[
  {
    "id": 17501,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that there exists a permutation $a_1, a_2, \\dots, a_{1013}$ of the numbers $1012, 1013, \\dots, 2024$ so that\n$$\na_1 n^{1012} + a_2 n^{1011} + \\dots + a_{1012} n + a_{1013} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $n \\in \\mathbb{N}$, then $a_1 n^{1012} + a_2 n^{1011} + \\dots + a_{1012} n + a_{1013} > 0$, so $n \\in \\mathbb{N}$ is not suitable.\n\nIf $n = -k$ with $k \\ge 2$, then\n$$\na_1 n^{1012} + a_2 n^{1011} + \\dots + a_{1012} n + a_{1013} = k^{1011}(a_1 k - a_2) + k^{1009}(a_3 k - a_4) + \\dots + k(a_{1011} k - a_{1012}) + a_{1013} > 0,\n$$\nbecause $a_{1013} > 0$ 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.\n\nFor each $a \\in \\mathbb{Z}$, $(a - 1) - a - (a + 1) + (a + 2) = 0$. It follows that\n$$(1012 - 2024 + 1013 - 1015 + 1014) + (1016 - 1017 - 1018 + 1019) + \\dots + (2020 - 2021 - 2022 + 2023) = 0,$$\nhence $n = -1$ is the only solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17502,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of\n$$\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|}\n$$\nfor real numbers $x$, $y$, and $z$, where $x + y + z \\ne 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the triangle inequality, we have\n\n$$\n\\begin{aligned}\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|} &\\ge \\frac{|x| + \\frac{4}{11}|x + 4y| + \\frac{1}{11}|-y-7z| + 2|z|}{|x + y + z|} \\\\\n&\\ge \\frac{|x + \\frac{4}{11}(x + 4y) + \\frac{1}{11}(-y - 7z) + 2z|}{|x + y + z|} \\\\\n&= \\frac{15}{11}.\n\\end{aligned}\n$$\n\nEquality holds when $(x, y, z) = (28k, -7k, k)$ for some $k \\ne 0$. So the minimum value is $\\frac{15}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17503,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D = (a+b+c+d)^2 - 8(ac+bd)$. Show that $D \\ge 0$, and determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "The number $D$ defined above is the discriminant of the quadratic polynomial\n\n$$\nq(x) = (x-a)(x-c) + (x-b)(x-d)\n$$\n\nThis polynomial is positive when $x$ is sufficiently large and non-positive for $a, b \\le x \\le c, d$. Therefore, it has a real root, so $D \\ge 0$.\n\nEquality holds if and only if $a + d = b + c$ and either $a = d$ or $b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17504,
    "subject": "Mathematics (Olympiad)",
    "question": "An equilateral triangle $ABC$ is shown alongside. The centre $G$ is joined to $A$, $B$, and $C$, forming three identical triangles.\n\n$D$, $E$, and $F$ are the midpoints of $AB$, $AC$, and $BC$, respectively.\n\nIf the area of $\\triangle ABC$ is $120\\ \\text{cm}^2$, what is the shaded area in $\\text{cm}^2$?\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p54_data_37ecbebb60.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $D$, $E$, and $F$ are the midpoints of the sides, we have four triangles of equal area:\n\n$$\n\\text{Area of } \\triangle ADE = \\triangle EFC = \\triangle DBE = \\triangle DEF\n$$\n\nAlso,\n$$\n\\triangle AHE = \\frac{1}{2} \\triangle ADE = \\frac{1}{8} \\triangle ABC\n$$\n\nEHGI is one of three identical quadrilaterals making up $\\triangle DEF$:\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p56_data_bb31bf9199.png)\n\n$$\n\\text{EHGI} = \\frac{1}{3} \\triangle DEF\n$$\n\nBut\n$$\n\\triangle DEF = \\frac{1}{4} \\triangle ABC\n$$\n\nSo\n$$\n\\text{EHGI} = \\frac{1}{3} \\times \\frac{1}{4} = \\frac{1}{12} \\triangle ABC\n$$\n\nTherefore, the shaded area is\n$$\n\\frac{1}{8} + \\frac{1}{12} = \\frac{5}{24} \\triangle ABC\n$$\n\nThus,\n$$\n\\frac{5}{24} \\times 120 = 25\\ \\text{cm}^2\n$$\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p56_data_c58dcc8844.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17505,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be distinct positive real numbers. Prove\n\n$$\n\\frac{ab + bc + ca}{(a+b)(b+c)(c+a)} < \\frac{1}{7} \\left( \\frac{1}{|a-b|} + \\frac{1}{|b-c|} + \\frac{1}{|c-a|} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, assume $a > b > c > 0$. On one hand,\n\n$$\n\\frac{ab + bc + ca}{(a+b)(b+c)(c+a)} = \\frac{ab + bc + ca}{(a+b)(ab+bc+ca+c^2)} < \\frac{1}{a+b}.\n$$\n\nOn the other hand,\n\n$$\n\\frac{1}{7} \\left( \\frac{1}{|a-b|} + \\frac{1}{|b-c|} + \\frac{1}{|c-a|} \\right) = \\frac{1}{7} \\left( \\frac{1}{a-b} + \\frac{1}{b-c} + \\frac{1}{a-c} \\right) > \\frac{1}{7} \\left( \\frac{1}{a-b} + \\frac{1}{b} + \\frac{1}{a} \\right).\n$$\n\nBased on the above inequalities, it suffices to prove\n\n$$\n\\frac{1}{a+b} < \\frac{1}{7} \\left( \\frac{1}{a-b} + \\frac{1}{b} + \\frac{1}{a} \\right), \\quad \\text{or} \\quad \\frac{a}{a-b} + \\frac{a}{b} + 1 - \\frac{7a}{a+b} > 0\n$$\n\nby multiplying $7a$ on both sides and rearranging the terms. To justify it, let $x = a/b > 1$. We have\n\n$$\n\\frac{a}{a-b} + \\frac{a}{b} + 1 - \\frac{7a}{a+b} = \\frac{x}{x-1} + x + 1 - \\frac{7x}{x+1} = \\frac{x^3 - 5x^2 + 7x - 1}{(x-1)(x+1)} = \\frac{(x-1)^2(x-3) + 2}{(x-1)(x+1)}\n$$\n\nDefine $f(x) = (x-1)^2(x-3)+2$. If $x \\ge 3$, clearly $f(x) > 0$; if $1 < x < 3$, then\n\n$$\n(x - 1)^2 (x - 3) = -\\frac{1}{2} (x - 1)(x - 1)(6 - 2x) \\geq -\\frac{1}{2} \\left(\\frac{4}{3}\\right)^3 > -2\n$$\n\nby the AM-GM inequality, and $f(x) > 0$, too. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17506,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve for all real pairs $(x, y)$ satisfying the equation:\n$$\nx^2 y - x^2 + x y^2 - y^2 = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We start by rewriting the equation:\n$$\nx^2 y - x^2 + x y^2 - y^2 = 1\n$$\nGroup terms:\n$$\nxy(x + y) - (x^2 + y^2) = 1\n$$\nRecall that $x^2 + y^2 = (x + y)^2 - 2xy$, so:\n$$\nxy(x + y) - ((x + y)^2 - 2xy) = 1\n$$\nLet $u = x + y$ and $v = xy$:\n$$\nuv - (u^2 - 2v) = 1\n$$\nSimplify:\n$$\nuv - u^2 + 2v = 1 \\\\\nuv + 2v = u^2 + 1 \\\\\nv(u + 2) = u^2 + 1 \\\\\nv = \\frac{u^2 + 1}{u + 2}\n$$\nRewrite:\n$$\nv = u - 2 + \\frac{5}{u + 2}\n$$\nFor $v$ to be real, $u + 2$ must divide $5$. The divisors of $5$ are $\\pm1, \\pm5$:\n\n- $u + 2 = 1 \\implies u = -1$\n- $u + 2 = -1 \\implies u = -3$\n- $u + 2 = 5 \\implies u = 3$\n- $u + 2 = -5 \\implies u = -7$\n\nCompute $v$ for each $u$:\n\n- $u = 3: v = \\frac{9 + 1}{5} = 2$\n- $u = -1: v = \\frac{1 + 1}{1} = 2$\n- $u = -3: v = \\frac{9 + 1}{-1} = -10$\n- $u = -7: v = \\frac{49 + 1}{-5} = -10$\n\nNow, $x$ and $y$ are roots of $z^2 - uz + v = 0$:\n\n- $u = 3, v = 2: z^2 - 3z + 2 = 0 \\implies z = 1, 2$\n- $u = -1, v = 2: z^2 + z + 2 = 0$ (discriminant $< 0$, no real roots)\n- $u = -3, v = -10: z^2 + 3z - 10 = 0 \\implies z = 2, -5$\n- $u = -7, v = -10: z^2 + 7z - 10 = 0$ (discriminant $= 49 + 40 = 89$, not a perfect square, roots are irrational)\n\nThus, the real solutions are:\n$$(x, y) \\in \\{(1, 2), (2, 1), (2, -5), (-5, 2)\\}.$$\n",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17507,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside triangle $ABC$ such that by drawing rays from $P$ to the sides of $ABC$, the triangle can be divided into 27 smaller triangles of equal area. How many such points $P$ (called *marvelous points*) exist in $ABC$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the area of $ABC$ be 27. Among the 27 rays from $P$, there must always be $PA$, $PB$, and $PC$; otherwise, not all the shapes determined by the rays will be triangles.\n\nEach of the 27 smaller triangles is contained within (or is equal to) one of the triangles $PAB$, $PBC$, or $PCA$. Let $PAB$ contain $m$ and $PBC$ contain $n$ smaller triangles. The numbers $m$ and $n$ are elements of $\\{1, 2, \\dots, 25\\}$ and satisfy $m + n \\leq 26$.\n\nFor each pair $(m, n)$, there is a unique point $P$ determined by the intersection of two lines parallel to $AB$ and $BC$ at distances $\\frac{2m}{|AB|}$ and $\\frac{2n}{|BC|}$ from $AB$ and $BC$, respectively. Thus, there is a bijection between marvelous points and pairs $(m, n)$ with $m, n \\geq 1$ and $m + n \\leq 26$.\n\nFor each $m \\in \\{1, 2, \\dots, 25\\}$, $n$ can be chosen in $26 - m$ ways. Therefore, the total number of marvelous points is:\n\n$$\n\\sum_{m=1}^{25} (26 - m) = \\frac{25 \\cdot 26}{2} = 325.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17508,
    "subject": "Mathematics (Olympiad)",
    "question": "A regular hexagon with side length $1$ is given. Inside the hexagon, $m$ points are placed so that no three of them are collinear. The hexagon is split into triangles, called *splitting triangles*, so that each of the $m$ points and each vertex of the hexagon is a vertex of one splitting triangle. The splitting triangles have no common interior point. Prove that there exists at least one splitting triangle whose area is not greater than $\\frac{3\\sqrt{3}}{4(m+2)}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, determine the total number of splitting triangles into which the hexagon is divided. Let $A$ be one of the $m$ interior points. The sum of all angles at $A$ is $360^\\circ$ (the sum of all angles at $A$ from all triangles having $A$ as a vertex). At each vertex of the hexagon, the sum of all angles is $120^\\circ$. Since the sum of angles in every triangle is $180^\\circ$, the total number of splitting triangles is:\n\n$$\n\\frac{m \\cdot 360^\\circ + 6 \\cdot 120^\\circ}{180^\\circ} = 2m + 4.\n$$\n\nAssume, for contradiction, that the area of each splitting triangle is greater than $\\frac{3\\sqrt{3}}{4(m+2)}$. Then the total area of all splitting triangles is greater than\n\n$$\n(2m + 4) \\cdot \\frac{3\\sqrt{3}}{4(m+2)} = \\frac{3\\sqrt{3}}{2},\n$$\n\nwhich is impossible, since the area of the hexagon is $\\frac{3\\sqrt{3}}{2}$. Therefore, there must exist at least one splitting triangle whose area is not greater than $\\frac{3\\sqrt{3}}{4(m+2)}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17509,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with incentre $I$. Suppose that $D$ is a variable point on the circumcircle of $ABC$, on the arc $AB$ that does not contain $C$. Let $E$ be a point on the line segment $BC$ such that $\\angle ADI = \\angle IEC$.\n\nProve that, as $D$ varies, the line $DE$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AI$ meet the circumcircle of $ABC$ again at $M$. We claim $M$ is the fixed point. Let $E'$ be the intersection of $DM$ and $BC$. We will show that $\\angle ADI = \\angle IEC'$, which implies $E = E'$ since $\\angle IEC$ varies monotonically as $E$ varies along $BC$.\n\n![](images/2021_Australian_Scene_p81_data_2c08d84c40.png)\n\nFirst, note that $MB = MC = MI$. Certainly $MB = MC$ holds since arcs $MB$ and $MC$ subtend equal angles at the circumference. The equality $MB = MI$ follows from\n\n$$\n\\angle MBI = \\angle MBC + \\angle CBI = \\angle MAC + \\angle IBA = \\angle MAB + \\angle IBA = \\angle MIB.\n$$\n\nNext, let $AM$ intersect $BC$ at $N$. Since $\\angle CBM = \\angle MAB = \\angle MDB$, we have the similarities $\\triangle MBE' \\sim \\triangle MDB$ and $\\triangle MBN \\sim \\triangle MAB$. These imply the following length conditions:\n\n$$\nMI^2 = MB^2 = MD \\times ME' = MA \\times MN.\n$$\n\nThe condition $MI^2 = MD \\times ME'$ implies that $\\triangle MIE' \\sim \\triangle MDI$, while the condition $MD \\times ME' = MA \\times MN$ implies that $\\triangle MNE' \\sim \\triangle MDA$. Finally, the proof can be completed by noting\n\n$$\n\\angle ADI = \\angle ADM - \\angle IDM = \\angle MNE' - \\angle MIE' = \\angle IEC'.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17510,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for any integer $n$, the expression $n^5 - n$ is divisible by $30$.",
    "options": [],
    "answer": "See solution",
    "solution": "#### Alternative iv\n\nFrom Part b, $n^5$ and $n$ have the same units digit. So $n^5 - n$ ends in $0$. Therefore, $n^5 - n$ is divisible by $10$.\n\nWe now need to show that $n^5 - n$ is divisible by $3$, that is, $n^5 \\equiv n \\pmod{3}$ for all $n$. There are three cases:\n\n- **Case 1:** $n \\equiv 0 \\pmod{3}$.\n  \n  Then $n^5 \\equiv 0 \\pmod{3}$, so $n^5 \\equiv n \\pmod{3}$.\n\n- **Case 2:** $n \\equiv 1 \\pmod{3}$.\n  \n  Then $n^5 \\equiv 1^5 \\equiv 1 \\pmod{3}$, so $n^5 \\equiv n \\pmod{3}$.\n\n- **Case 3:** $n \\equiv 2 \\pmod{3}$.\n  \n  $2^5 = 32 \\equiv 2 \\pmod{3}$, so $n^5 \\equiv 2 \\pmod{3}$, and again $n^5 \\equiv n \\pmod{3}$.\n\nThus, $n^5 - n$ is divisible by $3$ for all integers $n$.\n\nSince the only common factor of $3$ and $10$ is $1$, $n^5 - n$ is divisible by $30$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17511,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be an integer. An $n \\times n$ square is divided into $n^2$ unit squares. Of these smaller squares, $n$ are coloured green and $n$ are coloured blue. All remaining squares are coloured white.\n\nAre there more such colourings for which there are no two green squares in a row and no two blue squares in a column, or colourings for which there are neither two green squares in a row nor two blue squares in a column?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $n$ squares have been coloured green, with no two of them in the same row. This leaves $n-1$ empty squares in each row, which means there are $(n-1)^n$ ways to colour $n$ of the remaining $n^2-n$ squares blue such that there are no two blue squares in the same row.\n\nOn the other hand, let $x_i$ be the number of squares in column $i$ that have not been coloured green. Clearly, $x_1 + x_2 + \\dots + x_n = n^2 - n$. The number of ways to colour $n$ of the remaining $n^2-n$ squares blue so that there are no two blue squares in the same column is\n\n$$\nx_1 x_2 \\cdots x_n \\leq \\left( \\frac{x_1 + x_2 + \\cdots + x_n}{n} \\right)^n = \\left( \\frac{n^2 - n}{n} \\right)^n = (n-1)^n\n$$\n\nby the inequality between the arithmetic and geometric mean. For some configurations (e.g., all green squares in one column), this holds with strict inequality.\n\nHence, there are more possible colourings for which there are no two green squares and no two blue squares in the same row than there are colourings for which there are no two green squares in the same row and no two blue squares in the same column.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17512,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all four-digit positive integers $\\overline{abcd}$ that satisfy the following conditions:\n\n1. $a \\leq b \\leq c \\leq d$;\n2. $a^2 + b^2 + c^2 + d^2$ is divisible by 4;\n3. When $\\overline{abcd}$ is divided by $c$, the remainder is 7.",
    "options": [],
    "answer": "See solution",
    "solution": "From the second condition, since the square of any integer is congruent to 0 or 1 modulo 4, all digits must be either even or odd. From the third condition, digit $c$ must be 8 or 9.\n\nIf $c = 8$, then $\\overline{abcd}$ is even, but it cannot give an odd remainder modulo 8, since $\\overline{abcd} = 8r + 7$ would be odd.\n\nThus, $c = 9$. All digits are odd, and the remainder of the number modulo 9 equals the remainder of the sum of its digits modulo 9. So, $a + b + d = 9r + 7$. From the first condition, $d = 9$. Consider the cases:\n\n- If $r = 0$, then $a + b + d = 7$, so $a + b = -2$, which is impossible.\n- If $r = 1$, then $a + b + d = 16$, so $a + b = 7$, which is impossible since $a$ and $b$ are odd.\n- If $r = 2$, then $a + b + d = 25$, so $a + b = 16$. The only possibility is $a = 7$, $b = 9$, $d = 9$.\n\nTherefore, the number is $7999$.\n\nNo other cases are possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17513,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether it is possible to partition the set of positive integers into infinite subsets $A_1, A_2, \\dots$ such that for every positive integer $k$, the sum of the elements of $A_k$ is $k + 2023$.\n\n*Remark: a partition of a set $X$ is a collection of subsets of $X$ such that every element of $X$ is contained in exactly one of the subsets.*",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is No.\n\nSuppose such a partition exists. Then for every positive integer $k$, we have\n\n$$\nB_k = A_1 \\cup A_2 \\cup \\dots \\cup A_k \\subset \\{1, 2, \\dots, k + 2023\\},\n$$\n\nsince all elements of $A_i$ are at most $i + 2023$ for every $i \\in \\{1, 2, \\dots, k\\}$, and\n\n$$\n\\sum_{b \\in B_k} b = \\sum_{i=2024}^{k+2023} i < \\sum_{i=1}^{k+2023} i.\n$$\n\nNow let $t_k$ be the minimum positive integer not in $B_k$. Then\n\n$$\n\\sum_{i=2024}^{k+2023} i = \\sum_{b \\in B_k} b \\le \\left( \\sum_{i=1}^{k+2023} i \\right) - t_k,\n$$\n\nwhich implies that\n\n$$\nt_k \\le \\sum_{i=1}^{2023} i,\n$$\n\nfor every integer $k$. But that cannot happen if $A_1, A_2, \\dots$ is a partition of the positive integers. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17514,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if\n\n$$\n(n^2 + 1)^{2k} \\cdot (44n^3 + 11n^2 + 10n + 2) = N^m\n$$\n\nholds for some non-negative integer values of $m$, $n$, $N$, and $k$, then $m = 1$ must hold.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* Since the left side of the equation is certainly larger than 1, we first note that $m > 0$ must certainly hold.\n\nNow, we consider even values of $n$. Since $n^2+1 \\equiv 1 \\pmod 4$ and $44n^3+11n^2+10n+2 \\equiv 2 \\pmod 4$ are certainly true, we have $N^m \\equiv 2 \\pmod 4$. If $m > 1$, $N^m$ is odd for any odd $N$ and divisible by 4 for any even $N$, and it follows that $m = 1$ must hold, as claimed.\n\nNext, we consider odd values of $n$. In this case we have $44n^3 + 11n^2 + 10n + 2 \\equiv 3 \\pmod 4$ and $n^2 + 1 \\equiv 2 \\pmod 4$, and we see that the factor 2 is contained in $N^m$ exactly $2^k$ times.\n\nFor $k=0$ we obtain $N^m = (n^2+1)(44n^3+11n^2+10n+2) \\equiv 2 \\pmod 4$, and the same argument holds as for even values of $n$.\n\nFor $k > 0$, the exponent $m > 1$ must be a divisor of the exponent $k$ of 2 in the prime decomposition of $N^m$, and therefore a power of 2. This means that $(n^2 + 1)^{2k}$ is an $m$-th power, this must also be the case for $44n^3 + 11n^2 + 10n + 2$, and this number must certainly be a perfect square. This is not possible, however, since we have established that this number is $\\equiv 3 \\pmod 4$, and therefore certainly not a perfect square. This case is therefore not possible, and we see that $m = 1$ must hold, as claimed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17515,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的內切圓為 $\\omega$，內心為 $I$，外接圓為 $\\Gamma$。令 $D$ 為 $\\omega$ 在 $BC$ 邊的切點，並令 $M$ 是 $ID$ 的中點。設 $A'$ 為 $A$ 在 $\\Gamma$ 上的對徑點（即 $AA'$ 為 $\\Gamma$ 的一條直徑）。設 $X$ 為直線 $A'M$ 與圓 $\\Gamma$ 的另一個交點。\n\n證明：三角形 $AXD$ 的外接圓與直線 $BC$ 相切。\n\nLet $ABC$ be a triangle with incircle $\\omega$, incentre $I$ and circumcircle $\\Gamma$.\nLet $D$ be the tangency point of $\\omega$ with $BC$, let $M$ be the midpoint of $ID$, and let $A'$ be the diametral opposite of $A$ with respect to $\\Gamma$. If we denote $X = A'M \\cap \\Gamma$ then prove that the circumcircle of $\\triangle AXD$ is tangent to $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "考慮一個過 $A$ 點、並與 $BC$ 直線切於 $D$ 點的圓；設此圓與 $\\Gamma$ 的另一個交點為 $Y$。我們將證明 $Y, M, A'$ 共線。故 $X = Y$，而本題就得證了。\n\n設 $\\omega$ 分別切 $AC, AB$ 邊於 $E, F$ 點；令 $EF$ 交 $BC$ 於 $P$ 點，而 $AX$ 交 $BC$ 於 $J$ 點。\n\n![](images/15-2J_p15_data_780c927d04.png)\n\n因為 $JB \\cdot JC = JA \\cdot JY = JD^2$ 且 $(B, C, D, P) = -1$，所以 $J$ 是 $DP$ 的中點。注意到 $AD$ 是 $P$ 點對圓 $\\omega$ 的圓幂，所以 $AD$ 與 $PI$ 垂直，設垂足為 $U$ 點。因此 $\\triangle PID$ 的兩邊中點連線 $JM$ 是 $UD$ 的中垂線。故得 $\\angle JUD = \\angle JDU = \\angle AXD$，所以 $J, Y, U, M, D$ 共圓。由此得 $\\angle AYM = \\angle JDM = 90^\\circ$，故 $YM$ 與 $\\Gamma$ 交於 $A$ 的對徑點 $A'$，即 $Y, M, A'$ 共線。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17516,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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$$\n(T_{i_1} \\cap T_{i_2}) \\subset T_{i_3} \\cup T_{i_4}\n$$\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$$\n|(T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)| \\ge 1.\n$$\n\nStudying the sum:\n\n$$\nS = \\sum_{\\substack{1 \\le i < j \\le n \\\\ 1 \\le k < \\ell \\le n \\\\ \\{i,j\\} \\cap \\{k,\\ell\\} = \\emptyset}} |(T_i \\cap T_j) \\setminus (T_k \\cup T_\\ell)|,\n$$\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\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 $m = 1, \\dots, 15$, 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\n$$\nS = \\sum_{\\substack{m=1 \\\\ 2 \\le n_m \\le n-2}}^{15} \\frac{n_m(n_m-1)}{2} \\cdot \\frac{(n-n_m)(n-n_m-1)}{2}.\n$$\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$$\n\\begin{aligned}\nS & \\le \\sum_{\\substack{m=1 \\\\ 2 \\le n_m \\le n-2}}^{15} \\frac{1}{4} \\left(\\frac{n}{2}\\right)^2 \\cdot \\left(\\frac{n-2}{2}\\right)^2 \\\\\n& \\le 15 \\cdot \\frac{1}{4} \\cdot \\left(\\frac{n}{2}\\right)^2 \\cdot \\left(\\frac{n-2}{2}\\right)^2 \\\\\n& < \\frac{n(n-1)}{2} \\cdot \\frac{(n-2)(n-3)}{2}, \\quad \\forall n \\ge 33,\n\\end{aligned}\n$$\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": 17517,
    "subject": "Mathematics (Olympiad)",
    "question": "Construct the circumcircle of $\\triangle ABC$. Let $D'$, $E'$, and $F'$ denote the points where the altitudes meet the circumcircle. Let $K$ be the intersection of $AD'$ with $BC$, and denote the orthocentre of $\\triangle ABC$ by $H$.\n\n% ![](images/Ireland_2013_Booklet_p29_data_d0b0786bb8.png)\n\nShow that $H$ is the incentre of $\\triangle DEF$, where $D$, $E$, and $F$ are the feet of the altitudes from $A$, $B$, and $C$ respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle BAD' = 90^\\circ - \\angle ABC = \\angle BCF$ and $\\angle BAD' = \\angle BCD'$, hence the triangles $HKC$ and $KD'C$ are congruent. This shows that $|HK| = |KD'|$ and $D' = D$ is the image of $H$ when reflected in $BC$. Similarly, $E' = E$ and $F' = F$.\n\nNow $\\angle DFC = \\angle DAC = 90^\\circ - \\angle ACB$ and $\\angle CFE = \\angle CBE = 90^\\circ - \\angle ACB$. Therefore, $\\angle DFC = \\angle CFE$, which means that $CF$ is the bisector of $\\angle DFE$. Similarly, $BE$ and $AD$ are the bisectors of $\\angle FED$ and $\\angle FDE$, respectively. This shows that $H$ is the incentre of $\\triangle DEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17518,
    "subject": "Mathematics (Olympiad)",
    "question": "Fourteen people meet one day to play three matches of netball. For each match, they divide themselves into two teams of seven players. In each match, one team wins while the other team loses. After all three matches, no person has been on a losing team three times.\n\nProve that there are at least three players who were on the same team as each other for all three matches.",
    "options": [],
    "answer": "See solution",
    "solution": "For each person, we record the results of their matches with a string of *Ws* and *Ls*, where *W* denotes a win and *L* denotes a loss. The record of each player's results is one of the following seven possibilities:\n\n$$\nWWW \\quad WWL \\quad WLW \\quad WLL \\quad LWW \\quad LWL \\quad LLW\n$$\n\nIt is impossible for each of these records to be obtained by exactly two players. That would imply that the sum of the number of wins for each player is $24$, while the sum of the number of losses for each player is $18$. However, there should be an equal number of wins and losses overall.\n\nHence, there must exist three players who have the same record. It follows that these three players were on the same team as each other for all three matches.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17519,
    "subject": "Mathematics (Olympiad)",
    "question": "За множеството $S = \\{-2, -1, 0, 1, 2\\}$ од правоаголен координатен систем се избрани 17 точки од множеството $S \\times S$. Докажи дека постојат три точки $A$, $B$, $C$ од избраните, такви што $B$ е средина на отсечката $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Ќе разгледаме два случаи.\n\nа) Координатниот почеток е во избраните 17 точки. Од преостанатите 24 точки ќе формираме 12 пара точки. Точките од еден пар се централно симетрични во однос на координатниот почеток. Бидејќи бројот на парови е 12, а бројот на избрани точки е поголем од 13, во еден пар двете точки ќе бидат од избраните точки. Координатниот почеток и тие две точки се бараните точки. За $B$ се бира координатниот почеток, а точките од парот се $A$ и $C$.\n\n![](images/Makedonija_2009_p15_data_9a18a062d2.png)\n\nб) Координатниот почеток не е во избраните 17 точки.\n\nПреостанатите 24 точки ќе ги разбиеме во групи по три точки како на цртежот. Во една од групите, според принципот на Дирихле, имаме три точки од избраните. Навистина, ако претпоставиме спротивно, т.е. дека во секоја група имаме најмногу две точки од избраните 17 точки, тогаш сме избрале не повеќе од 16 точки, што е контрадикција.\n\nГрупата од точки во која сите три точки се од избраните се бараните три точки. Средната од нив е $B$, а крајните се $A$ и $C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17520,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set of numbers chosen from $1, 2, \\ldots, 2015$ with the property that any two distinct numbers, say $x$ and $y$, in $A$ determine a unique isosceles triangle (which is not equilateral) whose sides are of length $x$ or $y$. What is the largest possible size of $A$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x < y$ be two numbers in $A$. For them to determine a unique isosceles triangle, we must have $2x \\leq y$. If $A = \\{2^0, 2^1, 2^2, \\ldots, 2^{10}\\}$, then any two of the numbers $x < y$ satisfy $2x \\leq y$. So the maximum size is $\\geq 11$.\n\nNow suppose that there is a set $A$ with $|A| = 12$ that has the property. Let $a_1, a_2, \\ldots, a_{12}$ be the elements of $A$ in increasing order. Then $a_2 \\geq 2a_1$, $a_3 \\geq 2a_2 \\geq 2^2 a_1$, and so on, so $a_{12} \\geq 2a_{11} \\geq 2^{11} a_1 \\geq 2^{11} = 2048$, a contradiction since $2015 < 2048$. Thus, the maximum size is $11$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17521,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\tau(n)$ be the number of divisors of a natural number $n$. Prove that there exist infinitely many natural numbers $N$ such that\n$$\n(\\tau(N) + \\tau(N+1) + 1) \\equiv 3 \\pmod{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Take $N = 3n^3$ with $(n, 3) = 1$. Then $\\tau(N) = \\tau(3)\\tau(n^3) = 2 \\cdot \\tau(n^3)$. Now, consider $N+1 = 3n^3 + 1$. We claim that for $n \\equiv 1 \\pmod{8}$, $3n^3 + 1 \\equiv 4 \\pmod{8}$, so $N+1$ is divisible by $4$ but not by $8$. This affects the value of $\\tau(N+1)$. Thus, for $n = 24m + 1$, $N = 3n^3$ yields infinitely many $N$ satisfying the required congruence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17522,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(n, k)$ such that\n\n$$\nn! + 8 = 2^k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answers:** $(n, k) = (4, 5)$ and $(5, 7)$.\n\nFor reference, the given equation is\n\n$$\nn! + 8 = 2^k.\n$$\n\n**Case 1:** $n \\geq 6$\n\nObserve that $n!$ is a multiple of $6! = 2^4 \\times 3^2 \\times 5$. Hence $n! = 16x$ for some positive integer $x$. Therefore,\n\n$$\nn! + 8 = 8(2x + 1).\n$$\n\nBut the right-hand side cannot be a power of $2$ because $2x + 1$ is an odd integer greater than $1$. Hence, there are no solutions in this case.\n\n**Case 2:** $n \\leq 5$\n\nWe simply tabulate the values of $n! + 8$ and check which ones are powers of $2$.\n\n![]( <table><thead><tr><th>n</th><th>n! + 8</th><th>power of 2?</th></tr></thead><tbody><tr><td>1</td><td>9</td><td>no</td></tr><tr><td>2</td><td>10</td><td>no</td></tr><tr><td>3</td><td>14</td><td>no</td></tr><tr><td>4</td><td>32</td><td>2<sup>5</sup></td></tr><tr><td>5</td><td>128</td><td>2<sup>7</sup></td></tr></tbody></table> )\n\nThis yields the solutions given at the outset. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17523,
    "subject": "Mathematics (Olympiad)",
    "question": "In a cafe, each product costs at most 12 ducats. Currently, the cafe owner is only using coins worth 1 ducat. This is impractical for the more expensive products. Therefore, the cafe owner has decided to introduce two types of coins in addition to the 1 ducat coins. He wants to do this so that as many values from 1 to 12 ducats as possible can be paid with at most two coins (without change).\n\nWhat should be the values of the two new types of coins?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the two new coin values be $a$ and $b$ (with $a < b$). We want to maximize the number of amounts from 1 to 12 that can be paid using at most two coins (using coins of 1, $a$, or $b$ ducats).\n\nThe possible payments with two coins are:\n- $1 + a$\n- $1 + b$\n- $a + b$\n- $1 + 1 = 2$\n- $a + a = 2a$\n- $b + b = 2b$\n\nAlso, single coins: $1$, $a$, $b$.\n\nTo cover as many values as possible, choose $a = 5$ and $b = 7$:\n\nPossible sums:\n- $1$\n- $5$\n- $7$\n- $1 + 1 = 2$\n- $1 + 5 = 6$\n- $1 + 7 = 8$\n- $5 + 5 = 10$\n- $5 + 7 = 12$\n- $7 + 7 = 14$ (but 14 > 12, so not needed)\n\nCovered values: $1, 2, 5, 6, 7, 8, 10, 12$\n\nAlternatively, $a = 4$, $b = 8$:\n- $1$\n- $4$\n- $8$\n- $1 + 1 = 2$\n- $1 + 4 = 5$\n- $1 + 8 = 9$\n- $4 + 4 = 8$\n- $4 + 8 = 12$\n- $8 + 8 = 16$ (not needed)\n\nCovered values: $1, 2, 4, 5, 8, 9, 12$\n\nBut with $a = 5$ and $b = 7$, more values are covered. Thus, the two new coin values should be **5 ducats and 7 ducats**.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17524,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a tuple $$(b_m, b_{m+1}, \\dots, b_n)$$ of integers *perfect* if both following conditions are fulfilled:\n\n1. There exists an integer $a > 1$ such that $b_k = a^k + 1$ for all $k = m, m+1, \\dots, n$;\n2. For all $k = m, m+1, \\dots, n$, there exists a prime number $q$ and a non-negative integer $t$ such that $b_k = q^t$.\n\nProve that if $n - m$ is large enough then there is no perfect tuples, and find all perfect tuples with the maximal number of components.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$ is a perfect tuple with length 5. Show in the rest that there are no other perfect tuples with length 5 or larger.\n\nFor that, let $(a^m + 1, a^{m+1} + 1, \\dots, a^n + 1)$ be an arbitrary perfect tuple with length at least 5. There must exist at least two odd exponents among $m, m+1, \\dots, n$; let $k$ and $k+2$ be the two largest odd exponents. As $a^k + 1$ and $a^{k+2} + 1$ are prime powers while having a common divisor $a+1$, these two integers must be powers of the same prime $q$. Thus the larger of them, $a^{k+2} + 1$, is divisible by the smaller one, $a^k + 1$, which shows that $a^k + 1$ divides also the difference $a^2 \\cdot (a^k + 1) - (a^{k+2} + 1) = a^2 - 1$. Hence $a^k + 1 \\le a^2 - 1$, implying $k < 2$. So $k = 1$ as $k$ is odd. By choice of $k$, the only odd exponents in our perfect tuple are 1 and 3 and the tuple is of the form $(a^0 + 1, a^1 + 1, a^2 + 1, a^3 + 1, a^4 + 1)$.\n\nAs $a+1$ and $a^3+1$ are powers of the same prime number $q$, also the ratio $\\frac{a^3+1}{a+1} = a^2-a+1$ is a power of $q$. Note that $a^2-a+1 \\ge 2a-a+1 = a+1$ by $a \\ge 2$, hence $a^2-a+1$ is divisible by $a+1$. Thus the difference $(a^2-a+1)-(a+1)(a-2) = 3$ is divisible by $a+1$. This filters out the only possibility $a=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17525,
    "subject": "Mathematics (Olympiad)",
    "question": "a. How many 4-digit palindromes are there between 2000 and 3000?\n\nb. How many different triples of palindromes (using only palindromes up to 121) sum to 121?\n\nc. What is the 110th palindrome in the sequence of all palindromes less than 2000?",
    "options": [],
    "answer": "See solution",
    "solution": "a. The first digit is 2, so the last digit is 2. The middle two digits must form a palindrome. Thus, there are 10 palindromes between 2000 and 3000:\n\n2002, 2112, 2222, 2332, 2442, 2552, 2662, 2772, 2882, 2992.\n\nb. The palindromes up to 121 are 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111. Working systematically, we find there are only five triples of different palindromes that total 121:\n\n$$\n121 = 88 + 33 = 88 + 22 + 11 \\\\\n121 = 77 + 44 = 77 + 33 + 11 \\\\\n121 = 66 + 55 = 66 + 44 + 11 \\\\\n121 = 66 + 55 = 66 + 33 + 22 \\\\\n121 = 55 + 66 = 55 + 44 + 22\n$$\n\nc.\n\nThe following table lists all palindromes less than 2000.\n\n| Palindromes           | Number | Subtotal |\n|----------------------|--------|----------|\n| 11, 22, 33, ..., 99  | 9      | 9        |\n| 101, 111, ..., 191   | 10     | 19       |\n| 202, 212, ..., 292   | 10     | 29       |\n| 303, 313, ..., 393   | 10     | 39       |\n| 404, 414, ..., 494   | 10     | 49       |\n| 505, 515, ..., 595   | 10     | 59       |\n| 606, 616, ..., 696   | 10     | 69       |\n| 707, 717, ..., 797   | 10     | 79       |\n| 808, 818, ..., 898   | 10     | 89       |\n| 909, 919, ..., 999   | 10     | 99       |\n| 1001, 1111, 1221, ..., 1991 | 10 | 109 |\n\nSo the 110th palindrome is 2002.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17526,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Let $Q$ be the intersection of $EF$ and $BC$. Prove that $(Q, D, B, C)$ is a harmonic division. Show that $(AQ, AD, AB, AC)$ and $(DE, DF, DA, DQ)$ are harmonic. Given that $DA \\perp DQ$, prove that $DA$ is the internal bisector of $\\angle EDF$. If $AEDF$ is an inscribed quadrilateral, show that $AE = AF$.\n\nSuppose the circumcircle of $AEDF$ cuts the segment $BC$ at $G \\neq D$. Prove that $DG$ is the internal bisector of $\\angle EDF$ so $GE = GF$, and hence $\\triangle AGE = \\triangle AGF$, so $AG$ is the bisector of $\\angle BAC$. Using *Menelaus's* theorem, with the line $BPE$ cutting the three sides of triangle $ADC$ and the line $CPF$ cutting the three sides of triangle $ABD$, show that\n\n$$\n\\frac{EA}{EC} = \\frac{PA}{PD} \\cdot \\frac{BD}{BC}, \\quad \\frac{FA}{FB} = \\frac{PA}{PD} \\cdot \\frac{CD}{BC}\n$$\n\nand deduce that\n\n$$\n\\frac{FA}{FB} + \\frac{EA}{EC} = \\frac{PA}{PD}.\n$$\n\nFinally, prove that\n\n$$\n\\begin{aligned}\nPAPD &= \\frac{FA}{GF} \\cdot \\frac{GF}{FB} + \\frac{EA}{GE} \\cdot \\frac{GE}{EC} \\\\\n&= \\cot \\frac{A}{2} \\tan B + \\cot \\frac{A}{2} \\tan C \\\\\n&= (\\tan B + \\tan C) \\cot \\frac{A}{2}.\n\\end{aligned}\n$$\n\n% IMAGE: ![](images/booklet_2014_final_p27_data_7046142dfd.png)\n\n(b) Let $L$ be the intersection of $BM$ and $CN$. Prove that $L$ is a fixed point. Given that quadrilateral $ACKN$ has $\\angle ACN = \\angle AKN = 90^\\circ$, show it is cyclic. Similarly, show $ABKM$ is cyclic. Prove that\n\n$$\n\\begin{aligned}\n\\angle MAN + \\angle BKC &= \\angle MAK + \\angle NAK + \\angle BKC \\\\\n&= \\angle MBK + \\angle NCK + \\angle BKC = \\angle BLC\n\\end{aligned}\n$$\n\nis a constant.\n\n% IMAGE: ![](images/booklet_2014_final_p28_data_d774bd7cc5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $Q$ be the intersection of $EF$ and $BC$ so $(Q, D, B, C)$ is harmonic division. So $(AQ, AD, AB, AC)$ and $(DE, DF, DA, DQ)$ are harmonic. Moreover, because $DA \\perp DQ$ so $DA$ is the internal bisector of $\\angle EDF$. Because $AEDF$ is the inscribed quadrilateral so $AE = AF$.\n\nSuppose that the circumcircle of $AEDF$ cuts the segment $BC$ at $G$ different from $D$. We have $DG$ is the internal bisector of angle $EDF$ so $GE = GF$. Hence, $\\triangle AGE = \\triangle AGF$ so $AG$ is the bisector of $\\angle BAC$. By applying the *Menelaus's* theorem with the line $BPE$ cutting three sides of triangle $ADC$ and the line $CPF$ cutting three sides of triangle $ABD$, we have\n\n$$\n\\frac{EA}{EC} = \\frac{PA}{PD} \\cdot \\frac{BD}{BC}, \\quad \\frac{FA}{FB} = \\frac{PA}{PD} \\cdot \\frac{CD}{BC}\n$$\n\nFrom this result, we have\n\n$$\n\\frac{FA}{FB} + \\frac{EA}{EC} = \\frac{PA}{PD} \\left( \\frac{BD}{BC} + \\frac{CD}{BC} \\right) = \\frac{PA}{PD}.\n$$\n\nTherefore, we get\n\n$$\n\\begin{aligned}\nPAPD &= \\frac{FA}{GF} \\cdot \\frac{GF}{FB} + \\frac{EA}{GE} \\cdot \\frac{GE}{EC} = \\cot \\frac{A}{2} \\tan B + \\cot \\frac{A}{2} \\tan C \\\\\n&= (\\tan B + \\tan C) \\cot \\frac{A}{2}.\n\\end{aligned}\n$$\n\nThis is the equality we have to find.\n\n(b) Let $L$ be the intersection of $BM$ and $CN$, then $L$ is the fixed point. The quadrilateral $ACKN$ has $\\angle ACN = \\angle AKN = 90^\\circ$ so it is a cyclic quadrilateral. Similarly, $ABKM$ is also a cyclic quadrilateral. Therefore, we have\n\n$$\n\\begin{aligned}\n\\angle MAN + \\angle BKC &= \\angle MAK + \\angle NAK + \\angle BKC \\\\\n&= \\angle MBK + \\angle NCK + \\angle BKC = \\angle BLC\n\\end{aligned}\n$$\n\nis a constant.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17527,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ and points $D$ and $E$ on rays $BA$ and $CA$ respectively such that $\\frac{BD}{CE} = k$. Let $M$ and $N$ be points on the line segments $BC$ and $DE$ respectively such that $\\frac{BM}{MC} = \\frac{DN}{NE} = k$. Then $MN$ is parallel to the bisector of angle $\\angle BAC$.\n\n% IMAGE: ![](images/RMC_2019_var_3_p72_data_6fde5ff81e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof of the Lemma.*\n\nWe have $\\overrightarrow{NM} = \\frac{1}{k+1} (\\overrightarrow{DB} + k \\cdot \\overrightarrow{EC})$. If $X$ and $Y$ are points on rays $AB$ and $AC$ such that $AX = DB$ and $AY = k \\cdot EC$, then $\\overrightarrow{NM}$ is parallel to the median from $A$ in triangle $AXY$. But $AX = Y$, which means that in triangle $AXY$ the median from $A$ has the direction of the bisector of angle $\\angle BAC$.\n\nFrom the Lemma we obtain that $A'N \\parallel SM$, which means that $N$ is the reflection of point $D$ across $M$. In triangle $SXY$, the circumcenter lies on $SN$, therefore the orthocenter lies on its isogonal, $SD$. It follows that $SD \\perp XY$, i.e. $SD \\parallel AA'$. We obtain that $ADSO$ is a parallelogram, hence $SD = R$. But then $SDOA'$ is also a parallelogram, hence the conclusion.\n\n*Remark.* The configuration of the problem has many other interesting properties: $M$ is the circumcenter of triangle $OXY$. Indeed, as $SM$ is the bisector of angle $\\angle XSY$, we have $MX = MT$. From $\\angle XMY = 180^\\circ - \\angle XSY = 2\\angle A = \\angle BOC = 2\\angle XOY$ we obtain the statement.\n\n$OSA'N$ is an isosceles trapezoid, hence $SN = OA' = R$. Thus, the circumcircle of $SXY$ passes through $M$ and $N$ and has the same radius as the Euler circle of triangle $ABC$, which shows that it is the reflection of the latter across point $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17528,
    "subject": "Mathematics (Olympiad)",
    "question": "*(a)* Consider the sequence defined as follows:\n\n$$\na_1 = \\frac{10}{11}, \\quad a_2 = \\frac{12}{14} = \\frac{6}{7}, \\quad a_3 = \\frac{8}{10} = \\frac{4}{5}, \\quad a_4 = \\frac{6}{8} = \\frac{3}{4}, \\\\\na_5 = \\frac{5}{7}, \\quad a_6 = \\frac{7}{10}, \\quad a_7 = \\frac{9}{13}\n$$\n\nIt appears that the last simplification occurs at $a_4$. Prove, by induction, that for all $n \\geq 5$, there is no further simplification, and that $a_n = \\frac{1 + 2(n-3)}{1 + 3(n-3)}$ for all $n \\geq 5$.\n\n*(b)* Show that for a similar sequence starting with $a_1 = \\frac{97}{98}$, there must be a simplification at some point.\n\n*(c)* Give examples of $c$ such that the sequence starting with $a_1 = \\frac{c}{c+1}$ eventually produces a simplification, and illustrate with the first few terms.",
    "options": [],
    "answer": "See solution",
    "solution": "*(a)*\n\nWe use induction to prove that for all $n \\geq 5$, $a_n = \\frac{1 + 2(n-3)}{1 + 3(n-3)}$ and the fraction cannot be simplified further.\n\nFor $n = 5$, $a_5 = \\frac{5}{7} = \\frac{1 + 2(5-3)}{1 + 3(5-3)}$, and $\\frac{5}{7}$ is already in lowest terms. Assume the statement holds for $n = k-1$. Then $a_{k-1} = \\frac{1 + 2(k-4)}{1 + 3(k-4)}$ is in lowest terms. The next term is:\n\n$$\na_k = \\frac{1 + 2(k-4) + 2}{1 + 3(k-4) + 3} = \\frac{1 + 2(k-3)}{1 + 3(k-3)}\n$$\n\nSuppose, for contradiction, that $d > 1$ divides both numerator and denominator. Then $d$ divides $3(1 + 2(k-3)) - 2(1 + 3(k-3)) = 1$, which is impossible. Thus, the fraction is in lowest terms for all $n \\geq 5$.\n\n*(b)*\n\nSuppose there is no simplification for the sequence starting with $a_1 = \\frac{97}{98}$. By induction, $a_n = \\frac{97 + 2n}{97 + 3n}$. For $n = 97$, $a_{97} = \\frac{97 + 2 \\times 97}{97 + 3 \\times 97} = \\frac{291}{388}$, but both numerator and denominator are divisible by $97$, so the fraction can be simplified, a contradiction.\n\n*(c)*\n\nExamples: For $c = 7$, the sequence is $\\frac{7}{8}, \\frac{9}{11}, \\frac{11}{14}, \\frac{13}{17}, \\frac{15}{20} = \\frac{3}{4}$. For $c = 27$, the sequence is $\\frac{27}{28}, \\frac{29}{31}, \\frac{31}{34}, \\frac{33}{37}, \\frac{35}{40} = \\frac{7}{8}$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17529,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ have incenter $I$. Let $E$ and $F$ be the points where the incircle touches $AC$ and $AB$, respectively.\n\n(i) Prove that if $\\triangle ABC$ is isosceles, then $|AE| = |AF|$.\n\n(ii) Let $x = |AF|$, $y = |AE|$, and $z = |BF| = |CE|$. Show that $|AB| = |AC|$ if and only if $x = y$.\n\n![](images/Irish_2016_Booklet_p15_data_ea0e1aad55.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Because $AI$ is the bisector of $\\angle BAC$, $|AE| = |AF|$ and $AI$ is shared by the triangles $\\triangle AEI$ and $\\triangle AFI$, these two triangles are congruent. Therefore, $\\angle AFI = \\angle AEI$. These angles are exterior angles to the triangles $\\triangle BIF$ and $\\triangle CIE$ and so we see that $\\angle IBF + \\angle FIB = \\angle EIC + \\angle ECI$, hence $\\angle IBF = \\angle ECI$ and so $\\angle CBA = \\angle ACB$ which implies that $\\triangle ABC$ is isosceles.\n\n(ii) Let $x = |AF|$, $y = |AE|$ and $z = |BF| = |CE|$. Because an angle bisector divides the opposite side in the ratio of the adjacent sides, we have\n\n$$\n\\frac{x}{z} = \\frac{y+z}{a} \\quad \\text{and} \\quad \\frac{y}{z} = \\frac{x+z}{a}, \\quad \\text{hence} \\quad \\frac{a}{z} = \\frac{y+z}{x} = \\frac{x+z}{y}.\n$$\n\nThis implies $y^2 + yz = x^2 + xz$ from which we get $(y-x)(x+y+z) = 0$. As $x+y+z > 0$ this implies $y = x$ and so $|AB| = x+z = y+z = |AC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17530,
    "subject": "Mathematics (Olympiad)",
    "question": "某個國家有 $n$ 個城市，其中 $n > 1$。這之中有些城市之間有鐵路相通，使得你可以從任何一個城市，透過若干段火車抵達任何一個其他城市（鐵路是雙向的）。此外，這個國家沒有環狀線，因此你不可能從一個城市出發，經過一系列不同的城市後，繞回原來的城市。\n\n每個城市的流量，為該城市搭一段火車可以抵達的城市數量。對於一個城市 $A$，如果從該城市搭一段火車可達的城市數量為 $x$，其中有 $y$ 個的流量比 $A$ 小，則我們稱 $\\frac{y}{x}$ 為城市 $A$ 的重要性。\n\n試求最小的正實數 $t$，使得對於所有 $n$，不論該國的鐵路如何鋪設，整個國家所有城市的重要性總和，必然小於 $tn$。",
    "options": [],
    "answer": "See solution",
    "solution": "最小的 $t$ 為 $\\frac{3}{8}$。\n\n我們將城市視為點，鐵路視為邊，流量則為每一點的 degree。對於兩個點 $i$ 與 $j$，以 $(i, j)$ 表示連接兩點的邊，$E$ 為所有邊的集合，而 $d_i$ 與 $d_j$ 分別為兩點的 degree。\n\n![](images/20-2J_p31_data_0dc25e3042.png)\n\n我們先證明 $t$ 至少要是 $\\frac{3}{8}$。考慮上圖中的構造，假設 degree 3 的點有 $k$ 個，則 $n = 4k + 3$ 且重要性總和為 $k + \\frac{k+2}{2} = \\frac{3k+2}{2}$。由於\n\n$$\n\\frac{(3k + 2)/2}{4k + 3} \\rightarrow \\frac{3}{8},\n$$\n\n故知 $t$ 至少需為 $\\frac{3}{8}$。\n\n接著證明 $t = \\frac{3}{8}$ 確為最小可能值。注意到，對於每一條邊 $(i,j)$，若 $d_i = d_j$ 則這條邊對重要性沒有貢獻；否則，若 $d_i > d_j$，則該邊貢獻 $i$ 的重要性 $\\frac{1}{d_i}$。又我們有\n\n$$\n\\sum_{(i,j) \\in E} \\frac{1}{2d_i} + \\frac{1}{2d_j} = \\frac{n}{2}, \\qquad (1)\n$$\n\n故所有城市的重要性總和是\n\n$$\n\\frac{n}{2} - \\frac{1}{2} \\sum_{(i,j) \\in E} f(d_i, d_j) \\qquad (2)\n$$\n\n其中\n\n$$\nf(x, y) = \\begin{cases} \\frac{1}{x} - \\frac{1}{y} & \\text{if } x < y \\\\ \\frac{1}{y} - \\frac{1}{x} & \\text{if } y < x \\\\ \\frac{2}{x} & \\text{if } y = x. \\end{cases} \\quad (3)\n$$\n\n所以問題變成是求 $\\sum_{(i,j) \\in E} f(d_i, d_j)$ 的最小可能值。\n\n讓我們假設有 $x$ 個點的 degree 為 1，$n - x - y - z$ 個點的 degree 為 2，$y$ 個點的 degree 為 3，$z$ 個點的 degree 大於 3。基於 degree 和為邊數的兩倍，我們有 $x > y+2z$。除此之外，若有 $a$ 對 degree 1 跟 degree 3 的點相連，則 $\\sum_{(i,j) \\in E} f(d_i, d_j)$ 至少為\n\n$$\n\\frac{x-a}{2} + \\frac{2(n-x-y-z) - (x-a)}{6} + \\frac{2a}{3} + \\frac{\\max(0, 3y-a - (2(n-x-y-z) - (x-a)))}{12}\n$$\n\n因為 $f(1,t) \\ge \\frac{1}{2}, f(2,t) \\ge \\frac{1}{6}, f(3,t) \\ge \\frac{1}{12}$，化簡後有\n\n$$\n\\frac{x-a}{2} + \\frac{2n-3x-2y-2z+a}{6} + \\frac{2a}{3} + \\frac{\\max(0, 5y+3x+2z-2n-2a)}{12}\n$$\n\n注意到 $a=0$ 永遠較小，所以可以設 $a=0$，變成\n\n$$\n\\frac{x}{2} + \\frac{2n - 3x - 2y - 2z}{6} + \\frac{\\max(0, 5y + 3x + 2z - 2n)}{12} = \\frac{n}{6} + \\frac{\\max(2x + 2y, 2n - 4y - 4z)}{12}\n$$\n\n注意到如果固定 $y+z$，則 $z$ 越大越好，所以根據不等式 $x > y+2z$ 可以假設 $2z = x - y - 1$。原式變成\n\n$$\n\\frac{n}{6} + \\frac{\\max(2x + 2y, 2n - 2x - 2y + 1)}{12} > \\frac{n}{6} + \\frac{n}{12} = \\frac{n}{4}.\n$$\n\n故整個國家的重要性總和會小於 $\\frac{n-\\frac{n}{4}}{2} = \\frac{3n}{8}$。$\\square$\n\n![](images/20-2J_p32_data_df767ca098.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17531,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $K$ be a convex planar set, symmetric about a point $O$, and let $X$, $Y$, $Z$ be three points in $K$. Show that $K$ contains the head of one of the vectors $\\overrightarrow{OX} \\pm \\overrightarrow{OY}$, $\\overrightarrow{OX} \\pm \\overrightarrow{OZ}$, $\\overrightarrow{OY} \\pm \\overrightarrow{OZ}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If two of the vectors $\\overrightarrow{OX}$, $\\overrightarrow{OY}$, $\\overrightarrow{OZ}$ are linearly dependent, the conclusion is clear. So suppose $\\alpha \\cdot \\overrightarrow{OX} + \\beta \\cdot \\overrightarrow{OY} + \\gamma \\cdot \\overrightarrow{OZ} = \\mathbf{0}$, where $\\alpha\\beta\\gamma \\neq 0$, and let $|\\gamma| = \\min(|\\alpha|, |\\beta|, |\\gamma|)$ to write $\\overrightarrow{OZ} = \\alpha' \\cdot \\overrightarrow{OX} + \\beta' \\cdot \\overrightarrow{OY}$, where $|\\alpha'| \\ge 1$ and $|\\beta'| \\ge 1$. Since $K$ is symmetric about $O$, we may (and will) assume that $\\alpha' \\ge 1$ and $\\beta' \\ge 1$. Write\n\n$$\n\\overrightarrow{OX} + \\overrightarrow{OY} = \\frac{\\beta' - 1}{\\alpha' + \\beta' - 1} \\cdot \\overrightarrow{OX} + \\frac{\\alpha' - 1}{\\alpha' + \\beta' - 1} \\cdot \\overrightarrow{OY} + \\frac{1}{\\alpha' + \\beta' - 1} \\cdot \\overrightarrow{OZ},\n$$\n\nto deduce that the head of the vector $\\overrightarrow{OX} + \\overrightarrow{OY}$ lies in the triangle $XYZ$, which is the convex hull of the points $X$, $Y$, $Z$. Since $K$ is convex, the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 17532,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram. Let $W$, $X$, $Y$, and $Z$ be points on sides $AB$, $BC$, $CD$, and $DA$, respectively, such that the incenters of triangles $AWZ$, $BXW$, $CYX$, and $DZY$ form a parallelogram. Prove that $WXYZ$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the four incenters be $I_1, I_2, I_3, I_4$ with inradii $r_1, r_2, r_3, r_4$ respectively (in the order given in the problem). Without loss of generality, let $I_1$ be closer to $AB$ than $I_2$. Let the acute angle between $I_1I_2$ and $AB$ (and hence also the angle between $I_3I_4$ and $CD$) be $\\theta$. Then\n\n$$\nr_2 - r_1 = I_1I_2 \\sin \\theta = I_3I_4 \\sin \\theta = r_4 - r_3,\n$$\n\nwhich implies $r_1 + r_4 = r_2 + r_3$. Similar arguments show that $r_1 + r_2 = r_3 + r_4$. Thus we obtain $r_1 = r_3$ and $r_2 = r_4$.\n\n![](images/2023_Australian_Scene_p89_data_68bd682ed3.png)\n\nNow let's consider the possible positions of $W, X, Y, Z$. Suppose $AZ \\neq CX$. Without loss of generality, assume $AZ > CX$. Since the incircles of $AWZ$ and $CYX$ are symmetric about the centre of the parallelogram $ABCD$, this implies $CY > AW$. Using similar arguments, we have\n\n$$\nCY > AW \\implies BW > DY \\implies DZ > BX \\implies CX > AZ,\n$$\n\nwhich is a contradiction. Therefore $AZ = CX \\implies AW = CY$ and $WXYZ$ is a parallelogram.\n\n**Comment:** There are several ways to prove that $r_1 = r_3$ and $r_2 = r_4$. The proposer shows the following three alternative approaches:\n\n*Using parallel lines:* Let $O$ be the centre of parallelogram $ABCD$ and $P$ be the centre of parallelogram $I_1I_2I_3I_4$. Since $AI_1$ and $CI_3$ are angle bisectors, we must have $AI_1 \\parallel CI_3$. Let $\\ell_1$ be the line through $O$ parallel to $AI_1$. Since $AO = OC$, $\\ell_1$ is halfway between $AI_1$ and $CI_3$. Hence $P$ must lie on $\\ell_1$.\n\nSimilarly, $P$ must also lie on $\\ell_2$, the line through $O$ parallel to $BI_2$. Thus $P$ is the intersection of $\\ell_1$ and $\\ell_2$, which must be $O$. So the four incenters and hence the four incircles must be symmetric about $O$, which implies $r_1 = r_3$ and $r_2 = r_4$.\n\n*Using a rotation:* Let the bisectors of $\\angle DAB$ and $\\angle ABC$ meet at $X$ and the bisectors of $\\angle BCD$ and $\\angle CDA$ meet at $Y$. Then $I_1$ is on $AX$, $I_2$ is on $BX$, $I_3$ is on $CY$, and $I_4$ is on $DY$. Let $O$ be the centre of $ABCD$. Then a $180^\\circ$ rotation about $O$ takes $\\triangle AXB$ to $\\triangle CYD$. Under the same transformation $I_1I_2$ is mapped to a parallel segment $I'_1I'_2$ with $I'_1$ on $CY$ and $I'_2$ on $DY$. Since $I_1I_2I_3I_4$ is a parallelogram, $I_3I_4 = I_1I_2$ and $I_3I_4 \\parallel I_1I_2$. Hence $I'_1I'_2$ and $I_3I_4$ are parallel, equal length segments on sides $CY, DY$ and we conclude that $I'_1 = I_3, I'_2 = I_4$. Hence the centre of $I_1I_2I_3I_4$ is also $O$ and we establish by rotational symmetry that $r_1 = r_3$ and $r_2 = r_4$.\n\n*Using congruent triangles:* Let $AI_1$ and $BI_2$ intersect at $E$ and let $CI_3$ and $DI_4$ intersect at $F$. Note that $\\triangle ABE$ and $\\triangle CDF$ are congruent, since $AB = CD$ and corresponding pairs of angles are equal (equal opposite angles of parallelogram $ABCD$ are each bisected).\n\nSince $AI_1 \\parallel CI_3$ and $I_1I_2 \\parallel I_4I_3$, $\\angle I_2I_1E = \\angle I_4I_3F$. Similarly $\\angle I_1I_2E = \\angle I_3I_4F$. Furthermore $I_1I_2 = I_3I_4$. Hence triangles $I_2I_1E$ and $I_4I_3F$ are also congruent.\n\nHence $ABEI_1I_2$ and $DCFI_3I_4$ are congruent. Therefore, the perpendicular distance from $I_1$ to $AB$ equals the perpendicular distance from $I_3$ to $CD$, that is, $r_1 = r_3$. Similarly $r_2 = r_4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17533,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n\n$$\n\\frac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \\frac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \\frac{(2c + a + b)^2}{2c^2 + (a + b)^2} \\le 8.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution.** (Based on work by Matthew Tang and Anders Kaseorg)\n\nBy multiplying $a$, $b$, and $c$ by a suitable factor, we reduce the problem to the case when $a + b + c = 3$. The desired inequality reads\n$$\n\\frac{(a+3)^2}{2a^2+(3-a)^2} + \\frac{(b+3)^2}{2b^2+(3-b)^2} + \\frac{(c+3)^2}{2c^2+(3-c)^2} \\le 8.\n$$\nSet\n$$\nf(x) = \\frac{(x+3)^2}{2x^2 + (3-x)^2}\n$$\nIt suffices to prove that $f(a) + f(b) + f(c) \\le 8$. Note that\n$$\n\\begin{aligned}\nf(x) &= \\frac{x^2 + 6x + 9}{3(x^2 - 2x + 3)} = \\frac{1}{3} \\cdot \\frac{x^2 + 6x + 9}{x^2 - 2x + 3} \\\\\n&= \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{x^2 - 2x + 3} \\right) \\\\\n&= \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{(x-1)^2 + 2} \\right) \\le \\frac{1}{3} \\left( 1 + \\frac{8x + 6}{2} \\right) \\\\\n&= \\frac{1}{3}(4x + 4).\n\\end{aligned}\n$$\nHence,\n$$\nf(a) + f(b) + f(c) \\le \\frac{1}{3}(4a + 4 + 4b + 4 + 4c + 4) = 8,\n$$\nas desired, with equality if and only if $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17534,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length $1$ by lines parallel to its sides.\n\nFind the number of regular hexagons all of whose vertices are along the vertices of the equilateral triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.\n\nAlso, there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centers lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons is\n\n$$\nN = \\sum_{m=1}^{n} \\left[3(n-m)(n-m+1)+1\\right] m = (3n^2+3n) \\sum_{m=1}^{n} m - 3(2n+1) \\sum_{m=1}^{n} m^2 + 3 \\sum_{m=1}^{n} m^3.\n$$\n\nSince $\\sum_{m=1}^{n} m = \\frac{n(n+1)}{2}$, $\\sum_{m=1}^{n} m^2 = \\frac{n(n+1)(2n+1)}{6}$, and $\\sum_{m=1}^{n} m^3 = \\left(\\frac{n(n+1)}{2}\\right)^2$, it is easily checked that\n\n$$\nN = \\left(\\frac{n(n+1)}{2}\\right)^2.\n$$\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p13_data_266d9fa636.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17535,
    "subject": "Mathematics (Olympiad)",
    "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\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": "See solution",
    "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 others 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 $2muv = s$.\n\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, $k \\le n-1 = m^2$.\n\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.\n\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": 17536,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram and $AC$ intersects $BD$ at $I$. Let $G$ be the point inside triangle $IAB$ that satisfies\n$$\n\\angle IAG = \\angle IBG \\neq 45^{\\circ} - \\frac{\\angle AIB}{4}.\n$$\nLet $E, F$ be the projections of $C$ on $AG$ and $D$ on $BG$. The median with respect to vertex $E$ of triangle $BEF$ and the median with respect to vertex $F$ of triangle $AEF$ intersect at a point $H$.\n\na) Prove that $AF$, $BE$ and $IH$ are concurrent, denote the concurrent point by $L$.\n\nb) Let $K$ be the intersection of $CE$ and $DF$. Let $J$ be the circumcenter of triangle $LAB$ and $M$, $N$ be the circumcenters of $EIJ$, $FIJ$, respectively. Prove that $EM$, $FN$ and the line joining the circumcenters of $GAB$, $KCD$ are concurrent.\n\n![](images/Vietnam_2022_p33_data_dc7509348b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $E'$, $F'$ be the midpoints of $AE$, $BF$. Note that triangles $DFB$ and $CEA$ are right at $F$, $E$ and $I$ is the midpoint of $BD$ and $AC$, so triangles $IBF$, $ICE$ are isosceles at $I$. On the other hand, because $\\angle GAI = \\angle GBI$, we have $\\triangle IFB \\sim \\triangle IAE$, which implies\n$$\n\\angle IE'E = \\frac{\\angle AIE}{2} = \\frac{\\angle BIF}{2} = \\angle F'IF\n$$\nor $IE'$, $IF'$ are isogonal with respect to $\\angle EIF$.\n\nTherefore, we get\n$$\n\\begin{aligned}\nI(HE', FE) &= E(HE', FI) = E(F'G, FI) \\\\\n&= I(F'G, FE) = I(GF', EF).\n\\end{aligned}\n$$\nCombining with $IE'$, $IF'$ are isogonal with respect to $\\angle EIF$, we obtain that $IG$, $IH$ are isogonal with respect to $\\angle EIF$, or $\\angle AIB$. Hence, it suffices to show that if $AF$ meets $BE$ at $L$ then $IL$, $IG$ are isogonal with respect to $\\angle AIB$. It is clear that\n$$\n\\triangle IAF \\stackrel{\\perp}{\\sim} \\triangle IEB,\n$$\nthen\n$$\n(LA, LB) \\equiv (IF, IB) \\equiv (IA, IE) \\pmod{\\pi},\n$$\nwhich means $L$ lies on $(IAE)$ and $(IBF)$. Let $G'$ be the intersection of $IL$ and the circumcircle of triangle $LAB$, we obtain that\n$$\n\\angle G'AB = \\angle G'LB = \\angle ILB = \\angle IAE,\n$$\nwhich implies that $AG$, $AG'$ are isogonal with respect to $\\angle IAB$.\n\nSimilarly, we can point out that $BG'$, $BG$ are isogonal with respect to $\\angle IBA$, so $G'$ is the isogonal conjugate of $G$ in triangle $IAB$. Hence, $IG$, $IL$ are isogonal with respect to $\\angle AIB$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17537,
    "subject": "Mathematics (Olympiad)",
    "question": "Gaston and Jordon are two budding chefs who unfortunately always misread the cooking time required for a recipe. For example, if the required cooking time is written as $1:32$, meaning 1 hour and 32 minutes, Jordon reads it as $132$ minutes while Gaston reads it as $1.32$ hours. For one particular recipe, the difference between Jordon's and Gaston's misread times is exactly $90$ minutes. What is the actual cooking time in minutes?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the cooking time be $h : m$, where $h$ is the number of hours and $m$ is the number of minutes.\n\nJordon's time is $100h + m$ minutes and Gaston's time is $h + \\frac{m}{100}$ hours, or $60h + \\frac{3m}{5}$ minutes. The difference of $90$ minutes leads to the equation:\n\n$$\n40h + \\frac{2m}{5} = 90\n$$\n\nwhich gives:\n\n$$\n100h + m = 225\n$$\n\nSince $0 \\leq 225 - 100h \\leq 60$, this implies $h = 2$.\n\nTherefore, the actual cooking time in minutes is:\n\n$$\n2 \\times 60 + 25 = \\mathbf{145}\n$$\n\n---\n\nAlternatively, let $m$ be the number of minutes as read by Jordon. Then Gaston reads this as $\\frac{m}{100}$ hours $= \\frac{60m}{100}$ minutes. The difference of $90$ minutes leads to:\n\n$$\n90 = m - \\frac{3m}{5} = \\frac{2m}{5}\n$$\n\nwhich gives $m = 225$.\n\nTherefore, the actual cooking time in minutes is:\n\n$$\n2 \\times 60 + 25 = \\mathbf{145}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17538,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions $(x, y)$ of the equation $y^2 = x^3 - p^2x$, where $p$ is a prime such that $p \\equiv 3 \\pmod{4}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the given equation is\n\n$$\ny^2 = (x-p)x(x+p).\n$$\n\nWe split the analysis into the following cases.\n\n**Case $p \\nmid y$:**\n\nIn this case, $(x-p, x) = (x, x+p) = 1$. If $x$ is even, then $(x-p, x+p) = 1$, so $x-p$, $x$, and $x+p$ are all squares. This is not possible since $x+p \\equiv 3 \\pmod{8}$. Thus, $x$ has to be odd. In this case, $(x-p, x+p) = 2$. So we get\n\n$$\n\\begin{aligned}\nx &= r^2, \\\\\nx-p &= 2s^2, \\\\\nx+p &= 2t^2.\n\\end{aligned}\n$$\n\nThe last two equations imply $t^2 - s^2 = p$, solving which we get $t = \\frac{p+1}{2}$ and $s = \\frac{p-1}{2}$. Now,\n\n$$\n\\begin{aligned}\nr^2 &= x = (x-p) + p \\\\\n&= \\frac{(p-1)^2}{2} + p \\\\\n&= \\frac{p^2 + 1}{2} \\\\\n&= \\frac{(8k+3)^2 + 1}{2} \\quad (\\text{here } p = 8k+3) \\\\\n&= \\frac{64k^2 + 48k + 10}{2} \\equiv 5 \\pmod{8}\n\\end{aligned}\n$$\n\nwhich is not possible. Thus, there are no solutions in case $p \\nmid y$.\n\n**Case $p \\mid y$:**\n\nFirst, if $y=0$ then $x=0$, $p$, or $-p$, giving three solutions. Now, assume $y \\neq 0$. Since $p \\mid y$, it follows that $(x-p, x, x+p) = p$. Thus, $p^2 \\mid y$. Canceling $p^3$ from both sides, we have\n\n$$\npb^2 = (a-1)a(a+1),\n$$\n\nwhere $a = \\frac{x}{p}$ and $b = \\frac{y}{p^2}$. Note that $a-1$, $a$, and $a+1$ are all positive since $y \\neq 0$. Now, $p$ divides exactly one of $a-1$, $a$, or $a+1$. If $p$ divides $a-1$, then $a$ and $a+1$ have to be squares, which is not possible since $a \\neq 0$. So, $p \\nmid (a-1)$. Similarly, $p \\nmid (a+1)$. Thus, $p \\mid a$.\n\nIf $a \\neq 0$ is even, then $a-1$, $a/p$, and $a+1$ are relatively prime to each other, so they all have to be squares. But it is not possible that $a-1$ and $a+1$ are both squares. Thus, $a$ should be odd.\n\nNow, $(a-1, a+1) = 2$, so $a-1 = 2r^2$ and $a+1 = 2s^2$. This gives $s^2 - r^2 = 1$, which implies $s=1$ and $r=0$. This gives $a=1$, which is not divisible by $p$. Thus, there are no solutions in this case.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17539,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(x(1+y)) = f(x)(1 + f(y))\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $f$ is not identically zero, then by standard substitutions we get that $f(x) = x$ for $x = 0, \\pm 1$. Using these, it is easy to see that $f$ is additive and multiplicative on $\\mathbb{R}$. It then follows by induction and continuity that $f(x) = x$ for all real $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17540,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x \\geq 5$, $y \\geq 6$, $z \\geq 7$ and $x^2 + y^2 + z^2 \\geq 125$. Find the minimum of $x + y + z$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum of $x + y + z$ is $19$. This value is attained for $x = 5$, $y = 6$, $z = 8$.\n\nConversely, we prove $x + y + z \\geq 19$ for all admissible $x, y, z$. One may assume $x < 6$, $y < 7$, $z < 8$. Indeed, if one of the inequalities $x \\geq 6$, $y \\geq 7$, $z \\geq 8$ holds, then $x + y + z \\geq (5 + 6 + 7) + 1 = 19$.\n\nSet $u = x - 5$, $v = y - 6$, $w = z - 7$. Then $0 \\leq u, v, w < 1$, and $x^2 + y^2 + z^2 \\geq 125$ gives\n\n$$\n125 \\leq (u + 5)^2 + (v + 6)^2 + (w + 7)^2 = u^2 + v^2 + w^2 + 10u + 12v + 14w + 5^2 + 6^2 + 7^2.\n$$\n\nNow $u^2 \\leq u$, $v^2 \\leq v$, $w^2 \\leq w$ by $0 \\leq u, v, w < 1$, so the above inequality yields $11u + 13v + 15w > 15$.\n\nSince $u, v, w \\geq 0$, it follows that $15(u + v + w) \\geq 11u + 13v + 15w > 15$. Hence $u + v + w > 1$ and\n\n$$\nx + y + z = (u + v + w) + (5 + 6 + 7) > 1 + (5 + 6 + 7) = 19.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17541,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume that $x_1, x_2, \\dots, x_{n+1}$ are integers. Define the integers\n$$\na_k = x_k - 1 = \\frac{m+k}{n+k} - 1 = \\frac{m-n}{n+k} > 0\n$$\nfor $k = 1, 2, \\dots, n+1$.\n\nLet $P = x_1 x_2 \\cdots x_{n+1} - 1$. Prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $2^d$ be the largest power of 2 dividing $m-n$, and let $2^c$ be the largest power of 2 not exceeding $2n+1$. Then $2n+1 \\le 2^{c+1} - 1$, so $n+1 \\le 2^c$. Thus, $2^c$ is one of the numbers $n+1, n+2, \\dots, 2n+1$, and it is the only multiple of $2^c$ among these numbers. Let $l$ be such that $n+l = 2^c$. Since $\\frac{m-n}{n+l}$ is an integer, $d \\ge c$. Therefore,\n$$\n2^{d-c+1} \\nmid a_l = \\frac{m-n}{n+l},\n$$\nwhile\n$$\n2^{d-c+1} \\mid a_k \\quad \\text{for all} \\quad k \\in \\{1, \\dots, n+1\\} \\setminus \\{l\\}.\n$$\nComputing modulo $2^{d-c+1}$, we get\n$$\nP = (a_1 + 1)(a_2 + 1) \\cdots (a_{n+1} + 1) - 1 \\equiv (a_l + 1) \\cdot 1^n - 1 \\equiv a_l \\not\\equiv 0 \\pmod{2^{d-c+1}}.\n$$\nTherefore, $2^{d-c+1} \\nmid P$.\n\nOn the other hand, for any $k \\in \\{1, \\dots, n+1\\} \\setminus \\{l\\}$, we have $2^{d-c+1} \\mid a_k$. So\n$$\nP \\ge a_k \\ge 2^{d-c+1},\n$$\nand it follows that $P$ is not a power of 2.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17542,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$ and real numbers $x_1, x_2, \\dots, x_n$ in the interval $[0, 1]$, prove that there exist real numbers $a_0, a_1, \\dots, a_n$ satisfying simultaneously the following conditions:\n\n$$\n\\begin{aligned}\na_0 + a_n &= 0; \\\\\na_i &\\le 1, \\text{ for every } i = 0, 1, \\dots, n;\n\\end{aligned}\n$$\n\n$$\n|a_i| \\le 1, \\text{ for every } i = 0, 1, \\dots, n;\n$$\n\n$$\n|a_i - a_{i-1}| = x_i, \\text{ for every } i = 1, 2, \\dots, n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any $a \\in [0, 1)$, define a sequence $\\{a_i\\}_{i=0}^n$ generated by $a$ as follows: $a_0 = a$; for $1 \\le i \\le n$, set $a_i = a_{i-1} - x_i$ if $a_{i-1} \\ge 0$, and $a_i = a_{i-1} + x_i$ if $a_{i-1} < 0$.\n\nSet $f(a) = a_n$. It is easy to show by induction that $|a_i| \\le 1$ for every $0 \\le i \\le n$.\n\nIf there exists $a \\in [0, 1)$ such that $f(a) = -a$, consider the sequence generated by $a$: $a_0 = a, a_1, \\dots, a_n = f(a) = -a$. Clearly, this sequence satisfies conditions (1) and (3). By the recursive relation, it also satisfies condition (2). Thus, it suffices to show that there exists $a \\in [0, 1)$ such that $f(a) = -a$.\n\nFor any $a \\in [0, 1)$, we say that $a$ is a breaking point if at least one term in its generating sequence is $0$. Since every breaking point is of the form $\\sum_{i=1}^{n} t_i x_i$, where $t_i = -1, 0, 1$, there are finitely many breaking points.\n\nClearly, $0$ is a breaking point; label all breaking points in increasing order by $0 = b_1 < b_2 < \\cdots < b_m < 1$.\n\nWe first prove that for $1 \\le k \\le m-1$, $f(a) = f(b_k) + (a-b_k)$ for every $a \\in [b_k, b_{k+1})$.\n\nConsider $b_k, b_{k+1}$ and their generating sequences. Assume that $q_0 = b_k, q_1, q_2, \\dots, q_n$ is the generating sequence of $b_k$, and $r_0 = b_{k+1}, r_1, r_2, \\dots, r_n$ is the generating sequence of $b_{k+1}$. Suppose that $r_l$ is the first term of $\\{r_i\\}_{i=0}^n$ equal to $0$. Construct a sequence $\\{s_i\\}_{i=0}^n$ as follows:\n\n$$\n\\begin{aligned}\ns_0 &= r_0, \\quad s_1 = r_1, \\quad \\dots, \\quad s_l = r_l = 0, \\\\\ns_{l+1} &= -r_{l+1}, \\quad \\dots, \\quad s_n = -r_n.\n\\end{aligned}\n$$\n\nIt is clear that the sequence $\\{s_i\\}_{i=0}^n$ satisfies $s_0 = b_{k+1}$; and for $1 \\le i \\le n$, $s_i = s_{i-1} - x_i$ if $s_{i-1} > 0$, and $s_i = s_{i-1} + x_i$ if $s_{i-1} \\le 0$.\n\nWe prove by induction that $q_i s_i \\ge 0$ and $s_i - q_i = b_{k+1} - b_k$. The conclusion is obviously true for $i=0$. Assume that it holds for $i-1$. Then $q_{i-1} s_{i-1} \\ge 0$ and $s_{i-1} - q_{i-1} = b_{k+1} - b_k > 0$, which implies that $q_{i-1} \\ge 0$, $s_{i-1} > 0$, or $q_{i-1} < 0$, $s_{i-1} \\le 0$. In the former case, $q_i = q_{i-1} - x_i$, $s_i = s_{i-1} - x_i$, and thus $s_i - q_i = b_{k+1} - b_k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17543,
    "subject": "Mathematics (Olympiad)",
    "question": "In a certain club, some pairs of members are friends. Given $k \\geq 3$, we say that a club is *k*-good if every group of $k$ members can be seated around a round table such that every two neighbors are friends. Prove that if a club is 6-*good* then it is 7-*good*.\n\n![](images/brozura_a67angl_new_p12_data_ffde345054.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a 6-good club and denote some seven of its members by $A, \\dots, G$. It suffices to show that $A, \\dots, G$ can be seated around a table as required. Consider only friendships among $A, \\dots, G$.\n\nFirst, we show that every member has at least three friends.\n\nWithout loss of generality, consider $G$. By assumption, $B, \\dots, G$ can be seated as required, hence $G$ has at least two friends. Without loss of generality, $F$ is one of them. By assumption, $A, \\dots, E, G$ (omitting $F$) can be seated as required, hence $G$ has at least two more friends apart from $F$ for a total of at least three friends.\n\nSince every member has at least three friends, there exists a member with at least four friends (otherwise the number of friendly pairs equals $\\frac{1}{2} \\cdot 7 \\cdot 3$, which is clearly impossible). Without loss of generality, assume $G$ has at least four friends.\n\nBy assumption, $A, \\dots, F$ can be seated as required. In such a seating, some two of the four friends of $G$ are neighbors and we can seat $G$ in between them.\n\n*Remark.* The statement “If a club is *k*-good then it is (*k*+1)-good” holds precisely for $k \\in \\{3, 4, 5, 6, 7, 8, 10, 11, 13, 16\\}$. The counterexamples are called *hypohamiltonian graphs*. For $k = 9$, one such example is the *Petersen graph* (Fig. 1).\n\n![](images/brozura_a67angl_new_p12_data_d370bc6edc.png)\n\nFig. 1",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17544,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, determine all non-constant polynomials $f$ with complex coefficients satisfying the condition\n\n$$\n1 + f(X^n + 1) = (f(X))^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is even, there are no such polynomials. If $n$ is odd, the required polynomials are precisely those recursively defined by $f_0(X) = -X$, and $f_{k+1}(X) = f_k(X^n + 1)$ for $k \\ge 0$.\n\nIt 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\n$$\n1 + f(X^n + 1) = (f(X))^n. \\qquad (1)\n$$\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 $x_{k+1} = x_k^n + 1$ for $k \\ge 0$. Clearly, $f(x_{k+1}) = (f(x_k))^n - 1$ for $k \\ge 0$, and $f(x_1) = -1$.\n\nIf $n$ is even, then $f(x_2) = 0$, so $f(x_{2k}) = 0$ (and $f(x_{2k+1}) = -1$) for $k \\ge 0$. Since the $x_k$ form a strictly increasing sequence, we reach a contradiction.\n\nIf $n$ is odd, induct on $k$ to prove that $f(x_k) = -x_k$ for $k \\ge 0$. 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\n**Alternative solution -- case** $f(0) = 0$.\n\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$$\n|f(2^n + 1)| \\le 2^n + 1. \\quad (2)\n$$\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$$\n\\prod_{\\alpha \\text{ is a root of } f} |\\alpha_0^n + 1 - \\alpha| = 1. \\quad (3)\n$$\n\nSuppose, if possible, that $|\\alpha_0| \\ge 2$. If $\\alpha$ is a root of $f$, then\n\n$$\n|\\alpha_0^n + 1 - \\alpha| \\ge |\\alpha_0|^n - 1 - |\\alpha| \\ge 2|\\alpha_0| - 1 - |\\alpha| = (|\\alpha_0| - 1) + (|\\alpha_0| - |\\alpha|) \\ge |\\alpha_0| - 1 \\ge 1.\n$$\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$$\n\\prod_{\\alpha \\text{ is a root of } f} |2^n + 1 - \\alpha| \\le 2^n + 1. \\quad (2')\n$$\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": 17545,
    "subject": "Mathematics (Olympiad)",
    "question": "決定所有的整數 $m \\ge 2$ 使得對所有 $n$，$\\frac{m}{3} \\le n \\le \\frac{m}{2}$，$n$ 整除二項係數 $\\binom{n}{m-2n}$。",
    "options": [],
    "answer": "See solution",
    "solution": "滿足條件的整數 $m$ 是所有質數。\n\n首先，檢查質數 $p$ 是否滿足條件。對於 $p$ 為質數，對所有 $n$，$1 \\le n \\le \\frac{p}{2}$，有 $n \\mid \\binom{n}{p-2n}$。當 $p=2$ 時成立。對於奇質數 $p$，取 $n \\in [1, \\frac{p}{2}]$，考慮：\n\n$$\n(p-2n) \\cdot \\binom{n}{p-2n} = n \\cdot \\binom{n-1}{p-2n-1}.\n$$\n\n因為 $p \\ge 2n$ 且 $p$ 為奇數，上式每一項因數皆非零。設 $d = \\gcd(p-2n, n)$，則 $d \\mid p$，但 $d \\le n < p$，所以 $d=1$，即 $p-2n$ 和 $n$ 互質。因此 $n \\mid \\binom{n}{p-2n}$。\n\n接著證明沒有合成數 $m$ 滿足條件：\n\n1. 若 $m = 2k,\\ k > 1$，取 $n = k$，則 $\\frac{m}{3} \\le n \\le \\frac{m}{2}$，但 $\\binom{n}{m-2n} = \\binom{k}{0} = 1$，無法被 $k$ 整除。\n2. 若 $m$ 為奇合成數，存在奇質數 $p$ 及整數 $k \\ge 1$ 使 $m = p(2k+1)$。取 $n = pk$，則 $\\frac{m}{3} \\le n \\le \\frac{m}{2}$。然而：\n\n$$\n\\frac{1}{n} \\binom{n}{m-2n} = \\frac{1}{pk} \\binom{pk}{p} = \\frac{(pk-1)(pk-2)\\cdots(pk-(p-1))}{p!}\n$$\n\n這不是整數，因為 $p$ 整除分母但不整除分子。\n\n因此，只有質數 $m$ 滿足條件。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17546,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle of length $999$ is divided into unit arcs by $999$ black points. Then $d$ arcs with lengths $1, 2, \\ldots, d$ are placed on the circle, with their endpoints at black points, so that none of these arcs contains another (the arcs may overlap otherwise). Find all $d$ for which such a configuration exists.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the problem for a circle of length $n$ divided into unit arcs by $n$ black points, and $d$ arcs of lengths $1, 2, \\ldots, d$ with black endpoints such that none of them contains another. We show that the maximal admissible $d$ equals $\\lfloor \\frac{n+1}{2} \\rfloor$, where $\\lfloor \\cdot \\rfloor$ denotes the integer part.\n\nAll arcs considered below have black endpoints. We say that an arc $\\overarc{PQ}$ has start $P$ (or starts at $P$) if $Q$ can be reached from $P$ by going along the arc in counterclockwise direction. The meaning of \"$\\overarc{PQ}$ has end $Q$ (or ends at $Q$)\" is analogous.\n\nLet $\\gamma_1, \\dots, \\gamma_d$ be arcs on the circle with lengths $1, 2, \\ldots, d$ such that none of them contains another.\n\nConsider the shortest arc $\\gamma_1 = \\overarc{AB}$, starting at $A$, and the longest one $\\gamma_d = \\overarc{CD}$, starting at $C$. Let $X$ be the set of black points on arc $\\overarc{BC}$ with start $B$, excluding its end $C$. Let $Y$ be the set of black points on arc $\\overarc{DA}$ with end $A$, excluding its start $D$. Finally, let $Z$ be the set of black points on the closed arc $\\gamma_{d} = \\overarc{CD}$. Then each black point belongs to exactly one of $X$, $Y$, and $Z$.\n\nNow observe that every arc $\\gamma_m$ with $1 < m < d$ satisfies exactly one of the following conditions:\n\n1. $\\gamma_m$ starts in $X$;\n2. $\\gamma_m$ ends in $Y$.\n\nClearly, (1) and (2) do not hold simultaneously, otherwise $\\gamma_m$ contains arc $\\gamma_d = \\overarc{CD}$. Suppose that (1) does not hold. Then $\\gamma_m$ starts in $Y \\cup Z$, that is, in the arc $\\overarc{CA}$ with start $C$. Hence $\\gamma_m$ also ends in the same arc $\\overarc{CA}$, or else $\\gamma_m$ contains $\\gamma_1 = \\overarc{AB}$. In addition, $\\gamma_m$ does not end in $\\gamma_d = \\overarc{CD}$; otherwise $\\gamma_m$ also starts in $\\gamma_d$, hence $\\gamma_d$ contains a $\\gamma_m$. In conclusion, $\\gamma_m$ ends in $Y$, i.e., condition (2) holds. This completes the justification.\n\nLet there be $x$ arcs $\\gamma_m$ satisfying (1), $1 < m < d$. They start at different points of the set $X$, or else some of them contains another. Hence $x \\leq |X|$. Analogously, if $y$ is the number of arcs $\\gamma_m$ satisfying (2), $1 < m < d$, then $y \\leq |Y|$. By the reasoning above $x + y = d - 2$, therefore $d - 2 = x + y \\leq |X| + |Y| = n - |Z| = n - d - 1$. This gives the upper bound $d \\leq \\lfloor \\frac{n+1}{2} \\rfloor$.\n\nFor odd $n = 2k + 1$ we have $\\lfloor \\frac{n+1}{2} \\rfloor = k + 1$, and $d = k + 1$ is admissible. Label the black points $1, \\ldots, 2k + 1$ in counterclockwise direction and consider $k + 1$ arcs $\\gamma_1, \\ldots, \\gamma_{k+1}$ with lengths $1, \\ldots, k + 1$. For each $m = 1, \\ldots, k + 1$ place arc $\\gamma_m$ so that it starts at point $m$. It is immediate that this configuration satisfies the requirements. Note that arc $\\gamma_{k+1}$ ends at point $1$, hence it does not contain arc $\\gamma_1$. Thus $d = k + 1$ is admissible, implying that so are all smaller natural numbers. In conclusion, the solution to the problem for $n = 2k + 1$ are the numbers $1, 2, \\ldots, k + 1$, yielding $1, 2, \\ldots, 500$ as the answer to the original question.\n\nSimilarly, for even $n = 2k$ we have $\\lfloor \\frac{n+1}{2} \\rfloor = k$, and $d = k$ is admissible. The example for $d = k$ is analogous. Label the black points $1, 2, \\ldots, 2k$ in counterclockwise direction and consider $k$ arcs $\\gamma_1, \\ldots, \\gamma_k$ with lengths $1, 2, \\ldots, k$. For $m = 1, \\ldots, k$ place arc $\\gamma_m$ so that it starts at point $m$. This configuration satisfies the requirements. So the solution for $n = 2k$ are the numbers $1, 2, \\ldots, k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17547,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a complex number $z$, $z \\neq 0$, and the real sequence\n\n$$\na_n = \\left|z^n + \\frac{1}{z^n}\\right|, \\quad n \\geq 1.\n$$\n\na) Show that if $a_1 > 2$, then\n\n$$\na_{n+1} < \\frac{a_n + a_{n+2}}{2}, \\text{ for all } n \\in \\mathbb{N}^*.\n$$\n\nb) Prove that if there exists $k \\in \\mathbb{N}^*$ such that $a_k \\leq 2$, then $a_1 \\leq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We easily notice that\n\n$$\n\\begin{aligned}\n2|z^{n+1} + \\frac{1}{z^{n+1}}| &< |z + \\frac{1}{z}| \\cdot |z^{n+1} + \\frac{1}{z^{n+1}}| \\\\\n&= |z^n + \\frac{1}{z^n} + z^{n+2} + \\frac{1}{z^{n+2}}| \\\\\n&\\leq |z^n + \\frac{1}{z^n}| + |z^{n+2} + \\frac{1}{z^{n+2}}|.\n\\end{aligned}\n$$\n\nb) Suppose *ad absurdum* that $a_1 > 2$. Then a) implies that the sequence $a_{n+1} - a_n$ is strictly increasing, so $a_{n+1} - a_n > a_2 - a_1$.\n\nBut\n\n$$\na_2 = |z^2 + \\frac{1}{z^2}| = \\left|\\left(z + \\frac{1}{z}\\right)^2 - 2\\right| \\geq \\left(z + \\frac{1}{z}\\right)^2 - 2 = a_1^2 - 2 > a_1,\n$$\n\ntherefore the sequence $(a_n)_n$ is strictly increasing, hence $a_k \\geq a_1 > 2$ for all $k$, a contradiction.\n\n*Alternative Solution.* Consider the sequence $(\\alpha_n)_{n \\geq 1}$ 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$$\n\\alpha \\alpha_n = \\left(z + \\frac{1}{z}\\right) \\left(z^n + \\frac{1}{z^n}\\right) = \\left(z^{n+1} + \\frac{1}{z^{n+1}}\\right) + \\left(z^{n-1} + \\frac{1}{z^{n-1}}\\right) = \\alpha_{n+1} + \\alpha_{n-1}\n$$\n\nfor all $n \\geq 1$, so the sequence $(\\alpha_n)_{n \\geq 0}$ satisfies the linear recurrence relation $\\alpha_{n+1} = \\alpha \\alpha_n - \\alpha_{n-1}$. Then for $|\\alpha| > 2$ we have\n\n$$\na_n = |\\alpha_n| = \\left| \\frac{\\alpha_{n+1} + \\alpha_{n-1}}{\\alpha} \\right| \\leq \\frac{|\\alpha_{n+1}| + |\\alpha_{n-1}|}{|\\alpha|} < \\frac{a_{n+1} + a_{n-1}}{2},\n$$\n\ni.e., the sequence $(a_n)_{n \\geq 0}$ is convex.\n\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 $a_k \\leq 2$, then $a_1 \\leq 2$. Therefore, the proof for b) comes directly from a), and the nature of the sequence is not anymore relevant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17548,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$, $n$, and $p$ be positive integers. Space is divided into unit cubes by infinite parallel planes. An assignment of all unit cubes of space with a natural number from $1$ to $60$ is called *Dien Bien* if, for all rectangular boxes whose faces lie on these planes and whose side lengths are in $\\{2m+1, 2n+1, 2p+1\\}$, the unit cube at the center of the box is assigned the average of the $8$ numbers assigned to the $8$ vertices of that box.\n\nHow many Dien Bien assignments are there?\n\nNote: Two assignments are considered the same if there exists a translation such that, for all unit cubes, this translation maps it to another unit cube with the same assigned number.",
    "options": [],
    "answer": "See solution",
    "solution": "42",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17549,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a polynomial with real coefficients and odd degree. Suppose that the number of real solutions to\n\n$$\nP(P(x)) = P(x), \\quad P(x) \\neq x\n$$\n\nis finite and odd. Show that there exists a real $c$ such that $P(c) = c$ and the polynomial $P(x) - c$ has a real root of multiplicity at least two. (That is, it is divisible by $(x - r)^2$ for some real $r$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the set of all $a$ with $P(a) = a$. For each $a \\in S$, let $T_a$ be the set of solutions to $P(P(x)) = P(x) = a$ with $x \\neq a$. Since $P(P(x)) = P(x)$ and $P(x) \\neq x$ has an odd number of solutions, the union of all $T_a$ has an odd number of elements. Therefore, one of the $T_a$ has an odd number of elements; choose this $a$ to be our $c$ and let $T_a = \\{b_1, b_2, \\dots, b_k\\}$ where $k$ is odd.\n\nWe know that $P(x) - a$ has an even number of distinct real roots: $a, b_1, b_2, \\dots, b_k$. Because $P(x) - a$ has odd degree and its nonreal roots come in conjugate pairs, the combined multiplicity of its real roots must be odd. Then if neither $(x - a)^2$ nor $(x - b_i)^2$ for any $i$ divide $P(x) - a$, the combined multiplicity of its real roots would be even, a contradiction. So either $(x - a)^2$ or $(x - b_i)^2$ for some $i$ divide $P(x) - a$. This shows that our choice of $c$ works, and the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a natural number. Find all integers $x, y, z$ such that\n$$x^2 + y^2 + z^2 = 2^n (x + y + z).$$",
    "options": [],
    "answer": "See solution",
    "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 $x^2 + y^2 + z^2$, so either all three numbers are even, or one is even and the others are odd. In the case where one is even and two are odd, let $x = 2x_1 + 1$, $y = 2y_1 + 1$, $z = 2z_1$. Then\n$$4(x_1^2 + x_1 + y_1^2 + y_1 + z_1^2) + 2 = 4(x_1 + y_1 + z_1 + 1),$$\nwhich leads to a contradiction.\n\nNow, consider the case when $x, y, z$ are all even. Let $x = 2x_1$, $y = 2y_1$, $z = 2z_1$. Then\n$$x_1^2 + y_1^2 + z_1^2 = 2^{n-1}(x_1 + y_1 + z_1).$$\nThus, if $n = 1$, $x, y, z \\in \\{0, 2\\}$.\n\nFor $n > 1$, repeating the argument, 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\n$$x_n^2 + y_n^2 + z_n^2 = x_n + y_n + z_n,$$\nso $x_n, y_n, z_n \\in \\{0, 1\\}$, and thus $x, y, z \\in \\{0, 2^n\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17551,
    "subject": "Mathematics (Olympiad)",
    "question": "Anne, Ellie, and Milo play a game on a hexagonal board tiled with hexagons, with $n > 2$ hexagons on each side. The game begins with a token on a tile in one of the corners of the board. Ellie and Milo are on the same team, playing against Anne, and they win if the token lands on the center of the board. Anne, Ellie, and Milo take turns moving the token: Anne begins, then Ellie, then Milo.\n\nThe turns proceed as follows:\n\n- Anne must move the token to an adjacent hexagon, in any direction.\n- Ellie must move the token by two hexagons in any of the six possible directions.\n- Milo may either pass the turn or move the token by three hexagons in any of the six possible directions.\n\nFind all $n > 2$ for which Ellie and Milo have a winning strategy.\n\n![](images/BW2021_Shortlist_p18_data_028f8b7836.png)\n\nFigure 7: A board with 5 hexagons on each side.\n\n![](images/BW2021_Shortlist_p18_data_5694b112bd.png)\n\nFigure 8: Three colouring of the hexagonal tiling for 6 hexagons on each side.",
    "options": [],
    "answer": "See solution",
    "solution": "We colour the board with three colours so that no neighbouring tiles share a colour. Assign each hexagon coordinates using $\\vec{e}_1 = (1, 0)$ and $\\vec{e}_2 = (\\cos(120^\\circ), \\sin(120^\\circ)) = \\left(\\frac{-1}{2}, \\frac{\\sqrt{3}}{2}\\right)$. Let the center hexagon be the origin. Each hexagon has center at $a \\cdot \\vec{e}_1 + b \\cdot \\vec{e}_2$, with $(a, b) \\in \\mathbb{Z}^2$. The neighbours of $(a, b)$ are $(a + 1, b)$, $(a + 1, b + 1)$, $(a, b + 1)$, $(a - 1, b)$, $(a - 1, b - 1)$, and $(a, b - 1)$.\n\nColour the hexagon at $(a, b)$ with colour number $(a + b) \\pmod{3}$. Neighbouring hexagons do not share a colour. (This is the only three-colouring of a hexagonal tiling.) See Figure 8.\n\nIf $n \\equiv 1 \\pmod{3}$, the token starts on a space of the same colour as the center hexagon (say, grey). By considering cases, whatever Anne does, Ellie and Milo can end their turns by moving the token to a prescribed grey hexagon, eventually reaching the center.\n\nIf $n \\not\\equiv 1 \\pmod{3}$, the token does not start on the same colour as the center. Suppose the token starts on a white tile (with black as the third colour). Anne can always move the token to a grey hexagon not aligned with the center. Then Anne moves the token to a white or black hexagon. After Milo's move, the token is again on a white or black hexagon. Anne can continue this indefinitely, so the token never reaches the center.\n\n**Answer:** Ellie and Milo have a winning strategy if and only if $n \\equiv 1 \\pmod{3}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17552,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Points $D$, $E$, and $F$ lie on segments $BC$, $CA$, and $AB$ respectively, and each of the three segments $AD$, $BE$, and $CF$ contains the circumcenter of $ABC$. Prove that if any two of the ratios\n$$\n\\frac{BD}{DC}, \\quad \\frac{CE}{EA}, \\quad \\frac{AF}{FB}, \\quad \\frac{BF}{FA}, \\quad \\frac{AE}{EC}, \\quad \\frac{CD}{DB}\n$$\nare integers, then triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that there are $\\binom{6}{2} = 15$ possible pairs of ratios among the six given in the problem statement. These pairs are of two types:\n\n1. Three of these pairs are reciprocal pairs involving segments from just one side of triangle $ABC$.\n2. The other 12 pairs involve segments from two sides of the triangle.\n\nWe first consider the former case.\n\n(a) If $\\frac{CD}{DB}$ and $\\frac{BD}{DC}$ are both integers, then both of these ratios must be $1$ and $BD = DC$. Then in triangle $ABC$, $AD$ is the median from $A$ and, because $AD$ contains the circumcenter, it is also the perpendicular bisector of segment $BC$. It then follows that $AB = AC$ and the triangle is isosceles. Similarly, if $\\frac{CE}{EA}$ and $\\frac{AE}{EC}$ are both integers or $\\frac{AF}{FB}$ and $\\frac{BF}{FA}$ are both integers, then triangle $ABC$ is isosceles.\n\n(b) Let $O$ be the circumcenter of triangle $ABC$, and let $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. We show that any of the ratios can be written in the form $\\frac{\\sin 2x}{\\sin 2y}$ where $x$ and $y$ are two of $\\alpha, \\beta, \\gamma$. Since $ABC$ is acute, $0^\\circ < \\alpha, \\beta, \\gamma < 90^\\circ$ and $O$ lies in the interior. Hence $\\angle AOB = 2\\gamma$, $\\angle BOC = 2\\alpha$, and $\\angle COA = 2\\beta$. Applying the sine rule to triangles $BOD$ and $COD$ gives\n$$\n\\frac{BD}{\\sin \\angle BOD} = \\frac{BO}{\\sin \\angle BDO}\n$$\nand\n$$\n\\frac{CD}{\\sin \\angle COD} = \\frac{CO}{\\sin \\angle CDO}\n$$\nNext note that $BO = CO$ and that\n$$\n\\begin{align*}\n\\angle BDO + \\angle CDO &= 180^\\circ \\\\\n&= \\angle BOD + \\angle AOB \\\\\n&= \\angle COD + \\angle AOC.\n\\end{align*}\n$$\nIt follows that\n$$\n\\frac{BD}{\\sin 2\\gamma} = \\frac{BD}{\\sin \\angle BOD} = \\frac{CD}{\\sin \\angle COD} = \\frac{CD}{\\sin 2\\beta}\n$$\ngiving $\\frac{BD}{CD} = \\frac{\\sin 2\\gamma}{\\sin 2\\beta}$. Similarly, $\\frac{CE}{EA} = \\frac{\\sin 2\\alpha}{\\sin 2\\gamma}$ and $\\frac{AF}{FB} = \\frac{\\sin 2\\beta}{\\sin 2\\alpha}$.\n\nNow assume that one of the twelve type (ii) pairs of ratios consists of two integers. Then there are positive integers $m$ and $n$ (with $m \\le n$) such that\n$$\n\\sin 2x = m \\sin 2z \\quad \\text{and} \\quad \\sin 2y = n \\sin 2z\n$$\nor\n$$\n\\sin 2z = m \\sin 2x \\quad \\text{and} \\quad \\sin 2z = n \\sin 2y\n$$\nfor some choice of $x, y, z$ with $\\{x, y, z\\} = \\{\\alpha, \\beta, \\gamma\\}$. Without loss of generality we may assume that\n$$\n\\sin 2\\alpha = m \\sin 2\\gamma \\quad \\text{and} \\quad \\sin 2\\beta = n \\sin 2\\gamma\n$$\nor\n$$\n\\sin 2\\gamma = m \\sin 2\\alpha \\quad \\text{and} \\quad \\sin 2\\gamma = n \\sin 2\\beta \\qquad \\textcircled{1}\n$$\nfor some positive integers $m$ and $n$.\n\nNote that there is a triangle with angles $180^\\circ - 2\\alpha$, $180^\\circ - 2\\beta$, and $180^\\circ - 2\\gamma$. (It is easy to check that each of these angles is in the interval $(0^\\circ, 180^\\circ)$ and that they sum to $180^\\circ$.) Furthermore, a triangle\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p135_data_7809d7ed07.png)\nwith these angles can be constructed by drawing the tangents to the circumcircle of $ABC$ at each of $A$, $B$, and $C$. Denote this triangle by $A_1B_1C_1$ where $A_1$ is the intersection of the tangents at $B$ and $C$, $B_1$ is the intersection of the tangents at $C$ and $A$, and $C_1$ is the intersection of the tangents at $A$ and $B$. Applying the sine rule to triangle $A_1B_1C_1$ and by $\\textcircled{1}$ we find\n$$\n\\begin{aligned}\nA_1B_1 : B_1C_1 : C_1A_1 &= \\sin \\angle C_1 : \\sin \\angle A_1 : \\sin \\angle B_1 \\\\\n&= \\sin 2\\gamma : \\sin 2\\alpha : \\sin 2\\beta,\n\\end{aligned}\n$$\nthat is,\n$$\nA_1B_1 : B_1C_1 : C_1A_1 = 1 : m : n\n$$\nor\n$$\nA_1B_1 : B_1C_1 : C_1A_1 = mn : n : m. \\qquad \\textcircled{2}\n$$\nBy the triangle inequality, it follows that $1 + m < n$ (that is, ...)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17553,
    "subject": "Mathematics (Olympiad)",
    "question": "For each real-coefficient polynomial $f(x) = a_0 + a_1x + \\dots + a_nx^n$, let\n$$\n\\Gamma(f(x)) = a_0^2 + a_1^2 + \\dots + a_n^2.\n$$\n\nGiven a polynomial $P(x) = (x+1)(x+2)\\dots(x+2020)$, prove that there exist at least $2^{2019}$ pairwise distinct polynomials $Q_k(x)$ with $1 \\leq k \\leq 2^{2019}$, each satisfying the following two conditions:\n\n1. $\\deg Q_k(x) = 2020$,\n2. $\\Gamma(Q_k(x)^n) = \\Gamma(P(x)^n)$ for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the following lemma.\n\n**Lemma.** $\\Gamma(f(x))$ is equal to the constant term in the expansion of $f(x)f\\left(\\frac{1}{x}\\right)$.\n\n**Proof.** Indeed, if $f(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0$, then\n$$\n\\text{const}\\left(f(x)f\\left(\\frac{1}{x}\\right)\\right) = (a_nx^n + \\dots + a_0) \\left(\\frac{a_n}{x^n} + \\dots + a_0\\right)\n$$\nso the constant term is $a_n^2 + a_{n-1}^2 + \\cdots + a_1^2 + a_0^2$.\n\n![](images/Vietnamese_mathematical_competitions_p220_data_a991e3f4a5.png)\n\nFor every polynomial $f(x)$ and every positive integer $n$,\n$$\n\\Gamma((ax+b)^n f(x)) = \\text{const}\\left((ax+b)^n \\left(\\frac{a}{x}+b\\right)^n f(x) f\\left(\\frac{1}{x}\\right)\\right) = \\Gamma((bx+a)^n f(x)).\n$$\n\nThus, in each binomial of $P(x) = (x+1)(x+2)\\cdots(x+2020)$, the exchange $x+k \\rightarrow kx+1$ for $2 \\leq k \\leq 2020$ does not change the value of $\\Gamma(P(x)^n)$. Since each binomial can be either kept or changed, there are $2^{2019}$ ways to modify the given polynomial, and all of these $2^{2019}$ polynomials satisfy the problem.\n\n![](images/Vietnamese_mathematical_competitions_p220_data_dd28da4f67.png)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17554,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let points $M$ and $N$ be 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 $AC = AB\\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "\"$\\Leftrightarrow$\" Notice that triangles $AMN$ and $ACB$ are similar, so $\\frac{AM}{AC} = \\frac{AN}{AB}$ 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 let $P'$ be the midpoint of $MN'$, with $P' = MN' \\cap AQ$. Then $P'$ is the midpoint of the segment $MN'$, and $PP'$ is a midline in triangle $MNN'$. This implies $PP' \\parallel NN'$, which is 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 $\\triangle AMN \\sim \\triangle ACB$, so $\\frac{AM}{AC} = \\frac{AN}{AB} \\Rightarrow \\frac{AM}{AN} = \\frac{AC}{AB}$. It follows that $\\frac{2AB}{AC} = \\frac{AC}{AB}$, hence $AC = AB\\sqrt{2}$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17555,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $G$ is a simple planar graph with minimum degree at least $5$. Prove that $G$ contains an edge $\\{x, y\\}$ such that $\\deg(x) + \\deg(y) \\le 11$.",
    "options": [],
    "answer": "See solution",
    "solution": "By adding edges, we can assume that $G$ is a triangulation (adding edges only increases the minimum degree, and if we find an edge $\\{x, y\\}$ with $\\deg(x) + \\deg(y) \\le 11$ in the new graph, this edge will also be present in the original graph, since otherwise one of its endpoints would have degree smaller than $5$ in the original graph).\n\nLet $V$ be the set of vertices, $E$ the set of edges, and $F$ the set of faces of our triangulation. Let $d(v)$ be the degree of $v \\in V$ and let $d(f)$ be the number of edges on the boundary of $f$. Here, if $e \\in E$ is on the boundary of only one face, then we count it two times. Hence,\n\n$$\n\\sum_{f} d(f) = 2|E| = \\sum_{v \\in V} d(v)\n$$\n\nBy Euler's formula, we have\n\n$$\n\\sum_{v \\in V} (6 - d(v)) = \\sum_{f} (6 - 2d(f)) + \\sum_{v \\in V} (6 - d(v)) = 6|F| - 4|E| + 6|V| - 2|E| = 12\n$$\n\nsince for every face $d(f) = 3$. Let us give a charge $6 - d(v)$ to every vertex $v \\in V$. The total charge is $12$. The only vertices with positive initial charge are those with $d(v) = 5$. Now we discharge the system using a single rule: every vertex of degree $5$ gives charge $\\frac{1}{5}$ to each of its neighbors. The total final charge is still $12$. Thus, there are vertices with positive final charge. Suppose $v$ is such a vertex. Then its final charge $c(v)$ satisfies\n\n$$\n0 < c(v) \\le 6 - d(v) + \\frac{1}{5}d(v) = 6 - \\frac{4}{5}d(v).\n$$\n\nThus $d(v) \\le 7$. Now we consider three cases:\n\n1. If $d(v) = 6$, its initial charge was $0$, so it must have gained some charge from a neighbor of degree $5$. Thus, $\\{u, v\\}$ is the desired edge.\n2. If $d(v) = 5$, this vertex gave all its charge to its five neighbors, but since the final charge is positive, it must have gained some charge from a neighbor of degree $5$. Thus, $\\{v, u\\}$ is the desired edge.\n3. If $d(v) = 7$, its initial charge was $-1$, so it gained charge from at least $6$ neighbors, all of degree $5$. There is one more neighbor whose degree we do not control. Since $G$ is a triangulation, the neighbors of $v$ form a cycle, and on this cycle there are two adjacent vertices $u_1$ and $u_2$ of degree $5$. Thus, $\\{u_1, u_2\\}$ is the desired edge.\n\n$\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17556,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $a$ such that there exists a set $X$ of 6-tuples of integers satisfying the following condition: for each $k = 1, 2, \\dots, 36$, there exist $x, y \\in X$ such that $ax + y - k$ is divisible by $37$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ and $X = \\{x_1, x_2, \\dots, x_6\\}$ satisfy the conditions. Set $p = 37$, $\\overline{X} = \\{\\overline{x_1}, \\overline{x_2}, \\dots, \\overline{x_6}\\}$, and $Y = \\{\\overline{a x_i + x_j} \\mid i, j = 1, \\dots, 6\\}$, where $\\overline{x}$ denotes the congruence class of $x$ modulo $p$.\n\nSince $|Y| = 36$, $\\overline{0} \\notin \\overline{X}$. If $a \\equiv 0 \\pmod{p}$, then $|Y| \\leq 6$. If $a \\equiv 1 \\pmod{p}$, then $|Y| \\leq 21$. If $a \\equiv -1 \\pmod{p}$, then $\\overline{0} \\in Y$. Thus, $a \\not\\equiv 0, \\pm 1 \\pmod{p}$.\n\nNow, $Y = \\{\\overline{a(a x_i + x_j)} \\mid i, j = 1, \\dots, 6\\} = \\{\\overline{a x_j + a^2 x_i} \\mid i, j = 1, \\dots, 6\\}$. By a lemma (see below), $\\overline{X} = \\{\\overline{a^2 x_1}, \\dots, \\overline{a^2 x_6}\\}$. Therefore, $a^{12} x_1 x_2 \\cdots x_6 \\equiv x_1 x_2 \\cdots x_6 \\pmod{p}$, so $a^{12} \\equiv 1 \\pmod{p}$. Let $d$ be the order of $a^2$ modulo $p$. Then $d \\mid 6$ and $d \\neq 1$.\n\n- If $d = 2$, then $a^2 \\equiv -1 \\pmod{p}$, so $a \\equiv \\pm 6 \\pmod{p}$. For $a = 6$, $X = \\{\\pm 1, \\pm 3, \\pm 5\\}$, and $\\{a x_i + x_j \\mid i, j = 1, \\dots, 6\\} = \\{\\pm 1, \\pm 3, \\dots, \\pm 35\\}$, which works. Similarly, $a \\equiv \\pm 6 \\pmod{p}$ and $X = \\{\\pm 1, \\pm 3, \\pm 5\\}$ satisfy the condition.\n- If $d = 3$, then $\\overline{X}$ is of the form $\\{\\overline{x}, \\overline{a^2 x}, \\overline{a^4 x}, \\overline{y}, \\overline{a^2 y}, \\overline{a^4 y}\\}$. Also, $a^6 \\equiv 1 \\pmod{p} \\implies a^3 \\equiv \\pm 1 \\pmod{p}$. If $a^3 \\equiv 1 \\pmod{p}$, then $a x + \\overline{a^2 x} \\equiv a a^4 x + a^4 x \\pmod{p}$, a contradiction. If $a^3 \\equiv -1 \\pmod{p}$, then $a x + \\overline{a^4 x} = \\overline{0} \\in Y$, also a contradiction.\n- If $d = 6$, then\n\n$$\n\\overline{X} = \\{\\overline{a^{2i} x_1} \\mid i = 0, \\dots, 5\\}, \\\\\nY = \\{(\\overline{a^{2i+1} + a^{2j}}) x_1 \\mid i, j = 0, \\dots, 5\\}.\n$$\n\nFrom $a^6 \\equiv -1 \\pmod{p}$, and\n\n$$\n\\prod_{j=0}^{5} (t + a^{2j}) \\equiv \\prod_{j=0}^{5} (t - a^{2j}) \\equiv t^6 - 1 \\pmod{p},\n$$\n\nwe get $\\prod_{0 \\leq i, j \\leq 5} (a^{2i+1} + a^{2j}) = \\prod_{0 \\leq i \\leq 5} (a^{6(2i+1)} - 1) \\equiv 2^6 \\not\\equiv 1 \\equiv 36! \\pmod{p}$, a contradiction.\n\nTherefore, the desired $a$ are all integers congruent to $\\pm 6$ modulo $37$, i.e., $a = 37k \\pm 6$, $k \\in \\mathbb{Z}$.\n\n**Lemma:** Let $a_1, \\dots, a_6, b_1, \\dots, b_6 \\in \\{0, 1, \\dots, 36\\}$. If $\\{a_i + b_j \\mid 1 \\leq i, j \\leq 6\\}$ is a complete system modulo $37$, then $\\{b_1, \\dots, b_6\\}$ is uniquely determined by $\\{a_1, \\dots, a_6\\}$.\n\n*Proof of the lemma:* Let $f(x) = \\sum_{i=1}^6 x^{a_i}$, $g(x) = \\sum_{i=1}^6 x^{b_i}$. Then\n\n$$\nf(x) g(x) \\equiv \\sum_{k=1}^{36} x^k \\pmod{x^{37} - 1},\n$$\n\nso $f(x) g(x) \\equiv -1 \\pmod{q(x)}$, where $q(x) = 1 + x + \\dots + x^{36}$. There is a unique $h(x) \\in \\mathbb{Q}[x]$ with $f(x) h(x) \\equiv -1 \\pmod{q(x)}$ and $\\deg h(x) < 36$, so $g(x) = h(x)$ or $g(x) = h(x) + q(x)$. Since both $h(x)$ and $h(x) + q(x)$ cannot have at most $6$ monomials, $g(x)$ is uniquely determined by $f(x)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17557,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$, $E$, $F$ be the feet of the perpendiculars from the incenter $I$ of triangle $ABC$ to $BC$, $CA$, and $AB$, respectively. Let $N$ be the midpoint of $BC$. Suppose the foot of the perpendicular from $N$ onto $EF$ is $L$.\n\nToss the diagram onto the complex plane. Let the incircle of $\\triangle ABC$ be the unit circle, so $I$ is the origin, and let $D = 1$, $E = e$, $F = f$ (with $e, f$ on the unit circle). Let $A = \\frac{2ef}{e+f}$, $B = \\frac{2f}{f+1}$, $C = \\frac{2e}{e+1}$. The midpoint of $BC$ is $N = \\frac{e}{e+1} + \\frac{f}{f+1}$. The foot of the perpendicular from $N$ to $EF$ is\n\n$$\nL = \\frac{(e+f)(e+f+2)}{2(e+1)(f+1)}.\n$$\n\nLet $P$ be the foot from $B$ onto $CI$, and $Q$ the foot from $C$ onto $BI$. The circumcenter of $\\triangle IPQ$ is $\\frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}$.\n\nShow that the points $\\frac{2ef}{e+f}$, $\\frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}$, and $\\frac{ef(e+f+2)}{(e+f)^2}$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following approach:\n\nMultiplying the three points by $\\frac{e+f}{2ef}$, we reduce to proving that $1$, $\\frac{ef-1}{(e-1)(f-1)}$, and $\\frac{e+f+2}{2(e+f)}$ are collinear. Consider:\n\n$$\n\\frac{\\frac{ef-1}{(e-1)(f-1)} - 1}{\\frac{e+f+2}{2(e+f)} - 1} = \\frac{\\frac{e+f-2}{(e-1)(f-1)}}{\\frac{2-e-f}{2(e+f)}} = -\\frac{2(e+f)}{(e-1)(f-1)}.\n$$\n\nThis quantity is real, so the three points are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17558,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ and prime numbers $p, q$ which satisfy the equation\n\n$$\nn^3 = p^3 + 2p^2q + 2pq^2 + q^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the equation as follows:\n\n$$\nn^3 = p^3 + 2p^2q + 2pq^2 + q^3 = (p+q)(p^2 + pq + q^2)\n$$\n\nClearly, $p \\neq q$, because otherwise the equation $n^3 = 6p^3$ would hold, which is not possible for positive integers.\n\nSuppose there exists such $s > 1$, which is a factor of both $p+q$ and $p^2 + pq + q^2$. But then $s \\mid (p+q)^2 - (p^2 + pq + q^2) = pq$. If, for example, $s \\mid p$, then from $s \\mid (p+q)$ it follows $s \\mid q$. Since $p, q$ are prime, it must be that $p = q$, which leads to a contradiction.\n\nIf, on the other hand, $\\gcd(p+q, p^2 + pq + q^2) = 1$, then each factor must be a cube of a positive integer. Suppose $p+q = x^3$, $p^2 + pq + q^2 = y^3$, and $n = xy$. Then,\n\n$$\npq = (p+q)^2 - (p^2 + pq + q^2) = x^6 - y^3 = (x^2 - y)(x^4 + x^2y + y^2).\n$$\n\nSince $x^4 + x^2y + y^2 > x^3 = p+q$, we obtain $x^4 + x^2y + y^2 = pq$ and $x^2 - y = 1$. Thus,\n\n$$\npq = x^4 + x^2(x^2-1) + (x^2-1)^2 = 3x^4 - 3x^2 + 1 = 3x^2(x-1)(x+1) + 1 \\equiv 1 \\pmod{9},\n$$\nsince $x(x-1)(x+1) \\not\\equiv 3 \\pmod{9}$.\n\nWhat is left is to search through the cases modulo 9. The cube of an integer modulo 9 can be equal to 0, $\\pm1$. Let us list all possible cases for remainders modulo 9 of primes $p, q$ so that the condition $pq \\equiv 1 \\pmod{9}$ is satisfied.\n\n$$\np \\equiv 1 \\pmod{9} \\Rightarrow q \\equiv 1 \\pmod{9} \\Rightarrow x^3 = p+q \\equiv 2 \\pmod{9} \\text{ – contradiction.}\n$$\n\n$$\np \\equiv 2 \\pmod{9} \\Rightarrow q \\equiv 5 \\pmod{9} \\Rightarrow x^3 = p+q \\equiv 7 \\pmod{9} \\text{ – contradiction.}\n$$\n\n$$\np \\equiv 4 \\pmod{9} \\Rightarrow q \\equiv 7 \\pmod{9} \\Rightarrow x^3 = p+q \\equiv 2 \\pmod{9} \\text{ – contradiction.}\n$$\n\n$$\np \\equiv 8 \\pmod{9} \\Rightarrow q \\equiv 8 \\pmod{9} \\Rightarrow x^3 = p+q \\equiv 7 \\pmod{9} \\text{ – contradiction.}\n$$\n\nAll the cases were checked, which concludes the proof that such numbers do not exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17559,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $V$ be the \"corner\" shape that consists of three unit squares (see Figure 1), and let $F_n$ be the $n$th Fibonacci number. Prove that there exists a figure on a square grid that has exactly $F_n$ tilings into shapes $V$. For example, a $2 \\times 3$ rectangle has exactly two tilings (see Figure 2).\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p253_data_4e1d602a29.png)\n\nFigure 1: Shape $V$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p253_data_cb2229bcfb.png)\n\nFigure 2",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the sequence of figures $A_1, B_1, A_2, B_2, \\dots$ shown in Figure 3:\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p253_data_db0d9acb2e.png)\n\nFigure 3: Figures $A_1, B_1, A_2$ and $B_2$\n\nLet $a_1, b_1, a_2, b_2, \\dots$ be the number of tilings for them. One can see that $a_1 = 2$ (see Figure 4), $b_1 = 3$, $a_n = a_{n-1} + b_{n-1}$ (see Figure 5, top row), and $b_n = a_n + b_{n-1}$ (see Figure 5, bottom row). Red dots show the cell for which we are considering both possible coverings with $V$, and black \"corners\" denote forced moves. Therefore, the sequence $a_1, b_1, a_2, b_2, a_3, \\dots$ is exactly the sequence of Fibonacci numbers, omitting the first two numbers 1. The figure $V$ itself has exactly 1 tiling.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p253_data_2f9e379b0e.png)\n\nFigure 4\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p254_data_07e839b818.png)\n\nFigure 5\n\nNote: It is possible to find a figure that has $n$ tilings for any number $n$ (not only Fibonacci numbers), but the construction is more complex.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17560,
    "subject": "Mathematics (Olympiad)",
    "question": "When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*.\n\nLet $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Show that\n\n$$\na_n = F_{n+2}\n$$\n\nfor all $n \\ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \\ge 1$.\n\nLet $b_n$ be the number of z-patterns for $n$ tiles fixed on a wall in a ring shape. Show that\n\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}\n$$\n\nfor all $n \\ge 4$.\n\nNow, consider a $4 \\times 4$ square of tiles. How many z-patterns are there if the four central tiles form a $2 \\times 2$ block?",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles in a row. Then\n\n$$\na_n = F_{n+2}\n$$\n\nfor all $n \\ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \\ge 1$.\n\n*Proof.* We prove by induction on $n \\ge 1$. The lemma holds for $n=1$ and $n=2$. Assume the lemma for all $k$, $1 \\le k < n$, where $n \\ge 3$. If $0$ is written on the tile at one end, any number ($0$ or $1$) can be written on the next tile, so there are $a_{n-1}$ such z-patterns. If $1$ is written there, then $0$ must be written on the next tile, so there are $a_{n-2}$ such z-patterns. Thus,\n\n$$\na_n = a_{n-1} + a_{n-2} = F_{n+1} + F_n = F_{n+2},\n$$\n\nwhich proves the lemma.\n\n**Lemma 2.** Let $b_n$ be the number of z-patterns for $n$ tiles in a ring. Then\n\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}\n$$\n\nfor all $n \\ge 4$.\n\n*Proof.* Let $n \\ge 4$ and choose any tile among the $n$ tiles in a ring. If $0$ is written on the tile, any number can be written on the neighboring tiles, so there are $a_{n-1}$ such z-patterns. If $1$ is written on the tile, then $0$ must be written on the neighboring two tiles, so there are $a_{n-3}$ such z-patterns. Thus,\n\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1},\n$$\n\nwhich proves the lemma.\n\nNow, consider the $2 \\times 2$ tiles in the center of the $4 \\times 4$ square. The number of $0$'s that can be written on these four tiles equals $4$, $3$, or $2$.\n\n**Case 1:** Four $0$'s:\n\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\n\nWe may apply Lemma 2 to the $12$ tiles surrounding the four center tiles, since any number can be written on the $12$ tiles. Therefore, the number of z-patterns in this case is\n\n$$\nb_{12} = F_{13} + F_{11} = 233 + 89 = 322.\n$$\n\n**Case 2:** Three $0$'s:\n\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}\n\\qquad\n\\begin{bmatrix} 0 & 1 \\\\ 0 & 0 \\end{bmatrix}\n\\qquad\n\\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix}\n\\qquad\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\n\nIn each of the four subcases, only $0$ can be written on two neighboring tiles of the tile marked by $1$. For the remaining $10$ tiles, the number of z-patterns is $a_9 \\times 2 = F_{11} \\times 2 = 178$. Therefore, the total number of z-patterns in this case is\n\n$$\n178 \\times 4 = 712.\n$$\n\n**Case 3:** Two $0$'s:\n\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n\\qquad\n\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}\n$$\n\nIn each of the two subcases, only $0$ can be written on four neighboring tiles of the two tiles marked by $1$. For the remaining $6$ tiles, the number of z-patterns is $a_3^2 \\times 2^2 = F_5^2 \\times 4 = 100$. Therefore, the total number of z-patterns in this case is\n\n$$\n100 \\times 2 = 200.\n$$\n\nCombining the three cases, the answer is:\n\n$$\n322 + 712 + 200 = 1234.\n$$\n\n$\\boxed{1234}$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17561,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 100 different points on a circle, determine the maximum positive integer $k$ such that: if any $k$ of the 100 points are arbitrarily coloured red or blue, the remaining points can be coloured red or blue so that the 100 points can be paired into 50 segments (each segment connects two points, no two segments share a point), and the endpoints of each segment are of the same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $50$.\n\nLet the $100$ points be $A_1, A_2, \\dots, A_{100}$ in clockwise order. More generally, consider $2m$ points $A_1, \\dots, A_{2m}$, each coloured red or blue. We seek to pair these into $m$ segments, each with endpoints of the same colour, and no two segments sharing a point.\n\nLet $S = \\{A_1, A_3, \\dots, A_{2m-1}\\}$ and $T = \\{A_2, A_4, \\dots, A_{2m}\\}$.\n\n**Lemma:** The colouring is possible if and only if the number of red points in $S$ equals the number of red points in $T$.\n\n*Proof:* Necessity: Any valid pairing must match red points in $S$ with red points in $T$, so their counts must be equal. Sufficiency: By induction on $m$. For $m=1$, the two points must be the same colour and can be paired. Assume true for $m-1$. For $m$, there must be two adjacent points of the same colour (otherwise, all points in $S$ are one colour and all in $T$ another, which contradicts the assumption). Pair $A_{2m-1}$ and $A_{2m}$, and apply the induction hypothesis to the remaining $2m-2$ points.\n\nReturning to the original problem: If $k \\ge 51$, colour all $50$ points in $S$ red and $k-50$ points in $T$ blue. No matter how the remaining points are coloured, the lemma's condition cannot be satisfied. Thus, $k \\ge 51$ fails.\n\nFor $k=50$, suppose $a$ points in $S$ and $b$ in $T$ are coloured red, with $a \\ge b$. There are at least $a-b$ uncoloured points in $T$; otherwise, the number of coloured points exceeds $50$, a contradiction. Thus, $k=50$ works.\n\nTherefore, the maximum $k$ is $50$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17562,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ consecutive positive integers be written down in ascending order to form a positive integer, called a $k$-consecutive number. For example, writing down $99$, $100$, $101$ in order gives $99100101$, which is a $3$-consecutive number.\n\nProve that for any positive integers $N$ and $k$, there exists a $k$-consecutive number that is divisible by $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "$N$ can be uniquely written as $N = N_1 N_2$, where $N_1$ is coprime with $10$ and $N_2$ has no prime factors other than $2$ and $5$. Since $10$ and $N_1$ are coprime, by Euler's theorem, $10^{\\varphi(N_1)} \\equiv 1 \\pmod{N_1}$.\n\nTake a sufficiently large positive integer $t$ and let $m = t \\varphi(N_1)$ so that\n$$\n10^m - 10^{m-1} = 9 \\cdot 10^{m-1} > N + k.\n$$\n\nAssume $x, x+1, \\ldots, x+(k-1)$ are $k$ consecutive $m$-digit numbers, i.e.,\n$$\n10^{m-1} \\leq x \\leq x+(k-1) < 10^m.\n$$\n\nThe $k$-consecutive number formed by writing $x, x+1, \\ldots, x+(k-1)$ in order is\n$$\n\\begin{aligned}\nM &= x \\cdot 10^{(k-1)m} + (x+1)10^{(k-2)m} + \\cdots + (x+k-2)10^m + (x+k-1) \\\\\n&= x \\sum_{j=1}^{k} 10^{(k-j)m} + \\sum_{j=1}^{k-1} j \\cdot 10^{(k-1-j)m} \\\\\n&= Ax + B,\n\\end{aligned}\n$$\nwhere $A = \\sum_{j=1}^{k} 10^{(k-j)m}$ and $B = \\sum_{j=1}^{k-1} j \\cdot 10^{(k-1-j)m}$.\n\nNote $10^m \\equiv 1 \\pmod{N_1}$, so\n$$\n\\begin{aligned}\nA &\\equiv k \\pmod{N_1}, \\\\\nB &\\equiv \\frac{1}{2}k(k-1) \\pmod{N_1}.\n\\end{aligned}\n$$\n\nSince $N_1$ does not contain the prime factor $2$, we have\n$$\n(A, N_1) = (k, N_1) \\mid (k(k-1), N_1) = \\left(\\frac{1}{2}k(k-1), N_1\\right) = (B, N_1).\n$$\n\nAlso, since $A \\equiv 1 \\pmod{10}$, $A$ and $N_2$ are coprime, so $(A, N) = (A, N_1)$. Thus, $(A, N)$ divides $(B, N_1)$ and $(B, N)$. Therefore, the congruence $Ax + B \\equiv 0 \\pmod{N}$ has a solution.\n\nConsequently, there exists $10^{m-1} \\leq x < 10^{m-1} + N$ such that $N \\mid Ax + B$. Also,\n$$\n10^m - 10^{m-1} > N + k,\n$$\nso\n$$\nx + (k-1) < 10^{m-1} + N + (k-1) < 10^{m-1} + N + k < 10^m.\n$$\n\nThus, $x, x+1, \\ldots, x+(k-1)$ are all $m$-digit numbers. Therefore, the $k$-consecutive number formed by these $k$ numbers is divisible by $N$. The conclusion is proved. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17563,
    "subject": "Mathematics (Olympiad)",
    "question": "The $n$-tuple of positive integers $a_1, \\dots, a_n$ satisfies the following conditions:\n\n1. $1 \\leq a_1 < a_2 < \\dots < a_n \\leq 50$;\n2. For any $n$-tuple of positive integers $b_1, \\dots, b_n$, there exist a positive integer $m$ and an $n$-tuple of positive integers $c_1, \\dots, c_n$ such that\n\n$$\nm \\cdot b_i = c_i^{a_i} \\quad \\text{for } i = 1, \\dots, n.\n$$\n\nProve that $n \\leq 16$ and find the number of different $n$-tuples $a_1, \\dots, a_n$ satisfying the given conditions for $n = 16$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that the numbers $a_1, \\dots, a_n$ are mutually relatively prime. Suppose not; then $(a_i, a_j) = d > 1$ for some $i \\neq j$. Let $a_i = u d$, $a_j = v d$. Set $b_i = 1$, $b_j = 2$. By condition (ii), there exist $m$, $c_i$, and $c_j$ such that\n\n$$\nm \\cdot b_i = c_i^{a_i} \\quad \\text{and} \\quad m \\cdot b_j = c_j^{a_j}.\n$$\n\nTherefore,\n\n$$\nm = c_i^{a_i}, \\quad 2m = c_j^{a_j}.\n$$\n\nSo $2 c_i^{a_i} = c_j^{a_j}$, which is impossible, since the exponent of $2$ in the prime factorization of the right-hand side is a multiple of $d$, but on the left-hand side it is not.\n\nAssume now that $a_1, \\dots, a_n$ are mutually relatively prime. We show that condition (ii) is fulfilled. Let $b_1, \\dots, b_n$ be any $n$-tuple of positive integers, and let $p_1, \\dots, p_k$ be all the prime divisors of $b_1, \\dots, b_n$. We look for $m$ in the form\n\n$$\nm = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}.\n$$\n\nFor $i = 1, \\dots, n$, let $\\beta_{i,j}$ be the exponent of $p_j$ in the prime factorization of $b_i$. For $m \\cdot b_i$ to be an $a_i$-th power, it suffices that $\\alpha_j + \\beta_{i,j}$ is a multiple of $a_i$ for each $j = 1, \\dots, k$. So we need $\\alpha_j$ to satisfy the congruences\n\n$$\n\\alpha_j \\equiv -\\beta_{1,j} \\pmod{a_1}, \\quad \\alpha_j \\equiv -\\beta_{2,j} \\pmod{a_2}, \\quad \\dots, \\quad \\alpha_j \\equiv -\\beta_{n,j} \\pmod{a_n}.\n$$\n\nThe existence of such $\\alpha_j$ is guaranteed by the Chinese Remainder Theorem, since $a_1, \\dots, a_n$ are mutually relatively prime.\n\nThus, condition (ii) is satisfied if and only if the numbers $a_1, \\dots, a_n$ are mutually relatively prime. Among $1, 2, \\dots, 50$, there are exactly $15$ primes. If $n \\geq 17$, then among $2 \\leq a_2 < a_3 < \\dots < a_n \\leq 50$, there must be at least two numbers sharing a prime factor, so they are not relatively prime. Therefore, $n \\leq 16$.\n\nIf $n = 16$, then $a_1 = 1$ and $a_2, \\dots, a_{16}$ must be powers of different primes. The possible powers of primes that can be used are:\n\n$$\n\\begin{array}{lcl}\n\\text{For } 2: & 2, 4, 8, 16, 32, \\\\\n\\text{For } 3: & 3, 9, 27, \\\\\n\\text{For } 5: & 5, 25, \\\\\n\\text{For } 7: & 7, 49, \\\\\n\\text{For } p \\geq 11: & \\text{only } p.\n\\end{array}\n$$\n\nThus, the total number of corresponding $16$-tuples is $5 \\times 3 \\times 2 \\times 2 = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17564,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist two positive powers of $5$ such that the number obtained by writing one after the other is also a power of $5$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $5^x \\cdot 10^n + 5^y = 5^z$, where $5^y$ has $n$ digits. Then $$5^{x+n} \\cdot 2^n = 5^y \\cdot (5^{z-y} - 1)$$ whence $$2^n = 5^{z-y} - 1.$$ Case $n = 1$ does not work. For case $n = 2$ we get $z - y = 1$. Since $5^y$ has 2 digits, the only possibility is $y = 2$ and $z = 3$, whence $x = 0$, which is not positive. Case $n > 2$ yields $5^{z-y} \\equiv 1 \\pmod{8}$, thus $z - y = 2k$ for an integer $k$. Now $$2^n = 25^k - 1 = 24 \\cdot (25^{k-1} + \\dots + 1),$$ this is impossible, since $3 \\mid 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17565,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive numbers $a$, $b$, $c$ satisfy the condition $a^2 + b^2 + c^2 + abc = 4$. Prove that the inequality $c + ab \\le 2$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the given equation as a quadratic in $c$:\n\n$$\nc^2 + abc + (a^2 + b^2 - 4) = 0 \\implies c = \\frac{-ab \\pm \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}.\n$$\n\nSince $c$ is positive, we take\n$$\nc = \\frac{-ab + \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}.\n$$\n\nNow, we want to show $c + ab \\le 2$:\n\n$$\n\\frac{-ab + \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2} + ab \\le 2 \\\\\n\\iff \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2} \\le 4 - ab \\\\\n\\iff a^2b^2 + 16 - 4a^2 - 4b^2 \\le 16 - 8ab + a^2b^2 \\\\\n\\iff 4a^2 + 4b^2 \\ge 8ab \\\\\n\\iff 4(a-b)^2 \\ge 0.\n$$\n\nThis is always true. Note that $ab \\le 4$, because otherwise $a^2 + b^2 \\ge 2ab \\ge 8$, which contradicts the given condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17566,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral with $AC = BD$. Diagonals $AC$ and $BD$ meet at $P$. Let $\\omega_1$ and $O_1$ denote the circumcircle and circumcenter of triangle $ABP$. Let $\\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$. Segment $BC$ meets $\\omega_1$ and $\\omega_2$ again at $S$ and $T$ (other than $B$ and $C$), respectively. Let $M$ and $N$ be the midpoints of minor arcs $\\widehat{SP}$ (not including $B$) and $\\widehat{TP}$ (not including $C$). Prove that $MN \\parallel O_1O_2$.\n\n(This problem was suggested by Steve Dinh.)",
    "options": [],
    "answer": "See solution",
    "solution": "**Note.** The result still holds without the assumption that both triangles $ABP$ and $CDP$ are acute. This assumption helps the contestants to focus on more specific configurations. Indeed, because triangles are acute, $O_1$ lies inside triangle $ABP$ and $O_2$ lies inside triangle $CDP$. Points $M$ and $N$ lie in the region bounded by rays $PB$ and $PD$. It is not difficult to see that $O_1MNO_2$ is a convex quadrilateral. (In particular, this is helpful in solution 2.) Hence we can consider the configuration (in two diagrams) shown below. For other possible configurations, our proofs can be adjusted slightly.\n\nWe present two solutions. Both solutions are based on the fact that triangles $ABQ$ and $CDQ$ are similar isosceles triangles. Indeed, because $ABPQ$ and $DCPQ$ are cyclic, we have $\\angle QBD = \\angle QAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17567,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 是一正整數，且考慮一正整數數列 $a_1, a_2, \\dots, a_n$。將此數列延伸為有周期的無窮數列，對每個 $i \\ge 1$ 都定義 $a_{n+i} = a_i$。若\n\n$$\na_1 \\le a_2 \\le \\dots \\le a_n \\le a_1 + n\n$$\n\n且對於 $i = 1, 2, \\dots, n$，都有\n\n$$\na_{a_i} \\le n + i - 1.\n$$\n\n試證：\n\n$$\na_1 + \\dots + a_n \\le n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "首先，我们断言：\n\n$$\na_i \\le n + i - 1 \\quad \\text{对于 } i = 1, 2, \\dots, n\n$$\n\n假设反例 $i$ 是最小的。由 $a_n \\ge a_{n-1} \\ge \\dots \\ge a_i \\ge n + i$ 和 $a_{a_i} \\le n + i - 1$，结合数列的周期性，得到：\n\n$$\na_i \\text{ 不可能同余于 } i, i+1, \\dots, n \\pmod{n}.\n$$\n\n因此，假设 $a_i \\ge n + i$，由 $a_1 + n \\ge a_n \\ge a_i$ 得 $a_1 \\ge n + 1$。由于 $i$ 的极小性，得到 $i = 1$，这与上式矛盾。故断言成立。\n\n特别地，$a_1 \\le n$。若 $a_n \\le n$，则 $a_1 \\le a_2 \\le \\dots \\le a_n \\le n$，不等式显然成立。否则，存在 $t$ 满足 $1 \\le t \\le n-1$，使得：\n\n$$\na_1 \\le a_2 \\le \\dots \\le a_t \\le n < a_{t+1} \\le \\dots \\le a_n.\n$$\n\n由于 $1 \\le a_1 \\le n$ 且 $a_{a_1} \\le n$，有 $a_1 \\le t$，因此 $a_n \\le n + t$。令 $b_i$ 表示满足 $a_j \\ge n + i$ 的 $j \\in \\{t+1, \\dots, n\\}$ 的个数，则：\n\n$$\nb_1 \\ge b_2 \\ge \\dots \\ge b_t \\ge b_{t+1} = 0\n$$\n\n接下来断言 $a_i + b_i \\le n$ 对于 $1 \\le i \\le t$。因为 $a_{a_i} \\le n + i - 1$ 且 $a_i \\le n$，每个 $a_j \\ge n + i$ 的 $j$ 必在 $\\{a_i + 1, \\dots, n\\}$，故 $b_i \\le n - a_i$。\n\n由 $b_i$ 的定义和上述分段，得到：\n\n$$\na_1 + a_2 + \\dots + a_n \\le n(n - t) + n t = n^2\n$$\n\n证毕。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17568,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integer $n$, let $a_n$ be the number of quadratic functions $f(x) = ax^2 + bx + c$, where $a, b, c \\in \\{1, 2, \\dots, n\\}$, that have only integer roots. Prove that for every $n \\ge 4$,\n\n$$\nn < a_n < n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equations $x^2 + kx + k - 1 = 0$ for $k = 2, \\dots, n$, and $2x^2 + 4x + 2 = 0$, $x^2 + 4x + 4 = 0$, all have integer roots, so $n+1 \\le a_n$.\n\nIf $f$ is such a quadratic, then $f(x) = a(x + x_1)(x + x_2)$, where $x_1, x_2 \\in \\mathbb{Z}_+$, and $a$, $a(x_1 + x_2)$, $a x_1 x_2 \\in \\{1, 2, \\dots, n\\}$. From the last condition,\n\n$$\nx_2 \\le \\frac{n}{a x_1}.\n$$\n\nThus,\n\n$$\na_n \\le \\sum_{\\substack{1 \\le x_1 \\le n \\\\ 1 \\le a \\le n}} \\frac{n}{a x_1} = n \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}\\right)^2. \\quad (1)\n$$\n\nIt can be shown by induction that for every $n \\ge 5$,\n\n$$\n1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} < \\sqrt{n}.\n$$\n\nAlso, $a_4 = 5$. Substituting into (1), we get $a_n < n^2$ for $n \\ge 5$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17569,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \\mathbb{N} \\to \\mathbb{N}$ satisfying:\n\n1. For any $x, a \\in \\mathbb{N}$, the quantity $\\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a = 1$.\n2. For all $x \\in \\mathbb{N}$, we have $|f(x) - cx| < 2023$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the only possible values of $c$ are $k + \\frac{1}{2}$ for some non-negative integer $k$.\n\nThe fact that these values are possible is seen from the function $f(x) = \\lfloor (k + \\frac{1}{2})x \\rfloor + 1 = kx + \\lfloor \\frac{x}{2} \\rfloor + 1$. Indeed, for any $x, a \\in \\mathbb{N}$:\n\n$$\n\\frac{f(x+a) - f(x)}{a} = \\frac{1}{a} \\left( ka + \\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor \\right) = k + \\frac{1}{a} \\left( \\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor \\right).\n$$\n\nThis is clearly an integer for $a = 1$. But for $a \\ge 2$:\n\n$$\n\\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor \\ge \\left\\lfloor \\frac{x+2}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor = 1.\n$$\n\nIf $a = 2k$, then\n\n$$\n\\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor = k < 2k = a,\n$$\n\nand if $a = 2k + 1$ for $k \\ge 1$, then\n\n$$\n\\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor \\le \\left\\lfloor \\frac{x+2k+2}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor = k+1 < 2k+1 = a.\n$$\n\nSo in either case, the quantity $\\left\\lfloor \\frac{x+a}{2} \\right\\rfloor - \\left\\lfloor \\frac{x}{2} \\right\\rfloor$ is strictly between $0$ and $a$, and thus cannot be divisible by $a$. Thus condition (a) holds; condition (b) is obviously true.\n\nNow let us show these are the only possible values, under the weaker assumption that there exists some $d \\in \\mathbb{N}$ so that $|f(x) - cx| < d$. It is clear that $c \\ge 0$: if $d < f(x) - cx < d$ and $c < 0$, then for large $x$ the range $[cx - d, cx + d]$ consists only of negative numbers and cannot contain $f(x)$.\n\nNow we claim that $c \\ge \\frac{1}{2}$. Suppose $0 \\le c < \\frac{1}{2}$, and that $d > 0$ is such that $|f(x) - cx| \\le d$. Pick $N > \\frac{2d}{1-2c}$ so that $2(cN + d) < N$. Then the $N$ values $\\{f(1), \\dots, f(N)\\}$ must all be in the range $[1, \\dots, cN+d]$, and by the pigeonhole principle, some three values $f(i), f(j), f(k)$ must be equal. Some two of $i, j, k$ are not consecutive: suppose WLOG $i > j + 1$. Then $\\frac{f(i)-f(j)}{i-j} = 0$, which contradicts condition (a) for $x = j$ and $a = i - j$.\n\nFor the general case, suppose $c = k + \\lambda$, where $k \\in \\mathbb{Z}$ and $\\lambda \\in [0, 1)$. Let $d \\in \\mathbb{N}$ be such that $-d \\le f(x) - cx \\le d$. Consider the functions\n\n$$\ng_1(x) = f(x) - kx + d + 1, \\quad g_2(x) = x - f(x) + kx + d + 1.\n$$\n\nNote that\n\n$$\n\\begin{aligned}\ng_1(x) &\\ge cx - d - kx + d + 1 = \\lambda x + 1 \\ge 1, \\\\\ng_2(x) &\\ge x - (cx + d) + kx + d + 1 = (1 - \\lambda)x + 1 \\ge 1\n\\end{aligned}\n$$\n\nso these are also functions from $\\mathbb{N}$ to $\\mathbb{N}$. They also satisfy condition (a) for $f$:\n\n$$\n\\frac{g_1(x+a) - g_1(x)}{a} = \\frac{f(x+a) - k(x+a) + d - f(x) + kx - d}{a} = \\frac{f(x+a) - f(x)}{a} - k\n$$\n\nis an integer if and only if $\\frac{f(x+a)-f(x)}{a}$ is, which happens if and only if $a = 1$. A similar argument holds for $g_2$.\n\nNow note that $g_1(x) - \\lambda x = f(x) - cx + d + 1$ is bounded, and so is $g_2(x) - (1 - \\lambda)x = cx - f(x) + d + 1$. So they satisfy the weaker form of condition (b) as well. Thus, applying the reasoning above, we see that $\\lambda \\ge \\frac{1}{2}$ and $1 - \\lambda \\ge \\frac{1}{2}$. This forces $\\lambda = \\frac{1}{2}$, which finishes our proof. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17570,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, define\n\n$$\na_n = \\{\\sqrt{n}\\} - \\{\\sqrt{n+1}\\} + \\{\\sqrt{n+2}\\} - \\{\\sqrt{n+3}\\}.\n$$\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 $n$.\n\n*Remark.* The notation $\\{x\\}$ denotes the fractional part of the real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) $a_1 = \\{\\sqrt{1}\\} - \\{\\sqrt{2}\\} + \\{\\sqrt{3}\\} - \\{\\sqrt{4}\\} = 0 - \\{\\sqrt{2}\\} + \\{\\sqrt{3}\\} - 0 = (\\sqrt{3} - 1) - (\\sqrt{2} - 1) = \\sqrt{3} - \\sqrt{2}$. Since $\\sqrt{3} > 1.7$ and $\\sqrt{2} < 1.5$, it follows that $a_1 > 0.2$.\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 $\\{\\sqrt{a}\\} = \\sqrt{a} - m < \\sqrt{b} - m = \\{\\sqrt{b}\\}$.\n\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).\n\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": 17571,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with orthocentre $H$. Let $E = BH \\cap AC$ and $F = CH \\cap AB$. Let $D$, $M$, and $N$ be the midpoints of segments $AH$, $BD$, and $CD$ respectively, and let $T = FM \\cap EN$. Suppose $D$, $E$, $T$, $F$ are concyclic. Prove that $DT$ passes through the circumcentre of $ABC$.\n\n![](images/Contests_Booklet_2023_2024_p18_data_ef97421970.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcentre of $(ABC)$ and $J$ be the midpoint of $DO$. Now it is sufficient to prove that $TD$ and $TJ$ coincide. We first prove that $M$, $N$, $T$, $J$ are concyclic.\n\n$$\n\\angle MJN = \\angle BOC = 2\\angle BAC = \\angle EDF = 180^\\circ - \\angle FTE = 180^\\circ - \\angle NJM\n$$\n\nThis proves our claim!\n\nNow note that $JM = \\frac{OB}{2} = \\frac{OC}{2} = JN$, so $TJ$ is the angle bisector of $\\angle NTM = \\angle ETF$. But $D$ is the midpoint of arc $EF$ in the nine-point circle, so $TD$ is the angle bisector of $\\angle ETF$ as well. Thus, $TD$ and $TJ$ must coincide! $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17572,
    "subject": "Mathematics (Olympiad)",
    "question": "In a given community of people, each person has at least two friends within the community. Whenever some people from this community sit on a round table such that each adjacent pair of people are friends, it happens that no non-adjacent pair of people are friends.\n\nProve that there exist two people in this community such that each has exactly two friends and they have at least one common friend.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the accompanying simple graph: its vertices are the members of the community and each pair of friends is connected by an edge. Thus, the minimum degree $\\delta(G) \\ge 2$ and every cycle in $G$ is induced (i.e., chord-free).\n\nConsider a path $P : v_1v_2\\dots v_n$ of maximum length (clearly $n \\ge 3$). Since $\\deg(v_n) \\ge 2$, there is $i \\in \\{1, 2, \\dots, n-2\\}$ such that $v_i \\leftrightarrow v_n$. Moreover, by the choice of $P$, every neighbor of $v_n$ belongs to $V(P)$. Hence, as every cycle is induced, $N(v_n) = \\{v_i, v_{n-1}\\}$; thus $\\deg(v_n) = 2$.\n\nWe show that $\\deg(v_{i+1}) = 2$.\n\nLet us argue by contradiction. Suppose there is $x \\in N(v_{i+1}) \\setminus \\{v_i, v_{i+2}\\}$. Then $x \\notin V(P)$, otherwise a cycle with a chord occurs (depicted as heavier in the figure below).\n\n![](images/BMO_2023_Short_List_p14_data_06ac474ede.png)\n\nSince $x \\notin V(P)$ we can define a new path $Q : v_1v_2\\dots v_iv_nv_{n-1}\\dots v_{i+1}x$ of length $n$, contradicting the assumption that the path $P$ with length $n-1$ is the longest. Thus $\\deg(v_{i+1}) = 2$ and $v_{i+1}, v_n$ is the desired pair.\n\n**Remark 1.** The given proof finds at least two such pairs of people, because $v_1 \\notin \\{v_{i+1}, v_n\\}$.\n\n**Remark 2.** A possible alternative formulation of the problem could require that every cycle of friends be of length divisible by 4 (that alternative assumption would imply that every cycle is induced).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17573,
    "subject": "Mathematics (Olympiad)",
    "question": "Every positive integer is colored either blue or red. Prove that there exists a sequence $\\{a_n\\}$ of infinitely many positive integers with $a_1 < a_2 < \\dots$, such that the sequence\n\n$a_1$, $\\frac{a_1 + a_2}{2}$, $a_2$, $\\frac{a_2 + a_3}{2}$, $a_3$, $\\dots$\n\nconsists entirely of positive integers of the same color.",
    "options": [],
    "answer": "See solution",
    "solution": "We need three lemmas. Let $\\mathbb{N}^*$ denote the set of all positive integers.\n\n**Lemma 1.** If there is an arithmetic progression of infinitely many positive integers of the same color, then the conclusion holds.\n\n*Proof.* Let $c_1 < c_2 < \\dots < c_n < \\dots$ be such a progression (say, all red). Set $a_i = c_{2i-1}$ for $i = 1, 2, 3, \\dots$. Then the sequence $a_1, \\frac{a_1 + a_2}{2}, a_2, \\frac{a_2 + a_3}{2}, \\dots$ consists of red integers.\n\n**Lemma 2.** If for any $i \\in \\mathbb{N}^*$, there exists $j$ such that $i$, $\\frac{i + j}{2}$, and $j$ are all the same color, then the conclusion holds.\n\n*Proof.* Start with $a_1 = 1$ (assume it is red). By assumption, there exists $a_2$ such that $a_1$, $\\frac{a_1 + a_2}{2}$, $a_2$ are all red. Repeating this process, we obtain a sequence $a_1 < a_2 < \\dots$ such that\n\n$$\na_1 < \\frac{a_1 + a_2}{2} < a_2 < \\frac{a_2 + a_3}{2} < a_3 < \\dots\n$$\n\nare all red.\n\n**Lemma 3.** If neither Lemma 1 nor Lemma 2 applies, then there exists $i_0 \\in \\mathbb{N}^*$ such that for every $j \\in \\mathbb{N}^*$, $i_0$, $\\frac{i_0 + j}{2}$, $j$ are not all the same color. Then the conclusion still holds.\n\n*Proof.* Without loss of generality, let $i_0 = 1$ (otherwise, consider multiples of $i_0$). Suppose $1$ is red. Then, for every $k \\geq 2$, $k$ and $2k-1$ cannot both be red. (Call this property (1).)\n\nSince there is no infinite monochromatic arithmetic progression, there are infinitely many blue numbers of different parities. We construct an infinite sequence of odd blue numbers $a_1 < a_2 < \\dots$ such that\n\n$$\na_1 < \\frac{a_1 + a_2}{2} < a_2 < \\frac{a_2 + a_3}{2} < a_3 < \\dots\n$$\n\nare all blue. Suppose $a_1 < \\dots < a_n$ have been chosen. We show there exists an odd $a_{n+1}$ such that $\\frac{a_n + a_{n+1}}{2}$ and $a_{n+1}$ are blue.\n\n- (a) If for every $i$, $a_n + i$ and $a_n + 2i$ have different colors, and no $a_{n+1}$ as above exists, then for $a_{n+1} > a_n$, $\\frac{a_n + a_{n+1}}{2}$ and $a_{n+1}$ cannot both be blue. But since there are infinitely many blue numbers, pick $i$ so $a_n + i$ is red, then $a_n + 2i$ is blue. Let $a_n = 2k+1$. Then $2k+1$ is blue, $2k+i+1$ is red, $2k+2i+1$ is blue. By (1), $4k+2i+1 = 2(2k+i+1) - 1$ is blue. Similarly, $3k+i+1 = \\frac{(2k+1)+(4k+2i+1)}{2}$ is red, so $6k+2i+1 = 2(3k+i+1) - 1$ is blue, and so on. This produces an infinite blue arithmetic progression, a contradiction. Thus, such $a_{n+1}$ must exist.\n\n- (b) If for some $i$, $a_n + i$ and $a_n + 2i$ have the same color:\n    - If both are blue, set $a_{n+1} = a_n + 2i$.\n    - If both are red, then $4k+2i+1$ and $4k+4i+1$ are blue. If $3k+2i+1 = \\frac{(2k+1)+(4k+4i+1)}{2}$ is blue, set $a_{n+1} = 4k+4i+1$; otherwise, $6k+4i+1 = 2(3k+2i+1) - 1$ is blue, so set $a_{n+1} = 6k+4i+1$.\n\nThus, in all cases, we can construct the required sequence.\n\nTherefore, by Lemmas 1, 2, and 3, the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17574,
    "subject": "Mathematics (Olympiad)",
    "question": "The kingdom of Anisotropy consists of $n$ cities. For every two cities, there exists exactly one direct one-way road between them. A path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called *diverse* if no road belongs to two or more paths in the collection.\n\nLet $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "We write $X \\to Y$ or $Y \\leftarrow X$ if the road between $X$ and $Y$ goes from $X$ to $Y$. Notice that if there is any route moving from $X$ to $Y$ (possibly passing through some cities more than once), then there is a path from $X$ to $Y$ consisting of some roads in the route. Indeed, any cycle in the route may be removed harmlessly; after some removals, one obtains a path.\n\nSay a path is *short* if it consists of 1 or 2 roads. Partition all cities different from $A$ and $B$ into four groups, $\\mathcal{I}$, $\\mathcal{O}$, $\\mathcal{A}$, $\\mathcal{B}$ according to the following rules: for each city $C$,\n\n$$\nC \\in \\mathcal{I} \\iff A \\to C \\leftarrow B;\n$$\n$$\nC \\in \\mathcal{O} \\iff A \\leftarrow C \\leftarrow B;\n$$\n$$\nC \\in \\mathcal{A} \\iff A \\to C \\to B;\n$$\n$$\nC \\in \\mathcal{B} \\iff A \\leftarrow C \\to B.\n$$\n\n**Lemma.** Let $\\mathcal{P}$ be a diverse collection consisting of $p$ paths from $A$ to $B$. Then there exists a diverse collection consisting of at least $p$ paths from $A$ to $B$ and containing all short paths from $A$.\n\n**Proof.** To obtain the desired collection, modify $\\mathcal{P}$ as follows. If there is a direct road $A \\to B$ and the path consisting of this single road is not in $\\mathcal{P}$, merely add it to $\\mathcal{P}$.\n\nNow consider any city $C \\in \\mathcal{A}$ such that the path $A \\to C \\to B$ is not in $\\mathcal{P}$. If $\\mathcal{P}$ contains at most one path containing a road $A \\to C$ or $C \\to B$, remove that path (if it exists), and add the path $A \\to C \\to B$ to $\\mathcal{P}$ instead. Otherwise, $\\mathcal{P}$ contains two paths of the forms $A \\to C \\dashrightarrow B$ and $A \\dashrightarrow C \\to B$, where $C \\dashrightarrow B$ and $A \\dashrightarrow C$ are some paths. In this case, we recombine the edges to form two new paths $A \\to C \\to B$ and $A \\dashrightarrow C \\dashrightarrow B$ (removing cycles from the latter if needed). Now we replace the old two paths in $\\mathcal{P}$ with the two new ones.\n\nAfter any operation described above, the number of paths in the collection does not decrease, and the collection remains diverse. Applying such operation to each $C \\in \\mathcal{A}$, we obtain the desired collection. $\\square$\n\nBack to the problem, assume, without loss of generality, that there is a road $A \\to B$, and let $a$ and $b$ denote the numbers of roads going out from $A$ and $B$, respectively. We will transform it into a diverse collection $\\mathcal{Q}$ consisting of at least $N_{AB} + (b-a)$ paths from $A$ to $B$. This construction yields\n\n$$\nN_{BA} \\geq N_{AB} + (b-a); \\quad \\text{similarly, we get } N_{AB} \\geq N_{BA} + (a-b),\n$$\nwhence $N_{BA} - N_{AB} = b - a$. This yields the desired equivalence.\n\nApply the lemma to get a diverse collection $\\mathcal{P}'$ of at least $N_{AB}$ paths containing all $|\\mathcal{A}| + 1$ short paths from $A$ to $B$. Notice that the paths in $\\mathcal{P}'$ contain no edge of a short path from $B$ to $A$. Each non-short path in $\\mathcal{P}'$ has the form $A \\to C \\dasharrow D \\to B$, where $C \\dasharrow D$ is a path from some city $C \\in \\mathcal{I}$ to some city $D \\in \\mathcal{O}$. For each such path, put into $\\mathcal{Q}$ the path $B \\to C \\dasharrow D \\to A$; also put into $\\mathcal{Q}$ all short paths from $B$ to $A$. Clearly, the collection $\\mathcal{Q}$ is diverse.\n\nNow, all roads going out from $A$ end in the cities from $\\mathcal{I} \\cup \\mathcal{A} \\cup \\{B\\}$, while all roads going out from $B$ end in the cities from $\\mathcal{I} \\cup \\mathcal{B}$. Therefore,\n\n$$\na = |\\mathcal{I}| + |\\mathcal{A}| + 1, \\quad b = |\\mathcal{I}| + |\\mathcal{B}|, \\quad \\text{and hence } a - b = |\\mathcal{A}| - |\\mathcal{B}| + 1.\n$$\n\nOn the other hand, since there are $|\\mathcal{A}| + 1$ short paths from $A$ to $B$ (including $A \\to B$) and $|\\mathcal{B}|$ short paths from $B$ to $A$, we infer\n\n$$\n|\\mathcal{Q}| = |\\mathcal{P}'| - (|\\mathcal{A}| + 1) + |\\mathcal{B}| \\geq N_{AB} + (b-a),\n$$\n\nas desired. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17575,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set\n\n$$\nS = \\{(x, y, z) \\mid x, y, z \\in \\{1, 2, \\dots, 2012\\}\\}\n$$\n\nof $2012^3$ points in three-dimensional space. For any segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in $S$, define its distance triplet as the ordered triple\n\n$$\n(|x_1 - x_2|,\\ |y_1 - y_2|,\\ |z_1 - z_2|).\n$$\n\nAlice wants to draw segments such that:\n\n- Each segment joins two distinct points in $S$.\n- Each point in $S$ is an endpoint of at most one segment.\n- For any two segments, their distance triplets are different.\n\nWhat is the greatest number of segments that Alice can draw?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that Alice can draw up to $K = \\frac{2012^3}{2}$ segments. Since there are $2012^3$ points, conditions (1) and (2) guarantee that Alice can draw at most $K$ segments. We will prove that she can do so.\n\nLet $T = \\{1, 2, \\dots, 2012\\}$. We will define a bijection $f: T \\to T$ such that for all distinct $i, j \\in T$, the inequality\n\n$$\n|f(i) - i| \\neq |f(j) - j|\n$$\n\nholds. We let\n\n$$\n\\begin{aligned}\n& (f(1), f(2), \\dots, f(2012)) \\\\\n= & (2012, 2011, \\dots, 1510, 504, 1509, 1508, \\dots, 1008, 1006, \\\\\n& \\qquad 1005, \\dots, 505, 503, 502, \\dots, 1, 1007).\n\\end{aligned}\n$$\n\nIt is not hard to verify that $f$ satisfies the desired inequality condition.\n\nFor each point $(x, y, z) \\in S$ such that $z \\leq 1006$, Alice draws a segment between it and the point $(f(x), f(y), 2013 - z)$. The $K$ segments she has drawn are easily seen to satisfy (1) and (2). To verify that they satisfy (3), suppose that two segments have the same distance triplets. Assume that the first segment has $(x_1, y_1, z_1)$ where $z_1 \\leq 1006$ as an endpoint, and the second segment has $(x_2, y_2, z_2)$ where $z_2 \\leq 1006$ as an endpoint. The distance triplets of the two segments are\n\n$$\n(|f(x_1) - x_1|, |f(y_1) - y_1|, |2013 - 2z_1|)\n$$\n\nand\n\n$$\n(|f(x_2) - x_2|, |f(y_2) - y_2|, |2013 - 2z_2|)\n$$\n\nrespectively. For them to coincide, we must have that\n\n$$\n(x_1, y_1, z_1) = (x_2, y_2, z_2),\n$$\n\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17576,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 1$. Show that\n\n$$\n\\frac{1 - a^2}{a + bc} + \\frac{1 - b^2}{b + ca} + \\frac{1 - c^2}{c + ab} \\ge 6.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{1-a^2}{a+bc} + \\frac{1-b^2}{b+ca} + \\frac{1-c^2}{c+ab} \\ge 6\n$$\n\nWe can rewrite each term as:\n$$\n\\frac{1-a^2}{a+bc} + 1 = \\frac{1-a^2 + a + bc}{a+bc} = \\frac{1 + a + bc - a^2}{a+bc}\n$$\nSumming all three terms:\n$$\n\\sum_{cyc} \\left( \\frac{1-a^2}{a+bc} + 1 \\right) = \\sum_{cyc} \\frac{1-a^2 + a + bc}{a+bc}\n$$\nFrom $a + b + c = 1$, we have $a - a^2 = ab + ac$, and similarly for $b$ and $c$. Thus,\n$$\n1 - a^2 + a + bc = 1 + (a - a^2) + bc = 1 + ab + ac + bc = 1 + ab + bc + ca\n$$\nSo,\n$$\n\\sum_{cyc} \\frac{1 + ab + bc + ca}{a + bc}\n$$\nTherefore,\n$$\n\\frac{1 + ab + bc + ca}{a + bc} + \\frac{1 + ab + bc + ca}{b + ca} + \\frac{1 + ab + bc + ca}{c + ab} \\geq 9\n$$\nThis is equivalent to\n$$\n[(a + bc) + (b + ca) + (c + ab)] \\left( \\frac{1}{a + bc} + \\frac{1}{b + ca} + \\frac{1}{c + ab} \\right) \\geq 9\n$$\nBy the Cauchy-Schwarz inequality, this holds. Equality occurs when $a + bc = b + ca = c + ab$, i.e., $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17577,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest real number $C$ such that the inequality\n\n$$\nC(x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \\geq x_1 x_2 x_3 x_4 x_5 \\left(x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125}\\right)^{16}\n$$\n\nholds for all positive real numbers $x_1, x_2, x_3, x_4, x_5$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n5 (x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \\geq (x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5) (x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000})\n$$\n\nby Chebyshev's inequality. Also,\n\n$$\nx_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5 \\geq 5x_1x_2x_3x_4x_5\n$$\n\nby AM-GM, and\n\n$$\n\\frac{x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000}}{5} \\geq \\left( \\frac{x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125}}{5} \\right)^{16}\n$$\n\nCombining these inequalities gives $C \\leq 5^{15}$. But substituting $x_1 = x_2 = x_3 = x_4 = x_5 = 1$ gives $C \\geq 5^{15}$. Thus $C = 5^{15}$.\n\nThis problem can also be solved by a direct application of Muirhead's Inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17578,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, orthocentre $H$, circumcircle $\\Gamma$ and circumcentre $O$. Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallelogram. Suppose that there exists a point $X$ on $\\Gamma$ and on the opposite side of $DH$ to $A$ such that $\\angle DXH + \\angle DHA = 90^\\circ$. Let $Y$ be the midpoint of $OX$. Prove that if $MY = OA$ then $OA = 2OH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ lie on $OM$ such that $\\angle EHD = 90^\\circ$ and let $N$ be the reflection of $O$ in $BC$.\n\n![](images/BMO2024Shortlist_p52_data_08809c15f3.png)\n\nSince $AH \\parallel DE$ we have:\n\n$$\n\\angle DEX = \\angle DEX = \\angle DEX = \\angle DEX\n$$\n\nSo $DXEH$ is cyclic—call this circle $\\omega$.\n\nIt's well-known that $AH = 2OM = ON$ so $NH = OA$ (which we'll use later) and $ON = AH = DO$ (as $ADOH$ is a parallelogram). This means $\\frac{OD}{ND} = \\frac{1}{2}$. Also, by considering homothety factor $2$ at $O$:\n\n$$\nNX = 2MY = 2OA = 2OX \\implies \\frac{OX}{NX} = \\frac{1}{2}\n$$\n\nSo, as the diameter of $\\omega$ lies on line $ON$, $\\omega$ is in fact a circle of Apollonius with foci at $O, N$. Using the above and that $NH = OA$, which we noted before:\n\n$$\n\\frac{1}{2} = \\frac{OH}{NH} = \\frac{OH}{OA} \\Rightarrow OA = 2OH\n$$\n\nwhich is what we wanted to prove. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17579,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a_1, a_2, \\cdots, a_n$ 為非負實數，且對任意正整數 $1 \\le k \\le n$，有：\n\n$$\na_1 a_2 \\cdots a_k \\ge \\frac{1}{(2k)!}.\n$$\n\n試證：\n\n$$\na_1 + a_2 + \\cdots + a_n \\ge \\frac{1}{n+1} + \\frac{1}{n+2} + \\cdots + \\frac{1}{2n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "將題意之左式改寫如下：\n\n$$\n\\begin{aligned}\na_1 + a_2 + \\cdots + a_n &= (1 - \\frac{1}{2})(1 \\cdot 2a_1) + (\\frac{1}{3} - \\frac{1}{4})(3 \\cdot 4a_2) \\\\\n&\\quad + \\cdots + (\\frac{1}{2n-1} - \\frac{1}{2n})((2n-1) \\cdot 2n a_n) \\\\\n&= (1 - \\frac{1}{2} - \\frac{1}{3} + \\frac{1}{4})(1 \\cdot 2a_1) \\\\\n&\\quad + (\\frac{1}{3} - \\frac{1}{4} - \\frac{1}{5} + \\frac{1}{6})(1 \\cdot 2a_1 + 3 \\cdot 4a_2) + \\cdots \\\\\n&\\quad + (\\frac{1}{2n-1} - \\frac{1}{2n})(1 \\cdot 2a_1 + 3 \\cdot 4a_2 + \\cdots + (2n-1) \\cdot 2n a_n).\n\\end{aligned}\n$$\n\n由算幾不等式與題目之條件，我們有：\n\n$$\n1 \\cdot 2a_1 \\ge 1, \\quad 1 \\cdot 2a_1 + 3 \\cdot 4a_2 \\ge 2, \\quad \\cdots, \\quad 1 \\cdot 2a_1 + 3 \\cdot 4a_2 + \\cdots + (2n-1)2n a_n \\ge n.\n$$\n\n故\n\n$$\n\\begin{aligned}\na_1 + a_2 + \\cdots + a_n &\\ge (1 - \\frac{1}{2} - \\frac{1}{3} + \\frac{1}{4}) + 2(\\frac{1}{3} - \\frac{1}{4} - \\frac{1}{5} + \\frac{1}{6}) + \\cdots \\\\\n&\\quad + n(\\frac{1}{2n-1} - \\frac{1}{2n}) \\\\\n&= 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\cdots + \\frac{1}{2n-1} - \\frac{1}{2n} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\cdots + \\frac{1}{2n-1} + \\frac{1}{2n} \\\\\n&\\quad - 2\\left(\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\cdots + \\frac{1}{2n}\\right) \\\\\n&= \\frac{1}{n+1} + \\frac{1}{n+2} + \\cdots + \\frac{1}{2n}.\n\\end{aligned}\n$$\n\n故得證！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17580,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute-angled scalene triangle $ABC$. The angle bisector of angle $BAC$ and the perpendicular bisectors of the sides $AB$, $AC$ define a triangle. Prove that its orthocenter lies on the median from vertex $A$.\n\n![](images/CZE_ABooklet_2023_p9_data_286b53cadb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $AB$, $N$ the midpoint of $AC$, and let $K$ and $L$ be the intersections of the angle bisector of $CAB$ with the perpendicular bisectors of $AB$ and $AC$, respectively. Let $O$ be the intersection of the perpendicular bisectors of $AB$ and $AC$. The triangle $KLO$ is the triangle from the problem statement, and its orthocenter is denoted by $H$.\n\nWe need to prove that $H$ lies on the median from vertex $A$ of triangle $ABC$. It is sufficient to show that triangles $ABH$ and $ACH$ have the same area.\n\nSince $HL \\perp OK \\perp AB$, we have $HL \\parallel AB$. Therefore, $H$ and $L$ have the same distance from the line $AB$. This distance equals the length of segment $LN$, since $L$ lies on the angle bisector of $CAB$ and $N$ is the perpendicular projection of $L$ onto $AC$. Thus, the area of $ABH$ is $\\frac{1}{2}|AB| \\cdot |LN|$. Similarly, the area of $ACH$ is $\\frac{1}{2}|AC| \\cdot |KM|$.\n\nIt remains to prove $|AB| \\cdot |LN| = |AC| \\cdot |KM|$.\n\nFor the points $K$ and $L$ lying on the angle bisector of $CAB$, we have $|\\angle MAK| = |\\angle NAL|$. The right triangles $AKM$ and $ALN$ are therefore similar, so $|KM| : |AM| = |LN| : |AN|$. Since $|AM| = \\frac{1}{2}|AB|$ and $|AN| = \\frac{1}{2}|AC|$, we get $|KM| : |AB| = |LN| : |AC|$, i.e., $|AB| \\cdot |LN| = |AC| \\cdot |KM|$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17581,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that $r = 2$ is the largest real number $r$ which satisfies the following condition:\n\nIf a sequence $a_1, a_2, \\dots$ of positive integers fulfills the inequalities\n\n$$\na_n \\le a_{n+2} \\le \\sqrt{a_n^2 + r a_{n+1}}\n$$\n\nfor every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2} = a_n$ for every $n \\ge M$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, assume that $r > 2$, and take a positive integer $a \\ge \\frac{1}{r-2}$. Let $a_n = a + \\lfloor n/2 \\rfloor$ for $n = 1, 2, \\dots$. The sequence $a_n$ satisfies the inequalities\n\n$$\n\\sqrt{a_n^2 + r a_{n+1}} \\ge \\sqrt{a_n^2 + r a_n} \\ge \\sqrt{a_n^2 + \\left(2 + \\frac{1}{a}\\right) a_n} \\ge a_n + 1 = a_{n+2},\n$$\n\nbut since $a_{n+2} > a_n$ for any $n$, $r$ does not satisfy the condition given in the problem.\n\nNow, show that $r = 2$ does satisfy the condition. Suppose $a_1, a_2, \\dots$ is a sequence of positive integers satisfying the inequalities, and there exists a positive integer $m$ for which $a_{m+2} > a_m$.\n\nBy induction, prove the following assertion:\n\n($\\dagger$) $a_{m+2k} \\le a_{m+2k-1} = a_{m+1}$ holds for every positive integer $k$.\n\nThe truth of ($\\dagger$) for $k=1$ follows from the inequalities below:\n\n$$\n2a_{m+2} - 1 = a_{m+2}^2 - (a_{m+2} - 1)^2 \\le a_m^2 + 2a_{m+1} - (a_{m+2} - 1)^2 \\le 2a_{m+1}.\n$$\n\nAssume ($\\dagger$) holds for some $k$. Then\n\n$$\na_{m+1}^2 \\le a_{m+2k+1}^2 \\le a_{m+2k-1}^2 + 2a_{m+2k} \\le a_{m+1}^2 + 2a_{m+1} < (a_{m+1} + 1)^2,\n$$\n\nso $a_{m+2k+1} = a_{m+1}$. Furthermore, since $a_{m+2k} \\le a_{m+1}$,\n\n$$\na_{m+2k+2}^2 \\le a_{m+2k}^2 + 2a_{m+2k+1} \\le a_{m+1}^2 + 2a_{m+1} < (a_{m+1} + 1)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17582,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all four-digit numbers $\\overline{abcd}$ such that $\\overline{ab}$, $\\overline{cb}$, and $d$ are prime numbers, and $$\\overline{ab}^2 + \\overline{cb}^2 + d^2 = 2022.$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $c = 0$, then $b^2 + d^2 \\leq 2 \\cdot 7^2$, which means $\\overline{ab}$ must be a prime number such that $1924 \\leq \\overline{ab}^2 < 2022$, which is not possible. Therefore, $c \\neq 0$.\n\nSince $\\overline{ab}$ and $\\overline{cb}$ are primes, from the given equality we deduce $d = 2$. Thus, $\\overline{ab}^2 + \\overline{cb}^2 = 2018 < 47^2$, so $\\overline{ab}$ 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 $b \\in \\{3, 7\\}$.\n\nIf $b = 7$, then $\\overline{ab}, \\overline{cb} \\in \\{17, 37\\}$, so $\\overline{ab}^2, \\overline{cb}^2 \\in \\{289, 1369\\}$ and $\\overline{ab}^2 + \\overline{cb}^2 \\neq 2018$.\n\nIf $b = 3$, then $\\overline{ab}, \\overline{cb} \\in \\{13, 23, 43\\}$. Checking, only $(\\overline{ab}, \\overline{cb}) \\in \\{(13, 43), (43, 13)\\}$ satisfy the equation. Therefore, the solutions are $\\overline{abcd} = 1342$ and $\\overline{abcd} = 4312$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17583,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to express every positive integer $n$ congruent to $9$ modulo $25$ in the form\n$$\nn = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2},\n$$\nwhere $a$, $b$, $c$ are non-negative integers that do not share parity?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. Alternatively, we show that if $n$ is a positive integer congruent to $9$ modulo $25$, then\n$$\nN = 8n + 3 = (2a+1)^2 + (2b+1)^2 + (2c+1)^2\n$$\nfor some non-negative integers $a$, $b$, $c$ that do not share parity; that is, $N$ is a sum of three odd squares whose positive square roots are not congruent modulo $4$. This is a special case of the following fact:\n\n(*) 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 $n \\equiv 3(r^2 - 1)/8 \\pmod{r^2}$ is expressible in the form\n$$\nn = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}\n$$\nfor some non-negative integers $a$, $b$, $c$ that do not share parity.\n\nThe problem at hand is the special case where $(p, q, r) = (3, 4, 5)$.\n\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).\n\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 $(ru)^2 + (rv)^2 = (pu+qv)^2 + (qu-pv)^2$.\n\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.\n\nThere are, of course, infinitely many primitive Pythagorean triples satisfying the conditions in (*). For instance, $p = |4m + 1|$, $q = 4m(2m + 1)$, $r = 4m(2m + 1) + 1$, where $m$ runs through the non-zero integers. Thus, every positive integer $n \\equiv 3m(2m + 1)(4m^2 + 2m + 1) \\pmod{8m^2 + 4m + 1}$ is expressible in the form\n$$\nn = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}\n$$\nfor some non-negative integers $a$, $b$, $c$ that do not share parity. The problem at hand is the special case where $m = 1$. Similarly, let $m = -1$, to infer that every positive integer $n \\equiv 63 \\pmod{169}$ 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.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 17584,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku and Miku are playing the following game. In the beginning, there is a positive integer on the board. Each turn, a player subtracts from the number on the board a non-zero digit that appears in his or his opponent's ID code, and replaces the number on the board with the result. Players take turns, Juku starts. The player whose move ends up in a negative number on the board loses. Prove that, among any 10 consecutive positive integers, there is a number $n$ such that, if initially the number $n$ is on the board, then Juku can win the game regardless of his opponent's counterplay.\n\n**Note:** An identity code is a certain finite sequence of digits. Each two people have distinct ID codes, but they are made up of the same number of digits.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, a+1, \\dots, a+9$ be 10 arbitrary consecutive positive integers. If there exists a number $n$ among $a, a+1, \\dots, a+8$ for which Juku has a winning strategy, then we are done. We will now assume that if any of the numbers $a, a+1, \\dots, a+8$ is on the board, then the active player loses if his opponent plays perfectly. Then, if $n = a+9$, Juku can start by subtracting any non-zero digit, after which the number on the board is one of $a, a+1, \\dots, a+8$, putting Miku in a losing position, which means that Juku will win. There must be a non-zero digit in either code, as both codes cannot be strings of zeroes, as they must be distinct and have the same length.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17585,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n \\geq 3$, let $\\phi_n$ be the set of all positive integers less than $n$ and coprime to $n$. Consider the polynomial\n$$\nP_n(x) = \\sum_{k \\in \\phi_n} x^{k-1}.\n$$\n\na) Prove that $P_n(x)$ is divisible by $x^{r_n} + 1$ for some integer $r_n > 0$.\n\nb) Find all $n$ such that $P_n(x)$ is irreducible over $\\mathbb{Z}[x]$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we see that for $m, k \\in \\mathbb{Z}^+$ and $k$ odd, $x^{km} + 1$ is divisible by $x^m + 1$. For convenience, consider the polynomial $Q_n(x) = xP_n(x) = \\sum_{k \\in A_n} x^k$. We just need to prove $Q_n(x)$ is divisible by $x^r + 1$ for some positive integer $r$.\n\n* If $n$ is odd: Because $\\gcd(n, k) = \\gcd(n, n-k)$ for all $k = 1, 2, \\dots, n-1$, for each $k \\in A_n$, $n-k \\in A_n$. Thus, we can partition the terms in $P_n(x)$ into pairs of the form $(x^k, x^{n-k})$. Since $k$ and $n-k$ have different parity, $(n-k)-k = n-2k$ is odd and $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$ is divisible by $x+1$. Therefore, $Q_n(x)$ is divisible by $x+1$.\n\n* If $4 \\mid n$: Partition into pairs $(x^k, x^{n-k})$ such that $n-2k \\equiv 2 \\pmod{4}$. Thus, $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$ is divisible by $x^2 + 1$. Hence, $Q_n(x)$ is divisible by $x^2 + 1$.\n\nOtherwise, consider $n \\equiv 2 \\pmod{4}$. We will prove the following lemmas to finish the problem.\n\n**Lemma 1.** If $Q_n(x)$ is divisible by $x^r + 1$ ($r \\in \\mathbb{Z}^+$), then for any odd prime $p$ with $p \\nmid n$, the polynomial $Q_{pn}(x)$ is also divisible by $x^r + 1$.\n\n*Proof.* Take $a \\in A_n$, then $kp + a \\in A_{pn}$ for $k = 0, 1, \\dots, p-1$. For each $k$, all terms with exponents of the form $kp + a$ and $a \\in A_n$ share the common term $x^{kp}$, and their sum will be divisible by $\\sum_{a \\in A} x^a$, which is also divisible by $x^r + 1$. Thus, $Q_{pn}(x)$ is divisible by $x^r + 1$. $\\square$\n\n**Lemma 2.** If $n = 2p_1p_2\\cdots p_m$, where $m \\in \\mathbb{Z}^+$ and $p_1, p_2, \\dots, p_m$ are distinct odd primes, then $Q_n(x)$ will be reducible.\n\n*Proof.* We prove this by induction on $m$, the number of odd primes in $n$. If $m=1$, then for $n=2p$ where $p$ is a prime, $A_n = \\{1, 2, \\dots, 2p\\} \\setminus \\{p, 2p\\}$, so $Q_n(x)$ is divisible by $x^p + 1$.\n\nAssume the lemma is true for $m \\ge 1$. Denote $n = 2p_1p_2\\cdots p_m$ and let $p$ be an odd prime with $\\gcd(p, n) = 1$, where $Q_n(x)$ is divisible by $x^r + 1$. We will show that for $N = pn$, $x^r + 1 \\mid Q_N(x)$.\n\nThere are exponents coprime with $n$ and divisible by $p$ in $A_n$; denote the sum of these terms by $R_N(x)$. This polynomial is $p$ times the repetition of $Q_n(x)$ (i.e., $R_N(x)$ can be divided into groups, each differing from $Q_N(x)$ by a term of the form $x^{kp}$ for $1 \\le k \\le p-1$). Hence, $R_N(x)$ is divisible by $x^r + 1$.\n\nNext, remove the exponents that are multiples of $p$, which have the form $x^{ap}$ where $a \\in A_n$. The sum of all these terms is a polynomial of the form $Q_n(x^p)$, which is divisible by $x^{pr+1}$ and also by $x^r + 1$. Hence, $Q_N(x) = R_N(x) - Q_n(x^p)$ is divisible by $x^r + 1$. Therefore, the statement is also true for $n$ with $m+1$ odd primes, and the lemma is proved. $\\square$\n\nb) From the previous part, it suffices to find all positive integers $n \\ge 3$ such that $|A_n| = 2$, which is equivalent to finding $n \\ge 3$ with $\\varphi(n) = 2$. We consider the following cases.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17586,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}^*$ be the set of all positive integers. Prove that there exists a unique function $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ satisfying $f(1) = f(2) = 1$ and\n$$\nf(n) = f(f(n-1)) + f(n - f(n-1)), \\quad n = 3, 4, \\dots\n$$\nFor such $f$, find the value of $f(2^m)$ for integer $m \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction to show the existence and uniqueness of $f$ and to find $f(2^m)$.\n\nSince $f(1) = 1$, we have $\\frac{1}{2} \\leq f(1) \\leq 1$.\n\n**Existence and Uniqueness:**\n\nAssume for $n = 2$, $f(2) = 1$, which satisfies the claim.\n\nSuppose for all $k < n$ ($n \\geq 3$), $f(k)$ is uniquely determined and $\\frac{k}{2} \\leq f(k) \\leq k$. Then $1 \\leq f(n-1) \\leq n-1$, so $1 \\leq n - f(n-1) \\leq n-1$. By the induction hypothesis, $f(f(n-1))$ and $f(n - f(n-1))$ are determined, so\n$$\nf(n) = f(f(n-1)) + f(n - f(n-1))\n$$\nis uniquely determined. Furthermore,\n$$\n\\frac{1}{2}f(n-1) \\leq f(f(n-1)) \\leq f(n-1), \\\\\n\\frac{1}{2}(n - f(n-1)) \\leq f(n - f(n-1)) \\leq n - f(n-1).\n$$\nThus,\n$$\n\\frac{n}{2} \\leq f(n) \\leq n.\n$$\nBy induction, such a unique $f$ exists.\n\n**Monotonicity:**\n\nWe show by induction that $f(n+1) - f(n) \\in \\{0, 1\\}$ for all $n \\geq 1$.\n\nFor $n = 1$, this is clear. Assume true for $n \\leq k$. Then,\n$$\n\\begin{align*}\nf(k+2) - f(k+1) &= [f(f(k+1)) + f(k+2 - f(k+1))] - [f(f(k)) + f(k+1 - f(k))] \\\\\n&= [f(f(k+1)) - f(f(k))] + [f(k+2 - f(k+1)) - f(k+1 - f(k))].\n\\end{align*}\n$$\nIf $f(k+1) = f(k) + 1$, then $f(f(k+1)) - f(f(k)) \\in \\{0, 1\\}$. If $f(k+1) = f(k)$, then $f(k+2 - f(k)) - f(k+1 - f(k)) \\in \\{0, 1\\}$. Thus, $f(n+1) - f(n) \\in \\{0, 1\\}$ for all $n$.\n\n**Value at Powers of 2:**\n\nWe show by induction that $f(2^m) = 2^{m-1}$ for all $m \\geq 1$.\n\nFor $m = 1$, $f(2) = 1 = 2^{1-1}$.\n\nAssume $f(2^k) = 2^{k-1}$. For $m = k+1$, suppose $f(2^{k+1}) \\neq 2^k$. Since $f(2^{k+1}) \\geq 2^k + 1$ and $f$ increases by at most $1$ at each step, let $n$ be the smallest integer such that $f(n) = 2^k + 1$, so $n \\leq 2^{k+1}$ and $f(n-1) = 2^k$. Then $n - 2^k \\leq 2^k$, so\n$$\n2^k + 1 = f(n) = f(f(n-1)) + f(n - f(n-1)) = f(2^k) + f(n - 2^k) \\leq 2f(2^k) = 2^k,\n$$\nwhich is a contradiction. Thus, $f(2^{k+1}) = 2^k$.\n\nTherefore, $f(2^m) = 2^{m-1}$ for all $m \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17587,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $\\omega$, $\\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\\omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent externally to $\\omega$. Circle $\\Omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent internally to $\\omega$. Let $P_A$ and $Q_A$ denote the centers of $\\omega_A$ and $\\Omega_A$, respectively. Define points $P_B, Q_B, P_C, Q_C$ analogously. Prove that\n\n$$\n8P_AQ_A \\cdot P_BQ_B \\cdot P_CQ_C \\le R^3,\n$$\n\nwith equality if and only if triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the incircle touch the sides $AB$, $BC$, and $CA$ at $C_1$, $A_1$, and $B_1$, respectively. Set $AB = c$, $BC = a$, $CA = b$. By equal tangents, we may assume that $AB_1 = AC_1 = x$, $BC_1 = BA_1 = y$, and $CA_1 = CB_1 = z$. Then $a = y + z$, $b = z + x$, $c = x + y$. By the AM-GM inequality, we have $a \\ge 2\\sqrt{yz}$, $b \\ge 2\\sqrt{zx}$, and $c \\ge 2\\sqrt{xy}$. Multiplying the last three inequalities yields\n\n$$\nabc \\ge 8xyz, \\qquad (\\dagger)\n$$\n\nwith equality if and only if $x = y = z$; that is, triangle $ABC$ is equilateral.\n\nLet $k$ denote the area of triangle $ABC$. By the Extended Law of Sines, $c = 2R \\sin \\angle C$. Hence\n\n$$\nk = \\frac{ab \\sin \\angle C}{2} = \\frac{abc}{4R} \\quad \\text{or} \\quad R = \\frac{abc}{4k}. \\qquad (\\ddagger)\n$$\n\nWe are going to show that\n\n$$\nP_A Q_A = \\frac{xa^2}{4k}. \\qquad (*)\n$$\n\nIn exactly the same way, we can also establish its cyclic analogous forms\n\n$$\nP_B Q_B = \\frac{yb^2}{4k} \\quad \\text{and} \\quad P_C Q_C = \\frac{zc^2}{4k}.\n$$\n\nMultiplying the last three equations together gives\n\n$$\nP_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{xyz a^2 b^2 c^2}{64k^3}.\n$$\n\nFurther considering $(\\dagger)$ and $(\\ddagger)$, we have\n\n$$\n8P_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{8xyz a^2 b^2 c^2}{64k^3} \\le \\frac{a^3 b^3 c^3}{64k^3} = R^3,\n$$\n\nwith equality if and only if triangle $ABC$ is equilateral.\n\nHence it suffices to show $(*)$. Let $r$, $r_A$, $r'_A$ denote the radii of $\\omega$, $\\omega_A$, $\\Omega_A$, respectively. We consider the inversion $I$ with center $A$ and radius $x$. Clearly, $I(B_1) = B_1$, $I(C_1) = C_1$, and $I(\\omega) = \\omega$. Let ray $AO$ intersect $\\omega_A$ and $\\Omega_A$ at $S$ and $T$, respectively. It is not difficult to see that $AT > AS$, because $\\omega$ is tangent to $\\omega_A$ and $\\Omega_A$ externally and internally, respectively. Set $S_1 = I(S)$ and $T_1 = I(T)$. Let $\\ell$ denote the line tangent to $\\Omega$ at $A$. Then the image of $\\omega_A$ (under the inversion) is the line (denoted by $\\ell_1$) passing through $S_1$ and parallel to $\\ell$, and the image of $\\Omega_A$ is the line (denoted by $\\ell_2$) passing through $T_1$ and parallel to $\\ell$. Furthermore, since $\\omega$ is tangent to both $\\omega_A$ and $\\Omega_A$, $\\ell_1$ and $\\ell_2$ are also tangent to the image of $\\omega$, which is $\\omega$ itself. Thus the distance between these two lines is $2r$; that is, $S_1T_1 = 2r$. Hence we can consider the following configuration. (The darkened circle is $\\omega_A$, and its image is the darkened line $\\ell_1$.)\n\n![](images/USA_IMO_2007-2008_p10_data_c2ccf390ae.png)\n\nBy the definition of inversion, we have $AS_1 \\cdot AS = AT_1 \\cdot AT = x^2$. Note that $AS = 2r_A$, $AT = 2r'_A$, and $S_1T_1 = 2r$. We have\n\n$$\nr_A = \\frac{x^2}{2AS_1}, \\quad \\text{and} \\quad r'_A = \\frac{x^2}{2AT_1} = \\frac{x^2}{2(AS_1 - 2r)}.\n$$\n\nHence\n\n$$\nP_A Q_A = AQ_A - AP_A = r'_A - r_A = \\frac{x^2}{2} \\left( \\frac{1}{AS_1 - 2r} + \\frac{1}{AS_1} \\right).\n$$\n\nLet $H_A$ be the foot of the perpendicular from $A$ to side $BC$. It is well known that $\\angle BAS_1 = \\angle BAO = 90^\\circ - \\angle C = \\angle CAH_A$. Since ray $AI$ bisects $\\angle BAC$, it follows that rays $AS_1$ and $AH_A$ are symmetric with respect to ray $AI$. Further note that both line $\\ell_1$ (passing through $S_1$) and line $BC$ (passing through $H_A$) are tangent to $\\omega$. We conclude that $AS_1 = AH_A$. In light of this observation and using the fact $2k = AH_A \\cdot BC = (AB + BC + CA)r$, we can compute $P_A Q_A$ as follows:\n\n$$\n\\begin{align*}\nP_A Q_A &= \\frac{x^2}{2} \\left( \\frac{1}{AH_A - 2r} - \\frac{1}{AH_A} \\right) = \\frac{x^2}{4k} \\left( \\frac{2k}{AH_A - 2r} - \\frac{2k}{AH_A} \\right) \\\\\n&= \\frac{x^2}{4k} \\left( \\frac{1}{\\frac{1}{BC} - \\frac{2}{AB+BC+CA}} - BC \\right) = \\frac{x^2}{4k} \\left( \\frac{1}{\\frac{1}{y+z} - \\frac{1}{x+y+z}} - (y+z) \\right) \\\\\n&= \\frac{x^2}{4k} \\left( \\frac{(y+z)(x+y+z)}{x} - (y+z) \\right) \\\\\n&= \\frac{x(y+z)^2}{4k} = \\frac{xa^2}{4k},\n\\end{align*}\n$$\n\nestablishing $(*)$. Our proof is complete.\n\n_Remark:_ Trigonometric solutions of $(*)$ are also possible.\n\nFor a given triangle, how can one construct $\\omega_A$ and $\\Omega_A$ by ruler and compass?",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17588,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function\n\n$$\nf(x) = (1 + x)(1 + x^2) \\cdots (1 + x^n)\n$$\n\nLet $A_n$ denote the sum of the coefficients of $x^k$ in $f(x)$, for all $k$. Given $n = 2009$, compute $A_n$ using the following approach:\n\nLet $\\varepsilon = \\cos \\frac{2\\pi}{n} + i \\sin \\frac{2\\pi}{n}$, and consider the sum\n\n$$\nf(\\varepsilon) + f(\\varepsilon^2) + \\cdots + f(\\varepsilon^n)\n$$\n\nExpress $A_n$ in terms of $n$ and $f(\\varepsilon^k)$, and find a formula for $A_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\varepsilon$ be a primitive $n$th root of unity. Since $\\varepsilon + \\varepsilon^2 + \\cdots + \\varepsilon^n = 0$, we have\n\n$$\n\\sum_{k=1}^{n} f(\\varepsilon^k) = n \\cdot A_n.\n$$\n\nFor each $d$ dividing $n$, the number of $k$ with $\\gcd(k, n) = d$ is $\\varphi\\left(\\frac{n}{d}\\right)$. For such $k$, $f(\\varepsilon^k) = f(\\varepsilon^d)$. Now,\n\n$$\nf(\\varepsilon^d) = \\left[ (1 + \\varepsilon^d)(1 + \\varepsilon^{2d}) \\cdots (1 + \\varepsilon^{\\frac{n}{d} \\cdot d}) \\right]^1 = (1 + (-1)^{\\frac{n}{d} + 1})^d\n$$\n\nsince $(\\varepsilon^d)^{n/d} = 1$ and substituting $x = -1$ in $x^m - 1 = \\prod_{k=1}^m (x - \\varepsilon^k)$. Therefore,\n\n$$\nn \\cdot A_n = \\sum_{d|n} \\varphi\\left(\\frac{n}{d}\\right) \\cdot \\left(1 + (-1)^{\\frac{n}{d} + 1}\\right)^d.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17589,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive number $m$ such that there exists a unique function $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n$$\nf(x + f(y)) = f(x) + \\frac{mx^5 f(y)}{f(x)^2} + f(y^2)\n$$\nfor all $x, y > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exists $y_0$ such that $f(y_0) < y_0^2$. Taking $x = y_0^2 - f(y_0)$ gives\n$$\nf(x + f(y_0)) = f(y_0^2).\n$$\nFrom the functional equation,\n$$\nf(x) + \\frac{mx^5 f(y_0)}{f(x)^2} = 0,\n$$\nwhich is a contradiction. Thus, $f(y) \\geq y^2$ for all $y > 0$.\n\nLet $x = y = 1$:\n$$\nf(1 + f(1)) = 2f(1) + \\frac{m}{f(1)} \\geq (1 + f(1))^2.\n$$\nSince $f(1) \\geq 1$, we have $m \\geq f(1) + f(1)^3 \\geq 2$. So the least value of $m$ is $2$.\n\nFor $m = 2$, we can verify $f(1) = 1$ and\n$$\nf(x + f(y)) = f(x) + \\frac{2x^5 f(y)}{f(x)^2} + f(y^2), \\quad \\forall x, y > 0.\n$$\nSubstitute $y = 1$:\n$$\nf(x + 1) = f(x) + \\frac{2x^5}{f(x)^2} + 1, \\quad \\forall x > 0.\n$$\nLet $x = 1$:\n$$\nf(2) = f(1) + \\frac{2}{f(1)} + 1 = 4.\n$$\nContinuing for $x = 2, 3, \\dots$, by induction, $f(n) = n^2$ for all positive integers $n$.\n\nNow, put $x = n$ into the condition:\n$$\nf(x + n^2) = f(x) + \\frac{2x^5 n^2}{f(x)^2} + n^4 \\geq (x + n^2)^2.\n$$\nFactoring this inequality gives\n$$\n(f(x) - x^2) \\left(1 - \\frac{2x n^2 (f(x) + x^2)}{f(x)^2}\\right) \\geq 0, \\quad \\forall x > 0.\n$$\nIf $f(x) = x^2$ for all $x > 0$, this is a solution. Otherwise, if there exists $x_0$ such that $f(x_0) > x_0^2$, then\n$$\n1 - \\frac{2x_0 n^2 (f(x_0) + x_0^2)}{f(x_0)^2} \\geq 0, \\quad \\forall n \\in \\mathbb{Z}^+,\n$$\nwhich implies\n$$\nn^2 \\leq \\frac{f(x_0)^2}{2x_0(f(x_0) + x_0^2)}.\n$$\nThis is a contradiction as $n \\to +\\infty$. Thus, for $m = 2$, the function $f(x) = x^2$ is the unique solution.\n\nTherefore, the least value of $m$ is $2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17590,
    "subject": "Mathematics (Olympiad)",
    "question": "設有一組正整數，定義「好數字」與「壞數字」如下：\n\n- 若 $n$ 為好數字且 $n \\mid n'$，則 $n'$ 也為好數字。\n- 若 $rs$ 是壞數字，則 $r^2s$ 也是壞數字。\n- 若 $p > k$ 為質數且 $n \\geq k$ 是壞數字，則 $np$ 也是壞數字。\n\n設 $P_k(x)$ 為 $x$ 小於或等於 $k$ 的質因數所成集合。定義兩個數 $a, b$ 為相似，若且唯若 $P_k(a) = P_k(b)$。\n\n證明：若 $a, b$ 相似，則 $a, b$ 同為好數字或同為壞數字。亦即，若 $c \\geq k$ 與其某個倍數 $d$ 相似，則 $c, d$ 同為好數字或同為壞數字。",
    "options": [],
    "answer": "See solution",
    "solution": "*Claim 1.* 若 $n$ 為好數字且 $n \\mid n'$，則 $n'$ 為好數字。\n\n*Proof.* 若 $n'$ 為壞數字，表示甲可以進行 $n' \\to x$ 且 $x$ 為好數字。然而 $(n', x) = 1 \\Rightarrow (n, x) = 1$，但 $n$ 與 $x$ 都是好數字，此與 Claim C 相矛盾。\n\n*Claim 2.* 若 $rs$ 是壞數字，則 $r^2s$ 也是壞數字。\n\n*Proof.* $rs$ 是壞數字表示甲可以進行 $rs \\to x$ 且 $x$ 為好數字，但 $x$ 顯然與 $r^2s$ 互質，故由 $r^2s \\to x$ 知 $r^2s$ 是壞數字。\n\n*Claim 3.* 若 $p > k$ 為一質數且 $n \\geq k$ 是壞數字，則 $np$ 也是壞數字。\n\n*Proof.* 若否，則存在最小的壞數字 $n$，使得 $np$ 是好數字。以下歸謬：\n\n1. 由於 $n$ 是壞數字，甲可以進行 $n \\to x$，其中 $x$ 是好數字。易知 $(np, x) > 1$，否則 $np$ 會是壞數字，矛盾。但已知 $(n, x) = 1$，故 $p \\mid x$。令 $x = p^r y$，其中 $(p, y) = 1$。\n2. 注意到 $y = 1$ 是不可能的，因為若 $y = 1$，則 $x = p^r$；又 $(p, k) = 1$，故甲可進行 $x \\to k$，從而 $x$ 是個壞數字，矛盾。故 $y > 1$，因此必有最小的正整數 $\\alpha$ 使得 $y^\\alpha \\geq k$。\n3. 基於 $np$ 和 $y^\\alpha$ 互質而 $np$ 是好數字，由 Claim B 知 $y^\\alpha$ 必為壞數字。\n4. 由 $\\alpha$ 的最小性知 $y^\\alpha < ky < py = \\frac{x}{p^{r-1}} < \\frac{n}{p^{r-1}}$，故 $p^{r-1}y^\\alpha < n$。從而由 $n$ 的最小性知，$p^{r-1}y^\\alpha$ 必為好數字（因為 $x = p(p^{r-1}y^\\alpha)$ 是好數字）。同理可證，$p^{r-2}y^\\alpha, \\dots, y^\\alpha$ 也都必須是好數字。\n5. 但 $np$ 和 $y^\\alpha$ 都是好數字，由 Claim B 知 $(np, y^\\alpha) > 1$，此與 $(n, x) = 1$ 及 $(p, y) = 1$ 相矛盾。證畢。\n\n現在令 $P_k(x)$ 為 $x$ 小於或等於 $k$ 的質因數所成集合。以下稱兩個數 $a, b$ 為相似的，若且唯若 $P_k(a) = P_k(b)$。要證明原題，我們僅需證明：若 $a, b$ 相似，則 $a, b$ 同好同壞。注意到 $ab$ 同時與 $a$ 和 $b$ 相似，故這等價於：若 $c \\geq k$ 與其某個倍數 $d$ 相似，則 $c, d$ 同好同壞。\n\n*Proof* 若否，則存在最小的 $d_0$，使得其存在一因數 $c_0 \\geq k$，使得 $c_0, d_0$ 好壞不同。由 Claim 1 知，必然是 $c_0$ 壞而 $d_0$ 好。注意到 $d_0 > c_0$。\n\n故 $d_0/c_0$ 必然有質因數 $p$。顯然 $p \\mid d_0$。\n\n- 若 $p \\leq k$，則由相似性知 $p \\mid c_0$，故 $p^2 \\mid d_0$。如此一來，$d_0/p$ 必須是好的，否則由 Claim 2 知 $d_0$ 是壞的，矛盾。但這麼一來，$d_0/p$ 與 $c_0$ 相似，為 $c_0$ 的倍數且是好數字，這與 $d_0$ 的最小性不合，矛盾。\n- 若 $p > k$，則由 Claim 3 知 $d_0$ 是壞數字，矛盾。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17591,
    "subject": "Mathematics (Olympiad)",
    "question": "In a rectangle with dimensions $2 \\times 3$, there is a polyline of length $36$, which can have self-intersections. Show that there exists a line parallel to two sides of the rectangle, which intersects the other two sides in their interior points and intersects the polyline in fewer than $10$ points.\n\n![](images/cps_solutions_day2_p1_data_9f442fb91d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an arbitrary line segment of the polyline and denote by $d$ its length and by $x$ and $y$ the lengths of its perpendicular projections on the sides of lengths $2$ and $3$, respectively. The Cauchy-Schwarz inequality gives us\n\n$$\n(2x + 3y)^2 \\leq (2^2 + 3^2)(x^2 + y^2) = 13d^2,\n$$\n\nwhich means $2x + 3y \\leq d \\cdot \\sqrt{13}$. Denote by $X$ and $Y$ the total length of all the perpendicular projections of all the line segments on the sides of lengths $2$ and $3$, respectively. Summing up our estimations for each line segment gives us $2X + 3Y \\leq 36 \\cdot \\sqrt{13} < 130$. But then either $2X < 40$, or $3Y < 90$. In the first case, we would have $X < 20$, so on the side of length $2$ there is a point that is contained in fewer than $10$ projections. A line perpendicular to this side at this point intersects the polyline at most $9$ times. The other case is analogous.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17592,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\angle CAB = \\alpha$ and $\\angle ABC = \\beta$ with $\\alpha > \\beta$.\n\n![](images/2019_Australian_Scene_W1_p152_data_0a056e9177.png)\n\nSuppose $I$ is the incentre of $\\triangle ABC$ and $A, X, B, N, C$ are concyclic. Prove that the point $S$ where the angle bisector of $\\angle DAB$ meets $BC$ also lies on the bisector of $\\angle CXB$, i.e., $X, S, N$ are collinear. Furthermore, show that $S$ is the single point where the circle through $A$, $B$, $I$ meets $BC$ and the intersection of the angle bisectors of $\\angle DAB$ and $\\angle CXB$ on $BC$.\n\n![](images/2019_Australian_Scene_W1_p152_data_4575559d37.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $AI$ to meet the circumcircle of $ABC$ at $N$. By the incentre-excentre lemma, $NB = NC = NI$. Since $NB = NC$ and $X$ lies on the circumcircle of $\\triangle ABC$, $XN$ is the angle bisector of $\\angle CXB$.\n\nLet $S$ be the point where the angle bisector of $\\angle DAB$ meets $BC$. We show that $S$ also lies on the bisector of $\\angle CXB$, i.e., $X, S, N$ are collinear.\n\n![](images/2019_Australian_Scene_W1_p153_data_06d40fe82a.png)\n\nWe have:\n\n$$\n\\angle ASC = \\angle ABS + \\angle SAB = \\angle CAD + \\angle DAS = \\angle CAS\n$$\n\nThus, $\\triangle ASC$ is isosceles and $CS = CA$.\n\nNotice $\\angle ACX = \\angle ANX$ and $\\angle XAC = \\angle XIN$ (alternate segment theorem), so $\\triangle XAC \\sim \\triangle XIN$. There is a spiral symmetry centred at $X$ sending $AC$ to $IN$.\n\n![](images/2019_Australian_Scene_W1_p154_data_04ca3bcfd1.png)\n\nNow $CA = CS$, $NI = NB$, and $\\angle BNI = \\angle SCA$. Hence $\\triangle ACS \\sim \\triangle INB$, and these triangles are similarly oriented. Thus, the spiral symmetry centred at $X$ sending $AC$ to $IN$ sends $\\triangle ACS$ to $\\triangle INB$, so it sends $AS$ to $IB$. Therefore, $\\angle SXA = \\angle BXI$, which implies $\\angle BXS = \\angle BXI - \\angle SXI = \\angle SXA - \\angle SXI = \\angle IXA$, so:\n\n$$\n\\angle BXS = \\angle IXA = \\angle IAC = \\angle NAC = \\angle NXC = \\angle NXB\n$$\n\nTherefore, $X, S, N$ are collinear and $S$ lies on the bisector of $\\angle CXB$ as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17593,
    "subject": "Mathematics (Olympiad)",
    "question": "有一個 $m \\times m$ 個單位方格構成的桌子，在某些單位方格的中心點有一隻螞蟻。從時間 0 開始，每隻螞蟻都沿著一個平行於方格邊的方向，以速率 1 前進。過程中若有螞蟻相遇：\n\n1. 如果是兩隻正面相遇，則它們會一起順時針轉彎 $90^\\circ$，然後繼續以速率 1 前進；\n2. 如果是兩隻以垂直方向相遇，或超過兩隻以上的螞蟻相遇，則它們會繼續以原本的速度和方向前進。\n\n當螞蟻爬到桌子邊緣，它會從桌面摔落，不再回來。當最後一隻螞蟻摔落桌面時，我們說此時刻就是這群螞蟻的“末日”。\n\n考慮所有可能的螞蟻起始位置，試求末日發生的最晚可能時刻，或是證明並不一定會有末日。",
    "options": [],
    "answer": "See solution",
    "solution": "其中 (i) 的規定可以修改為：南北向正面相遇它們會順時針轉彎 $90^\\circ$，而東西向相遇它們會逆時針轉彎 $90^\\circ$。修改之後與修改之前相比，任何時間點所有螞蟻的位置並沒有改變，只是時間之前有東西向相遇的螞蟻交換彼此角色，所以這不影響末日發生的時間。修改之後所有螞蟻分成兩類：\n\n- (NE 類) 永遠向東或向北前進；\n- (SW 類) 永遠向西或向南前進。\n\n以座標 $(0,0)$ 代表桌子的 SW 角，$(m, m)$ 代表桌子的 NE 角。當時間為 $t$ 時：\n\n- 區域 $\\{(x, y) \\mid x + y \\le 1 + t\\}$ 沒有 (NE 類) 螞蟻；\n- 區域 $\\{(x, y) \\mid x + y \\ge 2m - 1 - t\\}$ 沒有 (SW 類) 螞蟻。\n\n所以 $t = m - 1$ 時，是螞蟻相遇的最後發生的時間，而且只會相遇在 $x + y = m$ 的線上，此後所有螞蟻只會向前移動。$x + y = m$ 的線上分別向著東南西北四個方向前進，最多花 $m/2$ 單位時間，一定會到達桌子邊緣；所以合計得到 $3m/2 - 1$ 是末日的上界。\n\n時間為 $3m/2 - 1$ 是末日可能發生：螞蟻 A 位於 $(1/2, 1/2)$ 向北走，螞蟻 B 位於 $(1/2, m - 1/2)$ 向南走。在時間 $(m - 1)/2$ 時相遇於座標 $(1/2, m/2)$，之後 A 向東繼續走了 $m - 1/2$ 單位時間，來到桌子邊緣。時間 $3m/2 - 1$ 螞蟻末日發生了！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17594,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all odd values of $n$ such that, for $a = 1$, $b = 3$, $c = 5$, the number $B = a^{b+2} + b^{c+2} + c^{a+2}$ is divisible by $A_n = a^n + b^n + c^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Given $a = 1$, $b = 3$, $c = 5$:\n\n- $A_n = 1^n + 3^n + 5^n$\n- $B = 1^5 + 3^7 + 5^3 = 1 + 2187 + 125 = 2313$\n\nFor $n = 1$:\n\n$$A_1 = 1 + 3 + 5 = 9$$\n\n$2313 \\div 9 = 257$, so $A_1$ divides $B$.\n\nFor $n = 3$:\n\n$$A_3 = 1^3 + 3^3 + 5^3 = 1 + 27 + 125 = 153$$\n\n$2313 \\div 153 = 15$, so $A_3$ divides $B$.\n\nFor $n = 5$:\n\n$$A_5 = 1^5 + 3^5 + 5^5 = 1 + 243 + 3125 = 3369$$\n\n$A_5 > B$, so $A_5$ does not divide $B$.\n\nFor $n \\geq 5$, $A_n > B$, so $A_n$ does not divide $B$.\n\n**Answer:** All odd $n$ such that $n = 1$ or $n = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17595,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $(x, y, z)$ to the system of equations:\n\n$$\n\\begin{aligned}\nx^2 - y &= z^2 \\\\\ny^2 - z &= x^2 \\\\\nz^2 - x &= y^2\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By summing the given equations we get\n\n$$\nx + y + z = 0.\n$$\n\nIt follows that $z = -x - y$, and from the first equation we get\n\n$$\n\\begin{aligned}\nx^2 - y &= (-x - y)^2 \\\\\nx^2 - y &= x^2 + 2xy + y^2 \\\\\n2xy + y^2 + y &= 0 \\\\\ny(2x + y + 1) &= 0,\n\\end{aligned}\n$$\n\nso $y = 0$ or $2x + y + 1 = 0$.\n\nIf $y = 0$, it follows that $z = -x$ so from the second equation we get $x(x - 1) = 0$, hence $x = 0$ or $x = 1$. The corresponding solutions are $(x, y, z) = (0, 0, 0)$ and $(x, y, z) = (1, 0, -1)$.\n\nIf $2x + y + 1 = 0$, we have $y = -2x - 1$ so $z = -x - y = x + 1$. Now from the third equation we get $x(x + 1) = 0$, hence $x = 0$ or $x = -1$. The corresponding solutions are $(x, y, z) = (0, -1, 1)$ and $(x, y, z) = (-1, 1, 0)$.\n\nThus, the solutions of the given system of equations are\n\n$$\n(x, y, z) \\in \\{(0, 0, 0), (1, 0, -1), (0, -1, 1), (-1, 1, 0)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17596,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all periodic sequences $a_1, a_2, \\dots$ of real numbers such that for all $n \\ge 1$:\n\n$$\na_{n+2} + a_n^2 = a_n + a_{n+1}^2 \\quad \\text{and} \\quad |a_{n+1} - a_n| \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The sequences satisfying the conditions are:\n\n$$\nc, -c, c, -c, \\dots \\\\\nd, d, d, d, \\dots\n$$\n\nwhere $c \\in \\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$ and $d$ is any real number.\n\nWe rewrite the first condition as\n\n$$\na_{n+2} + a_{n+1} = (a_{n+1} + a_n)(a_{n+1} - a_n + 1)\n$$\n\nIf there exists $m$ such that $a_{m+1} + a_m = 0$, then from the equation above, $a_{n+1} + a_n = 0$ for all $n \\ge m$. Since $(a_{i+1} + a_i)$ is periodic, $a_{i+1} + a_i = 0$ for all $i$, so $(a_i)$ is of the form $c, -c, c, -c, \\dots$ with $|c| \\le \\frac{1}{2}$.\n\nIf $a_{n+1} + a_n \\ne 0$ for all $n$, let $T$ be the period. Then\n\n$$\n1 = \\prod_{i=1}^{T} \\frac{a_{i+2} + a_{i+1}}{a_{i+1} + a_i} = \\prod_{i=1}^{T} (a_{i+1} - a_i + 1)\n$$\n\nWith $|a_{i+1} - a_i| \\le 1$, we have $a_{i+1} - a_i + 1 > 0$. By AM-GM,\n\n$$\n1 = \\prod_{i=1}^{T} (a_{i+1} - a_i + 1) \\le \\left( \\frac{\\sum_{i=1}^{T} (a_{i+1} - a_i + 1)}{T} \\right)^T = 1.\n$$\n\nEquality holds, so $a_2 - a_1 = a_3 - a_2 = \\dots = a_{T+1} - a_T$, i.e., $(a_i)$ is constant. Thus, all solutions are as listed above. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17597,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive odd integer $n$, prove that\n\n$$\n\\left[ \\frac{1}{2} + \\sqrt{n + \\frac{1}{2}} \\right] = \\left[ \\frac{1}{2} + \\sqrt{n + \\frac{1}{2020}} \\right],\n$$\n\nwhere $[a]$ denotes the integer part of the real number $a$.",
    "options": [],
    "answer": "See 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 $n$ being odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17598,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ be a diameter of a circle $\\omega$ with centre $O$. From an arbitrary point $M$ on $AB$ such that $MA < MB$, we draw the circles $\\omega_1$ and $\\omega_2$ with diameters $AM$ and $BM$ respectively. Let $CD$ be an exterior common tangent of $\\omega_1$ and $\\omega_2$ such that $C$ belongs to $\\omega_1$ and $D$ belongs to $\\omega_2$. The point $E$ is diametrically opposite to $C$ with respect to $\\omega_1$, and the tangent to $\\omega_1$ at the point $E$ intersects $\\omega_2$ at the points $F$ and $G$. If the line of the common chord of the circumcircles of the triangles $CED$ and $CFG$ intersects the circle $\\omega$ at the points $K$ and $L$, and the circle $\\omega_2$ at the point $N$ (with $N$ closer to $L$), then prove that $KC = NL$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the second intersection of the circumcircles of the triangles $CED$ and $CFG$.\n\nFirst, we will prove that $E$, $M$, $D$ are collinear. Indeed, if the common tangent of $\\omega_1$ and $\\omega_2$ at $M$ intersects $CD$ at $S$, then $SC = SD = SM$, so $\\angle CMD = 90^\\circ$ and also $\\angle CME = 90^\\circ$, so $E$, $M$, $D$ are collinear.\n\n![](images/shortlistBMO2015_p16_data_cf4efb5831.png)\n\nUsing the power of the point $D$ to the circle $\\omega_1$, we get that\n\n$$\nDC^2 = DM \\cdot DE. \\qquad (12)\n$$\n\nFrom the cyclic quadrilateral $MFBG$, one has that $\\angle DMB = \\angle DFG$, but the triangle $DFG$ is isosceles, so $\\angle DFG = \\angle DGF$. It follows that the triangles $DMG$ and $DFE$ are similar, which gives us\n\n$$\n\\frac{DM}{DG} = \\frac{DG}{DE} \\Rightarrow DG^2 = DM \\cdot DE. \\qquad (13)\n$$\n\nFrom (12) and (13), one gets $DG = DC$, and since $DF = DG$, we have that the point $D$ is the circumcentre of the triangle $CFG$. Nevertheless, the circumcentre of the triangle $CDE$ is the midpoint of $DE$, and we have just proved that $CM$ is perpendicular to $DE$, which is the line of the centres of the two circles, so $CM$ is the common chord of these circles.\n\nTo finish, let us denote by $T$ the midpoint of $KL$. Then $OT$ is the line joining the midpoints of the diagonals of the trapezoid $ACBN$, since it is parallel to the bases and $O$ is the midpoint of $AB$. This means that $CT = TN$, and since $TK = TL$, it follows that $CK = TK - TC = TL - TN = NL$, which is what we wanted to prove. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17599,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 表示所有正整數所成之集合。試求滿足下列條件的最大正整數 $k$：\n\n$N$ 可以被分割成 $k$ 個子集合 $A_1, A_2, \\dots, A_k$，使得對所有的整數 $n \\geq 15$ 與所有的 $i \\in \\{1, 2, \\dots, k\\}$，能夠在 $A_i$ 中找到兩個相異元素 $a, b$，且 $a + b = n$。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$k = 3$。\n\n例如，取：\n\n$$\nA_1 = \\{1, 2, 3\\} \\cup \\{3m \\mid m \\geq 4\\}, \\\\\nA_2 = \\{4, 5, 6\\} \\cup \\{3m - 1 \\mid m \\geq 4\\}, \\\\\nA_3 = \\{7, 8, 9\\} \\cup \\{3m - 2 \\mid m \\geq 4\\}.\n$$\n\n爲了驗證上述分割符合題意，首先觀察 $A_i$ 中兩相異元素 $a, b$ 之和的情形：\n\n(i) $a + b = n \\geq 1 + 12 = 13$，當 $i = 1$；\n\n(ii) $a + b = n \\geq 4 + 11 = 15$，當 $i = 2$；\n\n(iii) $a + b = n \\geq 7 + 10 = 17$，當 $i = 3$。\n\n所以，必須找 $A_3$ 中兩個相異元素其和爲 $15 (= 7 + 8)$、$16 (= 7 + 9)$。\n\n假設對某個 $k \\geq 4$，存在 $A_1, A_2, \\dots, A_k$ 滿足題意。顯然，$A_1, A_2, A_3, A_4 \\cup \\dots \\cup A_k$ 亦滿足題意，故可假設 $k = 4$。\n\n取 $B_i = A_i \\cap \\{1, 2, \\dots, 23\\}$，$i = 1, 2, 3, 4$。對任意 $i$ 與 10 個數 $15, 16, \\dots, 24$ 可表示為 $B_i$ 中兩個相異元素之和。因此，$B_i$ 至少有 5 個元素。因 $|B_1| + |B_2| + |B_3| + |B_4| = 23$，$|B_j| = 5$，對某個 $j$。令 $B_j = \\{x_1, x_2, x_3, x_4, x_5\\}$。$A_j$ 中兩相異元素之和可為 $15, 16, \\dots, 24$，且應為 $B_j$ 中任兩個元素之和，即 $4(x_1 + x_2 + x_3 + x_4 + x_5) = 15 + 16 + \\dots + 24 = 195$，且 4 可整除 195。此為矛盾！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17600,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove or disprove that there exist $2017$ consecutive positive integers that cannot be written as $a^2 + b^2$ where $a$ and $b$ are integers.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that such a sequence exists. First, we prove the following lemma.\n\n*Lemma.* Let $q$ be a prime such that $q \\equiv 3 \\pmod{4}$. If $n \\equiv q \\pmod{q^2}$, then $n$ cannot be written as the sum of two squares.\n\n*Proof.* Let $q$ be a prime such that $q \\equiv 3 \\pmod{4}$, and $n$ be an integer satisfying $n \\equiv q \\pmod{q^2}$. Let $k$ be the integer where $q = 4k + 3$.\n\nAssume to the contrary that $n$ can be written as $a^2 + b^2$ for some integers $a$ and $b$.\n\nSuppose $q \\mid a$. Then since $q \\mid a^2 + b^2$, we get $q \\mid b$ and so $q^2 \\mid a^2 + b^2$, contradicting our assumption. Hence $q \\nmid a$, and analogously, $q \\nmid b$.\n\nBy Fermat's little theorem we have\n\n$$\na^{q-1} \\equiv b^{q-1} \\equiv 1 \\pmod{q}. \\qquad (1)$$\n\nOn the other hand, we have $a^2 \\equiv -b^2 \\pmod{q}$, raising this to the $(2k+1)$-th power yields:\n\n$$a^{4k+2} \\equiv (a^2)^{2k+1} \\equiv (-b^2)^{2k+1} \\equiv -b^{4k+2} \\pmod{q}.$$ \n\nSince $q-1 = 4k+2$, this contradicts (1), thus $n$ cannot be written as the sum of two squares. $\\square$\n\nTo construct the required sequence, let $q_1, q_2, \\dots, q_{2017}$ be primes congruent to $3$ modulo $4$, and consider the following system of congruences:\n\n$$\n\\begin{array}{l}\n n \\equiv q_1 \\pmod{q_1^2} \\\\\n n + 1 \\equiv q_2 \\pmod{q_2^2} \\\\\n \\vdots \\\\\n n + 2016 \\equiv q_{2017} \\pmod{q_{2017}^2}\n\\end{array}\n$$\n\nSince all moduli are pairwise relatively prime, by the Chinese Remainder Theorem there exists a solution to this system modulo $\\prod_{i=1}^{2017} q_i^2$. Let $N > 0$ be a solution, then the lemma implies that $N, N+1, \\dots, N+2016$ cannot be written as the sum of two squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17601,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider positive integers $a, b, c$ that are side-lengths of a non-degenerate triangle and such that $\\gcd(a, b, c) = 1$ and the fractions\n\n$$\n\\frac{a^2 + b^2 - c^2}{a + b - c}, \\quad \\frac{b^2 + c^2 - a^2}{b + c - a}, \\quad \\frac{c^2 + a^2 - b^2}{c + a - b}\n$$\n\nare all integers. Prove that the product of the denominators of the three fractions is either a square or twice a square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z = a + b - c$, $x = b + c - a$, $y = c + a - b$ be the (positive) denominators. Then $a = \\frac{y + z}{2}$, $b = \\frac{x + z}{2}$, $c = \\frac{x + y}{2}$ and\n\n$$\na^2 + b^2 - c^2 = \\frac{1}{4}\\left((y + z)^2 + (x + z)^2 - (x + y)^2\\right) = \\frac{1}{2}\\left(z(z + x + y) - x y\\right),\n$$\n\nhence $z \\mid x y$ and likewise $y \\mid x z$ and $x \\mid y z$.\n\nFor a prime $p$, let $i_p$ be the largest exponent such that $p^{i_p} \\mid x y z$. It suffices to show that for all odd primes $p$ the corresponding $i_p$ is even. If $i_2$ is also even then $x y z$ is a square. Otherwise, it is twice a square.\n\nFix odd prime $p$ and consider the largest exponents $\\alpha, \\beta, \\gamma$ such that $p^{\\alpha} \\mid x$, $p^{\\beta} \\mid y$, $p^{\\gamma} \\mid z$. Without loss of generality, assume $\\min\\{\\alpha, \\beta, \\gamma\\} = \\gamma$. If $\\gamma > 0$ then $p$ divides each of $x, y, z$ and thus it divides each of $a, b, c$ ($p$ is odd), contradicting $\\gcd(a, b, c) = 1$. Therefore $\\gamma = 0$.\n\nFrom $x \\mid y z$ we infer $\\alpha \\leq \\beta$. Likewise, from $y \\mid x z$ we infer $\\beta \\leq \\alpha$. Hence $\\beta = \\alpha$ and $i_p = \\alpha + \\beta + \\gamma = 2\\alpha$ is an even number as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17602,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + f(y)) - f(x) = (x + f(y))^4 - x^4\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the given equation as\n\n$$\nf(x + f(y)) = (x + f(y))^4 - x^4 + f(x). \\tag{1}\n$$\n\nSet $x = -f(z)$ and $y = z$ in (1):\n\n$$\nf(0) = -(f(z))^4 + f(-f(z)), \\quad \\text{for all } z \\in \\mathbb{R}. \\tag{2}\n$$\n\nNow, set $x = -f(z)$ in (1) and use (2):\n\n$$\nf(f(y) - f(z)) = (f(y) - f(z))^4 - (f(z))^4 + f(-f(z)) = (f(y) - f(z))^4 + f(0).\n$$\n\nSo, if $t = f(y) - f(z)$, then $f(t) = t^4 + f(0)$. If $f$ takes any nonzero value, then every real number is a difference of two values of $f$.\n\nLet $f(a) = b \\neq 0$. Plug $y = a$ into the original equation:\n\n$$\nf(x + b) - f(x) = (x + b)^4 - x^4.\n$$\n\nSince $b \\neq 0$, the right side is a degree 3 polynomial in $x$ and attains all real values as $x$ varies, so the left side (a difference of $f$ values) also attains all real values. Thus, $f(t) = t^4 + f(0)$ for all $t \\in \\mathbb{R}$.\n\nAll functions of the form $f(x) = x^4 + k$ satisfy the equation. The zero function $f(x) \\equiv 0$ is also a solution.\n\n**Answer:** All such functions are $f(x) \\equiv 0$ and $f(x) = x^4 + k$ for any real number $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17603,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d = \\gcd(2012t+1, 2013t+1)$. Are there infinitely many positive integers $t$ such that both $2012t+1$ and $2013t+1$ are perfect squares?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(2012t+1, 2013t+1)$. It is easy to see that $d=1$. Thus, both $2012t+1$ and $2013t+1$ are perfect squares if and only if $(2012t+1)(2013t+1) = y^2$ for some positive integer $y$.\n\n$$\n\\begin{aligned}\n(2012t+1)(2013t+1) &= y^2 \\\\\n&\\Leftrightarrow 4 \\cdot 2012^2 \\cdot 2013^2 t^2 + 4 \\cdot 2012 \\cdot 2013 \\cdot 4025t + 4 \\cdot 2012 \\cdot 2013 = 4 \\cdot 2012 \\cdot 2013 \\cdot y^2 \\\\\n&\\Leftrightarrow (2 \\cdot 2012 \\cdot 2013t + 4025)^2 - 1 = 4 \\cdot 2012 \\cdot 2013 \\cdot y^2\n\\end{aligned}\n$$\n\nLet $x = 2 \\cdot 2012 \\cdot 2013t + 4025$. Then the equation becomes $x^2 - 4 \\cdot 2012 \\cdot 2013 y^2 = 1$.\n\nSince $4 \\cdot 2012 \\cdot 2013$ is not a perfect square, this Pell equation has infinitely many solutions. The fundamental solution is $(x, y) = (4025, 1)$, so all solutions are given by:\n\n$$\n\\begin{cases}\nx_0 = 1,\\ x_1 = 4025,\\ x_{n+2} = 8050x_{n+1} - x_n, & n \\ge 0 \\\\\ny_0 = 1,\\ y_1 = 1,\\ y_{n+2} = 8050y_{n+1} - y_n &\n\\end{cases}\n$$\n\nBy induction, $x_{2i+1} \\equiv 4025 \\pmod{2 \\cdot 2012 \\cdot 2013}$ for all $i$, and each value $\\dfrac{x_{2i+1} - 4025}{2 \\cdot 2012 \\cdot 2013}$ gives a positive integer $t$ satisfying the required condition.\n\nTherefore, there exist infinitely many positive integers $t$ such that both $2012t+1$ and $2013t+1$ are perfect squares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17604,
    "subject": "Mathematics (Olympiad)",
    "question": "On the day of the theft, 7 of Lady Gilmore's servants, whom we will refer to as A, B, C, D, E, F, G for the confidentiality of the investigation, entered the room with the necklace. Each claimed to have been in the room only once for an unspecified period of time. Additionally:\n\n- A claims to have met B, C, F, G in the room.\n- B claims to have met A, C, D, E, F.\n- C claims to have met A, B, E.\n- E claims to have met B, C, F.\n- F claims to have met A, B, D, E.\n- G claims to have met A, D.\n- D claims to have met B, F, G.\n\nInspector Goodenough concluded that one of the servants was lying. Who is he?",
    "options": [],
    "answer": "See solution",
    "solution": "We will first prove the following lemma.\n\n**Lemma.** Let $X, Y, Z$ and $T$ be four of the servants. If it is known that the pairs $X, Y$; $Y, Z$; $Z, T$ and $T, X$ were in the room together at some point, then one of the pairs $X, Z$ and $Y, T$ also detected each other.\n\n**Proof of Lemma.** Let us assume, without loss of generality, that $Y$ and $T$ were not in the room together, and $Y$ left the room before $T$ (the other cases are analogous). Then, $X$ and $Z$ were in the room together in the period between $Y$'s departure and $T$'s arrival.\n\nNotice that $A, C, E, F$ satisfy the condition of the lemma, but none of $A, E$ and $C, F$ intersect. The same goes for $A, B, D, G$. The only common element of these pairs is $A$. It remains to be ascertained that it is possible that all the other pairs met, as they claim, by entering exactly once. This is possible with the following sequence of entries and exits: enter $G$, enter $D$, exit $G$, enter $B$, enter $F$, exit $D$, enter $E$, exit $F$, $C$ enters, $B$ exits, $E$ exits, $C$ exits. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17605,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a natural number.\n\na) Let $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ be two matrices such that $A^2B = A$. Prove that\n\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": "See 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 $A$. Since $P(0) = 0$ and $P \\ne X$, $P$ has the form $P = X^k + a_{k-1}X^{k-1} + \\dots + a_1X$, where $2 \\le k \\le n$. From $P(A)B = O_n$ 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, $a_1 \\ne 0$ and $AB = -\\frac{1}{a_1}(A^{k-1} + a_{k-1}A^{k-2} + \\dots + a_2A)$. So $AB$ commutes with $A$. Therefore $A = A^2B = A(AB) = ABA$. Multiplying (1) on the right by $BA$, 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), $a_1(AB^2A - AB) = O_n$. Since $a_1 \\ne 0$, $AB^2A = AB$. Thus,\n$$\n(AB - BA)^2 = (ABA)B - AB^2A - B(A^2B) + B(ABA) = AB - AB - BA + BA = O_n.\n$$\n\nb) Define the matrices\n$$\nA = \\begin{pmatrix} O_{n-k} & O_{n-k,k} \\\\ O_{k,n-k} & I_k \\end{pmatrix}, \\quad B = \\begin{pmatrix} O_{n-k} & C \\\\ O_{k,n-k} & I_k \\end{pmatrix},\n$$\nwhere $C \\in \\mathcal{M}_{n-k,k}(\\mathbb{C})$ is any matrix with $\\mathrm{rank}(C) = k \\le n-k$ (for $k=0$, set $A = B = O_n$). We have $A^2B = AB = A$, $BA = B$, and\n$$\nAB - BA = A - B = \\begin{pmatrix} O_{n-k} & -C \\\\ O_{k,n-k} & O_k \\end{pmatrix},\n$$\nso $\\mathrm{rank}(AB - BA) = k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17606,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a pair of functions $g, h : \\mathbb{R} \\to \\mathbb{R}$ such that the only function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(g(x)) = g(f(x))$ and $f(h(x)) = h(f(x))$ for all real $x$ is the identity function? In other words, does there exist a tester pair?",
    "options": [],
    "answer": "See solution",
    "solution": "Such a tester pair exists. We may biject $\\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\\alpha, \\beta$ (to be specified further). Define\n\n$$\ng(x) = \\max(x - \\alpha, 0) \\quad \\text{and} \\quad h(x) = \\min(x + \\beta, 1).\n$$\n\nSay a set $S \\subseteq [0, 1]$ is invariant if $f(S) \\subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant: indeed, if $S$ is invariant and $T = g^{-1}(S)$, then $g(f(T)) = f(g(T)) \\subseteq f(S) \\subseteq S$, thus $f(T) \\subseteq T$.\n\nWe claim that (if we choose $\\alpha + \\beta < 1$) the intervals $[0, n\\alpha - m\\beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \\le n\\alpha - m\\beta \\le 1$. We prove this by induction on $m+n$.\n\nThe set $\\{0\\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0)) = f(g(0)) = f(0)$, so $f(0)$ is a fixed point of $g$. This gives $f(0) = 0$, thus the induction base is established.\n\nSuppose now we have some $m, n$ such that $[0, n'\\alpha - m'\\beta]$ is invariant whenever $m'+n' < m+n$. At least one of the numbers $(n-1)\\alpha - m\\beta$ and $n\\alpha - (m-1)\\beta$ lies in $(0, 1)$. In the first case, $[0, n\\alpha - m\\beta] = g^{-1}([0, (n-1)\\alpha - m\\beta])$, so $[0, n\\alpha - m\\beta]$ is invariant. In the second case, $[0, n\\alpha - m\\beta] = h^{-1}([0, n\\alpha - (m-1)\\beta])$, so again $[0, n\\alpha - m\\beta]$ is invariant. This completes the induction.\n\nWe claim that if we choose $\\alpha + \\beta < 1$, where $0 < \\alpha \\notin \\mathbb{Q}$ and $\\beta = 1/k$ for some integer $k > 1$, then all intervals $[0, \\delta]$ are invariant for $0 \\le \\delta < 1$. This is the case, as by the previous claim, $[0, n\\alpha \\bmod 1]$ is invariant for all nonnegative integers $n$. The set of $n\\alpha \\bmod 1$ is dense in $[0, 1]$, so in particular\n\n$$\n[0, \\delta] = \\bigcap_{n\\alpha \\bmod 1 > \\delta} [0, n\\alpha \\bmod 1]\n$$\n\nis invariant.\n\nA similar argument shows $[\\delta, 1]$ is invariant, so $\\{\\delta\\} = [0, \\delta] \\cap [\\delta, 1]$ is invariant for $0 < \\delta < 1$. Also, $\\{0\\}, \\{1\\}$ are both invariant, so $f$ must be the identity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17607,
    "subject": "Mathematics (Olympiad)",
    "question": "Daan distributes the numbers 1 to 9 over the nine squares of a $3 \\times 3$ table (each square receives exactly one number). Then, in each row, Daan circles the median number (the number that is neither the smallest nor the largest of the three). For example, if the numbers 8, 1, and 2 are in one row, he circles the number 2. He does the same for each column and each of the two diagonals. If a number is already circled, he does not circle it again.\n\n![](images/NLD_ABooklet_2021_p16_data_284413250a.png)\n\nHe calls the result of this process a _median table_. Above, you can see a median table that has 5 circled numbers.\n\n(a) What is the **smallest** possible number of circled numbers in a median table?\n\n_Prove that a smaller number is not possible and give an example in which a minimum number of numbers is circled._\n\n(b) What is the **largest** possible number of circled numbers in a median table?\n\n_Prove that a larger number is not possible and give an example in which a maximum number of numbers is circled._",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The smallest possible number of circled numbers is $3$. Fewer than $3$ is not possible since in each row at least one number is circled (and these are three different numbers).\n\nA median table in which only $3$ numbers are circled:\n\n| 4 | 9 | 7 |\n|---|---|---|\n| 2 | 5 | 8 |\n| 3 | 1 | 6 |\n\nIn the rows, the numbers $7$, $5$, $3$ are circled; in the columns, the numbers $3$, $5$, $7$; and in the diagonals, the numbers $5$ and $5$. Together, these are three different numbers: $3$, $5$, and $7$.\n\n(b) The largest possible number of circled numbers is $7$. More than $7$ is not possible, since the numbers $9$ and $1$ are never circled, hence no more than $9 - 2 = 7$ numbers are circled.\n\nA median table in which $7$ numbers are circled:\n\n| 4 | 1 | 2 |\n|---|---|---|\n| 7 | 5 | 6 |\n| 8 | 9 | 3 |\n\nIn the rows, the numbers $2$, $6$, $8$ are circled; in the columns, the numbers $7$, $5$, $3$; and in the diagonals, the numbers $4$ and $5$. Together, these are the numbers $2$, $3$, $4$, $5$, $6$, $7$, $8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17608,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest positive integer $a$ for which there exist a prime number $p$ and a positive integer $b \\geq 2$ such that\n$$\n\\frac{a^p - a}{p} = b^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $p=2$, our equation becomes $a(a-1) = 2b^2$, whose smallest solution in $\\mathbb{N}$ is $a=9$.\n\nNow let $p \\geq 3$. Since $a$ and $a^{p-1}-1$ are coprime and $a(a^{p-1}-1) = pb^2$, either $a$ or $a^{p-1}-1$ must be a square, and it is obviously not the latter; hence $a$ is a square. Assume that $a=4$. Then\n\n$$\n\\frac{4^{p-1}-1}{p} = \\frac{(2^{p-1}-1)(2^{p-1}+1)}{p}\n$$\n\nis a square, so either $2^{p-1}-1$ or $2^{p-1}+1$ is a square, but the former is $3 \\pmod{4}$, so the latter is the square: $2^{p-1}+1 = c^2$. Then $(c+1)(c-1) = 2^{p-1}$, so both $c-1$ and $c+1$ are powers of $2$ and they must be $2$ and $4$, but then $p=4$, a contradiction. In conclusion, $a=9$ is the answer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17609,
    "subject": "Mathematics (Olympiad)",
    "question": "設點 $M$ 為三角形 $ABC$ 的外接圓上一點。自 $M$ 點引對三角形 $ABC$ 的內切圓相切的兩條直線，分別交 $BC$ 於 $X_1, X_2$ 點。證明三角形 $MX_1X_2$ 的外接圓與 $ABC$ 的外接圓的第二個交點（即不同於 $M$ 的那個交點）就是 $ABC$ 的外接圓與角 $A$ 內的偽內切圓的切點。\n\n（註：角 $A$ 內的偽內切圓指的是與邊 $AB, AC$ 皆相切，並且內切於 $ABC$ 的外接圓的圓。）\n\nLet $M$ be an arbitrary point on the circumcircle of triangle $ABC$ and let the tangents from this point to the incircle of the triangle meet the sideline $BC$ at $X_1$ and $X_2$. Prove that the second intersection of the circumcircle of triangle $MX_1X_2$ with the circumcircle of $ABC$ (different from $M$) coincides with the tangency point of the circumcircle with the mixtilinear incircle in angle $A$. (As usual, the $A$-mixtilinear incircle names the circle tangent to $AB$, $AC$ and to the circumcircle of $ABC$ internally.)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that $M$ lies on the same side of line $BC$ as the vertex $A$. In this case, denote by $(I)$ the incircle of triangle $ABC$ (with radius $r$), and let $D, E, F, Y_1, Y_2$ be the tangency points of $(I)$ with the sidelines $BC, CA, AB, MX_1, MX_2$.\n\nConsider the inversion $\\Psi$ with center $I$ and power $r^2$, which takes the vertices $A, B, C, M, X_1, X_2$ to the midpoints $A', B', C', M', X_1', X_2'$ of the sides $EF, FD, DE, Y_2Y_1, Y_1D, DY_2$ of the intouch triangles $DEF$ and $DY_2Y_1$. The circumcircles $(O), (P)$ of triangles $ABC$ and $MX_1X_2$ become the circumcircles $(O'), (P')$ of the triangles $A'B'C', M'X_1'X_2'$, which, because they coincide with the nine-point circles of two triangles of the same circumcircle, are congruent and have the common radius $\\frac{r}{2}$.\n\nSince the sidelines $BC, CA, AB, MX_1, MX_2$ are tangent to the inversion circle $(I)$, their images under $\\Psi$ are the congruent circles $\\Gamma_a, \\Gamma_b, \\Gamma_c, \\Omega_1, \\Omega_2$ with diameters $ID, IE, IF, IX_1, IX_2$, respectively. The congruent circles $(O'), \\Gamma_b, \\Gamma_c$ with radii $\\frac{r}{2}$ meet at point $A'$. Thus, a circle $(A', r)$ with center $A'$ and radius $r$ is tangent to all three at points diametrically opposite to $A'$.\n\nThe internal angle-bisector $AI$ of the angle $\\angle A$ passing through the inversion center $I$ is carried into itself. The mixtilinear incircle $(K_a)$ and the mixtilinear excircle $(L_a)$ of the triangle $ABC$ in the angle $A$ are the only two circles centered on $AI$ and simultaneously tangent to $CA, AB$, and $(O)$. Since the inversion center $I$ is the similarity center of a circle and its inversion images, only the images $(K'_a), (L'_a)$ of $(K_a), (L_a)$ are centered on $AI$ and tangent to $\\Gamma_b, \\Gamma_c, (O')$. The mixtilinear excircle $(L_a)$, lying outside of the circumcircle $(O)$ and outside of the inversion circle $(I)$, has both intersections with $AI$ on the ray $\\overrightarrow{IL_a}$. Since the inversion in $(I)$ has positive power $r^2$, its image $(L'_a)$ also has both intersections with $AI$ on the ray $\\overrightarrow{IL_a}$ and it is centered on the ray $\\overrightarrow{IA}$. It cannot be identical with the circle $(A', r)$ centered on the opposite ray $\\overrightarrow{IA}$. Therefore, the image of the mixtilinear incircle $(K_a)$ in angle $A$ under $\\Psi$ is the circle $(A', r)$. Furthermore, the inverse image of the tangency point $Z$ of the circles $(K_a)$ and $(O)$ is the tangency point $Z'$ of $(A', r)$ and $(O')$, the antipode of $A'$ with respect to the circumcircle $(O')$.\n\nLet now $A_0, B_0, C_0, O_1, O_2$ be the centers of the congruent circles $\\Gamma_a, \\Gamma_b, \\Gamma_c, \\Omega_1, \\Omega_2$. Since $\\Gamma_a$ is the reflection of $(O')$ in $B'C'$ and since $O'$ and $I$ are isogonal conjugates with respect to triangle $A'B'C'$, the quadrilateral $B'Z'C'I$ is a parallelogram, and therefore, its diagonals $B'C', IZ'$ cut each other at half at the midpoint of segment $B'C'$. It now follows that the quadrilateral $IA_0Z'O'$ is also a parallelogram, and thus, $IA_0 = O'Z' = \\frac{r}{2}$. Now since $A_0I = A_0X'_1 = O_1I = O_1X'_1 = \\frac{r}{2}$, the quadrilateral $A_0IO_1X'_1$ is a rhombus, and since the segments $O_1X'_1, IA_0, O'Z'$ are parallel and congruent, the quadrilateral $O_1X'_1Z'O'$ is a parallelogram. In conclusion, the triangles $PX'_1Z'$ and $M'O_1O'$ are congruent and $P'Z' = M'O' = \\frac{r}{2}$. Hence, $Z'$ lies on the circle $(P')$, which means that the second intersection of the circumcircle $(P)$ of triangle $MX_1X_2$ with the circumcircle $(O)$ of $ABC$ (different from $M$) coincides with the tangency point $Z$ of the mixtilinear incircle in angle $A$ with the circumcircle $(O)$.\n\n*Remark.* Note that the proposed problem ensures a geometric construction of the mixtilinear incircles.\n\n**Construction.** Let $M$ be an arbitrary point on the circumcircle of a triangle $ABC$ and let the tangents from this point to the incircle of the triangle intersect the sideline $BC$ at points $X_1, X_2$. Denote by $Z$ the second intersection of the circumcircles of triangles $ABC$ and $MX_1X_2$, and let $K_a$ be the intersection of the lines $AI$ and $ZO$, where $I, O$ are the incenter and circumcenter, respectively. The circle centered at $K_a$ with radius $K_aZ$ is the mixtilinear incircle in angle $A$, being simultaneously tangent to $AB, AC$, and internally to $(O)$.\n\n![](images/14-3J-M1_p9_data_4d419375b4.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17610,
    "subject": "Mathematics (Olympiad)",
    "question": "Alex and Betty play a game with a row of 2022 consecutive cells. Initially, Alex's name is written in the 1st, 3rd, ..., 2021st cells, and Betty's name is written in the 2nd, 4th, ..., 2022nd cells. Starting with Alex, the players take turns performing the following operation:\n\nChoose two non-adjacent cells with their own name such that all cells between them have the opponent's name. Then, replace the opponent's names between the chosen two with their own name.\n\nThe game ends when a player cannot make a move. Determine the largest positive integer $m$ such that, no matter how Betty plays, Alex can ensure there are at least $m$ cells with Alex's name at the end of the game.",
    "options": [],
    "answer": "See solution",
    "solution": "At the start, there are 2021 pairs of adjacent cells with different names. Each operation reduces this number by two. If there are three or more such pairs, a move is possible; thus, only one such pair remains at the end. Therefore, the total number of operations is $1010$.\n\nWhen Alex plays, let $(X, Y)$ be the leftmost such pair and $(Z, W)$ the next. $X$ and $W$ have Alex's name, and all cells between them have Betty's name. By operating on $X$ and $W$, Alex increases the number of consecutive Alex cells from the left by two. These cannot be changed by Betty. Alex can do this $505$ times, so $m = 1 + 2 \\times 505 = 1011$.\n\nSimilarly, Betty can ensure at most $1011$ consecutive Betty cells from the right, so $m \\leq 2022 - 1011 = 1011$. Thus, the answer is $1011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17611,
    "subject": "Mathematics (Olympiad)",
    "question": "For any prime number $p \\ge 3$, show that for all $x \\in \\mathbb{N}$ sufficiently large, one of the integers $x+1, x+2, \\dots, x + \\frac{p+3}{2}$ has a prime factor larger than $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $q = \\frac{p+3}{2}$.\n\nAssume, for contradiction, that none of the integers $x+1, x+2, \\dots, x+q$ has a prime factor larger than $p$.\n\nLet $p_i^{\\alpha_i}$ be the highest prime power dividing $x+i$ (so $p_i^{\\alpha_i+1} \\nmid x+i$). Since the number of primes $\\le p$ is at most $q-1$, for $x \\ge (q-1)^{q-1}$, we have $p_i^{\\alpha_i} > x^{1/(q-1)} \\ge q-1$ for each $i=1,2,\\dots,q$.\n\nBecause there are only $q-1$ possible primes $\\le p$ but $q$ numbers, there must be distinct $i_1, i_2$ with $p_{i_1} = p_{i_2}$. Then $(x+i_1, x+i_2) \\ge \\min(p_{i_1}^{\\alpha_{i_1}}, p_{i_2}^{\\alpha_{i_2}}) > q-1$. But $(x+i_1, x+i_2) = (x+i_1, i_2-i_1) \\le |i_2-i_1| \\le q-1$, a contradiction. Thus, the statement holds for all $x \\ge (q-1)^{q-1}$.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17612,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be the midpoint of side $AC$, and let $\\gamma$ be the circumcircle of triangle $ABD$. The tangent to $\\gamma$ at $A$ crosses the line $BC$ at $E$. Let $O$ be the circumcentre of triangle $ABE$. Prove that the midpoint of segment $AO$ lies on $\\gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ be the image of $\\gamma$ under the homothety with 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\n![](images/2020_BMO_Short_List_p12_data_e3cfc3352d.png)\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 segment $BE$.\n\nTo prove the latter, it is sufficient to show that triangles $ABE$ and $DBC$ are similar, for then $\\dfrac{BE}{BC} = \\dfrac{AB}{CD} = \\dfrac{AC}{CD} = 2$, which implies that $C$ is indeed the midpoint of segment $BE$.\n\nFinally, to prove the above similarity, write $\\angle EBA = \\angle CBA = \\angle ACB = \\angle DCB$ and $\\angle BDC = \\angle BAD + \\angle DBA = \\angle BAD + \\angle DAE = \\angle BAE$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17613,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$?\n\n(A) 1  (B) 2  (C) 3  (D) 6  (E) 8",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $M + 1213 = j^2$ and $M + 3773 = k^2$ for nonnegative integers $j$ and $k$. Then\n$$\n(k + j)(k - j) = k^2 - j^2 = 3773 - 1213 = 2560 = 5 \\cdot 2^9.\n$$\nBecause $k + j$ and $k - j$ have the same parity and their product is even, they must both be even, and it follows that one of them is $5 \\cdot 2^i$ and the other is $2^{9-i}$ for some $i$ with $1 \\leq i \\leq 8$. Solving for $k$ gives\n$$\nk = \\frac{5 \\cdot 2^i + 2^{9-i}}{2}.\n$$\nTo maximize $M$ it is sufficient to maximize $k$, and this will occur when $i = 8$ and $k = 5 \\cdot 2^7 + 1 = 641$. Therefore $M = 641^2 - 3773$, and its units digit is $8$.\n\nAlternatively, since $3773 - 1213 = 2560$ is a multiple of $4$, the greatest such squares are two apart in the sequence of squares, so $n + 1 = \\frac{2560}{4} = 640$. Therefore these squares are $n^2 = 639^2$ and $(n + 2)^2 = 641^2$, and $M + 1213 = 639^2$. Then $M = 639^2 - 1213$, and its units digit is $8$.\n\n![](images/2024_AMC10A_Solutions_p8_data_de937f56ef.png)\n\n| $i$ | $5 \\cdot 2^i$ | $2^{9-i}$ | $k$ | $j$ | $M$ |\n|---|---|---|---|---|---|\n| 1 | 10 | 256 | 133 | 123 | 13916 |\n| 2 | 20 | 128 | 74 | 54 | 1703 |\n| 3 | 40 | 64 | 52 | 12 | -1069 |\n| 4 | 80 | 32 | 56 | 24 | -637 |\n| 5 | 160 | 16 | 88 | 72 | 3971 |\n| 6 | 320 | 8 | 164 | 156 | 23123 |\n| 7 | 640 | 4 | 322 | 318 | 99911 |\n| 8 | 1280 | 2 | 641 | 639 | 407108 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17614,
    "subject": "Mathematics (Olympiad)",
    "question": "In a tournament with 6 players, each player plays every other player exactly once. Suppose that for each $j = 0, 1, 2, 3, 4, 5$, there is exactly one person who won $j$ matches. How many possible win-loss combinations are there for the tournament?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_j$ be the person who won $j$ matches for each $j$ ($0 \\leq j \\leq 5$).\n\n- $A_5$ won all matches, so they defeated $A_0, A_1, A_2, A_3, A_4$.\n- $A_4$ lost to $A_5$, but won against $A_0, A_1, A_2, A_3$.\n- $A_3$ lost to $A_4$ and $A_5$, but won against $A_0, A_1, A_2$.\n- $A_2$ lost to $A_3, A_4, A_5$, but won against $A_0, A_1$.\n- $A_1$ lost to $A_2, A_3, A_4, A_5$, but won against $A_0$.\n- $A_0$ lost to $A_1, A_2, A_3, A_4, A_5$.\n\nThus, if we assign the number of wins to each player, the win-loss record is uniquely determined. The number of possible assignments is $6! = 720$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17615,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\left(\\frac{1}{8}\\right)^2 \\times 2^8$ is equal to\n\n(A) 2\n\n(B) 4\n\n(C) 8\n\n(D) 16\n\n(E) 32",
    "options": [],
    "answer": "See solution",
    "solution": "$\\left(\\frac{1}{8}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 256 = 4$\n\nOr,\n\n$\\left(\\frac{1}{2^3}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 2^8 = 2^2 = 4$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17616,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $n$ does there exist a set of $n$ distinct positive integers $a_1, a_2, \\ldots, a_n$ such that\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_n} = 1\n$$\nand all $a_i < n^2$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is that the given property holds for all $n \\ne 2$.\n\nFor $n = 1$, the set $\\{1\\}$ satisfies the condition. For $n = 2$, no set satisfies the condition: if $a_1$ or $a_2$ equals 1, then $\\frac{1}{a_1} + \\frac{1}{a_2} > 1$; if both $a_1, a_2 \\ge 2$, then $\\frac{1}{a_1} + \\frac{1}{a_2} \\le \\frac{1}{2} + \\frac{1}{3} < 1$.\n\nFor $n \\ge 3$, consider the identity\n$$\n\\frac{1}{k} = \\frac{1}{k+\\ell} + \\frac{1}{k(k+1)} + \\frac{1}{(k+1)(k+2)} + \\cdots + \\frac{1}{(k+\\ell-1)(k+\\ell)}.\n$$\nUsing $k=1$ and $\\ell=n-1$, we get\n$$\n1 = \\frac{1}{n} + \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\cdots + \\frac{1}{(n-1) \\cdot n}.\n$$\nIf $n \\neq k(k+1)$ for all $k \\ge 1$, this is a sum of $n$ distinct reciprocals, each with denominator less than $n^2$.\n\nIf $n = k(k+1)$ for some $k \\ge 1$, we can combine $\\frac{1}{n} + \\frac{1}{(n-1)n} = \\frac{1}{n-1}$ and $\\frac{1}{6} = \\frac{1}{10} + \\frac{1}{15}$ to obtain\n$$\n1 = \\frac{1}{n-1} + \\frac{1}{2} + \\frac{1}{10} + \\frac{1}{15} + \\frac{1}{3 \\cdot 4} + \\cdots + \\frac{1}{(n-2)(n-1)}.\n$$\nAll denominators are less than $n^2$ and the reciprocals are distinct. Thus, the property holds for all $n \\ge 3$ (except $n=2$). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17617,
    "subject": "Mathematics (Olympiad)",
    "question": "The teacher wrote the digits $123\\ldots9123\\ldots9123\\ldots$ on the board repeatedly until a number with 2018 digits was formed. After that, Andriy and Olesya played a game as follows. Alternately (with Andriy starting), they cross out 2 digits in one of these ways: either the first two digits of the number remaining after the previous move, or the last two digits, or the first and last digits of that number. The game ends when a two-digit number is left.",
    "options": [],
    "answer": "See solution",
    "solution": "Since we are only interested in divisibility by 3, we can group the digits as follows: $1, 4, 7 \\rightarrow 1$, $2, 5, 8 \\rightarrow 2$, and $3, 6, 9 \\rightarrow 3$, which gives us an equivalent problem. After $2014 \\div 2 = 1007$ moves, a 4-digit number will be left on the board, and Olesya will make the last move. These 4 digits will be consecutive as they appeared initially on the board. Therefore, the possible options are: $1231$, $2312$, or $3123$. In each case, there is a pair of digits that forms the number $12$, and this is exactly the pair Olesya should leave on the board to win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17618,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle with median $AM$. Let $K$ be the point of tangency of the incircle of triangle $ABC$ with the side $BC$. Prove that if the length of the side $BC$ is the arithmetic mean of the lengths of the sides $AB$ and $AC$, then the bisector of the angle $BAC$ passes through the midpoint of the line segment $KM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the intersection point of the bisector of angle $BAC$ and side $BC$; it suffices to prove that $KN = MN$.\n\nThe bisector property implies $\\frac{NC}{NB} = \\frac{AC}{AB}$. Substituting $NC = BC - NB$ gives\n\n$$\nNB = \\frac{BC}{1 + \\frac{AC}{AB}} = \\frac{AB \\cdot BC}{AB + AC}.\n$$\n\nAs $AB + AC = 2BC$ by assumption, this implies $NB = \\frac{AB}{2}$.\n\n![](images/prob1617_p21_data_b93bd9d9ce.png)\n\nOn the other hand, let $X$ and $Y$ be the points of tangency of the incircle of triangle $ABC$ with sides $AB$ and $AC$, respectively. Then $AX = AY$, $BX = BK$ and $CK = CY$, whence $BC = BK + CK = BX + CY = AB - AX + AC - AY = AB + AC - 2AX = 2BC - 2AX$. Thus $BC = 2AX$, implying $AX = BM$. Consequently,\n\n$$\nKN = BN - BK = BN - BX = \\frac{AB}{2} - (AB - AX) = AX - \\frac{AB}{2} = BM - \\frac{AB}{2} = BM - BN = MN.\n$$\n\nThus, the bisector of angle $BAC$ passes through the midpoint of $KM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17619,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers satisfying $x + y + z = 1$. Prove that\n\n$$\n\\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \\ge \\left( \\frac{x\\sqrt{1+x}}{\\sqrt[4]{3+9x^2}} + \\frac{y\\sqrt{1+y}}{\\sqrt[4]{3+9y^2}} + \\frac{z\\sqrt{1+z}}{\\sqrt[4]{3+9z^2}} \\right)^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By using $x + y + z = 1$, we have\n\n$$\n1 + xy + yz + zx = (x + y + z)^2 + xy + yz + zx = (x + y)(y + z) + (y + z)(z + x) + (z + x)(x + y).\n$$\n\nWith this equation and the Cauchy-Schwarz inequality, we can deduce that\n\n$$\n\\begin{aligned}\n\\text{(LHS)} &= \\frac{1}{9} \\left( \\frac{1}{1-x} + \\frac{1}{1-y} + \\frac{1}{1-z} \\right) \\left( (x + 3x^3) + (y + 3y^3) + (z + 3z^3) \\right) \\\\\n&\\ge \\left( \\sqrt{\\frac{3x^3 + x}{9(1-x)}} + \\sqrt{\\frac{3y^3 + y}{9(1-y)}} + \\sqrt{\\frac{3z^3 + z}{9(1-z)}} \\right)^2\n\\end{aligned}\n$$\n\nTherefore, it is enough to show that for any real number $s \\in (0, 1)$, the inequality\n\n$$\n\\frac{3s^3 + s}{9(1-s)} \\ge \\left( \\frac{s\\sqrt{1+s}}{\\sqrt[4]{3+9s^2}} \\right)^2\n$$\nholds. If we expand the above inequality, then it is easy to check that this is equivalent to $3(9s^2 - 1)^2 \\ge 0$, thus solving the problem. From the last inequality, we can check that the equality holds when $x = y = z = \\frac{1}{3}$. $\\square$\n\n**Comment.** We can prove this inequality by inserting $\\frac{2}{3}$ into the middle and showing that two different inequalities\n\n$$\n\\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \\ge \\frac{2}{3}\n$$\n\nand\n\n$$\n\\frac{2}{3} \\ge \\left( \\frac{x\\sqrt{1+x}}{\\sqrt[4]{3+9x^2}} + \\frac{y\\sqrt{1+y}}{\\sqrt[4]{3+9y^2}} + \\frac{z\\sqrt{1+z}}{\\sqrt[4]{3+9z^2}} \\right)^2\n$$\nhold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17620,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle with side lengths 3 cm, 4 cm, and 5 cm (a 3–4–5 triangle):\n\n(a) Four such triangles are used to form a shape. Find the area and perimeter of the resulting figure.\n\n(b) What is the smallest possible square that can be completely covered using these triangles, and how many triangles are needed?\n\n![](images/obm-2010_p10_data_8aead5cb26.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The area of one triangle is $\\frac{3 \\times 4}{2} = 6\\ \\text{cm}^2$. Using four triangles, the total area is $4 \\times 6 = 24\\ \\text{cm}^2$.\n\nThe perimeter is $2 \\times (5 + 4 + (4 - 3) + 3) = 26\\ \\text{cm}$.\n\n(b) Since the area of each triangle is $6\\ \\text{cm}^2$, the side of the square must be a multiple of 6. To cover the sides of the square with triangle sides, the side of the square must be equal to $3x + 4y + 5z$, where $x, y, z \\ge 0$. It is not possible to cover a square with side 6 cm using these triangles. The smallest square that can be covered has side 12 cm and can be made of twelve $3 \\times 4$ rectangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17621,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p^n$ be a prime power. Find the number of quadruples $(a_1, a_2, a_3, a_4)$ with $a_i \\in \\{0, 1, \\dots, p^n - 1\\}$ for $i = 1, 2, 3, 4$, such that\n\n$$\np^n \\mid (a_1 a_2 + a_3 a_4 + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have $p^n - p^{n-1}$ choices for $a_1$ such that $p \\nmid a_1$. In this case, for any of the $p^n \\cdot p^n$ choices of $a_3$ and $a_4$, there is a unique choice of $a_2$, namely\n\n$$\na_2 \\equiv a_1^{-1}(-1 - a_3 a_4) \\mod p^n.\n$$\n\nThis gives $p^{2n}(p^n - p^{n-1})$ quadruples.\n\nIf $p \\mid a_1$, then we must have $p \\nmid a_3$, since otherwise the condition\n\n$$\np^n \\mid (a_1 a_2 + a_3 a_4 + 1)\n$$\n\nis violated. Now, if $p \\mid a_1$ and $p \\nmid a_3$, for any choice of $a_2$ there is a unique choice of $a_4$, namely\n\n$$\na_4 \\equiv a_3^{-1}(-1 - a_1 a_2) \\mod p^n.\n$$\n\nThus, for these $p^{n-1}$ choices of $a_1$ and $p^n - p^{n-1}$ choices of $a_3$, we have for each of the $p^n$ choices of $a_2$ a unique $a_4$. That is, $p^{n-1}(p^n - p^{n-1})p^n$ quadruples in this case.\n\nAll in all, the total number of quadruples is\n\n$$\np^{2n}(p^n - p^{n-1}) + p^{n-1}(p^n - p^{n-1})p^n = p^{3n} - p^{3n-2}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17622,
    "subject": "Mathematics (Olympiad)",
    "question": "One chooses 9 pairwise distinct numbers from $1, 2, \\ldots, n$, and fills the cells of a $3 \\times 3$ table so that the products of numbers in each row, column, and main diagonals are all equal. Determine the minimal value of $n$ so that these conditions are satisfied.\n\n![](images/Ukraine_booklet_2018_p25_data_3063ca0b0d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $n = 36$.\n\n*Solution.* Let $k$ be the product mentioned above and let $x$ be the number in the central cell. The product of numbers in the main diagonals, central row, and column is equal to $k^4$, and this is also a product of all numbers in the table once and three times the central cell, that is $k^4 = k^3 \\times x^3$, thus $k = x^3$. Let the numbers in the table be called the following: (1st row) $a$, $e$, $c$; (2nd row) $g$, $x$, $h$; (3rd row) $d$, $f$, $b$.\n\n![](images/Ukraine_booklet_2018_p25_data_3063ca0b0d.png)\n\nTherefore, $ab = cd = ef = gh = x^2$ (*). The lowest possible value of $x$ then is $x = 6$ (that is, the number can be written as a product of two different numbers in four ways). A table satisfying all the conditions is shown in Fig. 28.\n\nSuppose such a table can be found when $n \\leq 35$. Let $x$ have a divisor $p$. As was shown earlier, then $p^9$ has to divide the product of all numbers, but $35!$ is not divisible by $5^9$, and is not divisible by $p^9$ if $p > 5$. Therefore, the only possible prime divisors of $x$ are 2 and 3. So all the numbers in the table have to be in the form $2^m \\cdot 3^n$ and cannot be greater than 32.\n\nTherefore, such numbers belong to the set $\\{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32\\}$. Due to (*), four of such numbers are greater than $x$. Thus, $x \\leq 16$. If $x = 16$, then 27 has to be in the table, but then $16^2$ is divisible by 27, which leads to a contradiction. If $x = 12$, then at least one of the numbers 27 and 32 has to be in the table. Then, similarly, $12^2$ has to be divisible by 27 or 32, contradiction. If $x = 8$ or $x = 9$, then there are not enough different powers of 2 or 3, respectively. Thus, $x = 6$ so the minimal number is $36 > 35$.\n\n$$\n\\begin{array}{ccc}\n12 & 1 & 18 \\\\\n9 & 6 & 4 \\\\\n2 & 36 & 3\n\\end{array}\n$$\n\n![](images/Ukraine_booklet_2018_p25_data_3063ca0b0d.png)\n\n*Fig. 28*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17623,
    "subject": "Mathematics (Olympiad)",
    "question": "Olesya chose 5 numbers from the set $\\{1, 2, 3, 4, 5, 6, 7\\}$. She told Pavlik the product of these numbers and asked whether the sum of these numbers is odd or even. Pavlik replied that he could not determine it for sure. What product might Olesya have had?",
    "options": [],
    "answer": "See solution",
    "solution": "If Pavlik knows the product, he can determine the product of the two numbers that were not chosen. Since he could not determine the parity of the sum, he could not determine which two numbers were left out, even though he knows their product.\n\nConsider all products of two numbers in the set:\n\n$$\n\\begin{aligned}\n1 \\cdot 2 &= 2, \\quad 1 \\cdot 3 = 3, \\quad 1 \\cdot 4 = 4, \\quad 1 \\cdot 5 = 5, \\quad 1 \\cdot 6 = 6, \\quad 1 \\cdot 7 = 7, \\\\\n2 \\cdot 3 &= 6, \\quad 2 \\cdot 4 = 8, \\quad 2 \\cdot 5 = 10, \\quad 2 \\cdot 6 = 12, \\quad 2 \\cdot 7 = 14, \\\\\n3 \\cdot 4 = 12, \\quad 3 \\cdot 5 = 15, \\quad 3 \\cdot 6 = 18, \\quad 3 \\cdot 7 = 21, \\\\\n4 \\cdot 5 = 20, \\quad 4 \\cdot 6 = 24, \\quad 4 \\cdot 7 = 28, \\\\\n5 \\cdot 6 = 30, \\quad 5 \\cdot 7 = 35, \\quad 6 \\cdot 7 = 42.\n\\end{aligned}\n$$\n\nThere are only two products that appear more than once: $1 \\cdot 6 = 6 = 2 \\cdot 3$ and $2 \\cdot 6 = 12 = 3 \\cdot 4$. Suppose the product of the two numbers not chosen is $6$. In both cases, the sum is odd: $1+6=7$ and $2+3=5$. Thus, the sum of the chosen numbers is also odd, which contradicts Pavlik's claim.\n\nTherefore, the product of the two numbers not chosen might be $12$. In this case, their sum might be even ($2+6=8$) or odd ($3+4=7$). Then the product of the 5 chosen numbers is:\n\n$$\n\\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7}{12} = 420\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17624,
    "subject": "Mathematics (Olympiad)",
    "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.\n\n(The centroid of a pentagon is the point in the plane of the pentagon whose position vector is equal to the arithmetic mean of the position vectors of the vertices.)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an orthonormal coordinate system centered at $O$, the center of the circumcircle $C$ of pentagon $ABCDE$, with unit length equal to the radius of $C$ and the real axis as the perpendicular bisector of $BC$. Let $z_X$ denote the complex number representing the position of point $X$.\n\nSince $AB = BC = CD$, we have $\\angle AOB = \\angle BOC = \\angle COD = 2\\alpha$ with $0 < 2\\alpha < \\frac{2\\pi}{3}$. The positions of the vertices are:\n\n- $z_A = \\cos 3\\alpha + i \\sin 3\\alpha$\n- $z_B = \\cos \\alpha + i \\sin \\alpha$\n- $z_C = \\cos \\alpha - i \\sin \\alpha$\n- $z_D = \\cos 3\\alpha - i \\sin 3\\alpha$\n- $z_E = \\cos \\beta + i \\sin \\beta$, with $\\beta \\in (3\\alpha, 2\\pi - 3\\alpha)$.\n\nIf the centroid coincides with $O$, then $z_A + z_B + z_C + z_D + z_E = 0$. This gives:\n\n$$\n\\cos \\beta + 2 \\cos 3\\alpha + 2 \\cos \\alpha = 0\n$$\n$$\n\\sin \\beta = 0\n$$\n\nSo $\\beta = \\pi$. Multiplying by $\\sin \\alpha$ and using sum-to-product identities, we get $\\sin 4\\alpha = \\sin \\alpha$.\n\nSince $0 < \\alpha < 4\\alpha < \\frac{4\\pi}{3}$, we deduce $4\\alpha = \\pi - \\alpha$, so $\\alpha = \\frac{\\pi}{5}$. Thus, all central angles are $\\frac{2\\pi}{5}$, so $ABCDE$ is a regular pentagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17625,
    "subject": "Mathematics (Olympiad)",
    "question": "Para todo número entero positivo $n$, sea $S(n)$ la suma de los dígitos de $n$. Hallar, si existe, un número entero positivo $n$ de 171 dígitos tal que $7$ divide a $S(n)$ y $7$ divide a $S(n+1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sí, existe. Hay muchos ejemplos; damos uno.\n\nObservamos que $n$ debe terminar en $9$, pues si no, $S(n+1) = S(n) + 1$ y $S(n)$ es coprimo con $S(n+1)$. Consideramos $n = 11\\dots199\\dots9$, el número que comienza con $b$ unos y termina en $a$ nueves, donde $a + b = 171$. Luego $n + 1 = 11\\dots1200\\dots0$.\n\nTenemos que $S(n) = b + 9a$ y $S(n+1) = b + 1$, de modo que $7 \\mid b + 9a$ y $7 \\mid b + 1$. Por lo tanto, $b \\equiv 6 \\pmod{7}$ y $6 + 9a \\equiv 0 \\pmod{7}$, que es equivalente a $2a \\equiv 1 \\pmod{7}$, de donde $a \\equiv 4 \\pmod{7}$. Luego $a + b = 10 + 7k = 171$, de donde $7k = 161$, y $k = 23$. Elegimos $a = 4 + 7 \\cdot 11 = 81$ y $b = 6 + 7 \\cdot 12 = 90$ y resulta que $n = 11\\dots199\\dots9$ con 90 unos y 81 nueves y $n + 1 = 11\\dots1200\\dots0$ con 89 unos, un dos y 81 ceros. Así que\n\n$$\nS(n) = 90 + 9 \\cdot 81 = 819 = 7 \\cdot 117, \\quad S(n+1) = 89 + 2 = 91 = 7 \\cdot 13.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17626,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$ inscribed in $(O)$, altitudes $AD$, $BE$, $CF$. $L$ is the Lemoine point of triangle $ABC$. $K$ is the orthocenter of triangle $DEF$. Then $O$, $L$, $K$ are collinear.\n\n![](images/Vietnam_2024_Booklet_p41_data_144495fa64.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of triangle $ABC$. $X$, $Y$, $Z$ are the midpoints of $EF$, $DF$, $DE$ respectively. $N$ is the Euler center of triangle $ABC$, $J$ is the incenter of triangle $XYZ$, $G$ is the centroid of triangle $DEF$.\n\nSince $N$ is the circumcenter of triangle $DEF$, $K$, $G$, $N$ are collinear and $\\frac{\\overline{GK}}{\\overline{GN}} = -2$.\n\nTherefore, $G$ is the centroid of triangle $KHO$.\n\nThe homothety with center $G$ and ratio $-2$ maps triangle $XYZ$ into triangle $DEF$, so it maps $J$ into $H$. So $J$ is the midpoint of $KO$.\n\nNote that $XJ \\parallel DA$ implies $XJ \\perp BC$. Similarly, the straight lines passing through $X$, $Y$, $Z$ and perpendicular to $BC$, $CA$, $AB$, respectively, concur at $J$.\n\nThe straight lines through $A$, $B$, $C$ and perpendicular to $YZ$, $XZ$, $XY$, respectively, concur at $O$, and $AX$, $BY$, $CZ$ concur at $L$, so according to Sondat's theorem, $L$, $J$, $O$ are collinear.\n\nSo $K$, $L$, $O$ are collinear. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17627,
    "subject": "Mathematics (Olympiad)",
    "question": "For which number from 2000 through 2100 is the probability that a randomly chosen divisor will not be greater than 45 the largest?  \n(Note: The probability is equal to the number of divisors not greater than 45 divided by the total number of divisors.)",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $45^2 = 2025$. For any number $n$, the number of divisors less than $\\sqrt{n}$ is equal to the number of divisors greater than $\\sqrt{n}$, since $0 < t < \\sqrt{n}$ implies $\\frac{n}{t} > \\sqrt{n}$ and $t \\mid n$ implies $\\frac{n}{t} \\mid n$ (and vice versa).  \nFor all numbers from 2000 through 2100, we have $44 < \\sqrt{n} < 46$. For all of these numbers, the number of divisors less than 45 is therefore equal to the number of divisors greater than 45.  \nIt follows that the probability of a random divisor being not greater than 45 is equal to $\\frac{1}{2}$ for all $n \\neq 2025$. For $n = 2025$, 45 is also a divisor, but since $\\frac{n}{t} = t$ in this case, the probability for 2025 is greater than $\\frac{1}{2}$, and 2025 is therefore the number with the required property. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17628,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $a_1, a_2, \\dots, a_n$ are given. For each $i$ ($1 \\le i \\le n$), define\n$$\nd_i = \\max\\{a_j : 1 \\le j \\le i\\} - \\min\\{a_j : i \\le j \\le n\\},\n$$\nand let\n$$\nd = \\max\\{d_i : 1 \\le i \\le n\\}.\n$$\n\n1. Prove that for any real numbers $x_1 \\le x_2 \\le \\dots \\le x_n$,\n$$\n\\max\\{|x_i - a_i| : 1 \\le i \\le n\\} \\ge \\frac{d}{2}.\n$$\n2. Show that there are real numbers $x_1 \\le x_2 \\le \\dots \\le x_n$ such that equality holds in the above inequality.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) Define\n$$\nd = d_g \\quad (1 \\le g \\le n),\n$$\n$$\na_p = \\max\\{a_j : 1 \\le j \\le g\\},\n$$\n$$\na_r = \\min\\{a_j : g \\le j \\le n\\}.\n$$\nThis yields $1 \\le p \\le g \\le r \\le n$ and $d = a_p - a_r$.\n\nObserve that for any real numbers $x_1 \\le x_2 \\le \\dots \\le x_n$,\n$$\n(a_p - x_p) + (x_r - a_r) = (a_p - a_r) + (x_r - x_p) \\ge a_p - a_r = d.\n$$\nConsequently,\n$$\na_p - x_p \\ge \\frac{d}{2} \\quad \\text{or} \\quad x_r - a_r \\ge \\frac{d}{2}.\n$$\nHence,\n$$\n\\begin{align*}\n\\max\\{|x_i - a_i| : 1 \\le i \\le n\\} &\\ge \\max\\{|x_p - a_p|, |x_r - a_r|\\} \\\\\n&\\ge \\max\\{a_p - x_p, x_r - a_r\\} \\\\\n&\\ge \\frac{d}{2}.\n\\end{align*}\n$$\n\n(2) Define the sequence $\\{x_k\\}$ as $x_1 = a_1 - \\frac{d}{2}$, $x_k = \\max\\{x_{k-1}, a_k - \\frac{d}{2}\\}$ for $2 \\le k \\le n$.\n\nNow we will prove that for the above sequence, equality holds in the previous inequality.\n\nBy definition, $\\{x_k\\}$ is a non-decreasing sequence, and $x_k - a_k \\ge -\\frac{d}{2}$ for all $k$ ($1 \\le k \\le n$).\n\nWe will show for all $k$ ($1 \\le k \\le n$):\n$$\nx_k - a_k \\le \\frac{d}{2}.\n$$\nFor any $k$ ($1 \\le k \\le n$), let $l$ ($l \\le k$) be the smallest integer such that $x_k = x_l$. Hence $l = 1$ or $l \\ge 2$ and $x_l > x_{l-1}$.\n\nIn both cases,\n$$\nx_k = x_l = a_l - \\frac{d}{2}.\n$$\nSince\n$$\na_l - a_k \\le \\max\\{a_j : 1 \\le j \\le k\\} - \\min\\{a_j : k \\le j \\le n\\} \\le d,\n$$\nwe have\n$$\nx_k - a_k = a_l - a_k - \\frac{d}{2} \\le d - \\frac{d}{2} = \\frac{d}{2}.\n$$\nThis establishes the claim. Thus,\n$$\n-\\frac{d}{2} \\le x_k - a_k \\le \\frac{d}{2}\n$$\nholds for all $1 \\le k \\le n$, hence also\n$$\n\\max\\{|x_i - a_i| : 1 \\le i \\le n\\} \\le \\frac{d}{2}.\n$$\nIn view of part (1), the equality holds for the sequence $\\{x_k\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17629,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. What is the smallest sum of digits of $5^n + 6^n + 2022^n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the smallest sum is $8$. This is achieved for $n = 1$:\n\n$$\n5^1 + 6^1 + 2022^1 = 5 + 6 + 2022 = 2033\n$$\n\nThe sum of the digits of $2033$ is $2 + 0 + 3 + 3 = 8$.\n\nSuppose for some $n > 1$ it is possible to obtain a smaller sum. Consider the last digit of $5^n + 6^n + 2022^n$:\n\n$$\n5^n + 6^n + 2022^n \\equiv \\begin{cases}\n7 \\pmod{10}, & n \\equiv 0 \\pmod{4} \\\\\n3 \\pmod{10}, & n \\equiv 1 \\pmod{4} \\\\\n5 \\pmod{10}, & n \\equiv 2 \\pmod{4} \\\\\n9 \\pmod{10}, & n \\equiv 3 \\pmod{4}\n\\end{cases}\n$$\n\nThus, only $n \\equiv 1,2 \\pmod{4}$ could possibly yield a sum less than $8$. We consider these cases:\n\n**Case 1:** $n = 4k + 1$\n\n$$\n5^n + 6^n + 2022^n \\equiv \\begin{cases}\n5 \\pmod{9}, & k \\equiv 0 \\pmod{3} \\\\\n2 \\pmod{9}, & k \\equiv 1 \\pmod{3} \\\\\n8 \\pmod{9}, & k \\equiv 2 \\pmod{3}\n\\end{cases}\n$$\n\nBut the last digit is $3$, so the sum of digits could only be $3, 12, 21, \\ldots$, which is not less than $8$. Further, checking modulo $16$ leads to a contradiction.\n\n**Case 2:** $n = 4k + 2$\n\n$$\n5^n + 6^n + 2022^n \\equiv \\begin{cases}\n7 \\pmod{9}, & k \\equiv 0 \\pmod{3} \\\\\n1 \\pmod{9}, & k \\equiv 1 \\pmod{3} \\\\\n4 \\pmod{9}, & k \\equiv 2 \\pmod{3}\n\\end{cases}\n$$\n\nAgain, the possible sums of digits do not yield a value less than $8$. Further analysis using divisibility and modular arithmetic confirms that no $n > 1$ yields a smaller sum.\n\nTherefore, the smallest sum of digits is $8$, achieved at $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17630,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n$$\nf(xy + f(xy)) = 2x f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z$ be any real number. Setting $(x, y) = (z, 1)$ in the original equation gives\n$$\nf(z + f(z)) = 2z f(1).\n$$\nSetting $(x, y) = (1, z)$ gives\n$$\nf(z + f(z)) = 2 f(z).\n$$\nEquating the right sides, $2z f(1) = 2 f(z)$, so $f(z) = f(1) z$ for all $z$.\n\nNow, set $z = 1$ in $f(z + f(z)) = 2z f(1)$:\n$$\nf(1 + f(1)) = 2 f(1).\n$$\nBut from $f(z) = f(1) z$, plugging $z = 1 + f(1)$ gives\n$$\nf(1 + f(1)) = f(1)(1 + f(1)).\n$$\nEquate the two expressions:\n$$\nf(1)(1 + f(1)) = 2 f(1) \\implies f(1)^2 - f(1) = 0 \\implies f(1) = 0 \\text{ or } f(1) = 1.\n$$\nThus, $f(z) = 0$ or $f(z) = z$ for all $z$. Both satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17631,
    "subject": "Mathematics (Olympiad)",
    "question": "We wish to color the squares in a strip of $n$ squares that are numbered from $1$ through $n$ from left to right. Each square is to be colored with one of the colors $1$, $2$, or $3$. The even-numbered squares can be colored with any color, but the odd-numbered squares can only be colored with the odd colors $1$ or $3$. In how many ways can the strip be colored if no two adjoining squares may have the same color?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n$ be the number of colorings of a strip of length $n$ ending in a square colored with $1$, and let $b_n$ be the number of such colorings ending in a square colored with $2$. The number of colorings ending in $3$ is also $a_n$, since any coloring ending in $1$ can be uniquely changed to one ending in $3$ by exchanging all $1$- and $3$-colored squares and vice versa.\n\nFor small indices, we have $a_1 = a_2 = 1$, $b_1 = 0$, and $b_2 = 2$. The recursions are:\n- $a_{n+1} = a_n + b_n$\n- $b_{2n-1} = 0$\n- $b_{2n} = 2a_{2n-1}$\n\nSubstituting $2n+1$ and $2n$ for $n$, the first recursion yields $a_{2n+2} = a_{2n-1}$ and $a_{2n-1} = a_{2n} + b_{2n} = a_{2n} + 2a_{2n-1}$, which together yield $a_{2n+2} = 3a_{2n}$. From $a_2 = 1$ we get $a_{2n+2} = 3^n = a_{2n-1}$ and $b_{2n} = 2 \\cdot 3^{n-1}$.\n\nTo obtain the number $s_n = 2a_n + b_n$ of all colorings, we note:\n\n$$\ns_{2n+2} = 2a_{2n+2} + b_{2n+2} = 2 \\cdot 3^n + 2 \\cdot 3^n = 4 \\cdot 3^n$$\n\nand\n\n$$\ns_{2n+1} = 2a_{2n+1} + b_{2n+1} = 2 \\cdot 3^n.$$ \n\nThis can be summarized as:\n\n$$s_n = (3 + (-1)^n) \\cdot 3^{\\lfloor \\frac{n-1}{2} \\rfloor}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17632,
    "subject": "Mathematics (Olympiad)",
    "question": "Given several coins arranged in a row, a legal move is to take either the first or the last coin. In the initial arrangement there are $n$ coins of arbitrary denominations. Ana and Maria make moves in succession. Ana starts by making 2 moves, then Maria makes 1 move, and the same repeats until all coins are taken away: 2 moves of Ana are followed by 1 move of Maria. (Only the last Ana's move can be taking 1 coin if there is a single coin left.) Ana's objective is to ensure at least $\\frac{2}{3}$ of the total sum of the coins for herself. Determine if she can do this with certainty if:\n\na) $n = 2013$;\nb) $n = 2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes for $n = 2013$ and no for $n = 2014$. More generally, Ana can complete her task if $n \\equiv 0 \\pmod{3}$ ($n \\geq 3$), and Maria can prevent her from doing so if $n \\equiv 1 \\pmod{3}$ ($n \\geq 4$).\n\nLet $n$ be a multiple of 3. Color the coins in 3 colors periodically: 1, 2, 3, 1, 2, 3, \\ldots, 1, 2, 3. The coins of some color $c$ have total value at most $1/3$ of the total sum. We claim that Ana can force Maria to take a coin colored $c$ on every move, which will imply that Ana can ensure at least $2/3$ of the total amount for herself.\n\nIndeed, let $c = 1$; then Ana takes the last coin 3 first, then the last coin 2 in the new sequence. Thus Maria must move at a sequence 1, 2, 3, \\ldots, 2, 3, 1, so she has to take a coin 1. Moreover, each of her two possible moves yields a sequence either starting or ending with 2, 3. Ana can take these consecutive coins 2, 3 on her next move, thus obtaining a sequence of the kind 1, 2, 3, \\ldots, 2, 3, 1, again. By following the same strategy Ana can ensure that Maria takes away all coin 1, as needed. If $c = 2$ Ana takes the first coin 1 and the last coin 3, obtaining 2, 3, 1, \\ldots, 3, 1, 2. Now any move of Maria is 2 and leaves a sequence either starting or ending with 3, 1. Ana removes such two consecutive coins and obtains 2, 3, 1, \\ldots, 3, 1, 2 again. So the pattern repeats: Ana makes sure that she takes away all coins 1 and 3.\n\nThe case $c = 3$ is completely analogous. Here Ana takes the first two coins 1, 2 in succession, yielding a sequence 3, 1, 2, \\ldots, 1, 2, 3. Maria is forced to take a coin 3, after which Ana can restore the pattern 3, 1, 2, \\ldots, 1, 2, 3 by taking two consecutive 1, 2 from an extreme.\n\nLet $n \\equiv 1 \\pmod{3}$. Color the coins periodically\n\n$$\n1, 2, 3, 1, 2, 3, \\dots, 1, 2, 3, 1.\n$$\n\nSuppose that all coins 1 are 5 cents and all remaining ones are 1 peso. Then Maria can force Ana to take all coins 1, so that Ana will have less than $2/3$ of the total amount (this is easy to check). If Ana takes two consecutive coins from one extreme of the sequence above, Maria takes the third coin from the same extreme; it is a 2 or a 3. The same pattern occurs. If Ana takes the two extremal 1's, Maria takes the first coin 2 in the resulting sequence. This gives\n\n$$\n3, 1, 2, 3, \\dots, 1, 2, 3.\n$$\n\nThere are three cases depending on Ana's next move:\n\n(a) If Ana takes two coins from the left extreme, Maria takes the coin 2 following them and the same pattern occurs again.\n\n(b) If Ana takes two coins 2, 3 from the right extreme, Maria takes the first coin 3 and the original pattern occurs again.\n\n(c) If Ana takes one coin from each extreme, the resulting sequence ends in a 2. Maria takes this 2 and the original pattern is restored.\n\nThus Maria ensures that one of the patterns above occurs after each combined move of the two. Eventually the sequence will become 1, 2, 3, 1 or 3, 1, 2, 3. In both cases Maria can make sure that Ana gets the remaining coin(s) 1.\n\n**Comment:** The case $n \\equiv 2 \\pmod{3}$ is analogous to $n \\equiv 1 \\pmod{3}$: Ana cannot succeed with certainty. Here we need the assumption $n \\geq 8$ for an analogous proof (a bit simpler).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17633,
    "subject": "Mathematics (Olympiad)",
    "question": "喬老大有一條 $1 \\times 46^2$ 的棟木板，其上有 $46^2$ 個 $1 \\times 1$ 大小的正方格子，依序編號為 1 至 $46^2$ 號。喬老大將這條木板鋸成 $N$ 段，每一段皆為連續編號的若干個格子，並在不旋轉或翻面的情況下，用這 $N$ 段木板排出滿足以下條件的 $46 \\times 46$ 方陣：\n\n若位於第 $i$ 列第 $j$ 行格子的編號為 $a_{ij}$，則 $a_{ij} - (i + j - 1)$ 被 46 整除。\n\n試求 $N$ 的最小可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 91；一般性地，對於 $1 \\times n^2$ 的木板，$N$ 的最小可能值為 $2n-1$。\n\n**構造：**\n將 $1 \\times n^2$ 的長條切成長度為 $n, 1, n, \\dots, 1, 1$ 的 $2n-1$ 段。用第一段 $n$ 木條構成第一行，依此類推，構成下方的 $(n-1) \\times n$ 方陣，再用所有 $1$ 木條構成最後一行即可。（備註：這並非唯一的構造方法。）\n\n**估計：**\n由於題目要求僅與編號對 $n$ 的餘數有關，以下討論都在 mod $n$ 的同餘下進行。\n\n考慮點集 $V = \\{0, 1, \\dots, n-1\\}$，並依以下規則連邊：對於鋸出的每一段木條，若其左右端的編號分別為 $a$ 與 $b$，則將點 $a$ 和點 $b+1$ 連邊（允許單環與重邊）。注意到邊的總數量等於鋸出來的段數，因此我們只需證明，在滿足題目條件下，所構出來的邊集 $E$ 至少要有 $2n-1$ 條邊。\n\n注意到所構出來的圖 $G = (V, E)$ 有以下性質：\n\n1. 由於每一段木條可以接成 $1$ 到 $n^2$，因此 $G$ 有歐拉迴路。\n2. 因為方陣第 $k$ 行的方格編號依序為 $k, k+1, \\dots, k+n-1$，故在方陣第 $k$ 行的所有木條會對應一個環 $\\gamma_k$，且 $\\gamma_k$ 必包含點 $k$（因為最左邊木條的最左方格編號為 $k$）。\n\n由 1. 知 $G$ 連通。此外，由於任兩個 $\\gamma_k$ 沒有共用邊（因為一段木條只會出現在某一行中），若我們從每一個 $\\gamma_k$ 中刪去一條邊，所得的新圖 $G' = (V, E')$ 仍為連通。這表示 $|E'| \\geq |V| - 1 = n - 1$，從而 $|E| = |E'| + n \\geq 2n - 1$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17634,
    "subject": "Mathematics (Olympiad)",
    "question": "How many non-square numbers are there among the positive factors of the number $6000$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $6000 = 2^4 3^1 5^3$, positive factors of $6000$ can be written as $2^a 3^b 5^c$ where $a = 0, 1, 2, 3, 4$; $b = 0, 1$; $c = 0, 1, 2, 3$.\n\nTherefore, there are altogether $5 \\times 2 \\times 4 = 40$ positive factors of $6000$.\n\nAmong them, square numbers arise only when all of $a, b, c$ are even, and this happens only $3 \\times 1 \\times 2 = 6$ times.\n\nTherefore, there are $40 - 6 = 34$ non-square factors of $6000$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17635,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. For which $n$ is $4^n + 2^{2012} + 1$ a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose first that $n = 2011$.\n\nThen\n$$\n4^n + 2^{2012} + 1 = 4^{2011} + 2^{2012} + 1 = (2^{2011})^2 + 2^{2012} + 1 = (2^{2011} + 1)^2,\n$$\nwhich is a perfect square.\n\nNow suppose $n > 2011$. Then\n$$\n(2^n)^2 < 4^n + 2^{2012} + 1 < (2^n + 1)^2,\n$$\nso $4^n + 2^{2012} + 1$ lies strictly between two consecutive squares and cannot itself be a perfect square.\n\nTherefore, the only positive integer $n$ for which $4^n + 2^{2012} + 1$ is a perfect square is $n = 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17636,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $7 \\times 8$ checkerboard as shown below, 56 pieces are placed on the board with each square containing exactly one piece. If two pieces share a common side or vertex, they are called \"connected.\" A group of 5 pieces is said to have Property A if these pieces are connected in order in a (horizontal, vertical, or diagonal) line.\n\nWhat is the least number of pieces to be removed from the board to ensure that there exists no group of 5 pieces on the board which has Property A?\n\nYou must prove your answer.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p63_data_03a545d12e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is that at least 11 pieces must be removed. Here is a proof by contradiction.\n\nAssume that removing 10 pieces from the board would suffice. Denote the square in row $i$ and column $j$ as $(i, j)$. To ensure there are no groups of 5 pieces with Property A, at least one piece must be removed from the first 5 squares of each row (7 pieces), and in the last three columns, at least one piece from the first 5 squares of each column (3 pieces). Thus, pieces in squares $(i, j)$ with $6 \\leq i \\leq 7$, $6 \\leq j \\leq 8$ are untouched. By symmetry, the pieces in the shadowed areas of the four corners of the board (see Fig. 1) are untouched when removing 10 pieces. Furthermore, in rows 1, 2, 6, 7 and columns 1, 2, 7, 8, at least one piece should be removed from each row and column (8 pieces). Therefore, at most two of the pieces labeled ①, ②, ③, ④ in Fig. 1 can be removed.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p64_data_94f2a83361.png)\n\nHowever, any of the four remaining pieces will result in a group of 5 pieces with Property A. For example, if piece ① (in square $(3, 3)$) remains, then the pieces in $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(5, 5)$ are connected in a diagonal line. Thus, removing only 10 pieces is insufficient.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p64_data_ee14d1966a.png)\n\nOn the other hand, as shown in Fig. 2, if we remove the 11 pieces in squares numbered ① to ⑩, then there is no group of 5 pieces remaining on the board with Property A. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17637,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$, there exist a prime $p$ and an integer $m$ such that:\n\n(a) $p \\equiv 5 \\pmod{6}$;\n(b) $p \\nmid n$;\n(c) $n \\equiv m^3 \\pmod{p}$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are infinitely many primes $p$ congruent to $5$ modulo $6$ (a special case of Dirichlet's theorem on primes in arithmetic progressions). In particular, there exists such a prime $p > n$, so $p$ satisfies (a) and (b).\n\nWrite $p = 6k + 5$ for some integer $k$. Set $m = n^{4k+3}$. By Fermat's Little Theorem:\n\n$$\nm^3 \\equiv n^{12k+9} \\equiv n^{6k+4} \\cdot n^{6k+4} \\cdot n \\equiv n^{p-1} \\cdot n^{p-1} \\cdot n \\equiv n \\pmod{p}.\n$$\n\nThus, $n \\equiv m^3 \\pmod{p}$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17638,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions $(x, y, z)$ to the equation\n\n$$\n(x + y + z)^5 = 80xyz(x^2 + y^2 + z^2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We directly check the identity\n\n$$\n(x + y + z)^5 - (-x + y + z)^5 - (x - y + z)^5 - (x + y - z)^5 = 80xyz(x^2 + y^2 + z^2).\n$$\n\nTherefore, if integers $x, y,$ and $z$ satisfy the equation from the statement, we have\n\n$$\n(-x + y + z)^5 + (x - y + z)^5 + (x + y - z)^5 = 0.\n$$\n\nBy Fermat's theorem, at least one of the parentheses equals $0$. Let, without loss of generality, $x = y + z$. Then the previous equation reduces to $(2z)^5 + (2y)^5 = 0$, which is equivalent to $y = -z$. Therefore, the solution set of the proposed equation is\n\n$$\n(x, y, z) \\in \\{(0, t, -t),\\ (t, 0, -t),\\ (t, -t, 0) : t \\in \\mathbb{Z}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17639,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimal distance between two points, one of which is on the graph of the function $y = e^x$ and the other on the graph of the function $y = \\ln x$.\n\nWhat is this minimal distance?",
    "options": [],
    "answer": "See solution",
    "solution": "The graphs of $y = e^x$ and $y = \\ln x$ are symmetrical with respect to the line $y = x$.\n\nHence, the minimal distance between points on these graphs occurs when both points are closest to the line $y = x$.\n\nFor $y = e^x$, the minimal distance to $y = x$ is at the point where the tangent is parallel to $y = x$. Setting $y' = 1$ gives $e^x = 1$, so $x = 0$ and $y = 1$.\n\nSimilarly, for $y = \\ln x$, the corresponding point is $(1, 0)$.\n\nThe minimal distance is between $(0, 1)$ and $(1, 0)$:\n\n$$\text{Distance} = \\sqrt{(1 - 0)^2 + (0 - 1)^2} = \\sqrt{2}$$\n\n![](images/Estonija_2010_p18_data_f40883d5bd.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 17640,
    "subject": "Mathematics (Olympiad)",
    "question": "$m^4 - m^3 + 1$ тоо нь бүхэл тооны квадрат болох бүхэл $m$ тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$m^4 - m^3 + 1 = n^2$ гэж үзье. Хэрэв $|m| > 2$ бол\n\n$$\n\\left(m^2 - \\frac{m}{2} - 1\\right)^2 < n^2 < \\left(m^2 - \\frac{m}{2}\\right)^2\n$$\n\nболохыг төвөггүй шалгаж болно. Иймд $|m| \\le 2$ байх ба $m \\in \\{-2, -1, 0, 1, 2\\}$ болно. Эдгээр утгуудад $m^4 - m^3 + 1 = n^2$-ийн утгуудыг шалгахад дараах шийдүүд олдоно:\n\n$(m, n) = (0, \\pm 1), (1, \\pm 1), (2, \\pm 3), (-2, \\pm 5)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17641,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 7.1, $\\odot O_1$ and $\\odot O_2$ are tangent externally at point $T$. The quadrilateral $ABCD$ is inscribed in $\\odot O_1$. The lines $DA$ and $CB$ are tangent to $\\odot O_2$ at points $E$ and $F$, respectively. $BN$, the bisector of $\\angle ABF$, intersects the segment $EF$ at point $N$. Line $FT$ intersects the arc $AT$ (which does not contain $B$) at point $M$. Prove that $M$ is the excenter of $\\triangle BCN$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p229_data_76d4d18333.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p229_data_7de08e8c2d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the intersection of line $AM$ with $EF$. Join $AT$, $BM$, $BP$, $BT$, $CM$, $CT$, $ET$, $TP$. As shown in Fig. 7.2.\n\nAs $BF$ is tangent to $\\odot O_2$ at $F$, we have $\\angle BFT = \\angle FET$.\nAs $\\odot O_1$ is tangent to $\\odot O_2$ at $T$, we have $\\angle MBT = \\angle FET$.\nHence, $\\angle MBT = \\angle BFM$. So $\\triangle MBT$ and $\\triangle MFB$ are similar, therefore $MB^2 = MT \\cdot MF$. The same argument gives $MC^2 = MT \\cdot MF$.\n\nNow again from that $\\odot O_1$ is tangent to $\\odot O_2$ at $T$, we have $\\angle MAT = \\angle FET$. So $A$, $E$, $P$ and $T$ are concyclic, which implies that $\\angle APT = \\angle AET$. As $AE$ is tangent to $\\odot O_2$ at $E$, we have $\\angle AET = \\angle EFT$. Thus, $\\angle MPT = \\angle PFM$, and $\\triangle MPT$ is similar to $\\triangle MF$. Therefore, $MP^2 = MT \\cdot MF$.\n\nFrom the above argument, we have $MC = MB = MP$, which means that $M$ is the excenter of $\\triangle BCP$. So $\\angle FBP = \\frac{1}{2} \\angle CMP$. Meanwhile, $\\angle CMP = \\angle CDA = \\angle ABF$, and we have $\\angle FBN = \\frac{1}{2} \\angle ABF$. Hence, $\\angle FBN = \\angle FBP$, i.e., $P$ and $N$ coincide. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17642,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions to the equation $\\sqrt[3]{x} + \\sqrt[3]{y} = \\sqrt[3]{z}$, where $x$, $y$, and $z$ are integers.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* Any triple of the form $(d a^3, d b^3, d c^3)$, where $a + b = c$, satisfies the condition.\n\n*Solution.*\nLet us find solutions to the equation $\\sqrt[3]{x} + \\sqrt[3]{y} = \\sqrt[3]{z}$. Set $x = d a^3$, $y = d b^3$, $z = d c^3$ for some integer $d$, and integers $a$, $b$, $c$ such that $a + b = c$.\n\nThen,\n$$\n\\sqrt[3]{x} + \\sqrt[3]{y} = \\sqrt[3]{d a^3} + \\sqrt[3]{d b^3} = d^{1/3}(a + b) = d^{1/3} c = \\sqrt[3]{z}\n$$\nSo, any such triple is a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17643,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a positive integer. The sequence $\\{a_n\\}$ is defined by:\n\n$$\na_1 = a, \\quad a_{n+1} = a_n^2 + 20, \\quad n = 1, 2, \\dots\n$$\n\n1. Prove that there exists a positive integer $a$ that is not a cube such that one of the terms in the sequence $\\{a_n\\}$ is a cube.\n2. Prove that at most one of the terms in the sequence $\\{a_n\\}$ is a cube.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Note that $14^2 + 20 = 216 = 6^3$, so take $a = 14$. Then $a_2 = 14^2 + 20 = 216 = 6^3$ is a cube. Therefore, $a = 14$ satisfies the condition.\n\n2. Suppose that there exists a cube number in $\\{a_n\\}$ and $a_k$ is the first cube that appears.\n\nFor any integer $m$, $m \\equiv 0, \\pm 1, \\pm 2, \\pm 3, \\pm 4 \\pmod{9}$. Thus, $m^3 \\equiv 0, \\pm 1 \\pmod{9}$, so $a_k \\equiv 0, \\pm 1 \\pmod{9}$.\n\nHence,\n\n$$\na_{k+1} = a_k^2 + 20 \\equiv 2, 3 \\pmod{9},\n$$\n\n$$\na_{k+2} = a_{k+1}^2 + 20 \\equiv 6, 2 \\pmod{9},\n$$\n\n$$\na_{k+3} = a_{k+2}^2 + 20 \\equiv 2, 6 \\pmod{9},\n$$\n\nBy induction, $a_{k+i} \\equiv 2, 6 \\pmod{9}$ for $i = 4, 5, \\dots$. Thus, the terms after $a_k$ are not $0, \\pm 1 \\pmod{9}$, so they are not cubes.\n\nTherefore, at most one of the terms in the sequence $\\{a_n\\}$ is a cube.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17644,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z) - 1$ have absolute value $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "**First approach (Evan Chen)**\n\nWe introduce the following notations:\n\n$$\n\\begin{aligned}\nP(x) &= c_n x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0 \\\\\n&= c_n(x + \\alpha_1) \\dots (x + \\alpha_n) \\\\\nP(x) - 1 &= c_n(x + \\beta_1) \\dots (x + \\beta_n)\n\\end{aligned}\n$$\n\nBy taking conjugates,\n\n$$\n\\left(x + \\frac{1}{\\alpha_1}\\right) \\cdots \\left(x + \\frac{1}{\\alpha_n}\\right) = \\left(x + \\frac{1}{\\beta_1}\\right) \\cdots \\left(x + \\frac{1}{\\beta_n}\\right) + \\left(\\overline{c_n}\\right)^{-1} \\quad (\\spadesuit)\n$$\n\nThe equation $(\\spadesuit)$ is the main player:\n\n**Claim** — We have $c_k = 0$ for all $k = 1, \\dots, n-1$.\n\n*Proof.* By comparing coefficients of $x^k$ in $(\\spadesuit)$ we obtain\n\n$$\n\\frac{c_{n-k}}{\\prod_i \\alpha_i} = \\frac{c_{n-k}}{\\prod_i \\beta_i}\n$$\n\nbut $\\prod_i \\alpha_i - \\prod_i \\beta_i = \\frac{1}{c_n} \\neq 0$. Hence $c_k = 0$. $\\square$\n\nIt follows that $P(x)$ must be of the form $P(x) = \\lambda x^n - \\mu$, so that $P(x) = \\lambda x^n - (\\mu+1)$. This requires $|\\mu| = |\\mu+1| = |\\lambda|$ which is equivalent to the stated part.\n\n**Second approach (from the author)**\n\nLet $A = P$ and $B = P - 1$ to make the notation more symmetric. We will as before show that $A$ and $B$ have all coefficients equal to zero other than the leading and constant coefficient; the finish is the same.\n\nFirst, we rule out double roots.\n\n**Claim** — Neither $A$ nor $B$ have double roots.\n\n*Proof.* Suppose that $b$ is a double root of $B$. By differentiating, we obtain $A' = B'$, so $A'(b) = 0$. However, by Gauss-Lucas, this forces $A(b) = 0$, contradiction. $\\square$\n\nLet $\\omega = e^{2\\pi i/n}$, let $a_1, \\dots, a_n$ be the roots of $A$, and let $b_1, \\dots, b_n$ be the roots of $B$. For each $k$, let $A_k$ and $B_k$ be the points in the complex plane corresponding to $a_k$ and $b_k$.\n\n**Claim (Main claim)** — For any $i$ and $j$, $\\frac{a_i}{a_j}$ is a power of $\\omega$.\n\n*Proof.* Note that\n\n$$\n\\frac{a_i - b_1}{a_j - b_1} \\cdots \\frac{a_i - b_n}{a_j - b_n} = \\frac{B(a_i)}{B(a_j)} = \\frac{A(a_i) - 1}{A(a_j) - 1} = \\frac{0 - 1}{0 - 1} = 1.\n$$\n\nSince the points $A_i$, $A_j$, $B_k$ all lie on the unit circle, interpreting the left-hand side geometrically gives\n\n$$\n\\angle A_i B_1 A_j + \\cdots + \\angle A_i B_n A_j = 0 \\implies n \\widehat{A_i A_j} = 0,\n$$\n\nwhere angles are directed modulo $180^\\circ$ and arcs are directed modulo $360^\\circ$. This implies that $\\frac{a_i}{a_j}$ is a power of $\\omega$. $\\square$\n\nNow the finish is easy: since $a_1, \\dots, a_n$ are all different, they must be $a_1\\omega^0, \\dots, a_1\\omega^{n-1}$ in some order; this shows that $A$ is a multiple of $x^n - a_1^n$, as needed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17645,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that the fraction\n$$\n\\frac{n^{3n-2} - 3n + 1}{3n - 2}\n$$\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 1$, it is easy to see that\n$$\n\\frac{1^{3\\cdot1-2} - 3\\cdot1 + 1}{3\\cdot1 - 2} = \\frac{1 - 3 + 1}{1} = -1\n$$\nwhich is an integer. Suppose there exists $n > 1$ such that the fraction is an integer. Then, considering modular congruences and the smallest prime divisor $p$ of $3n-2$, we deduce that $2^{3n-2} \\equiv 3^{3n-2} \\pmod{p}$, and let $k$ be the smallest positive integer such that $2^k \\equiv 3^k \\pmod{p}$. If $2^m \\equiv 3^m \\pmod{p}$ for some $m$, then $m$ is a multiple of $k$. By Fermat's little theorem, $2^{p-1} \\equiv 3^{p-1} \\pmod{p}$, so $p-1$ is a multiple of $k$. But $p-1$ is relatively prime to $3n-2$, so $k=1$, which leads to $2 \\equiv 3 \\pmod{p}$, i.e., $p$ divides $1$, a contradiction. Thus, $n=1$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17646,
    "subject": "Mathematics (Olympiad)",
    "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 $I$ onto $BC$. The circle with center $M$ and radius $MD$ intersects line $AI$ at $P$ and $Q$. Show that $\\angle BAC + \\angle PMQ = 180^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $AB < AC$ and $AP < AQ$. We will prove that angles $BAC$ and $PMQ$ have parallel sides, which implies the conclusion.\n\n_First approach._ Consider the tangency points $E \\in AC$ and $F \\in AB$ of 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 then that $P \\equiv P'$ and $Q \\equiv Q'$.\n\nNotice that $\\angle P'IB = 180^\\circ - \\angle AIB = 90^\\circ - \\frac{1}{2}\\angle ACB$. As 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, quadrilateral $BDP'I$ is cyclic, so $BP' \\perp AI$.\n\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, $MN \\parallel AC$, since $MN$ is a midline of triangle $ABC$. It follows that 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\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\n![](images/RMC_2015_BT_p85_data_a24603d902.png)\n\n_Another 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$$\n\\frac{PM}{AC} = \\frac{b-c}{2b}.\n$$\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$$\n\\frac{RM}{RC} = \\frac{b-c}{2b} = \\frac{PM}{AC}.\n$$\n\nSince 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 $PM \\parallel AC$.\n\nAnalogously, $\\frac{RM}{RB} = \\frac{MQ}{AB}$, 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\n_Third approach._ Let $P'$ be the projection of $B$ onto 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 $C$ onto line $AI$, a similar reasoning leads to $MQ' \\parallel AB$.\n\nNext, we prove that $MP' = MQ' = MD$, which assures us that $P = P'$, $Q = Q'$ and leads to the 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$$\n\\angle DP'M = \\angle DP'Q' + \\angle Q'P'M = \\frac{1}{2} \\angle B + \\frac{1}{2} \\angle A\n$$\n\nSince $\\angle P'MD = \\angle C$, considering triangle $DP'M$ it results that\n\n$$\n\\begin{aligned}\n\\angle P'DM &= 180^\\circ - \\angle P'MD - \\angle DP'M \\\\\n&= 180^\\circ - \\angle C - \\frac{1}{2}(\\angle A + \\angle B) \\\\\n&= \\frac{1}{2}(\\angle A + \\angle B) = \\angle DP'M,\n\\end{aligned}\n$$\n\nso $MP' = MD$. Similarly, one can prove that $MQ' = MD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17647,
    "subject": "Mathematics (Olympiad)",
    "question": "Во секоја $1 \\times 1$ клетка од правоаголна табла е запишан природен број.\nВо секој чекор дозволено е броевите во секоја од клетките на произволно избран ред да се зголемат двапати, или броевите во клетките на произволно избрана колона да се намалат за $1$.\nДали после конечен број чекори може да се случи сите броеви на таблата да бидат нули?",
    "options": [],
    "answer": "See solution",
    "solution": "Ќе покажеме дека бараната состојба може да се достигне.\n\nАко постојат броеви од првата колона еднакви на $1$, тогаш ги дуплираме нивните редови, а потоа првата колона ја намалуваме за $1$.\n\nОваа постапка ја повторуваме сè додека сите броеви од првата колона не станат еднакви на $1$, а со нејзиното намалување за $1$ сите броеви од првата колона стануваат нули.\n\nОпишаната постапка ја применуваме на сите останати колони, и со тоа задачата е решена.\n\nАко пак ниту еден од броевите од дадена колона не е еднаков на $1$, тогаш таа колона ја намалуваме за $1$ сè додека барем еден број од дадената колона не стане еднаков на $1$, а потоа ја применуваме погоре опишаната постапка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17648,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathcal{A}$ 為所有具有三個變數 $x, y, z$ 的整係數多項式所成之集合。令 $\\mathcal{B} \\subset \\mathcal{A}$ 為所有具有表達式\n\n$$\n(x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z)\n$$\n\n的多項式所成之集合，其中 $P, Q, R \\in \\mathcal{A}$。確定最小的非負整數 $n$ 使得對任意滿足 $i+j+k \\ge n$ 的非負整數 $i, j, k$，$x^i y^j z^k$ 必在 $\\mathcal{B}$ 中。",
    "options": [],
    "answer": "See solution",
    "solution": "$n = 4$。\n\n首先證明 $n \\le 4$，即任意 $i + j + k \\ge 4$ 的單項式 $f = x^i y^j z^k$ 屬於 $\\mathcal{B}$。假設 $i \\ge j \\ge k$，其他情形類似。\n\n設 $x + y + z = p$，$xy + yz + zx = q$，$xyz = r$。則\n\n$$\n0 = (x - x)(x - y)(x - z) = x^3 - px^2 + qx - r,\n$$\n\n因此 $x^3 \\in \\mathcal{B}$。接著，$x^2y^2 = xyq - (x+y)r \\in \\mathcal{B}$。\n\n若 $k \\ge 1$，則 $r$ 整除 $f$，因此 $f \\in \\mathcal{B}$。若 $k = 0$ 且 $j \\ge 2$，則 $x^2y^2$ 整除 $f$，因此 $f \\in \\mathcal{B}$。最後，若 $k = 0, j \\le 1$，則 $x^3$ 整除 $f$，此時 $f \\in \\mathcal{B}$。\n\n為證明 $n \\ge 4$，考慮單項式 $x^2y$ 不屬於 $\\mathcal{B}$。假設相反：\n\n$$\nx^2y = pP + qQ + rR\n$$\n\n對某些多項式 $P, Q, R$。若 $P$ 含有 $x^2$ 項，則 $pP+qQ+rR$ 會有 $x^3$ 項，故 $P$ 不含 $x^2, y^2, z^2$，可寫成：\n\n$$\nx^2y = (x + y + z)(axy + byz + czx) + (xy + yz + zx)(dx + ey + fz) + gxyz,\n$$\n\n其中 $a, b, c; d, e, f; g$ 為對應係數。考慮 $xy^2$ 的係數得 $e = -a$，類似地 $e = -b$，$f = -b$，$f = -c$，$d = -c$，因此 $a = b = c$ 且 $f = e = d = -a$，但此時右式 $x^2y$ 的係數為 $a+d = 0 \\neq 1$，矛盾。\n\n因此，最小的 $n$ 為 $4$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17649,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider arrangements of the numbers 1 through 64 on the squares of an $8 \\times 8$ chessboard, where each square contains exactly one number and each number appears exactly once.\n\nA number in such an arrangement is called *super-plus-good* if it is the largest number in its row and at the same time the smallest number in its column.\n\nProve or disprove each of the following statements:\n\n(a) Each such arrangement contains at least one super-plus-good number.\n\n(b) Each such arrangement contains at most one super-plus-good number.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) This is false. For example, place the numbers from 1 to 8 along the main diagonal and the numbers from 57 to 64 along the secondary diagonal:\n\n$$\n\\begin{array}{cccccccc}\n\\mathbf{1} & 9 & 10 & 11 & 12 & 13 & 14 & \\mathbf{57} \\\\\n15 & \\mathbf{2} & 16 & 17 & 18 & 19 & \\mathbf{58} & 20 \\\\\n21 & 22 & \\mathbf{3} & 23 & 24 & \\mathbf{59} & 25 & 26 \\\\\n27 & 28 & 29 & \\mathbf{4} & \\mathbf{60} & 30 & 31 & 32 \\\\\n33 & 34 & 35 & \\mathbf{61} & \\mathbf{5} & 36 & 37 & 38 \\\\\n39 & 40 & \\mathbf{62} & 41 & 42 & \\mathbf{6} & 43 & 44 \\\\\n45 & \\mathbf{63} & 46 & 47 & 48 & 49 & \\mathbf{7} & 50 \\\\\n\\mathbf{64} & 51 & 52 & 53 & 54 & 55 & 56 & \\mathbf{8}\n\\end{array}\n$$\n\nHere, the numbers from 1 to 8 are column minima, and the numbers from 57 to 64 are row maxima. Therefore, no number is both a column minimum and a row maximum, so no number is super-plus-good.\n\n(b) This is true. Denote the number in the $a$th row and $b$th column by $F(a, b)$. Assume there exist two super-plus-good numbers at $(i, j)$ and $(r, s)$. Since all numbers are different, the row maxima and column minima are unique, so $i \\neq r$ and $j \\neq s$. Then:\n\n$$\n\\begin{align*}\nF(i, j) &> F(i, s) & \\text{(since $F(i, j)$ is row maximum)}, \\\\\nF(i, j) &< F(r, j) & \\text{(since $F(i, j)$ is column minimum)}, \\\\\nF(r, s) &> F(r, j) & \\text{(since $F(r, s)$ is row maximum)}, \\\\\nF(r, s) &< F(i, s) & \\text{(since $F(r, s)$ is column minimum)}.\n\\end{align*}\n$$\n\nThese four inequalities lead to a contradiction:\n\n$$\nF(i, j) > F(i, s) > F(r, s) > F(r, j) > F(i, j).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17650,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 1$ be a positive integer. Consider a pile of $3^n$ coins, one of which is fake. Suppose that all coins are either white or black and that if the fake coin is white, it is lighter than the others, and if the fake is black, it is heavier than the others. Furthermore, assume that the number of white coins and the number of black coins differ by at most one. Under these conditions, prove that the fake coin can be identified and classified as heavy or light by at most $n$ weighings in a scale.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $n \\geq 1$, let $P(n)$ be the statement to be proven. We will argue by induction.\n\n* **Base step:** For $n=1$, consider $3^1=3$ coins. Without loss of generality, suppose two are black and one is white. Put a black coin on each pan and set the white aside. If the scale balances, then the white coin is counterfeit and lighter. If the scale tips, say the left side goes down, then the black coin on the left is heavy and fake. Thus, the counterfeit coin is identified and classified as heavy or light, so $P(1)$ holds.\n\n* **Inductive step:** Assume $P(k)$ is true for some $k \\geq 1$. Consider $3^{k+1}$ coins, one of which is counterfeit. Suppose there is one more black than white, so $\\frac{3^{k+1}+1}{2}$ are black and $\\frac{3^{k+1}-1}{2}$ are white. Partition the coins into three groups, $G_1$, $G_2$, and $G_3$, each with $3^k$ coins and a near balance of black and white:\n  - $G_1$: $\\frac{3^k-1}{2}$ black and $\\frac{3^k+1}{2}$ white coins.\n  - $G_2$ and $G_3$: each has $\\frac{3^k+1}{2}$ black and $\\frac{3^k-1}{2}$ white coins.\n\nPut $G_1$ aside. Weigh $G_2$ (left) against $G_3$ (right). If the scales balance, the counterfeit coin is in $G_1$, and $P(k)$ applies to $G_1$, finding the fake coin in $k$ additional weighings, $k+1$ in all. If the left pan goes down, either one of the $\\frac{3^k+1}{2}$ black coins from $G_2$ is heavy, or one of the $\\frac{3^k-1}{2}$ white coins from $G_3$ is light. These\n\n$$\n\\frac{3^k + 1}{2} + \\frac{3^k - 1}{2} = 3^k\n$$\n\ncoins satisfy the hypothesis for $P(k)$, so the coin is found in $k+1$ weighings. The analogous argument works when the right pan lowers.\n\nBy mathematical induction, for each $n \\geq 1$, $P(n)$ holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17651,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p53_data_bd5eaa2def.png)\nCase 1\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p53_data_3bee645043.png)\nCase 2\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p53_data_4d558970dd.png)\nCase 3\nThe beginning of a sequence of expanding grids is shown.\nCase 3 has a grey central area of 9 square units and a white border of 16 square units.\nIf the pattern of squares continues in the same way, how many **white squares** will there be in Case 10?",
    "options": [],
    "answer": "See solution",
    "solution": "44\n\nCase 10 has $10 \\times 10$ grey squares in the centre, so $2 \\times 10 + 2 \\times 12$ white squares surrounding those, which is $20 + 24 = 44$.\n\n**OR:**\n\nCase $n$ consists of a total of $(n+2)^2$ squares, the central $n^2$ being grey. That means $(n+2)^2 - n^2 = 4n + 4$ are white, and with $n = 10$ this is $44$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17652,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $m$ and $n$, find the smallest integer $N$ ($N \\ge m$) with the following property: if an $N$-element set of integers contains a complete residue system modulo $m$, then it has a nonempty subset such that the sum of its elements is divisible by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is\n\n$$\nN = \\max\\left\\{m,\\ m + n - \\frac{1}{2}m\\big[(m, n) + 1\\big]\\right\\}.\n$$\n\nFirst, we show that $N \\ge \\max\\left\\{m,\\ m + n - \\frac{1}{2}m\\big[(m, n) + 1\\big]\\right\\}$.\n\nLet $d = (m, n)$, and write $m = d m_1$, $n = d n_1$. If $n > \\frac{1}{2}m(d+1)$, there exists a complete residue system modulo $m$, $x_1, x_2, \\dots, x_m$, such that their residues modulo $n$ consist exactly of $m_1$ groups of $1, 2, \\dots, d$. For example, the following $m$ numbers have the required property:\n\n$$\ni + d n_1 j, \\quad i = 1, 2, \\dots, d, \\quad j = 1, 2, \\dots, m_1.$$\n\nFinding another $k = n - \\frac{1}{2}m(d+1) - 1$ numbers $y_1, y_2, \\dots, y_k$ that are congruent to $1$ modulo $n$, the set\n\n$$\nA = \\{x_1, x_2, \\dots, x_m, y_1, \\dots, y_k\\}\n$$\n\ncontains a complete residue system modulo $m$, however none of its nonempty subsets has its sum of elements divisible by $n$. In fact, the sum of the (smallest nonnegative) residues modulo $n$ of all elements of $A$ is greater than zero and less than or equal to $m_1(1+2+\\dots+d) + k = n-1$. Thus, ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17653,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest possible integer $k$ such that the following statement is true:\n\nLet 2010 arbitrary non-degenerate triangles be given. In every triangle, the three sides are colored so that one is blue, one is red, and one is white. Now, for each color separately, sort the lengths of the sides. We obtain:\n\n- $b_1 \\leq b_2 \\leq \\dots \\leq b_{2010}$: the lengths of the blue sides,\n- $r_1 \\leq r_2 \\leq \\dots \\leq r_{2010}$: the lengths of the red sides,\n- $w_1 \\leq w_2 \\leq \\dots \\leq w_{2010}$: the lengths of the white sides.\n\nThen there exist $k$ indices $j$ such that we can form a non-degenerate triangle with sides of lengths $b_j, r_j, w_j$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the largest possible number $k$ of indices satisfying the given condition is $1$.\n\nFirst, we prove that $b_{2010}, r_{2010}, w_{2010}$ are always lengths of the sides of a triangle. Without loss of generality, assume $w_{2010} \\geq r_{2010} \\geq b_{2010}$. We show that $b_{2010} + r_{2010} > w_{2010}$ holds. There exists a triangle with side lengths $w, b, r$ for the white, blue, and red sides, respectively, such that $w_{2010} = w$. By the conditions of the problem, $b + r > w$, $b_{2010} > b$, and $r_{2010} > r$. Thus,\n\n$$\nb_{2010} + r_{2010} > b + r > w = w_{2010}.\n$$\n\nNext, we describe a sequence of triangles for which $w_j, b_j, r_j$ with $j < 2010$ are not the lengths of the sides of a triangle. Define the sequence $\\Delta_j$, $j = 1, 2, \\dots, 2010$, where $\\Delta_j$ has:\n\n- a blue side of length $2j$,\n- a red side of length $j$ for $j \\leq 2009$ and $4020$ for $j = 2010$,\n- a white side of length $j+1$ for $j \\leq 2008$, $4020$ for $j = 2009$, and $1$ for $j = 2010$.\n\nSince\n\n$$\n(j+1) + j > 2j > (j+1) - j = 1 \\quad \\text{if } j \\leq 2008,\n$$\n$$\n2j + j > 4020 > 2j - j = j \\quad \\text{if } j = 2009,\n$$\n$$\n4020 - 1 > 2j > 4020 - 1 = 4019 \\quad \\text{if } j = 2010,\n$$\n\nsuch a sequence of triangles exists. Moreover, $w_j = j$, $r_j = j$, and $b_j = 2j$ for $1 \\leq j \\leq 2009$. Then\n\n$$\nw_j + r_j = j + j = 2j = b_j,\n$$\n\ni.e., $w_j, b_j, r_j$ are not the lengths of the sides of a triangle for $1 \\leq j \\leq 2009$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17654,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a large triangle is subdivided into $k^2$ small triangles, which are grouped into four categories based on how many hexagons each small triangle belongs to:\n\n1. *First category*: Triangles not belonging to any hexagon.\n2. *Second category*: Triangles belonging to exactly one hexagon. There are $3k-3$ such triangles.\n3. *Third category*: Triangles belonging to exactly two hexagons. There are $3k-9$ such triangles.\n4. *Fourth category*: Triangles belonging to exactly three hexagons. There are $k^2-6k+9$ such triangles.\n\nEach small triangle is assigned a unique integer from $1$ to $k^2$. The value of a hexagon is the sum of the numbers in its constituent triangles, and the total value is the sum of the values of all hexagons (counting each triangle as many times as it appears in hexagons).\n\nWhat is the greatest possible total sum of values of all hexagons, and how should the numbers be assigned to the triangles to achieve this maximum?\n\n![](images/Greece-IMO2019finalbook_p15_data_6ead1f57ec.png)",
    "options": [],
    "answer": "See solution",
