что $\\angle EBP = \\angle ECP = \\angle E'AH = \\angle E'DH$. Таким образом, $\\angle EBD = \\angle BDH$. Это означает, что трапеция $BHED$ — равнобокая, поэтому $BH = DE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39880,
    "subject": "Mathematics (Multi-modal)",
    "question": "100 distinct positive integers are arranged in a circle. Vasya divided each number by its clockwise neighbor with remainder, and he obtained only two distinct remainders. Petya divided each number by its counter-clockwise neighbor with remainder. Prove that Petya obtained 100 distinct remainders.\n(N. Agakhanov, S. Berlov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Пронумеруем числа по часовой стрелке $a_1, a_2, \\dots, a_{100}$ так, чтобы число $a_{100}$ было наименьшим. Тогда остаток от деления $a_{100}$ на $a_1$ будет равен $a = a_{100}$ (ибо $a_1 > a_{100}$), а остаток $b$ от деления $a_{99}$ на $a_{100}$ будет меньше, чем $a_{100}$. Значит, $a > b$ — единственные остатки, полученные Васей.\n\nПредположим, что $a_i < a_{i+1}$ при некотором $i < 100$. Тогда остаток от деления $a_i$ на $a_{i+1}$ равен $a_i$, что больше, чем $a$ (и тем более — чем $b$). Это невозможно. Значит, $a_1 > a_2 > \\dots > a_{100}$.\n\nИтак, Петя при делении $a_{i+1}$ на $a_i$ (при любом $i = 1, 2, \\dots, 99$) будет получать в остатке $a_{i+1}$, поскольку $a_{i+1} < a_i$. При делении же $a_1$ на $a_{100}$ он получит остаток $c$, меньший $a_{100}$. Значит, все его остатки $c < a_{100} < a_{99} < \\dots < a_2$ различны.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39881,
    "subject": "Mathematics (Multi-modal)",
    "question": "10,000 kids came to a camp; each has exactly 11 friends at the camp (friendship is always mutual). Each kid wears a T-shirt of one of the seven fixed colors, so that every two friends wear T-shirts of distinct colors. The guides requested that some kids (at least one) change their T-shirt colors (with some others among the same seven colors). A short poll showed that there are 100 kids which reject changing colors of their T-shirts. Prove that some remaining kids still can change their T-shirt colors so that the condition that every two friends wear distinct colors persists. (A. Magazinov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Перейдём к графу, вершины которого соответствуют детям, а рёбра — дружбам. Напомним, что раскраска вершин называется правильной, если цвета любых двух вершин, соединённые ребром, различны. Таким образом, граф правильно раскрашен в 7 цветов, в нём выделено 100 стабильных вершин, и требуется перекрасить часть остальных вершин так, чтобы раскраска осталась правильной.\n\nПредположим, что это невозможно. Пронумеруем цвета числами 1, 2, ..., 7. Рассмотрим любые два цвета $i < j$. Оставим в графе только вершины этих цветов и рёбра между ними; обозначим полученный граф через $G_{ij}$. Этот граф может распасться на несколько компонент связности; обозначим через $c_{ij}$ их количество. Если $c_{ij} > 100$, то в одной из компонент нет стабильных вершин; тогда можно изменить цвет каждой вершины в этой компоненте, заменив $i$ на $j$ и наоборот, и добиться требуемой альтернативной раскраски.\n\nЗначит, $c_{ij} \\le 100$ при всех $i$ и $j$. Заметим, что каждая компонента связности, содержащая $x$ вершин, содержит не менее $x - 1$ рёбер; значит, если в $G_{ij}$ есть $v_{ij}$ вершин, то количество рёбер в нём $e_{ij}$ не меньше, чем $v_{ij} - c_{ij}$, то есть\n$$\ne_{ij} \\ge v_{ij} - 100. \\quad (*)\n$$\nС другой стороны, нетрудно найти сумму $V$ всех чисел $v_{ij}$ и сумму $E$ всех чисел $e_{ij}$. Действительно, в исходном графе 10,000 вершин, и каждая из них участвует в 6 графах вида $G_{ij}$; поэтому $V = 6 \\cdot 10,000 = 60,000$. С другой стороны, в исходном графе $11 \\cdot 10,000/2 = 55,000$ рёбер, и каждое участвует ровно в одном графе $G_{ij}$; поэтому $E = 55,000$.\n\nНо, поскольку пар цветов всего $C_7^2 = 21$, неравенство (*) влечёт $E \\ge V - 21 \\cdot 100$, что не так для найденных значений. Значит, наше предположение было неверно, и требуемая перекраска возможна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39882,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes of an acute-angled triangle $ABC$ meet at $H$. The points $B_1$ and $C_1$ are chosen on the segments $BH$ and $CH$, respectively, so that $B_1C_1 \\parallel BC$. It appeared that the center of the circumcircle $\\omega$ of $\\triangle B_1HC_1$ lies on the line $BC$. Prove that $\\omega$ is tangent to the circumcircle $\\Gamma$ of $\\triangle ABC$.\n\n(A. Kuznetsov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Обозначим через $\\Gamma'$ окружность, описанную около треугольника $BHC$ (см. рис. 1). На касательной к $\\Gamma'$ в точке $H$ отметим точку $X$, лежащую внутри угла $BCH$. Тогда $\\angle BHX = \\angle BCH = \\angle B_1C_1H$ (последнее равенство следует из того, что $BC \\parallel B_1C_1$). Значит, окружность $\\omega$ касается прямой $HX$ и окружности $\\Gamma'$ в точке $H$.\n\n![](attached_image_1.png)\n\nРис. 1\n\nОбозначим через $H'$ точку, симметричную $H$ относительно прямой $BC$ (как известно, эта точка лежит на окружности $\\Gamma$). Итак, при симметрии относительно $BC$ окружность $\\Gamma'$ переходит в окружность $\\Gamma$, а окружность $\\omega$ — в себя, поскольку центр $\\omega$ лежит на прямой $BC$. Поскольку $\\omega$ касается $\\Gamma'$, она касается и $\\Gamma$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39883,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a kindergarden, a nurse took $n > 1$ congruent cardboard rectangles and gave them to $n$ kids, one per each. Each kid has cut its rectangle into congruent squares (the squares of different kids could be of different sizes). It turned out that the total number of the obtained squares is a prime number. Prove that all initial rectangles were in fact squares. (S. Berlov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Предположим противное: пусть прямоугольники имели размеры $k \\times l$, где $l < k$. Пусть $a_i$ — длина стороны квадрата у $i$-го ребёнка. Тогда каждая сторона прямоугольника составлена из нескольких отрезков длиной $a_i$, то есть $l = b_i a_i$ и $k = c_i a_i$, где $b_i$ и $c_i$ — натуральные числа. При этом у ребёнка получилось $b_i c_i$ квадратиков.\n\nЗаметим, что $1 < k/l = c_i/b_i$, то есть число $k/l$ рационально. Пусть $s/t$ — его несократимая запись; тогда $s > 1$, и $c_i$ делится на $s$ при всех $i$. Значит, и число квадратиков у $i$-го ребёнка делится на $s$. Тогда общее число квадратиков $Q$ также делится на $s > 1$, и притом $Q > s$ (поскольку количество детей больше 1). Значит, $Q$ — составное число; противоречие.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each face of a $1000 \\times 1000 \\times 1000$ cube is partitioned into $1000^2$ unit squares. Determine the maximal possible number of unit squares one can mark so that no two marked squares share a common side.",
    "options": [],
    "answer": "2998000",
    "solution": "Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into \"frames\" as shown in Fig. 2—500 frames around each of the eight vertices (one of the frames is shaded). Then, in the $k$-th frame, counting from the vertex, there are $S_k = 6k - 3$ cells. Since no two colored cells can be adjacent, in this frame there can be at most $\\left\\lfloor \\frac{S_k}{2} \\right\\rfloor = 3k - 2 = \\frac{S_k - 1}{2}$ colored cells. Summing over all 4000 frames and noting that their total area is $6 \\cdot 1000^2$, we get that the total number of colored cells does not exceed $\\frac{6 \\cdot 1000^2 - 4000}{2} = 3 \\cdot 10^6 - 2000$.\n\n![](attached_image_1.png)\n\nFig. 2\n\n![](attached_image_2.png)\n\nFig. 3\n\nIt remains to give an example showing that this many cells can indeed be colored. Call two opposite faces of the cube the top and bottom, and the others the side faces. On each side face, one can mark half the cells in a checkerboard pattern. After this, on the top and bottom faces, one can also color half the cells in all rows except the two extreme ones, leaving those empty—see Fig. 3, which shows two side faces and the top face. It is easy to see that with such coloring, each frame contains the maximum possible number of colored cells. (Instead of checking each frame, one can note that the entire surface is divided into $1 \\times 1000$ strips, four of which are empty, and in each of the others exactly half the cells are colored.)\n\n**Note.** There are other optimal examples. In particular, in the example given, the coloring of the top face can be changed as follows: divide the top face by diagonals into 4 triangles. In each of them, color the cells in a checkerboard pattern (so that the coloring of this triangle matches the coloring of the adjacent side face).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four consecutive integers greater than $100$ are given. Prove that the sum of some three of the given numbers can be represented as the product of three distinct positive integers greater than $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n$, $n+1$, $n+2$, $n+3$ be the given numbers. The sum of the three smallest of them is $3n + 3 = 3(n + 1)$, and the sum of the three largest numbers is $3(n + 2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of $2$ and $k$, where $k > 3$. Therefore, this sum can be represented as the product of three distinct natural numbers: $2$, $3$, and $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39886,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $D$ is chosen on the leg $AC$ of an isosceles triangle $ABC$ with base $BC$. A point $K$ is chosen on the smaller arc $CD$ of the circumcircle of the triangle $BCD$. The ray $CK$ meets the line through $A$ parallel to $BC$ at $T$. Let $M$ be the midpoint of $DT$. Prove that $\\angle AKT = \\angle CAM$.\n\nНа стороне $AC$ равнобедренного треугольника $ABC$ с основанием $BC$ выбрана точка $D$. На меньшей дуге $CD$ описанной окружности треугольника $BCD$ выбрана точка $K$. Луч $CK$ пересекает прямую, проходящую через $A$ параллельно $BC$, в точке $T$. Пусть $M$ — середина $DT$. Докажите, что $\\angle AKT = \\angle CAM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Продлим отрезок $AM$ на его длину за точку $M$, получим точку $N$ такую, что $ADNT$ — параллелограмм. Поскольку $\\angle ANT = \\angle CAM$, для решения задачи достаточно показать, что $\\angle AKT = \\angle ANT$, или что точки $A, T, N, K$ лежат на одной окружности (см. рис. 16).\n\n![](attached_image_1.png)\n\nПусть $ND$ пересекает $AB$ в точке $S$; тогда $DS \\parallel BC$, и $BSDC$ — равнобокая трапеция. Мы докажем, что точки $K$ и $N$ лежат на окружности $\\omega$, описанной около треугольника $AST$.\n\nИмеем $\\angle ATN = \\angle ADN = 180^\\circ - \\angle SDA = 180^\\circ - \\angle ASD$, значит, $N$ лежит на окружности $\\omega$. Из окружности, описанной около трапеции $BSDC$, имеем $\\angle SKT = \\angle SBC = 180^\\circ - \\angle SAT$, поэтому $K$ лежит на окружности $\\omega$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39887,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 16 coins — eight heavy ones of weight $11$ g each, and eight light ones of weight $10$ g each, but it is unknown which coin is of which type. One of the coins is commemorative. Determine whether the commemorative coin is light or heavy, by performing three weighings on a two-pan scales. (K. Knop)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the commemorative coin by $Y$. Set aside two non-commemorative coins $A$ and $B$, and distribute the remaining $14$ coins into two groups of $7$ each so that $Y$ is placed on the left pan. We will call a coin \"left\" or \"right\" if it was placed on the left or right pan, respectively, in this weighing.\n\nCase $(=)$: Suppose the pans are balanced.\nIn this case, either each pan contains $3$ heavy coins (and then $A$ and $B$ are both heavy), or $4$ (then $A$ and $B$ are light). In any case, both set-aside coins are of the same type. Now, take $Y$ and another coin $C$ from the left pan and compare this pair with the pair $A$ and $B$.\n\nSubcase $(=, =)$: Suppose again the scales are balanced.\nThen all $4$ coins $A$, $B$, $C$, $Y$ have the same weight. Compare them with any other four left coins. If the scales are balanced, then all $8$ coins in the weighing are of the same weight. Then among the left coins there were $6$ coins of that weight; this is impossible. Therefore, one pan will outweigh the other, and we will find out whether $A$, $B$, $C$, and $Y$ are heavy or light.\n\nSubcase $(=, <)$: The pan containing $Y$ in the second weighing is lighter.\nThen there cannot be two heavy coins on this pan. Comparing these two coins with each other, if they differ, we immediately find out the weight of $Y$, and if they are equal, we can conclude that both coins on this pan are light.\n\nSubcase $(=, >)$, when the pan with $Y$ is heavier, is analogous.\n\nCase $(<)$: Suppose the left pan in the first weighing is lighter.\nThen among the left coins there are at most three heavy coins. Comparing $Y$ with some left coin $C$, we either find out the weight of $Y$ (if they differ), or find two identical coins ($Y$ and $C$). Comparing this pair with another pair of left coins, again, we find out the weight of $Y$ if they differ. If they are equal, we find $4$ left coins of the same weight, one of which is $Y$. As already noted, they can only be light.\n\nCase $(>)$, when in the first weighing the pan with $Y$ is heavier, is analogous to the previous one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several cardboard rectangles, possibly of different sizes, are put onto a rectangular table; the sides of rectangles are parallel to those of the table. The rectangles may overlap, but there are no two rectangles with four common vertices. May it happen that each point which is a vertex of some rectangle is in fact a vertex of exactly three rectangles?\n(I. Bogdanov)",
    "options": [],
    "answer": "Yes; such configurations exist.",
    "solution": "![](attached_image_1.png)\nРис. 4\n\nНа рис. 4 показано, как можно положить три пары прямоугольников так, чтобы для каждой пары все точки $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ были вершинами ровно по разу. Одинаковыми точками отмечены вершины одного из прямоугольников пары.\n\n**Замечание.** Существует много других примеров; один из них указан на рис. 5.\n\n![](attached_image_2.png)\nРис. 5",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. A point $D$ is chosen on the external angle bisector of angle $ABC$ lying inside the angle $ABC$. Assume that $\\angle BCD = 60^\\circ$ and $CD = 2AB$. Let $M$ be the midpoint of $BD$. Prove that the triangle $AMC$ is isosceles.\n(A. Kuznetsov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Первое решение.** Обозначим $\\angle CBM = \\alpha$. Поскольку $BM$ — внешняя биссектриса угла $ABC$, то $\\angle ABM = 180^\\circ - \\alpha$ и $\\alpha < 90^\\circ$ (см. рис. 6).\n\nОпустим из точки $D$ перпендикуляр $DH$ на прямую $BC$. Так как в треугольнике $BCD$ углы при вершинах $B$ и $C$ острые, точка $H$ лежит на отрезке $BC$. Поскольку $DCH$ — прямоугольный треугольник с углом $60^\\circ$, имеем $CH = \\frac{CD}{2} = AB$.\n\nТреугольник $BHD$ — прямоугольный, $HM$ — его медиана, проведённая к гипотенузе, поэтому $HM = \\frac{BD}{2} = BM$. Следовательно, $\\angle MHB = \\angle HBM = \\alpha$, откуда мы получаем, что $\\angle MHC = 180^\\circ - \\alpha = \\angle ABM$. Поскольку также $AB = CH$ и $BM = MH$, треугольники $ABM$ и $CHM$ равны. Таким образом, $AM = MC$, что и требовалось доказать.\n\n\n**Второе решение.** Обозначим через $A'$ точку, симметричную $A$ относительно прямой $BD$ (см. рис. 7). Поскольку $BD$ — внешняя биссектриса угла $ABC$, точка $A'$ лежит на продолжении отрезка $CB$ за точку $B$. Пусть $K$ — середина отрезка $BC$. Тогда $MK$ — средняя линия треугольника $BCD$, поэтому $MK \\parallel CD$ и $\\angle MKB = \\angle BCD = 60^\\circ$, а также $MK = \\frac{CD}{2} = AB = A'B$.\n\nНа отрезке $A'K$ отметим точку $T$ так, что $MK = KT$. Тогда из того, что $TK = MK = A'B$, следует $A'T = BK = KC$. Заметим, что треугольник $MKT$ равносторонний, так как $MK = KT$ и $\\angle MKT = 60^\\circ$. Тогда $MT = MK$ и $\\angle MTA' = 120^\\circ = \\angle MKC$, а, как было доказано ранее, $A'T = KC$. Значит, треугольники $МКС$ и $МТА'$ равны, поэтому $CM = A'M = AM$.\n\n**Замечание.** Вместо того, чтобы отмечать точку $T$, можно записать теорему косинусов для треугольников $A'KM$ и $MSC$.\n\nВ них углы при вершине $K$ равны $60^\\circ$ и $120^\\circ$ соответственно. Пусть $MK = A'B = a$, $BK = KC = b$. Тогда\n$$\nA'M = \\sqrt{(a+b)^2 + a^2 - a(a+b)} = \\sqrt{a^2 + ab + b^2} = CM.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39890,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex $n$-gon is drawn on the blackboard, $n \\ge 4$. Paint each its vertex black or white. Say that a diagonal is *multicolor* if its endpoints are of different colors. We say that the painting is *good* if one can partition the $n$-gon into triangles by drawing some multicolor diagonals sharing no internal points. Find the number of good paintings.\n(S. Berlov)",
    "options": [],
    "answer": "n(n-1)",
    "solution": "Сразу же заметим, что раскраска всех вершин в один цвет хорошей не является; такие раскраски в дальнейшем решении не рассматриваются.\nНазовём сторону многоугольника разноцветной, если её концы окрашены в разные цвета (то есть расширим определение разноцветности на стороны). Назовём раскраску вершин упорядоченной, если все чёрные вершины на границе многоугольника идут подряд (иначе говоря, у многоугольника есть ровно две разноцветных стороны).\n**Лемма.** Раскраска вершин $n$-угольника (при $n \\ge 3$) является хорошей тогда и только тогда, когда она упорядочена.\n**Доказательство.** Индукция по $n$. При $n = 3$ доказывать нечего (напомним, что мы не рассматриваем одноцветные раскраски). Докажем теперь переход индукции. Пусть утверждение доказано для всех $m$ таких, что $3 \\le m < n$, где $n \\ge 4$.\n\nПредположим, что раскраска является хорошей. Разобьём многоугольник на треугольники непересекающимися разноцветными диагоналями; рассмотрим одну из этих диагоналей $AB$. Она делит $n$-угольник на два многоугольника $P_1$ и $P_2$ с меньшим количеством сторон, причём каждый из них раскрашен хорошо — а значит, по предположению индукции, и упорядоченно. Пусть $A$ — чёрный конец диагонали, а $B$ — белый. Все чёрные вершины в $P_1$ — это несколько последовательных вершин, начиная с $A$ (но не включая $B$). Аналогично с чёрными вершинами в $P_2$. Но тогда эти два блока чёрных вершин в объединении дают один связный блок в исходном многоугольнике, то есть раскраска вершин $n$-угольника также является упорядоченной.\n\nПусть теперь раскраска является упорядоченной. Нетрудно видеть, что тогда в многоугольнике есть разноцветная диагональ. Она делит многоугольник на два меньших, при этом, очевидно, каждый из них также раскрашен упорядоченно. По предположению индукции, каждый из них раскрашен хорошо, а значит, и исходный $n$-угольник — тоже. $\\square$\n\nВвиду леммы, осталось лишь посчитать число упорядоченных раскрасок $n$-угольника. Для каждого возможного количества чёрных вершин (от $1$ до $n-1$) можно $n$ способами выбрать расположение их блока среди всех $n$ вершин, то есть число способов равно $n(n-1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petr and Basil play a game on an initially empty checkered table $100 \\times 100$, making moves in turn. Petr starts. During his turn, the player writes any (capital) letter of the English alphabet into some empty cell (exactly one letter can be written in each cell). When all the cells are filled, Petr is declared the winner if there exist four consecutive cells horizontally in which the word «PETR» is written from left to right, or, if there exist four consecutive cells vertically in which the word «PETR» is written from top to bottom. Determine if Petr can win (regardless of Basil's actions).",
    "options": [],
    "answer": "No",
    "solution": "He won't be able to.\n\nLet's describe Vasya's winning strategy. Let Vasya always write the letter \"Y\" in a cell according to the following conditions; if the specified cell doesn't exist or is already occupied, or if Petya writes any letter other than \"P\" \"E\" \"T\" \"R\" then let Vasya write \"Y\" in any free cell.\n\nIf Petya writes the letter $P$ in some cell, then Vasya writes $Y$ in the cell to its right; if Petya writes the letter $E$ then Vasya writes $Y$ in the cell to its left; if Petya writes the letter $T$ then Vasya writes $Y$ in the cell below it; if Petya writes the letter $R$ then Vasya writes $Y$ in the cell above it.\n\nFrom the first two conditions it follows that the sequence \"PE\" read from left to right cannot appear in two horizontally adjacent cells. Indeed, suppose a horizontal \"PE\" appears; then after the first of these two letters appears, Vasya, according to the described strategy, will immediately occupy the position of the second letter — a contradiction. Similarly, the sequence \"TR\" read from top to bottom cannot appear in two vertically adjacent cells.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point lying inside triangle $ABC$. Let $Q$ be a point on the segment $AB$, and let $R$ be a point on the segment $AC$ such that both circles ($BPQ$) and ($CPR$) are tangent to line $AP$. Through $B$ and $C$ we draw the lines passing through the center of the circle ($BPC$), and through $Q$ and $R$ we draw the lines passing through the center of the circle ($PQR$). Prove that there exist a circle tangent to the four drawn lines.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB \\cdot AQ = AP^2 = AC \\cdot AR$, quadrilateral $BCRQ$ is cyclic. Let $O$ be the center of circle ($BCRQ$). Denote by $O_1$ and $O_2$ the centers of circles ($BPC$) and ($QPR$). We will show that lines $BO_1$, $CO_1$, $QO_2$, $RO_2$ are equidistant from $O$. Since $OB = OC = OQ = OR$, it suffices to establish the equality of (directed) angles $\\angle OCO_1 = \\angle O_1BO = \\angle OQO_2 = \\angle O_2RO$. Here the first and last equalities are obvious from symmetry about the perpendicular bisectors of $BC$ and $QR$.\n\n![](attached_image_1.png)\nРис. 5\n\nIt remains to prove the equality $\\angle O_1BO = \\angle OQO_2$ (*). By angle chasing we obtain $\\angle OQO_2 = \\angle OQR - \\angle O_2QR = (90^\\circ - \\angle RCQ) - (90^\\circ - \\angle RPQ) = \\angle RPQ - \\angle RCQ$. Similarly $\\angle O_1BO = \\angle BPC - \\angle BQC$. Thus, (*) is equivalent to the equality $\\angle RPQ - \\angle RCQ = \\angle BPC - \\angle BQC$ or $\\angle BQC - \\angle RCQ = \\angle BPC - \\angle RPQ$ (**). From the tangency of circles ($BPQ$) and ($CPR$) it follows that $\\angle RPQ = \\angle RCP + \\angle PBQ$, which equals (from the sum of angles in quadrilateral $BPCA$) $\\angle BPC - \\angle BAC$. Therefore, (**), transforms into $\\angle BQC - \\angle RCQ = \\angle BAC$, which holds true. The problem is solved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists an *even* positive integer $a$ such that $(a-1)(a^2-1)\\dots(a^n-1)$ is a perfect square.",
    "options": [],
    "answer": "1 and 2",
    "solution": "For $n=1$ and $n=2$.\n\nFor $n=1$, any even number $a$ of the form $m^2 + 1$ works, for example, $a = 2$.\n\nFor $n = 2$, any even number $a$ of the form $m^2 - 1$ works, for example, $a = 8$.\n\nAssume that for $n = 3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Since $a^2+a+1 = a(a+1)+1$, the numbers $a+1$ and $a^2+a+1$ are coprime. As $a+1$ is odd, the numbers $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a+1$ modulo $3$ can only be $0$ or $1$, and thus $a-1$ is not divisible by $3$. Hence,\n$$\n\\begin{aligned}\n\\gcd(a-1, a^2+a+1) &= \\gcd(a-1, (a+2)(a-1)+3) = \\\\\n&= \\gcd(a-1, 3) = 1,\n\\end{aligned}\n$$\nmeaning that both $a-1$ and $a^2+a+1$ must be perfect squares. However, the latter cannot be a square, since $a^2 < a^2+a+1 < (a+1)^2$. This is a contradiction.\n\nIt remains to prove that no such $a$ exists for $n \\ge 4$. Suppose such an $a$ exists. Take a natural number $k \\ge 2$ such that $2^k \\le n < 2^{k+1}$. Since $a^{2^k} - 1 = (a^{2^{k-1}} - 1)(a^{2^{k-1}} + 1)$, the number $(a-1)(a^2-1)\\dots(a^n-1)$ can be expressed as the product of $a^{2^{k-1}} + 1$ and several other factors of the form $a^m - 1$, where $1 \\le m \\le n$ and $m \\ne 2^k$.\n\nWe will show that the factor $a^{2^{k-1}} + 1$ is coprime with all other factors in this decomposition. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1)$ is divisible by $d$. But $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k,m)} - 1$. Since $m \\ne 2^k$ and $m \\le n < 2^{k+1}$, the number $m$ cannot be divisible by $2^k$. Thus, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Therefore, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k} - 1, a^m - 1)$, and hence also divides $d$. Because $a$ is even, the numbers $a^{2^{k-1}-1}$ and $a^{2^{k-1}} + 1$ have no common divisors other than $1$, so $d = 1$, as required.\n\nThe factor $a^{2^{k-1}} + 1$ is coprime with all other factors in the product, which is a perfect square, so it must itself be a perfect square. Then $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are perfect squares differing by $1$, which is impossible. Therefore, our assumption is false, and no such $a$ exists for $n \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39894,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a plane, $10^6$ points are marked no three of which are collinear; and all segments between these points are drawn. On each drawn segment Grisha placed a real number whose absolute value does not exceed $1$. Then for each six marked points, he calculated the sum of the numbers on all $15$ segments connecting these points. It turned out that the absolute value of each such sum is not less than $C$, and among such sums there exist both positive and negative sums. Find the greatest possible $C$.\n(Г. Р. Челноков)",
    "options": [],
    "answer": "15/4",
    "solution": "Consider a graph whose vertices are the marked points and edges are the drawn segments.\n\n*Upper bound.* We prove $C \\le \\frac{15}{4}$. The condition states that in our complete graph there exist both $6$-vertex subsets where the sum of edge weights is positive, and $6$-vertex subsets where it's negative. Therefore, there must exist two $6$-vertex subsets differing by exactly one vertex, such that one has positive sum and the other negative. Indeed, start with a positive $6$-vertex subset and transform it into a negative one by changing vertices one at a time - at some step the sign flips, giving the required pair.\n\nNow consider the complete subgraph on a $7$-vertex set $S$ - the union of this pair. Examine all seven $6$-vertex subsets obtained by removing one vertex from $S$. Suppose $k$ of them have negative sums (obtained by removing vertices $A_1, \\dots, A_k$, called *A-vertices*), and $7-k$ have positive sums (obtained by removing vertices $B_1, \\dots, B_{7-k}$, called *B-vertices*). We classify edges as:\n- *AA-edges* between two A-vertices\n- *BB-edges* between two B-vertices\n- *AB-edges* between an A-vertex and B-vertex\n\nWe modify the edge weights on $S$ by replacing:\n- All AA-edge weights with their average $x$\n- All AB-edge weights with their average $y$\n- All BB-edge weights with their average $z$\n\nClearly $|x|, |y|, |z| \\le 1$ since original weights were in $[-1, 1]$.\n\n**Lemma.** The subgraph $S$ with modified weights satisfies the condition with the same constant $C$.\n**Proof.** For any edge type, the number of $6$-vertex subsets containing both its endpoints is constant. Similarly, each $6$-vertex subset contains a fixed number of edges of each type. Therefore:\n- The sum $\\Sigma_A$ in any A-$6$-vertex subset is the average of original negative sums, so $\\Sigma_A \\le -C$\n- The sum $\\Sigma_B$ in any B-$6$-vertex subset is the average of original positive sums, so $\\Sigma_B \\ge C$\n\nWe analyze cases based on $k$:\n\n*Case $k = 1$:* One A-vertex, six B-vertices. The inequalities are:\n$$15z \\le -C$$\n(A-$6$-vertex subset)\n$$5y + 10z \\ge C$$\n(B-$6$-vertex subset)\nEliminating $z$ gives $15y \\ge 5C \\Rightarrow C \\le 3$\n\n*Case $k = 2$:* Two A-vertices, five B-vertices. Inequalities:\n$$5y + 10z \\le -C$$\n(A-$6$-vertex)\n$$x + 8y + 6z \\ge C$$\n(B-$6$-vertex)\nEliminating $z$ gives $8C \\le 5x + 25y \\le 30 \\Rightarrow C \\le \\frac{15}{4}$\n\n*Case $k = 3$:* Three A-vertices, four B-vertices. Inequalities:\n$$x + 8y + 6z \\le -C$$\n(A-$6$-vertex)\n$$3x + 9y + 3z \\ge C$$\n(B-$6$-vertex)\nEliminating $y$ gives $17C \\le 15x - 30z \\le 45 \\Rightarrow C \\le \\frac{45}{17} < \\frac{15}{4}$\n\nCases $k \\ge 4$ reduce to $k \\le 3$ by symmetry (multiplying all weights by $-1$). Thus $C \\le \\frac{15}{4}$ is proved.\n\n*Construction.* For any number of A-vertices from $2$ to $999,995$ (and remaining as B-vertices), set:\n- All BB-edge weights to $-\\frac{7}{8}$\n- All other edge weights to $1$\n\nThen any $6$-vertex subset with $\\ge 5$ B-vertices has sum $\\le -\\frac{15}{4}$, while others have sum $\\ge \\frac{15}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Positive integers $1, 2, \\dots, n$ are written in a row in some order. For any two neighboring numbers their GCD is written on the paper. Find the greatest possible number of distinct numbers among all $n-1$ numbers written on the paper.",
    "options": [],
    "answer": "⌊n/2⌋",
    "solution": "**Ответ.** $\\lfloor n/2 \\rfloor$.\n\n**Решение.** *Upper bound.* Assume one of the written numbers is greater than $\\lfloor n/2 \\rfloor$, say, $\\text{GCD}(a, b) = d > \\lfloor n/2 \\rfloor$. Then the larger of the numbers $a, b$ must be at least $2d$, which exceeds $n$ - a contradiction. Therefore, each written GCD cannot exceed $\\lfloor n/2 \\rfloor$, and thus the number of distinct GCDs cannot be greater than $\\lfloor n/2 \\rfloor$.\n\n*Example.* Let's partition all numbers from $1$ to $n$ into chains of the form $a, 2a, 4a, 8a, \\dots, 2^k a$, where $a$ is an odd number not exceeding $n$. Write these chains consecutively in a row. Then for any natural number $d \\le \\lfloor n/2 \\rfloor$, there exists a chain containing $d$ where the number following $d$ is $2d$. We see that every natural number $d \\le \\lfloor n/2 \\rfloor$ will appear on the sheet.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least $k$ satisfying the following condition: for any polynomial $f(x)$ of degree $100$ with real coefficients there exists a polynomial $g(x)$ of degree not greater than $k$ with real coefficients such that the graphs $y = f(x)$ and $y = g(x)$ have exactly $100$ common points.",
    "options": [],
    "answer": "98",
    "solution": "**Answer.** $98$.\n\nSet $n = 100$.\n\n1. We will show how to construct a polynomial $g$ of degree at most $n-2$ for a given polynomial $f$ of degree $n$. Multiplying $f$ by a nonzero constant doesn't change the condition (we can multiply $g$ by the same constant), so we assume the leading coefficient of $f$ is $1$, i.e., $f(x) = x^n + a_1 x^{n-1} + \\dots + a_n$.\n\nTake any set of $n$ distinct numbers $x_1, \\dots, x_n$ whose sum is $-a_1$, and define $h(x) = (x - x_1) \\dots (x - x_n)$, $g(x) = f(x) - h(x)$, so $h = f - g$.\n\nWe see that $f$ and $h$ have identical coefficients for $x^n$ and $x^{n-1}$, so the degree of $g$ doesn't exceed $n-2$. On the other hand, the $x$-coordinates of intersection points of $f$ and $g$ are exactly the roots of $h$, of which there are exactly $n$.\n\n2. We'll show that $k \\le n-3$ doesn't work. Let $f(x) = x^n$, and $g(x)$ be a polynomial of degree at most $n-3$. Suppose the graphs of $f$ and $g$ intersect at $n$ points with $x$-coordinates $x_1, \\dots, x_n$. Then the polynomial $f-g$ of degree $n$ has $n$ real roots $x_1, \\dots, x_n$.\n\nHowever, $f-g$ has zero coefficients for $x^{n-1}$ and $x^{n-2}$. By Vieta's formulas, the sum $s_1 = x_1 + \\dots + x_n$ and the sum of pairwise products $s_2 = x_1 x_2 + x_1 x_3 + \\dots + x_{n-1} x_n$ both equal $0$.\n\nThus $x_1^2 + \\dots + x_n^2 = s_1^2 - 2s_2 = 0$, which implies $x_1 = x_2 = \\dots = x_n = 0$, contradicting the distinctness of the $x_i$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 39897,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sport competition consists of 25 contests; in each contest there is exactly one winner who receives a gold medal. 25 athletes participate in this competition, each of them participates in all 25 contests. There are 25 sports experts. Each of 25 experts is to make his *prediction* how many gold medals each athlete will receive; and in his prediction the numbers of medals must be non-negative integers whose sum equals 25. An expert is considered *competent* if he correctly guesses the number of gold medals of at least one athlete. Find the greatest $k$ such that the experts can make their predictions so that at least $k$ of them will be considered competent, regardless of the results of the competition.",
    "options": [],
    "answer": "24",
    "solution": "**Ответ.** 24.\n\n**Решение.** *Upper bound.* We will show that $k \\le 24$, i.e., that any expert could be incompetent. If this expert believes that all athletes will receive one medal each, we can refute them with the result $(25, 0, 0, ..., 0)$. Otherwise, the expert believes that several (at least one) athletes will receive 0 medals. Then we can distribute all medals among these athletes so that each of them receives at least one medal. In this case, the expert won't guess any medal count correctly.\n\n*Example.* Let one expert's prediction be $(1, 1, 1, ..., 1)$, and the predictions of others be $(1, 0, ..., 0, 24)$, $(0, 1, ..., 0, 24)$, ..., $(0, 0, ..., 1, 24)$ (with 24 in the last position and one more 1).\n\nIf the first expert is incompetent, then the actual result must contain at least three zeros. Otherwise, at least 23 positions would have at least 2 medals, making the total medal count at least $23 \\cdot 2 > 30$ - a contradiction. But then all other experts must be competent.\n\nNow suppose two experts other than the first are incompetent. Then in two positions their predictions are 0 and 1 medals, meaning in the actual result these positions have at least 2 medals. Moreover, in 22 other positions both experts predicted zeros, so in reality these positions have at least 1 medal. Thus the total medal count would be at least $2 \\cdot 2 + 22 \\cdot 1 > 25$ - a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39898,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board 777 pairwise distinct *complex* numbers are written. It turned out that there are exactly 760 ways to choose a pair of numbers $a$ and $b$ written on the board so that\n$$\na^2 + b^2 + 1 = 2ab.\n$$\n\n(Here pairs considered unordered, i.e., $(a, b)$ and $(b, a)$ is the same pair.) Prove that one can choose numbers $c$ and $d$ written on the board such that\n$$\nc^2 + d^2 + 2025 = 2cd.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the condition $a^2 + b^2 + 1 = 2ab$ is equivalent to $(a-b)^2 = -1$ or $a-b = \\pm i$. Consider a graph whose vertices are the numbers written on the board, with an edge connecting two numbers if they differ by $i$. According to the problem's condition, this graph has exactly 760 edges.\n\nEach connected component of this graph forms a path consisting of numbers of the form $z, z+i, z+2i, \\dots, z+(n-1)i$. Suppose the graph has $k$ connected components. Then it contains $777-k$ edges, so $k=17$.\n\nSince $17 \\cdot 45 = 765 < 777$, at least one connected component must have at least 46 vertices. Therefore, there exist two numbers in this component, say $c$ and $d = c + 45i$. Then $(c-d)^2 = 45^2 \\cdot i^2 = -2025$, which implies $c^2 + d^2 + 2025 = 2cd$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39899,
    "subject": "Mathematics (Multi-modal)",
    "question": "A right prism $ABC A_1 B_1 C_1$ is given. It is known that triangles $A_1 BC$, $AB_1 C$, $ABC_1$, and $ABC$ are acute-angled. Prove that the orthocenters of these triangles, and the centroid of $ABC$ lie on a sphere.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ and $H$ denote the centroid and orthocenter of triangle $ABC$ respectively, and let $T$ be a point such that $3\\overrightarrow{MT} = \\overrightarrow{AA_1} = \\overrightarrow{BB_1} = \\overrightarrow{CC_1}$. Let $\\omega$ be the sphere with diameter $HT$. Since the line $MT$ is perpendicular to the plane $ABC$, the point $M$ lies on $\\omega$. We will show that the orthocenter $H_1$ of triangle $A_1 BC$ also lies on $\\omega$ (the proof for the other two triangles is analogous).\nLet $N$ be the midpoint of segment $BC$. Since $NA = 3NM$, the point $T$ lies on segment $A_1 N$, and consequently lies in the\n\n![](attached_image_1.png)\n\nplane $A_1BC$. Let $AA'$ be the altitude of triangle $ABC$. As $AA_1$ is perpendicular to the plane $ABC$, we have $\\angle A_1AA' = 90^\\circ$, and by the Three Perpendiculars Theorem, $A'A_1 \\perp BC$, meaning $H_1$ lies on segment $A_1A'$.\nSince the reflection of $H$ over $BC$ lies on the circumcircle of $ABC$, we have $A'H \\cdot A'A = A'B \\cdot A'C$. Applying the same reasoning to triangle $A_1BC$ gives $A'H_1 \\cdot A'A_1 = A'B \\cdot A'C = A'H \\cdot A'A$. Therefore, quadrilateral $AHH_1A_1$ is cyclic, so $\\angle HH_1A_1 = 180^\\circ - \\angle A_1AH = 90^\\circ$.\nFurthermore, since $HA' \\perp BC$ and $H_1A' \\perp BC$, applying the Three Perpendiculars Theorem again shows that line $HH_1$ is perpendicular to the plane $A_1BC$. Thus, $\\angle HH_1T = 90^\\circ$, proving that $H_1$ lies on sphere $\\omega$, as required.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 39900,
    "subject": "Mathematics (Multi-modal)",
    "question": "A pair of polynomials $F(x, y)$, $G(x, y)$ with integer coefficients is called *important*, if the following condition holds: if for some integers $a, b, c, d$ both $F(a, b) - F(c, d)$ and $G(a, b) - G(c, d)$ are divisible by 100, then both $a - c$ and $b - d$ are divisible by 100. Determine if there exist an important pair of polynomials $P(x, y)$, $Q(x, y)$ such that the pair $P(x, y) - xy$, $Q(x, y) + xy$ is also important.",
    "options": [],
    "answer": "Does not exist.",
    "solution": "**Ответ.** Does not exist.\n\n**Решение.** Let $F$ and $G$ be an important pair of polynomials. Consider pairs of residues modulo 100 of numbers $F(a,b)$ and $G(a,b)$, where $a,b$ range over all integer pairs from 0 to 99. According to the problem's condition, all such residue pairs are distinct. Since there are $100^2$ possible number pairs, each residue pair modulo 100 occurs exactly once. Therefore, all 4 possible parity combinations of $F(a,b)$ and $G(a,b)$ are achieved.\nSince the parity of a polynomial's value with integer coefficients at point $(a,b)$ depends only on the parity of $a$ and $b$, we conclude that the value pairs $(F(0,0); G(0,0))$, $(F(1,0); G(1,0))$, $(F(0,1); G(0,1))$, and $(F(1,1); G(1,1))$ must give all four possible parity combinations.\nHowever, observe that for both polynomial pairs $F = P, G = Q$ and $F(x,y) = P(x,y) - xy, G(x,y) = Q(x,y) + xy$, the first three parity pairs are identical, while the fourth pair differs. Consequently, both such polynomial pairs cannot be important.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. A cube $(2N+1) \\times (2N+1) \\times (2N+1)$ is made of $(2N+1)^3$ unit cubes; each unit cube is either black or white. It turned out that among any 8 unit cubes which form a cube $2 \\times 2 \\times 2$ the number of black unit cubes is not greater than 4. Find the greatest possible total number of black unit cubes.",
    "options": [],
    "answer": "(N+1)^2(4N+1)",