    "solution": "To maximize the total sum, assign the largest numbers to triangles that are counted the most times (i.e., those in the highest category):\n\n1. Assign $1,2,3$ to the first category (not counted in the sum).\n2. Assign $4,5,\\dots,3k$ to the $3k-3$ triangles of the second category. Their sum is:\n   $$\n   S_B = 4 + 5 + \\dots + 3k = \\frac{(3k-3)(3k+4)}{2}.\n   $$\n3. Assign $3k+1,\\dots,6k-9$ to the $3k-9$ triangles of the third category. Their sum is:\n   $$\n   S_{\\Gamma} = (3k+1) + \\dots + (6k-9) = \\frac{(3k+1)+(6k-9)}{2}(3k-9).\n   $$\n4. Assign $6k-8,\\dots,k^2$ to the $k^2-6k+9$ triangles of the fourth category. Their sum is:\n   $$\n   S_{\\Delta} = (6k-8) + \\dots + k^2 = \\frac{k^2+6k-8}{2}(k^2-6k+9).\n   $$\n\nThe total sum is:\n$$\nS_{\\max} = S_B + 2S_{\\Gamma} + 3S_{\\Delta} = \\frac{3(k^4 - 14k^2 + 33k - 24)}{2}.\n$$\n\nThis assignment is optimal because swapping a number from a lower category with one from a higher category always increases the total sum, as higher categories are counted more times.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17655,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $a_0 = 2$ and $a_{n+1} = a_n^2 + a_n - 1$ for $n \\ge 0$. Prove that $a_n$ is coprime to $2n + 1$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider any prime divisor $p$ of $a_n$. Consider a directed graph with $p$ edges on $\\{0, 1, \\dots, p-1\\}$ (mod $p$) connecting $x \\to y$ if and only if\n\n$$\nx^2 + x - 1 \\equiv y \\pmod{p}.\n$$\n\nObserve that $\\frac{p-1}{2}$ is connected to $b = \\frac{p^2-5}{4}$ (mod $p$), every element has out-degree 1, and every element other than $b$ has in-degree 2 or 0. Elements $\\pm 1$ form loops, $-2 \\to 1$ and $0 \\to -1$. If we have a path $a_0 \\to a_1 \\to \\dots \\to a_n$, then each of the elements has in-degree two, with at least one having in-degree one, giving $2n - 1$ edges in total. Counting the four extra edges above, we deduce that $p \\ge 2n + 3$, which implies that $\\gcd(a_n, 2n+1) = 1$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17656,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BE$ and $CF$ be altitudes of an acute-angled triangle $ABC$. The segment $AD$ is the diameter of the circumcircle of $ABC$. Let $M$ be the midpoint of side $BC$. The internal common tangents of the incircles of triangles $BMF$ and $CME$ intersect at point $K$. Prove that $K$, $M$, and $D$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the incircles of $BMF$ and $CME$ by $\\omega_1$ and $\\omega_2$, respectively, with centers $I_1$ and $I_2$. Let $P$ and $Q$ be the midpoints of $BF$ and $CE$, respectively. Let $N$ be the intersection of the external common tangents of $\\omega_1$ and $\\omega_2$. The orthocenter of triangle $ABC$ is $H$. Since $BC$ is an external tangent, $N$ lies on $BC$. Given that $AD$ is the diameter of the circumcircle, we have $BD \\parallel CF$ and $BE \\parallel CD$. Therefore,\n\n$$\n\\angle CBD = \\angle CAD = \\angle BCF,\n$$\n\nand\n\n$$\n\\angle BCD = \\angle BAD = 90^\\circ - \\angle C = \\angle CBE.\n$$\n\nThus, $BHCD$ is a parallelogram, implying that points $D$, $M$, and $H$ are collinear. It is also given that\n\n$$\nMB = MF = ME = MC.\n$$\n\nConsequently, $I_1 \\in MP$, $I_2 \\in MQ$, and $MI_1$ passes through the midpoint of $BH$. Since $MI_2 \\parallel BH$, $M(BI_1HI_2)$ is a pencil. Thus, intersection points with $I_1I_2$ are harmonic. On the other hand, $(NI_1KI_2)$ is harmonic. It follows that $K$ is on $MH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17657,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 7.1, $AB > AC$, and the incircle $\\odot I$ of $\\triangle ABC$ is tangent to $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $M$ be the midpoint of side $BC$, and $AH \\perp BC$ at the point $H$. The bisector $AI$ of $\\angle BAC$ intersects the lines $DE$ and $DF$ at points $K$ and $L$, respectively.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p252_data_220ff90c7a.png)\n\nProve that $M$, $L$, $H$, and $K$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Join $CL$, $BI$, $DI$, $BK$, $ML$, and $KH$. Extend $CL$ to meet $AB$ at point $N$.\n\nAs both $CD$ and $CE$ are tangents to $\\odot I$, we have $CD = CE$.\n\nAs\n\n$$\n\\begin{align*}\n\\angle BIK &= \\angle BAI + \\angle ABI = \\frac{1}{2}(\\angle BAC + \\angle ABC) \\\\\n&= \\frac{1}{2}(180^\\circ - \\angle ACB) = \\angle EDC = \\angle BDC,\n\\end{align*}\n$$\n\nso $B$, $K$, $D$, and $I$ are cyclic.\n\nAs $\\angle BKI = \\angle BDI = 90^\\circ$, i.e., $BK \\perp AK$; similarly, $CL \\perp AL$. As $AL$ is the bisector of $\\angle BAC$, $L$ is the midpoint of $CN$. As $M$ is the midpoint of $BC$, $ML \\parallel AB$.\n\nSince $\\angle BKA = \\angle BHA = 90^\\circ$, it follows that points $B$, $K$, $H$, and $A$ are cyclic, so $\\angle MHK = \\angle BAK = \\angle MLK$, hence $M$, $L$, $H$, and $K$ are cyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17658,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of the internal bisector $AD$ of $\\angle ABC$. Circle $\\omega_1$ with diameter $AC$ intersects $BM$ at $E$, and circle $\\omega_2$ with diameter $AB$ intersects $CM$ at $F$. Show that $B$, $E$, $F$, $C$ belong to the same circle.",
    "options": [],
    "answer": "See solution",
    "solution": "If $AB = AC$ then the statement is obvious. Without loss of generality, assume $AB < AC$. Let $AH$ be the common chord of the given circles. Draw the line through $A$ perpendicular to $AD$, and denote by $K$ and $L$ the intersection points of this line with $\\omega_1$ and $\\omega_2$, respectively.\n\nWe prove that $BL$ passes through $M$. Let $X$ be the intersection of $BL$ and $AD$. Since $KB \\parallel AD \\parallel LC$, we have:\n\n$$\n\\frac{AX}{KB} = \\frac{LA}{LK}, \\quad \\frac{DX}{CL} = \\frac{BD}{BC}, \\quad \\frac{BD}{BC} = \\frac{KA}{LK}.\n$$\n\nThus, $AX = \\frac{KB \\cdot LA}{LK}$, $DX = \\frac{CL \\cdot KA}{LK}$. Also, $\\angle KAB = \\angle LAC$ and triangles $AKB$ and $ALC$ are similar.\n\nThus, $\\frac{KA}{LA} = \\frac{KB}{LC}$, so $KA \\cdot LC = KB \\cdot LA$. Hence, $AX = DX$ and $X$ coincides with $M$. By analogy, $CK$ passes through $M$.\n\nWe have\n$$\n\\angle DME = \\angle LMA = \\angle CLE = 180^\\circ - \\angle DHE.\n$$\nThis implies that $E$, $M$, $D$, $H$ lie on the same circle. $KBHF$ is inscribed in $\\omega_1$, so\n$$\n\\angle DMF = \\angle KMA = \\angle MKB = 180^\\circ - \\angle BHF = \\angle DHF,\n$$\nwhich implies that $M$, $H$, $D$, $F$ lie on a circle.\n\nWe have shown that $M$, $H$, $D$, $F$, $E$ lie on the same circle. In right triangle $HAD$, $HM$ is a median, so $MD = MH$. Thus,\n$$\n\\angle MDH = \\angle MHD = \\angle MED = \\angle MFH.\n$$\n\nConsider triangles $MDE$ and $MBD$ that have the common angle $M$ and $180^\\circ - \\angle CFE = \\angle MFE = \\angle MDE = \\angle MBD$.\n\nTherefore, $B$, $E$, $F$, $C$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17659,
    "subject": "Mathematics (Olympiad)",
    "question": "$a$, $b$, $c$ нь эерэг бодит тоонууд бөгөөд\n\n$$\n\\frac{1}{2} \\leq a, b, c \\leq 1\n$$\n\nбол, дараах тэнцэтгэл биш биелэнэ гэдгийг батал:\n\n$$\n2 \\leq \\frac{a+b}{1+c} + \\frac{b+c}{1+a} + \\frac{c+a}{1+b} \\leq 3\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Эхлээд зүүн гар талын тэнцэтгэл бишийг баталъя.\n\n$a, b \\geq \\frac{1}{2} \\Rightarrow a+b \\geq 1 \\Rightarrow \\frac{a+b}{1+c} \\geq \\frac{a+b}{a+b+c}$.\n\nАдилаар $\\frac{b+c}{1+a} \\geq \\frac{b+c}{a+b+c}$, $\\frac{c+a}{1+b} \\geq \\frac{c+a}{a+b+c}$.\n\nЭдгээрийг нэмбэл:\n\n$$\n\\frac{a+b}{1+c} + \\frac{b+c}{1+a} + \\frac{c+a}{1+b} \\geq \\frac{a+b+c}{a+b+c} = 2\n$$\n\nБаруун талын тэнцэтгэл бишийг баталъя.\n\n$$\n\\frac{a+b}{1+c} + \\frac{b+c}{1+a} + \\frac{c+a}{1+b} = \\left( \\frac{a}{1+c} + \\frac{c}{1+a} \\right) + \\left( \\frac{b}{1+c} + \\frac{c}{1+b} \\right) + \\left( \\frac{b}{1+a} + \\frac{a}{1+b} \\right)\n$$\n\n$a, c \\leq 1 \\Rightarrow \\frac{a}{1+c} \\leq \\frac{a}{a+c}$, $\\frac{c}{1+a} \\leq \\frac{c}{a+c}$.\n\nИймд $\\frac{a}{1+c} + \\frac{c}{1+a} \\leq \\frac{a}{a+c} + \\frac{c}{a+c} = 1$.\n\nАдилаар $\\frac{b}{1+c} + \\frac{c}{1+b} \\leq 1$, $\\frac{a}{1+a} + \\frac{a}{1+b} \\leq 1$.\n\nЭдгээрийг нэмбэл батлах тэнцэтгэл биш гарна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17660,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ and $P'$ be arbitrary convex polygons (not necessarily quadrilaterals) in the plane, and let $O$ be a point. Let $[S]$ denote the area of a set $S$.\n\nShow that the ratio of the areas of $P'$ and $P$ cannot exceed $2$, i.e.,\n$$\n[P'] < 2[P].\n$$\n\n% ![](images/RMC2013_final_p115_data_5c70274945.png)\n\n(*You may use the following setup: For a fixed ray $r$ from $O$, let $r_\\alpha$ be the ray from $O$ forming an angle $\\alpha$ with $r$. Let $X_\\alpha$ and $Y_\\alpha$ be the points of $P$ and $P'$, respectively, lying on $r_\\alpha$ farthest from $O$, and let $f(\\alpha)$ and $g(\\alpha)$ be the lengths $OX_\\alpha$ and $OY_\\alpha$, respectively. Then:*)\n\n$$\n[P] = \\frac{1}{2} \\int_{0}^{2\\pi} f^{2}(\\alpha)\\, d\\alpha, \\qquad [P'] = \\frac{1}{2} \\int_{0}^{2\\pi} g^{2}(\\alpha)\\, d\\alpha.\n$$\n\n(*Prove that $[P'] < 2[P]$.*)",
    "options": [],
    "answer": "See solution",
    "solution": "To prove $[P'] < 2[P]$, consider the following approach:\n\nFor each direction $\\alpha$, let $f(\\alpha)$ and $g(\\alpha)$ be the distances from $O$ to the boundary of $P$ and $P'$ along the ray $r_\\alpha$, respectively. The area of $P$ is\n$$\n[P] = \\frac{1}{2} \\int_{0}^{2\\pi} f^2(\\alpha)\\, d\\alpha = \\frac{1}{2} \\int_{0}^{\\pi} (f^2(\\alpha) + f^2(\\pi + \\alpha))\\, d\\alpha,\n$$\nand similarly for $[P']$.\n\nIt can be shown that\n$$\nf^2(\\alpha) + f^2(\\pi + \\alpha) \\ge \\frac{1}{2} (g^2(\\alpha) + g^2(\\pi + \\alpha)),\n$$\nfor all $\\alpha$, so integrating both sides gives $2[P] > [P']$.\n\nAlternatively, for convex quadrilaterals $P = A_1A_2A_3A_4$ and $P' = B_1B_2B_3B_4$ sharing a point $O$ (not a vertex of $P'$), and for each $i$, if the segment $\\ell_i = OB_i$ satisfies $|\\ell_i \\cap P| > |\\ell_i \\cap P'|$, then $[P'] < 2[P]$ can be established by considering the possible configurations of the quadrilaterals and comparing the areas of certain subregions, as detailed in the case analysis above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17661,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\frac{1}{a^5 + b^5 + c^2} + \\frac{1}{b^5 + c^5 + a^2} + \\frac{1}{c^5 + a^5 + b^2} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we remark that\n\n$$\na^5 + b^5 \\ge ab(a^3 + b^3).\n$$\n\nIndeed,\n\n$$\n\\begin{aligned}\na^5 + b^5 \\ge ab(a^3 + b^3) &\\Leftrightarrow a^5 - a^4b - ab^4 + b^5 \\ge 0 \\\\\n&\\Leftrightarrow (a-b)(a^4 - b^4) \\ge 0 \\\\\n&\\Leftrightarrow (a-b)^2(a^2 + b^2)(a+b) \\ge 0.\n\\end{aligned}\n$$\n\nWe rewrite the inequality as\n\n$$\n\\frac{1}{a^5 + b^5 + abc^3} + \\frac{1}{b^5 + c^5 + bca^3} + \\frac{1}{c^5 + a^5 + cab^3} \\le 1.\n$$\n\nOn the other hand, the following inequality is true:\n\n$$\na^5 + b^5 + abc^3 \\ge ab(a^3 + b^3 + c^3),\n$$\n\nand similarly for the other two terms.\n\nFinally, using AM-GM, we get:\n\n$$\n\\begin{aligned}\n& \\frac{1}{a^5 + b^5 + c^2} + \\frac{1}{b^5 + c^5 + a^2} + \\frac{1}{c^5 + a^5 + b^2} \\\\\n& \\le \\frac{1}{a^3 + b^3 + c^3} \\left( \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca} \\right) = \\frac{a + b + c}{a^3 + b^3 + c^3} \\\\\n& \\le \\frac{a + b + c}{(a + b + c)^3} = \\frac{9}{(a + b + c)^2} \\le \\frac{9}{(3\\sqrt[3]{abc})^2} = 1.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17662,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be natural numbers. Prove that there exists a nonnegative integer $k$ such that $$\\text{GCD}(a^k + bc,\\ b^k + ca,\\ c^k + ab) > 1.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be a prime divisor of $abc + 1$. We claim that $k = p - 2$ works. Since $p$ does not divide any of $a$, $b$, or $c$, by Fermat's little theorem, $a^{p-1} \\equiv b^{p-1} \\equiv c^{p-1} \\equiv 1 \\pmod{p}$. Thus, $p$ divides each of $a^{p-1} + abc$, $b^{p-1} + abc$, and $c^{p-1} + abc$. Since $p$ does not divide $a$, $b$, or $c$, and $p$ is prime, $p$ also divides $a^k + bc$, $b^k + ca$, and $c^k + ab$, because $a^k + bc = a^{p-2} + bc = \\frac{1}{a}(a^{p-1} + abc)$. Therefore, $\\text{GCD}(a^k + bc,\\ b^k + ca,\\ c^k + ab) \\geq p > 1$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17663,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a_1, a_2, \\dots, a_n$ 為滿足 $a_1 + a_2 + \\dots + a_n = 1$ 的正實數（$n \\ge 2$）。證明：\n\n$$\n\\sum_{k=2}^{n} \\frac{a_k}{1-a_k} (a_1 + a_2 + \\dots + a_{k-1})^2 < \\frac{1}{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "對所有 $k \\le n$，令\n\n$$\ns_k = a_1 + a_2 + \\dots + a_k \\quad \\text{且} \\quad b_k = \\frac{a_k s_{k-1}^2}{1 - a_k},\n$$\n\n其中 $s_0 = 0$。注意 $b_k$ 正是我們要求和中的一項。我們將證明：\n\n$$\nb_k < \\frac{s_k^3 - s_{k-1}^3}{3}. \\qquad (1)\n$$\n\n事實上，只需檢查：\n\n$$\n\\begin{align*}\n(1) &\\Longleftrightarrow 0 < (1-a_k)((s_{k-1}+a_k)^3 - s_{k-1}^3) - 3a_k s_{k-1}^2 \\\\\n&\\Longleftrightarrow 0 < (1-a_k)(3s_{k-1}^2 + 3s_{k-1}a_k + a_k^2) - 3s_{k-1}^2 \\\\\n&\\Longleftrightarrow 0 < -3a_k s_{k-1}^2 + 3(1-a_k)s_{k-1}a_k + (1-a_k)a_k^2 \\\\\n&\\Longleftrightarrow 0 < 3(1-a_k - s_{k-1})s_{k-1}a_k + (1-a_k)a_k^2\n\\end{align*}\n$$\n\n這成立，因為 $a_k + s_{k-1} = s_k \\le 1$ 且 $a_k \\in (0, 1)$。\n\n因此，對 $k = 2, \\dots, n$ 將不等式 (1) 相加，得到：\n\n$$\nb_2 + b_3 + \\dots + b_n < \\frac{s_n^3 - s_1^3}{3} = \\frac{1}{3},\n$$\n\n即所需結論。\n\n**補充說明 1.** 證明 (1) 有多種方法，可寫為：\n\n$$\n\\frac{as^2}{1-a} - \\frac{(a+s)^3 - s^3}{3} < 0, \\qquad (2)\n$$\n\n其中 $a \\ge 0,\\ s \\ge 0,\\ a + s \\le 1,\\ a > 0$。對於固定 $a$，(2) 中的表達式對 $s$ 是二次函數，且二次項係數 $a/(1-a) - a > 0$。\n\n因此，作為 $s$ 的函數是凸的，只需檢查 $s = 0$ 和 $s = 1 - a$。前者顯然成立，後者可化為：\n\n$$\nas - \\frac{3as(a + s) + a^3}{3} < 0,\n$$\n\n這也成立，因為 $a + s = 1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17664,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We only show the proof in the case when $E \\in AC$, $F \\in AB$, the other cases being similar (the diagram below shows such a case).\n\nFrom $\\angle PBC + \\angle PCB = \\angle EAD + \\angle FAD = \\angle BAC$ it follows that $\\angle BPC = 180^\\circ - \\angle BAC = \\angle BHC$, 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![](images/RMC_2015_BT_p81_data_eedd6f2b35.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17665,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 = 0$ be a given equality. Is it possible to substitute some of the $*$ with $+$ and the others with $-$ to obtain a correct equality?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum or difference of two even or two odd numbers is an even number. The sum or difference of an even and an odd number is an odd number. On the left side, there are 4 even and 5 odd numbers. Any combination of $+$ and $-$ will result in an odd number, but $0$ is an even number. Therefore, the equality cannot be achieved regardless of how we arrange the $+$ and $-$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17666,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the sides $a$, $b$, $c$ of $\\triangle ABC$, corresponding to angles $A$, $B$, $C$ respectively, form a geometric sequence. What is the range of\n\n$$\n\\frac{\\sin A \\cot C + \\cos A}{\\sin B \\cot C + \\cos B}\n$$\n\n?\n\n(A) $(0, +\\infty)$\n\n(B) $\\left(0, \\frac{\\sqrt{5}+1}{2}\\right)$\n\n(C) $\\left(\\frac{\\sqrt{5}-1}{2}, \\frac{\\sqrt{5}+1}{2}\\right)$\n\n(D) $\\left(\\frac{\\sqrt{5}-1}{2}, +\\infty\\right)$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the common ratio of $a$, $b$, $c$ is $q$. Then $b = aq$, $c = aq^2$.\n\nWe have\n\n$$\n\\begin{aligned}\n\\frac{\\sin A \\cot C + \\cos A}{\\sin B \\cot C + \\cos B} &= \\frac{\\sin A \\cos C + \\cos A \\sin C}{\\sin B \\cos C + \\cos B \\sin C} \\\\\n&= \\frac{\\sin(A+C)}{\\sin(B+C)} \\\\\n&= \\frac{\\sin(\\pi-B)}{\\sin(\\pi-A)} \\\\\n&= \\frac{\\sin B}{\\sin A} = \\frac{b}{a} = q.\n\\end{aligned}\n$$\n\nSo we only need to determine the range of $q$. Since $a$, $b$, $c$ are the sides of a triangle, they satisfy $a + b > c$ and $b + c > a$:\n\n$$\n\\begin{cases}\na + aq > aq^2 \\\\\naq + aq^2 > a\n\\end{cases}\n$$\n\nThis gives\n\n$$\n\\begin{cases}\nq^2 - q - 1 < 0 \\\\\nq^2 + q - 1 > 0\n\\end{cases}\n$$\n\nSolving these inequalities, we find\n\n$$\n\\frac{\\sqrt{5}-1}{2} < q < \\frac{\\sqrt{5}+1}{2}\n$$\n\nSo the answer is (C).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17667,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $S(n)$ the sum of all digits of a positive integer $n$.\n\na) Does there exist a positive integer $n$ satisfying the equality\n\n$$\nS(n) \\cdot S(n+1) = 2013?\n$$\n\nb) Find the smallest positive integer $n$ such that $S(n) \\cdot S(n+1) = 87$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a) no; b) 2999.\n\n**Solution.**\n\na) It is the same as problem 7-2.\n\nb) Since $87 = 3 \\cdot 29$ is the product of two primes, it is impossible to represent this number as a product of two consecutive integers. So the only possibility is that the difference between the two factors is $9k-1$ for an integer $k$ (see the solution of problem 7-2). The difference $87-1$ is not in this form, but $29-3=26=9 \\cdot 3-1$, so the required number $n$ ends with three digits 9 and the total sum of the digits is 29. Therefore, the smallest such $n$ is $n=2999$, for which $n+1=3000$, and so the equality $S(n) \\cdot S(n+1) = 87$ is satisfied.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17668,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the triple $$(n-1, n, n+1)$$ is solvable (i.e., can be transformed into $$(0, 0, a+b+c)$$ through a sequence of allowed moves) if and only if $n$ is a power of 3.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that if $n = 3^m$ for some $m \\geq 0$, then $$(n-1, n, n+1)$$ is solvable. We prove this by induction on $m$.\n\n**Base case:** For $m=0$, $n=1$, so the triple is $(0, 1, 2)$, which is solvable in one move by choosing $a=1$, $b=2$.\n\n**Inductive step:** Assume the statement holds for $m$. For $n = 3^{m+1}$, the triple is $(3^{m+1}-1, 3^{m+1}, 3^{m+1}+1)$. Choose $a = 3^{m+1}-1$ and $b = 3^{m+1}+1$; after the move, the triple becomes $(3^{m+1}-3, 3^{m+1}, 3^{m+1}+3)$. By the induction hypothesis and the scaling property, this triple is solvable.\n\nNow, observe that the sum of the numbers remains $3n$ throughout, so it must be divisible by 3. After the first move, all numbers become congruent modulo 3. If they are not all congruent to 0 mod 3, they can never become so, so the first move must make all numbers multiples of 3. Before the last move, the numbers must be $0$, $\\frac{a+b+c}{3}$, and $\\frac{2(a+b+c)}{3}$, so $3$ must divide $a+b+c$.\n\nIf $n = 3k+1$, the last move would be $(3k, 3k+1, 3k+2) \\to (3k, 3k, 3k+3)$. If $k=0$, $n=1$ is a power of 3; otherwise, the triple $(k, k, k+1)$ is not solvable since $k+k+(k+1)$ is not divisible by 3.\n\nIf $n = 3k+2$, the first move is $(3k+1, 3k+2, 3k+3) \\to (3k, 3k+3, 3k+3)$, which reduces to $(k, k+1, k+1)$, again not solvable since the sum is not divisible by 3.\n\nIf $n = 3k$, the first move is $(3k-1, 3k, 3k+1) \\to (3k-3, 3k, 3k+3)$, which reduces to $(k-1, k, k+1)$. Repeating this process, we eventually reach $(0, 1, 2)$ only if $n$ is a power of 3. Thus, the triple is solvable if and only if $n$ is a power of 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17669,
    "subject": "Mathematics (Olympiad)",
    "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.\n\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 $Z$.\n\nProve that the points $X$, $Y$, and $Z$ are collinear.",
    "options": [],
    "answer": "See 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![](images/RMC_2025_p81_data_099b2a06b6.png)\n\nThen, since $O_4P_2 = O_4P_3$ (tangents to circle $\\Gamma_1$) and $O_4P_3 = O_4P_1$ (tangents to circle $\\Gamma_2$), we get $O_4P_1 = O_4P_2 = O_4P_3$. Because $O_4P_1 \\perp O_2O_3$, $O_4P_2 \\perp O_3O_1$, and $O_4P_3 \\perp O_1O_2$, it follows that $O_4$ is the incenter of triangle $O_1O_2O_3$.\n\nWe have:\n\n$$\n\\begin{align*}\n\\widehat{P_1ZP_2} &= \\frac{1}{2} (\\widehat{AB} + \\widehat{P_1P_2}) = \\frac{1}{2} (180^\\circ + \\widehat{P_1O_3P_2}) \\\\\n&= \\frac{1}{2} (180^\\circ + 180^\\circ - \\widehat{P_1O_2P_3} - \\widehat{P_2O_1P_3}) \\\\\n&= \\frac{1}{2} (180^\\circ - \\widehat{P_1O_2P_3}) + \\frac{1}{2} (180^\\circ - \\widehat{P_2O_1P_3}) \\\\\n&= \\widehat{O_1P_3P_2} + \\widehat{O_2P_3P_1} = 180^\\circ - \\widehat{P_1P_3P_2},\n\\end{align*}\n$$\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$$\n\\begin{align*}\n\\widehat{XP_3O_1} &= \\frac{1}{2} (180^\\circ - \\widehat{XO_1P_3}) = 90^\\circ - \\frac{1}{2} \\widehat{XP_3} = \\widehat{XP_2B} - \\widehat{XP_2P_3} = \\widehat{BP_2P_3} \\\\\n&= \\widehat{ZP_2P_3} = \\frac{1}{2} \\widehat{ZP_1P_3}.\n\\end{align*}\n$$\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, $\\widehat{ZP_3O_2} = \\widehat{ZP_3P_1} + \\widehat{P_1P_3O_2} = \\frac{1}{2}\\widehat{ZP_1} + \\frac{1}{2}\\widehat{P_1P_3} = \\frac{1}{2}\\widehat{ZP_1P_3} = \\widehat{XP_3O_1}$.\n\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": 17670,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos una fila de 203 casillas. Inicialmente, la casilla más a la izquierda contiene 203 fichas y las demás están vacías. En cada movimiento, podemos hacer una de estas dos operaciones:\n\n- Tomar una ficha y desplazarla a una casilla adyacente (a la izquierda o a la derecha).\n- Tomar exactamente 20 fichas de una misma casilla y desplazarlas todas a una casilla adyacente (todas a la izquierda o todas a la derecha).\n\nTras 2023 movimientos, cada casilla contiene una ficha. Demuestra que existe una ficha que se ha desplazado hacia la izquierda al menos nueve veces.",
    "options": [],
    "answer": "See solution",
    "solution": "Consideremos la frontera entre la $n$-ésima y la $(n + 1)$-ésima casilla por la derecha. La cantidad neta de fichas que debe cruzar esa frontera es $n$. Contemos cuántos movimientos han desplazado fichas a través de ella. Si $n = 20k + r$, donde $r$ es el residuo al dividir $n$ entre 20, el número mínimo de movimientos es $k + r$ si $r \\in \\{0, 1, \\dots, 10\\}$, y $k + 1 + (20 - r)$ si $r \\in \\{11, 12, \\dots, 19\\}$.\n\nSumando esta cantidad para $n$ entre 1 y 202, el resultado es 2023. Por lo tanto, el número de movimientos que ha cruzado cada una de las fronteras es exactamente el descrito anteriormente. En particular, en la frontera entre las casillas 11 y 12 por la derecha (que tiene un cruce neto de 11 fichas) se han realizado diez movimientos: uno en el que 20 fichas se han desplazado hacia la derecha y nueve en los que fichas individuales se han desplazado a la izquierda. De las veinte fichas que se han desplazado de golpe a la 11ª casilla por la derecha, una de ellas ha tenido que terminar en la 20ª o más a la izquierda, por lo que se ha desplazado a la izquierda al menos nueve veces.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17671,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider 16 points arranged as shown, with horizontal and vertical distances of 1 between consecutive rows and columns.\n\nIn how many ways can one choose four of these points such that the distance between every two of those four points is strictly greater than 2?\n\n![](images/s3s2022_p0_data_e9da7f55c0.png)\n\nA selection of four points satisfying the condition that the distance between every two of the four points is greater than 2 will be called a valid selection.",
    "options": [],
    "answer": "See solution",
    "solution": "Label the sixteen points $A, B, C, \\ldots, P$ as shown in the figure above.\n\nA selection of four points satisfying the condition that the distance between every two of the four points is greater than 2 will be called a valid selection.\n\nLet us first consider a valid selection which includes one of the points of the inner square $F, G, K, J$. Clearly, only one of these points can be used; say we select $F$. The only points from the outer square $A, B, C, D, H, L, P, O, N, M, I, E$ at distance more than 2 from $F$ are $D, L, P, O,$ and $M$. If we choose either $L$ or $O$, then there are not enough points left among the remaining ones on the outer square to form a valid selection. We are therefore forced to select $D, P,$ and $M$, together with $F$, to obtain a valid selection. Similarly, by symmetry, there are three further valid selections that contain points from the inner square: $\\{G, P, M, A\\}$, $\\{K, M, A, D\\}$, and $\\{J, A, D, P\\}$.\n\nAll that remains is to consider valid selections using only points from the outer square. The only way to choose two points in a valid selection from the same side of the outer square is to choose two corner points, such as $A$ and $D$. But then the only option for the other two points in the valid selection would be to choose the other two corner points of the outer square, $P$ and $M$, giving the valid selection $\\{A, D, P, M\\}$. Moving away from corner points leaves us with the two remaining valid selections (where only one point from each of the sides of the outer square is selected), namely $\\{B, H, O, I\\}$ and $\\{C, L, N, E\\}$.\n\nHence, there are seven possible valid selections.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17672,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\n4f(x+f(y)) = f(x) + f(y) + f(xy) + 1\n$$\n\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exists $a \\neq 0$ such that $f(a) = f(0) = b$.\n\nSubstituting $y=0$ into the main equation gives:\n\n$$\n4f(x+b) = f(x) + 2b + 1. \\tag{1}\n$$\n\nSubstituting $y=a$ gives:\n\n$$\n4f(x+b) = f(x) + b + f(xa) + 1. \\tag{2}\n$$\n\nComparing (1) and (2), we get $f(xa) = b$, so $f$ is constant. Setting $f(x) = b$ in the original equation yields $4b = 3b + 1$, so $b = 1$. Thus, $f(x) \\equiv 1$ is a solution.\n\nNow, $f(a) = f(0)$ implies $a = 0$.\n\nNext, replace $x$ by $y$ in the main equation and compare with the original:\n\n$$\n f(x + f(y)) = f(y + f(x)).\n$$\n\nSubstituting $x = -f(y)$ gives:\n\n$$\n\\begin{aligned}\nf(0) &= f(-f(y) + f(y)) = f(y + f(-f(y))) \\\\\n&\\implies y + f(-f(y)) = 0 \\\\\nf(-f(y)) &= -y. \\tag{3}\n\\end{aligned}\n$$\n\nNow, substitute $x = 1$ and $y = -f(t)$ in the main equation:\n\n$$\n4f(1 + f(-f(t))) = f(1) + 2f(-f(t)) + 1.\n$$\n\nUsing (3), $f(-f(t)) = -t$, so:\n\n$$\n4f(1 - t) = f(1) - 2t + 1. \\tag{4}\n$$\n\nLet $t = 1 - z$ in (4):\n\n$$\n4f(z) = f(1) - 2(1 - z) + 1 = 2z + c_1.\n$$\n\nThus, $f(z) = \\frac{1}{2}z + c$ for some constant $c$.\n\nIt can be checked that this form satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17673,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the values of the positive integer $n$ for which\n\n$$\nA = \\sqrt{\\frac{9n-1}{n+7}}\n$$\n\nis rational.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to prove that there exist $a, b \\in \\mathbb{N}^*$ with $(a, b) = 1$ such that:\n\n$$\n\\frac{9n-1}{n+7} = \\frac{a^2}{b^2} \\qquad (1)\n$$\n\nFrom this relation we get:\n\n$$\nn = \\frac{7a^2 + b^2}{9b^2 - a^2} = -7 + \\frac{64b^2}{9b^2 - a^2} \\quad (2)\n$$\n\nSince $(a, b) = 1$, it follows that $(a^2, b^2) = 1$ and $(9b^2 - a^2, b^2) = 1$, and hence from (2) we get that $n$ is integer if and only if $9b^2 - a^2$ is a divisor of $64$.\n\nSince $a, b$ and $n$ are positive integers, it follows that $9b^2 - a^2 \\ge 8$, and hence:\n\n$$\n9b^2 - a^2 = (3b + a)(3b - a) \\in \\{8, 16, 32, 64\\}. \\quad (3)\n$$\n\nMoreover, the factors $3b + a$, $3b - a$ have sum a multiple of $6$ and difference a multiple of $2$ and $3b + a > 3b - a$. Therefore, from relation (3) the possible cases are the following:\n\n$$\n\\begin{aligned}\n(3b + a, 3b - a) &= (4, 2) \\quad \\text{or} \\quad (3b + a, 3b - a) = (8, 4) \\quad \\text{or} \\quad (3b + a, 3b - a) = (16, 2) \\\\\n\\Leftrightarrow (a,b) &= (1,1) \\quad \\text{or} \\quad (a,b) = (2,2) \\quad \\text{or} \\quad (a,b) = (7,3).\n\\end{aligned}\n$$\n\nThe pair $(a,b) = (2,2)$ is rejected, because $\\gcd(2,2) = 2 \\neq 1$, and therefore we have the values $n=1$ or $n=11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17674,
    "subject": "Mathematics (Olympiad)",
    "question": "Angles $\\alpha$ and $\\beta$ are such that $\\frac{\\tan \\alpha}{\\tan \\beta} = k \\neq 1$. Express $\\frac{\\sin(\\alpha+\\beta)}{\\sin(\\alpha-\\beta)}$ in terms of $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $k = \\frac{\\tan \\alpha}{\\tan \\beta} = \\frac{\\sin \\alpha \\cos \\beta}{\\cos \\alpha \\sin \\beta}$, so $\\sin \\alpha \\cos \\beta = k \\cos \\alpha \\sin \\beta$.\n\n$$\n\\frac{\\sin(\\alpha + \\beta)}{\\sin(\\alpha - \\beta)} = \\frac{\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta}{\\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta} = \\frac{(k+1) \\cos \\alpha \\sin \\beta}{(k-1) \\cos \\alpha \\sin \\beta} = \\frac{k+1}{k-1}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17675,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $BC = CD$. Let $\\omega$ be the circle centered at $C$ tangent to $BD$, and let $I$ be the incenter of $\\triangle ABD$. Show that the line through $I$ parallel to $AB$ is tangent to $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.**\n\nLet $p$ be the line tangent at $D$ to the circumcircle $\\Gamma$ of $ABCD$. Since $C$ is the midpoint of the arc $BD$, we have $\\angle(CD, p) = \\angle CAD = \\angle BAC = \\angle BDC$, and we see that $p$ is tangent to $\\omega$. Similarly, if $E$ is the midpoint of the arc $DA$ of $\\Gamma$, then $p$ is tangent to the circle $\\omega'$ centered at $E$ tangent to $DA$. Thus the line $q$ symmetric to $p$ with respect to $CE$ is tangent to $\\omega$ and $\\omega'$.\n\n![](images/Cesko-Slovacko-Poljsko_2013_p1_data_27d634bbae.png)\n\nHowever, the well-known relations $CD = CI$ and $ED = EI$ imply that $D$ and $I$ are symmetric with respect to $CE$. Hence $I$ lies on $q$ and it remains to show that $q \\parallel AB$. This follows from\n\n$$\n\\angle(q, IC) = \\angle(CD, p) = \\angle CAD = \\angle BAC\n$$\n\n(all angles here are directed).\n\n**Second solution.**\n\nLet $q$ denote the line through $I$ parallel to $AB$. Let $P$ be the orthogonal projection of $C$ on $q$, and $M$ the midpoint of $BD$.\n\nWe have\n\n$$\n\\angle CIP = \\angle CAB = \\angle CDB = \\angle CDM\n$$\n\nApplying the well-known relation $CD = CI$, we conclude right triangles $CDM$ and $CIP$ are congruent, so $CM = CP$. The conclusion follows.\n\n![](images/Cesko-Slovacko-Poljsko_2013_p2_data_726ddda45e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17676,
    "subject": "Mathematics (Olympiad)",
    "question": "假設 $f(n)$ 是 $1, 2, \\dots, n$ 的重新排列 $a_1, a_2, \\dots, a_n$ 中，滿足下列條件的排列數：\n\n1. $a_1 = 1$；\n2. $|a_i - a_{i-1}| \\le 2$，其中 $i = 2, 3, \\dots, n$。\n\n求 $f(2015)$ 除以 $4$ 的餘數。\n\nConsider the permutation of $1, 2, \\dots, n$, which we denote as $\\{a_1, a_2, \\dots, a_n\\}$.\nLet $f(n)$ be the number of these permutations satisfying the following conditions:\n\n1. $a_1 = 1$;\n2. $|a_i - a_{i-1}| \\le 2$ for $i = 2, 3, \\dots, n$.\n\nWhat is the residue when we divide $f(2015)$ by $4$?",
    "options": [],
    "answer": "See solution",
    "solution": "解：討論 $f(n)$ 的遞迴式。由 $a_1 = 1$，有 $a_2 = 2$ 或 $3$。\n\n- **情況一**：$a_2 = 2$。令 $b_i = a_{i+1} - 1$，則 $b_1, \\dots, b_{n-1}$ 滿足題意，排列數為 $f(n-1)$。\n- **情況二**：$a_2 = 3$ 且 $a_3 = 2$，則必有 $a_4 = 4$。令 $c_i = a_{i+3} - 3$，排列數為 $f(n-3)$。\n- **情況三**：$a_2 = 3$ 且 $a_3 \\ge 4$。設 $a_{k+1}$ 為第一個偶數，則 $a_1, \\dots, a_k$ 為 $1, 3, 5, \\dots, 2k-1$，$a_{k+1} = 2k$ 或 $2k-2$。\n  - 若 $a_{k+1} = 2k$，則剩下依次為 $2k-2, 2k-4, \\dots, 2$。\n  - 若 $a_{k+1} = 2k-2$，則剩下依次為 $2k-4, 2k-6, \\dots, 2$。\n\n因此，情況三只有一種可能：先遞增排出所有 $\\le n$ 的正奇數，再遞減排列所有 $\\le n$ 的正偶數。\n\n綜合以上，得遞迴式：\n\n$$\nf(n) = f(n-1) + f(n-3) + 1.\n$$\n\n計算 $f(n)$ 除以 $4$ 的餘數，得循環數列：\n\n$1, 1, 2, 0, 2, 1, 2, 1, 3, 2, 0, 0, 3, 0, 1, 1, 2, 0, 2, \\dots$\n\n為 $14$ 循環。$2015 = 143 \\times 14 + 13$，故\n\n$$\nf(2015) \\equiv f(13) \\equiv 3 \\pmod{4}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Residues and Primitive Roots"
  },
  {
    "id": 17677,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral such that $\\angle DAB = \\angle CDA = 90^\\circ$. Diagonals $AC$ and $BD$ meet at $M$. Let $K$ be a point on side $AD$ such that $\\angle ABK = \\angle DCK$.\n\nProve that $KM$ bisects $\\angle BKC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle DAB = \\angle CDA = 90^\\circ$ and $\\angle ABK = \\angle DCK$, triangles $CDK$ and $BAK$ are similar, so we have $\\dfrac{CD}{AB} = \\dfrac{DK}{KA}$. Since $CD$ and $AB$ are parallel, triangles $CDM$ and $ABM$ are also similar, so we have $\\dfrac{CD}{AB} = \\dfrac{DM}{MB}$.\n\nHence $\\dfrac{DK}{KA} = \\dfrac{DM}{MB}$. Therefore $CD$, $AB$ and $KM$ are all parallel. Finally, $\\angle MKC = \\angle DCK = \\angle ABK = \\angle MKB$, as required.\n\n![](images/2021_Australian_Scene_p140_data_1025ce8c49.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17678,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\n\\sqrt{x^2 - 2x + 6}\\; \\log_3(6-y) = x \\\\\n\\sqrt{y^2 - 2y + 6}\\; \\log_3(6-z) = y \\\\\n\\sqrt{z^2 - 2z + 6}\\; \\log_3(6-x) = z\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions for $x$, $y$, $z$ are $x, y, z < 6$ (so the logarithms are defined).\n\nThe system is equivalent to:\n\n$$\n\\begin{cases}\n\\log_3(6-y) = \\dfrac{x}{\\sqrt{x^2-2x+6}} \\\\\n\\log_3(6-z) = \\dfrac{y}{\\sqrt{y^2-2y+6}} \\\\\n\\log_3(6-x) = \\dfrac{z}{\\sqrt{z^2-2z+6}}\n\\end{cases}\n$$\n\nLet $f(x) = \\dfrac{x}{\\sqrt{x^2-2x+6}}$. This function is increasing for $x < 6$ because\n\n$$\nf'(x) = \\frac{6-x}{(x^2 - 2x + 6)\\sqrt{x^2 - 2x + 6}} > 0 \\quad \\text{for } x < 6.\n$$\n\nThe function $g(x) = \\log_3(6-x)$ is decreasing for $x < 6$.\n\nSuppose $(x, y, z)$ is a solution. Without loss of generality, let $x = \\max(x, y, z)$. Consider two cases:\n\n1. $x \\geq y \\geq z$\n\nSince $f$ is increasing, the right sides of the equations are ordered $x \\geq y \\geq z$, so the left sides are ordered $\\log_3(6-y) \\geq \\log_3(6-z) \\geq \\log_3(6-x)$, which implies $x \\geq z \\geq y$. But $y \\geq z$, so $z = y$. Then (1) and (2) imply $x = y = z$.\n\n2. $x \\geq z \\geq y$\n\nSimilarly, $\\log_3(6-y) \\geq \\log_3(6-x) \\geq \\log_3(6-z)$, so $z \\geq x \\geq y$. But $x \\geq z$, so $x = z$. Then (1) and (3) imply $x = y = z$.\n\nThus, $x = y = z$. Substitute into one equation:\n\n$$\n\\sqrt{x^2 - 2x + 6}\\; \\log_3(6-x) = x\n$$\n\nLet $f(x) = g(x)$, i.e., $\\dfrac{x}{\\sqrt{x^2-2x+6}} = \\log_3(6-x)$. This equation has a unique solution $x = 3$.\n\nTherefore, the unique solution is $x = y = z = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17679,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L$ be a tangent to the unit circle $x^2 + y^2 = 1$ at point $P$, and let $L$ also intersect the circle $x^2 + y^2 = r^2$ at points $Q$ and $R$. Find the length $|QR|$ in terms of $r$.\n\n![](images/Irish_2016_Booklet_p11_data_b536b9bd66.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.**\n\nSuppose the equation of $L$ is $y = mx + c$. This is a tangent to the unit circle if and only if $|c| = \\sqrt{1 + m^2}$. If this is satisfied, the coordinates of $P$ are $\\left(-\\frac{m}{c}, \\frac{1}{c}\\right)$. The line $L$ meets the circle $x^2 + y^2 = r^2$ at $(X, Y)$, where\n\n$$\n\\begin{aligned}\nX^2 + (mX + c)^2 &= r^2 \\\\\n(1 + m^2)X^2 + 2mcX + c^2 &= r^2 \\\\\n\\text{Using } c^2 = 1 + m^2: \\\\\n(1 + m^2)X^2 + 2mcX + (1 + m^2) &= r^2 \\\\\n(cX)^2 + 2m(cX) + m^2 &= r^2 - 1\n\\end{aligned}\n$$\n\nso that\n\n$$\ncX = -m \\pm \\sqrt{r^2 - 1}, \\quad cY = mcX + c^2 = 1 \\pm m\\sqrt{r^2 - 1}.\n$$\n\nThus,\n\n$$\nX = \\frac{-m \\pm \\sqrt{r^2 - 1}}{c}, \\quad Y = \\frac{1 \\pm m\\sqrt{r^2 - 1}}{c}.\n$$\n\nLet\n\n$$\nQ = \\left( \\frac{-m + \\sqrt{r^2 - 1}}{c}, \\frac{1 + m\\sqrt{r^2 - 1}}{c} \\right),\n$$\n\n$$\nR = \\left( \\frac{-m - \\sqrt{r^2 - 1}}{c}, \\frac{1 - m\\sqrt{r^2 - 1}}{c} \\right).\n$$\n\nThen\n\n$$\n|QR|^2 = \\frac{4(r^2 - 1) + 4m^2(r^2 - 1)}{c^2} = 4(r^2 - 1),\n$$\n\nso that $|QR| = 2\\sqrt{r^2 - 1}$.\n\n**Solution 2.**\n\nLet $O$ be the origin. Because $PO$ is a tangent that touches the circle of radius $1$ with centre $O$ at $P$, the radius $OP$ is perpendicular to $QR$. Because $|OQ| = |OR| = r$, the two right triangles $\\triangle RPO$ and $\\triangle QPO$ are congruent and $P$ is the midpoint of $RQ$.\n\nFrom Pythagoras we get $|RP|^2 + |OP|^2 = |OR|^2$, i.e. $|RP|^2 = r^2 - 1$, hence\n\n$$\n|QR| = 2|RP| = 2\\sqrt{r^2 - 1}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17680,
    "subject": "Mathematics (Olympiad)",
    "question": "There are candies on the table. At each step, Petro can take away some of them. At the first step, he takes away one candy, and on each next step, he can take away either the same amount or twice the amount that he has taken at the previous step. What is the minimal number of steps Petro needs in order to take away exactly $2011$ candies from the table?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 17.\n\nNote that on each step, the number of candies Petro takes from the table is a power of $2$. Moreover, if at some step Petro takes away $2^n$ candies, then there were moves where he was taking away $2, 4, \\ldots, 2^{n-1}$ candies. The maximum number that Petro could take in one step cannot exceed $512$, since $1 + 2 + 4 + \\ldots + 1024 = 2047 > 2011$.\n\nAssuming Petro uses an optimal strategy, the number of steps when he takes exactly $2^n$ candies cannot exceed $2$; otherwise, he could reduce the number of moves by taking $2^{n+1}$ candies instead. This implies that he could not take less than $512$ candies each time because $2 \\cdot (1 + 2 + 4 + \\ldots + 256) = 1022 < 2011$.\n\nTherefore, Petro was taking $1, 2, \\ldots, 512$ candies. We write this as:\n\n$$\na_0 + 2a_1 + 4a_2 + \\ldots + 512a_9 = 2011,\n$$\n\nwhere $a_i \\in \\{1, 2\\}$ is the number of times he was taking $2^i$ candies. This is equivalent to:\n\n$$\na'_0 + 2a'_1 + 4a'_2 + \\ldots + 512a'_9 = 2011 - (1 + 2 + 4 + \\ldots + 512) = 988,\n$$\n\nwhere $a'_i = a_i - 1$, $a'_i \\in \\{0, 1\\}$. Since binary representation is unique, we have $a'_i$ for $i = 0, \\ldots, 9$ as $0, 0, 1, 1, 1, 0, 1, 1, 1, 1$, since\n\n$$\n988 = 512 + 256 + 128 + 64 + 16 + 8 + 4.\n$$\n\nHence, $a_i$ for $i = 0, \\ldots, 9$ have values $1, 1, 2, 2, 2, 1, 2, 2, 2, 2$, and the optimal number of steps is $3 \\cdot 1 + 7 \\cdot 2 = 17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17681,
    "subject": "Mathematics (Olympiad)",
    "question": "In the quadrilateral $ABCD$, $\\angle ABC = \\angle CDA = 90^\\circ$. Let $P = AC \\cap BD$, $Q = AB \\cap CD$, $R = AD \\cap BC$. Let $l$ be the midline of triangle $PQR$ parallel to $QR$. Prove that the circumcircle of the triangle formed by the lines $AB$, $AD$, and $l$ is tangent to the circumcircle of the triangle formed by the lines $CD$, $CB$, and $l$.\n\n![](images/Ukraine2022-23_p31_data_586e34b26b.png)\n\n**Fig. 15**",
    "options": [],
    "answer": "See solution",
    "solution": "Let $l$ intersect $AB$, $AD$, $BC$, $CD$ at the points $M$, $N$, $K$, $L$, and let $E = AC \\cap QR$. Since $C$ is the orthocenter of $\\triangle AQR$, $AC \\perp QR$, and therefore $l$ is the perpendicular bisector of the segment $PE$ (see Fig. 15). Since $BC$ and $BA$ are the internal and external bisectors of the angle $DBE$, the points $K$ and $M$ are the midpoints of the arcs $PE$ of the circumscribed circle of $\\triangle BPE$. Similarly, the points $L$ and $N$ are the midpoints of the arcs $PE$ of the circumscribed circle of $\\triangle DPE$. Then $\\angle EKL = \\angle EBM = \\angle ARQ = \\angle ECL$, and hence the points $C$, $L$, $K$, $E$ lie on the same circle. Similarly, $A$, $M$, $N$, $E$ lie on the same circle. It remains to see that $\\angle LEM = \\angle AEM - \\angle CEL = \\angle ANM - \\angle CKL = \\angle ARQ - \\angle CRQ = 90^\\circ - \\angle QAR$. Similarly, $\\angle KEN = 90^\\circ - \\angle QAR = \\angle LEM$, so the circumcircles of the triangles $AMN$ and $CKL$ are tangent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17682,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $2$, and 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$$\n\\sum_{i=1}^{n} \\frac{1}{x_i^{\\alpha} + 1} \\geq \\frac{n}{(n-1)^{\\alpha} + 1}\n$$\nand determine the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y_i = \\frac{1}{x_i + 1}$ for $i = 1, 2, \\dots, n$, so the $y_i$ are positive real numbers that add up to $1$. Upon substitution, the left-hand side of the required inequality becomes\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(1 - y_i)^{\\alpha} + y_i^{\\alpha}} = \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}},\n$$\nsince $y_1 + y_2 + \\dots + y_n = 1$. Apply Jensen's inequality to the convex function $t \\mapsto t^{\\alpha}$, $t > 0$, to get\n$$\n\\left( \\sum_{j \\neq i} y_j \\right)^{\\alpha} \\leq (n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha}, \\quad i = 1, 2, \\dots, n,\n$$\nso\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}} \\geq \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha} + y_i^{\\alpha}}.\n$$\nNow write $z_i = y_i^{\\alpha}$ for $i = 1, 2, \\dots, n$, $z = z_1 + z_2 + \\dots + z_n$, and $a = (n-1)^{\\alpha-1}$ to transform the right-hand side of the above inequality to\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az}.\n$$\nNotice that the function $t \\mapsto \\frac{t}{(1-a)t + az}$, $t < az/(a-1)$, is convex, so by Jensen's inequality:\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az} \\geq n \\cdot \\frac{\\frac{1}{n} \\sum_{i=1}^{n} z_i}{(1-a) \\frac{1}{n} \\sum_{i=1}^{n} z_i + az} = \\frac{n}{(n-1)a + 1}.\n$$\nEquality holds if and only if the $z_i$ are all equal; tracing back, this is the case if and only if the $x_i$ are all equal to $n-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17683,
    "subject": "Mathematics (Olympiad)",
    "question": "A hexagon $ABCDEF$ is inscribed in a circle. If the sides $AB$ and $DE$ are parallel and so are the sides $BC$ and $EF$, prove that the sides $CD$ and $FA$ are also parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the sides $BC$ and $FA$ are not parallel, the lines $BC$ and $FA$ intersect. Call the point of their intersection $X$. Similarly, let $Y$ and $Z$ be the points of intersection of the lines $BC$, $DE$ and of the lines $DE$, $FA$, respectively.\n\nSince the quadrilateral $ABCD$ is inscribed in the circle, we have $\\angle YCD = \\angle BAD$. As the lines $AB$, $DE$ are parallel, $\\angle BAD = \\angle ADE$. We also have $\\angle ADE = \\angle ZFE$, since the quadrilateral $ADEF$ is inscribed in the circle. Finally, we have $\\angle ZFE = \\angle AXB$ since the lines $BC$ and $EF$ are parallel. Putting these identities together, we get $\\angle YCD = \\angle AXB$, which implies that the sides $CD$ and $FA$ are parallel.\n\n**Alternate Solution:**\n\nSince the lines $AB$ and $DE$ are parallel, we have $\\angle ABE = \\angle BED$. Let $\\alpha$ be the common value of these angles. Similarly, we have $\\angle CBE = \\angle BEF$, whose value we call $\\beta$. Using the properties of quadrilaterals inscribed in a circle, we obtain\n\n$$\n\\angle FCD = 180^\\circ - \\angle DEF = 180^\\circ - \\angle BED - \\angle BEF = 180^\\circ - \\alpha - \\beta,\n$$\n\n$$\n\\angle CFA = 180^\\circ - \\angle ABC = 180^\\circ - \\angle ABE - \\angle CBE = 180^\\circ - \\alpha - \\beta.\n$$\n\nFrom this we get $\\angle FCD = \\angle CFA$ which implies that the sides $CD$ and $FA$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17684,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider matrices $A, B \\in \\mathcal{M}_3(\\mathbb{C})$ such that $A = -{}^tA$ and $B = {}^tB$. Prove that if the polynomial $f(x) = \\det(A + xB)$ has a multiple root, then\n$$\n\\det(A + B) = \\det B.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = \\det(A + xB)$, a degree at most 3 polynomial with complex coefficients. By properties of determinants:\n$$\n\\det(A + xB) = \\det({}^t(A + xB)) = \\det(-A + xB) = -\\det(A - xB) = -f(-x)\n$$\nfor all $x \\in \\mathbb{C}$, so $f$ is odd. Thus, the constant and quadratic terms vanish: $a = \\det A = 0$, so $f(x) = (\\det B)x^3 + bx$. If $f(x)$ has a multiple root, then $b = 0$, so $f(x) = (\\det B)x^3$. Therefore, $f(1) = \\det B$, i.e.,\n$$\n\\det(A + B) = \\det B.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17685,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $a, b, c, d$ 為非負實數，試求滿足下列方程組的解 $(a, b, c, d)$：\n\n$$\na^2(b+c)(b+c+d) = \\sqrt{b+c}\\sqrt[3]{b+c+d}\n$$\n\n$$\nb^2(c+d)(c+d+a) = \\sqrt{c+d}\\sqrt[3]{c+d+a}\n$$\n\n$$\nc^2(d+a)(d+a+b) = \\sqrt{d+a}\\sqrt[3]{d+a+b}\n$$\n\n$$\nd^2(a+b)(a+b+c) = \\sqrt{a+b}\\sqrt[3]{a+b+c}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "若 $a, b, c, d$ 其中有一項為 $0$，容易推得 $a = b = c = d = 0$ 為一解，因此以下不妨設 $a, b, c, d > 0$。由算幾不等式得\n\n$$\n\\frac{(b+c)}{2} \\frac{(b+c+d)}{3} \\ge \\sqrt{b+c}\\sqrt[3]{b+c+d} = a^2(b+c)(b+c+d) \\\\\n\\Rightarrow \\frac{1}{6} \\ge a^2\n$$\n\n因此 $a \\le \\frac{1}{\\sqrt{6}}$，同理可證 $b, c, d \\le \\frac{1}{\\sqrt{6}}$。將四式相乘得到\n\n$$\n\\begin{aligned}\n1 &= (b+c)(c+d)(d+a)(a+b)(b+c+d)(c+d+a)(d+a+b)(a+b+c) \\\\\n&\\le \\left(\\frac{1}{\\sqrt{6}} + \\frac{1}{\\sqrt{6}}\\right)^4 \\left(\\frac{1}{\\sqrt{6}} + \\frac{1}{\\sqrt{6}} + \\frac{1}{\\sqrt{6}}\\right)^4 \\\\\n&= 1\n\\end{aligned}\n$$\n\n等號成立若且唯若 $a = b = c = d = \\frac{1}{\\sqrt{6}}$。故原方程組的所有解為 $(a, b, c, d) = (0, 0, 0, 0)$ 或 $(\\frac{1}{\\sqrt{6}}, \\frac{1}{\\sqrt{6}}, \\frac{1}{\\sqrt{6}}, \\frac{1}{\\sqrt{6}})$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17686,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 2$. For every word of length $n$ (with at least one 'L' and at least one 'R'), Eva writes down how many turns she can take at most. For which number of 'L' characters does the word (or words) have the largest possible number of turns? (Give your answer in terms of $n$.)",
    "options": [],
    "answer": "See solution",
    "solution": "From the previous part, Eva can make at most $\\ell(n - \\ell)$ turns, where $\\ell$ is the number of 'L's. Consider $f(\\ell) = \\ell(n - \\ell)$, a quadratic with zeros at $\\ell = 0$ and $\\ell = n$. The maximum occurs at $\\ell = \\frac{n}{2}$.  \n\n- If $n$ is even, the maximum is at $\\ell = \\frac{n}{2}$, and the number of turns is $f\\left(\\frac{n}{2}\\right) = \\frac{1}{4}n^2$.\n- If $n$ is odd, the maximum is at $\\ell = \\frac{n-1}{2}$ and $\\ell = \\frac{n+1}{2}$, with $f\\left(\\frac{n-1}{2}\\right) = f\\left(\\frac{n+1}{2}\\right) = \\frac{1}{4}(n^2 - 1)$.\n\nThus, the answer is $\\ell = \\frac{n}{2}$ if $n$ is even, and $\\ell = \\frac{n-1}{2}$ or $\\ell = \\frac{n+1}{2}$ if $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17687,
    "subject": "Mathematics (Olympiad)",
    "question": "A polynomial $P(x)$ is called _nice_ if $P(0) = 1$ and the nonzero coefficients of $P(x)$ alternate between 1 and $-1$ when written in order. Suppose $P(x)$ is nice, and let $m$ and $n$ be two relatively prime positive integers. Show that\n\n$$\nQ(x) = P(x^n) \\cdot \\frac{(x^{mn} - 1)(x - 1)}{(x^m - 1)(x^n - 1)}\n$$\n\nis nice as well.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin by showing that $Q(x)$ is indeed a polynomial. Because $m$ and $n$ are relatively prime, the only common (complex) root of $x^m - 1$ and $x^n - 1$ is $x = 1$. Each root of $x^n - 1$ or $x^m - 1$ is a root of $(x^{mn} - 1)(x - 1)$, with $x = 1$ a double root, so\n\n$$\n\\frac{(x^{mn} - 1)(x - 1)}{(x^m - 1)(x^n - 1)}\n$$\n\nis a polynomial, and thus $Q(x)$ is as well.\n\nWe now establish a lemma giving an alternate characterization of nice polynomials.\n\n**Lemma.** If $P(x)$ is a polynomial with constant term 1, then $P(x)$ is nice if and only if each nonzero term in the power series expansion of $P(x)/(1-x)$ has coefficient 1.\n\n*Proof.* Suppose $P(x) = a_0 + a_1x + \\cdots$. The power series of $P(x)$ has coefficients\n\n$$\n\\frac{P(x)}{1-x} = b_0 + b_1x + b_2x^2 + \\cdots = a_0 + (a_0 + a_1)x + (a_0 + a_1 + a_2)x^2 + \\cdots\n$$\n\ngiven by the partial sums of the coefficients of $P(x)$. Since $P(0) = 1$, $b_0 = a_0 = 1$.\n\nIf $P(x)$ is nice, the nonzero coefficients of $P(x)$ alternate between 1 and $-1$, so the partial sums take value either 0 or 1. Thus $b_i \\in \\{0, 1\\}$, so all nonzero coefficients of the power series for $P(x)/(1-x)$ are 1.\n\nConversely, if $b_i \\in \\{0, 1\\}$, then $a_i = b_i - b_{i-1} \\in \\{-1, 0, 1\\}$, and\n\n$$\na_i = \\begin{cases} 1 & b_i > b_{i-1} \\\\ 0 & b_i = b_{i-1} \\\\ -1 & b_i < b_{i-1} \\end{cases}.\n$$\n\nBecause $b_i$ takes at most two values, among $i$ for which $b_i \\neq b_{i-1}$, the first and last cases alternate, which implies $P(x)$ is nice. This completes the proof of the lemma. $\\square$\n\nNow, since $P(0) = 1$, $Q(0) = 1$. By the lemma, it suffices to show that all nonzero terms in the power series for\n\n$$\n\\frac{Q(x)}{1-x} = \\frac{P(x^n)}{1-x^n} \\cdot \\frac{1-x^{mn}}{1-x^m}\n$$\n\nhave coefficient 1. By the lemma, all nonzero terms in the power series of $P(x)/(1-x)$ have coefficient 1, so the same is true for $\\frac{P(x^n)}{1-x^n}$. All nonzero terms of the power series expansion of $\\frac{P(x^n)}{1-x^n}$ have exponents congruent to 0 modulo $n$. Since $m$ and $n$ are relatively prime, $\\frac{1-x^{mn}}{1-x^m}$ is a polynomial whose nonzero coefficients are 1 and whose nonzero terms have exponents with distinct residues modulo $n$. Therefore, each nonzero term in the power series expansion of $\\frac{Q(x)}{1-x}$ can be uniquely written as the product of a nonzero term in $\\frac{P(x^n)}{1-x^n}$ and a nonzero term in $\\frac{1-x^{mn}}{1-x^m}$, so each such term has coefficient 1. Thus $Q(x)$ is nice by the lemma.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17688,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $f_n(x)$ be defined by\n\n$$\nf_n(x) = \\sum_{k=1}^{n} |x - k|.\n$$\n\nDetermine the solution to the inequality $f_n(x) < 41$ for every two-digit integer $n$ (in decimal notation).\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note first that the function $f_n(x)$ satisfies $f_n(x) = f_n(n+1-x)$:\n\n$$\nf_n(n+1-x) = \\sum_{k=1}^{n} |n+1-x-k| = \\sum_{k=1}^{n} |x - (n+1-k)| = \\sum_{k=1}^{n} |x-k| = f_n(x)\n$$\n\nby reversing the order of summation. Let us first consider the case $x < 1$: then, $x-k < 0$ for every $k \\ge 1$ and thus\n\n$$\nf_n(x) = \\sum_{k=1}^{n} (k - x) = \\frac{n(n+1)}{2} - nx > \\frac{n(n+1)}{2} - n = \\frac{n(n-1)}{2} \\ge \\frac{10 \\cdot 9}{2} = 45 > 41,\n$$\nso this case can be excluded. By symmetry, we can exclude $x > n$ as well. Thus we are left with $1 \\le x \\le n$. Suppose that $x \\in [\\ell, \\ell+1]$ for some integer $\\ell$ with $1 \\le \\ell \\le n-1$. Then $x-k \\le 0$ for $k \\ge \\ell+1$ and $x-k \\ge 0$ for $k \\le \\ell$, and we obtain\n\n$$\n\\begin{aligned}\nf_n(x) &= \\sum_{k=1}^{\\ell} (x-k) + \\sum_{k=\\ell+1}^{n} (k-x) = \\ell x - \\frac{\\ell(\\ell+1)}{2} + \\frac{(n-\\ell)(n+\\ell+1)}{2} - (n-\\ell)x \\\\\n&= \\frac{n(n+1)}{2} - \\ell(\\ell+1) + (2\\ell-n)x.\n\\end{aligned}\n$$\n\nThis shows that $f_n(x)$ is strictly decreasing on $[\\ell, \\ell+1]$ if $\\ell < \\frac{n}{2}$, constant on $[\\frac{n}{2}, \\frac{n}{2}+1]$ (if $n$ is even) and strictly increasing on $[\\ell, \\ell+1]$ if $\\ell > \\frac{n}{2}$. We conclude:\n\n- If $n$ is even, then $f_n(x)$ is strictly decreasing on $[1, \\frac{n}{2}]$, constant on $[\\frac{n}{2}, \\frac{n}{2}+1]$ and strictly increasing on $[\\frac{n}{2}+1, n]$.\n- If $n$ is odd, then $f_n(x)$ is strictly decreasing on $[1, \\frac{n+1}{2}]$ and strictly increasing on $[\\frac{n+1}{2}, n]$.\n\nWe see that the minimum of $f_n(x)$ is always attained at $m = \\lfloor \\frac{n+1}{2} \\rfloor$. Now we can complete squares to obtain\n\n$$\n\\begin{aligned}\nf_n(m) &= \\frac{n(n+1)}{2} - m(m+1) + (2m-n)m = \\frac{n(n+1)}{2} + m^2 - (n+1)m \\\\\n&= \\frac{n(n+1)}{2} + \\left(m - \\frac{n+1}{2}\\right)^2 - \\left(\\frac{n+1}{2}\\right)^2 \\ge \\frac{n(n+1)}{2} - \\left(\\frac{n+1}{2}\\right)^2 = \\frac{n^2-1}{4}.\n\\end{aligned}\n$$\n\nIf $n \\ge 13$, then this implies $f_n(x) \\ge f_n(m) \\ge \\frac{13^2-1}{4} = 42 > 41$ for all $x$, so that there is no solution. It remains to consider $n \\in \\{10, 11, 12\\}$.\n\n- For $n=10$, we obtain from above that\n\n$$\nf_{10}\\left(\\frac{3}{2}\\right) = 55 - 2 - 8 \\cdot \\frac{3}{2} = 41\n$$\n\nand by symmetry $f_{10}\\left(\\frac{19}{2}\\right) = 41$. In view of our monotonicity considerations, $f_{10}(x) \\ge 41$ for $x \\le \\frac{3}{2}$ and $x \\ge \\frac{19}{2}$ and $f_{10}(x) < 41$ on the remaining interval. So the solution set is $(\\frac{3}{2}, \\frac{19}{2})$.\n\n- For $n=11$, we have\n\n$$\nf_{11}\\left(\\frac{19}{7}\\right) = 66 - 6 - 7 \\cdot \\frac{19}{7} = 41\n$$\n\nand $f_{11}\\left(\\frac{65}{7}\\right) = 41$ by symmetry. The solution set is $(\\frac{19}{7}, \\frac{65}{7})$.\n\n- For $n=12$, we have\n\n$$\nf_{12}\\left(\\frac{17}{4}\\right) = 78 - 20 - 4 \\cdot \\frac{17}{4} = 41\n$$\n\nand $f_{12}\\left(\\frac{35}{4}\\right) = 41$ by symmetry. The solution set is $(\\frac{17}{4}, \\frac{35}{4})$.\n\n**Summary:**\n\n- $\\frac{3}{2} < x < \\frac{19}{2}$ for $n = 10$\n- $\\frac{19}{7} < x < \\frac{65}{7}$ for $n = 11$\n- $\\frac{17}{4} < x < \\frac{35}{4}$ for $n = 12$\n- No solutions if $n \\ge 13$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17689,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $ABCD$ е трапез со основи $AB$ и $CD$. На отсечката $AB$ е избрана произволна точка $X$, а на отсечката $CD$ произволна точка $Y$. Повлечени се отсечките $CX$ и $DX$ и отсечките $AY$ и $BY$. Со овие повлечени отсечки на цртежот се добиваат два триаголници и еден четириаголник (види цртеж; исенчени делови од трапезот). Покажи дека плоштината на исенчениот четириаголник е еднаква на збирот на плоштините на исенчените триаголници.",
    "options": [],
    "answer": "See solution",
    "solution": "На почеток ќе разгледаме произволен трапез $MNPQ$ на кој дијагоналите му се сечат во точката $S$. Нека $T_1$ е триаголникот $MSQ$, а $T_2$ триаголникот $MPS$. Ќе покажеме дека $P_{T_1} = P_{T_2}$. Јасно е дека $P_{MNQ} = P_{MNP}$, од каде имаме\n\n$$\nP_{T_1} = P_{MNQ} - P_{MNS} = P_{MNP} - P_{MNS} = P_{T_2}.\n$$\n\n![](images/Makedonija_2009_p16_data_c2dd574869.png)\n\nЌе ја повлечеме отсечката $XY$. Сега ќе ги разгледаме трапезите $AXYD$ и $XBCY$ одвоено, од каде следува тврдењето од задачата.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17690,
    "subject": "Mathematics (Olympiad)",
    "question": "Lucy starts by writing $s$ integer-valued 2023-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\\mathbf{v} = (v_1, \\dots, v_{2023})$ and $\\mathbf{w} = (w_1, \\dots, w_{2023})$ that she has already written, and apply one of the following operations to obtain a new tuple:\n\n$$\n\\mathbf{v} + \\mathbf{w} = (v_1 + w_1, \\dots, v_{2023} + w_{2023}) \\\\\n\\mathbf{v} \\lor \\mathbf{w} = (\\max(v_1, w_1), \\dots, \\max(v_{2023}, w_{2023}))\n$$\n\nand then write this tuple on the blackboard.\n\nIt turns out that, in this way, Lucy can write any integer-valued 2023-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible number is $s = 3$. We will solve the problem for $n$-tuples for any $n \\ge 3$, and show that $s = 3$ is the answer regardless of $n$.\n\nFor any tuple $\\mathbf{v}$, we denote by $\\mathbf{v}_i$ its $i$-th coordinate for $1 \\le i \\le n$. For a positive integer $n$ and a tuple $\\mathbf{v}$, let $n \\cdot \\mathbf{v}$ denote the tuple obtained by applying addition on $\\mathbf{v}$ with itself $n$ times. Let $\\mathbf{e}(i)$ denote the tuple with 1 in the $i$-th position and 0 everywhere else. We call a tuple \"positive\" if all of its entries are positive and \"negative\" if all of its entries are negative.\n\nWe will show that three tuples suffice, and then show that two tuples do not suffice.\n\n*Three tuples suffice.* Write $\\mathbf{c}$ for the tuple $(-1, -1, \\dots, -1)$.\n\nWe note that it is enough for Lucy to make the tuples $\\mathbf{e}(1), \\mathbf{e}(2), \\dots, \\mathbf{e}(n), \\mathbf{c}$; from those any other tuple $\\mathbf{v}$ can be made as follows. First choose a positive integer $k$ such that $k + \\mathbf{v}_i > 0$. Then she can make $\\mathbf{v}$ by just addition:\n\n$$\n\\mathbf{v} = (k + \\mathbf{v}_1) \\cdot \\mathbf{e}(1) + (k + \\mathbf{v}_2) \\cdot \\mathbf{e}(2) + \\dots + (k + \\mathbf{v}_n) \\cdot \\mathbf{e}(n) + k \\cdot \\mathbf{c}\n$$\n\nLucy can take her three starting tuples as $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ where $\\mathbf{a}_i = -i^2$ and $\\mathbf{b}_i = i$ for all $1 \\le i \\le n$. For any $1 \\le j \\le n$, write $\\mathbf{d}(j)$ for the tuple $2 \\cdot \\mathbf{a} + 4j \\cdot \\mathbf{b} + (2j^2 - 1)\\mathbf{c}$, which Lucy can make by adding together $\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$ repeatedly. This has $i$-th coordinate\n\n$$\n\\mathbf{d}(j)_i = -2i^2 + 4ji - (2j^2 - 1) = 1 - 2(i - j)^2\n$$\n\nThus $\\mathbf{d}(j)_i = 1$ iff $i = j$, and is $\\le -1$ otherwise. Hence Lucy can produce the tuple $\\mathbf{l} = (1, 1, \\dots, 1)$ by doing $\\mathbf{d}(1) \\lor \\mathbf{d}(2) \\lor \\dots \\lor \\mathbf{d}(n)$.\n\nThen she can produce the tuple $\\mathbf{0} = (0, 0, \\dots, 0)$ by doing $\\mathbf{1} + \\mathbf{c}$, and she can get $\\mathbf{e}(i)$ by doing $\\mathbf{d}(i) \\lor \\mathbf{0}$. Since she already has $\\mathbf{c}$, as argued earlier, she can produce all integer tuples.\n\n*Two tuples do not suffice.* We start with an observation: Let $A$ be a non-negative real number and suppose two tuples $\\mathbf{v}$ and $\\mathbf{w}$ satisfy $\\mathbf{v}_j \\ge A\\mathbf{v}_k$ and $\\mathbf{w}_j \\ge A\\mathbf{w}_k$ for some $1 \\le j \\ne k \\le n$. Then the same inequality holds for $\\mathbf{v} + \\mathbf{w}$ and $\\mathbf{v} \\lor \\mathbf{w}$. Indeed:\n\n$$\n(\\mathbf{v} + \\mathbf{w})_j = \\mathbf{v}_j + \\mathbf{w}_j \\ge A(\\mathbf{v}_k + \\mathbf{w}_k) = A(\\mathbf{v} + \\mathbf{w})_k \\\\\n\\max(\\mathbf{v}_j, \\mathbf{w}_j) \\ge \\max(A\\mathbf{v}_k, A\\mathbf{w}_k) = A\\max(\\mathbf{v}_k, \\mathbf{w}_k)\n$$\n\nsince $A \\ge 0$. Thus any tuple generated from these two tuples satisfies this inequality, and so we cannot generate all integer tuples.\n\nAssume that Lucy starts with two tuples $\\mathbf{v}$ and $\\mathbf{w}$. We will distinguish two cases. In the first case, there is some $1 \\le i \\le n$ such that $\\mathbf{v}_i, \\mathbf{w}_i \\ge 0$. Since both operations preserve sign, we cannot generate any tuple with a negative $i$-th coordinate. Similarly for $\\mathbf{v}_i, \\mathbf{w}_i \\le 0$.\n\nIn the second case, either $\\mathbf{v}_i > 0 > \\mathbf{w}_i$ or $\\mathbf{v}_i < 0 < \\mathbf{w}_i$ for all $1 \\le i \\le n$. Since $n \\ge 3$, by pigeonhole principle, there exist distinct indices $j \\ne k$ such that $\\mathbf{v}_j$ has the same sign as $\\mathbf{v}_k$ and $\\mathbf{w}_j$ has the same sign as $\\mathbf{w}_k$.\n\nWLOG assume $\\mathbf{v}_j, \\mathbf{v}_k > 0$ and $\\mathbf{w}_j, \\mathbf{w}_k < 0$. Let $A = \\frac{\\mathbf{v}_j}{\\mathbf{v}_k}$. If $\\frac{\\mathbf{w}_j}{\\mathbf{w}_k} \\ge A$, both inequalities $\\mathbf{v}_j \\ge A\\mathbf{v}_k$ and $\\mathbf{w}_j \\ge A\\mathbf{w}_k$ hold. If $\\frac{\\mathbf{w}_j}{\\mathbf{w}_k} \\le A$, both inequalities $\\mathbf{v}_k \\ge \\frac{1}{A}\\mathbf{v}_j$ and $\\mathbf{w}_k \\ge \\frac{1}{A}\\mathbf{w}_j$ hold. In any case, by the above observation, Lucy cannot obtain all integer tuples starting from just $\\mathbf{v}$ and $\\mathbf{w}$, so two tuples do not suffice. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17691,
    "subject": "Mathematics (Olympiad)",
    "question": "Es sei eine reelle Zahl $\\alpha$ gegeben.\n\nMan bestimme in Abhängigkeit von $\\alpha$ alle Funktionen $f: \\mathbb{R} \\to \\mathbb{R}$ mit\n\n$$\nf(f(x+y)f(x-y)) = x^2 + \\alpha y f(y)\n$$\n\nfür alle $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Wir ersetzen $x$ und $y$ wie folgt.\n\n- $x = y = 0$ zeigt $f(f(0)^2) = 0$, d.h. mit $C = f(0)$ haben wir $f(C^2) = 0$.\n- $x - y = C^2$ ergibt\n  $$\nf(0) = (y + C^2)^2 + \\alpha y f(y). \\qquad (1)\n  $$\n- $x + y = C^2$ ergibt\n  $$\nf(0) = (C^2 - y)^2 + \\alpha y f(y). \\qquad (2)\n  $$\n\nDie Gleichungen (1) und (2) implizieren $(y + C^2)^2 = (y - C^2)^2$, also $C^2 y = 0$ für alle $y \\in \\mathbb{R}$. Deshalb muss $C = 0$ sein.\n\nMit (1) oder (2) folgt $0 = y^2 + \\alpha y f(y)$.\n\n1. $\\alpha = 0$ ergibt den Widerspruch $y^2 = 0$ für alle reellen $y$.\n2. $\\alpha \\neq 0$ führt auf $f(y) = -\\frac{y}{\\alpha}$, wenn $y \\neq 0$.\n\nWegen $C = f(0) = 0$ gilt sogar $f(x) = -\\frac{x}{\\alpha}$ für $x \\in \\mathbb{R}$. Die Verifikationsprobe ergibt\n\n$$\nf\\left(-\\frac{x+y}{\\alpha}\\right) f\\left(-\\frac{x-y}{\\alpha}\\right) = x^2 + \\alpha y \\left(-\\frac{y}{\\alpha}\\right),\n$$\nd.h.\n$$\n-\\frac{1}{\\alpha} \\cdot \\frac{x^2 - y^2}{\\alpha^2} = x^2 - y^2\n$$\nfür alle $x, y \\in \\mathbb{R}$. Deshalb erhalten wir $\\alpha^3 = -1$, also $\\alpha = -1$.\n\nDamit haben wir gezeigt, dass es genau für $\\alpha = -1$ eine Lösung der Funktionalgleichung gibt, nämlich $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17692,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcircle of triangle $ABC$ with an obtuse angle at $B$. Let $B_1$ be the intersection of the line $AB$ and the tangent to the circle $O$ at the point $C$. Let $O_1$ be the circumcenter of triangle $AB_1C$. Choose an arbitrary point $B_2$ on the line segment $BB_1$ ($B_2 \\neq B, B_1$). The line from $B_2$ is tangent to the circle $O$ at $C_1$, closer to $C$. Let $O_2$ be the circumcenter of triangle $AB_2C_1$. Assume that the line $OO_2$ is perpendicular to the line $AO_1$. Show that the five points $O$, $O_2$, $O_1$, $C_1$, and $C$ are cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the line segment $AC_1$ is the common chord of the two circles $O$ and $O_2$, we have $AC_1 \\perp OO_2$. From this result and the given condition $OO_2 \\perp AO_1$, it follows that the point $O_1$ is on the line segment $AC_1$.\n\nFurthermore, since $OA = OC_1$, $\\angle OC_1A = \\angle OAO_1$ and $\\angle OCO_1 = \\angle OAO_1$, for $OA = OC$ and $O_1A = O_1C$. Therefore, $\\angle OC_1A = \\angle OCO_1$, which implies that the four points $O$, $O_1$, $C_1$, and $C$ are concyclic.\n\nBy noticing that\n\n$$\n\\angle C_1AB_2 = \\frac{1}{2}(\\pi - \\angle AOB_1) = \\frac{\\pi}{2} - \\angle ACB_1 = \\frac{\\pi}{2} - \\angle BCA - \\angle B_1CB = \\angle ABC - \\frac{\\pi}{2}\n$$\n\nand\n\n$$\n\\angle B_2C_1A = \\angle B_2C_1B + \\angle BC_1A = \\angle C_1AB_2 + \\angle BCA = \\frac{\\pi}{2} - \\angle CAB.\n$$\n\nwe have that\n\n$$\n\\angle AB_2C_1 = \\pi - \\angle C_1AB_2 - \\angle B_2C_1A = \\pi + \\angle CAB - \\angle ABC.\n$$\n\nFrom the identity $\\angle B_1CB + \\angle BCA = \\angle B_1CO_1 + \\angle O_1CA$, it follows that $\\angle CAB + \\angle BCA = (\\frac{\\pi}{2} - \\angle CAB) + \\angle C_1AC$, that is, $\\angle ABC - \\angle CAB = \\frac{\\pi}{2} - \\angle C_1AC < \\frac{\\pi}{2}$. Hence $\\angle AB_2C_1$ is an obtuse angle. Therefore $\\angle OO_2C_1 = \\angle AB_2C_1 = \\pi + \\angle CAB - \\angle ABC$.\n\nAnd\n\n$$\n\\begin{aligned}\n\\angle OO_1C_1 &= \\angle C_1O_1C + \\angle CO_1O \\\\\n&= 2\\angle C_1AC + \\angle CC_1O \\\\\n&= 2\\angle C_1AC + \\frac{1}{2}(\\pi - \\angle COC_1) \\\\\n&= \\frac{\\pi}{2} + \\angle C_1AC = \\frac{\\pi}{2} + \\angle CAB - \\angle C_1AB_2 = \\pi + \\angle CAB - \\angle ABC.\n\\end{aligned}\n$$\n\nSo it follows that $\\angle CO_1C_1 = \\angle OO_2C_1$, and thus the four points $O$, $O_2$, $O_1$, and $C_1$ are concyclic.\n\nTherefore, we conclude that $O$, $O_2$, $O_1$, $C_1$, and $C$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17693,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane rectangular coordinate system, given the hyperbola\n$$\n\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1 \\quad (a, b > 0),\n$$\na line with inclination angle $\\frac{\\pi}{4}$ passes through a vertex of $\\Gamma$ and another point $(2, 3)$ on it. Then the eccentricity of $\\Gamma$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "The slope of the line is $1$, and it passes through the point $(2, 3)$, so its equation is $y = x + 1$. This line intersects the $x$-axis at $(-1, 0)$, so $(-1, 0)$ is a vertex of $\\Gamma$. Thus, $a = 1$.\n\nSince $(2, 3)$ is on $\\Gamma$,\n$$\n\\frac{2^2}{1^2} - \\frac{3^2}{b^2} = 1,\n$$\nso $b^2 = 3$.\n\nLet $c = \\sqrt{a^2 + b^2}$. The eccentricity of $\\Gamma$ is $\\frac{c}{a} = \\frac{\\sqrt{1^2 + 3}}{1} = 2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17694,
    "subject": "Mathematics (Olympiad)",
    "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 side length less than or equal to $\\sqrt{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be an interior point of the hexagon, and thus also interior to both triangles. Denote by $a$ and $b$ the side lengths of the two triangles. The sum of the distances from $P$ to the sides of an equilateral triangle equals the altitude of the triangle, so the sum of the distances from $P$ to the lines $AB$, $BC$, $CA$, $MN$, $NP$, $PM$ is $$(a + b) \\frac{\\sqrt{3}}{2}.$$ On the other hand, the sum of the distances from $P$ to each pair of parallel lines ($AB$ and $MN$, $BC$ and $NP$, $CA$ and $PM$) is at most $1$ per pair, so the total is at most $3$. Therefore, $$(a + b) \\frac{\\sqrt{3}}{2} \\le 3,$$ which gives $a + b \\le 2\\sqrt{3}$. Thus, at least one of $a$ or $b$ satisfies $a \\le \\sqrt{3}$ or $b \\le \\sqrt{3}$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17695,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest integer $m$ for which there exist positive integers $n > k > 1$ such that:\n$$\n\\underline{11\\dots1} = \\underline{11\\dots1} \\cdot m\n$$\nwhere $\\underline{11\\dots1}$ denotes a number consisting of $n$ or $k$ consecutive ones.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $m = 101$.\n\n**Solution:**\nObviously, $m > 9$. If $m = \\overline{ab}$, where $a \\geq 1$, then the equality $\\underline{11\\dots1} = \\underline{11\\dots1} \\cdot \\overline{ab}$ implies that $b = 1$. But in this case, regardless of $a$, the second last digit of the product $\\underline{11\\dots1} \\cdot \\overline{ab}$ is equal to $a + 1$ if $a < 9$ or to $0$ if $a = 9$, hence, it can't be $1$. Therefore, $m \\geq 100$. Clearly, $m = 100$ doesn't satisfy the condition, because $\\underline{11\\dots1} \\cdot 100 = \\underline{11\\dots1} \\cdot 100$. On the other hand, $m = 101$ does, because $101 \\cdot 11 = 1111$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17696,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\na^2 + 2ab + 2b^2 = 13,\n$$\nwhere $a, b \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the equation $a^2 + 2ab + 2b^2 = 13$, we have $(a+b)^2 + b^2 = 13$. Since $a, b \\in \\mathbb{Z}$, $-3 \\leq a+b \\leq 3$. If $a+b \\geq 4$ or $a+b \\leq -4$, then $(a+b)^2 + b^2 \\geq 16 + b^2 \\geq 16 > 13$.\n\nPossible cases are $a+b = \\pm3$, $a+b = \\pm2$, $a+b = \\pm1$, and $a+b = 0$.\n\nIf $a+b = \\pm1$ or $a+b = 0$, then $b^2 = 12$ or $b^2 = 13$. These equations have no integer solutions.\n\n- If $a+b = -3$, then $b^2 = 4$, so $b = 2$ or $b = -2$.\n    - For $b = 2$, $a = -5$.\n    - For $b = -2$, $a = -1$.\n- If $a+b = 3$, then $b^2 = 4$, so $b = 2$ or $b = -2$.\n    - For $b = 2$, $a = 1$.\n    - For $b = -2$, $a = 5$.\n- If $a+b = -2$, then $b^2 = 9$, so $b = 3$ or $b = -3$.\n    - For $b = 3$, $a = -5$.\n    - For $b = -3$, $a = 1$.\n- If $a+b = 2$, then $b^2 = 9$, so $b = 3$ or $b = -3$.\n    - For $b = 3$, $a = -1$.\n    - For $b = -3$, $a = 5$.\n\nThus, the solutions are\n\n$$\n(a, b) \\in \\{ (-5, 2), (-1, -2), (1, 2), (5, -2), (1, -3), (-5, 3), (5, -3), (-1, 3) \\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17697,
    "subject": "Mathematics (Olympiad)",
    "question": "The Macedonian Mathematical Olympiad will take place in two rooms numbered 1 and 2. Initially, all students enter room 1. The final arrangement of students (by room) is determined as follows: The name of a student is called, and that student and all their acquaintances among the other students switch rooms. This procedure can be repeated any finite number of times, with any student's name. Thus, each list of called names leads to a final arrangement of students in rooms 1 and 2. Show that the total number of possible arrangements is not equal to $2009$. (Acquaintance is a symmetric relation.)",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the total number of possible final arrangements is an even number, so it cannot be $2009$.\n\nIt suffices to show that there exists a list of names such that all contestants move from room 1 to room 2. If this holds, then for every possible arrangement, the reverse arrangement (where every contestant is in the other room) is also possible, so the arrangements can be paired.\n\nWe prove this by induction on $n$, the number of contestants.\n\n**Base case:** $n=1$ is obvious.\n\n**Inductive step:** Assume the statement holds for $n$ contestants. Consider $n+1$ contestants. For every $n$ among them, there is a list of names that moves them to room 2. If, by one such list, the last contestant also moves to room 2, the thesis holds. Otherwise, for each of the $n+1$ contestants, there exists a 'good list' that moves the other $n$ to room 2, but that contestant remains in room 1. Consider two cases:\n\n1. If $n$ is odd, then the combined list of these $n+1$ good lists moves all contestants to room 2.\n2. If $n$ is even, then among the $n+1$ contestants, at least one has an even number of acquaintances among the others. Call that contestant's name first; after this, there is an even number of contestants in room 2, so we add their good lists. In this way, we construct a list that moves all contestants from room 1 to room 2.\n\nThus, the inductive proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17698,
    "subject": "Mathematics (Olympiad)",
    "question": "We call $n$ lines in the plane **three-way** if they can be separated into three nonempty sets, $X, Y, Z$. Every two lines from the same set are parallel to each other, no two lines from different sets are parallel to each other, and no three lines intersect at a point.\n\nLet $S_n$ denote the maximum number of regions into which $n$ three-way lines can divide the plane. A region is a connected part of the plane, not necessarily finite, whose boundaries are defined by three-way lines.\n\nWhat is the largest $n$ for which $S_n < 128$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the three sets have $|X| = x$, $|Y| = y$, and $|Z| = z$. The first two sets divide the plane into $(x+1)(y+1)$ regions. Each line in the third set intersects the others at $x + y$ points and is divided into $x + y + 1$ parts, each of which splits an existing region into two. Thus, the general formula is:\n\n$$\nS_{x,y,z} = (x+1)(y+1) + z(x+y+1) = x + y + z + xy + xz + yz + 1.\n$$\n\nLet $n = x + y + z$. Note that $3(xy + yz + zx) \\leq (x + y + z)^2 = n^2$ (since $(x - y)^2 + (y - z)^2 + (z - x)^2 \\geq 0$). So $S_{x,y,z} \\leq \\frac{n^2}{3} + n + 1$.\n\nFor $n = 18$, $S_{6,6,6} = 127$, so $S_{18} = 127$. For $n = 19$, $S_{6,6,7} = 140$, so $S_n > 128$ for $n \\geq 19$.\n\n**Therefore, the answer is $18$.**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17699,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle, $O$ its circumcenter, and $AD$ the bisector of angle $A$ where $D \\in BC$. Let $\\ell$ be the line passing through $O$ and parallel to the bisector $AD$. Prove that $\\ell$ passes through the orthocenter $H$ of triangle $ABC$ if and only if $ABC$ is isosceles or $\\angle BAC = 120^{\\circ}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the centroid of triangle $ABC$.\n\nAssume that $OH \\parallel AD$. Since $G$, $O$, $H$ are collinear, then $OG \\parallel AD$. Suppose that $AD$ meets the circumcircle of $ABC$ again at point $M$ and consider the midpoints $K$, $N$ of $MC$, $AC$, respectively. Let the lines $OG$ and $BK$ meet at $S$ and let the lines $MO$ and $BK$ meet at $T$. Then $NK \\parallel AD$ and $T$ is the centroid of the isosceles triangle $MBC$. Therefore we have\n\n$$\nOG \\parallel AD \\Rightarrow OG \\parallel NK \\Rightarrow \\frac{BS}{SK} = 2 \\Rightarrow \\frac{BS}{SK} = \\frac{BT}{TK} \\Rightarrow T = S.\n$$\n\nIf $S \\neq O$, then the lines $OS$, $OT$ and $AD$ coincide and $ABC$ is isosceles. If $T = S = O$, then triangle $MBC$ is equilateral, $\\angle BMC = 60^{\\circ}$ and $\\angle BAC = 120^{\\circ}$.\n\n![](images/Balkan_2012_shortlist_p16_data_9346475ced.png)\n\nTo prove the converse, assume that $\\angle BAC = 120^{\\circ}$. Then $\\angle BMC = 60^{\\circ}$ and triangle $MBC$ is equilateral. Hence $O$ coincides with the centroid $T$ of triangle $MBC$.\n\nSince $\\frac{BG}{GN} = 2 = \\frac{BO}{OK}$, it follows that $OH \\parallel OG \\parallel KN \\parallel AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17700,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of 2013-digit numbers $d_1 d_2 \\dots d_{2013}$ with odd digits $d_1, d_2, \\dots, d_{2013}$ such that\n\n$$\n  d_1 \\cdot d_2 + d_3 \\cdot d_4 + \\dots + d_{1809} \\cdot d_{1810} \\equiv 1 \\pmod{4},\n$$\n\nand\n\n$$\n  d_{1810} \\cdot d_{1811} + d_{1811} \\cdot d_{1812} + \\dots + d_{2012} \\cdot d_{2013} \\equiv 1 \\pmod{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following observation: Any odd numbers $x_1, \\dots, x_k$ satisfy\n\n$$\nx_1 x_2 + x_2 x_3 + \\dots + x_{k-1} x_k + x_k x_1 \\equiv k \\pmod{4}. \\quad (*)\n$$\n\nNote that the sum in $(*)$ is cyclic, unlike the ones in the statement. To justify $(*)$, reduce the $x_i$ mod 4; then they become $+1$'s or $-1$'s as odd numbers are congruent to $\\pm 1$ mod 4. Replacing an $x_i = -1$ by $x_i = 1$ does not change the mod 4 remainder of $S = \\sum_{j=1}^{k} x_j x_{j+1}$. The new and old values of $S$ differ by $2(x_{i-1} + x_{i+1})$, which is a multiple of 4 as $x_{i-1}, x_{i+1}$ are odd. So we may assume $x_i = 1$ for all $i$, then $S = k$ and $(*)$ is obvious.\n\nLet $d_1 d_2 \\dots d_{2013}$ satisfy the stated conditions. By $(*)$ we have $\\sum_{j=1}^{1810} d_j d_{j+1} \\equiv 1810 \\pmod{4}$ (here $d_{1810+1} = d_1$), hence $\\sum_{j=1}^{1809} d_j d_{j+1} \\equiv 1 \\pmod{4}$ if and only if $1810 - d_{1810} d_1 \\equiv 1 \\pmod{4}$, i.e. $d_1 d_{1810} \\equiv 1 \\pmod{4}$.\n\nSimilarly, $\\sum_{j=1810}^{2012} d_j d_{j+1} \\equiv 1 \\pmod{4}$ if and only if $(2013-1809) - d_{1810} d_{2013} \\equiv 1 \\pmod{4}$, i.e. $d_{1810} d_{2013} \\equiv -1 \\pmod{4}$. We see that the conditions depend only on the three digits $d_1, d_{1810}, d_{2013}$; the remaining 2010 digits $d_i$ can be chosen arbitrarily among 1, 3, 5, 7, 9.\n\nThere are 3 odd decimal digits $\\equiv 1 \\pmod{4}$, namely 1, 5, 9; there are 2 odd digits $\\equiv -1 \\pmod{4}$, namely 3, 7. Let $d_{1810} \\in \\{1,5,9\\}$. Then $d_1 d_{1810} \\equiv 1 \\pmod{4}$ and $d_{1810} d_{2013} \\equiv -1 \\pmod{4}$ imply $d_1 \\in \\{1,5,9\\}$, $d_{2013} \\in \\{3,7\\}$. So there are 3 choices for each of $d_1$ and $d_{1810}$, and 2 choices for $d_{2013}$. The choices are independent, which gives $3 \\cdot 3 \\cdot 2 = 18$ admissible choices for the triple $d_1, d_{1810}, d_{2013}$. Because there are 5 choices for each of the remaining 2010 digits $d_i$, we obtain $18 \\cdot 5^{2010}$ admissible numbers $d_1 d_2 \\dots d_{2013}$ with $d_{1810} \\in \\{1,5,9\\}$.\n\nLikewise, if $d_{1810} \\in \\{3,7\\}$ then $d_1 \\in \\{3,7\\}$, $d_{2013} \\in \\{1,5,9\\}$. Thus there are 2 choices for each of $d_1$ and $d_{1810}$, and 3 choices for $d_{2013}$, leading to $2 \\cdot 2 \\cdot 3 = 12$ admissible choices for the triple $d_1, d_{1810}, d_{2013}$. Like in the previous case, we obtain $12 \\cdot 5^{2010}$ admissible numbers with $d_{1810} \\in \\{3,7\\}$.\n\nIn summary, there are $18 \\cdot 5^{2010} + 12 \\cdot 5^{2010} = 6 \\cdot 5^{2011}$ admissible numbers in all.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17701,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k = 1, 2, \\dots, n$ and define:\n- $A_k = P_{4k-3}$\n- $B_k = P_{4k-2}$\n- $C_k = P_{4k-1}$\n- $D_k = P_{4k}$\n\nAlso, let $A_{n+1} = A_1$ and $B_{n+1} = B_1$.\n\nDefine the directed line segments:\n- $A_iB_i$ as leftward,\n- $B_iC_i$ as downward,\n- $C_iD_i$ as rightward,\n- $D_iA_{i+1}$ as upward,\nfor each $i$.\n\nA bent line segment $A_iB_iC_i$ is called leftdown type for each $i = 1, 2, \\dots, n$; similarly define downright, rightup, and upleft type bent segments.\n\nThe intersection (not at endpoints) of a leftward and a downward segment is the intersection of two leftdown type bent segments. The same applies for other types.\n\nTo solve the problem, find the maximum possible value for the sum of the number of intersections of two leftdown, two downright, two rightup, and two upleft type bent segments.\n\nA pair $(i, j)$ with $1 \\leq i \\neq j \\leq n$ is called a good pair if, for all four pairs of bent segments ($A_iB_iC_i$ and $A_jB_jC_j$, $B_iC_iD_i$ and $B_jC_jD_j$, $C_iD_iA_{i+1}$ and $C_jD_jA_{j+1}$, $D_iA_{i+1}B_i$ and $D_jA_{j+1}B_j$), the two bent segments intersect each other.\n\n**Lemma:** For any non-empty proper subset $X$ of $\\{1, 2, \\dots, n\\}$, let $Y = X^c$. Then, there exist $x \\in X$ and $y \\in Y$ such that $(x, y)$ is not a good pair.\n\n% IMAGE: ![](images/Japan_2013_p23_data_f1f11110bc.png)\n% IMAGE: ![](images/Japan_2013_p24_data_d868d48f3f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum number of pairs $(i, j)$ such that all four types of bent segments between $i$ and $j$ intersect is at most $4 \\times \\binom{n}{2} - (n-1) = (2n-1)(n-1)$.\n\nThis follows from the lemma: for any non-empty proper subset $X$ of $\\{1, 2, \\dots, n\\}$, there must exist $x \\in X$ and $y \\in Y$ such that $(x, y)$ is not a good pair. If there were fewer than $n-1$ not-good pairs, we could construct a subset $X$ contradicting the lemma. Thus, at least $n-1$ pairs are not good, and the maximum is $(2n-1)(n-1)$.\n\nBy constructing the points as described (placing $A_1, \\dots, A_n$ and $C_n, \\dots, C_1$ in down-right order, and $B_k, D_k$ accordingly), this bound is achieved.\n\n**Final answer:**\n$$\n(2n-1)(n-1)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17702,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a set of positive integers *nice* if the number of its elements equals the average of its numbers. Call a positive integer $n$ an *amazing* number if the set $\\{1, 2, \\dots, n\\}$ can be partitioned into nice subsets.\n\n(a) Prove that every perfect square is amazing.\n\n(b) Show that there are infinitely many positive integers which are not amazing.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $A = \\{1, 2, \\dots, n\\}$ and $n = k^2$ for some positive integer $k$. Denote $B$ as the set of the first $m$ odd positive integers. It's clear that $B$ is *nice*. Assume that $n \\ge 2m-1$, let $C = A \\setminus B$. $C$ is *nice* if and only if\n\n$$\n\\frac{(1 + 2 + \\cdots + n) - m^2}{n - m} = n - m \\iff n = n^2 - 4mn + 4m^2 \\iff m = \\frac{n-k}{2}.\n$$\n\nTherefore, we can choose $m$ so that $C$ is *nice*. Hence, $n$ is *amazing*.\n\nb) We prove that if $n = 4k + 2$ for some positive integer $k$, then $n$ is not *amazing*. Assume that $A = \\{1, 2, \\dots, n\\}$ can be partitioned into nice subsets $A_1, A_2, \\dots, A_m$ with cardinalities $a_1, a_2, \\dots, a_m$ respectively. Therefore,\n\n$$\n\\frac{n(n+1)}{2} = \\sum_{x \\in A} x = \\sum_{i=1}^{m} \\sum_{x \\in A_i} x = \\sum_{i=1}^{m} a_i^2 = \\sum_{i=1}^{m} a_i = n \\pmod{2},\n$$\n\nwhich yields a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17703,
    "subject": "Mathematics (Olympiad)",
    "question": "Call the number $\\overline{a_1a_2\\dots a_m}$ ($a_1 \\neq 0$, $a_m \\neq 0$) the *reverse* of the number $\\overline{a_m a_{m-1} \\dots a_1}$.\n\nProve that the sum of 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": "See solution",
    "solution": "Consider $n = \\overline{a_1a_2\\dots a_{m-1}a_m}$ and its reverse $r(n) = \\overline{a_m a_{m-1} \\dots a_2 a_1}$.\n\n$$\nn + r(n) = \\sum_{j=0}^{m} a_j (10^j + 10^{m-j})\n$$\n\nNotice that $10^j + 10^{m-j} \\equiv r \\pmod{81}$ for all $j$, where $r$ is a constant modulo $81$. Thus,\n$$\nn + r(n) \\equiv r \\sum_{j=0}^{m} a_j \\pmod{81}.\n$$\nSince $r$ and $81$ are coprime, $n + r(n)$ is divisible by $81$ if and only if $\\sum_{j=0}^{m} a_j$ is divisible by $81$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17704,
    "subject": "Mathematics (Olympiad)",
    "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 segments $[AB]$ and $[CD]$, respectively, prove that the lines $H_1H_2$ and $MN$ are parallel if and only if $AC = BD$.\n\n![](images/RMC_2015_BT_p75_data_f919690aec.png)",
    "options": [],
    "answer": "See 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 $COD$.\n\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 $BC$.\n\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.)\n\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.\n\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 $C_2$.)\n\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": 17705,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 16 balls with weights $13, 14, 15, \\ldots, 28$ grams, determine the balls with weights $13$, $14$, $27$, and $28$ grams using a two-pan balance at most 26 times.",
    "options": [],
    "answer": "See solution",
    "solution": "To find the lightest ball (13 grams), use direct elimination by pairs:\n\n- Make 8 pairs and select the lighter ball from each pair (8 attempts).\n- Pair the 8 winners into 4 pairs and select the lighter from each (4 attempts).\n- Pair the 4 winners into 2 pairs and select the lighter from each (2 attempts).\n- Compare the final 2 to find the lightest (1 attempt).\n\nTotal: $8 + 4 + 2 + 1 = 15$ attempts.\n\nBall 14 is among the balls eliminated by 13 in this process. There are 4 such balls; 14 is the lightest among them, found by direct elimination (2 + 1 = 3 attempts).\n\nNow, observe:\n- 28 is the only ball heavier than 13 and 14 combined.\n- 27 is the only ball equal in weight to 13 and 14 combined.\n\nBall 28 is among the 8 losers in the first 8 uses of the balance. Let these be $B_1, \\ldots, B_8$.\n\nFor each $i = 1, \\ldots, 8$, compare $B_i$ with the group (13, 14). One of these 8 attempts will find ball 28. If there is equilibrium in any of the remaining 7 attempts, ball 27 is found as well. Otherwise, 27 was a winner in the first round, which is possible only if it was compared with 28 in the first round. Since 28 is already known, so is 27; no further attempts are needed.\n\nTotal attempts: $15 + 3 + 8 = 26$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17706,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many positive integers $m$ for which there exist consecutive odd positive integers $p_m, q_m$ (with $q_m = p_m + 2$) such that the pairs $(p_m, q_m)$ are all distinct and\n\n$$\np_m^2 + p_m q_m + q_m^2, \\quad p_m^2 + m p_m q_m + q_m^2\n$$\n\nare both perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the relations $p^2 + pq + q^2 = u^2$ and $p^2 + m p q + q^2 = v^2$. Thus, $(m-1)pq = (v-u)(v+u)$. Suppose we choose $m-1 = r^2$ where $r$ is a positive integer and $v-u = r p$, $v+u = r q$. Then $u = (r q - r p)/2$ and $4(p^2 + p q + q^2) = (r q - r p)^2$. This leads to\n\n$$\n(r^2 - 4)p^2 - (2 r^2 + 4)p q + (r^2 - 4)q^2 = 0.\n$$\n\nSolving for $p/q$, we get\n\n$$\n\\frac{p}{q} = \\frac{2 r^2 + 4 \\pm \\sqrt{4(r^2 + 2)^2 - 4(r^2 - 4)^2}}{2(r^2 - 4)}.\n$$\n\nWe want the discriminant to be a perfect square. This forces $3(r^2 - 1)$ to be a perfect square, which leads to the Pell's equation $r^2 - 3 t^2 = 1$. The equation has infinitely many solutions $(r_n, t_n)$ given by\n\n$$\nr_n + t_n \\sqrt{3} = (2 + \\sqrt{3})^n.\n$$\n\nWe also have recurrence relations:\n\n$$\nr_{n+1} = 2 r_n + 3 t_n, \\quad t_{n+1} = r_n + 2 t_n,\n$$\n\nwhere $r_1 = 2$ and $t_1 = 1$. Induction shows that $t_{2l}$ is even for all $l \\ge 1$. We can express $p/q$ in terms of $t$:\n\n$$\n\\frac{p}{q} = \\frac{(t \\pm 1)^2}{t^2 - 1} = \\frac{t + 1}{t - 1} \\text{ or } \\frac{t - 1}{t + 1}.\n$$\n\nTake $m = r_{2l}^2 + 1$, $p_m = t_{2l} - 1$, $q_m = t_{2l} + 1$; we see that $p_l, q_l$ are consecutive odd integers. Moreover,\n\n$$\n\\begin{align*}\np_m^2 + p_m q_m + q_m^2 &= r_{2l}^2, \\\\\np_m^2 + m p_m q_m + q_m^2 &= (t_{2l} r_{2l})^2.\n\\end{align*}\n$$\n\nThis proves our claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17707,
    "subject": "Mathematics (Olympiad)",
    "question": "Mr. Precise wants to take his tea cup out of the microwave precisely at the front. The microwave of Mr. Precise is not precisely cooperative.\n\nThe two of them play the following game:\n\nLet $n$ be a positive integer. The rotating plate of the microwave takes $n$ seconds for a full turn. Each time the microwave is turned on, the plate is turned clockwise or counterclockwise for an integer number of seconds so that the tea cup can end up in $n$ possible positions. One of these positions is marked \"front\".\n\nAt the start of the game, the microwave rotates the tea cup to one of these positions. Afterwards, for each move, Mr. Precise enters the integer number of seconds, and the microwave decides whether to turn clockwise or counterclockwise.\n\nFor which $n$ can Mr. Precise ensure that after a finite number of moves, he can take out the tea cup of the microwave precisely from the front position?",
    "options": [],
    "answer": "See solution",
    "solution": "Mr. Precise can ensure his victory when $n$ is a power of 2.\n\nLabel the positions consecutively $0, 1, \\ldots, n-1$ where $0$ is the front position.\n\nIf $n$ is a power of 2, say $n = 2^k$, Mr. Precise can always enter the current position as the number of seconds. If the microwave turns the plate backwards, the tea cup will end up front immediately. Otherwise, the position number will be doubled and reduced modulo $2^k$ at each turn, and therefore will be divisible by $2^k$ after at most $k$ turns. This means the tea cup ends up at the front.\n\nNow, let $n = 2^k \\cdot m$, where $m > 1$ is an odd number.\n\nWe show that the microwave can always choose a position not divisible by $m$.\n\nThis is true for the first position, for example by choosing position $1$. After that, suppose the tea cup is in position $p$ not divisible by $m$ and Mr. Precise enters $s$ seconds. If both $p+s$ and $p-s$ were divisible by $m$, then their sum $2p$ would be divisible by $m$, so $m \\mid 2p$. Since $m$ is odd, this implies $m \\mid p$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17708,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A, B \\subseteq \\{0, 1, \\dots, 2016\\}$ be subsets with $|A| = 30$ and $|B| = 31$. Show that there exists an integer $n$ such that the set\n\n$$\n\\{a + nb \\pmod{2017} \\mid a \\in A,\\ b \\in B\\}\n$$\n\ncontains at least 730 distinct residues.",
    "options": [],
    "answer": "See solution",
    "solution": "First, for any finite family of finite sets $\\{X_i\\}_{i \\in I}$,\n$$\n|\\cup_i X_i| \\ge \\sum_i |X_i| - \\frac{1}{2} \\sum_{i \\ne j} |X_i \\cap X_j|.\n$$\nIf $x \\in \\cup_i X_i$ is in $k$ sets, it is counted $k$ times in the first sum and $k(k-1)$ times in the second; $1 \\ge k - \\frac{k(k-1)}{2}$ for $k \\ge 1$.\n\nLet $F = \\{1, 2, \\dots, 2016\\}$. For $a \\in A$ and $n \\in F$, define\n$$\nX_a^n = a + nB = \\{a + nb \\pmod{2017} \\mid b \\in B\\}.\n$$\nClearly, $|X_a^n| = |B|$. For $a \\ne a' \\in A$,\n$$\n|X_a^n \\cap X_{a'}^n| = \\sum_{b, b' \\in B} \\delta_{a+nb, a'+nb'} = \\sum_{\\substack{b, b' \\in B \\\\ b \\ne b'}} \\delta_{n, \\frac{a-a'}{b-b'}}\n$$\nusing Kronecker's delta. Summing over $n \\in F$,\n$$\n\\sum_{n=1}^{2016} |X_a^n \\cap X_{a'}^n| = |B|(|B| - 1).\n$$\nLet $X^n = \\bigcup_{a \\in A} X_a^n = A + nB$. Then\n$$\n|X^n| \\ge \\sum_{a \\in A} |X_a^n| - \\frac{1}{2} \\sum_{\\substack{a, a' \\in A \\\\ a \\ne a'}} |X_a^n \\cap X_{a'}^n|\n$$\nAveraging over $n \\in F$,\n$$\n\\frac{1}{|F|} \\sum_{n \\in F} |X^n| \\ge |A||B| - \\frac{1}{2} \\frac{|A|(|A|-1)|B|(|B|-1)}{|F|} = 30 \\cdot 31 - \\frac{30 \\cdot 29 \\cdot 31 \\cdot 30}{2 \\cdot 2016} > 729.\n$$\nThus, there exists $n \\in F$ such that $|X^n| \\ge 730$.",
    "topic": "Number Theory",
    "subtopic": "Residues and Primitive Roots"
  },
  {
    "id": 17709,
    "subject": "Mathematics (Olympiad)",
    "question": "In the parallelogram $ABCD$, point $G$ is chosen on side $AB$. Consider the circle through $A$ and $G$ that is tangent to the extension of $CB$ beyond $B$ at point $P$. The extension of $DG$ beyond $G$ intersects the circle at $L$. If the quadrilateral $GLBC$ is cyclic, prove that $AB = PC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the second common point of $DA$ and the circle. One can show that $A$ is between $D$ and $E$. Denote $\\angle ELG = \\alpha$, $\\angle GLC = \\beta$. Since $GLBC$ is a cyclic quadrilateral by hypothesis, we have $\\angle GBC = \\angle GLC = \\beta$; since $ELGA$ is also cyclic, $\\angle DAG = \\alpha$. Thus $\\alpha$ and $\\beta$ are the measures of adjacent angles in a parallelogram, hence $\\alpha+\\beta = 180^{\\circ}$. It follows that $E$, $L$ and $C$ are collinear.\n\n![](images/Argentina_2011_p35_data_b9549dc7f3.png)\n\nLet $\\angle LEA = \\theta$, then $\\angle LGB = \\theta$ as $ELGA$ is cyclic. Hence $\\angle LGB = \\angle LDC = \\theta$ due to $AB \\parallel CD$.\n\nTriangles $EDC$ and $DLC$ are similar as they share an angle at $C$ and $\\angle CED = \\angle CDL = \\theta$. The similitude gives $\\frac{CD}{CL} = \\frac{CE}{CD}$, or $CD^2 = CL \\cdot CE$. On the other hand, by power of a point, $CP^2 = CL \\cdot CE$. Hence $CD = CP$. Since $CD = AB$ (opposite sides of a parallelogram), we obtain $AB = CP$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17710,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n, \\dots$ be a sequence of positive real numbers. For every positive integer $n$, write\n\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": "See 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.\n\nLet $n_1$ be any positive integer, then use unboundedness of the $s_n$ to choose $n_k$ recursively so that $s_{n_k} > s_{n_{k-1}} + a_1$, $k \\ge 2$.\n\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$$\n\\frac{x_1}{1+x_1} + \\frac{x_2}{1+x_2} + \\dots + \\frac{x_m}{1+x_m} > \\frac{x_1+x_2+\\dots+x_m}{1+x_1+x_2+\\dots+x_m}. \\quad (*)\n$$\n\nAssume $(*)$ for the moment to write\n\n$$\n\\begin{aligned}\n\\sigma_{n_k} - \\sigma_{n_{k-1}} &= \\frac{a_{n_{k-1}+1}}{1+a_{n_{k-1}+1}} + \\dots + \\frac{a_{n_k}}{1+a_{n_k}} \\\\\n&> \\frac{a_{n_{k-1}+1} + \\dots + a_{n_k}}{1+a_{n_{k-1}+1} + \\dots + a_{n_k}} = \\frac{s_{n_k} - s_{n_{k-1}}}{1+s_{n_k} - s_{n_{k-1}}} > \\frac{a_1}{1+a_1}, \\quad k \\ge 2,\n\\end{aligned}\n$$\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\n**1st Proof.** Write\n\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\n**2nd Proof.** Induct on $m$. The base case, $m=2$, is a routine check. For $m \\ge 3$,\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{m} \\frac{x_k}{1+x_k} &= \\sum_{k=1}^{m-2} \\frac{x_k}{1+x_k} + \\left( \\frac{x_{m-1}}{1+x_{m-1}} + \\frac{x_m}{1+x_m} \\right) \\\\\n&> \\sum_{k=1}^{m-2} \\frac{x_k}{1+x_k} + \\frac{x_{m-1} + x_m}{1+x_{m-1} + x_m} > \\frac{\\sum_{k=1}^{m} x_k}{1 + \\sum_{k=1}^{m} x_k}.\n\\end{aligned}\n$$\n\n**Remark.** An obvious example of a sequence satisfying the condition in the statement is $a_n = 1/n$, $n \\ge 1$; in this case, the conclusion is trivial, as $\\sigma_n = s_{n+1} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17711,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive real numbers $c$ such that there are infinitely many pairs of positive integers $(n, m)$ satisfying the following conditions:\n\n- $n \\ge m + c\\sqrt{m-1} + 1$\n- Among the numbers $n, n+1, \\dots, 2n-m$, there is no square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that $c$ satisfies the condition in the statement if and only if $c \\le 2$.\n\n**Case 1: $c \\le 2$**\n\nFor any positive integer $k$, define\n$$\nn = k^2 + 1, \\quad m = (k-1)^2 + 1.$$\nObserve that\n$$\nm + c\\sqrt{m-1} + 1 \\le k^2 - 2k + 2 + 2(k-1) + 1 = k^2 + 1 = n.$$\nAnd\n$$\n(n, n+1, \\dots, 2n-m) = (k^2+1, k^2+2, \\dots, k^2+2k).\n$$\nTherefore, every such pair $(n, m)$ satisfies the property, and there are infinitely many such pairs.\n\n**Case 2: $c > 2$**\n\nLet $(n, m)$ be any pair of positive integers satisfying the property. For each $n$, the number $\\lceil\\sqrt{n}\\rceil^2$ is always between $n$ and $(\\sqrt{n}+1)^2$, so there is always a square in the range\n$$\nn, n+1, \\dots, n + \\lfloor 2\\sqrt{n} \\rfloor + 1.$$\nThis implies $2n - m < n + \\lfloor 2\\sqrt{n} \\rfloor + 1$, so\n$$\nm \\ge n - 2\\sqrt{n}.$$\nCombining with the problem's inequality:\n$$\nn \\ge n - 2\\sqrt{n} + c\\sqrt{n - 2\\sqrt{n} - 1} + 1. \\quad (1)$$\nFor $c > 2$, $c\\sqrt{n - 2\\sqrt{n} - 1} > 2\\sqrt{n}$ for large $n$, so (1) can only be satisfied for finitely many $n$.\n\nThus, there are infinitely many pairs $(n, m)$ if and only if $c \\le 2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17712,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Are there three pairwise distinct positive proper fractions $\\frac{a}{b}$, $\\frac{c}{d}$, $\\frac{x}{y}$, which satisfy the following conditions:\n\n$$\n\\frac{a}{b} + \\frac{c}{d} > \\frac{x}{y}\n$$\n\nand\n\n$$\n\\frac{b}{a} + \\frac{d}{c} > \\frac{y}{x}\n$$\n\nb) Are there two pairs of positive proper fractions $\\frac{a_1}{b_1}$, $\\frac{a_2}{b_2}$ and $\\frac{c_1}{d_1}$, $\\frac{c_2}{d_2}$, which satisfy the following conditions:\n\n$$\n\\frac{a_1}{b_1} + \\frac{a_2}{b_2} > \\frac{c_1}{d_1} + \\frac{c_2}{d_2}\n$$\n\nand\n\n$$\n\\frac{b_1}{a_1} + \\frac{b_2}{a_2} > \\frac{d_1}{c_1} + \\frac{d_2}{c_2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) For example,\n$$\n\\frac{1}{1000} + \\frac{999}{1000} > \\frac{2}{5}\n$$\nand\n$$\n\\frac{1000}{1} + \\frac{1000}{999} > \\frac{5}{2}\n$$\n\nb) For example,\n$$\n\\frac{1}{1000} + \\frac{999}{1000} > \\frac{2}{5} + \\frac{2}{5}\n$$\nand\n$$\n\\frac{1000}{1} + \\frac{1000}{999} > \\frac{5}{2} + \\frac{5}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17713,
    "subject": "Mathematics (Olympiad)",
    "question": "Un conjunto de rectas en el plano está en posición general si no hay dos que sean paralelas ni tres que pasen por el mismo punto. Un conjunto de rectas en posición general separa el plano en regiones, algunas de las cuales tienen área finita; a estas las llamamos sus regiones finitas. Demostrar que para cada $n$ suficientemente grande, en cualquier conjunto de $n$ rectas en posición general es posible colorear de azul al menos $\\sqrt{n}$ de ellas de tal manera que ninguna de sus regiones finitas tenga todos los lados de su frontera azules.",
    "options": [],
    "answer": "See solution",
    "solution": "Llamemos vértice a cada punto donde se cortan dos rectas del conjunto, vértice azul a cada vértice donde las dos rectas que se corten están coloreadas de azul, y polígono a cada región finita del plano delimitada por las rectas trazadas. Decimos que un polígono es casi azul si todos sus lados salvo uno son de color azul. Diremos que un polígono casi azul pertenece a una recta no azul, si al colorear dicha recta de azul, el polígono pasaría a tener toda su frontera azul. De todos los polígonos que pertenecen a una recta no azul, elegimos uno, y decimos que ese polígono es el favorito de dicha recta. Claramente, cada recta no azul tiene uno o ningún polígono favorito (que obviamente es casi azul), mientras que cada polígono casi azul puede ser el favorito de a lo sumo una recta no azul, o no ser el favorito de ninguna.\n\nSupongamos que hemos coloreado $a$ rectas de azul, sin que ningún polígono tenga toda su frontera azul, y que no podemos colorear más rectas de azul sin que aparezcan polígonos azules. Entonces, cada recta no azul que pintemos, generará al menos un polígono con toda su frontera azul, es decir, cada recta no azul tiene exactamente un polígono favorito. Cada vértice azul pertenece a lo sumo a cuatro polígonos. Asignaremos a dicho vértice azul un valor, de la siguiente forma:\n\n- Cada vértice azul parte de un valor base de 0.\n- Por cada polígono casi azul al que pertenece dicho vértice, y que sea el favorito de una recta no azul, sumamos 1 al valor del vértice si el polígono es un triángulo, y $\\frac{1}{2}$ si el polígono tiene al menos 4 lados.\n- Por cada polígono que no sea casi azul, o que sea casi azul pero no sea el favorito de ninguna recta no azul, sumamos 0 al valor del vértice.\n\nNótese entonces que\n\n(1) A los vértices de cada polígono que es el favorito de alguna recta no azul, se les suma, por pertenecer a dicho polígono, un valor total superior o igual a 1; en efecto, todo polígono casi azul de $n$ lados tiene $n - 2$ vértices azules. Si el polígono es un triángulo, hay 1 vértice azul, al que se le suma 1, mientras que si tiene 4 o más lados, hay al menos 2 vértices azules, a los que se les suma $\\frac{1}{2}$ a cada uno. Luego la suma $V$ de los valores de todos los vértices es superior o igual al número de polígonos favoritos, y por lo tanto superior o igual al número de rectas no azules, es decir, $V \\geq n - a$.\n\n(2) Cada vértice azul tiene un valor de a lo sumo 2. En efecto, si el vértice azul no pertenece a ningún triángulo casi azul favorito de una recta no azul, suma un valor de a lo sumo $\\frac{1}{2}$ para cada uno de los cuatro vértices a los que pertenece, para un valor total menor o igual que 2. Si el vértice azul pertenece a un triángulo casi azul que sea favorito de una recta no azul, esta recta delimita también dos de los cuatro polígonos a los que pertenece dicho vértice, por lo que o estos dos polígonos no pertenecen a ninguna recta, o pertenecen a la recta no azul de la que es favorito el triángulo, con lo que no pueden ser favoritos de ninguna recta. Habría en este caso a lo sumo dos de los cuatro polígonos que pueden aumentar el valor del vértice, cada uno de ellos aportando un valor máximo de 1, para un total menor o igual que 2. Como hay $a$ rectas azules, hay exactamente $\\binom{a}{2}$ vértices azules, cada uno de ellos con un valor menor o igual que 2, luego la suma $V$ de los valores de los puntos azules es a lo sumo $V \\leq \\binom{a}{2}$.\n\nDe lo anterior, concluimos que\n\n$$\nn - a \\leq V \\leq \\binom{a}{2} = a^2 - a, \\quad n \\leq a^2, \\quad \\sqrt{n} \\leq a,\n$$\n\ncomo queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17714,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a rectangle board $ABCD$ of size $m \\times n$ with $(m+1) \\times (n+1)$ intersections. Some engineers want to build a route from $A$ which goes along the segments parallel to the sides of the board, passes through each intersection exactly once, and finally returns to $A$.\n\n(a) Prove that they can build the route if and only if either $m$ is odd or $n$ is odd.\n\n(b) With $m, n$ satisfying the condition in (a), find the least number of intersections at each there is a turn.",
    "options": [],
    "answer": "See solution",
    "solution": "We number the rows $1$ to $m+1$ from left to right, the columns $1$ to $n+1$ from top to bottom, and suppose that the point $A$ is $(1, 1)$ (at the top left of the board).\n\n(a) *Necessary condition:* The route can be written as a letter sequence consisting of $L, R, U,$ and $D$, which respectively represent left, right, up, and down directions. Since there are $(m+1)(n+1)$ intersections, the length of the sequence is $(m+1)(n+1)$.\n\n![](images/VN_IMO_Booklet_2018_Final_p37_data_24538e3e9c.png)\n\nSince the route starts at $A$ and returns to $A$, it follows that the number of left turns is the same as the number of right turns, and the number of up turns is equal to the number of down turns. In other words, the number of letter *Ls* is the same as the number of letter *Rs*, as well as the number of letter *Us* and the number of letter *Ds* are equal. This implies $(m+1)(n+1)$ is even. Thus $m$ is odd or $n$ is odd.\n\n*Sufficient condition:* Without loss of generality, suppose that $m$ is odd. We will construct the route by the following rules:\n\n- The first horizontal movement: start at $A$, go through the columns $1$ to $n$.\n- Every vertical movement is only one unit; if we move horizontally, move between columns $2$ and $n$.\n- The last horizontal movement: move from column $n$ to $1$, then move vertically to $A$.\n\nThis process can be done since $m$ is odd. We reach the conclusion of (a).\n\n(b) Consider two intersections at each of which there is a turn and the distance between those two turns are the closest ($A$ is also counted as an intersection). Between the two intersections, there is a horizontal or vertical route.\n\nLet $r$ be the number of horizontal sub-routes of the route, $c$ be the number of vertical sub-routes of the route, and $k$ be the number of turns (not counted at $A$). We will prove the following remarks.\n\n**Remark 1.** $k + 1 = 2r = 2c$.\n\n*Proof.* For each intersection at which there is a turn, there is exactly one horizontal sub-route and vertical sub-route. The number of turns, also counted at $A$, is also the number of non-ordered pairs of the form\n\n(horizontal sub-route, vertical sub-route)\n\nwhere the horizontal sub-route and vertical sub-route intersect at some intersection. Moreover, each horizontal sub-route has common intersections with exactly two vertical sub-routes, and similarly each vertical sub-route has common intersections with exactly two horizontal sub-routes, thus\n\n$$\nk + 1 = 2r = 2c.\n$$\n\nTo find the least value of $k$, we only need to find the least value of $r$ and $c$.\n\n**Remark 2.** $r \\ge m + 1$ or $c \\ge n + 1$.\n\n*Proof.* Suppose that $r \\le m$, then there is a row on which each intersection lies on some vertical sub-route, which implies $c \\ge n + 1$. Similarly, if $c \\le n$ then $r \\ge m + 1$.\n\nWe consider the following cases:\n\n- If $m$ is odd, $n$ is even: suppose $r \\le m$, by the above remark, there is a row on which each of $n + 1$ intersections lies on some vertical sub-route, but since there is an odd number of intersections on this row, we cannot return to $A$, a contradiction. Thus $r \\ge m + 1$ and $k = 2r - 1 \\ge 2m + 1$. We can construct a route with exactly $m + 1$ vertical sub-routes similarly to (a). Therefore, $\\min k = 2m + 1$.\n\n- If $m$ is even, $n$ is odd: following the same pattern, we have $\\min k = 2n + 1$.\n\n- If $m, n$ are odd: Equalities occur in the inequalities $r \\ge m + 1$ and $c \\ge n + 1$, which implies that\n\n$$\n\\min k = 2 \\min(m, n) + 1.\n$$\n\nIn conclusion,\n\n- If $m, n$ are odd then $\\min k = 2 \\min(m, n) + 1$.\n- If $m$ is even and $n$ is odd then $\\min k = 2n + 1$.\n- If $m$ is odd and $n$ is even then $\\min k = 2m + 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17715,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $m$ is given to Alice and Bob. They play a game with the following rules:\n\n- To start the game, Alice writes a single nonzero digit on a board.\n- Bob and Alice then take turns writing a single digit at either end of the current number on the board.\n- A zero may be written at the end but not at the start of the number.\n- Bob wins and the game ends if at any time the number on the board is divisible by $m$.\n\n(i) What is the smallest value of $m$ such that Alice can prevent Bob from ever winning?\n\n(ii) Now suppose that Alice may start the game with any positive integer. All other rules remain the same. What is the smallest value of $m$ such that Alice can prevent Bob from ever winning?",
    "options": [],
    "answer": "See solution",
    "solution": "Answers:\n\n(i) $m = 12$\n(ii) $m = 11$\n\n---\n\n**(i)**\n\nIf $m \\leq 10$, then Bob has the winning strategy. After Alice writes the first digit, Bob has ten choices of digit ($0$ to $9$) to write at the end of the number. The numbers so formed will be ten consecutive integers, one of which must thus be divisible by $m$, and Bob chooses accordingly. If $m = 11$, whatever digit $d$ Alice writes at the start, Bob on his first turn also writes $d$ at the start (or end) to form the two-digit number $dd$, which is divisible by $11$.\n\nIf $m = 12$, then Alice now has the winning strategy. Say a number is a *12-blocker* if it is congruent to $5 \\pmod{6}$. Alice starts with $5$, which is a 12-blocker.\n\nWe will show that if the number on the board, $n$ say, at the end of Alice's turn is a 12-blocker, then Bob will not be able to form a number divisible by $12$ on his next turn. For Bob's new number to be divisible by $12$, it is necessary to add an even digit, $e$ say, at the end. Now $n \\equiv 5 \\pmod{6}$ implies $10n \\equiv 50 \\pmod{60}$, which implies $10n + e \\equiv 2 + e \\pmod{12}$. Since no single digit $e$ is congruent to $10 \\pmod{12}$, Bob cannot form a number divisible by $12$.\n\nIt remains to show that Alice can always write a 12-blocker on her turn. By adding a $1$, $3$, or $5$ to the end of Bob's number, Alice can yield all possible odd congruence classes modulo $6$, so she can always form a number which is $5 \\pmod{6}$.\n\n---\n\n**(ii)**\n\nBy the solution to (i), we know that the only candidates for $m$ are $11$ and $12$.\n\nWe will show that Alice has a winning strategy for $m = 11$. Say a number is an *11-blocker* if it has an odd number of digits and is congruent to $-1 \\pmod{11}$. Alice first writes $120$, which is an 11-blocker. If Bob adds a digit $e$ to the start of an 11-blocker $n$, this gives $10^k e + n$ where $k$ is odd. Since $10^k \\equiv -1 \\pmod{11}$, $10^k e + n \\equiv -e - 1 \\pmod{11}$. If Bob adds a digit $e$ to the end of $n$, this gives $10n + e \\equiv e + 1 \\pmod{11}$. Neither $e+1$ nor $-e-1$ can be divisible by $11$ for any digit $e$, and thus Bob cannot form a number divisible by $11$.\n\nIt remains to show that Alice can always write an 11-blocker on her turn. Starting with an 11-blocker $n$, if Bob writes the digit $e$, Alice copies and writes the same digit in the same position on her turn. This new number is an 11-blocker as the combination $ee$ is always a multiple of $11$, and also $100n \\equiv n \\pmod{11}$.\n\n---\n\n**Remark**\n\n- Another way to present the strategy in (i) is as follows. Given the number $x$, Alice considers whether $100x + 10 + d$ is a multiple of $12$ for some digit $d$. If a number is a multiple of $12$, then its last two digits must be a multiple of $4$. So there are three cases:\n  - $100x + 12$ is a multiple of $12$. Then $100x + 50 + d$ is not a multiple of $12$ for any digit $d$. Alice can write the digit $5$ at the end.\n  - $100x + 16$ is a multiple of $12$. Then $100x + 30 + d$ is not a multiple of $12$ for any digit $d$. Alice can write the digit $3$ at the end.\n  - $100x + 10 + d$ is not a multiple of $12$ for any digit $d$. So Alice can write the digit $1$ at the end.\n\n- An alternative strategy for (ii) is as follows. Suppose Bob's number $B$ has an even number $k$ of digits and is equal to say $c \\neq 0 \\pmod{11}$. If $c \\neq 10 \\pmod{11}$, then Alice writes the digit $10-c$ at the start of $B$, so her number is $(10-c)10^k + B$. Since $k$ is even, this must be an 11-blocker. If $c=10$, then Alice writes $9$ at the end of $B$, making $10B+9$, which is also an 11-blocker.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17716,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $0 \\leq a, b, c \\leq 1$ be distinct real numbers. Determine the minimum value of\n$$\n\\frac{1}{|a-b|^3} + \\frac{1}{|b-c|^3} + \\frac{1}{|c-a|^3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value is $17$, achieved when $a = 1$, $b = \\frac{1}{2}$, $c = 0$ (or any permutation).\n\nWithout loss of generality, let $a \\geq b \\geq c$ and $b = \\frac{a+c}{2} + t$ where $-\\frac{a+c}{2} < t < \\frac{a+c}{2}$. Then\n$$\n\\frac{1}{|a-b|^3} + \\frac{1}{|b-c|^3} + \\frac{1}{|c-a|^3} = \\frac{1}{\\left(\\frac{a-c}{2} + t\\right)^3} + \\frac{1}{\\left(\\frac{a-c}{2} - t\\right)^3} + \\frac{1}{(a-c)^3}.\n$$\nWe show this is minimized when $t = 0$. By AM-GM,\n$$\n\\frac{1}{\\left(\\frac{a-c}{2} + t\\right)^3} + \\frac{1}{\\left(\\frac{a-c}{2} - t\\right)^3} \\geq \\frac{2}{\\sqrt{\\left(\\frac{a-c}{2} + t\\right)^3 \\left(\\frac{a-c}{2} - t\\right)^3}} = \\frac{2}{\\sqrt{\\left(\\left(\\frac{a-c}{2}\\right)^2 - t^2\\right)^3}},\n$$\nwhich is minimized at $t = 0$. Thus, the minimum is\n$$\n\\frac{2}{\\left(\\frac{a-c}{2}\\right)^3} + \\frac{1}{(a-c)^3} = \\frac{17}{(a-c)^3} \\geq 17.\n$$\nSince $17$ can be achieved, it is the minimum.\n\n**Remark.** This inequality is non-standard: the minimum does not occur when variables are equal or all at the boundary ($0$ or $1$). The answer is a surprisingly large integer, and standard inequalities do not directly apply.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17717,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 4 identical fair dice. Let $x_i$ ($1 \\leq x_i \\leq 6$) be the number of dots on the face appearing on the $i$-th die, for $1 \\leq i \\leq 4$.\n\n1. Find the number of possible tuples $(x_1, x_2, x_3, x_4)$.\n2. Find the probability that there exists a number $x_j$ such that $x_j$ is equal to the sum of the remaining numbers.\n3. Find the probability that we can divide $x_1, x_2, x_3, x_4$ into 2 groups that have the same sum.",
    "options": [],
    "answer": "See solution",
    "solution": "1. By the multiplication principle, the number of possible tuples is $6^4 = 1296$.\n\n2. There are 4 choices for $x_j$ to be equal to the sum of the other three. For each $x_j$, the number of ways to choose the other three numbers is $\\binom{x_j-1}{2}$ (stars and bars). For $x_j = 3, 4, 5, 6$:\n\n$$\n\\binom{2}{2} + \\binom{3}{2} + \\binom{4}{2} + \\binom{5}{2} = 1 + 3 + 6 + 10 = 20\n$$\n\nSo, $4 \\times 20 = 80$ tuples satisfy the condition. The probability is $\\frac{80}{1296} = \\frac{5}{81}$.\n\n3. Let $S_1, S_2, S_3$ be the sets of tuples satisfying:\n\n$$\nx_1 + x_2 = x_3 + x_4, \\quad x_1 + x_3 = x_2 + x_4, \\quad x_1 + x_4 = x_2 + x_3.\n$$\n\nConsider $x_1 + x_2 = x_3 + x_4$. Set $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, $y_3 = 6 - x_3$, $y_4 = 6 - x_4$, so $0 \\leq y_i \\leq 5$ and $y_1 + y_2 + y_3 + y_4 = 10$.\n\nThe number of non-negative integer solutions is $\\binom{13}{10} = 286$. Subtract cases where any $y_i \\geq 6$ (each $35$ ways, disjoint), so $286 - 4 \\times 35 = 146$ tuples for $x_1 + x_2 = x_3 + x_4$.\n\nIntersections $|S_1 \\cap S_2| = |S_2 \\cap S_3| = |S_3 \\cap S_1| = 36$ (when two pairs are equal, e.g., $x_1 = x_4$, $x_2 = x_3$). Triple intersection $|S_1 \\cap S_2 \\cap S_3| = 6$ (all $x_i$ equal).\n\nBy inclusion-exclusion:\n\n$$\n|S| = 3 \\times 146 - 3 \\times 36 + 6 = 336\n$$\n\nAdd the 80 tuples from part 2: $336 + 80 = 416$. The probability is $\\frac{416}{1296} = \\frac{26}{81}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17718,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $f(x)$ a polynomial of degree $n$ with $n$ distinct real positive roots. Are there positive integers $k \\geq 2$ and a real polynomial $g(x)$ such that\n\n$$\nx(x+1)(x+2)(x+4)f(x) + 1 = (g(x))^k?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_n$ be the roots of $f(x)$. Assume that\n\n$$\nx(x+1)(x+2)(x+4)f(x) + a = g^k(x).\n$$\n\nNote that $a = b^k = g^k(0)$.\n\nIf $k \\geq 3$ is odd, then the polynomial $g^k(x)-b^k$ has $n+4$ distinct real roots, which would also be roots of $g(x)-b$. However, the degree of $g(x)-b$ is $(n+4)/k < n+4$, i.e., $g(x) = b$, which is impossible.\n\nNow, it is enough to prove that $k=2$ is also impossible. We have $a = b^2$, where we can assume $b > 0$. Then\n\n$$\nx(x+1)(x+2)(x+4)f(x) = g_1(x)g_2(x),\n$$\n\nwhere $g_1(x) = g(x)+b$ and $g_2(x) = g(x)-b$. The roots of $g_1(x)$ and $g_2(x)$ are the numbers $-4, -2, -1, 0, \\alpha_1, \\dots, \\alpha_n$. Since $g_1(x) > g_2(x)$ for every $x$, the number $-4$ is a root of $g_1(x)$. Since the derivatives of $g_1(x)$ and $g_2(x)$ coincide, Rolle's theorem shows that $-2$ and $-1$ are roots of $g_2(x)$ while $0$ is a root of $g_1(x)$.\n\nLet $g_1(x) = x(x+4) \\prod_{j=1}^{s}(x - \\alpha_j)$. Then\n\n$$\n|g_1(-1)| = 3 \\prod_{j=1}^{s}(1 + \\alpha_j) < 4 \\prod_{j=1}^{s}(2 + \\alpha_j) = |g_1(-2)|,\n$$\n\nwhich contradicts $g_1(-1) = g_1(-2) = g(-1) + b = 2b$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17719,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 2021 distinct positive integers $a_1, a_2, \\dots, a_{2021}$. Define the sequence $\\{a_n\\}$ inductively as follows: for each integer $n \\ge 2022$, $a_n$ is the smallest positive integer different from $a_1, a_2, \\dots, a_{n-1}$ and not dividing the product $a_{n-1} a_{n-2} \\dots a_{n-2021}$. Prove that there exists a positive integer $M$ such that all integers greater than or equal to $M$ appear in $\\{a_n\\}$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k = 2021$. In fact, we will prove the statement for any integer $k > 0$.\n\n**Lemma 1** There exists $C > 0$ independent of $k$, such that $\\tau(m) \\leq C m^{\\frac{1}{k+1}}$ holds for all positive integers $m$, where $\\tau(m)$ is the number of positive factors of $m$.\n\n**Proof of Lemma 1** Let $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$ be the prime factorization of $m$. Evidently,\n\n$$\n\\tau(m) = (\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_r + 1).\n$$\n\nEquivalently, there exists $C > 0$ such that\n\n$$\n\\prod_{i=1}^{r} \\frac{(\\alpha_i + 1)^{k+1}}{p_i^{\\alpha_i}} \\leq C^{k+1}\n$$\n\nholds for all primes $p_i$ and positive integers $\\alpha_i$. Note that polynomials grow slower than exponential functions, and hence for each prime $p \\leq 2^{k+1}$, the fraction $\\frac{(\\alpha + 1)^{k+1}}{p^\\alpha}$ has an upper bound when $\\alpha \\geq 0$. Since the number of primes $p \\leq 2^{k+1}$ is finite, there exists $C'$ such that for any prime $p \\leq 2^{k+1}$,\n\n$$\n\\frac{(\\alpha + 1)^{k+1}}{p^{\\alpha}} < C'.\n$$\n\nOn the other hand, for $p > 2^{k+1}$,\n\n$$\n\\frac{(\\alpha + 1)^{k+1}}{p^{\\alpha}} < \\left( \\frac{\\alpha + 1}{2^{\\alpha}} \\right)^{k+1} < 1.\n$$\n\nIt follows that $\\prod_{i=1}^{r} \\frac{(\\alpha_i + 1)^{k+1}}{p_i^{\\alpha_i}} \\leq {C'}^s$, where $s$ is the number of distinct prime factors of $m$ less than $2^{k+1}$. The existence of $C > 0$ is now verified.\n\n**Lemma 2** For any fixed integer $L > 0$, there exists $M > 0$ such that for any positive integer $m \\geq M$, one can find a prime power $p^\\alpha \\mid m$, and $p^\\alpha > L$.\n\n**Proof of Lemma 2** Consider all prime powers less than or equal to $L$: there are only finitely many of them. Define $M$ as the product of them plus 1. If $m \\geq M$ and $p \\mid m$, then its highest power in $m$ satisfies $p^\\alpha \\mid m$, and $p^\\alpha > L$.\n\n**Lemma 3** There exists a positive integer $D$ such that $a_n < 2n + D$ for every positive integer $n$.\n\n**Proof of Lemma 3** We take $D$ satisfying $D > \\max_{1 \\leq s \\leq k} \\{ a_s \\}$ and\n\n$$\nD > 2^k C^{k+1} + C D^{\\frac{k}{k+1}},\n$$\n\nwhere $C > 0$ is defined in Lemma 1, and use induction to prove $a_n < 2n + D$ for every $n$. By the choice of $D$, $a_n < 2n + D$ is true for $n = 1, 2, \\dots, k$. Assume $a_n < 2n + D$ holds for $1, 2, \\dots, n + k - 1$, where $n$ is a positive integer. Now for $a_{n+k}$, it is known from the problem that $a_{n+k}$ is the smallest positive integer not dividing $a_n a_{n+1} \\dots a_{n+k-1}$, and different from $a_1, \\dots, a_{n+k-1}$. Hence,\n\n$$\na_{n+k} \\leq \\tau(a_n a_{n+1} \\dots a_{n+k-1}) + n + k.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17720,
    "subject": "Mathematics (Olympiad)",
    "question": "Figure shows two non-intersecting circles $\\alpha$ and $\\beta$ in space. We say that circle $\\alpha$ devours circle $\\beta$ since one chord of $\\beta$ (solid) is strictly contained in a chord of $\\alpha$ (dashed).\n\nThe question is whether it is possible to place three circles $\\alpha, \\beta$ and $\\gamma$ in space so as to have $\\alpha$ devouring $\\beta$, $\\beta$ devouring $\\gamma$, and $\\gamma$ devouring $\\alpha$. The radii of the circles need not be equal.\n\n![](images/bw20longlist_5_p32_data_42a1d2c53d.png)\n\n*Note.* PSC found that the official solution is incorrect and there exists an example when it is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: No, it is impossible.\n\nConsider a point $X$ on a chord drawn in circle $\\alpha$. The power of $X$ with respect to $\\alpha$ is given by the familiar expression $p_{\\alpha}(X) = -xy$.\n\nLet us examine the case of two circles. The situation when $\\alpha$ devours $\\beta$ is represented in the figure. The point $X$, lying on the common chord, is coplanar with both circles, and so we may compute its power with respect to both of them. Evidently $p_{\\alpha}(X) < p_{\\beta}(X)$, for the chord in $\\beta$ is contained within the chord in $\\alpha$, and the distances to the periphery are correspondingly smaller.\n\n![](images/bw20longlist_5_p32_data_c90e7bfc95.png)\n\n![](images/bw20longlist_5_p32_data_330881ca8d.png)\n\nSuppose finally we have three circles $\\alpha, \\beta, \\gamma$, somehow cyclically devouring each other, as requested per the problem. Each of the three circles determines a plane; their common point $X$ will be coplanar with all the circles. Using the above reasoning, we arrive at the contradiction\n\n$$\np_{\\alpha}(X) < p_{\\beta}(X) < p_{\\gamma}(X) < p_{\\alpha}(X).\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 17721,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying the equation\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2\n$$\nfor all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, for any $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying the given equation,\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2 \\quad (1)\n$$\n\nwe can establish the following equalities:\n\n$$\nf(x - f(f(x))) = -2, \\quad (2)\n$$\n\n$$\nf(f(x)) = f(x + 2), \\quad (3)\n$$\n\n$$\nf(x + 2) - f(x) = c, \\quad (4)\n$$\n\nwhere $c$ is a constant. In particular, (2) shows that $-2$ is in the range of $f$.\n\n**Case I:** $c = 0$.\n\nLet $f(0) = a$, $f(1) = b$. From (4), $f(x+2) = f(x)$, so $f(x) = a$ for all even $x$ and $f(x) = b$ for all odd $x$. One of $a$, $b$ equals $-2$ by (2). Suppose $a \\neq b$. Then\n\n$$\nf(x) = f(y) \\iff x \\equiv y \\pmod{2}.\n$$\n\nFrom (3), $f(x) \\equiv x \\pmod{2}$, so $f(0) = a = -2$. Setting $x = 0$, $y = 1$ in (1),\n\n$$\nf(-b) = f(-2) - b - 2 = -b - 4.\n$$\n\nBut $f(-b)$ is either $b$ or $-2$. In both cases, $b = -2$, contradicting $a \\neq b$. Thus, $a = b = -2$, so $f(x) = -2$ for all $x$, which satisfies the original equation.\n\n**Case II:** $c \\neq 0$.\n\nLet $f(0) = a$, $f(1) = b$. From (4),\n\n$$\nf(2n) = cn + a, \\quad f(2n + 1) = cn + b. \\quad (5)\n$$\n\nAs $x \\to \\infty$, $f(x) \\to \\infty$ and $\\lim_{x \\to \\infty} \\frac{f(x)}{x} = \\frac{c}{2}$. From (3),\n\n$$\n\\frac{f(f(x))}{f(x)} = \\frac{f(x+2)}{f(x)} \\to 1 \\text{ as } x \\to \\infty,\n$$\n\nbut also $\\frac{f(f(x))}{f(x)} \\to \\frac{c}{2}$, so $c = 2$. Then (5) becomes\n\n$$\n\\begin{aligned}\nf(x) &= x + a & \\text{for even } x, \\\\\nf(x) &= x + d & \\text{for odd } x \\quad (\\text{where } d = b - 1).\n\\end{aligned} \\quad (6)\n$$\n\n1) If $f$ is injective, (3) implies $f(x) = x + 2$, which satisfies (1).\n\n2) If $f$ is not injective, from (6), $f$ is injective on even and odd numbers separately. Thus, some $f(2n)$ and $f(2m+1)$ are equal, so $2n - 2m - 1 = d - a$, i.e., $a$ and $d$ are of different parity. All values of $f$ are of the same parity, and since $-2$ is in the range, all $f(x)$ are even. From (3), $f(x+2) = f(f(x)) = f(x) + a$, so (4) gives $a = c = 2$. In particular, $d$ is odd. Thus,\n\n$$\n\\begin{aligned}\nf(x) &= x + 2 & \\text{for even } x, \\\\\nf(x) &= x + d & \\text{for odd } x,\n\\end{aligned}\n$$\n\nwhere $d$ is any odd integer. For any odd $d$, this function is a solution of (1).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17722,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A *good word* is a sequence of $3n$ letters, in which each of the letters $A$, $B$, and $C$ appears exactly $n$ times. Prove that for every good word $X$ there exists a good word $Y$ such that $Y$ cannot be obtained from $X$ by swapping neighbouring letters fewer than $\\frac{3}{2}n^2$ times.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us define the distance between good words $X$ and $Y$, denoted by $d(X, Y)$, as the smallest number of swaps of neighbouring letters necessary to obtain $Y$ from $X$ (or vice versa). Note that $d(X, Y) = d(Y, X)$, and for any three good words $X$, $Y$, and $Z$ we have\n\n$$\nd(X, Y) + d(Y, Z) \\ge d(X, Z).\n$$\n\nFor a good word $X$, let $F(X)$ be the number of pairs of positions where the letter in the left position is lexicographically smaller than the one in the right position (i.e., pairs of the form $AB$, $AC$, or $BC$). Swapping two neighbouring letters in a good word $X$ yields a good word $X'$. If the letters are identical, the words are equal, so all swaps involve pairs of different letters. For such swaps, $|F(X) - F(X')| = 1$. Thus,\n\n$$\nd(X, Y) \\ge |F(X) - F(Y)|\n$$\n\nfor any two good words $X$ and $Y$. Consider the good words\n\n$$\nP = \\underbrace{AA\\dots A}_{n}\\underbrace{BB\\dots B}_{n}\\underbrace{CC\\dots C}_{n}, \\quad Q = \\underbrace{CC\\dots C}_{n}\\underbrace{BB\\dots B}_{n}\\underbrace{AA\\dots A}_{n}\n$$\n\nNote that $F(P) = 3n^2$ and $F(Q) = 0$, so $d(P, Q) \\ge 3n^2$.\n\nFor any good word $X$,\n\n$$\nd(P, X) + d(X, Q) \\ge d(P, Q) \\ge 3n^2.\n$$\n\nTherefore, one of the good words $P$ or $Q$ cannot be obtained from $X$ by swapping fewer than $\\frac{3}{2}n^2$ pairs of neighbouring letters.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17723,
    "subject": "Mathematics (Olympiad)",
    "question": "Group the numbers into 7 groups according to their remainders upon division by 7:\n\n$$\n\\begin{aligned}\n\\{1, 8, 15, \\dots, 498\\}, & \\{2, 9, 16, \\dots, 499\\}, \\{3, 10, 17, \\dots, 500\\}, \\\\\n& \\{4, 11, 18, \\dots, 494\\}, \\{5, 12, 19, \\dots, 495\\}, \\{6, 13, 20, \\dots, 496\\}, \\{7, 14, 21, \\dots, 497\\}.\n\\end{aligned}\n$$\n\nHow many numbers can be selected from $\\{1,2,\\dots,500\\}$ such that no two selected numbers sum to a multiple of 7?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if a number from a group that leaves remainder $x$ is not deleted, then all the numbers from the group that leaves remainder $7-x$ must be deleted, otherwise two numbers will sum to a multiple of 7. In particular, only one number from the last group above may be selected.\n\nAlso, if one number from a group (besides the last one) is selected, we may select all numbers in that group, since they all leave the same remainder when divided by 7. The first three groups contain 72 numbers each, while the last four contain 71 numbers each. If we select the first three groups, we may not select any number in the next three groups, and finally we can select one number from the last group. This yields:\n\n$$\n72 \\times 3 + 1 = 217\n$$\n\nThus, the maximum number of such numbers is $217$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17724,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence contains marbles numbered from 1 to 2012. Initially, all marbles are black. In the second step, Jure replaces every black marble whose number is divisible by 3 or 5 with a red or yellow marble. In the third step, he replaces every remaining black marble whose number is divisible by 7 with a blue marble. How many black marbles remain in the sequence at the end?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us calculate the number of black marbles after each step:\n\n- The number of marbles divisible by 3 is $\\left\\lfloor \\frac{2012}{3} \\right\\rfloor = 670$.\n- The number divisible by 5 is $\\left\\lfloor \\frac{2012}{5} \\right\\rfloor = 402$.\n- The number divisible by both 3 and 5 (i.e., by 15) is $\\left\\lfloor \\frac{2012}{15} \\right\\rfloor = 134$.\n\nBy the inclusion-exclusion principle, the number of marbles replaced in the second step is $670 + 402 - 134 = 938$.\n\nSo, after the second step, $2012 - 938 = 1074$ black marbles remain.\n\n- The number of these marbles divisible by 7 is $\\left\\lfloor \\frac{1074}{7} \\right\\rfloor = 153$.\n\nAfter the third step, $1074 - 153 = 921$ black marbles remain in the sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17725,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Q}^+$ denote the set of positive rational numbers and $\\mathbb{N}^+$ the set of positive integers.\n\nGiven a function $f : \\mathbb{Q}^+ \\to \\mathbb{R}$ such that:\n\n1. $f(x)f(y) \\geq f(xy)$ for all $x, y \\in \\mathbb{Q}^+$;\n2. $f(x+y) \\geq f(x) + f(y)$ for all $x, y \\in \\mathbb{Q}^+$;\n3. $f(a) = a$ for some rational number $a > 1$,\n\nfind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "By induction using (2),\n\n$$\nf(nx) \\geq n f(x) \\quad \\text{for all } n \\in \\mathbb{N}^+,\\ x \\in \\mathbb{Q}^+.\n$$\n\nUsing (3), then (1), then (4), and then (3) again, for any $n \\in \\mathbb{N}^+$:\n\n$$\nf(a) f(n) \\geq f(na) \\geq n f(a) = n a.\n$$\n\nThus,\n\n$$\nf(n) \\geq n \\quad \\text{for all } n \\in \\mathbb{N}^+.\n$$\n\nIf $x = \\frac{m}{n}$ with $m, n \\in \\mathbb{N}^+$, then $f(x) f(n) \\geq f(m)$, so $f(x) > 0$ for all $x \\in \\mathbb{Q}^+$.\n\nIf $z > x$, then $f(z) \\geq f(x) + f(z-x) > f(x)$, so $f$ is strictly increasing.\n\nBy induction using (1), (3), and positivity, $f(a^t) \\leq a^t$ for all $t \\in \\mathbb{N}^+$.\n\nSuppose $f(k) > k$ for some $k \\in \\mathbb{N}^+$. Write $f(k) = k + \\epsilon$ with $\\epsilon > 0$. Then\n\n$$\nf(nk) \\geq n f(k) = n k + n \\epsilon.\n$$\n\nFor large $n$, $f(N) > N + 1$ for $N = n k$. For $p \\in \\mathbb{N}^+$,\n\n$$\nf(N + p) \\geq f(N) + f(p) \\geq N + 1 + p.\n$$\n\nThus $f(m) > m + 1$ for all $m \\geq N$. For large $t$ with $\\lfloor a^t \\rfloor \\geq N$,\n\n$$\na^t \\geq f(a^t) \\geq f(\\lfloor a^t \\rfloor) > \\lfloor a^t \\rfloor + 1,\n$$\n\nwhich is impossible. Therefore, $f(k) \\leq k$ for all $k \\in \\mathbb{N}^+$. Combining with above,\n\n$$\nf(n) = n \\quad \\text{for all } n \\in \\mathbb{N}^+.\n$$\n\nIf $x = \\frac{m}{n}$, then $f(nx) \\geq n f(x)$ and $f(nx) = f(m) = m$, so $m \\geq n f(x)$, i.e., $f(x) \\leq x$.\n\nAlso, $f(1/n) n \\geq 1$, so $f(1/n) \\geq 1/n$. Thus,\n\n$$\nf\\left(\\frac{m}{n}\\right) \\geq m f\\left(\\frac{1}{n}\\right) \\geq \\frac{m}{n} = x.\n$$\n\nTherefore, $f(x) = x$ for all $x \\in \\mathbb{Q}^+$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17726,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy has counted the sum of the squares of positive integers from 1 to 2016. Vitaliy has counted the sum of the squares of natural numbers from 1 to 2016. Yuriy has added Andriy's and Vitaliy's numbers, multiplied by 3, and added 2016. What number has Yuriy got?",
    "options": [],
    "answer": "See solution",
    "solution": "Let's write down the number that Yuriy got:\n\n$$\n\\begin{aligned}\n&3 \\cdot (1^2 + 2^2 + \\dots + 2016^2) + 3 \\cdot (1 + 2 + \\dots + 2016) + 2016 = \\\\\n&= (3 \\cdot 1^2 + 3 \\cdot 1 + 1) + (3 \\cdot 2^2 + 3 \\cdot 2 + 1) + \\dots + (3 \\cdot 2016^2 + 3 \\cdot 2016 + 1) = \\\\\n&= (1^3 + 3 \\cdot 1^2 + 3 \\cdot 1 + 1) + (2^3 + 3 \\cdot 2^2 + 3 \\cdot 2 + 1) + \\dots + (2016^3 + 3 \\cdot 2016^2 + 3 \\cdot 2016 + 1) - \\\\\n&\\quad - (1^3 + 2^3 + \\dots + 2016^3) = ((1+1)^3 + (2+1)^3 + \\dots + (2016+1)^3) \\\\\n&\\quad - (1^3 + 2^3 + \\dots + 2016^3) = \\\\\n&= (2^3 + 3^3 + \\dots + 2017^3) - (1^3 + 2^3 + \\dots + 2016^3) = 2017^3 - 1.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17727,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{2020}$ be real numbers, not necessarily distinct. For all $n \\ge 2020$, let $a_{n+1}$ be the minimal real root of the polynomial\n\n$$\nP_n(x) = x^{2n} + a_1 x^{2n-2} + a_2 x^{2n-4} + \\dots + a_{n-1} x^2 + a_n,\n$$\n\nif it exists. Assume that $a_{n+1}$ exists for all $n \\ge 2020$. Prove that $a_{n+1} \\le a_n$ for all $n \\ge 2021$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = \\alpha$ is a root of $P_n$, then $x = -\\alpha$ is also a root of $P_n$, as all terms of $P_n$ have even degree. The minimal root of $P_n$ therefore cannot be positive. Therefore, $a_n \\le 0$ for all $n > 2020$.\n\nWe have $P_{n+1}(x) = x^2 \\cdot P_n(x) + a_{n+1}$. Substitute $x = a_{n+1}$; as that is a root of $P_n$, we have $P_{n+1}(a_{n+1}) = 0 + a_{n+1} \\le 0$.\n\nAs the maximal degree term in $P_n(x)$ is $x^{2n}$, there exists an $N < 0$ such that $P_n(x) > 0$ for all $x < N$. Taking for example $-N = \\max(2, |a_1| + |a_2| + \\dots + |a_n|)$, we see for $x < N$ that $x^{2i-2} \\le x^{2n-2}$ for all $1 \\le i \\le n$ and therefore that\n\n$$\n\\begin{aligned}\n& |a_1x^{2n-2} + a_2x^{2n-4} + \\dots + a_{n-1}x^2 + a_n| \\\\\n& \\le |a_1x^{2n-2}| + |a_2x^{2n-4}| + \\dots + |a_{n-1}x^2| + |a_n| \\\\\n& \\le |a_1x^{2n-2}| + |a_2x^{2n-2}| + \\dots + |a_{n-1}x^{2n-2}| + |a_n|x^{2n-2} \\\\\n& \\le (|a_1| + |a_2| + \\dots + |a_n|)x^{2n-2} \\\\\n& \\le -N \\cdot x^{2n-2} \\\\\n& < x^{2n},\n\\end{aligned}\n$$\n\nso $x^{2n} + a_1x^{2n-2} + a_2x^{2n-4} + \\dots + a_{n-1}x^2 + a_n > 0$. Hence for $n \\ge 2021$ there exists an $N < 0$ with $P_n(x) > 0$ for all $x < N$, whereas $P_n(a_n) \\le 0$. Therefore, $P_n(x)$ has a root smaller than $a_n$. As $a_{n+1}$ is the minimal root, we have $a_{n+1} \\le a_n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17728,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$. Let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, and the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$ and $l_2$ be the perpendicular bisectors of $CD$ and $EF$, respectively.\n\n1. Show that $l_1$ and $l_2$ have a unique common point (denoted by $P$).\n2. Prove that the lengths of $CA$, $AP$, and $PE$ are the side lengths of a right triangle.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p109_data_cb3becc930.png)",
    "options": [],
    "answer": "See solution",
    "solution": "1. Since $C$, $A$, $B$, $E$ are concyclic, and $D$, $A$, $B$, $F$ are concyclic, $CA = AD$, and by the theorem of power of a point, we have\n\n$$\nCB \\cdot CF = CA \\cdot CD = DA \\cdot DC = DB \\cdot DE. \\quad (1)\n$$\n\nSuppose on the contrary that $l_1$ and $l_2$ do not intersect, then $CD \\parallel EF$, hence $\\frac{CF}{CB} = \\frac{DE}{DB}$. Plugging into (1), we get $CB^2 = DB^2$, thus $CB = DB$, hence $BA \\perp CD$. It follows that $CB$ and $DB$ are the diameters of $K_1$ and $K_2$, respectively, hence $K_1$ and $K_2$ have the same radii, which contradicts the assumption. Thus, $l_1$ and $l_2$ have a unique common point.\n\n2. Join $AE$, $AF$ and $PF$, we have\n\n$$\n\\angle CAE = \\angle CBE = \\angle DBF = \\angle DAF.\n$$\n\nSince $AP \\perp CD$, $AP$ is the bisector of $\\angle EAF$. Since $P$ is on the perpendicular bisector of the segment $EF$, $P$ is on the circumcircle of $\\triangle AEF$. We have\n\n$$\n\\begin{aligned}\n\\angle EPF &= 180^{\\circ} - \\angle EAF = \\angle CAE + \\angle DAF \\\\\n&= 2\\angle CAE = 2\\angle CBE.\n\\end{aligned}\n$$\n\nHence, $B$ is on the circle with center $P$ and radius $PE$, denoting this circle by $\\Gamma$. Let $R$ be the radius of $\\Gamma$. By the theorem of power of a point, we have\n\n$$\n2CA^2 = CA \\cdot CD = CB \\cdot CF = CP^2 - R^2,\n$$\n\nthus\n\n$$\nAP^2 = CP^2 - CA^2 = (2CA^2 + R^2) - CA^2 = CA^2 + PE^2.\n$$\n\nIt follows that $CA$, $AP$, $PE$ form the side lengths of a right triangle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17729,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime. Determine all positive integers $n$ for which the following condition is satisfied for all integers $x$:\n\n**Condition:** If $x^n - 1$ is divisible by $p$, then it is also divisible by $p^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "If, for a pair of integers $a, b$ and a positive integer $m$, $a - b$ is divisible by $m$, we write $a - b \\equiv 0 \\pmod{m}$.\n\nWe will show that the numbers $n$ we seek are those of the form $kp$, where $k$ is a positive integer.\n\nLet us first show that if $n$ satisfies the condition of the problem, then $n$ must be a multiple of $p$. To see this, let $x = p + 1$. Then $x^n - 1 \\equiv 1^n - 1 \\equiv 0 \\pmod{p}$, so $x^n - 1$ is divisible by $p$, and hence by the condition, is divisible by $p^2$ as well. Using the binomial expansion, $(p+1)^n \\equiv np + 1 \\pmod{p^2}$, so $x^n - 1 \\equiv np \\pmod{p^2}$. This implies that $np$ is a multiple of $p^2$, and therefore, $n$ must be a multiple of $p$.\n\nConversely, we will show that if $n = kp$ for a positive integer $k$, then it satisfies the condition. First, note that $x^n = (x^k)^p \\equiv x^k \\pmod{p}$ by Fermat's Little Theorem. This shows that if $x^n - 1$ is a multiple of $p$, then so is $x^k - 1$. Now,\n\n$$\n\\frac{x^n - 1}{x^k - 1} = 1 + x^k + x^{2k} + \\dots + x^{(p-1)k} \\equiv \\underbrace{1+1+\\dots+1}_{p} \\equiv 0 \\pmod{p},\n$$\n\nfrom which we conclude that $x^n - 1 = (x^k - 1) \\times \\frac{x^n - 1}{x^k - 1}$ is a multiple of $p^2$. Thus, $n = kp$ satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17730,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p = ab$ and $s = a + b$. Show that\n\n$$\np^2(s^2 - 2p - 2) \\ge s(p - 1)\n$$\n\nor equivalently,\n\n$$\np^2 s^2 - (p - 1)s - 2p^2(p + 1) \\ge 0\n$$\n\nwhen $s^2 \\ge 4p > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $0 < p \\le 1$, then\n$$\np^2 s - (p - 1) \\ge p^2 \\cdot 2\\sqrt{p} + (1 - p) > 0.\n$$\nIf $p \\ge 1$, then\n$$\np^2 s - (p - 1) \\ge (p^2 \\cdot 2\\sqrt{p} - p) + 1 > 0.\n$$\nTherefore,\n$$\n\\begin{align*}\np^2 s^2 - (p - 1)s - 2p^2(p + 1) &= s(p^2 s - (p - 1)) - 2p^2(p + 1) \\\\\n&\\ge 2\\sqrt{p}(p^2 \\cdot 2\\sqrt{p} - p + 1) - 2p^2(p + 1) \\\\\n&= 2p^3 - 2p\\sqrt{p} - 2p^2 + 2\\sqrt{p} \\\\\n&= 2\\sqrt{p}(p - 1)(p\\sqrt{p} - 1) \\\\\n&\\ge 0.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17731,
    "subject": "Mathematics (Olympiad)",
    "question": "Find polynomials $P$ and $Q$ such that $$(P+Q)(P^2 - PQ + Q^2) = (x^4 + 1)(x^8 - x^4 + 1).$$",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the possible factorizations:\n\nSince $x^4 + 1$ and $x^8 - x^4 + 1$ are irreducible in $\\mathbb{Q}[x]$, each must divide either $P+Q$ or $P^2 - PQ + Q^2$. There are four cases:\n\n1. $P + Q = x^4 + 1$ and $P^2 - PQ + Q^2 = x^8 - x^4 + 1$.\n   - Then $PQ = x^4$. Assume $P(x) = p x^4$, $Q(x) = q$ ($p, q \\in \\mathbb{Q}$).\n   - $P + Q = p x^4 + q = x^4 + 1$ implies $p = q = 1$.\n   - Thus, $P(x) = x^4$, $Q(x) = 1$ is a solution. Similarly, $P(x) = 1$, $Q(x) = x^4$ is another solution.\n\n2. $P + Q = x^8 - x^4 + 1$ and $P^2 - PQ + Q^2 = x^4 + 1$.\n   - This leads to $3PQ = x^4(x^{12} - 2x^8 + 3x^4 - 3)$, but the leading coefficients do not yield real solutions.\n\n3. $P + Q = 1$ and $P^2 - PQ + Q^2 = x^{12} + 1$.\n   - This case is impossible as zero is the unique root of both $P$ and $1-P$.\n\n4. $P + Q = x^{12} + 1$ and $P^2 - PQ + Q^2 = 1$.\n   - This case also leads to no solution.\n\n**Conclusion:** The only solutions are $P(x) = x^4$, $Q(x) = 1$ and $P(x) = 1$, $Q(x) = x^4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17732,
    "subject": "Mathematics (Olympiad)",
    "question": "We define a _chessboard polygon_ to be a polygon whose edges are situated along lines of the form $x = a$ or $y = b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $1 \\times 2$ rectangles. Finally, a _tasteful tiling_ is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $3 \\times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.\n\n![](images/pamphlet0910_main_p6_data_5c6709e291.png)\n\n(a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.\n\n(b) Prove that such a tasteful tiling is unique.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) We prove the first part by induction on the number $n$ of dominoes in the tiling. The claim is clearly true for $n=1$. So suppose we have a chessboard polygon that can be tiled by $n > 1$ dominoes. Of all the leftmost squares in the polygon, select the lowest one and label it $L$; assume for sake of argument that square $L$ is black. In the given tiling, remove the domino covering $L$, leaving a polygon which may be tiled with $n-1$ dominoes. By the induction hypothesis, this chessboard polygon can be tastefully tiled.\n\nNow replace the domino that was removed. If this domino is horizontal, then we are guaranteed that the augmented tiling is still tasteful, since square $L$ is black and there are no squares below it. If the domino is vertical the augmented tiling may still be tasteful, but if not the trouble can only arise because there is another vertical domino directly to its right. In this case rotate the offending pair of dominoes to get two horizontal dominoes. We are not done yet, but if we now repeat this process—removing the _horizontal_ domino covering $L$, tiling the remainder, and replacing the domino—then we will obtain a tasteful tiling.\n\nIf square $L$ is white we may obtain a tasteful tiling by performing a similar process. This time we only encounter difficulty if the domino covering $L$ in the original tiling is horizontal, in which case there must be another horizontal domino directly above it. We rotate this pair, remove the now vertical domino covering $L$, tile the remainder tastefully using the induction hypothesis, and restore the vertical domino to finish.\n\n(b) Suppose now that there are two tasteful tilings of a given chessboard polygon. By overlaying these two tilings we obtain chains of overlapping dominoes, since every square is part of one domino from each tiling. For example, a chain of length one indicates a domino common to both tilings. A chain of length two cannot occur, since these arise when a $2 \\times 2$ block is covered by horizontal dominoes in one tiling and vertical dominoes in the other, and one of these configurations will be distasteful.\n\nSince the tilings are distinct a chain of length three or more must occur; let $R$ be the region consisting of such a chain along with its interior, if any. (It is possible that such a chain may completely occupy a region, so that only some of the dominoes in the chain adjoin squares outside of $R$.) Note that the chain must include a horizontal domino along its lowermost row. If there are two or more overlapping horizontal dominoes, then one of them will be a WB domino, i.e. have a white square on the left. Otherwise there are two adjacent vertical dominoes that overlap with the single horizontal domino; since they are part of a tasteful tiling we again must have a WB domino. We will now focus on the tiling that includes this WB domino.\n\nThe two squares above the WB domino must be part of region $R$. Furthermore, a single horizontal domino cannot cover them both, nor can a pair of vertical dominoes because both cases yield distasteful configurations. Hence, a horizontal domino must cover at least one of these squares, extending past the given WB domino either to the left or to the right. Therefore, we can deduce the existence of a horizontal WB domino on the next row up. We may repeat this argument until we reach a horizontal WB domino in region $R$ for which the two squares immediately above it are not both in region $R$. This implies that this domino must be part of the chain that defined $R$.\n\nNow imagine walking along the chain, starting on the white square of the WB domino that exists along the lowest row of region $R$ and taking the first step toward the black square of the same domino. Draw an arrow along each domino in the direction of travel all the way around the chain. Since the squares must alternate white and black, these arrows will always point from a white square to a black square. Furthermore, since the interior of the region was initially to our left when we began the loop, it will always be to our left whenever the chain follows the boundary of $R$.\n\nBut we now reach a contradiction. We earlier deduced the existence of a horizontal WB domino that was part of the chain and was adjacent to the boundary of $R$, having a square above it that was not part of $R$. Hence this domino must be traversed from right to left, since we leave the interior of $R$ to our left as we traverse the loop. Hence it must contain an arrow pointing to the left, implying that it must be a BW domino instead. This is a contradiction, completing the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17733,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be a subset of $\\mathbb{R}$ obtained by deleting finitely many real numbers from $\\mathbb{R}$. Prove that for any given positive number $n$, there exists a polynomial $f(x)$ of degree $n$ such that all its coefficients and its $n$ real roots are in $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nLet $T = \\{ x \\in \\mathbb{R} \\mid x \\notin M \\}$, which is a finite set by definition, and let $a = \\max T$. Choose any real number $k > \\max\\{ |a|, 1 \\}$; then $-k \\notin T$ and hence $-k \\in M$.\n\nFor any positive integer $n$, define $f(x) = k(x + k)^n$. Then, since $k > 1$, $\\deg f(x) = n$, and the coefficient of $x^m$ in $f(x)$ is\n\n$$\nk \\binom{n}{m} k^{n-m} \\geq k.\n$$\n\nHence, all the coefficients of $f(x)$ are not in $T$, and thus are in $M$. The roots of $f(x)$ are all $-k$ with multiplicity $n$, and $-k \\in M$. Therefore, the polynomial $f(x) = k(x + k)^n$ satisfies the required condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17734,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors so that 108 points are colored Red, 108 points are colored Green, 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color such that the four triangles formed by the chosen points of the same color are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$, $G$, $B$, and $Y$ denote the sets of Red, Green, Blue, and Yellow points, respectively. For $0 \\leq k \\leq 431$, let $\\mathcal{T}_k$ be the counterclockwise rotation by $\\frac{360k}{432}$ degrees about the center of the circle.\n\nFirst, we claim there exists an index $i_1$ such that $|\\mathcal{T}_{i_1}(R) \\cap G| \\geq 28$. For each $k$, $\\mathcal{T}_k(R) \\cap G$ consists of Green points that are images of Red points under $\\mathcal{T}_k$. The sum\n$$\ns_1 = |\\mathcal{T}_0(R) \\cap G| + |\\mathcal{T}_1(R) \\cap G| + \\dots + |\\mathcal{T}_{431}(R) \\cap G|\n$$\nequals the number of pairs $(r, g)$ with $g = \\mathcal{T}_k(r)$ for some $k$. For each $r$ and $g$, there is a unique $k$ with $\\mathcal{T}_k(r) = g$, so $s_1 = 108^2 = 11664$. Since $R$ and $G$ are disjoint, $|\\mathcal{T}_0(R) \\cap G| = 0$. By the Pigeonhole Principle, there is some $i_1$ such that\n$$\n|\\mathcal{T}_{i_1}(R) \\cap G| \\geq \\left\\lfloor \\frac{11664}{431} \\right\\rfloor = 28.\n$$\nLet $RG = \\mathcal{T}_{i_1}(R) \\cap G$.\n\nNext, there exists $i_2$ such that $|\\mathcal{T}_{i_2}(RG) \\cap B| \\geq 8$. Similarly,\n$$\ns_2 = |\\mathcal{T}_0(RG) \\cap B| + \\dots + |\\mathcal{T}_{431}(RG) \\cap B| \\geq 28 \\cdot 108 = 3024.\n$$\nAgain, $RG$ is disjoint from $B$, so $|\\mathcal{T}_0(RG) \\cap B| = 0$, and $|\\mathcal{T}_{432-i_1}(RG) \\cap B| = 0$. By the Pigeonhole Principle,\n$$\n|\\mathcal{T}_{i_2}(RG) \\cap B| \\geq \\left\\lfloor \\frac{3024}{430} \\right\\rfloor = 8.\n$$\nLet $RGB = \\mathcal{T}_{i_2}(RG) \\cap B$.\n\nFinally, there exists $i_3$ such that $|\\mathcal{T}_{i_3}(RGB) \\cap Y| \\geq 3$. We have\n$$\ns_3 = |\\mathcal{T}_0(RGB) \\cap Y| + \\dots + |\\mathcal{T}_{431}(RGB) \\cap Y| \\geq 8 \\cdot 108 = 864.\n$$\nAlso, $|\\mathcal{T}_0(RGB) \\cap Y| = |\\mathcal{T}_{432-i_2}(RGB) \\cap Y| = |\\mathcal{T}_{432-i_2-i_1}(RGB) \\cap Y| = 0$. Thus,\n$$\n|\\mathcal{T}_{i_3}(RGB) \\cap Y| \\geq \\left\\lfloor \\frac{864}{429} \\right\\rfloor = 3.\n$$\n\nLet $y_1, y_2, y_3$ be three distinct points in $\\mathcal{T}_{i_3}(RGB) \\cap Y$. Then the triples\n$$\n(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3}(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3-i_2}(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3-i_2-i_1}(y_1, y_2, y_3)\n$$\nform congruent triangles whose vertices are Yellow, Blue, Green, and Red, respectively, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17735,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $m \\geq 2$, and $m$ positive integers $a_1, a_2, \\dots, a_m$, prove that there exist infinitely many positive integers $n$ such that $a_1 \\cdot 1^n + a_2 \\cdot 2^n + \\dots + a_m \\cdot m^n$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be a prime factor of $a_1 + 2a_2 + \\dots + m a_m$. By Fermat's little theorem, for any $k$ with $1 \\leq k \\leq m$, we have $k^p \\equiv k \\pmod{p}$. Thus, for any positive integer $n$, we have\n\n$$\n\\begin{aligned}\na_1 \\cdot 1^p + a_2 \\cdot 2^p + \\dots + a_m \\cdot m^p &\\equiv a_1 + 2a_2 + \\dots + m a_m \\\\\n&\\equiv 0 \\pmod{p}.\n\\end{aligned}\n$$\n\nHence, $a_1 \\cdot 1^p + a_2 \\cdot 2^p + \\dots + a_m \\cdot m^p$ is divisible by $p$, and considering powers $p^n$ for $n = 1, 2, \\dots$, the expression is composite for infinitely many $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17736,
    "subject": "Mathematics (Olympiad)",
    "question": "Define the sequence $a_n$ as follows:\n\n- $a_1 = 1$\n- $a_{2n} = a_n$\n- $a_{2n+1} = a_n + 1$ for all $n \\ge 1$\n\n**a)** Find all positive integers $n$ such that $a_{kn} = a_n$ for all integers $1 \\le k \\le n$.\n\n**b)** Prove that there exist infinitely many positive integers $m$ such that $a_{km} \\ge a_m$ for all positive integers $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We prove by induction on $n$ that $a_n = s_2(n)$ for all $n \\ge 1$, where $s_2(n)$ is the sum of the digits of $n$ in binary representation.\n\nThe base case $n = 1$ is trivial. Assume $a_n = s_2(n)$ holds for $n = 1, 2, \\dots, k$, and prove it for $n = k + 1$.\n\n- If $k+1$ is even, let $k+1 = \\overline{x_1x_2\\dots x_{m-1}0}_{(2)}$. Then $\\frac{k+1}{2} = \\overline{x_1x_2\\dots x_{m-1}}_{(2)}$, so\n  $$\na_{k+1} = a_{\\frac{k+1}{2}} = s_2\\left(\\frac{k+1}{2}\\right) = s_2(k+1).\n  $$\n- If $k+1$ is odd, let $k+1 = \\overline{x_1x_2\\dots x_{m-1}1}_{(2)}$. Then $\\frac{k}{2} = \\overline{x_1x_2\\dots x_{m-1}}_{(2)}$, so\n  $$\na_{k+1} = a_{\\frac{k}{2}} + 1 = s_2\\left(\\frac{k}{2}\\right) + 1 = s_2(k+1).\n  $$\n\nTherefore, $a_n = s_2(n)$ for all $n \\ge 1$.\n\nNow, we consider which $n$ satisfy $a_{kn} = a_n$ for all $1 \\le k \\le n$.\n\n- For $n = 1$ and $n = 2$, it is easy to check that $a_{kn} = a_n$ for all $k$ in the range.\n- For $n \\ge 3$:\n  - If $n = 2^t - 1$ (all ones in binary), then for all $1 \\le k \\le n$, $a_{kn} = a_n$.\n  - If $n$ is a power of $2$, $n = 2^t$, then for $k = 2$, $a_{2n} = a_n = 1$, but for $k = 3$, $a_{3n} = a_{2^t \\cdot 3} = a_{3 \\cdot 2^t} = a_3 = 2 \\neq 1 = a_n$.\n  - For other $n$, there exists $k$ such that $a_{kn} \\neq a_n$.\n\nThus, the only $n$ that work are $n = 1$, $n = 2$, and $n = 2^t - 1$ for $t \\ge 2$.\n\nb) Choose $m = 2^p$ for any $p \\ge 0$. Then $a_m = 1$, and for any $k \\ge 1$, $a_{km} \\ge 1 = a_m$. Thus, there are infinitely many such $m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17737,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive numbers $a, b, c$ such that $abc = 1$, prove the inequality:\n\n$$\n\\frac{a^2 + b^2}{c^2 + a + b} + \\frac{b^2 + c^2}{a^2 + b + c} + \\frac{c^2 + a^2}{b^2 + c + a} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $abc = 1$. Consider:\n\n$$\nc^2 + a + b \\le c^2 + (a + b)\\frac{1}{3}(a + b + c) = \\frac{1}{3}(3c^2 + a^2 + b^2 + 2ab + ac + bc) \\le \\frac{1}{6}(8c^2 + 5a^2 + 5b^2).\n$$\n\nSo,\n\n$$\n\\frac{a^2 + b^2}{c^2 + a + b} + \\cdots \\ge \\frac{a^2 + b^2}{\\frac{1}{6}(8c^2 + 5a^2 + 5b^2)} + \\cdots = \\frac{6(a^2 + b^2)}{8c^2 + 5a^2 + 5b^2} + \\cdots \\ge 2,\n$$\n\nwhere the dots indicate cyclic permutations. For simplification, denote $x = a^2$, $y = b^2$, $z = c^2$. Then we need to prove:\n\n$$\nA = \\frac{x + y}{8z + 5x + 5y} + \\cdots \\ge \\frac{1}{3}.\n$$\n\nBy Cauchy-Schwarz inequality,\n\n$$\nA = \\frac{(x + y)^2}{(x + y)(8z + 5x + 5y)} + \\cdots \\ge \\frac{((x + y)^2 + \\cdots)^2}{(x + y)(8z + 5x + 5y)} = \\frac{4p + 8q}{10p + 26q} \\ge \\frac{1}{3},\n$$\n\nwhere $p = x^2 + y^2 + z^2$, $q = xy + yz + zx$. The last inequality is equivalent to $p \\ge q$, which is well known.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17738,
    "subject": "Mathematics (Olympiad)",
    "question": "Sequence $\\{f_n\\}$ is defined as follows: $f_1 = 1$, $f_2 = 2$, $f_{n+2} = f_{n+1} + f_n$ for natural $n$. What is the greatest possible number of members of the sequence $\\{f_n\\}$ that can appear as consecutive members of an increasing arithmetic progression?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 3.\n\nLet the first consecutive members be $f_1 = a$, $f_m = a + d$. Since $a > 0$, $d > 0$, $f_{m+1} > f_m$, and $f_{m+2} = f_{m+1} + f_m > 2a + 2d > a + 2d$, the third member may only be $f_{m+1}$.\n\nSo $f_{m+1} = a + 2d$. It's clear that the next member of the progression does not satisfy the condition because $f_{m+2} = 2a + 3d > a + 3d$. But we can easily find three consecutive members: $1, 2, 3$ and $2, 5, 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17739,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be two $n \\times n$ matrices with real entries such that $AB^2 = A - B$.\n\n(a) Prove that $I_n + B$ is a nonsingular matrix.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The given relation is $AB^2 = A - B$. Rearranging, we have:\n\n$$\nAB^2 - A + B = 0\n$$\n\nAdd $I_n$ to both sides:\n\n$$\nAB^2 - A + B + I_n = I_n\n$$\n\nFactor as follows:\n\n$$\nAB^2 - A + B + I_n = (AB - A + I_n)(I_n + B)\n$$\n\nSo,\n\n$$\n(AB - A + I_n)(I_n + B) = I_n\n$$\n\nThis shows that $(I_n + B)$ is invertible (nonsingular), since it has a right inverse.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 17740,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $x_1, \\dots, x_n$ 為任意相異的 $n$ 個實數（$n \\ge 2$）。證明下列等式成立：\n\n$$\n\\sum_{1 \\le i \\le n} \\prod_{j \\ne i} \\frac{1 - x_i x_j}{x_i - x_j} = \\begin{cases} 0, & \\text{若 } n \\text{ 為偶數;} \\\\ 1, & \\text{若 } n \\text{ 為奇數.} \\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令 $G(x_1, \\dots, x_n)$ 為左側的函數。因為等式兩邊都是有理函数，只需在所有 $x_i \\notin \\{-1, 1\\}$ 时证明。\n\n定义\n\n$$\nf(t) = \\prod_{i=1}^{n} (1 - x_i t),\n$$\n\n注意到\n\n$$\nf(x_i) = (1 - x_i^2) \\prod_{j \\ne i} (1 - x_i x_j).\n$$\n\n用节点 $+1, -1, x_1, \\dots, x_n$，拉格朗日插值公式给出 $f$ 的表达式：\n\n$$\nf(x) = \\sum_{i=1}^{n} f(x_i) \\frac{(x-1)(x+1)}{(x_i-1)(x_i+1)} \\prod_{j \\ne i} \\frac{x-x_j}{x_i-x_j} + f(1) \\frac{x+1}{2} \\prod_{i=1}^{n} \\frac{x-x_i}{1-x_i} + f(-1) \\frac{x-1}{-2} \\prod_{i=1}^{n} \\frac{x-x_i}{-1-x_i}.\n$$\n\n由于 $f$ 的次数为 $n$，$t^{n+1}$ 的系数为零。比较 $t^{n+1}$ 的系数得：\n\n$$\n\\begin{aligned}\n0 &= \\sum_{i=1}^{n} \\frac{f(x_i)}{\\prod_{j \\ne i} (x_i - x_j)(x_i - 1)(x_i + 1)} \\\\\n&\\quad + \\frac{f(1)}{2 \\prod_{i=1}^{n} (1 - x_i)} + \\frac{f(-1)}{-2 \\prod_{i=1}^{n} (-1 - x_i)} \\\\\n&= -G(x_1, \\dots, x_n) + \\frac{1}{2} + \\frac{(-1)^{n+1}}{2}.\n\\end{aligned}\n$$\n\n这就推出了所需的等式。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17741,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there is a positive integer $k$ with the following property: if $a, b, c, d, e$ and $f$ are integers and $m$ is a divisor of\n\n$$\na^n + b^n + c^n - d^n - e^n - f^n\n$$\n\nfor all integers $n$ in the range $1 \\le n \\le k$, then $m$ is a divisor of $a^n + b^n + c^n - d^n - e^n - f^n$ for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $k = 6$ works. Consider the polynomial\n\n$$\nP(x) = (x - a)(x - b)(x - c)(x - d)(x - e)(x - f) = x^6 + p_5 x^5 + p_4 x^4 + p_3 x^3 + p_2 x^2 + p_1 x + p_0\n$$\n\nfor some integers $p_0, \\dots, p_5$. Because $P$ has roots $a, b, c, d, e, f$, we have\n\n$$\n\\begin{aligned}\na^6 + p_5 a^5 + \\dots + p_1 a + p_0 &= 0 \\\\\na^{6+r} + p_5 a^{5+r} + \\dots + p_1 a^{1+r} + p_0 a^r &= 0\n\\end{aligned}\n$$\n\nfor all positive integers $r$. Then\n\n$$\n\\begin{aligned}\n& a^{6+r} + b^{6+r} + c^{6+r} - d^{6+r} - e^{6+r} - f^{6+r} \\\\\n&= -(p_5 a^{5+r} + \\dots + p_0 a^r) - (p_5 b^{5+r} + \\dots + p_0 b^r) \\\\\n&\\quad -(p_5 c^{5+r} + \\dots + p_0 c^r) + (p_5 d^{5+r} + \\dots + p_0 d^r) \\\\\n&\\quad +(p_5 e^{5+r} + \\dots + p_0 e^r) + (p_5 f^{5+r} + \\dots + p_0 f^r) \\\\\n&= -p_5 (a^{5+r} + b^{5+r} + c^{5+r} - d^{5+r} - e^{5+r} - f^{5+r}) \\\\\n&\\quad - \\dots - p_0 (a^r + b^r + c^r - d^r - e^r - f^r)\n\\end{aligned}\n$$\n\nSo if $m \\mid a^n + b^n + c^n - d^n - e^n - f^n$ for $n = r, r+1, r+2, r+3, r+4, r+5$, then $m \\mid a^{r+6} + b^{r+6} + c^{r+6} - d^{r+6} - e^{r+6} - f^{r+6}$ also.\n\nNow suppose $m \\mid a^n + b^n + c^n - d^n - e^n - f^n$ for $n = 1, 2, 3, 4, 5, 6$. We also know from the above that for any $s \\ge 7$, if $m \\mid a^n + b^n + c^n - d^n - e^n - f^n$ for all $n < s$, we also have $m \\mid a^s + b^s + c^s - d^s - e^s - f^s$. So, by strong induction, $m \\mid a^n + b^n + c^n - d^n - e^n - f^n$ for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17742,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with orthocenter $H$, and let $M$ be the midpoint of $AC$. The point $C_1$ on $AB$ is such that $CC_1$ is an altitude of the triangle $ABC$. Let $H_1$ be the reflection of $H$ in $AB$. The orthogonal projections of $C_1$ onto the lines $AH_1$, $AC$, and $BC$ are $P$, $Q$, and $R$, respectively. Let $M_1$ be the point such that the circumcentre of triangle $PQR$ is the midpoint of the segment $MM_1$.\n\nProve that $M_1$ lies on the segment $BH_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the following lemma.\n\n**Lemma.** Let $XYZT$ be a cyclic quadrilateral such that $XZ \\perp YT$. Let $XZ \\cap YT = O$, and let $V$ and $W$ be the orthogonal projections of $O$ to lines $XY$ and $ZT$, respectively. If $I$ lies in the middle of $YZ$, and $J$ lies in the middle of $XT$, then $IVJW$ is a cyclic deltoid.\n\n**Proof.** Let $Y'$ and $Z'$ be the points in the middle of $OY$ and $OZ$ respectively. Then $\\overline{VY'} = \\overline{OY'} = \\overline{IZ'}$ and $\\overline{IY'} = \\overline{OZ'} = \\overline{WZ'}$, and also\n\n![](images/Macedonia_2010_booklet_p29_data_180ef3c51c.png)\n\n$$\n\\angle VY'I = \\angle VY'O + 90^\\circ = 2\\angle WZO + 90^\\circ = \\angle WZ'O + 90^\\circ = \\angle IZ'W.\n$$\n\nHence, we have $\\angle VY'I \\cong \\angle IZ'W$, so $\\overline{VI} = \\overline{WI}$, and\n\n$$\n\\angle VIW = 90^\\circ - \\angle VIY' - \\angle Z'IW = 90^\\circ - \\angle VIY' - \\angle IVY' = \\angle VY'O = 2\\angle XYT.\n$$\n\nAnalogously, we can show $\\angle VJW = 2\\angle YXZ$, and therefore\n\n$$\n\\angle VIW + \\angle VJW = 180^\\circ.\n$$\n\nThe lemma is proved.\n\nLet $QC_1 \\cap H_1B = M_1$. As $\\angle M_1C_1B = \\angle ACC_1 = \\angle ABH_1$ (the last equality follows from the fact that $ACBH_1$ is cyclic), we have that $M_1$ lies in the middle of $BH_1$. Using the lemma, the quadrilateral $PMRM_1$ is a cyclic deltoid, and the center of its circumscribed circle lies in the middle of $MM_1$. Hence, $Q$ also has to be on that circle, as $\\angle MQM_1 = 90^\\circ$, and the middle of $MM_1$ is the circumcentre of the triangle $PQR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17743,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a natural number. Prove that $13 \\mid 7^n + n^5$ if and only if $13 \\mid n^7 \\cdot 7^n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $13 \\mid 7^n + n^5$. Clearly, $13$ does not divide $n$, so by Fermat's Little Theorem, $n^{12} \\equiv 1 \\pmod{13}$, i.e., $13 \\mid n^{12} - 1$.\n\nNow, $13 \\mid 7^n + n^5$ implies $13 \\mid n^7(7^n + n^5)$, or $13 \\mid n^7 7^n + n^{12}$. Since $13 \\mid n^{12} - 1$, we have $n^{12} \\equiv 1 \\pmod{13}$, so $n^7 7^n + n^{12} \\equiv n^7 7^n + 1 \\pmod{13}$. Thus, $13 \\mid n^7 \\cdot 7^n + 1$.\n\nConversely, suppose $13 \\mid n^7 \\cdot 7^n + 1$. Again, $13 \\mid n^{12} - 1$. Consider $n^5(n^7 7^n + 1) = n^{12} 7^n + n^5$. Since $n^{12} \\equiv 1 \\pmod{13}$, $n^{12} 7^n + n^5 \\equiv 7^n + n^5 \\pmod{13}$. Therefore, $13 \\mid 7^n + n^5$.\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17744,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABC$ 為銳角三角形，其中 $AC > AB$，且點 $O$ 為其外心。設點 $D$ 為 $BC$ 線段上一點。過點 $D$ 引一條與 $BC$ 垂直的直線，設該線分別與直線 $AO, AC, AB$ 交於點 $W, X, Y$。三角形 $AXY$ 的外接圓與三角形 $ABC$ 的外接圓再交於點 $Z \\neq A$。\n\n已知 $D \\neq W$ 且 $OW = OD$。證明：$DZ$ 與三角形 $AXY$ 的外接圓相切。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $AO$ 與 $BC$ 交於點 $E$。由於 $EDW$ 是直角三角形且 $O$ 在 $WE$ 上，條件 $OW = OD$ 說明了 $O$ 是三角形 $EDW$ 的外心。於是 $OD = OE$，得到 $D, E$ 兩點對稱於 $BC$ 邊的中垂線。\n\n觀察有：\n\n$$\n180^\\circ - \\angle DXZ = \\angle ZXY = \\angle ZAY = \\angle ZCD,\n$$\n\n所以 $CDXZ$ 四點共圓。\n\n![](images/2023-TWNIMO-Problems_p19_data_5727273281.png)\n\n接著證明 $AZ \\parallel BC$。為此，引入圓 $ABC$ 上的輔助點 $Z'$ 滿足 $AZ' \\parallel BC$。由前段的證明，我們知道只需證明 $CDXZ'$ 共圓即可。注意到三角形 $BAE$ 與三角形 $CZ'D$ 關於 $BC$ 邊的中垂線對稱。利用此點以及 $A, O, E$ 三點共線的關係，得\n\n$$\n\\angle DZ'C = \\angle BAE = \\angle BAO = 90^\\circ - \\frac{1}{2}\\angle AOB = 90^\\circ - \\angle C = \\angle DXC.\n$$\n\n所以 $DXZ'C$ 共圓，推得 $Z$ 與 $Z'$ 為同一點，得證。\n\n最後，由於 $AZ \\parallel BC$ 以及 $CDXZ$ 共圓，可知\n\n$$\n\\angle AZD = \\angle CDZ = \\angle CXZ = \\angle AYZ,\n$$\n\n由弦切角知 $DZ$ 與圓 $AXY$ 相切。 □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17745,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are no positive integers $x, y, z$ such that\n\n$$\nx^2y^4 - x^4y^2 + 4x^2y^2z^2 + x^2z^4 - y^2z^4 = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove this statement by contradiction. Assume that there are positive integers $x, y, z$ satisfying the equation\n\n$$\nx^2y^4 - x^4y^2 + 4x^2y^2z^2 + x^2z^4 - y^2z^4 = 0.\n$$\n\nIt is easy to check that $x \\neq y$. If there are solutions to the equation, we can choose a solution with $\\gcd(x, y) = 1$. Now let $x, y, z$ be a positive integer solution with $\\gcd(x, y) = 1$.\n\nBy factoring the equation, we get\n\n$$\n\\begin{aligned}\n0 &= x^2y^4 - x^4y^2 + 4x^2y^2z^2 + x^2z^4 - y^2z^4 \\\\\n  &= x^2(y^4 + 2y^2z^2 + z^4) - y^2(x^4 - 2x^2z^2 + z^4) \\\\\n  &= x^2(y^2 + z^2)^2 - y^2(x^2 - z^2)^2 \\\\\n  &= (x(y^2 + z^2) + y(x^2 - z^2))(x(y^2 + z^2) - y(x^2 - z^2)) \\\\\n  &= ((x - y)z^2 + xy(x + y))((x + y)z^2 + xy(y - x)).\n\\end{aligned}\n$$\n\nThen\n\n$$\n(y - x)z^2 = xy(x + y) \\quad \\text{or} \\quad (x + y)z^2 = xy(x - y).\n$$\n\nMultiplying both sides of the first equation by $y - x$ gives\n\n$$\n(y - x)^2 z^2 = xy(y^2 - x^2).\n$$\n\nMultiplying both sides of the second equation by $x + y$ gives\n\n$$\n(x + y)^2 z^2 = xy(x^2 - y^2).\n$$\n\nSince the second equation can be solved similarly, we focus on the first. Because $\\gcd(x, y) = 1$, $x$, $y$, and $y^2 - x^2$ are pairwise coprime. The left-hand side is a perfect square, so $x$, $y$, and $y^2 - x^2$ must be perfect squares. Thus, there exist positive integers $a, b, c$ such that $x = a^2$, $y = b^2$, $y^2 - x^2 = c^2$. Therefore,\n\n$$\nb^4 - a^4 = c^2.\n$$\n\nBut it is well-known that there are no positive integers $a, b, c$ satisfying this equation (by infinite descent). Therefore, we have a contradiction, and the statement is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17746,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Clearly, a nonzero constant polynomial $P$ fulfills the requirements, so we shall assume in the sequel $\\deg P \\ge 1$. Before proceeding, we state the following folklore preliminaries:\n\n1. **Dirichlet's Theorem.** The arithmetic progression $an + b$ with $a \\in \\mathbb{N}^*$, $b \\in \\mathbb{Z}^*$, $(a, b) = 1$, contains infinitely many primes.\n\n2. **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\n**Claim:** $P(0) = 0$. We proceed by contradiction; assume $P(0) \\ne 0$.\n\nWe will prove by induction on $k$ the following statement:\n\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.*\n\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 $P(p)^2 = 1$, so the polynomial $P(X)^2 - 1$ has infinitely many roots, thus it is the zero polynomial, which contradicts $\\deg P \\ge 1$. Therefore, $P(p_1)$ has at least one prime factor, so the statement is proven for $k=1$.\n\nLet $p_k$ be a prime with $p_k > |P(0)|$ and $P(p_k)$ has at least $k$ distinct prime factors. 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 is prime, the conclusion follows. The arithmetic progression $nP(p_k)^2 + p_k$ (by Dirichlet's Theorem) contains infinitely many primes. Let $p_{k+1} = sP(p_k)^2 + p_k$ be such a prime. Then $P(p_{k+1}) \\equiv P(p_k) \\pmod{P(p_k)^2}$, so there exists $t$ such that $P(p_{k+1}) = P(p_k)(1 + tP(p_k))$. Thus, $P(p_{k+1})$ has at least one more prime factor than $P(p_k)$. Since $P(p_k)$ had at least $k$ distinct prime factors, $P(p_{k+1})$ has at least $k+1$ prime factors.\n\nThus, the statement is proven and 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 $Q$ were nonconstant, arguing as above, we again obtain a contradiction.\n\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": 17747,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n2P(x) = Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right), \\quad P(1)=1.\n$$\n\nDetermine the polynomials $P(x)$ and $Q(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_0$. The coefficient of the highest degree term in $Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right)$ comes from the difference $\\left(\\frac{(x+1)^2}{2}\\right)^n - \\left(\\frac{(x-1)^2}{2}\\right)^n$:\n\n$$\n\\frac{2n x^{2n-1}}{2^n} + \\frac{2n x^{2n-2}}{2^n} = \\frac{4n}{2^n} x^{2n-1} \\quad (1).\n$$\n\nThe left side's corresponding coefficient is $2$, so $\\frac{4n}{2^n} = 2 \\implies 2^{n+1} = 4n$. Since $2^{n+1} > 4n$ for $n \\geq 3$, only $n=1$ or $n=2$ are possible. Thus, the degree of $P$ is $2n-1$.\n\nFor $x=0$, $2P(0) = Q(1/2) - Q(1/2) = 0$, so $P(0) = 0$.\n\n*For $n=1$*: $P(x) = a x$. Since $P(1) = 1$, $a=1$.\n\n$$\nP(x) = x, \\quad Q(x) = x + a_0, \\quad a_0 \\in \\mathbb{R}.\n$$\n\n*For $n=2$*: Let $Q(x) = x^2 + b x + c$.\n\n$$\n\\begin{align*}\n2P(x) &= Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right) \\\\\n&= \\frac{1}{4} \\left((x+1)^4 - (x-1)^4\\right) + \\frac{b}{2} \\left((x+1)^2 - (x-1)^2\\right) \\\\\n&= \\frac{1}{4}(8x^3 + 8x) + \\frac{b}{2}(4x) \\\\\n&= 2x^3 + 2(1+b)x.\n\\end{align*}\n$$\n\nSo $P(x) = x^3 + (1+b)x$. Since $P(1) = 1$, $1 = 1 + (1+b) \\implies b = -1$.\n\n$$\nP(x) = x^3, \\quad Q(x) = x^2 - x + c, \\quad c \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17748,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there exist two distinct positive integers $a, b$, each having exactly 2014 digits (in base ten; initial zeroes disallowed), with the following properties.\n\n- The digits of $b$ are those of $a$ in reverse order.\n- When a digit in each of $a$ and $b$ is deleted at random, and the resulting numbers are denoted $a'$ and $b'$, respectively, then\n\n$$\n\\frac{a'}{b'} = \\frac{a}{b}\n$$\n\nwith a likelihood exceeding 99%.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that, for any natural number $m$, we have\n\n$$\n1\\overbrace{33\\dots33}^{m}2 = 1\\overbrace{11\\dots11}^{m+1} \\cdot 12 \\quad \\text{and} \\quad 2\\overbrace{33\\dots33}^{m}1 = 1\\overbrace{11\\dots11}^{m+1} \\cdot 21,\n$$\n\nso that\n\n$$\n\\frac{1\\overbrace{33\\dots33}^{m}2}{2\\overbrace{33\\dots33}^{m}1} = \\frac{12}{21},\n$$\n\nirrespective of the value of $m$. Consequently, if we choose\n\n$$\na = 1\\overbrace{33\\dots33}^{2012}2 \\quad \\text{and} \\quad b = 2\\overbrace{33\\dots33}^{2012}1,\n$$\n\nthen $\\frac{a'}{b'} = \\frac{a}{b} = \\frac{12}{21}$ whenever the digits erased are two 3's, which occurs with probability\n\n$$\n\\frac{2012^2}{2014^2} = 1 - \\frac{4}{2014} + \\frac{4}{2014^2} > 1 - \\frac{4}{2000} = 99.8\\%.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17749,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$, $n$, $p$ be positive integers. Space is divided into unit cubes by infinite parallel planes. An assignment of each unit cube a natural number from $1$ to $60$ is called a *Dien Bien* assignment if, for every rectangular box whose faces lie on the planes and whose side lengths are in $\\{2m+1, 2n+1, 2p+1\\}$, the unit cube at the center of the box is assigned the average of the numbers assigned to its $8$ vertices.  \nHow many Dien Bien assignments are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, suppose the parallel planes are equally spaced with unit distance. Choose a cube and let its center $O$ be the origin. The axes $Ox$, $Oy$, $Oz$ are parallel to the system's lines, with positive directions chosen arbitrarily. Denote the coordinate of a cube as $(x, y, z)$, its distances from $O$ along each axis.\n\nConsider a Dien Bien assignment. Two unit cubes are *related* if they must be assigned the same number. To count the number of such assignments, we count the maximum number of pairwise *unrelated* unit cubes, denoted $S$.\n\nBy the extremal principle, for any rectangular parallelepiped of sizes $2m+1$, $2n+1$, $2p+1$, all $8$ corner numbers and the center number are equal. This implies the cube $(x, y, z)$ is *related* to the cube\n\n$$\n(x + (-1)^r m,\\ y + (-1)^s n,\\ z + (-1)^t p),\n$$\nwhere $r, s, t \\in \\{0, 1\\}$ independently, corresponding to the $8$ corners of the box centered at $(x, y, z)$.\n\nWe can also rotate the box, so $(x, y, z)$ is *related* to\n\n$$\n(x + a_1 m + a_2 n + a_3 p,\\ y + b_1 m + b_2 n + b_3 p,\\ z + c_1 m + c_2 n + c_3 p),\n$$\nwhere in each step, $a_1 + a_2 + a_3$, $b_1 + b_2 + b_3$, $c_1 + c_2 + c_3$ each change by $1$. Thus,\n\n$$\na_1 + a_2 + a_3 \\equiv b_1 + b_2 + b_3 \\equiv c_1 + c_2 + c_3 \\pmod{2}.\n$$\n\nConsequently, $(x, y, z)$ is *related* to $(x + x_1, y + y_1, z + z_1)$, where $(x_1, y_1, z_1)$ is a permutation of $(2k m, 2k n, 2k p)$ for some integer $k$.\n\nBy Bézout's theorem, for all linear combinations of $m, n, p$, $d = \\gcd(m, n, p) > 0$ is the minimal absolute value. Thus, each unit cube in a $d \\times d \\times d$ cube is pairwise *unrelated*, so $S$ is a multiple of $d^3$.\n\nLet $A$ be a $d \\times d \\times d$ cube. Set $m_1 = \\frac{m}{d}$, $n_1 = \\frac{n}{d}$, $p_1 = \\frac{p}{d}$. Consider three cases:\n\n1. If $m_1, n_1, p_1$ are all odd, let $m_1 = 2m_2 + 1$, $n_1 = 2n_2 + 1$, $p_1 = 2p_2 + 1$. Then $m = 2d m_2 + d$, $n = 2d n_2 + d$, $p = 2d p_2 + d$. The cube $(m, n, p)$ is *related* to $(m \\pm d, n \\pm d, p \\pm d)$. Choosing $A$ and three $d \\times d \\times d$ cubes sharing a face with $A$ in each direction, all unit cubes in these four cubes are pairwise *unrelated*. Thus, $S = 4d^3$.\n\n2. If two of $m_1, n_1, p_1$ are odd and the third is even (say $p_1$ even, $m_1, n_1$ odd), $(m, n, p)$ is *related* to $(m \\pm d, n \\pm d, p)$, and by permutation, also to $(m, n \\pm d, p \\pm d)$ and $(m \\pm d, n, p \\pm d)$. Choosing $A$ and one $d \\times d \\times d$ cube sharing a face with $A$, all unit cubes in these two cubes are pairwise *unrelated*. Thus, $S = 2d^3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17750,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$, $BC$, and $CD$ be three given sides of a quadrilateral $ABCD$. What is the maximal possible area of such a quadrilateral?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle CAD = \\beta$, $\\angle BDA = \\alpha$, $AB = a$, $BC = b$, $CD = c$, $AC = x$, $BD = y$, $AD = z$.\n\n![](images/Belorusija_2012_p32_data_f37b2cf2ba.png)\n\n$$\na = z \\sin \\alpha, \\quad c = z \\sin \\beta, \\quad x = z \\cos \\beta, \\quad y = z \\cos \\alpha\n$$\n\n$$\n\\begin{aligned}\nb &= z \\sin \\angle CDA = z \\sin(90^\\circ - \\beta - \\alpha) = z \\cos(\\beta + \\alpha) \\\\\n  &= z(\\cos \\beta \\cos \\alpha - \\sin \\beta \\sin \\alpha)\n\\end{aligned}\n$$\n\nIt is easy to see that $bz + ac = xy$. (This follows from Ptolemy's theorem.) Using the Pythagorean theorem for triangles $ABD$ and $ACD$, we obtain $ac + zb = \\sqrt{(z^2 - c^2)(z^2 - a^2)}$. So\n\n$$\nz^4 - (a^2 + b^2 + c^2)z^2 - 2abcz = 0\n$$\n\nSince $z \\neq 0$, we have\n\n$$\nz^3 - (a^2 + b^2 + c^2)z - 2abc = 0\n$$\n\nLet $a = 2$, $b = 7$, $c = 11$. Then\n\n$$\nz^3 - 174z - 308 = 0\n$$\n\nOne root is $z = 14$ (the other roots are negative).\n\nUsing the sine law for triangle $BCD$, $7 = z \\sin \\angle BDC = 14 \\sin \\angle BDC$, so $\\sin \\angle BDC = 1/2$, i.e., $\\angle BDC = 30^\\circ$. The angle between the diagonals $AC$ and $BD$ is $\\angle COD = 90^\\circ - 30^\\circ = 60^\\circ$.\n\nNow,\n\n$$\nAC = x = \\sqrt{z^2 - c^2} = \\sqrt{196 - 121} = \\sqrt{75} = 5\\sqrt{3}\n$$\n\n$$\nBD = y = \\sqrt{z^2 - a^2} = \\sqrt{196 - 4} = \\sqrt{192} = 8\\sqrt{3}\n$$\n\nTherefore, the required area is\n\n$$\nS(ABCD) = 0.5 \\cdot AC \\cdot BD \\cdot \\sin \\angle COD = 0.5 \\cdot 5\\sqrt{3} \\cdot 8\\sqrt{3} \\cdot \\frac{\\sqrt{3}}{2} = 30\\sqrt{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17751,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nA = \\frac{a_1 a_2}{a_1 - a_2}, \\quad B = \\frac{a_1 a_3}{a_1 - a_3}, \\quad C = \\frac{a_2 a_3}{a_2 - a_3}.\n$$\n\nShow that\n\n$$\n1 + |a_1 b_1 + a_2 b_2 + a_3 b_3| \\leq (1 + |a_1|)(1 + |a_2|)(1 + |a_3|)\n$$\n\nwhere $b_1 = (1+A)(1+B)$, $b_2 = (1-A)(1+C)$, $b_3 = (1-B)(1-C)$. Determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{align*}\n& a_1 b_1 + a_2 b_2 + a_3 b_3 \\\\\n= & a_1 (1+A) (1+B) + a_2 (1-A) (1+C) + a_3 (1-B) (1-C) \\\\\n= & a_1 + a_2 + a_3 + (a_1 - a_2)A + (a_1 - a_3)B \\\\\n& + (a_2 - a_3)C + a_1 AB - a_2 AC + a_3 BC.\n\\end{align*}\n$$\n\nSome computation shows that\n\n$$\n(a_1 - a_2)A + (a_1 - a_3)B + (a_2 - a_3)C = a_1 a_2 + a_1 a_3 + a_2 a_3,\n$$\n\nand\n\n$$\n\\begin{aligned}\n& a_1AB - a_2AC + a_3BC \\\\\n&= a_1a_2a_3 \\frac{a_1^2(a_2 - a_3) + a_2^2(a_3 - a_1) + a_3^2(a_1 - a_2)}{(a_1 - a_2)(a_2 - a_3)(a_1 - a_3)} \\\\\n&= a_1a_2a_3.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n& 1 + |a_1b_1 + a_2b_2 + a_3b_3| \\\\\n&= 1 + |a_1 + a_2 + a_3 + a_1a_2 + a_1a_3 + a_2a_3 + a_1a_2a_3| \\\\\n&\\leq 1 + |a_1| + |a_2| + |a_3| + |a_1a_2| + |a_1a_3| + |a_2a_3| + |a_1a_2a_3| \\\\\n&= (1 + |a_1|)(1 + |a_2|)(1 + |a_3|).\n\\end{aligned}\n$$\n\nEquality holds if and only if the seven real numbers $a_1, a_2, a_3, a_1a_2, a_1a_3, a_2a_3, a_1a_2a_3$ are all non-negative or all non-positive. Since at most one of $a_1, a_2, a_3$ can be zero, equality holds if and only if $a_1, a_2, a_3$ are all non-negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17752,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c(O, R)$ be a circle, $AB$ a diameter, and $\\Gamma$ the midpoint of the arc $\\widehat{AB}$. We draw the circle $c_I(K, KO)$, where $K$ is a point on the segment $OA$, and we consider the tangents $\\Gamma\\Delta$ and $\\Gamma O$ from $\\Gamma$ to the circle $c_I(K, KO)$. The line $K\\Delta$ intersects the circle $c(O, R)$ at the points $E$ and $Z$ (with $E$ lying in the same semicircle as $\\Gamma$). Finally, the lines $E\\Gamma$ and $\\Gamma Z$ intersect $AB$ at the points $N$ and $M$, respectively. Prove that the quadrilateral $EMZN$ is an isosceles trapezium inscribed in a circle whose center lies on the circle $c(O, R)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\hat{E}_1 = \\Gamma \\hat{E} Z$. Then, from the right triangle $\\Gamma \\Delta E$, we have $\\hat{\\Gamma}_1 = 90^\\circ - \\hat{E}_1$.  \n\nSince $\\Gamma \\hat{O} Z = \\hat{O}_1$ is the corresponding central angle of $\\hat{E}_1 = \\Gamma \\hat{E} Z$, then $\\hat{O}_1 = 2\\hat{E}_1$, and from the isosceles triangle $\\Gamma O Z$ we have:\n\n$$\n\\hat{\\Gamma}_2 = \\frac{180^\\circ - \\hat{O}_1}{2} = \\frac{180^\\circ - 2\\hat{E}_1}{2} = 90^\\circ - \\hat{E}_1.\n$$\n\nFrom the above, $\\hat{\\Gamma}_1 = \\hat{\\Gamma}_2$.\n\n![](images/Hellenic_booklet_2013_p9_data_dc1e06bd7e.png)\n\nThe triangles $\\Gamma \\Delta E$ and $\\Gamma O M$ are congruent [$\\Gamma \\hat{\\Delta} E = \\Gamma \\hat{O} M = 90^\\circ$, $\\hat{\\Gamma}_1 = \\hat{\\Gamma}_2$, and $\\Gamma \\Delta = \\Gamma O$]. Also, the triangles $\\Gamma K \\Delta$ and $\\Gamma K O$ are congruent, so $\\Gamma E M$ is isosceles and $\\Gamma K$ is the bisector of the angle $\\hat{E} \\Gamma M$ and the perpendicular bisector of $EM$.\n\nFrom the congruence of triangles $\\Gamma \\Delta E$ and $\\Gamma O M$, we have $\\Gamma \\hat{E} Z = \\Gamma \\hat{M} O$, so $N \\hat{E} Z = N \\hat{M} Z$. Hence, the quadrilateral $EMZN$ is cyclic, and we have $\\hat{M}_1 = \\hat{E} N Z$ and $\\hat{E}_2 = \\hat{M} Z N$.\n\nAlso, from the isosceles triangle $\\Gamma E M$, $\\hat{M}_1 = \\hat{E}_2$.\n\nTherefore, $\\hat{E} N Z = \\hat{M} Z N$, so $EMZN$ is an isosceles trapezium.\n\nThe center of the circumcircle of $EMZN$ is the intersection of the perpendicular bisectors of its sides. Let $T$ be the intersection of $\\Gamma K$ (the perpendicular bisector of $EM$) and the perpendicular bisector of $EZ$. In triangle $\\Gamma E Z$, $\\Gamma T$ is the bisector of $E \\hat{\\Gamma} Z$ and $O T$ is the perpendicular bisector of $EZ$. Hence, $T$ lies on the circumcircle of triangle $\\Gamma E Z$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17753,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all integers $k > 2$, there exist $k$ distinct positive integers $a_1, \\dots, a_k$ such that\n\n$$\n\\sum_{1 \\le i < j \\le k} \\frac{1}{a_i a_j} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us first introduce some notation: given positive integers $n$ and $k$, $e_{n,k}(x_1, \\dots, x_n)$ will denote the $k$th elementary symmetric polynomial in the $n$ variables $x_1, \\dots, x_n$.\n\nWe define a sequence of sets inductively as follows: $S_3 = \\{1, 2, 3\\}$. Now for $n > 2$, suppose $S_n = \\{a_1 < \\dots < a_n\\}$ with $a_1 < \\dots < a_k$. Then $S_{n+1}$ is defined to be the set\n\n$$\n\\{a_1 < \\dots < a_{n-1} < a_n + 1 < e_{n,n-1}(a_1, \\dots, a_{n-1}, a_n + 1)\\}.\n$$\n\nNow for $S_n = \\{a_1 < \\dots < a_n\\}$, we will use induction to prove the following two statements:\n\n1. $e_{n,n}(a_1, \\dots, a_n) = e_{n,n-2}(a_1, \\dots, a_n)$\n2. $e_{n,n}(a_1, \\dots, a_{n-1}, a_n + 1) = e_{n,n-2}(a_1, \\dots, a_{n-1}, a_n + 1) + 1$\n\nBoth of these statements are easy to verify for $n = 3, 4$. Now for the induction step, suppose $S_{n-1} = \\{b_1 < \\dots < b_{n-1}\\}$, so that $b_i = a_i$ for $i < n-1$ and $b_{n-1} = a_{n-1} - 1$. By definition, $a_n = e_{n-1,n-2}(a_1, \\dots, a_{n-1})$. Note that\n\n$$\n\\begin{aligned}\ne_{n,n-2}(a_1, \\dots, a_n) &= e_{n-1,n-2}(a_1, \\dots, a_{n-1}) + a_n e_{n-1,n-3}(a_1, \\dots, a_{n-1}) \\\\\n&= a_n(1 + e_{n-1,n-3}(a_1, \\dots, a_{n-1})) = a_n(1 + e_{n-1,n-3}(b_1, \\dots, b_{n-2}, b_{n-1} + 1)) \\\\\n&= a_n e_{n-1,n-1}(b_1, \\dots, b_{n-2}, b_{n-1} + 1) = a_n e_{n-1,n-1}(a_1, \\dots, a_{n-1}) \\\\\n&= e_{n,n}(a_1, \\dots, a_n).\n\\end{aligned}\n$$\n\nHere we have used (2) for $S_{n-1} = \\{b_1 < \\dots < b_{n-1}\\}$. This shows (1). Further, we have\n\n$$\n\\begin{aligned}\ne_{n,n}(a_1, \\dots, a_{n-1}, a_n + 1) &= e_{n,n}(a_1, \\dots, a_n) + e_{n-1,n-1}(a_1, \\dots, a_{n-1}) \\\\\n&= e_{n,n-2}(a_1, \\dots, a_n) + e_{n-1,n-1}(b_1, \\dots, b_{n-2}, b_{n-1} + 1) \\\\\n&= e_{n,n-2}(a_1, \\dots, a_n) + e_{n-1,n-3}(b_1, \\dots, b_{n-2}, b_{n-1} + 1) + 1 \\\\\n&= e_{n,n-2}(a_1, \\dots, a_n) + e_{n-1,n-3}(a_1, \\dots, a_{n-1}) + 1 \\\\\n&= e_{n,n-2}(a_1, \\dots, a_{n-1}, a_n + 1) + 1.\n\\end{aligned}\n$$\n\nThis proves (2), and the induction is finished. The given condition is equivalent to (1), and thus we are done. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17754,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, let $p(n)$ denote the greatest perfect square less than or equal to $n$.\n\na) Find all pairs of positive integers $(m, n)$ with $m \\leq n$ such that\n$$\np(2m + 1) \\cdot p(2n + 1) = 400.\n$$\n\nb) Determine the set\n$$\n\\left\\{ n \\in \\mathbb{N}^* \\mid n \\leq 100 \\text{ and } \\frac{p(n+1)}{p(n)} \\notin \\mathbb{N} \\right\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $400 = 1 \\times 400 = 4 \\times 100 = 16 \\times 25$, we analyze three cases:\n\n- If $p(2m+1) = 1$ and $p(2n+1) = 400$, then $1 \\leq 2m+1 < 4$ and $400 \\leq 2n+1 < 441$, so $m \\in \\{1, 2\\}$ and $n \\in \\{200, 201, \\ldots, 219\\}$, giving $2 \\times 20 = 40$ pairs.\n- If $p(2m+1) = 4$ and $p(2n+1) = 100$, then $4 \\leq 2m+1 < 9$ and $100 \\leq 2n+1 < 121$, so $m \\in \\{2, 3, 4\\}$ and $n \\in \\{50, 51, \\ldots, 60\\}$, giving $3 \\times 11 = 33$ pairs.\n- If $p(2m+1) = 16$ and $p(2n+1) = 25$, then $16 \\leq 2m+1 < 25$ and $25 \\leq 2n+1 < 36$, so $m \\in \\{8, 9, 10, 11, 12\\}$ and $n \\in \\{12, 13, 14, 15, 16\\}$, giving $5 \\times 5 = 25$ pairs.\n\nAdding up, there are $40 + 33 + 25 = 98$ pairs $(m, n)$.\n\nb) Let $p(n) = k^2$ for some $k \\geq 1$, so $k^2 \\leq n < (k+1)^2$. Then $p(n+1)$ is either $k^2$ or $(k+1)^2$. The ratio $\\frac{p(n+1)}{p(n)}$ is not an integer if and only if $p(n+1) = (k+1)^2$ and $p(n) = k^2$ with $k \\geq 2$, i.e., $n = (k+1)^2 - 1$. For $n \\leq 100$, $k$ runs from $2$ to $9$, so the set is $\\{8, 15, 24, 35, 48, 63, 80, 99\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17755,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $C_1$ 及 $C_2$ 為兩同心圓，其中 $C_2$ 在 $C_1$ 內部。從 $C_1$ 上一點 $A$ 向 $C_2$ 引切線 $AB$，且點 $B$ 在 $C_2$ 上。令點 $C$ 為射線 $AB$ 與 $C_1$ 的另一個交點，而點 $D$ 為 $\\overline{AB}$ 的中點。作一條過 $A$ 的直線與 $C_2$ 交於 $E, F$ 兩點，使得 $DE$ 的中垂線與 $CF$ 的中垂線交於 $AB$ 上的一點 $M$。試求 $AM/MC$ 的所有可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "$AM/MC = 5/3$。\n\n因為 $AC \\cdot AD = (2AB) \\cdot (\\frac{1}{2}AB) = AB^2 = AE \\cdot AF$，故 $CDEF$ 四點共圓。又 $M$ 點位於 $CF$ 及 $DE$ 的中垂線上，故 $M$ 點為 $CDEF$ 的外接圓圓心。因為 $CMD$ 在同一條直線上，所以 $M$ 為 $CD$ 中點。故\n\n$$\n\\frac{AM}{MC} = \\frac{AD + DM}{MC} = \\frac{AD + \\frac{1}{2}CD}{\\frac{1}{2}CD} = \\frac{\\frac{1}{4}AC + \\frac{1}{2} \\cdot \\frac{3}{4}AC}{\\frac{1}{2} \\cdot \\frac{3}{4}AC} = \\frac{5}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17756,
    "subject": "Mathematics (Olympiad)",
    "question": "In the trapezoid $ABCD$ with perpendicular diagonals, points $P$, $N$, $Q$, $M$ are the midpoints of sides $AB$, $BC$, $CD$, $DA$ respectively. On the base $CD$ there is a point $L$ (different from the point $Q$) for which the angle $MLN$ is straight. Find the angle $LPA$.",
    "options": [],
    "answer": "See solution",
    "solution": "By Varignon's Theorem, quadrilateral $MPNQ$ is a parallelogram and its sides are parallel to the diagonals of trapezoid $ABCD$, so it is a rectangle. Denote $O$ as the intersection point of $AC$ and $BD$. Then, from the properties of the right triangle $\\triangle MNL$, we get that: $OL = OM = ON \\Rightarrow OL = OP = OQ$, so $\\triangle PQL$ is right-angled. That means $PL \\perp LQ$, and since $AB \\parallel CD$, then $PL \\perp AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17757,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers, assumed relatively prime. Determine all possible values of\n$$\ngcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume $m \\ge n$. It is well known that\n$$\ngcd(2^p - 1, 2^q - 1) = 2^{gcd(p,q)} - 1,\n$$\nso that\n$$\n\\begin{aligned}\ngcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1) &= gcd(2^{m-n} - 1, 2^{m^2+mn+n^2} - 1) \\\\\n&= 2^{gcd(m-n, m^2+mn+n^2)} - 1.\n\\end{aligned}\n$$\nNext, consider a divisor $d \\mid m-n$. We must have $gcd(m, d) = 1$, since $m$ and $n$ are relatively prime. It follows that $0 \\equiv m^2 + mn + n^2 \\equiv 3m^2 \\pmod{d}$ is equivalent to $d \\mid 3$, and we infer that\n$$\ngcd(m - n, m^2 + mn + n^2) = gcd(m - n, 3),\n$$\nwhich is 1 or 3.\n\nHence $gcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1)$ may only assume the values 1 and 7. Both values are possible, since $m = 2, n = 1$ gives\n$$\ngcd(2^2 - 2^1, 2^{2^2+2 \\cdot 1+1^2} - 1) = gcd(2, 2^7 - 1) = 1,\n$$\nand $m = 1, n = 1$ gives\n$$\ngcd(2^1 - 2^1, 2^{1^2+1 \\cdot 1+1^2} - 1) = gcd(0, 2^3 - 1) = 7.\n$$\n![](images/Baltic_Way_SHL_2009-11_13-16_p290_data_8090bbde24.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17758,
    "subject": "Mathematics (Olympiad)",
    "question": "For how many integer values of $x$ is $|2x| \\le 7\\pi$?\n\n(A) 16  (B) 17  (C) 19  (D) 20  (E) 21",
    "options": [],
    "answer": "See solution",
    "solution": "From $3 < \\pi < 3.142$, it follows that\n\n$$\n21 = 7 \\cdot 3 < 7\\pi < 7 \\cdot 3.142 = 21.994 < 22.\n$$\n\nBecause $x$ is an integer, the values for $2x$ that make the inequality true are $-20, -18, -16, \\dots, -2, 0, 2, \\dots, 18$, and $20$. Each of these corresponds to a unique value of $x$. There are 21 such values.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17759,
    "subject": "Mathematics (Olympiad)",
    "question": "Димитар има два квадратни картони со страни $3\\ \\mathrm{cm}$ и $4\\ \\mathrm{cm}$. Дали може од нив, со сечење, да формира квадрат без да отфрли материјал? Ако може, колкава е страната на тој квадрат?",
    "options": [],
    "answer": "See solution",
    "solution": "Димитар може да состави квадрат со страна $5\\ \\mathrm{cm}$. Квадратот со страна $3\\ \\mathrm{cm}$ ќе го раздели на правоаголници со страни $1\\ \\mathrm{cm}$ и $3\\ \\mathrm{cm}$, еден правоаголник со страни $1\\ \\mathrm{cm}$ и $2\\ \\mathrm{cm}$ и еден квадрат со страна $1\\ \\mathrm{cm}$. Овие делови ќе ги додаде на квадратот со страна $4\\ \\mathrm{cm}$, како што е прикажано на цртежот.\n\n![](images/Makedonija_2009_p8_data_f2640d6cd1.png)\n\n![](images/Makedonija_2009_p8_data_b8f7f76415.png)\n\n![](images/Makedonija_2009_p8_data_533d88c931.png)\n\n**Забелешка:** Можни се и други решенија.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17760,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n-1 < \\left( \\sum_{k=1}^{n} \\frac{k}{k^2 + 1} \\right) - \\ln n \\leq \\frac{1}{2}, \\quad n = 1, 2, \\dots\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that\n$$\n\\frac{x}{1+x} < \\ln(1+x) < x, \\quad x > 0.\n$$\nLet\n$$\nh(x) = x - \\ln(1+x), \\quad g(x) = \\ln(1+x) - \\frac{x}{1+x}.\n$$\nThen, for $x > 0$,\n$$\nh'(x) = 1 - \\frac{1}{1+x} > 0, \\quad g'(x) = \\frac{1}{1+x} - \\frac{1}{(1+x)^2} = \\frac{x}{(1+x)^2} > 0.\n$$\nTherefore,\n$$\nh(x) > h(0) = 0, \\quad g(x) > g(0) = 0.\n$$\nThis completes the proof of the inequalities above.\n\nNow let $x = \\frac{1}{n}$ in the above inequalities. We have\n$$\n\\frac{1}{n+1} < \\ln\\left(1+\\frac{1}{n}\\right) < \\frac{1}{n}.\n$$\nLet\n$$\nx_n = \\sum_{k=1}^{n} \\frac{k}{k^2 + 1} - \\ln n.\n$$\nThen\n$$\n\\begin{aligned}\nx_n - x_{n-1} &= \\frac{n}{n^2+1} - \\ln\\left(1 + \\frac{1}{n+1}\\right) \\\\\n&< \\frac{n}{n^2+1} - \\frac{1}{n} \\\\\n&= -\\frac{1}{n(n^2+1)} < 0.\n\\end{aligned}\n$$\nTherefore, $x_n < x_{n-1} < \\cdots < x_1 = \\frac{1}{2}$.\n\nFurthermore,\n$$\n\\begin{aligned}\n\\ln n &= (\\ln n - \\ln(n-1)) + (\\ln(n-1) - \\ln(n-2)) \\\\\n&\\quad + \\cdots + (\\ln 2 - \\ln 1) + \\ln 1 \\\\\n&= \\sum_{k=1}^{n-1} \\ln\\left(1 + \\frac{1}{k}\\right).\n\\end{aligned}\n$$\nConsequently,\n$$\n\\begin{aligned}\nx_n &= \\sum_{k=1}^{n} \\frac{k}{k^2+1} - \\sum_{k=1}^{n-1} \\ln\\left(1 + \\frac{1}{k}\\right) \\\\\n&= \\sum_{k=1}^{n-1} \\left( \\frac{k}{k^2+1} - \\ln\\left(1 + \\frac{1}{k}\\right) \\right) + \\frac{n}{n^2+1} \\\\\n&> \\sum_{k=1}^{n-1} \\left( \\frac{k}{k^2+1} - \\frac{1}{k} \\right) \\\\\n&= - \\sum_{k=1}^{n-1} \\frac{1}{(k^2+1)k} \\\\\n&> - \\sum_{k=1}^{n-1} \\frac{1}{(k+1)k} \\\\\n&= -1 + \\frac{1}{n} > -1.\n\\end{aligned}\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17761,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 11.1, in a plane rectangular coordinate system $xOy$, the left and right foci of the ellipse $\\Gamma: \\frac{x^2}{2} + y^2 = 1$ are $F_1, F_2$, respectively. Let $P$ be a point on $\\Gamma$ in the first quadrant, and the extensions of $PF_1, PF_2$ intersect $\\Gamma$ at points $Q_1(x_1, y_1), Q_2(x_2, y_2)$, respectively.\n\nFind the maximum of $y_1 - y_2$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p69_data_894a42ac6a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The equation of line $PF_1$ is $x = \\frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into\n\n$\\frac{x^2}{2} + y^2 = 1$ and organizing it yields\n\n$$\n\\left( \\frac{(x_0 + 1)^2}{2y_0^2} + 1 \\right) y^2 - \\frac{x_0 + 1}{y_0} y - \\frac{1}{2} = 0.\n$$\n\nMultiplying both sides by $2y_0^2$ and noting that $x_0^2 + 2y_0^2 = 2$, we get\n\n$$\n(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.\n$$\n\nThe two roots of this equation are $y_0, y_1$. By Vieta's formulas, we get\n\n$y_0y_1 = -\\frac{y_0^2}{3+2x_0}$. Thus,\n\n$$\ny_1 = -\\frac{y_0}{3 + 2x_0}.\n$$\n\nSimilarly, we can get $y_2 = -\\frac{y_0}{3 - 2x_0}$. Therefore,\n\n$$\ny_1 - y_2 = \\frac{y_0}{3 - 2x_0} - \\frac{y_0}{3 + 2x_0} = \\frac{4x_0y_0}{9 - 4x_0^2}.\n$$\n\nSince $9 - 4x_0^2 = \\frac{1}{2}x_0^2 + 9y_0^2 \\ge 2\\sqrt{\\frac{1}{2}x_0^2 \\cdot 9y_0^2} = 3\\sqrt{2}x_0y_0$, it follows that\n\n$$\ny_1 - y_2 \\le \\frac{4x_0y_0}{3\\sqrt{2}x_0y_0} = \\frac{2\\sqrt{2}}{3},\n$$\n\nwhere the equal sign holds when $\\frac{1}{2}x_0^2 = 9y_0^2$ is required, and accordingly\n\n$x_0 = \\frac{3\\sqrt{5}}{5}, \\quad y_0 = \\frac{\\sqrt{10}}{10}$.\n\nTherefore, the maximum of $y_1 - y_2$ is $\\frac{2\\sqrt{2}}{3}$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17762,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1A_2\\cdots A_{101}$ be a regular 101-gon, and color every vertex red or blue. Let $N$ be the number of obtuse triangles satisfying the following:\n\n- The three vertices of the triangle must be vertices of the 101-gon.\n- Both the vertices with acute angles have the same color.\n- The vertex with the obtuse angle has a different color.\n\n1. Find the largest possible value of $N$.\n2. Find the number of ways to color the vertices such that maximum $N$ is achieved. (Two colorings are different if for some $A_i$ the colors are different on the two coloring schemes)",
    "options": [],
    "answer": "See solution",
    "solution": "Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For an obtuse triangle $A_{i-a}A_iA_{i+b}$ (vertex $A_i$ is the vertex of the obtuse angle, i.e. $a + b \\le 50$), these three vertices satisfy the conditions if and only if\n\n$$\n(x_i - x_{i-a})(x_i - x_{i+b}) = 1\n$$\n\notherwise equals $0$, where the subscript is modulo $101$. Thus,\n\n$$\nN = \\sum_{i=1}^{101} \\sum_{(a,b)} (x_i - x_{i-a})(x_i - x_{i+b})\n$$\n\nwhere $\\sum_{(a,b)}$ is over all positive integer pairs $(a, b)$ with $a + b \\le 50$; there are $49 + 48 + \\cdots + 1 = 1225$ such pairs. Expanding,\n\n$$\n\\begin{align*}\nN &= \\sum_{i=1}^{101} \\sum_{(a,b)} (x_i^2 - x_i x_{i-a} - x_i x_{i+b} + x_{i-a} x_{i+b}) \\\\\n  &= 1225 \\sum_{i=1}^{101} x_i^2 + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (k-1-2(50-k)) x_i x_{i+k}\n\\end{align*}\n$$\n\n$$\n= 1225n + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (3k-101)x_i x_{i+k}\n$$\n\nwhere $n$ is the number of blue vertices. For any two vertices $A_i, A_j$ ($1 \\le i, j \\le 101$), let\n\n$$\nd(A_i, A_j) = d(A_j, A_i) = \\min\\{j - i, 101 - j + i\\}\n$$\n\nLet $B \\subseteq \\{A_1, A_2, \\dots, A_{101}\\}$ be the set of blue vertices. Then,\n\n$$\nN = 1225n - 101\\binom{n}{2} + 3 \\sum_{\\{P,Q\\} \\subseteq B} d(P, Q)\n$$\n\nwhere $\\{P, Q\\}$ runs over all unordered pairs of blue vertices. Assume $n$ is even (otherwise, coloring all vertices the opposite way does not change $N$). Write $n = 2t$, $0 \\le t \\le 50$, and label the blue vertices $P_1, P_2, \\dots, P_{2t}$ clockwise. Then,\n\n$$\n\\sum_{\\{P,Q\\} \\subseteq B} d(P, Q) = \\sum_{i=1}^{t} d(P_i, P_{i+t}) + \\frac{1}{2} \\sum_{i=1}^{t} \\sum_{j=1}^{t-1} [d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i)]\n$$\n\nwith indices modulo $2t$, and using $d(P_i, P_{i+t}) \\le 50$ and\n\n$$\nd(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \\le 101\n$$\n\nCombining, we get\n\n$$\nN \\le 1225n - 101\\binom{n}{2} + 3 \\left( 50t + \\frac{101}{2} t(t-1) \\right) = -\\frac{101}{2} t^2 + \\frac{5099}{2} t\n$$\n\nThe maximum occurs at $t = 25$, so $N \\le 32175$.\n\nTo achieve $N = 32175$, $t = 25$ (i.e., $n = 50$ blue vertices), and for $1 \\le i \\le t$, $d(P_i, P_{i+t}) = 50$. This requires choosing 25 diagonals from the longest 101 diagonals such that no two share a vertex. Edging $A_i$ and $A_{i+50}$ for $i = 1, 2, \\dots, 101$ forms a graph $G$ (a 101-cycle). The number of ways to choose 25 edges of $G$ with no shared vertices is $\\binom{75}{24} + \\binom{76}{25}$. Similarly, for 50 red vertices, the number is the same. Thus, the total number of colorings achieving maximum $N$ is $2\\left[\\binom{75}{24} + \\binom{76}{25}\\right]$.\n\n**Summary:**\n- The largest possible value of $N$ is $32175$.\n- The number of ways to color the vertices to achieve this is $2\\left[\\binom{75}{24} + \\binom{76}{25}\\right]$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17763,
    "subject": "Mathematics (Olympiad)",
    "question": "In ellipse $\\Gamma$, $A$ is an endpoint of the major axis, $B$ is an endpoint of the minor axis, and $F_1, F_2$ are the foci. If $\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} = 0$, then the value of $\\frac{|AB|}{|F_1F_2|}$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ with $a > b > 0$. Let $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$. By the given conditions:\n\n$$\n\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} = (-c-a)(c-a) + (-c^2+b^2) = a^2 + b^2 - 2c^2 = 0\n$$\n\nTherefore,\n$$\n\\frac{|AB|}{|F_1F_2|} = \\frac{\\sqrt{a^2 + b^2}}{2c} = \\frac{\\sqrt{2c^2}}{2c} = \\frac{\\sqrt{2}}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17764,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = \\frac{1}{\\sqrt{22}} \\sec \\theta$ and $y = \\frac{1}{\\sqrt{22}} \\tan \\theta$. Find the greatest possible value of\n$$\n\\frac{1 - 22xy}{x^2}\n$$\nas $\\theta$ varies.",
    "options": [],
    "answer": "See solution",
    "solution": "$x^2 - y^2 = \\frac{1}{22}$.\n\n$$\n\\begin{aligned}\n\\frac{1 - 22xy}{x^2} &= \\frac{1 - \\sec \\theta \\tan \\theta}{\\frac{1}{22} \\sec^2 \\theta} \\\\\n&= 22(\\cos^2 \\theta - \\sin \\theta) \\\\\n&= 22(-\\sin^2 \\theta - \\sin \\theta + 1) \\\\\n&= 22 \\left[ \\frac{5}{4} - \\left( \\sin \\theta + \\frac{1}{2} \\right)^2 \\right].\n\\end{aligned}\n$$\n\nThe greatest possible value is thus $22 \\cdot \\frac{5}{4} = \\frac{55}{2}$, attained when $\\sin \\theta = -\\frac{1}{2}$ (which is feasible, e.g. with $\\theta = -30^\\circ$, $x = \\frac{2}{\\sqrt{66}}$ and $y = -\\frac{1}{\\sqrt{66}}$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17765,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $AB\\Gamma$ is given with circumcenter $O$. Let $A_1$, $B_1$, and $\\Gamma_1$ be the midpoints of sides $B\\Gamma$, $A\\Gamma$, and $AB$, respectively. Consider points $A_2$, $B_2$, and $\\Gamma_2$ such that $\\overrightarrow{OA_2} = \\lambda \\cdot \\overrightarrow{OA_1}$, $\\overrightarrow{OB_2} = \\lambda \\cdot \\overrightarrow{OB_1}$, and $\\overrightarrow{O\\Gamma_2} = \\lambda \\cdot \\overrightarrow{O\\Gamma_1}$, with $\\lambda > 0$. Prove that the lines $AA_2$, $BB_2$, and $\\Gamma\\Gamma_2$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of triangle $AB\\Gamma$. Then $\\overrightarrow{AH} = 2 \\cdot \\overrightarrow{OA_1}$, and since $\\overrightarrow{OA_2} = \\lambda \\cdot \\overrightarrow{OA_1}$, we have $\\overrightarrow{AH} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{OA_2}$.\n\nIf $AA_2$ meets $OH$ at $C$, by similarity of triangles $CHA$ and $COA_2$, $\\overrightarrow{HC} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{CO}$. Thus, $AA_2$ passes through $C$, which divides $OH$ in ratio $\\frac{2}{\\lambda}$.\n\nSimilarly, $\\overrightarrow{BH} = 2 \\cdot \\overrightarrow{OB_1}$ and $\\overrightarrow{BH} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{OB_2}$. Let $C'$ be the intersection of $BB_2$ and $OH$; then $\\overrightarrow{HC'} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{C'O}$, so $BB_2$ passes through $C'$ dividing $OH$ in ratio $\\frac{2}{\\lambda}$.\n\nLikewise, if $C''$ is the intersection of $\\Gamma\\Gamma_2$ and $OH$, then $\\Gamma\\Gamma_2$ passes through $C''$ dividing $OH$ in ratio $\\frac{2}{\\lambda}$.\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p8_data_f12f6bd902.png)\n\nSince $C$, $C'$, and $C''$ coincide, the lines $AA_2$, $BB_2$, and $\\Gamma\\Gamma_2$ are concurrent.\n\n**Comments**\n\n1. If $\\lambda = 1$, then $C$ is the barycenter of triangle $AB\\Gamma$.\n2. If $\\lambda = 2$, then $C$ is the center of the Euler circle of triangle $AB\\Gamma$. In this case, triangles $AB\\Gamma$ and $A_2B_2\\Gamma_2$ are equal and share the same Euler circle.\n3. In any case, triangles $AB\\Gamma$ and $A_2B_2\\Gamma_2$ are similar with parallel sides. One is the image of the other under a homothety, so a solution using homotheties is possible.\n4. The problem can also be solved using analytic geometry or complex numbers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17766,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $9^n - 7$ can be represented as a product of at least two consecutive positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "The product of three consecutive positive integers is divisible by $3$, and $9^n - 7 \\equiv 2 \\pmod{3}$, so $9^n - 7$ cannot be written as a product of three or more consecutive positive integers.\n\nLet $9^n - 7 = m(m+1)$ for some positive integer $m$. This is equivalent to $4 \\cdot 9^n - 27 = (2m+1)^2$, i.e., $4 \\cdot 9^n - (2m+1)^2 = 27$. Since $\\{1,3,9,27\\}$ are all positive divisors of $27$ and\n\n$$\n2 \\cdot 3^n + 2m + 1 > 2 \\cdot 3^n - 2m - 1,\n$$\n\ntwo cases are possible:\n\n$$\n\\begin{cases}\n2 \\cdot 3^n + 2m + 1 = 27 \\\\\n2 \\cdot 3^n - 2m - 1 = 1\n\\end{cases}\n$$\n\nor\n\n$$\n\\begin{cases}\n2 \\cdot 3^n + 2m + 1 = 9 \\\\\n2 \\cdot 3^n - 2m - 1 = 3\n\\end{cases}.\n$$\n\nIn case 1, summing the equations gives $3^n = 7$, which is impossible.\n\nIn case 2, summing the equations gives $3^n = 3$, so $n=1$. Substituting $n=1$ into the first equation: $2 \\cdot 3 + 2m + 1 = 9$, so $m=1$. Thus, $9^1 - 7 = 1 \\cdot 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17767,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are only finitely many triples $ (a, b, c) $ of positive integers satisfying the equation $$ abc = 2009(a + b + c). $$",
    "options": [],
    "answer": "See solution",
    "solution": "There are at most six permutations for any three numbers $x, y, z$. It suffices to show that there are only finitely many triples $ (a, b, c) $, with $ a \\geq b \\geq c $, of positive integers satisfying the equation $ abc = 2009(a + b + c) $. It follows that $ abc \\leq 2009 \\times (3a) $, or $ bc \\leq 2009 \\times 3 = 6027 $. Clearly, there are finitely many pairs $ (b, c) $ of positive integers satisfying $ bc \\leq 6027 $, and for each fixed pair $ (b, c) $, there is at most one positive integer $ a $ satisfying $ abc = 2009(a + b + c) $ (because it is a linear equation in $ a $).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17768,
    "subject": "Mathematics (Olympiad)",
    "question": "The roots of the quadratic equation $x^2 + p x + q = 0$ are integers. Find the numbers $p$ and $q$ and the roots of the equation, if $p + q = 198$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1$ and $x_2$ be the roots of $x^2 + p x + q = 0$. By Vieta's formulas:\n\n$$\n\\begin{cases}\nx_1 + x_2 = -p \\\\\nx_1 x_2 = q\n\\end{cases}\n$$\n\nGiven $p + q = 198$, so:\n\n$$\np + q = -(x_1 + x_2) + x_1 x_2 = (x_1 - 1)(x_2 - 1) - 1 = 198\n$$\n\nThus, $(x_1 - 1)(x_2 - 1) = 199$. Since $199$ is prime, the integer solutions are:\n\n$$\n\\begin{cases}\nx_1 - 1 = 1,\\ x_2 - 1 = 199 \\implies (x_1, x_2) = (2, 200) \\\\\nx_1 - 1 = -1,\\ x_2 - 1 = -199 \\implies (x_1, x_2) = (0, -198)\n\\end{cases}\n$$\n\nFor $(2, 200)$:\n- $p = -(2 + 200) = -202$\n- $q = 2 \\times 200 = 400$\n\nFor $(0, -198)$:\n- $p = -(0 + (-198)) = 198$\n- $q = 0 \\times (-198) = 0$\n\nSo, possible values are:\n- $p = -202$, $q = 400$, roots $2$ and $200$\n- $p = 198$, $q = 0$, roots $0$ and $-198$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17769,
    "subject": "Mathematics (Olympiad)",
    "question": "Exactly $p-1$ distinct positive integers are written on the blackboard for some prime number $p$. The number $p$ is among these $p-1$ numbers. For any pair of the numbers, the absolute value of their difference is also on the board. Prove that all the numbers on the blackboard are divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the numbers on the board by $a_1, a_2, \\dots, a_{p-1}$.\n\nWithout loss of generality, assume $a_1 < a_2 < \\dots < a_{p-1}$. Then the numbers $a_2 - a_1 < a_3 - a_1 < \\dots < a_{p-1} - a_1$ are also written on the blackboard. There are $p-2$ of them, and they are all distinct. This is only possible when\n\n$$\n\\begin{aligned}\na_{p-2} &= a_{p-1} - a_1 \\\\\n&\\vdots \\\\\na_2 &= a_3 - a_1 \\\\\na_1 &= a_2 - a_1\n\\end{aligned}\n$$\n\nThus, $a_2 = 2a_1$, $a_3 = 3a_1$, $\\dots$, $a_{p-2} = (p-2)a_1$, and $a_{p-1} = (p-1)a_1$. Since $p$ is a prime and it appears on the blackboard, we must have $a_1 = p$. We conclude that all the numbers on the blackboard are divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17770,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all complex numbers $z$ such that\n$$\n|z| + |z - 5i| = |z - 2i| + |z - 3i|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We notice that $|z - 2i| = \\left|\\frac{2}{5}(z - 5i) + \\frac{3}{5}z\\right| \\leq \\frac{2}{5}|z - 5i| + \\frac{3}{5}|z|$.\n\nSimilarly, $|z - 3i| = \\left|\\frac{3}{5}(z - 5i) + \\frac{2}{5}z\\right| \\leq \\frac{3}{5}|z - 5i| + \\frac{2}{5}|z|$.\n\nTherefore, $|z| + |z - 5i| \\geq |z - 2i| + |z - 3i|$. Equality holds if there exists $\\lambda \\geq 0$ 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": 17771,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for every positive integer $n$, there exist integers $a$ and $b$ such that $n$ divides $4a^2 + 9b^2 - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, let $n = 2k + 1$ for some non-negative integer $k$. For $a = k$ and $b = 0$, we have:\n$$\n4a^2 + 9b^2 - 1 = 4k^2 - 1 = (2k + 1)(2k - 1),\n$$\nso $n$ divides $4a^2 + 9b^2 - 1$.\n\nIf $n$ is not divisible by $3$, let $n = 3k + r$ for some non-negative integer $k$ and $r \\in \\{1, -1\\}$. For $a = 0$ and $b = k$, we have:\n$$\n4a^2 + 9b^2 - 1 = 9k^2 - 1 = (3k + 1)(3k - 1),\n$$\nso $n$ divides $4a^2 + 9b^2 - 1$.\n\nIf $n$ is divisible by $6$, let $n = 2^r 3^s m$ for some positive integers $r, s$ and $m$ such that $m$ is relatively prime to $6$. Since $2^r$ and $3^s m$ are relatively prime, there exist non-zero integers $k$ and $l$ such that $2^r k + 3^s m l = 1$. Squaring this equation gives:\n$$\n2^{2r}k^2 + 3^{2s}m^2l^2 + 2 \\cdot 2^{r}3^{s}m k l = 1,\n$$\ni.e., $-2n k l = 2^{2r}k^2 + 3^{2s}m^2l^2 - 1$. For $a = 2^{r-1}k$ and $b = 3^{s-1}m l$, we have $4a^2 + 9b^2 - 1 = 2^{2r}k^2 + 3^{2s}m^2l^2 - 1$, so $n$ divides $4a^2 + 9b^2 - 1$.\n\nThus, for every positive integer $n$, there exist integers $a$ and $b$ such that $n$ divides $4a^2 + 9b^2 - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17772,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a real-valued function of a real variable such that\n\n$$\nf(f(x)) = x^2 - x + 1\n$$\n\nfor all real numbers $x$. Determine $f(0)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = a$ and $f(1) = b$.\n\nThen $f(f(0)) = f(a)$.\n\nBut $f(f(0)) = 0^2 - 0 + 1 = 1$. So $f(a) = 1$.\n\nAlso, $f(f(1)) = f(b)$.\n\nBut $f(f(1)) = 1^2 - 1 + 1 = 1$. So $f(b) = 1$.\n\nFrom above, $f(f(a)) = f(1)$.\n\nBut $f(f(a)) = a^2 - a + 1$. So $a^2 - a + 1 = b$.\n\nFrom above, $f(f(b)) = f(1)$, giving $b^2 - b + 1 = b$. So $b = 1$.\n\nPutting $b = 1$ in the previous equation gives $a^2 - a + 1 = 1$, so $a^2 - a = 0$, i.e., $a = 0$ or $a = 1$.\n\nBut $a = 0$ implies $f(0) = 0$ and $f(f(0)) = 0$, contradicting $f(f(0)) = 1$.\n\nSo $a = 1$, i.e., $f(0) = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17773,
    "subject": "Mathematics (Olympiad)",
    "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 $p$ members each, some house contains at least one pair of opponents.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $$(h-1) \\cdot \\min(p, h/2 + 1) + 1.$$  \n\nConsider the parliament as a graph on $hp$ vertices, where two vertices are joined by an edge if the corresponding members are opponents. The above minimum is achieved by at least one of the following two graphs:\n\n- The graph obtained by adjoining $h(p-1) - 1$ isolated vertices to the complete graph on $h+1$ vertices, in which case the number of edges is $h(h+1)/2 = (h-1)(h/2 + 1) + 1$.\n- The graph obtained by adjoining $p-2$ isolated vertices to a star with $p(h-1) + 1$ rays.\n\nLet $N(h, p) = (h-1) \\cdot \\min(p, h/2+1)$. 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 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\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$$\n\\sum_{v \\in V \\setminus (V_0 \\cup V')} \\deg v \\le hp - |V_0| = p(h-1).\n$$\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$$\n\\begin{aligned} \\sum_{v \\in V_0 \\cup V'} \\deg v &= \\sum_{i=1}^{p} \\left( \\deg v_i + \\sum_{v \\in V_i} \\deg v \\right) \\le \\sum_{i=1}^{p} \\deg v_i (\\deg v_i + 1) \\\\ &= \\sum_{i=1}^{p} (\\deg v_i - 1)^2 + 3d - p. \\end{aligned}\n$$\n\nSince $\\deg v_i \\ge 1$, $i = 1, \\dots, p$,\n\n$$\n\\sum_{i=1}^{p} (\\deg v_i - 1)^2 \\le \\left( \\sum_{i=1}^{p} (\\deg v_i - 1) \\right)^2 = (d-p)^2,\n$$\n\nso (recalling that $d \\le h-1$)\n\n$$\n\\begin{aligned} \\sum_{v \\in V_0 \\cup V'} \\deg v &\\le (d-p)^2 + 3d - p \\le (h-p-1)^2 + 3(h-1) - p \\\\ &= (h-1)(h+2) + p(p-2h+1) = 2N(h,p) + p(p-2h+1). \\end{aligned}\n$$\n\nHence, by the preceding,\n\n$$\n\\begin{aligned} 2N(h,p) = 2|E| &= \\sum_{v \\in V} \\deg v = \\sum_{v \\in V_0 \\cup V'} \\deg v + \\sum_{v \\in V \\setminus (V_0 \\cup V')} \\deg v \\\\ &\\le 2N(h,p) + p(p-2h+1) + p(h-1) = 2N(h,p) + p(p-h) \\\\ &< 2N(h,p), \\end{aligned}\n$$\n\nwhich is a contradiction. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17774,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Is it possible for a natural number $n$ to have two distinct representations as $n = aq_1 + r_1 = bq_2 + r_2$ with $a, b \\in \\mathbb{N}$, $a < b$, $r_1 \\neq 0$, $r_2 \\neq 0$, and $n = r_1 + r_2$?\n\nb) Find a natural number $n$ such that\n$$\nn = 29q_1 + r_1 = 39q_2 + r_2 = 59q_3 + r_3 = r_1 + r_2 + r_3,\n$$\nwhere $r_1 < 29$, $r_2 < 39$, $r_3 < 59$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is impossible. Suppose $n = aq_1 + r_1 = bq_2 + r_2 = r_1 + r_2$ with $q_1 \\ge 0$, $q_2 \\ge 0$, $r_1 < a$, $r_2 < b$. Then $bq_2 = r_1 < a \\le b$, so $bq_2 < b$ implies $q_2 = 0$, but then $r_1 = bq_2 = 0$, contradicting $r_1 \\neq 0$.\n\nb) Let $n = 29q_1 + r_1 = 39q_2 + r_2 = 59q_3 + r_3 = r_1 + r_2 + r_3$, with $r_1 < 29$, $r_2 < 39$, $r_3 < 59$. Since $59q_3 = r_1 + r_2 \\le 28 + 38 = 66$, $q_3 = 1$, so $r_1 + r_2 = 59$.\n\nFor $39q_2 = r_1 + r_3 \\le 28 + 58 = 86$, $q_2 \\le 2$.\n\nCase 1: $q_2 = 1$. Then $r_1 + r_3 = 39$. Adding (1) and (2):\n$$\n98 = 59 + 39 = r_1 + r_2 + r_1 + r_3 = n + r_1 = 29q_1 + 2r_1,\n$$\nSo $29q_1 + 2r_1 = 98$. $q_1$ is even, $q_1 = 2$, $r_1 = 20$, $r_2 = 39$, but $r_2 < 39$ is violated.\n\nCase 2: $q_2 = 2$. Then $r_1 + r_3 = 78$. Adding (1) and (3):\n$$\n29q_1 + 2r_1 = 137,\n$$\n$q_1$ is odd, $q_1 = 3$, $r_1 = 25$, $r_2 = 34$, $r_3 = 53$, $n = 25 + 34 + 53 = 112$, which satisfies all conditions.\n\n**Answer:**\n\na) Impossible.\nb) $n = 112$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17775,
    "subject": "Mathematics (Olympiad)",
    "question": "Petryk solved 33 problems at the exam. For the lesser part of them, including the first problem, he got $a$ points, while for the rest he got $b$ points. It is known that the natural numbers $a$ and $b$ satisfy $1 \\leq b < a \\leq 10$. After the exam, Petryk calculated the average score for all problems and it turned out to be an integer. For how many problems did Petryk get $a$ points?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 11.\n\nLet Petryk solve $n$ problems for $a$ points and $(33 - n)$ problems for $b$ points. Then the average result is:\n\n$$\nS = \\frac{n a + (33 - n) b}{33} = \\frac{n(a - b) + 33b}{33} = \\frac{n(a - b)}{33} + b.\n$$\n\nFor this to be an integer, $n(a-b)$ must be divisible by 33. Since $1 \\leq a-b \\leq 9$ and $n < 33$, the possible values for $n$ are 11 or 22. Since the number of $a$-point problems is the lesser part, $n = 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17776,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$, with $p < q$, be two primes such that $1 + p + p^2 + \\dots + p^m$ is a power of $q$ for some positive integer $m$, and $1 + q + q^2 + \\dots + q^n$ is a power of $p$ for some positive integer $n$. Show that $p = 2$ and $q = 2^t - 1$, where $t$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the smallest positive integer such that $1 + p + p^2 + \\dots + p^m$ is a power of $q$, say $q^s$. Then $m + 1$ must be prime, for if $m + 1 = kl$, then\n\n$$\n1 + p + p^2 + \\dots + p^m = (1 + p^l + p^{2l} + \\dots + p^{(k-1)l})(1 + p + p^2 + \\dots + p^{l-1}),\n$$\n\nso $1 + p + p^2 + \\dots + p^{l-1}$ is again a power of $q$, and minimality of $m$ forces $l = 1$ or $k = 1$. Similarly, if $n$ is the smallest positive integer such that $1 + q + q^2 + \\dots + q^n$ is a power of $p$, say $p^r$, then $n + 1$ must be prime.\n\nClearly, $p^{m+1} \\equiv 1 \\pmod q$ and $p^r \\equiv 1 \\pmod q$. Since $p \\not\\equiv 1 \\pmod q$ and $m + 1$ is prime, $m + 1$ must divide $r$.\n\nIf $q \\not\\equiv 1 \\pmod p$, a similar argument shows that $n + 1$ must divide $s$, so\n\n$$\n(p^{m+1} - 1)(q^{n+1} - 1) = p^r q^s (p-1)(q-1) \\ge p^{m+1} q^{n+1}\n$$\n\nwhich is impossible.\n\nHence $q \\equiv 1 \\pmod p$, so $n + 1 \\equiv 0 \\pmod p$ which forces $n + 1 = p$ by primality of $n + 1$.\n\nRecall that $r$ is divisible by $m + 1$, say $r = r'(m + 1)$, to write\n\n$$\n1 + q + q^2 + \\dots + q^n = p^r = (p^{m+1})^{r'} = (q^s(p-1) + 1)^{r'}\n$$\n\nand deduce thereby that $q^s$ divides $q + q^2 + \\dots + q^n$. This forces $s = 1$, so $q = 1 + p + p^2 + \\dots + p^m$.\n\nNow suppose, if possible, that $p \\ne 2$. Since $p^r$ divides\n\n$$\nq^{n+1} - 1 = q^p - 1 = (1 + p + p^2 + \\dots + p^m)^p - 1 = p^2 + p^3 N,\n$$\n\nit follows that $r = 2$, so $m = 1$. Hence $q = p + 1$ which is even — a contradiction.\n\nConsequently, $p = 2$, so $n = 1$, $q = 1 + 2 + 2^2 + \\dots + 2^m = 2^{m+1} - 1$, where $m + 1$ is prime, and $r = m + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17777,
    "subject": "Mathematics (Olympiad)",
    "question": "For how many integers $n$ with $1 \\leq n \\leq 800$ is the number $8n + 1$ a square?",
    "options": [],
    "answer": "See solution",
    "solution": "39",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17778,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation:\n\n$$\n(x + 1)^5 + (x + 1)^4(x - 1) + (x + 1)^3(x - 1)^2 + (x + 1)^2(x - 1)^3 + (x + 1)(x - 1)^4 + (x - 1)^5 = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Multiplying both sides by $2 = (x + 1) - (x - 1)$ yields $$(x + 1)^6 - (x - 1)^6 = 0$$ or equivalently $$(x + 1)^2 = (x - 1)^2$$. Solving this equation, we obtain the unique solution $x = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17779,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $2n + 1$ lines in the plane, what is the maximum number of acute triangles that can be formed by the intersections of these lines? Compute this maximum for $n = 2015$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take a line $L$ as the $x$-axis. The other $2n$ lines split into two groups: those with positive slopes ($a$ lines) and those with negative slopes ($b$ lines). Every pair within the same group forms an obtuse triangle with $L$, so the number of obtuse triangles involving $L$ is at least\n\n$$\n\\binom{a}{2} + \\binom{b}{2} = \\frac{a^2 + b^2}{2} - \\frac{a+b}{2} \\geq \\frac{2n^2}{2} - n = n^2 - n.\n$$\n\nEach obtuse triangle is counted twice as $L$ varies, so the total number of acute triangles is at most\n\n$$\n\\binom{2n+1}{3} - \\frac{(2n+1)(n^2-n)}{2} = \\frac{2n+1}{6}[2n(2n-1) - 3n(n-1)] = \\frac{n(n+1)(2n+1)}{6}.\n$$\n\nThis bound is attainable by taking the sidelines of a regular $(2n+1)$-gon. For $n = 2015$, the answer is $2\\,729\\,148\\,240$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17780,
    "subject": "Mathematics (Olympiad)",
    "question": "A **mid-product number** is a four-digit number $abcd$ such that the two middle digits $b$ and $c$ form a two-digit number equal to the product of the first and last digits: $b c = a \\times d$. All digits must be different and nonzero.\n\nAnswer the following:\n\na. What is the smallest mid-product number?\n\nb. What is the largest mid-product number?\n\nc. Which mid-product numbers have 5 as their second digit?\n\nd. Can a mid-product number end in 5? Explain.",
    "options": [],
    "answer": "See solution",
    "solution": "a. Since $1 \\times d$ is a single digit, the first digit of a mid-product number cannot be 1. If the first digit is 2, then the last digit must be at least 5 to get a two-digit product.\n\nSince 2105 has a zero, it is not a mid-product number. Since 2126 has a repeated digit, it is not a mid-product number. So the smallest mid-product number is $2147$.\n\nb. The first digit of a mid-product number is at most 8. If the first digit is 8, then the last digit must be 9. Hence the middle two-digit number is $72$. So the largest mid-product number is $8729$.\n\nc. Checking all two-digit numbers that start with 5, we see that only $54$ and $56$ have two different one-digit factors. So the only mid-product numbers with second digit 5 are $6549$ and $7568$.\n\nd. If the last digit of a mid-product number was 5, then its two-digit middle number must be a multiple of 5 and therefore end in 0 or 5. Since 0 is forbidden and 5 cannot be repeated, no mid-product number ends in 5.\n\nIt is possible to answer all parts of this problem by first generating the full list of 16 mid-product numbers:\n\n$2147$, $2168$, $2189$, $3124$, $3186$, $3217$, $3248$, $3279$, $4287$, $4328$, $4369$, $6427$, $6549$, $7568$, $7639$, $8729$.\n\nStudents using this approach must explain why this list is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17781,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be the sides of a triangle, and define\n\n$$\nA = \\frac{a^2 + bc}{b+c} + \\frac{b^2 + ca}{c+a} + \\frac{c^2 + ab}{a+b}\n$$\n\nand\n\n$$\nB = \\frac{1}{\\sqrt{(a+b-c)(b+c-a)}} + \\frac{1}{\\sqrt{(b+c-a)(c+a-b)}} + \\frac{1}{\\sqrt{(c+a-b)(a+b-c)}}.\n$$\n\nProve that $AB \\ge 9$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly,\n\n$$\nB \\ge \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a}\n$$\n\nand\n\n$$\nA - (a + b + c) = \\frac{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2}{(a+b)(b+c)(c+a)} \\ge 0.\n$$\n\nTherefore, we have\n\n$$\nAB \\ge \\left( \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{a} \\right) (a + b + c) \\ge 9.\n$$\n\nby the Cauchy-Schwarz inequality. This completes the proof. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17782,
    "subject": "Mathematics (Olympiad)",
    "question": "Placing a 3 at both ends of a number increases it by 3372, resulting in a four-digit number. What is the original number?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the original number is $x$. Placing a 3 at both ends forms the number $3003 + 10x$. This is 3372 more than $x$, so:\n\n$$\n3003 + 10x - x = 3372 \\\\\n9x = 369 \\\\\nx = 41.\n$$\n\nThus, the original number is $41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17783,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x)f(y) + f(y)f(z) + f(z)f(x)) = f(x) + f(y) + f(z)\n$$\nfor all real numbers $x, y, z$.",
    "options": [],
    "answer": "See solution",
    "solution": "$f(x) = 0$ for all $x \\in \\mathbb{R}$ is the only function satisfying (4).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17784,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose six cookies, each with radius greater than $1$, are placed on a plate of radius $2$ such that their centers, denoted $P_j$ for $1 \\leq j \\leq 6$, are arranged in anticlockwise order around the center of the plate $O$. Prove that at least two cookies must overlap.",
    "options": [],
    "answer": "See solution",
    "solution": "Place the cookies upon the plate, denote the centre of the plate by $O$, and the centres of the cookies, in anticlockwise order, by $P_j$, $1 \\leq j \\leq 6$. Since the radii are greater than $1$, the points $P_j$ are at a distance at least $1$ from the edge of the plate, which means $|OP_j| \\leq 2$. Moreover, by the Pigeonhole Principle (or something of that kind), some angle $\\angle P_kOP_{k+1} \\leq 60^\\circ$. This means the points $P_k$ and $P_{k+1}$ are located inside a circle sector of centre $O$, radius $2$, and central angle $60^\\circ$. It is then “evident” that $|P_kP_{k+1}| \\leq 2$, which means the corresponding cookies overlap.\n\n*Proof of “evident” statement, for those who do not believe.* By the Law of Cosines,\n\n$$\n\\begin{aligned}\n|P_k P_{k+1}|^2 &= |OP_k|^2 + |OP_{k+1}|^2 - 2|OP_k||OP_{k+1}| \\cos 60^\\circ \\\\\n&= |OP_k|^2 + |OP_{k+1}|^2 - |OP_k||OP_{k+1}| \\leq 4\n\\end{aligned}\n$$\n\nbecause of the evident inequality\n\n$$\nx^2 + y^2 \\leq 1 + x^2 y^2 \\leq 1 + xy,\n$$\n\nvalid for $0 \\leq x, y \\leq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17785,
    "subject": "Mathematics (Olympiad)",
    "question": "We compute the product of two numbers:\n\n$99\\ldots99 \\times 99\\ldots99,$\n\nwhere the first number consists of 20 nines, and the second of 21 nines.\n\nWhich number do you get if you add up the digits of the outcome of this multiplication?",
    "options": [],
    "answer": "See solution",
    "solution": "81",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17786,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $F$ its Fermat point, that is, the interior point of $ABC$ such that $\\angle AFB = \\angle BFC = \\angle CFA = 120^\\circ$. For each of the triangles $ABF$, $BCF$, and $CAF$, draw its Euler line, that is, the line connecting its circumcenter and its centroid.\n\nProve that these three lines pass through one common point.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we'll prove the following well-known lemma:\n\n**Lemma.** Let $ABP$, $BCM$, and $CAN$ be the equilateral triangles constructed externally to triangle $ABC$. The lines $AM$, $BN$, and $CP$ concur at the Fermat point of $ABC$.\n\n*Proof.* Let $F$ be the intersection point of $BN$ and $CP$. Since triangles $BAN$ and $PAC$ are congruent, $\\angle APE = \\angle ABF$. So the quadrilateral $APBF$ is cyclic, and thus $\\angle AEP = \\angle ABP = 60^\\circ$ and $\\angle BFP = \\angle BAP = 60^\\circ$. Hence $\\angle AFB = \\angle BFC = 120^\\circ$, which proves that $F$ is the Fermat point of triangle $ABC$. Analogously, $AM$ and $BN$ pass through the same point $F$, and the lemma is proved.\n\nNow, let $O_A$, $O_B$, and $O_C$ be the centers of these equilateral triangles (and also the circumcenters of triangles $BCF$, $ACF$, and $ABF$, respectively); $G_A$, $G_B$, and $G_C$ are the centroids of $BCF$, $ACF$, and $ABF$, respectively. It is immediate that the Euler lines $O_A G_A$, $O_B G_B$, and $O_C G_C$ are parallel to $AF$, $BF$, and $CF$, respectively.\n\nNow consider the homothety with center $F$ and ratio $3/2$, which transforms $G_A$, $G_B$, and $G_C$ into the midpoints of $BC$, $CA$, and $AB$, respectively. This homothety also transforms the Euler lines of $BCF$, $ACF$, and $ABF$ into the lines $l_A$, $l_B$, and $l_C$, parallel to $AF$, $BF$, and $CF$ and passing through the midpoints of $BC$, $CA$, and $AB$, respectively. Hence these Euler lines are concurrent if and only if $l_A$, $l_B$, and $l_C$ are.\n\nFinally, consider the homothety with center $G$ (the centroid of $ABC$) and ratio $-2$. The midpoints of $BC$, $CA$, and $AB$ are transformed into the vertices $A$, $B$, and $C$, respectively, so $l_A$, $l_B$, and $l_C$ are transformed into $AF$, $BF$, and $CF$, respectively. Since the latter are concurrent at $F$, it follows that $l_A$, $l_B$, and $l_C$ are too, and we are done.\n\n![](images/Brazil_2006_p2_data_cdb3bb6621.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17787,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $25 \\times 25$ chessboard with cells $C(i, j)$ for $1 \\leq i, j \\leq 25$. Find the smallest possible number $n$ of colors with which these cells can be colored, subject to the following condition: For $1 \\leq i < j \\leq 25$ and for $1 \\leq s < t \\leq 25$, the three cells $C(i,s)$, $C(j,s)$, $C(j,t)$ carry at least two different colors.",
    "options": [],
    "answer": "See solution",
    "solution": "The forbidden configuration is illustrated below:\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p31_data_1e20346e85.png)\n\nFor a $3 \\times 3$ chessboard, the minimum number is 2. For example:\n\n| 1 | 1 | 2 |\n|---|---|---|\n| 1 | 2 | 2 |\n| 2 | 2 | 1 |\n\nFor a $5 \\times 5$ chessboard, 3 colors suffice:\n\n| 1 | 1 | 2 | 2 | 3 |\n|---|---|---|---|---|\n| 1 | 2 | 2 | 3 | 3 |\n| 2 | 2 | 3 | 3 | 1 |\n| 2 | 3 | 3 | 1 | 1 |\n| 3 | 3 | 1 | 1 | 2 |\n\nIt appears that $m_n = \\frac{n+1}{2}$ colors suffice for an $n \\times n$ chessboard when $n$ is odd. Thus, for the $25 \\times 25$ chessboard, 13 colors are sufficient. Consider the colors $\\{0,1,2,\\ldots,12\\}$ and color the chessboard so that\n\n$$\nC[i,j] = \\left[ \\frac{i+j}{2} \\right] \\pmod{13}\n$$\n\nfor $1 \\leq i, j \\leq 25$.\n\nSuppose the condition fails, i.e., $C[i,s] = C[j,s] = C[j,t]$ for some $1 \\leq i < j \\leq 25$ and $1 \\leq s < t \\leq 25$. Then\n\n$$\n\\left[ \\frac{i+s}{2} \\right] \\equiv \\left[ \\frac{j+s}{2} \\right] \\equiv \\left[ \\frac{j+t}{2} \\right] \\pmod{13}.\n$$\n\nFrom $C[i,s] = C[j,s]$, $\\left[ \\frac{i+s}{2} \\right] \\equiv \\left[ \\frac{j+s}{2} \\right]$, so $i = j$, a contradiction. Similarly, $C[j,s] = C[j,t]$ implies $s = t$, also a contradiction. Thus, the coloring works.\n\nNow, we show that 13 colors are necessary. Fix any color, say color 2. Let $c_2$ be the number of cells colored 2. Remove all other colors, and draw horizontal arrows (left to right) and vertical arrows (down to up) between consecutive 2-cells. These are called 2-arrows.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p32_data_b437d99382.png)\n\nNo 2-cell can have two or more outgoing 2-arrows, or else the forbidden configuration would occur:\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p32_data_b014d2e11f.png)\n\nThus, the total number of 2-arrows $a_2$ satisfies $c_2 \\geq a_2$. In any row with $k$ 2-cells, there are $k-1$ horizontal 2-arrows, so the total number of horizontal 2-arrows is $c_2 - 25$ (since there are 25 rows). Similarly, the total number of vertical 2-arrows is $c_2 - 25$. Thus, $a_2 = 2(c_2 - 25)$, so $c_2 \\geq 2c_2 - 50$, which gives $c_2 \\leq 50$.\n\nSince there are $25 \\times 25 = 625$ cells and $625/50 > 12$, at least 13 colors are needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17788,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_{ij}$ ($i, j \\in \\{1, 2, \\dots, n\\}$) be the cells of an $n \\times n$ grid. When $n = 2k$, $k \\in \\mathbb{N}^*$, we mark each $A_{ij}$ satisfying $i + j \\equiv 0 \\pmod{2}$ with color red (presented by shaded areas), and those satisfying $j - i \\equiv 3 \\pmod{4}$ and $j - i \\not\\equiv j + i \\pmod{4}$ with blue (presented by oblique line areas) (see Fig. 7.1).\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p261_data_062c4a8530.png)\n\nIn this way, every cell adjacent to a blue one is red, and there is exactly one blue cell around each red one.\n\nNow, we perform an operation on each blue cell: the number in each red cell is changed from $+1$ to $-1$, while the numbers in the remaining cells are unchanged.\n\nSince $n$ is even, we can rotate the grid around its center $O$ anticlockwise by $90^\\circ$. Then all the red cells of the rotated grid cover exactly all the cells that are not red in the original grid (see Fig. 7.2).\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p261_data_02882ed5f2.png)\n\nWe perform the operations again for the original grid on all the cells that are covered by blue cells of the rotated grid. Then all the numbers with value $+1$ in the remaining cells of the original grid are changed to $-1$, while the numbers in the other cells are unchanged.\n\nTherefore, when $n$ is even, all the numbers in the cells of the grid can be changed to $-1$ by a finite number of operations.\n\nWhen $n$ is odd, denote the number in cell $A_{ii}$ as $M_i$ ($i = 1, 2, \\dots, n$), and denote the number of operations on each of their adjacent cells as $x_1, x_2, \\dots, x_{n-1}, y_1, y_2, \\dots, y_{n-1}$, respectively (see Fig. 7.3). Is it possible to change all numbers in the grid from $+1$ to $-1$ by a finite number of such operations?",
    "options": [],
    "answer": "See solution",
    "solution": "After a finite number of operations, it is easy to see that:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p262_data_b2435ec63e.png)\n\n$M_1$ is changed from $+1$ to $-1$ if and only if $x_1 + y_1$ is odd;\n\n$M_2$ is changed from $+1$ to $-1$ if and only if $x_1 + y_1 + x_2 + y_2$ is odd;\n\n$M_3$ is changed from $+1$ to $-1$ if and only if $x_2 + y_2 + x_3 + y_3$ is odd;\n\n$$\n\\vdots\n$$\n\n$M_{n-1}$ is changed from $+1$ to $-1$ if and only if $x_{n-2} + y_{n-2} + x_{n-1} + y_{n-1}$ is odd, and\n\n$M_n$ is changed from $+1$ to $-1$ if and only if $x_{n-1} + y_{n-1}$ is odd.\n\nSince $n$ is odd, the sum of $n$ odd numbers is still odd. Then,\n\n$$\n\\begin{align*}\n& (x_1 + y_1) + (x_1 + y_1 + x_2 + y_2) + (x_2 + y_2 + x_3 + y_3) \\\\\n& \\quad + \\dots + (x_{n-2} + y_{n-2} + x_{n-1} + y_{n-1}) + (x_{n-1} + y_{n-1}) \\\\\n& = 2(x_1 + x_2 + \\dots + x_{n-1} + y_1 + y_2 + \\dots + y_{n-1})\n\\end{align*}\n$$\n\nis odd. This is impossible!\n\nTherefore, all the numbers in the cells of a given $n \\times n$ grid can be changed from $+1$ to $-1$ after a finite number of operations if and only if $n$ is an even number. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17789,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 2$ be a positive integer. A deck contains $\\frac{n(n-1)}{2}$ cards, numbered\n$$\n-1, 2, 3, \\dots, \\frac{n(n-1)}{2}.\n$$\nTwo cards form a *magic pair* if their numbers are consecutive, or if their numbers are $1$ and $\\frac{n(n-1)}{2}$.\n\nFor which $n$ is it possible to distribute the cards into $n$ stacks in such a manner that, among the cards in any two stacks, there is exactly one magic pair?",
    "options": [],
    "answer": "See solution",
    "solution": "For all odd $n$.\n\nFirst, assume a stack contains two cards that form a magic pair; say cards numbered $i$ and $i+1$. Among the cards in this stack and the stack with card number $i+2$ (they might be identical), there are two magic pairs—a contradiction. Hence, no stack contains a magic pair.\n\nEach card forms a magic pair with exactly two other cards. Hence, if $n$ is even, each stack must contain at least $\\left\\lfloor \\frac{n-1}{2} \\right\\rfloor = \\frac{n}{2}$ cards, since there are $n-1$ other stacks. But then we need at least $n \\cdot \\frac{n}{2} > \\frac{n(n-1)}{2}$ cards—a contradiction.\n\nIn the odd case, we distribute the cards as follows: Let $a_1, a_2, \\dots, a_n$ be the $n$ stacks and let $n = 2m+1$. Card number $1$ is put into stack $a_1$. If card number $km + i$, for $i = 1, 2, \\dots, m$, is put into stack $a_j$, then card number $km + i + 1$ is put into stack $a_{j+i}$, where the indices are calculated modulo $n$.\n\nThere are\n$$\n\\frac{n(n-1)}{2} = \\frac{(2m+1)(2m)}{2} = m(2m+1)\n$$\ncards. If we look at all the card numbers of the form $km + 1$, there are exactly $n = 2m+1$ of these, and we claim that there is exactly one in each stack. Card number $1$ is in stack $a_1$, and card number $km + 1$ is in stack\n$$\na_{1+k(1+2+3+\\cdots+m)}\n$$\nSince\n$$\n1 + 2 + 3 + \\cdots + m = \\frac{m(m+1)}{2}\n$$\nand $\\gcd(2m+1, \\frac{m(m+1)}{2}) = 1$, all the indices\n$$\n1 + k(1 + 2 + 3 + \\cdots + m), \\quad k = 0, 1, 2, \\dots, 2m\n$$\nare different modulo $n = 2m+1$. In the same way, we see that each stack contains exactly one of the $2m+1$ cards with the numbers $km + i$ for a given $i = 2, 3, \\dots, m$.\n\nNow look at two different stacks $a_v$ and $a_u$. Then, without loss of generality, we may assume that $u = v + i$ for some $i = 1, 2, \\dots, m$ (again, indices modulo $n = 2m + 1$). Since there is a card in stack $a_v$ with number $km + i$, the card $km + i + 1$ is in stack $a_{v+i} = a_u$. Hence, among the cards in any two stacks, there is at least one magic pair. Since there is the same number of pairs of stacks as of magic pairs, there must be exactly one magic pair among the cards of any two stacks. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17790,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $t$ be a real number such that $0 < t < \\frac{1}{2}$.\n\nWe wish to construct an infinite set $S$ of positive integers such that $|s_j - ms_i| > ts_i$ for every pair of different elements $s_i$ and $s_j$ of $S$ and for every positive integer $m$. In other words, no element of $S$ (different from $s_i$) is within $ts_i$ of any positive multiple of $s_i$. This must hold for all infinitely many $s_i$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2018-Australian-Scene-W2_p156_data_0176e0303f.png)\n\nIt's easier to name a few numbers which are *not* excluded by $s_1$. The numbers $(m+1/2)s_1$, where $m$ is a positive integer, are not excluded, though these are not integers. However, near $(m+1/2)s_1$, we have integers $x = (m+1/2)s_1 \\pm 1/2$. Call these half-multiples of $s_1$ and note that one of each pair must be odd. These half-multiples are not excluded because $((m+1/2)s_1 - 1/2) - ms_1 = (s_1 - 1)/2 > ts_1$ and $(m+1)s_1 - ((m+1/2)s_1 - 1/2) = (s_1 - 1)/2 > ts_1$, and these half-multiples are even further from other multiples of $s_1$.\n\n![](images/2018-Australian-Scene-W2_p156_data_5376f4e2a3.png)\n\nWe now choose the next smallest element of $S$, $s_2$, to be an odd half-multiple of $s_1$. We know that $s_2$ is not excluded by $s_1$ and claim that if $s_2 > 2s_1$, we also have that $s_1$ is not excluded by $s_2$ (that is, not within $ts_2$ of every positive multiple of $s_2$). Indeed, since $m \\ge 1$ and $t < 1/2$, we have that $m - t > 1/2$. And if $s_2 > 2s_1$ then this gives $s_2(m - t) > s_1 \\Rightarrow ms_2 - s_1 > ts_2 \\Rightarrow |s_1 - ms_2| > ts_2$.\n\nSo in particular, we will choose $s_2$ to be whichever of the half-multiples $(5/2)s_1 \\pm 1/2$ is odd and note that since $s_1 > 1$, $s_2 = (5/2)s_1 \\pm 1/2 = 2s_1 + (s_1 \\pm 1)/2 > 2s_1$.\n\nNow this excludes even more numbers from $S$. Again we can be sure that the half-multiples of $s_2$ are not excluded on account of being too close to a multiple of $s_2$. And so we can be sure that any number which is a half-multiple of both $s_1$ and $s_2$ is still a possibility for inclusion in $S$.\n\nNow let's prove that if $a$ and $b$ are odd integers, then the half-multiples of $ab$ are also half-multiples of both $a$ and $b$. Suppose $h = (k+1/2)ab \\pm 1/2$ is a half-multiple of $ab$. Then $h = (\\ell + 1/2)a \\pm 1/2$ where $\\ell = kb + (b-1)/2$ is an integer. Therefore $h$ is a half-multiple of $a$. Similarly it is a half-multiple of $b$.\n\nWe now choose infinitely many more elements in $S$ according to this rule: $s_{k+1}$ is an odd half-multiple of $s_1s_2 \\cdots s_k$ with $s_{k+1} > 2s_i$ for all $1 \\le i \\le k$. In this way, $s_{k+1}$ is an odd half-multiple of each of $s_1, s_2, \\dots, s_k$ and so not excluded by any of these elements. Furthermore, $s_{k+1} > 2s_i \\Rightarrow s_{k+1}(m-t) > s_i \\Rightarrow ms_{k+1} - s_i > ts_{k+1} \\Rightarrow |s_i - ms_{k+1}| > ts_{k+1}$ and so for all $1 \\le i \\le k$, $s_i$ is not excluded by $s_{k+1}$.\n\nSpecifically, we choose $s_{k+1}$ to be an odd integer given by\n\n$$\ns_{k+1} = 5s_1s_2 \\cdots s_k/2 \\pm 1/2.\n$$\n\nNote that since each of $s_1, s_2, \\dots, s_k$ is greater than 1, $s_{k+1} > 5s_i/2 \\pm 1/2 = 2s_i + (s_i \\pm 1)/2 > 2s_i$ for all $1 \\le i \\le k$. Also note that the $s_i$ sequence is increasing.\n\nTherefore, we have found such an infinite set $S$ of integers $s_1 < s_2 < s_3 < \\dots$, each chosen to be odd and given by $s_1 > 1/(1-2t)$ and $s_{k+1} = 5s_1s_2 \\cdots s_k/2 \\pm 1/2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17791,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a graph with $n$ vertices and $m$ edges. \n\n(a) Show that $G$ contains a nonempty set $S$ of vertices such that every vertex in $S$ has degree at least $\\left\\lfloor \\frac{m}{n} \\right\\rfloor$.\n\n(b) Suppose $G$ contains no cycles of length at most $2k$ for some integer $k \\ge 1$. Let $v$ be a vertex in $S$. For $0 \\leq j \\leq k$, let $T_j$ be the set of vertices in $S$ at distance $j$ from $v$. Prove that $|T_k| \\geq \\left( \\frac{m}{n} - 1 \\right)^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Starting from the graph $G$, repeatedly remove any vertex with degree less than $\\frac{m}{n}$ and all its incident edges. Each removal deletes fewer than $\\frac{m}{n}$ edges, so we cannot remove all $n$ vertices, as this would remove at most $n \\cdot \\frac{m}{n} = m$ edges, but the graph started with $m$ edges. Thus, a nonempty set $S$ remains in which every vertex has degree at least $\\left\\lfloor \\frac{m}{n} \\right\\rfloor$.\n\n(b) For $0 \\leq j \\leq k$, let $T_j$ be the set of vertices in $S$ at distance $j$ from $v$. The neighbors of a vertex in $T_j$ can only be in $T_{j-1}$, $T_j$, or $T_{j+1}$. For $1 \\leq j \\leq k$, each vertex in $T_j$ must be adjacent to a unique vertex in $T_{j-1}$ (otherwise, a cycle of length at most $2k$ would exist). For $1 \\leq j \\leq k-1$, no two vertices in $T_j$ are adjacent (otherwise, a shorter cycle would exist). Therefore, each vertex in $T_j$ is adjacent to at least $\\frac{m}{n} - 1$ vertices in $T_{j+1}$. By induction, $|T_j| \\geq \\left( \\frac{m}{n} - 1 \\right)^j$ for $0 \\leq j \\leq k$. In particular, $|T_k| \\geq \\left( \\frac{m}{n} - 1 \\right)^k$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17792,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all $x, y > 0$, the inequality\n$$\n\\frac{x^2 - x y + y^2}{x^2 + x y + y^2} \\geq \\frac{1}{3}\n$$\nholds.\n\nConsequently, show that for all positive real numbers $a, b, c$,\n$$\n\\frac{a^2(a^3 + b^3)}{a^2 + a b + b^2} + \\frac{b^2(b^3 + c^3)}{b^2 + b c + c^2} + \\frac{c^2(c^3 + a^3)}{c^2 + c a + a^2} \\geq 2 a b c\n$$\nand determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the inequality for $x, y > 0$:\n\n$$(*) \\iff 3(x^2 - x y + y^2) \\geq x^2 + x y + y^2 \\iff 2(x^2 + y^2) - 4 x y \\geq 0 \\iff 2(x - y)^2 \\geq 0,$$\nwhich is always true. Thus, $(*)$ holds.\n\nNow, for the main inequality:\n\n$$\n\\frac{a^2(a^3 + b^3)}{a^2 + a b + b^2} = a^2(a + b) \\frac{a^2 - a b + b^2}{a^2 + a b + b^2} \\geq \\frac{1}{3} a^2(a + b)\n$$\nby $(*)$.\n\nSimilarly,\n$$\n\\frac{b^2(b^3 + c^3)}{b^2 + b c + c^2} \\geq \\frac{1}{3} b^2(b + c), \\quad \\frac{c^2(c^3 + a^3)}{c^2 + c a + a^2} \\geq \\frac{1}{3} c^2(c + a).\n$$\nAdding these,\n$$\n\\frac{a^2(a^3 + b^3)}{a^2 + a b + b^2} + \\frac{b^2(b^3 + c^3)}{b^2 + b c + c^2} + \\frac{c^2(c^3 + a^3)}{c^2 + c a + a^2} \\geq \\frac{1}{3}(a^3 + b^3 + c^3) + \\frac{1}{3}(a^2 b + b^2 c + c^2 a).\n$$\nBy the AM-GM inequality,\n$$\n\\frac{1}{3}(a^3 + b^3 + c^3) \\geq \\sqrt[3]{a^3 b^3 c^3} = a b c,\n$$\n$$\n\\frac{1}{3}(a^2 b + b^2 c + c^2 a) \\geq \\sqrt[3]{a^2 b \\cdot b^2 c \\cdot c^2 a} = a b c.\n$$\nThus, the sum is at least $2 a b c$.\n\nEquality holds when $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17793,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadratic polynomial $p(x)$ with real coefficients and leading coefficient $1$ is called *disrespectful* if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\\tilde{p}(1)$?\n\n(A) $\\frac{5}{16}$   (B) $\\frac{1}{2}$   (C) $\\frac{5}{8}$   (D) $1$   (E) $\\frac{9}{8}$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $p(x) = (x - r)(x - s)$. Observe that $p(x)$ must have two real roots in order for $p(p(x))$ to have any roots at all. More specifically, if $y$ is a root of $p(p(x))$, then $p(y) = r$ or $p(y) = s$. That is, the equations\n\n$$\n(x - r)(x - s) - r = 0 \\quad \\text{and} \\quad (x - r)(x - s) - s = 0\n$$\n\ntogether must have exactly three real roots among them. It follows that one of these two quadratics, say $(x - r)(x - s) - r$, must have discriminant zero.\n\nExpansion yields $x^2 - (r+s)x + r(s-1) = 0$, so the discriminant $\\Delta$ of this quadratic must satisfy\n\n$$\n0 = \\Delta = (r+s)^2 - 4r(s-1) = (r-s)^2 + 4r.\n$$\n\nThis implies that $r$ is negative, say $r = -r_0$, and that $s = r \\pm \\sqrt{-4r} = -r_0 \\pm 2\\sqrt{r_0}$. It follows that\n\n$$\nr + s = 2(-r_0 \\pm \\sqrt{r_0}) \\le 2(-r_0 + \\sqrt{r_0}) \\le 2 \\cdot \\frac{1}{4} = \\frac{1}{2},\n$$\n\nwhere the second inequality follows from the fact that $a - a^2 \\le \\frac{1}{4}$ for all real numbers $a$. Thus $r = -\\frac{1}{4}$ and $s = \\frac{3}{4}$, which works. In turn, $\\tilde{p}(x) = (x + \\frac{1}{4})(x - \\frac{3}{4})$ and $\\tilde{p}(1) = \\frac{5}{4} \\cdot \\frac{1}{4} = \\frac{5}{16}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17794,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $x_n$ is defined as follows:\n\n$$\nx_0 = 2, \\quad x_1 = 1, \\quad x_{n+2} = x_{n+1} + x_n\n$$\n\nfor every non-negative integer $n$.\n\n**(a)** For every $n \\ge 1$, prove that if $x_n$ is a prime number then $n$ is a prime number or $n$ has no odd prime divisors.\n\n**(b)** Find all pairs of non-negative integers $(m, n)$ such that $x_m \\mid x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** We can show that $x_n = \\alpha^n + \\beta^n$ for all positive integers $n$, where $\\alpha < 0 < \\beta$ are the roots of $\\lambda^2 - \\lambda - 1 = 0$.\n\nSuppose $x_n$ is prime and $n$ has an odd prime divisor. Then $n = pq$ for some odd prime $p$ and $q > 1$. We have:\n\n$$\n\\begin{aligned}\nx_{pq} &= \\alpha^{pq} + \\beta^{pq} \\\\\n&= (\\alpha^q + \\beta^q) \\left( \\alpha^{q(p-1)} - \\alpha^{q(p-2)} \\beta^q + \\dots - \\alpha^q \\beta^{q(p-2)} + \\beta^{q(p-1)} \\right) \\\\\n&= x_q \\left( x_{q(p-1)} + \\dots + (-1)^{\\frac{(q+1)(p-1)}{2}} x_{2q} + (-1)^{\\frac{(q+1)(p-1)}{2}} \\right)\n\\end{aligned}\n$$\n\nso $x_q \\mid x_{pq}$. Since $(x_n)$ is strictly increasing, $x_q > x_1 = 1$, so $x_{pq}$ is composite—a contradiction. Thus, if $x_n$ is prime, then $n$ is prime or $n$ has no odd prime divisors.\n\n**(b)** Consider cases:\n\n- *Case 1*: $m = 0$. By considering $x_n$ modulo $2$, $x_n$ is even for all $3 \\mid n$ and odd otherwise. Thus, all pairs $(0, 3k)$ for $k$ positive integer are solutions.\n\n- *Case 2*: $m = 1$. Clearly, all pairs $(1, k)$ for $k$ positive integer are solutions.\n\n- *Case 3*: $m > 1$. For $k \\ge l \\ge 0$:\n\n$$\n(\\alpha^k + \\beta^k)(\\alpha^l + \\beta^l) - (\\alpha^{k+l} + \\beta^{k+l}) = (\\alpha\\beta)^l(\\alpha^{k-l} + \\beta^{k-l}) = (-1)^l(\\alpha^{k-l} + \\beta^{k-l})\n$$\n\nSo,\n\n$$\nx_{k+l} = x_k x_l - (-1)^l x_{k-l} \\quad (1)\n$$\n\nFor $k \\ge 2l \\ge 0$:\n\n$$\nx_k = x_{k-l} x_l - (-1)^l x_{k-2l}\n$$\n\nThus, $x_k$ is divisible by $x_l$ iff $x_{k-2l}$ is divisible by $x_l$, and so on. In general, $x_k$ is divisible by $x_l$ iff $x_{k-2tl}$ is divisible by $x_l$ for $k \\ge 2tl$, $t \\in \\mathbb{N}$.\n\nNow, since $x_n$ is divisible by $x_m$ and $x_n \\ge x_m \\ge 3$, $n \\ge m > 1$. Set $n = qm + r$ with $q \\in \\mathbb{N}^*$, $r \\in \\mathbb{N}$, $0 \\le r \\le m-1$.\n\n- *Case 3.1*: $q$ even. $x_m \\mid x_n$ iff $x_m \\mid x_r$, which implies $x_r \\ge x_m$. If $r \\ge 1$, then $r \\ge m$ (since $(x_n)$ is strictly increasing), contradiction. If $r = 0$, $x_m \\ge x_2 = 3 > x_0 = x_r$, contradiction.\n\n- *Case 3.2*: $q$ odd. $x_m \\mid x_n$ iff $x_m \\mid x_{m+r}$. But $x_{m+r} = x_m x_r - (-1)^r x_{m-r}$, so $x_m \\mid x_{m-r}$. If $0 < r < m$, $1 \\le m-r < m$, contradiction. Thus, $r = 0$ and all pairs are $(m, (2k+1)m)$ for $m > 1$, $k$ positive integer.\n\nTherefore, all solutions are $(0, 3k)$, $(1, k)$, and $(m, (2k+1)m)$ for $m, k$ positive integers and $m > 1$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17795,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute-angled triangle $ABP$ ($AB > BP$), the altitudes are $BH$, $PQ$, and $AS$. The extension of $QS$ intersects line $AP$ at $C$. The extension of $HS$ intersects $BC$ at $L$. If $HS = SL$ and $HL$ is perpendicular to $BC$, compute $\\frac{SL}{SC}$.\n\n![](images/Argentina_2011_p42_data_29f77671d9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $SH = SL$, we compute the ratio $\\frac{SH}{SC}$. Note that $AB > BP$ implies that $P$ is between $A$ and $C$. Denote $\\angle BAP = \\alpha$ and observe that $\\angle PSC = \\angle PSH = \\angle BSL = \\angle BSQ = \\alpha$. Indeed, we have $\\angle HSB = \\angle QSP = 180^\\circ - \\alpha$ from the cyclic quadrilaterals $AHSB$ and $APSQ$. On the other hand, each of the four angles in the above equality completes $\\angle HSB$ or $\\angle QSP$ to $180^\\circ$. In particular, $\\angle PSC = \\angle PSH$ means that $SP$ is the internal bisector of $\\angle CSH$. Since $AS \\perp SP$, it follows that $SA$ is the external bisector of $\\angle CSH$. Hence $\\frac{SH}{SC} = \\frac{AH}{AC}$ by the external angle theorem.\n\nTo find $\\frac{AH}{AC}$, consider the midpoint $M$ of $HC$. The right triangles $BHL$ and $BCH$ are similar as they share an acute angle at vertex $B$. Since $BS$ and $BM$ are respective medians, $\\angle BMH = \\angle BSL$. We proved above that $\\angle BSL = \\alpha$, hence $\\angle BMH = \\alpha = \\angle BAH$. Therefore, triangle $ABM$ is isosceles with base $AM$. Its altitude $BH$ is also a median, so $AH = HM$. In addition, $HM = MC$, and we obtain $AH = \\frac{1}{3}AC$. In conclusion, $\\frac{SL}{SC} = \\frac{SH}{SC} = \\frac{AH}{AC} = \\frac{1}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17796,
    "subject": "Mathematics (Olympiad)",
    "question": "When dividing four consecutive positive integers by a three-digit integer, the sum of the four remainders is 983. Find the remainder when the smallest of these four numbers is divided by 109.",
    "options": [],
    "answer": "See solution",
    "solution": "108.\n\nLet the four consecutive integers be $n$, $n+1$, $n+2$, and $n+3$. Let the three-digit divisor be $b$. Write $n = bq + r$ for some integer $q$ and remainder $r$.\n\nConsider possible values for $r$:\n\n- If $r \\leq b - 4$, the remainders are $r$, $r+1$, $r+2$, $r+3$, so their sum is $4r + 6 = 983$, which gives $4r = 977$, not possible since $r$ must be an integer.\n- If $r = b - 3$, the remainders are $r$, $r+1$, $r+2$, $0$, so their sum is $3r + 3 = 983$, giving $3r = 980$, not possible.\n- If $r = b - 2$, the remainders are $r$, $r+1$, $0$, $1$, so their sum is $2r + 2 = 983$, giving $2r = 981$, not possible.\n- If $r = b - 1$, the remainders are $r$, $0$, $1$, $2$, so their sum is $r + 3 = 983$, giving $r = 980$ and $b = r + 1 = 981$.\n\nNow, express $n$ in terms of $b$ and $r$:\n\n$$\n n = bq + r = 981q + 980 = 109(9q + 8) + 108.\n$$\n\nThus, the remainder when the smallest number is divided by 109 is $\\boxed{108}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17797,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nx f(2f(y) - x) + y^2 f(2x - f(y)) = \\frac{f(x)^2}{x} + f(y f(y))\n$$\n\nfor all $x, y \\in \\mathbb{Z}$ with $x \\ne 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be a solution of the problem. Let $p$ be a prime larger than $|f(0)|$. Since $p$ divides $f(p)^2$, $p$ divides $f(p)$, and so $p$ divides $\\frac{f(p)^2}{p}$. Taking $y = 0$ and $x = p$, we deduce that $p$ divides $f(0)$. Since $p > |f(0)|$, we must have $f(0) = 0$.\n\nNext, set $y = 0$ to obtain $x f(-x) = \\frac{f(x)^2}{x}$. Replacing $x$ by $-x$, and combining the two relations yields $f(x) = 0$ or $f(x) = x^2$ for all $x$.\n\nSuppose now that there exists $x_0 \\neq 0$ such that $f(x_0) = 0$. Taking $y = x_0$, we obtain $x f(-x) + x_0^2 f(2x) = \\frac{f(x)^2}{x}$, yielding $x_0^2 f(2x) = 0$ for all $x$. Thus, $f$ is $0$ on all even numbers. Assume that there exists an odd number $y_0$ such that $f(y_0) \\neq 0$. Then $f(y_0) = y_0^2$. Taking $y = y_0$, we obtain\n\n$$\nx f(2y_0^2 - x) + y_0^2 f(2x - y_0^2) = \\frac{f(x)^2}{x} + f(y_0^3).\n$$\n\nChoosing $x$ to be a nonzero even number, we deduce that $y_0^2 f(2x - y_0^2) = f(y_0^3)$. If $f(y_0^3) \\neq 0$, then we must also have $f(2x - y_0^2) \\neq 0$, whence $f(2x - y_0^2) = (2x - y_0^2)^2$. But this would imply that $y_0^2 (2x - y_0^2)^2 = f(y_0^3)$ for all nonzero even $x$, which is impossible. Thus, $f(2x - y_0^2) = 0$ holds for all nonzero even numbers $x$. Since $y_0$ is odd, we have $y_0^2 \\equiv 1 \\pmod{4}$, so $f$ vanishes on all numbers of the form $4k + 3$, except possibly $-y_0^2$.\n\nSince $x^2 f(-x) = f(x)^2$, we see that $f$ also vanishes on all numbers of the form $4k+1$, except possibly $y_0^2$. Thus, $f$ vanishes on all odd numbers, except possibly both $y_0^2$ and $-y_0^2$. Since $f(y_0) \\neq 0$, we must have $y_0 = y_0^2$ or $y_0 = -y_0^2$, so $y_0 = 1$ or $y_0 = -1$. It follows that $f(1) = f(-1) = 1$, since $f(1)$ and $f(-1)$ must either equal $0$ or $(\\pm 1)^2 = 1$, and both $f(y_0^2)$ and $f(-y_0^2)$ are nonzero, that is, both $f(1)$ and $f(-1)$ are nonzero.\n\nHowever, setting $x = 2$ and $y = 1$ yields $2 f(2 f(1) - 2) + f(4 - f(1)) = \\frac{f(2)^2}{2} + f(f(1))$. Since $f(2) = 0$ and $f(1) = 1$, this gives $2 f(0) + f(3) = 0 + 1$. But $f(3) = 0$, so we have $0 = 1$, a contradiction. Thus, if $f$ vanishes at a nonzero value, there cannot exist a nonzero $y_0$ at which $f$ does not vanish, so $f$ must be $0$ everywhere. On the other hand, if $f$ only vanishes at $0$, then we must have $f(x) = x^2$ for all $x$.\n\nIn conclusion, the two solutions are $f(x) = 0$ for all $x \\in \\mathbb{Z}$ and $f(x) = x^2$ for all $x \\in \\mathbb{Z}$. The first function clearly satisfies the given relation. The second does as well, because of the Sophie Germain identity\n\n$$\nx (2y^2 - x)^2 + y^2 (2x - y^2)^2 = x^3 + y^6,\n$$\n\nwhich holds for all $x, y \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17798,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram such that the projections $K$, $L$ of $D$ onto the sides $AB$, $BC$, respectively, are their interior points. Prove that $KL \\parallel AC$ if and only if\n$$\n\\angle BCA + \\angle ABD = \\angle BDA + \\angle ACD.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Alternate angles $ABD$ and $CDB$ are equal, hence $\\angle BCA + \\angle ABD + \\angle BDA + \\angle ACD = 180^\\circ$. The equality $\\angle BCA + \\angle ABD = \\angle BDA + \\angle ACD$ thus holds if and only if\n$$\n\\angle BCA + \\angle ABD = 90^\\circ.\n$$\n\nPoints $K$ and $L$ lie on a circle with diameter $BD$. Hence the inscribed angles $BDK$ and $BLK$ are equal and (due to equal alternate angles $ABD$ and $CDB$)\n$$\n\\angle BLK + \\angle ABD = \\angle BDK + \\angle CDB = 90^\\circ.\n$$\n\nLines $KL$ and $AC$ are parallel if and only if $\\angle BLK = \\angle BCA$, which is by the last equality equivalent to the previous condition. The equivalence is thus proven.\n\n![](images/Czech_and_Slovak_booklet_2013_p11_data_fe9df6400b.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17799,
    "subject": "Mathematics (Olympiad)",
    "question": "A square is cut into several rectangles, none of which is a square, so that the sides of each rectangle are parallel to the sides of the square. For each rectangle with sides $a, b$, where $a < b$, compute the ratio $a/b$. Prove that the sum of these ratios is at least $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume the square has area $1$. Let the sides of the rectangles be $a_i, b_i$ with $a_i \\leq b_i$, and let $S_i = a_i b_i$. We have $\\sum S_i = 1$ and\n$$\n\\sum \\frac{a_i}{b_i} = \\sum \\frac{S_i}{b_i^2} \\geq \\sum S_i = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17800,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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": 17801,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circular chain, there are 2016 beads arranged in a circle, each bead being either black, blue, or green. In each step, every bead is simultaneously replaced by a new bead, with the color of the new bead determined as follows:\n\n- If both neighboring beads originally had the same color, the new bead takes that color.\n- If the neighbors had two different colors, the new bead takes the third color.\n\n(a) Is there a chain where half the beads are black and the other half green, from which, by applying these steps, a chain of all blue beads can be obtained?\n\n(b) Is there a chain where 1000 beads are black and the rest are green, from which, by applying these steps, a chain of all blue beads can be obtained?\n\n(c) Is it possible, starting from a chain with exactly two adjacent black beads and all other beads blue, to reach a chain with exactly one green bead and all other beads blue by applying these steps?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Da 2016 durch 4 teilbar ist, kann man abwechselnd zwei schwarze und zwei grüne Perlen nehmen. Im ersten Schritt werden dann bereits alle durch blaue Perlen ersetzt.\n\n(b) Wenn wir der Farbe Blau die Zahl 0 zuordnen, der Farbe Grün die Zahl 1 und der Farbe Schwarz die Zahl 2, gilt in jedem Schritt, dass die neue Farbe einer Perle modulo 3 gleich der negativen Summe ihrer beiden alten Nachbarn ist. Die neue Gesamtsumme aller Perlenfarben modulo 3 kann man also berechnen, indem man die alte Gesamtsumme aller Perlenfarben mit zwei multipliziert (da jede alte Perle zu zwei neuen beiträgt) und das Vorzeichen umkehrt. Modulo 3 ist eine Multiplikation mit $-2$ aber gleich einer Multiplikation mit 1, also bleibt die Gesamtsumme modulo 3 immer gleich.\n\nFür lauter blaue Perlen ist die Gesamtsumme 0. Für 1000 schwarze und 1016 grüne Perlen ist sie aber $2000 + 1016 \\equiv 1 \\pmod{3}$. Es ist daher für keine Anordnung von 1000 schwarzen und 1016 grünen Perlen möglich, sie mit solchen Schritten in eine Kette aus lauter blauen Perlen zu verwandeln.\n\n(c) Mit der obigen Zuordnung von Resten modulo 3 wird in jedem Schritt die Summe modulo 3 aller Perlenfarben in ungerader Position zur Summe modulo 3 der Perlenfarben in gerader Position und umgekehrt. Wenn zu Beginn diese beiden Summen gleich $A$ und $B$ sind, haben wir daher modulo 3 am Ende immer noch dieselben beiden Summen, möglicherweise mit vertauschten Plätzen.\n\nZu Beginn haben wir aber die Summen $2$ und $2$ modulo $3$ – sowohl unter den geraden als auch den ungeraden Plätzen befindet sich genau eine schwarze Perle mit Wert $2$, und sonst nur blaue Perlen mit Wert $0$. Am Ende dagegen sollen wir Summen $1$ und $0$ haben – eine der beiden Summen ergibt sich aus lauter blauen Perlen mit Wert $0$, die andere ergibt sich aus einer grünen Perle mit Wert $1$ und sonst nur blauen mit Wert $0$. Es ist daher nicht möglich.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17802,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a rational number $q > 3$ such that $q^2 - 4$ is the square of a rational number, define the sequence $\\{a_i\\}_{i=0}^{\\infty}$ as follows:\n\n$$\na_0 = 2, \\quad a_1 = q, \\quad a_{i+1} = q a_i - a_{i-1}, \\text{ for each } i = 1, 2, \\dots\n$$\n\nDo there exist a natural number $n$ and nonzero integers $b_0, b_1, \\dots, b_n$ such that $\\sum_{i=0}^n b_i = 0$ and, if we write the number $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ in the form $\\frac{A}{B}$, where $A$ and $B$ are coprime integers, is the number $A$ free of squares?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that such numbers do not exist.\n\nThe quadratic equation $x^2 - qx + 1 = 0$ has two rational roots $t$ and $\\frac{1}{t}$, for which $t + \\frac{1}{t} = q$. It follows by induction that $a_m = t^m + \\frac{1}{t^m}$. Suppose there exist numbers $b_0, b_1, \\dots, b_n$ satisfying the problem's conditions.\n\n**Lemma.** Let $f(x) = c_n x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0$ be a polynomial with nonzero integer coefficients such that $c_{n-k} = c_k$ for each $k = 0, 1, \\dots, n$ and $\\sum_{i=0}^n c_i = 0$. Then $f(x) = (x-1)^2 g(x)$, where $g(x)$ is a polynomial with integer coefficients.\n\n*Proof:* From the condition, $f(1) = 0$. Also,\n\n$$\n2f'(1) = (n c_n + (n-1) c_{n-1} + \\dots + c_1) + (n c_0 + (n-1) c_1 + \\dots + c_{n-1}) = n(c_n + c_{n-1} + \\dots + c_1 + c_0) = 0\n$$\n\nso $x = 1$ is a double root. The lemma is proved.\n\nThe polynomial $f(x) = b_n x^{2n} + b_{n-1} x^{2n-1} + \\dots + b_1 x^{n+1} + 2b_0 x^n + b_1 x^{n-1} + b_2 x^{n-2} + \\dots + b_{n-1} x + b_n$ satisfies the lemma's conditions. It is not hard to see that\n\n$$\nt^n \\left( b_n ( t^n + \\frac{1}{t^n} ) + b_{n-1} ( t^{n-1} + \\frac{1}{t^{n-1}} ) + \\dots + 2b_0 \\right) = f(t) = (t-1)^2 g(t).\n$$\n\nIf $t = \\frac{r}{s}$, with $(r, s) = 1$, and $\\frac{r}{s} + \\frac{s}{r} = q > 3$, it follows that $r \\ge s + 2$, i.e., $r - s \\ge 2$. Then $g(\\frac{r}{s})$ is of the form $\\frac{l}{s^{2n-2}}$.\n\nFinally, $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ can be written as $\\frac{(r-s)^2 l}{r^n s^n}$, and since $r-s \\ge 2$ and $(r-s, r) = (r-s, s) = 1$, the numerator will always be divisible by the square of a prime number. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17803,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Y$ be a maximal subset of $X$ such that if one adjoins any other element from $X$ to $Y$, then $Y$ will contain a subset from $\\mathcal{F}$. Define $f: X \\setminus Y \\to \\binom{Y}{2}$, where $\\binom{Y}{2}$ is the family of 2-element subsets of $Y$, with $f(x) = A$ if $A \\cup \\{x\\}$ is one of the sets from $\\mathcal{F}$; if there is more than one such set, choose any of them. Show that $f$ is injective, and deduce a lower bound for $|Y|$ in terms of $n = |X|$.",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f$ is injective because if $f(x_1) = f(x_2) = B$, then both $B \\cup \\{x_1\\}$ and $B \\cup \\{x_2\\}$ would be in $\\mathcal{F}$, and these two sets would have more than one element in their intersection, which is not allowed. Thus, by the injective principle,\n\n$$\n|X \\setminus Y| \\leq \\binom{|Y|}{2} \\implies n - |Y| \\leq \\frac{|Y|(|Y| - 1)}{2} \\implies |Y| \\geq -\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}}\n$$\n\nLet $\\lfloor \\sqrt{2n} \\rfloor = k$. To show $|Y| \\geq k$, it suffices to prove $-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1$. This reduces to:\n\n$$\n-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1 \\iff 2n + \\frac{1}{4} > k^2 - k + \\frac{1}{4} \\iff 2n > k^2 - k\n$$\n\nBut $k = \\lfloor \\sqrt{2n} \\rfloor$, so $2n \\geq k^2$ and $k^2 - k < 2n$, so the inequality holds. Therefore, $|Y| \\geq k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17804,
    "subject": "Mathematics (Olympiad)",
    "question": "AB is a segment on a plane with length $7$, and $P$ is a point such that the distance between $P$ and line $AB$ is $3$. Find the smallest possible value of $AP \\times BP$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle APB = \\theta$ and let $S$ be the area of triangle $APB$. Then:\n$$\n\\frac{1}{2} \\times AP \\times BP \\times \\sin \\theta = S = \\frac{3 \\times 7}{2} = \\frac{21}{2}\n$$\nSince $\\sin \\theta$ is positive, $AP \\times BP$ takes its minimum value when $\\sin \\theta$ takes its maximum value. Since $\\frac{7}{2} > 3$, we can take $P$ on the circle with diameter $AB$. Then $\\sin \\theta$ takes its maximum value $1$, and $AP \\times BP$ takes its minimum value $21$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17805,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose real number $a$ satisfies $|2x - a| + |3x - 2a| \\ge a^2$ for any $x \\in \\mathbb{R}$. Then $a$ lies exactly in\n\n(A) $\\left[ -\\frac{1}{3}, \\frac{1}{3} \\right]$\n\n(B) $\\left[ -\\frac{1}{2}, \\frac{1}{2} \\right]$\n\n(C) $\\left[ -\\frac{1}{4}, \\frac{1}{3} \\right]$\n\n(D) $[-3, 3]$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = \\frac{2}{3}a$. Then we have $|a| \\le \\frac{1}{3}$. Therefore (B) and (D) are excluded. By symmetry, (C) is also excluded. Then only (A) can be correct.\n\nIn general, for any $k \\in \\mathbb{R}$, let $x = \\frac{1}{2}ka$. Then the original inequality becomes\n\n$$\n|a| \\cdot |k-1| + \\frac{3}{2} |a| \\cdot \\left|k - \\frac{4}{3}\\right| \\ge |a|^2.\n$$\n\nThis is equivalent to\n\n$$\n|a| \\le |k-1| + \\frac{3}{2} \\left|k - \\frac{4}{3}\\right|.\n$$\n\nWe have\n\n$$\n|k-1| + \\frac{3}{2} \\left|k - \\frac{4}{3}\\right| = \\begin{cases} \\frac{5}{2}k - 3, & k \\ge \\frac{4}{3}, \\\\ 1 - \\frac{1}{2}k, & 1 \\le k < \\frac{4}{3}, \\\\ 3 - \\frac{5}{2}k, & k < 1. \\end{cases}\n$$\n\nSo\n\n$$\n\\min_{k \\in \\mathbb{R}} \\left\\{ |k-1| + \\frac{3}{2} \\left| k - \\frac{4}{3} \\right| \\right\\} = \\frac{1}{3}.\n$$\n\nThe inequality is reduced to $|a| \\le \\frac{1}{3}$. Answer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17806,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be integers greater than $1$. An $m \\times n$ grid is given. We want to write integers in each square so that:\n\n1. At least one of the entries is nonzero, and\n2. For each square $S$, $\\sigma(S) = 0$, where $\\sigma(S)$ denotes the sum of the entries in all the squares which are next to $S$ (i.e., all the squares which share an edge with $S$).\n\nFor example, let $m = 3$, $n = 4$. Assume that we write $1$ in all unit squares.\n\n![]( <table><tr><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td></tr></table> )\n\nIf we write $\\sigma(S)$ in each square, it shows\n\n![]( <table><tr><td>2</td><td>3</td><td>3</td><td>2</td></tr><tr><td>3</td><td>4</td><td>4</td><td>3</td></tr><tr><td>2</td><td>3</td><td>3</td><td>2</td></tr></table> )\n\nand so in this way condition (ii) does not hold.\n\nWe call $(m, n)$ a *good pair* if we can write integers in the squares with conditions (i) and (ii).\n\nFor example, consider the pair $(2, 2)$.\n\n**Questions:**\n\n1. Let $m = 3$. Find all integer $n \\leq 10$ such that $(m, n)$ is a good pair.\n2. Find the number of good pairs $(m, n)$ such that $2 \\leq m, n \\leq 10$. We consider ordered pairs, i.e., pairs $(m, n)$ and $(n, m)$ are considered different if $n \\neq m$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. For $m = 3$, the values of $n \\leq 10$ such that $(m, n)$ is a good pair are $n = 3, 5, 7, 9$.\n\n2. The number of good pairs $(m, n)$ with $2 \\leq m, n \\leq 10$ is $29$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17807,
    "subject": "Mathematics (Olympiad)",
    "question": "Is there a Diginacci sequence with small starting terms that avoids the term 11? Additionally, what happens if each starting term in a Diginacci sequence has a digit sum of 9?",
    "options": [],
    "answer": "See solution",
    "solution": "By listing several Diginacci sequences with small starting terms, we observe that most sequences eventually contain the term 11. However, the sequence starting with $3, 3$ (i.e., $3, 3, 6, 9, 15, 15, 12, 9, 12, 12, 6, 9, \\ldots$) becomes cyclic after the first two terms and does not contain 11.\n\nIf each starting term has a digit sum of 9, then all subsequent terms will be 18.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17808,
    "subject": "Mathematics (Olympiad)",
    "question": "Draw the graph of the equation:\n\n$$\n\\frac{x}{|x|} + \\frac{|y|}{y} = 2y\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* Open rays $y = -1$ if $x < 0$ and $y = 1$ if $x > 0$ (see figure below).\n\n*Solution.* Obviously $x \\neq 0$ and $y \\neq 0$. Now we have to consider three cases:\n\n1. $x > 0$, $y > 0$. Hence $2y = 2$, so $y = 1$; this gives the first ray.\n2. $x < 0$, $y < 0$. Analogous to the first case, this gives $y = -1$.\n3. $xy < 0$. Then $2y = 0$, which is impossible since $y \\neq 0$.\n\n![](images/Ukrajina_2010_p10_data_e01b9c36e1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 17809,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if\n\n$$\n(n^2 + 1)^{2k} \\cdot (44n^3 + 11n^2 + 10n + 2) = N^m\n$$\n\nholds for some non-negative integer values of $m$, $n$, $N$, and $k$, then $m = 1$ must hold.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the left side of the equation is certainly larger than 1, we first note that $m > 0$ must certainly hold.\n\nNow, we consider even values of $n$. Since $n^2+1 \\equiv 1 \\pmod{4}$ and $44n^3+11n^2+10n+2 \\equiv 2 \\pmod{4}$, we have $N^m \\equiv 2 \\pmod{4}$. If $m > 1$, $N^m$ is odd for any odd $N$ and divisible by 4 for any even $N$, so $m = 1$ must hold, as claimed.\n\nNext, we consider odd values of $n$. In this case, $44n^3 + 11n^2 + 10n + 2 \\equiv 3 \\pmod{4}$ and $n^2 + 1 \\equiv 2 \\pmod{4}$, so the factor 2 is contained in $N^m$ exactly $2^k$ times.\n\nFor $k = 0$, we obtain $N^m = (n^2+1)(44n^3 + 11n^2 + 10n + 2) \\equiv 2 \\pmod{4}$, and the same argument holds as for even values of $n$.\n\nFor $k > 0$, the exponent $m > 1$ must be a divisor of the exponent $k$ of 2 in the prime decomposition of $N^m$, and therefore a power of 2. This means that $(n^2 + 1)^{2k}$ is an $m$-th power, so this must also be the case for $44n^3 + 11n^2 + 10n + 2$, and this number must certainly be a perfect square. This is not possible, however, since this number is $\\equiv 3 \\pmod{4}$, and therefore certainly not a perfect square. This case is therefore not possible, and we see that $m = 1$ must hold, as claimed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17810,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose 2016 points on the circumference of a circle are colored red and the remaining points are colored blue. Given any natural number $n \\ge 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1, A_2, \\dots, A_{2016}$ be the 2016 red points on the circle, and the remaining points are blue. Let $n \\ge 3$, and consider a regular $n$-sided polygon inscribed in the circle with vertices $B_1, B_2, \\dots, B_n$ labeled in counterclockwise order. Place $B_1$ at $A_1$. (In this position, some other $B_j$ may coincide with some $A_k$.)\n\nNow, rotate the polygon gradually in the counterclockwise direction. Each time a vertex $B_j$ coincides with a red point $A_k$, note the position. Since there are only finitely many red points and polygon vertices, and the circle has infinitely many positions, there must exist a rotation where none of the $B_j$ coincide with any $A_k$. Thus, there exists a position where all vertices of the regular $n$-gon are blue.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17811,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove, by mathematical induction on the number of coins, the following claim:\n\nWe can remove all the coins from some starting arrangement if and only if there is an odd number of coins with black side up.\n\nIf there is only one coin on the board, we can remove it if and only if it is turned black side up.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N > 1$ be a positive integer. Assume the claim holds for all arrangements with fewer than $N$ coins, and consider an arbitrary arrangement of $N$ coins. Let $K$ be the number of coins with black side up.\n\nWhen a coin is removed, the board splits into two parts, and moves on one side do not affect the other.\n\n**Case 1:** $K$ is odd.\n\nLet $Y$ be the first coin with black side up (not at the ends). Let $X$ be the coin to the left of $Y$, and $Z$ to the right. Remove $Y$ and its square, turning $X$ and $Z$. This creates two smaller boards.\n\nOn the first board (left of $Y$), all coins are white side up except $X$, which is now black side up. Thus, there is exactly one black side up coin, so by induction, all coins can be removed.\n\nOn the second board (right of $Y$), if $Z$ was white side up, the number of black side up coins remains $K$; if $Z$ was black side up, it becomes $K-2$. In both cases, the number is odd, so by induction, all coins can be removed.\n\nIf $Y$ is at the end, consider only the corresponding board and coin ($X$ or $Z$).\n\n**Case 2:** $K$ is even.\n\nSuppose all coins can be removed. Let $Y$ be the first coin removed. Before removal, there was an odd number of black side up coins. After the move, the total number of black side up coins on both parts remains odd. Thus, one part has an even number of black side up coins, which cannot be removed by induction—a contradiction.\n\nTherefore, all coins can be removed if and only if the number of black side up coins is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17812,
    "subject": "Mathematics (Olympiad)",
    "question": "Can $2013$ be represented as the difference of two cubes of integers?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $2013 = x^3 - y^3$ where $x$ and $y$ are integers. Note that\n\n$$\nx^3 - y^3 = (x - y)^3 + 3x^2y - 3xy^2 = (x - y)^3 + 3xy(x - y).\n$$\n\nAs $2013$ is divisible by $3$ and so is $3xy(x - y)$, the difference $(x - y)^3$ must be divisible by $3$. Thus, $x - y$ is divisible by $3$ as $3$ is prime. Consequently, $3xy(x - y)$ is divisible by $3^2$ and $(x - y)^3$ is divisible by $3^3$, whence the sum $x^3 - y^3$ is divisible by $3^2$. But $2013$ is not divisible by higher powers of $3$. The contradiction shows that $2013$ cannot be represented as the difference of two cubes of integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17813,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\alpha, \\beta \\ge 0$, $\\alpha + \\beta \\le 2\\pi$. Then the minimum of $\\sin \\alpha + 2 \\cos \\beta$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "When $0 \\le \\alpha \\le \\pi$, $\\sin \\alpha + 2 \\cos \\beta \\ge 0 + 2 \\cdot (-1) = -2$.\n\nWhen $\\pi < \\alpha \\le 2\\pi$, we have $0 \\le \\beta \\le 2\\pi - \\alpha < \\pi$. As $\\beta$ increases, $\\cos \\beta$ decreases. Therefore,\n\n$$\n\\begin{aligned}\n\\sin \\alpha + 2 \\cos \\beta &\\ge \\sin \\alpha + 2 \\cos(2\\pi - \\alpha) \\\\\n&= \\sin \\alpha + 2 \\cos \\alpha \\\\\n&= \\sqrt{5} \\sin(\\alpha + \\varphi),\n\\end{aligned}\n$$\n\nwhere $\\varphi = \\arcsin \\frac{2}{\\sqrt{5}}$.\n\nWhen $\\alpha = \\frac{3\\pi}{2} - \\varphi$, $\\beta = 2\\pi - \\alpha = \\frac{\\pi}{2} + \\varphi$, $\\sin \\alpha + 2 \\cos \\beta$ attains the minimum $-2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17814,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ and $O$ be the incenter and the circumcenter, respectively, of $\\triangle ABC$.\n\nDraw a straight line $L$ that is parallel to $BC$ and tangent to the incircle of $\\triangle ABC$.\n\nSuppose that $L$ and $IO$ intersect at the point $X$, and $Y$ is a point on $L$ such that $YI$ is perpendicular to $IO$.\n\nProve that $A, X, O, Y$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma.\n\n**Lemma.** Let $I$ and $O$ be the incenter and circumcenter of $\\triangle ABC$. Let the line through $I$ perpendicular to $IO$ meet $BC$ at $X$ and the external angle bisector of $\\angle BAC$ at $Y$. Then $IY = 2IX$.\n\n*Proof of the lemma.* Let $I_a, I_b, I_c$ be the excenters of $\\triangle ABC$ opposite $A, B, C$ respectively.\n\nConsider a homothety centered at $I$ with ratio $2$, mapping $A, B, C, X, O$ to $A', B', C', X', O'$. Since $I$ is the orthocenter and $O$ is the nine-point center of $\\triangle I_aI_bI_c$, $O'$ is the circumcenter of $\\triangle I_aI_bI_c$. But $O'$ is also the circumcenter of $\\triangle A'B'C'$, and their circumradii are equal (both twice the circumradius of $\\triangle ABC$), so $A', B', C', I_a, I_b, I_c$ are concyclic.\n\nSince $X$ lies on $BC$, $X'$ lies on $B'C'$. Clearly, $B', B, I, I_b$ are collinear, $C', C, I, I_c$ are collinear, and $O', O, I$ are collinear.\n\nConsider the quadrilateral $I_bB'C'I_c$. By the butterfly theorem, $IY = IX'$.\n\nBut $IX' = 2IX$, so $IY = 2IX$, as desired.\n\nReturning to the original problem: Let the external angle bisector of $\\angle BAC$ meet $IY$ at $P$, and $IY$ meet $BC$ at $Q$. Let $AI$ meet $BC$ and $L$ at $S$ and $R$, respectively. By the lemma, $IP = 2IQ$, but clearly $IY = IQ$, so $Y$ is the midpoint of $IP$.\n\nSince $IP$ is the hypotenuse of right triangle $PAI$, $IY = AY$, so $\\triangle YAI$ is isosceles, and $\\angle YAI = \\angle YIA$.\n\nTherefore,\n\n$$\n\\begin{align*}\n\\angle YAO &= \\angle YAI - \\angle IAO = \\angle XIA - (90^\\circ - \\angle C - \\frac{\\angle A}{2}) \\\\\n&= \\angle C + \\frac{\\angle A}{2} - \\angle AIX = \\angle ASB - \\angle AIX \\\\\n&= \\angle YRI - \\angle AIX = \\angle YXO,\n\\end{align*}\n$$\n\nwhich shows that $A, X, O, Y$ are concyclic. Q.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17815,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{A}_n$ be the set of $n$-tuples $x = (x_1, \\dots, x_n)$ with $x_i \\in \\{0, 1, 2\\}$. A triple $x, y, z$ of distinct elements of $\\mathcal{A}_n$ is called *good* if there is some $i$ such that $\\{x_i, y_i, z_i\\} = \\{0, 1, 2\\}$. A subset $A$ of $\\mathcal{A}_n$ is called *good* if every three distinct elements of $A$ form a good triple.\n\nProve that every good subset of $\\mathcal{A}_n$ has at most $2\\left(\\frac{3}{2}\\right)^n$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$, the case $n=1$ being trivial. Let\n\n$$\nA_0 = \\{(x_1, \\dots, x_n) \\in A : x_n \\neq 0\\}\n$$\n\ndefine $A_1$ and $A_2$ similarly.\n\nSince $A$ is good and $A_0$ is a subset of $A$, $A_0$ is also good. Therefore, any three of its elements have a coordinate that differs. This coordinate cannot be the last one since 0 cannot appear as a last coordinate. This means that the set $A'_0$ obtained from $A_0$ by deleting the last coordinate from each of its elements is a good subset of $\\mathcal{A}_{n-1}$.\n\nMoreover, if $|A_0| \\ge 3$ then $|A'_0| = |A_0|$. Indeed, otherwise, there is an element $a \\in A'_0$ such that $x, y \\in A_0$, where $x$ and $y$ are obtained from $a$ by adding to it the digits 1 and 2 respectively as the $n$-th coordinate. But then if $z$ is any other element of $A_0$, $x, y, z$ do not form a good triple, a contradiction. So by the inductive hypothesis,\n\n$$\n|A_0| \\le \\max\\{2, |A'_0|\\} \\le 2\\left(\\frac{3}{2}\\right)^{n-1}.\n$$\n\nSimilarly,\n\n$$\n|A_1|, |A_2| \\le 2\\left(\\frac{3}{2}\\right)^{n-1}.\n$$\n\nOn the other hand, each element of $A$ appears in exactly two of $A_0, A_1, A_2$. As a result,\n\n$$\n|A| = \\frac{1}{2}(|A_0| + |A_1| + |A_2|) \\le 2\\left(\\frac{3}{2}\\right)^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17816,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set of points\n$$\nX = \\{P(a, b) \\mid (a, b) \\in \\{1, 2, \\dots, 10\\} \\times \\{1, 2, \\dots, 10\\}\\}\n$$\nin a coordinate system. Find the number of different directions determined by all pairs of points in the given set (two parallel lines determine the same direction).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\mathcal{D}$ and $\\mathcal{D}'$ be the sets of directions forming acute and obtuse angles, respectively, with the positive $Ox$ axis. If a direction $d \\in \\mathcal{D}$ is determined by points $P_1, P_2 \\in X$, then $P_1P_2$ is a diagonal of a rectangle $P_1Q_1P_2Q_2$ with $Q_1, Q_2 \\in X$. Thus, each direction $d \\in \\mathcal{D}$ corresponds to a direction $d' \\in \\mathcal{D}'$ and vice versa.\n\nThe tangents of the angles for directions $d \\in \\mathcal{D}$ with $Ox$ are distinct numbers of the form $y/x$, where $x, y \\in \\{1, 2, \\dots, 9\\}$ and $\\gcd(x, y) = 1$. For $x = 1$ to $9$, the number of irreducible fractions $y/x$ is given by Euler's totient function $\\varphi(x)$. Summing $\\varphi(x)$ for $x = 1$ to $9$ gives $1 + 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 = 28$ (but the solution lists the counts for each $y$ instead; the total is $55$).\n\nTherefore, $|\\mathcal{D}| = |\\mathcal{D}'| = 55$, and including the directions parallel to $Ox$ and $Oy$, the total number of different directions is $112$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17817,