    "solution": "**Ответ.** $(N+1)^2(4N+1)$.\n\nLet $k = (N+1)^2(4N+1)$. Introduce a coordinate system where all vertices of unit cubes have integer coordinates from $0$ to $2N+1$.\n\nFirst, we present an example showing that the number of black cubes can indeed be $k$. For each cube, consider its vertex closest to the origin (its coordinates range from $0$ to $2N$). Let the cube be black if at least two coordinates of this vertex are even, and white otherwise. Clearly, in any $2 \\times 2 \\times 2$ cube there will be exactly $4$ black and $4$ white cubes. The number of black cubes with all three corresponding coordinates even is $(N+1)^3$, and the number with exactly two even coordinates is $3(N+1)^2N$, making the total number of black cubes equal to $k$.\n\nNow we prove this example is optimal. Consider a cube partitioned into black and white unit cubes satisfying the problem's conditions. We'll call a cube *dark* or *light* if it's respectively black or white in the above example.\n\nFor each point $(a, b, c)$ in the large cube, define its $x$-, $y$-, and $z$-rank as $r_x = \\min(a, 2N + 1 - a)$, $r_y = \\min(b, 2N + 1 - b)$, and $r_z = \\min(c, 2N+1-c)$. The *rank* is $r = \\min(r_x, r_y, r_z)$ - the distance to the nearest face of the large cube.\n\nMark all unit cube vertices with odd ranks. For each marked vertex, consider the difference between the number of black and white cubes meeting at it. Since these vertices are centers of $2 \\times 2 \\times 2$ cubes, this difference is non-positive, making the total sum $\\Sigma$ of such differences non-positive.\n\nDefine the *multiplicity* of a unit cube as the number of its marked vertices. Then $\\Sigma$ equals the difference between the sum of multiplicities of black cubes and white cubes. We need to show that if this difference is non-positive, the number $\\ell$ of black cubes doesn't exceed $k$.\n\nLet $r_x \\le r_y \\le r_z$ be the ranks of a cube's center. Then:\n- If $r_x < r_y$, its multiplicity is $4$\n- If $r_x = r_y = \\frac{1}{2} + d$ with even $d$, multiplicity is $< 4$ and it's dark\n- If $r_x = r_y = \\frac{1}{2} + d$ with odd $d$, multiplicity is $> 4$ and it's light\n\nThus, dark cubes have multiplicity $\\le 4$, light cubes $\\ge 4$.\n\nLet $s_1 \\le \\dots \\le s_{(2N+1)^3}$ be the multiplicities in order. From our example where $\\Sigma = 0$, we have $s_1 + \\dots + s_k - s_{k+1} - \\dots - s_{(2N+1)^3} = 0$. If $\\ell > k$:\n$$\n\\begin{aligned}\n0 \\ge \\Sigma &\\ge s_1 + \\dots + s_\\ell - s_{\\ell+1} - \\dots - s_{(2N+1)^3} > \\\\\n&> s_1 + \\dots + s_k - s_{k+1} - \\dots - s_{(2N+1)^3} = 0,\n\\end{aligned}\n$$\nsince $s_{k+1} \\ge 4$. This contradiction proves $\\ell \\le k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially $100$ numbers $1$ are arranged on a circle. Petya and Vasya play the following game, taking turns; each boy performs $10^{10}$ moves; Petya starts. By his move, Petya chooses $9$ consecutive numbers and decreases each of them by $2$. By his move, Vasya chooses $10$ consecutive numbers and increases each of them by $1$. Prove that Vasya can play so that after each his move among the numbers on the circle there will be at least $5$ positive numbers (regardless of Petya's moves).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the numbers written in a circle be denoted as $a_1, a_2, \\dots, a_{100}$. Vasya will track only ten numbers, which he will pair as follows: $(a_9, a_{18})$, $(a_{27}, a_{36})$, $\\dots$, $(a_{90}, a_{99})$. In one move, Petya can decrease at most one of these $10$ numbers. If Petya decreases one number in a pair $(a_i, a_{i+1})$, Vasya will respond by adding $1$ to each of $a_i, a_{i+1}, \\dots, a_{i+9}$. If Petya doesn't decrease any of these $10$ numbers, Vasya will make any allowed move.\n\nThus, after each pair of moves (Petya's and Vasya's), the sum of numbers in each of Vasya's five pairs will not decrease. Since initially all five pair sums are positive, after each of Vasya's moves the sum in each pair will remain positive, meaning each pair will contain at least one positive number. Therefore, after any of Vasya's moves there will be at least $5$ positive numbers, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral having no two parallel sides, inscribed into the circle $\\Omega$. Let $\\omega_a, \\omega_b, \\omega_c, \\omega_d$ be circles inscribed into triangles $DAB, ABC, BCD, CDA$, respectively. Let us draw common external tangents $t_1, t_2, t_3, t_4$ to pairs of circles $\\omega_a$ and $\\omega_b$, $\\omega_b$ and $\\omega_c$, $\\omega_c$ and $\\omega_d$, $\\omega_d$ and $\\omega_a$, respectively, so that lines $t_1, t_2, t_3, t_4$ do not contain the sides of $ABCD$. The quadrilateral whose consecutive sides lie on $t_1, t_2, t_3, t_4$ (in this order) is inscribed into the circle $\\Gamma$. Prove that the three lines connecting centers of $\\omega_a$ and $\\omega_c$, $\\omega_b$ and $\\omega_d$, $\\Omega$ and $\\Gamma$, are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, let rays $AB$ and $DC$ intersect at point $P$, and rays $AD$ and $BC$ intersect at point $Q$. Denote the center of circle $\\omega_a$ as $I_a$, and define points $I_b, I_c, I_d$ similarly. Let the quadrilateral formed by the four tangent lines be $A'B'C'D'$ (line $A'B'$ is the common external tangent to $\\omega_a$ and $\\omega_b$, and similarly for the other three sides).\n\n![](attached_image_1.png)\n\nFor quadrilateral $ABCD$, denote $\\angle A = \\alpha$, $\\angle D = \\delta$. By the cyclic properties of quadrilaterals $ABCD$ and $AI_aI_dD$, we have angle relations: $\\angle QCD = \\alpha$, $\\angle UI_aA = \\angle ADI_d = \\delta/2$, $\\angle VI_dD = \\angle DAI_a = \\alpha/2$. Consequently, $\\angle CQD = \\delta - \\alpha$ and $\\angle PUV = \\angle PVU = (\\alpha+\\delta)/2$. In particular, $\\alpha < \\delta$, so point $X_{ad}$ lies on ray $AD$ and $\\angle DX_{ad}I_d = \\angle ADV - \\angle PVU = (\\delta - \\alpha)/2$. This equality implies that line $I_aI_d$ is parallel to the angle bisector of $\\angle CQD$.\nSince line $A'D'$ is symmetric to line $AD$ with respect to the line of centers $I_aI_d$, we obtain that $A'D' \\parallel BC$, and line $A'D'$ passes through point $X_{ad}$. Defining points $X_{ab}, X_{bc}, X_{cd}$ similarly, we find these points lie on lines $A'B', B'C', C'D'$ which are parallel to the sides of quadrilateral $ABCD$.\nAs established, line $I_aI_d$ is parallel to the bisector of $\\angle CQD$. Similarly, line $I_bI_c$ is parallel to this bisector, while lines $I_aI_b$\n\nand $I_c I_d$ are parallel to the bisector of $\\angle BPC$. Since $\\angle PUV = \\angle PVU$, the bisector of $\\angle BPC$ is perpendicular to line $I_a I_d$. Thus, adjacent sides of quadrilateral $I_a I_b I_c I_d$ are perpendicular, making it a rectangle. Therefore, it is cyclic - denote its circumcircle by $\\omega$, with center at the intersection of its diagonals.\nIt remains to prove that the centers of circles $\\omega$, $\\Omega$ and $\\Gamma$ are collinear. We'll show these three circles share a radical axis containing points $X_{ab}, X_{bc}, X_{cd}, X_{da}$ ($\\star$).\nLet $\\gamma$ be the circumcircle of quadrilateral $AI_a I_d D$. Then point $X_{ad}$ is the radical center of circles $\\gamma$, $\\omega$ and $\\Omega$, lying on two of their radical axes. Hence, the radical axis of $\\omega$ and $\\Omega$ passes through $X_{ad}$, and similarly through $X_{ab}, X_{bc}, X_{cd}$. Thus these four points are collinear.\nLet line $B'C'$ intersect side $AB$ at $S$ and side $CD$ at $T$. Since $B'C' \\parallel AD$, we have $\\angle BST = \\angle BAD = 180^{\\circ} - \\angle BCT$, making quadrilateral $BCTS$ cyclic. As $C'D' \\parallel AB$ and $A'B' \\parallel CD$, by Thales' theorem:\n$$\n\\frac{X_{bc}B'}{X_{bc}T} = \\frac{X_{bc}X_{ab}}{X_{bc}X_{cd}} = \\frac{X_{bc}S}{X_{bc}C'}\n$$\nFrom these ratios and the cyclicity of $BCTS$, we obtain $X_{bc}B' \\cdot X_{bc}C' = X_{bc}S \\cdot X_{bc}T = X_{bc}B \\cdot X_{bc}C$, showing equal power of $X_{bc}$ with respect to $\\Omega$ and $\\Gamma$. Analogous reasoning applies to $X_{ab}, X_{ad}, X_{cd}$, proving claim ($\\star$) as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function. Let us call a chord a segment of integer length parallel to the axis $Ox$ whose endpoints belong to the graph $y = f(x)$. It is known that the graph $y = f(x)$ has exactly $N$ chords, and moreover, among them there is a chord of length $2025$. Find the least possible value of $N$.",
    "options": [],
    "answer": "4049",
    "solution": "**Ответ.** 4049.\n\nFor a natural number $n$, define $g_n(x) = f(x+n) - f(x)$. Then the number of chords of length $n$ equals the number of zeros of the function $g_n(x)$.\n\nAs an *example*, consider the following piecewise linear function $f$: $f(x) = x$ for $x \\le 2024\\frac{9}{10}$ and $f(x) = 20249 \\cdot |2025 - x|$ for $x \\ge 2024\\frac{9}{10}$. Note that for $a \\notin [0; 2025\\frac{1}{10}]$, the function $f(x)$ takes the value $f(a)$ only at point $a$. Therefore, if $g_n(x) = 0$, then both points $x$ and $x+n$ must lie in the interval $[0, 2025 \\frac{1}{10}]$. In particular, $n \\le 2025$, and the zeros of $g_n(x)$ lie in $[0; 2025 \\frac{1}{10} - n]$.\n\nFor $n = 2025$ and $x \\in [0, \\frac{1}{10}]$, we have $g_{2025}(x) = 20248x$. Thus, $g_{2025}(x)$ has a unique zero at $x = 0$, meaning the function $f$ has exactly one chord of length $2025$. For natural $n \\le 2024$, the function $g_n(x)$ is monotonically decreasing on $[0, 2025 - n]$ and monotonically increasing on $[2025 - n, 2025 \\frac{1}{10} - n]$, with $g_n(0) > 0$, $g_n(2025 - n) < 0$, and $g_n(2025 \\frac{1}{10} - n) > 0$. Therefore, this function has exactly two zeros, meaning $f$ has two chords of each length $n = 1, 2, \\dots, 2024$. In total, it has $4049$ distinct chords.\n\n$g_k(0) = 0$. The differences $d_j = f(r_{j+1}) - f(r_j)$ satisfy: if $r_{j+1} > r_j$ then $d_j = g_k(r_j) \\ge 0$; if $r_{j+1} < r_j$ then $d_j = -g_m(r_{j+1})$.\nThe sum $\\sum d_j = 0$ must contain both positive and negative terms, implying $g_m(t) > 0$ for some $t \\in (0, k)$. Since $g_k(0) + g_m(k) = g_m(0) + g_k(m) = 0$ and $g_k \\ge 0$, we get $g_m(0) \\le 0$ and $g_m(k) \\le 0$. This leads to $g_m$ having at least $3$ zeros (at $0$, $k$, and between them), plus $g_k$ having at least $1$ zero, totaling $4$ zeros as required. The case when $g_k(m) = 0$ is similar.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 39905,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a rectangular sheet of paper, several segments were drawn parallel to its sides. These segments divide the sheet into several rectangles (so that there are no parts of drawn segments inside rectangles). Petya wants to draw one of two diagonals in each of these rectangles dividing it into two triangles, and then color all triangles, each triangle either black or white. Determine if Petya can always do this so that no two triangles of the same color have a common boundary segment.",
    "options": [],
    "answer": "Yes, always possible.",
    "solution": "Let Petya draw a diagonal in each of the rectangles from the bottom-left corner to the top-right corner. After this, he will color all the triangles adjacent to the top-left corners of the rectangles black, and the rest — white.\n\nLet us prove that such a coloring will work. Consider a common boundary segment of two triangles. If this segment is a diagonal, then a black triangle adjoins it from above, and a white one from below. If the segment is horizontal, then a white triangle adjoins it from above, and a black one — from below; the case of a vertical segment is similar. Therefore, such a coloring is suitable.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Diagonals of a convex quadrilateral *ABCD* intersect at *E*. The four points of tangency of the circles (*ABE*) and (*CDE*) with their external common tangents lie on a circle $\\omega$. Analogously, the four points of tangency of the circles (*ADE*) and (*BCE*) with their external common tangents lie on a circle $\\gamma$. Prove that the centers of $\\omega$ and $\\gamma$ coincide.\n\n*(А. Д. Терёшин)*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the centers of the circumscribed circles of triangles *ABE*, *BCE*, *CDE*, *ADE* by $O_{AB}$, $O_{BC}$, $O_{CD}$, $O_{AD}$ respectively. Let $T_1, T_2$ be the points of tangency of one of the common tangents with the circumscribed circles of triangles *ABE* and *CDE*, respectively; denote by *O* and *T* the midpoints of segments $O_{AB}O_{CD}$ and $T_1T_2$, respectively (см. рис. 1). Then, in the right trapezoid $O_{AB}T_1T_2O_{CD}$, the line $OT$ is the midline, hence it is the perpendicular bisector of segment $T_1T_2$. Note that the circle $\\omega$ is symmetric with respect to the line $O_{AB}O_{CD}$, on which the point $O$ also lies, meaning that $O$ is the center of $\\omega$.\n\n![](attached_image_1.png)\nРис. 1\n\n![](attached_image_2.png)\nРис. 2\n\nSimilarly, we find that the midpoint of segment $O_{AD}O_{BC}$ is the center of $\\gamma$. Therefore, the statement of the problem is equivalent to the fact that $O_{AB}O_{BC}O_{CD}O_{AD}$ is a parallelogram. To prove this, it suffices to note that $O_{AB}O_{BC}$ and $O_{CD}O_{AD}$ are the perpendicular bisectors of segments $EB$ and $ED$, respectively, hence $O_{AB}O_{BC} \\parallel O_{CD}O_{AD}$; similarly, $O_{AB}O_{AD} \\parallel O_{BC}O_{CD}$, from which the required conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exists an *even* positive integer $a$ such that $(a-1)(a^2-1)\\dots(a^n-1)$ is a perfect square.",
    "options": [],
    "answer": "n = 1 and n = 2",
    "solution": "**Ответ.** For $n=1$ and $n=2$.\n\nFor $n=1$, any even $a$ of the form $m^2+1$ works, for example, $a=2$.\n\nFor $n=2$, any even $a$ of the form $m^2-1$ works, for example, $a=8$.\n\nAssume that for $n=3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Since $a^2+a+1 = a(a+1)+1$, the numbers $a+1$ and $a^2+a+1$ are coprime. Because $a+1$ is odd, the numbers $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a + 1$ modulo 3 can only be 0 or 1, and thus $a - 1$ is not divisible by 3. Hence,\n$$\n\\begin{align*}\n\\gcd(a - 1, a^2 + a + 1) &= \\gcd(a - 1, (a + 2)(a - 1) + 3) = \\\\\n&= \\gcd(a - 1, 3) = 1,\n\\end{align*}\n$$\nmeaning that both $a-1$ and $a^2+a+1$ must be perfect squares.\nHowever, the latter cannot be a square, since $a^2 < a^2+a+1 < (a+1)^2$. This is a contradiction.\n\nIt remains to prove that no such $a$ exists for $n \\geq 4$. Suppose such an $a$ exists. Take a natural number $k \\geq 2$ such that $2^k \\leq n < 2^{k+1}$. Since $a^{2^k} - 1 = (a^{2^k} - 1)(a^{2^k} + 1)$, the number $(a - 1)(a^2 - 1) \\dots (a^n - 1)$ can be expressed as the product of $a^{2^k} - 1 + 1$ and several other factors of the form $a^m - 1$, where $1 \\leq m \\leq n$ and $m \\neq 2^k$.\nWe will show that the factor $a^{2^{k-1}} + 1$ is coprime with all other factors in this decomposition. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1)$ is divisible by $d$. But $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k,m)} - 1$. Since $m \\neq 2^k$ and $m \\leq n < 2^{k+1}$, the number $m$ cannot be divisible by $2^k$. Thus, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Therefore, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k}-1, a^m-1)$, and hence also divides $d$. Because $a$ is even, the numbers $a^{2^{k-1}-1}$ and $a^{2^{k-1}}+1$ have no common divisors other than 1, so $d=1$, as required.\nThe factor $a^{2^{k-1}} + 1$ is coprime with all other factors in the product, which is a perfect square, so it must itself be a perfect square. Then $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are perfect squares differing by 1, which is impossible. Therefore, our assumption is false, and no such $a$ exists for $n \\geq 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39908,
    "subject": "Mathematics (Multi-modal)",
    "question": "The chess King was placed on a cell of the board $8 \\times 8$, and then he made $64$ moves so that he visited all the cells and returned back to the initial cell. At each moment, we calculate the distance from the center of the cell occupied by the King to the center of the board. We call the move *pleasant* if after the move this distance became less than it was before the move. Find the greatest possible number of pleasant moves.\n\n(By one move the chess King moves to a cell adjacent by a side or by a vertex.)",
    "options": [],
    "answer": "44",
    "solution": "Ответ. $44$ moves.\n\nLet us prove that there must have been at least $20$ unpleasant moves (and thus the number of pleasant moves cannot exceed $44$). Let's place numbers in the cells as shown in рис. 3; cells with the same numbers are equidistant from the center, and cells with smaller numbers are closer to the center than those with larger numbers.\n\n![](attached_image_1.png)\nРис. 3\n\n![](attached_image_2.png)\nРис. 4\n\nEvery move from a cell with number $1$ does not decrease the distance to the center and is therefore unpleasant — there are $4$ such moves. A move from a cell with number $2$ can be pleasant only if it goes to a cell with number $1$. But there are eight cells with number $2$ and only four with number $1$, so at least four moves from cells with number $2$ will be unpleasant.\n\nNow consider moves leading to the $32$ cells with numbers not less than $6$. Note that these moves cannot originate from cells with numbers $1$ or $2$, meaning they weren't accounted for in the previous reasoning. Such a move can only be pleasant if it comes from a cell with a number not less than $7$; however, there are only $20$ such cells. Therefore, among these moves, there are at least $32 - 20 = 12$ unpleasant ones, bringing the total number of unpleasant moves to no fewer than $4 + 4 + 12 = 20$.\n\nAn example of a traversal with $44$ pleasant moves is shown in рис. 4.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 39909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_1(x)$ and $P_2(x)$ be monic quadratic polynomials (i.e., quadratic polynomials with leading coefficient $1$). Let points $A_1$ and $A_2$ be vertices of parabolas $y = P_1(x)$ and $y = P_2(x)$, respectively. By $m(g(x))$ denote the minimal value of the function $g(x)$. It happens that the differences $m(P_1(P_2(x))) - m(P_1(x))$ and $m(P_2(P_1(x))) - m(P_2(x))$ are equal positive real numbers. Find the angle between line $A_1A_2$ and the coordinate axis $Ox$. (Н. Х. Агаханов)",
    "options": [],
    "answer": "45 degrees",
    "solution": "Let the given trinomials be $P_1(x) = (x - x_1)^2 + y_1$ and $P_2(x) = (x - x_2)^2 + y_2$, where $A_1(x_1; y_1)$ and $A_2(x_2; y_2)$ are the coordinates of the vertices of the parabolas. Then $m(P_1(x)) = y_1$, and $P_1(P_2(x)) = ((x - x_2)^2 + y_2 - x_1)^2 + y_1$. If $y_2 \\le x_1$, then the minimal value of the expression $((x - x_2)^2 + y_2 - x_1)^2$ is zero, whence $m(P_1(P_2(x))) - m(P_1(x)) = y_1 - y_1 = 0$. The latter contradicts the fact that $m(P_1(P_2(x))) - m(P_1(x))$ is a positive number. Thus, $y_2 > x_1$, from which it follows that $m(P_1(P_2(x))) = (y_2 - x_1)^2 + y_1$ and $m(P_1(P_2(x))) - m(P_1(x)) = (y_2 - x_1)^2$.\n\nSimilarly, $y_1 > x_2$ and $m(P_2(P_1(x))) - m(P_2(x)) = (y_1 - x_2)^2$. Now, the condition of equality of the differences can be rewritten as $(y_1 - x_2)^2 = (y_2 - x_1)^2$. Hence, since $y_2 > x_1$ and $y_1 > x_2$, we obtain $y_1 - x_2 = y_2 - x_1$, that is, $y_2 - y_1 = -(x_2 - x_1)$. Therefore, the desired angle is $45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Petya chose 100 pairwise distinct positive real numbers, each less than $1$, and arranged these numbers on a circle. Then he performs the following moves. By one move he takes some three consecutive numbers $a, b, c$ (in this order) and replaces the middle number $b$ by $a - b + c$. Find the greatest possible $k$ such that Petya can choose initial numbers and perform several moves so that in the resulting arrangement there are $k$ integers.",
    "options": [],
    "answer": "50",
    "solution": "Estimate. We will show that the number of integers never exceeds $50$.\nWe will track the differences between each number and the next one in clockwise order. If three consecutive numbers were $a$, $b$, and $c$, their differences were $a-b$ and $b-c$. After applying the operation to $b$, the numbers become $a$, $a-b+c$, and $c$, with differences $a - (a-b+c) = b-c$ and $(a-b+c)-c = a-b$. Thus, the operation simply swaps two adjacent differences. Initially, all differences were non-integer, so they remain non-integer at all times. Therefore, two integers can never appear consecutively, meaning their count cannot exceed $50$.\n\nExample. First, arrange the numbers $0.1$ and $0.2$ alternately in a circle. If we perform the operation on each $0.2$, it will be replaced by $0.1 - 0.2 + 0.1 = 0$, making every other number an integer.\n\nNow, we need to adjust the example so that all numbers become distinct. To do this, it is sufficient to add a small unique value to each $0.1$ and to each $0.2$ the sum of the values added to its neighbors. For instance, choosing $t = 0.001$, we can add to consecutive $0.1$s the numbers $0, t, 2t, \\dots, 47t, 48t, 50t$; then, to the $0.2$s, we add $t, 3t, 5t, \\dots, 95t, 98t, 50t$. As a result, all numbers will become distinct.\n\nThe explicitly constructed example looks as follows:\n$$\n0.1 \\ 0.2001 \\ 0.1001 \\ 0.2003 \\ 0.1002 \\ 0.2005 \\ 0.1003 \\ \\dots \\\\\n\\dots \\ 0.1047 \\ 0.2095 \\ 0.1048 \\ 0.2098 \\ 0.105 \\ 0.205\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $1, 2, 3, \\ldots, 60$ are written in a row (in this order). Igor and Ruslan take turns in making moves; Igor starts; by one move the player puts one of the signs of operation $+$, $-$, $\\times$ between some pair of adjacent numbers. When a sign is placed between every two adjacent numbers we calculate the value of the obtained expression. Igor wins if this value is divisible by $3$, otherwise Ruslan wins. Determine who of the two players has a winning strategy.",
    "options": [],
    "answer": "Igor",
    "solution": "Let us replace all numbers in the sequence with their remainders modulo $3$; this will not change the game's outcome.\n\nWe obtain the sequence $1, 2, 0, \\ldots, 1, 2, 0$. The gaps between the numbers are numbered from left to right from $1$ to $59$.\nOn his first move, Igor places a \"$-$\" sign in the $30$th gap, while pairing all other gaps as $(i, 30+i)$. If Ruslan places a \"$+$\" or \"$-$\" sign in some gap, Igor responds by placing a \"$-$\" or \"$+$\" sign, respectively, in the paired gap. If Ruslan places a \"$\\times$\" sign, Igor also places a \"$\\times$\" sign in the paired gap.\nOnce all signs are placed, the resulting expression splits into several terms. The sets of terms in the left and right halves of the expression are identical but have opposite signs. Consequently, the expression's value will be congruent to $0$ modulo $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39912,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the boundary of triangle $ABC$ points $D_1, D_2, E_1, E_2, F_1, F_2$ are chosen so that when going around the perimeter, the points are encountered in the following order: $A, F_1, F_2, B, D_1, D_2, C, E_1, E_2$. Given that $AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2$, prove that two triangles formed by the triples of lines $AD_1, BE_1, CF_1$, and $AD_2, BE_2, CF_2$, have equal perimeters.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's begin with the following useful lemma.\n\n**Lemma.** Let points $F$ and $E$ be chosen on sides $AB$ and $AC$ of parallelogram $ABKC$ respectively, such that $BE = CF$. Then point $K$ is equidistant from lines $BE$ and $CF$ (see Fig. 1).\n\n![](attached_image_1.png)\n\nРис. 1\n\n**Proof.** Since $BK \\parallel EC$ and $CK \\parallel FB$, we have $S_{KBE} = S_{KBC} = S_{KFC}$. As $BE = CF$, it follows that the distances from point $K$ to lines $BE$ and $CF$ are equal. $\\square$\n\nNow let's proceed to the solution. Suppose the lines given in the condition form triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ (points are labeled as in Fig. 2).\n\nChoose point $K$ such that $ABKC$ is a parallelogram; according to the lemma, point $K$ is equidistant from lines $BE_1, CF_1, BE_2$ and $CF_2$; therefore, there exists a circle centered at $K$ that is tangent to these lines at points $P_1, Q_1, P_2$ and $Q_2$ respectively. Then from the equality of tangent segments we obtain:\n\n![](attached_image_2.png)\n\nРис. 2\n\n$$\n\\begin{aligned}\nBX_1 - CX_1 &= BP_1 + X_1P_1 - X_1Q_1 + CQ_1 = BP_2 + CQ_2 = \\\\\n&= BP_2 - X_2P_2 + X_2Q_2 + CQ_2 = CX_2 - BX_2.\n\\end{aligned}\n$$\n\nSimilarly, we get $CY_1 - AY_1 = AY_2 - CY_2$ and $AZ_1 - BZ_1 = BZ_2 - AZ_2$. Adding these three equalities yields the required equality of perimeters.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of positive integers such that\n$$\nx + y + z = 2010 \\text{ and } x^2 + y^2 + z^2 - x y - y z - z x = 3.\n$$",
    "options": [],
    "answer": "All permutations of (669, 670, 671)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Using each of the first eight primes exactly once and several algebraic operations, obtain the result $2010$.",
    "options": [],
    "answer": "2×3×5×(7+11+13+17+19) = 2010",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_{1} A_{2} \\ldots A_{2010}$ be a regular $2010$-gon. Find the number of obtuse triangles whose vertices are among $A_{1}, A_{2}, \\ldots, A_{2010}$.",
    "options": [],
    "answer": "1012047060",
    "solution": "We will solve the problem for a regular $n$-gon $A_{1} A_{2} \\ldots A_{n}$, $n \\geq 3$.\n\nSolution 1.\nLet $P(n)$ be the desired number of obtuse triangles and let $P_{1}(n)$ be the number of obtuse angles $A_{1} A_{i} A_{j}$, where $1 < i < j \\leq n$. Clearly $P(n) = n \\cdot P_{1}(n)$. Any of the considered angles $A_{1} A_{i} A_{j}$ is obtuse if and only if $j \\leq \\frac{n+1}{2}$, hence $P_{1}(n)$ equals the number of all two-elements of $\\{2,3, \\ldots, \\left[\\frac{n+1}{2}\\right]\\}$. Thus\n\n$$\n\\begin{gathered}\nP(n) = n \\cdot P_{1}(n) = n\\left(\\left[\\frac{n+1}{2}\\right] - 1\\right)\\left(\\left[\\frac{n+1}{2}\\right] - 2\\right) \\\\\n= \\frac{n}{2} \\cdot \\left[\\frac{n-1}{2}\\right] \\cdot \\left[\\frac{n-3}{2}\\right] .\n\\end{gathered}\n$$\n\n\nSolution 2.\nLet $\\alpha = \\widehat{A_{i} A_{j}}$ be the largest arc of the circumcircle that contains the vertices of an obtuse triangle, say $\\triangle A_{i} A_{k} A_{j}$. The size of $\\alpha$ is less than $180^{\\circ}$ and the number of all obtuse triangles $A_{i} A_{k} A_{j}$ with the common longest side $A_{i} A_{j}$ is identical with the numbers $v(\\alpha)$ of those vertices $A_{k}$ of the $n$-gon that are inner points of the arc $\\alpha$. Clearly, $v(\\alpha) \\in \\{1,2, \\ldots, \\left[\\frac{n-3}{2}\\right]\\}$. Since there are exactly $n$ arcs $\\alpha$ with the same value $v(\\alpha)$, the desired number is\n\n$$\nn\\left(1+2+\\ldots+\\left[\\frac{n-3}{2}\\right]\\right) = \\frac{n}{2} \\cdot \\left[\\frac{n-3}{2}\\right] \\cdot \\left[\\frac{n-1}{2}\\right] .\n$$\n\nIn our problem $n = 2010$, hence the desired result is\n\n$$\n\\frac{2010}{2} \\cdot \\left[\\frac{2007}{2}\\right] \\cdot \\left[\\frac{2009}{2}\\right] = 1003 \\cdot 1004 \\cdot 1005\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of positive integers such that\n$$\n3^{x} + 4^{y} = 5^{z}\n$$",
    "options": [],
    "answer": "(x, y, z) = (2, 2, 2)",
    "solution": "Looking $\\bmod 4$ we get $(-1)^{x} \\equiv (\\bmod 4)$, hence $x$ is even. Looking $\\bmod 3$ we get $1 \\equiv (-1)^{z} (\\bmod 3)$, hence $z$ is even. Let $x = 2m$, $z = 2n$. We have\n$$\n4^{y} = (5^{n} - 3^{m})(5^{n} + 3^{m})\n$$\nhence $5^{n} + 3^{m} = 2^{a}$, $5^{n} - 3^{m} = 2^{b}$, $a + b = 2y$, $y > 0$. Then\n$$\n2 \\cdot 3^{m} = 2^{a} - 2^{b} = 2^{b}(2^{a-b} - 1),\n$$\nso $b = 1$ and obtain\n$$\n3^{m} = 2^{a-1} - 1.\n$$\nLooking mod 3 it follows $(-1)^{a-1} \\equiv 1 (\\bmod 3)$, hence $a-1$ is even. Let $a-1 = 2\\alpha$. We obtain\n$$\n3^{m} = 2^{2\\alpha} - 1 = (2^{\\alpha} - 1)(2^{\\alpha} + 1),\n$$\nand obtain\n$$\n2^{\\alpha} - 1 = 3^{t}, \\quad 2^{\\alpha} + 1 = 3^{s}, \\quad t + s = m, \\quad t < s.\n$$\nWe get\n$$\n3^{s} - 3^{t} = 2\n$$\nor\n$$\n3^{t}(3^{s-t} - 1) = 2,\n$$\nhence $t = 0$, $s = 1$, $\\alpha = 1$, $a = 3$.\nWe get $m = 1$, $5^{n} = 5$, so $n = 1$. Finally, $x = 2$, $z = 2$, and $y = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-empty sets $S$ of nonzero real numbers such that\n\na) $S$ has at most 5 elements;\n\nb) If $x$ is in $S$, then so are $1-x$ and $\\frac{1}{x}$.",
    "options": [],
    "answer": "{-1, 1/2, 2}",
    "solution": "Let $x \\in S$. Then\n$$\n1-x, \\frac{1}{x}, \\frac{1}{1-x}, 1-\\frac{1}{x}, 1-\\frac{1}{1-x} \\in S.\n$$\n(1) Since $|S| \\leq 5$, it follows that at least two of these numbers are equal. Considering all 15 possible cases it follows $x=-1$ or $x=\\frac{1}{2}$. We get\n$$\nS=\\left\\{-1, \\frac{1}{2}, 2\\right\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\geq b \\geq c > 0$. Prove that\n$$\n(a - b + c)\\left(\\frac{1}{a} - \\frac{1}{b} + \\frac{1}{c}\\right) \\geq 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ for which $9 n+16$ and $16 n+9$ are both perfect squares.",
    "options": [],
    "answer": "0, 1, 52",
    "solution": "It is clear that $n=0$ and $n=1$ are solutions. Let $9 n+16 = x^2$, $16 n+9 = y^2$. Then\n$$\n16 x^2 - 9 y^2 = 16^2 - 9^2\n$$\nthat is\n$$\n(4 x - 3 y)(4 x + 3 y) = 7 \\cdot 5^2.\n$$\nWe can assume that $x, y > 0$, hence we have $4 x - 3 y > 0$. Considering the following possibilities\n$$\n\\left\\{\n\\begin{array}{l}\n4 x - 3 y = 1 \\\\\n4 x + 3 y = 175\n\\end{array}\n\\right.\n\\quad\n\\left\\{\n\\begin{array}{l}\n4 x - 3 y = 5 \\\\\n4 x + 3 y = 35\n\\end{array}\n\\right.\n\\quad\n\\left\\{\n\\begin{array}{l}\n4 x - 3 y = 7 \\\\\n4 x + 3 y = 25\n\\end{array}\n\\right.\n$$\nwe get the solutions $x = 22, 5, 8$, respectively. Hence $n = 52$, $n = 1$, and $n = 0$ are the desired values of $n$.\nLet $9 n+16 = x^2$, $16 n+9 = y^2$. Then\n$$\n(x y)^2 = (9 n+16)(16 n+9) = (12 n)^2 + 337 n + 12^2,\n$$\nhence\n$$\n(12 n + 15)^2 > (x y)^2 \\geq (12 n + 12)^2\n$$\nIt follows that there are only three possibilities for $(x y)^2$, that is $(12 n + 12)^2$, $(12 n + 13)^2$, and $(12 n + 14)^2$. Solving the corresponding equation we get $n = 0, 1$ and $n = 52$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of positive integers such that\n$$\nx^{2}+y^{2}+33^{2}=2010 \\sqrt{x-y}\n$$",
    "options": [],
    "answer": "(54, 45)",
    "solution": "From $3 \\mid x^{2}+y^{2}$ it follows $3 \\mid x$ and $3 \\mid y$, hence $x=3u$ and $y=3v$, for some positive integers $u$ and $v$. Replace in the equation and get $3\\left(u^{2}+v^{2}+11^{2}\\right)=670 \\sqrt{3(u-v)}$, hence $u-v=3k^{2}$, for some positive integer $k$, i.e. $u^{2}+v^{2}+11^{2}=670k$. We have $u^{2}+v^{2}>(u-v)^{2}=9k^{4}$, and obtain $9k^{4}+121<670k$, hence $k<5$.\n\nCase 1. If $k=2$ or $k=4$, then $u^{2}+v^{2}+11^{2} \\equiv 0\\pmod{4}$, not possible since we have $11^{2} \\equiv 1\\pmod{4}$ and $u^{2}, v^{2} \\equiv 0$ or $1\\pmod{4}$.\n\nCase 2. If $k=1$, then we get the system\n$$\n\\left\\{\\begin{array}{l}\nu^{2}+v^{2}=549 \\\\\nu-v=3\n\\end{array}\\right.\n$$\nhaving integer solutions $u=18$ and $v=15$. In this case it follows $x=54$ and $y=45$.\n\nCase 3. If $k=3$, then we get the system\n$$\n\\left\\{\\begin{array}{l}\nu^{2}+v^{2}=1882 \\\\\nu-v=27\n\\end{array}\\right.\n$$\nhaving no solutions in integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39921,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $A B C$ the circumcircle has radius $R$ and center $O$ and the incircle has radius $r$ and center $I \\neq O$. Let $G$ denote the centroid of triangle $A B C$. Prove that $I G \\perp B C$ if and only if $A B=A C$ or $A B+A C=3 B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = B C$, $\\beta = C A$, $\\gamma = A B$. Without loss of generality we may assume that $\\beta \\geq \\gamma$. Let $M$ be the midpoint of $B C$, and let $A'$, $I'$, and $G'$ be the orthogonal projections of $A$, $I$ and $G$ on $M B$.\n\n![](attached_image_1.png)\n\nWe have\n$$\nM I' = M B - B I' = \\frac{\\alpha}{2} - (s - \\beta) = \\frac{\\beta - \\gamma}{2},\n$$\nand because\n$$\n\\gamma^2 - A' B^2 = \\beta^2 - (\\alpha - A' B)^2,\n$$\nwe have\n$$\n\\begin{aligned}\n& M G' = \\frac{1}{3} M A' = \\frac{1}{3}(M B - A' B) \\\\\n= & \\frac{1}{3}\\left(\\frac{\\alpha}{2} - \\frac{\\alpha^2 - \\beta^2 + \\gamma^2}{2 \\alpha}\\right) = \\frac{\\beta^2 - \\gamma^2}{6 \\alpha} .\n\\end{aligned}\n$$\nWe conclude that $I G \\perp B C$ is equivalent to $I'$ coincides to $G'$, that is $M I' = M G'$. The last relation is equivalent to\n$$\n\\frac{\\beta - \\gamma}{2} = \\frac{\\beta^2 - \\gamma^2}{6 \\alpha}\n$$\nor\n$$\n(\\beta - \\gamma)(\\beta + \\gamma - 3 \\alpha) = 0\n$$\nand the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ that can be written as\n$$\nx = \\frac{a_{0}}{a_{1} a_{2} \\ldots a_{n}} + \\frac{a_{1}}{a_{2} a_{3} \\ldots a_{n}} + \\frac{a_{2}}{a_{3} a_{4} \\ldots a_{n}} + \\ldots + \\frac{a_{n-2}}{a_{n-1} a_{n}} + \\frac{a_{n-1}}{a_{n}},\n$$\nwhere $n, a_{1}, a_{2}, \\ldots, a_{n}$ are positive integers and\n$$\n1 = a_{0} \\leq a_{1} < a_{2} < \\ldots < a_{n}.\n$$",
    "options": [],
    "answer": "All rational numbers in the interval (0, 1], i.e., Q ∩ (0, 1].",
    "solution": "It is clear that $x \\in \\mathbb{Q}$ and $x > 0$.\n\nOn the other hand, if $x$ has such a representation, then from the inequalities\n$$\n\\begin{aligned}\n& \\frac{1}{a_{1} a_{2} \\ldots a_{n}} \\leq \\frac{1}{a_{2} a_{3} \\ldots a_{n}} \\\\\n& \\frac{a_{1}}{a_{2} a_{3} \\ldots a_{n}} \\leq \\frac{a_{2} - 1}{a_{2} a_{3} \\ldots a_{n}} = \\frac{1}{a_{3} a_{4} \\ldots a_{n}} - \\frac{1}{a_{2} a_{3} \\ldots a_{n}} \\\\\n& \\frac{a_{n-2}}{a_{n-1} a_{n}} \\leq \\frac{a_{n-1} - 1}{a_{n-1} a_{n}} = \\frac{1}{a_{n}} - \\frac{1}{a_{n-1} a_{n}} \\\\\n& \\frac{a_{n-1}}{a_{n}} \\leq \\frac{a_{n} - 1}{a_{n}} = 1 - \\frac{1}{a_{n}},\n\\end{aligned}\n$$\nwe get $x \\leq 1 - \\frac{1}{a_{n}}$, hence $x \\in (0, 1]$. We note that $x = 1$ can be represented in this form by considering $n = 1$ and $a_{1} = 1$.\n\nIf $x = \\frac{p}{q} \\in (0, 1)$, where $p$ and $q$ are positive integers, $p < q$, then we can write\n$$\nx = x \\cdot 1 = \\frac{p}{q} \\left( \\frac{1}{1 \\cdot 2 \\ldots p} + \\frac{1}{2 \\cdot 3 \\ldots p} + \\frac{2}{3 \\cdot 4 \\ldots p} + \\frac{p-1}{p} \\right),\n$$\nand we can consider $n = p$, $a_{1} = 1$, $a_{2} = 2$, $\\ldots$, $a_{p-1} = p-1$, $a_{p} = q$.\n\nThe desired numbers are the elements of the set $\\mathbb{Q} \\cap (0, 1]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of positive integers for which\n$$\na + b c = 2010 \\text{ and } b + c a = 250.\n$$",
    "options": [],
    "answer": "(a, b, c) = (3, 223, 9)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs ($m, n$) of integers, $m, n \\geq 2$ such that $m n - 1$ divides $n^{3} - 1$.",
    "options": [],
    "answer": "All pairs are (k, k^2) and (k^2, k) with k ≥ 2.",
    "solution": "The solutions are $\\left(k, k^{2}\\right)$ and $\\left(k^{2}, k\\right)$, with $k \\geq 2$.\nWe have $m n - 1 \\mid \\left(n^{3} - 1\\right) m - n^{2}(m n - 1) = n^{2} - m$, hence\n$$\nm n - 1 \\mid m\\left(n^{2} - m\\right) - (m n - 1) n = n - m^{2}\n$$\nIf $n > m^{2}$, then $m n - 1 \\leq n - m^{2} \\leq n - 1$, so $m n \\leq n$, not possible.\nIf $n = m^{2}$, then obviously $m^{3} - 1 \\mid m^{6} - 1$, so all pairs ($m, m^{2}$), $m \\geq 2$ are solutions.\nIf $n < m^{2}$, from $m n - 1 \\leq n^{3} - 1$ we obtain that $\\sqrt{n} < m \\leq n^{2}$. Then $m n - 1 \\leq m^{2} - n < m^{2} - 1$, hence $n < m$. If $n^{2} - m > 0$, we get $m n - 1 \\leq n^{2} - m < n^{2} - 1$, so $m < n$, a contradiction. It follows $n = m^{2}$, satisfying the condition in the problem since $m^{3} - 1 \\mid m^{3} - 1$, so all pairs $\\left(n^{2}, n\\right), n \\geq 2$, are also solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\widehat{B} \\geq 2 \\widehat{C}$. Denote by $D$ the foot of the altitude from $A$ and by $M$ the midpoint of $BC$. Prove that $DM \\geq \\frac{AB}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nDenote by $a, b, c$ the length sides of triangle $ABC$. In triangle $ADM$ we have\n$$\n\\begin{gathered}\nDM^{2} = AM^{2} - AD^{2} = \\frac{2\\left(b^{2} + c^{2}\\right) - a^{2}}{4} - \\frac{4K^{2}}{a^{2}} \\\\\n= \\frac{2\\left(b^{2} + c^{2}\\right) - a^{2}}{4} - \\frac{16K^{2}}{4a^{2}}\n\\end{gathered}\n$$\n$$\n\\begin{gathered}\n= \\frac{2\\left(b^{2} + c^{2}\\right) - a^{2}}{4} - \\frac{1}{4a^{2}}\\left(2 \\sum a^{2}b^{2} - \\sum a^{4}\\right) \\\\\n= \\frac{1}{4a^{2}}\\left(-2b^{2}c^{2} + b^{4} + c^{4}\\right) = \\left(\\frac{b^{2} - c^{2}}{2a}\\right)^{2}\n\\end{gathered}\n$$\nSince $\\widehat{B} \\geq 2\\widehat{C} > \\widehat{C}$, it follows $b > c$, hence\n$$\n\\begin{equation*}\nDM = \\frac{b^{2} - c^{2}}{2a} \\tag{1}\n\\end{equation*}\n$$\nThe inequality $DM \\geq \\frac{AB}{2}$ is equivalent to $\\frac{b^{2} - c^{2}}{2a} \\geq \\frac{c}{2}$, that is $b^{2} \\geq c^{2} + ac$. Using the cosine law, the last inequality becomes $a \\geq 2c \\cos B + c$, or $\\sin A \\geq 2 \\sin C \\cos B + \\sin C$. We can write $\\sin(B + C) \\geq 2 \\sin C \\cos B + \\sin C$, hence $\\sin(B - C) \\geq \\sin C$, and we get $2 \\sin \\frac{B - 2C}{2} \\cos \\frac{B}{2} \\geq 0$. This inequality is true because $\\widehat{B} \\geq 2\\widehat{C}$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers such that\n$$\nx^{2}+y^{2}+z^{2}+1=xy+yz+zx+|x-2y+z| .\n$$",
    "options": [],
    "answer": "All triples of the form (a, a − 1, a) or (a, a + 1, a) for any real a.",
    "solution": "We can write\n$$\nx^{2}+y^{2}+z^{2}+1=xy+yz+zx+|x-y+z-y|\n$$\nhence\n$$\n(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2=2|x-y+z-y| .\n$$\nIt follows\n$$\n(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2 \\leq 2|x-y|+2|y-z| .\n$$\nThe last relation is equivalent to\n$$\n(|x-y|-1)^{2}+(|y-z|-1)^{2}+(z-x)^{2} \\leq 0 .\n$$\nWe get $|x-y|=1$, $|y-z|=1$ and $x=z$. The desired triples $(x, y, z)$ are $(a, a-1, a)$, $(a, a+1, a)$, where $a \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39927,
    "subject": "Mathematics (Multi-modal)",
    "question": "1) Prove that there is a triangle with side lengths\n$$\n\\sqrt{a^{2}-a+1},\\ \\sqrt{a^{2}+a+1},\\ \\text{ and }\\ \\sqrt{4 a^{2}+3}.\n$$\n\n2) Prove that the area of this triangle does not depend on $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and let $MNPQ$ be a square inscribed in the triangle such that $M, N \\in BC$, $P \\in AC$, $Q \\in AB$. Prove that\n$$\n\\operatorname{area}[MNPQ] \\leq \\frac{1}{2} \\operatorname{area}[ABC]\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nDenote by $x$ the length of sides of square $MNPQ$, $a = BC$, $h_a = AA'$, where $AA' \\perp BC$. The triangle $AQP$ and $ABC$ are similar, hence we have $\\frac{x}{a} = \\frac{h_a - x}{h_a}$. We get\n$$\nx = \\frac{a h_a}{a + h_a} \\leq \\frac{a h_a}{2 \\sqrt{a h_a}} = \\frac{1}{2} \\sqrt{a h_a} = \\frac{1}{2} \\sqrt{2 \\operatorname{area}[ABC]}\n$$\nand the desired inequality follows. The equality holds if and only if $h_a = a$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that in any triangle $ABC$ with $\\widehat{A}=90^{\\circ}$ the following inequality holds:\n$$\n(AB-AC)^{2}\\left(BC^{2}+4 AB \\cdot AC\\right)^{2} \\leq 2 BC^{6} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$$\n\\left(\\frac{AB}{BC}-\\frac{AC}{BC}\\right)^{2}\\left(1+4 \\frac{AB}{BC} \\cdot \\frac{AC}{BC}\\right)^{2} \\leq 2,\n$$\nhence\n$$\n(\\cos B-\\sin B)^{2}(1+4 \\cos B \\sin B)^{2} \\leq 2,\n$$\nthat is\n$$\n(1-\\sin 2B)\\left(1+4 \\sin 2B+4 \\sin ^{2} 2B\\right) \\leq 2 .\n$$\nAfter easy computations we get the equivalent inequality\n$$\n(1+\\sin 2B)(2 \\sin 2B-1)^{2} \\geq 0 .\n$$\nWe have equality if and only if $\\sin 2B=\\frac{1}{2}$, i.e. $\\widehat{B}=\\frac{\\pi}{12}$.\nLet $\\alpha=\\frac{AB}{BC}, \\beta=\\frac{AC}{BC}$. We have $\\alpha^{2}+\\beta^{2}=1$, and the inequality is equivalent to:\n$$\n(\\alpha-\\beta)^{2}(1+4 \\alpha \\beta)^{2} \\leq 2,\n$$\nhence\n$$\n(1-2 \\alpha \\beta)(1+4 \\alpha \\beta)^{2} \\leq 2 .\n$$\nPut $x=2 \\alpha \\beta$ and obtain\n$$\n(1-x)(1+2 x)^{2} \\leq 2 \\text{ or } 4 x^{2}-3 x+1 \\geq 0,\n$$\nwhich is equivalent to $(2 x-1)^{2}(x+1) \\geq 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle. Altitude $DE$ in triangle $ABD$ intersects diagonal $AC$ in $F$. Prove that $FB = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n$F$ is the orthocenter of $\\triangle ABD$, that is $BF \\perp AD$. We have $\\widehat{BCA} \\equiv \\widehat{BDA}$ ($ABCD$ is cyclic) and $\\widehat{BDA} = \\frac{\\pi}{2} - \\widehat{PAD}$. Also, $\\widehat{BFP} = \\widehat{AFB'} = \\frac{\\pi}{2} - \\widehat{FAB'} = \\frac{\\pi}{2} - \\widehat{PAD}$. It follows, $\\widehat{BCA} = \\widehat{BFC}$, i.e. $FB = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $n$ for which there are $n$ consecutive integers whose sum of squares is a prime.",
    "options": [],
    "answer": "n = 2, 3, 6",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\\widehat{MAB} = \\widehat{NAC}$ and $\\widehat{MBA} = \\widehat{NBC}$. Prove that\n$$\n\\frac{AM \\cdot AN}{AB \\cdot AC} + \\frac{BM \\cdot BN}{BA \\cdot BC} + \\frac{CM \\cdot CN}{CA \\cdot CB} = 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $K$ be a point on the ray $BN$ such that $\\widehat{BMA} = \\widehat{BCK}$. It is clear that $K$ is outside of triangle because $\\widehat{BMA} > \\widehat{ACB}$. We have $\\triangle ABM \\sim \\triangle KBC$, hence\n$$\n\\frac{AB}{BK} = \\frac{BM}{BC} = \\frac{AM}{CK} . \\tag{1}\n$$\nNow, we have $\\triangle ABK \\sim \\triangle MBC$ (since $\\widehat{ABK} = \\widehat{MBC}$ and $\\frac{AB}{BK} = \\frac{BM}{BC}$), hence\n$$\n\\frac{AB}{BM} = \\frac{BK}{BC} = \\frac{AK}{CM} . \\tag{2}\n$$\nWe obtained $\\widehat{CKN} = \\widehat{MAB} = \\widehat{NAC}$. It follows that quadrilateral $ANCK$ is cyclic, and by using Ptolemy's theorem we get\n$$\nAC \\cdot (BK - BN) = AN \\cdot CK + CN \\cdot AK . \\tag{3}\n$$\nFrom (1) and (2) it follows\n$$\nCK = \\frac{AM \\cdot BC}{BM}, \\quad AK = \\frac{AB \\cdot CM}{BM}, \\quad BK = \\frac{AB \\cdot BC}{BM} .\n$$\nReplacing these relations in (3) we get\n$$\nAC \\cdot \\left( \\frac{AB \\cdot BC}{BM} - BN \\right) = \\frac{AM \\cdot AN \\cdot BC}{BM} + \\frac{CM \\cdot CN \\cdot AB}{BM},\n$$\nhence\n$$\n\\frac{AM \\cdot AN}{AB \\cdot AC} + \\frac{BM \\cdot BN}{BA \\cdot BC} + \\frac{CM \\cdot CN}{CA \\cdot CB} = 1 .\nUsing Ceva's theorem in trigonometric form for points $M$ and $N$ we get\n$$\n\\frac{\\sin \\widehat{MAB}}{\\sin \\widehat{MAC}} \\cdot \\frac{\\sin \\widehat{MBC}}{\\sin \\widehat{MBA}} \\cdot \\frac{\\sin \\widehat{MCA}}{\\sin \\widehat{MCB}} = 1,\n$$\nand\n$$\n\\frac{\\sin \\widehat{NAB}}{\\sin \\widehat{NAC}} \\cdot \\frac{\\sin \\widehat{NBC}}{\\sin \\widehat{NBA}} \\cdot \\frac{\\sin \\widehat{NCA}}{\\sin \\widehat{NCB}} = 1\n$$\nMultiplying these relations, it follows\n$$\n\\frac{\\sin \\widehat{MCA}}{\\sin \\widehat{MBC}} \\cdot \\frac{\\sin \\widehat{NCA}}{\\sin \\widehat{NCB}} = 1,\n$$\nhence\n$$\n\\frac{\\sin \\widehat{MCA}}{\\sin \\widehat{MCB}} = \\frac{\\sin \\widehat{NCB}}{\\sin \\widehat{NCA}} . \\tag{2}\n$$\nDenote $\\widehat{MCB} = \\alpha$ and $\\widehat{MCA} = \\beta$ and get\n$$\n\\frac{\\sin (C - \\alpha)}{\\sin \\alpha} = \\frac{\\sin (C - \\beta)}{\\sin \\beta} . \\tag{3}\n$$\nor\n$$\n\\sin C \\cdot \\cot \\alpha - \\cos C = \\sin C \\cdot \\cot \\beta - \\cos C\n$$\nhence $\\alpha = \\beta$.\n\nUsing complex coordinates, relation $\\widehat{MCB} = \\widehat{ACN}$ shows that\n$$\n\\arg \\frac{m-c}{b-c} = \\arg \\frac{a-c}{n-c},\n$$\nhence the number $\\frac{m-c}{b-c} : \\frac{a-c}{n-c}$ is real. We have\n$$\n\\frac{m-c}{b-c} \\cdot \\frac{n-c}{a-c} = \\left| \\frac{m-c}{b-c} \\cdot \\frac{n-c}{a-c} \\right| = \\frac{CM \\cdot CN}{CB \\cdot CA} .\n$$\nFinally, we get\n$$\n\\sum \\frac{AM \\cdot AN}{AB \\cdot AC} = \\sum \\frac{(m-a)(n-a)}{(b-a)(c-a)} \\\\\n= \\frac{\\sum (m-a)(n-a)(b-c)}{(a-b)(b-c)(c-a)} = 1\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among any nine divisors of $30^{2010}$ there are two whose product is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us factor $30^{2010} = 2^{2010} \\cdot 3^{2010} \\cdot 5^{2010}$.\n\nAny divisor $d$ of $30^{2010}$ can be written as $d = 2^a 3^b 5^c$ where $0 \\leq a, b, c \\leq 2010$.\n\nThe product of two divisors $d_1 = 2^{a_1} 3^{b_1} 5^{c_1}$ and $d_2 = 2^{a_2} 3^{b_2} 5^{c_2}$ is a perfect square if and only if $a_1 + a_2$, $b_1 + b_2$, and $c_1 + c_2$ are all even.\n\nThis is equivalent to $a_1 \\equiv a_2 \\pmod{2}$, $b_1 \\equiv b_2 \\pmod{2}$, $c_1 \\equiv c_2 \\pmod{2}$.\n\nThus, for each divisor, consider the triple $(a \\bmod 2, b \\bmod 2, c \\bmod 2)$. There are $2 \\times 2 \\times 2 = 8$ possible such triples.\n\nIf we select 9 divisors, by the pigeonhole principle, at least two of them must have the same triple. For these two divisors, the exponents of $2$, $3$, and $5$ have the same parity, so their sum is even for each prime, and thus their product is a perfect square.\n\nTherefore, among any nine divisors of $30^{2010}$, there are two whose product is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ for which $n(n+2010)$ is a perfect square.",