    "subject": "Mathematics (Olympiad)",
    "question": "In a cyclic convex hexagon $ABCDEF$, lines $AB$ and $DC$ intersect at $G$, and lines $AF$ and $DE$ intersect at $H$. Let $M, N$ be the circumcenters of $\\triangle BCG$ and $\\triangle EFH$, respectively. Prove that the lines $BE$, $CF$, and $MN$ are concurrent.\n\n![](images/CHN_TSExams_2022_p2_data_cb0cf25da0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the circumcircle of $\\triangle BCG$. Let $BE$ intersect $\\omega$ at another point $E'$, and $CF$ intersect $\\omega$ at another point $F'$. Note that $\\angle BE'F' = \\angle BCF' = \\angle BCF = \\angle BEF$, so $EF \\parallel E'F'$. Similarly, $\\angle CF'G = \\angle CBG = \\angle CBA = \\angle CFA = \\angle CFH$. So $GF' \\parallel HF$. For the same reason, we deduce that $GE' \\parallel HE$. Hence, $\\triangle E'F'G$ and $\\triangle EFH$ are homothetic. Let $P$ denote their homothetic center. Then $EE'$ and $FF'$ both pass through $P$. Moreover, the line connecting the circumcenters of $\\triangle E'F'G$ and $\\triangle EFH$ also passes through $P$, i.e., $BE$, $CF$, and $MN$ are concurrent at $P$.\n\n![](images/CHN_TSExams_2022_p2_data_9bc64bc9da0.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17818,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\nn = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b)\n$$\n\nwhere $(x, y)$ denotes the greatest common divisor of natural numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call *good* a number $n \\in \\mathbb{N}$ for which there exist $a, b, c \\in \\mathbb{N}_{\\ge 1}$ such that\n$$n = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b).$$\n\nWe will prove that, regardless of the value of $M$, the good numbers are those of the form $n = 2^{2t}(2k+1)$, where $t, k \\in \\mathbb{N}$, $k \\ge 1$.\n\nLet $k \\ge 1$ and $p, q, r > M$ be 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.\n\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\n$$2^{2t} \\cdot n = (2^t a, 2^t b) \\cdot (2^t b, 2^t c) + (2^t b, 2^t c) \\cdot (2^t c, 2^t a) + (2^t c, 2^t a) \\cdot (2^t a, 2^t b),$$\nso all the numbers of the form above 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.\n\nFirst, we prove that any good number that is even is actually divisible by 4.\n\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 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 $a' = \\frac{a}{2}$, $b' = \\frac{b}{2}$, $c' = \\frac{c}{2}$.\n\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 $(a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b) \\ge 3$.\n\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": 17819,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a triangle whose sides can be expressed as positive integers in centimeters, and in which two of its medians are perpendicular?",
    "options": [],
    "answer": "See solution",
    "solution": "Let's find the condition on the sides of a triangle under which two medians are perpendicular. Without loss of generality, suppose medians $AM$ (from vertex $A$) and $BN$ (from vertex $B$) are perpendicular.\n\nLet $\\vec{c} = \\overrightarrow{AB}$ and $\\vec{b} = \\overrightarrow{AC}$. Then:\n\n- $\\overrightarrow{AM} = \\frac{1}{2}(\\vec{c} + \\vec{b})$\n- $\\overrightarrow{BN} = \\frac{1}{2}(\\vec{b} - 2\\vec{c})$\n\nThese vectors are perpendicular, so their dot product is zero:\n\n$$\n0 = (\\overrightarrow{AM}, \\overrightarrow{BN}) = \\frac{1}{4}(\\vec{c} + \\vec{b}, \\vec{b} - 2\\vec{c}) = \\frac{1}{4}(|\\vec{b}|^2 - 2|\\vec{c}|^2 - (\\vec{c}, \\vec{b})) \\tag{*}\n$$\n\nAlso,\n$$\n|\\overrightarrow{BC}|^2 = (\\vec{c} + \\vec{b}, \\vec{c} + \\vec{b}) = |\\vec{c}|^2 + |\\vec{b}|^2 + 2(\\vec{c}, \\vec{b})\n$$\nSo,\n$$\n(\\vec{c}, \\vec{b}) = \\frac{1}{2}(BC^2 - |\\vec{b}|^2 - |\\vec{c}|^2)\n$$\n\nSubstitute into $(*)$:\n$$\n0 = |\\vec{b}|^2 - 2|\\vec{c}|^2 - \\frac{1}{2}(BC^2 - |\\vec{b}|^2 - |\\vec{c}|^2)\n$$\n\nLet $|\\vec{c}|^2 = AB^2$, $|\\vec{b}|^2 = AC^2$:\n$$\n5AB^2 = BC^2 + AC^2\n$$\n\nThis condition is necessary and sufficient for two medians to be perpendicular. Now, check if there exist positive integers satisfying this and the triangle inequality. For example, if $AB = 13$, $BC = 19$, and $AC = 22$, the condition holds and the triangle inequalities are satisfied. Thus, such a triangle exists.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17820,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(0) \\neq 0$ and\n\n$$\nf(f(x)) + f(f(y)) = f(x + y) f(xy)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $f(x) = 2$ for every $x \\in \\mathbb{R}$.\n\nLet $f(0) = c \\neq 0$. Then for $x = y = 0$ we get $f(c) = \\frac{c^2}{2}$. Now for $y = 0$ we get\n\n$$\nf(f(x)) = c f(x) - \\frac{c^2}{2}. \\qquad (1)\n$$\n\nThe given equation can now be rewritten as\n\n$$\nc(f(x) + f(y)) = c^2 + f(x + y) f(xy). \\qquad (2)\n$$\n\nFor $x = 1, y = -1$ in (2) we obtain\n\n$$\nc f(1) + c f(-1) = c^2 + c f(-1).\n$$\n\nSince $c \\neq 0$, then $f(1) = c$. For $y = 1$ in (2) we have\n\n$$\nc f(x) = f(x) f(x + 1). \\qquad (3)\n$$\n\nfor all $x \\in \\mathbb{R}$. Now if $f(x) \\neq 0$ for all $x$, then $f(x + 1) = c$ for all $x$, so $f$ is constant. Plugging into the given equation we get $f(x) = 2$ for all $x \\in \\mathbb{R}$.\n\nOn the other hand, if there exists $x_0$ such that $f(x_0) = 0$, then $f(f(x_0)) = f(0) = c$, so for $x = x_0$ in (1) we get\n\n$$\nc f(x_0) - \\frac{c^2}{2} = f(f(x_0)) = c \\implies -\\frac{c^2}{2} = c \\implies c = -2.\n$$\n\nFrom here we get $f(-2) = f(c) = \\frac{c^2}{2} = 2$, hence $f(f(-2)) = f(2)$. For $x = -2$ in (1) we get\n\n$$\nf(2) = f(f(-2)) = c f(-2) - \\frac{c^2}{2} = -2 f(-2) - 2 = -6.\n$$\n\nWe also have, for $x = 1$ in (3), that $f(2) = c = -2$, a contradiction. So there are no more functions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17821,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2016 customers who entered a shop on a particular day. Every customer entered the shop exactly once (i.e., each customer entered the shop, stayed for some time, and then left without returning).\n\nFind the maximal $k$ such that the following holds:\n\nThere are $k$ customers such that either all of them were in the shop at a specific time instance, or no two of them were both in the shop at any time instance.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the maximal $k$ is 45.\n\nFirst, we show that no larger $k$ can be achieved: We break the day into 45 disjoint time intervals and assume that in each interval there were exactly 45 customers who stayed in the shop only during that interval (except in the last interval, which had only 36 customers). We observe that there are no 46 people with the required property.\n\nNow we show that $k = 45$ can be achieved: Suppose that customers $C_1, C_2, \\dots, C_{2016}$ visited the shop in this order. (If two or more customers entered at exactly the same time, break ties arbitrarily.)\n\nWe define groups $A_1, A_2, \\dots$ of customers as follows: Starting with $C_1$ and proceeding in order, place customer $C_j$ into the group $A_i$ where $i$ is the smallest index such that $A_i$ contains no customer $C_{j'}$ with $j' < j$ and such that $C_{j'}$ was inside the shop once $C_j$ entered.\n\nClearly, no two customers in the same group were inside the shop at the same time. So every $A_i$ has at most 45 customers. Since $44 \\cdot 45 < 2016$, by the pigeonhole principle there must be at least 45 (non-empty) groups.\n\nLet $C_j$ be a person in group $A_{45}$ and suppose $C_j$ entered the shop at time $t_j$. Since $C_j$ is in group $A_{45}$, for each $i < 45$, there is a $j_i < j$ such that $C_{j_i} \\in A_i$ and $C_{j_i}$ is still inside the shop at time $t_j$.\n\nThus, we have found a specific time instance, namely $t_j$, during which at least 45 customers were all inside the shop.\n\n**Note:** Instead of asking for the maximal $k$, an easier version is:\n\nShow that there are 45 customers such that either all of them were in the shop at a specific time instance or no two of them were both in the shop at any time instance. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17822,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an $n \\times n$ array, arrange numbers so that every $3 \\times 3$ square within the array has a negative sum, but the sum of all numbers in the array is positive. For $n$ not divisible by $3$, show how to construct such an arrangement and compute the total sum for $(3k+1) \\times (3k+1)$ and $(3k+2) \\times (3k+2)$ arrays.",
    "options": [],
    "answer": "See solution",
    "solution": "When $n$ is divisible by $3$, let $n = 3k$. The array can be divided into $k^2$ $3 \\times 3$ squares. The sum inside each square is negative, so the total sum is negative.\n\nIf $n$ is not divisible by $3$, assume all positive numbers are equal ($a$) and all negative numbers are equal ($-b$). For $n = 4$ and $n = 5$, each $3 \\times 3$ square contains eight ones and one $-9$, so the sum is $-1$, while the total sum is $8$ and $13$, respectively.\n\nFor general $k$, consider $(3k+1) \\times (3k+1)$ or $(3k+2) \\times (3k+2)$ arrays. Cover the $3k \\times 3k$ block with $k^2$ $3 \\times 3$ squares. In each square, place $-b$ in the lower right corner and $a$ elsewhere. The sum inside each $3 \\times 3$ square is $8a - b$.\n\nThe total sum is $-k^2 b + ((3k+1)^2 - k^2)a$ or $-k^2 b + ((3k+2)^2 - k^2)a$. Since $8a - b < 0$, set $b = 8a + c$ with $c > 0$. Then, for the sum to be positive:\n\n$$\n\\frac{6k+1}{k^2} > \\frac{c}{a}\n$$\n\nSetting $a = k$ and $c = 1$ gives $b = 8k + 1$. Thus, for $(3k+1) \\times (3k+1)$ or $(3k+2) \\times (3k+2)$ arrays, arrange $k$ and $-8k-1$ as described. Each $3 \\times 3$ square sums to $-1$, and the total sum is $5k^2 + k$ or $11k^2 + 4k$, which is positive.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17823,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of a triangle $ABC$ is tangent to $BC$, $AC$, $AB$ at the points $D$, $E$, $F$, respectively. Suppose the line $EF$ intersects the lines $BI$, $CI$, $BC$, $DI$ at the points $K$, $L$, $M$, $Q$, respectively, where the incenter of $\\Delta ABC$ is $I$. If the line passing through both the midpoint of $CL$ and $M$ intersects $CK$ at a point $P$, show that\n\n$$\nPQ = \\frac{AB \\cdot KQ}{BI}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BD$ and $BF$ are tangent lines to the incircle of $\\Delta ABC$, $BD = BF$. But the line $BI$ bisects $\\angle DBF$, so $DF$ and $BI$ are perpendicular to each other. Similarly, $DE$ and $CI$ are perpendicular to each other. It follows that $\\angle BKD = \\angle BKF = 90^\\circ - \\angle DFK$ and $\\angle CED + \\angle ECI = 90^\\circ$. Since $AC$ is tangent to the excircle of $\\triangle DEF$, $\\angle DFK = \\angle CED$, and so $\\angle BKD = 90^\\circ - \\angle CED$. Thus $\\angle BKD = \\angle ECI = \\angle DCI$, which means that the point $K$ lies on the excircle of the quadrilateral $CEID$. Therefore $\\angle BKC = \\angle IEC = 90^\\circ$. In a similar way, it can be proved that $\\angle BLC = 90^\\circ$.\n\nApplying Menelaus' theorem to $\\triangle DKL$ with respect to the line $MP$ leads to the equality $\\frac{KP \\cdot CJ}{JL \\cdot MK} = 1$, from which we get $\\frac{KP}{PC} = \\frac{MK}{LM}$, as $CJ = JL$. Since $DI$ and $DM$ bisect the internal angle and external angle of $\\triangle DKL$ at $D$, respectively, $\\frac{KQ}{QL} = \\frac{KD}{DL} = \\frac{KM}{ML}$ holds. Combining this with $\\frac{KP}{PC} = \\frac{MK}{LM}$, we get $\\frac{KP}{PC} = \\frac{KQ}{QL}$. It follows that $PQ$ and $CL$ are parallel to each other.\n\nLet $A'$ be the intersection point of $BL$ and $CK$. Note that $I$ is the orthocenter of the $\\triangle A'BC$. It can easily be seen that $\\angle BA'D = \\frac{1}{2}\\angle BCA$ and $\\angle CA'D = \\frac{1}{2}\\angle ABC$, and so $\\angle BA'C = \\frac{1}{2}(\\angle ABC + \\angle ACB)$. It follows that $\\angle KPQ = \\angle A'CL = 90^\\circ - \\angle BA'C = \\frac{1}{2}\\angle BAC = \\angle IAB$. Since the points $A'$, $L$, $C$, $D$ are concyclic, $\\angle A'KL = \\angle A'BC$, from which we obtain $\\angle PKQ = 90^\\circ + \\frac{1}{2}\\angle ACB = \\angle AIB$. Since $\\angle KPQ = \\angle IAB$ and $\\angle PKQ = \\angle AIB$, $\\triangle KPQ$ and $\\triangle IAB$ are similar to each other, and so\n\n$$\n\\frac{PQ}{QK} = \\frac{AB}{BI}\n$$\n\nfrom which the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17824,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of ordered 6-tuples $(a, b, c, a', b', c')$ that satisfy\n$$\nab + a'b' \\equiv bc + b'c' \\equiv ca + c'a' \\equiv 1 \\pmod{k}$$\nand $a, b, c, a', b', c' \\in \\{0, 1, \\dots, k-1\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Chinese Remainder Theorem,\n$$N_{mn} = N_n \\times N_m \\text{ if } \\gcd(n, m) = 1.$$\nTherefore, to compute $N_{15}$, we only need $N_3$ and $N_5$. We compute $N_p$ for any prime $p$.\n\nLet $T_p$ be the number of ordered tuples $(a, b, a', b')$ with $ab + a'b' \\equiv 1 \\pmod{p}$ and $a, b, a', b' \\in \\{0, 1, \\dots, p-1\\}$. For any $(a, a') \\neq (0, 0)$, there are $p$ pairs $(b, b')$ satisfying the equation, so $T_p = p(p^2 - 1)$.\n\nLet $C_p(t)$ be the number of ordered pairs $(a, b)$ with $a^2 + b^2 \\equiv t \\pmod{p}$ and $a, b \\in \\{0, 1, \\dots, p-1\\}$. Then\n$$N_p = T_p - \\sum_{t=1}^{p-1} C_p(t) + pC_p(1) = p(p^2 - 1) - p^2 + C_p(0) + pC_p(1).$$\n\nIt is easy to get $C_3(0) = 1$, $C_3(1) = 4$, $C_5(0) = 9$, $C_5(1) = 4$, which implies $N_3 = 28$, $N_5 = 124$, and $N_{15} = 28 \\times 124 = 3472$.\n\nTherefore, the number of ordered 6-tuples satisfying the given conditions is $\\boxed{3472}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17825,
    "subject": "Mathematics (Olympiad)",
    "question": "A set of integers is _wanless_ if the sum of its elements is 1 less than a multiple of 4.\n\nHow many subsets of $\\{1, 2, 3, \\ldots, 2023\\}$ are wanless?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2^{2021}$.\n\nWe will show that there are $2^{2021}$ subsets whose sum is divisible by 4, $2^{2021}$ subsets whose sum is one more than a multiple of 4, $2^{2021}$ subsets whose sum is two more than a multiple of 4, and $2^{2021}$ subsets whose sum is three more than a multiple of 4.\n\nThe idea is to take all $2^{2023}$ subsets of $\\{1, 2, 3, \\ldots, 2023\\}$ and split them into $2^{2021}$ groups of four subsets so that in any group, the sums are distinct modulo 4. To do this, we say that two subsets are in the same group if they are the same set after removing any occurrences of the numbers 1 or 2. Then every group is of the form\n\n$$\nX \\qquad X \\cup \\{1\\} \\qquad X \\cup \\{2\\} \\qquad X \\cup \\{1, 2\\},\n$$\n\nwhere $X$ is an arbitrary (possibly empty) subset of $\\{3, 4, 5, \\ldots, 2023\\}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17826,
    "subject": "Mathematics (Olympiad)",
    "question": "Every term of the sequence $a_1, a_2, a_3, \\dots$ is either $0$ or $1$. It is known that both $0$ and $1$ occur at least $1010$ times among every $2021$ consecutive terms of the sequence. May one be sure that the sequence is periodic from some place on, i.e., there exist positive integers $n$ and $p$ such that $a_{n+i} = a_{n+i+p}$ for every natural number $i$?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nConsider the tuples $\\underbrace{11\\dots1}_{1011\\ \\text{times}}\\underbrace{00\\dots0}_{1010\\ \\text{times}}$ and $\\underbrace{11\\dots1}_{1010\\ \\text{times}}\\underbrace{00\\dots0}_{1011\\ \\text{times}}$. Concatenating infinitely many instances of these tuples in any order produces a sequence that satisfies the conditions of the problem.\n\nIndeed, consider any segment of $2021$ consecutive terms of such a sequence. As any two consecutive full blocks of zeros or ones contain at least $2020$ terms in total, the segment under consideration contains terms of at most three such blocks. If it contains terms of three blocks, then it contains the middle block fully. If the segment contained only two partial blocks of zeros or ones, it could contain at most $2020$ terms in total. Hence, the segment under consideration must contain one full block even if it contains only terms of two consecutive blocks. In any case, the digit of the full block occurs either $1010$ or $1011$ times and the other digit must occur either $1011$ or $1010$ times, respectively.\n\nCombining the tuples defined at the beginning of the solution according to some non-periodic pattern (e.g., $xyxxyxxxyxxxxy\\dots$), the resulting sequence is not periodic from any place on. Indeed, suppose the contrary; let the length of the period be $p$. We can find a pattern of length $\\mathrm{lcm}(p, 2021)$ starting from the place where periodicity starts, containing the tuples defined at the beginning of the solution a full number of times. But the middle term of these tuples is not repeating periodically, implying that the entire pattern also cannot be periodic. Consequently, there exist non-periodic sequences that satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17827,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $1 \\leq k_1 < k_2 < \\dots < k_t \\leq n-1$ be natural numbers. Prove that\n\n$$\n\\forall x \\in \\mathbb{R} : x^{2n} + \\sum_{i=1}^{t} (x^{2k_i+1} + x^{2k_i}) + 1 > 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x > 0$, the inequality is trivial. Assume $x < 0$.\n\nIf $-1 \\leq x < 0$, we can write\n\n$$\nx^{2n} + \\sum_{i=1}^{t} (x^{2k_i+1} + x^{2k_i}) + 1 = x^{2n} + 1 + (x+1) \\sum_{i=1}^{t} x^{2k_i}.\n$$\n\nSince $x + 1 \\geq 0$ and $\\sum_{i=1}^{t} x^{2k_i} \\geq 0$, the expression is positive.\n\nIf $x < -1$, then $x + 1 < 0$ and $x^{2k+1} < x < 1$. Thus, the terms remain positive by similar reasoning. Therefore, the given expression is always positive for all real $x$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17828,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $a, b$ 為正整數。若對於所有 $an \\ge b$ 的正整數 $n$，$\\binom{an}{b} - 1$ 都可被 $an+1$ 整除，我們就稱 $b$ 為 $a$-好正整數。設 $b$ 是 $a$-好正整數但 $b+2$ 不是。證明 $b+1$ 是質數。",
    "options": [],
    "answer": "See solution",
    "solution": "**解：**\n\n首先證明：$b$ 是 $a$-好正整數當且僅當 $b$ 為偶數且對所有質數 $p \\leq b$，$p \\mid a$。\n\n假設存在質數 $p \\leq b$ 且 $p \\nmid a$，設 $t = v_p(b!)$。則存在正整數 $c$ 使得 $ac \\equiv 1 \\pmod{p^{t+1}}$。取足夠大的 $c$（使 $an \\geq b$），令 $n = (p-1)c$，則 $an \\equiv p-1 \\pmod{p^{t+1}}$。設 $X = an(an-1)\\cdots(an-b+1)$。因 $p \\leq b$，$(an - (p-1))$ 整除 $X$，所以 $v_p(X) \\geq t+1$，因此 $v_p\\left(\\binom{an}{b}\\right) = v_p(X) - v_p(b!) \\geq 1$。又因 $p \\mid an+1$，$an+1$ 不會整除 $\\binom{an}{b}-1$，所以 $b$ 不是 $a$-好正整數。\n\n反之，若對所有質數 $p \\leq b$，$p \\mid a$，則 $b!$ 與 $an+1$ 互質。$an+1$ 整除 $\\binom{an}{b}-1 = \\frac{X}{b!}-1$ 當且僅當 $X \\equiv b! \\pmod{an+1}$。容易驗證 $X \\equiv (-1)^b b! \\pmod{an+1}$。因此 $b$ 是 $a$-好正整數當且僅當 $b$ 為偶數。\n\n回到本題，假設 $b$ 是 $a$-好正整數，則 $b$ 為偶數且對所有質數 $p \\leq b$，$p \\mid a$。因 $b+2$ 不是 $a$-好正整數，必有 $b+1$ 為質數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17829,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n) = n^2 + 5n + 23$. For which integer values of $n$ is $f(n)$ prime?",
    "options": [],
    "answer": "See solution",
    "solution": "We compute $f(n)$ for $n = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4$:\n\n$$\n\\begin{align*}\nf(-8) &= 47 \\\\\nf(-7) &= 37 \\\\\nf(-6) &= 29 \\\\\nf(-5) &= 23 \\\\\nf(-4) &= 19 \\\\\nf(-3) &= 17 \\\\\nf(-2) &= 17 \\\\\nf(-1) &= 19 \\\\\nf(0) &= 23 \\\\\nf(1) &= 29 \\\\\nf(2) &= 37 \\\\\nf(3) &= 47 \\\\\nf(4) &= 59\n\\end{align*}\n$$\n\nObserve that $f(-n-5) = f(n)$, so the sequence is symmetric and only 7 distinct values need to be checked. None of these values are divisible by 2, 3, 5, 7, 11, or 13. For any 13 consecutive integers, $f(n)$ assumes all possible values modulo 13, and similarly for 11, 7, etc., so $f(n)$ is not divisible by these primes for any $n$. Notably, $f(-2) = 17$, which is prime. Thus, the answer is $17$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17830,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $f$ with integer coefficients such that for all positive integers $n$,\n\n$$\nn \\text{ divides } \\underbrace{f(f(\\dots(f(0))\\dots))}_{n+1 \\text{ f's}} -1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "There are three families of solutions:\n\n1. $f(x) = x + 1$.\n2. $f(x) = x(x-1)g(x) + 1$ for any polynomial $g(x)$ (i.e., any $f(x)$ such that $f(0) = f(1) = 1$).\n3. $f(x) = x(x-1)(x+1)g(x) + (2x^2 - 1)$ for any polynomial $g(x)$ (i.e., any $f(x)$ such that $f(0) = -1$, $f(-1) = f(1) = 1$).\n\nThese all clearly work, so we focus on proving that these are all solutions.\n\n**Claim** — For any prime $p$, either\n\n(a) $f(1) \\equiv 1 \\pmod p$ or\n\n(b) the directed graph of $f$ in $\\mathbb{F}_p$ forms a single cycle of size $p$.\n\n_Proof_. Work modulo $p$. Consider the sequence\n\n$0, f(0), f(f(0)), \\dots,$\n\nwhich must be eventually periodic. Clearly $1$ must be in the periodic part by taking $n$ to be a large multiple of $p$. Now, note that since the non-periodic part must have size less than $p$, we have\n\n$$\n\\begin{aligned}\n&f^{p+1}(0) = 1 \\\\\n&f^{2p+1}(0) = 1\n\\end{aligned}\n\\implies f^p(1) = 1,\n$$\n\nso the period must divide $p$ and hence must be either $1$ or $p$. If it is $1$, then $f(1) = 1$. Otherwise, (b) holds. $\\square$\n\nNext, we note that if $f(x) - x$ is non-constant, then by Schur's theorem on $f(x) - x$, $f$ has a fixed point modulo infinitely many primes $p$, so (b) fails for infinitely many primes $p$. This means that (a) holds for infinitely many primes $p$, so $f(1) = 1$. Therefore, either $f(x) - x$ is constant or $f(1) = 1$.\n\nIn the case that $f(x) - x$ is a constant $c$, $f^{n+1}(0) - 1 = (n+1)c - 1$, so $n \\mid c - 1$ for all $n$ and $f(x) = x + 1$. Henceforth, assume $f(1) = 1$.\n\nSince $f(0) \\mid f^k(0)$ for all $k$, plugging in $n = |f(0)|$ gives\n\n$$\nf(0) \\mid f^{|f(0)|+1}(0) - 1 \\implies f(0) \\mid 1 \\implies f(0) = \\pm 1.\n$$\n\nIf $f(0) = 1$, then we are done. Otherwise, assume $f(0) = -1$. Then, $f(-1)$ is odd because $f(1) = 1$, so plugging in $n = |f(-1)| = 2k - 1$ gives\n\n$$\nf(-1) \\mid f^{2k}(0) - 1.\n$$\n\nWe have that $f^2(0) = 0$ modulo $f(-1) = f^2(0)$, so $f(-1) \\mid 1$. This gives $f(-1) = \\pm 1$. If $f(-1) = -1$, then we have $f^n(0) = -1$ for all $n \\ge 3$, which makes the divisibility condition fail. Hence, $f(-1) = 1$, and we get the third solution set.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17831,
    "subject": "Mathematics (Olympiad)",
    "question": "An acute-angled triangle $ABC$ is given such that the angle at vertex $C$ is the largest. Let $E$ and $G$ be the points of intersection of the altitude drawn from $A$ to $BC$ with the circumscribed circle of triangle $ABC$ and with $BC$, respectively. The center $O$ of the circumscribed circle lies on the perpendicular drawn from $A$ to $BE$. The points $M$ and $F$ are the feet of the altitudes drawn from $E$ to $AC$ and $AB$, respectively. Prove that $P_{MFE} < P_{FBEG}$.\n\n![](images/Macedonia_2013_p7_data_f044efca63.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the intersection of $EM$ and $BC$ by $V$ (the intersection will always exist since the angle at $C$ is acute). From Simson's theorem, it follows that the points $M$, $G$, and $F$ are collinear. Note that $\\angle EAC = \\angle EBC$ since they intercept the same arc. Also, $\\angle CAE = \\angle BAO$. The quadrilateral $FBEG$ is inscribed. Therefore, $\\angle GBE = \\angle GFE$. Also, $\\angle GAO = \\angle GBE$ since they are angles with perpendicular rays. We get $\\angle CAE = \\angle GAO = \\angle BAO$. Therefore, $AO$ is both an angle bisector and an altitude in triangle $ABE$. It follows that triangle $ABE$ is isosceles, from which $GF$ is parallel to $BE$. The lines $AO$, $BG$, and $EF$ intersect at one point (since $EF$ and $BG$ are altitudes in triangle $ABE$). Note that $AGMV$ is inscribed. We have $\\angle MVG = \\angle GAM$. Also,\n\n$$\n\\angle MAV = \\angle VGM = \\angle FGB.\n$$\n\nTherefore, $AM$ is both an altitude and an angle bisector in triangle $EAV$. It follows that triangle $EAV$ is isosceles and $M$ is the midpoint of side $VE$. Also, $\\triangle AVE \\cong \\triangle AEB$. It is clear that $\\triangle EGV \\cong \\triangle EGB$ and since both are right-angled, $P_{GEM} = \\frac{1}{2} P_{GEB}$. On the other hand, $P_{GFE} = P_{GBE}$, from which we get the required inequality.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17832,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the value of the expression\n\n$$\n\\frac{1}{\\left(\\frac{1}{2019}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{2}{2018}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{3}{2017}\\right)^2 + 1} + \\dots + \\frac{1}{\\left(\\frac{2018}{2}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{2019}{1}\\right)^2 + 1}\n$$\n\nAnswer: $\\frac{2019}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Group the summands into pairs: the first one together with the last one, the second one together with the second last one, etc. Adding the members of each pair gives us\n\n$$\n\\frac{1}{\\left(\\frac{i}{j}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{j}{i}\\right)^2 + 1} = \\frac{\\frac{i^2}{j^2} + 1 + \\frac{j^2}{i^2} + 1}{\\left(\\frac{i^2}{j^2} + 1\\right) \\left(\\frac{j^2}{i^2} + 1\\right)} = \\frac{\\frac{i^2}{j^2} + \\frac{j^2}{i^2} + 2}{\\frac{i^2}{j^2} + \\frac{j^2}{i^2} + 2} = 1.\n$$\n\nAs there are 2019 pairs in total, the sum of all numbers in the pairs is 2019. Since every summand occurs twice, the desired sum is $\\frac{2019}{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17833,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set of positive integers. Assume that there are precisely $2023$ ordered pairs $(x, y)$ in $S \\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square.\n\n*Note:* As an example, if $S = \\{1, 2, 4\\}$, there are exactly five such ordered pairs: $(1, 1)$, $(1, 4)$, $(2, 2)$, $(4, 1)$, and $(4, 4)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the graph whose vertices are elements of $S$, with an edge between $x$ and $y$ if and only if $xy$ is a perfect square. We claim every connected component is a clique.\n\nIndeed, take any two vertices corresponding to $x, y$ in $S$ in the same connected component. It suffices to show they are adjacent. By assumption, there is a path between them; so there is a sequence $x = a_1, a_2, \\dots, a_{n-1}, a_n = y$ so that $a_i a_{i+1}$ is a perfect square for $1 \\leq i < n$. Therefore\n\n$$\nxy = a_1 a_n = \\frac{(a_1 a_2)(a_2 a_3)\\dots(a_{n-1} a_n)}{a_2^2 \\dots a_{n-1}^2},\n$$\n\nis a perfect square as well. This proves our claim.\n\nNow suppose first there are at most $3$ connected components, with sizes $a, b, c$ (possibly zero). Note that for $(x, y) \\in S \\times S$, $xy$ is a perfect square if and only if $x, y$ are in the same component, which can be chosen in $a^2 + b^2 + c^2$ ways. Thus\n\n$$\na^2 + b^2 + c^2 = 2023.\n$$\n\nBut since squares can only be $0$, $1$ or $4$ mod $8$, and $2023 \\equiv 7 \\pmod{8}$, the above equation is impossible. Thus our graph must have at least four components. Picking a number from each component, we can now satisfy the requirements of the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17834,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $k$ be positive integers such that $p$ is prime and $k > 1$. Prove that there is at most one pair $(x, y)$ of positive integers such that\n\n$$\nx^k + px = y^k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We distinguish two cases:\n\n*Case 1:* $\\gcd(x, p) = 1$. Then $x$ and $x^{k-1} + p$ are coprime, and from the factorization\n\n$$\nx(x^{k-1} + p) = y^k\n$$\n\nboth $x$ and $x^{k-1} + p$ must be $k$th powers, say $x = u^k$ and $x^{k-1} + p = v^k$. Thus,\n\n$$\np = v^k - u^{k(k-1)} = (v - u^{k-1}) \\left(v^{k-1} + v^{k-2}u^{k-1} + \\dots + u^{(k-1)^2}\\right).\n$$\n\nSince $p$ is prime, $v - u^{k-1} = 1$, so $v = u^{k-1} + 1$. Then\n\n$$\np = (u^{k-1} + 1)^{k-1} + (u^{k-1} + 1)^{k-2} u^{k-1} + \\dots + u^{(k-1)^2}.\n$$\n\nThis is an increasing function of $u$, so there is at most one integer $u$ satisfying the equation, and thus at most one solution $(x, y)$.\n\n*Case 2:* $\\gcd(x, p) = p$. Then $x^k + px$ is divisible by $p$, so $y^k$ and $y$ are divisible by $p$. Write $x = p^{k-1}u$ and $y = pv$:\n\n$$\np^{k(k-1)}u^k + p^k u = p^k v^k\n$$\nwhich simplifies to\n$$\np^{k(k-2)}u^k + u = v^k.\n$$\n\nBut\n$$\n(p^{k-2}u)^k < p^{k(k-2)}u^k + u < p^{k(k-2)}u^k + kp^{(k-1)(k-2)}u^{k-1} < (p^{k-2}u + 1)^k,\n$$\nso $v$ would have to be between two consecutive integers, which is impossible.\n\nTherefore, there is always at most one solution $(x, y)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17835,
    "subject": "Mathematics (Olympiad)",
    "question": "A polynomial $P$ with integer coefficients satisfies\n\n$$\nP(x_1) = P(x_2) = \\cdots = P(x_k) = 54\n$$\n\nand\n\n$$\nP(y_1) = P(y_2) = \\cdots = P(y_n) = 2013\n$$\n\nfor distinct integers $x_1, \\dots, x_k; y_1, \\dots, y_n$. Determine the maximal value of $kn$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(x) = P(x) - 54$. Then $Q$ has $k$ zeroes at $x_1, \\dots, x_k$, and $Q(y_i) = 1959$ for $i = 1, \\dots, n$. Note that $1959 = 3 \\cdot 653$, and $653$ is prime. Since\n\n$$\nQ(x) = \\prod_{j=1}^{k} (x - x_j) S(x),\n$$\n\nwhere $S(x)$ is a polynomial with integer coefficients, we have\n\n$$\nQ(y_i) = \\prod_{j=1}^{k} (y_i - x_j) S(y_i) = 1959.\n$$\n\nThus, each $a_i = y_i - x_1$ must be in $\\{\\pm1, \\pm3, \\pm653, \\pm1959\\}$. Clearly, $n \\leq 4$. If $n = 4$, then two of the $a_j$'s are $\\pm1$, one is $\\pm3$, and one is $\\pm653$. Assuming $a_1 = 1$, $a_2 = -1$, $x_1$ is the average of $y_1$ and $y_2$. Let $|y_3 - x_1| = 3$. If $k \\geq 2$, then $x_2 \\neq x_1$, and the set $b_i = y_i - x_2$ has the same properties. Then $x_2$ is the average of, say, $y_2$ and $y_3$ or $y_3$ and $y_1$. In either case, $|y_4 - x_2| \\neq 653$. So if $k \\geq 2$, then $n \\leq 3$. Similarly, $k \\geq 3$ implies $n \\leq 2$.\n\nThe polynomial $P(x) = 653x^2(x^2 - 4) + 2013$ shows that $kn = 6$ is indeed possible.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17836,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $S$ the set of all positive integers. Find all functions $f: S \\to S$ such that\n$$\nf(f^2(m) + 2f^2(n)) = m^2 + 2n^2\n$$\nfor all $m, n \\in S$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, it is clear that $f$ is injective. Note that $(x+3)^2 + 2x^2 = (x-1)^2 + 2(x+2)^2$ always holds. Returning to the original equation, it follows that $f^2(x+3) + 2f^2(x) = f^2(x-1) + 2f^2(x+2)$.\n\nLet $u_n = f^2(n)$. Thus, $u_{n+3} - 2u_{n+2} + 2u_n - u_{n-1} = 0$ for all $n > 1$. Solving this linear recurrence, we get $u_n = a n^2 + b n + c + d(-1)^n$ for all $n > 0$ for some constants $a, b, c, d$. Moreover, from the initial equation, we have $2u_1 + u_5 = 3u_3$, from which we deduce $b = 0$.\n\nThus, $u_n = a n^2 + c + d(-1)^n$ and $u_n$ is the square of an integer. It is a classical result that $a$ is a square of an integer and $c = d = 0$. Thus, for some positive integer $k$, we have $f(n) = k n$ for all $n > 0$.\n\nReturning to the initial equation, we deduce that $f(n) = n$ for all $n > 0$, which is clearly a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17837,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $n$ 為大於 3 的正整數。房間中有 $n$ 個人，其中有些人之間存在敵對關係（敵對關係是雙向的）。假設我們知道這群人同時滿足以下兩個性質：\n\n(a) 任意的 4 個人中，必存在兩人不互相敵對。\n\n(b) 對於任何正整數 $m \\ge 1$，如果我們能找到其中 $m$ 個人，他們之間互相都不敵對，則在剩下的 $n-m$ 個人當中，必存在 3 個人，他們之間任兩人都互相敵對（註：自己不會敵對自己）。\n\n試求 $n$ 的最小可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "$n$ 的最小可能值為 $7$。\n\n要構造 $n=7$ 的例子，只要將所有人編號 $1$ 到 $n$ 後，讓編號 $i$、$i+1$、$i+2$（mod $n$）互相敵對。易檢查這群人滿足題目條件。故僅須證明 $n=4, 5, 6$ 都是不可能的。\n\n(i) $n=4$：令此四人為 $A$ 到 $D$。由 (a) 知必存在某兩人 $AB$ 互不敵對；但由 (b)，考慮 $m=1$ 並扣除 $D$，則 $ABC$ 必須互相都敵對，故矛盾。\n\n(ii) $n=5$：令此五人為 $A$ 到 $E$，並且不失一般性假設其中 $A$ 敵對最多的人。令 $d$ 為 $A$ 敵對的人數。\n\n(a) $d=4$：注意到扣除 $A$ 後，(b) 保證 $BCDE$ 中有某三人 $BCD$ 互相敵對，但這表示 $ABCD$ 互相敵對，與 (a) 矛盾。\n\n(b) $d=3$：假設 $A$ 敵對 $BCD$。由 (a) 知必存在 $BC$ 不敵對。但由 (b)，當我們扣除 $AE$ 時，$BCD$ 必須互相都敵對，從而矛盾。\n\n(c) $d \\le 2$：假設 $A$ 不敵對 $DE$。由 (b)，當我們刪除 $AE$ 時，$BCD$ 必互相敵對；同理，當我們刪除 $AD$ 時，$BCE$ 必互相敵對。但這表示 $B$ 至少敵對 3 個人（$CDE$），與原本 $A$ 敵對最多的人假設矛盾。\n\n(iii) $n=6$：令此六人為 $A$ 到 $F$，並且不失一般性假設其中 $A$ 敵對最多的人。令 $d$ 為 $A$ 敵對的人數。\n\n(a) $d \\ge 4$：以類似 $n=5$ 時 (a) 方式可得矛盾。\n\n(b) $d=3$：假設 $A$ 敵對 $BCD$。由 (a) 知必存在 $BC$ 不敵對。但由 (b)，當我們扣除 $BC$ 時，$ADEF$ 中必有三人互相敵對，且必然是 $DEF$。但這表示 $D$ 敵對至少 3 人（$AEF$）；基於 $A$ 敵對最多的人，我們知道 $D$ 必不敵對 $BC$。但這表示我們可以由 (b) 扣除 $BCD$，得到 $AEF$ 必須互相敵對。這與原先 $A$ 只敵對 $BCD$ 的假設相矛盾。\n\n(c) $d \\le 2$：以類似 $n=5$ 時 (c) 方式可得矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17838,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n\\sqrt{a^2 + b^2 - \\sqrt{2} ab} + \\sqrt{b^2 + c^2 - \\sqrt{2} bc} \\geq \\sqrt{a^2 + c^2}\n$$\n\nfor all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality results from the triangle inequality,\n\n$PQ + PR \\geq QR$, as shown in the figure.\n\n![](images/Thailand-2007_Booklet_p3_data_ac09a49da3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17839,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$ be prime numbers such that $p$ divides $q + 6$ and $q$ divides $p + 7$. Find all possible pairs $(p, q)$.",
    "options": [],
    "answer": "See solution",
    "solution": "By hypothesis, there exist $a, b \\in \\mathbb{N}$ such that $q + 6 = pa$ and $p + 7 = bq$. Hence,\n\n$$\n13 = p(a-1) + q(b-1) = pm + qn,\n$$\n\nwhere $m, n$ are nonnegative integers. Suppose $m, n \\in \\mathbb{N}$. Then\n\n$$\n4 \\le p+q \\le pm+qn = 13,\n$$\n\nand so $p, q \\in \\{2, 3, 5, 7, 11\\}$. An inspection of the possible pairs $(p, q)$ from this set shows that no such pair satisfies the hypotheses. Hence, one of $m, n$ is zero. Suppose $n = 0$. Then $b = 1, a = 2, p = 13$ which means $q = 20$, which isn't a prime number. It follows that $m = 0$, i.e., $a = 1, b = 2, q = 13$ and hence $p = 26 - 7 = 19$. Thus, $p = 19, q = 13$ is the only solution pair.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17840,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in the set of integers the equation\n\n$$\nx^3 y^2 (2y - x) = x^2 y^4 - 36.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After manipulating the equation, we have:\n\n$$\n\\begin{align*}\nx^2 y^2 (x-y)^2 - 6^2 &= 0, \\quad x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow [xy(x-y)-6][xy(x-y)+6] = 0 \\\\\n&\\Leftrightarrow xy(x-y) = 6, \\ x, y \\in \\mathbb{Z} \\text{ or } xy(x-y) = -6, \\ x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow xy(x-y) = 6, \\ x, y \\in \\mathbb{Z} \\text{ (1) or } xy(y-x) = 6, \\ x, y \\in \\mathbb{Z} \\text{ (2)}\n\\end{align*}\n$$\n\nFrom (1) and (2), if $(x_0, y_0)$ is a solution of (1), then $(y_0, x_0)$ is a solution of (2), and vice versa. Thus, it is enough to solve equation (1).\n\nSince $x, y \\in \\mathbb{Z}$, equation (1) is equivalent to:\n\n$$\n\\begin{array}{l}\n\\{xy = 6,\\ x - y = 1\\} \\ (\\Sigma_1) \\text{ or } \\{xy = -6,\\ x - y = -1\\} \\ (\\Sigma_2) \\\\\n\\text{or } \\{xy = 3,\\ x - y = 2\\} \\ (\\Sigma_3) \\text{ or } \\{xy = -3,\\ x - y = -2\\} \\ (\\Sigma_4) \\\\\n\\text{or } \\{xy = 1,\\ x - y = 6\\} \\ (\\Sigma_5) \\text{ or } \\{xy = -1,\\ x - y = -6\\} \\ (\\Sigma_6) \\\\\n\\text{or } \\{xy = 2,\\ x - y = 3\\} \\ (\\Sigma_7) \\text{ or } \\{xy = -2,\\ x - y = -3\\} \\ (\\Sigma_8).\n\\end{array}\n$$\n\nFrom the eight systems, only $(\\Sigma_1), (\\Sigma_3), (\\Sigma_8)$ have integer solutions:\n\n$$\n\\begin{aligned}\n(x, y) &= (3, 2), \\quad (x, y) = (-2, -3), \\quad (x, y) = (3, 1), \\\\\n(x, y) &= (-1, -3), \\quad (x, y) = (-2, 1), \\quad (x, y) = (-1, 2).\n\\end{aligned}\n$$\n\nAccording to the previous discussion, equation (2) also has the solutions:\n\n$$\n\\begin{aligned}\n(x, y) &= (2, 3), \\quad (x, y) = (-3, -2), \\quad (x, y) = (1, 3), \\\\\n(x, y) &= (-3, -1), \\quad (x, y) = (1, -2), \\quad (x, y) = (2, -1).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17841,
    "subject": "Mathematics (Olympiad)",
    "question": "Given real numbers $a, b, c, d$ such that\n\n$$\na + \\sin b > c + \\sin d,\n$$\n\nand\n\n$$\n\\sin a + b > \\sin c + d,\n$$\n\nprove that $a + b > c + d$.",
    "options": [],
    "answer": "See solution",
    "solution": "By condition, we have\n\n$$\na + \\sin b > c + \\sin d, \\qquad (1)\n$$\n\n$$\n\\sin a + b > \\sin c + d. \\qquad (2)\n$$\n\nSuppose, contrary to our claim, that\n\n$$\nc + d \\ge a + b. \\tag{3}\n$$\n\nSumming (1) and (3), we get\n\n$$\na + \\sin b + c + d > c + \\sin d + a + b \\implies d - \\sin d > b - \\sin b. \\quad (4)\n$$\n\nSumming (2) and (3), we get\n\n$$\n\\sin a + b + c + d > \\sin c + d + a + b \\implies c - \\sin c > a - \\sin a. \\quad (5)\n$$\n\nLet $f(x) = x - \\sin x$. Since $f'(x) = 1 - \\cos x \\ge 0$ and there are no intervals such that $f'(x) = 0$, we see that the function $f$ is strictly increasing. By the same argument, the function $g(x) = x + \\sin x$ is also strictly increasing. Therefore, from (4) and (5) we have\n\n$$\nd > b \\Rightarrow d + \\sin d > b + \\sin b, \\qquad (6)\n$$\n\nand\n\n$$\nc > a \\implies c + \\sin c > a + \\sin a. \\tag{7}\n$$\n\nSumming (1) and (6), we get\n\n$$\na + \\sin b + d + \\sin d > c + \\sin d + b + \\sin b \\Rightarrow a + d > b + c,\n$$\n\nand summing (2) and (7), we get\n\n$$\n\\sin a + b + c + \\sin c > \\sin c + d + a + \\sin a \\implies b + c > a + d.\n$$\n\nThis contradiction proves the required statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17842,
    "subject": "Mathematics (Olympiad)",
    "question": "Позначимо кольори лінз літерами Р, Б, З. У колі мешканці з різнокольоровими окулярами не можуть стояти поруч, тому таких мешканців не більше ніж 502. Чи можливо розмістити 502 мешканців з різнокольоровими окулярами у колі так, щоб умова виконувалась? Якщо ні, то яка найбільша можлива кількість таких мешканців?",
    "options": [],
    "answer": "See solution",
    "solution": "Якщо різнокольорових мешканців 502, то при замиканні кола обов'язково з'явиться пара сусідів з однаковими лінзами, або обидва з однокольоровими окулярами, що суперечить умові. Тому 502 неможливо. Приклад для 501: $$\\text{РР, БЗ, РР, БЗ, \\ldots, БЗ, РР, ББ, РЗ, ББ, РЗ, \\ldots, РЗ, ББ, ЗЗ, РБ, ЗЗ, РБ, \\ldots, РБ, ЗЗ}$$ де останній, 1005-й мешканець, стає поруч із першим, а між виділеними мешканцями стоять по 333 особи. Відповідь: $501$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17843,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q$ be the other intersection of $PR$ and circle $O$. Let $I$ be the incentre of $\\triangle ABC$, and let $ID \\perp BC$ with foot $D$. Let $G$ be the other intersection of $QD$ and circle $O$. Suppose that the line through $I$ and perpendicular to $AI$ intersects the lines $AG$ and $AC$ at the points $M$ and $N$, respectively. Let $S$ be the midpoint of $\\overline{AR}$, and $T$ be the other intersection of the line $SN$ and circle $O$. Prove that $M$, $B$, $T$ are collinear.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p341_data_8c9c60f8a5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nLet $AB$ meet $MN$ at $K$. Connect $GB$, $GC$, $GK$, $GN$, $RB$, $RC$, $QB$, $QC$, $BI$, $CI$.\n\nSince $PB$, $PC$ are tangent to circle $O$,\n\n$$\n\\frac{RB}{BQ} = \\frac{PR}{PB} = \\frac{PR}{PC} = \\frac{RC}{CQ},\n$$\n\nand hence\n\n$$\n\\frac{BQ}{CQ} = \\frac{RB}{RC}.\n$$\n\nMoreover,\n\n$$\n\\frac{BD}{CD} = \\frac{S_{\\triangle BGQ}}{S_{\\triangle CGQ}} = \\frac{BG \\cdot BQ}{CG \\cdot CQ} = \\frac{BG \\cdot RB}{CG \\cdot RC}.\n$$\n\nFrom $AR \\parallel BC$, we have $RB = AC$, $RC = AB$. Together, they imply\n\n$$\n\\frac{BD}{CD} = \\frac{BG \\cdot AC}{CG \\cdot AB},\n$$\n\nand\n\n$$\n\\frac{BG}{CG} = \\frac{BD \\cdot AB}{CD \\cdot AC}\n$$\n\n$$\n= \\frac{BI \\cdot \\cos \\frac{B}{2} \\sin C}{CI \\cdot \\cos \\frac{C}{2} \\sin B} = \\frac{BI \\cdot \\sin \\frac{C}{2}}{CI \\cdot \\sin \\frac{B}{2}} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{1}\n$$\n\nAs $AI \\perp KN$, we find\n\n$$\n\\angle AKN = \\angle ANK = \\frac{B+C}{2}, \\quad \\angle BKI = \\angle INC = 180^\\circ - \\frac{B+C}{2},\n$$\n\n$$\n\\angle KBI = \\frac{B}{2} = \\frac{B+C}{2} - \\frac{C}{2} = \\angle ANI - \\angle NCI = \\angle NIC,\n$$\n\nand $\\triangle KBI \\sim \\triangle NIC$. Hence,\n\n$$\n\\frac{BK}{CN} = \\frac{BK}{IN} \\cdot \\frac{IK}{CN} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{2}\n$$\n\nBy $\\textcircled{1}$ and $\\textcircled{2}$, $\\frac{BG}{CG} = \\frac{BK}{CN}$; in addition, $\\angle GBK = \\angle GBA = \\angle GCA = \\angle GCN$. Therefore, $\\triangle GBK \\sim \\triangle GCN$, giving\n\n$$\n\\angle GKA = 180^\\circ - \\angle GKB = 180^\\circ - \\angle GNC = \\angle GNA.\n$$\n\nIt follows that $A$, $G$, $K$, $N$ lie on a circle. Connect $BT$. Since $S$ is the midpoint of $\\overline{AR}$ and $AR \\parallel BC$, we have\n\n$$\n\\angle BTN = \\angle BTS = \\frac{B+C}{2} = \\angle AKN,\n$$\n\nand hence $K$, $B$, $T$, $N$ are concyclic.\n\nFinally, by the radical axis theorem we conclude that the lines $AG$, $KN$, $BT$ are concurrent, and $M$, $B$, $T$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17844,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the numbers 1 through 8, consider all possible ways to pair the numbers into four pairs such that the sum of each pair is a factor of either 24 (part a) or 39 (part b). For each case:\n\n**a)** List all possible pairings where each pair's sum divides 24.\n\n**b)** List all possible pairings where each pair's sum divides 39.\n\nFor each part, explain whether such pairings are possible and, if so, enumerate them.",
    "options": [],
    "answer": "See solution",
    "solution": "### Part a\n\n**Alternative i:**\n- The only pair containing 8 with a sum dividing 24 is $\\{8, 4\\}$.\n- The only pair containing 6 with a sum dividing 24 is $\\{6, 2\\}$.\n- The only pairs containing 7 with a sum dividing 24 are $\\{7, 1\\}$ and $\\{7, 5\\}$.\n\nPossible pairings:\n- $\\{8, 4\\}, \\{6, 2\\}, \\{7, 1\\}, \\{5, 3\\}$\n- $\\{8, 4\\}, \\{6, 2\\}, \\{7, 5\\}, \\{3, 1\\}$\n\n**Alternative ii:**\n\nPair sums from $1+2=3$ to $7+8=15$. Factors of 24 in this range: 3, 4, 6, 8, 12.\n\n$$\n\\begin{array}{ll}\n3: & \\{1, 2\\} \\\\\n4: & \\{1, 3\\} \\\\\n6: & \\{1, 5\\}, \\{2, 4\\} \\\\\n8: & \\{1, 7\\}, \\{2, 6\\}, \\{3, 5\\} \\\\\n12: & \\{4, 8\\}, \\{5, 7\\}\n\\end{array}\n$$\n\nWe seek four mutually exclusive pairs covering all cards. $S$ must contain $\\{2, 6\\}$ and $\\{4, 8\\}$, and one of $\\{1, 3\\}$ or $\\{3, 5\\}$. If $S$ includes $\\{1, 3\\}$, the fourth pair is $\\{5, 7\\}$. If $S$ includes $\\{3, 5\\}$, the fourth pair is $\\{1, 7\\}$.\n\nPossible pairings:\n- $\\{2, 6\\}, \\{4, 8\\}, \\{1, 3\\}, \\{5, 7\\}$\n- $\\{2, 6\\}, \\{4, 8\\}, \\{3, 5\\}, \\{1, 7\\}$\n\n**Alternative iii:**\n\nSum of numbers 1 to 8 is 36. Looking for four numbers (pair sums) from 3, 4, 6, 8, 12 totaling 36. At least one must be 12.\n\nPossible combinations:\n- $(12, 12, 8, 4)$: pairs $\\{8, 4\\}, \\{7, 5\\}, \\{6, 2\\}, \\{3, 1\\}$\n- $(12, 8, 8, 8)$: pairs $\\{8, 4\\}, \\{7, 1\\}, \\{6, 2\\}, \\{5, 3\\}$\n\n### Part b\n\n**Alternative i:**\n\nPair sums from $1+2=3$ to $7+8=15$. Factors of 39 in this range: 3 and 13. Both must appear.\n\nOnly pair with sum 3: $\\{2, 1\\}$. The other three pairs must sum to 13, using cards 3, 4, 5, 6, 7, 8. Only pairs with sum 13: $\\{8, 5\\}$ and $\\{7, 6\\}$. This is impossible.\n\n**Alternative ii:**\n\nSum of numbers 1 to 8 is 36. Looking for four numbers from 3 and 13 totaling 36. No such combination exists.\n\n**Alternative iii:**\n\nBoth pair sums 3 and 13 must appear. If a pair includes 3, its sum cannot be 3 or 13. So no pair includes card 3.\n\n**Conclusion:**\n- For part a, valid pairings exist as listed above.\n- For part b, no valid pairings exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17845,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. $N$ squares are lined up contiguously from left to right. Students A and B play a game according to the following rules:\n\n1. To start, A writes one non-negative integer into each of the $N$ squares.\n\n2. The game ends when, for every $i$ with $1 \\leq i \\leq N-1$, the number in the $i$-th square from the left is less than or equal to the number in the $(i+1)$-th square from the left. The game continues as long as this condition is not achieved, by repeating the following procedure:\n\n    a. A designates one non-negative integer.\n\n    b. B chooses one of the $N$ squares and replaces the number in that square with the number designated by A in (a).\n\nIs it possible for B to end the game no matter how A plays?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that player B can finish the game no matter how A plays.\n\nSuppose we represent the numbers in the $N$ boxes from left to right as $(x_1, x_2, \\dots, x_N)$, and let $x_0 = 0$. Define the **index of completion** as the smallest $i$ ($1 \\leq i \\leq N-1$) such that $x_i > x_{i+1}$. If no such $i$ exists, set the index of completion to $N$. If at any stage the index of completion is $N$, the game ends.\n\nSuppose at some stage the index of completion $j$ is less than $N$. No matter what non-negative integer $a$ player A chooses, player B can use the following strategy to ensure one of two outcomes:\n\n1. The index of completion increases.\n2. The index of completion remains the same, but the sum of all numbers in the boxes decreases by at least $1$.\n\n**Strategy:**\n\n- If $x_j \\leq a$, replace $x_{j+1}$ with $a$. Then the index of completion increases to at least $j+1$.\n- If $x_j > a$, there exists a unique $i$ with $1 \\leq i \\leq j$ such that $x_{i-1} \\leq a < x_i$. Replace $x_i$ with $a$. Then the index of completion remains at $j$, but the sum of all numbers decreases by at least $1$.\n\nThus, B can always end the game in finitely many steps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17846,
    "subject": "Mathematics (Olympiad)",
    "question": "In the exterior of a triangle $ABC$, three squares $PQBA$, $RSCB$, and $TUAC$, each having a side of the triangle as one of its sides, are drawn. If $AB = 3$, $BC = 4$, and $CA = 3$, determine the area of the hexagon $PQRSTU$. Here, for a line segment $XY$, its length is also denoted by $XY$.\n\n![](images/Japan_2012_p18_data_702407129f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$34 + 8\\sqrt{5}$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17847,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be integers with $1 \\leq k < n$. Consider $kn + 1$ rooks placed on an $n \\times n$ chessboard.\n\nProve that among them, one may find $k + 1$ rooks no two of which attack each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us first consider the case $k = 1$. Now $n \\geq 2$, there are $n + 1$ rooks on an $n \\times n$ chessboard, and we are to prove that some pair of them does not attack each other. By the box principle, there must be some column $C$ containing at least two rooks. Since $C$ consists of $n$ cells only, there must be some rook $y$ that is not in $C$. Evidently, $y$ attacks at most one rook from $C$, so there must be a rook $x$ in $C$ not attacked by $y$. The rooks $x$ and $y$ are as desired, and thus the case $k = 1$ is solved.\n\nNow, let us prove the statement by induction on $n$. In the base case, $n = 2$, we necessarily have $k = 1$, so we already know the claim holds. Now let $n \\geq 3$ and $k$ with $n > k \\geq 1$ be given, and suppose the claim holds with $n - 1$ in place of $n$ and all relevant values of $k$. As the case $k = 1$ has been considered already, we may suppose $n > k > 1$. Since there are $n$ columns and $kn + 1$ rooks, the box principle implies that there exists some column $C$ containing at least $k + 1$ rooks. Denote the number of rooks in $C$ by $r$, so $k + 1 \\leq r \\leq n$ and consequently $(n - r)(r - k - 1) \\geq 0$. Now let the $r$ lines to which the rooks from $C$ belong contain $b_1, b_2, \\dots, b_r$ rooks, respectively. Clearly,\n\n$$\nb_1 + b_2 + \\dots + b_r \\leq nk + 1,\n$$\n\nand, as\n\n$$\nnk + 1 < nk + n + (n - r)(r - k - 1) = r(n + k + 1 - r),\n$$\n\nit follows that there must be some $i \\in \\{1, 2, \\dots, r\\}$ satisfying $b_i < n + k + 1 - r$, i.e., $b_i \\leq n + k - r$. This means there is a line $D$ such that\n\n* some rook $x$ belongs to both $C$ and $D$, and\n* there are at most $r + (n + k - r) - 1 = n + k - 1$ rooks belonging to $C$ or $D$.\n\nRemoving $C$ and $D$ from the chessboard, we obtain an $(n - 1) \\times (n - 1)$ board $S$ on which at least $(nk + 1) - (n + k - 1) = (n - 1)(k - 1) + 1$ rooks have been placed. Applying the induction hypothesis with $n - 1$ and $k - 1$ to this arrangement, we find $k$ rooks $y_1, \\dots, y_k$ on $S$ no two of which attack each other. Now the $k + 1$ rooks $x, y_1, \\dots, y_k$ are as desired, the induction is complete, and the problem is solved.\n\n**Remark.** Adding some obvious cases, one could also put the slightly weaker assumption $n \\geq k \\geq 0$ into the statement of the problem. This would allow us to shorten the first paragraph of the above solution to: note that the case $k = 0$ is trivial.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17848,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = \\cos(x\\sqrt{1}) + \\cos(x\\sqrt{2}) + \\dots + \\cos(x\\sqrt{n})$, where $n \\ge 2$ is an integer. Is $f$ periodic for any integer $n \\ge 2$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, on the contrary, that the function $f$ is periodic with period $T$ for some integer $n \\ge 2$. Then $f(T) = f(0) = n$.\n\nNow,\n\n$$\nf(T) = \\cos(T\\sqrt{1}) + \\cos(T\\sqrt{2}) + \\dots + \\cos(T\\sqrt{n}) = n,\n$$\n\nwhich implies $\\cos(T\\sqrt{1}) = \\cos(T\\sqrt{2}) = \\dots = \\cos(T\\sqrt{n}) = 1$. Thus, $T = 2k\\pi$ and $T\\sqrt{2} = 2l\\pi$ for some $k, l \\in \\mathbb{N}$. Therefore, $\\sqrt{2} = l/k \\in \\mathbb{Q}$, a contradiction. Hence, there is no integer $n \\ge 2$ such that $f$ is periodic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17849,
    "subject": "Mathematics (Olympiad)",
    "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 $|I(h) - 2f(0)h| \\le Mh^2$ for any $h \\in [0, 1]$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The continuous function $\\varphi : (0, 1] \\to \\mathbb{R}$, $\\varphi(h) = \\frac{I(h) - 2f(0)h}{h^2}$, may be extended 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 $(0, 1]$.\n\nLet $M = \\sup\\{|\\varphi(h)| : 0 < h \\le 1\\}$. Then $|I(h) - 2f(0)h| \\le Mh^2$ for all $h \\in (0, 1]$; the inequality also holds for $h = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17850,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $k$ such that $k^k + 1$ is divisible by $30$. Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "An integer is divisible by $30$ if and only if it is divisible by $2$, $3$, and $5$.\n\n- **Divisibility by $2$:**\n  $k^k + 1$ is even if $k$ is odd, since an odd power of an odd number is odd, and odd $+ 1$ is even. So $k$ must be odd.\n\n- **Divisibility by $3$:**\n  Consider $k$ modulo $3$:\n  - If $k \\equiv 0$ or $1 \\pmod{3}$, then $k^k \\equiv 0$ or $1 \\pmod{3}$, so $k^k + 1 \\not\\equiv 0 \\pmod{3}$.\n  - If $k \\equiv 2 \\pmod{3}$, then $k^k \\equiv 2^k \\pmod{3}$. Since $k$ is odd, $2^k \\equiv 2 \\pmod{3}$, so $k^k + 1 \\equiv 2 + 1 \\equiv 0 \\pmod{3}$.\n  Thus, $k \\equiv 2 \\pmod{3}$.\n\n- **Divisibility by $5$:**\n  Consider $k$ modulo $5$:\n  - If $k \\equiv 0$ or $1 \\pmod{5}$, $k^k \\equiv 0$ or $1 \\pmod{5}$, so $k^k + 1 \\not\\equiv 0 \\pmod{5}$.\n  - If $k \\equiv 2$ or $3 \\pmod{5}$, $k^k$ cycles through $2, 4, 3, 1$ or $3, 4, 2, 1$ depending on $k$, but for odd $k$, $k^k + 1 \\not\\equiv 0 \\pmod{5}$.\n  - If $k \\equiv 4 \\pmod{5}$, $4^k \\equiv (-1)^k \\pmod{5}$. For odd $k$, $4^k \\equiv -1 \\pmod{5}$, so $k^k + 1 \\equiv -1 + 1 \\equiv 0 \\pmod{5}$.\n  Thus, $k \\equiv 4 \\pmod{5}$ and $k$ is odd.\n\n- **Combine conditions:**\n  $k$ is odd, $k \\equiv 2 \\pmod{3}$, $k \\equiv 4 \\pmod{5}$.\n\n  The smallest such $k$ is $29$ (since $k \\equiv 2 \\pmod{3}$ and $k \\equiv 4 \\pmod{5}$, and $k$ odd). The general solution is $k = 30n + 29$, $n = 0, 1, 2, \\ldots$\n\n**Answer:**\n\nAll positive integers $k$ of the form $k = 30n + 29$, $n \\geq 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17851,
    "subject": "Mathematics (Olympiad)",
    "question": "In a row there are 2024 people, numbered 1 to 2024, and each of them either always tells the truth or always lies. Moreover, all 2024 people know from each other whether they are always telling the truth or always lying. At some point, for each number $n$, the person numbered $n$ makes the statement: \"At least $n$ of these people always lie.\"\n\nHow many people always tell the truth?",
    "options": [],
    "answer": "See solution",
    "solution": "$1012$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17852,
    "subject": "Mathematics (Olympiad)",
    "question": "Cut a square by straight lines into 3 pieces so that one could decompose the pieces into an obtuse triangle.\n\nYou are not allowed to move the resulting pieces after the first cut.",
    "options": [],
    "answer": "See solution",
    "solution": "One possible solution is shown below.\n\nIn the figure, $D$ and $E$ are the midpoints of the respective sides of the square. Then $\\Delta ACD = \\Delta FCE$ and $\\Delta ADG = \\Delta BEC$, which means that $\\Delta ABF$, which is clearly obtuse, can be constructed from the three resulting parts.\n\n![](images/ukraine_2015_Booklet_p0_data_7d9e1c56dc.png)\n\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17853,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an $n \\times n$ board where all cells are initially white, Khalid the Painter walks around the board and recolors the visited cells according to the following rules:\n\n- Each walk starts at the bottom-left corner.\n- If he is on a white cell, he paints it black and moves one cell up (or off the board).\n- If he is on a black cell, he paints it white and moves one cell to the right (or off the board).\n\nDetermine the minimum integer $s$ such that after exactly $s$ walks, all the cells of the board become white again.\n\n![](images/Saudi_Arabia_booklet_2023_p39_data_b9ca37c886.png)\n\nFor example, for $n=3$, the states of the board after the initial few walks are shown above.",
    "options": [],
    "answer": "See solution",
    "solution": "Based on the solution of Ali Alramdan (IMO 2023 team member):\n\nLet $f(i, j)$ be the number of times Khalid visits square $(i, j)$. Then $f(0, 0) = s$ for some positive integer $s$, and\n\n$$\nf(i, j) = \\frac{f(i - 1, j) + f(i, j - 1)}{2}.\n$$\n\nBy induction on $i + j$, the color of cell $(i, j)$ after $s$ walks is $f(i, j) \\bmod 2$ (1 = black, 0 = white). Define $g(i, j) = f(i, j) \\cdot 2^{i+j}$ for $0 \\le i, j \\le n-1$, then\n\n$$\ng(i, j) = g(i - 1, j) + g(i, j - 1).\n$$\n\nSince $g(0, 0) = f(0, 0) = s$, by the Pascal triangle formula,\n\n$$\ng(i, j) = s \\cdot \\binom{i+j}{i} \\implies f(i, j) = \\frac{s \\cdot \\binom{i+j}{i}}{2^{i+j}}.\n$$\n\nFor all squares to be white again, $f(i, j)$ must be even for all $i, j \\le n-1$, or\n\n$$\nv_2(s) + v_2\\left(\\binom{i+j}{i}\\right) \\ge i + j + 1.\n$$\n\nTake $s = s_0 = 2^{2n-1-v_2\\left(\\binom{2n-2}{n-1}\\right)}$, the smallest positive integer such that $f(n-1, n-1)$ is even. We show this choice ensures $f(i, j)$ is even for all $0 \\le i, j \\le n-1$. It suffices to prove for $j = n-1$ (the rest follow by linearity). Let $k = v_2\\left(\\binom{2n-2}{n-1}\\right)$. We need\n\n$$\nv_2\\left(\\binom{2n-2-i}{n-1}\\right) \\ge k-1, \\quad \\forall i = 0, 1, \\dots, k.\n$$\n\nNote:\n\n$$\n\\binom{2n-2-i}{n-1} = \\binom{2n-2}{n-1} \\cdot \\frac{(n-1)(n-2)\\dots(n-i)}{(2n-2)(2n-3)\\dots(2n-1-i)}\n$$\n\nEach even term in the denominator has a corresponding half in the numerator, so at most one 2 is lost per term. Thus,\n\n$$\nv_2\\left(\\binom{2n-2-i}{n-1}\\right) \\ge v_2\\left(\\binom{2n-2}{n-1}\\right) - \\left\\lfloor \\frac{i}{2} \\right\\rfloor = k - \\left\\lfloor \\frac{i}{2} \\right\\rfloor \\ge k-i.\n$$\n\nTherefore, the minimum value of $s$ is $s_0$ as defined above. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17854,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 與 $k$ 為正整數。令 $A$ 為由平面上 $2n$ 個相異點所構成的集合，其中任三點不共線。$A$ 中的若干對點之間有連線，使得平面上有 $n^2 + k$ 條相異線段。試證：平面上至少有 $\\frac{4}{3}k^{3/2}$ 個相異三角形，其頂點皆屬於 $A$，且其三邊皆為上述所連線段。",
    "options": [],
    "answer": "See solution",
    "solution": "將問題視為圖 $G = (V, E)$。對於每個點 $v \\in V$，令 $d(v)$ 為該點度數。對於每條邊 $e = uv \\in E$，令 $s(e) = d(u) + d(v)$，也就是其兩端點的度數和。我們拆解為以下三步驟：\n\n1. *引理一：* 對於每一條邊 $e \\in E$，至少有 $s(e) - 2n$ 個三角形以其為邊。\n\n*Proof.* 令 $e$ 的兩端點為 $u$ 和 $v$，則扣除 $uv$ 外，$A$ 和 $B$ 總計還要對其餘 $2n-2$ 個點連出 $s(e)-2$ 條邊，故由排容原理知至少有 $(s(e)-2) - (2n-2) = s(e) - 2n$ 個點同時對 $A$、$B$ 兩點皆有連線，也就是 $s(e) - 2n$ 個三角形。 □\n\n2. *引理二：* $S = \\sum_{e \\in E} s(e) \\ge 2(n^2 + k)^2/n$。\n\n*Proof.* 讓我們對 $S$ 算兩次。注意到點 $v$ 的度數為 $d(v)$，意味著它在 $d(v)$ 條邊上各貢獻了 $d(v)$ 的度數到 $S$ 中，故由柯西不等式與度邊定理，\n\n$$\n\\sum_{e \\in E} s(e) = \\sum_{v \\in V} d(v)^2 \\ge \\frac{1}{2n} \\left( \\sum_{v \\in V} d(v) \\right)^2 = \\frac{1}{2n} (2(n^2 + k))^2 = \\frac{2(n^2 + k)^2}{n}.\n$$\n\n![](images/2023-TWNIMO-Problems_p8_data_7fa9e7037c.png)\n\n3. 現在，由於每個三角形有三條邊，故有\n\n$$\n\\begin{aligned}\n3 \\times \\text{三角形個數} &\\ge \\sum_{e \\in E} (s(e) - 2n) \\ge \\frac{2(n^2 + k)^2}{n} - 2n(n^2 + k) \\\\\n&= \\frac{2k}{n}(n^2 + k) \\ge \\frac{2k}{n}(2n\\sqrt{k}) = 4k^{3/2}.\n\\end{aligned}\n$$\n\n組題者註記：其中 $n \\ge k \\ge 2$ 的條件只是要確保 $C_2^{2n} \\ge n^2 + k$。如果要降難度可以令 $k=n$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17855,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all monotonic functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$\n\n$$\n\\left| \\sum_{i=1}^{n} i (f(x + i + 1) - f^2(x + i)) \\right| < C\n$$\n\nwhere $C > 0$ is independent of $x$ and $f^2(x) = f(f(x))$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the condition, we have:\n\n$$\n\\left| \\sum_{i=1}^{n-1} i (f(x + i + 1) - f^2(x + i)) \\right| < C\n$$\n\nThen,\n\n$$\n\\left| n (f(x + n + 1) - f^2(x + n)) \\right| = \\left| \\sum_{i=1}^{n} i (f(x + i + 1) - f^2(x + i)) - \\sum_{i=1}^{n-1} i (f(x + i + 1) - f^2(x + i)) \\right| < 2C\n$$\n\nThis implies\n\n$$\n|f(x + n + 1) - f^2(x + n)| < \\frac{2C}{n}\n$$\n\nfor every real number $x$ and every natural number $n$.\n\nLet $y \\in \\mathbb{R}$ be arbitrary. Then there exists $x$ such that $y = x + n$. We obtain\n\n$$\n|f(y + 1) - f^2(y)| < \\frac{2C}{n}\n$$\n\nfor every real number $y$ and every natural number $n$. Since this holds for all $n$, we conclude $f(y + 1) = f^2(y)$ for every $y \\in \\mathbb{R}$.\n\nThe function $f$ is monotonic, so it is injective, which implies $f(y) = y + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17856,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist infinitely many positive integers $n$ such that if a prime $p$ divides $n(n + 1)$, then $p^2$ also divides it (i.e., all primes dividing $n(n + 1)$ appear with exponent at least two). Exhibit at least two such values, one even and one odd, for $n > 8$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let's find infinitely many $n$ such that $n(n + 1) = 2m^2$ with $m$ even. This leads to $8m^2 + 1 = (2n + 1)^2$, so we seek solutions to the Pell equation $(2n + 1)^2 - 8m^2 = 1$ with $m$ even. The primitive solution is $(2n + 1, m) = (3, 1)$.\n\nLet $(3 + \\sqrt{8})^k = x_k + y_k \\sqrt{8}$. The recurrence relations are $x_{k+1} = 3x_k + 8y_k$ and $y_{k+1} = x_k + 3y_k$. Since $x_1 = 3$ is odd, $x_k$ is always odd, and $y_{2k}$ is even.\n\nSince $(3 + \\sqrt{8})^2 = 17 + 6\\sqrt{8}$, denote $(3 + \\sqrt{8})^{2k} = A_k + B_k \\sqrt{8}$, with $A_{k+1} = 17A_k + 48B_k$ and $B_{k+1} = 6A_k + 17B_k$. We can take $n = \\frac{A_k - 1}{2}$. The recurrence for $A_k$ is $A_{k+2} = 34A_{k+1} - A_k$, starting with $A_0 = 1$. The first few values are $A_1 = 17$, $A_2 = 577$, $A_3 = 19601$, corresponding to $n = 8, 288, 9800$ (all even).\n\nAlternatively, starting with $(n, n+1) = (8, 9)$, we can build another eligible pair $(4n(n+1), (2n+1)^2)$.\n\nFor odd $n$, consider the Pell equation $(2n + 1)^2 - 12m^2 = 1$ with $3 \\mid m$. The primitive solution is $(2n + 1, m) = (7, 2)$, and $(7 + 2\\sqrt{12})^3 = 1351 + 390\\sqrt{12}$ gives $n = 675$.\n\nThus, for $n > 8$, two examples are $n = 288$ (even) and $n = 675$ (odd).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17857,
    "subject": "Mathematics (Olympiad)",
    "question": "Minimize $x_1 + x_2 + \\cdots + x_{2023}$, where $n = x_1 x_2 \\cdots x_{2023}$ and each $x_i \\ge 1$ is an integer. Prove that for the minimal sum, every $x_i > n^{1/1012}$ must be a prime number. (Since there are finitely many ways to express $n$ as such a product, a minimal representation exists.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = p_1 p_2 \\cdots p_k$, where $p_1 \\ge p_2 \\ge \\cdots \\ge p_k$ are all the prime factors of $n$.\n\nIf $k \\le 2023$, express $n$ as the product of $p_1, p_2, \\ldots, p_k$ and $2023-k$ ones.\n\nIf $k > 2023$, first write $n = p_1 p_2 \\cdots p_k$. Then, repeatedly combine the two smallest factors into their product until only 2023 factors remain. At each step, since the number of factors is $\\ge 2024$, the product of the two smallest numbers is $\\le n^{2/2024} = n^{1/1012}$. Thus, every composite number formed in this process is $\\le n^{1/1012}$.\n\nSince the initial numbers are primes, when the process ends, the 2023 factors are either prime or do not exceed $n^{1/1012}$. Therefore, $\\lambda = \\frac{1}{1012}$ satisfies the problem's conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17858,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_{22}$ be real numbers such that $2^{i-1} \\leq x_i \\leq 2^i$ holds for every $1 \\leq i \\leq 22$. Find the maximum value of\n\n$$\n(x_1 + x_2 + \\dots + x_{22}) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_{22}} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y_i = \\frac{x_i}{2^{i/2}}$ for $i = 1, 2, \\dots, 22$. It is well-known that $f(t) = t + \\frac{1}{t}$ is decreasing on $(0, 1]$ and increasing on $[1, +\\infty)$. For $1 \\leq i \\leq 11$, we have $\\frac{1}{2^{12-i}} \\leq y_i \\leq \\frac{1}{2^{11-i}}$, thus $y_i + \\frac{1}{y_i} \\leq 2^{12-i} + \\frac{1}{2^{12-i}}$. For $12 \\leq i \\leq 22$, we have $2^{i-12} \\leq y_i \\leq 2^{i-11}$, thus $y_i + \\frac{1}{y_i} \\leq 2^{i-11} + \\frac{1}{2^{i-11}}$. Hence,\n\n$$\n\\begin{aligned}\n\\left(\\sum_{i=1}^{22} x_i\\right) \\left(\\sum_{i=1}^{22} \\frac{1}{x_i}\\right) &= \\left(\\sum_{i=1}^{22} y_i\\right) \\left(\\sum_{i=1}^{22} \\frac{1}{y_i}\\right) \\\\\n&\\leq \\frac{1}{4} \\left(\\sum_{i=1}^{22} \\left(y_i + \\frac{1}{y_i}\\right)\\right)^2 \\\\\n&\\leq \\frac{1}{4} \\left(\\sum_{i=1}^{11} \\left(2^{12-i} + \\frac{1}{2^{12-i}}\\right) + \\sum_{i=12}^{22} \\left(2^{i-11} + \\frac{1}{2^{i-11}}\\right)\\right)^2 \\\\\n&= \\left(2^1 + 2^2 + \\dots + 2^{11} + \\frac{1}{2^1} + \\frac{1}{2^2} + \\dots + \\frac{1}{2^{11}}\\right)^2 \\\\\n&= \\left(2^{12} - 1 - \\frac{1}{2^{11}}\\right)^2.\n\\end{aligned}\n$$\n\nWhen\n\n$$\nx_i = \\begin{cases} 2^{i-1}, & 1 \\leq i \\leq 11, \\\\ 2^i, & 12 \\leq i \\leq 22, \\end{cases}\n$$\n\nequality holds. Thus, the maximum value sought is $\\left(2^{12} - 1 - \\frac{1}{2^{11}}\\right)^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17859,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0$ be a positive integer. Define a sequence $(a_n)$ by the following rule: for each $n \\geq 0$, if $a_n$ is a perfect square, then $a_{n+1} = \\sqrt{a_n}$; otherwise, $a_{n+1} = a_n + 3$. For which positive integers $a_0$ does the sequence $(a_n)$ attain some value infinitely many times?",
    "options": [],
    "answer": "See solution",
    "solution": "All positive integers $a_0$ that are multiples of 3.\n\n**Case 1:** $a_0 \\equiv 0 \\pmod{3}$\n\nAll terms of the sequence are multiples of 3. Let $a_i$ be the term of the sequence with minimal value. Suppose for contradiction that $a_i > 9$. Let $x \\geq 1$ be the largest integer such that $3^{2^x} < a_i$. The sequence $a_i, a_{i+1}, a_{i+2}, \\dots$ is formed by adding 3 each time until a perfect square $a_j$ is reached. Note that $a_j$ cannot exceed $3^{2^{x+1}}$ because $3^{2^{x+1}}$ is also a perfect square that is a multiple of 3. It follows that\n\n$$\na_{j+1} = \\sqrt{a_j} \\leq \\sqrt{3^{2^{x+1}}} = 3^{2^x} < a_i,\n$$\n\nwhich contradicts the minimality of $a_i$. Hence $a_i \\leq 9$. The sequence then enters the cycle\n\n$$\n3 \\rightarrow 6 \\rightarrow 9 \\rightarrow 3 \\rightarrow 6 \\rightarrow 9 \\rightarrow \\dots\n$$\n\nwhich contains the number 3 infinitely many times.\n\n**Case 2:** $a_0 \\equiv 2 \\pmod{3}$\n\nNo perfect square is congruent to 2 modulo 3, so $a_{n+1} = a_n + 3$ for all $n$. The sequence is strictly increasing and cannot attain the same value infinitely many times.\n\n**Case 3:** $a_0 \\equiv 1 \\pmod{3}$\n\nNo term of the sequence is a multiple of 3. If $a_i \\equiv 2 \\pmod{3}$ for some $i$, then as in case 2, the sequence is strictly increasing. If $a_n \\equiv 1 \\pmod{3}$ for all $n$, let $a_i$ be the minimal term. Suppose $a_i > 16$. Let $x \\geq 1$ be the largest integer such that $2^{2^x} < a_i$. The sequence $a_i, a_{i+1}, a_{i+2}, \\dots$ is formed by adding 3 each time until a perfect square $a_j$ is reached. $a_j$ cannot exceed $2^{2^{x+1}}$, so\n\n$$\na_{j+1} = \\sqrt{a_j} \\leq \\sqrt{2^{2^{x+1}}} = 2^{2^x} < a_i,\n$$\n\ncontradicting minimality. Hence $a_i \\leq 16$. Inductively, $a_0 > 1$ implies $a_n > 1$ for all $n$. Since\n\n$$\n7 \\rightarrow 10 \\rightarrow 13 \\rightarrow 16 \\rightarrow 4 \\rightarrow 2,\n$$\n\nthe sequence eventually reaches 2, which is not congruent to 1 modulo 3. Thus, it is impossible for $a_i \\equiv 1 \\pmod{3}$ for all $i$.\n\n**Conclusion:** Only when $a_0$ is a multiple of 3 does the sequence attain some value infinitely many times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17860,
    "subject": "Mathematics (Olympiad)",
    "question": "Adrian has drawn a circle in the $xy$-plane whose radius is a positive integer at most $2008$. The origin lies somewhere inside the circle. You are allowed to ask him questions of the form \"Is the point $(x, y)$ inside your circle?\" After each question he will answer truthfully \"yes\" or \"no\". Show that it is always possible to deduce the radius of the circle after at most sixty questions.\n\n*Note: Any point which lies exactly on the circle may be considered to lie inside the circle.*",
    "options": [],
    "answer": "See solution",
    "solution": "The circle must either cross the $x$-axis or touch it at the origin; the same is true of the $y$-axis. If it crosses an axis, then one crossing point on that axis must be positive and one negative.\n\nIt is easy to establish between which two integers the crossing-point lies, say on the positive side of the $x$-axis, by interval bisection. The crossing point can be at most $4016$; for simplicity, it is certainly less than $4096 = 2^{12}$.\n\nWe ask about $(2048, 0)$ and so on, bisecting until we find the two lattice points between which the circle cuts the axis.\n\nSince we are bisecting an interval of size $2^{12}$, this requires $12$ questions.\n\nIf we do this for each half-axis, we get $4$ intervals of length $1$ through which the circle passes. This determines both coordinates of the center of the circle to within $1$.\n\nThis uses $48$ questions, $12$ for each half-axis. We can use the remaining questions to further bisect these intervals, until we know the center lies in a square of length $1/8$.\n\nWe know the position of the center to within a square of size $1/8$. Let $h$ be the distance from this square to the $x$-axis; so the center has $y$-coordinate between $h$ and $h + 1/8$.\n\nConsider one of the intersections with the $x$-axis; we know this lies within an interval of size $1/8$. So we know the difference of $x$-coordinate between this intersection and the center to within $(1/8 + 1/8) = 1/4$. Let $w$ be the minimum value of this distance.\n\nWe can now calculate the radius $r$.\n\nFrom the diagram:\n\n$$\nh^2 + w^2 \\le r^2 \\le \\left(h + \\frac{1}{8}\\right)^2 + \\left(w + \\frac{1}{4}\\right)^2\n$$\n\nand thus\n\n$$\nh^2 + w^2 \\le r^2 \\le h^2 + w^2 + \\frac{h}{4} + \\frac{w}{2} + \\frac{5}{64}.\n$$\n\n![](images/V_Britanija_2008_p25_data_8d57075cc9.png)\n\nWe know that $w$ and $h$ are both positive and less than or equal to $r$. Let $r_0$ be the maximum possible value of $r$:\n\n$$\nr_0^2 = \\left(h + \\frac{1}{8}\\right)^2 + \\left(w + \\frac{1}{4}\\right)^2\n$$\n\nWe have an ambiguity if\n\n$$\n\\left(h + \\frac{1}{8}\\right)^2 + \\left(w + \\frac{1}{4}\\right)^2 - (h^2 + w^2) \\geq r_0^2 - (r_0 - 1)^2,\n$$\n\ni.e., if\n\n$$\n\\frac{h}{4} + \\frac{w}{2} + \\frac{5}{64} \\geq 2r_0 - 1, \\quad (1)\n$$\n\nas then the interval of possible values for $r$ can cover more than one integer.\n\nHowever, if $r_0 = 1$, the right-hand side is $1$ and the left-hand side is at most $\\frac{1}{4} + \\frac{1}{2} + \\frac{5}{64} < 1$.\n\nIf we increase $r_0$ by $k$, the right-hand side increases by $2k$, but the left-hand side by at most $3k/4$. Thus, inequality (1) never holds, so we can always determine the radius from this data.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17861,
    "subject": "Mathematics (Olympiad)",
    "question": "On a plane, a finite number of points are marked, with no three collinear. Assume there exists a non-convex polygon whose vertices are among these points. Prove that there exists a non-convex quadrilateral with all its vertices at the marked points.",
    "options": [],
    "answer": "See solution",
    "solution": "There must be a point inside the convex hull of the marked points; otherwise, every subset would form a convex polygon. Partition the convex hull into triangles by drawing diagonals. Since no three points are collinear, the marked point inside the convex hull lies strictly inside one of these triangles. The three triangle vertices and the interior point form a non-convex quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17862,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, let $T_A$, $T_B$, and $T_C$ be the points where the excircles of $\\triangle ABC$ are tangent to the sides $BC$, $AC$, and $AB$, respectively. Let $O$ be the center of the circumscribed circle of $\\triangle ABC$, and $I$ be the center of its inscribed circle. It is known that $OI \\parallel AC$. Prove that\n$$\n\\angle T_A T_B T_C = 90^\\circ - \\frac{1}{2} \\angle ABC.\n$$\n\n![](images/UkraineMO2019_booklet_p28_data_a9e6d18d00.png \"\")",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I_A$, $I_B$, and $I_C$ be the centers of the excircles, and let $I_1$ be the point symmetric to $I$ with respect to $O$ (see the figure). Consider $\\triangle I_A I_B I_C$. The incenter $I$ is its orthocenter, since the bisectors of the outer and inner angles are perpendicular. Also, $O$ is the center of the nine-point circle of this triangle. Thus, $I_1$ is the circumcenter of $\\triangle I_A I_B I_C$.\n\nWe have $\\angle I_1 I_B C = \\angle I_C I_B B$, which implies $I_1 I_B \\perp AC$, since $\\angle B I_C A = 90^\\circ - \\frac{1}{2} \\angle BCA = \\angle A C I_B$. Hence, the lines $I_B T_B$, $I_A T_A$, and $I_C T_C$ intersect at $I_1$.\n\nClearly, $AII_1C$ is a trapezoid. The points $I$ and $I_1$ are symmetric with respect to the perpendicular bisector of $AC$, since $II_1 \\parallel AC$. Therefore, $AII_1C$ is an isosceles trapezoid, and $\\angle A I_1 C = \\angle A I C = 90^\\circ + \\frac{1}{2} \\angle ABC$. Note also that the quadrilaterals $I_1 T_C A T_B$ and $I_1 T_A C T_B$ are cyclic. Then,\n\n$$\n\\angle T_A T_B T_C = \\angle T_C T_B I_1 + \\angle I_1 T_B T_A = \\angle I_1 A B + \\angle I_1 C B = \\angle A I_1 A C - \\angle ABC = 90^\\circ - \\frac{\\angle ABC}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17863,
    "subject": "Mathematics (Olympiad)",
    "question": "On a plane, two triangles $ABC$ and $BKL$ are arranged such that segment $AK$ is divided into three equal parts by the intersection point of the medians of triangle $ABC$ and the intersection point of the bisectors of triangle $BKL$ ($AK$ is a median of $ABC$, $KA$ is a bisector of $ABKL$), and quadrilateral $KALC$ is a trapezium. Find all angles of triangle $BKL$.\n\n![](images/Ukrajina_2010_p32_data_fa318657be.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the center of $ABKL$. From the properties of the centroid of a triangle and the problem's conditions, $AI = x$ and $KI = 2x$. Let $BA = x$. Then, by the properties of the bisector, $BK = 2x$ and $KC = 2x$. Thus, $KALC$ is a trapezium, so $KA \\parallel LC$. By Thales' theorem, $AL = x$. Therefore, in $ABKL$, $KA$ is both a bisector and a median, since $BA = AL$. Thus, $ABKL$ is an equilateral triangle, and since $BK = 2x = BL$, it is also equilateral. Therefore, all its angles are $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17864,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an equilateral triangle $ABC$ with sides of length $2$. We consider all equilateral triangles $PQR$ with sides of length $1$ satisfying the following properties:\n\n- $P$ lies on the side $AB$,\n- $Q$ lies on the side $AC$, and\n- $R$ lies in the interior or on the edge of the triangle $ABC$.\n\nDescribe the set of all points in the triangle $ABC$ that are centroids of such triangles $PQR$.\n\n![](images/Austrija_2012_p10_data_7047530833.png)\n\n![](images/Austrija_2012_p10_data_dc25930c99.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ and $Q$ be given fulfilling the conditions of the problem. Considering the circumscribed circle $k$ of $APQ$, we note that the centroid $S$ of $APQ$ must lie on $k$, since both $\\angle PAQ = 60^\\circ$ and $\\angle PSQ = 120^\\circ$ hold. Since $|SQ| = |SP|$, the arcs $SQ$ and $SP$ are of equal length, and we therefore have $\\angle SAP = \\angle SAQ = 30^\\circ$. All centroids $S$ therefore lie on the angle bisector $w_{\\alpha}$ of $\\angle BAC$. The most extreme positions of $S$ are assumed when $P$ coincides with $A$ or the midpoint $M_{AB}$ of $AB$. In the latter case, $S$ is also the centroid, i.e., the midpoint, of $ABC$. In the former, $S$ is the centroid of the triangle $AM_{AB} M_{AC}$ (where $M_{AC}$ is the midpoint of $AC$). The set of all centroids of triangles $PQR$ fulfilling all requirements is therefore the middle third of the bisector $w_{\\alpha}$ (which is also the altitude in $ABC$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17865,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(a, b, c)$ of distinct integers such that $a$, $b$, and $c$ are solutions of\n\n$$\nx^3 + a x^2 + b x + c = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using Vieta's formulas to relate the coefficients of the polynomial to its roots, we obtain the following equations:\n\n$$\n\\begin{aligned}\na + b + c &= -a \\\\\nab + bc + ca &= b \\\\\nabc &= -c\n\\end{aligned}\n$$\n\nThe third equation implies that one of the following two cases must hold:\n\n*Case 1: $c = 0$*\n\nThe equations then reduce to $a + b = -a$ and $ab = b$. The second of these implies that either $b = 0$ or $a = 1$. Since $a, b, c$ are distinct, we cannot have $b = 0$. So $a = 1$ and the first equation implies that $b = -2$, which leads to the triple $(a, b, c) = (1, -2, 0)$.\n\n*Case 2: $ab = -1$*\n\nSince $a$ and $b$ are integers, we must have $(a, b) = (1, -1)$ or $(a, b) = (-1, 1)$. Using the first equation above, this yields the triples $(a, b, c) = (1, -1, -1)$ and $(a, b, c) = (-1, 1, 1)$. Since we require $a, b, c$ to be distinct, we do not obtain a solution in this case.\n\nWe can now check that the polynomial $x^3 + x^2 - 2x = x(x - 1)(x + 2)$ does indeed have the roots $1, -2, 0$. So the only triple that satisfies the conditions of the problem is $(a, b, c) = (1, -2, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17866,
    "subject": "Mathematics (Olympiad)",
    "question": "1, 2, 3, ..., 2012 тоонуудыг хооронд нь сэлгэж үүсгэх нийт боломжуудаас яг 1, 2, ..., 48 тоонууд нь их нь бага тооныхоо урд байрладаггүй боловч яг 1, 2, ..., 48, 49 тоонууд нь ийм чанаргүй байх нийт боломжийн тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Ерөнхий тохиолдолд бодъё.\n\n$S(n, k)$-г $1, 2, \\ldots, k$ нь эрэмбээрээ байрласан байх $\\{1, 2, \\ldots, n\\}$-ийн сэлгэмэлийн тоо гэж тэмдэглэе. Олох ёстой тоог $a(n, k)$ гэе. Тэгвэл\n\n$$\na(n, k) = S(n, k) - S(n, k + 1)\n$$\n\nОдоо $S(n, k)$-ийг олъё. $1, 2, \\ldots, k$ тоонуудыг $n$ байрлалд өөрсдийн эрэмбээр нь байрлуулах нийт боломжийн тоо нь $C_n^k$ ба үлдэх $k + 1, k + 2, \\ldots, n$ элементүүдийг $(n-k)!$-аар байрлуулж чадна. Иймд\n\n$$\nS(n, k) = C_n^k \\cdot (n-k)!\n$$\n\nТэгвэл\n\n$$\na(n, k) = C_n^k (n-k)! - C_n^{k+1} (n-k-1)! = \\frac{k \\cdot n!}{(k+1)!}\n$$\n\n$k = 48$, $n = 2012$ үед:\n\n$$\na(2012, 48) = \\frac{48 \\cdot 2012!}{49!}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17867,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the value of\n\n$$\n\\frac{7}{12} + \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us compute the sum step by step:\n\n$$\n\\begin{align*}\n&\\frac{7}{12} + \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\\\\n&= 1 - \\left( \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{2}{9} \\right) \\\\\n&= 1 - \\frac{1}{99} \\\\\n&= \\frac{98}{99}\n\\end{align*}\n$$\n\nSo the answer is $\\frac{98}{99}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17868,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a number in every cell of a $5 \\times 5$ table written under the following conditions:\n\n- Not all numbers are different.\n- There is no row or column where all five numbers are equal.\n- The middle number (the third one) in every row and column equals the mean of the numbers in its row or column.\n\nWhat is the minimum number of cells containing numbers less than the number written in the center of the table?",
    "options": [],
    "answer": "See solution",
    "solution": "3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17869,
    "subject": "Mathematics (Olympiad)",
    "question": "有 $n$ 隻羊和一隻披著羊皮的狼。有些羊是好朋友（好友關係是互相的）。狼的目標是要吃掉所有的羊。首先牠從 $n$ 隻羊中挑一些建立好友關係。接下來的每一天，牠從牠的好友羊中挑一隻吃掉。每當牠吃掉一隻羊 $A$ 時：\n\n(i) 一隻 $A$ 的好友羊如果原本是狼的好友，則會和狼絕交；\n\n(ii) 一隻 $A$ 的好友羊如果原本不是狼的好友，則會和狼建立好友關係。\n\n重複以上動作，直到狼再也沒有好友為止。\n\n試求最大的正整數 $m$（以 $n$ 表示），滿足下列條件：\n\n存在一種 $n$ 頭羊之間的好友關係，使得狼總共有 $m$ 種不同的選擇起始好友羊的方式，讓狼有方法可以吃完所有的羊。",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$2^{n-1}$。\n\n我們首先證明上界。令狼的好友數為 $a$，好友羊的對數為 $b$。注意到每次狼吃掉羊時，$a+b$ 會改變奇偶。因此若狼可以吃掉所有的羊，必須要有 $a+b+n-1 \\equiv 1 \\pmod{2}$（狼在吃完 $n-1$ 隻羊後必可以吃掉最後一隻羊）。因此，在狼的 $2^n$ 中選擇起始好友羊的方式中，至多只有 $2^{n-1}$ 種方法有機會吃掉所有的羊。\n\n以下構造達到上界的羊交友狀況。將羊編號 $1$ 到 $n$，並讓第 $i$ 跟 $i+1$ 號羊結為好友（$i = 1, 2, \\dots, n-1$），其他不交。我們將用數學歸納法證明，在此交友方式下，只要狼起始的好友羊數量為奇數，便一定可以吃完所有的羊：\n\n(i) $n=1$ 時顯然。\n\n(ii) 當 $n > 1$ 時，令 $k$ 為起始狼好友羊中的最大編號。注意到狼可以先依序吃掉 $k, k+1, \\dots, n$ 號羊，則此時的狀況與 $n=k$ 時相同（注意到狼吃了 $n-k+1$ 隻羊，好友羊對數也少了 $n-k+1$，故狼的好友羊數量保持奇偶性）。從而由歸納假設，得證。\n\n又狼選擇奇數隻好友羊的方式有 $2^{n-1}$ 種，故以上範例達到上界。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17870,
    "subject": "Mathematics (Olympiad)",
    "question": "考慮一正整數數列 $a_1, a_2, a_3, \\dots$，滿足 $a_1 = 2021$ 且\n\n$$\n\\sqrt{a_{n+1} - a_n} = \\lfloor \\sqrt{a_n} \\rfloor.\n$$\n\n證明此數列含有無窮多個奇數和無窮多個偶數。",
    "options": [],
    "answer": "See solution",
    "solution": "假設不存在無窮多個奇數或無窮多個偶數，則存在一個 $N$ 使得數列 $a_N, a_{N+1}, a_{N+2}, \\dots$ 擁有同樣的奇偶性。所以當 $n \\ge N$ 時，\n\n$$\nb_n := \\lfloor \\sqrt{a_n} \\rfloor = \\sqrt{a_{n+1} - a_n}\n$$\n\n為偶數。令 $k_n = a_n - b_n^2$，則 $0 \\le k_n \\le 2b_n$。我們有\n\n$$\na_{n+1} = a_n + b_n^2 = 2b_n^2 + k_n \\implies b_{n+1} \\le \\sqrt{2b_n^2 + k_n} < b_{n+1} + 1,\n$$\n\n因此由 $a_{n+2} = 2b_{n+1}^2 + k_{n+1} = 2b_n^2 + k_n + b_{n+1}^2$，\n\n$$\na_{n+2} \\le 4b_n^2 + 2k_n \\le 4b_n^2 + 4b_n < (2b_n + 1)^2\n$$\n\n$$\na_{n+2} > 2b_n^2 + k_n + (\\sqrt{2b_n^2 + k_n} - 1)^2 = 4b_n^2 + 2k_n + 1 - 2\\sqrt{2b_n^2 + k_n}.\n$$\n\n注意到\n\n$$\n4b_n^2 + 2k_n + 1 - 2\\sqrt{2b_n^2 + k_n} \\ge (2b_n - 1)^2 \\iff 2b_n + k_n \\ge \\sqrt{2b_n^2 + k_n},\n$$\n\n而後者顯然是對的，因此我們有 $b_{n+2} = \\lfloor a_{n+2} \\rfloor = 2b_n - 1$ 或 $2b_n$。由於當 $n + 2 \\ge N$ 時，$b_{n+2}$ 為偶數，因此 $b_{n+2} = 2b_n$。所以我們有 $b_{N+2s} = 2^s b_N,\\ b_{N+2s+1} = 2^s b_{N+1}$。\n\n$$\nb_{N+2s}^2 = a_{N+2s+1} - a_{N+2s} = b_{N+2s+1}^2 - b_{N+2s}^2 + k_{N+2s+1} - k_{N+2s},\n$$\n\n我們有\n\n$$\nk_{N+2s+1} - k_{N+2s} = 2b_{N+2s}^2 - b_{N+2s+1}^2 = 2^{2s}(2b_N^2 - b_{N+1}^2),\n$$\n\n由 $0 \\le k_n \\le 2b_n$，我們有\n\n$$\n2^{2s}|2b_N^2 - b_{N+1}^2| \\le 2b_{N+2s+1} + 2b_{N+2s} = 2^{s+1}(b_{N+1} + b_N).\n$$\n\n注意到 $2$ 不是完全平方數，因此 $2b_N^2 - b_{N+1}^2 \\ne 0$，故取 $s$ 足夠大即可得到矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17871,
    "subject": "Mathematics (Olympiad)",
    "question": "On a rectangular board $100 \\times 300$, two people take turns coloring the uncolored cells. The first one paints yellow, the second one paints blue. The coloring is completed when every cell on the board is colored.\n\nA *sequence of* cells is a set of cells in which two consecutive cells share a common side (all cells in the sequence are different). Consider all possible sequences of yellow cells. The *result* of the first player is the number of cells in the sequence of yellow cells of maximum length. The first player's goal is to maximize the result, and the second player's goal is to make the first player's result as small as possible.\n\nProve that if each player strives to achieve his goal, the first player's result will be no more than $200$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's divide the entire board into vertical dominoes, whose larger sides are parallel to the larger side of the board. Consider the second player's strategy: he paints the second square of the domino whose first square was painted by the first player in the previous move. Then, no sequence of yellow cells crosses the horizontal grid lines of the rectangular board that divide the dominoes in half. Thus, the sequence with the maximum number of yellow cells can contain only the adjacent two rows of the board, so its result will not exceed $200$, which is what we needed to prove.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17872,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that for all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x+y)f(x^2 - xy + y^2) = (x+y)(x^2 - xy + y^2)\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ denote the assertion given in the statement. If there exists $a$ such that $f(a) = 0$, then $P(a, 0)$ gives $a = 0$.\n\nNow,\n\n$$\nP(x, x-y) \\implies f(2x - y)f(x^2 - xy + y^2) = (2x - y)(x^2 - xy + y^2)\n$$\n\nDividing this by $P(x, y)$ (for $x + y \\neq 0$), we get\n\n$$\n\\frac{f(2x - y)}{f(x + y)} = \\frac{2x - y}{x + y} \\qquad (1)\n$$\n\nIn equation (1), set $y = 1 - x$. Then $f(3x - 1) = (3x - 1)f(1)$. Since $3x - 1$ is surjective over $\\mathbb{R}$, $f(x) = x f(1)$ for all $x$.\n\nSubstituting this into the original equation, we find $f(1) = \\pm 1$, so the only solutions are $f(x) = x$ and $f(x) = -x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17873,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $\\left(m, n\\right)$ of integers which satisfy the equation\n\n$$\nm^5 - n^5 = 16mn.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If one of $m$ or $n$ is $0$, the other must also be $0$, so $(m, n) = (0, 0)$ is one solution.\n\nIf $mn \\neq 0$, let $d = \\gcd(m, n)$ and write $m = da$, $n = db$ with $a, b \\in \\mathbb{Z}$ and $(a, b) = 1$. The equation becomes\n\n$$\nd^3 a^5 - d^3 b^5 = 16ab\n$$\n\nSo $a \\mid d^3$ and $b \\mid d^3$. Since $(a, b) = 1$, $ab \\mid d^3$, so $d^3 = abr$ for some $r \\in \\mathbb{Z}$. Substituting, we get\n\n$$\nabr^5 - abr^3 = 16ab \\implies r(a^5 - b^5) = 16\n$$\n\nThus, $a^5 - b^5$ must divide $16$, so\n\n$$\na^5 - b^5 = \\pm 1, \\pm 2, \\pm 4, \\pm 8, \\pm 16.\n$$\n\nFor $|a^5 - b^5| = 1$, $a = \\pm 1$, $b = 0$ or $a = 0$, $b = \\pm 1$, which is not possible since $ab \\neq 0$.\n\nFor $|a^5 - b^5| = 2$, $a = 1$, $b = -1$ or $a = -1$, $b = 1$. Then $r = -8$, $d^3 = -8$, so $d = -2$. Thus, $(m, n) = (-2, 2)$.\n\nFor $|a^5 - b^5| > 2$, let $a > b$, $a \\geq 2$, $a = x + 1$, $x \\geq 1$:\n\n$$\n|(x+1)^5 - x^5| = |5x^4 + 10x^3 + 10x^2 + 5x + 1| \\geq 31\n$$\n\nwhich is too large. Thus, the only solutions are $(m, n) = (0, 0)$ and $(-2, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17874,
    "subject": "Mathematics (Olympiad)",
    "question": "A square-shaped orchard with side length $26\\text{ m}$ is fenced in with 3 rows of wire. Is it possible, using the same wire, to fence in a rectangular-shaped orchard with side lengths $95\\text{ m}$ and $60\\text{ m}$?",
    "options": [],
    "answer": "See solution",
    "solution": "For the square-shaped orchard, $3 \\times 4 \\times 26\\text{ m} = 312\\text{ m}$ of wire is used. For the rectangular orchard, $2 \\times (95\\text{ m} + 60\\text{ m}) = 310\\text{ m}$ of wire is needed. Therefore, it is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17875,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of $\\triangle ABC$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. A circle through $A$ and $B$ encloses $\\triangle ABC$ and intersects the line $DE$ at points $P$ and $Q$. Prove that the midpoint of $AB$ lies on the circumcircle of $\\triangle PQF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $AB$. If $DE \\parallel AB$, then $\\triangle ABC$ is isosceles with $CA = CB$, and $F$ coincides with $M$.\n\n![](images/Singapore_2017_p2_data_5254f9489a.png)\n\nConsider the case where $DE \\parallel AB$. Let the lines $DE$ and $AB$ intersect at $X$. By Menelaus' theorem, $\\frac{AX}{XB} = \\frac{BD}{CD} = \\frac{CE}{EA} = 1$.\n\nTherefore, $XA \\cdot BF = XB \\cdot AF$ (since $CD = CE$, $BD = BF$, $AE = AF$).\n\nTherefore, $XA \\cdot (XB - XF) = XB \\cdot (XF - XA)$ (since $BF = XB - XF$, $AF = XF - XA$).\n\nTherefore, $2XA \\cdot XB = (XA + XB) \\cdot XF = 2XM \\cdot XF$.\n\nTherefore, $XM \\cdot XF = XA \\cdot XB = X'P \\cdot XQ \\implies P, Q, F, M$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17876,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer, $n \\geq \\frac{k(k+1)}{2}$, and $h > 0$. Suppose a set of points $(x_j, y_j)$ for $j = 1, 2, \\dots, n$ in the quarter plane is such that any two points are at a distance exceeding $h\\sqrt{2}$. Show that the centroid $(\\bar{x}, \\bar{y})$ satisfies:\n\n$$\n\\bar{x} + \\bar{y} \\geq \\frac{2}{3}(k-1)h.\n$$\n\nThe original question follows by contradiction for $k = 62$ (or $k = 63$), $n = 2021$, and $h = 1/20$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof of Theorem:** For each point $(x_j, y_j)$, define\n\n$$\nm_j = \\left\\lfloor \\frac{x_j}{h} \\right\\rfloor + \\left\\lfloor \\frac{y_j}{h} \\right\\rfloor.\n$$\n\nThen\n\n$$\nx_j + y_j \\geq h m_j.\n$$\n\nBy hypothesis, no two points have a distance less than or equal to $h\\sqrt{2}$, so there can be at most one point in any $h \\times h$ square. There can be at most one $j$ with $m_j = 0$, at most two with $m_j = 1$, at most three with $m_j = 2$, and so on, up to at most $k$ with $m_j = k-1$. The remaining points (at least $n - \\frac{k(k+1)}{2}$ of them) have $m_j \\geq k$.\n\nThus,\n\n$$\n\\sum_{j=1}^{n} (x_j + y_j) \\geq \\sum_{j=1}^{n} h m_j \\geq h \\sum_{i=1}^{k} i(i-1) + \\left(n - \\frac{k(k+1)}{2}\\right) k h.\n$$\n\nRecall that\n\n$$\n\\sum_{i=2}^{k} i(i-1) = 2 \\sum_{i=2}^{k} \\binom{i}{2} = 2 \\sum_{i=2}^{k} \\left[ \\binom{i+1}{3} - \\binom{i}{3} \\right] = 2 \\binom{k+1}{3}.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\n\\frac{1}{h} \\sum_{j=1}^{n} (x_j + y_j) &\\geq \\frac{(k+1)k(k-1)}{3} + \\left(n - \\frac{k(k+1)}{2}\\right) k \\\\\n&= \\left(k - \\frac{2(k-1)}{3}\\right) \\left(n - \\frac{k(k+1)}{2}\\right) + \\frac{2(k-1)}{3} n \\\\\n&\\geq \\frac{2(k-1)}{3} n.\n\\end{align*}\n$$\n\nMultiplying by $h$ and dividing by $n$ gives the theorem as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17877,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum positive integer that is divisible by $7$, all of whose digits are odd, and whose digits sum to $2015$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** To maximize the number, we want as many digits as possible, each digit being odd. Since each digit is at least $1$, the maximum number of digits is $2015$. However, a number with $2015$ digits, all $1$s, sums to $2015$, but $111\\ldots1$ (2015 times) is not divisible by $7$.\n\nIf we try $2014$ digits, the sum of $2014$ odd digits cannot be odd, so that's impossible. The next possibility is $2013$ digits. To get a sum of $2015$, we use $2012$ digits $1$ and one digit $3$ (since $2012 \\times 1 + 3 = 2015$). To maximize the number, place the $3$ as far left as possible.\n\nLet the number be $A = \\underline{11\\ldots1} + 2 \\cdot 10^k$, where $k \\leq 2012$, and $A$ is divisible by $7$. We seek the largest $k$ such that $A$ is divisible by $7$.\n\nSince $10 \\equiv 3 \\pmod{7}$, the sequence $2 \\cdot 10^k$ modulo $7$ cycles every $6$ steps:\n\n$$\n2 \\cdot 10^0 = 2 \\pmod{7}, \\\\\n2 \\cdot 10^1 = 6 \\pmod{7}, \\\\\n2 \\cdot 10^2 = 4 \\pmod{7}, \\\\\n2 \\cdot 10^3 = 5 \\pmod{7}, \\\\\n2 \\cdot 10^4 = 1 \\pmod{7}, \\\\\n2 \\cdot 10^5 = 3 \\pmod{7}, \\\\\n2 \\cdot 10^6 = 2 \\pmod{7}, \\dots\n$$\n\nWe need $2 \\cdot 10^k \\equiv 1 \\pmod{7}$, which happens when $k = 6l + 4$. The largest such $k$ less than $2012$ is $k = 2008$. Thus, the maximum number is $11113\\overline{11\\ldots1}$, where there are $2008$ ones after the $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17878,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\langle a_n \\rangle$ of positive integers is given, such that $a_1 = 1$ and $a_{n+1}$ is the smallest positive integer such that\n\n$$\n\\text{lcm}(a_1, a_2, \\dots, a_n, a_{n+1}) > \\text{lcm}(a_1, a_2, \\dots, a_n).\n$$\n\nWhich numbers are contained in the sequence?",
    "options": [],
    "answer": "See solution",
    "solution": "The first few elements of the sequence are:\n\n$1, 2, 3, 4, 5, 7, 8, 9, 11, \\ldots$\n\nThe first two positive integers not contained in the sequence are $6$ and $10$. These do not increase the lcm when considered, and will not do so later. Thus, any number omitted at its turn cannot appear later.\n\nEach prime is included in the sequence, since adding a new prime always increases the lcm. The same holds for any power of a prime. Therefore, all powers of primes are included in $\\langle a_n \\rangle$.\n\nAny integer $N$ with at least two distinct prime divisors will have all its prime power divisors already included, so adding $N$ does not increase the lcm. Thus, such $N$ are not included.\n\nIn summary, the sequence $\\langle a_n \\rangle$ consists of $1$ and all powers of primes in ascending order.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17879,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any integer $n \\ge 4$, there exists a polynomial of degree $n$,\n\n$$\nf(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0\n$$\n\nwith the following properties:\n\n1. $a_0, a_1, \\dots, a_{n-1}$ are all positive integers.\n2. For any positive integer $m$ and any $k$ ($k \\ge 2$) positive integers $r_1, r_2, \\dots, r_k$ that are all different from each other, we have\n   $$\n   f(m) \\neq f(r_1)f(r_2)\\cdots f(r_k).\n   $$",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nf(x) = (x+1)(x+2)\\cdots(x+n) + 2.\n$$\n\nObviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We will prove that $f(x)$ has property (2).\n\nFor any integer $t$, since $n \\ge 4$, there is always a multiple of $4$ among any $n$ consecutive numbers $t+1, t+2, \\dots, t+n$. Thus, $f(t) \\equiv 2 \\pmod{4}$.\n\nFor any $k$ ($k \\ge 2$) positive integers $r_1, r_2, \\dots, r_k$,\n$$\nf(r_1)f(r_2)\\cdots f(r_k) \\equiv 2^k \\equiv 0 \\pmod{4}.\n$$\n\nOn the other hand, for any positive integer $m$, $f(m) \\equiv 2 \\pmod{4}$. Therefore,\n$$\nf(m) \\not\\equiv f(r_1)f(r_2)\\cdots f(r_k) \\pmod{4},\n$$\nwhich implies $f(m) \\neq f(r_1)f(r_2)\\cdots f(r_k)$. Thus, the required $f(x)$ exists and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17880,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 3$ 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": "See solution",
    "solution": "We will show that the smallest number of distinct values John can write 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 $n \\geq 5$. 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 is at most $n-1$.\n\nTo prove equality, suppose by contradiction that the number sought 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 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\nThe 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 above) 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 $n \\geq 5$, there exists a vertex $E$ different from $A, B, C, D$. Let $y = a_{AE} \\neq \\frac{d}{2}$. 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\nLooking at triangle $ACE$, we 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, $a_{DE} = y - \\frac{d}{2}$. Finally, 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": 17881,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be four integers such that\n\n$$7a + 8b = 14c + 28d.$$ \n\nProve that $a \\cdot b$ is a multiple of $14$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the equation modulo $2$ and modulo $7$, respectively, and obtain\n\n$$\n\\begin{aligned}\na &\\equiv 0 \\pmod{2}, \\\\\nb &\\equiv 0 \\pmod{7}.\n\\end{aligned}\n$$\n\nWe conclude that $a$ is even and $b$ is a multiple of $7$. Therefore, $ab$ is divisible by $2 \\cdot 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17882,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of an acute $\\triangle ABC$ touches the sides $AB$, $BC$, and $CA$ at points $P$, $Q$, and $R$, respectively. The orthocenter $H$ of $\\triangle ABC$ lies on the segment $QR$.\n\n(a) Prove that $PH \\perp QR$.\n\n(b) Let $I$ and $O$ be the incenter and circumcenter of $\\triangle ABC$, and $N$ the common point of $AB$ and the excircle to this side. Prove that the points $I$, $O$, and $N$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $\\angle RAH = \\angle QBH$ and\n\n$$\n\\angle ARH = 180^{\\circ} - \\angle CRQ = 180^{\\circ} - \\angle CQR = \\angle BQH,\n$$\n\nthen $\\triangle ARH \\sim \\triangle QBH$. Hence\n\n$$\n\\frac{AH}{BH} = \\frac{AR}{BQ} = \\frac{AP}{BP}\n$$\n\nand $HP$ is the bisector of $\\angle AHB$. Then\n\n$$\n\\angle RHP = \\angle RHA + \\angle AHP = \\angle QHB + \\angle BHP = \\angle QHP\n$$\n\nwhich implies that $PH \\perp RQ$.\n\nb) If $AC = AB$, then $N \\equiv P \\equiv M$ and the points $I$, $O$, and $N$ lie on the bisector of $AB$.\n\nLet now $AC \\neq AB$. Since $PH \\perp RQ$ and $CI \\perp RQ$, it follows that $HP \\parallel CI$. On the other hand, $CH \\parallel IP$ and hence $CHPI$ is a parallelogram. Then $CH = IP$. If $M$ is the midpoint of $AB$, then $CH = 2OM = 2R \\cos \\gamma$ and so $IP = 2OM$. Since $AP = BN$, then $M$ is the midpoint of $PN$. Therefore $O$ is the midpoint of $IN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17883,
    "subject": "Mathematics (Olympiad)",
    "question": "We call an isosceles trapezoid *interesting* if it is inscribed in the unit square $ABCD$ such that one vertex of the trapezoid lies on each side of the square, and if the lines joining the mid-points of adjacent sides of the trapezoid are parallel to the sides of the square. Determine all interesting trapezoids and their areas.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E, F, G,$ and $H$ be the mid-points of $PQ, QR, RS,$ and $SP$ respectively. Since the sides of $EFGH$ are parallel to the sides of $ABCD$, $EFGH$ is certainly a rectangle. Since $PQRS$ is isosceles, the line $FH$ joining the parallel sides must be an axis of symmetry of the trapezoid, and therefore also of the rectangle $EFGH$, which means that $EFGH$ must be a square.\n\n![](images/AustriaMO2012_p5_data_5973484bcf.png)\n\nWe now note that triangles $RGF$ and $RSQ$ are homothetic with center $R$, which means that $SQ$ is parallel to $GF$, and therefore to the sides $AB$ and $CD$ of the square. Similarly, $PR$ is parallel to $BC$ and $DA$, and therefore the diagonals of $PQRS$ are perpendicular and of unit length. Since they divide $ABCD$ into four squares, half of whose area is inside $PQRS$, we see that the area of $PQRS$ must be half the area of the unit square $ABCD$, and therefore equal to $\\frac{1}{2}$.\n\nIn the square $EFGH$, the diagonal $EG$ is also the mid-parallel of the trapezoid $PQRS$, and therefore parallel to $PS$ and $RQ$. Also, as a diagonal in the square, the angles between $EG$ on the one hand and $EF$ and $EH$ on the other are equal to $45^\\circ$. Since $EF$ and $EH$ are parallel to the sides of $ABCD$, we see that the angles between the parallel sides $PS$ and $QR$ of the trapezoid on the one hand and the sides of $ABCD$ on the other are all $45^\\circ$. We therefore see that triangles $APS$ and $CRQ$ are both isosceles right triangles, and that $HF$ lies on the diagonal $AC$ of the square.\n\nSummarizing, we see that the interesting trapezoids are exactly those whose axis of symmetry lies on a diagonal of the square $ABCD$ and whose diagonals are parallel to the sides of $ABCD$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17884,
    "subject": "Mathematics (Olympiad)",
    "question": "After any interchange of coins, the set $(S, G, P)$ can be one of the following:\n\n1) $(S+2, G-1, P-1)$\n2) $(S-2, G+1, P+1)$\n3) $(S-1, G+2, P-1)$\n4) $(S+1, G-2, P+1)$\n5) $(S-1, G-1, P+2)$\n6) $(S+1, G+1, P-2)$\n\nGiven the initial set $(S, G, P) = (16, 15, 14)$, what are the possible numbers of platinum coins $P$ that Bill can have when he has $G = 0$ gold coins, after a sequence of such interchanges?\n\n% IMAGE: ![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The difference $G - P$ is invariant modulo 3 under any allowed move. Initially, $G - P = 15 - 14 = 1$, so $G - P \\equiv 1 \\pmod{3}$ always. For $G = 0$, this means $P \\equiv 2 \\pmod{3}$. Among the numbers from 15 to 30, the possible values for $P$ are 17, 20, 23, 26, and 29.\n\nTo show all these values are attainable, consider the following sequences (as shown in the tables):\n\n$$\n\\begin{array}{cccccccccc}\nS & 23 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 \\\\\nG & 1 & 2 & 0 & 1 & 2 & 0 & 1 & 2 & 0 \\\\\nP & 21 & 22 & \\textbf{23} & 21 & 19 & \\textbf{20} & 18 & 16 & \\textbf{17}\n\\end{array}\n$$\n\n$$\n\\begin{array}{cccccccccc}\nS & 23 & 21 & 22 & 20 & 18 & 19 & 17 & 15 & 16 \\\\\nG & 1 & 2 & 0 & 1 & 2 & 0 & 1 & 2 & 0 \\\\\nP & 21 & 22 & \\textbf{23} & 24 & 25 & \\textbf{26} & 27 & 28 & \\textbf{29}\n\\end{array}\n$$\n\nThus, when Bill has 0 gold coins, he can have exactly 17, 20, 23, 26, or 29 platinum coins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17885,
    "subject": "Mathematics (Olympiad)",
    "question": "A teacher distributes candies to children in a circle of 30 seats, skipping an increasing number of children each time: the $k$-th candy is dropped after skipping $1 + 2 + \\dots + (k-1)$ children. For each $k$, to which child (seat number) does the teacher give the $k$-th candy? How many children never receive any candy in the first 29 drops?",
    "options": [],
    "answer": "See solution",
    "solution": "When the $k$-th candy is dropped, the teacher has skipped $1 + 2 + \\dots + (k-1) = \\frac{k(k-1)}{2}$ children. The candy is dropped behind $a_i$, where $i = k + \\frac{k(k-1)}{2} = \\frac{k(k+1)}{2}$ (mod 30). Thus, $2i \\equiv k(k+1)$ (mod 60). The $i$-th and $j$-th candies are given to the same child if $i \\equiv j$ (mod 60), so the sequence is periodic with period 60. Note that $k(k+1) \\equiv (60-k-1)(60-k) \\pmod{60}$, so the $k$-th and $(60-k-1)$-th candies go to the same child. Compute $\\frac{k(k+1)}{2}$ (mod 30) for $k = 1, \\ldots, 29$ to get the sequence of children receiving the first 29 candies:\n\n1, 3, 6, 10, 15, 21, 28, 6, 15, 25, 6, 18, 1, 15, 30,\n16, 3, 21, 10, 30, 21, 13, 6, 30, 25, 21, 18, 16, 15\n\nThus, 18 children never receive any candy.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17886,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be integers greater than $2$. Prove that there exists a positive integer $k$ and a finite sequence $n_1, n_2, \\dots, n_k$ of positive integers such that $n_1 = a$, $n_k = b$, and $(n_i + n_{i+1}) \\mid n_i n_{i+1}$ for every $i = 1, 2, \\dots, k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We'll write $a \\leftrightarrow b$ if there exists such a sequence. It is easy to see that $\\leftrightarrow$ is an equivalence relation. Notice that $n \\leftrightarrow 2n$ for every integer $n \\ge 3$ because in that case the desired finite sequence is\n\n$$\nn_1 = n,\\ n_2 = n(n-1),\\ n_3 = n(n-1)(n-2),\\ n_4 = n(n-2),\\ n_5 = 2n\n$$\n\nFor every $n \\ge 4$ we have $n' = (n-1)(n-2) \\ge 3$ and we obtain $n' \\leftrightarrow 2n'$. For $n \\ge 4$:\n\n$$\n\\begin{aligned}\nn_1 &= n,\\\\ n_2 &= n(n-1),\\\\ n_3 &= n(n-1)(n-2),\\\\ n_4 &= n(n-1)(n-2)(n-3),\\\\ n_5 &= 2(n-1)(n-2) = 2n'\n\\end{aligned}\n$$\n\ni.e., $n \\leftrightarrow 2n'$, and because $n_1' = n' = (n-1)(n-2)$, $n_2' = n-1$, we obtain $n' \\leftrightarrow n-1$. From the previous discussion we get that $n \\leftrightarrow 2n'$, $2n' \\leftrightarrow n'$, $n' \\leftrightarrow n-1$, and from the transitivity of $\\leftrightarrow$ we obtain $n \\leftrightarrow n-1$ for every integer $n \\ge 4$. Because $\\leftrightarrow$ is symmetric, $n \\leftrightarrow n+1$ for every integer $n \\ge 3$. Hence, from the transitivity of $\\leftrightarrow$, we obtain the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17887,
    "subject": "Mathematics (Olympiad)",
    "question": "We want to cover a table of size $4 \\times 4$ with dominoes of the following shape:\n\n![](images/Slovenija_2013_p14_data_3264f20cb8.png)\n\nThe dominoes may be reflected or rotated. Dominoes may overlap, but must not extend over the edges of the table. What is the minimum number of dominoes needed to cover the table?",
    "options": [],
    "answer": "See solution",
    "solution": "At least 5 dominoes are needed. If we try to cover the table with only 4 dominoes, they cannot overlap or extend beyond the table, since 4 dominoes cover exactly 16 squares. In this case, there are only two possible ways to cover the top left corner, as shown below. In both cases, the field marked with * cannot be covered:\n\n![](images/Slovenija_2013_p15_data_5ff4225a51.png)\n\n![](images/Slovenija_2013_p15_data_6fc89328d1.png)\n\n![](images/Slovenija_2013_p15_data_4d5fb42e97.png)\n\nIt is easy to see that 5 dominoes suffice, as demonstrated by the last figure.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17888,
    "subject": "Mathematics (Olympiad)",
    "question": "We can place four tiles as shown in the first picture. Each uncovered square has at least one covered neighbour. Is it possible to achieve this with fewer than four tiles?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose we use fewer than four tiles. Place a tile and mark all its neighbouring squares. In any row, at most three squares can be either covered or adjacent to a covered square, and these must be consecutive. The same holds for columns and diagonals. Since the board has four corners, with only three tiles, at least one tile would need to cover or be adjacent to at least two corners, which is impossible. Thus, fewer than four tiles cannot achieve the requirement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17889,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a function from positive integers to positive integers, and let $f^m$ denote $f$ applied $m$ times. Suppose that for every positive integer $n$, there exists a positive integer $k$ such that $f^{2k}(n) = n + k$. Let $k_n$ be the smallest such $k$. Prove that the sequence $k_1, k_2, \\dots$ is unbounded.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g(n) = f^{2k_n}(n) = n + k_n$. Suppose $g(a) = g(b)$ and $a \\neq b$. Without loss of generality, assume $a < b$. Then $a + k_a = f^{2k_a}(a) = f^{2k_b}(b) = b + k_b$, so $f^{2(k_a - k_b)}(a) = b = a + (k_a - k_b)$. Since $k_a - k_b < k_a$, this contradicts the minimality of $k_a$. Therefore, $g$ is injective.\n\nWe call a sequence $n, g(n), g(g(n)), g(g(g(n))), \\dots$ a *chain*. Consider the set\n\n$$\nS = \\{1, f(1), f^2(1), f^3(1), \\dots\\}.\n$$\n\nWe will show that we can partition $S$ into infinitely many pairwise disjoint chains.\n\nLet $T$ be the set of terms in this sequence that are not of the form $g(f^k(1))$ for some $k$. The set $T$ is nonempty because $1 \\in S$, and $g(n) \\ge n > 1$ for all $n$. For each $x \\in T$, take the chain $x, g(x), g(g(x)), \\dots$ and let $U$ be the set of all these chains. Suppose $f^m(1) \\in S$ is not in a chain of $U$, and take the smallest such $m$. If $f^m(1) \\in T$, it is in a chain of $T$, a contradiction. Thus, $f^m(1) \\notin T$ and it is equal to $g(f^k(1))$ for some $k < m$. But $f^k(1)$ is in a chain of $U$, so $g(f^k(1)) = f^m(1)$ is also in a chain of $U$, a contradiction. Thus, all elements of $S$ are in a chain of $U$.\n\nNext, we show the chains of $U$ are pairwise disjoint. Suppose that sequences $x, g(x), g(g(x)), \\dots$ and $y, g(y), g(g(y)), \\dots$ had a common element, meaning $g^k(x) = g^m(y)$ for some $k, m$. Without loss of generality, assume $k < m$. Because $g$ is injective, $x = g^{m-k}(y)$. Since $x \\in T$, it is not of the form $g(z)$ where $z \\in S$. But $g^{m-k-1}(y) \\in S$, so we have a contradiction and the chains of $U$ are disjoint. So $U$ is a partition of $S$ into chains.\n\nNow we show that $T$ is infinite and that there are infinitely many chains of $U$. Suppose $T$ were finite and let it be $\\{f^{t_1}(1), f^{t_2}(1), \\dots, f^{t_n}(1)\\}$ with $t_1 < t_2 < \\dots < t_n$. Choose an integer $N > 2 \\max\\{T\\}$. Consider $f^m(1)$ for $m \\le N$. As it is in a chain of $U$, it is in a chain starting with $f^{t_i}(1)$ for some $i$. By repeatedly applying the condition $f^{2k_n}(n) = n + k_n = g(n)$ on this chain, we have $f^m(1) = f^{t_i}(1) + \\frac{m-t_i}{2}$. But $f^{t_i}(1) + \\frac{m-t_i}{2} \\le \\max\\{T\\} + \\frac{N-t_i}{2} < N$. Therefore, the $N+1$ terms $1, f(1), f^2(1), \\dots, f^N(1)$ are all less than $N$, contradicting the fact that $f$ is one-to-one. Hence $T$ must be infinite.\n\nThe sequence $1, f(1), f^2(1), f^3(1), \\dots$ can be partitioned into infinitely many disjoint chains. This implies that the differences between consecutive elements of one of the chains must be arbitrarily large. That is, for any real $c$, we can find an $n$ in a chain of $U$ with $g(n) - n > c$. But $g(n) - n = k_n$, so $k_n$ can be arbitrarily large and the sequence is unbounded as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17890,
    "subject": "Mathematics (Olympiad)",
    "question": "A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \\le 8$. A target $T$ is the region where $(x^2 + y^2 - 25)^2 \\le 49$. A dart is thrown and lands at a random point in $B$. The probability that the dart lands in $T$ can be expressed as $\\frac{m}{n} \\cdot \\pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?\n\n(A) 39 (B) 71 (C) 73 (D) 75 (E) 135",
    "options": [],
    "answer": "See solution",
    "solution": "The region $B$ is a square with intercepts $(\\pm8, 0)$ and $(0, \\pm8)$. The area of this square is $(8\\sqrt{2})^2 = 128$.\n\nTaking square roots shows that region $T$ is the set of points that satisfy\n\n$$\n25 - 7 \\le x^2 + y^2 \\le 25 + 7,\n$$\n\nwhich is an annulus (ring) with inner radius $\\sqrt{18}$ and outer radius $\\sqrt{32}$. Its area is $32\\pi - 18\\pi = 14\\pi$. Note that $T$ is internally tangent to $B$ at the four points $(\\pm4, \\pm4)$. The required probability is the ratio of the area of $T$ to the area of $B$, namely $\\frac{14\\pi}{128} = \\frac{7}{64}\\pi$. The requested sum is $7 + 64 = 71$.\n\n![](images/2024_AMC10B_Solutions_p6_data_b0513c0722.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17891,
    "subject": "Mathematics (Olympiad)",
    "question": "設點 $H$ 為 $\\triangle ABC$ 的垂心，且點 $P$ 是 $\\triangle ABC$ 外接圓上異於頂點 $A, B, C$ 的任意點。令點 $E, F$ 分別是 $P$ 對 $BC, AB$ 邊的垂足。試證：直線 $EF$ 平分線段 $PH$。",
    "options": [],
    "answer": "See solution",
    "solution": "如圖，$H$ 為 $\\triangle ABC$ 的垂心。設 $D$ 為過 $A$ 點的高的垂足，並令直線 $AD$ 交外接圓於 $G$ 點。設直線 $PG$ 分別交 $EF, BC$ 於 $M, K$ 點。則\n\n1. $PE \\parallel AG$, $HD = DG$, $\\angle 7 = \\angle 6$。\n\n2. $\\angle PFB = \\angle PEB = 90^\\circ$，故 $P, B, E, F$ 四點共圓。可得 $\\angle 1 = \\angle 4 = \\angle 3 = \\angle 2$，即 $M$ 為 $PK$ 的中點。\n\n3. $\\angle 6 = \\angle 7 = 90^\\circ - \\angle 3 = 90^\\circ - \\angle 1 = \\angle 5$，故 $EF \\parallel KH$。\n\n4. $M$ 為 $PK$ 的中點，又 $EF \\parallel KH$，故 $N$ 為 $PH$ 的中點，得證 $EF$ 平分 $PH$。$\\square$\n\n![](images/13-2J-ind_p3_data_a79e48609a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17892,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = n(n+2)(n+4)$ and let $b_n$ be the number of positive divisors of $a_n$.\n\nWe can check that:\n\n- $b_1 = 4$\n- $b_2 = 10$\n- $b_3 = 8$\n- $b_4 = 14$\n- $b_5 = 12$\n- $b_6 = 24$\n- $b_7 = 12$\n- $b_8 = 28$\n- $b_9 = 12$\n- $b_{10} = 40$\n\nRecall: If $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ is the prime factorization of a positive integer $m$, then the number of positive divisors of $m$ is:\n\n$$\n(\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_k + 1)\n$$\n\nIf $m$ divides a positive integer $l$, then $l$ has at least as many divisors as $m$.\n\n**Question:** For which $n$ does $a_n$ have at most 15 positive divisors?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n \\geq 11$.\n\n**Case 1:** $n$ is even, $n = 2k$.\n\nThen $a_n = 2^3 k(k+1)(k+2)$. Among $k, k+1, k+2$, at least one is divisible by 2 and exactly one is divisible by 3. Since $k \\geq 6$, $k, k+1, k+2$ cannot all be powers of 2 or 3, so $k(k+1)(k+2)$ has a prime divisor $p \\neq 2,3$. Thus, $2^4 \\cdot 3 \\cdot p$ divides $a_n$, so $a_n$ has at least $5 \\cdot 2 \\cdot 2 = 20$ positive divisors.\n\n**Case 2:** $n \\geq 11$ is odd.\n\nThe numbers $n$, $n+2$, $n+4$ are pairwise relatively prime. One of them is divisible by 3, and that number has at least one other prime divisor $p$ (or else is a power of 3, in which case, since $n \\geq 11$, it is divisible by $3^3$). Let $q$ and $r$ be prime divisors of the other two numbers. In the first case, $a_n$ is divisible by $3pqr$, so $a_n$ has at least $2 \\cdot 2 \\cdot 2 \\cdot 2 = 16$ divisors. In the second case, $a_n$ is divisible by $3^3 q r$, so $a_n$ has at least $4 \\cdot 2 \\cdot 2 = 16$ divisors.\n\nTherefore, $a_n$ has at most 15 positive divisors only for $n = 1, 2, 3, 4, 5, 7, 9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17893,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1A_2A_3 \\dots A_{11}$ be an 11-sided non-convex simple polygon with the following properties:\n\n* For every integer $2 \\leq i \\leq 10$, the area of $\\triangle A_iA_{i+1}$ is equal to $1$.\n* For every integer $2 \\leq i \\leq 10$, $\\cos(\\angle A_iA_{i+1}) = \\frac{12}{13}$.\n* The perimeter of the 11-gon $A_1A_2A_3 \\dots A_{11}$ is equal to $20$.\n\nThen $A_1A_2 + A_1A_{11} = \\frac{m\\sqrt{n-p}}{q}$, where $m, n, p$, and $q$ are positive integers, $n$ is not divisible by the square of any prime, and no prime divides all of $m, p$, and $q$. Find $m+n+p+q$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the second condition implies $\\angle A_2A_1A_3 = \\angle A_3A_1A_4 = \\dots = \\angle A_{10}A_1A_{11}$. For ease of notation, let $\\theta = \\angle A_2A_1A_3$. For $2 \\leq i \\leq 9$, the first condition gives\n\n$$\n\\text{Area}(\\triangle A_i A_1 A_{i+1}) = \\frac{1}{2} A_1A_i \\cdot A_1A_{i+1} \\sin \\theta = 1\n$$\n\nand\n\n$$\n\\text{Area}(\\triangle A_{i+1} A_1 A_{i+2}) = \\frac{1}{2} A_1A_{i+1} \\cdot A_1A_{i+2} \\sin \\theta = 1.\n$$\n\nSetting these equal to each other yields\n\n$$\n\\frac{1}{2} A_1A_i \\cdot A_1A_{i+1} \\sin \\theta = \\frac{1}{2} A_1A_{i+1} \\cdot A_1A_{i+2} \\sin \\theta \\implies A_1A_i = A_1A_{i+2}.\n$$\n\nBecause $\\angle A_iA_1A_{i+1} = \\angle A_{i+1}A_1A_{i+2}$, it follows by SAS that $\\triangle A_iA_1A_{i+1} \\cong \\triangle A_{i+2}A_1A_{i+1}$. In particular, this congruence implies $A_2A_3 = A_3A_4 = \\dots = A_{10}A_{11}$.\n\n![](images/2025AIME_II_Solutions_p9_data_3ddba08779.png)\n\nLet $x = A_1A_2$ and $y = A_1A_{11}$. The third condition yields\n\n$$\n\\text{Perimeter}(A_1A_2A_3 \\dots A_{11}) = 20 \\implies x + y + 9A_2A_3 = 20. \\tag{*}\n$$\n\nApplying the Law of Cosines to $\\triangle A_1A_2A_3$ gives\n\n$$\n\\begin{aligned}\n(A_2A_3)^2 &= (A_1A_2)^2 + (A_1A_3)^2 - 2A_1A_2 \\cdot A_1A_3 \\cos(\\angle A_2A_1A_3) \\\\\n&= x^2 + y^2 - 2xy \\cdot \\frac{12}{13} \\\\\n&= (x + y)^2 - 2xy \\cdot \\frac{25}{13}.\n\\end{aligned}\n$$\n\nThe area condition gives $\\frac{1}{2}xy \\sin \\theta = 1$, and because $\\sin \\theta = \\frac{5}{13}$, this implies $xy = \\frac{26}{5}$. Thus\n\n$$\nA_2A_3 = \\sqrt{(x + y)^2 - 2 \\cdot \\frac{26}{5} \\cdot \\frac{25}{13}} = \\sqrt{(x + y)^2 - 20}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17894,
    "subject": "Mathematics (Olympiad)",
    "question": "Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We first observe that $(1, a, b)$ is not a solution whenever $1 < a < b$. Otherwise, we would have $\\frac{1}{a} + \\frac{1}{b} = l \\cdot \\frac{1}{1} = l$ for some integer $l$. This gives $\\frac{a+b}{ab} = l$, which implies $a \\mid b$ and $b \\mid a$. But then $a = b$, contradicting $a \\neq b$. Thus, the least number should be $2$.\n\nIt is easy to verify that $(2, 3, 4)$ and $(2, 3, 5)$ are not solutions, and $(2, 3, 6)$ satisfies all the conditions. (We may observe $(2, 4, 5)$ is also not a solution.) Since $3 + 4 + 5 = 12 > 11 = 2 + 3 + 6$, it follows that $(2, 3, 6)$ has the required minimality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17895,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 1000$ athletes, labeled $A_1, A_2, \\dots, A_{1000}$, stand equally spaced around a circular track of length $1000$ metres.\n\n(a) In how many ways can the athletes be divided into $500$ pairs so that the members of each pair are $335$ metres apart?\n\n(b) In how many ways can the athletes be divided into $500$ pairs so that the members of each pair are $336$ metres apart?\n\nFor the general case of $n$ (where $n$ is an even positive integer) athletes standing equally spaced around a circular track of length $n$ metres, show that the number of ways of dividing the athletes into $\\frac{n}{2}$ pairs such that the members of each pair are $k$ metres apart is\n\n$$\n\\begin{cases} 2^{\\text{gcd}(k,n)}, & \\text{if } \\frac{n}{\\text{gcd}(k,n)} \\text{ is even,} \\\\ 0, & \\text{if } \\frac{n}{\\text{gcd}(k,n)} \\text{ is odd.} \\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Answers: (a) $32$ (b) $0$\n\nLet the athletes be $A_1, A_2, \\dots, A_{1000}$ in that order, clockwise around the track.\n\n(a) Starting from $A_1$ and proceeding clockwise around the track in intervals of $335$ metres, we meet athletes in the order:\n\n$$\nA_1, A_{336}, A_{671}, A_6, A_{341}, A_{676}, A_{11}, \\dots, A_{331}, A_{666}, (A_1).\n$$\n\nSince $335$ and $1000$ are both divisible by $5$, only athletes $A_i$ with $i \\equiv 1 \\pmod{5}$ can occur in the above list. Furthermore, the directed distance between every third athlete in the above list is $5$ metres. Thus, the above list contains precisely all $A_i$ with $i \\equiv 1 \\pmod{5}$, so there are exactly $200$ different athletes.\n\nAthlete $A_1$ can either be paired with $A_{336}$ or $A_{666}$. Once this pairing is chosen, all other pairings are forced. Thus, there are exactly two ways of pairing up all the athletes $A_i$ for $i \\equiv 1 \\pmod{5}$.\n\nWe could have started four other lists with $A_2, A_3, A_4$, and $A_5$, respectively. Analogous arguments show that there are exactly two ways of pairing up the athletes in each such list. Since the five lists are independent, the total number of pairings is $2^5 = 32$.\n\n(b) Starting from $A_1$ and proceeding clockwise around the track in intervals of $336$ metres, we meet athletes in the order:\n\n$$\nA_1, A_{337}, A_{673}, A_9, A_{345}, A_{681}, A_{17}, \\dots, A_{329}, A_{665}, (A_1).\n$$\n\nSince $336$ and $1000$ are both divisible by $8$, only athletes $A_i$ with $i \\equiv 1 \\pmod{8}$ can occur in the above list. The directed distance between every third athlete in the above list is $8$ metres. Thus, the above list contains precisely all $A_i$ with $i \\equiv 1 \\pmod{8}$, so there are exactly $125$ different athletes. However, since $125$ is odd, it is not possible to pair everyone up from the above list. Hence, no such pairing is possible.\n\nFor the general case, the number of ways of dividing $n$ athletes into $\\frac{n}{2}$ pairs such that the members of each pair are $k$ metres apart is\n\n$$\n\\begin{cases} 2^{\\text{gcd}(k,n)}, & \\text{if } \\frac{n}{\\text{gcd}(k,n)} \\text{ is even,} \\\\ 0, & \\text{if } \\frac{n}{\\text{gcd}(k,n)} \\text{ is odd.} \\end{cases}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17896,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $\\mathbb{Z}$ the set of integers. Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a, b, c$ that satisfy $a + b + c = 0$, the following equality holds:\n\n$$\nf(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "There are three classes of solutions:\n\n* For a fixed integer $m$, the function $f(n) = mn^2$.\n\n* For a fixed nonzero integer $m$, the function\n\n$$\nf(n) = \\begin{cases} 0 & \\text{if } n \\text{ is even} \\\\ m & \\text{if } n \\text{ is odd} \\end{cases}\n$$\n\n* For a fixed nonzero integer $m$, the function\n\n$$\nf(n) = \\begin{cases} 0 & \\text{if } n \\equiv 0 \\pmod{4} \\\\ 4m & \\text{if } n \\equiv 2 \\pmod{4} \\\\ m & \\text{if } n \\text{ is odd} \\end{cases}\n$$\n\nAll of these functions satisfy the given identity. It remains to show they are the only solutions. Setting $a = b = c = 0$ gives $3f(0)^2 = 6f(0)^2$, so $f(0) = 0$. Setting $a = 0$ and $c = -b$ gives $f(b)^2 + f(-b)^2 = 2f(b)f(-b)$, so $(f(b) - f(-b))^2 = 0$, hence $f(b) = f(-b)$ for all $b \\in \\mathbb{Z}$.\n\nSetting $c = -a - b$ and using $f(-a - b) = f(a + b)$, we get\n\n$$\nf(a)^2 + f(b)^2 + f(a+b)^2 = 2f(a)f(b) + 2f(b)f(a+b) + 2f(a+b)f(a).\n$$\n\nRearranging and factoring yields\n\n$$\n(f(a+b) - f(a) - f(b))^2 = 4f(a)f(b).\n$$\n\nIf $f(a) = 0$ for some $a$, then $f(a+b) = f(b)$ for all $b$, so $f$ has period $a$. If $f(1) = 0$, $f$ is identically zero (first class). If $f(2) = 1$, $f$ is in the second class. Otherwise, assume $f(1) = m \\neq 0$ and $f(2) \\neq 0$.\n\nApplying the above with $a = b = 1$ gives $f(2) = 4m$, and with $a = 2$, $b = 1$ gives $f(3) = m$ or $f(3) = 9m$. If $f(3) = m$, then with $a = b = 2$, $f(4) \\in \\{0, 16m\\}$. Since $m \\neq 0$, $f(4) = 0$, so $f$ is in the third class.\n\nIf $f(0) = 0$, $f(1) = m$, $f(2) = 4m$, $f(3) = 9m$, we claim by induction that $f(n) = mn^2$ for all $n \\geq 0$. The base cases hold. Suppose $f(l) = ml^2$ for all $l \\leq n$ with $n \\geq 3$. Then with $a = n$, $b = 1$, $f(n+1) \\in \\{(n+1)^2m, (n-1)^2m\\}$, and with $a = n-1$, $b = 2$, $f(n+1) \\in \\{(n+1)^2m, (n-3)^2m\\}$. For $n \\geq 3$, $(n-3)^2m \\neq (n-1)^2m$, so $f(n+1) = (n+1)^2m$, completing the induction and placing $f$ in the first class.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17897,
    "subject": "Mathematics (Olympiad)",
    "question": "設銳角三角形 $ABC$ 的外接圓為 $\\Omega$。令 $D, E, F$ 分別為 $\\Omega$ 上劣弧 $BC, CA, AB$ 的中點，$G$ 為 $D$ 關於 $\\Omega$ 的對徑點。令 $X$ 為 $GE$ 與 $AB$ 的交點，$Y$ 為 $FG$ 與 $CA$ 的交點。設三角形 $BEX$ 的外心為 $S$，三角形 $CFY$ 的外心為 $T$。試證 $D, S, T$ 共線。",
    "options": [],
    "answer": "See solution",
    "solution": "解法一：令 $I$ 為 $\\triangle ABC$ 的內心，我們有 $\\angle IFG = \\angle CBG = 90^\\circ - \\frac{1}{2}\\angle A = \\angle CIE$，即 $FG \\parallel IE$。同理有 $GE \\parallel IF$。\n\n我們證明 $BY$ 與 $CX$ 的交點 $W$ 位於 $\\Omega$ 上：考慮折線 $BWCFGE$，由帕斯卡定理知 $W$ 位於 $\\Omega$ 上等價於 $X, Y, I$ 共線，而這是因為\n\n$$\nA(X,Y; I, G) = (B,C; D, G) = -1 = G(E,F; I, \\infty_{EF}) = G(X,Y; I, A).\n$$\n\n令 $U$ 為 $BY$ 與 $\\odot(BEX)$ 的第二個交點，則由 $\\angle XUB = \\angle XEB = \\angle XGY$ 知 $U$ 位於 $\\odot(GXY)$ 上。同理，$CX$ 與 $\\odot(CFY)$ 的第二個交點 $V$ 也位於 $\\odot(GXY)$ 上。由於 $W$ 位於 $\\Omega$ 上，$\\angle GEB = \\angle GWB$，即 $XU \\parallel GW$。同理，$YV \\parallel GW$。因此 $XYVU$ 是等腰梯形。\n\n由於 $\\angle BWC$ 的內角平分線 $WD$ 會是 $\\overline{XU}$ 的中垂線 $WS$，也是 $\\overline{YV}$ 的中垂線 $WT$，因此 $D, S, T$ 共線，證畢。 □\n\n![](images/2023-TWNIMO-Problems_p49_data_a45ee56570.png)\n\n註：要證明 $X, Y, I$ 共線，我們也有如下的算長度作法：注意到 $X, Y, I$ 共線等價於 $\\frac{FY}{YG} = \\frac{IF}{GX}$，而\n\n$$\n\\frac{IF}{GX} = \\frac{EG}{GX} = \\frac{EX}{GX} - 1 = \\frac{AE \\cdot BE}{AG \\cdot BG}, \\quad \\frac{FY}{YG} = \\frac{AF \\cdot CF}{AG \\cdot CG}.\n$$\n\n因此，由 $BG = CG$，又等價於\n\n$$\nAE \\cdot BE = AF \\cdot CF + AG \\cdot CG\n$$\n\n$$\n\\Leftrightarrow \\sin \\frac{B}{2} \\sin\\left(A + \\frac{B}{2}\\right) = \\sin \\frac{C}{2} \\sin\\left(A + \\frac{C}{2}\\right) + \\sin \\frac{B-C}{2} \\sin \\frac{B+C}{2},\n$$\n\n而這只是簡單的和角公式。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17898,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of 2018 children each holds a token with a number. Each child holds hands with the two children who share the same number as them, forming several closed rounds (possibly more than one). The entertainer can make an exchange between two children, swapping their tokens. If the exchange is between children in different rounds, the two rounds merge into one. If the exchange is between children in the same round, the round either splits into two or is simply reordered. What is the minimum number of exchanges the entertainer must make to guarantee that, regardless of the initial distribution, all children can be arranged into a single round?",
    "options": [],
    "answer": "See solution",
    "solution": "Since every child holds hands with the two children sharing a number, the children form several rounds. An exchange between two children in different rounds merges the rounds, reducing the number of rounds by one. An exchange within the same round either splits the round or reorders it, so after each exchange, the number of rounds decreases by at most one. If there are $K$ rounds initially, at least $K-1$ exchanges are needed.\n\nEach round must have at least 3 children. Since $2018 = 3 \\times 672 + 2$, the maximum number of rounds is 672 (with 670 rounds of 3 children and 2 rounds of 4 children). Therefore, the minimum number of exchanges required is $672 - 1 = 671$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17899,
    "subject": "Mathematics (Olympiad)",
    "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\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "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\nGiven a configuration of exactly $n$ red cells, we show that $N \\ge (n + 6)/5$. Consider the minimal rectangular grid array $A$ containing the $n$ red cells. By minimality, $A$ contains some (not necessarily pairwise distinct) red cells $a$, $b$, $c$ and $d$ on the bottom row, the rightmost column, the top row, and the leftmost column, respectively; if, for instance, $a$ is the lower-left corner cell, then the list reads $a$, $b$, $c$, $a$.\n\nLetting $[xy]$ denote the (unique) rectangular grid array whose opposite corner cells are $x$ and $y$, notice that the (not necessarily pairwise distinct) special rectangular grid arrays $[ab]$, $[bc]$, $[cd]$, $[da]$ and $[ac]$ cover $A$: the first two cover the part of $A$ to the right of $[ac]$, and the next two cover the part of $A$ to the left of $[ac]$.\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": 17900,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ with $A$ a right angle. Consider points $D \\in (AC)$ and $E \\in (BD)$ so that $\\angle ABC = \\angle ECD = \\angle CED$. Show that $BE = 2AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Reflect $D$ across $A$ to $D'$, and denote $x = \\angle ABC = \\angle ECD = \\angle CED$. Then $\\angle ADB = 2x$ and $\\angle ABD' = \\angle ABD = 90^\\circ - 2x$, so $\\angle CBD' = x + 90^\\circ - 2x = 90^\\circ - x$. In triangle $BCD'$, from $\\angle BD'C = 2x$ it follows that $\\angle BCD' = 90^\\circ - x = \\angle CBD'$, so triangle $D'BC$ is isosceles and $D'C = D'B = DB$. Finally, $DC = DE$ yields $BE = BD - DE = CD' - CD = DD' = 2AD$.\n\n![](images/RMC2012_p21_data_65103ba1d6.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17901,
    "subject": "Mathematics (Olympiad)",
    "question": "The vertices $A$ and $B$ of an equilateral triangle $ABC$ lie on a circle $k$ of radius $1$, and the vertex $C$ is in the interior of circle $k$. A point $D$, different from $B$, lies on $k$ so that $AD = AB$. The line $DC$ intersects $k$ for the second time at point $E$. Find the length of the line segment $CE$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $\\overline{AD} = \\overline{AC}$, $\\triangle CDA$ is isosceles. If $\\angle ADC = \\angle ACD = \\alpha$ and $\\angle BCE = \\beta$ then $\\beta = 120^\\circ - \\alpha$. The quadrilateral $ABED$ is cyclic, so $\\angle ABE = 180^\\circ - \\alpha$. Then $\\angle CBE = 120^\\circ - \\alpha$ so $\\angle CBE = \\beta$. Thus $\\triangle CBE$ is isosceles, so $AE$ is the perpendicular bisector of $BC$, so it bisects $\\angle BAC$. Now the arc $BE$ is intercepted by a $30^\\circ$ inscribed angle, so it measures $60^\\circ$. Then $\\overline{BE}$ equals the radius of $k$, namely $1$. Hence $\\overline{CE} = \\overline{BE} = 1$.\n\n![](images/Makedonija_2008_p25_data_eed0df2e83.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17902,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a polynomial of degree $n > 1$ with integer coefficients and let $k$ be a positive integer. Consider the polynomial\n\n$$\nQ(x) = \\underbrace{P(P(\\dots(P(x)\\dots)))}_{k \\text{ terms}}\n$$\n\nProve that there are at most $n$ integers $t$ such that $Q(t) = t$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\mathbb{N}$ denote the set of integers. Define\n\n$$\nS_P = \\{t \\mid t \\in \\mathbb{N} \\text{ and } P(t) = t\\}, \\quad S_Q = \\{t \\mid t \\in \\mathbb{N} \\text{ and } Q(t) = t\\}.\n$$\n\nClearly, $S_P \\subseteq S_Q$. There are at most $n$ elements in $S_P$ because $t \\in S_P$ if and only if $t$ is a root of $P(x) - x = 0$, a degree $n$ polynomial, which has at most $n$ integer roots. If $S_Q = S_P$, we are done. Assume $S_P$ is a proper subset of $S_Q$, and $t \\in S_Q$ but $t \\notin S_P$.\n\nConsider the sequence $\\{t_i\\}_{i=0}^{\\infty}$ with $t_0 = t$, $t_{i+1} = P(t_i)$. Since $t \\in S_Q$, $t_k = Q(t) = t = t_0$.\n\nFor polynomials with integer coefficients, $a-b$ divides $P(a)-P(b)$. Thus, for our sequence,\n\n$$\n(t_{i+1} - t_i) \\mid (t_{i+2} - t_{i+1})\n$$\n\nfor all $i$. Since $t_{k+1} - t_k = t_1 - t_0 = P(t) - t \\neq 0$, each difference in the chain $t_1 - t_0, t_2 - t_1, \\dots, t_k - t_{k-1}, t_{k+1} - t_k$ is a nonzero divisor of the next, and all have equal absolute values. Let $t_i = \\max\\{t_0, t_1, \\dots, t_k\\}$. Then $t_{i-1} - t_i = -(t_i - t_{i+1})$, so $t_{i-1} = t_{i+1}$, and $t_{i+2} = t_i$ for all $i$; that is,\n\n$$\nt_1 = P(t_0), \\quad t_0 = P(t_1), \\quad \\text{so} \\quad P(P(t_0)) = t_0.\n$$\n\nTherefore,\n\n$$\nS_Q = \\{t \\mid t \\in \\mathbb{N},\\ P(P(t)) = t\\}.\n$$\n\nAssume $t_0 < t_1$. If $s_0$ is another element in $S_Q$, let $s_1 = P(s_0)$. Assume $s_0 < s_1$ and $t_0 < s_0 \\leq s_1$, $t_0 < t_1$. Note $s_1 - t_0$ divides $P(s_1) - P(t_0) = s_0 - t_1$. We must have $t_0 < s_0 < s_1 < t_1$. Also, $s_0 - t_1$ divides $P(s_0) - P(t_1) = s_1 - t_0$, so $s_0 - t_1 = -(s_1 - t_0)$, i.e.,\n\n$$\nt_0 + t_1 = s_0 + s_1 = s_0 + P(s_0).\n$$\n\nThus, $s_0$ is a root of $P(x) + x = t_0 + t_1$. Since $P(x) + x$ has degree $n$, there are at most $n$ integer roots (including $t_0$). Hence, there are at most $n$ elements in $S_Q$, as required.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17903,
    "subject": "Mathematics (Olympiad)",
    "question": "$a_n = a + a^{2n+1}$, $n \\geq 1$ дараалал өгөгдөв.\n\n$a_1, a_2, \\dots, a_{2012}$ нь бүгд $a$-тай харилцан анхны хоёр бүхэл тооны квадратуудын нийлбэрт байх $a$ натурал тоо төгсгөлгүй олон олдохыг харуул.",
    "options": [],
    "answer": "See solution",
    "solution": "$a = p$ гэж авъя, энд $p$ нь $4k + 1$ хэлбэрийн анхны тоо.\n\n1. $4k+1$ хэлбэрийн анхны тоо төгсгөлгүй олон байдаг.\n\n2. Аливаа $4k + 1$ хэлбэрийн анхны тоог хоёр бүтэн квадратын нийлбэрт задалж болно, $p = u^2 + v^2$ гэж бичигдэнэ ($u, v \\in \\mathbb{Z}$).\n\n$$\na_k = a + a^{2k+1} = a(a^{2k} + 1) = (u^2 + v^2)((a^k)^2 + 1^2)\n$$\n\nЭнэ нь\n\n$$\n((u a^k) + v)^2 + (u - v a^k)^2 = c^2 + d^2\n$$\n\nболно. Иймд $\\forall n \\in \\mathbb{N}$, $a_1, a_2, \\dots, a_n$ тоонууд бүгд хоёр бүтэн квадратын нийлбэрт тавигдаж байх $a$ ($a = p$, $p$ нь $4k + 1$ хэлбэрийн анхны тоо) төгсгөлгүй олон олдоно. Мөн $(a, c) = (a, d) = 1$ байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17904,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $a_1, a_2, a_3, \\dots$ 為無窮正整數數列，且對所有正整數 $n, m$，都有 $a_{n+2m}$ 整除 $a_n + a_{n+m}$ 這個性質。證明這個數列最終有週期性，也就是說，存在正整數 $N$ 和 $d$，使得對於所有 $n > N$，都有 $a_n = a_{n+d}$。",
    "options": [],
    "answer": "See solution",
    "solution": "我們將重複使用以下簡單觀察：\n\n**Lemma 1.** 如果正整數 $d$ 整除 $a_n$ 和 $a_{n-m}$（對某些 $m$ 且 $n > 2m$），則 $d$ 也整除 $a_{n-2m}$。\n\n*證明*：兩部分都很明顯，因為 $a_n$ 整除 $a_{n-2m} + a_{n-m}$。\n\n**Claim.** 數列 $(a_n)$ 是有界的。\n\n*證明*：假設相反。則存在無窮多個指標 $n$，使得 $a_n$ 大於所有前面的項 $a_1, a_2, \\dots, a_{n-1}$。令 $a_n = k$ 為其中一項，$n > 10$。對每個 $s < \\frac{n}{2}$，$a_n = k$ 整除 $a_{n-s} + a_{n-2s} < 2k$，因此\n\n$$\na_{n-s} + a_{n-2s} = k.\n$$\n\n特別地，\n\n$$\na_n = a_{n-1} + a_{n-2} = a_{n-2} + a_{n-4} = a_{n-4} + a_{n-8},\n$$\n\n即 $a_{n-1} = a_{n-4}$ 且 $a_{n-2} = a_{n-8}$。由 Lemma 1 可知 $a_{n-1}$ 整除 $a_{n-1-3s}$（對 $3s < n-1$），$a_{n-2}$ 整除 $a_{n-2-6s}$（對 $6s < n-2$）。由於 $a_{n-1}$ 和 $a_{n-2}$ 至少有一個不小於 $\\frac{a_n}{2}$，因此某個 $a_i$（$i \\le 6$）也如此。然而 $a_n$ 可以任意大，矛盾。\n\n因此 $(a_n)$ 有界，只有有限多個 $i$ 使得 $a_i$ 在數列中只出現有限次。換句話說，存在 $N$，使得若 $a_i = t$ 且 $i > N$，則 $a_j = t$ 對無窮多個 $j$ 成立。\n\n顯然，數列 $(a_{n+N})_{n>0}$ 也滿足可整除性質，只需證明此數列最終週期即可。因此，若有需要可截去前面部分，假設每個數在數列中都出現無窮多次。令 $k$ 為數列中出現的最大數。\n\n**Lemma 2.** 若正整數 $d$ 整除某項 $a_n$，則所有 $d$ 整除 $a_i$ 的 $i$ 構成一個差為奇數的等差數列。\n\n*證明*：令 $i_1 < i_2 < i_3 < \\dots$ 為所有 $d$ 整除 $a_i$ 的指標。若 $i_s + i_{s+1}$ 為偶數，則由 Lemma 1，$d$ 也整除 $a_{\\frac{i_s + i_{s+1}}{2}}$，但 $i_s < \\frac{i_s + i_{s+1}}{2} < i_{s+1}$，矛盾。因此 $i_s$ 和 $i_{s+1}$ 必為不同奇偶性，故 $i_s + i_{s+2}$ 為偶數。再用 Lemma 1，$d$ 整除 $a_{\\frac{i_s + i_{s+2}}{2}}$，即 $\\frac{i_s + i_{s+2}}{2} = i_{s+1}$。\n\n現在可以解決問題了。\n\n所有項的正因數個數有限。令 $d_s$ 為對應於 $s$ 的等差數列的差，即 $s$ 整除 $a_n$ 當且僅當 $s$ 整除 $a_{n + t d_s}$（任意正整數 $t$）。令 $d$ 為所有 $d_s$ 的乘積。則每個 $s$ 整除某項時，$s$ 整除 $a_n$ 當且僅當 $s$ 整除 $a_{n+D}$。這表示 $a_n$ 和 $a_{n+D}$ 的因數集合相同，且 $a_{n+D} = a_n$。因此 $D$ 為數列的週期。\n\n**補充說明**：上述解法未求出週期部分的具體結構。稍加補充可知，數列的週期有以下三種形式：\n\n(i) $t$（此時數列最終常數）；\n\n(ii) $t, 2t, 3t$ 或 $2t, t, 3t$（週期為 3）；\n\n(iii) $t, t, \\dots, 2t$（週期可為任意奇數）。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17905,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. What proportion of the non-empty subsets of $\\{1, 2, \\dots, 2n\\}$ has a smallest element that is odd?",
    "options": [],
    "answer": "See solution",
    "solution": "The number of subsets of $\\{1, 2, \\dots, 2n\\}$ that have $k$ as smallest element is $2^{2n-k}$ for $1 \\leq k \\leq 2n$, since each element bigger than $k$ is either contained in the subset or not.\n\nThe number $O$ of subsets with an odd smallest element is therefore equal to\n\n$$\nO = 2^{2n-1} + 2^{2n-3} + \\dots + 2^3 + 2^1 = 2 \\cdot (4^{n-1} + 4^{n-2} + \\dots + 4^1 + 4^0).\n$$\n\nThe number $E$ of subsets with an even smallest element is equal to\n\n$$\nE = 2^{2n-2} + 2^{2n-4} + \\dots + 2^2 + 2^0 = 4^{n-1} + 4^{n-2} + \\dots + 4^1 + 4^0.\n$$\n\nThis implies $O = 2E$ and consequently the desired proportion is $2/3$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17906,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a$ is an integer, and that $n! + a$ divides $(2n)!$ for infinitely many positive integers $n$. Prove that $a = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n(2n)! = \\binom{2n}{n} \\cdot n!^2 \\equiv \\binom{2n}{n} \\cdot (-a)^2 \\pmod{n! + a},\n$$\n\nso if $n! + a$ divides $(2n)!$, then it also divides $a^2 \\binom{2n}{n}$. We will show that when $n$ is large, $n! + a$ is greater than $a^2 \\binom{2n}{n}$ and therefore does not divide it (unless $a = 0$). Assume in the following that $a \\neq 0$ and $n > \\frac{4^5}{12}a^2$.\n\nFirst, by the binomial theorem,\n\n$$\n\\binom{2n}{n} \\leq \\sum_{k=0}^{2n} \\binom{2n}{k} = 2^{2n} = 4^n.\n$$\n\nIf $a > 0$, we have\n\n$$\n0 < \\frac{a^2}{n! + a} \\binom{2n}{n} < \\frac{a^2}{n!} \\binom{2n}{n} \\leq \\frac{a^2}{n!} 4^n = \\left( \\frac{4^{n-5}}{5 \\cdot 6 \\cdots (n-1)} \\right) \\frac{4^5 a^2}{24n} < 1.\n$$\n\nIf $a < 0$, then $n! > n > 2|a|$, so\n\n$$\n0 < \\frac{a^2}{n! + a} \\binom{2n}{n} = \\frac{a^2}{n! - |a|} \\binom{2n}{n} < \\frac{a^2}{\\frac{1}{2}n!} \\binom{2n}{n} \\leq \\frac{2a^2}{n!} 4^n = \\left( \\frac{4^{n-5}}{5 \\cdot 6 \\cdots (n-1)} \\right) \\frac{4^5 a^2}{12n} < 1.\n$$\n\nThus $\\frac{a^2 \\binom{2n}{n}}{n!+a}$ is not an integer for all such $n$, which contradicts our assumption that $n! + a$ divides $(2n)!$ (and thus $a^2 \\binom{2n}{n}$) for infinitely many $n$.\n\nSo $a = 0$ is the only possibility (and indeed $n!$ divides $(2n)!$ for all $n$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17907,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, the internal bisectors through $A$ and $B$ meet the opposite sides at $D$ and $E$, respectively. Prove that\n\n$$\nDE \\leq (3 - 2\\sqrt{2})(AB + BC + CA)\n$$\n\nand determine the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$, $b$, $c$ denote the lengths of the sides $BC$, $CA$, $AB$, respectively. The barycentric coordinates of $D$ and $E$ are $(0, \\frac{b}{b+c}, \\frac{c}{b+c})$ and $(\\frac{a}{a+c}, 0, \\frac{c}{a+c})$, respectively. The components of the vector $\\vec{DE}$ are\n\n$$\n\\frac{a}{a+c},\\quad -\\frac{b}{b+c},\\quad \\frac{c(b-a)}{(a+c)(b+c)}.\n$$\n\nUsing the standard formula and simple manipulations,\n\n$$\n\\begin{aligned}\nDE^2 &= \\frac{abc}{(a+c)(b+c)} \\left( c - \\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)} \\right) \\\\\n&\\leq \\frac{abc^2}{(a+c)(b+c)} \\leq \\frac{1}{4} \\left( \\frac{ac}{a+c} + \\frac{bc}{b+c} \\right)^2.\n\\end{aligned}\n$$\n\nSo $DE \\leq \\frac{1}{2}\\left( \\frac{ac}{a+c} + \\frac{bc}{b+c} \\right)$. Since $(a+c)(2a+c) \\geq (3+2\\sqrt{2})ac$ and $(b+c)(2b+c) \\geq (3+2\\sqrt{2})bc$, it follows that\n\n$$\nDE \\leq \\frac{1}{2} \\left( \\frac{2a+c}{3+2\\sqrt{2}} + \\frac{2b+c}{3+2\\sqrt{2}} \\right) = (3-2\\sqrt{2})(a+b+c).\n$$\n\nEquality holds if and only if $c = a\\sqrt{2} = b\\sqrt{2}$, i.e., when triangle $ABC$ is an isosceles right-angled triangle with the apex at $C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17908,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega$ be the circumcircle of an acute-angled triangle $ABC$. Point $D$ lies on the arc $BC$ of $\\omega$ not containing point $A$. Point $E$ lies in the interior of triangle $ABC$, does not lie on the line $AD$, and satisfies $\\angle DBE = \\angle ACB$ and $\\angle DCE = \\angle ABC$. Let $F$ be a point on the line $AD$ such that lines $EF$ and $BC$ are parallel, and let $G$ be a point on $\\omega$ different from $A$ such that $AF = FG$. Prove that points $D$, $E$, $F$, $G$ lie on one circle.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\alpha = \\angle CAB$, $\\beta = \\angle ABC$, and $\\gamma = \\angle BCA$. Let $K$ and $L$ be the second intersections of lines $BE$ and $CE$ with $\\omega$, respectively, different from $B$ and $C$. Observe that\n\n$$\n\\angle BAK = \\angle BAD + \\angle DAK = \\angle BAD + \\angle DBE = \\angle BAD + \\gamma = \\angle ACD\n$$\n\nand symmetrically $\\angle CAL = \\angle ABD$. It follows that arcs $AD$, $BK$, and $CL$ of $\\omega$ have equal lengths, so chords $AD$, $BK$, and $CL$ also have equal lengths. In particular, since $E = BK \\cap CL$ does not lie on $AD$, these chords are not diameters. It follows that if $O$ is the center of $\\omega$, then $O$ does not lie on any of the chords $AD$, $BK$, $CL$, and in particular $O \\ne E$. Moreover, $O$ and $E$ lie on the same side of line $AD$. Suppose without loss of generality that $O$ and $E$ lie in triangle $ACD$, for the second case is symmetric.\n\nObserve that\n\n$$\n\\angle BEC = 360^{\\circ} - \\angle DBE - \\angle DCE - \\angle BDC = 360^{\\circ} - \\beta - \\gamma - (180^{\\circ} - \\alpha) = 2\\alpha = \\angle BOC,\n$$\n\nwhich implies that $B$, $O$, $E$, $C$ are concyclic. Further, if we denote $P = AD \\cap BC$, then we have\n\n$$\n\\begin{aligned}\n\\angle DOE &= \\angle BOE - \\angle BOD = 180^{\\circ} - \\angle BCE - 2\\angle BAD \\\\\n&= 180^{\\circ} - \\angle DCE - \\angle BAD = 180^{\\circ} - \\beta - \\angle BAD = \\angle APB = \\angle EFD,\n\\end{aligned}\n$$\n\nwhich implies that $D$, $F$, $O$, $E$ are also concyclic.\n\nConsider triangles $AFO$ and $GFO$. We have $AF = FG$ by assumption, also $OA = OG$ since $G$ lies on $\\omega$, hence these two triangles are congruent. In particular $\\angle FAO = \\angle FGO$. Triangle $AOD$ is isosceles, hence $\\angle FAO = \\angle FDO$. This implies that $F$, $O$, $G$, $D$ are concyclic as well.\n\nSince points $F$, $O$, $D$ are pairwise distinct, this implies that all five points $D$, $F$, $O$, $E$, $G$ lie on the circumference of triangle $FOD$, so in particular $D$, $E$, $F$, $G$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17909,
    "subject": "Mathematics (Olympiad)",
    "question": "A three-digit natural number $n$ is initially written on the board. Two players, A and B, take turns, with A going first. On each turn, the player reduces the number on the board by some proper divisor of the current number (i.e., a divisor other than 1 and the number itself). For example, if the number on the board is 6, it can be reduced by 2, resulting in 4. Whoever cannot make a move loses, and the other wins. It is known that both player A and player B have a way to win. What are all possible values of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "If a prime number is on the board, the player loses, since there are no proper divisors. If the number is even and not a power of 2, the player can always reduce it by an odd divisor, leaving an odd number. If the number is odd and is reduced by its (odd) divisor $a$, i.e., a number of the form $ab$ is replaced by $a(b - 1)$, the result is even and not a power of 2. Thus, if the starting number is even and not a power of 2, the player can always make a move to ensure the next number is of the same kind. Therefore, an even number that is not a power of 2 is a winning position, and an odd number is a losing position.\n\nNow, consider when the number is $2^m$ for some natural $m$. The only proper divisors are $2^k$ for $k < m$. If $k < m - 1$, the resulting number $2^k(2^{m-k} - 1)$ is even and not a power of 2, which would be a winning move for the opponent, so this is not optimal. Thus, we need $k = m - 1$. Playing this way, one player will get even powers of 2, the other odd. Since 2 is a losing position, even powers of 2 are winning, odd powers are losing.\n\nTherefore, the suitable $n$ are the even numbers that are not odd powers of 2. Among the three-digit numbers, there are $900 \\div 2 - 2 = 448$ such numbers (excluding $128 = 2^7$ and $512 = 2^9$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17910,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ for which it is possible to cut a square into $2n$ squares of two different sizes: $n$ squares of one size and $n$ squares of another size.\n\n![](images/Ukraine_2021-2022_p17_data_949248e8e8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "An example of cutting into $2n = 18$ squares is provided in Fig. 10.\n\nSuppose there exists an example for $n \\leq 8$. Denote the sides of the small squares by $a$ and $b$, and the side of the large square by $N$. We first show that it is possible to represent $N$ as the sum $ka + lb$, where $k, l \\leq n$, in at least two different ways.\n\nClearly, at least one such representation must exist. Let $N = k_1 a + l_1 b$. Any vertical or horizontal line not passing through any side of any square intersects precisely $k_1$ squares of side $a$ and $l_1$ squares of side $b$.\n\nChoose any $x$ and draw vertical lines at distances $x, x+a, x+2a, \\ldots$ from the left side of the large square (with $x$ chosen so that no line passes through any side of any square). Each such line will intersect exactly $k_1$ squares of size $a$, so the large square is intersected by precisely $\\frac{n}{k_1}$ lines for any choice of $x$, which is only possible if $N = \\frac{n}{k_1} a$. Similarly, $N = a \\frac{n}{l_1}$. Thus, we obtain two representations of $N$, which is a contradiction.\n\nNow, such $k_2, l_2$ exist, so $a(k_1 - k_2) = b(l_2 - l_1)$, and we can rescale and let $a = l$, $b = k$ for some $l, k \\leq n$. Thus, $1 \\leq a, b \\leq n$, $a \\neq b$, $a, b, N$ are positive integers, and $\\gcd(a, b) = 1$.\n\nThe area condition is $n(a^2 + b^2) = N^2$.\n\nFor $n = 3, 6, 7$, this equation has no solution: since $n$ has a prime divisor of the form $p = 4k-1$, we get $N^2 \\mid p$, so $(a^2 + b^2) \\mid p$, so $a$ and $b$ are divisible by $p$, but they are coprime.\n\nFor $n = 1, 2$, we can check all pairs $(a, b)$ satisfying the above conditions and see that for none of them is $n(a^2 + b^2)$ a perfect square.\n\nFor $n = 4$, the only such pair is $(a, b) = (3, 4)$. Then $N = 10$, but $10$ can be represented as $3a + 4b$ in a unique way, so it does not work.\n\nFor $n = 5$, the only such pair is $(a, b) = (1, 2)$, then $N = 5$, but from the square $5 \\times 5$ we cannot cut out $5$ squares $2 \\times 2$, as each contains at least one of the cells $(2,2), (2,4), (4,2), (4,4)$.\n\nFor $n = 8$, the only such pair is $(1, 7)$, then $N = 20$, but from the square $20 \\times 20$ we cannot cut out $8$ squares $7 \\times 7$, as each contains at least one of the cells $(7,7), (7,14), (14,7), (14,14)$.\n\nTherefore, the smallest $n$ for which such a division is possible is $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17911,
    "subject": "Mathematics (Olympiad)",
    "question": "Every day, Maurits bikes to school. He can choose between two different routes. Route B is 1.5 km longer than route A. However, because he encounters fewer traffic lights, his average speed along route B is 2 km/h higher than along route A. This means that travelling along the two routes takes exactly the same amount of time.\n\nHow long does it take for Maurits to bike to school?",
    "options": [],
    "answer": "See solution",
    "solution": "45 minutes",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17912,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, set $x_n = \\binom{2n}{n}$.\n\n**a)** Prove that if $\\dfrac{2017^k}{2} < n < 2017^k$ for some positive integer $k$, then $2017$ divides $x_n$.\n\n**b)** Find all positive integers $h > 1$ such that there exist positive integers $N, T$ so that $(x_n)_{n>N}$ is periodic modulo $h$ with period $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)**\nWe prove the statement for all odd primes $p$ (including $2017$). Suppose there exists a positive integer $k$ such that $\\dfrac{p^k}{2} < n < p^k$. We have\n\n$$\nv_p(x_n) = v_p\\left(\\binom{2n}{n}\\right) = v_p((2n)!) - 2v_p(n!).\n$$\n\nSince $\\dfrac{p^k}{2} < n < p^k$, it follows that $p^k < 2n < 2p^k < p^{k+1}$. Thus,\n\n$$\nv_p((2n)!) = \\left\\lfloor \\frac{2n}{p} \\right\\rfloor + \\left\\lfloor \\frac{2n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2n}{p^k} \\right\\rfloor.\n$$\n\nFor every $x \\in \\mathbb{R}$, $\\lfloor 2x \\rfloor \\ge 2\\lfloor x \\rfloor$, with equality if $\\{x\\} < \\frac{1}{2}$. Given $\\dfrac{p^k}{2} < n < p^k$, we have\n\n$$\nv_p((2n)!) > 2 \\left( \\left\\lfloor \\frac{n}{p} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{p^k} \\right\\rfloor \\right) = 2v_p(n!).\n$$\n\nTherefore, $v_p(x_n) > 0$, so $p \\mid x_n$.\n\n**b)**\nSuppose $h > 1$ satisfies the periodicity condition. For any odd prime $p$ dividing $h$, the sequence $x_n$ modulo $p$ would also be periodic. By part (a), for $\\dfrac{p^k}{2} < n < p^k$,\n\n$$\nx_n \\equiv 0 \\pmod{p}.\n$$\n\nLet $k$ be such that $\\dfrac{p^k}{2} > T + 1$. Then for all $n \\ge n_0$ (for some large $n_0$), $x_n \\equiv 0 \\pmod{p}$. However, for large $t$ with $p^t - 1 > 2n_0$ and $n = \\dfrac{p^t - 1}{2}$, we have $v_p(x_n) = 0$, so $x_n$ is not divisible by $p$, a contradiction.\n\nThus, $2$ is the only possible prime divisor of $h$, so $h = 2^k$ for some $k$. If $k > 1$, choose $r = k - 1$ and $n = 2^{a_1} + \\dots + 2^{a_r}$ with $a_1 > \\max\\{T, N\\}$. Then\n\n$$\nv_2(x_n) = 2S_2(n) - S_2(2n) = r,\n$$\n\nwhere $S_2(x)$ is the sum of the binary digits of $x$. Thus, $x_n \\equiv 2^{k-1} \\pmod{h}$. For $i < 2^{a_1}$, $x_{n+i} \\equiv 0 \\pmod{h}$. Since $a_1 > \\max\\{T, N\\}$, $x_n \\equiv x_{n+T} \\equiv 0 \\pmod{h}$, a contradiction.\n\nTherefore, $k = 1$ and $h = 2$. This works because $x_n$ is even for all $n$: if $n = 2^{a_1} + \\cdots + 2^{a_r}$ with $0 \\le a_1 < \\cdots < a_r$,\n\n$$\n2n = 2^{a_1+1} + \\cdots + 2^{a_r+1},\n$$\n\nso\n\n$$\nv_2(x_n) = 2S_2(n) - S_2(2n) = r \\ge 1.\n$$\n\nHence, $x_n$ is always even, so $h = 2$ is the only possibility.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 17913,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $n, n+1, n+2, n+3$ — четыре последовательных натуральных числа. Докажите, что сумма трёх наименьших из них и сумма трёх наибольших из них могут быть представлены в виде произведения трёх различных натуральных чисел.",
    "options": [],
    "answer": "See solution",
    "solution": "Сумма трёх наименьших чисел: $n + (n+1) + (n+2) = 3n + 3 = 3(n + 1)$. Сумма трёх наибольших чисел: $(n+1) + (n+2) + (n+3) = 3n + 6 = 3(n + 2)$. Среди $n+1$ и $n+2$ хотя бы одно число чётно, то есть оно делится на 2, и его можно записать как $2k$, где $k > 3$. Следовательно, сумма представляется в виде произведения трёх различных натуральных чисел: $2$, $3$ и $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17914,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with real coefficients such that\n\n$$\nP(a) \\in \\mathbb{Z} \\implies a \\in \\mathbb{Z}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x) = a_n x^n + \\cdots + a_1 x + a_0$. Define $Q(x) = P(x+1) - P(x)$. Then $Q(x)$ is a polynomial of degree $n-1$.\n\nSuppose, for contradiction, that $|Q(a)| > 3$ for some $a \\in \\mathbb{R}$. Then $|P(a+1) - P(a)| > 3$, so there are at least three integers between $P(a)$ and $P(a+1)$. Thus, there exist three values $b_1, b_2, b_3$ in $(a, a+1)$ such that $P(b_i) \\in \\mathbb{Z}$ for $i=1,2,3$. But then $b_i \\notin \\mathbb{Z}$, contradicting the condition.\n\nTherefore, $|Q(x)| \\leq 3$ for all $x$. But $Q(x)$ is a polynomial, so it must be constant. Thus, $P(x)$ is at most degree $1$.\n\nLet $P(x) = ax + b$. If $P(a) \\in \\mathbb{Z}$, then $a a + b \\in \\mathbb{Z}$. For $a \\notin \\mathbb{Z}$, $P(a)$ is not always integer unless $a = 0$ and $b \\in \\mathbb{Z}$. Thus, the only polynomials are $P(x) = c$, where $c \\notin \\mathbb{Z}$, or $P(x) = kx + c$ with $k \\neq 0$, $k \\in \\mathbb{R}$, and $c \\notin \\mathbb{Z}$, but only $P(x) = kx + c$ with $k \\neq 0$ and $k$ irrational, $c$ irrational, so that $P(a) \\in \\mathbb{Z}$ only when $a \\in \\mathbb{Z}$.\n\nBut in fact, the only polynomials that satisfy the condition are $P(x) = kx + c$ with $k \\neq 0$, $k \\in \\mathbb{Z}$, $c \\notin \\mathbb{Z}$, or $P(x) = c$ with $c \\notin \\mathbb{Z}$.\n\nThus, all constant polynomials $P(x) = c$ with $c \\notin \\mathbb{Z}$, and all linear polynomials $P(x) = kx + c$ with $k \\in \\mathbb{Z} \\setminus \\{0\\}$ and $c \\notin \\mathbb{Z}$, are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17915,
    "subject": "Mathematics (Olympiad)",
    "question": "En la circunferencia circunscrita al triángulo $ABC$, sea $A_1$ el punto diametralmente opuesto al vértice $A$. Sea $A'$ el punto en el que la recta $AA_1$ corta al lado $BC$. La perpendicular a la recta $AA'$ trazada por $A'$ corta a los lados $AB$ y $AC$ (o a sus prolongaciones) en $M$ y $N$, respectivamente. Demostrar que los puntos $A$, $M$, $A_1$ y $N$ están en una circunferencia cuyo centro se encuentra en la altura desde $A$ en el triángulo $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $O$ el circuncentro de $ABC$, y $D$ el pie de la altura desde $A$. Es conocido que $AO$ y la altura desde $A$ son rectas isogonales en cualquier triángulo. En nuestro caso, lo son en los dos triángulos $ABC$ y $AMN$, por la manera como se construye el triángulo $AMN$.\n\nEn $ABC$, $AA_1$ es diámetro de la circunferencia circunscrita $\\Gamma$ y la recta $AD$ es altura. En $AMN$, $AO$ es altura, así que el centro de la circunferencia circunscrita $\\Gamma'$ estará en la recta $AD$ (isogonal de $AO$ en $AMN$).\n\nPara terminar el problema hay que probar que $A_1$ pertenece a $\\Gamma'$. En primer lugar, es fácil demostrar que los triángulos $ABC$ y $AMN$ son semejantes ($\\widehat{AMN} = 90^\\circ - \\alpha = \\hat{C}$). Escribiendo la proporcionalidad entre sus lados, obtenemos:\n\n$$\n\\frac{AB}{AC} = \\frac{AN}{AM},\n$$\n\ny esto quiere decir que las rectas $MC$ y $BN$ son antiparalelas y el cuadrilátero $BMCN$ es cíclico. Sea $\\Gamma''$ la circunferencia que pasa por los vértices de dicho cuadrilátero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17916,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest constant $C$ such that\n\n$$\n(x_1 + x_2 + \\cdots + x_6)^2 \\geq C \\cdot (x_1(x_2 + x_3) + x_2(x_3 + x_4) + \\cdots + x_6(x_1 + x_2))\n$$\n\nholds for all real numbers $x_1, x_2, \\dots, x_6$.\n\nFor this $C$, determine all $x_1, x_2, \\dots, x_6$ such that equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the right-hand side:\n\n$$\nx_1x_2 + x_1x_3 + x_2x_3 + x_2x_4 + x_3x_4 + x_3x_5 + x_4x_5 + x_4x_6 + x_5x_6 + x_1x_5 + x_1x_6 + x_2x_6\n$$\n\nas\n\n$$\n(x_1 + x_4)(x_2 + x_5) + (x_2 + x_5)(x_3 + x_6) + (x_3 + x_6)(x_1 + x_4).\n$$\n\nLet $X = x_1 + x_4$, $Y = x_2 + x_5$, and $Z = x_3 + x_6$. The inequality becomes\n\n$$\n(X + Y + Z)^2 \\geq C \\cdot (XY + YZ + ZX),\n$$\n\nwhere $X, Y, Z$ are arbitrary real numbers.\n\nFor $X = Y = Z = 1$, we get $9 \\geq 3C$, so $C \\leq 3$.\n\nWe now prove that\n\n$$\n(X + Y + Z)^2 \\geq 3(XY + YZ + ZX).\n$$\n\nExpanding gives\n\n$$\nX^2 + Y^2 + Z^2 \\geq XY + YZ + ZX.\n$$\n\nThis is equivalent to\n\n$$\n(X - Y)^2 + (Y - Z)^2 + (Z - X)^2 \\geq 0,\n$$\n\nwith equality when $X = Y = Z$, i.e., $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17917,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the hyperboloid with equation $3x^2 + 3y^2 - z^2 - 1 = 0$. \n\n(a) Show that every point of $H$ belongs to at least two lines contained in $H$, and that these lines are parallel to either $(0, \\frac{\\sqrt{3}}{3}, 1)$ or $(0, -\\frac{\\sqrt{3}}{3}, 1)$. Prove that $H$ does not contain any other lines.\n\n(b) Compute the angle that the vectors $(0, \\pm\\frac{\\sqrt{3}}{3}, 1)$ make with the $xy$-plane.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write the hyperboloid as $x^2 + y^2 - \\frac{1}{3}z^2 = \\frac{1}{3}$. Consider the point $(x_0, y_0, 0) = (\\frac{\\sqrt{3}}{3}, 0, 0)$ and the direction vector $(a, b, 1) = (0, \\frac{\\sqrt{3}}{3}, 1)$.\n\n(a) The lines $\\ell_1: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(0, \\frac{\\sqrt{3}}{3}, 1)$ and $\\ell_2: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(0, -\\frac{\\sqrt{3}}{3}, 1)$ are both contained in $H$. To show these are the only lines, suppose a line $r: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(c, d, 1)$ is contained in $H$. For all $t \\in \\mathbb{R}$,\n\n$$\n3\\left(\\frac{\\sqrt{3}}{3} + tc\\right)^2 + 3(td)^2 - t^2 - 1 = 0\n$$\n\nExpanding and collecting terms, we get:\n\n$$\n(3c^2 + 3d^2 - 1)t^2 + \\frac{2\\sqrt{3}}{3}tc = 0\n$$\n\nFor this to hold for all $t$, both coefficients must vanish:\n\n$$\n\\begin{cases}\n\\frac{2\\sqrt{3}}{3}c = 0 \\\\\n3c^2 + 3d^2 - 1 = 0\n\\end{cases}\n$$\n\nSo $c = 0$ and $d = \\pm \\frac{\\sqrt{3}}{3}$. Thus, the only possible direction vectors are $(0, \\frac{\\sqrt{3}}{3}, 1)$ and $(0, -\\frac{\\sqrt{3}}{3}, 1)$.\n\n(b) The angle $\\theta$ between $(0, \\pm\\frac{\\sqrt{3}}{3}, 1)$ and the $xy$-plane is:\n\n$$\n\\theta = \\arctan\\left(\\frac{1}{\\sqrt{3}/3}\\right) = \\arctan(3/\\sqrt{3}) = \\arctan(\\sqrt{3}) = 60^\\circ\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17918,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $g$ be a primitive root of $p$. Show that there exists a permutation $(k_1, k_2, \\dots, k_{p-1})$ of $\\{1, 2, \\dots, p-1\\}$ such that\n\n$$\n\\sum_{i=1}^{p-1} g^{i k_i} \\equiv 0 \\pmod{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Choose $k_i = \\frac{p+1}{2} - i$ and $k_{i+\\frac{p-1}{2}} = p - i$ for $i = 1, 2, \\dots, \\frac{p-1}{2}$. Then,\n\n$$\n\\sum_{i+j = \\frac{p+1}{2}} \\left(g^{i j} + g^{(p-i)(p-j)}\\right).\n$$\n\nSince the congruence $g^{p^2 - p(i + j)} \\equiv -1 \\pmod{p}$ always holds (independent of the choice of $g$), the sum $S$ is divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Residues and Primitive Roots"
  },
  {
    "id": 17919,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nS = \\frac{2^2 + 1}{2^2 - 1} + \\frac{3^2 + 1}{3^2 - 1} + \\dots + \\frac{100^2 + 1}{100^2 - 1}.\n$$\n\nFind the value of $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can rewrite each term:\n\n$$\n\\frac{k^2 + 1}{k^2 - 1} = \\frac{k^2 - 1 + 2}{k^2 - 1} = 1 + \\frac{2}{k^2 - 1}\n$$\n\nSo,\n\n$$\nS = \\sum_{k=2}^{100} \\left(1 + \\frac{2}{k^2 - 1}\\right) = 99 + 2 \\sum_{k=2}^{100} \\frac{1}{k^2 - 1}\n$$\n\nNote that\n\n$$\n\\frac{2}{k^2 - 1} = \\frac{2}{(k-1)(k+1)} = \\frac{1}{k-1} - \\frac{1}{k+1}\n$$\n\nThus,\n\n$$\n\\sum_{k=2}^{100} \\left(\\frac{1}{k-1} - \\frac{1}{k+1}\\right) = \\sum_{j=1}^{99} \\frac{1}{j} - \\sum_{j=3}^{101} \\frac{1}{j} = \\frac{1}{1} + \\frac{1}{2} - \\frac{1}{100} - \\frac{1}{101}\n$$\n\nTherefore,\n\n$$\nS = 99 + 1 + \\frac{1}{2} - \\frac{1}{100} - \\frac{1}{101} = \\frac{1014849}{10100}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17920,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle BCA = 90^\\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK = ML$.",
    "options": [],
    "answer": "See solution",
    "solution": "Construct segments $FK$ and $GL$. Let $\\angle FMK = \\angle GML = \\theta$. Also let $D = MK \\cap BJ$ and $E = ML \\cap CJ$.\n\nWe have $BM = BK$ due to equal tangents from $B$ to the excircle. Furthermore, since $BJ$ bisects $\\angle KBM$, this means that the line through $B$ and $J$ is a line of symmetry for $\\triangle KBM$. Since $F$ also lies on this line, we deduce that $\\angle FKG = \\angle FKM = \\angle FMK = \\theta$. Similarly, we show that $\\angle FLG = \\angle GML = \\theta$. Also, due to symmetry, we have $MK \\perp BJ$, and $ML \\perp CJ$. Hence $MDJE$ is cyclic. Thus $\\angle FJG = \\angle DJE = \\angle FMK = \\theta$. Since $\\angle FKG$, $\\angle FJG$, and $\\angle FLG$ are all equal (to $\\theta$), we conclude that $FKJLG$ is cyclic.\n\nA radius is perpendicular to its tangent, hence we have $AK \\perp JK$ and $AL \\perp JL$. So, $AKJL$ is cyclic. Combining this with $FKJLG$, we deduce that $AFKJLG$ is cyclic.\n\nFrom circle $AFKJLG$ we find $\\angle AFJ = \\angle AKJ = 90^\\circ$. But now in $\\triangle ASB$ the angle bisector $BF$ at $B$ is also the altitude. This implies that $AB = BS$. Thus $MS = BS + BM = AB + BK = AK$. Similarly, $MT = AL$. Since $AK = AL$, we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17921,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $A = \\{1, 2, 3\\}$, $B = \\{4x - y \\mid x, y \\in A\\}$, $C = \\{4x + y \\mid x, y \\in A\\}$. Then the sum of all the elements of $B \\cap C$ is ____.",
    "options": [],
    "answer": "See solution",
    "solution": "When $x$ and $y$ take all the elements of $A$, respectively, $4x - y$ gets exactly the values $1, 2, 3, 5, 6, 7, 9, 10, 11$, and $4x + y$ gets exactly the values $5, 6, 7, 9, 10, 11, 13, 14, 15$.\n\nHence, $B \\cap C = \\{5, 6, 7, 9, 10, 11\\}$. Then the sum of all the elements of $B \\cap C$ is $48$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17922,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $P$ is chosen inside triangle $ABC$ so that $BC = AP$ and $\\angle APC = 180^\\circ - \\angle ABC$. On side $AB$, there exists a point $K$ such that $AK = KB + PC$. Prove that $\\angle AKC = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We extend the ray $AB$ beyond $B$ to find the point $T$ such that $BT = PC$ (see the figure below). Then, $\\triangle TBC \\cong \\triangle CPA$ because they have two equal sides and the included angle is the same. Hence, $TC = CA$.\n\nSimilarly,\n\n$$\nAK = KB + PC = KB + BT = KT.\n$$\n\nTherefore, in the isosceles triangle $ATC$, the segment $KC$ is a median, which also makes it the altitude. Thus, $KC \\perp AB$, as required.\n\n![](images/Ukraine_booklet_2018_p7_data_482d671a23.png)\n\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17923,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an odd positive integer, and let $x_1, x_2, \\dots, x_n$ be non-negative real numbers. Show that\n$$\n\\min_{i=1,2,\\dots,n} (x_i^2 + x_{i+1}^2) \\le \\max_{j=1,2,\\dots,n} (2x_j x_{j+1}),\n$$\nwhere $x_{n+1} = x_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1} - x_k$, $k = 1, 2, \\dots, n$. 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 are non-negative, so $x_j \\le x_{j+1} \\le x_{j+2}$. Consequently,\n$$\n\\min_{i=1,2,\\dots,n} (x_i^2 + x_{i+1}^2) \\le x_j^2 + x_{j+1}^2 \\le 2x_j^2 + 2x_{j+1}^2 \\le 2x_{j+1}x_{j+2} \\le \\max_{k=1,2,\\dots,n} 2x_k x_{k+1}.\n$$\n\n**Remark.** If $n \\ge 3$ is odd, and one of the $x_k$ is negative, then the conclusion may no longer hold. This is the case if, for instance, $x_1 = -b$, and $x_{2k} = a$, $x_{2k+1} = b$, $k = 1, 2, \\dots, \\frac{n-1}{2}$, where $0 \\le a < b$, so the string of numbers is $-b, a, b, a, b, \\dots, b, a$.\n\nIf $n$ is even, the conclusion may again no longer hold, as shown by any string of alternate real numbers: $a, b, a, b, \\dots, a, b$, where $a \\ne b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17924,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a scalene triangle whose incenter lies on its Euler line?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is **No**.\n\nLet $\\triangle ABC$ be a triangle with orthocenter $H$, circumcenter $O$, and incenter $I$. Assume $I$ lies on $OH$. We'll show that $\\triangle ABC$ must be isosceles, contradicting the scalene property of the statement.\n\nFirst, a well-known claim:\n\n**Claim.** $O$ and $H$ are isogonal conjugates in $\\triangle ABC$.\n\n*Proof.* Let $AH$ intersect $BC$ at $D$. Then\n\n$$\n\\angle BAH = \\angle BAD = 90^\\circ - \\angle CBA = 90^\\circ - \\frac{1}{2} \\angle AOC = \\angle OAC,\n$$\n\nso $AH$ and $AO$ are isogonal in $\\angle BAC$. Similarly, $BH$ and $BO$ are isogonal in $\\angle CBA$, so the conclusion follows. $\\Box$\n\nNote that at most one of the vertices can lie on the Euler line because the centroid lies in the interior of $\\triangle ABC$. Therefore, we may assume that $OH$ passes through neither $B$ nor $C$.\n\nNow from the claim, $BI$ bisects $\\angle HBO$ and $CI$ bisects $\\angle HCO$. And since $I$ lies on $OH$, we get from the Angle Bisector Theorem that\n\n$$\n\\frac{|BH|}{|BO|} = \\frac{|IH|}{|IO|} = \\frac{|CH|}{|CO|}\n$$\n\nso $|BH| = |CH|$ since $|BO| = |CO|$. Now this implies that $OH$ is the perpendicular bisector of $BC$. In particular, $OH \\perp BC$ so $A$ lies on $OH$. That is, $A$ lies on the perpendicular bisector of $BC$ so $|AB| = |AC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17925,
    "subject": "Mathematics (Olympiad)",
    "question": "Label the $5 \\times 5$ grid as shown:\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p73_data_71a68eda9b.png)\n\nComplete the grid so that each row and column contains the letters A, H, M, S, and T exactly once, and each diagonal also contains each letter exactly once.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the diagonal $A5 - E1$: the letter T must appear in the diagonal, but it cannot appear in D2 or C3, since columns C and D already contain a T. Hence $A5 = T$. The letter A must also appear in the diagonal, but cannot be in C3, since column C already contains an A, so $D2 = A$ and hence $C3 = H$.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p73_data_4ca993b467.png)\n\nConsidering the other diagonal, $B2$ cannot be an A since $B1 = A$, hence $E5 = A$ which forces $B2 = S$. Looking at row 2 and letter T, columns A and C already contain Ts, which implies that $E2 = T$. This then forces $C2 = M$ and $A2 = H$.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p73_data_07a58e0e10.png)\n\nThe only place in row 3 where A can occur is in $A3$ and T can only occur in $B3$. The rest of the grid can now be easily completed in a similar fashion.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p73_data_104c9b027f.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17926,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a white square $8 \\times 8$. In one move, Dmitry can choose a totally white square $2 \\times 2$ and paint in black any two cells of this square, located on the diagonal.\n\n![](images/ukraine_2015_Booklet_p15_data_2a77b1b97f.png)\n\nWhat is the maximum number of cells Dmitry can paint according to these rules?",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show how to achieve the required number of painted cells. In each $4 \\times 4$ box, paint all four $2 \\times 2$ squares by coloring their diagonals, as shown in Fig. 31 (black squares). After this, there remains a completely white $2 \\times 2$ square inside. Paint any of its diagonals (gray squares). After painting all four $4 \\times 4$ squares, there is a completely white $2 \\times 2$ square in the center of the $8 \\times 8$ square, where we can paint two more cells. In total: $10 \\cdot 4 + 2 = 42$ squares of size $1 \\times 1$.\n\nNow, we show that a greater amount cannot be achieved. Along with the given $8 \\times 8$ square (call it \"initial\"), consider a $7 \\times 7$ square formed from the centers of the squares of the initial square (call it \"central\"). Coloring the diagonal of a $2 \\times 2$ square in the initial square corresponds to drawing a diagonal in the central square (Fig. 32).\n\n![](images/ukraine_2015_Booklet_p16_data_8cae1813d1.png)\n\nIn two adjacent $1 \\times 1$ squares of the central square, the diagonals can't be drawn. Consider a $3 \\times 4$ rectangle of the central square. We show it is impossible to draw diagonals in exactly half of all squares. Assume the contrary: then the diagonals are drawn in six $1 \\times 1$ squares. Choose four squares A, B, C, D that contain diagonals and have no corners among them (Fig. 33). It is impossible to draw diagonals in all of them. WLOG, the first diagonal is drawn in square A in the direction shown in Fig. 8. Then, in square C, it is impossible to draw a diagonal. So, in the $3 \\times 4$ rectangle, it is possible to draw at most 5 diagonals. Divide the $7 \\times 7$ square into four rectangles $3 \\times 4$ and one $1 \\times 1$ square, as shown in Fig. 34. The maximum number is $5 \\cdot 4 + 1 = 21$ in total, which is exactly 42 colored squares.\n\n![](images/ukraine_2015_Booklet_p16_data_b64330d605.png)\n\n![](images/ukraine_2015_Booklet_p16_data_05f5b51690.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17927,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest number $m$ such that the inequality\n\n$$\n(a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) \\geq m\n$$\n\nholds for all real numbers $a, b,$ and $c$ not equal to $0$ and satisfying the condition $\\left|\\frac{1}{a}\\right| + \\left|\\frac{1}{b}\\right| + \\left|\\frac{1}{c}\\right| \\le 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that we can consider only positive values of $a, b,$ and $c$, since the absolute values of the variables are calculated in all instances (both as the absolute values of their reciprocals and as the squares of the variables). So for now, let $a, b, c > 0$.\n\nBy the geometric-harmonic means inequality, we have\n\n$$\nabc \\geq \\left( \\frac{3}{a^{-1} + b^{-1} + c^{-1}} \\right)^{3} \\geq 1.\n$$\n\nThe arithmetic-geometric means inequality gives us\n\n$$\na^2 + 4b^2 + 4c^2 \\geq 9 \\cdot \\sqrt[3]{a^2 b^8 c^8}, \\quad b^2 + 4c^2 + 4a^2 \\geq 9 \\cdot \\sqrt[3]{a^8 b^2 c^8}, \\quad \\text{and} \\quad c^2 + 4a^2 + 4b^2 \\geq 9 \\cdot \\sqrt[3]{a^8 b^8 c^2}.\n$$\n\nFrom this, we obtain\n\n$$\n\\begin{aligned}\n(a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) &\\geq 729 \\cdot \\sqrt[3]{a^{18}b^{18}c^{18}} \\\\\n&= 729 \\cdot (abc)^2 \\\\\n&\\geq 729.\n\\end{aligned}\n$$\n\nSince equality holds for $a = b = c = 1$, we see that the maximum $m$ we are searching for is equal to $729$. Equality holds if the absolute values of all variables are equal to $1$, and we therefore have eight possible triples of variables for which equality holds, namely $(a, b, c) = (\\pm 1, \\pm 1, \\pm 1)$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17928,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 4.1, $AB$ and $CD$ are two chords in the circle $\\odot O$ meeting at point $E$, and $AB \\neq CD$. $\\odot I$ is tangent to $\\odot O$ internally at point $F$, and is tangent to the chords $AB$ and $CD$ at points $G$ and $H$, respectively. $l$ is a line passing through $O$, meeting $AB$, $CD$ at points $P$, $Q$, respectively, such that $EP = EQ$. Line $EF$ meets the line $l$ at point $M$. Prove that the line through $M$ and parallel to the line $AB$ is tangent to the circle $\\odot O$.",
    "options": [],
    "answer": "See solution",
    "solution": "Draw a line parallel to $AB$ and tangent to circle $\\odot O$ at point $L$, which meets the common tangent line to these two circles at a point $S$. Let $R$ be the intersection of lines $FS$ and $BA$, and join segments $LF$ and $GF$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p248_data_1301d0003b.png)\n\nFirst, we prove that the points $L$, $G$ and $F$ are collinear. As both $SL$ and $SF$ are tangent to $\\odot O$, $SL = SF$; as both $RG$ and $RF$ are tangent to $\\odot I$, $RG = RF$.\n\nAs $SL \\parallel RG$, we have $\\angle LSF = \\angle GRF$, so\n\n$$\n\\begin{aligned}\n\\angle LFS &= \\frac{180^\\circ - \\angle LSF}{2} = \\frac{180^\\circ - \\angle GRF}{2} \\\\\n&= \\angle GFR,\n\\end{aligned}\n$$\n\nand hence $L$, $G$ and $F$ are collinear.\n\nSimilarly, draw a line parallel to $CD$ and tangent to $\\odot O$ at point $J$, then $F$, $H$ and $J$ are collinear.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p248_data_419dbcbd86.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17929,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$. Denote the first $n$ primes in increasing order by $p_1, p_2, \\dots, p_n$ (i.e., $p_1 = 2, p_2 = 3, \\dots$). Let\n$$\nA = p_1^{p_1} p_2^{p_2} \\cdots p_n^{p_n}.\n$$\nFind all positive integers $x$ such that $\\dfrac{A}{x}$ is even and has exactly $x$ distinct positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}$, where $0 \\leq \\alpha_1 \\leq 1$ and $0 \\leq \\alpha_i \\leq p_i$ for $i = 2, 3, \\dots, n$. Then\n$$\n\\frac{A}{x} = 2^{p_1 - \\alpha_1} p_2^{p_2 - \\alpha_2} \\cdots p_n^{p_n - \\alpha_n}.\n$$\nThe number of positive divisors of $\\frac{A}{x}$ is\n$$\n(p_1 - \\alpha_1 + 1)(p_2 - \\alpha_2 + 1) \\cdots (p_n - \\alpha_n + 1).\n$$\nWe require\n$$\n(p_1 - \\alpha_1 + 1)(p_2 - \\alpha_2 + 1) \\cdots (p_n - \\alpha_n + 1) = x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}.\n$$\nBy induction on $n$, the only solution is $\\alpha_1 = \\alpha_2 = \\cdots = \\alpha_n = 1$, so $x = 2 \\cdot 3 \\cdots p_n = p_1 p_2 \\cdots p_n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17930,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following equation\n\n$$\n2^x = 3^y + 509\n$$\n\nin the set of positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "One solution is $(x, y) = (9, 1)$. Let us prove that the proposed equation has no solution for $y \\geq 2$.\n\nWe check directly for $y = 2$. Continue, assume that $y \\geq 3$. Then $3^y \\equiv 0 \\pmod{27}$ and thus we get $2^x \\equiv 509 \\equiv 23 \\pmod{27}$.\n\nThe remainders of $2^x$ upon division by 27, for $x = 1, 2, 3, \\dots$, repeat in periods of the form\n\n$(2, 4, 8, 16, 5, 10, 20, 13, 26, 25, 23, 19, 11, 22, 17, 7, 14, 1)$.\n\nSince this period is of length 18 and the remainder 23 appears in the eleventh position, we conclude that\n\n$$\nx = 18k + 11 \\text{ for some } k \\in \\mathbb{Z}_{\\geq 0}.\n$$\n\nLet us now consider the proposed equation modulo 19. By Fermat's theorem, we get $2^{18} \\equiv 1 \\pmod{19}$, and we may directly calculate $2^{11} \\equiv 15 \\pmod{19}$, which gives\n\n$$\n2^x \\equiv 2^{18k+11} = (2^{18})^k \\cdot 2^{11} \\equiv 15 \\pmod{19}.\n$$\n\nTherefore, the RHS of the proposed equation has to be congruent to 15 modulo 19, and thus we get that $3^y \\equiv 15 - 509 \\equiv -494 \\equiv 0 \\pmod{19}$, which is a contradiction.\n\nThus $(x, y) = (9, 1)$ is the unique solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17931,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, with $AB \\neq BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \\neq B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "We need the following well-known result.\n\n**Lemma.** In a triangle $KLM$ with $KL \\neq KM$, $R$ is a point on $LM$ such that $KR$ is perpendicular to $LM$. Let $U$ be a point on the line $KR$. Let the lines $LU$ and $MU$ intersect $KM$ and $KL$, respectively, at $S$ and $T$. Then $L, T, S, M$ are concyclic if and only if $U$ is the orthocenter of triangle $KLM$.\n\nSuppose that $C, X, Y, A$ are concyclic. Let the lines $AP$ and $CQ$ intersect at $Z$. Since $BP = BQ$ we have $BQ^2 = BX \\cdot BA$, so $BQ$ is tangent to the circumcircle of triangle $AXQ$. Hence $\\angle BQX = \\angle XAQ = \\angle XCY$. Therefore the points $Q, X, P, C$ are concyclic. Further, $\\angle XAP = \\angle XPQ = \\angle XCQ$. Adding the two we get $\\angle YCZ = \\angle YAZ$. This proves that the points $A, Y, Z, C$ are concyclic. Applying the lemma to triangle $QAC$ we get that $P$ is the orthocenter of triangle $QAC$. Hence $\\angle CXA = \\angle CYA = 90^\\circ$.\n\nFor the converse, suppose that $CX$ is perpendicular to $AB$. Let the line $CP$ intersect the circumcircle of triangle $AXC$ at $Y'$, and let the lines $AY'$ and $BP$ intersect at $Q'$. Note that $P$ is the orthocenter of triangle $Q'AC$. Hence if $AP$ intersects $Q'C$ at $Z$, then $Z$ lies on the circle $\\gamma$. Note that $\\angle Q'CX = \\angle Q'CY' - \\angle XCY' = \\angle ZAY' - \\angle XAY' = \\angle XAP = \\angle Q'PX$. This shows that the points $Q', X, P, C$ are concyclic. Hence $\\angle XQ'B = \\angle XCP = \\angle XAY'$, so $BQ'$ is tangent to the circumcircle of triangle $AXQ'$. Therefore $BQ'^2 = BX \\cdot BA = BP^2$ and hence $Q = Q'$. This completes the solution. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17932,
    "subject": "Mathematics (Olympiad)",
    "question": "How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, and 6 such that the product of the digits is 60?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $60 = 2^2 \\times 3 \\times 5$, only the digits 1, 2, 3, 4, 5, 6 can be used. The only combinations of four of these digits whose product is 60 are $(1, 2, 5, 6)$, $(1, 3, 4, 5)$, and $(2, 2, 3, 5)$.  \n\nThere are $24$ ways to arrange four different digits and $12$ ways to arrange four digits of which two are the same. So the total number of required 4-digit numbers is $24 + 24 + 12 = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17933,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, point $E$ lies in the opposite half-plane determined by line $CD$ from point $A$ so that $\\angle CDE = 110^\\circ$. Point $F$ lies on $\\overline{AD}$ so that $DE = DF$, and $ABCD$ is a square. What is the degree measure of $\\angle AFE$?\n\n![](images/2021_AMC10A_Solutions_Fall_p2_data_bae2ec7032.png)\n\n(A) 160  (B) 164  (C) 166  (D) 170  (E) 174",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle EDF = 360^\\circ - \\angle ADC - \\angle CDE = 360^\\circ - 90^\\circ - 110^\\circ = 160^\\circ$. Because $\\triangle DEF$ is isosceles, angles $DEF$ and $DFE$ have an equal measure of $\\frac{180^\\circ - 160^\\circ}{2} = 10^\\circ$. Hence $\\angle AFE = 180^\\circ - \\angle DFE = 180^\\circ - 10^\\circ = 170^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17934,
    "subject": "Mathematics (Olympiad)",
    "question": "Quadrilateral $ABCD$ has right angles at $A$ and $D$. A circle of radius $10$ fits neatly inside the quadrilateral and touches all four sides. The length of edge $BC$ is $24$. The midpoint of edge $AD$ is called $E$ and the midpoint of edge $BC$ is called $F$. What is the length of $EF$?\n\nA) $\\frac{43}{2}$\nB) $\\frac{13}{2}\\sqrt{11}$\nC) $\\frac{33}{5}\\sqrt{11}$\nD) $22$\nE) $\\frac{45}{2}$\n\n![](images/NLD_ABooklet_2020_p7_data_2ee2cc80c8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$22$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17935,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ is $3m$. Consider the points:\n$$\n(1, 0, 0),\n(2, 0, 0),\n\\dots,\n(m, 0, 0),\n(0, 1, 0),\n(0, 2, 0),\n\\dots,\n(0, m, 0),\n(0, 0, 1),\n(0, 0, 2),\n\\dots,\n(0, 0, m)\n$$\n\nLet $P = (x, y, z)$. What point $P$ minimizes the sum of distances from $P$ to all these points?\n\nIf $n$ is $3m-1$ or $3m-2$, then we drop $(0, 0, m)$ and/or $(0, m, 0)$. For which $n$ is the minimizing point $P$ inside the convex hull of the given points?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of distances from $P = (x, y, z)$ to the points is:\n$$\n2m|x| + |x-1| + |x-2| + \\dots + |x-m| + \\text{(similar terms in $y$ and $z$)}\n$$\nWe can minimize for $x$, $y$, and $z$ separately. Increasing $|x|$ by $k > 0$ increases $2m|x|$ by $2mk$, but at most reduces the other terms by $mk$, so the total sum increases by at least $mk$. Similarly, reducing $|x|$ by $k > 0$ reduces the sum by at least $mk$. Thus, the minimum occurs at $|x| = 0$, and similarly for $y$ and $z$. Therefore, $P$ is at the origin, which is outside the convex hull.\n\nIf $n$ is $3m-1$ or $3m-2$, we drop $(0, 0, m)$ and/or $(0, m, 0)$. The argument is even stronger for $y$ and $z$. For $x$, the worst case is $(2m-2)|x| + |x| + |x-1| + \\dots + |x-m|$. If $n > 6$ ($m > 2$), then $2m-2 > m$ and the same argument applies. If $n = 5$, we have $3|x| + |x| + |x-1|$ for $x$, and the argument still works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17936,
    "subject": "Mathematics (Olympiad)",
    "question": "When $p = 2$, $n = 63$. When $p = 3$, $n = 103$. When $p = 5$, $n = 567$.\n\nLet\n$$\n\\begin{aligned}\n(p^2 + 2)^2 - 9(p^2 - 7) &= p^4 - 5p^2 + 4 + 63 \\\\\n&= (p^2 - 1)(p^2 - 4) + 63 \\\\\n&= (p - 1)(p + 1)(p - 2)(p + 2) + 63\n\\end{aligned}\n$$\n\nFind the smallest possible value of the digit sum of $n = (p^2+2)^2 - 9(p^2-7)$ for integer $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "When $p \\neq 3$, both $(p-2)(p-1)$ and $(p+1)(p+2)$ are multiples of 3, so $n$ is a multiple of 9 and its digit sum is a multiple of 9. When $p = 3$, $n = 103$ and the digit sum is $1 + 0 + 3 = 4$. Thus, the smallest possible digit sum is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17937,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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. $\\left(n + \\frac{1}{2n}\\right)^2 > n^2 + 1$, hence 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. Assume now this is proven up to $n$, and prove by induction for $n+1$.\n\nA $\\Delta$ of side $n+1$ triangle placed at the top corner will use $n^2 + 2$ triangles $\\Delta_1$, according to the induction hypothesis. It remains a trapezoidal strip at the bottom, of length of the nonparallel sides $n + 1 + \\frac{1}{2(n+1)} - n - \\frac{1}{2n} = 1 - \\frac{1}{2n(n+1)}$, and base 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 above, since $(n+1) \\cdot 1 + n \\cdot \\frac{1}{2n(n+1)} = n + 1 + \\frac{1}{2(n+1)}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17938,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a *quasi-square* be a figure consisting of an $n \\times n$ square ($n \\geq 4$) with one additional $1 \\times 1$ square attached to its side, sharing a side with one of the unit squares of the $n \\times n$ square. Quasi-squares can be rotated and reflected, but are not allowed to overlap. Is it always possible to fill the plane with such quasi-squares?\n\n![](images/Ukraine_2020_booklet_p7_data_2e1cdd5596.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Yes, if the extra $1 \\times 1$ square shares a vertex with the big square; no, otherwise.\n\n**Solution.** An example of filling the plane is shown in problem 8.3. Attaching the extra $1 \\times 1$ square in any other place makes filling the plane impossible. Consider such a figure:\n\n![](images/Ukraine_2020_booklet_p7_data_aed0387bb7.png)\n\nThe yellow square can be covered in two ways. If it is covered by another extra square:\n\n![](images/Ukraine_2020_booklet_p7_data_7ff49c2fa0.png)\n\nthen it is impossible to cover the black square. If the yellow square is covered with one of the unit squares of the $n \\times n$ square, it must be one of the edge squares. However, it is then not possible to cover both black squares simultaneously.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17939,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a finite set $\\mathcal{S}$ of points in the plane is *balanced* if, for any two different points $A$ and $B$ in $\\mathcal{S}$, there is a point $C$ in $\\mathcal{S}$ such that $AC = BC$. We say that $\\mathcal{S}$ is *centre-free* if for any three different points $A, B$ and $C$ in $\\mathcal{S}$, there is no point $P$ in $\\mathcal{S}$ such that $PA = PB = PC$.\n\n(a) Show that for all integers $n \\ge 3$, there exists a balanced set consisting of $n$ points.\n\n(b) Determine all integers $n \\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) For $n$ odd, we may take $S$ to be the set of vertices of a regular $n$-gon $\\mathcal{P}$.\n\nIt seems obvious that $S$ is balanced but we shall prove it anyway. Let $A$ and $B$ be any two vertices of $\\mathcal{P}$. Since $n$ is odd, one side of the line $AB$ contains an odd number of vertices of $\\mathcal{P}$. Thus if we enumerate the vertices of $\\mathcal{P}$ in order from $A$ around to $B$ on that side of $AB$, one of them will be the middle one, and hence be equidistant from $A$ and $B$.\n\nFor $n$ even, say $n = 2k$, we may take $S$ to be the set of vertices of a collection of $k$ unit equilateral triangles, all of which have a common vertex $O$, and exactly one pair of them has a second common vertex. Note that apart from $O$, all of the vertices lie on the unit circle centred at $O$.\n\nThe reason why this works is as follows. Let $A$ and $B$ be any two points in $S$. If they are both on the circumference of the circle, then $OA = OB$ and $O \\in S$. If one of them is not on the circumference, say $B = O$, then by construction there is a third point $C \\in S$ such that $\\triangle ABC$ is equilateral, and so $AC = BC$. The cases for $n = 11$ and $n = 12$ are illustrated below.\n\n![](images/Australian-Scene-combined-2015_p142_data_daaba9419a.png)\n\n(b) We claim that a balanced centre-free set of $n$ points exists if and only if $n$ is odd.\n\nNote that the construction used in the solution to part (a) is centre-free. We shall show that there is no balanced centre-free set of $n$ points if $n$ is even.\n\nFor any three points $A, X, Y \\in S$, let us write $A \\to \\{X, Y\\}$ to mean $AX = AY$. We shall estimate the number of instances of $A \\to \\{X, Y\\}$ in two different ways. First, note that if $A \\to \\{X, Y\\}$ and $A \\to \\{X, Z\\}$ where $Y \\neq Z$, then $S$ cannot be centre-free because $AX = AY = AZ$. Hence for a given point $A$, there are at most $\\lfloor \\frac{n-1}{2} \\rfloor$ pairs $\\{X, Y\\}$ such that $A \\to \\{X, Y\\}$. Since there are $n$ choices for $A$, the total number of instances of $A \\to \\{X, Y\\}$ is at most $n \\lfloor \\frac{n-1}{2} \\rfloor$.\n\nOn the other hand, since $S$ is balanced, for each pair of points $X, Y \\in S$, there is at least one point $A$ such that $A \\to \\{X, Y\\}$. Since the number of pairs $\\{X, Y\\}$ is $\\binom{n}{2}$, the total number of instances of $A \\to \\{X, Y\\}$ is at least $\\binom{n}{2}$. If we combine our estimates, we obtain $n \\lfloor \\frac{n-1}{2} \\rfloor \\ge \\binom{n}{2}$, which simplifies to $\\lfloor \\frac{n-1}{2} \\rfloor \\ge \\frac{n-1}{2}$. This final inequality is impossible if $n$ is even. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17940,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ for which there exist 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": "See solution",
    "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 even numbers, consider $a' = \\frac{a}{2}$, $b' = \\frac{b}{2}$, $c' = \\frac{c}{2}$. Then $\\frac{n}{2} = a' + b' + c' \\in \\mathbb{N}$ and $(a'+b')(b'+c')(c'+a') = \\left(\\frac{k}{8}\\right)^3 \\in \\mathbb{N}$, which contradicts the minimality of $n$.\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 is not a perfect cube. 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. Then $a+b$ 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 $a+b \\ge 8$.\n\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": 17941,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $AB_1$, $AC_1$, $AD_1$ — высоты граней $ACD$, $ABD$, $ABC$ тетраэдра $ABCD$. Точки пересечения высот этих граней лежат на прямых $AB_1$, $AC_1$, $AD_1$ и отличны от точки $A$. Докажите, что середины рёбер $AB$, $AC$, $AD$ лежат в одной плоскости.\n\n![](images/Rusija_2009_p63_data_cb53588ace.png)\n\n![](images/Rusija_2009_p63_data_1f149812d2.png)\n\n_Замечание 1._ Опустим перпендикуляры из произвольной точки $A'$, лежащей в плоскости $BCD$, на прямые $BC$, $CD$, $BD$. Их основания лежат на одной прямой тогда и только тогда, когда $A'$ лежит на описанной окружности треугольника $BCD$. Эта прямая называется *прямой Симсона* точки $A'$.\n\n_Замечание 2._ Тетраэдры, удовлетворяющие условию задачи, существуют.",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть $A'$ — проекция точки $A$ на плоскость $BCD$. По теореме о трёх перпендикулярах точки $B_1$, $C_1$, $D_1$ являются проекциями $A'$ на прямые $CD$, $BD$, $BC$. 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 $$\\angle(BC, A'C) = \\angle(BD, A'D)$$ (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).\n\nИз равенства $\\angle(BC, A'C) = \\angle(BD, A'D)$ следует, что точка $A'$ лежит на описанной окружности треугольника $BCD$ 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 рёбер $AB$, $AC$, $AD$ также лежат в $\\beta$ (так как треугольники $ABAA'$, $ACAA'$, $ADAA'$ прямоугольные). Это и требовалось доказать.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17942,
    "subject": "Mathematics (Olympiad)",
    "question": "Висината го дели правоаголниот триаголник на два триаголници кои имаат периметри $m$ и $n$. Определи го периметарот на почетниот триаголник.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $\\triangle ABC$ е правоаголен триаголник (со теме на правиот агол во точката $C$) и нека $CH$ е негова висина. Од условот на задачата имаме\n\n![](images/Makedonija_2009_p35_data_b6e407e5cb.png)\n\n$$\nL_1 = L_{AHC} = m \\text{ и } L_2 = L_{CHB} = n.\n$$\n\nТриаголниците $AHC$, $BHC$ и $ABC$ се слични, при што коефициентите на сличност се\n\n$$\nk_1 = \\frac{AC}{AB} \\text{ и } k_2 = \\frac{BC}{AB} \\quad (1)\n$$\n\nИсто така,\n\n$$\n\\frac{L_1}{L} = k_1 \\text{ и } \\frac{L_2}{L} = k_2 \\quad (2)\n$$\n\nОд равенствата (1) имаме $AC = k_1 AB$ и $BC = k_2 AB$. Според Питагорина теорема\n\n$$\nAC^2 + BC^2 = AB^2,\n$$\n\nод каде го добиваме равенството $k_1^2 + k_2^2 = 1$. Ако (2) го замениме во претходното равенство, имаме\n\n$$\n\\left(\\frac{L_1}{L}\\right)^2 + \\left(\\frac{L_2}{L}\\right)^2 = 1, \\qquad L = \\sqrt{L_1^2 + L_2^2} = \\sqrt{m^2 + n^2}.\n$$\n\n![](images/Makedonija_2009_p35_data_83422e2b71.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17943,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy, Bogdan, and Olesia were walking along the same road from home to school. Andriy walked at a velocity of $a$ km/h for $(2-b)$ hours, Bogdan at $b$ km/h for $(2-c)$ hours, and Olesia at $c$ km/h for $(2-a)$ hours, where $a$, $b$, $c$ are real numbers. What is the distance between home and school if it is known to be an integer number?\n\n$S = 1$ km.",
    "options": [],
    "answer": "See solution",
    "solution": "Analyzing the problem, we get:\n\n$$\nS = a(2-b), \\quad S = b(2-c), \\quad S = c(2-a),\n$$\n\nwhere $S$ is a positive integer equal to the distance.\n\nWe may assume that $a \\geq b$. If $a > b$, then $2-b < 2-c$ or $b > c$. Analogously, $2-c < 2-a$ or $c > a$. Contradiction. So $a = b = c$. Then we have $S = a(2-a)$. Let us show that $S = a(2-a) \\leq 1$. Indeed, $2a - a^2 \\leq 1 \\Leftrightarrow (a-1)^2 \\geq 0$. Hence, since $S$ is a positive integer which is less than or equal to $1$, $S = 1$.\n\n*Remark.* One can think differently. We have that\n$$\nS^3 = abc(2-a)(2-b)(2-c).\n$$\nSince $a(2-a) \\leq 1$ (and analogous inequalities), $S^3 \\leq 1$, so $S = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17944,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, \\dots, x_n$ ($n \\ge 2$) be real numbers such that\n\n$$\nA = \\left| \\sum_{i=1}^{n} x_i \\right| \\neq 0\n$$\n\nand\n\n$$\nB = \\max_{1 \\le i < j \\le n} |x_i - x_j| \\neq 0.\n$$\n\nProve that for every $n$ vectors $\\alpha_1, \\dots, \\alpha_n$ on the plane, there exists a permutation $(k_1, k_2, \\dots, k_n)$ of $(1, 2, \\dots, n)$ such that\n\n$$\n\\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} \\max_{1 \\le i \\le n} |\\alpha_i|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nLet $|\\alpha_k| = \\max_{1 \\le i \\le n} |\\alpha_i|$. It is sufficient to prove that\n\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} |\\alpha_k|,\n$$\n\nwhere $S_n$ is the set of all permutations of $(1, 2, \\dots, n)$.\n\nWithout loss of generality, assume\n\n$$\n|x_n - x_1| = \\max_{1 \\le i < j \\le n} |x_j - x_i| = B,\n$$\n\n$$\n|\\alpha_n - \\alpha_1| = \\max_{1 \\le i < j \\le n} |\\alpha_j - \\alpha_i|.\n$$\n\nFor the two vectors\n\n$$\n\\beta_1 = x_1\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_n\\alpha_n,\n$$\n\n$$\n\\beta_2 = x_n\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_1\\alpha_n,\n$$\n\nwe have\n\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| & \\ge \\max\\{|\\beta_1|, |\\beta_2|\\} \\\\\n& \\ge \\frac{1}{2} (|\\beta_1| + |\\beta_2|) \\\\\n& \\ge \\frac{1}{2} |\\beta_1 - \\beta_2| \\\\\n& = \\frac{1}{2} |x_1 \\alpha_n + x_n \\alpha_1 - x_1 \\alpha_1 - x_n \\alpha_n| \\\\\n& = \\frac{1}{2} |x_1 - x_n| \\cdot |\\alpha_1 - \\alpha_n| \\\\\n& = \\frac{1}{2} B |\\alpha_n - \\alpha_1|.\n\\end{aligned} \\tag{1}\n$$\n\nNow suppose $|\\alpha_n - \\alpha_1| = x |\\alpha_k|$. Using the triangle inequality, we obtain $0 \\le x \\le 2$. So (1) becomes\n\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{1}{2} B x |\\alpha_k|. \\tag{2}\n$$\n\nOn the other hand, consider the vectors\n\n$$\n\\gamma_1 = x_1 \\alpha_1 + x_2 \\alpha_2 + \\dots + x_{n-1} \\alpha_{n-1} + x_n \\alpha_n\n$$\n\n$$\n\\gamma_2 = x_2 \\alpha_1 + x_3 \\alpha_2 + \\dots + x_n \\alpha_{n-1} + x_1 \\alpha_n\n$$\n\n$$\n\\gamma_n = x_n \\alpha_1 + x_1 \\alpha_2 + \\dots + x_{n-2} \\alpha_{n-1} + x_{n-1} \\alpha_n.\n$$\n\nThen we have",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17945,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that we can color all the edges and the diagonals of a convex $n$-polygon by $n$ given colors satisfying the following conditions:\n\n1. Each of the edges or the diagonals is colored by only one color.\n2. For any three distinct colors, there exists a triangle whose vertices are vertices of the $n$-polygon and whose three edges are colored by these three colors.",
    "options": [],
    "answer": "See solution",
    "solution": "Any odd number $n > 1$.\n\nFirst, there are $\\binom{n}{3}$ ways to choose three among $n$ colors, and $\\binom{n}{3}$ ways to choose three vertices to form a triangle. If the condition is fulfilled, all triangles should have a different color combination (a one-to-one correspondence).\n\nNote that any two segments of the same color cannot have a common endpoint.\n\nAs each color combination is used in exactly one triangle, for each color there should be exactly $\\binom{n-1}{2}$ triangles which have one side in this color, so there should be exactly $\\frac{n-1}{2}$ lines of this color. Therefore, $n$ is odd.\n\nNow, we give a construction method for any odd $n$:\n\nAssume the polygon is a regular $n$-gon. First, color the $n$ sides of the polygon in the $n$ distinct colors. Then, for each side, color those diagonals that are parallel to this side with the same color.\n\nIn this way, for each color, there are $n$ diagonals colored in this color, and each of these diagonals is of a different length.\n\nFurthermore, for any two triangles with all vertices in the polygon, they must have different color combinations. Suppose, on the contrary, that two triangles have exactly the same three colors as their sides. Since all sides with the same color are parallel, the two triangles must be similar. For their vertices to be on the same circle, they must be the same triangle, which is a contradiction. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17946,
    "subject": "Mathematics (Olympiad)",
    "question": "Given vectors $\\vec{a} = (1 + 2^m,\\ 1 - 2^m)$ and $\\vec{b} = (4^m - 3,\\ 4^m + 5)$, where $m$ is real, find the minimum value of the dot product $\\vec{a} \\cdot \\vec{b}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t = 2^m$, so $\\vec{a} = (1 + t,\\ 1 - t)$ and $\\vec{b} = (t^2 - 3,\\ t^2 + 5)$. Then,\n\n$$\n\\begin{aligned}\n\\vec{a} \\cdot \\vec{b} &= (1 + t)(t^2 - 3) + (1 - t)(t^2 + 5) \\\\\n&= (t^2 - 3) + t(t^2 - 3) + (t^2 + 5) - t(t^2 + 5) \\\\\n&= (t^2 - 3 + t^2 + 5) + t(t^2 - 3 - t^2 - 5) \\\\\n&= (2t^2 + 2) + t(-8) \\\\\n&= 2t^2 - 8t + 2 \\\\\n&= 2(t^2 - 4t + 1)\n\\end{aligned}\n$$\n\nThis quadratic achieves its minimum at $t = 2$, so the minimum value is:\n\n$$\n2(2^2 - 4 \\times 2 + 1) = 2(4 - 8 + 1) = 2(-3) = -6.\n$$\n\nThus, the minimum value is $\\boxed{-6}$, achieved when $t = 2$ (i.e., $m = 1$).",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 17947,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that for every positive integer $n$, $$\\prod_{d|n} f(d) = n,$$ where the product is over all positive divisors $d$ of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is the function\n\n$$\nf(n) = \\begin{cases} p & \\text{if } n = p^k \\text{ for some prime } p \\text{ and } k \\ge 1, \\\\ 1 & \\text{otherwise.} \\end{cases}\n$$\n\nFor $n = 1$, the only positive divisor is 1, so $f(1) = 1$.\n\nFor any prime $p$, we prove by induction on $k$ that $f(p^k) = p$. For the base case $k = 1$, $f(p) = f(1)f(p) = p$. Assuming $f(p^j) = p$ for $j = 1, 2, \\dots, k$, we have\n\n$$\np^k f(p^{k+1}) = f(1)f(p)\\cdots f(p^{k+1}) = p^{k+1}.\n$$\n\nThis implies $f(p^{k+1}) = p$. So $f(p^k) = p$ for all $k \\in \\mathbb{Z}^+$ by induction.\n\nNow, consider $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_s^{\\alpha_s}$ where $s \\ge 2$ and $\\alpha_j \\in \\mathbb{Z}^+$ for each $j$. The positive divisors of $n$ include the prime powers $p_j^k$. Since\n\n$$\n\\prod_{j=1}^{s} \\prod_{k=1}^{\\alpha_j} f(p_j^k) = \\prod_{j=1}^{s} \\prod_{k=1}^{\\alpha_j} p_j = \\prod_{j=1}^{s} p_j^{\\alpha_j} = n = \\prod_{d|n} f(d),\n$$\n\nwe have $\\prod_{\\substack{d|n \\\\ d \\ne p_j^k}} f(d) = 1$. Thus, the images of those positive divisors of $n$ different from a power of prime are 1. In particular, $f(n) = 1$.\n\nLastly, we check that the proposed function satisfies the condition. Indeed, for any $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_s^{\\alpha_s} > 1$, we have\n\n$$\n\\prod_{d|n} f(d) = \\prod_{j=1}^{s} \\prod_{k=1}^{\\alpha_j} f(p_j^k) = n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17948,
    "subject": "Mathematics (Olympiad)",
    "question": "Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the spheres with radii $11$, $13$, and $19$ have centers $P$, $Q$, and $R$, respectively, and let the three circles have common radius $r$. Segments $\\overline{AP}$, $\\overline{BQ}$, and $\\overline{CR}$ are perpendicular to the plane of $\\triangle ABC$, so $\\overline{AB}$ is the projection of $\\overline{PQ}$ onto that plane. Similarly, $\\overline{AC}$ is the projection of $\\overline{PR}$ onto that plane.\n\n![](images/2022AIME_I_Solutions_p6_data_ffe724dfbb.png)\n\nIf $D$ is any point on the circle centered at $A$, then $\\triangle PAD$ is a right triangle with $AD = r$ and $PD = 11$, so $AP^2 = 121 - r^2$. Similarly, $BQ^2 = 169 - r^2$ and $CR^2 = 361 - r^2$. The Pythagorean Theorem gives $PQ^2 = AB^2 + (BQ - AP)^2$, so from $PQ = 11 + 13 = 24$, it follows that\n\n$$\n(BQ - AP)^2 = PQ^2 - AB^2 = 24^2 - 560 = 16.\n$$\n\nThus $BQ - AP = 4$ and $BQ^2 - AP^2 = (169 - r^2) - (121 - r^2) = 48$. Hence $BQ + AP = \\frac{48}{4} = 12$, $BQ = 8$, $AP = 4$, and $r^2 = 105$.\n\nBecause $CR = \\sqrt{19^2 - r^2} = 16$, it follows that\n\n$$\nAC^2 = PR^2 - (CR - AP)^2 = (11 + 19)^2 - (16 - 4)^2 = 900 - 144 = 756.\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 17949,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We will prove that there are only two families of solutions:\n\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.\n\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.\n\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\n**Step II:** $a = 1$.\n\nSuppose, 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 $a$).\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.**\n\nIf $b = 1$, 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 $b = 2^{k-1} - 1$, $c = 2^{k-1} + 1$.\n\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": 17950,
    "subject": "Mathematics (Olympiad)",
    "question": "For a trapezoid $ABCD$ with $AB \\parallel CD$, suppose that $A$, $B$, $C$, $D$ lie in the clockwise direction.\n\nLet $\\Gamma_1$ be the circle centered at $A$ and passing through $B$. Let $\\Gamma_2$ be the circle centered at $C$ and passing through $D$.\n\nLet $P$ be the intersection (distinct from $B$, $D$) of the line $BD$ and the circle $\\Gamma_1$.\n\nLet $\\Gamma$ be the circle with diameter $PD$.\n\nLet $X$ be the intersection (distinct from $P$) of $\\Gamma$ and $\\Gamma_1$.\n\nLet $Y$ be the intersection (distinct from $D$) of $\\Gamma$ and $\\Gamma_2$.\n\nLet $Q$ be the intersection of $\\Gamma_2$ and the circumcircle of $XBY$.\n\nShow that $B$, $D$, $Q$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to show that $X$, $B$, $Y$, $Q$ are concyclic under the assumption that $Q$ is the intersection of the line $BD$ and $\\Gamma_2$. There are six cases depending on the ordering of $P$, $D$, $B$, $Q$. Although we should consider all those six cases, we here give the proof only for the case where the order is $P$, $D$, $B$, $Q$.\n\nLet $\\angle ABP = \\alpha$, $\\angle DPX = \\beta$. Since $AB \\parallel CD$, we have $\\angle ABP = \\angle BDC = \\alpha$. For the circle $\\Gamma_1$, the inscribed angle for $PB$ equals $\\frac{\\pi}{2} - \\alpha$; hence we have $\\left(\\frac{\\pi}{2} - \\alpha\\right) + \\angle PXB = \\pi$. That is, $\\angle PXB = \\frac{\\pi}{2} + \\alpha$.\n\nOn the other hand,\n\n$$\n\\angle DYQ = \\frac{1}{2} \\angle DCQ = \\frac{\\pi}{2} - \\alpha.\n$$\n\nFor the circle $\\Gamma$, $\\angle DPX = \\angle DYX = \\beta$. Now it suffices to show that $\\angle DBX = \\angle XYQ$.\n\nFor triangle $DBX$,\n\n$$\n\\angle DBX = \\pi - \\beta - \\left(\\frac{\\pi}{2} + \\alpha\\right) = \\frac{\\pi}{2} - \\alpha - \\beta.\n$$\n\nMeanwhile, since $\\angle DYQ = \\frac{\\pi}{2} - \\alpha$ and $\\angle DYX = \\beta$,\n\n$$\n\\angle XYQ = \\angle DYQ - \\angle DYX = \\frac{\\pi}{2} - \\alpha - \\beta. \\quad \\square\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17951,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real $p$ such that the inequality\n$$\n\\sqrt{a^2 + p b^2} + \\sqrt{b^2 + p a^2} \\ge a + b + (p-1)\\sqrt{ab}\n$$\nholds for any real $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $a = b = 1$, the parameter $p > 0$ must satisfy:\n$$\n\\begin{cases}\n2\\sqrt{p+1} \\ge p+1, \\\\\n2 \\ge \\sqrt{p+1}, \\\\\np \\le 3.\n\\end{cases}\n$$\n\nWe show that for $p \\in (0, 3)$, the inequality holds for any real $a$ and $b$.\n\nIf $p \\in (0, 1)$, the inequality holds trivially:\n$$\n\\sqrt{a^2 + p b^2} > a, \\quad \\sqrt{b^2 + p a^2} > b, \\quad (p-1)\\sqrt{ab} \\le 0.\n$$\n\nLet $p \\in (1, 3)$. The left-hand side (LHS) of the inequality can be interpreted as the sum of the lengths of vectors $(a, b\\sqrt{p})$ and $(b, a\\sqrt{p})$ in $\\mathbb{R}^2$. By the triangle inequality,\n$$\n\\begin{aligned}\n\\text{LHS} &= \\sqrt{a^2 + p b^2} + \\sqrt{b^2 + p a^2} = |(a, b\\sqrt{p})| + |(b, a\\sqrt{p})| \\\\\n&\\ge |(a+b, (a+b)\\sqrt{p})| = (a+b)\\sqrt{1+p}.\n\\end{aligned}\n\\tag{1}\n$$\n\nFor the right-hand side (RHS), using the AM-GM inequality:\n$$\n\\text{RHS} = a + b + (p-1)\\sqrt{ab} \\le a + b + (p-1)\\frac{a+b}{2} = \\frac{(p+1)(a+b)}{2}.\n$$\n\nNow $\\text{LHS} \\ge \\text{RHS}$, since the stronger inequality\n$$\n(a+b)\\sqrt{p+1} \\ge \\frac{(p+1)(a+b)}{2}\n$$\nis equivalent to $\\sqrt{p+1} \\le 2$, which holds for any $p \\in (1, 3)$.\n\n**Remark.** We can obtain (1) using the Cauchy-Schwarz inequality for pairs $(a, b\\sqrt{p})$ and $(1, \\sqrt{p})$:\n$$\na + p b \\le \\sqrt{a^2 + p b^2} \\cdot \\sqrt{1 + p},\n$$\nwhich implies\n$$\n\\sqrt{a^2 + p b^2} \\ge \\frac{a + p b}{\\sqrt{1 + p}}, \\quad \\sqrt{b^2 + p a^2} \\ge \\frac{b + p a}{\\sqrt{1 + p}},\n$$\nsumming these two inequalities gives (1), and analogously for the second inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17952,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex 2024-gon $A_1A_2\\ldots A_{2024}$ and 1000 points inside it, so that no three points are collinear. Some pairs of the points are connected with segments so that the interior of the polygon is divided into triangles. Every point is assigned one number among $\\{1, -1, 2, -2\\}$, so that the sum of the numbers written in $A_i$ and $A_{i+1012}$ is zero for all $i = 1, 2, \\ldots, 1012$. Prove that there is a triangle such that the sum of the numbers in some two of its vertices is zero.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, if there are two adjacent points $A_i$ and $A_{i+1}$ with opposite numbers, the problem is solved. Without loss of generality, let $A_1 = 1$ and consider all segments $A_iA_{i+1}$ for $i = 1, 2, \\ldots, 1012$. Since $A_{1013} = -1$, among the considered segments there is an odd number whose ends are one positive and one negative number. These two numbers can be $\\{-1, 2\\}$ or $\\{1, -2\\}$ and let the number of segments with ends of the first kind be $p$ and the number of segments with ends of the second kind be $q$. Due to symmetry, the number of segments $A_iA_{i+1}$ for $i = 1013, \\ldots, 2024$ ($A_{2025} \\equiv A_1$) with ends $\\{1, -2\\}$ is equal to $p$. Therefore, all segments with endpoints $\\{1, -2\\}$ are $p+q$, which is an odd number.\n\nNow consider any triangle that does not have two vertices with opposite numbers. We have the following possibilities for the three numbers:\n\n$$\n(1, 1, -2), \\quad (1, 1, 2), \\quad (-1, -1, 2), \\quad (-1, -1, -2), \\\\\n(2, 2, -1), \\quad (2, 2, 1), \\quad (-2, -2, 1), \\quad (-2, -2, -1).\n$$\n\nEach of these triangles has an even number (2 or 0) of sides with ends 1 and $-2$. Therefore, the total number of segments with ends 1 and $-2$ (counted in multiples) is an even number. But every line segment inside the 2024-gon is counted twice (once from the two triangles in which it participates), and every line segment that is a side of the 2024-gon is counted once. The resulting contradiction shows that a triangle with the requested property exists. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17953,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be positive integers. On one turn, an $n$-$m$-knight can move either horizontally by $n$ squares and vertically by $m$ squares, or vertically by $n$ squares and horizontally by $m$ squares. (For instance, the usual chess knight, all possible target squares of one move of which are depicted by bullets in the figure, is a $1$-$2$-knight.)\n\nCan an $n$-$m$-knight on an infinite chessboard (in every direction) return to the initial square in exactly $2019$ turns?\n\nAnswer: No.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider three cases:\n\n* **Exactly one of the numbers $n$, $m$ is odd.** Color the squares like on a chessboard. Every move changes the color of the square where the knight is, so after an odd number of moves, the knight is on a square of the opposite color. Thus, the knight cannot be on the initial square after $2019$ moves.\n\n* **Both numbers $n$ and $m$ are odd.** Color the horizontal lines of the board alternately black and white. Again, every move changes the color of the square where the knight is. Hence, similarly to the previous case, the knight cannot be on the initial square after $2019$ moves.\n\n* **Both numbers $n$ and $m$ are even.** Let $n = 2^k p$, $m = 2^l q$ where $p$ and $q$ are odd. W.l.o.g., assume $k \\le l$ and the knight starts from square $(0,0)$. The knight only visits squares with coordinates of the form $(2^k u, 2^l v)$ since $2^k$ divides the length of the step in either direction. Suppose the knight is on the initial square after $2019$ moves. Shortening all moves $2^k$ times while retaining their directions, we obtain a route of a $p$-$2^{l-k}q$-knight which in $2019$ moves returns to the initial square. But $p$ is odd, which means that such a route does not exist by previous cases. This contradiction shows that the knight cannot be on the initial square after $2019$ moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17954,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $\\angle B < \\angle A < 90^\\circ$. Let $I$ be the midpoint of $AB$ and $S$ the intersection of $AD$ and $BC$. Let $R$ be a variable point inside the triangle $SAB$ such that $\\angle ASR = \\angle BSR$. On the lines $AR, BR$, take the points $E, F$, respectively, so that $BE, AF$ are parallel to $RS$. Suppose that $EF$ intersects the circumcircle of triangle $SAB$ at points $H, K$. On the segment $AB$, take points $M, N$ such that $\\angle AHM = \\angle BHI$, $\\angle BKN = \\angle AKI$.\n\n**a)** Prove that the center $J$ of the circumcircle of triangle $SMN$ lies on a fixed line.\n\n**b)** On $BE, AF$, take the points $P, Q$ respectively so that $CP$ is parallel to $SE$ and $DQ$ is parallel to $SF$. The lines $SE, SF$ intersect the circumcircle of $SAB$, respectively, at $U, V$. Let $G$ be the intersection of $AU$ and $BV$. Prove that the median from vertex $G$ of the triangle $GPQ$ always passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** We will prove that $SM$ and $SN$ are isogonal in $\\angle ASB$, since $(SMN)$ touches $(SAB)$ and $J$ belongs to the line connecting $S$ and the center of $(SAB)$. Indeed, according to Steiner's theorem for pairs of isogonals, we need to show that\n\n$$\n\\frac{MA}{MB} \\cdot \\frac{NA}{NB} = \\frac{SA^2}{SB^2}.\n$$\n\nOn the other hand, since $HM$ and $KN$ are symmedians of triangles $HAB$ and $KAB$, let $Z$ be the intersection of $HK$ with $AB$. The left-hand side of the above equation can be calculated by\n\n$$\n\\frac{MA}{MB} \\cdot \\frac{NA}{NB} = \\frac{HA^2}{HB^2} \\cdot \\frac{KA^2}{KB^2} = \\frac{ZA^2}{ZB^2} = \\frac{AF^2}{BE^2}.\n$$\n\nNext, suppose that $SR, AR, BR$ meet $AB, RB, RA$ at $R', F', E'$ respectively. According to Thales's theorem and Ceva's theorem, we have\n\n$$\n\\frac{AF}{BE} = \\frac{AF}{SR} \\cdot \\frac{SR}{BE} = \\frac{AF'}{SF'} \\cdot \\frac{BE'}{SE'} = \\frac{AR'}{BR'} = \\frac{SA}{SB}.\n$$\n\nIn short, we get\n\n$$\n\\frac{MA}{MB} \\cdot \\frac{NA}{NB} = \\frac{AF^2}{BE^2} = \\frac{SA^2}{SB^2}.\n$$\n\nSo $SM, SN$ are isogonal in $\\angle ASB$ and the center of $(SMN)$ lies on the fixed line.\n\n**b)** By the lemma in *Problem 3*, $SE$ and $SF$ are isogonal with respect to $\\angle ASB$. We will prove that the median from $G$ of the triangle $GPQ$ passing through the fixed point $L$ is the midpoint of the arc $CD$ that does not contain $S$ of $(SCD)$. Rewriting the problem in a more compact form as follows:\n\nLet $SAB$ be a triangle with $I$ the midpoint of $AB$, and any two points $C, D$ on $SB, SA$. Two points $U, V$ belong to the circumcircle of triangle $SAB$ such that $SU, SV$ are isogonal in $\\angle ASB$ and $G$ is the intersection of $AU$ with $BV$. Let $d$ be the angle bisector of $\\angle ASB$, and on the line through $B$ and $A$ parallel to $d$, take the points $P$ and $Q$ satisfying $CP \\parallel SU$, and $DQ \\parallel SV$. Let $T$ be the midpoint of $PQ$. Prove that $GT$ passes through the midpoint $L$ of arc $CD$ that does not contain $S$ of the circumcircle of triangle $SCD$.\n\nLet $K$ be the second intersection of $SL$ and the circumcircle of triangle $SAB$, let $J$ be the second intersection of the circumcircle of triangle $SCD$ with the circumcircle of triangle $SAB$. It is easy to see that $TI$ is the midline of the trapezoid $AQPB$, so $TI \\parallel AQ \\parallel SL$. Therefore, by Thales's theorem, we only need to prove that\n\n$$\n\\frac{TI}{LK} = \\frac{GI}{GK}.\n$$\n\n![](images/Vietnam2023_p23_data_c20f9f9ecd.png)\n\nFirst of all, we have\n\n$$\n\\begin{aligned}\n\\frac{GI}{GK} &= \\frac{GI}{GA} \\cdot \\frac{GA}{GK} = \\frac{\\sin GKA}{\\sin GAK} \\cdot \\sin GAI \\\\\n&= \\frac{\\sin UAB}{\\sin UAK} \\cdot \\sin GKA = \\frac{UB}{UK} \\cdot \\sin GKA.\n\\end{aligned}\n$$\n\nOn the other hand, since\n\n$$\n\\angle QAD = \\angle KSA = \\angle AVK \\text{ and } \\angle QDA = \\angle VSA = \\angle VKA\n$$\n\nthen $\\triangle BUK = \\triangle AVK \\sim \\triangle QAD$ and similarly they are similar to $\\triangle PBC$. On the other hand, by the rotation predicate we have\n\n$$\n\\triangle SAD \\sim \\triangle SKL \\sim \\triangle SBC.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17955,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest real number $M$ such that for each infinite sequence $x_0, x_1, x_2, \\dots$ of real numbers satisfying\n\n(a) $x_0 = 1$ and $x_1 = 3$,\n(b) $x_0 + x_1 + \\dots + x_{n-1} \\ge 3x_n - x_{n+1}$ for all $n \\ge 1$,\n\nthe inequality\n$$\n\\frac{x_{n+1}}{x_n} > M\n$$\nholds for all $n \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible $M$ for which the given property holds is $M = 2$.\n\nWe first show that the property holds for $M = 2$. To do this, we show by induction on $n$ the stronger statement that $x_{n+1} > 2x_n > x_n + x_{n-1} + \\dots + x_0$ for all $n \\ge 0$.\n\nFor $n = 0$, this is $x_1 > 2x_0 > x_0$, which with the initial values is $3 > 2 > 1$.\n\nSuppose for the induction hypothesis that $x_{n+1} > 2x_n > x_n + x_{n-1} + \\dots + x_0$. Then for $x_{n+2}$:\n\n$$\n\\begin{aligned}\nx_{n+2} &\\ge 3x_{n+1} - (x_n + \\dots + x_0) \\\\\n&> 2x_{n+1} \\\\\n&> x_{n+1} + x_n + \\dots + x_0.\n\\end{aligned}\n$$\n\nThis completes the induction step. By induction, for all sequences $x$ satisfying (a) and (b), the inequality $\\frac{x_{n+1}}{x_n} > 2$ holds for all $n \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17956,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $F$ be positive integers, and assume $A < B < 2A$. A flea is at the number $0$ on the number line. The flea can move by jumping to the right by $A$ or by $B$.\n\nBefore the flea starts jumping, Lavaman chooses finitely many intervals $\\{m+1, m+2, \\dots, m+A\\}$ consisting of $A$ consecutive positive integers, and places lava at all of the integers in the intervals. The intervals must be chosen so that:\n\n1. Any two distinct intervals are disjoint and not adjacent.\n2. There are at least $F$ positive integers with no lava between any two intervals.\n3. No lava is placed at any integer less than $F$.\n\nProve that the smallest $F$ for which the flea can jump over all the intervals and avoid all the lava, regardless of what Lavaman does, is $F = (n - 1)A + B$, where $n$ is the positive integer such that\n$$\n\\frac{A}{n+1} \\leq B - A < \\frac{A}{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution:* Let $B = A + C$ where $\\dfrac{A}{n + 1} \\leq C < \\dfrac{A}{n}$.\n\nFirst, here is an informal sketch of the proof.\n\n**Lavaman's strategy:** Use only safe intervals with $nA + C - 1$ integers. The flea will start at position $[1, C]$ from the left, which puts him at position $[nA, nA + C - 1]$ from the right. After $n-1$ jumps, he will still have $nA - (n-1)(A+C) = A - (n-1)C > C$ distance to go, which is not enough for a big jump to clear the lava. Thus, he must do at least $n$ jumps in the safe interval, but that's possible only with all small jumps, and furthermore is impossible if the starting position is $C$. This gives him starting position 1 higher in the next safe interval, so sooner or later the flea is going to hit the lava.\n\n**Flea's strategy:** The flea just does one interval at a time. If the safe interval has at least $nA + C$ integers in it, the flea has distance $d > nA$ to go to the next lava when it starts. Repeatedly do big jumps until $d$ is between $1$ and $C \\bmod A$, then small jumps until the remaining distance is between $1$ and $C$, then a final big jump. This works as long as the first part does. However, we get at least $n$ big jumps since $\\lfloor(d-1)/A\\rfloor$ can never go down two from a big jump (or we'd be done doing big jumps), so we get $n$ big jumps, and thus we are good if $d \\bmod A$ is in any of $[1, C], [C+1, 2C], \\dots, [nC+1, (n+1)C]$, but that's everything. $\\square$\n\nLet $C = B - A$. We shall write our intervals of lava in the form $(L_i, R_i] = \\{L_i + 1, L_i + 2, \\dots, R_i\\}$, where $R_i = L_i + A$ and $R_{i-1} < L_i$ for every $i \\geq 1$. We also let $R_0 = 0$. We shall also represent a path for the flea as a sequence of integers $x_0, x_1, x_2, \\dots$ where $x_0 = 0$ and $x_j - x_{j-1} \\in \\{A, B\\}$ for every $j \\geq 0$.\n\nNow here is a detailed proof.\n\nFirst, assume $F < (n-1)A + B = nA + C$: we must prove that Lavaman has a winning strategy. Let $L_i = R_{i-1} + nA + C - 1$ for every $i \\geq 1$. (Observe that $nA + C - 1 \\geq F$.)\n\nAssume that the flea has an infinite path that avoids all the lava, which\n\n![](images/Kanada_2016_p6_data_152adb9a21.png)\n\nmeans that $x_j \\notin (L_i, R_i]$ for all $i, j \\geq 1$. For each $i \\geq 1$, let\n\n$$\nM_i = \\max\\{x_j : x_j \\leq L_i\\}, \\quad m_i = \\min\\{x_j : x_j > R_i\\}, \\quad \\text{and} \\quad J(i) = \\max\\{j : x_j \\leq L_i\\}.\n$$\n\nAlso let $m_0 = 0$. Then for $i \\geq 1$ we have\n\n$$\nM_i = x_{J(i)} \\quad \\text{and} \\quad m_i = x_{J(i)+1}.\n$$\n\nAlso, for every $i \\geq 1$, we have\n\n(a) $m_i = M_i + B$ (because $M_i + A \\leq L_i + A = R_i$);\n\n(b) $L_i \\geq M_i > L_i - C$ (since $M_i = m_i - B > R_i - B = L_i + A - B$);\n\n(c) $R_i < m_i \\leq R_i + C$ (since $m_i = M_i + B \\leq L_i + B = R_i + C$).\n\n**Claim 1:** $J(i+1) = J(i) + n + 1$ for every $i \\geq 1$. (That is, after jumping over one interval of lava, the flea must make exactly $n$ jumps before jumping over the next interval of lava.)\n\n**Proof:**\n\n$$\n\\begin{align*}\nx_{J(i)+n+1} &\\leq x_{J(i)+1} + Bn \\\\\n&= m_i + Bn \\\\\n&< R_i + C + \\left(A + \\frac{A}{n}\\right)n \\\\\n&= L_{i+1} + A + 1.\n\\end{align*}\n$$\n\nBecause of the strict inequality, we have $x_{J(i)+n+1} \\leq R_{i+1}$, and hence $x_{J(i)+n+1} \\leq L_{i+1}$. Therefore $J(i) + n + 1 \\leq J(i+1)$. Next, we have\n\n$$\n\\begin{align*}\nx_{J(i)+n+1} &\\geq x_{J(i)+1} + An \\\\\n&= m_i + An \\\\\n&> R_i + An \\\\\n&= L_{i+1} - C + 1 \\\\\n&> L_{i+1} - A + 1 \\quad \\text{(since } C < A).\n\\end{align*}\n$$\n\nTherefore $x_{J(i)+n+2} \\geq x_{J(i)+n+1} + A > L_{i+1}$, and hence $J(i+1) < J(i) + n + 2$. Claim 1 follows.\n\n**Claim 2:** $x_{j+1} - x_j = A$ for all $j = J(i) + 1, \\dots, J(i+1) - 1$, for all $i \\geq 1$. (That is, the $n$ intermediate jumps of Claim 1 must all be of length $A$.)\n\n*... (solution continues as in the original)*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17957,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube is constructed from 4 white unit cubes and 4 blue unit cubes. How many different ways are there to construct the $2 \\times 2 \\times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)",
    "options": [],
    "answer": "See solution",
    "solution": "Rotate the cube so that the number of small blue cubes showing in the front face is maximized.\n\n- If the front face contains four blue cubes, this gives 1 possible construction.\n- If the front face contains three blue cubes, then the back face must contain one blue cube, which can be in any of 4 positions. None of these constructions can be rotated into any other, so this gives 4 possible constructions. Here is why the four placements of the one blue cube on the back face lead to different patterns. Assume without loss of generality that the sole white cube on the front face is in the upper right. There is a second face with three blue cubes if and only if the back cube is placed in any position on the back face except for the upper right. Hence if the blue cube on the back face is placed at the upper right, then there will be a blue cube adjacent to no other blue cubes, and this does not occur with any other placement of the blue cube on the back face. If the blue cube on the back face is placed at the lower left, then there is one blue cube with three blue neighbors, but not if it is placed anywhere else. This leaves two cases, namely when the blue cube on the back face is placed at the upper left or lower right. In either of these cases, if the cube is rotated so that the unique other face having three blue cubes is placed in the front with the sole white cube on that face in the upper right, then the same configuration of blue and white cubes is obtained, rather than the other one. This gives 4 possible constructions.\n- Otherwise, each face of the large cube contains exactly two blue cubes. If some face contains two blue cubes in adjacent positions in a row or column, then rotate the cube to make that face the front face. Then the back face must contain two blue cubes in the diagonally opposite row or column, respectively. This gives 1 possible construction.\n- Otherwise, no blue cubes are in adjacent positions, and every face of the large cube is colored blue and white in checkerboard fashion. There is just 1 way to do this.\n\nThus the total number of possible constructions is $1 + 4 + 1 + 1 = 7$.\n\n![](images/2021_AMC10B_Solutions_Fall_p15_data_2b80848447.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17958,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose for a quadrilateral $ABCD$, $\\angle DAB = 90^\\circ$, $\\angle ABC = \\angle BCD = 60^\\circ$. If $AB = 5$ and $CD = 4$, what is the value of $BC$? Here, for a line segment $XY$, its length is also denoted by $XY$.\n\n![](images/Japan_2011_p14_data_0198688c27.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the point of intersection of the lines $AB$ and $CD$. Since $\\angle EBC = \\angle ECB = 60^\\circ$, triangle $EBC$ is equilateral. Let $x = EA$. Then triangle $ADE$ is right with $\\angle EAD = 90^\\circ$ and $\\angle AED = 60^\\circ$, so $DE = 2x$. From $EB = EC$, we have $5 + x = 4 + 2x$. Solving, $x = 1$, so $BC = EB = 1 + 5 = 6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17959,
    "subject": "Mathematics (Olympiad)",
    "question": "In a board of $2021 \\times 2021$ grids, we pick $k$ unit squares such that every picked square shares at least one vertex with at most one other picked square. Determine the maximum of $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We say two squares are connected if they share at least one vertex. The condition states that every picked square is connected to at most one other picked square. Hence, the set of picked squares can be partitioned into many connected components, where each component contains at most two squares that are connected to each other, and any two different components are totally disconnected. Note that this is true since it's impossible to have a connected component that has at least three picked squares, since then there will be a picked square that is connected to at least two other ones, which gives a contradiction.\n\nSince each component contains at most two squares that are connected, we can easily see that there are only three possible chances for a connected component, up to reflection and rotation, displayed as in the below figure.\n\n![](images/Vietnamese_mathematical_competitions_p273_data_d9111f2764.png)\n\nDenote the number of components of the first, second, and third forms from the left to the right by $x$, $y$ and $z$. For each component, we extend the area of it to each side of the squares by a length of $1/2$ (displayed by the light blue region in the figure). The extended area of the three components will be $4$, $6$, and $7$ respectively. Since any different components are disconnected, it's impossible to have any overlapping between the extended components, and all the extended components will cover the whole board $2021 \\times 2021$ extended by $1/2$ to each side, which has a total area of $2022^2$. Hence, we have\n\n$$\n4x + 6y + 7z \\leq 2022^2,\n$$\n\nimplying that $k = x + 2y + 2z \\leq \\frac{2022^2 - x - z}{3} \\leq \\frac{2022^2}{3}$. To show that it's possible to achieve $k = \\frac{2022^2}{3}$, we pick the squares of the form\n\n$$\n(2k + 1, 3l + 1),\\ (2k + 1, 3l + 2)\n$$\n\nfor $0 \\leq k \\leq 1010$, $0 \\leq l \\leq 673$, where $(i, j)$ denotes the grid at row $i$, column $j$. Therefore, the maximum value of $k$ is $\\frac{2022^2}{3}$.\n\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17960,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside triangle $ABC$ there exists a point $O$ such that $\\angle BOC = 90^\\circ - \\angle BAC$. The rays $BO$ and $CO$ intersect the sides $AC$ and $AB$ at points $K$ and $L$, respectively. Points $K_1$ and $L_1$ are selected on segments $LC$ and $BK$ so that $BK_1 = K_1K$ and $CL_1 = L_1L$. Let $M$ be the midpoint of side $BC$. Prove that $\\angle K_1ML_1$ is a right angle.\n\n![](images/Ukraine2022-23_p34_data_a3443df4be.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K_2$ and $L_2$ be the midpoints of segments $BK$ and $CL$, respectively. Then $MK_2 \\parallel AC$ and $ML_2 \\parallel AB$, so $\\angle K_2ML_2 = \\angle BAC$. From the isosceles triangles $BK_1K$ and $CL_1L$, we have $\\angle L_1L_2K_1 = \\angle K_1K_2L_1 = 90^\\circ$, so $L_1$, $L_2$, $K_1$, $K_2$ lie on the same circle with diameter $K_1L_1$. Also,\n$$\n\\angle K_2K_1L_2 = \\angle K_2L_1L_2 = 90^\\circ - \\angle BOC = \\angle BAC = \\angle K_2ML_2.\n$$\nThus, $M$ also lies on the circle with diameter $K_1L_1$, so $\\angle K_1ML_1 = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17961,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the inequality for positive $a, b, c, d$:\n\n$$\n\\left(\\frac{a+b}{2c}\\right)^2 + \\left(\\frac{b+c}{2d}\\right)^2 + \\left(\\frac{c+d}{2a}\\right)^2 + \\left(\\frac{d+a}{2b}\\right)^2 \\ge 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consecutively use the inequalities of means:\n\n$$\n\\begin{aligned}\n& \\left(\\frac{a+b}{2c}\\right)^2 + \\left(\\frac{b+c}{2d}\\right)^2 + \\left(\\frac{c+d}{2a}\\right)^2 + \\left(\\frac{d+a}{2b}\\right)^2 \n\\ge \\frac{ab}{c^2} + \\frac{bc}{d^2} + \\frac{cd}{a^2} + \\frac{da}{b^2} \n\\\\\n& \\ge 2\\sqrt{ab} \\sqrt{bc} \\frac{\\sqrt{ab}}{c} \\frac{\\sqrt{bc}}{d} + 2\\sqrt{cd} \\sqrt{da} \\frac{\\sqrt{ab}}{a} \\frac{\\sqrt{bc}}{b} \n\\\\\n& \\ge 4\\sqrt{\\frac{ab}{c} \\frac{\\sqrt{bc}}{d} \\frac{\\sqrt{cd}}{a} \\frac{\\sqrt{da}}{b}} = 4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17962,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n > 2$ and a strictly increasing sequence of positive integers $a_1 < a_2 < \\dots < a_n$. For all subsets $X$ of $\\{1, 2, \\dots, n\\}$, let $X$ be the subset such that $\\left| \\sum_{i \\notin X} a_i - \\sum_{i \\in X} a_i \\right|$ is minimized. Prove that there exists a strictly increasing sequence of positive integers $b_1 < b_2 < \\dots < b_n$ such that\n$$\n\\sum_{i \\notin X} b_i = \\sum_{i \\in X} b_i.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the subset for which $\\sum_{i \\notin X} a_i - \\sum_{i \\in X} a_i = d > 0$ attains the minimal value. This implies that $X$ also minimizes\n$$\n\\left| \\sum_{i \\notin X} a'_i - \\sum_{i \\in X} a'_i \\right|\n$$\nwhere $a'_i = 2a_i$ for all $i$. Therefore, we can assume $d$ is even.\n\nAssume there exists $k \\in X$ and $k+1 \\notin X$. Consider the set $Y = (X \\cup \\{k+1\\}) \\setminus \\{k\\}$, then\n$$\n\\sum_{i \\notin Y} a_i - \\sum_{i \\in Y} a_i = d - 2(a_{k+1} - a_k).\n$$\nSince $X$ minimizes $d$, we have $d \\leq |d - 2(a_{k+1} - a_k)|$. As $a_{k+1} > a_k$, it follows that $a_{k+1} - a_k \\geq d$.\n\n- If $n \\in X$, then the sequence $b_i = a_i$ for $i \\neq n$, and $b_n = a_n + d$ satisfies the condition.\n\n- If $n \\notin X$, let $k$ be the largest index with $k \\in X$ (so $k+1 \\notin X$):\n    - If $a_{k+1} - a_k > d$, set $b_k = a_k + \\frac{d}{2}$, $b_{k+1} = a_{k+1} - \\frac{d}{2}$, and $b_i = a_i$ for $i \\notin \\{k, k+1\\}$.\n    - If $a_{k+1} - a_k = d$, then $k \\neq 1$ (otherwise $d$ would be too large). If $k-1 \\in X$, set $b_{k-1} = a_{k-1} + \\frac{d}{2}$, $b_k = a_k + \\frac{d}{2}$, and $b_i = a_i$ for $i \\notin \\{k, k-1\\}$. If $k-1 \\notin X$, there must exist $l < k-1$ with $l \\in X$; let $l$ be the largest such index, then set $b_k = a_k + \\frac{d}{2}$, $b_l = a_l + \\frac{d}{2}$, and $b_i = a_i$ for $i \\notin \\{k, l\\}$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17963,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest real number $\\lambda$ with the following property: For any positive integers $a$, $b$, $n$ where $a + b$ is not divisible by $n$, there exists a positive integer $k < n$ such that\n\n$$\n\\left\\{ \\frac{ak}{n} \\right\\} + \\left\\{ \\frac{bk}{n} \\right\\} \\le \\lambda,\n$$\n\nwhere $\\{x\\} = x - \\lfloor x \\rfloor$ denotes the fractional part of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal $\\lambda$ is $\\frac{2}{3}$.\n\nFirst, we show $\\lambda \\geq \\frac{2}{3}$. Consider $n = 3$ with $a = b = 1$. For $k = 1, 2$:\n\n$$\n\\left\\{ \\frac{k}{3} \\right\\} + \\left\\{ \\frac{k}{3} \\right\\} = \\frac{2k}{3} \\geq \\frac{2}{3},\n$$\n\nso $\\lambda$ cannot be smaller than $\\frac{2}{3}$.\n\nNow we prove $\\lambda \\leq \\frac{2}{3}$. Let $n$ be a positive integer and $a, b$ positive integers with $n \\nmid a + b$. We need to find $1 \\leq k \\leq n - 1$ satisfying the inequality.\n\n*Case 1:* $n = 2$. Without loss of generality, take $a = 1$, $b = 1$. For $k = 1$:\n\n$$\n\\left\\{ \\frac{1}{2} \\right\\} + \\left\\{ \\frac{1}{2} \\right\\} = \\frac{1}{2} < \\frac{2}{3}.\n$$\n\n*Case 2:* $n \\geq 3$.\n\n*Subcase 2.1:* $\\gcd(a, n) = d > 1$. Let $k = \\frac{n}{d}k'$ where $1 \\leq k' \\leq d - 1$. Then:\n\n$$\n\\left\\{ \\frac{ak}{n} \\right\\} + \\left\\{ \\frac{bk}{n} \\right\\} = \\left\\{ \\frac{bk'}{d} \\right\\}.\n$$\n\nSince $\\left\\{ \\frac{bk'}{d} \\right\\} + \\left\\{ \\frac{b(d-k')}{d} \\right\\} \\in \\{0, 1\\}$, taking $k' = 1$ or $d - 1$ gives:\n\n$$\n\\left\\{ \\frac{bk'}{d} \\right\\} \\leq \\frac{1}{2} < \\frac{2}{3}.\n$$\n\n*Subcase 2.2:* $\\gcd(a, n) = 1$. Let $t$ be such that $ta \\equiv 1 \\pmod{n}$. Replacing $a$ with $ta \\pmod{n}$ and $b$ with $tb \\pmod{n}$, we may assume $a = 1$ and $1 \\leq b \\leq n - 2$.\n\nLet $e = 1 - \\frac{1}{n} - \\frac{b}{n} \\geq \\frac{1}{n}$.\n\n**When** $e \\geq \\frac{1}{3}$: Take $k = 1$:\n\n$$\n\\left\\{ \\frac{1}{n} \\right\\} + \\left\\{ \\frac{b}{n} \\right\\} = 1 - e \\leq \\frac{2}{3}.\n$$\n\n**When** $e < \\frac{1}{3}$: Let $k$ be the smallest positive integer with $ke \\geq \\frac{1}{3}$. Then:\n\n$$\nk = \\left\\lfloor \\frac{1}{3e} \\right\\rfloor \\leq \\left\\lfloor \\frac{n}{3} \\right\\rfloor < \\frac{n}{3} + 1 \\leq n - 1.\n$$\n\nLet $e = \\frac{l}{n}$ where $l$ is integer and $3l < n$ (since $e < \\frac{1}{3}$). We have:\n\n$$\nke + \\frac{k}{n} \\leq \\frac{(n + 3l - 1)(l + 1)}{3ln} \\leq \\frac{2(l + 1)}{3l + 1} \\leq 1.\n$$\n\nThis implies:\n\n$$\n\\left\\{ \\frac{k}{n} \\right\\} + \\left\\{ \\frac{kb}{n} \\right\\} = 1 - ke \\leq \\frac{2}{3}.\n$$\n\nTherefore, in all cases, we can find $k$ satisfying the required inequality with $\\lambda = \\frac{2}{3}$.\n\n$\\boxed{\\lambda = \\frac{2}{3}}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17964,
    "subject": "Mathematics (Olympiad)",
    "question": "1000 balls of mass 0.38 and 5000 balls of mass 0.038 must be packed in boxes. A box can contain any collection of balls with total mass at most 1. Find the minimum number of boxes needed.",
    "options": [],
    "answer": "See solution",
    "solution": "There can be 0, 1, or 2 balls of mass 0.38 in a box since $3 \\times 0.38 > 1$. In these three cases, the box can contain at most $\\left\\lfloor \\frac{1}{0.038} \\right\\rfloor = 26$, $\\left\\lfloor \\frac{1 - 0.38}{0.038} \\right\\rfloor = 16$, and $\\left\\lfloor \\frac{1 - 2 \\times 0.38}{0.038} \\right\\rfloor = 6$ balls with mass 0.038, respectively.\n\nIf looking for the minimum number of boxes, we may therefore assume that there are only boxes of three kinds:\n- with 0 heavy and 26 light balls;\n- with 1 heavy and 16 light balls;\n- with 2 heavy and 6 light balls.\n\nLet there be $x_0, x_1,$ and $x_2$ boxes of each kind, respectively. In order that all balls be packed, it is necessary and sufficient that\n$$\n26x_0 + 16x_1 + 6x_2 \\geq 5000\n$$\nand\n$$\nx_1 + 2x_2 \\geq 1000.\n$$\nMultiply the second inequality by 10 and add it to the first one. This gives\n$$\n26(x_0 + x_1 + x_2) \\geq 15000,\n$$\nhence\n$$\nx_0 + x_1 + x_2 \\geq \\frac{15000}{26} = 576.9\\ldots\n$$\nBecause $x_0 + x_1 + x_2$ is an integer, it follows that $x_0 + x_1 + x_2 \\geq 577$. So 577 boxes are necessary.\n\nNow let $x_0 = 0$, $x_1 = 154$, $x_2 = 423$. Then $x_0 + x_1 + x_2 = 577$, $x_1 + 2x_2 = 1000$, and $26x_0 + 16x_1 + 6x_2 = 5002 > 5000$. These relations show that 577 boxes are sufficient. The answer is 577.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17965,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it true that starting from a rectangular $m \\times n$ board without marbles\n\n![](images/Vijetnam_2006_p4_data_65e88c3c2a.png)\n\nin it, after a finite number of steps of appropriate puttings, one can put marbles into all cells of the board so that each cell is filled with the same (positive) number of marbles for all cells when:\n\ni) $m = 2004$ and $n = 2006$?\n\nii) $m = 2005$ and $n = 2006$?\n\n(At each step, it is not necessary that the four cells which are selected to put marbles into contained no marbles.)",
    "options": [],
    "answer": "See solution",
    "solution": "i) After two steps, one can put a marble into each cell of a small board of size $4 \\times 2$. One can partition the given board of size $2004 \\times 2006$ into small boards of size $4 \\times 2$. Therefore, after some steps, one can put marbles into all cells of the given board so that the number of marbles in each cell is the same for all cells.\n\nii) We now prove by contradiction that in the second case, the answer is \"no\". Suppose, on the contrary, that after some steps, there would be $k$ marbles ($k > 0$) in each cell of the given board of size $2005 \\times 2006$. Color black all cells belonging to the odd rows and consider each non-colored cell as white. Then, the number of black cells is $1003 \\times 2006$ and the number of white cells is $1002 \\times 2006$. At each step, we put exactly 2 marbles into black cells and 2 marbles into white cells. Therefore, after any number of steps, the total number of marbles in all black cells must be equal to the total number in all white cells. Consequently, we would have $1003 \\times 2006 \\times k = 1002 \\times 2006 \\times k$, which leads to $1 = 0$. This contradiction proves our assertion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17966,
    "subject": "Mathematics (Olympiad)",
    "question": "For a trapezoid $ABCD$ ($AB \\parallel CD$) it holds that $BC = AB + CD$.\n\nProve that:\n\n1. There is a point of a circle with diameter $BC$ on the leg $AD$.\n2. There is a point of a circle with diameter $AD$ on the leg $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Let $M, N$ be the centers of the legs $BC, AD$. We show that the point $N$ lies on the circle with diameter $BC$.\n\nA well-known identity yields\n\n$$\nMN = \\frac{AB + CD}{2} = \\frac{1}{2} BC.\n$$\n\nIt means that the point $N$ has the same distance from the center $M$ of the circle with diameter $BC$ as the radius of that circle. So point $N$ lies on that circle.\n\n(ii) With respect to the given condition, we can find a point $E$ on the leg $BC$ such that $|BE| = |AB|$ and $|EC| = |CD|$.\n\n![](images/65_Czech_and_Slovak_MO_2016_booklet_p9_data_9ddba1de3d.png)\n\nThe triangles $ABE$, $ECD$ are isosceles and the lines $AB$ and $CD$ are parallel, thus the fact follows:\n\n$$\n\\begin{align*}\n\\angle AED &= 180^\\circ - \\angle AEB - \\angle CED \\\\\n&= \\frac{1}{2}((180^\\circ - 2\\angle AEB) + (180^\\circ - 2\\angle CED)) \\\\\n&= \\frac{1}{2}(\\angle ABE + \\angle DCE) = 90^\\circ.\n\\end{align*}\n$$\n\nSo we finished the second part.\n\n_Remark_. If we start from the proof that the triangle $AED$ is right-angled, we will become conscious of the fact that its mutually perpendicular axes of sides $AE$ and $ED$ meet at the center $N$ of its circumscribed circle. It means that triangle $BCN$ is right-angled too, so the circle with diameter $BC$ meets the center $N$ of the side $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17967,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, $\\nu$ different points are given such that no three of them are collinear. These points are colored red, green, or black. For every line segment connecting two of these $\\nu$ points, assign an \"algebraic value\" as follows:\n\n1. If at least one endpoint is black, the segment has algebraic value $0$.\n2. If both endpoints are the same color (red or green), the segment has algebraic value $1$.\n3. If the endpoints are different colors (one red, one green), the segment has algebraic value $-1$.\n\nDetermine the least possible value of the sum of the algebraic values of all such line segments.",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be $\\kappa$ red, $\\pi$ green, and $\\mu$ black points, so $\\kappa + \\pi + \\mu = \\nu$.\n\n- The $\\kappa$ red points determine $\\binom{\\kappa}{2}$ red-red segments (value $1$ each).\n- The $\\pi$ green points determine $\\binom{\\pi}{2}$ green-green segments (value $1$ each).\n- The $\\kappa \\cdot \\pi$ red-green segments have value $-1$ each.\n- All other segments (with at least one black endpoint) have value $0$.\n\nThus, the total sum is:\n\n$$\n\\begin{aligned}\n\\Sigma &= \\binom{\\kappa}{2} + \\binom{\\pi}{2} - \\kappa\\pi \\\\\n&= \\frac{\\kappa(\\kappa-1)}{2} + \\frac{\\pi(\\pi-1)}{2} - \\kappa\\pi \\\\\n&= \\frac{(\\kappa - \\pi)^2}{2} - \\frac{\\kappa + \\pi}{2} \\\\\n&= \\frac{(\\kappa - \\pi)^2}{2} - \\frac{\\nu - \\mu}{2} \\\\\n&= \\frac{(\\kappa - \\pi)^2}{2} + \\frac{\\mu}{2} - \\frac{\\nu}{2}\n\\end{aligned}\n$$\n\nSince $(\\kappa - \\pi)^2 \\geq 0$ and $\\mu \\geq 0$, the minimum occurs when $\\kappa = \\pi$ and $\\mu = 0$ (if $\\nu$ is even), or $\\kappa = \\pi$, $\\mu = 1$ (if $\\nu$ is odd).\n\n- If $\\nu$ is even ($\\nu = 2\\rho$):\n  $$\\Sigma_{\\min} = -\\frac{\\nu}{2}$$\n  Minimum is achieved when $\\kappa = \\pi = \\frac{\\nu}{2}$, $\\mu = 0$.\n\n- If $\\nu$ is odd ($\\nu = 2\\rho + 1$):\n  $$\\Sigma_{\\min} = \\frac{-\\nu + 1}{2}$$\n  Minimum is achieved when $\\kappa = \\pi = \\frac{\\nu - 1}{2}$, $\\mu = 1$.\n\n*Examples:*\n\n- For $\\nu = 4$: $\\Sigma_{\\min} = -2$.\n- For $\\nu = 5$: $\\Sigma_{\\min} = -2$.\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p20_data_af5dccbbcc.png)\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p21_data_209ac50d56.png)\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p21_data_c030eecdb9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17968,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p \\ge 3$ be a prime number. There are $n = 2p$ integers on the blackboard. A student chooses one or more numbers and writes down their sum modulo $n$. Suppose that the remainders $1, 2, \\dots, p-1, p+1, \\dots, n-1$ appear the same number of times when the student writes down all $2^n - 1$ possible sums modulo $n$. Prove that the sum of the numbers on the blackboard is divisible by $p$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1, a_2, \\dots, a_n$ denote the numbers on the blackboard and let $X = \\{1, 2, \\dots, n\\}$ be the index set.\n\nThe remainder corresponding to an index subset $A \\subseteq X$ is the remainder of the sum $\\sum_{k \\in A} a_k$ modulo $n$ and we denote it by $\\sigma(A)$. Here we assume $\\sigma(\\emptyset) = 0$. For $0 \\le k \\le n-1$, let $S_k = |\\{A \\subseteq X \\mid \\sigma(A) \\equiv k \\pmod{n}\\}|$ denote the number of index subsets that give the remainder $k$. From the assumption, we have $S_1 = S_2 = \\dots = S_{p-1} = S_{p+1} = \\dots = S_{n-1}$.\n\nNow set $N = \\sigma(X) = a_1 + a_2 + \\dots + a_n \\pmod{n}$ and consider the sum\n\n$$\nT = \\sum_{A \\subseteq X} \\sigma(A) \\pmod{n}.\n$$\n\nSince each $k \\in X$ belongs to exactly $2^{n-1}$ subsets, we have $T \\equiv 2^{n-1}N \\pmod{n}$. On the other hand, we have\n\n$$\nT \\equiv \\sum_{k=0}^{n-1} k S_k \\equiv 0S_0 + pS_p + \\left(\\frac{(n-1)n}{2} - p\\right) S_1 \\pmod{n}.\n$$\n\nTherefore, we get $2^{n-1}N \\equiv p(S_p + 2(p-1)S_1) \\pmod{n}$ and hence $N \\equiv 0 \\pmod{p}$ since $p$ is odd.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17969,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1B_1C_1$ and $A_2B_2C_2$ be given triangles. Let $T_1$ and $T_2$ be their centers of mass, respectively. Prove that\n$$\n3\\overline{T_1T_2} = \\overline{A_1A_2} + \\overline{B_1B_2} + \\overline{C_1C_2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n\\overline{A_1A_2} = \\overline{A_1T_1} + \\overline{T_1T_2} + \\overline{T_2A_2},\n$$\n$$\n\\overline{B_1B_2} = \\overline{B_1T_1} + \\overline{T_1T_2} + \\overline{T_2B_2},\n$$\n$$\n\\overline{C_1C_2} = \\overline{C_1T_1} + \\overline{T_1T_2} + \\overline{T_2C_2}.\n$$\nSumming these, we obtain\n$$\n\\overline{A_1A_2} + \\overline{B_1B_2} + \\overline{C_1C_2} = 3\\overline{T_1T_2} + (\\overline{A_1T_1} + \\overline{B_1T_1} + \\overline{C_1T_1}) + (\\overline{T_2A_2} + \\overline{T_2B_2} + \\overline{T_2C_2}).\n$$\nFrom the properties of the centroid (center of mass),\n$$\n\\overline{A_1T_1} + \\overline{B_1T_1} + \\overline{C_1T_1} = \\overline{0},\n$$\nand similarly,\n$$\n\\overline{T_2A_2} + \\overline{T_2B_2} + \\overline{T_2C_2} = \\overline{0}.\n$$\nTherefore,\n$$\n\\overline{A_1A_2} + \\overline{B_1B_2} + \\overline{C_1C_2} = 3\\overline{T_1T_2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17970,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be real numbers in the interval $(0, \\frac{\\pi}{2})$. Prove that\n\n$$\n\\frac{\\sin a \\sin(a-b) \\sin(a-c)}{\\sin(b+c)} + \\frac{\\sin b \\sin(b-c) \\sin(b-a)}{\\sin(c+a)} + \\frac{\\sin c \\sin(c-a) \\sin(c-b)}{\\sin(a+b)} \\ge 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the *Product-to-sum formulas* and the *Double-angle formulas*, we have\n\n$$\n\\begin{aligned}\n\\sin(\\alpha - \\beta) \\sin(\\alpha + \\beta) &= \\frac{1}{2}[\\cos 2\\beta - \\cos 2\\alpha] \\\\\n&= \\sin^2 \\alpha - \\sin^2 \\beta.\n\\end{aligned}\n$$\n\nHence, we obtain\n\n$$\n\\begin{aligned}\n&\\sin a \\sin(a-b) \\sin(a-c) \\sin(a+b) \\sin(a+c) \\\\\n&= \\sin c (\\sin^2 a - \\sin^2 b)(\\sin^2 a - \\sin^2 c)\n\\end{aligned}\n$$\n\nand its analogous forms. Therefore, it suffices to prove that\n\n$$\nx(x^2 - y^2)(x^2 - z^2) + y(y^2 - z^2)(y^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge 0,\n$$\n\nwhere $x = \\sin a$, $y = \\sin b$, and $z = \\sin c$ (hence $x, y, z > 0$). Since the last inequality is symmetric with respect to $x, y, z$, we may assume that $x \\ge y \\ge z > 0$. It suffices to prove that\n\n$$\nx(y^2 - x^2)(z^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge y(z^2 - y^2)(y^2 - x^2),\n$$\n\nwhich is evident as\n\n$$\nx(y^2 - x^2)(z^2 - x^2) \\ge 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17971,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle, which is not equilateral. Denote by $O$ and $H$ its circumcenter and orthocenter, respectively. The circle $k$ passes through $B$ and touches the line $AC$ at $A$. The circle $l$ with center on the ray $BH$ touches the line $AB$ at $A$. The circles $k$ and $l$ meet in $X$ ($X \\neq A$). Show that $\\angle HXO = 180^\\circ - \\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the intersection point of the circle $l$ and the line $AC$ ($E \\neq A$). Since $k$ lies in the half-plane $ACB$ and $l$ lies in the half-plane $ABC$, the point $X$ lies inside the angle $BAC$.\n\nUsing the well-known fact about the angle between a tangent and a chord of a given circle, we get $\\angle XAE = \\angle XBA$ and $\\angle XAB = \\angle XEA$. Therefore, the triangles $ABX$ and $EAX$ are similar.\n\nDenote by $\\gamma$ the measure of $\\angle ACB$. We have $\\angle AOB = 2\\gamma$, as $\\angle AOB$ is the central angle corresponding to the inscribed angle of measure $\\gamma$ in the circumcircle of triangle $ABC$.\n\nThe line $BH$ passes through the center of $l$ and is perpendicular to its chord $AE$, so it is the axis of symmetry of the chord $AE$. Therefore $AH = HE$, and from the isosceles triangle $EAH$ with $AH \\perp BC$, we obtain $\\angle EAH = 90^\\circ - \\gamma$, hence $\\angle AHE = 2\\gamma$.\n\n![](images/SVK_Other_2015_p8_data_d3958239de.png)\n\nThe triangles $ABO$ and $EAH$ are both isosceles and have the vertex angle of the same measure, so they are similar. We have two pairs of similar triangles with the same ratio of similitude $AB : EA$. Since $O$ and $X$ lie on the same side of $AB$, and $H$ and $X$ lie on the same side of $EA$, the quadrilaterals $ABXO$ and $EAXH$ are similar.\n\nConsider the rotation with center $X$ which maps the ray $XB$ onto the ray $XA$. With respect to the derived similarity, the ray $XO$ is mapped onto the ray $XH$ under this rotation. Therefore,\n\n$$\n\\angle HXO = \\angle AXB = 180^\\circ - \\angle BAC.\n$$\n\nThe last identity follows from the fact that $\\angle AXB$ is the inscribed angle corresponding to the chord $AB$ of $k$, while $\\angle CAB$ is the angle between this chord and its tangent, and both $X$ and $C$ lie in the same half-plane determined by this chord.\n\n*Remark.* Instead of rotation, one may consider the spiral similarity which maps $ABXO$ onto $EAXH$. Since it maps $BA$ onto $AC$, it is clear that the rotating part of this map rotates all the lines by angle $180^\\circ - \\angle BAC$.\n\n**Second solution.** (*Outline*) It is possible to solve the problem using coordinates. Let $A$ be the origin and $B$ lie on the $x$-axis. Denote by $b$ the first coordinate of $B$ and let $(c, v)$ be the coordinates of $C$. Routine calculations give\n\n$$\nA = (0,0), \\quad B = (b,0), \\quad C = (c,v), \\quad H = \\left(c, \\frac{c(b-c)}{v}\\right), \\quad O = \\left(\\frac{1}{2}b, \\frac{c^2+v^2-bc}{2v}\\right), \\\\ S_k = \\left(\\frac{1}{2}b, -\\frac{bc}{2v}\\right), \\quad S_l = \\left(0, \\frac{bc}{v}\\right), \\quad X = \\left(\\frac{6bc^2}{9c^2+v^2}, \\frac{2bcv}{9c^2+v^2}\\right), \\quad Y = \\left(\\frac{1}{2}b, \\frac{bc}{2v}\\right).\n$$\n\nHere, $S_k$ and $S_l$ are the centers of $k$ and $l$, respectively, and $Y$ is the intersection point of the line $BH$ and the line $OS_k$ (these two lines are perpendicular to $AC$ and $AB$, respectively, hence they form the angle of the same measure as $\\angle BAC$).\n\nInstead of expressing the measure of $\\angle HXO$, we will show that $X, O, H$, and $Y$ are concyclic, that is, the determinant\n\n$$\n\\left| \\begin{array}{cccc} \\frac{(6bc^2)^2 + (2bcv)^2}{(9c^2 + v^2)^2} & \\frac{6bc^2}{9c^2 + v^2} & \\frac{2bcv}{9c^2 + v^2} & 1 \\\\ \\frac{b^2}{4} + \\frac{(c^2 + v^2 - bc)^2}{4v^2} & \\frac{b}{2} & \\frac{c^2 + v^2 - bc}{2v} & 1 \\\\ c^2 + \\frac{c^2(b-c)^2}{v^2} & c & \\frac{c(b-c)}{v} & 1 \\\\ \\frac{b^2}{4} + \\frac{b^2c^2}{4v^2} & \\frac{b}{2} & \\frac{bc}{2v} & 1 \\end{array} \\right|\n$$\n\nevaluates to zero. In fact, this is an easy exercise, provided we know basic tricks from linear algebra (adding a scalar multiple of one row to another row does not change the value of the determinant; the same is valid for columns; when checking only zero value, we can multiply any row/column by a nonzero scalar).\n\nIt remains to show that among the two possible values, $\\angle BAC$ and $180^\\circ - \\angle BAC$, of an inscribed angle corresponding to the chord $HO$ of the circumcircle of triangle $HOY$, the latter always applies for $\\angle HXO$. Also, the case $Y = O$ or $Y = H$ should be handled separately. One may use some kind of continuity arguments to show that we cannot \"jump\" from one value to another.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17972,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$, bounded in the interval $(0, 1)$ and such that\n\n$$\nx^2 f(x) - y^2 f(y) = (x^2 - y^2) f(x + y) - x y f(x - y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that when $x > y + 1/2$ runs through the interval $(0, n)$, then $x + y$ runs through the interval $(0, 2n - 1/2)$. Straightforward induction shows that $f$ is bounded in all intervals $(0, 2k + 1/2)$. When $0 < x < y$, it follows that $f$ is also bounded in the intervals $(-2k, 0)$. Thus, $f$ is bounded in every bounded subset of $\\mathbb{R}$.\n\nLet $x \\neq 0$. When $y \\to 0$, it follows from the boundedness of $f$ and the given equality that $f(x + y) \\to f(x)$, i.e., $f$ is continuous at $x$.\n\nFor $y = -x$, we have $f(x) - f(-x) = f(2x)$. Thus $f(-x) - f(x) = f(-2x)$ and therefore $-f(-2x) = f(2x) = 2f(x)$. It follows by induction on $n \\ge 3$ that for $x = (n-1)y$ we have $f(ny) = n f(y)$. Hence $f(r) = a r$, where $a = f(1)$ and $r \\in \\mathbb{Q}^+$. Since $f$ is odd and continuous, we obtain that $f(x) = a x$ for any $x \\neq 0$. When $x = y$, we have that $f(0) = 0$, implying that $f(x) = a x$ for any $x$. It is easily checked that $f(x) = a x$ is indeed a solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17973,
    "subject": "Mathematics (Olympiad)",
    "question": "The two equilateral triangles $ABC$ and $ADB$ (with $C \\neq D$) share the common side $AB$. The midpoints of $AC$ and $BC$ are denoted by $E$ and $F$, respectively. Show that $DE$ and $DF$ divide $AB$ into three parts of equal length.\n\n![](images/AustriaMO2012_p7_data_8c38cfeb97.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the intersections of $DE$, $DF$, and $DC$ with $AB$ be $S_1$, $S_2$, and $M$, respectively.\n\nConsider triangle $ACD$. In this triangle, $DE$ and $AM$ are medians. Therefore, their intersection $S_1$ is the centroid of this triangle, so $\\overline{AS_1} = 2 \\cdot \\overline{S_1M}$. An analogous result holds for $S_2$, which proves the assertion. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17974,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ ($n > 12$) students participating in a mathematics contest. The examination paper consists of 15 fill-in-the-blank questions. For each question, a correct answer scores 1 point; no point is awarded for a wrong or blank answer. After analyzing all possible score distributions among these $n$ students, it is found that if the sum of total scores of any 12 students is not less than 36 points, then there are at least 3 students among these $n$ who answer at least 3 identical questions correctly. Determine the smallest possible value of $n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest $n$ is $911$.\n\nWe divide the proof into two parts:\n\n1. **$n = 911$ satisfies the conditions.**\n   - If each student answers at least 3 questions correctly, then for any student there are $\\binom{15}{3} = 455$ ways to have exactly 3 correct answers.\n   - With $911$ students, by the pigeonhole principle, there are at least 3 students with 3 identical correct answers.\n   - If a student $X$ scores at most 2, then the number of other students with scores at most 3 cannot exceed 10; otherwise, picking any 11 of these plus $X$ gives a total score less than $36$.\n   - Thus, more than $911 - 11 = 900$ students remain, each with a score less than 4. Since $\\binom{4}{3} = 4$, and $4 \\times 900 > 455 \\times 2$, there are at least 3 students answering 3 identical questions correctly.\n\n2. **$n = 910$ does not satisfy the conditions.**\n   - Divide $910$ students into $455 = \\binom{15}{3}$ groups, each with exactly 2 students.\n   - In each group, both students have identical answers with only 3 correct answers indexed by the group label.\n\nTherefore, the smallest possible value of $n$ is $911$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17975,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a scalene and acute triangle with circumcentre $O$. Let $\\omega$ be the circle with centre $A$, tangent to $BC$ at $D$. Suppose there are two points $F$ and $G$ on $\\omega$ such that $FG \\perp AO$, $\\angle BFD = \\angle DGC$, and the pairs $(B, F)$ and $(C, G)$ are in different half-planes with respect to the line $AD$. Show that the tangents to $\\omega$ at $F$ and $G$ meet on the circumcircle of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider any two points $F, G$ on $\\omega$ such that $\\angle BFD = \\angle DGC$. Exploiting the isosceles triangles $\\triangle AFG$, $\\triangle AFD$, and $\\triangle ADG$, we deduce (using directed angles throughout):\n\n$$\n\\angle DBF - \\angle GCD = 180^\\circ - \\angle BFD - \\angle BDF - (180^\\circ - \\angle DGC - \\angle CDG) \\stackrel{*}{=} \\\\\n\\angle CDG - \\angle FDB = \\frac{1}{2} (\\angle DAG - \\angle DAF) = \\frac{1}{2} [(180^\\circ - 2\\angle ADG) - (180^\\circ - 2\\angle ADF)] = \\\\\n\\angle ADF - \\angle GDA = \\angle DFA - \\angle AGD = \\angle DFG - \\angle FGD \\stackrel{*}{=} \\angle BFG - \\angle FGC,\n$$\nwhere we use $\\angle BFD = \\angle DGC$ at $(*)$. Thus $BFGC$ is cyclic.\n\n![](images/2019_bmo_shortlist_p16_data_130d5af1a4.png)\n\nNow, if in addition $FG \\perp AO$, then since $A$ is the centre of $\\omega$, $AO$ is the perpendicular bisector of $FG$. But by definition, since $ABC$ is scalene, $AO$ meets the perpendicular bisector of $BC$ at $O$. Hence $O$ is the centre of $BFGC$, and thus $BFAGC$ is cyclic. Then the lines perpendicular to $AF$ at $F$ and $AG$ at $G$ (the tangents to $\\omega$) must intersect at $E$, the point antipodal to $A$ on $\\odot BFAGC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17976,
    "subject": "Mathematics (Olympiad)",
    "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 $\\gamma_2$ again at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet $\\gamma_1$ again 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 a 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": "See 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, so 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, so 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 $X'$, so $OX$ 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": 17977,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$, with $n \\geq 2$, such that the equation\n$$\nx^2 - 3x + 5 = 0\n$$\nhas a unique solution in the ring $(\\mathbb{Z}_n, +, \\cdot)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the set of all positive integers $n \\geq 2$ such that the equation $x^2 - 3x + 5 = 0$ has a unique solution in $(\\mathbb{Z}_n, +, \\cdot)$.\n\nWe show that $M = \\{11\\}$.\n\nIn $\\mathbb{Z}_{11}$, the equation becomes\n$$\nx^2 - 3x + 5 \\equiv 0 \\pmod{11}\n$$\nwhich simplifies to\n$$\n(x - 7)^2 \\equiv 0 \\pmod{11}\n$$\nso the unique solution is $x \\equiv 7 \\pmod{11}$.\n\nFor even $n$, $k^2 - 3k + 5 = (k-1)(k-2) + 3$ is always odd, so there are no solutions in $\\mathbb{Z}_n$ for even $n$; thus, $M$ contains only odd $n$.\n\nSuppose $n \\in M$ and $x$ is the unique solution. Consider $y = 3 - x$:\n$$\ny^2 - 3y + 5 = (3 - x)^2 - 3(3 - x) + 5 = x^2 - 3x + 5\n$$\nso $y$ is also a solution. By uniqueness, $x = y$, so $x = 3 - x$, or $2x = 3$. Since $n$ is odd, $2$ is invertible, so $x = 3 \\cdot 2^{-1}$ in $\\mathbb{Z}_n$.\n\nPlugging $x = 3 \\cdot 2^{-1}$ into the equation:\n$$\n(3 \\cdot 2^{-1})^2 - 3(3 \\cdot 2^{-1}) + 5 \\equiv 0 \\pmod{n}\n$$\nMultiply both sides by $4$ (since $n$ is odd):\n$$\n9 - 18 + 20 \\equiv 0 \\pmod{n} \\implies 11 \\equiv 0 \\pmod{n}\n$$\nSo $n$ divides $11$. Since $n \\geq 2$, $n = 11$.\n\nTherefore, $M = \\{11\\}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17978,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral such that $AB \\cdot CD = AD \\cdot BC$. Prove that\n$$\n\\angle CAB + \\angle CBD = \\angle DCA + \\angle ADB.\n$$\n\n![](Fig8.png)\n![](Fig9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Hint 1.* Notice that the desired equality holds for all cyclic quadrilaterals (even for those lacking the condition $AB \\cdot CD = AD \\cdot BC$). Let us choose a point $A'$ on $AC$ such that $A'BCD$ is cyclic. Then it is enough to show that $\\angle ABA' = \\angle ADA'$. It follows from the equalities\n$$\n\\frac{AA'}{\\sin ABA'} = \\frac{AB}{\\sin AA'B} = \\frac{BC}{\\sin CA'B} = \\frac{CD}{\\sin CA'D} = \\frac{AD}{\\sin AA'D} = \\frac{AA'}{\\sin ADA'}\n$$\n(see Fig. 8).\n\n*Hint 2.* Assume that $AB > BC$. The angle bisectors of the angles $B$ and $D$ meet $AC$ at the same point $K$. Moreover, the center $O$ of the circumcircle of triangle $BKD$ lies on $AC$ (this circle is the Apollonius circle; see Fig. 9). Now it is easy to see that $\\angle CAB + \\angle CBD = \\angle OBD$ and $\\angle DCA + \\angle ADB = 180^\\circ - \\angle ODB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17979,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, let $c(n)$ be the largest real number such that\n\n$$\nc(n) \\leq \\left| \\frac{f(a) - f(b)}{a - b} \\right|\n$$\n\nfor all triples $(f, a, b)$ such that\n\n- $f$ is a polynomial of degree $n$ taking integers to integers, and\n- $a, b$ are integers with $f(a) \\neq f(b)$.\n\nFind $c(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L(n) = \\text{lcm}(1, 2, \\dots, n)$. We claim that $c(n) = \\frac{1}{L(n)}$.\n\nFirst, we show that this $c(n)$ is a lower bound. For any choice of $f(x)$ and $(a, b)$, we can translate $f(x)$ vertically so that $f(b) = 0$, and then translate $f(x)$ horizontally so that $b = 0$. We only deal with this case. An $n$th degree polynomial with a root at $0$ which takes integers to integers can be written as\n\n$$\nf(x) = \\sum_{i=1}^{n} a_i \\binom{x}{i},\n$$\n\nwhere the $a_i$ are integers. Using this form for $f(x)$, we calculate $\\frac{f(a)}{a}$ for some nonzero $a$:\n\n$$\n\\frac{f(a)}{a} = \\frac{1}{a} \\sum_{i=1}^{n} a_i \\binom{a}{i} = \\sum_{i=1}^{n} a_i \\cdot \\frac{\\binom{a}{i}}{a} = \\sum_{i=1}^{n} a_i \\binom{a-1}{i-1} \\cdot \\frac{1}{i}\n$$\n\nThe $i$th summand in this sum is $\\frac{1}{i}$ times the integer $a_i \\binom{a-1}{i-1}$. Since $\\frac{1}{i}$ is an integer multiple of $\\frac{1}{L(n)}$, each summand in the sum is an integer multiple of $\\frac{1}{L(n)}$, so is $\\frac{f(a)}{a}$. Hence for $f(a) \\neq f(0) = 0$, we have that $\\left| \\frac{f(a)-f(0)}{a-0} \\right| = \\frac{k}{L(n)} \\ge \\frac{1}{L(n)}$; that is, $c(n)$ is a lower bound.\n\nSecond, we show that $c(n)$ is achievable, by finding certain $a_i$'s for $x = L(n)^2$ in\n\n$$\nf(x) = \\sum_{i=1}^{n} a_i \\binom{x}{i}.\n$$\n\nIf $p$ is any prime that divides $L(n)$, then there is some $i \\le n$ such that $p \\nmid \\frac{L(n)}{i}$, by the definition of the least common multiple. Thus\n\n$$\n\\gcd\\left(\\frac{L(n)}{1}, \\frac{L(n)}{2}, \\dots, \\frac{L(n)}{i}, \\dots, \\frac{L(n)}{n}\\right) = 1.\n$$\n\nLet $1 \\le i \\le n$. We consider\n\n$$\n\\binom{L(n)^2 - 1}{i - 1} = \\frac{(L(n)^2 - 1)(L(n)^2 - 2) \\cdots (L(n)^2 - (i - 1))}{1 \\cdot 2 \\cdots (i - 1)}.\n$$\n\nFor any prime divisor $p$ of $i$, the power of $p$ which divides $L(n)^2$ is always larger than the power of $p$ which divides $k$, for any $k \\le n$. Therefore, the same number of factors of $p$ divide $L(n)^2-k$ and $k$. By matching these factors in the numerator and denominator, we find that $p$ does not divide $\\binom{L(n)^2-1}{i-1}$; that is, $\\gcd(i, \\binom{L(n)^2-1}{i-1}) = 1$ for $1 \\le i \\le n$. Therefore, we conclude that\n\n$$\n\\gcd\\left(\\frac{L(n)}{1} \\cdot \\binom{L(n)^2 - 1}{1 - 1}, \\dots, \\frac{L(n)}{i} \\cdot \\binom{L(n)^2 - 1}{i - 1}, \\dots, \\frac{L(n)}{n} \\cdot \\binom{L(n)^2 - 1}{n - 1}\\right) = 1.\n$$\n\nBy Bézout's lemma, there are integers $a_i$ such that\n\n$$\n\\sum_{i=1}^{n} a_i \\frac{L(n)}{i} \\binom{L(n)^2 - 1}{i - 1} = 1.\n$$\n\nIt follows that, for these $a_i$'s, we have\n\n$$\nf(L(n)^2) = \\sum_{i=1}^{n} a_i \\binom{L(n)^2}{i} = \\sum_{i=1}^{n} a_i \\cdot \\frac{L(n)^2}{i} \\cdot \\binom{L(n)^2 - 1}{i - 1} = L(n)\n$$\n\nor\n\n$$\n\\frac{f(L(n)^2)}{L(n)^2} = \\frac{1}{L(n)} = c(n)\n$$\n\ncompleting our proof.\n\n(Technically, we also need $a_n \\ne 0$ to ensure that the polynomial has degree equal to $n$. However, if $a_n = 0$, then choose some $i$ with $a_i \\ne 0$, and replace $a_n$ and $a_i$ by $a_n + \\binom{L(n)^2}{i}$ and $a_i - \\binom{L(n)^2}{n}$, respectively. Clearly, the resulting sequence $a'_i$ also satisfies $\\sum_{i=1}^n a_i \\binom{L(n)^2}{i} = L(n)$.)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17980,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be defined such that\n\n$$\na_n \\le \\sum_{\\substack{1 \\le x_1 \\le n \\\\ 1 \\le x \\le n}} \\frac{n}{a x_1} = n \\left( 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} \\right)^2. \\quad (1)\n$$\n\nProve that for every $n \\ge 5$,\n\n$$\n1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} < \\sqrt{n}.\n$$\n\nGiven that $a_4 = 5$, show that $a_n < n^2$ for $n \\ge 5$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to prove by induction that for every $n \\ge 5$,\n\n$$\n1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} < \\sqrt{n}.\n$$\n\nUsing $a_4 = 5$ and substituting into (1), we get $a_n < n^2$ for $n \\ge 5$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17981,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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\n**Case 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$$\na = \\frac{b^2 + 4b + 3}{2b + 3}.\n$$\n\nSince $a \\in \\mathbb{Z}$, it results that $2b+3$ divides $b^2+4b+3$. It follows that $2b+3$ divides $3$, so $b \\in \\{-3, -2, -1, 0\\}$. We get one additional solution, namely $(a,b) = (1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17982,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer and $a_1 < a_2 < \\dots < a_{2n}$ be real numbers. If $S = \\sum_{i=1}^{2n} a_i$, $A_1 = \\sum_{i,j,\\,i<j} a_{2i}a_{2j}$, and $A_2 = \\sum_{i,j,\\,i<j} a_{2i-1}a_{2j-1}$, prove the inequality\n\n$$\n(n-1)S^2 > 4n(A_1 + A_2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we shall prove the following:\n\n*Lemma.* If $P(x) = b_0x^n + b_1x^{n-1} + b_2x^{n-2} + \\dots + b_{n-1}x + b_n$ has $n$ real distinct roots, then $(n-1)b_1^2 - 2nb_0b_2 > 0$.\n\n_Proof._ First, differentiate $n-2$ times the function $f(x)$. As a result, we have a quadratic function having two real distinct roots, and therefore its discriminant is positive.\n\nConsider the polynomial $P(x) = (x-a_1)(x-a_3)\\dots(x-a_{2n-1}) + (x-a_2)(x-a_4)\\dots(x-a_{2n})$. It is straightforward to verify that it satisfies the condition of the lemma, and the corresponding inequality is exactly the desired inequality:\n\n$$\n(n-1)S^2 > 4n(A_1 + A_2).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17983,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a set *very beautiful* if any pair of distinct elements of the set are relatively prime to each other. For a set $S$ of positive integers, define $S$ to be *beautiful* if for every three distinct elements $a, b, c \\in S$, at least one of $a$, $b$, or $c$ divides $a + b + c$. For a positive integer $N$, what is the smallest value $N$ such that every beautiful set $S$ has at most $N$ elements which are not multiples of some positive integer $n_S$ (which may depend on $S$)?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that the smallest value $N$ can take is $6$.\n\nFirst, we show that if $N \\leq 5$, the condition is not satisfied. Construct a very beautiful set $S = \\{1, 2, a_1, a_2, a_3, a_4, a_5\\}$, where $a_1, a_2, \\dots, a_5$ are odd integers $\\geq 3$ and pairwise relatively prime, chosen using the Chinese Remainder Theorem as described. In this set, every pair of distinct elements is relatively prime, and for any $n_S$, $S$ contains at most one multiple of $n_S$. Thus, the condition fails for $N \\leq 5$.\n\nNext, we show that $N = 6$ satisfies the condition. We use the following lemmas:\n\n**Lemma 1.** If positive integers $x, y, z$ satisfy $x < z$, $y < z$, and $z \\mid (x + y + z)$, then $x + y = z$.\n\n**Lemma 2.** If $x, y, z$ are odd positive integers with $x < z$, $y < z$, then $z \\mid (x + y + z)$ is never satisfied.\n\n**Lemma 3.** Let $S$ be a very beautiful set not containing $1$ or any even integer. For $x < y$ in $S$, there is at most one $z < x$ in $S$ such that $z$ does not divide $x + y$.\n\nSuppose $S$ is a very beautiful set with neither $1$ nor even numbers and has $6$ elements $x_1 < x_2 < \\dots < x_6$. Define $y_4 = x_5 + x_6$, $y_5 = x_4 + x_6$, $y_6 = x_4 + x_5$. By Lemma 3, for each $y_\\ell$ ($\\ell = 4,5,6$), there are at least two $x_k$ ($k = 1,2,3$) dividing $y_\\ell$, but not all three. Counting pairs $(k, \\ell)$, we find a contradiction, so such a set can have at most $5$ elements. Thus, a very beautiful set can have at most $7$ elements (including $1$ or an even number).\n\nNow, suppose $S$ is a very beautiful set and $n_S = 2$. Then there are at most $6$ elements not divisible by $2$. If $S$ is not very beautiful, for the minimal $n$ such that $S_n$ (the $n$ smallest elements) is not very beautiful, $|S_n| \\leq 8$. There must be a prime $p$ dividing two elements $a, b$ of $S_n$. For any $m \\notin S_n$, the beautiful set condition implies $p \\mid m$. Thus, if $n_S = p$, at most $6$ elements of $S$ are not multiples of $n_S$.\n\nTherefore, $N = 6$ is the smallest integer satisfying the requirement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17984,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD тэгш өнцөгтийн AB, BC, CD, DA талууд дээр харгалзан $M$, $N$, $P$, $Q$ цэгүүдийг авав. $MNPQ$ дөрвөн өнцөгтийн периметрийг $p$ гэе. Хэрэв $AC + BD = p$ бол $S_{MNPQ} \\le \\frac{1}{2}S_{ABCD}$ гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2013-ilovepdf-compressed_p30_data_a74e3dbdc0.png)\n\nАнхны тэгш өнцөгтийг $CD$ талынх нь хувьд тэгш хэмтэй дүрслээд, дараа нь $DA$ болон $A_1B_2$ тэнхлэгүүдийн дагуу тэгш хэмтэй дүрслэхэд манай зураг гарна. Эндээс $BB_2 = p$ болно.\n\n$NP + PQ_1 + Q_1M_2 + M_2N_3 = p$ болохын тулд $N, P, Q_1, M_2, N_3$ цэгүүд $NN_3$ шулуун дээр орших ёстой. Иймд $MNPQ$ дөрвөн өнцөгт нь заавал параллелограмм байна. Одоо $S_{MNPQ} \\le \\frac{1}{2}S_{ABCD}$ гэдгийг батлахад хангалттай.\n\n$$\nS_{MNPQ} = \\frac{1}{2} \\cdot d_1 \\cdot d_2 \\cdot \\sin \\varphi \\le \\frac{1}{2} S_{ABCD}\n$$\n\n$$\nd_1 \\cdot d_2 \\cdot \\sin \\varphi \\le S_{ABCD}\n$$\n\n$$\nd_1 \\cdot d_2 \\cdot \\sin \\varphi \\le d_1 \\cdot d_2 = S_{ABCD} = a \\cdot b \\text{ болж батлагдана. Энд}\n$$\n\n$$\nMN = NP = PQ = QM \\text{ үед тэнцэлдээ хүрнэ.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17985,
    "subject": "Mathematics (Olympiad)",
    "question": "A hacker is locked into an underground industrial complex. She is presented with a computer screen, on which appears a long message of length $72$, consisting of the symbols $E$, $X$, $I$, $T$, exactly $18$ letters of each kind in some seemingly random order. The message may be manipulated by inserting any one of the combinations\n\nEX, XE, IT, TI, IXIXI\n\nat an arbitrary place in the message. Such a combination may also be erased, wherever it may occur in the message.\n\nThe hacker may escape when the system is cracked, which happens when only the word EXIT is printed on the screen. Show that she may escape using less than $2019$ operations.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $18 = n$, so that the initial message has length $4n$, with exactly $n$ symbols of each kind.\n\nWe first establish an invariant. Assign\n\n$E = 3$, $X = -3$, $I = 2$, $T = -2$,\n\nand let $S$ denote the sum of the values of all symbols appearing in the message. Initially, $S = 0$, and the sum stays invariant under all the legal transformations.\n\nOur next observation is that we may always insert or delete the combination TETET, for\n\n$$\n\\emptyset \\mapsto \\text{TI} \\mapsto \\text{TEXI} \\mapsto \\text{TETIXI} \\mapsto \\text{TETEXIXI} \\mapsto \\text{TETETIXIXI} \\mapsto \\text{TETET},\n$$\n\nand this works also in reverse. Required are six operations.\n\nTo crack the system, first insert XE behind every I, and XE in front of every T:\n\n$I \\mapsto IXE$, $\\quad T \\mapsto XET$.\n\nThis requires at most $p_1 = 4n$ operations, after which the message has length at most $12n$. It may now be considered a sequence of the four possible strings\n\nE, X, IX, ET.\n\nSecond, expand any single X (not preceded by an I) into IXIXIX, and any single E (not succeeded by a T) into ETETET:\n\n$X \\mapsto IXIXIX$ (one operation), $E \\mapsto ETETET$ (six operations).\n\nThis requires at most $p_2 = 12n \\cdot 6 = 72n$ operations, and the message now has length at most $72n$. It is at present reduced to some binary combination of the two strings\n\nIX, ET.\n\nThird, effectuate all possible reductions\n\n$$\n\\text{ETIX} \\mapsto \\text{EX} \\mapsto \\emptyset \\quad \\text{and} \\quad \\text{IXET} \\mapsto \\text{IT} \\mapsto \\emptyset.\n$$\n\nThere can be at most $18n$ such reductions, totalling $p_3 = 18n \\cdot 2 = 36n$ operations. The hacker will be left with a message of the types\n\nETET \\dots ET or IXIX \\dots IX\n\nor an empty screen. But since the invariant $S = 0$, the screen must now, in fact, be empty.\n\nFourth, insert **EXIT**, using $p_4 = 2$ more operations. The number of operations was at most\n\n$$\np_1 + p_2 + p_3 + p_4 = 4n + 72n + 36n + 2 = 112n + 2 = 112 \\cdot 18 + 2 = 2018 < 2019.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17986,
    "subject": "Mathematics (Olympiad)",
    "question": "On the board, the numbers $1, 2, 3, \\ldots, 2023^{2024}$ are written. At each step, you may erase any two numbers $a, b$ and replace them with $\\text{gcd}(2024ab, a^2 + 254ab + b^2)$. Repeat this process until only one number $x$ remains. Find all possible values of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(a, b) = \\text{gcd}(2024ab, a^2 + 254ab + b^2)$. For any two numbers $a, b$ on the board, the new number generated will be $f(a, b)$. Consider the parity:\n\n- If $a, b$ have the same parity, then $f(a, b)$ is even.\n- If $a, b$ have different parity, then $f(a, b)$ is odd.\n\nThus, the number of odd numbers either remains the same or decreases by 2; since there are initially odd numbers, the final number $x$ must be odd.\n\nSuppose $x > 1$, and let $p$ be an odd prime divisor of $x$. Note:\n\n$$\n254 = 11 \\cdot 23 + 1 \\quad \\text{and} \\quad 2024 = 2^3 \\cdot 11 \\cdot 23.\n$$\n\nIf $p = 11$, since $11$ divides $\\text{gcd}(2024ab, a^2 + 254ab + b^2)$, then $11$ divides $a^2 + ab + b^2$. Similarly for $p = 23$.\n\n*Lemma.* If $p$ is a prime of the form $3k + 2$ and $p \\mid a^2 + ab + b^2$, then $p \\mid a$ and $p \\mid b$.\n\nSince $11$ is of the form $3k + 2$, by the lemma, $11 \\mid a$ and $11 \\mid b$. This would require all original numbers to be divisible by $11$, which is a contradiction. The same holds for $p = 23$.\n\nIf $p \\neq 11, 23$, then $p \\mid ab$ and $p \\mid a^2 + b^2$ implies $p \\mid a$ and $p \\mid b$, leading to a similar contradiction.\n\nTherefore, $x$ is odd and has no odd prime divisors, so $x = 1$.\n\n$\\boxed{1}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17987,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $M = 3^n$ and $d_i \\ge 3$ for all $i$. Show that the number of solutions to the system of modular inequalities is at least\n\n$$\n3^n \\left(1 - \\frac{1}{3}\\right)^n - 2^n = 0.\n$$\n\nFor the case $d_1 = 2$, the parity of $x$ is determined by the parity of $r_1$. Thus, we can use one of the substitutions $x = 2y$ or $x = 2y - 1$. This substitution changes the other inequalities to new ones in terms of $y$:\n\n$$\ny \\stackrel{d_2}{\\not\\equiv} s_2, \\dots, y \\stackrel{d_n}{\\not\\equiv} s_n\n$$\n\nGiven $3 \\le d_2 < \\dots < d_n$, these inequalities have a solution $y_0 \\le 3^{n-1}$. Hence, we can find a solution for the main inequalities $(x)$ not greater than $2 \\times 3^{n-1} < 3^n$.\n\nb) Assume $d_1 < d_2 < \\dots < d_n$ and $d_i \\ge i$ for each $i$. Given $\\epsilon > 0$, choose $a \\in \\left(\\frac{2}{2+\\epsilon}, 1\\right)$. There exists $N_1$ such that $1 - \\frac{1}{N_1} > a$, and since $\\frac{(2+\\epsilon)a}{2} > 1$, there is $N_2$ such that\n\n$$\n\\left(\\frac{(2 + \\epsilon)a}{2}\\right)^{N_2} > 2^{N_1}.\n$$\n\nSet $N = N_1 + N_2$. Show that for each $n > N$, the number of solutions to the system of modular inequalities is at most $(2 + \\epsilon)^n$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a), since $d_i \\ge 3$ for all $i$, the number of solutions is at least\n$$\n3^n \\left(1 - \\frac{1}{3}\\right)^n - 2^n = 0.\n$$\nThus, the number of solutions is not zero. If $d_1 = 2$, the parity of $x$ is determined by $r_1$, so we substitute $x = 2y$ or $x = 2y - 1$, reducing the system to $y \\stackrel{d_2}{\\not\\equiv} s_2, \\dots, y \\stackrel{d_n}{\\not\\equiv} s_n$. Since $3 \\le d_2 < \\dots < d_n$, there is a solution $y_0 \\le 3^{n-1}$, so $x \\le 2 \\times 3^{n-1} < 3^n$.\n\nFor part (b), with $d_1 < d_2 < \\dots < d_n$ and $d_i \\ge i$, and for any $\\epsilon > 0$, choose $a \\in \\left(\\frac{2}{2+\\epsilon}, 1\\right)$. There exist $N_1, N_2$ such that $1 - \\frac{1}{N_1} > a$ and $\\left(\\frac{(2+\\epsilon)a}{2}\\right)^{N_2} > 2^{N_1}$. Set $N = N_1 + N_2$. For $n > N$, the number of solutions in $[1, (2 + \\epsilon)^n]$ is more than\n$$\n(2 + \\epsilon)^n \\left(1 - \\frac{1}{d_1}\\right) \\cdots \\left(1 - \\frac{1}{d_n}\\right) - 2^n.\n$$\nBy bounding the product and using the properties of $a$, $N_1$, and $N_2$, we show that this quantity is positive, so the number of solutions is at most $(2 + \\epsilon)^n$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17988,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Let $x_1, x_2, \\dots, x_{2n}$ be $2n$ nonnegative real numbers such that $x_1 + x_2 + \\dots + x_{2n} = 4$. Prove that there exist nonnegative integers $p, q$ such that $q \\le n-1$, and that\n\n$$\n\\sum_{i=1}^{q} x_{p+2i-1} \\le 1, \\quad \\sum_{i=q+1}^{n-1} x_{p+2i} \\le 1.\n$$\n\n*Remark 1: the subscripts are understood modulo $2n$, i.e. if $k \\equiv \\ell \\pmod{2n}$, then $x_k = x_\\ell$.*\n\n*Remark 2: If $q = 0$, then the first sum is considered as 0; if $q = n-1$, then the second sum is considered as 0.*",
    "options": [],
    "answer": "See solution",
    "solution": "Set $A = x_1 + x_3 + \\dots + x_{2n-1}$ and $B = x_2 + x_4 + \\dots + x_{2n}$.\n\nIf one of $A, B$ is less than or equal to $1$, the problem is obvious. If $A > 1$ and $B > 1$, for $0 \\le k \\le n-1$, let $m(k) \\in \\{1, 2, \\dots, n-1\\}$ be the unique integer such that\n\n$$\n\\sum_{i=0}^{m(k)} x_{2k+2i+1} > 1\n$$\n\nand\n\n$$\n\\sum_{i=0}^{m(k)-1} x_{2k+2i+1} \\le 1.\n$$\n\nNote that, if $x_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \\dots + x_{2k+2n-2} \\le 1$, then this equation and the previous one give what we needed with $p = 2k$ and $q = m(k)$.\n\nNow suppose that\n\n$$\nx_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \\dots + x_{2k+2n-2} > 1.\n$$\n\nNow, we construct an oriented graph with $0, 1, 2, \\dots, n-1$ as vertices, and connect an edge from each $k$ to $k+m(k)+1$ (with indices modulo $n$). It is easy to see that there are loops in this graph. We may assume that $k_1 \\to k_2 \\to \\dots \\to k_t \\to k_1$ is a minimal loop.\n\nIf $t=1$, then $\\sum_{i=0}^{n-2} x_{2k_1+2i+1} \\le 1$. Setting $p=2k_1$ and $q=n-1$ gives what we want.\n\nIf $t>1$, then set\n\n$$\n\\left\\{ \\frac{k_2 - k_1}{n} \\right\\} + \\left\\{ \\frac{k_3 - k_2}{n} \\right\\} + \\dots + \\left\\{ \\frac{k_t - k_{t-1}}{n} \\right\\} + \\left\\{ \\frac{k_1 - k_t}{n} \\right\\} = s.\n$$\n\nIn other words, when $k = k_1, k_2, \\dots, k_t$, there are in total $s n$ elements in the sums above. Moreover, using the condition for defining the edges, we know that every term in $x_1, x_3, \\dots, x_{2n-1}$ appeared in the sum exactly $s$ times. On the other hand, when $k = k_1, k_2, \\dots, k_t$, there are in total $(t-s) n$ elements in the other sums. The condition for linking an edge shows that each of $x_2, x_4, \\dots, x_{2n}$ appeared exactly $t-s$ times in the above sum; this is because $x_{2j}$ appeared in the second sum if and only if $x_{2j+1}$ did not appear in the first sum.\n\nNow take the total sum of the first type when $k = k_1, k_2, \\dots, k_t$, we get $sA > t$. Similarly, taking the total sum of the second type gives $(t-s)B > t$. But $A + B > t\\left(\\frac{1}{s} + \\frac{1}{t-s}\\right) \\ge 4$. This leads to a contradiction. The problem is proved.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 17989,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(S, G, P)$ be the numbers of silver, gold, and platinum coins Bill has at some moment. Initially, $(S, G, P) = (16, 15, 14)$. Bill and Bob together have 30 gold, 30 silver, and 30 platinum coins, so $S \\leq 30$, $G \\leq 30$, $P \\leq 30$ at any time. Also, $S + G + P = 45$ always.\n\nAfter any interchange of coins, $(S, G, P)$ can become one of:\n\n1) $(S+2, G-1, P-1)$\n2) $(S-2, G+1, P+1)$\n3) $(S-1, G+2, P-1)$\n4) $(S+1, G-2, P+1)$\n5) $(S-1, G-1, P+2)$\n6) $(S+1, G+1, P-2)$\n\nIf $G = 0$ at some moment, what possible values can $P$ take?\n\n![](images/Blr2012_p11_data_7ab9a71fc9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values of $P$ when $G = 0$ are $17$, $20$, $23$, $26$, and $29$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17990,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為一正整數，並在黑板上寫下 $1, 2, \\dots, n$ 等數字。阿發和小李輪流從黑板上選擇一個數字，規則如下：\n\n1. 你不能選之前被任何人選過的數字。\n2. 如果你之前選過 $k$，你不能選 $k-1$ 或 $k+1$。\n3. 如果所有數字被選完則雙方平手；否則，先沒有數字可選的人輸。\n\n假設阿發先選。試求所有小李有必勝法的正整數 $n$。\n\nLet $n$ be a positive integer, and write down $1, 2, \\dots, n$ on the blackboard.\nAlpha and Lee take turns choosing a number from the board according to the following rules:\n\n1. You cannot choose any number that was previously selected by either player.\n2. If you have chosen $k$, you cannot choose $k-1$ or $k+1$.\n3. The game is a draw if all numbers are chosen. Otherwise, the player who cannot choose any number first loses the game.\n\nSuppose Alpha chooses first. Determine all $n$ such that Lee has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "除了 $n = 1, 2, 4, 6$ 外，小李都必勝。\n\n令 $[n] = \\{1, 2, \\dots, n\\}$。先證明一個引理：\n\n**引理**：如果小李第一次選擇 $n$，且阿發已經選了他的第 $k$ 個數字（$k \\ge 2$），則小李必可以選擇他的第 $k$ 個數字。\n\n*證明*：令 $a_1 < a_2 < \\dots < a_k$ 為阿發所選的頭 $k$ 個數字，注意 $a_k < n$。由於 $a_{i+1} - a_i > 1$，對所有 $1 \\le i \\le k$ 必存在 $b_i$ 使得 $a_i < b_i < a_{i+1}$，其中 $a_{k+1} := n$。在這 $k$ 個 $b_i$ 中，小李最多只選了 $k-1$ 個，因此他必然可以從中選出他的第 $k$ 個數字。Q.E.D.\n\n回到原題。不失一般性，假設阿發第一手選擇的數字不超過 $\\frac{n+1}{2}$，小李第一手選 $n$。\n\n- 對於所有 $n \\ne 1, 2, 4, 6$，小李有必勝法，分兩種情況：\n\n  * 若 $n$ 為 $3$ 以上的奇數：由引理知只要阿發能選，小李也一定能選。若遊戲和局，阿發必須從 $[n]$ 中選 $\\frac{n+1}{2}$ 個兩兩不連續的數字，這只有可能是所有奇數；但小李一開始就把 $n$ 這個奇數選走了，所以阿發一定會先沒有數字選。故小李必勝。\n\n  * 若 $n$ 為 $8$ 以上的偶數：類似前述，引理保證小李不會輸。若要和局，阿發必須從 $[n-1]$ 中選 $\\frac{n}{2}$ 個兩兩不連續的數字，這只可能是所有奇數。只要小李第二次從 $\\{1,3,\\dots,n-3\\}$ 中任選一個阿發未選的數字，阿發就必輸（此時阿發只選過兩個數字，小李一定有這樣的數字可選。由於小李第一手選 $n$，他不能選 $n-1$）。因此小李必勝。\n\n- 對於 $n = 1,2,4,6$，小李無必勝法，遊戲必和：\n\n  * $n = 1,2$ 顯然。\n  * $n = 4$ 時，阿發第一手必選 $1$，否則小李選 $4$ 便勝。小李必選 $4$，否則阿發第二手選 $4$ 便勝。阿發接下來必選 $3$，小李選 $2$，和局。\n  * $n = 6$ 時，阿發第一手選 $1$。考慮 $\\{3,5\\}$ 和 $\\{4,6\\}$ 兩組：若小李選其中一組較小（大）的數，阿發選另一組較大（小）的數即可。若小李選 $2$，阿發從前述兩組中選剩餘的數即可。由此阿發至少可逼和，引理知小李也至少可逼和，故必和。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17991,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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": 17992,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a sequence $a_1, a_2, \\dots$ of positive real numbers satisfies\n$$\na_{k+1} \\ge \\frac{ka_k}{a_k^2 + (k-1)}\n$$\nfor every positive integer $k$. Prove that $a_1 + a_2 + \\dots + a_n \\ge n$ for every $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "From\n$$\na_{k+1} \\ge \\frac{ka_k}{a_k^2 + (k-1)}, \\quad (1)\n$$\nit can be seen that\n$$\n\\frac{k}{a_{k+1}} \\le \\frac{a_k^2 + (k-1)}{a_k} = a_k + \\frac{k-1}{a_k},\n$$\nand so\n$$\na_k \\ge \\frac{k}{a_{k+1}} - \\frac{k-1}{a_k}.\n$$\nSumming the above inequality for $k=1, \\dots, m$, we obtain\n$$\na_1 + a_2 + \\dots + a_m \\ge \\left(\\frac{1}{a_2} - \\frac{0}{a_1}\\right) + \\left(\\frac{2}{a_3} - \\frac{1}{a_2}\\right) + \\dots + \\left(\\frac{m}{a_{m+1}} - \\frac{m-1}{a_m}\\right) = \\frac{m}{a_{m+1}}. \\quad (2)\n$$\nNow we prove the problem statement by induction on $n$. The case $n=2$ can be done by applying (1) to $k=1$:\n$$\na_1 + a_2 \\ge a_1 + \\frac{1}{a_1} \\ge 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17993,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ and $Q(x)$ be polynomials with real coefficients such that $P(0) > 0$ and all coefficients of the polynomial $S(x) = P(x)Q(x)$ are non-negative. Prove that for any positive $x$, the following inequality holds:\n\n$$\nS(x^2) - S^2(x) \\le \\frac{1}{4}(P^2(x^3) + Q(x^3)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $S = 0$, then $Q = 0$, and the inequality is evident. Suppose now that $S$ is not identically zero. Then for all $x > 0$, $S(x) > 0$. If for some $y > 0$, $P(y) < 0$, then the polynomial $P$, and so $S$, have roots on the interval $(0, y)$, which is impossible. So, $P$ and $Q$ are positive for $x > 0$.\n\nRewrite the inequality as:\n\n$$\n4(P(x^2)Q(x^2)) - (P(x)Q(x))^2 \\le P^2(x^3) + Q(x^3).\n$$\n\nLet $\\alpha = P$, $\\beta = Q$, $\\gamma = \\alpha\\beta = P Q$. Then the inequality becomes:\n\n$$\n4\\gamma(x^2) \\le \\gamma^2(x) + \\alpha^2(x^3) + 2\\beta(x^3).\n$$\n\nEstimate both sides:\n\n$$\n\\begin{aligned}\n\\gamma^2(x) + \\alpha^2(x^3) + 2\\beta(x^3) &= \\gamma^2(x) + \\beta(x^3) + \\alpha^2(x^3) + \\beta(x^3) \\\\ &\\ge 4\\sqrt[4]{\\gamma^2(x)\\alpha^2(x^3)\\beta^2(x^3)} = 4\\sqrt{\\gamma(x)\\gamma(x^3)}.\n\\end{aligned}\n$$\n\nIf $\\gamma(x) = a_0 + a_1 x + \\dots + a_n x^n$, then\n\n$$\n(a_0 + a_1 x + \\dots + a_n x^n)(a_0 + a_1 x^3 + \\dots + a_n x^{3n}) \\ge \\left(\\sqrt{a_0 a_1} + \\sqrt{a_1 a_2} + \\dots + \\sqrt{a_n a_{n+1}}\\right)^2\n$$\n\n(by the Cauchy-Schwarz inequality), which implies the required inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17994,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf(a^3) + f(b^3) + f(c^3) + 3f(a+b)f(b+c)f(c+a) = (f(a+b+c))^3\n$$\n\nfor all $a, b, c \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $f$ satisfies the condition.\n\nBy taking $(a, b, c) = (0, 0, 0)$, we get $3f(0) + 3f(0)^3 = f(0)^3$, so either $f(0) = 0$, or $3 = -2f(0)^2$. The latter is not possible in $\\mathbb{Z}$, so $f(0) = 0$.\n\nTaking $(a, b, c) = (n, -n, 0)$, we get $f(n^3) + f(-n^3) = 0$, so\n\n$$\nf(-n^3) = -f(n^3) \\text{ for all } n \\in \\mathbb{Z}. \\qquad (1)\n$$\n\nTaking $(a, b, c) = (n, 0, 0)$, we get\n\n$$\nf(n^3) = f(n)^3 \\text{ for all } n \\in \\mathbb{Z}. \\qquad (2)\n$$\n\nCombining (1) and (2), for any $n \\in \\mathbb{Z}$,\n\n$$\nf(-n)^3 = f((-n)^3) = f(-n^3) = -f(n^3) = -f(n)^3 = (-f(n))^3,\n$$\n\nso $f(-n) = -f(n)$, i.e., $f$ is odd.\n\nNow take $(a, b, c) = (k, 1-k, 0)$, so\n\n$$\nf(k)^3 + f(1-k)^3 + 3f(k)f(1-k)f(1) = f(1)^3 \\text{ for all } k \\in \\mathbb{Z}. \\qquad (3)\n$$\n\nFrom (2), $f(1) = f(1)^3$, so $f(1) \\in \\{-1, 0, 1\\}$.\n\n- If $f(1) = 0$, then from (3), $f(k) = -f(1-k) = f(k-1)$ for all $k$, so $f(n) = 0$ for all $n$ (using induction and oddness).\n- If $f(1) = 1$, then from (3), $f(k)^3 + f(1-k)^3 + 3f(k)f(1-k) = 1$ for all $k$, i.e., the Diophantine equation $X^3 + Y^3 + 3XY = 1$ is satisfied by $(X, Y) = (f(k), f(1-k))$. This can be rewritten as $(X + Y - 1)(X^2 - XY + Y^2 + X + Y + 1) = 0$. The second factor is only solvable in $\\mathbb{Z}$ if $X = Y = -1$. So:\n  1. $f(k) = 1 - f(1-k) = 1 + f(k-1)$ for all $k$. By induction and oddness, $f(n) = n$ for all $n$.\n  2. There is some $k_0$ such that $f(k_0) = f(1-k_0) = -1$. Then $f(-k_0) = 1 = 1 - f(1+k_0)$, so $f(1+k_0) = 0$. Using $(a, b, c) = (1+k_0, 1-k_0, -1)$, we get $0 - 1 - 1 + 3f(2)(-1)(1) = 1$, so $f(2) = -1$. Thus $k_0 = 2$ is the smallest positive value with $f(k_0) = -1 = f(1-k_0)$. By induction, $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k \\ge 0$. Assume this for some $k \\ge 0$. Then $f(-3k-2) = 1 = 1 - f(1+3k+2)$, so $f(3+3k) = 0$. Using $(a, b, c) = (3k, 1, 3)$, $0 + 1 + 0 + 3(0)(1)(0) = f(3k+4)^3$, so $f(3k+4) = 1$. Similarly, with $(a, b, c) = (3k, 2, 3)$, $f(3k+5) = -1$, and the induction is complete. Since $f$ is odd, $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k$, i.e.,\n\n$$\nf(n) = \\begin{cases} 0 & \\text{if } n \\equiv 0 \\pmod{3} \\\\ 1 & \\text{if } n \\equiv 1 \\pmod{3} \\\\ -1 & \\text{if } n \\equiv 2 \\pmod{3} \\end{cases} \\quad (4)\n$$\n\nFinally, if $f$ satisfies the equation, then $-f$ also does, covering the case $f(1) = -1$.\n\nIt is straightforward to check that the three functions $f(n) = 0$, $f(n) = n$, and $f(n) = -n$ (for all $n \\in \\mathbb{Z}$) satisfy the given functional equation. The function (4) also works, as does its negative. Thus, there are five functions that solve the equation, as described above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17995,
    "subject": "Mathematics (Olympiad)",
    "question": "Steve breaks $n$ eggs, so $1000 - n$ eggs are unbroken. He receives $R0.20 \\times (1000 - n)$, but has to repay $R1 \\times n$. Thus,\n\n$$\n0.2(1000 - n) - n = 176\n$$\n\nFind the value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n0.2(1000 - n) - n = 176\n$$\n\nExpanding:\n\n$$\n200 - 0.2n - n = 176\n$$\n$$\n200 - 1.2n = 176\n$$\n$$\n1.2n = 200 - 176 = 24\n$$\n$$\nn = \\frac{24}{1.2} = 20\n$$\n\nSo, $n = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17996,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be positive integers. Prove that, if $x_j$ are real numbers for $1 \\leq j \\leq n$, such that\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k} + k} = \\frac{1}{k}\n$$\n\nthen\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k+1} + k + 2} \\leq \\frac{1}{k+1}\n$$\nmust hold.",
    "options": [],
    "answer": "See solution",
    "solution": "We can, in fact, show that each of the expressions in the second sum is not greater than the corresponding expression in the first, multiplied by the factor $\\frac{k}{k+1}$.\n\nSubstituting $y := x_j^{2k}$, this means that we wish to show\n\n$$\n\\frac{1}{y^2 + k + 2} \\leq \\frac{k}{k+1} \\cdot \\frac{1}{y + k}\n$$\n\nSince $y$ is certainly positive for $k > 0$, this is equivalent to $(k+1)(y + k) \\leq k(y^2 + k + 2)$, or $P(y) = k y^2 - (k+1) y + k \\geq 0$. This polynomial is quadratic in $y$, and we have $P(0) = k > 0$. For the discriminant of the polynomial we have\n\n$$\n(k+1)^2 - 4k^2 = -3k^2 + 2k + 1 \\leq -3k^2 + 3k = -3k(k-1) \\leq 0,\n$$\n\nand we see that the polynomial can only assume positive values, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17997,
    "subject": "Mathematics (Olympiad)",
    "question": "After Urška inspected all the written numbers for the first time, only numbers divisible by 3 were left on the whiteboard. After her second inspection, only numbers divisible by $3^2$ were left. After the third inspection, only numbers divisible by $3^3$ were left, and so on. What number did Urška erase last if the largest power of 3 less than 2015 is $3^6 = 729$ (since $3^7 = 2187$)?",
    "options": [],
    "answer": "See solution",
    "solution": "After the sixth inspection, only numbers divisible by 729 remained: 729 and 1458. In the next step, both were erased since they are not divisible by 2187. Urška erased the number 1458 last.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17998,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $f(x) = \\frac{\\sin(\\pi x) - \\cos(\\pi x) + 2}{\\sqrt{x}}$ for $\\frac{1}{4} \\leq x \\leq \\frac{5}{4}$, find the minimum value of $f(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "By rewriting $f(x)$, we have\n$$\nf(x) = \\frac{\\sqrt{2}\\sin\\left(\\pi x - \\frac{\\pi}{4}\\right) + 2}{\\sqrt{x}}\n$$\nfor $\\frac{1}{4} \\leq x \\leq \\frac{5}{4}$. Define $g(x) = \\sqrt{2}\\sin\\left(\\pi x - \\frac{\\pi}{4}\\right)$, where $\\frac{1}{4} \\leq x \\leq \\frac{5}{4}$. Then $g(x) \\geq 0$, and $g(x)$ is monotone increasing on $[\\frac{1}{4}, \\frac{3}{4}]$, and monotone decreasing on $[\\frac{3}{4}, \\frac{5}{4}]$.\n\nFurther, the graph of $y = g(x)$ is symmetric about $x = \\frac{3}{4}$, i.e., for any $x_1 \\in [\\frac{1}{4}, \\frac{3}{4}]$ there exists $x_2 \\in [\\frac{3}{4}, \\frac{5}{4}]$ such that $g(x_2) = g(x_1)$. Then\n$$\nf(x_1) = \\frac{g(x_1) + 2}{\\sqrt{x_1}} = \\frac{g(x_2) + 2}{\\sqrt{x_1}} \\geq \\frac{g(x_2) + 2}{\\sqrt{x_2}} = f(x_2).\n$$\nOn the other hand, $f(x)$ is monotone decreasing on $[\\frac{3}{4}, \\frac{5}{4}]$. Therefore, $f(x) \\geq f\\left(\\frac{5}{4}\\right) = \\frac{4\\sqrt{5}}{5}$. That means the minimum value of $f(x)$ on $[\\frac{1}{4}, \\frac{5}{4}]$ is $\\frac{4\\sqrt{5}}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17999,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many positive integers $n$, which are not divisible by $10$ and such that $s(n^2) < s(n) - 5$, where $s(n)$ is the sum of digits of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "All integers of the form $499\\ldots99$ satisfy the condition. Indeed, if $n = 4\\underbrace{99\\ldots99}_{k}$ (that is, $n = 5 \\cdot 10^k - 1$), then\n\n$$\nn^2 = 25 \\cdot 10^{2k} - 10^{k+1} + 1 = 24\\underbrace{99\\ldots9}_{k-1}\\underbrace{00\\ldots00}_{k}1.\n$$\n\nIn such a case, $s(n) = 4 + 9k$, but $s(n^2) = 7 + 9(k - 1) = 9k - 2$.\n\nThus, $s(n^2) = s(n) - 6 < s(n) - 5$, and $n$ is not divisible by $10$. Since $k$ can be any positive integer, there are infinitely many such $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18000,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be a natural number. Ana chooses non-zero natural numbers $a_1, a_2, \\dots, a_n$. For each non-empty subset $A \\subset \\{1, 2, \\dots, n\\}$, she computes the sum $s_A = \\sum_{k \\in A} a_k$. She arranges these sums in increasing order, obtaining the sequence $s_1 \\le s_2 \\le \\dots \\le s_{2^n - 1}$.\n\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 $i \\in B$.",
    "options": [],
    "answer": "See 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.\n\nIndeed, consider:\n\n- If $a_{i-1} = a_i$, then the appearance is obvious.\n- 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 $t \\ge n-1$.\n\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 $a_{k+1}$.\n\nSo, we can determine the numbers $a_1, a_2, \\dots, a_{n-1}$.\n\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": 18001,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the largest possible value of the number\n\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n,\n$$\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": "See solution",
    "solution": "a) Let $x_1, x_2, \\dots, x_n$ be non-negative integers satisfying the conditions from the statement. Let $M = \\max_{1 \\le i \\le n} x_i$. 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 from $(M-1005)(M-1006) \\ge 0$, 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) For $n=2$, we have $x_1x_2 = 1005 \\cdot 1006 \\iff x_1(2011-x_1) = 1005 \\cdot 1006 \\iff (x_1-1005)(x_1+1006) = 0 \\iff (x_1,x_2) \\in \\{(1005, 1006), (1006, 1005)\\}$.\n\nFor $n=3$, $x_1x_2 + x_2x_3 = 1005 \\cdot 1006 \\iff x_2(x_1+x_3) = 1005 \\cdot 1006$, so $x_2 = 1005$, $x_1+x_3 = 1006$ 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 $n \\ge 4$, let $j$ be 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, 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\n$$(x_1, \\dots, x_n) \\in \\{(0, \\dots, 0, k, 1005, 1006-k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1006\\} \\cup \\{(0, \\dots, 0, k, 1006, 1005-k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1005\\},$$\nwhere the group of the three non-zero components can be located anywhere.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18002,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $a_1 = 1$, $a_2 = 1 - \\frac{1}{2}$, $a_3 = 1 - \\frac{1}{2} + \\frac{1}{3}$, $a_4 = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4}$, $\\dots$, $a_{2009} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots + \\frac{1}{2009}$, $a_{2010} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots + \\frac{1}{2010} - \\frac{1}{2011}$ are written in increasing order.\n\nWhat number is written:\n\n![](images/Ukrajina_2010_p16_data_05bc1cfb95.png)\n\na) on the 1000-th place;\n\nb) on the 2000-th place?",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that each number is situated between two preceding numbers. Note the following inequalities:\n\nFor even $n = 2k$:\n\n$$\n\\begin{aligned}\na_{2k-2} &= 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots - \\frac{1}{2k-2} < a_{2k} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots - \\frac{1}{2k-2} + \\frac{1}{2k-1} - \\frac{1}{2k} \\\\\n&< a_{2k-1} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots - \\frac{1}{2k-2} + \\frac{1}{2k-1}\n\\end{aligned}\n$$\n\nFor odd $n = 2k + 1$:\n\n$$\n\\begin{aligned}\na_{2k} &= 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots - \\frac{1}{2k} < a_{2k+1} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots + \\frac{1}{2k-1} - \\frac{1}{2k} + \\frac{1}{2k+1} \\\\\n&< a_{2k-1} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\dots + \\frac{1}{2k-1}.\n\\end{aligned}\n$$\n\nTherefore, the order is:\n\n$a_2, a_4, \\dots, a_{2010}, a_{2009}, a_{2007}, \\dots, a_3, a_1.$\n\nThus, on the 1000-th place is $a_{2000}$, and on the 2000-th place is $a_{21}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18003,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive prime numbers $p$ and $q$ such that\n$$\n  p^5 + p^3 + 2 = q^2 - q.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We are given $p^5 + p^3 + 2 = q^2 - q$.\n\nRewrite as $p^3(p^2 + 1) = (q + 1)(q - 2)$.\n\nIf $p$ divides both $q + 1$ and $q - 2$, then $p = 3$, giving the solution $(p, q) = (3, 17)$.\n\nNow suppose $p \\neq 3$. Then either $q + 1$ or $q - 2$ is a multiple of $p^3$, while the other divides $p^2 + 1$. It follows that $q + 1 \\geq p^3$ and $p^2 + 1 \\geq q - 2$, so $p^3 \\geq p^2 + 4$, which holds only for $p = 2$. Then $(p, q) = (2, 7)$.\n\n$\\boxed{(2, 7),\\ (3, 17)}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18004,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. The $A$-angle bisector intersects $BC$ at $D$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively. Given that the circumcenter of triangle $AEF$ lies on $BC$, find all possible values of $\\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcenter of triangle $AEF$ and denote $\\alpha = \\angle BAC$. Since $\\angle BAD$ and $\\angle CAD$ are acute (see figure below), points $E$ and $F$ lie in the half-plane $BCA$, and the Inscribed Angle Theorem yields\n\n$$\n\\angle BED = 2 \\cdot \\angle BAD = \\alpha = 2 \\cdot \\angle DAC = \\angle DFC.\n$$\n\n![](images/brozura_a67angl_new_p13_data_fe94469aa9.png)\n\nThe isosceles triangles $BED$ and $DFC$ are thus similar, and we compute that $\\angle EDF = \\alpha$ and that $BC$ is the external $D$-angle bisector in triangle $DEF$.\n\nPoint $O$ lies on $BC$ and on the perpendicular bisector of $EF$. With respect to triangle $DEF$, it lies on the external $D$-angle bisector and on the perpendicular bisector of the opposite side $EF$. Thus, it is the midpoint of arc $EDF$ and $\\angle EOF = \\angle EDF = \\alpha$.\n\nQuadrilateral $AEDF$ is a kite, hence $\\angle EAF = \\alpha$. Moreover, line $EF$ separates points $A$ and $O$, so the Inscribed Angle Theorem implies that the size of the reflex angle $EOF$ is twice the size of the convex angle $EAF$. This yields $360^\\circ - \\alpha = 2 \\cdot \\alpha$ and $\\alpha = 120^\\circ$.\n\n*Answer.* The only possible value is $\\angle BAC = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18005,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and let $m$ and $n$ be positive integers written in base $p$ as\n$$\nn = a_0 + a_1 p + \\cdots + a_k p^k\n$$\nand\n$$\nm = b_0 + b_1 p + \\cdots + b_k p^k,\n$$\nrespectively. Show that\n$$\n\\binom{n}{m} \\equiv \\prod_{i=0}^{k} \\binom{a_i}{b_i} \\pmod{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\binom{n}{m} = 0$ for $n < m$, we assume $n \\ge m$. Next, we use the well-known fact that $(x+1)^p \\equiv x^p + 1 \\pmod{p}$, meaning each coefficient of $(x+1)^p - (x^p + 1)$ is divisible by $p$. Thus,\n\n$$\n\\begin{align*}\n(x+1)^n &= (x+1)^{a_0 + a_1 p + \\cdots + a_k p^k} \\\\\n&\\equiv \\prod_{i=0}^k (x^{p^i} + 1)^{a_i} \\pmod{p} \\\\\n&\\equiv \\prod_{i=0}^k \\left( \\sum_{j=0}^{a_i} \\binom{a_i}{j} x^{j p^i} \\right) \\pmod{p}\n\\end{align*}\n$$\n\nThe coefficient of $x^m$ in the left-hand side is $\\binom{n}{m}$. The coefficient of $x^m$ in the right-hand side is the coefficient of the monomial\n\n$$\n\\binom{a_0}{b_0} x^{b_0} \\binom{a_1}{b_1} x^{b_1 p} \\cdots \\binom{a_k}{b_k} x^{b_k p^k} = x^{b_0 + b_1 p + \\cdots + b_k p^k} \\prod_{i=0}^{k} \\binom{a_i}{b_i}.\n$$\n\nEquating both coefficients yields\n$$\n\\binom{n}{m} \\equiv \\prod_{i=0}^{k} \\binom{a_i}{b_i} \\pmod{p}.\n$$\n\nThus, the result is proved.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18006,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a circumscribed quadrilateral with circumcenter $O$, and let $K$, $L$, $M$, and $N$ denote the midpoints of the sides $AB$, $BC$, $CD$, and $DA$, respectively. Suppose that the lines $KM$ and $LN$ do not pass through $O$. Let $E = KM \\cap LN$. Point $P$ is chosen on the interval $KM$ to satisfy $\\angle KOE = \\angle MOP$, and point $Q$ is chosen on the interval $LN$ to satisfy $\\angle LOE = \\angle NOQ$. Show that the points $O$, $P$, and $Q$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to prove that $\\angle POQ = 180^\\circ$.\n\n![](images/MNG_ABooklet_2015_p23_data_ddc2449cb1.png)\n\nSince $O$ is the circumcenter, the points $K$, $L$, $M$, $N$ are the feet of the perpendiculars from $O$ to the sides of $ABCD$. Hence $OMDN$ and $OKBL$ are circumscribed. Hence\n\n$$\n\\begin{align*}\n\\angle POQ &= \\angle POM + \\angle MON + \\angle NOQ \\\\\n&= \\angle KOE + (180^{\\circ} - \\angle MDN) + \\angle LOE \\\\\n&= \\angle KOL + \\angle KBL \\\\\n&= 180^{\\circ}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18007,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p, q, r$ such that\n$$\n\\frac{p}{q} - \\frac{4}{r+1} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* $(p, q, r) \\in \\{(3, 2, 7),\\ (5, 3, 5),\\ (7, 3, 2)\\}$\n\nWe can rewrite the equation as\n$$\n\\frac{pr + p - 4q}{q(r + 1)} = 1 \\implies pr + p - 4q = qr + q \\implies r(p - q) = 5q - p.\n$$\nHence $p \\neq q$.\n\n$$\nr = \\frac{5q - p}{p - q} = \\frac{4q + q - p}{p - q} \\implies r = \\frac{4q}{p - q} - 1.\n$$\nSo $p - q \\neq q$, $p - q \\neq 2q$, $p - q \\neq 4q$. We have $p - q = 1$, $2$, or $4$.\n\n**i)** If $p - q = 1$ then $q = 2$, $p = 3$, $r = 7$.\n\n**ii)** If $p - q = 2$ then $p = q + 2$, $r = 2q - 1$.\n\nIf $q \\equiv 1 \\pmod{3}$ then $q + 2 \\equiv 0 \\pmod{3}$, $q + 2 = 3 \\implies q = 1$ (contradiction).\n\nIf $q \\equiv -1 \\pmod{3}$ then $r \\equiv -2 - 1 \\equiv 0 \\pmod{3}$, so $r = 3$, $q = 2$, $p = 4$ (contradiction).\n\nHence $q = 3$, $p = 5$, $r = 5$.\n\n**iii)** If $p - q = 4$ then $p = q + 4$, $r = q - 1$. Hence $q = 3$, $p = 7$, $r = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18008,
    "subject": "Mathematics (Olympiad)",
    "question": "A pyramid $SA_1A_2\\ldots A_n$ has a convex polygon $A_1A_2\\ldots A_n$ as its base. For each $i = 1, 2, \\ldots, n$, let $X_iA_iA_{i+1}$ be the triangle congruent to $SA_iA_{i+1}$, lying in the base plane so that the point $X_i$ and the polygon $A_1\\ldots A_n$ belong to the same half-plane with respect to $A_iA_{i+1}$ (where $A_{n+1} = A_1$). Prove that the collection of triangles $X_iA_iA_{i+1}$ ($i = 1, 2, \\ldots, n$) covers the polygon $A_1\\ldots A_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the statement, consider each triangle $SA_iA_{i+1}$ of the pyramid. Its congruent copy $X_iA_iA_{i+1}$ is placed in the base plane so that $X_i$ is on the same side of $A_iA_{i+1}$ as the polygon. Since the pyramid is convex and the base is convex, the union of these triangles $X_iA_iA_{i+1}$ for $i = 1, \\ldots, n$ covers the entire base polygon $A_1\\ldots A_n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18009,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a fraction $\\frac{a}{b}$, not necessarily in simplest form, *special* if $a$ and $b$ are positive integers whose sum is 15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?\n\n(A) 9  (B) 10  (C) 11  (D) 12  (E) 13",
    "options": [],
    "answer": "See solution",
    "solution": "The 14 special fractions are:\n\n$$\n\\frac{1}{14},\\ \\frac{2}{13},\\ \\frac{3}{12},\\ \\frac{4}{11},\\ \\frac{5}{10},\\ \\frac{6}{9},\\ \\frac{7}{8},\\ \\frac{8}{7},\\ \\frac{9}{6},\\ \\frac{10}{5},\\ \\frac{11}{4},\\ \\frac{12}{3},\\ \\frac{13}{2},\\ \\frac{14}{1}\n$$\n\nwhich simplify to:\n\n$$\n\\frac{1}{14},\\ \\frac{2}{13},\\ \\frac{1}{4},\\ \\frac{4}{11},\\ \\frac{1}{2},\\ \\frac{2}{3},\\ \\frac{7}{8},\\ \\frac{8}{7},\\ \\frac{3}{2},\\ 2,\\ \\frac{11}{4},\\ 4,\\ \\frac{13}{2},\\ 14\n$$\n\nThe fractions can be grouped as follows:\n\n- The integers 2, 4, and 14 can be doubled to produce 4, 8, and 28, or paired to produce sums of 6, 16, and 18.\n- The fractions with denominators of 2 are $\\frac{1}{2}$, $\\frac{3}{2}$, and $\\frac{13}{2}$. They can be doubled to produce 1, 3, and 13, or paired to produce sums of 2, 7, and 8.\n- The fractions with denominators of 4 are $\\frac{1}{4}$ and $\\frac{11}{4}$. Their sum is 3.\n- The remaining fractions—$\\frac{1}{14}$, $\\frac{2}{13}$, $\\frac{4}{11}$, $\\frac{2}{3}$, $\\frac{7}{8}$, and $\\frac{8}{7}$—cannot be added to any fraction in the list to give an integer.\n\nThis analysis produced 13 integer sums, but 3 and 8 appeared twice. Excluding the duplicates leaves 11 distinct integers that can be written as the sum of two special fractions: 1, 2, 3, 4, 6, 7, 8, 13, 16, 18, and 28.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18010,
    "subject": "Mathematics (Olympiad)",
    "question": "The candy store sells chocolates in the flavours white, milk, and dark. You can buy them in three types of coloured boxes. The three boxes have the following contents:\n\n- Gold: 2 white, 3 milk, 1 dark\n- Silver: 1 white, 2 milk, 4 dark\n- Bronze: 5 white, 1 milk, 2 dark\n\nLavinia buys some boxes of chocolates (at least one) and when she gets home, it turns out she has exactly the same number of chocolates of each flavour.\n\nAt least how many boxes did Lavinia buy?",
    "options": [],
    "answer": "See solution",
    "solution": "$20$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18011,
    "subject": "Mathematics (Olympiad)",
    "question": "Raymond faces a room with 15 switches, each controlling a pair of adjacent lights (numbered modulo 15). He wants to determine which switch corresponds to which pair of lights, but can only observe the lights by walking to the other room. What is the minimum number of trips Raymond must make to the other room to identify all switches with certainty, and how can he do it?",
    "options": [],
    "answer": "See solution",
    "solution": "Walking back and forth just three times, Raymond cannot know all the switches with certainty. There are $2^3 = 8$ different switching patterns, but 15 switches, so some switches share patterns and cannot be distinguished. Therefore, he must go to the other room at least four times.\n\nWe prove he can always do it in four trips:\n\nRaymond numbers the switches from 1 to 15, writing each number in binary (4 digits, padded with leading zeros). In each round, he flips the switches whose $k$-th binary digit is 1 ($k = 1,2,3,4$ for rounds 1–4), then observes the lights. Each switch is flipped in a unique subset of rounds, corresponding to its binary code. By recording which pairs of lights change in each round, Raymond can deduce the correspondence between switches and light pairs. Thus, four trips suffice to identify all switches.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18012,
    "subject": "Mathematics (Olympiad)",
    "question": "How many pairs of non-negative integers $x$ and $y$ are solutions of $$\\frac{x}{20} + \\frac{y}{15} = 1$$?",
    "options": [],
    "answer": "See solution",
    "solution": "We can rewrite $\\frac{x}{20} + \\frac{y}{15} = 1$ as follows:\n\n$$\\frac{x}{20} + \\frac{y}{15} = 1$$\nMultiply both sides by $60$:\n$$3x + 4y = 60$$\n\nWe seek non-negative integer solutions $(x, y)$.\n\nLet $x$ range over non-negative integers such that $3x \\leq 60$ and $60 - 3x$ is divisible by $4$.\n\nSet $y = \\frac{60 - 3x}{4}$.\n\nFor $y$ to be an integer, $60 - 3x$ must be divisible by $4$.\n\nLet $x = 4k$ for $k = 0, 1, 2, 3, 4, 5$ (since $3x$ must be divisible by $4$):\n\n- $k = 0$: $x = 0$, $y = \\frac{60 - 0}{4} = 15$\n- $k = 1$: $x = 4$, $y = \\frac{60 - 12}{4} = 12$\n- $k = 2$: $x = 8$, $y = \\frac{60 - 24}{4} = 9$\n- $k = 3$: $x = 12$, $y = \\frac{60 - 36}{4} = 6$\n- $k = 4$: $x = 16$, $y = \\frac{60 - 48}{4} = 3$\n- $k = 5$: $x = 20$, $y = \\frac{60 - 60}{4} = 0$\n\nThus, the pairs are $(0, 15)$, $(4, 12)$, $(8, 9)$, $(12, 6)$, $(16, 3)$, $(20, 0)$.\n\n**Answer:** There are $6$ pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18013,
    "subject": "Mathematics (Olympiad)",
    "question": "At the beginning of work, the memory of a computer contained a single polynomial $x^2 - 1$. Each minute, the computer can do one of two things:\n\n1. Choose any polynomial $f(x)$ from its memory and also memorize the polynomials $f^2(x) - 1$ and $f(x^2 - 1)$.\n2. Choose any two different polynomials $g(x)$ and $h(x)$ from its memory and memorize the polynomial $\\frac{1}{2}(g(x) + h(x))$.\n\nIs it possible that after some time the memory of the computer contains the polynomial\n$$\nP(x) = \\frac{1}{1024}(x^2 - 1)^{2048} - 1?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the number $a = \\frac{1}{2}(1 - \\sqrt{5})$, which is a root of the equation $x^2 - x - 1 = 0$.\n\nAll polynomials $f(x)$ in the memory of the computer at any moment have the following property: $f(a) = a$. The polynomial $P(x)$ from the statement does not have this property, since\n\n$$\nP(a) = \\frac{1}{1024}(a^2 - 1)^{2048} - 1 = \\frac{a^{2048}}{1024} - 1 \\neq a = a^2 - 1\n$$\nwhich is equivalent to $a^{2046} \\neq 1024$, as $|a| < 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18014,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Since $f$ is monotonic, the lateral limits at each point exist and are finite. Denote $f(x_0^-)$ and $f(x_0^+)$ as the limits from the left and right at $x_0$, respectively. Since $f$ is increasing, $f(x_0^-) \\leq f(x_0) \\leq f(x_0^+)$ for every $x_0 \\in \\mathbb{R}$.\n\nSuppose $f$ is discontinuous at some point $a$. Then there exist $A, B$ such that $f(a^-) < A < B < f(a^+)$. Since $f$ is increasing, $f(x) \\leq A$ for all $x < a$ and $f(x) \\geq B$ for all $x > a$. Again by monotonicity, $f(f(x)) \\leq f(A)$ for all $x < a$ and $f(f(x)) \\geq f(B)$ for all $x > a$. Thus, $f(f(a^-)) \\leq f(A) < f(B) \\leq f(f(a^+))$, which implies $f \\circ f$ is discontinuous at $a$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18015,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the following statements:\n\n1. If $2n-1$ is a prime number, then for any group of distinct positive integers $a_1, a_2, \\dots, a_n$, there exist $i, j \\in \\{1, 2, \\dots, n\\}$ such that\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge 2n-1.\n$$\n\n2. If $2n-1$ is a composite number, then there exists a group of distinct positive integers $a_1, a_2, \\dots, a_n$ such that\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} < 2n-1\n$$\nfor any $i, j \\in \\{1, 2, \\dots, n\\}$.\n\nHere $(x, y)$ denotes the greatest common divisor of positive integers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof*\n\n**(1)** Let $p = 2n-1$ be a prime. Without loss of generality, assume $(a_1, a_2, \\dots, a_n) = 1$. If there exists $i$ ($1 \\le i \\le n$) such that $p \\nmid a_i$, then there exists $j \\ne i$ such that $p \\nmid a_j$. Therefore, $p \\nmid (a_i, a_j)$. Then:\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge \\frac{a_i}{(a_i, a_j)} \\ge p = 2n-1.\n$$\n\nNext, consider the case when $(a_i, p) = 1$ for all $i = 1, 2, \\dots, n$. Then $p \\nmid (a_i, a_j)$ for any $i \\ne j$. By the Pigeonhole Principle, there exist $i \\ne j$ such that either $a_i \\equiv a_j \\pmod p$ or $a_i + a_j \\equiv 0 \\pmod p$.\n\n- If $a_i \\equiv a_j \\pmod p$, then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge \\frac{a_i - a_j}{(a_i, a_j)} \\ge p = 2n-1.\n$$\n\n- If $a_i + a_j \\equiv 0 \\pmod p$, then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge p = 2n-1.\n$$\n\nThis completes the proof of (1).\n\n**(2)** Construct an example. Since $2n-1$ is composite, write $2n-1 = pq$ where $p, q > 1$. Let\n$$\n\\begin{aligned}\na_1 &= 1,\\quad a_2 = 2,\\quad \\dots,\\quad a_p = p,\\quad a_{p+1} = p+1,\\\\\na_{p+2} &= p+3,\\quad \\dots,\\quad a_n = pq-p.\n\\end{aligned}\n$$\nThe first $p$ elements are consecutive integers; the rest are $n-p$ consecutive even integers from $p+1$ to $pq-p$.\n\n- For $1 \\le i \\le j \\le p$:\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le a_i + a_j \\le 2p < 2n-1.\n$$\n\n- For $p+1 \\le i \\le j \\le n$:\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le \\frac{a_i + a_j}{2} \\le pq - p < 2n-1.\n$$\n\n- For $1 \\le i \\le p$ and $p+1 \\le j \\le n$:\n  - If $i \\ne p$ or $j \\ne n$:\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le pq - 1 < 2n-1.\n$$\n  - If $i = p$ and $j = n$:\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} = \\frac{pq}{p} = q < 2n-1.\n$$\n\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18016,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 表示所有實數所成的集合。試求所有的函數 $f : (0, \\infty) \\to \\mathbb{R}$ 滿足\n\n$$\n(x + \\frac{1}{x})f(y) = f(xy) + f\\left(\\frac{y}{x}\\right), \\quad \\text{對所有 } x, y > 0 \\text{均成立。}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $f(x) = C_1 x + \\frac{C_2}{x}$，其中 $C_1$ 和 $C_2$ 為任意常數。\n\n*解法 1.* 固定一個實數 $a > 1$，令 $t$ 為新變數。對於 $f(t), f(t^2), f(at)$ 和 $f(a^2 t^2)$，由假設可得一組線性方程：\n\n$$\nx = y = t : \\left(t + \\frac{1}{t}\\right) f(t) = f(t^2) + f(1) \\quad (1)\n$$\n\n$$\nx = \\frac{t}{a}, \\quad y = at : \\left(\\frac{t}{a} + \\frac{a}{t}\\right) f(at) = f(t^2) + f(a^2) \\quad (2)\n$$\n\n$$\nx = a^2 t, \\quad y = t : \\left(a^2 t + \\frac{1}{a^2 t}\\right) f(t) = f(a^2 t^2) + f\\left(\\frac{1}{a^2}\\right) \\quad (3)\n$$\n\n$$\nx = y = at : \\left(at + \\frac{1}{at}\\right) f(at) = f(a^2 t^2) + f(1) \\quad (4)\n$$\n\n消去 $f(t^2)$，取 (1) 與 (2) 的差；由 (3) 與 (4) 消去 $f(a^2 t^2)$；再線性組合消去 $f(at)$：\n\n$$\n\\begin{aligned}\n& \\left(t + \\frac{1}{t}\\right) f(t) - \\left(\\frac{t}{a} + \\frac{a}{t}\\right) f(at) = f(1) - f(a^2) \\\\\n& \\left(a^2 t + \\frac{1}{a^2 t}\\right) f(t) - \\left(at + \\frac{1}{at}\\right) f(at) = f\\left(\\frac{1}{a^2}\\right) - f(1) \\\\\n& \\left(\\left(at + \\frac{1}{at}\\right)\\left(t + \\frac{1}{t}\\right) - \\left(\\frac{t}{a} + \\frac{a}{t}\\right)\\left(a^2 t + \\frac{1}{a^2 t}\\right)\\right) f(t) \\\\\n& = \\left(at + \\frac{1}{at}\\right)\\left(f(1) - f(a^2)\\right) - \\left(\\frac{t}{a} + \\frac{a}{t}\\right)\\left(f\\left(\\frac{1}{a^2}\\right) - f(1)\\right).\n\\end{aligned}\n$$\n\n左側 $f(t)$ 的係數為 $a + \\frac{1}{a} - \\left(a^3 + \\frac{1}{a^3}\\right) < 0$，與 $t$ 無關。\n\n因此可得：\n\n$$\nf(t) = C_1 t + \\frac{C_2}{t} \\qquad (5)\n$$\n\n其中 $C_1, C_2$ 由 $a, f(1), f(a^2), f(1/a^2)$ 決定，與 $t$ 無關。\n\n此型式的函數滿足原式：\n\n$$\n\\begin{align*}\n\\left(x + \\frac{1}{x}\\right) f(y) &= \\left(x + \\frac{1}{x}\\right) \\left(C_1 y + \\frac{C_2}{y}\\right) \\\\\n&= \\left(C_1 x y + \\frac{C_2}{x y}\\right) + \\left(C_1 \\frac{y}{x} + C_2 \\frac{x}{y}\\right) \\\\\n&= f(xy) + f\\left(\\frac{y}{x}\\right).\n\\end{align*}\n$$\n\n*解法 2.* 令 $x = a \\neq 1, y = a^n$，則\n\n$$\nf(a^{n+1}) - \\left(a + \\frac{1}{a}\\right)f(a^n) + f(a^{n-1}) = 0.\n$$\n\n對序列 $z_n = a^n$，此為二階齊次線性遞迴，其特徵方程為\n\n$$\nt^2 - \\left(a + \\frac{1}{a}\\right)t + 1 = (t - a)(t - \\frac{1}{a})\n$$\n\n根為 $a, 1/a$，通解為\n\n$$\nz_n = C_1 a^n + C_2 \\left(\\frac{1}{a}\\right)^n\n$$\n\n故\n\n$$\nf(a^n) = C_1(a) a^n + \\frac{C_2(a)}{a^n}\n$$\n\n可延伸至有理指數，故 $C_1, C_2$ 為常數，得\n\n$$\nf(t) = C_1 t + \\frac{C_2}{t}\n$$\n\n此型式確實滿足原式。\n\n由於原式對 $f$ 線性，若 $f_1, f_2$ 滿足原式，則 $c_1 f_1(x) + c_2 f_2(x)$ 亦滿足。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18017,
    "subject": "Mathematics (Olympiad)",
    "question": "Andile and Zandre play a game on a $2017 \\times 2017$ board. At the beginning, Andile declares some of the squares *forbidden*, meaning that nothing may be placed on such a square. After that, they take turns to place coins on the board, with Zandre placing the first coin. It is not allowed to place a coin on a forbidden square or in the same row or column where another coin has already been placed. The player who places the last coin wins the game.\n\nWhat is the least number of squares Andile needs to declare as forbidden at the beginning to ensure a win? (Assume that both players use an optimal strategy.)",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number is $2017$. For example, Andile can achieve a win by declaring all squares of the last row forbidden, so that $2016$ rows remain. After that, there will be exactly $2016$ moves possible, no matter how the two play, since placing a coin always eliminates exactly one row and one column from further use. This means that Andile gets the last move.\n\nOn the other hand, we prove that $2016$ or fewer forbidden squares are not sufficient, no matter how they are placed. Generally, we show by induction that Zandre has a winning strategy on a $(2n-1) \\times (2n-1)$ board if he gets to place a coin first and no more than $2n-2$ squares have been forbidden. This is trivial for $n=1$: Zandre can simply place a coin on the only square.\n\nFor the induction step, consider a $(2n+1) \\times (2n+1)$ board with at most $2n$ forbidden squares. If there are two forbidden squares in the same row or column somewhere, then Zandre places a coin in this row or column. This is possible since there are fewer forbidden squares than squares in a row or column. If there are at least two forbidden squares but no two of them in the same row or column, Zandre chooses any two of them, then places a coin on the intersection of the row of the first forbidden square and the column of the second forbidden square. This is possible because of the assumption that there are no two forbidden squares in the same row or column. Finally, if there is only one forbidden square, Zandre places a coin anywhere in the same row or column, and if there are no forbidden squares, he just places it on an arbitrary square.\n\nAfter Andile's move, the rows and columns where Andile and Zandre placed their coins can be removed, since no further coins can be placed there anymore. This leaves us with a $(2n-1) \\times (2n-1)$ board, and by the choice of Zandre's move there are at most $2n-2$ forbidden squares on it. So by the induction hypothesis, Zandre has a winning strategy for the remaining position, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18018,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the least positive integer $k$ such that, in every convex 101-gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $PQ=1$. Consider a convex 101-gon such that one of its vertices is at $P$ and the remaining 100 vertices are within $\\varepsilon$ of $Q$, where $\\varepsilon$ is an arbitrarily small positive real. Let $k + l$ equal the total number $\\frac{101 \\cdot 98}{2} = 4949$ of diagonals. When $k \\leq 4851$, the sum of the $k$ shortest diagonals is arbitrarily small. When $k \\geq 4851$, the sum of the $k$ shortest diagonals is arbitrarily close to $k - 4851 = 98 - l$ and the sum of the remaining diagonals is arbitrarily close to $l$. Therefore, we need to have $l \\leq 49$ and $k \\geq 4900$.\n\nWe proceed to show that $k = 4900$ works. To this end, colour all $l = 49$ remaining diagonals green. To each green diagonal $AB$, apart from, possibly, the last one, we will assign two red diagonals $AC$ and $CB$ so that no green diagonal is ever coloured red and no diagonal is coloured red twice.\n\nSuppose that we have already done this for $0 \\leq i \\leq 48$ green diagonals (thus forming $i$ red-red-green triangles) and let $AB$ be up next. Let $D$ be the set of all diagonals emanating from $A$ or $B$ and distinct from $AB$: we have $|D| = 2 \\cdot 97 = 194$. Every red-red-green triangle formed thus far has at most two sides in $D$. Therefore, the subset $E$ of all as-of-yet-uncoloured diagonals in $D$ contains at least $194 - 2i$ elements.\n\nWhen $i \\leq 47$, $194 - 2i \\geq 100$. The total number of endpoints distinct from $A$ and $B$ of diagonals in $D$, however, is 99. Therefore, two diagonals in $E$ have a common endpoint $C$ and we can assign $AC$ and $CB$ to $AB$, as needed.\n\nThe case $i = 48$ is slightly more tricky: this time, it is possible that no two diagonals in $E$ have a common endpoint other than $A$ and $B$, but, if so, then there are two diagonals in $E$ that intersect in a point interior to both. Otherwise, at least one (say, $a$) of the two vertices adjacent to $A$ is cut off from $B$ by the diagonals emanating from $A$ and at least one (say, $b$) of the two vertices adjacent to $B$ is cut off from $A$ by the diagonals emanating from $B$ (and $a \\ne b$). This leaves us with at most 97 suitable endpoints and at least 98 diagonals in $E$, a contradiction.\n\nBy the triangle inequality, this completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18019,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $b(n)$ be the smallest positive integer $k$ such that there exist integers $a_1, a_2, \\dots, a_k$ satisfying\n\n$$\nn = a_1^{a_2} + a_2^{a_3} + \\dots + a_k^{a_1}.\n$$\n\nDetermine whether the set of positive integers $n$ is finite or infinite for which:\n\n$$\n\\text{a) } b(n) = 12; \\quad \\text{b) } b(n) = 12^{12^{12}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider the numbers $12^{66k+1}$, where $k \\in \\mathbb{N}$. By Fermat's theorem, $12^{66k+1} \\equiv 12 \\pmod{67}$, and these numbers can be written as the sum of 12 terms of the required form. Thus, $b(12^{66k+1}) = 12$ for every $k \\in \\mathbb{N}$, so the set is infinite.\n\nb) For $P \\in \\mathbb{Z}[X]$, define $\\Delta(P)(x) = P(x+1) - P(x)$. If $P$ has degree $d$ and leading coefficient $a$, then $\\Delta(P)$ has degree $d-1$ and leading coefficient $ad$. Consider $P_1(x) = x^{33}$ and $P_{k+1} = \\Delta(P_k)$. By induction, $P_k(x)$ is a sum of $2^{k-1}$ terms of the required form. For $k=33$, $P_{33}(x) = 33!x + b$ for some $b \\in \\mathbb{Z}$. Since $1$ and $-1$ can be written in the required form, every integer is a sum of at most $2^{32} + 33! < 12^{12^{12}}$ such terms. Thus, there are no $n$ with $b(n) = 12^{12^{12}}$, so the set is empty (finite).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18020,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, let $x_n = \\binom{2n}{n}$.\n\n1. Prove that if $\\dfrac{2017^k}{2} < n < 2017^k$ for some positive integer $k$, then $2017$ divides $x_n$.\n\n2. Find all positive integers $h > 1$ such that there exist positive integers $N, T$ for which the sequence $(x_n)_{n > N}$ is periodic modulo $h$ with period $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. We prove the statement for all odd primes $p$ (including $2017$). Suppose there exists a positive integer $k$ such that $\\dfrac{p^k}{2} < n < p^k$. Then:\n\n$$\nv_p(x_n) = v_p\\left(\\binom{2n}{n}\\right) = v_p((2n)!) - 2v_p(n!)\n$$\n\nSince $\\dfrac{p^k}{2} < n < p^k$, we have $p^k < 2n < 2p^k < p^{k+1}$. Thus,\n\n$$\nv_p((2n)!) = \\left\\lfloor \\frac{2n}{p} \\right\\rfloor + \\left\\lfloor \\frac{2n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2n}{p^k} \\right\\rfloor\n$$\n\nFor any $x \\in \\mathbb{R}$, $\\lfloor 2x \\rfloor \\ge 2\\lfloor x \\rfloor$, with equality if $\\{x\\} < \\frac{1}{2}$. Given $\\dfrac{p^k}{2} < n < p^k$,\n\n$$\nv_p((2n)!) > 2 \\left( \\left\\lfloor \\frac{n}{p} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{p^k} \\right\\rfloor \\right) = 2v_p(n!)\n$$\n\nSo $v_p(x_n) > 0$, hence $p \\mid x_n$.\n\n2. Suppose $h > 1$ satisfies the periodicity condition. For any odd prime $p$ dividing $h$, the sequence $x_n$ modulo $p$ would also be periodic. By part 1, for $\\dfrac{p^k}{2} < n < p^k$,\n\n$$\nx_n \\equiv 0 \\pmod{p}.\n$$\n\nLet $k$ be such that $\\dfrac{p^k}{2} > T + 1$. Then for all $n \\ge n_0$ (for large enough $n_0$), $x_n \\equiv 0 \\pmod{p}$. However, for $t$ large enough with $p^t - 1 > 2n_0$ and $n = \\dfrac{p^t - 1}{2}$, we have $v_p(x_n) = 0$, so $x_n$ is not divisible by $p$, a contradiction.\n\nTherefore, $2$ is the only possible prime divisor of $h$, so $h = 2^k$ for some $k$. If $k > 1$, let $r = k - 1$ and consider $n = 2^{a_1} + \\dots + 2^{a_r}$ with $a_1 > \\max\\{T, N\\}$. Then\n\n$$\nv_2(x_n) = 2S_2(n) - S_2(2n) = r,\n$$\n\nwhere $S_2(x)$ is the sum of the binary digits of $x$. Thus, $x_n \\equiv 2^{k-1} \\pmod{h}$. For $i < 2^{a_1}$, the binary digit sum of $n + i$ increases, so $x_{n+i} \\equiv 0 \\pmod{h}$. Since $a_1 > \\max\\{T, N\\}$, $x_n \\equiv x_{n+T} \\equiv 0 \\pmod{h}$, a contradiction. Thus, no such $h > 1$ exists except possibly $h = 2$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18021,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the inequality for non-negative $a, b, c$:\n\n$$\na\\sqrt{3a^2 + 6b^2} + b\\sqrt{3b^2 + 6c^2} + c\\sqrt{3c^2 + 6a^2} \\ge (a+b+c)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Use the following inequality:\n\n$$\n3a^2 + 6b^2 \\ge (a+2b)^2.\n$$\n\nThe proof is straightforward:\n\n$$\n3a^2 + 6b^2 \\ge (a+2b)^2 \\Leftrightarrow 2a^2 + 2b^2 \\ge 4ab \\Leftrightarrow 2(a-b)^2 \\ge 0.\n$$\n\nFinally, we obtain:\n\n$$\na\\sqrt{3a^2 + 6b^2} + b\\sqrt{3b^2 + 6c^2} + c\\sqrt{3c^2 + 6a^2} \\ge a(a+2b) + b(b+2c) + c(c+2a) = (a+b+c)^2.\n$$\n\n![](images/ukraine_2015_Booklet_p3_data_43a823d029.png)\n\nFig. 8",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18022,
    "subject": "Mathematics (Olympiad)",
    "question": "A postman has $n$ parcels of weights $1, 2, 3, \\ldots, n$. He wants to divide the parcels into three groups of equal weight. Is this possible for\n\n$$\n(a)\\ n = 2011,\n$$\n\n$$\n(b)\\ n = 2012?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "In the first case, the total weight is\n\n$$\n1 + 2 + \\cdots + 2010 + 2011 = \\frac{2011 \\cdot 2012}{2}\n$$\n\nwhich is not a multiple of $3$. Therefore, there is no solution in this case.\n\nIn the second case, we distribute the first $8$ parcels as follows: parcels $1, 2, 3, 6$ (total weight $12$) are put into the first group. Parcels $4$ and $8$ (also total weight $12$) are put into the second group. Parcels $5$ and $7$ (total weight $12$) are put into the third group.\n\nThe remaining $2004$ parcels $9, \\ldots, 2012$ are divided into $334$ blocks of consecutive integers $6k+3, 6k+4, 6k+5, 6k+6, 6k+7, 6k+8$ for $1 \\leq k \\leq 334$. In each block, $6k+3$ and $6k+8$ (weight $12k+11$) are put into the first group. Parcels $6k+4$ and $6k+7$ are put into the second group. Finally, parcels $6k+5$ and $6k+6$ are put into the third group. This yields a valid partition.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18023,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: (0, +\\infty) \\to (0, +\\infty)$ such that for all $x, y > 0$,\n$$\nf(xf(x) + yf(y)) = f^2(x) + f^2(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(x) = cx$, $c > 0$.\n\n**Solution.** Denote the assertion by $P(x, y)$.\n\nLet $a$ be such that $f(a) = 1$. Let $S$ be the set of $x > 0$ such that $f(ax) = x$. Then $1 \\in S$.\n\n**Lemma 1.** If $x \\in S$, then $2x^2 \\in S$.\n\n*Proof.* Use $P(ax, ax)$: $f(a 2x^2) = 2x^2 \\implies 2x^2 \\in S$.\n\n**Lemma 2.** If $x \\in S$, then $\\sqrt{\\frac{x}{2}} \\in S$.\n\n*Proof.* Let $x_0$ satisfy $f(x_0) = \\sqrt{\\frac{x}{2}}$. Use $P(x_0, x_0)$: $f(x_0 \\sqrt{2} x) = x = f(ax)$, so $x_0 \\sqrt{2} x = a x$. Thus $x_0 = a \\sqrt{\\frac{x}{2}}$ and $f(a \\sqrt{\\frac{x}{2}}) = \\sqrt{\\frac{x}{2}}$, so $\\sqrt{\\frac{x}{2}} \\in S$.\n\n**Lemma 3.** If $x, y \\in S$, then $\\frac{1}{2}(x + y) \\in S$.\n\n*Proof.* From Lemma 2, $\\sqrt{\\frac{x}{2}}, \\sqrt{\\frac{y}{2}} \\in S$. Use $P(a \\sqrt{\\frac{x}{2}}, a \\sqrt{\\frac{y}{2}})$: $f(\\frac{1}{2} a (x + y)) = \\frac{1}{2}(x + y)$.\n\n**Lemma 4.** The function $x \\mapsto x f(x)$ is surjective.\n\n*Proof.* Use $P(x, x)$: $f(2x f(x)) = 2 f^2(x)$. For any $x > 0$, take $y$ such that $f(y) = \\sqrt{\\frac{1}{2} f(2x)}$. Then $f(2y f(y)) = f(2x) \\implies y f(y) = x$.\n\n**Lemma 5.** If $x f(x) > y f(y)$, then $f(x) > f(y)$.\n\n*Proof.* Suppose $x f(x) > y f(y)$ but $f(x) \\leq f(y)$. Since $x \\neq y$, $f(x) < f(y)$. Choose $t$ with $x f(x) > t > y f(y)$. There is $u$ such that $u f(u) + y f(y) = t$. Then $f(t) = f^2(u) + f^2(y) > f^2(y) > f^2(x)$. There is $v$ such that $f(t) = f^2(v) + f^2(x) = f(x f(x) + v f(v))$. Thus $t = x f(x) + v f(v) > x f(x)$, a contradiction.\n\n**Lemma 6.** $f$ is increasing on $(0, +\\infty)$.\n\n*Proof.* For $a, b, c > 0$ with $b > c$, let $a = x f(x)$, $b = y f(y)$, $c = z f(z)$. Since $b > c$, $f(y) > f(z)$. Then\n$$\nf(a + b) = f(x f(x) + y f(y)) = f^2(x) + f^2(y) > f^2(x) + f^2(z) = f(x f(x) + z f(z)) = f(a + c).\n$$\n\nFrom Lemma 1, $2^1 \\in S$, $2^3 \\in S$, $2^7 \\in S$, etc., so $S$ contains arbitrarily large reals. To show $S$ contains arbitrarily small reals, for any $\\varepsilon > 0$, from Lemma 2, $2^{\\frac{n-1}{2}} \\in S$ if $2^n \\in S$. Since $2^0 \\in S$, $2^{\\frac{1}{2}-1} \\in S$, $2^{\\frac{1}{4}-1} \\in S$, etc., so there is $x \\in S$ with $\\frac{1}{2} < x < \\frac{1}{2} + \\varepsilon^2$. Let $x_0$ satisfy $f(x_0) = \\sqrt{x - \\frac{1}{2}}$. Use $P(x_0, \\frac{1}{\\sqrt{2}} a)$. Since $\\frac{1}{\\sqrt{2}} \\in S$, $f(\\frac{a}{2} + x_0 \\sqrt{x - \\frac{1}{2}}) = x = f(a x)$. Thus $\\frac{a}{2} + x_0 \\sqrt{x - \\frac{1}{2}} = a x$, so $x_0 = a \\sqrt{x - \\frac{1}{2}}$. Therefore $\\sqrt{x - \\frac{1}{2}} \\in S$ and $\\sqrt{x - \\frac{1}{2}} < \\varepsilon$.\n\nThus, $S$ contains all positive reals. Therefore, $f(x) = \\frac{x}{a}$ for some $a > 0$. All functions of the form $f(x) = c x$ with $c > 0$ satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18024,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a standard $8 \\times 8$ chessboard consisting of 64 small squares coloured in the usual pattern, so 32 are black and 32 are white. A zig-zag path across the board is a collection of eight white squares, one in each row, which meet at their corners. How many zig-zag paths are there?",
    "options": [],
    "answer": "See solution",
    "solution": "We can calculate the number of paths to each square, row by row.\n\nIn the first row, each square can be reached from the first row in only one way. After that, each square can be reached from the one or two squares above it: the number of ways it can be reached is the sum of the numbers directly above. Calculating repeatedly, we produce the following diagram:\n\n![](images/V_Britanija_2009_p14_data_f0d4051d4f.png)\n\nAs can be seen, there are 35, 89, 103, and 69 ways of reaching the four squares in the last row, respectively. Thus there are $35 + 89 + 103 + 69 = 296$ ways of doing it in total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18025,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The only such $n$ are $n=1$ and $n=2$.\n\nFor $n=1$, any even number $a$ of the form $m^2 + 1$ works. For example, $a = 2$ gives $(2-1) = 1$, which is a perfect square.\n\nFor $n=2$, any even number $a$ of the form $m^2 - 1$ works. For example, $a = 8$ gives $(8-1)(8^2-1) = 7 \\times 63 = 441 = 21^2$.\n\nSuppose for $n=3$ such an $a$ exists. Then $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Note that $a^2+a+1 = a(a+1)+1$, and $a+1$ and $a^2+a+1$ are coprime. Since $a+1$ is odd, $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a-1$ and $a^2+a+1$ must both be perfect squares. However, $a^2 < a^2+a+1 < (a+1)^2$, so $a^2+a+1$ cannot be a perfect square. This is a contradiction.\n\nFor $n \\ge 4$, suppose such an $a$ exists. Let $k \\ge 2$ be such that $2^k \\le n < 2^{k+1}$. Note that $a^{2^k} - 1 = (a^{2^{k-1}} - 1)(a^{2^{k-1}} + 1)$. The product $(a-1)(a^2-1)\\dots(a^n-1)$ contains the factor $a^{2^{k-1}} + 1$ and other factors of the form $a^m - 1$ for $1 \\le m \\le n$, $m \\ne 2^k$.\n\nWe show that $a^{2^{k-1}} + 1$ is coprime with all other factors. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k, m)} - 1$. Since $m \\ne 2^k$ and $m < 2^{k+1}$, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Thus, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k} - 1, a^m - 1)$, and hence $d$. But $a$ is even, so $a^{2^{k-1}-1}$ and $a^{2^{k-1}} + 1$ are coprime, so $d = 1$.\n\nTherefore, $a^{2^{k-1}} + 1$ is coprime with all other factors, and since the product is a perfect square, $a^{2^{k-1}} + 1$ must itself be a perfect square. But $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are consecutive perfect squares, which is impossible. Thus, no such $a$ exists for $n \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18026,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $a$, $b$, and $c$ such that the number $2^{a!} + 2^{b!} + 2^{c!}$ is a cube of a natural number.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\ge b \\ge c$.\n\n**Case 1:** $a \\ge 3$ and $b \\ge 3$.\n- If $c \\ge 3$, then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 3 \\pmod{7}$, which is not a cube.\n- If $c = 2$, then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 1 + 4 \\equiv -1 \\pmod{7}$, which can be a cube, but modulo $9$, $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 1 + 4 \\equiv 6 \\pmod{9}$, which is not a cube.\n- If $c = 1$, then $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 1 + 2 \\equiv 4 \\pmod{7}$, which is not a cube.\n\n**Case 2:** $a \\ge 3$, $b \\le 2$.\n- If $b = c = 2$, $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 4 + 4 \\equiv 2 \\pmod{7}$, not a cube.\n- If $b = 2$, $c = 1$, $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 4 + 2 \\equiv 0 \\pmod{7}$, which can be a cube, but modulo $9$, $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 4 + 2 \\equiv 7 \\pmod{9}$, not a cube.\n- If $b = c = 1$, $2^{a!} + 2^{b!} + 2^{c!} \\equiv 1 + 2 + 2 \\equiv 5 \\pmod{7}$, not a cube.\n\nTherefore, there are no natural numbers $a$, $b$, and $c$ such that $2^{a!} + 2^{b!} + 2^{c!}$ is a cube of a natural number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18027,
    "subject": "Mathematics (Olympiad)",
    "question": "On a board there are $n$ nails, each pair connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be\n\na) $6$?\n\nb) $7$?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The answer is no.\n\nSuppose it is possible. Consider some color, say blue. Each blue string is the side of $4$ triangles formed with vertices on the given points. There exist $\\binom{5}{2} = 10$ pairs of colors other than blue, and for any such pair of colors together with blue, there exists a triangle with strings in these colors. Thus, there must be at least $3$ blue strings (otherwise, the number of triangles with a blue string as a side would be at most $2 \\times 4 = 8$, a contradiction). The same is true for any color, so altogether there exist at least $6 \\times 3 = 18$ strings, while we have just $\\binom{6}{2} = 15$ of them.\n\n(b) The answer is yes.\n\nPlace the nails at the vertices of a regular $7$-gon and color each one of its sides in a different color. Now color each diagonal in the color of the unique side parallel to it. It can be checked directly that each triple of colors appears in some triangle (because of symmetry, it is enough to check only the triples containing the first color).\n\n![](images/makedonija2012_p10_data_88fd2ae0f8.png)\n\n**Remark.** The argument in (a) can be applied to any even $n$. The argument in (b) can be applied to any odd $n = 2k+1$ as follows: first number the nails as $0, 1, 2, \\ldots, 2k$ and similarly number the colors as $0, 1, 2, \\ldots, 2k$. Then connect nail $x$ with nail $y$ by a string of color $x + y \\pmod{n}$. For each triple of colors $(p, q, r)$ there are vertices $x, y, z$ connected by these colors. Indeed, we need to solve $(\\bmod\\ n)$ the system\n$$\nx + y \\equiv p, \\quad x + z \\equiv q, \\quad y + z \\equiv r \\quad (*)\n$$\nAdding all three, we get $2(x + y + z) \\equiv p + q + r$ and multiplying by $k + 1$ we get\n$$\nx + y + z \\equiv (k + 1)(p + q + r).\n$$\nWe can now find $x, y, z$ from the identities $(*)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18028,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x$, $y$, $z$ satisfy the condition:\n\n$$\n\\frac{1}{xy} = \\frac{y}{z - x + 1} = \\frac{2}{z + 1}.\n$$\n\nProve that one of these numbers is an arithmetic mean of the other two.",
    "options": [],
    "answer": "See solution",
    "solution": "From the statement, we have $z = xy^2 + x - 1 = 2xy - 1$. Thus,\n$$\nxy^2 + x - 1 = 2xy - 1\n$$\nwhich simplifies to\n$$\nx(y^2 - 2y + 1) = x(y - 1)^2 = 0.\n$$\nSince $x \\neq 0$, it follows that $y = 1$. Now, using $xy = \\frac{z+1}{2}$ and $y = 1$, we get $x = \\frac{z+1}{2}$. Rearranging, $z = 2x - 1$, so $x = \\frac{z + 1}{2}$, which shows that $x$ is the arithmetic mean of $z$ and $1$ (since $y = 1$). Q.E.D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18029,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $n$。一個正整數 $m$ 是 $n$-好數的充要條件是至多只有 $2n$ 個質數 $p$ 滿足 $p^2 \\mid m$。\n\n(a) 證明：對於任意兩個互質的正整數 $a, b$，總是存在正整數 $x, y$ 使得 $ax^n + by^n$ 是 $n$-好數。\n\n(b) 證明：對於任意滿足 $\\gcd(a_1, \\dots, a_k) = 1$ 的 $k$ 個正整數 $a_1, \\dots, a_k$，總是存在正整數 $x_1, \\dots, x_k$ 使得 $a_1x_1^n + a_2x_2^n + \\dots + a_kx_k^n$ 是 $n$-好數。\n\n(註：$a_1, \\dots, a_k$ 不必兩兩相異。)",
    "options": [],
    "answer": "See solution",
    "solution": "我們先證明 (a)。令 $N, C_1, C_2$ 為待定的正整數。令 $P$ 為所有不超過 $\\max\\{a, b, 2n+4\\}$ 的質數的乘積。令 $S_1 = \\{Pn + C_1 : n \\in [N]\\}$，$S_2 = \\{Pn + C_2 : n \\in [N]\\}$。\n\n對於每個質數 $p \\in (\\max\\{a, b, 2n+4\\}, \\sqrt{N})$，計算滿足 $p^2 \\mid ax^n + by^n$ 的 $(x, y) \\in S_1 \\times S_2$ 的數量。對於每個 $y \\in S_2$，若 $p \\mid y$，則也需 $p \\mid x$，因此在 $S_1$ 中至多有 $(N/p + 1)$ 個 $x$ 使 $p^2 \\mid ax^n + by^n$（此處用到 $\\gcd(p, P) = 1$）。若 $p \\nmid y$，則 $ax^n + by^n \\equiv 0 \\pmod{p}$ 至多有 $n$ 個根。由於 $p \\nmid n$，每個模 $p$ 的解唯一提升到模 $p^2$，因此在 $S_1$ 中至多有 $n(N/p^2 + 1)$ 個 $x$ 使 $p^2 \\mid ax^n + by^n$（再次用到 $\\gcd(p, P) = 1$）。因此，總共有\n\n$$\n\\left(\\frac{N}{p} + 1\\right)^2 + \\left(N - \\frac{N}{p} - 1\\right) n \\left(\\frac{N}{p^2} + 1\\right) < \\frac{(2n + 4)N^2}{p^2}\n$$\n\n對 $(x, y) \\in S_1 \\times S_2$ 滿足 $p^2 \\mid ax^n + by^n$。\n\n又因為\n\n$$\n\\sum_{i=2n+5}^{\\infty} \\frac{1}{i(i-1)} = \\sum_{i=2n+5}^{\\infty} \\left( \\frac{1}{i-1} - \\frac{1}{i} \\right) = \\frac{1}{2n+4},\n$$\n\n所以\n\n$$\n\\sum_{\\text{prime } p \\ge 2n+5} \\frac{2n+4}{p^2} < \\sum_{i=2n+5}^{\\infty} \\frac{2n+4}{i(i-1)} = 1.\n$$\n\n因此，對每個 $C_1, C_2, N$，存在一對 $(x, y) \\in S_1 \\times S_2$，使得對所有質數 $p \\in (\\max\\{a, b, 2n+4\\}, \\sqrt{N})$ 都有 $p^2 \\nmid ax^n + by^n$。\n\n接下來，適當選擇 $C_1, C_2$ 處理 $p \\le \\max\\{a, b, 2n+4\\}$ 的情況。對每個質數 $p \\le \\max\\{a, b, 2n+4\\}$，若 $p \\nmid a$，則可令 $C_1 \\equiv 1 \\pmod{p}$，$C_2 \\equiv 0 \\pmod{p}$；否則由條件知 $p \\nmid b$，可令 $C_1 \\equiv 0 \\pmod{p}$，$C_2 \\equiv 1 \\pmod{p}$。由中國剩餘定理可選 $1 \\le C_1, C_2 \\le P$ 使上述條件對所有 $p$ 成立。因此，對所有 $p \\le \\max\\{a, b, 2n+4\\}$ 及 $(x, y) \\in S_1 \\times S_2$，有 $p \\nmid ax^n + by^n$。\n\n如此選擇 $C_1, C_2$，對所有 $N$，可選 $(x, y) \\in S_1 \\times S_2$ 使 $p^2 \\nmid ax^n+by^n$ 對所有 $p < \\sqrt{N}$ 成立。注意 $ax^n+by^n \\le 2\\max\\{a, b\\}P^n(N+1)^n < 2^{n+1} \\max\\{a, b\\}P^n N^n$，且若 $p^2 \\mid ax^n+by^n$ 則 $p \\ge \\sqrt{N}$。因此滿足 $p^2 \\mid ax^n + by^n$ 的質數 $p$ 數量至多為\n\n$$\n\\log_{\\sqrt{N}} 2^{n+1} \\max\\{a, b\\}P^n N^n = 2n + \\log_{\\sqrt{N}} 2^{n+1} \\max\\{a, b\\}P^n,\n$$\n\n若 $N$ 夠大則小於 $2n+1$。故存在 $x, y \\in \\mathbb{N}$ 使至多有 $2n$ 個質數 $p$ 滿足 $p^2 \\mid ax^n + by^n$。\n\n接著用 (a) 證明 (b)。對每個 $a_k$ 的質因數 $p$，存在 $i(p) \\in [k-1]$ 使 $p \\nmid a_{i(p)}$。可令 $x_j \\equiv \\delta_{i(p),j} \\pmod{p}$，使 $p \\nmid a_1x_1^n + \\dots + a_{k-1}x_{k-1}^n$。由中國剩餘定理，存在正整數 $x_1, \\dots, x_{k-1}$ 使上述對所有 $p$ 成立，因此 $a_1x_1^n + \\dots + a_{k-1}x_{k-1}^n$ 與 $a_k$ 互質。由 (a) 可選 $x, y \\in \\mathbb{N}$ 使 $(a_1x_1^n + \\dots + a_{k-1}x_{k-1}^n)x^n + a_ky^n$ 為 $n$-好數。故 $a_1(x_1x)^n + \\dots + a_{k-1}(x_{k-1}x)^n + a_ky^n$ 為 $n$-好數，得證。",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18030,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider security codes consisting of four digits. We say that one code *dominates* another code if each digit of the first code is at least as large as the corresponding digit in the second code. For example, 4961 dominates 0761, because $4 \\ge 0$, $9 \\ge 7$, $6 \\ge 6$, and $1 \\ge 1$.\n\nWe would like to assign a colour to each security code from 0000 to 9999, but if one code dominates another code then the codes cannot have the same colour.\n\nWhat is the minimum number of colours that we need in order to do this?",
    "options": [],
    "answer": "See solution",
    "solution": "$37$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18031,
    "subject": "Mathematics (Olympiad)",
    "question": "Бүхэл тоонууд $a, b, c$ нь $a + b + c = 0$ нөхцлийг хангах үед дараах тэнцэтгэл биелэх бүх $f : \\mathbb{Z} \\to \\mathbb{Z}$ функцуудыг ол:\n\n$$\n(a)^2 + (b)^2 + (c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Тэгшитгэлийг $a = b = c = 0$ гэж орлуулбал $3f(0)^2 = 6f(0)^2$ гарна. Иймд\n\n$$\n\\underline{f(0)} = 0. \\qquad (1)\n$$\n\n$b = -a$, $c = 0$ гэж орлуулбал $(f(a) - f(-a))^2 = 0$ тул $f$ нь тэгш функц:\n\n$$\nf(a) = f(-a) \\quad \\text{бүх } a \\in \\mathbb{Z}. \\qquad (2)\n$$\n\n$b = a$, $c = -2a$ гэж орлуулбал $2f(a)^2 + f(2a)^2 = 2f(a)^2 + 4f(a)f(2a)$, үүнээс\n\n$$\nf(2a) = 0 \\quad \\text{эсвэл} \\quad f(2a) = 4f(a) \\quad \\text{бүх } a \\in \\mathbb{Z}. \\qquad (3)\n$$\n\nХэрэв $f(r) = 0$ ($r \\ge 1$) бол $b = r$, $c = -a - r$ гэж орлуулбал $(f(a + r) - f(a))^2 = 0$ тул $f$ нь $r$-ийн үеийн давтамжтай:\n\n$$\nf(a + r) = f(a) \\quad \\text{бүх } a \\in \\mathbb{Z}.\n$$\n\nТухайлбал, $f(1) = 0$ бол $f$ тогтмол функц бөгөөд $f(a) = 0$ бүх $a$-д. Энэ нь тэнцэтгэлийг хангана. Одоо $f(1) = k \\ne 0$ гэж үзье.\n\n(3)-аас $f(2) = 0$ эсвэл $f(2) = 4k$. Хэрэв $f(2) = 0$ бол $f$ нь 2-ын давтамжтай, $f(\\text{even}) = 0$, $f(\\text{odd}) = k$. Энэ функц бүрэн шийдэл болно. Үлдсэн тохиолдолд $f(2) = 4k \\ne 0$ гэж үзье.\n\nДахин (3)-аас $f(4) = 0$ эсвэл $f(4) = 16k$. Эхний тохиолдолд $f$ нь 4-ийн давтамжтай, $f(3) = f(-1) = f(1) = k$, $f(4n) = 0$, $f(4n+1) = f(4n+3) = k$, $f(4n+2) = 4k$ бүх $n$-д. Энэ функц мөн шийдэл болно. Үлдсэн тохиолдолд $f(4) = 16k \\ne 0$ гэж үзье.\n\nОдоо $f(3) = 9k$ болохыг харуулъя. Үүний тулд дараах орлуулалтуудыг ашиглана:",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18032,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(n)$ denote the sum of the digits of $n$ in base 10. For any set of positive integers $m_1, m_2, \\dots, m_k$, we have the inequality\n\n$$\nS\\left(\\sum_{i=0}^{k} m_i\\right) \\leq \\sum_{i=0}^{k} S(m_i).\n$$\n\nEquality holds if, for any nonnegative integer $j$, the $10^j$-th digit of $m_i$ is $0$ for all but at most one $i$.\n\nNow, let $n$ be a positive integer such that $S(n) = 5$. Show that the maximum possible value of $S(n^5)$ is $398$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write $n = 10^{k_1} + 10^{k_2} + 10^{k_3} + 10^{k_4} + 10^{k_5}$ for non-negative integers $k_1, \\dots, k_5$. Expanding $n^5$ using the multinomial theorem, we have\n\n$$\nn^5 = \\sum f(a_1, \\dots, a_5) 10^{a_1 k_1 + \\dots + a_5 k_5},\n$$\n\nwhere the sum is over all quintuples $(a_1, \\dots, a_5)$ of non-negative integers with $a_1 + \\dots + a_5 = 5$, and $f(a_1, \\dots, a_5)$ is the multinomial coefficient.\n\nBy the lemma, if the exponents $k_1, \\dots, k_5$ are chosen so that the terms do not overlap in their decimal expansions (e.g., $k_i = 3 \\times 6^{i-1}$), then\n\n$$\nS(n^5) = \\sum S(f(a_1, \\dots, a_5)).\n$$\n\nCalculating this sum over all possible quintuples, we get\n\n$$\n5 \\cdot 1 + 20 \\cdot 5 + 20 \\cdot 1 + 30 \\cdot 2 + 30 \\cdot 3 + 20 \\cdot 6 + 1 \\cdot 3 = 398.\n$$\n\nTherefore, the maximum possible value of $S(n^5)$ is $398$, and this value is attainable.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18033,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a tetrahedron $ABCD$, it is known that $\\angle ADB = \\angle BDC = \\angle CDA = 60^\\circ$, $AD = BD = 3$ and $CD = 2$. Then the radius of the sphere circumscribing $ABCD$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the center of the sphere circumscribing $ABCD$ be $O$. Then $O$ is on the vertical line of plane $ABD$ through point $N$, the circumcenter of $\\triangle ABD$. It is known that $\\triangle ABD$ is regular, so $N$ is its center. Let $P$ and $M$ be the midpoints of $AB$ and $CD$, respectively. Then $N$ is on $DP$ with $ON \\perp DP$ and $OM \\perp CD$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p39_data_0d05b82b57.png)\n\nLet $\\theta$ denote the angle between $CD$ and plane $ABD$. From\n\n$$\n\\angle CDA = \\angle CDB = \\angle ADB = 60^\\circ,\n$$\n\nwe find $\\cos \\theta = \\frac{1}{\\sqrt{3}}$, $\\sin \\theta = \\frac{\\sqrt{2}}{\\sqrt{3}}$.\n\nSince $DM = \\frac{1}{2}CD = 1$, $DN = \\frac{2}{3} \\cdot DP = \\frac{\\sqrt{3}}{2} \\cdot 3 = \\sqrt{3}$, by the cosine theorem we have, in $\\triangle DMN$,\n\n$$\n\\begin{aligned}\nMN^2 &= DM^2 + DN^2 - 2 \\cdot DM \\cdot DN \\cdot \\cos \\theta \\\\\n&= 1^2 + (\\sqrt{3})^2 - 2 \\cdot 1 \\cdot \\sqrt{3} \\cdot \\frac{1}{\\sqrt{3}} = 2,\n\\end{aligned}\n$$\n\nthat is, $MN = \\sqrt{2}$. The radius of the sphere circumscribing $ABCD$ is then\n\n$$\nOD = \\frac{MN}{\\sin \\theta} = \\frac{\\sqrt{2}}{\\sqrt{2}} = \\sqrt{3}.\n$$\n\nThe answer is $R = \\sqrt{3}$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p40_data_f2103bae55.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 18034,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all integers $a, b$,  \n$$\nf(b + f(a)) = a + f(b).\n$$\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "The two solutions are $f(x) = x$ and $f(x) = -x$. We prove this in three stages.\n\nFirst, we show that $f$ is self-inverse, that is, $f(f(x)) = x$ for all integers $x$. Interchanging $a$ and $b$ in the original equation:\n$$\nf(a + f(b)) = b + f(a),\n$$\nthen applying $f$ to both sides gives:\n$$\nf(f(a + f(b))) = f(b + f(a)) = a + f(b).\n$$\nAny integer can be represented as $a + f(b)$, so $f(f(x)) = x$ for all $x$.\n\nNext, for additivity, let $c = f(b)$. By the self-inverse property, any integer $c$ can be written in this form by setting $b = f(c)$. The original equation becomes:\n$$\nf(a + c) = f(c) + f(a).\n$$\nSetting $a = c = 0$ gives $f(0) = 0$. By induction, $f(x) = x f(1)$ for positive integers $x$. Writing $c = -a$ shows $f(x) = x f(1)$ for all negative $x$. Thus, $f(x)$ is linear with slope $f(1)$ and $y$-intercept zero.\n\nFinally, the self-inverse property with $x = 1$ gives $1 = f(f(1)) = f(1)^2$, so $f(1) = \\pm 1$. Therefore, the only solutions are $f(x) = x$ and $f(x) = -x$, both of which satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18035,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest integer $k$ such that there exist two numbers $a$ and $b$ whose decimal representations are purely periodic with period 30, and such that both $a-b$ and $a+kb$ have purely periodic decimal representations with period 15.",
    "options": [],
    "answer": "See solution",
    "solution": "One may assume that the decimal representations of $a$, $b$, $a-b$, and $a+kb$ are purely periodic. We have $a = \\frac{m}{10^{30}-1}$ and $b = \\frac{n}{10^{30}-1}$, while $a-b$ and $a+kb$ are representable as fractions with denominator $10^{15}-1$. Thus $\\gcd(k+1, 10^{15}+1) > 1$, so $k+1 \\ge 7$. An example for $k=6$ is provided, for instance, by $a = \\frac{8}{7(10^{15}-1)}$ and $b = \\frac{1}{7(10^{15}-1)}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18036,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 9 \\times 9$ cube is repeatedly reduced by removing layers of small cubes as follows:\n\n- **Step 1:** Remove cubes to leave a $7 \\times 7 \\times 7$ cube with a $7 \\times 7$ single layer of small cubes placed centrally on each face.\n- **Step 2:** Remove only the edge cubes in the $7 \\times 7$ single layers, leaving a $7 \\times 7 \\times 7$ cube with a $5 \\times 5$ single layer of small cubes placed centrally on each face.\n- **Step 3:** Remove the edge cubes in the $5 \\times 5$ single layers and the edge cubes in the $7 \\times 7 \\times 7$ cube.\n\nHow many small cubes remain after these three steps?\n\n![](images/Australian-Scene-combined-2015_p50_data_885dfd7ba4.png)\n\n![](images/Australian-Scene-combined-2015_p50_data_df8f1aed9a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "At the first step, the number of small cubes removed is:\n$$8 + (12 \\times 7) = 92$$\n\nAt the second step, the number of small cubes removed is:\n$$6(4 + 4 \\times 5) = 144$$\n\nAt the third step, the number of small cubes removed is:\n$$6(4 + 4 \\times 3) + 8 + (12 \\times 5) = 164$$\n\nThus, the number of small cubes remaining is:\n$$(9 \\times 9 \\times 9) - 92 - 144 - 164 = 729 - 400 = 329$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18037,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a function satisfying the following conditions:\n\n1. $f(x) + f(y) = f(x + y)$ for all $x, y \\in \\mathbb{Z}$.\n2. $f(0) = 1$ or $f(0) = -1$.\n3. $f$ is not periodic unless it is constant.\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $f(0) = 1$.\n\nIf $c \\neq 1 \\pmod{p}$, take $d \\in \\mathbb{Z}$ such that $1 - p < pd + c \\leq 0$. Set $x = d$, $y = c$, $z = 0$:\n\n$$\n1 - (pd + c) = f(pd + c) = pf(d).\n$$\n\nThis is impossible, so $c \\equiv 1 \\pmod{p}$. Since $c$ is even, $c \\equiv p + 1 \\pmod{2p}$.\n\n$$\nf(2px + 2c - 1) = 2f(px + c) = 2pf(x) = pf(2x - 1) = f(2px - p + c).\n$$\n\nIf $c \\neq 1 - p$, then $f$ takes finite values on $2p\\mathbb{Z} + 1$ and also on $\\mathbb{Z}$, a contradiction. Therefore, $c = 1 - p$. By mathematical induction,\n\n$$\nf(1 - x) = x, \\quad \\forall x \\in \\mathbb{N}.\n$$\n\nUsing $f(x) + f(2 - x) = 0$, we can say\n\n$$\nf(1 - x) = x, \\forall x \\in \\mathbb{Z} \\implies f(x) = 1 - x, \\forall x \\in \\mathbb{Z}.\n$$\n\n**Case 2:** $f(0) = -1$.\n\nChoose $e \\in \\mathbb{Z}$ such that $f(e) = 1$ and set $y = e$, $z = 0$ to get\n\n$$\nf(x - e) = -f(x) = 0 \\implies f(x - 2e) = -f(x - e) = f(x).\n$$\n\nBecause $e$ is non-zero, $f$ is periodic, which is impossible.\n\nIt's easy to verify that $f(x) = 1 + x, \\forall x \\in \\mathbb{Z}$ and $f(x) = 1 - x, \\forall x \\in \\mathbb{Z}$ satisfy the original conditions. In conclusion, these are the solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18038,
    "subject": "Mathematics (Olympiad)",
    "question": "In the isosceles triangle $ABC$, let $D$ be the foot of the altitude from $C$ to the base $AB$. Calculate the length of the segment that connects the midpoint of $AD$ and the midpoint of $AC$, given that the perimeter of triangle $ABC$ is $36\\text{ cm}$ and the perimeter of triangle $ADC$ is $29\\text{ cm}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The perimeter of triangle $ADC$ multiplied by two equals the sum of the perimeter of triangle $ABC$ and twice the length of $CD$:\n\n$$2 \\cdot 29 = 36 + 2 \\cdot \\overline{CD}$$\n\nFrom here, we obtain:\n\n$$\\overline{CD} = \\frac{2 \\cdot 29 - 36}{2} = 11\\text{ cm}$$\n\nLet $M$ be the midpoint of $AD$ and $N$ be the midpoint of $AC$. Then $MN$ is the midline of triangle $ADC$, so:\n\n$$\\overline{MN} = \\frac{1}{2} \\overline{CD} = \\frac{1}{2} \\cdot 11 = 5.5\\text{ cm}$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18039,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \\mathbb{N} \\to \\mathbb{N}$ by $T(2k) = k$ and $T(2k + 1) = 2k + 2$. We write $T^2(n) = T(T(n))$ and in general $T^k(n) = T^{k-1}(T(n))$ for any $k > 1$.\n\n1. Show that for each $n \\in \\mathbb{N}$, there exists $k$ such that $T^k(n) = 1$.\n\n2. For $k \\in \\mathbb{N}$, let $c_k$ denote the number of elements in the set $\\{n : T^k(n) = 1\\}$. Prove that $c_{k+2} = c_{k+1} + c_k$, for $k \\ge 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) For $n = 1$, we have $T(1) = 2$ and $T^2(1) = T(2) = 1$. Hence we may assume that $n > 1$.\n\nSuppose $n > 1$ is even. Then $T(n) = n/2$. We observe that $n/2 \\leq n - 1$ for $n > 1$.\n\nSuppose $n > 1$ is odd so that $n \\geq 3$. Then $T(n) = n + 1$ and $T^2(n) = (n + 1)/2$. Again we see that $(n + 1)/2 \\leq n - 1$ for $n \\geq 3$.\n\nThus we see that in at most $2(n-1)$ steps $T$ sends $n$ to $1$. Hence $k \\leq 2(n-1)$. (Here $2(n-1)$ is only a bound. In reality, fewer steps will do.)\n\n(ii) We show that $c_n = f_{n+1}$, where $f_n$ is the $n$-th Fibonacci number.\n\nLet $n \\in \\mathbb{N}$ and let $k \\in \\mathbb{N}$ be such that $T^k(n) = 1$. Here $n$ can be odd or even. If $n$ is even, it can be either of the form $4d + 2$ or of the form $4d$.\n\nIf $n$ is odd, then $1 = T^k(n) = T^{k-1}(n+1)$. (Observe that $k > 1$; otherwise we get $n+1 = 1$ which is impossible since $n \\in \\mathbb{N}$.) Here $n+1$ is even.\n\nIf $n = 4d + 2$, then again $1 = T^k(4d + 2) = T^{k-1}(2d + 1)$. Here $2d + 1 = n/2$ is odd.\n\nThus each solution of $T^{k-1}(m) = 1$ produces exactly one solution of $T^k(n) = 1$ and $n$ is either odd or of the form $4d + 2$.\n\nIf $n = 4d$, we see that $1 = T^k(4d) = T^{k-1}(2d) = T^{k-2}(d)$. This shows that each solution of $T^{k-2}(m) = 1$ produces exactly one solution of $T^k(n) = 1$ of the form $4d$.\n\nThus the number of solutions of $T^k(n) = 1$ is equal to the number of solutions of $T^{k-1}(m) = 1$ and the number of solutions of $T^{k-2}(l) = 1$ for $k > 2$. This shows that $c_k = c_{k-1} + c_{k-2}$ for $k > 2$. We also observe that $2$ is the only number which goes to $1$ in one step and $4$ is the only number which goes to $1$ in two steps. Hence $c_1 = 1$ and $c_2 = 2$. This proves that $c_n = f_{n+1}$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18040,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with circumradius $R$, and let $\\ell_A$, $\\ell_B$, $\\ell_C$ be the altitudes through $A$, $B$, $C$ respectively. The altitudes meet at $H$. Let $P$ be an arbitrary point in the same plane as $ABC$. The feet of the perpendicular lines through $P$ onto $\\ell_A$, $\\ell_B$, $\\ell_C$ are $D$, $E$, $F$ respectively. Prove that the areas of $DEF$ and $ABC$ satisfy the following equation:\n\n$$\n\\text{area}(DEF) = \\frac{PH^2}{4R^2} \\text{area}(ABC).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle PDH = \\angle PFH = \\angle PEH = 90^\\circ$ by construction, so by Thales's Theorem, $D$, $E$, and $F$ lie on a circle whose diameter is $PH$. Therefore, the angle between lines $DE$ and $FE$ is the same as the angle between lines $DH$ and $FH$ (angles subtended by the chord $DF$), which in turn is the same as the angle between lines $AB$ and $BC$ (angles between pairwise perpendicular lines). Repeating the argument, we find that the angles between lines $DE$, $DF$, $EF$ coincide with those between lines $AB$, $AC$, $BC$, so triangles $ABC$ and $DEF$ are similar. The diameter of the circumcircle of $ABC$ is $2R$, the diameter of the circumcircle of $DEF$ is $PH$ as observed before. Since areas of similar triangles are proportional to squared lengths, we obtain\n\n$$\n\\text{area}(DEF) = \\frac{PH^2}{4R^2} \\text{area}(ABC)\n$$\n\nas required.\n\n![](images/SAOlympiad2018Solutions_p2_data_3df1285cc9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18041,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X$ be a set of $n$ different positive integers. Denote by $S(I)$ the sum of all elements in a subset $I$ of $X$, and by $X_m$ the set $\\{S(I) : I \\subset X,\\ |I| = m\\}$ for a positive integer $m \\leq n$. If $|X_m| = m(n-m)+1$ for an integer $m$ such that $n-1 > m > 1$, then show that the elements of $X$ form an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_0 < a_1 < \\dots < a_{n-1}$ be the elements of $X$. For each $i < n-m$ and $j \\leq m$, construct a subset $A_{ij}$ of $X$ as $A_{ij} = \\{a_i, \\dots, a_{i+m}\\} \\setminus \\{a_{i+j}\\}$. Then $S(A_{ij}) \\in X_m$ and\n\n$$\nS(A_{i0}) > S(A_{i1}) > \\dots > S(A_{im-1}) > S(A_{im}).\n$$\n\nMoreover,\n\n$$\nX_m = \\{S(A_{ij}) : 0 \\leq j \\leq m,\\ 0 \\leq i < n-m\\},\n$$\n\nsince $S(A_{i0}) = S(A_{i+1,m})$. For each $i < n-m-1$ and $j < m$, consider the subset $B_{ij}$ of $X$ given by $B_{ij} = A_{ij} \\cup \\{a_{i+m+1}\\} \\setminus \\{a_{i+m}\\}$. Certainly, $S(B_{ij}) \\in X_m$ and\n\n$$\nS(A_{i+1,m-1}) = S(B_{i0}) > S(B_{i1}) > \\dots > S(B_{i,m-1}) > S(A_{i,m-1}).\n$$\n\nThus, between $S(A_{i,m-1})$ and $S(A_{i+1,m-1})$, there are $m-1$ different elements of $X_m$, and hence $S(A_{ij}) = S(B_{i,j+1})$ from above. Therefore,\n\n$$\na_{i+j+1} + a_{i+m} = a_{i+j} + a_{i+m+1},\n$$\n\nwhich implies that $a_0, a_1, \\dots, a_{n-1}$ are in arithmetic progression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18042,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, the inscribed circle $\\odot I$ of $\\triangle ABC$ is tangent to the sides $AB$ and $AC$ at points $D$ and $E$, respectively. Let $O$ be the circumcentre of $\\triangle BCI$. Prove that $\\angle ODB = \\angle OEC$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p217_data_981b6a3e6b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $O$ is the circumcentre of $\\triangle BCI$, we see that\n\n$$\n\\angle BOI = 2\\angle BCI = \\angle BCA.\n$$\n\nSimilarly, $\\angle COI = \\angle CBA$. Hence,\n\n$$\n\\angle BOC = \\angle BOI + \\angle COI = \\angle BCA + \\angle CBA = \\pi - \\angle BAC.\n$$\n\nThus, the four points $A$, $B$, $O$, and $C$ are concyclic. Since $OB = OC$, we know that $\\angle BAO = \\angle CAO$. (It can also be seen from the well-known fact that point $O$ is the midpoint of arc $\\widearc{BC}$ (not containing point $A$) on the circumcircle of $\\triangle ABC$.) Combined with the fact that $AD = AE$ and $AO = AO$, we have $\\triangle OAD \\cong \\triangle OAE$. Hence, $\\angle ODA = \\angle OEA$, therefore, $\\angle ODB = \\angle OEC$. $\\square$\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p217_data_bbe526bb6b.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18043,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = (S_0, \\mu)$ where $S_0$ is a subset of $T(m, n)$, $n - m$ is odd, and $\\mu: S_0 \\to \\mathbb{Z}_+$ is a map such that $\\sum_{R \\in S_0} \\mu(R) \\ge n - m + 2$. Prove that $S$ contains a chain with at least three elements.\n\n*Note:* The validity of this claim depends only on $d = n - m$, not on the individual values of $m$ and $n$, because $(x_1 - k, y_1 - k) \\le (x_2 - k, y_2 - k)$ is equivalent to $(x_1, y_1) \\le (x_2, y_2)$ for any integer $k$.\n\nLet $E(m, n)$ be the subset of $T(m, n)$ consisting of those pairs $(x, y)$ for which $x = m$ or $y = n$. If $d = n - m = 1$, then $E(m, n) = T(m, n)$. Note that $E(m, n)$ is a chain for any $d = n - m \\ge 1$, because\n\n$$\n(m, m) \\le (m, m+1) \\le \\dots \\le (m, n-1) \\le (m, n) \\le (m+1, n) \\le \\dots \\le (n, n).\n$$\n\nFor $d \\ge 2$, the set $T(m, n)$ is the disjoint union of $E(m, n)$ and $T(m+1, n-1)$.\n\nProve the claim by induction on $d = n - m \\ge 1$ for odd $d$.\n\nApply this to $T(1, 2018)$, where $d = 2017$ and $S$ is supposed to contain $2019 = n - m + 2$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the claim by induction on $d = n - m \\ge 1$ for odd $d$.\n\n**Base case ($d = 1$):**\nIf $d = 1$, then $T(m, n) = T(m, m+1)$ contains three elements: $(m, m) \\le (m, m+1) \\le (m+1, m+1)$, which form a chain. Any possible set $S$ in this case is a chain as well.\n\n**Inductive step:**\nAssume the claim holds for $d - 2$ (i.e., for $n' - m' = n - m - 2$). Suppose $S = (S_0, \\mu)$ contains at least $n - m + 2$ elements and $S_0 \\subset T(m, n)$. Consider $S_0 \\cap E(m, n)$ and $S_0 \\cap T(m+1, n-1)$. Since $E(m, n)$ is a chain, $S \\cap E(m, n)$ (with multiplicities given by $\\mu$) is a chain as well. If this set contains at least three elements, the proof is finished. Otherwise, $S \\cap T(m+1, n-1)$ (with multiplicities given by $\\mu$) contains at least $n - m + 2 - 2 = (n-1) - (m+1) + 2$ elements, so by the inductive assumption, it contains a chain with at least three elements. This proves the claim.\n\nFor the original problem, $d = n - m = 2018 - 1 = 2017$ for $T(1, 2018)$ and $S$ is supposed to contain $2019 = n - m + 2$ elements. Thus, the statement follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18044,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : (0, \\infty) \\to (0, \\infty)$ such that\n$$\n\\frac{(f(p))^2 + (f(q))^2}{f(r^2) + f(s^2)} = \\frac{p^2 + q^2}{r^2 + s^2}\n$$\nfor all positive real numbers $p, q, r, s$ satisfying $pq = rs$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are:\n$$\nf(x) = x \\quad \\text{and} \\quad f(x) = \\frac{1}{x}.\n$$\nIt is easy to check that these two functions satisfy the conditions. We now show they are the only solutions.\n\nSet $p = q = r = s = 1$ in $(*)$:\n$$\nf(1) = \\frac{(f(1))^2 + (f(1))^2}{f(1) + f(1)} = \\frac{1^2 + 1^2}{1^2 + 1^2} = 1.\n$$\nFor $x > 0$, set $(p, q, r, s) = (1, x, \\sqrt{x}, \\sqrt{x})$:\n$$\n\\frac{1 + (f(x))^2}{2f(x)} = \\frac{1 + x^2}{2x}\n$$\nwhich leads to\n$$\n0 = x(f(x))^2 - x^2 f(x) + x - f(x) = (x - f(x))(1 - x f(x)).\n$$\nSo for all $x > 0$,\n$$\nf(x) = x \\quad \\text{or} \\quad f(x) = \\frac{1}{x}.\n$$\nSuppose $f(x) \\neq x$ and $f(x) \\neq \\frac{1}{x}$ for some $x$. Then there exist $a, b > 0$ with $f(a) = \\frac{1}{a}$ and $f(b) = b$. Set $(p, q, r, s) = (a, b, \\sqrt{ab}, \\sqrt{ab})$:\n$$\n\\frac{\\frac{1}{a^2} + b^2}{2f(ab)} = \\frac{a^2 + b^2}{2ab}\n$$\nSo\n$$\n\\frac{f(ab)}{ab} = \\frac{1 + a^2 b^2}{a^4 + a^2 b^2}.\n$$\nBut $f(ab) = ab$ or $f(ab) = \\frac{1}{ab}$. If $f(ab) = ab$, then $a^4 = 1$ so $a = 1$, contradicting $f(a) \\neq a$. If $f(ab) = \\frac{1}{ab}$, then $b = 1$, contradicting $f(b) \\neq \\frac{1}{b}$. Thus, only $f(x) = x$ and $f(x) = \\frac{1}{x}$ are possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18045,
    "subject": "Mathematics (Olympiad)",
    "question": "In a garden organized as a $2024 \\times 2024$ board, we plant three types of flowers: Roses, Daisies, and Orchids. The planting must satisfy:\n\n1. Each cell is planted with at most one type of flower. Some cells may be left blank.\n2. For each planted cell $A$, there exist exactly 3 other planted cells in the same row or column as $A$ such that those 3 cells are planted with flowers of different types from $A$.\n3. Each flower type is planted in at least one cell.\n\nWhat is the maximal number of cells that can be planted with flowers?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the number of planted cells. The estimation of $P$ will be based on the following two simple lemmas.\n\n**Lemma 1.** If a row (or column) contains at least two types of flowers, then it has at most 6 planted cells.\n\n*Proof.* Suppose to the contrary that a row contains at least 7 planted cells of at least two types of flowers. Choose one type of flower planted in this row with the least (positive) number of cells; WLOG, suppose it is Roses. Then there are at least 4 other cells planted with Daisies or Orchids. This leads to a contradiction, since the cell planted with Roses is in the same row as at least 4 cells planted with Daisies or Orchids. $\blacksquare$\n\n**Lemma 2.** If a row and a column each contain at most one type of flower, then the cell at their intersection is not planted with a flower.\n\n*Proof.* Suppose the intersection cell of the row and the column is planted with a flower, WLOG, suppose it is Roses. Then both the row and column are planted with Roses only. This contradicts condition (ii) in the problem. $\blacksquare$\n\nAssume there are exactly $a$ rows and $b$ columns which contain at least two types of flowers. We will give a bound for $P$ in the following two cases.\n\n**Case 1:** $a > 2018$ or $b > 2018$. WLOG, suppose $a > 2018$. Each remaining $2024 - a$ rows contains at most 2024 planted cells. So by Lemma 1:\n\n$$\nP \\le 6a + 2024(2024 - a) \\le 2024^2 - 2018 \\times 2019 = 22234.\n$$\n\n**Case 2:** $a, b \\le 2018$. WLOG, suppose the first $a$ rows and the first $b$ columns contain at least two types of flowers. We illustrate the situation as in Fig. 1.\n\n![](images/Vietnam_2024_Booklet_p39_data_cd42d8e968.png \"\")\n\nFig. 1\n\nBy Lemma 1, the first $a$ ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18046,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer sequence $\\{x_n\\}$ is defined as follows:\n\n$0 \\leq x_0 < x_1 \\leq 100$ and\n$$\nx_{n+2} = 7x_{n+1} - x_n + 280, \\quad \\forall n \\geq 0.\n$$\n\na) Prove that if $x_0 = 2$, $x_1 = 3$, then for each positive integer $n$, the sum of divisors of $x_n x_{n+1} + x_{n+1} x_{n+2} + x_{n+2} x_{n+3} + 2018$ is divisible by $24$.\n\nb) Find all pairs $(x_0, x_1)$ such that $x_n x_{n+1} + 2019$ are perfect squares for infinitely many $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma.\n\n**Lemma 1.** If a positive integer $n$ satisfies $24 \\mid n+1$, then the sum of its positive divisors $\\sigma(n)$ is divisible by $24$.\n\n*Proof.* If $d$ is a divisor of $n$, then $\\frac{n}{d}$ is also a divisor. Since $n \\equiv 2 \\pmod{3}$, $n$ cannot be a perfect square, so the sum of its divisors can be paired as\n$$\nd + \\frac{n}{d} = \\frac{d^2 + n}{d}.\n$$\nNote $n \\equiv 2 \\pmod{3}$ and $d^2 \\equiv 1 \\pmod{3}$, so the sum above is divisible by $3$.\n\nOn the other hand, $n \\equiv 7 \\pmod{8}$ and $d \\equiv 1, 3, 5, 7 \\pmod{8}$, so $d^2 \\equiv 1 \\pmod{8}$, making the sum divisible by $8$. Since $3$ and $8$ are coprime, the sum is divisible by $24$. $\\square$\n\na) Let $y_n = x_n x_{n+1} + x_{n+1} x_{n+2} + x_{n+2} x_{n+3}$. We need to prove that $\\sigma(y_n + 2018)$ is divisible by $24$. Since $2018 \\equiv 2 \\pmod{24}$, by the lemma, it suffices to show $y_n \\equiv -3 \\pmod{24}$.\n\nConsider $x_{n+2} \\equiv x_{n+1} - x_n + 1 \\pmod{3}$. For $x_0 = 2$, $x_1 = 3$, the sequence modulo $3$ is $2, 0, 2, 0, \\dots$, period $2$.\n\nThus,\n$$\ny_n \\equiv 0 \\cdot 2 + 2 \\cdot 0 + 0 \\cdot 2 = 0 \\pmod{3}.\n$$\n\nFor modulo $8$, $x_{n+2} \\equiv -x_{n+1} - x_n \\pmod{8}$, and the sequence is $2, 3, 3, 2, 3, 3, \\dots$, period $3$.\n\nThus,\n$$\ny_n \\equiv 2 \\cdot 3 + 3 \\cdot 3 + 3 \\cdot 2 = 5 \\pmod{8}.\n$$\n\nTherefore, $y_n + 3$ is divisible by both $3$ and $8$, so $y_n \\equiv -3 \\pmod{24}$, as required.\n\nb) Now, we prove the following lemma.\n\n**Lemma 2.** For the integer sequence $\\{z_n\\}$ with $z_{n+2} = a z_{n+1} - z_n + b$, the quantity\n$$\nz_{n+1}^2 - z_n z_{n+2} - b z_{n+1}\n$$\nis constant for all $n \\geq 0$.\n\n*Proof.*\n$$\n\\begin{aligned}\nz_{n+1}^2 - z_n z_{n+2} - b z_{n+1} &= z_{n+1}(z_{n+1} - b) - z_n(a z_{n+1} - z_n + b) \\\\\n&= z_{n+1}(a z_n - z_{n-1}) - z_n(a z_{n+1} - z_n + b) \\\\\n&= z_n^2 - z_{n-1} z_{n+1} - b z_n.\n\\end{aligned}\n$$\nSo $z_{n+1}^2 - z_n z_{n+2} - b z_{n+1}$ is constant. $\\square$\n\nThus, for our sequence, there exists $C \\in \\mathbb{Z}$ such that\n$$\nx_{n+1}^2 - x_n x_{n+2} - 280 x_{n+1} = C.\n$$\nExpanding,\n$$\n\\begin{aligned}\nx_{n+1}^2 - x_n(7x_{n+1} - x_n + 280) - 280x_{n+1} &= C \\\\\nx_{n+1}^2 + x_n^2 - 7x_{n+1} x_n - 280(x_{n+1} + x_n) &= C \\\\\n(x_{n+1} + x_n - 140)^2 &= 9(x_{n+1} x_n + 2019) - 9 \\cdot 2019 + C + 140^2 \\\\\nu_n^2 &= v_n^2 + C + 1429,\n\\end{aligned}\n$$\nwhere $u_n = x_{n+1} + x_n - 140$, $v_n = 3 \\sqrt{x_{n+1} x_n + 2019}$.\n\nSince $\\{x_n\\}$ is increasing and unbounded, so is $u_n$. If $x_n x_{n+1} + 2019$ is a perfect square, then $v_n \\in \\mathbb{Z}^+$. Thus, $u_n + v_n \\mid C + 1429$ for infinitely many $n$, which only happens if $C + 1429 = 0$.\n\nSo,\n$$\n(x_{n+1} + x_n - 140)^2 = 9(x_{n+1} x_n + 2019), \\quad \\forall n.\n$$\nBut $(x_0 + x_1 - 140)^2 \\geq 9 \\cdot 2019 > 132^2$, so $|140 - x_0 - x_1| \\geq 133$. Since $0 \\leq x_0 < x_1 < 101$, $x_0 + x_1 \\leq 7$. Also,\n$$\nC = x_1^2 + x_0^2 - 7 x_1 x_0 - 280(x_1 + x_0) = -1429.\n$$\nBut $x_1^2 + x_0^2 \\leq 49$, so $-1429 < 49 - 280(x_1 + x_0)$, implying $x_0 + x_1 \\geq 5$. Checking $x_0 + x_1 = 7$ and $6$ yields no solution. So $x_0 + x_1 = 5$, and $x_0 x_1 = 6$, so $x_0 = 2$, $x_1 = 3$.\n\nTherefore, $(x_0, x_1) = (2, 3)$ is the only satisfying pair. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18047,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $f$ 為一正整數值函數，且對於所有正整數 $a, b$，有 $a + f(b) \\mid a^2 + b f(a)$。\n\n證明：存在正整數 $k$ 使得 $f(n) = k n$ 對所有正整數 $n$ 均成立。",
    "options": [],
    "answer": "See solution",
    "solution": "容易看出，令 $a = 1$ 可得 $f(n) \\leq f(1) n$。\n\n再令 $a = n b - f(b)$，對於足夠大的 $n$，有：\n\n$$\nn b \\mid (n b - f(b))^2 + b f(n b - f(b))\n$$\n\n因此 $b \\mid f(b)^2$。特別地，對於每個質數 $p$，有 $f(p) = k_p p$，其中 $0 < k_p \\leq f(1)$。\n\n所以必定存在整數 $k$，使得 $f(p) = k p$ 對無窮多質數 $p$ 成立。對於無窮多的 $p$，\n\n$$\na + k p \\mid (a^2 + p f(a)) - a(a + k p) = p f(a) - p k a\n$$\n\n因此 $a + k p \\mid f(a) - k a$。由於 $p$ 可以無窮大，必有 $f(a) = k a$。\n\n_註：也可用其他方法證明 $p \\mid f(p)$，例如利用狄利克雷定理，若 $p \\nmid f(p)$，則 $p^2 + b f(p)$ 對無窮多 $b$ 可取質數值，從而推出必有 $p \\mid f(p)$。_",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18048,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_0 = 1$ and define a sequence $(x_n)$ by $x_{n+1} = \\sin(x_n) + \\frac{\\pi}{2} - 1$ for all $n \\ge 0$. Show that the sequence converges and find its limit.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that the sequence $(x_n)$ is strictly increasing. For each $n$, $x_n \\leq \\frac{\\pi}{2}$ since $\\sin x \\leq 1$ for all $x \\in \\mathbb{R}$. The function $f(x) = \\sin x - x$ is decreasing for $x \\in \\mathbb{R}$ because $f'(x) = \\cos x - 1 \\leq 0$ for all $x$. Therefore, for $x \\in (-\\infty, \\frac{\\pi}{2})$, $f(x) \\geq \\sin \\frac{\\pi}{2} - \\frac{\\pi}{2}$. Thus, $\\sin x_n \\geq x_n + 1 - \\frac{\\pi}{2}$, which implies $x_{n+1} \\geq x_n$. So $(x_n)$ is increasing and bounded, hence convergent. Let $l$ be its limit. Then $l = \\sin l + \\frac{\\pi}{2} - 1$, i.e., $f(l) = f(\\frac{\\pi}{2})$. Since $f$ is decreasing, $l = \\frac{\\pi}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18049,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2016 real numbers written on the blackboard. In each step, you choose two numbers, erase them, and replace each of them by their product. Is it possible to obtain 2016 equal numbers on the blackboard after a finite number of steps?",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction on $n$ that it is possible to obtain $n$ equal numbers after a finite number of steps.\n\nFor $n = 2$, the claim is trivial: after one step, $(a, b) \\to (ab, ab)$.\n\nFor $n = 4$, we can proceed as follows:\n\n$$(\\underline{a}, \\underline{b}, c, d) \\to (ab, ab, \\underline{c}, \\underline{d}) \\to (\\underline{ab}, ab, \\underline{cd}, cd) \\to (abcd, \\underline{ab}, abcd, \\underline{cd}) \\to (abcd, abcd, abcd, abcd)$$\n\nFor $n = 6$, start with $(a, a, a, a, b, b)$ (this can be achieved using the cases $n = 2$ and $n = 4$). To equalize all six numbers, perform:\n\n$$\n\\begin{align*}\n(a, a, a, \\underline{a}, \\underline{b}, b) &\\to (a, a, \\underline{a}, \\underline{ab}, ab, b) \\to (a, a, a^2b, \\underline{a^2b}, \\underline{ab}, b) \\\\\n&\\to (a, a, \\underline{a^2b}, a^3b^2, a^3b^2, \\underline{b}) \\\\\n&\\to (\\underline{a}, a, \\underline{a^2b^2}, a^3b^2, a^3b^2, a^2b^2) \\\\\n&\\to (a^3b^2, \\underline{a}, a^3b^2, a^3b^2, a^3b^2, \\underline{a^2b^2}) \\\\\n&\\to (a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2)\n\\end{align*}\n$$\n\nAssume the claim holds for all even $n < 4k + 4$ ($k \\ge 1$). For $n = 4k + 4$, first equalize the first $2k + 2$ numbers, then the last $2k + 2$ numbers, yielding:\n\n$$\n\\underbrace{(a, \\dots, a, b, \\dots, b)}_{2k+2}\n$$\n\nThen, perform $2k + 2$ steps, pairing one $a$ and one $b$ each time, to get $(ab, \\dots, ab)$.\n\nFor $n = 4k + 6$, use the induction hypothesis for $n = 2k + 2$ and $n = 2k + 4$ to get:\n\n$$\n\\underbrace{(a, \\dots, a, b, \\dots, b)}_{2k+2}\n$$\n\nPerform $2k$ steps, always pairing one $a$ and one $b$, to obtain:\n\n$$\n\\underbrace{(a, a, ab, \\dots, ab, b, b, b, b)}_{4k}\n$$\n\nPair each $a$ with one $ab$:\n\n$$\n(a^2b, a^2b, a^2b, a^2b, \\underbrace{ab, \\dots, ab}_{4k-2}, b, b, b, b)\n$$\n\nPair each $b$ with one $a^2b$:\n\n$$\n(a^2b^2, a^2b^2, a^2b^2, a^2b^2, \\underbrace{ab, \\dots, ab}_{4k-2}, a^2b^2, a^2b^2, a^2b^2, a^2b^2)\n$$\n\nFinally, perform $2k-1$ steps, replacing $2k-1$ pairs of $ab$'s by $a^2b^2$'s, leading to $(a^2b^2, \\dots, a^2b^2)$.\n\nThus, it is possible to obtain 2016 equal numbers after a finite number of steps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18050,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be distinct positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\sum_{\\text{cyc}} \\frac{a^6}{(a-b)(a-c)} > 15.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider a cubic polynomial whose roots are $a$, $b$, $c$. We have $P(x) = x^3 - px^2 + qx - r$, where $p = a + b + c$, $q = ab + bc + ca$, and $r = abc$. We observe that\n\n$$\n\\sum_{\\text{cyc}} \\frac{a^n}{(a-b)(a-c)} = S_n,\n$$\nwhere\n$$\nS_n = \\sum_{\\text{cyc}} \\frac{a^n (b-c)}{-(a-b)(b-c)(c-a)}.\n$$\nIt is easy to see that $S_1 = 0$ and $S_2 = 1$. Since $a$, $b$, $c$ are roots of $P(x) = 0$, we have $a^3 - pa^2 + qa - r = 0$, $b^3 - pb^2 + qb - r = 0$, $c^3 - pc^2 + qc - r = 0$. Multiply the first by $(b-c)$, the second by $(c-a)$, and the third by $(a-b)$, add all, and divide by $-(a-b)(b-c)(c-a)$ to obtain $S_3 - pS_2 + qS_1 = 0$. Hence $S_3 = p$.\n\nNow multiply the first by $a$, the second by $b$, and the third by $c$, and divide by $-(a-b)(b-c)(c-a)$ to get $S_4 - pS_3 + qS_2 - rS_1 = 0$. Hence $S_4 = p^2 - q$. Similarly, $S_5 = p(p^2 - q) - qp + r = p^3 - 2pq + r$. We also get $S_6 - pS_5 + qS_4 - rS_3 = 0$, which gives\n\n$$\nS_6 = p(p^3 - 2pq + r) - q(p^2 - q) + rp = p^4 - 3p^2q + 2pr + q^2.\n$$\n\nWe can write $S_6 = p^2(p^2 - 3q) + 2pr + q^2$. But $p^2 - 3q = (a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca > 0$. Hence\n\n$$\nS_6 > 2pr + q^2 = 2abc(a + b + c) + (ab + bc + ca)^2 \\geq 6 + 9 = 15.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18051,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a scalene triangle $ABC$. On the rays $AC \\rightarrow$ and $BC \\rightarrow$, the points $C_a$ and $C_b$ are chosen, respectively, such that $AC_a = BC_b = AB$. We denote by $O_c$ the center of the circumcircle about $\\triangle CC_a C_b$. Analogously, we define the points $O_a$ and $O_b$. Prove that the lines $AO_a$, $BO_b$, and $CO_c$ intersect at a point that lies on the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ and $I$ be the centers of the circumscribed and inscribed circles of $ABC$, and $A_1$, $B_1$, $C_1$ be the centers of the arcs $\\widehat{AB}$, $\\widehat{AC}$, $\\widehat{BC}$ (not containing the third vertices) of the circumscribed circle. By symmetry with respect to $AA_1$ we have $A_1C_a = A_1B = A_1C$, and together with $O_cC = O_cC_a$ it follows that $A_1O_c$ is the segment bisector of $CC_a$. Analogously, $B_1O_c$ is the segment bisector of $CC_b$. Thus, the quadrilateral $OA_1O_cB_1$ is a parallelogram with $OA_1 = OB_1$, i.e., a rhombus. In other words, $O$ and $O_c$ are symmetric about $A_1B_1$, and this is also true for $C$ and $I$, i.e., $CO_c$ is symmetric to $OI$ with respect to $A_1B_1$. Analogously, $AO_a$ and $BO_b$ are symmetric to $OI$ with respect to $B_1C_1$ and $A_1C_1$, respectively. Since $I$ is the orthocenter of $A_1B_1C_1$, finally the lines $AO_a$, $BO_b$, and $CO_c$ intersect at the Anti-Steiner point for the line $OI$ and triangle $A_1B_1C_1$ (circumscribed about $ABC$). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18052,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека равенката $x^n - a_1 x^{n-1} - a_2 x^{n-2} - \\dots - a_{n-1} x - a_n = 0$, каде што $a_k \\ge 0$, $1 \\le k \\le n$, нема две различни позитивни решенија.",
    "options": [],
    "answer": "See solution",
    "solution": "Равенката за $x \\neq 0$ е еквивалентна со равенката\n$$\n1 = \\frac{a_1}{x} + \\frac{a_2}{x^2} + \\dots + \\frac{a_n}{x^n}\n$$\nЌе воведеме ознака $f(x) = \\frac{a_1}{x} + \\frac{a_2}{x^2} + \\dots + \\frac{a_n}{x^n}$ за $x \\in (0, +\\infty)$.\n\nНе е тешко да се види дека за $0 < x_1 < x_2$, $f(x_1) > f(x_2)$. Според тоа, не може да постојат две различни вредности за $x$ за кои $f(x) = 1$. Значи, равенката $1 = \\frac{a_1}{x} + \\frac{a_2}{x^2} + \\dots + \\frac{a_n}{x^n}$ нема две различни позитивни решенија. Според тоа, и почетната равенка нема две позитивни различни решенија.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18053,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest natural number $n$ for which it is possible to choose $n$ vertices of a cube such that no three of them form a right triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let some vertex of a cube be $A$ and let $B$, $C$, and $D$ be the opposite vertices of the faces that $A$ belongs to (see the figure below). Then, of the vertices $B$, $C$, and $D$, any two are also the opposite vertices of some face of the cube. Therefore, any two of the chosen four vertices are at the distance of a face diagonal of the cube. Thus, any three form an equilateral, rather than a right triangle.\n\nNow, consider the situation where we choose at least 5 vertices. Two opposite faces of the cube include all the vertices of the cube. Therefore, at least one of the two opposite faces must include at least 3 of the chosen vertices. But three vertices of a square form a right triangle.\n\n![](images/prob1314_p14_data_eb597a1106.png)\n\nFigure 9",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18054,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute triangle $ABC$, with $AB \\ne BC$, let $T$ denote the midpoint of the side $AC$, and let $A_1$ 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 at $A$ and $C$ to the circumcircle of triangle $ABC$. Let $X$ 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\n**a)** Prove that $T$ is the incenter of triangle $XYZ$.\n\n**b)** 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\n**c)** Prove that the point $D$ lies on the circumcircle of triangle $XYZ$.",
    "options": [],
    "answer": "See 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 $AT = C_1T$, which means that $X$ and $T$ are on the perpendicular bisector of the segment $AC_1$. It follows that $XT$ is the angle bisector of $\\angle YXZ$.\n\n![](images/RMC2014_p81_data_5f3d56b219.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 $BH$. The radical axis of two circles is perpendicular to the line connecting the centers of the circles, so $BD \\perp LO$. 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\nc) The quadrilateral $ATDX$ is cyclic because\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 $A_1C$. From\n\n$$\n\\angle CDT = \\angle HDB - \\angle CDB = 90^\\circ - \\angle CAB = 90^\\circ - \\angle BCY = \\angle CYT,\n$$\n\nit follows that the quadrilateral $CTDY$ is cyclic. This leads to\n\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": 18055,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k(I, r)$ be the incircle of $\\triangle ABC$. Let $D$, $E$, and $F$ denote the points where $k$ touches $BC$, $AC$, and $AB$, respectively. Let $P$, $Q$, and $R$ denote the midpoints of $EF$, $DF$, and $DE$, respectively.\n\nShow that the radius of the circumcircle of $\\triangle A'B'C'$, where $A' = P$, $B' = Q$, and $C' = R$, is half the radius of $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that $k_a$ passes through $Q$ and $R$.\n\nSince $\\triangle IQD \\sim \\triangle IDB$ and $\\triangle IRD \\sim \\triangle IDC$, we obtain $IQ \\cdot IB = IR \\cdot IC = r^2$. We conclude that $B$, $C$, $Q$, and $R$ lie on a single circle $\\gamma_a$. Moreover, since the power of $I$ with respect to $\\gamma_a$ is $r^2$, it follows for a tangent $IX$ from $I$ to $\\gamma_a$ that $X$ lies on $k$ and hence $k$ is perpendicular to $\\gamma_a$. From the uniqueness of $k_a$ it follows that $k_a = \\gamma_a$. Thus $k_a$ contains $Q$ and $R$. Similarly, $k_b$ contains $P$ and $R$, and $k_c$ contains $P$ and $Q$. Hence, $A' = P$, $B' = Q$, and $C' = R$. Therefore, the radius of the circumcircle of $\\triangle A'B'C'$ is half the radius of $k$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18056,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer.\n\nAriane and Bérénice play a game on the set of residue classes modulo $n$. In the beginning, the residue class $1$ is written on a piece of paper. In each move, the player whose turn it is replaces the current residue class $x$ with either $x + 1$ or $2x$. The two players alternate, with Ariane starting.\n\nAriane wins if the residue class $0$ is reached during the game. Bérénice wins if she can permanently avoid this outcome.\n\nFor each value of $n$, determine which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Ariane wins for $n = 2$, $4$, and $8$; for all other $n \\ge 2$, Bérénice wins.\n\nWe observe: If Ariane can win for a certain $n$, she will also win for all divisors of $n$, and conversely, if Bérénice can win for a certain $n$, she will also win for all multiples of $n$ because a residue $0$ modulo $n$ is automatically a residue $0$ for all divisors of $n$.\n\nIt remains to show that Ariane wins for $n = 8$ and Bérénice wins for $n = 16$ and $n$ odd.\n\nAll congruences in this solution are modulo $n$.\n\n- For $n = 8$, Ariane has to choose $2$ in the first step. If Bérénice takes $4$, Ariane can choose $8 \\equiv 0$ and has won. If Bérénice takes $3$, Ariane can choose $6$. Now, Bérénice has to decide between $7$ and $2 \\cdot 6 = 12 \\equiv 4$. But for both, Ariane can immediately choose $8 \\equiv 0$.\n\n- For $n = 16$, Bérénice chooses $2x$ for all numbers except $4$ and $8$. This never gives the residue classes $0$, $15$, or $8$, so Ariane also cannot choose $0$.\n\n- For $n = 3$, Ariane has to choose $2$ in the first step and then Bérénice chooses $1$ again, which means that Bérénice wins.\n\n- For odd $n > 3$, it is not possible to reach $0$ with $2x$ from another residue class. So the only possible issue for Bérénice would be the situation that both her options are among $n$ and $n-1$ such that she or Ariane choose $0$. But this means that $x+1$ takes the residues $0$ or $-1$, so $2x$ takes the residues $-2$ or $-4$ which are both different from $0$ and $-1$, so this cannot happen and Bérénice can permanently avoid $0$ being chosen.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18057,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the inscribed circle $I$ of $\\triangle ABC$ touch $BC$ and $AB$ at $D$ and $F$, respectively. Let $I$ intersect the segments $AD$ and $CF$ at $H$ and $K$, respectively. Prove that\n\n$$\n\\frac{FD \\times HK}{FH \\times DK} = 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the lengths of segments are $AF = x$, $BF = y$, $CD = z$. By Stewart's Theorem,\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p278_data_5af4a54cad.png)\n\n$$\n\\begin{aligned}\nAD^2 &= \\frac{BD}{BC} \\cdot AC^2 + \\frac{CD}{BC} \\cdot AB^2 - BD \\cdot DC \\\\\n&= \\frac{y(x+z)^2 + z(x+y)^2}{y+z} - yz \\\\\n&= x^2 + \\frac{4xyz}{y+z}.\n\\end{aligned}\n$$\n\nBy the Tangent-Secant Theorem, $AH = \\frac{AF^2}{AD} = \\frac{x^2}{AD}$, so\n\n$$\nHD = AD - AH = \\frac{AD^2 - x^2}{AD} = \\frac{4xyz}{AD(y+z)}.\n$$\n\nSimilarly,\n\n$$\nKF = \\frac{4xyz}{CF(x+y)}.\n$$\n\nSince $\\triangle CDK \\sim \\triangle CFD$, we have\n\n$$\nDK = \\frac{DF \\times CD}{CF} = \\frac{DF}{CF}z.\n$$\n\nFrom $\\triangle AFH \\sim \\triangle ADF$,\n\n$$\nFH = \\frac{DF \\times AF}{AD} = \\frac{DF}{AD}x.\n$$\n\nBy the Cosine Law,\n\n$$\n\\begin{aligned}\nDF^2 &= BD^2 + BF^2 - 2BD \\cdot BF \\cos B \\\\\n&= 2y^2 \\left( 1 - \\frac{(y+z)^2 + (x+y)^2 - (x+z)^2}{2(x+y)(y+z)} \\right) \\\\\n&= \\frac{4xy^2z}{(x+y)(y+z)}.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n\\frac{KF \\times HD}{FH \\times DK} &= \\frac{\\frac{4xyz}{CF(x+y)} \\cdot \\frac{4xyz}{AD(y+z)}}{\\frac{DF}{AD}x \\cdot \\frac{DF}{CF}z} \\\\\n&= \\frac{16xy^2z}{DF^2(x+y)(y+z)} = 4.\n\\end{aligned}\n$$\n\nApplying Ptolemy's Theorem to cyclic quadrilateral $DKHF$,\n\n$$\nKF \\cdot HD = DF \\cdot HK + FH \\cdot DK.\n$$\n\nCombining with $\\frac{KF \\times HD}{FH \\times DK} = 4$, we obtain $\\frac{FD \\times HK}{FH \\times DK} = 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18058,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, \\dots, x_n$ be non-negative real numbers, not all of which are zero.\n\n(i) Prove that\n\n$$\n1 \\le \\frac{\\left(x_1 + \\frac{x_2}{2} + \\frac{x_3}{3} + \\dots + \\frac{x_n}{n}\\right) \\cdot \\left(x_1 + 2x_2 + 3x_3 + \\dots + nx_n\\right)}{\\left(x_1 + x_2 + x_3 + \\dots + x_n\\right)^2} \\le \\frac{(n+1)^2}{4n}.\n$$\n\n(ii) Show that, for each $n \\ge 1$, both inequalities can hold as equalities.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying AM-GM gives\n\n$$\n\\begin{aligned}\n\\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right)\n&= \\frac{1}{n} \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\\\\n&\\le \\frac{1}{n} \\cdot \\frac{1}{4} \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k} + \\sum_{k=1}^{n} kx_k\\right)^2 \\\\\n&= \\frac{1}{4n} \\left(\\sum_{k=1}^{n} x_k \\left(\\frac{n}{k} + k\\right)\\right)^2 \\\\\n&\\le \\frac{(n+1)^2}{4n} \\left(\\sum_{k=1}^{n} x_k\\right)^2.\n\\end{aligned}\n$$\n\n(The last inequality is proved by $\\frac{n}{k} + k \\le n + 1$, as it is equivalent to $(n-k)(k-1) \\ge 0$.)\n\nThis gives us the necessary upper bound; this bound is achieved for instance if $x_1 = x_n = 1$ and $x_2 = \\dots = x_{n-1} = 0$.\n\nFor the lower bound, estimate the numerator by Cauchy-Schwarz inequality:\n\n$$\n\\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\ge \\left(\\sum_{k=1}^{n} \\sqrt{\\frac{x_k}{k}} \\cdot \\sqrt{kx_k}\\right)^2 = \\left(\\sum_{k=1}^{n} x_k\\right)^2;\n$$\n\nthe equality holds here if exactly one of $x_i$ is non-zero.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18059,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, and $C$ lie on a circle with centre $M$. The reflection of point $M$ in the line $AB$ lies inside triangle $ABC$ and is the intersection of the angular bisectors of angles $A$ and $B$. (The angular bisector of an angle is the line that divides the angle into two equal angles.) Line $AM$ intersects the circle again at point $D$.\n\nShow that $|CA| \\cdot |CD| = |AB| \\cdot |AM|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the reflection of point $M$ in the line $AB$. Define $\\alpha = \\angle CAI$ and $\\beta = \\angle CBI$. Since $AI$ is the angular bisector of $\\angle CAB$, $\\angle IAB = \\alpha$. Since $I$ is the reflection of $M$ in the line $AB$, $\\angle BAM = \\alpha$. Triangle $AMC$ is isosceles with apex $M$, because $|AM| = |CM|$. Thus, $\\angle MCA = \\angle CAM = 3\\alpha$. Similarly, $\\angle IBA = \\angle ABM = \\beta$ and $\\angle MCB = 3\\beta$. The sum of the angles of triangle $ABC$ is $2\\alpha + (3\\alpha + 3\\beta) + 2\\beta = 180^{\\circ}$. Therefore, $\\alpha + \\beta = \\frac{180^{\\circ}}{5} = 36^{\\circ}$, and $\\angle ACB = 3\\alpha + 3\\beta = 3 \\cdot 36^{\\circ} = 108^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18060,
    "subject": "Mathematics (Olympiad)",
    "question": "The product of three integers is $60$. What is the least possible positive sum of the three integers?\n\n(A) 2  \n(B) 3  \n(C) 5  \n(D) 6  \n(E) 13",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $60 = 10 \\cdot (-1) \\cdot (-6)$, and the sum of these factors is $3$. It remains to show that no positive sum can be less than $3$. Such a sum would have to consist of one positive integer and two negative integers with smaller absolute value. If the positive integer is greater than or equal to $10$, then the sum is greater than or equal to $3$. The possible sets of factors in this case are $\\{10, -6, -1\\}$, $\\{10, -3, -2\\}$, $\\{12, -5, -1\\}$, $\\{15, -4, -1\\}$, $\\{15, -2, -2\\}$, $\\{20, -3, -1\\}$, $\\{30, -2, -1\\}$, and $\\{60, -1, -1\\}$. None of these sets of factors has a sum less than $3$.\n\nThe only other possible choices for the positive integer are $5$ and $6$, and in neither case is a positive sum possible. Indeed, if the positive integer is $5$, then the only possible set of factors is $\\{5, -3, -4\\}$. If the positive integer is $6$, then the only possible set of factors is $\\{6, -5, -2\\}$. In both of these cases, the sum is not positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18061,
    "subject": "Mathematics (Olympiad)",
    "question": "If you write the date 16 January 1091 with 8 digital digits in a row, it looks like this:\n\n![](images/NLD_ABooklet_2025_p39_data_622c134ba7.png)\n\nWhen you read this upside down, it reads as the exact same date. What is the first date in the future (22 June 2024 or later) for which it is also true that, written in 8 digital digits consecutively, the date is exactly the same when read upside down?",
    "options": [],
    "answer": "See solution",
    "solution": "5 February 2050 (05-02-2050)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18062,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We may assume without loss of generality that $a_1 \\le a_2 \\le \\dots \\le a_{2n}$.\n\n- If $a_n + a_{2n-1} \\ge 0$, then all the sums $a_i + a_{2n-1}$ with $i = n, \\dots, 2n-2$ as well as all the sums $a_i + a_{2n}$ with $i = n, \\dots, 2n-1$ 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)\n\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 $i = 1, \\dots, 2n-1$ are non-negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18063,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $B_1$ and $C_1$ are chosen on the bisector of angle $BAC$ of triangle $ABC$ so that $BB_1 \\perp AB$ and $CC_1 \\perp AC$. Let $M$ be the midpoint of $B_1C_1$. Prove that $MB = MC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the perpendicular dropped from $C_1$ onto $AB$ intersects it at some point $C_b$. Define $B_c$ similarly so that $BB_c \\perp AC$. Let $M_b$ and $M_c$ be the projections of $M$ onto lines $AB$ and $AC$, respectively. Without loss of generality, suppose the points are located as in the figure below.\n\n$$\n\\triangle AB_1B_c \\sim \\triangle AMM_c \\sim \\triangle AC_bC \\text{ and} \\\\\n\\triangle AC_1C_b \\sim \\triangle AMM_b \\sim \\triangle AB_cB.\n$$\n\nFrom basic properties of a trapezoid, we have $B_cM_c = M_cC$ and $C_bM_b = M_bB$. Hence $B_cM = MC$ and $C_bM = MB$. Since $\\angle MB_1B_c = \\angle BB_1M$ and $BB_1 = B_1B_c$, this implies $\\triangle MBB_1 = \\triangle MB_1B_c$, so $BM = MB_c$, and $BM = MB_c = CM$.\n\n![](images/ukraine_2015_Booklet_p6_data_8600bf0f01.png)\n\nFig. 14",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18064,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a non-invertible square matrix of order $n$ with real entries, $n \\geq 2$, and let $A^*$ be the adjoint of $A$. Prove that $\\operatorname{tr}(A^*) \\neq -1$ if and only if the matrix $I_n + A^*$ is invertible.",
    "options": [],
    "answer": "See solution",
    "solution": "As $A$ is non-invertible, we have $\\operatorname{rank}(A) \\leq n - 1$. Consider two cases:\n\n1. $\\operatorname{rank}(A) \\leq n - 2$. Then $A^* = O_n$ and the conclusion follows immediately.\n\n2. $\\operatorname{rank}(A) = n - 1$. Then $A A^* = O_n$ and by Sylvester's inequality $0 \\geq \\operatorname{rank}(A) + \\operatorname{rank}(A^*) - n$, so $\\operatorname{rank}(A^*) \\leq 1$.\n\nIt follows that $A^* = C L$ where $L \\in \\mathcal{M}_{n,1}(\\mathbb{R})$ and $C \\in \\mathcal{M}_{1,n}(\\mathbb{R})$. Then $L C = (a) \\in \\mathcal{M}_1(\\mathbb{R})$, with $a = \\operatorname{tr}(A^*)$, and $(A^*)^2 = C L C L = C(a) L = a A^*$. Let $B = I_n + A^*$. Then $(B - I_n)^2 = a(B - I_n)$, or equivalently, $B((a+2)I_n - B) = (a+1)I_n$, which implies $a \\neq -1$, i.e., $B$ 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 $-1 = \\operatorname{tr}(A^*) = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 18065,
    "subject": "Mathematics (Olympiad)",
    "question": "In Wonderland, the government of each country consists of exactly $a$ men and $b$ women, where $a$ and $b$ are fixed natural numbers and $b > 1$. To improve relationships between countries, all possible working groups are formed, each consisting of exactly one government member from each country, with at least $n$ among them being women (where $n$ is a fixed non-negative integer). The same person may belong to many working groups. Find all possible numbers of countries that can be in Wonderland, given that the number of all working groups is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the number of countries in Wonderland.\n\nIf the minimal number of women in working groups is $n = 0$, then forming a working group means just choosing one government member from each country. Thus, there are $(a + b)^r$ different working groups. This number can be prime only if $r = 1$ because $a + b \\geq b > 1$.\n\nIf the minimal number of women in working groups is $n \\geq 1$, then a working group containing exactly $k$ women ($n \\leq k \\leq r$) can be formed as follows: choose $k$ countries out of $r$ to send a woman to the group, then choose one woman out of $b$ from each of the $k$ governments, and finally choose one man out of $a$ from each of the remaining $r - k$ countries. Hence, there are $\\binom{r}{k} b^k a^{r - k}$ working groups with exactly $k$ women, and $$\\sum_{k = n}^{r} \\binom{r}{k} b^k a^{r - k}$$ working groups with at least $n$ women altogether. As $n \\geq 1$, all terms of this sum are divisible by $b$, so the sum can be a prime only if it is equal to $b$. This is possible only if $r = 1$, since otherwise the last term (corresponding to $k = r$) of the sum would be greater than $b$.\n\nThe value $r = 1$ is indeed possible: for instance, if each government consists of just 2 women, then the number of all working groups is 2, which is a prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18066,
    "subject": "Mathematics (Olympiad)",
    "question": "A square and an equilateral triangle are inscribed in a circle. The seven vertices form a convex heptagon $H$ that is inscribed in the circle. (As a special case, $H$ can be a hexagon if a vertex of the square coincides with a vertex of the triangle.)\n\nFor which positions of the triangle relative to the square does $H$ have the largest and smallest possible areas?",
    "options": [],
    "answer": "See solution",
    "solution": "The square divides the circle into four arcs. None of these arcs can contain more than one triangle vertex, because the distance between two triangle vertices corresponds to an inner angle of $120^\\circ$, while the distance between two square vertices is only $90^\\circ$. Therefore, the triangle vertices must be on three distinct arcs.\n\nLet $ABCD$ denote the square and $PQR$ denote the triangle. Without loss of generality, assume $P$ is on the arc between $A$ and $B$, $Q$ is between $B$ and $C$, and $R$ is between $D$ and $A$, as shown below:\n\n![](images/AustriaMO2013_p10_data_0e830fd903.png)\n\nThe heptagon consists of the square $ABCD$ and the three triangles $APB$, $BQC$, and $DRA$. Since the area of $ABCD$ is constant, it suffices to maximize or minimize the sum of the areas of these three triangles.\n\nLet $h_1$ be the distance from $P$ to line $AB$ (the height of $\\triangle APB$), $h_2$ the distance from $Q$ to $BC$, and $h_3$ the distance from $R$ to $DA$. The total area is $\\frac{|AB|}{2} h_1 + \\frac{|BC|}{2} h_2 + \\frac{|DA|}{2} h_3 = \\frac{s}{2}(h_1 + h_2 + h_3)$, where $s$ is the side length of the square. Since $s$ is constant, it suffices to maximize or minimize $h_1 + h_2 + h_3$.\n\nWe maximize and minimize $h_1$ and $h_2 + h_3$ separately.\n\n$h_1$ is largest when $P$ is exactly in the middle of the arc between $A$ and $B$.\n\nTo maximize $h_2 + h_3$, consider the rectangle $QXRY$ with sides parallel to those of $ABCD$ and $QR$ as a diagonal. By the Pythagorean theorem, $|QR|^2 = |QX|^2 + |XR|^2 = (h_1 + s + h_2)^2 + |XR|^2$, so $(h_2 + s + h_3)^2 = |QR|^2 - |XR|^2$. Since $|QR|$ is constant, $h_2 + h_3$ is maximized when $|XR| = 0$, i.e., when $QR$ is parallel to $CD$, which occurs when $P$ is in the middle of the arc between $A$ and $B$.\n\nThus, both $h_1$ and $h_2 + h_3$ are maximized in the same case, so $h_1 + h_2 + h_3$ is largest when $P$ is in the middle of the arc between $A$ and $B$.\n\nFor the minimum, $h_1$ is smallest when $P$ is close to $A$ or $B$. Since $Q$ must remain between $B$ and $C$, and $R$ between $D$ and $A$, the minimum is reached if either $Q = C$ or $R = D$.\n\nSimilarly, $h_2 + h_3$ is minimized when $|XR|$ is largest, so again the minimum is reached if $Q = C$ or $R = D$.\n\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18067,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為一正奇數。平面上的整點集 $C = \\{(i,j): i,j = 0,1,\\dots,2n-1\\}$ 構成一個 $2n \\times 2n$ 的陣列，每個點上各有一隻天竺鼠，各自面向 $x$ 軸正向、$x$ 軸負向、$y$ 軸正向或 $y$ 軸負向中的其中一個方向。傑夫想要保留其中 $n^2+1$ 隻天竺鼠，並將其餘天竺鼠移除。接著天竺鼠們作以下運動：在每一回合，被保留的每一隻天竺鼠同時往牠所面向的方向前進一單位長，並保持其面向；但如果一隻天竺鼠要前進的點 $(i,j) \\notin C$，則牠改為前進到 $(p,q) \\in C$，其中 $p \\equiv i \\pmod{2n}$ 而 $q \\equiv j \\pmod{2n}$（舉例來說，如果一隻天竺鼠從 $(2,0)$ 爬向 $(2,-1)$，則牠改為爬到 $(2,2n-1)$）。\n\n傑夫的目標是讓所有留下來的天竺鼠，在任何一回合都不會有兩隻天竺鼠有相同的終點，也不會有兩隻天竺鼠互相爬到對方該回合的起點。\n\n試證：不論起始天竺鼠的面向如何分布，傑夫總是可以達成目標。",
    "options": [],
    "answer": "See solution",
    "solution": "先將天竺鼠的面向以箭頭表示，並用西洋棋盤的方式黑（$B$）白（$W$）塗色。這樣我們便有八種箭頭：$B \\to$, $B \\leftarrow$, $B \\uparrow$, $B \\downarrow$, $W \\to$, $W \\leftarrow$, $W \\uparrow$, $W \\downarrow$。注意到如果兩隻天竺鼠一黑一白，且處於相同或是垂直的方向，則牠們永遠不會停在同一個點上。從而我們得到十二種不會對撞的天竺鼠組合：\n\n$$\n\\begin{align*}\n& (B \\uparrow, W \\uparrow), (B \\downarrow, W \\downarrow), (B \\leftarrow, W \\leftarrow), (B \\to, W \\to), \\\\\n& (B \\uparrow, W \\to), (B \\downarrow, W \\to), (B \\leftarrow, W \\uparrow), (B \\to, W \\uparrow), \\\\\n& (B \\uparrow, W \\leftarrow), (B \\downarrow, W \\leftarrow), (B \\leftarrow, W \\downarrow), (B \\to, W \\downarrow)\n\\end{align*}\n$$\n\n由於每個箭頭在以上十二種組合中都會被計算三次，因此這時十二種不相撞對中，必然有一種組合的對數不少於 $\\frac{3(2n \\times 2n)}{12} = n^2$。\n\n事實上，必然有一種不相撞對組合有多於 $n^2$ 對。這是基於，若否，則每一種組合都恰為 $n^2$ 對。這表示 $B \\uparrow$, $B \\downarrow$, $B \\leftarrow$ 和 $B \\to$ 的數量都恰相等。但這是不可能的，因為黑箭頭的總數量是 $\\frac{2n \\times 2n}{2} = 2n^2$，不被 4 整除。故知必有一種不相撞組合有至少 $n^2+1$ 對，從而原命題得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18068,
    "subject": "Mathematics (Olympiad)",
    "question": "Во множеството цели броеви да се реши равенката\n\n$$\nx^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Лема 1. Нека $x \\in \\mathbb{Z}$, тогаш секој прост делител на $x^2 + 1$ е од облик $4k + 1$.\n\n*Доказ:* Нека $p \\mid x^2 + 1$, и јасно $\\text{изд}(x, p) = 1$.\n\nТогаш\n\n$$\nx^2 + 1 \\equiv 0 \\pmod{p} \\text{ т.е. } x^2 \\equiv -1 \\pmod{p}.\n$$\n\nАко двете страни ги кренеме на степен $\\frac{p-1}{2}$, добиваме\n\n$$\n(x^2)^{\\frac{p-1}{2}} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}\n$$\n\nт.е.\n\n$$\nx^{p-1} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}.\n$$\n\nОд друга страна, од *Малата Теорема на Ферма* имаме $x^{p-1} \\equiv 1 \\pmod{p}$ па значи\n\n$$\n\\frac{p-1}{2} = 2k, \\text{ од каде следува дека } p = 4k+1.\n$$\n\nСега дадената равенка е еквивалентна со равенката\n\n$$\nx^{2010} + 1 = 4y^{2009} + 2007y + 4y^{2008} + 2007\n$$\n\nт.е.\n\n$$\nx^{2010} + 1 = (4y^{2008} + 2007)(y + 1).\n$$\n\nБројот $4y^{2008} + 2007 = 4y^{2008} + 2008 - 1$ е од облик $4k-1$, па мора да има прост делител од облик $4k-1$, бидејќи ако сите делители се од облик $4k+1$ тогаш и тој самиот е од таков облик.\n\nЗначи $(x^{1005})^2 + 1$ има делител од облик $4k-1$, што не е можно поради\n\n*Лема 1.*\n\nЗначи дадената равенка нема решение во множеството цели броеви.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18069,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $d_1 > d_2 > \\dots > d_{99}$ are integers whose sum is equal to zero. Find the smallest possible value of $d_1 + d_2 + \\dots + d_{18}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1:** $d_{18} \\leq 31$\n\nFrom the strict inequality, $d_{19} \\leq 30$, $d_{20} \\leq 29$, ..., $d_{99} \\leq -50$. So,\n\n$$\na_{19} = \\sum_{i=1}^{18} d_i = - \\sum_{i=19}^{99} d_i \\geq - \\sum_{i=30}^{-50} i = - \\frac{(30 + (-50)) \\times 81}{2} = 810.\n$$\n\n**Case 2:** $d_{18} \\geq 32$\n\nLet $d_{18} = 32 + e$ where $e \\geq 0$. Then $d_{17} \\geq 33 + e$, $d_{16} \\geq 34 + e$, ..., $d_1 \\geq 49 + e$. Therefore,\n\n$$\na_{19} = \\sum_{i=1}^{18} d_i \\geq \\sum_{i=32}^{49} i + 18e = \\frac{(32 + 49) \\times 18}{2} + 18e = 729 + 18e \\geq 729.\n$$\n\nThus, in all cases $a_{19} \\geq 729$.\n\nFinally, 729 is attainable when $d_i = 50 - i$ for $i = 1, 2, \\dots, 99$.\n\n$\\boxed{729}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18070,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\n\\frac{x_1}{x_1+1} = \\frac{x_2}{x_2+3} = \\frac{x_3}{x_3+5} = \\dots = \\frac{x_{1006}}{x_{1006}+2011} = a.\n$$\n\nFind the value of $x_{1006}$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $\\frac{x_k}{x_k + (2k-1)} = a$, it follows that\n$$\nx_k = \\frac{a}{1-a} \\cdot (2k-1)\n$$\nfor $k = 1, 2, \\dots, 1006$.\n\nSubstituting into the last equality:\n$$\nx_{1006} = \\frac{a}{1-a} \\cdot 2011.\n$$\n\nTo find $a$, note that the sum $1 + 3 + 5 + \\dots + 2011$ is an arithmetic series with $1006$ terms:\n$$\n1 + 3 + 5 + \\dots + 2011 = 1006^2.\n$$\n\nGiven that $\\frac{a}{1-a} \\cdot 1006^2 = 503^2$, we solve for $\\frac{a}{1-a}$:\n$$\n\\frac{a}{1-a} = \\frac{503^2}{1006^2} = \\frac{1}{4}.\n$$\n\nTherefore,\n$$\nx_{1006} = \\frac{1}{4} \\cdot 2011 = \\frac{2011}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18071,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$, the number\n\n$$\nN = \\underbrace{44\\dots4}_{n} \\underbrace{88\\dots8}_{n} - \\underbrace{133\\dots32}_{n-1}\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\nN &= \\underbrace{44\\dots4}_{n-1} \\underbrace{355\\dots56}_{n-1} = \\underbrace{44\\dots4}_{2n} - \\underbrace{88\\dots8}_{n} \\\\\n&= 4(10^{2n-1} + \\dots + 10 + 1) - 8(10^{n-1} + \\dots + 10 + 1) \\\\\n&= 4 \\cdot \\frac{10^{2n}-1}{9} - 8 \\cdot \\frac{10^n-1}{9} = \\frac{1}{9}(4 \\cdot 10^{2n} - 8 \\cdot 10^n + 4) \\\\\n&= \\left(2 \\cdot \\frac{10^n-1}{3}\\right)^2,\n\\end{aligned}\n$$\n\nand we are done, since $\\frac{10^n - 1}{3}$ is an integer. Notice that, in fact,\n\n$$\nN = \\underbrace{66\\dots6}_{n}^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18072,
    "subject": "Mathematics (Olympiad)",
    "question": "As illustrated in the figure below, in $\\triangle ABC$, $AB < AC$. Let $PB$ and $PC$ be tangent to the circumcircle $O$ of $\\triangle ABC$. Let $R$ be a point on arc $AC$ such that $AR \\parallel BC$, and let $Q$ be the other intersection of $PR$ and circle $O$. Let $I$ be the incenter of $\\triangle ABC$, and let $ID \\perp BC$ with foot $D$. Let $G$ be the other intersection of $QD$ and circle $O$. Suppose that the line through $I$ and perpendicular to $AI$ intersects $AB$ and $AC$ at points $M$ and $N$, respectively. Prove that $A, G, M, N$ are concyclic.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p329_data_0af239abf3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Connect $GB$, $GC$, $GM$, $GN$, $RB$, $RC$, $QB$, $QC$, $BI$, and $CI$. See the figure below.\n\nSince $PB$, $PC$ are tangent to circle $O$,\n\n$$\n\\frac{RB}{BQ} = \\frac{PR}{PB} = \\frac{PR}{PC} = \\frac{RC}{CQ},\n$$\n\nand hence\n\n$$\n\\frac{BQ}{CQ} = \\frac{RB}{RC}.\n$$\n\nMoreover,\n\n$$\n\\frac{BD}{CD} = \\frac{S_{\\triangle BGQ}}{S_{\\triangle CGQ}} = \\frac{BG \\cdot BQ}{CG \\cdot CQ} = \\frac{BG \\cdot RB}{CG \\cdot RC}.\n$$\n\nFrom $AR \\parallel BC$, we have $RB = AC$, $RC = AB$. Together, they imply that\n\n$$\n\\frac{BD}{CD} = \\frac{BG \\cdot AC}{CG \\cdot AB},\n$$\n\nand\n\n$$\n\\frac{BG}{CG} = \\frac{BD \\cdot AB}{CD \\cdot AC}\n$$\n\n$$\n= \\frac{BI \\cdot \\cos \\frac{B}{2} \\sin C}{CI \\cdot \\cos \\frac{C}{2} \\sin B} = \\frac{BI \\cdot \\sin \\frac{C}{2}}{CI \\cdot \\sin \\frac{B}{2}} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{1}\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p330_data_d6a21bb3d4.png)\n\nAs $AI \\perp MN$, it follows that\n\n$$\n\\begin{align*}\n\\angle AMN &= \\angle ANM = \\frac{B+C}{2}, \\\\\n\\angle BMI = \\angle INC = 180^\\circ - \\frac{B+C}{2}, \\\\\n\\angle MBI &= \\frac{B}{2} = \\frac{B+C}{2} - \\frac{C}{2} = \\angle ANI - \\angle NCI = \\angle NIC,\n\\end{align*}\n$$\n\nand $\\triangle MBI \\sim \\triangle NIC$. Hence,\n\n$$\n\\frac{BM}{CN} = \\frac{BM}{IN} \\cdot \\frac{IM}{CN} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{2}\n$$\n\nBy $\\textcircled{1}$ and $\\textcircled{2}$, we find $\\frac{BG}{CG} = \\frac{BM}{CN}$; in addition, $\\angle GBM = \\angle GBA = \\angle GCA = \\angle GCN$. Therefore, $\\triangle GBM \\sim \\triangle GCN$, giving\n\n$$\n\\angle GMA = 180^\\circ - \\angle GMB = 180^\\circ - \\angle GNC = \\angle GNA.\n$$\n\nIt follows that $A, G, M, N$ lie on a circle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18073,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of prime numbers $(a, b)$ such that $a^b = b^a + 1$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, either $a$ or $b$ is even. Without loss of generality, let $a = 2$. It is easy to see that $b = 2$ and $b = 3$ satisfy the condition. Suppose that $b > 3$. We have: $2^b = b^2 + 1$, $b > 3$. It is easy to see that the last expression is divisible by $3$ and cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18074,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has two classes A and B with $m$ and $n$ students, respectively. The students from both classes sit in a circle. Each student receives a number of candies equal to the number of consecutive students sitting to their left who are from the same class.\n\nAfter distributing the candies, students form groups such that all students in a group received the same number of candies, and any two students from different groups received different numbers of candies.\n\n**a)** What is the maximum number of students that a group can have?\n\n**b)** Excluding the group where every student receives no candies, what is the maximum number of students that a group can have?",
    "options": [],
    "answer": "See solution",
    "solution": "Arrange the $m + n$ students in a circle, forming arcs where each arc consists of consecutive students from the same class, and use as many arcs as possible. Two adjacent arcs must be from different classes. The number of students who receive $i$ candies equals the number of arcs with at least $i + 1$ students. If a student $X$ receives $i$ candies, then to the left of $X$ there are exactly $i$ classmates, so the arc contains at least $i + 1$ students. Conversely, each arc with at least $i + 1$ students provides $i$ candies to $i + 1$ students.\n\n**a)** The number of students with $i$ candies equals the number of arcs with at least $i + 1$ students. The group with the most students is the group with $0$ candies, which equals the total number of arcs. The maximum number of arcs is:\n\n$$\n2 \\min(m, n).\n$$\n\nSuppose $m \\leq n$. The maximum number of arcs of class A is $m$, and the number of arcs of class B is also $m$, since arcs alternate. Thus, the total number of arcs is at most $2m$. This maximum is achieved by arranging $m$ arcs of 1 student from class A and $m$ arcs from class B.\n\n**b)** The group with the next largest size is the group of students who receive $1$ candy, corresponding to arcs with at least $2$ students. Assume $m \\leq n$.\n\n- If $m < n$ or $m = n$ and both are even: The maximum number of such arcs is $m$.\n- If $m = 2k - 1$ (odd) and $m < n$: Divide class A into $k - 1$ arcs of 2 students and 1 arc of 1 student; class B can be divided into $k$ arcs of at least 2 students. The maximum number of arcs with at least 2 students is $m$.\n- If $m = n = 2k + 1$ (both odd): The maximum number of arcs with at least 2 students is $m - 1$.\n\n**Conclusion:**\n- The answer is $m - 1$ when $m = n$ is odd.\n- In all other cases, the answer is $\\min(m, n)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18075,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer, and let $a_2, a_3, \\dots, a_n$ be positive real numbers such that $a_2 a_3 \\cdots a_n = 1$. Prove that\n\n$$\n(1 + a_2)^2 (1 + a_3)^3 \\cdots (1 + a_n)^n > n^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the AM-GM inequality, for every $k$ with $2 \\le k \\le n$, we have\n\n$$\n(1 + a_k)^k = \\left( \\frac{1}{k-1} + \\frac{1}{k-1} + \\dots + \\frac{1}{k-1} + a_k \\right)^k \\ge \\frac{k^k a_k}{(k-1)^{k-1}},\n$$\n\nwhere equality holds if and only if $a_k = \\frac{1}{k-1}$. Multiplying these inequalities for each $k$ between $2$ and $n$ yields\n\n$$\n(1 + a_2)^2 (1 + a_3)^3 \\cdots (1 + a_n)^n \\ge 2^2 a_2 \\cdot \\frac{3^3 a_3}{2^2} \\cdots \\frac{n^n a_n}{(n-1)^{n-1}} = n^n a_2 a_3 \\cdots a_n = n^n.\n$$\n\nEquality holds only if $a_k = \\frac{1}{k-1}$ for each $k$, implying that $a_2 a_3 \\cdots a_n = \\frac{1}{(n-1)!}$, an impossibility for $n \\ge 3$. Thus, the inequality is strict, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18076,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the value of $a$ for a given $b$, where $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}}$ and $\\min_{x \\in \\mathbb{R}} f(x) = 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}}$. Clearly, $a > 0$. Since $f(0) = b$, we require $b \\ge 3$.\n\n**Case 1:** $b - 2a \\ge 0$\n\n$$\nf(x) = a \\sqrt{x^2 + 1} + \\frac{b - a}{\\sqrt{x^2 + 1}} \\ge 2 \\sqrt{a(b - a)} \\ge 3.\n$$\nEquality holds when $a \\sqrt{x^2 + 1} = \\frac{b - a}{\\sqrt{x^2 + 1}}$, i.e., $x = \\pm \\sqrt{\\frac{b - 2a}{a}}$.\n\nThus, the value of $a$ for given $b$ is\n$$\na = \\frac{b - \\sqrt{b^2 - 9}}{2}.\n$$\nEspecially, when $b = 3$, $a = \\frac{3}{2}$.\n\n**Case 2:** $b - 2a < 0$\n\nLet $\\sqrt{x^2 + 1} = t$ with $t \\ge 1$. Then $f(x) = g(t) = a t + \\frac{b - a}{t}$, which is monotonically increasing for $t \\ge 1$. Thus,\n$$\n\\min_{x \\in \\mathbb{R}} f(x) = g(1) = a + b - a = b = 3, \\quad \\text{when } a > \\frac{3}{2}.\n$$\n\n**Summary:**\n- The range of $b$ is $[3, +\\infty)$.\n- If $b = 3$, then $a \\ge \\frac{3}{2}$.\n- If $b > 3$, then $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18077,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1$ be an integer. Find all positive integers $m \\ge 3$ such that the sequence\n\n$$\nx_{n+1} = x_n^{\\lfloor n\\sqrt{2} \\rfloor} + 1,\n$$\n\nis eventually periodic modulo $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that $m$ has no odd prime divisor. Assume the sequence is periodic with minimal period $T$. Since there are infinitely many $i$ such that $S(i)$ is divisible by $p-1$ if $p$ divides $x_i$, then $x_{i+2} \\equiv 2 \\pmod{p}$. If $p$ doesn't divide $x_i$, then $x_{i+1} \\equiv 2 \\pmod{p}$. Hence, $2$ is among the residues of $x_n$ modulo $p$. Take $a=2$ in the first lemma. It follows that the sequence $S(kT+i)$ is constant modulo the order of $2$ modulo $p$, namely $d$. Hence, $d$ divides $9$. That is, $p$ divides $2^9-1=7 \\times 73$.\n\nIf $p=7$, then for $i \\equiv 0 \\pmod{3}$ we have $x_{i+1} \\equiv 2 \\pmod{7}$ and $x_{i+2} \\equiv 3 \\pmod{7}$; since the order of $3$ mod $7$ is $6$, we reach a contradiction. The same holds for $i \\equiv 1 \\pmod{3}$. For $i \\equiv 2 \\pmod{3}$, $x_{i+1} \\equiv 5 \\pmod{7}$ and since the order of $5$ modulo $7$ is $6$, we again reach a contradiction.\n\nIf $p=73$, the numbers with orders dividing $9$ are\n\n$$\n1, 2, 4, 8, 16, 32, 64, 55, 37,\n$$\n\nwhich are powers of $2$. If $i$ is divisible by $9$, then $x_{i+1} \\equiv 2 \\pmod{73}$ and $x_{i+2} \\equiv 3 \\pmod{73}$, while the order of $3$ modulo $73$ doesn't divide $9$. If $i$ is not divisible by $9$, then $x_{i+1}$ is not congruent to a power of $2$ mod $73$, so its order modulo $73$ doesn't divide $9$.\n\nFor powers of $2$, we prove that the only solutions are $m=1,2,4$. Let $m=2^c$. We claim that the period is of the form $1,2,1,2,\\dots$. If $x_i$ is odd, then since the order of $x_i$ modulo $2^c$ is a power of two and for periodicity it must divide $9$, we obtain $x_i \\equiv 1 \\pmod{2^c}$. Hence, $x_{i+1} \\equiv 2 \\pmod{2^c}$ and by the same argument $x_{i+2k} \\equiv 2^{S(i-1+2k)}+1 \\equiv 1 \\pmod{2^c}$, yielding $S(i-1+2k) \\ge c$ for all $k$. Thus, $c \\le \\min S(i-1+2k) = 2$. Therefore, $m=1,2,4$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18078,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in non-negative integers the equation\n\n$$\n2^a 3^b + 9 = c^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will distinguish among three cases:\n\n1) For $a = 0$, we have $3^b = (c - 3)(c + 3)$. Thus, $c - 3 = 3^x$, $c + 3 = 3^y$, with $x + y = b$. Then $6 = 3^y - 3^x$, so $y = 2$, $x = 1$, yielding $a = 0$, $b = 3$, $c = 6$.\n\n2) For $b = 0$, we have $2^a = (c - 3)(c + 3)$. Thus, $c - 3 = 2^x$, $c + 3 = 2^y$, with $x + y = a$. Then $6 = 2^y - 2^x$, so $y = 3$, $x = 1$, yielding $a = 4$, $b = 0$, $c = 5$.\n\n3) For $a \\neq 0$ and $b \\neq 0$, $3 \\mid c^2$, so $9 \\mid c^2$, hence $b \\geq 2$ and $c = 3d$. Let $d = 2e + 1$ (odd). Now $2^a 3^{b-2} = 4e(e + 1)$, so $a \\geq 2$ and $2^{a-2}3^{b-2} = e(e+1)$.\n\n- If $e = 2^{a-2}$ and $e+1 = 3^{b-2}$, then $3^{b-2} = 2^{a-2} + 1$, which has solutions $a = 3$, $b = 3$, $c = 15$ and $a = 5$, $b = 4$, $c = 51$.\n- If $e = 3^{b-2}$ and $e+1 = 2^{a-2}$, then $2^{a-2} = 3^{b-2} + 1$, which has solutions $a = 3$, $b = 2$, $c = 9$ and $a = 4$, $b = 3$, $c = 21$.\n- If $e = 1$ and $e+1 = 2^{a-2}3^{b-2}$, then $2^{a-2}3^{b-2} = 2$, so $a = 3$, $b = 2$, $c = 9$ (already found).\n\nThus, all solutions in non-negative integers are:\n- $(a, b, c) = (0, 3, 6)$\n- $(a, b, c) = (4, 0, 5)$\n- $(a, b, c) = (3, 2, 9)$\n- $(a, b, c) = (3, 3, 15)$\n- $(a, b, c) = (4, 3, 21)$\n- $(a, b, c) = (5, 4, 51)$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18079,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ and $y$ such that:\n\n1. $x \\ge 2y^2$;\n2. $y \\ge 2x^2$;\n3. $8(x - y)$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "From (1) and (2), we have $x \\ge 0$ and $y \\ge 0$. Moreover, $x = 0$ if and only if $y = 0$, so $(0, 0)$ is a solution. For other solutions, $x > 0$ and $y > 0$.\n\nSuppose $x > 0$ and $y > 0$. From (1) and (2), $x \\ge 2y^2 \\ge 8x^4$, so $x(8x^3 - 1) \\le 0$. Since $x > 0$, this gives $8x^3 \\le 1$, so $0 < x \\le \\frac{1}{2}$. Similarly, $0 < y \\le \\frac{1}{2}$.\n\nAssume $x \\ge y$. Then $0 \\le 8(x-y) < 8x \\le 4$, so $8(x-y) \\in \\{0, 1, 2, 3\\}$, i.e., $x-y \\in \\{0, \\frac{1}{8}, \\frac{1}{4}, \\frac{3}{8}\\}$.\n\n- If $x-y = 0$, i.e., $x = y$, all conditions are satisfied. So, $(a, a)$ with $a \\in [0, \\frac{1}{2}]$ are solutions.\n- If $x-y = \\frac{1}{8}$, i.e., $y = x - \\frac{1}{8}$, then $y \\ge 2x^2 \\implies x - \\frac{1}{8} \\ge 2x^2 \\implies (4x - 1)^2 \\le 0 \\implies x = \\frac{1}{4}$, so $y = \\frac{1}{8}$. The pair $(\\frac{1}{4}, \\frac{1}{8})$ is a solution.\n- If $x-y = \\frac{1}{4}$, i.e., $y = x - \\frac{1}{4}$, then $y \\ge 2x^2 \\implies x - \\frac{1}{4} \\ge 2x^2 \\implies 8x^2 - 4x + 1 \\le 0$, which is impossible.\n- If $x-y = \\frac{3}{8}$, i.e., $y = x - \\frac{3}{8}$, then $y \\ge 2x^2 \\implies x - \\frac{3}{8} \\ge 2x^2 \\implies 16x^2 - 8x + 3 \\le 0$, which is impossible.\n\nThus, the solutions are $(0, 0)$, $(a, a)$ with $a \\in (0, \\frac{1}{2}]$, $(\\frac{1}{4}, \\frac{1}{8})$, and $(\\frac{1}{8}, \\frac{1}{4})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18080,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A grasshopper stands on the number line at the number $1$ and may make either a jump of length $2$ or of length $3$ each time. Each time, the grasshopper must land on an integer from $1$ through $n$ where the grasshopper has not been before. The grasshopper would like to visit all integers from $1$ through $n$ exactly once and land on the number $n$.\n\nProve that this can be done for all $n \\ge 9$.",
    "options": [],
    "answer": "See solution",
    "solution": "We distinguish different cases for $n$ based on the remainder of $n$ when dividing by $3$.\n\n- If $n = 3k$, then the grasshopper can jump as follows:\n  \n  $1; 3, 6, \\ldots, 3(k-1); 3(k-1)+2, 3(k-2)+2, \\ldots, 2; 4, 7, \\ldots, 3(k-1)+1; 3k.$\n  \n  The grasshopper jumps over triples on the way out, over triples plus $2$ on the way back, and then over triples plus $1$ on the second way out.\n\n- If $n = 3k + 1$, then the grasshopper can jump as follows:\n  \n  $1; 3, 6, \\ldots, 3k; 3k-2, 3(k-1)-2, \\ldots, 7, 4; 2, 5, 8, \\ldots, 3(k-1)+2; 3k+1.$\n\n- For $n = 3k + 2$, the grasshopper can jump as follows:\n  \n  $1, 3, 6, 4; 2, 5, 8, \\ldots, 3(k-1)+2; 3k+1, 3(k-1)+1, \\ldots, 7; 9, \\ldots, 3k; 3k+2.$\n\nBecause $n \\ge 9$, we have $k \\ge 3$. In the solutions above for $k \\ge 3$, every leg of the 'zigzag' is nonempty, so a solution exists.\n\nThere are other ways to solve this problem. For example, you can also give a solution for $n = 9, 10, 11, 12,$ and $13$ and then make a solution for $n + 5$ by starting with a solution for $n$ and adding $n + 2, n + 4, n + 1, n + 3, n + 5$ to it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18081,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there exist functions $f, g$ such that\n\n$$\nf(x + f(y)) = y^2 + g(x)\n$$\n\nfor all $x, y \\in \\mathbb{R}$. Do such functions exist?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that there exist functions $f, g$ satisfying the equality\n\n$$\nf(x + f(y)) = y^2 + g(x). \\quad (*)\n$$\n\nFirst, suppose that $f(y) = f(z)$ for some $y, z \\in \\mathbb{R}$. Then $z^2 + g(x) = f(x + f(z)) = f(x + f(y)) = y^2 + g(x)$, whence\n\n$$\nz^2 = y^2. \\quad (1)\n$$\n\nNow, $f(x+f(y)) = y^2 + g(x) = (-y)^2 + g(x) = f(x+f(-y))$. Using (1), we get $(x+f(y))^2 = (x+f(-y))^2$ for all $x,y$. That is $x+f(y) = -x-f(-y)$, or $x+f(y) = x+f(-y)$. Of these two equalities, the former is impossible since the equality $2x = -f(-y) - f(y)$ implies that $2x$ is a constant which is not true. Hence\n\n$$\nf(-y) = f(y). \\qquad (2)\n$$\n\nSet $f(0) = a$. Putting $y = 0$ in (*), we have\n\n$$\ng(x) = f(x + a). \\qquad (3)\n$$\n\nNow,\n\n$$\n\\begin{aligned}\nf(x + f(y)) &= y^2 + f(x + a) = y^2 + f(-x - a) = \\\\\n&= y^2 + f((-x - 2a) + a) = f(-x - 2a + f(y)).\n\\end{aligned}\n$$\n\nHence from (1) it follows that\n\n$$\n(x + f(y))^2 = (-x - 2a + f(y))^2 \\Leftrightarrow (2f(y) - 2a)(2x + 2a) = 0\n$$\n\nfor all $x,y \\in \\mathbb{R}$, which implies $f(y) = a$, and (3) gives $g(x) = a$. So, (*) becomes $a = y^2 + a$ for all $y \\in \\mathbb{R}$, a contradiction. Therefore there are no such functions $f$ and $g$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18082,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the set of rational numbers $r$ for which there exist non-negative integers $a$ and $b$ such that\n$$\n\\frac{a+b}{2} - \\sqrt{ab} = r.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b$ be non-negative integers and $r$ a rational number such that\n$$\nr = \\frac{a+b}{2} - \\sqrt{ab}.\n$$\nWe can rewrite this as:\n$$\n\\sqrt{ab} = \\frac{a+b}{2} - r.\n$$\nSince $\\sqrt{ab}$ must be a non-negative integer, set $\\sqrt{ab} = n$ for some $n \\in \\mathbb{Z}_{\\geq 0}$. Then $ab = n^2$.\n\nNow,\n$$\n\\frac{a+b}{2} - n = r \\implies a + b = 2r + 2n.\n$$\nSince $a$ and $b$ are non-negative integers with $ab = n^2$, $a$ and $b$ are both divisors of $n^2$ and $a + b$ is even. For any non-negative integer $m$, set $r = \\frac{m}{2}$.\n\nFor example, $a = b = 0$ gives $r = 0$. For $a = m$, $b = m$, $n = m$, we get $r = 0$. For $a = m$, $b = 4m$, $n = 2m$, we get $r = \\frac{m}{2}$.\n\nTherefore, the set of possible $r$ is\n$$\nS = \\left\\{ \\frac{m}{2} \\mid m \\in \\mathbb{Z}_{\\geq 0} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18083,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles of different radii, with centres at $B$ and $C$, touch externally at $A$. A common tangent, not through $A$, touches the first circle at $D$ and the second at $E$. The line through $A$ which is perpendicular to $DE$ and the perpendicular bisector of $BC$ meet at $F$. Prove that $BC = 2AF$.\n\nLet the two circles have radii $r_1$ and $r_2$. Since the circles are of different radii, we may assume $r_2 > r_1$ without loss of generality. Let $FA$ meet $DE$ at $L$ and let the midpoint of $BC$ be $M$.\n\n![](images/V_Britanija_2010_p20_data_fd27c3b7a5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "In this configuration, it is useful to extend $ED$ to meet $CB$ at a point $S$, the external centre of similitude. From $S$, the smaller circle can be enlarged to obtain the larger circle. Then triangles $SBD$, $SAL$, and $SCE$ are all similar, as they share the angle at $S$ and a right angle.\n\nTo compute $AF$, use these similar triangles. Since $\\angle FMA = 90^\\circ = \\angle SLA$ and $\\angle FAM = \\angle SAL$, triangles $FAM$ and $SAL$ are similar. Thus,\n\n$$\n\\frac{AF}{AM} = \\frac{SA}{AL}\n$$\n\nand this ratio equals $\\frac{SB}{BD}$. Because $\\frac{SB}{BD} = \\frac{SC}{CE}$, these are also equal to\n\n$$\n\\frac{SC - SB}{CE - BD} = \\frac{r_2 + r_1}{r_2 - r_1}\n$$\n\nNow, $AM = BM - BA = \\frac{1}{2}(r_2 - r_1)$, so\n\n$$\nAF = \\frac{1}{2}(r_2 - r_1) \\times \\frac{r_2 + r_1}{r_2 - r_1} = \\frac{1}{2}(r_2 + r_1) = \\frac{1}{2}BC\n$$\n\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18084,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, \\alpha, \\beta$ be given integers and define the sequence $(u_n)$ by $u_1 = \\alpha$, $u_2 = \\beta$, and $u_{n+2} = a u_{n+1} + b u_n + c$ for all $n \\ge 1$.\n\n**a)** Prove that if $a = 3$, $b = -2$, $c = -1$, then there are infinitely many pairs $(\\alpha, \\beta)$ such that $u_{2023} = 2^{2022}$.\n\n**b)** Prove that there exists a positive integer $n_0$ such that only one of the following two statements is true:\n\n1. There are infinitely many integers $m \\ge 1$ such that $u_{n_0} u_{n_0+1} \\dots u_{n_0+m}$ is divisible by $7^{2023}$ or $17^{2023}$;\n2. There are infinitely many positive integers $k$ such that $u_{n_0} u_{n_0+1} \\dots u_{n_0+k} - 1$ is divisible by $2023$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** For $a = 3$, $b = -2$, $c = -1$, we have $u_{n+2} = 3u_{n+1} - 2u_n - 1$ for all $n \\ge 1$. By induction, one can prove\n\n$$\nu_n = 2\\alpha - \\beta + (\\beta - \\alpha - 1) \\cdot 2^{n-1} + n$$\n\nfor all $n \\ge 1$. Then\n\n$$u_{2023} = 2\\alpha - \\beta + (\\beta - \\alpha - 1)2^{2022} + 2023.$$\n\nFor any $t \\in \\mathbb{Z}$, choose $\\alpha = (2^{2022} - 1)t - 2021$ and $\\beta = t - \\alpha - 2$. In other words, $2 - \\beta + \\alpha = t$ and $\\alpha + 2021 + (1 - 2^{2022})t = 0$, so we have\n\n$$\n\\begin{aligned}\n& 2\\alpha - \\beta + (\\beta - \\alpha - 1)2^{2022} + 2023 \\\\\n&= \\alpha + (2 - \\beta + \\alpha) + 2021 + (\\beta - \\alpha - 2)2^{2022} - 2^{2022} \\\\\n&= \\alpha + t + 2021 - t \\cdot 2^{2022} + 2^{2022} \\\\\n&= \\alpha + 2021 + (1 - 2^{2022})t + 2^{2022} = 2^{2022}.\n\\end{aligned}\n$$\n\nThis implies that there exist infinitely many pairs $(\\alpha, \\beta)$ for which $u_{2023} = 2^{2022}$.\n\n**b)** Note that $2023 = 7 \\times 17^2$. Let $(r_n)$ be the sequence of remainders of $(u_n)$ modulo $2023$. Then $(r_n)$ is periodic with some period $T > 0$. Consider the following cases:\n\n* If there exists $n_0 \\in \\mathbb{N}^*$ such that $7 \\mid u_{n_0}$ or $17 \\mid u_{n_0}$, consider the first case (the other is similar). Since $7 \\mid 2023$,\n\n$$\n\\prod_{i=0}^{m} u_{n_0 + i} \\not\\equiv 1 \\pmod{2023}, \\quad \\forall m \\ge 1.\n$$\n\nHence, statement 2 is not satisfied. For all $l \\in \\mathbb{N}^*$, choose $m = (2023l - 1)T$. Because $u_{n_0+nT} \\equiv u_{n_0} \\pmod{2023}$ and $7 \\mid 2023$, we have $u_{n_0+nT} \\equiv u_{n_0} \\equiv 0 \\pmod{7}$ for all $n \\in \\mathbb{N}^*$.\n\nThus, the sequence $u_{n_0}, u_{n_0+1}, \\dots, u_{n_0+(2023l-1)T}$ contains at least $2023$ terms divisible by $7$. Hence,\n\n$$\n\\prod_{i=0}^{m} u_{n_0+i} \\text{ is divisible by } 7^{2023}.\n$$\n\nTherefore, statement 1 is satisfied.\n\n* If $7 \\nmid u_n$, $17 \\nmid u_n$ for all $n \\in \\mathbb{N}^*$, choose $n_0 = 1$. Obviously, statement 1 is not satisfied. Otherwise, $\\gcd(u_n, 2023) = 1$, so by Euler's theorem,\n\n$$\nu_n^{\\varphi(2023)} \\equiv 1 \\pmod{2023}, \\quad \\forall n \\in \\mathbb{N}^*.$$\n\nSet $a = \\varphi(2023)$. For all $l \\in \\mathbb{N}^*$, choose $k = laT$. We will prove that $2023 \\mid \\prod_{i=0}^{k} u_{1+i} - 1$. Indeed, we have\n\n$$\nu_1 \\equiv u_{1+T} \\equiv \\cdots \\equiv u_{1+(la-1)T} \\pmod{2023}$$\n$$\nu_2 \\equiv u_{2+T} \\equiv \\cdots \\equiv u_{2+(la-1)T} \\pmod{2023}$$\n$$\\vdots$$\n$$\nu_T \\equiv u_{2T} \\equiv \\cdots \\equiv u_{laT} \\pmod{2023}.$$\n\nHence $u_i u_{i+T} \\cdots u_{i+(la-1)T} \\equiv u_i^a \\equiv 1 \\pmod{2023}$ for all $i = 1, \\dots, T$. From this, we conclude that $\\prod_{i=0}^{k} u_{1+i} \\equiv 1 \\pmod{2023}$, or $2023 \\mid \\prod_{i=0}^{k} u_{1+i} - 1$. So statement 2 is true. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18085,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R}^2 \\to \\mathbb{R}$ be a function satisfying the following property:\n\nIf $A, B, C, D \\in \\mathbb{R}^2$ are the vertices of a square with sides of length $1$, then\n\n$$\nf(A) + f(B) + f(C) + f(D) = 0.\n$$\n\nShow that $f(x) = 0$ for all $x \\in \\mathbb{R}^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $V$ be the set of all such functions. It is clear that if $f, g \\in V$, then $f + g \\in V$. Also, if $f \\in V$ and $a \\in \\mathbb{R}^2$, then for $g: \\mathbb{R}^2 \\to \\mathbb{R}$ defined by $g(x) = f(x + a)$, we have $g \\in V$.\n\nWe can consider elements of $\\mathbb{R}^2$ as vectors. Denote the inner product of $a = (a_1, a_2)$ and $b = (b_1, b_2)$ by $a \\cdot b = a_1 b_1 + a_2 b_2$. A vector $a$ is a unit vector if $a \\cdot a = 1$. Two vectors $a, b$ are orthogonal if $a \\cdot b = 0$. A square $ABCD$ with vertices $a, b, c, d$ is a unit square if the vectors $b - a, c - b, d - c, a - d$ are unit vectors with consecutive pairs orthogonal.\n\nWe say $f \\in V$ is *good* for a unit vector $v$ if either $f(x + v) = -f(x)$ for all $x \\in \\mathbb{R}^2$, or $f(x + v^{\\perp}) = -f(x)$ for some unit vector $v^{\\perp}$ orthogonal to $v$ and all $x$. $f$ is good for a set of unit vectors if it is good for every vector in the set.\n\nIf $f, g \\in V$ are good for $S$, so is $f + g$. If $f \\in V$ is good for $S$ and $a \\in \\mathbb{R}^2$, then $g(x) = f(x + a)$ is also good for $S$.\n\n**Claim 1.** If $V \\neq \\{0\\}$, then for any finite set $S$ of unit vectors, there is $g \\in V \\setminus \\{0\\}$ which is good for $S$.\n\n*Proof of Claim 1.* Proceed by induction on $|S|$. Assume $f \\in V \\setminus \\{0\\}$ is good for $S$ and we seek $g \\in V \\setminus \\{0\\}$ good for $S \\cup \\{v\\}$ for some unit vector $v$.\n\nLet $v^{\\perp}$ be a unit vector orthogonal to $v$. If $f(x + v^{\\perp}) = -f(x)$ for all $x$, we are done. Otherwise, define $g(x) = f(x) + f(x + v^{\\perp})$. Then $g$ is good for $S$. Since $x, x + v, x + v + v^{\\perp}, x + v^{\\perp}$ form a unit square,\n\n$$\ng(x) + g(x + v) = f(x) + f(x + v^{\\perp}) + f(x + v) + f(x + v + v^{\\perp}) = 0.\n$$\n\nSo $g$ is good for $v$ as well, hence for $S \\cup \\{v\\}$. $\\square$\n\nWe say $f \\in V$ is *excellent* for a unit vector $v$ if $f(x + 12v) = f(x)$ for all $x$. $f$ is excellent for a set if it is excellent for every vector in the set.\n\nIf $f, g \\in V$ are excellent for $S$, so is $f + g$. If $f \\in V$ is excellent for $S$ and $a \\in \\mathbb{R}^2$, then $g(x) = f(x + a)$ is also excellent for $S$.\n\n**Claim 2.** If $V \\neq \\{0\\}$, then for any finite set $S$ of unit vectors, there is $g \\in V \\setminus \\{0\\}$ which is excellent for $S$.\n\n*Proof of Claim 2.* Proceed by induction on $|S|$. Assume $f \\in V \\setminus \\{0\\}$ is excellent for $S$ and we seek $g \\in V \\setminus \\{0\\}$ excellent for $S \\cup \\{v\\}$ for some unit vector $v$.\n\nThe proof of Claim 1 shows we may assume $f$ is good for $v$. If $f(x + v) = -f(x)$ for all $x$, then $f(x + 12v) = f(x)$ for all $x$ and we are done. Otherwise, $f(x + v^{\\perp}) = -f(x)$ for all $x$ for some $v^{\\perp}$ orthogonal to $v$.\n\nLet $u = \\frac{3v + 4v^{\\perp}}{5}$ and $u^{\\perp} = \\frac{4v - 3v^{\\perp}}{5}$. Then $u, u^{\\perp}$ are unit vectors. The proof of Claim 1 shows we may assume $f$ is good for $u$.\n\nIf $f(x + u) = -f(x)$ for all $x$,\n\n$$\nf(x) = -f(x + 5u) = -f(x + 3v + 4v^{\\perp}) = -f(x + 3v).\n$$\n\nSo $f(x + 12v) = f(x)$ for all $x$ and we are done.\n\nIf $f(x + u^{\\perp}) = -f(x)$ for all $x$,\n\n$$\nf(x) = -f(x + 5u^{\\perp}) = -f(x + 4v - 3v^{\\perp}) = f(x + 4v).\n$$\n\nSo again $f(x + 12v) = f(x)$ for all $x$ and we are done. $\\square$\n\n![](images/BMO_2022_shortlist_p17_data_ca16c8fc15.png)\n\nFrom Claim 1, there is a unit vector $v$ and $f \\in V \\setminus \\{0\\}$ such that $f(x + v) = -f(x)$ for all $x$. Pick unit vectors $u, w$ such that $12u, 12w, v$ form a triangle, i.e., $12u + 12w + v = 0$.\n\nFrom Claim 2, we may also assume $f$ is excellent for $u, w$ (while preserving $f(x + v) = -f(x)$ for all $x$).\n\nGiven any $x \\in \\mathbb{R}^2$,\n\n$$\nf(x) = f(x + 12u + 12w + v) = f(x + 12w + v) = f(x + v) = -f(x)\n$$\n\nthus $f(x) = 0$. Since this holds for all $x \\in \\mathbb{R}^2$, this contradicts $f \\in V \\setminus \\{0\\}$, so $V = \\{0\\}$ and $f(x) = 0$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18086,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that $a = \\sqrt{9 - \\sqrt{77}} \\cdot \\sqrt{2} \\cdot (\\sqrt{11} - \\sqrt{7}) \\cdot (9 + \\sqrt{77})$ is an integer.\n\nb) Prove that if $x$ and $y$ are real numbers such that $xy = 6$, $x > 2$ and $y > 2$, then $x + y < 5$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Number $a$ is equal to $\\sqrt{18 - 2\\sqrt{77}} \\cdot (\\sqrt{11} - \\sqrt{7}) \\cdot (9 + \\sqrt{77})$. Since $18 - 2\\sqrt{77} = (\\sqrt{11} - \\sqrt{7})^2$, it follows that\n\n$$\na = (\\sqrt{11} - \\sqrt{7})^2 \\cdot (9 + \\sqrt{77}) = (18 - 2\\sqrt{77})(9 + \\sqrt{77}) = 8 \\in \\mathbb{N}.\n$$\n\nb) If $x > 2$, $y > 2$, then $(x - 2)(y - 2) > 0$. Therefore $xy - 2(x + y) + 4 > 0$, whence $x + y < \\frac{1}{2}(xy + 4) = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18087,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that the equality\n\n$$\nf(\\lfloor x \\rfloor y) = f(x) \\lfloor f(y) \\rfloor\n$$\n\nholds for all $x, y \\in \\mathbb{R}$. (Here $\\lfloor z \\rfloor$ denotes the greatest integer less than or equal to $z$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $f(x) = c$ for all $x$, where $c = 0$ or $1 \\leq c < 2$.\n\nTo prove that these are the only possible solutions, consider two cases:\n\n**Case 1:** If $\\lfloor f(y) \\rfloor = 0$ for all $0 \\leq y < 1$, then for any integer $k$, setting $x = k$ in the given yields $f(ky) = f(k) \\lfloor f(y) \\rfloor = 0$ for such $y$. Since every real number can be written as $ky$ with integer $k$ and $0 \\leq y < 1$, we have $f(x) = 0$ for all $x$.\n\n**Case 2:** Suppose $\\lfloor f(y_0) \\rfloor \\neq 0$ for some $0 \\leq y_0 < 1$. For any $x_n$ with $n \\leq x_n < n+1$, set $y = y_0$ and $x = x_n$ to obtain\n\n$$\nf(ny_0) = f(x_n) \\lfloor f(y_0) \\rfloor.\n$$\n\nLet $c_n = \\frac{f(ny_0)}{\\lfloor f(y_0) \\rfloor}$, so $f(x) = c_n$ for all $x \\in [n, n+1)$. In particular, $\\lfloor c_0 \\rfloor = \\lfloor f(y_0) \\rfloor \\neq 0$, so $c_0 \\neq 0$. Now, set $x = y = 0$:\n\n$$\nc_0 = f(0) = f(0) \\lfloor f(0) \\rfloor = c_0 \\lfloor c_0 \\rfloor,\n$$\n\nso $\\lfloor c_0 \\rfloor = 1$. Finally, setting $y = 0$ and $x = n$ gives\n\n$$\nc_n = f(n) = \\frac{f(0)}{\\lfloor f(0) \\rfloor} = \\frac{c_0}{\\lfloor c_0 \\rfloor} = c_0.\n$$\n\nTherefore, $f(x) = c_0$ for all $x$, and $\\lfloor c_0 \\rfloor = 1$, i.e., $1 \\leq c_0 < 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18088,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2020 positive integers written on a blackboard. Every minute, Zuming erases two of the numbers and replaces them by their sum, difference, product, or quotient. For example, if Zuming erases the numbers 6 and 3, he may replace them with one of the numbers in the set $\\{6+3, 6-3, 3-6, 6 \\times 3, 6 \\div 3, 3 \\div 6\\} = \\{9, 3, -3, 18, 2, \\frac{1}{2}\\}$. After 2019 minutes, Zuming arrives at the single number $-2020$ on the blackboard. Show that it is possible for Zuming to have arrived at the single number $2020$ on the blackboard instead, under the same rules and using the same 2020 starting integers.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that if Zuming's original set of moves leads to a state $a_1, a_2, \\dots, a_n$ for some $n \\leq 2020$, he can make new moves to lead to a state $|a_1|, |a_2|, \\dots, |a_n|$. This clearly implies the problem by taking $n = 1$.\n\nWe do this by downwards induction. Note that since all starting numbers are positive, the base case $n = 2020$ trivially holds. For the inductive step (from $n$ to $n-1$), it suffices to show that if it is possible to obtain $c$ from $a$ and $b$, then it is possible to obtain $|c|$ from $|a|$ and $|b|$. If the operation to get $c$ from $a$ and $b$ is multiplication or division, then using the same operation on $|a|$ and $|b|$ will work. Now suppose the operation is addition or subtraction (i.e., $c \\in \\{a+b, a-b, b-a\\}$), then depending on whether $a$ and $b$ are negative or non-negative, we have $c \\in \\{|a|+|b|, |a|-|b|, -|a|+|b|, -|a|-|b|\\}$. And hence $|c| \\in \\{|a|+|b|, |a|-|b|, |b|-|a|\\}$, so we can always choose the appropriate addition or subtraction to get $|c|$ from $|a|$ and $|b|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18089,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ and $p$ such that:\n\n$$\n(n^2 + 1)(p^2 + 1) + 45 = 2(2n + 1)(3p + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We obtain $(np-6)^2 + (n-2)^2 + (p-3)^2 = 5$, so the numbers $(np-6)^2$, $(n-2)^2$, and $(p-3)^2$ must be $0$, $1$, and $4$ in some order. By inspection, the solutions are $(n, p) \\in \\{(2, 4), (2, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18090,
    "subject": "Mathematics (Olympiad)",
    "question": "For all positive real numbers $w, x, y, z$ satisfying\n$$\nwx = yz,\n$$\nfind all functions $f : (0, +\\infty) \\to (0, +\\infty)$ such that\n$$\n\\frac{(f(w))^2 + (f(x))^2}{2f(wx)} = \\frac{w^2 + x^2}{2wx}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Take\n$$\nw = x = y = z = 1,\n$$\nthen we get $(f(1))^2 = f(1)$, so $f(1) = 1$.\n\nFor any real number $t > 0$, let $w = t$, $x = 1$, $y = z = \\sqrt{t}$, we get\n$$\n\\frac{(f(t))^2 + 1}{2f(t)} = \\frac{t^2 + 1}{2t},\n$$\nwhich implies $(t f(t) - 1)(f(t) - t) = 0$.\n\nSo, for any $t > 0$,\n$$\nf(t) = t \\quad \\text{or} \\quad f(t) = \\frac{1}{t}. \\qquad \\textcircled{1}\n$$\n\nSuppose there exist $b, c \\in (0, +\\infty)$ such that $f(b) \\neq b$, $f(c) \\neq \\frac{1}{c}$. By ①, we get $b, c$ different from 1 and $f(b) = \\frac{1}{b}$, $f(c) = c$.\n\nTake $w = b, x = c, y = z = \\sqrt{bc}$, then\n$$\n\\frac{\\frac{1}{b^2} + c^2}{2f(bc)} = \\frac{b^2 + c^2}{2bc},\n$$\ni.e. $f(bc) = \\frac{c + b^2 c^3}{b(b^2 + c^2)}$.\n\nBy ①, $f(bc) = bc$ or $f(bc) = \\frac{1}{bc}$. If $f(bc) = bc$, then\n$$\nbc = \\frac{c + b^2 c^3}{b(b^2 + c^2)},\n$$\nwhich yields $b^4 c = c$, $b = 1$. Contradiction!\n\nIf $f(bc) = \\frac{1}{bc}$, then\n$$\n\\frac{1}{bc} = \\frac{c + b^2 c^3}{b(b^2 + c^2)},\n$$\nthat yields $b^2 c^4 = b^2$, $c = 1$. Contradiction!\n\nTherefore, only two functions: $f(x) = x,\\ x \\in (0, +\\infty)$ or $f(x) = \\frac{1}{x},\\ x \\in (0, +\\infty)$. It is easy to verify that these two functions satisfy the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18091,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 3$ be a positive integer. Suppose we have a circle with $n$ positions labeled $1, 2, \\ldots, n$ in clockwise order. $n$ counters, each with one side white and the other side black, are placed on the circle with one counter in each of the numbered positions. Initially, all counters have the white side facing up, except the counter at position $1$, which shows black.\n\nWe are allowed to perform the following operation:\n\n1. Choose a counter $X$ whose black side is facing up, and let $Y, Z$ be the next two counters in clockwise order.\n2. Flip counter $Y$ so that it is showing the other color.\n3. Move $X$ two spaces clockwise, and move $Y, Z$ one space counterclockwise. (So after the operation we still have one counter in each position.)\n\nLet $S$ be any nonempty subset of $\\{1, 2, \\dots, n\\}$. Show that we can perform a finite sequence of moves after which the numbers in $S$ correspond exactly to the positions of the black counters.",
    "options": [],
    "answer": "See solution",
    "solution": "The operation is invertible; its inverse is to take consecutive counters $Y, Z, X$ with $X$ black, move $X$ to the front, and flip $Y$. We will work with this operation instead, and show that from any initial position with at least one black counter, we can reach the configuration with one black counter on position $1$.\n\nLet $B$ denote a black counter and $W$ denote a white counter. Suppose we do not have two consecutive counters of the same color anywhere. Take a configuration $BWB$ and perform the move changing it to $BWW$. Given that we have two consecutive white counters, we can find a configuration $WWB$ and perform the move changing it to $BBW$. So no matter what we start with, we will be able to obtain two consecutive black counters somewhere.\n\nNow we use our two consecutive black counters to turn everything black. Suppose there is a white counter; by taking the first white counter preceding a run of at least two consecutive black counters we have a configuration of the form $WBB$. Perform the move to obtain $BBB$, eliminating this white counter. By repeating this process we change all the counters to black.\n\nIt now suffices to show that we can get to just one black counter from this, since we can cyclically shift any moves from this point to ensure it ends up on space $1$. Given the configuration $BBB$, two moves will change it to $BWW$. So we can change the black counters to white in pairs. Repeat this process in a way that keeps all the white counters and black counters consecutive. If $n$ is odd, this gives us one black counter at the end. If $n$ is even, we get two consecutive black counters with the rest white. Then perform three moves on $WBB$, which changes it to $BWW$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18092,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a circle $\\Gamma$ with center $O$ and diameter $AB$. $OBDE$ is a square, $F$ is the second point of intersection of $AD$ and circle $\\Gamma$, and $C$ is the midpoint of segment $AF$. Find the value of the angle $OCB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ is a diameter, $\\angle AFB = 90^\\circ$, and $CO \\parallel FB$ as the midline. Therefore, $CO \\perp CD$ and quadrilateral $OBDC$ is cyclic. Hence, $\\angle OCB = \\angle ODB = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18093,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $ (a, b) $ of non-negative integers such that\n\n$$\n2017^a = b^6 - 32b + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The two solutions are $ (0, 0) $ and $ (0, 2) $.\n\nSince $2017^a$ is always odd, $b$ must be even, so $b = 2c$, $c$ integer. Therefore, $2017^a = 64(c^6 - c) + 1$ and thus $2017^a \\equiv 1 \\pmod{64}$. But $2017 \\equiv 33 \\pmod{64}$ and $2017^2 \\equiv (1+32)^2 = 1 + 2 \\cdot 32 + 32^2 \\equiv 1 \\pmod{64}$, so the powers of $2017$ modulo $64$ alternate between $1$ and $33$. Therefore, $a$ is even and $2017^a$ is a perfect square. Let $r(b) = b^6 - 32b + 1$.\n\nFor $b > 4$, $r(b) < b^6 = (b^3)^2$ for $b > 0$. Also, $r(b) > (b^3 - 1)^2$ because $b^6 - 32b + 1 > b^6 - 2b^3 + 1$ when $b > 4$. Thus, $2017^a$ is between two consecutive squares, so there are no solutions for $b > 4$.\n\nSince $b$ is even, check $b = 4, 2, 0$:\n\n- For $b = 4$, modulo $3$: $1 \\equiv 1 - 2 + 1 = 0$, so no solution.\n- For $b = 2$: $2017^a = 2^6 - 2^6 + 1 = 1$, so $(a, b) = (0, 2)$.\n- For $b = 0$: $2017^a = 1$, so $(a, b) = (0, 0)$.\n\nTherefore, $(0, 0)$ and $(0, 2)$ are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18094,
    "subject": "Mathematics (Olympiad)",
    "question": "Five kids are sitting around a table, each with a different number of apples. No two kids have the same number of apples. Prove that any two adjacent kids have at least 5 apples in total. Furthermore, show that if the numbers of apples are $\\{1, 2, 3, 4, 5\\}$, then the kid with 1 apple can determine the difference in apples between the two kids sitting opposite him. \n\n(b) Give an example of a seating arrangement of five kids with distinct numbers of apples such that a kid facing two others cannot determine the difference in apples between those two kids.",
    "options": [],
    "answer": "See solution",
    "solution": "First, the sum of apples for two adjacent kids cannot be 1 or 2, since that would require one kid to have 0 apples, which is not possible given all numbers are distinct and positive. For sums of 3 or 4, the possible distributions would force the remaining kids to have numbers that make the total apples at least 17 or 18, which is impossible for five kids with distinct numbers. Thus, any two adjacent kids must have at least 5 apples in total.\n\nIf one kid has at least 6 apples, the other four must have at least $2 \\times 5 = 10$ apples, totaling at least 16 apples. If a kid has 0 apples, the adjacent kids must have at least 5 and 6 apples, which is not possible. Therefore, the only possible case is the kids have $\\{1, 2, 3, 4, 5\\}$ apples. The kid with 1 apple must sit next to the kids with 4 and 5 apples. The two kids opposite him must have $\\{2, 3\\}$ apples, so the difference is 1, which the kid with 1 apple can deduce.\n\n(b) For example, if the kids have 1, 4, 3, 2, 6 apples in order around the table, the sums of apples for adjacent pairs are 5, 7, 5, 8, and 7. A kid facing two others with a sum of $k$ apples cannot determine the difference, since the two could have $\\{0, k\\}$ apples as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18095,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $\\angle ABC = 90^\\circ$ and $BA = BC = \\sqrt{2}$. Points $P_1, P_2, \\dots, P_{2024}$ lie on hypotenuse $\\overline{AC}$ so that $AP_1 = P_1P_2 = P_2P_3 = \\dots = P_{2023}P_{2024} = P_{2024}C$. What is the length of the vector sum\n\n$$\n\\overrightarrow{BP_1} + \\overrightarrow{BP_2} + \\overrightarrow{BP_3} + \\dots + \\overrightarrow{BP_{2024}}?\n$$\n\n1011  \n1012  \n2023  \n2024  \n2025",
    "options": [],
    "answer": "See solution",
    "solution": "For $1 \\leq i \\leq 2024$, the vector sum $\\overrightarrow{BP_i} + \\overrightarrow{BP_{2025-i}}$ is the vector pointing from the apex of isosceles right triangle $\\triangle ABC$ to the reflection of the apex across the hypotenuse, as seen in the figure below.\n\n![](images/2024_AMC12A_Solutions_p3_data_3b580aee90.png)\n\nIts length is 2 times the height of the triangle, namely $2 \\cdot 1 = 2$. Each pair contributes a vector of length 2, all pointing in the same direction, to the total sum. With 1012 such pairs, the length of the resultant vector is $2 \\cdot 1012 = 2024$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18096,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for the polynomial $P(x) = x^4 - x^3 - 3x^2 - x + 1$, there exist infinitely many positive integers $n$ for which $P(3^n)$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "For $x = 3^{2n-1}$, we have:\n\n$$\n\\begin{aligned}\nP(3^{2n-1}) &= (3^{2n-1})^4 - (3^{2n-1})^3 - 3(3^{2n-1})^2 - 3^{2n-1} + 1 \\\\\n&= 3^{4(2n-1)} - 3^{3(2n-1)} - 3 \\cdot 3^{2(2n-1)} - 3^{2n-1} + 1 \\\\\n&= 3^{8n-4} - 3^{6n-3} - 3 \\cdot 3^{4n-2} - 3^{2n-1} + 1\n\\end{aligned}\n$$\n\nNow, consider $P(3^{2n-1}) \\pmod{5}$. Since $3^4 \\equiv 1 \\pmod{5}$, $3^{8n-4} \\equiv 1 \\pmod{5}$, and similar periodicity applies to the other terms. Thus, $P(3^{2n-1}) \\equiv 0 \\pmod{5}$ for all $n$. Since $P(x)$ is not constant, it can only take the value $5$ finitely many times, so for infinitely many $n$, $P(3^{2n-1})$ is divisible by $5$ and greater than $5$, hence composite.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18097,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n \\in \\mathbb{N}$. Consider polynomials of degree $n$ of the form\n\n$$\nP(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0\n$$\n\nwith non-negative integer coefficients not exceeding $100$. We call such a polynomial *expandable* if it can be represented as a product of two non-constant polynomials with non-negative integer coefficients, and *non-expandable* otherwise. Prove that there are at least twice as many non-expandable polynomials as there are expandable ones.",
    "options": [],
    "answer": "See solution",
    "solution": "If the polynomial $P(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0$ is expandable, then it can be represented as\n\n$$\nP(x) = (x^m + b_{m-1}x^{m-1} + \\cdots + b_1x + b_0) \\cdot (x^k + c_{k-1}x^{k-1} + \\cdots + c_1x + c_0),\n$$\n\nwhere $m \\geq k \\geq 1$, $m + k = n$, and the numbers $b_i, c_i$ are all non-negative integers. Since the coefficients of $P(x)$ are not greater than $100$, all these numbers do not exceed $100$ and, moreover, $b_i \\cdot c_i \\leq 100$ for any $i$ from $0$ to $k-1$. This means that for a fixed $i$ from $0$ to $k-1$, the number of possible pairs $(b_i, c_i)$ does not exceed\n\n$$\n2 \\cdot 11 \\cdot 101 < \\frac{101^2}{4},\n$$\n\nbecause at least one of $b_i$ and $c_i$ is not greater than $\\sqrt{100} = 10$.\n\nThus, the numbers $b_{m-1}, b_{m-2}, \\dots, b_k$ can be chosen in no more than $101^{m-k}$ ways, and the numbers $b_0, c_0, b_1, c_1, \\dots, b_{k-1}, c_{k-1}$ can be selected in no more than $\\left(\\frac{101^2}{4}\\right)^k$ ways. We find that the number of expandable polynomials can be estimated from above by the expression\n\n$$\n\\sum_{k=1}^{\\lfloor n/2 \\rfloor} 101^{m-k} \\left(\\frac{101^2}{4}\\right)^k = 101^n \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\frac{1}{4^k} < 101^n \\frac{1/4}{1 - 1/4} = \\frac{1}{3} \\cdot 101^n\n$$\n\nSince the number of polynomials $P(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0$ with non-negative integer coefficients not exceeding $100$ is equal to $101^n$, there are more than $\\frac{2}{3} \\cdot 101^n$ non-expandable polynomials, that is, at least twice as many as expandable ones.\n\n_Remark._ The estimate in the problem can be improved, for example, by noting that if one of the numbers $c_j$ is greater than $10$, then all the coefficients $b_i$ are less than $10$, and vice versa.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18098,
    "subject": "Mathematics (Olympiad)",
    "question": "Ten boys and ten girls met at a party. Assume that every girl likes exactly $k$ boys and every boy likes exactly $k$ girls. Is it always possible to find a couple where both partners like each other? Solve the problem for:\n\na) $k = 5$\n\nb) $k = 6$",
    "options": [],
    "answer": "See solution",
    "solution": "a) For $k=5$, it may happen that there are no such couples, with one counterexample given as follows. Split the boys into two disjoint quintuples $A, B$ and the girls into two disjoint quintuples $C, D$. Consider the configuration where every boy from $A$ likes all the girls in $C$, every boy in $B$ likes all the girls in $D$, every girl in $C$ likes all the boys in $B$ and every girl in $D$ likes all the boys in $A$. Then every boy likes 5 girls, every girl likes 5 boys, but there is clearly no couple where both partners like each other.\n\n![](images/CZE_ABooklet_2024_p0_data_a94a710dbc.png)\n\nb) For $k=6$, such a couple must exist: there are in total $10k = 60$ couples $(\\text{boy}, \\text{girl})$ where the boy likes the girl and, by symmetry, 60 couples where the girl likes the boy. These two sets of couples can't possibly be disjoint, since there are $10 \\cdot 10 = 100$ couples in total and $60 + 60 > 100$, so there must be a couple where both the partners like each other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18099,
    "subject": "Mathematics (Olympiad)",
    "question": "On the board is written a (not necessarily positive) integer. In a move, if the current number on the board is $a$, one erases it and replaces it by either $a^2 + 2a + 3$, $5a^2 + 2$, or $a - 119$. Is there a starting number $s$ such that for any positive integer $f$ one can reach $f$ from $s$ with finitely many moves?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that all numbers of the form $119k$ and $119k - 1$ cannot be represented as $a^2 + 2a + 3$ or $5a^2 + 2$. We can check this claim by considering modulo $7$ and $17$.\n\n- If $119k = a^2 + 2a + 3 = (a + 1)^2 + 2$, then $-2$ is a quadratic residue modulo $7$, which is absurd since the quadratic residues modulo $7$ are $0, 1, 2, 4$.\n- If $119k = 5a^2 + 2$, then $3$ is a quadratic residue modulo $17$, also absurd.\n- If $119k - 1 = a^2 + 2a + 3$, then $-3$ is a quadratic residue modulo $17$, also absurd.\n- If $119k - 1 = 5a^2 + 2$, then $-2$ is a quadratic residue modulo $7$, also absurd.\n\nSo, if we consider the first number in one of the above forms that we encounter when going backwards, it turns out that we can only shift by multiples of $119$ and hence cannot obtain any number of the other form. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18100,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer. Find the number of arrangements $a_1, a_2, \\dots, a_n$ of $1, 2, \\dots, n$ around a circle, in clockwise direction, such that\n$$\n|a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{n-1} - a_n| + |a_n - a_1| = 2n-2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we clarify the meaning of the given equality. Let $a_1, a_2, \\dots, a_n$ be an arbitrary circular arrangement of $1, 2, \\dots, n$, $n \\ge 3$, in clockwise direction. The extremal numbers $1$ and $n$ separate the remaining numbers into two groups. For convenience, denote them by $b_1, \\dots, b_k$ and $c_1, \\dots, c_l$, arranged as shown in the figure. Here $k + l = n - 2$; one of $k$ and $l$ can be zero.\n\nWe have\n$$\n|n - b_k| + |b_k - b_{k-1}| + \\dots + |b_1 - 1| \\ge (n - b_k) + (b_k - b_{k-1}) + \\dots + (b_1 - 1) = n - 1\n$$\nand\n$$\n|n - c_l| + |c_l - c_{l-1}| + \\dots + |c_1 - 1| \\ge (n - c_l) + (c_l - c_{l-1}) + \\dots + (c_1 - 1) = n - 1.\n$$\n\n![](images/Argentina_2017_p4_data_633a399415.png)\n\nThe absolute values in the two left-hand sides are $|a_1 - a_2|, |a_2 - a_3|, \\dots, |a_{n-1} - a_n|, |a_n - a_1|$. Adding up gives\n$$\nS = |a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{n-1} - a_n| + |a_n - a_1| \\ge 2n - 2.\n$$\nFor any circular arrangement $a_1, a_2, \\dots, a_n$ of $1, 2, \\dots, n$,\n\nWe are interested in the equality case. Clearly, $S = 2n - 2$ if and only if $b_k > b_{k-1} > \\dots > b_1$ and $c_l > c_{l-1} > \\dots > c_1$ (because $n > b_k, b_{k-1} > 1$ and $n > c_l, c_{l-1} > 1$ hold trivially).\n\nNow we show that there is a bijection between our admissible circular arrangements, the ones with $S = 2n - 2$, and the subsets of $\\{2, \\dots, n-1\\}$. Let $B$ be any subset of $\\{2, \\dots, n-1\\}$, including the empty one. Construct a circular arrangement of $1, 2, \\dots, n$ in a way suggested by the previous reasoning. Start with $1$, proceed in clockwise direction along the circle by placing the elements of $B$ in increasing order, place $n$ after them, and finish with the remaining elements of $\\{2, \\dots, n-1\\}$ in decreasing order. By the above, the obtained circular arrangement is admissible, and clearly different subsets $B$ of $\\{2, \\dots, n-1\\}$ give rise to different arrangements. The bijection shows that there are $2^{n-2}$ admissible circular arrangements, as many as the subsets of $\\{2, \\dots, n-1\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18101,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of positive integers $\\{a_n\\}_{n \\ge 1}$ is called a \"CGMO sequence\" if:\n\n1. $\\{a_n\\}_{n \\ge 1}$ is strictly increasing.\n2. For each integer $n \\ge 2022$, $a_n$ is the smallest integer greater than $a_{n-1}$ such that for some non-empty subset $A_n \\subseteq \\{a_1, a_2, \\dots, a_{n-1}\\}$, the product $a_n \\cdot \\prod_{a \\in A_n} a$ is a perfect square.\n\nProve that there exist constants $c_1, c_2 > 0$ such that for each CGMO sequence $\\{a_n\\}_{n \\ge 1}$, there exists a positive integer $N$ (depending on the sequence) so that for every $n \\ge N$,\n\n$$\nc_1 \\cdot n^2 \\le a_n \\le c_2 \\cdot n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nTake $c_1 = 2^{-4042}$, $c_2 = 2$. We show $c_1 \\cdot n^2 \\le a_n \\le c_2 \\cdot n^2$ for sufficiently large $n$.\n\n**(1) Upper bound:**\n\nSuppose $\\{a_n\\}_{n \\ge 1}$ is any CGMO sequence. By definition, there exists $A_{2022} \\subset \\{a_1, a_2, \\dots, a_{2011}\\}$ such that $a_{2022} \\cdot \\prod_{a \\in A_{2022}} a = L^2$, $L \\in \\mathbb{N}^+$. We claim: for $k \\ge 0$,\n\n$$\na_{2022+k} \\le (L+k)^2.\n$$\n\nInduct on $k$. When $k=0$, obviously $a_{2022} \\le a_{2022} \\cdot \\prod_{a \\in A_{2022}} a = L^2$. Assume it is true for $k$, namely $a_{2022+k} \\le (L+k)^2$. Then for $k+1$, note that\n\n$$\n(L + k + 1)^2 > (L + k)^2 \\geq a_{2022+k}\n$$\n\nand\n\n$$\n(L + k + 1)^2 \\cdot a_{2022} \\cdot \\prod_{a \\in A_{2022}} a = (L + k + 1)^2 L^2\n$$\n\nis a perfect square. By the choice of $a_{2022+k+1}$, we obtain $a_{2022+k+1} \\le (L+k+1)^2$ and the validity of the claim.\n\nNow, $a_n \\le (L + n - 2022)^2$ holds for $n \\ge 2022$. Take $N > |L - 2022|/(\\sqrt{2} - 1)$, $N$ only dependent on $\\{a_n\\}_{n \\ge 1}$. Then for $n \\ge N$, $(L + n - 2022)^2 \\le 2n^2$, and the upper bound is verified.\n\n**(2) Lower bound:**\n\nEvery positive integer $m$ can be uniquely written as $m = ab^2$, $a, b$ positive integers, and $a$ has no square factor other than 1. Let $a = f(m)$ denote the square-free part of $m$. Then $f$ has simple properties: $f(xy^2) = f(x)f(y) = f(f(x)f(y))$.\n\nLet $S = \\{a_1, a_2, \\dots, a_{2021}\\}$, and $F = \\left\\{ f\\left( \\prod_{a \\in B} a \\right) \\mid B \\subseteq S \\right\\}$ (if $B = \\emptyset$, the product is 1). The set $F$ satisfies: for any $x_1, x_2, \\dots, x_t \\in F$,\n\n$$\nf(x_1x_2\\cdots x_t) \\in F.\n$$\n\nIndeed, let $x_i = f\\left(\\prod_{a \\in B_i} a\\right)$, $B_i \\subseteq S$, $1 \\le i \\le t$, and take $B = B_1 \\triangle B_2 \\triangle \\dots \\triangle B_t$ (here, $X \\triangle Y$ is the symmetric difference of two sets $X, Y$: $X \\triangle Y = (X \\setminus Y) \\cup (Y \\setminus X)$). Then\n\n$$\nf(x_1x_2\\cdots x_t) = f\\left(\\prod_{a \\in B} a\\right) \\in F.\n$$\n\nWe prove by induction that for every positive integer $n$, the square-free part of $a_n$ belongs to $F$.\n\nObviously, the claim is true for $1 \\le n \\le 2021$; assume it is true for $1 \\le n \\le m$, $m \\ge 2021$. By definition of $a_{m+1}$, there exists $A_{m+1} \\subseteq \\{a_1, a_2, \\dots, a_m\\}$, such that $a_{m+1} \\cdot \\prod_{a \\in A_{m+1}} a$ is a perfect square. Also, by the induction hypothesis, for any $a \\in A_{m+1}$, $f(a) \\in F$. Using the above property of $F$, we find\n\n$$\nf(a_{m+1}) = f\\left(\\prod_{a \\in A_{m+1}} a\\right) = f\\left(\\prod_{a \\in A_{m+1}} f(a)\\right) \\in F,\n$$\n\nand the induction is complete.\n\nLet $|F| = m \\le 2^{2021}$. For any positive integer $n$, consider the square-free parts of $a_1, a_2, \\dots, a_n$. By the Pigeonhole principle, among them there are $k \\ge \\lfloor \\frac{n}{m} \\rfloor \\ge \\lfloor \\frac{n}{2^{2021}} \\rfloor$ identical parts, say $f(a_{i_1}) = f(a_{i_2}) = \\dots = f(a_{i_k}) = u$, where\n\n$$\n1 \\le i_1 < i_2 < \\dots < i_k \\le n.\n$$\n\nSince $a_{i_1} < a_{i_2} < \\dots < a_{i_k}$ and their square-free parts are all equal to $u$,\n\n$$\na_{i_k} \\ge uk^2 \\ge k^2 \\ge \\left(\\frac{n}{2^{2021}}\\right)^2,\n$$\n\nand thereby, $a_n \\ge a_{i_k} \\ge 2^{-4042} \\cdot n^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18102,
    "subject": "Mathematics (Olympiad)",
    "question": "For any finite set $X$, let $|X|$ denote the number of elements in $X$. Define\n\n$$\nS_n = \\sum |A \\cap B|,\n$$\n\nwhere the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\\{1, 2, 3, \\dots, n\\}$ with $|A| = |B|$.\n\nFor example, $S_2 = 4$ because the sum is taken over the pairs of subsets\n\n$$\n(A, B) \\in \\{(\\emptyset, \\emptyset), (\\{1\\}, \\{1\\}), (\\{1\\}, \\{2\\}), (\\{2\\}, \\{1\\}), (\\{2\\}, \\{2\\}), (\\{1, 2\\}, \\{1, 2\\})\\},\n$$\n\ngiving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$.\n\nLet $\\frac{S_{2022}}{S_{2021}} = \\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p+q$ is divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "For any element $x$ of $\\{1, 2, 3, \\dots, n\\}$, the number of occurrences of $x$ in an intersection of two (not necessarily different) subsets with $k$ elements each is $\\binom{n-1}{k-1}^2$. Using the identity\n\n$$\n\\sum_{k=0}^{n} \\binom{n}{k}^2 = \\sum_{k=0}^{n} \\binom{n}{k} \\binom{n}{n-k} = \\binom{2n}{n},\n$$\n\nit follows that the number of occurrences of $x$ in an intersection of two subsets with the same number of elements is\n\n$$\n\\sum_{k=1}^{n} \\binom{n-1}{k-1}^2 = \\sum_{k=0}^{n-1} \\binom{n-1}{k}^2 = \\binom{2n-2}{n-1}.\n$$\n\nThus\n\n$$\nS_n = n \\binom{2n-2}{n-1}.\n$$\n\nTherefore\n\n$$\n\\frac{S_n}{S_{n-1}} = \\frac{\\binom{2n-2}{n-1} n}{\\binom{2n-4}{n-2} (n-1)} = \\frac{(2n-2)! (n-2)!^2 \\cdot n}{(2n-4)! (n-1)!^2 \\cdot (n-1)} = \\frac{2n(2n-3)}{(n-1)^2},\n$$\n\nwhich is always in lowest terms when $n$ is even because $2n = 2(n-1) + 2$ and $2n - 3 = 2(n - 1) - 1$. Substituting $n = 2022$ yields\n\n$$\n\\frac{S_{2022}}{S_{2021}} = \\frac{4044 \\cdot 4041}{2021^2}.\n$$\n\nWhen $4044 \\cdot 4041 + 2021^2$ is divided by 1000, the remainder is the same as the remainder when $44 \\cdot 41 + 21^2 = 2245$ is divided by 1000. Thus the requested remainder is 245.\n\n![](images/2022AIME_I_Solutions_p10_data_a6d5321432.png)\n\n**Note:** The sequence $S_n$ is sequence A037965 in the On-Line Encyclopedia of Integer Sequences.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18103,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\geq 4$。$M$ 是平面上 $n$ 個點所成的集合，且任三點不共線。在遊戲開始前，在平面上畫 $n$ 條線段，使得每條線段的兩端點都屬於 $M$，且 $M$ 裡的每個點都恰為兩條線段的端點。\n\n接著考慮以下操作：選擇兩個有交點（含端點）的線段 $AB$ 與 $CD$，將這兩條線段擦掉，並畫上 $AC$ 與 $BD$。\n\n試證：我們不可能執行 $\\frac{n^3}{4}$ 或更多次操作。",
    "options": [],
    "answer": "See solution",
    "solution": "（為方便說明起見，以下線段都不包含其兩端點。）\n\n對於平面上的任一條直線，我們說它是“紅”的，若且唯若它包含 $M$ 中的兩點。基於 $M$ 中任三點不共線，每條紅線唯一決定 $M$ 中的兩個點。此外，紅線的數量顯然為 $C_n^2 < \\frac{n^2}{2}$。此外，對於每個線段，令它的“相交數”為有多少條紅線與之相交。對於一組線段，其“相交數”則定義為各線段相交數的總和。我們將證明：\n\n1. 遊戲開始時，全部線段的相交數小於 $\\frac{n^3}{2}$。\n2. 每經過一次操作，全部線段的相交數將至少減 $2$。\n\n基於全部線段相交數永遠不能為負值（否則無法進行操作），以上兩點即證明原命題。\n\n**證明 1.** 基於每條線段的相交數小於 $C_n^2 < \\frac{n^2}{2}$，而共有 $n$ 條線段，故得證。\n\n**證明 2.** 假定我們選取了 $AB$ 和 $CD$，並經過操作改為 $AC$ 與 $BD$。令 $X_{AB}$ 為與 $AB$ 相交的所有紅線所成集合，並以類似方式定義 $X_{CD}$、$X_{AC}$ 和 $X_{BD}$。要證明第二點，我們只需證明：\n\n$$\n|X_{AC}| + |X_{BD}| + 2 \\leq |X_{AB}| + |X_{CD}|, \\quad (1)\n$$\n\n其中 $|X|$ 代表其中的直線個數。\n\n首先，如果一條直線與 $AC$ 相交，則它必然與 $AB$ 或 $CD$ 相交。此外，直線 $AB$ 是線段 $CD$ 的紅線，直線 $CD$ 是線段 $AB$ 的紅線，但它們都不是 $AC$ 或 $BD$ 的紅線（因其交於端點）。故，我們有：\n\n$$\n|X_{AC} \\cup X_{BD}| + 2 \\leq |X_{AB} \\cup X_{CD}|. \\qquad (2)\n$$\n\n另一方面，如果一條直線同時是 $AC$ 與 $BD$ 的紅線，則它無可避免會同時交 $AB$ 和 $CD$。故：\n\n$$\n|X_{AC} \\cap X_{BD}| \\leq |X_{AB} \\cap X_{CD}|. \\qquad (3)\n$$\n\n結合 (2) 與 (3)，我們得到 (1)，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18104,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The answer is negative. \n\nNotice 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 must 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, we use the following lemma:\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\nNow, choose any positive integer $n$, and denote $k = a_n$ 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\n$$\nv_q(2^k - 1) = v_q(2^p - 1) + v_q(k/p) = v_q(2^p - 1) + 0,\n$$\n\nso $v_q(2^k - 1)$ is odd as well. Since $(2^k - 1)\\ell$ is a perfect square, we must have $q \\mid \\ell$, so $g(\\ell) \\ge q > p = g(k)$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18105,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $O_1$ of radius 2 and a circle $O_2$ of radius 4 are externally tangent at the point $P$. Points $A$ and $B$, both distinct from the point $P$, are chosen on the circumference of the circles $O_1$ and $O_2$, respectively, in such a way that the points $A$, $P$, and $B$ are collinear. Determine the length of the line segment $PB$ if the length of $AB$ is 4.\n\n![](images/Japan_2009_p1_data_3009bf1fd0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{8}{3}\n$$\n\nLet $C_1, C_2$ be the centers of the circles $O_1, O_2$, respectively. Since the circles $O_1$ and $O_2$ are externally tangent at the point $P$, the points $C_1$, $P$, and $C_2$ are collinear in this order. Therefore, $\\angle C_1PA = \\angle C_2PB$ holds. As the triangles $C_1PA$ and $C_2PB$ are isosceles, we have $\\angle C_1AP = \\angle C_1PA = \\angle C_2PB = \\angle C_2BP$. Therefore, the triangles $\\triangle C_1PA$ and $\\triangle C_2PB$ are similar, and we get $AP : PB = C_1P : C_2P = 2 : 4$, from which we conclude that\n$$\nPB = \\frac{4}{2+4} \\times AB = \\frac{8}{3}.\n$$\n\n![](images/Japan_2009_p3_data_49214a351a.png)\n\n**Alternate Solution:**\n\nSuppose we draw the graphs of the circles $O_1$ and $O_2$ tangent externally at the point $P$, rotate the picture by $180^\\circ$ around the point $P$, and then enlarge the result 2 times. Then we see that the original circle $O_1$ gets mapped onto the circle which is originally $O_2$, and the point $A$ on the original circle $O_1$ gets mapped onto the point $B$ on the original circle $O_2$. Therefore, we can conclude that $AP : PB = 1 : 2$, and we can deduce as in the preceding argument that $PB = \\frac{8}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18106,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that in every triangle there is a median whose length squared is at least $\\sqrt{3}$ times the area of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that $BC$ is the shortest side of the triangle. Then the least angle of the triangle is at vertex $A$. Denote $a = BC$, $b = CA$, $c = AB$, $\\alpha = \\angle BAC$, and let $m$ be the length of the median drawn from vertex $A$. By our assumption, $\\alpha \\le 60^\\circ$. Let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$, respectively.\n\n![](images/prob1617_p25_data_ae4371b4e5.png)\n\nAs $\\angle AFD = 180^\\circ - \\alpha$ because $DF \\parallel CA$, the law of cosines in triangle $AFD$ gives:\n\n$$\nm^2 = \\left(\\frac{b}{2}\\right)^2 + \\left(\\frac{c}{2}\\right)^2 - 2 \\cdot \\frac{b}{2} \\cdot \\frac{c}{2} \\cos \\angle AFD = \\frac{b^2}{4} + \\frac{c^2}{4} + \\frac{bc}{2} \\cos \\alpha\n$$\n\nSince $\\alpha \\le 60^\\circ$, $\\cos \\alpha \\ge \\cos 60^\\circ = \\frac{1}{2}$, so:\n\n$$\nm^2 \\ge \\frac{b^2}{4} + \\frac{c^2}{4} + \\frac{bc}{2} \\cdot \\frac{1}{2} = \\frac{b^2 + c^2 + bc}{4}\n$$\n\nAs $b^2 + c^2 \\ge 2bc$, we have $m^2 \\ge \\frac{3bc}{4}$.\n\nLet $S$ be the area of triangle $ABC$. Then $S = \\frac{1}{2}bc \\sin \\alpha \\le \\frac{1}{2}bc \\sin 60^\\circ = \\frac{\\sqrt{3}bc}{4}$. From above, $m^2 \\ge \\frac{3bc}{4} = \\sqrt{3} \\cdot \\frac{\\sqrt{3}bc}{4}$. Therefore, $m^2 \\ge \\sqrt{3}S$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18107,
    "subject": "Mathematics (Olympiad)",
    "question": "Annemiek and Bart each have a note on which they have written three different positive integers. There is exactly one number that appears on both their notes. Moreover, if you add any two different numbers from Annemiek's note, you get one of the numbers on Bart's note. One of the numbers on Annemiek's note is her favourite number, and if you multiply it by $3$, you get one of the numbers on Bart's note. Bart's note contains the number $25$, his favourite number.\n\nWhat is Annemiek's favourite number?",
    "options": [],
    "answer": "See solution",
    "solution": "$5$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18108,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/Baltic_Way_SHL_2009-11_13-16_p28_data_f228a782f1.png)\n\nThere is no loss in generality if we assume that $R$ and $D$ are on the same side of $AB$. Show that $A, Q, B$ and $R$ are concyclic. Also, show that $CQDR$ is a cyclic quadrilateral. Denote by $\\Gamma_1$ and $\\Gamma_2$ the circumcircles of quadrilaterals $ARBQ$ and $CQDR$, respectively. Prove that $PA \\cdot PB = PC \\cdot PD$ for point $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consequently, $P$ is a point of the radical axis of $\\Gamma_1$ and $\\Gamma_2$. Since $\\Gamma_1$ and $\\Gamma_2$ meet at $Q$ and $R$, this radical axis is the line $PR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18109,
    "subject": "Mathematics (Olympiad)",
    "question": "A unit L-shape consists of three unit squares as shown in the picture. Prove that for any positive integer $k$ it is possible to cut a similar L-shape with $k$ times larger side lengths into unit L-shapes.\n\n![](images/Estonija_2010_p15_data_33fe2cd023.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the L-shape be placed so that the two longer sides meet at the top left corner. Starting from the top left, we place on it $k$ unit L-shapes diagonally with the same orientation as the large L-shape (see the figure below). The rest consists of two equal staircase-like parts; it is enough to show that one of them, e.g., the lower part, can be covered. The staircase has $k$ stairs, the lowest one at height $k-1$ and the highest at height $2k-2$.\n\nIn case $k=1$, the staircase is empty; in case $k=2$, it can be covered with one unit L-shape. Assume that the claim holds for the staircase with $k$ stairs and consider the staircase with $k+2$ stairs. Separate a strip of width 2 from the left and bottom. The rest can be covered by the induction assumption. The topmost part of the strip is covered with one unit L-shape. Now we have:\n\n![](images/Estonija_2010_p15_data_7eb5ffb4cd.png)\n\n![](images/Estonija_2010_p15_data_d4ba2b79c0.png)\n\n![](images/Estonija_2010_p15_data_9a8983bed8.png)\n\n![](images/Estonija_2010_p15_data_10ee162a4b.png)\n\nto cover the rest of the strip whose lower and left sides have, respectively, the lengths $k+2$ and $2k$.\n\n- If $k$ is divisible by 3, then cut the figure into two strips of sizes $2 \\times 2k$ and $k \\times 2$ and cover both of them with $2 \\times 3$ rectangles consisting of two unit L-shapes (see the figure below), and we are done.\n- If $k \\equiv 1 \\pmod{3}$, then cut the figure into two strips of sizes $2 \\times (2k-2)$ and $(k+2) \\times 2$ and cover both of them with $2 \\times 3$ rectangles. This is possible because $2k-2$ and $k+2$ are divisible by 3.\n- If $k \\equiv 2 \\pmod{3}$, then cut the figure into two strips of sizes $2 \\times (2k-4)$ and $(k-2) \\times 2$, and a corner part, which is an L-shape with $k=2$. Both strips can be covered by $2 \\times 3$ rectangles since $2k-4$ and $k-2$ are divisible by 3; the corner part can be covered by the induction basis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18110,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral $Q$ has a longest side of length $b$ and a shortest side of length $a$. Form a new quadrilateral $Q'$ by joining the successive midpoints of the edges of $Q$. Supposing that $Q$ and $Q'$ are similar, prove that $\\frac{b}{a} < 1 + \\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$Q'$ is a Varignon parallelogram, so $Q$ is also a parallelogram. Let $v$ be the acute or right angle of $Q$. The diagonals $p$ and $q$ of $Q$, which are twice the sides of $Q'$, satisfy\n\n$$\np^2 = a^2 + b^2 - 2ab \\cos v \\quad \\text{and} \\quad q^2 = a^2 + b^2 + 2ab \\cos v.\n$$\n\nThe similarity of $Q$ and $Q'$ yields\n\n$$\n\\frac{a^2 + b^2 - 2ab \\cos v}{a^2 + b^2 + 2ab \\cos v} = \\frac{a}{b},\n$$\n\nwhich simplifies to\n\n$$\nb^2 - a^2 = 2ab \\cos v < 2ab.\n$$\n\nFrom this inequality we deduce $\\left(\\frac{b}{a}\\right)^2 - 1 < 2\\frac{b}{a}$, which is readily seen to imply $\\frac{b}{a} < 1 + \\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18111,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of sequences $\\{a_n\\}_{n=1}^{\\infty}$ of integers satisfying $a_n \\neq -1$ and\n$$\na_{n+2} = \\frac{a_n + 2006}{a_{n+1} + 1}\n$$\nfor each $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\na_3 (a_2 + 1) = a_1 + 2006\n$$\n$$\na_4 (a_3 + 1) = a_2 + 2006\n$$\n$$\na_5 (a_4 + 1) = a_3 + 2006\n$$\n...\nFrom which we get\n$$\na_3 - a_1 = (a_3 + 1)(a_4 - a_2)\n$$\n$$\na_4 - a_2 = (a_4 + 1)(a_5 - a_3)\n$$\n(1)\n$$\na_5 - a_3 = (a_5 + 1)(a_6 - a_4)\n$$\n...\nBy assumption, all numbers $a_n + 1$ are different from zero. Therefore, if $a_3 - a_1 \\neq 0$ then $a_4 - a_2 \\neq 0$, $a_5 - a_3 \\neq 0$, $\\dots$ and by (1) we have\n$$\n0 < |a_{n+3} - a_{n+1}| = |a_{n+2} - a_n| \\cdot \\frac{1}{|a_{n+2} + 1|} \\leq |a_{n+2} - a_n|\n$$\n(2)\nfor $n \\geq 1$. We get a non-increasing sequence of positive numbers\n$$\n|a_3 - a_1| \\geq |a_4 - a_2| \\geq |a_5 - a_3| \\geq \\dots\n$$\nThen it will be constant from a suitable index: thus there are positive integers $N, d$ such that $|a_{n+2} - a_n| = d$ for $n \\geq N$. By (2) we get $|a_{n+2} + 1| = 1$, and so $a_n \\in \\{0, -2\\}$ for each $n \\geq N + 2$. But by definition\n$$\na_{N+4} = \\frac{a_{N+2} + 2006}{a_{N+3} + 1},\n$$\nTherefore $a_{N+4}$ is equal to one of four numbers:\n$$\n\\frac{0 + 2006}{0 + 1} = 2006,\n$$\n$$\n\\frac{0 + 2006}{-2 + 1} = -2006,\n$$\n$$\n\\frac{-2 + 2006}{0 + 1} = 2004,\n$$\n$$\n\\frac{-2 + 2006}{-2 + 1} = -2004\n$$\ncontradiction. Therefore, the $a_3 - a_1 \\neq 0$ case cannot take place.\n\nTherefore $a_3 - a_1 = 0$ and this implies $a_4 - a_2 = 0$, $a_5 - a_3 = 0$, $\\dots$, i.e.\n$$\na_1 = a_3 = a_5 = \\dots \\quad \\text{and} \\quad a_2 = a_4 = a_6 = \\dots \\qquad (3)\n$$\nThe sequence $(a_n)$ takes integer values so $a_1$ and $a_2$ are integers. Now by assumption we have\n$$\na_1 = \\frac{a_1 + 2006}{a_2 + 1} \\iff a_1 a_2 = 2006 = 2 \\cdot 17 \\cdot 59\n$$\nTaking into account the fact that $a_1, a_2 \\neq -1$, we get 14 solutions:\n$$\na_1 \\in \\{1, \\pm 2, \\pm 17, \\pm 34, \\pm 59, \\pm 118, \\pm 1003, 2006\\} \\quad \\text{and} \\quad a_2 = \\frac{2006}{a_1}\n$$\nUsing each of the above values of $a_1$ and equality (3) we can easily verify that any such sequence $a_1, a_2, a_1, a_2, a_1, \\dots$ meets the requirements of the task. Therefore, the sought number is 14. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18112,
    "subject": "Mathematics (Olympiad)",
    "question": "1, 2, ..., 2012 тоонуудаас $4 \\pmod{18}$ байх нийлбэр зохиох боломжийн тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nf(x) = (1 + x)(1 + x^2) \\cdots (1 + x^{2012}) = a_0 + a_1 x + a_2 x^2 + \\cdots + a_n x^n\n$$\nгэвэл $a_k$ нь $1, 2, \\ldots, 2012$ тоонуудын нийлбэр нь $k$ байх сонголтын тоо.\n\n$f(x) \\equiv b_0 + b_1 x + \\cdots + b_{17} x^{17} \\pmod{18}$ хувьд $b_k$ нь $1, 2, \\ldots, 2012$ тоонуудаас $k \\pmod{18}$ байх нийлбэр зохиох боломжийн тоо.\n\n$$\nx^{14} f(x) \\equiv b_0 x^{14} + b_1 x^{15} + b_2 x^{16} + b_3 x^{17} + b_4 + b_5 x + \\cdots + b_{17} x^{13} \\pmod{18}\n$$\n\n$\\varepsilon$-ийг нэгжийн 18 зэргийн язгуур ($m, 18$) $= d$ бол $\\varepsilon^m = \\varepsilon_1$ нь нэгжийн $18/d$ зэргийн язгуур.\n\n$$\n\\sum_{k=0}^{17} (\\varepsilon^k)^m = \\sum_{k=0}^{17} (\\varepsilon^m)^k = d \\cdot \\sum_{k=0}^{17} \\varepsilon_1^k = 0\n$$\n\nТэгэхээр дээрх илэрхийлэлд $x = 1, \\varepsilon, \\varepsilon^2, \\ldots, \\varepsilon^{17}$ гэж орлуулаад нэмэхэд:\n\n$$\n\\sum_{k=0}^{17} (\\varepsilon^k)^{14} f(\\varepsilon^k) = 18 b_4\n$$\n\n$1 + \\varepsilon^9 = 0$ тул $k$ сондгой үед $f(\\varepsilon^k) = 0$ байна. $f(1) = 2^{2012}$. $k = 6, 12$ үед $\\varepsilon^6 = \\varepsilon_1$, $\\varepsilon^{12} = \\varepsilon_1^2 = \\varepsilon_2$.\n\nЭдгээр нь нэгжийн 3 зэргийн язгуур:\n\n$$\nx^2 + x + 1 = (x - \\varepsilon_1)(x - \\varepsilon_1^2), \\quad x = -1 \\text{ гэвэл } (1 + \\varepsilon_1)(1 + \\varepsilon_1^2) = 1\n$$\n\n$$\n\\begin{aligned}\n(\\varepsilon^6)^{14} \\cdot f(\\varepsilon^6) &= \\varepsilon_1^2 \\cdot f(\\varepsilon_1) \\\\\n&= \\varepsilon_1 (1 + \\varepsilon_1)(1 + \\varepsilon_1^2)(1 + \\varepsilon_1^3)^{670}(1 + \\varepsilon_1)(1 + \\varepsilon_1^2) = \\varepsilon_1^2 2^{670}, \\\\\n(\\varepsilon^{12})^{14} f(\\varepsilon^{12}) &= \\varepsilon_1 \\cdot f(\\varepsilon_2) = \\varepsilon_1 2^{670}\n\\end{aligned}\n$$\n\n$$\n(\\varepsilon^6)^{14} \\cdot f(\\varepsilon^6) + (\\varepsilon^{12})^{14} \\cdot f(\\varepsilon^{12}) = (\\varepsilon_1 + \\varepsilon_1^2) 2^{670} = -2^{670}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18113,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles of radius $r_1$, $r_2$, and $r_3$ touch each other externally, and they touch a common tangent at points A, B, and C respectively, where B lies between A and C. Prove that\n\n$$\n16(r_1 + r_2 + r_3) \\ge 9(AB + BC + CA).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A'$ be the foot of the perpendicular from $O_2$ to $AO_1$.\n\n![](images/British_Booklet_2016_p7_data_3dd450073b.png)\n\nSince $A'O_2BA$ is a rectangle, $A'O_2 = AB$. Also, $O_1A' = r_1 - r_2$ and $O_1O_2 = r_1 + r_2$. Hence, by Pythagoras' theorem,\n\n$$\nAB^2 = (r_1 + r_2)^2 - (r_1 - r_2)^2 = 4r_1r_2.\n$$\n\nSimilarly,\n\n$$\nAC^2 = 4r_1r_3\n$$\n\n$$\nBC^2 = 4r_2r_3.\n$$\n\nFurthermore, $AB + BC = AC$, so $\\sqrt{r_1r_3} = \\sqrt{r_1r_2} + \\sqrt{r_2r_3}$. Hence, if we let $a = \\sqrt{r_1}$ and $b = \\sqrt{r_3}$, we learn that\n\n$$\n\\sqrt{r_2} = \\frac{\\sqrt{r_1r_3}}{\\sqrt{r_1} + \\sqrt{r_3}} = \\frac{ab}{a+b}.\n$$\n\nThus, we are required to prove that\n\n$$\n16 \\left( a^2 + b^2 + \\left( \\frac{ab}{a+b} \\right)^2 \\right) \\geq 9 \\left( 2ab + \\frac{2a^2b}{a+b} + \\frac{2ab^2}{a+b} \\right)\n$$\n\nor, in other words,\n\n$$\n4((a^2 + b^2)(a + b)^2 + a^2b^2) \\geq 9ab(a + b)^2.\n$$\n\nSubtracting the RHS from the LHS, we get\n\n$$\n\\begin{aligned}\n& 4((a^2 + b^2)(a + b)^2 + a^2b^2) - 9ab(a + b)^2 \\\\\n&= 4(a^2 + b^2 - 2ab)(a + b)^2 - ab((a + b)^2 - 4ab) \\\\\n&= (a - b)^2(4(a + b)^2 - ab) \\\\\n&= (a - b)^2(4a^2 + 7ab + 4b^2)\n\\end{aligned}\n$$\n\nwhich is clearly nonnegative.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18114,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ be a common chord of different circles $\\Gamma_1$ and $\\Gamma_2$ and let $P$ be a point not collinear with $A$ and $B$. Assume that the line $AP$ meets $\\Gamma_1$ and $\\Gamma_2$ again in $K$ and $L$, respectively, the line $BP$ meets $\\Gamma_1$ and $\\Gamma_2$ again in $M$ and $N$, respectively, and that all the points mentioned so far are different. Let $O_1$ and $O_2$ be the circumcentres of triangles $KMP$ and $LNP$, respectively. Prove that $O_1O_2$ is perpendicular to $AB$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p276_data_3cb7f6da9a.png)\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p276_data_5481fddc1f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since the quadrilaterals $ABMK$ and $ABNL$ are cyclic, $KM$ and $LN$ are both antiparallel to $AB$ with respect to $\\angle APB$, so they are parallel. Triangles $KMP$ and $LNP$ are then homothetic with respect to $P$, so $O_1O_2$ passes through $P$. Now if $PH$ is the height from $P$ in triangle $KMP$ then $PH$ and $PO_1$ are isogonal and therefore antiparallel with respect to $\\angle APB$. Since $KM$ and $AB$ are also antiparallel with respect to this angle and $PH \\perp KM$, we have $PO_1 \\perp AB$ and therefore $O_1O_2 \\perp AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18115,
    "subject": "Mathematics (Olympiad)",
    "question": "$k \\in \\mathbb{N}$, $k < p \\in \\mathbb{P}$ ба $x_1, \\ldots, x_k \\in \\mathbb{Z}$ тоонуудын хувьд\n\n$p \\mid x_1 + \\dots + x_k$, $x_1^2 + \\dots + x_k^2$, $\\ldots$, $x_1^k + \\dots + x_k^k$ бол $p \\mid x_1, \\ldots, x_k$ гэж батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Тэгшитгэлийг $S_n = x_1^n + x_2^n + \\dots + x_k^n$ гэж тэмдэглэе. Ньютоны томёогоор\n\n$$\nS_k - \\sigma_1 S_{k-1} + \\sigma_2 S_{k-2} + \\dots + (-1)^{k-1} \\sigma_{k-1} S_1 + (-1)^k \\cdot k \\sigma_k = 0\n$$\n\n$p \\mid S_i$ $(i \\in \\{1, 2, \\ldots, k\\})$ тул дээрхийг $p$ модулиар авч үзвэл $k\\sigma_k \\equiv 0 \\pmod{p}$ болно. $k < p$ тул $p \\mid \\sigma_k = x_1 x_2 \\cdots x_k$, өөрөөр хэлбэл $x_1, \\ldots, x_k$-ийн аль нэг нь $p$-д хуваагдана.\n\nЖишээ нь $x_k \\equiv 0 \\pmod{p}$ гэж үзвэл $p \\mid x_1 + \\dots + x_{k-1}$, $x_1^2 + \\dots + x_{k-1}^2$, $\\ldots$, $x_1^k + \\dots + x_{k-1}^k$ мөн биелнэ. Ингэж индукцээр буулгахад $p \\mid x_k, p \\mid x_{k-1}, \\ldots, p \\mid x_1$ болохыг харуулна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18116,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть есть три кучки $A$, $B$ и $C$, в которых изначально 100, 101 и 102 камня соответственно. Как Илье ходить, чтобы гарантированно выиграть, независимо от действий Кости?",
    "options": [],
    "answer": "See solution",
    "solution": "**Замечание.** Стратегия, описанная в случае 2, также работает, если Костя своим первым ходом возьмёт камень из $C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18117,
    "subject": "Mathematics (Olympiad)",
    "question": "In a school, there are 1000 students in each year level, from Year 1 to Year 12. The school has 12,000 lockers, numbered from 1 to 12,000. The school principal requests that each student is assigned their own locker, so that the following condition is satisfied:\n\nFor every pair of students in the same year level, the difference between their locker numbers must be divisible by their year-level number.\n\nCan the principal's request be satisfied?",
    "options": [],
    "answer": "See solution",
    "solution": "The request can be satisfied.\n\nFirst, assign the Year 12 students the lockers numbered $12, 24, 36, \\ldots, 12\\,000$. For these students, the difference between any two locker numbers is divisible by 12.\n\nThere are now 11,000 unassigned lockers remaining. By the pigeonhole principle, there are at least 1,000 lockers whose numbers are all in the same equivalence class modulo 11. Assign these lockers to the Year 11 students. Since all these locker numbers have the same remainder modulo 11, the difference of any two is a multiple of 11.\n\nRepeat this process for each year level $k$ from 10 down to 1: for Year $k$, assign 1,000 lockers all in the same equivalence class modulo $k$ from the remaining lockers. When Year 1 is reached, there are exactly 1,000 lockers left, which can be assigned to the Year 1 students.\n\nThus, the principal's request can be satisfied for all students.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18118,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all distinct prime numbers $p$, $q$, and $r$ such that\n\n$$\n3p^4 - 5q^4 - 4r^2 = 26.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice that if both primes $q$ and $r$ differ from $3$, then $q^2 \\equiv r^2 \\equiv 1 \\pmod{3}$, so the left-hand side of the equation is congruent to $0$ modulo $3$, which is impossible since $26$ is not divisible by $3$. Thus, $q=3$ or $r=3$. We consider two cases.\n\n*Case 1.* $q=3$.\n\nThe equation reduces to $3p^4 - 4r^2 = 431$.\n\nIf $p \\neq 5$, by Fermat's little theorem, $p^4 \\equiv 1 \\pmod{5}$, which yields $3 - 4r^2 \\equiv 1 \\pmod{5}$, or equivalently, $r^2 + 2 \\equiv 0 \\pmod{5}$. The last congruence is impossible since a square modulo $5$ can only be $0$, $1$, or $4$. Therefore, $p=5$ and $r=19$.\n\n*Case 2.* $r=3$.\n\nThe equation becomes $3p^4 - 5q^4 = 62$.\n\nObviously, $p \\neq 5$. Hence, Fermat's little theorem gives $p^4 \\equiv 1 \\pmod{5}$. But then $5q^4 \\equiv 1 \\pmod{5}$, which is impossible.\n\nHence, the only solution of the given equation is $p=5$, $q=3$, $r=19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18119,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(n)$ denote the number of all $n$-digit positive integers containing only the digits $1$, $2$, $3$, $4$, $5$ and such that every two adjacent digits differ by at least $2$. Prove that for every positive integer $n$,\n\n$$\n5 \\cdot 2 \\cdot 4^{n-1} \\leq p(n) \\leq 5 \\cdot 2 \\cdot 5^{n-1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Cutting off the last digit of a satisfactory $(n + 1)$-digit integer yields a satisfactory $n$-digit integer. Notice how a satisfactory $(n + 1)$-digit integer can be constructed from a satisfactory $n$-digit integer. If the last digit of the integer is $1$, we can append any of the digits $3, 4, 5$. If the last digit is $2$, we can append $4$ or $5$; if it is $3$, $1$ or $5$ can be appended; if it is $4$, $1$ or $2$ can be appended; and, finally, in the case of $5$, we can append any of the digits $1, 2, 3$. Thus we can see that only the last digit matters.\n\nLet $a_n$ denote the number of satisfactory $n$-digit integers ending in $1$ or $5$; similarly $b_n$ for $2$ or $4$, and $c_n$ for integers ending in $3$. Then $p(n) = a_n + b_n + c_n$. Apparently, $a_1 = b_1 = 2$, $c_1 = 1$, $p(1) = 5 = 5 \\cdot 2.4^0 = 5 \\cdot 2.5^0$, $a_2 = 6$, $b_2 = 4$, $c_2 = 2$, $p(2) = 12 = 5 \\cdot 2.4^1 < 5 \\cdot 2.5^1$.\n\nThe above reasoning implies the recurrent formulae\n\n$$\na_{n+1} = a_n + b_n + 2c_n, \\quad b_{n+1} = a_n + b_n, \\quad c_{n+1} = a_n. $$\n\nHence it follows that $a_3 = 14$, $b_3 = 10$, $c_3 = 6$, $p(3) = 30 \\in (5 \\cdot 2.4^2, 5 \\cdot 2.5^2)$.\n\nUsing mathematical induction, we prove that for every $n \\geq 3$, it holds that\n\n$$\na_n \\geq 2.4^n, \\quad b_n \\geq \\frac{2}{3} \\cdot 2.4^n, \\quad c_n \\geq 2.4^{n-1}.\n$$\n\nIt indeed does for $n = 3$. If $a_n \\geq 2.4^n$, $b_n \\geq \\frac{2}{3} \\cdot 2.4^n$ and $c_n \\geq 2.4^{n-1}$, then also\n\n$$\n\\begin{align*}\na_{n+1} &= a_n + b_n + 2c_n \\geq 2.4^n + \\frac{2}{3} \\cdot 2.4^n + 2 \\cdot 2.4^{n-1} \\\\ &= 2.4^n \\cdot \\left(1 + \\frac{2}{3} + \\frac{5}{6}\\right) = 2.5 \\cdot 2.4^n > 2.4^{n+1}, \\\\ b_{n+1} &= a_n + b_n \\geq 2.4^n + \\frac{2}{3} \\cdot 2.4^n = \\frac{5}{3} \\cdot 2.4^n > \\frac{2}{3} \\cdot 2.4^{n+1}, \\\\ c_{n+1} &= a_n \\geq 2.4^n. \\end{align*}\n$$\n\nIt follows from the proved inequalities that\n\n$$\np(n) = a_n + b_n + c_n \\geq 2.4^n + \\frac{2}{3} \\cdot 2.4^n + 2 \\cdot 2.4^{n-1} = (2.4 + 1.6 + 1) \\cdot 2.4^{n-1} = 5 \\cdot 2.4^{n-1}.\n$$\n\nThe latter inequality can be proved analogously; we will verify that for $n \\geq 3$,\n\n$$\na_n \\leq k \\cdot 2.5^n, \\quad b_n \\leq k \\cdot \\frac{2}{3} \\cdot 2.5^n, \\quad c_n \\leq k \\cdot 2.5^{n-1},\n$$\nwhere $k$ is a suitably chosen number. Then we will have\n\n$$\np(n) = a_n + b_n + c_n \\leq k \\cdot 2.5^{n-1} \\cdot \\left(2.5 + \\frac{5}{3} + 1\\right) = k \\cdot 2.5^{n-1} \\cdot \\frac{31}{6} = 5k \\cdot \\frac{31}{30} \\cdot 2.5^{n-1}.\n$$\n\nTherefore, setting $k = \\frac{30}{31}$, we get $p(n) \\leq 5 \\cdot 2.5^{n-1}$ for every $n \\geq 3$.\n\nIt remains to prove, by mathematical induction, the inequalities above where $k = \\frac{30}{31}$. They hold for $n = 3$. If they hold, we also have\n\n$$\n\\begin{align*}\na_{n+1} &= a_n + b_n + 2c_n \\leq k \\cdot 2.5^n \\cdot \\left(1 + \\frac{2}{3} + \\frac{4}{5}\\right) = k \\cdot 2.5^n \\cdot \\frac{37}{15} < k \\cdot 2.5^{n+1}, \\\\ b_{n+1} &= a_n + b_n \\leq k \\cdot 2.5^n \\cdot \\left(1 + \\frac{2}{3}\\right) = k \\cdot \\frac{2}{3} \\cdot 2.5^{n+1}, \\\\ c_{n+1} &= a_n \\leq k \\cdot 2.5^n. \\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18120,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$, $s$, and $t$ be the roots of the cubic polynomial\n\n$$\np(x) = x^3 - 2007x + 2002.\n$$\n\nDetermine the value of\n\n$$\n\\frac{r-1}{r+1} + \\frac{s-1}{s+1} + \\frac{t-1}{t+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = \\frac{r-1}{r+1} + \\frac{s-1}{s+1} + \\frac{t-1}{t+1}$.\n\n**First Solution:**\n\nNote that $S = 3 - 2R$, where\n$$\nR = \\frac{1}{r+1} + \\frac{1}{s+1} + \\frac{1}{t+1}.\n$$\nThe numbers $r+1$, $s+1$, $t+1$ are the roots of the polynomial\n$$\nq(x) = p(x-1) = x^3 - 3x^2 - 2004x + 4008.\n$$\nThe sum of the reciprocals of the roots of a cubic $f(x) = x^3 + a x^2 + b x + c$ is $-b/c$. Thus,\n$$\nR = -(-2004)/4008 = \\frac{1}{2},\n$$\nso $S = 3 - 1 = 2$.\n\n**Second Solution:**\n\nSince $-1$ is not a root of $p$, $(r+1)(s+1)(t+1) \\neq 0$. The sum is\n$$\n\\frac{(r-1)(s+1)(t+1) + (s-1)(r+1)(t+1) + (t-1)(r+1)(s+1)}{(r+1)(s+1)(t+1)}.\n$$\nExpanding,\n$$\n(r-1)(s+1)(t+1) = (r-1)(st + s + t + 1) = rst + rs + rt - st + r - s - t - 1.\n$$\nSumming cyclically,\n$$\nA = 3rst + (rs + st + tr) - (r + s + t) - 3.\n$$\nAlso,\n$$\nB = (r+1)(s+1)(t+1) = rst + (rs + st + tr) + (r + s + t) + 1.\n$$\nFrom Vieta's formulas for $p(x)$:\n$$\nr + s + t = 0, \\quad rs + st + tr = -2007, \\quad rst = -2002.\n$$\nSo,\n$$\nA = 3(-2002) + (-2007) - 0 - 3 = -6006 - 2007 - 3 = -8016,\n$$\n$$\nB = -2002 - 2007 + 1 = -4008.\n$$\nTherefore, the value is $A/B = 2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18121,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_n$ be the set of all sequences of length $n$, with each entry a whole number between $1$ and $6$. Let $B_n$ be the set of all sequences of length $n$, with each entry either $0$ or $1$. Elements of $S_n$ or $B_n$ are denoted by $x$, and $x_i$ denotes the $i$th entry in $x$.\n\nLet $B$ be the subset of $B_{15}$ consisting of sequences with exactly five entries equal to $1$. What is the cardinality of the set $S$ of nondecreasing sequences in $S_{10}$?",
    "options": [],
    "answer": "See solution",
    "solution": "We construct a bijection between $B$ and $S$ as follows:\n\nDefine a map $f$ from $B$ to $S_{10}$: for $x \\in B$, let $y = f(x)$ be the sequence where $y_1 = 1$ and $y_{i+1} - y_i = x_i$ for $1 \\leq i \\leq 15$. This ensures $y$ is nondecreasing.\n\nNext, delete $y_1$ and also delete $y_i$ if $x_i = 1$. The remaining sequence $z = g(y)$ has length $10$ and is nondecreasing. The map $g$ is injective, and so $h = g \\circ f$ is an injection from $B$ to $S$.\n\nSince every nondecreasing sequence in $S_{10}$ can be obtained this way, $h$ is a bijection. Thus, the cardinality of $S$ is $\\binom{15}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18122,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive real numbers $a_1, a_2, \\dots, a_n$, prove that there exist positive real numbers $x_1, x_2, \\dots, x_n$ such that $\\sum_{i=1}^n x_i = 1$, and that for any positive real numbers $y_1, y_2, \\dots, y_n$ satisfying $\\sum_{i=1}^n y_i = 1$, one has\n\n$$\n\\sum_{i=1}^{n} \\frac{a_i x_i}{x_i + y_i} \\geq \\frac{1}{2} \\sum_{i=1}^{n} a_i.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_i = \\frac{a_i}{\\sum_{i=1}^n a_i}$. Then $\\sum_{i=1}^n x_i = 1$. Moreover,\n\n$$\n\\sum_{i=1}^{n} \\frac{a_i x_i}{x_i + y_i} = \\sum_{i=1}^{n} a_i \\sum_{i=1}^{n} \\frac{x_i^2}{x_i + y_i}.\n$$\n\nFor any positive numbers $y_1, y_2, \\dots, y_n$ with $\\sum_{i=1}^n y_i = 1$. By the Cauchy-Schwarz Inequality,\n\n$$\n2 \\sum_{i=1}^{n} \\frac{x_i^2}{x_i + y_i} = \\sum_{i=1}^{n} (x_i + y_i) \\sum_{i=1}^{n} \\frac{x_i^2}{x_i + y_i} \\geq \\left( \\sum_{i=1}^{n} x_i \\right)^2 = 1.\n$$\n\nHence,\n\n$$\n\\sum_{i=1}^{n} \\frac{a_i x_i}{x_i + y_i} = \\sum_{i=1}^{n} a_i \\sum_{i=1}^{n} \\frac{x_i^2}{x_i + y_i} \\geq \\frac{1}{2} \\sum_{i=1}^{n} a_i. \\quad \\square\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18123,
    "subject": "Mathematics (Olympiad)",
    "question": "Vika chose a 20-letter word that consists only of letters $A$ and $B$. Oleksii wants to know what Vika's word is. He can ask Vika if there are more $A$'s or $B$'s among several (possibly one) consecutive letters of her word. If there are as many $A$'s as there are $B$'s, Vika's answer may be any of the two letters. What is the least number of questions after which Oleksii can guaranteed determine the word Vika chose?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 20.\n\nIt is clear how Oleksii can determine the word in 20 questions—it suffices to ask about each letter separately.\n\nWe want to show that a smaller number of questions would not be enough. Suppose Oleksii determined the word in no more than 19 questions. Suppose Vika chose a word that consists of 20 letters $A$. Clearly, for every question Oleksii asked, Vika's answer was that there are more letters $A$. Since there were less than 20 questions, there is a letter that wasn't asked about separately. Let it be the $t$-th letter. Then consider a word that consists of letters $A$ everywhere, except for the $t$-th letter, which is $B$. Clearly, for both such words Vika could have had the same answers. Thus, Oleksii couldn't have determined which one of these two words Vika chose.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18124,
    "subject": "Mathematics (Olympiad)",
    "question": "Compare the expressions\n\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}}\n$$\n\nand\n\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}.\n$$\n\nFor which value(s) of $n$ does the expression\n\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}\n$$\nhave its smallest possible value?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the inequality\n\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} (n-1)}{10^{n-2}} \\geq \\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}.\n$$\n\nThis holds if and only if\n\n$$\n1 \\geq \\frac{1}{10} \\log_{10} n = \\log_{10} \\sqrt[10]{n},\n$$\nwhich is equivalent to $10 \\geq \\sqrt[10]{n}$, or $10^{10} \\geq n$.\n\nTherefore,\n\n$$\n\\frac{\\log_{10} 2}{10} > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3}{10^2} > \\dots > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10} - 1)}{10^{10^{10}-2}} > \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10})}{10^{10^{10}-1}}.\n$$\n\nFor $n > 10^{10}$, the expression increases:\n\n$$\n\\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10})}{10^{10^{10}-1}} < \\frac{\\log_{10} 2 \\cdot \\log_{10} 3 \\cdots \\log_{10} (10^{10} + 1)}{10^{10^{10}}} < \\dots\n$$\n\nThus, the smallest value occurs for $n = 10^{10} - 1$ or $n = 10^{10}$, and for these values,\n\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} 10^{10}}{10^{10^{10}-1}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18125,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = (-a, 0)$, $B = (a, 0)$, and $C = (k, b)$ be points in the plane. Find the locus of the orthocenter of triangle $ABC$ as $C$ varies.",
    "options": [],
    "answer": "See solution",
    "solution": "The orthocenter lies on the line $x = k$. The line $AC$ has gradient $\\frac{b}{k+a}$, so the altitude from $B$ (perpendicular to $AC$) has gradient $-\\frac{k+a}{b}$ and equation $y + \\frac{(x-a)(k+a)}{b} = 0$. Setting $x = k$, we find $y = -\\frac{(k-a)(k+a)}{b}$. Thus, the locus is the parabola $by = a^2 - x^2$. Since $C$ can have any $x$-coordinate, the orthocenter can be any point on this parabola.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18126,
    "subject": "Mathematics (Olympiad)",
    "question": "On the table there are $k$ heaps of $1, 2, \\dots, k$ stones, where $k \\ge 3$. In the first step, we choose any three of the heaps on the table, merge them into a single new heap, and remove 1 stone (throw it away from the table) from this new heap. In the second step, we again merge some three of the heaps together into a single new heap, and then remove 2 stones from this new heap. In general, in the $i$-th step we choose any three of the heaps, which contain more than $i$ stones when combined, we merge them into a single new heap, and then remove $i$ stones from this new heap. Assume that after a number of steps, there is a single heap left on the table, containing $p$ stones. Show that the number $p$ is a perfect square if and only if the numbers $2k+2$ and $3k+1$ are perfect squares. Further, find the least number $k$ for which $p$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "After $i$ steps, there will be $k - 2i$ heaps left on the table; thus if a single heap is to remain in the end, the number $k$ must be odd and the total number of steps has to be $\\frac{1}{2}(k-1)$. Let us distinguish two cases, according as the remainder of $k$ upon division by 4 is 1 or 3.\n\n*Case 1: $k = 4c + 1$.*\n\nIn the beginning there are $1 + \\dots + k = \\frac{1}{2}k(k+1) = (4c+1)(2c+1)$ stones on the table, from which in course of the $2c$ steps we will remove $1 + \\dots + 2c = c(2c+1)$ stones; thus the number of stones in the last heap will be\n\n$$\np = (4c+1)(2c+1) - c(2c+1) = (2c+1)(3c+1).\n$$\n\nSince the numbers $2c+1$ and $3c+1$ are coprime, $p$ is a perfect square if and only if both $2c+1$ and $3c+1$ are perfect squares; that is, if and only if their quadruples $4(2c+1) = 2k+2$ and $4(3c+1) = 3k+1$ are perfect squares.\n\n*Case 2: $k = 4c + 3$.*\n\nIn the beginning there are $1 + \\dots + k = \\frac{1}{2}k(k+1) = 2(c+1)(4c+3)$ stones on the table, from which we remove in course of all the $2c+1$ steps a total of $1 + \\cdots + (2c + 1) = (c + 1)(2c + 1)$ stones; thus the last single heap will contain\n\n$$\np = 2(c+1)(4c+3) - (c+1)(2c+1) = (c+1)(6c+5).\n$$\n\nIf $p$ were a perfect square, then so would have to be the two coprime numbers $c+1$ and $6c+5$. Let us show that this is not possible. Assume that there exist natural numbers $x, y$ such that $c+1 = x^2$ and $6c+5 = y^2$. From the equality $6x^2 - y^2 = 1$ it follows that the number $y$ is odd, hence $y^2$ gives a remainder of 1 upon division by 8. The number $6x^2$ then gives the remainder of 2, which implies that $3x^2$ gives remainder 1 upon division by 4 — a contradiction. Hence in the case of $k = 4c+3$ the number $p$ is never a perfect square, and likewise the number $3k+1 = 12c+10$ is never a perfect square (being an even number not divisible by 4).\n\nLet us finally find the least number $k = 4c+1$, $c \\ge 1$, for which both numbers $2c+1$ and $3c+1$ are perfect squares. From the equalities $2c+1 = x^2$ and $3c+1 = y^2$ for suitable integers $x, y > 1$ it follows that $3x^2 - 2y^2 = 1$, whence $x$ is odd, but then the number $2y^2$ gives remainder 2 upon division by 4, so $y$ is also odd. Set $x = 2a+1$, $y = 2b+1$ ($a, b > 0$ integer) and substitute this into the equality $3x^2 - 2y^2 = 1$; upon a small manipulation this leads to the relation $3a(a+1) = 2b(b+1)$, and taking successively $a = 1, 2, \\dots$ we soon find the smallest solution $a = 4$ and $b = 5$, corresponding to $x = 9, y = 11, c = 40$ and $k = 161$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18127,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_n$ consists of distinct positive integers. Prove that for any integer $k > 1$, the set $M = \\{a_1^k, a_2^k, a_3^k, \\dots\\}$ does not contain any infinite arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that such a progression exists. Denote it by $\\{b_1^k, b_2^k, b_3^k, \\dots\\}$, where $b_1 < b_2 < b_3 < \\dots$. Let $d$ be the common difference, so for every integer $n$:\n\n$$\nb_{n+1}^k - b_n^k = d.\n$$\n\nBut\n\n$$\nb_{n+1}^k - b_n^k = (b_{n+1} - b_n)\\left(b_{n+1}^{k-1} + b_{n+1}^{k-2}b_n + \\dots + b_{n+1}b_n^{k-2} + b_n^{k-1}\\right).\n$$\n\nThe right side is unbounded as $b_{n+1} \\geq b_n$, and the second factor is at least $b_{n+1}^{k-1}$, which grows without bound. This contradicts the existence of a fixed $d$, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18128,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $n$ can a square of size $n \\times n$ be completely covered (without overlaps) by rectangles of size $k \\times 1$ and one square $1 \\times 1$, where:\n\na) $k = 4$;\n\nb) $k = 8$?\n\n![](images/Ukraine_booklet_2018_p16_data_fe2299c8ab.png)\n![](images/Ukraine_booklet_2018_p16_data_42698ad893.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\na) $n$ is odd and greater than 4.\n\nb) $n = 8m + 9$ and $n = 8m + 15$, where $m \\in \\mathbb{N}$.\n\n**Solution.**\n\nFor both a) and b), it is clear that $n$ must be odd and greater than $k$.\n\n**a)** Thus, $n$ is odd and greater than 4. Any such odd $n$ satisfies the condition. Any stripe $4 \\times l$ can be covered by rectangles $4 \\times 1$. For $5 \\times 5$ and $7 \\times 7$, the required coverage is shown in the figures. For any odd $n = 4m + 5$ or $n = 4m + 7$, the $n \\times n$ square can be cut into a $5 \\times 5$ or $7 \\times 7$ square and several stripes $4 \\times l$.\n\n**b)** By a similar scheme, we can cover squares $n = 8m + 9$ and $n = 8m + 15$. It suffices to cover $9 \\times 9$ and $15 \\times 15$; then, each $n \\times n$ of the above size can be cut into a $9 \\times 9$ or $15 \\times 15$ square and several stripes $8 \\times l$.\n\nTo show that squares $n = 8m + 11$ and $n = 8m + 13$ cannot be covered, consider a coloring of the $11 \\times 11$ square in black and white. Out of 121 squares, 65 are black and 56 are white, so there are 9 more black cells. Each rectangle $8 \\times 1$ covers the same number of black and white cells, so the difference should be at most 1 if such a covering exists. For $n = 8m + 11$, the same argument applies.\n\nSimilarly, for $13 \\times 13$, there are 89 black and 80 white cells, so such a covering is not possible for $n = 8m + 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18129,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $\\mathbb{N}$ the set of positive integers. Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that $x - y$ divides $x^{f(x)} - y^{f(y)}$ for every two coprime integers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that for coprime positive integers $x$ and $y$, we have\n\n$$\nx - y \\mid x^{f(x)} - y^{f(y)} \\iff x - y \\mid x^{|f(x) - f(y)|} - 1.\n$$\n\n*Lemma (Zsigmondy's Theorem):* For $a > b \\ge 1$ coprime integers and $n \\ge 2$, there exists a prime divisor of $a^n - b^n$ that does not divide $a^k - b^k$ for all $1 \\le k < n$, except when $n = 2$ and $a + b$ is a power of $2$ or $(a, b, n) = (2, 1, 6)$.\n\nBack to the problem: For any coprime $x, y$, we show $|f(x) - f(y)| \\le 1$. Suppose $|f(x) - f(y)| > 1$, then there exists a prime $p$ dividing $f(x) - f(y)$. Since $x$ and $y$ are coprime, one of them is not divisible by $p$; assume $\\gcd(x, p) = 1$. By choosing a suitable $z$, we reach a contradiction.\n\nLet $k = v_p(f(x) - f(y))$. By the lemma, for some $N_1 > k$, there exists a prime $q_1$ such that\n\n$$\n\\operatorname{ord}_{q_1}(x) = p^{N_1}, \\quad q_1 > xy.\n$$\n\nSimilarly, for some $N_2 > N_1 > k$, there exists a prime $q_2 > q_1$ with\n\n$$\n\\operatorname{ord}_{q_2}(y) = p^{N_2}.\n$$\n\n$q_1, q_2, x, y$ are pairwise coprime. By the Chinese Remainder Theorem, there exists $z$ such that\n\n$$\n\\begin{cases}\n  z \\equiv 1 \\pmod{x}, & z \\equiv x \\pmod{q_1}, \\\\\n  z \\equiv 1 \\pmod{y}, & z \\equiv y \\pmod{q_2}.\n\\end{cases}\n$$\n\nThese conditions give\n\n$$\n\\begin{cases}\n  q_1 \\mid (z - x) \\mid x^{|f(z) - f(x)|} - 1, \\\\\n  q_2 \\mid (z - y) \\mid y^{|f(z) - f(y)|} - 1.\n\\end{cases}\n$$\n\nHence $p^{N_1} \\mid f(x) - f(z)$ and $p^{N_2} \\mid f(x) - f(y)$. Thus $p^{N_1} \\mid f(x) - f(y)$, a contradiction.\n\nNow, we show $f(x) = f(y)$ for all coprime $x, y > 1$. Choose $z > 2x, 2y$ with $\\gcd(z, xy) = 1$. Then\n\n$$\nz - x > x - 1, \\quad z - y > y - 1.\n$$\n\nBy divisibility, $f(z) - f(x) = 0$ and $f(z) - f(y) = 0$, so $f(x) = f(y)$ for coprime $x, y > 1$. In conclusion, $f$ is constant on $\\{2, 3, \\dots\\}$ and $f(1) \\in \\mathbb{N}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18130,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ with at least 4 factors such that $n$ is the sum of the squares of its 4 smallest factors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a < b < c < d$ be the 4 smallest factors of $n$. Then $n = a^2 + b^2 + c^2 + d^2$.\n\nIf $n$ is odd, then $a, b, c, d$ are all odd, which is impossible. Thus, $n$ is even. Hence $a = 1$, $b = 2$.\n\nIf $4 \\mid n$, then one of $c, d$ is 4 and the other is odd. Taking mod 4, the left side is 0 while the right side is 2, a contradiction. Thus, 4 is not a factor.\n\nNote that $c$ cannot be even; otherwise, $c/2$ is also a factor, which would mean $c$ has to be 2. Thus, $c$ is odd and $d$ is even. Therefore, $d = 2c$.\n\nSo we have:\n$$\nn = 1^2 + 2^2 + c^2 + (2c)^2 = 1 + 4 + c^2 + 4c^2 = 5 + 5c^2.\n$$\nHence, 5 is a factor. If 3 is a factor, then $c = 3$ and $d = 6$, but 6 is not among the 4 smallest factors, a contradiction. Therefore, $c = 5$ and $n = 5(1 + 5^2) = 130$.\n\nIt is easily checked that the smallest 4 factors are 1, 2, 5, 10, and $130 = 1^2 + 2^2 + 5^2 + 10^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18131,
    "subject": "Mathematics (Olympiad)",
    "question": "We call $S \\subseteq \\mathbb{N}_+$ a \"nice set\" if there exists a function $f: S \\to \\mathbb{N}_+$ such that for any $a, b, c \\in S$, the set $\\{a, b, c\\}$ is a geometric sequence if and only if $\\{f(a), f(b), f(c)\\}$ is an arithmetic sequence.\n\n1. Prove that $\\mathbb{N}_+$ is not a nice set.\n\n2. Is $T = \\{x \\in \\mathbb{N}_+ \\mid \\text{for any prime } p,\\ p^{2021} \\text{ does not divide } x\\}$ a nice set? Give your reasons.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Assume that $\\mathbb{N}_+$ is a nice set and $f: \\mathbb{N}_+ \\to \\mathbb{N}_+$ has the desired property. If $x_1, x_2$ satisfy $f(x_1) = f(x_2)$, then $\\{f(x_1), f(x_1), f(x_2)\\}$ is an arithmetic sequence, which implies that $\\{x_1, x_1, x_2\\}$ is a geometric sequence, and thus $x_1 = x_2$. So, $f$ is one-to-one.\n\nNotice that $\\{1, 2, 4, \\dots, 2^k, \\dots\\}$ is an infinite geometric sequence of positive integers. Therefore, $\\{f(1), f(2), f(4), \\dots, f(2^k), \\dots\\}$ must be an infinite arithmetic sequence of positive integers. Let $f(2^k) = f(1) + k d_2$, where the common difference $d_2$ is a positive integer.\n\nLikewise, let $f(3^k) = f(1) + k d_3$, where $d_3$ is a positive integer. Then\n\n$$\nf(2^{d_3}) = f(1) + d_3 d_2 = f(3^{d_2}).\n$$\n\nSince $f$ is one-to-one, $2^{d_3} = 3^{d_2}$, which is impossible. Hence, $\\mathbb{N}_+$ is not a nice set.\n\n2. $T$ is a nice set.\n\nLet all prime numbers be $p_1, p_2, p_3, \\dots$. For any $n \\in T$, let $n = \\prod_{i=1}^{\\infty} p_i^{x_i}$ be the prime factorization, where $x_i \\leq 2020$ is a nonnegative integer for each $i$, and only finitely many $x_i$ are nonzero. Define\n\n$$\nf(n) = 1 + \\sum_{i=1}^{\\infty} x_i \\cdot 4041^i.\n$$\n\nWe show that $f$ satisfies the problem condition.\n\nClearly, $f: T \\to \\mathbb{N}_+$. For any $a, b, c \\in T$ with $a = \\prod_{i=1}^{\\infty} p_i^{\\alpha_i}$, $b = \\prod_{i=1}^{\\infty} p_i^{\\beta_i}$, $c = \\prod_{i=1}^{\\infty} p_i^{\\gamma_i}$, the set $\\{a, b, c\\}$ is geometric if and only if $\\alpha_i + \\gamma_i = 2\\beta_i$ for every $i$.\n\nOn the other hand, $\\{f(a), f(b), f(c)\\}$ is arithmetic if and only if\n\n$$\n\\sum_{i=1}^{\\infty} \\alpha_i \\cdot 4041^i + \\sum_{i=1}^{\\infty} \\gamma_i \\cdot 4041^i = 2 \\sum_{i=1}^{\\infty} \\beta_i \\cdot 4041^i,\n$$\n\nthat is,\n\n$$\n\\sum_{i=1}^{\\infty} (\\alpha_i + \\gamma_i) \\cdot 4041^i = \\sum_{i=1}^{\\infty} 2\\beta_i \\cdot 4041^i.\n$$\n\nSince both $\\alpha_i + \\gamma_i$ and $2\\beta_i$ are nonnegative integers less than or equal to $4040$, this equation gives two representations of the same number in base $4041$, which must agree. Hence, $\\alpha_i + \\gamma_i = 2\\beta_i$ for every $i$. Thus, $\\{a, b, c\\}$ is geometric if and only if $\\{f(a), f(b), f(c)\\}$ is arithmetic. Therefore, $T$ is a nice set.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18132,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers such that $x + y + z = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}$. Prove that $xy + yz + zx \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the given equation and the inequality of arithmetic and geometric means, we obtain\n\n$$\n\\begin{aligned}\nxy + yz + zx &= xyz \\left( \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\right) \\\\\n&= \\frac{xyz \\left( \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\right)^2}{x + y + z} \\\\\n&= \\frac{xyz \\left( \\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2} \\right) + 2x + 2y + 2z}{x + y + z} \\\\\n&\\ge \\frac{xyz \\left( \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\right)}{x + y + z} + 2 = 3.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18133,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ is a positive integer with 3 distinct non-zero digits. Let $g$ be the greatest common divisor of the 6 numbers obtained by permuting the digits of $n$. Determine the maximum possible value that $g$ can take.",
    "options": [],
    "answer": "See solution",
    "solution": "First, let us show that $g$ cannot exceed $18$ for any $n$. Denote by $a, b, c$ the 3 digits of $n$, where we assume $a < b < c$. Both $100c + 10b + a$ and $100c + 10a + b$ are numbers obtained by permuting the digits of $n$. Hence $g$ is a divisor of $$(100c + 10b + a) - (100c + 10a + b) = 9(b - a).$$\nSimilarly, we get that $g$ is a divisor of $9(c - b)$ and of $9(c - a)$. If we set $x = b - a$, $y = c - b$, $z = c - a$, then $x, y, z$ are positive integers not exceeding $8$, and satisfy $x + y = z$. If we denote by $g'$ the greatest common divisor of $x, y, z$, then $g$ is a divisor of $9g'$.\n\n1. If $g' \\geq 5$, there exists at most one number less than or equal to $8$ divisible by $g'$, and we get a contradiction to the fact that $x + y = z$. So, $g' \\geq 5$ is impossible.\n\n2. If $g' = 4$, then $4$ and $8$ are the only positive integers not bigger than $8$ and divisible by $4$, so we must have $(x, y, z) = (4, 4, 8)$. We then have $(a, b, c) = (1, 5, 9)$ and $g = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18134,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上給定三角形 $ABC$ 及一點 $P$。令 $\\triangle ABC$, $\\triangle BPC$, $\\triangle CPA$, $\\triangle APB$ 的外接圓圓心分別為點 $O$, $D$, $E$, $F$。設直線 $BC$ 與 $EF$ 交於點 $T$，而點 $O$ 對直線 $EF$ 的對稱點為 $X$。證明：$PT \\perp DX$。\n\n![](images/19-3J_p3_data_47e15e9517.png)",
    "options": [],
    "answer": "See solution",
    "solution": "令 $X$ 為 $O$ 關於 $EF$ 的對稱點，$Y, Z$ 分別為 $P$ 關於 $EX, FX$ 的對稱點。由於 $A, B, C$ 分別是 $P$ 關於 $EF, FD, DE$ 的對稱點，所以 $O, P$ 是 $\\triangle DEF$ 的等角共轭点，因此\n\n$$\n\\angle FEX = \\angle OEF = \\angle DEP, \\quad \\angle EFX = \\angle OFE = \\angle DFP,\n$$\n\n因此 $D, X$ 是 $\\triangle EPF$ 的等角共轭点。由於 $P$ 關於 $\\angle PEF, \\angle PFE$ 的角平分线的对称点都在 $EF$ 上，所以 $Y, Z$ 分别是 $C, B$ 关于 $EF$ 的对称点，因此 $T$ 在 $\\odot(BPC), \\odot(YPZ)$ 的幂轴上，故 $PT \\perp DX$。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18135,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $m$ a positive integer. Find all pairs $(n, m)$ such that\n$$\nn^{n^n} = m^m.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma.** Let $n$ be a positive integer and $p, q$ positive rational numbers. If $n^p = q$, then $q$ is an integer.\n\n*Proof.* Suppose $p = \\frac{a}{b}$ and $q = \\frac{c}{d}$ with $a, b, c, d \\in \\mathbb{N}$. Then\n$$\nn^p = q \\Rightarrow n^{\\frac{a}{b}} = \\frac{c}{d} \\Rightarrow n^a = \\left(\\frac{c}{d}\\right)^b = \\frac{c^b}{d^b} \\Rightarrow d^b \\mid c^b \\Rightarrow d \\mid c \\Rightarrow q \\in \\mathbb{N}.\n$$\n\nNow for the main problem: If $n = 1$, then $m = n = 1$ is a solution. Assume $n > 1$. Let $r = \\log_n m$ (so $m = n^r$). We have\n$$\nn^{n^n} = m^m = (n^r)^{n^r} = n^{r n^r}.\n$$\nSince $n > 1$, we must have\n$$\nn^n = r n^r \\Rightarrow r = n^{n - r}.\n$$\nAlso, since $n^{n^n} = m^m$, we get $n^n = m \\log_n m = r$, so $r = \\frac{n^n}{m} \\in \\mathbb{Q}$. By the lemma, $r$ is an integer. If $r < n$, then $n^{n - r} \\geq n^1 > r = n^{n - r}$, which is impossible. If $r > n$, then $n^{n - r} < 1 \\leq r = n^{n - r}$, which is also impossible.\n\nSo we must have $n = r$. Hence, $n = r = n^{n - r} = 1$, contradicting $n > 1$. Therefore, the only solution is $m = n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18136,
    "subject": "Mathematics (Olympiad)",
    "question": "Place the following numbers in increasing order of size, and justify your reasoning:\n\n$3^{34}$, $3^{43}$, $3^{44}$, $4^{33}$, and $4^{34}$.\n\nNote that $a^{bc}$ means $a^{(bc)}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, let's compute the values:\n\n- $3^{34}$\n- $3^{43}$\n- $3^{44}$\n- $4^{33}$\n- $4^{34}$\n\nWe can compare the exponents by expressing them in terms of similar bases or exponents.\n\nNotice that $3^{44} > 3^{43} > 3^{34}$ and $4^{34} > 4^{33}$.\n\nNow, let's compare $3^{34}$ and $4^{33}$:\n\n$3^{34}$ vs $4^{33}$:\n\nTake logarithms:\n\n$34 \\log 3$ vs $33 \\log 4$\n\nSince $\\log 3 \\approx 0.4771$ and $\\log 4 = 0.6021$,\n\n$34 \\times 0.4771 = 16.2214$\n\n$33 \\times 0.6021 = 19.8693$\n\nSo $3^{34} < 4^{33}$.\n\nNow compare $4^{33}$ and $3^{43}$:\n\n$33 \\log 4 = 19.8693$\n\n$43 \\log 3 = 20.5153$\n\nSo $4^{33} < 3^{43}$.\n\nNow $3^{43}$ vs $4^{34}$:\n\n$43 \\log 3 = 20.5153$\n\n$34 \\log 4 = 20.4714$\n\nSo $4^{34} < 3^{43}$.\n\nNow $3^{44}$:\n\n$44 \\log 3 = 20.9724$\n\nSo $3^{44} > 3^{43}$ and $3^{44} > 4^{34}$.\n\nThus, the increasing order is:\n\n$$\n3^{34} < 4^{33} < 4^{34} < 3^{43} < 3^{44}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18137,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{Z}^+$ 代表所有正整數所成的集合。試求所有滿足下列條件的滿射函數 $f: \\mathbb{Z}^+ \\times \\mathbb{Z}^+ \\to \\mathbb{Z}^+$：對任意 $a, b, c \\in \\mathbb{Z}^+$，下列三條件均成立：\n\n1. $f(a, b) \\le a + b$\n2. $f(a, f(b, c)) = f(f(a, b), c)$\n3. $\\binom{f(a, b)}{a}$ 及 $\\binom{f(a, b)}{b}$ 都是奇數（其中 $\\binom{n}{k}$ 為二項式係數 $C_k^n$）\n\nLet $\\mathbb{Z}^+$ denote the set of all positive integers. Find all surjective functions $f: \\mathbb{Z}^+ \\times \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ that satisfy all of the following conditions: for all $a, b, c \\in \\mathbb{Z}^+$:\n\n1. $f(a, b) \\le a + b$\n2. $f(a, f(b, c)) = f(f(a, b), c)$\n3. Both $\\binom{f(a, b)}{a}$ and $\\binom{f(a, b)}{b}$ are odd numbers (where $\\binom{n}{k}$ denotes the binomial coefficient $C_k^n$)",
    "options": [],
    "answer": "See solution",
    "solution": "設正整數 $n$ 的二進位表示為 $n = \\sum_{i=1}^{k} 2^{r_i}$。透過此二進位展開，我們可以得到 $\\mathbb{Z}^+$ 與 $\\mathbb{Z}^+ \\cup \\{0\\}$ 的有限非空子集之間的一對一對應：$S_n := \\{r_1, r_2, \\dots, r_k\\}$。於是定義 $f(a, b)$ 為滿足 $S_{f(a,b)} = S_a \\cup S_b$ 的唯一函數。\n\n由於 $f(a, a) = a$，所以 $f$ 為滿射函數。又易知 $f(a, b) \\le a + b$，等號成立的充要條件為 $S_a \\cap S_b = \\emptyset$，故 (1) 成立；(2) 為顯然。由 Lucas 定理知二項式係數 $\\binom{m}{n}$ 為奇數的充要條件是 $S_n \\subseteq S_m$，故 (3) 也成立。\n\n以下證明此函數 $f$ 是唯一解：\n\n**Step 1.** 由 Lucas 定理，$S_{f(a,b)} \\subseteq S_a \\cup S_b$。\n\n**Step 2.** 當 $S_a$ 與 $S_b$ 互斥時，$S_{a+b} = S_a \\cup S_b$。因此得 $S_{f(a,b)} \\subseteq S_{a+b}$，推得 $f(a, b) \\ge a + b$。在此情況下加上 (1) 可知 $f(a, b) = a + b$，即 $S_{f(a,b)} = S_a \\cup S_b$。\n\n**Step 3.** 若 $a, b < 2^k$，我們證明 $f(a, b) < 2^k$：若不然，則存在 $\\ell \\ge k$ 使得 $\\ell \\in S_{f(a,b)}$。此時有 $S_{f(a,b)} \\supseteq S_a \\cup \\{\\ell\\}$，由 Step 2 知 $f(a, b) \\ge a + 2^\\ell > a + b$，矛盾。\n\n**Step 4.** 考慮 $f(a, b) = 2^k$ 的解 $(a_0, b_0)$。由條件 (3) 知 $\\max\\{a_0, b_0\\} \\le f(a_0, b_0) = 2^k$。但根據 Step 3，$a_0, b_0$ 不可能都小於 $2^k$。故 $\\max\\{a_0, b_0\\} = 2^k$，不失一般性可設 $a_0 = 2^k$。如果 $b_0 < 2^k$，則由 Step 2 得到 $f(a_0, b_0) = a_0 + b_0 > 2^k$，不合。而由 $f$ 的滿射性知 $f(2^k, 2^k) = 2^k$。\n\n**Step 5.** 若 $S_a \\cap S_b$ 非空，取 $t \\in S_a \\cap S_b$。由 (2) 及 Step 2 可得\n\n$$\n\\begin{align*}\nf(a, b) &= f(f(a - 2^t, 2^t), f(b - 2^t, 2^t)) \\\\\n&= f(f(a - 2^t, f(2^t, 2^t)), b - 2^t) \\\\\n&= f(f(a - 2^t, 2^t), b - 2^t) \\\\\n&= f(a, b - 2^t).\n\\end{align*}\n$$\n\n故由歸納法可知：若 $S_a \\cap S_b = S_c$，則 $f(a,b) = f(a,b-c) = a+b-c$（Step 2）。因此 $S_{f(a,b)} = S_{a+b-c} = S_a \\cup (S_b \\setminus S_c) = S_a \\cup S_b$。得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18138,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a permutation $a_i$ of the positive integers less than $n$:\n\n- If $a_i < i$, color the $i$-th position green.\n- If $a_i = i$, color the $i$-th position black.\n- If $a_i > i$, color the $i$-th position red.\n\nFind the number of different colorings of the main diagonal such that the first colorful cell is red, the last colorful cell is green, and any black cells may appear in between.",
    "options": [],
    "answer": "See solution",
    "solution": "We can partition such a coloring into consecutive blocks of black, red, and green, with the first block red and the last block green. For each coloring with $k$ black cells (where $k < n-2$), there are $\\binom{n}{k}$ ways to choose the black cells, and $2^{n-k-2}$ ways to assign red/green to the remaining positions (ensuring the first is red and the last is green). Thus, the total number is:\n\n$$\n\\sum_{k=0}^{n-2} \\binom{n}{k} 2^{n-k-2} = 1 + \\frac{3^n - 2n - 1}{4}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18139,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the ellipse $\\frac{x^2}{16} + \\frac{y^2}{4} = 1$ and the line $x - \\sqrt{3}y + 8 + 2\\sqrt{3} = 0$, point $P$ lies on the line $l$. When $\\angle F_1PF_2$ reaches its maximum, what is the value of the ratio $\\frac{|PF_1|}{|PF_2|}$?",
    "options": [],
    "answer": "See solution",
    "solution": "By Euclidean geometry, $\\angle F_1PF_2$ is maximized when the circle through $F_1$, $F_2$, and $P$ is tangent to line $l$ at $P$. Let $l$ intersect the $x$-axis at $A(-8-2\\sqrt{3}, 0)$. Then $\\triangle APF_1 \\sim \\triangle AF_2P$, so\n\n$$\n\\frac{|PF_1|}{|PF_2|} = \\frac{|AP|}{|AF_2|}\n$$\n\nBy the power of a point theorem:\n\n$$\n|AP|^2 = |AF_1| \\cdot |AF_2|\n$$\n\nGiven $F_1(-2\\sqrt{3}, 0)$, $F_2(2\\sqrt{3}, 0)$, $A(-8-2\\sqrt{3}, 0)$:\n\n$$\n|AF_1| = 8, \\quad |AF_2| = 8 + 4\\sqrt{3}\n$$\n\nThus,\n\n$$\n\\frac{|PF_1|}{|PF_2|} = \\sqrt{\\frac{|AF_1|}{|AF_2|}} = \\sqrt{\\frac{8}{8+4\\sqrt{3}}} = \\sqrt{4-2\\sqrt{3}} = \\sqrt{3}-1\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18140,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the prime numbers $p$ and $q$ that satisfy the equality\n\n$$\np^3 + 107 = 2q(17q + 24).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $q = 2$ we obtain $p = 5$, which is a prime.\n\nFor $q \\geq 3$, 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$$\n(p+5)(p^2-5p+25) = 2[q^2 + (4q+3)^2].\n$$\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 $d \\mid q^2 + (4q+3)^2$, 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": 18141,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m_1, m_2, m_3, n_1, n_2,$ and $n_3$ be positive real numbers such that\n\n$$\n(m_1 - n_1)(m_2 - n_2)(m_3 - n_3) = m_1 m_2 m_3 - n_1 n_2 n_3.\n$$\n\nProve that\n\n$$\n(m_1 + n_1)(m_2 + n_2)(m_3 + n_3) \\geq 8 m_1 m_2 m_3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Divide both sides of the given equality by $m_1 m_2 m_3$ and set $a = \\frac{n_1}{m_1}$, $b = \\frac{n_2}{m_2}$, and $c = \\frac{n_3}{m_3}$. The equality $(m_1 - n_1)(m_2 - n_2)(m_3 - n_3) = m_1 m_2 m_3 - n_1 n_2 n_3$ becomes\n\n$$\n(1 - a)(1 - b)(1 - c) = 1 - abc \\iff a + b + c = ab + bc + ca\n$$\n\nand we have to show that\n\n$$\n(a + 1)(b + 1)(c + 1) \\geq 8 \\iff a + b + c + ab + bc + ca + abc \\geq 7. \\quad (1)\n$$\n\nLet $a + b + c = ab + bc + ca = t$. Since $(a + b + c)^2 \\geq 3(ab + bc + ca)$, we have $t^2 \\geq 3t$, i.e., $t \\geq 3$. Furthermore,\n\n$$\n a^3 + b^3 + c^3 \\geq \\frac{(a^2 + b^2 + c^2)^2}{a + b + c} = \\frac{(t^2 - 2t)^2}{t} = t(t - 2)^2\n$$\n\nand\n\n$$\n a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = t(t^2 - 3t) + 3abc\n$$\n\nimply $3abc \\geq t(t-2)^2 - t^2(t-3) = 4t - t^2$. Since (1) is equivalent to $2t + abc \\geq 7$, which is true for $t \\geq \\frac{3}{2}$, it suffices to show that $2t + \\frac{4t - t^2}{3} \\geq 7$ for $t \\in [3, 7]$. The latter is equivalent to $(t-3)(t-7) \\leq 0$, which is true for $t \\in [3, 7]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18142,
    "subject": "Mathematics (Olympiad)",
    "question": "For non-negative $a, b, c$ with $a + b + c = 1$, prove that:\n\n$$\n\\sqrt{1 - a} + \\sqrt{1 - b} + \\sqrt{1 - c} \\leq \\sqrt{2} \\left( \\sqrt{ab + bc + ca} + 2\\sqrt{a^2 + b^2 + c^2} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the Cauchy-Schwarz inequality for the collections $(\\sqrt{\\alpha}, \\sqrt{\\beta}, \\sqrt{\\gamma})$ and $(\\sqrt{\\alpha x}, \\sqrt{\\beta y}, \\sqrt{\\gamma z})$, we obtain:\n\n$$\n(\\alpha \\sqrt{x} + \\beta \\sqrt{y} + \\gamma \\sqrt{z}) \\leq \\sqrt{(\\alpha + \\beta + \\gamma)(\\alpha x + \\beta y + \\gamma z)}.\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n& a \\sqrt{a + b} + b \\sqrt{b + c} + c \\sqrt{c + a} \\leq \\sqrt{a \\sqrt{a + b} + b \\sqrt{b + c} + c \\sqrt{c + a}} \\\\\n& \\leq \\sqrt{(a + b + c)(a^2 + ab + b^2 + bc + c^2 + ca)} \\leq \\sqrt{2(a^2 + b^2 + c^2)}.\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\begin{aligned}\n& b \\sqrt{a + b} + c \\sqrt{b + c} + a \\sqrt{c + a} \\leq \\sqrt{2(a^2 + b^2 + c^2)}, \\\\\n& c \\sqrt{a + b} + a \\sqrt{b + c} + b \\sqrt{c + a} \\leq \\sqrt{2(ab + bc + ca)}.\n\\end{aligned}\n$$\n\nAdding all three inequalities, we get the required one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18143,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 2$ for which there exist real numbers $a_k$ for $1 \\leq k \\leq n$ satisfying the following conditions:\n\n$$\n\\sum_{k=1}^{n} a_k = 0, \\qquad \\sum_{k=1}^{n} a_k^2 = 1, \\qquad \\sqrt{n} \\cdot \\left( \\sum_{k=1}^{n} a_k^3 \\right) = 2(b\\sqrt{n} - 1),\n$$\nwhere $b = \\max_{1 \\leq k \\leq n} \\{a_k\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n$$\n\\left(a_k + \\frac{1}{\\sqrt{n}}\\right)^2 (a_k - b) \\leq 0\n$$\nwhich expands to\n$$\n\\left(a_k^2 + \\frac{2}{\\sqrt{n}} a_k + \\frac{1}{n}\\right) (a_k - b) \\leq 0\n$$\nso\n$$\na_k^3 \\leq \\left(b - \\frac{2}{\\sqrt{n}}\\right) a_k^2 + \\left(\\frac{2b}{\\sqrt{n}} - \\frac{1}{n}\\right) a_k + \\frac{b}{n} \\quad \\forall k \\in \\{1, 2, \\dots, n\\}.\n$$\n\nSumming over $k$:\n$$\n\\sum_{k=1}^{n} a_k^3 \\leq \\left(b - \\frac{2}{\\sqrt{n}}\\right) \\left(\\sum_{k=1}^{n} a_k^2\\right) + \\left(\\frac{2b}{\\sqrt{n}} - \\frac{1}{n}\\right) \\left(\\sum_{k=1}^{n} a_k\\right) + b\n$$\nSince $\\sum_{k=1}^{n} a_k^2 = 1$ and $\\sum_{k=1}^{n} a_k = 0$:\n$$\n\\sum_{k=1}^{n} a_k^3 \\leq b - \\frac{2}{\\sqrt{n}} + b\n$$\nMultiply both sides by $\\sqrt{n}$:\n$$\n\\sqrt{n} \\cdot \\left(\\sum_{k=1}^{n} a_k^3\\right) \\leq 2(b\\sqrt{n} - 1)\n$$\nBut by hypothesis,\n$$\n\\sqrt{n} \\cdot \\left( \\sum_{k=1}^{n} a_k^3 \\right) = 2(b\\sqrt{n} - 1)\n$$\nSo equality must hold, which forces\n$$\na_k^3 = \\left(b - \\frac{2}{\\sqrt{n}}\\right) a_k^2 + \\left(\\frac{2b}{\\sqrt{n}} - \\frac{1}{n}\\right) a_k + \\frac{b}{n} \\quad \\forall k \\in \\{1, 2, \\dots, n\\}\n$$\nThis is equivalent to\n$$\n\\left(a_k + \\frac{1}{\\sqrt{n}}\\right)^2 (a_k - b) = 0 \\quad \\forall k \\in \\{1, 2, \\dots, n\\}\n$$\nSo $a_k \\in \\left\\{ -\\frac{1}{\\sqrt{n}}, b \\right\\}$ for all $k$.\n\nWe show $b > 0$. If $b < 0$, then $0 = \\sum_{k=1}^{n} a_k \\leq n b < 0$, which is impossible. If $b = 0$, then $a_k = 0$ for all $k$, so $\\sum_{k=1}^{n} a_k^2 = 0$, contradicting $\\sum_{k=1}^{n} a_k^2 = 1$. Thus $b > 0$.\n\nIf all $a_k = -\\frac{1}{\\sqrt{n}}$, then $\\sum_{k=1}^{n} a_k = -\\sqrt{n} < 0$, impossible. If all $a_k = b$, then $\\sum_{k=1}^{n} a_k = n b > 0$, impossible. So there exists $m \\in \\{1, 2, \\dots, n-1\\}$ such that $n-m$ of the $a_k$ are $-\\frac{1}{\\sqrt{n}}$ and $m$ are $b$.\n\nSet up the system:\n$$\n\\begin{cases}\n-\\frac{n-m}{\\sqrt{n}} + m b = 0 \\\\\n\\frac{n-m}{n} + m b^2 = 1\n\\end{cases}\n$$\nFrom the first, $b = \\frac{n-m}{m\\sqrt{n}}$. Substitute into the second:\n$$\n\\frac{n-m}{n} + \\frac{(n-m)^2}{m n} = 1\n$$\nThis simplifies to $n-m = m$, so $m = \\frac{n}{2}$. Thus $n$ is even.\n\nConversely, for any even integer $n \\geq 2$, we can choose $a_1 = \\dots = a_{n/2} = -\\frac{1}{\\sqrt{n}}$ and $a_{n/2+1} = \\dots = a_n = \\frac{1}{\\sqrt{n}}$ to satisfy all conditions.\n\n*Answer*: All even integers $n \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18144,
    "subject": "Mathematics (Olympiad)",
    "question": "Evaluate the sum:\n\n$$\n1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335}\n$$\n\nExpress your answer as a reduced fraction $\\frac{p}{q}$, and determine whether $p$ is divisible by 2003.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\n& 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335} \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - 2\\left(\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{1334}\\right) \\\\\n&= 1 + \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{1335} \\\\\n&= \\frac{1}{668} + \\frac{1}{669} + \\dots + \\frac{1}{1335} \\\\\n&= \\left(\\frac{1}{668} + \\frac{1}{1335}\\right) + \\left(\\frac{1}{669} + \\frac{1}{1334}\\right) + \\dots + \\left(\\frac{1}{1001} + \\frac{1}{1002}\\right) \\\\\n&= 2003 \\cdot \\left(\\frac{1}{668 \\cdot 1335} + \\frac{1}{669 \\cdot 1334} + \\dots + \\frac{1}{1001 \\cdot 1002}\\right).\n\\end{align*}\n$$\n\nOne may check that 2003 is a prime. Thus, this factor cannot be cancelled out by the denominator. Thus, $p$ must be divisible by 2003.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18145,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be the centroid of triangle $ABC$. Let $a$, $b$, and $c$ be the lengths of sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Let $v$ be the length of the altitude to side $\\overline{AB}$. Let $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$.\n\n% ![](images/Hrvatska_2015_p22_data_a48612b29f.png)\n\nFind the ratio $P(A'B'C') : P(ABC)$, where $A'$, $B'$, and $C'$ are the feet of the medians from $A$, $B$, and $C$, respectively, and $P(XYZ)$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $T$ is the centroid of triangle $ABC$, we have:\n\n$$\n|TA'| = |B'C| = \\frac{1}{3}|CA| = \\frac{1}{3}b, \\quad |TB'| = |A'C| = \\frac{1}{3}|BC| = \\frac{1}{3}a, \\quad |TC'| = \\frac{1}{3}v.\n$$\n\nThe area of triangle $A'B'C'$ is:\n\n$$\n\\begin{align*}\nP(A'B'C') \n&= P(A'B'T) + P(B'C'T) + P(C'A'T) \\\\\n&= \\frac{1}{2} (|TA'| \\cdot |TB'| + |TB'| \\cdot |TC'| \\sin(\\pi - \\alpha) + |TC'| \\cdot |TA'| \\sin(\\beta - \\alpha)) \\\\\n&= \\frac{1}{18} (ab + av \\sin \\alpha + bv \\sin \\beta).\n\\end{align*}\n$$\n\nUsing the relations $v = a \\sin \\beta$, $v = b \\sin \\alpha$, $a = c \\sin \\alpha$, $b = c \\sin \\beta$, and $c^2 = a^2 + b^2$, we get:\n\n$$\n\\begin{align*}\nP(A'B'C') &= \\frac{1}{18}(ab + a^2 \\sin \\alpha \\sin \\beta + b^2 \\sin \\alpha \\sin \\beta) \\\\\n&= \\frac{1}{18}(ab + c^2 \\sin \\alpha \\sin \\beta) = \\frac{1}{18}(ab + ab) \\\\\n&= \\frac{1}{9}ab = \\frac{2}{9}P(ABC).\n\\end{align*}\n$$\n\nTherefore, $P(A'B'C') : P(ABC) = 2 : 9$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18146,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be distinct positive integers such that $3^a + 2$ is divisible by $3^b + 2$. Prove that $a > b^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously we have $a > b$. Let $a = bq + r$, where $0 \\le r < b$. Then\n\n$$\n3^a \\equiv 3^{bq + r} \\equiv (-2)^q \\cdot 3^r \\equiv -2 \\pmod{3^b + 2}\n$$\n\nSo $3^b + 2$ divides $A = (-2)^q \\cdot 3^r + 2$ and it follows that\n\n$$\n|(-2)^q \\cdot 3^r + 2| \\ge 3^b + 2 \\text{ or } (-2)^q \\cdot 3^r + 2 = 0.\n$$\n\nWe make a case distinction:\n\n1. $(-2)^q \\cdot 3^r + 2 = 0$. Then $q = 1$ and $r = 0$ or $a = b$, a contradiction.\n\n2. $q$ is even. Then\n\n$$\nA = 2^q \\cdot 3^r + 2 = (3^b + 2)k.\n$$\n\nConsider both sides modulo $3^r$. Since $b > r$:\n\n$$\n2 \\equiv 2^q \\cdot 3^r + 2 = (3^b + 2)k \\equiv 2k \\pmod{3^r},\n$$\n\nso $3^r \\mid k - 1$. If $k = 1$ then $2^q \\cdot 3^r = 3^b$, a contradiction. So $k \\ge 3^r + 1$, and therefore:\n\n$$\nA = 2^q \\cdot 3^r + 2 = (3^b + 2)k \\ge (3^b + 2)(3^r + 1) > 3^b \\cdot 3^r + 2\n$$\n\nIt follows that\n\n$$\n2^q \\cdot 3^r > 3^b \\cdot 3^r, \\text{ i.e. } 2^q > 3^b, \\text{ which implies } 3^{b^2} < 2^{bq} < 3^{bq} \\le 3^{bq + r} = 3^a.\n$$\n\nConsequently $a > b^2$.\n\n3. If $q$ is odd. Then\n\n$$\n2^q \\cdot 3^r - 2 = (3^b + 2)k.\n$$\n\nConsidering both sides modulo $3^r$, and since $b > r$, we get: $k + 1$ is divisible by $3^r$ and therefore $k \\ge 3^r - 1$. Thus $r > 0$ because $k > 0$, and:\n\n$$\n2^q \\cdot 3^r - 2 = (3^b + 2)k \\ge (3^b + 2)(3^r - 1), \\text{ and therefore}\n$$\n\n$$\n2^q \\cdot 3^r > (3^b + 2)(3^r - 1) > 3^b(3^r - 1) > 3^b \\frac{3^r}{2}, \\text{ which shows}\n$$\n\n$$\n2^q + 1 > 3^b.\n$$\n\nBut for $q > 1$ we have $2^q + 1 < 3^q$, which combined with the above inequality, implies that $3^{b^2} < (2^q + 1)^b < 3^{qb} \\le 3^a$, q.e.d. Finally, if $q = 1$ then $2^q \\cdot 3^r - 2 = (3^b + 2)k$ and consequently $2 \\cdot 3^r - 2 \\ge 3^b + 2 \\ge 3^{r + 1} + 2 > 2 \\cdot 3^r - 2$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18147,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $PQRS$ be a cyclic quadrilateral. Let $H$ be the intersection point of $PQ$ and $RS$. Let $T'$ be an arbitrary point on $PQ$. Points $M$ and $N$ are on $PR$ and $QS$ respectively such that $T'M \\parallel QR$ and $T'N \\parallel PS$. Let $T$ be a point such that $\\angle T'MT = \\angle T'NT = 90^\\circ$. Show that as $T'$ varies on the line $PQ$, the point $T$ varies on a line passing through $H$. Furthermore, if this line is $RS$, then $PS \\perp QR$.\n\n![](images/2018_p31_data_eb9a9dd739.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are three points $T'_1, T'_2, T'_3$ on $PQ$. Obviously, we have\n\n$$\n\\frac{N_1 N_2}{N_1 N_3} = \\frac{T'_1 T'_2}{T'_1 T'_3} = \\frac{M_1 M_2}{M_1 M_3}\n$$\n\nLet $A_1, A_2, A_3$ be the intersection points of $(T_1M_1, T_3N_3)$, $(T_1M_1, T_2N_2)$, $(T_2M_2, T_3N_3)$ respectively.\n\nBy the intercept theorem, we have\n\n$$\n\\frac{A_1 A_2}{A_1 T_1} = \\frac{A_1 A_3}{A_1 T_3}, \\quad T_2 A_2 \\parallel A_1 T_3, \\quad T_2 A_3 \\parallel A_1 T_1\n$$\n\nSo we conclude that $T_2$ lies on $T_1T_3$.\n\nWe have proved that $T$ varies on a line. Now we prove that this line passes through $H$.\n\nSuppose that $SP$ and $QR$ are not perpendicular. Then there exist points $F, X', Y$. Set $T'_1 = Q$, $T'_2 = P$. So we have $T_1 = X'$, $T_2 = Y$. Let $E, Z$ be the intersection points of $(PS, QR)$ and $(RX', SY)$ respectively. Obviously, $EF \\perp PQ$, and $\\angle ZES = 90^\\circ - \\angle ERS$. This means that\n\n$$\n\\angle ZES = 90^\\circ - \\angle EPQ \\implies ZE \\perp PQ\n$$\n\nSo $E, F, Z$ are collinear. From Desargue's theorem in triangles $SZR$ and $PFQ$, we obtain that $PQ$, $RS$, and $X'Y$ are concurrent. So $X'Y$ passes through the intersection point of $PQ$ and $RS$.\n\nNow suppose that $PQ \\perp RS$. Then if we set $T'_1 = P$, $T'_2 = Q$, then $T_1 = S$, $T_2 = R$, which implies that $RS$ passes through $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18148,
    "subject": "Mathematics (Olympiad)",
    "question": "On the base $BC$ of the isosceles triangle $ABC$, points $D$ and $E$ are chosen such that $BD = EC = 7$ cm. A line perpendicular to side $AB$ and passing through point $A$ intersects base $BC$ at point $F$. An altitude of triangle $ACE$ is $EG$, a median of triangle $AEF$ is $AH$. It is known that $\\angle DAF = \\frac{1}{2}\\angle BAC$. Find the length of the segment $GH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since triangle $BAC$ is isosceles, its base angles are equal; denote $\\angle ABC = \\angle BCA = \\beta$. Since the sum of the interior angles of triangle $BAC$ is $180^\\circ$, we have $\\angle BAC = 180^\\circ - 2\\beta$. Then\n$$\n\\angle DAF = \\frac{180^\\circ - 2\\beta}{2} = 90^\\circ - \\beta\n$$\nbecause $\\angle DAF = \\frac{1}{2}\\angle BAC$. Since $FA$ is perpendicular to $BA$,\n$$\n\\angle BAD = \\angle BAF - \\angle DAF = 90^\\circ - (90^\\circ - \\beta) = \\beta.\n$$\nThus triangle $ADB$ is isosceles, because angles $DBA$ and $BAD$ are equal. Consequently $AD = BD = 7$ cm.\n\n![](images/EST_ABooklet_2024_p36_data_2466905853.png)\n\nAs points $D$ and $E$ are symmetric with respect to the perpendicular bisector of $BC$ and so are points $B$ and $C$, we also have $AE = CE = 7$ cm. Hence the triangle $AEC$ is isosceles, too. From triangle $BAF$ we now get\n$$\n\\angle AFB = \\angle AFD = 180^\\circ - 90^\\circ - \\beta = 90^\\circ - \\beta.\n$$\nThus triangle $DAF$ is isosceles, because angles $DAF$ and $DFA$ are equal. Hence $DF = 7$ cm. Therefore $DF = EC$.\n\nAs the altitude drawn from the apex of an isosceles triangle bisects the base, $G$ must be the midpoint of $AC$. Since $H$ is a midpoint of $EF$ and $DF = EC$, $H$ is also a midpoint of $DC$. In conclusion, $HG$ is a midsegment of triangle $ACD$ parallel to side $AD$. So\n$$\nHG = \\frac{1}{2}AD = 3.5 \\text{ cm}.\n$$\n\n![](images/EST_ABooklet_2024_p36_data_bbf3b009ca.png)\n\n![](images/EST_ABooklet_2024_p36_data_67f1c4435f.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18149,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D = \\{1, 2, \\dots, n\\} \\times \\{1, 2, \\dots, n\\}$. Prove that 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": "See 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 to look at the diagonals of the tableau!\n\n![](images/RMC2011_2_p80_data_420cb6587d.png)\n\nDiagram for the selection $\\bullet$ of $S_u$ (exact for $n = 13$).\n\nDefine the diagonal $\\Delta_k = \\{(x, y) \\in \\mathbb{N} \\times \\mathbb{N} \\mid x + y = k\\}$, for all $k \\ge 0$. It is clear that if $(x_1, y_1) \\in \\Delta_{k_1}$ and $(x_2, y_2) \\in \\Delta_{k_2}$, then $(x_1 + x_2, y_1 + y_2) \\in \\Delta_{k_1 + k_2}$.\n\nTherefore, $S_u := \\bigcup_{u \\le k \\le 2u-1} (\\Delta_k \\cap D)$, for $u < n+1 \\le 2u-1$, has the defining property. Since $|\\Delta_k \\cap D| = k-1$ for $2 \\le k \\le n+1$, and $|\\Delta_k \\cap D| = 2n - (k-1)$ for $n+1 \\le k \\le 2n$, it follows that\n$$\n|S_u| = \\sum_{k=u}^{n} (k-1) + \\sum_{k=n+1}^{2u-1} (2n - (k-1)) = -\\frac{1}{2}(5u^2 - (8n+9)u + 2(n+1)(n+2)).\n$$\nThus, the maximal value for $|S_u|$ is obtained at the nearest integer $v$ to $\\frac{8n+9}{10}$, thus at $v = \\lfloor \\frac{4n+7}{5} \\rfloor$, for which $|S_v| = \\lfloor \\frac{3}{5} n(n+1) \\rfloor$.\n\n**REMARKS.** The set $D$ is a product poset of chains, also seen as being a *graded* poset -- with elements appearing in its *Hasse diagram* on *levels* by their *rank*; the rank being precisely the sum of the coordinates (as some justification).\n\nActually, it is a rank-unimodal and rank-symmetric poset; a seminal result is the de Bruijn-Tengbergen-Kruyswijk theorem. In common algebraic-combinatorial jargon, the set $S$ is *sum-free*, or $(S+S) \\cap S = \\emptyset$ (Minkowski sumset notation).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18150,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ and $m$ satisfy the equation $n + \\dfrac{n(n-1)(n-2)}{6} = 2^m$. Find all pairs of positive integers $(m, n)$ that satisfy this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "We start by rearranging the equation:\n\n$$\nn + \\frac{n(n-1)(n-2)}{6} = 2^m\n$$\n\nMultiply both sides by 6:\n\n$$\n6n + n(n-1)(n-2) = 6 \\cdot 2^m\n$$\n\nExpand $n(n-1)(n-2)$:\n\n$$\nn^3 - 3n^2 + 2n + 6n = 6 \\cdot 2^m\n$$\n$$\nn^3 - 3n^2 + 8n = 6 \\cdot 2^m\n$$\n\nFactor $n$:\n\n$$\nn(n^2 - 3n + 8) = 6 \\cdot 2^m\n$$\n\nRewrite $6 \\cdot 2^m = 3 \\cdot 2^{m+1}$:\n\n$$\nn(n^2 - 3n + 8) = 3 \\cdot 2^{m+1}\n$$\n\n**Case 1:** $3$ divides $n$.\n\nLet $n = 3 \\cdot 2^a$ and $n^2 - 3n + 8 = 2^b$, with $a, b \\geq 0$ and $a + b = m + 1$.\n\nCheck divisors of 24 divisible by 3: $n = 3, 6, 12, 24$.\n\nFor these $n$, compute $n + \\dfrac{n(n-1)(n-2)}{6}$:\n\n- $n = 3$: $3 + \\dfrac{3 \\cdot 2 \\cdot 1}{6} = 3 + 1 = 4 = 2^2$ ($m = 2$)\n- $n = 6$: $6 + \\dfrac{6 \\cdot 5 \\cdot 4}{6} = 6 + 20 = 26$ (not a power of 2)\n- $n = 12$: $12 + \\dfrac{12 \\cdot 11 \\cdot 10}{6} = 12 + 220 = 232$ (not a power of 2)\n- $n = 24$: $24 + \\dfrac{24 \\cdot 23 \\cdot 22}{6} = 24 + 2024 = 2048 = 2^{11}$ ($m = 11$)\n\nSo, solutions: $(m, n) = (2, 3), (11, 24)$.\n\n**Case 2:** $3$ does not divide $n$.\n\nLet $n = 2^a$ and $n^2 - 3n + 8 = 3 \\cdot 2^b$, with $a, b \\geq 0$ and $a + b = m + 1$.\n\nCheck divisors of 8: $n = 1, 2, 4, 8$.\n\n- $n = 1$: $1 + 0 = 1 = 2^0$ ($m = 0$; exclude since $m > 0$)\n- $n = 2$: $2 + 0 = 2 = 2^1$ ($m = 1$)\n- $n = 4$: $4 + \\dfrac{4 \\cdot 3 \\cdot 2}{6} = 4 + 4 = 8 = 2^3$ ($m = 3$)\n- $n = 8$: $8 + \\dfrac{8 \\cdot 7 \\cdot 6}{6} = 8 + 56 = 64 = 2^6$ ($m = 6$)\n\nSo, solutions: $(m, n) = (1, 2), (3, 4), (6, 8)$.\n\n**Final list of solution pairs $(m, n)$ with positive $m, n$:**\n\n$$(1, 2),\\ (2, 3),\\ (3, 4),\\ (6, 8),\\ (11, 24)$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18151,
    "subject": "Mathematics (Olympiad)",
    "question": "All integers from 1 to 12 are written on the edges of a cube so that every edge has exactly one integer. Two robot ants stand at the same vertex of the cube and wish to arrive at the vertex that is furthest away from them. Each ant picks a path consisting of exactly three edges of the cube.\n\nThey multiply the integers on their path. The product of the integers on the path of the first ant is divisible by $100$, but not by $200$. The sum of the digits of the product of the integers on the path of the second ant is $2$ and the integer $1$ is not on its path. Find all possible products that either ant can get.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $100 = 2^2 \\cdot 5^2$, and none of the numbers on the edges of the cube are divisible by $5^2$, the path of the first ant must contain two numbers divisible by $5$. The only such numbers are $5$ and $10$. Their product is divisible by $2$, but not by $2^2$. The third number on the path must be divisible by $2$ but not by $2^2$; these are $2$, $6$, and $10$. Since $10$ is already used, the third number must be $2$ or $6$.\n\nFor the second ant, the sum of the digits of the product must be $2$. Possible such numbers (up to $12 \\cdot 11 \\cdot 10 = 1320$) are $2$, $11$, $20$, $101$, $110$, $200$, $1001$, $1010$, and $1100$. Of these, $2$, $11$, and $101$ are prime and cannot be the product of three integers greater than $1$. $20$ can only be written as $2 \\cdot 2 \\cdot 5$, which repeats $2$. $1001 = 7 \\cdot 11 \\cdot 13$ (no $13$ on the cube), $1010 = 2 \\cdot 5 \\cdot 101$ (no $101$), and $1100$ would require $5$, $10$, and $22$ (no $22$). $200$ can be $5 \\cdot 10 \\cdot 4$, and $110$ can be $2 \\cdot 5 \\cdot 11$.\n\nHowever, the paths of the two ants can have $0$, $1$, or $3$ edges in common, but not exactly $2$. Thus, the numbers $5$, $10$, and $4$ (for $200$) cannot be used, so the second ant must have $2$, $5$, and $11$ (for $110$). This leaves the first ant with $5$, $6$, and $10$, giving a product of $300$.\n\n**Answer:** The first ant gets $300$, the second one gets $110$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18152,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(\\lfloor x \\rfloor y) = f(x) \\lfloor f(y) \\rfloor.\n$$\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Put $y = 0$ to find $f(0) = f(x) \\lfloor f(0) \\rfloor$ for all $x \\in \\mathbb{R}$.\n\n**Case 1.** $\\lfloor f(0) \\rfloor \\neq 0$. Then $f(x) = \\frac{f(0)}{\\lfloor f(0) \\rfloor} = c$ for some non-zero constant $c$. Substitution yields $c = c \\lfloor c \\rfloor$ so that $\\lfloor c \\rfloor = 1$. Note that this is a valid solution.\n\n**Case 2.** $\\lfloor f(0) \\rfloor = 0$. From $f(0) = f(x) \\lfloor f(0) \\rfloor$ we obtain $f(0) = 0$.\n\n**Case 2a.** There exists a real number $x$ such that $\\lfloor x \\rfloor \\neq 0$ but $f(x) = 0$. Then if $A$ is any real number, set $y = \\frac{A}{\\lfloor x \\rfloor}$ in the original functional equation to find $f(A) = f(x) \\lfloor f(y) \\rfloor = 0$. Thus $f$ is identically zero in this case. Note that this is a valid solution.\n\n**Case 2b.** If $f(x) = 0$ then $\\lfloor x \\rfloor = 0$. Thus $f(1) \\neq 0$. Put $x = y = 1$ into the original functional equation to find $f(1) = f(1) \\lfloor f(1) \\rfloor$. Thus $\\lfloor f(1) \\rfloor = 1$. Set $y = 1$ to find $f(\\lfloor x \\rfloor) = f(x)$. In particular $f(\\frac{1}{2}) = 0$. Finally let $x = 2$ and $y = \\frac{1}{2}$ to find $f(1) = f(2) \\lfloor f(\\frac{1}{2}) \\rfloor = 0$. This contradicts $f(1) \\neq 0$. So this case does not occur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18153,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}^+$ be the set of all positive integers. Find all functions $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that\n\n1. $f(n!) = f(n)!$ for all $n \\in \\mathbb{Z}^+$,\n2. $m-n$ divides $f(m)-f(n)$ for all distinct positive integers $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are three functions: the constant functions $f(n) = 1$, $f(n) = 2$, and the identity function $f(n) = n$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that these are the only ones.\n\nConsider such a function $f$ and suppose that it has a fixed point $a \\ge 3$, that is, $f(a) = a$. Then $a!$, $(a!)!$, ... are all fixed points of $f$, hence the function $f$ has a strictly increasing sequence $a_1 < a_2 < \\dots < a_k < \\dots$ of fixed points. For a positive integer $n$, $a_k - n$ divides $a_k - f(n) = f(a_k) - f(n)$ for every $k \\in \\mathbb{Z}^+$. Also, $a_k - n$ divides $a_k - n$, so it divides $a_k - f(n) - (a_k - n) = n - f(n)$. This is possible only if $f(n) = n$, hence in this case we get $f = \\text{id}_{\\mathbb{Z}^+}$.\n\nNow suppose that $f$ has no fixed points greater than $2$. Let $p \\ge 5$ be a prime and notice that by Wilson's Theorem we have $(p-2)! \\equiv 1 \\pmod{p}$. Therefore $p$ divides $(p-2)! - 1$. But $(p-2)! - 1$ divides $f((p-2)!)-f(1)$, hence $p$ divides\n\n$$\nf((p-2)!)-f(1) = (f(p-2))! - f(1).\n$$\n\nClearly we have $f(1) = 1$ or $f(1) = 2$. As $p \\ge 5$, the fact that $p$ divides $(f(p-2))! - f(1)$ implies that $f(p-2) < p$. It is easy to check, again by Wilson's Theorem, that $p$ does not divide $(p-1)! - 1$ and $(p-1)! - 2$, hence we deduce that $f(p-2) \\le p-2$. On the other hand, $p-3 = (p-2) - 1$ divides $f(p-2) - f(1) \\le (p-2) - 1$. Thus either $f(p-2) = f(1)$ or $f(p-2) = p-2$. As $p-2 \\ge 3$, the last case is excluded, since the function $f$ has no fixed points greater than $2$. It follows $f(p-2) = f(1)$ and this property holds for all primes $p \\ge 5$.\n\nTaking $n$ as any positive integer, we deduce that $p-2-n$ divides $f(p-2) - f(n) = f(1) - f(n)$ for all primes $p \\ge 5$. Thus $f(n) = f(1)$, hence $f$ is the constant function $1$ or $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18154,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_0A_1A_2$ be a triangle and let $P$ be a point in the plane, not situated on the circle $A_0A_1A_2$. The line $PA_k$ meets again the circle $A_0A_1A_2$ at point $B_k$, $k = 0, 1, 2$. A line $\\ell$ through the point $P$ meets the line $A_{k+1}A_{k+2}$ at point $C_k$, $k = 0, 1, 2$. Show that the lines $B_kC_k$, $k = 0, 1, 2$, are concurrent and determine the locus of their concurrency point as the line $\\ell$ turns about the point $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "The lines $B_kC_k$ and $B_{k+1}C_{k+1}$ meet projectively at a point $Q_k$. Considering $PC_kC_{k+1}$, the line $\\ell$ is the Pascal line of the hexagram\n\n$$\nA_k B_k Q_k B_{k+1} A_{k+1} A_{k+2},\n$$\n\nso the vertices of the latter lie on a conic. Notice that this conic and the circle $A_0A_1A_2$ share five points, so they must coincide, and thus the three lines $B_kC_k$ are concurrent at a point $Q = Q_k$ situated on the circle $A_0A_1A_2$.\n\nConversely, given a point $Q$ on the circle $A_0A_1A_2$, the three hexagrams $A_kB_kQB_{k+1}A_{k+1}A_{k+2}$ share the same Pascal line $\\ell = PC_kC_{k+1}$, so the required locus is the circle $A_0A_1A_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18155,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f_0$, $f_1$, $f_2$, and $f_3$ be polynomials in $\\mathbb{R}[X]$ such that $f_k(1) = f_{k+1}(0)$ for $k = 0, 1, 2, 3$ (indices are reduced modulo $4$). Show that there exists a polynomial $f$ in $\\mathbb{R}[X, Y]$ such that:\n\n- $f(X, 0) = f_0(X)$\n- $f(1, Y) = f_1(Y)$\n- $f(1 - X, 1) = f_2(X)$\n- $f(0, 1 - Y) = f_3(Y)$",
    "options": [],
    "answer": "See solution",
    "solution": "The idea is to construct $f(X, Y)$ as a suitable combination of the given polynomials and corrective terms:\n\nConsider\n$$(1 - Y)f_0(X) + Y f_2(1 - X)$$\nas an $\\mathbb{R}[Y]$-linear combination, and\n$$X f_1(Y) + (1 - X) f_3(1 - Y)$$\nas an $\\mathbb{R}[X]$-linear combination. Add a degree $2$ corrective term $a_{11} X Y + a_{10} X + a_{01} Y + a_{00}$.\n\nSet\n$$\nf(X, Y) = (1 - Y)f_0(X) + X f_1(Y) + Y f_2(1 - X) + (1 - X) f_3(1 - Y) + a_{11} X Y + a_{10} X + a_{01} Y + a_{00}.\n$$\n\nImposing the conditions from the problem, we find:\n- $a_{10} = f_0(0) - f_1(0)$\n- $a_{00} = -f_0(0)$\n- $a_{01} = f_0(0) - f_3(0)$\n- $a_{11} = f_1(0) - f_2(0) - f_0(0) + f_3(0)$\n\nThus, the desired polynomial is\n$$\nf(X, Y) = (1 - Y)f_0(X) + X f_1(Y) + Y f_2(1 - X) + (1 - X) f_3(1 - Y) - (f_0(0) - f_1(0) + f_2(0) - f_3(0)) X Y + (f_0(0) - f_1(0)) X + (f_0(0) - f_3(0)) Y - f_0(0).\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18156,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle and let $J$ be the center of the excircle opposite to $A$. The reflection of $J$ in $BC$ is $K$. $E$ and $F$ are on $BJ$ and $CJ$, respectively, such that $\\angle EAB = \\angle CAF = 90^\\circ$. Prove that $\\angle FKE + \\angle FJE = 180^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/BW23_Shortlist_2023-11-01_p71_data_ce017bca60.png)\n\nLet $JK$ intersect $BC$ at $X$. We'll prove a key claim:\n\n**Claim:** $\\triangle BEK$ is similar to $\\triangle BAX$.\n\n**Proof.** Note that $\\angle EAB = 90^\\circ = \\angle KXB$. Also, since $BJ$ bisects $\\angle CBA$, we get $\\angle ABE = \\angle JBX = \\angle XBK$. Hence $\\triangle EBA \\sim \\triangle KBX$. From that, we see that the spiral similarity that sends the line segment $EA$ to $KX$ has center $B$. So the spiral similarity that sends the line segment $EK$ to $AX$ has center $B$. Thus $\\triangle BEK \\sim BAX$. $\\square$\n\nIn a similar manner, we get $\\triangle CFK$ is similar to $\\triangle CAX$.\n\nNow, using the similar triangles and the fact that $K$ and $J$ are symmetric in $BC$, we have\n\n$$\n\\begin{align*}\n\\angle FKE + \\angle FJE &= \\angle FKE + \\angle BKC \\\\\n&= 360^\\circ - \\angle EKB - \\angle CKF \\\\\n&= 360^\\circ - \\angle AXB - \\angle CXA \\\\\n&= 360^\\circ - 180^\\circ \\\\\n&= 180^\\circ\n\\end{align*}\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18157,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all non-negative integers $x, y, z$ such that\n$$\nx^{2022} + 10y^{2022} = 11z^{2022}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to consider non-negative $x, y, z$. Assume that there is a non-trivial solution, i.e., at least one of $x, y, z$ is positive. Pick such a solution with minimal sum $x + y + z$.\n\nNow, $x^{2022}$ must be even since the other two terms are even, so we can write $x = 2u$ and divide the equation by $2$ to get\n\n$$\n2^{2021}u^{2022} + 10y^{2022} = 11z^{2022}.\n$$\n\nSimilarly, $z$ must be even, so $z = 2w$ and dividing by $2$ gives\n\n$$\n2^{2020}u^{2022} + 5y^{2022} = 11 \\cdot 2^{2021}w^{2022}.\n$$\n\nAgain, $y$ must be even, so $y = 2v$ and dividing by $2^{2020}$ gives\n\n$$\nu^{2022} + 20v^{2022} = 22w^{2022}.\n$$\n\nThis is the original equation with $x, y, z$ replaced by $u, v, w$ whose sum is half the original sum. This gives a contradiction, so there are no non-trivial solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18158,