    "options": [],
    "answer": "All integers n of the form n = -1005 ± (d + 2010^2/d)/4, where d ranges over positive divisors of 2010^2 such that d + 2010^2/d is divisible by 4.",
    "solution": "Let $n(n+2010) = k^2$ for some integer $k$.\n\nWe can write:\n$$\nn(n+2010) = k^2\n$$\n$$\nn^2 + 2010n - k^2 = 0\n$$\nThis is a quadratic in $n$:\n$$\nn^2 + 2010n - k^2 = 0\n$$\nThe discriminant must be a perfect square for $n$ to be integer:\n$$\n\\Delta = 2010^2 + 4k^2 = m^2\n$$\nfor some integer $m$.\n\nSo:\n$$\nm^2 - 4k^2 = 2010^2\n$$\n$$\n(m - 2k)(m + 2k) = 2010^2\n$$\nLet $d$ be a positive divisor of $2010^2$, and set:\n$$\nm - 2k = d, \\quad m + 2k = \\frac{2010^2}{d}\n$$\nThen:\n$$\nm = \\frac{d + \\frac{2010^2}{d}}{2} = \\frac{d + 2010^2/d}{2}\n$$\n$$\n2k = m + 2k - m = \\frac{2010^2}{d} - d\n$$\n$$\nk = \\frac{2010^2 - d^2}{4d}\n$$\nNow, $n$ is given by the quadratic formula:\n$$\nn = \\frac{-2010 \\pm m}{2}\n$$\nSo:\n$$\nn = \\frac{-2010 \\pm \\frac{d + 2010^2/d}{2}}{2} = \\frac{-2010}{2} \\pm \\frac{d + 2010^2/d}{4}\n$$\nThus, all integers $n$ for which $n(n+2010)$ is a perfect square are:\n$$\nn = \\frac{-2010}{2} \\pm \\frac{d + 2010^2/d}{4}\n$$\nwhere $d$ is a positive divisor of $2010^2$ such that $d + 2010^2/d$ is divisible by $4$ (so $n$ is integer).\n\nTherefore, all such integers $n$ are given by the above formula for all positive divisors $d$ of $2010^2$ with $d + 2010^2/d$ divisible by $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$ and a point $P$ in its interior. Lines $PA$, $PB$, $PC$ intersect $BC$, $CA$, $AB$ at $A'$, $B'$, $C'$, respectively. Prove that\n$$\n\\frac{BA'}{BC} + \\frac{CB'}{CA} + \\frac{AC'}{AB} = \\frac{3}{2}\n$$\nif and only if at least two of the triangles $PAB$, $PBC$, $PCA$ have the same area.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $AB = c$, $BC = a$, $CA = b$, $BA' = x$, $CB' = y$, $AC' = z$. The condition in the problem $\\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} = \\frac{3}{2}$ is equivalent to\n$$\n\\begin{equation*}\nbcx + cay + abz = \\frac{3}{2}abc \\tag{1}\n\\end{equation*}\n$$\nFrom Ceva's Theorem,\n$$\n\\frac{x}{a-x} \\cdot \\frac{y}{b-y} \\cdot \\frac{z}{c-z} = 1,\n$$\nso\n$$\n\\begin{gathered}\nxyz = (a-x)(b-y)(c-z) = abc - (bcx + cay + abz) \\\\\n+ (axy + byz + czx) - xyz\n\\end{gathered}\n$$\nor\n$$\n2xyz = \\frac{1}{4}abc + \\frac{3}{4}abc - (bcx + cay + abz) + (axy + byz + czx).\n$$\nThe (1) is equivalent to\n$$\n2xyz = \\frac{1}{4}abc - \\frac{1}{2}(bcx + cay + abz) + (axy + byz + czx),\n$$\nwhich can be written as\n$$\n8xyz - 4(axy + byz + czx) + 2(bcx + cay + abz) - abc = 0\n$$\nso $(2x - a)(2y - b)(2z - c) = 0$.\n\nThis means that at least one of the segments $AA'$, $BB'$, $CC'$ is a median in triangle $ABC$ and the conclusion follows.\n\n\nSolution 2:\n\nDenote\n$$\nx = \\frac{A'C}{A'B}, \\quad y = \\frac{B'A}{B'C}, \\quad z = \\frac{C'B}{C'A}.\n$$\nFrom Ceva's Theorem, we have $xyz = 1$.\nThe condition in the problem is equivalent to\n$$\n\\frac{1}{1+x} + \\frac{1}{1+y} + \\frac{1}{1+z} = \\frac{3}{2}\n$$\nThat is\n$$\n\\begin{gathered}\n2[(x+1)(y+1) + (y+1)(z+1) + (z+1)(x+1)] \\\\\n= 3(x+1)(y+1)(z+1)\n\\end{gathered}\n$$\nWe obtain\n$$\n\\begin{gathered}\n2(xy + yz + zx) + 4(x + y + z) + 6 = 3xyz + 3(xy + yz + zx) \\\\\n+ 3(x + y + z) + 3\n\\end{gathered}\n$$\nhence\n$$\n3xyz - (x + y + z) + (xy + yz + zx) - 3 = 0\n$$\nthat is\n$$\n1 - (x + y + z) + (xy + yz + zx) - 1 = 0.\n$$\nIt follows\n$$\nxyz - (x + y + z) + (xy + yz + zx) - 1 = 0\n$$\nhence\n$$\n(x - 1)(y - 1)(z - 1) = 0\n$$\nand the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39936,
    "subject": "Mathematics (Multi-modal)",
    "question": "The squares $OAB C$ and $OA_{1}B_{1}C_{1}$ are situated in the same plane and are directly oriented. Prove that the lines $AA_{1}$, $BB_{1}$, and $CC_{1}$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AA_{1} \\cap CC_{1} = \\{M\\}$. We have $\\triangle AOA_{1} \\equiv \\triangle COC_{1}$ (S.A.S.), then $\\widehat{AA_{1}O} \\equiv \\widehat{MA_{1}O} \\equiv \\widehat{CC_{1}O}$. It follows that $A_{1}C_{1}MO$ is cyclic, hence $\\widehat{A_{1}MC} \\equiv \\widehat{A_{1}OC_{1}} = 90^{\\circ}$, that is $A_{1}M \\perp CC_{1}$.\n\nThe quadrilateral $A_{1}B_{1}C_{1}M$ is cyclic, then $\\widehat{A_{1}MB_{1}} \\equiv \\widehat{A_{1}C_{1}B_{1}} = 45^{\\circ}$.\n\nThe quadrilateral $AOCM$ is cyclic, hence we have $\\widehat{OMC} \\equiv \\widehat{OAC} = 45^{\\circ}$ and $\\widehat{AMO} \\equiv \\widehat{ACO} = 45^{\\circ}$.\n\nThe quadrilateral $OBMC$ is cyclic, since $\\widehat{OMC} \\equiv \\widehat{OBC} = 45^{\\circ}$. It follows $\\widehat{OMB} = 90^{\\circ}$.\n\n![](attached_image_1.png)\n\nWe have\n$$\n\\widehat{A_{1}MB} = \\widehat{OMB} - \\widehat{AMO} = 90^{\\circ} - 45^{\\circ} = 45^{\\circ},\n$$\n\nhence $\\widehat{A_{1}MB_{1}} = \\widehat{AMB} = 45^{\\circ}$, that is points $M, B, B_{1}$ are collinear. It follows that the lines $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.\n\nSolution 2:\n\nWe use complex coordinates. Assume that the origin of the complex plane is at $O$ and we have $A(1)$, $B(1+i)$, $C(i)$. If $A_{1}(a_{1})$, then $C_{1}(a_{1}i)$ and $B_{1}(a_{1}(1+i))$. As in the previous solution, consider $AA_{1} \\cap CC_{1} = \\{M\\}$, and $M(m)$. The points $A_{1}, A, M$ are collinear if and only if\n$$\n\\frac{m-a_{1}}{1-a_{1}} \\in \\mathbb{R}^{*}\n$$\nThat is equivalent to\n$$\n\\begin{equation*}\n\\frac{m-a_{1}}{1-a_{1}} = \\frac{\\bar{m}-\\bar{a}_{1}}{1-\\bar{a}_{1}} . \\tag{1}\n\\end{equation*}\n$$\nThe points $C_{1}, C, M$ are collinear if and only if\n$$\n\\frac{m-a_{1}i}{i-a_{1}i} \\in \\mathbb{R}^{*}\n$$\nhence\n$$\n\\begin{equation*}\n\\frac{m-a_{1}i}{i-a_{1}i} = \\frac{\\bar{m}+\\bar{a}_{1}i}{-i+\\bar{a}_{1}i} \\tag{2}\n\\end{equation*}\n$$\nFrom (1) and (2) we get\n$$\nm = \\frac{1}{2}\\left[-\\frac{(1+i)\\bar{a}_{1}(1-a_{1})}{1-\\bar{a}_{1}} + (1+i)a_{1}\\right] = \\frac{1+i}{2} \\cdot \\frac{a_{1}-\\bar{a}_{1}}{1-\\bar{a}_{1}} .\n$$\nWe have only to show that the points $B_{1}, B, M$ are collinear. That is\n$$\n\\frac{m-a_{1}(1+i)}{1+i-a_{1}(1+i)} \\in \\mathbb{R}^{*}\n$$\nIndeed,\n$$\n\\begin{gathered}\n\\frac{m-a_{1}(1+i)}{1+i-a_{1}(1+i)} = \\frac{\\frac{1}{2} \\cdot \\frac{a_{1}-\\bar{a}_{1}}{1-\\bar{a}_{1}} - a_{1}}{1-a_{1}} \\\\\n= -\\frac{a_{1}+\\bar{a}_{1}-a_{1}\\bar{a}_{1}}{2(1-a_{1})(1-\\bar{a}_{1})} \\in \\mathbb{R}^{*}\n\\end{gathered}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Evaluate the sum\n$$\n1+2+3-4-5+6+7+8-9-10+\\ldots-2010\n$$\nwhere each three consecutive signs $+$ are followed by two signs $-$.",
    "options": [],
    "answer": "401799",
    "solution": "We can write the sum as follows\n$$\n\\begin{gathered}\n\\sum_{k=0}^{401}[5k+1+5k+2+5k+3-(5k+4)-(5k+5)] \\\\\n=\\sum_{k=0}^{401}(5k-3)=5\\sum_{k=1}^{401}k-3\\cdot 402 \\\\\n=5 \\cdot \\frac{401 \\cdot 402}{2}-3 \\cdot 402 \\\\\n=402 \\cdot\\left(5 \\cdot \\frac{401}{2}-3\\right)=201 \\cdot 1999=401799\n\\end{gathered}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39938,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that for each positive integer $n$ there is a unique positive integer $a_{n}$ such that\n$$\n(1+\\sqrt{5})^{n}=\\sqrt{a_{n}}+\\sqrt{a_{n}+4^{n}}\n$$\n\nb) Prove that $a_{2010}$ is divisible by $5 \\cdot 4^{2009}$ and find the quotient.",
    "options": [],
    "answer": "F_{2010}^2",
    "solution": "(a) Let $(1+\\sqrt{5})^{n}=x_{n}+y_{n} \\sqrt{5}$, where $x_{n}, y_{n}$ are positive integers, $n=1,2, \\ldots$ Then\n$$\n(1-\\sqrt{5})^{n}=x_{n}-y_{n} \\sqrt{5}, \\quad n=1,2, \\ldots\n$$\nhence\n$$\n\\begin{equation*}\nx_{n}^{2}-5 y_{n}^{2}=(-4)^{n}, \\quad n=1,2, \\ldots \\tag{1}\n\\end{equation*}\n$$\nIf $n$ is even, consider $a_{n}=x_{n}^{2}-4^{n}$ and we have\n$$\n\\begin{gathered}\n\\sqrt{a_{n}}+\\sqrt{a_{n}+4^{n}}=\\sqrt{x_{n}^{2}-4^{n}}+\\sqrt{x_{n}^{2}}=\\sqrt{5 y_{n}^{2}}+\\sqrt{x_{n}^{2}} \\\\\n=y_{n} \\sqrt{5}+x_{n}=(1+\\sqrt{5})^{n}\n\\end{gathered}\n$$\nIf $n$ is odd, consider $a_{n}=5 y_{n}^{2}-4^{n}$ and we have\n$$\n\\begin{gathered}\n\\sqrt{a_{n}}+\\sqrt{a_{n}+4^{n}}=\\sqrt{5 y_{n}^{2}-4^{n}}+\\sqrt{5 y_{n}^{2}}=\\sqrt{x_{n}^{2}}+\\sqrt{5 y_{n}^{2}} \\\\\n=x_{n}+y_{n} \\sqrt{5}=(1+\\sqrt{5})^{n}\n\\end{gathered}\n$$\n\n(b) If $n$ is even, then we have $a_{n}=x_{n}^{2}-4^{n}=5 y_{n}^{2}$, where\n$$\n\\begin{gathered}\ny_{n}=\\frac{1}{2 \\sqrt{5}}\\left[(1+\\sqrt{5})^{n}-(1-\\sqrt{5})^{n}\\right] \\\\\n=\\frac{2^{n}}{2 \\sqrt{5}}\\left[\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}\\right]=2^{n-1} F_{n}\n\\end{gathered}\n$$\nwhere $F_{n}$ is the $n^{\\text{th}}$ Fibonacci number. In this case we get $a_{n}= 5 \\cdot 4^{n-1} F_{n}^{2}$, hence $5 \\cdot 4^{n-1} \\mid a_{n}$ and the quotient is $F_{n}^{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{0}$ be a positive integer and $a_{n+1} = \\sqrt{a_{n}^{2} + 1}$, for all $n \\geq 0$.\n\n1) Prove that for all $a_{0}$ the sequence contains infinitely many integers and infinitely many irrational numbers.\n\n2) Is there an $a_{0}$ for which $a_{2010}$ is an integer?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEFG$ be a regular heptagon. If $AC = m$ and $AD = n$, prove that $AB = \\frac{mn}{m + n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > 0$ be a real number and let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function satisfying\n$$\nf\\left(x_{1}\\right) + f\\left(x_{2}\\right) \\geq a f\\left(x_{1} + x_{2}\\right), \\forall x_{1}, x_{2} \\in \\mathbb{R} .\n$$\nProve that\n$$\nf\\left(x_{1}\\right) + f\\left(x_{2}\\right) + f\\left(x_{3}\\right) \\geq \\frac{3 a^{2}}{a+2} f\\left(x_{1} + x_{2} + x_{3}\\right), \\forall x_{1}, x_{2}, x_{3} \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the inequality in the hypothesis we get successively:\n$$\n\\begin{aligned}\n& f\\left(x_{1}\\right) + f\\left(x_{2}\\right) + a f\\left(x_{3}\\right) \\geq a^{2} f\\left(x_{1} + x_{2} + x_{3}\\right) \\\\\n& f\\left(x_{1}\\right) + a f\\left(x_{2}\\right) + f\\left(x_{3}\\right) \\geq a^{2} f\\left(x_{1} + x_{2} + x_{3}\\right) \\\\\n& a f\\left(x_{1}\\right) + f\\left(x_{2}\\right) + f\\left(x_{3}\\right) \\geq a^{2} f\\left(x_{1} + x_{2} + x_{3}\\right)\n\\end{aligned}\n$$\nAdd all and get:\n$$\n(a+2)\\left(f\\left(x_{1}\\right) + f\\left(x_{2}\\right) + f\\left(x_{3}\\right)\\right) \\geq 3 a^{2} f\\left(x_{1} + x_{2} + x_{3}\\right)\n$$\nand the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $a_{1} = 1$ and $a_{n+1} = \\frac{3 a_{n}^{2} + 1}{2} - a_{n}$ for $n = 1, 2, \\ldots$. Prove that if $n$ is a power of $3$ then $n$ divides $a_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The recursive relation is equivalent to\n$$\n3 a_{n+1} - 1 = \\frac{1}{2} \\left(9 a_{n}^{2} - 6 a_{n} + 1\\right), \\quad n = 1, 2, \\ldots\n$$\nLet $x_{n} = 3 a_{n} - 1$, $n = 1, 2, \\ldots$, and get $x_{1} = 8 = 2^{3}$,\n$$\nx_{n+1} = \\frac{1}{2} x_{n}^{2}, \\quad n = 1, 2, \\ldots\n$$\nLet $x_{n} = 2^{y_{n}}$, $y_{1} = 3$, $n = 1, 2, \\ldots$, and obtain\n$$\ny_{n+1} = 2 y_{n} - 1, \\quad n = 1, 2, \\ldots\n$$\nIt follows $y_{n} = 2^{n} + 1$, $n = 1, 2, \\ldots$, hence $x_{n} = 2^{2^{n} + 1}$, and we get\n$$\na_{n} = \\frac{1}{3} \\left(2^{2^{n}} + 1\\right), \\quad n = 1, 2, \\ldots\n$$\nIf $n = 3^{k}$, then\n$$\n2^{2 \\cdot 3^{k}} = 4^{3^{k}} = (3 + 1)^{3^{k}} \\equiv 1 \\quad (\\bmod\\ 3^{k+1})\n$$\nhence\n$$\n\\left(2^{3^{k}} - 1\\right)\\left(2^{3^{k}} + 1\\right) \\equiv 0 \\quad (\\bmod\\ 3^{k+1})\n$$\nBut, we have\n$$\n2^{3^{k}} - 1 = (3 - 1)^{3^{k}} - 1 \\equiv -2 \\quad (\\bmod\\ 3^{k+1})\n$$\nand we obtain\n$$\n2^{3^{k}} + 1 \\equiv 0 \\quad (\\bmod\\ 3^{k+1})\n$$\nhence\n$$\n2^{3^{k}} \\equiv -1 \\quad (\\bmod\\ 3^{k+1}) .\n$$\nIt follows $2^{3^{k}} = 3^{k+1} m - 1$, for some odd integer $m$, hence\n$$\n\\begin{gathered}\n2^{2^{3^{k}+1} + 1} = 2^{3^{k+1} m} + 1 = (3 - 1)^{3^{k+1} m} + 1 \\\\\n\\equiv (-1)^{3^{k+1} m} + 1 \\quad (\\bmod\\ 3^{k+1}) \\equiv 0 \\quad (\\bmod\\ 3^{k+1})\n\\end{gathered}\n$$\nWe obtain $3^{k+1} \\mid 2^{2^{n}} + 1$, hence $3^{k} \\mid a_{n}$, that is $n \\mid a_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the arithmetic sequence $8, 21, 34, 47, \\ldots$.\n\na) Prove that this sequence contains infinitely many integers written only with digit $9$.\n\nb) How many such integers less than $2010^{2010}$ are in the sequence?",
    "options": [],
    "answer": "1106",
    "solution": "(a) We are looking for integers $m$ such that\n$$\n10^{m}-1=13 n+8 \\text{ for some positive integer } n\n$$\nThe last relation is equivalent to $10^{m}-9 \\equiv 0 \\pmod{13}$.\nWe will prove that $\\operatorname{ord}_{13}(10)=6$, that is, the smallest positive integer $s$ such that $10^{s} \\equiv 1 \\pmod{13}$ is $s=6$. Indeed, we have\n$$\n10^{6}=(13-3)^{6} \\equiv 3^{6}=(3^{3})^{2}=(26+1)^{2} \\equiv 1 \\quad (\\bmod\\ 13).\n$$\nMoreover, $10^{2} \\equiv 9 \\pmod{13}$ and $10^{3} \\equiv -1 \\pmod{13}$, hence\n$$\n\\operatorname{ord}_{13}(10)=6.\n$$\nFrom the previous property, it follows that for any integer $k \\geq 0$, we have $10^{6k} \\equiv 1 \\pmod{13}$. Therefore,\n$$\n10^{6k+2}=10^{6k} \\cdot 10^{2} \\equiv 9 \\quad (\\bmod\\ 13),\n$$\nhence the integers\n$$\n10^{6k+2}-1=\\underbrace{99\\ldots 9}_{6k+2},\\ k=0,1,\\ldots,\n$$\nbelong to the sequence, and all integers in the sequence containing only digit $9$ are of this form.\n\n(b) It is clear that we have to find the greatest integer $k$ such that $10^{6k+2}-1<2010^{2010}$, that is $10^{6k+2}-1<10^{6k+2}<2010^{2010}$. It follows $6k+2<2010 \\log 2010$, hence\n$$\n\\begin{gathered}\n3k+1<1005 \\log 201 \\cdot 10=1005+1005 \\log 201 \\\\\n=1005+2010+1005 \\log \\left(2+\\frac{1}{100}\\right) \\\\\n=3015+1005 \\log \\left(2+\\frac{1}{100}\\right)\n\\end{gathered}\n$$\nhence\n$$\nk=1004+\\left[\\frac{2}{3}+335 \\log \\left(2+\\frac{1}{100}\\right)\\right]=1105.\n$$\nThe desired number is $1106$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ be a strictly increasing function such that $f(f(n))=3 n$, for all $n \\in \\mathbb{N}$. Find $f(2010)$.\n(Note: $\\mathbb{N}=\\{0,1,2, \\ldots\\}$ ).",
    "options": [],
    "answer": "3843",
    "solution": "It follows $f(f(f(n)))=f(3 n)$, hence for all $n \\in \\mathbb{N}$ we have\n$$\nf(3 n)=3 f(n) \\tag{1}\n$$\nFor $n=0$ we find $f(0)=0$. We have $f(1) \\neq 1$. Indeed, if $f(1)=1$, then we get $3=f(f(1))=f(1)=1$, not possible. Hence $f(1)>1$, so $3=f(f(1))>f(1)>1$. It follows $f(1)=2$, and consequently $f(2)=f(f(1))=3$.\nNow we prove by induction that for all $n \\in \\mathbb{N}$, we have\n$$\nf\\left(3^{n}\\right)=2 \\cdot 3^{n} \\text{ and } f\\left(2 \\cdot 3^{n}\\right)=3^{n+1} \\tag{2}\n$$\nIndeed, the relations (2) hold for $n=0$. If they hold for a positive integer $n$, then we get\n$$\nf\\left(3^{n+1}\\right)=f\\left(3 \\cdot 3^{n}\\right)=3 f\\left(3^{n}\\right)=2 \\cdot 3^{n+1}\n$$\nand\n$$\nf\\left(2 \\cdot 3^{n+1}\\right)=3 f\\left(2 \\cdot 3^{n}\\right)=3^{n+1}\n$$\nThere are exactly $3^{n}-1$ positive integers $k$ such that $3^{n}<k< 2 \\cdot 3^{n}$. Also, there are exactly $3^{n}-1$ positive integers $k'$ such that\n$$\nf\\left(3^{n}\\right)=2 \\cdot 3^{n}<k'<3^{n+1}=f\\left(2 \\cdot 3^{n}\\right)\n$$\nThe function $f$ is strictly increasing, then we have\n$$\nf\\left(3^{n}+k\\right)=2 \\cdot 3^{n}+k, \\quad 0 \\leq k \\leq 3^{n}\n$$\nhence $f\\left(2 \\cdot 3^{n}+k\\right)=f\\left(f\\left(3^{n}+k\\right)\\right)=3\\left(3^{n}+k\\right)$.\nIn our case, we have $2010=3 \\cdot 670$ and it follows\n$$\nf(2010)=3 f(670)=3 f\\left(2 \\cdot 3^{5}+184\\right)=9\\left(3^{5}+184\\right)=3843 .\n$$\nLet $n=\\overline{a_{1} a_{2} \\ldots a_{t}}$ the representation of $n$ in base 3. We have, if $a_{1}=1$, then $f(n)=\\overline{2 a_{2} \\ldots a_{t}}$. If $a_{1}=2$, then $f(n)= \\overline{1 a_{2} a_{3} \\ldots a_{t} 0}$. These relations are direct consequences from (2). As in the previous solution we have $f(2010)=3 f(670)$. But $670=220211$, hence\n$$\n\\begin{gathered}\nf(670)=f\\left(220211_{3}\\right)=1202110_{3} \\\\\n=3^{6}+2 \\cdot 3^{5}+2 \\cdot 3^{3}+3^{2}+3=1281,\n\\end{gathered}\n$$\nand we get $f(2010)=3 \\cdot 1281=3843$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ for which $p^{2}-p+1$ is a perfect cube.",
    "options": [],
    "answer": "19",
    "solution": "Write the equation $p^{2}-p+1 = x^{3}$ as\n$$\np(p-1) = (x-1)\\left(x^{2}+x+1\\right).\n$$\nBecause $p > x$, $p$ divides $x^{2}+x+1$ so $x^{2}+x+1 = k p$ and $k(x-1) = p-1$, for some positive integer $k$. It follows that\n$$\nk^{2}(x-1) + k = x^{2} + x + 1\n$$\nand consider the following cases:\n\nCase 1. If $k^{2} \\leq x+1$, then\n$$\nk^{2}(x-1) + k \\leq x^{2} - 1 + k \\leq x^{2} - 1 + \\sqrt{x+1} < x^{2} + x + 1,\n$$\nnot possible.\n\nCase 2. If $k^{2} \\geq x+3$, then\n$$\n\\begin{aligned}\nk^{2}(x-1) + k & \\geq (x-1)(x+3) + k \\geq (x-1)(x+3) + \\sqrt{x+3} \\\\\n& = x^{2} + 2x - 3 + \\sqrt{x+3} > x^{2} + x + 1\n\\end{aligned}\n$$\nnot possible.\n\nTherefore $x+1 < k^{2} < x+3$, hence $k^{2} = x+2$. This implies\n$$\n(x+2)(x-1) + k = x^{2} + x + 1,\n$$\nthat is $k = 3$. It follows $x = 7$ and $p = 19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a circle of center $O$ and a chord $AB$ of it (not a diameter). Take a point $T$ on the ray $OB$. The perpendicular at $T$ onto $OB$ meets the chord $AB$ at $C$ and the circle at $D$ and $E$. Denote by $S$ the orthogonal projection of $T$ onto the chord $AB$. Prove that $AS \\cdot BC = TE \\cdot TD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nDenote $BB'$ the diameter corresponding to point $B$ and consider $\\alpha$ the angle $\\widehat{ABB'}$. Writing the power of $T$ with respect to the circle, we get\n$$\n\\begin{gathered}\nTE \\cdot TD = TB \\cdot TB' = TB \\cdot (2R - TB) = 2R \\cdot \\frac{BS}{\\cos \\alpha} - TB^2 \\\\\n= 2R \\cdot \\frac{BS}{\\cos \\alpha} - BC \\cdot BS\n\\end{gathered}\n$$\nwhere $R$ is the radius of the circle.\n\nWe have $AS \\cdot BC = TE \\cdot TD$ if and only if\n$$\nAS \\cdot BC = 2R \\cdot \\frac{BS}{\\cos \\alpha} - BC \\cdot BS,\n$$\nthat is equivalent to\n$$\n2R \\cdot \\frac{BS}{\\cos \\alpha} = AS \\cdot BC + BC \\cdot BS = BC \\cdot (AS + BS) = BC \\cdot AB.\n$$\nIt follows that the desired relation is equivalent to\n$$\nBC \\cdot \\frac{AB}{2R} = \\frac{BS}{\\cos \\alpha},\n$$\nthat is $BC \\cdot \\cos \\alpha = \\frac{BS}{\\cos \\alpha}$, hence $\\frac{BS}{BC} = \\cos^2 \\alpha$.\n\nOn the other hand, it is clear that\n$$\n\\cos \\alpha = \\frac{BS}{BT} \\text{ and } \\cos \\alpha = \\frac{BT}{BC}.\n$$\nMultiplying these relations we get $\\cos^2 \\alpha = BS / BC$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39947,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integers in the set $\\{1,2, \\ldots, 2010\\}$ divide $5^{2010!}-3^{2010!}$?",
    "options": [],
    "answer": "1072",
    "solution": "Let $k \\in \\{1,2, \\ldots, 2010\\}$. If $3 \\mid k$, then $k \\nmid 5^{2010!}-3^{2010!}$. Also, if $5 \\mid k$, then $k \\nmid 5^{2010!}-3^{2010!}$. It follows that any multiple of $3$ or $5$ in the set $\\{1,2, \\ldots, 2010\\}$ is not a divisor of $5^{2010!}-3^{2010!}$. Any number $k$ in $\\{1,2, \\ldots, 2010\\}$ which is not divisible by $3$ or $5$ is also a divisor of $5^{2010!}-3^{2010!}$. We have $\\varphi(k)<k<2010$, and from Euler's Theorem we get\n$$\n5^{2010!}-3^{2010!}=a^{\\varphi(k)}-b^{\\varphi(k)} \\equiv 1-1 \\quad(\\bmod k) .\n$$\nThe desired number is\n$$\n2010-\\left[\\frac{2010}{3}\\right]-\\left[\\frac{2010}{5}\\right]+\\left[\\frac{2010}{15}\\right]=1072 .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a b c = 8$. Prove that\n$$\n\\frac{a-2}{a+1} + \\frac{b-2}{b+1} + \\frac{c-2}{c+1} \\leq 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\left(a_{n}\\right)_{n \\geq 0}$ and $\\left(b_{n}\\right)_{n \\geq 0}$ be sequences defined by $a_{n+2}= a_{n+1}+a_{n}$, $n=0,1, \\ldots$, $a_{0}=1$, $a_{1}=2$, and $b_{n+2}=b_{n+1}+b_{n}$, $n=0,1, \\ldots$, $b_{0}=2$, $b_{1}=1$. How many integers do the sequences have in common?",
    "options": [],
    "answer": "3",
    "solution": "We have $a_{2}=3$, $a_{3}=5$, $a_{4}=8$, $\\ldots$ and $b_{2}=3$, $b_{3}=4$, $b_{4}=7$, $\\ldots$ It follows $a_{0}=b_{0}$, $a_{1}=b_{1}$, $a_{2}=b_{2}$, and $b_{3}<a_{3}<b_{4}<a_{4}<b_{5}$. We prove by induction of step 2 that for $m \\geq 3$ we have $b_{m}<a_{m}<b_{m+1}$. The basis cases $m=3$, $m=4$ are verified above. Assume that\n$$\nb_{k}<a_{k}<b_{k+1} \\quad \\text{ and } \\quad b_{k+1}<a_{k+1}<b_{k+2}\n$$\nAdding these inequalities we get $b_{k}+b_{k+1}<a_{k}+a_{k+1}<b_{k+1}+ b_{k+2}$, that is $b_{k+2}<a_{k+2}<b_{k+3}$.\nThe sequences have in common only three integers: $1,2,3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right angled triangle with $\\widehat{A}=90^{\\circ}$ and $BC=a$, $AC=b$, $AB=c$. Let $d$ be a line passing through the incenter of triangle and intersecting the sides $AB$ and $AC$ in $P$ and $Q$, respectively.\n\na. Prove that\n$$\nb \\cdot \\frac{PB}{PA} + c \\cdot \\frac{QC}{QA} = a ;\n$$\n\nb. Find the minimum of\n$$\n\\left(\\frac{PB}{PA}\\right)^{2} + \\left(\\frac{QC}{QA}\\right)^{2} .\n$$",