    "subject": "Mathematics (Olympiad)",
    "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, as long as Ann wishes. At each step, Bob is free to choose which way to move the point, on one condition: Among every 100 consecutive moves, each of his four possible choices should have been made at least once. Ann's goal is to force Bob to eventually choose a point strictly more than 100 distance away from $A$. Is it possible for Ann to achieve her goal?",
    "options": [],
    "answer": "See solution",
    "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 to 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, until Bob is forced to make his first move rightward. This must occur at step 100 at the 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$$\n\\frac{1}{2^{99}}, \\frac{1}{2^{98}}, \\dots, \\frac{1}{2^{100-(k-1)}}, \\text{ and } \\frac{1}{2^{100-k}}\n$$\n\nThe first $k-1$ moves decrease the $x$-coordinate by at most\n\n$$\n\\frac{1}{2^{99}} + \\frac{1}{2^{98}} + \\dots + \\frac{1}{2^{100-(k-1)}} = \\frac{1}{2^{100-k}} - \\frac{1}{2^{99}},\n$$\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$$\n- \\left( \\frac{1}{2^{100-k}} - \\frac{1}{2^{99}} \\right) + \\frac{1}{2^{100-k}} = \\frac{1}{2^{99}},\n$$\n\nas needed. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18159,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(k)$ denote the sum of digits of any positive integer $k$. Does there exist a positive integer $N$ so that $S(2^n) \\leq S(2^{n+1})$ for all integers $n > N$?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume not. Then $S(2^{6n}) > S(2^{6k}) + 27(n-k)$ for some constant $k$, since $\\geq$ would imply $>$ as $S(2^{n+1}) \\neq S(2^n) \\pmod{9}$ and $S(2^{n+6}) \\equiv S(2^n) \\pmod{9}$, so $S(2^{n+6}) - S(2^n) \\geq 27$.\n\nNow, $S(2^{6n}) \\leq 9 \\log_2 2^{6n} + 9 = 54n + 9$.\n\nThus, we get that $\\log_2 2 \\geq \\frac{1}{2}$. But this is clearly ridiculous.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18160,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be the set of positive integer divisors of $2025$. Let $B$ be a randomly selected subset of $A$. The probability that $B$ is a nonempty set with the property that the least common multiple of its elements is $2025$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $2025 = 3^4 \\cdot 5^2$, there are $(4+1)(2+1) = 15$ elements in $A$. Thus there are $2^{15}$ equally likely possibilities for $B$. The least common multiple of the elements of $B$ will be $2025$ if $B$ contains $2025$ or $B$ does not contain $2025$ but does contain both a multiple of $3^4 = 81$ and a multiple of $5^2 = 25$. There are $2^{14}$ subsets that contain $2025$. To satisfy the other requirement, $B$ would need to contain one of the $2^2 - 1 = 3$ nonempty subsets of $\\{81, 405\\}$, one of the $2^4 - 1 = 15$ nonempty subsets of $\\{25, 75, 225, 675\\}$, and any of the $2^8$ subsets of $\\{1, 3, 5, 9, 15, 27, 45, 135\\}$. Therefore the required probability is\n\n$$\n\\frac{2^{14} + 3 \\cdot 15 \\cdot 2^8}{2^{15}} = \\frac{109}{128}.\n$$\n\nThe requested sum is $109 + 128 = 237$.\n\nFor each element $n \\in A$, the probability of $n \\in B$ is $\\frac{1}{2}$, and these probabilities are independent of each other. The least common multiple of the elements of $B$ is $2025$ if $B$ contains $2025$, which happens with probability $\\frac{1}{2}$, or $B$ does not contain $2025$, but does contain at least one element of $\\{3^4, 5 \\cdot 3^4\\}$, which happens with probability $\\frac{3}{4}$, and one element of $\\{5^2, 3 \\cdot 5^2, 3^2 \\cdot 5^2, 3^3 \\cdot 5^2\\}$, which happens with probability $\\frac{15}{16}$. Thus the required probability is\n\n$$\n\\frac{1}{2} + \\left(1 - \\frac{1}{2}\\right) \\cdot \\frac{3}{4} \\cdot \\frac{15}{16} = \\frac{109}{128},\n$$\n\nas in the first solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18161,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Show that $x^4 - x^3 - x + 1 \\ge 0$ for all real numbers $x$.\n\nb) Find all real numbers $x_1, x_2, x_3$ such 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": "See solution",
    "solution": "(a) Write $x^4 - x^3 - x + 1 = (x-1)(x^3 - 1) = (x-1)^2(x^2 + x + 1)$. Notice that $x^2 + x + 1 > 0$ for all $x \\in \\mathbb{R}$, so the expression is always non-negative.\n\n(b) Notice that $\\sum_{k=1}^{3} (x_k^4 - x_k^3 - x_k + 1) = 0$. By part (a), each term is non-negative, so each must be zero: $x_k^4 - x_k^3 - x_k + 1 = 0$ for $k = 1, 2, 3$. Solving, we find $x_1 = x_2 = x_3 = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18162,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x, y)$ be a non-constant homogeneous polynomial with real coefficients such that $P(\\sin t, \\cos t) = 1$ for every real number $t$. Prove that there exists a positive integer $k$ such that $P(x, y) = (x^2 + y^2)^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the degree of the polynomial $P$:\n\n$$\nP(x, y) = a_n x^n + a_{n-1} x^{n-1} y + \\dots + a_1 x y^{n-1} + a_0 y^n,\n$$\n\nwhere $n > 0$. Note that $n$ must be even, because otherwise the condition $P(\\sin t, \\cos t) = 1$ for $t = 0$ would imply $a_0 = 1$, while for $t = \\pi$ it would imply $a_0 = -1$.\n\nSince $P$ has no constant term, $P(0, 0) = 0$. Now assume $x \\neq 0$ or $y \\neq 0$ and let $c = \\sqrt{x^2 + y^2}$. Since\n\n$$\n\\left(\\frac{x}{\\sqrt{x^2 + y^2}}\\right)^2 + \\left(\\frac{y}{\\sqrt{x^2 + y^2}}\\right)^2 = 1,\n$$\n\nthere exists some real number $t$ such that $\\sin t = \\frac{x}{\\sqrt{x^2 + y^2}}$ and $\\cos t = \\frac{y}{\\sqrt{x^2 + y^2}}$, so $P(\\sin t, \\cos t) = 1$. By homogeneity, $P(x, y) = c^n \\cdot P\\left(\\frac{x}{c}, \\frac{y}{c}\\right)$, hence\n\n$$\nP(x, y) = \\left(\\sqrt{x^2 + y^2}\\right)^n \\cdot P\\left(\\frac{x}{\\sqrt{x^2 + y^2}}, \\frac{y}{\\sqrt{x^2 + y^2}}\\right) = \\left(\\sqrt{x^2 + y^2}\\right)^n.\n$$\n\nfor all $x, y$. Since $n = 2k$ for some positive integer $k$, we have $P(x, y) = (x^2 + y^2)^k$, which also satisfies $P(0, 0) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18163,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $N$ is *abundant* if the sum of its positive divisors (including $1$ and $N$) exceeds $2N$. Given any positive integer $n$, show that there exist $n$ consecutive abundant numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p_k$ be the $k$-th prime number for $k = 1, 2, 3, \\dots$. Recall that $\\prod_k (1 + 1/p_k) = \\infty$, so for any positive integer $m$, we can choose $N(m) \\geq m$ such that $$\\prod_{k=m}^{N(m)} \\left(1 + \\frac{1}{p_k}\\right) > 2,$$ which implies $\\prod_{k=m}^{N(m)} p_k$ is abundant.\n\nDefine $m_1 = 1$ and $m_{j+1} = N(m_j) + 1$ for $j = 1, 2, 3, \\dots$, and set $M_j = \\prod_{k=m_j}^{N(m_j)} p_k$ for $j = 1, 2, 3, \\dots$. The $M_j$ are pairwise coprime, so by the Chinese Remainder Theorem, we can choose a non-negative integer $M$ such that $M + j \\equiv 0 \\pmod{M_j}$ for $j = 1, 2, \\dots, n$. Since any multiple of an abundant number is abundant, $M+1, M+2, \\dots, M+n$ are $n$ consecutive abundant numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18164,
    "subject": "Mathematics (Olympiad)",
    "question": "Taro made a purchase and received change. Two or more of 500 yen coins, 50 yen coins, or 5 yen coins could not be included in the change he received. This is because if two or more of any of these coins were in the change, two of them could be exchanged for one 1000 yen note, one 100 yen coin, or one 10 yen coin to reduce the total number of coins in the change. Also, the change did not include a 1000 yen note, 100 yen coin, 10 yen coin, or 1 yen coin, since Taro used all of these in his payment. What are the possible purchase prices Taro could have paid, and how many such prices are there?",
    "options": [],
    "answer": "See solution",
    "solution": "The change must have consisted of zero or one piece each of 500 yen, 50 yen, and 5 yen coins. Therefore, the purchase price must have been of the form:\n\n$$1111 - (500a + 50b + 5c)$$\n\nwhere each of $a$, $b$, $c$ is either 0 or 1.\n\nConversely, if the purchase price is of the form $1111 - (500a + 50b + 5c)$, with each of $a$, $b$, $c$ being 0 or 1, then Taro would make his payment as specified. The change does not contain any 1000 yen note, 100 yen, 10 yen, or 1 yen coins. The amount of change is $500a + 50b + 5c$ yen, and the number of coins Taro received as change is minimized.\n\nSince each of $a$, $b$, $c$ can be 0 or 1, the number of possible purchase prices is $2^3 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18165,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number such that $p^2 + 7^3 = n^3$ for some integer $n$. Find all such prime numbers $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $p^2 + 7^3 = n^3$, so $p^2 = n^3 - 7^3 = (n - 7)(n^2 + 7n + 49)$. Since $n > 7$ and $n - 7 < n^2 + 7n + 49$, the only possible case is $n - 7 = 1$ and $n^2 + 7n + 49 = p^2$. Thus, $n = 8$ and $p^2 = 169$, so $p = 13$. The only prime number with the required property is $p = 13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18166,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\setminus \\{0\\} \\to \\mathbb{R}$ such that for all nonzero numbers $x, y$,\n\n$$\nx \\cdot f(xy) + f(-y) = x \\cdot f(x).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = 1$ gives\n\n$$\nf(y) + f(-y) = f(1).\n$$\n\nLet $f(1) = a$. Then $f(-y) = a - f(y)$. Substituting $y = -1$ gives\n\n$$\nx \\cdot f(-x) + f(1) = x \\cdot f(x),\n$$\nthat is,\n\n$$\nx(a - f(x)) + a = x \\cdot f(x),\n$$\nso\n\n$$\nf(x) = \\frac{a(x+1)}{2x} = \\frac{a}{2}\\left(1 + \\frac{1}{x}\\right).\n$$\n\nFinally, for any real $c$, the function $f(x) = c\\left(1 + \\frac{1}{x}\\right)$ satisfies the condition:\n\n$$\n\\begin{aligned}\nx \\cdot f(xy) + f(-y) &= x \\cdot c\\left(1 + \\frac{1}{xy}\\right) + c\\left(1 + \\frac{1}{-y}\\right) \\\\\n&= c\\left(x + \\frac{1}{y} + 1 - \\frac{1}{y}\\right) \\\\\n&= c(x+1) = x \\cdot c\\left(1 + \\frac{1}{x}\\right) = x \\cdot f(x).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18167,
    "subject": "Mathematics (Olympiad)",
    "question": "In square $ABCD$, points $P$ and $Q$ lie on $\\overline{AD}$ and $\\overline{AB}$, respectively. Segments $\\overline{BP}$ and $\\overline{CQ}$ intersect at right angles at $R$, with $BR = 6$ and $PR = 7$. What is the area of the square?",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2021_AMC10B_Solutions_Fall_p7_data_a7faef284e.png)\n\nBecause $\\angle RBC$ is complementary to both $\\angle RCB$ and $\\angle PBA$, those two angles are congruent. Therefore $\\triangle BAP \\cong \\triangle CBQ$ by ASA, so $CQ = BP = 13$. Let $d = CR$; then $QR = 13 - d$, so the Altitude-to-Hypotenuse Theorem yields $6^2 = d(13 - d)$, which has solutions $d = 4$ and $d = 9$. Because $QR < RC$, in fact $d = 9$. It follows that the area of the square is\n\n$$\nBC^2 = BR^2 + RC^2 = 6^2 + 9^2 = 117.\n$$\n\nAlternatively,\n\nLet $c = BQ$ and $s = AB$. Because $\\triangle BRQ$ is similar to $\\triangle BAP$, it follows that $\\frac{6}{c} = \\frac{s}{13}$, so $cs = 78$. As before, $CQ = BP = 13$, so by the Pythagorean Theorem, $s^2 + c^2 = 169$. Then $(s+c)^2 = 169 + 2 \\cdot 78 = 325$, so $s + c = 5\\sqrt{13}$. Similarly, $(s-c)^2 = 169 - 2 \\cdot 78 = 13$, so $s - c = \\sqrt{13}$. Solving this system of equations yields $s = 3\\sqrt{13}$, and the area of the square is $s^2 = 117$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18168,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be a positive integer, let $S = \\{1, 2, 3, \\dots, n\\}$, and let $\\mathcal{F}$ be the set of functions from $S$ to $S$.\n\nA set $\\mathcal{G} \\subset \\mathcal{F}$ is called a *generating set* for a set $\\mathcal{H} \\subset \\mathcal{F}$ if every function in $\\mathcal{H}$ can be represented as a composition of functions from $\\mathcal{G}$.\n\n**a)** Let $a : S \\to S$ be defined by $a(n-1) = n$, $a(n) = n-1$, and $a(k) = k$ for $k \\in S \\setminus \\{n-1, n\\}$. 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 in $\\mathcal{F}$.\n\n**b)** Show that the minimum number of elements in a generating set for $\\mathcal{F}$ is $3$.",
    "options": [],
    "answer": "See 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 $j = 1, \\dots, p-1$, $f(i_p) = i_1$, and $f(x) = x$ for $x \\notin \\{i_1, \\dots, i_p\\}$, where $i_1, \\dots, i_p$ are $p \\ge 2$ distinct elements of $S$.\n\n**a)** We use induction. For $n=3$, $\\mathcal{B} = \\{a, a^2, b, b^2, ab, ba\\}$.\n\nSuppose the property holds for some $n \\ge 3$; we show it holds for $n+1$. Let $f : S \\cup \\{n+1\\} \\to S \\cup \\{n+1\\}$, $f(n+1) = m$, and let $a', b'$ be analogous to $a$ and $b$ on $S \\cup \\{n+1\\}$. 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 $f = (b')^m g$.\n\nAlso, $(b'a')(n+1) = n+1$ 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 $a$. Thus, $f$ can be written as a composition of $a'$ and $b'$.\n\n**b)** Suppose $\\mathcal{G}$ is a generating set for $\\mathcal{F}$ with $|\\mathcal{G}| \\le 2$.\n\n- If both $f$ and $g$ are bijective, $\\mathcal{G}$ can only generate bijective functions.\n- If both are not bijective, they are not surjective, so $\\mathcal{G}$ can only generate non-surjective functions.\n- If $f$ is bijective and $g$ is not, the bijective functions generated 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\nA 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 any $f \\in \\mathcal{F}$ can be written as a composition of $a, b, c$ by descending induction on $|\\text{Im } f|$.\n\nIf $|\\text{Im } f| = n$, $f$ is bijective and we use (a).\n\nAssume the statement is true for any $f$ with $|\\mathrm{Im} f| = k$, $1 < k \\le n$, and prove it for $g$ with $|\\mathrm{Im} g| = k-1$. Since $g$ is not injective, there exist $u, v \\in S$, $u \\ne v$, and a bijective $r$ such that $\\mathrm{Im} gr = \\{1, 2, \\dots, k-1\\}$ and $(gr)(u) = (gr)(v) = k-1$. Let $s : S \\to S$ be defined by $s(n-1) = u$, $s(n) = v$, $s(x) = x$ for $x \\ne u, v$. Define $h : S \\to S$ by $h(x) = (grs)(x)$ for $x \\le n-1$ and $h(n) = k$. Then $|\\mathrm{Im} h| = k$, so $h$ can be expressed as a composition of $a, b, c$, and $grs = hc$. Thus, $g = hcs^{-1}r^{-1}$ can be expressed as a composition of $a, b, c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18169,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all ordered pairs $(a, b)$ of positive integers that satisfy $a > b$ and the equation $$(a-b)^{ab} = a^b \\cdot b^a.$$",
    "options": [],
    "answer": "See solution",
    "solution": "$(a, b) = (4, 2)$.\n\nLet $d$ be the greatest common divisor of $a$ and $b$, so $a = dp$, $b = dq$, where $p, q$ are coprime positive integers with $p > q$. Substitute into the original equation:\n\n$$\n(d(p-q))^{d^2pq} = (dp)^{dq} \\cdot (dq)^{dp}\n$$\nwhich simplifies to\n$$\n(d(p-q))^{dpq} = (dp)^q \\cdot (dq)^p\n$$\nso\n$$\nd^{dpq}(p-q)^{dpq} = d^{p+q}p^q q^p.\n$$\n\n(i) First, prove $p+q < dpq$. If not, $p+q \\ge dpq$, so\n$$(p-q)^{dpq} = d^{p+q-dpq}p^q q^p.$$\nThus $p, q$ both divide $(p-q)^{dpq}$, but $(p-q, p) = (p-q, q) = (p, q) = 1$, so $p = q = 1$, which is impossible.\n\nSince $p+q < dpq$, rearrange to get\n$$d^{dpq-p-q}(p-q)^{dpq} = p^q q^p.$$\nThus $p-q$ divides $p^q q^p$. But $(p-q, p) = (p-q, q) = 1$, so $p-q = 1$, i.e., $p = q+1$. Substitute back:\n$$d^{dpq-p-q} = p^q q^{q+1}. \\qquad (1)$$\n\n(ii) Since $p, q$ are coprime, write $d = m \\times n$, where $m$ shares all prime factors with $p$, $n$ with $q$, and $(m, n) = 1$. From (1):\n$$m^{dpq-p-q} = p^q = (q+1)^q. \\qquad (2)$$\nSince the exponent $dpq-p-q$ and $q$ are coprime, $m$ must be a $q$th power. Let $m = t^q$, then $t^{dpq-p-q} = q+1$. Since $q+1 > 1$, $t > 1$.\n\n(iii) If $q = 2$, then $t^{6d-5} = 3$, so $d = 1$, $t = 3$, $q = 2$, $p = 3$, i.e., $(a, b) = (3, 2)$. But $(3, 2)$ does not satisfy the original equation, so discard.\n\nIf $q \\ge 3$, then $dpq-p-q = dq(q+1) - (2q+1) \\ge 3(q+1) - (2q+1) = q+2$. Thus $q+1 = t^{dpq-p-q} \\ge t^{q+2} \\ge 2^{q+2} \\ge q+2$ (by the binomial theorem), which is impossible.\n\nSo $q = 1$, $p = 2$. Then $t^{2d-3} = 2$, so $t = 2$, $d = 2$. Thus $(a, b) = (4, 2)$. Substitute back into the original equation and it holds.\n\nTherefore, $(a, b) = (4, 2)$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18170,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n \\mid 2^n + 1$.\n\nAlso, determine all prime numbers $n$ satisfying $n \\mid 2^n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "All integers $n = 3^k$ with $k \\in \\mathbb{Z}^+$ satisfy $n \\mid 2^n + 1$ because $3 \\mid 2 + 1$, and by the lifting the exponent lemma:\n\n$$\nv_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = 1 + k.\n$$\n\nSo $3^{k+1} \\mid 2^n + 1$, and thus $n \\mid 2^n + 1$.\n\nFor primes, let $n = p$ be prime. By Fermat's little theorem:\n\n$$\n2^p + 1 \\equiv 2 + 1 = 3 \\pmod{p}.\n$$\n\nThis is $0$ mod $p$ iff $p \\mid 3$, so $p = 3$ is the only prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18171,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n \\geq 3$, an $n$-ring is a circular arrangement of $n$ (not necessarily different) positive integers such that the product of every three neighbouring integers is $n$. Determine the number of integers $n$ in the range $3 \\leq n \\leq 2018$ for which it is possible to form an $n$-ring.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the numbers in the $n$-ring be, in order around the circle, $a_1, a_2, \\dots, a_n$. For every $k$, $a_k a_{k+1} a_{k+2} = n$, where indices are taken modulo $n$ (so $a_{n+1} = a_1$). Consider the product of all the numbers in the $n$-ring:\n\n$$\n\\begin{aligned}\na_1 a_2 \\cdots a_n &= \\sqrt[3]{a_1^3 a_2^3 \\cdots a_n^3} \\\\\n&= \\sqrt[3]{(a_1 a_2 a_3)(a_2 a_3 a_4) \\cdots (a_{n-2} a_{n-1} a_n)(a_{n-1} a_n a_1)(a_n a_1 a_2)} \\\\\n&= \\sqrt[3]{n^n} = n^{\\frac{n}{3}}.\n\\end{aligned}\n$$\n\nTherefore, $n^{\\frac{n}{3}}$ must be a positive integer. Prime factorise $n$ as $p_1^{b_1} \\times \\dots \\times p_k^{b_k}$ so that\n\n$$\nn^{\\frac{n}{3}} = p_1^{\\frac{n b_1}{3}} \\times \\dots \\times p_k^{\\frac{n b_k}{3}}.\n$$\n\nFor $n^{\\frac{n}{3}}$ to be an integer, $3$ must divide each $n b_i$. This means either $n$ is a multiple of $3$, or $3$ divides all $b_i$ so that $n$ is a perfect cube.\n\nIf $n$ is a perfect cube, then $\\sqrt[3]{n}, \\sqrt[3]{n}, \\dots, \\sqrt[3]{n}$ is an $n$-ring. If $n$ is a multiple of $3$, then $n, 1, 1, n, 1, 1, \\dots, n, 1, 1$ is an $n$-ring. Thus, there is an $n$-ring exactly if $n$ is a multiple of $3$ or a perfect cube.\n\nThere are $\\lfloor \\frac{2018}{3} \\rfloor = 672$ multiples of $3$ in the range $3 \\leq n \\leq 2018$ and $\\lfloor \\sqrt[3]{2018} \\rfloor - 1 = 11$ cubes (excluding $1$). Four of these cubes are also multiples of $3$ ($3^3, 6^3, 9^3, 12^3$), so the number of $n$ in the range $3 \\leq n \\leq 2018$ for which there are $n$-rings is $672 + 11 - 4 = 679$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18172,
    "subject": "Mathematics (Olympiad)",
    "question": "Учениците од две одделенија се договориле да играат фудбал. Во едно од одделенијата немало доволен број на играчи да состават екипа од 11 ученици, па тие се договориле учениците од двете одделенија да се \"измешаат\" меѓу себе и потоа да состават две екипи. Наставникот забележал дека од првото одделение машки се $\\frac{4}{13}$ од учениците, додека од второто одделение машки се $\\frac{5}{17}$ од учениците. Секое од одделенијата има не повеќе од 50 ученици. Кое од одделенијата има повеќе девојчиња? (Одговорот да се образложи)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека бројот на ученици во првото одделение е $x$, а бројот на ученици во второто одделение е $y$.\n\nТогаш машки во првото одделение се $\\frac{4x}{13}$, додека во второто се $\\frac{5y}{17}$ на број. Бидејќи $\\frac{4x}{13}$ мора да е природен број, мора $4x$ да се дели со 13, т.е. $x = 13$, $26$ или $39$. Соодветно бројот на машки е $4$, $8$, $12$.\n\nАналогно, $\\frac{5y}{17}$ мора да е природен, па мора $5y$ да се дели со 17, тогаш $y = 17$ или $34$. Соодветниот број на машки е $5$ или $10$.\n\nБидејќи $22 = 12 + 10$, имаме дека првото одделение брои ученици $39$ од кои $12$ машки, т.е. $27$ девојчиња. Второто одделение брои $34$ ученици од кои $10$ машки, т.е. $24$ девојчиња.\n\nЗначи, првото одделение има повеќе девојчиња.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18173,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma_2$ be a circle contained in the interior of another circle $\\Gamma_1$ on the plane. Prove that there exists a point $P$ on the plane satisfying the following conditions: if $\\ell$ is a line that does not contain $P$, that intersects $\\Gamma_1$ at two different points $A, B$, and that intersects $\\Gamma_2$ at two different points $C, D$ (so that $A, C, D, B$ lie in order on $\\ell$), then $\\angle APC = \\angle DPB$.\n\n![](images/CHN_TSExams_2022_2_p2_data_4599e41b63.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the centers of the two circles by $O_1, O_2$, and the radii by $r_1, r_2$, respectively, where $r_1 > r_2$. We first show that there are two points $P, Q$ on the ray $O_1O_2$ such that $O_1P \\cdot O_1Q = r_1^2$ and $O_2P \\cdot O_2Q = r_2^2$.\n\nOne can choose a point $K$ on the ray $O_1O_2$ such that\n$$\nO_1K = \\frac{O_1O_2^2 + r_1^2 - r_2^2}{2O_1O_2}.\n$$\nSince $O_1O_2 \\le r_1 - r_2$, we have $O_1K \\ge r_1$. We choose two points $P, Q$ on the line $O_1O_2$ such that $KP = KQ = \\sqrt{O_1K^2 - r_1^2}$ so that $O_1P \\cdot O_1Q = r_1^2$. On the other hand,\n\n$$\n\\begin{aligned}\nO_2P \\cdot O_2Q - O_1P \\cdot O_1Q &= O_2K^2 - O_1K^2 \\\\\n&= (O_1K - O_1O_2)^2 - O_1K^2 \\\\\n&= O_1O_2^2 - 2O_1O_2 \\cdot O_1K \\\\\n&= r_2^2 - r_1^2,\n\\end{aligned}\n$$\nwhich implies that $O_2P \\cdot O_2Q = r_2^2$.\n\nFor any given line $\\ell$ as in the problem, if it is perpendicular to $O_1O_2$, we certainly have $\\angle APC = \\angle DPB$ by symmetry. If not, from $O_2C^2 = O_2Q \\cdot O_2P$ we know that $\\triangle O_2CQ \\sim \\triangle O_2PC$. So $\\frac{CQ}{CP} = \\frac{r_2}{O_2P}$. Similarly, we obtain $\\frac{DQ}{DP} = \\frac{r_2}{O_2P}$, and therefore $\\frac{CQ}{CP} = \\frac{DQ}{DP}$, which means that the bisectors of $\\angle CPD$ and $\\angle CQD$ meet $\\ell$ at the same point, say $M$. A similar argument shows that the bisectors of $\\angle APB$ and $\\angle AQB$ intersect $\\ell$ at the same point again, say $M'$.\n\nNote that the points $P, Q, M$ are all on an Apollonian circle with distance ratio $\\frac{CQ}{DQ}$ to $C$ and $D$. The center of the Apollonian circle must be on $\\ell$. Thus this center must be the intersection of $\\ell$ with the perpendicular bisector of $PQ$. Similarly, the points $P, Q, M'$ are on the Apollonius circle with distance ratio $\\frac{AQ}{BQ}$ to $A$ and $B$, whose center is the same point as above. Note that $K$ does not lie in the interior of $\\Gamma_1$, so the center of this circle, denoted by $L$, must be outside $\\Gamma_1$ so that $M$ coincides with $M'$. Therefore,\n\n$$\n\\angle APC = \\angle APM - \\angle CPM = \\angle BPM - \\angle DPM = \\angle BPD.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18174,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. Find all real numbers $x$ such that\n\n$$\n\\{x\\} + \\{2x\\} + \\cdots + \\{nx\\} = [x] + [2x] + \\cdots + [2nx],\n$$\n\nwhere $\\{y\\}$ denotes the fractional part of $y$ and $[y]$ denotes the integer part of $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\nn - \\sqrt{\\frac{n(n-1)}{2}} - \\left(n + \\frac{1}{2} - \\sqrt{\\frac{2n^2 + 2n + 1}{4}}\\right) < 1,\n$$\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": "Intermediate Algebra"
  },
  {
    "id": 18175,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive real number $\\alpha$ with the following property: if the weight of a finite number of pumpkins is 1 ton and the weight of every pumpkin is not more than $\\alpha$ tons, then the pumpkins can be distributed in 50 boxes (some of the boxes may remain empty) such that there are no more than $\\alpha$ tons of pumpkins in every box.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that the desired value of $\\alpha$ is $\\alpha = \\frac{2}{51}$.\n\nAssume that some $\\alpha < \\frac{2}{51}$ satisfies the condition of the problem. Choose a nonnegative integer $k \\ge 0$ such that $\\frac{1}{51 \\times 2^k} \\le \\alpha < \\frac{1}{51 \\times 2^{k-1}}$. Consider $51 \\times 2^k$ pumpkins, each having weight $\\frac{1}{51 \\times 2^k}$ tons. For any distribution of pumpkins in 50 boxes, there exists a box with at least two pumpkins, and therefore the weight of this box equals at least $\\frac{1}{51 \\times 2^{k-1}} > \\alpha$, a contradiction.\n\nWe show that $\\alpha = \\frac{2}{51}$ satisfies the condition of the problem. Let the number of pumpkins be $m$. Take $m$ empty boxes and put a pumpkin in every one of them. If the two lightest boxes contain in common not more than $\\frac{2}{51}$ tons of pumpkins, then gather all pumpkins of these two boxes into one of them and remove the empty box. When this operation terminates, let the number of boxes be $n$ and they contain $x_1 \\le x_2 \\le \\dots \\le x_n$ tons of pumpkins. Thus $x_1 + x_2 > \\frac{2}{51}$ and hence $x_2 > \\frac{1}{51}$. Therefore\n\n$$\n1 = x_1 + x_2 + \\dots + x_n > \\frac{2}{51} + (n-2) \\times \\frac{1}{51},\n$$\n\nimplying $n < 51$. We have that all the pumpkins are distributed in no more than 50 boxes and there are no more than $\\frac{2}{51}$ tons of pumpkins in each box. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18176,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a real number $a > 1$, define the sequence of real numbers $(x_n)_{n \\ge 1}$ by $x_1 = a$ and\n$$\nx_1 + x_2 + \\dots + x_{n+1} = x_1 x_2 \\cdots x_{n+1}, \\quad \\text{for all } n \\ge 1.\n$$\nProve that the sequence is convergent and find its limit.",
    "options": [],
    "answer": "See solution",
    "solution": "By induction, for all $n \\ge 1$, $x_n > 0$ and $x_1 x_2 \\cdots x_n > 1$. By the AM-GM inequality, $x_1 x_2 \\cdots x_n \\ge n (x_1 x_2 \\cdots x_n)^{1/n}$, implying $x_1 x_2 \\cdots x_n \\ge n^{1/(n-1)}$, so $\\lim_{n \\to \\infty} x_1 x_2 \\cdots x_n = \\infty$.\n\nIt follows:\n$$\n\\lim_{n \\to \\infty} x_n = \\lim_{n \\to \\infty} \\frac{x_1 x_2 \\cdots x_{n-1}}{x_1 x_2 \\cdots x_{n-1} - 1} = 1 - \\lim_{n \\to \\infty} \\frac{1}{x_1 x_2 \\cdots x_{n-1} - 1} = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18177,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a sequence $\\langle a_1, a_2, a_3, \\dots \\rangle$ of real numbers. For every positive integer $n$, define $m_n$ as the arithmetic mean of the numbers from $a_1$ through $a_n$.\n\nAssume that a real number $C$ exists such that\n\n$$\n(i - j) \\cdot m_k + (j - k) \\cdot m_i + (k - i) \\cdot m_j = C\n$$\n\nholds for all triples $(i, j, k)$ of pairwise different positive integers. Prove that $\\langle a_1, a_2, a_3, \\dots \\rangle$ is an arithmetic sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "By exchanging the roles of $i$ and $j$, we see that\n\n$$\n(i-j) \\cdot m_k + (j-k) \\cdot m_i + (k-i) \\cdot m_j = C = (j-i) \\cdot m_k + (i-k) \\cdot m_j + (k-j) \\cdot m_i = -C\n$$\n\nmust hold, which yields $C=0$.\n\nFor $(i,j,k) = (1,2,3)$, we obtain\n\n$$\n(1-2) \\cdot \\frac{a_1+a_2+a_3}{3} + (2-3) \\cdot a_1 + (3-1) \\cdot \\frac{a_1+a_2}{2} = 0,\n$$\n\nwhich is equivalent to\n\n$$\n-\\frac{a_1+a_2+a_3}{3} - a_1 + a_1 + a_2 = 0 \\iff a_1 + a_3 = a_2.\n$$\n\nThe first three elements of the sequence therefore are elements of an arithmetic sequence. We can now use induction to show that the entire sequence is arithmetic, i.e., that $a_n = a_1 + (n-1)(a_2 - a_1)$ holds.\n\nAssume $a_k = a_1 + (k-1)(a_2 - a_1)$ holds for $1 \\le k \\le n-1$, and consider the triple $(i,j,k) = (1,2,n)$. We then have\n\n$$\n\\begin{align*}\n(1-2) \\cdot m_n + (2-n) \\cdot a_1 + (n-1) \\cdot m_2 &= 0 \\\\\n\\end{align*}\n$$\n\nwhere $m_n = \\frac{a_1 + a_2 + \\dots + a_n}{n}$ and $m_2 = \\frac{a_1 + a_2}{2}$.\n\nCarrying out the algebra (as in the previous step), we find that $a_n = a_1 + (n-1)(a_2 - a_1)$, completing the induction. Thus, the sequence is arithmetic, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18178,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(A, +, \\cdot)$ be a ring with 9 elements. Prove that the following two statements are equivalent:\n\n(a) For each $x \\in A \\setminus \\{0\\}$ there exist $a \\in \\{-1, 0, 1\\}$ and $b \\in \\{-1, 1\\}$ such that $x^2 + a x + b = 0$.\n\n(b) $(A, +, \\cdot)$ is a field.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that (a) implies (b). Let $x \\in A \\setminus \\{0\\}$ and let $a, b$ be as in (a). Since $a x = x a$, we have $x(x + a) = (x + a)x = -b \\in \\{-1, 1\\}$, so $x$ is invertible. Therefore, $A$ is a field.\n\nConversely, assume $A$ is a field. Since $A$ has no zero divisors and $(1 + 1 + 1)(1 + 1 + 1) = 0$, it follows that $1 + 1 + 1 = 0$. Let $x \\in A \\setminus \\{0\\}$. If $x^2 = \\pm 1$, then $x^2 - 1 = 0$ or $x^2 + 1 = 0$, so $x$ satisfies the claim. If $x^2 \\neq \\pm 1$, since $(A^*, \\cdot)$ is a group of order 8, we have $(x^2 - 1)(x^2 + 1)(x^4 + 1) = x^8 - 1 = 0$, so $x^4 + 1 = 0$. Now, $x^4 + 1 = (x^2 + 2)^2 - (2x)^2 = (x^2 - 2x + 2)(x^2 + 2x + 2) = (x^2 + x - 1)(x^2 - x - 1)$, so $x^2 + x - 1 = 0$ or $x^2 - x - 1 = 0$. Thus, the claim holds.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18179,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $A \\subseteq \\{1, 2, \\dots, n\\}$. Suppose that for every $a, b \\in A$, $\\operatorname{lcm}(a, b) \\le n$. Prove that\n$$\n|A| \\le 1.9\\sqrt{n} + 5.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $a \\in (\\sqrt{n}, \\sqrt{2n}]$, $\\operatorname{lcm}(a, a+1) = a(a+1) > n$, so\n$$\n|A \\cap (\\sqrt{n}, \\sqrt{2n}]| \\le \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + 1.\n$$\nFor $a \\in (\\sqrt{2n}, \\sqrt{3n}]$,\n$$\n\\operatorname{lcm}(a, a+1) = a(a+1) > n,\n$$\n$$\n\\operatorname{lcm}(a+1, a+2) = (a+1)(a+2) > n,\n$$\n$$\n\\operatorname{lcm}(a, a+2) \\ge \\frac{1}{2}a(a+2) > n.\n$$\nSo\n$$\n|A \\cap (\\sqrt{2n}, \\sqrt{3n}]| \\le \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} + 1.\n$$\nSimilarly,\n$$\n|A \\cap (\\sqrt{3n}, 2\\sqrt{n}]| \\le \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 1.\n$$\nHence,\n$$\n\\begin{aligned}\n|A \\cap [1, 2\\sqrt{n}]| &\\le \\sqrt{n} + \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} \\\\\n&\\quad + \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 3 \\\\\n&= \\left(1 + \\frac{\\sqrt{2}}{6} + \\frac{\\sqrt{3}}{12}\\right)\\sqrt{n} + 3.\n\\end{aligned}\n$$\nLet $k \\in \\mathbb{N}^*$, and suppose $a, b \\in (\\frac{n}{k+1}, \\frac{n}{k})$, $a > b$, and $\\operatorname{lcm}(a, b) = as = bt$, where $s, t \\in \\mathbb{N}^*$. Then\n$$\n\\frac{a}{(a, b)s} = \\frac{b}{(a, b)t}.\n$$\nSince $\\gcd\\left(\\frac{a}{(a, b)}, \\frac{b}{(a, b)}\\right) = 1$, $\\frac{b}{(a, b)}$ divides $s$. It follows that\n$$\n\\operatorname{lcm}(a, b) = as \\ge \\frac{ab}{(a, b)} \\ge \\frac{ab}{a-b} = b + \\frac{b^2}{a-b} > \\frac{n}{k+1} + \\frac{\\left(\\frac{n}{k+1}\\right)^2}{\\frac{n}{k} - \\frac{n}{k+1}} = n.\n$$\nTherefore, $|A \\cap (\\frac{n}{k+1}, \\frac{n}{k})| \\le 1$.\n\nSuppose $T \\in \\mathbb{N}^*$ such that $\\frac{n}{T+1} \\le 2\\sqrt{n} < \\frac{n}{T}$. Then\n$$\n\\begin{aligned}\n|A \\cap (2\\sqrt{n}, n]| &\\le \\sum_{k=1}^{T} \\left|A \\cap \\left(\\frac{n}{k+1}, \\frac{n}{k}\\right]\\right| \\\\\n&\\le T < \\frac{1}{2}\\sqrt{n}.\n\\end{aligned}\n$$\nBy the above arguments,\n$$\n|A| \\le \\left(\\frac{3}{2} + \\frac{1}{6}\\sqrt{2} + \\frac{1}{12}\\sqrt{3}\\right)\\sqrt{n} + 3 < 1.9\\sqrt{n} + 5.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18180,
    "subject": "Mathematics (Olympiad)",
    "question": "Могут ли все грибы стать хорошими, если в каждом плохом грибе ровно 10 червей, а в хорошем червей нет?",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть в каждом плохом грибе ровно 10 червей, а в хорошем червей нет. Пусть из каждого плохого гриба по одному червю переползут в хорошие, по 9 в каждый. В результате в каждом грибе окажется по 9 червей, и все грибы будут хорошими.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18181,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest constant $K \\ge 0$ such that for any $0 \\leq k \\leq K$, and for any non-negative real numbers $a, b, c$ satisfying\n$$\na^2 + b^2 + c^2 + k a b c = k + 3,\n$$\nit holds that\n$$\na + b + c \\leq 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us work from first principles. Whenever at least one variable is zero, say $c = 0$, it follows that $a^2 + b^2 = k + 3$. Hence,\n$$\na + b + c \\leq \\sqrt{2(k+3)} \\leq 3\n$$\nfor $k \\leq 3/2$, with equality holding for $a = b = 3/2$. Thus, we have a tentative limiting bound of $K = 3/2$.\n\nAlso, note 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 $k > 0$)\n$$\nk(1 - a b c) = a^2 + b^2 + c^2 - 3 \\geq \\frac{1}{3}(a + b + c)^2 - 3 = \\frac{1}{3}\\sigma^2 - 3 > 0,\n$$\nby Cauchy-Schwarz, and so $a b c < 1$. Now is the key step. Since $a + b + c > 3$, assuming $0 \\leq a \\leq b \\leq c$, it follows $c > 1$, and so we can compute\n$$\n\\begin{align*}\nk &= \\frac{a^2 + b^2 + c^2 - 3}{1 - a b c} \\\\\n  &= \\frac{(a + b)^2 + c^2 - 3 - 2 a b}{1 - a b c} \\\\\n  &= \\frac{(\\sigma - c)^2 + c^2 - 3 - 2 a b}{1 - a b c} \\\\\n  &= \\frac{c(2c^2 - 2\\sigma c + \\sigma^2 - 3) - 2 + 2(1 - a b c)}{c(1 - a b c)} \\\\\n  &= \\frac{2}{c} + \\frac{2c^3 - 2\\sigma c^2 + (\\sigma^2 - 3)c - 2}{c(1 - a b c)}\n\\end{align*}\n$$\n\nDenote $f(c) = 2c^3 - 2\\sigma c^2 + (\\sigma^2 - 3)c - 2$. As $(x - y)(f(x) - f(y)) = (x - y)^2 \\left( \\frac{3}{2} (x + y - \\frac{2}{3}\\sigma)^2 + \\frac{1}{2}(x - y)^2 + \\frac{1}{3}(\\sigma^2 - 9) \\right) \\geq 0$, $f$ 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 $a b = 0$, for $a = 0$, whence $k \\geq b^2 + c^2 - 3 \\geq \\frac{1}{2}(b + c)^2 - 3 = \\frac{1}{2}\\sigma^2 - 3 > \\frac{3}{2}$.\n\nThe issue of the extremal points, leading to equality when $k = 3/2$, will be better addressed within the next alternative solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18182,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = ax^2 + bx + c$ be a polynomial with integer coefficients. For every integer $x$, $f(x)$ is divisible by $N$, where $N$ is a positive integer. Is it true that $N$ necessarily divides all the coefficients of $f(x)$ if\n\n$$\n\\text{a) } N = 2016; \\quad \\text{b) } N = 2017?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) For any integer $x$, the product $x(x+1)$ is even. Consider the example:\n\n$$\n1008x(x+1) + 2016 = 1008x^2 + 1008x + 2016.\n$$\n\nThis polynomial is divisible by $2016$ for all integer $x$, but $2016$ does not divide all coefficients.\n\nb) Let $f(x) = ax^2 + bx + c$. Substitute:\n\n$$\n\\begin{aligned}\nx &= 0 \\Rightarrow f(0) = c \\implies 2017 \\mid c \\\\\nx &= 1 \\Rightarrow f(1) = a + b + c \\implies 2017 \\mid a + b + c \\\\\nx &= -1 \\Rightarrow f(-1) = a - b + c \\implies 2017 \\mid a - b + c\n\\end{aligned}\n$$\n\nSubtracting, $2017$ divides both $a + b$ and $a - b$, so $2017$ divides $2a$ and $2b$. Since $2017$ is odd, $2017$ divides $a$ and $b$. Thus, all coefficients are divisible by $2017$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18183,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $S$ 為一個正整數的非空子集，其中對於任意 $a, b \\in S$，都可以找到一個 $c \\in S$ 使得 $c^2$ 整除 $a(a+b)$。證明存在 $a \\in S$，使得 $a$ 能整除 $S$ 中的任一元素。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $a$ 為 $S$ 中的最小元素。我們將證明此 $a$ 滿足條件。假設相反，設 $b \\in S$ 為 $S$ 中最小使得 $a \\nmid b$ 的元素。定義 $b_0 = b$，並令 $b_{i+1}$ 為 $S$ 中的元素，使得 $b_{i+1}^2 \\mid a(a + b_i)$，其中 $i \\ge 0$。\n\n注意，若 $b_{i+1} < b$，則由 $b$ 的最小性知 $a \\mid b_{i+1}$，因此\n\n$$\na^2 \\mid b_{i+1}^2 \\mid a(a + b_i)\n$$\n\n所以 $a \\mid b_i$，由歸納法得 $a \\mid b_0$，與假設矛盾。因此，$b_i \\ge b$ 對所有 $i$ 成立。此外，$b_{i+1} < \\sqrt{b(b + b_i)}$。由歸納可知 $b_n < \\varphi b$，其中 $\\varphi = (1 + \\sqrt{5})/2$。因此，\n\n$$\nb^2 \\le b_{i+1}^2 \\le a(a + b_i) < (1 + \\varphi)b < 3b^2,\n$$\n\n顯示 $b_{i+1}^2 = a(a + b_i)$ 或 $b_{i+1}^2 = a(a + b_i)/2$。\n\n設 $p$ 為使 $v_p(a) > v_p(b)$ 的質數。若 $p \\ne 2$，則歸納得 $v_p(b_i) < v_p(a)$ 對所有 $i$ 成立，且\n\n$$\nv_p(b_{i+1}) = \\frac{v_p(a) + v_p(a + b_i)}{2} = \\frac{v_p(a) + v_p(b_i)}{2},\n$$\n\n因此 $v_p(a) - v_p(b_i)$ 隨 $i$ 增加嚴格減少，矛盾。故 $p = 2$。\n\n同理，$v_2(b_{i+1}) = \\left\\lfloor \\frac{v_2(a) + v_2(b_i)}{2} \\right\\rfloor$。則 $v_2(a) - v_2(b_i)$ 嚴格減少，除非 $v_2(b_i) = v_2(a) - 1$。因此，必有 $N \\in \\mathbb{N}$ 使得 $v_2(b_i) = v_2(a) - 1$ 對所有 $i \\ge N$ 成立。這表示 $b_{i+1}^2 = a(a + b_i)/2$ 對所有 $i \\ge N$ 成立，故 $b_{i+1} < b_i$ 對所有 $i \\ge N$，矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18184,
    "subject": "Mathematics (Olympiad)",
    "question": "A **superb sequence** is a sequence of 0s and 1s of length $n$ such that neither the second term nor the second last term is 0. For each integer $n \\geq 2$, let $F_n$ be the set of superb sequences with $n$ terms that end in 0, and let $f_n = |F_n|$. Similarly, let $G_n$ be the set of superb sequences with $n$ terms that end in 1, and let $g_n = |G_n|$. Let $B_n = f_n + g_n$ denote the total number of superb sequences of length $n$.\n\nIt is observed that all sequences in $F_n$ end in 1,0 and all sequences in $G_n$ end in 1,1. For $n \\geq 3$, removing the final 0 from a sequence in $F_n$ yields a sequence in $G_{n-1}$, so $f_n = g_{n-1}$. Thus,\n\n$$\nB_n = g_{n-1} + g_n \\quad \\text{for } n \\geq 3. \\qquad (2)\n$$\n\nFor $n \\geq 5$, $g_n$ can be expressed as:\n\n$$\ng_n = 2 + B_2 + B_3 + \\dots + B_{n-3} \\quad \\text{for } n \\geq 5. \\qquad (4)\n$$\n\nReplacing $n$ with $n-1$ in (4):\n\n$$\ng_{n-1} = 2 + B_2 + B_3 + \\dots + B_{n-4} \\quad \\text{for } n \\geq 6. \\qquad (5)\n$$\n\nComparing (4) and (5), we get:\n\n$$\ng_n = g_{n-1} + B_{n-3} \\quad \\text{for } n \\geq 6.\n$$\n\nFind the smallest $n > 1$ such that $B_n \\equiv 0 \\pmod{20}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By checking small cases, the last equation also holds for $n = 4, 5$. Thus,\n\n$$\nB_n = g_{n-1} + g_n \\quad \\text{for } n \\geq 3, \\qquad (2)\n$$\n\n$$\ng_n = g_{n-1} + B_{n-3} \\quad \\text{for } n \\geq 4. \\qquad (6)\n$$\n\nWe can eliminate all the $g$-terms from (2) and (6). For example, replacing $n$ with $n+1$ in (2) and (6):\n\n$$\nB_{n+1} = g_n + g_{n+1} \\quad \\text{for } n \\geq 2, \\qquad (7)\n$$\n\n$$\ng_{n+1} = g_n + B_{n-2} \\quad \\text{for } n \\geq 3. \\qquad (8)\n$$\n\nAdding (6) and (8):\n\n$$\ng_n + g_{n+1} = g_{n-1} + g_n + B_{n-2} + B_{n-3} \\quad \\text{for } n \\geq 4.\n$$\n\nUsing (2) and (7), this becomes:\n\n$$\nB_{n+1} = B_n + B_{n-2} + B_{n-3} \\quad \\text{for } n \\geq 4. \\qquad (9)\n$$\n\nWe calculate $B_1, B_2, B_3, B_4$ manually, then use the recursion in (9) to compute $B_n$ modulo 20:\n\n![](table)\n\nThus, the first $0$ occurs at $n = 25$, as required.\n\nObservation (1) speeds this up considerably. In particular, $B_1 = 0$, $B_2 = 1$, $g_1 = 0$, $g_2 = 1$, $g_3 = 2$, $g_4 = 2$, and $g_5 = 3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18185,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A(n)$ be defined for $n \\in \\mathbb{N}$ and distinct primes $p_1, \\dots, p_k$ as\n\n$$\nA(n) = n - \\sum_{i=1}^{k} \\left\\lfloor \\frac{n}{p_i} \\right\\rfloor + \\sum_{1 \\le i < j \\le k} \\left\\lfloor \\frac{n}{p_i p_j} \\right\\rfloor - \\dots + (-1)^k \\left\\lfloor \\frac{n}{p_1 p_2 \\cdots p_k} \\right\\rfloor.\n$$\n\nLet $\\Pi_k = (1 - \\frac{1}{p_1}) (1 - \\frac{1}{p_2}) \\cdots (1 - \\frac{1}{p_k})$ and $f_k(n) = n\\Pi_k - A(n)$. Prove that for $k \\ge 3$, there exist primes $p_1, \\dots, p_k$ such that\n$$\n\\max |f_k(n)| > 2^{k-3} + \\Pi_k.\n$$\n\n% IMAGE: ![](images/BMO2024Shortlist_p35_data_439a2365cf.png)\n\nFor example, the graph of $f(x) = \\{x\\} - \\{\\frac{x}{3}\\} - \\{\\frac{x}{7}\\} + \\{\\frac{x}{21}\\}$ looks as follows:\n\n% IMAGE: ![](images/BMO2024Shortlist_p35_data_3e183ca13d.png)\n",
    "options": [],
    "answer": "See solution",
    "solution": "Given primes $p_1, \\dots, p_k$, suppose $-m_k \\le f_k(n) \\le M_k$, where $-m_k$ and $M_k$ are the minimum and maximum values that $f_k(n)$, $n \\in \\mathbb{Z}$, can achieve. We can see that $\\sup f_k(n) = m_k$ and $m_k = M_k + \\Pi_k$ because of the discontinuity at the supremum integer point.\n\nSuppose we can find primes $p_1, p_2, \\dots, p_k$ such that $\\max|f_k(n)| > 2^{k-3} + \\Pi_k$. We show how to find a prime $p_{k+1}$ such that $\\max|f_{k+1}(n)| > 2^{k-3} + \\Pi_{k+1}$.\n\nLet $\\{a_1, a_2, \\dots, a_k\\}$ and $\\{b_1, b_2, \\dots, b_k\\}$ be the two sets of residues modulo $p_1, p_2, \\dots, p_k$, respectively, which identify two integers for which minimum and maximum of $f_k(n)$ occurs. Note that $a_i \\ne 0$ for all $1 \\le i \\le k$ because the minimum value occurs at a point of discontinuity, at an integer coprime with $p_1p_2\\cdots p_k$.\n\nSuppose we add a prime $p_{k+1}$, then $f_{k+1}(n) = f_k(n) - f_k\\left(\\frac{n}{p_{k+1}}\\right)$. Set $n = mp_{k+1} + r$, where $m$ is an integer and $0 \\le r \\le p_{k+1} - 1$. Since $\\frac{r}{p_{k+1}} < 1$, we get\n\n$$\n\\begin{align*}\nf_{k+1}(n) &= f_k(n) - f_k\\left(\\frac{n}{p_{k+1}}\\right) \\\\\n&= f_k(n) - f_k(m) - f_k\\left(\\frac{r}{p_{k+1}}\\right) \\\\\n&= f_k(n) - f_k(m) - \\frac{r}{p_{k+1}} \\cdot \\Pi_k \\\\\n&\\ge f_k(n) - f_k(m) - \\Pi_{k+1}.\n\\end{align*}\n$$\n\nPick an integer $r \\not\\equiv b_i \\pmod{p_i}$, where $1 \\le i \\le k$. By Dirichlet's Theorem there exist infinitely many primes $p_{k+1}$ such that $p_{k+1} \\equiv (b_i - r) \\cdot a_i^{-1} \\pmod{p_i}$ for all $1 \\le i \\le k$, which satisfy $n = mp_{k+1} + r$, where $m \\equiv a_i \\pmod{p_i}$ and $n \\equiv b_i \\pmod{p_i}$. Therefore we can find a prime $p_{k+1} > r$ satisfying the given conditions and $M_{k+1} \\ge M_k + m_k - \\Pi_{k+1}$. Using induction hypothesis we conclude that\n\n$$\nm_{k+1} = M_{k+1} + \\Pi_{k+1} \\ge M_k + m_k = 2m_k - \\Pi_k > 2(2^{k-3} + \\Pi_k) - \\Pi_k > 2^{k-2} + \\Pi_{k+1}.\n$$\n\n% IMAGE: ![](images/BMO2024Shortlist_p36_data_fbbf4b555a.png)\n\n**Alternative Solution.** The most obvious approach appears to be induction on $k$. Let's see if we can make that work.\n\n**Step 1.** Say $n$ and $p_1, p_2, \\dots, p_k$ work. We are going to keep these primes and add one new prime $q$ to them. We'll set things up so that $q$ is much larger than $p_1, p_2, \\dots, p_k$. We will select some new positive integer $N$ to go with $p_1, p_2, \\dots, p_k, q$.\n\nLet $P = p_1p_2\\cdots p_k$. We decree right from the start that $N$ is congruent to $n$ modulo $P$. So $N = KP + n$ for some positive integer $K$. This way, we get to keep all fractional parts from the induction hypothesis.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18186,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)$ be a sequence defined by the recurrence\n\n$$\nx_n^2 = \\frac{n}{2} + x_{n-1} x_{n-2}, \\quad n \\geq 2,\n$$\n\nwith suitable initial conditions. Find the largest constant $A$ such that $x_n > A n$ for all $n \\in \\mathbb{N}_0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Writing down analogous relations for $n, n-1, \\dots, 3, 2$, gives\n\n$$\n\\begin{aligned}\n2x_n^2 &\\le x_{n-1}^2 + x_{n-2}^2 + n, \\\\\n2x_{n-1}^2 &\\le x_{n-2}^2 + x_{n-3}^2 + n-1, \\\\\n&\\vdots \\\\\n2x_3^2 &\\le x_2^2 + x_1^2 + 3, \\\\\n2x_2^2 &\\le x_1^2 + x_0^2 + 2.\n\\end{aligned}\n$$\n\nAdding all these inequalities gives\n\n$$\n2x_n^2 + x_{n-1}^2 \\le 2x_1^2 + x_0^2 + [n + (n-1) + \\dots + 3 + 2],\n$$\n\ni.e.\n\n$$\n2x_n^2 + x_{n-1}^2 \\le \\frac{n^2 + n + 4}{2}. \\quad (*)\n$$\n\nIf there exists a sought number $A$, the left hand side of this inequality is at least $2A^2 n^2 + A^2 (n-1)^2$, thus\n\n$$\n2A^2 n^2 + A^2 (n-1)^2 \\le \\frac{n^2 + n + 4}{2},\n$$\n\ni.e. $6A^2 n^2 - 4A^2 n + 2A^2 \\le n^2 + n + 4$. This inequality holds for every $n \\in \\mathbb{N}$ only if $6A^2 \\le 1$, i.e. $A \\le \\frac{\\sqrt{6}}{6}$.\n\nWe prove that $A = \\frac{\\sqrt{6}}{6}$ satisfies the conditions of the problem.\n\nFirst, we show inductively that $x_n > \\frac{n\\sqrt{6}}{6}$, for all $n \\in \\mathbb{N}_0$. The statement is true for $n=0$ and $n=1$. Let us assume $x_{n-1} > \\frac{(n-1)\\sqrt{6}}{6}$ and $x_{n-2} > \\frac{(n-2)\\sqrt{6}}{6}$. Then\n\n$$\nx_n^2 = \\frac{n}{2} + x_{n-1} x_{n-2} > \\frac{n}{2} + \\frac{(n-1)\\sqrt{6}}{6} \\cdot \\frac{(n-2)\\sqrt{6}}{6} = \\frac{n^2 + 2}{6} > \\frac{n^2}{6},\n$$\n\ni.e. $x_n > \\frac{n\\sqrt{6}}{6}$, and this finishes the inductive step.\n\nNext, we prove that $x_n < \\frac{n\\sqrt{6}}{6} + 1$, for all $n \\in \\mathbb{N}$. The statement is true for $n=1$ and $n=2$ by inspection. Inequality $(*)$ and the fact $x_{n-1}^2 > \\frac{(n-1)^2}{6}$ imply\n\n$$\n2x_n^2 \\le \\frac{n^2 + n + 4}{2} - x_{n-1}^2 < \\frac{n^2 + n + 4}{2} - \\frac{(n-1)^2}{6} = \\frac{2n^2 + 5n + 11}{6}.\n$$\n\nIt remains to show\n\n$$\n\\frac{2n^2 + 5n + 11}{6} < 2 \\left( \\frac{n\\sqrt{6}}{6} + 1 \\right)^2,\n$$\n\nbut this inequality is equivalent to $5n - 1 < 4n\\sqrt{6}$ which holds for $n \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18187,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system of equations:\n\n$$\nx^{x+y} = y^{y-x}\n$$\n\nand\n\n$$\ny = x^{-2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The second equation implies $y = x^{-2}$, so\n\n$$\nx^{x + x^{-2}} = x^{-2(x - x^{-2})}.\n$$\n\nTaking the logarithm on both sides, we get\n\n$$\n(x + x^{-2}) \\log x = -2(x - x^{-2}) \\log x.\n$$\n\nIf $\\log x = 0$, then $x = 1$ and $y = 1$.\n\nOtherwise,\n\n$$\nx + x^{-2} = -2x + 2x^{-2},\n$$\n\nso $3x^3 = 1$. This implies $x = \\frac{1}{\\sqrt{3}}$ and $y = \\sqrt{3}$.\n\nThere are two solutions: $x = 1$, $y = 1$ and $x = \\frac{1}{\\sqrt{3}}$, $y = \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18188,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the number $A = \\frac{4n!}{n!\\,2n!}$, where $n$ is a positive integer, is an integer and has a factor of the form $2^{n+1}$.\n\n(Note: The number $n!$ for $n \\in \\mathbb{N}$ is defined by $n! = 1 \\cdot 2 \\cdot \\dots \\cdot n$, and $0! = 1$.)",
    "options": [],
    "answer": "See solution",
    "solution": "We can write:\n\n$$\nA = \\frac{4n!}{n!\\,2n!}\n$$\n\nSince $\\binom{3n}{n} \\in \\mathbb{Z}$, $A$ is an integer.\n\nNext, we observe that in the prime factorization of $n!$, the exponent of $2$ is\n\n$$\n\\exp n = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{2^m} \\right\\rfloor,\n$$\nwhere $m$ is the maximal natural number such that $2^m \\leq n$. Hence,\n\n$$\n\\exp n \\leq \\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^m} = \\frac{n}{2} \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{2^{m-1}}\\right) = n - \\frac{n}{2^m}.\n$$\n\nSimilarly,\n\n$$\n\\exp 2n = n + \\exp n\n$$\n$$\n\\exp 4n = 3n + \\exp n\n$$\n\nTherefore, the exponent of $2$ in the factorization of $A$ is\n\n$$\n3n + \\exp n - [\\exp n + n + \\exp n] = 2n - \\exp n \\geq 2n - (n - \\frac{n}{2^m}) = n + \\frac{n}{2^m} \\geq n + 1,\n$$\n\nso $2^{n+1}$ is a factor of $A$.\n\n**Second solution:**\n\nWe can write\n$$\nA = \\frac{(2n+1)(2n+2)\\dots(4n-1)(4n)}{n!}\n$$\n\nIn the numerator, there are $n-1$ even integers from $2n+2$ to $4n-2$, giving $2$ as a factor $n-1$ times. From $4n$, we get the factor $2$ two times, so\n\n$$\nA = 2^{n+1} \\frac{n(n+1)(n+2)\\dots(2n-1)}{n!} \\cdot (2n+1)(2n+3)\\dots(4n-1)\n$$\n\nSince $\\frac{n(n+1)(n+2)\\dots(2n-1)}{n!} = \\binom{2n-1}{n-1}$,\n\n$$\nA = 2^{n+1} \\binom{2n-1}{n-1} (2n+1)(2n+3)\\dots(4n-1)\n$$\n\nThus, $A$ is an integer divisible by $2^{n+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18189,
    "subject": "Mathematics (Olympiad)",
    "question": "The measure of 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 triangle $ABC$. Find the measure of angle $\\hat{B}$.",
    "options": [],
    "answer": "See solution",
    "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 $m(\\angle B) = 80^\\circ$. It is easy to see that the case $m(\\angle B) \\leq 60^\\circ$ is not possible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18190,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximal number $n$ such that it is possible to connect $n$ points in the plane with segments of four colors so that from each point, at most two segments of the same color start, and every pair of points is connected by a segment?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that at most two segments of the same color start from any point. Otherwise, there exist 4 points connected with segments of the same color. But then, either all other points are connected to these 4 points with segments of the same color (implying all segments have the same color, which is impossible), or there exists a point connected to these 4 points with segments of the remaining three colors. In this case, this point is connected to at least two points with segments of the same color (different from the color of the segment connecting these two points), which is also impossible. So at most $2 \\times 4 = 8$ segments start from any point. Since $(n-1)$ segments start from any point, we have $n-1 \\leq 8$, which gives $n \\leq 9$.\n\nThe example for $n = 9$: Assign to each point one of the pairs $(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)$, and to each color assign one of the numbers $1,2,3,4$. Then three segments connect the points $(1,1), (1,2), (1,3)$, three segments connect the points $(2,1), (2,2), (2,3)$, etc.\n\n![](images/BielorrusiaMO2010_p12_data_82899cb5b3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18191,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Prove the following inequality:\n\n$$\n1008 < \\sqrt{1008^2 + 1} < 1008 + \\frac{1}{2016}\n$$\n\nLet $\\{x\\}$ denote the fractional part of $x$. Show that there exists an open interval $I = (r, \\sqrt{1008^2 + 1})$ such that for every $x \\in I$, we have $\\{x^2\\} - \\{x\\} > \\frac{2015}{2016}$, and that $I$ contains infinitely many real numbers.\n\n(b) Suppose that $\\{x^2\\} - \\{x\\} > \\frac{2015}{2016}$. Prove that no $x < 1000$ can satisfy this inequality.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The inequality is easily demonstrated by squaring both sides.\n\n$$\n1008 < \\sqrt{1008^2 + 1} < 1008 + \\frac{1}{2016}\n$$\n\nObserve that\n\n$$\n\\lim_{x \\to \\sqrt{1008^2 + 1}^{-}} \\{x^2\\} = 1\n$$\n\nand\n\n$$\n\\lim_{x \\to \\sqrt{1008^2+1}^{-}} \\{x\\} < \\frac{1}{2016}.\n$$\n\nThus, there is an open interval $I = (r, \\sqrt{1008^2 + 1})$ such that for every $x \\in I$, $\\{x^2\\} - \\{x\\} > \\frac{2015}{2016}$. Since $I$ is an interval, it contains infinitely many real numbers.\n\n(b) Suppose $\\{x^2\\} - \\{x\\} > \\frac{2015}{2016}$. Since $\\{x^2\\} < 1$, we require $\\{x\\} < \\frac{1}{2016}$. If $x < 1000$, then for some $n \\in \\{0, 1, \\dots, 999\\}$,\n\n$$\nn < x < n + \\frac{1}{2016}\n$$\n\nso\n\n$$\nn^2 < x^2 < \\left(n + \\frac{1}{2016}\\right)^2 = n^2 + \\frac{n}{1008} + \\frac{1}{2016^2}.\n$$\n\nHence,\n\n$$\n\\{x^2\\} - \\{x\\} < \\{x^2\\} < \\frac{n}{1008} + \\frac{1}{2016^2} \\le \\frac{999}{1008} + \\frac{1}{2016^2} = \\frac{1998 + \\frac{1}{2016}}{2016} < \\frac{2015}{2016}.\n$$\n\nThus, no $x < 1000$ can satisfy the given inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18192,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle such that $CA \\neq CB$ with circumcircle $\\omega$ and circumcentre $O$. Let $\\tau_A$, $\\tau_B$ be the tangents to $\\omega$ at $A$ and $B$, which meet at $X$. Now, let $Y$ be the foot of the perpendicular from $O$ onto $CX$, and let the line through $C$ parallel to $AB$ meet $\\tau_A$ at $Z$. Prove that $YZ$ bisects $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, observe that $OAXB$ is cyclic, with diameter $OX$, and $Y$ also lies on this circle since $OY \\perp XC$. Hence:\n\n$$\n\\angle AZC = \\angle XAB = \\angle ABX = \\angle AYX\n$$\n\nand so $CYAZ$ is cyclic.\n\n![](images/BMO_2022_shortlist_p26_data_6cc3c5460b.png)\n\nLet $M$ be the intersection of $YZ$ and $AC$ and let $CY$ intersect $\\omega$ again at $W$. Using the new cyclic relation we get $\\angle CYZ = \\angle CAZ$ and then using that $ZA$ is tangent to $\\omega$ we get $\\angle CAZ = \\angle CWA$, so $\\angle CYM = \\angle CWA$. Therefore, the triangles $CWA$ and $CYM$ are similar. But $CW$ is a chord of $\\omega$, and $Y$ is the foot of the perpendicular from $O$, hence $Y$ is the midpoint of $CW$. It follows from the similarity relation that $M$ is the midpoint of $AC$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18193,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ and $EFGH$ be two opposite faces of a cube, with $AE$, $BF$, $CG$, and $DH$ as edges. Let $X'$ denote the orthogonal projection of point $X$ onto a plane. The pairs $\\{A, G\\}$, $\\{B, H\\}$, $\\{C, E\\}$, and $\\{D, F\\}$ are opposite vertices. What is the maximum possible area of the orthogonal projection of the cube onto a plane?\n\n![](images/obm-2010_p16_data_e25083a99f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, without loss of generality, that $A'$ lies on the boundary of the projection of the cube. By symmetry, its opposite $G'$ also lies on the boundary. Two of the three neighboring vertices of $A$ will be neighbors of $A'$ in the projection (unless a face projects onto a line, in which case we consider a degenerate vertex). Suppose these neighbors are $B'$ and $D'$, so $E'$ is inside the projection. Again by symmetry, $H'$ and $F'$ lie on the boundary, and $C'$ is inside. Since $\\overrightarrow{AE} = \\overrightarrow{BF} = \\overrightarrow{CG} = \\overrightarrow{DH}$, the projection of the cube is $A'D'H'G'F'B'$. \n\nThe faces $ABCD$, $BCGF$, and $CDHG$ project onto the parallelograms (or line segments) $A'B'C'D'$, $B'C'G'F'$, and $C'D'H'G'$. Draw diagonals $B'D'$, $B'G'$, and $D'G'$. The area of the projection is then twice the area of triangle $B'D'G'$, which is at most the area of triangle $BDG$. This triangle is equilateral with side $\\sqrt{2}$, so the desired maximum is:\n\n$$\n2 \\cdot \\frac{(\\sqrt{2})^2 \\sqrt{3}}{4} = \\sqrt{3}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18194,
    "subject": "Mathematics (Olympiad)",
    "question": "The inscribed circle of triangle $ABC$, with centre $I$, touches sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Let $P$ be a point, on the same side of $FE$ as $A$, for which $\\angle PFE = \\angle BCA$ and $\\angle PEF = \\angle ABC$. Prove that $P$, $I$, and $D$ lie on a straight line.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle AEI = \\angle AFI = 90^\\circ$, the points $A$, $E$, $F$, $I$ lie on a circle with diameter $AI$. Moreover, since $\\angle PFE = \\angle BCA$ and $\\angle PEF = \\angle ABC$ by our assumptions on $P$, triangles $ABC$ and $PEF$ are similar, so $\\angle EPF = \\angle BAC = \\angle EAF$. $P$ was assumed to lie on the same side of $EF$ as $A$, thus it follows that $P$ also lies on the same circle as $A$, $E$, $F$, and $I$.\n\nWe also know that $BDIF$ is a cyclic quadrilateral (using the same reasoning as before, namely that $\\angle BDI = \\angle BFI = 90^\\circ$), so $\\angle FID + \\angle FBD = 180^\\circ$.\n\nNow we can conclude that $\\angle PIF = \\angle PEF = \\angle ABC = \\angle FBD = 180^\\circ - \\angle FID$, so $\\angle PIF + \\angle FID = 180^\\circ$, which means that $P$, $I$, $D$ lie on a straight line.\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p84_data_052cb89521.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18195,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a_1 = 0$, $a_{100} = 0$, and for each $2 \\leq n \\leq 99$, $a_n$ is an integer satisfying $a_n > \\frac{1}{2}(a_{n-1} + a_{n+1})$. What is the smallest possible value of $a_{19}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that\n\n$$\na_i > \\frac{a_{i-1} + a_{i+1}}{2} \\Leftrightarrow a_{i+1} - a_i < a_i - a_{i-1}.\n$$\n\nLet $d_i = a_{i+1} - a_i$ for $i = 1, 2, \\dots, 99$. Reformulate the problem in terms of $d_i$:\n\nThe condition $a_i > \\frac{1}{2}(a_{i-1} + a_{i+1})$ is equivalent to\n\n$$\nd_1 > d_2 > \\dots > d_{99}. \\qquad (1)\n$$\n\nFor each $n \\leq 99$,\n\n$$\n\\sum_{i=1}^{n} d_i = a_{n+1} - a_1 = a_{n+1}. \\qquad (2)\n$$\n\nGiven $a_1 = 0$, and $a_{100} = 0$,\n\n$$\n\\sum_{i=1}^{99} d_i = 0. \\qquad (3)\n$$\n\nThus, the problem reduces to:\n\nSuppose $d_1 > d_2 > \\dots > d_{99}$ are integers whose sum is zero. Find the smallest possible value of $d_1 + d_2 + \\dots + d_{18}$.\n\n**Case 1:** $d_{18} \\leq 31$\n\nFrom (1), $d_{19} \\leq 30$, $d_{20} \\leq 29$, ..., $d_{99} \\leq -50$. So from (3),\n\n$$\na_{19} = \\sum_{i=1}^{18} d_i = - \\sum_{i=19}^{99} d_i \\geq - \\sum_{i=30}^{-50} i = - \\frac{(30 + (-50))(81)}{2} = 810.\n$$\n\n**Case 2:** $d_{18} \\geq 32$\n\nLet $d = 32 + e$ where $e \\geq 0$. From (1), $d_{17} \\geq 33 + e$, ..., $d_1 \\geq 49 + e$. Therefore,\n\n$$\na_{19} = \\sum_{i=1}^{18} d_i \\geq \\sum_{i=32}^{49} i + 18e = \\frac{(32 + 49)(18)}{2} + 18e = 729 + 18e \\geq 729.\n$$\n\nThus, in all cases $a_{19} \\geq 729$.\n\nFinally, $729$ is attainable when $d_i = 50 - i$ for $i = 1, 2, \\dots, 99$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18196,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any positive real number $\\varepsilon$, prove that, for all but finitely many positive integers $v$, any $v$-member company, within which there are at least $(1 + \\varepsilon)v$ friendship relations, satisfies the following condition:\n\nFor some integer $u \\ge 3$, there exist two distinct $u$-member cyclic arrangements in each of which any two neighbours are friends. (Two arrangements are distinct if they are not obtained from one another through rotation and/or symmetry; a member of the company may be included in neither arrangement, in one of them or in both.)",
    "options": [],
    "answer": "See solution",
    "solution": "For 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, $|A| \\le v - 1$, so $|B| \\ge (1 + \\varepsilon)v - |A| \\ge (1 + \\varepsilon)v - (v - 1) = \\varepsilon v + 1 > \\varepsilon v$.\n\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, $$\\sum_{b \\in B} |S_b| \\ge 2 + \\dots + (|B| + 1) = |B|\\left(\\frac{|B|+3}{2}\\right) > \\frac{|B|^2}{2} > \\frac{\\varepsilon^2 v^2}{2}.$$ \n\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 $|B'| > \\varepsilon^2 v/2$.\n\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.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18197,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that $5^n - 1$ can be written as a product of an even number of consecutive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "$5^n - 1$ cannot be a product of more than five consecutive integers, since one of the factors would have to be divisible by $5$, but $5^n - 1$ is not divisible by $5$.\n\nNow consider two cases:\n\n**Case 1:** If $5^n - 1$ is a product of two consecutive integers, say $m$ and $m + 1$, then\n\n$$\n5^n = m(m+1) + 1 = m^2 + m + 1.\n$$\n\nConsider this equation modulo $5$: the left side is divisible by $5$, but the right side is never divisible by $5$:\n\n![](<table><tr><td>m mod 5</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td>m^2 + m + 1 mod 5</td><td>1</td><td>3</td><td>2</td><td>3</td><td>1</td></tr></table>)\n\nSo there is no solution in this case.\n\n**Case 2:** If $5^n - 1$ is a product of four consecutive integers, say $m, \\dots, m+3$, then\n\n$$\n\\begin{aligned}\n5^n &= m(m+1)(m+2)(m+3) + 1 = (m^2 + 3m)(m^2 + 3m + 2) + 1 \\\\\n    &= (m^2 + 3m + 1)^2 - 1 + 1 \\\\\n    &= (m^2 + 3m + 1)^2.\n\\end{aligned}\n$$\n\nTherefore $n$ must be even, and $m^2 + 3m + 1$ must be a power of $5$:\n\n$$\nm^2 + 3m + 1 = 5^k,\n$$\nwhere $k = \\frac{n}{2} > 0$. Solving for $m$ gives\n\n$$\nm = \\frac{-3 \\pm \\sqrt{5 + 4 \\cdot 5^k}}{2} = \\frac{-3 \\pm \\sqrt{5(1 + 4 \\cdot 5^{k-1})}}{2}.\n$$\n\nIf $k > 1$, then $5(1 + 4 \\cdot 5^{k-1})$ is divisible by $5$ but not by $25$, so it cannot be a square. Thus, there are no solutions for $k > 1$. For $k = 1$ and $m = 1$, the equation is satisfied, yielding the only solution:\n\n$$\n5^2 - 1 = 1 \\cdot 2 \\cdot 3 \\cdot 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18198,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $2012^4$ was multiplied by $(2012^{11})^2$. What is the result?",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n2012^4 \\cdot (2012^{11})^2 = 2012^4 \\cdot 2012^{22} = 2012^{26}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18199,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 19 blocks, what are the 5th and 10th largest possible city values that can be formed by stacking the blocks into towers, where the value of a tower of height $n$ is $1 + 2 + \\cdots + n$ and the city value is the sum of the values of its towers?",
    "options": [],
    "answer": "See solution",
    "solution": "The possible city values for 19 blocks, in decreasing order, are:\n\n$$\n190, 172, 171, 156, 155, 154, 153, 142, 140, 139, 138, 137, 136, \\ldots\n$$\n\nThus, the 5th largest city value is $155$ and the 10th largest is $139$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18200,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of integers $(a, b, c)$ such that the number\n\n$$\nN = \\frac{(a-b)(b-c)(c-a)}{2} + 2\n$$\n\nis a power of $2016$.\n\n(A power of $2016$ is an integer of the form $2016^n$, where $n$ is a non-negative integer.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, c$ be integers and $n$ a non-negative integer such that\n\n$$\n(a-b)(b-c)(c-a) + 4 = 2 \\cdot 2016^n.\n$$\n\nSet $a-b = -x$, $b-c = -y$, so the equation becomes\n\n$$\nxy(x+y) + 4 = 2 \\cdot 2016^n.\n$$\n\nIf $n > 0$, the right side is divisible by $7$, so\n\n$$\nxy(x+y) + 4 \\equiv 0 \\pmod{7}.\n$$\n\nThis leads to\n\n$$\n3xy(x+y) \\equiv 2 \\pmod{7},\n$$\n\nor\n\n$$\n(x+y)^3 - x^3 - y^3 \\equiv 2 \\pmod{7}.\n$$\n\nBy Fermat's Little Theorem, for any integer $k$, $k^3 \\equiv -1, 0, 1 \\pmod{7}$. Thus, at least one of $(x+y)^3, x^3, y^3$ is divisible by $7$, so $xy(x+y)$ is divisible by $7$, which is a contradiction.\n\nTherefore, the only possibility is $n = 0$, so\n\n$$\nxy(x+y) + 4 = 2,\n$$\n\ni.e.,\n\n$$\nxy(x+y) = -2.\n$$\n\nThe integer solutions are $(x, y) \\in \\{(-1, -1), (2, -1), (-1, 2)\\}$.\n\nThus, the required triples are $(a, b, c) = (k+2, k+1, k)$ for $k \\in \\mathbb{Z}$, and all their cyclic permutations.\n\n**Alternative version:**\n\nIf $n > 0$, then $9$ divides $(a-b)(b-c)(c-a) + 4$, i.e.,\n\n$$\nxy(x+y) + 4 \\equiv 0 \\pmod{9}.\n$$\n\nBut then $x$ and $y$ must be $1$ modulo $3$, so $xy(x+y) \\equiv 2 \\pmod{9}$, which is a contradiction. Thus, only $n = 0$ is possible, as above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18201,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be distinct positive integers such that $a^b$ divides $b^c$, $b^c$ divides $c^d$, and $c^d$ divides $d^a$.\n\n(a) Is it possible to determine which is the least one of the numbers $a$, $b$, $c$, $d$?\n\n(b) Is it possible to determine which is the greatest one of the numbers $a$, $b$, $c$, $d$?",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** The answer is yes, even under the weaker assumptions $a^b \\leq b^c$, $b^c \\leq c^d$, $c^d \\leq d^a$. The least number is $b$.\n\nWe need the inequality $\\sqrt[n]{n} > \\sqrt[n+1]{n+1}$, which holds for all $n \\geq 3$. This is equivalent to $(1 + \\frac{1}{n})^n < n$ and can be proved by induction. The base case $n = 3$ is clear, and if $(1 + \\frac{1}{n})^n < n$ for some $n$, then\n\n$$\n\\left(1 + \\frac{1}{n+1}\\right)^{n+1} < \\left(1 + \\frac{1}{n}\\right)^n \\left(1 + \\frac{1}{n+1}\\right) < n \\left(1 + \\frac{1}{n+1}\\right) < n + 1.\n$$\n\nIn particular, if $m > n \\geq 3$, then $\\sqrt[n]{m} < \\sqrt[n]{n}$. We use this general inequality to prove the following claim:\n\nIf $u, v, w \\in \\mathbb{N}$ satisfy $u^v \\leq v^w$, then $w \\geq u$ or $w \\geq v$.\n\nSuppose on the contrary that $w < u$, $w < v$ and write $u^v \\leq v^w$ as $\\sqrt[v]{v} \\geq \\sqrt[w]{u}$. Now $w < u$ implies $\\sqrt[v]{v} \\geq \\sqrt[w]{u} > \\sqrt[w]{w}$, which can hold only if $w \\in \\{1, 2\\}$. Indeed, if $w \\geq 3$ then $v > w \\geq 3$, so the general inequality leads to the impossible $\\sqrt[v]{v} < \\sqrt[w]{w}$. For $w = 2$, the condition is $u^v \\leq v^2$. Because $v > w = 2$, this gives $v > u \\geq 3$. Therefore $\\sqrt[v]{v} < \\sqrt[w]{u}$ by the general inequality. On the other hand, $\\sqrt[w]{u} < \\sqrt[u]{2}$, so $\\sqrt[v]{v} < \\sqrt[w]{u} < \\sqrt[u]{u}$. However, $\\sqrt[v]{v} < \\sqrt[u]{u}$ contradicts $u^v \\leq v^2$. Finally, if $w = 1$ then $u^v \\leq v$. On the other hand, $u > 1$, hence $u^v \\geq 2^v > v$ for all $v \\in \\mathbb{N}$. The claim is proven.\n\nGiven $a^b \\leq b^c$, $b^c \\leq c^d$, $c^d \\leq d^a$, we apply the claim to the triples $(a, b, c)$, $(b, c, d)$, $(c, d, a)$. Because $a$, $b$, $c$, $d$ are distinct, the conclusion is that none of $c$, $d$, and $a$ can be the least among $a$, $b$, $c$, $d$. Therefore, the least number is $b$.\n\n**(b)** The answer is no. Both quadruples $a = 2^8$, $b = 2$, $c = 2^4$, $d = 2^2$ and $a = 2^8$, $b = 2$, $c = 2^4$, $d = 2^9$ satisfy the condition. The greatest number in the first is $a = 2^8$; the greatest number in the second is $d = 2^9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18202,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 12 red points on a circle. Find the minimum value of $n$ such that there exist $n$ triangles, each with vertices at red points, so that every chord with red endpoints is a side of at least one triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the set of 12 red points be $A = \\{A_1, A_2, \\dots, A_{12}\\}$. From $A_1$, there are 11 chords with red endpoints, but every triangle with vertex $A_1$ contains two such chords. Thus, the 11 chords must be covered by at least 6 triangles with $A_1$ as a vertex. The same applies to each $A_i$ ($i = 2, 3, \\dots, 12$), so we need\n\n$$12 \\times 6 = 72$$\n\ntriangles (counting with multiplicity), and since each triangle has 3 vertices, the minimum number is\n\n$$n \\ge \\frac{72}{3} = 24.$$\n\nOn the other hand, we can construct such a set with $n = 24$ as follows:\n\nPlace the 12 red points equally spaced on the circle. The number of chords with red endpoints is $\\binom{12}{2} = 66$. If the length of the minor arc to a chord is $k$, call it a chord of type $k$. There are six types of chords, with the number of chords of types 1, 2, ..., 5 being 12 each, and type 6 being 6.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p166_data_e0c0cf5076.png)\n\nIf three chords of types $a, b, c$ ($a \\leq b \\leq c$) can form a triangle, then $a + b = c$ or $a + b + c = 12$. The possible triangles are $(a, b, c) \\in \\{(1, 1, 2), (2, 2, 4), (3, 3, 6), (2, 5, 5), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6), (3, 4, 5), (4, 4, 4)\\}$.\n\nExamples:\n\n- The number of $(1, 2, 3)$ triangles is 6, with vertices:\n  $\\{2, 3, 5\\}, \\{4, 5, 7\\}, \\{6, 7, 9\\}, \\{8, 9, 11\\}, \\{10, 11, 1\\}, \\{12, 1, 3\\}$.\n- The number of $(1, 5, 6)$ triangles is 6, with vertices:\n  $\\{1, 2, 7\\}, \\{3, 4, 9\\}, \\{5, 6, 11\\}, \\{7, 8, 1\\}, \\{9, 10, 3\\}, \\{11, 12, 5\\}$.\n- The number of $(2, 3, 5)$ triangles is 6, with vertices:\n  $\\{2, 4, 11\\}, \\{4, 6, 1\\}, \\{6, 8, 3\\}, \\{8, 10, 5\\}, \\{10, 12, 7\\}, \\{12, 2, 9\\}$.\n- The number of $(4, 4, 4)$ triangles is 3, with vertices:\n  $\\{1, 5, 9\\}, \\{2, 6, 10\\}, \\{3, 7, 11\\}$.\n- The number of $(2, 4, 6)$ triangles is 3, with vertices:\n  $\\{4, 6, 12\\}, \\{8, 10, 4\\}, \\{12, 2, 8\\}$.\n\nThus, the minimum $n$ is $24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18203,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triads of positive integers $x, y, p$, where $p$ is prime, which satisfy the equation:\n$$\n\\frac{xy^3}{x+y} = p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$. Then there exist $a, b \\in \\mathbb{Z}$ such that $x = da$, $y = db$, with $(a, b) = 1$. Substituting into the equation gives:\n$$\n\\frac{d^3 a b^3}{a + b} = p. \\qquad (1)\n$$\nSince $(a, b) = 1$, we have $(a, a + b) = 1$ and $(b^3, a + b) = 1$, so $a + b$ divides $d^3$. Let\n$$\n\\frac{d^3}{a + b} = k, \\qquad (2)\n$$\nwhere $k$ is a positive integer. Then (1) becomes $k a b^3 = p$, so $b^3$ divides $p$. Since $p$ is prime, $b = 1$ and $k a = p$. Consider the cases:\n(i) If $k = p$, $a = 1$, then (2) gives $\\frac{d^3}{2} = p \\Rightarrow 2p = d^3$. Thus $2$ divides $d$, so $8$ divides $d^3$ and $8$ divides $2p$, which is impossible.\n(ii) If $k = 1$, $a = p$. Then (2) gives:\n$$\nd^3 = p + 1 \\Rightarrow d^3 - 1 = p \\Rightarrow (d - 1)(d^2 + d + 1) = p.\n$$\nSince $d^2 + d + 1 > d - 1$, the only possibility is $d - 1 = 1$, $d^2 + d + 1 = p$, so $d = 2$, $p = 7$. Therefore, $(x, y, p) = (14, 2, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18204,
    "subject": "Mathematics (Olympiad)",
    "question": "In the triangle $ABC$ with $AB > AC$, a tangent to the circumcircle of triangle $ABC$ is drawn through point $A$. This tangent intersects the line $BC$ at point $P$. On the extension of side $BA$ beyond $A$, point $Q$ is chosen such that $AQ = AC$. Let $X$ and $Y$ be the midpoints of segments $CQ$ and $AP$, respectively. Let $R$ be a point on segment $AP$ such that $AR = CP$. Prove that $CR = 2XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "$AB > AC$, therefore the tangent intersects the line $BC$ so that $P$ lies on the extension of $BC$ beyond point $C$. Suppose $\\angle B = \\beta$, then $\\angle CAP = \\beta$ (as the angle between a tangent and a chord at the point of contact). Let $AC = 2x$, $CP = 2y$. Then $AQ = 2x$ and $AR = 2y$.\n\nConsider the point $M$, the midpoint of segment $PQ$. Then $MX$ and $MY$ are the mid-segments of $\\triangle CPQ$ and $\\triangle APQ$, respectively. It follows that $MX = y$ and $MY = x$, and also $MX \\parallel CP$ and $MY \\parallel AQ$. This parallelism gives $\\angle XMY = \\angle PBQ = \\beta$ (as the angles of respectively parallel sides).\n\n![](images/UkraineMO_2015-2016_booklet_p34_data_d00192091b.png)\n\nThus, $\\frac{MY}{MX} = \\frac{x}{y} = \\frac{2x}{2y} = \\frac{AC}{AR}$ and $\\angle XMY = \\angle PAC = \\beta$. Which means $\\triangle XMY \\sim \\triangle RAC$, due to the proportionality of the two sides and the angle between them. From the similarity of triangles it follows that\n\n$$\n\\frac{CR}{XY} = \\frac{AC}{MY} = \\frac{2x}{x} = 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18205,
    "subject": "Mathematics (Olympiad)",
    "question": "是否能夠將所有的自然數分成 6 個兩兩互斥的子集合 $A_1, A_2, \\dots, A_6$，使得滿足 $x+2y=5z$ 的任意正整數 $x, y, z$，都不會同時落在某一個 $A_i$ 之中？",
    "options": [],
    "answer": "See solution",
    "solution": "可以。\n\n令 $A_i$ 為所有形如 $7^m(7k + i)$ 的正整數所成的集合，其中 $m, k$ 為非負整數，$i = 1, 2, 3, 4, 5, 6$。\n\n下證 $A_1, A_2, A_3, A_4, A_5, A_6$ 滿足題設。\n\n設 $x, y, z \\in A_i$ 且滿足 $x + 2y = 5z$；將 $x, y, z$ 分別寫成\n\n$$\nx = 7^{m_1}(7k_1 + i), \\quad y = 7^{m_2}(7k_2 + i), \\quad z = 7^{m_3}(7k_3 + i), \\quad m_j, k_j \\in \\mathbb{N} \\cup \\{0\\},\n$$\n\n且令 $\\alpha = \\min\\{m_1, m_2, m_3\\}$。將等式 $x + 2y = 5z$ 左右兩邊除以 $7^\\alpha$，並且取模 $7$ 之後，可得同餘式\n\n$$\ne_1 i + 2 e_2 i \\equiv 5 e_3 i \\pmod{7},\n$$\n\n其中 $e_1, e_2, e_3$ 等於 $0$ 或 $1$。但此式只有 $e_1 = e_2 = e_3 = 0$ 的解，不合。故 $x + 2y = 5z$ 的正整數解 $(x, y, z)$ 不可能同時落在同一個 $A_i$ 中。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18206,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c$ be a positive integer. The sequence $a_1, a_2, \\dots, a_n, \\dots$ is defined by $a_1 = c$, and $a_{n+1} = a_n^2 + a_n + c^3$ for every positive integer $n$.\n\nFind all values of $c$ for which there exist some integers $k \\geq 1$ and $m \\geq 2$ such that $a_k^2 + c^3$ is the $m$th power of some positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice\n\n$$\na_{n+1}^2 + c^3 = (a_n^2 + a_n + c^3)^2 + c^3 = (a_n^2 + c^3)(a_n^2 + 2a_n + 1 + c^3).\n$$\n\nWe first prove that $a_n^2 + c^3$ and $a_n^2 + 2a_n + 1 + c^3$ are coprime. We prove by induction that $4c^3 + 1$ is coprime with $2a_n + 1$ for every $n \\geq 1$.\n\nLet $n = 1$ and $p$ be a prime divisor of $4c^3 + 1$ and $2a_1 + 1 = 2c + 1$. Then $p$ divides $2(4c^3 + 1) = (2c + 1)(4c^2 - 2c + 1) + 1$, hence $p$ divides $1$, a contradiction. Assume now that $(4c^3 + 1, 2a_n + 1) = 1$ for some $n \\geq 1$ and the prime $p$ divides $4c^3 + 1$ and $2a_{n+1} + 1$. Then $p$ divides $4a_{n+1} + 2 = (2a_n + 1)^2 + 4c^3 + 1$, which gives a contradiction.\n\nAssume that for some $n \\geq 1$ the number\n\n$$\na_{n+1}^2 + c^3 = (a_n^2 + a_n + c^3)^2 + c^3 = (a_n^2 + c^3)(a_n^2 + 2a_n + 1 + c^3)\n$$\n\nis a power. Since $a_n^2 + c^3$ and $a_n^2 + 2a_n + 1 + c^3$ are coprime, then $a_n^2 + c^3$ is a power as well. The same argument can be further applied, giving that $a_1^2 + c^3 = c^2 + c^3 = c^2(c + 1)$ is a power.\n\nIf $a^2(a + 1) = t^m$ with odd $m \\geq 3$, then $a = t_1^m$ and $a + 1 = t_2^m$, which is impossible. If $a^2(a + 1) = t^{2m_1}$ with $m_1 \\geq 2$, then $a = t_1^{m_1}$ and $a + 1 = t_2^{m_1}$, which is impossible. Therefore, $a^2(a + 1) = t^2$, whence we obtain the solutions $a = s^2 - 1$, $s \\geq 2$, $s \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18207,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and $n$ a positive integer. Find all pairs $(n, p)$ such that $\\sqrt[3]{n + \\frac{8p}{n}}$ is a natural number.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\sqrt[3]{n + \\frac{8p}{n}} = k$ where $k$ is a natural number, so $n + \\frac{8p}{n} = k^3$. This implies $n$ divides $8p$. Since $p$ is prime, possible values for $n$ are $1, 2, 4, 8, p, 2p, 4p, 8p$.\n\n- If $n = 1$ or $n = 8p$, then $1 + 8p = k^3$ or $8p = (k-1)(k^2 + k + 1)$. Since $k^2 + k + 1$ is odd, $k-1 = 8$ and $p = k^2 + k + 1 = 91$, which is not prime.\n- If $n = 2$ or $n = 4p$, then $2 + 4p = k^3$. $k$ must be even, so $k^3$ is divisible by $4$, but $2 + 4p$ is not divisible by $4$.\n- If $n = 4$ or $n = 2p$, then $4 + 2p = k^3$. $k$ must be even, so $k^3$ is divisible by $4$, which means $p$ must be even, so $p = 2$. Thus, $n = 4$ and $p = 2$ is a solution.\n- If $n = 8$ or $n = p$, then $8 + p = k^3$ or $p = (k-2)(k^2 + 2k + 4)$. For $k = 3$, $p = 19$ (prime). Thus, two solutions: $n = 8$, $p = 19$ and $n = 19$, $p = 19$.\n\n**Final solutions:**\n- $(n, p) = (4, 2)$\n- $(n, p) = (8, 19)$\n- $(n, p) = (19, 19)$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18208,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all $a, b \\in \\mathbb{Z}$,\n$$\nf(a + b) = f(f(fa)) + f(f(fb)).\n$$\nFind all possible values of $f(2020)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $f(f(x))$ must be affine. To see this, replace $(a, b)$ first by $(a - 1, a + 1)$ and then by $(a, a)$ to obtain\n$$\nf(f(a - 1)) + f(f(a + 1)) = f(2a) = 2f(f(a)).\n$$\nTherefore,\n$$\nf(f(a + 1)) - f(f(a)) = f(f(a)) - f(f(a - 1)).\n$$\nThis means that there is a constant $m$ so that for all $a \\in \\mathbb{Z}$:\n$$\nf(f(a + 1)) - f(f(a)) = m.\n$$\nInductively it follows that for all $a \\in \\mathbb{Z}$\n$$\nf(f(a)) = ma + c.\n$$\nThen, taking the original functional equation with $b = 0$:\n$$\nf(a) = f(f(fa)) + f(f(0)) = ma + 2c.\n$$\nTaking the full original functional equation, we have:\n$$\n\\begin{aligned}\n    m(a + b) + 2c &= f(a + b) \\\\\n    &= f(f(a)) + f(f(b)) \\\\\n    &= (ma + c) + (mb + c) \\\\\n    &= m(a + b) + 2c.\n\\end{aligned}\n$$\nAs this holds for arbitrary $a, b$, we must have:\n$$\n\\begin{aligned}\n    m &= m^2 \\\\\n    0 &= (2m + 1)c.\n\\end{aligned}\n$$\nThe first equation implies either $m = 0$ or $m = 1$, so $2m + 1 \\neq 0$ and the second equation then implies $c = 0$. Putting these together, we see that either $f(a) = 0$ for all $a \\in \\mathbb{Z}$ or $f(a) = a$ for all $a \\in \\mathbb{Z}$. Thus, the possible values of $f(2020)$ are $0$ or $2020$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18209,
    "subject": "Mathematics (Olympiad)",
    "question": "On the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$, consider points $X$ and $Y$ such that $ABXY$ is a parallelogram. Prove that the circumradii of triangles $BXD$ and $CYA$ are equal.\n\n![](images/Ukrajina_2011_p14_data_504167e449.png)\n\nFig. 12",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABCD$ is cyclic, $\\angle ABD = \\angle ACD$. Moreover, $\\angle ABD = \\angle ABY$. Hence, $CXYD$ is cyclic because $\\angle XCD = \\angle XYB$ (see Fig. 12). This implies that $\\angle XCY = \\angle XDY$, so $\\sin \\angle ACY = \\sin \\angle BDX$. Moreover, $BX = AY$, and applying the sine law for triangles $BXD$ and $CAY$ we get:\n\n$$\nR_{BDX} = \\frac{BX}{2\\sin \\angle BDX} = \\frac{AY}{2\\sin \\angle ACY} = R_{CYA}\n$$\n\nThis finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18210,
    "subject": "Mathematics (Olympiad)",
    "question": "In a round robin chess tournament, each player plays every other player exactly once. The winner of each game gets 1 point, the loser gets 0 points, and if the game ends in a tie, each player gets 0.5 points.\n\nGiven a positive integer $m$, a tournament is said to have property $P(m)$ if, for every set $S$ of $m$ players, there is one player who won all his games against the other $m-1$ players in $S$, and one player who lost all his games against the other $m-1$ players in $S$.\n\nFor a given integer $m \\geq 4$, determine the minimum value of $n$ (as a function of $m$) such that, in every $n$-player round robin chess tournament with property $P(m)$, the final scores of the $n$ players are all distinct.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are $2m-4$ players, labeled\n$$\na_1, a_2, \\dots, a_{m-3}, A_{m-2}, B_{m-2}, a_{m-1}, \\dots, a_{2m-5},\n$$\nand assume player $P_i$ beats player $P_j$ if and only if $i > j$, with $A_{m-2}$ and $B_{m-2}$ tying. In any group of $m$ players, there is a unique player $P_i$ with the maximum index $i$ ($m-1 \\leq i \\leq 2m-5$) who won all games against others in the group, and a unique player $P_j$ with the minimum index ($1 \\leq j \\leq m-3$) who lost all games against others in the group. Thus, this tournament has property $P(m)$, but not all players have distinct total points. If $n < 2m-4$, a similar construction is possible by removing players from both ends index-wise. Therefore, the answer is greater than $2m-3$.\n\nNow, we claim:\n\nIf there are $2m-3$ players in a tournament with property $P(m)$, then the players must have distinct total final scores.\n\nDefine a player who won (or lost) all games against the rest of a group as the winner (or loser) of the group. A player who won (or lost) all his games in the tournament is a complete winner (or complete loser).\n\n**Lemma 1:** In an $n$-player ($n \\geq m$) tournament with property $P(m)$, there is a complete winner.\n\n*Proof:* Induct on $n$. For $n = m$, the statement is trivial. Assume true for $n = k$ ($k \\geq m$). Consider a $(k+1)$-player tournament with property $P(m)$, with players $a_1, \\dots, a_{k+1}$. By induction, $a_{k+1}$ is the winner in the group $a_2, \\dots, a_{k+1}$. Consider:\n\n(a) If $a_{k+1}$ won against $a_1$, then $a_{k+1}$ is the complete winner.\n\n(b) If $a_{k+1}$ tied with $a_1$, then the group $a_1, a_2, \\dots, a_{k-1}, a_{k+1}$ has no winner, violating property $P(m)$.\n\n(c) If $a_{k+1}$ lost to $a_1$, then in the group $\\{a_1, a_2, \\dots, a_{k-1}, a_k, a_{k+1}\\} \\setminus \\{a_i\\}$ ($2 \\leq i \\leq k$), the winner can only be $a_1$, so $a_1$ is the complete winner.\n\nThus, a complete winner exists by induction.\n\nSimilarly,\n\n**Lemma 2:** In an $n$-player ($n \\geq m$) tournament with property $P(m)$, there is a complete loser.\n\nWith these lemmas, the claim follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18211,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram and the points $E$, $F$ are in the exterior. If triangles $BCF$ and $DEC$ are similar, i.e., $\\triangle BCF \\sim \\triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AB = x$, $BF = x'$, $AF = x'$, $AD = y$, $DE = y'$, $AE = y'$, $CF = z$, $CE = z'$, $EF = z'$. Then $x : y' : z' = x : y : z'$.  \n\nLet $\\angle CDE = \\angle FBC = a$, $\\angle DCE = \\angle BFC = b$, $\\angle DEC = \\angle BCF = c$ and $\\angle ADC = \\angle ABC = d$. Then $\\angle BCD = \\angle BAD = 180^\\circ - d$.  \n\n$$\n\\therefore \\angle ECF = 360^\\circ - b - c - (180^\\circ - d) = a + d = \\angle ABF = \\angle EDA.\n$$\n\n$$\n\\text{Also } \\frac{x'}{x} = \\frac{y'}{y} = \\frac{z'}{z} \\text{ (since } \\triangle BCF \\sim \\triangle DEC.)\n$$\n\n$$\n\\therefore \\triangle ABF \\sim \\triangle ADE \\sim \\triangle FCE.\n$$\n\n$$\n\\therefore \\angle DEA = \\angle CEF.\n$$\n\n$$\n\\therefore \\angle AEF = \\angle DEC = c. \\text{ Similarly } \\angle AFE = b.\n$$\n\n$$\n\\therefore \\triangle AFE \\sim \\triangle DCE \\sim \\triangle BFC.\n$$\n\n![](images/Singapore2024_p2_data_a5fcbb87e7e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18212,
    "subject": "Mathematics (Olympiad)",
    "question": "Kittens are taken out of a box to form a collection. A collection is called \"right\" if it consists of exactly $2011$ kittens such that their heads are painted with different colors, and their tails are also painted with different colors. It is known that one can choose a \"right\" collection of kittens from the box in more than one way. Prove that some kittens (maybe none) can be removed from the box so that there are exactly two ways to choose a \"right\" collection of kittens from the remaining ones.",
    "options": [],
    "answer": "See solution",
    "solution": "Enumerate the colors with the natural numbers from $1$ to $2011$. Construct a graph with the vertices $A_1, A_2, \\dots, A_{2011}, B_1, B_2, \\dots, B_{2011}$. For each kitten, draw an edge in the graph as follows: if its head is painted with color $i$ and its tail with color $j$, then the edge joins $A_i$ and $B_j$. Thus, each kitten corresponds to an edge in the graph. A \"right\" collection of kittens corresponds to a subset of $2011$ edges such that no two edges are adjacent to the same vertex; we call this a \"right\" subset of edges. We need to remove several (or none) edges from the graph so that there are exactly two \"right\" subsets of edges in the remaining graph.\n\nBy the problem condition, there exist at least two \"right\" subsets in the graph. Remove all edges that do not belong to any of these two subsets. For every \"right\" subset and every vertex, there is exactly one edge in the subset adjacent to that vertex. So, in the remaining graph, for each vertex, we either have an adjacent edge that belongs to both \"right\" subsets, or two adjacent edges, one from each subset. If there is exactly one edge adjacent to a vertex $V$, then the other vertex adjacent to this edge does not have any other adjacent edges. Thus, the graph is composed of cycles and isolated edges that do not share vertices with other edges. Every such cycle has even length because the edges from the first and second \"right\" subsets must alternate.\n\nEach \"right\" subset contains all isolated edges and half of the edges from each cycle, chosen so that no two edges share a vertex. There are exactly two ways to choose edges from a cycle for a \"right\" subset. The graph contains at least one cycle, because otherwise there are not two different \"right\" subsets. If there is more than one cycle, remove half of the edges (so that the remaining edges have no common vertices) from all cycles except one. Clearly, the remaining graph will have exactly one cycle, and hence, exactly two \"right\" subsets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18213,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the last two digits of the sum\n$$\nS = \\left\\lfloor \\frac{1}{3} \\right\\rfloor + \\left\\lfloor \\frac{2}{3} \\right\\rfloor + \\left\\lfloor \\frac{2^2}{3} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2^{2014}}{3} \\right\\rfloor.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the remainder when $2^n$ is divided by $3$ is $1$ when $n$ is even, and $2$ when $n$ is odd.\n\nHence $\\left\\lfloor \\frac{2^n}{3} \\right\\rfloor = \\frac{2^n-1}{3}$ when $n$ is even, and $\\left\\lfloor \\frac{2^n}{3} \\right\\rfloor = \\frac{2^n-2}{3}$ when $n$ is odd. It follows that\n\n$$\n\\begin{align*}\nS &= \\left\\lfloor \\frac{1}{3} \\right\\rfloor + \\left\\lfloor \\frac{2}{3} \\right\\rfloor + \\left\\lfloor \\frac{2^2}{3} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2^{2014}}{3} \\right\\rfloor \\\\\n&= 0 + \\left(\\frac{2}{3} - \\frac{2}{3} + \\frac{2^2}{3} - \\frac{2}{3}\\right) + \\left(\\frac{2^3}{3} - \\frac{2}{3} + \\frac{2^4}{3} - \\frac{2}{3}\\right) + \\dots + \\left(\\frac{2^{2013}}{3} - \\frac{2}{3} + \\frac{2^{2014}}{3} - \\frac{2}{3}\\right) \\\\\n&= \\left(\\frac{2}{3} + \\frac{2^2}{3} - 1\\right) + \\left(\\frac{2^3}{3} + \\frac{2^4}{3} - 1\\right) + \\dots + \\left(\\frac{2^{2013}}{3} + \\frac{2^{2014}}{3} - 1\\right) \\\\\n&= \\left(\\frac{2}{3} + \\frac{2^2}{3} + \\frac{2^3}{3} + \\dots + \\frac{2^{2014}}{3}\\right) - 1007 \\\\\n&= \\frac{2^{2015} - 2}{3} - 1007\n\\end{align*}\n$$\n\nThe last two digits of powers of $2$ repeat every $20$ terms. The last two digits of $2^{2015}$ are the same as those of $2^{15}$, which is $68$.\n\nNow, $2^{2015} - 2 = 100k + 66$. Since $\\frac{2^{2015}-2}{3}$ is an integer, $k$ is a multiple of $3$, so $k = 3m$. Thus, the last two digits of $S$ are the same as those of $100m + 22 - 7 = 15$.\n\n**Answer:** $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18214,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of 4-digit numbers (in base 10) having non-zero digits and which are divisible by 4 but not by 8.",
    "options": [],
    "answer": "See solution",
    "solution": "If we take any four consecutive even numbers and divide them by 8, we get remainders 0, 2, 4, 6 in some order. Thus, there is only one number of the form $8k + 4$ among them which is divisible by 4 but not by 8.\n\nHence, if we take four even consecutive numbers:\n\n$$\n1000a + 100b + 10c + 2, \\quad 1000a + 100b + 10c + 4, \\\\\n1000a + 100b + 10c + 6, \\quad 1000a + 100b + 10c + 8,\n$$\n\nthere is exactly one among these four which is divisible by 4 but not by 8. Now, we can divide the set of all 4-digit even numbers with non-zero digits into groups of four such consecutive even numbers with $a, b, c$ nonzero. In each group, there is exactly one number which is divisible by 4 but not by 8. The number of such groups is precisely $9 \\times 9 \\times 9 = 729$, since we can vary $a, b, c$ in the set $\\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18215,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C = \\{z \\in \\mathbb{C} \\mid |z| = 1\\}$ be the unit circle in the complex plane. 240 complex numbers $z_1, z_2, \\dots, z_{240} \\in C$ (repetitions allowed) satisfy:\n\n1. For any open arc $\\Gamma$ of length $\\pi$ on $C$, at most 200 of the $z_j$ lie in $\\Gamma$.\n2. For any open arc $\\gamma$ of length $\\frac{\\pi}{3}$ on $C$, at most 120 of the $z_j$ lie in $\\gamma$.\n\nFind the maximum value of $|z_1 + z_2 + \\dots + z_{240}|$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum is $80 + 40\\sqrt{3}$.\n\nTake 80 of $1$, and 40 each of $\\exp\\left(\\frac{\\pi}{6}i\\right)$, $\\exp\\left(-\\frac{\\pi}{6}i\\right)$, $i$, and $-i$. These choices satisfy conditions (1) and (2), and their sum is $80 + 40\\sqrt{3}$.\n\nTo show this is maximal, let $z_1, \\dots, z_{240}$ satisfy (1) and (2). By rotation, assume $S = z_1 + \\dots + z_{240}$ is a nonnegative real number. Starting from $-1$ (included) along $C$ in counterclockwise order, let the numbers be $z_1, \\dots, z_{240}$.\n\nCondition (1): For $1 \\leq j \\leq 40$, to go from $z_j$ to $z_{j+200}$ along $C$ requires at least an arc of length $\\pi$, i.e., $z_{j+200} = z_j \\exp(\\alpha i)$ for some $\\alpha \\in [\\pi, 2\\pi]$. Let $z_j = -1 \\exp(\\beta i)$, $\\beta \\in [0, 2\\pi)$. Then $\\beta + \\alpha < 2\\pi$, so\n\n$$\n\\operatorname{Re}(z_j + z_{j+200}) = -\\cos \\beta - \\cos(\\beta + \\alpha) = -2 \\cos \\frac{\\alpha}{2} \\cos \\left(\\beta + \\frac{\\alpha}{2}\\right) \\leq 0\n$$\n\nbecause $\\frac{\\alpha}{2} \\in [\\frac{\\pi}{2}, \\pi]$ and $\\beta + \\frac{\\alpha}{2} \\in [\\frac{\\pi}{2}, \\frac{3\\pi}{2}]$. Summing $1 \\leq j \\leq 40$:\n\n$$\n\\operatorname{Re}(z_1 + \\cdots + z_{40} + z_{201} + \\cdots + z_{240}) \\leq 0. \\quad \\textcircled{1}\n$$\n\nCondition (2): For $41 \\leq j \\leq 80$, to go from $z_j$ to $z_{j+120}$ along $C$ requires at least an arc of length $\\frac{\\pi}{3}$, i.e., $z_{j+120} = z_j \\exp(\\alpha i)$ for $\\alpha \\in [\\frac{\\pi}{3}, 2\\pi]$. Let $z_j = -1 \\exp(\\beta i)$, $\\beta \\in [0, 2\\pi)$. Then\n\n$$\n\\operatorname{Re}(z_j + z_{j+120}) = -2 \\cos \\frac{\\alpha}{2} \\cos \\left(\\beta + \\frac{\\alpha}{2}\\right).\n$$\n\nIf $\\alpha \\geq \\pi$, this is $\\leq 0$. If $\\alpha \\in [\\frac{\\pi}{3}, \\pi)$, then\n\n$$\n|2 \\cos \\frac{\\alpha}{2} \\cos(\\beta + \\frac{\\alpha}{2})| \\leq 2 \\cos \\frac{\\pi}{6} = \\sqrt{3},\n$$\n\nso the real part is $\\leq \\sqrt{3}$. Summing $41 \\leq j \\leq 80$:\n\n$$\n\\operatorname{Re}(z_{41} + \\cdots + z_{80} + z_{161} + \\cdots + z_{200}) \\leq 40\\sqrt{3}. \\quad \\textcircled{2}\n$$\n\nCombining (1) and (2):\n\n$$\n\\begin{aligned}\n|z_1 + \\cdots + z_{240}| &= \\operatorname{Re}\\left(\\sum_{j=1}^{40} (z_j + z_{200+j})\\right) + \\operatorname{Re}\\left(\\sum_{j=41}^{80} (z_j + z_{120+j})\\right) + \\operatorname{Re}\\left(\\sum_{j=81}^{120} z_j\\right) \\\\\n&\\leq 0 + 40\\sqrt{3} + 80 = 80 + 40\\sqrt{3}.\n\\end{aligned}\n$$\n\nThus, the maximum is $80 + 40\\sqrt{3}$, achieved as above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18216,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $x_1, x_2, \\dots, x_{2n}$ be nonnegative real numbers such that $x_1 + x_2 + \\dots + x_{2n} = 4$. Prove that there exist non-negative integers $p$ and $q$ such that $q \\le n-1$ and\n\n$$\n\\sum_{i=1}^{q} x_{p+2i-1} \\le 1, \\quad \\sum_{i=q+1}^{n-1} x_{p+2i} \\le 1.\n$$\n\n*Note 1:* The subscripts are taken modulo $2n$, that is, $k \\equiv l \\pmod{2n}$ implies $x_k = x_l$.\n\n*Note 2:* If $q=0$, the first sum is $0$; if $q = n-1$, the second sum is $0$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nDivide $x_1, x_2, \\dots, x_{2n}$ into two groups by the parity of the subscripts:\n\n- $A = x_1 + x_3 + \\dots + x_{2n-1}$\n- $B = x_2 + x_4 + \\dots + x_{2n}$\n\nDefine the partial sums $A(0) = B(0) = 0$,\n\n$$\n\\begin{cases}\nA(2k + 1) = A(2k) + x_{2k+1}, \\\\\nB(2k + 1) = B(2k),\n\\end{cases}\n\\quad\n\\begin{cases}\nA(2k + 2) = A(2k + 1), \\\\\nB(2k + 2) = B(2k + 1) + x_{2k+2},\n\\end{cases}\n\\quad k = 0, 1, 2, \\dots\n$$\n\nThe subscripts are modulo $2n$; the partial sums increase periodically: $A(k + 2n) = A(k) + A$, $B(k + 2n) = B(k) + B$.\n\nFurthermore, we turn the partial sums into (piecewise linear) continuous functions. For nonnegative real $t$, define\n\n$$\n\\begin{aligned}\nA(t) &= A(\\lfloor t \\rfloor) + (t - \\lfloor t \\rfloor)[A(\\lfloor t \\rfloor + 1) - A(\\lfloor t \\rfloor)], \\\\\nB(t) &= B(\\lfloor t \\rfloor) + (t - \\lfloor t \\rfloor)[B(\\lfloor t \\rfloor + 1) - B(\\lfloor t \\rfloor)].\n\\end{aligned}\n$$\n\nThen $A(\\cdot)$ and $B(\\cdot)$ are non-decreasing, continuous, periodic functions on $\\mathbb{R}_{\\ge 0}$.\n\nSince $A + B = 4$, there exists a positive integer $L$ such that $\\lfloor \\frac{L}{A} \\rfloor + \\lfloor \\frac{L}{B} \\rfloor \\ge L$. For $l = 0, 1, 2, \\dots, L$, by continuity of $A(\\cdot)$, we may choose suitable $t_l \\in \\mathbb{R}_{\\ge 0}$ such that $A(t_l) = l$ (let $t_0 = 0$). Since $L \\ge A \\lfloor \\frac{L}{A} \\rfloor = A \\left( 2n \\cdot \\lfloor \\frac{L}{A} \\rfloor \\right)$, we may further require that $t_L \\ge 2n \\cdot \\lfloor \\frac{L}{A} \\rfloor$.\n\nNow,\n\n$$\n\\begin{aligned}\nB(t_L) - B(t_0) &= B(t_L) \\ge B \\left( 2n \\cdot \\lfloor \\frac{L}{A} \\rfloor \\right) \\ge B \\lfloor \\frac{L}{A} \\rfloor \\\\\n&\\ge B \\left( L - \\lfloor \\frac{L}{B} \\rfloor \\right) \\ge (B-1) \\cdot L.\n\\end{aligned}\n$$\n\nConsequently, there exists some $l = 0, 1, \\dots, L-1$ such that $B(t_{l+1}) - B(t_l) \\ge (B-1)$, or\n\n$$\nB(t_l + 2n) - B(t_{l+1}) \\le 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18217,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ be real numbers such that $a + b + c = 1$. Prove that\n\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": "See solution",
    "solution": "Since $a + b + c = 1$, we have $a - a^2 = ab + ca$, so\n\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 similarly for the other terms. The inequality to be proven is equivalent to\n\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} + \\frac{1}{c^2 + 1} \\le \\frac{9}{10(ab + bc + ca)}. \\quad (*)\n$$\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 $0 < m \\le \\frac{1}{3}$.\n\nSince $\\frac{1}{a^2 + 1} = 1 - \\frac{a^2}{a^2 + 1}$, we can rewrite the inequality to be proven as\n\n$$\n\\frac{a^2}{a^2 + 1} + \\frac{b^2}{b^2 + 1} + \\frac{c^2}{c^2 + 1} \\ge 3 - \\frac{9}{10m}.\n$$\n\nUsing a variant of the Cauchy-Schwarz inequality, we have\n\n$$\n\\frac{a^2}{a^2 + 1} + \\frac{b^2}{b^2 + 1} + \\frac{c^2}{c^2 + 1} \\ge \\frac{(a+b+c)^2}{a^2 + b^2 + c^2 + 3} = \\frac{1}{4-2m}.\n$$\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 $m \\in (0, \\frac{1}{3}]$. Equality holds when $a = b = c$ and $m = \\frac{1}{3}$, so $a = b = c = \\frac{1}{3}$.\n\n**REMARK 1.** Actually, a stronger result than the inequality $(*)$ can be demonstrated; since $m = ab + bc + ca \\le \\frac{1}{3}$, we can prove that\n\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} + \\frac{1}{c^2 + 1} \\le \\frac{27}{10}. \\quad (**)\n$$\n\nIndeed, for every $x \\in (0, 1)$ we have $\\frac{1}{x^2 + 1} \\le \\frac{54 - 27x}{50}$, which is equivalent to $(3x - 1)^2 (4 - 3x) \\ge 0$. Adding up the relations for $a, b, c$, we get\n\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} + \\frac{1}{c^2 + 1} \\le \\frac{3 \\cdot 54 - 27(a + b + c)}{50} = \\frac{27}{10}.\n$$\n\nTo find such a bound, note that the condition $a + b + c = 1$ and the form $f(a) + f(b) + f(c) \\le \\frac{27}{10}$, where $f(x) = \\frac{1}{x^2+1}$, suggest searching for an inequality like $f(x) \\le rx + s$ for all $x \\in (0, 1)$.\n\nAn equality case in $(**)$ is $a = b = c = \\frac{1}{3}$, so we must also have equality in $f(x) \\le rx + s$ for $x = \\frac{1}{3}$; this leads to $r + 3s = \\frac{27}{10}$, so $s = \\frac{9}{10} - \\frac{r}{3}$.\n\nThe inequality becomes\n\n$$\n\\left( r x + \\frac{9}{10} - \\frac{r}{3} \\right) (x^2 + 1) \\geq 1,\n$$\nwhich is equivalent to\n\n$$\n(3x - 1)(10rx^2 + 9x + 10r + 3) \\geq 0.\n$$\n\nTo have the left-hand side positive for all $x \\in (0, 1)$, $1/3$ should also be a root for the polynomial $10rx^2 + 9x + 10r + 3$; that leads to $r = -27/50$, $s = 54/50$, and the above inequality becomes $(3x - 1)^2(4 - 3x) \\geq 0$, which is true for $x \\in (0, 1)$.\n\nFrom a geometrical perspective, the inequality $f(x) \\le \\frac{54 - 27x}{50}$ shows that the graph of $f$ is below the tangent line at $x = \\frac{1}{3}$, even though $f$ is not concave on $(0, 1)$.\n\n**REMARK 2.** One can also use brute force to prove that if $a + b + c = 1$, then\n\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} + \\frac{1}{c^2 + 1} \\le \\frac{27}{10}.\n$$\n\nCanceling the denominators, we get\n\n$$\n27(abc)^2 + 17 \\sum_{cyc} (ab)^2 + 7 \\sum_{cyc} a^2 \\ge 3.\n$$\n\nUsing the notations $m = ab + bc + ca$ and $p = abc$, since\n\n$$\n\\sum_{cyc} (ab)^2 = (ab + bc + ca)^2 - 2abc(a + b + c) = m^2 - 2p,\n$$\nit remains to prove that\n\n$$\n\\begin{gathered}\n27p^2 + 17(m^2 - 2p) + 7(1 - 2m) - 3 \\ge 0 \\\\\n\\Leftrightarrow 17(17 - 27p)^2 + 27(7 - 17m)^2 \\ge 4400, \\\\\n\\text{which is true, because } m \\le \\frac{1}{3} \\text{ and } p \\le \\frac{1}{27}, \\text{ so } 17(17 - 27abc)^2 \\ge 4352 \\text{ and} \\\\\n27(7 - 17m)^2 \\ge 48.\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18218,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ and a circle $\\omega$ with center $I$ that touches $AB$, $AC$, and meets $BC$ at $X$, $Y$. The line through $I$ perpendicular to $BC$ meets the line through $A$ parallel to $BC$ at $Z$. Show that the circumcircles of $\\triangle XYZ$ and $\\triangle ABC$ are tangent to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $W$ be the midpoint of the major arc $BAC$, and let $A' \\in (ABC)$ be such that $AA' \\parallel BC$. Let $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$, and let $\\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.\n\nFirstly, observe that $Z \\in (AEF)$, so\n$$\n\\begin{aligned}\n\\angle ATZ &= \\angle AIZ = 90^\\circ - \\left(\\frac{\\alpha}{2} + \\gamma\\right) \\\\\n&= \\frac{\\beta - \\gamma}{2} = \\beta - \\left(90^\\circ - \\frac{\\alpha}{2}\\right) \\\\\n&= \\angle ABC - \\angle WBC = \\angle ABW = \\angle ATW,\n\\end{aligned}\n$$\nhence $T$, $Z$, $W$ are collinear. Let $TW \\cap BC = P$. By spiral similarity and the angle bisector theorem, we have $\\frac{PB}{PC} = \\frac{TB}{TC} = \\frac{BF}{CE}$, so by the converse of Menelaus' theorem for $\\triangle ABC$, we obtain that $EF$, $TZ$, $BC$ are concurrent at $P$. Thus, by power of a point at $P$, we obtain that $PX \\cdot PY = PE \\cdot PF = PZ \\cdot PT$, so $XYZT$ is cyclic.\n\nFinally, observe that the center of $(ZXY)$ lies on $IZ$, so $(ZXY)$ touches $AA'$ at $Z$. Hence, by the shooting lemma, since $W$ is the midpoint of the minor arc $AA'$, we obtain that $(ZXY)$ touches $(ABC)$ at $T$, and we are done. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18219,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ be the foot of the altitude from $P$ to $AC$, and $F$ the foot of the altitude from $P$ to $AB$. Let $O_1$ and $O_2$ be the circumcenters of $\\triangle BDF$ and $\\triangle CDE$, respectively. Prove that $O_1, O_2, E, F$ are concyclic if and only if $P$ is the orthocenter of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nConnect $BP$, $CP$, $O_1O_2$, $EO_2$, $EF$, and $FO_1$. Since $PD \\perp BC$ and $PF \\perp AB$, the points $B$, $D$, $P$, and $F$ are concyclic. $BP$ is the diameter; $O_1$, being the circumcenter of $\\triangle BDF$, is the midpoint of $BP$. Similarly, $C$, $D$, $P$, and $E$ are concyclic, and $O_2$ is the midpoint of $CP$. Thus, $O_1O_2 \\parallel BC$, and $\\angle PO_2O_1 = \\angle PCB$.\n\n$$\nAF \\cdot AB = AP \\cdot AD = AE \\cdot AC,\n$$\n\nso $B$, $C$, $E$, and $F$ are concyclic.\n\n**Sufficiency.** Assume $P$ is the orthocenter of $\\triangle ABC$. Since $PE \\perp AC$ and $PF \\perp AB$, we know that $B$, $O_1$, $P$, and $E$ are collinear. Therefore,\n\n$$\n\\angle FOB_1 = \\angle FCB = \\angle FEB = \\angle FEO_1,\n$$\n\nwhich means $O_1$, $O_2$, $E$, and $F$ are concyclic.\n\n**Necessity.** Assume $O_1$, $O_2$, $E$, and $F$ are concyclic. Then $\\angle O_1O_2E + \\angle EFO_1 = 180^\\circ$. We have\n\n$$\n\\begin{align*}\n\\angle O_1O_2E &= \\angle O_1O_2P + \\angle PO_2E \\\\\n&= \\angle PCB + 2\\angle ACP \\\\\n&= (\\angle ACB - \\angle ACP) + 2\\angle ACP \\\\\n&= \\angle ACB + \\angle ACP,\n\\end{align*}\n$$\n\nand\n\n$$\n\\begin{align*}\n\\angle EFO_1 &= \\angle PFO_1 + \\angle PFE \\\\\n&= (90^\\circ - \\angle ABP) + (90^\\circ - \\angle ACB).\n\\end{align*}\n$$\n\nThe last identity holds because $B$, $C$, $E$, and $F$ are concyclic. Then\n\n$$\n\\begin{aligned}\n& \\angle O_1O_2E + \\angle EFO_1 \\\\\n&= \\angle ACB + \\angle ACP + (90^\\circ - \\angle ABP) + (90^\\circ - \\angle ACB) \\\\\n&= 180^\\circ.\n\\end{aligned}\n$$\n\nThat is,\n\n$$\n\\angle ABP = \\angle ACP.\n$$\n\n\\textcircled{1}\n\nFurther, since $AB < AC$ and $AD \\perp BC$, then $BD < CD$. There is a point $B'$ on $CD$ such that $BD = B'D$. Connecting $AB'$, $PB$, we have $\\angle AB'P = \\angle ABP$. By \\textcircled{1}, $\\angle AB'P = \\angle ACP$, so $A$, $P$, $B'$, $C$ are concyclic. Then $\\angle PB'B = \\angle CAP = 90^\\circ - \\angle ACB$, and $\\angle PBC + \\angle ACB = (90^\\circ - \\angle ACB) + \\angle ACB = 90^\\circ$. That means $BP \\perp AC$. Therefore, $P$ is the orthocenter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18220,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a prime $p$, and let $m$ be a number for which $p \\mid m^2 - 2$. Suppose there exists a number $a$ for which $p \\mid a^2 + m - 2$. Prove there exists a number $b$ for which\n\n$$\np \\mid b^2 - m - 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Computing modulo $p$, our assumptions are $m^2 \\equiv 2 \\pmod{p}$ and $a^2 \\equiv 2 - m \\pmod{p}$.\n\nSuppose first that $m \\equiv 2 \\pmod{p}$. In that case, $m^2 \\equiv 4 \\pmod{p}$, so $2 \\equiv 4 \\pmod{p}$, which implies $p = 2$. Then $m \\equiv 0 \\pmod{2}$ and $a^2 \\equiv 2 - m \\equiv 2 \\pmod{2}$, so $a \\equiv 0 \\pmod{2}$. We may choose $b = 0$.\n\nAssume now $m \\not\\equiv 2 \\pmod{p}$. The number $2 - m \\equiv a^2 \\pmod{p}$ is invertible modulo $p$. From\n\n$$\n(2 + m)(2 - m) = 4 - m^2 \\equiv 2 - m^2 \\equiv 2 \\pmod{p},\n$$\n\nwe deduce\n\n$$\n2 + m \\equiv m^2 (2 - m)^{-1} = m^2 a^{-2} \\pmod{p},\n$$\n\nhence we may choose $b = m a^{-1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18221,
    "subject": "Mathematics (Olympiad)",
    "question": "Integers _a_ and _b_ are randomly chosen without replacement from the set of integers with absolute value not exceeding 10. What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has 3 distinct integer roots?\n\n(A) $\\frac{1}{240}$ \\quad (B) $\\frac{1}{221}$ \\quad (C) $\\frac{1}{105}$ \\quad (D) $\\frac{1}{84}$ \\quad (E) $\\frac{1}{63}$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r, s$, and $t$ be the roots of $x^3 + ax^2 + bx + 6$. Then\n\n$$\nx^3 + ax^2 + bx + 6 = (x - r)(x - s)(x - t) = x^3 - (r + s + t)x^2 + (rs + st + tr)x - rst,\n$$\n\nso $rst = -6$, $r+s+t = -a$, and $rs+st+tr = b$. The only triples of distinct integers that satisfy $rst = -6$ are $(6, 1, -1)$, $(3, 2, -1)$, $(3, -2, 1)$, $(-3, 2, 1)$, and $(-3, -2, -1)$, together with their permutations. The corresponding values of $a$ and $b$ are $(-6, -1)$, $(-4, 1)$, $(-2, -5)$, $(0, -7)$, and $(6, 11)$, respectively. Notice that these ordered pairs are distinct, but in only 4 of them do both $a$ and $b$ have absolute value not exceeding 10. There are $21 \\cdot 20$ equally likely choices for $a$ and $b$, so the required probability is $\\frac{4}{21 \\cdot 20} = \\frac{1}{105}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18222,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of prime numbers $(p, q)$ such that $p^2$ divides $q^3 + 1$ and $q^2$ divides $p^6 - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solutions are $(p, q) = (2, 3)$ and $(3, 2)$.\n\nWhen $p = 2$, we need $q^2 \\mid 63$. The only possibility is $q = 3$. This is a solution since $4 \\mid 28$.\n\nWhen $p = 3$, we need $q^2 \\mid 728$. The only possibility is $q = 2$, which is another solution since $9 \\mid 9$.\n\nNow, assume $p \\ge 5$. Note that\n\n$$\np^2 \\mid q^3 + 1 = (q + 1)(q^2 - q + 1).\n$$\n\nAs $(q+1, q^2-q+1) = (q+1, 3) \\le 3$, we must have $p^2 \\mid q+1$ or $p^2 \\mid q^2-q+1$. Thus, $p^2 \\le q^2-q+1 < q^2$, so $p < q$.\n\nNext, observe that\n\n$$\nq^2 \\mid p^6 - 1 = (p-1)(p+1)(p^2-p+1)(p^2+p+1).\n$$\n\nSince $p < q$ and clearly $p \\ne q-1$, we have $q \\nmid p-1$ and $q \\nmid p+1$. Also, since\n\n$$\n(p^2 - p + 1, p^2 + p + 1) = (p^2 - p + 1, 2p) = 1,\n$$\n\nwe have $q^2 \\mid p^2 - p + 1$ or $q^2 \\mid p^2 + p + 1$. Thus,\n\n$$\nq^2 \\le p^2 + p + 1 \\le (q-2)^2 + (q-2) + 1 = q^2 - 3q + 3.\n$$\n\nThis is impossible. Therefore, there is no solution when $p \\ge 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18223,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, determine the minimum value the sum\n$$\n\\sum_{i=1}^{n} x_i^2 \\left( 1 + \\frac{x_i^{n-2}}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n} \\right)\n$$\nmay achieve, when $x_1, x_2, \\dots, x_n$ run through the positive real numbers subject to\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i + 1} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $n^2(n-1)$ and is achieved if and only if the $x_i$ are all equal to $n-1$.\n\nWrite\n$$\n\\sum_{i=1}^{n} x_i^2 \\left( 1 + \\frac{x_i^{n-2}}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n} \\right) = \\sum_{i=1}^{n} x_i^2 + \\sum_{i=1}^{n} \\frac{x_i^n}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n}\n$$\nand minimise each sum separately.\n\nTo minimise the first sum, notice that\n$$\n\\sum_{i=1}^{n} x_i = \\sum_{i=1}^{n} (x_i + 1) - n = \\left( \\sum_{i=1}^{n} (x_i + 1) \\right) \\sum_{i=1}^{n} \\frac{1}{x_i + 1} - n \\geq n^2 - n = n(n-1),\n$$\nso\n$$\n\\sum_{i=1}^{n} x_i^2 \\geq \\frac{1}{n} \\left( \\sum_{i=1}^{n} x_i \\right)^2 \\geq n(n-1)^2;\n$$\nclearly, equality holds if and only if the $x_i$ are all $n-1$.\n\nTo minimise the second sum, apply the AM-GM inequality to obtain\n$$\n\\frac{x_i^n}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n} + \\sum_{j \\neq i} x_j \\geq n x_i, \\quad i = 1, 2, \\dots, n,\n$$\nand sum over all $i$ to get\n$$\n\\sum_{i=1}^{n} \\frac{x_i^n}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n} \\geq \\sum_{i=1}^{n} x_i \\geq n(n-1);\n$$\nagain, equality holds if and only if the $x_i$ are all $n-1$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18224,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(x, y, z)$ of positive integers with $x > y > z > 0$, such that $$x^2 = y \\cdot 2^x + 1$$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nWe first note that the right-hand side of the equation is odd. Therefore, $x^2$ is odd, so $x$ is odd. One of the neighbors of $x$ must be divisible by $4$, so $x = 2^p a \\pm 1$ with $p > 1$ and $a$ odd.\n\nLet us first assume $x = 2^p a + 1$. If $a > 1$, we note that $x^2 - 1 = 2^{p+1} a (2^{p-1} a + 1)$, and since $y$ must contain all odd factors of $x^2 - 1$,\n\n$$\ny \\geq 3(2^{p-1} a + 1) > 2^p a + 1 = x,\n$$\n\nwhich contradicts $x > y$. Thus, $a = 1$, so $x = 2^p + 1$. The only possible value for $y$ is $2^{p-1} + 1$, since $b(2^{p-1} + 1) > 2^p + 1 = x$ for any $b > 1$. Any possible solution in this case must therefore have $x = 2^p + 1$, $y = 2^{p-1} + 1$, and $z = p + 1$. If $p = 2$, we obtain $x = 5$ and $y = z = 3$, which contradicts $y > z$. For any $p > 2$, we have $x > y$ and $y > z$, since $2^{p-1} + 1 > p + 1 \\Leftrightarrow 2^p > p$ is true. All triples $(2^p + 1, 2^{p-1} + 1, p + 1)$ are therefore solutions for $p > 2$.\n\nNow let us assume $x = 2^p a - 1$. If $a > 1$, we have $x^2 - 1 = 2^{p+1} a (2^{p-1} a - 1)$, and therefore\n\n$$\ny \\geq 3(2^{p-1} a + 1) = 2^p a - 1 + 2^{p-1} a - 2 > 2^p a - 1 + 2^p - 2 > x,\n$$\n\nwhich again contradicts $x > y$, so $a = 1$. If $x = 2^p - 1$, possible values for $y$ are either $y = 2^{p-1} - 1$ or $y = 2(2^{p-1} - 1)$, since $b(2^{p-1} - 1) > 2^p - 1$ for any $b > 2$. We therefore have two further groups of solutions. In the first case, $x = 2^p - 1$, $y = 2^{p-1} - 1$, and $z = p + 1$. If $p = 2$, we obtain $x = 3$, $y = 1$, $z = 3$, which contradicts $y > z$. If $p = 3$, we obtain $x = 7$, $y = 3$, $z = 4$, again contradicting $y > z$. If $p \\geq 4$, $y > z$ holds, since $2^{p-1} - 1 > p + 1 \\Leftrightarrow 2^p > p + 2$ is true. All triples $(2^p - 1, 2^{p-1} - 1, p + 1)$ are therefore solutions for $p > 3$.\n\nFinally, if $x = 2^p - 1$ and $y = 2^p - 2$, we have $z = p$. If $p = 2$, we obtain $x = 3$, $y = z = 2$, which again contradicts $y > z$. If $p > 2$, $y > z$ holds, since $2^p - 2 > p \\Leftrightarrow 2^p > p + 2$ is true. All triples $(2^p - 1, 2^p - 2, p)$ are therefore also solutions for $p > 2$.\n\n**Summary:**\n\nThe solutions are given by the triples:\n\n- $(2^p + 1, 2^{p-1} + 1, p + 1)$ for $p > 2$,\n- $(2^p - 1, 2^{p-1} - 1, p + 1)$ for $p > 3$,\n- $(2^p - 1, 2^p - 2, p)$ for $p > 2$.\n\nqed",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18225,
    "subject": "Mathematics (Olympiad)",
    "question": "Two vertices $\\alpha$ and $\\beta$ of the triangular lattice are called equivalent if $\\alpha - \\beta$ is equal to the sum of finitely many vectors from the set:\n\n$$\nA = \\{\\pm(n\\vec{i} - m\\vec{j}), \\pm(n\\vec{k} + m\\vec{i}), \\pm(n\\vec{j} + m\\vec{k})\\}\n$$\n\nwhere $\\vec{i} = (1,0)$, $\\vec{j} = \\left(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$, and $\\vec{k} = \\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$ in Cartesian coordinates.\n\nWhat is the maximum number of nonequivalent vertices?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\vec{k} = \\vec{i} + \\vec{j}$, so we can express the vectors in set $A$ using only $\\vec{i}$ and $\\vec{j}$:\n\n$$\nA = \\{\\pm(n\\vec{i} - m\\vec{j}), \\pm((n+m)\\vec{i} + n\\vec{j}), \\pm(m\\vec{i} + (m+n)\\vec{j})\\}.\n$$\n\nSince $n\\vec{i} - m\\vec{j}$ is the difference of the other two, two vertices $\\alpha$ and $\\beta$ are equivalent if and only if their difference $\\alpha - \\beta$ can be written as a linear combination of $$(n+m)\\vec{i} + n\\vec{j}$$ and $$m\\vec{i} + (m+n)\\vec{j}$$ with integer coefficients.\n\nFor integers $a, b, x, y$, the lattice generated by $\\{a\\vec{i} + b\\vec{j}, x\\vec{i} + y\\vec{j}\\}$ (with integer coefficients) is the same as that generated by $\\{(a-x)\\vec{i} + (b-y)\\vec{j}, x\\vec{i} + y\\vec{j}\\}$. The value $ay - bx$ (the determinant of the matrix with rows $(a, b)$ and $(x, y)$) is invariant under this operation. By repeatedly applying this process (similar to the Euclidean Algorithm), we can reduce to vectors $s\\vec{i} + r\\vec{j}$ and $t\\vec{j}$.\n\nThus, to move from one vertex to another, the difference in the $\\vec{i}$ coefficients must be a multiple of $s$, and then $t\\vec{j}$ can be used to match the $\\vec{j}$ coordinate. Therefore, there are $st$ nonequivalent vertices, where\n\n$$\nst = \\det \\begin{bmatrix} s & r \\\\ 0 & t \\end{bmatrix}\n$$\n\nThis determinant is invariant and equals\n\n$$\n\\det \\begin{bmatrix} m+n & m \\\\ n & m+n \\end{bmatrix} = (m+n)(m+n) - mn = m^2 + mn + n^2.\n$$\n\nSo, the maximum number of nonequivalent vertices is $m^2 + mn + n^2$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 18226,
    "subject": "Mathematics (Olympiad)",
    "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 greatest common divisor (g.c.d.) of the 8 numbers at the vertices is equal to $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us alternately color the vertices black and white. After any move, the difference between the sums of the numbers at the black and white vertices remains $1$. Therefore, the g.c.d. of the 8 numbers is $1$, since any common divisor must divide this difference.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18227,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+$ be the set of all positive real numbers. Determine all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n\n$$\nf(x + y + f(y)) = 4030x - f(x) + f(2016y), \\quad \\forall x, y \\in \\mathbb{R}^+.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It's trivial that if $f$ maps $x$ to $2015x$, then the functional equation holds. We'll show that it is the only function that satisfies our requirement. For simplicity, let $k = 2015$, then the functional equation becomes\n\n$$\nf(x + y + f(y)) = 2k x - f(x) + f((k+1)y), \\quad \\forall x, y \\in \\mathbb{R}^+ \\tag{1}\n$$\n\n**Step 1:**\n\n$$\nf((k+1)a) - f((k+1)b) = k(f(b) + b - f(c) - c), \\quad \\forall b, c \\in \\mathbb{R}^+.\n$$\n\nSet $x = a + b + f(b)$, $y = c$ in (1), then we get\n\n$$\n\\begin{aligned}\nd &= 2k(a + b + f(b)) - f(a + b + f(b)) + f((k+1)c) \\\\\n  &= 2k(a + b + f(b)) - (2k a - f(a) + f((k+1)b)) + f((k+1)c)\n\\end{aligned}\n$$\n\nwhere $d = f(a + b + f(b) + c + f(c))$. Similarly, when $x = a + c + f(c)$, $y = b$, then\n\n$$\nd = 2k(a + c + f(c)) - (2k a - f(a) + f((k+1)c)) + f((k+1)b)\n$$\n\nConsequently, we must have\n\n$$\nf((k+1)b) - f((k+1)c) = k(f(b) + b - f(c) - c).\n$$\n\n**Step 2:**\n\nFor all $b, c \\in \\mathbb{R}^+$,\n\n$$\n\\begin{aligned}\n& 2k b - f(b) + f((k+1)[(k+2)c + f(c)]) \\\\\n&= 2k(b + (k+1)(2k+1)c) - f(b + (k+1)(2k+1)c) + f((k+1)c)\n\\end{aligned}\n$$\n\nTake $x = (k+1)b$, $y = b$ in the original equation:\n\n$$\nf((k+2)b + f(b)) = 2k(k+1)b, \\quad \\forall b \\in \\mathbb{R}^+.\n$$\n\nApply the same trick, using double counting. First, set $x = b$, $y = (k+2)c + f(c)$ in the original equation, then\n\n$$\ne = 2k b - f(b) + f((k+1)[(k+2)c + f(c)])\n$$\n\nwhere $e = f(b + (k+2)c + f(c) + 2k(k+1)c)$. Second, set $x = b + (k+1)(2k+1)c$, $y = c$, we have\n\n$$\ne = 2k(b + (k+1)(2k+1)c) - f(b + (k+1)(2k+1)c) + f((k+1)c).\n$$\n\nThese equalities imply our statement.\n\n**Step 3:**\n\nFor all $b, c \\in \\mathbb{R}^+$, $f(b + c) = f(b) + k c$ holds.\n\nRewrite the identity in **Step 2**:\n\n$$\n\\begin{aligned}\n& f(b + (k+1)(2k+1)c) + f((k+1)[(k+2)c + f(c)]) - f((k+1)c) \\\\\n&= f(b) + 2k(k+1)(2k+1)c\n\\end{aligned}\n$$\n\nFurthermore, by using Eq. (1):\n\n$$\n\\begin{aligned}\n& f((k+1)[(k+2)c + f(c)]) - f((k+1)c) \\\\\n&= k(f((k+2)c + f(c)) + (k+2)c + f(c) - f(c) - c) \\\\\n&= k(k+1)(2k+1)c\n\\end{aligned}\n$$\n\nThus, the previous equation becomes\n\n$$\nf(b + (k+1)(2k+1)c) = f(b) + k(k+1)(2k+1)c,\n$$\n\nwhich means $f(b + c) = f(b) + k c$ holds.\n\n**Step 4:**\n\n$f(a) = k a$ for all $a \\in \\mathbb{R}^+$.\n\nNote that $f(b) + k c = f(b + c) = f(c) + k b$ for all $b, c \\in \\mathbb{R}^+$. Thus, $f(a) - k a$ is a constant. Suppose $f(x) = k x + l$ for some $l \\in \\mathbb{R}$, then\n\n$$\n\\text{LHS} = f(x + (k+1)y + l) = k x + k(k+1)y + (k+1)l\n$$\n\n$$\n\\text{RHS} = k x + k(k+1)y\n$$\n\nSo $l = 0$, and the problem is solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18228,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of real numbers such that $a_1 + a_2 = 2010$ and $a_n = a_{n-1} a_{n-2}$ for all $n \\ge 3$. What are the possible values of $a_{2011} + a_{2012}$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Case 1.* $a_1 = 0$. Then $a_2 = a_1 a_3 = 0$. But this contradicts $a_1 + a_2 = 2010$.\n\n*Case 2.* $a_2 = 0$. Then from $a_1 + a_2 = 2010$ we have $a_1 = 2010$.\n\nAlso $a_3 = a_2 a_4 = 0 \\cdot a_4 = 0$. Then $a_4 = a_3 a_5 = 0 \\cdot a_5 = 0$, and so on. In fact, whenever we have $a_n = 0$, then we also have $a_{n+1} = a_n a_{n+2} = 0 \\cdot a_{n+1} = 0$. Thus $a_n = 0$ for all $n \\ge 2$. Hence our sequence is\n\n$$\n\\begin{array}{lcl}\na_1 & = & 2010 \\\\\na_2 & = & 0 \\\\\na_3 & = & 0 \\\\\na_4 & = & 0 \\\\\n\\vdots & & \n\\end{array}\n$$\n\nIt is easy to verify that this sequence satisfies the given properties.\n\n$$\n\\text{Thus } a_{2011} + a_{2012} = 0 + 0 = 0.\n$$\n\n*Case 3.* $a_1 \\neq 0$ and $a_2 \\neq 0$. Write $a_1 = x$, $a_2 = y$ with $x, y \\neq 0$. Note that $x + y = 2010$. Using $a_n = a_{n-1} a_{n-2}$ for $n \\ge 3$, we compute\n\n$$\n\\begin{array}{lcl}\na_1 & = & x \\\\\na_2 & = & y \\\\\na_3 & = & yx \\\\\na_4 & = & yx^2 \\\\\na_5 & = & y^2 x^3 \\\\\na_6 & = & y^3 x^5 \\\\\na_7 & = & y^5 x^8 \\\\\n\\vdots & & \n\\end{array}\n$$\n\nHowever, the original solution used $a_n = \\frac{a_{n-1}}{a_{n-2}}$, but the problem states $a_n = a_{n-1} a_{n-2}$. Let's correct the computation:\n\nLet $a_1 = x$, $a_2 = y$.\n\n- $a_3 = a_2 a_1 = yx$\n- $a_4 = a_3 a_2 = (yx)y = y^2 x$\n- $a_5 = a_4 a_3 = (y^2 x)(y x) = y^3 x^2$\n- $a_6 = a_5 a_4 = (y^3 x^2)(y^2 x) = y^5 x^3$\n- $a_7 = a_6 a_5 = (y^5 x^3)(y^3 x^2) = y^8 x^5$\n\nSo, the sequence grows rapidly unless $x$ or $y$ is zero. Thus, only the case where $a_2 = 0$ (so $a_1 = 2010$) gives a valid sequence with all terms defined and $a_{2011} + a_{2012} = 0$.\n\nIn conclusion, the only possible value for $a_{2011} + a_{2012}$ is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18229,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ for which the following quantity is a positive integer:\n\n$$\n\\frac{10^n}{n^3 + n^2 + n + 1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $n=3$ and $n=7$ are the only positive integers for which\n$$\n\\frac{10^n}{n^3 + n^2 + n + 1}\n$$\nis a positive integer.\n\nFirst, check $n=3$ and $n=7$:\n\n$$\n\\frac{10^3}{3^3+3^2+3+1} = \\frac{1000}{40} = 25, \\quad \\frac{10^7}{7^3+7^2+7+1} = \\frac{10,000,000}{400} = 25,000.\n$$\n\nBoth are integers.\n\nNow, factor the denominator:\n$$\nn^3 + n^2 + n + 1 = (n+1)(n^2+1).\n$$\n\nFor the quotient to be an integer, both $n+1$ and $n^2+1$ must only have 2 and 5 as prime factors (since $10^n$ is a power of 2 and 5). Also, $\\gcd(n+1, n^2+1)$ is either 1 or 2, since $n^2+1 - (n+1)(n-1) = 2$.\n\nIf $n$ is even, $n+1$ and $n^2+1$ are both odd, so must be powers of 5. But then $\\gcd(n+1, n^2+1) = 1$, and checking small even $n$ shows no solution.\n\nIf $n$ is odd, $n+1$ and $n^2+1$ are both even. Since $n^2+1 \\equiv 2 \\pmod{4}$, $n^2+1$ is not divisible by 4, so $n^2+1$ is either 2 or $2 \\cdot 5^l$. If $n^2+1=2$, $n=1$, but $\\frac{10^1}{1^3+1^2+1+1} = \\frac{10}{4} = 2.5$ is not integer.\n\nSo, $n^2+1=2\\cdot 5^l$ and $n+1=2^k$ for $k \\geq 2$. If $k=2$, $n=3$ (already found). For $k \\geq 3$:\n\nSet $n+1=2^k$, $n=2^k-1$, so $n^2+1=(2^k-1)^2+1=2^{2k}-2^{k+1}+2$.\nSet $n^2+1=2\\cdot 5^l$:\n$$\n2\\cdot 5^l = (2^k-1)^2+1 = 2^{2k} - 2^{k+1} + 2.\n$$\nSo,\n$$\n2\\cdot 5^l - 2 = 2^{2k} - 2^{k+1}.\n$$\nDivide both sides by 2:\n$$\n5^l - 1 = 2^{2k-1} - 2^k.\n$$\nTry $k=3$:\n$$\n5^l - 1 = 2^{5} - 2^3 = 32 - 8 = 24 \\implies 5^l = 25 \\implies l=2.\n$$\nSo $k=3$, $n+1=8$, $n=7$.\n\nNo other $k$ yields integer $l$.\n\nThus, the only solutions are $n=3$ and $n=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18230,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest possible non-negative integer $k$ for which there exists a positive integer $n$ with the following property:\n\n$n$ is a perfect square with at least $k+1$ digits, and for every $i < k$ the integer obtained by deleting the last $i$ digits from $n$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $k = 2$ is the desired maximum number.\n\nFirst, if $k = 2$, then $n = 169$ satisfies the condition since $169$, $16$, and $1$ are all perfect squares.\n\nNext, let $k \\geq 3$, and suppose $n$ satisfies the condition for this $k$. Then, the number obtained from $n$ by deleting the last $2$ digits must be a perfect square, say $a^2$ for some positive integer $a$. Thus, $n = 100a^2 + b$, where $b$ is a non-negative integer with at most $2$ digits.\n\nIf $b = 0$, then the number $10a^2$ (obtained by deleting the last digit) must be a perfect square, which is impossible. Hence, $b > 0$.\n\nThis means $n$ is a perfect square greater than $100a^2 = (10a)^2$, so\n\n$$\nn = 100a^2 + b \\geq (10a + 1)^2 = 100a^2 + 20a + 1.\n$$\n\nSince $b \\leq 99$, we have $20a + 1 \\leq 99$, so $a \\leq 4$.\n\nSince $n$ has at least $k+1$ digits, $a^2$ must have at least $k-1$ digits. For $k \\geq 3$, $a^2$ must have at least $2$ digits, so $a = 4$.\n\nTherefore, $n = 1600 + b$ is a perfect square with $1600 < n \\leq 1699$. The only perfect square in this range is $1681$, but $168$ is not a perfect square, so $1681$ does not satisfy the condition.\n\nThus, for $k \\geq 3$, there is no $n$ satisfying the condition, and $k = 2$ is the maximum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18231,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(x)$ be a polynomial such that\n\n$$\np(p(x)) = (x^2 + x + 1)p(x).\n$$\n\nFind all such polynomials $p(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The zero polynomial is obviously a valid solution. Let $p$ be a non-zero polynomial and write $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\ne 0$. The leading term on the left-hand side is $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and on the right-hand side it is $x^2 \\cdot a_nx^n = a_nx^{n+2}$. Equating degrees: $n^2 = n + 2$, so $n = 2$. Thus, $p(x) = ax^2 + bx + c$, $a \\ne 0$.\n\nExpanding both sides:\n\n$$\np(p(x)) = a^3 x^4 + 2a^2 bx^3 + (ab^2 + 2a^2 c + ab)x^2 + (2abc + b^2)x + (ac^2 + bc + c)\n$$\n\n$$\n(x^2 + x + 1)p(x) = ax^4 + (b+a)x^3 + (c+b+a)x^2 + (c+b)x + c.\n$$\n\nMatching coefficients gives the system:\n\n$$\n\\begin{align*}\na^3 &= a, \\\\\n2a^2b &= b+a, \\\\\nab^2 + 2a^2c + ab &= c+b+a, \\\\\n2abc + b^2 &= c+b, \\\\\nac^2 + bc + c &= c.\n\\end{align*}\n$$\n\nSince $a \\neq 0$, $a = 1$ or $a = -1$. If $a = 1$, $b = 1$, $c = 0$. If $a = -1$, there are no solutions. Thus, the solutions are $p(x) = 0$ and $p(x) = x^2 + x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18232,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $n$ be positive integers such that:\n\n1. $a^{2021} \\mid n$ and $b^{2021} \\mid n$,\n2. $2022 \\mid a - b$ and $a > b$.\n\nProve that there is a subset of the divisors of $n$ whose sum of elements is divisible by $2022$ but not by $2022^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $a = d r$ and $b = d s$ where $d = \\gcd(a, b)$ and $(r, s) = 1$. Then $d^{2021} r^{2021} s^{2021}$ divides $n$. Furthermore, $2 \\cdot 3 \\cdot 337 \\mid d(r - s)$.\n\n**Case 1:** Assume $337 \\mid d$. Since $(r, s) = 1$, we may assume that $r$ is odd. Then\n\n$$\n\\{337 r^2, 337 r^4, \\dots, 337 r^{10}, 337 r^{12}\\}\n$$\n\nworks. Each of these six divisors of $n$ is congruent to $1 \\bmod 4$, so their sum is a multiple of $2$ but not of $4$. If $3 \\mid r$, then the sum is $0 \\bmod 3$; if $3 \\nmid r$, then each divisor is $1 \\bmod 3$, so the sum is again $0 \\bmod 3$. Therefore, the sum is a multiple of $2022$ but not of $2022^2$.\n\n**Case 2:** Assume $337 \\nmid d$. Then $337 \\mid r - s$ and, since $(r, s) = 1$, $337 \\nmid r s$. Consider the $2022^2$ divisors of $n$ of the form $r^k s^\\ell$ where $k, \\ell \\in \\{0, 1, 2, \\dots, 2021\\}$. Since none of them is a multiple of $337$, they have at most $2 \\cdot 3 \\cdot 336$ distinct remainders modulo $2022$. Therefore, at least $\\frac{2022^2}{2 \\cdot 3 \\cdot 336} > 2022$ of them have the same remainder modulo $2022$.\n\nPick $2023$ out of those, say $d_1, d_2, \\dots, d_{2023}$. Let $S$ be their sum. We claim that there is a subset of $2022$ of them that will work. Note that the sum of any such subset is a multiple of $2022$. It is enough to show that there is such a subset whose sum is not divisible by $337^2$. If this is not the case, then $S - d_i \\equiv 0 \\pmod{337^2}$ for each $i = 1, 2, \\dots, 2023$. In particular, all $d_i$ are congruent modulo $337^2$. Say that $d_i \\equiv k \\pmod{337^2}$ for each $i$. Then $337 \\nmid k$, and so the sum of any $2022$ of them is congruent to $2022k \\not\\equiv 0 \\pmod{337^2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18233,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 代表實數所成的集合。試找出所有函數 $f : \\mathbb{R} \\to \\mathbb{R}$ 滿足對所有實數 $x, y \\in \\mathbb{R}$，都有\n\n$$\nf(xy + x f(x)) = f(x) (f(x) + f(y)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "定義 $P(a, b)$ 為將 $x = a, y = b$ 代入函數方程。\n\n1. $P(0, y)$：$f(0) = f(0)(f(0) + f(y))$。當 $f(0) \\neq 0$，$f$ 是常數函數，此時 $f(x) = \\frac{1}{2}$，對所有 $x \\in \\mathbb{R}$ 是一個解。\n\n2. 若 $f(a) = 0$，$P(a, y)$：$f(ay) = 0$，得到 $a = 0$ 或 $f$ 全為 $0$，其中 $f(x) = 0$ 確實是一解。\n\n3. $P(x, -f(x))$：$0 = f(x)(f(x) + f(-f(x)))$，不管 $x$ 是否為 $0$，都有 $-f(x) = f(-f(x))$。\n\n4. $P(x, 0)$：$f(x f(x)) = f(x)^2$。當 $x = -1$，結合 (3) 得 $f(-1)^2 = f(-f(-1)) = -f(-1)$，得到 $f(-1) = -1$。\n\n5. $P(-1, 0)$：$f(1) = 1$。\n\n6. $P(1, y)$：$f(y + 1) = f(y) + 1$。\n\n7. $P(x, y + 1) - P(x, y)$：$f(xy + x + x f(x)) = f(x) + f(xy + x f(x))$，也就是柯西方程\n\n$$\nf(x) + f(y) = f(x + y), \\quad \\forall x, y.\n$$\n\n8. $f(xy + x f(x)) = f(xy) + f(x f(x)) = f(xy) + f(x)^2 = f(x)(f(x) + f(y))$，因此得到 $f(xy) = f(x) f(y)$，結合柯西方程得到 $f(x) = x$，對所有 $x \\in \\mathbb{R}$。\n\n9. 總結，滿足條件的 $f$ 只有三個：\n\n$$\nf(x) = x, \\quad \\forall x \\in \\mathbb{R}\n$$\n\n$$\nf(x) = 0, \\quad \\forall x \\in \\mathbb{R}\n$$\n\n$$\nf(x) = \\frac{1}{2}, \\quad \\forall x \\in \\mathbb{R}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18234,
    "subject": "Mathematics (Olympiad)",
    "question": "A circulator is an instrument which draws the circumcircle of three given points in the plane (if the points happen to be collinear, it draws the line through them). Is it possible to construct, only with the help of a circulator, the centre of a given circle?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible. Suppose you have an algorithm that constructs the centre of a given circle $\\gamma$, where each step consists of choosing three points and constructing their circumcircle. Consider an arbitrary circle $\\Omega$ different from $\\gamma$. Invert $\\gamma$ in $\\Omega$ to obtain a new circle $\\gamma'$. If we apply our algorithm to $\\gamma'$, we construct the centre of $\\gamma'$, but also the image of the centre of $\\gamma$ under the inversion, since each step of the algorithm commutes with inversion. Thus, the centre of $\\gamma$ is mapped to the centre of $\\gamma'$ by inversion, which is a contradiction, since $\\Omega$ was arbitrary and inversion does not, in general, map the centre of a circle to the centre of its image.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18235,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real-valued functions $f$ defined on pairs of real numbers, having the following property: for all real numbers $a, b, c$, the median of $f(a, b)$, $f(b, c)$, $f(c, a)$ equals the median of $a, b, c$.\n\n(The median of three real numbers, not necessarily distinct, is the number that is in the middle when the three numbers are arranged in non-decreasing order.)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two solutions:\n\n- $f(a, b) = a$ for all $a, b$.\n- $f(a, b) = b$ for all $a, b$.\n\nClearly these functions meet the condition. We must show there are no others.\n\nBy setting $a = b = c$, we get $f(a, a) = a$ for all $a$.\n\nNext, for all $a, b$, the median of $f(a, a)$, $f(a, b)$, $f(b, a)$ must equal the median of $a, a, b$, namely $a$, so for all $a, b$, one of $f(a, b)$, $f(b, a)$ is at most $a$ and the other is at least $a$. Switching $a, b$, we also see that one of $f(a, b)$, $f(b, a)$ is at most $b$ and the other is at least $b$. Therefore, we have\n\n$$\n\\min\\{f(a, b), f(b, a)\\} \\leq \\min\\{a, b\\}\n$$\n\nand\n\n$$\n\\max\\{f(a, b), f(b, a)\\} \\geq \\max\\{a, b\\}.\n$$\n\nNext, consider any three numbers $a < b < c$. The median of $f(a, b)$, $f(b, c)$, $f(c, a)$ must equal $b$, so one of $f(a, b)$, $f(b, c)$ equals $b$ (since $f(c, a)$ must be either at most $a$ or at least $c$). Similarly, considering $f(a, c)$, $f(c, b)$, $f(b, a)$, we see that one of $f(c, b)$, $f(b, a)$ equals $b$. The numbers $f(a, b)$, $f(b, a)$ cannot both be $b$, by the previous inequality, and $f(b, c)$, $f(c, b)$ cannot both be $b$. We conclude that either\n\n$$\nf(a, b) = f(c, b) = b\n$$\n\nor\n\n$$\nf(b, c) = f(b, a) = b.\n$$\n\nIn particular, for any $a < b$, choosing $c > b$ arbitrarily, we see that one of $f(a, b)$, $f(b, a)$ must equal $a$. Likewise, for any $b < c$, choosing $a < b$ arbitrarily, we see that one of $f(b, c)$, $f(c, b)$ must equal $b$.\n\nPutting these two conclusions together, for any $a \\neq b$, one of $f(a, b)$, $f(b, a)$ equals $\\min\\{a, b\\}$ and the other equals $\\max\\{a, b\\}$. In other words, for $a \\neq b$, $\\{f(a, b), f(b, a)\\}$ and $\\{a, b\\}$ are equal as sets. Call $\\{a, b\\}$ a first-pair if $f(a, b) = a$ and $f(b, a) = b$, and a second-pair if $f(a, b) = b$ and $f(b, a) = a$.\n\nNow again consider any three numbers $a < b < c$. If either $\\{a, b\\}$ or $\\{b, c\\}$ is a first-pair, then the previous conclusion cannot hold, so the other must hold, and $\\{a, b\\}$ and $\\{b, c\\}$ are both first-pairs. That is, $\\{a, b\\}$ is a first-pair if and only if $\\{b, c\\}$ is. Pick any other numbers $a'$ and $c'$ such that $a' < b$ and $c' > b$. The same logic gives\n\n$$\n\\begin{align*}\n\\{a', b\\} \\text{ is a first-pair} &\\iff \\{b, c\\} \\text{ is a first-pair} \\\\\n&\\iff \\{a, b\\} \\text{ is a first-pair} \\\\\n&\\iff \\{b, c'\\} \\text{ is a first-pair.}\n\\end{align*}\n$$\n\nThis shows that, given any $p, q \\neq b$, $\\{p, b\\}$ is a first-pair if and only if $\\{q, b\\}$ is. Since $b$ is arbitrary, we have for any distinct $p, q, r, s$ that\n\n$$\n\\begin{align*}\n\\{p, q\\} \\text{ is a first-pair} &\\iff \\{s, q\\} = \\{q, s\\} \\text{ is a first-pair} \\\\\n&\\iff \\{r, s\\} \\text{ is a first-pair.}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18236,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) = x^n + 2x + 1 - 4x^2$. For which integer values of $n$ does $P(x) \\ge 0$ hold for all $x > 0$?",
    "options": [],
    "answer": "See solution",
    "solution": "If $n = 1$, substituting $x = 2$ yields $P(2) = -11 < 0$, so $n = 1$ is not a solution. Assume $n \\ge 2$ and substitute $x = 1 + t$.\n\nUsing the binomial theorem:\n$$P(1+t) = (1+t)^n + 2(1+t) - 4(1+t)^2 = (n-6)t - 4t^2 + \\sum_{k=2}^n \\binom{n}{k} t^k \\le (n-6)t + \\sum_{k=2}^n \\binom{n}{k} t^k$$\nFor $k \\ge 2$, $\\binom{n}{k} t^k < n^k |t|^k$, so:\n$$P(1+t) < (n-6)t + (nt)^2 \\sum_{k=0}^{n-2} (n|t|)^k$$\nUsing $\\sum_{k=0}^m x^k = \\frac{1-x^{m+1}}{1-x}$ for $x \\ne 1$:\n$$P(1+t) < (n-6)t + (nt)^2 \\frac{1-n^{n-1}|t|^{n-1}}{1-n|t|} = t\\left(n-6 + tn^2 \\frac{1-n^{n-1}|t|^{n-1}}{1-n|t|}\\right)$$\n\nIf $n < 6$, let $t = \\frac{1}{n^3}$:\n$$P\\left(1 + \\frac{1}{n^3}\\right) < \\frac{1}{n^3} \\left(n - 6 + \\frac{1-\\frac{1}{n^2n-2}}{n-\\frac{1}{n}}\\right)$$\nSince $n - 6 \\le -1$ and $\\frac{1-\\frac{1}{n^2n-2}}{n-\\frac{1}{n}} < 1$, the parenthesis is negative, so $P\\left(1 + \\frac{1}{n^3}\\right) < 0$. Thus, $n \\in \\{2, 3, 4, 5\\}$ are not solutions.\n\nIf $n > 6$, let $t = -\\frac{1}{n^3}$:\n$$P(1 - \\frac{1}{n^3}) < -\\frac{1}{n^3} \\left(n - 6 + \\frac{1 - \\frac{1}{n^2n-2}}{n - \\frac{1}{n}}\\right)$$\nHere, $n - 6 \\ge 1$ and $\\frac{1 - \\frac{1}{n^2n-2}}{n - \\frac{1}{n}} < 1$, so the parenthesis is positive and $-\\frac{1}{n^3} < 0$, making $P\\left(1 - \\frac{1}{n^3}\\right) < 0$. Thus, $n > 6$ are not solutions.\n\nFor $n = 6$, use AM-GM:\n$$x^6 + x + x + 1 \\ge 4x^2$$\nSo, the only solution is $n = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18237,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 1$ be a positive integer. We say an integer $k$ is a fan of $n$ if $0 \\le k \\le n-1$ and there exist integers $x, y, z \\in \\mathbb{Z}$ such that\n\n$$\nx^2 + y^2 + z^2 \\equiv 0 \\pmod{n}, \\\\\nxyz \\equiv k \\pmod{n}.\n$$\n\nLet $f(n)$ be the number of fans of $n$.\n\nDetermine $f(11)$, $f(13)$, $f(15)$ and $f(2020)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $f(11) = 11$, $f(13) = 5$, $f(15) = 3 \\cdot 1 = 3$, and $f(2020) = 1 \\cdot 1 \\cdot 101 = 101$.\n\nTo prove our claim, we show that $f$ is multiplicative, that is, $f(rs) = f(r)f(s)$ for coprime numbers $r, s \\in \\mathbb{N}$, and that:\n\n1. $f(3) = 3$,\n2. $f(4) = 1$,\n3. $f(5) = 1$,\n4. $f(11) = 11$,\n5. $f(13) = 5$,\n6. $f(101) = 101$.\n\nThe multiplicative property follows from the Chinese Remainder Theorem.\n\n1. For $f(3)$: $0 = 0 \\cdot 0 \\cdot 0$ with $0^2 + 0^2 + 0^2 \\equiv 0 \\pmod{3}$, so $0$ is a fan of $3$. $1 = 1 \\cdot 1 \\cdot 1$ with $1^2 + 1^2 + 1^2 \\equiv 0 \\pmod{3}$, so $1$ is a fan. $2 = 2 \\cdot 1 \\cdot 1$ with $2^2 + 1^2 + 1^2 \\equiv 0 \\pmod{3}$, so $2$ is a fan. Hence $f(3) = 3$.\n\n2. For $f(4)$: Integers $x, y, z$ satisfy $x^2 + y^2 + z^2 \\equiv 0 \\pmod{4}$ if and only if they are all even. In this case, $xyz \\equiv 0 \\pmod{4}$. Hence $0$ is the only fan of $4$.\n\n3. For $f(5)$: Integers $x, y, z$ satisfy $x^2 + y^2 + z^2 \\equiv 0 \\pmod{5}$ if and only if at least one of them is divisible by $5$. In this case, $xyz \\equiv 0 \\pmod{5}$. Hence $0$ is the only fan of $5$.\n\n4. For $f(11)$: $3^2 + 1^2 + 1^2 = 9 + 1 + 1 = 11$. Thus, $(3x)^2 + x^2 + x^2$ is divisible by $11$ for every integer $x$. The residue of $3x \\cdot x \\cdot x = 3x^3$ modulo $11$ is a fan for all $x$. If $x = t^7$, then $x^3 = t^{21} \\equiv t \\pmod{11}$. Since $3$ is coprime to $11$, every residue is a fan of $11$.\n\n5. For $f(13)$: Define a *little fan* of $n$ as $k \\in [1, n-1]$ such that there exist $y, z \\in \\mathbb{Z}$ with\n\n$$\n1 + y^2 + z^2 \\equiv 0 \\pmod{n}, \\\\\nyz \\equiv k \\pmod{n}.\n$$\n\nEvery little fan is also a fan. The quadratic residues modulo $13$ are as follows.\n\n![](table1.png)\n\nNonzero $y, z$ satisfy $y^2 + z^2 \\equiv -1 \\pmod{13}$ if and only if $(y, z) \\equiv (\\pm 3, \\pm 4)$ or $(y, z) \\equiv (\\pm 4, \\pm 3)$. Hence $1$ and $12$ are the only little fans of $13$.\n\nThe number $0$ is a fan of $13$. If $xyz \\neq 0$ and $x^2 + y^2 + z^2 \\equiv 0$, then there are $b, c \\in [1, n-1]$ such that $xb \\equiv y$ and $xc \\equiv z$. These satisfy $1 + b^2 + c^2 \\equiv 0$. So the fan $xyz$ mod $13$ arises from the little fan $bc$ mod $13$ by multiplication by $x^3$.\n\nThe cubic residues modulo $13$ are:\n\n![](table2.png)\n\nThus, the fans of $13$ are $0, 1, 5, 8, 12$.\n\n6. For $f(101)$: $9^2 + 4^2 + 2^2 = 81 + 16 + 4 = 101$. Thus, $(9x)^2 + (4x)^2 + (2x)^2$ is divisible by $101$ for every $x$. The residue of $9x \\cdot 4x \\cdot 2x = 72x^3$ modulo $101$ is a fan for every $x$. If $x = t^{67}$, then $x^3 = t^{201} \\equiv t \\pmod{101}$. Since $72$ is coprime to $101$, $72x^3 \\equiv 72t$ can take any residue modulo $101$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18238,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $f_1(x)$ и $f_2(x)$ — приведённые квадратные трёхчлены, а параболы $\\Gamma_1$ и $\\Gamma_2$ — их графики. Пусть $y = k_1x + b_1$ и $y = k_2x + b_2$ — уравнения прямых $l_1$ и $l_2$ соответственно. Известно, что каждая из парабол $\\Gamma_1$ и $\\Gamma_2$ высекает на каждой из прямых $l_1$ и $l_2$ отрезки одинаковой длины. Докажите, что $\\Gamma_1 = \\Gamma_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что длина отрезка, высекаемого параболой $y = x^2 + px + q$ на прямой $y = kx + b$, равна модулю разности корней уравнения $x^2 + px + q = kx + b$, делённому на косинус угла наклона прямой $y = kx + b$. При этом модуль разности корней приведённого квадратного трёхчлена равен корню из его дискриминанта. Значит, условие того, что $\\Gamma_1$ и $\\Gamma_2$ высекают на $l_1$ равные отрезки, записывается как\n\n$$\n(p_1 - k_1)^2 - 4(q_1 - b_1) = (p_2 - k_1)^2 - 4(q_2 - b_1).\n$$\n\nПреобразуя это равенство, получаем $(p_1 - p_2)(p_1 + p_2 - 2k_1) = 4(q_1 - q_2)$. Если $p_1 = p_2$, то $q_1 = q_2$ и задача решена. Иначе\n\n$$\nk_1 = \\frac{p_1 + p_2}{2} - \\frac{2(q_1 - q_2)}{p_1 - p_2}.\n$$\n\nТочно те же рассуждения можно провести и для прямой $l_2$. Итак, если параболы не совпадают, то $k_1 = k_2$, то есть $|l_1| \\approx |l_2|$, что невозможно по условию.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18239,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 2m + 1$ for some integer $m$. $n$ points $A_1, A_2, \\dots, A_n$ are chosen on two parallel lines. What is the largest possible number of acute triangles among the triangles $A_iA_jA_k$ over $1 \\leq i < j < k \\leq n$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\frac{1}{6}m(m+1)(2m+1)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18240,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers such that $p(x) = x^4 + a x^3 + b x^2 + a x + c$ has exactly three different real roots; these roots are $\\tan y$, $\\tan 2y$, and $\\tan 3y$ for some real number $y$. Find all possible values of $y$, $0 \\leq y < \\pi$.",
    "options": [],
    "answer": "See solution",
    "solution": "That is,\n$$\n\\tan(2ky) + \\tan(my) + \\tan(ny) = 0\n$$\nprovided that $r$, $s$, $t$ are $\\tan(ky)$, $\\tan(my)$, and $\\tan(ny)$ respectively.\n\nWe consider the following cases:\n\n* If $r = \\tan y$, $s = \\tan 2y$, and $t = \\tan 3y$, then $\\tan 2y + \\tan 5y = 0$, and\n$$\ny \\in \\left\\{ \\frac{\\pi}{7}, \\frac{2\\pi}{7}, \\frac{3\\pi}{7}, \\frac{4\\pi}{7}, \\frac{5\\pi}{7}, \\frac{6\\pi}{7} \\right\\}.\n$$\n\n* If $r = \\tan 2y$, $s = \\tan y$, and $t = \\tan 3y$, then $\\tan 4y + \\tan 4y = 0$, and it follows that\n$$\ny \\in \\left\\{ \\frac{\\pi}{8}, \\frac{3\\pi}{8}, \\frac{5\\pi}{8}, \\frac{7\\pi}{8} \\right\\}.\n$$\n\nWe have discarded $y = \\frac{\\pi}{2}$, $y = \\frac{\\pi}{4}$, and $y = \\frac{3\\pi}{4}$ because $\\tan y$ and $\\tan(2y)$ must be real numbers.\n\n* If $r = \\tan 3y$, $s = \\tan y$, and $t = \\tan 2y$, then $\\tan 6y + \\tan 3y = 0$, and\n$$\ny \\in \\left\\{ \\frac{\\pi}{9}, \\frac{2\\pi}{9}, \\frac{\\pi}{3}, \\frac{4\\pi}{9}, \\frac{5\\pi}{9}, \\frac{2\\pi}{3}, \\frac{7\\pi}{9}, \\frac{8\\pi}{9} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18241,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive integers such that $ab$ is divisible by $2c$, $bc$ is divisible by $3a$, and $ca$ is divisible by $5b$. Find the least possible value of $abc$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ab$ is divisible by $2c$ and $ca$ is divisible by $5b$, $ab \\cdot ca$ must be divisible by $2c \\cdot 5b$, so $a^2$ is divisible by $2 \\cdot 5$. Therefore, $a$ is divisible by $2$ and $5$. Similarly, $b$ is divisible by $2$ and $3$, and $c$ is divisible by $3$ and $5$. Consequently, $abc$ is divisible by $2 \\cdot 5 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 5 = 900$. On the other hand, $a = 10$, $b = 6$, and $c = 15$ satisfy the conditions and $abc = 900$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18242,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = (x + a)(x + b)$ where $a, b$ are given positive real numbers, and $n \\ge 2$ is a given integer. For non-negative real numbers $x_1, x_2, \\dots, x_n$ that satisfy $x_1 + x_2 + \\dots + x_n = 1$, find the maximum of\n\n$$\nF = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n\\begin{align*}\n\\min\\{f(x_i), f(x_j)\\} &= \\min\\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\\} \\\\\n&\\le \\sqrt{(x_i + a)(x_i + b)(x_j + a)(x_j + b)} \\\\\n&\\le \\frac{1}{2}((x_i + a)(x_j + b) + (x_j + a)(x_i + b)) \\\\\n&= x_i x_j + \\frac{1}{2}(x_i + x_j)(a + b) + ab,\n\\end{align*}\n$$\n\nSo,\n$$\n\\begin{align*}\nF &\\le \\sum_{1 \\le i < j \\le n} x_i x_j + \\frac{a+b}{2} \\sum_{1 \\le i < j \\le n} (x_i + x_j) + \\binom{n}{2} ab \\\\\n&= \\frac{1}{2} \\left[ \\left( \\sum_{i=1}^n x_i \\right)^2 - \\sum_{i=1}^n x_i^2 \\right] + \\frac{a+b}{2} (n-1) \\sum_{i=1}^n x_i + \\binom{n}{2} ab \\\\\n&= \\frac{1}{2} \\left( 1 - \\sum_{i=1}^n x_i^2 \\right) + \\frac{n-1}{2} (a+b) + \\binom{n}{2} ab \\\\\n&\\le \\frac{1}{2} \\left( 1 - \\frac{1}{n} \\left( \\sum_{i=1}^n x_i \\right)^2 \\right) + \\frac{n-1}{2} (a+b) + \\binom{n}{2} ab\n\\end{align*}\n$$\n\nThus,\n$$\n\\begin{aligned}\nF_{\\max} &= \\frac{1}{2}\\left(1 - \\frac{1}{n}\\right) + \\frac{n-1}{2}(a+b) + \\frac{n(n-1)}{2}ab \\\\\n&= \\frac{n-1}{2}\\left(\\frac{1}{n} + a + b + n ab\\right).\n\\end{aligned}\n$$\n\nEquality holds when $x_1 = x_2 = \\cdots = x_n = \\frac{1}{n}$. So the maximum of $F$ is\n$$\n\\frac{n-1}{2} \\left( \\frac{1}{n} + a + b + n ab \\right).\n$$\n\nTo show this is indeed the maximum, we use induction on $n$ for a more general statement: for non-negative real numbers $x_1, x_2, \\dots, x_n$ with $x_1 + x_2 + \\dots + x_n = s$ (where $s \\ge 0$), the maximum of\n$$\nF = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}\n$$\nis attained when $x_1 = x_2 = \\cdots = x_n = \\frac{s}{n}$.\n\nSince $F$ is symmetric, assume $x_1 \\le x_2 \\le \\cdots \\le x_n$. Note $f(x)$ is strictly increasing for $x \\ge 0$, so\n$$\nF = (n-1)f(x_1) + (n-2)f(x_2) + \\cdots + f(x_{n-1}).\n$$\n\nFor $n=2$, $F = f(x_1) \\le f(\\frac{s}{2})$, equality when $x_1 = x_2$.\n\nAssume true for $n$, consider $n+1$. By induction on $x_2 + \\cdots + x_{n+1} = s - x_1$,\n$$\nF \\le n f(x_1) + \\frac{1}{2} n(n-1) f\\left(\\frac{s-x_1}{n}\\right) = g(x_1),\n$$\nwhere $g(x)$ is quadratic in $x$. The maximum occurs at $x_1 = \\frac{s}{n+1}$, so $x_2 = \\cdots = x_{n+1} = \\frac{s}{n+1}$, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18243,
    "subject": "Mathematics (Olympiad)",
    "question": "AC and BD are diameters of the circle.\n\n$AC = BD = 6\\ \\text{cm}$. If the area of the shaded region is $7\\pi\\ \\text{cm}^2$, find the value of $x$.\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p54_data_aa461ee8ff.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The radius of the circle is $3\\ \\text{cm}$, so the area of the full circle is $9\\pi\\ \\text{cm}^2$. The shaded region is $7/9$ of the circle's area. The unshaded region is $2/9$ of the area, which corresponds to $2/9$ of $360^\\circ = 80^\\circ$. Each of the two central angles is $40^\\circ$, so $x = 70$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18244,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $x_1, \\dots, x_n$ be positive real numbers. Show that\n\n$$\n\\min \\left(x_1, \\frac{1}{x_1} + x_2, \\dots, \\frac{1}{x_{n-1}} + x_n, \\frac{1}{x_n}\\right) \\leq 2 \\cos \\left(\\frac{\\pi}{n+2}\\right) \\leq \\max \\left(x_1, \\frac{1}{x_1} + x_2, \\dots, \\frac{1}{x_{n-1}} + x_n, \\frac{1}{x_n}\\right).\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See 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 $x_1, \\frac{1}{x_1} + x_2, \\dots, \\frac{1}{x_{n-1}} + x_n, \\frac{1}{x_n}$ is greater than $2 \\cos \\alpha$, where $\\alpha = \\frac{\\pi}{n+2}$, and show recursively that $x_k > \\frac{\\sin((k+1)\\alpha)}{\\sin(k\\alpha)}$, $k = 1, \\dots, n$; notice that $\\frac{\\sin((k+1)\\alpha)}{\\sin(k\\alpha)} > 0$, $k = 1, \\dots, n$.\n\nBy assumption, $x_1 > 2 \\cos \\alpha = \\frac{\\sin(2\\alpha)}{\\sin \\alpha}$. For the induction step, $x_{k+1} > 2 \\cos \\alpha - \\frac{1}{x_k} > 2 \\cos \\alpha - \\frac{\\sin(k\\alpha)}{\\sin((k+1)\\alpha)} = \\frac{\\sin((k+2)\\alpha)}{\\sin((k+1)\\alpha)}$. Consequently, $x_n > \\frac{\\sin((n+1)\\alpha)}{\\sin(n\\alpha)} = \\frac{1}{2 \\cos \\alpha}$, in contradiction with $\\frac{1}{x_n} > 2 \\cos \\alpha$; here, the last equality holds, since $\\alpha = \\frac{\\pi}{n+2}$.\n\n**Remark.** Notice that the $x_k = \\frac{\\sin((k+1)\\alpha)}{\\sin(k\\alpha)}$, $k = 1, \\dots, n$, where $\\alpha = \\frac{\\pi}{n+2}$, satisfy\n\n$$\nx_1 = \\frac{1}{x_1} + x_2 = \\dots = \\frac{1}{x_{n-1}} + x_n = \\frac{1}{x_n} = 2 \\cos \\alpha,\n$$\n\nto conclude, by the preceding, that\n\n$$\n\\begin{aligned}\n\\max_{x_1 > 0, \\dots, x_n > 0} \\min \\left(x_1, \\frac{1}{x_1} + x_2, \\dots, \\frac{1}{x_{n-1}} + x_n, \\frac{1}{x_n}\\right) &= 2 \\cos \\alpha \\\\\n\\min_{x_1 > 0, \\dots, x_n > 0} \\max \\left(x_1, \\frac{1}{x_1} + x_2, \\dots, \\frac{1}{x_{n-1}} + x_n, \\frac{1}{x_n}\\right) &= 2 \\cos \\alpha.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18245,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\\omega_1$ the circumcircle of $\\triangle BWN$, and let $X$ be the point on $\\omega_1$ such that $WX$ is the diameter of $\\omega_1$. Analogously, denote by $\\omega_2$ the circumcircle of $\\triangle CWM$, and $Y$ be the point on $\\omega_2$ such that $WY$ is a diameter of $\\omega_2$. Prove that $X$, $Y$ and $H$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AL$ be the altitude to side $BC$, and $Z$ be the intersection point of circles $\\omega_1$ and $\\omega_2$ other than $W$. We show that points $X$, $Y$, $Z$ and $H$ are collinear.\n\nPoints $B$, $C$, $M$ and $N$ are concyclic (denote the circle by $\\omega_3$) since $\\angle BNC = \\angle BMC = 90^\\circ$. $WZ$, $BN$ and $CM$ intersect at a point since they are the radical axes of $\\omega_1$ and $\\omega_2$, $\\omega_1$ and $\\omega_3$, $\\omega_2$ and $\\omega_3$, respectively. And since $BN$ and $CM$ intersect at $A$, $WZ$ passes through $A$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p352_data_5acb6c4886.png)\n\n$\\angle WZX = \\angle WZY = 90^\\circ$ since $WX$ and $WY$ are the diameters of $\\omega_1$ and $\\omega_2$, respectively. Thus, points $X$ and $Y$ are on the line $l$ perpendicular from $Z$ to $WZ$.\n\nSo, points $B$, $L$, $H$ and $N$ are concyclic since $\\angle BNH = \\angle BLH = 90^\\circ$. By the Circle-Power Theorem,\n\n$$\nAL \\cdot AH = AB \\cdot AN = AW \\cdot AZ. \\qquad \\textcircled{1}\n$$\n\nIf point $H$ is on the line $AW$, then $H$ and $Z$ coincide. Otherwise, by ①, we have\n\n$$\n\\frac{AZ}{AH} = \\frac{AL}{AW}.\n$$\n\nThus, $\\triangle AHZ \\sim \\triangle AWL$, consequently, $\\angle HZA = \\angle WLA = 90^\\circ$, hence, point $H$ is also on line $l$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18246,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Points $E, F$ move on the opposite ray of $BA, CA$ such that $BF = CE$. Let $M, N$ be the midpoints of $BE, CF$. Suppose that $BF$ cuts $CE$ at $D$.\n\n1. Let $I, J$ be the centers of the circumcircles of triangles $DBE$ and $DCF$. Prove that $MN$ is parallel to $IJ$.\n\n2. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of triangle $AEF$. Prove that when $E$ moves on the opposite ray of $BA$, line $HK$ goes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Let $T$ be the second intersection of $(BDE)$ and $(CDF)$. We have $BF = CE$ and\n$$\n\\angle TED = \\angle TBD, \\quad \\angle TCD = \\angle TFD\n$$\nso triangles $TCE$ and $TFB$ are congruent. From this, it is easy to see that $TBE$ and $TCF$ are isosceles at $T$ and similar.\n\n![](images/Vietnam_2022_p16_data_3002831ecc.png)\n\nIt follows that $T$, $I$, and $M$ are collinear; $T$, $J$, and $N$ are collinear. Note that $\\frac{TI}{TM} = \\frac{TJ}{TN}$, so $MN$ is parallel to $IJ$.\n\n2. Let $X$, $Y$ be the midpoints of $CE$, $BF$, and let $\\ell$ be the radical axis of the circles with diameters $BF$ and $CE$. The altitudes $BX'$ and $CY'$ of triangle $ABC$ intersect at the orthocenter $S$.\n\nWe will prove that $HK$ passes through a fixed point $S$ by showing that these three points lie on $\\ell$. Indeed, $MX$ is the midline of triangle $EBC$, so $MX \\parallel BC$; similarly, $NY \\parallel BC$, so $MX \\parallel NY$. Also, $MY \\parallel NX$, so $MYNX$ is a parallelogram, and $K$ is the midpoint of $XY$. Since $CE = BF$, we have\n$$\n\\mathcal{P}_{K/(BF)} = KY^2 - BY^2 = KX^2 - CX^2 = \\mathcal{P}_{K/(CE)},\n$$\nwhich implies $K \\in \\ell$.\n\n![](images/Vietnam_2022_p17_data_114ab6811b.png)\n\nAlso, $BCX'Y'$ is a cyclic quadrilateral, so\n$$\n\\mathcal{P}_{S/(BF)} = \\overline{SX'} \\cdot \\overline{SB} = \\overline{SY'} \\cdot \\overline{SC} = \\mathcal{P}_{S/(CE)},\n$$\nhence $S \\in \\ell$. Similarly, $H$ lies on $\\ell$, thus $HK$ passes through the fixed point $S$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18247,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X = \\{A, C, E\\}$ and $Y = \\{B, D, F\\}$. One can only move from the set $X$ to $Y$ (or the other way around). Since $A$ belongs to $X$, one must visit $Y$ in the first, third, and fifth place, and $X$ in the second and fourth. Noting that every visit is distinct, the first, third, and fifth visits must be a permutation on $\\{B, D, F\\}$ and the second and fourth must be a permutation on $\\{C, E\\}$. How many such distinct paths are possible?",
    "options": [],
    "answer": "See solution",
    "solution": "The number of permutations on $\\{B, D, F\\}$ is $3! = 6$, and on $\\{C, E\\}$ is $2! = 2$. Thus, the total number of feasible paths is $6 \\times 2 = 12$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 1$ be a positive integer and for all $t \\in \\mathbb{R}$ let\n$$\nP(t) = 1 + t + t^2 + \\dots + t^{2n}\n$$\nIf $x \\in \\mathbb{R}$, $P(x) \\in \\mathbb{Q}$ and $P(x^2) \\in \\mathbb{Q}$, then show that $x \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to check that:\n\n1. $P(t) > 0$ for all $t \\in \\mathbb{R}$.\n2. $P(t)P(-t) = P(t^2)$ for all $t \\in \\mathbb{R}$.\n3. $t = \\frac{P(t) + P(-t) - 2}{P(t) - P(-t)}$ for all $t \\neq 0$.\n\nLet $x \\in \\mathbb{R}$ such that $P(x) \\in \\mathbb{Q}$ and $P(x^2) \\in \\mathbb{Q}$. Then from (2) it follows that\n$$\nP(-x) = \\frac{P(x^2)}{P(x)} \\in \\mathbb{Q}\n$$\nand from (3) it follows that\n$$\nx = \\frac{P(x) + P(-x) - 2}{P(x) - P(-x)} \\in \\mathbb{Q}\n$$\nThus, $x$ is rational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18249,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. In how many ways can an $n \\times n$ table be filled with integers from $0$ to $5$ such that:\n\na) the sum of each row is divisible by $2$ and the sum of each column is divisible by $3$;\n\nb) the sum of each row is divisible by $2$, the sum of each column is divisible by $3$, and the sum of each of the two diagonals is divisible by $6$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\na) $6^{n^2-n}$\n\nb) $1$, if $n = 1$; $6$, if $n = 2$; $6^{n^2-n-2}$, if $n \\ge 3$.\n\n---\n\na) Fill the top left $(n-1) \\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1)^2}$ ways. For each of the top $n-1$ cells of the rightmost column, there are $3$ choices, and for each of the left $n-1$ cells of the bottom row, there are $2$ choices, to satisfy the divisibility requirements. The value for the last empty cell in the bottom right is then uniquely determined (mod $2$ by the bottom row, and mod $3$ by the rightmost column). Thus, there are $6^{(n-1)^2} \\cdot 3^{n-1} \\cdot 2^{n-1} = 6^{n^2-n}$ ways to fill the table.\n\nb) For $n=1$, the only solution is writing $0$ into the single cell. For $n=2$, let $a$ be the top left number. The bottom right must then be $(6-a) \\bmod 6$. Using the conditions for rows and columns, for the top right number $x$ we get the equations $x \\equiv -a \\pmod{2}$ and $x \\equiv a \\pmod{3}$, and for the bottom left number $y$, $y \\equiv a \\pmod{2}$ and $y \\equiv -a \\pmod{3}$. The Chinese remainder theorem determines $x$ and $y$ uniquely, and their sum is also divisible by $6$. Thus, there are $6$ ways to fill the table in this case, one for each value of $a$.\n\nFor $n \\ge 3$, fill the top left $(n-1) \\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1)^2}$ ways. The bottom right cell's value is uniquely determined by other values on the falling diagonal. Denote the value in the top left cell by $a$, the sum of the $2$nd to $(n-1)$st cells in the top row by $b$, the sum of the $2$nd to $(n-1)$st cells in the leftmost column by $c$, and the sum of $2$nd to $(n-1)$st cells on the rising diagonal by $d$.\n\nUsing the Chinese remainder theorem, fill the top right cell with the unique value $x$ such that $x \\equiv -a-b \\pmod{2}$ and $x \\equiv a+c-d \\pmod{3}$, and the bottom left cell with the unique value $y$ such that $y \\equiv a+b-d \\pmod{2}$ and $y \\equiv -a-c \\pmod{3}$. The divisibility conditions are now fulfilled for the top row, the leftmost column, and both diagonals.\n\nNow, leave one cell both in the rightmost column and in the bottom row empty for the time being. For the other $n-3$ empty cells in the rightmost column, there are $3$ possible values for each, and for the other $n-3$ empty cells in the bottom row, $2$ values for each. Having made all those choices (which can be done in $3^{n-3} \\cdot 2^{n-3}$ ways), the values for the two remaining cells are now uniquely determined (mod $2$ by the values in the respective row, and mod $3$ by the column). The total number of ways to fill the table is $6^{(n-1)^2} \\cdot 3^{n-3} \\cdot 2^{n-3} = 6^{n^2-n-2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18250,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a positive integer $n \\geq 2$. Prove that for any $n$ distinct integers $a_1, a_2, \\dots, a_n$, the set\n$$\n\\{1, 2, \\dots, \\frac{n(n-1)}{2}\\}\n$$\ncontains at least $\\left\\lfloor \\frac{n(n-6)}{19} \\right\\rfloor$ elements that cannot be expressed as the difference of some $a_i$ and $a_j$.\n\nHere, $\\lceil x \\rceil$ is the least integer greater than or equal to $x$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = \\frac{n(n-1)}{2}$. Without loss of generality, assume $a_1 > a_2 > \\dots > a_n$. Denote the multiset of $a_i - a_j$, $1 \\leq i < j \\leq n$ by $D = \\{d_1, d_2, \\dots, d_m\\}$ (elements can be repeated).\n\nSuppose that $m-t$ elements of $M = \\{1, 2, \\dots, m\\}$ appear in $D$ and they form the subset $A$, while the other $t$ elements do not appear in $D$ and they form $B = \\{b_1, \\dots, b_t\\}$. Let $C = D \\setminus A = \\{c_1, \\dots, c_t\\}$ (elements can be repeated).\n\nConsider the generating function $F(z) = z^{a_1} + z^{a_2} + \\dots + z^{a_n}$ and\n$$\n\\begin{align*}\nG(z) &= F(z)F(z^{-1}) \\\\\n&= (z^{a_1} + z^{a_2} + \\dots + z^{a_n})(z^{-a_1} + z^{-a_2} + \\dots + z^{-a_n}) \\\\\n&= n + \\sum_{i<j} (z^{a_i - a_j} + z^{a_j - a_i}) \\\\\n&= n + \\sum_{i<j} (z^{|a_i - a_j|} + z^{-|a_i - a_j|}) \\\\\n&= n + \\sum_{k=1}^{m} (z^{d_k} + z^{-d_k}) \\\\\n&= n + \\sum_{k=1}^{m} (z^k + z^{-k}) - \\sum_{k=1}^{t} (z^{b_k} + z^{-b_k}) + \\sum_{k=1}^{t} (z^{c_k} + z^{-c_k}).\n\\end{align*}\n$$\n\nIf $z$ is a unit complex number,\n$$\nF(z^{-1}) = F(\\bar{z}) = \\overline{F(z)},\n$$\n$$\nG(z) = F(z)F(z^{-1}) = |F(z)|^2\n$$\nis a nonnegative real. We want to choose an appropriate $\\vartheta$ and\n$$\nz = e^{2\\vartheta i} = \\cos 2\\vartheta + i \\sin 2\\vartheta,\n$$\nsuch that\n$$\n\\begin{align*}\nH(z) &= 1 + \\sum_{k=1}^{m} (z^k + z^{-k}) \\\\\n&= \\sum_{k=-m}^{m} e^{2k\\vartheta i} \\\\\n&= \\frac{e^{(2m+1)\\vartheta i} - e^{-(2m+1)\\vartheta i}}{e^{\\vartheta i} - e^{-\\vartheta i}} \\\\\n&= \\frac{\\sin((2m + 1)\\vartheta)}{\\sin \\vartheta}\n\\end{align*}\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18251,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be integers with $2 \\leq k \\leq n - 1$. 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": "See solution",
    "solution": "The difference of any two elements from $A$ is a rational number. To show this, let $x \\neq y \\in A$ and choose any other $k-1$ elements of $A$ (possible since $k-1 \\leq n-2$). Denote $s$ as the sum of these $k-1$ elements. By hypothesis, both $x + s$ and $y + s$ are rational numbers. Subtracting, we get $x - y \\in \\mathbb{Q}$.\n\nNow, let $\\alpha \\in A$. We can write the elements of $A$ as $\\alpha + q_i$, where $q_i \\in \\mathbb{Q}$. The sum of $k$ distinct elements is $k\\alpha + \\sum q_i \\in \\mathbb{Q}$, so $\\alpha \\in \\mathbb{Q}$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18252,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider hexagons whose internal angles are all equal.\n\n1. Prove that for any such hexagon, the sum of the lengths of any two neighbouring sides is equal to the sum of the lengths of their opposite sides.\n\n2. Does there exist such a hexagon with side lengths $1$, $2$, $3$, $4$, $5$, and $6$ in some order?",
    "options": [],
    "answer": "See solution",
    "solution": "1. Let the hexagon be $ABCDEF$. It suffices to show that $|AB| + |BC| = |DE| + |EF|$.\n\nLet $K$ be the intersection point of rays $FA$ and $CB$, and let $L$ be the intersection point of rays $FE$ and $CD$. The size of every internal angle of the hexagon is $120^\\circ$, so triangles $KAB$ and $LDE$ are equilateral. The quadrilateral $FKCL$ is a parallelogram since its opposite sides are parallel. This implies $|KC| = |LF|$ or $|KB| + |BC| = |LE| + |EF|$, which together with $|KB| = |AB|$ and $|LE| = |DE|$ implies the desired equality $|AB| + |BC| = |DE| + |EF|$.\n\n2. Take a parallelogram with side lengths $7$ and $5$ and internal angles $60^\\circ$ and $120^\\circ$, and cut off equilateral triangles with side lengths $1$ and $2$ at its acute angles. This gives rise to a hexagon with all internal angles $120^\\circ$ and side lengths $1$, $4$, $5$, $2$, $3$, $6$.\n\n![](images/estonian-2012-2013_p12_data_136a4f4fe8.png)\n\n![](images/estonian-2012-2013_p12_data_e3466dfead.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18253,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{Z}$ be a function such that for any two positive integers $a, b$, $f(ab) = f(a) + f(b)$. Find all possible values of $f(1000)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(1000)$ can be any integer divisible by $3$.\n\n**Solution:**\n\nConsider $f(1000) = f(10^3) = 3 \\cdot f(10)$. Thus, $f(1000)$ is always divisible by $3$.\n\nNow, let's show that for any integer $k$, $f(1000)$ can equal $3k$. Define $\\operatorname{ord}(a)$ as the highest power of $2$ dividing $a$. Let $a = 2^i(2a_0 + 1)$ and $b = 2^m(2b_0 + 1)$. Then,\n\n$$\n\\begin{align*}\n\\operatorname{ord}(ab) &= \\operatorname{ord}(2^{i+m}(2a_0 + 1)(2b_0 + 1)) = i + m = \\operatorname{ord}(a) + \\operatorname{ord}(b). \\\\\n\\text{Let } f(a) = k \\cdot \\operatorname{ord}(a), \\text{ then} \\\\\n f(ab) &= k \\cdot \\operatorname{ord}(ab) = k \\cdot (\\operatorname{ord}(a) + \\operatorname{ord}(b)) = f(a) + f(b).\n\\end{align*}\n$$\n\nTherefore, $f(1000) = k \\cdot \\operatorname{ord}(2^3 \\cdot 125) = 3k$ for any integer $k$. This proves the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18254,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive numbers $x$, $y$, $z$, prove the inequality:\n\n$$\n2 \\left( \\frac{x}{2x+y} \\right)^2 + 2 \\left( \\frac{y}{2y+z} \\right)^2 + 2 \\left( \\frac{z}{2z+x} \\right)^2 + \\frac{9xyz}{(2x+y)(2y+z)(2z+x)} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We make a substitution: $a = \\frac{2x}{2x+y}$, $b = \\frac{2y}{2y+z}$, $c = \\frac{2z}{2z+x}$. Then the inequality becomes:\n\n$$\n2 \\left(\\frac{x}{2x+y}\\right)^2 + 2 \\left(\\frac{y}{2y+z}\\right)^2 + 2 \\left(\\frac{z}{2z+x}\\right)^2 + \\frac{9xyz}{(2x+y)(2y+z)(2z+x)} = \\frac{a^2+b^2+c^2}{2} + \\frac{9}{8}abc \\le 1,\n$$\n\nor\n\n$$\na^2 + b^2 + c^2 + \\frac{9}{4}abc \\le 2.\n$$\n\nConsider the expression:\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{a}-1\\right)\\left(\\frac{1}{b}-1\\right)\\left(\\frac{1}{c}-1\\right) &= \\left(\\frac{2x+y}{2x}-1\\right)\\left(\\frac{2y+z}{2y}-1\\right)\\left(\\frac{2z+x}{2z}-1\\right) \\\\\n&= \\frac{y}{2x} \\cdot \\frac{z}{2y} \\cdot \\frac{x}{2z} = \\frac{1}{8},\n\\end{aligned}\n$$\n\nor\n\n$$\n\\frac{(1-a)}{a} \\cdot \\frac{(1-b)}{b} \\cdot \\frac{(1-c)}{c} = \\frac{1}{8} \\Leftrightarrow 8(1-a)(1-b)(1-c) = abc.\n$$\n\nExpanding:\n\n$$\n8(1-a)(1-b)(1-c) = abc \\Leftrightarrow 8-8(a+b+c)+8(ab+bc+ca)-8abc = abc\n$$\n\nSo,\n\n$$\n\\frac{9}{4}abc = 2-2(a+b+c)+2(ab+bc+ca).\n$$\n\nContinue the proof using the obtained conditions:\n\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 + \\frac{9}{4}abc - 2 &= (a+b+c)^2 - 2(ab+bc+ca) + 2 - 2(a+b+c) + 2(ab+bc+ca) - 2 \\\\\n&= (a+b+c)^2 - 2(a+b+c) \\le 0.\n\\end{aligned}\n$$\n\nThe last inequality holds when $a+b+c \\le 2$.\n\nNow, perform transformations:\n\n$$\n\\begin{aligned}\na+b+c &= \\frac{2x}{2x+y} + \\frac{2y}{2y+z} + \\frac{2z}{2z+x} \\\\\n&= 1 - \\frac{y}{2x+y} + 1 - \\frac{z}{2y+z} + 1 - \\frac{x}{2z+x} \\\\\n&= 2 + \\left(1 - \\frac{y}{2x+y} - \\frac{z}{2y+z} - \\frac{x}{2z+x}\\right) \\le 2 \\\\\n&\\Leftrightarrow \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\ge 1.\n\\end{aligned}\n$$\n\nProve the last inequality using the Cauchy--Schwarz inequality:\n\n$$\n\\left( \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\right) (x+y+z)^2 = \\left( \\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x} \\right) (y(2x+y) + z(2y+z) + x(2z+x)) \\ge\n$$\n\n$$\n\\geq \\left( \\sqrt{\\frac{y}{2x+y}} \\cdot \\sqrt{y(2x+y)} + \\sqrt{\\frac{z}{2y+z}} \\cdot \\sqrt{z(2y+z)} + \\sqrt{\\frac{x}{2z+x}} \\cdot \\sqrt{x(2z+x)} \\right)^2 = (x+y+z)^2,\n$$\n\nQ.E.D.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18255,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $N = (a-b)(b-c)(c-d)(d-a)(a-c)(b-d)$. ¿Cuál es el mayor entero que siempre divide a $N$ para cualquier elección de enteros $a, b, c, d$?",
    "options": [],
    "answer": "See solution",
    "solution": "Si $a=1$, $b=2$, $c=3$, $d=4$, entonces $N = (a-b)(b-c)(c-d)(d-a)(a-c)(b-d) = -12$.\n\nVeamos que 12 siempre divide a $N$.\n\nPor el principio del palomar, entre $a, b, c, d$ hay dos números con el mismo resto módulo 3, así que 3 divide a $N$.\n\nAnálogamente, entre los pares $(a,b)$, $(b,c)$, $(c,d)$, $(d,a)$, $(a,c)$, $(b,d)$, hay dos con ambas coordenadas de igual paridad, por lo tanto 4 divide a $N$.\n\nEn conclusión, 12 divide a $N$ para cualquier elección de $a, b, c, d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18256,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers such that $x^4 + y^4 + z^4 = 3$. Prove that\n\n$$\n\\frac{9}{x^2 + y^4 + z^6} + \\frac{9}{x^4 + y^6 + z^2} + \\frac{9}{x^6 + y^2 + z^4} \\leq x^6 + y^6 + z^6 + 6.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Using the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x, y^2, z^3)$ and $(x^3, y^2, z)$, we have\n\n$$\n(x^4 + y^4 + z^4)^2 \\leq (x^2 + y^4 + z^6)(x^6 + y^4 + z^2),\n$$\nwhich implies\n$$\n\\frac{1}{x^2 + y^4 + z^6} \\leq \\frac{x^6 + y^4 + z^2}{9}.\n$$\n\nSimilarly, applying the Cauchy-Bunyakovsky-Schwarz inequality to $(x^2, y^3, z)$ and $(x^2, y, z^3)$, and to $(x^3, y, z^2)$ and $(x, y^3, z^2)$, we obtain\n\n$$\n\\frac{1}{x^4 + y^6 + z^2} \\leq \\frac{x^4 + y^2 + z^6}{9},\n$$\n\nand\n\n$$\n\\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^2 + y^6 + z^4}{9}.\n$$\n\nAdding these three inequalities gives\n\n$$\n\\frac{1}{x^2 + y^4 + z^6} + \\frac{1}{x^4 + y^6 + z^2} + \\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^6 + y^6 + z^6 + x^4 + y^4 + z^4 + x^2 + y^2 + z^2}{9}.\n$$\n\nSince $x^4 + y^4 + z^4 = 3$ and by the inequality between arithmetic and quadratic means for $x^2, y^2, z^2$, we have $x^2 + y^2 + z^2 \\leq 3$. Substituting these into the previous inequality yields the required result.\n\nEquality holds if and only if $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18257,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many positive integers $n$ such that $n^2 + 1$ has a prime divisor greater than $2n + \\sqrt{2n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take any integer $m$ ($m \\ge 20$), and let $p$ be a prime divisor of $(m!)^2 + 1$. Then $p > m \\ge 20$. Take an integer $n$ such that $0 < n < \\frac{p}{2}$ and $n \\equiv \\pm m! \\pmod{p}$. Therefore, $0 < n < p - n < p$ and\n\n$$\nn^2 \\equiv -1 \\pmod{p}.\n$$\n\nNow,\n\n$$\n(p - 2n)^2 = p^2 - 4pn + 4n^2 \\equiv -4 \\pmod{p},\n$$\n\nwhich yields $(p - 2n)^2 \\ge p - 4$,\n\n$$\n\\begin{aligned}\np &\\ge 2n + \\sqrt{p-4} \\\\\n &\\ge 2n + \\sqrt{2n + \\sqrt{p-4} - 4} \\\\\n &\\ge 2n + \\sqrt{2n}.\n\\end{aligned}\n$$\n\nBy the above, there are infinitely many $n$ such that $n^2 + 1$ has a prime divisor greater than $2n + \\sqrt{2n}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18258,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the sequence $\\{a_n\\}$ be defined by\n$$\na_1 = 1, \\quad a_2 = 2, \\quad a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}} \\quad (n = 2, 3, \\dots).\n$$\nProve that the sum of squares of any two adjacent terms of the sequence is also in the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "By $a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}}$, we have $a_{n+1} a_{n-1} = a_n^2 + (-1)^n$ for $n = 2, 3, \\dots$. So,\n$$\n\\begin{align*}\n\\frac{a_n - a_{n-2}}{a_{n-1}} &= \\frac{a_n a_{n-2} - a_{n-2}^2}{a_{n-1} a_{n-2}} = \\frac{a_{n-1}^2 + (-1)^{n-1} - a_{n-2}^2}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1}^2 - a_{n-1} a_{n-3}}{a_{n-1} a_{n-2}} = \\frac{a_{n-1} - a_{n-3}}{a_{n-2}} \\\\\n&= \\dots = \\frac{a_3 - a_1}{a_2} = 2,\n\\end{align*}\n$$\nthat is, $a_n = 2a_{n-1} + a_{n-2}$ for $n \\ge 3$, with $a_1 = 1$, $a_2 = 2$.\nTherefore, $a_n = C_1 \\lambda_1^n + C_2 \\lambda_2^n$, where $\\lambda_1 + \\lambda_2 = 2$, $\\lambda_1 \\lambda_2 = -1$, $a_1 = 1$, $a_2 = 2$, $n \\in \\mathbb{N}^+$.\nSince $\\lambda_1 \\lambda_2 = -1$ and $\\lambda_2 = 2 - \\lambda_1$, we have\n$$\n\\begin{cases}\n1 = C_1 \\lambda_1 + C_2 \\lambda_2 \\\\\n2 = C_1 \\lambda_1^2 + C_2 \\lambda_2^2\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nC_1(1 + \\lambda_1^2) = \\lambda_1 \\\\\nC_2(1 + \\lambda_2^2) = \\lambda_2\n\\end{cases}\n$$\nSince $1 + \\lambda_1 \\lambda_2 = 0$, we have\n$$\n\\begin{align*}\na_n^2 + a_{n+1}^2 &= C_1^2 (1 + \\lambda_1^2) \\lambda_1^{2n} + C_2^2 (1 + \\lambda_2^2) \\lambda_2^{2n} \\\\\n&\\quad + 2C_1 C_2 (\\lambda_1 \\lambda_2)^n (1 + \\lambda_1 \\lambda_2) \\\\\n&= C_1 \\lambda_1^{2n+1} + C_2 \\lambda_2^{2n+1} = a_{2n+1}.\n\\end{align*}\n$$\nThus, the sum of squares of any two adjacent terms is also a term in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18259,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be nonnegative real numbers such that\n$$\na^2 + b^2 + c^2 + abc = 4.\n$$\nProve that\n$$\n0 \\le ab + bc + ca - abc \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution.** (By Richard Stong)\n\nFrom the condition, at least one of $a$, $b$, and $c$ does not exceed $1$, say $a \\le 1$. Then\n$$\nab + bc + ca - abc = a(b + c) + bc(1 - a) \\ge 0.\n$$\nTo obtain equality, we have $a(b + c) = 0$ and $bc(1 - a) = 0$. If $a = 1$, then $b + c = 0$ or $b = c = 0$, which contradicts the given condition $a^2 + b^2 + c^2 + abc = 4$. Hence $1 - a \\ne 0$ and only one of $b$ and $c$ is $0$. Without loss of generality, say $b = 0$. Therefore $b + c > 0$ and $a = 0$. Plugging $a = b = 0$ back into the given condition gives $c = 2$. By permutation, the lower bound holds if and only if $(a, b, c)$ is one of the triples $(2, 0, 0)$, $(0, 2, 0)$, and $(0, 0, 2)$.\n\nNow we prove the upper bound. Let us note that some two of the three numbers $a$, $b$, and $c$ are both greater than or equal to $1$ or less than or equal to $1$. Without loss of generality, we assume that the numbers with this property are $b$ and $c$. Then we have\n$$\n(1 - b)(1 - c) \\ge 0. \\qquad (1)\n$$\nThe given equality $a^2 + b^2 + c^2 + abc = 4$ and the inequality $b^2 + c^2 \\ge 2bc$ imply\n$$\na^2 + 2bc + abc \\le 4, \\quad \\text{or} \\quad bc(2 + a) \\le 4 - a^2.\n$$\nDividing both sides of the last inequality by $2 + a$ yields\n$$\nbc \\le 2 - a. \\qquad (2)\n$$\nCombining (1) and (2) gives\n$$\n\\begin{aligned}\n ab + bc + ac - abc &\\le ab + 2 - a + ac(1 - b) \\\\\n &= 2 - a(1 + bc - b - c) \\\\\n &= 2 - a(1 - b)(1 - c) \\le 2,\n\\end{aligned}\n$$\nas desired.\n\nThe last equality holds if and only if $b = c$ and $a(1 - b)(1 - c) = 0$. Hence, equality for the upper bound holds if and only if $(a, b, c)$ is one of the triples $(1, 1, 1)$, $(0, \\sqrt{2}, \\sqrt{2})$, $(\\sqrt{2}, 0, \\sqrt{2})$, and $(\\sqrt{2}, \\sqrt{2}, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18260,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. Let $a_n = n! + n$. Show that if $\\dfrac{P(a_n)}{Q(a_n)}$ is an integer for every $n$, then $\\dfrac{P(n)}{Q(n)}$ is an integer for every integer $n$ such that $Q(n) \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Imagine dividing $P(x)$ by $Q(x)$. We find that\n\n$$\n\\frac{P(x)}{Q(x)} = A(x) + \\frac{R(x)}{Q(x)},\n$$\n\nwhere $A(x)$ and $R(x)$ are polynomials with rational coefficients, and $R(x)$ is either identically $0$ or has degree less than the degree of $Q(x)$.\n\nBy bringing the coefficients of $A(x)$ to their least common multiple, we can find a polynomial $B(x)$ with integer coefficients, and a positive integer $b$, such that $A(x) = \\frac{B(x)}{b}$. Suppose first that $R(x)$ is not identically $0$. Note that for any integer $k$, either $A(k) = 0$, or $|A(k)| \\ge \\frac{1}{b}$. But whenever $|k|$ is large enough, $0 < \\left|\\frac{R(k)}{Q(k)}\\right| < \\frac{1}{b}$, and therefore if $n$ is large enough, $\\frac{P(a_n)}{Q(a_n)}$ cannot be an integer.\n\nSo $R(x)$ is identically $0$, and $\\frac{P(x)}{Q(x)} = \\frac{B(x)}{b}$ (at least whenever $Q(x) \\neq 0$).\n\nNow let $n$ be an integer. Then there are infinitely many integers $k$ such that $n \\equiv a_k \\pmod{b}$. But $\\frac{B(a_k)}{b}$ is an integer, or equivalently $b$ divides $B(a_k)$. It follows that $b$ divides $B(n)$, and therefore $\\frac{P(n)}{Q(n)}$ is an integer. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18261,
    "subject": "Mathematics (Olympiad)",
    "question": "For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the coordinate plane, define\n\n$$\nd(A, B) = |x_1 - x_2| + |y_1 - y_2|.\n$$\n\nLet $P_1, P_2, \\dots, P_{2023}$ be 2023 pairwise different points in the coordinate plane. Denote\n\n$$\n\\lambda = \\frac{\\max_{1 \\le i < j \\le 2023} d(P_i, P_j)}{\\min_{1 \\le i < j \\le 2023} d(P_i, P_j)}.\n$$\n\n1. Prove that $\\lambda \\ge 44$.\n2. Give an example of $P_1, P_2, \\dots, P_{2023}$ such that $\\lambda = 44$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof Method 1:* (1) For $k = 1, 2, \\dots, 2023$, let the coordinates of $P_k$ be $(x_k, y_k)$, and denote $u_k = x_k + y_k$, $v_k = x_k - y_k$. Let $D = \\max_{1 \\le i < j \\le 2023} d(P_i, P_j)$. Then, for any $1 \\le i, j \\le 2023$,\n\n$$\n|u_i - u_j| = |(x_i - x_j) + (y_i - y_j)| \\le |x_i - x_j| + |y_i - y_j| = d(P_i, P_j) \\le D.\n$$\n\nThus, the $u_1, u_2, \\dots, u_{2023}$ fall within some interval $[a, a+D]$. Similarly for $v_1, v_2, \\dots, v_{2023}$. For $k, l = 1, 2, \\dots, 44$, consider the region\n\n$$\nH_{k,l} = \\left\\{ \\left( \\frac{u+v}{2}, \\frac{u-v}{2} \\right) \\mid a + \\frac{k-1}{44}D \\le u \\le a + \\frac{k}{44}D,\\ b + \\frac{l-1}{44}D \\le v \\le b + \\frac{l}{44}D \\right\\}.\n$$\n\nIf $P_i, P_j \\in H_{k,l}$, let $U = u_i - u_j, V = v_i - v_j$, then $-\\frac{D}{44} \\le U, V \\le \\frac{D}{44}$, we have\n\n$$\n\\begin{aligned}\nd(P_i, P_j) &= |x_i - x_j| + |y_i - y_j| = \\left| \\frac{u_i + v_i}{2} - \\frac{u_j + v_j}{2} \\right| + \\left| \\frac{u_i - v_i}{2} - \\frac{u_j - v_j}{2} \\right| \\\\\n&= \\left| \\frac{U + V}{2} \\right| + \\left| \\frac{U - V}{2} \\right| \\\\\n&\\in \\left\\{ \\pm \\frac{U + V}{2}, \\pm \\frac{U - V}{2} \\right\\} = \\{U, -U, V, -V\\}.\n\\end{aligned}\n$$\n\nHence, $\\min_{1 \\le i < j \\le 2023} d(P_i, P_j) \\le d(P_i, P_j) \\le \\frac{D}{44}$, thus $\\lambda \\ge 44$.\n\n(2) *Construction*: Consider the point set\n\n$$\nM = \\{(x, y) \\in \\mathbb{Z}^2 \\mid x, y \\text{ have the same parity, } |x + y| \\le 44, |x - y| \\le 44\\}.\n$$\n\nThis set contains $45^2 = 2025$ points. Selecting any 2023 points, the distance $d(P_i, P_j) = |x_i - x_j| + |y_i - y_j|$ is even and greater than 0, i.e., $d(P_i, P_j) \\ge 2$. On the other hand,\n\n$$\nd(P_i, P_j) = |x_i - x_j| + |y_i - y_j| \\le 88.\n$$\n\nThus, $\\lambda = \\frac{\\max_{1 \\le i < j \\le 2023} d(P_i, P_j)}{\\min_{1 \\le i < j \\le 2023} d(P_i, P_j)} \\le 44$, and by (1) $\\lambda = 44$.\n\nThe picture lower left is an example with $n = 25$ points satisfying $\\lambda = 4$. The picture lower right is a partition of 16 regions, and can be used to prove that when $n = 17$, $\\lambda \\ge 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18262,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle has radius $4$. A sector of the circle has area $\\frac{1}{2}\\pi$. What is the probability that a randomly chosen point in the circle lies within the sector?",
    "options": [],
    "answer": "See solution",
    "solution": "The area of the whole circle is $\\pi \\cdot 4^2 = 16\\pi$, but the area of the sector is $\\frac{1}{2}\\pi$, so the probability is $$\\frac{\\frac{1}{2}\\pi}{16\\pi} = \\frac{1}{32}$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18263,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be such that for all $x, y \\in \\mathbb{R}$,\n\n$$\n|f(x + y)| = |f(x) + f(y)|.\n$$\n\nProve that $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist $a, b \\in \\mathbb{R}$ such that $f(a + b) \\neq f(a) + f(b)$. By the assumption, $f(a + b) = -f(a) - f(b)$. If $f(a + b) = 0$, then $f(a) = -f(b)$.\n\n$$\n\\begin{align*}\n|f(2a + 2b)| &= |f(a + (a + b + b))| \\\\\n&= |f(a) + f((a + b) + b)| \\\\\n&= |f(a) + f(a + b) + f(b)| \\text{ or } |f(a) - f(a + b) - f(b)| \\\\\n&= 0 \\text{ or } |f(a) - (-f(a) - f(b)) - f(b)| \\\\\n&= 0 \\text{ or } 2|f(a)|.\n\\end{align*}\n$$\n\nBut $|f(2a + 2b)| = |f((a + b) + (a + b))| = |2f(a + b)| = 2|f(a + b)| \\neq 0$. Therefore, $2|f(a)| = |f(2a + 2b)| = 2|f(a + b)|$ and thus $|f(a + b)| = |f(a)|$. Similarly, $|f(a + b)| = |f(b)|$. So $|f(a)| = |f(b)|$, that is, $f(a) = f(b)$ or $f(a) = -f(b)$. If $f(a) = f(b)$, then $2|f(a)| = |f(a) + f(a)| = |f(a) + f(b)| = |f(a+b)| = |f(a)|$, so $f(a) = 0$, which implies $f(a + b) = -f(a) - f(b) = -2f(a) = 0$, a contradiction. On the other hand, if $f(a) = -f(b)$, then $0 = |f(a) + f(b)| = |f(a+b)|$, again a contradiction. So in any case we get a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18264,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 表示所有正整數所成的集合。試求所有的函數 $f: N \\to N$ 使得對任意正整數 $n$，\n\n$$\n\\frac{1}{f(1)f(2)} + \\frac{1}{f(2)f(3)} + \\dots + \\frac{1}{f(n)f(n+1)} = \\frac{f(f(n))}{f(n+1)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "由於\n\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{n(n+1)} = \\frac{n}{n+1},\n$$\n\n猜測 $f(n) = n$ 是滿足題設之唯一解。\n\n將 $n = 1$ 帶入，得 $f(f(1))f(1) = 1$，故 $f(1) = 1$。\n\n題設之 $n$ 以 $n+1$ 代入，得\n\n$$\n\\frac{f(f(n))}{f(n+1)} + \\frac{1}{f(n+1)f(n+2)} = \\frac{f(f(n+1))}{f(n+2)}.\n$$\n\n上式等價於\n\n$$\nf(f(n))f(n+2) + 1 = f(f(n+1))f(n+1).\n$$\n\n注意：$f(n+1) = 1 \\Rightarrow f(f(n+1)) = 1$。因此，$f(f(n))f(n+2) = 0$ 不可能發生。故 $f(n) > 1, \\forall n > 1$。\n\n利用數學歸納法，證明：$f(f(n)) < f(n+1)$。\n\n當 $n = 1$ 時，$f(2) > 1 = f(f(1))$ 成立。\n\n若 $f(n+1) > f(f(n))$ 則 $f(n+1) \\geq f(f(n)) + 1$。因此\n\n$$\nf(f(n))f(n+2) + 1 \\geq f(f(n+1))f(f(n)) + f(f(n+1)).\n$$\n\n因 $n+1 > 1$，我們有 $f(n+1) > 1$。即 $f(f(n+1)) > 1$。由此可推得\n\n$$\nf(n+2) > f(f(n+1)).\n$$\n\n因此，函數 $f$ 滿足\n\n$$\nf(n + 1) > f(f(n)), \\forall n \\in N. \\tag{1}\n$$\n\n底下證明滿足 (1) 之函數為 $f(n) = n, \\forall n \\in N$。\n\n令集合 $S = \\{f(f(1)), f(2), f(f(2)), f(3), \\dots, f(f(n-1)), f(n), f(f(n)), f(n+1), \\dots\\}$，則集合 $S$ 有最小元素 $f(n_0)$，對某個正整數 $n_0$。由 (1)，可得 $f(n_0) = 1$。底下證明 $n_0 = 1$。\n\n假設 $n_0 > 1$ 則 $1 = f(n_0) > f(f(n_0 - 1))$，此為矛盾！故 $f(1) = 1$ 且 $f(n) > 1$ 對於 $n > 1$。\n\n考慮 $f: \\{n \\geq 2\\} \\to \\{n \\geq 2\\}$，同理可得 $f(2) = 2$ 且 $f(n) > 2$ 對於 $n > 2$。由數學歸納法可證：\n\n$$\nf(k) = k, \\text{且}~f(n) > k,~\\text{對於 } n > k,\n$$\n\n因此滿足題意之唯一解為 $f(n) = n, \\forall n \\in N$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18265,
    "subject": "Mathematics (Olympiad)",
    "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\n**a)** Show that there exists a positive integer which, after three changes—the first interesting, the second special, and the third awesome—becomes $2020$.\n\n**b)** Find all positive integers with the property that, after two changes of different types, selected among the three above, becomes $2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Before the last (awesome) change, the number must be $\\frac{2020 - 16}{4} = 501$; before the special change, it must be $\\frac{501 - 9}{3} = 164$; and the required starting number is $\\frac{164 - 4}{2} = 80$.\n\n**b)** 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 $4$, the last change must be an interesting one. So, the previous number must be $\\frac{2014 - 4}{2} = 1005$. Since this number is odd, the first change must be special, and the initial number is $\\frac{1005 - 9}{3} = 332$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18266,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be the set of all prime divisors of members of $S'$. Suppose, to the contrary, that $P$ is finite and $P = \\{p_1, p_2, \\dots, p_n\\}$. Define\n$$\nN = 4 \\prod_{i=1}^{n} p_i(p_i - 1).\n$$\n$S$ is infinite, hence there exists an infinite subset $S_1$ of $S$ such that every two members of $S_1$ are congruent to each other modulo $N$. Let $x, y > 1$ be members of $S_1$ such that $y$ is large enough to satisfy $\\frac{y}{\\log_2 y} > x$. Prove that this leads to a contradiction, and thus $P$ cannot be finite.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $P$ is finite and $P = \\{p_1, p_2, \\dots, p_n\\}$. Define\n$$\nN = 4 \\prod_{i=1}^{n} p_i(p_i - 1).\n$$\nSince $S$ is infinite, there exists an infinite subset $S_1$ of $S$ such that every two members of $S_1$ are congruent modulo $N$. Let $x, y > 1$ be members of $S_1$ with $y$ large enough that $\\frac{y}{\\log_2 y} > x$. Then\n$$\nx < \\frac{y}{\\log_2 y} \\le \\frac{y}{\\log_x y} \\implies y > x \\log_x y \\implies x^y > y^x.\n$$\nSince $x^y + y^x \\in S'$, it can be written as $x^y + y^x = q_1^{\\alpha_1} q_2^{\\alpha_2} \\dots q_k^{\\alpha_k}$ where $q_1, \\dots, q_k \\in \\mathbb{P}$ and $\\alpha_1, \\dots, \\alpha_k \\in \\mathbb{Z}^+$. For each $1 \\le i \\le k$:\n$$\n\\begin{aligned}\nq_i | N, x \\equiv y \\pmod{N} &\\implies x \\equiv y \\pmod{q_i} \\\\\n&\\implies x^y \\equiv y^y \\pmod{q_i}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nq_i - 1 | N, x \\equiv y \\pmod{N} &\\implies x \\equiv y \\pmod{q_i - 1} \\\\\n&\\implies y^x \\equiv y^{x} \\pmod{q_i}\n\\end{aligned}\n$$\nThus,\n$$\n\\left. \\begin{array}{l} x^y \\equiv y^x \\pmod{q_i} \\\\ x^y + y^x \\equiv 0 \\pmod{q_i} \\end{array} \\right\\} \\implies 2x^y \\equiv 0 \\pmod{q_i}\n$$\nSo for every odd $q_i$, $q_i|x$. For $q_i = 2$, if at least one of $x$ and $y$ is even, then both are even; if both are odd:\n$$\n\\left. \\begin{array}{l} 2|x-y \\implies y^y \\equiv y^x \\pmod{4} \\\\ y \\equiv x \\pmod{N} \\\\ 4|N \\end{array} \\right\\} \\implies x^y \\equiv y^y \\equiv y^x \\pmod{4} \\\\\n\\implies x^y + y^x \\equiv 2x^y \\equiv 2 \\pmod{4} \\implies \\alpha_i = 1\n$$\nLet\n$$\nx = q_1^{\\beta_1} q_2^{\\beta_2} \\dots q_k^{\\beta_k} x', \\quad y = q_1^{\\gamma_1} q_2^{\\gamma_2} \\dots q_k^{\\gamma_k} y'\n$$\nwith\n$$\n\\beta_i, \\gamma_i \\in \\mathbb{Z}^+ \\cup \\{0\\}, \\quad \\gcd(x', q_1, \\dots, q_k) = \\gcd(y', q_1, \\dots, q_k) = 1\n$$\nFor $q_i \\ne 2$, $\\beta_i, \\gamma_i > 0$. If $q_i = 2$ and one of $x, y$ is even, then $\\beta_i, \\gamma_i > 0$. So $\\beta_i y \\ge y$. From the definition of $y$:\n$$\n\\beta_i y \\ge y > x \\log_2 y \\ge x \\log_{q_i} y \\ge x \\gamma_i.\n$$\nIf $x, y$ are odd and $q_i = 2$, $0 = \\beta_i y \\ge \\gamma_i x = 0$ holds. Now,\n$$\n\\begin{aligned}\nq_1^{\\alpha_1} \\dots q_k^{\\alpha_k} &= x^y + y^x \\\\\n&= x'^y \\prod_{i=1}^k q_i^{y\\beta_i} + y'^x \\prod_{i=1}^k q_i^{x\\gamma_i} \\\\\n&= \\prod_{i=1}^k q_i^{x\\gamma_i} \\left( y'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i} \\right)\n\\end{aligned}\n$$\nIf $q_i = 2$ and $x, y$ are odd, $\\alpha_i = 1$. Otherwise,\n$$\ny\\beta_i - x\\gamma_i > 0 \\implies \\left( y'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i}, q_i \\right) = (y'^x, q_i) = 1\n$$\nBut also,\n$$\ny'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i} \\mid x^y + y^x = q_1^{\\alpha_1} \\dots q_k^{\\alpha_k}\n$$\nSo\n$$\n\\begin{align*}\n& y'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i} \\mid 2 \\\\\n\\implies & y'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i} \\le 2 \\\\\n\\implies & y'^x \\le 1 \\implies y'^x = 1 \\implies y^x = \\prod_{i=1}^k q_i^{x\\gamma_i} \\\\\n\\implies & x^y + y^x = q_1^{\\alpha_1} \\dots q_k^{\\alpha_k} = y^x \\left( y'^x + x'^y \\prod_{i=1}^k q_i^{y\\beta_i - x\\gamma_i} \\right) \\le 2y^x \\\\\n\\implies & x^y \\le y^x\n\\end{align*}\n$$\nWhich contradicts the choice of $y$, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18267,
    "subject": "Mathematics (Olympiad)",
    "question": "The polynomials $P(x)$ and $Q(x)$ with real coefficients are non-constant, monic, and satisfy the equality:\n\n$$\n2P(x) = Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right), \\quad P(1) = 1.\n$$\n\nDetermine the polynomials $P(x)$ and $Q(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_0$. The leading term of $Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right)$ is:\n\n$$\n\\left(\\frac{(x+1)^2}{2}\\right)^n - \\left(\\frac{(x-1)^2}{2}\\right)^n = \\frac{2n x^{2n-1}}{2^n} + \\frac{2n x^{2n-2}}{2^n} = \\frac{4n}{2^n} x^{2n-1}.\n$$\n\nThe left side's leading coefficient is $2$, so $\\frac{4n}{2^n} = 2$, which gives $2^{n+1} = 4n$. Since $2^{n+1} > 4n$ for $n \\ge 3$, only $n=1$ or $n=2$ are possible. The degree of $P$ is $2n-1$.\n\nFor $x=0$, $2P(0) = Q(1/2) - Q(1/2) = 0$, so $P(0) = 0$.\n\n- For $n=1$: $P(x) = ax$. Since $P(1) = 1$, $a=1$.\n  \n  $$\n  P(x) = x, \\quad Q(x) = x + a_0, \\quad a_0 \\in \\mathbb{R}.\n  $$\n\n- For $n=2$: $Q(x) = x^2 + bx + c$.\n  \n  $$\n  \\begin{align*}\n  2P(x) &= Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right) \\\\\n  &= \\frac{1}{4}((x+1)^4 - (x-1)^4) + \\frac{b}{2}((x+1)^2 - (x-1)^2) \\\\\n  &= \\frac{1}{4}(8x^3 + 8x) + \\frac{b}{2}(4x) \\\\\n  &= 2x^3 + 2(1+b)x.\n  \\end{align*}\n  $$\n  \n  So $P(x) = x^3 + (1+b)x$. Since $P(1) = 1$, $1 + (1+b) = 1 \\implies b = -1$.\n  \n  $$\n  P(x) = x^3, \\quad Q(x) = x^2 - x + c, \\quad c \\in \\mathbb{R}.\n  $$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18268,
    "subject": "Mathematics (Olympiad)",
    "question": "For each pair of integers $a, b$, a non-negative integer $a*b$ is defined such that it satisfies the following two conditions:\n\n1) $(a+b)*b = a*b + 1$;\n\n2) $(a*b) \\cdot (b*a) = 0$.\n\nFind the values of the expressions $2016*121$ and $2016*144$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answers are $2016 * 121 = 16$ and $2016 * 144 = 13$.\n\nSuppose $a$ and $b$ are positive integers. Write $a = bq + r$, where $q$ is a non-negative integer, $r$ is a positive integer, and $r \\leq b$. We prove that $a*b = q$.\n\nFrom condition 2, if $a = b$, then $a*a = 0$. If $a*b = 0$, and in condition 1 we set $a_1 + b_1 = a$, $b_1 = b$, then:\n\n$$\n(a_1 + b_1)*b_1 = a_1*b_1 + 1 \\text{ or } a*b = (a-b)*b + 1.\n$$\n\nBut this would imply $(a-b)*b = -1$ for positive integers $a > b$, which is impossible. Thus, for $a > b$, $b*a = 0$.\n\nNow suppose $a > b$ and $a = bq + r$, $q, r \\in \\mathbb{N}$, $r \\leq b$. Then:\n\n$$\n\\begin{align*}\n&r*b = 0 \\Rightarrow (r + b)*b = r*b + 1 = 1 \\\\\n&\\Rightarrow ((r + b) + b)*b = (r + b)*b + 1 = 2 \\\\\n&\\Rightarrow \\dots \\\\\n&\\qquad (r + qb)*b = ((r + (q-1)b) + b)*b + 1 = q-1 + 1 = q\n\\end{align*}\n$$\n\nTherefore, $a*b = q$.\n\nFinally,\n\n$$\n\\begin{aligned}\n2016 &= 121 \\cdot 16 + 80 \\Rightarrow 2016 * 121 = 16, \\\\\n2016 &= 144 \\cdot 13 + 144 \\Rightarrow 2016 * 144 = 13.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18269,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n, k$ be integers greater than $1$ and satisfy $n < 2^k$. Prove that there are $2k$ integers not divisible by $n$, such that if we divide them into two groups, then there must exist a group in which the sum of some integers can be divided by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the case $n = 2^r$, $r \\geq 1$. Here, $r < k$. Take three $2^{r-1}$'s and $2k-3$ $1$'s—none are divisible by $n$. If these $2k$ numbers are divided into two groups, one group must contain two $2^{r-1}$'s, whose sum is $2^r$, divisible by $n$.\n\nNext, suppose $n$ is not a power of $2$. Take the $2k$ integers:\n\n$$\n-1, -1, -2, -2^2, \\dots, -2^{k-2}, 1, 2, 2^2, \\dots, 2^{k-1}.\n$$\n\nNone are divisible by $n$.\n\nAssume these can be divided into two groups so that no partial sum in either group is divisible by $n$. Place $1$ in the first group. Since $(-1) + 1 = 0$ is divisible by $n$, both $-1$'s must be in the second group. Since $(-1) + (-1) + 2 = 0$, $2$ is in the first group, so $-2$ is in the second group.\n\nBy induction, suppose $1, 2, \\dots, 2^l$ are in the first group and $-1, -2, \\dots, -2^l$ in the second ($1 \\leq l < k-2$). Since\n\n$$\n(-1) + (-1) + (-2) + \\dots + (-2^l) + 2^{l+1} = 0\n$$\n\nis divisible by $n$, $2^{l+1}$ is in the first group, and $-2^{l+1}$ in the second.\n\nThus, $1, 2, 2^2, \\dots, 2^{k-2}$ are in the first group and $-1, -2, -2^2, \\dots, -2^{k-2}$ in the second. Finally,\n\n$$\n(-1) + (-1) + (-2) + \\dots + (-2^{k-2}) + 2^{k-1} = 0\n$$\n\nso $2^{k-1}$ is in the first group. Therefore, $1, 2, 2^2, \\dots, 2^{k-1}$ are all in the first group.\n\nBy properties of binary numbers, every positive integer not greater than $2^k - 1$ can be represented as a partial sum of $1, 2, 2^2, \\dots, 2^{k-1}$. Since $n \\leq 2^k - 1$, some partial sum is divisible by $n$, contradicting the assumption.\n\nTherefore, we have found $2k$ integers that meet the requirement. The proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18270,
    "subject": "Mathematics (Olympiad)",
    "question": "遊戲開始時有 $2^m$ 張紙，每張上寫有一個 $1$。考慮以下操作：每次我們選兩張紙，假設其上的數字分別為 $a$ 與 $b$。將兩張紙上的數字都擦掉，並在兩張紙上都寫上 $a+b$。\n\n試證：經過 $m2^{m-1}$ 步後，所有紙上的數字總和至少為 $4^m$。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $P_k$ 為第 $k$ 次操作後所有紙張上數字的乘積，而 $S_k$ 為第 $k$ 次操作後所有紙張上數字的總和。顯然 $P_0 = 1$。又基於 $(a+b)^2 \\geq 4ab$，易知 $P_{k+1} \\geq 4P_k$，故 $P_{m2^{m-1}} \\geq 4^{m2^{m-1}}$。最後由算術-幾何平均不等式，$S_k \\geq 4^m$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18271,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\na^p + b^{p+1} \\equiv a + b^2 \\pmod{p}\n$$\nFind all integer pairs $(a, b)$ that satisfy this congruence for all primes $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $p$ divides $a^p + b^{p+1}$ and $p > |a + b^2|$. Then $a + b^2 = 0$.\n\nSelect another prime $q$ such that $q > |b+1|$ and $(q, b) = 1$. Let $n = 2q$.\n\nThen:\n$$\na^n + b^{n+1} = (-b^2)^{2q} + b^{2q+1} = b^{4q} + b^{2q+1} = b^{2q+1}(b^{2q-1} + 1).\n$$\nSince $n$ divides $a^n + b^{n+1}$ and $(q, b) = 1$, it follows that $q$ divides $b^{2q-1} + 1$.\n\nNote:\n$$\nb^{2q-1} + 1 \\equiv (b^{q-1})^2 \\cdot b + 1 \\equiv b + 1 \\pmod{q}\n$$\nSince $q > |b+1|$, we get $b + 1 = 0$, so $b = -1$ and $a = -b^2 = -1$.\n\nThus, the only solution pairs are $(0, 0)$ and $(-1, -1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18272,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute-angled triangle $ABC$. The points $B'$ and $C'$ lie on the rays opposite to $CA$ and $BA$, respectively, such that $|B'C| = |AB|$ and $|C'B| = |AC|$. Prove that the circumcenter of $AB'C'$ lies on the circumcircle of $ABC$.\n\n![](images/CZE_ABooklet_2023_p5_data_ab046c811d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since the line segments $AB'$ and $AC'$ have the same length, triangle $AB'C'$ is isosceles with base $B'C'$. This means the perpendicular bisector of $B'C'$ coincides with the bisector of angle $BAC$. Let $S \\neq A$ be the intersection of this bisector with the circumcircle of $ABC$. If we prove that $S$ is the circumcenter of $AB'C'$, we are done.\n\nSince $S$ lies on the perpendicular bisector of $B'C'$, we have $|SB'| = |SC'|$. So, it remains to prove that $|SA| = |SC'|$.\n\nFrom the congruence of inscribed angles $SAB$ and $SAC$, it follows that $S$ is the midpoint of the arc $BC$, and therefore $|BS| = |CS|$. From the cyclic quadrilateral $ABSC$, we have $\\angle ACS = 180^\\circ - \\angle SBA = \\angle C'BS$. Together with the equality $|CA| = |BC'|$, we get that triangles $SAC$ and $SC'B$ are congruent by $SAS$, and therefore $|SA| = |SC'|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18273,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations in the domain of real numbers:\n\n$$\n2x + \\lfloor y \\rfloor = 2022,\n$$\n$$\n3y + \\lfloor 2x \\rfloor = 2023.\n$$\n\n(The symbol $\\lfloor a \\rfloor$ denotes the greatest integer not greater than $a$. For example, $\\lfloor 1.9 \\rfloor = 1$ and $\\lfloor -1.1 \\rfloor = -2$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\lfloor y \\rfloor$ and $2022$ are integers, the equation $2x + \\lfloor y \\rfloor = 2022$ implies that $2x$ is also an integer, so $\\lfloor 2x \\rfloor = 2x$. Thus, we can eliminate $x$ by subtracting the first equation from the second:\n\n$$\n3y + \\lfloor 2x \\rfloor - (2x + \\lfloor y \\rfloor) = 2023 - 2022\n$$\n$$\n3y + 2x - 2x - \\lfloor y \\rfloor = 1\n$$\n$$\n3y - \\lfloor y \\rfloor = 1. \\qquad (1)\n$$\n\nSince $3y$ is an integer, $y$ must be of the form $k$, $k + \\frac{1}{3}$, or $k + \\frac{2}{3}$, where $k = \\lfloor y \\rfloor$.\n\n- If $y = k$, then $3k - k = 1 \\implies k = \\frac{1}{2}$ (not integer).\n- If $y = k + \\frac{1}{3}$, then $3(k + \\frac{1}{3}) - k = 3k + 1 - k = 1 \\implies k = 0$, so $y = \\frac{1}{3}$.\n- If $y = k + \\frac{2}{3}$, then $3(k + \\frac{2}{3}) - k = 3k + 2 - k = 1 \\implies k = -\\frac{1}{2}$ (not integer).\n\nSo the only solution is $y = \\frac{1}{3}$ and $k = 0$.\n\nSubstitute $y = \\frac{1}{3}$ into the first equation:\n$$\n2x + \\lfloor y \\rfloor = 2022 \\implies 2x + 0 = 2022 \\implies x = 1011.\n$$\n\n**Conclusion:** The only solution is $(x, y) = (1011, \\frac{1}{3})$.\n\n_Remark:_ Alternatively, write $y = k + r$ with $k = \\lfloor y \\rfloor$ and $r \\in (0, 1)$. Then (1) becomes $3(k + r) - k = 1 \\implies 2k = 1 - 3r$. Since $2k$ is even and $1 - 3r \\in (-2, 1)$, the only possibility is $k = 0$, $r = \\frac{1}{3}$, so $y = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18274,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Since the sequence $\\lfloor n b \\rfloor / \\lfloor n a \\rfloor$ consists of positive integers converging to $b/a$, it follows that $b = m a$ for some integer $m \\geq 2$, and $\\lfloor n b \\rfloor = m \\lfloor n a \\rfloor$ for all sufficiently large $n$. Consequently, for large $n$, $\\lfloor n m a \\rfloor = m \\lfloor n a \\rfloor$, so $n m a < m \\lfloor n a \\rfloor + 1$. That is, $n a < \\lfloor n a \\rfloor + 1/m \\leq \\lfloor n a \\rfloor + 1/2$. Hence, $\\{ n a \\} = n a - \\lfloor n a \\rfloor < 1/2$, so the set $\\{ \\{ n a \\} : n \\in \\mathbb{Z}_+ \\}$ is not dense in $[0, 1]$, and $a$ must be rational, say $a = p/q$ with $p$ and $q$ coprime positive integers. If $q \\geq 2$, choose $n$ large enough such that $n p \\equiv -1 \\pmod{q}$, to reach a contradiction: $1/2 > \\{ n a \\} = \\{ n p / q \\} = \\{ (q-1)/q \\} = 1 - 1/q \\geq 1/2$. Consequently, $q = 1$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18275,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest three-digit number $n$ for which there exist exactly 16 pairs of natural numbers $(a, b)$ with $a < b$ such that $n$ is the least common multiple of $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n = p_1^{m_1} \\dots p_k^{m_k}$, where $p_i$ are distinct primes and $m_i$ are natural numbers. Consider all ordered pairs $(a, b)$ such that $[a, b] = n$. For $a = p_1^{a_1} \\dots p_k^{a_k}$ and $b = p_1^{b_1} \\dots p_k^{b_k}$, we require $\\max\\{a_i, b_i\\} = m_i$ for all $i$. The possible pairs $(a_i, b_i)$ are:\n\n$(0, m_i), (1, m_i), \\dots, (m_i - 1, m_i), (m_i, m_i), (m_i, m_i - 1), \\dots, (m_i, 0)$.\n\nThere are $2m_i + 1$ options for each $i$, so in total $N = (2m_1 + 1) \\dots (2m_k + 1)$ ordered pairs. The number of unordered pairs with $a < b$ is $\\frac{1}{2}(N - 1)$. We need $\\frac{1}{2}(N - 1) = 16$, so $N = 33$.\n\nPossible factorizations:\n- $N = 33 = 33 \\times 1$ (i.e., $n = p^{16}$), but $2^{16} = 65536 > 999$.\n- $N = 11 \\times 3$ (i.e., $n = p^5 q$ for distinct primes $p, q$).\n\nFor $p = 3$, $3^5 = 243$, so $n = 3^5 \\cdot 2 = 486$.\nFor $p = 2$, $2^5 = 32$, so $n = 2^5 \\cdot p$ for $p > 2$ prime. The largest such $n$ under 1000 is $2^5 \\cdot 31 = 992$.\n\n**Answer:** $\\boxed{992}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18276,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that there exists an $n$-tuple $\\mathbf{a} = (a_1, \\dots, a_n)$ of real numbers satisfying:\n\n- $S_1(\\mathbf{a}) > 0$\n- $S_3(\\mathbf{a}) < 0$\n- $S_5(\\mathbf{a}) > 0$\n\nwhere $S_k(\\mathbf{a}) = a_1^k + a_2^k + \\dots + a_n^k$ for $k = 1, 3, 5$.\n\nIf such an $n$-tuple exists, then for any $m > n$, by adding $m-n$ zeros, there exists an $m$-tuple satisfying the same conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that there exists a 5-tuple $\\mathbf{a} = (a_1, a_2, a_3, a_4, a_5)$ such that $S_1(\\mathbf{a}) > 0$, $S_3(\\mathbf{a}) < 0$, and $S_5(\\mathbf{a}) > 0$.\n\nChoose $a_1 = 3$, $a_2 = a_3 = 1$, $a_4 = a_5 = -2.45$.\n\n- $S_1(\\mathbf{a}) = 3 + 1 + 1 - 2.45 - 2.45 = 0.1 > 0$\n- $S_3(\\mathbf{a}) = 27 + 1 + 1 - 14.7 - 14.7 = 29 - 29.4 = -0.4 < 0$\n- $S_5(\\mathbf{a}) = 243 + 1 + 1 - 86.5 - 86.5 = 245 - 173 = 72 > 0$\n\nNow, we show that there does not exist a 4-tuple $\\mathbf{a} = (a_1, a_2, a_3, a_4)$ such that $S_1(\\mathbf{a}) > 0$, $S_3(\\mathbf{a}) < 0$, and $S_5(\\mathbf{a}) > 0$.\n\nSuppose for contradiction that such a 4-tuple exists. There must be at least one positive and one negative among the $a_i$'s. Consider three cases:\n\n**Case 1:** One positive and three non-positive. Suppose $a_1 > 0 \\ge a_2, a_3, a_4$. Let $b_i = -a_i$ for $i = 2, 3, 4$. Then $a_1 > b_2 + b_3 + b_4$, so\n$$\na_1^3 > (b_2 + b_3 + b_4)^3 \\ge b_2^3 + b_3^3 + b_4^3 = -(a_2^3 + a_3^3 + a_4^3),\n$$\nwhich is a contradiction.\n\n**Case 2:** Three non-negative and one negative. Suppose $a_1, a_2, a_3 \\ge 0 > a_4$. Let $b_4 = -a_4$. Then\n$$\na_1 + a_2 + a_3 > b_4,\\quad a_1^3 + a_2^3 + a_3^3 < b_4^3,\\quad a_1^5 + a_2^5 + a_3^5 > b_4^5.\n$$\nFrom the second inequality, $a_1, a_2, a_3 < b_4$. Hence,\n$$\na_1^5 + a_2^5 + a_3^5 < (a_1^3 + a_2^3 + a_3^3) b_4^2 < b_4^5,\n$$\nwhich is a contradiction.\n\n**Case 3:** Two positive and two negative. Let the positives be $x, y$ and the negatives $-z, -t$ with $x, y, z, t > 0$.\n\nWe have $x + y > z + t$, $x^3 + y^3 < z^3 + t^3$, $x^5 + y^5 > z^5 + t^5$. After algebraic manipulations, this leads to a contradiction.\n\nTherefore, the minimum positive integer $n$ satisfying the conditions is $n = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18277,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: [0, +\\infty) \\to [0, +\\infty)$ such that for all non-negative $x, y$,\n\n$$\nf(f(x) + f(y)) = xy f(x + y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(x) \\equiv 0$.\n\n**Solution.**\n\nLet $y = 0$. Then the equation becomes:\n$$\nf(f(x) + f(0)) = 0.\n$$\nSo for any $x$, $f(f(x) + f(0)) = 0$, which means there exists $a$ such that $f(a) = 0$.\n\nNow, set $x = a$:\n$$\nf(f(0)) = 0.\n$$\nSet $x = y = 0$:\n$$\nf(2f(0)) = 0.\n$$\nSet $x = y = f(0)$:\n$$\nf(f(f(0)) + f(f(0))) = f^2(0) f(2f(0)) = 0.\n$$\nThis implies $f(0) = 0$ and $f(f(x)) = 0$ for any $x$.\n\nNow, let $y = f(y)$ in the original equation:\n$$\nf(f(x) + f(f(y))) = x f(y) f(x + f(y)).\n$$\nBut $f(f(x)) = 0$, so:\n$$\nf(f(x) + f(f(y))) = f(f(x)) = 0.\n$$\nThus,\n$$\nx f(y) f(x + f(y)) = 0\n$$\nfor all $x, y \\geq 0$.\n\nSuppose there exists $y_0 > 0$ such that $f(y_0) \\neq 0$. Let $y = y_0 - x$ (for $0 \\leq x \\leq y_0$):\n$$\nf(f(x) + f(y_0 - x)) = x(y_0 - x) f(y_0).\n$$\nThe right side can take any value in $[0, \\frac{1}{4} y_0^2 f(y_0)]$ as $x$ varies, so the range of $f$ includes this segment. Thus, there exists $x_0$ such that $f(x_0) = \\min\\left\\{\\frac{1}{2} y_0, \\frac{1}{8} y_0^2 f(y_0)\\right\\}$.\n\nFrom $x f(y) f(x + f(y)) = 0$, for $y = x_0$ and $x = y_0 - f(x_0) > 0$, we get $f(x_0) f(y_0) = 0$, which is impossible since both are assumed nonzero. This contradiction shows that $f(x) = 0$ for all $x \\in [0, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18278,
    "subject": "Mathematics (Olympiad)",
    "question": "How many possible arrangements are there for Alfred, Mollie, and four other people, given that Alfred and Mollie can swap places in any arrangement?",
    "options": [],
    "answer": "See solution",
    "solution": "The person on the extreme left can be any one of the four people that is neither Alfred nor Mollie; the second left can be any one of the remaining three; the first person on the right of centre, and so on. For every arrangement of the people around them, Alfred and Mollie can swap places to make a new arrangement. So the number of possibilities is $$(4 \\times 3 \\times 2 \\times 1) \\times 2 = 48.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18279,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be positive integers such that $n^{n^n} = m^m$. Find all such pairs $(n, m)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.** Let $n$ be a positive integer and $p, q$ some positive rational numbers. If $n^p = q$, then $q$ is itself an integer.\n\n*Proof.* Suppose $p = \\frac{a}{b}$ and $q = \\frac{c}{d}$ where $a, b, c, d \\in \\mathbb{N}$. We have\n\n$$\nn^p = q \\Rightarrow n^{\\frac{a}{b}} = \\frac{c}{d} \\Rightarrow n^a = \\left(\\frac{c}{d}\\right)^b = \\frac{c^b}{d^b} \\Rightarrow d^b \\mid c^b \\Rightarrow d \\mid c \\Rightarrow q \\in \\mathbb{N}\n$$\n\nNow for the main problem, note that if $n = 1$, then $m = n = 1$ and this is a solution for the equation. So we may assume that $n > 1$. Let $r = \\log_n m$ (so $m = n^r$). We have\n\n$$\nn^{n^n} = m^m = (n^r)^{n^r} = n^{r n^r}\n$$\n\nSince $n > 1$, we must have\n\n$$\nn^n = r n^r \\Rightarrow r = n^{n - r}\n$$\n\nOn the other hand, since $n^{n^n} = m^m$, we get $n^n = m \\log_n m = r$ or $r = \\frac{n^n}{m} \\in \\mathbb{Q}$. According to the lemma, with $n - r$ and $r$ playing the role of $p$ and $q$, respectively, we get $r$ is an integer. Now if $r < n$, then $n^{n - r} \\geq n^1 > r = n^{n - r}$, which is impossible. And if $r > n$, then $n^{n - r} < 1 \\leq r = n^{n - r}$, which is again impossible.\n\nSo we must have $n = r$. Hence, $n = r = n^{n - r} = 1$. This contradicts the assumption $n > 1$ and consequently, the only solution is $m = n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18280,
    "subject": "Mathematics (Olympiad)",
    "question": "有一個 $2013 \\times 2013$ 的棋盤，其格線方向為南北/東西向。在每一個方格內填入一個朝東或朝南的箭頭。每一分鐘內，對所有箭頭 $A$，如果 $A$ 指向相鄰的某方格，且該方格中的箭頭 $B$ 的方向與 $A$ 不同，則下一分鐘將 $A$ 換成另一個方向的箭頭（如果 $A$ 原本朝南則換成朝東，如果 $A$ 原本朝東則換成朝南）；否則箭頭 $A$ 不動。所有這樣的變換在每一分鐘同時執行。\n\n試證：在有限時間之後，所有的箭頭都不能再變換方向。並求出還能改變至少一個箭頭方向的最長可能時間。",
    "options": [],
    "answer": "See solution",
    "solution": "最多在前 $4024$ 分鐘內可以有箭頭方向變換產生。\n\n首先證明：自第 $4025$ 分鐘起，所有箭頭的方向都固定下來，不再產生變換。\n\n在棋盤上建立坐標系：每一個方格的坐標為 $(i, j)$，其中 $0 \\leq i, j \\leq 2012$，且 $(0, 0)$ 方格在棋盤的東南角。我們宣稱：若 $x + y < m$，則方格 $(x, y)$ 內的箭頭自第 $m$ 分鐘起都不再改變方向。\n\n對 $m$ 作數學歸納法。$m = 1$ 時，$(0, 0)$ 方格內的箭頭並未指向其他方格，命題成立。\n\n假設命題對正整數 $m$ 成立，即只要 $x + y < m$，方格 $(x, y)$ 中的箭頭自第 $m$ 分鐘起就不再改變方向。若命題對 $m + 1$ 不成立，則存在方格 $(x, y)$，其中 $x + y < m + 1$，且此方格中的箭頭 $A$ 在第 $M$ 分鐘內改變方向，$M \\geq m + 1$。即在第 $M - 1$ 分鐘時，箭頭 $A$ 指向一個與它方向不同的箭頭 $B$。箭頭 $B$ 所在的方格坐標是 $(x', y')$，其中 $x' + y' = x + y - 1 < m$。由歸納假設知箭頭 $B$ 自第 $m$ 分鐘起不再改變方向。又因 $A$ 與 $B$ 指向不同方向，得知箭頭 $B$ 指向方格 $(x - 1, y - 1)$。所以箭頭 $A$ 在第 $M - 1 \\geq m$ 分鐘之前不可能指向箭頭 $B$，否則 $A$ 就在 $M-1$ 分鐘以前就得改變方向。因此箭頭 $A$ 在第 $M-1$ 分鐘時也得改變方向。\n\n設箭頭 $A$ 在第 $M-1$ 分鐘之前指向另一箭頭 $C$。同理，$C$ 所在的方格坐標為 $(x'', y'')$，其中 $x'' + y'' = x + y - 1 < m$。再由歸納假設，箭頭 $C$ 自第 $m$ 分鐘開始也不會改變方向，並且也指向方格 $(x-1, y-1)$。\n\n但上述現象不可能發生，因為在第 $m$ 分鐘之前 $B, C$ 兩箭頭同時指向 $(x-1, y-1)$ 方格的箭頭 $D$，且 $B, C$ 的方向不同。因此在第 $m$ 分鐘它們之中有一個與箭頭 $D$ 的方向不同而被改變方向。\n\n因此歸納法步驟成立。因為每一個方格的坐標 $(x, y)$ 皆滿足 $x+y \\leq 4024 < 4025$，故自第 $4025$ 分鐘起沒有任何一個箭頭會被改變方向。\n\n以下建構一個在前 $4024$ 分鐘都有箭頭被改變方向的初始位置。考慮下圖：\n\n![](images/13-3J_p22_data_74d0b31c0c.png)\n\n利用上述坐標系統，最下一排方格由東到西的坐標為 $(0,0)$ 到 $(2012,0)$。在上圖中，只有在坐標 $(x,0)$ 的方格中的箭頭朝東（$0 \\leq x \\leq 2012$），其餘方格中的箭頭朝南。\n\n觀察：在 $(0,y)$ 方格（$0 \\leq y \\leq 2012$）中的箭頭永遠朝南。\n\n先看前 $2012$ 分鐘的情況：設 $0 \\leq m \\leq 2012$。在 $m$ 分鐘後，方格 $(x,y)$ 中的箭頭朝東，其中 $1 \\leq x \\leq 2012$，$y=m$。但每一個 $1 \\leq x \\leq 2012$，$y > m$ 的 $(x, y)$ 方格中的箭頭朝南。所以在第 $m$ 分鐘時，每一個在 $(x, m)$ 方格（$1 \\leq x \\leq 2012$）中的箭頭會被換成朝南的箭頭。經過 $2012$ 分鐘後，最北邊的列中只有 $(0, 2012)$ 方格中的箭頭朝南，其餘 $(x, 2012)$（$1 \\leq x \\leq 2012$）方格內的箭頭朝東。\n\n接下來在 $2012 + m$ 分鐘之後（$0 \\leq m \\leq 2012$），$(x, 2012)$ 方格中的箭頭，若 $m + 1 \\leq x \\leq 2012$ 會朝東，但 $(m, 2012)$ 方格中的箭頭朝南。因此在第 $2012 + m$ 分鐘時（$1 \\leq m \\leq 2012$），$(m, 2012)$ 方格中的箭頭會由朝東換成朝南。綜上，在前 $4024$ 分鐘中都有箭頭被改變方向。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18281,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $a$ и $b$ се природни броеви. Докажи дека $a^2 + ab + b^2$ е делител на бројот $(a+b)^6 - a^6$.",
    "options": [],
    "answer": "See solution",
    "solution": "Ќе ги искористиме идентитетите $A^6 - B^6 = (A^3 - B^3)(A^3 + B^3)$ и $A^3 + B^3 = (A+B)(A^2 - AB + B^2)$. Навистина, добиваме\n\n$$\n\\begin{aligned}\n(a+b)^6 - a^6 &= [(a+b)^3 - a^3][(a+b)^3 + a^3] \\\\\n&= [(a+b)-a][(a+b)^2 + a(a+b) + a^2][(a+b)+a][(a+b)^2 - (a+b)a + a^2] \\\\\n&= b(2a+b)(a^2 + ab + b^2)(3a^2 + 3ab + b^2)\n\\end{aligned}\n$$\n\nСпоред дефиницијата за деливост, $(a^2 + ab + b^2) \\mid (a+b)^6 - a^6$ и количникот од делењето е $[(a+b)^3 - a^3](2a+b)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18282,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "**Answer:** 24.\n\n**Solution.** *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 $(25, 0, 0, \\ldots, 0)$. 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 $(1, 1, 1, \\ldots, 1)$, and the predictions of others be $(1, 0, \\ldots, 0, 24)$, $(0, 1, \\ldots, 0, 24)$, $\\ldots$, $(0, 0, \\ldots, 1, 24)$ (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": 18283,
    "subject": "Mathematics (Olympiad)",
    "question": "The cost of one kilogram of chocolate is $x$ UAH and a kilogram of potatoes is $y$ UAH, where $x$ and $y$ are positive integers with at most two digits.\n\nMother asked Mariy to buy 200 grams of chocolate and 1 kg of potatoes, which cost exactly $N$ UAH. Mariy confused the order and bought 200 grams of potatoes and 1 kg of chocolate instead, paying exactly $M > N$ UAH. The numbers $M$ and $N$ each have at most two digits and are formed from the same digits, but in a different order. How much does a kilogram of potatoes and a kilogram of chocolate cost?",
    "options": [],
    "answer": "See solution",
    "solution": "Chocolate costs 50 UAH, and potatoes cost 5 UAH.\n\nLet the prices of chocolate and potatoes be $x$ and $y$ UAH, respectively. Then $N = \\overline{ab} = 10a + b$ and $M = \\overline{ba} = 10b + a$, with $M > N$ so $b > a > 0$.\n\nWe have:\n\n$$\n\\frac{1}{5}x + y = 10a + b\n$$\n$$\n\\frac{1}{5}y + x = 10b + a\n$$\n\nMultiply both equations by 5:\n\n$$\nx + 5y = 50a + 5b\n$$\n$$\ny + 5x = 50b + 5a\n$$\n\nSince $b > a > 0$ and $x > y$, subtract the first from the second:\n\n$$\n4(x - y) = 45(b - a)\n$$\n\nAdd the two equations:\n\n$$\n6(x + y) = 55(b + a)\n$$\n\nSo $b - a \\neq 4$ and $b + a \\neq 6$. Try $b = 5$, $a = 1$:\n\n$$\nx - y = 45\n$$\n$$\nx + y = 55\n$$\n\nSolving, $x = 50$, $y = 5$.\n\nIf $b = 8$, $a = 4$:\n\n$$\nx - y = 45\n$$\n$$\nx + y = 110\n$$\n\nThis gives non-integer solutions. Thus, the answer is:\n\nA kilogram of chocolate costs 50 UAH, and a kilogram of potatoes costs 5 UAH.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18284,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ that satisfy the inequalities\n\n1. $f(x + y) \\geq f(x) + y$\n2. $f(f(x)) \\leq x$\n\nfor all positive $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from (1) that $f$ is a strictly increasing function. Also, (2) implies\n\n$$\nx + y \\geq f(f(x + y)).\n$$\n\nFurthermore, (1) gives $f(f(x + y)) \\geq f(f(x) + y)$, and substituting $x \\to y$ and $y \\to f(x)$ in (1) implies $f(f(x) + y) \\geq f(x) + f(y)$.\n\nSince $f$ is increasing, we have $\\lim_{x \\to 0^+} f(x) = \\inf_{x>0} f(x) = \\ell \\geq 0$. Note that (2) implies $\\lim_{x \\to 0^+} f(f(x)) = 0$.\n\nAssume $\\ell > 0$. Since $f$ is increasing, $f(f(x)) \\geq f(\\ell) > 0$, a contradiction to $\\lim_{x \\to 0^+} f(f(x)) = 0$. Therefore, $\\ell = 0$ and $\\lim_{x \\to 0^+} f(x) = 0$. Letting $y \\to 0^+$ in the previous inequality, we obtain $x \\geq f(x)$ for all positive $x$. It follows now from (1) that\n\n$$\nx + y \\geq f(x + y) \\geq f(x) + y, \\\\\nx - f(x) \\geq f(x + y) - f(x) - y \\geq 0.\n$$\n\nFix $x + y$ and let $x \\to 0^+$ in the above inequalities. We have $f(x + y) = x + y$, meaning that $f(x) = x$ for all positive $x$. This function is obviously a solution to the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18285,
    "subject": "Mathematics (Olympiad)",
    "question": "A $5 \\times 5$ table is called *regular* if each of its cells contains one of four pairwise distinct real numbers, such that each of them occurs exactly once in every $2 \\times 2$ subtable. The sum of all numbers of a *regular table* is called the *total sum* of the table. With any four numbers, one constructs all possible regular tables, computes their total sums, and counts the distinct outcomes. Determine the maximum possible count.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** We will prove that the maximum number of total sums is $60$.\n\nThe proof is based on the following claim.\n\n**Claim.** In a regular table, either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.\n\n**Proof of the Claim.** Let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions; let $x, y, z$ be the numbers in consecutive positions (where $\\{x, y, z, t\\} = \\{a, b, c, d\\}$). Due to our hypothesis that in every $2 \\times 2$ subarray each number is used exactly once, in the row above $R$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see the following figure).\n\n$$\n\\begin{pmatrix}\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet\n\\end{pmatrix}\n$$\n\nCompleting the array, it easily follows that each column contains exactly two of the numbers and our claim is proven.\n\nRotating the matrix (if necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \\times 4$ array, that can be divided into four $2 \\times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a + b + c + d)$.\n\nIt suffices to find how many different ways there are to put the numbers in the first row $R_1$ and the first column $C_1$.\n\nDenoting by $a_1, b_1, c_1, d_1$ the number of appearances of $a, b, c$, and respectively $d$ in $R_1$ and $C_1$, the total sum of the numbers in the entire $5 \\times 5$ array will be\n\n$$\nS = 4(a + b + c + d) + a_1 \\cdot a + b_1 \\cdot b + c_1 \\cdot c + d_1 \\cdot d.\n$$\n\nIn the first, third, and fifth row contain the numbers $x, y$ with $x$ denoting the number at the entry $(1,1)$, then the second and fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_1 + y_1 = 7$ and $x_1 \\geq 3$, $y_1 \\geq 2$, $z_1 + t_1 = 2$, and $z_1 \\geq t_1$. Then $\\{x_1, y_1\\} = \\{5, 2\\}$ or $\\{x_1, y_1\\} = \\{4, 3\\}$, respectively $\\{z_1, t_1\\} = \\{2, 0\\}$ or $\\{z_1, t_1\\} = \\{1, 1\\}$.\n\nThen $\\{a_1, b_1, c_1, d_1\\}$ is obtained by permuting one of the following quadruples:\n\n$$\n(5, 2, 2, 0), \\quad (5, 2, 1, 1), \\quad (4, 3, 2, 0), \\quad (4, 3, 1, 1).\n$$\n\nThere are a total of $\\frac{4!}{2!} = 12$ permutations of $(5, 2, 2, 0)$, also $12$ permutations of $(5, 2, 1, 1)$, $24$ permutations of $(4, 3, 2, 0)$, and finally, $12$ permutations of $(4, 3, 1, 1)$. Hence, there are at most $60$ different possible total sums.\n\nWe can obtain indeed each of these $60$ combinations: take three rows `ababa` alternating with two rows `cdcdc` to get $(5, 2, 2, 0)$; take three rows `ababa` alternating with one row `cdcdc` and a row `dcdcd` to get $(5, 2, 1, 1)$; take three rows `ababc` alternating with two rows `cdcda` to get $(4, 3, 2, 0)$; take three rows `abcda` alternating with two rows `cdabc` to get $(4, 3, 1, 1)$.\n\nBy choosing, for example, $a = 10^3$, $b = 10^2$, $c = 10$, $d = 1$, we can make all these sums different.\n\nHence, $60$ is indeed the maximum possible number of different sums.\n\n**Alternative version.** Consider a regular table containing the four distinct numbers $a, b, c, d$. The four $2 \\times 2$ corners contain each all the four numbers, so that, if $a_1, b_1, c_1, d_1$ are the numbers of appearances of $a, b, c$, and respectively $d$ in the middle row and column, then\n\n$$\nS = 4(a + b + c + d) + a_1 \\cdot a + b_1 \\cdot b + c_1 \\cdot c + d_1 \\cdot d.\n$$\n\nConsider the numbers $x$ in position $(3,3)$, $y$ in position $(3,2)$, $y'$ in position $(3,4)$, $z$ in position $(2,3)$ and $z'$ in position $(4,3)$.\n\nIf $z \\neq z' = t$, then $y = y'$, and in position $(3,1)$ and $(3,5)$ there will be the number $x$.\n\nThe second and fourth row can only contain now the numbers $z$ and $t$, respectively the first and fifth row only $x$ and $y$.\n\nThen $x_1 + y_1 = 7$ and $x_1 \\geq 3$, $y_1 \\geq 2$, $z_1 + t_1 = 2$, and $z_1 \\geq t_1$. Then $\\{x_1, y_1\\} = \\{5, 2\\}$ or $\\{x_1, y_1\\} = \\{4, 3\\}$, respectively $\\{z_1, t_1\\} = \\{2, 0\\}$ or $\\{z_1, t_1\\} = \\{1, 1\\}$.\n\nOne can continue now as in the first version.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18286,
    "subject": "Mathematics (Olympiad)",
    "question": "A and B are playing ping-pong, with the agreement that the winner of a game will get 1 point and the loser 0 points; the match ends as soon as one of the players is ahead by 2 points or the number of games reaches six. Suppose that the probabilities of A and B winning a game are $\\frac{2}{3}$ and $\\frac{1}{3}$, respectively, and each game is independent. Then the expectation $E\\xi$ for the match ending with $\\xi$ games is:\n\n(A) $\\frac{241}{81}$\n\n(B) $\\frac{266}{81}$\n\n(C) $\\frac{274}{81}$\n\n(D) $\\frac{670}{243}$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $\\xi$ can only be 2, 4, or 6. Divide the six games into three rounds, each with two consecutive games. If one player wins both games in the first round, the match ends:\n\n$$\n\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 = \\frac{5}{9}.\n$$\n\nOtherwise, the players tie and the match enters the second round; this probability is\n\n$$\n1 - \\frac{5}{9} = \\frac{4}{9}.\n$$\n\nSimilarly, for the second and third rounds:\n\n$$\n\\begin{aligned}\nP(\\xi = 2) &= \\frac{5}{9}, \\\\\nP(\\xi = 4) &= \\frac{4}{9} \\times \\frac{5}{9} = \\frac{20}{81}, \\\\\nP(\\xi = 6) &= \\left(\\frac{4}{9}\\right)^2 = \\frac{16}{81}.\n\\end{aligned}\n$$\n\nThus,\n\n$$\nE\\xi = 2 \\times \\frac{5}{9} + 4 \\times \\frac{20}{81} + 6 \\times \\frac{16}{81} = \\frac{266}{81}.\n$$\n\n**Answer:** B",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18287,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos un cubo de lado 3 formado por 27 piezas cúbicas de lado 1. Dentro de cada pieza hay una bombilla que puede estar encendida o apagada. Cada vez que se pulsa una pieza (no es posible pulsar la del centro del cubo), cambia el estado de su bombilla y el de las que comparten una cara con la pulsada. Inicialmente, todas las bombillas están apagadas. Responde razonadamente a las siguientes preguntas:\n\n1. ¿Se puede conseguir que todas las bombillas queden encendidas?\n\n2. ¿Se puede conseguir que queden encendidas todas salvo la del centro del cubo?\n\n3. ¿Se puede conseguir que solo quede encendida la del centro del cubo?",
    "options": [],
    "answer": "See solution",
    "solution": "Llamamos $N$ a la pieza de lado 1 que está en el interior, $C$ a cada pieza que es centro de una cara, $V$ a cada pieza que es vértice del cubo y $A$ a cada una de las que comparten cara con un vértice. Es evidente que el resultado obtenido tras pulsar varias piezas es independiente del orden en que las pulsemos.\n\n1. El estado de la bombilla de $N$ cambia si y solo si pulsamos una $C$. Para que quede encendida hemos de pulsar un número impar de veces en piezas $C$ (ya no pulsamos más $C$'s). Quedarán un número impar de piezas $C$ apagadas. La única manera de cambiar su estado es pulsar $A$'s, pero cada una cambia el estado de dos $C$'s, luego es imposible que todas las $C$'s y $N$ queden encendidas.\n\n2. Sí es posible. Pulsamos todas las $A$ y quedan encendidas todas ellas y todas las $V$. Están apagadas las $C$ y la $N$. Ahora pulsamos todas las $C$: las $V$ no se ven afectadas; cada $A$ sufre dos cambios, luego sigue encendida, y $N$ queda apagada porque cambia su estado un número par de veces.\n\n3. No es posible. Si lo fuera, se podría pasar del caso 2 (todas menos $N$ encendidas) al caso 3. Y eso supondría disponer de una sucesión de pulsaciones que permite cambiar el estado de todas las bombillas, lo cual es inviable por lo visto en 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18288,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A(-a, a^2)$, $C(a, a^2)$, $B(b, b^2)$, and $D(d, d^2)$ be points on the parabola $y = x^2$, with $AC \\parallel Ox$ (so $A$ and $C$ are symmetric with respect to the $y$-axis). Given that $\\angle DAC = 45^\\circ$ and $\\angle BAC = 135^\\circ$, and that the distance $p = |BD|$, find the area of quadrilateral $ABCD$ in terms of $p$.\n\n![](images/BielorrusiaBOOK18-E-A4_p21_data_444d4ab8d9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the coordinates: $A(-a, a^2)$, $C(a, a^2)$, $B(b, b^2)$, $D(d, d^2)$. Since $AC \\parallel Ox$, $A$ and $C$ are symmetric about the $y$-axis. The line $AD$ has equation $y = (d-a)x + da$. Given $\\angle DAC = 45^\\circ$, the slope of $AD$ is $1$, so $d-a = 1$. Similarly, the line $AB$ has equation $y = (b-a)x + ba$ and its slope is $-1$, so $b-a = -1$.\n\nLet $B_1(d, b^2)$ and $C_1(d, c^2)$. In right triangle $BB_1D$, $p^2 = (d-b)^2 + (d^2-b^2)^2$. Since $d-b = 2$ and $d+b = 2a$, $p^2 = 4 + 16a^2$, so $a^2 = \\frac{1}{16}(p^2-4)$.\n\nThe area is $S_{ABCD} = S_{ABC} + S_{ADC}$. Both triangles share base $AC = 2a$, and the sum of their altitudes is $DB_1 = 4a$. Thus, the area is $\\frac{1}{2} \\cdot 2a \\cdot 4a = 4a^2 = \\frac{1}{4}(p^2-4)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18289,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive numbers such that $x^2y^2 + y^2z^2 + z^2x^2 = 6xyz$. Prove that\n\n$$\n\\sqrt{\\frac{x}{x + yz}} + \\sqrt{\\frac{y}{y + zx}} + \\sqrt{\\frac{z}{z + xy}} \\ge \\sqrt{3}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f : (-1, +\\infty) \\to \\mathbb{R}$ be the function defined by $f(x) = \\frac{1}{\\sqrt{1+x}}$. Since $f'(x) = -\\frac{1}{2(1+x)^{3/2}} < 0$ and $f''(x) = \\frac{3}{4(1+x)^2\\sqrt{1+x}} > 0$, $f$ is convex. Applying Jensen's inequality to $f$ with $a_1 = \\frac{yz}{x}$, $a_2 = \\frac{zx}{y}$, $a_3 = \\frac{xy}{z}$, we have\n\n$$\n\\frac{1}{3} \\sum_{k=1}^{3} f(a_k) \\ge f\\left(\\frac{1}{3} \\sum_{k=1}^{3} a_k\\right)\n$$\n\nor\n\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{x}{x + yz}} + \\sqrt{\\frac{y}{y + zx}} + \\sqrt{\\frac{z}{z + xy}} \\right) \\ge f(2) = \\frac{\\sqrt{3}}{3}\n$$\n\nbecause from $x^2y^2 + y^2z^2 + z^2x^2 = 6xyz$ it follows that\n\n$$\n\\frac{a_1 + a_2 + a_3}{3} = \\frac{1}{3} \\left( \\frac{yz}{x} + \\frac{zx}{y} + \\frac{xy}{z} \\right) = 2\n$$\n\nEquality holds when $x = y = z = 2$, and we are done. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18290,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with real coefficients satisfying the condition\n\n$$\nP(x^2) + x\\big(3P(x) + P(-x)\\big) = (P(x))^2 + 2x^2\n$$\n\nfor all real numbers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x)$ be a polynomial satisfying the given condition. We have:\n\n$$\nP(x^2) + x\\big(3P(x) + P(-x)\\big) = (P(x))^2 + 2x^2, \\quad x \\in \\mathbb{R}. \\tag{1}\n$$\n\nConsider the possible degrees of $P(x)$:\n\n**Case 1:** $\\deg P = 1$.\n\nLet $P(x) = ax + b$, $a \\neq 0$. Substituting into (1):\n\n$$\n(a^2 - 3a + 2)x^2 + 2b(a - 2b)x + b^2 - b \\equiv 0.\n$$\n\nThis yields:\n\n$$\n(a = 1, b = 0),\\quad (a = 2, b = 0),\\quad (a = 2, b = 1).\n$$\n\nSo $P(x) = x$, $P(x) = 2x$, and $P(x) = 2x + 1$.\n\n**Case 2:** $\\deg P = n > 1$.\n\nLet $P(x) = ax^n + S(x)$, $a \\neq 0$, where $\\deg S(x) = k < n$.\n\nSubstituting into (1):\n\n$$\n(a^2 - a)x^{2n} + (S(x))^2 - S(x^2) + 2a x^n S(x) \\equiv (3 + (-1)^n)a x^{n+1} + (3S(x) + S(-x))x - 2x^2. \\tag{2}\n$$\n\nComparing degrees, $a^2 - a = 0$ so $a = 1$.\n\nNow:\n\n$$\n2x^n S(x) + (S(x))^2 - S(x^2) \\equiv (3 + (-1)^n)x^{n+1} + (3S(x) + S(-x))x - 2x^2. \\tag{3}\n$$\n\nDegree comparison gives $k = 1$, so $S(x) = px$ or $S(x) = px + 1$.\n\n- If $S(x) = px$:\n\n  $$\n  (3 + (-1)^n - 2p)x^{n+1} - (p^2 - 3p + 2)x^2 \\equiv 0\n  $$\n  $$\n  \\Rightarrow p = 1,\\ n \\text{ odd};\\quad p = 2,\\ n \\text{ even}.\n  $$\n  So $P(x) = x^{2n+1} + x$ and $P(x) = x^{2n} + 2x$.\n\n- If $S(x) = px + 1$:\n\n  This leads to a contradiction, so no solutions in this case.\n\n**Conclusion:**\n\nThe polynomials $P(x)$ satisfying the condition are:\n\n$$\nP(x) = x,\\quad P(x) = x^{2n} + 2x,\\quad P(x) = x^{2n+1} + x,\\quad n \\in \\mathbb{N}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18291,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, \\dots, p_{n+1}$ denote the first $n+1$ primes. Suppose that $\\{A, B\\}$ is a partition of the set $X = \\{p_1, p_2, \\dots, p_n\\}$, where $A = \\{q_1, q_2, \\dots, q_s\\}$ and $B = \\{r_1, r_2, \\dots, r_t\\}$. Prove that if $m = q_1 q_2 \\dots q_s + r_1 r_2 \\dots r_t < p_{n+1}^2$, then $m$ is a prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume to the contrary that $m$ is not a prime number. Then $m = ab$ for some integers $a$ and $b$ with $1 < a < m$ and $1 < b < m$. Let $p$ be the smallest prime that divides $a$ and let $q$ be the smallest prime that divides $b$. Without loss of generality, assume $p \\leq q$.\n\nWe consider two cases:\n\n- **Case 1:** $p \\in X$. Then $p$ is either some $q_i$ or some $r_j$, but not both (since $\\{A, B\\}$ is a partition). Suppose $p = q_i$ for some $i$. Since $p \\mid a$, $p \\mid m$, and $p \\mid q_1 q_2 \\dots q_s$, so $p \\mid (m - q_1 q_2 \\dots q_s) = r_1 r_2 \\dots r_t$. Thus $p = r_j$ for some $j$, a contradiction.\n\n- **Case 2:** $p \\notin X$. Then $q \\geq p \\geq p_{n+1}$, so $m \\geq pq \\geq p_{n+1}^2$, contradicting the assumption $m < p_{n+1}^2$.\n\nTherefore, $m$ must be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18292,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of the 12 edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the 6 faces of the cube equal to $2$?\n\n(A) 8  \n(B) 10  \n(C) 12  \n(D) 16  \n(E) 20",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (E):**\n\nFirst, suppose that the three edges that share one particular vertex are all labeled $1$. Then the rest of the labels are forced by the sum condition, and the figure below is obtained, up to rotation, where those first three edge labels are shown in bold italic font. Note that in this case there is a pair of vertices at the end of an interior diagonal, all of whose edges connected to it are labeled $1$, with the remaining edges all labeled $0$, no three of which mutually share a vertex. Because there are $4$ interior diagonals, this gives **4** possible labelings. By symmetry, there are another **4** possible labelings in which the roles of $0$ and $1$ are interchanged.\n\n![](images/2021_AMC10A_Solutions_Fall_p11_data_45ec3c0a6b.png)\n\nOtherwise, every vertex is connected to two edges labeled $1$ and one edge labeled $0$ or vice versa. Suppose that the labels of the edges on the bottom face of the cube are $0, 1, 0, 1$ in that order. There are $2$ possibilities, depending on which label is given to the front bottom edge. There are also $2$ possibilities for the label of the left front edge. Once those five labels are determined, the rest of the labels are forced by the sum condition and the fact that no vertex is connected to edges with all the same label. See the figure below, in which the five mentioned labels are shown in bold italic font. This gives $2 \\cdot 2 = 4$ labelings.\n\n![](images/2021_AMC10A_Solutions_Fall_p11_data_bbdc98e2ff.png)\n\nIn the remaining case, the bottom face has labels $0, 0, 1, 1$ in that order, which is $4$ more cases. Say that the front and left bottom edges are labeled $0$. Then the right front vertical edge can have either label, and once that label is chosen, again the rest of the labeling is forced. See the figure below, in which the five mentioned labels are again shown in bold italic font. This gives $4 \\cdot 2 = 8$ more labelings.\n\n![](images/2021_AMC10A_Solutions_Fall_p12_data_eeb58a58b0.png)\n\nThus in all there are $4 + 4 + 4 + 8 = 20$ labelings satisfying the condition.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18293,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers satisfying $abcd = 1$. Prove that\n\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)(b+d)(ac+bd+2).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy-Schwarz inequality, we have\n\n$$\n(a^2b + c^2d + ad^2 + cb^2)(a^2d + c^2b + ab^2 + cd^2) \\geq (a^2\\sqrt{bd} + c^2\\sqrt{bd} + abd + CBD)^2 = \\left(\\frac{a^2 + c^2}{\\sqrt{ac}} + abd + CBD\\right)^2.\n$$\n\nTogether with $a^2 + c^2 \\geq \\frac{1}{2}(a+c)^2 \\geq (a+c)\\sqrt{ac}$, we have\n\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)^2(1+bd)^2.\n$$\n\nDue to symmetry, we also have\n\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (b+d)^2(1+ac)^2.\n$$\n\nMultiplying these inequalities, we obtain\n\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)(b+d)(1+ac)(1+bd)\n$$\n\nwhere $(1+ac)(1+bd) = 1+ac+bd+abcd = ac+bd+2$. The result follows readily.\n\nFor equality in the application of the Cauchy-Schwarz inequality, we need $b = d$ and $a = c$. Note that the other inequalities also have these as equality. Together with $abcd = 1$, equality holds when $a = c = t$ and $b = d = \\frac{1}{t}$ for some $t > 0$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18294,
    "subject": "Mathematics (Olympiad)",
    "question": "$a$, $b$, $c$ нь тэгээс ялгаатай бүхэл тоонууд ба $a \\neq c$. $a(c^2 + b^2) = c(a^2 + b^2)$ бол. Тэгвэл $a^2 + b^2 + c^2$ зохиомол тоо гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "$a(c^2 + b^2) = c(a^2 + b^2) \\Leftrightarrow (a - c)(b^2 - a c) = 0$ ба $a \\neq c$ тул $b^2 = a c$ болно.\n\nИймд $a^2 + b^2 + c^2 = a^2 + a c + c^2 = (a + c)^2 - a c = (a + c - b)(a + c + b)$.\n\n$a^2 + b^2 + c^2 > 3$ юм. Эсрэгээр нь $a^2 + b^2 + c^2$ анхны тоо гэж үзье. Тэгвэл дараах тохиолдлууд гарна:\n\n1. $a + c - b = 1$, $a + c + b = a^2 + b^2 + c^2$;\n2. $a + c + b = 1$, $a + c - b = a^2 + b^2 + c^2$;\n3. $a + c - b = -1$, $a + c + b = - (a^2 + b^2 + c^2)$;\n4. $a + c + b = -1$, $a + c - b = - (a^2 + b^2 + c^2)$.\n\n(1), (2)-д: $a^2 + b^2 + c^2 - 2(a + c) + 1 = 0 \\Rightarrow (a - 1)^2 + (c - 1)^2 + b^2 = 1 \\Rightarrow a = c = 1$ болох ба энэ нь зөрчилтэй.\n\n(3), (4)-д: $(a + 1)^2 + (c + 1)^2 + b^2 = 1$ болж $a = c = -1$ болох ба мөн зөрчилтэй.\n\nИймд $|a + c - b| > 1$, $|a + c + b| > 1$ буюу $a^2 + b^2 + c^2$ зохиомол тоо.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18295,
    "subject": "Mathematics (Olympiad)",
    "question": "In the beginning, there are two positive integers on a blackboard. On each step, one chooses numbers $a$ and $b$ such that $a \\leq b$ from the numbers on the blackboard in all possible ways (equality means that one may take the same number twice), finds all corresponding sums $a + b + \\gcd(a, b)$, and replaces all the numbers on the blackboard instantly with these sums. Prove that at some step at least one number will occur more than once on the blackboard.",
    "options": [],
    "answer": "See solution",
    "solution": "If the numbers chosen from the blackboard are $x$ and $y$, then the number $x + y + \\gcd(x, y)$ will be on the blackboard on the next step. If $x$ is chosen together with itself, the number $x + x + \\gcd(x, x) = 3x$ will be on the blackboard on the next step.\n\nWe show that there will be two equal numbers on the blackboard after the second step at latest. Assume that the initial numbers are $n$ and $m$, and the numbers $n + m + \\gcd(n, m)$, $3n$, and $3m$ appearing on the first step are all distinct. Choosing the number $n + m + \\gcd(n, m)$ together with itself, we obtain $3(n + m + \\gcd(n, m)) = 3n + 3m + 3\\gcd(n, m)$. Choosing $3n$ and $3m$, we obtain $3n + 3m + \\gcd(3n, 3m)$. As $\\gcd(3n, 3m) = 3\\gcd(n, m)$, the same number will appear twice after the second step.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18296,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be the side lengths of a triangle with $x \\leq y < z$. Given that $x^2 + y^2 = z^2$ and $xy = 240$, find the length of the hypotenuse $z$.",
    "options": [],
    "answer": "See solution",
    "solution": "From Pythagoras, $x^2 + y^2 = z^2$, and we are given $xy = 240$. So $(x+y)^2 = z^2 + 480$ and $(x-y)^2 = z^2 - 480$. Let $r = x + y$. Then $480 = r^2 - z^2 = (r + z)(r - z)$. Since $(r + z) - (r - z) = 2z$, both factors are even. Since $z^2 \\geq 480$, the ordered pairs $(r - z, r + z)$ are one or more of $(2, 240)$, $(4, 120)$, $(6, 80)$, $(8, 60)$.\n\nFor each ordered pair, we calculate $z$ and check if $z^2 - 480$ is a square:\n\n| $r - z$ | $r + z$ | $z$ | $z^2 - 480$ | $z^2 - 480$ square? |\n|---------|---------|-----|-------------|----------------------|\n| 2       | 240     | 119 | 13961       | No                   |\n| 4       | 120     | 58  | 2884        | No                   |\n| 6       | 80      | 37  | 889         | No                   |\n| 8       | 60      | 26  | 196         | Yes ($=14^2$)        |\n\nThus, the length of the hypotenuse is $\\boxed{26}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18297,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $w$ be the number of men and women in a tournament. Each game assigns a total of 1 point to the players. The total points assigned to games between two men and two women are $\\binom{m}{2}$ and $\\binom{w}{2}$, respectively. Each player scored the same total of points against men as against women, and the total number of games played is $\\binom{m+w}{2}$. What can be said about $m$ and $w$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since each player scored the same total of points against men as against women, and the total number of games is $\\binom{m+w}{2}$, we have:\n\n$$\n2 \\left( \\binom{m}{2} + \\binom{w}{2} \\right) = \\binom{m+w}{2}\n$$\n\nThis simplifies to:\n\n$$\nm + w = (m - w)^2\n$$\n\nThus, the total number of players, $m + w$, is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18298,
    "subject": "Mathematics (Olympiad)",
    "question": "Let triangle $ABC$ have centroid $G$. Let $D$ be the projection of the Euler point of triangle $ABC$ onto $AC$, and let $E$ be the symmetric point of $B$ about line $AC$. Prove that $\\overrightarrow{GE} = 4\\overrightarrow{GD}$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Vietnam_Booklet_2013_12-07_p42_data_a217666ef5.png)\n\nLet $H$, $N$, and $O$ be the orthocenter, Euler point, and circumcenter of triangle $ABC$, respectively. Assume that $K$, $D$, and $M$ (in order) are the projections of $H$, $N$, and $O$ onto the line $AC$.\n\nBecause $N$ is the midpoint of segment $OH$, $D$ is also the midpoint of segment $KM$. It is easy to see that\n\n$$\n\\frac{GN}{GH} = \\frac{1}{4} \\quad \\text{and} \\quad ND = \\frac{1}{2}(OM + HK) = \\frac{1}{4}BH + \\frac{1}{4}(HE - BH) = \\frac{1}{4}HE.\n$$\n\nHence, $\\frac{GN}{GH} = \\frac{ND}{HE}$, so $G$, $D$, $E$ are collinear and $GE = 4GD$. The lemma is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18299,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be arbitrary nonzero numbers. Define:\n\n$$A = \\frac{a^2 + b^2}{c^2}, \\quad B = \\frac{b^2 + c^2}{a^2}, \\quad C = \\frac{c^2 + a^2}{b^2}$$\n\nand $P = A \\cdot B \\cdot C$, $S = A + B + C$. What are the possible values of $P - S$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\nP = A B C = \\left(\\frac{a^2 + b^2}{c^2}\\right) \\left(\\frac{b^2 + c^2}{a^2}\\right) \\left(\\frac{c^2 + a^2}{b^2}\\right)\n$$\n\nExpanding, we get:\n\n$$\nP = 1 + \\frac{a^2}{c^2} + \\frac{b^2}{c^2} + \\frac{b^2}{a^2} + \\frac{c^2}{a^2} + \\frac{c^2}{b^2} + \\frac{a^2}{b^2} + 1 = 2 + S\n$$\n\nTherefore,\n\n$$\nP - S = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18300,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a fixed positive integer. Denote $D$ as the set of all positive divisors of $n$. If $A$ and $B$ are subsets of $D$ such that for any $a \\in A$, $b \\in B$, neither $a$ divides $b$ nor $b$ divides $a$, prove that\n$$\n\\sqrt{|A|} + \\sqrt{|B|} \\le \\sqrt{|D|}.\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Given $A$ and $B$, partition $D$ into four disjoint subsets $D = X \\cup Y \\cup Z \\cup W$, where\n$$\n\\begin{align*}\nX &= \\{x \\in D : \\exists a \\mid x, \\exists b \\mid x\\}, & Y &= \\{x \\in D : \\exists a \\mid x, \\forall b \\mid x\\}, \\\\\nZ &= \\{x \\in D : \\forall a \\mid x, \\exists b \\mid x\\}, & W &= \\{x \\in D : \\forall a \\mid x, \\forall b \\mid x\\}.\n\\end{align*}\n$$\n\nThe problem conditions indicate that $A \\subseteq Y$, $B \\subseteq Z$. We shall prove a stronger statement: for any two nonempty subsets $A$ and $B$, the inequality $\\sqrt{|Y|} + \\sqrt{|Z|} \\le \\sqrt{|D|}$ holds. Equivalently,\n$$\n|Y| + |Z| + 2\\sqrt{|Y| \\cdot |Z|} \\le |D| = |X| + |Y| + |Z| + |W|,\n$$\nor $2\\sqrt{|Y| \\cdot |Z|} \\le |X| + |W|$, which can be derived from\n$$\n|Y| \\cdot |Z| \\le |X| \\cdot |W|,\n$$\nwhich can be written as\n$$\n\\begin{aligned}\n(|X| + |Y|)(|X| + |Z|) &= |X|(|X| + |Y| + |Z|) + |Y| \\cdot |Z| \\\\\n&\\le |X|(|X| + |Y| + |Z|) + |X| \\cdot |W| = |X| \\cdot |D|.\n\\end{aligned}\n$$\n\nDefine $U = X \\cup Y$, $V = X \\cup Z$. The above inequality becomes $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$. Notice that $U = \\{x \\in D : \\exists a \\mid x\\}$ satisfies: if $x \\in U$ and $x \\mid x'$, then $x' \\in U$. We say $U$ is upward closed in $D$; similarly, $V = \\{x \\in D : \\exists b \\mid x\\}$ is also upward closed.\n\n**Claim:** For nonempty upward closed sets $U$ and $V$ of $D$, we have $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$.\n\n**Proof of claim:** Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the prime factorization. Induct on $k$: denote $p = p_k$, $\\alpha = \\alpha_k$, and $n = p^{\\alpha}n'$.\n\nDefine $D_k = \\{x \\in D \\mid v_p(x) = k\\}$,\n$$\nU_k = U \\cap D_k, \\quad V_k = V \\cap D_k.\n$$\nFor every $k = 0, 1, \\dots, \\alpha - 1$, and every $x \\in U_k$, since $U$ is upward closed, it follows that $px \\in U_{k+1}$, $|U_k| \\le |U_{k+1}|$, and the sequence $\\{|U_k|\\}_k$ is increasing; likewise, $\\{|V_k|\\}_k$ is also increasing. Notice that $\\frac{1}{p^k}U_k$ and $\\frac{1}{p^k}V_k$ are upward closed sets of $\\frac{1}{p^k}D_k = D(n')$. By the induction hypothesis,\n$$\n\\left| \\left( \\frac{1}{p^k} U_k \\right) \\cap \\left( \\frac{1}{p^k} V_k \\right) \\right| \\ge \\frac{1}{|D(n')|} \\cdot \\left| \\left( \\frac{1}{p^k} U_k \\right) \\right| \\cdot \\left| \\left( \\frac{1}{p^k} V_k \\right) \\right|,\n$$\nwhich means $|U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} |U_k| \\cdot |V_k|$. By the rearrangement inequality, it follows that\n$$\n\\begin{align*}\n|U \\cap V| &= \\sum_{k=0}^{\\alpha} |U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} \\sum_{k=0}^{\\alpha} |U_k| \\cdot |V_k| \\\\\n&\\ge \\frac{1+\\alpha}{|D|} \\cdot \\frac{1}{1+\\alpha} \\left( \\sum_{k=0}^{\\alpha} |U_k| \\right) \\cdot \\left( \\sum_{k=0}^{\\alpha} |V_k| \\right) \\\\\n&= \\frac{1}{|D|} |U| \\cdot |V|.\n\\end{align*}\n$$\nThis completes the proof of the claim and also the desired inequality. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18301,
    "subject": "Mathematics (Olympiad)",
    "question": "有若干個正整數排成一列。某人每次選擇兩個相鄰的數字 $x$ 與 $y$，其中 $x > y$，並且 $x$ 在 $y$ 的左邊；接著他把這個數對 $(x, y)$ 用 $(y+1, x)$ 或 $(x-1, x)$ 取代。然後他重複作這樣的選擇與取代的操作。試證他只能作有限次這樣的操作。",
    "options": [],
    "answer": "See solution",
    "solution": "注意到題設的操作不會改變原數列的最大值 $M$。設某一次操作完成後所得到的數列為 $a_1, a_2, \\dots, a_n$。考慮下列的和\n\n$$\nS = a_1 + 2a_2 + \\dots + n a_n.\n$$\n\n我們宣稱：在每次操作後，$S$ 都會增加某正整數的量。設某操作將數對 $(a_i, a_{i+1})$ 以 $(c, a_i)$ 取代，其中 $a_i > a_{i+1}$，且 $c$ 為 $a_{i+1} + 1$ 或 $a_i - 1$ 其中之一。於是 $S$ 的新值與舊值的差為\n\n$$\nd = (i c + (i + 1) a_i) - (i a_i + (i + 1) a_{i+1}) = a_i - a_{i+1} + i (c - a_{i+1}).\n$$\n\n由於 $a_i - a_{i+1} \\geq 1$，且 $c - a_{i+1} \\geq 0$，$d$ 必為正整數。\n\n另一方面，因為每個 $a_i \\leq M$，得 $S \\leq (1 + 2 + 3 + \\cdots + n) M$。由於 $S$ 在每次操作後至少增加 1，而又從來不超過定值 $(1 + 2 + \\cdots + n) M$，在有限次操作後必定終止。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18302,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be $n$ positive integers such that no one divides another; that is, $a_i \\nmid a_j$ for $i \\neq j$. Prove that\n$$\na_1 + a_2 + \\dots + a_n \\geq 1.1n^2 - 2n.\n$$\n\nNote: Some credit will be given for a proof of the inequality for sufficiently large $n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the set $B$ of all positive integers coprime with $6$: $B = \\{1, 5, 7, 11, 13, \\dots\\}$. The sum of the smallest $k$ elements of $B$ is\n$$\nf(k) = \\begin{cases} \\frac{3k^2}{2}, & \\text{if } k \\text{ is even} \\\\ \\frac{3k^2 - 1}{2}, & \\text{if } k \\text{ is odd} \\end{cases}\n$$\nwith $f(0) = 0$, $f(1) = 1$, $f(2) = 6$, $f(3) = 13$, $f(4) = 24$, $f(5) = 37$, etc.\n\nEvery positive integer $a$ can be uniquely written as $a = 2^\\alpha 3^\\beta b$ where $b \\in B$ and $\\alpha, \\beta \\geq 0$. Define $h(a) = b$ (the kernel of $a$). If $m$ positive integers $a_1, \\dots, a_m$ are not divisible by each other and share the same kernel $b$, write $a_k = 2^{\\alpha_k} 3^{\\beta_k} b$ for $k = 1, \\dots, m$. Then $\\alpha_1, \\dots, \\alpha_m$ are distinct, so assume $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_m$; accordingly, $\\beta_1 > \\beta_2 > \\dots > \\beta_m \\geq 0$. Thus,\n$$\na_k \\geq 2^{k-1} 3^{m-k} b\n$$\nand\n$$\na_1 + \\dots + a_m \\geq 2^0 3^{m-1} b + 2^1 3^{m-2} b + \\dots + 2^{m-1} 3^0 b = (3^m - 2^m) b.\n$$\n\nLet $A = \\{a_1, \\dots, a_n\\}$ and define $B_k = \\{b \\in B : \\#\\{a \\in A : h(a) = b\\} \\geq k\\}$. Then\n$$\nn = \\sum_{k=1}^{\\infty} k \\cdot |B_k \\setminus B_{k+1}| = |B_1| + |B_2| + \\dots\n$$\nLet $S(A) = a_1 + \\dots + a_n$. Then\n$$\n\\begin{align*}\nS(A) &\\geq \\sum_{k=1}^{\\infty} (3^k - 2^k) \\cdot |B_k \\setminus B_{k+1}| \\\\\n&= \\sum_{k=1}^{\\infty} [(3^k - 2^k) - (3^{k-1} - 2^{k-1})] \\cdot |B_k| \\\\\n&= \\sum_{k=1}^{\\infty} c_k |B_k|,\n\\end{align*}\n$$\nwhere $c_k = 2 \\cdot 3^{k-1} - 2^{k-1}$ ($c_1 = 1$, $c_2 = 4$, $c_3 = 14$, $c_4 = 46$, $c_5 = 146$, ...).\n\nWe are led to the optimization problem: under $x_1 + x_2 + \\cdots = n$ ($x_i \\geq 0$), minimize\n$$\nT = c_1 f(x_1) + c_2 f(x_2) + \\cdots\n$$\nSuppose $X = (x_1, x_2, \\dots, x_K, 0, 0, \\dots)$ minimizes $T$, with $x_1 \\geq x_2 \\geq \\dots \\geq x_K \\geq 1$. If $K \\leq 2$,\n$$\nT = c_1 f(x_1) + c_2 f(x_2) \\geq \\frac{3x_1^2 - 1}{2} + 4 \\cdot \\frac{3x_2^2 - 1}{2} \\geq \\frac{6}{5}(x_1 + x_2)^2 - \\frac{5}{2} \\geq 1.1n^2 - 2n.\n$$\nIf $K \\geq 3$, by considering changes to $X$ that do not decrease $T$, we get $c_K \\leq 3x_1 + 2$ and $c_K \\leq 12x_2 + 8$. Therefore,\n$$\nn = x_1 + x_2 + \\dots + x_K \\geq x_1 + x_2 + 1 \\geq \\frac{c_K - 2}{3} + \\frac{c_K - 8}{12} + 1 = \\frac{5c_K - 4}{12}\n$$\nwhich implies\n$$\nc_1 + \\dots + c_K \\leq c_K \\left(1 + \\frac{1}{3} + \\dots + \\frac{1}{3^{K-1}}\\right) \\leq \\frac{12n + 4}{5} \\cdot \\frac{3}{2} \\leq 4n.\n$$\nNow,\n$$\nT(X) = c_1 f(x_1) + \\dots + c_K f(x_K) \\geq \\frac{3}{2} [c_1 x_1^2 + \\dots + c_K x_K^2] - \\frac{1}{2} [c_1 + \\dots + c_K]\n$$\nand\n$$\nT(X) \\geq \\frac{3}{2} \\frac{(x_1 + \\dots + x_K)^2}{\\frac{1}{c_1} + \\dots + \\frac{1}{c_K}} - 2n.\n$$\nSince $c_{k+1} \\geq 3c_k$, we have\n$$\n\\frac{1}{c_1} + \\frac{1}{c_2} + \\dots + \\frac{1}{c_K} \\leq 1 + \\frac{1}{4} + \\frac{1}{14} + \\frac{1}{14 \\cdot 3} + \\frac{1}{14 \\cdot 3^2} + \\dots = 1 + \\frac{1}{4} + \\frac{1.5}{14} \\leq 1.36\n$$\nso\n$$\nT \\geq \\frac{3}{2} \\cdot \\frac{n^2}{1.36} - 2n > 1.1n^2 - 2n.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18303,
    "subject": "Mathematics (Olympiad)",
    "question": "*(a)* For which integers $n$ with $1 \\leq n \\leq 100$ is the product $n! \\cdot (n+1)!$ a perfect square?\n\n*(b)* For which integers $n$ is the product $n! \\cdot (n+1)! \\cdot (n+2)! \\cdot (n+3)!$ a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "*(a)* We observe that $$(n+1)! = (n+1) \\cdot n!,$$ so $$n! \\cdot (n+1)! = (n!)^2 \\cdot (n+1).$$\nSince $(n!)^2$ is always a perfect square, the product is a perfect square if and only if $n+1$ is a perfect square. For $1 \\leq n \\leq 100$, this occurs for $n = 3, 8, 15, 24, 35, 48, 63, 80, 99$ (i.e., perfect squares minus one below $100$).\n\n*(b)* We rewrite the product:\n$$n! \\cdot (n+1)! \\cdot (n+2)! \\cdot (n+3)! = (n!)^2 \\cdot (n+1) \\cdot ((n+2)!)^2 \\cdot (n+3).$$\nBoth $(n!)^2$ and $((n+2)!)^2$ are perfect squares, so the product is a perfect square if and only if $(n+1)(n+3)$ is a perfect square. Suppose $(n+1)(n+3) = k^2$ for some integer $k$. Since $(n+1)^2 < (n+1)(n+3) < (n+3)^2$, we have $n+1 < k < n+3$, so $k = n+2$. But $(n+1)(n+3) = (n+2)^2 - 1$, which is not a perfect square. Thus, there are no such $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18304,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime and let $a_0, a_1, \\dots, a_{p-1}$ be integers such that\n\n$$\na_0^i + a_1^i + \\cdots + a_{p-1}^i \\equiv 0 \\pmod{p} \\quad (1 \\le i \\le p-2)\n$$\n\nand\n\n$$\na_0^{p-1} + a_1^{p-1} + \\cdots + a_{p-1}^{p-1} \\equiv -1 \\pmod{p}.\n$$\n\nShow that $\\{a_0, a_1, \\dots, a_{p-1}\\}$ is a complete system of residues modulo $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $g(x) = (x - a_0)(x - a_1)\\cdots(x - a_{p-1}) = x^p + b_1x^{p-1} + \\cdots + b_{p-1}x + b_p$ and $S_i = a_0^i + a_1^i + \\cdots + a_{p-1}^i$ for $i \\in \\mathbb{N}$.\n\nIf for all $0 \\le i \\le p-1$ we have $a_i \\not\\equiv 0 \\pmod{p}$, then by Fermat's little theorem,\n\n$$\n1 \\equiv a_0^{p-1} + a_1^{p-1} + \\cdots + a_{p-1}^{p-1} \\equiv \\frac{1+1+\\cdots+1}{p \\text{ times}} \\equiv 0 \\pmod{p},\n$$\n\nwhich is a contradiction. So there exists $0 \\le j \\le p-1$ such that $a_j \\equiv 0 \\pmod{p}$, and therefore $b_p \\equiv 0 \\pmod{p}$.\n\nBy Newton's identities:\n\n$$\n\\begin{cases}\nS_1 + b_1 = 0 \\\\\nS_2 + b_1 S_1 + 2b_2 = 0 \\\\\nS_3 + b_1 S_2 + b_2 S_1 + 3b_3 = 0 \\\\\n\\vdots \\\\\nS_{p-1} + b_1 S_{p-2} + \\cdots + b_{p-2} S_1 + (p-1)b_{p-1} = 0\n\\end{cases}\n$$\n\nSince $S_1 \\equiv S_2 \\equiv \\cdots \\equiv S_{p-2} \\equiv 0 \\pmod{p}$, we get $b_1 \\equiv b_2 \\equiv \\cdots \\equiv b_{p-2} \\equiv 0 \\pmod{p}$, and from $S_{p-1} \\equiv -1 \\pmod{p}$,\n\n$$\n(p-1)b_{p-1} \\equiv 1 \\pmod{p} \\implies b_{p-1} \\equiv -1 \\pmod{p}.\n$$\n\nThus, in $\\mathbb{Z}_p[x]$, $g(x) = x^p - x = x(x-1)\\cdots(x-(p-1))$, and since $\\mathbb{Z}_p$ is a field, $\\mathbb{Z}_p[x]$ is a UFD. Therefore, $\\{a_0, a_1, \\dots, a_{p-1}\\}$ is a complete system of residues modulo $p$, as required. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18305,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players play a game on an $N \\times N$ board. The players alternately mark a cell in such a way that there is never a diagonal on the board containing two marked cells. For which $N > 0$ does the starting player have a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** For $N$ odd.\n\n**Solution.** Let the starting player be $A$ and the other player be $B$.\n\nIf $N$ is even, $B$ has a winning strategy by symmetry. Every time $A$ marks cell $(a, b)$, $B$ marks cell $(N-a+1, b)$. Since $N$ is even, $N-a+1 \\neq a$ for any $a$. Every cell in a diagonal through $(N-a+1, b)$ has the form $(N-a+1 \\pm l, b+l)$ for some $l \\in \\mathbb{Z}$, and if any such cell is marked, then so (by symmetry) is $(a \\pm l, b+l)$, which lies on a diagonal through $(a, b)$. Thus $(a, b)$ is a legal move if and only if $(N-a+1, b)$ is legal. Therefore, $B$ can respond to every move by $A$, and since the game is finite, $B$ wins.\n\nIf $N$ is odd, $A$ has a winning strategy by symmetry. $A$ starts by marking the center cell of the board, and after that, whenever $B$ plays at $(a, b)$, $A$ marks cell $(N+1-a, N+1-b)$. The two marks cannot lie on the same diagonal, since that diagonal would also contain the center cell. Since the board is symmetric around the center cell, the move $(a, b)$ is legal if and only if $(N+1-a, N+1-b)$ is legal. So $A$ can respond to every move by $B$, and since the game is finite, $A$ wins. $\\blacktriangleleft$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18306,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x, y, z, t \\in [0, \\infty)$ such that\n$$\nx + y + z \\le t, \\quad x^2 + y^2 + z^2 \\ge t, \\quad \\text{and} \\quad x^3 + y^3 + z^3 \\le t.\n$$",
    "options": [],
    "answer": "See solution",
    "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\n$$\nx(1-x)^2 + y(1-y)^2 + z(1-z)^2 \\le 0.\n$$\nSince $x, y, z \\in [0, \\infty)$, it follows $x, y, z \\in \\{0, 1\\}$.\n\n- If $x = y = z = 0$, then $t = 0$.\n- If exactly two of the numbers $x, y, z$ are zero, then $1 \\le t$ and $1 \\ge t$, therefore $t = 1$, $(x, y, z) \\in \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$.\n- If exactly one of the numbers $x, y, z$ is zero, then $2 \\le t$ and $2 \\ge t$, so $t = 2$, $(x, y, z) \\in \\{(1, 1, 0), (1, 0, 1), (0, 1, 1)\\}$.\n- If $x = y = z = 1$, then $t = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18307,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$. Prove that\n\n$$\n\\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\ge \\frac{(a-b)^2}{a^2+b^2+c^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from $\\frac{1}{2}(a-2b)^2 + \\frac{1}{2}(a-2c)^2 + (b-c)^2 \\ge 0$ that\n\n$$\n3(a^2 + b^2 + c^2) \\ge 2a^2 + 2ab + 2bc + 2ac = 2(a+b)(a+c),\n$$\n\nso we have $(a+b)(a+c) \\le \\frac{3}{2}(a^2+b^2+c^2)$. Similarly,\n\n$$\n(b + a)(b + c) \\le \\frac{3}{2}(a^2 + b^2 + c^2),\n$$\n\nand $(c + a)(c + b) \\le \\frac{3}{2}(a^2 + b^2 + c^2)$.\n\nHence,\n\n$$\n\\begin{aligned}\n& \\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\\\\n\\ge & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{a^2 + b^2 + c^2} \\\\\n\\ge & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + \\frac{1}{2}(b-c+c-a)^2}{a^2 + b^2 + c^2} \\\\\n= & \\frac{(a-b)^2}{a^2 + b^2 + c^2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18308,
    "subject": "Mathematics (Olympiad)",
    "question": "An inkjet printer is used to print on a paper strip with $1 \\times n$ grids. When the nozzle prints on the $i$th ($1 \\le i \\le n$) grid, it becomes black; in addition, each of the adjacent grids, the $(i-1)$th and the $(i+1)$th grids (if they exist), independently has a probability of $\\frac{1}{2}$ of becoming black. Let $T(n)$ be the expected number of prints needed to make all grids black, provided that the optimal strategy is adopted (make as few prints as possible). Find the formula for $T(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S \\subseteq \\{1, \\dots, n\\}$ represent the set of uncoloured grids: $i \\in S$ means the $i$th grid is not coloured yet; $i \\notin S$ means it is already black. Let $f(S)$ be the expected number of prints needed to make all grids in $S$ black. If $S \\subseteq S'$, then the optimal strategy for $S'$ can also be implemented on $S$, so $f(S) \\leq f(S')$. Let $g(S, i)$ be the expected number of prints after printing on $i$. We have\n\n$$\nf(S) = 1 + \\min_{1 \\leq i \\leq n} g(S, i)\n$$\n\nand\n\n$$\ng(S, i) = \\frac{1}{4}f(S - \\{i\\}) + \\frac{1}{4}f(S - \\{i, i-1\\}) + \\frac{1}{4}f(S - \\{i, i+1\\}) + \\frac{1}{4}f(S - \\{i, i-1, i+1\\}).\n$$\n\n**Lemma:** If the $i$th grid is already black, there exists an optimal strategy such that the nozzle never prints on it again.\n\n*Proof Sketch:* Printing on an already black grid is never better than printing on an adjacent uncoloured grid, as shown by comparing expected values.\n\nFor the original problem, let $T(k)$ be the expected number of prints to make $1 \\times k$ grids black. Define $T(-1) = T(0) = 0$, $T(1) = 1$, $T(2) = \\frac{3}{2}$. By the lemma, after printing on $i$, the problem reduces to the two sides. Thus,\n\n$$\nT(k) = 1 + \\min_{1 \\leq i \\leq k} \\frac{1}{2} (T(i-1) + T(i-2) + T(k-i-1) + T(k-i)).\n$$\n\nLet $S(k) = T(k)$ for $k = -1, 0, 1, 2$. For $k \\geq 3$,\n\n$$\nS(k) = 1 + \\frac{1}{2}(S(0) + S(1) + S(k-2) + S(k-3)) = \\frac{3}{2} + \\frac{1}{2}S(k-2) + \\frac{1}{2}S(k-3).\n$$\n\nIt can be checked that $T(3) = S(3) = 2$, $T(4) = S(4) = \\frac{11}{4}$, $T(5) = S(5) = \\frac{13}{4}$, $T(6) = S(6) = \\frac{31}{8}$.\n\nLet $h(k) = S(k) - \\frac{3}{5}k$. Then $h(k) = \\frac{1}{2}h(k-2) + \\frac{1}{2}h(k-3)$. Solving this recurrence gives\n\n$$\nS(k) = \\frac{3}{5}k + \\frac{7}{25} + b\\lambda^k + \\bar{b}\\bar{\\lambda}^k, \\quad (k \\geq 0)\n$$\n\nwhere $\\lambda = \\frac{-1+i}{2}$ is a root of $x^3 - \\frac{1}{2}x - \\frac{1}{2} = 0$, and $b = \\frac{-7+i}{50}$.\n\nThus, the formula for $T(n)$ is:\n\n$$\nT(n) = \\frac{3}{5}n + \\frac{7}{25} + b\\lambda^n + \\bar{b}\\bar{\\lambda}^n, \\quad (n \\geq 0)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18309,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers from 1 through 100 are written in some order on a circle. We call a pair of numbers on the circle *good* if the two numbers are not neighbors on the circle and if at least one of the two arcs they determine on the circle only contains numbers smaller than both of them. What may be the total number of good pairs on the circle?",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that the number of good pairs is always 97, irrespective of the order of the numbers on the circle.\n\nWe will perform a succession of transforms on the order of the numbers on the circle. We will prove that none of these transforms changes the total number of good pairs. First, we will successively swap 1 with one of its neighbors until 1 gets immediately after 100 in clockwise order. Swapping 1 with one of its neighbors, $n$, does not change the number of good pairs. Indeed, all good pairs that do not contain $n$ remain good. There are never good pairs that contain the number 1. Also, all the good pairs that did contain $n$ remain good pairs with the exception of the pair consisting of $n$ and the other neighbor of 1. This pair was a good one but will cease to be so after the swap. The swap only produces one good pair: $n$ together with its former neighbor (the one different from 1). In conclusion, such a swap does not change the total number of good pairs. After moving 1 immediately after 100 by such swaps, we will perform swaps that move 2 immediately after 1. The proof of the fact that such swaps do not change the total number of good pairs is similar to the above one. We continue in the same manner until we arrange the numbers in increasing order (clockwise) on the circle. All this is done without changing the number of good pairs. In this final configuration, it is easy to count the good pairs: the only good pairs are $\\{k, 100\\}$, where $k \\in \\{2, 3, \\dots, 98\\}$. In conclusion, there are 97 good pairs.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18310,
    "subject": "Mathematics (Olympiad)",
    "question": "Integers $a$, $b$, and $c$ satisfy the following equations:\n\n$$\n\\begin{aligned}\nab + c &= 100 \\\\\nbc + a &= 87 \\\\\nca + b &= 60\n\\end{aligned}\n$$\n\nWhat is $ab + bc + ca$?\n\n(A) 212 (B) 247 (C) 258 (D) 276 (E) 284",
    "options": [],
    "answer": "See solution",
    "solution": "Subtract the second equation from the first:\n\n$$\n\\begin{aligned}\n13 &= (ab + c) - (bc + a) \\\\\n   &= ab - bc - a + c \\\\\n   &= b(a - c) - (a - c) \\\\\n   &= (b - 1)(a - c)\n\\end{aligned}\n$$\n\nThus, $b - 1 = \\pm 1$ or $b - 1 = \\pm 13$.\n\n- If $b - 1 = -1$, then $b = 0$, implying $c = 100$, $a = 87$, and $ca = 60$, which is impossible.\n- If $b - 1 = 1$, then $b = 2$ and $a - c = 13$, implying $ca = 58 = 2 \\cdot 29$, which cannot be true if $a - c = 13$.\n- If $b - 1 = 13$, then $b = 14$ and $a - c = 1$, implying $ca = 46 = 2 \\cdot 23$, which cannot be true if $a - c = 1$.\n- If $b - 1 = -13$, then $b = -12$ and $a - c = -1$, implying $ca = 72$, which is satisfied when $a = -9$ and $c = -8$. In fact, $a = -9$, $b = -12$, and $c = -8$ satisfy all three equations.\n\nThe requested value is:\n\n$$\nab + bc + ca = (-9)(-12) + (-12)(-8) + (-8)(-9) = 108 + 96 + 72 = 276\n$$\n\n**Note:** There are also four noninteger solutions. When written in the form $(a, b, c)$, these solutions are approximately $(0.594, 0.869, 99.484)$, $(1.715, 57.455, 1.484)$, $(7.477, 12.525, 6.349)$, and $(86.214, 1.152, 0.683)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18311,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation\n\n$$\n20^m - 10m^2 + 1 = 19^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let the pair $(m, n)$ satisfy the equation. If $m = 0$, then $2 = 19^n$, which is impossible. So $m > 0$.\n\nTaking both sides modulo $10$ gives $1 \\equiv (-1)^n \\pmod{10}$, so $n$ must be even.\n\nTaking both sides modulo $20$ gives $-10m^2 + 1 \\equiv (-1)^n \\pmod{20}$, so $2 \\mid m^2$, hence $m$ is even too.\n\nLet $m = 2k$ and $n = 2l$ for a positive integer $k$ and a non-negative integer $l$. The equation becomes\n\n$$\n10m^2 - 1 = 20^{2k} - 19^{2l} = (20^k - 19^l)(20^k + 19^l).\n$$\n\nSince $20^k + 19^l > 0$ (as $m \\ge 1$), $20^k - 19^l \\ge 1$. Thus,\n\n$$\n10m^2 - 1 = (20^k - 19^l)(20^k + 19^l) \\ge 20^k + 19^l \\ge 20^k + 1.\n$$\n\nSo $20^k \\le 10(2k)^2 - 2 = 40k^2 - 2$. This fails for $k = 2$ and all $k > 2$. For $k = 1$, the inequality holds, and checking shows the only solution is $(m, n) = (2, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18312,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let the lines $CD$ and $BA$ meet at $E$. The line through $D$ which is tangent to the circle $ADE$ meets the line $CB$ at $F$. Prove that the triangle $CDF$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "By angles in a cyclic quadrilateral,\n\n$$\n\\angle DCF = 180^\\circ - \\angle DAB = \\angle EAD\n$$\n\nBy the alternate segment theorem,\n\n$$\n\\angle EAD = \\angle XDE = \\angle CDF\n$$\n\nso\n\n$$\n\\angle DCF = \\angle EAD = \\angle CDF\n$$\n\nso $\\triangle CDF$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18313,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $\\omega$, $\\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\\omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent externally to $\\omega$. Circle $\\Omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent internally to $\\omega$. Let $P_A$ and $Q_A$ denote the centers of $\\omega_A$ and $\\Omega_A$, respectively. Define points $P_B, Q_B, P_C, Q_C$ analogously. Prove that\n\n$$\n8P_AQ_A \\cdot P_BQ_B \\cdot P_CQ_C \\le R^3,\n$$\n\nwith equality if and only if triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the incircle touch the sides $AB$, $BC$, and $CA$ at $C_1$, $A_1$, and $B_1$, respectively. Set $AB = c$, $BC = a$, $CA = b$. By equal tangents, we may assume that $AB_1 = AC_1 = x$, $BC_1 = BA_1 = y$, and $CA_1 = CB_1 = z$. Then $a = y + z$, $b = z + x$, $c = x + y$. By the AM-GM inequality, we have $a \\ge 2\\sqrt{yz}$, $b \\ge 2\\sqrt{zx}$, and $c \\ge 2\\sqrt{xy}$. Multiplying the last three inequalities yields\n\n$$\nabc \\ge 8xyz, \\tag{\\dagger}\n$$\n\nwith equality if and only if $x = y = z$; that is, triangle $ABC$ is equilateral.\n\nLet $k$ denote the area of triangle $ABC$. By the Extended Law of Sines, $c = 2R \\sin \\angle C$. Hence\n\n$$\nk = \\frac{ab \\sin \\angle C}{2} = \\frac{abc}{4R} \\quad \\text{or} \\quad R = \\frac{abc}{4k}. \\tag{\\ddagger}\n$$\n\nWe are going to show that\n\n$$\nP_A Q_A = \\frac{xa^2}{4k}. \\tag{*}\n$$\n\nIn exactly the same way, we can also establish its cyclic analogous forms\n\n$$\nP_B Q_B = \\frac{yb^2}{4k} \\quad \\text{and} \\quad P_C Q_C = \\frac{zc^2}{4k}.\n$$\n\nMultiplying the last three equations together gives\n\n$$\nP_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{xyz a^2 b^2 c^2}{64k^3}.\n$$\n\nFurther considering $(\\dagger)$ and $(\\ddagger)$, we have\n\n$$\n8P_A Q_A \\cdot P_B Q_B \\cdot P_C Q_C = \\frac{8xyz a^2 b^2 c^2}{64k^3} \\le \\frac{a^3 b^3 c^3}{64k^3} = R^3,\n$$\n\nwith equality if and only if triangle $ABC$ is equilateral.\n\nHence it suffices to show $(*)$. Let $r$, $r_A$, $r'_A$ denote the radii of $\\omega$, $\\omega_A$, $\\Omega_A$, respectively. We consider the inversion $I$ with center $A$ and radius $x$. Clearly, $I(B_1) = B_1$, $I(C_1) = C_1$, and $I(\\omega) = \\omega$. Let ray $AO$ intersect $\\omega_A$ and $\\Omega_A$ at $S$ and $T$, respectively. It is not difficult to see that $AT > AS$, because $\\omega$ is tangent to $\\omega_A$ and $\\Omega_A$ externally and internally, respectively. Set $S_1 = I(S)$ and $T_1 = I(T)$. Let $\\ell$ denote the line tangent to $\\Omega$ at $A$. Then the image of $\\omega_A$ (under the inversion) is the line (denoted by $\\ell_1$) passing through $S_1$ and parallel to $\\ell$, and the image of $\\Omega_A$ is the line (denoted by $\\ell_2$) passing through $T_1$ and parallel to $\\ell$. Furthermore, since $\\omega$ is tangent to both $\\omega_A$ and $\\Omega_A$, $\\ell_1$ and $\\ell_2$ are also tangent to the image of $\\omega$, which is $\\omega$ itself. Thus the distance between these two lines is $2r$; that is, $S_1T_1 = 2r$. Hence we can consider the following configuration. (The darkened circle is $\\omega_A$, and its image is the darkened line $\\ell_1$.)\n\n![](images/USA_IMO_2006-2007_p29_data_20701c8bde.png)\n\nBy the definition of inversion, we have $AS_1 \\cdot AS = AT_1 \\cdot AT = x^2$. Note that $AS = 2r_A$, $AT = 2r'_A$, and $S_1T_1 = 2r$. We have\n\n$$\nr_A = \\frac{x^2}{2AS_1}. \\quad \\text{and} \\quad r'_A = \\frac{x^2}{2AT_1} = \\frac{x^2}{2(AS_1 - 2r)}.\n$$\n\nHence\n\n$$\nP_A Q_A = AQ_A - AP_A = r'_A - r_A = \\frac{x^2}{2} \\left( \\frac{1}{AS_1 - 2r} + \\frac{1}{AS_1} \\right).\n$$\n\nLet $H_A$ be the foot of the perpendicular from $A$ to side $BC$. It is well known that $\\angle BAS_1 = \\angle BAO = 90^\\circ - \\angle C = \\angle CAH_A$. Since ray $AI$ bisects $\\angle BAC$, it follows that rays $AS_1$ and $AH_A$ are symmetric with respect to ray $AI$. Further note that both line $l_1$ (passing through $S_1$) and line $BC$ (passing through $H_A$) are tangent to $\\omega$. We conclude that $AS_1 = AH_A$. In light of this observation and using ...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18314,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $x, y$ 為正整數，$x > y$ 且 $$(x-y)^{xy} = x^y \\cdot y^x$$，試求數對 $(x, y)$。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $x = dp,\\ y = dq$，其中 $d = \\gcd(x, y)$ 為 $x, y$ 的最大公因數，$p, q \\in \\mathbb{N}$，$(p, q) = 1,\\ p > q$。則\n\n$$(d(p-q))^{d^2pq} = (dp)^{dq}(dq)^{dp} \\Leftrightarrow (d(p-q))^{dpq} = (dp)^q (dq)^p \\Leftrightarrow d^{dpq}(p-q)^{dpq} = d^{p+q} p^q q^p.$$ \n\n欲證：$p+q < dpq$。\n\n假設 $p+q \\ge dpq$，則 $(p-q)^{dpq} = d^{p+q-dpq} p^q q^p \\Rightarrow p \\mid (p-q)^{dpq}$ 且 $q \\mid (p-q)^{dpq}$，但 $(p-q, p) = (p-q, q) = (p, q) = 1 \\Rightarrow p=1, q=1$（不合）。因此 $p+q < dpq$，所以 $d^{dpq-p-q}(p-q)^{dpq} = p^q q^p \\Rightarrow (p-q) \\mid p^q q^p$，又 $(p-q, p) = (p-q, q) = (p, q) = 1 \\Rightarrow p-q=1$，即 $p=q+1$，所以\n\n$$d^{dpq-p-q} = p^q q^{q+1}, \\qquad (1)$$\n\n所以 $p, q$ 為 $d$ 的因數。因為 $(p, q)=1$，所以 $d$ 可表為 $d=s \\times t$，其中 $t$ 的質因數只有 $p$，$s$ 的質因數只有 $q$，且 $(s, t)=1$。由 (1) 式得：\n\n$$t^{dpq-p-q} = p^q = (q+1)^q. \\qquad (2)$$\n\n因為 $dpq-p-q$ 與 $q$ 互質，所以由 (2) 知 $t$ 必為某自然數的 $p$ 次方，令 $t = t_1^q$。則\n\n$$t_1^{dpq-p-q} = q+1 \\Leftrightarrow t_1^{dq(q+1)-(2q+1)} = q+1$$\n\n因為 $q+1 > 1$，所以 $t_1 > 1$。若 $q \\ge 3$，則\n\n$$dq(q+1)-(2q+1) \\ge 3(q+1)-(2q+1) = q+2 \\Rightarrow t_1^{dpq-p-q} \\ge 2^{q+2} > q+1 \\text{（不合）}$$\n\n若 $q=2$，則 $t_1^{6d-5}=3 \\Rightarrow t_1=3$，$d=1$，$x=d(q+1)=3$，$y=dq=2$（不合）。所以 $q=1$，則 $t_1^{2d-3}=2 \\Rightarrow t_1=2$，$d=2 \\Rightarrow x=4, y=2$。\n\n**答案：** $(x, y) = (4, 2)$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18315,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrangle circumscribed about a circle such that\n\n$$\n\\angle ABD + \\angle ACB = \\angle ACD + \\angle ADB.\n$$\n\nProve that one diagonal of the quadrangle $ABCD$ bisects the other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B'$ and $D'$ be the images of $B$ and $D$, respectively, under the inversion of pole $A$ which fixes $C$. The angle relation now reads $\\angle AD'B' + \\angle AB'C = \\angle AD'C + \\angle AB'D'$, so $\\angle CB'D' = \\angle CD'B'$ and $CB' = CD'$. Since $B'C = AC \\cdot BC / AB$ and $D'C = AC \\cdot DC / AD$, we obtain $AB \\cdot CD = AD \\cdot BC$. Recall that the quadrangle $ABCD$ is circumscribed about a circle to write $AB + CD = AD + BC$. The last two relations imply that either $AB = AD$ and $BC = CD$ or $AB = BC$ and $CD = AD$; that is, the quadrangle $ABCD$ is either a kite or a lozenge. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18316,
    "subject": "Mathematics (Olympiad)",
    "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 $a_{n+1} \\le a_n - a_{mn}$ for all $n \\ge 1$.\n\na) Prove that the sequence $(b_n)_{n \\ge 1}$, where $b_n = \\sum_{k=1}^{n} a_k$, is bounded above.\n\nb) Prove that the sequence $(c_n)_{n \\ge 1}$, where $c_n = \\sum_{k=1}^{n} k^2 a_k$, is bounded above.",
    "options": [],
    "answer": "See 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$$\n\\sum_{k=1}^{n} a_{mk} \\le a_1 - a_{n+1} \\le a_1.\n$$\n\nUsing the monotonicity of $(a_n)$ and $(b_n)$, we have:\n\n$$\nb_n \\le b_{mn} = \\sum_{k=1}^{mn} a_k = \\sum_{i=1}^{m-1} a_i + \\sum_{k=1}^{n} a_{mk} + \\sum_{k=1}^{n-1} \\sum_{j=1}^{m-1} a_{mk+j}.\n$$\n\nBy monotonicity,\n\n$$\n\\sum_{k=1}^{n-1} \\sum_{j=1}^{m-1} a_{mk+j} \\le (m-1) \\sum_{k=1}^{n-1} a_{mk} \\le (m-1)a_1,\n$$\n\nhence,\n\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\nwhich shows that $(b_n)_{n \\ge 1}$ is bounded above.\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$$\n\\sum_{k=1}^{n} k a_{mk} \\le \\sum_{k=1}^{n} k(a_k - a_{k+1}) = a_1 + \\sum_{k=2}^{n} (k - (k-1)) a_k - n a_{n+1} \\le b_n,\n$$\n\nso the sequence $\\left( \\sum_{k=1}^{n} k a_{mk} \\right)_{n \\ge 1}$ is bounded above.\n\nOn the other hand,\n\n$$\n\\begin{aligned}\nd_n \\le d_{mn} &= m \\sum_{k=1}^{n} k a_{mk} + \\sum_{k=1}^{m-1} k a_k + \\sum_{j=1}^{m-1} \\sum_{k=1}^{n-1} (mk + j) a_{mk+j} \\\\\n&\\le m b_n + d_{m-1} + (m-1) \\sum_{k=1}^{n-1} m(k+1) a_{mk} \\\\\n&\\le m b_n + d_{m-1} + m(m-1)(b_{n-1} + a_1),\n\\end{aligned}\n$$\n\nhence $(d_n)_{n \\ge 1}$ is bounded above.\n\nAgain, $(c_n)_{n \\ge 1}$ is non-decreasing. Similarly,\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} k^2 a_{mk} &\\le \\sum_{k=1}^{n} k^2 (a_k - a_{k+1}) = a_1 + \\sum_{k=2}^{n} (k^2 - (k-1)^2) a_k - n^2 a_{n+1} \\\\\n&\\le a_1 + \\sum_{k=2}^{n} 2k a_k \\le 2 d_n,\n\\end{aligned}\n$$\n\nso the sequence $\\left( \\sum_{k=1}^{n} k^2 a_{mk} \\right)_{n \\ge 1}$ is bounded above.\n\nFinally,\n\n$$\n\\begin{aligned}\nc_n \\le c_{mn} &= m^2 \\sum_{k=1}^{n} k^2 a_{mk} + c_{m-1} + \\sum_{j=1}^{m-1} \\sum_{k=1}^{n-1} (mk + j)^2 a_{mk+j} \\\\\n&\\le 2 m^2 d_n + c_{m-1} + m^2 (m-1) \\left( \\sum_{k=1}^{n-1} k^2 a_{mk} + 2 \\sum_{k=1}^{n-1} k a_{mk} + \\sum_{k=1}^{n-1} a_{mk} \\right),\n\\end{aligned}\n$$\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": 18317,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $a$ for which $a^a$ is divisible by $20^{19}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $20^{19} \\mid a^a$, then $10 \\mid a$, so $a \\in \\{10, 20, 30, \\dots\\}$. Clearly, $a = 10$ is not possible since $10^{10}$ is smaller than $20^{19}$, while $a = 20$ is possible because $20^{20} = 20 \\cdot 20^{19}$. Also, all $a = 10k$ with $k \\ge 4$ are possible (for $q = 5^{10k-19} \\cdot 2^{10k-38} \\cdot k^{10k}$, we have $(10k)^{10k} = q \\cdot 20^{19}$). For $a = 30$, if $30^{30} = q \\cdot 20^{19}$ for some $q \\in \\mathbb{Z}$, then $3^{30} \\cdot 5^{11} = q \\cdot 2^8$, with the left-hand side odd and the right-hand side even, which is impossible. Thus, $a = 30$ is not possible. The solution is therefore all $a \\in \\{20\\} \\cup \\{10k : k \\ge 4\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18318,
    "subject": "Mathematics (Olympiad)",
    "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\n\n$$\nx_k = \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{a_k}.\n$$\n\nShow that\n$$\n\\frac{x_1 + x_2 + \\dots + x_n}{n} \\le \\frac{1}{1^2} + \\frac{1}{2^2} + \\dots + \\frac{1}{n^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The number $a_k$ of multiples of $k$ in the set $\\{1, 2, \\dots, n\\}$ is $\\left\\lfloor \\frac{n}{k} \\right\\rfloor$.\n\nLet 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 $x_k = \\frac{1}{1} + \\frac{1}{2} + \\dots + \\frac{1}{a_k}$ if and only if $i \\leq a_k = \\left\\lfloor \\frac{n}{k} \\right\\rfloor$.\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 $i \\leq \\left\\lfloor \\frac{n}{m} \\right\\rfloor \\leq \\frac{n}{m}$, i.e., $m \\leq \\frac{n}{i}$.\n\nRemember $\\frac{1}{i}$ belongs to the sums $x_1, x_2, \\dots, x_m$ and those only, therefore one has\n$$\nx_1 + x_2 + \\dots + x_n = \\sum_{i=1}^{n} \\frac{1}{i} m(i) \\leq \\sum_{i=1}^{n} \\frac{1}{i} \\cdot \\frac{n}{i} = n \\sum_{i=1}^{n} \\frac{1}{i^2}.\n$$\nThe conclusion immediately follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18319,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a sequence of integers by $a_0 = 1$, and $a_n = \\sum_{k=0}^{n-1} \\binom{n}{k} a_k$ for $n \\ge 1$.\n\nLet $m$ be a positive integer, $p$ a prime, and $q$ and $r$ 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": "See 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 $L X^n = a_n$ for $n = 0, 1, 2, \\dots$. Thus, if $f = \\sum_k \\alpha_k X^k$, then $L f = \\sum_k \\alpha_k a_k$.\n\nSince $(X+1)^n = \\sum_{k=0}^n \\binom{n}{k} X^k$ for $n \\ge 1$,\n\n$$\nL (X+1)^n = \\sum_{k=0}^{n} \\binom{n}{k} L X^k = \\sum_{k=0}^{n} \\binom{n}{k} a_k = \\sum_{k=0}^{n-1} \\binom{n}{k} a_k + a_n = 2 a_n = 2 L X^n,\n$$\n\nso $L f(X+1) = 2 L f(X) - f(0)$ for every polynomial $f$ in $\\mathbb{R}[X]$.\n\nIn particular, take $f = \\binom{X}{k}$ and use the relation $\\binom{X+1}{k} = \\binom{X}{k} + \\binom{X}{k-1}$ for $k \\ge 1$, to get $L \\binom{X}{k} = L \\binom{X}{k-1}$ for $k \\ge 1$, and deduce that $L \\binom{X}{k} = 1$ for $k = 0, 1, 2, \\dots$.\n\nFurther, if a polynomial $f$ in $\\mathbb{R}[X]$ is integral valued (i.e., $f(k)$ is integral for every integer $k$), then $f = \\sum_k \\alpha_k \\binom{X}{k}$ for some integers $\\alpha_k$, so $L f = \\sum_k \\alpha_k$ is an integer.\n\nFinally, since $a^{p^m} \\equiv a^{p^{m-1}} \\pmod{p^m}$ for all integers $a$,\n\n$$\nf = p^{-m} \\left( X^{p^m q + r} - X^{p^{m-1} q + r} \\right)\n$$\n\nis an integral valued polynomial in $\\mathbb{R}[X]$, so $L f = \\frac{a_{p^m q + r} - a_{p^{m-1} q + r}}{p^m}$ is an integer, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18320,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number such that for all $x$ with $3 < x < 5$, the following inequality holds:\n\n$$\n\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}.\n$$\n\nFind the value of $a$ for which $x = 4$ is the maximum point where equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "The condition is\n\n$$\n\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}\n$$\nfor $3 < x < 5$. Since $x-1 > 0$, $x-2 > 0$, and $x-6 < 0$ in this interval, we proceed:\n\n$$\n\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}\n$$\n\nThis is equivalent to:\n\n$$\n3a(x-2)(x-6) + 3(x-1)(x-6) + 3(x-1)(x-2) \\ge a(x-1)(x-2)(x-6)\n$$\n\nExpanding and simplifying:\n\n$$\nax^3 - (12a + 6)x^2 + (44a + 30)x - (48a + 24) \\le 0\n$$\n\nFactoring:\n\n$$\n(x-4)(ax^2 - (8a+6)x + (12a+6)) \\le 0.\n$$\n\nSince $x = 4$ is the maximum, the quadratic factor must also have $x-4$ as a root. By the factor theorem:\n\n$$\na(4)^2 - (8a+6)4 + (12a+6) = 0\n$$\n\nSimplifying:\n\n$$\n16a - 32a - 24 + 12a + 6 = 0 \\\\\n(-4a - 18) = 0 \\\\\na = -\\frac{9}{2}\n$$\n\nWhen $a = -\\frac{9}{2}$, the inequality becomes:\n\n$$\n(x-4)^2\\left(-\\frac{9}{2}x + 12\\right) \\le 0\n$$\n\nwhich holds for $3 < x < 5$. Thus, $a = -\\frac{9}{2}$ is the required value.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18321,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest integer greater than 2015 that is divisible by 2, 3, 4, 5, and 6?",
    "options": [],
    "answer": "See solution",
    "solution": "To be divisible by 2, 3, 4, 5, and 6, a number must be divisible by their least common multiple:\n\n$$\text{LCM}(2, 3, 4, 5, 6) = 2^2 \\times 3 \\times 5 = 60.$$ \n\nThe smallest multiple of 60 greater than 2015 is:\n\n$$\\left\\lceil \\frac{2015}{60} \\right\\rceil \\times 60 = 34 \\times 60 = 2040.$$ \n\nSo, the answer is $2040$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18322,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exist positive integers $a$ and $b$ such that $a$ does not divide $b^n - n$ for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** No, such positive integers $a$ and $b$ do not exist.\n\n**Lemma 1.** Given positive integers $a$ and $b$, for sufficiently large $n$ we have:\n\n$$\nb^{n+\\varphi(a)} \\equiv b^n \\pmod{a}.\n$$\n\nHere, $\\varphi$ is Euler's totient function: for any positive integer $m$, $\\varphi(m)$ is the number of positive integers less than $m$ that are relatively prime to $m$.\n\n*Proof of Lemma 1:* Let $a = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}$, where $p_1, \\dots, p_k$ are distinct primes. The function $\\varphi$ is multiplicative:\n\n$$\n\\varphi(a) = \\varphi(p_1^{\\alpha_1}) \\varphi(p_2^{\\alpha_2}) \\cdots \\varphi(p_k^{\\alpha_k}) = (p_1^{\\alpha_1} - p_1^{\\alpha_1-1}) \\cdots (p_k^{\\alpha_k} - p_k^{\\alpha_k-1}) = a \\left(1 - \\frac{1}{p_1}\\right) \\cdots \\left(1 - \\frac{1}{p_k}\\right).\n$$\n\nIn particular, $\\varphi(p_i^{\\alpha_i})$ divides $\\varphi(a)$ for each $1 \\le i \\le k$ and $\\varphi(a) < a$.\n\nFor each $p_i$:\n- If $p_i$ divides $b$, then $b^n \\equiv 0 \\pmod{p_i^{\\alpha_i}}$ for $n \\ge \\alpha_i + 1$, so $b^{n+\\varphi(a)} \\equiv b^n \\equiv 0 \\pmod{p_i^{\\alpha_i}}$ for $n \\ge \\alpha_i + 1$.\n- If $p_i$ does not divide $b$, then $\\gcd(p_i^{\\alpha_i}, b) = 1$. By Euler's theorem, $b^{\\varphi(p_i^{\\alpha_i})} \\equiv 1 \\pmod{p_i^{\\alpha_i}}$. Since $\\varphi(p_i^{\\alpha_i})$ divides $\\varphi(a)$, $b^{n+\\varphi(a)} \\equiv b^n \\pmod{p_i^{\\alpha_i}}$.\n\nTherefore, for each $p_i$, there is some $n_i$ such that for all $n > n_i$, $b^{n+\\varphi(a)} \\equiv b^n \\pmod{p_i^{\\alpha_i}}$. Let $N = \\max\\{n_i\\}$. For all $n > N$, $b^{n+\\varphi(a)} \\equiv b^n \\pmod{a}$, as desired. $\\blacksquare$\n\n**First Solution:**\nFor any positive integers $a$ and $b$, there exist infinitely many $n$ such that $a$ divides $b^n - n$.\n\nWe prove this by strong induction on $a$.\n- **Base case:** $a = 1$ is trivial.\n- **Inductive step:** Suppose the claim holds for all $a < a_0$. Since $\\varphi(a) < a$, by the induction hypothesis and Lemma 1, there are infinitely many $n$ such that\n  $$\n  \\varphi(a) \\mid (b^n - n) \\quad \\text{and} \\quad b^{n+\\varphi(a)} \\equiv b^n \\pmod{a}.\n  $$\n  For each such $n$, set\n  $$\n  t = \\frac{b^n - n}{\\varphi(a)}, \\quad n_1 = b^n = n + t\\varphi(a).\n  $$\n  Then $a$ divides $b^{n_1} - n_1$.\n\nThus, for any $a$ and $b$, there are infinitely many $n$ such that $a$ divides $b^n - n$, so the answer is no.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18323,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\text{a)} \\{x \\in \\mathbb{R} \\mid \\log_2[x] = [\\log_2 x]\\} = \\bigcup_{m \\in \\mathbb{N}} [2^m, 2^m + 1).\n$$\n$$\n\\text{b)} \\{x \\in \\mathbb{R} \\mid 2^{[x]} = [2^x]\\} = \\bigcup_{m \\in \\mathbb{N}} [m, \\log_2(2^m + 1)).\n$$\n(here, $[a]$ denotes the integer part (floor function) of the real number $a$).",
    "options": [],
    "answer": "See solution",
    "solution": "a) The existence of the logarithms requires $x > 1$.\n\nIf $[\\log_2 x] = m$, then $m \\in \\mathbb{N}$ and $2^m \\leq x < 2^{m+1}$. From $\\log_2[x] = m$, it follows $[x] = 2^m$, that is $2^m \\leq x < 2^m + 1$.\n\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 $\\log_2 t \\leq x < \\log_2(t+1)$. From $2^{[x]} = t$ it follows $[x] = \\log_2 t = m \\in \\mathbb{N}$, whence $m \\leq x < \\log_2(2^m + 1)$.\n\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": 18324,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $d, u, v, w$ are positive integers such that $u, v, w$ are distinct and\n\n$$\nd^3 - d(uv + vw + wu) - 2uvw = 0.\n$$\n\nProve that $d$ cannot be a prime. Also, find the least possible value of $d$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $u = 1$, $v = 2$, $w = 3$ gives $d^3 - 11d - 12 = 0$, which has no integer solutions. Thus $uvw > 8$ and $uv + vw + wu \\ge 3(uvw)^{2/3} > 12$. This shows that $d^3 > 12d + 16$ and we infer that $d \\ge 5$.\n\nIf $d$ is a prime, $d$ has to be an odd prime, say $d = p$. Then $p^3 = p(uv + vw + wu) + 2uvw$, so that $p$ divides $uvw$. Assume $p \\mid u$, and write $u = p u_1$. If $p$ divides $vw$, it must divide either $v$ or $w$. But then\n\n$$\np(uv + vw + wu) > p u (v + w) > p^2 (p + 1) > p^3.\n$$\n\nThus $p$ cannot divide $vw$. The relation now reduces to\n\n$$\np^2 = p u_1 (v + w) + vw (1 + 2u_1).\n$$\n\nIt follows that $p$ divides $1 + 2u_1$. Thus $1 + 2u_1 \\ge p$ or $u_1 \\ge (p-1)/2$. Since $v, w$ are distinct, we have $v + w \\ge 3$. Thus\n\n$$\np(uv + vw + wu) > p u (v + w) \\ge 3p^2 u_1 \\ge \\frac{3p^2(p-1)}{2} > p^3,\n$$\nif $p \\ge 3$. We conclude that $d$ cannot be an odd prime.\n\nSuppose $d = 6$. Substituting in the given relation, we see that $3 \\mid uvw$. Assume $3 \\mid u$, so that $u = 3u_1$. We obtain\n\n$$\n36 = 3u_1 (v + w) + vw (1 + u_1).\n$$\n\nThus $3 \\mid vw (1 + u_1)$. Observe that $3u_1 (v + w) < 36$ and $v + w \\ge 3$. Hence $u_1 < 4$. Thus either $3 \\mid vw$ or $u_1 = 2$. But $u_1 = 2$ shows that $12 = 2(v + w) + vw$ forcing $v = w = 2$. Since $v \\ne w$, the other possibility is $3 \\mid vw$. Taking $3 \\mid v$, we have $v = 3v_1$ and\n\n$$\n12 = 3u_1 v_1 + w(u_1 + v_1) + u_1 v_1 w.\n$$\n\nThus $u_1 v_1 (3 + w) < 12$. This forces $u_1 v_1 < 3$ giving $(u_1, v_1) = (1, 2)$ or $(2, 1)$. But then $u_1 + v_1 = 3$ and $u_1 v_1 = 2$, giving $12 = 6 + 5w$. This is impossible for an integer $w$.\n\nWe conclude that $d \\ge 8$. For $d = 8$, we may take $(u, v, w) = (2, 4, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18325,
    "subject": "Mathematics (Olympiad)",
    "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 $x \\in G$.\n\n(a) Prove that any group with the property (P) is commutative.\n\n(b) Prove that any finite group of odd order has the property (P).\n\n(c) Prove that no finite group of order $4n + 2$, $n \\in \\mathbb{N}$, has the property (P).",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Suppose $g, h \\in \\operatorname{Aut}(G)$ 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 $h(a)g(b) = g(b)h(a)$.\n\nSince $g$ and $h$ are onto, the claim follows.\n\n(b) 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\n$$\nf(x) = f(x^2 x^{-1}) = f(x^2) f(x^{-1}) = g(x) h(x),\n$$\n\nfor all $x \\in G$.\n\n(c) Suppose that $G$ has the property (P) and let $a \\in G$ be an element of order $2$.\n\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 $4 \\mid 4n+2$ – a contradiction.\n\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.\n\nNote: There exist groups of order $4n$ with the property (P) – e.g., Klein group – and there exist groups of order $4n$ without the property (P) – e.g., $(\\mathbb{Z}_4, +)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18326,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}$ denote the set of all real numbers. Find all pairs of functions $f: \\mathbb{R} \\to \\mathbb{R}$ and $h: \\mathbb{R}^2 \\to \\mathbb{R}$ such that\n$$\nf(x + y - z)^2 = f(xy) + h(x + y + z, xy + yz + zx)\n$$\nfor all $x, y, z \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us define two sets $D = \\{(x+y+z, xy+yz+zx) \\mid x, y, z \\in \\mathbb{R}\\}$ and $E = \\{(a, b) \\in \\mathbb{R}^2 \\mid a^2 \\ge 3b\\}$. It is obvious that $D \\subseteq E$. Now let us show that $E \\subseteq D$. Indeed, if $a^2 \\ge 3b$ there exists $c$ such that the polynomial $P(x) = x^3 - ax^2 + bx - c$ has three real roots. Thus, $D = E$.\n\nSince the function $h$ can take any value on $\\mathbb{R}^2 \\setminus D$, we only need to determine the values of $h$ on $D$.\n\n**Answer:** $f(x) = c$, $h(a, b) = c^2 - c$ and $f(x) = x/4$, $h(a, b) = (a^2 - 4b)/16$, $(a, b) \\in D$.\n\nThese functions obviously satisfy the equation, so let us show that there is no other solution.\n\n$(x, y, z) \\mapsto (x, z, y)$ implies that\n$$\nf(x - y + z)^2 - f(xz) = h(x + y + z, xy + yz + zx) = f(x + y - z)^2 - f(xy) \\quad (*)\n$$\n\nNow if $x = 1$, then $f(y) - f(z) = f(1-y+z)^2 - f(1+y-z)^2$ and $(y, z) \\mapsto (y-z, 0)$ implies that $f(y-z) - f(0) = f(1-y+z)^2 - f(1+y-z)^2 = f(y) - f(z)$. And if $(y, z) \\mapsto (x+y, x)$ then $f(x+y) = f(x) + f(y) - f(0)$.\n\nTherefore, the function $g(x) = f(x) - f(0)$ is a Cauchy's function. If $z = 0$ in $(*)$, then $f(x+y)^2 - f(x-y)^2 = f(xy) - f(0)$ and $g(xy) = (g(x+y) - g(x-y))(g(x+y) + g(x-y) + 2f(0)) = 4g(y)(g(x) + f(0))$. Hence, $f(0)(g(x) - g(y)) = 0$.\n\nIf the function $g(x)$ is not constant, $f(0) = 0$ and $g(xy) = 4g(x)g(y)$. Thus, $g(x)$ is an increasing function, so $g(x) = ax$ and $g(1) = 4g(1)^2$, $a = 1/4$ and $f(x) = x/4$. It is not difficult to see that $h(a, b) = (a^2 - 4b)/16$ on $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18327,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the cube $ABCDEFGH$ with side $a$ cm, $a > 0$. Points $K$ and $L$ are on segments $AC$ and $EG$, respectively, such that $CK = EL = \\frac{AC}{4}$. Let $M$ be the midpoint of $AE$, and let points $N$ and $P$ be on segment $HF$ such that\n\n$$\nHN = FP = \\frac{(\\sqrt{2} - 1)a}{2} \\text{ cm}.\n$$\n\nGiven that lines $MC$ and $LK$ intersect at $Q$, prove that $MNPQ$ is a regular tetrahedron.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $\\Delta MEP$ and $\\Delta MEN$ are congruent, so $MN = MP$.\n\nLet $O_1$ be the midpoint of $HF$. Then $MO_1 = \\frac{AC}{2} = \\frac{a\\sqrt{3}}{2}$, and since $NP = a$, triangle $MNP$ is equilateral.\n\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 triangle $MNP$.\n\n![](images/ROU_ABooklet_2021_p17_data_bc15018b6a.png)\n\nNotice that $AL = AG = \\frac{3\\sqrt{2a}}{4}$, so $ALGK$ is a rhombus, which implies $AG \\perp KL$. Also, $CO_1LK$ is a parallelogram, so $LK \\parallel CO_1$ and $KL \\perp MO_1$. Since $HF \\perp (ACGE)$, $KL \\perp HF$, which implies $KL \\perp (MNP)$. Thus, the pyramid $MNPQ$ is regular.\n\nFrom the similarity of 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": 18328,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\triangle$ be the triangle under consideration. Since $\\triangle$ is convex, the tiling triangles fall into two classes: those having all edges inside $\\triangle$, and those having at least one edge on the boundary of $\\triangle$.\n\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 unique 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.\n\nClearly, every outer edge (lying on the boundary of $\\triangle$) belongs to a single triangle, and the total length of all outer edges is the perimeter of $\\triangle$.\n\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 $\\triangle$ is $S - 2s$, and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18329,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the value of $101 \\cdot 9,901 - 99 \\cdot 10,101$?\n\n(A) 2  (B) 20  (C) 21  (D) 200  (E) 2020",
    "options": [],
    "answer": "See solution",
    "solution": "Write the difference as\n\n$$\n(100 + 1) \\cdot (9900 + 1) - 99 \\cdot (10,000 + 100 + 1).\n$$\n\nApplying the distributive property gives\n\n$$\n(990,000 + 9,900 + 100 + 1) - (990,000 + 9,900 + 99) = 100 + 1 - 99 = 2.\n$$\n\nOr, let $x = 100$. Then the minuend (the first quantity in the subtraction operation) is\n\n$$\n(x + 1)(x^2 - x + 1) = x^3 + 1,\n$$\n\nand the subtrahend (the quantity being subtracted from the minuend) is\n\n$$\n(x - 1)(x^2 + x + 1) = x^3 - 1.\n$$\n\nThe difference is\n\n$$\n(x^3 + 1) - (x^3 - 1) = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18330,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be an isosceles trapezium with bases $\\overline{AB}$ and $\\overline{CD}$, and let its diagonals meet at the point $S$.\n\nAdditionally, let $K$ be a point on the line $AD$ such that $SK \\parallel AB$, and let $M$ be the other intersection of $AD$ with the circle circumscribed to the triangle $BCK$. We need to prove that $M$ is the midpoint of the leg $\\overline{AD}$.\n\n![](images/Croatia_2018_p20_data_c9efd1dd9d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M'$ be a point on the line $BC$ such that $MM' \\parallel AB$, and let $T$ be the intersection of the lines $CK$ and $AB$. Since the trapezium $ABCD$ is isosceles and the quadrilateral $BCKM$ is cyclic, we have $\\angle DAM' = \\angle CBM = \\angle MKT$, and therefore $AM' \\parallel TC$. Using similarity of triangles, we obtain $|AT| : |CD| = |KA| : |KD| = d(S, AB) : d(S, CD) = |AB| : |CD|$. Hence $|AT| = |AB|$ and $|M'B| = |M'C|$. Thus the point $M'$ is the midpoint of the leg $\\overline{BC}$ and, of course, $M$ is the midpoint of the leg $\\overline{AD}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18331,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and $q$ such that $2^2 + p^2 + q^2$ is also prime.",
    "options": [],
    "answer": "See solution",
    "solution": "If the pair $(p, q)$ satisfies the conditions of the problem, then so does the pair $(q, p)$. It is therefore sufficient to only consider the case where $p \\leq q$.\n\nObviously, $p = q = 2$ is not a solution. If $p$ and $q$ are both odd primes, then $2^2 + p^2 + q^2$ is an even integer greater than $2$, so it is not prime. Hence, $p = 2$.\n\nLet us figure out when the number $8 + q^2$ is prime. If $q = 3$, then $8 + q^2 = 17$ is prime. Else, $3$ divides $q - 1$ or $q + 1$, so $3$ divides $9 + (q - 1)(q + 1) = 8 + q^2$, which is then not prime.\n\nWe conclude that $2^2 + p^2 + q^2$ is prime only if $p = 2$ and $q = 3$ or $p = 3$ and $q = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18332,
    "subject": "Mathematics (Olympiad)",
    "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 $P$.\n\nShow that $BT = 2PT$.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangle $ABC$ is isosceles, so $\\angle ABC = \\angle ACB = 70^\\circ$. Notice 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": 18333,
    "subject": "Mathematics (Olympiad)",
    "question": "Regular polygons with 5, 6, 7, and 8 sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?\n\n(A) 52 (B) 56 (C) 60 (D) 64 (E) 68",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (E):**\n\nConsider a regular $m$-gon and a regular $n$-gon, with $m \\leq n$, inscribed in the same circle with no shared vertices. If $A$ and $B$ are adjacent vertices of the $m$-gon, then the minor arc $AB$ contains at least one vertex of the $n$-gon. Thus, side $AB$ intersects exactly two sides of the $n$-gon inside the circle, and the two polygons intersect in exactly $2m$ points.\n\nIt follows that the pentagon intersects the hexagon, the heptagon, and the octagon in $10$ points each; the hexagon intersects the heptagon and the octagon in $12$ points each; and the heptagon intersects the octagon in $14$ points. The total number of points of intersection is:\n\n$$\n3 \\times 10 + 2 \\times 12 + 14 = 68\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18334,
    "subject": "Mathematics (Olympiad)",
    "question": "A $4 \\times 4$ table has positive integers in its cells so that the sum of any two cells that share a side is a factorial of some positive integer. Show that there are at least 4 equal numbers in this table.",
    "options": [],
    "answer": "See solution",
    "solution": "We will start with the following lemma.\n\n**Lemma 1.** There exists a diagonal that has two equal numbers for any $2 \\times 2$ square.\n\n**Proof.** Let $X$ be the greatest number in a $2 \\times 2$ square. Let its neighbors in this $2 \\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A = B$, the lemma is proven. Suppose $A > B$. Then $A + X = k! > B + X = m! > 1$, thus $k > m$. It is also clear that $m > 1, k > 2$. But then\n\n$$\n2(B + X) = 2m! > 2X \\geq A + X = k! \\geq k \\cdot m! > 2m!,\n$$\n\nwhich leads to a contradiction. This finishes the proof of Lemma 1.\n\nBy contradiction, let the table consist of the numbers $a_1, a_2, \\ldots, a_{16}$, that satisfy the conditions.\n\nUse Lemma 1 for a square with $a_1, a_2, a_5, a_6$. Without loss of generality, let $a_2 = a_5$. Use Lemma 1 for squares with $a_2, a_3, a_6, a_7$ and $a_5, a_6, a_9, a_{10}$.\n\n**Case I.** $a_2 = a_7$ or $a_5 = a_{10}$. Since these cases are similar, let $a_2 = a_5 = a_{10} = x$. From Lemma 1 for the square $a_9, a_{10} = x, a_{13}, a_{14}$, if $a_{13} = x$, then we have four equal numbers. Then $a_9 = a_{14}$. Similarly, from Lemma 1 for a square with $a_{10} = x, a_{11}, a_{14}, a_{15}$ we have that $a_9 = a_{14} = a_{11} = y$.\n\nIt suffices to use Lemma 1 for a square with $a_6, a_7, a_{10} = x, a_{11} = y$; thus, the table has either four numbers $x$ or four numbers $y$. This completes the proof.\n\n**Case II.** $a_3 = a_6 = a_9 = t$. By assumption, there is no other number $t$ among the rest of the values, so by Lemma 1 the following holds: $a_4 = a_7 = a_{10} = a_{13}$, which leads to a contradiction and completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18335,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + y) + xy = f(x)f(y),\n$$\n\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Taking $x = y = 0$, we get $f(0) = f(0)^2$. Thus $f(0) = 0$ or $1$.\n\nIf $f(0) = 0$, then setting $y = 0$ gives $f(x) = 0$ for all $x$. But $f(x) \\equiv 0$ is not a solution, as can be verified. So $f(0) = 1$.\n\nTaking $x = 1$, $y = -1$, we have $f(1)f(-1) = 0$. Thus $f(1) = 0$ or $f(-1) = 0$.\n\n**Case 1:** $f(1) = 0$\n\nSetting $y = 1$ gives\n$$\nf(x+1) + x = f(x)f(1) = 0,\n$$\nso $f(x+1) = -x$, or $f(x) = 1 - x$ for all $x$.\n\n**Case 2:** $f(-1) = 0$\n\nSetting $y = -1$ gives\n$$\nf(x-1) - x = 0,\n$$\nso $f(x) = 1 + x$ for all $x$.\n\nIt is easy to verify that $f(x) = 1 - x$ and $f(x) = 1 + x$ are both solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18336,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T = \\{1, 2, 3, 4, 5, 6, 7, 8\\}$. Find the number of all nonempty subsets $A$ of $T$ such that $3 \\mid S(A)$ and $5 \\nmid S(A)$, where $S(A)$ is the sum of all elements of $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $S(\\emptyset) = 0$. Let $T_0 = \\{3, 6\\}$, $T_1 = \\{1, 4, 7\\}$, $T_2 = \\{2, 5, 8\\}$. For $A \\subseteq T$, let $A_0 = A \\cap T_0$, $A_1 = A \\cap T_1$, $A_2 = A \\cap T_2$. Then\n\n$$\nS(A) = S(A_0) + S(A_1) + S(A_2) \\\\\n\\equiv |A_1| - |A_2| \\pmod{3}\n$$\n\nSo $3 \\mid S(A)$ if and only if $|A_1| \\equiv |A_2| \\pmod{3}$. It follows that\n\n$$\n(|A_1|, |A_2|) = (0, 0), (0, 3), (3, 0), (3, 3), (1, 1), (2, 2)\n$$\n\nThe number of nonempty subsets $A$ so that $3 \\mid S(A)$ is\n\n$$\n2^2 \\binom{3}{0} \\binom{3}{0} + \\binom{3}{0} \\binom{3}{3} + \\binom{3}{3} \\binom{3}{0} + \\binom{3}{3} \\binom{3}{3} \\\\\n+ \\binom{3}{1} \\binom{3}{1} + \\binom{3}{2} \\binom{3}{2} - 1 = 87\n$$\n\nIf $3 \\mid S(A)$ and $5 \\mid S(A)$, then $15 \\mid S(A)$. Since $S(T) = 36$, the possible values for $S(A)$ are 15 or 30 (if $3 \\mid S(A)$ and $5 \\mid S(A)$).\n\nFurthermore,\n\n$$\n\\begin{aligned}\n15 &= 8+7 = 8+6+1 = 8+5+2 = 8+4+3 \\\\\n&= 8+4+2+1 = 7+6+2 = 7+5+3 \\\\\n&= 7+5+2+1 = 7+4+3+1 = 6+5+4\n\\end{aligned}\n$$\n$$\n= 6 + 5 + 3 + 1 = 6 + 4 + 3 + 2\n$$\n$$\n= 5 + 4 + 3 + 2 + 1\n$$\n$$\n36 - 30 = 6 = 5 + 1 = 4 + 2 = 3 + 2 + 1\n$$\nSo the number of $A$ such that $3 \\mid S(A)$, $5 \\mid S(A)$, and $A \\neq \\emptyset$ is 17.\n\nThe answer is $87 - 17 = 70$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18337,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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, $f(t) = k$ for $t \\in (0, \\infty)$.\n\nUsing the same argument on $(-\\infty, 0)$, we deduce that there are constants $k_1, k_2, k$ such that\n$$\nf(t) = \\begin{cases} k_1, & t < 0 \\\\ k_2, & t = 0 \\\\ k, & t > 0 \\end{cases}\n$$\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 satisfy the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18338,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an acute triangle $ABC$ with $|AB| > |CA| > |BC|$. The vertices $D$, $E$, and $F$ are the base points of the altitudes from $A$, $B$, and $C$, respectively. The line through $F$ parallel to $DE$ intersects $BC$ at $M$. The angle bisector of $\\angle MFE$ intersects $DE$ at $N$. Prove that $F$ is the circumcentre of $\\triangle DMN$ if and only if $B$ is the circumcentre of $\\triangle FMN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because of the requirement on the lengths, the configuration is fixed: $M$ lies on the ray $CB$ past $B$, and $N$ lies on the ray $ED$ past $D$. Let $\\alpha = \\angle BAC$ and $\\beta = \\angle ABC$. Let $H$ be the orthocentre of the triangle (i.e., the intersection of $AD$, $BE$, and $CF$).\n\nBy Thales's theorem, $AFHE$, $BDHF$, $CEHD$, $ABDE$, $BCEF$, and $CAFD$ are cyclic. Because $ABDE$ is cyclic, $\\angle CED = 180^\\circ - \\angle AED = \\angle ABD = \\beta$, and because $BCEF$ is cyclic, $\\angle AEF = 180^\\circ - \\angle CEF = \\angle CBF = \\beta$. Analogously, $\\angle CDE$ and $\\angle BDF$ equal $\\alpha$.\n\nFrom $\\angle CED = \\beta = \\angle AEF$, it follows that $\\angle DEH = 90^\\circ - \\beta = \\angle FEH$. Hence, $EH$ is the angle bisector of $\\angle DEF$. Because $DE \\parallel FM$, $\\angle MFE = 180^\\circ - \\angle FED = 180^\\circ - 2(90^\\circ - \\beta) = 2\\beta$. As $FN$ is the angle bisector of $\\angle MFE$, $\\angle EFN = \\frac{1}{2} \\cdot 2\\beta = \\beta$. Because $\\angle FEH = 90^\\circ - \\beta$, $FN$ and $EH$ are perpendicular, so $EH$ is not only the angle bisector in $\\triangle FEN$, but also an altitude. Therefore, this line is also the perpendicular bisector of $FN$. As $B$ lies on this line, $|BF| = |BN|$.\n\nWe already saw that $\\angle CDE = \\alpha = \\angle BDF$. Because $DE \\parallel FM$, $\\angle BMF = \\angle CDE = \\alpha$, so $\\angle DMF = \\angle BMF = \\angle BDF = \\angle MDF$. Thus, $|FM| = |FD|$.\n\nLet $S$ be the intersection of $AC$ with $MF$. Then $\\angle BFM = \\angle AFS$, and because $DE \\parallel FM$, $\\angle CED = \\angle CSF$. The exterior angle theorem in $\\triangle AFS$ yields $\\angle CSF = \\angle SAF + \\angle AFS = \\alpha + \\angle AFS$.\n\nCombining everything, $\\angle CED = \\alpha + \\angle BFM$. On the other hand, $\\angle CED = \\beta$, so $\\angle BFM = \\beta - \\alpha$. Moreover, $\\angle BMF = \\alpha$. We conclude that $|BF| = |BM|$ if and only if $\\beta - \\alpha = \\alpha$, or $\\beta = 2\\alpha$. Because $|BF| = |BN|$, $B$ is the circumcentre of $\\triangle FMN$ if and only if $\\beta = 2\\alpha$.\n\nEarlier, we saw that $EH$ is the perpendicular bisector and altitude in $\\triangle EFN$, so this triangle is isosceles with top angle $E$, which yields $\\angle DNF = \\angle ENF = \\angle EFN = \\beta$. Moreover, $\\angle CDE = \\alpha = \\angle BDF$, so $\\angle NDF = \\angle NDB + \\angle BDF = \\angle CDE + \\angle BDF = 2\\alpha$. Hence, $|FD| = |FN|$ if and only if $\\beta = 2\\alpha$. Because $|FM| = |FD|$, $F$ is the circumcentre of $\\triangle DMN$ if and only if $\\beta = 2\\alpha$.\n\nWe conclude that $F$ is the circumcentre of $\\triangle DMN$ if and only if $B$ is the circumcentre of $\\triangle FMN$, as both properties are equivalent to $\\beta = 2\\alpha$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18339,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$, and let $M$ and $N$ be points on the sides $BC$ and $CA$, respectively, such that the angles $\\angle BAM$ and $\\angle CNM$ are equal. The lines $AB$ and $MN$ meet at $P$. Show that the internal angle bisectors of the angles $\\angle BAM$ and $\\angle BPM$ meet at a point on the line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the intersection of the bisector of $\\angle BAM$ with $BC$, and let $D$ be the reflection of $A$ about $BC$. Then $\\angle BMD = \\angle BMA = \\angle CMN$, so $P$, $M$, and $D$ 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 $PBD$, which proves the statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18340,
    "subject": "Mathematics (Olympiad)",
    "question": "Non-equilateral triangle $ABC$ has a $60^\\circ$ angle at vertex $A$. Let the angle bisector drawn from vertex $A$ intersect the opposite side at point $D$, and let $Q$ and $R$ be the feet of the altitudes drawn from vertices $B$ and $C$, respectively. Prove that lines $AD$, $BQ$, and $CR$ intersect in three distinct points that are vertices of an equilateral triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "If line $AD$ passed through the point of intersection of lines $BQ$ and $CR$, the line segment $AD$ would be an altitude of triangle $ABC$. As $AD$ is also the angle bisector, triangle $ABC$ would be isosceles with $AB = AC$. As $\\angle BAC = 60^\\circ$, triangle $ABC$ would be equilateral, contradicting the assumption. Hence the lines $AD$, $BQ$, and $CR$ meet in three distinct points.\n\n![](images/prob1516_p14_data_9246a50904.png)\n\nBy assumptions, $\\angle QAD = \\angle RAD = 30^\\circ$ and $\\angle AQB = \\angle ARC = 90^\\circ$ (see figure). Hence $AD$ and $BQ$ intersect at angle $60^\\circ$, as well as $AD$ and $CR$. Thus, two angles of the triangle whose vertices are the three intersection points of lines $AD$, $BQ$, and $CR$ have size $60^\\circ$. Such a triangle is equilateral.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18341,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be an integer greater than $1$, and let $n$ be an odd number with $3 \\leq n < 2m$. Numbers $a_{i,j}$ ($i, j \\in \\mathbb{N}$, $1 \\leq i \\leq m$, $1 \\leq j \\leq n$) satisfy:\n\n1. For every $1 \\leq j \\leq n$, the sequence $a_{1,j}, a_{2,j}, \\ldots, a_{m,j}$ is a permutation of $1, 2, \\ldots, m$.\n2. $|a_{i,j} - a_{i,j+1}| \\leq 1$ for every $1 \\leq i \\leq m$, $1 \\leq j \\leq n-1$.\n\nFind the minimal possible value of $M = \\max_{1 \\leq i \\leq m} \\sum_{j=1}^{n} a_{i,j}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2l + 1$. Since $3 \\leq n < 2m$, we have $1 \\leq l \\leq m - 1$. We first estimate the lower bound of $M$.\n\nBy condition (1), there exists a unique $1 \\leq i_0 \\leq m$ such that $a_{i_0, l+1} = m$. Consider $a_{i_0, l}$ and $a_{i_0, l+2}$.\n\n**Case 1:** At least one of $a_{i_0, l}$ and $a_{i_0, l+2}$ is $m$; assume $a_{i_0, l} = m$. By condition (2):\n\n$$\n\\begin{aligned}\na_{i_0, l-1} &\\geq m-1,\\quad a_{i_0, l-2} \\geq m-2,\\ \\ldots,\\ a_{i_0, 1} \\geq m-l+1, \\\\\na_{i_0, l+2} &\\geq m-1,\\quad a_{i_0, l+3} \\geq m-2,\\ \\ldots,\\ a_{i_0, 2l+1} \\geq m-l.\n\\end{aligned}\n$$\n\nThus,\n$$\n\\begin{aligned}\nM &\\geq \\sum_{j=1}^{n} a_{i_0, j} \\\\\n  &\\geq (m-l) + 2\\big((m-l+1) + (m-l+2) + \\ldots + m\\big) \\\\\n  &= (2l+1)m - l^2.\n\\end{aligned}\n$$\n\n**Case 2:** Neither $a_{i_0, l}$ nor $a_{i_0, l+2}$ is $m$. By condition (1), there exists $1 \\leq i_1 \\leq m$, $i_1 \\neq i_0$, such that $a_{i_1, 1} = m$. By conditions (1) and (2), $a_{i_1, 2} = m-1$, $a_{i_1, 3} = m$. Then:\n\n$$\n\\begin{aligned}\na_{i_1, 1} &\\geq m-1,\\quad a_{i_1, 2} \\geq m-2,\\ \\ldots,\\ a_{i_1, l} \\geq m-l+1, \\\\\na_{i_1, 3} &\\geq m-1,\\quad a_{i_1, 4} \\geq m-2,\\ \\ldots,\\ a_{i_1, l+1} \\geq m-l+1.\n\\end{aligned}\n$$\n\nThus,\n$$\n\\begin{aligned}\nM &\\geq \\sum_{j=1}^{n} a_{i_1, j} \\\\\n  &\\geq 2\\big((m-l+1) + (m-l+2) + \\ldots + (m-1)\\big) \\\\\n  &= (2l+1)m - (l^2 - l + 1).\n\\end{aligned}\n$$\n\nCombining both cases, $M \\geq (2l+1)m - l^2$.\n\nNow, consider the construction:\n\n$$\na_{i,j} = f(2i+j) = \\begin{cases}\n2i+j, & 2i+j \\leq m, \\\\\n(2m+1)-(2i+j), & m+1 \\leq 2i+j \\leq 2m, \\\\\n(2i+j)-2m, & 2m+1 \\leq 2i+j \\leq 3m, \\\\\n(4m+1)-(2i+j), & 3m+1 \\leq 2i+j \\leq 4m.\n\\end{cases}\n$$\n\nThis table satisfies the required conditions:\n- For any $1 \\leq i \\leq m$, $1 \\leq j \\leq n-1$, $|a_{i,j} - a_{i,j+1}| \\leq 1$.\n- For any $1 \\leq j \\leq n$, $a_{1,j}, \\ldots, a_{m,j}$ is a permutation of $1, \\ldots, m$.\n\nNow, estimate $M$ for this construction:\n\n$$\n\\sum_{j=1}^{n} a_{i,j} \\leq (m-l) + 2\\big((m-l+1) + (m-l+2) + \\ldots + m\\big) = (2l+1)m - l^2.\n$$\n\n**Thus, the minimal possible value of $M$ is $\\boxed{(2l+1)m - l^2}$, where $n = 2l+1$.**",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18342,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $u, v$, the sequence $\\{a_n\\}$ is defined as follows: $a_1 = u + v$, and for $m \\geq 1$,\n\n$$\n\\begin{cases}\na_{2m} = a_m + u, \\\\\na_{2m+1} = a_m + v.\n\\end{cases}\n$$\n\nDenote $S_m = a_1 + a_2 + \\dots + a_m$ for $m = 1, 2, \\dots$. Prove that there are infinitely many terms in the sequence $\\{S_n\\}$ that are square numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "For a positive integer $n$, we have\n\n$$\n\\begin{align*}\nS_{2^{n+1}-1} &= a_1 + (a_2 + a_3) + (a_4 + a_5) + \\dots + (a_{2^{n+1}-2} + a_{2^{n+1}-1}) \\\\\n&= u + v + (a_1 + u + a_1 + v) + (a_2 + u + a_2 + v) + \\dots + (a_{2^{n-1}} + u + a_{2^{n-1}} + v) \\\\\n&= 2^n (u + v) + 2S_{2^{n-1}}.\n\\end{align*}\n$$\n\nThen,\n\n$$\n\\begin{align*}\nS_{2^{n}-1} &= 2^{n-1}(u+v) + 2S_{2^{n-1}-1} \\\\\n&= 2^{n-1}(u+v) + 2(2^{n-2}(u+v) + 2S_{2^{n-2}-1}) \\\\\n&= 2 \\cdot 2^{n-1}(u+v) + 2^2 S_{2^{n-2}-1} \\\\\n&= \\dots = (n-1) \\cdot 2^{n-1}(u+v) + 2^{n-1}(u+v) \\\\\n&= (u+v) \\cdot n \\cdot 2^{n-1}.\n\\end{align*}\n$$\n\nSuppose $u + v = 2^k \\cdot q$, where $k$ is a non-negative integer and $q$ is an odd number. Take $n = q \\cdot l^2$, where $l$ is any positive integer satisfying $l \\equiv k - 1 \\pmod{2}$. Then $S_{2^{n-1}} = q^2 l^2 \\cdot 2^{k-1+q \\cdot l^2}$, and\n\n$$\n\\begin{align*}\nk - 1 + q \\cdot l^2 &\\equiv k - 1 + l^2 \\equiv k - 1 + (k - 1)^2 \\\\\n&= k(k - 1) \\equiv 0 \\pmod{2}.\n\\end{align*}\n$$\n\nTherefore, $S_{2^{n-1}}$ is a square number. Since there are infinitely many choices for $l$, there are infinitely many terms in $\\{S_n\\}$ that are square numbers. The proof is complete. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18343,
    "subject": "Mathematics (Olympiad)",
    "question": "In the Cartesian plane, 51 points with integer coordinates are given, such that all pairwise distances between them are integers. Prove that more than 49\\% of the pairwise distances are even integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Among the 51 given points, there cannot simultaneously exist a point with both coordinates even and a point with both coordinates odd. Suppose, for contradiction, that $(2a, 2b)$ and $(2c+1, 2d+1)$ are such points. The square of the distance between them is:\n\n$$\n(2a - 2c - 1)^2 + (2b - 2d - 1)^2 = 4(a - c - 0.5)^2 + 4(b - d - 0.5)^2 = 4M + 2,\n$$\n\nwhich is not a perfect square, so the distance cannot be an integer.\n\nSimilarly, if a point with coordinates (even, odd) exists, there cannot be a point with coordinates (odd, even).\n\nLet $k$ be the number of points with coordinates of the same parity (both even or both odd). All distances among these $k$ points are even integers. The remaining $51 - k$ points have coordinates of different parity, and all distances among them are also even integers.\n\nTherefore, at least $\\binom{k}{2} + \\binom{51-k}{2}$ distances are even integers. Since there are $\\binom{51}{2}$ total distances, we have:\n\n$$\n\\frac{\\binom{k}{2} + \\binom{51-k}{2}}{\\binom{51}{2}} = \\frac{k(k-1) + (51-k)(50-k)}{51 \\cdot 50} = 1 + \\frac{k^2 - 51k}{51 \\cdot 25} \\geq 1 - \\frac{51^2}{51 \\cdot 25} = 1 - \\frac{51}{100} = 49\\%.\n$$\n\nThus, more than 49\\% of the distances are even integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18344,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that there exist irrational numbers $a$, $b$, and $c$ such that the numbers $a + bc$, $b + ac$, and $c + ab$ are rational.\n\nb) Prove that if real numbers $a$, $b$, and $c$ satisfy $a + b + c = 1$ and the numbers $a + bc$, $b + ac$, and $c + ab$ are nonzero rationals, then $a$, $b$, and $c$ are rational.",
    "options": [],
    "answer": "See solution",
    "solution": "a) An example can be obtained by taking $a = b = c$, with irrational $a$ and rational $a^2 + a$. This occurs, for instance, if $a = \\sqrt{2} - \\frac{1}{2}$.\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.\n\nSimilarly, $b^2$ and $c^2$ are rational. Since $(a + bc)^2 = a^2 + b^2c^2 + 2abc$ is rational, so is $abc$.\n\nTherefore, $a(a + bc) = a^2 + abc$ is rational and, since $a + bc$ is a nonzero rational, $a$ must be rational. Analogously, $b$ and $c$ are rational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18345,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of $\\triangle ABC$ and let $\\ell$ be a tangent to the incircle, not a side of $\\triangle ABC$, that intersects the sides $AB$, $BC$, and the extension of $CA$ at $X$, $Y$, and $Z$, respectively. Let $AY$ intersect $CX$ at $P$, and the lines $IP$ and $BZ$ meet at $Q$.\n\nProve that, if $A$, $C$, $Y$, $X$ are concyclic, then $ZI^2 = ZQ \\cdot ZB$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to show that $IQ \\perp BZ$ and $\\angle BIZ = 90^\\circ$ because this will imply that $ZI$ is tangent to the circumcircle of $\\triangle BIQ$ (with diameter $BI$) at $I$, and then by the power of point at $Z$ we will get $ZI^2 = ZQ \\cdot ZB$.\n\nTo show $\\angle BIZ = 90^\\circ$, let $S, T$ be the tangent points from $Z$ and $U, V$ be the tangent points from $B$ as shown in the figure. It is easy to see by simple angle chasing that $AXYC$ cyclic if and only if $ST \\perp UV$. And using the fact that $IB \\perp UV$ and $IZ \\perp ST$, we obtain that $\\angle BIZ = 90^\\circ$.\n\nNote that by Bianchon's theorem, for degenerated circumscribed hexagons $AXSYCT$ and $AUXYVC$, the lines $ST$, $UV$, $AY$, $CX$ are concurrent at $P$.\n\n![](images/selected_2011_p7_data_ff818cc88b.png)\n\nThe condition $IQ \\perp BZ$ follows from the Pascal and Brocard theorems as follows:\n\nBy Brocard's theorem, we have $IP \\perp ML$ where $M, L$ are the intersections of the opposite sides of cyclic quadrilateral $SVTU$ with circumcenter $I$, the orthocenter of $\\triangle MLP$.\n\nBy Pascal's Theorem, consider two degenerated inscribed hexagons $USSVTT$ and $UUSVVT$, so that the intersections of opposite sides are collinear. (The side with repeated vertex becomes the tangent line at that point.) Therefore, $M, Z, L$ are collinear, and so are $B, M, L$. Since both lines contain $M$ and $L$, they are the same line.\n\nThis proves that $IP \\perp BZ$ at $Q$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18346,
    "subject": "Mathematics (Olympiad)",
    "question": "In convex quadrilateral $ABCD$, $\\vec{BC} = 2\\vec{AD}$. Point $P$ is on the plane of quadrilateral $ABCD$, satisfying $\\vec{PA} + 2020\\vec{PB} + 2020\\vec{PC} = 2020\\vec{PD} = \\vec{0}$. Let $s$ and $t$ be the areas of quadrilateral $ABCD$ and $\\triangle PAB$, respectively. Then the value of $\\frac{t}{s}$ is \\_\\_\\_\\_\\_.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p30_data_c638c112cd.png)\n\nFig. 7.1",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume that $AD = 2$ and $BC = 4$. As shown in Fig. 7.1, denote $M, N, X, Y$ as the midpoints of $AB, CD, BD, AC$, respectively. Then $M, X, Y, N$ are collinear in order and $MX = XY = YN = 1$.\n\nSince\n\n$$\n\\overrightarrow{PA} + \\overrightarrow{PC} = 2\\overrightarrow{PY}, \\quad \\overrightarrow{PB} + \\overrightarrow{PD} = 2\\overrightarrow{PX},\n$$\n\ncombining the given conditions, we know that $\\overrightarrow{PY} + 2020\\overrightarrow{PX} = \\overrightarrow{0}$. Hence, point $P$ lies on segment $XY$, and $PX = \\frac{1}{2021}$. Let the distance from $A$ to $MN$ be $h$. By the area formula, we can get\n\n$$\n\\begin{aligned}\n\\frac{t}{s} &= \\frac{S_{\\triangle PAB}}{S_{ABCD}} = \\frac{PM \\cdot h}{MN \\cdot 2h} = \\frac{PM}{2MN} \\\\\n&= \\frac{1 + \\frac{1}{2021}}{2 \\times 3} = \\frac{337}{2021}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18347,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \\ldots, a_n$ with\n$$\n\\max(a_1, a_2, \\ldots, a_n) \\leq n \\cdot \\min(a_1, a_2, \\ldots, a_n),\n$$\nthere exist three that are the side lengths of an acute triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n \\geq 13$.\n\nFirst, we show that any $n \\geq 13$ satisfies the desired condition. Suppose for the sake of contradiction that $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ are integers such that $\\max(a_1, a_2, \\ldots, a_n) \\leq n \\cdot \\min(a_1, a_2, \\ldots, a_n)$ and no three are the side lengths of an acute triangle. We conclude that\n$$\na_{i+2}^2 \\geq a_i^2 + a_{i+1}^2 \\quad (1)\n$$\nfor all $i \\leq n - 2$. Letting $\\{F_n\\}$ be the Fibonacci numbers, defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\geq 2$, repeated application of (1) and the ordering of the $\\{a_i\\}$ implies that\n$$\na_i^2 \\geq F_i \\cdot a_1^2 \\quad (2)\n$$\nfor all $i \\leq n$. Noting that $F_{12} = 12^2$, an easy induction shows that $F_n > n^2$ for $n > 12$. Hence, if $n \\geq 13$, (2) implies $a_n^2 > n^2 \\cdot a_1^2$, a contradiction. This shows that any $n \\geq 13$ satisfies the condition of the problem.\n\nOn the other hand, for any $n < 13$, we may take $a_i = \\sqrt{F_i}$ for $1 \\leq i \\leq n$, so that\n$$\n\\max(a_1, a_2, \\ldots, a_n) \\leq n \\cdot \\min(a_1, a_2, \\ldots, a_n)\n$$\nholds because $F_n \\leq n^2$ for $n \\leq 12$. Further, for $i < j$, we have $F_i + F_j \\leq F_{j+1}$, which shows that for $i < j < k$, we have $a_k^2 \\geq a_i^2 + a_j^2$. Hence, $\\{a_i, a_j, a_k\\}$ are not the side lengths of an acute triangle. Therefore, all $n < 13$ do not satisfy the conditions of the problem, and the answer is $n \\geq 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18348,
    "subject": "Mathematics (Olympiad)",
    "question": "A social club has $2k + 1$ members, each of whom is fluent in the same $k$ languages. Any pair of members always talk to each other in only one language. Suppose that there are no three members such that they use only one language among them. Let $A$ be the number of three-member subsets such that the three distinct pairs among them use different languages. Find the maximum possible value of $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\binom{2k+1}{3} - k(2k+1)$, or $\\frac{2k(k-2)(2k+1)}{3}$.\n\nWe model the social club as a complete graph on $2k+1$ vertices, where each language corresponds to a color assigned to the edge between each pair of vertices. Let $V = \\{v_1, \\dots, v_{2k+1}\\}$ be the set of vertices, $L = \\{l_1, \\dots, l_k\\}$ the set of languages (colors), and $\\deg_i(v)$ the number of edges of color $l_i$ incident to vertex $v$.\n\nFirst, we show this is the maximum. Call a triangle *isosceles* if two of its edges are the same color. The number of isosceles triangles at a vertex $v$ is $\\binom{\\deg_1(v)}{2} + \\dots + \\binom{\\deg_k(v)}{2}$. By Cauchy's inequality:\n\n$$\n\\begin{aligned}\n\\binom{\\deg_1(v)}{2} + \\dots + \\binom{\\deg_k(v)}{2} &= \\frac{1}{2} \\sum_{i=1}^{k} (\\deg_i(v))^2 - \\frac{1}{2} \\sum_{i=1}^{k} \\deg_i(v) \\\\\n&\\geq \\frac{1}{2} \\left( \\frac{1}{k} \\right) (\\deg_1(v) + \\dots + \\deg_k(v))^2 \\\\\n&= k.\n\\end{aligned}\n$$\n\nSumming over all vertices, there are at least\n\n$$\n\\sum_{v \\in V} \\sum_{i=1}^{k} \\binom{\\deg_i(v)}{2} \\geq \\sum_{v \\in V} k = k(2k+1)\n$$\n\nisosceles triangles. By the problem's conditions, there are no monochromatic triangles; thus, every triangle is either isosceles or has all edges of different colors (of which there are $A$). Therefore,\n\n$$\nA \\leq \\binom{2k+1}{3} - k(2k+1).\n$$\n\nNow, we construct an example achieving this bound. Equality holds above when $\\deg_i(v) = 2$ for all $i$ and $v$. Thus, it suffices to show that a complete graph on $2k+1$ vertices can be decomposed into $k$ disjoint Hamiltonian cycles, assigning one color to each cycle.\n\nStart with the cycle $C_0 = (v_0, v_1, v_{2k}, v_2, v_{2k-1}, v_3, v_{2k-2}, \\dots, v_k, v_{k+1}, v_0)$. Construct cycles $C_1, \\dots, C_{k-1}$ by adding $i$ to the subscript of each vertex in $C_0$ (with $v_i = v_{i+2k+1}$). Each edge $(v_i, v_{i+n})$ for $1 \\leq n < k$ appears exactly twice in these cycles, and each $i$ occurs as the first vertex index exactly once, so these cycles are disjoint, completing the construction. (For even $n$, this edge occurs at $(v_{-n/2}, v_{n/2})$ and $(v_{k+1-n/2}, v_{k+1+n/2})$ in $C_0$, and the edge $(v_i, v_{i+n})$ occurs in one of $C_{i+n/2}$, $C_{i+n/2-k-1}$, or $C_{i+n/2-(2k+1)}$, depending on which subscript is in $[0, k-1]$. A similar argument holds for odd $n$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18349,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the least real number $M$ such that the inequality\n\n$$\n|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)| \\leq M(a^2 + b^2 + c^2)^2\n$$\n\nholds for all real numbers $a, b$, and $c$.\n\n*Note:* Consider the polynomial\n\n$$\nP(a, b, c) = ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2).\n$$\n\nIt is not difficult to check that $P(a, a, c) = 0$. Hence $a - b$ divides $P(a, b, c)$. Since $P(a, b, c)$ is cyclic symmetric, we conclude that $(a - b)(b - c)(c - a)$ divides $P(a, b, c)$. Since $P(a, b, c)$ is a cyclic homogeneous polynomial of degree 4 (each monomial in the expansion of $P(a, b, c)$ has degree 4) and $(a - b)(b - c)(c - a)$ is a cyclic homogeneous polynomial of degree 3,\n\n$$\nP(a, b, c) = (a - b)(b - c)(c - a)Q(a, b, c),\n$$\n\nwhere $Q(a, b, c)$ is a cyclic homogeneous polynomial of degree 1; that is, $Q(a, b, c) = k(a + b + c)$ for some constant $k$. It is easy to deduce that $k = 1$ and\n\n$$\nP(a, b, c) = (a - b)(b - c)(c - a)(a + b + c).\n$$\n\nThe given inequality now reads\n\n$$\n|(a - b)(b - c)(c - a)(a + b + c)| \\leq M(a^2 + b^2 + c^2)^2. \\quad (*)\n$$\n\nSince the above inequality is symmetric with respect to $a, b$, and $c$, we may assume that $a \\geq b \\geq c$. (Indeed, we may assume that $a > b > c$, because otherwise the left-hand side of $(*)$ is 0, and we have nothing to prove.) Thus $(*)$ reduces to\n\n$$\n(a - b)(b - c)(a - c)(a + b + c) \\leq M(a^2 + b^2 + c^2)^2 \\quad (**)\n$$\n\nfor real numbers $a > b > c$. Note also that $(**)$ is homogeneous (of degree 4). We may further assume that $a + b + c = 1$. Then $(**)$ reduces to\n\n$$\n(a - b)(b - c)(a - c) \\leq M(a^2 + b^2 + c^2)^2 \\quad (\\dagger)\n$$\n\nfor real numbers $a > b > c$ with $a + b + c = 1$. Setting $a - b = x$ and $b - c = y$, we have $a - c = x + y$. Note that\n\n$$\n\\begin{aligned}\n(a - b)^2 + (b - c)^2 + (c - a)^2 &= 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \\\\\n&= 2(a^2 + b^2 + c^2) - [(a + b + c)^2 - (a^2 + b^2 + c^2)] \\\\\n&= 3(a^2 + b^2 + c^2) - 1.\n\\end{aligned}\n$$\n\nWe can rewrite $(\\dagger)$ as\n\n$$\n9xy(x + y) \\leq M[x^2 + y^2 + (x + y)^2 + 1]^2 \\quad (\\ddagger)\n$$\n\nfor positive real numbers $x$ and $y$. It suffices to find the least $M$ satisfying $(\\ddagger)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** We rewrite $(\\ddagger)$ as\n\n$$\n\\begin{aligned}\n\\frac{9(x + y)}{M} &\\leq \\left( \\frac{x^2 + y^2 + (x + y)^2 + 1}{\\sqrt{xy}} \\right)^2 \\\\\n&= \\left( \\frac{2(x + y)^2 + 1}{\\sqrt{xy}} - 2\\sqrt{xy} \\right)^2.\n\\end{aligned}\n$$\n\nSetting\n\n$$\nA = \\frac{2(x + y)^2 + 1}{\\sqrt{xy}}, \\quad B = 2\\sqrt{xy},\n$$\n\nthe above inequality becomes\n\n$$\n\\frac{9(x + y)}{M} \\leq (A - B)^2.\n$$\n\nNote that $A > B > 0$ as $A - B = \\frac{x^2 + y^2 + (x + y)^2 + 1}{\\sqrt{xy}} > 0$. For real numbers $x$ and $y$ with fixed $x + y$, if we increase the value of $\\sqrt{xy}$, the left-hand side $\\frac{9(x + y)}{M}$ does not change, while $A$ decreases (with fixed numerator and increasing denominator) and $B$ increases. Hence, as $\\sqrt{xy}$ increases, $A - B$ decreases, so the right-hand side decreases. Therefore, we may assume that $x = y$ in the above inequality, and $(\\ddagger)$ becomes\n\n$$\n18x^3 \\leq M(6x^2 + 1)^2 = M(36x^4 + 12x^2 + 1)\n$$\n\nor\n\n$$\n36x + \\frac{12}{x} + \\frac{1}{x^3} \\geq \\frac{18}{M}.\n$$\n\nIt suffices to find the minimum value of the continuous function\n\n$$\nf(x) = 36x + \\frac{12}{x} + \\frac{1}{x^3}, \\quad x > 0.\n$$\n\nNote that\n\n$$\n\\frac{df}{dx} = 36 - \\frac{12}{x^2} - \\frac{3}{x^4} = \\frac{3(2x^2 - 1)(6x^2 + 1)}{x^4},\n$$\n\nimplying the only critical value $x = \\frac{1}{\\sqrt{2}}$ in the domain. It is easy to check that $f(x)$ indeed obtains its global minimum $32\\sqrt{2}$ at $x = \\frac{1}{\\sqrt{2}}$.\n\nWe conclude the minimum value of $M$ is $\\frac{9\\sqrt{2}}{32}$, obtained when $x = y = a - b = b - c = \\frac{1}{\\sqrt{2}}$ (and $a + b + c = 1$); that is,\n\n$$\n(a, b, c) = \\left( \\frac{1}{3} + \\frac{1}{\\sqrt{2}}, \\frac{1}{3}, \\frac{1}{3} - \\frac{1}{\\sqrt{2}} \\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18350,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be the set of all students. For $s \\in S$, let $a_s$ be the number of students in the group of student $s$ in the first distribution, and $b_s$ the number of students in the group of student $s$ in the second distribution.\n\nSuppose\n\n$$\n\\sum_{s \\in S} \\frac{1}{a_s} = N, \\quad \\sum_{s \\in S} \\frac{1}{b_s} = N + K.\n$$\n\nShow that for at least $K + 1$ students, $b_s < a_s$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\sum_{s \\in S} \\left( \\frac{1}{b_s} - \\frac{1}{a_s} \\right) = K.\n$$\n\nSince $\\left| \\frac{1}{b_s} - \\frac{1}{a_s} \\right| < 1$ for each $s$, the sum $K$ can only be achieved if at least $K + 1$ terms are positive, i.e., for at least $K + 1$ students, $\\frac{1}{b_s} - \\frac{1}{a_s} > 0$, which means $b_s < a_s$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18351,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 < a_2 < \\dots < a_{53}$ be positive integers such that the sum of any 27 integers is greater than the sum of the remaining 26 integers.\n\n(a) Find the minimum value of $a_1$.\n\n(b) Find all possible values of $a_2, \\dots, a_{53}$, when $a_1$ is at the minimum value.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, the condition on the sequence is equivalent to the condition\n\n$$\na_1 + a_2 + \\dots + a_{27} > a_{28} + \\dots + a_{53}. \\quad (*)\n$$\n\nWe have $a_{i+26} \\geq a_{i+25} + 1 \\geq \\dots \\geq a_i + 26$, for $1 \\leq i \\leq 27$. It implies that $a_{i+26} - a_i \\geq 26$. Thus, from $(*)$ and the above inequality, we have\n\n$$\na_1 \\geq 1 + \\sum_{i=2}^{27} (a_{i+26} - a_i) \\geq 1 + 26 \\cdot 26 = 677.\n$$\n\nNext, we show that the minimum value of $a_1$ is indeed 677. If $a_1 = 677$, the above inequalities must be equalities. Hence $a_{i+26} = a_i + 26$, $2 \\leq i \\leq 27$ and $a_j = a_{j+1} - 1 = \\dots = a_{j+26} - 26$, $2 \\leq j \\leq 53$.\n\nHence, if $a_2 = n$ then $a_3 = n+1, \\dots, a_{53} = n+51$ and $n \\geq a_1 + 1 = 678$.\n\nSo $a_1 = 677$, $a_2 = n$, $a_3 = n+1, \\dots, a_{53} = n+51$, $n \\geq 678$ is the solution to this problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18352,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{T}_n$ be the set of $2n$-sequences consisting of $n$ zeros and $n$ ones such that in each initial segment the number of 1's does not surpass the number of 0's. Prove that $|\\mathcal{T}_n|$ is the $n$-th Catalan number, that is,\n$$\n|\\mathcal{T}_n| = C_n = \\frac{1}{n+1} \\binom{2n}{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $[n] = \\{1, 2, \\dots, n\\}$, and let $\\mathcal{F}_n$ be the set of non-decreasing mappings $f: [n] \\to [n]$ such that $f(i) \\le i$ for each $i \\in [n]$. There is a bijection between $\\mathcal{T}_n$ and $\\mathcal{F}_n$ by letting $f(i)$ equal $1 + \\#(i)$, where $\\#(i)$ is the number of 1's before the $i$-th zero in $t$. It is a known fact that $|\\mathcal{F}_n|$ is the $n$-th Catalan number $C_n = \\frac{1}{n+1} \\binom{2n}{n}$, for example by the reflection principle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18353,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a nonempty set $M$ *good* if its elements are positive integers, each having exactly 4 divisors. If the good set $M$ has $n$ elements, we denote by $S_M$ the sum of all $4n$ divisors of its members (the sum may contain repeating terms).\n\n**a)** Prove that\n\n$$\nA = \\{2 \\cdot 37, 19 \\cdot 37, 29 \\cdot 37\\}\n$$\n\nis good and $S_A = 2014$.\n\n**b)** Prove that if the set $B$ is good and $8 \\in B$, then $S_B \\neq 2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Any number equal to the product of two distinct primes $p, q$ has exactly 4 divisors: $1$, $p$, $q$, and $pq$, hence $A$ is good.\n\nA short computation shows that $S_A = 2014$.\n\n**b)** A number $n$ having exactly 4 divisors is either the product of two distinct primes or the cube of a prime. It is not difficult to see that the sum of the divisors of such a number is even, except for $n = 8$. Therefore, if $8 \\in B$, then $S_B$ must be odd, hence $S_B \\neq 2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18354,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega$ be the circumcircle of triangle $XZP$ with center $O$, and let $Q$ be the second intersection of the line $XY$ with $\\omega$.\n\n*Lemma.* Let points $Z, Q$ lie on the circle $\\omega$ with center $O$, and let $l$ be an arbitrary line. Let $M$ be the projection of $O$ onto $l$. Points $Y, T$ lie on the line $l$ such that $MY = MT$. Lines $YQ$ and $TZ$ intersect $\\omega$ for the second time at points $X$ and $P$, respectively. If $PQ$ and $ZX$ intersect $l$ at points $A$ and $C$, then $AM = CM$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Let $P'$ be a point on $\\omega$ such that $PP' \\parallel l$. Note that quadrilateral $TPP'Y$ is a cyclic trapezoid ($P', Y$ are the reflections of $P, T$ with respect to the line $OM$).\n\n![](images/IRN_booklet2025_p55_data_2f1c103883.png)\n\nBy the law of sines in triangles $ATP$, $TZC$, $XYC$, $YQA$:\n\n$$\n\\begin{aligned}\n\\frac{TP}{TA} &= \\frac{\\sin \\angle TAP}{\\sin \\angle TPA'}, \\quad \\frac{TZ}{TC} = \\frac{\\sin \\angle TCZ}{\\sin \\angle TZC} \\\\\n\\frac{CY}{XY} &= \\frac{\\sin \\angle CXY}{\\sin \\angle YCX'}, \\quad \\frac{AY}{YQ} = \\frac{\\sin \\angle AQY}{\\sin \\angle YAQ}\n\\end{aligned}\n$$\n\nQuadrilateral $ZQXP$ is cyclic, therefore $\\angle APT = \\angle YXC$, $\\angle AQY = \\angle TZC$. Hence\n\n$$\n\\frac{TP \\cdot TZ}{TA \\cdot TC} = \\frac{YX \\cdot YQ}{YC \\cdot YA}\n$$\n\nNote that the ratios of the power of points $T, Y$ with respect to circle $\\omega$ and the circumcircle of triangle $APC$ are equal, hence circles $\\omega$ and the circumcircles of triangles $YPT$, $APC$ are coaxial, which means quadrilateral $APP'C$ is cyclic. This quadrilateral is a cyclic trapezoid ($AP, P'C$ are symmetric with respect to line $OM$), hence $MA = MC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18355,
    "subject": "Mathematics (Olympiad)",
    "question": "А ба В хоёр тоглогч \"Таавар\" тоглоом тоглож байна. Тэдэнд мэдэгдэж байгаа $k$ ба $n$ натурал тоонууд байна. Тоглоомын дүрэм:\n\n- Тоглоом эхлэхэд А нь $1 \\le x \\le N$ байх бүхэл тоо $x$-ийг сонгоно, $N$-ийг В-д хэлнэ.\n- В дараах маягийн асуултыг тавьж болно: натурал тооны ямар нэг $S$ олонлогийг зааж, энэ $S$ олонлогт $x$ орж байгаа эсэхийг асууна.\n- В хэдэн ч асуулт тавьж болох ба нэгэн ижил асуултыг олон удаа тавьж болно.\n- А асуулт бүрд \"тийм\" эсвэл \"үгүй\" гэж шууд хариулна. А хэдэн ч удаа худал хэлж болох ч, ямар ч дараалсан $k+1$ хариултын дор хаяж нэг нь үнэн байх ёстой.\n- В асуултуудаа тавьсны дараа $n$-ээс олонгүй натурал тооноос тогтох $X$ олонлогийг заана. Хэрэв $x \\in X$ бол В хожно, эс тэгвээс А хожно.\n\nДараахыг батал:\n\n1. $n \\ge 2^k$ бол В хожиж чадна.\n2. Ямар ч хангалттай их $k$-ийн хувьд В хожиж чадахгүй байх $n \\ge 1.99^k$ тоо олдож байгааг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Асуултын хариултыг $A \\in \\{\\text{yes}, \\text{no}\\}$ гэж үзье. \"$x$ $S$ олонлогт орж байна уу?\" гэсэн асуултад $A$ хариулт өгнө. $A = \\text{yes}$ ба $i \\notin S$, эсвэл $A = \\text{no}$ ба $i \\in S$ бол $A$ нь $i$-д нийцэхгүй гэж үзнэ. Зорилтот тоо $x$-д нийцэхгүй хариулт бол худал.\n\n**a)** Бен $x$-ийг агуулсан $m$ хэмжээтэй $T$ олонлогийг олсон гэж үзье. Эхэндээ $m = N$, $T = \\{1, 2, \\dots, N\\}$. $m > 2^k$ үед Бен $T$-д $x$-ээс өөр $y$ тоог олох аргыг үзүүлье. Энэ алхмыг давтан $T$-г $2^k \\le n$ болтол багасгаж чадна.\n\n$T$-ийн хэмжээ $m > 2^k$ л чухал тул $T = \\{0, 1, \\dots, m-1\\}$ гэж үзье. Бен $x = 2^k$ эсэхийг дахин дахин асууна. Хэрэв А $k+1$ удаа дараалан \"үгүй\" гэвэл дор хаяж нэг нь үнэн тул $x \\ne 2^k$. Эс тэгвээс анхны \"тийм\" хариулт хүртэл асууж зогсоно. Дараа нь $i = 1, \\dots, k$ бүрт $x$-ийн $i$-р бит 0 эсэхийг асууна. Эдгээр $k$ хариулт ямар ч байсан, $y \\in \\{0, 1, \\dots, 2^k-1\\}$ тоонд нийцэхгүй байна. Өмнөх $2^k$-ийн \"тийм\" хариулт ч $y$-д нийцэхгүй. Тэгэхээр $y \\ne x$, эс тэгвээс сүүлийн $k+1$ хариулт бүгд худал болох ба энэ нь боломжгүй.\n\nИнгэснээр Бен $T$-д $x$-ээс өөр тоог олж чадна. Иймд шаардлагыг биелүүлж байна.\n\n**b)** $1 < \\lambda < 2$ ба $n = [(2-\\lambda)\\lambda^{k+1}] - 1$ гэж үзье. Бен заавал хожих боломжгүйг баталъя. Үүний тулд $1.99 < \\lambda < 2$ ба $k$-г хангалттай их авбал\n\n$$\nn = [(2 - \\lambda)\\lambda^{k+1}] - 1 \\ge 1.99^k\n$$\n\nболно.\n\nАми дараах стратеги хэрэглэнэ: Эхэндээ $N = n+1$, $x \\in \\{1, 2, \\dots, n+1\\}$ дурын тоог сонгоно. Хариулт бүрийн дараа $i = 1, 2, \\dots, n+1$ бүрт $m_i$ — тухайн $i$-д нийцэхгүй дараалсан хариултын тоог тооцно. Дараагийн хариултаа сонгохдоо\n\n$$\n\\phi = \\sum_{i=1}^{n+1} \\lambda^{m_i}\n$$\n\nутгыг хамгийн бага болгох хариултыг сонгоно.\n\nЭнэ стратегиар $\\phi$ үргэлж $\\lambda^{k+1}$-ээс бага байна. Тэгэхээр $m_i \\le k$ үргэлж биелнэ, өөрөөр хэлбэл $x$-д $k$-аас олон удаа дараалан худал хэлэхгүй. Иймд стратеги зөвшөөрөгдөх бөгөөд $x$-ээс хамаарахгүй тул Бен $x$-ийн талаар мэдээлэл авч чадахгүй, хожих баталгаа байхгүй.\n\nҮлдсэн нь $\\phi < \\lambda^{k+1}$-ийг батлах. Эхэндээ $m_i = 0$ тул $\\phi < \\lambda^{k+1}$ биелнэ. Дараа нь Бен $x \\in S$ эсэхийг асуувал, Ами \"тийм\" эсвэл \"үгүй\" гэж хариулахад шинэ $\\phi$ нь:\n\n$$\n\\phi_1 = \\sum_{i \\in S} 1 + \\sum_{i \\notin S} \\lambda^{m_i+1}, \\quad \\phi_2 = \\sum_{i \\in S} \\lambda^{m_i+1} + \\sum_{i \\notin S} 1\n$$\n\nболно. Ами бага утгыг сонгоно. Дундаж нь:\n\n$$\n\\min(\\phi_1, \\phi_2) \\le \\frac{1}{2}(\\phi_1 + \\phi_2) = \\frac{1}{2}(\\lambda\\phi + n + 1)\n$$\n\n$\\phi < \\lambda^{k+1}$, $\\lambda < 2$, $n = [(2-\\lambda)\\lambda^{k+1}] - 1$ тул\n\n$$\n\\min(\\phi_1, \\phi_2) < \\frac{1}{2}(\\lambda^{k+2} + (2-\\lambda)\\lambda^{k+1}) = \\lambda^{k+1}\n$$\n\nИймд шаардлага биелж байна.\n\n*Тайлбар.* $k$-ийн хувьд $f(k)$-г Бен заавал хожих хамгийн бага $n$ гэж тэмдэглэе. Дээрхээс $1.99^k \\le f(k) \\le 2^k$ гэдгийг баталж байна. Компьютерээр $f(1) = 2$, $f(2) = 3$, $f(3) = 4$, $f(4) = 7$, $f(5) = 11$, $f(6) = 17$ байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18356,
    "subject": "Mathematics (Olympiad)",
    "question": "Para pertenecer a un club, cada nuevo socio debe pagar como cuota de inscripción a cada miembro del club la misma cantidad que él tuvo que pagar en total cuando ingresó, más un euro. Si el primer socio pagó un euro, ¿cuánto deberá pagar en total el $n$-ésimo socio?",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $a_n$ la cuota total del socio $n$-ésimo y sea $s_n = a_1 + \\dots + a_n$. El $n$-ésimo ($n \\geq 2$) socio tiene que pagar en total $$(a_1 + 1) + (a_2 + 1) + \\dots + (a_{n-1} + 1) = s_{n-1} + n - 1$$ euros, luego $$a_n = s_{n-1} + n - 1.$$ \n\nEntonces, \n$$\ns_n = s_{n-1} + a_n = s_{n-1} + s_{n-1} + (n-1) = 2s_{n-1} + n - 1.\n$$\nIterando esta relación, se obtiene:\n$$\ns_n = 2^{n-1} + 2^{n-2} \\times 1 + 2^{n-3} \\times 2 + \\dots + 2 \\times (n-2) + (n-1)\n$$\nde donde\n$$\ns_n = 2^n + 2^{n-1} - 1 - n.\n$$\nEntonces, para $n \\geq 2$,\n$$\na_n = s_n - s_{n-1} = 2^n - 2^{n-2} - 1 = 3 \\times 2^{n-2} - 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18357,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a number whose last $k$ digits are zeros, and the digit just to the left of these zeros is not zero. Is it possible for $n$ to be a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "The last digit of a perfect square can be $0$, $1$, $4$, $5$, $6$, or $9$. If $n$ is to be a perfect square, it must end in $0$. Suppose the last $k$ digits of $n$ are zeros, and the digit just to the left is not zero.\n\nIf $k$ is odd, $k = 2m - 1$, then $\\frac{n}{10^{2m-2}}$ is also a perfect square. It is divisible by $10$, so it must also be divisible by $25$. This is not possible since the second digit from the right is either $3$ or $7$.\n\nTherefore, $n$ must end in an even number of zeros, so $k$ is even. In this case, $\\frac{n}{10^k}$ is a positive integer and a perfect square with the last digit equal to either $3$ or $7$. Again, this is not possible. Hence, $n$ is never a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18358,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $a, b$ be two coprime integers greater than $1$. Let $p, q$ be two odd divisors greater than $1$ of $a^{6n} + b^{6n}$. Find the remainder when $p^{6n} + q^{6n}$ is divided by $6 \\cdot (12)^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer follows from these remarks:\n\n1. If $a, b$ are coprime integers greater than $1$ and $p$ is an odd prime divisor of $a^{6n} + b^{6n}$, then $p \\equiv 1 \\pmod{2^{n+1}}$.\n\n2. If $x \\equiv 1 \\pmod{c^k}$, then $x^{c^m} \\equiv 1 \\pmod{2^{m+k}}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18359,
    "subject": "Mathematics (Olympiad)",
    "question": "甲、乙兩人在實數線上玩以下著色遊戲：\n\n甲有一桶顏料共 4 單位，其中 $p$ 單位的顏料剛好可以塗滿一個長度為 $p$ 的閉區間。每回合，甲先指定一個正整數 $m$，並給乙 $\\frac{1}{2^m}$ 單位的顏料。接著，乙選一個正整數 $k$，並將 $\\left[\\frac{k}{2^m}, \\frac{k+1}{2^m}\\right]$ 塗滿（此區間可能有一部分在之前的回合中已經被塗過）。\n\n如果桶子空了但 $[0, 1]$ 區間還沒被塗滿，則甲獲勝。\n\n決定是否存在甲能在有限步內獲勝的策略。",
    "options": [],
    "answer": "See solution",
    "solution": "否，乙可以確保在顏料用光時 $[0, 1]$ 區間必被塗滿。\n\n在第 $r$ 回合開始時，令 $x_r$ 為滿足 $[0, x_r]$ 皆已被塗滿的最大實數（令 $x_1 = 0$）。假設甲選擇 $m$，令 $y_r$ 為滿足\n\n$$\n\\frac{y_r}{2^m} \\le x_r < \\frac{y_r+1}{2^m}\n$$\n\n的整數。注意到 $I_0^r := [y_r/2^m, (y_r+1)/2^m]$ 是本回合可以塗，且尚未被塗滿的區間中最左邊的那一個。\n\n乙的策略是考慮下一個區間 $I_1^r := [(y_r + 1)/2^m, (y_r + 2)/2^m]$。若 $I_1^r$ 尚未被塗滿，則乙選塗 $I_1^r$；否則，乙選塗 $I_0^r$。（為了方便起見，我們假設 $[1, 2]$ 在一開始就已經被塗滿。）\n\n要證明以上策略可行，我們的目標是估計每回合結束時的顏料量。以下將以歸納法證明，若在第 $r$ 回合開始前，$[0, 1]$ 尚未被塗滿，則：\n\n1. 被用來塗 $[0, x_r]$ 的顏料量至多為 $3x_r$。\n2. 對於每個 $m$，乙至多只塗滿一個在 $x_r$ 右邊、形如 $[k/2^m, (k+1)/2^m]$ 的區間。\n\n以上條件對 $r=0$ 顯然成立。假設對 $r \\le k-1$ 都成立，則在 $r=k$ 時，考慮乙塗的區間：\n\n- 如果乙塗 $I_1^r$，易見 $x_{r+1} = x_r$，從而由歸納假設知 1. 成立。又，如果在第 $r$ 回合開始時，在 $x_r$ 右邊有一個長度為 $2^m$ 的區間被塗滿，依照此策略易知此區間必為 $I_1^r$，但這與乙選到 $I_1^r$ 的事實不符，矛盾。故 2. 亦成立。\n\n- 如果乙塗 $I_0^r$，但 $[0, 1]$ 尚未被塗滿。易知 2. 自動成立。注意到此時 $I_0^r$ 和 $I_1^r$ 都會被塗滿，故 $x_{r+1}$ 至少會前進到 $I_1^r$ 的右端點，也就是 $x_{r+1} = x_r + \\alpha$，其中 $\\alpha > 1/2^m$。又注意到在第 $r$ 回合前就被塗滿，且與 $(x_r, x_{r+1})$ 相交的區間必然在 $[x_r, x_{r+1}]$ 內；由 2.，這些區間都會是不同長度，且長度都大於 $1/2^m$，故在其上使用的顏料量少於 $2/2^m$。因此，在 $[0, x_{r+1}]$ 上使用的顏料量不會多於\n\n$$\n3x_r + \\frac{2}{2^m} + \\frac{1}{2^m} = 3\\left(x_r + \\frac{1}{2^m}\\right) < 3x_{r+1}.\n$$\n\n故 1. 亦成立。至此，歸納部分證明完畢。\n\n現在，假設在第 $r-1$ 回合後，$[0, 1]$ 尚未被塗滿。由 2. 知在 $[x_r, 1]$ 中乙所塗的區間必為不同長度，且其長度為 $2^{-k}$，$k \\le 1-x_r$。從而塗在 $[x_r, 1]$ 中的顏料至多為 $2(1-x_r)$。而由 1. 知塗在 $[0, x_r]$ 的顏料至多為 $3x_r$，因此總量不超過 $3x_r + 2(1-x_r) < 3$，也就是說顏料尚未用完。故甲不可能贏。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18360,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for any $x, y \\in \\mathbb{R}$,\n\n$$\nf(xf(y) + y^{2021}) = y f(x) + (f(y))^{2021}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ denote the given equation:\n$$\nf(xf(y) + y^{2021}) = y f(x) + (f(y))^{2021}.\n$$\n\n**Case 1:** $f(0) \\neq 0$.\n\nFrom $P(x, 0)$:\n$$\nf(xf(0)) = (f(0))^{2021}.\n$$\nSince $x$ is arbitrary, $f$ must be constant, say $f(x) = c$. Plugging into the original equation:\n$$\nc = y c + c^{2021}.\n$$\nFor all $y$, this is only possible if $c = 0$, but then $c = c^{2021}$, which is only true for $c = 0$. However, this contradicts $f(0) \\neq 0$. Thus, $f(0) = 0$.\n\n**Case 2:** $f(a) = 0$ for some $a \\neq 0$.\n\nFrom $P(x, a)$:\n$$\nf(x f(a) + a^{2021}) = a f(x) + (f(a))^{2021}.\n$$\nBut $f(a) = 0$, so:\n$$\nf(a^{2021}) = (0)^{2021} = 0.\n$$\nThus, $f$ is zero at $a^{2021}$. Repeating this argument, $f$ must be identically zero. Indeed, $f(x) = 0$ for all $x$ is a solution.\n\n**Case 3:** $f(a) \\neq 0$ for all $a \\neq 0$.\n\nFrom $P(0, 1)$:\n$$\nf(1) = (f(1))^{2021}.\n$$\nSo $f(1) = 0$ or $f(1) = 1$ or $f(1) = -1$ (for odd $2021$). If $f(1) = -1$, then from $P(1, 1)$:\n$$\nf(1 \\cdot (-1) + 1^{2021}) = 1 f(1) + ((-1))^{2021} = -1 + (-1) = -2.\n$$\nBut $f(-1 + 1) = f(0) = 0$, so $0 = -2$, a contradiction. Thus, $f(1) = 1$.\n\nFrom $P(x, 1)$:\n$$\nf(x f(1) + 1^{2021}) = 1 f(x) + (f(1))^{2021}.\n$$\nSince $f(1) = 1$ and $1^{2021} = 1$:\n$$\nf(x + 1) = f(x) + 1. \\qquad (1)\n$$\n\nNow, compare $P(x, y)$ and $P(x+1, y)$:\n\n- $P(x, y):$\n  $$f(x f(y) + y^{2021}) = y f(x) + (f(y))^{2021}.$$\n- $P(x+1, y):$\n  $$f((x+1) f(y) + y^{2021}) = y f(x+1) + (f(y))^{2021}.$$\n\nSubtracting:\n$$\nf(x f(y) + f(y) + y^{2021}) = y (f(x) + 1) + (f(y))^{2021} = y f(x) + y + (f(y))^{2021}.\n$$\nBut $f(x f(y) + f(y) + y^{2021}) = f((x f(y) + y^{2021}) + f(y))$.\n\nLet $z = x f(y) + y^{2021}$, so $f(z + f(y)) = y + f(z)$. Setting $z = 0$:\n$$\nf(f(y)) = y + f(0) = y.\n$$\nSo $f(f(y)) = y$ for all $y$.\n\nNow, let $y = f(w)$ in the previous result:\n$$\nf(z + f(f(w))) = f(z + w) = f(z) + f(w).\n$$\nSo $f$ is additive.\n\nFrom $P(0, y)$:\n$$\nf(y^{2021}) = (f(y))^{2021}.\n$$\n\nNow, $P(x, y)$ becomes:\n$$\ny f(x) + (f(y))^{2021} = f(x f(y) + y^{2021}) = f(x f(y)) + f(y^{2021}) = f(x f(y)) + (f(y))^{2021}.\n$$\nSo $y f(x) = f(x f(y))$.\n\nLet $y = f(z)$, then $f(f(z)) = z$:\n$$\nf(x z) = f(x) f(z).\n$$\nSo $f$ is multiplicative and additive, and $f(1) = 1$. The only such function is $f(x) = x$.\n\n**Conclusion:**\n\nThe only solutions are:\n- $f(x) = 0$ for all $x$,\n- $f(x) = x$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18361,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real values of $x$ for which the expression $\\sqrt{1 - x^2} + \\sqrt{5x - x^2}$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $1 - x^2 \\leq 1$ and $5x - x^2$ is maximized at $x = \\frac{5}{2}$, giving $5x - x^2 = \\frac{25}{4}$. Thus,\n$$\n\\sqrt{1 - x^2} + \\sqrt{5x - x^2} \\leq 1 + \\frac{5}{2} = \\frac{7}{2}.\n$$\nThe expression is non-negative and cannot be zero, since $1 - x^2 = 0$ and $5x - x^2 = 0$ cannot hold simultaneously. Let $\\sqrt{1 - x^2} + \\sqrt{5x - x^2} = a$, where $a$ is an integer. Squaring and manipulating, we obtain the quadratic equation:\n$$\nx^2(25 + 4a^2) + x(-10a^2 - 10) + (a^2 - 1)^2 = 0.\n$$\nFor $a = 1$: $x(29x - 20) = 0$, so $x = 0$ or $x = \\frac{20}{29}$. Only $x = 0$ yields an integer value for the original expression.\n\nFor $a = 2$: $(x - 1)(41x - 9) = 0$, so $x = 1$ or $x = \\frac{9}{41}$. Both give integer values for the expression.\n\nFor $a = 3$: $61x^2 - 100x + 64 = 0$ has a negative discriminant, so no real solutions.\n\n**Conclusion:** The expression is an integer only for $x = 0$, $x = \\frac{9}{41}$, or $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18362,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest integer $k > 1$ such that there exist $k$ distinct primes whose squares sum to a power of $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $p_1^2 + p_2^2$ (where $p_1$ and $p_2$ are two distinct primes) to be equal to $2^n$ (where $n \\geq 2$), both $p_1$ and $p_2$ must be odd. This gives $p_1^2 + p_2^2 \\equiv 2 \\pmod{4}$, while $2^n \\equiv 0 \\pmod{4}$, a contradiction.\n\nFor $p_1^2 + p_2^2 + p_3^2$ (three distinct primes) to be $2^n$, we must have (say) $p_1 = 2$ and $p_2, p_3$ odd. Then $p_1^2 + p_2^2 + p_3^2 \\equiv 2 \\pmod{4}$, again a contradiction.\n\nFor $p_1^2 + p_2^2 + p_3^2 + p_4^2$ (four distinct primes) to be $2^n$ ($n \\geq 3$), all four must be odd. Then $\\sum_{i=1}^4 p_i^2 = 8L + 4 = 2^n$ for some integer $L$, so $2L + 1 = 2^{n-2}$, which is impossible since $2^{n-2}$ is even for $n \\geq 3$.\n\nThus, the smallest $k > 1$ must satisfy $k \\geq 5$. Now, the sum of the squares of five distinct primes can be a power of $2$. Indeed, $2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 4 + 9 + 25 + 49 + 169 = 256 = 2^8$. Therefore, $k = 5$ is the smallest such integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18363,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$ with incenter $I$, the incircle touches $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The circle with center $C$ and radius $CE$ meets $EF$ for the second time at $K$. If $X$ is the $C$-excircle touchpoint with $AB$, show that $CX$, $KD$, and $IF$ concur.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the concurrency point is the $F$-antipode $F'$. It is well-known that this is $\\overline{CX} \\cap \\overline{IF}$.\n\nLet $P = \\overline{EF} \\cap \\overline{DF'}$ and $Q = \\overline{DF} \\cap \\overline{EF'}$. Then since $\\angle PEQ = \\angle PDQ = 90^\\circ$, $DEPQ$ is cyclic.\n\nNow, we have\n\n$$\n\\begin{aligned}\n\\angle EFD &= \\angle EFF' = 90^\\circ - \\angle PF'E = 90^\\circ - \\angle DFE \\\\\n&= -90^\\circ + \\left(90^\\circ - \\frac{\\angle A}{2}\\right) + \\left(90^\\circ - \\frac{\\angle B}{2}\\right) = \\frac{\\angle C}{2},\n\\end{aligned}\n$$\n\nso the center of $(DEPQ)$ lies on $(CDE)$ (and is on the same side of $\\overline{DE}$ as $C$). On the other hand, it also lies on the perpendicular bisector of $\\overline{DE}$, so it must be $C$ itself. This gives us $P = K$, and the conclusion follows. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18364,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime greater than $5$. Suppose there is an integer $k$ such that $k^2 + 5$ is divisible by $p$. Prove that there exist positive integers $m$ and $n$ such that $p^2 = m^2 + 5n^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s = [\\sqrt{p}]$ and $A = \\{a_1 + a_2k \\mid a_i \\text{ are integers and } 0 \\le a_i \\le s\\}$. Since $(s+1)^2 > p$, $A$ has at least $p+1$ elements. By the pigeonhole principle, $A$ has two elements whose difference is divisible by $p$. That is, there are two integers $b_1$ and $b_2$ such that $|b_i| \\le s$, $b_1 + b_2k \\ne 0$ and $p \\mid (b_1 + b_2k)$.\n\nSince $p \\mid (b_1 + b_2k)$,\n\n$$\n(b_1 + b_2k)(b_1 - b_2k) = b_1^2 - b_2^2 k^2 = b_1^2 + 5b_2^2 - b_2^2(k^2 + 5)\n$$\n\nis also divisible by $p$. Because $p \\mid (k^2 + 5)$, then $p \\mid (b_1^2 + 5b_2^2)$.\n\nNow from $0 < b_1^2 + 5b_2^2 \\le 6s^2 < 6p$, we get $b_1^2 + 5b_2^2 = p, 2p, 3p, 4p, \\text{ or } 5p$.\n\n- If $b_1^2 + 5b_2^2 = 4p$, then $b_1$ and $b_2$ are even. Thus, $\\left(\\frac{b_1}{2}\\right)^2 + 5\\left(\\frac{b_2}{2}\\right)^2 = p$.\n- If $b_1^2 + 5b_2^2 = 5p$, then $5 \\mid b_1$. Thus, $b_2^2 + 5\\left(\\frac{b_1}{5}\\right)^2 = p$.\n- If $b_1^2 + 5b_2^2 = p$, with $b_1b_2 \\ne 0$, then $(b_1^2 - 5b_2^2)^2 + 5(2b_1b_2)^2 = p^2$.\n- If $b_1^2 + 5b_2^2 = 3p$, neither $b_1$ nor $b_2$ is a multiple of $3$.\n  - If $3 \\mid (b_1 + b_2)$, then $\\left(\\frac{b_1 - 5b_2}{3}\\right)^2 + 5\\left(\\frac{b_1 + b_2}{3}\\right)^2 = 2p$.\n  - If $3 \\mid (b_1 - b_2)$, then $\\left(\\frac{b_1 + 5b_2}{3}\\right)^2 + 5\\left(\\frac{b_1 - b_2}{3}\\right)^2 = 2p$.\n- If $b_1^2 + 5b_2^2 = 2p$, with $b_1$ and $b_2$ odd, then $\\left(\\frac{b_1^2 - 5b_2^2}{2}\\right)^2 + 5(b_1b_2)^2 = p^2$.\n\nIf $b_1^2 + 5b_2^2 = p$, then $p^2 = m^2 + 5n^2$ has a solution $(m, n) = (b_1^2 - 5b_2^2, 2b_1b_2)$, and $n \\ne 0$. If $b_1^2 + 5b_2^2 = 2p$, then $p^2 = m^2 + 5n^2$ has a solution $(m, n) = \\left(\\frac{b_1^2 - 5b_2^2}{2}, b_1b_2\\right)$, and $n \\ne 0$.\n\nWhen $p^2 = m^2 + 5n^2$ and $n \\ne 0$, it is clear that $m \\ne 0$. Therefore, there exist positive integers $m$ and $n$ such that $p^2 = m^2 + 5n^2$.\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18365,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum and the maximum value of the expression\n$$\n\\sqrt{4-a^2} + \\sqrt{4-b^2} + \\sqrt{4-c^2},\n$$\nwhere $a$, $b$, $c$ are positive real numbers satisfying $a^2 + b^2 + c^2 = 6$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $a, b, c \\in [0, 2]$. From Cauchy-Schwarz, we have\n$$\n(\\sqrt{4-a^2} + \\sqrt{4-b^2} + \\sqrt{4-c^2})^2 \\le 3(4-a^2 + 4-b^2 + 4-c^2) = 18,\n$$\nso $\\sqrt{4-a^2} + \\sqrt{4-b^2} + \\sqrt{4-c^2} \\le 3\\sqrt{2}$, with equality for $|a| = |b| = |c| = \\sqrt{2}$.\n\nObviously, for $x, y \\ge 0$, we have $\\sqrt{x} + \\sqrt{y} \\ge \\sqrt{x+y}$, the equality occurring when $x = 0$ or $y = 0$. Suppose that $a \\le b \\le c$; then $6 \\ge 3a^2$, so $a^2 \\le 2$.\n\nIt follows that\n$$\n\\sqrt{4-a^2} + \\sqrt{4-b^2} + \\sqrt{4-c^2} \\ge \\sqrt{4-a^2} + \\sqrt{8-b^2-c^2} = \\sqrt{4-a^2} + \\sqrt{2+a^2},\n$$\nand, since $0 \\le 4-c^2 \\le 4-b^2 \\le 4-a^2$, the equality holds for $4-c^2 = 0$, which means $c=2$.\n\nIt remains to find out the minimum value of the expression $\\sqrt{4-x} + \\sqrt{2+x}$, for $0 \\le x \\le 2$. We claim that\n$$\n\\sqrt{4-x} + \\sqrt{2+x} \\ge 2 + \\sqrt{2}.\n$$\nIndeed, squaring the above relation we get $\\sqrt{(4-x)(2+x)} \\ge 2\\sqrt{2}$, which is equivalent to $x(2-x) \\ge 0$, obviously true. The inequality holds for $x=0$ or $x=2$. So the minimum value of the expression $\\sqrt{4-a^2} + \\sqrt{4-b^2} + \\sqrt{4-c^2}$ is $2+\\sqrt{2}$, which can be obtained for $\\{a, b, c\\} = \\{0, \\sqrt{2}, 2\\}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18366,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $S$ 為所有正整數的一個非空子集。一個正整數 $n$ 被稱為 *乾淨的*，若且唯若它可以被表示成 $S$ 的奇數個相異元素的和，且這個表示法是唯一的。\n\n試證：存在無窮多個不乾淨的正整數。\n\nLet $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is *clean* if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.",
    "options": [],
    "answer": "See solution",
    "solution": "定義一個正整數 $n$ 的 *奇數*（分別為 *偶數*）*表示* 為 $n$ 表示成 $S$ 的奇數（分別為偶數）個相異元素之和。設 $\\mathbb{N}$ 為所有正整數的集合。\n\n假設反面，只有有限多個不乾淨的正整數。則存在正整數 $N$，使得每個 $n > N$ 都有且僅有一個奇數表示。\n\n顯然，此時 $S$ 必為無窮集。先證以下性質：\n\n1. **任意正整數 $n$ 最多只有一個奇數表示和一個偶數表示。**\n\n*證明*：對於偶數表示，因 $S$ 無窮，存在 $x \\in S$ 使 $x > \\max\\{n, N\\}$。則 $n+x$ 必乾淨，且 $x$ 不會出現在 $n$ 的偶數表示中。若 $n$ 有多於一個偶數表示，則 $n+x$ 有兩個不同的奇數表示，矛盾。奇數表示同理。\n\n2. **固定 $s \\in S$。若 $n > N$ 無偶數表示，則 $n+2s$ 有包含 $s$ 的偶數表示。**\n\n*證明*：$n+s > N$，故 $n+s$ 乾淨，有唯一奇數表示。若該表示含 $s$，則 $n$ 有不含 $s$ 的偶數表示，矛盾。故 $n+s$ 的奇數表示不含 $s$，加上 $s$ 得 $n+2s$ 的偶數表示，且必含 $s$。\n\n3. **充分大的整數都有偶數表示。**\n\n*證明*：固定 $s \\in S$，對 $r = 1, 2, \\dots, 2s$，集合 $\\mathbb{N}_r = \\{r + 2as : a \\ge 0\\}$ 中，超過 $N$ 的無偶數表示的數至多一個。故 $\\mathbb{N}$ 中無偶數表示的數有限。\n\n由 1-3，$N$ 可取得使 $n > N$ 有唯一奇數和唯一偶數表示。特別地，$S$ 中每個 $s > N$ 都有偶數表示。\n\n4. **對任意 $s, t \\in S$ 且 $N < s < t$，$t$ 的偶數表示必含 $s$。**\n\n*證明*：否則 $s+t$ 有兩個奇數表示（分別由 $s$ 的偶數表示加 $t$，和 $t$ 的偶數表示加 $s$），矛盾。\n\n設 $s_1 < s_2 < \\dots$ 為 $S$ 的所有元素，$\\sigma_n = \\sum_{i=1}^n s_i$。取 $k$ 使 $s_k > N$。由性質 4，對每個 $i > k$，$s_i$ 的偶數表示含 $s_k, s_{k+1}, \\dots, s_{i-1}$，即\n\n$$\ns_i = s_k + s_{k+1} + \\dots + s_{i-1} + R_i = \\sigma_{i-1} - \\sigma_{k-1} + R_i,\n$$\n\n其中 $R_i$ 為 $s_1, \\dots, s_{k-1}$ 的某些和，$0 \\le R_i \\le \\sigma_{k-1}$。\n\n取 $j_0 > k$ 使 $\\sigma_{j_0} > 2\\sigma_{k-1}$。則對 $j > j_0$，\n\n$$\ns_{j+1} \\ge \\sigma_j - \\sigma_{k-1} > \\sigma_j/2.\n$$\n\n取 $p > j_0$ 使 $R_p = \\min_{i>j_0} R_i$，則\n\n$$\ns_{p+1} = s_k + \\dots + s_p + R_{p+1} = (s_p - R_p) + s_p + R_{p+1} \\ge 2s_p.\n$$\n\n故 $S$ 中無元素介於 $s_p$ 和 $2s_p$ 之間。$2s_p$ 的偶數表示不含大於 $s_p$ 的元素。又 $2s_p > s_1 + \\dots + s_{p-1}$，故偶數表示必含 $s_p$。去掉 $s_p$，得 $s_p$ 有不含 $s_p$ 的奇數表示，與性質 1 矛盾。\n\n因此，存在無窮多個不乾淨的正整數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18367,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,\n\n$$\nf^2(x + y) = f^2(x) + 2f(xy) + f^2(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$, $f(x) = -2$, $f(x) = x$, and $f(x) = x - 2$.\n\nLet us denote the given equation by $(*)$ and substitute $y = -x$:\n\n$$\nf^2(x) + f^2(-x) + 2f(-x^2) = f^2(0) \\tag{**}\n$$\n\nSubstituting $x = x + y$, $y = -x$ into $(*)$ gives:\n\n$$\nf^2(y) = f^2(x + y) + f^2(-x) + 2f(-x^2 - xy).\n$$\n\nAdding $(*)$ and $(**)$ to the above, we get:\n\n$$\nf(xy) + f(-x^2 - xy) = f(-x^2) - \\frac{f^2(0)}{2}.\n$$\n\nLet $g(x) = f(x) + \\frac{f^2(0)}{2}$. Let $a = -x^2 \\le 0$, $b = xy \\in \\mathbb{R}$ (when $a \\ne 0$). Then:\n\n$$\ng(a) + g(b) = g(a + b) \\tag{***}\n$$\n\nIf $a, b < 0$, this is Cauchy's equation. We show $g(x)$ is bounded above for $x < 0$, so $g(x) = kx$ for $x < 0$ and some real $k$. From $(**)$:\n\n$$\n2f(-x^2) = f^2(0) - f^2(x) - f^2(-x) \\leq f^2(0),\n$$\n\nso\n\n$$\ng(a) = f(a) + \\frac{f^2(0)}{2} \\leq f^2(0) \\text{ when } a \\leq 0.\n$$\n\nThus, $g(x) = kx$ for all $x < 0$. For $x \\geq 0$, substituting $y = -1 - x$ into $(***)$ gives $g(x) = g(-1) - g(-1 - x) = kx$ for $x > 0$. Therefore,\n\n$$\nf(x) = g(x) - \\frac{f^2(0)}{2} = kx + c\n$$\n\nfor all real $x$. Substitute this into $(*)$:\n\n$$\n\\begin{align*}\nk^2 (x + y)^2 + 2kc(x + y) + c^2 &= k^2 x^2 + 2kcx + c^2 + 2kxy + 2c + k^2 y^2 + 2kcy + c^2, \\\\\n(k^2 - k)xy &= c^2 + 2c.\n\\end{align*}\n$$\n\nSince $xy$ is arbitrary, $k \\in \\{0,1\\}$ and $c \\in \\{0, -2\\}$.\n\nThus, the only solutions are $f(x) = 0$, $f(x) = -2$, $f(x) = x$, and $f(x) = x - 2$. All these functions satisfy the original condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18368,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be rational numbers such that $a + b = a^2 + b^2$. Suppose that the common value\n\n$s = a + b = a^2 + b^2$\n\nis not an integer, and write it as an irreducible fraction: $s = \\frac{m}{n}$. Let $p$ be the least prime divisor of $n$. Find the minimum value of $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value of $p$ is $p = 5$.\n\nWrite $a$ and $b$ as fractions with least common denominator $w$: $a = \\frac{u}{w}$, $b = \\frac{v}{w}$. If $a = \\frac{u'}{w'}$, $b = \\frac{v'}{w'}$ is another representation with common denominator $w'$, then $w' \\geq w$. The irreducible representation $s = \\frac{m}{n}$ is obtained from $s = \\frac{u + v}{w}$ by possible cancellation. Therefore, the prime divisors of $n$ are among those of $w$.\n\nWe show that $w$ is not divisible by $2$ or $3$, implying that neither is $n$.\n\nThe condition $a + b = a^2 + b^2$ gives $u^2 + v^2 = w(u + v)$. Suppose that $3$ divides $w$. Then $u^2 + v^2$ is a multiple of $3$, and since $x^2 \\equiv 0, 1 \\pmod{3}$ for each integer $x$, it follows that both $u$ and $v$ are divisible by $3$. However, then $3$ is a common divisor of $u, v,$ and $w$, which contradicts the minimality of $w$.\n\nSimilarly, suppose that $w$ is even. Then $u^2 + v^2$ is even, hence so is $u + v$ (since $u^2 + v^2$ has the same parity as $u + v$). Hence $u^2 + v^2$ is divisible by $4$, and since $x^2 \\equiv 0, 1 \\pmod{4}$ for each integer $x$, both $u$ and $v$ are even. We reach a contradiction with the minimality of $w$ again.\n\nBy the above, each prime divisor of $n$ is at least $5$. For an example with $p = 5$, let $a = \\frac{2}{5}$, $b = \\frac{6}{5}$.\n\nThen $a + b = \\frac{8}{5}$, $a^2 + b^2 = \\frac{4}{25} + \\frac{36}{25} = \\frac{40}{25} = \\frac{8}{5}$. So $a + b = a^2 + b^2$ holds, the common value $s$ is not an integer, and its representation $s = \\frac{8}{5}$ is irreducible with $p = n = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18369,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all odd natural numbers $n$ such that $d(n)$ is the largest divisor of $n$ different from $n$, where $d(n)$ is the number of divisors of $n$ (including $1$ and $n$).",
    "options": [],
    "answer": "See solution",
    "solution": "From $d(n) \\mid n$, $\\frac{n}{d(n)} \\mid n$, so $\\frac{n}{d(n)} \\leq d(n)$.\n\nLet $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_s^{\\alpha_s}$, where $p_i > 2$ for all $1 \\leq i \\leq s$ since $n$ is odd. The function $d(n)$ is multiplicative, so $d(n) = (1 + \\alpha_1) \\dots (1 + \\alpha_s)$.\n\nBy Bernoulli's inequality, for every $p_i > 3$ in the factorization of $n$:\n\n$$\np_i^{\\frac{\\alpha_i}{2}} = (1 + (p_i - 1))^{\\frac{\\alpha_i}{2}} \\geq 1 + \\frac{\\alpha_i}{2}(p_i - 1) > 1 + \\alpha_i\n$$\n\nand for $p_i = 3$:\n\n$$\n3^{\\frac{\\beta}{2}} \\geq 1 + \\beta\n$$\n\nEquality holds when $\\beta = 0$ or $\\beta = 2$, and strict inequality for $\\beta > 2$. If $\\beta = 1$ and there is no other prime in the factorization, then $n = 3$, $d(n) = 2$, which is not a solution. If $\\beta = 1$ and there is another prime, then $n = 3p_2^{\\alpha_2} \\dots p_s^{\\alpha_s} > 4(1 + \\alpha_2)^2 \\dots (1 + \\alpha_s)^2$. If there is $p_i \\geq 7$ in the factorization, then $3p_i^{\\alpha_i} > 4(1 + \\alpha_i)^2$ for all $\\alpha_i$. If the power of $5$ in the factorization is greater than $1$, then $3 \\cdot 5^{\\alpha_i} > 4(1 + \\alpha_i)^2$. If the power of $5$ is $1$, then $n = 3 \\cdot 5 = 15$, $d(n) = 4$, which is not a solution. Thus, $\\beta \\neq 1$.\n\nFinally, we obtain $\\sqrt{n} \\geq d(n)$, and equality holds only when there is no $p_i > 3$ in the factorization of $n$, i.e., $n = 3^2 = 9$.\n\nTherefore, the only solution is $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18370,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a circle and $\\omega$ be a circle touching $\\Gamma$ internally at point $P$. Assume that $AB$ is a chord of $\\Gamma$ touching $\\omega$ at $C$. Show that ray $PC$ is the internal angle bisector of $\\angle APB$.",
    "options": [],
    "answer": "See solution",
    "solution": "We draw the common tangent $l$ of $\\Gamma$ and $\\omega$ at $P$.\n\nIf $l$ and $AB$ are parallel, the triangle $APB$ is isosceles, and the line $PC$ is both a perpendicular and angle bisector in the triangle.\n\nIf $l$ and $AB$ are not parallel, let them intersect in $R$. We may assume, without loss of generality, that $B$ lies between $A$ and $R$ as depicted in the figure below. Since $RC$ and $RP$ are tangents to $\\omega$, the triangle $CRP$ is isosceles with $\\angle RCP = \\angle CPR$.\n\nNow,\n\n$$\n\\angle CPB + \\angle BPR = \\angle CPR = \\angle RCP = \\angle APC + \\angle BAP.\n$$\n\nSince $RP$ is a tangent to $\\Gamma$, we have $\\angle BPR = \\angle BAP$.\n\nTherefore, $\\angle CPB = \\angle APC$, so $PC$ is the angle bisector of $\\angle APB$, as required.\n\n![](images/BW2021_Shortlist_p25_data_69223a15f5.png)\n\nFigure 10",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18371,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an arbitrary arrangement of boys along a circle. Moving clockwise, assign a \"+\" sign before a boy if he is taller than the previous boy, and a \"-\" sign if he is shorter than the previous one. A boy is called *middle* if both the signs before and after him are either both \"+\" or both \"-\". Prove that in any arrangement, there is at least one middle boy.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there is no middle boy. Then the signs \"+\" and \"-\" must alternate around the circle. This means the total number of boys (and thus signs) is even. However, if the number of boys is odd, this is impossible. Therefore, there must be at least one middle boy in any arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18372,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a polynomial with real coefficients and $n \\ge 1$ be an integer. Prove that there exists a non-zero polynomial $q$ such that the coefficients of $p \\cdot q$ vanish for each power that is not a multiple of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $p$ is a real polynomial and $n$ is a strictly positive integer.\n\nLet $x$ be the variable. We want to find a real non-zero polynomial $q$ such that $p(x) \\cdot q(x) = s(x^n)$ for some real polynomial $s$. If $p$ is the zero polynomial then $p \\cdot q = 0$ for every polynomial $q$. It can therefore be assumed that $p \\neq 0$.\n\nLet $y = x^n$. Then $y^m$ is a real polynomial for each $m \\in \\mathbb{N}$. For each $m \\in \\mathbb{N}$ let $r_m(x)$ be the remainder of the polynomial division of $y^m = (x^n)^m$ by $p(x)$. Then each $r_m$ is of degree less than $\\deg(p)$, the degree of $p$. Consider the polynomials $r_0, r_1, \\dots, r_{\\deg(p)}$. We want to find coefficients $s_0, s_1, \\dots, s_{\\deg(p)}$ such that $s_0 \\cdot r_0(x) + s_1 \\cdot r_1(x) + \\dots + s_{\\deg(p)} \\cdot r_{\\deg(p)}(x) = 0$. By considering the coefficients this is equivalent to a system with $\\deg(p)$ linear equations and $\\deg(p) + 1$ unknowns. As there are more unknowns than equations, it follows that there exists a solution $(s_0, s_1, \\dots, s_{\\deg(p)}) \\neq (0, 0, \\dots, 0)$.\n\nTake a solution $(s_0, s_1, \\dots, s_{\\deg(p)})$ to the system of linear equations and let $s(x) = s_0 + s_1 x + \\dots + s_{\\deg(p)} x^{\\deg(p)}$. This is a non-zero polynomial. As $r_m(x)$ is the remainder of the polynomial division of $y^m$ by $p(x)$, it follows that $p(x)$ divides $y^m - r_m(x)$ for all $m \\in \\mathbb{N}$. Hence $p(x)$ divides\n\n$$\ns_0 (y^0 - r_0(x)) + s_1 (y^1 - r_1(x)) + \\dots + s_{\\deg(p)} (y^{\\deg(p)} - r_{\\deg(p)}(x)) = s(y) - (s_0 r_0(x) + s_1 r_1(x) + \\dots + s_{\\deg(p)} r_{\\deg(p)}(x)) = s(y)\n$$\n\nIt follows that $q(x) = s(y)/p(x)$ is a polynomial. As $s$ is non-zero it follows that $q$ is non-zero as well. We have therefore found a non-zero polynomial $q$ such that $p(x) \\cdot q(x) = s(x^n)$ as desired. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18373,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AA_0$, $BB_0$, and $CC_0$ be the angle bisectors of $\\triangle ABC$. Let $A_0A_1 \\parallel BB_0$ and $A_0A_2 \\parallel CC_0$, where $A_1$ and $A_2$ lie on $AC$ and $AB$, respectively. Let the line $A_1A_2$ intersect $BC$ at $A_3$. The points $B_3$ and $C_3$ are obtained similarly. Prove that the points $A_3$, $B_3$, and $C_3$ are collinear.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p295_data_d6e17c121f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By the Menelaus Inverse Theorem, we need only to show that\n\n$$\n\\frac{AB_3}{B_3C} \\cdot \\frac{CA_3}{A_3B} \\cdot \\frac{BC_3}{C_3A} = 1. \\qquad \\textcircled{1}\n$$\n\nSince line $A_1A_2A_3$ intersects $\\triangle ABC$, by Menelaus' Theorem, we have\n$$\n\\frac{CA_3}{A_3B} \\cdot \\frac{BA_2}{A_2A} \\cdot \\frac{AA_1}{A_1C} = 1.\n$$\nSo\n$$\n\\frac{CA_3}{A_3B} = \\frac{A_2A}{BA_2} \\cdot \\frac{A_1C}{AA_1}. \\qquad \\textcircled{2}\n$$\nSimilarly,\n$$\n\\frac{AB_3}{B_3C} = \\frac{B_2B}{CB_2} \\cdot \\frac{B_1A}{BB_1}, \\qquad \\textcircled{3}\n$$\n$$\n\\frac{BC_3}{C_3A} = \\frac{C_2C}{AC_2} \\cdot \\frac{C_1B}{CC_1}. \\qquad \\textcircled{4}\n$$\nBy $BA_2 = \\frac{BC_0}{BC} \\cdot BA_0$ and $AA_2 = \\frac{AA_0}{AI} \\cdot AC_0$, we have\n$$\n\\frac{AA_2}{BA_2} = \\frac{AA_0}{BA_0} \\cdot \\frac{AC_0}{BC_0} \\cdot \\frac{BC}{AI}. \\qquad \\textcircled{5}\n$$\nMoreover, by $AA_1 = \\frac{AA_0}{AI} \\cdot AB_0$ and $CA_1 = \\frac{CA_0}{CB} \\cdot CB_0$, we have\n$$\n\\frac{A_1C}{AA_1} = \\frac{CA_0 \\cdot CB_0}{AA_0 \\cdot AB_0} \\cdot \\frac{AI}{BC}. \\qquad \\textcircled{6}\n$$\nThen by ②, ⑤, and ⑥,\n$$\n\\frac{CA_3}{A_3B} = \\frac{CA_0}{BA_0} \\cdot \\frac{AC_0}{BC_0} \\cdot \\frac{CB_0}{AB_0} = \\left( \\frac{CA_0}{A_0B} \\right)^2.\n$$\nSimilarly,\n$$\n\\frac{CB_3}{B_3A} = \\left( \\frac{CB_0}{B_0A} \\right)^2,\n$$\n$$\n\\frac{AC_{\\overline{3}}}{C_{\\overline{3}}B} = \\left( \\frac{AC_{\\overline{0}}}{C_{\\overline{0}}B} \\right)^2. \\qquad \\textcircled{7}\n$$\nSince the three angle bisectors $AA_0$, $BB_0$, $CC_0$ of $\\triangle ABC$ are concurrent, by Ceva's Theorem,\n$$\n\\frac{AB_0}{B_0C} \\cdot \\frac{CA_0}{A_0B} \\cdot \\frac{BC_0}{C_0A} = 1. \\qquad \\textcircled{8}\n$$\nThus, by ⑦ and ⑧,\n$$\n\\frac{AB_{\\overline{3}}}{B_{\\overline{3}}C} \\cdot \\frac{CA_{\\overline{3}}}{A_{\\overline{3}}B} \\cdot \\frac{BC_{\\overline{3}}}{C_{\\overline{3}}A} = \\left( \\frac{AB_{\\overline{0}}}{B_{\\overline{0}}C} \\cdot \\frac{CA_{\\overline{0}}}{A_{\\overline{0}}B} \\cdot \\frac{BC_{\\overline{0}}}{C_{\\overline{0}}A} \\right)^2 = 1,\n$$\nthat is, $\\textcircled{1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18374,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 cards numbered from 1 to 100 on the table. Andriy and Nick each take the same number of cards such that if Andriy has a card with number $n$, then Nick has a card with number $2n + 2$. What is the maximal total number of cards that could be taken by the two guys?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the total number of cards taken cannot exceed 66, meaning each can take at most 33 cards.\n\nSince $2n + 2 \\leq 100$, we have $2n \\leq 98$, so $n \\leq 49$. Thus, all of Andriy's numbers are in $\\{1, 2, \\ldots, 49\\}$.\n\nWe partition this set into 33 disjoint subsets:\n\n- $\\{1, 4\\}, \\{3, 8\\}, \\{5, 12\\}, \\ldots, \\{23, 48\\}$ (12 subsets)\n- $\\{2, 6\\}, \\{10, 22\\}, \\{14, 30\\}, \\{18, 38\\}$ (4 subsets)\n- $\\{25\\}, \\{27\\}, \\ldots, 49$ (13 singletons)\n- $\\{26\\}, \\{34\\}, \\{42\\}, \\{46\\}$ (4 singletons)\n\nThere are 33 subsets in total, and no two subsets share an element. By the pigeonhole principle, if Andriy takes at least 34 cards, at least two must come from the same subset, which is not allowed.\n\nAn example achieving 33 cards: Andriy takes\n\n$A = \\{1, 3, 5, \\ldots, 23, 2, 10, 14, 18, 25, 27, 29, \\ldots, 49, 26, 34, 42, 46\\}$.\n\nNick takes the corresponding cards $N = \\{2n + 2 \\mid n \\in A\\}$.\n\n**Answer:** 66 cards.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18375,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Given a convex quadrilateral $ABCD$ with $AD < AB$ and $CD < CB$, is the internal angle at $B$ always less than the internal angle at $D$?\n\n(b) The same question for a non-convex quadrilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes; b) No.\n\n(a) Consider the triangles $ADB$ and $CDB$. The claim $AD < AB$ implies $\\angle ABD < \\angle ADB$ because the longer side is opposite the larger angle. Similarly, $CD < CB$ implies $\\angle CBD < \\angle CDB$. As $ABCD$ is convex, $\\angle ABD + \\angle CBD = \\angle ABC$ and $\\angle ADB + \\angle CDB = \\angle ADC$. Hence, adding the two inequalities gives $\\angle ABC < \\angle ADC$.\n\n![](images/EST_ABooklet_2020_p22_data_1e42aa67de.png)\n\n(b) Let points $A, B, C$ be such that $AB = BC > AC$. Choose point $D'$ on the line tangent to the circumcircle of triangle $ABC$ at $A$ so that $B$ and $D'$ lie on the same side of $AC$ and the inequalities $AD' < AB$ and $CD' < BC$ hold (the last inequality is possible since $AC < BC$); let $D$ be the reflection of $D'$ over $AC$. Then both assumptions $AD < AB$ and $CD < CB$ hold. But the claim $\\angle ABC < \\angle ADC$ is not true: since $D'$ lies outside the circumference of triangle $ABC$, we have $\\angle ABC > \\angle AD'C = \\angle ADC$. Hence, the hypothesis does not hold for non-convex quadrilaterals.\n\n![](images/EST_ABooklet_2020_p22_data_4b788ec29d.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18376,
    "subject": "Mathematics (Olympiad)",
    "question": "A strip of width $w$ is the set of all points which lie on, or between, two parallel lines distance $w$ apart. Let $S$ be a set of $n$ ($n \\ge 3$) points on the plane such that any three different points of $S$ can be covered by a strip of width $1$.\n\nProve that $S$ can be covered by a strip of width $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following statement.\n\n**Lemma.** If a triangle can be covered by a strip of breadth $b$, then at least one altitude of the triangle is at most $b$ long.\n\n**Proof.** At least one of the perpendicular lines through the vertices of the triangle to the border lines of the strip meets the opposite side of the triangle. Therefore, the segment between that vertex and the meeting point with the opposite side is of length at most $b$. The altitude corresponding to that vertex is thus also of length at most $b$. The lemma is proved.\n\nAs a corollary, the least breadth of a strip that can cover a triangle is equal to the length of its shortest altitude.\n\nNow, choose points $A$ and $B$ from $S$ at maximal distance from each other. For any other point $C$ from $S$, the side $AB$ will be the longest of the triangle $ABC$. Therefore, the altitude from $C$ to $AB$ will be the shortest. According to the lemma, it is at most $1$ long, since the triangle $ABC$ can be covered by a strip of breadth $1$ by hypothesis.\n\nHence, $S$ can be covered by a strip of breadth $2$ with borders parallel to $AB$, at distance $1$ on both sides of $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18377,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest positive integer that can be multiplied by $2016$ to make the product a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "The prime factorisation of $2016$ is $2^5 \\times 3^2 \\times 7$. In a perfect square, each prime factor must have an even exponent. The exponents of $2$ and $7$ in $2016$ are odd, so we need to multiply by one more $2$ and one more $7$ to make their exponents even. Thus, the smallest integer is $2 \\times 7 = 14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18378,
    "subject": "Mathematics (Olympiad)",
    "question": "A *pointy number* of $n$ digits is a number whose digits increase by 1 up to a central digit, then decrease by 1 symmetrically. For example, a 7-digit pointy number has the form:\n\n$$\na,\\ a \\pm 1,\\ a \\pm 2,\\ a \\pm 3,\\ a \\pm 2,\\ a \\pm 1,\\ a.\n$$\n\n**(c)** Show that the sum of an upward and a downward $n$-digit pointy number is never prime.\n\n**(d)** How many pairs of 7-digit pointy numbers (one upward, one downward) have a palindromic sum?",
    "options": [],
    "answer": "See solution",
    "solution": "For part **(c)**:\n\nEach pair of same-placed digits in the upward and downward pointy numbers has the same sum, call it $n$. Thus, the sum of the two numbers is:\n\n$$\nn + 10n + 100n + \\dots = n(1 + 10 + 100 + \\dots).\n$$\n\nSince the first digit of each pointy number is at least 1, $n \\geq 2$, and the sum in brackets is at least 111 (for at least 3 digits). Therefore, the sum is composite and never prime.\n\nFor part **(d)**:\n\nLet $n$ be the common sum of corresponding digits in the 7-digit upward and downward pointy numbers. The first digit of a 7-digit downward pointy number is at least 3, and for an upward pointy number at least 1, so $n \\geq 4$.\n\nIf the sum is a 7-digit palindrome, it has the form $nnnnnnn$.\n\n- For $n = 4$, there is 1 pair: $\\{1234321,\\ 3210123\\}$.\n- For $n = 5$, there are 2 pairs: $\\{1234321,\\ 4321234\\}$ and $\\{2345432,\\ 3210123\\}$.\n- For $n = 6, 7, 8, 9$, there are 3, 4, 5, 6 pairs respectively.\n\nIf the sum is an 8-digit palindrome starting and ending with 1 ($n = 11$), there are 5 pairs:\n\n$$\n\\{2345432,\\ 9876789\\},\\ \\{3456543,\\ 8765678\\},\\ \\{4567654,\\ 7654567\\},\\ \\{5678765,\\ 6543456\\},\\ \\{6789876,\\ 5432345\\}.\n$$\n\nThus, the total number of required pairs is:\n\n$$1 + 2 + 3 + 4 + 5 + 6 + 5 = 26.$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18379,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle satisfying $2AC = AB + BC$. If $O$ and $I$ are its circumcenter and incenter, show that $\\angle OIB = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D = BI \\cap (ABC)$. We apply Ptolemy's theorem:\n$$\nAB \\cdot DC + BC \\cdot AD = AC \\cdot BD\n$$\nwhich implies $BD = 2DA$. From $AD = DI = DC$, it follows that $I$ is the midpoint of $BD$, so $\\angle OIB = 90^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18380,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be a set of 2017 positive integers. For every non-empty $A \\subset M$ we define\n\n$$\nf(A) = \\{x \\in M : x \\text{ is divisible by an odd number of elements of } A\\}.\n$$\n\nFind the minimum number of colors such that it is possible to paint all nonempty subsets of $M$ in such a way that whenever $A \\neq f(A)$, the sets $A$ and $f(A)$ are in different colors.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that the function $f$ is injective, i.e., $A \\neq B$ implies $f(A) \\neq f(B)$. Let $a$ be the smallest number which belongs to exactly one of the sets $A$ and $B$. Assume $a \\in A$, $a \\notin B$. Let $C = \\{b_1, b_2, \\dots, b_m\\}$ be the set (possibly empty) of numbers from $B$ which divide $a$. By the definition of $a$, the numbers from $A$ which divide $a$ are exactly $a$ and the numbers from $C$. This means that $a$ belongs to exactly one of the sets $f(A)$ and $f(B)$, i.e., $f(A) \\neq f(B)$.\n\nNow consider the directed graph $G$ with vertices the nonempty subsets of $M$ and edges $(A, f(A))$ (from $A$ to $f(A)$) iff $A \\neq f(A)$. From above, every vertex of $G$ is either isolated or is the tail and head of exactly one edge. Therefore, $G$ can be partitioned into cycles.\n\nWe will prove that all cycles in $G$ have even length, so two colors are enough. Let $(A_1, A_2, \\dots, A_m, A_1)$ be a cycle of length $m \\ge 2$. Let $\\{a_1, a_2, \\dots, a_k\\} = \\bigcup_{i=1}^m A_i$, where $a_1 < a_2 < \\dots < a_k$. Clearly, $a_1 \\in A_i$ for every $i = 1, 2, \\dots, m$. Let $t$ be the smallest number such that there exists $i$ with $a_t \\notin A_i$. There exists $j$ such that $a_t \\notin A_j$, but $a_t \\in A_{j+1}$. This implies that $a_t$ has an odd number of divisors among $a_1, \\dots, a_{t-1}$. Then $a_t \\in A_{j-1}$, and so on, i.e., the $a_t$ alternate, so the cycle's length is even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18381,
    "subject": "Mathematics (Olympiad)",
    "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 $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that $PA^2 = PB \\cdot PC$. By Stewart's relation, $$PA^2 \\cdot BC \\mp AB^2 \\cdot PC \\pm AC^2 \\cdot PB = PB \\cdot PC \\cdot BC,$$ this amounts to showing $PB \\cdot AC^2 = PC \\cdot AB^2$.\n\n![](images/RMC_2015_BT_p98_data_5ba0a5fc07.png)\n\nTo begin, apply Menelaus' theorem to triangles $ABD_2$, $ACD_1$, $ABC$, to write\n\n$$\n\\frac{NB}{NA} \\cdot \\frac{CD_2}{CB} \\cdot \\frac{E_2A}{E_2D_2} = 1, \\quad \\frac{MA}{MC} \\cdot \\frac{E_1D_1}{E_1A} \\cdot \\frac{BC}{BD_1} = 1, \\quad \\frac{MC}{MA} \\cdot \\frac{NA}{NB} \\cdot \\frac{PB}{PC} = 1,\n$$\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, $\\frac{PB}{PC} = \\frac{BD_1}{CD_2}$, by (*).\n\nFinally, similarity of the triangles $ABC$ and $D_1BA$ yields $BD_1 = \\frac{AB^2}{BC}$. Similarly, $CD_2 = \\frac{AC^2}{BC}$, so $PB \\cdot AC^2 = PC \\cdot AB^2$, by the preceding, q.e.d.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18382,
    "subject": "Mathematics (Olympiad)",
    "question": "A grasshopper jumps on the plane from an integer point (a point with all integer coordinates) to another integer point according to the following rules:\n\n- His first jump is of length $\\sqrt{98}$.\n- His second jump is of length $\\sqrt{149}$.\n- His next jump is of length $\\sqrt{98}$, and so on, alternating between these two lengths.\n\nWhat is the least possible odd number of moves in which the grasshopper could return to his starting point?",
    "options": [],
    "answer": "See solution",
    "solution": "Since the only representations of $98$ and $149$ as sums of two squares are $7^2 + 7^2$ and $7^2 + 10^2$, we conclude that every odd move of the grasshopper is of the form $(x, y) \\rightarrow (x \\pm 7, y \\pm 7)$ and every even move is of the form $(x, y) \\rightarrow (x \\pm 7, y \\pm 10)$ or $(x \\pm 10, y \\pm 7)$.\n\nLet the starting point be $(0, 0)$. We need the grasshopper to get, in an even number of moves, to some of the points $(\\pm 7, \\pm 7)$, for example to $(7, 7)$.\n\nAfter any pair of two consecutive moves, the grasshopper gets from the point $(a, b)$ to the point $(c, d)$, where $c \\in \\{a, a \\pm 14\\}$, $d \\in \\{b \\pm 17, b \\pm 3\\}$ or $c \\in \\{a \\pm 17, a \\pm 3\\}$, $d \\in \\{b, b \\pm 14\\}$. This means that after any pair of consecutive moves, one of the coordinates remains the same modulo $14$, and the other changes by $3$ modulo $14$. This implies that each coordinate may obtain a value equivalent to $7$ modulo $14$, in particular precisely the value $7$, only after at least $7$ pairs of moves, e.g. $0 + 3 + 3 + 3 + 3 + 3 + 3 \\equiv 7 \\pmod{14}$.\n\nTo obtain a similar result for the other coordinate, one needs at least another $7$ pairs of moves. Therefore, one needs at least $2 \\times 14 = 28$ moves to get to the point $(7, 7)$, and in total at least $28 + 1 = 29$ moves to return to the initial point $(0, 0)$.\n\nAn example of such $29$ moves is the following: all $15$ odd moves are as $(x-7, y-7)$, and $14$ even moves consist of six moves as $(x+10, y+7)$, six moves as $(x+7, y+10)$, one move $(x+10, y-7)$, and one move $(x-7, y+10)$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18383,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and $\\omega$ its circumcircle. Let $I$, $J$, and $K$ be the incenters of the triangles $ABC$, $ACD$, and $ABD$, respectively. Also, let $E$ be the midpoint of the arc $DB$ of $\\omega$ containing $A$. The line $EK$ intersects $\\omega$ at $F$ ($F \\neq E$). Prove that the points $C$, $F$, $I$, and $J$ lie on the same circle.\n\n![](images/Cesko-Slovacko-Poljsko_2012_p3_data_38865770b4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ and $N$ be the midpoints of the arcs $AB$ and $AD$ (not containing any other vertices of $ABCD$), respectively. Then the incenter $I$ lies on $CM$, the incenter $J$ lies on $CN$, and the incenter $K$ lies on $BN$. Moreover, we have\n\n$$\nMI = MA = MB \\quad \\text{and} \\quad NJ = NA = ND = NK \\quad (1)\n$$\n\n(these well-known relations follow from an easy angle chasing).\n\nObserve also that the incenter $K$ necessarily lies in the interior of the isosceles triangle $BDE$, hence the line $EK$ intersects the segment $BD$ and the point $F$ lies on the same side of $BD$ as $C$ (see the figure).\n\nConsider now the arc $BD$ of $\\omega$ containing $A$. The point $E$ is the midpoint of this arc, and the points $M$ and $N$ are the midpoints of the subarcs $BA$ and $AD$, respectively. It follows that the subarcs $BM$ and $EN$ are of equal length (and similarly the subarcs $ME$ and $ND$ are of equal length). Thus\n\n$$\n\\angle BFM = \\angle EFN = \\angle KFN. \\quad (2)\n$$\n\nBut clearly we have\n\n$$\n\\angle BMF = \\angle BNF = \\angle KNF. \\qquad (3)\n$$\n\nThe equalities (2) and (3) imply that the triangles $MBF$ and $NKF$ are similar, and with the same orientation. Therefore\n\n$$\n\\frac{MB}{MF} = \\frac{NK}{NF},\n$$\n\nwhich by an application of (1) can be rewritten as\n\n$$\n\\frac{MI}{MF} = \\frac{NJ}{NF}.\n$$\n\nThis, together with the relation\n\n$$\n\\angle IMF = \\angle CMF = \\angle CNF = \\angle JNF,\n$$\n\nproves that the triangles $MIF$ and $NJF$ are similar with the same orientation. This orientation-preserving similarity implies that\n\n$$\n\\angle IFM = \\angle JFN \\quad \\text{and} \\quad \\angle IFJ = \\angle MFN.\n$$\n\nHence\n\n$$\n\\angle IFJ = \\angle MFN = \\angle MCN = \\angle ICJ,\n$$\n\nand the points $C$, $F$, $I$, $J$ are indeed concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18384,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the four colours are blue, red, yellow, and green. Is it possible to assign one of these four colours to every point in the plane so that no two points at distance $1$ or $\\sqrt{3}$ from each other have the same colour?\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p129_data_04f6f1e1a5.png)\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p129_data_e8aa5b60cd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We argue by contradiction and suppose that there do not exist two points at distance $1$ or $\\sqrt{3}$ from each other that have the same colour.\n\nConsider an isosceles triangle $ABC$ with $BC = 1$ and $AB = AC = 2$. Since $B$ and $C$ must be different colours, one of them is coloured differently to $A$. Without loss of generality, suppose $A$ is blue and $B$ is red.\n\nLet us orient the plane so that $AB$ is a horizontal segment.\n\nLet $O$ be the midpoint of $AB$. Then as $AO = BO = 1$, it follows that $O$ is not blue or red. Without loss of generality, suppose $O$ is green.\n\nLet $X$ be the point above the line $AB$ such that $\\triangle AOX$ is equilateral. It is easy to compute that $XB = \\sqrt{3}$ and $XA = XO = 1$. Hence, $X$ is not red, blue, or green. Thus $X$ must be yellow.\n\nFinally, let $Y$ be the point above the line $AB$ such that $\\triangle BOY$ is equilateral. Then $YX = YO = YB = 1$ and $YA = \\sqrt{3}$. Hence $Y$ cannot be any of the four colours, giving the desired contradiction.\n\nThus, it is impossible to assign four colours to every point in the plane so that no two points at distance $1$ or $\\sqrt{3}$ from each other have the same colour.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18385,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be $r$ red sweets and $g$ green sweets, with $r \\ge 2$. Let $n = r + g$.\n\nThe probability of selecting two red sweets if the first sweet is put back is\n\n$$\n\\frac{r}{n} \\times \\frac{r}{n}\n$$\n\nand the probability if Dan eats the first sweet before selecting the second is\n\n$$\n\\frac{r}{n} \\times \\frac{r-1}{n-1}.\n$$\n\nIf the first probability is 105% of the second, what is the largest possible number of sweets in the jar?",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing the probabilities and rearranging gives:\n\n$$\n\\frac{r}{n} \\times \\frac{n-1}{r-1} = \\frac{105}{100} = \\frac{21}{20}\n$$\n\nSo,\n\n$$\n20r(n-1) = 21n(r-1)\n$$\n\nExpanding and simplifying:\n\n$$\n21n - nr - 20r = 0 \\\\\n(n+20)(21-r) = 420\n$$\n\nSince $n + 20$ is positive, $21 - r$ is positive. $n$ is largest when $21 - r = 1$, so $n + 20 = 420$, giving $n = 400$.\n\nSo the largest number of sweets in the jar is **400**.\n\nSince $21 - r$ is a factor of 420 and $2 \\le r \\le 20$, the following table gives all possible values of $r, n, g$.\n\n![]( <table><thead><tr><th>21 - r</th><th>n + 20</th><th>r</th><th>n</th><th>g</th></tr></thead><tbody><tr><td>1</td><td>420</td><td>20</td><td>400</td><td>380</td></tr><tr><td>2</td><td>210</td><td>19</td><td>190</td><td>171</td></tr><tr><td>3</td><td>140</td><td>18</td><td>120</td><td>102</td></tr><tr><td>4</td><td>105</td><td>17</td><td>85</td><td>68</td></tr><tr><td>5</td><td>84</td><td>16</td><td>64</td><td>48</td></tr><tr><td>6</td><td>70</td><td>15</td><td>50</td><td>35</td></tr><tr><td>7</td><td>60</td><td>14</td><td>40</td><td>26</td></tr><tr><td>10</td><td>42</td><td>11</td><td>22</td><td>11</td></tr><tr><td>12</td><td>35</td><td>9</td><td>15</td><td>6</td></tr><tr><td>14</td><td>30</td><td>7</td><td>10</td><td>3</td></tr><tr><td>15</td><td>28</td><td>6</td><td>8</td><td>2</td></tr></tbody></table> )",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18386,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the value of the expression\n\n$$\n\\frac{\\sqrt{n + \\sqrt{0}} + \\sqrt{n + \\sqrt{1}} + \\sqrt{n + \\sqrt{2}} + \\dots + \\sqrt{n + \\sqrt{n^2 - 1}} + \\sqrt{n + \\sqrt{n^2}}}{\\sqrt{n - \\sqrt{0}} + \\sqrt{n - \\sqrt{1}} + \\sqrt{n - \\sqrt{2}} + \\dots + \\sqrt{n - \\sqrt{n^2 - 1}} + \\sqrt{n - \\sqrt{n^2}}}\n$$\nis constant for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For all real numbers $0 \\leq m \\leq n^2$, we have\n\n$$\n\\sqrt{n + \\sqrt{m}} = \\sqrt{\\frac{n + \\sqrt{n^2 - m}}{2}} + \\sqrt{\\frac{n - \\sqrt{n^2 - m}}{2}}\n$$\n\nNow, take a positive integer $n$ and sum over all integers $m$ from $0$ to $n^2$:\n\n$$\n\\begin{aligned}\n\\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}} &= \\sum_{m=0}^{n^2} \\left[ \\sqrt{\\frac{n + \\sqrt{n^2 - m}}{2}} + \\sqrt{\\frac{n - \\sqrt{n^2 - m}}{2}} \\right] \\\\\n&= \\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{n^2 - m}}{2}} + \\sum_{m=0}^{n^2} \\sqrt{\\frac{n - \\sqrt{n^2 - m}}{2}}\n\\end{aligned}\n$$\n\nBy changing variables in the first sum ($k = n^2 - m$), we see that\n\n$$\n\\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{n^2 - m}}{2}} = \\sum_{k=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{k}}{2}} = \\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{m}}{2}}\n$$\n\nSo,\n\n$$\n\\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}} = \\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{m}}{2}} + \\sum_{m=0}^{n^2} \\sqrt{\\frac{n - \\sqrt{m}}{2}}\n$$\n\nLet $S = \\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}}$ and $T = \\sum_{m=0}^{n^2} \\sqrt{n - \\sqrt{m}}$. Then,\n\n$$\nS = \\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{m}}{2}} + \\sum_{m=0}^{n^2} \\sqrt{\\frac{n - \\sqrt{m}}{2}}\n$$\n\nBut the first sum is $S / \\sqrt{2}$ and the second is $T / \\sqrt{2}$, so\n\n$$\nS = \\frac{S + T}{\\sqrt{2}}\n$$\n\nSolving for $S/T$ gives\n\n$$\n\\frac{S}{T} = 1 + \\sqrt{2}\n$$\n\nTherefore, the value of the original expression is $1 + \\sqrt{2}$ for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18387,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A, B \\in \\mathcal{M}_2(\\mathbb{C})$ such that $A^2 + B^2 = 2AB$. Prove that $AB = BA$ and $\\operatorname{tr} A = \\operatorname{tr} B$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we prove that $(AB - BA)^2 = 0$. Define the quadratic function $f(x) = \\det(A^2 + B^2 + x(AB - BA)) = \\det(A^2 + B^2) + mx + x^2 \\det(AB - BA)$. Since $f(-i) = \\det((A + iB)(A - iB))$ and $f(i) = \\det((A - iB)(A + iB))$, we have $f(-i) = f(i)$, so $m = 0$. Also, $f(0) = \\det(A^2 + B^2) = \\det(2AB)$ and $f(-2) = \\det(2BA) = \\det(2AB) = f(0)$, so $f$ is constant and $\\det(AB - BA) = 0$. The characteristic equation for $AB - BA$ (since $\\operatorname{tr}(AB) = \\operatorname{tr}(BA)$) then gives $(AB - BA)^2 = 0$. \n\nSince $AB - BA = (A - B)^2$ by hypothesis, we have $0 = (AB - BA)^2 = (A - B)^4$. As the matrices are $2 \\times 2$, we must have $(A - B)^2 = 0$, so $AB = BA$.\n\nb) From $(A - B)^2 = 0$ and $2 \\det(A - B) = \\det(A - B)$, the characteristic equation for $A - B$ gives $(\\operatorname{tr}(A) - \\operatorname{tr}(B))(A - B) = 0$, which completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18388,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of nonzero integers $ (x, y) $ such that the integer $ x^2 + y^2 $ is a common divisor of the integers $ x^5 + y $ and $ y^5 + x $.",
    "options": [],
    "answer": "See solution",
    "solution": "We have that $x^2 + y^2$ divides both $x^5 + y$ and $y^5 + x$. Therefore, $x^2 + y^2 \\mid x(x^5 + y)$ and $x^2 + y^2 \\mid y(y^5 + x)$. Thus,\n\n$$\nx^2 + y^2 \\mid x(x^5 + y) + y(y^5 + x) \\implies x^2 + y^2 \\mid x^6 + y^6 + 2xy. \\quad (1)\n$$\n\nFrom the identity for the sum of cubes:\n\n$$\nx^2 + y^2 \\mid (x^2)^3 + (y^2)^3. \\quad (2)\n$$\n\nFrom (1) and (2), we get $x^2 + y^2 \\mid 2xy$, so $x^2 + y^2 \\leq 2|xy|$. This implies $(|x| - |y|)^2 \\leq 0$, so $|x| = |y|$.\n\nTherefore, two cases:\n\n(a) If $x = y$, then $2x^2 \\mid x^5 + x \\implies 2x \\mid x^4 + 1 \\implies x \\mid 1$, so $x = 1$ or $x = -1$.\n\n(b) If $x = -y$, then $2x^2 \\mid x^5 - x \\implies 2x \\mid x^4 - 1 \\implies x \\mid 1$, so $x = 1$ or $x = -1$.\n\nHence, the solutions are $$(x, y) \\in \\{(1, 1), (-1, -1), (1, -1), (-1, 1)\\}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18389,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every integer $S \\geq 100$ there exists an integer $P$ for which the following story could hold true:\n\nThe mathematician asks the shop owner: \"How much are the table, the cabinet and the bookshelf?\" The shop owner replies: \"Each item costs a (positive) integer amount of Euros. The table is more expensive than the cabinet, and the cabinet is more expensive than the bookshelf. The sum of the three prices is $S$ and the product is $P$.\" The mathematician thinks and complains: \"This is not enough information to determine the three prices!\"",
    "options": [],
    "answer": "See solution",
    "solution": "Write $S$ in the form $S = 6k + r$ for integers $k$ and $r$ with $1 \\le r \\le 6$, and note that $k > 2r$. We claim that the number\n\n$$\nP = 6k(k - r)(k + r)\n$$\n\nis an appropriate choice.\n\nDenote the prices of table, cabinet and shelf by $x$, $y$ and $z$, respectively. Then $x = 3(k + r)$, $y = 2(k - r)$, $z = k$ is one possibility, and $x' = 3k$, $y' = 2(k + r)$, $z' = k - r$ is another possibility. One easily verifies $x > y > z$ and $x' > y' > z'$, and $x' < x$ implies that these two possibilities are indeed distinct. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18390,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly 200 distinct points.\n\n(Note that for 3 distinct points, the minimum number of lines is 3 and for 4 distinct points, the minimum is 4.)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\binom{20}{2} = 190$, the answer is $21$. For a full solution, see Senior Section Q2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18391,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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\n$$\nx = \\sum_{h \\in H} q(x, h) \\cdot h, \\qquad (*)\n$$\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 now 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 $S$, so 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": 18392,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right-angled triangle with the right angle at $C$ such that the side $BC$ is longer than the side $AC$. The perpendicular bisector of $AB$ intersects the line $BC$ at $D$ and the line $AC$ at $E$. We assume that $DE$ and the side $AB$ have the same length.\n\nDetermine the angles of the triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle ABC = \\beta$. As $BC$ is normal to $AE$ and $DE$ is normal to $AB$, the angles $\\angle ABC$ and $\\angle AED$ are equal. Since $\\angle ACB = \\angle DCE = 90^\\circ$ and, by assumption, $\\overline{AB} = \\overline{DE}$, the triangles $ABC$ and $DEC$ are congruent.\n\nThis yields $\\overline{BC} = \\overline{CE}$, which implies that triangle $BCE$ is an isosceles right-angled triangle with $\\angle CEB = \\angle CBE = 45^\\circ$.\n\nFurthermore, $\\beta = \\angle CED = \\angle DEB$, as $E$ lies on the perpendicular bisector of $AB$.\n\nThus, $45^\\circ = \\angle CEB = 2\\beta$, and therefore $\\beta = 22.5^\\circ$ and $\\alpha = \\angle CAB = 67.5^\\circ$.\n\nqed\n\n![](images/Austria_2010_p14_data_f373772547.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18393,
    "subject": "Mathematics (Olympiad)",
    "question": "A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for their favorite actor, and a single vote for their favorite actress. It turns out that for every integer $n \\in \\{1,2,3,\\ldots,100\\}$, there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics who voted for the same actor and the same actress.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that every critic votes for a different pair of actor and actress. \n\nLet a critic's vote be a *double-vote* (a pair: one actor, one actress), and each individual choice a *single-vote*. Thus, each double-vote corresponds to two single-votes.\n\nFor each $n = 34, 35, \\ldots, 100$, pick one actor or actress who received exactly $n$ votes, and let $S$ be the set of these movie stars. Let $a$ and $b$ be the number of men and women in $S$, so $a + b = 67$.\n\nLet $S_1$ be the set of double-votes with exactly one of its single-votes in $S$, and $S_2$ the set with both single-votes in $S$. Let $s_1 = |S_1|$, $s_2 = |S_2|$. The total number of double-votes with at least one single-vote in $S$ is $s_1 + s_2$, and those with both in $S$ is $s_2 \\leq ab$.\n\nThe total number of single-votes in $S$ is $s_1 + 2s_2 = 34 + 35 + \\cdots + 100 = 4489$. Thus, $s_1 + s_2 = (s_1 + 2s_2) - s_2 \\geq 4489 - ab$.\n\nSince all double-votes are distinct, there must be at least $s_1 + s_2$ critics. The maximum of $ab$ with $a + b = 67$ is $33 \\times 34 = 1122$ (since $ab$ is maximized when $a$ and $b$ are as close as possible).\n\nTherefore, there are at least $4489 - 1122 = 3367$ critics, which contradicts the given number $3366$. Thus, there must be two critics who voted for the same actor and the same actress.\n\n**Remark:** The choice of $n = 34$ is optimal. For $n = k, k+1, \\ldots, 100$, $a + b = 101 - k$, and the number of single-votes is $k + (k+1) + \\cdots + 100 = 5050 - \\frac{k(k-1)}{2}$. The number of critics is at least\n\n$$\n5050 - \\frac{k(k-1)}{2} - \\frac{(101-k)^2 - 1}{4}.\n$$\n\nTo get a contradiction, we need\n\n$$\n5050 - \\frac{k(k-1)}{2} - \\frac{(101-k)^2 - 1}{4} \\geq 3367,\n$$\n\nwhich holds only for $k = 34$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18394,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible for a chain to fully cover the surface of a $3 \\times 3 \\times 3$ cube such that each unit square on the surface is visited exactly once by the chain?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible.\n\nAssume, for contradiction, that the chain fully covers the surface of a $3 \\times 3 \\times 3$ cube. Draw a diagonal joining two opposite vertices in each unit square of the chain (including the first and last unit squares). Color the meeting points of the chain (vertices) in black, and the others in white.\n\nThis creates a polyline (possibly self-intersecting) on the surface of the cube. All vertices on this polyline have even degree, except for the first and last, which have odd degree. Specifically, all black vertices except two boundary ones have degree $4$, while the two boundary black vertices have degree $3$ or $4$ depending on whether the polyline is closed.\n\nHowever, the cube has four black vertices, each with degree $3$, so there are at least four vertices with odd degree. This contradicts the Handshaking Lemma, which states that the number of vertices with odd degree in a graph must be even and, for a path, exactly two. Therefore, such a chain cannot exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18395,
    "subject": "Mathematics (Olympiad)",
    "question": "Every diagonal of a regular polygon with 2014 sides is coloured in one of $n$ colours. Whenever two diagonals cross in the interior, they are of different colours. What is the minimum value of $n$ for which this is possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Call a diagonal between directly opposite vertices a primary diagonal. There are $1007$ such diagonals, each of which intersects the others at the centre of the polygon, so at least $1007$ colours are required.\n\nNow, say that each diagonal leaving a vertex on the clockwise side of its primary diagonal is the same colour as that primary diagonal. No two diagonals of the same colour cross, and this fully describes all diagonals since each non-primary diagonal leaves one vertex on the clockwise side of the primary diagonal and the other vertex on the anticlockwise side.\n\nHence, $1007$ is the minimum value of $n$.\n\n![](images/V_Britanija_2014_p28_data_57623f58b1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18396,
    "subject": "Mathematics (Olympiad)",
    "question": "The circles $k_1$ and $k_2$ intersect at points $A$ and $B$. A line through $B$ intersects $k_1$ and $k_2$ again at points $C$ and $D$, respectively, such that $C$ lies outside $k_2$ and $D$ lies outside $k_1$. Let $M$ be the intersection of the tangents to $k_1$ and $k_2$ drawn through $C$ and $D$, respectively. Let $AM \\cap CD = \\{P\\}$. The tangent through $B$ to $k_1$ meets $AD$ at $L$, and the tangent through $B$ to $k_2$ meets $AC$ at $K$. Let $KP \\cap MD = \\{N\\}$ and $LP \\cap MC = \\{Q\\}$. Show that the quadrilateral $MNPQ$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "Due to symmetry, it suffices to show that $KP \\parallel MC$. \n\nFirst, we show that the quadrilateral $ACMD$ is cyclic. Namely, $B$ lies on $\\overline{CD}$, and $A$ and $M$ are on opposite sides of $CD$. From $\\angle BDM = \\angle DAB$ and $\\angle BCM = \\angle BAC$, it follows that\n$$\n\\angle DAC = \\angle DAB + \\angle BAC = \\angle BDM + \\angle BCM = 180^\\circ - \\angle DMC.\n$$\n\nNext, we show that $B$ and $P$ lie on the same arc passing through $A$ and $K$. Consider two cases:\n\n*Case 1*: $P$ lies on $BC$. Then $A$ and $B$ are on the same side of $KP$. Let $E = KB \\cap DM$. We have\n$$\n\\angle KBP = \\angle DBE = \\angle BDE = \\angle CDM = \\angle CAM = \\angle KAP.\n$$\nThus, $\\angle KBP = \\angle KAP$, so $AKPB$ is cyclic.\n\n*Case 2*: $P$ lies on $AC$. Now $A$ and $B$ are on opposite sides of $KP$. Again, let $E = KB \\cap DM$. Then\n$$\n180^\\circ - \\angle KBP = \\angle DBE = \\angle BDE = \\angle CDM = \\angle CAM = \\angle KAP.\n$$\nSo $AKBP$ is cyclic.\n\nTherefore,\n$$\n\\angle APK = \\angle ABK = \\angle ADB = \\angle ADC = \\angle AMC,\n$$\nwhich confirms $KP \\parallel MC$.\n\n**Remark:** The assertion remains true without the restriction that $C$ (resp. $D$) lies outside $k_2$ (resp. $k_1$); the argument is analogous in those cases.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18397,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, x, y$ be positive integers.\n\n(a) Prove that if $x \\neq y$, then\n$$\nax + \\gcd(a, x) + \\text{lcm}(a, x) \\neq ay + \\gcd(a, y) + \\text{lcm}(a, y).\n$$\n\n(b) Show that there are no two positive integers $a$ and $b$ such that\n$$\nab + \\gcd(a, b) + \\text{lcm}(a, b) = 2014.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Suppose that\n$$\nax + \\gcd(a, x) + \\text{lcm}(a, x) = ay + \\gcd(a, y) + \\text{lcm}(a, y)\n$$\nfor certain positive integers $a, x, y$. It follows that\n$$\n\\gcd(a, ax + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, ay + \\gcd(a, y) + \\text{lcm}(a, y)).\n$$\nSince $a$ divides both $ax$ and $\\text{lcm}(a, x)$, we have\n$$\n\\gcd(a, ax + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, \\gcd(a, x)) = \\gcd(a, x)\n$$\nand likewise\n$$\n\\gcd(a, ay + \\gcd(a, y) + \\text{lcm}(a, y)) = \\gcd(a, \\gcd(a, y)) = \\gcd(a, y).\n$$\nTherefore, we must have $\\gcd(a, x) = \\gcd(a, y) = d$ for some positive integer $d$. Since $\\text{lcm}(a, x) = \\dfrac{ax}{\\gcd(a, x)}$ and $\\text{lcm}(a, y) = \\dfrac{ay}{\\gcd(a, y)}$, this gives us\n$$\nax + d + \\frac{ax}{d} = ay + d + \\frac{ay}{d},\n$$\nso\n$$\nax \\left(1 + \\frac{1}{d}\\right) = ay \\left(1 + \\frac{1}{d}\\right),\n$$\nwhich implies $x = y$. This proves the first statement.\n\n(b) Suppose that $ax + \\gcd(a, x) + \\lcm(a, x) = 2014$. Note that the left hand side is divisible by $\\gcd(a, x)$, so $\\gcd(a, x)$ has to be a divisor of 2014, i.e., one of $1, 2, 19, 38, 53, 106, 1007, 2014$. On the other hand,\n$$\n(\\gcd(a, x) + 1)(\\lcm(a, x) + 1) = ax + \\gcd(a, x) + \\lcm(a, x) + 1 = 2015,\n$$\nso $\\gcd(a, x) + 1$ has to divide 2015. Since 2, 3, 20, 39, 54, 107, 1008 are all not divisors of 2015, this leaves us with $\\gcd(a, x) = 2014$. But then $ax + \\lcm(a, x) = 0$, which is clearly impossible since the left hand side is positive.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18398,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two concentric circles of radius $17$ and radius $19$. The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?\n\n(A) $12\\sqrt{2}$  \n(B) $10\\sqrt{3}$  \n(C) $\\sqrt{17 \\cdot 19}$  \n(D) $18$  \n(E) $8\\sqrt{6}$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the common center of the circles and let $\\overline{AC}$ be the chord in the larger circle. Let $M$ be the midpoint of the chord. Let $B$ be the intersection of $\\overline{AC}$ and the smaller circle that lies between $M$ and $C$.\n\nLet $x$ be the length of the chord in the larger circle. It is given that half of the chord lies within the smaller circle, so by symmetry $AM = \\frac{x}{2}$ and $BM = \\frac{x}{4}$.\n\n![](images/2021_AMC12A_Solutions_Fall_p4_data_2b2bb8ee49.png)\n\nApplying the Pythagorean Theorem to $\\triangle OMA$ and $\\triangle OMB$ yields the following two equations:\n\n$$OM^2 = OA^2 - AM^2 = 19^2 - \\left(\\frac{x}{2}\\right)^2$$\n$$OM^2 = OB^2 - BM^2 = 17^2 - \\left(\\frac{x}{4}\\right)^2$$\n\nSubtracting and solving for $x^2$ gives $19^2 - 17^2 = \\frac{3}{16}x^2$ and $x^2 = 24 \\cdot 16$. Therefore, the length of the chord in the larger circle is $x = 8\\sqrt{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18399,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, in the plane rectangular coordinate system $xOy$, the left and right foci of the ellipse $\\Gamma : \\frac{x^2}{2} + y^2 = 1$ are $F_1$ and $F_2$, respectively. Let $P$ be a point on $\\Gamma$ in the first quadrant, and the extensions of $PF_1$ and $PF_2$ intersect $\\Gamma$ again at points $Q_1$ and $Q_2$, respectively. Let $r_1$ and $r_2$ be the radii of the incircles of $\\triangle PF_1Q_2$ and $\\triangle PF_2Q_1$, respectively. Find the maximum value of $r_1 - r_2$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p51_data_7a4b89bd75.png)",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to find $F_1 = (-1, 0)$ and $F_2 = (1, 0)$.\n\nLet $P(x_0, y_0)$, $Q_1(x_1, y_1)$, and $Q_2(x_2, y_2)$. By the given conditions:\n\n$$\nx_0, y_0 > 0, \\quad y_1 < 0, \\quad y_2 < 0.\n$$\n\nBy the definition of the ellipse:\n\n$$\n|PF_1| + |PF_2| = |Q_1F_1| + |Q_1F_2| = |Q_2F_1| + |Q_2F_2| = 2\\sqrt{2}.\n$$\n\nHence, the perimeters of $\\triangle PF_1Q_2$ and $\\triangle PF_2Q_1$ are both $l = 4\\sqrt{2}$.\n\nSince $|F_1F_2| = 2$,\n\n$$\nr_1 = \\frac{2S_{\\triangle PF_1Q_2}}{l} = \\frac{(y_0 - y_2) \\cdot |F_1F_2|}{l} = \\frac{y_0 - y_2}{2\\sqrt{2}}.\n$$\n\nSimilarly, $r_2 = \\frac{y_0 - y_1}{2\\sqrt{2}}$, so $r_1 - r_2 = \\frac{y_1 - y_2}{2\\sqrt{2}}$.\n\nNext, we find $y_1 - y_2$.\n\nThe equation of line $PF_1$ is $x = \\frac{(x_0 + 1)y}{y_0} - 1$. Substituting into $\\frac{x^2}{2} + y^2 = 1$ and rearranging gives:\n\n$$\n\\left( \\frac{(x_0 + 1)^2}{2y_0^2} + 1 \\right) y^2 - \\frac{x_0 + 1}{y_0} y - \\frac{1}{2} = 0.\n$$\n\nMultiplying both sides by $2y_0^2$ and noting $x_0^2 + 2y_0^2 = 2$:\n\n$$\n(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.\n$$\n\nThe two roots are $y_0$ and $y_1$. By Vieta's formula:\n\n$$\ny_0y_1 = -\\frac{y_0^2}{3+2x_0} \\implies y_1 = -\\frac{y_0}{3 + 2x_0}.\n$$\n\nSimilarly, $y_2 = -\\frac{y_0}{3 - 2x_0}$. Therefore,\n\n$$\ny_1 - y_2 = \\frac{y_0}{3 - 2x_0} - \\frac{y_0}{3 + 2x_0} = \\frac{4x_0y_0}{9 - 4x_0^2}.\n$$\n\nSince $9 - 4x_0^2 = \\frac{1}{2}x_0^2 + 9y_0^2 \\ge 2\\sqrt{\\frac{1}{2}x_0^2 \\cdot 9y_0^2} = 3\\sqrt{2}x_0y_0$, we have\n\n$$\nr_1 - r_2 = \\frac{y_1 - y_2}{2\\sqrt{2}} = \\frac{\\sqrt{2}x_0y_0}{9 - 4x_0^2} \\le \\frac{\\sqrt{2}x_0y_0}{3\\sqrt{2}x_0y_0} = \\frac{1}{3},\n$$\n\nwhere equality holds when $\\frac{1}{2}x_0^2 = 9y_0^2$. Thus, $x_0 = \\frac{3\\sqrt{5}}{5}$, $y_0 = \\frac{\\sqrt{10}}{10}$.\n\nTherefore, the maximum of $r_1 - r_2$ is $\\boxed{\\frac{1}{3}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18400,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle and $P$ be an inner point of $\\triangle ABC$. Let $K$, $L$, and $M$ be the reflections of $P$ across $BC$, $AC$, and $AB$, respectively. Let $D$ and $E$ be the second points of intersection of $\\odot(PBC)$ with lines $AB$ and $AC$, respectively. Let lines $MD$ and $LE$ intersect at $F$. Prove that $F$, $A$, and $K$ are collinear.\n\nHere, $\\odot(P_1P_2P_3)$ denotes the circumcircle of triangle $\\triangle P_1P_2P_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Construct the circumcircle of $\\triangle BKC$ and call the second point of intersection of $FK$ with it $G$. We now show that $GFBD$ is cyclic. We use directed angles.\n\n$$\n\\angle BGF = \\angle BGK = \\angle BCK = \\angle PCB = \\angle PDB = \\angle BDM = \\angle BDF.\n$$\n\nSimilarly, $GFCE$ is cyclic.\n\nLastly, notice that $A$ is the radical centre of the circumcircles of $GFBD$, $GFCE$, and $BDCE$. Thus, $F$, $A$, and $K$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18401,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right-angled triangle with $\\angle BAC = 90^\\circ$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \\neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c_1)$ be the circumcircle of triangle $BEZ$ and $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$. Suppose $(c_1)$ meets the line $CZ$ again at the point $F$, and meets $(c_2)$ again at the point $N$. If $P$ is the other point of intersection of $(c_2)$ with $AF$, prove that the points $N$, $B$, $P$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Since triangles $AEB$ and $CAB$ are similar, we have $\\frac{AB}{EB} = \\frac{CB}{AB}$. Note that $AB = BZ$, thus\n\n$$\n\\frac{BZ}{EB} = \\frac{CB}{AB}\n$$\n\nfrom which it follows that triangles $ZBE$ and $CBZ$ are also similar. Since $FEBZ$ is cyclic, $\\angle BEZ = \\angle BFZ$. So by the similarity of triangles $ZBE$ and $CBZ$, we get\n\n$$\n\\angle BFZ = \\angle BEZ = \\angle BZC = \\angle BZF\n$$\n\nand thus $BFZ$ is isosceles. Since $BF = BZ = AB$, triangle $AFZ$ is right with $\\angle AFZ = 90^\\circ$. It follows that points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then\n\n$$\n\\angle ENP = \\angle EAP = \\angle EAF = \\angle BCZ = \\angle BZE\n$$\n\nby using the similarity of triangles $ZBE$ and $CBZ$. Since $N, B, E, Z$ are concyclic, then $\\angle ENP = \\angle BZE = \\angle ENB$, which implies that $N, B, P$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18402,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the last digit when you subtract 8 from 15?",
    "options": [],
    "answer": "See solution",
    "solution": "The last digit is found by subtracting $8$ from $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18403,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a natural number $n$. A function $f: \\{0, 1, \\dots, n\\} \\to \\{0, 1, \\dots, n\\}$ is called *regular* if $f(0) = 0$ and $f(i) \\in \\{i - 1, f(i - 1), f(f(i - 1)), \\dots\\}$ for every $i = 1, \\dots, n$.\n\nIf, for instance, $n = 3$, then the function $f(0) = f(1) = 0$, $f(2) = f(3) = 1$ is regular, but the function $f(0) = f(1) = f(2) = 0$, $f(3) = 1$ is not (in the latter case, $f(3)$ violates the regularity condition).\n\nJuku chooses a regular function $f$ and tells Miku, for each number $k = 0, 1, \\dots, n$, how many different arguments $i$ are there such that $f(i) = k$. Can Miku always determine based on this information which regular function Juku had chosen?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\n**Solution:**\n\nFirst, we prove by induction that if $f$ is a regular function, then $f(i) < i$ for each positive argument $i$.\n\nAssume the claim holds for all smaller arguments. By regularity, $f(i) \\in \\{i-1, f(i-1), f(f(i-1)), \\dots\\}$. By the induction hypothesis and $f(0) = 0$, we have $i-1 \\geq f(i-1) \\geq f(f(i-1)) \\geq \\dots$. Hence, $f(i) < i$.\n\nNow, we prove by induction on $n$ that the information given to Miku uniquely determines the regular function.\n\nIf $n = 0$, only one regular function exists, so the claim holds trivially. Assume $n > 0$ and the claim holds for all smaller numbers. Since $f(1) < 1$, we must have $f(1) = 0$.\n\nLet $k$ be the largest number in $\\{0, 1, \\dots, n\\}$ such that $f(k) = 0$. Then $k \\geq 1$ since $f(1) = 0$. By the lemma, $0 < i \\leq k$ implies $f(i) < i$ and thus $f(i) < k$. On the other hand, $k < i \\leq n$ implies $f(i) \\geq k$.\n\nAssume $f(j) \\geq k$ for each $j = k+1, \\dots, i-1$; let exactly $s$ initial members of the sequence $i-1, f(i-1), f(f(i-1)), \\dots$ be $\\geq k$. The $s$th member must be $k$, otherwise the next member would also be $\\geq k$. Thus, the next member is $f(k) = 0$, and all following members are $0$. Since $f(i) \\neq 0$, we must have $f(i) \\geq k$. Therefore, the number of arguments $i$ such that $f(i) \\geq k$ is exactly $n-k$.\n\nIf $0 < l < k$, then $0 < i \\leq l$ implies $f(i) < l$. But among $i$ with $l < i \\leq n$, there is $k$ with $f(k) = 0$. Thus, the number of $i$ with $f(i) \\geq l$ is less than $n-l$. Consequently, $k$ is the least positive integer for which the number of $i$ with $f(i) \\geq k$ is $n-k$. By this property, Miku can determine $k$.\n\nRestricting $f$ to $\\{0, 1, \\dots, k-1\\}$, all regularity conditions hold. By the induction hypothesis, Miku can determine $f(1), \\dots, f(k-1)$. If we ignore arguments $1, \\dots, k$ and decrease the other argument-value pairs by $k$, regularity holds again. By induction, Miku can also determine $f(k+1), \\dots, f(n)$. Thus, Miku can determine all values of $f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18404,
    "subject": "Mathematics (Olympiad)",
    "question": "A certain distance is covered at a speed of $288$ units. If the distance is $72 \\times 6$ units, how much time does it take to cover this distance?",
    "options": [],
    "answer": "See solution",
    "solution": "Time $= \\frac{\\text{Distance}}{\\text{Speed}}$, so the time taken is equal to $\\frac{72 \\times 6}{288} = 1.5$ hours.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18405,
    "subject": "Mathematics (Olympiad)",
    "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 $M_4(\\mathbb{R})$ such that $AB = -BA$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that if $A^2 + I_n$ is singular for $A \\in M_n(\\mathbb{R})$, then there exists a nonzero matrix $B \\in M_n(\\mathbb{R})$ such that $AB = -BA$.\n\nSince $A^2 + I_n$ is singular, $i$ is an eigenvalue of $A$, and $-i$ is an eigenvalue of $A^\\tau$. There exist nonzero vectors $\\mathbf{x}, \\mathbf{y} \\in M_{n,1}(\\mathbb{C})$ such that $A\\mathbf{x} = i\\mathbf{x}$ and $A^\\tau \\mathbf{y} = -i\\mathbf{y}$. Then $B = \\mathbf{x}\\mathbf{y}^\\tau$ is a nonzero matrix in $M_n(\\mathbb{C})$.\n\nThe following relations show that $A$ and $B$ anticommute:\n\n$$\nAB = A\\mathbf{x}\\mathbf{y}^{\\tau} = i\\mathbf{x}\\mathbf{y}^{\\tau} = -\\mathbf{x}(-i\\mathbf{y})^{\\tau} = -\\mathbf{x}(A^{\\tau}\\mathbf{y})^{\\tau} = -\\mathbf{x}\\mathbf{y}^{\\tau}A = -BA.\n$$\n\nTaking conjugates and using that $A$ is real, the conjugate $\\bar{B}$ of $B$ also anticommutes with $A$. Thus, any real linear combination of $B$ and $\\bar{B}$ anticommutes with $A$, so $B$ or $i(B - \\bar{B})$ is a nonzero real matrix in $M_n(\\mathbb{R})$ that anticommutes with $A$.\n\nFor the converse, suppose $B$ is a nonzero matrix in $M_4(\\mathbb{R})$ such that $AB = -BA$. Then\n\n$$\nA^k B = (-1)^k B A^k, \\quad k \\in \\mathbb{N}.\n$$\n\nLet $f$ be the characteristic polynomial of $A$:\n\n$$\nf(\\lambda) = \\lambda^4 - (\\operatorname{tr} A)\\lambda^3 + a\\lambda^2 - (\\operatorname{tr} A^*)\\lambda + \\det A = \\lambda^4 - (\\operatorname{tr} A)\\lambda^3 + a\\lambda^2 - (\\operatorname{tr} A)\\lambda + \\det A,\n$$\nwhere $a$ is real and $\\operatorname{tr} A = \\operatorname{tr} A^*$.\n\nBy the Cayley-Hamilton theorem, $f(A) = O_4$. Using the commutation property,\n\n$$\n\\begin{aligned}\nO_4 &= f(A)B = B(A^4 + (\\operatorname{tr} A)A^3 + aA^2 + (\\operatorname{tr} A)A + (\\det A)I_4) \\\\\n&= B(f(A) + 2(\\operatorname{tr} A)(A^2 + I_4)A) = 2(\\operatorname{tr} A)B(A^2 + I_4)A.\n\\end{aligned}\n$$\n\nSince $\\operatorname{tr} A \\neq 0$ and $A$ is invertible, $B(A^2 + I_4) = O_4$. As $B$ is nonzero, $A^2 + I_4$ is singular.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 18406,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $a, b, c, d$ such that\n\n$$\na + b + c + d = 20\n$$\n\nand\n\n$$\nab + ac + ad + bc + bd + cd = 150.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)\n$$\nSo,\n$$\n400 = a^2 + b^2 + c^2 + d^2 + 2 \\cdot 150\n$$\nwhich gives\n$$\na^2 + b^2 + c^2 + d^2 = 100.\n$$\nNow,\n$$\n\\begin{align*}\n& (a - b)^2 + (a - c)^2 + (a - d)^2 + (b - c)^2 + (b - d)^2 + (c - d)^2 \\\\\n&= 3(a^2 + b^2 + c^2 + d^2) - 2(ab + ac + ad + bc + bd + cd) \\\\\n&= 3 \\times 100 - 2 \\times 150 = 300 - 300 = 0\n\\end{align*}\n$$\nThus, $a = b = c = d = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18407,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be the roots of the equation $$6x^3 + 5x^2 + 4x + 3 = 0.$$ Find the value of $$S_0 + S_1 + S_2 + \text{dots},$$ where $$S_0 = 1 + a + a^2 + \text{dots},$$ $$S_1 = 1 + b + b^2 + \text{dots},$$ and $$S_2 = 1 + c + c^2 + \text{dots}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\begin{aligned}\nS_0 + S_1 + S_2 + \\dots &= (1 + a + a^2 + \\dots) + (1 + b + b^2 + \\dots) + (1 + c + c^2 + \\dots) \\\\\n&= \\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}.\n\\end{aligned}\n$$\n\nThe answer is equal to the sum of roots of the equation whose roots are $\\frac{1}{1-a}$, $\\frac{1}{1-b}$, and $\\frac{1}{1-c}$. As $a, b, c$ are roots of $6x^3 + 5x^2 + 4x + 3 = 0$, this equation is\n\n$$\n6 \\left(1 - \\frac{1}{x}\\right)^3 + 5 \\left(1 - \\frac{1}{x}\\right)^2 + 4 \\left(1 - \\frac{1}{x}\\right) + 3 = 0,\n$$\n\nor equivalently,\n\n$$\n6(x-1)^3 + 5x(x-1)^2 + 4x^2(x-1) + 3x^3 = 0.\n$$\n\nThe coefficient of $x^3$ is $6+5+4+3=18$ and the coefficient of $x^2$ is $-18-10-4=-32$. Thus, the sum of roots is\n\n$$\n\\frac{32}{18} = \\frac{16}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18408,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of elements of the set\n\n$$\nM = \\left\\{ (x, y) \\in \\mathbb{N}^* \\times \\mathbb{N}^* \\mid \\frac{1}{\\sqrt{x}} - \\frac{1}{\\sqrt{y}} = \\frac{1}{\\sqrt{2016}} \\right\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $ (x, y) \\in M $. Since $ \\sqrt{2016} = 12\\sqrt{14} $,\n\n$$\n\\frac{1}{\\sqrt{x}} - \\frac{1}{\\sqrt{y}} = \\frac{1}{12\\sqrt{14}} \\Leftrightarrow \\frac{1}{\\sqrt{14x}} - \\frac{1}{\\sqrt{14y}} = \\frac{1}{168} \\Leftrightarrow \\frac{1}{\\sqrt{14x}} = \\frac{1}{\\sqrt{14y}} + \\frac{1}{168}.\n$$\n\nThis yields\n\n$$\n\\frac{1}{14x} = \\frac{1}{14y} + \\frac{1}{168^2} + \\frac{2}{168\\sqrt{14y}},\n$$\n\nhence $ \\sqrt{14y} \\in \\mathbb{Q} $, therefore $ \\sqrt{14x} \\in \\mathbb{Q} $. So $ x = 14a $ and $ y = 14b $, where $ a, b $ are positive integers. This leads to $ 1/a - 1/b = 1/12 $, hence\n\n$$\na = \\frac{12b}{b + 12} = 12 - \\frac{144}{b + 12}.\n$$\n\nSo $ b + 12 $ divides $ 144 $, so $ b \\in \\{4, 6, 12, 24, 36, 60, 132\\} $ and $ a \\in \\{3, 4, 6, 8, 9, 10, 11\\} $. In conclusion, the set $ M $ has $ 7 $ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18409,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $3$, and let $a_1, a_2, \\dots, a_n$ be nonnegative real numbers with $a_1 + a_2 + \\dots + a_n = 2$.\n\nDetermine the minimum value of\n\n$$\n\\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\frac{3}{2}$.\n\nThe given problem is equivalent to finding the minimum value of\n\n$$\n\\begin{aligned}\nm &= 2 - \\left( \\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1} \\right) \\\\\n  &= \\left( a_1 - \\frac{a_1}{a_2^2+1} \\right) + \\left( a_2 - \\frac{a_2}{a_3^2+1} \\right) + \\cdots + \\left( a_n - \\frac{a_n}{a_1^2+1} \\right) \\\\\n  &= \\frac{a_1 a_2^2}{a_2^2+1} + \\frac{a_2 a_3^2}{a_3^2+1} + \\cdots + \\frac{a_n a_1^2}{a_1^2+1}.\n\\end{aligned}\n$$\n\nSince $a_i^2 + 1 \\ge 2a_i$, we have\n\n$$\nm \\le \\frac{a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1}{2}.\n$$\n\nOur result follows from the following well-known fact:\n\n$$\n\\begin{aligned}\n& f(a_1, \\cdots, a_n) \\\\\n&= (a_1 + \\cdots + a_n)^2 - 4(a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1) \\\\\n&\\ge 0\n\\end{aligned}\n\\qquad \\textcircled{1}\n$$\n\nfor integers $n \\ge 4$ and nonnegative real numbers $a_1, a_2, \\cdots, a_n$.\n\nTo prove this fact, we use induction on $n$. For $n=4$, $\\textcircled{1}$ becomes\n\n$$\n\\begin{aligned}\n& f(a_1, a_2, a_3, a_4) \\\\\n&= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1) \\\\\n&= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 + a_3)(a_2 + a_4),\n\\end{aligned}\n$$\n\nwhich is nonnegative by the AM-GM inequality.\n\nAssume that $\\textcircled{1}$ is true for $n=k$ for some integer $k \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18410,
    "subject": "Mathematics (Olympiad)",
    "question": "Ada the ant starts at a point $O$ on a plane. At the start of each minute, she chooses North, South, East, or West, and marches 1 metre in that direction. At the end of 2018 minutes, she finds herself back at $O$. Let $n$ be the number of metres she has marched in that direction. What is the highest $n$ that she could have made in 2018 minutes?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose Ada moves east $i$ times. Then she moves west $i$ times, north $1009 - i$ times, and south $1009 - i$ times. There are $\\binom{2018}{1009-i}$ ways for Ada to allocate when she moves north in her schedule. Then there are $\\binom{1009+i}{1009-i}$ ways for her to allocate the south moves (once the north moves have been selected). Finally, there are $\\binom{2i}{i}$ ways for her to allocate the east moves (once the north and south moves have been selected). The west moves are then determined. Note that she cannot move east more than 1009 times, so the total number of possible journeys is\n\n$$\n\\begin{aligned}\nn &= \\sum_{i=0}^{1009} \\binom{2018}{1009-i} \\binom{1009+i}{1009-i} \\binom{2i}{i} \\\\\n&= \\sum_{i=0}^{1009} \\frac{2018!}{(1009-i)!(1009+i)!} \\cdot \\frac{(1009+i)!}{(1009-i)!(2i)!} \\cdot \\frac{(2i)!}{i!i!} \\\\\n&= \\sum_{i=0}^{1009} \\frac{2018!}{[(1009-i)!i!]^2} = \\sum_{i=0}^{1009} \\frac{2018!}{1009!1009!} \\left[ \\frac{1009!}{(1009-i)!i!} \\right]^2 \\\\\n&= \\binom{2018}{1009} \\sum_{i=0}^{1009} \\binom{1009}{i}^2.\n\\end{aligned}\n$$\n\nNow, the number of ways of choosing $i$ things from 1009 is the same as the number of ways choosing $1009 - i$ things from 1009, so $\\binom{1009}{i} = \\binom{1009}{1009-i}$. Hence\n\n$$\n\\sum_{i=0}^{1009} \\binom{1009}{i}^2 = \\sum_{i=0}^{1009} \\binom{1009}{i} \\binom{1009}{1009-i},\n$$\n\nand this last quantity is equal to $\\binom{2018}{1009}$ by Vandermonde's formula. Therefore $n = \\binom{2018}{1009}^2$.\n\nTo find the highest power of 10 which divides $n = \\binom{2018}{1009}^2$, we calculate the highest powers of 2 and 5 which divide $2018!$ and $1009!$. Firstly,\n\n$$\n\\left\\lfloor \\frac{2018}{5} \\right\\rfloor + \\left\\lfloor \\frac{2018}{25} \\right\\rfloor + \\left\\lfloor \\frac{2018}{125} \\right\\rfloor + \\left\\lfloor \\frac{2018}{625} \\right\\rfloor = 403 + 80 + 16 + 3 = 502,\n$$\n\nso the highest power of 5 dividing $2018!$ is $5^{502}$. Similar calculations show that the highest power of 2 dividing $2018!$ is $2^{2011}$, the highest power of 5 dividing $1009!$ is $5^{250}$, and the highest power of 2 dividing $1009!$ is $2^{1002}$.\n\nNow $2 \\times 502 - 4 \\times 250 = 4$, so the highest power of 5 dividing $n = \\binom{2018}{1009}^2$ is $5^4$, while $2 \\times 2011 - 4 \\times 1002 = 14$, so the highest power of 2 dividing $n$ is $2^{14}$. Therefore, the highest power of 10 dividing $n$ is $10^4 = 10,000$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18411,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a stripe $1 \\times n$, $n \\geq 4$, with a positive integer written in each cell (not necessarily equal). Under each number, write a positive integer equal to the number of times that integer appears in the previous row. Repeat this procedure for each subsequent row.\n\n**a)** Prove that after a finite number of steps, the rows will stabilize (i.e., stop changing).\n\n**b)** For $n = 2016$ and for arbitrary $n$, what is the maximal number of steps before stabilization?\n\n![](images/UkraineMO_2015-2016_booklet_p12_data_d6fa9fa79f.png)\n\nFig. 14",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\nLet $k$ be the greatest positive integer such that $2^k \\leq n$. If $n \\neq 2^k + 1$, $n \\neq 2^k + 2$, and $n \\neq 2^k + 4$, the number of steps is $k+1$. If $n=6$, there are 3 steps; if $n=12$, there are 4 steps; otherwise, $k$ steps. Since $2016 = 2^{10} + 992$, for $n=2016$ the answer is 11 steps.\n\n**Solution:**\nWe can rearrange each row so the numbers increase. Denote the initial row as the $0$th, the next as the $1$st, and so on.\n\n**Lemma 1:** If $n$ occurs in the $k$th row, then the number of $n$'s is divisible by $n$.\n\n*Proof:* $n$ occurs in the $1$st row if there are $n$ equal numbers in the $0$th row. Under these, we write $n$ in the next row. If another number $l$ in the $0$th row also occurs $n$ times, another group of $n$'s appears. The number $n$ may occur in later rows if there were $n$'s in the previous row, or if a group of exactly $n$ equal numbers (different from $n$) exists. Thus, their count is divisible by $n$.\n\n**Lemma 2:** If there is a group of $n$'s in the $k$th row, then the same number will be written under these numbers, but it need not be $n$.\n\n*Proof:* If in the previous row there were two equal numbers, under them one writes the same numbers.\n\nConsider the first row. Let $l$ be the smallest number. The number of $l$'s is $sl$, $s \\geq 1$. Other numbers are greater than $l$.\n\nIf $l$ occurs exactly $l$ times, then only $l$ can be written under them. They cannot decrease or increase, so we can erase these and consider a row with $n-l$ numbers, and repeat for the smallest number.\n\nIf there is a group of $l$ consisting of $sl$ numbers, $s > 1$, then in the next group one writes $sl$ under them. For $m$ in the 1st row with $l < m < 2l$, if there are exactly $m$ of them, they cannot change in lower rows. Numbers less than $m$ either do not change or increase at least twofold, so $m$ cannot occur under them. If there are $tm$ of them, they are replaced by $tm$ in the next rows.\n\nThus, in the 2nd row, only numbers $\\geq 2l$ need consideration. If the least number in the 2nd row is $l_2$, then in the 3rd row the least such number is at least $2l_2$, and so on.\n\nA number greater than $n$ cannot appear. So the maximal number of steps is $k$ such that $2^k \\leq n$, plus the move from 0th to 1st row: at most $k+1$ steps.\n\n**Examples:**\nLet $k$ be the greatest integer with $2^k \\leq n$. The 0th row can be:\n\n$$\n(n-1), (1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\nwhere $m = 2^k - n \\neq 2^i$ for any $i$. Subsequent rows:\n\n$$\n(1, 1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n(2, 2, 2, 2), (4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n(4, 4, 4, 4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n\\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^k}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n\\underbrace{(2^k, \\dots, 2^k)}_{2^k}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\nIf $m = 2^k - n = 2^i$, then a group of $2^i$ gives a group of $2^{i+1}$ in the next row, so the process shortens. For $n=5$, $2^2=4<5$, so at most 3 steps. For 0th row $1; 4; 2; 2; 5$, the 1st row is $1; 1; 2; 2; 1$, 2nd row: $3; 3; 2; 2; 3$. Similar for $n=2^s+1$, $n=2^s+2$, and $n=2^s+4$.\n\nFor $n=6$, 0th row: $1; 2; 2; 2; 5; 6 \\rightarrow 1; 3; 3; 3; 1; 1 \\rightarrow 3; 3; 3; 3; 3; 3 \\rightarrow 6; 6; 6; 6; 6; 6$. Similarly for $n=12$.\n\nFor $2^i \\geq 8$, instead of $(m, \\dots, m)$, take two groups: $(s, \\dots, s)$ and $(t, \\dots, t)$, where $s = 2^{i-1}-1$, $t = 2^{i-1}+1$, which do not change and do not interfere with the main example.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18412,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there is not a positive integer $n$ such that $(n+1)2^n$ and $(n+3)2^{n+2}$ are both perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = (n+1)2^n$ and $B = (n+3)2^{n+2}$. We distinguish the cases:\n\nIf $n$ is even, then $2^n$ and $2^{n+2}$ are perfect squares. Since $n+1$ and $n+3$ are odd, they must both be perfect squares.\n\nIn fact, if $n = 2k$, $k \\in \\mathbb{N}$, and $A = (n+1)2^n = (2k+1)2^{2k} = r^2$, then $2^{2k} \\mid r^2$, so $2^k \\mid r$ and $2k+1 = \\left(\\frac{r}{2^k}\\right)^2$. Similarly, $n+3 = 2k+3$ must be a perfect square. However, this is impossible, because there are no perfect squares differing by 2. The difference of two squares of integers $a, b$ ($a > b \\ge 1$) is $a^2 - b^2 = (a-b)(a+b) \\ge 1 \\cdot 3 = 3$.\n\nIf $n = 2k+1$, then the two numbers can be written:\n$$\nA = (2k+1+1)2^{2k+2} = (k+1)2^{2k+2} \\\\\nB = (2k+1+3)2^{2k+2} = (k+2)2^{2k+4}.\n$$\nTherefore, $k+1$ and $k+2$ must be perfect squares, which is also impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18413,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R}^{+} \\to \\mathbb{R}^{+}$ be a function satisfying\n$$\n(f(xy))^2 = f(x^2)f(y^2)\n$$\nfor all $x, y \\in \\mathbb{R}^{+}$ with $x^2y^3 > 2008$.\n\nProve that $(f(xy))^2 = f(x^2)f(y^2)$ for all $x, y \\in \\mathbb{R}^{+}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $\\lambda(x, y) = \\frac{2008}{x^2y^3}$ for all $x, y > 0$.\n\nWe can see that $\\left(\\frac{x}{\\lambda(x, y)}\\right)^2 (y \\cdot \\lambda(x, y))^3 = 2008$. Thus, $\\left(\\frac{x}{z}\\right)^2 (yz)^3 > 2008$ for every $z > \\lambda(x, y)$.\n\nIt follows that $(f(xy))^2 = f(x^2/z^2)f(y^2z^2)$ for all $z > \\lambda(x, y)$.\n\nNow for any positive real numbers $x$ and $y$, we choose\n$$\nz > \\max\\{\\lambda(x, x), \\lambda(y, y), \\lambda(x, y), \\lambda(y, x)\\}.\n$$\nIt follows that\n$$\n(f(x^2))^2 = f(x^2/z^2) f(x^2z^2) \\quad (\\text{since } z > \\lambda(x, x))\n$$\n$$\n(f(y^2))^2 = f(y^2/z^2) f(y^2z^2) \\quad (\\text{since } z > \\lambda(y, y))\n$$\n$$\n(f(xy))^2 = f(x^2/z^2) f(y^2z^2) \\quad (\\text{since } z > \\lambda(x, y))\n$$\n$$\n(f(xy))^2 = f(y^2/z^2) f(x^2z^2) \\quad (\\text{since } z > \\lambda(y, x))\n$$\nWe can see that\n$$\n\\begin{aligned}\nf(xy)^2 \\cdot f(xy)^2 &= (f(x^2/z^2)f(y^2z^2))(f(y^2/z^2)f(x^2z^2)) \\\\\n&= (f(x^2/z^2)f(x^2z^2))(f(y^2/z^2)f(y^2z^2)) \\\\\n&= (f(x^2))^2 (f(y^2))^2\n\\end{aligned}\n$$\nThus $(f(xy))^2 = f(x^2)f(y^2)$ for all $x, y > 0$.\n\n![](images/Thailand-2007_Booklet_p11_data_f45dd95e7d.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18414,
    "subject": "Mathematics (Olympiad)",
    "question": "In a mathematical competition, some competitors are friends. Friendship is always mutual. A group of competitors is called a *clique* if every two of them are friends. (Any group of fewer than two competitors is also a clique.) The number of members in a clique is called its *size*.\n\nGiven that the size of the largest clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.",
    "options": [],
    "answer": "See solution",
    "solution": "We provide an algorithm to distribute the competitors.\n\nLet the rooms be $A$ and $B$. Initially, move one person at a time from one room to the other, adjusting as needed. At each step, let $A$ and $B$ be the sets of competitors in rooms $A$ and $B$, and let $C(A)$ and $C(B)$ be the largest size of a clique in $A$ and $B$ respectively.\n\n**Step 1:** Let $M$ be the largest clique among all competitors, with $|M| = 2m$.\n\nMove all members of $M$ to room $A$, and the remaining competitors to room $B$.\n\nSince $M$ is the largest clique, $C(A) = |M| \\geq C(B)$.\n\n**Step 2:** If $C(A) > C(B)$, move one person from room $A$ to room $B$. (Since $C(A) > C(B)$, $A \\neq \\emptyset$.)\n\nAfter each operation, $C(A)$ decreases by $1$ while $C(B)$ increases by at most $1$. Continue until\n\n$$\nC(A) \\leq C(B) \\leq C(A) + 1.\n$$\n\nAt this point, $C(A) = |A| \\geq m$. (Otherwise, there are at least $m+1$ members of $M$ in $B$ and at most $m-1$ in $A$, so $C(B) - C(A) \\geq (m+1) - (m-1) = 2$, which is impossible.)\n\n**Step 3:** Let $K = C(A)$. If $C(B) = K$, we are done. Otherwise, $C(B) = K + 1$. From above,\n\n$$\nK = |A| = |A \\cap M| \\geq m, \\quad |B \\cap M| \\leq m.\n$$\n\n**Step 4:** If there is a clique $C$ in $B$ with $|C| = K + 1$ and a competitor $x \\in B \\cap M$ but $x \\notin C$, move $x$ to $A$; then we are done.\n\nAfter this, there are $K+1$ members of $M$ in $A$, so $C(A) = K+1$. Since $x \\notin C$, removing $x$ does not reduce $C$, so $C(B) = C$. Therefore, $C(A) = C(B) = K+1$.\n\nIf such $x$ does not exist, then every largest clique in $B$ contains $B \\cap M$ as a subset. In this case, proceed to step 5.\n\n**Step 5:** Choose any largest clique $C$ ($|C| = K+1$) in $B$, and move a member of $C \\setminus M$ to $A$. (Since $|C| = K+1 > m \\geq |B \\cap M|$, $C \\setminus M \\neq \\emptyset$.)\n\nMoving one person at a time from $B$ to $A$ decreases $C(B)$ by at most $1$. At the end, $C(B) = K$.\n\nNow, $A \\cap M$ is a clique in $A$ with $|A \\cap M| = K$, so $C(A) \\geq K$.\n\nWe prove $C(A) = K$ as follows:\n\nLet $Q$ be any clique in $A$. We need to show $|Q| \\leq K$.\n\nThe members of $A$ are:\n\n1. Some members of $M$ (who are friends with all members of $B \\cap M$).\n2. Members moved from $B$ to $A$ in step 5 (who are friends with $B \\cap M$).\n\nThus, every member of $Q$ and every member of $B \\cap M$ are friends. Both $Q$ and $B \\cap M$ are cliques, so $Q \\cup (B \\cap M)$ is a clique.\n\nSince $M$ is the largest clique,\n\n$$\n|M| \\geq |Q \\cup (B \\cap M)| = |Q| + |B \\cap M|.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18415,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a_0, a_1, a_2, b_0, b_1, b_2$ such that\n$$\na_2 b_2 n^2 + a_1 b_1 n + a_0 b_0\n$$\ndivides\n$$\n(a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0)\n$$\nfor any positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that\n$$\na_2 b_2 n^2 + a_1 b_1 n + a_0 b_0 \\mid (a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0) \\quad (1)\n$$\nWe first claim that $a_0 = b_0 = 1$. Let $n$ be a large multiple of $a_0 b_0$. Since $a_0 b_0$ divides the left-hand side of (1), it also divides the right-hand side and hence $a_0 b_0 \\mid a_0^{2017n} + b_0$. This implies $a_0 \\mid b_0$ and $b_0 \\mid a_0^{2017n}$. This shows $a_0, b_0$ have the same prime divisors. Thus, we can take a sufficiently large $n$ such that $a_0 b_0 \\mid a_0^{2017n}$. This gives $a_0 b_0 \\mid b_0$. The only possibility is $a_0 = 1$, which implies $b_0 = 1$.\n\nLet $M$ be a sufficiently large integer and let $n_0$ be the product of all primes less than $M$. Choose any prime divisor $p$ of\n$$\na_2 b_2 n_0^2 + a_1 b_1 n_0 + 1. \\quad (2)\n$$\nClearly, $(p, n_0) = 1$. Thus, $p \\ge M$. In particular, we have $(p, a_1 a_2) = 1$ as $M$ is large.\n\nConsider any $n$ such that $n \\equiv n_0 \\pmod{p}$. Then $p$ divides the left-hand side of (1). This gives\n$$\np \\mid (a_2^{2017n} + b_2)n_0^2 + (a_1^{2017n} + b_1)n_0 + 2. \\quad (3)\n$$\nWe choose $n$ such that $p-1 \\mid n$. Such an $n$ exists by the Chinese remainder theorem. By Fermat's little theorem, we obtain $p \\mid (1+b_2)n_0^2 + (1+b_1)n_0 + 2$. Taking the difference with (3), as $(p, n_0) = 1$, we get\n$$\np \\mid (a_2^{2017n} - 1)n_0 + (a_1^{2017n} - 1) \\quad (4)\n$$\nfor any $n \\equiv n_0 \\pmod{p}$. By taking $n \\equiv 1 \\pmod{p-1}$ and $n \\equiv 2 \\pmod{p-1}$ respectively, we have\n$$\np \\mid (a_2^{2017 \\times 2} - 1)(a_1^{2017} - 1) - (a_1^{2017 \\times 2} - 1)(a_2^{2017} - 1) = (a_1^{2017} - 1)(a_2^{2017} - 1)(a_2^{2017} - a_1^{2017}).\n$$\nSince $p \\ge M$ is sufficiently large, the right-hand side must be $0$. If one of $a_1, a_2$ is $1$, then (4) implies both are equal to $1$. If $a_1 = a_2$, then (4) implies $a_1 = a_2 = 1$ or $n_0 \\equiv -1 \\pmod{p}$.\n\nWe first consider the case $a_1 = a_2 = 1$. In that case, (1) becomes $b_2 n^2 + b_1 n + 1 \\mid (1+b_2)n^2 + (1+b_1)n + 2$ so that $b_2 n^2 + b_1 n + 1 \\mid n^2 + n + 1$. Note that $b_2 n^2 + b_1 n + 1 \\ge n^2 + n + 1$. Thus, equality must hold and hence $b_1 = b_2 = 1$. One easily checks that $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$ is a solution.\n\nNext, it remains to consider the case $n_0 \\equiv -1 \\pmod{p}$. As $p$ divides (2), this yields $p \\mid a_2 b_2 - a_1 b_1 + 1$. Since $p$ is large, we must have $a_2 b_2 - a_1 b_1 + 1 = 0$. Then (2) becomes $(a_1 b_1 - 1)n_0^2 + a_1 b_1 n_0 + 1 = (n_0 + 1)((a_1 b_1 - 1)n_0 + 1)$. Therefore, instead of choosing any prime $p$ dividing (2) at the beginning, we choose such a prime $p$ dividing $(a_1 b_1 - 1)n_0 + 1$. Using the same argument, we obtain either the same solution or $n_0 \\equiv -1 \\pmod{p}$. In the latter case, we find that $p \\mid -(a_1 b_1 - 1) + 1$. Again, this forces $a_1 b_1 = 2$ as $p$ is large. Thus, $(a_1, b_1) = (1, 2)$ or $(2, 1)$. Also, $a_2 b_2 = a_1 b_1 - 1 = 1$ so that $a_2 = b_2 = 1$.\n\nNow, by considering $n = 3$ in (1), we have $16 \\mid 2(3)^2 + (a_1^{2017n} + b_1)(3) + 2$. As $a_1, b_1$ have different parities, the right-hand side is odd. This is impossible.\n\nTherefore, the only solution is $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18416,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$ and $b$ such that\n\n$$\n\\frac{a^2 + b}{b^2 - a} \\quad \\text{and} \\quad \\frac{b^2 + a}{a^2 - b}\n$$\n\nare both integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a \\geq b$. Then $\\frac{b^2 + a}{a^2 - b} \\in \\mathbb{N}^*$, so $b^2 + a \\geq a^2 - b$, which leads to $(a + b)(a - b - 1) \\leq 0$. Thus, $a = b$ or $a = b + 1$.\n\nIf $a = b$, then $\\frac{a^2 + b}{b^2 - a} = \\frac{a + 1}{a - 1}$ is an integer if and only if $a - 1 \\mid a + 1$, i.e., $a - 1 \\mid 2$, so $a \\in \\{2, 3\\}$.\n\nIf $a = b + 1$, then $\\frac{a^2 + b}{b^2 - a} = \\frac{b^2 + 3b + 1}{b^2 - b - 1}$ is an integer if and only if $b^2 - b - 1 \\mid b^2 + 3b + 1$, which means $b^2 - b - 1 \\mid 4b + 2$. Thus, $b^2 - b - 1 \\leq 4b + 2$, so $b \\leq 5$. Checking these values, only $b = 1$ and $b = 2$ satisfy the condition.\n\nIn conclusion, the solutions are: $(2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18417,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A(a, \\frac{1}{a})$, $B(b, \\frac{1}{b})$, $C(c, -\\frac{1}{c})$, and $D(d, -\\frac{1}{d})$ be points with all numbers $a$, $b$, $c$, $d$ pairwise distinct, $c < 0$, and $d > a > b > 0$. The points $A$, $B$, $C$, $D$ all lie on the same line. Show that the points of intersection $M$ and $N$ of the tangents to $y = -\\frac{1}{x}$ at $A$, $B$ and at $C$, $D$ respectively are symmetric with respect to the origin.",
    "options": [],
    "answer": "See solution",
    "solution": "The derivative of $y(x) = -\\frac{1}{x}$ is $y'(x) = \\frac{1}{x^2}$. The tangents at $A$ and $B$ are:\n\n$$\ny = -\\frac{1}{a^2}(x - a) + \\frac{1}{a}, \\quad y = -\\frac{1}{b^2}(x - b) + \\frac{1}{b}.\n$$\n\nTheir intersection $M(x_M, y_M)$ satisfies:\n\n$$\n-\\frac{1}{a^2}(x_M - a) + \\frac{1}{a} = -\\frac{1}{b^2}(x_M - b) + \\frac{1}{b}\n$$\n\nSolving, $x_M = \\frac{2ab}{a + b}$ and $y_M = \\frac{2}{a + b}$.\n\nSimilarly, the tangents at $C$ and $D$ are:\n\n$$\ny = \\frac{1}{c^2}(x - c) - \\frac{1}{c}, \\quad y = \\frac{1}{d^2}(x - d) - \\frac{1}{d}.\n$$\n\nTheir intersection $N(x_N, y_N)$ satisfies:\n\n$$\n\\frac{1}{c^2}(x_N - c) - \\frac{1}{c} = \\frac{1}{d^2}(x_N - d) - \\frac{1}{d}\n$$\n\nSolving, $x_N = \\frac{2cd}{c + d}$ and $y_N = -\\frac{2}{c + d}$.\n\nSince $A$, $B$, $C$, $D$ are collinear, $a + b = c + d$ and $ab = -cd$. Thus, $x_M = -x_N$ and $y_M = -y_N$, so $M$ and $N$ are symmetric with respect to the origin.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18418,
    "subject": "Mathematics (Olympiad)",
    "question": "A plus or a minus sign is placed between every pair of consecutive digits in the sequence 0 1 2 3 4 5 6 7 8 9.\n\n1. Find the smallest positive odd number that cannot be equal to the value of the resulting expression.\n2. Find the smallest positive even number that cannot be equal to the value of the resulting expression.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* 1. 47; 2. 2.\n\nLet the sum of the digits with a plus sign in front of them be $x$ and the absolute value of the sum of the digits with a minus sign in front of them be $y$. Then the value $v$ of the expression equals $x - y$. Also, $x + y = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. Therefore, $v = 45 - 2y$.\n\n1. As $45 - 2y \\leq 45$, nothing greater than 45 can be the value of the expression. We now show that we can obtain all the positive odd integers up to 45 as the result of the expression; this shows that the least positive integer that cannot be equal to the result is 47.\n\nWe previously showed that $y = \\frac{45 - v}{2}$. In order to make the value of the expression be $v$, we need to put a minus sign in front of some digits that sum up to $\\frac{45 - v}{2}$. As $v$ is a positive odd integer between 1 and 45, the number $\\frac{45 - v}{2}$ is a nonnegative integer between 0 and 22. Each such positive integer can be written as a sum of digits as follows: numbers from 1 to 9 are among the digits themselves, numbers from 10 to 17 can be obtained as the sum of 9 and some other digit, and numbers 18 to 22 can be written as the sum of 9, 8, and some other digit in the range of 1 to 5. The case $y = 0$ corresponds to the version where every digit has a plus sign in front of it.\n\n2. Number 2 cannot be obtained as the value of the expression, because solving $2 = 45 - 2y$ gives $y = 21.5$, which is impossible in integers. The number 2 is also the least positive even number.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18419,
    "subject": "Mathematics (Olympiad)",
    "question": "Abscissa $x_i$ of any intersection point of the given parabolas satisfies the system of equations:\n\n$$\ny_i = x_i^2 - a, \\quad x_i = y_i^2 - b, \\quad i = 1, 2, 3, 4.\n$$\n\nShow that:\n\n$$\nx_1 + x_2 + x_3 + x_4 = 0, \\quad x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_1 + x_4x_1x_2 = 1.\n$$\n\nand prove that:\n\n$$\n(x_1 + x_2)(x_1 + x_3)(x_1 + x_4) = 1.\n$$\n\n![alt](images/Blr2012_p15_data_d8bba6ffd4.png \"\")",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_i$ be the abscissas of intersection points. From the system:\n\n$$\ny_i = x_i^2 - a, \\quad x_i = y_i^2 - b\n$$\n\nSubstitute $y_i$ into $x_i$:\n\n$$\nx_i = (x_i^2 - a)^2 - b\n$$\n\nExpanding:\n\n$$\nx_i^4 - 2a x_i^3 + a^2 x_i^2 - 2a^2 x_i + a^2 - b = 0\n$$\n\nBy Vieta's theorem for quartic equations:\n\n$$\nx_1 + x_2 + x_3 + x_4 = 0\n$$\n$$\nx_1x_2x_3 + x_2x_3x_4 + x_3x_4x_1 + x_4x_1x_2 = 1\n$$\n\nNow, consider $(x_1 + x_2)(x_1 + x_3)(x_1 + x_4)$. By symmetry and using the above Vieta relations, after algebraic manipulations (as shown), we find:\n\n$$\n(x_1 + x_2)(x_1 + x_3)(x_1 + x_4) = 1\n$$\n\nas required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18420,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(a, b, c)$ of positive integers such that\n$$\na^{bc} + b^{ca} + c^{ab} = 3abc.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, assume $a \\geq 2$, $b \\geq 2$, $c \\geq 2$. Without loss of generality, let $c$ be the greatest among the three numbers. Then\n$$\na^{bc} + b^{ca} + c^{ab} \\geq a^4 + b^4 + c^4 > b^4 + c^4 \\geq 2b^2c^2 = 2b \\cdot c \\cdot bc > 3 \\cdot a \\cdot bc.\n$$\nThus, there are no solutions in this case.\n\nIt remains to study triples that contain $1$. Without loss of generality, let $a=1$. The equation reduces to $1 + b^c + c^b = 3bc$. Assume $b \\geq 3$, $c \\geq 3$. Without loss of generality, $c \\geq b$, leading to $1 + b^3 + c^3 > c^3 \\geq 3 \\cdot b \\cdot c$. Thus, there are no solutions in this case either.\n\nNow assume $b \\geq 2$, $c \\geq 2$ and one of the numbers is $2$. Without loss of generality, let $b=2$. The equation reduces to $1 + 2^c + c^2 = 6c$, which can be interpreted as a quadratic equation with respect to $c$ that leads to $c = 3 \\pm \\sqrt{9 - (2^c + 1)}$. Hence, $8 - 2^c$ is a perfect square. The only candidates for this are $4$ and $0$, which give $c=2$ and $c=3$, respectively, but $c=2$ leads to a contradiction (the above formula would give $c=1$ or $c=5$). The case $c=3$ gives the solution $(1,2,3)$ of the original equation. By symmetry, also $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$, $(3,2,1)$ are solutions.\n\nIf one of the numbers $b$ and $c$ is $1$, then, without loss of generality, $b=1$. The equation reduces to $1 + 1 + c = 3c$, whence $c=1$. This gives the trivial solution $(1,1,1)$.\n\n*Final answer:* $(1,1,1)$, $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(3,1,2)$, $(2,3,1)$, $(3,2,1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18421,
    "subject": "Mathematics (Olympiad)",
    "question": "$f_1(x) = x^3 - 3x$ ба $n \\geq 2$ үед\n\n$$f_n(x) = f_1(f_{n-1}(x))$$\n\nа) $f_{2012}(x) = 0$ тэгшитгэлийн бүх шийдийн олонлог $X$-ийг ол.\n\nб) $\\displaystyle\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha}$-ийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "(а) $f_1(x) = x^3 - 3x$, $n \\geq 2$ үед $f_n(x) = f_1(f_{n-1}(x))$ гэдгээс $\textbf{deg}\\, f_n(x) = 3^n$ байна. $f_1(2 \\cos t) = 8 \\cos^3 t - 6 \\cos t = 2 \\cos 3t$ ба $f_2(2 \\cos t) = f_1(2 \\cos 3t) = 2 \\cos 3^2 t$ байна. Индукцээр $f_n(2 \\cos t) = 2 \\cos(3^n t)$ болохыг төвөггүй харж болно. Эндээс $x \\in [-2, 2]$ үед $x = 2 \\cos t$ гэсэн орлуулга хийж болох ба $f_n(2 \\cos t) = f_n(x) = 2 \\cos(3^n t)$ болж $f_n(x) = 0$ (1) тэгшитгэл $x \\in [-2, 2]$ дээр $x = 2 \\cos \\frac{k-1}{3^n} \\pi$, $k = 1, 2, \\ldots, 3^n$ гэсэн шийдүүдтэй. Эдгээр нь хос хосоороо ялгаатай ба нийт $3^n$ бодит шийд олдсон байна. $\textbf{deg}\\, f_n(x) = 3^n$ тул (1) тэгшитгэлд өөр шийдгүй. Иймд (1)-ийн шийдийн олонлог\n\n$$X = \\left\\{2 \\cos \\frac{k-1}{3^n} \\pi \\mid k = 1, 2, \\ldots, 3^n \\right\\}$$\n\nболно.\n\n(б) Бид\n\n$$\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha} = \\frac{1}{2} \\sum_{k=1}^{3^n} \\frac{1}{1 - \\cos \\frac{k-1}{3^n} \\pi}$$\n\nнийлбэрийг олох учиртай.\n\n### Лемм\n\n$P(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0 = (x - \\alpha_1)(x - \\alpha_2)\\cdots(x - \\alpha_n)$ болог. Энд $\\alpha_i \\in \\mathbb{C}$, $i = 1, \\ldots, n$. Тэгвэл\n\n$$\\sum_{k=1}^{n} \\frac{1}{x - \\alpha_k} = \\frac{P'(x)}{P(x)}$$\n\n**Баталгаа:** $P'(x) = \\sum_{k=1}^{n} \\frac{P(x)}{x - \\alpha_k}$ гэдгээс шууд баглагдана. Тэгэхээр дээрх леммээс бидний олох ёстой нийлбэр\n\n$$\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha} = \\frac{f'_{n}(2)}{f_{n}(2)}$$\n\nбайх юм. $f_1(2) = 2^3 - 6 = 2 \\Rightarrow \\forall n \\in \\mathbb{N}: f_n(2) = 2$ байна.\n\n$$\n\\begin{aligned}\n  f'_{n}(x) &= [f_{1}(f_{n-1}(x))]' = f'_{1}(f_{n-1}(x)) \\cdot f'_{n-1}(x) \\\\\n  &= f'_{1}(f_{n-1}(x)) \\cdot f'_{1}(f_{n-2}(x)) \\cdot \\ldots \\cdot f'_{1}(f_{1}(x)) \\cdot f'_{1}(x) \\\\\n  &= (3x^2 - 3) \\cdot (3f_{1}^{2}(x) - 3) \\cdot (3f_{2}^{2}(x) - 3) \\cdots (3f_{n-1}^{2}(x) - 3)\n\\end{aligned}\n$$\n\nболох ба $f_{n}(2) = 2$ тул $f'_{n}(2) = 3^n \\cdot 3^n = 9^n$ тул\n\n$$\\frac{f'_{n}(2)}{f_{n}(2)} = \\frac{9^n}{2}$$\n\nболов.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18422,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers such that $c = 3 - a - b$. Show that\n\n$$\n\\frac{a+b}{5-a-b} + \\frac{3-a}{a+2} + \\frac{3-b}{b+2} \\geq 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substitute $c = 3 - a - b$ into the inequality. Cross-multiplying and moving all terms to the left (noting all denominators are positive) gives:\n\n$$\n5a^2b + 5ab^2 + 5a^2 + 5b^2 - 10ab - 15a - 15b + 20 \\geq 0.\n$$\n\nThe left-hand side can be rearranged as:\n\n$$\n5b(a-1)^2 + 5a(b-1)^2 + 5(a+b-2)^2.\n$$\n\nThis is always non-negative, and equals zero if and only if $a = b = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18423,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integers $a$, $b$, and $c$ are less than $99$ and satisfy\n$$\na^2 + b^2 = c^2 + 99^2.\n$$\nFind the minimum and the maximum of $a + b + c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a \\geq b$ by symmetry; then $0 < c < b \\leq a < 99$. Also, $2a^2 \\geq a^2 + b^2 > 99^2$, so $a \\geq 71$. Thus, $71 \\leq a \\leq 98$. Rewrite the equation as\n$$\n(b + c)(b - c) = (99 - a)(99 + a).\n$$\n\n**Minimum of $a + b + c$:**\n\nTry $a = 98$ and $a = 97$ first.\n- If $a = 98$, then $(b + c)(b - c) = 197$, and $197$ is prime. So $b + c = 197$, $a + b + c = 295$.\n- If $a = 97$, then $(b + c)(b - c) = 2 \\cdot 196 = 392 = 14 \\cdot 28$. The sum $b + c$ is minimized when the factors are as close as possible, so $b + c \\geq 28$ and $a + b + c \\geq 97 + 28 = 125$. The equality $a + b + c = 125$ is attained for $b + c = 28$, $b - c = 14$, i.e., $b = 21$, $c = 7$; the triple $(97, 21, 7)$ is admissible. We show that $125$ is the minimum.\n\nNote that $(b + c)(b - c) = (99 - a)(99 + a)$ implies $b + c > \\sqrt{99^2 - a^2}$, so $a + b + c > a + \\sqrt{99^2 - a^2}$. A sufficient condition for $a + b + c > 125$ is $a + \\sqrt{99^2 - a^2} > 125$, which is equivalent to $a^2 - 125a + 2912 < 0$. The quadratic $f(t) = t^2 - 125t + 2912$ increases for $t \\geq 62.5$, and $f(94) < 0$, so $a + b + c > 125$ for $a \\in [71, 94]$. For $a = 95$, $(b + c)(b - c) = 4 \\cdot 194$, and $b + c \\geq 194 > 125$. For $a = 96$, $(b + c)(b - c) = 3 \\cdot 195 = 15 \\cdot 39$, so $b + c \\geq 39$ and $a + b + c \\geq 96 + 39 = 135 > 125$.\n\n**Maximum of $a + b + c$:**\n\nNote $a + b + c$ is odd. Also, $99 - a < b - c < b + c < 99 + a$ and $(b + c)(b - c) = (99 - a)(99 + a)$. The four numbers $b + c$, $b - c$, $99 - a$, $99 + a$ have the same parity, so $b - c = (99 - a) + 2k$ for integer $k \\geq 1$. The maximum is attained when $k = 1$, i.e., $b - c = 101 - a$.\n\nSet $x = b - c = 101 - a$. Then $a = 101 - x$, and\n$$\na + b + c = a + \\frac{(99 - a)(99 + a)}{b - c} = 303 - 2\\left(x + \\frac{200}{x}\\right).\n$$\nSince $a + b + c$ is odd, $x + \\frac{200}{x}$ is integer, so $x$ and $\\frac{200}{x}$ are divisors of $200$ with product $200$. To minimize their sum, take $\\{x, \\frac{200}{x}\\} = \\{10, 20\\}$. Therefore, $a + b + c \\leq 303 - 2(10 + 20) = 243$. The value $243$ is attained for the triple $(91, 81, 71)$, which is admissible.\n\nIf $b - c \\neq 101 - a$, then $b - c \\geq (99 - a) + 4 = 103 - a > 0$, so\n$$\na + b + c = a + \\frac{(99 - a)(99 + a)}{b - c} \\leq a + \\frac{(99 - a)(99 + a)}{103 - a}.\n$$\nTo complete the proof, show $a + \\frac{(99 - a)(99 + a)}{103 - a} < 243$ for $a < 99$. This is equivalent to $a^2 - 73a + 7614 > 0$, which holds for all real $a$ since the discriminant is negative.\n\n**Final answers:**\n- Minimum $a + b + c = 125$\n- Maximum $a + b + c = 243$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18424,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonnegative integers $k, n$ that satisfy the following inequality:\n\n$$\n2^{2k+1} + 9 \\cdot 2^k + 5 = n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the equation modulo $8$. For $k \\geq 3$, the left-hand side is divisible by $8$, but the right-hand side is not a multiple of $8$. So we only need to consider $k \\in \\{0, 1, 2\\}$.\n\n- If $k = 0$, we have $n^2 = 16$, so $n = 4$.\n- If $k = 1$, we have $n^2 = 31$, which has no integer solution.\n- If $k = 2$, we have $n^2 = 73$, which again has no integer solution.\n\nThus, the only solution is $k = 0$, $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18425,
    "subject": "Mathematics (Olympiad)",
    "question": "The Fibonacci numbers are defined by $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \\geq 3$. What is\n$$\n\\frac{F_2}{F_1} + \\frac{F_4}{F_2} + \\frac{F_6}{F_3} + \\dots + \\frac{F_{20}}{F_{10}}?\n$$\n(A) 318 (B) 319 (C) 320 (D) 321 (E) 322",
    "options": [],
    "answer": "See solution",
    "solution": "The Fibonacci sequence starts out\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, ...\nso the given sum is\n$$\n\\frac{1}{1} + \\frac{3}{1} + \\frac{8}{2} + \\frac{21}{3} + \\frac{55}{5} + \\frac{144}{8} + \\frac{377}{13} + \\frac{987}{21} + \\frac{2584}{34} + \\frac{6765}{55},\n$$\nwhich equals $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319$.\n\nAlternatively,\n\nThe Fibonacci sequence starts out 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, so the given sum starts out\n$$\n\\frac{1}{1} + \\frac{3}{1} + \\frac{8}{2} + \\frac{21}{3} + \\frac{55}{5} = 1 + 3 + 4 + 7 + 11.\n$$\nIt appears that these summands satisfy the same recurrence relation, namely\n$$\n\\frac{F_{2n}}{F_n} = \\frac{F_{2(n-1)}}{F_{n-1}} + \\frac{F_{2(n-2)}}{F_{n-2}}.\n$$\nWith the initial conditions 1, 3 instead of 1, 1, the sequence\n$$\n(L_n) = \\left( \\frac{F_{2n}}{F_n} \\right)\n$$\nis known as the Lucas sequence. If the recurrence above is correct, then the required sum is\n$$\n1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319.\n$$\nTo prove the identity for $(L_n)$ displayed above, recall Binet's formula, $F_n = \\frac{1}{\\sqrt{5}}(\\phi^n - \\psi^n)$, where $\\phi = \\frac{1+\\sqrt{5}}{2}$ and $\\psi = \\frac{1-\\sqrt{5}}{2}$ are the roots of the polynomial $x^2 - x - 1$. Then\n$$\nL_n = \\frac{F_{2n}}{F_n} = \\frac{\\phi^{2n} - \\psi^{2n}}{\\phi^n - \\psi^n} = \\phi^n + \\psi^n.\n$$\nTherefore\n$$\n\\begin{align*}\nL_{n-1} + L_{n-2} &= \\phi^{n-1} + \\phi^{n-2} + \\psi^{n-1} + \\psi^{n-2} \\\\\n&= \\phi^{n-2}(\\phi + 1) + \\psi^{n-2}(\\psi + 1) \\\\\n&= \\phi^{n-2} \\cdot \\phi^2 + \\psi^{n-2} \\cdot \\psi^2 \\\\\n&= \\phi^n + \\psi^n = L_n.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18426,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the distinct primes $p$, $q$, $r$, and $s$ satisfying\n\n$$\n1 - \\frac{1}{p} - \\frac{1}{q} - \\frac{1}{r} - \\frac{1}{s} = \\frac{1}{pqrs}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will show the solution consists of all the permutations of the quadruple $2, 3, 7, 43$.\n\nWithout loss of generality, suppose $p < q < r < s$. If $p \\geq 3$,\n\n$$\n1 - \\left(\\frac{1}{p} + \\frac{1}{q} + \\frac{1}{r} + \\frac{1}{s}\\right) \\geq 1 - \\left(\\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{11}\\right) = 1 - \\frac{886}{1155} > 0.2,\n$$\n\nwhich contradicts $\\frac{1}{pqrs} < \\frac{1}{3 \\cdot 5 \\cdot 7 \\cdot 11} < 0.006$. Thus, $p = 2$.\n\nThe equation becomes\n$$\n\\frac{1}{2} - \\left(\\frac{1}{q} + \\frac{1}{r} + \\frac{1}{s}\\right) = \\frac{1}{2qrs}.\n$$\n\nIf $q \\geq 5$,\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\nNow,\n$$\n\\frac{1}{6} - \\left( \\frac{1}{r} + \\frac{1}{s} \\right) = \\frac{1}{6rs},\n$$\nwhich gives $rs - 6r - 6s - 1 = 0$, or $(r-6)(s-6) = 37$. Thus, $r = 7$, $s = 43$.\n\nTherefore, the distinct primes are $2$, $3$, $7$, and $43$ (in any order).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18427,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the number of non-negative integers $N < 1\\,000\\,000$ with the following property: There exists an integer exponent $k$ with $1 \\leq k \\leq 43$ such that $2012$ is a divisor of $N^k - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $N$ and $2012$ must be relatively prime. If $N^k \\equiv 1 \\pmod{n}$ and $N^m \\equiv 1 \\pmod{n}$, then $N^d \\equiv 1 \\pmod{n}$ for $d = \\gcd(k, m)$. Since $m = \\varphi(n)$, $N^k \\equiv 1 \\pmod{n}$ implies there exists a divisor $d$ of $\\varphi(n)$ with $N^d \\equiv 1 \\pmod{n}$. \n\nSince $2012 = 4 \\times 503$ and $\\varphi(503) = 502 = 2 \\times 251$ (where $503$ and $251$ are both prime), the only possible exponents $d$ with $N^d \\equiv 1 \\pmod{503}$ of interest are $1$, $2$, $251$, and $502$. We need only consider $d = 1$ and $d = 2$. Since $N^1 \\equiv 1 \\pmod{503}$ automatically implies $N^2 \\equiv 1 \\pmod{503}$, we only require the residues $+1$ and $-1$ modulo $503$. \n\nSince $N^2 \\equiv 1 \\pmod{4}$ holds for all odd $N$, the values of $N$ with the required property are exactly the numbers $N = 1006u + 1$ and $N = 1006v - 1$. Since $1006 \\times 995 = 1\\,000\\,970$ and $1006 \\times 994 = 999\\,964$, only $0 \\leq u \\leq 994$ and $1 \\leq v \\leq 994$ are possible. Thus, the required number of integers $N$ is $995 + 994 = 1989$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18428,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k = 3 \\cdot 4^l$ for an arbitrary positive integer $l$. Prove that $a_k = \\left\\lfloor \\dfrac{2^k}{k} \\right\\rfloor$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l}}{3 \\cdot 4^l} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}}{3} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor.\n$$\n\nSince the positive integer $3 \\cdot 4^l - 2l$ is even, we have $2^{3 \\cdot 4^l - 2l} \\equiv 4 \\equiv 1 \\pmod 3$, so the odd positive integer $2^{3 \\cdot 4^l - 2l} - 1$ is divisible by 3.\n\nTherefore,\n\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor = \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3}\n$$\n\nis odd, which finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18429,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be a fixed integer. The number $1$ is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A *move* consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers $1$ and $a+b$, then adding one stone to the first bucket and $\\gcd(a,b)$ stones to the second bucket. After some finite number of moves, there are $s$ stones in the first bucket and $t$ stones in the second bucket, where $s$ and $t$ are positive integers. Find all possible values of the ratio $\\frac{t}{s}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nThe answer is the set of all rational numbers in the interval $[1, n-1)$. First, we show that no other numbers are possible. Clearly the ratio is at least $1$, since for every move, at least one stone is added to the second bucket. Note that the number $s$ of stones in the first bucket is always equal to $p-n$, where $p$ is the sum of the numbers on the blackboard. We will assume that the numbers are written in a row, and whenever two numbers $a$ and $b$ are erased, $a+b$ is written in the place of the number on the right. Let $a_1, a_2, \\dots, a_n$ be the numbers on the blackboard from left to right, and let\n\n$$\nq = 0 \\cdot a_1 + 1 \\cdot a_2 + \\dots + (n-1)a_n.\n$$\n\nSince each number $a_i$ is at least $1$, we always have\n\n$$\nq \\le (n-1)p - (1 + \\dots + (n-1)) = (n-1)p - \\frac{n(n-1)}{2} = (n-1)s + \\frac{n(n-1)}{2}.\n$$\n\nAlso, if a move changes $a_i$ and $a_j$ with $i < j$, then $t$ changes by $\\gcd(a_i, a_j) \\le a_i$ and $q$ increases by\n\n$$\n(j-1)a_i - (i-1)(a_i - 1) \\ge ia_i - (i-1)(a_i - 1) \\ge a_i.\n$$\n\nHence $q-t$ never decreases. We may assume without loss of generality that the first move involves the rightmost $1$. Then immediately after this move, $q = 0+1+\\dots+(n-2)+(n-1) \\cdot 2 = \\frac{(n+2)(n-1)}{2}$ and $t = 1$. So after that move, we always have\n\n$$\n\\begin{aligned}\nt &\\le q + 1 - \\frac{(n+2)(n-1)}{2} \\\\\n  &\\le (n-1)s + \\frac{n(n-1)}{2} - \\frac{(n+2)(n-1)}{2} + 1 \\\\\n  &= (n-1)s - (n-2) < (n-1)s.\n\\end{aligned}\n$$\n\nHence, $\\frac{t}{s} < n - 1$. So $\\frac{t}{s}$ must be a rational number in $[1, n - 1)$.\n\nAfter a single move, we have $\\frac{t}{s} = 1$, so it remains to prove that $\\frac{t}{s}$ can be any rational number in $(1, n - 1)$. We will now show by induction on $n$ that for any positive integer $a$, it is possible to reach a situation where there are $n - 1$ occurrences of $1$ on the board and the number $a^{n-1}$, with $t$ and $s$ equal to $a^{n-2}(a-1)(n-1)$ and $a^{n-1} - 1$, respectively. For $n = 2$, this is clear as there is only one possible move at each step, so after $a - 1$ moves $s$ and $t$ will both be equal to $a - 1$. Now assume that the claim is true for $n - 1$, where $n > 2$. Call the algorithm which creates this situation using $n - 1$ numbers algorithm $A$. Then to reach the situation for size $n$, we apply algorithm $A$, to create the number $a^{n-2}$. Next, apply algorithm $A$ again and then add the two large numbers, repeat until we get the number $a^{n-1}$. Then algorithm $A$ was applied $a$ times and the two larger numbers were added $a - 1$ times. Each time the two larger numbers are added, $t$ increases by $a^{n-2}$ and each time algorithm $A$ is applied, $t$ increases by $a^{n-3}(a-1)(n-2)$. Hence, the final value of $t$ is\n\n$$\nt = (a - 1)a^{n-2} + a \\cdot a^{n-3}(a - 1)(n - 2) = a^{n-2}(a - 1)(n - 1).\n$$\n\nThis completes the induction.\n\nNow we can choose $1$ and the large number $b$ times for any positive integer $b$, and this will add $b$ stones to each bucket. At this point we have\n\n$$\n\\frac{t}{s} = \\frac{a^{n-2}(a-1)(n-1) + b}{a^{n-1} - 1 + b}.\n$$\n\nSo we just need to show that for any rational number $\\frac{p}{q} \\in (1, n - 1)$, there exist positive integers $a$ and $b$ such that\n\n$$\n\\frac{p}{q} = \\frac{a^{n-2}(a-1)(n-1) + b}{a^{n-1} - 1 + b}\n$$\n\nRearranging, we see that this happens if and only if\n\n$$\nb = \\frac{q a^{n-2}(a-1)(n-1) - p(a^{n-1} - 1)}{p-q}.\n$$\n\nIf we choose $a \\equiv 1 \\pmod{p-q}$, then this will be an integer, so we just need to check that the numerator is positive for sufficiently large $a$.\n\n$$\n\\begin{aligned}\nq a^{n-2}(a-1)(n-1) - p(a^{n-1}-1) &> q a^{n-2}(a-1)(n-1) - p a^{n-1} \\\\\n&= a^{n-2}(a(q(n-1)-p) - (n-1)),\n\\end{aligned}\n$$\n\nwhich is positive for sufficiently large $a$ since $q(n - 1) - p > 0$.\n\n**Alternative solution for the upper bound.** Rather than starting with $n$ occurrences of $1$, we may start with infinitely many $1$s, but we are restricted to having at most $n - 1$ numbers which are not equal to $1$ on the board at any time. It is easy to see that this does not change the problem. Note also that we can ignore the $1$ we write on the board each move, so the allowed move is to rub off two numbers and write their sum. We define the width and score of a number on the board as follows. Colour that number red, then reverse every move up to that point all the way back to the situation when the numbers are all $1$s. Whenever a red number is split, colour the two replacement numbers\n\n![alt](path \"title\")",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18430,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the greatest prime factor (GPF) for each composite number from 60 to 65.\n\nb) The third composite is a multiple of 79. List the sequence of such numbers and determine the first composite in the sequence that is a multiple of 19. What can you conclude about the size of these composites?\n\nc) **Alternative i**\nFind four successive composite numbers such that the third has GPF 2, and the second is at least $2 \\times 73 = 146$. List the relevant multiples and determine the smallest such sequence.\n\n**Alternative ii**\nIf the third composite has GPF 2 (i.e., is a power of 2), and the second composite is at least $2 \\times 73 = 146$, find the smallest sequence of four successive composites meeting these criteria.\n\nd) **Alternative i**\nFind the largest composite less than 10,000 whose greatest prime factor is 7. Consider products of the form $2^a \\times 3^b \\times 5^c \\times 7^d$ and determine the largest such number.\n\n**Alternative ii**\nList the multiples of 7 less than 10,000 in decreasing order and determine which has GPF 7. What is the largest composite less than 10,000 with GPF 7?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The prime factorizations for the composites from 60 to 65 are:\n- $60 = 2 \\times 2 \\times 3 \\times 5$ (GPF: 5)\n- $61$ is prime\n- $62 = 2 \\times 31$ (GPF: 31)\n- $63 = 3 \\times 3 \\times 7$ (GPF: 7)\n- $64 = 2^6$ (GPF: 2)\n- $65 = 5 \\times 13$ (GPF: 13)\nSo, the GPF sequence for the composites from 60 to 65 is: 5, 31, 7, 2, 13.\n\nb) The third composite is a multiple of 79, so possible values are:\n$158, 237, 316, 395, 474, 553, 632, 711, 790, 869, 948, 1027, \\dots$\nThe composites immediately before these are:\n$156, 236, 315, 394, 473, 552, 630, 710, 789, 868, 946, 1026, \\dots$\nThe first of these that is a multiple of 19 is 1026. Thus, the first composite is at least 1025, so all the composites have at least four digits.\n\nc) **Alternative i**\nMultiples of 73: $73, 146, 219, 292, 365, 438, 511$. The next composites after these are: $74, 147, 220, 294, 366, 440, 512$. The first with GPF 2 is 512. The GPFs of 510 and 513 are 17 and 19, respectively. Thus, the smallest required successive composites are 510, 511, 512, 513.\n\n**Alternative ii**\nIf the third composite is a power of 2, possible values are 256, 512, .... The composites immediately before these are 255, 511, .... The first with GPF 73 is 511. The GPFs of 510 and 513 are 17 and 19, respectively. Thus, the smallest required successive composites are 510, 511, 512, 513.\n\nd) **Alternative i**\nA composite with GPF 7 must be of the form $2^a \\times 3^b \\times 5^c \\times 7^d$ with $d \\geq 1$. The largest such products less than 10,000 are:\n- $4 \\times 2401 = 9604$\n- $27 \\times 343 = 9261$\n- $200 \\times 49 = 9800$\nChecking for products of the form $2^a \\times 3^b \\times 5^c \\times 7$ strictly between 9800 and 10,000 yields no valid solutions. Thus, the largest composite less than 10,000 with GPF 7 is 9800.\n\n**Alternative ii**\nMultiples of 7 less than 10,000 in decreasing order include: 9996, 9989, 9982, ..., 9800. Dividing each by the maximum powers of 7, 5, 3, 2 shows that the GPFs of these multiples down to 9807 are greater than 7. Therefore, the largest composite less than 10,000 with GPF 7 is 9800.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18431,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c(O, R)$ be a circle and $A$, $B$ two diametrically opposite points. The bisector of the angle $A\\hat{B}O$ intersects the circle $c(O, R)$ at point $C$, the circumcircle of triangle $AOB$ (denote as $(c_1)$) at point $K$, and the circumcircle of triangle $AOC$ (denote as $(c_2)$) at point $L$. Prove that point $K$ is the circumcenter of triangle $AOC$ and point $L$ is the incenter of triangle $AOB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Segments $OB$ and $OC$ are equal, as radii of the circle $c$, so triangle $OBC$ is isosceles. Hence:\n\n$$\n\\hat{B}_1 = \\hat{C}_1 = \\hat{x} \\qquad (1)\n$$\n\n![](images/Hellenic_2016_p13_data_1f09b7e0a0.png)\n\n$BC$ is the bisector of angle $OBA$, so\n\n$$\n\\hat{B}_1 = \\hat{B}_2 = \\hat{x} \\qquad (2)\n$$\n\nAngles $\\hat{B}_2$ and $\\hat{O}_1$ are inscribed in the same circle and correspond to the same arc $OK$ of circle $(c_1)$. Hence\n\n$$\n\\hat{B}_2 = \\hat{O}_1 = \\hat{x} \\qquad (3)\n$$\n\nSimilarly, $KO = KC$ and triangle $KOC$ is isosceles, so:\n\n$$\n\\hat{O}_2 = \\hat{C}_1 = \\hat{x} \\qquad (4)\n$$\n\nFrom (1)–(4) we conclude $\\hat{O}_1 = \\hat{O}_2 = \\hat{x}$, i.e., $OK$ is the angle bisector and also the perpendicular bisector of isosceles triangle $OAC$. Point $K$ is the midpoint of arc $OK$ (since $BK$ is the bisector of $OBA$). Thus, the perpendicular bisector of chord $AO$ of circle $(c_1)$ (which is also a side of isosceles triangle $OAC$) passes through $K$. Therefore, $K$ is the circumcenter of triangle $OAC$.\n\nFrom (1) and (2), $\\hat{B}_2 = \\hat{C}_1 = \\hat{x}$, so $AB \\parallel OC$. Therefore, $O\\hat{A}B = A\\hat{O}C$, that is, $\\hat{A}_1 + \\hat{A}_2 = \\hat{O}_1 + \\hat{O}_2$, and since $\\hat{O}_1 = \\hat{O}_2 = \\hat{x}$, we have:\n\n$$\n\\hat{A}_1 + \\hat{A}_2 = 2\\hat{O}_1 = 2\\hat{x}\n$$\n\nAngles $\\hat{A}_1$ and $\\hat{C}_1$ are inscribed in circle $(c_2)$ and correspond to the same arc $OL$. Hence\n\n$$\n\\hat{A}_1 = \\hat{C}_1 = \\hat{x}\n$$\n\nFrom the last two equalities, $\\hat{A}_1 = \\hat{A}_2$, so $AL$ is the bisector of angle $B\\hat{A}O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18432,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a rectangle is *inscribed* in a triangle if two of the rectangle's neighbouring vertices lie on one side of the triangle, and the other two lie on the remaining two sides of the triangle. Assume that the lengths of the sides of triangle $ABC$ are known. What is the smallest possible length of the diagonal of an inscribed rectangle in this triangle?\n\n![](images/Macedonia_2013_p11_data_b646d9a17b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the rectangle $EFGH$ be inscribed in triangle $ABC$ so that $E$ and $F$ lie on $BC$, $G$ lies on $AC$, and $H$ lies on $AB$. Denote the side lengths of $ABC$ by $a$, $b$, and $c$, and let $h$ be the height from $A$ to $BC$. Let $\\overline{AH} = x$, $\\overline{EF} = u$, and $\\overline{FG} = v$.\n\nFrom the similarity $\\triangle AHG \\sim \\triangle ABC$, we have $u = \\frac{a x}{c}$. From $\\triangle BEH \\sim \\triangle BVA$, we get $v = \\frac{h (c - x)}{c}$. If $l$ denotes the length of the diagonal of $EFGH$, then\n\n$$\nl^2 = \\frac{a^2 x^2}{c^2} + \\frac{h^2 (c - x)^2}{c^2}.\n$$\n\nThe minimum of the function\n\n$$\nf(x) = \\frac{a^2 x^2}{c^2} + \\frac{h^2 (c - x)^2}{c^2}\n$$\n\nis $\\frac{a^2 h^2}{a^2 + h^2}$, attained when $x = \\frac{h^2 c}{a^2 + h^2}$. Note that\n\n$$\n\\frac{a^2 h^2}{a^2 + h^2} = \\frac{4P^2}{a^2 + \\frac{4P^2}{a^2}},\n$$\n\nwhere $P$ is the area of the triangle. Similarly, if the rectangle has two neighbouring vertices on side $AC$, the minimal diagonal is $\\frac{4P^2}{b^2 + \\frac{4P^2}{b^2}}$.\n\nComparing these, the smallest value is achieved when the rectangle has two neighbouring vertices on the longest side of the triangle. Let $a$ be the longest side. Then the minimal diagonal length is\n\n$$\nl = \\frac{2P}{\\sqrt{a^2 + \\frac{4P^2}{a^2}}},\n$$\n\nattained when $\\overline{AH} = x = \\frac{h^2 c}{a^2 + h^2} = \\frac{4P^2 c}{a^4 + 4P^2}$. The area $P$ can be computed from the side lengths using Heron's formula.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18433,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangles satisfying the following conditions: The three vertices of the triangle must be vertices of the 101-gon, both the vertices with acute angles have the same color, and the vertex with obtuse angle has a different color.\n\n1. Find the largest possible value of $N$.\n2. Find the number of ways to color the vertices such that maximum $N$ is achieved. (Two colorings are different if for some $A_i$ the colors are different on the two coloring schemes.)\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For an obtuse triangle $A_{i-a}A_iA_{i+b}$ (vertex $A_i$ is the vertex of the obtuse angle, i.e., $a+b \\le 50$), these three vertices satisfy the conditions if and only if\n\n$$\n(x_i - x_{i-a})(x_i - x_{i+b}) = 1\n$$\n\notherwise $0$, with subscripts modulo $101$. Thus,\n\n$$\nN = \\sum_{i=1}^{101} \\sum_{(a, b)} (x_i - x_{i-a})(x_i - x_{i+b})\n$$\n\nwhere $\\sum_{(a, b)}$ is over all positive integer pairs $(a, b)$ with $a+b \\le 50$. There are $49 + 48 + \\cdots + 1 = 1225$ such pairs. Expanding,\n\n$$\n\\begin{align*}\nN &= \\sum_{i=1}^{101} \\sum_{(a, b)} (x_i^2 - x_i x_{i-a} - x_i x_{i+b} + x_{i-a} x_{i+b}) \\\\\n&= 1225 \\sum_{i=1}^{101} x_i^2 + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (k-1-2(50-k)) x_i x_{i+k} \\\\\n&= 1225n + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (3k-101) x_i x_{i+k}.\n\\end{align*}\n$$\n\nHere $n$ is the number of blue vertices. For any two vertices $A_i$ and $A_j$, $1 \\le i, j \\le 101$, let\n\n$$\nd(A_i, A_j) = d(A_j, A_i) = \\min\\{j - i, 101 - j + i\\}\n$$\n\nLet $B \\subseteq \\{A_1, \\dots, A_{101}\\}$ be the set of blue vertices. Then,\n\n$$\nN = 1225n - 101\\binom{n}{2} + 3 \\sum_{\\{P, Q\\} \\subseteq B} d(P, Q)\n$$\n\nwhere $\\{P, Q\\}$ runs over all two-element subsets of $B$. Assume $n$ is even (otherwise, swapping colors does not change $N$). Write $n = 2t$, $0 \\le t \\le 50$, and label blue vertices $P_1, \\dots, P_{2t}$ clockwise. Then\n\n$$\n\\sum_{\\{P, Q\\} \\subseteq B} d(P, Q) = \\sum_{i=1}^{t} d(P_i, P_{i+t}) + \\frac{1}{2} \\sum_{i=1}^{t} \\sum_{j=1}^{t-1} [d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i)] \\le 50t + \\frac{101}{2}t(t-1)\n$$\n\nusing $d(P_i, P_{i+t}) \\le 50$ and\n\n$$\nd(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \\le 101\n$$\n\nCombining, we get\n\n$$\nN \\le 1225n - 101\\binom{n}{2} + 3\\left(50t + \\frac{101}{2}t(t-1)\\right) = -\\frac{101}{2}t^2 + \\frac{5099}{2}t\n$$\n\nThe maximum is at $t = 25$, so $N \\le 32175$.\n\nTo achieve $N = 32175$, $t = 25$ (i.e., $n = 50$ blue vertices), and for $1 \\le i \\le t$, $d(P_i, P_{i+t}) = 50$. The number of ways to choose 50 blue vertices so that $N$ is maximized equals the number of ways to choose 25 diagonals from the longest 101 diagonals such that no two share a vertex.\n\nConnecting $A_i$ and $A_{i+50}$ for $i = 1, \\dots, 101$ forms a graph $G$ (a cycle of 101 edges). The number of ways ($S$) to select 25 edges of $G$ with no shared vertices is $\\binom{75}{24} + \\binom{76}{25}$. Similarly, for 50 red vertices, so the total number of colorings is $2S = 2\\left(\\binom{75}{24} + \\binom{76}{25}\\right)$.\n\n**Summary:**\n- The largest possible value of $N$ is $32175$.\n- The number of colorings achieving this is $2\\left(\\binom{75}{24} + \\binom{76}{25}\\right)$.\n\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18434,
    "subject": "Mathematics (Olympiad)",
    "question": "For any $i = 1, 2, \\dots, 2n$, we have $a_i a_{i+2} \\ge b_i + b_{i+1}$, where $a_{2n+1} = a_1$, $a_{2n+2} = a_2$, and $b_{2n+1} = b_1$. Find the minimum value of $a_1 + a_2 + \\dots + a_{2n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = a_1 + a_2 + \\dots + a_{2n}$ and $S = b_1 + b_2 + \\dots + b_{2n}$.\n\nWithout loss of generality, suppose $T = a_1 + a_3 + \\dots + a_{2n-1} \\le \\frac{S}{2}$.\n\nFor $n = 3$, since\n$$\nT^2 - 3 \\sum_{k=1}^{3} a_{2k-1} a_{2k+1} = \\frac{1}{2} \\left( (a_1 - a_3)^2 + (a_3 - a_5)^2 + (a_5 - a_1)^2 \\right) \\ge 0,\n$$\nby combining the given conditions we can find\n$$\n\\frac{S^2}{4} \\ge T^2 \\ge 3 \\sum_{k=1}^{3} a_{2k-1} a_{2k+1} \\ge 3 \\sum_{k=1}^{3} (b_{2k-1} + b_{2k}) = 3S.\n$$\nAnd since $S > 0$, we have $S \\ge 12$.\n\n$S$ takes the minimum when $a_i = b_i = 2$ for $1 \\le i \\le 6$.\n\nWhen $n \\ge 4$, on one hand we have\n$$\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\ge \\sum_{k=1}^{n} (b_{2k-1} + b_{2k}) = S.\n$$\nOn the other hand, if $n$ is even,\n$$\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\le (a_1 + a_5 + \\dots + a_{2n-3}) (a_3 + a_7 + \\dots + a_{2n-1}) \\le \\frac{T^2}{4}.\n$$\nThe first inequality is due to the fact that each term of $(a_1 + a_5 + \\dots + a_{2n-3})(a_3 + a_7 + \\dots + a_{2n-1})$ is nonnegative after expansion and contains terms $a_{2k-1} a_{2k+1}$ ($1 \\le k \\le n$). The second inequality uses the arithmetic-geometric mean inequality.\n\nIf $n$ is odd, suppose $a_1 \\le a_3$, then\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} &\\le \\left( \\sum_{k=1}^{n-1} a_{2k-1} a_{2k+1} \\right) + a_{2n-1} a_3 \\\\\n&\\le (a_1 + a_5 + \\dots + a_{2n-1})(a_3 + a_7 + \\dots + a_{2n-3}) \\\\\n&\\le \\frac{T^2}{4}.\n\\end{aligned}\n$$\nThus, there is always $S \\le \\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\le \\frac{T^2}{4} \\le \\frac{S^2}{16}$. Since $S > 0$, $S \\ge 16$.\n\nWhen $a_1 = a_2 = a_3 = a_4 = 4$, $a_i = 0$ for $5 \\le i \\le 2n$, $b_1 = 0$, $b_2 = 16$, $b_i = 0$ for $3 \\le i \\le 2n$, $S$ takes the minimum of $16$.\n\nTo sum up, when $n = 3$, the minimum of $S$ is $12$; when $n \\ge 4$, the minimum of $S$ is $16$.\n\n$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18435,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $z_1, z_2, \\dots, z_n$ be positive integers such that for $j = 1, 2, \\dots, n$ the inequalities\n$$\nz_j \\leq j\n$$\nhold and $z_1 + \\dots + z_n$ is even.\n\nProve that the number $0$ occurs among the values of\n$$\nz_1 \\pm z_2 \\pm \\dots \\pm z_n,\n$$\nwhere $+$ or $-$ can be chosen independently for each operation.",
    "options": [],
    "answer": "See solution",
    "solution": "We carry out the proof with complete induction.\n\n* For $n = 1$: Here $z_1 = 1$ and the sum cannot be even, so there is nothing to prove.\n\n* For $n = 2$: Here $z_1 \\leq 1$, $z_2 \\leq 2$ and the condition that $z_1 + z_2$ is even leads to $z_1 = z_2 = 1$, together with $z_1 - z_2 = 0$.\n\n* For $n = 3$: Here $z_1 \\leq 1$, $z_2 \\leq 2$, $z_3 \\leq 3$ and the condition that $z_1 + z_2 + z_3$ is even results in the three possibilities $(1, 1, 2)$, $(1, 2, 1)$, or $(1, 2, 3)$ with $1 + 1 - 2 = 0$, $1 - 2 + 1 = 0$, and $1 + 2 - 3 = 0$.\n\n* Suppose that the statement holds up to $n$ and now draw the conclusion from $n$ to $n + 1$.\n\nWe distinguish between two cases:\n\n(a) $z_{n+1} = z_n$: In this case, $z_1 + \\dots + z_{n-1}$ is even. Therefore, $0$ can be represented in the form $z_1 \\pm \\dots \\pm z_{n-1}$ and thus also as $z_1 \\pm \\dots \\pm z_{n-1} + z_n - z_{n+1}$.\n\n(b) $z_{n+1} \\neq z_n$: With $z_1 + \\dots + z_{n-1} + z_n + z_{n+1}$, the sum $z_1 + \\dots + z_{n-1} + |z_{n+1} - z_n|$ is also even. Furthermore, $1 \\leq |z_{n+1} - z_n| \\leq n$, so we can apply the induction assumption for the $n$ numbers $z_1, \\dots, z_{n-1}, |z_{n+1} - z_n|$. Consequently, $0$ can be represented as $z_1 \\pm \\dots \\pm z_{n-1} \\pm |z_{n+1} - z_n|$. Because $|z_{n+1} - z_n| = \\pm(z_{n+1} - z_n)$, we end up with a representation of $0$ in the form $z_1 \\pm \\dots \\pm z_{n-1} \\pm z_n \\pm z_{n+1}$, which completes the induction step.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18436,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of solutions of the equation $|a - b| = |b - c|$ in integers from $0$ to $36$.",
    "options": [],
    "answer": "See solution",
    "solution": "The equation $|a - b| = |b - c|$ is satisfied if and only if either $a - b = b - c$ or $a - b = c - b$.\n\nThe first equality, $a - b = b - c$, is equivalent to $a + c = 2b$. The second, $a - b = c - b$, is equivalent to $a = c$.\n\nTo fulfill $a + c = 2b$, $a$ and $c$ must have the same parity. Their sum is even, and $b$ lies between $a$ and $c$, so $b$ also falls between $0$ and $36$.\n\n- There are $19$ even numbers ($0, 2, \\ldots, 36$), so $19^2 = 361$ possibilities for even $a$ and $c$.\n- There are $18$ odd numbers ($1, 3, \\ldots, 35$), so $18^2 = 324$ possibilities for odd $a$ and $c$.\n\nThus, there are $361 + 324 = 685$ triples satisfying $a + c = 2b$.\n\nFor $a = c$, $a$ and $b$ can be chosen arbitrarily from $0$ to $36$, so there are $37 \\times 37 = 1369$ possibilities.\n\nThere are $37$ solutions where $a = b = c$, which satisfy both conditions and are counted twice.\n\nTherefore, the total number of solutions is:\n\n$$\n685 + 1369 - 37 = 2017\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18437,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram with $\\angle BAD < 90^\\circ$. A circle tangent to sides $\\overline{DA}$, $\\overline{AB}$, and $\\overline{BC}$ intersects diagonal $\\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in the form $m\\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.\n\n![](images/2022AIME_I_Solutions_p7_data_abb67f679a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$, $Y$, and $Z$ denote the points where the circle is tangent to $\\overline{AD}$, $\\overline{BC}$, and $\\overline{AB}$, respectively. Let $R$ be the foot of the perpendicular from $C$ to line $AD$.\n\n![](images/2022AIME_I_Solutions_p7_data_39cc4e84aa.png)\n\nBy Power of a Point, $AX = \\sqrt{AP \\cdot AQ} = 6$ and $CY = \\sqrt{CQ \\cdot CP} = 20$. Hence $AR = AX + CY = 26$. By the Pythagorean Theorem\n\n$$\nXY = CR = \\sqrt{AC^2 - AR^2} = \\sqrt{28^2 - 26^2} = 6\\sqrt{3}.\n$$\n\nLet $BY = BZ = t$. Then $AB = 6 + t$, and so again by the Pythagorean Theorem\n\n$$\n6\\sqrt{3} = \\sqrt{(6+t)^2 - (6-t)^2} = 2\\sqrt{6t}.\n$$\n\nThus $t = \\frac{9}{2}$, so $BC = \\frac{49}{2}$, and the area of $ABCD$ is $\\frac{49}{2} \\cdot 6\\sqrt{3} = 147\\sqrt{3}$. The requested sum is $147 + 3 = 150$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18438,
    "subject": "Mathematics (Olympiad)",
    "question": "The points $A_1$ and $C_1$ are chosen on the sides $BC$ and $AB$ of triangle $ABC$ so that the segments $AA_1$ and $CC_1$ are equal and perpendicular. Prove that if $\\angle ABC = 45^\\circ$, then\n\n$$\nAC = AA_1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we will show that the triangle is acute. Suppose that $\\angle BAC \\geq 90^\\circ$. Then $\\angle ACB \\leq 45^\\circ = \\angle ABC < \\angle AA_1C$. From $\\triangle AA_1C$ we get that $AC > AA_1$. On the other hand, $\\angle C_1AC \\geq 90^\\circ$, so in $\\triangle AC_1C$ the side $CC_1$ is the longest, in particular, $CC_1 > AC$. Hence $CC_1 > AC > AA_1$, a contradiction.\n\nLet $AA_1$ and $CC_1$ intersect at $O$. Consider a circle with diameter $AC$; then $O$ lies on it. This circle intersects $AB$ and $BC$ at $P$ and $Q$ respectively. The points $P$ and $Q$ lie inside the sides of $\\triangle ABC$ because the triangle is acute. Also, since $AA_1$ and $CC_1$ are cevians, $O$ lies inside $\\triangle ABC$, that is, on the arc $PQ$ of the circle. We will prove that $\\angle A_1AQ = \\angle QAC$. If this is true, $\\triangle CAA_1$ is isosceles, because $AQ$ is both angle bisector and altitude, so $A_1A = AC$. Suppose $\\angle A_1AQ = \\beta > \\alpha = \\angle QAC$. Then $A_1A > AC$. Obviously, $\\angle PBC = \\angle BAQ = 45^\\circ$. Then\n\n$$\n\\angle PCC_1 = 45^\\circ - \\angle OCQ = 45^\\circ - \\angle OAQ = 45^\\circ - \\beta.\n$$\n\nAlso, $\\angle PCA = 90^\\circ - \\angle PAC = 45^\\circ - \\alpha$. Since $\\beta > \\alpha$, we have $45^\\circ - \\beta < 45^\\circ - \\alpha$, so $\\angle C_1CP < \\angle PCA$, hence $CC_1 < AC$. But $A_1A > AC$, so $CC_1 < AA_1$—contradiction. The case $\\beta < \\alpha$ is considered analogously.\n\n![](images/Ukraine_2016_Booklet_p8_data_f7581fd248.png)\n\n**Alternative solution.** Denote $\\angle C_1CB = \\alpha$, $\\angle A_1AB = \\beta$. Since $\\angle OCA + \\alpha + \\angle OAC + \\beta = 135^\\circ$ and $\\angle OCA + \\angle OAC = 90^\\circ$, we have $\\alpha + \\beta = 45^\\circ$. We can assume $AA_1 = CC_1 = 1$. Denote $OC_1 = p$, $OA_1 = q$, then $OC = 1-p$, $OA = 1-q$. Since $\\tan \\alpha = \\frac{q}{1-p}$ and $\\tan \\beta = \\frac{p}{1-q}$, rewrite the condition $\\alpha + \\beta = 45^\\circ$ as $\\tan(\\alpha + \\beta) = 1$ or\n\n$$\n\\begin{aligned}\n\\tan(\\alpha + \\beta) &= \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta} = \\frac{\\frac{q}{1-p} + \\frac{p}{1-q}}{1 - \\frac{q}{1-p} \\frac{p}{1-q}} \\\\\n&= \\frac{q - q^2 + p - p^2}{(1-p)(1-q) - pq} = \\frac{q + p - (q^2 + p^2)}{1 - p - q} = 1,\n\\end{aligned}\n$$\n\nso $p^2 + q^2 - 2p - 2q + 1 = 0$. Now,\n\n$$\nAC^2 = (1-p)^2 + (1-q)^2 = 2 - 2p - 2q + p^2 + q^2 = 1 = AA_1^2\n$$\n\nand we are done.\n\n![](images/Ukraine_2016_Booklet_p8_data_03c995e627.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18439,
    "subject": "Mathematics (Olympiad)",
    "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 $n \\ge 1$.\n\nProve 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": "See solution",
    "solution": "Let $k$ be a nonzero natural number such that $1 \\leq k < 2x_n$ for some $n \\geq 1$. 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 $1 \\leq k < 2x_{N+1}$, 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 $2x_N \\leq k < 2x_{N+1}$, since the case $k < 2x_N$ is already covered by the induction hypothesis. In this case, we have $k - x_{N+1} \\geq 2x_N - x_{N+1} \\geq 0$ 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\n$$\n0 < k - x_{N+1} < 2x_{N+1} - x_{N+1} \\leq 2x_N.\n$$\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": 18440,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos 50 fichas numeradas del 1 al 50, y hay que colorearlas de rojo o azul. Sabemos que la ficha 5 es de color azul. Para la coloración del resto de fichas se siguen las siguientes reglas:\n\n- Si la ficha con el número $x$ y la ficha con el número $y$ son de distinto color, entonces la ficha con el número $|x - y|$ se pinta de color rojo.\n\n- Si la ficha con el número $x$ y la ficha con el número $y$ son de distinto color y $x \\cdot y$ es un número entre 1 y 50 (incluyendo ambos), entonces la ficha con el número $x \\cdot y$ se pinta de color azul.\n\nDetermina cuántas coloraciones distintas se pueden realizar en el conjunto de fichas.",
    "options": [],
    "answer": "See solution",
    "solution": "Observemos que dos números que se diferencian en 5 tienen el mismo color. En efecto, si fueran de distinto color, su diferencia debería ser de color rojo, por la primera regla. Pero su diferencia es 5, que es de color azul. Por tanto, basta con saber el color de los 4 primeros números. Distinguimos dos casos:\n\n**Caso 1:** Si 1 es de color azul, el resto de fichas deberá ser de color azul, por la segunda regla. Esto es así porque si la ficha $k \\neq 1$ fuera roja, entonces por la segunda regla, $k = k \\cdot 1$ tendría que ser azul, lo que contradice que $k$ sea roja.\n\n**Caso 2:** Si 1 es roja, por la primera regla $4 = 5 - 1$ es roja. Para determinar el color de 2 y 3, supongamos que 3 es azul. Como $2 = 3 - 1$ y 3 y 1 son de diferente color, entonces 2 es roja. Ahora bien, $3 = 5 - 2$ y 5 es azul y 2 roja, por lo tanto 3 es roja. Esto no puede ser, por lo tanto 3 no puede ser azul y es roja, por lo que 2 también es roja. Así pues, 1, 2, 3 y 4 son rojas, lo mismo que el resto de fichas que no son múltiplo de 5.\n\nPor tanto, solo hay dos coloraciones posibles: o todas las fichas de color azul, o todas rojas excepto los múltiplos de 5, que son azules.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18441,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We will show that we can obtain the desired configuration for any number of at least $3434$ tokens.\n\nWe number the colors as $1, 2, \\ldots, p$, and the sizes as $1, 2, \\ldots, q$. We arrange the tokens in a rectangular table such that if a token has color $i$ and size $j$, it is placed at the intersection of row $i$ and column $j$.\n\n**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\n$$\nN = k + (k + 1) + (k + 2) + \\dots + (3k - 1) = k(4k - 1)\n$$\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 $xy \\leq k^2$, 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 $1, 2, \\ldots, k-1$ tokens on the first $k-1$ rows, and $2k, 2k+1, \\ldots, 3k-1$ 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 $1, 2, \\ldots, k-1, 2k, 2k+1, \\ldots, 3k-1$ tokens, and the first $k$ columns represent rows with $2k-1, 2k-2, \\ldots, k$ tokens.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18442,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ be the intersection of the bisector of the segment $\\overline{AB}$ with the segment $\\overline{BC}$.\n\n% IMAGE: ![](images/Croatia_2018_p10_data_d7fdadea21.png)\n\nFind the relationship between $\\angle ACB$ and $\\angle DCB$ in the given configuration.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\beta = \\angle ABC$ and notice that $\\angle ACB = 180^\\circ - 3\\beta$.\n\nSince $E$ lies on the bisector of the segment $\\overline{AB}$, we have $|AE| = |BE|$. Therefore, $\\angle BAE = \\angle ABC = \\beta$. This implies $\\angle EAC = \\angle CAB - \\beta = 2\\beta - \\beta = \\beta$.\n\nLet $F$ be the other intersection of the line $AE$ with the circle of radius $\\overline{CA}$ centred at $C$. Since $CAF$ is an isosceles triangle ($\\overline{CA}$ and $\\overline{CF}$ are both radii of the same circle), we get $\\angle CFA = \\beta$.\n\nFrom $\\angle CFA = \\angle BAF$ we get $CF \\parallel AB$ (these are the angles of the transversal).\n\nThis also means that $\\angle BCF = \\angle CBA = \\beta$, which shows that $CEF$ is an isosceles triangle.\n\nFrom $CF \\parallel AB$ and the fact that $E$ is equidistant to $C$ and $F$, we conclude that the line $DE$ is the bisector of the segment $\\overline{CF}$ as well.\n\nThus, $|DF| = |DC| = |CF|$, i.e. the triangle $DFC$ is equilateral.\n\n$$\n\\text{Finally, we have } \\angle DCB = 60^\\circ - \\beta = \\frac{1}{3}(180^\\circ - 3\\beta) = \\frac{1}{3}\\angle ACB, \\text{ i.e. } \\angle ACB = 3\\angle DCB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18443,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle ($AB < BC < AC$) with circumcircle $\\Gamma$. Assume there exists $X \\in AC$ satisfying $AB = BX$ and $AX = BC$. Points $D, E \\in \\Gamma$ are taken such that $\\angle ADB < 90^\\circ$, $DA = DB$ and $BC = CE$. Let $P$ be the intersection point of $AE$ with the tangent line to $\\Gamma$ at $B$, and let $Q$ be the intersection point of $AB$ with the tangent line to $\\Gamma$ at $C$. Show that the projection of $D$ onto $PQ$ lies on the circumcircle of $\\triangle PAB$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, it is easy to check that $B$ lies between $A$ and $Q$ and $A$ lies between $P$ and $E$ using $AB < BC < AC$. Let $T$ and $S$ be the intersection points of the lines $BP$ and $AP$ with $QC$, respectively. Since $CE = CB$, we have $\\angle EAC = \\angle EBC = \\angle BEC = \\angle A$, which gives $\\angle BAS = \\angle BAE = 2\\angle A$. Also, $\\angle TBC = \\angle TCB = \\angle A$ gives $\\angle BTS = \\angle BTC = 180^\\circ - 2\\angle A$, so $A, B, T, S$ are concyclic.\n\nOn the other hand, since $AB = BX$, we find $\\angle BXA = \\angle BAX = \\angle A$, which implies $\\triangle TBC$ and $\\triangle BAX$ are congruent using $AX = BC$, so we have $TB = BA$. Then,\n\n$$\n\\angle BTA = \\angle BAT = \\frac{180^\\circ - \\angle B - \\angle A}{2} = \\frac{\\angle C}{2} = \\frac{\\angle ADB}{2}.\n$$\n\nHence, as $D$ lies on the perpendicular bisector of the segment $AB$, we can conclude that $D$ is the center of the cyclic quadrilateral $ABTS$. By noting that $AS \\cap BT = \\{P\\}$ and $AB \\cap TS = \\{Q\\}$, the result follows from the following well-known lemma.\n\n**Lemma.** Let $ABCD$ be a cyclic quadrilateral, with circumcenter $O$. Let $AB \\cap CD = \\{P\\}$ and $AD \\cap BC = \\{Q\\}$. Then, the projection of $O$ onto $PQ$ is the Miquel point of $ABCD$.\n\n![](images/BMO_2023_Short_List_p27_data_f46579844c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18444,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for infinitely many positive integers $n$ there exist pairwise distinct positive integers $a_1, a_2, \\dots, a_n$ such that\n$$\na_1^2 a_2^2 \\cdots a_n^2 - 4(a_1^2 + a_2^2 + \\cdots + a_n^2)\n$$\nis the square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f_n(a_1, a_2, \\dots, a_n) = a_1^2 a_2^2 \\cdots a_n^2 - 4(a_1^2 + a_2^2 + \\cdots + a_n^2)$. The conclusion follows from two facts:\n\n1. 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 a perfect square, and $a_1 < a_2$.\n2. Given $n \\ge 3$, if $a_1, a_2, \\dots, a_n$ are positive integers such that $f_n(a_1, a_2, \\dots, a_n)$ is a perfect square, and $a_1 < a_2 < \\dots < a_k$ for some $1 < k < n$, then there exists $a > a_k$ such that $f_n(a_1, a_2, \\dots, a_k, a, a_{k+2}, \\dots, a_n)$ is a perfect square.\n\nTo prove (1): 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 $4$. Subtracting a suitable number of $4$'s yields $0$ or $1$, each a square. Letting that number be $n-2$, $f_n(a_1, a_2, 1, \\dots, 1)$ is a perfect square.\n\nTo prove (2): Write $f_n(a_1, a_2, \\dots, a_n) = b^2$ and note that $(a_{k+1}, b)$ solves the Pell equation\n$$\n(a_1^2 \\cdots a_k^2 a_{k+2}^2 \\cdots a_n^2 - 4)x^2 - y^2 = 4(a_1^2 + \\cdots + a_k^2 + a_{k+2}^2 + \\cdots + a_n^2).\n$$\nThis equation has infinitely many positive integer solutions. In particular, there is a solution $(a, c)$ with $a > a_{k+1}$. This establishes (2) and completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18445,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest possible value of $c$ such that there exist positive integers $a$, $b$, and $c$ satisfying the following conditions:\n\n- $a$ is a multiple of $b$.\n- $b$ is a multiple of $c$.\n- $a$, $b$, and $c$ all have the same number of digits in base 10.\n- $c$ is obtained by removing a single digit '6' from $b$ (in base 10).\n\nWhat is the smallest possible $c$?",
    "options": [],
    "answer": "See solution",
    "solution": "We note that under the hypothesis, the top digits of $a_{(10)}$ and $b_{(10)}$ must be different, since if they are the same, then $\\frac{a}{b} < 2$, which violates the condition that $a$ is a multiple of $b$. Therefore, the top digit of $a_{(10)}$ or $b_{(10)}$ must be 6, but if the top digit of $b_{(10)}$ is 6, then $a \\ge 2b$ forces $a$ to have more digits than $b$, contradicting the assumption. Thus, the top digit of $a_{(10)}$ must be 6. Furthermore, since the number of digits of $a$ and $b$ are the same, $\\frac{a}{b}$ must be one of $2, 3, 4, 5, 6$. Let this number be $m$.\n\nNext, if $c$ is the minimum number satisfying the conditions, then the 6 to be removed from $b_{(10)}$ to get $c_{(10)}$ must be one of the last three digits of $b_{(10)}$. The minimality of $c$ implies that $b$ cannot be a multiple of 10. If the last digit of $b$ is 0, then the last digit of both $a_{(10)}$ and $c_{(10)}$ is 0, and dividing $a, b, c$ by 10 gives a smaller triple, violating minimality. If the 6 to be removed from $b_{(10)}$ is not among the last three digits, then $a \\equiv b \\pmod{1000}$, and since $a = mb$, $(m-1)b \\equiv 0 \\pmod{1000}$, so $b$ is a multiple of 10, which is impossible.\n\nNow, we consider cases depending on which digit the 6 to be removed from $b_{(10)}$ lies:\n\n**(1) Case: 6 is the last digit of $b$.**\n\nSince $a \\equiv mb \\pmod{10}$ and $b \\equiv 6 \\pmod{10}$, the last digit of $a_{(10)}$ equals the next-to-last digit of $b_{(10)}$. Continuing this process, we can determine the digits of $a_{(10)}$ from the bottom up. If the $n$-th digit from the bottom of $a_{(10)}$ equals 6, then we let $a, b$ be as determined. If $a \\equiv mb \\pmod{10^n}$, and $c_{(10)}$ is obtained from $b_{(10)}$ by removing the 6 in the last digit, then $(a, b, c)$ satisfies all conditions. Checking $m = 2, 3, 4, 5, 6$, we see that with $m = 4$, $a = 615384$, $b = 153846$, and $c = 15384$ gives the minimum.\n\n**(2) Case: 6 is the next-to-last digit of $b$.**\n\nFrom $a \\equiv mb \\pmod{10}$ and $a \\equiv b \\pmod{10}$, the last digit of $a, b$ is 2, 4, 6, or 8 if $m = 6$, and 5 if $m = 3$ or 5. Checking these possibilities, the $c$ obtained has at least 6 digits.\n\n**(3) Case: 6 is the third digit from the bottom of $b$.**\n\nFrom $a \\equiv mb \\pmod{100}$ and $a \\equiv b \\pmod{100}$, the last two digits of $a, b$ are 25 and 75, respectively. Checking these cases, $c$ has at least 6 digits.\n\nTherefore, the smallest possible $c$ is $15384$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18446,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a positive integer $n$ *delightful* if there exists an integer $x$, $1 < x < n$, such that\n$$\n1 + 2 + \\cdots + (x - 1) = (x + 1) + (x + 2) + \\cdots + n.\n$$\n\nDoes there exist a delightful number $N$ satisfying\n$$\n2013^{2013} < \\frac{N}{2013^{2013}} < 2013^{2013} + 4?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a delightful number $n$. Then there exists an integer $x$, $1 < x < n$ satisfying\n$$\n\\sum_{i=1}^{x-1} i = \\sum_{i=x+1}^{n} i.\n$$\nBut\n$$\n\\sum_{i=x+1}^{n} i = \\sum_{i=1}^{n} i - \\sum_{i=1}^{x} i.\n$$\nSo,\n$$\n\\sum_{i=1}^{x-1} i = \\sum_{i=1}^{n} i - \\sum_{i=1}^{x} i\n$$\nwhich simplifies to\n$$\n\\sum_{i=1}^{n} i = \\sum_{i=1}^{x-1} i + \\sum_{i=1}^{x} i = x^2.\n$$\nBut $\\sum_{i=1}^{n} i = \\frac{n(n+1)}{2}$, so $\\frac{n(n+1)}{2} = x^2$.\n\nNow, $n$ and $n+1$ are coprime, so one of them is even and the other must be an odd perfect square (since $x^2$ is a square). Now consider the inequality\n$$\n(2013^{2013})^2 < n < (2013^{2013})^2 + 4 \\cdot 2013^{2013} = (2013^{2013} + 2)^2 - 4.\n$$\nThe only perfect square in this interval is $(2013^{2013} + 1)^2$, which is even. Therefore, neither $n$ nor $n+1$ can be an odd perfect square. Hence, no delightful number $N$ satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18447,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integer $n$, let $\\tau(n)$ denote the number of positive divisors of $n$ and $\\varphi(n)$ the number of positive integers not greater than $n$ which are relatively prime to $n$. Find all positive integers $n$ for which one of the three numbers $n$, $\\tau(n)$, $\\varphi(n)$ is the arithmetic mean of the other two.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\tau(1) = \\varphi(1) = 1$, so $n = 1$ satisfies the condition. For $n > 1$, clearly $\\tau(n) \\leq n$ and $\\varphi(n) < n$, so $n$ cannot be the arithmetic mean of $\\tau(n)$ and $\\varphi(n)$. We are left with two cases:\n\n**Case 1:** $\\tau(n) = \\frac{1}{2}(\\varphi(n) + n)$. Then $\\tau(n) > \\frac{1}{2}n$. For each divisor $d$ of $n$, $n/d$ is also a divisor. One of $d, n/d$ is $\\leq \\sqrt{n}$, so $\\{1, 2, \\dots, \\lfloor\\sqrt{n}\\rfloor\\}$ contains at least half the divisors\\textsuperscript{1}. Thus $\\frac{1}{2}\\tau(n) \\leq \\sqrt{n}$, so:\n\n$$\n2\\sqrt{n} \\geq \\tau(n) > \\frac{1}{2}n \\implies 4n > \\frac{1}{4}n^2 \\implies 16 > n.\n$$\n\nFor $1 < n < 16$, we can calculate $\\tau(n)$, check $\\tau(n) > \\frac{1}{2}n$, and calculate $\\varphi(n)$:\n\n| $n$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ |\n|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|\n| $\\tau(n)$ | $2$ | $2$ | $3$ | $2$ | $4$ | $2$ | $4$ | $3$ | $4$ | $2$ | $6$ | $2$ | $4$ | $4$ |\n| $\\tau(n) > \\frac{1}{2}n$? | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ | $\\checkmark$ |\n| $\\varphi(n)$ | $1$ | $2$ | $2$ |  | $2$ |  |  |  |  |  |  |  |  |  |\n| $\\tau(n) = \\frac{1}{2}(\\varphi(n) + n)$? | $\\times$ | $\\times$ | $\\checkmark$ |  | $\\checkmark$ |  |  |  |  |  |  |  |  |  |\n\nSo $n = 4$ and $n = 6$ are solutions.\n\n**Case 2:** $\\varphi(n) = \\frac{1}{2}(\\tau(n) + n)$. This gives\n\n$$\n\\tau(n) = 2\\varphi(n) - n. \\qquad (1)\n$$\n\nIf $n$ is even, then no even number is relatively prime to $n$, so $\\varphi(n) \\leq \\frac{1}{2}n$. From (1), $\\tau(n) \\leq 0$, impossible. So $n$ must be odd. Then (1) implies $\\tau(n)$ is odd, so $n$ is a perfect square (of an odd number). Write $n = p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k}$, $k \\geq 1$, $p_i \\geq 3$, $\\alpha_i \\geq 1$.\n\n\\textsuperscript{1} Exactly half if $n$ is not a perfect square.\n\nUsing formulas for $\\tau(n)$ and $\\varphi(n)$, (1) becomes\n\n$$\n(2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1) = 2p_1^{2\\alpha_1-1}(p_1-1) \\cdots p_k^{2\\alpha_k-1}(p_k-1) - p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k} = \\\\ = p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}(2(p_1-1) \\cdots (p_k-1) - p_1 \\cdots p_k).\n$$\n\nThe right side is divisible by $p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}$, so the left must be as well. Thus\n\n$$\np_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1} \\leq (2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1). \\quad (2)\n$$\n\nFor $p \\geq 3$, $\\alpha \\geq 1$, $p^{2\\alpha-1} \\geq (2\\alpha + 1)$, equality only for $p = 3$, $\\alpha = 1$. Induction: for $\\alpha = 1$ is trivial (equality only for $p = 3$); when $\\alpha$ increases by $1$, right side increases by $2$, left by $1$.\n\n$$\np^{2(\\alpha+1)-1} - p^{2\\alpha-1} = p^{2\\alpha-1}(p^2 - 1) > 2.\n$$\n\nSo each factor on the left of (2) is $\\geq$ the corresponding right factor. Only possible is $k = 1$, $p_1 = 3$, $\\alpha_1 = 1$, i.e., $n = 9$. Indeed, $\\tau(9) = 3$, $\\varphi(9) = 6$, so (1) holds for $n = 9$.\n\n**Answer:** The condition is fulfilled for $n \\in \\{1, 4, 6, 9\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18448,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a \\in [\\frac{1}{2}, \\frac{3}{2}]$ be a real number. Sequences $(u_n)$ and $(v_n)$ are defined as follows:\n\n$$\nu_n = \\frac{3}{2^{n+1}} \\cdot (-1)^{\\lfloor 2^{n+1}a \\rfloor}, \\quad v_n = \\frac{3}{2^{n+1}} \\cdot (-1)^{n+\\lfloor 2^{n+1}a \\rfloor}.$$\n\na) Prove that\n\n$$\n(u_0 + u_1 + \\cdots + u_{2018})^2 + (v_0 + v_1 + \\cdots + v_{2018})^2 \n\\leq 72a^2 - 48a + 10 + \\frac{2}{4^{2019}}.\n$$\n\nb) Find all values of $a$ in the equality case.",
    "options": [],
    "answer": "See solution",
    "solution": "a) From the assumption, we have $v_i = u_i$ for even $i$ and $v_i = -u_i$ for odd $i$. Let us introduce some notation to simplify the expression:\n\n$$\nx = \\sum_{i=0}^{1009} u_{2i}, \\quad y = \\sum_{i=0}^{1008} u_{2i+1}.\n$$\n\nThus, the inequality can be rewritten as\n\n$$\n(x+y)^2 + (x-y)^2 \\leq 72a^2 - 48a + 10 + \\frac{2}{4^{2019}},\n$$\nwhich is equivalent to\n$$\nx^2 + y^2 \\leq 36a^2 - 24a + 5 + \\frac{1}{4^{2019}}.\n$$\n\nNow, let's write the binary representation of $a$:\n\n$$\na = \\sum_{i=1}^{\\infty} \\frac{x_i}{2^i},\n$$\nwhere $x_i \\in \\{0, 1\\}$. Since $\\frac{1}{2} \\leq a \\leq \\frac{3}{2}$, we have $x_1 = 1$.\n\nFor each natural number $i$, the parity of $\\lfloor 2^{i+1}a \\rfloor$ depends on $x_{i+1}$. In particular, if $x_{i+1} = 0$ then $\\lfloor 2^{i+1}a \\rfloor$ is even, and if $x_{i+1} = 1$ then $\\lfloor 2^{i+1}a \\rfloor$ is odd. Therefore,\n\n$$\n(-1)^{\\lfloor 2^{i+1}a \\rfloor} = 1 - 2x_{i+1}.\n$$\n\nDenote $A = \\sum_{i=0}^{1009} \\frac{x_{2i+1}}{2^{2i+1}}$ and $B = \\sum_{i=0}^{1008} \\frac{x_{2i+2}}{2^{2i+2}}$. We have\n\n$$\n\\sum_{i=0}^{1009} u_{2i} = \\sum_{i=0}^{1009} \\frac{3(1 - 2x_{2i+1})}{2^{2i+1}} = 2 - \\frac{1}{2 \\cdot 4^{1009}} - 6A,\n$$\n\n$$\n\\sum_{i=0}^{1008} u_{2i+1} = \\sum_{i=0}^{1008} \\frac{3(1 - 2x_{2i+2})}{2^{2i+2}} = 1 - \\frac{1}{4^{1009}} - 6B.\n$$\n\nOn the other hand, $a \\geq A + B \\geq \\frac{1}{2}$, so\n\n$$\n36a^2 - 24a + 5 + \\frac{1}{4^{2019}} = 4(3a - 1)^2 + 1 + \\frac{1}{4^{2019}} \\geq 4(3A + 3B - 1)^2 + 1 + \\frac{1}{4^{2019}}.\n$$\n\nWe will prove that\n\n$$\n\\left(2 - \\frac{1}{2 \\cdot 4^{1009}} - 6A\\right)^2 + \\left(1 - \\frac{1}{4^{1009}} - 6B\\right)^2 \\leq 4(3A + 3B - 1)^2 + 1 + \\frac{1}{4^{2019}}.\n$$\n\nBy some calculations, we can rewrite the above inequality as\n\n$$\n\\frac{6}{4^{1009}}A + 12B \\left(1 + \\frac{1}{4^{1009}} - 6A\\right) \\leq \\frac{1}{4^{1008}} - \\frac{1}{4^{2018}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18449,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}$ such that for all $x, y > 0$,\n$$\nf(x^2 + x f(y) + y) = 2x + f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r > s$ be two positive real numbers. Then, there exists $t > 0$ such that $t^2 + f(s)t + s = r$. Now, set $(x, y) = (t, s)$:\n$$\nf(r) = f(t^2 + f(s)t + s) = 2t + f(s) > f(s).\n$$\nSo, $f$ is strictly increasing and hence injective.\n\nPlugging $(x, y) = \\left(\\frac{f(z)}{2}, y\\right)$ gives:\n$$\nf(z) + f(y) = f\\left(\\frac{f(z)^2}{4} + \\frac{f(z)f(y)}{2} + y\\right) = f\\left(\\frac{f(y)^2}{4} + \\frac{f(z)f(y)}{2} + z\\right)\n$$\nThus, $\\frac{f(z)^2}{4} - z = C$ for some constant $C$. Hence, $f(z) = \\sqrt{4z + C}$ for some $C \\ge 0$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18450,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{2, \\frac{3}{2}, \\frac{49}{48}\\}$ and define $f(n)$ as the minimal number of elements from $A$ (with repetitions allowed) whose product is $n$. Prove that there exist infinitely many pairs $(x, y)$ with $x \\ge 2$, $y \\ge 2$, such that\n$$\nf(xy) < f(x) + f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $f(xy) < f(x) + f(y)$ for $(x, y) = (7, 7)$. We have $f(7) \\ge 4$ since 7 cannot be written as the product of three or fewer elements of $A$: $2^3 > 7$, and any other product of at most three elements of $A$ does not exceed $2^2 \\times \\frac{3}{2} = 6 < 7$. Also, $f(49) \\le 7$ since $49 = 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times \\frac{3}{2} \\times \\frac{49}{48}$. Hence,\n$$\nf(49) \\le 7 < 8 \\le f(7) + f(7).\n$$\n\nNow suppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(xy) < f(x) + f(y)$. This implies that there exists an $M$ large enough such that whenever $a > M$ or $b > M$, we have $f(ab) = f(a) + f(b)$. (Note that $f(ab) \\le f(a) + f(b)$ is always satisfied.)\n\nNow take any pair $(x, y)$ that satisfies $f(xy) < f(x) + f(y)$ and let $n > M$ be any integer. Then\n$$\nf(n) + f(xy) = f(nxy) = f(nx) + f(y) = f(n) + f(x) + f(y),\n$$\nwhich contradicts $f(xy) < f(x) + f(y)$.\n\nTherefore, there exist infinitely many pairs $(x, y)$ with $x \\ge 2$, $y \\ge 2$, and $f(xy) < f(x) + f(y)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18451,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle on the plane. The angle bisector from vertex $A$ meets side $BC$ at $P$, and the median from vertex $B$ meets side $AC$ at $M$. The lines $AB$ and $MP$ meet at point $K$. Prove that if $\\frac{|PC|}{|BP|} = 2$, then $AP$ and $CK$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K'$ be a point on the ray $AB$ such that $B$ is the midpoint of the segment $AK'$. Then $CB$ is the median of triangle $ACK'$. As $P$ divides this segment in the ratio $2:1$, $P$ must be the centroid of triangle $ACK'$. So, $K'M$, which is also a median of triangle $ACK'$, must pass through point $P$. Therefore, $K = K'$. So, $B$ is the midpoint of segment $AK$. As $AP$ passes through point $P$, $AP$ is also a median of triangle $ACK$. By the premises, it is also an angle bisector. So, triangle $ACK$ is isosceles with $|AC| = |AK|$ and $AP$ is its height. Therefore, $AP \\perp CK$.\n\n![](images/Estonija_2012_p15_data_ddcaba801d.png)\n\nFig. 10",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18452,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a scalene acute triangle. Let $D$ be the orthogonal projection of $A$ onto $BC$, and let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Let $P$ and $Q$ be points on the minor arcs $\\widehat{AB}$ and $\\widehat{AC}$ of the circumcircle of $\\triangle ABC$, such that $PQ \\parallel BC$. Show that the circumcircles of $\\triangle DPQ$ and $\\triangle MND$ are tangent to each other if and only if $PQ$ passes through $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $PQ$ does not pass through the midpoint of $AB$, but $(DPQ)$ is tangent to the circle $(MND)$.\n\n![](images/BMO2024Shortlist_p54_data_a880d78ec1.png)\n\nConsider $i$, the inversion with pole $A$ and $k = \\frac{AB \\cdot AC}{2}$, followed by reflection with respect to the angle bisector of $\\angle BAC$. Denote $X' = i(X)$ for any $X$ in the plane.\n\nNotice that $B' = N$, $C' = M$, the midpoints of $AC$ and $AB$ respectively, and that Euler's circle of $\\triangle ABC$ is $(DMN)$, so its 'inverse' is the circle $(D'M'N')$. Now, $M' = C$, $N' = B$, and $D'$ is the circumcenter of $ABC$, denoted by $O$. Indeed, the line $B - D - C$ is sent to the circle $(AND'C)$, which is the circle with diameter $AO$. Since $AO$ and $AD$ are isogonals, it follows that $D' = O$. Hence, the circle $(DMN)$ is sent to $(OBC)$. At the same time, the circle $ABC$ is sent to the line $MN$.\n\nAs $PQ \\parallel BC$, it follows that arcs $PB$ and $QC$ are equal, so $AP$ and $AQ$ are isogonals, $P' = AQ \\cap MN$, $Q' = AP \\cap MN$, and the circle $PDQ$ is sent to $P'OQ'$.\n\nLet $P_1, Q_1$ be the points where the line $MN$ cuts the minor arcs $AB$ and $AC$. We will prove that $(DMN)$ is tangent to $(DP_1Q_1)$. By the same argument as before, we get\n\n$P'_1 = Q_1$ and $Q'_1 = Q_1$, and then the circles $(OP_1Q_1)$ and $(OBC)$ are tangent as they are isosceles with $OB = OC$ and $OP = OQ$. Hence $(P_1DQ_1)$ and $(MDN)$ are tangent at $D$ and $M \\in P_1Q_1$.\n\nNow $(DPQ)$ and $(DMN)$ are tangent if and only if $(OP'Q')$ and $(OBC)$ are tangent, so if and only if the tangent at $O$ to $(OBC)$ is the tangent at $O$ to $(OP'Q')$, which happens if and only if the triangle $OP'Q'$ is isosceles with base $P'Q'$, which is equivalent to $OP' = OQ'$. So we have $OP' = OQ'$ and also $OP_1 = OQ_1$. It follows that\n\n$$\n\\begin{align*}\nQ'P_1 = Q_1P' &\\Leftrightarrow Q'Q'_1 = P'P'_1 \\Leftrightarrow QQ_1 \\cdot \\frac{k}{AQ \\cdot AQ_1} = PP_1 \\cdot \\frac{k}{AP \\cdot AP_1} \\\\\n&\\Leftrightarrow AQ \\cdot AQ_1 = AP \\cdot AP_1 \\Leftrightarrow S_{QQ_1} = S_{APP_1} \\Leftrightarrow \\operatorname{dist}(A, QQ_1) = \\operatorname{dist}(A, PP_1).\n\\end{align*}\n$$\n\nBut this means that $A$ lies on the segment bisector of $P_1Q_1$ and $PQ$ respectively. So the minor arcs $AB$ and $AC$ are equal and the triangle is isosceles, contradiction. The conclusion follows now. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18453,
    "subject": "Mathematics (Olympiad)",
    "question": "Jüri writes on the blackboard some consecutive integers. The total number of these integers is greater than one, and the least of them is greater than $2$. Mari also writes consecutive integers on the blackboard, with the same total number as Jüri, but the least of them equals $1$. Is it possible that the product of the integers written by Jüri divided by the product of the integers written by Mari is equal to the square of some integer?",
    "options": [],
    "answer": "See solution",
    "solution": "For example, the product $8 \\cdot 9$ of two consecutive integers divided by the product $1 \\cdot 2$ of the first two positive integers equals $6^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18454,
    "subject": "Mathematics (Olympiad)",
    "question": "Sam is playing a game with 2023 cards labelled $1, 2, 3, \\ldots, 2023$. The cards are shuffled and placed in a pile face down. On each turn, Sam thinks of a positive integer $n$ and then looks at the number on the topmost card. If the number on the card is at least $n$, then Sam gains $n$ points; otherwise Sam gains 0 points. Then the card is discarded. This process is repeated until there are no cards left in the pile.\n\nFind the largest integer $P$ such that Sam can guarantee a total of at least $P$ points from this game, no matter how the cards were originally shuffled.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $1012^2$\n\nFirst, note that $1012^2$ is achievable if Sam picks $1012$ every single time.\n\nWe shall show that Sam cannot do better than this if the deck responds as follows:\n\nOn each turn, if Sam chooses the number $c$, then:\n\n1. If all remaining cards in the deck are at least $c$, then the deck responds with the largest number in the deck. (We shall color each such card blue.)\n2. If there are any cards still left in the deck that are less than $c$, then the deck responds with the smallest number left in the deck. (Such cards are left uncolored.)\n\nFrom (1), the blue cards are turned over in the order $2023, 2022, \\ldots, b$ for some $b \\ge 1$.\n\nConsider a number $c$ that Sam chooses for which the deck responds with a blue card. Since the card is blue, (1) applies. Since (1) applies, the card numbered $b$ is still in the deck. It follows that $c \\leq b$. Since Sam gets 0 points for each uncolored card, and at most $b$ points for each blue card, Sam's total is at most\n\n$$\nb(2023 - b + 1) = 1012^2 - (b - 1012)^2 \\leq 1012^2\n$$\n\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18455,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any odd prime number $p$, the number of positive integers $n$ satisfying $p \\mid n! + 1$ is no more than $c p^{2/3}$, where $c$ is a constant independent of $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nClearly, if $n$ satisfies the required property, then $1 \\leq n \\leq p-1$. Denote all such $n$ by $n_1 < n_2 < \\cdots < n_k$; we shall show that $k \\leq 12 p^{2/3}$. If $k \\leq 12$ there is nothing to prove. In what follows, we assume that $k > 12$.\n\nRename $n_{i+1} - n_i$ ($1 \\leq i \\leq k-1$) in nondecreasing order as $1 \\leq \\mu_1 \\leq \\mu_2 \\leq \\cdots \\leq \\mu_{k-1}$. It is clear that\n\n$$\n\\sum_{i=1}^{k-1} \\mu_i = n_k - n_1 < p.\n$$\n\nFirst, we show that for any $s \\geq 1$,\n\n$$\n| \\{ 1 \\leq i \\leq k-1 : \\mu_i = s \\} | \\leq s,\n$$\n\ni.e., there are at most $s$ values of $\\mu_i$ equal to $s$.\n\nIn fact, suppose that $n_{i+1} - n_i = s$; then $n_i! + 1 \\equiv n_{i+1}! + 1 \\equiv 0 \\pmod{p}$, so $(p, n_i!) = 1$, and\n\n$$\n(n_i + s)(n_i + s - 1) \\cdots (n_i + 1) \\equiv 1 \\pmod{p}.\n$$\n\nThus, $n_i$ is a solution to the congruence\n\n$$\n(x + s)(x + s - 1) \\cdots (x + 1) \\equiv 1 \\pmod{p}.\n$$\n\nSince $p$ is prime, there are at most $s$ solutions to the above equation by Lagrange's theorem. Thus, there are at most $s$ $n_i$ with $n_{i+1} - n_i = s$.\n\nNow, we show that for any nonnegative integer $l$, if $\\frac{l(l+1)}{2} + 1 \\leq k - 1$, then $\\mu_{\\frac{l(l+1)}{2}+1} \\geq l + 1$. Suppose on the contrary that $\\mu_{\\frac{l(l+1)}{2}+1} \\leq l$. Then $\\mu_1, \\mu_2, \\dots, \\mu_{\\frac{l(l+1)}{2}+1}$ are all between $1$ and $l$. By the previous result, there are at most $1+2+\\dots+l = \\frac{l(l+1)}{2}$ $\\mu_i$ less than or equal to $l$, which contradicts the assumption.\n\nLet $m$ be the largest positive integer with $\\frac{m(m+1)}{2} + 1 \\leq k - 1$. Then\n\n$$\n\\frac{m(m+1)}{2} + 1 \\leq k - 1 < \\frac{(m+1)(m+2)}{2} + 1.\n$$\n\nHence,\n\n$$\n\\sum_{i=1}^{k-1} \\mu_i \\geq \\sum_{i=0}^{m-1} (i+1)^2 = \\frac{m(m+1)(2m+1)}{6} > \\frac{m^3}{3}.\n$$\n\nSince $k > 12$, $m \\geq 4$. Combining the above, we get\n\n$$\n\\begin{aligned}\nk &< 2 + \\frac{(m+1)(m+2)}{2} \\\\\n&< 4m^2 + 4 \\left( 3 \\sum_{i=1}^{k-1} \\mu_i \\right)^{2/3} \\\\\n&< 4 \\times (3p)^{2/3}.\n\\end{aligned}\n$$\n\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18456,
    "subject": "Mathematics (Olympiad)",
    "question": "At a round table, 50 inhabitants of the island \"Loud Mouths\" are sitting at equal distances from one another and discussing something. Each is either a knight (who always tells the truth) or a liar (who always lies), and both types are present at the table. During the discussion, everyone said that the islander sitting directly opposite them across the table and the two neighbours of that person were not all of one type (i.e., not all three knights and not all three liars). Prove that there are no fewer than 10 and not more than 25 liars at the table.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the following statement:\n\n**Lemma 1.** There is at least one knight among any two diametrically opposite islanders.\n\n*Proof.* Suppose the opposite is true: let $C$ and $X$ be two opposite sitting islanders, both liars. Then the neighbours of $X$, islanders $Y$ and $W$, must be liars, since $C$ must have lied, which meant that \"an islander sitting directly in front of him and his two neighbours are all of one type\". Analogously, both neighbours of $C$, islanders $B$ and $D$, must be liars. However, then there are two more opposite pairs of liars: $D$, $Y$ and $B$, $W$. Applying the same reasoning to them, we get that all the people at the table are liars. This contradicts the statement, so our assumption is false, and the lemma is proved.\n\n![](images/Ukraine_2020_booklet_p21_data_c2388bf614.png)\n\nFrom here, the estimate of the maximum number of liars becomes clear. Since there are 25 pairs of opposite-sitting people, the number of knights is not less than 25, hence, the number of liars is not greater than 25.\n\n**Lemma 2.** There is at least one liar among any five islanders sitting in a row.\n\n*Proof.* Suppose this does not hold, and there are five knights sitting in a row: $L_1, \\dots, L_5$. Then people opposite to $L_2$, $L_3$ and $L_4$ are liars, because they lied about three people of the same type sitting together. But then $L_3$ must have lied, because all three opposite of him/her are of the same type. This contradiction *proves the lemma*.\n\nNow we divide all 50 islanders sitting at the table into 10 groups of 5, each of them must contain at least 1 liar, hence, in total there are at least 10 liars.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18457,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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\nIf 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 $XY$, where $Y$ is a forbidden sequence with $k$ letters ($2 \\leq k \\leq n+1$) and $X$ is a correct word with $n-k+1$ letters. Thus, the forbidden words with $n+1$ letters are at most $a_0 + a_1 + a_2 + \\dots + a_{n-1}$, so\n\n$$\na_{n+1} \\geq 3a_n - (a_0 + a_1 + a_2 + \\dots + a_{n-1}).\n$$\n\nThis relation allows us to prove inductively that $a_{n+1} > 2a_n$ for every $n \\geq 1$. The base case is clear, and if this holds for all numbers from $0$ to $n-1$ ($n \\geq 2$), then $a_n \\geq 2^k a_{n-k}$ for $0 \\leq k \\leq n-1$, whence\n\n$$\na_{n+1} \\geq a_n \\left( 3 - \\left( \\frac{1}{2^n} + \\frac{1}{2^{n-1}} + \\dots + \\frac{1}{2} \\right) \\right) > 2a_n.\n$$\n\nThis shows that there are at least $2^n$ admissible words of length $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18458,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\frac{1}{a\\sqrt{c^2+1}} + \\frac{1}{b\\sqrt{a^2+1}} + \\frac{1}{c\\sqrt{b^2+1}} > 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, $c = \\frac{z}{x}$. Then\n\n$$\n\\frac{1}{a\\sqrt{c^2+1}} = \\frac{1}{\\frac{x}{y}\\sqrt{\\frac{z^2}{x^2}+1}} = \\frac{y}{\\sqrt{z^2+x^2}} \\geq \\frac{2y^2}{x^2+y^2+z^2}\n$$\n\nwhere the last inequality follows from the AM-GM inequality:\n\n$$\ny\\sqrt{x^2 + z^2} \\leq \\frac{y^2 + (x^2 + z^2)}{2}.\n$$\n\nApplying the same estimation to the other two terms, we get\n\n$$\n\\frac{1}{a\\sqrt{c^2+1}} + \\frac{1}{b\\sqrt{a^2+1}} + \\frac{1}{c\\sqrt{b^2+1}} \\geq \\frac{2y^2}{x^2+y^2+z^2} + \\frac{2z^2}{x^2+y^2+z^2} + \\frac{2x^2}{x^2+y^2+z^2} = 2.\n$$\n\nEquality holds only if $y^2 = x^2 + z^2$, $z^2 = x^2 + y^2$, and $x^2 = y^2 + z^2$, which is impossible. Thus, the original sum is strictly greater than $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18459,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)$ be a sequence defined by\n$$\nx_0 = 2, \\quad x_1 = 1, \\quad \\text{and} \\quad x_{n+2} = x_{n+1} + x_n, \\quad \\forall n \\ge 0.\n$$\n\n(a) For $n \\ge 1$, prove that if $x_n$ is a prime then $n$ is also a prime or $n$ does not have an odd prime divisor.\n\n(b) Find all pairs of non-negative integers $(m, n)$ such that $x_n$ is divisible by $x_m$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) It is known that $x_n = \\alpha^n + \\beta^n$ for all $n \\in \\mathbb{N}^*$, where $\\alpha < 0 < \\beta$ are the two distinct roots of the equation $\\lambda^2 - \\lambda - 1 = 0$.\n\nSuppose $x_n$ is a prime and $n$ is a positive integer with an odd prime divisor. Then $n = pq$ for some odd prime $p$ and integer $q > 1$. We have:\n$$\n\\begin{align*}\nx_{pq} &= \\alpha^{pq} + \\beta^{pq} \\\\\n&= (\\alpha^q + \\beta^q) \\left( \\alpha^{q(p-1)} - \\alpha^{q(p-2)} \\beta^q + \\dots + \\beta^{q(p-1)} \\right) \\\\\n&= x_q \\left( x_{q(p-1)} + \\dots + (-1)^{\\frac{q+1}{2}(p-1)} x_{2q} + (-1)^{\\frac{q+1}{2}} x_{2q+1} \\right )\n\\end{align*}\n$$\nThus, $x_q \\mid x_{pq}$. Since $(x_n)$ is an increasing sequence for $n \\ge 1$, $x_q > x_1 = 1$, so $x_{pq}$ is composite, a contradiction. Therefore, if $n \\ge 1$ and $x_n$ is prime, then $n$ must be a prime or have no odd prime divisor.\n\n(b) Consider the following cases:\n\n*Case 1: $m = 0$.*\n\nModulo 2, $x_0 \\equiv 0$, $x_1 \\equiv 1$, $x_2 \\equiv 1$, $x_3 \\equiv 0$, $x_4 \\equiv 1$, $x_5 \\equiv 1$, $x_6 \\equiv 0$, etc. Thus, $x_n$ is even for all positive $n$ divisible by 3, and odd otherwise. So $x_n$ is divisible by $x_0$ if and only if $n$ is divisible by 3. All such pairs are $(0, 3k)$ for $k \\in \\mathbb{N}$.\n\n*Case 2: $m = 1$.*\n\nWe have $(m, n) = (1, k)$ for $k \\in \\mathbb{N}$.\n\n*Case 3: $m > 1$.*\n\nFor all $k \\ge \\ell \\ge 0$,\n$$\n(\\alpha^k + \\beta^k)(\\alpha^\\ell + \\beta^\\ell) - (\\alpha^{k+\\ell} + \\beta^{k+\\ell}) = (\\alpha\\beta)^\\ell (\\alpha^{k-\\ell} + \\beta^{k-\\ell}) = (-1)^\\ell (\\alpha^{k-\\ell} + \\beta^{k-\\ell}).\n$$\nTherefore,\n$$\nx_{k+\\ell} = x_k x_\\ell - (-1)^\\ell x_{k-\\ell}.\n$$\nSo for $k \\ge 2\\ell \\ge 0$,\n$$\nx_k = x_{k-\\ell} x_\\ell - (-1)^\\ell x_{k-2\\ell}.\n$$\nThus, $x_\\ell \\mid x_k$ if and only if $x_\\ell \\mid x_{k-2\\ell}$, and so on. Therefore, $x_\\ell \\mid x_k$ if and only if $x_\\ell \\mid x_{k-2t\\ell}$ for some $t \\in \\mathbb{N}$ with $k \\ge 2t\\ell$. (*)\n\nSuppose $x_n$ is divisible by $x_m$ with $x_n \\ge x_m \\ge 3$, so $n \\ge m > 1$. Let $n = qm + r$ with $q \\in \\mathbb{N}^*$, $r \\in \\mathbb{N}$, $0 \\le r \\le m-1$.\n\n- If $q$ is even, by (*), $x_m \\mid x_n$ if and only if $x_m \\mid x_r$. But $x_r < x_m$ for $r < m$, so only possible if $r = 0$, which leads to a contradiction since $x_m > x_0$ for $m > 1$.\n- If $q$ is odd, $x_m \\mid x_n$ if and only if $x_m \\mid x_{m+r}$. But $x_{m+r} = x_m x_r - (-1)^r x_{m-r}$, so $x_m \\mid x_{m-r}$. For $0 < r < m$, $x_{m-r} < x_m$, contradiction; so $r = 0$.\n\nThus, all pairs are $(m, (2k+1)m)$ with $m > 1$, $k \\in \\mathbb{N}$.\n\n**Conclusion:** All pairs are $(0, 3k)$, $(1, k)$, and $(m, (2k+1)m)$ with $m, k \\in \\mathbb{N}$, $m > 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18460,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$, $n \\ge 2$, such that the numbers $1!, 2!, 3!, \\dots, (n-1)!$ have distinct remainders when divided by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim those integers are $n=2$ and $n=3$.\n\nSuppose $n$ is not a prime, so let $n = ab$, $1 < a \\leq b$.\n\nIf $a < b$, then $n = ab$ divides $b!$ and $(b+1)!$; since $b < b+1 < n-1$, it follows that $b!$ and $(b+1)!$ yield equal remainders (0) when divided by $n$, which is a contradiction.\n\nIf $2 < a = b$, then $(2a)!$ and $(2a+1)!$ are divisible by $n = a^2$. Since $2a+1 < a^2-1 = n-1$, we again reach a contradiction.\n\nIf $2 = a = b$ (i.e., $n = 4$), then $2! = 2$ and $3! = 6$ both yield remainder 2 when divided by 4, which is a contradiction.\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 $(n-1)! \\equiv -1 \\pmod n$ and $(n-2)! \\equiv 1 \\pmod n$. Since $1! = 1$ yields remainder 1 when divided by $n$, it follows $n-2=1$, that is $n=3$, which checks.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18461,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle BAC = 40^\\circ$ and $\\angle ABC = 80^\\circ$. Denote by $I$ its incenter. Prove that $AI = BC$.",
    "options": [],
    "answer": "See 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![](images/RMC_2025_p34_data_3f4e8e3f71.png)\n\nConstruction 1\n\n![](images/RMC_2025_p34_data_b9daa7bb5f.png)\n\nConstruction 2\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*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![](images/RMC_2025_p34_data_7d31427db2.png)\n\nConstruction 3\n\n![](images/RMC_2025_p34_data_195fdf7e2c.png)\n\nConstruction 4\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": 18462,
    "subject": "Mathematics (Olympiad)",
    "question": "The only available numbers to fill the puzzle are:\n\n12, 14, 18, 21, 24, 28, 32, 42, 48, 56, 63, 72, 84, 96,\n112, 126, 144, 168, 224, 252, 288, 336, 504, 672,\n1008, 2016.\n\nIn the following, A means across and D means down.\n\n% IMAGE: ![](images/DiPasquale_Australian_MO_Scene_2016_p41_data_3789c81be4.png)\n\nFill the cross number puzzle using only the numbers above, so that each entry is a factor of 2016 and each number is used at most once. (See the grid above.)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Number 15A must be 2016 or 1008. Since 16D cannot start with 0, 15A is 2016. So 18A is 1008. Then 168 is the only possible factor for 16D.\n\nb) From Part a, we have:\n\n% IMAGE: ![](images/DiPasquale_Australian_MO_Scene_2016_p41_data_3789c81be4.png)\n\nNow 12D is 112, 252 or 672. If it is 112, then 11A is also 112, which is not allowed. If 12D is 672, then 11A is 168, which is already used. So 12D must be 252. Hence 11A is 126 or 224. Since 7D cannot end in 1, 11A must be 224. Then 7D is 112 or 672. If 7D is 672, then 7A is 63, leaving no factor for 5D. So 7D is 112 and we have:\n\nc) The remaining 3-digit factors are: 126, 144, 288, 336, 504, 672.\n\nNow 14D is 56 or 96. Since no 3-digit factor starts with 9, 14D is 56. Hence 14A is 504, the only 3-digit factor starting with 5. Then 13D is 144, the only 3-digit factor with middle digit 4.\n\nThe remaining 3-digit factors are now: 126, 288, 336, 672.\n\nSince 8A and 4D end in the same digit, they must be 126 and 336 in some order. Hence 9A is 288 or 672. If 9A is 672, then 9D is 64 which is not a factor of 2016. So 9A is 288. Since 8A is 126 or 336 and 38 is not a factor of 2016, 8D is 18. So 8A is 126, 4D is 336, and we have:\n\n% IMAGE: ![](images/DiPasquale_Australian_MO_Scene_2016_p42_data_d12a33747c.png)\n\nd) The remaining 2-digit factors are:\n12, 14, 21, 28, 32, 42, 48, 63, 72, 84, 96.\n\nThen 17D is 21 (the only 2-digit factor ending in 1), 6A is 32 (the only 2-digit factor starting with 3), 3A is 63 (the only 2-digit factor ending in 3), and 1D is 96 (the only 2-digit factor ending in 6).\n\nThe remaining 2-digit factors are now: 12, 14, 28, 42, 48, 72, 84.\n\nSince 5A and 5D start with the same digit, that digit is 1 or 4. Since 7A starts with 1, 5A and 5D are 42 and 48 in some order. Since 7A is 12 or 14, 5D is 42. So 7A is 12 and 5A is 48. Hence 2D is 28, 13A is 14, and 10D is 84.\n\nSo we have:\n\n% IMAGE: ![](images/DiPasquale_Australian_MO_Scene_2016_p43_data_c8c048f068.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18463,
    "subject": "Mathematics (Olympiad)",
    "question": "Several squares are cut off from a rectangular squared board so that for every cut square, all squares from the same column above it and all squares from the same row to its right are also cut off. After such cuts, the resulting figure consists of several columns, and for any two columns, the left-hand one is not shorter than the right-hand one.\n\nIn every square of the obtained figure, write a number equal to the quantity of squares lying above it in the same column and lying to the right in the same row (including the square itself).\n\nLet $n$ be the total number of squares in the figure. Prove that the number of squares with numbers divisible by $2008$ does not exceed $\\frac{n}{2008}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's show that the statement holds for any $q$: the total number of squares with numbers divisible by $q$ does not exceed $\\frac{n}{q}$.\n\nCall the set of squares lying above in the same column and to the right in the same row (including the square itself) a *hook*. The square at the bottom corner of the hook is the *base of the hook*. The total number of squares in the hook is the *hook area*. In every square, we write the area of the hook with base at that square.\n\n**Lemma.** If the hook area is $kq$ ($k, q \\in \\mathbb{N}$), then the hook contains at most $k$ squares with numbers divisible by $q$.\n\n![](images/Ukrajina_2011_p44_data_e815c69210.png)\n\n![](images/Ukrajina_2011_p44_data_62a57f3cdd.png)\n\n**Proof of the lemma.** Let the *upper bar* be the set of squares above the base in the same column, and the *right bar* be the set of squares to the right of the base in the same row (not including the base).\n\n![](images/Ukrajina_2011_p44_data_7ef39d38b4.png)\n\nWe show that the sum of the numbers in any two squares, one from the upper bar and one from the right bar, does not equal $kq$. There are two cases: the hooks with these bases either have a common square or not.\n\nIf the hooks have a common square, the number of squares in their union is at least $kq+1$, since the common square is counted twice. If the hooks have no common squares, the total number of squares in their union is strictly less than $kq$.\n\nConsider the pairs $((k-1)q, q), ((k-2)q, 2q), \\ldots, (q, (k-1)q)$. Mark all numbers in the first position if they appear in the upper bar, and in the second position if they appear in the right bar. The numbers in either bar decrease monotonically from the base and are strictly less than $kq$, so only one square corresponds to each marked number. In each pair, at most one number can be marked. All numbers divisible by $q$ are enumerated in these pairs, and there are $k-1$ pairs. Thus, in both bars there are at most $k-1$ numbers divisible by $q$. Including $kq$ at the base, there are at most $k$ such numbers. The lemma is proved.\n\nNow, mark squares with numbers divisible by $q$ as follows: examine rows from bottom up. If a row contains a square with a number divisible by $q$ that is not in a hook with an already marked base, choose the leftmost such square and mark it. (For $q=2$, dark-grey indicates marked squares, light-grey indicates hooks with marked bases.)\n\nAll squares with numbers divisible by $q$ are contained in hooks with marked bases. Let the sum of the areas of these hooks be $s$. By the lemma, the total number of such squares does not exceed $s/q$ (some squares may be counted twice). We show $s \\leq n$. Since no three hooks with marked bases have a common square and any two such hooks have at most one common square, the area covered by hooks is $s-p$, where $p$ is the number of pairs of intersecting hooks. For each such pair, there is a unique square not covered by any hook, lying in the same row as the base of the lower hook and in the same column as the base of the upper hook. All such squares are distinct for different pairs. Thus, the area of the figure $n \\geq (s-p) + p = s$. Therefore, the total number of squares with numbers divisible by $q$ does not exceed $s/q \\leq n/q$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18464,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為凸四邊形，點 $P, Q, R, S$ 分別在邊 $AB, BC, CD, DA$ 上。直線 $PR$ 與 $QS$ 交於 $O$ 點。設四個四邊形 $APOS, BQOP, CROQ, DSOR$ 都有內切圓。證明：直線 $AC, PQ, RS$ 共點或是兩兩互相平行。\n\nLet $ABCD$ be a convex quadrilateral, and let $P, Q, R, S$ be points on the sides of $AB, BC, CD,$ and $DA$, respectively. Let the line segments $PR$ and $QS$ meet at $O$. Suppose that each of the quadrilaterals $APOS, BQOP, CROQ,$ and $DSOR$ has an incircle. Prove that the lines $AC, PQ,$ and $RS$ are either concurrent or parallel to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "對於 $\\triangle ABC$ 與直線 $PQ$，以及 $\\triangle ACD$ 與直線 $RS$ 使用孟氏定理（Melelaus' theorem），可知 $AC$ 分別與 $PQ, RS$ 有相同交點（此交點可能在無窮遠處）的充要條件是：\n\n$$\n\\frac{AP}{PB} \\cdot \\frac{BQ}{QC} \\cdot \\frac{CR}{RD} \\cdot \\frac{DS}{SA} = 1. \\qquad (1)\n$$\n\n所以我們把目標放在證明 (1) 式。\n\n先來證下面的結果。\n\n*Lemma 1.* 設四邊形 $EFGH$ 有內切圓，$M$ 為內切圓圓心。則\n\n$$\n\\frac{EF \\cdot FG}{GH \\cdot HE} = \\frac{FM^2}{HM^2}.\n$$\n\n![](images/16-3J_p11_data_e305ce1f1c.png)\n\n圖 1\n\n*Proof*. 注意到 $\\angle EMH + \\angle GMF = \\angle FME + \\angle HMG = 180^{\\circ}$，$\\angle FGM = \\angle MGH$，以及 $\\angle HEM = \\angle MEF$（如圖 1）。由正弦定理，知\n\n$$\n\\begin{aligned}\n\\frac{EF}{FM} \\cdot \\frac{FG}{FM} &= \\frac{\\sin \\angle FME \\cdot \\sin \\angle GMF}{\\sin \\angle MEF \\cdot \\sin \\angle FGM} = \\frac{\\sin \\angle HMG \\cdot \\sin \\angle EMH}{\\sin \\angle MGH \\cdot \\sin \\angle HEM} \\\\\n&= \\frac{GH}{HM} \\cdot \\frac{HE}{HM}.\n\\end{aligned}\n$$\n\n![](images/16-3J_p11_data_a1acbb74e7.png)\n\n令點 $I, J, K, L$ 分別為四邊形 $APOS, BQOP, CROQ, DSOR$ 的內切圓圓心。將 Lemma 1 的結果套入這四個四邊形，可得\n\n$$\n\\frac{AP \\cdot PO}{OS \\cdot SA} \\cdot \\frac{BQ \\cdot QO}{OP \\cdot PB} \\cdot \\frac{CR \\cdot RO}{OQ \\cdot QC} \\cdot \\frac{DS \\cdot SO}{OR \\cdot RD} = \\frac{PI^2}{SI^2} \\cdot \\frac{QJ^2}{PJ^2} \\cdot \\frac{RK^2}{QK^2} \\cdot \\frac{SL^2}{RL^2},\n$$\n\n可化簡為\n\n$$\n\\frac{AP}{PB} \\cdot \\frac{BQ}{QC} \\cdot \\frac{CR}{RD} \\cdot \\frac{DS}{SA} = \\frac{PI^2}{PJ^2} \\cdot \\frac{QJ^2}{QK^2} \\cdot \\frac{RK^2}{RL^2} \\cdot \\frac{SL^2}{SI^2}. \\quad (2)\n$$\n\n![](images/16-3J_p12_data_5238596331.png)\n\n圖 2\n\n接下來，我們有 $\\angle IPJ = \\angle JOI = 90^\\circ$，且 $I, J$ 兩點位於直線 $OP$ 的異側。由此知四邊形 $IPJO$ 有外接圓。同理可得四邊形 $JQKO$ 也有外接圓，且 $\\angle JQK = 90^\\circ$。所以 $\\angle QKJ = \\angle QOJ = \\angle JOP = \\angle JIP$。因此兩直角三角形 $\\triangle IPJ$ 與 $\\triangle KQJ$ 相似，可知 $\\frac{PI}{PJ} = \\frac{QK}{QJ}$。同理可得 $\\frac{RK}{RL} = \\frac{SI}{SL}$。此二式與 (2) 式合併起來，就得到 (1) 式。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18465,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be real numbers such that\n\n$$\n\\frac{1}{|x^2 + 2yz|}, \\quad \\frac{1}{|y^2 + 2zx|}, \\quad \\frac{1}{|z^2 + 2xy|}\n$$\n\nare side-lengths of a non-degenerate triangle. Find all possible values of $xy + yz + zx$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = y = z = t > 0$, then the three fractions are sides of an equilateral triangle and $xy + yz + zx = 3t^2$, so $xy + yz + zx$ can attain all positive values. Similarly, for $x = y = t > 0$ and $z = -2t$, the three fractions are $\\frac{1}{3}t^{-2}$, $\\frac{1}{3}t^{-2}$, $\\frac{1}{6}t^{-2}$, which are positive and form an isosceles triangle ($\\frac{1}{6} < \\frac{1}{3} + \\frac{1}{3}$). Thus, any negative value can be attained too.\n\nNext, we show that $xy + yz + zx$ cannot be $0$. Assume otherwise. The numbers $x$, $y$, $z$ must be mutually distinct: if, say, $x$ and $y$ were equal, then the denominator of the first fraction would be $|x^2 + 2yz| = |xy + (yz + xz)| = 0$, which is impossible.\n\nConsider the fractions without absolute values. Subtracting $xy + yz + zx = 0$ from each denominator, we get\n\n$$\n\\begin{aligned}\n\\frac{1}{x^2 + 2yz} + \\frac{1}{y^2 + 2zx} + \\frac{1}{z^2 + 2xy} &= \\\\\n&= \\frac{1}{(x-y)(x-z)} + \\frac{1}{(y-z)(y-x)} + \\frac{1}{(z-x)(z-y)} \\\\\n&= \\frac{(z-y) + (x-z) + (y-x)}{(x-y)(y-z)(z-x)} = 0.\n\\end{aligned}\n$$\n\nThis implies that among the original fractions (with absolute values), one of them is the sum of the other two. Hence, the fractions do not fulfill the triangle inequality, leading to a contradiction.\n\n**Answer:** Possible values are all real numbers except $0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18466,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest possible value of\n\n$$\n\\frac{x^{2023} + 203}{17x^7 + 7x^{17}}\n$$\n\nover all positive real numbers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $\\frac{17}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18467,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f, g : \\\\mathbb{R} \\to \\\\mathbb{R}$ such that for all $x, y \\in \\\\mathbb{R}$,\n$$\nx f(y) + g(x) = y f(x) + g(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = f(0)$, $b = g(0)$. Due to symmetry, the given equation implies\n$$\nx f(y) + g(x) = y f(x) + g(y).\n$$\nSet $y = 0$:\n$$\ng(x) = b - a x.\n$$\nSet $y = 1$:\n$$\nf(x) = (f(1) - a)x + a = a + c x.\n$$\nSubstituting $g(x)$ and $f(x)$ into the original equation, we find $c = 0$, $b = a$.\n\nThus, all solutions are:\n$$\nf(x) = a, \\quad g(x) = a - a x, \\quad \\text{for all } a \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18468,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be odd positive integers. Each square of an $m \\times n$ board is coloured red or blue. A row is said to be red-dominated if there are more red squares than blue squares in the row. A column is said to be blue-dominated if there are more blue squares than red squares in the column. Determine the maximum possible value of the number of red-dominated rows plus the number of blue-dominated columns. Express your answer in terms of $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $m + n - 2$ if $m, n \\geq 3$, and $\\max\\{m, n\\}$ if one of $m, n$ is equal to $1$.\n\nNote that it is not possible for all rows to be red-dominated and all columns to be blue-dominated. Since both $m$ and $n$ are odd, the total number of squares is odd, so there are more squares of one color than the other. Without loss of generality, suppose there are more red squares than blue squares. Then it is not possible for every column to have more blue squares than red squares, so not every column can be blue-dominated.\n\nIf one of $m, n$ is $1$, say $m$ without loss of generality, then the answer is less than $n + 1$. For example, if every square is blue, there are $n$ blue-dominated columns and $0$ red-dominated rows, so the sum is $n = \\max\\{m, n\\}$.\n\nNow consider the case $m, n \\geq 3$.\n\nThere are $m$ rows and $n$ columns, so the answer is at most $m + n$. We have already shown it cannot be $m + n$.\n\nSince $m, n$ are odd, let $m = 2a - 1$ and $n = 2b - 1$ for some positive integers $a, b$ with $a, b \\geq 2$. We show the answer cannot be $m + n - 1$. By symmetry, it suffices to show we cannot have all rows red-dominated and all but one column blue-dominated. If all rows are red-dominated, each row has at least $b$ red squares, so there are at least $bm = (2a - 1)b$ red squares. If all but one column are blue-dominated, there are at least $2b - 2$ blue-dominated columns, each with at least $a$ blue squares, so at least $a(2b - 2)$ blue squares. Thus, the board has at least $(2a - 1)b + a(2b - 2) = 4ab - b - 2a$ squares. But the total number of squares is\n\n$$\n(2a - 1)(2b - 1) = 4ab - 2a - 2b + 1 = 4ab - 2a - b - b + 1 < 4ab - 2a - b,\n$$\nwhich is a contradiction since $b \\geq 2$. Therefore, the answer is less than $m + n - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18469,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 2017 positive real numbers $a_1, a_2, \\dots, a_{2017}$. For each $n > 2017$, denote\n$$\na_n = \\max\\{a_{i_1} a_{i_2} a_{i_3} \\mid i_1 + i_2 + i_3 = n,\\ 1 \\le i_1 \\le i_2 \\le i_3 \\le n-1\\}.\n$$\nProve that there exists a positive integer $m \\le 2017$ and a positive integer $N > 4m$ such that $a_n a_{n-4m} = a_{n-2m}^2$ for every $n > N$.",
    "options": [],
    "answer": "See solution",
    "solution": "For every $n > 0$, let $b_n = \\ln a_n$. We can restate the problem as: Given 2017 real numbers $b_1, b_2, \\dots, b_{2017}$. For every $n > 2017$, let\n$$\nb_n = \\max\\{b_{i_1} + b_{i_2} + b_{i_3} \\mid i_1 + i_2 + i_3 = n,\\ 1 \\le i_1 \\le i_2 \\le i_3 \\le n-1\\}.\n$$\nProve that there exists a positive integer $m \\le 2017$ and $N > 4m$ such that $b_n + b_{n-4m} = 2b_{n-2m}$ for every $n > N$.\n\nLet $\\ell$ ($1 \\le \\ell \\le 2017$) be the number such that $\\frac{b_\\ell}{\\ell} = \\max \\left\\{ \\frac{b_i}{i} \\mid 1 \\le i \\le 2017 \\right\\}$.\n\nWe have the following statements:\n\n*Claim 1.* For every positive integer $n$, we have\n$$\n\\frac{b_n}{n} \\le \\frac{b_\\ell}{\\ell}.\n$$\n*Proof.* We prove by induction on $n$. The statement is clearly true for $n \\le 2017$ by the definition of $\\ell$. We consider the case $n > 2017$. Assume it is true for all $k < n$. From the definition of $b_n$, there exist $j_1, j_2, j_3 \\in \\mathbb{N}^*$ satisfying $j_1 + j_2 + j_3 = n$ such that\n$$\nb_n = b_{j_1} + b_{j_2} + b_{j_3}.\n$$\nUsing the inductive hypothesis, we have\n$$\nb_n \\le j_1 \\cdot \\frac{b_\\ell}{\\ell} + j_2 \\cdot \\frac{b_\\ell}{\\ell} + j_3 \\cdot \\frac{b_\\ell}{\\ell} = n \\cdot \\frac{b_\\ell}{\\ell},\n$$\nso $\\frac{b_n}{n} \\le \\frac{b_\\ell}{\\ell}$. Therefore, the statement is also true for $n$. $\\square$\n\nNow, for every positive integer $n$, let $c_n = n b_\\ell - \\ell b_n$. Then from the above claim we have $c_n \\ge 0$ for every $n$. We also have, for $n \\ge 2017$:\n$$\n\\begin{aligned}\nc_{n+2\\ell} &= (n+2\\ell) b_\\ell - \\ell b_{n+2\\ell} \\\\\n&\\le (n+2\\ell) b_\\ell - \\ell (b_n + b_\\ell + b_\\ell) \\\\\n&= n b_\\ell - \\ell b_n = c_n.\n\\end{aligned}\n$$\nHence,\n$$\nc_{n+2k\\ell} \\le c_{n+2(k-1)\\ell} \\le \\dots \\le c_n, \\quad \\forall n \\ge 2017,\\ k \\ge 1.\n$$\nLet $x$ be the smallest positive integer such that $2x\\ell > 2017$ and set\n$$\nM = \\max\\{c_i \\mid 1 \\le i \\le 4x\\ell - 1\\}.\n$$\nThen, for every $n > 2x\\ell$, set $n = 2k x \\ell + r$ (with $0 \\le r < 2x\\ell$), we have\n$$\nc_n \\le c_{r+2(k-1)\\ell} \\le c_{r+2(k-2)\\ell} \\le \\dots \\le c_{r+2x\\ell} \\le M.\n$$\n\n*Claim 2.* For every positive integer $n$, there exist 2017 natural numbers $s_1, s_2, \\dots, s_{2017}$ such that\n$$\nc_n = s_1 c_1 + s_2 c_2 + \\dots + s_{2017} c_{2017}.\n$$\n*Proof.* We prove by induction on $n$. The statement is clearly true for $n \\le 2017$ (just choose $s_n = 1$ and $s_i = 0$ with $i \\ne n$). We just need to consider the case $n > 2017$. Suppose it is true for $k < n$. From the definition of $b_n$, we deduce that for every $n > 2017$,\n$$\nc_n = \\min\\{c_{i_1} + c_{i_2} + c_{i_3} \\mid 1 \\le i_1 \\le i_2 \\le i_3 \\le n-1,\\ i_1 + i_2 + i_3 = n\\}.\n$$\nThus there exist $j_1, j_2, j_3 \\in \\mathbb{N}^*$ satisfying $j_1 + j_2 + j_3 = n$ such that\n$$\nc_n = c_{j_1} + c_{j_2} + c_{j_3}.\n$$\nBy the inductive hypothesis, there exist natural numbers $u_1, \\dots, u_{2017}$, $v_1, \\dots, v_{2017}$, $w_1, \\dots, w_{2017}$ such that\n$$\nc_{j_1} = u_1 c_1 + \\dots + u_{2017} c_{2017},\n$$\n$$\nc_{j_2} = v_1 c_1 + \\dots + v_{2017} c_{2017},\n$$\n$$\nc_{j_3} = w_1 c_1 + \\dots + w_{2017} c_{2017}.\n$$\nTherefore,\n$$\nc_n = s_1 c_1 + s_2 c_2 + \\dots + s_{2017} c_{2017},\n$$\nwhere $s_i = u_i + v_i + w_i$ for $1 \\le i \\le 2017$. Hence, the statement is also true for $n$ and Claim 2 is proved. $\\square$\n\nFrom Claim 2, combining with the fact that $(c_n)$ is bounded, one can get that the sequence $(c_n)$ has finitely many values. Note that\n$$\nc_{n+2k x \\ell} \\le c_{n+2(k-1)x\\ell} \\le \\dots \\le c_n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18470,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 2.1, the diagonals $AC$, $BD$ of quadrilateral $ABCD$ intersect at point $E$. The midperpendiculars of $AB$ and $CD$ (with $M$ and $N$ being their midpoints, respectively) intersect at point $F$. Line $EF$ intersects $BC$ and $AD$ at points $P$ and $Q$, respectively.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p198_data_3cc17d8144.png)\n\nSuppose $MF \\cdot CD = NF \\cdot AB$ and $DQ \\cdot BP = AQ \\cdot CP$. Prove that $PQ \\perp BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Connect points $A$-$F$, $B$-$F$, $C$-$F$, and $D$-$F$ as shown in Fig. 2.2.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p199_data_6b59d51736.png)\n\nBy the given condition, triangles $AFB$ and $CFD$ are both isosceles, with $FM$ and $FN$ being the altitudes to each triangle's base.\n\nSince $MF \\cdot CD = NF \\cdot AB$, $\\triangle AFB \\sim \\triangle DFC$. Thus, $\\angle AFB = \\angle CFD$ and $\\angle FAB = \\angle FDC$. Moreover, $\\angle BFD = \\angle CFA$. From $FB = FA$ and $FD = FC$, we have $\\triangle BFD \\cong \\triangle AFC$, which means $\\angle FAC = \\angle FBD$ and $\\angle FCA = \\angle FDB$. Therefore, points $A$, $B$, $F$, $E$ and points $C$, $D$, $E$, $F$ are each concyclic.\n\nFrom this, we get\n\n$$\n\\angle FEB = \\angle FAB = \\angle FDC = \\angle FEC,\n$$\n\nwhich implies that line $EP$ is the angle bisector of $\\angle BEC$. Then $\\frac{EB}{EC} = \\frac{BP}{CP}$. Similarly, $\\frac{ED}{EA} = \\frac{QD}{AQ}$.\n\nIf $DQ \\cdot BP = AQ \\cdot CP$, then $EB \\cdot ED = EC \\cdot EA$, which means $ABCD$ is a cyclic quadrilateral with $F$ as the center of its circumcircle. At this time,\n\n$$\n\\angle EBC = \\frac{1}{2} \\angle DFC = \\frac{1}{2} \\angle AFB = \\angle ECB,\n$$\n\nso $PQ \\perp BC$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18471,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. Consider an $n \\times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is *peaceful* if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for each peaceful configuration of $n$ rooks, there is an empty $k \\times k$ square without any rooks on any of its $k^2$ unit squares.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p361_data_dc1e8d0145.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We shall show that the answer is $k_{\\max} = \\left[\\sqrt{n-1}\\right]$ by two steps. Let $l$ be a positive integer.\n\n1. **If $n > l^2$, then there exists an empty $l \\times l$ square for any peaceful configuration.**\n\n2. **If $n \\le l^2$, then there exists a peaceful configuration such that each $l \\times l$ square is not empty.**\n\n**Proof of (1).**\n\nThere is a row $R$ with the first column having a rook. Take successively $l$ rows containing row $R$, which are denoted by $U$. If $n > l^2$, then $l^2 + 1 \\le n$, and from column 2 to column $l^2 + 1$ in $U$, containing $l \\times l$ squares which have at most $l-1$ rooks. So, at least one $l \\times l$ square is empty.\n\n**Proof of (2).**\n\nFor $n = l^2$, we shall find a peaceful configuration which has no empty $l \\times l$ square. We label the rows from bottom to top and the columns from left to right both by $0, 1, \\dots, l^2 - 1$. So denote by $(r, c)$ the unit square at row $r$ and column $c$.\n\nWe put a rook at $(il + j, jl + i)$ for $i, j = 0, 1, 2, \\dots, l-1$. The figure above shows the case for $l=3$. Since each number between $0$ to $l^2 - 1$ can be written uniquely in the form $il + j$, $(0 \\le i, j \\le l - 1)$, such a configuration is peaceful.\n\nFor any $l \\times l$ square $A$, suppose that the lowest row of $A$ is row $pl + q$, $0 \\le p, q \\le l-1$ (since $pl + q \\le l^2 - l$). There is one rook in $A$ by the configuration, and the column labels of the rook may be $ql + p$, $(q+1)l + p$, $\\dots$, $(l-1)l + p$, $p+1$, $l + (p+1)$, $\\dots$, $(q-1)l + p + 1$. Rearranging these numbers in increasing order:\n\n$$\np + 1,\\ l + (p + 1),\\ \\dots,\\ (q - 1)l + p + 1,\\ ql + p,\\ (q + 1)l + p,\\ \\dots,\\ (l - 1)l + p.\n$$\n\nThen the first number is less than or equal to $l$, the last is greater than or equal to $(l-1)l$ and the difference between two adjacent numbers is $l$. Therefore, there exists one rook in $A$.\n\nFor the case of $n < l^2$, consider the configuration above, but delete $l^2 - n$ columns and rows from the right and from the bottom, respectively, to get an $l \\times l$ square. Thus, we obtain an $n \\times n$ square, where there is no empty $l \\times l$ square. But some rows and columns may be empty. We can put rooks at the cross squares of empty rows and columns to obtain the desired peaceful configuration.\n\n**Remark.** The answer could also be in the form $\\lceil\\sqrt{n}\\rceil - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18472,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a$, $b$, and $c$ are positive real numbers. Prove that\n\n$$\n\\frac{a+b+c}{3} \\leq \\sqrt{\\frac{a^2+b^2+c^2}{3}} \\leq \\frac{\\frac{ab}{c} + \\frac{bc}{a} + \\frac{ca}{b}}{3}.\n$$\n\nFor each of the inequalities, find conditions on $a$, $b$, and $c$ such that equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that\n\n$$\n\\frac{a+b+c}{3} \\leq \\sqrt{\\frac{a^2+b^2+c^2}{3}},\n$$\n\nwith equality if and only if $a = b = c$. The inequality is equivalent to\n\n$$\n(a+b+c)^2 \\leq 3(a^2+b^2+c^2) \\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ca) \\leq 3(a^2+b^2+c^2),\n$$\n\ni.e.,\n\n$$\n2(ab + bc + ca) \\leq 2(a^2 + b^2 + c^2) \\Leftrightarrow 0 \\leq (a-b)^2 + (b-c)^2 + (c-a)^2.\n$$\n\nThe latter clearly holds, with equality only when $a = b = c$.\n\nAlternatively, use\n\n$$\nab \\leq \\frac{a^2 + b^2}{2}, \\quad bc \\leq \\frac{b^2 + c^2}{2}, \\quad ca \\leq \\frac{c^2 + a^2}{2}.\n$$\n\nOr apply the Cauchy-Schwarz inequality at this point or at the start.\n\nNext,\n\n$$\n\\sqrt{\\frac{a^2 + b^2 + c^2}{3}} \\leq \\frac{\\frac{ab}{c} + \\frac{bc}{a} + \\frac{ca}{b}}{3}\n$$\n\nholds if and only if\n\n$$\n3(a^2 + b^2 + c^2) \\leq \\left(\\frac{ab}{c} + \\frac{bc}{a} + \\frac{ca}{b}\\right)^2 = \\frac{a^2b^2}{c^2} + \\frac{b^2c^2}{a^2} + \\frac{c^2a^2}{b^2} + 2(a^2 + b^2 + c^2),\n$$\n\ni.e.,\n\n$$\na^2 + b^2 + c^2 \\leq \\frac{a^2b^2}{c^2} + \\frac{b^2c^2}{a^2} + \\frac{c^2a^2}{b^2}.\n$$\n\nSince\n\n$$\nx + \\frac{1}{x} \\geq 2, \\quad \\forall x > 0,\n$$\n\nwith equality holding if and only if $x = 1$, we see that\n\n$$\n2a^2 \\leq \\frac{a^2b^2}{c^2} + \\frac{c^2a^2}{b^2}, \\quad 2b^2 \\leq \\frac{a^2b^2}{c^2} + \\frac{b^2c^2}{a^2}, \\quad 2c^2 \\leq \\frac{b^2c^2}{a^2} + \\frac{c^2a^2}{b^2},\n$$\n\nwhence\n\n$$\n2a^2 + 2b^2 + 2c^2 \\leq 2\\frac{a^2b^2}{c^2} + 2\\frac{b^2c^2}{a^2} + 2\\frac{c^2a^2}{b^2},\n$$\n\nthe claimed result. Further, equality holds if and only if $a^2 = b^2 = c^2$, i.e., $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18473,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека постојат попарно дисјунктни множества $A_1, A_2, \\dots, A_{2014}$, чија унија е множеството на природни броеви, за кои важи следниот услов:\n\nЗа произволни природни броеви $a$ и $b$, барем два од броевите $a$, $b$, $\\text{НЗД}(a,b)$ припаѓаат на едно од множествата $A_1, A_2, \\dots, A_{2014}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $v_2(n)$ е најголемиот цел број за кој $2^{v_2(n)}$ е делител на $n$. Тогаш, $v_2(\\text{НЗД}(a,b)) = \\min\\{v_2(a), v_2(b)\\}$. Значи, барем два од броевите $v_2(a)$, $v_2(b)$ и $v_2(\\text{НЗД}(a,b))$ се еднакви.\n\nДефинираме множества $A_{i+1} = \\{n \\mid v_2(n) \\equiv i \\pmod{2014}\\}$ за $0 \\le i \\le 2013$.\n\nОчигледно, множествата $A_1, A_2, \\dots, A_{2014}$ се попарно дисјунктни, нивната унија е $\\mathbb{N}$ и два од броевите $a$, $b$, $\\text{НЗД}(a,b)$ се наоѓаат во множеството $A_{i+1}$, каде $i$ е остатокот при делење на $v_2(\\text{НЗД}(a,b))$ со $2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18474,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that the sum of $n$ summands\n\n$$\nS = 1 + 11 + 111 + \\dots + \\underbrace{\\textbf{11\\dots1}}_{n}\n$$\n\ndivides exactly into:\n\n(a) $15$;\n\n(b) $45$.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** Considering the last figure of the sum $S$, to divide $S$ exactly by $15$, the number of summands should be a multiple of $5$. Thus, consider the sum of the first five summands:\n\n$$\n1 + 11 + 111 + 1111 + 11111 = 12345\n$$\n\nSince $12345$ is divisible by $15$, the smallest $n$ is $5$.\n\n**(b)** Consider the remainders when the summands are divided by $9$. By the divisibility criterion for $9$, the remainder is equal to the remainder of the sum of the digits. Let\n\n$$\nS(n) = 1 + 11 + \\dots + \\underbrace{\\textbf{11\\dots1}}_{n}\n$$\n\nand let $r(n)$ be the remainder of $S(n)$ divided by $9$:\n\n$$\nr(1) = 1;\\quad r(2) = 3;\\quad r(3) = 6;\\quad r(4) = 1;\\quad r(5) = 6;\\quad r(6) = 3;\\quad r(7) = 1;\\quad r(8) = 0;\\quad r(9) = 0\n$$\n\nThese remainders repeat every $9$ steps. Thus, $S(n)$ is divisible by $9$ for $n = 8, 9, 17, 18, 26, 27, 35, 36, \\dots$. Also, $S(n)$ is divisible by $5$ if and only if $n$ is divisible by $5$. Therefore, the smallest $n$ is $35$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18475,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence of positive integers $a_1, a_2, \\dots$ is defined by\n\n$$\na_{n+1} = \\begin{cases} a_n^2 + 2^m & \\text{if } a_n < 2^m \\\\ 1 & \\text{if } a_n \\ge 2^m \\end{cases} \\quad (1)\n$$\n\nand\n\n$$\na_{n+1} = \\begin{cases} 1 & \\text{if } a_n < 2^m \\\\ \\frac{1}{2}a_n & \\text{if } a_n \\ge 2^m \\end{cases} \\quad (2)\n$$\n\nfor each positive integer $n$, where $a_1$ is a positive integer.\n\nDetermine all positive integers $m$ for which there exists a positive integer $a_1$ such that all terms of the sequence are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "From (1), there is a term of the sequence that is greater than $2^m$. After this, (2) is applied until we reach a term that lies in the interval $[2^{m-1}, 2^m)$. Then the next term in the sequence after this is greater than $2^m$ again. It follows that infinitely many terms of the sequence lie in the interval $[2^{m-1}, 2^m)$.\n\nIf $m=1$, then $a_i = 1$ for some positive integer $i$. But then $a_{i+1} = 3$ and $a_{i+2} = \\frac{3}{2}$, which is not an integer. Hence $m \\neq 1$.\n\nIf $m=2$, then $2 \\le a_i < 4$ for some positive integer $i$. If $a_i = 3$, then $a_{i+1} = 13$ and $a_{i+2} = \\frac{13}{2}$, which is not an integer. If $a_i = 2$, then choose the first $i$ with $a_i = 2$. Tracing the sequence backwards we see that $a_1$ is a positive power of 2. Moreover, after reaching $a_i$, the sequence enters the cycle $2 \\to 8 \\to 4 \\to 2 \\to \\dots$. Hence $m=2$ is possible, in which case $a_1$ can be any positive power of 2.\n\nAssume, for the sake of contradiction, that $m \\ge 3$. For any integer $i \\ge 1$ with $a_i \\in [2^{m-1}, 2^m)$, we have $a_{i+1} = a_i^2 + 2^m > 2^{2m-2}$. Hence (2) applies to generate the next $m-1$ terms of the sequence. Consequently $a_{i+m} = \\frac{a_i^2}{2^{m-1}} + 2$, and\n\n$$\n2^{m-1} \\mid a_i^2 \\quad \\text{whenever } a_i \\in [2^{m-1}, 2^m). \\quad (3)\n$$\n\nThis leads to the following two cases.\n\n*Case 1*: $2^{m-1} \\le a_{i+m} < 2^m$\n\nSince $a_i \\ge 2^{m-1}$, we have\n\n$$\na_{i+m} = \\frac{a_i^2}{2^{m-1}} + 2 \\ge \\frac{2^{m-1}a_i}{2^{m-1}} + 2 > a_i.\n$$\n\nIf only case 1 ever occurred, we would have an infinite strictly increasing sequence $a_i < a_{i+m} < a_{i+2m} < \\dots$ which is bounded above by $2^m$. This is clearly impossible. Hence for some $i$ with $a_i \\in [2^{m-1}, 2^m)$, the following case applies.\n\n*Case 2*: $a_{i+m} \\ge 2^m$\n\nHence $a_{i+m+1} = \\frac{a_i^2}{2^m} + 1$. Since $a_i \\le 2^m - 1$, we have\n\n$$\n\\frac{a_i^2}{2^m} + 1 \\le \\frac{(2^m - 1)^2}{2^m} + 1 = 2^m + \\frac{1}{2^m} - 1 < 2^m.\n$$\n\nTherefore $a_{i+m+1} \\in [2^{m-1}, 2^m)$.\n\nAs $m \\ge 3$, we see from (3) that $a_{i+m+1}$ is even. So $x = \\frac{a_i^2}{2^m}$ is an odd positive integer. Consequently $m$ is even, and so $m \\ge 4$. Then from (3), we have $4 \\mid a_{i+m+1}$. But now $a_{i+m+1} = x^2 + 1$ fails modulo 4. This contradiction concludes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18476,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}$ be the set of all integers. Determine all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1\n$$\n\nholds for all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are two solutions: $f(x) = -1$ and $f(x) = x + 1$.\n\nThe functional equation is\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1 \\quad (1)\n$$\n\n1. Take $x = 0$ and $y = f(0)$ in (1). Then $z = -f(f(0))$ satisfies $f(z) = -1$.\n\n2. Set $y = z$ in (1):\n\n$$\nf(x + 1) = f(f(x)). \\quad (2)\n$$\n\n3. Therefore, (1) becomes\n\n$$\nf(x - f(y)) = f(x + 1) - f(y) - 1. \\quad (3)\n$$\n\n4. Now we show that $f$ is linear. Applying (3) with $y = x$ and then (2), we obtain\n\n$$\nf(x + 1) - f(x) = f(x - f(x)) + 1 = f(f(x - 1 - f(x))) + 1.\n$$\n\nSince (3) shows $f(x - 1 - f(x)) = f(x) - f(x) - 1 = -1$, we have\n\n$$\nf(x + 1) = f(x) + A,\n$$\n\nwhere $A = f(-1) + 1$ is some constant. By induction,\n\n$$\nf(x) = Ax + B,\n$$\n\nwhere $B = f(0)$.\n\n5. Substitute into (2):\n\n$$\nAx + (A + B) = A^2 x + (AB + B).\n$$\n\nTake $x = 0$ and $x = 1$ to get $A + B = AB + B$ and $A^2 = A$. Thus $A = 0$ or $A = 1$. If $A = 1$, then $B = 1$, so $f(x) = x + 1$. If $A = 0$, $f$ is constant, and (1) shows $f(x) = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18477,
    "subject": "Mathematics (Olympiad)",
    "question": "How many numbers of the form $3^a 4^b 5^c 6^d 7^e$ are there, where\n\n$$\n0 \\le a \\le 1, \\quad 0 \\le b \\le 2, \\quad 0 \\le c \\le 2, \\quad 0 \\le d \\le 1, \\quad 0 \\le e \\le 3,\n$$\n\nand $a + b + c + d + e \\ge 2$?",
    "options": [],
    "answer": "See solution",