    "options": [],
    "answer": "1",
    "solution": "(a) Assume that the origin of the coordinates system is at $A$. Let $r$ be the inradius of $\\triangle ABC$, and $I$ the incenter. Then $I(r, r)$ and the line $d$ has equation\n$$\nd: y - r = m(x - r)\n$$\nthat is $y = r + m(x - r)$. We get\n$$\nP\\left(\\frac{r(m-1)}{m}, 0\\right), \\quad Q(0, r(m-1)) .\n$$\n![](attached_image_1.png)\nIt follows\n$$\n\\begin{aligned}\nb \\cdot \\frac{PB}{PA} & + c \\cdot \\frac{QC}{QA} = b \\cdot \\frac{c - \\frac{r(m-1)}{m}}{\\frac{r(m-1)}{m}} + c \\cdot \\frac{b + r(m-1)}{-r(m-1)} \\\\\n& = b \\cdot \\frac{m c - r(m-1)}{r(m-1)} + c \\cdot \\frac{b + r(m-1)}{-r(m-1)} \\\\\n& = b \\cdot \\frac{m c}{r(m-1)} - c \\cdot \\frac{b}{r(m-1)} - (b + c) \\\\\n& = \\frac{b c}{r(m-1)}(m-1) - (b + c) = \\frac{b c}{r} - (b + c) .\n\\end{aligned}\n$$\nBut $r = \\frac{2K}{a + b + c} = \\frac{b c}{a + b + c}$, hence $\\frac{b c}{r} = a + b + c$, and we get:\n$$\nb \\cdot \\frac{PB}{PA} + c \\cdot \\frac{QC}{QA} = (a + b + c) - (b + c) = a .\n$$\n\n(b) From the previous relation we have\n$$\n\\frac{b}{a} \\cdot \\frac{PB}{PA} + \\frac{c}{a} \\cdot \\frac{QC}{QA} = 1,\n$$\nso from Cauchy-Schwarz inequality it follows\n$$\n\\left[\\left(\\frac{PB}{PA}\\right)^{2} + \\left(\\frac{QC}{QA}\\right)^{2}\\right]\\left(\\frac{b^{2}}{a^{2}} + \\frac{c^{2}}{a^{2}}\\right) \\geq \\left(\\frac{b}{a} \\cdot \\frac{PB}{PA} + \\frac{c}{a} \\cdot \\frac{QC}{QA}\\right)^{2} = 1\n$$\nhence\n$$\n\\left(\\frac{PB}{PA}\\right)^{2} + \\left(\\frac{QC}{QA}\\right)^{2} \\geq 1\n$$\nThe minimum value is $1$ and it is obtained for the line $d$ satisfying the property\n$$\n\\frac{PB}{PA} \\cdot \\frac{a}{b} = \\frac{QC}{QA} \\cdot \\frac{a}{c} .\n$$\n(a) (Abdullah Al-Saeed). The relation is equivalent to\n$$\nb \\cdot \\frac{c - PA}{PA} + c \\cdot \\frac{b - QA}{QA} = a\n$$\nthat is\n$$\n\\begin{equation*}\nbc\\left(\\frac{1}{PA} + \\frac{1}{QA}\\right) = a + b + c \\tag{1}\n\\end{equation*}\n$$\nWe know that $K[ABC] = \\frac{b c}{2} = s r$, where $s$ is the semiperimeter of triangle $ABC$, hence\n$$\nr = \\frac{b c}{a + b + c},\n$$\nand by replacing in (1) we obtain\n$$\n\\begin{equation*}\n\\frac{r}{PA} + \\frac{r}{QA} = 1 \\tag{2}\n\\end{equation*}\n$$\nFrom the similarity we have $\\frac{r}{PA} = \\frac{QI}{QP}$ and $\\frac{r}{QA} = \\frac{PI}{QP}$, and a relation (2) follows.\n\n(b) From the previous relation we have\n$$\n\\frac{b}{a} \\cdot \\frac{PB}{PA} + \\frac{c}{a} \\cdot \\frac{QC}{QA} = 1,\n$$\nso from Cauchy-Schwarz inequality it follows\n$$\n\\left[\\left(\\frac{PB}{PA}\\right)^{2} + \\left(\\frac{QC}{QA}\\right)^{2}\\right]\\left(\\frac{b^{2}}{a^{2}} + \\frac{c^{2}}{a^{2}}\\right) \\geq \\left(\\frac{b}{a} \\cdot \\frac{PB}{PA} + \\frac{c}{a} \\cdot \\frac{QC}{QA}\\right)^{2} = 1\n$$\nhence\n$$\n\\left(\\frac{PB}{PA}\\right)^{2} + \\left(\\frac{QC}{QA}\\right)^{2} \\geq 1\n$$\nThe minimum value is $1$ and it is obtained for the line $d$ satisfying the property\n$$\n\\frac{PB}{PA} \\cdot \\frac{a}{b} = \\frac{QC}{QA} \\cdot \\frac{a}{c} .\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of integers such that\n$$\nx^{2} y + y^{2} z + z^{2} x = 2010^{2} \\text{ and } x y^{2} + y z^{2} + z x^{2} = -2010.\n$$",
    "options": [],
    "answer": "(1, 0, -2010), (-2010, 1, 0), (0, -2010, 1)",
    "solution": "We replace $2010$ by $p-1$, for some odd prime. Subtracting the first equation from the second, we obtain\n$$\n(x-y)(y-z)(z-x) = -p(p-1)\n$$\nWe have $(x-y)+(y-z)+(z-x)=0$ and $(x-y)(y-z)(z-x)<0$, so precisely two of them are positive. Assume that $x-y>0$ and $y-z>0$. Without loss of generality, suppose $x-y \\leq y-z$. Because $p$ is a prime, the only possibility is\n$$\nx-y=1, \\quad y-z=p-1, \\quad z-x=-p\n$$\nThen $x=y+1$, $z=y-(p-1)$, and the first equation reduces to\n$$\ny\\left[3y^{2} + 3(p-2)y + (p-2)^{2}\\right]=0\n$$\nThe only solution is $y=0$, implying $x=1$ and $z=p-1$.\nThe solutions $(x, y, z)$ are $(1,0,1-p)$, $(1-p, 1,0)$ and $(0,1-p, 1)$. We have $p=2011$, hence the desired triples are $(1,0,-2010)$, $(-2010,1,0)$, and $(0,-2010,1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers such that $a+b \\neq 0$. Solve the equation\n$$\n\\frac{1}{(x+a)^2-b^2}+\\frac{1}{(x+b)^2-a^2}=\\frac{1}{x^2-(a+b)^2}+\\frac{1}{x^2-(a-b)^2}\n$$",
    "options": [],
    "answer": "x = (a^2 + b^2) / (a + b) (with no solution if ab = 0 due to domain restrictions)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a>0$. If the system\n$$\n\\left\\{\\begin{array}{l}\na^{x}+a^{y}+a^{z}=14-a \\\\\nx+y+z=1\n\\end{array}\\right.\n$$\nhas a solution in real numbers, prove that $a \\leq 8$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, by contradiction, that $a>8$.\nThen $14-a<6$. Applying AM-GM inequality, we get\n$$\n6>14-a=a^{x}+a^{y}+a^{z} \\geq 3 a^{\\frac{x+y+z}{3}}>3 \\cdot 8^{\\frac{1}{3}}=3 \\cdot 8^{\\frac{1}{3}}=6,\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of real numbers that satisfy the system of equations\n$$\n\\begin{cases}\nx^{4} + 2x^{3} - y = \\sqrt{3} - \\dfrac{1}{4} \\\\\ny^{4} + 2y^{3} - x = -\\sqrt{3} - \\dfrac{1}{4}\n\\end{cases}\n$$",
    "options": [],
    "answer": "(x, y) ≈ (0.3842, -1.3468)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39955,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ with centroid $G$ let $M \\in (AB)$ and $N \\in (AC)$ be points on two of its sides. Prove that points $M$, $G$, $N$ are collinear if and only if\n$$\n\\frac{MB}{MA} + \\frac{NC}{NA} = 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\rightarrow [0, \\infty)$ be a function satisfying the following conditions:\n\na) $f(4) = 2$;\n\nb) $\\frac{1}{f(0) + f(1)} + \\frac{1}{f(1) + f(2)} + \\ldots + \\frac{1}{f(n) + f(n+1)} = f(n+1)$, for all integers $n \\geq 0$.\n\nFind $f(n)$ in closed form.",
    "options": [],
    "answer": "f(n) = sqrt(n)",
    "solution": "For $k \\geq 0$ we get\n$$\nf(k+2) - f(k+1) = \\frac{1}{f(k+1) + f(k+2)}\n$$\nhence $f^{2}(k+2) = f^{2}(k+1) + 1$. It follows\n$$\n4 = f^{2}(4) = f^{2}(3) + 1\n$$\nhence $f(3) = \\sqrt{3}$. Also, $3 = f^{2}(3) = f^{2}(2) + 1$ implies\n$$\nf(2) = \\sqrt{2} \\quad \\text{and} \\quad 2 = f^{2}(2) = f^{2}(1) + 1\n$$\ngives $f(1) = 1$. Finally, $f(1) = \\frac{1}{f(0) + f(1)}$ implies\n$$\n1 = \\frac{1}{f(0) + 1}\n$$\nhence $f(0) = 0$.\n\nNow we prove by induction that for any $n \\geq 0$, we have $f(n) = \\sqrt{n}$. Assume that $f(k+1) = \\sqrt{k+1}$, and get\n$$\nf^{2}(k+2) = f^{2}(k+1) + 1 = k+1+1 = k+2\n$$\nhence $f(k+2) = \\sqrt{k+2}$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39957,
    "subject": "Mathematics (Multi-modal)",
    "question": "In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $O$. Denote by $P, Q, R, S$ the orthogonal projections of $O$ onto $AB, BC, CD, DA$, respectively. Prove that\n$$\nPA \\cdot AB + RC \\cdot CD = \\frac{1}{2}\\left(AD^2 + BC^2\\right)\n$$\nif and only if\n$$\nQB \\cdot BC + SD \\cdot DA = \\frac{1}{2}\\left(AB^2 + CD^2\\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nDenote by $1,2, \\ldots, 8$ the right angled triangles as in the above figure. Applying Pythagoras theorem successively in triangles $1,2, \\ldots, 8$, we get:\n$$\n\\begin{aligned}\n& PA^2 + PO^2 - OA^2 = 0 \\\\\n& PB^2 - PO^2 - OB^2 = 0 \\\\\n& QB^2 + QO^2 - OB^2 = 0 \\\\\n& QC^2 + QO^2 - OC^2 = 0 \\\\\n& RC^2 + RO^2 - OC^2 = 0 \\\\\n& RD^2 + RO^2 - OD^2 = 0 \\\\\n& SD^2 + SO^2 - OD^2 = 0 \\\\\n& SA^2 + SO^2 - OA^2 = 0\n\\end{aligned}\n$$\nIt follows\n$$\n\\begin{equation*}\nPA^2 - PB^2 + QB^2 - QC^2 + RC^2 - RD^2 + SD^2 - SA^2 = 0. \\tag{1}\n\\end{equation*}\n$$\nThe last relation is equivalent to:\n$$\n\\begin{gathered}\nPA^2 - (AB - PA)^2 + QB^2 - (BC - QB)^2 + RC^2 - (CD - RC)^2 \\\\\n+ SD^2 - (DA - SD)^2 = 0\n\\end{gathered}\n$$\nhence\n$$\n\\begin{gathered}\n2(PA \\cdot AB + QB \\cdot BC + RC \\cdot CD + SD \\cdot DA) \\\\\n= AB^2 + BC^2 + CD^2 + DA^2\n\\end{gathered}\n$$\nand the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p, q$ satisfying the equation $2 p^{q} - q^{p} = 7$.",
    "options": [],
    "answer": "(2, 3), (5, 3)",
    "solution": "It is clear that $q$ must be odd, hence $q \\geq 3$.\n\nCase 1. If $p = 2$, then we get $2^{q+1} = q^{2} + 7$. For any $n \\geq 4$ we have $2^{n+1} > n^{2} + 7$ (by induction). Hence $q = 3$, and we get solution $(2, 3)$.\n\nCase 2. If $p \\geq 3$, then we can write, and using Fermat Little Theorem,\n$$\nq^{p} = 2 p^{q} - 7 = 2\\left(p^{q} - p\\right) + (2p - 7) \\Rightarrow q \\mid 2p - 7\n$$\nand\n$$\n2 p^{q} = q^{p} + 7 = \\left(q^{p} - q\\right) + (q + 7) \\Rightarrow p \\mid q + 7.\n$$\nSet $q + 7 = k p$.\nIf $2p - 7 \\leq 0 \\Rightarrow p = 3$, hence $q \\mid -1$, not possible.\nIf $2p - 7 > 0 \\Rightarrow 2p - 7 \\geq q$, hence $2p \\geq q + 7 = k p$, and we get $k \\leq 2$. Therefore, we have only two possibilities for $k$: $k = 1$, $k = 2$.\nIf $k = 1 \\Rightarrow q + 7 = p \\Rightarrow q \\mid 2q + 7 \\Rightarrow q = 7$ and $p = 14$, not a prime, contradiction.\nIf $k = 2 \\Rightarrow q + 7 = 2p$. If $q > p$, then we have $p, q \\geq 3$. Hence $7 = 2 p^{q} - q^{p} = p^{q} \\geq 27$, since $p^{q} > q^{p}$, not possible. Then $p > q$, and we get $q + 7 = 2p > 2q \\Rightarrow q < 7 \\Rightarrow q = 3$ or $q = 5$. In first case we obtain $p = 5$ and in the second case contradiction. Finally, the solutions are $(p, q) = (2, 3), (5, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AMNB$ be a quadrilateral inscribed in a semicircle of diameter $AB = x$. Denote $AM = a$, $MN = b$, $NB = c$. Prove that\n$$\nx^{3} - (a^{2} + b^{2} + c^{2})x - 2abc = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$, $n \\geq 2$, such that the numbers $1!$, $2!$, $\\ldots$, $(n-1)!$ give distinct remainders when divided by $n$.",
    "options": [],
    "answer": "2 and 3",
    "solution": "We claim that those integers are $n=2$ and $n=3$.\n\nSuppose $n$ is a composite integer, $n=a b$, $1<a \\leq b$. If $a<b$, then $n=a b$ divides $b!$ and $(b+1)!$. Since $b<b+1<n-1$, it follows that $b!$ and $(b+1)!$ yield equal remainders (both $0$) at division by $n$. If $2<a=b$ then $(2 a)!$ and $(2 a+1)!$ are divisible by $n=a^{2}$. Since $2 a+1<a^{2}-1=n-1$, it follows that $n$ does not satisfy the desired property.\n\nIt remains to settle the case of prime $n$. The value $n=2$ checks; let then $n \\geq 3$. From Wilson's theorem we have\n$$\n(n-1)!\\equiv-1 \\quad(\\bmod n)\n$$\nand then\n$$\n(n-2)!\\equiv 1 \\quad(\\bmod n) .\n$$\nSince $1!=1$ yields remainder $1$ at division by $n$, it follows $n-2=1$, that is $n=3$, which checks.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer.\nFind all real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ such that\n$$\n\\prod_{k=1}^{n}\\left(x_{k}^{2}+(k+2) x_{k}+k^{2}+k+1\\right)=\\left(\\frac{3}{4}\\right)^{n}(n!)^{2}\n$$",
    "options": [],
    "answer": "x_k = -k/2 - 1 for k = 1, 2, ..., n",
    "solution": "We have $t^{2}+(k+2) t+k^{2}+k+1 \\geq \\frac{3}{4} k^{2}, \\forall t \\in \\mathbb{R}$.\nIndeed, this inequality is equivalent to\n$$\nt^{2}+(k+2) t+\\frac{1}{4} k^{2}+k+1 \\geq 0\n$$\nhence\n$$\nt^{2}+(k+2) t+\\left(\\frac{1}{2} k+1\\right)^{2} \\geq 0\n$$\nthat is\n$$\n\\left(t+\\frac{1}{2} k+1\\right)^{2} \\geq 0\n$$\nWe have equality if and only if $t=-\\frac{1}{2} k-1$.\nIt follows that for any real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ we have\n$$\n\\prod_{k=1}^{n}\\left(x_{k}^{2}+(k+2) x_{k}+k^{2}+k+1\\right) \\geq \\prod_{k=1}^{n} \\frac{3}{4} k^{2}=\\left(\\frac{3}{4}\\right)^{n}(n!)^{2}\n$$\nwith equality if and only if $x_{k}=-\\frac{1}{2} k-1, k=1,2, \\ldots, n$, giving the unique solution to the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$ with real coefficients such that for all $x, y, z \\in \\mathbb{R}$,\n$$\nP(x) + P(y) + P(z) + P(x + y + z) = P(x + y) + P(y + z) + P(z + x).\n$$",
    "options": [],
    "answer": "P(t) = a t^2 + b t for real a, b",
    "solution": "Assume that $P(t) = a_n t^n + \\ldots + a_1 t + a_0$, where $a_n \\neq 0$. Taking $x = y = z$ in the given relation, it follows\n$$\n3 P(x) + P(3x) = 3 P(2x).\n$$\nLooking for the coefficient of $x^n$ in this equation, we obtain the relation $3 + 3^n = 3 \\cdot 2^n$, that is $3^{n-1} = 2^n - 1$. This relation holds for $n = 1$ and $n = 2$. For $n \\geq 3$ we prove by induction that $3^{n-1} > 2^n - 1$.\n\nIf $n = 0$, that is $P$ is a constant polynomial $P(t) = c$, then it follows, for example from the above, that $c = 0$.\n\nIf $n = 1$, that is $P(t) = b t$, $b \\neq 0$, then an easy checking shows that it satisfies the relation in the problem.\n\nIf $n = 2$, we can choose $P(t) = a t^2$, $a \\neq 0$, which also satisfies the relation.\n\nFinally, we find that all desired polynomials are\n$$\nP(t) = a t^2 + b t,\n$$\nwhere $a, b \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39963,
    "subject": "Mathematics (Multi-modal)",
    "question": "The shape of a military base is an equilateral triangle of side $10$ kilometers. Security constraints make cellular phone communication possible only within $2.5$ kilometers. Each of $17$ soldiers patrols the base randomly and tries to contact all others. Prove that at each moment at least two soldiers can communicate.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Divide any side of the equilateral triangle into four equal segments as follows. Then construct through the interior points to the sides parallel lines to the equilateral triangle.\n\n![](attached_image_1.png)\n\nIn this way we get a net of the military base consisting in $16$ equilateral triangles of side $2.5$ kilometers each. Since there are $17$ soldiers, according to the Pigeonhole Principle, at each moment at least two of them are in a triangle of the previous net, hence they can communicate.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$ and let $A'$, $B'$, $C'$ be midpoints of sides $BC$, $CA$, $AB$, respectively. If $IA' = IB' = IC'$, then prove that triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively.\n\n![](attached_image_1.png)\n\nSince $IA' = IB' = IC'$ it follows that triangles $IA_1A'$, $IB_1B'$, and $IC_1C'$ are congruent. We get $A_1A' = B_1B' = C_1C'$, hence\n$$\n\\begin{equation*}\n\\left|(s-b)-\\frac{a}{2}\\right| = \\left|(s-c)-\\frac{b}{2}\\right| = \\left|(s-a)-\\frac{c}{2}\\right|, \\tag{1}\n\\end{equation*}\n$$\nwhere $a$, $b$, $c$ are the length sides of triangle $ABC$, and $s$ its semiperimeter. The relations (1) are equivalent to\n$$\n\\begin{equation*}\n|c-b| = |a-c| = |b-a|. \\tag{2}\n\\end{equation*}\n$$\nFrom (2), considering all 6 possible orders for $a$, $b$, $c$, it follows $a = b = c$.\n\n![](attached_image_1.png)\nThe relations $IA' = IB' = IC'$ imply that $I = O_9$, the center of the Euler nine-point circle of triangle $ABC$. Hence $HI = IO$, where $H$ is the orthocenter and $O$ the circumcenter of triangle $ABC$.\n\n![](attached_image_2.png)\n\nBut $\\widehat{BAH} = \\widehat{OAC} = \\frac{\\pi}{2} - \\widehat{B}$, that is $\\widehat{HAI} = \\widehat{OAI}$. That is triangle $HAO$ is isosceles, hence $AH = AO$.\n\nIn similar way, we get $BH = BO$ and $CH = CO$. Since $OA = OB = OC$, it follows $HA = HB = HC$, that is $H = O$, hence triangle $ABC$ is equilateral.\n\n![](attached_image_2.png)\nWe have\n$IA^2 = r^2 + (s-a)^2$, $IB^2 = r^2 + (s-b)^2$, $IC^2 = r^2 + (s-c)^2$,\nwhere $a$, $b$, $c$ are the length sides of triangle $ABC$, $s$ the semiperimeter, and $r$ the inradius.\n\n![](attached_image_3.png)\n\nApplying the Median Theorem in triangle $BIC$, we get\n$IA'^2 = \\frac{1}{2}(IB^2 + IC^2) - \\frac{1}{4}BC^2 = r^2 + \\frac{1}{2}[(s-b)^2 + (s-c)^2] - \\frac{a^2}{4}$, and similarly\n$$\n\\begin{aligned}\n& IB'^2 = r^2 + \\frac{1}{2}[(s-a)^2 + (s-c)^2] - \\frac{b^2}{4} \\\\\n& IC'^2 = r^2 + \\frac{1}{2}[(s-a)^2 + (s-b)^2] - \\frac{c^2}{4}\n\\end{aligned}\n$$\nIt follows that $IA' = IB'$ if and only if $IA'^2 = IB'^2$, that is\n$$\n\\begin{equation*}\n(a-b)\\left[c - \\frac{1}{2}(a+b)\\right] = 0 \\tag{1}\n\\end{equation*}\n$$\nAlso, $IB' = IC'$ if and only if\n$$\n\\begin{equation*}\n(b-c)\\left[a - \\frac{1}{2}(b+c)\\right] = 0 \\tag{2}\n\\end{equation*}\n$$\nRelations (1) and (2) hold if and only if $a = b = c$.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$ there is an equiangular hexagon whose side-lengths are $n+1, n+2, \\ldots, n+6$ in some order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that the equiangular hexagon has the side-lengths $a_{1}, a_{2}, \\ldots, a_{6}$. Since all angles of the hexagon are $120^{\\circ}$, extending its sides we get an equilateral triangle.\nIt is clear that\n$$\na_{1}+a_{2}+a_{6}=a_{2}+a_{3}+a_{4}=a_{4}+a_{5}+a_{6}\n$$\nthat is\n$$\na_{1}+a_{6}=a_{3}+a_{4} \\quad \\text{ and } \\quad a_{2}+a_{3}=a_{5}+a_{6} \\tag{1}\n$$\nwhich are the necessary and sufficient conditions for a hexagon with side-lengths $a_{1}, a_{2}, \\ldots, a_{6}$ to be equiangular.\n\n![](attached_image_1.png)\n\nIf $a_{1}=n+1$, $a_{2}=n+6$, $a_{3}=n+2$, $a_{4}=n+4$, $a_{5}=n+3$, $a_{6}=n+5$, then (1) is verified and we are done.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be integers such that $a-b=a^{2} c-b^{2} d$ for some consecutive integers $c$ and $d$. Prove that $|a-b|$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d = c + 1$. The equality $a-b = a^{2} c - b^{2} (c + a)$ implies\n$$\n(a-b)[c(a+b)-1] = b^{2}.\n$$\nBut $c(a+b)-1$ and $a-b$ are relatively prime. Indeed, if $p$ is a prime dividing $a-b$, then the above equality shows that $p$ also divides $b$, so $p$ will divide $a+b = (a-b) + 2b$. Hence $p$ cannot divide $c(a+b)-1$. It follows that $|a-b|$ is a perfect square as well. A first nontrivial example is $18-22 = 18^{2}(-3) - 22^{2}(-2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{\\sin^{3} a}{\\sin b} + \\frac{\\cos^{3} a}{\\cos b} \\geq \\frac{1}{\\cos(a-b)},\n$$\nfor all $a$ and $b$ in the interval $\\left(0, \\frac{\\pi}{2}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a, b \\in \\left(0, \\frac{\\pi}{2}\\right)$ we have $\\sin a, \\sin b, \\cos a, \\cos b > 0$.\nThen\n$$\n\\begin{aligned}\n\\frac{\\sin^{3} a}{\\sin b} + \\frac{\\cos^{3} a}{\\cos b} &= \\frac{\\sin^{4} a}{\\sin a \\sin b} + \\frac{\\cos^{4} a}{\\cos a \\cos b} \\\\\n&= \\frac{(\\sin^{2} a)^2}{\\sin a \\sin b} + \\frac{(\\cos^{2} a)^2}{\\cos a \\cos b} \\\\\n&\\geq \\frac{(\\sin^{2} a + \\cos^{2} a)^2}{\\cos a \\cos b + \\sin a \\sin b} = \\frac{1}{\\cos(a-b)}\n\\end{aligned}\n$$\nwhere we have used the Cauchy-Schwarz inequality. We have equality if and only if\n$$\n\\frac{\\sin^{2} a}{\\sin a \\sin b} = \\frac{\\cos^{2} a}{\\cos a \\cos b}\n$$\nthat is $\\sin a \\cos b - \\sin b \\cos a = 0$, hence $\\sin(a-b) = 0$ yielding $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39968,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a semicircle of diameter $A B$ and center $C$ consider variable points $M$ and $N$ such that $M C \\perp N C$. The circumcircle of triangle $M N C$ intersects $A B$ for the second time at $P$. Prove that $\\frac{|P M-P N|}{P C}$ is a constant and find its value.",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Consider the case when point $P$ is between $C$ and $B$. Quadrilateral $M C P N$ is cyclic and from Ptolemy's relation it follows\n$$\nP M \\cdot C N = P C \\cdot M N + C M \\cdot P N \\tag{1}\n$$\n\n![](attached_image_1.png)\n\nWe have $C M = C N = R$, $M N = R \\sqrt{2}$, and replacing in (1) we obtain $P M = P C \\sqrt{2} + P N$, hence\n$$\n\\frac{P M - P N}{P C} = \\sqrt{2} .\n$$\n\nIf $P$ is between $C$ and $A$, then in similar way we get\n$$\n\\frac{P N - P M}{P C} = \\sqrt{2} .\n$$\n\nCombining these two cases we obtain\n$$\n\\frac{|P M - P N|}{P C} = \\sqrt{2} .\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n8(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\leq 9\\left(1+\\frac{a}{b}\\right)\\left(1+\\frac{b}{c}\\right)\\left(1+\\frac{c}{a}\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$$\n8(a+b+c)(ab+bc+ca) \\leq 9(a+b)(b+c)(c+a)\n$$\nUsing the identity\n$$\n(a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a) + abc\n$$\nwe get\n$$\n8(a+b)(b+c)(c+a) + 8abc \\leq 9(a+b)(b+c)(c+a),\n$$\nhence $8abc \\leq (a+b)(b+c)(c+a)$. Applying AM-GM inequality, it follows\n$$\n2\\sqrt{ab} \\leq a+b, \\quad 2\\sqrt{bc} \\leq b+c, \\quad 2\\sqrt{ca} \\leq c+a,\n$$\nand by multiplication $8abc \\leq (a+b)(b+c)(c+a)$. The equality holds if and only if $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle A = 90^\\circ$ and let $P$ be a point on the hypotenuse $BC$. Prove that\n$$\n\\frac{AB^2}{PC} + \\frac{AC^2}{PB} \\geq \\frac{BC^3}{PA^2 + PB \\cdot PC}\n$$",
    "options": [],
    "answer": "null",