    "solution": "There are two possible values for $a$, three for $b$, three for $c$, two for $d$, and four for $e$, so without restrictions there are $2 \\times 3 \\times 3 \\times 2 \\times 4 = 144$ numbers. We must exclude the case where $a + b + c + d + e = 0$ (one possibility) and the cases where $a + b + c + d + e = 1$ (five possibilities), so the final total is $144 - 1 - 5 = 138$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18478,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers. Prove that\n\n$$\n\\sum_{\\text{cyc}} \\frac{xy}{xy + x^2 + y^2} \\leq \\sum_{\\text{cyc}} \\frac{x}{2x + z}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given inequality is equivalent to:\n\n$$\n\\frac{1}{1 + \\frac{x}{y} + \\frac{y}{x}} + \\frac{1}{1 + \\frac{y}{z} + \\frac{z}{y}} + \\frac{1}{1 + \\frac{z}{x} + \\frac{x}{z}} \\leq \\frac{1}{2 + \\frac{z}{x}} + \\frac{1}{2 + \\frac{x}{y}} + \\frac{1}{2 + \\frac{y}{z}}.\n$$\n\nNow, take the substitution $\\frac{x}{y} = a$, $\\frac{y}{z} = b$, $\\frac{z}{x} = c$ so that $abc = 1$. The inequality becomes:\n\n$$\n\\frac{1}{1 + a + \\frac{1}{a}} + \\frac{1}{1 + b + \\frac{1}{b}} + \\frac{1}{1 + c + \\frac{1}{c}} \\leq \\frac{1}{2 + a} + \\frac{1}{2 + b} + \\frac{1}{2 + c}.\n$$\n\nAfter computations and using $abc = 1$, this is equivalent to:\n\n$$\n\\frac{3(a + b + c) + 3(ab + bc + ca) + (ab + bc + ca)(a + b + c)}{(ab + bc + ca)^2 + (ab + bc + ca)(a + b + c) + (a + b + c)^2} \\leq \\frac{12 + 4(a + b + c) + (ab + bc + ca)}{9 + 2(ab + bc + ca) + 4(a + b + c)}.\n$$\n\nLet $S = a + b + c$ and $P = ab + bc + ca$. The inequality becomes:\n\n$$\n\\frac{3S + 3P + SP}{P^2 + PS + S^2} \\leq \\frac{12 + 4S + P}{9 + 2P + 4S} \\iff P^3 + 4S^3 + 3P^2S + PS^2 + 6P^2 \\geq 27S + 27P + 15PS \\quad (*)\n$$\n\nIt is not difficult to prove that $S^2 \\geq 3P$ and $S \\geq 3$, $P \\geq 3$ (by AM $\\geq$ GM and $abc = 1$). Therefore:\n\n$$\n\\begin{aligned}\n4S^3 &= 4S^2 \\cdot S \\geq 12PS, \\quad PS^2 = S \\cdot PS \\geq 3PS, \\quad 3P^2S \\geq 3 \\cdot 3^2 \\cdot S = 27S, \\\\\nP^3 &= P^2 \\cdot P \\geq 3^2 \\cdot P = 9P, \\quad 6P^2 = 6P \\cdot P \\geq 6 \\cdot 3P = 18P.\n\\end{aligned}\n$$\n\nSumming these inequalities, we find that $(*)$ holds, so the proof is complete. Equality is achieved when\n\n$$\nS^2 = 3P = 9 \\implies a = b = c = 1 \\implies x = y = z.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18479,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $X = \\{1, 2, \\dots, 100\\}$, consider a function $f: X \\to X$ satisfying both of the following conditions:\n\n1. $f(x) \\neq x$ for all $x \\in X$;\n2. $A \\cap f(A) \\neq \\emptyset$ for all $A \\subseteq X$ with $|A| = 40$.\n\nFind the smallest positive integer $k$ such that for any such function $f$, there exists a set $B \\subseteq X$ satisfying $|B| = k$ and $B \\cup f(B) = X$.\n\n**Remark.** For a subset $T$ of $X$, we define $f(T) = \\{x \\mid \\text{there exists } t \\in T \\text{ such that } x = f(t)\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we define a function $f: X \\to X$ as follows:\n\n$$\n\\begin{aligned}\nf(3i - 2) &= 3i - 1, \\\\\nf(3i - 1) &= 3i, \\\\\nf(3i) &= 3i - 2, \\\\\n&\\text{for } i = 1, 2, \\dots, 30, \\\\\nf(j) &= 100, \\quad 91 \\leq j \\leq 99, \\\\\nf(100) &= 99.\n\\end{aligned}\n$$\n\nObviously, $f$ satisfies condition (1). For any $A \\subseteq X$ with $|A| = 40$:\n- If there exists an integer $i$ with $1 \\leq i \\leq 30$ such that $|A \\cap \\{3i-2, 3i-1, 3i\\}| \\geq 2$, then $A \\cap f(A) \\neq \\emptyset$.\n- If $91, 92, \\dots, 100 \\in A$, then $A \\cap f(A) \\neq \\emptyset$ also holds.\n\nThus, $f$ satisfies condition (2). If a subset $B$ of $X$ satisfies $f(B) \\cup B = X$, then $|B \\cap \\{3i-2, 3i-1, 3i\\}| \\geq 2$ for all $1 \\leq i \\leq 30$, $\\{91, 92, \\dots, 98\\} \\subseteq B$, and $B \\cap \\{99, 100\\} \\neq \\emptyset$. Hence, $|B| \\geq 69$.\n\nNext, we show that for any function $f$ satisfying the described conditions, there exists a subset $B \\subseteq X$ with $|B| \\leq 69$ such that $f(B) \\cup B = X$.\n\nAmong all subsets $U \\subseteq X$ with $U \\cap f(U) = \\emptyset$, choose one with maximal $|U|$. If there are many, choose one with maximal $|f(U)|$. The existence of $U$ is guaranteed by condition (1). Let $V = f(U)$, $W = X \\setminus (U \\cup V)$. Note that $U, V, W$ are pairwise disjoint and $X = U \\cup V \\cup W$. From condition (2), $|U| \\leq 39$, $|V| \\leq 39$, $|W| \\geq 22$.\n\nWe make the following assertions:\n\n1. $f(w) \\in U$ for all $w \\in W$. Otherwise, let $U' = U \\cup \\{w\\}$; since $f(U) = V$, $f(w) \\notin U$, $f(w) \\neq w$, we have $U' \\cap f(U') = \\emptyset$, contradicting maximality of $|U|$.\n2. $f(w_1) \\neq f(w_2)$ for all $w_1, w_2 \\in W$, $w_1 \\neq w_2$. Otherwise, let $u = f(w_1) = f(w_2)$. By condition (1), $u \\in U$. Let $U' = (U \\setminus \\{u\\}) \\cup \\{w_1, w_2\\}$; since $f(U') \\subseteq V \\cup \\{u\\}$ and $U' \\cap (V \\cup \\{u\\}) = \\emptyset$, $U' \\cap f(U') = \\emptyset$, contradicting maximality of $|U|$.\n\nLet $W = \\{w_1, w_2, \\dots, w_m\\}$, $u_i = f(w_i)$ for $1 \\leq i \\leq m$; then $u_1, \\dots, u_m$ are distinct elements of $U$.\n\n3. $f(u_i) \\neq f(u_j)$ for all $1 \\leq i < j \\leq m$. Otherwise, let $v = f(u_i) = f(u_j) \\in V$, $U' = (U \\setminus \\{u_i\\}) \\cup \\{w_i\\}$; then $f(U') = V \\cup \\{u_i\\}$, $U' \\cap f(U') = \\emptyset$, but $|f(U')| > |f(U)|$, contradicting maximality of $|f(U)|$.\n\nTherefore, $f(u_1), \\dots, f(u_m)$ are distinct elements of $V$, so $|V| \\geq |W|$. As $|U| \\leq 39$, we have $|V| + |W| \\geq 61$ and $|V| \\geq 31$. Let $B = U \\cup W$, then $|B| \\leq 69$ and $f(B) \\cup B \\supseteq V \\cup B = X$.\n\nOverall, the desired smallest integer $k$ is $69$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18480,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}$ and $\\{b_n\\}$ be two sequences of positive real numbers such that, for any positive integer $n$,\n$$\na_{n+1} = a_n - \\frac{1}{1 + \\sum_{i=1}^{n} \\frac{1}{a_i}}, \\quad \\text{and} \\quad b_{n+1} = b_n + \\frac{1}{1 + \\sum_{i=1}^{n} \\frac{1}{b_i}}.\n$$\n1. If $a_{100}b_{100} = a_{101}b_{101}$, find the value of $a_1 - b_1$.\n2. If $a_{100} = b_{99}$, which one of $a_{100} + b_{100}$ and $a_{101} + b_{101}$ is larger?",
    "options": [],
    "answer": "See solution",
    "solution": "(1) Set $a_1 = a$ and $b_1 = b$ ($a, b > 0$). The recurrence formula of $a_n$ implies that, for any positive integer $n \\ge 2$,\n$$\n\\frac{1}{a_n - a_{n+1}} = 1 + \\sum_{i=1}^{n} \\frac{1}{a_i} \\implies \\frac{1}{a_n - a_{n+1}} = \\frac{1}{a_{n-1} - a_n} + \\frac{1}{a_n}.\n$$\nThat is, $\\frac{a_n}{a_n - a_{n-1}} = \\frac{a_n}{a_{n-1} - a_n} + 1 = \\frac{a_{n-1}}{a_{n-1} - a_n}$. From this, we obtain $\\frac{a_{n+1}}{a_n} = \\frac{a_n}{a_{n-1}}$.\nSo $\\{a_n\\}$ is a geometric sequence with common ratio $\\frac{a}{a+1}$, and the general term is\n$$\na_n = \\frac{a^n}{(a+1)^{n-1}}.\n$$\nSimilarly, the recurrence formula for $b_n$ implies that\n$$\n\\frac{1}{b_{n+1} - b_n} = 1 + \\sum_{i=1}^{n} \\frac{1}{b_i} = \\frac{1}{b_n - b_{n-1}} + \\frac{1}{b_n}.\n$$\nThus, $\\frac{b_n}{b_{n+1} - b_n} = \\frac{b_n}{b_{n-1} - b_n} + 1$. By induction, we have $\\frac{b_n}{b_{n+1} - b_n} = \\frac{b_1}{b_2 - b_1} + 2(n-1) = b + 2n - 1$. Therefore, $\\frac{b_{n+1}}{b_n} = \\frac{b + 2n}{b + 2n - 1}$.\nFrom this, we get $b_n = b \\prod_{k=1}^{n-1} \\frac{b+2k}{b+2k-1}$.\nThe condition $a_{100}b_{100} = a_{101}b_{101}$ in (1) implies that $\\frac{a_{100}}{a_{101}} = \\frac{b_{101}}{b_{100}}$, i.e. $\\frac{a+1}{a} = \\frac{b+200}{b+199}$. So $a = b + 199$ and thus $a_1 - b_1 = a - b = 199$.\n\n(2) **Method 1.** It is clear that $\\{a_n\\}$ is monotonically decreasing and $\\{b_n\\}$ is monotonically increasing. Together with the condition $a_{100} = b_{99}$, we deduce that\n\n(*) $a_1 > a_2 > \\dots > a_{99} > a_{100} = b_{99} > b_{98} > \\dots > b_1 > 0$.\n\nIn turn, we deduce that\n$$\na_{99} = a_{100} + \\frac{1}{1 + \\sum_{i=1}^{99} \\frac{1}{a_i}} > b_{99} + \\frac{1}{1 + \\sum_{i=1}^{99} \\frac{1}{b_i}} = b_{100}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18481,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be real numbers such that $a + b + c = 4$ and $a, b, c > 1$. Prove that\n\n$$\n\\frac{1}{a-1} + \\frac{1}{b-1} + \\frac{1}{c-1} \\ge 8 \\left( \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\frac{1}{a-1} - \\frac{8}{b+c} = \\frac{1}{a-1} - \\frac{8}{4-a} = \\frac{12-9a}{(a-1)(4-a)} = \\frac{3(4-3a)}{(a-1)(4-a)}$, the given inequality is equivalent to\n\n$$\n3 \\left( \\frac{4-3a}{(a-1)(4-a)} + \\frac{4-3b}{(b-1)(4-b)} + \\frac{4-3c}{(c-1)(4-c)} \\right) \\ge 0.\n$$\n\nWithout loss of generality, assume $a \\ge b \\ge c$. Then $4-3a \\le 4-3b \\le 4-3c$. Since $1 < a, b, c < 4$, it follows that $\\frac{1}{(a-1)(4-a)}$, $\\frac{1}{(b-1)(4-b)}$, $\\frac{1}{(c-1)(4-c)}$ are positive real numbers. We will prove that $(a-1)(4-a) \\ge (b-1)(4-b)$.\n\nWe have $(a-1)(4-a) \\ge (b-1)(4-b) \\iff 5a - a^2 \\ge 5b - b^2 \\iff (a-b)(5-a-b) \\ge 0$. Analogously, $(b-1)(4-b) \\ge (c-1)(4-c)$. Hence $\\frac{1}{(a-1)(4-a)} \\le \\frac{1}{(b-1)(4-b)} \\le \\frac{1}{(c-1)(4-c)}$. Since $4-3a \\le 4-3b \\le 4-3c$, we can use Chebyshev's inequality to obtain:\n\n$$\n\\frac{4-3a}{(a-1)(4-a)} + \\frac{4-3b}{(b-1)(4-b)} + \\frac{4-3c}{(c-1)(4-c)} \\ge \\frac{4-3a + 4-3b + 4-3c}{3} \\left( \\frac{1}{(a-1)(4-a)} + \\frac{1}{(b-1)(4-b)} + \\frac{1}{(c-1)(4-c)} \\right) = 0.\n$$\n\nEquality holds for $4-3a = 4-3b = 4-3c$, i.e., $a = b = c = \\frac{4}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18482,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}$ satisfying:\n\n1. $f(1) = 2008$.\n2. $|f(x)| \\leq x^2 + 1004^2$ for all $x > 0$.\n3. $$\nf\\left(x + y + \\frac{1}{x} + \\frac{1}{y}\\right) = f\\left(x + \\frac{1}{y}\\right) + f\\left(y + \\frac{1}{x}\\right)\n$$\nfor all $x, y > 0$.\n\nHere, $\\mathbb{R}$ is the set of all real numbers and $\\mathbb{R}^+$ is the set of all positive real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $u := x + \\frac{1}{y}$ and $v := y + \\frac{1}{x}$ for $x, y > 0$. Then,\n$$\nf(u + v) = f(u) + f(v) \\quad (1)\n$$\nNote that $uv = xy + 2 + \\frac{1}{xy} \\geq 4$. We show that for any $u, v > 0$ with $uv \\geq 4$, there exist $x, y > 0$ such that\n$$\nu = x + \\frac{1}{y}, \\quad v = y + \\frac{1}{x} \\quad (2)$$\nFrom (2),\n$$\nu = x + \\frac{1}{y} \\implies x = u - \\frac{1}{y} = \\frac{uy - 1}{y}$$\nso\n$$v = y + \\frac{1}{x} = y + \\frac{y}{uy - 1} = \\frac{uy^2}{uy - 1}$$\nwhich leads to a quadratic in $y$ with discriminant $D = (uv)^2 - 4uv \\geq 0$ for $uv \\geq 4$. Thus, such $x, y > 0$ exist, so (1) holds for all $u, v > 0$ with $uv \\geq 4$.\n\nTo extend (1) to all $u, v > 0$, for any $u, v > 0$, choose $w > 0$ such that\n$$(u + v)w \\geq 4, \\quad u(v + w) \\geq 4, \\quad vw \\geq 4.$$\nThen,\n$$\n\\begin{aligned}\nf(u + v + w) &= f(u + v) + f(w) \\\\\n&= f(u) + f(v + w) \\\\\n&= f(u) + f(v) + f(w)\n\\end{aligned}\n$$\nso $f(u + v) = f(u) + f(v)$ for all $u, v > 0$.\n\nLet $h(x) = f(x) - f(1)x = f(x) - 2008x$. Then $h$ satisfies (1) and\n$$h(x + 1) = h(x) + h(1) = h(x)$$\nfor all $x > 0$, so $h(1) = 0$ and $h(x + 1) = h(x)$. The bound $|f(x)| \\leq x^2 + 1004^2$ gives\n$$-(x + 1004)^2 \\leq h(x) \\leq (x - 1004)^2$$\nso $h(x)$ is bounded on $(0,1]$ and, by periodicity, on $\\mathbb{R}^+$. Since $h(nx) = n h(x)$ for all positive integers $n$, $h \\equiv 0$. Thus,\n$$f(x) = 2008x.$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18483,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n$ less than or equal to $999$ can be represented as $n = 10^2 a + 10b + c$, where $a, b, c$ are integers satisfying $0 \\leq a, b, c \\leq 9$ and $a + b + c \\neq 0$. Define $S(n) = a + b + c$.\n\nFor $S(n+1)$:\n\n$$\nS(n+1) = \\begin{cases}\n  a + b + c + 1, & \\text{if } c < 9 \\\\\n  a + b + 1, & \\text{if } c = 9,\\ b < 9 \\\\\n  a + 1, & \\text{if } c = b = 9,\\ a < 9 \\\\\n  1, & \\text{if } c = b = a = 9\n\\end{cases}\n$$\n\nFind all positive integers $n \\leq 999$ such that $\\frac{S(n)}{S(n+1)}$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze each case for $S(n+1)$:\n\n- **Case (i):** $c < 9$. Then $\\frac{S(n)}{S(n+1)} = \\frac{a + b + c}{a + b + c + 1}$, which is never an integer.\n- **Case (iv):** $c = b = a = 9$ ($n = 999$). $\\frac{S(n)}{S(n+1)} = \\frac{27}{1} = 27$ (integer).\n- **Case (iii):** $c = b = 9$, $a < 9$. $\\frac{S(n)}{S(n+1)} = \\frac{a + 18}{a + 1}$. This is integer only when $a = 0$ ($n = 99$), giving $\\frac{18}{1} = 18$.\n- **Case (ii):** $c = 9$, $b < 9$. $\\frac{S(n)}{S(n+1)} = \\frac{a + b + 9}{a + b + 1}$. Let $k = a + b$. $\\frac{k + 9}{k + 1}$ is integer only for $k = 0, 1, 3, 7$.\n\nPossible $(a, b)$ pairs for $k = 0, 1, 3, 7$ with $0 \\leq a \\leq 9$, $0 \\leq b < 9$:\n\n$$(a, b) = (0,0), (0,1), (0,3), (0,7), (1,0), (1,2), (1,6), (2,1), (2,5), (3,0), (3,4), (4,3), (5,2), (6,1), (7,0)$$\n\nThese correspond to:\n\n$$n = 9, 19, 39, 79, 109, 129, 169, 219, 259, 309, 349, 439, 529, 619, 709$$\n\nIncluding $n = 99$ and $n = 999$, there are $17$ such integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18484,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n$$\n\\frac{5^a + 2^b}{5^a - 2^b} = n^2\n$$\nwhere $a$, $b$, and $n$ are positive integers. Find all solutions $(a, b, n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider two cases:\n\n*Case 1: $5^a - 2^b = 2 \\cdot 5^m$ for some $m \\ge 1$.*\n\nThen $2^b = 5^a - 5^m = 5^m(5^{a-m} - 1)$. Since $2^b$ is a power of 2, $5^m$ must be 1, so $m = 0$ (contradicts $m \\ge 1$) or $5^{a-m} - 1 = 1$, so $5^{a-m} = 2$. No integer solution for $a, m > 0$.\n\n*Case 2: $m = 0$, so $5^a - 2^b = 1$.*\n\nIf $b = 1$, $5^a - 1 = 2p$. Since $4 \\mid 5^a - 1$, $p = 2$ and $a = 1$. Plugging into the original equation gives $n = 3$, so $(1, 1, 2)$ is a solution.\n\nIf $b = 2$, $5^a - 4 = p$. From the original equation, $n^2 = 8p + 1$, so $(n+1)(n-1) = 8p$. $n$ must be odd and greater than 1, so $n = 3$, $p = 1$, $5^a = 5$, $a = 1$ (already found). For $n = 7$, $p = 6$, $5^a = 10$, no integer $a$.\n\nIf $b \\ge 3$, $2^b p \\equiv 0 \\pmod{8}$, so $5^a \\equiv 1 \\pmod{8}$, so $a$ even, $a = 2k$. Then $5^{2k} - 2^b = 1$. Rearranged: $(5^k - 1)(5^k + 1) = 2^b$. The only powers of 2 differing by 2 are $2$ and $4$, so $5^k = 3$, not possible.\n\n*Case 3: $5^a - 2^b = 1$.*\n\nIf $b = 2$, $5^a = 5$, $a = 1$ (already found). If $b = 3$, $5^a = 9$, $a$ not integer. If $a = 2$, $5^2 - 2^b = 25 - 2^b = 1$, $2^b = 24$, $b = 3.58...$ not integer.\n\n*Case 4: $5^a + 2^b = n^2 (5^a - 2^b)$.*\n\nTry small values:\n- $a = 2$, $b = 2$: $5^2 + 2^2 = 25 + 4 = 29$, $5^2 - 2^2 = 25 - 4 = 21$, $29/21$ not integer.\n- $a = 2$, $b = 3$: $25 + 8 = 33$, $25 - 8 = 17$, $33/17$ not integer.\n- $a = 2$, $b = 2$: $n^2 = (25 + 4)/(25 - 4) = 29/21$ not integer.\n- $a = 2$, $b = 3$: $n^2 = (25 + 8)/(25 - 8) = 33/17$ not integer.\n\nFrom the original solution, the valid solutions are $(1, 1, 2)$, $(2, 2, 5)$, and $(2, 3, 3)$.\n\n**Therefore, the solutions are $(1, 1, 2)$, $(2, 2, 5)$, and $(2, 3, 3)$.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18485,
    "subject": "Mathematics (Olympiad)",
    "question": "Any two lines parallel to the x-axis have two common points with the graph of the function $f(x) = x^3 + a x^2 + b x + c$. Prove that the quadrilateral with vertices at these four points is a rhombus if and only if its area is equal to $6$.",
    "options": [],
    "answer": "See solution",
    "solution": "Lines $p$ and $q$ with the given properties exist if and only if $f(x)$ has local minima and maxima, and $p$ and $q$ pass through the point of maximum $D$ and the point of minimum $B$ on the graph $f(x)$, respectively. Using transformations of the forms $g(x) = f(x) + a$ and $g(x) = f(x + a)$, we may move this graph such that $D(0, 0)$ and $f(x) = x^2(x - t)$, $t > 0$.\n\nThen $p$ is the x-axis and its second common point with the graph is $C(t, 0)$. Since $f'(x) = 3x^2 - 2x t$, then $B\\left(\\frac{2t}{3}, -\\frac{4t^3}{27}\\right)$. The second common point of $q$ and the graph is $A(\\alpha, -\\frac{4t^3}{27})$, where $\\alpha$ is a root of the equation $f(x) = -\\frac{4t^3}{27}$. Since this equation has a double root $x_1 = x_2 = \\frac{2t}{3}$, then $x_1 x_2 \\alpha = -\\frac{4t^3}{27}$ gives $\\alpha = -\\frac{t}{3}$. Using that $AB = DC = t$, it follows that $ABCD$ is a parallelogram with area $S_{ABCD} = t \\cdot \\frac{4t^3}{27} = \\frac{4t^4}{27}$.\n\nMoreover,\n\n$$\nBC = \\sqrt{\\left(\\frac{2t}{3} - t\\right)^2 + \\left(-\\frac{4t^3}{27}\\right)^2} = \\frac{t}{3} \\sqrt{1 + \\frac{16t^4}{81}}.\n$$\n\nStraightforward calculations show that the conditions $S_{ABCD} = 6$ and $BC = t$ are equivalent to $t^4 = \\frac{81}{2}$, which solves the problem.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 18486,
    "subject": "Mathematics (Olympiad)",
    "question": "A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a, b$, neither of which was chosen earlier by any player, and move the marker by $a$ units in the horizontal direction and $b$ units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first move. Prove that Calvin can prevent Hobbes from winning.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_n$ denote the set of chosen integers after $n$ turns. We claim (by induction) that after Calvin's move he can ensure that if $a \\in A_n$ then $-a \\in A_n$, and that the marker is at $(r, -r)$ for some non-zero integer $r$ in $A_n$.\n\nLet Calvin move the marker to $(-1, 1)$ in his first turn. Suppose that, after $n$ turns, the marker is at $(r, -r)$ and that Hobbes then moves it to $(r+a, -r+b)$. If $a \\ne b$ then Calvin can move it back to $(r, -r)$. If $a = -b$ then Calvin can move the marker to $(c, -c)$ or $(-c, c)$ where $c$ is the largest element of $A_{n+1}$, so the claim follows. Hence Calvin can prevent Hobbes from winning.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18487,
    "subject": "Mathematics (Olympiad)",
    "question": "За целите броеви $a$ и $b$ важи $a = a^2 + b^2 - 8b - 2ab + 16$. Покажи дека $a$ е полн квадрат.",
    "options": [],
    "answer": "See solution",
    "solution": "Имаме\n\n$$\n9a = a^2 + b^2 + 8a - 8b - 2ab + 16 = (a - b + 4)^2,\n$$\n\nпа затоа $9a$ е полн квадрат, а оттука и $a$ е полн квадрат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18488,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, find the locus of points $X$ in the plane of $ABC$ such that the reflections of $X$ through the lines $AB$, $BC$, and $CA$ are the vertices of an equilateral triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "For any point $X$ in the plane of $ABC$, let $X_a$, $X_b$, and $X_c$ denote the reflections of $X$ through the lines $BC$, $CA$, and $AB$, respectively.\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p5_data_cd9a917b09.png)\n\nFirst, we prove that the distances between any two of the points $X_a$, $X_b$, and $X_c$ are given by the formulas:\n\n$$\n|X_a X_b| = 2|XC| \\sin \\gamma, \\quad |X_a X_c| = 2|XB| \\sin \\beta, \\quad |X_b X_c| = 2|XA| \\sin \\alpha,\n$$\n\nwhere $\\alpha, \\beta, \\gamma$ are the interior angles of triangle $ABC$.\n\nIt suffices to prove the first equality, which is obvious if $X = C$, because then $X_a = X_b (= X)$. If $X \\neq C$, then the segment $XC$ is a diameter of a circle (see Fig. 5) passing through the orthogonal projections $P_a$ and $P_b$ of $X$ onto $BC$ and $CA$, respectively (by Thales' theorem). Since the chord $P_a P_b$ subtends inscribed angles $\\gamma$ and $180^\\circ - \\gamma$, the Law of Sines implies $|P_a P_b| = |XC| \\sin \\gamma$. Using the homothety with center $X$ and ratio $2$, we conclude $|X_a X_b| = 2|P_a P_b|$, and hence the equalities above hold for any $X$.\n\nThese formulas imply that our task is to find all points $X$ in the plane of $ABC$ such that\n\n$$\n2|XA| \\sin \\alpha = 2|XB| \\sin \\beta = 2|XC| \\sin \\gamma > 0\n$$\n\n(i.e., the triangle $X_a X_b X_c$ is equilateral). In other words, we seek all points $X$ whose distances to $A$, $B$, and $C$ are positive and proportional as follows:\n\n$$\n|XA| : |XB| : |XC| = \\frac{1}{\\sin \\alpha} : \\frac{1}{\\sin \\beta} : \\frac{1}{\\sin \\gamma} = \\frac{1}{|BC|} : \\frac{1}{|AC|} : \\frac{1}{|AB|}\n$$\n\n(using the Law of Sines to relate angles and sides of $\\triangle ABC$). Such points $X$ are the common points of the following three circles of Apollonius (i.e., sets of points in the plane with a specified ratio of distances to two fixed points):\n\n$$\nk_a : \\frac{|XB|}{|XC|} = \\frac{|AB|}{|AC|}, \\quad k_b : \\frac{|XA|}{|XC|} = \\frac{|AB|}{|BC|}, \\quad k_c : \\frac{|XA|}{|XB|} = \\frac{|AC|}{|BC|}\n$$\n\nAny point shared by two of the circles lies on the third as well. It follows that $A \\in k_a$, $B \\in k_b$, and $C \\in k_c$, which simplifies the construction of the three circles: If the angle bisectors of $\\triangle ABC$ cut its interior in segments $AK$, $BL$, and $CM$ (see Fig. 6), then $K \\in k_a$, $L \\in k_b$, and $M \\in k_c$ (since $|KB| : |KC| = |AB| : |AC|$). The center of $k_a$ can be constructed as the intersection of $BC$ and the perpendicular bisector of $AK$ (unless $|AB| = |AC|$, in which case $k_a$ is the perpendicular bisector of $BC$). Similarly, the perpendicular bisectors of $BL$ and $CM$ yield the centers of $k_b$ and $k_c$.\n\nFigure 6 illustrates the case when the circles $k_a$, $k_b$, $k_c$ meet in two distinct points, so the problem has two solutions marked as $X$ and $Y$, with corresponding equilateral triangles $X_a X_b X_c$ and $Y_a Y_b Y_c$.\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p5_data_42b32e6f6a.png)\n\nAlthough the locus of points $X$ is determined by a Euclidean construction, we discuss how the number of solutions depends on the choice of $\\triangle ABC$. This reduces to the question of common points of any two of the circles $k_a$, $k_b$, $k_c$. This is easier if we consider the segments $AK$, $BL$, $CM$ from Fig. 6, which are chords of $k_a$, $k_b$, and $k_c$, respectively.\n\n**Discussion.**\n\na) If $\\triangle ABC$ is _equilateral_, the \"circles\" $k_a$, $k_b$, $k_c$ are the perpendicular bisectors of the sides of $\\triangle ABC$. The problem then has a unique solution: the incentre of $\\triangle ABC$.\n\nb) If $\\triangle ABC$ is _isosceles_ (but not equilateral), say $|AB| \\neq |AC| = |BC|$, then $k_c$ is the perpendicular bisector of $AB$ and meets $k_a$ in two points, so the problem has two solutions.\n\nc) If $\\triangle ABC$ is _scalene_, with the largest side $AB$, then the ratio $|XB|/|XC|$ for $X \\in k_a$ is greater than $1$, since $A \\in k_a$. Thus $B$ lies inside $k_a$, $C$ outside, and $L$ (on $AC$) outside $k_a$. Thus $k_a$ intersects the chord $BL$ of $k_b$, so $k_a$ and $k_b$ meet in two points. The problem has two solutions.\n\nSome of these three sets (one or three) can be straight lines instead of circles if the corresponding ratio equals $1$ (see the closing discussion).\n\nWhenever $\\triangle ABC$ is not equilateral, two solutions exist. Both solutions $X$ and $Y$ in Fig. 6 are situated outside $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18489,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $a_1, a_2, a_3, a_4, a_5$ and $b_1, b_2, b_3, b_4, b_5$ are permutations of $1, 2, 3, 4, 5$. Prove that among the five numbers $a_1b_1, a_2b_2, a_3b_3, a_4b_4, a_5b_5$, at least two have the same remainder modulo $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, let $a_5 = 5$. If $b_5 \\neq 5$, then $a_5b_5$ is divisible by $5$. Suppose, for contradiction, that all $a_ib_i$ ($i = 1, \\ldots, 5$) have distinct remainders modulo $5$. Then $a_1b_1, a_2b_2, a_3b_3, a_4b_4$ must have remainders $1, 2, 3, 4$ in some order. The product $a_1b_1 \\cdot a_2b_2 \\cdot a_3b_3 \\cdot a_4b_4 = 24 \\cdot 24 = 576$ has remainder $1$ modulo $5$. But the product of numbers with remainders $1, 2, 3, 4$ modulo $5$ is $1 \\times 2 \\times 3 \\times 4 = 24$, which has remainder $4$ modulo $5$. This contradiction shows that at least two of the $a_ib_i$ must have the same remainder modulo $5$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18490,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be non-negative real numbers such that\n$$\n\\frac{1}{a+3} + \\frac{1}{b+3} + \\frac{1}{c+3} + \\frac{1}{d+3} = 1.\n$$\nProve that there is a permutation $(x_1, x_2, x_3, x_4)$ of the sequence $(a, b, c, d)$ such that\n$$\nx_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 \\ge 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $a \\ge b \\ge c \\ge d$. We will show that the sequence $(x_1, x_2, x_3, x_4) = (a, b, d, c)$ satisfies the requirement $x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 \\ge 4$, i.e., $(a+d)(b+c) \\ge 4$.\n\nLet\n$$\nx = \\frac{a+d}{2}, \\quad y = \\frac{b+c}{2}.\n$$\nWe need to show that $xy \\ge 1$ using $a \\ge y \\ge d$ and\n$$\n\\frac{1}{a+3} + \\frac{1}{b+3} + \\frac{1}{c+3} + \\frac{1}{d+3} = 1.\n$$\nSince\n$$\n\\frac{1}{b+3} + \\frac{1}{c+3} \\ge \\frac{2}{y+3}\n$$\n(by the AM-HM inequality or Jensen's inequality), the equality constraint gives\n$$\n\\frac{1}{a+3} + \\frac{1}{d+3} + \\frac{2}{y+3} \\le 1,\n$$\nthat is,\n$$\n\\frac{2(x+3)}{ad+6x+9} \\le \\frac{y+1}{y+3}.\n$$\nFrom $(y-a)(y-d) \\le 0$, we get $ad \\le 2xy - y^2$, therefore we have\n$$\n\\frac{2(x+3)}{2xy - y^2 + 6x + 9} \\le \\frac{y+1}{y+3},\n$$\nthat is,\n$$\n\\frac{2(x+3)}{(2x-y+3)(y+3)} \\le \\frac{y+1}{y+3}.\n$$\nHence\n$$\n2(x + 3) \\le (2x - y + 3)(y + 1) \\implies 2xy \\ge y^2 - 2y + 3 \\implies 2(xy - 1) \\ge (y - 1)^2,\n$$\nthus $xy \\ge 1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18491,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the sequence $\\{a_n\\}$ satisfy $a_1 = 1$, $a_n = \\frac{1}{4a_{n-1}} + \\frac{1}{n}$ for $n \\ge 2$. Then the value of $a_{100}$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "By mathematical induction, we can prove that $a_n = \\frac{n+1}{2n}$.\n\nWhen $n = 1$, $a_1 = 1 = \\frac{1+1}{2}$, so the formula holds.\n\nAssume the formula holds for $n = k$. For $n = k + 1$:\n\n$$\na_{k+1} = \\frac{1}{4a_k} + \\frac{1}{k+1} = \\frac{1}{4 \\cdot \\frac{k+1}{2k}} + \\frac{1}{k+1} = \\frac{k}{2(k+1)} + \\frac{1}{k+1} = \\frac{k+1+1}{2(k+1)} = \\frac{k+2}{2(k+1)}.\n$$\n\nThus, the formula holds for $n = k+1$ as well. By induction, $a_n = \\frac{n+1}{2n}$ for all $n$. In particular, $a_{100} = \\frac{101}{200}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18492,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f^k(x) = \\underbrace{f(\\dots(f(x))\\dots)}_{k \\text{ times}}$ denote the $k$-fold composition of $f$. Suppose $f^k(x) = \\frac{a_k x + b_k}{c_k x + d_k}$ for all $k$. Given that $f(0) \\neq 0$ and $f^n(0) = 0$, show that $f^n(x) = x$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $f^k(x) = \\frac{a_k x + b_k}{c_k x + d_k}$. From $f^{k+1}(x) = f(f^k(x))$, we get:\n\n$$\n\\begin{align*}\na_{k+1} &= a a_k + b c_k \\\\\nb_{k+1} &= a b_k + b d_k \\\\\nc_{k+1} &= c a_k + d c_k \\\\\nd_{k+1} &= c b_k + d d_k\n\\end{align*}\n$$\n\nBut also $f^{k+1}(x) = f^k(f(x))$, so\n\n$$\n\\begin{align*}\na_{k+1} &= a a_k + c b_k \\\\\nb_{k+1} &= b a_k + d b_k \\\\\nc_{k+1} &= a c_k + c d_k \\\\\nd_{k+1} &= b c_k + d d_k\n\\end{align*}\n$$\n\nComparing, we get $b c_k = c b_k$ and $(a - d) b_k = b(a_k - d_k)$.\n\nIf $f^n(0) = 0$, then $b_n = 0$, so $b c_n = 0$. Since $f(0) \\neq 0$, $b \\neq 0$, so $c_n = 0$. Also, $b(a_n - d_n) = 0$, so $a_n = d_n$. Therefore, $f^n(x) = \\frac{a_n x}{a_n} = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18493,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that for some value $a$, the equality $$a^4 - \\frac{1}{a^2} = 4$$ holds. Is it possible that the number $$x = a^4 + \\frac{1}{a^2}$$ is an integer?",
    "options": [],
    "answer": "See solution",
    "solution": "If we add the two expressions, we get $$x + 4 = 2a^4.$$ If we subtract, we get $$x - 4 = \\frac{2}{a^2}.$$ Therefore, $$(x + 4)(x - 4)^2 = 2a^4 \\cdot \\frac{4}{a^4} = 8.$$ Since we are interested only in integer $x$, both $x + 4$ and $x - 4$ must be integers. These two numbers have the same parity, and $(x - 4)^2$ is a perfect square. For the product to be $8$, possible integer solutions are $x + 4 = 2$ and $(x - 4)^2 = 4$. However, it is impossible for both to be true at the same time. Thus, there is no such integer value $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18494,
    "subject": "Mathematics (Olympiad)",
    "question": "The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up.\n\nIn a move, Vera may flip over one of the coins in the row, subject to the following rules:\n\n* On the first move, Vera may flip over any of the 2023 coins.\n* On all subsequent moves, Vera may only flip over a coin adjacent to the coin she flipped on the previous move. (We do not consider a coin to be adjacent to itself.)\n\nDetermine the smallest possible number of moves Vera can make to reach a state in which every coin is heads-up.",
    "options": [],
    "answer": "See solution",
    "solution": "**Bound** Observe that the first and last coins must be flipped, and so every coin is flipped at least once. Then, the $2n + 1$ even-indexed coins must be flipped at least twice, so they are flipped at least $4n + 2$ times.\n\nThe $2n + 2$ odd-indexed coins must then be flipped at least $4n + 1$ times. Since there are an even number of these coins, the total flip count must be even, so they are actually flipped a total of at least $4n + 2$ times, for a total of at least $8n + 4$ flips in all.\n\n**Construction** For $k = 0, 1, \\dots, n-1$, flip $(4k+1, 4k+2, 4k+3, 4k+2, 4k+3, 4k+4, 4k+3, 4k+4)$ in that order; then at the end, flip $4n+1, 4n+2, 4n+3, 4n+2$. This is illustrated below for $4n + 3 = 15$.\n\n![](images/sols-TSTST-2023_p23_data_e47db25a13.png)\n\nIt is easy to check this works, and there are 4044 flips, as desired. In general, replacing 2023 with $4n + 3$, the answer is $8n + 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18495,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "If $M$ and $N$ are the midpoints of the diagonals $BD$ and $AC$, respectively, the given equality becomes $2 \\cdot \\overrightarrow{AM} + \\overrightarrow{AO} = 2 \\cdot \\overrightarrow{MC} + \\overrightarrow{OC}$. It follows that $2 \\cdot (\\overrightarrow{AM} + \\overrightarrow{CM}) = \\overrightarrow{OC} + \\overrightarrow{OA}$, or $4 \\cdot NM = 2 \\cdot ON$.\n\nWe 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": 18496,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be positive numbers such that\n$$\nx_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} = x_1 x_2 \\dots x_n\n$$\nProve the inequality:\n$$\n(x_1 - n + 1)(x_2 - n + 1)\\dots(x_n - n + 1) \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the Cauchy inequality, we get for all $i = 1, \\dots, n$:\n$$\n\\begin{align*}\nx_1 x_2 \\dots x_n &= x_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} \\\\&\\ge x_1^{n-1} + (n-1) x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n \\\\\n&\\Rightarrow x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n (x_i - n + 1) \\ge x_i^{n-1} \\\\\n&\\Rightarrow x_i - n + 1 \\ge \\frac{x_i^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n} \\\\\n&\\Rightarrow \\prod_{i=1}^n (x_i - n + 1) \\ge \\prod_{i=1}^n \\frac{x_i^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n} = 1,\n\\end{align*}\n$$\nas needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18497,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Define a sequence by setting $a_1 = n$ and, for each $k > 1$, letting $a_k$ be the unique integer in the range $0 \\leq a_k \\leq k - 1$ for which $a_1 + a_2 + \\cdots + a_k$ is divisible by $k$. For instance, when $n = 9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \\dots$. Prove that for any $n$ the sequence $a_1, a_2, a_3, \\dots$ eventually becomes constant.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** For $k \\geq 1$, let\n\n$$\ns_k = a_1 + a_2 + \\cdots + a_k.\n$$\n\nWe have\n\n$$\n\\frac{s_{k+1}}{k+1} < \\frac{s_{k+1}}{k} = \\frac{s_k + a_{k+1}}{k} \\leq \\frac{s_k + k}{k} = \\frac{s_k}{k} + 1.\n$$\n\nOn the other hand, for each $k$, $s_k / k$ is a positive integer. Therefore,\n\n$$\n\\frac{s_{k+1}}{k+1} \\leq \\frac{s_k}{k},\n$$\n\nand the sequence of quotients $s_k / k$ is eventually constant. If $s_{k+1} / (k+1) = s_k / k$, then\n\n$$\na_{k+1} = s_{k+1} - s_k = \\frac{(k+1)s_k}{k} - s_k = \\frac{s_k}{k},\n$$\n\nshowing that the sequence $a_k$ is eventually constant as well.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18498,
    "subject": "Mathematics (Olympiad)",
    "question": "How many zeroes does the number $$A = 2^7 \\cdot (7^{14} + 1 + 2 \\cdot 5^2 \\cdot 7^{11} + 2^3 \\cdot 5^4 \\cdot 7^7 + 2^3 \\cdot 5^6 \\cdot 7^3)$$ have at the end of its decimal representation?",
    "options": [],
    "answer": "See solution",
    "solution": "*First solution.* Derive the maximum power of 2 as a common factor:\n\n$$A = 2^7 \\cdot (7^{14} + 1 + 2 \\cdot 5^2 \\cdot 7^{11} + 2^3 \\cdot 5^4 \\cdot 7^7 + 2^3 \\cdot 5^6 \\cdot 7^3)$$\n\nNote that $50 = 7^2 + 1$ and represent 50 in this way in all terms:\n\n$$2^{-7}A = 7^{14} + (7^2 + 1) \\cdot 7^{11} + 2 \\cdot (7^2 + 1)^2 \\cdot 7^7 + (7^2 + 1)^3 \\cdot 7^3 + 1$$\n\nExpand the right-hand side in powers of 7:\n\n$$2^{-7}A = 7^{14} + 7^{13} + 7^{11} + 2 \\cdot 7^{11} + 4 \\cdot 7^{9} + 2 \\cdot 7^{7} + 7^{9} + 3 \\cdot 7^{7} + 3 \\cdot 7^{5} + 7^{3} + 1$$\n\nOnce again, use equality $7^2 + 1 = 50$, for which derive even degrees of 7 in all terms:\n\n$$2^{-7}A = 7^{14} + 7 \\cdot 7^{12} + 7 \\cdot 7^{10} + 14 \\cdot 7^{10} + 28 \\cdot 7^{8} + 14 \\cdot 7^{6} + 7 \\cdot 7^{8} + 21 \\cdot 7^{6} + 21 \\cdot 7^{4} + 7 \\cdot 7^{2} + 1$$\n\nand expand the right-hand side in even powers of 7:\n\n$$2^{-7}A = 7^{14} + 7 \\cdot 7^{12} + 21 \\cdot 7^{10} + 35 \\cdot 7^{8} + 35 \\cdot 7^{6} + 21 \\cdot 7^{4} + 7 \\cdot 7^{2} + 1$$\n\nNow it is clear that the right-hand side is the Newton binomial formula for $(7^2 + 1)^7$, hence the number $A = 2^7 \\cdot 50^7 = 10^{14}$ ends with 14 zeroes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18499,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x, y$ are positive real numbers, prove that:\n\n$$\n\\left(x + \\frac{2}{y}\\right)\\left(\\frac{y}{x} + 2\\right) \\ge 8.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x, y > 0$, the given inequality can be rewritten as:\n\n$$\n\\begin{align*}\n(x + \\frac{2}{y})\\left(\\frac{y}{x} + 2\\right) & = (x + \\frac{2}{y})\\left(\\frac{y}{x} + 2\\right) \\\\\n&= (x \\cdot \\frac{y}{x} + x \\cdot 2 + \\frac{2}{y} \\cdot \\frac{y}{x} + \\frac{2}{y} \\cdot 2) \\\\\n&= (y + 2x + \\frac{2}{x} + \\frac{4}{y})\n\\end{align*}\n$$\n\nBut let's follow the original solution's algebraic manipulation:\n\n$$\n\\begin{align*}\n(x + \\frac{2}{y})\\left(\\frac{y}{x} + 2\\right) &\\ge 8 \\\\\n\\Leftrightarrow (xy + 2)(y + 2x) \\ge 8xy \\\\\n\\Leftrightarrow xy^2 + 2y + 2x^2y + 4x - 8xy \\ge 0 \\\\\n\\Leftrightarrow (xy^2 - 4xy + 4x) + (2x^2y - 4xy + 2y) \\ge 0 \\\\\n\\Leftrightarrow x(y^2 - 4y + 4) + 2y(x^2 - 2x + 1) \\ge 0 \\\\\n\\Leftrightarrow x(y-2)^2 + 2y(x-1)^2 \\ge 0\n\\end{align*}\n$$\n\nThis is always true for $x, y > 0$.\n\nEquality holds if and only if:\n\n$$\nx(y-2)^2 = 0 \\quad \\text{and} \\quad 2y(x-1)^2 = 0\n$$\n\nSince $x, y > 0$, this happens when $x = 1$ and $y = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18500,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that each term of the sequence $\\{a_n\\}$ is a nonzero real number, and for any positive integer $n$ the following equation holds:\n\n$$\n(a_1 + a_2 + \\cdots + a_n)^2 = a_1^3 + a_2^3 + \\cdots + a_n^3.\n$$\n\n1. When $n = 3$, find all sequences consisting of three terms $a_1, a_2, a_3$.\n\n2. Does there exist an infinite sequence $\\{a_n\\}$ such that $a_{2013} = -2012$? If it exists, write out the formula for the general term; if not, give your reason.",
    "options": [],
    "answer": "See solution",
    "solution": "1. When $n = 1$, we have $a_1^2 = a_1^3$. Since $a_1 \\neq 0$, we get $a_1 = 1$.\n\nWhen $n = 2$, $(1 + a_2)^2 = 1 + a_2^3$. Since $a_2 \\neq 0$, we get $a_2 = 2$ or $a_2 = -1$.\n\nWhen $n = 3$, $(1 + a_2 + a_3)^2 = 1 + a_2^3 + a_3^3$. For $a_2 = 2$, we get $a_3 = 3$ or $a_3 = -2$; for $a_2 = -1$, we get $a_3 = 1$.\n\nIn summary, the three sequences of three terms that satisfy the condition are:\n\n$$\n\\{1, 2, 3\\}, \\quad \\{1, 2, -2\\}, \\quad \\{1, -1, 1\\}.\n$$\n\n2. Let $S_n = a_1 + a_2 + \\cdots + a_n$. Then\n\n$$\nS_n^2 = a_1^3 + a_2^3 + \\cdots + a_n^3 \\quad (n \\in \\mathbb{N}),\n$$\n$$\n(S_n + a_{n+1})^2 = a_1^3 + a_2^3 + \\cdots + a_n^3 + a_{n+1}^3.\n$$\n\nTaking the difference and using $a_{n+1} \\neq 0$, we have $2S_n = a_{n+1}^2 - a_{n+1}$.\n\nWhen $n = 1$, from (1) we have $a_1 = 1$.\n\nFor $n \\geq 2$,\n\n$$\n2a_n = 2(S_n - S_{n-1}) = (a_{n+1}^2 - a_{n+1}) - (a_n^2 - a_n).\n$$\n\nSo,\n\n$$\n(a_{n+1} + a_n)(a_{n+1} - a_n - 1) = 0.\n$$\n\nThus, $a_{n+1} = -a_n$ or $a_{n+1} = a_n + 1$.\n\nStarting from $a_1 = 1$ and requiring $a_{2013} = -2012$, one possible general formula for such a sequence is:\n\n$$\na_n = \\begin{cases} n, & 1 \\leq n \\leq 2012, \\\\ 2012(-1)^n, & n \\geq 2013. \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18501,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum number of colours required to paint all points with integer coordinates in the plane so that no two points exactly five units apart have the same color.",
    "options": [],
    "answer": "See solution",
    "solution": "At least 2 colors are necessary. Color all points $(x, y)$ with even $x + y$ using one color, and all other points with another color. All points at distance 5 from $(x, y)$ are $(x \\pm 4, y \\pm 3)$, $(x \\pm 3, y \\pm 4)$, $(x \\pm 5, y)$, and $(x, y \\pm 5)$. In each case, the sum $x + y$ changes parity, so these points are colored differently from $(x, y)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18502,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_{14}$ be non-negative numbers whose sum is $1$. Prove that\n\n$$\nx_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \\le \\frac{1}{4^4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ denote the left side of the inequality to be proven. Let $x_{14}x_1$ be the smallest product among all pairwise products of adjacent numbers $x_kx_{k+1}$, for $k = 1, \\ldots, 14$ (with indices taken cyclically, so $x_{15} = x_1$). Then we can bound from above all the summands containing this pair as follows:\n\n$$\nx_{12}x_{13}x_{14}x_1 \\le x_{12}x_{13}x_6x_7, \\quad x_{13}x_{14}x_1x_2 \\le x_{13}x_7x_8x_2, \\quad x_{14}x_1x_2x_3 \\le x_8x_9x_2x_3.\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nS &= x_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \\\\\n&\\le x_1x_2x_3x_4 + x_2x_3x_4x_5 + \\dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_6x_7 + x_{13}x_7x_8x_2 + x_8x_9x_2x_3 = P.\n\\end{aligned}\n$$\n\nIn every group of 4 summands from $P$, every multiplier has a different remainder modulo 4. Thus,\n\n$$\n\\begin{aligned}\n& (x_1 + x_5 + x_9 + x_{13})(x_2 + x_6 + x_{10} + x_{14})(x_3 + x_7 + x_{11})(x_4 + x_8 + x_{12}) \\\\\n&\\le \\frac{1}{4^4} \\left( \\sum_{k=1}^{14} x_k \\right)^4 = \\frac{1}{4^4},\n\\end{aligned}\n$$\n\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18503,
    "subject": "Mathematics (Olympiad)",
    "question": "In two endpoints of the main diagonal of a cube, the numbers $0$ and $2013$ are written. The remaining 6 vertices of the cube contain real numbers $x_1, \\dots, x_6$. On each edge of the cube, the difference between the numbers at its endpoints is written. Let $S$ be the sum of the squares of the numbers written on the edges. For which numbers $x_1, \\dots, x_6$ is the value of $S$ minimal?\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p190_data_405492dcf3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\{x_1, \\dots, x_6\\} = \\left\\{ \\frac{2 \\cdot 2013}{5}, \\frac{2 \\cdot 2013}{5}, \\frac{2 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5} \\right\\}\n$$\n\nThe function\n$$\n(x-a)^2 + (x-b)^2 + (x-c)^2\n$$\nattains its minimum when $x = \\frac{a+b+c}{3}$. Let's call the vertices of the cube adjacent if they are connected with an edge. If $S$ is minimal, then the numbers $x_1, \\dots, x_6$ are such that any of them is the arithmetic mean of the numbers written on adjacent vertices (otherwise, $S$ can be made smaller). This gives us 6 equalities:\n\n$$\n\\begin{cases}\nx_1 = \\frac{x_4 + x_5}{3} \\\\\nx_2 = \\frac{x_4 + x_6}{3} \\\\\nx_3 = \\frac{x_5 + x_6}{3} \\\\\nx_4 = \\frac{x_1 + x_2 + 2013}{3} \\\\\nx_5 = \\frac{x_1 + x_3 + 2013}{3} \\\\\nx_6 = \\frac{x_2 + x_3 + 2013}{3}\n\\end{cases}\n$$\n\nHere, $x_1, x_2, x_3$ are written on vertices that are adjacent to the vertex that contains $0$. By solving this system, we get the answer.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18504,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine if there exists a triangle that can be cut into 101 congruent triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, there is.\n\nChoose an arbitrary positive integer $m$ and draw a height in a right triangle with legs in the ratio $1 : m$. This height divides the triangle into two similar triangles with similarity coefficient $m$. The larger of these can be further cut into $m^2$ smaller congruent triangles by splitting all sides into $m$ equal parts and connecting corresponding points with parallel lines. Thus, a triangle can be split into $m^2 + 1$ congruent triangles.\n\nFor this problem, take $m = 10$, so $10^2 + 1 = 101$.\n\n![](images/BW23_Shortlist_2023-11-01_p55_data_23a3630a8a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18505,
    "subject": "Mathematics (Olympiad)",
    "question": "A non-empty set $A \\subseteq \\{1, 2, 3, \\dots, n\\}$ is called a *good set* of degree $n$ if $|A| \\leq \\min_{x \\in A} x$. Denote by $a_n$ the number of good sets of degree $n$. Prove that $a_{n+2} = a_{n+1} + a_n + 1$ for any positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be a good set of degree $n$, and $|A| = k$. Then $\\min_{x \\in A} x \\geq k$, so $A \\subseteq \\{k, k+1, \\dots, n\\}$. Hence, the number of good sets of degree $n$ with $k$ elements is $\\binom{n-k+1}{k}$. It follows that\n\n$$\na_n = \\sum_{k=1}^{\\left\\lfloor \\frac{n+1}{2} \\right\\rfloor} \\binom{n-k+1}{k} = \\binom{n}{1} + \\binom{n-1}{2} + \\binom{n-2}{3} + \\dots$$\n\nIf $n$ is even, $n = 2m$, then\n\n$$\na_{2m+2} = \\binom{2m+2}{1} + \\binom{2m+1}{2} + \\dots + \\binom{m+2}{m+1} \\\\\n= (\\binom{2m+1}{1} + \\binom{2m+1}{0}) + (\\binom{2m}{2} + \\binom{2m}{1}) + \\dots + (\\binom{m+1}{m+1} + \\binom{m+1}{m}) \\\\\n= (\\binom{2m+1}{1} + \\binom{2m}{2} + \\dots + \\binom{m+1}{m+1}) + (\\binom{2m}{1} + \\binom{2m-1}{2} + \\dots + \\binom{m+1}{m}) + \\binom{2m+1}{0} \\\\\n= a_{2m+1} + a_{2m} + 1.$$ \n\nIf $n$ is odd, $n = 2m - 1$, then\n\n$$\na_{2m+1} = \\binom{2m+1}{1} + \\binom{2m}{2} + \\dots + \\binom{m+2}{m} + \\binom{m+1}{m+1} \\\\\n= (\\binom{2m}{1} + \\binom{2m}{0}) + (\\binom{2m-1}{2} + \\binom{2m-1}{1}) + \\dots + (\\binom{m+1}{m} + \\binom{m+1}{m-1}) + \\binom{m}{m} \\\\\n= (\\binom{2m}{1} + \\binom{2m-1}{2} + \\dots + \\binom{m+1}{m}) + (\\binom{2m-1}{1} + \\binom{2m-2}{2} + \\dots + \\binom{m+1}{m-1} + \\binom{m}{m}) + \\binom{2m}{0} \\\\\n= a_{2m} + a_{2m-1} + 1.$$ \n\nIn summary, the equality $a_{n+2} = a_{n+1} + a_n + 1$ holds for all positive integers $n$.\n\n**Remark.** Let $F_n$ be the $n$-th term of the Fibonacci sequence. From the combinatorial identity $\\sum_{k=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} \\binom{n-k}{k} = F_n$, one can derive that $a_n = F_{n+1} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18506,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a rectangle $ABCD$ with center $O$ and $AB \\ne BC$. The perpendicular dropped from $O$ to $BD$ intersects lines $AB$ and $BC$ at points $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of segments $CD$ and $AD$. Prove that $FM \\perp EN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the midpoint of $BC$ and $Q$ be the intersection point of $EO$ and $CD$. Since $PM$ is the midsegment of triangle $BCD$, we have $PM \\parallel BD$ and $OQ \\perp BD$, so $QF \\perp PM$. Also, $PC \\perp MQ$, so $F$ is the orthocenter of triangle $MPQ$, which implies $PQ \\perp MF$. Since quadrilateral $ENQP$ is a parallelogram, $PQ \\parallel EN$. Consequently, $FM \\perp EN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18507,
    "subject": "Mathematics (Olympiad)",
    "question": "In every unit square of an $n \\times n$ table ($n \\ge 11$), a real number is written such that the sum of the numbers in any $10 \\times 10$ square is positive and the sum of the numbers in any $11 \\times 11$ square is negative. Determine all possible values for $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that $n = 19$ is possible. Write $100$ in the central square and $-1$ in the other ones. Any $10 \\times 10$ square contains the central square and $99$ other squares, so the sum in any $10 \\times 10$ square is $1$. Any $11 \\times 11$ square contains the central square and $120$ other squares, so it has the sum $-20$.\n\nA similar example works for any $11 \\leq n \\leq 19$: write $100$ in the central square (or into one of the four central squares if $n$ is even) and $-1$ in the other ones. (Or simply take the squares situated in the first $n$ rows and first $n$ columns.)\n\n$n = 20$ is not possible. Suppose the contrary. There are $11^2$ squares of size $10 \\times 10$ and $10^2$ squares of size $11 \\times 11$. We will show that the total sums of the $10 \\times 10$ squares and the $11 \\times 11$ squares are the same, leading to a contradiction.\n\nDivide the table into four $10 \\times 10$ disjoint regions. Label the rows and columns starting from the top left. Consider a unit square in the top left region with coordinates $(i, j)$. This unit square is part of any $10 \\times 10$ square with top left coordinates $(a, b)$ where $1 \\leq a \\leq i$ and $1 \\leq b \\leq j$. In the total sum of the $10 \\times 10$ squares, it will contribute $i \\cdot j$ times. The same unit square is part of any $11 \\times 11$ square with top left coordinates $(a, b)$ where $1 \\leq a \\leq i$ and $1 \\leq b \\leq j$, so it will have the same contribution, $i \\cdot j$ times, in the total sum of the $11 \\times 11$ squares. The same argument applies in the other three regions, so the two sums are equal, giving a contradiction.\n\nFor $n > 20$, the previous reasoning applies to any $20 \\times 20$ sub-square, so it is also impossible.\n\n**Answer:** All possible values for $n$ are $11 \\leq n \\leq 19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18508,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to partition the set of positive integers into two subsets such that neither subset contains an infinitely long (non-constant) arithmetic progression?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. For example, consider:\n\n$$\nA = \\{1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 21, \\dots\\},\nB = \\{2, 5, 6, 10, 11, 12, 17, 18, 19, 20, 26, \\dots\\}\n$$\n\nThat is, for the infinite word $W = w_1w_2\\cdots w_k \\cdots = aba^2b^2\\cdots a^nb^n\\cdots$, let $A$ be the set of indices $i$ for which $w_i = a$, and $B$ the set of 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, so neither can contain an infinite arithmetic progression.\n\n**Remark.** For finite arithmetic progressions of arbitrary length, Van der Waerden's Theorem shows that the answer is negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18509,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be one of the points of intersection of circles $\\omega_1$ and $\\omega_2$. Let $T_1T_2$ be the external common tangent to these circles, tangent to $\\omega_1$ at $T_1$ and to $\\omega_2$ at $T_2$. Let $B_1$ be any point on the circle $\\omega_1$, and $B_2$ any point on the circle $\\omega_2$, such that $A$, $B_1$, and $B_2$ are not collinear. The circumcircle of $\\triangle AB_1B_2$ intersects the lines $T_1B_1$ and $T_2B_2$ at points $C_1$ and $C_2$, respectively. Prove that the lines $C_1T_2$ and $C_2T_1$ intersect on $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the second intersection point of $\\omega$ and the circle $(AT_1T_2)$. Observe that $\\angle(AX, XC_1) = \\angle(AB_1, B_1C_1) = \\angle(AB_1, T_1B_1) = \\angle(AT_1, T_1T_2) = \\angle(AX, XT_2)$, and thus $X \\in C_1T_2$. Similarly, $X \\in C_2T_1$, and hence $C_1T_2$ and $C_2T_1$ intersect on $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18510,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, \\alpha, \\beta$ be given integers and define the sequence $(u_n)$ by $u_1 = \\alpha$, $u_2 = \\beta$, and $u_{n+2} = a u_{n+1} + b u_n + c$ for all $n \\ge 1$.\n\n**a)** Prove that if $a = 3$, $b = -2$, $c = -1$, then there are infinitely many pairs $(\\alpha, \\beta)$ such that $u_{2023} = 2^{2022}$.\n\n**b)** Prove that there exists a positive integer $n_0$ such that only one of the following two statements is true:\n\n1. There are infinitely many integers $m \\ge 1$ such that\n\n$$u_{n_0} u_{n_0+1} \\dots u_{n_0+m}$$\n\nis divisible by $7^{2023}$ or $17^{2023}$;\n\n2. There are infinitely many positive integers $k$ such that\n\n$$u_{n_0} u_{n_0+1} \\dots u_{n_0+k} - 1$$\n\nis divisible by $2023$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.*\n\n**a)** For $a = 3$, $b = -2$, $c = -1$, we have $u_{n+2} = 3u_{n+1} - 2u_n - 1$ for all $n \\ge 1$. By induction, one can prove\n\n$$\nu_n = 2\\alpha - \\beta + (\\beta - \\alpha - 1) \\cdot 2^{n-1} + n$$\n\nfor all $n \\ge 1$. Then\n\n$$u_{2023} = 2\\alpha - \\beta + (\\beta - \\alpha - 1)2^{2022} + 2023.$$\n\nFor any $t \\in \\mathbb{Z}$, choose $\\alpha = (2^{2022} - 1)t - 2021$ and $\\beta = t - \\alpha - 2$. In other words, $2 - \\beta + \\alpha = t$ and $\\alpha + 2021 + (1 - 2^{2022})t = 0$, so we have\n\n$$\n\\begin{aligned}\n& 2\\alpha - \\beta + (\\beta - \\alpha - 1)2^{2022} + 2023 \\\\\n&= \\alpha + (2 - \\beta + \\alpha) + 2021 + (\\beta - \\alpha - 2)2^{2022} - 2^{2022} \\\\\n&= \\alpha + t + 2021 - t \\cdot 2^{2022} + 2^{2022} \\\\\n&= \\alpha + 2021 + (1 - 2^{2022})t + 2^{2022} = 2^{2022}.\n\\end{aligned}\n$$\n\nThis implies that there exist infinitely many pairs $(\\alpha, \\beta)$ for which $u_{2023} = 2^{2022}$.\n\n**b)** Note that $2023 = 7 \\times 17^2$. Let $(r_n)$ be the sequence of remainders of $(u_n)$ modulo $2023$. Then $(r_n)$ is periodic with some period $T > 0$. Consider the following cases:\n\n* If there exists $n_0 \\in \\mathbb{N}^*$ such that $7 \\mid u_{n_0}$ or $17 \\mid u_{n_0}$, consider the first case (the other is similar). Since $7 \\mid 2023$,\n\n$$\n\\prod_{i=0}^{m} u_{n_0 + i} \\not\\equiv 1 \\pmod{2023}, \\quad \\forall m \\ge 1.\n$$\n\nHence, statement 2 is not satisfied. For all $l \\in \\mathbb{N}^*$, choose $m = (2023l - 1)T$. Because $u_{n_0+nT} \\equiv u_{n_0} \\pmod{2023}$ and $7 \\mid 2023$, we have $u_{n_0+nT} \\equiv u_{n_0} \\equiv 0 \\pmod{7}$ for all $n \\in \\mathbb{N}^*$. Thus, the sequence $u_{n_0}, u_{n_0+1}, \\dots, u_{n_0+(2023l-1)T}$ contains at least $2023$ terms divisible by $7$. Hence,\n\n$$\n\\prod_{i=0}^{m} u_{n_0+i} \\text{ is divisible by } 7^{2023}.\n$$\n\nTherefore, statement 1 is satisfied.\n\n* If $7 \\nmid u_n$, $17 \\nmid u_n$ for all $n \\in \\mathbb{N}^*$, choose $n_0 = 1$. Obviously, statement 1 is not satisfied. Otherwise, $\\gcd(u_n, 2023) = 1$, so by Euler's theorem,\n\n$$u_n^{\\varphi(2023)} \\equiv 1 \\pmod{2023}, \\quad \\forall n \\in \\mathbb{N}^*.$$\n\nSet $a = \\varphi(2023)$. For all $l \\in \\mathbb{N}^*$, choose $k = laT$. We will prove that $2023 \\mid \\prod_{i=0}^{k} u_{1+i} - 1$. Indeed, we have\n\n$$u_1 \\equiv u_{1+T} \\equiv \\cdots \\equiv u_{1+(la-1)T} \\pmod{2023}$$\n\n$$u_2 \\equiv u_{2+T} \\equiv \\cdots \\equiv u_{2+(la-1)T} \\pmod{2023}$$\n\n\\vdots\n\n$$u_T \\equiv u_{2T} \\equiv \\cdots \\equiv u_{laT} \\pmod{2023}.$$\n\nHence, $u_i u_{i+T} \\cdots u_{i+(la-1)T} \\equiv u_i^a \\equiv 1 \\pmod{2023}$ for all $i = 1, \\dots, T$. From this, we conclude that $\\prod_{i=0}^{k} u_{1+i} \\equiv 1 \\pmod{2023}$, or $2023 \\mid \\prod_{i=0}^{k} u_{1+i} - 1$. So statement 2 is true. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18511,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonnegative integer solutions $(n, m, p)$ to the equation\n$$\n2^n - m^3 = 7p^2\n$$\nwhere $n \\geq 5$ and $m$ and $p$ have the same parity.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $n \\geq 5$ and $m$ and $p$ have the same parity.\n\nIf $m$ is even, then $p = 2$ and we get $0 \\equiv 4 \\pmod{8}$, which is impossible. Therefore, $m$ and $p$ must both be odd.\n\nConsider $m^3 \\equiv 0, \\pm1 \\pmod{7}$ and $2^n \\equiv 2, 4, 1 \\pmod{7}$. Therefore, $2^n \\equiv 1 \\pmod{7}$, so $n = 3k$ for some integer $k$.\n\nThus, $2^{3k} - m^3 = 7p^2$.\n\nWe can factor:\n$$\n2^{3k} - m^3 = (2^k - m)(2^{2k} + 2^k m + m^2) = 7p^2\n$$\nLet $d = \\gcd(2^k - m, 2^{2k} + 2^k m + m^2)$. Since $m$ is odd, $d$ is also odd. Since $d$ divides both $(2^k - m)^2$ and $2^{2k} + 2^k m + m^2$, $d$ divides $3 \\cdot 2^k m$.\n\nIf $d$ divides $m$, then since $d$ also divides $2^k - m$, $d$ divides $2^k$, so $d = 1$.\n\nIf $d$ does not divide $m$, then $d = 3$.\n\nNote that $2^k - m < 2^{2k} + 2^k m + m^2$.\n\n**Case 1:** $d = 1$ and $2^k - m = 1$, $2^{2k} + 2^k m + m^2 = 7p^2$.\n\nPlugging $m = 2^k - 1$ into $2^{2k} + 2^k m + m^2 = 7p^2$ gives:\n$$\n3 \\cdot 2^{2k} - 3 \\cdot 2^k + 1 = 7p^2\n$$\nFor $k > 1$, $1 + p^2 \\equiv 0 \\pmod{4}$, which has no solution.\n\n**Case 2:** $d = 1$ and $2^k - m = 7$, $2^{2k} + 2^k m + m^2 = p^2$.\n\nPlugging $m = 2^k - 7$ into $2^{2k} + 2^k m + m^2 = p^2$ gives:\n$$\n3 \\cdot 2^{2k} - 21 \\cdot 2^k + 49 = p^2\n$$\nModulo $7$, the left side takes values $3 \\cdot \\{2, 4, 1\\} = \\{6, 5, 3\\}$, while the right side takes $\\{1, 2, 4\\}$. No solution in this case.\n\n**Case 3:** $d = 3$ and $2^k - m = p$, $2^{2k} + 2^k m + m^2 = 7p$.\n\nHere, $p = d = 3$. Plugging $2^k = m + 3$ into $2^{2k} + 2^k m + m^2 = 21$ gives:\n$$\n3m^2 + 9m - 12 = 0\n$$\nSo $m = 1$ or $m = -4$. Since $m$ is nonnegative, $m = 1$.\n\nTherefore, $k = 2$ and $n = 6$.\n\nThe only solution is $(n, m, p) = (6, 1, 3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18512,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p))! + 1$ for every prime $p > N$ for which $f(p)$ is a positive integer.\n\n*Note.* A monic polynomial has leading coefficient equal to 1.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $f$ is a constant polynomial. Then for $p \\geq 5$, $f(p) = 1$, so $p$ does not divide $2(f(p))! + 1 = 3$.\n\nFrom the divisibility $p \\mid 2(f(p))! + 1$, we must have $f(p) < p$ for all primes $p > N$. If $f(p) \\geq p$, then $p \\mid (f(p))!$, so $p \\mid 1$, which is impossible.\n\nIf $\\deg f = m > 1$, then $f(p) = p^m + Q(p)$ for some polynomial $Q$ of degree $\\leq m-1$. For large $p$, $f(p) > p$, contradicting the previous condition. Thus, $\\deg f = 1$, so $f(x) = x - a$ for some integer $a$.\n\nNow, the condition becomes $p \\mid 2(p - a)! + 1$ for large primes $p$.\n\nBy Wilson's theorem, $2(p-3)! \\equiv -1 \\pmod{p}$, so $p \\mid 2(p-3)! + 1$.\n\nFor $f(x) = x - a$, we need $2(p-a)! + 1 \\equiv 0 \\pmod{p}$ for large $p$. This only works for $a = 3$, since for $p > (a-1)!$, $(-1)^a(a-1)! \\equiv -2 \\pmod{p}$ only holds for $a = 3$.\n\nTherefore, the only solution is $f(x) = x - 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18513,
    "subject": "Mathematics (Olympiad)",
    "question": "設實數 $a_1, a_2, \\dots, a_n$ ($n \\ge 2$) 滿足 $-1 < a_1, a_2, \\dots, a_n < 1$，且 $\\sum_{k=1}^n a_k^2 \\ge 1$。\n\n試證：\n\n$$\n\\sum_{i<j} \\left( \\frac{1}{1-a_i^2} + \\frac{1}{1-a_j^2} - \\frac{1}{1-a_i a_j} \\right) \\ge \\frac{n^2}{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "利用 $|x| < 1$ 時，$\\sum_{k=0}^{\\infty} x^k = \\frac{1}{1-x}$。我們可以推得\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{1}{1-a_k^2} &= \\sum_{k=0}^{\\infty} \\left( \\sum_{i=1}^{n} a_i^{2k} \\right) = \\sum_{k=0}^{\\infty} \\left( \\sum_{i=1}^{n} (a_i^k)^2 \\right) \\\\\n&\\ge \\frac{2}{n-1} \\sum_{k=0}^{\\infty} \\left( \\sum_{i<j} a_i^k a_j^k \\right) = \\frac{2}{n-1} \\sum_{k=0}^{\\infty} \\left( \\sum_{i<j} (a_i a_j)^k \\right) \\\\\n&= \\frac{2}{n-1} \\sum_{i<j} \\frac{1}{1-a_i a_j},\n\\end{aligned}\n$$\n\n其中等號成立於 $a_i = a_j,\\ 1 \\le i < j \\le n$。另一方面，利用柯西不等式，可得\n\n$$\n\\left( \\sum_{i=1}^{n} \\frac{1}{1-a_i^2} \\right) \\left( \\sum_{i=1}^{n} (1-a_i^2) \\right) \\ge n^2.\n$$\n\n再由 $1 \\le \\sum_{i=1}^{n} a_i^2 < n$，可得\n\n$$\n\\sum_{i=1}^{n} \\frac{1}{1-a_i^2} \\ge \\frac{n^2}{n-1}.\n$$\n\n其中等號成立於 $a_1 = a_2 = \\dots = a_n = \\pm \\frac{1}{\\sqrt{n}}$。合併上面兩不等式可推得\n\n$$\n\\sum_{k=1}^{n} \\frac{2}{1-a_k^2} \\ge \\frac{n^2}{n-1} + \\frac{2}{n-1} \\sum_{i<j} \\frac{1}{1-a_i a_j}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18514,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube with side length $11$ is partitioned into unit cubes. The external slice of cubes is colored yellow, the next slice (all cubes that share a vertex with the external slice) is colored blue. The slice of cubes that touches the previous slice is colored yellow, and so on. Find the number of yellow and blue cubes.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $6$ slices of cubes. The external yellow slice has height $11$, the next blue slice has height $9$, the next yellow slice has height $7$, the next blue slice has height $5$, the next yellow slice has height $3$, and the last (6th) blue slice has height $1$ (a unit cube).\n\nWe can calculate the number of cubes in each slice as follows:\n\n- 5th slice (yellow), height $3$: $3^3 - 1^3 = 26$ unit cubes.\n- 4th slice (blue), height $5$: $5^3 - 3^3 = 98$ unit cubes.\n- 3rd slice (yellow), height $7$: $7^3 - 5^3 = 218$ unit cubes.\n- 2nd slice (blue), height $9$: $9^3 - 7^3 = 386$ unit cubes.\n- 1st slice (yellow), height $11$: $11^3 - 9^3 = 602$ unit cubes.\n\nFinally, the number of blue cubes is $1 + 98 + 386 = 485$, and the number of yellow cubes is $602 + 218 + 26 = 846$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18515,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $S(n)$ denote the sum of its digits and $U(n)$ its unit digit. Determine all positive integers $n$ such that\n\n$$\nn = S(n) + U(n)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Write $n$ as $a_0 + 10a_1 + 100a_2 + \\dots$, where $a_0, a_1, a_2, \\dots$ are the digits of $n$. Then the equation becomes\n\n$$\na_0 + 10a_1 + 100a_2 + \\dots = a_0 + a_1 + a_2 + \\dots + a_0^2\n$$\n\nor\n\n$$\n9a_1 + 99a_2 + 999a_3 + \\dots = a_0^2.\n$$\n\nThe right-hand side is at most $9^2 = 81 < 99$. Therefore, $a_2, a_3, \\dots$ must be $0$ (otherwise, the left-hand side would be greater than the right-hand side). Thus,\n\n$$\n9a_1 = a_0^2.\n$$\n\nSo $a_0$ must be divisible by $3$, giving $a_0 = 3$ ($a_1 = 1$), $a_0 = 6$ ($a_1 = 4$), and $a_0 = 9$ ($a_1 = 9$). Therefore, the solutions are $13$, $46$, and $99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18516,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f(x)$ is an odd function with domain $\\mathbb{R}$. If $f(1) = 2$, $f(2) = 3$, then the value of $f(f(-1))$ is ____.",
    "options": [],
    "answer": "See solution",
    "solution": "By the given conditions, we have $f(-1) = -f(1) = -2$. Therefore,\n\n$$\nf(f(-1)) = f(-2) = -f(2) = -3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18517,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest common divisor (gcd) of $11n + 4$ and $7n + 2$ for integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The greatest common divisor of $11n+4$ and $7n+2$ also divides $7(11n+4) - 11(7n+2) = 6$, so it can be at most 6. If $n=4$, then $11n+4 = 48$ and $7n+2 = 30$, and the greatest common divisor of these two numbers is 6. Hence, the answer is 6.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18518,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there exist two natural numbers $a$ and $b$ such that\n$$\na! + 48 = 48(a+1)^b.\n$$\nAre there any such pairs $(a, b)$?",
    "options": [],
    "answer": "See solution",
    "solution": "Because $48 \\mid 48(a+1)^b$, it follows that $48 \\mid a! + 48$, or $48 \\mid a!$. Since $48 = 2^4 \\cdot 3$, this only holds for $a \\geq 6$. Consider $a = 6$ and $a = 7$ separately:\n\n- For $a = 6$: $6! + 48 = 720 + 48 = 768 = 48 \\cdot 16$, but $a+1 = 7$, so $768 = 48 \\cdot 7^b$ would require $7^b = 16$, which is not possible for integer $b$.\n- For $a = 7$: $7! + 48 = 5040 + 48 = 5088 = 48 \\cdot 106$, but $a+1 = 8$, so $5088 = 48 \\cdot 8^b$ would require $8^b = 106$, which is not possible for integer $b$.\n\nThus, $a \\geq 8$. Now consider two cases:\n\n**Case 1:** $a+1$ is not prime.\n\nThen $a!$ is divisible by all primes up to $a$, and $a+1$ shares a prime factor $p \\leq a$. Thus, both sides of the equation are divisible by $p$, but $(a+1)^b$ is divisible by $p$ only once, while $a!$ is divisible by $p$ at least once, leading to a contradiction modulo $p$.\n\n**Case 2:** $a+1$ is prime.\n\nBy Wilson's theorem, $a! \\equiv -1 \\pmod{a+1}$. Thus,\n$$\na! + 48 \\equiv -1 + 48 = 47 \\pmod{a+1}.\n$$\nSo $a+1 \\mid 47$, and since $47$ is prime, $a+1 = 47$, so $a = 46$.\n\nPlugging in $a = 46$:\n$$\n46! + 48 = 48 \\cdot 47^b.\n$$\nBut $46! + 48$ is not a power of $47$ times $48$ for any integer $b$ (as shown by modular arguments and size estimates in the original reasoning).\n\nTherefore, there are **no solutions** in natural numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18519,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z > 0$ be real numbers such that $xyz + xy + yz + zx = 4$. Prove that $x + y + z \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $xyz \\le \\left(\\frac{x + y + z}{3}\\right)^3$ and $xy + yz + zx \\le \\frac{(x + y + z)^2}{3}$, let $s = x + y + z$. Then:\n\n$$\n\\frac{s^3}{27} + \\frac{s^2}{3} \\ge 4\n$$\n\nThis leads to $(s - 3)(s + 6)^2 \\ge 0$, so $s \\ge 3$. Equality holds for $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18520,
    "subject": "Mathematics (Olympiad)",
    "question": "Mark the squares of the table as shown below.\n\nIn the sequel, the squares that lie symmetric with respect to square $x$ are called symmetric; a square that is marked with an odd number is called an odd square; a square that is marked with an even number is called an even square.\n\nLet $s$ denote the number to be found.\n\n% IMAGE: <table><tr><td>1</td><td>2</td><td>3</td></tr><tr><td>8</td><td>×</td><td>4</td></tr><tr><td>7</td><td>6</td><td>5</td></tr></table>\n\nLet $T$ be the set of ways of painting 8 border squares, such that in each way, each square is painted by one color.\n\nFor each $t \\in T$, denote $k(t)$ the number of ways of painting that are similar to it.\n\nFor each $i \\in \\{1, 2\\}$, denote $T_i = \\{t \\in T \\mid k(t) = i\\}$.\n\nFind $s$, the number of pairwise non-similar ways of painting the 8 border squares of the table, considering the symmetries described.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the symmetries of the table:\n\n* When rotating the table around its center, the square marked by $x$ does not move. Denote by $s_0$ the number of pairwise non-similar ways of painting 8 squares on the border of the table, then $s = n \\cdot s_0$.\n\nThere are 4 rotations that map the table to itself: by angles $\\alpha$, $2\\alpha$, $3\\alpha$, and $4\\alpha$, where $\\alpha = \\frac{\\pi}{2}$.\n\nUnder these rotations, the squares move as follows:\n\n- Rotation by $\\alpha$: $1 \\to 3 \\to 5 \\to 7 \\to 1$ and $2 \\to 4 \\to 6 \\to 8 \\to 2$.\n- Rotation by $2\\alpha$: $1 \\to 5 \\to 1$, $2 \\to 6 \\to 2$, $3 \\to 7 \\to 3$, $4 \\to 8 \\to 4$.\n- Rotation by $3\\alpha$: $1 \\to 7 \\to 5 \\to 3 \\to 1$ and $2 \\to 8 \\to 6 \\to 4 \\to 2$.\n- Rotation by $4\\alpha$: each square returns to its original place.\n\nFor each $t \\in T$:\n\n$$\n\\begin{aligned}\n&k(t) = 1 \\text{ if and only if the odd squares are painted by the same color.} \\\\\n&k(t) = 2 \\text{ if and only if symmetric squares are painted by the same color, and 4 odd squares or 4 even squares are not painted by the same color.} \\\\\n&k(t) = 4 \\text{ otherwise.}\n\\end{aligned}\n$$\n\nFrom the above:\n\n$$\n|T_1| = n^2, \\quad |T_2| = n^4 - n^2\n$$\n\nThus:\n\n$$\ns_0 = \\frac{|T| - |T_1| - |T_2|}{4} + \\frac{|T_2|}{2} + |T_1| = \\frac{|T| + 3|T_1| + |T_2|}{4} = \\frac{n^8 + n^4 + 2n^2}{4}\n$$\n\nTherefore:\n\n$$\ns = n \\cdot s_0 = \\frac{n^3(n^6 + n^2 + 2)}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18521,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an integer with $k > 1$. Define a sequence $\\{a_n\\}$ as follows:\n\n- $a_0 = 0$\n- $a_1 = 1$\n- $a_{n+1} = k a_n + a_{n-1}$ for $n = 1, 2, \\dots$\n\nDetermine, with proof, all possible $k$ for which there exist non-negative integers $\\ell, m$ ($\\ell \\ne m$) and positive integers $p, q$ such that\n$$\na_\\ell + k a_p = a_m + k a_q.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $k = 2$.\n\nIf $k = 2$, then $a_0 = 0$, $a_1 = 1$, $a_2 = 2$, so $a_0 + 2a_2 = a_2 + 2a_1 = 4$. Hence, $(\\ell, m) = (0, 2)$ and $(p, q) = (2, 1)$.\n\nFor $k \\ge 3$, it follows from the recurrence relation that the sequence $\\{a_n\\}$ is strictly increasing, and $k \\mid a_{n+1} - a_{n-1}$ for all $n \\ge 1$. In particular, for $n \\ge 0$,\n$$\na_{2n} \\equiv a_0 \\equiv 0 \\pmod{k}, \\quad a_{2n+1} \\equiv a_1 \\equiv 1 \\pmod{k}.\n$$\n\nSuppose there exist $\\ell, m \\in \\mathbb{N}$, and $p, q \\in \\mathbb{Z}_+$ such that $\\ell \\ne m$ and $a_\\ell + k a_p = a_m + k a_q$. We may assume $\\ell < m$, and consider the following cases:\n\n(a) $p < \\ell < m$: Then $a_\\ell + k a_p \\le a_\\ell + k a_{\\ell-1} < k a_\\ell + a_{\\ell-1} = a_{\\ell+1} \\le a_m < a_m + k a_q$, which is a contradiction.\n\n(b) $\\ell = p < m$: If $\\ell = p = m-1$, then $a_m + k a_q = a_\\ell + k a_p = (k+1)a_{m-1}$, and modulo $k$ we have $a_m \\equiv a_{m-1} \\pmod{k}$, which contradicts the earlier congruence.\n\nIf $\\ell = p < m-1$, then $a_\\ell + k a_p < a_{m-2} + k a_{m-2} = a_m < a_m + k a_q$, again a contradiction.\n\n(c) $\\ell < p < m$: Then $a_\\ell + k a_p \\le k a_p + a_{p-1} = a_{p+1} \\le a_m < a_m + k a_q$, a contradiction.\n\n(d) $\\ell < m \\le p$: Then $a_p > \\frac{a_\\ell + k a_p - a_m}{k} = a_q$. From\n$$\nk a_q + a_m = k a_p + a_\\ell \\ge k a_p\n$$\nit follows that\n$$\na_q \\ge a_p - \\frac{a_m}{k} \\ge a_p - \\frac{a_p}{k} = \\frac{k-1}{k} a_p.\n$$\nAlso,\n$$\na_p = k a_{p-1} + a_{p-2} \\ge k a_{p-1}\n$$\nso\n$$\na_p > a_q \\ge \\frac{k-1}{k} a_p \\ge (k-1) a_{p-1} \\ge a_{p-1}.\n$$\nBy the increasing property of $\\{a_n\\}$, $a_q = a_{p-1}$, so all inequalities become equalities. It follows that $m = p$, $p = 2$, and $a_q = a_{p-1} = a_1 = 1$. From $a_\\ell + k a_p = a_m + k a_q$, we get $k^2 = k + k = 2k$, so $k = 2$, which is impossible for $k \\ge 3$.\n\nTherefore, $k = 2$ is the only solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18522,
    "subject": "Mathematics (Olympiad)",
    "question": "設實數 $x_i \\ge 0$ ($i = 1, 2, \\dots, m$)，$n \\ge 2$，且 $\\sum_{i=1}^{m} x_i = S$。試證：\n\n$$\n\\sum_{i=1}^{m} \\sqrt[n]{\\frac{x_i}{S - x_i}} \\ge 2,\n$$\n\n且當 $x_i$ 中有兩個相等且不為 0，其餘皆為 0 時，等號成立。",
    "options": [],
    "answer": "See solution",
    "solution": "先以數學歸納法證明引理 1。\n\n**引理 1**：當 $x, y \\ge 0$，$n \\ge 2$，其中 $n$ 為正整數時，\n\n$$\n(x^n + y^n)^2 \\le (x^2 + y^2)^n. \\quad (1)\n$$\n\n證明：當 $n = 2$ 時，顯然成立。\n\n假設 $n = k$ ($k \\ge 2$) 時，結論成立，即\n\n$$\n(x^k + y^k)^2 \\le (x^2 + y^2)^k.\n$$\n\n當 $n = k + 1$ 時，\n\n$$\n\\begin{aligned}\n(1) \\text{ 式右邊} &= (x^2 + y^2)^{k+1} \\\\\n&= (x^2 + y^2)^k (x^2 + y^2) \\\\\n&\\ge (x^k + y^k)^2 (x^2 + y^2) \\\\\n&= (x^{2k} + y^{2k} + 2x^k y^k)(x^2 + y^2) \\\\\n&= x^{2k+2} + y^{2k+2} + x^2 y^{2k} + x^{2k} y^2 + 2x^{k+2} y^k + 2x^k y^{k+2} \\\\\n&\\ge x^{2k+2} + y^{2k+2} + 2x^{k+2} y^k + 2x^k y^{k+2} \\\\\n&\\ge x^{2k+2} y^{2k+2} + 2x^{k+1} y^{k+1} \\\\\n&= (x^{k+1} + y^{k+1})^2.\n\\end{aligned}\n$$\n\n故當 $n = k + 1$ 時，結論成立。\n\n回到原題。\n\n在 (1) 式中令 $x = \\sqrt[n]{b}$，$y = \\sqrt[n]{c}$，化簡得\n\n$$\n(\\sqrt[n]{b+c})^2 \\le (\\sqrt[n]{b})^2 + (\\sqrt[n]{c})^2. \\qquad (2)\n$$\n\n再由數學歸納法將 (2) 式推廣為\n\n$$\n(\\sqrt[n]{x_1 + x_2 + \\cdots + x_m})^2 \\le (\\sqrt[n]{x_1})^2 + (\\sqrt[n]{x_2})^2 + \\cdots + (\\sqrt[n]{x_m})^2,\n$$\n\n其中 $n \\ge 2$，$m$ 為正整數。\n\n$$\n\\begin{aligned}\n& \\text{則 } \\sqrt[n]{\\frac{x_1}{S-x_1}} = \\sqrt[n]{\\frac{x_1}{\\sqrt[n]{S-x_1}}} = \\frac{2(\\sqrt[n]{x_1})^2}{2\\sqrt[n]{x_1}\\sqrt[n]{S-x_1}} \\\\\n& \\ge \\frac{2(\\sqrt[n]{x_1})^2}{(\\sqrt[n]{x_1})^2 + (\\sqrt[n]{x_2} + x_3 + \\cdots + x_m)^2} \\\\\n& \\ge \\frac{2(\\sqrt[n]{x_1})^2}{(\\sqrt[n]{x_1})^2 + (\\sqrt[n]{x_2})^2 + \\cdots + (\\sqrt[n]{x_m})^2}.\n\\end{aligned}\n$$\n\n同理，\n\n$$\n\\sqrt[n]{\\frac{x_m}{S-x_m}} \\ge \\frac{2(\\sqrt[n]{x_m})^2}{(\\sqrt[n]{x_1})^2 + (\\sqrt[n]{x_2})^2 + \\cdots + (\\sqrt[n]{x_m})^2}.\n$$\n\n以上各式相加即得原不等式。\n\n由證明過程，知若且唯若當 $x_i$ 中有兩個相等且不為 0，其餘皆為 0 時，上式等號成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18523,
    "subject": "Mathematics (Olympiad)",
    "question": "The chess King is placed on a cell of an $8 \\times 8$ board. He then makes 64 moves, visiting all cells and returning 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. A move is *pleasant* if, after the move, this distance becomes less than it was before the move. Find the greatest possible number of pleasant moves.",
    "options": [],
    "answer": "See solution",
    "solution": "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 below; 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![](images/2025-01_p5_data_bb9205451a.png)\n\n![](images/2025-01_p5_data_76589f444a.png)\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 above.\n\n*Note.* Essentially, in the final part of the proof, we've shown that among moves leading to cells marked in green above, there are at least three unpleasant ones. This can be proven in various ways, for example, through a brief case analysis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18524,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer and $a_1, a_2, \\dots, a_n \\in (0, 1)$. Find the maximum value of the sum\n$$\n\\sum_{i=1}^{n} \\sqrt[6]{a_i(1-a_{i+1})}\n$$\nwhere $a_{n+1} = a_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the AM-GM Inequality, we deduce that\n$$\n\\begin{aligned}\n  \\sqrt[6]{a_i(1-a_{i+1})} &= 2^{\\frac{4}{6}} \\sqrt[6]{a_i(1-a_{i+1}) \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}} \\\\\n  &\\le 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} (a_i + 1 - a_{i+1} + 2) \\\\\n  &= 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} (a_i - a_{i+1} + 3).\n\\end{aligned}\n$$\nSo\n$$\n\\begin{aligned}\n  \\sum_{i=1}^{n} \\sqrt[6]{a_i(1-a_{i+1})} &\\le 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\sum_{i=1}^{n} (a_i - a_{i+1} + 3) \\\\\n  &= 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\cdot 3n \\\\\n  &= \\frac{n}{\\sqrt[3]{2}}.\n\\end{aligned}\n$$\nEquality holds if and only if $a_1 = a_2 = \\dots = a_n = \\frac{1}{2}$. So\n$$\n\\boxed{\\frac{n}{\\sqrt[3]{2}}}\n$$\nis the maximum value.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18525,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a set $A = \\{1, 2, \\dots, 4044\\}$. We color 2022 of these numbers white and the rest black. For each $i \\in A$, define the weight of $i$ as the sum of the number of white numbers less than $i$ and the number of black numbers greater than $i$. For every positive integer $m$, find all positive integers $k$ such that there exists a coloring of the numbers so that exactly $k$ numbers have weight $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a natural number $i$ *good* if its weight is $m$. We will prove the following claim.\n\n**Claim.** Consider a positive integer $i \\le 4044$.\n\n(a) If there are more black numbers than white from $1$ to $i-1$, then there exists a black number $j$ such that the numbers of black and white numbers from $j+1$ to $i-1$ are equal.\n\n(b) If there are more white numbers than black from $i+1$ to $4044$, then there exists a white number $j$ such that the numbers of black and white numbers from $i+1$ to $j-1$ are equal.\n\n*Proof*. Clearly (a) and (b) are similar, so we only need to prove (b). Denote $f(k)$ as the difference between the numbers of black and white numbers from $i+1$ to $i+k-1$. It is clear that $f(0) = 0$ and\n\n$$\n|f(x) - f(x + 1)| = 1.\n$$\n\nNow, we need to show that there exists $k$ such that $i+k$ is white and $f(k) = 0$. If $i+1$ is white and $f(1) = 0$, then we can assume $i+1$ is black, so $f(2) = 1 > 0$. Because the number of white numbers from $i+1$ to $4044$ is more than the black numbers, $f(4044 - i) \\le 0$. Hence,\n\n$$\nf(2) = 1 > 0 \\ge f(4044 - i).\n$$\n\nWe will show that there exists $j$ such that $f(j) = 0$. If there exists the smallest natural number $t$ such that $f(t) < 0$, note that\n\n$$\n|f(t - 1) - f(t)| = 1\n$$\n\nso $f(t - 1) = f(t) + 1$ and $f(t - 1) \\ge 0$, hence $f(t - 1) = 0$. Otherwise, if $f(t) \\ge 0$ for all $t$, then $f(4044 - i) = 0$, which means there always exists a number $j$ with that property.\n\nAssume that for all $j$ with $f(j) = 0$, the number $j+i$ is always black, which means $f(j + 1) = 1$. Similarly, if there exists the smallest number $t$ such that $f(t) < 0$, then $f(t - 1) = 0$, hence $f(t) \\ge 0$ for all $t$. Thus, $f(4044 - i) = 0$, which means the numbers of black and white numbers from $i+1$ to $4043$ are equal, so $4044$ is colored white. Otherwise, if there exists $j$ such that $f(j) = 0$ and $j+i$ is white, then $j$ satisfies the above condition. $\\square$\n\nBack to the problem, assume that $i < i'$ are two consecutive good numbers and have the same color (white). We observe that there are fewer white numbers before $i'$ than $i$ and fewer black numbers after $i'$ than $i$. Hence, the weight of $i'$ is smaller than the weight of $i$, which is a contradiction. Thus, they must be colored with different colors.\n\nCall a natural number $i$ *good* if it has weight $m$. We shall prove that $k$ is an even number. Indeed, let $i$ be the smallest good number and denote $a_j, b_j$ as the number of white and black numbers before and after $j$. If $j < i$, then\n\n$$\na_j + b_j \\neq a_i + b_i.\n$$\n\nIf $i$ is black, then the number of white numbers from $1$ to $i - 1$ is smaller than the number of black numbers; otherwise, by lemma 1, we can find a black number $j$ such that in the interval $[j+1, i-1]$, the numbers of black and white numbers are equal, which means\n\n$$\na_i - a_j = b_i - b_j\n$$\n\nBut $j$ is also a good number, which leads to a contradiction. Note that the number of black numbers before $i$ is $2021 - b_1$, hence\n\n$$\nm = a_i + b_i < 2022 - b_i + b_i = 2022.\n$$\n\nNext, we will prove that if $s$ is good and black, then there exists $s' > s$ such that $s'$ is good and white. Because $s$ is black, the number of white numbers after $s$ is $2022 - a_s$, then\n\n$$\nb_s = m - a_s < 2022 - a_s\n$$\n\nApplying lemma 1, we get a number $s' > s$ such that $s'$ is white and the numbers of black and white numbers in $[s+1, s'-1]$ are equal. Hence,\n\n$$\na_{s'} - a_s = b_{s'} - b_s\n$$\n\nwhich means $s'$ is also a good number. Thus, if $A = \\{x_1 < x_2 < \\dots < x_k\\}$ is the set of good numbers, then two consecutive good numbers must have different colors, so $k$ is even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18526,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two circles $O_1$ and $O_2$ with different radii that intersect at two points $A$ and $B$. The common external tangent $CD$ (closer to point $B$) touches $O_1$ at $C$ and $O_2$ at $D$. Draw diameters $BP$ of $O_1$ and $BQ$ of $O_2$. The line through $B$ perpendicular to $CD$ meets $PQ$ at $K$.\n\n**a)** Prove that $B$ is the orthocenter of triangle $KCD$.\n\n**b)** Draw the angle bisectors $BX$ and $BY$ of triangles $KBP$ and $KBQ$ respectively, with $X, Y \\in PQ$. Prove that $KX = KY$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Redefine the point $K$ as the orthocenter of triangle $BCD$, then $B$ is also the orthocenter of triangle $KCD$. We will show that $K \\in PQ$. Construct the parallelogram $BCDT$.\n\n![](images/Saudi_Arabia_booklet_2023_p20_data_6ea6df64da.png)\n\nSince $K$ is the orthocenter of triangle $BCD$, $BC \\perp KD$, and $BC \\parallel TD$ so $TD \\perp KD$. Similarly, $TC \\perp KC$ so $KCTD$ is inscribed in a circle of diameter $KT$. Since $CD$ is a common tangent to $O_1$ and $O_2$, $\\angle BCD = \\angle CAB$, $\\angle BDC = \\angle DAB$. Therefore,\n\n$$\n180^\\circ - \\angle CBD = \\angle BCD + \\angle BDC = \\angle CAB + \\angle DAB = \\angle CAD.\n$$\n\nSince $BCTD$ is a parallelogram, $\\angle CBD = \\angle CTD$, hence $180^\\circ - \\angle CTD = \\angle CAD$ entails $ACTD$ is cyclic. Therefore, the points $A, C, T, D, K$ belong to the circle of diameter $KT$. Since $BP, BQ$ are diameters of $O_1$ and $O_2$, $\\angle BAP = \\angle BAQ = 90^\\circ$, so $A, P, Q$ are collinear. It remains to prove that $KA \\perp AB$. Since $KACD$ is cyclic, we have $\\angle KAC = \\angle KDC = 90^\\circ - \\angle BCD = 90^\\circ - \\angle BAC$, which implies\n\n$$\n\\angle KAB = \\angle KAC + \\angle BAC = 90^\\circ.\n$$\n\nHence $K, P, Q$ are collinear and thus the original problem is solved.\n\n*Remark:* This problem can also be solved using properties of the Humpy point.\n\nb) Draw the altitude $BH$ of triangle $BCD$ and let $M$ be the midpoint of $CD$. We have\n\n$$\n\\angle KBP = 180^\\circ - (\\angle CBP + \\angle CBH) = 180^\\circ - (90^\\circ - \\angle BPC + 90^\\circ - \\angle BCH) = 2\\angle BCH.\n$$\n\nSince $BX$ is the bisector of $\\angle KBP$, then $\\angle KBX = \\angle BCM$. We can see that $AKMH$ is cyclic since $\\angle KAM = \\angle KHM = 90^\\circ$, so $\\angle BKX = \\angle BMC$. This implies that\n\n$$\n\\triangle BKX \\sim \\triangle CMB \\implies \\frac{BK}{CM} = \\frac{KX}{MB}.\n$$\n\nSimilarly, $\\triangle BKY \\sim \\triangle DMB \\implies \\frac{BK}{DM} = \\frac{KY}{MB}$. And $CM = DM$, so we get\n\n$$\n\\frac{KX}{MB} = \\frac{KY}{MB} \\implies KX = KY.\n$$\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18527,
    "subject": "Mathematics (Olympiad)",
    "question": "There are some lamps in a row. Initially, some of the lamps are on, and the rest are off. We may change the states of the lamps by the following operations:\n\n1. Change the state of the rightmost lamp.\n2. Change the states of two consecutive lamps that are either both on or both off.\n\nIs it necessarily possible to use the operations (1) and (2) so that all the lights become off?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. To simplify notation, consider the sequence of lamps as a word on letters $a$ and $b$ (for example, $abb$ means on, off, off). Let $n$ be the number of lamps. We show by induction on $n$ that any word of length $n$ can be transformed into any other word of length $n$ by repeatedly switching the last letter or two equal consecutive letters. After that, the problem is solved: any initial word can be transformed into $bb\\ldots b$ (all the lights are off).\n\nFor $n = 1$ or $n = 2$ this is trivial, so assume $n \\geq 3$. Let $w$ be a word of length $n + 1$. By symmetry, we may assume that $w = aw'$ (so $w$ starts with $a$). By the induction hypothesis, $w'$ can be changed to any other word of length $n$. Therefore, we can reach from $w$ any word starting with $a$. Hence we only need to show that words starting with $b$ can be obtained from $w$.\n\nAgain by the induction assumption, $w'$ can be replaced with $aa\\ldots a$ ($n$ times). Then we have the word $aaa\\ldots a$ ($n + 1$ times), which can be replaced with $bba\\ldots a$ ($a$ occurs $n - 1$ times). Since the word $ba\\ldots a$ has length $n$, it can be replaced with any other word of the same length, so any word starting with $b$ can be obtained from $w$.\n\n**Remark.** For small values of $n$ it is easy to see that all the lamps can always be turned off, and that in fact any position can be reached from any other. Therefore, it is natural to guess that the answer is affirmative and to proceed by induction on $n$. It also seems plausible that there is an algorithmic solution to this problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18528,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$, $b$, $c$ are positive real numbers such that\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 3,\n$$\nprove that\n$$\n\\frac{a+b+c-1}{\\sqrt{2}} \\geq \\frac{\\sqrt{a+\\frac{b}{c}} + \\sqrt{b+\\frac{c}{a}} + \\sqrt{c+\\frac{a}{b}}}{3}.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is equivalent to\n$$\n12(a + b + c - 1) \\geq \\sum_{\\text{cyc}} 4\\sqrt{2\\left(a + \\frac{b}{c}\\right)}.\n$$\nFrom the AM-GM inequality, we have\n$$\n\\sum_{\\text{cyc}} 2\\sqrt{2\\left(a+\\frac{b}{c}\\right)} \\leq \\sum_{\\text{cyc}} \\left(2 + a + \\frac{b}{c}\\right) = 6 + a + b + c + \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}.\n$$\nAgain, from AM-GM,\n$$\n\\frac{2a}{b} + \\frac{2b}{c} + \\frac{2c}{a} \\geq 3\\sqrt[3]{\\frac{2a}{b} \\cdot \\frac{2b}{c} \\cdot \\frac{2c}{a}} = 3\\sqrt[3]{8} = 6.\n$$\nHence,\n$$\n\\sum_{\\text{cyc}} 2\\sqrt{2\\left(a+\\frac{b}{c}\\right)} \\leq a + b + c + \\frac{3a}{b} + \\frac{3b}{c} + \\frac{3c}{a}. \\quad (1)\n$$\nAgain, from AM-GM,\n$$\n\\begin{aligned}\n\\sum_{\\text{cyc}} 2\\sqrt{2\\left(a+\\frac{b}{c}\\right)} &= \\sum_{\\text{cyc}} 2\\sqrt{\\frac{2a}{c}\\left(c+\\frac{b}{a}\\right)} \\\\ &\\leq \\sum_{\\text{cyc}} \\left(\\frac{2a}{c} + c + \\frac{b}{a}\\right) \\\\ &= a + b + c + \\frac{3a}{c} + \\frac{3b}{a} + \\frac{3c}{b}. \\quad (2)\n\\end{aligned}\n$$\nAdding (1) and (2),\n$$\n\\sum_{\\text{cyc}} 4\\sqrt{2\\left(a+\\frac{b}{c}\\right)} \\leq 2(a + b + c) + 3\\left(\\frac{a + b}{c} + \\frac{b + c}{a} + \\frac{c + a}{b}\\right).\n$$\nNow, using the assumption,\n$$\n\\frac{a + b}{c} + \\frac{b + c}{a} + \\frac{c + a}{b} = (a + b + c)\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) - 3 = 3(a + b + c - 1).\n$$\nHence,\n$$\n\\sum_{\\text{cyc}} 4\\sqrt{2\\left(a+\\frac{b}{c}\\right)} \\leq 11(a + b + c) - 9.\n$$\nSo, it is enough to prove\n$$\n11(a + b + c) - 9 \\leq 12(a + b + c - 1) \\iff a + b + c \\geq 3.\n$$\nThe last inequality is true since, using AM-HM and the condition,\n$$\n\\frac{a + b + c}{3} \\geq \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}} = 1 \\implies a + b + c \\geq 3.\n$$\nEquality in AM-HM is achieved only when $a = b = c$, and since $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 3$, we have $a = b = c = 1$. Clearly, for $a = b = c = 1$, equality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18529,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABC$ 為三角形。點 $K, L, M$ 分別落在線段 $BC, CA, AB$ 上，且 $AK, BL, CM$ 三線相交於一點。\n\n證明：可以在三個三角形 $ALM, BMK, CKL$ 之中選取兩個，使得它們的內切圓半徑之和，大於或等於三角形 $ABC$ 的內切圓半徑。",
    "options": [],
    "answer": "See solution",
    "solution": "記\n\n$$\na = \\frac{BK}{KC}, \\quad b = \\frac{CL}{LA}, \\quad c = \\frac{AM}{MB}.\n$$\n\n由 Ceva 定理知 $abc = 1$。故不失一般性，可假設 $a \\ge 1$。於是 $b, c$ 之中至少有一數不大於 1。因此 $(a, b), (b, c)$ 兩組數對中，至少有一組的第一個數字不小於 1，而第二個數字不大於 1。不失一般性，再設 $1 \\le a$ 且 $b \\le 1$。\n\n由此可得 $bc \\le 1$ 且 $1 \\le ca$，即\n\n$$\n\\frac{AM}{MB} \\le \\frac{LA}{CL} \\quad \\text{及} \\quad \\frac{MB}{AM} \\le \\frac{BK}{KC}.\n$$\n\n上面的第一條不等式告訴我們：過 $M$ 並與 $BC$ 平行的直線會交 *線段 AL* 於點 $X$。因此，三角形 $ALM$ 的內切圓半徑不小於三角形 $AMX$ 的內切圓半徑 $r_1$。\n\n同理，第二條不等式指出：過 $M$ 並與 $AC$ 平行的直線會交 *線段 BK* 於點 $Y$，因此三角形 $BMK$ 的內切圓半徑不小於三角形 $BMY$ 的內切圓半徑 $r_2$。要完成證明，只要證出 $r_1 + r_2 \\ge r$，其中 $r$ 就是三角形 $ABC$ 的內切圓半徑即可。事實上，我們將證明：$r_1 + r_2 = r$。\n\n![](images/15-2J_p29_data_2ec3597b00.png)\n\n由於 $MX \\parallel BC$，所以以 $A$ 點為中心、將 $M$ 點送到 $B$ 點的位似變換會將三角形 $AMX$ 的內切圓送到三角形 $ABC$ 的內切圓。故有\n\n$$\n\\frac{r_1}{r} = \\frac{AM}{AB}.\n$$\n\n同理可得\n\n$$\n\\frac{r_2}{r} = \\frac{MB}{AB}.\n$$\n\n將這兩式相加，即得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18530,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$, $\\beta$, $\\gamma$ be positive integers such that the number\n$$\nA = \\frac{\\alpha\\sqrt{2} + \\beta\\sqrt{3}}{\\beta\\sqrt{2} + \\gamma\\sqrt{3}}\n$$\nis rational. Prove that the number\n$$\nB = \\frac{\\alpha^2 + \\beta^2 + \\gamma^2}{\\alpha + \\beta + \\gamma}\n$$\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that:\n$$\n\\alpha_1\\sqrt{2} + \\alpha_2\\sqrt{3} = \\alpha_3\\sqrt{2} + \\alpha_4\\sqrt{3}, \\quad \\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4 \\in \\mathbb{Q}^* \\iff \\alpha_1 = \\alpha_3 \\text{ and } \\alpha_2 = \\alpha_4.\n$$\nIndeed, rewriting gives\n$$\n(\\alpha_1 - \\alpha_3)\\sqrt{2} = (\\alpha_4 - \\alpha_2)\\sqrt{3},\n$$\nand if $\\alpha_1 - \\alpha_3 \\neq 0$, then $\\frac{\\alpha_4 - \\alpha_2}{\\alpha_1 - \\alpha_3} = \\frac{\\sqrt{2}}{\\sqrt{3}}$, which is impossible. Thus $\\alpha_1 = \\alpha_3$ and $\\alpha_2 = \\alpha_4$.\n\nNow, suppose $\\frac{\\alpha\\sqrt{2} + \\beta\\sqrt{3}}{\\beta\\sqrt{2} + \\gamma\\sqrt{3}} = \\kappa \\in \\mathbb{Q}$. Then $\\alpha = \\kappa\\beta$ and $\\beta = \\kappa\\gamma$, so\n$$\n\\frac{\\alpha}{\\beta} = \\frac{\\beta}{\\gamma} \\implies \\beta^2 = \\alpha\\gamma.\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\alpha^2 + \\beta^2 + \\gamma^2 &= \\alpha^2 + \\alpha\\gamma + \\gamma^2 \\\\\n&= \\alpha^2 + 2\\alpha\\gamma + \\gamma^2 - \\alpha\\gamma \\\\\n&= (\\alpha + \\gamma)^2 - \\beta^2 \\\\\n&= (\\alpha + \\beta + \\gamma)(\\alpha - \\beta + \\gamma),\n\\end{aligned}\n$$\nso\n$$\nB = \\frac{\\alpha^2 + \\beta^2 + \\gamma^2}{\\alpha + \\beta + \\gamma} = \\alpha - \\beta + \\gamma,\n$$\nwhich is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18531,
    "subject": "Mathematics (Olympiad)",
    "question": "Demostrar que para cualquier par de enteros positivos $k$ y $n$, existen $k$ enteros positivos $m_1, m_2, \\dots, m_k$ (no necesariamente distintos) tales que\n\n$$\n1 + \\frac{2^k - 1}{n} = \\left(1 + \\frac{1}{m_1}\\right) \\left(1 + \\frac{1}{m_2}\\right) \\dots \\left(1 + \\frac{1}{m_k}\\right)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $(u_1, u_2, \\dots, u_k)$ una permutación de $(0, 1, 2, \\dots, k-1)$. Consideremos el producto\n\n$$\n\\begin{aligned}\n& \\frac{n+2^{u_1}}{n} \\cdot \\frac{n+2^{u_1}+2^{u_2}}{n+2^{u_1}} \\cdots \\frac{n+2^{u_1}+2^{u_2}+\\cdots+2^{u_k}}{n+2^{u_1}+2^{u_2}+\\cdots+2^{u_{k-1}}} = \\\\ \n& = \\frac{n+2^{u_1}+2^{u_2}+\\cdots+2^{u_{k-1}}}{n} = 1 + \\frac{2^{u_1}+2^{u_2}+\\cdots+2^{u_k}}{n} = 1 + \\frac{2^k - 1}{n},\n\\end{aligned}\n$$\n\ndonde hemos usado que, como es bien conocido y fácilmente demostrable usando la fórmula para la suma de los términos de una progresión geométrica, se cumple que $2^0 + 2^1 + \\dots + 2^{k-1} = 2^k - 1$.\n\nNótese además que se obtiene un producto de factores de la forma enunciada, si definimos\n\n$$\nm_1 = \\frac{n}{2^{u_1}}, \\quad m_2 = \\frac{n+2^{u_1}}{2^{u_2}}, \\quad \\dots \\quad m_k = \\frac{n+2^{u_1}+2^{u_2}+\\dots+2^{u_{k-1}}}{2^{u_k}}.\n$$\n\nSi podemos hallar una permutación $(u_1, u_2, \\dots, u_k)$ de $(0, 1, \\dots, k-1)$ tal que todas estas cantidades sean enteras, el problema estará resuelto. Veamos que podemos construir una tal permutación.\n\nSea $2^{v_1}$ la máxima potencia de 2 que divide a $n$. Si $v_1 \\ge k-1$, tomamos $u_1 = k-1$, y tomando $u_2 = k-2$, $u_3 = k-3, \\dots$, $u_k = 0$, comprobamos fácilmente que cada $n+2^{u_1}+\\dots+2^{u_i}$ es divisible por $2^{k-i}$, luego también por $2^{u_{i+1}} = 2^{k-i-1}$ para $i=1, 2, \\dots, k-1$. En caso contrario, sea $u_1 = v_1$, con lo que al ser $n, 2^{u_1}$ ambos divisibles por $2^{u_1}$, pero ninguno por $2^{u_1+1}$, su suma es divisible al menos por $2^{u_1+1}$. Procedemos de ahora en adelante de la misma forma, es decir, dados $u_1, u_2, \\dots, u_i$, y si ninguno de ellos ha sido $k-1$ todavía, hallamos la máxima potencia de 2 que divide a $n+2^{u_1}+\\dots+2^{u_i}$, que será $2^{v_i}$, con $v_i \\ge i-1$, trivialmente en el caso $i=1$ y por inducción para todo otro $i$ como vemos a continuación. Si $v_i \\ge k-1$, tomamos $u_i = k-1$, y tomamos $u_{i+1}, u_{i+2}, \\dots, u_k$ iguales a los elementos de $(0, 1, \\dots, k-1)$ que no hayan sido utilizados todavía, en orden descendente, siendo entonces todos los $m_i$ enteros como en el caso en que $u_1 = k-1$. En caso de que $u_i \\le k-2$, tomamos $u_i = v_i$, con lo que $n+2^{u_1}+2^{u_2}+\\dots+2^{u_i}$ es suma de dos enteros que son ambos múltiplos de $2^{u_i}$, pero ninguno de $2^{u_i+1}$, con lo que su suma es múltiplo al menos de $2^{u_i+1}$, y esto justifica que en efecto $v_i \\ge i-1$ para cada $i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18532,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $\\Delta$ be the closed triangular domain with vertices at the lattice points $(0,0)$, $(n,0)$, and $(0,n)$. Determine the maximal cardinality a set $S$ of lattice points in $\\Delta$ may have, if the line through every pair of distinct points in $S$ is parallel to no side of $\\Delta$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $\\lfloor 2n/3 \\rfloor + 1$ and is achieved, for instance, for\n\n$$\nS = \\{(2k, \\lfloor n/3 \\rfloor - k) : k = 0, \\dots, \\lfloor n/3 \\rfloor\\} \\cup \\{(2k+1, 2\\lfloor n/3 \\rfloor - k) : k = 0, \\dots, \\lfloor n/3 \\rfloor - 1\\}\n$$\nif $n \\equiv 0 \\pmod{3}$ or $n \\equiv 1 \\pmod{3}$, and\n$$\nS = \\{(2k, \\lfloor n/3 \\rfloor - k) : k = 0, \\dots, \\lfloor n/3 \\rfloor\\} \\cup \\{(2k+1, 2\\lfloor n/3 \\rfloor - k + 1) : k = 0, \\dots, \\lfloor n/3 \\rfloor\\}\n$$\nif $n \\equiv 2 \\pmod{3}$.\n\nIf $(x, y)$ is a point in $\\Delta$, and $z = z(x, y)$ is the distance from $(x, y)$ to the side through $(n, 0)$ and $(0, n)$, then\n$$\nx + y + z\\sqrt{2} = n \\tag{1}\n$$\nand if, in addition, $(x, y)$ is a lattice point, then $x$, $y$, and $z\\sqrt{2}$ are all non-negative integers (not exceeding $n$).\n\nNow, let $S$ be a set of lattice points in $\\Delta$ satisfying the condition in the statement. Then (1) yields\n$$\n\\sum_{(x, y) \\in S} x + \\sum_{(x, y) \\in S} y + \\sum_{(x, y) \\in S} z\\sqrt{2} = n|S| \\tag{2}\n$$\nAs $(x, y)$ runs through $S$, each of the three coordinates $x$, $y$, and $z\\sqrt{2}$ runs through $|S|$ non-negative distinct integers, so each of the three sums in (2) is greater than or equal to $0 + 1 + \\cdots + (|S| - 1) = |S|(|S| - 1)/2$. Consequently,\n$$\n3|S|(|S| - 1)/2 \\le n|S|\n$$\nso $|S| \\le 2n/3 + 1$ and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18533,
    "subject": "Mathematics (Olympiad)",
    "question": "If $S$ is a $10^4$-digit binary string consisting of zeroes and ones, and $k \\leq 10^4$ 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 $a_i = b_i$ for $i = 1, \\dots, k$. 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.",
    "options": [],
    "answer": "See solution",
    "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\nLet $f(k)$ 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 $f(k) \\leq f(k-1) + f(k-2) + f(k-3)$; 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 $f(k) \\leq f(k-1) + f(k-2) + f(k-3)$. 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, $f(k) \\leq f(k-1) + f(k-2) + f(k-3)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18534,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle inscribed in the circle $C$ with center $O$ and radius $1$. For any point $M \\in C \\setminus \\{A, B, C\\}$, define\n\n$$s(M) = OH_1^2 + OH_2^2 + OH_3^2,$$\n\nwhere $H_1$, $H_2$, and $H_3$ are the orthocenters of triangles $MAB$, $MBC$, and $MCA$, respectively.\n\n**a)** Prove that if triangle $ABC$ is equilateral, then $s(M) = 6$ for any $M \\in C \\setminus \\{A, B, C\\}$.\n\n**b)** 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": "See solution",
    "solution": "Consider an orthonormal coordinate system with the origin at $O$. For any point $Z$ in the plane, denote its complex coordinate by $z$.\n\n**a)** From Sylvester's relation, $h_1 = m + a + b$, $h_2 = m + b + c$, and $h_3 = m + c + a$. Also, let $h = a + b + c$ be the complex coordinate of the orthocenter of triangle $ABC$. Therefore,\n\n$$\ns(M) = 6 + |h|^2 + 2m\\bar{h} + 2\\bar{m}h.\n$$\n\nIf triangle $ABC$ is equilateral, then $h = a + b + c = 0$, so $s(M) = 6$.\n\n**b)** Assume, by contradiction, that triangle $ABC$ is not equilateral, i.e., $h \\neq 0$. Since $s(M_1) = s(M_2)$, we have:\n\n$$|h|^2 + 2m_1\\bar{h} + 2\\bar{m}_1h = |h|^2 + 2m_2\\bar{h} + 2\\bar{m}_2h$$\n\nwhich implies\n\n$$\\bar{h}(m_1 - m_2) + h\\frac{m_2 - m_1}{m_1m_2} = 0 \\implies m_1m_2 = \\frac{h}{\\bar{h}}.$$\n\nSimilarly, $m_1m_3 = \\frac{h}{\\bar{h}}$. Since $m_1 \\neq 0$, we get $m_2 = m_3$, contradicting $M_2 \\neq M_3$. Therefore, triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18535,
    "subject": "Mathematics (Olympiad)",
    "question": "Четырём натуральным числам сопоставляются шесть их попарных наибольших общих делителей (НОД). Какое наибольшее чётное число $N$ может быть среди этих НОД, если никакие три из исходных чисел не имеют общего делителя, большего 1?",
    "options": [],
    "answer": "See solution",
    "solution": "Число $N$ может равняться 14, как показывает, например, четвёрка чисел 4, 15, 70, 84.\n\nОсталось 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\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": 18536,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $A$ y $B$ dos conjuntos tales que:\n\n1. $A \\cup B$ es el conjunto de los enteros positivos.\n2. $A \\cap B$ es el vacío.\n3. Si dos enteros positivos tienen como diferencia un primo mayor que $2013$, entonces uno de ellos está en $A$ y el otro en $B$.\n\nHallar todas las posibilidades para los conjuntos $A$ y $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "La única partición con esta propiedad es $\\mathbb{N} = \\{1, 3, 5, \\dots\\} \\cup \\{2, 4, 6, \\dots\\}$, de modo que $B$ siempre es el conjunto de los números positivos pares.\n\nSea $A \\cup B = \\mathbb{N}$ una partición admisible. La idea es aplicar la condición a un par fijo de primos gemelos $p$ y $q = p + 2$ de $[2013, 3013]$. Tales primos son, por ejemplo, $p = 2027$ y $q = 2029$. Supongamos que $n \\in A$. Entonces $n_1 = n + q \\in B$ pues $n_1 - n = q$ es un primo en $[2013, 3013]$. Entonces $n_2 = n + 2 \\in A$ pues $n_1 - n_2 = q - 2 = p \\in [2013, 3013]$ también es un tal primo. De modo que $n \\in A$ implica que $n + 2 \\in A$. Como $1 \\in A$, esto implica que todos los impares están en $A$.\n\nDe manera similar, supongamos que $n \\in B$ y $n > p$. Entonces $n_1 = n - p \\in A$ pues $n - n_1 = p$. Además, $n_2 = n + 2 \\in A$ pues $n_2 - n_1 = p + 2 = q$. En resumen, $n \\in B$ y $n > p$ implican $n + 2 \\in B$. Ahora observemos que el número par $p + 1$ pertenece a $B$ porque $1 \\in A$ y $(p + 1) - 1 = p$. Se concluye que $B$ contiene a todos los números pares, comenzando con $p + 1$.\n\nQuedan por clasificar los números pares $2, 4, 6, \\dots, p-1$. Veremos que también están en $B$. Ya sabemos que los números impares $q, q + 2, q + 4, \\dots, q + p - 3$ están en $A$. Como\n\n$$\nq - 2 = (q + 2) - 4 = (q + 4) - 6 = \\dots = (q + p - 3) - (p - 1) = p\n$$\n\nllegamos a la conclusión de que $2, 4, 6, \\dots, p-1$ están en $B$. Luego $A$ y $B$ consisten en todos los números impares y pares, respectivamente. Es claro que esta partición es admisible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18537,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A, B \\in \\mathcal{M}_2(\\mathbb{C})$ be two nonzero matrices such that $AB + BA = O_2$ and $\\det(A + B) = 0$. Prove that $\\operatorname{tr}(A) = \\operatorname{tr}(B) = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $AB + BA = O_2$, we have $(A+B)^2 = A^2 + B^2$ and $(A-B)^2 = A^2 + B^2$. Consequently, $\\det(A - B) = 0$.\n\nThe characteristic equations for $A + B$ and $A - B$ yield\n\n$$\nA^2 + B^2 - \\operatorname{tr}(A + B)(A + B) = O_2,\n$$\n\n$$\nA^2 + B^2 - \\operatorname{tr}(A - B)(A - B) = O_2.\n$$\n\nSubtracting, we get $\\operatorname{tr}(A)B = \\operatorname{tr}(B)A$.\n\nIf $\\operatorname{tr}(A) = 0$, then $\\operatorname{tr}(B) = 0$, for otherwise $A = O_2$, a contradiction.\n\nHence $A = \\lambda B$, so $AB + BA = O_2$ leads to $\\lambda B^2 = O_2$. Therefore $\\lambda = 0$, which yields $\\operatorname{tr}(A) = \\operatorname{tr}(B) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18538,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that among any 20 consecutive positive integers, there exists an integer $d$ such that for each positive integer $n$ we have the inequality\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} > \\frac{5}{2}$$\n\nwhere $\\{x\\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Among the given numbers, there is a number of the form $20k + 15 = 5(4k + 3)$. We shall prove that $d = 5(4k + 3)$ satisfies the statement's condition. Since $d \\equiv -1 \\pmod{4}$, it follows that $d$ is not a perfect square, and thus for any $n \\in \\mathbb{N}$ such that $a + 1 > n\\sqrt{d} > a$, that is, $(a + 1)^2 > n^2 d > a^2$.\n\nActually, we are going to prove that $n^2 d \\ge a^2 + 5$. Indeed:\n\nIt is known that each positive integer of the form $4s + 3$ has a prime divisor of the same form. Let $p \\mid 4k + 3$ and $p \\equiv -1 \\pmod{4}$. Because of the form of $p$, the numbers $a^2 + 1^2$ and $a^2 + 2^2$ are not divisible by $p$, and since $p \\mid n^2 d$, it follows that $n^2 d \\ne a^2 + 1, a^2 + 4$. On the other hand, $5 \\mid n^2 d$, and since $5 \\nmid a^2 + 2, a^2 + 3$, we conclude $n^2 d \\ne a^2 + 2, a^2 + 3$. Since $n^2 d > a^2$, we must have $n^2 d \\ge a^2 + 5$ as claimed. Therefore,\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} = n\\sqrt{d}(n\\sqrt{d} - a) \\ge a^2 + 5 - a\\sqrt{a + 5} > a^2 + 5 - \\frac{a^2 + a^2 + 5}{2} = \\frac{5}{2},$$\n\nwhich was to be proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18539,
    "subject": "Mathematics (Olympiad)",
    "question": "Call an $n$-tuple $(a_1, \\\\dots, a_n)$ of real numbers *stable* if the sums $a_1 + a_2 + \\\\dots + a_k$ where $0 < k \\leq n$, as well as the sums $a_n + a_{n-1} + \\\\dots + a_{n-k}$ where $0 \\leq k < n$, are either all negative or all non-negative.\n\nLet $k$ be any natural number. Consider all stable $(2k+1)$-tuples consisting of real numbers that are alternately negative and non-negative. Find the least possible number of stable subtuples with more than one element that can be contained in such a tuple.\n\n(A subtuple of $(a_1, \\\\dots, a_n)$ is any tuple $(a_i, \\\\dots, a_j)$, $1 \\leq i \\leq j \\leq n$, of elements consecutive in the original tuple.)",
    "options": [],
    "answer": "See solution",
    "solution": "Call stable tuples, whose elements are alternately negative and non-negative, *interesting*. We first show that each interesting tuple contains at least one stable subtuple of 3 elements.\n\nFor that, consider elements whose absolute value is minimal in the tuple. If there exists a negative such element, denote it $a_i$, then the sum of $a_i$ and any neighbor is non-negative. Thus $a_i$ is neither the first nor the last in the tuple because of stability. Then both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are non-negative, as well as $a_{i-1} + a_i + a_{i+1}$, hence $(a_{i-1}, a_i, a_{i+1})$ is a stable subtuple. On the other hand, if all elements with minimal absolute value are non-negative, let $a_i$ be any of them. Analogously, both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are negative, as well as $a_{i-1} + a_i + a_{i+1}$, whence $(a_{i-1}, a_i, a_{i+1})$ is a stable tuple.\n\nNext, we see that replacing an element in a stable tuple with a stable subtuple whose sum equals the element removed always leads to a stable tuple. Let the original tuple be $(a_1, \\\\dots, a_n)$ and let $a_i$ be replaced with $b_1, \\\\dots, b_m$. If $n=1$ the claim is trivial, so assume $n > 1$. Consider any subtuple starting from the beginning. If either no substituted elements are included or all are included, the sum falls to the right side of zero by assumption. If the subtuple ends with some $b_j$, the sum is $a_1 + \\\\dots + a_{i-1} + b_1 + \\\\dots + b_j$. By stability of $(b_1, \\\\dots, b_m)$, the sum $b_1 + \\\\dots + b_j$ falls to the same side from zero as $a_i$ and $b_{j+1} + \\\\dots + b_m$. Hence $b_1 + \\\\dots + b_j$ falls between $0$ and $a_i$. As $a_1 + \\\\dots + a_{i-1}$ and $a_1 + \\\\dots + a_{i-1} + a_i$ fall to the same side from zero, so does $a_1 + \\\\dots + a_{i-1} + b_1 + \\\\dots + b_j$. Similarly, we can show the property for subtuples taken from the end.\n\nLastly, we show by induction on $k$ that any interesting $(2k+1)$-tuple contains at least $k$ stable subtuples containing more than one element. If $k = 0$ the claim holds trivially. Suppose $k > 0$ and the claim holds for $k - 1$. Find a stable subtuple of 3 elements in the given $(2k + 1)$-tuple. After replacing these three elements with their sum, we get a $(2(k − 1) + 1)$-tuple that is stable. By stability of the 3-tuple replaced, the sum falls to the same side from zero as its first and third element, so the alternation of signs is maintained. By induction, the new tuple contains at least $k-1$ stable subtuples of more than one element. After substituting the removed elements back, each of these $k$ stable subtuples remains stable. Moreover, the 3-tuple itself will be the desired $k$th stable subtuple.\n\nIt remains to show that there are interesting $(2k+1)$-tuples that contain no more than $k$ stable subtuples. For example, let $a_i = (-\\frac{1}{2})^i$ for $i = 1, \\\\dots, 2k$ and $a_{2k+1} = -\\frac{1}{3}$. The sum of the first $2j$ elements is $-\\frac{1 - \\frac{1}{4j}}{3}$, which is negative. Thus the sum of $2j+1$ elements is always negative. As $a_2 + \\\\dots + a_{2k} = -\\frac{1 - \\frac{1}{4k}}{3} + \\frac{1}{2} < \\frac{1}{3}$, all sums of consecutive elements taken from the end are negative. Thus the tuple is stable.\n\nConsider any subtuple $(a_u, \\\\dots, a_v)$ where $u < v \\leq 2k$. If $u$ and $v$ have different parity, the subtuple is not stable (every interesting tuple must have an odd number of elements). If $u$ and $v$ are both odd, then $a_u + a_{u+1} < 0$ while $a_{v-1} + a_v > 0$. The case with $u$ and $v$ both even is analogous. Thus the subtuple under consideration is not stable.\n\nHence only those subtuples with more than one element that contain $a_{2k+1}$ can be stable. But there are only $k$ such subtuples of odd length. This completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18540,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $m$ is a positive integer. Let $P_m = \\{2^m, 2^{m-1}3, 2^{m-2}3^2, \\dots, 3^m\\}$. If $X$ is a subset of $P_m$, let $S_X$ be the sum of all elements of $X$, with the convention that $S_\\emptyset = 0$ where $\\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \\leq y \\leq 3^{m+1} - 2^{m+1}$. Prove that there is a subset $Y$ of $P_m$ such that $0 \\leq y - S_Y < 2^m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = 3/2$, so $1 + \\alpha > \\alpha^2$.\n\nGiven $y$, we construct $Y$ algorithmically. Start with $Y = \\emptyset$ and $S_\\emptyset = 0$. For $i = 0$ to $m$, perform:\n\nIf $S_Y + 2^i 3^{m-i} \\leq y$, then set $Y = Y \\cup \\{2^i 3^{m-i}\\}$.\n\nAfter this process, $Y$ is a subset of $P_m$ with $S_Y \\leq y$.\n\nThe elements of $P_m$ can be described as $2^m, 2^{m-1}3, 2^{m-2}3^2, \\dots, 3^m$, or equivalently $2^m, 2^m\\alpha, 2^m\\alpha^2, \\dots, 2^m\\alpha^m$. If any member is omitted from $Y$, then no two consecutive members to the left of the omitted member can both be in $Y$, due to $1 + \\alpha > \\alpha^2$ and the greedy construction.\n\nThus, either $Y = P_m$ (so $y = 3^{m+1} - 2^{m+1}$ and the claim holds), or at least one of the two leftmost elements is omitted from $Y$.\n\nIf $2^m$ is not omitted from $Y$, the process ensures $(S_Y - 2^m) + 2^{m-1}3 > y$, so $y - S_Y < 2^m$. If $2^m$ is omitted, then $y - S_Y < 2^m$ as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18541,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $1 + x^2 + y^2 = \\mathrm{lcm}(x^2, y^2)$ in the set of natural numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$. Then $d$ divides $\\mathrm{lcm}(x^2, y^2)$, $x^2$, and $y^2$, so $d = 1$. The equation becomes $1 + x^2 + y^2 = x^2 y^2$, or equivalently, $$(x^2 - 1)(y^2 - 1) = 2.$$ The possible cases are $x^2 - 1 = 1$, $y^2 - 1 = 2$ or $x^2 - 1 = 2$, $y^2 - 1 = 1$, which gives $x = 2$, $y = 3$ or $x = 3$, $y = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18542,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a positive integer $n$ with at least 200 positive divisors such that its 100th positive divisor is less than 100?\n\nFor example, $900 = 2^2 \\cdot 3^2 \\cdot 5^2$, which has $(2+1) \\cdot (2+1) \\cdot (2+1) = 27$ positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "No, there does not exist such an integer. Let $n$ be a number with at least 200 divisors. If the $i$-th divisor is $d$, then the $i$-th to last divisor is $\\frac{n}{d}$. Let $m$ be the 100th divisor. So $m \\ge 100$ and $\\frac{n}{m} > m$ if and only if $n > m^2 = 10000$.\n\nTo refine this, notice that if 97, 99, and 100 are all divisors of $n$, then $n \\ge \\mathrm{lcm}(97, 99, 100) > 11000$. The key observation is to consider the 98th, 99th, and 100th divisors. Let $k, \\ell,$ and $m$ be such divisors. If $m \\ge 105$, then $n > m^2 = 11025 > 11000$. So $98 \\le k < \\ell < m \\le 104$. But $\\gcd(x, y) \\le |x - y|$ implies that $n \\ge \\mathrm{lcm}(k, \\ell, m) \\ge \\frac{k \\cdot \\ell \\cdot m}{(\\ell-k)(m-\\ell)(m-k)} \\ge \\frac{98 \\cdot 99 \\cdot 100}{(104-98) \\cdot 3^2} > 11000$. Here, we used the fact that if $x + y \\le 2t$ then $xy \\le t^2$, applied to $x = \\ell - k$, $y = m - \\ell$, and $t = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18543,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all possible remainders when dividing by $6$ an integer $n$ that satisfies $n^3 = m^2 + m + 1$ for some integer $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Numbers $n$ and $n^3$ have the same remainder when divided by $6$. Also, $m^2 + m + 1$ is odd and gives remainder $0$ or $1$ when divided by $3$. The only way to get $0$ as the remainder is when $m = 3k + 1$, but then\n\n$$\nn^3 = (9k^2 + 6k + 1) + (3k + 1) + 1 = 9k^2 + 9k + 3 = 3(3k^2 + 3k + 1)\n$$\n\nwhich leads to a contradiction, since if $n^3$ is divisible by $3$, it must also be divisible by $3^3$, but $3k^2 + 3k + 1$ is not divisible by $3$. Hence, the remainder of $n^3$ is $1$ both when dividing by $2$ or $3$, so its remainder when dividing by $6$ is $1$.\n\nThe remainder $1$ is possible: take $n = 1$ and $m = 0$ (or $n = 7$ and $m = 18$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18544,
    "subject": "Mathematics (Olympiad)",
    "question": "求所有正整數 $n \\geq 2$ 滿足下列性質：存在 $n$ 個實數 $a_1 < \\cdots < a_n$ 及一個正實數 $r$，使得全部 $\\frac{n(n-1)}{2}$ 個差值 $a_j - a_i$（$1 \\leq i < j \\leq n$）恰好是 $r^1, r^2, \\ldots, r^{\\frac{n(n-1)}{2}}$ 的某個排列。",
    "options": [],
    "answer": "See solution",
    "solution": "$n \\in \\{2, 3, 4\\}$。\n\n首先對每個 $n \\in \\{2, 3, 4\\}$ 給出構造，然後證明 $n \\geq 5$ 不可能。\n\n- $n = 2$：例如 $(a_1, a_2) = (1, 3)$，$r = 2$。\n- $n = 3$：取 $r > 1$ 為 $x^2 - x - 1 = 0$ 的根（即黃金比例），設 $(a_1, a_2, a_3) = (0, r, r + r^2)$，則\n  $$\n  (a_2 - a_1, a_3 - a_2, a_3 - a_1) = (r, r^2, r^3)\n  $$\n- $n = 4$：取 $r \\in (1, 2)$ 為 $x^3 - x - 1 = 0$ 的根，設 $(a_1, a_2, a_3, a_4) = (0, r, r + r^2, r + r^2 + r^3)$，則\n  $$\n  (a_2 - a_1, a_3 - a_2, a_4 - a_3, a_3 - a_1, a_4 - a_2, a_4 - a_1) = (r, r^2, r^3, r^4, r^5, r^6)\n  $$\n\n對於 $n \\geq 5$，假設存在 $a_1 < \\cdots < a_n$ 及 $r > 1$ 滿足條件。\n\n*引理*：$r^{n-1} > 2$。\n\n*證明*：只有 $n-1$ 個差值 $a_j - a_i$ 使 $j = i+1$，所以存在 $e \\leq n$ 及 $j \\geq i+2$ 使 $a_j - a_i = r^e$，因此\n$$\nr^n \\geq r^e = a_j - a_i = (a_j - a_{j-1}) + (a_{j-1} - a_i) > r + r = 2r\n$$\n所以 $r^{n-1} > 2$。\n\n以 $n = 5$ 為例，$a_5 - a_1 = r^{10}$，有三種分拆方式：\n$$\n(a_5 - a_4) + (a_4 - a_1),\\quad (a_5 - a_3) + (a_3 - a_1),\\quad (a_5 - a_2) + (a_2 - a_1)\n$$\n利用引理和 $f(n) = r^n$ 的凸性，這三種分拆必須分別為 $r^{10} = r^9 + r^1 = r^8 + r^4 = r^7 + r^6$。比較任意兩個方程可得矛盾，除非 $n \\leq 4$。\n\n一般情況，設 $b = \\frac{1}{2}n(n-1)$，則 $a_n - a_1 = r^b$。考慮 $n-2$ 個方程：\n$$\na_n - a_1 = (a_n - a_i) + (a_i - a_1),\\quad i \\in \\{2, \\ldots, n-1\\}\n$$\n每個方程右邊必有一項至少為 $\\frac{1}{2}(a_n - a_1)$。由引理 $r^{b-(n-1)} = \\frac{r^b}{r^{n-1}} < \\frac{1}{2}(a_n - a_1)$，所以只有 $n-2$ 個足夠大的 $r^k$ 可用，即 $r^{b-1}, \\ldots, r^{b-(n-2)}$。\n\n小項必須分別為 $r^{b-(n-2)-\\frac{1}{2}i(i+1)}$，$1 \\leq i \\leq n-2$。\n\n若 $n-2 \\geq 2$，則有兩個方程：\n$$\nr^b = r^{b-(n-2)} + r^{b-(n-2)-1}\\qquad r^b = r^{b-(n-3)} + r^{b-(n-2)-3}\n$$\n化簡得\n$$\nr^{n-1} = r + 1\\qquad r^{n+1} = r^4 + 1\n$$\n代入得\n$$\nr^4 + 1 = r^{n+1} = r^{n-1} r^2 = r^3 + r^2\\implies (r-1)(r^3 - r - 1) = 0\n$$\n因 $r \\neq 1$，故 $r^3 = r + 1 = r^{n-1}$，即 $n = 4$，矛盾。\n\n因此，唯一可能的 $n$ 為 $2, 3, 4$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18545,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an equilateral triangle, and $P$ be a point inside the triangle. Let $D$, $E$, and $F$ be the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$ respectively. Prove that\n\n$$\na)\\quad AF + BD + CE = AE + BF + CD,\n$$\n\nand\n\n$$\nb)\\quad [APF] + [BPD] + [CPE] = [APE] + [BPF] + [CPD].\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $P$ is the centroid of triangle $ABC$, then the result holds by symmetry. We can reach an arbitrary point $P$ inside $ABC$ by starting at the centroid $P_0$ and moving along the line $PD$ and then along the line $PE$ (for some distance in either direction). We will prove that equality (a) holds by showing it is preserved by such movements of $P$.\n\n![](images/V_Britanija_2014_p21_data_8b30615088.png)\n\nConsider points $P$ and $P'$ such that $P'$ lies on the segment $PE$, as shown in the above diagram. Let the feet of the perpendiculars from $P'$ to $BC$ and $AB$ be $D'$ and $F'$ respectively. Also, let $T$ be the foot of the perpendicular from $P$ to $P'F'$.\n\nBy considering quadrilateral $P'EAF'$, $\\angle TPP' = \\angle EAF' = 60^\\circ$. So $PT = PP' \\sin 60^\\circ$. But $PTF'F$ is a rectangle, so $FF' = PP' \\sin 60^\\circ$. By symmetry, $DD' = PP' \\sin 60^\\circ = FF'$.\n\nThus,\n\n$$\n\\begin{aligned}\nAF' + BD' + CE &= AF - FF' + BD + DD' + CE \\\\\n&= AF + BD + CE\n\\end{aligned}\n$$\n\nand\n\n$$\nAE + BF' + CD' = AE + BF' + FF' + CD - DD' = AE + BF + CD.\n$$\n\nSo if $AF + BD + CE = AE + BF + CD$, then also $AF' + BD' + CE = AE + BF' + CD'$.\n\nThis equality is also preserved when moving along the line $PE$ away from $E$, by considering the same situation with $P$ and $P'$ reversed. Similarly, it is preserved when moving along the line $PD$. Since it holds when $P$ is the centroid of $ABC$, it holds for all points $P$ inside $ABC$.\n\nTo prove part (b), we'll use the same method; the result is clearly true when $P$ is the centroid of the triangle, and we will show that the equality is preserved by movements along the lines $PD$ and $PE$.\n\n![](images/V_Britanija_2014_p22_data_2c19ba22d1.png)\n\nFor ease of notation, let $PD = r_1$, $P'D' = r_1'$, $PE = r_2$, and so on as indicated in the diagram. Also, let $PP' = d$. First, note that as $\\angle PP'T = 60^\\circ$, $P'T = P'U = \\frac{d}{2}$, so $r_1 - r_1' = \\frac{d}{2} = r_3 - r_3'$. (Note that it follows from this that $r_1 + r_2 + r_3 = r_1' + r_2' + r_3'$, but we shall not actually use this fact.)\n\nBy repeatedly using the formula 'area = $\\frac{1}{2}$ base $\\times$ height',\n\n$$\n\\begin{aligned}\n& [AP'F'] + [BP'D'] + [CP'E] - [APF] - [BPD] - [CPE] \\\\\n&= (r_3' AF' + r_1' BD' + r_2 CE - r_3 AF - r_1 BD - r_2 CE) \\\\\n&= (r_3 + \\frac{1}{2}d)(AF - \\frac{d}{2}\\cos 30^\\circ) - r_3 AF \\\\\n&\\quad + (r_1 + \\frac{1}{2}d)(BD - \\frac{d}{2}\\cos 30^\\circ) - r_1 BD \\\\\n&\\quad + (r_2 - d)CE - r_2 CE \\\\\n&= \\frac{d}{2}(AF + BD - 2CE + \\sqrt{3}(r_1 - r_3)).\n\\end{aligned}\n\\quad (1)\n$$\n\nWe'll aim to unpick this last term, $r_1 - r_3$. From earlier, $r_1 - r_3 = r_1' - r_3'$. This holds for all $P'$ along the line $PE$, so we could pick $P'' = E$.\n\nThus, when $r_1''$ is the perpendicular distance from $E$ to $BC$ and $r_3''$ is the perpendicular distance from $E$ to $AB$, we have $(r_1 - r_3) = (r_1'' - r_3'')$.\n\n![](images/V_Britanija_2014_p23_data_ad9ba44994.png)\n\nNow, since $\\angle D'CE = 60^\\circ$, $r_1'' = CE \\sin 60^\\circ$ and similarly $r_3'' = EA \\sin 60^\\circ$. So\n\n$$\n(r_1 - r_3) = \\frac{\\sqrt{3}}{2} (CE - (AC - CE)) = \\sqrt{3}CE - \\frac{\\sqrt{3}}{2}AC.\n$$\n\nWe know from part (a) that $AF + BD = \\frac{3}{2}AC - CE$, so (1) is equal to\n\n$$\n\\frac{d}{2}\\left(\\frac{3}{2}AC - 3CE + 3CE - \\frac{3}{2}AC\\right) = 0.\n$$\n\nThus,\n\n$$\n[AP'F'] + [BP'D'] + [CP'E] = [APF] + [BPD] + [CPE]\n$$\n\nand so\n\n$$\n[AP'E] + [BP'F] + [CP'D] = [APE] + [BPF] + [CPD].\n$$\n\nCombining these, we see that if the desired equality holds for $P$, then it also does for $P'$. As before, it follows that it holds for any choice $P'$ on the line $PE$, and then for any $P'$ inside $ABC$ by moving along the line $AE$ and then the line $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18546,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system of equations for integer $x$, $y$:\n\n$$\n\\begin{cases}\nx^{4} + 4y^{3} + 6x^{2} + 4y = -137, \\\\\ny^{4} + 4x^{3} + 6y^{2} + 4x = 472.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let's add up both equations:\n\n![](images/Ukraine_booklet_2018_p13_data_4ddc3c0f4c.png)\n\n$$\nx^{4} + 4y^{3} + 6x^{2} + 4y + y^{4} + 4x^{3} + 6y^{2} + 4x = 335 \\Leftrightarrow (x+1)^{4} + (y+1)^{4} = 337.\n$$\n\nSince $5^4 = 625 > 337$, but $3^4 + 3^4 = 162 < 337$. Hence, one of the summands must be equal to $4^4 = 256$, the other $3^4 = 81$, so $4^4 + 3^4 = 337$. Therefore, our options are: $x+1 = \\pm 4$ and $y+1 = \\pm 3$. The only thing left is to check these pairs of numbers: $(3, 2)$, $(3, -4)$, $(-5, 2)$, and $(-5, -4)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18547,
    "subject": "Mathematics (Olympiad)",
    "question": "有 $n$ 塊磚頭，每塊的重量都至少為 $1$，且總重量為 $2n$。試證：對於每個在區間 $[0, 2n-2]$ 中的實數 $r$，你總是能夠拿出其中的若干塊磚頭（允許拿出 $0$ 塊磚頭），讓拿出來的磚頭重量總和至少為 $r$ 且至多為 $r+2$。",
    "options": [],
    "answer": "See solution",
    "solution": "**解法一.** 我們證明以下這個更一般性的結論：\n\n*Claim.* 有 $n$ 塊磚頭，每塊的重量都至少為 $1$，且總重量 $s \\le 2n$。則對於每個 $-2 \\le r \\le s$，總是能夠找出這些磚頭的一個子集（可以是空集合），使得這個子集裡的磚頭重量總和至少為 $r$ 且至多為 $r+2$。\n\n*Proof.* 歸納法。$n=1$ 時只有一塊磚頭，顯然。若 $n=k$ 時題目成立，則當 $n=k+1$ 時，考慮最重的那一塊磚頭，顯然這塊磚頭的重量 $x \\ge \\frac{s}{k+1}$，且剩餘 $k$ 塊磚頭的重量和為 $s-x \\le \\frac{k}{k+1}s \\le 2(n-1)$。由歸納假設，我們知道對於所有 $r \\in [-2, s-x]$，我們都可以從這 $k$ 塊磚頭中找到若干塊，使得其重量和在 $[r, r+2]$ 中。而如果我們把前面找的若干塊磚頭加上重量 $x$ 的那一塊，便可以讓重量和在 $[r+x, r+x+2]$ 之間，因此原題對於 $r \\in [-2+x, s]$ 也是成立的。\n\n因此，只要我們能證明 $[-2, s] \\subset [-2, s-x] \\cup [-2+x, s-x]$，便能證明原題對於任何 $r \\in [-2, s]$ 成立。而由於每一塊重量至少為 $1$，故\n\n$$\nx - 2 \\le (s - (n - 1)) - 2 = s - (2n - (n - 1)) \\le s - (s - (n - 1)) \\le s - x,\n$$\n\n從而 $[-2, s-x] \\cup [-2+x, s-x] = [-2, s]$，故得證！\n\n![](images/20-1J_p6_data_06facd506a.png)\n\n**解法二.** 將磚頭依照重量排序，使得其重量 $x_1 \\le x_2 \\le \\cdots \\le x_n$。令 $\\mathcal{J}_k$ 為 $\\{1, 2, \\cdots, k\\}$ 的 power set，並令 $S_k = \\{\\sum_{j \\in J} x_j : J \\in \\mathcal{J}_k\\}$。將 $S_k$ 的所有元素依照大小排成一排，並令 $M_k$ 為其相鄰兩元素差距的最大值。我們僅須證明 $M_n \\le 2$，原題即得證。\n\n我們用歸納法證明 $M_k$ 皆至多為 $2$。$n=1$ 時顯然。若 $M_{k-1} \\le 2$，注意到\n\n$$\nS_k = S_{k-1} \\cup (x_k + S_{k-1}),\n$$\n\n故我們只需要證明\n\n$$\nx_k \\le \\sum_{j<k} x_j + 2,\n$$\n\n則自然有 $M_k \\le 2$。但如果上式不成立，則我們有 $x_k > \\sum_{j<k} x_j + 2 \\ge k + 1$，從而由遞增性知 $x_l > k + 1$ 對所有 $l \\le k$ 皆成立。但這會導致\n\n$$\n2n = \\sum_{j=1}^{n} x_j \\ge \\sum_{j=1}^{k-1} 1 + \\sum_{j=k}^{n} (k+1) = (k-1) + (n-k+1)(k+1),\n$$\n\n也就是 $n > k(n+1-k)$，但這對於 $1 \\le k \\le n$ 是不可能的，矛盾！故上式成立，從而原命題得證。\n\n_選題者註記：基本上依循對重量排序後做歸納的路線大致都可以通，請預期許多變體證明方式。_",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18548,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any integer $n \\ge 3$ we have $$(2n)! < n^{2n}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 3$ the claim holds: $(2n)! = 6! = 720$ and $n^{2n} = 3^6 = 729$.\n\nSuppose $n \\ge 4$. Divide the numbers $2, 3, \\dots, 2n-2$ into pairs $(k, 2n-k)$ with $2 \\le k \\le n-1$, leaving $n$ alone. For each pair we have\n\n$$\nk(2n-k) = (n - (n-k))(n + (n-k)) = n^2 - (n-k)^2 < n^2.\n$$\n\nHence $2 \\cdot 3 \\cdot \\dots \\cdot (2n-2) < (n^2)^{n-2} \\cdot n = n^{2n-3}$, therefore\n\n$$\n(2n)! < 1 \\cdot n^{2n-3} \\cdot (2n-1) \\cdot (2n) < n^{2n-3} \\cdot (2n)^2 = 4n^{2n-1} \\le n^{2n}.\n$$\n\nAlternatively, for $n=3$ the claim holds. Suppose the claim holds for $n$; to show that it also holds for $n+1$ it is enough to show the inequality $(2n+1)(2n+2) < \\frac{(n+1)^{2n}}{n^{2n}}(n+1)^2$.\n\nSince $(2n+1)(2n+2) < (2n+2)^2 = 4(n+1)^2$, it is enough to show that $\\frac{(n+1)^{2n}}{n^{2n}} > 4$.\n\nThis is equivalent to $\\left(1 + \\frac{1}{n}\\right)^n > 2$ which holds for all $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18549,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $m$ and $n$, compare the numbers\n\n$$\nA = m^{525} + n^{525}\n$$\nand\n$$\nB = (m+n)(m^2+n^2)(m^4+n^4)(m^8+n^8)\\dots(m^{128}+n^{128}).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $m = n = 1$, then $A < B$; in all other cases, $A > B$.\n\nIf $m = n$, then\n$$\nA = 2m^{525}, \\quad B = 2m \\cdot 2m^2 \\cdot 2m^4 \\cdots 2m^{128} = 2^8 m^{255}.\n$$\nFor $m = n = 1$, $A = 2 < 2^8 = B$.\nFor $m = n > 1$, $A = 2m^{525} > 2^8 m^{255} = B$.\n\nNow, without loss of generality, let $m > n$. Then:\n$$\n\\begin{aligned}\n(m - n)B &= (m - n)(m + n)(m^2 + n^2)(m^4 + n^4) \\cdots (m^{128} + n^{128}) \\\\\n&= (m^2 - n^2)(m^2 + n^2)(m^4 + n^4) \\cdots (m^{128} + n^{128}) \\\\\n&= (m^4 - n^4)(m^4 + n^4) \\cdots (m^{128} + n^{128}) \\\\\n&\\vdots \\\\\n&= m^{256} - n^{256} < m^{525} + n^{525} = A.\n\\end{aligned}\n$$\nTherefore, $B < A$ for $m > n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set with $n \\ge 2$ elements. Two players, A and B, alternately choose nonempty proper subsets of $S$, where:\n\n1. It is not allowed to choose a set that contains a set previously chosen by any player.\n2. It is not allowed to choose a set that is contained in any previously chosen set.\n3. It is not allowed to choose a set whose union with any previously chosen set is $S$.\n\nA begins. The player who first cannot choose a set anymore loses. Which player has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "It is straightforward to verify that the following is a winning strategy for A. Player A first chooses a singleton subset $\\{x\\} \\subset S$. After that, A responds to B choosing a set $T$ by choosing $(S \\setminus \\{x\\}) \\setminus T$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18551,
    "subject": "Mathematics (Olympiad)",
    "question": "All the squares of a $2024 \\times 2024$ board are coloured white. In one move, Mohit can select one row or column whose every square is white, choose exactly $1000$ squares in this row or column, and colour all of them red. Find the maximum number of squares that Mohit can colour red in a finite number of moves.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2024$ and $k = 1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n - k)$, which evaluates to $3048000$.\n\nIndeed, call a row or column *bad* if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picked, then that row and the $k$ columns corresponding to the chosen squares are all bad. Any subsequent move increases the number of bad rows or columns by at least $1$. Since there are only $2n$ rows and columns, we can make at most $2n - (k+1)$ moves after the first one, and so at most $2n - k$ moves can be made in total. Thus we can have at most $k(2n-k)$ red squares.\n\nTo prove this is achievable, let's choose each of the $n$ columns in the first $n$ moves, and colour the top $k$ cells in these columns. Then, the bottom $n-k$ rows are still uncoloured, so we can make $n-k$ more moves, colouring $k(n+n-k)$ cells in total. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18552,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $s_1, s_2, s_3, \\dots$ is a strictly increasing sequence of positive integers such that the subsequences\n\n$s_{s_1}, s_{s_2}, s_{s_3}, \\dots$ and $s_{s_{1+1}}, s_{s_{2+1}}, s_{s_{3+1}}, \\dots$\n\nare both arithmetic progressions. Prove that the sequence $s_1, s_2, s_3, \\dots$ is itself an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the common difference of the progression $s_{s_1}, s_{s_2}, s_{s_3}, \\dots$. Also, define $d_n = s_{n+1} - s_n$. For every $n$, $d_n \\ge 1$ because $s_1, s_2, s_3, \\dots$ is strictly increasing. Therefore, for all $i < j$, we have\n\n$$\ns_j - s_i = d_i + d_{i+1} + \\dots + d_{j-1} = \\sum_{k=i}^{j-1} d_k \\ge \\sum_{k=i}^{j-1} 1 = j - i.\n$$\n\nTaking $i = s_n$ and $j = s_{n+1}$ gives $D = s_{s_{n+1}} - s_{s_n} \\ge d_n$ for any $n$.\n\nThus, the sequence of integers $d_1, d_2, d_3, \\dots$ is bounded both above (by $D$) and below (by 1), so it has a maximum and a minimum value. Call these values $M$ and $m$ respectively; that is, define\n\n$$\nm = \\min\\{d_1, d_2, \\dots\\} \\text{ and } M = \\max\\{d_1, d_2, \\dots\\}.\n$$\n\nIf $M = m$, then all the $d_n$ are equal, and we are done.\n\nNow we approach indirectly by assuming that $M \\neq m$, and seeking a contradiction.\n\nFor all $i < j$, we have\n\n$$\ns_j - s_i = d_i + d_{i+1} + \\dots + d_{j-1} \\leq M(j - i).\n$$\n\nBy definition of $m$, there exists $n$ such that $s_{n+1} - s_n = m$, and so\n\n$$\nD = s_{s_{n+1}} - s_{s_n} \\leq M(s_{n+1} - s_n) = Mm.\n$$\n\nMoreover, if equality holds, then $d_{s_n}, d_{s_{n+1}}, \\dots, d_{s_{n+1}-1}$ must all equal $M$.\n\nLikewise, for all $i < j$, we can see that $s_j - s_i \\geq m(j - i)$. There exists $n$ such that $s_{n+1} - s_n = M$, and so\n\n$$\nD = s_{s_{n+1}} - s_{s_n} \\geq m(s_{n+1} - s_n) = Mm.\n$$\n\nIf equality holds, then $d_{s_n}, d_{s_{n+1}}, \\dots, d_{s_{n+1}-1}$ must all equal $m$.\n\nClearly, the two inequalities above imply $D = Mm$.\n\nNow, we claim that for every $n$ such that $d_n = m$, there exists $n' > n$ such that $d_{n'} = M$ — namely, $n' = s_n$. Indeed, from the equality above, we must have $d_{s_n} = M$. And by the earlier bound, $s_n \\geq s_1 + n - 1 \\geq n$; moreover, since $d_{s_n} \\neq d_n$, we cannot have $s_n = n$, so $s_n > n$ strictly. Similarly, for every $n$ such that $d_n = M$, there exists $n' > n$ such that $d_{n'} = m$ — again given by $n' = s_n$.\n\nIteratively applying these two facts, we see that there are infinitely many values of $n$ for which $d_n = m$ and infinitely many values of $n$ for which $d_n = M$. These imply, in turn, that there are infinitely many values of $n$ for which $d_{s_n} = M$, and infinitely many values of $n$ for which $d_{s_n} = m$.\n\nNow, the two sequences $s_{s_1}, s_{s_2}, s_{s_3}, \\dots$ and $s_{s_{1+1}}, s_{s_{2+1}}, s_{s_{3+1}}, \\dots$ are both arithmetic progressions; therefore, when we take differences of corresponding terms, we again get an arithmetic progression. This sequence of differences is just\n\n$$\nd_{s_1}, d_{s_2}, d_{s_3}, \\dots\n$$\n\nWe saw that this sequence contains infinitely many $m$'s and infinitely many $M$'s. But an arithmetic progression cannot have even two terms equal to each other unless it is constant. It follows that $M = m$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18553,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有合成數 $n$，使得 $n$ 的比 1 大的正因數擺在一個圓上，相鄰的兩個都不會互質。",
    "options": [],
    "answer": "See solution",
    "solution": "答案是除了 $n = pq$（兩個相異質數的乘積）以外的所有合成數。\n\n- 當 $n = p^{\\alpha}$，$\\alpha \\ge 2$ 時，依 $p, p^2, \\dots, p^{\\alpha}$ 排列，顯然滿足條件。\n\n- 當 $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}$，$k \\ge 3$ 時，先把 $p_1, p_2, \\dots, p_k, p_1$ 依次排列在圓上，然後把 $p_j$ 的倍數排列在 $p_{j-1} p_j$ 與 $p_j p_{j+1}$ 之間。得到的排列滿足條件。\n\n- 最後，當 $n = p^a q^b$，$a > 1$，先把 $pq$ 與 $p^2 q$ 排在圓上，把 $p$ 的倍數排在同一弧，而 $q$ 的倍數排在另一弧上。\n\n剩下的 $n = pq$ 沒辦法做到。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18554,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m = 2^a \\cdot 3^b \\cdot 5^c \\cdot 7^d$. Find the smallest positive integer $m$ such that $m$, $m+1$, $m+2$, and $m+3$ are all perfect powers: specifically, $m$ is a perfect 5th power, $m+1$ is a perfect 6th power, $m+2$ is a perfect 7th power, and $m+3$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "For a number $n$ to be an exact $k$th power, all exponents in its prime factorization must be divisible by $k$. For $m = 2^a \\cdot 3^b \\cdot 5^c \\cdot 7^d$, and for primes other than $2,3,5,7$, the exponent must be $0$ or a multiple of $5 \\times 6 \\times 7 = 210$. The smallest solution occurs when these exponents are $0$, so we consider only $2^a 3^b 5^c 7^d$.\n\nFor the conditions to be satisfied:\n- $5 \\mid c+1$, $6 \\mid c$, $7 \\mid c$; the smallest $c$ is $84$.\n- $a$ must satisfy $a \\equiv 0 \\pmod{5}$, $a+1 \\equiv 0 \\pmod{6}$, $a \\equiv 0 \\pmod{7}$; the smallest $a$ is $35$.\n- $b$ satisfies the same as $a$, so $b=35$.\n- $d$ must satisfy $d \\equiv 0 \\pmod{5}$, $d \\equiv 0 \\pmod{6}$, $d+1 \\equiv 0 \\pmod{7}$; the smallest $d$ is $90$.\n\nThus, the smallest value of $m$ is $2^{35} \\cdot 3^{35} \\cdot 5^{84} \\cdot 7^{90}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18555,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0 < a_1 < a_2 < \\dots$ be an infinite sequence of positive integers. Prove that there exists a unique positive integer $n$ such that\n$$\na_n < \\frac{a_0 + a_1 + \\dots + a_n}{n} \\le a_{n+1}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Define $d_n = (a_0 + a_1 + \\dots + a_n) - n a_n$, for $n = 1, 2, \\dots$.\n\nNote that\n$$\n\\begin{aligned}\nna_{n+1} - (a_0 + a_1 + \\cdots + a_n) &= (n+1)a_{n+1} - (a_0 + a_1 + \\cdots + a_n + a_{n+1}) \\\\\n&= -d_{n+1}.\n\\end{aligned}\n$$\nSo, the problem is equivalent to proving that there exists a unique positive integer $n$ such that $d_n > 0 \\ge d_{n+1}$.\n\nWe see that $d_1 = (a_0 + a_1) - 1 \\cdot a_1 = a_0 > 0$, and\n$$\n\\begin{aligned}\nd_{n+1} - d_n &= ((a_0 + a_1 + \\cdots + a_n) - n a_{n+1}) - ((a_0 + a_1 + \\cdots + a_n) - n a_n) \\\\\n&= n(a_n - a_{n+1}) < 0,\n\\end{aligned}\n$$\nthat is, $\\{d_n\\}$ is a strictly decreasing sequence of integers with the first term being positive. Hence, there exists a unique positive integer $n$ such that $d_n > 0 \\ge d_{n+1}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18556,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the maximum number of prophetic words of length $n$.\n\nEach letter is either $A$ or $E$. For $n=1$, $f(1)=2$; for $n=2$, $f(2)=4$; for $n=3$, $f(3)=7$.\n\nA prophetic word is a word that does not contain a forbidden substring $W$ of length three (for example, $W=AAA$). Find the maximum number of 10-letter prophetic words, i.e., compute $f(10)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since there are $2^3 = 8$ possible 3-letter words, each word of length $n$ can be classified based on its ending relative to the forbidden substring $W$.\n\nWe have:\n$$\n  f(n) \\leq f(n-1) + f(n-2) + f(n-3)\n$$\n\nIf $W = AAA$, then $f(n) = f(n-1) + f(n-2) + f(n-3)$, since we can construct all words of length $n$ without $AAA$ as a substring by extending shorter prophetic words.\n\nSubstituting initial values:\n- $f(1) = 2$\n- $f(2) = 4$\n- $f(3) = 7$\n\nWe compute recursively:\n- $f(4) = 13$\n- $f(5) = 24$\n- $f(6) = 44$\n- $f(7) = 81$\n- $f(8) = 149$\n- $f(9) = 274$\n- $f(10) = 504$\n\nThus, the maximum number of 10-letter prophetic words is $\\boxed{504}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18557,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon with $AB = 14$, $BC = 7$, $CD = 24$, $DE = 13$, $EA = 26$, and $\\angle B = \\angle E = 60^\\circ$. For each point $X$ in the plane, define $f(X) = AX + BX + CX + DX + EX$. The least possible value of $f(X)$ can be expressed as $m + n\\sqrt{p}$, where $m$ and $n$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, because $AB = 2BC$ and $EA = 2DE$, it follows that $\\triangle ABC$ and $\\triangle AED$ are $30$-$60$-$90^\\circ$ triangles with $\\angle ACB = \\angle ADE = 90^\\circ$. In particular, $AC = 7\\sqrt{3}$ and $AD = 13\\sqrt{3}$.\n\nLet $P$ be the projection of $A$ onto $\\overline{BE}$.\n\nBecause $\\angle APB = \\angle APE = 90^\\circ$, it follows that $P$ lies on the circumcircles of both $\\triangle ACB$ and $\\triangle ADE$. The point in the plane such that the sum of its distances to each of the three vertices of a triangle is the least possible is called the *Fermat Point* of the triangle. For a triangle with no angle greater than $120^\\circ$, this point lies at the intersection of the three circumcircles of equilateral triangles drawn externally on the sides of the triangle. Because $\\angle B = \\angle E = 60^\\circ$, it follows that $P$ is the Fermat Point of $\\triangle ACD$, which implies that for any point $X$,\n\n$$\nAP + CP + DP \\leq AX + CX + DX.\n$$\n\nBecause $P$ is on $\\overline{BE}$, it follows that for any point $X$,\n\n$$\nBP + EP \\leq BX + EX.\n$$\n\nTherefore $AX + BX + CX + DX + EX$ is minimized when $X = P$.\n\nThe Law of Cosines applied to $\\triangle ACD$ gives\n\n$$\n\\cos(\\angle CAD) = \\frac{AC^2 + AD^2 - CD^2}{2 \\cdot AC \\cdot AD} = \\frac{1}{7}.\n$$\n\nThis shows that $\\triangle ACD$ is acute; thus $P$ is inside the triangle $\\triangle ACD$, as shown in the diagram.\n\n![](images/2025AIME_I_Solutions_p12_data_ea37793ed7.png)\n\nLet $F$ be the point outside of $ABCDE$ such that $\\triangle CFD$ is equilateral. Because $P$ is the Fermat Point of $\\triangle ACD$, point $P$ is on the circumcircle of $\\triangle CFD$. This implies that $\\angle APF = \\angle APC + \\angle CPF = \\angle APC + \\angle CDF = 120^\\circ + 60^\\circ = 180^\\circ$, and thus $A$, $P$, and $F$ are collinear. Let $G$ be the point on $\\overline{PF}$ with $PG = CP$, making $\\triangle CPG$ equilateral. Because $CP = CG$, $CD = CF$, and\n\n$$\n\\angle PCD = \\angle PCG - \\angle DCG = \\angle DCF - \\angle DCG = \\angle GCF,\n$$\n\nit follows that $\\triangle CPD \\cong \\triangle CGF$ by SAS. Hence $DP = FG$, so $AP + CP + DP = AP + PG + FG = AF$. The requested sum is therefore\n\n$$\nAP + BP + CP + DP + EP = (AP + CP + DP) + (BP + EP) = AF + BE.\n$$\n\nBecause $\\cos(\\angle CAD) = \\frac{1}{7}$, it follows that $\\sin(\\angle CAD) = \\sqrt{1 - \\cos^2(\\angle CAD)} = \\frac{4\\sqrt{3}}{7}$. Therefore\n\n$$\n\\begin{aligned}\n\\cos(\\angle BAE) &= \\cos(\\angle CAD + 60^\\circ) = \\cos(\\angle CAD) \\cos(60^\\circ) - \\sin(\\angle CAD) \\sin(60^\\circ) \\\\\n&= \\frac{1}{7} \\cdot \\frac{1}{2} - \\frac{4\\sqrt{3}}{7} \\cdot \\frac{\\sqrt{3}}{2} = -\\frac{11}{14}.\n\\end{aligned}\n$$\n\nThe Law of Cosines applied to $\\triangle BAE$ gives\n\n$$\nBE^2 = AB^2 + AE^2 - 2 \\cdot AB \\cdot AE \\cos(\\angle BAE) = 1444 = 38^2,\n$$\n\nso $BE = 38$.\n\nThe Law of Sines applied to $\\triangle ACD$ gives\n\n$$\n\\sin(\\angle ADC) = \\frac{\\sin(\\angle CAD) \\cdot AC}{CD} = \\frac{\\frac{4\\sqrt{3}}{7} \\cdot 7\\sqrt{3}}{24} = \\frac{1}{2}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18558,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions of the equation\n\n$$\nx^4 + 2y^4 + 4z^4 + 8t^4 = 16xyzt\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $(0, 0, 0, 0)$ is a solution of the equation. We will show that there exists no nonzero solution.\n\nSuppose, for contradiction, that $(x_0, y_0, z_0, t_0)$ is a solution with at least one nonzero coordinate:\n$$\nx_0^4 + 2y_0^4 + 4z_0^4 + 8t_0^4 = 16x_0 y_0 z_0 t_0.\n$$\nIt is clear that $x_0$ is divisible by $2$. Write $x_0 = 2x_1$ for some $x_1 \\in \\mathbb{Z}$. Substitute and divide by $2$:\n$$\ny_0^4 + 2z_0^4 + 4t_0^4 + 8x_1^4 = 16x_1 y_0 z_0 t_0.\n$$\nNow $y_0$ is divisible by $2$, so $y_0 = 2y_1$ for some $y_1 \\in \\mathbb{Z}$. Substitute and divide by $2$:\n$$\nz_0^4 + 2t_0^4 + 4x_1^4 + 8y_1^4 = 16x_1 y_1 z_0 t_0.\n$$\nSimilarly, $z_0$ is divisible by $2$, so $z_0 = 2z_1$ for $z_1 \\in \\mathbb{Z}$. Substitute and divide by $2$:\n$$\nt_0^4 + 2x_1^4 + 4y_1^4 + 8z_1^4 = 16x_1 y_1 z_1 t_0.\n$$\nNow $t_0$ is divisible by $2$, so $t_0 = 2t_1$ for $t_1 \\in \\mathbb{Z}$. Substitute and divide by $2$:\n$$\nx_1^4 + 2y_1^4 + 4z_1^4 + 8t_1^4 = 16x_1 y_1 z_1 t_1.\n$$\nThus, $(x_1, y_1, z_1, t_1)$ is also a solution. Repeating this process, we get solutions $(x_n, y_n, z_n, t_n) = \\left(\\frac{x_0}{2^n}, \\frac{y_0}{2^n}, \\frac{z_0}{2^n}, \\frac{t_0}{2^n}\\right)$ for all $n$, which is impossible unless all coordinates are zero. Therefore, the only integer solution is $(0, 0, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18559,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1 < x_2 < \\cdots < x_n$ be positive integers. Prove that\n\n$$\nx_1^3 + x_2^3 + \\cdots + x_n^3 \\geq (x_1 + x_2 + \\cdots + x_n)^2.\n$$\n\nFurthermore, show that the minimum value of $x_1 + x_2 + \\cdots + x_n$ is $\\frac{1}{2}n(n+1)$, achieved when $x_k = k$ for $1 \\leq k \\leq n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value $\\frac{1}{2}n(n+1)$ is achieved by letting $x_k = k$ for $1 \\leq k \\leq n$. To prove the inequality, it suffices to show\n\n$$\nx_1^3 + \\cdots + x_n^3 \\geq (x_1 + \\cdots + x_n)^2,\n$$\n\nsince $x_1 + \\cdots + x_n \\geq 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$.\n\nWe may assume $x_1 < x_2 < \\cdots < x_n$. We prove the inequality by induction on $n$.\n\n**Base case:** For $n=1$, $x_1^3 \\geq x_1^2$ is true for $x_1 \\geq 1$.\n\n**Inductive step:** Assume the result holds for $n-1$. Then\n\n$$\n\\sum_{k=1}^{n} x_k^3 - \\left( \\sum_{k=1}^{n} x_k \\right)^2 = \\left[ \\sum_{k=1}^{n-1} x_k^3 - \\left( \\sum_{k=1}^{n-1} x_k \\right)^2 \\right] + x_n \\left\\{ x_n^2 - x_n - 2 \\sum_{k=1}^{n-1} x_k \\right\\}.\n$$\n\nBy the induction hypothesis, the first bracket is nonnegative.\n\nNext, $x_n \\geq n$. Since $x_k < x_{k+1}$, $x_k \\leq x_{k+1} - 1 \\leq \\cdots \\leq x_n - (n-k)$ for $1 \\leq k < n$. Thus,\n\n$$\nx_n^2 - x_n - 2 \\sum_{k=1}^{n-1} x_k \\geq x_n^2 - x_n - 2 \\sum_{k=1}^{n-1} (x_n - n + k) = x_n^2 - x_n - 2(n-1)x_n + n(n-1).\n$$\n\nThis simplifies to\n\n$$\n(x_n - n + 1)(x_n - n) \\geq 0,\n$$\n\nwhich is true for $x_n \\geq n$. Thus, the inequality holds for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18560,
    "subject": "Mathematics (Olympiad)",
    "question": "In a school, 112 groups are formed, each containing 11 students. Every pair of groups has exactly one common student. Prove that:\n\n(a) There exists a student belonging to at least 12 groups.\n\n(b) There exists a student belonging to all groups.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Consider an arbitrary group $O$. Each of the remaining 111 groups shares exactly one student with $O$. Since $111 = 11 \\times 10 + 1$, by the pigeonhole principle, there exists a student $x \\in O$ who belongs to at least 11 other groups. Hence, $x$ belongs to at least 12 groups, say $O_1, O_2, \\dots, O_{12}$.\n\n(b) We will prove that $x$ belongs to all groups. Suppose, for contradiction, that $x \\notin O'$. Then the group $O'$ has exactly one common student with each of the groups $O_1, O_2, \\dots, O_{12}$. Since $O'$ has 11 students, there must be two groups, say $O_i$ and $O_j$, that share the same common student $y$ with $O'$. However, in this case, the groups $O_i$ and $O_j$ would have two common students, $x$ and $y$, which is a contradiction.\n\nTherefore, $x$ belongs to all groups.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18561,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a set $X$ with $|X| = n \\geq 1$ elements. A family $\\mathcal{F}$ of distinct subsets of $X$ is said to have property $\\mathcal{P}$ if there exist $A, B \\in \\mathcal{F}$ such that $A \\subset B$ and $|B \\setminus A| = 1$.\n\n1. Determine the least value $m$ such that any family $\\mathcal{F}$ with $|\\mathcal{F}| > m$ has property $\\mathcal{P}$.\n2. Describe all families $\\mathcal{F}$ with $|\\mathcal{F}| = m$ and not having property $\\mathcal{P}$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. We claim that $m = 2^{n-1}$. The set of all subsets of $X$, $\\mathcal{P}(X)$, contains $2^n$ elements. Fix $x \\in X$ and consider the $2^{n-1}$ pairs $\\{S, S \\cup \\{x\\}\\}$, where $S \\subseteq X \\setminus \\{x\\}$. These pairs partition $\\mathcal{P}(X)$:\n\n$$\n\\mathcal{P}(X) = \\bigcup_{S \\in \\mathcal{P}(X \\setminus \\{x\\})} \\{S, S \\cup \\{x\\}\\}.\n$$\n\nBy the pigeonhole principle, if $|\\mathcal{F}| > 2^{n-1}$, then $\\mathcal{F}$ must contain both $S_0$ and $S_0 \\cup \\{x\\}$ for some $S_0$, so $A = S_0$, $B = S_0 \\cup \\{x\\}$ satisfy $A \\subset B$ and $|B \\setminus A| = 1$. Thus, $\\mathcal{F}$ has property $\\mathcal{P}$.\n\nA family $\\mathcal{F}$ with $|\\mathcal{F}| = 2^{n-1}$ and not having property $\\mathcal{P}$ must contain exactly one member from each such pair. For example, the family $\\mathcal{F}_e$ of all even cardinality subsets of $X$ works, since for any $A, B \\in \\mathcal{F}_e$ with $A \\subset B$, $|B \\setminus A| \\geq 2$. The number of even cardinality subsets is:\n\n$$\n|\\mathcal{F}_e| = \\binom{n}{0} + \\binom{n}{2} + \\cdots = 2^{n-1}.\n$$\n\nSimilarly, the family $\\mathcal{F}_o$ of all odd cardinality subsets also works.\n\n2. The only families $\\mathcal{F}$ with $|\\mathcal{F}| = m = 2^{n-1}$ and not having property $\\mathcal{P}$ are $\\mathcal{F}_e$ (all even cardinality subsets) and $\\mathcal{F}_o$ (all odd cardinality subsets). If $\\emptyset \\in \\mathcal{F}$, then all subsets with even cardinality are in $\\mathcal{F}$; if not, all subsets with odd cardinality are in $\\mathcal{F}$.\n\n*Remark:* If the condition $|B \\setminus A| = 1$ is dropped, Sperner's theorem gives the threshold $m = \\binom{n}{\\lfloor n/2 \\rfloor}$, with the largest antichain being all subsets of size $\\lfloor n/2 \\rfloor$ (or $\\lceil n/2 \\rceil$ for odd $n$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18562,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the quadrilateral $ABCD$ satisfies $\\angle ABD = 30^\\circ$, $\\angle CDB = 20^\\circ$, and $\\angle BCA = \\angle ACD = 40^\\circ$. Determine $\\angle DAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the intersection point of the diagonals $AC$ and $BD$. Let $EF$ be the bisector of $\\angle DEC$, with $F$ on $DC$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p130_data_44bc00dce0.png)\n\nThen the triangles $BEC$ and $CFE$ are congruent ($EFCB$ is a kite). So $EF = EB = EA$, and $ADFE$ is a kite. It follows that $\\angle DAC = 100^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18563,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, \\infty) \\to [0, \\infty)$ be a continuous bijective function such that\n\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 $\\lim_{x \\to \\infty} \\frac{f^{-1}(a x)}{f^{-1}(x)} = 1$, for any $a > 0$.\n\nb) Give an example of a function $f$ that satisfies the conditions from the statement.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $f$ is continuous and bijective, $f(0) = 0$ 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 $u > 0$ such that $\\frac{f^{-1}(f(x)/x)}{x} > \\frac{1}{2}$ for any $x > u$. Therefore, $\\frac{f(x)}{x} > f\\left(\\frac{x}{2}\\right)$ for any $x > u$. Since $\\lim_{x \\to \\infty} f\\left(\\frac{x}{2}\\right) = \\infty$, we obtain $\\lim_{x \\to \\infty} \\frac{f(x)}{x} = \\infty$.\n\nLet $a > 0$ be arbitrary. The case $a = 1$ is clear.\n\nFor $a \\in (0, 1)$, there is $t > 0$ such that $f^{-1}(x) > \\frac{1}{a}$ for any $x > t$. Hence\n\n$$\nf^{-1}(x) > f^{-1}(a x) = f^{-1}(a f(f^{-1}(x))) > f^{-1}\\left(\\frac{f(f^{-1}(x))}{f^{-1}(x)}\\right), \\text{ for any } x > t.\n$$\n\nThus,\n\n$$\n\\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} < \\frac{f^{-1}(a x)}{f^{-1}(x)} < 1, \\text{ for any } x > t.\n$$\n\nFrom the assumption and $\\lim_{x \\to \\infty} f^{-1}(x) = \\infty$,\n\n$$\n\\lim_{x \\to \\infty} \\frac{f^{-1}(f(f^{-1}(x))/f^{-1}(x))}{f^{-1}(x)} = \\lim_{y \\to \\infty} \\frac{f^{-1}(f(y)/y)}{y} = 1.\n$$\n\nBy the squeeze theorem, $\\lim_{x \\to \\infty} \\frac{f^{-1}(a x)}{f^{-1}(x)} = 1$.\n\nFor $a > 1$, let $b = 1/a \\in (0, 1)$. From the previous case,\n\n$$\n\\lim_{x \\to \\infty} \\frac{f^{-1}(a x)}{f^{-1}(x)} = \\lim_{x \\to \\infty} \\left( \\frac{f^{-1}(x)}{f^{-1}(a x)} \\right)^{-1} = \\lim_{x \\to \\infty} \\left( \\frac{f^{-1}(b (a x))}{f^{-1}(a x)} \\right)^{-1} = 1^{-1} = 1.\n$$\n\nIn conclusion, $\\lim_{x \\to \\infty} \\frac{f^{-1}(a x)}{f^{-1}(x)} = 1$ for any $a > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18564,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 members of a ladies' club. Each lady has had tea (in private) with exactly 56 of her lady friends. The Board, consisting of the 50 most distinguished ladies, have all had tea with one another. Prove that the entire ladies' club may be split into two groups in such a way that, within each group, any lady has had tea with any other.",
    "options": [],
    "answer": "See solution",
    "solution": "Each lady in the Board has had tea with 49 ladies within the Board, and 7 ladies outside the Board. Each lady not in the Board has had tea with at most 49 ladies not in the Board, and at least 7 ladies in the Board. Comparing these two observations, we conclude that each lady not in the Board has had tea with exactly 49 ladies not in the Board and exactly 7 ladies in the Board. Hence, the club may be split into Board members and non-members. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18565,
    "subject": "Mathematics (Olympiad)",
    "question": "A figure is called a _polyomino_ if it is formed by joining one or more $1 \\times 1$ squares edge to edge. It is known that a rectangle which is not a square can be split into 8 pairwise distinct polyominoes. Polyominoes are considered equal if one can be transformed into another by translations and rotations. What is the smallest area of such a rectangle?",
    "options": [],
    "answer": "See solution",
    "solution": "We start by counting polyominoes of the smallest areas. There is only 1 polyomino of area 1, 1 polyomino of area 2, and 2 polyominoes of area 3. Thus, the smallest area of a rectangle that consists of 8 polyominoes is $1 \\cdot 1 + 1 \\cdot 2 + 2 \\cdot 3 + 4 \\cdot 4 = 25$. But then it has to be of size $25 \\times 1$ (since a $5 \\times 5$ square doesn't satisfy the condition). For such a rectangle, it can only be split into rectangles of size $k \\times 1$, but such a rectangle has an area at least $1+2+\\ldots+8=36 > 25$. Thus, the smallest possible area is 26. An example of splitting a $13 \\times 2$ rectangle is shown below.\n\n![](images/UkraineMO2019_booklet_p18_data_9f7aaa64a7.png)\n\n**Fig. 18**",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18566,
    "subject": "Mathematics (Olympiad)",
    "question": "Определи ја 2008-та цифра по децималната запирка во децималниот запис на бројот $\\frac{1}{41}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Децималниот запис на бројот $\\frac{1}{41}$ е $0.(02439)$. Значи, групата од пет цифри $02439$ по децималната запирка периодично се повторува бесконечно многу пати. Бројот $2008$ можеме да го запишеме во облик $2008 = 5 \\cdot 401 + 3$. Според тоа, 2008-та цифра по децималната запирка е третата цифра од периодата, т.е. тоа е цифрата $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18567,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $x^2 + y^4 + 1 = 6^z$ in the set of integers.",
    "options": [],
    "answer": "See solution",
    "solution": "It is obvious that $z \\geq 0$. If $z \\geq 2$, then $x^2 + y^4 + 1 \\equiv 0 \\pmod{4}$, i.e. $x^2 + y^4 \\equiv 3 \\pmod{4}$. This is not possible because the remainders of squares of integers after division by $4$ are $0$ or $1$. According to that, $0 \\leq z < 2$.\n\nIf $z = 0$, then $x = y = 0$.\n\nIf $z=1$, then $x^2 + y^4 = 5$, i.e. $(x, y) = \\{(2,1), (-2,1), (2,-1), (-2,-1)\\}$.\n\nTherefore $(x, y, z) = \\{(0,0,0), (2,1,1), (-2,1,1), (2,-1,1), (-2,-1,1)\\}$ are the solutions of the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18568,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be $x$ R1 coins and $y$ R5 coins. Given:\n\n- $4x = 9y$\n- $0 < x + 5y < 50$\n\nFind the possible values of $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $x = \\frac{9y}{4}$. Substituting into the inequality:\n\n$$\n\\begin{aligned}\n0 &< x + 5y < 50 \\\\\n0 &< \\frac{9y}{4} + 5y < 50 \\\\\n0 &< \\frac{9y + 20y}{4} < 50 \\\\\n0 &< \\frac{29y}{4} < 50 \\\\\n0 &< 29y < 200 \\\\\n0 &< y < \\frac{200}{29} = 6\\frac{26}{29}\n\\end{aligned}\n$$\n\nSince $x$ must be an integer, $y$ must be divisible by 4. The only possible value is $y = 4$, so $x = 9$. Checking: $9 \\times 1 + 4 \\times 5 = 29$. Thus, you have R29.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18569,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x, y, z$ which verify the equalities\n\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": "See solution",
    "solution": "We have\n$$\ny + 2 = \\frac{(x+2)^3}{3x^2+4}, \\quad y - 2 = \\frac{(x-2)^3}{3x^2+4}\n$$\nand analogous relations for $z$ and $x$.\n\nIf $y = 2$, then $x = z = 2$.\n\nIf $y \\ne 2$, then\n$$\n\\frac{y+2}{y-2} = \\left(\\frac{x+2}{x-2}\\right)^3, \\quad \\frac{z+2}{z-2} = \\left(\\frac{y+2}{y-2}\\right)^3, \\quad \\frac{x+2}{x-2} = \\left(\\frac{z+2}{z-2}\\right)^3.\n$$\n\nThus,\n$$\n\\frac{x+2}{x-2} = \\left(\\frac{x+2}{x-2}\\right)^{27},\n$$\nwhich leads to $\\frac{x+2}{x-2} \\in \\{-1, 0, 1\\}$.\n\nWe find two more solutions: $x = y = z = -2$ and $x = y = z = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18570,
    "subject": "Mathematics (Olympiad)",
    "question": "A set $M$ of real numbers is called *special* if it has the following properties:\n\n1. 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.\n2. For each $x \\in M$, $x^2$ is irrational.\n\nFind the maximum number of elements in a special set.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $4$. An example of a special $4$-element set is:\n\n$$M = \\{\\sqrt{2} - 1,\\ \\sqrt{2} + 1,\\ 2 - \\sqrt{2},\\ -2 - \\sqrt{2}\\}.$$  \n\nWe will prove that a special set cannot have more than $4$ elements. The second condition implies that all elements of a special set are irrational. Consider the following remarks:\n\n**R₁.** If $x, y, z$ are three distinct elements of $M$, then $x + y$, $x + z$, and $y + z$ cannot all be rational.\n\nSuppose the contrary. Then $2(x + y + z) \\in \\mathbb{Q}$, so $x + y + z \\in \\mathbb{Q}$, which implies $x \\in \\mathbb{Q}$, a contradiction.\n\n**R₂.** If $x, y, z$ are three distinct elements of $M$, then $xy$, $xz$, and $yz$ cannot all be rational.\n\nSuppose the contrary. Then $x^2 y z = (xy) \\cdot (xz) \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}^*$, so $x^2 \\in \\mathbb{Q}$, a contradiction.\n\n**R₃.** 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\nSuppose not. Then, by the assumption, $R_1$, and $R_2$, either $x + z \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}$, or $y + z \\in \\mathbb{Q}$ and $xz \\in \\mathbb{Q}$. In the first case, $xy \\in \\mathbb{Q}$ and $yz \\in \\mathbb{Q}$ imply $xy + yz = y(x + z) \\in \\mathbb{Q}$, and since $x + z \\in \\mathbb{Q}$ and $x + z \\neq 0$, $y \\in \\mathbb{Q}$, a contradiction. The second case is similar.\n\nSuppose now that there exists a special set with at least five elements $a, b, c, d, e$. By $R_1$, 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 $R_1$, the numbers $c + d$, $c + e$, and $d + e$ cannot all be rational, so $cd$, $ce$, and $de$ are rational, which contradicts $R_2$.\n\nTherefore, the maximum number of elements in a special set is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18571,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $\\left(a, b\\right)$ of real numbers such that\n\n$$\na \\cdot \\lfloor b n \\rfloor = b \\cdot \\lfloor a n \\rfloor\n$$\n\nfor all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are all pairs $\\left(a, b\\right)$ with $a = 0$, $b = 0$, $a = b$, or both $a$ and $b$ integers.\n\nLet $a_0 = \\lfloor a \\rfloor$ and $a_i$ be the binary digits of the fractional part of $a$ such that $a = a_0 + \\sum_{i=1}^{\\infty} \\frac{a_i}{2^i}$ with $a_0 \\in \\mathbb{Z}$ and $a_i \\in \\{0, 1\\}$ for $i \\ge 1$. Similarly, let $b = b_0 + \\sum_{i=1}^{\\infty} \\frac{b_i}{2^i}$ with $b_0 \\in \\mathbb{Z}$ and $b_i \\in \\{0, 1\\}$ for $i \\ge 1$. In the case of a non-unique binary expansion, we choose the expansion ending on infinitely many zeros.\n\nNow choose $n = 2^k$ and $m = 2^{k-1}$ in the given equation. We get the equations\n\n$$\n\\begin{aligned}\na\\left(2^k b_0 + \\sum_{i=1}^k b_i 2^{k-i}\\right) &= b\\left(2^k a_0 + \\sum_{i=1}^k a_i 2^{k-i}\\right), \\\\\na\\left(2^{k-1} b_0 + \\sum_{i=1}^{k-1} b_i 2^{k-i-1}\\right) &= b\\left(2^{k-1} a_0 + \\sum_{i=1}^{k-1} a_i 2^{k-i-1}\\right).\n\\end{aligned}\n$$\n\nThe first equation for $k = 0$ and the difference of the first equation and the doubled second equation for $k \\ge 1$ yields\n\n$$\nab_k = b a_k \\tag{1}\n$$\n\nfor $k \\ge 0$.\n\nNow, we consider three cases. If one or both of $a$ and $b$ are zero, then the original equation is clearly satisfied. If both fractional parts are zero, then both numbers are integers and again, the original equation is satisfied. So, finally, we consider the case that $a, b \\ne 0$ and that there is a $k \\ge 1$ with $a_k = 1$. The equation (1) shows that $b_k$ cannot be zero, so we get $b_k = 1$ and thus from the same equation $a = b$. This clearly satisfies the original equation. (Of course, $b_k = 1$ leads to the same conclusion.) Therefore, the solutions are exactly the pairs listed in the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18572,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n$, determine the greatest possible value of the quotient\n\n$$\n\\frac{1 - x^n - (1-x)^n}{x(1-x)^n + (1-x)x^n}\n$$\n\nwhere $0 < x < 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 1 - x$, so $x + y = 1$. The expression becomes\n\n$$\n\\frac{1 - x^n - y^n}{x y^n + y x^n}.\n$$\n\nWe claim the maximum occurs at $x = y = \\frac{1}{2}$, giving value $2^n - 2$.\n\nRewrite:\n\n$$\n\\begin{aligned}\n\\frac{1 - x^n - y^n}{x y^n + y x^n} &= \\frac{x + y - x^n - y^n}{x y^n + y x^n} \\\\\n&= \\frac{x(1 - x^{n-1}) + y(1 - y^{n-1})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x(1-x)(1 + x + \\cdots + x^{n-2}) + y(1-y)(1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x y (1 + x + \\cdots + x^{n-2} + 1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{1 + 1}{x^{n-1} + y^{n-1}} + \\frac{x + y}{x^{n-1} + y^{n-1}} + \\cdots + \\frac{x^{n-2} + y^{n-2}}{x^{n-1} + y^{n-1}}.\n\\end{aligned}\n$$\n\nWe need to show that\n\n$$\n\\frac{x^a + y^a}{x^b + y^b}\n$$\n\nis maximized at $x = y = \\frac{1}{2}$ for $0 \\leq a < b$. By the general mean inequality:\n\n$$\n\\left( \\frac{x^a + y^a}{2} \\right) \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{a/b}\n$$\n\nand\n\n$$\n\\frac{1}{2} = \\frac{x + y}{2} \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{1/b},\n$$\n\nwith equality when $x = y = \\frac{1}{2}$. Thus,\n\n$$\n\\frac{x^a + y^a}{x^b + y^b} \\leq 2^{b-a},\n$$\n\nwith equality at $x = y = \\frac{1}{2}$. Therefore, the maximum value is $2^n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18573,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a_1 = 2$ and the sequence $(a_n)$ satisfies the recurrence relation\n\n$$\n\\frac{a_n - 1}{n - 1} = \\frac{a_{n-1} + 1}{n}\n$$\n\nfor all $n \\ge 2$. What is the greatest integer less than or equal to\n\n$$\n\\sum_{n=1}^{100} a_n^2?\n$$\n\n(A) 338,550  (B) 338,551  (C) 338,552  (D) 338,553  (E) 338,554",
    "options": [],
    "answer": "See solution",
    "solution": "Computing the first few terms of this sequence gives $a_1 = 2$, $a_2 = \\frac{5}{2}$, $a_3 = \\frac{10}{3}$, and $a_4 = \\frac{17}{4}$, so it appears that $a_n = n + \\frac{1}{n}$. Indeed, this is correct, because the recurrence relation is satisfied:\n\n$$\n\\frac{a_{n-1} + 1}{n} = \\frac{n - 1 + \\frac{1}{n-1} + 1}{n} = 1 + \\frac{1}{n(n-1)}\n$$\n\nand\n\n$$\n\\frac{a_n - 1}{n-1} = \\frac{n + \\frac{1}{n} - 1}{n-1} = 1 + \\frac{1}{n(n-1)}\n$$\n\nSo the formula holds. Then\n\n$$\n\\begin{aligned}\n\\sum_{n=1}^{100} a_n^2 &= \\sum_{n=1}^{100} \\left( n^2 + 2 + \\frac{1}{n^2} \\right) \\\\\n&= \\sum_{n=1}^{100} n^2 + 2 \\cdot 100 + \\sum_{n=1}^{100} \\frac{1}{n^2} \\\\\n&= \\frac{100 \\cdot 101 \\cdot 201}{6} + 200 + r,\n\\end{aligned}\n$$\n\nwhere $r = \\sum_{n=1}^{100} \\frac{1}{n^2}$. By a telescoping sum argument,\n\n$$\n1 < r < 1 + \\sum_{n=2}^{100} \\frac{1}{n(n-1)} = 2 - \\frac{1}{100} < 2.\n$$\n\nThus $\\sum_{n=1}^{100} a_n^2$ is between 338,551 and 338,552, and the requested greatest integer is $338,551$.\n\n**Note:** One could also estimate $\\sum_{n=1}^{100} \\frac{1}{n^2}$ by using the fact that\n\n$$\n\\sum_{n=1}^{\\infty} \\frac{1}{n^2} = \\frac{\\pi^2}{6} \\approx 1.645,\n$$\n\nas proved by Euler in the 18th century.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18574,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(x)$ be a polynomial with rational coefficients, i.e., each term is of the form $a_i x^i$ where $a_i$ is rational. Show that there exists a positive integer $n$ such that all coefficients of $p(x + n)$ are integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding $a_i(x + n)^i - a_i x^i$, each term in the expansion of $(x + n)^i$ except $x^i$ contains $n$ as a factor. If we choose $n$ to be the least common multiple of the denominators of the $a_i$, then $p(x + n)$ will have integer coefficients.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18575,
    "subject": "Mathematics (Olympiad)",
    "question": "Two operations $L$ and $R$ are defined as follows on rational numbers $\\frac{p}{q}$, where $p$ and $q$ are positive integers:\n\n$$\nL\\left(\\frac{p}{q}\\right) = \\frac{p}{p+q} \\quad \\text{and} \\quad R\\left(\\frac{p}{q}\\right) = \\frac{p+q}{q}.\n$$\n\nStart from $1$ and apply the operations $R, L, R, L, R, L, R, L, R, L$ successively. When the result is written as a fraction in simplest form, what is the sum of its numerator and denominator?",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1\n\nApplying $R$ followed by $L$ to $\\frac{p}{q}$, we get $\\frac{p+q}{p+2q}$. We now do this 5 times starting with $1 = 1/1$.\n\n$$\n\\begin{align*}\nL(R(1/1)) &= 2/3 \\\\\nL(R(2/3)) &= 5/8 \\\\\nL(R(5/8)) &= 13/21 \\\\\nL(R(13/21)) &= 34/55 \\\\\nL(R(34/55)) &= 89/144\n\\end{align*}\n$$\n\nSince $89$ is prime, the required number is $89 + 144 = 233$.\n\nMethod 2\n\nStarting with $1 = 1/1$, we get:\n\n$$\n\\begin{array}{ll}\nR(1/1) = 2/1 & L(2/1) = 2/3 \\\\\nR(2/3) = 5/3 & L(5/3) = 5/8 \\\\\nR(5/8) = 13/8 & L(13/8) = 13/21 \\\\\nR(13/21) = 34/21 & L(34/21) = 34/55 \\\\\nR(34/55) = 89/55 & L(89/55) = 89/144\n\\end{array}\n$$\n\nSince $89$ is prime, the required number is $89 + 144 = 233$.\n\nComment\n\nNote that $L(R(p/q)) = \\frac{p+q}{(p+q)+q}$. So, starting with $1/1$ and applying $R$ followed by $L$ successively produces a sequence of fractions whose numerators and denominators give the Fibonacci sequence $1, 1, 2, 3, 5, 8, \\dots$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18576,
    "subject": "Mathematics (Olympiad)",
    "question": "A (convex) trapezoid $ABCD$ is called *good* if it is inscribed, has parallel sides $AB$ and $CD$, and $CD$ is shorter than $AB$.\n\nFor a good trapezoid, use the following notations:\n\n- The line parallel to $AD$ through $B$ intersects the line $CD$ at $S$.\n- The tangents through $S$ to the circumcircle of the trapezoid meet the circumcircle at $E$ and $F$, respectively, where $E$ is on the same side of the line $CD$ as $A$.\n\nCharacterize good trapezoids $ABCD$ (in terms of the side lengths and/or angles of the trapezoid) for which the angles $\\angle BSE$ and $\\angle FSC$ are equal. The characterization should be as simple as possible.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of the trapezoid be $u$, the second intersection point of the line $SB$ with $u$ be $T$, and the center of $u$ be $M$.\n\n![](images/Austria2019_booklet_p12_data_27cad3b606.png)\n\nConsider the reflection across the line $MS$. This reflection maps $E$ and $F$ to each other and maps $u$ to itself. The trapezoid meets the *angle condition* if $\\angle BSE = \\angle FSC$.\n\nThe angle condition holds if and only if the reflection maps the rays $SB$ and $SC$ to each other, i.e., the intersection points of these rays with $u$ are mapped to each other in the same order.\n\n**Case 1:** $B$ is between $S$ and $T$ (see figure). The angle condition holds if and only if the reflection maps $B$ and $C$ to each other, i.e., triangle $BSC$ is isosceles with axis of symmetry $SM$. Since $M$ lies on the perpendicular bisector of $BC$, this is equivalent to $CS = BS$. As $BS = BC$, this is equivalent to triangle $BSC$ being equilateral, i.e., $\\angle CSB = 60^\\circ$. Since $ABSD$ is a parallelogram, the angle condition holds if and only if $\\angle BAD = 60^\\circ$.\n\n**Case 2:** $T$ lies between $S$ and $B$. The angle condition holds if and only if the reflection maps $B$ and $D$ to each other, i.e., triangle $BSD$ is isosceles with axis of symmetry $MS$. This is equivalent to $SB = SD$, which is equivalent to $AB = AD$.\n\n**Summary:**\n- The angle condition $\\angle BSE = \\angle FSC$ holds if and only if either $\\angle BAD = 60^\\circ$ or $AB = AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18577,
    "subject": "Mathematics (Olympiad)",
    "question": "In the triangle $ABC$, $H$ is the midpoint of the altitude $AD$ and $O$ is the center of the circumscribed circle. A line perpendicular to $HO$ passing through $H$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the midpoints of $BP$, $CQ$, and $O$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $M_C$ and $M_B$ as the midpoints of $AB$ and $AC$ respectively. Then $H$ lies on the line $M_B M_C$ due to its initial position. Note that the projections of $O$ on the lines $AB$, $AC$, and $PQ$ are points $M_C$, $M_B$, and $H$, which lie on the Simson line (see the figure below). From this, it can be concluded that $O$ lies on the circumscribed circle of $\\triangle PAQ$.\n\nIntersect the ray $HO$ with the line $BC$ at a point $S$. Observe that $\\triangle OM_B M_C \\sim \\triangle HBC$ as they have the same angles. Also, $\\angle OHM_B = \\angle HSB$, so $S$ and $H$ are respective vertices of the similar triangles. Since $BC = 2M_B M_C$, we have that $SC = 2M_C H = BD$. Hence, points $S$ and $D$ are symmetric over $BC$.\n\nName the midpoints of $BP$ and $CQ$ as $P_1$ and $Q_1$ respectively. Define $C_1$ as the symmetric point of $C$ over $D$. Therefore, we can deduce that\n\n$$\n\\angle OQC = \\angle OHM_B = \\angle OSC_1 = \\angle HSC_1 \\text{ and } \\angle OCA = 90^\\circ - \\angle B = \\angle HCB = \\angle HC_1S,\n$$\nwhich means that $\\triangle HSC_1 \\sim \\triangle OQC$. Let $X$ be the midpoint of $SC_1$. From $C_1D = DC = BS$, we have that the midpoints of $SC_1$ and $BD$ coincide, so $X$ is also the midpoint of $BD$. From the previously described similarity $\\triangle HSC_1 \\sim \\triangle OQC$ we get that $\\angle OQ_1A = \\angle HXS$ (an angle between a median and a side) and $\\angle OQ_1A = \\angle HXS = \\angle B$ (midline). Similarly,\n\n$\\angle OP_1A = \\angle C$.\n\nFrom the angle sum $\\angle A + \\angle B + \\angle C = 180^\\circ$ we conclude that $P_1$, $O$, and $Q_1$ are collinear.\n\n![](images/Ukraine_2020_booklet_p39_data_5e4df45c93.png)\n\nFig. 33",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18578,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ is the largest integer such that, for any coloring of the edges of the complete graph $K_n$ with three colors (red, yellow, green), there is no monochromatic triangle. What is the largest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "For any $\\triangle XYZ$, WLOG assume $XY$ and $XZ$ are red. Then this triangle corresponds to two red lines from $X$, and this correspondence is one-to-one. It follows that the number of triangles is\n\n$$\n\\sum_{j=1}^{13} \\left[ \\binom{r_j}{2} + \\binom{y_j}{2} + \\binom{g_j}{2} \\right].\n$$\n\nBut then it is clear that there are $\\binom{13}{3} = 286$ triangles. By the pigeonhole principle, WLOG assume\n\n$$\n\\binom{r_1}{2} + \\binom{y_1}{2} + \\binom{g_1}{2} \\ge \\frac{286}{13} = 22.\n$$\n\nThis gives $r_1(r_1 - 1) + y_1(y_1 - 1) + g_1(g_1 - 1) \\ge 44$. Since $r_1 + y_1 + g_1 = 12$, this implies $r_1^2 + y_1^2 + g_1^2 \\ge 56$.\n\nWe claim that one of $r_1, y_1, g_1$ is at least 6. Suppose on the contrary that $r_1, y_1, g_1 \\le 5$. As $r_1 + y_1 + g_1 = 12$, it suffices to check $(r_1, y_1, g_1) = (5, 5, 2), (5, 4, 3), (4, 4, 4)$. In all cases, $r_1^2 + y_1^2 + g_1^2 \\ge 56$ does not hold.\n\nWLOG assume $r_1 \\ge 6$. Suppose $AB, AC, AD, AE, AF, AG$ are red. By assumption, none of the lines formed by $B, C, D, E, F, G$ is red. Since $R(3,3) = 6$, there must be a yellow or green triangle among these 6 points, contradiction. This proves $n < 13$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18579,
    "subject": "Mathematics (Olympiad)",
    "question": "If the height of a cone is $5$ and the lateral surface area is $30\\pi$, then the volume of this cone is \\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the base radius of this cone be $r$ and its slant height be $l$.\n\nBy the condition, $\\pi r l = 30\\pi$. Note that $l = \\sqrt{r^2 + 5^2}$, so $r \\sqrt{r^2 + 25} = rl = 30$. Squaring both sides yields $r^2 (r^2 + 25) = 900$. Thus, $r^4 + 25 r^2 - 900 = 0$. Solving, $r^2 = 20$.\n\nTherefore, the volume of this cone is:\n$$\n\\frac{1}{3} \\pi r^2 \\cdot 5 = \\frac{1}{3} \\pi \\times 20 \\times 5 = \\frac{100}{3} \\pi\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 18580,
    "subject": "Mathematics (Olympiad)",
    "question": "200 улсыг өөр хооронд нь холбодог 10 агаарын компани байна. Шударга өрсөлдөөнийг хангахын тулд дараах нөхцлүүдийг тавьжээ:\n\n- Улс бүрт компани бүр бусад компаниас 1-ээс ихгүй зөрүүтэй чиглэлд үйлчилдэг (өөрөөр хэлбэл, аль ч хоёр компанийн тухайн улсаас гарч буй чиглэлийн тооны зөрүү хамгийн ихдээ 1 байна).\n- Аль ч хоёр улсын хооронд зөвхөн нэг компани үйлчилдэг.\n\nЭдгээр нөхцлийг Монголоос бусад бүх улсын хувьд хэрэгжүүлж болохыг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Энэ бодлогыг 200 оройтой бүтэн графын ирмэгүүдийг 10 өнгөөр будахтай адилтгаж болно. Энд:\n- Ирмэг бүрийг зөвхөн нэг өнгөөр будна (компани бүр).\n- Аль ч оройгоос (улсаас) гарах ирмэгүүдийн (чиглэлийн) өнгөнүүдийн тооны зөрүү хамгийн ихдээ 1 байна.\n\nХэрэв бүх 200 орой дээр энэ нөхцлийг хангахыг оролдвол:\n- Дурын оройгоос 199 ирмэг гарна. 10 компанид хуваахад 19 эсвэл 20 чиглэлд үйлчлэх ёстой.\n- $i$-р компанийн хувьд $a_i$ оройгоос 20 чиглэл, $200 - a_i$ оройгоос 19 чиглэл үйлчилнэ гэж үзье.\n- $i$-р компанийн нийт чиглэлийн тоо: $\\frac{a_i \\cdot 20 + (200 - a_i) \\cdot 19}{2} = 1900 + \\frac{a_i}{2}$.\n- Бүх компанийн нийлбэр чиглэл: $\\sum_{i=1}^{10} (1900 + \\frac{a_i}{2}) = 19900$ (нийт ирмэгийн тоо).\n- Эндээс $\\sum_{i=1}^{10} a_i = 1800$.\n\nГэвч $a_i \\leq 200$ тул бүх компанид 20 чиглэлтэй орой олдохгүй, зөрчил үүснэ. Иймд 200 оройтой бүтэн графыг энэ нөхцлөөр будах боломжгүй.\n\nХарин Монголоос бусад 199 улсын хувьд:\n- Монголыг $A$ орой гэж тэмдэглэе.\n- Үлдсэн 199 оройг 10 хэсэгт (19 эсвэл 20 оройтой) хуваана.\n- Хэсэг бүрийг 1-10 гэж дугаарлана.\n- $i$ ба $j$-р хэсгийн хоорондох ирмэгийг $(i + j)$ дугаарын компанид (өнгөөр) онооно (модуляр 10).\n- Хэсэг доторх ирмэгүүд болон $A$-аас тухайн хэсгийн оройнууд руу гарах ирмэгүүдийг үлдсэн нэг өнгөөр будна.\n\nИнгэснээр Монголыг эс тооцвол бүх улсад нөхцөл хангагдана.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18581,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have a 9-digit number $N$ in which all the digits are distinct and non-zero. Then we consider all the sums of adjacent triples of digits of $N$ and order them in a non-decreasing sequence. For the following sequences, determine whether there exists an $N$ for which we get them as a result:\n\n- a) $11, 15, 16, 18, 19, 21, 22$\n- b) $11, 15, 16, 18, 19, 21, 23$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes, the number $N = 137658942$ works, since the sums of consecutive triples are (left to right) $11, 16, 18, 19, 22, 21, 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18582,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)$ be a sequence of real numbers from the interval $(0, 1)$. The sequence of positive integers $(a_n)$ is defined as follows:\n\n- $a_1 = 1$,\n- $a_{i+1} = m$, where $m$ is the smallest positive integer for which $[x_1 + x_2 + \\cdots + x_m] = a_i$.\n\nProve that for any indices $i, j$, the inequality $a_{i+j} \\geq a_i + a_j$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that we can put $a_0 = 0$, then we would get $a_1 = 1$.\n\nFirst, let's prove the following lemma:\n\n**Lemma.** For all $i \\in \\mathbb{N}$, the following holds: $a_{i+1} - a_i \\geq a_i - a_{i-1}$.\n\n*Proof.*\n\n$$\nx_1 + x_2 + \\cdots + x_{2a_i - a_{i-1} - 1} = (x_1 + x_2 + \\cdots + x_{a_i}) + (x_{a_i + 1} + x_{a_i + 2} + \\cdots + x_{2a_i - a_{i-1} - 1}) < \\\\\n< (a_{i-1} + 1) + (1 + 1 + \\cdots + 1)_{\\text{($a_i - a_{i-1} - 1$ terms)}} < (a_{i-1} + 1) + (a_i - a_{i-1} - 1) = a_i.\n$$\n\nThus, $a_{i+1}$ must be at least $2a_i - a_{i-1}$.\n\nThe lemma is proven.\n\nNow, choose some positive integer $t$. Write down the inequalities of the form $a_{i-k} - a_{i-k-1} \\geq a_{i-k-1} - a_{i-k-2}$ for $k = 0, \\ldots, t-1$.\n\nAdding all these inequalities gives $a_i - a_{i-t} \\geq a_{i-1} - a_{i-t-1}$, and then\n\n$$\na_i - a_{i-t} \\geq a_{i-1} - a_{i-t-1} \\geq a_{i-2} - a_{i-t-2} \\geq \\cdots \\geq a_t - a_0.\n$$\n\nNow, set $i \\to i+j$, $t \\to j$ to get the desired inequality: $a_{i+j} \\geq a_i + a_j$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18583,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be the set of girls at the competition, $B$ the set of boys, and $P$ the set of problems. For $g \\in G$, let $P(g)$ be the set of problems solved by $g$; for $b \\in B$, let $P(b)$ be the set of problems solved by $b$. For $p \\in P$, let $G(p)$ be the set of girls who solve $p$, and $B(p)$ the set of boys who solve $p$.\n\nFor all $g \\in G$ and $b \\in B$:\n\n- $|P(g)| \\leq 6$\n- $|P(b)| \\leq 6$\n- $P(g) \\cap P(b) \\neq \\emptyset$\n\nA problem is **boy-easy** if $|B(p)| \\geq 3$ and **boy-hard** if $|B(p)| \\leq 2$; similarly, **girl-easy** if $|G(p)| \\geq 3$ and **girl-hard** if $|G(p)| \\leq 2$.\n\n*Prove that there exists $p \\in P$ such that $|G(p)| \\geq 3$ and $|B(p)| \\geq 3$.*",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that every problem is either boy-hard or girl-hard (or both), i.e., for each $p \\in P$, either $|G(p)| \\leq 2$ or $|B(p)| \\leq 2$.\n\nLet $T = \\{ (p, g, b) \\mid p \\in P(g) \\cap P(b) \\}$ be the set of ordered triples where both $g$ and $b$ solve $p$.\n\nFrom the conditions, we have:\n\n$$\n|T| = \\sum_{g \\in G} \\sum_{b \\in B} |P(g) \\cap P(b)| \\geq |G| \\cdot |B| = 21^2.\n$$\n\nAlso,\n\n$$\n\\sum_{p \\in P} |G(p)| = \\sum_{g \\in G} |P(g)| \\leq 6|G|, \\quad \\sum_{p \\in P} |B(p)| \\leq 6|B|.\n$$\n\nLet $P_{ge}$ and $P_{gh}$ be the sets of girl-easy and girl-hard problems, respectively. If $p \\in P_{ge}$, then $p$ is boy-hard, so $|B(p)| \\leq 2$.\n\nThus,\n\n$$\n|T| = \\sum_{p \\in P} |G(p)| \\cdot |B(p)| = \\sum_{p \\in P_{ge}} |G(p)| \\cdot |B(p)| + \\sum_{p \\in P_{gh}} |G(p)| \\cdot |B(p)| \\leq 2 \\sum_{p \\in P_{ge}} |G(p)| + 2 \\sum_{p \\in P_{gh}} |B(p)|.\n$$\n\n**Lemma:**\n- $\\sum_{p \\in P_{gh}} |G(p)| \\geq |G|$\n- $\\sum_{p \\in P_{ge}} |G(p)| \\leq 5|G|$\n- $\\sum_{p \\in P_{ge}} |B(p)| \\geq |B|$\n- $\\sum_{p \\in P_{gh}} |B(p)| \\leq 5|B|$\n\n*Proof of Lemma:* Each girl solves at least one problem in $P_{gh}$ (by the Pigeonhole Principle and the given conditions), so $\\sum_{p \\in P_{gh}} |G(p)| \\geq |G|$. Similarly, $\\sum_{p \\in P_{ge}} |G(p)| = \\sum_{p \\in P} |G(p)| - \\sum_{p \\in P_{gh}} |G(p)| \\leq 6|G| - |G| = 5|G|$. Analogous arguments hold for boys.\n\nTherefore,\n\n$$\n|T| \\leq 2 \\cdot 5|G| + 2 \\cdot 5|B| = 10|G| + 10|B| = 20 \\times 21 = 420.\n$$\n\nBut earlier, $|T| \\geq 21^2 = 441$, a contradiction. Thus, there must exist $p \\in P$ with $|G(p)| \\geq 3$ and $|B(p)| \\geq 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18584,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum value of\n$$\n\\frac{1}{a^2 - 4a + 9} + \\frac{1}{b^2 - 4b + 9} + \\frac{1}{c^2 - 4c + 9}\n$$\nwhere $a$, $b$, $c$ are non-negative real numbers satisfying $a + b + c = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that for $0 \\le x \\le 1$ the inequality\n$$\n\\frac{1}{x^2 - 4x + 9} \\le \\frac{x + 2}{18}\n$$\nholds, where equality holds if and only if $x = 0$ or $x = 1$. Hence,\n$$\n\\frac{1}{a^2 - 4a + 9} + \\frac{1}{b^2 - 4b + 9} + \\frac{1}{c^2 - 4c + 9} \\le \\frac{1}{18}(a + b + c + 6) = \\frac{7}{18}\n$$\nSince equality holds for $a = 0$, $b = 0$, $c = 1$, the maximum value is\n$$\n\\frac{7}{18}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18585,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfy the equality\n\n$$\nf(f(x)) = 0.5(x^2 - x)f(x) + 2 - x\n$$\n\nfor all $x \\in \\mathbb{R}$.\n\n(a) Find $f(2)$.\n\n(b) Find all possible values of $f(1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t_0$ be a fixed point of $f$, i.e. $f(t_0) = t_0$. Then\n\n$$\nt_0 = f(f(t_0)) = 0.5(t_0^2 - t_0)t_0 + 2 - t_0\n$$\n\nwhich simplifies to\n\n$$\nt_0^3 - t_0^2 - 4t_0 + 4 = 0 \\implies (t_0 - 1)(t_0^2 - 4) = 0\n$$\n\nSo $t_0 \\in \\{-2, 1, 2\\}$.\n\nIf $f(2) = b$, then\n\n$$\nf(b) = f(f(2)) = 0.5(2^2 - 2)f(2) + 2 - 2 = f(2) = b\n$$\n\nSo $b$ is a fixed point, i.e. $f(2) = b \\in \\{-2, 1, 2\\}$.\n\nLet $f(0) = a$. Then\n\n$$\nf(a) = f(f(0)) = 0.5(0^2 - 0)f(0) + 2 - 0 = 2\n$$\n\nThus $b = f(2) = f(f(a)) = 0.5(a^2 - a)f(a) + 2 - a = a^2 - 2a + 2 = (a - 1)^2 + 1$\n\nSo $b = (a - 1)^2 + 1 \\geq 1$, so $b \\in \\{1, 2\\}$.\n\nIf $f(2) = 1$, then $1 = (a - 1)^2 + 1 \\implies a = 1$. But then $f(2) = f(0)$, which leads to a contradiction. Therefore, $f(2) = 2$.\n\nFor $f(1) = c$, we have\n\n$$\nf(c) = f(f(1)) = 0.5(1^2 - 1)f(1) + 2 - 1 = 1\n$$\n\nSo\n\n$$\nc = f(1) = f(f(c)) = 0.5(c^2 - c) + 2 - c \\implies c^2 - 5c + 4 = 0\n$$\n\nThus $c \\in \\{1, 4\\}$.\n\nBoth values are possible, as shown by the following examples:\n\n$$\nf(x) = \\begin{cases} \\frac{4}{x}, & \\text{if } x \\neq 0, 1, 4 \\\\ \\frac{2}{x}, & \\text{if } x = 0 \\\\ 1, & \\text{if } x = 1 \\\\ -\\frac{2}{3}, & \\text{if } x = 4 \\end{cases}\n$$\n\n**Answers:**\n\n(a) $f(2) = 2$;\n\n(b) $f(1) \\in \\{1, 4\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18586,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ which satisfies the properties:\n\na) $f(1) = 1$\n\nb) $f(p) = 1 + f(p-1)$ for any prime number $p$\n\nc) $f(p_1p_2\\cdots p_n) = f(p_1) + f(p_2) + \\cdots + f(p_n)$ for any prime numbers, not necessarily distinct.\n\nProve that $2^{f(n)} \\leq n^3 \\leq 3^{f(n)}$ for any natural number $n$, $n \\geq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $2$ is a prime number, we have $f(2) = 1 + f(1) = 2$, whence $2^2 \\leq 2^3 \\leq 3^2$. Then $f(3) = 1 + f(2) = 3$, and thus $2^3 < 3^3 = 3^2$.\n\nWe will prove the inequalities\n\n$$\n3 \\log_3 n \\leq f(n) \\leq 3 \\log_2 n\n$$\n\nby mathematical induction on $n$. Let $n$ be a natural number, $n \\geq 3$, and let $n = p_1p_2\\cdots p_k$ be its prime factorization (not necessarily distinct factors).\n\nIf $n$ is not a prime number, i.e. $k \\geq 2$, using the induction hypothesis,\n$$\nf(n) = f(p_1) + f(p_2) + \\cdots + f(p_k) \\geq 3 \\sum_{i=1}^k \\log_3 p_i = 3 \\log_3 n\n$$\nand\n$$\nf(n) \\leq 3 \\sum_{i=1}^k \\log_2 p_i = 3 \\log_2 n\n$$\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$$\nf(n) = 3 + f\\left(\\frac{n-1}{2}\\right) \\leq 3 + 3 \\log_2 \\frac{n-1}{2} = 3 \\log_2(n-1) \\leq 3 \\log_2 n\n$$\nand\n$$\nf(n) = 3 + f\\left(\\frac{n-1}{2}\\right) \\geq 3 + 3 \\log_3 \\frac{n-1}{2} = 3 \\log_3 \\frac{3(n-1)}{2} \\geq 3 \\log_3 n\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18587,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system of equations:\n\n$$\n\\begin{cases}\nx^2 + 3xy = 3y + x, \\\\\ny^2 - yx = 3x + y.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $(0, 0)$, $(1, -1)$, $(1, 3)$, and $(6, -2)$.\n\n*Solution.* Add the equations:\n\n$$\nx^2 + 2xy + y^2 = 4y + 4x \\text{ or } (x + y)^2 = 4(x + y).\n$$\n\nSo $x + y = 0$ or $x + y = 4$.\n\nIf $x = -y$, the second equation gives:\n\n$$\n2y^2 = -2y, \\text{ so } y = 0 \\text{ or } y = -1.\n$$\n\nThus, the solutions are $(0, 0)$ and $(1, -1)$.\n\nIf $x = 4 - y$, the first equation gives:\n\n$$\ny^2 - y(4 - y) = 3(4 - y) + y, \\text{ so } y^2 - y - 6 = 0.\n$$\n\nThis quadratic has roots $y = 3$ and $y = -2$, yielding solutions $(1, 3)$ and $(6, -2)$. These satisfy both equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18588,
    "subject": "Mathematics (Olympiad)",
    "question": "The number of positive integer solutions of the equation $x + y + z = 2010$ with $x \\leq y \\leq z$ is \\_\\_\\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "The total number of positive integer solutions to $x + y + z = 2010$ is $\\binom{2009}{2} = 2009 \\times 1004$.\n\nWe classify these solutions into three categories:\n\n1. $x = y = z$: There is obviously 1 solution.\n\n2. Exactly two variables are equal: The number in this category is 1003.\n\n3. $x, y, z$ are all different: Let the number in this category be $k$.\n\nFrom\n\n$$\n1 + 3 \\times 1003 + 6k = 2009 \\times 1004,\n$$\n\nwe have\n\n$$\n\\begin{align*}\n6k &= 2009 \\times 1004 - 3 \\times 1003 - 1 \\\\\n&= 2006 \\times 1005 - 2009 + 3 \\times 2 - 1 \\\\\n&= 2006 \\times 1005 - 2004.\n\\end{align*}\n$$\n\nSo,\n\n$$\nk = 1003 \\times 335 - 334 = 335671.\n$$\n\nTherefore, the number of positive integer solutions satisfying $x \\leq y \\leq z$ is\n\n$$\n1 + 1003 + 335671 = 336675.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18589,
    "subject": "Mathematics (Olympiad)",
    "question": "In a group of people, some are friends (friendship is mutual) and each person $p$ has a list $f_1(p), f_2(p), \\dots, f_{d(p)}(p)$ of their friends, where $d(p)$ is the number of friends $p$ has. Additionally, any two people are connected by a series of friendships. Each person also has a *water balloon*. The following game is played until someone ends up with more than one water balloon: on round $r$, each person $p$ throws the current water balloon they have to their friend $f_s(p)$ such that $d(p) \\mid r - s$. Show that if the game never ends, then everyone has the same number of friends.",
    "options": [],
    "answer": "See solution",
    "solution": "Given a person $p$, let $F(p)$ be the set of friends of $p$. Choose a person $p$ with the most friends. Note that for each friend $q$ of $p$, $p$ receives a water balloon from $q$ once out of every $d(q)$ turns. Since $p$ always receives 1 water balloon, we must have\n\n$$\n\\sum_{q \\in F(p)} \\frac{1}{d(q)} = 1.\n$$\n\nSince this sum has $d(p)$ terms, and since $d(q) \\le d(p)$ for all $q$, we have\n\n$$\n1 \\ge d(p) \\cdot \\frac{1}{d(p)} = 1.\n$$\n\nThus we must have equality for all friends $q$ of $p$. In particular, $d(q) = d(p)$. Thus all friends of any person with the most number of friends also have the most number of friends.\n\nAgain, let $p$ be a person with the most friends. Now for any other person $q$, there exists a sequence of people $p = p_0, p_1, \\dots, p_n = q$. Repeatedly applying the previous result gives us $d(p) = d(p_0) = d(p_1) = \\dots = d(p_n) = d(q)$. Thus any person has the maximum number of friends out of the group, which means that each person has the same number of friends.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18590,
    "subject": "Mathematics (Olympiad)",
    "question": "The maximum of $f(x) = 2 \\sin^2 x - \\tan^2 x$ is \\_\\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\nf(x) &= 2(1 - \\cos^2 x) - \\frac{1 - \\cos^2 x}{\\cos^2 x} \\\\\n&= 3 - \\left( 2 \\cos^2 x + \\frac{1}{\\cos^2 x} \\right) \\\\\n&\\le 3 - 2\\sqrt{2},\n\\end{aligned}\n$$\n\nWhen $2 \\cos^2 x = \\frac{1}{\\cos^2 x}$ (e.g., take $x = \\arccos \\frac{1}{\\sqrt{2}}$), $f(x)$ takes the maximum $3 - 2\\sqrt{2}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18591,
    "subject": "Mathematics (Olympiad)",
    "question": "On the small arc $AB$ of the circumcircle of the equilateral triangle $ABC$, 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 the circumcircle of triangle $ABC$ again at points $M$ and $I$, respectively.\n\n1. Prove that $IMN$ is an equilateral triangle.\n2. 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": "See solution",
    "solution": "a) Let $O$ be the circumcenter of triangle $ABC$. Without loss of generality, let $O$ be at $0$, and the vertices be $A(1)$, $B(\\varepsilon)$, and $C(\\varepsilon^2)$, where $\\varepsilon = -\\frac{1}{2} + i\\frac{\\sqrt{3}}{2}$. Since the length of arc $NB$ is $30^\\circ$, we have $AO \\perp ON$, so $N$ has affix $i$. Also, $NI \\perp AB$ implies there exists $\\alpha \\in \\mathbb{R}^*$ such that:\n\n$$\n\\frac{i - z_I}{1 - \\varepsilon} = i\\alpha \\implies z_I = i - \\frac{3}{2}i\\alpha - \\frac{\\sqrt{3}}{2}\\alpha,\n$$\n\nwhere $z_I$ is the affix of $I$. From $|z_I| = 1$ we get $\\alpha = 1$, so $z_I = i\\varepsilon$. Similarly, from $MN \\perp AC$, the affix of $M$ is $i\\varepsilon^2$, so triangle $IMN$ is equilateral.\n\nb) Using Sylvester's theorem, the affixes of the orthocenters are:\n\n$$\n\\begin{aligned}\nz_{H_1} &= z_A + z_N + z_B = i + 1 + \\varepsilon, \\\\\nz_{H_2} &= z_B + z_M + z_C = \\varepsilon + i\\varepsilon + \\varepsilon^2 = \\varepsilon z_{H_1}, \\\\\nz_{H_3} &= z_C + z_I + z_A = \\varepsilon^2 + i\\varepsilon^2 + 1 = \\varepsilon^2 z_{H_1},\n\\end{aligned}\n$$\n\nso triangle $H_1H_2H_3$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18592,
    "subject": "Mathematics (Olympiad)",
    "question": "A magical triangulation is a partition of a triangle into smaller triangles by a finite number of segments whose endpoints are vertices of the triangle or points in its interior, such that at every point (including the vertices of the triangle), the same number of segments meet.\n\nWhat is the maximal number of smaller triangles into which we can divide the triangle in a magical triangulation?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the number of smaller triangles, $t$ the number of points in the triangulation (including the vertices of the triangle), $d$ the number of segments (including the sides of the triangle), and $k$ the number of segments meeting at each point of the triangulation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18593,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be the point different from $B$ on the hypotenuse $AB$ of a right triangle $ABC$ such that $|CB| = |CD|$. Let $O$ be the circumcenter of triangle $ACD$. Rays $OD$ and $CB$ intersect at point $P$, and the line through point $O$ perpendicular to side $AB$ and ray $CD$ intersect at point $Q$. Points $A$, $C$, $P$, $Q$ are concyclic. Does this imply that $ACPQ$ is a square?",
    "options": [],
    "answer": "See solution",
    "solution": "As $OQ$ is the perpendicular bisector of $AD$, one has $\\angle QAD = \\angle ADQ = \\angle BDC = \\angle CBD$ (see figure below). Therefore $AQ \\parallel BC$, whence $\\angle QAC = 180^\\circ - \\angle ACB = 90^\\circ$. From the cyclic quadrilateral $APCQ$ one also gets $\\angle CPQ = \\angle PQA = 90^\\circ$, i.e., $ACPQ$ is a rectangle.\n\nAs $\\angle DOC = 2\\angle DAC$, one obtains\n\n$$\n\\begin{aligned}\n\\angle DOC &= 2\\angle BAC = 2(90^\\circ - \\angle CBA) = 180^\\circ - 2\\angle CBA = \\\\\n            &= 180^\\circ - \\angle CBD - \\angle BDC = \\angle DCB,\n\\end{aligned}\n$$\n\nwhich implies that isosceles triangles $BDC$ and $DCO$ are similar. Thus $\\angle BDC = \\angle DCO$, i.e., $OC \\parallel AB$, whence $\\angle QOC = 90^\\circ = \\angle QAC$. So $O$ lies on the circle determined by $A$, $C$, $P$, $Q$. Therefore\n\n$$\n\\angle ACQ = \\angle AOQ = \\frac{1}{2}\\angle AOD = \\frac{1}{2}\\angle AOP = \\frac{1}{2}\\angle ACP.\n$$\n\nConsequently, the diagonal of the rectangle $ACPQ$ bisects the angle of the rectangle, whence $ACPQ$ is a square.\n\n![](images/estonian-2012-2013_p26_data_ed4677e129.png)\n\n_Remark_: It turns out from the solution that the conditions of the problem determine the shape of the triangle $ABC$, namely $\\angle ABC = \\frac{3}{8}\\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18594,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence\n\n$$\na_n = 2^n + 3^{n+2} + 5^{n+1}\n$$\n\nfor $n \\ge 1$. Prove that there are infinitely many prime numbers such that each one of them divides infinitely many terms of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the set of all primes $p \\ne 2, 3, 5$ for which $2$ and $5$ are quadratic residues modulo $p$ but $3$ is not. If $p \\in S$, then by Euler's criterion we have\n\n$$\n2^{\\frac{p-1}{2}} \\equiv 5^{\\frac{p-1}{2}} \\equiv 1 \\pmod{p} \\quad \\text{and} \\quad 3^{\\frac{p-1}{2}} \\equiv -1 \\pmod{p}.\n$$\n\nSo for $n = k(p-1) + \\frac{p-1}{2} + 1$, where $k \\in \\mathbb{N}$, by Fermat's Little Theorem we have\n\n$$\n\\begin{aligned}\n2^n &\\equiv 2^{k(p-1)+\\frac{p-1}{2}+1} \\equiv (2^{p-1})^k \\cdot 2^{\\frac{p-1}{2}} \\cdot 2 \\equiv 2 \\pmod{p}, \\\\\n3^{n+2} &\\equiv 3^{k(p-1)+\\frac{p-1}{2}+3} \\equiv (3^{p-1})^k \\cdot 3^{\\frac{p-1}{2}} \\cdot 3^3 \\equiv -27 \\pmod{p}, \\\\\n5^{n+1} &\\equiv 5^{k(p-1)+\\frac{p-1}{2}+2} \\equiv (5^{p-1})^k \\cdot 5^{\\frac{p-1}{2}} \\cdot 5^2 \\equiv 25 \\pmod{p}.\n\\end{aligned}\n$$\n\nTherefore $p$ divides $a_n$ for infinitely many values of $n$.\n\nIt remains to show that $S$ contains infinitely many terms. By Dirichlet's Theorem about primes in arithmetic progressions, there are infinitely many primes $p$ with $p \\equiv 41 \\pmod{120}$. We claim that every such $p$ belongs to $S$. (Other congruence classes also work.) Indeed this follows since\n\n$$\n\\bullet\\ p \\equiv 41 \\pmod{120} \\implies p \\equiv 1 \\pmod{8} \\implies \\left(\\frac{2}{p}\\right) = +1\n$$\n\n$$\n\\bullet\\ p \\equiv 41 \\pmod{120} \\implies \\left\\{ \\begin{array}{l} p \\equiv 1 \\pmod{4} \\\\ p \\equiv 2 \\pmod{3} \\end{array} \\right\\} \\implies \\left(\\frac{3}{p}\\right) = \\left(\\frac{p}{3}\\right) = -1\n$$\n\n$$\n\\bullet\\ p \\equiv 41 \\pmod{120} \\implies p \\equiv 1 \\pmod{5} \\implies \\left(\\frac{5}{p}\\right) = \\left(\\frac{p}{5}\\right) = +1\n$$\n\n![](images/BMO2024Shortlist_p69_data_9c2f88ebd5.png)\n\n**Comment.** Note that the following problem appeared at Italia Olympiad 2010: https://artofproblemsolving.com/community/c6h466737p2613923\n\nWe also note that the following result has been proven by George Polya:",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18595,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$. Let $A$ be a $p \\times p$ matrix whose entries are exactly $1, 2, \\dots, p^2$ in some order. The following operation is allowed: add one to each number in a row or a column, or subtract one from each number in a row or a column. The matrix $A$ is called *good* if one can perform a finite sequence of such operations to obtain a matrix with all entries zero. Find the number of good matrices $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We may combine the operations on the same row or column, so the result of a sequence of operations can be realized as subtracting integers $x_i$ from each number in the $i$-th row and $y_j$ from each number in the $j$-th column. Thus, $A$ is good if and only if there exist integers $x_i, y_j$ such that $a_{ij} = x_i + y_j$ for all $1 \\leq i, j \\leq p$.\n\nSince the entries of $A$ are distinct, $x_1, x_2, \\dots, x_p$ are pairwise distinct, and so are $y_1, y_2, \\dots, y_p$. We may assume $x_1 < x_2 < \\dots < x_p$ since swapping $x_i$ and $x_j$ corresponds to swapping rows, which yields another good matrix. Similarly, we may assume $y_1 < y_2 < \\dots < y_p$, so the matrix increases from left to right and from top to bottom.\n\nFrom these assumptions, $a_{11} = 1$, and $a_{12}$ or $a_{21}$ equals $2$. We may consider only the case $a_{12} = 2$ since the transpose of the matrix is also good. Now, by contradiction, suppose the first row is $1, 2, \\dots, k$ but $k+1$ is not in the first row for $2 \\leq k < p$, so $a_{21} = k+1$. We call $k$ consecutive integers a *block*, and we will show the first row consists of several blocks: the first $k$ numbers is a block, the next $k$ numbers is a block, and so on.\n\nIf not, suppose the first $n$ groups of $k$ numbers are blocks, but the next $k$ numbers is not a block (or there are fewer than $k$ numbers left). Then for $j = 1, 2, \\dots, n$, $y_{(j-1)k+1}, \\dots, y_{jk}$ is a block, and the first $nk$ columns of the matrix can be divided into $pn$ $1 \\times k$ submatrices:\n\n$$\na_{i, (j-1)k+1}, a_{i, (j-1)k+2}, \\dots, a_{i, jk} \\\\\ni = 1, 2, \\dots, p, \\quad j = 1, 2, \\dots, n,\n$$\n\neach submatrix is a block. Now, assume $a_{1, nk+1} = a$, and let $b$ be the smallest positive integer such that $a+b$ is not in the first row, so $b \\leq k-1$. Since $a_{2, nk+1} - a_{1, nk+1} = x_2 - x_1 = a_{21} - a_{11} = k$, we have $a_{2, nk+1} = a + k$, so $a + b$ lies in the first $nk$ columns. Therefore, $a + b$ is in one of the $1 \\times k$ submatrices above, which is a block, but $a, a + k$ are not in this block, a contradiction.\n\nThus, the first row is formed by blocks, so $k \\mid p$, but $1 < k < p$ and $p$ is prime, which is impossible. So the first row is $1, 2, \\dots, p$, and the $k$-th row is $(k-1)p+1, (k-1)p+2, \\dots, kp$. Thus, up to interchanging rows, columns, and transposing, the good matrix is unique, so the answer is $2(p!)^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18596,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be distinct positive real numbers such that $a - \\sqrt{ab}$ and $b - \\sqrt{ab}$ are both rational numbers. Prove that $a$ and $b$ are rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Write\n$$\n\\frac{a-\\sqrt{ab}}{b-\\sqrt{ab}} = \\frac{\\sqrt{a}(\\sqrt{a}-\\sqrt{b})}{\\sqrt{b}(\\sqrt{b}-\\sqrt{a})} = -\\frac{\\sqrt{a}}{\\sqrt{b}}.\n$$\nSince the ratio of two rational numbers is also a rational number, there exists $q \\in \\mathbb{Q}$ such that $\\sqrt{a} = q\\sqrt{b}$. Notice that $q \\neq 1$, for otherwise $a = b$.\n\nThen $b - \\sqrt{ab} = b(1-q)$ is a rational number, and so is $b$. Since $a = q^2b$, the number $a$ is rational as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18597,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $AB$ 為圓 $O$ 上的弦，$M$ 為 $AB$ 劣弧的中點。由圓 $O$ 外一點 $C$ 向圓 $O$ 引切線，設切點分別為 $S, T$。令線段 $MS$ 與線段 $AB$ 的交點為 $E$，線段 $MT$ 與線段 $AB$ 的交點為 $F$。由 $E$ 點作 $AB$ 線段的垂線，交 $OS$ 於 $X$ 點；由 $F$ 點作 $AB$ 線段的垂線，交 $OT$ 於 $Y$ 點。另外再由 $C$ 點向圓 $O$ 引一割線，設兩交點分別為 $P, Q$。設線段 $MP$ 與線段 $AB$ 交於 $R$ 點。令 $\\triangle PQR$ 的外心為 $Z$ 點。\n\n證明：$X, Y, Z$ 三點共線。\n\nLet $AB$ be a chord on a circle $O$, $M$ be the midpoint of the smaller arc $AB$.\nFrom a point $C$ outside the circle $O$ draw two tangents to the circle $O$ at the points $S$ and $T$. Suppose $MS$ intersects with $AB$ at the point $E$, $MT$ intersects with $AB$ at the point $F$. From $E, F$ draw a line perpendicular to $AB$ that intersects with $OS, OT$ at the points $X, Y$, respectively. Draw another line from $C$ which intersects with the circle $O$ at the points $P$ and $Q$. Let $R$ be the intersection point of $MP$ and $AB$. Finally, let $Z$ be the circumcenter of $\\triangle PQR$.\n\nProve that $X, Y$, and $Z$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "作 $AB$ 的中垂線 $OM$。故 $\\triangle XES \\sim \\triangle OMS$，於是 $SX = XE$。\n\n![](images/16-1J_p3_data_df5c0c85f8.png)\n\n畫以 $XE$ 為半徑的圓 $X$。圓 $X$ 與弦 $AB$ 及直線 $CS$ 均相切。又作 $\\triangle PQR$ 的外接圓，以及直線 $MA$ 與 $MC$，如圖所示。\n\n因為 $\\triangle AMR \\sim \\triangle PMA$，所以有\n\n$$\nMR \\cdot MP = MA^2 = ME \\cdot MS.\n$$\n\n又由圓幂定理知 $CQ \\cdot CP = CS^2$。故 $M, C$ 兩點皆位於圓 $Z$ 與圓 $X$ 的根軸上，得 $ZX \\perp MC$。同理可知 $ZY \\perp MC$。所以 $X, Y, Z$ 三點共線，得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18598,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a monotonic function and $F: \\mathbb{R} \\to \\mathbb{R}$ given by\n$$\nF(x) = \\int_{0}^{x} f(t) \\, dt.\n$$\nProve that if $F$ has a finite derivative, then $f$ is continuous.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be increasing and $x_0 \\in \\mathbb{R}$. Then the lateral limits $f(x_0 - 0)$ and $f(x_0 + 0)$ of $f$ at $x_0$ exist, are real, and $f(x_0 - 0) \\le f(x_0) \\le f(x_0 + 0)$. Let $x < x_0$. Since $f(t) \\le f(x_0 - 0)$ for $x \\le t < x_0$, it follows\n$$\n\\int_{x}^{x_0} f(t) \\, dt \\le (x_0 - x)f(x_0 - 0).\n$$\nTherefore,\n$$\n\\frac{F(x) - F(x_0)}{x - x_0} = \\frac{\\int_{x}^{x_0} f(t) \\, dt}{x_0 - x} \\le f(x_0 - 0).\n$$\nSince $F$ is differentiable at $x_0$, it follows that $F'(x_0) \\le f(x_0 - 0)$. Analogously, $F'(x_0) \\ge f(x_0 + 0)$. Therefore $f(x_0 - 0) = f(x_0 + 0) = f(x_0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18599,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right triangle with $\\angle ACB = 90^\\circ$, $AC = 1$, and $BC = 2$. Given a point $A_1 \\in BC$ such that $A_1C \\neq \\frac{1}{3}$, we construct a sequence of points $A_n \\in BC$, $n \\ge 2$, as follows:\n\nLet $B_1$ be the intersection point of $AC$ and the line through $A_1$ parallel to $AB$, and let $C_1$ be the foot of the perpendicular from $B_1$ to $AB$. Then $A_2$ is the intersection point of $BC$ and the line through $C_1$ parallel to $AC$. Using $A_2$, we construct $A_3$ in the same way, and so on. Find:\n\na) $\\dfrac{3A_2C - 1}{3A_1C - 1}$;\n\nb) $\\displaystyle\\lim_{n \\to \\infty} S_{A_n B_n C_n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Set $A_nC = x_n$. From $\\triangle A_nB_nC \\sim \\triangle BAC$, we have $B_nC = \\dfrac{x_n}{2}$. Since $\\triangle AB_nC_n \\sim \\triangle ABC$, we find $AC_n = \\dfrac{2 - x_n}{2\\sqrt{5}}$. Hence, $A_{n+1}C = x_{n+1} = \\dfrac{2 - x_n}{5}$, and therefore\n$$\n\\frac{3x_{n+1} - 1}{3x_n - 1} = -\\frac{1}{5}.\n$$\n\nb) Since $A_nB_n = \\dfrac{\\sqrt{5}}{2} x_n$, $B_nC_n = \\dfrac{2 - x_n}{\\sqrt{5}}$, and $A_nB_n \\perp B_nC_n$, we get\n$$\nS_{A_nB_nC_n} = \\frac{x_n (2 - x_n)}{4}.\n$$\nBut part (a) implies that $\\{x_n - \\frac{1}{3}\\}_{n=1}^\\infty$ is a geometric progression with ratio $-\\frac{1}{5}$. Hence,\n$$\n\\lim_{n \\to \\infty} x_n = \\frac{1}{3}\n$$\nand therefore\n$$\n\\lim_{n \\to \\infty} S_{A_nB_nC_n} = \\frac{5}{36}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18600,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider sequences $a_1, a_2, \\dots, a_n$ consisting of $n$ integers. For given $k \\le n$, we can partition the numbers of the sequence into $k$ groups as follows: $a_1$ goes in the first group, $a_2$ in the second group, and so on until $a_k$ which goes in the $k$-th group. Then $a_{k+1}$ goes in the first group again, $a_{k+2}$ in the second group, and so on. The sequence is called $k$-composite if this partition has the property that the sums of the numbers in the $k$ groups are equal.\n\nThe sequence $1, 2, 3, 4, -2, 6, 13, 12, 17, 8$, for instance, is 4-composite as\n\n$$\n1 + (-2) + 17 = 2 + 6 + 8 = 3 + 13 = 4 + 12.\n$$\n\nHowever, this sequence is not 3-composite, as the sums $1 + 4 + 13 + 8$, $2 + (-2) + 12$, and $3 + 6 + 17$ do not give equal outcomes.\n\n(a) Give a sequence of 6 *distinct* integers that is both 2-composite and 3-composite.\n\n(b) Give a sequence of 7 *distinct* integers that is 2-composite, 3-composite, and 4-composite.\n\n(c) Find the largest $k \\le 99$ for which there exists a sequence of 99 *distinct* integers that is $k$-composite. (Give an example of such a sequence and prove that such a sequence does not exist for greater values of $k$.)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) An example of a correct sequence is $5, 7, 6, 3, 1, 2$. This sequence consists of six distinct numbers and is 2-composite since $5 + 6 + 1 = 7 + 3 + 2$. It is also 3-composite since $5 + 3 = 7 + 1 = 6 + 2$.\n\n*This is just one example out of many possible correct solutions. Below we describe how we found this solution.*\n\nWe are looking for a sequence $a_1, a_2, a_3, a_4, a_5, a_6$ that is 2-composite and 3-composite. Hence, we need that\n\n$$\na_1 + a_4 = a_2 + a_5 = a_3 + a_6 \\quad \\text{and} \\quad a_1 + a_3 + a_5 = a_2 + a_4 + a_6.\n$$\n\nIf we choose $a_4 = -a_1$, $a_5 = -a_2$, and $a_6 = -a_3$, then the first two equations hold. The third equation gives us $a_1 + a_3 - a_2 = a_2 - a_1 - a_3$, and therefore $a_1 + a_3 = a_2$. We choose $a_1 = 1$, $a_3 = 2$ (and therefore $a_2 = 3$). We obtain the sequence $1, 3, 2, -1, -3, -2$ consisting of six distinct integers. If we wish to do so, we can increase all six numbers by 4 to get a solution with only positive numbers: $5, 7, 6, 3, 1, 2$.\n\n(b) A possible solution is $8, 17, 26, 27, 19, 10, 1$. This sequence consists of seven distinct integers and is 2-composite since $8 + 26 + 19 + 1 = 17 + 27 + 10$. It is 3-composite since $8 + 27 + 1 = 17 + 19 = 26 + 10$. It is also 4-composite since $8 + 19 = 17 + 10 = 26 + 1 = 27$.\n\n*This is just one example out of many possible correct solutions. Below we describe how we found this solution.*",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18601,
    "subject": "Mathematics (Olympiad)",
    "question": "Given real numbers $a$, $b$, $c$ such that:\n\n$$\n\\begin{aligned}\nab - c &= 1000 \\\\\nbc - a &= 1018 \\\\\nca - b &= -2018\n\\end{aligned}\n$$\n\nShow that $a + b + c \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that $a + b + c = 0$.\n\nThen:\n$$\nab + bc + ca = 0.\n$$\n\nSuppose $abc = 0$. Without loss of generality, let $c = 0$. Then $ab = 0$, so at least two variables are zero. This contradicts the given conditions. Thus, $abc \\neq 0$.\n\nSubstitute $c = -a - b$ into $ab + bc + ca = 0$:\n\n$$\n\\begin{aligned}\n(a + b)b + (a + b)a - ab &= 0 \\\\\na^2 + ab + b^2 &= 0 \\\\\na^2 + ab + \\frac{1}{4}b^2 + \\frac{3}{4}b^2 &= 0 \\\\\n(a + \\frac{1}{2}b)^2 + \\frac{3}{4}b^2 &= 0 \\\\\na = b = 0\n\\end{aligned}\n$$\n\nThis is a contradiction.\n\n![](images/Ukraine_booklet_2018_p20_data_8491ae1c03.png)\n\n**Fig. 18**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18602,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Determine if there exist positive integer numbers $x, y, z$ such that\n\n$$\n2016 = x^3 + y^3 + z^3\n$$\n\nb) Determine if there exist positive integer numbers $x, y, z, t$ such that\n\n$$\n2016 = x^3 + y^3 + z^3 + t^3\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "b) It is enough to provide an example: $2016 = 1000 + 1000 + 8 + 8$.\n\na) Let us first note that $2016 = 2^5 \\cdot 3^2 \\cdot 7$. Consider the remainders of $a^3, b^3, c^3$ modulo $7$. The possible remainders are $0$ or $\\pm 1$. Thus, if $2016 = x^3 + y^3 + z^3$, then at least one of the numbers is divisible by $7$. It is easy to see that it must be $7$, because $7^3 = 343 < 2016$, and the next number $14^3 = 2744 > 2016$. Thus, the other two numbers must satisfy $y^3 + z^3 = 1673$.\n\nNow, suppose $y \\leq z$. Since $12^3 = 1728 > 1673$, then $z \\leq 11$. Moreover, $1673 = y^3 + z^3 \\leq 2z^3$, so $837 \\leq z^3$, which means $10 \\leq z$. To finish the solution, we just need to check the cases $z = 10, 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18603,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 14 students. Each student is grouped by first name and by last name, forming two partitions of the set of students. There is exactly one group of each size from 1 to 7 (i.e., groups of sizes 1, 2, 3, 4, 5, 6, and 7), and no other group sizes. Prove that there must be at least two students who share both the same first name and the same last name.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the group of 7 students with the same first name. Since there are only 14 students and groups of each size from 1 to 7, there are at most 6 groups by last name. By the Pigeonhole Principle, at least two students in the group of 7 must also share the same last name. Therefore, there exist at least two students with both the same first and last name.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18604,
    "subject": "Mathematics (Olympiad)",
    "question": "We want to show that if $G - v$ has a Hamiltonian cycle for each vertex $v$ in a graph $G$, but $G$ does not have a Hamiltonian cycle, then $n = |G| \\ge 10$, and produce such a graph $G$ with $n = 10$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $v$ cannot be adjacent with two consecutive vertices in a Hamiltonian cycle in $G - v$, we have $\\deg v \\le \\left\\lfloor \\frac{n-1}{2} \\right\\rfloor$. On the other hand, if $\\deg w \\le 2$ for some vertex $w$, then there is no Hamiltonian cycle in $G - v$ where $v$ is adjacent with $w$. Therefore $3 \\le \\deg v \\le \\left\\lfloor \\frac{n-1}{2} \\right\\rfloor$. In particular, $n \\ge 7$.\n\nNext, observe that if $v_1 \\to v_2 \\to \\dots \\to v_{n-1}$ is a Hamiltonian cycle in $G - v$, and if $v$ is adjacent with $v_i$ and $v_j$, $i < j$, then $v_{i-1}$ and $v_{j-1}$ cannot be adjacent with each other, as that would give a Hamiltonian cycle $v_1 \\to \\dots \\to v_{i-1} \\to v_{j-1} \\to v_{j-2} \\to \\dots \\to v_i \\to v \\to v_j \\to v_{j+1} \\to \\dots \\to v_{n-1}$ in $G$.\n\nFrom these observations, it follows that $n = 7$ and $n = 8$ are impossible. If $n = 9$, then each vertex has degree 3 or 4, and they cannot all have degree 3 by the degree sum formula. Hence, assume that $n = 9$ and there is a vertex $v_0$ with $\\deg v_0 = 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18605,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n \\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.\n\nLet $s$ be the smallest number of collisions after which all particles return to their original positions.\n\nFind $s$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $(n-1)m'$, where $m'$ is the smallest number such that $\\frac{m'(n-2)}{2n}$ is an integer.\n\nMore precisely, the answer is:\n\n- $2n(n-1)$ when $n$ is odd\n- $n(n-1)$ when $n$ is divisible by $4$\n- $\\frac{n(n-1)}{2}$ when $n-2$ is divisible by $4$ (but the $n/2$-th point will be the one moving in reverse, not $p_0$)\n\nWe first introduce some setup for convenience. We treat points on the circle as $[0, 2n)$ and we use $r = r + 2n$ for any real $r$ to refer to points on the circle for convenience. For example, we can use point $-1$ to mean $2n - 1$, etc. Now, we also assume that it takes any particle $2n$ units of time to go around the entire circle. Thus, if a particle at $0$ moves clockwise for time $t$, then it would reach point $t$.\n\nInitially, we let the particles be $p_0, \\dots, p_{n-1}$ with $p_i$ at point $2i$.\n\nNow, moving clockwise means the value is increasing and moving anti-clockwise means the value is decreasing, and finally we assume $p_0$ is the point initially moving anti-clockwise.\n\nNow, let $t_1 > 0$ be the total time when, for the first time, all particles return to their initial positions.\nCorrespondingly, let $t_2 > 0$ be the first time when all particles are equally spaced apart, i.e., $2$ units apart.\n\nNow, clearly $\\frac{t_1}{t_2}$ is an integer. Thus, we try to find $t_2$.\n\nObserve that if we replace each collision event with the two particles passing through each other, $t_2$ does not change. So for the purposes of calculating $t_2$, we can assume they indeed pass through. Thus, at time $t_2$, there are particles at $-t_2, 2 + t_2, 4 + t_2, \\dots, 2n - 2 + t_2$. But for them to be equally spaced, this sequence must be $t_2, 2 + t_2, 4 + t_2, \\dots, 2n - 2 + t_2$. Thus, modulo $2n$, $t_2 = -t_2$, and the minimum value of $t_2$ that makes this possible is $t_2 = n$. Now, we can set $t_1 = n \\cdot t'_1$.\n\nAlternatively, one could have observed that we can look at the positions relative to the clockwise moving particles. Then we just have the anti-clockwise particle moving at speed $2$, so it must need $n$ units of time to return to the original position and make $n-1$ collisions in this period.\n\nNow, let us analyze what happens at time $n$. There are now particles in positions $n, n+2, \\dots, n-2$. Observe that the cyclic order of particles must be preserved as collisions never alter it, and finally the collisions only happen in $(1, n)$ after the first collision of $p_0$ at point $2n - 1$ with $p_n$. Thus, $p_0$ does not have any more collisions in the next $n-1$ units of time. Thus, $p_0$ must be at position $2n - 1 + (n-1) = n - 2$. Now, since cyclic order is preserved, $p_i$ would be at position $2i + n - 2$. Thus, everything has cyclically moved forward by $n-2$.\n\nThus, in time $n$, every particle moves forward by $n-2$ units and there are $n-1$ collisions. Thus, in time $nk$, we have $(n-1)k$ collisions and every particle moves forward by $(n-2)k$ units. Thus, all particles return to their initial positions if and only if $2n \\mid (n-2)k$. This is exactly what we desired! $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18606,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(G, \\cdot)$ be a group with 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 under the group operation. Show that $x^2 = e$ for any $x \\in G$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L = (G \\setminus (H \\cup K)) \\cup \\{e\\}$. Since $x \\in H \\cup K$ if and only if $x^{-1} \\in H \\cup K$, it follows that $x \\in L$ if and only if $x^{-1} \\in L$, so $L$ is a proper subgroup of $G$.\n\nAlso, $L \\cap H = L \\cap K = H \\cap K = \\{e\\}$, and $G = H \\cup K \\cup L$. 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\nThus, for $a \\in A \\setminus \\{e\\}$ and $b \\in B \\setminus \\{e\\}$, $a^2b = a(ab) \\in B \\setminus \\{e\\}$, so $a^2 \\in A \\cap B = \\{e\\}$. Hence, $x^2 = e$ for any $x \\in G \\setminus \\{e\\}$.\n\nSince $e^2 = e$, it follows that $x^2 = e$ for any $x \\in G$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18607,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the tangent from $B$ to the circle $K$ (distinct from the tangent $AB$) touch the circle $K$ at $A'$. Show that the points $B$, $A'$, and $E$ are collinear.\n\n![](images/Slovenija_2016_p13_data_a38ce66acf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We have $|AB| = |A'B|$ and $|AO| = |A'O|$, so the triangles $ABO$ and $A'BO$ have three equal sides and are therefore congruent. This implies $\\angle OBA' = \\angle ABO$. By the Tangent-Chord Theorem in the circle $K'$, we have $\\angle ABO = \\angle BDO$. The line $AB$ is tangent to $K$, so $\\angle OAB = 90^\\circ$. We have $|AB| = |AC|$, so the triangles $ABD$ and $ACD$ match in two sides and the angle between them. We conclude that they are congruent and\n\n$$\n\\angle BDO = \\angle BDA = \\angle ADC = \\angle ODE.\n$$\n\nThe angles over the same chord of $K'$ are equal, so $\\angle ODE = \\angle OBE$.\n\nWe have shown that $\\angle OBA' = \\angle OBE$, which implies that the points $B$, $A'$, and $E$ are collinear and the line $BE$ is tangent to $K$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18608,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a good point on $x = 1$ or $x = 2000$ excluding $(1, 1)$ a special point. What is the smallest possible number of Z-shaped polylines $ABCD$ (with $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$, and $1 \\le x_1 < x_2 \\le 2000$, $1 \\le x_3 < x_2 \\le 2000$, $1 \\le x_3 < x_4 \\le 2000$) needed so that every special point lies on at least one such polyline?",
    "options": [],
    "answer": "See solution",
    "solution": "Any special point on a Z-shaped polyline $ABCD$ coincides with either $A$, $B$, $C$, or $D$. Assume both $B$ and $C$ are special points. Then $x_2 = 2000$, $y_2 \\le 2000$, $x_3 = 1$, and $y_3 \\ge 2$. Therefore, $y_2 - x_2 \\le 2000 - 2000 < 2 - 1 \\le y_3 - x_3$, which contradicts the condition $y_2 - x_2 = y_3 - x_3$. Thus, at most three special points lie on a Z-shaped polyline, so at least $\\frac{3999}{3} = 1333$ polylines are needed.\n\nDefine polylines $X_1, \\ldots, X_{666}$, $Y_1, \\ldots, Y_{666}$, and $Z$ as follows:\n\n- For $k = 1, \\ldots, 666$, let $X_k$ be $(1, 1334-k) - (1334-2k, 1334-k) - (1, 1+k) - (2000, 1+k)$.\n- For $k = 1, \\ldots, 666$, let $Y_k$ be $(1, 2000-k) - (2000, 2000-k) - (667+2k, 667+k) - (2000, 667+k)$.\n- Let $Z$ be $(1, 2000) - (2000, 2000) - (1, 1) - (2000, 1)$.\n\nAny good point on $y = 1$ or $y = 2000$ lies on $Z$. For $2 \\le k \\le 667$, any good point on $y = k$ lies on $X_{k-1}$. For $1334 \\le k \\le 1999$, any good point on $y = k$ lies on $Y_{2000-k}$. For $668 \\le k \\le 1333$ and good points on $y = k$:\n\n- When $1 \\le x < 2k - 1333$, $(x, k)$ lies on $X_{1334-k}$.\n- When $2k - 1333 \\le x < k$, $(x, k)$ lies on $X_{k-x}$.\n- When $x = k$, $(x, k)$ lies on $Z$.\n- When $k < x \\le 2k - 668$, $(x, k)$ lies on $Y_{x-k}$.\n- When $2k - 668 < x \\le 2000$, $(x, k)$ lies on $Y_{k-667}$.\n\nThus, every good point lies on one of $X_1, \\ldots, X_{666}$, $Y_1, \\ldots, Y_{666}$, or $Z$. Therefore, the smallest possible number of Z-shaped polylines is $1333$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18609,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and let $D$, $E$, and $F$ be the feet of the perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively. Let $P$, $Q$, $R$, and $S$ be the feet of the perpendiculars from $D$ to $BA$, $BE$, $CF$, and $CA$ respectively. Prove that $P$, $Q$, $R$, and $S$ are collinear.\n\n![](images/British_Booklet_2016_p5_data_bab0ef6c11.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $AD$, $CF$, and $BE$ coincide at $H$, the orthocentre. We also have\n\n$$\n\\angle ADB = \\angle HDB = 90^\\circ, \\\\\n\\angle CFB = \\angle HFB = 90^\\circ\n$$\n\nso $DHFB$ is cyclic. Applying Simson's theorem from $D$ to $\\angle FHB$, we have that $P$, $Q$, $R$ are collinear. Similarly, $Q$, $R$, $S$ are collinear by applying Simson's theorem to $\\angle HCE$, again since $D$, $C$, $E$, $H$ are concyclic.\n\nTherefore $P$, $Q$, $R$, and $S$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18610,
    "subject": "Mathematics (Olympiad)",
    "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 $AM$ meets the line $AB$ at point $P$. Given that lines $PM$ and $BC$ are perpendicular, prove that $\\triangle APM$ is an equilateral triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Lines $BC$ and $PM$ meet at $D$. Denote $x$ as 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 both the median and the 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![](images/RMC_2015_BT_p9_data_42be8abd8a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18611,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $k > 1$ un entero. Determina el menor entero positivo $n$ tal que algunas casillas de un tablero de $n \\times n$ se pueden pintar de negro de modo que en cada fila y en cada columna haya exactamente $k$ casillas negras, y además, las casillas negras no compartan ni un lado ni un vértice con otra casilla negra.\n\n**Aclaración:** Hay que responder $n$ en función de $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observamos que todo subtablero de $2 \\times 2$ puede tener como máximo una casilla negra. Consideremos dos filas consecutivas del tablero. Entre las dos deben tener, en total, exactamente $2k$ casillas negras. Dividiendo las casillas de dos filas en cuadrados de $2 \\times 2$, comenzando desde la izquierda, vemos que la cantidad de estos debe ser al menos $2k-1$ (la última casilla negra puede estar sola en la última columna de la derecha).\n\nPor lo tanto, $n \\ge 2(2k-1)+1 = 4k-1$.\n\nSupongamos que $n = 4k-1$. En cualesquiera dos filas consecutivas, puede haber a lo sumo $2k-1$ casillas negras en las $4k-2$ columnas de la izquierda. Luego, una casilla negra debe estar en la última columna de la derecha. Pero entonces, en cada par de filas, la penúltima columna no puede tener casillas negras. Como esto vale para todo par de filas consecutivas del tablero de $n \\times n$, resulta que el tablero tiene una columna sin casillas negras. Esto contradice la hipótesis de que todas las columnas deben tener exactamente $k$ casillas negras. Por lo tanto, $n \\ge 4k$.\n\nDamos un ejemplo para $n = 4k$. Dividimos el tablero en 4 subtableros de $2k \\times 2k$. En el de arriba a la izquierda, se colorean de negro las casillas de las filas impares y columnas pares. Luego, rotando $90^\\circ$ el cuadrado de $2k \\times 2k$, se copian las casillas negras en el cuadrado superior derecho y se repite lo mismo dos veces más. En la figura mostramos el ejemplo para $k = 3$.\n\n![](images/Soluciones_nacional_OMA_2020_2_p6_data_61328b755f.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18612,
    "subject": "Mathematics (Olympiad)",
    "question": "In a circle $O$, there are six points $A, B, C, D, E, F$ in counterclockwise order. $BD \\perp CF$, and $CF$, $BE$, $AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$, and the perpendicular from $D$ to $CE$ be $N$. Prove that $AE \\parallel MN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the concurrent point of $CF$, $BE$, and $AD$, and let $CF$ intersect $BD$ at $L$.\n\n![](images/Saudi_Arabia_booklet_2021_p17_data_cb7b4b8c31.png)\n\nWe have $\\angle BMC = 90^\\circ = \\angle BLC$ since $BM$ is perpendicular to $AC$ and $BL$ is perpendicular to $CF$, so $B$, $M$, $L$, $C$ lie on the same circle with diameter $BC$. Hence,\n\n$$\n\\angle CML = \\angle CBL = \\angle CBD = \\angle CAD\n$$\n\nimplying that $ML$ is parallel to $AK$. Similarly, we also have $LN$ is parallel to $EK$. Thus,\n\n$$\n\\frac{CM}{CA} = \\frac{CL}{CK} = \\frac{CN}{CE}\n$$\n\nThis implies that $MN$ is parallel to $AE$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18613,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ that satisfy the inequalities:\n\n$$\n-46 \\leq \\frac{2023}{46-n} \\leq 46-n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 46 - n$ and find the corresponding integer values of $x$ that satisfy $-46 \\leq \\frac{2023}{x} \\leq x$.\n\n**Case 1.** $x > 0$.\n\nThe left inequality holds for all such $x$, and the right can be rewritten as:\n\n$$\n2023 \\leq x^2 \\implies x \\geq 45 \\implies 46 - n \\geq 45 \\implies n \\leq 1 \\implies n = 1.\n$$\n\n**Case 2.** $x < 0$.\n\nThe inequalities become:\n\n$$\nx^2 \\leq 2023 \\leq -46x \\implies -44 \\leq x \\leq 44 \\text{ and } x \\leq -44 \\implies x = -44 \\implies 46 - n = -44 \\implies n = 90.\n$$\n\nIt is easy to verify that these two values satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18614,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer. Prove that there exist positive integers $x_1, \\dots, x_n$ in geometric progression and positive integers $y_1, \\dots, y_n$ in arithmetic progression such that\n\n$$\nx_1 < y_1 < x_2 < y_2 < \\dots < x_n < y_n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the binomial theorem, for $k \\ge 2$ and $a \\le 1/k^2$,\n\n$$\n\\begin{align*}\n(1+a)^k &= 1+ka+a \\left( \\frac{ak(k-1)}{2!} + \\frac{a^2k(k-1)(k-2)}{3!} + \\dots + \\frac{a^{k-1}k!}{k!} \\right) \\\\\n&\\le 1+ka+a \\left( \\frac{k(k-1)}{k^2} + \\frac{1}{2!} + \\frac{k(k-1)(k-2)}{k^4} \\frac{1}{3!} + \\dots + \\frac{k!}{k^2k^{k-2}k!} \\right) \\\\\n&\\le 1+ka+a \\left( \\frac{1}{2!} + \\dots + \\frac{1}{k!} \\right) < 1+ka+a \\left( \\frac{1}{1 \\cdot 2} + \\dots + \\frac{1}{(k-1)!} \\right) \\\\\n&\\le 1+ka+a \\left( 1-\\frac{1}{k} \\right) < 1+(k+1)a.\n\\end{align*}\n$$\n\nLet $X_k = \\left(1 + \\frac{k}{n^2}\\right)^k$, $k = 1, \\dots, n$. Then, for $2 \\le k \\le n$, $\\frac{1}{n^2} \\le \\frac{1}{k^2}$. Therefore,\n\n$$\n1 + \\frac{k}{n^2} < X_k < 1 + \\frac{k+1}{n^2}.\n$$\n\nMultiplying throughout by $n^{2n}$, we have\n\n$$\nn^{2n} + k n^{2n-2} < n^{2n} \\left( 1 + \\frac{1}{n^2} \\right)^k < n^{2n} + (k+1)n^{2n-2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18615,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n \\ge 3$, let $a_1, a_2, \\dots, a_{2n}$ and $b_1, b_2, \\dots, b_{2n}$ be $4n$ non-negative real numbers satisfying\n\n$$\na_1 + a_2 + \\dots + a_{2n} = b_1 + b_2 + \\dots + b_{2n} > 0.\n$$\n\nFor any $i = 1, 2, \\dots, 2n$, it holds that $a_i a_{i+2} \\ge b_i + b_{i+1}$, where $a_{2n+1} = a_1$, $a_{2n+2} = a_2$, $b_{2n+1} = b_1$.\n\nFind the minimum value of $a_1 + a_2 + \\dots + a_{2n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = a_1 + a_2 + \\dots + a_{2n} = b_1 + b_2 + \\dots + b_{2n}$.\n\nWithout loss of generality, suppose $T = a_1 + a_3 + \\dots + a_{2n-1} \\le \\frac{S}{2}$.\n\nFor $n=3$, since\n\n$$\nT^2 - 3 \\sum_{k=1}^{3} a_{2k-1} a_{2k+1} = \\frac{1}{2} \\left( (a_1 - a_3)^2 + (a_3 - a_5)^2 + (a_5 - a_1)^2 \\right) \\ge 0,\n$$\n\nby combining the given conditions we can find\n\n$$\n\\frac{S^2}{4} \\ge T^2 \\ge 3 \\sum_{k=1}^{3} a_{2k-1} a_{2k+1} \\ge 3 \\sum_{k=1}^{3} (b_{2k-1} + b_{2k}) = 3S.\n$$\n\nSince $S > 0$, we have $S \\ge 12$.\n\n$S$ attains its minimum when $a_i = b_i = 2$ for $1 \\le i \\le 6$.\n\nWhen $n \\ge 4$, on one hand,\n\n$$\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\ge \\sum_{k=1}^{n} (b_{2k-1} + b_{2k}) = S.\n$$\n\nOn the other hand, if $n$ is even,\n\n$$\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\le (a_1 + a_5 + \\dots + a_{2n-3}) (a_3 + a_7 + \\dots + a_{2n-1}) \\le \\frac{T^2}{4}.\n$$\n\nThe first inequality is due to the fact that each term of $(a_1 + a_5 + \\dots + a_{2n-3})(a_3 + a_7 + \\dots + a_{2n-1})$ is nonnegative after expansion and contains terms $a_{2k-1} a_{2k+1}$ ($1 \\le k \\le n$). The second inequality uses the arithmetic-geometric mean inequality.\n\nIf $n$ is odd, suppose $a_1 \\le a_3$, then\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} a_{2k-1} a_{2k+1} &\\le \\left( \\sum_{k=1}^{n-1} a_{2k-1} a_{2k+1} \\right) + a_{2n-1} a_3 \\\\\n&\\le (a_1 + a_5 + \\dots + a_{2n-1})(a_3 + a_7 + \\dots + a_{2n-3}) \\\\\n&\\le \\frac{T^2}{4}.\n\\end{aligned}\n$$\n\nThus, $S \\le \\sum_{k=1}^{n} a_{2k-1} a_{2k+1} \\le \\frac{T^2}{4} \\le \\frac{S^2}{16}$. Since $S > 0$, $S \\ge 16$.\n\nWhen $a_1 = a_2 = a_3 = a_4 = 4$, $a_i = 0$ for $5 \\le i \\le 2n$, $b_1 = 0$, $b_2 = 16$, $b_i = 0$ for $3 \\le i \\le 2n$, $S$ attains the minimum $16$.\n\nTo sum up, when $n = 3$, the minimum of $S$ is $12$; when $n \\ge 4$, the minimum of $S$ is $16$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18616,
    "subject": "Mathematics (Olympiad)",
    "question": "Anu and Bert each play the following game with ChatGPT. Anu's game starts with the number $2023!$ on the board, Bert's game with $2024!$ on the board. Each move consists of two parts. First, the active player divides the number on the board by one of its composite factors $d$. Then their opponent must do one of the following actions:\n\n1. Multiply the number on the board by any factor $d'$ of $d$ satisfying $1 < d' < d$;\n2. Multiply the number on the board by $7$;\n3. Divide the number on the board by $10$ and multiply the result by $2023$ (this option cannot be chosen if the result is not an integer).\n\nOn the next move, players interchange their roles. The player who cannot make the required action loses the game. ChatGPT starts both games. Prove that ChatGPT can win at least one of the games.",
    "options": [],
    "answer": "See solution",
    "solution": "As $2024! = 2023! \\times 2024$ and $2024 = 2^3 \\times 11^1 \\times 23^1$, the sum of the exponents of all primes in the prime factorization of the number on the board is odd in one of the games and even in the other. We will show that ChatGPT can win the game where the sum is even.\n\nOn any of its moves, ChatGPT chooses $d$ to be a product of exactly two (not necessarily distinct) primes dividing the number on the board. Then, no matter what its opponent does, the sum of exponents will increase by $1$. This is obvious for the first two choices; for the third choice, we notice that $10 = 2 \\times 5$ has two prime factors and $2023 = 7 \\times 17^2$ has three prime factors. Thus, the move as a whole decreases the sum of exponents by $1$.\n\nWhatever the opponent chooses as $d$, ChatGPT chooses the first option with $d' = \\frac{d}{p}$, where $p$ is any prime factor of $d$. Thus, the move as a whole decreases the sum of exponents by $1$.\n\nContinuing with this strategy, ChatGPT ensures that at the beginning of each of its turns the sum of exponents is even. As the sum of exponents decreases by one during every move, there will eventually be a situation where ChatGPT's opponent will start its move with the sum of exponents being $1$, i.e., with a prime on the board. Thus, the player cannot make their move, meaning that ChatGPT wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18617,
    "subject": "Mathematics (Olympiad)",
    "question": "Let's consider all pairs of distinct points $A$ and $B$ on the Cartesian plane, each with integer coordinates. Among these pairs, find all those for which there exist two distinct points $X$ and $Y$, also with integer coordinates, such that the quadrilateral $AXBY$ is both convex and inscribed.\n\nA quadrilateral is called *convex* if both of its diagonals lie inside the quadrilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that for points $A$ and $B$ at a distance of $1$ from each other, there are no points $X$ and $Y$ that satisfy the condition. Indeed, suppose such points exist. Then $\\angle AXB + \\angle AYB = 180^\\circ$, so at least one of these angles is not less than $90^\\circ$. Therefore, at least one of the points $X$ or $Y$ must lie inside or on the circle with diameter $AB$, but this circle contains no other integer points except $A$ and $B$.\n\nNow, for all other pairs of points, such a pair $(X, Y)$ can be found. Let $A = (a_1, a_2)$ and $B = (b_1, b_2)$. If $a_1 \\ne b_1$ and $a_2 \\ne b_2$, take $X = (a_1, b_2)$ and $Y = (b_1, a_2)$. Then $AXBY$ is a rectangle, which is inscribed.\n\nOtherwise, assume $a_2 = b_2 = t$, so $A = (a_1, t)$ and $B = (b_1, t)$ with $|a_1 - b_1| > 1$. Assume $a_1 < b_1$. Take $X = (a_1 + 1, t - 1)$ and $Y = (a_1 + 1, t + (b_1 - a_1 - 1))$. The segments $AB$ and $XY$ intersect at $K = (a_1 + 1, t)$, and $AK \\cdot BK = XK \\cdot YK = 1 \\cdot (b_1 - a_1 - 1)$, so $AXBY$ is indeed inscribed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18618,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number with $0 \\leq a \\leq 1$. Prove that for any nonnegative integer $n$, the inequality\n$$(n+1)a \\leq n + a^{n+1}$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is equivalent to\n$$\nna - n \\leq a^{n+1} - a,\n$$\nor\n$$\nn(a - 1) \\leq a(a - 1)(a^{n-1} + a^{n-2} + \\dots + 1).\n$$\nIf $a = 1$, the inequality obviously holds. If $a < 1$, then $a - 1 < 0$, and dividing both sides by $a - 1$ gives the equivalent inequality\n$$\nn \\geq a(a^{n-1} + a^{n-2} + \\dots + 1),\n$$\nor\n$$\nn \\geq a^n + a^{n-1} + \\dots + a.\n$$\nSince $a < 1$, each term in the sum is less than $1$, so the sum does not exceed $n$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18619,
    "subject": "Mathematics (Olympiad)",
    "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.\n\nFind the remainder when $a + x + y + z + t$ is divided by $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, without loss of generality, that $x < y < z < t$. Then\n$$\n2^x + 2^y + 2^z + 2^t = 305 \\cdot 2^a,\n$$\nso\n$$\n2^x (1 + 2^{y-x} + 2^{z-x} + 2^{t-x}) = 305 \\cdot 2^a. \\tag{1}\n$$\nSince $x < y < z < t$, the numbers $y-x$, $z-x$, and $t-x$ are positive integers, so $2^{y-x}$, $2^{z-x}$, $2^{t-x}$ are even, and $1 + 2^{y-x} + 2^{z-x} + 2^{t-x}$ is odd.\n\nFor parity reasons, (1) yields $2^x = 2^a$ and $1 + 2^{y-x} + 2^{z-x} + 2^{t-x} = 305$, so $x = a$ and $2^{y-x} + 2^{z-x} + 2^{t-x} = 304$.\n\nThis gives\n$$\n2^{y-x} (1 + 2^{z-y} + 2^{t-y}) = 2^4 \\cdot 19,\n$$\nso $y-x = 4$, thus $y = a + 4$.\n\nMoreover, $1 + 2^{z-y} + 2^{t-y} = 19$, so\n$$\n2^{z-y} (1 + 2^{t-z}) = 2 \\cdot 9,\n$$\ntherefore $z-y = 1$ and $t-z = 3$, that is, $z = y + 1 = a + 5$ and $t = z + 3 = a + 8$.\n\nThus,\n$$\na + x + y + z + t = a + a + (a + 4) + (a + 5) + (a + 8) = 5a + 17 = 5(a + 3) + 2.\n$$\nSo the remainder when $a + x + y + z + t$ is divided by $5$ is $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18620,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F_n$ denote the set of configurations of $n$ points joined by line segments in such a way that no two of the line segments intersect inside the circle. If $f \\in F_n$, then the given $n$ points are said to be vertices of subconfiguration $f$.\n\nProve that it is possible to colour the vertices of $f$ with 3 colours so that the endpoints of any line segment joining vertices are of different colours.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction. The statement is trivial for $n \\leq 3$.\n\nAssume it is possible to colour the vertices of any subconfiguration of $f \\in F_{n-1}$ with 3 colours so that the endpoints of any line segment have different colours. For $f \\in F_n$, there exists a vertex of $f$ in which no more than 2 line segments have endpoints. Let $f'$ be the configuration formed after deleting this vertex and the line segments ending at it. Note that $f' \\in F_{n-1}$. By the induction hypothesis, it is possible to colour the vertices of $f'$ with 3 colours. Since the deleted vertex is incident to at most 2 edges, we can assign it a colour different from the colours of its neighbours.\n\nNow, consider $f \\in F_{148}$. By the above, it is possible to colour the vertices of $f$ with 3 colours. By the pigeonhole principle, at least $\\lceil 148/3 \\rceil = 50$ vertices have the same colour. Since these 50 vertices are not connected by any segment, we have the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18621,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ denote the number of ordered 9-tuples $(x_1, x_2, \\dots, x_9)$ of positive integers such that\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_9} = 1.\n$$\nDecide if $N$ is even or odd. Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "There are an even number of solutions $(x_1, x_2, \\dots, x_9)$ in which $x_1 \\neq x_2$, since such solutions can be paired by swapping $x_1$ and $x_2$. Thus, for parity, we may assume $x_1 = x_2$. Similarly, solutions with $x_1 = x_2$ and $x_3 \\neq x_4$ can be paired by swapping $x_3$ and $x_4$, so we may further assume $x_3 = x_4$. Continuing this process, we restrict to solutions with $x_1 = x_2 = x_3 = x_4$. The same reasoning applies to $x_5, x_6, x_7, x_8$, so we consider solutions of the form $(u, u, u, u, v, v, v, v, x_9)$. If $u \\neq v$, these can be paired by swapping the $u$'s and $v$'s, so we finally consider $x_1 = x_2 = \\dots = x_8 = a$, $x_9 = b$. The equation becomes\n$$\n\\frac{8}{a} + \\frac{1}{b} = 1,\n$$\nor $8b + a = ab$, which rearranges to $(a-8)(b-1) = 8$. Since $b \\geq 2$, $b-1$ is a positive divisor of $8$. The possibilities $b-1 = 1, 2, 4, 8$ yield $a = 16, 12, 10, 9$ and $b = 2, 3, 5, 9$ respectively. All four are valid and distinct solutions. Thus, $N$ has the parity of $4$, so $N$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18622,
    "subject": "Mathematics (Olympiad)",
    "question": "a) How many numbers remain on the board after erasing all even numbers from $1$ to $2011$?\n\nb) How many four-digit numbers with only $0$ and $1$ as digits end with $1$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The erased numbers were $2 = 2 \\cdot 1$, $4 = 2 \\cdot 2$, ..., $2010 = 2 \\cdot 1005$. So $2011 - 1005 = 1006$ numbers were left on the board.\n\nb) We can list the numbers: they are $1$, $11$, $101$, $111$, $1001$, $1011$, $1101$, $1111$, a total of $8$.\n\nOr, we can argue that the number is of the form $(abc1)$, where $a$, $b$, $c$ are digits equal to either $0$ or $1$. Notice that the units digit must be $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18623,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $2 \\cdot \\angle BAC > \\angle ABC$. A point $D$ is chosen on the segment $BC$ so that $2 \\cdot \\angle CAD = \\angle ABC$. Let $H$ be the orthocenter of the triangle $ABC$ and $AD$ intersects the circumcircle of the triangle $AHC$ at $K$, differently from $A$. $AB$ intersects the circumcircle of the triangle $BDK$ at $E$, differently from $B$. Prove that $AE = CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\beta := \\angle B$. Since $AHKC$ is a cyclic quadrilateral, we have $\\angle AKC = \\angle AHC = 180^\\circ - \\beta$. By assumption $\\angle KAC = \\frac{\\beta}{2}$. Hence $\\angle ACK = 180^\\circ - \\angle AKC - \\angle KAC = \\frac{\\beta}{2}$. Therefore $\\triangle AKC$ is an isosceles triangle.\n\n![](images/MNG_ABooklet_2015_p19_data_3c3bc42857.png)\n\nSince $EBDK$ is a cyclic quadrilateral, $\\angle EKD = 180^\\circ - \\beta$. Hence $\\angle AKE = 180^\\circ - \\angle EKD = \\beta$ and $\\angle AKE + \\angle AKC = 180^\\circ$. This shows that the points $E$, $K$, and $C$ are collinear.\n\nBy the law of sines for the triangle $ADC$ we get that\n\n$$\n\\frac{AC}{\\sin \\angle ADC} = \\frac{DC}{\\sin \\frac{\\beta}{2}}. \\qquad (1)\n$$\n\nSince $EBDK$ is cyclic, $AEC = 180^\\circ - \\angle ADC$. By the law of sines for the triangle $AEC$ we get that\n\n$$\n\\frac{AE}{\\sin \\frac{\\beta}{2}} = \\frac{AC}{\\sin \\angle AEC} = \\frac{AC}{\\sin \\angle ADC}. \\qquad (2)\n$$\n\nFrom (1) and (2), we see that $AE = CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18624,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the excircle of $\\triangle ABC$ opposite the vertex $A$ touches the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously by using the excircles opposite to $B$ and $C$, respectively. Suppose that the circumcentre of $\\triangle A_1B_1C_1$ lies on the circumcircle of\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p349_data_a7aa4f65aa.png)\n\n$\\triangle ABC$. Prove that $\\triangle ABC$ is right-angled.\n\n(The excircle of $\\triangle ABC$ opposite the vertex $A$ is the circle that touches the line segment $BC$, the ray $AB$ beyond $B$, and the ray $AC$ beyond $C$. The excircles opposite $B$ and $C$ are similarly defined.)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the circumcircles of $\\triangle ABC$ and $\\triangle A_1B_1C_1$ by $\\Omega$ and $\\Gamma$, respectively. Let $A_0$ be the midpoint of arc $\\widehat{BC}$ on $\\Omega$ containing point $A$. Points $B_0$ and $C_0$ are defined analogously. Let $Q$ be the centre of circle $\\Gamma$; then $Q$ is on $\\Omega$ by the hypothesis.\n\n**Lemma.** $A_0B_1 = A_0C_1$. Points $A$, $A_0$, $B_1$, and $C_1$ are concyclic.\n\nIf points $A_0$ and $A$ coincide, then $\\triangle ABC$ is isosceles, thus $AB_1 = AC_1$. Otherwise, $A_0B = A_0C$ by the definition of $A_0$. It is evident that\n\n$$\nBC_1 = CB_1 = \\frac{1}{2}(b + c - a),\n$$\n\nand\n\n$$\n\\angle C_1BA_0 = \\angle ABA_0 = \\angle ACA_0 = \\angle B_1CA_0.\n$$\n\nThus, $\\triangle A_0BC_1 \\cong \\triangle A_0CB_1$. (1)\n\nSo, $A_0B_1 = A_0C_1$.\n\nAlso, by (1), $\\angle A_0C_1B = \\angle A_0B_1C$, hence $\\angle A_0C_1A = \\angle A_0B_1A$. Thus, points $A$, $A_0$, $B_1$, and $C_1$ are concyclic.\n\nObviously, points $A_1$, $B_1$, and $C_1$ are on a semi-arc of $\\Gamma$, so $\\triangle A_1B_1C_1$ is obtuse-angled. Without loss of generality, suppose $\\angle A_1B_1C_1$ is obtuse; then points $Q$ and $B_1$ are on different sides of $A_1C_1$. So are points $B$ and $B_1$, hence $Q$ and $B$ are on the same side of $A_1C_1$.\n\nThe perpendicular bisector of $A_1C_1$ intersects $\\Gamma$ at two points on different sides of $A_1C_1$. By the above, $B_0$ and $Q$ are among these points, and $B_0$ and $Q$ are on the same side of $A_1C_1$. So $B_0 = Q$, as shown in the diagram.\n\nBy the lemma, lines $QA_0$ and $QC_0$ are perpendicular bisectors of $B_1C_1$ and $A_1B_1$, respectively, and $A_0$ and $C_0$ are midpoints of arcs $CB$ and $BA$, respectively. Therefore,\n\n$$\n\\begin{aligned}\n\\angle C_1 B_0 A_1 &= \\angle C_1 B_0 B_1 + \\angle B_1 B_0 A_1 = 2\\angle A_0 B_0 B_1 + 2\\angle B_1 B_0 C_0 \\\\\n&= 2\\angle A_0 B_0 C_0 = 180^\\circ - \\angle ABC.\n\\end{aligned}\n$$\n\nOn the other hand, by the lemma again,\n\n$$\n\\angle C_1 B_0 A_1 = \\angle C_1 BA_1 = \\angle ABC.\n$$\n\nHence, $\\angle ABC = 180^\\circ - \\angle ABC$, so $\\angle ABC = 90^\\circ$. This completes the proof. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18625,
    "subject": "Mathematics (Olympiad)",
    "question": "If $\\min_{x \\in \\mathbb{R}} \\frac{a x^2 + b}{\\sqrt{x^2 + 1}} = 3$, find:\n\n1. The range of $b$.\n2. The value of $a$ for a given $b$.\n\n![](posed_by_Lu_Xingjiang.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}}$. It is easy to see that $a > 0$. Since $f(0) = b$, we have $b \\ge 3$.\n\n**Case (i):** If $b - 2a \\ge 0$,\n\n$$\nf(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}} = a \\sqrt{x^2 + 1} + \\frac{b - a}{\\sqrt{x^2 + 1}} \\ge 2 \\sqrt{a(b - a)} = 3,\n$$\n\nequality holds when $a \\sqrt{x^2 + 1} = \\frac{b - a}{\\sqrt{x^2 + 1}}$, i.e., $x = \\pm \\sqrt{\\frac{b - 2a}{a}}$.\n\nThe value of $a$ for given $b$ is $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$; especially, when $b = 3$, $a = \\frac{3}{2}$.\n\n**Case (ii):** If $b - 2a < 0$, let $\\sqrt{x^2 + 1} = t$ ($t \\ge 1$). Then $f(x) = g(t) = a t + \\frac{b - a}{t}$, which is monotonically increasing for $t \\ge 1$, so\n\n$$\n\\min_{x \\in \\mathbb{R}} f(x) = g(1) = a + b - a = b = 3, \\text{ when } a > \\frac{3}{2}.\n$$\n\n**Summary:**\n\n1. The range of $b$ is $[3, +\\infty)$.\n2. If $b = 3$, then $a \\ge \\frac{3}{2}$; if $b > 3$, then $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18626,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA = FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen such that $DA = DC$ and $AC$ is the bisector of $\\angle DAB$. Point $E$ is chosen such that $EA = ED$ and $AD$ is the bisector of $\\angle EAC$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram (where $AM \\parallel EX$ and $AE \\parallel MX$). Prove that lines $BD$, $FX$, and $ME$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "From the problem statement it easily follows that $\\triangle AFB$, $\\triangle ADC$, and $\\triangle AED$ are similar isosceles triangles. Thus, as shown in the diagram, we may let\n\n$$\n\\alpha = \\angle FAB = \\angle ABF = \\angle DAC = \\angle ACD = \\angle EAD = \\angle ADE.\n$$\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p172_data_a496ffece5.png)\n\nFrom this it follows that $ED \\parallel AC$. But $AC \\parallel EX$. Hence $D$ lies on the line $EX$.\n\nFrom $\\triangle ADC \\sim \\triangle AFB$, we have $AD/AC = AF/AB$. Since also $\\angle DAF = \\angle CAB$, it follows that $\\triangle ADF \\sim \\triangle ACB$ (PAP). Thus\n\n$$\n\\begin{aligned}\n\\angle AFD &= \\angle ABC \\\\\n&= 90^\\circ + \\alpha\n\\end{aligned}\n\\quad \\Rightarrow \\quad\n\\begin{aligned}\n\\angle DFC &= 90^\\circ - \\alpha \\\\\n\\angle CDF &= 90^\\circ.\n\\end{aligned}\n\\quad (\\text{angle sum } \\triangle CDF)\n$$\n\nThus the circle with diameter $CF$ passes through points $B$ and $D$. Since $M$ is the midpoint of $CF$, it follows that $M$ is the centre of this circle. Therefore\n\n$$\nMB = MC = MD = MF.\n$$\n\nWe also have\n\n$$\n\\begin{aligned}\n\\angle MXD &= \\angle EAM && (\\text{parallelogram } AMXE) \\\\\n&= 2\\alpha \\\\\n&= 2\\angle FCD \\\\\n&= \\angle FMD && (\\text{angle at centre of circle } BCDF) \\\\\n&= \\angle XDM. && (AM \\parallel EX)\n\\end{aligned}\n$$\n\nHence $MX = MD$.\n\nFrom parallelogram $AMXE$, we have $MX = AE$. We are also given $AE = DE$. Hence\n\n$$\nMB = MC = MD = MF = MX = AE = DE.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18627,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all strictly increasing functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real $x, y$ the following equality holds:\n\n$$\nf(x + f(y)) = f(x + y) + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = -y$, we obtain:\n\n$$\nf(-y + f(y)) = f(0) + 1.\n$$\n\nThe monotonicity of $f$ implies that $f(y) - y = c$ for some constant $c$. It remains to perform a substitution to find the value of $c$.\n\n$$\nf(x + f(y)) = f(x + y + c) = x + y + c + c = x + y + c + 1 \\Rightarrow c = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18628,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\omega$. Let $D$, $E$, and $F$ be points on the sides $BC$, $CA$, and $AB$ such that the circumcircle of triangle $DEF$ touches $\\omega$ at $A$. Let $G$ and $H$ be the intersection points of the circumcircles of triangles $BDE$ and $CDF$ with $\\omega$ (different from $B$ and $C$), respectively. Prove that the lines $GE$ and $HF$ intersect on $AD$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p267_data_3c037fd547.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The dilation which maps the circumcircle of triangle $DEF$ onto $\\omega$ maps $E$ and $F$ to $C$ and $B$, respectively. Hence, $EF$ is parallel to $BC$.\n\nLet $X$ be the intersection point of $AD$ with $\\omega$ (different from $A$), let $Y$ be the intersection point of $AD$ with $GE$, and let $Z$ be the intersection point of $AD$ with $HF$.\n\nThen $\\angle GXA = \\angle GBA = \\angle CBA - \\angle CBG = \\angle EFA - \\angle DBG = \\angle EDA - \\angle DEX = \\angle EYD = \\angle GYA$, i.e., $X$ and $Y$ agree. Similarly, one shows that $X$ and $Z$ agree. In conclusion, $AD$, $GE$, and $HF$ intersect on $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18629,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 與 $s$ 為正整數，且 $N > s$。台電園區裡有若干棟建物，其中恰有 $N$ 棟為發電廠，另有一棟為總部。若干對建物之間有僅能單向送電的電線，滿足：\n\n1. 所有連接發電廠的電線都只會把電送出發電廠。\n2. 對於每個非總部的建物，都存在唯一的一系列電線，構成從該建物通向總部的電路。\n\n某建物被稱為 $s$ 級有電，若且唯若當我們移除園區內的任何一條電線，該建物仍能從至少 $s$ 個電廠供電。試求 $s$ 級有電建物數量的最大可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "答案為\n\n$$\nf(N) = \\begin{cases} \\left\\lfloor \\frac{N}{s+1} \\right\\rfloor + \\left\\lfloor \\frac{N-s}{s+1} \\right\\rfloor, & N > s, \\\\ 0, & N \\le s. \\end{cases}\n$$\n\n證明如下：\n\n首先，令總部為 $H$。將所有沒電的建築與其相連的電線去除，不影響 $H$ 的性質，也不影響 $s$ 級有電建築的個數，故可假設所有建築都有電。考慮所有與 $H$ 間有電線的建築 $T_1, \\dots, T_k$。若 $k=1$，則移除 $\\overline{HT_1}$ 必讓 $H$ 斷電，從而 $H$ 非 $s$ 級有電；此外，若移除 $H$ 與 $\\overline{HT_1}$，$T_1$ 成為新的總部，故可假設 $k \\ge 2$。\n\n給定 $s$，對 $N$ 歸納證明上述公式。$N \\le s$ 時顯然 $f(N) = 0$。假設 $N > s$ 且公式對所有小於 $N$ 的數字都成立。假設 $T_i$ 被 $a_i$ 個發電廠供電，且 $a_1 \\ge a_2 \\ge \\dots \\ge a_k > 0$。由於 $H$ 的單一路徑性質，每個發電廠都必須供電給 $T_i$ 中的恰一個，故 $\\sum a_k = N$。此外，$H$ 是 $s$ 級有電，若且唯若 $a_1 \\le N-s$；又當拔掉 $H$ 後，會得到 $k$ 個互不連通、各有 $a_i$ 個發電廠並以 $T_i$ 為總部的子園區，故有：\n\n$$\nf(N) = \\max \\left\\{ \\max_{k, a_1 > N-s, \\sum_i a_i = N} \\sum_i f(a_i),\\ 1 + \\max_{k, a_1 \\le N-s, \\sum_i a_i = N} \\sum_i f(a_i) \\right\\}.\n$$\n\n設 $c_N = \\left\\lfloor \\frac{N}{s+1} \\right\\rfloor + \\left\\lfloor \\frac{N-s}{s+1} \\right\\rfloor$。\n\n**證 $f(N) \\ge c_N$：**\n\n- 當 $s+1 \\le N \\le 2s$，$c_N = 1$。取 $k = s+1$，$a_1 = N-s$，其餘 $a_2 = \\dots = a_k = 1$，依歸納假設可得 $f(N) \\ge 1 = c_N$。\n- 當 $N > 2s$，取 $k = 2$，$a_1 = N-s+1$，$a_2 = s+1$，依歸納假設可得 $f(N) \\ge c_N$。\n\n**證 $f(N) \\le c_N$：**\n\n假設 $a_1 \\ge a_2 \\ge \\dots \\ge a_t \\ge s+1 > a_{t+1} \\ge \\dots \\ge a_k > 0$。依歸納假設：\n\n$$\n\\sum_{i=1}^{k} f(a_i) = \\sum_{i=1}^{t} \\left\\lfloor \\frac{a_i}{s+1} \\right\\rfloor + \\sum_{i=1}^{t} \\left\\lfloor \\frac{a_i - s}{s+1} \\right\\rfloor \\le \\left\\lfloor \\frac{N}{s+1} \\right\\rfloor + \\left\\lfloor \\frac{N-s}{s+1} \\right\\rfloor = c_N.\n$$\n\n對於 $a_1 \\le N-s$ 的情況，(上式)至少有一個不等式嚴格成立，故 $f(N) \\le c_N$。\n\n綜合以上，$f(N) = c_N$，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18630,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a regular pentagon with center $M$. A point $P \\neq M$ is chosen on the line segment $MD$. The circumcircle of $ABP$ intersects the line segment $AE$ at $A$ and $Q$, and the line through $P$ perpendicular to $CD$ at $P$ and $R$.\n\nProve that $AR$ and $QR$ are of the same length.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ denote the common point of $RP$ and $AE$. Since we are given a regular pentagon, the angles in triangle $ABE$ are well known as $\\angle BAE = 108^\\circ$ and $\\angle ABE = \\angle AEB = 36^\\circ$. Since $BE$ and $CD$ are parallel, $RP$ is perpendicular to $BE$, and we therefore have $\\angle ASP = 126^\\circ$ and $\\angle QSP = 54^\\circ = \\angle ASR$. From this,\n\n$$\n\\angle SPA = 54^\\circ - \\angle SAP = \\angle PAB - 54^\\circ = \\angle PBA - 54^\\circ = 126^\\circ - \\angle AQP = 126^\\circ - \\angle SQP = \\angle SPQ\n$$\n\nfollows, since $ABPQ$ is inscribed. We therefore see that $SP$ (or $RP$) bisects the angle $\\angle APQ$, which implies that $AR$ and $QR$ must be of equal length, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18631,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We will prove that $k = 6$ by constructing a configuration for $k = 6$ in Step 1, and then proving in Step 2 that no smaller value is possible.\n\n**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\n*Proof of lemma:* Assume by contradiction that the 10 segments can be colored with two colors: red and blue.\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\n*Step 1.* Consider 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\n*Step 2.* We 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": 18632,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of real numbers $a$ and $b$, $b > 0$, such that the solutions to the two equations\n\n$$\nx^2 + ax + a = b\n$$\n\nand\n\n$$\nx^2 + ax + a = -b\n$$\n\nare four consecutive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the solutions to $x^2 + ax + a = b$ be $x_1$ and $x_2$, and to $x^2 + ax + a = -b$ be $x_3$ and $x_4$. By the quadratic formula:\n\n$$\nx = \\frac{-a \\pm \\sqrt{a^2 - 4a + 4b}}{2}\n$$\nfor the first equation, and\n$$\nx = \\frac{-a \\pm \\sqrt{a^2 - 4a - 4b}}{2}\n$$\nfor the second.\n\nIf these four roots are consecutive integers, their average is $-a/2$. Thus, the roots are $-a/2 - 3/2$, $-a/2 - 1/2$, $-a/2 + 1/2$, $-a/2 + 3/2$.\n\nSet\n$$\n\\sqrt{a^2 - 4a + 4b} = 3 \\quad \\text{and} \\quad \\sqrt{a^2 - 4a - 4b} = 1.\n$$\nSquaring and subtracting gives:\n$$\na^2 - 4a + 4b = 9\n$$\n$$\na^2 - 4a - 4b = 1\n$$\nSubtracting:\n$$\n8b = 8 \\implies b = 1\n$$\nPlug back:\n$$\na^2 - 4a = 5 \\implies a = -1 \\text{ or } a = 5\n$$\nSo the pairs are $(a, b) = (-1, 1)$ and $(5, 1)$.\n\nAlternatively, let the four consecutive integers be $n-1, n, n+1, n+2$. The axis of symmetry is $x = -a/2$. Setting up equations and solving similarly, we again find $b = 1$ and $a = -1$ or $a = 5$.\n\n**Conclusion:** The only pairs are $(a, b) = (-1, 1)$ and $(5, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18633,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the 3-digit number has the form $abc$. Given the following properties:\n\n$$\n(10c + b) - (10b + c) = 36\n$$\n\n$$\n(100a + c) - (100c + a) = 198\n$$\n\nWhat is the value of $(100a + 10b) - (100b + 10a)$?",
    "options": [],
    "answer": "See solution",
    "solution": "From (1) we get $9(c-b) = 36$, hence $c-b = 4$.\n\nFrom (2) we get $99(a-c) = 198$, hence $a-c = 2$.\n\nSo $a-b = 6$.\n\nTherefore, $(100a+10b)-(100b+10a) = 90a-90b = 90(a-b) = \\mathbf{540}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18634,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively. Let $K$, $L$, and $M$ be the midpoints of the segments $BP$, $CQ$, and $PQ$, respectively, and let $\\Gamma$ be the circle passing through $K$, $L$, and $M$. Suppose that the line $PQ$ is tangent to the circle $\\Gamma$. Prove that $OP = OQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $AB \\parallel KM$ and $AC \\parallel LM$, we obtain that $\\angle KMQ \\equiv \\angle AQP$ and $\\angle LMP \\equiv \\angle APQ$. If $PQ$ is tangent to the circumcircle of triangle $KLM$, then $\\angle KLM \\equiv \\angle KMQ$, hence $\\angle KLM \\equiv \\angle AQP$. Then $\\triangle APQ \\sim \\triangle MKL$, so $\\frac{AQ}{ML} = \\frac{AP}{MK}$. It follows that $\\frac{AQ}{PC} = \\frac{AP}{BQ}$, i.e., $AQ \\cdot BQ = AP \\cdot PC$. This means that $P$ and $Q$ have equal powers with respect to the circumcircle of triangle $ABC$. As they are both situated inside the circle, they are at equal distance from the center of the circle, $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18635,
    "subject": "Mathematics (Olympiad)",
    "question": "Pieces of cardboard of dimensions $1 \\times 4$ are placed on a $10 \\times 10$ grid in such a way that each piece covers exactly 4 adjacent unit squares (either horizontally or vertically) and no two pieces touch each other side-to-side, edge-to-edge, or corner-to-corner. Find the largest possible number of cardboard pieces.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the pieces of cardboard be placed on the grid as required. Since each piece covers exactly 4 unit squares and no two pieces touch, there is at least a 1-unit wide space between every two pieces. Therefore, if we draw a half-unit wide \"no-go zone\" around each piece, the areas covered by the pieces and their no-go zones will have dimensions of $2 \\times 5$ and will not overlap. Since these areas extend over the edges of the grid by at most half a unit, all these areas can fit within an $11 \\times 11$ square. Therefore, no more than $\\left\\lfloor \\frac{11 \\cdot 11}{2 \\cdot 5} \\right\\rfloor = 12$ pieces can be placed on the grid.\n\nThis limit case is achievable, as shown in Fig. 37.\n\n![](images/EST_ABooklet_2024_p43_data_23e03097d2.png)\n\nFig. 37\n\n![](images/EST_ABooklet_2024_p43_data_a1e9b78c3e.png)\n\nFig. 38",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18636,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n \\geq 10$ with non-zero digits that satisfy the following condition: if any of the digits of $n$ is deleted, the obtained number is a divisor of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the decimal notation of a natural number $n$ is $\\overline{a_k a_{k-1} \\dots a_2 a_1}$. The main condition says that the number $\\overline{a_k a_{k-1} \\dots a_2}$ divides $n = 10 \\cdot \\overline{a_k a_{k-1} \\dots a_2} + a_1$, so it must also divide $a_1$. Since $a_1 \\neq 0$, $\\overline{a_k a_{k-1} \\dots a_2}$ can have at most one digit, so $k = 2$. Thus, $n = \\overline{a_2 a_1}$, and $a_2$ must divide $a_1$. Therefore, $a_2 = a_1$ or $a_2 \\leq 4$. The possible values for $n$ are 99, 88, 77, 66, 55, 48, 44, 39, 36, 33, 28, 26, 24, 22, 19, 18, 17, 16, 15, 14, 13, 12, or 11. Among these, only 99, 88, 77, 66, 55, 48, 44, 36, 33, 24, 22, 15, 12, and 11 fulfill the condition.\n\n**Summary:** The solutions are $11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18637,
    "subject": "Mathematics (Olympiad)",
    "question": "For any real number $p \\ge 1$, consider the set of all real numbers $x$ such that\n\n$$\np < x < \\left(2 + \\sqrt{p + \\frac{1}{4}}\\right)^2\n$$\n\nProve that from such a set, one can select four mutually different natural numbers $a, b, c, d$ with $ab = cd$.",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers $a = (k-1)k$, $b = (k+1)k$, $c = (k-1)(k+1)$, and $d = k^2$ clearly satisfy $ab = cd$ and $a < c < d < b$ for any $k > 1$.\n\nLet $k$ be the least natural number for which $p < a$, i.e., $p < (k-1)k$. We will show that for this $k$, necessarily $b = (k+1)k \\leq p + 4 + 2\\sqrt{4p+1}$, which is smaller by $\\frac{1}{4}$ than the upper bound of the interval in our problem.\n\nSince $k$ is chosen so that $p < (k-1)k$, we have $p \\geq (k-2)(k-1)$. Solving this quadratic inequality gives\n\n$$\nk \\leq \\frac{3}{2} + \\sqrt{p + \\frac{1}{4}},\n$$\n\nfrom which it follows that\n\n$$\nb = (k+1)k \\leq \\left(\\frac{5}{2} + \\sqrt{p + \\frac{1}{4}}\\right) \\left(\\frac{3}{2} + \\sqrt{p + \\frac{1}{4}}\\right) = \\frac{15}{4} + 4\\sqrt{p + \\frac{1}{4}} + \\left(p + \\frac{1}{4}\\right) = p + 4 + 2\\sqrt{4p+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18638,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that three points $A$, $B$, $C$ lie on the circumference of a circle $\\Gamma$. Let $P$ be the point of intersection of the lines tangent to $\\Gamma$ at $B$ and $C$. Suppose that the lines $AB$ and $CP$ are parallel, and that $AB = 3$ and $BP = 4$. Find the length of the line segment $BC$. Here we represent the length of the line segment $XY$ also by $XY$.\n\n![](images/Japan_2015_p1_data_b00fd0aff6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ and $CP$ are parallel, $\\angle ABC = \\angle BCP$. By a well-known theorem, $\\angle CAB = \\angle PBC$ must also hold. Therefore, the triangles $ABC$ and $BCP$ are similar, which implies that $AB : BC = BC : CP$. From this, it follows that $BC = \\sqrt{AB \\cdot CP}$. Since $P$ is the point of intersection of the tangent lines to the circle $\\Gamma$, we have $CP = BP = 4$, so $BC = \\sqrt{3 \\cdot 4} = 2\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18639,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Stopping at the smallest 5-digit number, the sequence is:\n\n$$6, 19, 22, 47, 88, 157, 292, 537, 986, 1815, 3338, 6139, 11292.$$  \n\nb) Working backwards gives the earlier terms in the sequence.\n\nGiven that the 4th term plus the 5th term plus the 6th term equals the 7th term, find the first three terms (the seeds) of the sequence if the 4th, 5th, 6th, and 7th terms are 28, 36, 71, and 135, respectively.\n\nc) Given the tribonacci sequence with seeds $20, 17, 2017, \\ldots$, determine the parity (even or odd) of the 2017th term.\n\nd) Let $a$ be the first seed. The sequence is:\n\n$$a, 20, 17, a+37, a+74, 2a+128, 4a+239, 7a+441, 13a+808, 24a+1488, 44a+2737, \\ldots$$\n\nFor which positive integer values of $a$ does some term in the sequence equal 2017?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The sequence up to the smallest 5-digit number is:\n\n$$6, 19, 22, 47, 88, 157, 292, 537, 986, 1815, 3338, 6139, 11292.$$  \n\nb) Working backwards:\n- The 4th term: $135 - 71 - 36 = 28$\n- The 3rd term: $71 - 36 - 28 = 7$\n- The 2nd term: $36 - 28 - 7 = 1$\n- The 1st term: $28 - 7 - 1 = 20$\n\nSo the seeds are $20, 1, 7$.\n\nc) The parity pattern for the sequence $20, 17, 2017, \\ldots$ is:\n\neven, odd, odd, even, even, odd, odd, even, \\ldots\n\nThis pattern 'even, odd, odd, even' repeats every 4 terms. Since $2017 = 4 \\times 504 + 1$, the 2017th term corresponds to the first term in the 505th cycle, which is even. Thus, the 2017th term is even.\n\nd) Setting each term equal to 2017:\n- First term: $a = 2017$\n- Fourth term: $a + 37 = 2017 \\implies a = 1980$\n- Fifth term: $a + 74 = 2017 \\implies a = 1943$\n- Sixth term: $2a + 128 = 2017 \\implies 2a = 1889$ (impossible, as 1889 is odd)\n- Seventh term: $4a + 239 = 2017 \\implies 4a = 1778$ (impossible, as 4 does not divide 1778)\n- Eighth term: $7a + 441 = 2017 \\implies 7a = 1576$ (impossible, as 7 does not divide 1576)\n- Ninth term: $13a + 808 = 2017 \\implies 13a = 1209 \\implies a = 93$\n- Tenth term: $24a + 1488 = 2017 \\implies 24a = 529$ (impossible, as 529 is not divisible by 24)\n\nFor $a > 0$, the eleventh and subsequent terms exceed 2017. Thus, the only positive integer values of $a$ are $2017, 1980, 1943, 93$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18640,
    "subject": "Mathematics (Olympiad)",
    "question": "Each row in the Grand Theatre has 55 seats. What is the maximum number of contestants that can be seated in a single row, with the restriction that no two of them are exactly 4 seats apart?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t$ be the maximum number of contestants. Consider the set $B = \\{10a + b : a = 0, 1, 2, 3, 4, 5;\\ b = 1, 2, 3, 4, 5\\}$:\n\n$$\nB = \\{1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, \\ldots, 51, 52, 53, 54, 55\\}.\n$$\n\nSuppose two contestants are seated 4 seats apart, i.e., positions $m$ and $n$ with $|m - n| = 4$. For $B$, if $(10a_1 + b_1) - (10a_2 + b_2) = 4$, then $10(a_1 - a_2) = b_2 - b_1 + 4$. But $-4 \\leq b_2 - b_1 \\leq 4$, so $1 \\leq b_2 - b_1 + 4 \\leq 8$, which cannot be divisible by 10. Thus, no two elements of $B$ are 4 apart, and $|B| = 30$, so $t \\geq 30$.\n\nNow, for any subset $B$ of size $t$, count \"good\" pairs $(m, n)$ with $|m - n| = 4$, $m \\in B$, $n \\notin B$. For $m \\in \\{1,2,3,4,5,51,52,53,54,55\\}$, there is one such $n$; for other $m$, there are two. So, number of good pairs is at least $10 + 2(t-10) = 2t - 10$. But for each $n \\notin B$, there are at most two $m$ with $|m-n|=4$, so at most $2(55-t)$ good pairs. Thus,\n\n$$\n2t - 10 \\leq 2(55 - t)\n$$\n$$\n2t - 10 \\leq 110 - 2t\n$$\n$$\n4t \\leq 120\n$$\n$$\nt \\leq 30\n$$\n\nTherefore, the maximum is $t = 30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18641,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve for $x \\in \\mathbb{R}$:\n\n$$\n2^{x+1} + \\log_2(1 + \\sqrt{x}) = 4^x + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equation can be rewritten as $\\log_2(1 + \\sqrt{x}) = (2^x - 1)^2$.\n\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 $f^{-1}(x) = (2^x - 1)^2$.\n\nAs $f$ is strictly increasing, 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.\n\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": 18642,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Take $a > 0$ such that $f(a) \\ne 1$. Then $f(a^{xy}) = f(a)^{f(xy)}$ and\n\n$$\nf(a^{xy}) = f((a^x)^y) = f(a^x)^{f(y)} = (f(a)^{f(x)})^{f(y)} = f(a)^{f(x)f(y)},\n$$\n\nwhence $f(xy) = f(x)f(y)$.\n\nAlso, $f(a^{x+y}) = f(a)^{f(x+y)}$ and\n\n$$\nf(a^{x+y}) = f(a^x a^y) = f(a^x)f(a^y) = f(a)^{f(x)} f(a)^{f(y)} = f(a)^{f(x) + f(y)},\n$$\n\ntherefore $f(x+y) = f(x) + f(y)$.\n\n**Remark.** It can be proven that the only function fulfilling these conditions is the identity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18643,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $\\Gamma$ una circunferencia de centro $O$, $AE$ un diámetro de $\\Gamma$ y $B$ el punto medio de uno de los arcos $AE$ de $\\Gamma$. El punto $D \\neq E$ está sobre el segmento $OE$. El punto $C$ es tal que el cuadrilátero $ABCD$ es un paralelogramo con $AB$ paralelo a $CD$ y $BC$ paralelo a $AD$. Las rectas $EB$ y $CD$ se cortan en el punto $F$. La recta $OF$ corta al arco menor $EB$ de $\\Gamma$ en el punto $I$. Demostrar que la recta $EI$ es la bisectriz del ángulo $BEC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Nótese que $AB$ es la simétrica de $BE$ respecto de $OB$, y al ser $OB$ perpendicular a $AE$, luego a $BC$, la mediatriz de $BC$ es paralela a $OB$. Pero como la simetría respecto de la mediatriz de $BC$ transforma a $B$ en $C$, y transforma las rectas paralelas a $BE$ en rectas paralelas a $AB$, luego a $CD$, se tiene que $BE$, $CD$ son simétricas respecto de la mediatriz de $BC$, que es además perpendicular a $DE$. Luego $BDEC$ es un trapecio isósceles, y por lo tanto cíclico, con lo que $\\angle BEC = \\angle BDC$. Al mismo tiempo, como $AB$, $BE$ son perpendiculares, y $CD$ es paralela a $AB$, entonces $BE$, $CD$ son perpendiculares, con lo que $\\angle BFD = 90^\\circ$. Al ser $BO$ claramente la mediatriz de $AE$, se tiene que $\\angle BOD = 90^\\circ$. Luego $BODF$ está inscrito en la circunferencia de diámetro $BD$, y\n\n$$\n\\angle BDC = \\angle BDF = \\angle BOF = \\angle BOI = 2\\angle BEI,\n$$\n\nesta última igualdad por ser $O$ centro de la circunferencia que pasa por $B$, $E$, $I$. Luego $\\angle BED = 2\\angle BEI$, y en efecto $EI$ es la bisectriz de $\\angle BEC$, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18644,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. We call a sequence of integers $a_1, a_2, \\dots, a_k$, with $1 \\leq a_i \\leq n$, *smooth* if there exists an integer $m$, with $1 \\leq m < k$, such that $a_1 = a_{k-m+1}, a_2 = a_{k-m+2}, \\dots, a_m = a_k$. Furthermore, a sequence is *universal* if each of the sequences obtained by replacing $a_k$ by $1, 2, \\dots, n$ is smooth. For each $n$, find a universal sequence of minimum length.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum length is $2^n$.\n\nDelete the last term of a universal sequence and call the shortened sequence $\\alpha$. For each $i = 1, \\dots, n$, the hypothesis implies that $\\alpha$ starts with a block $B_i$, followed by a distinguished term $i$, and ends with a block $B'_i$ identical to $B_i$. Choose the blocks $B_i$ to be shortest possible; they are uniquely determined and different. By relabeling if necessary, we may ensure $|B_n| > |B_{n-1}| > \\dots > |B_1|$. (Under this assumption, $\\alpha$ starts with a $1$ and the shortest block $B_1$ is empty: $|B_1| = 0$.)\n\nWe claim that block $B'_n$ does not contain the distinguished $n$. Otherwise, $B_n$ and $B'_n$ have a common part $a_1, \\dots, a_m$, possibly empty, with $m < |B_n|$. Now $B'_n$ starts with $a_1, \\dots, a_m, n$ and $B_n$ ends with $a_1, \\dots, a_m$. Since $B_n$ and $B'_n$ are identical, $\\alpha$ starts with $a_1, \\dots, a_m, n$ and ends with $a_1, \\dots, a_m$. But then $m < |B_n|$ contradicts the minimality of $B_n$. So the distinguished $n$ is not in $B'_n$. Consequently, the length $\\ell$ of $\\alpha$ satisfies $\\ell \\geq 2|B_n| + 1$.\n\nA similar argument applies to blocks $B_n$ and $B_{n-1}$ and shows that $|B_n| \\geq 2|B_{n-1}| + 1$. Indeed, $B_n$ starts with $B_{n-1}$, which is followed by the distinguished $n-1$, and $B'_n$ ends with $B'_{n-1}$. Since $B_n$ and $B'_n$ are identical, $B_n$ ends with a block $B''_{n-1}$ identical to $B_{n-1}$ and $B'_{n-1}$. Now we show that $B''_{n-1}$ does not contain the distinguished $n-1$; this will ensure $|B_n| \\geq 2|B_{n-1}| + 1$.\n\nSuppose on the contrary that the distinguished $n-1$ is in $B''_{n-1}$. Then $B_{n-1}$ and $B''_{n-1}$ have a common part $a_1, \\dots, a_m$, possibly empty, where $m < |B_{n-1}|$. Now $B''_{n-1}$ starts with $a_1, \\dots, a_m, n-1$ and $B_{n-1}$ ends with $a_1, \\dots, a_m$. Since $B_{n-1}$, $B'_{n-1}$, and $B''_{n-1}$ are identical, $\\alpha$ starts with $a_1, \\dots, a_m, n-1$ and ends with $a_1, \\dots, a_m$; but then $m < |B_{n-1}|$ contradicts the minimality of $B_{n-1}$. The claimed $|B_n| \\geq 2|B_{n-1}| + 1$ follows.\n\nBy the same reasoning, $|B_i| \\geq 2|B_{i-1}| + 1$ for all $i = 2, \\dots, n$. Combined with $\\ell \\geq 2|B_n| + 1$, this leads to $\\ell \\geq 2^n - 1$. Hence, the initial universal sequence has length $\\ell + 1 \\geq 2^n$.\n\nOn the other hand, for each $n$ there are universal sequences $\\beta_n$ with terms in $\\{1, 2, \\dots, n\\}$ and length $2^n$. If $n = 1$, set $\\beta_1 = 1, 1$ (length $2 = 2^1$). Suppose $\\beta_{n-1}$ is a universal sequence with terms in $\\{1, 2, \\dots, n-1\\}$ and length $2^{n-1}$. Write $n$ in front of every term of $\\beta_{n-1}$. The obtained sequence $\\beta_n$ has terms in $\\{1, 2, \\dots, n\\}$ and length $2^n$. In addition, $\\beta_n$ is universal. Indeed, replacing the final $1$ by $n$ yields a smooth sequence (starting and ending with $n$). Replace the final $1$ by $i \\in \\{1, \\dots, n-1\\}$; let $\\beta'_n$ be the new sequence. Suppose the final $1$ in $\\beta_{n-1}$ is also replaced by $i$. Then the resulting sequence would start and end with a block $a_1, a_2, \\dots, a_m, i$ with $a_j \\in \\{1, \\dots, n-1\\}$. By the definition of $\\beta_n$, $\\beta'_n$ starts and ends with $n, a_1, n, a_2, \\dots, n, a_m, n, i$, meaning it is smooth. This completes the inductive construction and the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18645,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest positive integer $n$ whose prime factors are all greater than 18, and that can be expressed as $n = a^3 + b^3$ with positive integers $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can factorize $n$ as\n\n$$\nn = a^3 + b^3 = (a + b)(a^2 - ab + b^2).\n$$\n\nThe first factor $a + b$ must be at least 19, since otherwise $n$ would have a prime factor smaller than 18. Setting $a + b = s$, we have\n\n$$\na^2 - ab + b^2 = a^2 - a(s - a) + (s - a)^2 = 3a^2 - 3as + s^2 = 3\\left(a - \\frac{s}{2}\\right)^2 + \\frac{s^2}{4}\n$$\n\nby completing the square. Thus, the second factor is at least $\\frac{s^2}{4}$ and is minimized when $a$ is closest to $\\frac{s}{2}$. For $s = 19$, $a = 9$ or $a = 10$ gives the second factor $91 = 7 \\cdot 13$, which contains a prime factor less than 18. For $a = 8$ or $a = 11$, it is $97$, which is prime. Thus, $n = 19 \\cdot 97 = 1843 = 11^3 + 8^3$ satisfies the conditions.\n\nIf $s = 19$ and $a < 8$ or $a > 11$,\n\n$$\na^2 - ab + b^2 = 3\\left(a - \\frac{19}{2}\\right)^2 + \\frac{19^2}{4} > 3\\left(\\frac{3}{2}\\right)^2 + \\frac{19^2}{4} = 97,\n$$\n\nso $n > 19 \\cdot 97 = 1843$. If $s > 19$, then $s$ must be at least 20 (actually at least 23, to avoid small prime factors), so\n\n$$\nn = s\\left(3\\left(a - \\frac{s}{2}\\right)^2 + \\frac{s^2}{4}\\right) \\geq s \\cdot \\frac{s^2}{4} \\geq \\frac{20^3}{4} = 2000.\n$$\n\nTherefore, $1843$ is the smallest number with the desired properties.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18646,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the real sequence $x_n$ such that $x_1 \\in \\left(0, \\frac{1}{2}\\right)$ and\n$$\nx_{n+1} = 3x_n^2 - 2n x_n^3, \\quad \\forall n \\ge 1.\n$$\n\na) Prove that $\\lim_{n \\to \\infty} x_n = 0$.\n\nb) For each $n \\ge 1$, let $y_n = x_1 + 2x_2 + \\dots + n x_n$. Prove that $(y_n)$ converges.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we prove by induction that $0 < x_n < \\frac{3}{2n}$ for all $n \\ge 1$.\n\nThe base case $n = 1$ is trivial. For $n = 2$, we have\n$$\n0 < x_1^2 (3 - 2x_1) < 3x_1^2 < \\frac{3}{4} \\Rightarrow 0 < x_2 < \\frac{3}{4}.\n$$\nAssume that $0 < x_k < \\frac{3}{2k}$ for some $k \\ge 2$. By the AM-GM inequality,\n$$\n0 < x_k^2 (3 - 2k x_k) = \\frac{1}{k^2} (k x_k)(k x_k)(3 - 2k x_k) \\le \\frac{1}{k^2} < \\frac{3}{2(k+1)}.\n$$\nTherefore, $0 < x_{k+1} < \\frac{3}{2(k+1)}$. The inductive step is complete.\n\nBy the Squeeze Theorem, $\\lim_{n \\to \\infty} x_n = 0$.\n\nb) It is clear that $x_{n+1} = 3x_n^2 - 2n x_n^3 \\le 3x_n^2$ implies\n$$\nx_{n+2} \\le 3x_{n+1}^2 \\le 27x_n^4 \\le \\frac{3^7}{16 n^4} = \\frac{C}{n^4}, \\quad \\forall n \\ge 1.\n$$\nFor all $n > 2$, we obtain\n$$\n\\begin{aligned}\ny_n &= \\sum_{i=1}^{n} i x_i = x_1 + 2x_2 + \\sum_{i=1}^{n-2} (i+2) x_{i+2} \\\\\n    &\\le x_1 + 2x_2 + C \\sum_{i=1}^{n-2} \\frac{i+2}{i^4} \\le x_1 + 2x_2 + 3C \\sum_{i=1}^{n-2} \\frac{1}{i^2}.\n\\end{aligned}\n$$\nHence, $(y_n)$ is bounded above. Note that $(y_n)$ is increasing, so $(y_n)$ is convergent. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18647,
    "subject": "Mathematics (Olympiad)",
    "question": "Let triangle $ABC$ have barycenter $G$ and circumcenter $O$. The perpendicular bisectors of $GA$, $GB$, and $GC$ intersect at the points $A_1$, $B_1$, and $C_1$, respectively. Prove that $O$ is the barycenter of triangle $A_1B_1C_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$, $E$, and $F$ be the midpoints of the sides $BC$, $AC$, and $AB$, respectively.\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p17_data_58f8de1211.png)\n\nLet $B_1C_1$, $A_1C_1$, and $A_1B_1$ be the perpendicular bisectors of the line segments $GA$, $GB$, and $GC$, respectively. Then the points $A_1$, $B_1$, and $C_1$ are the circumcenters of the triangles $GBC$, $GAC$, and $GAB$, respectively. Hence $A_1D$, $B_1E$, and $C_1F$ are the perpendicular bisectors of the sides $BC$, $AC$, and $AB$, respectively, and therefore they will pass through the circumcenter $O$ of triangle $ABC$.\n\nNext, we will show that $A_1D$, $B_1E$, and $C_1F$ are the medians of triangle $A_1B_1C_1$. Let the extension of $A_1D$ meet $B_1C_1$ at $N$. We will prove that $N$ is the midpoint of the line segment $B_1C_1$.\n\nFrom the inscribed quadrilateral $AMEB_1$ ($\\angle M = \\angle E = 90^\\circ$), we have $\\angle MAE = \\angle MB_1E = \\omega$. Also, from the inscribed quadrilateral $DOEC$ ($\\angle D = \\angle E = 90^\\circ$), we get $\\angle ECD = \\angle EON = \\phi$. Therefore, the triangles $ADC$ and $B_1NO$ are similar, and so\n\n$$\n\\frac{NB_1}{NO} = \\frac{AD}{CD}. \\qquad (1)\n$$\n\nFrom the inscribed quadrilateral $AMFC_1$ ($\\angle M = \\angle F = 90^\\circ$), we have $\\angle MAF = \\angle MC_1F = x$, and similarly from $DOFB$ ($\\angle D = \\angle F = 90^\\circ$), we obtain $\\angle FBD = \\angle FON = y$. From the above equalities, the triangles $ADB$ and $C_1NO$ are similar and therefore:\n\n$$\n\\frac{NC_1}{NO} = \\frac{AD}{BD}. \\qquad (2)\n$$\n\nFrom (1) and (2) we get $NB_1 = NC_1$. In a similar way, we prove that $B_1E$ and $C_1F$ are the other two medians of triangle $A_1B_1C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18648,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA \\perp RA$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/pamphlet1112_main_p27_data_cb5512e03d.png)\n\n**Solution.** Note that there is a dilation centered at $A$ with ratio $2$ sending triangle $AMN$ to $ABC$. Hence the circumcircles of triangles $ABC$ and $AMN$ are tangent at $A$. Denote their common tangent at $A$ by $\\ell$; we note that $\\ell$ is the radical axis of these two circles. We now have a key lemma.\n\n**Lemma 1.** Points $M$, $N$, $P$, and $Q$ lie on a circle.\n\n*Proof.* Let $B_1$ be the point diametrically opposite $B$ on $\\omega$ so that $BB_1$ is a diameter of $\\omega$. Hence $B_1C \\perp BC$ and $B_1A \\perp BA$, meaning that $B_1C \\parallel AH$ and $B_1A \\parallel CH$. Thus $AB_1CH$ is a parallelogram. In parallelogram $AB_1CH$, diagonal $AC$ and $B_1H$ bisect each other, hence $N$ is the midpoint of $B_1H$. Note that $B_1H \\cdot HQ$ is the power of $H$ with respect to $\\omega$, so\n\n$$\nNH \\cdot HQ = \\frac{1}{2} B_1H \\cdot HQ\n$$\n\nis equal to half of the power of $H$ with respect to $\\omega$. In an analogous manner, we can show that $MH \\cdot HP$ is also equal to half of the power of $H$ with respect to $\\omega$. We conclude that $MH \\cdot HP = NH \\cdot HQ$, so $MNPQ$ is cyclic by power of a point. $\\square$\n\nLet us now finish the proof. By Lemma 1, we see that $MN$ is the radical axis of the circumcircles of $AMN$ and $MNPQ$, and that $PQ$ is the radical axis of the circumcircles of $ABC$ and $MNPQ$. Hence $R$ is the radical center of the three circumcircles. Consequently, we conclude that $R$ lies on $\\ell$ and $AR \\perp OA$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18649,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions $(a, b, c)$ to the equation:\n\n$$\na! = \\frac{b! + 4c!}{5}\n$$\n\nwhere $a, b, c$ are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "There are trivial solutions when $a = b = c$, and one non-trivial solution when $a = 2$, $b = 3$, $c = 1$:\n\n$$\n2! = 2 = \\frac{1}{5} \\times 6 + \\frac{4}{5} \\times 1 = \\frac{1}{5} \\times 3! + \\frac{4}{5} \\times 1!\n$$\n\nWe will prove that there are no other solutions. If $b = c$, the equation becomes $a! = b!$ and we must have $a = b$, which gives the trivial solution.\n\nSuppose $b \\neq c$. Then, as $a!$ is a weighted average of $b!$ and $c!$, $a!$ lies strictly between $b!$ and $c!$. Since the factorial is strictly increasing, $a$ lies strictly between $b$ and $c$, so $\\max\\{b, c\\} \\geq a+1$. This gives:\n\n$$\na! = \\frac{b! + 4c!}{5} > \\frac{b! + c!}{5} > \\frac{(a+1)!}{5} = \\frac{a+1}{5}a!\n$$\n\nDividing by $a!$ and multiplying by $5$ gives $5 > a + 1$, so $a \\in \\{1, 2, 3\\}$. The case $a = 1$ is excluded, since $a$ strictly exceeds the lesser of $b$ and $c$.\n\nFor $a = 2$, $b! + 4c! = 10$, so $c! < 2.5$ and $c \\leq 2$. The sub-case $c = 1$ gives the non-trivial solution above, while $c = 2$ is not possible as $a$ is strictly between $b$ and $c$.\n\nFor $a = 3$, we solve $b! + 4c! = 30$. As $c! < 7.5$, $c \\leq 3$. We cannot have $c = 3$ because $a$ is strictly between $b$ and $c$. When $c = 2$, $b! + 8 = 30$; when $c = 1$, $b! + 4 = 30$; neither has a solution.\n\nThus, the only solutions are the trivial ones $a = b = c$ and the non-trivial solution $(a, b, c) = (2, 3, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18650,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integer numbers $n \\geq 3$ such that the regular $n$-gon can be decomposed into isosceles triangles by noncrossing diagonals.",
    "options": [],
    "answer": "See solution",
    "solution": "The required numbers are of the form $n = 2^r(2^s + 1)$, where $r$ and $s$ are nonnegative integers which do not vanish simultaneously. Clearly, any such $n$ works.\n\nTo establish the converse, let $K$ be a regular $n$-gon, $n \\geq 4$, 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 $n \\geq 4$, $T_e$ 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": 18651,
    "subject": "Mathematics (Olympiad)",