    "solution": "Applying Cauchy-Schwarz inequality we have\n$$\n\\begin{gathered}\n\\frac{AB^2}{PC} + \\frac{AC^2}{PB} = \\frac{AB^4}{AB^2 \\cdot PC} + \\frac{AC^4}{AC^2 \\cdot PB} \\geq \\frac{(AB^2 + AC^2)^2}{AB^2 \\cdot PC + AC^2 \\cdot PB} \\\\\n= \\frac{BC^4}{AB^2 \\cdot PC + AC^2 \\cdot PB}\n\\end{gathered}\n$$\n![](attached_image_1.png)\nFrom Stewart's Theorem we have\n$$\nAB^2 \\cdot PC + AC^2 \\cdot PB = PA^2 \\cdot BC + PB \\cdot PC \\cdot BC\n$$\nand we get the inequality.\nWe have equality if and only if\n$$\n\\frac{AB^2}{PC \\cdot AB^2} = \\frac{AC^2}{PB \\cdot AC^2},\n$$\nthat is $PB = PC$, hence $P$ is the midpoint of segment $BC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that the interval\n$$\nI_{n} = \\left( \\frac{1 + \\sqrt{8n + 1}}{2}, \\frac{1 + \\sqrt{8n + 9}}{2} \\right)\n$$\ndoes not contain any integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that we have\n$$\n\\frac{1 + \\sqrt{8n + 1}}{2} < x < \\frac{1 + \\sqrt{8n + 9}}{2}\n$$\nfor some positive integer $x$. Then\n$$\n\\sqrt{8n + 1} < 2x - 1 < \\sqrt{8n + 9}\n$$\nhence $8n + 1 < 4x^{2} - 4x + 1 < 8n + 9$. It follows $2n < x^{2} - x < 2n + 2$, that is $x^{2} - x = 2n + 1$, not possible since $x^{2} - x$ is an even integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square of center $O$. The parallel to $AD$ through $O$ intersects $AB$ and $CD$ at $M$ and $N$ and a parallel to $AB$ intersects diagonal $AC$ at $P$. Prove that\n$$\nOP^{4} + \\left(\\frac{MN}{2}\\right)^{4} = MP^{2} \\cdot NP^{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AB = 2a$ and let $Q$ be the intersection point of the parallel to $AB$ with $MN$. Let $x = OQ$. Then $PQ = x$. We have\n$$\n\\begin{gathered}\nOP^{2} = 2x^{2}, \\quad MP^{2} = x^{2} + (a - x)^{2} \\\\\nNP^{2} = x^{2} + (a + x)^{2}, \\quad MN = 2a\n\\end{gathered}\n$$\n\n![](attached_image_1.png)\n\nThe relation is equivalent to\n$$\n4x^{4} + a^{4} = \\left[x^{2} + (a - x)^{2}\\right]\\left[x^{2} + (a + x)^{2}\\right]\n$$\nthat is\n$$\n4x^{4} + a^{4} = \\left(2x^{2} + a^{2} - 2ax\\right)\\left(2x^{2} + a^{2} + 2ax\\right)\n$$\nhence\n$$\n4x^{4} + a^{4} = \\left(2x^{2} + a^{2}\\right)^{2} - 4a^{2}x^{2}\n$$\nand we are done.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39973,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Geostationary Earth Orbit is situated directly above the equator and has a period equal to the Earth's rotational period. It is at the precise distance of $22,236$ miles above the Earth that a satellite can maintain an orbit with a period of rotation around the Earth exactly equal to $24$ hours. Because the satellites revolve at the same rotational speed of the Earth, they appear stationary from the Earth surface. That is why most station antennas (satellite dishes) do not need to move once they have been properly aimed at a target satellite in the sky. In an international project, a total of ten stations were equally spaced on this orbit (at the precise distance of $22,236$ miles above the equator). Given that the radius of the Earth is $3960$ miles, find the exact straight distance between two neighboring stations. Write your answer in the form $a+b \\sqrt{c}$, where $a, b, c$ are integers and $c>0$ is square-free.",
    "options": [],
    "answer": "-13098 + 13098 sqrt 5",
    "solution": "Let $A$ and $B$ be two neighboring stations. We have $\\widehat{A O B}=\\frac{\\pi}{5}$, hence $A B=2 R \\sin \\frac{\\pi}{10}$, where $R=22236+3960=26196$. We will prove that $\\sin \\frac{\\pi}{10}=\\frac{\\sqrt{5}-1}{4}$.\n\n![](attached_image_1.png)\n\nSince $\\sin \\pi=0$, then $\\sin \\left(\\frac{2 \\pi}{5}+\\frac{3 \\pi}{5}\\right)=0$. We have:\n$$\n\\begin{gathered}\n\\sin \\frac{2 \\pi}{5} \\cos \\frac{3 \\pi}{5}+\\cos \\frac{2 \\pi}{5} \\sin \\frac{3 \\pi}{5}=0 \\Rightarrow \\\\\n2 \\sin \\frac{\\pi}{5} \\cos \\frac{\\pi}{5}\\left(4 \\cos ^{3} \\frac{\\pi}{5}-3 \\cos \\frac{\\pi}{5}\\right) \\\\\n+\\left(2 \\cos ^{2} \\frac{\\pi}{5}-1\\right)\\left(3 \\sin \\frac{\\pi}{5}-4 \\sin ^{3} \\frac{\\pi}{5}\\right)=0\n\\end{gathered}\n$$\nWe may divide by $\\sin \\frac{\\pi}{5}$ and we get\n$$\n\\begin{gathered}\n2 \\cos \\frac{\\pi}{5}\\left(4 \\cos ^{3} \\frac{\\pi}{5}-3 \\cos \\frac{\\pi}{5}\\right) \\\\\n+\\left(2 \\cos ^{2} \\frac{\\pi}{5}-1\\right)\\left(-1+4 \\cos ^2 \\frac{\\pi}{5}\\right)=0\n\\end{gathered}\n$$\nDenote $y=\\cos \\frac{\\pi}{5}$. Then, our equation becomes\n$$\n16 y^{4}-12 y^{2}+1=0\n$$\nwith the solutions given by $y^{2}=\\frac{3 \\pm \\sqrt{5}}{8}$. Since $\\frac{\\pi}{6}<\\frac{\\pi}{5}<\\frac{\\pi}{4}$, then $\\cos \\frac{\\pi}{5} \\in\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{3}}{2}\\right)$, so that\n$$\ny^{2}=\\frac{3+\\sqrt{5}}{8} \\text{ and } \\cos \\frac{\\pi}{5}=\\sqrt{\\frac{3+\\sqrt{5}}{8}}=\\frac{\\sqrt{5}+1}{4}\n$$\nNow, $\\sin \\frac{\\pi}{10}=\\sqrt{\\frac{1-\\cos \\frac{\\pi}{5}}{2}}=\\frac{\\sqrt{5}-1}{4}$. It follows\n$$\nA B=26196 \\cdot \\frac{\\sqrt{5}-1}{2}=-13098+13098 \\sqrt{5}\n$$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 39974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer such that $2011^{2011}$ divides $n!$. Prove that $2011^{2012}$ divides $n!$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $2011$ is a prime and $2011^{2011}$ divides $n!$, it follows that in $n!$ we have at least $2011$ multiples of $2011$. These are\n$$\n2011, 2 \\cdot 2011, 3 \\cdot 2011, \\ldots, 2011 \\cdot 2011.\n$$\n\nTherefore $2011! \\cdot 2011^{2011}$ divides $n!$, hence $2011^{2012}$ divides $n!$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Set $A$ consists of 7 consecutive positive integers less than $2011$, while set $B$ consists of 11 consecutive positive integers. If the sum of the numbers in $A$ is equal to the sum of the numbers in $B$, what is the maximum possible element that $A$ could contain?",
    "options": [],
    "answer": "2005",
    "solution": "Let\n$A=\\{x+1, x+2, \\ldots, x+7\\}$ and $B=\\{y+1, y+2, \\ldots, y+11\\}$, where $x<2004$. We have\n$$(x+1)+(x+2)+\\ldots+(x+7)=(y+1)+(y+2)+\\ldots+(y+11)$$\nhence\n$$\n7x+\\frac{7 \\cdot 8}{2}=11y+\\frac{11 \\cdot 12}{2}\n$$\nThis equation is equivalent to\n$$\n7x-11y=38.\n$$\nThe smallest solution in positive integers to this is $(7,1)$ and all solutions are $x=7+11n$, $y=1+7n$, where $n$ is an arbitrary positive integer.\nSince $x<2004$, it follows $7+11n<2004$, that is $n<\\frac{1997}{11}$. We get $n \\leq 181$, hence the maximum possible element of $A$ is $x+7=14+11 \\cdot 181=2005$.\nChoose\n$$\nA=\\{a-3, a-2, a-1, a, a+1, a+2, a+3\\}\n$$\nand\n$$\nB=\\{b-5, b-4, \\ldots, b-1, b, b+1, \\ldots, b+4\\}\n$$\nWe have $7a=11b$, hence $11 \\mid a$ and $7 \\mid b$. It follows $a=11m$ and $b=7m$, for some positive integer $m$. But $a+3 \\leq 2010$ implies $11m \\leq 2007$, that is $m \\leq \\frac{2007}{11}$, hence $m=182$. The desired number is $11 \\cdot 182+3=2005$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB \\neq AC$. Its incircle has center $I$ and touches the side $BC$ at point $D$. Line $AI$ intersects the circumcircle $\\mathcal{C}$ of triangle $ABC$ at $M$ and $DM$ intersects again $\\mathcal{C}$ at $P$. Prove that $\\widehat{API}=90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $AB < AC$. We have $MB = MC = MI$. Indeed, from the hypothesis it follows\n$$\n\\widehat{IBM} = \\frac{1}{2}(\\widehat{A} + \\widehat{B}) = \\widehat{BIM},\n$$\nhence triangle $BMI$ is isosceles.\n\n![](attached_image_1.png)\n\nFurthermore, since\n$$\n\\widehat{BPM} = \\frac{1}{2} \\widehat{A} = \\widehat{MBC}\n$$\none has $\\triangle DBM \\sim \\triangle BPM$, hence $MB^2 = MD \\cdot MP$. It follows $MI^2 = MD \\cdot MP$, so $\\triangle MDI \\sim \\triangle MIP$, and consequently\n$$\n\\begin{equation*}\n\\widehat{MPI} = \\widehat{MID}. \\tag{1}\n\\end{equation*}\n$$\nNotice now that $ID \\parallel OM$, implying $\\widehat{MID} = \\widehat{AMO}$. Let $N$ be the antipodal point of $M$ in circle $\\mathcal{C}$. Finally, we have\n$$\n\\begin{gathered}\n\\widehat{API} = \\widehat{APM} - \\widehat{MPI} = \\widehat{APM} - \\widehat{AMN} \\\\\n= \\frac{1}{2}(\\overparen{AM} - \\overparen{AN}) = \\frac{1}{2} \\overparen{MN} = 90^{\\circ}\n\\end{gathered}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1} + a_{2} + \\ldots + a_{n} = 0$ and $|a_{1}| + |a_{2}| + \\ldots + |a_{n}| = 1$.\nProve that\n$$\n\\left|a_{1} + 2 a_{2} + \\ldots + n a_{n}\\right| \\leq \\frac{n-1}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\sum_{k=1}^{n} a_{k} = 0$, it follows that\n$$\n\\sum_{k=1}^{n} k a_{k} = \\sum_{k=1}^{n} (k - x) a_{k}\n$$\nfor every $x \\in \\mathbb{R}$. We obtain\n$$\n\\begin{aligned}\n\\left|\\sum_{k=1}^{n} k a_{k}\\right| & = \\left|\\sum_{k=1}^{n} (k - x) a_{k}\\right| \\leq \\sum_{k=1}^{n} |k - x| \\left|a_{k}\\right| \\\\\n& \\leq M(x) \\sum_{k=1}^{n} \\left|a_{k}\\right| = M(x)\n\\end{aligned}\n$$\nwhere $M(x) = \\max \\{|k - x| : k = 1, 2, \\ldots, n\\}$.\nFor $x = \\frac{n+1}{2}$ one has $M(x) = \\frac{n-1}{2}$, and the conclusion follows.\nFor each $k = 1, \\ldots, n$, we have\n$$\n\\left|a_{1} + \\ldots + a_{k}\\right| + \\left|a_{k+1} + \\ldots + a_{n}\\right| \\leq |a_{1}| + \\ldots + |a_{n}| = 1.\n$$\nThe relation $\\left|a_{1} + \\ldots + a_{k}\\right| = \\left|a_{k+1} + \\ldots + a_{n}\\right|$ implies\n$$\n\\left|a_{1} + \\ldots + a_{k}\\right| \\leq \\frac{1}{2}, \\quad k = 1, \\ldots, n\n$$\nNow we can write\n$$\n\\begin{aligned}\n\\left|a_{1} + 2 a_{2} + \\ldots + n a_{n}\\right| & \\leq |a_{1} + \\ldots + a_{n}| + |a_{2} + \\ldots + a_{n}| + \\ldots + |a_{n}| \\\\\n& \\leq \\frac{n-1}{2},\n\\end{aligned}\n$$\nand we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pentagon $A B C D E$ is inscribed in a circle. Distances from point $E$ to lines $A B$, $B C$ and $C D$ are equal to $a$, $b$ and $c$, respectively. Find the distance from point $E$ to line $A D$.",
    "options": [],
    "answer": "ac/b",
    "solution": "Let $K$, $L$, $M$ and $N$ be the feet of perpendiculars dropped from point $E$ to lines $A B$, $B C$, $C D$ and $D A$, respectively.\n\n\n\nPoints $K$ and $N$ lie on the circle with diameter $A E$, hence $\\widehat{E K N} \\equiv \\widehat{E A N}$. Similarly, $\\widehat{E L M} \\equiv \\widehat{E C M} \\equiv \\widehat{E A N}$, hence\n$$\n\\begin{equation*}\n\\widehat{E K N} \\equiv \\widehat{E L N} . \\tag{1}\n\\end{equation*}\n$$\nIn the same way we obtain\n$$\n\\begin{equation*}\n\\widehat{E N K} \\equiv \\widehat{E M L} . \\tag{2}\n\\end{equation*}\n$$\nFrom (1) and (2) it follows that $\\triangle E K N \\sim \\triangle E L M$, therefore\n$$\n\\frac{E K}{E N}=\\frac{E L}{E M}, \\text{ hence } E N=\\frac{E K \\cdot E M}{E L}=\\frac{a c}{b} .\n$$\n\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39979,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$ let $a_{n}$ be the number of pairs $(x, y)$ of integers satisfying $|x^{2}-y^{2}|=n$.\n\na. Find $a_{1432}$ and $a_{1433}$.\n\nb. Find $a_{n}$.",
    "options": [],
    "answer": "a_1432 = 16, a_1433 = 8; and in general\n- a_n = 4·tau(n) if n is odd,\n- a_n = 0 if n ≡ 2 mod 4,\n- a_n = 4·tau(n/4) if n is divisible by 4.",
    "solution": "Let us consider the following cases.\n\nCase 1: $n$ is odd. The equation $|x^{2}-y^{2}|=n$ is equivalent to\n$$\n|x-y||x+y|=n,\n$$\nhence $|x-y|=d$ and $|x+y|=\\frac{n}{d}$, where $d$ is a divisor of $n$. The system\n$$\n\\left\\{\\begin{array}{l}\n|x-y|=d  \\\\\n|x+y|=\\frac{n}{d}\n\\end{array}\\right.\n$$\nhas four solutions, hence we have $a_{n}=4 \\tau(n)$, where $\\tau(n)$ is the number of divisors of $n$.\n\nCase 2: $n=4k+2$. In this case $x^{2} \\equiv y^{2} (\\bmod 4)$, that is\n$$\nx^{2}-y^{2} \\equiv 0 \\quad(\\bmod 4),\n$$\ngiving $a_{n}=0$.\n\nCase 3: $n=2^{s}(2k+1),\\ s \\geq 2$. The system (1) has no solutions when $d \\mid 2k+1$ or $\\left.\\frac{n}{d} \\right\\rvert\\, 2k+1$. It follows that\n$$\na_{n}=4(\\tau(n)-2 \\tau(2k+1)) .\n$$\nAssume that $n=2^{s} p_{1}^{\\alpha_{1}} \\ldots p_{t}^{\\alpha_{t}}$, where $p_{1}, \\ldots, p_{t}$ are odd primes. Then,\n$$\n\\begin{aligned}\na_{n} & =4\\left((s+1)\\left(\\alpha_{1}+1\\right) \\ldots\\left(\\alpha_{t}+1\\right)-2\\left(\\alpha_{1}+1\\right) \\ldots\\left(\\alpha_{t}+1\\right)\\right) \\\\\n& =4(s-1)\\left(\\alpha_{1}+1\\right) \\ldots\\left(\\alpha_{t}+1\\right) \\\\\n& =4 \\tau\\left(\\frac{n}{4}\\right)\n\\end{aligned}\n$$\nFinally, we get\n$$\na_{n}= \\begin{cases}4 \\tau(n) & \\text{ if } n \\text{ is odd } \\\\ 0 & \\text{ if } n=4k+2 \\\\ 4 \\tau\\left(\\frac{n}{4}\\right) & \\text{ if } n=2^{s}(2k+1),\\ s \\geq 2 .\\end{cases}\n$$\n\nFor $n=1432=2^{3} \\cdot 179$, we have $a_{1432}=4 \\tau(2 \\cdot 179)=16$. For $n=1433$, we have $a_{1433}=4 \\tau(1433)=8$, since $1433$ is a prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39980,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $A B C$, let $I_{a}, I_{b}, I_{c}$ be the centers of the excircles tangent to sides $B C, C A, A B$, respectively. Let $P$ and $Q$ be the tangency points of the excircle of center $I_{a}$ with lines $A B$ and $A C$. Line $P Q$ intersects $I_{a} B$ and $I_{a} C$ at $D$ and $E$. Let $A_{1}$ be the intersection of $D C$ and $B E$. In an analogous way we define points $B_{1}$ and $C_{1}$. Prove that $A A_{1}, B B_{1}, C C_{1}$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove that $A_{1}$ is the orthocenter of triangle $I_{a} B C$. Indeed, we have\n$$\n\\begin{aligned}\n\\widehat{P E I_{a}} & =180^{\\circ}-\\widehat{P Q A}-\\widehat{E C Q} \\\\\n& =180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\widehat{A}\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\widehat{C}\\right) \\\\\n& =\\frac{1}{2}(\\widehat{A}+\\widehat{C})=90^{\\circ}-\\frac{1}{2} \\widehat{B}=\\widehat{P B I_{a}}\n\\end{aligned}\n$$\nhence quadrilateral $B E I_{a} P$ is cyclic. Since $\\widehat{B P I_{a}}=90^{\\circ}$, it follows $B E \\perp C I_{a}$. In an analogous way we get $C D \\perp B I$, hence $A_{1}$ is the orthocenter of triangle $I_{a} B C$.\n\n![](attached_image_1.png)\n\nIt follows that $B A_{1} \\parallel C I$, where $I$ is the incenter of triangle $A B C$. Also, $C A_{1} \\parallel B I$, hence $B I C A_{1}$ is a parallelogram.\nSimilarly, $A I B C_{1}$ is a parallelogram, hence $A C_{1} A_{1} C$ is a parallelogram. We obtain that the segments $A A_{1}$ and $C C_{1}$ have the same midpoint. In an analogous way, the segments $B B_{1}$ and $A A_{1}$ have the same midpoint, and the conclusion follows.\n\n![](attached_image_1.png)\n\nRemark. It is clear that $I_{b}, A, I_{c}$ and $I_{c}, B, I_{a}$, and $I_{a}, C, I_{b}$ are collinear.\n![](attached_image_2.png)\nAs in the previous solution, $A_{1}$ is the orthocenter of triangle $I_{a} B C$, hence $I_{a} A_{1} \\perp B C$. Similarly, $I_{b} B_{1} \\perp C A$ and $I_{c} C_{1} \\perp A B$.\nThe triangles $A B C$ and $I_{a} I_{b} I_{c}$ are orthological, that is the perpendicular lines through $A, B, C$ on $I_{b} I_{c}, I_{c} I_{a}$, and $I_{a} I_{b}$, respectively, are concurrent (as internal bisectors of triangle $A B C$ ). It follows that also, $I_{a} A_{1}, I_{b} B_{1}, I_{c} C_{1}$ are concurrent.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for each $n \\geq 4$ a parallelogram can be dissected in $n$ cyclic quadrilaterals.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ABCD$ be a parallelogram. If $ABCD$ is a rectangle, then it is clear that we can dissect it into $n$ rectangles by parallel lines to its sides.\n\nAssume that $ABCD$ is not a rectangle. Without loss of generality we can assume that $AB \\geq AD$. Let $E$ and $F$ be the midpoints of $BC$ and $AD$, respectively. There are two cases: $\\angle A < 90^{\\circ}$ and $\\angle A > 90^{\\circ}$. Because the second case is analogous to the first one, we shall study only the situation $\\angle A < 90^{\\circ}$.\n\n![](attached_image_1.png)\n\nConstruct the isosceles triangle $AFG$, with $AF = FG$ and vertex $G$ on line $AB$. Because $\\angle A < 90^{\\circ}$ we have\n$$\nAG = 2AF \\cos A < 2AF = AD \\leq AB,\n$$\nhence $G$ belongs to the interior of the segment $AB$. Let $G'$ be any point in the interior of segment $GB$. Construct the parallelogram $GG'F'F$. Then $AFF'G'$ and $G'BEF'$ are isosceles trapezoids, hence they are cyclic quadrilaterals. Similarly, we divide $F E C D$ into two isosceles trapezoids, hence $ABCD$ is dissected into 4 isosceles trapezoids.\n\nIn order to pass from $n$ to $n+1$ cyclic quadrilaterals it is sufficient to notice that an isosceles trapezoid is divided into isosceles trapezoids by a parallel line to the basis.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find positive integers $a_{1} < a_{2} < \\ldots < a_{2010}$ such that\n$$\na_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \\ldots + a_{2010}(2010!)^{2010} = (2011!)^{2010}.\n$$",
    "options": [],
    "answer": "a1 = 2^{2010}, a2 = 3^{2010} − 1, a3 = 4^{2010} − 1, ..., a2010 = 2011^{2010} − 1",
    "solution": "Consider\n$a_{1} = 2^{2010},\\ a_{2} = 3^{2010} - 1,\\ a_{3} = 4^{2010} - 1,\\ \\ldots,\\ a_{2010} = 2011^{2010} - 1$\nand get\n$$\n\\begin{gathered}\na_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \\ldots + a_{2010}(2010!)^{2010} \\\\\n= (2!)^{2010} + (3!)^{2010} - (2!)^{2010} + (4!)^{2010} - (3!)^{2010} + \\ldots + \\\\\n+ (2011!)^{2010} - (2010!)^{2010} = (2011!)^{2010}\n\\end{gathered}\n$$\nMoreover, it is clear that $a_{1} < a_{2} < a_{3} < \\ldots < a_{2010}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the set $S=\\{(a+b)^{7}-a^{7}-b^{7}: a, b \\in \\mathbb{Z}\\}$. Find the greatest common divisor of all members in $S$.",
    "options": [],
    "answer": "42",
    "solution": "Let $d$ be the greatest common divisor of the numbers in $S$. Since $2^{7}-2=126=2 \\cdot 3^{2} \\cdot 7 \\in S$ for $x=y=1$, one gets that $d$ divides $126$. Now, for $x=2$ and $y=1$, one gets $(2+1)^{7}-2^{7}-1=3^{7}-2^{7}-1 \\in S$, therefore $d$ will divide the number $(3^{7}-2^{7}-1)+(2^{7}-2)=3^{7}-3$. Since $9$ does not divide $3^{7}-3$, it follows that $d$ divides $126 / 3=42$.\n\nLet us check that $d=42$, i.e. the prime numbers $2$, $3$ and $7$ divide $(x+y)^{7}-x^{7}-y^{7}$, for all integers $x$ and $y$. Notice it is enough to show that $42$ divides $a^{7}-a$ for all integers $a$, since one can write\n$$(x+y)^{7}-x^{7}-y^{7}=\\left((x+y)^{7}-(x+y)\\right)-\\left(x^{7}-x\\right)-\\left(y^{7}-y\\right).$$\nBut $a^{7}-a=a(a-1)(a+1)\\left(a^{2}-a+1\\right)\\left(a^{2}+a+1\\right)$. Now, $2$ divides $a(a-1)$, $3$ divides $a(a-1)(a+1)$, while $7$ divides $a^{7}-a$, by Fermat's Little Theorem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39984,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ let the set $A_{n}$ consist of all numbers $\\pm 1 \\pm 2 \\pm \\ldots \\pm n$. For example,\n$$\n\\begin{gathered}\nA_{1}=\\{-1,1\\}, \\quad A_{2}=\\{-3,-1,1,3\\} \\\\\nA_{3}=\\{-6,-4,-2,0,2,4,6\\}\n\\end{gathered}\n$$\nFind the number of elements in $A_{n}$.",
    "options": [],
    "answer": "n(n+1)/2 + 1",
    "solution": "The greatest element of set $A_{n}$ is\n$$\n1+2+\\ldots+n=\\frac{n(n+1)}{2}\n$$\nand the smallest element of $A_{n}$ is\n$$\n-1-2-\\ldots-n=-\\frac{n(n+1)}{2} .\n$$\nAlso, the difference of any two elements of $A_{n}$ is even, hence all elements of $A_{n}$ are of the same parity.\nLet us prove that all integers between $-\\frac{n(n+1)}{2}$ and $\\frac{n(n+1)}{2}$, and of the same parity with $\\frac{n(n+1)}{2}$, belong to $A_{n}$, and these are all the elements of $A_{n}$. Indeed, let $x \\in A_{n}$ be an element such that $-\\frac{n(n+1)}{2} \\leq x<\\frac{n(n+1)}{2}$.\nCase 1. If the writing of $x$ begins with -1 , then by changing -1 by +1 , we get $x+2 \\in A_{n}$.\nCase 2. If the writing of $x$ begins with +1 , then consider the first term with sign -. A such term exists, otherwise\n$$\nx=1+2+\\ldots+n=\\frac{n(n+1)}{2} .\n$$\nIn this case we have\n$$\nx=+1+2+\\ldots+(k-1)-k \\pm \\ldots \\pm n,\n$$\nwhere $-k$ is the considered term. By changing the signs of terms $k-1$ and $k$, it follows $x+2 \\in A_{n}$.\nWe obtain\n$A_{n}=\\left\\{-\\frac{n(n+1)}{2},-\\frac{n(n+1)}{2}+2, \\ldots, \\frac{n(n+1)}{2}-2, \\frac{n(n+1)}{2}\\right\\}$, that is the number of elements in $A_{n}$ is $\\frac{n(n+1)}{2}+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $q_{1}, q_{2}, q_{3}, q_{4}, q_{5}$ such that $q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4}$ is the product of two consecutive even integers.",
    "options": [],
    "answer": "q1 = q2 = q3 = q4 = q5 = 2",
    "solution": "Assume that $q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 2k(2k+2)$ for some positive integer $k$. That is\n$$\nq_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 4k(k+1)\n$$\nIf $p$ is a prime, $p \\neq 2$, then $p^{4} \\equiv 1 \\pmod{8}$.\nAssume that $q_{1} \\leq q_{2} \\leq q_{3} \\leq q_{4} \\leq q_{5}$. A simple parity argument shows that $q_{1} = 2$. If $q_{2} \\neq 2$, then we have\n$$\nq_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} \\equiv 4 \\pmod{8}\n$$\nSince $8 \\mid 4k(k+1) - 16$, we get a contradiction. Therefore $q_{2} = 2$. In similar way it follows $q_{3} = q_{4} = q_{5} = 2$.\nThe only solution is $q_{1} = q_{2} = q_{3} = q_{4} = q_{5} = 2$, and $k = 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 39986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle with circumcenter $O$, incenter $I$, and orthocenter $H$. Prove that angle $\\widehat{OIH}$ is obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Recall some preliminary facts. The nine-point circle of a triangle $ABC$ passes through the midpoints of the sides, the midpoints of the segments joining its vertices to the orthocenter $H$ and the pedal point (i.e., the feet of its altitudes to the sides). Its center is the midpoint $O_{9}$ of the segment joining the circumcenter $O$ and the orthocenter $H$ of the triangle. Its radius $\\frac{1}{2} R$ is equal to half the circumradius $R$ of the triangle $ABC$ and it touches internally the incircle with radius $r$ (as well as all three excircles). The square of the length of the segment $OI$ is\n$$\n|OI|^{2}=R^{2}-2Rr=R(R-2r).\n$$\n![](attached_image_1.png)\nAlso, it is clear that $O_{9}I=\\frac{1}{2}R-r$.\nLet $M$ be the symmetric point of $O$ with respect to $I$. It follows that\n$$\nHM=2O_{9}I=2\\left(\\frac{1}{2}R-r\\right)=R-2r.\n$$\nSince\n$$\nIM=OI=\\sqrt{R(R-2r)}>R-2r=HM,\n$$\nwe get that $\\widehat{IH M}>\\widehat{MIH}$. Hence $\\widehat{MIH}<90^{\\circ}$, so that $\\widehat{OIH}>90^{\\circ}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of integers such that $a+b+c=2010 \\cdot 2011$ and the solutions to the equation $2011 x^{3}+a x^{2}+b x+c=0$ are all nonzero integers.",
    "options": [],
    "answer": "(2006*2011, -4*2009*2011, 4*2010*2011)",
    "solution": "For a prime $p$ consider the equation\n$$\np x^{3}+a x^{2}+b x+c=0\n$$\nwhere $a+b+c=p(p-1)$. Let $x_{1}, x_{2}, x_{3}$ be its roots. From Viète's relation,\n$$\n\\begin{gathered}\nx_{1}+x_{2}+x_{3}=-\\frac{a}{p} \\\\\nx_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=\\frac{b}{p} \\\\\nx_{1} x_{2} x_{3}=-\\frac{c}{p}\n\\end{gathered}\n$$\nWe have\n$$\n\\begin{gathered}\n\\left(x_{1}-1\\right)\\left(x_{2}-1\\right)\\left(x_{3}-1\\right)=-\\frac{c}{p}-\\frac{b}{p}-\\frac{a}{p}-1 \\\\\n=-\\frac{a+b+c}{p}-1=1-p-1=-p\n\\end{gathered}\n$$\nBecause $x_{1}, x_{2}, x_{3} \\neq 0$, we must have $x_{1}=2, x_{2}=2, x_{3}=1-p$, up to permutation. It follows that\n$$\n5-p=-\\frac{a}{p}, \\quad 4(2-p)=\\frac{b}{p}, \\quad 4(1-p)=-\\frac{c}{p}\n$$\nhence all triples are $(p(p-5), 4 p(2-p), 4 p(1-p))$.\nIn our case, $p=2011$ and we obtain\n$$\n(a, b, c)=(2006 \\cdot 2011,-4 \\cdot 2009 \\cdot 2011,4 \\cdot 2010 \\cdot 2011)\n$$\nAs in the previous solution assume that $p$ is a prime and $a+b+1=p(p-1)$. Consider the polynomial\n$$\nP(x)=p x^{3}+a x^{2}+b x+c\n$$\nand let $x_{1}, x_{2}, x_{3} \\in \\mathbb{Z}^{*}$ be its roots. We have\n$$\nP(x)=p\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)\\left(x-x_{3}\\right)\n$$\nhence\n$P(1)=p\\left(1-x_{1}\\right)\\left(1-x_{2}\\right)\\left(1-x_{3}\\right)=p+a+b+c=p+p(p-1)=p^{2}$.\nIt follows\n$$\n\\begin{equation*}\n\\left(1-x_{1}\\right)\\left(1-x_{2}\\right)\\left(1-x_{3}\\right)=p \\tag{1}\n\\end{equation*}\n$$\nSince $x_{1}, x_{2}, x_{3} \\neq 0$, we get $1-x_{1}=-1,1-x_{2}=-1,1-x_{3}=p$, hence $x_{1}=x_{2}=2$, and $x_{3}=1-p$, up to a permutation of $x_{1}, x_{2}, x_{3}$. Therefore\n$$\n\\begin{aligned}\nP(x) & =p(x-2)^{2}(x+p-1)=p\\left(x^{2}-4 x+4\\right)(x+p-1) \\\\\n& =p x^{3}+p(p-5) x^{2}+4 p(2-p) x+4 p(1-p),\n\\end{aligned}\n$$\nhence $(a, b, c)=(p(p-5), 4 p(2-p), 4 p(1-p))$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive real numbers $a, b, c$,\n$$\n2\\left(a^{3}+b^{3}+c^{3}+a b c\\right) \\geq (a+b)(b+c)(c+a).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$2 a^{3} + 2 b^{3} + 2 c^{3} + 2 a b c \\geq a^{2}(b+c) + b^{2}(c+a) + c^{2}(a+b) + 2 a b c$,\nhence\n$$\n2 a^{3} + 2 b^{3} + 2 c^{3} \\geq a b(a+b) + b c(b+c) + c a(c+a) \\tag{1}\n$$\nFor any positive real numbers $x$ and $y$ we have\n$$\nx^{3} + y^{3} \\geq x y(x+y)\n$$\nIndeed, this inequality is equivalent to $x^{2} - x y + y^{2} \\geq x y$, that is $(x-y)^{2} \\geq 0$. We have equality in this inequality if and only if $x = y$.\nApplying the inequality above we get\n$a^{3} + b^{3} \\geq a b(a+b)$, $b^{3} + c^{3} \\geq b c(b+c)$, $c^{3} + a^{3} \\geq c a(c+a)$.\nAdding these inequalities we obtain (1). We have equality in our inequality if and only if $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x$ and $y$ such that\n$$\n\\binom{x}{y}=1432\n$$",
    "options": [],
    "answer": "[(1432, 1), (1432, 1431)]",
    "solution": "We have $1432=2^{3} \\cdot 179$. The equation is equivalent to\n$$\n\\frac{x!}{y!(x-y)!}=2^{3} \\cdot 179\n$$\nor $y!(x-y)! \\cdot 2^{3} \\cdot 179 = x!$. It follows $179 \\mid x!$, hence $x \\geq 179$.\n\nIt is clear that $(x, y) = (1432, 1), (1432, 1431)$ are solutions. We shall prove that there are no other solutions. Because of the symmetry of the binomial coefficients, we can assume that $y \\leq \\left\\lfloor \\frac{x}{2} \\right\\rfloor$. Then\n$$\n\\binom{x}{1} < \\binom{x}{2} < \\ldots < \\binom{x}{\\left\\lfloor \\frac{x}{2} \\right\\rfloor}.\n$$\nIf $y=1$, then we get $x=1432$. If $y \\geq 2$, then we have\n$$\n\\binom{x}{y} \\geq \\binom{x}{2} = \\frac{x(x-1)}{2} \\geq \\frac{179 \\cdot (179-1)}{2} > 1432,\n$$\nnot possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of nonnegative integers $(a, b)$ such that\n$$\na + 2b - b^{2} = \\sqrt{2a + a^{2} + |2a + 1 - 2b|}\n$$",
    "options": [],
    "answer": "[(0,1), (1,1)]",
    "solution": "It is clear that $2a + 1 - 2b$ is an odd integer, hence\n$$\n|2a + 1 - 2b| \\geq 1\n$$\nIf $(a, b)$ is a pair of nonnegative integers satisfying\n$$\na + 2b - b^{2} = \\sqrt{2a + a^{2} + |2a + 1 - 2b|}\n$$\nthen we get\n$$\na + 2b - b^{2} \\geq \\sqrt{2a + a^{2} + 1} = \\sqrt{(a + 1)^{2}} = a + 1\n$$\nTherefore $0 \\geq (b - 1)^{2}$, that is $b = 1$.\n\nFor $b = 1$, the equation becomes\n$$\na + 1 = \\sqrt{2a + a^{2} + |2a - 1|} \\tag{1}\n$$\nEquation (1) is equivalent to $|2a - 1| = 1$. We get $a = 0$ or $a = 1$.\n\nThe desired pairs are $(a, b) = (0, 1), (1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $x_{n} = 2^{n} - n$, $n = 0, 1, 2, \\ldots$. Find all integers $m \\geq 0$ such that $s_{m} = x_{0} + x_{1} + x_{2} + \\ldots + x_{m}$ is a power of $2$.",
    "options": [],
    "answer": "m = 0, 1, 2",
    "solution": "We have\n$$\ns_{m} = \\sum_{k=0}^{m} \\left(2^{k} - k\\right) = 2^{m+1} - 1 - \\frac{m(m+1)}{2}\n$$\nWe prove that for $m \\geq 3$, we have $2^{m} < s_{m} < 2^{m+1}$. This inequality is equivalent to\n$$\n2^{m} < 2^{m+1} - 1 - \\frac{m(m+1)}{2} < 2^{m+1}\n$$\nThe right inequality is obvious. The left inequality is equivalent to\n$$\n1 + \\frac{m(m+1)}{2} < 2^{m}\n$$\nthat is\n$$\n2 + 2 \\frac{m(m+1)}{2} < 2^{m+1}\n$$\nor\n$$\n2\\left(\\binom{m+1}{0} + \\binom{m+1}{2}\\right) < 2^{m+1}\n$$\nand we are done.\n\nFor $m=0$, we have $s_{0} = 1 = 2^{0}$.\nFor $m=1$, we have $s_{1} = 2 = 2^{1}$.\nFor $m=2$, we have $s_{2} = 4 = 2^{2}$.\nThe solutions are $m \\in \\{0, 1, 2\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that all roots of the equation\n$$\nx(x+2)(x+4) \\ldots(x+2n) + (x+1)(x+3) \\ldots(x+2n-1) = 0\n$$\nare real and irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the polynomial function\n$$\nf(x) = x(x+2)(x+4) \\ldots(x+2n) + (x+1)(x+3) \\ldots(x+2n-1)\n$$\nWe have $\\deg f = n+1$ and the leading coefficient is $1$.\n\nObserve that for $0 \\leq k < n$,\n$$\nf(-2k) = (-1)^k (2k-1)(2k-3) \\ldots 1 \\cdot 3 \\ldots (-2k+2n-1)\n$$\nand\n$$\n\\begin{gathered}\nf(-2k-2) = f(-2(k+1)) \\\\\n= (-1)^{k+1} (2k+1)(2k-1) \\ldots 1 \\cdot 3 \\ldots (-2k+2n-3) .\n\\end{gathered}\n$$\nIt is clear that $f(-2k-2) f(-2k) < 0$, that is, the interval $I_k = (-2k-2, -2k)$ contains a root of $f$ for any $k = 0, 1, \\ldots, n-1$. Because $f$ is of degree $n+1$ and it has $n$ real roots, it follows that $f$ has $n+1$ real roots.\n\nIt is clear that $f$ has integer coefficients. If $f$ has a rational root $a$, then $a$ must be integer since the leading coefficient is $1$.\n\nIf $a$ is even, then from $f(a) = 0$ we get\n$$\na(a+2)(a+4) \\ldots(a+2n) + (a+1)(a+3) \\ldots(a+2n+1) = 0,\n$$\nthat is, an odd number is equal to $0$, not possible.\n\nIf $a$ is odd, then from $f(a) = 0$, it follows again that an odd number is equal to $0$, not possible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 39993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f = a X^{2} + b X + c \\in \\mathbb{Z}[X]$ be a polynomial such that for every positive integer $n$, $f(n)$ is a perfect square. Prove that $f = g^{2}$ for some polynomial $g \\in \\mathbb{Z}[X]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sequence $x_{n} = \\sqrt{f(n+1)} - \\sqrt{f(n)}$, $n \\geq 1$, contains only integers. We have\n$$\n\\lim_{n \\rightarrow \\infty} x_{n} = \\lim_{n \\rightarrow \\infty} \\frac{f(n+1) - f(n)}{\\sqrt{f(n+1)} + \\sqrt{f(n)}} = \\frac{2a}{2\\sqrt{a}} = \\sqrt{a}\n$$\nIt follows that $\\sqrt{a}$ is an integer, hence\n$$\ny_{n} = \\sqrt{f(n)} - n \\sqrt{a} \\in \\mathbb{Z}\n$$\nfor every $n \\geq 1$. We have\n$$\n\\lim_{n \\rightarrow \\infty} y_{n} = \\frac{b n + c}{\\sqrt{f(n)} + n \\sqrt{a}} = \\frac{b}{2\\sqrt{a}}\n$$\nhence $\\frac{b}{2\\sqrt{a}} \\in \\mathbb{Z}$, and $y_{n} = \\frac{b}{2\\sqrt{a}}$ for any $n \\geq n_{0}$. It follows $f(n) = (\\sqrt{a} n + d)^{2}$ for any $n \\geq n_{0}$, and we get $f = g^{2}$, where $g = \\sqrt{a} X + d$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ be distinct positive integers. Prove that the number\n$$\n\\frac{(x+y)^2}{x^3 + x y^2 - x^2 y - y^3}\n$$\nis not an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $E(x, y)$ the expression in the problem. We have\n$$\nE(x, y) = \\frac{(x+y)^2}{x(x^2 + y^2) - y(x^2 + y^2)} = \\frac{(x+y)^2}{(x-y)(x^2 + y^2)} \\tag{1}\n$$\nand we can assume that $x > y$.\n\nIf $x-y = 1$, then $E(x, y) \\in \\mathbb{Z}$ if and only if $x^2 + y^2 \\mid 2 x y$. This is not possible because $x^2 + y^2 > 2 x y$.\n\nIf $x-y > 1$, then $x-y \\geq 2$. We have\n$$\n0 < E(x, y) \\leq \\frac{(x+y)^2}{2(x^2 + y^2)} < 1\n$$\nhence $E(x, y) \\notin \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a triangle $ABC$. Let $A_{1}$ be the symmetric point of $A$ with respect to the line $BC$, $B_{1}$ the symmetric point of $B$ with respect to the line $CA$, and $C_{1}$ the symmetric point of $C$ with respect to the line $AB$. Determine the possible set of angles of triangle $ABC$ for which $A_{1}B_{1}C_{1}$ is equilateral.",
    "options": [],
    "answer": "(60°, 60°, 60°), (30°, 75°, 75°), (150°, 15°, 15°)",
    "solution": "We will use the following relation: For any angle $\\theta$,\n$$\n\\cos 3\\theta = \\cos \\theta - 4 \\cos \\theta \\sin^2 \\theta\n$$\nLet $a, b, c$ be the sides of the triangle and $\\alpha, \\beta, \\gamma$ be the respective angles opposite these sides. Since the triangles $A_{1}BC$, $AB_{1}C$ and $ABC_{1}$ are all congruent to the triangle $ABC$, we have that $\\widehat{C_{1}AB_{1}} = 3\\alpha$ or $|2\\pi - 3\\alpha|$, $\\widehat{A_{1}BC_{1}} = 3\\beta$ or $|2\\pi - 3\\beta|$, and $\\widehat{B_{1}CA_{1}} = 3\\gamma$ or $|2\\pi - 3\\gamma|$.\n\nApplying the Cosine Law to triangle $B_{1}CA_{1}$ yields that\n$$\nA_{1}B_{1}^2 = a^2 + b^2 - 2ab \\cos 3\\gamma\n$$\nwhere $R$ is the circumradius of triangle $ABC$, the Cosine and the Sine Laws applied to that triangle yields that\n$$\n2ab \\cos \\gamma = a^2 + b^2 - c^2\n$$\nand\n$$\n\\sin \\gamma = \\frac{c}{2R}\n$$\nApplying the above to the previous equation yields that\n$$\n\\begin{gathered}\nA_{1}B_{1}^2 = a^2 + b^2 - 2ab \\cos 3\\gamma = a^2 + b^2 - 2ab \\cos \\gamma (1 - 4 \\sin^2 \\gamma) \\\\\n= a^2 + b^2 - (a^2 + b^2 - c^2)(1 - 4 \\sin^2 \\gamma) \\\\\n= (a^2 + b^2)(1 - 1 + 4 \\sin^2 \\gamma) + c^2(1 - 4 \\sin^2 \\gamma) \\\\\n= (a^2 + b^2 - c^2)(4 \\sin^2 \\gamma) + c^2 = \\frac{(a^2 + b^2 - c^2)c^2}{R^2} + c^2 \\\\\n= \\frac{c^2}{R^2}(a^2 + b^2 - c^2 + R^2)\n\\end{gathered}\n$$\nSimilarly,\n$$\nB_{1}C_{1}^2 = \\frac{a^2}{R^2}(b^2 + c^2 - a^2 + R^2)\n$$\nand\n$$\nC_{1}A_{1}^2 = \\frac{b^2}{R^2}(c^2 + a^2 - b^2 + R^2)\n$$\nIt follows that $A_{1}B_{1} = B_{1}C_{1}$ if and only if\n$$\nc^2(a^2 + b^2 - c^2 + R^2) - a^2(b^2 + c^2 - a^2 + R^2) = 0\n$$\nFactoring the left side yields that\n$$\n(c^2 - a^2)(R^2 + b^2 - a^2 - c^2) = 0\n$$\nThe equality of other pairs of sides can be similarly handled. Thus, triangle $A_{1}B_{1}C_{1}$ is equilateral if and only if the following system of three equations is valid:\n$$\n\\begin{aligned}\n& (c^2 - a^2)(R^2 + b^2 - a^2 - c^2) = 0 \\\\\n& (a^2 - b^2)(R^2 + c^2 - b^2 - a^2) = 0 \\\\\n& (b^2 - c^2)(R^2 + a^2 - b^2 - c^2) = 0\n\\end{aligned}\n$$\nIf $a$ is unequal to both $b$ and $c$, then\n$$\nR^2 = a^2 + c^2 - b^2 = b^2 + a^2 - c^2\n$$\nso that $b = c$. Hence, the triangle is isosceles in any case.\n\nAssume that $b = c$. Then from the middle equation, we obtain that\n$$\n(a^2 - b^2)(R^2 - a^2) = 0\n$$\nTherefore, either $a = b$, in which case the triangle is equilateral, or $R = a$ and $\\alpha = a / 2R = 1/2$. Therefore, $\\alpha = 30^\\circ$ or $\\alpha = 150^\\circ$. Thus, there are three possible sets of angles for the triangle $ABC$: $(60^\\circ, 60^\\circ, 60^\\circ)$, $(30^\\circ, 75^\\circ, 75^\\circ)$ and $(150^\\circ, 15^\\circ, 15^\\circ)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f \\in \\mathbb{Z}[X], f = X^{2} + a X + b$, be a quadratic polynomial. Prove that $f$ has integer zeros if and only if for each positive integer $n$ there is an integer $u_{n}$ such that $n \\mid f\\left(u_{n}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $f = (X - x_{1})(X - x_{2}),\\ x_{1}, x_{2} \\in \\mathbb{Z}$, then take $u_{n} = n + x_{1}$ and get $f\\left(u_{n}\\right) = n\\left(n + x_{1} - x_{2}\\right)$, hence $n \\mid f\\left(u_{n}\\right),\\ n \\geq 1$.\n\nConversely, assume that $f\\left(u_{n}\\right) = k_{n} \\cdot n$, for some integer $k_{n}$, $n = 1, 2, \\ldots$ Then, the quadratic equation\n$$\nu^{2} + a \\nu + b - k_{n} \\cdot n = 0$$\nhas integer zeros, hence its discriminant $\\Delta_{n}$ is a perfect square, that is $\\Delta_{n} = t_{n}^{2},\\ n = 1, 2, \\ldots$. This is equivalent to\n$$a^{2} - 4\\left(b - k_{n} \\cdot n\\right) = t_{n}^{2}$$\nhence we have\n$$\n\\Delta + 4 k_{n} \\cdot n = t_{n}^{2}, \\quad n = 1, 2, \\ldots \\tag{1}\n$$\nwhere $\\Delta = a^{2} - 4b$ is the discriminant of equation $f(n) = 0$. In (1) we take $n = \\Delta^{2}$ and obtain\n$$\n\\Delta \\cdot \\left(1 + 4 k_{\\Delta^{2}} \\cdot \\Delta\\right) = t_{\\Delta_{2}}^{2} . \\tag{2}\n$$\nSince $\\Delta$ and $1 + 4 k_{\\Delta^{2}} \\cdot \\Delta$ are relatively prime, from (2) it follows that $\\Delta$ is a perfect square, hence the zeros of $f$, $x_{1}, x_{2}$ are rational. We have $\\Delta = a^{2} - 4b = c^{2},\\ c \\in \\mathbb{Z}$, and\n$$\nx_{1,2} = \\frac{1}{2}(-a \\pm c) . \\tag{3}\n$$\nIt is clear that $a$ and $c$ are of the same parity, and from (3) we get that $x_{1}, x_{2} \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 39997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive real numbers for which\n$$\n\\frac{1}{1+x_{1}}+\\frac{1}{1+x_{2}}+\\ldots+\\frac{1}{1+x_{n}}=1\n$$\nProve that\n$$\nx_{1} x_{2} \\ldots x_{n} \\geq (n-1)^{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $y_{k}=\\frac{1}{1+x_{k}},\\ k=1,2, \\ldots, n$. Then\n$$\nx_{k}=\\frac{1}{y_{k}}-1=\\frac{y_{1}+\\ldots+y_{n}}{y_{k}}-1=\\frac{S_{k}}{y_{k}}\n$$\nwhere\n$$\nS_{k}=x_{1}+\\ldots+x_{k-1}+x_{k+1}+\\ldots+x_{n}, \\quad k=1,2, \\ldots, n.\n$$\nApplying AM-GM inequality it follows\n$$\n\\frac{S_{k}}{n-1} \\geq \\sqrt[n-1]{y_{1} \\ldots y_{k-1} y_{k+1} \\ldots y_{n}}\n$$\nfor $k=1,2, \\ldots, n$. We obtain\n$$\nx_{1} x_{2} \\ldots x_{n}=\\frac{S_{1} S_{2} \\ldots S_{n}}{y_{1} y_{2} \\ldots y_{n}} \\geq (n-1)^{n} \\frac{y_{1} y_{2} \\ldots y_{n}}{y_{1} y_{2} \\ldots y_{n}}=(n-1)^{n}.\n$$\nThe equality holds if and only if $y_{1}=\\ldots=y_{n}=\\frac{1}{n}$ that is $x_{1}=\\ldots=x_{n}=n-1$.\nWrite the condition in the hypothesis as\n$$\n\\frac{1}{1+x_{1}}+\\ldots+\\frac{1}{1+x_{n-1}}=\\frac{x_{n}}{1+x_{n}}\n$$\nApplying AM-GM inequality we get\n$$\n(n-1) \\sqrt[n-1]{\\frac{1}{1+x_{1}} \\cdots \\frac{1}{1+x_{n-1}}} \\leq \\frac{x_{n}}{1+x_{n}} \\tag{1}\n$$\nIn similar way we obtain other $n-1$ inequalities of the form. Therefore, we have\n$$\n(n-1) \\sqrt[n-1]{\\frac{1}{1+x_{1}} \\cdots \\frac{1}{1+x_{k-1}} \\frac{1}{1+x_{k+1}} \\cdots \\frac{1}{1+x_{n}}} \\leq \\frac{x_{k}}{1+x_{k}} \\tag{2}\n$$\nfor $k=1,2, \\ldots, n$. Multiplying inequalities (2) and simplifying by $\\frac{1}{1+x_{1}} \\ldots \\frac{1}{1+x_{n}}$ it follows $(n-1)^{n} \\leq x_{1} \\ldots x_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 39998,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ the angle bisector $AL$, $L \\in BC$, intersects its circumcircle at $N$. Let $K$ and $M$ be the projections of $L$ onto sides $AB$ and $AC$. Prove that triangle $ABC$ and quadrilateral $AKNM$ have equal areas.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the second point of intersection of segment $BC$ and the circle circumscribed about quadrilateral $AKLM$. Denote by $E$ the intersection point of the lines $KN$ and $BC$ and by $F$ the intersection point of the lines $MN$ and $BC$.\n\nThen $\\widehat{BCN} = \\widehat{BAN}$ and $\\widehat{MAL} = \\widehat{MPL}$, as angles on the same arc. Since $AL$ is a bisector, $\\widehat{BCN} = \\widehat{BAL} = \\widehat{MAL} = \\widehat{MPL}$, and consequently $PM \\parallel NC$. Similarly we prove $KP \\parallel BN$. Then the quadrilaterals $BKPN$ and $NPMC$ are trapezoids; hence\n$$\nS_{BKE} = S_{NPE} \\quad \\text{and} \\quad S_{NPF} = S_{CMF}.\n$$\nTherefore $S_{ABC} = S_{AKNM}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 39999,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the isosceles triangle $ABC$, with $AB = AC$, the angle bisector of $\\widehat{B}$ intersects side $AC$ at $B'$. Suppose that $BB' + B'A = BC$. Find the angles of the triangle.",
    "options": [],
    "answer": "∠B = ∠C = 40°, ∠A = 100°",
    "solution": "On the side $BC$ take point $M$ such that $CM = AB'$. The angle bisector theorem implies\n$$\n\\frac{AB'}{B'C} = \\frac{AB}{BC} = \\frac{AC}{BC} \\text{, hence } \\frac{MC}{B'C} = \\frac{AC}{BC} \\text{.}\n$$\n![](attached_image_1.png)\nIt follows $\\frac{MC}{AC} = \\frac{B'C}{BC}$, that is $\\triangle MCB' \\sim \\triangle ACB$, and we get $MC = MB'$. Moreover, $\\widehat{C} = \\widehat{MCB'} = \\widehat{MB'C}$ and $MC = MB'$.\nFrom $BB' + B'A = BC$ it follows $BB' = BC - B'A = BC - MC = BM$, hence $\\triangle B'BM$ is isosceles.\nIn $\\triangle BB'M$ we have $180^{\\circ} = 2\\widehat{C} + 2\\widehat{C} + \\frac{\\widehat{C}}{2} = \\frac{9\\widehat{C}}{2}$, implying $\\widehat{C} = 40^{\\circ}$. It follows\n$$\n\\widehat{B} = \\widehat{C} = 40^{\\circ} \\text{ and } \\widehat{A} = 100^{\\circ} .\n$$\n\n![](attached_image_1.png)\nIn $\\triangle ABB'$, we have\n$$\n\\frac{BB'}{\\sin 4x} = \\frac{AB'}{\\sin x} = \\frac{AB}{\\sin 3x} .\n$$\nIn $\\triangle BB'C$ we have\n$$\n\\frac{BB'}{\\sin 2x} = \\frac{B'C}{\\sin x} = \\frac{BC}{\\sin 3x}\n$$\nand in $\\triangle ABC$ we can write\n$$\n\\frac{BC}{\\sin 4x} = \\frac{AC}{\\sin 2x} .\n$$\n\nWe get\n$$\nBB' + BB' \\frac{\\sin x}{\\sin 4x} = BB' \\frac{\\sin 3x}{\\sin 2x}\n$$\nhence\n$$\n\\begin{equation*}\n1 + \\frac{\\sin x}{\\sin 4x} = \\frac{\\sin 3x}{\\sin 2x} \\tag{1}\n\\end{equation*}\n$$\nRelation (1) is equivalent to\n$$\n\\frac{\\sin x}{\\sin 4x} = \\frac{\\sin 3x - \\sin 2x}{\\sin 2x},\n$$\nthat is\n$$\n\\frac{2 \\sin \\frac{x}{2} \\cos \\frac{x}{2}}{2 \\sin 2x \\cos 2x} = \\frac{2 \\sin \\frac{x}{2} \\cos \\frac{5x}{2}}{\\sin 2x} .\n$$\nWe get\n$$\n\\cos \\frac{x}{2} = 2 \\cos 2x \\cos \\frac{5x}{2},\n$$\nor\n$$\n\\cos \\frac{x}{2} = \\cos \\frac{9x}{2} + \\cos \\frac{x}{2} .\n$$\nIt follows $\\cos \\frac{9x}{2} = 0$, that is $\\frac{9x}{2} = \\frac{\\pi}{2}$, and we obtain $x = \\frac{\\pi}{9}$.\nFinally,\n$$\n\\widehat{B} = \\widehat{C} = \\frac{2\\pi}{9}, \\quad \\widehat{A} = \\frac{5\\pi}{9} .\n$$\n\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 40000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that at least one of the integers\n$$\n\\left[2^{n} \\cdot \\sqrt{2}\\right],\\left[2^{n+1} \\cdot \\sqrt{2}\\right], \\ldots,\\left[2^{2 n} \\cdot \\sqrt{2}\\right]\n$$\nis even, where $[a]$ denotes the integer part of $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume by contradiction that all integers are odd. Then, we have\n$$\n2a-1 < 2^{n} \\sqrt{2} < 2a\n$$\nfor some positive integer $a$. Multiplying inequalities (1) by $2$, we get $4a-2 < 2^{n+1} \\sqrt{2} < 4a$. Since the integer $\\left[2^{n+1} \\sqrt{2}\\right]$ is odd, it follows\n$$\n4a-1 < 2^{n+1} \\sqrt{2} < 4a.\n$$\nIn analogous way, after $n$ steps, we get\n$$\n2^{n+1} a - 1 < 2^{2n} \\sqrt{2} < 2^{n+1} a\n$$\nFrom the left inequality above it follows\n$$\na - 2^{n-1} \\sqrt{2} < \\frac{1}{2^{n+1}}\n$$\nhence\n$$\n\\frac{a^{2} - 2^{2n-1}}{a + 2^{n-1} \\sqrt{2}} < \\frac{1}{2^{n+1}}\n$$\nFrom the right inequality in the first step we have $2a > 2^{n} \\sqrt{2}$ hence $a^{2} > 2^{2n-1}$, that is $a^{2} - 2^{2n-1} \\geq 1$. Using this inequality, from above it follows\n$$\n\\frac{1}{a + 2^{n-1} \\sqrt{2}} < \\frac{1}{2^{n+1}}\n$$\nhence\n$$\n2^{n+1} < 2^{n-1} \\sqrt{2} + a < 2^{n-1} \\sqrt{2} + \\frac{1}{2} (2^{n} \\sqrt{2} + 1)\n$$\nFrom this we get $2^{n+2} < 2^{n+1} \\sqrt{2} + 1$, that is\n$$\n2 < \\sqrt{2} + \\frac{1}{2^{n+1}} < \\frac{3}{2} + \\frac{1}{2} = 2\n$$\ncontradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  }
]