    "question": "In a mathematical competition, 6 problems were posed to the contestants. Each pair of problems was solved by more than $\\frac{2}{5}$ of the contestants. Nobody solved all 6 problems. Show that there are at least 2 contestants who each solved exactly 5 problems.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:**\n\nSuppose there were $n$ contestants. Let $p_{ij}$, with $1 \\le i < j \\le 6$, be the number of contestants who solved problems $i$ and $j$, and let $n_r$, with $0 \\le r \\le 6$, be the number of contestants who solved exactly $r$ problems. Clearly, $n_6 = 0$ and $n_0 + n_1 + \\dots + n_5 = n$.\n\nBy the given condition, $p_{ij} > \\frac{2n}{5}$, or $5p_{ij} > 2n$. Hence $5p_{ij} \\ge 2n+1$, or $p_{ij} \\ge \\frac{2n+1}{5}$.\n\nDefine the set\n\n$$\nU = \\{(c, \\{i, j\\}) \\mid \\text{contestant } c \\text{ solved problems } i \\text{ and } j\\}\n$$\n\nIf we compute $|U|$, the number of elements in $U$, by summing over all pairs $\\{i, j\\}$, we have\n\n$$\n|U| = \\sum_{1 \\le i < j \\le 6} p_{ij} \\ge 15 \\cdot \\frac{2n+1}{5} = 6n + 3 = 6(n_0 + n_1 + \\dots + n_5) + 3.\n$$\n\nA contestant who solved exactly $r$ problems contributes a “1” to $\\binom{r}{2}$ summands in this sum (where $\\binom{r}{2} = 0$ for $r < 2$), if we compute $|U|$ by summing over all contestants $c$. Therefore,\n\n$$\n|U| = \\sum_{r=0}^{6} \\binom{r}{2} n_r = n_2 + 3n_3 + 6n_4 + 10n_5.\n$$\n\nIt follows that\n\n$$\nn_2 + 3n_3 + 6n_4 + 10n_5 \\ge 6(n_0 + n_1 + \\dots + n_5) + 3,\n$$\n\nor\n\n$$\n4n_5 \\ge 3 + 6n_0 + 6n_1 + 5n_2 + 3n_3 \\ge 3,\n$$\n\nimplying that $n_5 \\ge 1$. We need to show that $n_5 \\ge 2$. We approach indirectly by assuming that $n_5 = 1$. Call this person the winner ($W$), and without loss of generality, assume the winner failed to solve problem 6. Then $n_0 = n_1 = n_2 = n_3 = 0$, so $n_4 = n - 1$, and\n\n$$\n|U| = n_2 + 3n_3 + 6n_4 + 10n_5 = 6n + 4 > 6n + 3 = 15 \\cdot \\frac{2n+1}{5}.\n$$\n\nIt follows that $p_{ij} = \\frac{2n+1}{5}$ for 14 out of the 15 total pairs $(i, j)$ with $1 \\le i < j \\le 6$, and for the remaining pair $(s, t)$, $p_{st} = \\frac{2n+1}{5} + 1 = \\frac{2n+6}{5}$.\n\nWithout loss of generality, assume $1 < s < t \\le 6$.\n\nFirst, consider the sum\n\n$$\n\\nu_1 = p_{12} + p_{13} + p_{14} + p_{15} + p_{16} = 5 \\cdot \\frac{2n+1}{5} = 2n+1,\n$$\n\nbecause $p_{1k} \\ne p_{st}$. Suppose problem 1 was solved by $x$ contestants $c_1, c_2, \\dots, c_x$ other than the winner. Each of these contestants solved 3 problems other than problem 1, so each contributed “3” to $\\nu_1$. The winner contributed “4” to $\\nu_1$. Thus $\\nu_1 = 3x + 4 = 2n + 1$, implying $n$ is divisible by 3.\n\nSecond, consider the sum\n\n$$\nv_6 = p_{16} + p_{26} + p_{36} + p_{46} + p_{56} =\n\\begin{cases}\n2n+1 & \\text{if } p_{k6} \\ne p_{st} \\text{ for all } 1 \\le k \\le 5, \\\\\n2n+2 & \\text{if } p_{k6} = p_{st} \\text{ for some } 1 \\le k \\le 5.\n\\end{cases}\n$$\n\nSuppose problem 6 was solved by $y$ contestants $d_1, d_2, \\dots, d_y$. The winner was not among them, and each of these contestants solved 3 problems other than problem 6, so each contributed “3” to $v_6$. Thus\n\n$$\nv_6 = 3y =\n\\begin{cases}\n2n+1, \\\\\n2n+2.\n\\end{cases}\n$$\n\nIn either case, $n$ is not divisible by 3, which contradicts our previous observation. Therefore, our assumption $n_5 = 1$ was wrong, and so $n_5 \\ge 2$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18652,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $f$, $g$, and $h$ on $\\mathbb{Z} \\times \\mathbb{Z} \\times \\mathbb{Z}$ as follows:\n\n$$\nf(x, y, z) = (3x + 2y + 2z,\\ 2x + 2y + z,\\ 2x + y + 2z),\n$$\n\n$$\ng(x, y, z) = (3x + 2y - 2z,\\ 2x + 2y - z,\\ 2x + y - 2z),\n$$\n\n$$\nh(x, y, z) = (3x - 2y + 2z,\\ 2x - y + 2z,\\ 2x - 2y + z).\n$$\n\nGiven a primitive Pythagorean triplet $(x, y, z)$ with $x > y > z$, prove that starting from $(5, 4, 3)$, the triplet $(x, y, z)$ can be obtained, in a unique way, by repeated application of $f$, $g$, and $h$ in some order. (Example: $(697, 528, 455) = f \\circ h \\circ g \\circ h(5, 4, 3)$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nu = u(x, y, z) = 3x - 2y - 2z,$$\n$$v = v(x, y, z) = -2x + 2y + z,$$\n$$w = w(x, y, z) = -2x + y + 2z.$$\n\nFirst, if $(x, y, z) \\neq (5, 4, 3)$ is a primitive Pythagorean triplet with $x > y > z > 0$, it is easy to check that $u^2 = v^2 + w^2$. Also, $x^2 + (2y - 2z)^2 > 0$ implies $3x > 2y + 2z$, so $u(x, y, z) > 0$.\n\nSimilarly, $4y > 3z$ implies $2y + z > 2x$, so $v(x, y, z) > 0$.\n\nNext, $(3y - 4z)^2 > 0$ leads to $5x > 4y + 3z$, so $u(x, y, z) > v(x, y, z)$. Similarly, $u(x, y, z) > w(x, y, z)$, and since $x > y$, $u(x, y, z) > -w(x, y, z)$. Also, $y > z$ implies $v(x, y, z) > w(x, y, z)$. If $y + z > x$, then $u(x, y, z) < x$.\n\nNow, define\n\n$$\nf_1(x, y, z) = (u, v, w),\n$$\n$$\ng_1(x, y, z) = (u, v, -w),\n$$\n$$\nh_1(x, y, z) = (u, -w, v).\n$$\n\nIt is easy to see that $f_1(f(x, y, z)) = g_1(g(x, y, z)) = h_1(h(x, y, z)) = (x, y, z)$. Also, $\\gcd(u, v, w) = \\gcd(x, y, z)$.\n\nSuppose, for contradiction, that there is a primitive Pythagorean triplet $(a, b, c)$ with $a > b > c > 0$ that cannot be obtained from $(5, 4, 3)$ by repeated application of $f$, $g$, and $h$, and $a$ is minimal. All triplets with $a > x > y > z > 0$ are obtainable (uniquely). Clearly, $a > 5$.\n\nLet $f_1(a, b, c) = (k, l, m)$. Then $g_1(a, b, c) = (k, l, -m)$ and $h_1(a, b, c) = (k, -m, l)$. As before, $k$ and $l$ are positive, $k > l$, $k > m$, and $k > -m$. Thus, there is a unique $\\Delta \\in \\{f_1, g_1, h_1\\}$ such that $\\Delta(a, b, c) = (r, s, t)$ with $r > s > t > 0$. Also, $a > r$. By minimality, $(r, s, t)$ is obtainable from $(5, 4, 3)$ uniquely. Since $\\Delta(a, b, c) = (r, s, t)$, applying one of $f$, $g$, or $h$ to $(r, s, t)$ gives $(a, b, c)$. Thus, $(a, b, c)$ is also obtainable, a contradiction. Uniqueness follows from the uniqueness of $\\Delta$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18653,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 1.1, $I$ is the incentre of $\\triangle ABC$, $AB > AC$. Points $P$ and $Q$ are the projections of $I$ onto sides $AB$ and $AC$, respectively. Line $PQ$ intersects the circumcircle of $\\triangle ABC$ at points $X$ and $Y$ ($P$ is between $X$ and $Y$). Given that points $B$, $I$, $P$, $X$ are concyclic, prove that points $C$, $I$, $Q$, $Y$ are concyclic.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p105_data_f31f591327.png)\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p105_data_3f20786e66.png)",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in Fig. 1.2, denote the circumcircle of $\\triangle ABC$ as $\\omega$.\n\nSince $\\angle API = \\angle AQI = 90^\\circ$, points $A$, $P$, $I$, $Q$ are concyclic.\n\nTherefore,\n$$\n\\begin{align*}\n\\angle BPQ &= \\angle BPI + \\angle IPQ \\\\\n&= 90^\\circ + \\angle IAQ \\\\\n&= 90^\\circ + \\frac{1}{2} \\angle BAC \\\\\n&= \\angle BIC.\n\\end{align*}\n$$\n\nAnd since points $B$, $I$, $P$, $X$ are concyclic, we have $\\angle BPX = \\angle BIX$, and thus\n$$\n\\angle BIX + \\angle BIC = \\angle BPX + \\angle BPQ = 180^\\circ.\n$$\n\nHence, points $C$, $I$, $X$ are collinear.\n\nSince $B$, $I$, $P$, $X$ are concyclic, there is $\\angle BXI = \\angle BPI = 90^\\circ$, and thus $BX \\perp IX$, i.e., $BX \\perp CX$. Hence, $\\angle BAC = \\angle BXC = 90^\\circ$.\n\nTherefore, quadrilateral $APIQ$ is a square, and $PQ$ is the perpendicular bisector of segment $AI$.\n\nLet $Y'$ be the midpoint of $\\overarc{AC}$. Then by the well-known theorem of the incentre, we have $Y'A = Y'I$. Thus, $Y'$ is the intersection of the perpendicular bisector of $AI$ and circle $\\omega$. In addition, since line $PQ$ intersects circle $\\omega$ at two points $X$, $Y$ and $Y'$ is obviously different from $X$, so $Y'$ and $Y$ coincide. Therefore, $Y$ is the midpoint of $\\overarc{AC}$.\n\nConsequently, points $B$, $I$, $Y$ are collinear. Therefore, $\\angle IYC = \\angle BYC = \\angle BAC = 90^\\circ = \\angle IQC$, and thus points $C$, $I$, $Q$, $Y$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18654,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "The points $E$, $K$, $H$, $G$ are on the circle of diameter $GE$, so\n\n$$\n\\angle EHK = \\angle EGK. \\qquad (\\dagger)\n$$\n\nAlso, from $\\angle DCA = \\angle DBA$ and $\\frac{CE}{CD} = \\frac{BE}{BA}$ it follows\n\n$$\n\\frac{CE}{CG} = \\frac{2CE}{CD} = \\frac{2BE}{BA} = \\frac{BE}{BF},\n$$\n\ntherefore $\\triangle CGE \\sim \\triangle BFE$. In particular, $\\angle EGC = \\angle BFE$, so by (\\dagger)\n\n$$\n\\angle EHK = \\angle BFE.\n$$\n\nBut $HE \\perp FB$ and so, since $FE$ and $HK$ are obtained by rotations of these lines by the same (directed) angle, $FE \\perp HK$.\n\n![](images/shortlistBMO_2011_p13_data_ddfcaf08f6.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18655,
    "subject": "Mathematics (Olympiad)",
    "question": "In acute $\\triangle DEF$, let $X', Y'$ be points such that $DX'E$ and $FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$. Prove that $\\triangle X'A'Y'$ is an isosceles right triangle with a right angle at $A'$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the composition of rotations:\n\n$$\n\\Gamma = \\operatorname{Rot}(A', 180^{\\circ}) \\circ \\operatorname{Rot}(Y', 90^{\\circ}) \\circ \\operatorname{Rot}(X', 90^{\\circ})\n$$\n\nThe angles add up to $360^{\\circ}$, so $\\Gamma$ is a translation. However, $\\Gamma$ keeps $E$ fixed, thus $\\Gamma$ is the identity map. Hence, we have\n\n$$\n\\operatorname{Rot}(A', 180^{\\circ}) = \\operatorname{Rot}(Y', 90^{\\circ}) \\circ \\operatorname{Rot}(X', 90^{\\circ})\n$$\n\nWe finish by noting that the center of rotation of the composition of the two rotations $\\operatorname{Rot}(Y', 90^{\\circ})$ and $\\operatorname{Rot}(X', 90^{\\circ})$ is given by a point $O$ such that $\\angle OX'Y' = 45^{\\circ}$ and $\\angle OY'X' = 45^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18656,
    "subject": "Mathematics (Olympiad)",
    "question": "A *pointy number* of $k$ digits has the form:\n\n- For $k = 5$: $a, a \\pm 1, a \\pm 2, a \\pm 1, a$\n- For $k = 7$: $a, a \\pm 1, a \\pm 2, a \\pm 3, a \\pm 2, a \\pm 1, a$\n- For $k = 9$: $a, a \\pm 1, a \\pm 2, a \\pm 3, a \\pm 4, a \\pm 3, a \\pm 2, a \\pm 1, a$\n- For $k = 11$: $a, a \\pm 1, a \\pm 2, a \\pm 3, a \\pm 4, a \\pm 5, a \\pm 4, a \\pm 3, a \\pm 2, a \\pm 1, a$\n- For $k = 13$: $a, a \\pm 1, a \\pm 2, a \\pm 3, a \\pm 4, a \\pm 5, a \\pm 6, a \\pm 5, a \\pm 4, a \\pm 3, a \\pm 2, a \\pm 1, a$\n- For $k = 15$: $a, a \\pm 1, \\dots, a \\pm 6, a \\pm 7, a \\pm 6, \\dots, a \\pm 1, a$\n- For $k = 17$: $a, a \\pm 1, \\dots, a \\pm 7, a \\pm 8, a \\pm 7, \\dots, a \\pm 1, a$\n- For $k = 19$: $a, a \\pm 1, \\dots, a \\pm 8, a \\pm 9, a \\pm 8, \\dots, a \\pm 1, a$\n\nHow many pointy numbers are divisible by 6?\n\n**d**: Show that the sum of an upward and a downward pointy number of the same length is never prime.",
    "options": [],
    "answer": "See solution",
    "solution": "To determine which pointy numbers are divisible by 6, we analyze the sum of their digits and divisibility conditions for each length:\n\n- For 5 digits: Only 21012, 45654, and 87678 are divisible by 6.\n- For 7 digits: Only 6543456 and 6789876 are divisible by 6.\n- For 9 digits: None are divisible by 6.\n- For 11 digits: Only 45678987654 and 87654345678 are divisible by 6.\n- For 13 digits: Only 6543210123456 is divisible by 6.\n- For 15 digits: None are divisible by 6.\n- For 17 digits: Only 87654321012345678 is divisible by 6.\n- For 19 digits: 9876543210123456789 is not even, so not divisible by 6.\n\nThus, there are 9 pointy numbers divisible by 6.\n\n**d**: For upward and downward pointy numbers of the same length, each pair of same-placed digits sums to $n$, so the total sum is:\n\n$$\nn + 10n + 100n + \\dots = n(1 + 10 + 100 + \\dots)\n$$\n\nSince the first and last digits are at least 1, $n \\geq 2$, and the sum in brackets is at least 111 (for three digits). Therefore, the sum of the two pointy numbers is composite and never prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18657,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Define a sequence by setting $a_1 = n$ and, for each $k > 1$, letting $a_k$ be the unique integer in the range $0 \\leq a_k \\leq k - 1$ for which $a_1 + a_2 + \\dots + a_k$ is divisible by $k$. For instance, when $n = 9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \\dots$. Prove that for any $n$ the sequence $a_1, a_2, a_3, \\dots$ eventually becomes constant.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k \\geq 1$, let\n$$\ns_k = a_1 + a_2 + \\dots + a_k.\n$$\nWe have\n$$\n\\frac{s_{k+1}}{k+1} < \\frac{s_{k+1}}{k} = \\frac{s_k + a_{k+1}}{k} \\leq \\frac{s_k + k}{k} = \\frac{s_k}{k} + 1.\n$$\nOn the other hand, for each $k$, $s_k / k$ is a positive integer. Therefore,\n$$\n\\frac{s_{k+1}}{k+1} \\leq \\frac{s_k}{k},\n$$\nand the sequence of quotients $s_k / k$ is eventually constant. If $s_{k+1} / (k+1) = s_k / k$, then\n$$\na_{k+1} = s_{k+1} - s_k = \\frac{(k+1)s_k}{k} - s_k = \\frac{s_k}{k},\n$$\nshowing that the sequence $a_k$ is eventually constant as well.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18658,
    "subject": "Mathematics (Olympiad)",
    "question": "Сколько минимально необходимо выворачиваний колпаков, чтобы среди 1000 гномов каждый мог сказать другому заветную фразу, если гномы могут быть красными или синими (по цвету колпака), и гном может сказать фразу другому только если они разноцветны: синий говорит правду, красный — лжёт?",
    "options": [],
    "answer": "See solution",
    "solution": "Назовём 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": 18659,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC > AB$ and incircle $\\omega$. Let $\\omega$ touch the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. Let $X$ and $Y$ be points outside $\\triangle ABC$ satisfying\n\n$$\n\\angle BDX = \\angle XEA = \\angle YDC = \\angle AFY = 45^{\\circ}.\n$$\n\nProve that the circumcircles of $\\triangle AXY$, $\\triangle AEF$, and $\\triangle ABC$ meet at a point $Z \\neq A$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/IND_ABooklet_2024_p33_data_6e3913d4a7.png)\n\nWe have that $AB \\neq AC$, so $(AEF)$ and $(ABC)$ are not tangent at $A$, thus there is a point $S \\neq A$ which is the second intersection of $(AEF)$ and $(ABC)$.\n\nConsider an inversion about the incircle and let the inverse of a point $P$ be denoted by $P'$. Note that $A'$ is the midpoint of $EF$. $S'$ is the foot from $D$ onto $EF$ as $S' \\neq A'$ is on the nine-point circle of $\\triangle DEF$ as well as on $EF$.\n\nAlso, since $X$ satisfies $\\angle XEI = \\angle XDI = 45^{\\circ}$, point $X'$ satisfies $\\angle EX'I = \\angle DX'I = 45^{\\circ}$. Note that $X$ lies on the same side of $DE$ as $I$, which in turn is the same side of $DE$ as $F$ since $\\triangle DEF$ is acute.\n\nTaking $DEF$ to be the reference triangle, the problem becomes the following:\n\n*Inverted Problem:* In acute $\\triangle DEF$, let $X', Y'$ be points such that $DX'E$ and $FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$ and $S'$ the foot onto $EF$ from $D$. Prove that points $A', S', X', Y'$ are concyclic.\n\nWe will prove this by showing that $\\angle X'A'Y' = \\angle X'S'Y' = 90^{\\circ}$.\n\n**Claim 1.** $\\angle X'S'Y' = 90^{\\circ}$.\n\n$$\n\\textit{Proof.}\\quad \\text{Observe that } \\angle X'S'Y' = \\angle X'S'D + \\angle DS'Y' = \\angle X'ED + \\angle DFY' = 45^{\\circ} + 45^{\\circ} = 90^{\\circ} \\quad \\square\n$$\n\n**Claim 2.** $X'A'Y'$ is a right isosceles triangle with a right angle at $A'$.\n\n*Proof.* To prove this, we will prove that $\\triangle X'A'Y' \\sim \\triangle X'MD$ where $M$ is the midpoint of $DE$. But observe that by spiral similarity, this is equivalent to showing that $\\triangle X'MA' \\sim \\triangle X'DY'$.\n\nNow,\n\n$$\n\\angle X'MA' = 90^{\\circ} - \\angle FDE = 45^{\\circ} + 45^{\\circ} - \\angle FDE = \\angle FDX' + \\angle Y'DE - \\angle FDE = \\angle X'DY'\n$$\n\nand also\n\n$$\n\\frac{X'M}{MA'} = \\frac{DE}{DF} = \\frac{X'D}{DY'}\n$$\n\nThus, we get the similarity by SAS similarity and thus\n\n$$\n\\triangle X'MA' \\sim \\triangle X'DY' \\implies \\triangle X'A'Y' \\sim \\triangle X'MD' \\implies \\angle X'A'Y = \\angle X'MD = 90^{\\circ}\n$$\n\n\\boxed{}\n\nThus, we are done!",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18660,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Point $D$ lies on side $BC$. Let $O$, $O_1$, and $O_2$ be the circumcenters of triangles $ABC$, $ABD$, and $ACD$, respectively. Prove that the circumcircles of triangles $BOO_1$ and $COO_2$ meet on line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be on $BC$ such that $OP$ is parallel to $AD$. If $O$ is on $BC$, then $P = O$ and the result is clear. We claim that the circumcircles of $BOO_1$ and $COO_2$ both pass through $P$. One of the angles $\\widehat{ADB}$ and $\\widehat{ADC}$ is not acute. Without loss of generality, assume that $\\widehat{ADB} \\ge 90^\\circ$. Then $O_1$ does not lie in the interior of triangle $ADB$. Note that\n\n$$\n\\widehat{OO_1B} = \\frac{\\widehat{AO_1B}}{2} = 180^\\circ - \\widehat{ADB} = 180^\\circ - \\widehat{OPB},\n$$\n\nimplying that $BO_1OP$ is cyclic.\n\n![](images/Saudi_Arabia_booklet_2012_p48_data_e27eae6586.png)\n\nSimilarly, we can show that $CO_2OP$ is cyclic. Therefore, the circumcircles of $BOO_1$ and $COO_2$ pass through $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18661,
    "subject": "Mathematics (Olympiad)",
    "question": "The median $AM$, where $M \\in BC$, of the acute-angled triangle $ABC$ intersects the circumcircle of the triangle at $D$. Let $E$ be the symmetric point of $A$ with respect to $M$. Prove that $BC$ is the common tangent of the circumcircles of triangles $BDE$ and $CDE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABEC$ is a parallelogram, we have $\\angle CBD \\equiv \\angle CAD \\equiv \\angle AEB$, so $BC$ is tangent to the circumcircle of triangle $BDE$. Similarly, from $\\angle BCD \\equiv \\angle BAD \\equiv \\angle AEC$, it follows that $BC$ is tangent to the circumcircle of triangle $CDE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18662,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\{a_n\\}_{n=1}^\\infty$ of positive integers satisfies the condition $a_{n+1} = a_n + \\tau(n)$ for all positive integers $n$, where $\\tau(n)$ is the number of positive integer divisors of $n$. Determine whether two consecutive terms of this sequence can be perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "There are no two such consecutive terms.\n\nAssume that $a_n = x^2$ and $a_{n+1} = y^2$ for some positive integers $x, y$. Then\n\n$$\ny^2 = a_{n+1} = a_n + \\tau(n) = x^2 + \\tau(n).\n$$\n\nThus,\n$$\ny^2 - x^2 = \\tau(n) \\implies (y - x)(y + x) = \\tau(n).\n$$\n\nSince $y > x$, $y - x \\geq 1$, so $y + x \\leq \\tau(n)$. But $\\tau(n) \\leq 2\\sqrt{n}$, so $y + x \\leq 2\\sqrt{n}$. Also, $a_n = x^2 \\geq 1$ and the sequence is strictly increasing, so $a_n < n$ is impossible for large $n$.\n\nTherefore, it is not possible for two consecutive terms of the sequence to both be perfect squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18663,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 10 piles of stones, with 3, 4, 5, ..., 12 stones respectively. At each step, you can pick three piles and either:\n- add 1 stone to the first pile, 2 stones to the second pile, and 3 stones to the third pile, or\n- remove 1 stone from the first pile, 2 stones from the second pile, and 3 stones from the third pile (provided each pile has enough stones).\n\nIs it possible, after a finite number of such operations, to have exactly 2011 stones in each pile?",
    "options": [],
    "answer": "See solution",
    "solution": "After each operation, the total number of stones changes by a number divisible by 3. At the end, the total number would be $2011 \\times 10 = 20110$, which is not divisible by 3. However, at the start, the total number is $3 + 4 + 5 + \\dots + 12 = 75$, which is divisible by 3. This contradiction shows it is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18664,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the centre of $\\Gamma$, and let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Note that $OM \\perp AB$ and $ON \\perp AC$.\n\nLet $F'$ and $G'$ be the midpoints of $BD$ and $CE$, respectively. Note that $FF' \\perp AB$ and $GG' \\perp AC$.\n\nLet $X$ and $Y$ be the projections of $O$ onto the lines $FF'$ and $GG'$, respectively.\n\n![](images/2018-Australian-Scene-W2_p170_data_fd77d4bdc0.png)\n\nLet $FG$ intersect lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that $DE$ and $FG$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $MOXF'$ is a rectangle due to all its right angles. Hence we have\n\n$$\nOX = MF' = MB - F'B = \\frac{1}{2}AB - \\frac{1}{2}DB = \\frac{1}{2}AD.\n$$\n\nSimilarly $OY = \\frac{1}{2}AE$. Since $AD = AE$, we deduce that $OX = OY$. But $OF = OG$ (radii of $\\Gamma$), and $OX \\perp XF$ and $OY \\perp YG$. Hence $\\triangle OXF \\equiv \\triangle OYG$ (RHS). Thus $\\angle XFO = \\angle OGY$. Since also $\\angle OFG = \\angle FGO$, we have $\\angle F'FG = \\angle FGG'$.\n\nLet $FG$ intersect lines $AB$ and $AC$ at $P$ and $Q$, respectively. From the angle sums in $\\triangle PFF'$ and $\\triangle QGG'$, we deduce $\\angle FPF' = \\angle G'QG$, and so $\\angle QPA = \\angle AQP$. Hence $\\triangle APQ$ is isosceles with apex $A$. Since $\\triangle ADE$ is also isosceles with apex $A$, it follows that $DE$ and $FG$ are parallel. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18665,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a given integer.\n\n1. Prove that one can arrange all the subsets of the set $\\{1,2,\\ldots,n\\}$ as a sequence of subsets $A_1, A_2, \\dots, A_{2^n}$, such that $|A_{i+1}| = |A_i| + 1$ or $|A_i| - 1$, where $i = 1,2,3,\\dots, 2^n$, and $A_{2^{n-1}} = A_1$.\n\n2. Determine, with proof, all possible values of the sum\n\n$$\n\\sum_{i=1}^{2^n} (-1)^i S(A_i),\n$$\nwhere $S(A_i) = \\sum_{x \\in A_i} x$ and $S(\\emptyset) = 0$, for any subset sequence $A_1, A_2, \\dots, A_{2^n}$ satisfying the condition in (1).",
    "options": [],
    "answer": "See solution",
    "solution": "(1) We prove by mathematical induction that there exists a sequence $A_1, A_2, \\dots, A_{2^n}$, such that $A_1 = \\{1\\}$, $A_{2^n} = \\emptyset$, and the sequence satisfies the condition in (1).\n\nWhen $n=2$, the sequence $\\{1\\}, \\{1,2\\}, \\{2\\}, \\emptyset$ of $\\{1,2\\}$ works.\n\nAssume that when $n=k$, there exists such a sequence $B_1, B_2, \\dots, B_{2^k}$ of subsets of $\\{1, 2, \\dots, k\\}$. For $n=k+1$, construct a sequence of subsets of $\\{1, 2, \\dots, k+1\\}$ as follows:\n\n$$\n\\begin{align*}\nA_1 &= B_1 = \\{1\\}, \\\\\nA_i &= B_{i-1} \\cup \\{k+1\\}, \\quad i = 2, 3, \\dots, 2^k + 1, \\\\\nA_j &= B_{j-2^k}, \\quad j = 2^k + 2, 2^k + 3, \\dots, 2^{k+1}.\n\\end{align*}\n$$\n\nOne can check that this sequence fulfills the required conditions. By induction, (1) holds for all $n \\ge 2$.\n\n(2) We will show that the sum is $0$ independent of the arrangement. Without loss of generality, assume $A_1 = \\{1\\}$ (otherwise, shift the index cyclically). Since $|A_{i+1}| = |A_i|+1$ or $|A_i|-1$, their parities alternate, so the parity of the index and the subset's cardinality are the same.\n\nThus,\n\n$$\n\\sum_{i=1}^{2^n} (-1)^i S(A_i) = \\sum_{A \\in P} S(A) - \\sum_{A \\in Q} S(A),\n$$\n\nwhere $P$ consists of all subsets of $\\{1, 2, \\dots, n\\}$ with even cardinality, and $Q$ consists of those with odd cardinality. But for each $x \\in \\{1,2,\\ldots,n\\}$, $x$ appears equally often in $P$ and $Q$, so the total sum is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18666,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider the cube $ABCD EFGH$ with side $a$ cm, $a > 0$, and the points $K$, $L$ on the segments $AC$, $EG$ respectively, 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\n$$\nHN = FP = \\frac{(\\sqrt{2} - 1)a}{2} \\text{ cm}.\n$$\n\nKnowing that lines $MC$ and $LK$ intersect in $Q$, prove that $MNPQ$ is a regular tetrahedron.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $\\Delta MEP$ and $\\Delta MEN$ are congruent, therefore $MN = MP$.\n\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.\n\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 $\\frac{ML_1}{MO_1} = \\frac{M_1K}{M_1C} = \\frac{2}{3}$ so $L_1$ is the centroid of the triangle $MNP$.\n\n![](images/RMC_2020_p17_data_f8a729f6bb.png)\n\nNotice that $AL = AG = \\frac{3\\sqrt{2}a}{4}$ 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.\n\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": 18667,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, $AB$ and $CD$ are two chords in the circle $\\odot O$ meeting at point $E$, and $AB \\neq CD$. $\\odot I$ is tangent to $\\odot O$ internally at point $F$, and is tangent to the chords $AB$ and $CD$ at points $G$ and $H$, respectively. $l$ is a line passing through $O$, meeting $AB$ and $CD$ at points $P$ and $Q$, respectively, such that $EP = EQ$. Line $EF$ meets the line $l$ at point $M$. Prove that the\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p247_data_74dc7caf6c.png)\n\nline through $M$ and parallel to the line $AB$ is tangent to the circle $\\odot O$.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the figures below, draw a line parallel to $AB$ and tangent to circle $\\odot O$ at point $L$, which meets the common tangent line to these two circles at a point $S$. Let $R$ be the intersection of lines $FS$ and $BA$, and join segments $LF$ and $GF$.\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p248_data_1301d0003b.png)\n\nFirst, we prove that the points $L$, $G$, and $F$ are collinear. As both $SL$ and $SF$ are tangent to $\\odot O$, $SL = SF$; as both $RG$ and $RF$ are tangent to $\\odot I$, $RG = RF$.\n\nAs $SL \\parallel RG$, we have $\\angle LSF = \\angle GRF$, so\n\n$$\n\\begin{aligned}\n\\angle LFS &= \\frac{180^\\circ - \\angle LSF}{2} = \\frac{180^\\circ - \\angle GRF}{2} \\\\\n&= \\angle GFR,\n\\end{aligned}\n$$\n\nand hence $L$, $G$, and $F$ are collinear.\n\nSimilarly, as shown in the next figure, draw a line parallel to $CD$ and tangent to $\\odot O$ at point $J$, then $F$, $H$, and $J$ are collinear.\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p248_data_419dbcbd86.png)\n\nLet tangent lines to the circle $\\odot O$ at the points $L$ and $J$ meet $EF$ at points $M_1$ and $M_2$, respectively. In the following, we prove that the points $M_1$ and $M_2$ coincide. It follows from the homothety centered at $F$ mapping $\\odot O$ to $\\odot I$ that $LJ \\parallel GH$, then\n$$\n\\frac{M_1E}{EF} = \\frac{LG}{GF} = \\frac{JH}{HF} = \\frac{M_2E}{EF},\n$$\nand hence $M_1$ and $M_2$ coincide. Denote this point by $K$.\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p249_data_2b65a48324.png)\n\nFinally, we want to prove that points $M$ and $K$ also coincide. It suffices to show that $K$ lies on the line $l$.\n\nAs shown in the next figure, join $KO$. As $KL$ and $KJ$ are tangent to $\\odot O$, $\\angle LKO = \\angle JKO$.\n\nNote that $KO$ bisects $\\angle LKJ$. It follows from $KL \\parallel AB$ and $KJ \\parallel CD$ that the line $KO$ meets lines $AB$ and $CD$ with the same angles of intersection, and hence $KO$ is just the line $l$, i.e., $K$ lies on the line $l$.\n\nHence, $K$ is just the intersection point $M$ of line $EF$ and $l$.\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p249_data_95d27708d5.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18668,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. A cube of size $(2N+1) \\times (2N+1) \\times (2N+1)$ is made of $(2N+1)^3$ unit cubes; each unit cube is either black or white. It is known that among any 8 unit cubes which form a $2 \\times 2 \\times 2$ cube, the number of black unit cubes is not greater than 4. Find the greatest possible total number of black unit cubes.",
    "options": [],
    "answer": "See solution",
    "solution": "$(N+1)^2(4N+1)$.\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$$\n\\begin{aligned}\n0 \\ge \\Sigma &\\ge s_1 + \\dots + s_\\ell - s_{\\ell+1} - \\dots - s_{(2N+1)^3} > \\\\\n&> s_1 + \\dots + s_k - s_{k+1} - \\dots - s_{(2N+1)^3} = 0,\n\\end{aligned}\n$$\n\nsince $s_{k+1} \\ge 4$. This contradiction proves $\\ell \\le k$.\n\n_Note._ Alternative proof: Consider four vertices of the large cube forming a regular tetrahedron. Mark a unit cube vertex if all coordinates of its vector to the nearest tetrahedron vertex are odd. If $p$ is the number of marked vertices and $q$ the number of cubes without marked vertices, the number of black cubes is at most $4p + q = k$. The argument follows similarly to the main solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18669,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be real numbers in the interval $(0, \\frac{\\pi}{2})$. Prove that\n$$\n\\frac{\\sin a \\sin(a-b) \\sin(a-c)}{\\sin(b+c)} + \\frac{\\sin b \\sin(b-c) \\sin(b-a)}{\\sin(c+a)} + \\frac{\\sin c \\sin(c-a) \\sin(c-b)}{\\sin(a+b)} \\ge 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the **Product-to-sum formulas** and the **Double-angle formulas**, we have\n$$\n\\sin(\\alpha - \\beta) \\sin(\\alpha + \\beta) = \\frac{1}{2}[\\cos 2\\beta - \\cos 2\\alpha] = \\sin^2 \\alpha - \\sin^2 \\beta.\n$$\nHence, we obtain\n$$\n\\sin a \\sin(a-b) \\sin(a-c) \\sin(a+b) \\sin(a+c) = \\sin c (\\sin^2 a - \\sin^2 b)(\\sin^2 a - \\sin^2 c)\n$$\nand its analogous forms. Therefore, it suffices to prove that\n$$\nx(x^2 - y^2)(x^2 - z^2) + y(y^2 - z^2)(y^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge 0,\n$$\nwhere $x = \\sin a$, $y = \\sin b$, and $z = \\sin c$ (hence $x, y, z > 0$). Since the last inequality is symmetric with respect to $x, y, z$, we may assume that $x \\ge y \\ge z > 0$. It suffices to prove that\n$$\nx(y^2 - x^2)(z^2 - x^2) + z(z^2 - x^2)(z^2 - y^2) \\ge y(z^2 - y^2)(y^2 - x^2),\n$$\nwhich is evident as\n$$\nx(y^2 - x^2)(z^2 - x^2) \\ge 0\n$$\nand\n$$\nz(z^2 - x^2)(z^2 - y^2) \\geq z(y^2 - x^2)(z^2 - y^2) \\geq y(z^2 - y^2)(y^2 - x^2).\n$$\n**Note.** The key step of the proof is an instance of **Schur's Inequality** with $r = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18670,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n \\ge 2$ is *lucky* if $n^2$ can be represented as a sum of $n$ consecutive positive integers.\n\nProve 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": "See solution",
    "solution": "a) We seek seven consecutive integers $a, a+1, a+2, a+3, a+4, a+5, a+6$ such that $7^2 = a + (a+1) + (a+2) + (a+3) + (a+4) + (a+5) + (a+6)$. This gives $49 = 7a + 21$, so $a = 4$. Thus, $7^2 = 4 + 5 + 6 + 7 + 8 + 9 + 10$, so $7$ is lucky.\n\nb) Suppose there exist ten consecutive integers $a, a+1, \\dots, a+9$ such that $10^2 = a + (a+1) + \\dots + (a+9)$. This gives $100 = 10a + 45$, so $a = 5.5$, which is not an integer. Thus, $10$ is not lucky.\n\nc) We claim that a number $m$ is lucky if and only if $m$ is odd. Suppose $m$ is lucky, so $m^2 = (a+1) + (a+2) + \\dots + (a+m)$ for some $a$. Then\n\n$$\nm^2 = m a + (1 + 2 + \\dots + m) = m a + \\frac{m(m+1)}{2}\n$$\n\nSo $m^2 - \\frac{m(m+1)}{2} = m a$, so $a = \\frac{m^2 - \\frac{m(m+1)}{2}}{m} = m - \\frac{m+1}{2}$. For $a$ to be integer, $m$ must be odd. Conversely, if $m$ is odd, $a$ is integer and positive. The product of two odd numbers is odd, so the product of two lucky numbers is lucky.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18671,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $10 \\times 10$ table and number its cells sequentially and row by row from 1 to 100. If we remove two opposite corners of this table, determine the number of dominoes that can be placed in the table without any overlap.",
    "options": [],
    "answer": "See solution",
    "solution": "Using coloring, it follows that at least two cells cannot be covered, and one can easily cover all but two cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18672,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, \\dots, p_n$ be the first $n$ primes in increasing order (i.e., $p_1 = 2, p_2 = 3, \\dots$). Let\n$$\nA = p_1^{p_1} p_2^{p_2} \\cdots p_n^{p_n}.\n$$\nFind all positive integers $x$ such that $\\dfrac{A}{x}$ is even and has exactly $x$ distinct positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}$, where $0 \\leq \\alpha_1 \\leq 1$ and $0 \\leq \\alpha_i \\leq p_i$ for $i = 2, 3, \\dots, n$. Then\n$$\n\\frac{A}{x} = 2^{p_1 - \\alpha_1} p_2^{p_2 - \\alpha_2} \\cdots p_n^{p_n - \\alpha_n}.\n$$\nThe number of positive divisors of $\\frac{A}{x}$ is\n$$\n(p_1 - \\alpha_1 + 1)(p_2 - \\alpha_2 + 1) \\cdots (p_n - \\alpha_n + 1).\n$$\nWe require\n$$\n(p_1 - \\alpha_1 + 1)(p_2 - \\alpha_2 + 1) \\cdots (p_n - \\alpha_n + 1) = x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}.\n$$\nBy induction, the only solution is $\\alpha_1 = \\alpha_2 = \\cdots = \\alpha_n = 1$, so $x = 2 \\cdot 3 \\cdots p_n = p_1 p_2 \\cdots p_n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18673,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $H$ its orthocenter. Point $D$ lies on segment $AC$ and $E$ is the foot of the perpendicular from $D$ onto the line $BC$. Prove that $EH \\perp BD$ if and only if $BD$ bisects $AE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BD \\cap AH = X$. Then $XH \\perp BE$, so $EH \\perp BD$ if and only if $H$ is the orthocenter of $BXE$, which is equivalent to $BH \\perp EX$, which in turn is equivalent to $EX \\parallel AC$. This happens if and only if $AXED$ is a parallelogram, which is equivalent to $BD$ bisecting $AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18674,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有的實係數多項式 $P(x)$，使得對滿足 $2xyz = x + y + z$ 的非零實數，皆有\n\n$$\n\\frac{P(x)}{yz} + \\frac{P(y)}{zx} + \\frac{P(z)}{xy} = P(x - y) + P(y - z) + P(z - x)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "定\n\n$$\nQ(x, y, z) = xP(x) + yP(y) + zP(z) - xyz[P(x - y) + P(y - z) + P(z - x)]\n$$\n\n則 $Q(x, y, z)$ 也是實係數多項式，且當 $xyz \\neq 0$ 時\n\n$$\n2xyz = x + y + z \\Rightarrow Q(x, y, z) = 0\n$$\n\n上面的性質可延伸到複數上面，即 $x, y, z$ 也可用複數帶入。當 $(x, y, z) = (t, -t, 0)$ 帶入得出 $P(t) = P(-t)$，知 $P(x)$ 是偶函數。又帶入\n\n$$\n(x, y, z) = \\left(x, \\frac{i}{\\sqrt{2}}, -\\frac{i}{\\sqrt{2}}\\right)\n$$\n\n得到\n\n$$\n\\begin{aligned}\n& xP(x) + \\frac{i}{\\sqrt{2}}\\left(P\\left(\\frac{i}{\\sqrt{2}}\\right) - P\\left(-\\frac{i}{\\sqrt{2}}\\right)\\right) \\\\\n& = \\frac{1}{2}x\\left(P\\left(x - \\frac{i}{\\sqrt{2}}\\right) + P\\left(x + \\frac{i}{\\sqrt{2}}\\right) + P(\\sqrt{2}i)\\right)\n\\end{aligned}\n$$\n\n推出\n\n$$\nP\\left(x + \\frac{i}{\\sqrt{2}}\\right) + P\\left(x - \\frac{i}{\\sqrt{2}}\\right) - 2P(x) = P(\\sqrt{2}i)\n$$\n\n看出 $\\deg P(x) \\leq 2$，$P(x)$ 的一般形式是 $ax^2 + b$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18675,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote the number of arrangements of $n$ envelopes as $K_n$.\n\nHow many different arrangements are there for 100 envelopes, where each envelope can be placed inside a larger one, but not inside itself or inside an envelope that is already inside it?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume we are given an arrangement of $n$ envelopes. If we remove the smallest envelope, we obtain a possible arrangement of $n-1$ remaining envelopes. Conversely, given an arrangement of $n-1$ envelopes, we can put the smallest one directly in the biggest one, or in any of the remaining $n-2$ envelopes. Therefore, $K_n$ is exactly $n-1$ times larger than $K_{n-1}$.\n\nWe conclude that\n$$\nK_n = (n-1) \\cdot K_{n-1} = (n-1) \\cdot (n-2) \\cdots 2 \\cdot K_2.\n$$\nFor $n=2$, the only possible arrangement is putting the smaller envelope inside the larger one, so $K_2=1$. Therefore,\n$$\nK_n = (n-1) \\cdot (n-2) \\cdots 1 = (n-1)!\n$$\nFor 100 envelopes, the number of different arrangements is\n$$\nK_{100} = 99 \\cdot 98 \\cdots 1 = 99!\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18676,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be real numbers in the interval $(-1, 1)$. Prove that\n\n$$\n\\frac{1}{(1-x^2)(1-y^2)(1-z^2)} + \\frac{2}{(1-xy)(1-yz)(1-zx)} \\geq \\frac{1}{(1-x^2)(1-yz)^2} + \\frac{1}{(1-y^2)(1-zx)^2} + \\frac{1}{(1-z^2)(1-xy)^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First we prove a lemma.\n\n**Lemma.** Let $a$, $b$, $c$, $d$, $e$, and $f$ be real numbers. If for all real numbers $\\lambda$ and $\\mu$, we have\n\n$$\na + b\\lambda^2 + c\\mu^2 + 2(d\\lambda + e\\lambda\\mu + f\\mu) \\geq 0, \\qquad (1)$$\n\nthen we have\n\n$$\nabc + 2def \\geq ae^2 + bf^2 + cd^2. \\qquad (2)\n$$\n\n*Proof.* Clearly, $b \\geq 0$. If $b=0$, then it is easy to see that $d=e=0$, thus (2) holds. Thus we may assume $b > 0$.\n\nSince the left-hand side of (1) is quadratic in the variable $\\lambda$, its discriminant is non-positive, i.e., $(d+e\\mu)^2 - b(a+2f\\mu+c\\mu^2) \\leq 0$. Hence\n\n$$\n(bc - e^2)\\mu^2 + 2(bf - de)\\mu + (ab - d^2) \\geq 0.\n$$\n\nAgain, its left-hand side is quadratic in $\\mu$ and so its discriminant is non-positive:\n\n$$\nb \\cdot (abc + 2def - ae^2 - bf^2 - cd^2) \\geq 0.\n$$\n\nThis proves the claim.\n\n![](images/MNG_ABooklet_2015_p27_data_03ff7ab1b3.png)\n\nBy the Lemma, it suffices to show that for any real numbers $\\lambda$ and $\\mu$, we have\n\n$$\n\\frac{1}{1-x^2} + \\frac{\\lambda^2}{1-y^2} + \\frac{\\mu^2}{1-z^2} + 2 \\left( \\frac{\\lambda}{1-xy} + \\frac{\\lambda\\mu}{1-yz} + \\frac{\\mu}{1-zx} \\right) \\geq 0.\n$$\n\nUsing the summation formula of geometric series we obtain that the left side of the inequality equals\n\n$$\n\\sum_{n=0}^{\\infty} (x^n + \\lambda y^n + \\mu z^n)^2\n$$\n\nwhich is clearly non-negative.\n\nThe equality holds if and only if there exist real numbers $\\lambda$, $\\mu$ such that $x^n + \\lambda y^n + \\mu z^n = 0$ for any $n \\geq 0$. So the equality holds if and only if two of $x$, $y$, $z$ are equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18677,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be a set of numbers, and let $t$ be the smallest element in $T$ (so $t \\le 201$). For each integer $i$ with $0 \\le i \\le 2211 - 11t$, consider the sum of the following 11 elements of $T$, all of which are greater than or equal to $t$:\n\n$t, t, \\dots, t$ (10 times), $t + i$.\n\nThese sums, $11t + i$, are in a set $S$.\n\nProve that there is at least one $i$ with $0 \\le i \\le 2211 - 11t$ for which both $t + i$ and $11t + i$ are in $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "Of the numbers in $T$, let $t$ be the smallest. Note that $t \\le 201$.\n\nFor each $i$ with $0 \\le i \\le 2211 - 11t$, consider the sum of the following 11 elements of $T$ all of which are greater than or equal to $t$:\n\n$t, t, \\dots, t$ (10 times), $t + i$.\n\nThese sums, $11t + i$, are in $S$.\n\nIt suffices to prove that there is at least one $i$ with $0 \\le i \\le 2211 - 11t$, for which both $t + i$ and $11t + i$ are in $T$.\n\nThere are exactly $2211 - (t - 1) - 2011 = 201 - t$ numbers in $S$ that are greater than $t$ and are not in $T$. Hence, there are at most $201 - t$ values of $i$ for which $t + i$ is not in $T$.\n\nLikewise, there are at most $201 - t$ values of $i$ for which $11t + i$ is not in $T$. So, there are at most $2(201 - t) = 402 - 2t$ values of $i$ for which either $t + i$ or $11t + i$ (or both) is not in $T$.\n\nIt follows that there are at least\n\n$$\n(2211 - 11t + 1) - (402 - 2t) = 1810 - 9t\n$$\n\nvalues of $i$ for which both $t + i$ and $11t + i$ are in $T$. Since $t \\le 201$, we have $1810 - 9t \\ge 1$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18678,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle. A line parallel to $BC$ meets the side $AB$ at $P$ and the side $AC$ at $Q$. The line through $C$ that is parallel to $AB$ meets the line $PQ$ at $R$. Let $D$ be the reflection of $C$ in the line $BR$.\n\nProve that $D$ lies on the circumcircle of triangle $APQ$ if and only if $AB = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the intersection of lines $BR$ and $AC$. The conclusion shall follow from the following equivalences.\n\n$$\nADEB \\text{ cyclic} \\Leftrightarrow AB = BC \\quad (1)\n$$\n\n$$\nADEB \\text{ cyclic} \\Leftrightarrow APQD \\text{ cyclic} \\quad (2)\n$$\n\n*Proof of (1).* From the reflection, we have $\\angle BDE = \\angle ECB$. Hence, $\\angle BAE = \\angle BDE$ if and only if $\\angle BAE = \\angle ECB$. That is, $ADEB$ is cyclic if and only if $AB = BC$.\n\n![](images/2019_Australian_Scene_W1_p77_data_36c0556e4f.png)\n\n*Proof of (2).* From the reflection and the parallelogram $BCRP$, we have\n\n$$\n\\triangle BDR \\equiv \\triangle BCR \\equiv \\triangle RPB.\n$$\n\nIt follows that $BPDR$ is an isosceles trapezium and hence cyclic.$^1$ Thus, $\\angle QPD = \\angle RBD$. It follows that $\\angle EAD = \\angle EBD$ if and only if $\\angle EAD = \\angle QPD$. That is, $ADEB$ is cyclic if and only if $APQD$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18679,
    "subject": "Mathematics (Olympiad)",
    "question": "By replacing each * in the expression $1 * 2 * 3 * 4 * 5 * \\cdots * 2019 * 2020$ by a $+$ or a $-$ sign, we get a long calculation. Put the $+$ and $-$ signs in such a way that the outcome is a positive number (greater than $0$) which is as small as possible.\n\nWhat is this outcome?",
    "options": [],
    "answer": "See solution",
    "solution": "$2$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18680,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations in the set of rational numbers:\n\n$$\n\\begin{aligned}\n(x^2 + 1)^3 &= y + 1 \\\\\n(y^2 + 1)^3 &= z + 1 \\\\\n(z^2 + 1)^3 &= x + 1.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $(0, 0, 0)$ is obviously a solution of the system of equations. We will now show that there are no others.\n\nLet $x = \\frac{p}{q}$ with relatively prime integers $p$ and $q$ and $q > 0$. We then have\n\n$$\ny = \\left( \\left( \\frac{p}{q} \\right)^2 + 1 \\right)^3 - 1 = \\frac{(p^2 + q^2)^3 - q^6}{q^6} = \\frac{p^6 + qQ}{q^6} = \\frac{r}{q^6},\n$$\n\nand this fraction cannot be simplified, since $p$ and $q$ are relatively prime. Further substitutions then yield $z = \\frac{s}{q^{36}}$ and $x = \\frac{t}{q^{216}}$, and since these fractions similarly cannot be simplified, $q^{216} = q = 1$ follows. We see that $x$ (and also $y$ and $z$) must be integers. For integer values not equal to $0$, we have $(x^2 + 1)^3 > x^2 + 1 \\ge x + 1$, and since equality must hold if the three equations are multiplied, this yields a contradiction. We see that $(0, 0, 0)$ is indeed the only solution, as claimed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18681,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_{2014}$ be real numbers such that each $x_i \\in \\{-1, 0, 1\\}$. What is the minimal possible value of the sum of all products $x_i x_j$ for $1 \\leq i < j \\leq 2014$?",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the double sum of all the products $x_i x_j$ for $1 \\leq i < j \\leq 2014$ equals\n$$\n(x_1 + \\cdots + x_{2014})^2 - (x_1^2 + \\cdots + x_{2014}^2).\n$$\nLet $A = (x_1 + \\cdots + x_{2014})^2$ and $B = x_1^2 + \\cdots + x_{2014}^2$.\n\nWe want to minimize $A$ and maximize $B$ at the same time.\n\nObviously, $A \\geq 0$. The minimum $A = 0$ is attained when among $x_i$ there is an equal number of $1$'s and $-1$'s, e.g., when $x_1 = x_2 = \\cdots = x_{1007} = 1$, $x_{1008} = x_{1009} = \\cdots = x_{2014} = -1$.\n\nClearly, $B = x_1^2 + \\cdots + x_{2014}^2 \\leq 2014$. The maximal value $B = 2014$ is attained if none of the $x_i$ is $0$, e.g., $x_1 = x_2 = \\cdots = x_{1007} = 1$, while $x_{1008} = x_{1009} = \\cdots = x_{2014} = -1$.\n\nSince the minimum of $A$ and the maximum of $B$ can be attained for the same values of $x_1, \\ldots, x_{2014}$, we conclude that the minimal possible value of the sum of all products of pairs $x_i x_j$ ($1 \\leq i < j \\leq 2014$) is\n$$\n\\frac{A - B}{2} = \\frac{0 - 2014}{2} = -1007.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18682,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a_i = \\min\\left\\{k + \\frac{i}{k} \\mid k \\in \\mathbb{N}^*\\right\\}$. Find the value of\n$$S_n^2 = [a_1] + [a_2] + \\cdots + [a_{n^2}]$$\nwhere $n \\ge 2$, and $[x]$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$a_{i+1} = \\min\\left\\{k + \\frac{i+1}{k} \\mid k \\in \\mathbb{N}^*\\right\\} = k_1 + \\frac{i+1}{k_1} \\quad (k_1 \\in \\mathbb{N}^*)$$\nThen,\n$$a_i \\leq k_1 + \\frac{i}{k_1} < k_1 + \\frac{i+1}{k_1} = a_{i+1}$$\nwhich means $\\{a_n\\}$ is a monotonic increasing sequence. Since $k + \\frac{m^2}{k} \\geq 2m$ (with equality if and only if $k = m$), we have $a_{m^2} = 2m$ for $m \\in \\mathbb{N}^*$. On the other hand,\n$$k + \\frac{m(m+1)}{k} = 2m + 1$$\nwhen $k = m$ or $k = m + 1$. For $k \\leq m$ or $k \\geq m + 1$, we get $(k - m)(k - m - 1) \\geq 0$, which means\n$$k^2 - (2m+1)k + m(m+1) \\geq 0$$\nso\n$$k + \\frac{m(m+1)}{k} \\geq 2m + 1.$$ \nThus, $a_{m^2+m} = 2m+1$. Moreover, due to the monotonicity of $\\{a_n\\}$, we have $2m+1 \\leq a_i < 2(m+1)$ when $m^2+m \\leq i < (m+1)^2$. Hence,\n$$[a_i] = \\begin{cases} 2m, & m^2 \\leq i < m^2 + m \\\\ 2m+1, & m^2 + m \\leq i < (m+1)^2 \\end{cases}$$\nTherefore,\n$$\\sum_{i=m^2}^{m^2+2m} [a_i] = 2m \\cdot m + (2m+1) \\cdot (m+1) = 4m^2 + 3m + 1$$\nand\n$$\\begin{align*}\nS_n^2 &= \\sum_{m=1}^{n-1} (4m^2 + 3m + 1) + 2n \\\\\n&= 4 \\cdot \\frac{n(n-1)(2n-1)}{6} + 3 \\cdot \\frac{n(n-1)}{2} + (n-1) + 2n \\\\\n&= \\frac{8n^3 - 3n^2 + 13n - 6}{6}\n\\end{align*}$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18683,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute the sum\n\n$$\n\\Sigma = \\sqrt{1+\\frac{8 \\cdot 1^2-1}{1^2 \\cdot 3^2}} + \\sqrt{1+\\frac{8 \\cdot 2^2-1}{3^2 \\cdot 5^2}} + \\dots + \\sqrt{1+\\frac{8 \\cdot 1003^2-1}{2005^2 \\cdot 2007^2}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $v = 1, 2, \\dots, 1003$, we have\n\n$$\n1 + \\frac{8v^2 - 1}{(2v-1)^2 (2v+1)^2} = \\frac{(4v^2 - 1)^2 + 8v^2 - 1}{(4v^2 - 1)^2} = \\frac{16v^4}{(4v^2 - 1)^2} = \\left( \\frac{4v^2}{4v^2 - 1} \\right)^2\n$$\n\nHence we can write\n\n$$\n\\sqrt{1+\\frac{8v^2-1}{(2v-1)^2(2v+1)^2}} = \\sqrt{\\left(\\frac{4v^2}{4v^2-1}\\right)^2} = \\frac{4v^2}{4v^2-1} = \\frac{4v^2-1+1}{4v^2-1} = 1+\\frac{1}{(2v-1)(2v+1)} = 1+\\frac{2v+1-(2v-1)}{2(2v-1)(2v+1)} = 1+\\frac{1}{2}\\left(\\frac{1}{2v-1}-\\frac{1}{2v+1}\\right).\n$$\n\nSo,\n\n$$\n\\Sigma = 1+\\frac{1}{2}\\left(1-\\frac{1}{3}\\right)+1+\\frac{1}{2}\\left(\\frac{1}{3}-\\frac{1}{5}\\right)+\\dots+1+\\frac{1}{2}\\left(\\frac{1}{2005}-\\frac{1}{2007}\\right)\n$$\n\nThis telescopes to\n\n$$\n\\Sigma = 1003+\\frac{1}{2}\\left(1-\\frac{1}{2007}\\right) = 1003+\\frac{1003}{2007} = \\frac{1003 \\cdot 2008}{2007} = \\frac{2014024}{2007}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18684,
    "subject": "Mathematics (Olympiad)",
    "question": "The minimum of $y = (a \\cos^2 x - 3) \\sin x$ is $-3$. Then the range of real number $a$ is \\underline{\\hspace{2cm}}.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\sin x = t$. The expression becomes:\n\n$$\ng(t) = (a \\cos^2 x - 3) \\sin x = (a (1 - t^2) - 3)t = -a t^3 + (a - 3)t.\n$$\n\nGiven that the minimum of $g(t)$ is $-3$ for $t \\in [-1, 1]$, we require:\n\n$$\n-at^3 + (a - 3)t \\geq -3, \\quad \\forall t \\in [-1, 1].\n$$\n\nRewriting:\n\n$$\n-at^3 + (a - 3)t + 3 \\geq 0\n$$\n\nor\n\n$$\n-a t^3 + (a - 3)t + 3 = -a t^3 + a t - 3t + 3 = -a t^3 + a t - 3t + 3.\n$$\n\nAlternatively, factor as:\n\n$$\n-at^3 + (a - 3)t + 3 = -a t^3 + a t - 3t + 3 = -a t^3 + a t - 3t + 3.\n$$\n\nTo find the range of $a$, analyze the critical points and endpoints. After checking the values at $t = -1, 0, 1$ and considering the behavior in $[-1, 1]$, we find:\n\n$$\n-\\frac{3}{2} \\leq a \\leq 12.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18685,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive integers with $x \\mid y^3$, $y \\mid z^3$, and $z \\mid x^3$. What is the smallest positive integer $n$ such that $xyz \\mid (x + y + z)^n$ for all such $x$, $y$, $z$?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible integer with that property is $n = 13$.\n\nWe note that $xyz \\mid (x + y + z)^n$ if and only if for each prime $p$, the inequality $v_p(xyz) \\leq v_p((x + y + z)^n)$ holds, where $v_p(m)$ denotes the exponent of $p$ in the prime factorization of $m$.\n\nLet $x$, $y$, and $z$ be positive integers with $x \\mid y^3$, $y \\mid z^3$, and $z \\mid x^3$. Let $p$ be an arbitrary prime, and without loss of generality, let the multiplicity of $p$ be lowest in $z$, that is, $v_p(z) = \\min\\{v_p(x), v_p(y), v_p(z)\\}$.\n\nThen $v_p(x + y + z) \\geq v_p(z)$, and from the divisibility constraints we get $v_p(x) \\leq 3v_p(y) \\leq 9v_p(z)$. It follows that\n\n$$\n\\begin{aligned}\nv_p(xyz) &= v_p(x) + v_p(y) + v_p(z) \\\\\n&\\leq 9v_p(z) + 3v_p(z) + v_p(z) = 13v_p(z) \\\\\n&\\leq 13v_p(x + y + z) = v_p((x + y + z)^{13}),\n\\end{aligned}\n$$\n\nwhich proves that for $n = 13$ the desired property is satisfied.\n\nIt remains to show that this is indeed the smallest possible integer with this property. For this, let $n$ be a number that has the desired property. By setting $(x, y, z) = (p^9, p^3, p^1)$ with an arbitrary prime $p$ (so both inequalities above become equalities), we get\n\n$$\n\\begin{aligned}\n13 &= v_p(p^{13}) = v_p(p^9 \\cdot p^3 \\cdot p^1) = v_p(xyz) \\\\\n  &\\leq v_p((x + y + z)^n) = v_p((p^9 + p^3 + p^1)^n) = n \\cdot v_p(p(p^8 + p^2 + 1)) = n,\n\\end{aligned}\n$$\n\nwhich yields $n \\geq 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18686,
    "subject": "Mathematics (Olympiad)",
    "question": "Taro has coins and notes of denominations 1000 yen, 100 yen, 10 yen, and 1 yen, one of each. He uses all of them to pay for an item costing less than 1111 yen, and receives change in coins of denominations 500 yen, 50 yen, and 5 yen. How many possible values are there for the purchasing price?",
    "options": [],
    "answer": "See solution",
    "solution": "To begin with, we note that 2 or more of 500 yen coins, or of 50 yen coins, or of 5 yen coins could not be included in the change Taro received. This is because if 2 or more of any of these coins were in the change, 2 of them could be exchanged for 1 1000 yen note, or 1 100 yen coin or 1 10 yen coin to reduce the total number of coins in the change. Also, the change did not include 1000 yen note, 100 yen coin, 10 yen coin, or 1 yen coin, since Taro used all of these in his initial possession in his payment. Therefore the change must have consisted of 0 or 1 piece each of 500 yen, 50 yen and 5 yen coins. Consequently, the purchasing price must have been of the form:  \n$$1111 - (500a + 50b + 5c)$$  \nwhere each of $a, b, c$ is either 0 or 1.\n\nConversely, let us show that if the purchasing price is of the form $1111 - (500a + 50b + 5c)$ yen, with each one of $a, b, c$ being either 0 or 1, Taro would make his payment in the way specified in the statement of the problem. Note that $a$ pieces of 500 yen coin, $b$ pieces of 50 yen coin and $c$ pieces of 5 yen coin together would give the method of paying $500a+50b+5c$ yen with the minimum number of coins. As was remarked above, the change will not involve any of 1000 yen note, 100 yen coin, 10 yen coin or 1 yen coin. Therefore, it suffices to show that the number of coins in the change received must be the minimum to represent the actual amount of the change. But the amount of the change is $500a + 50b + 5c$ yen regardless of the method of payment, and hence again by the remark made above, the number of coins received by Taro for the change is the minimum.\n\nSince each of $a, b, c$ can take either 0 or 1, the number of possible values for the purchasing price is $2^3 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18687,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ and $a < b + c$. Prove that\n$$\n\\frac{a}{1+a} < \\frac{b}{1+b} + \\frac{c}{1+c}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, if $x, y > -1$, then\n$$\n\\frac{x}{1+x} < \\frac{y}{1+y} \\Leftrightarrow x + xy < y + yx,\n$$\ni.e., if and only if $-1 < x < y$. Since $a < b + c$ and $b, c > 0$,\n$$\n\\frac{a}{1+a} < \\frac{b+c}{1+b+c} = \\frac{b}{1+b+c} + \\frac{c}{1+b+c} < \\frac{b}{1+b} + \\frac{c}{1+c},\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18688,
    "subject": "Mathematics (Olympiad)",
    "question": "$f: \\mathbb{N} \\rightarrow \\mathbb{R}$ функц дээр\n$$\n\\forall n \\in \\mathbb{N}: \\sqrt{f(n+2)+2} \\leq f(n) \\leq 2\n$$\nэнэхүү тэнцэтгэл бишийг хангах бүх $f$ функцийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$0 \\leq f(n) \\leq 2$, $\\forall n \\in \\mathbb{N}$.\n\n$$\n\\Rightarrow f(n) = 2 \\cos g(n),\\quad g(n) \\in [0, \\frac{\\pi}{2}] \\text{ гэж үзэж болно.}\n$$\n\n$$\n\\sqrt{f(n+2)+2} \\leq f(n) \\text{ ба } \\cos 2\\alpha + 1 = 2\\cos^2 \\alpha\n$$\n\n$$\n\\Rightarrow \\cos \\frac{g(n+2)}{2} \\leq \\cos g(n) \\text{ ба } \\cos t \\text{ нь } t \\in [0, \\frac{\\pi}{2}] \\text{ дээр буурна.\n}\n$$\n\nЭндээс\n\n$$\n\\frac{g(n+2)}{2} \\geq g(n),\\quad \\forall n \\in \\mathbb{N}.\n$$\n\nИндукцээр $\\forall k, n \\in \\mathbb{N}: g(n) \\leq \\frac{g(n+2k)}{2k}$ болох ба $k \\to \\infty$ үед хязгаарт шилжвэл $g(n) = 0 \\Rightarrow f(n) = 2 \\cos 0 = 2$ болж өгөгдсөн нөхдөлмийг хангах функц $f(n) = 2, \\forall n \\in \\mathbb{N}$-ээс өөр байхгүй.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18689,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral and let $P$ be the point of intersection of $AC$ and $BD$. Suppose that $AC + AD = BC + BD$. Prove that the internal angle bisectors of $\\angle ACB$, $\\angle ADB$, and $\\angle APB$ meet at a common point.",
    "options": [],
    "answer": "See solution",
    "solution": "I. Construct $A'$ on $CA$ so that $AA' = AD$ and $B'$ on $CB$ such that $BB' = BD$. Then we have three angle bisectors that correspond to the perpendicular bisectors of $A'B'$, $A'D$, and $B'D$. These perpendicular bisectors are concurrent, so the angle bisectors are also concurrent. This tells us that the external angle bisectors at $A$ and $B$ meet at the excentre of $PDB$. A symmetric argument for $C$ finishes the problem.\n\nII. Note that the angle bisectors $\\angle ACB$ and $\\angle APB$ intersect at the excentres of $\\triangle PBC$ opposite $C$ and the angle bisectors of $\\angle ADB$ and $\\angle APB$ intersect at the excentres of $\\triangle PAD$ opposite $D$. Hence, it suffices to prove that these two excentres coincide.\n\nLet the excircle of $\\triangle PBC$ opposite $C$ touch side $PB$ at a point $X$, line $CP$ at a point $Y$ and line $CB$ at a point $Z$. Hence, $CY = CZ$, $PX = PY$ and $BX = BZ$. Therefore, $CP + PX = CB + BX$. Since $CP + PX + CB + BX$ is the perimeter of $\\triangle CBP$, $CP + PX = CB + BX = s$, where $s$ is the semi-perimeter of $\\triangle CBP$. Therefore,\n\n$$\nPX = CB + BX - CP = \\frac{s}{2} - CP = \\frac{CB + BP + PC}{2} - CP = \\frac{CB + BP - PC}{2}.\n$$\n\nSimilarly, if we let the excircle of $\\triangle PAD$ opposite $D$ touch side $PA$ at a point $X'$, then\n\n$$\nPX' = \\frac{DA + AP - PD}{2}.\n$$\n\nSince both excircles are tangent to $AC$ and $BD$, if we show that $PX = PX'$, then we would show that the two excircles are tangent to $AC$ and $BD$ at the same points, i.e. the two excircles are identical. Hence, the two excentres coincide.\n\nWe will use the fact that $AC + AD = BC + BD$ to prove that $PX = PX'$. Since $AC + AD = BC + BD$, $AP + PC + AD = BC + BP + PD$. Hence, $AP + AD - PD = BC + BP - PC$. Therefore, $PX = PX'$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18690,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider how many squares have their vertices on the perimeter of a given $ (n+1) \\times (n+1) $ array whose sides are parallel to the sides of the array.\n\n![](images/obm-book_p59_data_d6f8ad6267.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The diagram shows that there are $n$ such squares. There are $ (k+1-n)^2 $ such arrays. So the total number of squares is:\n\n$$\n  k \\cdot 1^2 + (k-1)^2 \\cdot 2 + \\cdots + 1 \\cdot k^2 = \\sum_{i=1}^{k} (k+1-i)i^2\n$$\n\nThis sum simplifies to:\n\n$$\n  \\frac{(k+1)k(k+1)(2k+1)}{6} - \\frac{k^2(k+1)^2}{4} = \\frac{k(k+1)2(k+2)}{12}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18691,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $k$ ($k \\ge 2$) and $k$ nonzero real numbers $a_1, a_2, \\dots, a_k$, prove that there are at most finitely many $k$-element integer arrays $(n_1, n_2, \\dots, n_k)$ such that $n_1, n_2, \\dots, n_k$ are pairwise distinct and\n\n$$\na_1 \\cdot n_1! + a_2 \\cdot n_2! + \\dots + a_k \\cdot n_k! = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Take a positive integer $N \\ge \\dfrac{|a_1| + |a_2| + \\dots + |a_k|}{\\min_{1 \\le i \\le k} |a_i|}$ (note that $a_1, a_2, \\dots, a_k \\ne 0$).\n\nWe will show that when positive integers $n_1, n_2, \\dots, n_k$ satisfy the condition, we must have $\\max_{1 \\le i \\le k} n_i \\le N$.\n\nAssume otherwise and set $\\max_{1 \\le i \\le k} n_i = n_1 > N$. For $i = 2, \\dots, k$, since $n_i < n_1$, it follows that\n\n$$\nn_i! \\le (n_1 - 1)! = \\frac{n_1!}{n_1} < \\frac{n_1!}{N}.\n$$\n\nHence,\n\n$$\n\\begin{align*}\n|a_2 \\cdot n_2! + \\dots + a_k \\cdot n_k!| &\\le \\sum_{i=2}^k |a_i| \\cdot n_i! < \\frac{n_1!}{N} \\sum_{i=2}^k |a_i| < \\frac{n_1!}{N} \\sum_{i=1}^k |a_i| \\\\\n&\\le \\min_{1 \\le i \\le k} |a_i| \\cdot n_1! \\le |a_1| \\cdot n_1!.\n\\end{align*}\n$$\n\nHowever, $|a_2 \\cdot n_2! + \\dots + a_k \\cdot n_k!| = |-a_1 \\cdot n_1!| = |a_1| \\cdot n_1!$, a contradiction.\n\nTherefore, there are at most $N^k$ sets of positive integers $(n_1, n_2, \\dots, n_k)$ that satisfy the condition. The proof is complete. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18692,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive numbers $a$, $b$, $c$ satisfy $a^2 + b^2 + c^2 = 3$. Show that the following inequality holds:\n\n$$\n\\frac{1}{a} + \\frac{3}{b} + \\frac{5}{c} \\ge 4a^2 + 3b^2 + 2c^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the inequality as:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{2}{b} + b^2 + \\frac{4}{c} + 2c^2 \\ge 4(a^2 + b^2 + c^2) = 12.\n$$\n\nSince $a^2 + b^2 + c^2 = 3$, by the AM-GM inequality:\n\n$$\n3 = a^2 + b^2 + c^2 \\ge 3 \\sqrt[3]{(abc)^2} \\implies abc \\le 1.\n$$\n\nThus,\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\ge 3 \\cdot \\frac{1}{\\sqrt[3]{abc}} \\ge 3.\n$$\n\nApply the AM-GM inequality to other terms:\n\n$$\n\\frac{2}{b} + b^2 = \\frac{1}{b} + \\frac{1}{b} + b^2 \\ge 3 \\sqrt[3]{\\frac{1}{b} \\cdot \\frac{1}{b} \\cdot b^2} = 3.\n$$\n\nSimilarly,\n\n$$\n\\frac{4}{c} + 2c^2 = \\frac{2}{c} + \\frac{2}{c} + 2c^2 \\ge 3 \\sqrt[3]{\\frac{2}{c} \\cdot \\frac{2}{c} \\cdot 2c^2} = 6.\n$$\n\nAdding these inequalities proves the rewritten inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18693,
    "subject": "Mathematics (Olympiad)",
    "question": "Exhibit a set of 199 points in the grid such that no three distinct points $A$, $B$, $C$ satisfy: $A$ is stronger than $B$ and $B$ is stronger than $C$ (i.e., the relation is transitive and all three pairs satisfy the relation in one direction). Additionally, for any $n$ with $2 < n < 199$, construct a similar set of $n$ points with the same property.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nT_1 = \\{(x, y) : 1 \\le x, y \\le 100 \\text{ and } x + y = 101\\},\n$$\n\n$$\nT_2 = \\{(x, y) : 2 \\le x, y \\le 100 \\text{ and } x + y = 102\\}\n$$\n\nand $T = T_1 \\cup T_2$. Note that $T$ has exactly 199 points. By the pigeonhole principle, any three points in $T$ must have at least two in either $T_1$ or $T_2$. There are no two points in $T_1$ (or $T_2$) such that one is stronger than the other. Therefore, by transitivity, no three distinct points $A$, $B$, $C$ in $T$ satisfy the requirement. For $2 < n < 199$, a similar set can be constructed by deleting sufficient elements from $T$. For $n = 1$ or $2$, no set contains three distinct points.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18694,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for positive real numbers $a, b, c$, the following inequality holds:\n\n$$\n(16a^2 + 8b + 17)(16b^2 + 8c + 17)(16c^2 + 8a + 17) \\geq 2^{12}(a+1)(b+1)(c+1).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By twice using the inequality between the arithmetic mean and geometric mean, we get\n\n$$\n\\begin{aligned}\n(16a^2 + 8b + 17) &= (16a^2 + 1 + 8b + 16) \\\\ &\\ge 8a + 8b + 16 = 8(a+b+2) = 8(a+1+b+1) \\\\ &\\ge 8 \\cdot 2\\sqrt{(a+1)(b+1)} = 2^4\\sqrt{(a+1)(b+1)}.\n\\end{aligned} \\quad (1)\n$$\n\nAnalogously,\n\n$$\n(16b^2 + 8c + 17) \\ge 2^4 \\sqrt{(b+1)(c+1)} \\quad (2)\n$$\n\n$$\n(16c^2 + 8a + 17) \\ge 2^4 \\sqrt{(c+1)(a+1)} \\quad (3)\n$$\n\nMultiplying the three inequalities, we get\n\n$$\n(16a^2 + 8b + 17)(16b^2 + 8c + 17)(16c^2 + 8a + 17) \\ge 2^{12}(a+1)(b+1)(c+1).\n$$\n\nIn (1), equality is obtained when $16a^2 = 1$ and $a = b$, i.e., $a = b = \\frac{1}{4}$. By an analogous argument for (2) and (3), we get $a = b = c = \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18695,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a natural number $n \\ge 3$, find the smallest real number $k > 0$ such that for any connected graph $G$ with $n$ vertices and $m$ edges, it is always possible to delete no more than $k \\cdot \\left(m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor\\right)$ edges so that the remaining graph can be colored with two colors, and every undeleted edge connects vertices of different colors.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma:\n\n*Lemma*: Let $G$ be a connected graph with at least 3 vertices. Then either there exist two vertices connected by an edge whose removal (along with their incident edges) leaves $G$ connected, or there exist two vertices of degree 1 (leaves).\n\nConsider a spanning tree of $G$ and choose a root that is not a leaf. Let $v$ be the farthest vertex from the root, and $u$ its parent. Let $v_1, v_2, \\dots, v_k$ be the children of $u$. These are all leaves in the tree.\n\n- *Case 1*: Among $v_1, \\dots, v_k$, two are connected by an edge in $G$. Removing these two vertices leaves the tree (and $G$) connected.\n- *Case 2*: Among $v_1, \\dots, v_k$, two are leaves in $G$. Then $G$ has at least two leaves.\n- *Case 3*: At most one of $v_1, \\dots, v_k$ is a leaf in $G$. For $v_2, \\dots, v_k$, connect each to a vertex in $G$ other than $u$ (such edges are not in the tree). Removing $u$ and $v_1$ leaves a spanning tree, so $G$ remains connected.\n\nThis proves the lemma.\n\nNow, we prove:\n\n*Assertion*: For any connected graph $G$ with $n \\ge 2$ vertices, we can 2-color its vertices so that if $x$ is the number of multicolored edges and $y$ is the number of single-colored edges, then $x - y \\ge \\left\\lfloor \\frac{n}{2} \\right\\rfloor$.\n\n*Proof*: For $n = 2, 3$, the statement is clear. For $n \\ge 4$, let $u$ and $v$ be the two vertices from the lemma. Remove $u$ and $v$ and color $G \\setminus \\{u, v\\}$ by induction. We can color $u$ and $v$ so that the difference $x - y$ increases by at least 1. Thus, by induction, the assertion holds.\n\nFor a connected graph $G$ with $n \\ge 3$ vertices and $m$ edges, coloring as above gives $x - y \\ge \\left\\lfloor \\frac{n}{2} \\right\\rfloor$ and $x + y = m$, so\n\n$$\ny \\le \\frac{1}{2} \\left( m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\right)\n$$\n\nDeleting $y$ edges suffices, so $k \\le \\frac{1}{2}$.\n\nTo show $k \\ge \\frac{1}{2}$, consider the complete graph on $n$ vertices. To make it bipartite (so all edges are multicolored), we must delete all edges within each part. The minimum is when the parts are as equal as possible.\n\n- If $n = 2n_1$, $m = \\binom{n}{2}$, and we must delete at least $2\\binom{n_1}{2} = n_1^2 - n_1$ edges. Thus,\n  $$\nn_1^2 - n_1 \\le k \\left( \\binom{2n_1}{2} - n_1 \\right) \\implies k \\ge \\frac{1}{2}.$$\n- If $n = 2n_1 + 1$, we must delete at least $\\binom{n_1+1}{2} + \\binom{n_1}{2} = n_1^2$ edges, so\n  $$\nn_1^2 \\le k \\left( \\binom{2n_1+1}{2} - n_1 \\right) \\implies k \\ge \\frac{1}{2}.$$\n\nTherefore, the smallest such $k$ is $\\boxed{\\frac{1}{2}}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18696,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AH_A$, $BH_B$, $CH_C$ be the altitudes in $\\triangle ABC$. Draw perpendiculars $p_A$, $p_B$, $p_C$ through the vertices $A$, $B$, $C$ to $H_BH_C$, $H_CH_A$, $H_AH_B$, respectively. Prove that $p_A$, $p_B$, $p_C$ pass through the same point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the circumscribed circle around $\\triangle ABC$. We will show that each of the lines $p_A$, $p_B$, $p_C$ passes through $O$. Because of symmetry, it is enough to show that $OC \\perp H_A H_B$. Let $D$ be the point of intersection of these two lines. We restrict ourselves to the case where $\\triangle ABC$ is acute (since in the case of $\\triangle ABC$ being obtuse the argument is analogous). It is enough to use the fact that $\\angle H_A CD = \\angle BCO = 90^\\circ - \\alpha$ and $\\angle DH_A C = \\angle H_B H_A C = \\alpha$ (the last equality follows from the fact that $ABH_A H_B$ is inscribed).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18697,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to colour the squares of a given array using only 7 colours so that every L-shaped tile (covering four squares) contains four different colours? If not, how many colours are needed?\n\n![](images/Slovenija_2009_p22_data_051d4921c4.png)\n\n![](images/Slovenija_2009_p22_data_3c6f3468f0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We can show that 8 colours are enough. Number the colours from 1 to 8 and colour the array as shown in the first figure. Any L-shaped tile contains four different colours since no two squares of the same colour can be covered with a single tile.\n\nAssume that we could do this using only 7 colours. We say that two squares are *connected* if they can both be covered with one tile. Label some of the squares in the array as shown in the second figure. Any two of the squares marked $A$, $B$, $C$, $D$, $E$, $F$, $G$ are connected, so all of these squares have to be coloured differently (and there are exactly 7 of them).\n\nThe square marked as $u$ is connected to the squares $A$, $B$, $C$, $D$, $E$, $G$, so it must have the same colour as $F$. The square marked as $x$ is connected to the squares $B$, $C$, $D$, $E$, $F$, $G$, so it must have the same colour as $A$. The square marked as $v$ is connected to $B$, $u$, $D$, $E$, $G$, $x$. Since $u$ and $F$ have the same colour and $x$ and $A$ have the same colour, the colour of $v$ must be the same as that of $C$. The square marked as $z$ is connected to $C$, $D$, $E$, $F$, $G$, $x$, so it must have the same colour as $B$. The square marked as $y$ is connected to the squares $u$, $D$, $E$, $v$, $G$, $x$, which share their colours with $F$, $D$, $E$, $C$, $G$, $A$, so $y$ has to have the colour of $B$. Now, $G$, $x$, $y$, $z$ can be covered with the L-shaped tile but $y$ and $z$ have the same colour (that of $B$). Hence, 7 colours is not enough.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18698,
    "subject": "Mathematics (Olympiad)",
    "question": "In the nation of Onewaynia, certain pairs of cities are connected by roads. Every road connects exactly two cities (roads are allowed to cross each other, e.g., via bridges). Some roads have a traffic capacity of $1$ unit and other roads have a traffic capacity of $2$ units. However, on every road, traffic is only allowed to travel in one direction. It is known that for every city, the sum of the capacities of the roads connected to it is always odd. The transportation minister needs to assign a direction to every road. Prove that he can do it in such a way that for every city, the difference between the sum of the capacities of roads entering the city and the sum of the capacities of roads leaving the city is always exactly one.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be a graph representing the setup, where each vertex represents a city, and each edge represents a road connecting two cities. Every edge is labeled $1$ or $2$. Initially, all edges are unoriented. For each orientation of edges, define the *surplus* at vertex $v$ to be the sum of labels of edges entering the vertex minus the sum of labels of edges leaving the vertex. A *good orientation* of $G$ is an orientation of edges so that every vertex has surplus $\\pm 1$. An *odd graph* is a graph whose sum of edge-labels at each vertex is odd. We are asked to show that every odd graph has a good orientation.\n\nAllow multiple edges between the same pair of vertices in our graphs. Let us consider a number of possible transformations on odd graphs, reducing the number of edges at each step, so that existence of a good orientation in the new graph implies that of a good orientation in the old graph (with the same surplus at each vertex).\n\n- If $G$ has two edges $AB$ and $BC$, where $AB$ and $BC$ have the same label $x$ and $A \\neq C$, then construct $G'$ from $G$ by removing $AB$ and $BC$ and then adding a new edge $AC$ with label $x$. A good orientation in $G'$ gives a good orientation in $G$ by orienting $AB$ and $BC$ in $G$ in the same direction as $AC$ in $G'$, and keeping the orientation of all other vertices the same.\n- If $G$ has two edges between $A$ and $B$ of the same label, then construct $G'$ from $G$ by deleting the two parallel edges. A good orientation in $G'$ gives a good orientation in $G$ by directing the two deleted edges in opposite directions and keeping the other orientations the same.\n\nWe can repeatedly perform the above transformations until it is no longer possible to do any more. Let $G'$ be the graph we obtain at the end. It suffices to show that $G'$ has a good orientation. Note that no vertex of $G'$ is adjacent to two edges of the same label. Since $G'$ is an odd graph, we deduce that $G'$ must be a disjoint union of subgraphs of the following forms:\n\n- Two edges of different label between $A$ and $B$. We can orient the two edges in opposite directions.\n- A path $P_1P_2\\cdots P_{2n}$, where the edges alternate in value: $1$, $2$, $1$, ..., $2$, $1$. We can orient all the edges along one direction of the path.\n- A cycle with edges of alternating value. We can orient all the edges along one direction of the cycle.\n\nThis shows that $G'$ has a good orientation, and therefore so does $G$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18699,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Let $P$ be the point on the line $CD$ such that $MP$ is perpendicular to $CD$. Let $Q$ be the point on the line $AB$ such that $NQ$ is perpendicular to $AB$. Prove that $AD$ is parallel to $BC$ if and only if $$\\frac{AB}{CD} = \\frac{MP}{NQ}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the area of $A$ by $[A]$. Observe that\n![](images/Tajland_2013_p4_data_631d25c080.png)\n\n$$\n[ABCD] = [ADM] + [DMC] + [MCB] = \\frac{1}{2}[ABD] + \\frac{1}{2}CD \\cdot MP + \\frac{1}{2}[ABC]\n$$\n\n$$\n[ABCD] = [ADN] + [ANB] + [NBC] = \\frac{1}{2}[ADC] + \\frac{1}{2}AB \\cdot NQ + \\frac{1}{2}[BCD]\n$$\n\nThus,\n\n$$\n\\begin{aligned}\nCD \\cdot MP - AB \\cdot NQ &= [ADC] + [BCD] - [ABD] - [ABC] \\\\\n&= [ABCD] + [DOC] - [AOB] - \\{[ABCD] + [AOB] - [DOC]\\} \\\\\n&= 2\\{[DOC] - [AOB]\\} = 2\\{[ADC] - [ADB]\\},\n\\end{aligned}\n$$\n\ni.e., $CD \\cdot MP - AB \\cdot NQ = 0 \\iff [ADC] = [ADB]$. Thus,\n\n$$\nAD \\parallel BC \\iff [ADC] = [ADB] \\iff \\frac{AB}{CD} = \\frac{MP}{NQ}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18700,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the sum of 74 real numbers lying in the interval $[4, 10]$ is 356. Find the maximal possible value of the sum of their squares.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the desired maximum is equal to 2024.\n\nDenote the numbers $x_1, \\dots, x_{74}$. The assumption $4 \\le x_i \\le 10$ guarantees that $(x_i-4)(10-x_i) \\ge 0$ holds for all $i$. Expanding this to $x_i^2 \\le 14x_i - 40$ and summing over $i$ gives us an upper bound\n\n$$\nx_1^2 + x_2^2 + \\dots + x_{74}^2 \\le 14(x_1 + \\dots + x_{74}) - 40 \\cdot 74 = 14 \\cdot 356 - 40 \\cdot 74 = 2024.\n$$\n\nTherefore, it is now sufficient to show that there exists a set of 74 real numbers that attains this bound. Our approach above shows that the bound is attained if and only if $x_i \\in \\{4, 10\\}$ for all $i$, so it suffices to check that such a 74-tuple with sum 356 exists.\n\nSuppose that the tuple contains $a$ fours and $b$ tens, then the non-negative integers $a, b$ must satisfy the system of equations:\n\n$$\n\\begin{aligned}\na+b &= 74, \\\\\n4a+10b &= 356.\n\\end{aligned}\n$$\n\nIt is easy to see that $(a, b) = (64, 10)$ is the only solution, so a set consisting of 64 fours and 10 tens proves that the upper bound is tight.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18701,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum possible number of distinct real roots of a polynomial $P(x)$ of degree $2012$ with real coefficients satisfying the condition\n\n$$\nP(a)^3 + P(b)^3 + P(c)^3 \\geq 3P(a)P(b)P(c)\n$$\n\nfor all real numbers $a, b, c$ with $a + b + c = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that there exists a polynomial $P(x)$ which satisfies the given condition and has $2012$ distinct real roots.\n\nFirst, note that the given inequality is equivalent to\n\n$$\n(P(a) + P(b) + P(c))\\big((P(a) - P(b))^2 + (P(b) - P(c))^2 + (P(c) - P(a))^2\\big) \\geq 0,\n$$\n\nso it is enough to find a polynomial $P$ such that $P(a) + P(b) + P(c) \\geq 0$ whenever $a + b + c = 0$.\n\nFor positive numbers $M$ and $\\varepsilon$, let\n\n$$\nP_{M,\\varepsilon}(x) = (x - M)(x - M - \\varepsilon) \\cdots (x - M - 2011\\varepsilon).\n$$\n\n$P_{M,\\varepsilon}$ is positive and decreasing on $(-\\infty, M)$, and positive and increasing on $(M + 2011\\varepsilon, \\infty)$. We have $P_{M,\\varepsilon}(x) \\geq M^{2012}$ for $x \\leq 0$ and\n\n$$\n|P_{M,\\varepsilon}(x)| = |(x - M) (x - M - \\varepsilon) \\cdots (x - M - 2011\\varepsilon)| \\leq (2011\\varepsilon)^{2012}\n$$\n\nfor $x \\in [M, M + 2011\\varepsilon]$. Therefore, $P_{M,\\varepsilon}(x) \\geq -(2011\\varepsilon)^{2012}$ for $x \\geq 0$.\n\nLet $a, b, c$ be real numbers with $a + b + c = 0$. Without loss of generality, assume $a \\leq 0$. From the previous inequalities,\n\n$$\nP_{M,\\varepsilon}(a) + P_{M,\\varepsilon}(b) + P_{M,\\varepsilon}(c) \\geq M^{2012} + 2\\big(- (2011\\varepsilon)^{2012}\\big).\n$$\n\nSince the right-hand side is positive for $M = 2$ and $\\varepsilon = 1/2011$, we can take $P(x) = P_{2,1/2011}(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18702,
    "subject": "Mathematics (Olympiad)",
    "question": "Two cyclists leave towns A and B and travel towards each other with speeds $v_1$ and $v_2$, where $v_1 \\ge v_2$. They meet for the first time after 1 hour. After meeting, both continue to their destination towns without stopping. Upon reaching their destination, each cyclist turns around and heads back in the opposite direction.\n\nHow much time after their first meeting will it take for them to meet a second time?",
    "options": [],
    "answer": "See solution",
    "solution": "If $v_1 < 2v_2$, then the second meeting occurs 2 hours after the first meeting; otherwise, it occurs after $\\frac{2v_2}{v_1 - v_2}$ hours.\n\nUntil the first meeting, the first cyclist travels $S_1 = v_1$ and the second $S_2 = v_2$, so the distance between towns is $v_1 + v_2$.\n\nThere are two cases:\n\n![](images/UkraineMO2019_booklet_p22_data_b2cb529b61.png)\n\n*Fig. 22*\n\n**Case 1:** Both cyclists reach their destination towns and turn around before the second meeting. Suppose the second meeting occurs $S_3$ from town B and $t_2$ hours after the first meeting (see Fig. 22). Then:\n\n![](images/UkraineMO2019_booklet_p22_data_bef5cc6d28.png)\n\n*Fig. 23*\n\n$$\nv_2 + S_3 = v_1 t_2 \\quad \\text{and} \\quad v_1 + (v_1 + v_2 - S_3) = v_2 t_2.\n$$\n\nAdding these equations:\n\n$$\nv_2 + S_3 + 2v_1 + v_2 - S_3 = (v_2 + v_1)t_2 \\implies t_2 = 2.\n$$\n\n**Case 2:** The first cyclist reaches town B, turns around, and meets the second cyclist before the latter reaches town A. Suppose the second meeting occurs $S_3$ from town B and $t_2$ hours after the first meeting. Then:\n\n$$\nS_3 - v_2 = v_2 t_2 \\quad \\text{and} \\quad v_2 + S_3 = v_1 t_2.\n$$\n\nSubtracting the first from the second:\n\n$$\nv_2 + S_3 + v_2 - S_3 = (v_1 - v_2)t_2 \\implies t_2 = \\frac{2v_2}{v_1 - v_2}.\n$$\n\nTo determine which case applies: Case 1 occurs when the first cyclist reaches A later than the second cyclist, i.e.,\n\n$$\n\\frac{v_2 + v_2 + v_1}{v_1} > \\frac{v_1}{v_2} \\implies 2v_2^2 + v_1 v_2 > v_1^2.\n$$\n\nLet $x = \\frac{v_1}{v_2}$, then:\n\n$$\nx^2 - x - 2 < 0 \\implies (x + 1)(x - 2) < 0 \\implies x < 2, \\text{ i.e., } v_1 < 2v_2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18703,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be a sequence defined by the recurrence relation\n$$\na_n - \\beta a_{n-1} = \\alpha (a_{n-1} - \\beta a_{n-2}).\n$$\nFind the general term of $a_n$ for the following cases:\n1. When $\\Delta = p^2 - 4q = 0$ (i.e., $\\alpha = \\beta \\neq 0$).\n2. When $\\Delta > 0$ (i.e., $\\alpha \\neq \\beta$).\n\nAdditionally, for $p = 1$, $q = \\frac{1}{4}$, compute the sum $S_n = a_1 + a_2 + \\dots + a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_n = a_{n+1} - \\beta a_n$. Then $b_{n+1} = \\alpha b_n$ for $n = 1, 2, \\dots$. Thus, $\\{b_n\\}$ is a geometric sequence with common ratio $\\alpha$.\n\nThe first term is\n$$\nb_1 = a_2 - \\beta a_1 = (\\alpha + \\beta)^2 - \\alpha\\beta - \\beta(\\alpha + \\beta) = \\alpha^2.\n$$\nSo $b_n = \\alpha^2 \\alpha^{n-1} = \\alpha^{n+1}$, and $a_{n+1} - \\beta a_n = \\alpha^{n+1}$, or\n$$\na_{n+1} = \\alpha^{n+1} + \\beta a_n.\n$$\n\n**Case 1:** $\\Delta = p^2 - 4q = 0$, $\\alpha = \\beta \\neq 0$.\nThen $a_{n+1} = \\alpha^{n+1} + \\alpha a_n$, so\n$$\n\\frac{a_{n+1}}{\\alpha^{n+1}} - \\frac{a_n}{\\alpha^n} = 1.\n$$\nThus, $\\left\\{\\frac{a_n}{\\alpha^n}\\right\\}$ is arithmetic with difference $1$ and first term $2$, so\n$$\n\\frac{a_n}{\\alpha^n} = n + 1 \\implies a_n = (n + 1)\\alpha^n.\n$$\n\n**Case 2:** $\\Delta > 0$, $\\alpha \\neq \\beta$.\nWe have\n$$\na_{n+1} = \\alpha^{n+1} + \\beta a_n.\n$$\nRewriting,\n$$\na_{n+1} + \\frac{\\alpha^{n+2}}{\\beta - \\alpha} = \\beta \\left(a_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha}\\right).\n$$\nSo $\\left\\{a_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha}\\right\\}$ is geometric with ratio $\\beta$ and first term $\\frac{\\beta^2}{\\beta - \\alpha}$, so\n$$\na_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha} = \\frac{\\beta^2}{\\beta - \\alpha} \\beta^{n-1}.\n$$\nThus,\n$$\na_n = \\frac{\\beta^{n+1} - \\alpha^{n+1}}{\\beta - \\alpha}.\n$$\n\n**Sum for $p = 1$, $q = \\frac{1}{4}$:**\nHere $\\Delta = 0$, $\\alpha = \\beta = \\frac{1}{2}$, so\n$$\na_n = (n+1)\\left(\\frac{1}{2}\\right)^n = \\frac{n+1}{2^n}.\n$$\nThe sum is\n$$\nS_n = \\frac{2}{2} + \\frac{3}{2^2} + \\dots + \\frac{n+1}{2^n}.\n$$\nConsider\n$$\n\\frac{1}{2} S_n = \\frac{2}{2^2} + \\frac{3}{2^3} + \\dots + \\frac{n+1}{2^{n+1}}.\n$$\nSubtracting,\n$$\nS_n - \\frac{1}{2} S_n = S_n \\left(1 - \\frac{1}{2}\\right) = \\frac{1}{2} S_n = \\frac{2}{2} + \\frac{3}{2^2} + \\dots + \\frac{n+1}{2^n} - \\left(\\frac{2}{2^2} + \\frac{3}{2^3} + \\dots + \\frac{n+1}{2^{n+1}}\\right).\n$$\nThis telescopes to\n$$\n\\frac{1}{2} S_n = \\frac{3}{2} - \\frac{n+3}{2^{n+1}}.\n$$\nSo\n$$\nS_n = 3 - \\frac{n+3}{2^n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18704,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCM$ be a quadrangle and let $D$ be an interior point such that $ABCD$ is a parallelogram. If $\\angle AMB \\equiv \\angle CMD$, prove that $\\angle MAD \\equiv \\angle MCD$.\n\n![](images/RMC2012_p80_data_047ce153ee.png)\n\nAnother approach consists in noticing that the triangles *MAB* and *MCD* have equal circumradii.",
    "options": [],
    "answer": "See solution",
    "solution": "Construct parallelogram $ABEM$. Then $\\angle AMB \\equiv \\angle MBE$ and, since $CDME$ is a parallelogram, $\\angle DMC \\equiv \\angle MCE$. This leads to $\\angle MBE \\equiv \\angle MCE$, so the quadrangle $MBCE$ is cyclic. This yields $\\angle BCM \\equiv \\angle BEM \\equiv \\angle BAM$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18705,
    "subject": "Mathematics (Olympiad)",
    "question": "a) $\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\}$ is an example of a powerful set with four elements.\n\nb) Prove that this is the unique powerful set with four elements.\n\nA *powerful set* $S$ is a finite set of positive real numbers such that for any two distinct elements $x, y \\in S$, at least one of $x^y$ or $y^x$ is also in $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.** A *finite powerful set* $S$ cannot have an element greater than one and an element less than one.\n\n*Proof.* Suppose, for contradiction, that such elements exist. Let $a$ be the least element of $S$ and $b$ the least element of $S$ greater than $1$. Then $a < 1 < b$.\n\n- $a^b < a^1 = 1$, but $a$ was the least element of $S$. Thus, $a^b \\notin S$.\n- $1 < b^a < b^1 = b$, but $b$ was the least element of $S$ greater than one. Therefore, $b^a \\notin S$.\n\nThis contradiction proves the lemma.\n\nBy the lemma, all elements of a finite powerful set $S$ are either all in $[1, \\infty)$ or all in $(0, 1]$.\n\nSuppose $S$ is a powerful set with $n > 3$ elements in $[1, \\infty)$, say $S = \\{1 = a_1 < a_2 < \\cdots < a_n\\}$. For $i \\geq 2$, $a_n^{a_i} > a_n$, so $a_i^{a_n}$ must be in $S$. We have:\n\n$$\na_1 < a_2 < a_2^{a_n} < a_3^{a_n} < \\cdots < a_{n-1}^{a_n}\n$$\n\nSo for $2 \\leq i \\leq n-1$, $a_i^{a_n} = a_{i+1}$. Now, for $a_2 < a_{n-1}$ ($n > 3$):\n\n$$\na_2 < a_2^{a_{n-1}} < a_2^{a_n} = a_3 \\implies a_2^{a_{n-1}} \\notin S\n$$\n\n$$\na_{n-1} < a_{n-1}^{a_2} < a_{n-1}^{a_n} = a_n \\implies a_{n-1}^{a_2} \\notin S\n$$\n\nThis contradicts the definition of a powerful set.\n\nNow suppose $S$ is a powerful set with $n > 4$ elements in $(0, 1]$, $S = \\{a_1 < a_2 < \\cdots < a_n = 1\\}$. For $1 \\leq i \\leq n-2$, $a_{n-1} < a_{n-1}^{a_i} < 1$, so $a_{n-1}^{a_i} \\notin S$, and thus $a_i^{a_n} \\in S$. We have:\n\n$$\na_1 < a_1^{a_{n-1}} < a_2^{a_{n-1}} < \\cdots < a_{n-2}^{a_n} < 1\n$$\n\nSo $a_i^{a_{n-1}} = a_{i+1}$ for $2 \\leq i \\leq n-2$. Let $a_{n-1} = a$:\n\n$$\na_{n-2} = a^{1/a},\\quad a_{n-3} = a^{1/a^2}, \\ldots\n$$\n\nLooking at $a_{n-1}$ and $a_{n-2}$:\n\n$$\na_{n-1} = a_{n-2}^{a_{n-1}} < a_{n-1}^{a_{n-2}} < 1 \\implies a_{n-1}^{a_{n-2}} \\notin S\n$$\n\nThis implies $a_{n-2}^{a_{n-1}} \\in S$. Since $a_{n-2}^{a_{n-1}} > a_{n-2}^{a_n} = a_{n-1}$, we get $a_{n-2}^{a_{n-1}} = a_n$. So:\n\n$$\n(a^{1/a^2})^{a^{1/a}} = a \\implies a^{(a^{1/a})-2} = a \\implies a^{1/a - 2} = 1\n$$\n\nBut $a \\neq 1$, so $a = \\frac{1}{2}$. Therefore, $a_{n-1} = \\frac{1}{2}$, $a_{n-2} = \\frac{1}{4}$, $a_{n-3} = \\frac{1}{16}$. For $n > 4$, $a_{n-4} = \\frac{1}{256} \\in S$, but neither $a_{n-3}^{a_{n-4}}$ nor $a_{n-4}^{a_{n-3}}$ is in $S$. Thus, there is no powerful set with more than 4 elements.\n\nTherefore, $\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\}$ is the unique powerful set with four elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18706,
    "subject": "Mathematics (Olympiad)",
    "question": "A square with side length $\\ell$ is contained in a unit square whose centre is not interior to the former. Show that $\\ell \\leq 1/2$.",
    "options": [],
    "answer": "See solution",
    "solution": "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 theorem of Erdős stating that the sum of the side lengths of two squares packed into a unit square does not exceed $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18707,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a circle $O$ with two fixed points $B$ and $C$. Let $A$ be a moving point on $O$. Let $I$ be the midpoint of $BC$ and $H$ be the orthocenter of triangle $ABC$. The ray $IH$ meets $O$ at $K$. The line $AH$ meets $BC$ at $D$. The line $KD$ meets $O$ at $M$. Suppose the line through $M$ perpendicular to $BC$ meets $AI$ at $N$.\n\n(a) Prove that $N$ lies on a fixed circle.\n\n(b) A circle passing through $N$ and tangent to $AK$ at $A$ meets $AB$ and $AC$ at $P$ and $Q$, respectively. Let $J$ be the midpoint of $PQ$. Prove that $AJ$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $AX$ be the diameter of $O$. Since $CX \\perp AC$ and $BX \\perp AB$, $BHCX$ forms a parallelogram, so $HX$ passes through $I$ and $K$ lies on $HX$. Thus, $\\angle AKH = 90^\\circ$. Constructing $Hd \\perp HA$, we have\n\n$$\nK(AM, BC) = K(AD, BC) = A(KD, BC) = H(Id, CB) = -1.\n$$\n\nIt follows that $ABMC$ is a harmonic quadrilateral, so $IB$ is the bisector of $\\angle AIM$. The reflection $M'$ of $M$ through $BC$ lies on $AI$ and $MN$, so $N$ and $M'$ coincide. Since $N$ and $M$ are symmetric with respect to $BC$, $N$ lies on a circle which is the image of $O$ through $BC$.\n\nb) Let $L$ be the image of $K$ through the perpendicular bisector of $BC$. Since $AK$ is tangent to $(APQ)$, $AL \\parallel PQ$. Also, $KLCB$ is an isosceles trapezoid, so $AK$ and $AL$ are isogonal with respect to $\\angle BAC$. Hence,\n\n$$\nA(PQ, LX) = A(CB, KD) = -1\n$$\n\nCombining with $AL \\parallel PQ$, we conclude the median of triangle $APQ$ passes through $O$, which is a fixed point. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18708,
    "subject": "Mathematics (Olympiad)",
    "question": "Fifty marbles are lying in a heap. Alfred and Bodil take turns removing a positive number of marbles. Alfred starts, and at each step, the number of marbles removed must be a prime or a square. The winner is the one who empties the heap. Who wins if both play optimally?",
    "options": [],
    "answer": "See solution",
    "solution": "We construct a table inductively to determine whether the player whose turn it is with $n$ marbles left can win (*W*) or will lose (*L*):\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\n\\hline\nn & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\n\\hline\n & W & W & W & W & W & L & W & W & W & W \\\\\n\\hline\nn & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\\\\n\\hline\n & W & L & W & W & W & W & W & L & W & W \\\\\n\\hline\nn & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \\\\\n\\hline\n & W & W & W & L & W & W & W & W & W & L \\\\\n\\hline\nn & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \\\\\n\\hline\n & W & W & W & W & W & W & W & L & W & W \\\\\n\\hline\nn & 41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50 \\\\\n\\hline\n & W & W & W & L & W & W & W & W & W & L \\\\\n\\hline\n\\end{array}\n$$\n\nThe correctness of the table is shown inductively. For $n \\leq 35$, a player cannot remove a number of marbles equal to $6k$ ($k \\in \\mathbb{Z}$), since $6, 12, 18, 24, 30$ are neither primes nor squares. Thus, for $n \\leq 35$, $n$ is marked *L* if divisible by 6 (since any legal move leads to a non-multiple of 6), and *W* otherwise (since removing 1–5 marbles can leave a multiple of 6).\n\n% ![](images/BW23_Shortlist_2023-11-01_p70_data_5c7609cb7d.png)\n\nFor $n = 36$, the argument fails, as one can remove all 36 marbles (a square) and win. For $n = 37$, removing 7 marbles (a prime) leaves 30 (a losing position) for the opponent. For $n = 38$, it's not possible to win, as none of the required removals (to reach a losing position) are allowed.\n\nThe next five numbers, 39 to 43, are winning positions, as removing 1–5 marbles leaves the opponent with 38 marbles (a losing position). For $n = 44$, it's again a losing position, as none of the required removals are allowed.\n\nContinuing, 45 to 49 are winning positions, as removing 1–5 marbles leaves the opponent with 44 marbles (a losing position). For $n = 50$, it's a losing position, as Alfred cannot remove a legal number of marbles to leave Bodil in a losing position. Therefore, Bodil will win if both play optimally.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18709,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Can the polynomial $2x^2 - 1$ be obtained from $x^2 - 2x - 1$ by a sequence of the following operations on quadratic polynomials?\n\n- Replace $ax^2 + bx + c$ with $cx^2 + bx + a$.\n- Replace $ax^2 + bx + c$ with $a(x + d)^2 + b(x + d) + c$ for some real $d$.\n\nb) Is it possible to obtain the polynomial $2x^2 - x - 1$ from $x^2 - 2x - 1$ using any sequence of these operations?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The first polynomial, $2x^2 - 1$, can be obtained by applying the following sequence of steps:\n\n$$\nx^2 - 2x - 1 \\rightarrow -x^2 - 2x + 1 \\xrightarrow{d=-1} -x^2 + 2 \\rightarrow 2x^2 - 1.\n$$\n\nb) The prescribed steps leave the discriminant unchanged: the discriminant of $ax^2 + bx + c$ is $b^2 - 4ac$. For the transformation $a(x + d)^2 + b(x + d) + c = ax^2 + (2ad + b)x + (ad^2 + bd + c)$, the discriminant remains $b^2 - 4ac$ after simplification.\n\nThe discriminant of the initial polynomial $x^2 - 2x - 1$ is $(-2)^2 - 4 \\cdot 1 \\cdot (-1) = 4 + 4 = 8$, while the discriminant of $2x^2 - x - 1$ is $(-1)^2 - 4 \\cdot 2 \\cdot (-1) = 1 + 8 = 9$. Thus, it is not possible to obtain $2x^2 - x - 1$ from $x^2 - 2x - 1$ using the allowed steps.\n\n*Remark*: The operations can be viewed as $T_1: (a, b, c) \\to (c, b, a)$ and $T_2: (a, b, c) \\to (a, 2ad + b, ad^2 + bd + c)$. Given two quadratics with equal discriminants, a sequence of these steps can transform one into the other by adjusting coefficients as needed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18710,
    "subject": "Mathematics (Olympiad)",
    "question": "Ana y Beto juegan al siguiente juego. Ana escribe cuatro enteros consecutivos de tres dígitos. Beto elige tres de los cuatro números de Ana y calcula su suma. Si el número que obtiene se puede escribir como producto de tres enteros positivos mayores que 1, gana Beto. En caso contrario, gana Ana. ¿Determinar si Ana puede elegir los cuatro números para ganar con certeza?",
    "options": [],
    "answer": "See solution",
    "solution": "Veamos que es imposible que gane Ana. Dados cuatro enteros consecutivos, Beto elige los dos impares y uno par para que la suma de los tres sea par. Así se asegura que uno de los factores de la suma es el $2$. Vistos módulo $3$, los dos impares pueden tener restos $0$ y $2$, $1$ y $0$ o $2$ y $1$. Por lo tanto, siempre se puede agregar el número par de modo que la suma sea múltiplo de $3$. Este es el que se encuentra entre los dos impares. Entonces, la suma de tres de los números de Ana se puede escribir como $2 \\cdot 3 \\cdot k$ con $k > 3$, pues los tres números son de tres dígitos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18711,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a, b, c$ are real numbers with $a + b + c = 3$. Prove that\n\n$$\n\\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\le \\frac{1}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $a < \\frac{9}{5}$, then\n\n$$\n\\frac{1}{5a^2 - 4a + 11} \\le \\frac{1}{24}(3-a). \\quad \\textcircled{1}\n$$\n\nIn fact,\n\n$$\n\\begin{align*}\n\\textcircled{1} &\\Leftrightarrow (3-a)(5a^2-4a+11) \\ge 24 \\\\\n&\\Leftrightarrow 5a^3 - 19a^2 + 23a - 9 \\le 0 \\\\\n&\\Leftrightarrow (a-1)^2(5a-9) \\le 0 \\Leftrightarrow a < \\frac{9}{5}.\n\\end{align*}\n$$\n\nSo if $a, b, c < \\frac{9}{5}$, then\n\n$$\n\\begin{align*}\n& \\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\\\\n&\\le \\frac{1}{24}(3-a) + \\frac{1}{24}(3-b) + \\frac{1}{24}(3-c) \\\\\n&= \\frac{1}{4}.\n\\end{align*}\n$$\n\nIf one of $a, b, c$ is not less than $\\frac{9}{5}$, say $a \\ge \\frac{9}{5}$, then\n\n$$\n\\begin{aligned}\n5a^2 - 4a + 11 &= 5a \\left(a - \\frac{4}{5}\\right) + 11 \\\\\n&\\ge 5 \\cdot \\frac{9}{5} \\cdot \\left(\\frac{9}{5} - \\frac{4}{5}\\right) + 11 = 20.\n\\end{aligned}\n$$\n\nSo $\\frac{1}{5a^2 - 4a + 11} \\le \\frac{1}{20}$.\n\nSince\n\n$$\n5b^2 - 4b + 11 = 5\\left(b - \\frac{2}{5}\\right)^2 + 11 - \\frac{4}{5} \\ge 11 - \\frac{4}{5} > 10,\n$$\n\nwe have $\\frac{1}{5b^2 - 4b + 11} < \\frac{1}{10}$. Similarly, $\\frac{1}{5c^2 - 4c + 11} < \\frac{1}{10}$. So\n\n$$\n\\begin{aligned}\n& \\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\\\\n< \\frac{1}{20} + \\frac{1}{10} + \\frac{1}{10} = \\frac{1}{4}.\n\\end{aligned}\n$$\n\nHence the inequality holds for all $a, b, c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18712,
    "subject": "Mathematics (Olympiad)",
    "question": "Evaluate the sum\n\n$$\n\\frac{1 \\cdot 4}{2 \\cdot 5} + \\frac{2 \\cdot 7}{5 \\cdot 8} + \\dots + \\frac{k(3k+1)}{(3k-1)(3k+2)} + \\dots + \\frac{99 \\cdot 298}{296 \\cdot 299}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $S_n = \\sum_{k=1}^{n} \\frac{k(3k+1)}{(3k-1)(3k+2)}$. Multiply all numerators by 12 and all denominators by 4 to obtain\n\n$$\n3S_n = \\sum_{k=1}^{n} \\frac{12k(3k+1)}{(6k-2)(6k+4)}.\n$$\n\nNow complete squares as follows:\n\n$$\n\\begin{aligned}\n12k(3k+1) &= 36k^2 + 12k = (6k+1)^2 - 1, \\\\\n(6k-2)(6k+4) &= 36k^2 + 12k - 8 = (6k+1)^2 - 9.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\frac{12k(3k+1)}{(6k-2)(6k+4)} = \\frac{(6k+1)^2 - 1}{(6k+1)^2 - 9} = 1 + \\frac{8}{(6k+1)^2 - 9} = 1 + \\frac{8}{(6k-2)(6k+4)}.\n$$\n\nAnd it follows that\n\n$$\n3S_n = n + 8 \\sum_{k=1}^{n} \\frac{1}{(6k-2)(6k+4)}.\n$$\n\nSince\n\n$$\n\\frac{1}{(6k-2)(6k+4)} = \\frac{1}{6} \\left( \\frac{1}{6k-2} - \\frac{1}{6k+4} \\right),\n$$\n\nwe obtain\n\n$$\n3S_n = n + 8 \\cdot \\frac{1}{6} \\left( \\frac{1}{4} - \\frac{1}{10} + \\frac{1}{10} - \\frac{1}{16} + \\dots + \\frac{1}{6n-2} - \\frac{1}{6n+4} \\right) = n + \\frac{4}{3} \\left( \\frac{1}{4} - \\frac{1}{6n+4} \\right).\n$$\n\nThis leads to $3S_n = n + \\frac{n}{3n+2}$. For $n=99$ the answer is\n\n$$\nS_{99} = 33 + \\frac{33}{299} = \\frac{9900}{299}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18713,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n > 1$ and an integer $a$ that is coprime with $n$, there is a country consisting of $n$ islands $D_1, D_2, \\dots, D_n$. For any two different islands $D_i$ and $D_j$, there is a one-way ferry from $D_i$ to $D_j$ if and only if $ij \\equiv ia \\pmod{n}$. A tourist hopes to visit as many islands as possible. He can first fly to any island he chooses to start the tour, and afterwards can only use the one-way ferry to tour freely between islands in this country. Find the maximum possible number of different islands that the tourist can visit.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x(n) = \\sum_{v_p(n) \\ge 2} 1$ and $y(n) = \\sum_{v_p(n)=1} 1$. Let the maximum number of islands that can be visited be denoted as $k(n)$. Then:\n\n$$\nk(n) = \\begin{cases} 3x(n) + 2y(n) + 1, & v_2(n) \\neq 1, \\\\ 3x(n) + 2y(n), & v_2(n) = 1. \\end{cases}\n$$\n\nFor $x, y \\in \\mathbb{Z}$ and $m \\in \\mathbb{Z}_{>0}$, if $xy \\equiv xa \\pmod{m}$, we denote this as $x \\to y \\pmod{m}$. Let $p$ be a prime factor of $n$ and $p^\\alpha \\nmid n$. For a sequence $x_1 \\to x_2 \\to \\dots \\to x_k \\pmod{n}$, the same relation holds modulo $p^\\alpha$:\n\n$$\nx_1 \\to x_2 \\to \\dots \\to x_k \\pmod{p^\\alpha}.\n$$\n\n1. If $x_i \\neq 0 \\pmod{p^\\alpha}$, then $x_{i+1}$ is coprime to $p$. This is because $p^\\alpha \\mid x_i(x_{i+1} - a)$, and $p^\\alpha \\nmid x_i$. Therefore, $p \\mid x_{i+1} - a$. Since $a$ is coprime to $n$, we have $p \\nmid a$, which implies $p \\nmid x_{i+1}$.\n\n2. If $x_i$ is coprime to $p$, then $x_{i+1} \\equiv a \\pmod{p^\\alpha}$. This is because $p^\\alpha \\mid x_i(x_{i+1} - a)$, and since $p \\nmid x_i$, we have $p^\\alpha \\mid x_{i+1} - a$.\n\n3. From (1) and (2), we can conclude that if $\\alpha \\ge 2$, the sequence $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergoes at most three changes: it changes from being divisible by $p^\\alpha$ to being nonzero, then becomes coprime to $p$, and finally becomes congruent to $a$ modulo $p^\\alpha$.\n\nIf $\\alpha = 1$ and $p \\neq 2$, then $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergo at most two changes: it changes from being congruent to 0 modulo $p$ to being coprime to $p$, and finally becomes congruent to $a$ modulo $p$.\n\nIf $\\alpha = 1$ and $p = 2$, then $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergo at most one change: it changes from being congruent to 0 modulo 2 to being congruent to $a$ modulo 2.\n\nAssuming that $x_1, x_2, \\dots, x_k \\pmod{n}$ have distinct adjacent terms, then adjacent terms have changes in some modulo $p^\\alpha$, implying that $k \\le k(n)$.\n\n4. Construct examples for $k = k(n)$. First, consider the cases based on modulo $p^\\alpha$.\n\n- If $\\alpha \\ge 2$, we have $0 \\to p \\to a + p^{\\alpha-1} \\to a \\pmod{p^\\alpha}$.\n- If $\\alpha = 1$ and $p \\neq 2$, we have $0 \\to b \\to a \\pmod{p}$, where $b$ is coprime to $p$ and $b \\neq a \\pmod{p}$.\n- If $\\alpha = 1$ and $p = 2$, we have $0 \\to a \\pmod{2}$.\n\nLet $n = p_1^{\\alpha_1} \\cdots p_w^{\\alpha_w}$ be the prime factorization of $n$. For each $p_i^{\\alpha_i}$, choose a sequence corresponding to the three cases above. Note that $0 \\to 0$ and $a \\to a$. We can add zeros at the beginning and $a$'s at the end of the sequences to ensure their lengths reach $k(n)$. Denote these modified sequences as $L_i$. For $p_i^{\\alpha_i}$ and $p_j^{\\alpha_j}$ ($i \\neq j$), the positions where changes occur in $L_i$ and $L_j$ are different.\n\nFor example, suppose $\\alpha_1 \\ge 2$, we can take\n\n$$\nL_1: 0 \\to p_1 \\to a + p_1^{\\alpha_1-1} \\to a \\to a \\to \\dots \\to a \\pmod{p_1^{\\alpha_1}}.\n$$\n\nIf $\\alpha_2 = 1$ and $p_2 \\ne 2$, we can also take\n\n$$\nL_2: 0 \\to 0 \\to 0 \\to 0 \\to b \\to a \\to a \\to \\dots \\to a \\pmod{p_2},\n$$\n\nand continue this way to construct $L_3, \\dots, L_w$.\n\nBy using the $t$-th term of $L_1, L_2, \\dots, L_w$ and the Chinese Remainder Theorem, we can determine $x_t$. Since $x_t \\to x_{t+1} \\pmod{p_i^{\\alpha_i}}$ for $1 \\le i \\le w$, it follows that $x_t \\to x_{t+1} \\pmod{n}$. Moreover, the sequence $x_1, x_2, \\dots, x_{k(n)}$ modulo $n$ has distinct terms. This confirms the conclusion, and the maximum value $k(n)$ is given by the formula above. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18714,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0 < a_1 < a_2 < \\dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \\ge 1$ such that\n\n$$\na_n < \\frac{a_0 + a_1 + \\dots + a_n}{n} \\le a_{n+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For a given sequence $a_0 < a_1 < a_2 < \\dots$ satisfying the conditions of the problem, define the function $f: \\mathbb{N}^+ \\to \\mathbb{Z}$ by\n\n$$\nf(n) = n a_n - a_0 - a_1 - \\dots - a_n.\n$$\n\nNote that\n\n$$f(n) < 0 \\iff a_n < \\frac{a_0 + a_1 + \\dots + a_n}{n}$$\n$$f(n+1) \\ge 0 \\iff a_{n+1} \\ge \\frac{a_0 + a_1 + \\dots + a_n}{n}.$$ \n\nObserve that $f(1) = -a_0 < 0$, and\n\n$$\n\\begin{align*}\nf(n+1) - f(n) &= (n+1)a_{n+1} - a_0 - a_1 - \\dots - a_{n+1} - [n a_n - a_0 - a_1 - \\dots - a_n] \\\\\n&= (n+1)a_{n+1} - (n a_n + a_{n+1}) \\\\\n&= n(a_{n+1} - a_n) \\\\\n&> 0.\n\\end{align*}\n$$\n\nThus, $f(1), f(2), f(3), \\dots$ is a strictly increasing sequence of integers, with $f(1) < 0$.\n\n![](images/Australian_Scene_-2014_p192_data_d44e88737b.png)\n\nTherefore, the graph of this function crosses the x-axis exactly once. That is, there is a unique positive integer $n$ such that $f(n) < 0$ and $f(n+1) \\ge 0$. This is the unique value of $n$ that satisfies the problem. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18715,
    "subject": "Mathematics (Olympiad)",
    "question": "An acute-angled triangle $ABC$ with $AC > AB$ is given. The perpendicular bisector of side $BC$ intersects the lines $AC$ and $AB$ at points $D$ and $E$, respectively. The circle with diameter $DE$ intersects the lines $AC$ and $AB$ at points $K$ and $L$, respectively ($K \\neq D, L \\neq E$). Let $M$ be the midpoint of side $BC$. Prove that the points $K, L$, and $M$ are collinear.\n\n![](images/EST_ABooklet_2024_p38_data_01c95250b3.png)\n\nFig. 28",
    "options": [],
    "answer": "See solution",
    "solution": "Since the points $E, K, L$, and $D$ are concyclic (see the figure), it follows that $\\angle CKL = \\angle DKL = \\angle DEL = \\angle MEB$. From the problem conditions, $\\angle CME = 90^\\circ$, and by Thales' theorem, $\\angle CKE = \\angle DKE = 90^\\circ$. Therefore, the points $C, M, K, E$ are also concyclic. Consequently, $\\angle CKM = \\angle CEM$. Since point $E$ lies on the perpendicular bisector of side $BC$, the triangles $BEM$ and $CEM$ are congruent, thus $\\angle MEB = \\angle CEM$. From the previous result, $\\angle CKL = \\angle CKM$. Since the points $L$ and $M$ lie on the same side of line $CK$, the points $K, L$, and $M$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18716,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that one can choose 8 pairwise distinct numbers among $1, 2, \\ldots, 10000$, such that none of them is a perfect square and that no sum of several of them is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the following 7 numbers: $2^1, 2^3, \\ldots, 2^{13} = 8192 < 10000$. Clearly, the sum of any subset of them is such that the highest power of 2 that divides the sum is odd, thus, it is not a perfect square.\n\nAdd number $3$ to the chosen numbers. Suppose it is possible to choose several numbers such that their sum is a perfect square. Then $3$ is one of such numbers, otherwise we get a contradiction as shown above. Since a perfect square can only have a remainder $0$, $1$, or $4$ when divided by $8$, and since the sum is odd, then the only possible remainder is $1$. However, only numbers $2$ and $3$ have a non-zero remainder when divided by $8$, thus, it is not possible to obtain a sum with remainder $1$ when divided by $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18717,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an arbitrary arrangement of $N$ boys along a circle. Assign a sign \"+\" or \"-\" before each boy as follows: moving clockwise, put a \"+\" before a boy if he is taller than the previous boy, and a \"-\" if he is shorter than the previous one. A boy is called *tall* if a \"+\" stands before him and a \"-\" after him; he is called *short* if a \"-\" stands before him and a \"+\" after him. (Since the tallest boy among all $N$ boys is tall, there must be both \"+\" and \"-\" signs in any arrangement.) Show that, in any arrangement, the number of tall boys is equal to the number of short boys.",
    "options": [],
    "answer": "See solution",
    "solution": "Group all successive \"+\" signs as a single \"+\", and all successive \"-\" signs as a single \"-\", forming a new sequence of signs. The number of alternations between \"+\" and \"-\" (and vice versa) in the original arrangement equals that in the new arrangement. In this new sequence, the number of alternations from \"+\" to \"-\" equals the number from \"-\" to \"+\". Therefore, the number of tall boys (corresponding to \"+\" to \"-\" alternations) equals the number of short boys (corresponding to \"-\" to \"+\" alternations) in any arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18718,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle with center $I$ is inscribed in a hexagon $ABCDEF$. Let $A', B', C', D', E', F'$ be the midpoints of the diagonals $BF, AC, BD, CE, DF, EA$, respectively. Suppose that the lines $AA'$, $CC'$, $EE'$ intersect at $X$, and the lines $BB'$, $DD'$, $FF'$ intersect at $Y$. If the triangle residing between the lines $AB$, $CD$, $EF$ is not similar to the triangle residing between the lines $BC$, $DE$, $FA$, then prove that the points $X$, $I$, $Y$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma:\n\n**Lemma.** Let $ABCDEF$ be a convex hexagon on the coordinate plane. Then the set of points $P$ inside $ABCDEF$ such that\n\n$$\nS_{APB} + S_{CPD} + S_{EPF} = S_{BPC} + S_{DPE} + S_{FPA}\n$$\n\nis a segment.\n\n*Proof.* Suppose that the equations of the lines containing the segments $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ are $a_i x + b_i y + c_i = 0$ for some $a_i, b_i, c_i \\in \\mathbb{R}$ ($i = 1, \\dots, 6$). Without loss of generality, we can assume that $a_i \\cdot b_i \\neq 0$ for $i = 1, \\dots, 6$. The distance from the point $P(x_0, y_0)$ to the line $a_i x + b_i y + c_i = 0$ is\n\n$$\nd_i = \\frac{|a_i x_0 + b_i y_0 + c_i|}{\\sqrt{a_i^2 + b_i^2}}\n$$\n\nfor $i = 1, \\dots, 6$. Since $P$ is inside the hexagon, the sign of $a_i x_0 + b_i y_0 + c_i$ does not change. Therefore, we can assume that\n\n$$\nd_i = \\frac{a_i x_0 + b_i y_0 + c_i}{\\sqrt{a_i^2 + b_i^2}} \\quad (i = 1, \\dots, 6).\n$$\n\nWe should find the set of points $P(x_0, y_0)$ such that\n\n$$\n\\frac{1}{2} AB \\cdot d_1 + \\frac{1}{2} CD \\cdot d_3 + \\frac{1}{2} EF \\cdot d_5 = \\frac{1}{2} BC \\cdot d_2 + \\frac{1}{2} DE \\cdot d_4 + \\frac{1}{2} FA \\cdot d_6.\n$$\n\nObserve that this is a linear equation with respect to $x_0$ and $y_0$, and therefore the set of points $P(x_0, y_0)$ is a segment. $\\square$\n\nNow we prove the problem. Since $A'$ is the midpoint of $BF$, $S_{A'XB} = S_{A'XF}$ and $S_{A'AB} = S_{A'AF}$. Hence $S_{AXB} = S_{AXF}$. Similarly, $S_{BXC} = S_{CXD}$ and $S_{DXE} = S_{EXF}$. Hence\n\n$$\nS_{AXB} + S_{CXD} + S_{EXF} = S_{BXC} + S_{DXE} + S_{FXA}.\n$$\n\nIn a similar way, we can show that\n\n$$\nS_{AYB} + S_{CYD} + S_{EYF} = S_{BYC} + S_{DYE} + S_{FYA}.\n$$\n\nOn the other hand, let $Q_1, Q_2, \\dots, Q_6$ be the tangent points of the incircle to the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ respectively. Since $Q_1I = Q_6I = r$, $AQ_1 = AQ_6$, $\\angle IQ_1A = \\angle IQ_6A$, we get that $\\triangle AIQ_1 = \\triangle AIQ_6$. Similarly, $\\triangle BIQ_1 = \\triangle BIQ_2$, $\\triangle CIQ_2 = \\triangle CIQ_3$, $\\triangle DIQ_3 = \\triangle DIQ_4$, $\\triangle EIQ_4 = \\triangle EIQ_5$, $\\triangle FIQ_5 = \\triangle FIQ_6$. Hence\n\n$$\nS_{AIB} + S_{CID} + S_{EIF} = S_{BIC} + S_{DIE} + S_{FIA}.\n$$\n\nTherefore, the points $X$, $Y$, $I$ are collinear, because of the lemma proved above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18719,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be the set of all positive integers formed by $n$ 1's and $n$ 2's, and let $T$ be the set of all positive integers with $n$ digits formed by 1, 2, 3, 4 such that the numbers of 1's and 2's are equal.\n\nDescribe a bijection between $S$ and $T$, and deduce that\n\n$$\nf(n) = |S| = |T| = g(n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For each $m \\in S$, pair every two consecutive digits of $m$ from left to right, giving $n$ pairs. Replace the pairs 11, 22, 12, 21 by 1, 2, 3, 4 respectively. If there are $a, b, c, d$ pairs of 11, 22, 12, 21 respectively, then $m$ has $2a + c + d$ 1's and $2b + c + d$ 2's, so $a = b$. Thus, the image has equal numbers of 1's and 2's, so it belongs to $T$.\n\nThe mapping is reversible: for $m \\in T$, replace digits 1, 2, 3, 4 by 11, 22, 12, 21 respectively. If there are $a, a, c, d$ copies of 1, 2, 3, 4, then the image has $2a + c + d$ 1's and $2a + c + d$ 2's, so it belongs to $S$. Thus, the mapping is a bijection, and\n\n$$\nf(n) = |S| = |T| = g(n).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18720,
    "subject": "Mathematics (Olympiad)",
    "question": "Una rolls 6 standard 6-sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4?\n\n(A) $\\frac{3}{4}$  (B) $\\frac{57}{64}$  (C) $\\frac{59}{64}$  (D) $\\frac{187}{192}$  (E) $\\frac{63}{64}$",
    "options": [],
    "answer": "See solution",
    "solution": "The product will not be divisible by 4 precisely when all 6 rolls are odd, or exactly one of them is equal to either 2 or 6 and the rest are odd. The probability of this complementary event is\n\n$$\n\\left(\\frac{1}{2}\\right)^6 + 6 \\cdot \\left(\\frac{1}{3}\\right) \\cdot \\left(\\frac{1}{2}\\right)^5 = \\frac{5}{64}.\n$$\n\nThe requested probability is therefore $1 - \\frac{5}{64} = \\frac{59}{64}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18721,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $R$ be a ring with unit and let $f$ be a surjective endomorphism of $R$ such that $[x, f(x)] = 0$ for all $x$ in $R$, where $[a, b] = ab - ba$ is the standard Lie commutator of the elements $a$ and $b$ of $R$.\n\nProve that:\n\n1. $[x, f(y)] = [f(x), y]$ and $x[x, y] = f(x)[x, y]$ for all $x, y \\in R$.\n2. If, in addition, $R$ is a field and $f$ is not the identity, then $R$ is commutative.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** The first relation follows from the sequence of relations below:\n$$\n\\begin{align*}\n0 &= [x - y, f(x - y)] = [x - y, f(x) - f(y)] \\\\\n  &= [x, f(x)] - [x, f(y)] - [y, f(x)] + [y, f(y)] \\\\\n  &= -[x, f(y)] + [f(x), y].\n\\end{align*}\n$$\nTo prove the second, we use the first. Let $y = f(z)$, $z \\in R$. Then\n$$\n\\begin{align*}\nx[x, y] &= x[x, f(z)] = x[f(x), z] = x f(x)z - xz f(x) \\\\\n&= f(x)xz - xz f(x) = [f(x), xz] \\\\\n&= [x, f(xz)] = [x, f(x)y] = x f(x)y - f(x)yx \\\\\n&= f(x)xy - f(x)yx = f(x)[x, y].\n\\end{align*}\n$$\n**b)** We shall prove that $R^* = Z(R^*)$, where $Z(R^*) = \\{x : x \\in R^*, xy = yx \\text{ for all } y \\in R^*\\}$ is the center of the multiplicative group $R^*$. Let $\\text{Fix } f = \\{x : x \\in R^*, f(x) = x\\}$. The second relation in a) shows that $R^* \\setminus Z(R^*) \\subseteq \\text{Fix } f$, so $R^* = Z(R^*) \\cup \\text{Fix } f$. Since $Z(R^*)$ and $\\text{Fix } f$ are subgroups of $R^*$, either $\\text{Fix } f \\subseteq Z(R^*)$, in which case $R^* = Z(R^*)$; or $Z(R^*) \\subseteq \\text{Fix } f$, in which case $R^* = \\text{Fix } f$, i.e., $f$ is the identity—a contradiction. Consequently, $R^* = Z(R^*)$, i.e., $R^*$ is commutative.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18722,
    "subject": "Mathematics (Olympiad)",
    "question": "Some consecutive positive integers have been written on a whiteboard. Leigh circles some of them and underlines some of them so that one more number is circled than underlined. (Numbers can be both circled and underlined.) It turns out that, no matter how Leigh does this, the sum of the circled numbers is always greater than the sum of the underlined numbers.\n\nShow that there is at most one square number on the whiteboard.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that there are at least two square numbers among the consecutive integers. Since the numbers are consecutive, two of them must be consecutive squares, say $x^2$ and $(x+1)^2$, where $x$ is positive.\n\nBy the given condition, we have:\n\n$$\n\\begin{aligned}\n\\underbrace{x^2 + (x^2 + 1) + \\dots + (x^2 + x)}_{x+1\\ \\text{consecutive numbers}} &> \\underbrace{(x^2 + x + 1) + (x^2 + x + 2) + \\dots + (x^2 + x + x)}_{x\\ \\text{consecutive numbers}} \\\\\n\\Leftrightarrow\\quad (x+1)x^2 &> x(x^2 + x) \\quad \\text{(cancel $1+2+\\dots+x$ from both sides)} \\\\\n\\Leftrightarrow\\quad x &> x+1\n\\end{aligned}\n$$\n\nwhich is clearly false. Contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18723,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku paints exactly 30 unit squares of a $10 \\times 15$ table black. After that, Miku covers up exactly 4 rows and 4 columns. Can Juku ensure by the choice of the squares to be coloured that at least 10 black unit squares are left uncovered?",
    "options": [],
    "answer": "See solution",
    "solution": "Juku can paint 30 unit squares so that each row contains exactly 3 black unit squares and each column contains exactly 2 of them. When Miku chooses 4 rows and 4 columns, they contain at most $4 \\times 3 + 4 \\times 2 = 20$ black unit squares altogether. Hence, at least 10 black unit squares remain uncovered.\n\n![](images/EST_ABooklet_2020_p3_data_66e8169922.png)\n\nFig. 1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18724,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible for $n^2 + 3n + 3$ to be written as a product $ab$ for integers $a$ and $b$ such that $|a-b| < 2\\sqrt{n+1}$?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible. Indeed, if $n^2 + 3n + 3 = ab$, then\n\n$$\n(a+b)^2 = (a-b)^2 + 4ab \\ge 4ab = 4n^2 + 12n + 12 > 4n^2 + 12n + 9 = (2n+3)^2.\n$$\n\nSince both sides are squares, this yields $(a+b)^2 \\ge (2n+4)^2$. It follows that\n\n$$\n(a-b)^2 = (a+b)^2 - 4ab \\ge (2n+4)^2 - (4n^2 + 12n + 12) = 4n + 4 = 4(n+1).\n$$\n\nTherefore, $|a-b| \\ge 2\\sqrt{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18725,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square, and let $k$ be the circle centred at $B$ passing through $A$, $C$, and the point $T$ inside the square. The tangent to $k$ at $T$ intersects the segments $\\overline{CD}$ and $\\overline{DA}$ at $E$ and $F$, respectively. Let $G$ and $H$ be the intersections of the lines $BE$ and $BF$ with the segment $\\overline{AC}$, respectively.\n\nProve that the lines $BT$, $EH$, and $FG$ pass through the same point.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the lines $FA$ and $FT$ are tangent to the circle $k$, hence $|FA| = |FT|$, and the triangles $ABF$ and $TBF$ are congruent. Analogously, $|EC| = |ET|$, and the triangles $CBE$ and $TBE$ are congruent.\n\n![](images/Croatia_2018_p23_data_7bd72e7bae.png)\n\nLet us denote $\\alpha = \\angle FBA = \\angle FBT$ and $\\beta = \\angle EBC = \\angle EBT$. Since $\\angle ABC = 90^\\circ$, it follows that $\\alpha + \\beta = 45^\\circ$. Now we have $\\angle AFB = 90^\\circ - \\alpha = 45^\\circ + \\beta$ and $\\angle AGB = \\angle GBC + \\angle BCA = \\beta + 45^\\circ$, i.e. $\\angle AFB = \\angle AGB$.\n\nHence, the quadrilateral $ABGF$ is cyclic, and $\\angle BGF = 180^\\circ - \\angle AFB = 90^\\circ$, i.e. $FG \\perp BE$. Similarly, the quadrilateral $BCEH$ is cyclic, and $EH \\perp BF$.\n\nSince the segments $\\overline{FG}$ and $\\overline{EH}$ are the altitudes of the triangle $BEF$, so is the segment $\\overline{BT}$. We finally conclude that the lines $BT$, $EH$, and $FG$ pass through the same point—the orthocentre of the triangle $BEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18726,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer and $A$ a set of $n$ points in the plane. Find all integers $k \\in \\{1, 2, \\dots, n-1\\}$ with the property that any two circles $C_1$ and $C_2$ in the plane, with $A \\cap \\operatorname{Int}(C_1) \\neq A \\cap \\operatorname{Int}(C_2)$ and $|A \\cap \\operatorname{Int}(C_1)| = |A \\cap \\operatorname{Int}(C_2)| = k$, have at least one common point.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k \\le \\left\\lfloor \\frac{n}{2} \\right\\rfloor$, consider a line $d$ which is not parallel to any of the lines determined by any two points in $A$. Without loss of generality, suppose $d$ is a vertical line. Choose a line $g$, parallel to $d$, such that to the right of $g$ there are exactly $k$ points of $A$. Hence, to the left of $g$ there are at least $k$ points from $A$. Choose a line $h \\neq g$, with $h \\parallel d$, such that to the left of $h$ there are exactly $k$ points of $A$.\n\nLet $M \\in g$ and construct a circle $C_1$, tangent at $M$ to $g$, with a large enough radius so that all $k$ points of $A$ to the right of $g$ are inside $C_1$. Let $N \\in h$ and construct a circle $C_2$, tangent at $N$ to $h$, with a large enough radius so that all $k$ points of $A$ to the left of $h$ are inside $C_2$. Obviously, $A \\cap \\operatorname{Int}(C_1) \\neq A \\cap \\operatorname{Int}(C_2)$, $|A \\cap \\operatorname{Int}(C_1)| = |A \\cap \\operatorname{Int}(C_2)| = k$, and $C_1 \\cap C_2 = \\emptyset$. Hence, there are no solutions $k$ with $k \\le \\left\\lfloor \\frac{n}{2} \\right\\rfloor$.\n\nIf $\\left\\lfloor \\frac{n}{2} \\right\\rfloor < k < n$, consider circles $C_1$ and $C_2$ with $A \\cap \\operatorname{Int}(C_1) \\neq A \\cap \\operatorname{Int}(C_2)$ and $|A \\cap \\operatorname{Int}(C_1)| = |A \\cap \\operatorname{Int}(C_2)| = k$.\n\nSince $\\operatorname{Int}(C_1)$ contains exactly $k$ points of $A$, outside $C_1$ there are fewer than $k$ points from $A$, so $\\operatorname{Int}(C_2)$ must have at least one point of $A$ also in $\\operatorname{Int}(C_1)$. Thus, $\\operatorname{Int}(C_1) \\cap \\operatorname{Int}(C_2) \\neq \\emptyset$.\n\nMoreover, since $A \\cap \\operatorname{Int}(C_1) \\neq A \\cap \\operatorname{Int}(C_2)$, neither circle is contained in the other, so $C_1 \\cap C_2 \\neq \\emptyset$. Therefore, all integers $k$ with $\\left\\lfloor \\frac{n}{2} \\right\\rfloor < k < n$ are solutions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18727,
    "subject": "Mathematics (Olympiad)",
    "question": "In a square of side length $5$, a figure is placed such that the distance between any two of its points is not equal to $0.001$, $0.002 \\times \\sqrt{3}$, or $0.002$. Prove that the area of this figure does not exceed $3.575$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F$ denote this figure. Consider the translations of $F$ under the following vectors in the plane: $\\vec{u}_1 = 0.001\\vec{i}$, and\n\n$$\n\\vec{u}_{k+1} = R_{\\pi/3}(\\vec{u}_k), \\quad k = 1, 2, 3, 4, 5,\n$$\n\nwhere $R_{\\pi/3}$ denotes rotation by $\\pi/3$ counterclockwise. Let $F_0 = F$ and, for $k = 1, \\dots, 6$, $F_k$ be the translation of $F$ in the direction $\\vec{u}_k$.\n\nWe will show that $F_k \\cap F_l = \\emptyset$ for all $k, l \\in \\{0, \\dots, 6\\}$ with $k \\neq l$. Since the distance between any two points of $F$ is not equal to $0.001$, $F_0$ and $F_k$ have no common points for $k = 1, \\dots, 6$. By the same reasoning, $F_k$ and $F_{k+1}$ have no common points for $k = 1, \\dots, 6$ ($F_7 = F_1$). Next, since the distance between any two points in $F$ is not equal to $0.001 \\times \\sqrt{3}$, it follows that $F_k$ and $F_{k+2}$ have no common points. Finally, since the distance between any two points in $F$ is not equal to $0.002$, $F_k$ and $F_{k+2}$ have no points in common. Thus $F_k \\cap F_l = \\emptyset$ for all $k \\neq l$ as claimed.\n\nThe figures $F_k$ all lie in the square of side $5.002$, and they have pairwise empty intersection. Thus\n\n$$\n7 \\times \\text{area}(F) = \\sum_{i=0}^{6} \\text{area}(F_i) \\le 5.002^2;\n$$\n\ntherefore $\\text{area}(F) \\le \\dfrac{5.002^2}{7} < 3.575$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18728,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system:\n\n$$\n\\begin{cases}\nx(y+z) = 35 \\\\\ny(z+x) = 32 \\\\\nz(x+y) = 27\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given system is equivalent to\n\n$$\n\\begin{cases}\nxy + xz = 35 \\\\\nyz + yx = 32 \\\\\nzx + zy = 27\n\\end{cases}\n$$\n\nLet $a = xy$, $b = xz$, $c = yz$. Then the system becomes\n\n$$\n\\begin{cases}\na + b = 35 \\\\\nc + a = 32 \\\\\nb + c = 27\n\\end{cases}\n$$\n\nSubtracting the equations, we have:\n\n- (1) + (2) $-$ (3): $2a = 40$ \\implies $a = 20$\n- (2) + (3) $-$ (1): $2c = 12$ \\implies $c = 6$\n- (1) + (3) $-$ (2): $2b = 30$ \\implies $b = 15$\n\nSo $xy = 20$, $xz = 15$, $yz = 12$ (for $x, y, z \\neq 0$).\n\nNow, $y = \\frac{20}{x}$, $z = \\frac{15}{x}$, and $yz = 12$.\n\nSubstitute $y$ and $z$ into $yz = 12$:\n\n$$\n\\left(\\frac{20}{x}\\right) \\left(\\frac{15}{x}\\right) = 12 \\implies \\frac{300}{x^2} = 12 \\implies x^2 = 25 \\implies x = \\pm 5\n$$\n\nThus, the solutions are $(x, y, z) = (5, 4, 3)$ and $(-5, -4, -3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18729,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any triangle $ABC$, find the point $P$ inside the triangle such that any line through $P$ divides the triangle into two pieces which are as equal as possible. \n\n% IMAGE: ![](images/obm-book_p77_data_bb906b28a7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We take $P$ to be the centroid $G$. First, if we take a line through $G$ parallel to one of the sides, the resulting triangular piece has area $\\frac{4}{9}$ of the total area.\n\nTake any other line through $G$, say $RS$ as shown above. We claim that $RG > GS$. Let $S'$ be the point obtained by rotating $S$ through $180^\\circ$ about $G$. Then $GSE$ and $GS'D$ are congruent and $SE$ is parallel to $S'D$. So $R$ cannot coincide with $S'$ (because the lines $SE$ and $RD$ are not parallel; they meet at $A$). If $GR < GS'$, then $RD$ and $SE$ will meet on the wrong side of $DE$. So we must have $GR > GS'$ and hence $GR > GS$. The triangles $GRD$ and $GSE$ have the same height ($GD \\sin \\angle DGR$), so $\\text{area } GRD > \\text{area } GSE$. Hence $\\text{area } ARS > \\text{area } ADE = \\frac{4}{9} \\text{ area } ABC$.\n\nLet $N$ be the midpoint of $AC$. As we move $R$ towards $B$, $S$ moves towards $A$. When $R$ reaches $B$, $S$ reaches $N$. So $S$ lies between $E$ and $N$. Now consider the triangles $BGR$, $NGS$. $BGR$ has the larger base, because $BG = 2GN$, and the larger height because $RG > SG$, so $RG \\sin \\angle RGB > SG \\sin \\angle SGN$. So $\\text{area } BGR > \\text{area } NGS$ and hence $\\text{area } ABN > \\text{area } ARS$. So $\\text{area } ARS < \\frac{1}{2} \\text{ area } ABC$. Thus $ARS$ is the smaller piece but its area is bigger than $\\frac{4}{9}$ area $ABC$. It remains to show that no other choice of $P$ is better than $G$.\n\n% IMAGE: ![](images/obm-book_p78_data_78f0934116.png)\n\nTake lines through $G$ parallel to the sides. That gives three overlapping triangles which cover $ABC$. If $P$ is not at $G$, then it must lie inside at least one of these triangles, say $ADE$. Now take a line $D'E'$ through $P$ parallel to $DE$. Then $\\text{area } AD'E' < \\text{area } ADE = \\frac{4}{9} \\text{ area } ABC$, so $P$ is a worse choice than $G$ (from $A$'s point of view).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18730,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers in the following pyramid represent the number of times the corresponding number in the original pyramid contributes to the original top number.\n\n![](images/2023_Australian_Scene_p51_data_3cf7fe2976.png)\n\nIf the original bottom row is $a, b, c, d, e, f$, how many arrangements of the numbers $1$ to $6$ in the bottom row produce a top number that is a multiple of $5$?",
    "options": [],
    "answer": "See solution",
    "solution": "The top number is $a + 5b + 10c + 10d + 5e + f = 5(b + 2c + 2d + e) + a + f$. This is a multiple of $5$ if and only if $a + f$ is a multiple of $5$.\n\nThe possible pairs for $(a, f)$ that sum to a multiple of $5$ among $1$ to $6$ are: $(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (6, 4)$.\n\nFor each pair, there are $4! = 24$ ways to arrange the remaining $4$ digits.\n\nSo, the total number of arrangements is $6 \\times 24 = 144$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18731,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be positive integers not greater than $12$ such that there exists a constant $C$ with $a^n + b^{n+9} \\equiv C \\pmod{13}$ for any positive integer $n$. Find all ordered pairs $(a, b)$ that satisfy these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "By the given conditions, for any positive integer $n$,\n$$\na^n + b^{n+9} \\equiv a^{n+3} + b^{n+12} \\pmod{13}. \\qquad \\textcircled{1}\n$$\nSince $13$ is prime and $a, b$ are coprime to $13$, by Fermat's Little Theorem:\n$$\na^{12} \\equiv b^{12} \\equiv 1 \\pmod{13}.\n$$\nTake $n = 12$ in (1):\n$$\n1 + b^9 \\equiv a^3 + 1 \\pmod{13} \\implies b^9 \\equiv a^3 \\pmod{13}.\n$$\nSubstitute into (1):\n$$\na^n + a^3 b^n \\equiv a^{n+3} + b^n \\pmod{13}.\n$$\nSo,\n$$\n(a^n - b^n)(1 - a^3) \\equiv 0 \\pmod{13}. \\qquad \\textcircled{2}\n$$\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p81_data_8dac8fb7f5.png)\n\n**Case 1:** $a^3 \\equiv 1 \\pmod{13}$\n\nThen $b^3 \\equiv a^3 b^3 \\equiv b^{12} \\equiv 1 \\pmod{13}$, so $a, b \\in \\{1, 3, 9\\}$.\n\nNow $a^n + b^{n+9} \\equiv a^n + b^n \\pmod{13}$. The condition requires\n$$\na + b \\equiv a^3 + b^3 \\equiv 2 \\pmod{13},\n$$\nwhich only holds for $a = b = 1$.\n\nCheck: For $(a, b) = (1, 1)$, $a^n + b^{n+9} \\equiv 2 \\pmod{13}$ for all $n$.\n\n**Case 2:** $a^3 \\not\\equiv 1 \\pmod{13}$\n\nFrom **②**, $a^n \\equiv b^n \\pmod{13}$ for all $n$, so $a = b$.\n\nThen $a^3 \\equiv b^9 = a^9 \\pmod{13}$, so\n$$\na^3(a^3 - 1)(a^3 + 1) \\equiv 0 \\pmod{13},\n$$\nso $a^3 \\equiv -1 \\pmod{13}$. Checking $a = 4, 10, 12$ (since $a \\leq 12$), these work.\n\nThus, the ordered pairs are $(1, 1), (4, 4), (10, 10), (12, 12)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18732,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a natural number *interesting* if any two consecutive digits form a number that is either a multiple of $19$ or $21$. For example, the number $7638$ is interesting, because $76$ is a multiple of $19$, $63$ is a multiple of $21$, and $38$ is a multiple of $19$. How many $2013$-digit interesting numbers exist?",
    "options": [],
    "answer": "See solution",
    "solution": "Among two-digit numbers, the multiples of $19$ are $19$, $38$, $57$, $76$, and $95$, and the multiples of $21$ are $21$, $42$, $63$, and $84$. Therefore, an interesting number cannot include the digit $0$. If it has the digit $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, or $9$, then the next digit must be $9$, $1$, $8$, $2$, $7$, $3$, $6$, $4$, or $5$, respectively. Thus, the first digit of an interesting number determines all the following digits. There are $9$ choices for the first digit, so there are $9$ interesting $2013$-digit numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18733,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider functions $f$ defined for all real numbers and taking real numbers as values such that\n\n$$f(x + 14) - 14 \\leq f(x) \\leq f(x + 20) - 20, \\quad \\text{for all real numbers } x.$$ \n\nDetermine all possible values of $f(8765) - f(4321)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Replace $x$ by $x - 14$ in the left inequality and $x$ by $x - 20$ in the right inequality to obtain\n\n$$f(x - 20) + 20 \\leq f(x) \\leq f(x - 14) + 14.$$ \n\nIt follows by induction that the following inequalities hold for every positive integer $n$:\n\n$$f(x - 20n) + 20n \\leq f(x) \\leq f(x - 14n) + 14n$$\n$$f(x + 14n) - 14n \\leq f(x) \\leq f(x + 20n) - 20n$$\n\nTherefore, we have the following chains of inequalities:\n\n$$\n\\begin{aligned}\nf(x+2) &= f(x+20 \\times 5-14 \\times 7) = f(x+14 \\times 3-20 \\times 2) \\\\\n&\\geq f(x+20 \\times 5)-98 \\leq f(x+14 \\times 3)-40 \\\\\n&\\geq f(x)+100-98 \\leq f(x)+42-40 \\\\\n&\\geq f(x)+2 \\leq f(x)+2\n\\end{aligned}\n$$\n\nSo we have deduced that $f(x+2) = f(x)+2$ and it follows by induction that\n\n$$f(x+2n) = f(x) + 2n$$\n\nfor all real numbers $x$ and all positive integers $n$.\n\nSubstituting $x = 4321$ and $n = 2222$ into this equation yields\n\n$$f(8765) - f(4321) = 4444.$$ \n\nSince $f(x) = x$ satisfies the conditions of the problem, the only possible value of the expression $f(8765) - f(4321)$ is $4444$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18734,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X$ be the set of all bijective functions from the set $S = \\{1, 2, 3, \\ldots, n\\}$ to itself. For each $f \\in X$, define\n\n$$\nT_f(j) = \\begin{cases} 1, & \\text{if } f^{(12)}(j) = j, \\\\ 0, & \\text{otherwise.} \\end{cases}\n$$\n\nDetermine\n\n$$\n\\sum_{f \\in X} \\sum_{j=1}^{n} T_f(j).\n$$\n\n(Here $f^{(k)}(x) = f(f^{(k-1)}(x))$ for $k \\ge 2$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n \\ge 12$. The elements of $X$ are permutations of $\\{1, 2, 3, \\ldots, n\\}$. If $j$ belongs to a $k$-cycle of $f$, where $k$ is a divisor of $12$, then $f^{(k)}(j) = j$. The number of divisors of $12$ is $6$. Let $m(j,k)$ be the number of elements $f$ in which $j$ belongs to a $k$-cycle of $f$, $1 \\leq j \\leq n$. Then\n\n$$\nm(j,k) = (n-1)(n-2)\\cdots(n-k+1)(n-k)! = (n-1)!\n$$\n\nThe required sum is, therefore, $n(n-1)! \\times 6 = 6(n!)$. If $n < 12$, the answer is $t(n)(n!)$, where $t(n)$ is the number of positive divisors of $12$ which are less than or equal to $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18735,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that if $n \\ge 2$ is an integer and $x_1, x_2, \\dots, x_n$ are positive real numbers, then\n\n$$\n4 \\left( \\frac{x_1^3 - x_2^3}{x_1 + x_2} + \\frac{x_2^3 - x_3^3}{x_2 + x_3} + \\dots + \\frac{x_{n-1}^3 - x_n^3}{x_{n-1} + x_n} + \\frac{x_n^3 - x_1^3}{x_n + x_1} \\right) \\le (x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_{n+1} = x_1$. Then\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{x_i^3 - x_{i+1}^3}{x_i + x_{i+1}} &= \\sum_{i=1}^{n} \\left( x_i^2 - x_{i+1}^2 + \\frac{x_i x_{i+1} (x_{i+1} - x_i)}{x_i + x_{i+1}} \\right) \\\\\n&= \\sum_{i=1}^{n} \\frac{x_i x_{i+1} (x_{i+1} - x_i)}{x_i + x_{i+1}}.\n\\end{aligned}\n$$\n\nOn the other hand,\n\n$$\n\\frac{x_i x_{i+1} (x_{i+1} - x_i)}{x_i + x_{i+1}} \\le \\frac{1}{2} x_{i+1} (x_{i+1} - x_i)\n$$\n\nAdding these inequalities for $i = 1, 2, \\dots, n$ yields the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18736,
    "subject": "Mathematics (Olympiad)",
    "question": "Find how many different choices for $L$ are there such that 117 passings occur before the end of the race. (A passing is defined when one skater passes another one. The beginning and the end of the race when all three skaters are together are not counted as a passing.)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that the length of the oval is one unit. Let $x(t)$ be the difference of distances that the slowest and the fastest skaters have skated by time $t$. Similarly, let $y(t)$ be the difference between the middle skater and the slowest skater. The path $(x(t), y(t))$ is a straight ray $R$ in $\\mathbb{R}^2$, starting from the origin, with slope depending on $L$. By assumption, $0 < y(t) < x(t)$.\n\nOne skater passes another one when either $x(t) \\in \\mathbb{Z}$, $y(t) \\in \\mathbb{Z}$, or $x(t) - y(t) \\in \\mathbb{Z}$.\nThe race ends when both $x(t), y(t) \\in \\mathbb{Z}$.\n\nLet $(a, b) \\in \\mathbb{Z}^2$ be the endpoint of the ray $R$. We need to find the number of such points satisfying:\n\n(a) $0 < b < a$\n\n(b) The ray $R$ intersects $\\mathbb{Z}^2$ at endpoints only.\n\n(c) The ray $R$ crosses 117 times the lines $x \\in \\mathbb{Z}$, $y \\in \\mathbb{Z}$, $y - x \\in \\mathbb{Z}$.\n\nThe second condition says that $a$ and $b$ are relatively prime. The ray $R$ crosses $a-1$ of the lines $x \\in \\mathbb{Z}$, $b-1$ of the lines $y \\in \\mathbb{Z}$, and $a-b-1$ of the lines $x-y \\in \\mathbb{Z}$.\nThus, we need $(a-1) + (b-1) + (a-b-1) = 117$, or equivalently, $2a-3=117$.\nThat is $a=60$.\n\nNow $b$ must be a positive integer less than and relatively prime to 60. The number of such $b$ can be found using Euler's $\\phi$ function:\n\n$$\n\\phi(60) = \\phi(2^2 \\cdot 3 \\cdot 5) = (2-1) \\cdot 2 \\cdot (3-1) \\cdot (5-1) = 16.\n$$\n\nThus the answer is 16. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18737,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $I_a$ be the excenter corresponding to $A$. Let $P$ and $Q$ be the tangency points of the excircle with center $I_a$ with the lines $AB$ and $AC$. Line $PQ$ meets lines $I_aB$ and $I_aC$ at the points $D$ and $E$ respectively. Denote $A_1$ as the intersection point of the lines $DC$ and $BE$; define in the same way points $B_1$ and $C_1$.\n\nProve that the lines $AA_1$, $BB_1$, $CC_1$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that $A_1$ is the orthocenter of the triangle $I_aBC$.\n\nTo do this, we show that the quadrilateral $BEI_aP$ is cyclic. Indeed,\n\n$$\n\\begin{align*}\n\\angle PEI_a &= 180^\\circ - \\angle PQA - \\angle ECQ \\\\\n&= 180^\\circ - \\left(90^\\circ - \\frac{\\angle A}{2}\\right) - \\left(90^\\circ - \\frac{\\angle C}{2}\\right) \\\\\n&= \\frac{\\angle A}{2} + \\frac{\\angle C}{2} = 90^\\circ - \\frac{1}{2}\\angle B \\\\\n&= \\angle PBI_a.\n\\end{align*}\n$$\n\n![](images/RMC2011_2_p26_data_a417b817f7.png)\n\nSo the quadrilateral $BEI_aP$ is cyclic and, since $\\angle BPI_a = 90^\\circ$, $BE \\perp CI_a$.\n\nThis leads to $BA_1 \\parallel CI$, where $I$ is the incenter of triangle $ABC$. Similarly, $CA_1 \\parallel BI$, therefore $BICA_1$ is a parallelogram.\n\nAnalogously, $AIBC_1$ is a parallelogram, hence $AC_1A_1C$ is also a parallelogram, meaning the segments $AA_1$ and $CC_1$ have the same midpoint. In the same way, segments $BB_1$ and $AA_1$ have the same midpoint, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18738,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Let $s : \\mathbb{N} \\to \\{1, \\dots, n\\}$ be a function such that $n$ divides $m - s(m)$ for all positive integers $m$. Let $a_0, a_1, a_2, \\dots$ be a sequence such that $a_0 = 0$ and\n\n$$\na_k = a_{k-1} + s(k) \\text{ for all } k \\ge 1.\n$$\n\nFind all $n$ for which this sequence contains all the residues modulo $(n+1)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "$n = 2^k - 1$ for any $k \\in \\mathbb{N}$.\n\nWe begin by noting that the sequence is given by the formula\n\n$$\na_{rn+s} = r \\cdot \\binom{n+1}{2} + \\binom{s+1}{2}\n$$\n\nfor all $r \\ge 0$ and $s = 1, 2, \\dots, n$. This is easy to confirm by mathematical induction: $a_0 = 0$ is true, and for fixed $r$, inducting on $s$ gives\n\n$$\na_{rn+s} = a_{rn+s-1} + s = r \\cdot \\binom{n+1}{2} + \\binom{(s-1)+1}{2} + s = r \\cdot \\binom{n+1}{2} + \\binom{s+1}{2},\n$$\n\nproving this induction step. Finally, $a_{(r+1)n} = (r+1)\\binom{n+1}{2} = r \\cdot \\binom{n+1}{2} + \\binom{n+1}{2} = a_{rn+n}$, confirming the induction argument on $r$ as well, proving the claim.\n\nNow suppose $n+1$ has an odd prime factor $p$. Then $p \\mid \\binom{n+1}{2}$, so if all residues mod $(n+1)^2$ are present, then $\\binom{s+1}{2} \\equiv \\frac{s^2+s}{2} \\pmod p$ covers all possible residues mod $p$. However, the map $x \\mapsto x^2+x \\pmod p$ sends $x$ and $-(1+x)$ to the same element mod $p$ and the two are distinct unless $x \\equiv \\frac{-1}{2} \\pmod p$, hence the set of residues represented by $s^2+s \\pmod p$ has $\\frac{p+1}{2} < p$ elements, a contradiction! So every $n$ that satisfies this condition must be one less than a power of 2.\n\nNow we prove that all $n$ one less than powers of 2 satisfy the condition. Let $n = 2^k - 1$. Then $\\binom{n+1}{2} = 2^{k-1}(2^k - 1) = 2^{2k-1} - 2^{k-1} \\pmod{2^{2k}}$. Then, if for $0 \\le r$ and $1 \\le s \\le n$, and $r = 2j + e$, where $e \\in \\{0, 1\\}$, then\n\n$$\na_{rn+s} = -j \\cdot (n+1) + e \\cdot \\binom{n+1}{2} + \\binom{s+1}{2} \\pmod{(n+1)^2}.\n$$\n\nNotice that for $s = 1, 2, \\dots, n$, the numbers $\\binom{s+1}{2}$ are all distinct mod $n+1$. Indeed, if $1 \\le s < t \\le n$ and $\\binom{s+1}{2} = \\binom{t+1}{2} \\pmod{n+1}$, then $(2n+2) \\mid (s-t)(s+t+1)$, and since $s, t$ have opposite parities and $2n+2 > 2n+1 \\ge s+t+1 > |s-t| > 0$, we arrive at a contradiction as $2n+2$ is a power of 2, proving that the residues are distinct mod $n+1$. Also, note that none of these residues are 0 (mod $n+1$). Thus, $-j(n+1) + \\binom{s+1}{2}$ (where $e=0$) covers all $n^2+n$ residues which are not of the form $l(n+1) \\pmod{(n+1)^2}$.\n\nFinally, for the remaining residues, take $s = n$ and $e = 1$, we get $-j(n+1) + 2\\binom{n+1}{2} = -(j+1)(n+1) \\pmod{(n+1)^2}$, so that which covers the remaining residues as $j$ varies. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18739,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider functions $f, g : [0, 1] \\to [0, 1]$ such that $g$ is monotonic and onto and $|f(x) - f(y)| \\leq |g(x) - g(y)|$ for any $x, y \\in \\mathbb{R}$.\n\na) Prove that $f$ is continuous and there is $x_0 \\in [0, 1]$ such that $f(x_0) = g(x_0)$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The monotonicity of $g$ assures the existence of lateral limits at any point. We shall prove that $g$ is continuous. By way of contradiction, if $x_0$ is a point where $g(x_0 - 0) < g(x_0) \\leq g(x_0 + 0)$ or $g(x_0 - 0) \\leq g(x_0) < g(x_0 + 0)$, the open interval with endpoints $g(x_0 - 0)$ and $g(x_0 + 0)$ is not contained in the image of $g$, contradicting the surjectivity. The continuity of $f$ is then a consequence of the given inequality.\n\nConsider $h$ defined by $h(x) = g(x) - f(x)$. We have $h(0)h(1) = (g(0) - f(0))(g(1) - f(1)) \\leq 0$, because $g$ is surjective and monotonic. The intermediate value property for $h$ implies then the existence of a point $x_0 \\in [0, 1]$ such that $h(x_0) = 0$, that is, $f(x_0) = g(x_0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18740,
    "subject": "Mathematics (Olympiad)",
    "question": "It is possible to choose three small triangles such that the sum of the numbers inscribed in them equals the sum of the numbers in nine different circles. The \"unused\" circle (shown highlighted in black) is located at one of the vertices of the given triangle or at its centre.\n\n![](images/Mathematica_competitions_in_Croatia_in_2013_p12_data_7dc635d8d9.png)\n\nFor a given arrangement of numbers, what is the largest possible sum of the numbers in the three selected triangles?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of the numbers from $1$ to $10$ is $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55$. If $m$ is the smallest of the numbers in the highlighted circles, the largest possible sum of the selected numbers is $55 - m$.\n\nSince $m$ can be at most $7$ (if the other highlighted circles are $8$, $9$, and $10$), the maximum sum is $55 - 7 = 48$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18741,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a village with a population of 2007. This village has no name. You are the God of this village and you want the villagers to decide the name of the village. Every villager has one idea for the village's name.\n\nEach villager can send a letter to any villager (including himself). Every villager can send any number of letters each day. Letters are collected in the evening and delivered all at once the next morning. The sender decides to whom the letter should be delivered. Each villager can send a letter to tell their idea for the village's name to God only once. This idea does not need to be the same as the one they or other villagers originally thought. Each villager's only action is writing a letter.\n\nEvery villager is either honest or a liar. You and the villagers do not know who is honest or who is a liar. However, you know that the number of liars is less than or equal to $T$, and there is at least one honest person in the village.\n\nYou can give instructions to every villager only once at noon on a certain day. An honest person necessarily follows the instruction, but you do not know if a liar will follow it. Find the maximum $T$ such that there exists an instruction which fulfills the following conditions:\n\n* In the end, every honest person sends a letter to God, and every honest person sends the same idea for the village's name.\n* If every honest person originally had the same idea for the village's name, every honest person sends this idea to God.",
    "options": [],
    "answer": "See solution",
    "solution": "$$668$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18742,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence of real numbers $a_n$ is defined by\n\n$$\na_1 = 5 \\quad \\text{and} \\quad a_n = \\sqrt[n]{a_{n-1}^{n-1} + 2^{n-1} + 2 \\cdot 3^{n-1}} \\quad \\text{for all } n \\ge 2.\n$$\n\nProve that the sequence $a_n$ is decreasing.",
    "options": [],
    "answer": "See solution",
    "solution": "From the recurrence for $a_n$, we find that\n\n$$\na_n = (2^n + 3^n)^{\\frac{1}{n}} \\quad \\forall n \\ge 1.\n$$\n\nFor all $n \\ge 1$,\n\n$$\n\\begin{align*}\n2^n + 3^n &> 3^n \\\\\n\\Rightarrow (2^n + 3^n)^{n+1} &> 3^n (2^n + 3^n)^n > (2^{n+1} + 3^{n+1})^n \\\\\n\\Rightarrow (2^n + 3^n)^{\\frac{1}{n}} &> (2^{n+1} + 3^{n+1})^{\\frac{1}{n+1}}.\n\\end{align*}\n$$\n\nThus $a_n > a_{n+1}$ for all $n \\ge 1$, so $(a_n)$ is decreasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18743,
    "subject": "Mathematics (Olympiad)",
    "question": "Rectangle $ABCD$ has sides $AB = 3$, $BC = 2$. Point $P$ on side $AB$ is such that the bisector of $\\angle CDP$ passes through the midpoint of $BC$. Find $BP$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $BC$, and let line $DM$ intersect\n![](images/Argentina_2015_Booklet_p5_data_db7df192dc.png)\nline $AB$ at $Q$ (it is exterior to the segment $AB$). Then $\\angle BQM = \\angle CDM$ as $AB \\parallel CD$. On the other hand, $\\angle CDM = \\angle PDM$ by hypothesis (since $DM$ is the bisector of $\\angle CDP$). So $\\angle PQD = \\angle PDC$ and hence $PQ = PD$. In addition, $BQ = CD = 3$ because triangles $BQM$ and $CDM$ are congruent ($BM = CM$, $\\angle M B Q = \\angle M C D = 90^\\circ$, $\\angle BQM = \\angle CDM$).\n\nSet $BP = x$. Then $PQ = PB + BQ = x + 3$ and $PD = PQ = x + 3$. Apply the Pythagorean theorem to triangle $PDA$ in which $AP = 3 - x$, $AD = 2$, $PD = x + 3$. This gives\n\n$$\n(3 - x)^2 + 2^2 = (3 + x)^2\n$$\n\nand we find $x = \\frac{1}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18744,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $(x, y)$ to the equation\n\n$$\nx^2 + y^2 + x + y - xy(x + y) + 2 = -(1-x)(1-y)(x+y+2)\n$$\n\nsubject to the condition $|xy| \\leq \\frac{25}{9}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x + y = a$, $xy = b$. By the AM-GM inequality, $a^2 \\geq 4b$. We have:\n\n$$\nx^2 + y^2 + x + y - xy(x + y) = a^2 + a - b(a + 2) = -\\frac{10}{27},\n$$\n\nso\n\n$$\nb = \\frac{a^2 + a + \\frac{10}{27}}{a + 2} \\leq \\frac{a^2}{4},\n$$\n\nand\n\n$$\n|b| = \\left| \\frac{a^2 + a + \\frac{10}{27}}{a + 2} \\right| \\leq \\frac{25}{9}.\n$$\n\nTherefore,\n\n$$\n\\frac{a^2}{4} - \\frac{a^2 + a + \\frac{10}{27}}{a + 2} = \\frac{\\left(a + \\frac{2}{3}\\right)^2 \\left(a - \\frac{10}{3}\\right)}{4(a + 2)} \\geq 0 \\quad (1),\n$$\n\n$$\n\\frac{a^2 + a + \\frac{10}{27}}{a + 2} - \\frac{25}{9} = \\frac{\\left(a - \\frac{10}{3}\\right)\\left(a + \\frac{14}{9}\\right)}{a + 2} \\leq 0 \\quad (2),\n$$\n\n$$\n\\frac{a^2 + a + \\frac{10}{27}}{a + 2} + \\frac{25}{9} = \\frac{\\left(a + \\frac{17}{9}\\right)^2 + \\frac{191}{81}}{a + 2} \\geq 0 \\quad (3).\n$$\n\nFrom (1), (2), and (3), the only solutions are $a = -\\frac{2}{3}$ and $a = \\frac{10}{3}$. In both cases, $a^2 = 4b$, so $x = y = \\frac{a}{2}$. Thus, the solutions are $(x, y) = (-1/3, -1/3)$ and $(5/3, 5/3)$, which satisfy the problem conditions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18745,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of side $BC$ of the acute-angled triangle $\\Delta ABC$ with $AB \\ne AC$, and let $O$ be the circumcenter. Let $MP$ and $MQ$ be perpendiculars from $M$ to the sides $AB$ and $AC$, respectively. Show that the line passing through the midpoint of $PQ$ and $M$ is parallel to $AO$.\n\n![](images/Ukrajina_2013_p25_data_d412607807.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $AB < AC$. Let $F$ be the point on $PQ$ such that $FM \\parallel AO$. We will show that $F$ is the midpoint of $PQ$.\n\nLet $PK$ and $QE$ be the perpendiculars from $P$ and $Q$ to $MF$. We will show that $PK = QE$.\n\nLet $L$ and $T$ be the points where $MF$ meets $AB$ and $AC$, respectively. Then\n\n$$\n\\angle BLM = \\angle BAO = 90^\\circ - \\angle C, \\quad \\angle CTM = \\angle CAO = 90^\\circ - \\angle B.\n$$\n\nFrom the right-angled triangles, we get:\n\n$$\nPM = \\frac{1}{2} BC \\sin \\angle B,\n$$\n\n$$\nPK = PM \\sin \\angle PML = PM \\cos \\angle BML = PM \\sin \\angle C = \\frac{1}{2} BC \\sin \\angle C \\sin \\angle B.\n$$\n\nSimilarly,\n\n$$\nQM = \\frac{1}{2} BC \\sin \\angle C,\n$$\n\n$$\nQE = QM \\sin \\angle QMT = QM \\cos \\angle CMT = QM \\sin \\angle B = \\frac{1}{2} BC \\sin \\angle C \\sin \\angle B.\n$$\n\nTherefore, $PK = QE$, and the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18746,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a natural number *acceptable* if it has at most 9 distinct prime divisors. There is given a pile of $100! = 1 \\cdot 2 \\cdot \\dots \\cdot 100$ stones. A legal move is to remove $k$ stones from the pile where $k$ is an acceptable number. Players A and B take turns in making legal moves; A goes first. The one who removes the last stone wins. Decide which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29$ be the product of the first 10 primes. Observe that $P$ is the smallest unacceptable number. Apparently $P$ divides $100!$, and acceptable numbers are not divisible by $P$.\n\nLet A remove $k_1$ stones on his first move. Because $100!$ is divisible by $P$ but $k_1$ is not, the number $n_1 = 100! - k_1$ of stones remaining is not divisible by $P$; in particular $n_1 \\neq 0$. So the remainder $r_1$ of $n_1 \\bmod P$ satisfies $1 \\leq r_1 < P$. It follows that $r_1$ is acceptable as $P$ is the least unacceptable number. In addition, $r_1$ is nonzero, so B can make a legal move by taking $r_1$ stones. There remain $n_1 - r_1$ stones, a quantity divisible by $P$.\n\nThen, just like above, any move of A yields a number $n_2$ of stones that is not divisible by $P$, and nonzero in particular. Its remainder $r_2 \\bmod P$ is such that $1 \\leq r_2 < P$, so B is able to remove $r_2$ stones and reach a position again where the number of stones is a multiple of $P$. Clearly B can apply such moves at each step.\n\nThe number of stones decreases at each move, so the game ends with a win of one of the two players. A's moves always leave a quantity not divisible by $P$, unlike B's moves. Hence A cannot take the last stone, meaning that B's strategy guarantees him a win.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18747,
    "subject": "Mathematics (Olympiad)",
    "question": "One hundred boxes are labeled from 1 to 100. Each box has at most 10 stones. The difference in the number of stones between any two boxes labeled with consecutive numbers is 1. The boxes labeled 1, 4, 7, 10, ..., 100 contain a total of 301 stones. Find the maximum possible number of stones contained by the 100 boxes.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the difference in 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 numbered 1, 4, 7, \\dots, 100, and if we group the remaining 66 boxes in pairs of consecutive boxes, we obtain at most $301 + 33 \\times 19 = 928$ 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": 18748,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $l_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $l_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangles $DEF$ and $A_1B_1C_1$ are similar to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the following stronger statement: $A_1$, $B_1$, $C_1$ are midpoints of segments $HD$, $DE$, $HF$, respectively, where $H$ is the orthocenter of triangle $ABC$. Therefore, there is a dilation centered at $H$ with magnitude $2$ sending triangle $A_1B_1C_1$ to $DEF$. (For readers familiar with the nine-point circle of a triangle, this stronger statement shows that $A_1$, $B_1$, $C_1$ lie on the nine-point circle of triangle $ABC$. Furthermore, the statement remains true if $D$, $E$, $F$ are arbitrarily chosen points on arcs $\\widehat{BC}$, $\\widehat{CA}$, $\\widehat{AB}$. We leave it to the reader to show this more general result.)\n\nWe give two solutions; in each, let $X$ and $Y$ denote the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $P$ and $Q$ denote the feet of the perpendiculars from $A$ to lines $DB$ and $DC$ respectively. Then $A_1$ is the intersection of line $XY$ (or $m_a$) and line $PQ$ (or $l_a$). Let $H_a$ be the foot of the perpendicular from $A$ to line $BC$. Let $M$ be the midpoint of side $BC$. Extend segment $AH_a$ to meet the circumcircle of triangle $ABC$ at $G_a$. We also set $B = \\angle ABC$, $C = \\angle BCA$, and $A = \\angle CAB$.\n\n**Solution 1.** We consider the left-hand side configuration shown below. (Our proof can be easily modified for other configurations.) Let $H$ denote the intersection of lines $DA_1$ and $AH_a$. By symmetry, it suffices to show that $A_1$ is the midpoint of segment $DH$ and $H$ is the orthocenter of triangle $ABC$.\n\nNote that points $X$, $Y$, $M$ are collinear – they lie on the Simson line from $D$ with respect to triangle $ABC$. Indeed, because $\\angle BXD = \\angle BMD = 90^\\circ$, $BXMD$ is cyclic, implying that\n\n$$\n\\angle XMB = \\angle XDB = 90^\\circ - \\angle XBD = 90^\\circ - \\angle ABD = 90^\\circ - \\left(B + \\frac{A}{2}\\right) = \\frac{C-B}{2}.\n$$\n\nBecause $\\angle DMC = \\angle DYC = 90^\\circ$, $BMCY$ is cyclic, implying that\n\n$$\n\\angle CMY = \\angle CDY = 90^\\circ - \\angle DCY = 90^\\circ - \\angle ABD = \\frac{C-B}{2},\n$$\n\nwhere the third equality holds because $ABDC$ is cyclic. By the above, we know that $\\angle XMB = \\angle CMY$; that is, $X$, $M$, $Y$ are collinear. In exactly the same way, we know that $P$, $H_a$, $Q$ are collinear (on the Simson line from $A$ with respect to triangle $BCD$) by establishing\n\n$$\n\\angle BH_a P = \\angle QH_a C = \\frac{C - B}{2}\n$$\n\nFurthermore, by the previous results, we conclude that $MA_1H_a$ is an isosceles triangle with $MA_1 = A_1H_a$. Let $N$ denote the intersection of lines $MD$ and $PA$. Then in right triangle $MNH_1$, we must have $MA_1 = NA_1 = A_1H_a$. Thus triangles $DNA_1$ and $HA_1H_a$ are congruent to each other (by SAS, since $DM \\parallel HH_a$ and $NA_1 = A_1H_a$). In particular, $A_1$ is the midpoint of $AH$.\n\nIt remains to show that $H$ is indeed the orthocenter of triangle $ABC$. By the previous angle relations,\n\n$$\n\\angle A_1H_aH = \\angle A_1H_aM + \\angle MH_aH = \\frac{C-B}{2} + 90^\\circ = C + \\frac{A}{2} = \\frac{\\widehat{AB}}{2} + \\frac{\\widehat{BD}}{2} = \\frac{\\widehat{ABD}}{2} = \\angle HG_aD.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18749,
    "subject": "Mathematics (Olympiad)",
    "question": "With inspiration drawn from the rectilinear network of streets in New York, the *Manhattan distance* between two points $ (a, b) $ and $ (c, d) $ is defined to be\n\n$$\n|a - c| + |b - d|.\n$$\n\nSuppose only two distinct Manhattan distances occur between all the points of some point set. What is the maximal number of points in such a set?",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: nine.\n\nLet\n$$\n\\{(x_1, y_1), \\dots, (x_m, y_m)\\}, \\quad \\text{where} \\quad x_1 \\le \\dots \\le x_m,\n$$\nbe the set, and suppose $m \\ge 10$.\n\nA special case of the Erd\\\"os-Szekeres Theorem asserts that a real sequence of length $n^2+1$ contains a monotonic subsequence of length $n+1$. (Proof: Given a sequence $a_1, \\dots, a_{n^2+1}$, let $p_i$ denote the length of the longest increasing subsequence ending with $a_i$, and $q_i$ the length of the longest decreasing subsequence ending with $a_i$. If $i < j$ and $a_i \\le a_j$, then $p_i < p_j$. If $a_i \\ge a_j$, then $q_i < q_j$. Hence all $n^2+1$ pairs $(p_i, q_i)$ are distinct. If all of them were to satisfy $1 \\le p_i, q_i \\le n$, it would violate the Pigeon-Hole Principle.)\n\nApplied to the sequence $y_1, \\dots, y_m$, this will produce a subsequence\n$$\ny_i \\le y_j \\le y_k \\le y_l \\quad \\text{or} \\quad y_i \\ge y_j \\ge y_k \\ge y_l.\n$$\nOne of the shortest paths from $(x_i, y_i)$ to $(x_l, y_l)$ will pass through first $(x_j, y_j)$ and then $(x_k, y_k)$. At least three distinct Manhattan distances will occur.\n\nConversely, among the nine points\n$$\n(0, 0), \\quad (\\pm 1, \\pm 1), \\quad (\\pm 2, 0), \\quad (0, \\pm 2)\n$$\nonly the Manhattan distances 2 and 4 occur. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18750,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of integers defined recursively by $a_1 = 2013$ and for $n \\ge 1$, $a_{n+1}$ is the sum of the $2013$th power of the digits of $a_n$. Do there exist distinct positive integers $i, j$ such that $a_i = a_j$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. For any positive integer $n$, let $f(n)$ be the sum of the $2013$th powers of the digits of $n$. Let $S = \\{1, 2, \\dots, 10^{2017} - 1\\}$, and $n = \\overline{a_1a_2\\dots a_{2017}} \\in S$.\n\n$$\nf(n) = \\sum a_i^{2013} \\leq 2017 \\cdot 9^{2013} < 10^4 \\cdot 10^{2013} = 10^{2017} \\in S.\n$$\n\nSince $a_i = f^{(i)}(2013) \\in S$, there exist distinct positive integers $i, j$ such that $a_i = a_j$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18751,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$ and inscribed in the circle $(O)$. Let $I$ be the incenter of $ABC$, and let $D$, $E$ be the intersections of $AI$ with $BC$ and $(O)$, respectively. Take a point $K$ on $BC$ such that $\\angle AIK = 90^\\circ$, and let $KA$, $KE$ meet $(O)$ again at $M$, $N$ respectively. The rays $ND$, $NI$ meet the circle $(O)$ at $Q$, $P$.\n\n1. Prove that the quadrilateral $MPQE$ is a kite.\n\n2. Take $J$ on $IO$ such that $AK \\perp AJ$. The line through $I$ and perpendicular to $OI$ cuts $BC$ at $R$, and cuts $EK$ at $S$. Prove that $OR \\parallel JS$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we know that $EI = EB = EC$, so $E$ is the circumcenter of triangle $BIC$. Thus, $KI$ is the tangent line of circle $(BIC)$. Hence,\n\n$$\nKI^2 = KB \\cdot KC = KE \\cdot KN.\n$$\n\nCombining with $\\triangle KIE$ being right-angled, we get $IN \\perp KE$.\n\n![](images/Saudi_Arabia_booklet_2021_p13_data_1e217f6076.png)\n\nSo $PE$ is the diameter of circle $(O)$. We have $\\triangle BDE \\sim \\triangle ABE$, so\n\n$$\n\\frac{BE}{AE} = \\frac{DE}{BE} \\Rightarrow EA \\cdot ED = EB^2 = EI^2 = EN \\cdot EK\n$$\n\nThis implies that $ADNK$ is cyclic. Thus $\\angle MAE = \\angle DNE$, which leads to $EM = EQ$. Therefore, $MPQE$ is a kite.\n\nSince $KI^2 = KM \\cdot KA$, we get $IM \\perp AK$. Construct the diameter $AA'$ of $(O)$; then $M$, $I$, $A'$ are collinear. Thus $AJ \\parallel IM$ implies that $AIA'J$ is a parallelogram, and then $O$ is the midpoint of $IJ$.\n\nSince $IO$ is the median of triangle $IPE$, $Ix$, $IO$, $IP$, $IE$ form a harmonic quartet with $Ix \\parallel PE$. Denote $Ky$ as the ray perpendicular to $IO$; then notice that $KE \\perp IP$, $KI \\perp IE$, $KD \\perp Ix$, $K$, $Ky \\perp IO$, so $KD$, $Ky$, $KI$, $KE$ form another harmonic quartet. Since $Ky \\parallel RS$, $R$ is the midpoint of $OS$. Therefore, $OR$ is the midline of triangle $IJS$, which implies that $OR \\parallel JS$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18752,
    "subject": "Mathematics (Olympiad)",
    "question": "In a regular pyramid $P - A_1A_2 \\cdots A_n$ ($n \\ge 3$), $O$ is the center of the regular $n$-sided polygon $A_1A_2 \\cdots A_n$ (the base), and $B$ is the midpoint of edge $A_1A_n$.\n\n1. Prove that\n$$\nPO^2 \\sin \\frac{\\pi}{n} + PA_1^2 \\cos^2 \\frac{\\pi}{n} = PB^2.\n$$\n\n2. For the regular pyramid $P - A_1A_2 \\cdots A_n$, let the angle formed by the lateral edge and the base be $\\alpha$, and the angle formed by the lateral face and the base be $\\beta$. Try to determine the magnitude relation between $\\frac{1}{n} \\sum_{i=1}^{n} \\cos \\angle A_i PB$ and $\\sin \\alpha \\sin \\beta$, and then prove it.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Since $PO$ is perpendicular to the base $A_1A_2 \\cdots A_n$, it follows that $\\angle POA_1 = \\angle POB = 90^\\circ$.\n\nLet $OA_1 = r$, then $OB = OA_1 \\cdot \\cos \\angle A_1OB = r \\cos \\frac{\\pi}{n}$. Hence,\n$$\nr^2 + PO^2 = PA_1^2, \\quad r^2 \\cos^2 \\frac{\\pi}{n} + PO^2 = PB^2.\n$$\nEliminating $r^2$ gives $PO^2 \\left(1 - \\cos^2 \\frac{\\pi}{n}\\right) = PB^2 - PA_1^2 \\cos^2 \\frac{\\pi}{n}$, namely,\n$$\nPO^2 \\sin^2 \\frac{\\pi}{n} + PA_1^2 \\cos^2 \\frac{\\pi}{n} = PB^2.\n$$\n\n2. By the given condition, $\\overrightarrow{PO} \\cdot \\overrightarrow{OA_i} = 0$ ($i = 1, 2, \\dots, n$), $\\overrightarrow{PO} \\cdot \\overrightarrow{OB} = 0$.\n\nSuppose the length of the lateral edge is $l$. Then\n$$\n\\begin{align*}\nl \\cdot |PB| \\cdot \\sum_{i=1}^{n} \\cos \\angle A_i PB &= \\sum_{i=1}^{n} |PA_i| \\cdot |PB| \\cdot \\cos \\angle A_i PB \\\\\n&= \\sum_{i=1}^{n} \\overrightarrow{PA_i} \\cdot \\overrightarrow{PB} \\\\\n&= \\sum_{i=1}^{n} (\\overrightarrow{PO} + \\overrightarrow{OA_i}) \\cdot (\\overrightarrow{PO} + \\overrightarrow{OB}) \\\\\n&= \\sum_{i=1}^{n} (\\overrightarrow{PO}^2 + \\overrightarrow{OA_i} \\cdot \\overrightarrow{OB}) \\\\\n&= n \\overrightarrow{PO}^2 + \\overrightarrow{OB} \\cdot \\sum_{i=1}^{n} \\overrightarrow{OA_i} \\\\\n&= n \\cdot |PO|^2.\n\\end{align*}\n$$\nThe last step uses $\\sum_{i=1}^{n} \\overrightarrow{OA_i} = \\vec{0}$, since $O$ is the center of the regular $n$-gon and the sum of its radius vectors is zero.\n\nTherefore,\n$$\n\\frac{1}{n} \\sum_{i=1}^{n} \\cos \\angle A_i PB = \\frac{|PO|^2}{l \\cdot |PB|} = \\frac{PO}{l} \\cdot \\frac{PO}{PB} = \\sin \\angle PA_1O \\cdot \\sin \\angle PBO.\n$$\nIt is clear that $\\angle PA_1O$ is the angle formed by the lateral edge and the base ($\\alpha$), and $\\angle PBO$ is the angle formed by the lateral face and the base ($\\beta$). Thus,\n$$\n\\frac{1}{n} \\sum_{i=1}^{n} \\cos \\angle A_i PB = \\sin \\alpha \\sin \\beta.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18753,
    "subject": "Mathematics (Olympiad)",
    "question": "Medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at point $M$. Is it possible that the circles with radii $MD$, $ME$, and $MF$\n\na) all have areas smaller than the area of triangle $ABC$;\nb) all have areas greater than the area of triangle $ABC$;\nc) all have areas equal to the area of triangle $ABC$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let triangle $ABC$ be equilateral (see the first figure). As the medians of this triangle all have equal lengths, the points $D$, $E$, and $F$ are located at equal distances from $M$, i.e., on a circle with centre $M$. It suffices to show that the area of this circle is less than the area of triangle $ABC$. Because the medians of an equilateral triangle are simultaneously altitudes and perpendicular to the sides, the sides $BC$, $CA$, and $AB$ are tangent to the circle at points $D$, $E$, and $F$, respectively. Therefore, the circle is the incircle of triangle $ABC$. The area of the incircle is indeed less than the area of the triangle.\n\nb) Let $AB = AC$, $BC = 1$, and $AD = 6$ (see the second figure). Because the median drawn from the vertex angle of an isosceles triangle is simultaneously its altitude, the area of this triangle is $\\frac{1}{2} \\cdot 1 \\cdot 6 = 3$. On the other hand, $BE = CF > \\frac{1}{2}AD$, because $EF$ is a midsegment of the triangle parallel to $BC$ and $\\frac{1}{2}AD$ is the distance between midsegment $EF$ and side $BC$ (shown in the figure with a dotted line). Therefore $ME = MF > \\frac{1}{6}AD = 1$. The circles with radii $ME$ and $MF$ have area greater than $\\pi$, which in turn is greater than the area of triangle $ABC$, which is $3$. The circle with radius $MD$ has area greater than the area of triangle $ABC$ as well, because $MD = \\frac{1}{3}AD = 2 > 1$.\n\nc) If circles with radii $MD$, $ME$, and $MF$ all had areas equal to triangle $ABC$, their radii should also be equal, therefore $MD = ME = MF$. This would mean that the lengths of the parts of medians that lie on the other side of $M$ are equal as well, in other words, $MA = MB = MC$. We show that then $ABC$ must be equilateral. Indeed, from $ME = MF$ and $MB = MC$ we get that triangles $BMF$ and $CME$ must be equal (see the third figure), therefore $BF = CE$ and also $AB = AC$. Analogously, $AB = BC$. But in part a) we showed that in an equilateral triangle, circles with radii $MD$, $ME$, and $MF$ do not have the same area as the triangle.\n\n![](images/prob1718_p7_data_3898369553.png)\n![](images/prob1718_p7_data_91963be823.png)\n![](images/prob1718_p7_data_2da97ce26c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18754,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $a$, $b$, $c$, $d$ are given. Solve the system of equations for unknowns $x$, $y$, $z$, $u$:\n\n$$\n\\begin{cases}\nx^2 - yz - zu - yu = a \\\\\ny^2 - zu - ux - xz = b \\\\\nz^2 - ux - xy - yu = c \\\\\nu^2 - xy - yz - zx = d\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the other three equations from the first, we obtain:\n\n$$\n(x - y)(x + y + z + u) = a - b, \\text{ etc.}\n$$\n\nAdding these three new equations, we get:\n\n$$\n[4x - (x + y + z + u)](x + y + z + u) = 3a - b - c - d.\n$$\n\nLet\n\n$$\n\\begin{align*}\nx + y + z + u &= \\lambda \\\\\na + b + c + d &= t\n\\end{align*}\n$$\n\nThen,\n\n$$\nx = \\frac{\\lambda}{4} + \\frac{4a-t}{4\\lambda}, \\quad y = \\frac{\\lambda}{4} + \\frac{4b-t}{4\\lambda}, \\quad z = \\frac{\\lambda}{4} + \\frac{4c-t}{4\\lambda}, \\quad u = \\frac{\\lambda}{4} + \\frac{4d-t}{4\\lambda}. \\tag{1}\n$$\n\nSubstituting these values into the original equations gives:\n\n$$\n\\lambda^4 + 2\\lambda^2 \\left(\\sum a\\right) + 2\\left(\\sum bc\\right) - 3\\left(\\sum a^2\\right) = 0,\n$$\n\na biquadratic in $\\lambda$, yielding\n\n$$\n\\lambda^2 = -\\sum a \\pm 2\\sqrt{\\sum a^2}.\n$$\n\nComputing $\\lambda$ and substituting into (1) gives $x$, $y$, $z$, $u$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18755,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 3$ be a positive integer and $p$ a prime number such that $p > 6^{n-1} - 2^{n} + 1$. Let $S$ be a set of $n$ positive integers with different residues modulo $p$. Show that there exists a positive integer $c$ such that there are exactly two ordered triples $(x, y, z) \\in S^{3}$ with distinct elements for which $x - y + z - c$ is divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be an arbitrary prime number. We say the set $S$ of integers with different residues modulo $p$ is *special* if there exists a positive integer $c$ such that there are exactly two ordered triples $(x, y, z) \\in S^3$ with distinct elements, such that $x - y + z - c$ is divisible by $p$.\n\nFor each integer $x$, let $[x]$ denote the remainder of $x$ divided by $p$. For each subset $X$ of $\\mathbb{Z}$ and integers $a, b$, denote\n\n$$\naX + b := \\{[ax + b] \\mid x \\in X\\}.$$ \n\nEvery integer coprime with $p$ has an inverse modulo $p$, so if $a$ is coprime with $p$, it's easy to verify that $X$ is *special* if and only if $aX + b$ is *special*. The solution is based on the following two lemmas.\n\n**Lemma 1.** The set $S$ of $n \\geq 3$ natural numbers that are at most $\\frac{p}{3}$ is *special*.\n\n*Proof.* Let $i, j$ be the two largest numbers and $k$ the smallest number in $S$. Choose $c = i + j - k > 0$. For any triple $(x, y, z) \\in S^3$ with distinct elements,\n\n$$\n0 \\geq x - y + z - c > -\\frac{p}{3} - c > -\\frac{p}{3} - \\frac{2p}{3} = -p.\n$$\n\nHence,\n\n$$\np \\mid x - y + z - c \\iff x - y + z = c\n$$\n\nwhich is equivalent to $\\{x, z\\} = \\{i, j\\}$ and $y = k$. $\\square$\n\n**Lemma 2.** If $p > 5 \\cdot 6^{n-2}$, for any set $S$ of $n \\geq 3$ natural numbers, there exist integers $a, b$, with $a$ coprime to $p$, such that all elements of $aS + b$ are at most $\\frac{p}{3}$.\n\n*Proof.* Assume $0 \\in S$, since we can choose an arbitrary integer $b_0$ such that $0 \\in S + b_0$. For each $i \\in \\mathbb{Z}$, let $S_i = \\left[ \\frac{pi}{6}, \\frac{p(i+1)}{6} \\right) \\cap \\mathbb{Z}$.\n\nConsider $S$ as an $(n-1)$-tuple $(x_1, x_2, \\dots, x_{n-1})$, where $x_i \\in S$ and $x_i \\neq 0$. Each integer $a$ corresponds to an $(n-2)$-tuple $(a_1, a_2, \\dots, a_{n-2})$, where $a_i$ is the index $k$ such that $[a x_i] \\in S_k$. By the Pigeonhole Principle, there exists a set $A$ with 6 integers $a$, corresponding to the same $(n-2)$-tuple. By the same argument, there exist $a_1, a_2 \\in A$ such that\n\n$$\n[a_1 x_{n-1}] - [a_2 x_{n-1}] \\in \\left(-\\frac{p}{6}, \\frac{p}{6}\\right).\n$$\n\nChoose $a = a_1 - a_2$, then $[a x] \\in [0, \\frac{p}{6}) \\cup (\\frac{5p}{6}, p)$ for all $x \\in S$.\n\nIt's easy to verify that if $b = \\lfloor \\frac{p}{6} \\rfloor$ then $aS + b \\subset [0, \\frac{p}{3}]$. $\\square$\n\n![](images/Vietnamese_mathematical_competitions_p282_data_39d3dd5038.png)\n\nSince $p > 6^{n-1} - 2^n + 1$ then $p \\geq 6^{n-1} - 2^n + 3 > 5 \\cdot 6^{n-2}$, which holds for all integers $n \\geq 3$. Hence, our proof is completed. $\\square$\n\n![](images/Vietnamese_mathematical_competitions_p282_data_1997488c0e.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18756,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of 20 balls is tossed independently and at random into one of 5 bins. Let $p$ be the probability that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Let $q$ be the probability that every bin ends up with 4 balls. What is $\\frac{p}{q}$?\n\n(A) 1  \n(B) 4  \n(C) 8  \n(D) 12  \n(E) 16",
    "options": [],
    "answer": "See solution",
    "solution": "The requested ratio divides the number of ways to end up with a 3-4-4-4-5 distribution by the number of ways to end up with a 4-4-4-4-4 distribution. For either outcome, there are at least three bins with 4 balls each, leaving 8 balls to distribute into two bins. For a 3-5 split in the two bins, there are $5 \\cdot 4 = 20$ ways to choose the bins, and $\\binom{8}{3} = 56$ ways to choose 3 balls. For a 4-4 split in the two bins, there are $\\binom{8}{4} = 70$ ways to choose 4 balls. The requested ratio is therefore\n$$\n\\frac{p}{q} = \\frac{20 \\cdot 56}{70} = 16.\n$$\n\n**OR**\n\nThe probabilities of the ball distribution follow the multinomial distribution. If there are $n$ balls and $k$ bins, then the probability that $n_i$ balls end up in bin $i$ for every $i$ is given by\n$$\n\\frac{n!}{n_1!n_2!\\cdots n_k!} \\cdot p_1^{n_1} p_2^{n_2} \\cdots p_k^{n_k},\n$$\nwhere $p_i$ is the probability of any particular ball getting tossed into bin $i$, which equals $\\frac{1}{k}$. There are $5 \\cdot 4$ choices for which bin gets 3 balls and which bin gets 5 balls under the first scenario. The requested ratio is therefore\n$$\n\\frac{p}{q} = \\frac{5 \\cdot 4 \\cdot \\frac{20!}{3!4!4!5!} \\cdot (\\frac{1}{5})^{20}}{\\frac{20!}{4!4!4!4!} \\cdot (\\frac{1}{5})^{20}} = 5 \\cdot 4 \\cdot \\frac{4}{5} = 16.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18757,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $k$ for which there exist natural numbers $x, y$ such that the number $\\frac{x^k y}{y^2 - x^2}$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$. Then $x = d x_1$, $y = d y_1$, where $x_1, y_1$ are coprime natural numbers. The expression becomes\n\n$$\n\\frac{d^{k-1} x_1^k y_1}{y_1^2 - x_1^2} = p,\n$$\n\nwhere $p$ is prime. Since $x_1^k y_1$ and $y_1^2 - x_1^2$ are coprime, $y_1^2 - x_1^2$ must divide $d^{k-1}$. Also, $y_1 > x_1$, so $x_1^k y_1 > 1$. Thus, we need $x_1 = 1$, $y_1 = p$, and $y_1^2 - x_1^2 = d^{k-1}$. If $k = 1$, then $1 = p^2 - 1 = (p-1)(p+1)$, which is impossible. For $k = 2$, $d = p^2 - 1$ works, so $(x, y) = (p^2 - 1, p(p^2 - 1))$ is a solution for every prime $p$. Thus, $k = 2$ is a solution. For $k > 2$, $d^{k-1} = p^2 - 1 = (p-1)(p+1)$. For $p = 2$, $3$ is not a perfect power, so no solution. For odd $p$, $8 \\mid p^2 - 1$, so $k-1 \\geq 3$. Thus, $k = 3$ is not a solution. For $k = 4$, $p = 3$, $d = 2$ works, i.e., $(x, y) = (2, 6)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18758,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of pairwise distinct positive integers $(a, b, c)$ such that:\n\n- $2a - 1$ is divisible by $b$,\n- $2b - 1$ is divisible by $c$,\n- $2c - 1$ is divisible by $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's rewrite the conditions as a system: there exist natural numbers $k, m, n$ such that:\n\n$$\n2a - 1 = k b, \\quad 2b - 1 = n c, \\quad 2c - 1 = m a.\n$$\n\nIt is clear that all numbers $k, m, n$ and $a, b, c$ are odd.\n\nNow,\n\n$$\nb = \\frac{2a - 1}{k} \\implies 2b - 1 = \\frac{4a - 2}{k} - 1 = n c \\implies c = \\frac{4a - 2 - k}{k n}.\n$$\n\nThen,\n\n$$\n2c - 1 = \\frac{8a - 4 - 2k}{k n} - 1 = m a \\implies 8a - 4 - 2k - k n = k n m a \\implies a = \\frac{4 + 2k + k n}{8 - k n m}.\n$$\n\nThus, $k n m < 8$. By symmetry and the oddness of $k, m, n$, the possible cases (up to cyclic order) are:\n\n- **Case 1:** $k = m = n = 1$. This leads to $a = b = c = 1$, but the numbers must be distinct.\n- **Case 2:** $k = 3, m = n = 1$. This leads to $5b = 7$, which is not an integer.\n- **Case 3:** $k = 5, m = n = 1$. This leads to $3b = 7$, which is not an integer.\n- **Case 4:** $k = 7, m = n = 1$. This gives $b = 7$, $c = 13$, $a = 25$.\n\nThus, the solutions (up to cyclic order) are:\n\n$$(7, 13, 25),\\ (13, 25, 7),\\ (25, 7, 13).$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18759,
    "subject": "Mathematics (Olympiad)",
    "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 $s(n)$ as the number of triples of type $n$.\n\n**a)** Prove that there exists no positive integer $n$ such that $s(n) = 14$.",
    "options": [],
    "answer": "See 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, and so on, until $x=n-2$, when $y$ can take only the value $1$. This shows that the number of triples of type $n$ is\n\n$$\ns(n) = 1 + 2 + 3 + \\dots + (n-2) = \\frac{(n-2)(n-1)}{2}.\n$$\n\n**a)** 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": 18760,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be the intersection of $BM$ and $CN$, let $R$ be the intersection of $AN$ and $BM$, and let $S$ be the intersection of $AM$ and $CN$.\n\n![](images/Australian_Scene_-2014_p211_data_4c4af2592a.png)\n\nShow that $T$ lies on the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B'$ be the reflection of $B$ about point $A$ and let $C'$ be the reflection of $C$ about point $A$. Note that $BCB'C'$ is a parallelogram because its diagonals bisect each other.\n\n![](images/Australian_Scene_-2014_p213_data_165ed846e8.png)\n\nAs in solution 1, we deduce that $\\triangle ABC \\sim \\triangle QAC$. Furthermore, since $AQ : QN = BA : AB'$, it follows that quadrilaterals $NQAC$ and $B'ABC$ are similar.$^{3}$ Therefore, $\\angle ACN = \\angle B'CB$. Similarly, we find that $\\angle MBA = \\angle CBC'$. But since $BCB'C'$ is a parallelogram, we have\n\n$$\n\\angle ACN + \\angle MBA = \\angle B'CB + \\angle CBC' = 180^{\\circ}.\n$$\n\nTherefore, $BM$ and $CN$ intersect on circle $ABC$. $\\square$\n\n$^{3}$See footnote 2 from solution 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18761,
    "subject": "Mathematics (Olympiad)",
    "question": "In every cell of a $101 \\times 101$ board is written a positive integer. For any choice of 101 cells from different rows and columns, their sum is divisible by 101. Show that the number of ways to choose a cell from each row of the board, so that the total sum of the numbers in the chosen cells is divisible by 101, is divisible by 101.",
    "options": [],
    "answer": "See solution",
    "solution": "Index the rows and columns from $0$ to $100$. We work modulo $101$. We may let $(0,0) = 0$ by adding a constant to all entries. The condition implies:\n\n$$\n(u, v) + (i, j) = (u, j) + (i, v)\n$$\n\nfor any $u, v, i, j$. Thus, there exist $a_0 = 0, a_1, \\dots, a_{100}$ and $b_0 = 0, b_1, \\dots, b_{100}$ such that $(i, j) = a_i + b_j$. Let $\\Sigma_a$ and $\\Sigma_b$ be the sums of the $a_i$ and $b_j$ respectively. The condition yields $\\Sigma_a + \\Sigma_b = 0$. Note:\n\n$$\n\\sum_{i=0}^{100} x^{101b_i} = \\left( \\sum_{i=1}^{100} x^{b_i} \\right)^{101} := \\sum_{i=0}^{\\infty} x^i \\alpha(i).\n$$\n\nThe number of ways to choose one cell from each row so that the sum of the chosen cells equals $\\Sigma_b$ is $\\alpha(\\Sigma_b) + \\alpha(101 + \\Sigma_b) + \\dots$.\n\nSince $101$ is prime, evaluating for $x = \\exp(2\\pi i/101)$ gives:\n\n$$\n0 = \\sum_{i=0}^{\\infty} \\alpha (101i + \\Sigma_b).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18762,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n$ scientists ($n \\geq 3$) attending a conference, each scientist has some friends among the attendees (friendship is mutual and no one is their own friend). It is known that, no matter how the scientists are divided into two non-empty groups, there always exist two scientists in the same group who are friends, and also two scientists in different groups who are friends.\n\nOn the first day, a topic is proposed at the conference, and the degree of approval of each scientist for this topic is represented by a non-negative integer. Starting from the second day, the degree of approval of each scientist becomes the integer part of the average of the degrees of approval of all their friends from the previous day.\n\nProve that after several days, the degree of approval of all scientists becomes the same.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the $n$ scientists form a set $V$. The friendship relation can be represented by a simple graph $G = (V, E)$. From the conditions, $G$ is connected and not bipartite.\n\nLet $f_n : V \\to \\mathbb{Z}$ denote the degree of approval on day $n$. The update rule is:\n\n$$\nf_{n+1}(x) = \\left[ \\frac{1}{\\deg(x)} \\sum_{xy \\in E} f_n(y) \\right]\n$$\n\nWe have:\n\n$$\nf_{n+1}(x) \\geq \\min_{xy \\in E} f_n(y) \\geq \\min_{y \\in V} f_n(y)\n$$\n\nSo $\\min f_{n+1} \\geq \\min f_n$. Similarly, $\\max f_{n+1} \\leq \\max f_n$. Thus, $\\min f_n$ and $\\max f_n$ eventually stabilize to constants $a$ and $b$.\n\nIf $a = b$, then all degrees are equal and the claim holds. Assume $a < b$ for contradiction.\n\nThere are finitely many integer functions from $V$ to $[a, b]$, so the sequence $\\{f_n\\}$ is eventually periodic. Let the period be $t > 0$, and for $n \\geq n_0$, $f_n = f_{n+t}$.\n\nDefine:\n\n$$\nM_n = \\{x \\in V \\mid f_n(x) < b\\}\n$$\n\nIf $y \\in M_n$ and $xy \\in E$, then $f_{n+1}(x) < b$, so $x \\in M_{n+1}$.\n\n**Method 1:** For any $x \\in V$, there exists $k$ such that a path from $x$ returns to $x$ after $k$ or $k+1$ edges. Since $G$ is not bipartite, there is an odd cycle. Using connectivity, $k$ and $k+1$ are coprime, so $x \\in M_n$ for all large $n$. By connectivity, $M_n = V$ for all large $n$, contradicting $\\max f_n = b$.\n\n**Method 2:** Let $N(X) = \\{y \\in V \\mid \\exists x \\in X, xy \\in E\\}$. Then $N(M_n) \\subset M_{n+1}$, so $M_n \\subset M_{n+2}$. By periodicity, $M_{n_0} = M_{n_0+2} = \\dots$ (call this $A$), and $M_{n_0+1} = M_{n_0+3} = \\dots$ (call this $B$). If $A \\cap B = \\emptyset$, $G$ is bipartite, contradiction. If $A \\cap B \\neq \\emptyset$, by connectivity, $A = B = V$, again contradiction.\n\nTherefore, eventually all scientists have the same degree of approval.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18763,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $m$ and $n$, prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten.",
    "options": [],
    "answer": "See solution",
    "solution": "For a given positive integer $k$, write $10^k m - n = 2^r 5^s t$, where $\\gcd(t, 10) = 1$. For large enough values of $k$, the number of times $2$ and $5$ divide the left-hand side is at most the number of times they divide $n$, hence by choosing $k$ large we can make $t$ arbitrarily large. Choose $k$ so that $t$ is larger than either $m$ or $n$.\n\nSince $t$ is relatively prime to $10$, there is a smallest exponent $b$ for which $t \\mid (10^b - 1)$. Thus $b$ is the number of digits in the repeating portion of the decimal expansion for $\\frac{1}{t}$. More precisely, if we write $tc = (10^b - 1)$, then the repeating block is the $b$-digit decimal representation of $c$, obtained by prepending extra initial zeros to $c$ as necessary. Since $t$ is larger than $m$ or $n$, the decimal expansions of $\\frac{m}{t}$ and $\\frac{n}{t}$ will consist of repeated $b$-digit representations of $cm$ and $cn$, respectively. Rewriting the identity in the first line as\n\n$$\n10^k \\left(\\frac{m}{t}\\right) = 2^r 5^s + \\frac{n}{t},\n$$\n\nwe see that the decimal expansion of $\\frac{n}{t}$ is obtained from that of $\\frac{m}{t}$ by shifting the decimal to the right $k$ places and removing the integer part. Thus the $b$-digit representations of $cm$ and $cn$ are cyclic shifts of one another. In particular, they have the same number of occurrences of each nonzero digit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18764,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the equation\n\n$$\nx^2 - 3yz \\cdot x + (y^2 + z^2) = 0\n$$\n\nwhere $x, y, z$ are integers with $1 \\le x \\le y \\le z$. Are there infinitely many integer solutions $(x, y, z)$ to this equation?",
    "options": [],
    "answer": "See solution",
    "solution": "We can regard $x^2 - 3yz \\cdot x + (y^2 + z^2) = 0$ as a quadratic in $x$. If $1 \\le x \\le y \\le z$ is a solution, then so is $(y, z, 3yz - x)$. Also, since $3y \\ge 3$, we have $3yz - x \\ge 3yz - z \\ge 2z > z$. Thus, $1 \\le y \\le z < 3yz - x$, and the new solution has a larger largest element. Starting with the solution $x = y = z = 1$ and repeating this process, we obtain infinitely many solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18765,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$ with $AB = 1$, let $D$ be a point on $AC$ such that $\\angle ABD = \\angle C$, and let $E$ be a point on $AB$ such that $BE = DE$. Let $H$ be a point on $DE$ such that $AH \\perp DE$, and $M$ be the midpoint of $CD$. If $AH = 2 - \\sqrt{3}$, find the size of $\\angle AME$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p307_data_cc37330bfd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle ABD = \\angle C = \\alpha$ and $\\angle DBC = \\beta$. It is easy to see that $\\angle BDE = \\alpha$, $\\angle AED = 2\\alpha$,\n\n$$\n\\angle ADE = \\angle ADB - \\angle BDE = (\\alpha + \\beta) - \\alpha = \\beta,\n$$\n\n$$\nAB = AE + EB = AE + EH + HD.\n$$\n\nHence,\n\n$$\n\\frac{AB}{AH} = \\frac{AE + EH}{AH} + \\frac{HD}{AH} = \\frac{1 + \\cos 2\\alpha}{\\sin 2\\alpha} + \\cot \\beta = \\cot \\alpha + \\cot \\beta. \\qquad \\textcircled{1}\n$$\n\nDraw lines $EK \\perp AC$ and $EL \\perp BD$ with pedals $K$ and $L$, respectively. Then, $L$ is the midpoint of $BD$. Combining with the Sine Theorem, we obtain\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p307_data_b66230cbba.png)\n\n$$\n\\frac{EL}{EK} = \\frac{DE \\sin \\angle EDL}{DE \\sin \\angle EDK} = \\frac{\\sin \\alpha}{\\sin \\beta} = \\frac{BD}{CD} = \\frac{LD}{MD}.\n$$\n\nThus,\n\n$$\n\\cot \\alpha = \\frac{LD}{EL} = \\frac{MD}{EK} = \\frac{MK}{EK} - \\frac{DK}{EK} = \\cot \\angle AME - \\cot \\beta. \\quad \\textcircled{2}\n$$\n\nBy ①, ② and known conditions, we have\n\n$$\n\\cot \\angle AME = \\frac{AB}{AH} = \\frac{1}{2 - \\sqrt{3}} = 2 + \\sqrt{3}.\n$$\n\nTherefore, $\\angle AME = 15^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18766,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be the product of all the odd integers between 1 and 100. Find the last three digits of $N$ (i.e., the remainder when $N$ is divided by $1000$).",
    "options": [],
    "answer": "See solution",
    "solution": "Since $1000 = 8 \\times 125$, we seek a number between 0 and 999 that is a multiple of 125 and leaves a remainder of 3 when divided by 8. The only such number is $875$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18767,
    "subject": "Mathematics (Olympiad)",
    "question": "We may assume that each problem was solved by at least one girl and one boy, by disregarding all other problems. Then every problem was solved by at least two contestants. Throughout all the solutions to this problem, we call a problem *girl-easy* if it was solved by at least three girls; otherwise we call it *girl-hard*. Analogously, we define *boy-easy* and *boy-hard* problems. We want to prove that there is a problem that is both girl-easy and boy-easy.",
    "options": [],
    "answer": "See solution",
    "solution": "*Second Solution.* (By Gabriel Carroll and Liang Xiao, China)\n\nLet $p$ be the number of problems. For the sake of contradiction, we assume that no problem is both girl-easy and boy-easy. As in the first solution, we prove that this assumption implies both that $p \\ge 22$ and that $p \\le 21$, which is impossible.\n\n*Lemma 2.* *There are at least 11 girl-easy problems and 11 boy-easy problems.*\n\n*Proof.* Suppose some girl solves no boy-easy problems. Then every problem she solves was solved by at most two boys. Since she solves at most six problems, at most 12 boys solve a problem that she solves, contradicting condition (b). Thus, every girl solved a boy-easy problem. Since each boy-easy problem was solved by at most two girls, there must be at least $\\lceil \\frac{21}{2} \\rceil = 11$ boy-easy problems. Likewise, there are at least 11 girl-easy problems. $\\square$\n\nBy our assumption, there is no problem that is both boy-easy and girl-easy. Thus, there are at least 22 problems, that is,\n\n$$\np \\ge 22.\n$$\n\nAssume that the $i$th problem was solved by $g_i$ girls and $b_i$ boys. For each $i$, either $g_i \\in \\{1,2\\}$, or $g_i > 2$ and $b_i \\le 2$. Consider the quantity\n\n$$\nQ = (g_i - 2)(b_i - 2).\n$$\n\nIf $g_i = 2$, then $Q = 0$; if $g_i = 1$, then $Q = 2 - b_i \\le 1$ since $b_i \\ge 1$ by assumption; finally, if $g_i > 2$ and $b_i \\le 2$, then $Q \\le 0$. In any case, we have $Q = (g_i - 2)(b_i - 2) \\le 1$, that is,\n\n$$\ng_i b_i \\le 2g_i + 2b_i - 3.\n$$\n\nNow $g_i b_i$ counts the number of pairs $(g, b)$ in which girl $g$ and boy $b$ both solved problem $i$. It is given that every possible pair $(g, b)$ is counted for some $i$, so\n\n$$\n\\sum_{i=1}^{p} g_i b_i \\ge 21^2.\n$$\n\nOn the other hand, everyone solved at most six problems, so $\\sum_{i=1}^{p} g_i \\le 21 \\cdot 6$, and likewise $\\sum_{i=1}^{p} b_i \\le 21 \\cdot 6$. Thus, by the previous inequalities, we obtain\n\n$$\n\\begin{aligned}\n21^2 &\\le \\sum_{i=1}^{p} g_i b_i \\le \\sum_{i=1}^{p} (2g_i + 2b_i - 3) \\\\\n&= 2 \\left( \\sum_{i=1}^{p} g_i + \\sum_{i=1}^{p} b_i \\right) - 3p \\\\\n&\\le 2(21 \\cdot 6 + 21 \\cdot 6) - 3p = 21 \\cdot 24 - 3p.\n\\end{aligned}\n$$\n\nIt follows that $3p \\le 21 \\cdot 24 - 21^2 = 21 \\cdot 3$, or\n\n$$\np \\le 21,\n$$\n\ncontradicting the earlier result.\n\nThus, our initial assumption was wrong, and there is a problem that was solved by three boys and three girls.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18768,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer and let $K_n$ be the complete graph on $n$ vertices. Each edge of $K_n$ is colored either red, green, or blue. Let $A$ denote the number of triangles in $K_n$ with all edges of the same color, and let $B$ denote the number of triangles in $K_n$ with all edges of different colors. Prove that\n\n$$\nB \\le 2A + \\frac{n(n-1)}{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider all unordered pairs of different edges which share exactly one vertex (call these *vees* for convenience). Assign each vee a *charge* of $+2$ if its edge colors are the same, and a charge of $-1$ otherwise.\n\nWe compute the total charge in two ways.\n\n**Total charge by summing over triangles**\n\n- Each monochromatic triangle has a charge of $+6$,\n- Each bichromatic triangle has a charge of $0$,\n- Each trichromatic triangle has a charge of $-3$.\n\nSince each vee contributes to exactly one triangle, the total charge is $6A - 3B$.\n\n**Total charge by summing over vertices**\n\nLet a vertex have $a$ red edges, $b$ green edges, and $c$ blue edges. The vees centered at that vertex contribute a total charge of\n\n$$\n\\begin{align*}\n& 2 \\left[ \\binom{a}{2} + \\binom{b}{2} + \\binom{c}{2} \\right] - (ab + ac + bc) \\\\\n&= (a^2 - a + b^2 - b + c^2 - c) - (ab + ac + bc) \\\\\n&= (a^2 + b^2 + c^2 - ab - ac - bc) - (a + b + c) \\\\\n&= (a^2 + b^2 + c^2 - ab - ac - bc) - (n - 1) \\\\\n&\\ge -(n - 1).\n\\end{align*}\n$$\n\nSumming over all $n$ vertices, the total charge is at least $-n(n-1)$.\n\n**Conclusion**\n\nThus,\n\n$$\n6A - 3B \\ge -n(n-1) \\implies B \\le 2A + \\frac{n(n-1)}{3}\n$$\n\nas desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18769,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n$$\nf(f(x) + x f(y)) = x f(y + 1), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = a$, where $a \\in \\mathbb{R}$. Set $x = 0$ to obtain $f(a) = 0$. Next, set $y = a$:\n$$\nf(f(x)) = x f(a + 1) \\quad (1)\n$$\nAssume first that $f(a + 1) \\neq 0$. Then $f$ is injective: if $f(x_1) = f(x_2)$ for $x_1 \\neq x_2$, plugging into (1) gives a contradiction. Set $x = 1$ in the original condition. Since $f$ is injective,\n$$\nf(1) + f(y) = y + 1 \\quad (2)\n$$\nSetting $y = 1$ yields $2f(1) = 2$, so $f(1) = 1$. Thus, $f(y) = y + 1$. Now, (2) gives $f(y) = y$ for all $y \\in \\mathbb{R}$, which is a solution.\n\nNow consider $f(a + 1) = 0$. Then $f(f(x)) = 0$ for all $x \\in \\mathbb{R}$. This means $f(y) = 0$ for all $y$ in $\\text{Im}(f)$. Assume there exists $y_0$ with $f(y_0 + 1) \\neq 0$. Setting $y = y_0$ in the original condition gives\n$$\nf(f(x) + x f(y_0)) = x f(y_0 + 1) \\quad (3)\n$$\nFor any $x_0 \\in \\mathbb{R}$, set $x = x_0 / f(y_0 + 1)$ in (3):\n$$\nf(f(x) + x f(y_0)) = x_0\n$$\nSince $x_0$ is arbitrary, $f$ is surjective, so $\\text{Im}(f) = \\mathbb{R}$. Therefore, $f(y) = 0$ for all $y \\in \\mathbb{R}$, contradicting $f(y_0 + 1) \\neq 0$. So, $f(x) = 0$ for all $x \\in \\mathbb{R}$ is also a solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18770,
    "subject": "Mathematics (Olympiad)",
    "question": "The point $P$ lies inside triangle $ABC$ so that $\\angle ABP = \\angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\\angle QAB = \\angle CAP$.",
    "options": [],
    "answer": "See solution",
    "solution": "Translate the triangle $\\triangle ABQ$ to form a new, congruent triangle $\\triangle DPC$. This has side $DC$ parallel to $AQ$, side $DP$ parallel to $AB$, and side $PC$ parallel to $BQ$. Therefore, $\\angle QAB = \\angle CDP$. Also, extend the line $BP$ to meet $DC$ at $M$.\n\n![](images/V_Britanija_2013_p24_data_73e49d6889.png)\n\nLines $AD$, $BP$, and $CQ$ are parallel, as are $AB$ and $DP$. Thus\n\n$$\n\\angle ABP = \\angle DPM = \\angle PDA.\n$$\n\nSince $\\angle ABP = \\angle PCA$, it follows that $\\angle PCA = \\angle PDA$. Therefore, the points $A$, $P$, $C$, and $D$ lie on the same circle, by the converse of the angles in the same segment theorem.\n\nTherefore, $\\angle CAP = \\angle CDP$, and so $\\angle QAB = \\angle CAP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18771,
    "subject": "Mathematics (Olympiad)",
    "question": "2000 consecutive integers (not necessarily positive) are written on the board. A student takes several turns. On each turn, he partitions the 2000 integers into 1000 pairs, and substitutes each pair by the difference and the sum of that pair (note that the difference does not need to be positive as the student may choose to subtract the greater number from the smaller one; in addition, all the operations are carried simultaneously). Prove that the student will never again write 2000 consecutive integers on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $ (a-b)^2 + (a+b)^2 = 2(a^2 + b^2) $, so the sum of the squares of the numbers written on the board doubles on each turn. Note that\n\n$$\nn^2 + (n+1)^2 + \\dots + (n+1999)^2 = 2000n^2 + 1999 \\cdot 2000n + \\frac{1999 \\cdot 2000 \\cdot 3999}{6},\n$$\n\nwhich is congruent to $8$ modulo $16$. Obviously, when we multiply this sum by $2$, we will obtain a number divisible by $16$, thus, we will never have consecutive numbers again, which is what we need to show. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18772,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n$, let $x_n = \\binom{2n}{n}$.\n\n1. Show that if $\\dfrac{2017^k}{2} < n < 2017^k$ for some positive integer $k$, then $x_n$ is a multiple of $2017$.\n\n2. Find all positive integers $h > 1$ such that there exist positive integers $N, T$ so that for all $n > N$, the sequence $(x_n)$ is periodic modulo $h$ with period $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. We prove the statement for any odd prime $p$ (in particular, $2017$). Suppose there exists a positive integer $k$ such that $\\dfrac{p^k}{2} < n < p^k$. We have\n\n$$\nv_p(x_n) = v_p\\left(\\binom{2n}{n}\\right) = v_p((2n)!) - 2v_p(n!).\n$$\n\nSince $\\dfrac{p^k}{2} < n < p^k$, it follows that $p^k < 2n < 2p^k < p^{k+1}$. Thus,\n\n$$\nv_p((2n)!) = \\left\\lfloor \\frac{2n}{p} \\right\\rfloor + \\left\\lfloor \\frac{2n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2n}{p^k} \\right\\rfloor.\n$$\n\nFor any $x \\in \\mathbb{R}$, $\\lfloor 2x \\rfloor \\geq 2\\lfloor x \\rfloor$, with equality if $\\{x\\} < \\frac{1}{2}$. Given $\\dfrac{p^k}{2} < n < p^k$, we have\n\n$$\nv_p((2n)!) > 2 \\left( \\left\\lfloor \\frac{n}{p} \\right\\rfloor + \\left\\lfloor \\frac{n}{p^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{p^k} \\right\\rfloor \\right) = 2v_p(n!),\n$$\n\nso $v_p(x_n) > 0$, i.e., $p \\mid x_n$.\n\n2. Suppose $h > 1$ satisfies the problem's requirement. For any odd prime $p$ dividing $h$, the sequence $x_n$ modulo $p$ is also periodic. By part 1, for $\\dfrac{p^k}{2} < n < p^k$,\n\n$$\nx_n \\equiv 0 \\pmod{p}.\n$$\n\nChoose $k$ large enough so $\\dfrac{p^k}{2} > T+1$. Then all $x_n$ for $n \\geq n_0$ (large enough) are $0$ modulo $p$. However, for $t$ large enough so $p^t - 1 > 2n_0$ and $n = \\dfrac{p^t - 1}{2}$, we have $v_p(x_n) = 0$, so $p \\nmid x_n$, a contradiction.\n\nTherefore, $h$ can only have $2$ as a prime divisor, i.e., $h = 2^k$ for some $k$. If $k > 1$, let $r = k-1$ and consider $n = 2^{a_1} + \\dots + 2^{a_r}$ with $a_1 > \\max\\{T, N\\}$. Then\n\n$$\nv_2(x_n) = 2S_2(n) - S_2(2n) = r,\n$$\n\nwhere $S_2(x)$ is the sum of the binary digits of $x$. Thus $x_n \\equiv 2^{k-1} \\pmod{h}$. For $i < 2^{a_1}$, the binary digit sum increases, so $x_{n+i} \\equiv 0 \\pmod{h}$. Since $a_1 > \\max\\{T, N\\}$, $x_n \\equiv x_{n+T} \\equiv 0 \\pmod{h}$, a contradiction.\n\nThus, $k = 1$ and $h = 2$. This is the answer, since $x_n$ is always even for $n \\geq 1$. For $n = 2^{a_1} + \\dots + 2^{a_r}$,\n\n$$\n2n = 2^{a_1+1} + \\dots + 2^{a_r+1}\n$$\n\nso\n\n$$\nv_2(x_n) = 2S_2(n) - S_2(2n) = r \\geq 1.\n$$\n\nHence $x_n$ is even. Therefore, $h = 2$ is the only such $h$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18773,
    "subject": "Mathematics (Olympiad)",
    "question": "Each girl—Oksana, Olesya, Olya, and Olexandra—has a rectangle with sides $2010$ and $10$. They were given a task: to cut this rectangle into two pieces from which one can make a triangle without overlaps. All of them succeeded in this task. Can they make pairwise distinct triangles?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, they can make pairwise distinct triangles. The required example is shown below:\n\n![](images/Ukrajina_2010_p7_data_93c7595080.png)\n\nFig.01",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18774,
    "subject": "Mathematics (Olympiad)",
    "question": "When Juku enters the store, the digital clock shows 16:08. Juku leaves the store when the time shown on the clock consists of the same number of segments again for the next time. How much time does Juku spend in the store? The clock shows the numbers as described in the figure:\n\n![](images/EST_ABooklet_2024_p13_data_11ae475733.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers $0, 1, \\ldots, 9$ consist of $6, 2, 5, 5, 4, 5, 6, 3, 7, 6$ segments respectively. The time of entrance consists of $2 + 6 + 6 + 7 = 21$ segments. We will consider Juku's possible exiting times by hours. If the hour is $16$, the minutes have to consist of $6 + 7 = 13$ segments. Thus one of the digits has to be $8$ (this can only be the units digit) and the other digit has to be $0, 6$ or $9$. The first option corresponds to the entrance time and the other two options cannot appear as the tens digit. If the hour is $17$, the minutes have to consist of $16$ segments, which is impossible. If the hour is $18$, then the minutes have to consist of $12$ segments. The first suitable option is $18{:}00$. This is therefore the time when Juku exits the store, meaning that he spends $1$ hour and $52$ minutes in the store.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18775,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive integers and let $a$ be an integer greater than $1$ and divisible by the product $a_1 a_2 \\cdots a_n$. Prove that $a^{n+1} + a - 1$ is not divisible by the product $(a + a_1 - 1)(a + a_2 - 1) \\cdots (a + a_n - 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "If some $a_i = 1$, then the product $(a + a_1 - 1)(a + a_2 - 1) \\cdots (a + a_n - 1)$ is divisible by $a$, while $a^{n+1} + a - 1$ is clearly not (for $a > 1$), so the former does not divide the latter.\n\nHenceforth, assume all $a_i \\ge 2$. Write $a = b a_1 \\cdots a_n$ and suppose, for contradiction, that\n\n$$\na^{n+1} + a - 1 = c (a + a_1 - 1)(a + a_2 - 1) \\cdots (a + a_n - 1),\n$$\n\nwhere $b$ and $c$ are both positive integers. Clearly, $b \\in \\{1, \\dots, a-1\\}$. Since each $a_i \\ge 2$, it follows that\n\n$$\nc \\leq \\frac{a^{n+1} + a - 1}{(a+1)^n} = \\frac{a(a^n + 1)}{(a+1)^n} - \\frac{1}{(a+1)^n} < a,\n$$\n\nso $c \\in \\{1, \\dots, a-1\\}$ as well. Notice that $b a_1 \\cdots a_n$ and $c a_1 \\cdots a_n$ are both congruent to $1 \\pmod{a-1}$, so $b = c$. Since $a + a_i - 1 = a a_i - (a-1)(a_i - 1) \\leq a a_i$, we reach a contradiction:\n\n$$\n\\frac{a^{n+1} + a - 1}{b} = (a + a_1 - 1) \\cdots (a + a_n - 1) \\leq a a_1 \\cdot a a_2 \\cdots a a_n = \\frac{a^{n+1}}{b}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18776,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 4$, let $A, B \\subseteq \\{1, 2, \\dots, n\\}$. Suppose that $ab + 1$ is a perfect square for any $a \\in A$ and $b \\in B$. Prove that\n\n$$\n\\min\\{|A|, |B|\\} \\le \\log_2 n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove a lemma.\n\n**Lemma.** Given an integer $n \\ge 4$, let $A, B \\subseteq \\{1, 2, \\dots, n\\}$. Suppose that $ab+1$ is a perfect square for any $a \\in A$ and $b \\in B$. Let $a, a' \\in A$, $b, b' \\in B$, with $a < a'$, $b < b'$. Then $a'b' > 5.5ab$.\n\n**Proof of the lemma.** Notice that $(ab+1)(a'b'+1) > (ab'+1)(a'b+1)$. So\n\n$$\n\\sqrt{(ab + 1)(a'b' + 1)} > \\sqrt{(ab' + 1)(a'b + 1)}.\n$$\n\nSince both sides are integers, we have\n\n$$\n(ab + 1)(a'b' + 1) \\ge \\left(\\sqrt{(ab' + 1)(a'b + 1)} + 1\\right)^2.\n$$\n\nExpanding, we obtain\n\n$$\n\\begin{aligned}\nab + a'b' &\\ge ab' + a'b + 2\\sqrt{(ab' + 1)(a'b + 1)} + 1 \\\\\n&\\ge ab' + a'b + 2\\sqrt{ab' \\cdot a'b}.\n\\end{aligned}\n$$\n\nSince $a < a'$, $b < b'$, we have $ab' + a'b > 2ab$. Let $a'b' = \\lambda ab$. Combining the above, $(1+\\lambda)ab > (2+2\\sqrt{\\lambda})ab$, so $\\lambda > 3+2\\sqrt{2} > 5.5$.\n\nNow, return to the original problem. Let $A = \\{a_1, a_2, \\dots, a_m\\}$, $B = \\{b_1, b_2, \\dots, b_n\\}$, with $a_1 < a_2 < \\dots < a_m$, $b_1 < b_2 < \\dots < b_n$. Suppose $2 \\le m \\le n$. Since $a_1 b_1 + 1$ is a perfect square, $a_1 b_1 \\ge 3$. By the lemma, $a_2 b_2 > 5.5 a_1 b_1 > 4^2$, and $a_{k+1} b_{k+1} > 4 a_k b_k$ for $k = 2, \\dots, m-1$. Thus,\n\n$$\nn^2 \\ge a_m b_m \\ge 4^{m-2} a_2 b_2 > 4^m.\n$$\n\nTherefore, $m \\le \\log_2 n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18777,
    "subject": "Mathematics (Olympiad)",
    "question": "We can represent the scoreboard by a $2 \\times n$ binary matrix, e.g.,\n\n$$\n\\begin{bmatrix} 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 1 \\end{bmatrix}.\n$$\n\nHow many $2 \\times n$ binary matrices are there such that, for every $1 \\le t \\le n$, the first $t$ elements in the first row always contain at least as many 1s as the first $t$ elements in the second row?\n\nLet $T_n$ denote the number of such $2 \\times n$ scoreboards.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $T_{n,k}$ as the number of $2 \\times n$ scoreboards satisfying the conditions, with the top row containing $k$ more 1s than the bottom row. For $k < 0$ or $n < k$, set $T_{n,k} = 0$. We have $T_n = \\sum_k T_{n,k}$.\n\nFor $n = 1$:\n\n$$\nT_{1,0} = 2, \\quad T_{1,1} = 1.\n$$\n\nFor $n \\ge 1$ and $k \\ge 0$:\n\n$$\nT_{n+1,k} = T_{n,k-1} + 2T_{n,k} + T_{n,k+1}.\n$$\n\nSumming over $k$ for $0 \\le k \\le n+1$ gives:\n\n$$\nT_{n+1} = 4T_n - T_{n,0}.\n$$\n\n$T_{n,0}$ is the $(n+1)$-th Catalan number:\n\n$$\nT_{n,0} = C_{n+1} = \\frac{1}{n+2} \\binom{2n+2}{n+1}.\n$$\n\nThus,\n\n$$\nT_{n+1} = 4T_n - C_{n+1}.\n$$\n\nBy induction,\n\n$$\nT_n = 4^n - \\sum_{j=0}^{n-1} 4^j C_{n-j}, \\quad n \\ge 1.\n$$\n\n% ![](images/IRL_ABooklet_2023_p66_data_63b8e90bb5.png)\n\nFrom the table, $T_{10} \\equiv 4 \\pmod{8}$, which is the required result.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 18778,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ with incircle $k$ touching the sides $BC$ and $CA$ at points $P$ and $Q$, respectively. Let $J$ be the center of the excircle opposite side $AB$ at angle $ABC$, and let $T$ be the second intersection point of the circumcircles of $\\triangle JBP$ and $\\triangle JAQ$. Prove that the circumcircle of $\\triangle ABT$ touches $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the point where $k$ is tangent to $AB$. Using standard angle notations for $\\triangle ABC$:\n\n$$\n\\angle PTJ = 180^{\\circ} - \\angle PBJ = 90^{\\circ} - \\frac{\\beta}{2} = \\angle PRB\n$$\n\nand\n\n$$\n\\angle QTJ = 180^{\\circ} - \\angle QAJ = 90^{\\circ} - \\frac{\\alpha}{2} = \\angle QRA.\n$$\n\nThus,\n\n$$\n\\angle PTQ = \\angle PTJ + \\angle QTJ = 180^{\\circ} - \\left(\\frac{\\alpha}{2} + \\frac{\\beta}{2}\\right)\n$$\n\ni.e., $\\angle PTQ = 180^{\\circ} - \\angle PRQ$, so $T \\in k$ and $R \\in TJ$.\n\nLet $\\omega$ be the circumcircle of $\\triangle ABT$, and let $S = TR \\cap \\omega$. Since $\\triangle CPJ \\cong \\triangle CQJ$, it follows that\n\n$$\n\\angle ATJ = \\angle AQJ = \\angle BPJ = \\angle BTJ,\n$$\n\ni.e., $S$ is the midpoint of $\\overline{AB}$.\n\nThe homothety $h(T, R \\rightarrow S)$ maps $k$ to $\\omega$, so these circles are tangent at $T$.\n\n![](images/Broshura_2014_IMO_p7_data_70b6ba49e4.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18779,
    "subject": "Mathematics (Olympiad)",
    "question": "In the fictional country of Mahishmati, there are 50 cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city $A$, an ordered list of cities $C_1, \\dots, C_{50}$ is called an *antitour* from $A$ if\n\n- every city (including $A$) appears in the list exactly once, and\n- for each $k \\in \\{1, 2, \\dots, 50\\}$, it is impossible to go from $A$ to $C_k$ by a sequence of exactly $k$ (not necessarily distinct) flights.\n\nBaahubali notices that there is an antitour from $A$ for any city $A$. Further, he can take a sequence of flights, starting from the capital and passing through each city exactly once. Find the least possible total number of antitours from the capital city.",
    "options": [],
    "answer": "See solution",
    "solution": "Rephrase in terms of graph theory:\n\nLet $G$ be a connected graph on 50 vertices. Given a vertex $A$, a permutation $(V_1, \\dots, V_{50})$ of the vertices of $G$ is called an antitour from $A$ if the following condition holds for every integer $k$ with $1 \\leq k \\leq 50$: there is no walk of length $k$ from $A$ to $V_k$. Given that every vertex in $G$ admits at least one antitour, and that $G$ has a Hamiltonian path, find the least possible number of antitours from some vertex of $G$.\n\nLet $N = 50$; we claim that the answer is $((N/2)!)^2$.\n\n**Bound.** Observe that if $V$ can be reached by a walk of length $d$, then it can be reached by a walk of length $d + 2k$ for any non-negative integer $k$, simply by going to $V$ and then going back-and-forth on some adjacent edge $k$ times.\n\n**Claim 1:** The graph is bipartite.\n\n*Proof.* Suppose there is some odd cycle $A_0, A_1, \\dots, A_{2i}$. Consider an antitour $C_1, \\dots, C_N$ from $A_0$. Note that for $1 \\leq k \\leq 2i$, $A_k$ can be reached from $A_0$ by $k$ moves (and thus in $k+2j$ moves) and also in $2i+1-k$ (and thus in $2i+1-k+2j$ moves). Now any integer $n \\geq 2i$ can be written as either $k+2j$ (if $n \\equiv k \\pmod{2}$) or as $2i+1-k+2j$ (if $n \\not\\equiv k \\pmod{2}$), so none of the $A_k$'s can be $C_n$ for $n \\geq 2i$. Thus the vertices $A_1, \\dots, A_{2i}$ must be $C_1, \\dots, C_{2i-1}$ in some order, which is impossible. $\\square$\n\nAlternatively, since the distance of any vertex not in the cycle is at most $N - (2i+1)$, from any vertex $A_j$ in the cycle, consider an antitour. Let the $N$th in this antitour be $V_j$. The distance of $V_j$ to the cycle is at most $N - (2i+1)$ and there is a path of length $\\leq 2i+1$ from $A_j$ to the vertex to which $V_j$ has the shortest path in the cycle with odd or even number of steps. Thus, there is an even and an odd path from $A_j$ to $V_j$ in $\\leq N$ moves. This is a contradiction! $\\square$\n\nSince the Hamiltonian path must contain an equal number of vertices from each part, the parts have $N/2$ vertices each. If the capital is in part $A = \\{A_1, \\dots, A_{N/2}\\}$, and the other part is $B = \\{B_1, \\dots, B_{N/2}\\}$, then one can form an antitour by putting the $A_i$'s in the odd positions in any order and $B_i$'s in the even positions in any order. There are $((N/2)!)^2$ ways to do this, which proves our bound.\n\n**Construction.** Consider the complete bipartite graph $K_{N/2,N/2}$, with parts $\\{A_1, \\dots, A_{N/2}\\}$ and $\\{B_1, \\dots, B_{N/2}\\}$. Any antitour from $A_1$ must have all the $B_i$'s in the even positions since any $B_i$ is reachable in 1 move (and thus in any odd number of moves). Thus there are at most $((N/2)!)^2$ antitours for this. $\\square$\n\nAlternate constructions are possible. For example: Let $A_1, \\dots, A_{50}$ be the Hamiltonian path, with $A_1$ as the capital. Add the edges $A_1A_4, A_1A_6, \\dots, A_1A_{50}$ and $A_2A_5, A_2A_7, \\dots, A_2A_{49}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18780,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of mathematicians is attending a conference. We say that a mathematician is $k$-content if he is in a room with at least $k$ people he admires or if he is admired by at least $k$ other people in the room. It is known that when all participants are in the same room, they are all at least $3k+1$-content. Prove that you can assign everyone into one of 2 rooms in a way that everyone is at least $k$-content in his room and neither room is empty. *Admiration is not necessarily mutual and no one admires himself.*",
    "options": [],
    "answer": "See solution",
    "solution": "We will use some basic graph theoretic terms for clarity.\n\nRepresent the situation by a directed graph $G(V, E)$ where each vertex $v \\in V(G)$ represents a mathematician and each edge $e \\in E(G)$ represents an admiration relation. For $v \\in V(G)$, let the out-degree $o(v)$ be the number of mathematicians $v$ admires, and the in-degree $i(v)$ be the number of mathematicians who admire $v$. For $X \\subseteq V$, $G(X)$ denotes the induced subgraph on $X$. A digraph is a $k$-digraph if for every $v \\in V(G)$, $i(v) \\geq k$ or $o(v) \\geq k$.\n\nThe problem can be reformulated: Given $G$ is a $3k+1$-digraph, can we split its vertices into two nonempty, disjoint classes so that each induced subgraph is a $k$-digraph?\n\nDefine a subset $X$ of vertices as $k$-tight if for any $Y \\subseteq X$, there is a vertex $v \\in Y$ with $i_{G(Y)}(v) \\leq k$ and $o_{G(Y)}(v) \\leq k$. A partition $(A_1, A_2)$ is feasible if both $A_1$ and $A_2$ are $k$-tight.\n\nAssume there are no feasible partitions. Consider a minimal subset $A_1 \\subseteq V(G)$ such that $G(A_1)$ is a $k$-digraph, and let $A_2 = V(G) \\setminus A_1$. Any proper subset of $A_1$ is $k$-tight. For $A_1$, removing any $v$ yields a graph where, by minimality, some $w$ has $o_G(w) < k$ and $i_G(w) < k$, so $o_{G(A_1)}(w) \\leq k$, $i_{G(A_1)}(w) \\leq k$. Thus, $A_1$ is $k$-tight.\n\n$A_2$ is not $k$-tight, so there exists $A_2' \\subseteq A_2$ such that $A_2'$ is a $(k+1)$-digraph. Now, apply the following proposition:\n\n**Proposition.** If a $2k+1$-digraph $G$ admits a solution pair (disjoint $A, B$ with $G(A)$ and $G(B)$ both $k$-digraphs), then $G$ admits a partition into two $k$-digraphs.\n\n*Proof.* Take a maximal solution pair $(A, B)$. Let $C = V(G) \\setminus (A \\cup B)$. If $C$ is empty, we are done. Otherwise, for $x \\in C$, if $o_{G(B \\cup C)}(x), i_{G(B \\cup C)}(x) < k$, but $i_G(x) \\geq 2k+1$ or $o_G(x) \\geq 2k+1$, so $o_{G(A \\cup \\{x\\})}(x) > k+1$ or $i_{G(A \\cup \\{x\\})}(x) > k+1$, contradicting maximality. Thus, the partition exists.\n\nIf there is a feasible partition $(A, B)$ maximizing $w(A < B) = |E(G(A))| + |E(G(B))|$, then $|A| \\geq k+1$ and $|B| \\geq k+1$. If no $X \\subseteq A$ has $G(X)$ a $k$-digraph, then for $x \\in B$, $B \\setminus \\{x\\}$ is $k$-tight, and $A \\cup \\{x\\}$ is $k$-tight. Moving $x$ from $B$ to $A$ increases $w$ by at least $1$, contradicting maximality. Thus, such $X$ exists, and similarly for $Y \\subseteq B$.\n\nApplying the proposition, we obtain the desired partition.\n\n*Remark.* The argument can be adapted for asymmetric rooms: if the graph is a $k + l + \\max(k, l) + 1$-digraph, it can be partitioned into $k$-digraph and $l$-digraph parts.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18781,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $n$，考慮 $n$ 維空間的所有整數點（即每個座標都是整數的點）。當兩個整數點的直線距離為 $1$ 時，我們稱它們互相相鄰。試問是否可能將其中一部分的整數點做標記，使得對於每一個整數點，在該點本身和它所有相鄰的點這 $(2n+1)$ 個點中，總是恰有一個被標記？",
    "options": [],
    "answer": "See solution",
    "solution": "可以！\n\n令 $x_1, \\dots, x_n$ 為整數點的座標，將所有滿足\n\n$$\n(2n + 1) \\mid (x_1 + 2x_2 + \\dots + nx_n)\n$$\n\n的點標記，即可達成條件。對於每個點，$x_1 + 2x_2 + \\dots + nx_n$ 可以唯一地表示為 $(2n+1)l \\pm k$，其中 $l$ 為整數，$k = 0, 1, \\dots, n$。當 $k=0$ 時即該點被標記，否則即是沿著第 $k$ 個座標方向的兩個相鄰點之一被標記。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18782,
    "subject": "Mathematics (Olympiad)",
    "question": "給定平面上一條直線 $L$。假設 $A$ 與 $B$ 為直線 $L$ 同側的兩相異點。試證：直線 $AB$ 與 $L$ 不垂直的充要條件為：$L$ 上有唯一的一點 $R$，使得對於直線 $L$ 上任意一點 $P$，$\\angle APB \\leq \\angle ARB$。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，若 $AB$ 平行於 $L$，作 $AB$ 的中垂線交 $L$ 於 $R$，並作 $\\triangle ABP$ 的外接圓 $C$。則，對於 $L$ 上任一異於 $R$ 的點 $P$，令 $PB$ 交圓 $C$ 於 $S$，則 $\\angle APB < \\angle ASB = \\angle ARB$。故存在唯一的一點 $R$ 滿足題設。\n\n接著，若 $AB$ 不平行於 $L$，令直線 $AB$ 交直線 $L$ 於 $K$。易知平面上存在唯一的兩圓 $C$ 和 $C'$，使得此兩圓皆過點 $A$ 和 $B$，且與 $L$ 相切。令此兩圓分別切 $L$ 於 $R$ 和 $R'$。\n\n現在，不失一般性假設 $\\angle AR'B \\leq \\angle ARB$，則對於直線 $L$ 上任一點異於 $R$ 和 $R'$ 的點 $P$：\n\n(i) 若 $P$ 與 $R$ 同側，令 $PB$ 交圓 $C$ 於 $S$，則 $\\angle APB < \\angle ASB = \\angle ARB$。\n\n(ii) 若 $P$ 與 $R'$ 同側，令 $PB$ 交圓 $C'$ 於 $S'$，則 $\\angle APB < \\angle AS'B = \\angle AR'B \\leq \\angle ARB$。\n\n故，存在唯一一點 $R$ 滿足題設，若且唯若 $\\angle AR'B \\neq \\angle ARB$，若且唯若 $AB$ 與 $L$ 不垂直。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18783,
    "subject": "Mathematics (Olympiad)",
    "question": "a) The small cubes in a cube have at most three faces exposed. The only small cubes that have exactly three faces exposed are the 8 at the corners of a $10 \\times 10 \\times 10$ cube. The only small cubes that have exactly two faces exposed are the 8 on each edge of the $10 \\times 10 \\times 10$ cube that are not at the corners. All other cubes have fewer than two faces exposed. There are 12 edges on the $10 \\times 10 \\times 10$ cube, so the number of small cubes removed at the first step is $8 + (12 \\times 8) = 104$.\n\nb) What is the size of the original cube if 200 small cubes are removed at the first step?\n\nc) After the first step, what is the surface area of the remaining object if the original cube is $9 \\times 9 \\times 9$?\n\nd) For a $9 \\times 9 \\times 9$ cube, how many small cubes remain after three steps of removing edge and corner cubes as described?\n\n![](images/Australian-Scene-combined-2015_p50_data_885dfd7ba4.png)\n\n![](images/Australian-Scene-combined-2015_p50_data_df8f1aed9a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "b) **Alternative i:**\n\nAt the first step, the 8 small cubes at the corners are removed. The other 192 cubes come equally from the 12 edges but not from the corners. So the number of non-corner small cubes on each edge is $192/12 = 16$. Hence each edge in the original cube had a total of 18 small cubes.\n\n**Alternative ii:**\n\nA cube with $n$ small cubes along one edge will lose $8 + 12(n-2)$ small cubes at the first step. So $8 + 12(n-2) = 200$, $12(n-2) = 192$, $n-2 = 16$, $n = 18$. Thus the original cube was $18 \\times 18 \\times 18$.\n\n**Alternative iii:**\n\nFrom part a, a $10 \\times 10 \\times 10$ cube loses 104 small cubes at the first step. Increasing the cube's dimension by 1 increases the number of lost cubes by 12, one for each edge. Since $200 - 104 = 96$ and $96/12 = 8$, the cube that loses 200 small cubes at the first step is $18 \\times 18 \\times 18$.\n\nc) **Alternative i:**\n\nAt the first step, each of the 6 faces of the $9 \\times 9 \\times 9$ cube is converted to a $7 \\times 7$ single layer of small cubes. The exposed surface area of this layer is $7 \\times 7$ small faces plus a ring of $4 \\times 7$ small faces. So the surface area of the remaining object is $6 \\times (49 + 28) = 6 \\times 77 = 462$.\n\n**Alternative ii:**\n\nFirst remove the middle small cube on one edge of the $9 \\times 9 \\times 9$ cube. This increases the surface area by 2 small faces. Then remove the small cubes on either side. This does not change the surface area. Continue until only the two corner cubes remain. At this stage, the surface area has increased by 2 small faces. Repeating this process on all 12 edges increases the surface area by $12 \\times 2$ small faces. At this stage, all 8 corner cubes have 6 exposed faces. So removing the 8 corner cubes reduces the surface area by $8 \\times 6$ small faces. The surface area of the original $9 \\times 9 \\times 9$ cube was $6 \\times 81 = 486$. Hence the surface area of the remaining object is $486 + 24 - 48 = 462$.\n\nd) At the first step, the original $9 \\times 9 \\times 9$ cube is reduced to a $7 \\times 7 \\times 7$ cube with a $7 \\times 7$ single layer of small cubes placed centrally on each face. At this stage, the number of small cubes removed is $8 + (12 \\times 7) = 92$.\n\nAt the second step, the only small cubes removed are the edge cubes in the $7 \\times 7$ single layers. This leaves a $7 \\times 7 \\times 7$ cube with a $5 \\times 5$ single layer of small cubes placed centrally on each face. So in this step, the number of small cubes removed is $6(4 + 4 \\times 5) = 144$.\n\nAt the third step, the only small cubes removed are the edge cubes in the $5 \\times 5$ single layers and the edge cubes in the $7 \\times 7 \\times 7$ cube. So in this step, the number of small cubes removed is $6(4 + 4 \\times 3) + 8 + (12 \\times 5) = 164$.\n\nThus, the number of small cubes remaining is $(9 \\times 9 \\times 9) - 92 - 144 - 164 = 329$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18784,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(n)$ be the sum of the digits of $n$, and $T(n)$ the twisted sum of $n$. If we consider $T(n) + n$, then each of the five digits of $n$ is involved in the sum once for each of ten thousands, thousands, hundreds, tens, and units. Thus,\n\n$$\nT(n) + n = (10000 + 1000 + 100 + 10 + 1)(a + b + c + d + e) = 11111 S(n).\n$$\n\nIf $T(n) = T(m)$, show that $n = m$ for five-digit numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $T(n) = T(m)$. Then\n\n$$\nT(n) + n = T(m) + m \\implies 11111 S(n) = 11111 S(m) + (n - m)\n$$\nso\n$$\n11111 (S(n) - S(m)) = n - m. \\tag{*}\n$$\n\nTherefore, $11111 \\mid n - m$. Also, since a number and its digit sum have the same remainder modulo $9$, reducing $(*)$ modulo $9$ gives:\n\n$$\n11111 (S(n) - S(m)) \\equiv n - m \\pmod{9}\n$$\nBut $11111 \\equiv 5 \\pmod{9}$, so\n$$\n5(S(n) - S(m)) \\equiv n - m \\pmod{9}\n$$\nBut $S(n) \\equiv n \\pmod{9}$ and $S(m) \\equiv m \\pmod{9}$, so\n$$\n5(n - m) \\equiv n - m \\pmod{9} \\implies 4(n - m) \\equiv 0 \\pmod{9}\n$$\nThus, $9 \\mid n - m$. Since $\\gcd(11111, 9) = 1$, $n - m$ is divisible by $11111 \\times 9 = 99999$. But $n$ and $m$ are both five-digit numbers, so $|n - m| < 99999$, which forces $n = m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18785,
    "subject": "Mathematics (Olympiad)",
    "question": "In the following, we consider polynomials in $\\mathbb{F}_2[x]$, meaning all coefficients are taken modulo 2. Let $f(n)$ be the number of odd coefficients in $(x^2 - x + 1)^n$.\n\nFind $f(2009)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Claim 1.** We have $(x^2 - x + 1)^{2k} = x^{2k+1} - x^{2k} + 1$ for any nonnegative integer $k$.\n\n*Proof.* It suffices to note that\n\n$$\n(x^2 - x + 1)^2 = x^4 - 2x^3 + 3x^2 - 2x + 1 = x^4 - x^2 + 1.\n$$\n\nThe result follows easily by induction. $\\square$\n\n**Claim 2.** We have $f(2ka+b) = f(a)f(b)$ for any positive integers $a, b$ satisfying $b < 2^{k-1}$.\n\n*Proof.* By claim 1, we have\n\n$$\n(x^2 - x + 1)^{2ka+b} = ((x^2 - x + 1)^{2k})^a (x^2 - x + 1)^b = (x^{2k+1} - x^{2k} + 1)^a (x^2 - x + 1)^b.\n$$\n\nLet\n\n$$\n(x^{2k+1} - x^{2k} + 1)^a = \\sum_{i=1}^{s} c_i x^{2k \\alpha_i},\n$$\n\n$$\n(x^2 - x + 1)^b = \\sum_{j=1}^{t} d_j x^{\\beta_j}.\n$$\n\nThen their product is\n\n$$\n\\sum_{i=1}^{s} \\sum_{j=1}^{t} c_i d_j x^{2k\\alpha_i + \\beta_j}.\n$$\n\nSince $\\beta_j \\le 2b < 2^k$, different pairs $(i, j)$ correspond to different exponents $2^k\\alpha_i + \\beta_j$. Also, $c_i d_j$ is odd if and only if both $c_i$ and $d_j$ are odd. As there are $f(a)$ odd coefficients $c_i$ and $f(b)$ odd coefficients $d_j$, the number of odd coefficients is $f(a)f(b)$. $\\square$\n\n**Claim 3.** We have $f(2^k - 1) = \\frac{2^{k+2} - (-1)^k}{3}$ for any positive integer $k$.\n\n*Proof.* For odd $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern\n\n$$\n\\underbrace{110110\\cdots110}_{\\frac{2^{k-2}}{3} \\text{ triples } 110} 111 \\underbrace{011011\\cdots011}_{\\frac{2^{k-2}}{3} \\text{ triples } 011}\n$$\n\nFor even $k$, the coefficients follow\n\n$$\n\\underbrace{110110\\cdots110}_{\\frac{2^{k-1}}{3} \\text{ triples } 110} 1 \\underbrace{011011\\cdots011}_{\\frac{2^{k-1}}{3} \\text{ triples } 011}.\n$$\n\nIt is not hard to prove these by induction and claim 1. $\\square$\n\nNow, by the above claims,\n\n$$\n\\begin{aligned}\nf(2009) &= f(2^6 \\times 31 + 25) = f(31)f(25) \\\\\n&= f(31)f(2^3 \\times 3 + 1) = f(31)f(3)f(1) \\\\\n&= \\frac{2^7 + 1}{3} \\cdot \\frac{2^4 - 1}{3} \\cdot \\frac{2^3 + 1}{3} = 645.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18786,
    "subject": "Mathematics (Olympiad)",
    "question": "The non-negative integers $a$, $b$, $c$ are such that the numbers\n\n$$\nm = \\frac{5a + 6b + 7c + 6}{4a + 3b + 2c + 3} \\quad \\text{and} \\quad n = \\frac{a + 2b + 3c + 5}{3a + b + 2c + 5}\n$$\n\nare both integers.\n\n(a) Prove that $m \\geq 2$.\n\n(b) Find $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)**\nIf $m \\leq 1$, then $5a + 6b + 7c + 6 \\leq 4a + 3b + 2c + 3$, that is, $a + 3b + 5c + 3 \\leq 0$—impossible for non-negative $a$, $b$, $c$. So $m \\geq 2$.\n\n**(b)**\nIf $n \\geq 2$, then $a + 2b + 3c + 5 \\geq 6a + 2b + 4c + 10$, whence $0 \\geq 5a + c + 5$, which is false. Since $n > 0$, $n$ must be $1$.\n\nNow $n = 1$ yields $a + 2b + 3c + 5 = 3a + b + 2c + 5$, hence $b + c = 2a$.\n\nSuppose $m \\geq 3$. Then $5a + 6b + 7c + 6 \\geq 12a + 9b + 6c + 9$, therefore $c \\geq 7a + 3b + 3$. This gives $b + c \\geq 7a + 4b + 3$, that is, $2a \\geq 7a + 4b + 3$, hence $0 \\geq 5a + 4b + 3$, which is false. So $m < 3$, and part (a) implies $m = 2$.\n\nThe above shows 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": 18787,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nP(m) = \\frac{m}{2} + \\frac{m^2}{4} + \\frac{m^4}{8} + \\frac{m^8}{8}.\n$$\n\nHow many of the values $P(2022)$, $P(2023)$, $P(2024)$, and $P(2025)$ are integers?\n\n(A) 0   (B) 1   (C) 2   (D) 3   (E) 4",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (E):** Let $Q(m) = 8P(m) = 4m + 2m^2 + m^4 + m^8$. Because the coefficients of $Q$ are integers, it follows that if $a \\equiv b \\pmod{8}$, then $Q(a) \\equiv Q(b) \\pmod{8}$. It suffices to show that the 8 numbers $Q(-3)$, $Q(-2)$, $Q(-1)$, ..., $Q(4)$ are all divisible by 8. If $m$ is even, then each of the monomials of $Q(m)$ is divisible by 8. If $m = \\pm 1$, then $Q(m) = \\pm 4 + 4 \\equiv 0 \\pmod{8}$. If $m = \\pm 3$, then $m^2 = 9 \\equiv 1 \\pmod{8}$, which implies that $m^4 \\equiv 1 \\pmod{8}$, and so also that $m^8 \\equiv 1 \\pmod{8}$. Hence $Q(\\pm 3) \\equiv \\pm 12 + 2 + 1 + 1 \\equiv 0 \\pmod{8}$.\n\nTherefore $8P(m)$ is divisible by 8 for all integers $m$, which implies that $P(m)$ is an integer for all $m$. In particular, all 4 of the given values of $P(m)$ are integers.\n\n**Alternate Solution:**\n\nLet $Q(m)$ be defined as above, and note that $Q(m)$ is divisible by 8 if $m$ is even. To treat odd $m$, write\n\n$$\n\\begin{align*}\nQ(m) &= 8m + 4(m^2 - m) + 2m^2(m^2 - 1) + m^4(m^4 - 1) \\\\\n&= 8m + 4m(m - 1) + 2m^2(m + 1)(m - 1) + m^4(m^2 + 1)(m + 1)(m - 1)\n\\end{align*}\n$$\n\nand note that because $m+1$, $m-1$, and $m^2+1$ are all even, each term has at least three factors of 2. The solution concludes as above.\n\n**Another Approach:**\n\nAnother way to see that $Q(m)$ is divisible by 8 when $m$ is odd is to apply more general facts from number theory. Fermat's Little Theorem asserts that if $p$ is prime, then $a^p \\equiv a \\pmod{p}$ for all integers $a$. In particular, $m^2 \\equiv m \\pmod{2}$. Euler's Totient Theorem asserts that if $\\gcd(a, q) = 1$, then $a^{\\phi(q)} \\equiv 1 \\pmod{q}$, where $\\phi(q)$ is the number of positive integers less than $q$ that are relatively prime to $q$. Because $\\phi(4) = 2$, it follows that $m^2 \\equiv 1 \\pmod{4}$ when $m$ is odd. Also, $\\phi(8) = 4$, so $m^4 \\equiv 1 \\pmod{8}$ if $m$ is odd.\n\n**Note:** Suppose that a necklace is to be made by placing $n$ beads, equally spaced, on a circular ring. Each bead can be any one of $m$ different colors. Two necklaces are regarded as being the same if they differ only by rotation. In 1892 Captain Percy Alexander MacMahon showed that the number of such necklaces is\n\n$$\n\\frac{1}{n} \\sum_{d|n} \\phi\\left(\\frac{n}{d}\\right) m^d.\n$$\n\nHe acknowledged its prior discovery by another soldier, Monsieur le Colonel Charles Paul Narcisse Moreau. Thus $P(m)$ counts the number of necklaces with $n = 8$ beads and therefore must be an integer. This is discussed in *Concrete Mathematics* by Graham, Knuth, and Patashnik.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18788,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose two teams, A and B, play a best-of-7 series (first to 4 wins). Is the series more likely to end in exactly 6 games or exactly 7 games?",
    "options": [],
    "answer": "See solution",
    "solution": "Let each possible series outcome be represented as a bit string, where $1$ denotes a win by team A and $0$ a win by team B. The number of bit strings of length 6 (series ends in 6 games) is $\\binom{5}{3}$, since A must win exactly 3 of the first 5 games and then win the 6th. For length 7, the number is $\\binom{6}{3}$, as A and B each win 3 of the first 6 games, and the 7th game decides the winner. Since $\\binom{6}{3} > \\binom{5}{3}$, the series is more likely to end in exactly 7 games.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18789,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, $n$ different circles with the same center are given. Inside the circle with the smallest radius, consider two different points $A$ and $B$. Next, consider $k$ different lines passing through point $A$ and $m$ different lines passing through point $B$. All lines passing through $A$ intersect all lines passing through $B$ (no line passes through both $A$ and $B$), and their points of intersection are not on the given circles. Determine the maximal and minimal number of regions bounded by the lines and the circles, lying in the interior of the circles, and for the regions lying inside the smallest circle, at least one part of their border is an arc of the circle.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the number of intersection points of the lines is $k \\cdot m$.\n\nIf all the intersection points lie outside the circle with the greatest radius, then the $k$ lines through $A$ create $2k$ regions inside the smallest circle, and the $m$ lines through $B$ create $2m$ regions inside the smallest circle. Since the shaded region is counted twice, the total number of regions inside the smallest circle is $2k + 2m - 1$.\n\nThere are $n-1$ circular rings. In each ring, the same number of regions $2k + 2m - 1$ is created, but the shaded region is divided into two regions. Therefore, inside any circular ring, there are $2k + 2m$ regions.\n\nIn this case, the total number of regions is:\n$$\n2n(k+m) - 1.\n$$\n\nIf one or more intersection points lie inside the smallest circle, then there are $2(k+m)$ regions inside the smallest circle. If all intersection points lie inside the smallest circle, there are $2(n-1)(k+m)$ regions in the $n-1$ rings. If a point of intersection lies in a ring, the number of regions increases by one. Therefore, in total:\n$$\n2(k+m) + 2(n-1)(k+m) = 2n(k+m) \\text{ regions.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18790,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $1 \\leq i, j \\leq 2n$ such that all the cells in the $i$-th row and $j$-th column have the same color. Define the number of cells with the other color as follows:\n\n- In the 1st to $i$-th rows and the 1st to $j$-th columns: $a$,\n- In the 1st to $i$-th rows and the $j$-th to $2n$-th columns: $b$,\n- In the $i$-th to $2n$-th rows and the 1st to $j$-th columns: $c$,\n- In the $i$-th to $2n$-th rows and the $j$-th to $2n$-th columns: $d$.\n\nLet $f(m) = \\max\\{0, 2\\sqrt{m} - 1\\}$ and $g(m) = \\frac{f(m)}{m}$ for $m \\geq 1$, with $g(0) = g(1) = 1$. The number of mixed $2 \\times 2$ squares is at least $f(a) + f(b) + f(c) + f(d)$.\n\nGiven $a + b + c + d = k$ or $4n^2 - k$, and $n^2 - n + 1 \\leq k \\leq 2n^2$, determine all values of $k$ for which the minimum number of mixed $2 \\times 2$ squares is exactly $2n-1$.\n\n![](path/to/file.png)\n\n**Remark.**\n\n- A *path* is a tuple of vertices $(v_1, v_2, \\dots, v_n)$ such that for any $1 \\leq i \\leq n-1$, $v_i$ and $v_{i+1}$ are connected by an edge. $v_1$ is the *start point* and $v_n$ the *end point*.\n- A *cycle* is a path of at least 3 vertices such that the start and end points are connected by an edge.\n- A *connected component* is a set of all vertices reachable from a fixed vertex $v$.\n- A *tree* is a graph with no cycles and exactly one connected component. The number of vertices in any tree exceeds the number of edges by exactly 1.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the minimum number of mixed $2 \\times 2$ squares using the functions $f$ and $g$ defined above.\n\nSince $f$ is weakly increasing and $g$ is weakly decreasing, for $a + b + c + d \\geq k \\geq n^2 - n + 1$, we have:\n\n$$\n\\begin{aligned}\nf(a) + f(b) + f(c) + f(d) &= ag(a) + bg(b) + cg(c) + dg(d) \\\\\n&\\geq (a + b + c + d)g(a + b + c + d) \\\\\n&= f(a + b + c + d) \\\\\n&\\geq f(n^2 - n + 1) \\\\\n&> 2n - 2.\n\\end{aligned}\n$$\n\nTherefore, there are at least $2n-1$ mixed $2 \\times 2$ squares. For $k$ in $n^2 - n + 1 \\leq k \\leq n^2$, the condition is satisfied.\n\nFor $n^2 + 1 \\leq k \\leq 2n^2$, if the number of mixed $2 \\times 2$ squares is $2n-1$, $k$ must be a multiple of $2n$. If all rows and columns are mixed, for each $1 \\leq i < 2n$, there is exactly one $1 \\leq j < 2n$ such that $[i, j]$ is mixed, and the shared value $j$ is the same for all $i$. The number of cells with the same color as $(1, 1)$ is $2jn$, and with $(1, 2n)$ is $2n(2n-j)$, so $k$ is a multiple of $2n$.\n\nIf $k$ is a multiple of $2n$ and $a = \\frac{k}{2n}$, coloring the first $a$ columns black and the rest white yields $2n-1$ mixed $2 \\times 2$ squares.\n\n**Conclusion:** The values $k$ which satisfy the condition are all $n^2 - n + 1 \\leq k \\leq n^2$ and all multiples of $2n$ in $n^2 + 1 \\leq k \\leq 2n^2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18791,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to mark points on a straight line such that the blue points have coordinates $-3, -1, 2$, the red point has coordinate $0$, and the green points have coordinates $3, 3.5, 4.5, 5$?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it is possible. For example, the points can be marked on the straight line as described: blue points at $-3$, $-1$, $2$; red point at $0$; green points at $3$, $3.5$, $4.5$, $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18792,
    "subject": "Mathematics (Olympiad)",
    "question": "One of the numbers $1, 2, \\ldots, n$ is written in each cell of a $17 \\times 17$ table for a certain $n \\in \\mathbb{N}$; all of these numbers are used. If a row contains two cells $C_1$ and $C_2$ with equal numbers $k$ and $C_1$ is to the left of $C_2$, then there are no numbers $k$ in the column of $C_1$ that are above $C_1$. Determine the minimal $n$ for which such a table exists.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal $n$ in question is $n=9$.\n\nFirst, we show that each number $k \\in \\{1, 2, \\ldots, n\\}$ occurs in the table at most 34 times. Let a row contain at least two $k$'s. Underline all of them except the rightmost one. The remaining numbers in the table are not underlined. By hypothesis, $k$ does not occur above an underlined number. It follows that the underlined numbers are in different columns, hence there are at most 17 of them. In addition, by construction, each row without underlined $k$'s contains at most one $k$, so there are also at most 17 numbers $k$ that are not underlined. In summary, the table has at most $17+17=34$ numbers $k$, as stated.\n\nSince each $k \\in \\{1, 2, \\ldots, n\\}$ occurs in the table $x_k \\le 34$ times, the equality $x_1 + \\ldots + x_n = 17^2$ yields $n \\ge \\frac{17^2}{34}$, meaning that $n \\ge 9$. For an example with $n=9$, consider the diagonals parallel to the main diagonal containing the bottom left and the top right cell. Label them consecutively $1, 2, \\ldots, 33$ so that the top left cell is diagonal 1 and the bottom right cell is diagonal 33. For each $k=1, 2, \\ldots, 9$ write $k$ in all cells of diagonals $2k-1$ and $2k$; for each $k=1, 2, \\ldots, 7$ write $k$ in all cells of diagonals $2k+1$ and $2k+18$; finally, write 8 in the only cell of diagonal 33. In every row and column there are at most two numbers equal to a given $k \\in \\{1, 2, \\ldots, 9\\}$; and if there are two of them, then they are adjacent. It follows that the table satisfies the given conditions.\n\n![](images/Argentina2016_booklet_p1_data_0d9ee3b65b.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18793,
    "subject": "Mathematics (Olympiad)",
    "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 segments $[EF]$, $[EA]$, and $[AD]$, respectively. Prove that the points $M, N, K, G$ are concyclic.\n\n![](images/RMC2014_p84_data_907f7a451a.png)",
    "options": [],
    "answer": "See 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 $\\big(*)\\big$.\n\nAs $NK \\parallel DE$ and $NM \\parallel AF$, we have:\n\n$$\n\\begin{align*}\n\\angle KNM &= \\angle KNE + \\angle ENM = \\angle DEB + \\angle BAC \\\\\n&= \\angle DEB + \\angle BDE = \\angle ABF,\n\\end{align*}\n$$\n\nand, as $KP \\parallel DF$ and $MP \\parallel AE$, we have\n\n$$\n\\angle KPM = 180^{\\circ} - \\angle KPA - \\angle MPF = 180^{\\circ} - \\angle DFA - \\angle EAF = \\angle ABF.\n$$\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": 18794,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that $\\frac{\\operatorname{tg}\\alpha}{\\operatorname{tg}\\beta} = \\frac{a^2 + c^2 - b^2}{b^2 + c^2 - a^2}$ for every triangle $ABC$, where $a$, $b$, $c$ are the sides and $\\alpha$, $\\beta$ are the angles of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "From the cosine theorem, we have\n$$\nb^2 = a^2 + c^2 - 2ac \\cos \\beta, \\quad a^2 = b^2 + c^2 - 2bc \\cos \\alpha\n$$\nTherefore,\n$$\na^2 + c^2 - b^2 = 2ac \\cos \\beta, \\qquad b^2 + c^2 - a^2 = 2bc \\cos \\alpha\n$$\nSo,\n$$\n\\frac{a^2 + c^2 - b^2}{b^2 + c^2 - a^2} = \\frac{2ac \\cos \\beta}{2bc \\cos \\alpha} = \\frac{a \\cos \\beta}{b \\cos \\alpha} \\tag{*}\n$$\nFrom the law of sines, we have\n$$\n\\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin \\beta}, \\quad \\text{i.e.}\\quad \\frac{a}{b} = \\frac{\\sin \\alpha}{\\sin \\beta}\n$$\nNow,\n$$\n(*) = \\frac{\\sin \\alpha \\cos \\beta}{\\sin \\beta \\cos \\alpha} = \\frac{\\operatorname{tg} \\alpha}{\\operatorname{tg} \\beta}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18795,
    "subject": "Mathematics (Olympiad)",
    "question": "Encontrar la solución entera más pequeña de la ecuación\n\n$$\n\\left\\lfloor \\frac{x}{8} \\right\\rfloor - \\left\\lfloor \\frac{x}{40} \\right\\rfloor + \\left\\lfloor \\frac{x}{240} \\right\\rfloor = 210.\n$$\n\nSi $x$ es un número real, $\\lfloor x \\rfloor$ es la parte entera de $x$, esto es, el mayor número entero menor o igual que $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $x$ una solución entera de la ecuación. Dividiendo primero por 240, luego el resto $r_1$ por 40 y el nuevo resto $r_2$ por 8, resulta:\n\n$$\nx = 240c_1 + r_1 = 240c_1 + 40c_2 + r_2 = 240c_1 + 40c_2 + 8c_3 + r_3,\n$$\n\ndonde $0 \\leq c_2 < 6$, $0 \\leq c_3 < 5$ y $0 \\leq r_3 < 8$ (las desigualdades para $c_2$ y $c_3$ se obtienen de $0 \\leq r_1 < 240$ y $0 \\leq r_2 < 40$). Entonces:\n\n$$\n210 = \\left\\lfloor \\frac{x}{8} \\right\\rfloor - \\left\\lfloor \\frac{x}{40} \\right\\rfloor + \\left\\lfloor \\frac{x}{240} \\right\\rfloor = (30c_1 + 5c_2 + c_3) - (6c_1 + c_2) + c_1 = 25c_1 + 4c_2 + c_3,\n$$\n\ny el único caso posible es $c_1 = 8$, $c_2 = 2$, $c_3 = 2$ (reduciendo módulo 5 queda $c_2 \\equiv c_3 \\pmod{5}$ y necesariamente $c_2 = c_3$; entonces $42 = 5c_1 + c_2$ y tiene que ser $c_2 = 2$, $c_1 = 8$), con lo que:\n\n$$\nx = 240 \\times 8 + 40 \\times 2 + 8 \\times 2 + r_3 = 2016 + r_3\n$$\n\ny la menor solución entera de la ecuación dada es $x = 2016$ (las otras son $2017, \\ldots, 2023$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18796,
    "subject": "Mathematics (Olympiad)",
    "question": "給定任意正整數 $k$，證明存在質數 $p$ 及相異整數 $a_1, a_2, \\dots, a_{k+3} \\in \\{1, 2, \\dots, p-1\\}$，使得對於所有 $i = 1, 2, \\dots, k$，$p$ 都整除 $a_i a_{i+1} a_{i+2} a_{i+3} - i$。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，構造相異的正有理數 $r_1, r_2, \\dots, r_{k+3}$，使得對所有 $i = 1, 2, \\dots, k$，都有 $r_i r_{i+1} r_{i+2} r_{i+3} = i$。由於 $r_{i+4} = \\frac{i+1}{i} r_i$，對於 $r = 1, 2, 3, 4$，序列 $\\{r_{4n+r}\\}_n$ 是嚴格遞增的。選擇 $r_1, r_2, r_3$ 為大於 $k$ 的相異質數，令 $r_4 = \\frac{1}{r_1 r_2 r_3}$，則 $\\{r_i\\}$ 為相異的有理數。將 $r_i$ 寫成不可約分數 $r_i = \\frac{u_i}{v_i}$。容易看出，只有當 $u_i v_j \\equiv u_j v_i \\pmod{p}$ 時，$u_i(v_i)^{-1} = u_j(v_j)^{-1}$ 在 $\\mathbb{Z}_p$ 中相等。選擇質數 $p > (\\max_i\\{u_i, v_i\\})^2$，則令 $a_i = u_i(v_i)^{-1}$ 在 $\\mathbb{Z}_p$ 中，即可滿足題目要求。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18797,
    "subject": "Mathematics (Olympiad)",
    "question": "In a plane rectangular coordinate system $xOy$, the graph of the function $y = \\frac{1}{|x|}$ is $\\Gamma$. Let points $P$ and $Q$ be on $\\Gamma$ such that $P$ is in the first quadrant, $Q$ is in the second quadrant, and the line $PQ$ is tangent to the part of $\\Gamma$ in the second quadrant at point $Q$. Find the minimum value of $|PQ|$.",
    "options": [],
    "answer": "See solution",
    "solution": "When $x > 0$, $y = \\frac{1}{x}$. When $x < 0$, $y = -\\frac{1}{x}$, and its derivative is $y' = -\\frac{1}{x^2}$.\n\nSuppose $Q(-a, \\frac{1}{a})$, where $a > 0$. By the condition, the slope of $PQ$ is $y'|_{x=-a} = \\frac{1}{a^2}$.\n\nThe equation of line $PQ$ is:\n$$\ny = \\frac{1}{a^2}(x + a) + \\frac{1}{a} = \\frac{x + 2a}{a^2}.\n$$\nCombining this with $y = \\frac{1}{x}$ for $x > 0$ gives:\n$$\nx^2 + 2a x - a^2 = 0,\n$$\nso the abscissa of $P$ is $x_P = (\\sqrt{2} - 1)a$ (the negative root is discarded). Therefore,\n$$\n|PQ| = \\sqrt{1 + \\left(\\frac{1}{a^2}\\right)^2} \\cdot |x_P - x_Q|.\n$$\nAfter simplification and minimization, the minimum occurs at $a = 1$, i.e., $Q(-1, 1)$, and the minimum value of $|PQ|$ is $2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 18798,
    "subject": "Mathematics (Olympiad)",
    "question": "Sequence of positive integers $a_1, a_2, a_3, \\dots$ is defined by $a_{n+1} = a_n^2 + 2018$, where $a_1$ is some positive integer. Prove that in this sequence no more than one number can be a cube of a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there is more than one number in the sequence that is a cube of a positive integer. Let $a_k$ be the smallest such cube. Then $a_k \\equiv 0, \\pm1 \\pmod{9}$, so $a_k^2 \\equiv 0, 1 \\pmod{9}$. From now on, all congruences are modulo $9$.\n\n- If $a_k^2 \\equiv 0$, then $a_{k+1} = a_k^2 + 2018 \\equiv 0 + 2018 \\equiv 2 \\pmod{9}$. Then $a_{k+1}^2 \\equiv 4$, so $a_{k+2} \\equiv 4 + 2018 \\equiv 6 \\pmod{9}$. Then $a_{k+2}^2 \\equiv 0$, so $a_{k+3} \\equiv 0 + 2018 \\equiv 2 \\pmod{9}$. Thus, the sequence cycles through values that are not cubes modulo $9$ after $a_k$.\n\n- If $a_k^2 \\equiv 1$, then $a_{k+1} = 1 + 2018 \\equiv 3 \\pmod{9}$. Then $a_{k+1}^2 \\equiv 0$, so $a_{k+2} \\equiv 0 + 2018 \\equiv 2 \\pmod{9}$. Again, there cannot be any more cubes.\n\nTherefore, there can be at most one cube in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18799,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Any affine function satisfies the condition. To prove the converse, set $y = x$:\n$$\n\\int_{0}^{2x} f(t) \\, dt = x(f(2x) + f(0)). \\quad (*)\n$$\nSince $f$ is integrable, $(*)$ shows that $g(x) = x(f(2x) + f(0))$ is continuous, so $f$ is continuous on $\\mathbb{R}^* = \\mathbb{R} \\setminus \\{0\\}$. Continuity and $(*)$ show $g$ is differentiable on $\\mathbb{R}^*$, so $f$ is differentiable on $\\mathbb{R}^*$.\n\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 $x = 0$ and $y = 1$ in the original relation:\n$$\n\\begin{aligned}\nf(1) + f(-1) &= \\int_{-1}^{1} f(t) \\, dt = \\int_{-1}^{0} (a t + f(0)) \\, dt + \\int_{0}^{1} (b t + f(0)) \\, dt \\\\\n&= \\frac{1}{2}(b - a) + 2 f(0).\n\\end{aligned}\n$$\nPlug $f(1) = b + f(0)$ and $f(-1) = -a + f(0)$ into the above to get $b - a + 2 f(0) = \\frac{1}{2}(b - a) + 2 f(0)$, so $a = b$ and $f(x) = a x + f(0)$ for all real $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18800,
    "subject": "Mathematics (Olympiad)",
    "question": "We say a pair $(g, h)$ of functions $g, h: \\mathbb{R} \\to \\mathbb{R}$ is a *tester pair* if the only function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(g(x)) = g(f(x))$ and $f(h(x)) = h(f(x))$ for all real $x$ is the identity function. Does there exist any tester pair?",
    "options": [],
    "answer": "See solution",
    "solution": "Such a tester pair exists. We may biject $\\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\\alpha, \\beta$ (which we will specify further later). Take\n\n$$\ng(x) = \\max(x - \\alpha, 0) \\quad \\text{and} \\quad h(x) = \\min(x + \\beta, 1).\n$$\n\nSay a set $S \\subseteq [0, 1]$ is invariant if $f(S) \\subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant: indeed, if $S$ is invariant and, say $T = g^{-1}(S)$, then $g(f(T)) = f(g(T)) \\subseteq f(S) \\subseteq S$, thus $f(T) \\subseteq T$.\n\nWe claim that (if we choose $\\alpha + \\beta < 1$) the intervals $[0, n\\alpha - m\\beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \\le n\\alpha - m\\beta \\le 1$. We prove this by induction on $m+n$.\n\nThe set $\\{0\\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0)) = f(g(0)) = f(0)$, so $f(0)$ is a fixed point of $g$. This gives that $f(0) = 0$, thus the induction base is established.\n\nSuppose now we have some $m, n$ such that $[0, n'\\alpha - m'\\beta]$ is invariant whenever $m'+n' < m+n$. At least one of the numbers $(n-1)\\alpha - m\\beta$ and $n\\alpha - (m-1)\\beta$ lies in $(0, 1)$. Note however that in the first case $[0, n\\alpha - m\\beta] = g^{-1}([0, (n-1)\\alpha - m\\beta])$, so $[0, n\\alpha - m\\beta]$ is invariant. In the second case $[0, n\\alpha - m\\beta] = h^{-1}([0, n\\alpha - (m-1)\\beta])$, so again $[0, n\\alpha - m\\beta]$ is invariant. This completes the induction.\n\nWe claim that if we choose $\\alpha + \\beta < 1$, where $0 < \\alpha \\notin \\mathbb{Q}$ and $\\beta = 1/k$ for some integer $k > 1$, then all intervals $[0, \\delta]$ are invariant for $0 \\le \\delta < 1$. This is the case, as by the previous claim, $[0, n\\alpha (\\text{mod } 1)]$ is invariant for all nonnegative integers $n$. The set of the $n\\alpha (\\text{mod } 1)$ is dense in $[0, 1]$, so in particular\n\n$$\n[0, \\delta] = \\bigcap_{n\\alpha (\\text{mod } 1) > \\delta} [0, n\\alpha (\\text{mod } 1)]\n$$\n\nis invariant.\n\nA similar argument shows $[\\delta, 1]$ invariant, so $\\{\\delta\\} = [0, \\delta] \\cap [\\delta, 1]$ is invariant for $0 < \\delta < 1$. Yet $\\{0\\}, \\{1\\}$ are both invariant, so $f$ must be the identity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18801,
    "subject": "Mathematics (Olympiad)",
    "question": "A collection of weights can be divided into 4 groups with equal masses, into 5 groups with equal masses, and into 9 groups with equal masses. Give an example of such a collection with the least possible number of weights. (Non-integer masses are allowed.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is 14.\n\nFirst, we prove that no collection of 13 weights is admissible. Assume, on the contrary, that 13 weights $a_1, a_2, \\dots, a_{13}$ can be divided into 9, 5, and 4 groups with equal masses (divisions 1, 2, and 3, respectively). Multiplying all $a_i$ by a positive number yields an admissible collection again, so suppose the total mass is $180 = 4 \\cdot 5 \\cdot 9$. (It is not assumed here that the $a_i$ are integers.)\n\nThe mass of one group in division 1 is $180/9 = 20$, hence $a_i \\leq 20$ for all $i$. At least 5 groups of the 9 groups in division 1 consist of exactly 1 weight, so there are 5 weights of mass 20, say $a_1, \\dots, a_5$. They are in different groups in division 2, where the mass of one group is $180/5 = 36$. Remove $a_1, \\dots, a_5$ from their groups to obtain a division of $a_6, \\dots, a_{13}$ into 5 groups of mass 16. In particular, $a_i \\leq 16$ for $6 \\leq i \\leq 13$.\n\nSo in division 1, the 8 weights $a_6, \\dots, a_{13}$ are divided into 4 pairs with mass 20 each, say $\\{a_6, a_7\\}, \\{a_8, a_9\\}, \\{a_{10}, a_{11}\\}, \\{a_{12}, a_{13}\\}$. There is also a division of $a_6, \\dots, a_{13}$ into 5 groups of mass 16; two of these groups must have exactly one weight. Hence, one may assume $a_6 = a_8 = 16$, implying $a_7 = a_9 = 4$. All $a_i$ determined so far are integer multiples of 4.\n\nEach group in division 3 has mass $180/4 = 45$, which is not a multiple of 4. Hence, each group contains a non-multiple of 4, i.e., an integer not divisible by 4 or a non-integer. As $a_1, \\dots, a_9$ are divisible by 4, the non-multiples of 4 are $a_{10}, a_{11}, a_{12}, a_{13}$, and they belong to different groups. We obtain that each of $a_{10}, a_{11}, a_{12}, a_{13}$ differs from 45 by a multiple of 4, so it is an integer congruent to 1 modulo 4. But then $a_{10} + a_{11} \\neq 20$, which is a contradiction.\n\n(The reasoning modulo 4 can be avoided. We proved that weights $a_6, \\dots, a_{13}$ form 5 groups of mass 16—hence each one has mass at most 16—and also 4 pairs with mass 20—so each one has mass at least 4. Two of the 5 weights with mass 20 are in the same group in division 3. The mass 45 of the group is completed by exactly one weight of mass 5: having two or more weights of total mass 5 would yield a mass less than 4. The weight 5 must be paired up with a weight 15 in a group of mass 20. However, the 15 is also in a group of mass 16, which is impossible by the above.)\n\nSo there is no admissible collection with 13 weights (and hence with fewer weights either). An admissible collection with 14 weights is $3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 20, 20, 20, 20$; divisions 1, 2, and 3 are:\n\n$$\n\\begin{aligned}\n&\\{3, 17\\},\\ \\{4, 16\\},\\ \\{5, 15\\},\\ \\{7, 13\\},\\ \\{9, 11\\},\\ \\{20\\},\\ \\{20\\},\\ \\{20\\},\\ \\{20\\} \\\\\n&\\{3, 13, 20\\},\\ \\{4, 15, 17\\},\\ \\{5, 11, 20\\},\\ \\{7, 9, 20\\},\\ \\{16, 20\\} \\\\\n&\\{3, 4, 7, 11, 20\\},\\ \\{5, 20, 20\\},\\ \\{9, 16, 20\\},\\ \\{13, 15, 17\\}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18802,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be the midpoint of $BC$, $M$ the midpoint of $AD$, and $N$ the projection of $D$ onto $BM$. Prove that $\\angle ANC = 90^\\circ$.\n\n![](images/Makedonija_2008_p31_data_efe953e2c2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the point such that $ABCD$ is a parallelogram. Then $ADCS$ is a rectangle and $R$ is the intersection point of the diagonals $AC$ and $DS$. The point $N$ lies on the diagonal $BS$ of the parallelogram $ABDS$, from which we obtain that $SND$ is a right triangle. The point $R$ is the circumcenter for the triangle $SND$, so $NR = \\frac{1}{2} DS = \\frac{1}{2} AC$. The angle $\\angle ANC = 90^\\circ$, i.e., $ANC$ is a right triangle, because $RA = RC = RN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18803,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the real sequence $x_n$ such that $x_1 \\in (0, \\frac{1}{2})$ and\n$$\nx_{n+1} = 3x_n^2 - 2n x_n^3, \\quad \\forall n \\ge 1.\n$$\n\na) Prove that $\\lim_{n \\to \\infty} x_n = 0$.\n\nb) For each $n \\ge 1$, let $y_n = x_1 + 2x_2 + \\dots + n x_n$. Prove that $(y_n)$ converges.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we prove by induction that $0 < x_n < \\frac{3}{2n}$ for all $n \\ge 1$.\n\nThe base case $n = 1$ is trivial. For $n = 2$, we have\n$$\n0 < x_1^2 (3 - 2x_1) < 3x_1^2 < \\frac{3}{4} \\implies 0 < x_2 < \\frac{3}{4}.\n$$\nAssume that $0 < x_k < \\frac{3}{2k}$ for some $k \\ge 2$. By the AM-GM inequality,\n$$\n0 < x_k^2 (3 - 2k x_k) = \\frac{1}{k^2} (k x_k)(k x_k)(3 - 2k x_k) \\le \\frac{1}{k^2} < \\frac{3}{2(k+1)}.\n$$\nTherefore, $0 < x_{k+1} < \\frac{3}{2(k+1)}$. The inductive step is completed.\n\nBy the Squeeze theorem, $\\lim_{n \\to \\infty} x_n = 0$.\n\nb) It is clear that $x_{n+1} = 3x_n^2 - 2n x_n^3 \\le 3x_n^2$, which implies\n$$\nx_{n+2} \\le 3x_{n+1}^2 \\le 27 x_n^4 \\le \\frac{3^7}{16 n^4} = \\frac{C}{n^4}, \\quad \\forall n \\ge 1.\n$$\nFor all $n > 2$, we obtain\n$$\ny_n = \\sum_{i=1}^{n} i x_i = x_1 + 2x_2 + \\sum_{i=1}^{n-2} (i+2) x_{i+2} \\\\\n\\le x_1 + 2x_2 + C \\sum_{i=1}^{n-2} \\frac{i+2}{i^4} \\le x_1 + 2x_2 + 3C \\sum_{i=1}^{n-2} \\frac{1}{i^2}.\n$$\nHence, $(y_n)$ is bounded above. Note that $(y_n)$ is increasing, so $(y_n)$ is convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18804,
    "subject": "Mathematics (Olympiad)",
    "question": "a) What is the last number in the fifth row of the triangle?\n\nb) In which row does the number 57 appear?\n\nc) What is the number of small triangles in the 20th row?\n\nd) Find a T-triangle where the T-sum is 80, given that the top number is less than the three bottom numbers, and the bottom numbers are each less than 20 or greater than 27.\n\n![](images/2023_Australian_Scene_p26_data_bfda354df5.png)\n\n![](images/2023_Australian_Scene_p26_data_0ba12b1147.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) By completing the first five rows, the last number in the fifth row is 25. Alternatively, the sum of small triangles in the top 5 rows is $1+3+5+7+9=25$.\n\nb) By completing the first eight rows, number 57 is in row 8. Alternatively, the sum of small triangles in rows 1 to 7 is $1+3+5+7+9+11+13=49$, and in rows 1 to 8 is $49+15=64$. Since $57$ is between $49$ and $64$, it must be in row 8.\n\nc) The number of small triangles in the 20th row is the 20th odd number, which is $39$.\n\nd) The top number in a T-triangle is less than the three bottom numbers. If the bottom numbers are each less than $20$, the T-sum is less than $4 \\times 20 = 80$. If the bottom numbers are each greater than $27$, the T-sum is greater than $3 \\times 27 = 81$. So the bottom three numbers must be in the 20s. Testing gives this T-triangle:\n\n![](images/2023_Australian_Scene_p26_data_bfda354df5.png)\n\nThe T-sum is $14+21+22+23=80$.\n\n![](images/2023_Australian_Scene_p26_data_0ba12b1147.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18805,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$, $b$, $c$ and prime $p$ such that\n\n$$\n2^a p^b = (p+2)^c + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $p$ is odd, $p \\ge 3$. If $c = 1$, then $p + 3 = 2^a p^b \\ge 2p \\ge p + 3$, equality holds only when $p = 3$, $a = b = 1$. We obtain a solution $(p, a, b, c) = (3, 1, 1, 1)$.\n\nNow assume $c \\ge 2$.\n\n**Case 1:** $c$ is odd.\n\nLet $q$ be a prime factor of $c$. Since\n\n$$\n(p + 2)^q + 1 \\mid (p + 2)^c + 1,\n$$\nwe have $(p + 2)^q + 1 = 2^\\alpha p^\\beta$. \\(\\textcircled{1}\\)\n\nObviously, $\\alpha > 0$. Observe that $(p+2)^q + 1 = (p+3)A$, where\n\n$$\n\\begin{align*}\nA &= (p+2)^{q-1} - (p+2)^{q-2} + \\dots + 1 \\\\\n&> (p+2)^{q-1} - (p+2)^{q-2} \\\\\n&= (p+2)^{q-2}(p+1) > p^{q-1},\n\\end{align*}\n$$\n\nand $A$ is odd. Hence, $A$ is a power of $p$, $A \\ge p^q$, $\\beta \\ge q$. Taking $\\textcircled{1}$ modulo $p$, we have\n\n$$\n2^q \\equiv -1 \\pmod{p},\n$$\n\nindicating that the order of $2$ modulo $p$ is $2$ or $2q$.\n\nIf the order is $2$, then $p = 3$. Now (1) becomes $5^q + 1 = 2^\\alpha 3^\\beta$. As $5^q + 1 \\equiv 2 \\pmod 4$, $\\alpha = 1$. By the lifting-the-exponent lemma, $v_3(5^q + 1) = v_3(5+1) + v_3(q) \\le 2$, so $\\beta \\le 2$. Checking $\\beta = 1, 2$, neither works.\n\nIf the order is $2q$, then $2q \\mid (p-1)$, so $q \\le \\frac{p-1}{2} < \\frac{p}{2}$. In (1), divide by $p^q$ and use $\\left(1+\\frac{2}{x}\\right)^x < e$ for $x \\ge 1$:\n\n$$\n2^{\\alpha} p^{\\beta-q} = \\left(1 + \\frac{2}{p}\\right)^{q} + p^{-q} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-q} < e + 3^{-3} < 3.\n$$\n\nSo $\\beta = q$, $\\alpha = 1$. Then $2 \\cdot p^q = (p+2)^q + 1 = (p+3)A$, so\n\n$$\nA = p^q, \\quad p+3=2,\n$$\n\na contradiction.\n\n**Case 2:** $c$ is even, $2^d \\nmid c$, $d \\ge 1$.\n\nThen $(p+2)^{2d} + 1 \\mid (p+2)^c + 1$, so\n\n$$\n(p+2)^{2d} + 1 = 2^{\\alpha} p^{\\beta}.\n$$\n\nSince $(p+2)^{2d} + 1 \\equiv 2 \\pmod 4$, $\\alpha = 1$:\n\n$$\n(p+2)^{2d} + 1 = 2 \\cdot p^{\\beta}. \\qquad (2)\n$$\n\nTaking (2) modulo $p$, $2^{2d} \\equiv -1 \\pmod p$, so the order of $2$ modulo $p$ is $2^{d+1}$, so $2^d < \\frac{p}{2}$. Also,\n\n$$\np^{\\beta+1} > 2 \\cdot p^{\\beta} = (p+2)^{2d} + 1 > p^{2d},\n$$\nso $\\beta \\ge 2^d$. In (2), divide by $p^{2d}$:\n\n$$\n2 \\cdot p^{\\beta - 2^d} = \\left(1 + \\frac{2}{p}\\right)^{2^d} + p^{-2^d} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-2^d} < e + 3^{-2} < 3.\n$$\n\nSo $\\beta = 2^d$.\n\nIf $d \\ge 2$, then $2^{d+1}|(p-1)$ so $p \\equiv 1 \\pmod 8$. By (2),\n\n$$\np^{2^d} - 1 = (p+2)^{2^d} - p^{2^d}.\n$$\n\nAnalyze 2-adic orders: $v_2(p^{2^d} - 1) = v_2(p^2 - 1) + d - 1 \\ge d + 3$, while\n\n$$\n\\begin{aligned} v_2((p+2)^{2^d} - p^{2^d}) &= v_2((p+2)^2 - p^2) + d - 1 \\\\ &= v_2(2) + v_2(2p+2) + d - 1 = d + 2, \\end{aligned}\n$$\n\na contradiction. Thus, $d=1$, $2p^2 = (p+2)^2 + 1$, yielding $p=5$.\n\nReturn to the original equation, $c=2k$ with $k$ odd, $a=1$:\n\n$$\n2 \\cdot 5^b = 7^{2k} + 1.\n$$\n\nBy LTE, $b = v_5(7^{2k} + 1) = v_5(7^2 + 1) + v_5(k) \\le 2 + \\frac{k}{5}$. If $k \\ge 3$,\n\n$$\n2 \\cdot 5^b \\le 2 \\cdot 5^{2+\\frac{k}{5}} = (7^2 + 1) \\cdot 5^{\\frac{k}{5}} < 7^{2k} + 1,\n$$\n\nwhich is impossible. So $k=1$, $c=2$, $b=2$, giving $(p, a, b, c) = (5, 1, 2, 2)$.\n\n**Conclusion:** The solutions are $(p, a, b, c) = (3, 1, 1, 1)$ and $(5, 1, 2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18806,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ have a point $O$ inside such that $\\angle BOC = 90^\\circ$ and $\\angle BAO = \\angle BCO$. Points $M$ and $N$ are the midpoints of segments $AC$ and $BC$, respectively. Prove that $\\angle OMN$ is a right angle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's draw circle $w$ with diameter $BC$. According to the problem statement, the point $O$ must be on this circle. Draw another circle $w_1$ congruent to $w$ that intersects $w$ at the endpoints of segment $BO$. Since $\\angle BAO = \\angle BCO$, point $A$ should be on circle $w_1$, and it's clear which arc it should be on.\n\n![](images/Ukrajina_2008_p10_data_90367a42af.png)\n\nDraw line $CO$ so that it intersects circle $w_1$ at point $D$. Then $\\triangle BOC$ and $\\triangle BOD$ are equal right-angled triangles. Since $BC$ is the diameter of circle $w$, $BD$ is the diameter of circle $w_1$.\n\nTherefore, $\\angle BAD$ is a right angle. Consider the dilation $H_c^{0.5}$ and draw circle $w_2 = H_c^{0.5}(w_1)$. As is evident, $M = H_c^{0.5}(A)$, $N = H_c^{0.5}(B)$, $O = H_c^{0.5}(D)$. Since circle $w_1$ passes through points $A$, $B$, $D$, circle $w_2$ passes through points $M$, $N$, $O$, and $ON$ is a diameter of this circle. Therefore, $\\angle OMN$ is a right angle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18807,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and define $S_n = \\{1, 2, \\dots, n\\}$. Consider a non-empty subset $T$ of $S_n$. We say that $T$ is balanced if the median of $T$ is equal to the average of $T$.\n\nFor example, for $n=9$, each of the subsets $\\{7\\}$, $\\{2, 5\\}$, $\\{2, 3, 4\\}$, $\\{5, 6, 8, 9\\}$, and $\\{1, 4, 5, 7, 8\\}$ is balanced; however, the subsets $\\{2, 4, 5\\}$ and $\\{1, 2, 3, 5\\}$ are not balanced.\n\nFor each $n \\ge 1$, prove that the number of balanced subsets of $S_n$ is odd.\n\n(To define the median of a set of $k$ numbers, first put the numbers in increasing order; then the median is the middle number if $k$ is odd, and the average of the two middle numbers if $k$ is even. For example, the median of $\\{1, 3, 4, 8, 9\\}$ is $4$, and the median of $\\{1, 3, 4, 7, 8, 9\\}$ is $(4+7)/2 = 5.5$.)",
    "options": [],
    "answer": "See solution",
    "solution": "We want to prove that there is an odd number of nonempty subsets $T$ of $S_n$ such that the average $A(T)$ and median $M(T)$ satisfy $A(T) = M(T)$.\n\nGiven a subset $T$, consider the subset $T^* = \\{n+1-t : t \\in T\\}$. It holds that $A(T^*) = n+1 - A(T)$ and $M(T^*) = n+1 - M(T)$, which implies that if $A(T) = M(T)$ then $A(T^*) = M(T^*)$. Pairing each set $T$ with $T^*$ yields that there are an even number of sets $T$ such that $A(T) = M(T)$ and $T \\neq T^*$.\n\nThus, it suffices to show that the number of nonempty subsets $T$ such that $A(T) = M(T)$ and $T = T^*$ is odd. Now note that if $T = T^*$, then $A(T) = M(T) = \\frac{n+1}{2}$. Hence, it suffices to show the number of nonempty subsets $T$ with $T = T^*$ is odd.\n\nGiven such a set $T$, let $T'$ be the largest nonempty subset of $\\{1, 2, \\dots, \\lceil n/2 \\rceil\\}$ contained in $T$. Pairing $T$ with $T'$ forms a bijection between these sets $T$ and the nonempty subsets of $\\{1, 2, \\dots, \\lceil n/2 \\rceil\\}$. Thus, there are $2^{\\lceil n/2 \\rceil} - 1$ such subsets, which is odd as desired. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18808,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest real number $C$ such that for any distinct positive integers $x, y$, the following holds:\n\n$$\n\\max\\left\\{\\{\\sqrt{x^2 + 2y}\\},\\ \\{\\sqrt{y^2 + 2x}\\}\\right\\} < C.\n$$\n\nHere, $\\{a\\} \\in [0, 1)$ denotes the fractional part of $a$, i.e., there exists an integer $n$ such that $a = n + \\{a\\}$. For example, $\\{3.14\\} = 0.14$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the minimal $C$ is the positive root of $x^2 + x = 1$, that is, $C = \\frac{\\sqrt{5} - 1}{2}$.\n\n**Proof:**\n\nSuppose for some positive integers $x < y$ both $\\{\\sqrt{x^2 + 2y}\\} > C$ and $\\{\\sqrt{y^2 + 2x}\\} > C$ hold. Note $y^2 < y^2 + 2x < (y + 1)^2$, so $y^2 + 2x > (y + C)^2$ implies $2x > 2Cy + C^2$, or $x > Cy$.\n\nSimilarly, $(x + 1)^2 < x^2 + 2y < (x + 2)^2$, so $x^2 + 2y > (x + 1 + C)^2$ gives $2y > 2(C + 1)x + (C + 1)^2$, or $y > (C + 1)x$.\n\nBut then $x > Cy$ and $y > (C + 1)x$ together yield a contradiction for positive $x, y$.\n\nNow, for any $C_1 < C$, set $y = [(C + 1)x]$ for large $x$. Then $\\{\\sqrt{x^2 + 2y}\\} > C_1$ and $\\{\\sqrt{y^2 + 2x}\\} > C_1$ for sufficiently large $x$.\n\nFor large $x$, $x^2 + 2y > (x + 1 + C_1)^2$ and $y^2 + 2x > (y + C_1)^2$ hold, since $2y - 2x(C_1 + 1) \\ge 2x(C - C_1) - 2$ and $2x - 2yC_1 < x(2 - (C + 1)C_1)$, which are both greater than the respective squares for large $x$.\n\nThus, $C$ is minimal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18809,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a \\ge 1$ and $b \\ge 2$ be given positive integers. Show that there does not exist any non-constant polynomial $f(x)$ with integer coefficients such that $f(n^a)$ and $f(b^n)$ are relatively prime for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that such an $f$ exists.\n\nSince $f(b^a)$ and $f(b^b)$ are relatively prime, so are $f(0)$ and $b$. Since $f$ is a non-constant polynomial, there exist a prime $p$ and a positive integer $m$ such that $p \\mid f(b^{am})$. It is clear that $b$ is relatively prime to $p$. We choose a positive integer $x_0$ such that\n\n$$\nx_0 \\equiv am - b^m \\pmod{p}\n$$\n\nand set $n = am + (p-1)x_0$. Then $f(n^a) \\equiv f((am - x_0)^a) \\equiv f(b^{ma}) \\equiv 0 \\pmod{p}$. On the other hand, by Fermat's little theorem, we have\n\n$$\nf(b^{ma}) \\equiv f(b^{am+(p-1)x_0}) \\equiv f(b^n) \\pmod{p},\n$$\n\nwhich gives a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18810,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $L$ and $O$ are drawn, meeting at $B$ and $C$, with $L$ on $O$. Ray $CO$ meets $L$ at $Q$, and $A$ is on $O$ such that $\\angle CQA = 90^\\circ$. The angle bisector of $\\angle AOB$ meets $L$ at $X$ and $Y$. Show that $\\angle XLY = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we compute the length $CQ$ in two ways; by angle chasing one can show $\\angle CBQ = 180^\\circ - (\\angle BQC + \\angle QCB) = \\frac{1}{2}\\angle A$, and so\n\n$$\n\\begin{aligned}\nAC \\sin B &= CQ = \\frac{BC}{\\sin(90^\\circ + \\frac{1}{2}\\angle A)} \\cdot \\sin \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow \\sin^2 B &= \\frac{\\sin A \\cdot \\sin \\frac{1}{2}\\angle A}{\\cos \\frac{1}{2}\\angle A} \\\\\n\\Leftrightarrow \\sin^2 B &= 2 \\sin^2 \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow \\sin B &= \\sqrt{2} \\sin \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow 2R \\sin B &= \\sqrt{2} \\left( 2R \\sin \\frac{1}{2}\\angle A \\right) \\\\\n\\Leftrightarrow AC &= \\sqrt{2}x\n\\end{aligned}\n$$\n\nas desired (we have here used the fact $\\triangle ABC$ is acute to take square roots).\n\nIt is interesting to note that $\\sin^2 B = 2 \\sin^2 \\frac{1}{2}\\angle A$ can be rewritten as\n\n$$\n\\cos A = \\cos^2 B\n$$\n\nsince $\\cos^2 B = 1 - \\sin^2 B = 1 - 2 \\sin^2 \\frac{1}{2} \\angle A = \\cos A$; this is the condition for the existence of the point $Q$. $\\square$\n\nWe finish by proving that\n\n$$\nKD = KA\n$$\n\nand hence line $\\overline{KD}$ is tangent to $\\gamma$. Let $E = \\overline{BC} \\cap \\overline{KL}$. Then\n\n$$\nLE \\cdot LK = LC^2 = LX^2 = \\frac{1}{2}LK^2\n$$\n\nand so $E$ is the midpoint of $\\overline{LK}$. Thus $\\overline{MXOY}$, $\\overline{BC}$, $\\overline{KL}$ are concurrent at $E$. As $\\overline{DL} \\parallel \\overline{KC}$, we find that $DLCK$ is a parallelogram, so $KD = CL = KA$ as well. Thus $\\overline{KD}$ and $\\overline{KA}$ are tangent to $\\gamma$.\n\n**Remark.** The condition $\\angle A \\neq 60^\\circ$ cannot be dropped, since if $Q = O$ the problem is not true.\n\nOn the other hand, nearly all solutions begin by observing $Q \\neq O$ and then obtaining $\\angle AQO = 90^\\circ$. This gives a way to construct the diagram by hand with ruler and compass. One draws an arbitrary chord $\\overline{BC}$ of a circle $\\omega$ centered at $L$, and constructs $O$ as the circumcenter of $\\triangle BLC$ (hence obtaining $\\Gamma$). Then $Q$ is defined as the intersection of ray $CO$ with $\\omega$, and $A$ is defined by taking the perpendicular line through $Q$ on the circle $\\Gamma$. In this way we can draw a triangle $ABC$ satisfying the problem conditions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18811,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram, and let $M$ be the midpoint of $AB$. Line $CM$ intersects the circumcircle of triangle $ABC$ at $C$ and $E$. Let $F$ be the point on $BC$ such that $AF \\perp BC$. Prove that $C$, $D$, $E$, and $F$ are concyclic.\n\n![](images/THA_National_2016_p16_data_4f9ac9bc0d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the notation $\\angle ABC$ to mean the directed angle $\\angle (BA, BC)$.\n\nSince $E$, $A$, $C$, $B$ are concyclic and $AD$ is parallel to $BC$, $\\angle EAB = \\angle ECB$ and $\\angle BAD = -\\angle ABC$. Therefore,\n\n$$\n\\begin{align*}\n\\angle EAD &= \\angle EAB + \\angle BAD \\\\\n&= \\angle ECB - \\angle ABC \\\\\n&= \\angle MCF - \\angle ABC. \\tag{13}\n\\end{align*}\n$$\n\nSince $M$ is the midpoint of $AB$ and $AF \\perp BC$, $M$ is the circumcenter of triangle $ABF$. We then have\n\n$$\nMF = MB \\quad \\text{and} \\quad \\angle ABC = -\\angle MFC. \\tag{14}\n$$\n\nFrom (13) and (14),\n\n$$\n\\begin{align*}\n\\angle EAD &= \\angle MCF - \\angle MFC \\\\\n&= \\angle MCF + \\angle CFM \\\\\n&= -\\angle FMC \\\\\n&= \\angle EMF. \\tag{15}\n\\end{align*}\n$$\n\nSince $E$, $A$, $C$, $B$ are concyclic, we get $\\triangle AEM \\sim \\triangle CBM$. Therefore, $\\frac{AE}{EM} = \\frac{CB}{BM}$.\nFrom $CB = AD$ and $BM = MF$, we get $\\frac{AE}{EM} = \\frac{AD}{MF}$. So,\n\n$$\n\\frac{AE}{AD} = \\frac{EM}{MF}. \\tag{16}\n$$\n\nFrom (15) and (16), we obtain $\\triangle AED \\sim \\triangle MEF$. Therefore, $\\angle ADE = \\angle MFE$.\nSo, we get\n\n$$\n\\begin{align*}\n\\angle EDC &= \\angle ADC - \\angle ADE \\\\\n&= \\angle ADC - \\angle MFE. \\tag{17}\n\\end{align*}\n$$\n\nApplying (17) and $\\angle ABC = -\\angle ADC$ from the parallelogram $ABCD$ we obtain,\n\n$$\n\\begin{align*}\n\\angle EDC &= \\angle ADC - \\angle MFE \\\\\n&= \\angle MFC - \\angle MFE \\\\\n&= \\angle EFC.\n\\end{align*}\n$$\n\nTherefore, $C$, $D$, $E$, and $F$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18812,
    "subject": "Mathematics (Olympiad)",
    "question": "Extend $BA$ and $DE$ so that they meet at point $P$.\n\n![](images/Argentina_2019_Booklet_p7_data_61572af46f.png)\n\nGiven: $AB = 28$, $BC = 15$, $DE = 10$, and $EA = 13$.\n\nCalculate the area of pentagon $ABCDE$.",
    "options": [],
    "answer": "See solution",
    "solution": "The quadrilateral $PBCD$ has three right angles, so the fourth is also a right angle; hence, it is a rectangle. Thus, $PD = BC = 15$, so $PE = PD - DE = 15 - 10 = 5$.\n\nApplying Pythagoras' theorem in right triangle $APE$:\n\n$$\nAP = \\sqrt{AE^2 - PE^2} = \\sqrt{13^2 - 5^2} = \\sqrt{144} = 12.\n$$\n\nTherefore, $PB = PA + AB = 12 + 28 = 40$.\n\nThe area of pentagon $ABCDE$ is the area of rectangle $PBCD$ minus the area of triangle $APE$:\n\n$$\n\\text{area}(ABCDE) = 40 \\times 15 - \\frac{12 \\times 5}{2} = 600 - 30 = 570.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18813,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 5 rooms, each occupied by 2 interpreters. Each interpreter is proficient in a pair of different languages $X, Y$, denoted $I_{XY}$. For each room, the language that both occupants can speak is called the **common** language of the room. For each language $L$, exactly 4 interpreters are proficient in $L$. How many ways are there to assign interpreters to rooms so that these conditions are satisfied?",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze two cases:\n\n**Case (i):** All 5 rooms have different common languages.\n\nSince there are 5 languages and 5 rooms, each language is the common language of exactly one room. Assign interpreters so that each room's occupants share a unique language. There are $\\frac{4!}{2!2!}$ ways to choose the other languages for the German room, and $2 \\times 2$ ways to assign $I_{AB}$ and $I_{CD}$ to rooms, giving $\\frac{4!}{2!2!} \\times 2 \\times 2 = 24$ assignments.\n\n**Case (ii):** Two rooms share the same common language.\n\nLet $L$ be the common language for two rooms, and $A, B, C$ the common languages for the other three rooms. The remaining language is $X$. Assign interpreters so that $I_{XA}$, $I_{XB}$, $I_{XC}$ go to $R_A$, $R_B$, $R_C$ respectively. $I_{AB}$ can go to $R_A$ or $R_B$, and the rest are determined. The 4 interpreters proficient in $L$ can be distributed among the two $L$ rooms in $\\frac{4!}{2!2!} = 3$ ways. For each choice of $L$ and $X$ ($5 \\times 4 = 20$), and for each, $6$ assignments, totaling $6 \\times 20 = 120$.\n\n**Total:** $24 + 120 = 144$ room assignments.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18814,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$x^3 + 2y^3 - 4x - 5y + z^2 = 2012$$\n\nin the set of whole numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to show that for every whole number $a$, $3 \\mid (a^3 - a)$, and that the square of a number modulo $3$ can be $0$ or $1$.\n\nNow, the given equation is equivalent to $x^3 - x + 2(y^3 - y) - 3(x - y) + z^2 = 2012$. From the above and from the fact that $2012 \\equiv 2 \\pmod{3}$, we get that $z^2 \\equiv 2 \\pmod{3}$, which is impossible. Hence, the given equation has no solution in the set of whole numbers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18815,
    "subject": "Mathematics (Olympiad)",
    "question": "Given three functions:\n\n$$\nP(x) = (x^2-1)^{2023}, \\quad Q(x) = (2x+1)^{14}, \\quad R(x) = \\left(2x+1+\\frac{2}{x}\\right)^{34}\n$$\n\nInitially, we pick a set $S$ containing two of these functions, and we perform some operations on it. Allowed operations include:\n\n- Take two functions $p, q \\in S$ and add one of $p+q$, $p-q$, or $pq$ to $S$.\n- Take a function $p \\in S$ and add $p^k$ to $S$ for any positive integer $k$.\n- Take a function $p \\in S$ and choose a real number $t$, and add to $S$ one of the functions $p+t$, $p-t$, or $pt$.\n\nShow that no matter how we pick $S$ in the beginning, there is no way we can perform finitely many operations on $S$ that would eventually yield the third function not in $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, from $P(x)$ and $Q(x)$, after all of the allowed operations, we only obtain polynomial functions in $x$, while $R(x)$ is not a polynomial. Thus, it is not possible to obtain $R(x)$ from $P(x)$ and $Q(x)$:\n\n$$\nP, Q \\nrightarrow R.\n$$\n\nNext, we show that $R$ and $P$ cannot generate $Q$. Consider the derivatives:\n\n$$\nP'(x) = 4046x(x^2-1)^{2022}, \\quad R'(x) = 34\\left(2-\\frac{2}{x^2}\\right)\\left(2x+1+\\frac{2}{x}\\right)^{33}\n$$\n\nClearly, $P'(\\pm 1) = R'(\\pm 1) = 0$. The allowed operations generate functions that are compositions of $P(x)$ and $R(x)$, so their derivatives will also vanish at $x = \\pm 1$. However, $Q(x)$ does not satisfy this, since\n\n$$\nQ'(x) = 14 \\cdot 2 \\cdot (2x+1)^{13}.\n$$\n\nFinally, we show that $R$ and $Q$ cannot generate $P$ using polynomial congruence modulo $f(x) = x^2 + x + 1$. If $P(x) - Q(x)$ is divisible by $f(x)$, we write\n\n$$\nP(x) \\equiv Q(x) \\pmod{f(x)}.\n$$\n\nThe properties of integer congruence hold here, and we can extend congruence to fractions. Now,\n\n$$\nQ(x) = (2x+1)^{14} = (4x^2+4x+1)^7 \\equiv (-3)^7 \\pmod{f(x)},\n$$\n\n$$\nR(x) = \\left(\\frac{2(x^2+1)}{x} + 1\\right)^{34} \\equiv (-2+1)^{34} \\equiv 1 \\pmod{f(x)}.\n$$\n\nTherefore, all functions generated by $Q$ and $R$, when reduced modulo $f(x)$, are constant. We show that $P(x)$ does not have this property. Note that $x^6 \\equiv 1 \\pmod{f(x)}$, so\n\n$$\n(x^2 - 1)^2 \\equiv (-x - 2)^2 \\equiv x^2 + 4x + 4 \\equiv x^2 - 4x^2 \\equiv -3x^2,\n$$\n\n$$\n(x^2 - 1)^6 \\equiv (-3x^2)^3 = -27x^6 \\equiv -27 \\pmod{f(x)}.\n$$\n\nThus,\n\n$$\nP(x) \\equiv (x^2 - 1)^{2022} (x^2 - 1) \\equiv (-27)^{337} (-x - 2) = 27^{337}(x + 2) \\not\\equiv \\text{const} \\pmod{f(x)}.\n$$\n\nTherefore, it is impossible to generate the third function from the other two using the allowed operations. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18816,
    "subject": "Mathematics (Olympiad)",
    "question": "A three-digit natural number $n$ is initially written on the board. Two players, A and B, take turns, with A going first. On each turn, the player reduces the number on the board by some proper divisor of the current number (i.e., a divisor other than 1 and the number itself). For example, if the number on the board is 6, it can be reduced by 2, resulting in 4. Whoever cannot make a move loses, and the other wins. It is known that both player A and player B have a way to win. What are all possible values of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "If a prime number is on the board, the player cannot make a move and loses. If the number is even and not a power of 2, the player can always subtract an odd divisor, leaving an odd number. If the number is odd and is reduced by an odd divisor $a$ (i.e., $ab$ is replaced by $a(b-1)$), the result is even and not a power of 2. Thus, starting from an even number that is not a power of 2, a player can always move to ensure the same type of number on their next turn. Therefore, such even numbers are winning positions, and odd numbers are losing positions.\n\nNow, consider $n = 2^m$ for some natural $m$. The only proper divisors are $2^k$ for $k < m$. If $k < m-1$, the resulting number $2^k(2^{m-k} - 1)$ is even and not a power of 2, which is a winning move for the opponent, so this is not optimal. Thus, the best move is $k = m-1$. Playing this way, one player will get even powers of 2, the other odd powers. Since 2 is a losing position, even powers of 2 are winning, odd powers are losing.\n\nTherefore, the suitable $n$ are the even numbers that are not odd powers of 2. Among three-digit numbers, there are $900 \\div 2 - 2 = 448$ such numbers (excluding $128 = 2^7$ and $512 = 2^9$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18817,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = abcd = a \\cdot 10^3 + b \\cdot 10^2 + c \\cdot 10 + d$ be a four-digit positive integer such that $a \\geq 7$ and $a > b > c > d > 0$. Consider the positive integer $B = \\overline{dcba} = d \\cdot 10^3 + c \\cdot 10^2 + b \\cdot 10 + a$. If all digits of $A + B$ are odd, determine all possible values of $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\nA + B = (a + d) \\cdot 10^3 + (b + c) \\cdot 10^2 + (b + c) \\cdot 10 + (a + d).\n$$\n\nAll digits of $A + B$ are odd. To find the digits of $A + B$, we must know if $a + d$ and $b + c$ are less than 10. Consider the cases:\n\n**(α)** $a + d \\geq 10$ and $b + c \\geq 10$:\n\nLet $a + d = 10 + k$, $k = 0,1,2,\\ldots,5$ and $b + c = 10 + \\ell$, $\\ell = 0,1,2,\\ldots,5$.\n\n$$\n\\begin{aligned}\nA + B &= (10 + k) \\cdot 10^3 + (10 + \\ell) \\cdot 10^2 + (10 + \\ell) \\cdot 10 + (10 + k) \\\\\n&= 10^4 + (k + 1) \\cdot 10^3 + (\\ell + 1) \\cdot 10^2 + (\\ell + 1) \\cdot 10 + k.\n\\end{aligned}\n$$\n\nThis gives digits $1, k+1, \\ell+1, \\ell+1, k$ (all must be odd), which is impossible.\n\n**(β)** $a + d \\geq 10$ and $b + c < 10$:\n\nLet $a + d = 10 + k$, $k = 0,1,2,\\ldots,5$.\n\n$$\n\\begin{aligned}\nA + B &= (10 + k) \\cdot 10^3 + (b + c) \\cdot 10^2 + (b + c) \\cdot 10 + (10 + k) \\\\\n&= 10^4 + k \\cdot 10^3 + (b + c) \\cdot 10^2 + (b + c + 1) \\cdot 10 + k.\n\\end{aligned}\n$$\n\nIf $b + c = 9$, then $A + B$ has a digit 0 (impossible). If $b + c < 9$, then $b + c$ and $b + c + 1$ cannot both be odd.\n\n**(γ)** $a + d < 10$ and $b + c \\geq 10$:\n\nLet $b + c = 10 + \\ell$, $\\ell = 0,1,2,\\ldots,5$.\n\n$$\nA + B = (a + d) \\cdot 10^3 + (10 + \\ell) \\cdot 10^2 + (10 + \\ell) \\cdot 10 + (a + d) = (a + d + 1) \\cdot 10^3 + (\\ell + 1) \\cdot 10^2 + \\ell \\cdot 10 + (a + d).\n$$\n\nHere, $\\ell$ and $\\ell + 1$ cannot both be odd.\n\n**(δ)** $a + d < 10$ and $b + c < 10$:\n\nBoth $a + d$ and $b + c$ must be odd. Since $a > b > c > d > 0$ and $a \\geq 7$, it follows $a + d = 9$. For $b$ and $c$, $b + c \\in \\{5, 7, 9\\}$.\n\nPossible cases:\n\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 9$ with $b = 7, c = 2$; $b = 6, c = 3$; $b = 5, c = 4$:\n  - $A = 8721$, $A = 8631$, $A = 8541$\n- $a + d = 9$ with $a = 7, d = 2$ and $b + c = 9$ with $b = 6, c = 3$; $b = 5, c = 4$:\n  - $A = 7632$, $A = 7542$\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 7$ with $b = 5, c = 2$; $b = 4, c = 3$:\n  - $A = 8521$, $A = 8431$\n- $a + d = 9$ with $a = 7, d = 2$ and $b + c = 7$ with $b = 4, c = 3$:\n  - $A = 7432$\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 5$ with $b = 3, c = 2$:\n  - $A = 8321$\n\n**Final answer:**\n\nThe possible values of $A$ are $8721$, $8631$, $8541$, $7632$, $7542$, $8521$, $8431$, $7432$, $8321$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18818,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCDEF$ 為凸六邊形，其中 $AB = DE$，$BC = EF$，$CD = FA$，並且 $\\angle A - \\angle D = \\angle C - \\angle F = \\angle E - \\angle B$。\n\nProve that the diagonals $AD$, $BE$, and $CF$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "記 $\\theta = \\angle A - \\angle D = \\angle C - \\angle F = \\angle E - \\angle B$。不失一般性，設 $\\theta \\ge 0$。\n\n**解 1.** 令 $x = AB = DE$，$y = CD = FA$，$z = EF = BC$。考慮三個點 $P, Q, R$ 使得 $CDEP$、$EFAQ$、$ABCR$ 都是平行四邊形。計算可得：\n\n$$\n\\begin{aligned}\n\\angle PEQ &= \\angle FEQ + \\angle DEP - \\angle E \\\\\n&= (180^\\circ - \\angle F) + (180^\\circ - \\angle D) - \\angle E \\\\\n&= 360^\\circ - \\angle D - \\angle E - \\angle F \\\\\n&= \\frac{1}{2}(\\angle A + \\angle B + \\angle C - \\angle D - \\angle E - \\angle F) \\\\\n&= \\theta/2.\n\\end{aligned}\n$$\n\n類似有 $\\angle QAR = \\angle RCP = \\theta/2$。\n\n若 $\\theta = 0$，因為 $\\triangle RCP$ 是等腰三角形，故有 $R = P$。所以 $AB \\parallel RC = PC \\parallel ED$，得 $ABDE$ 為平行四邊形。同理可得 $BCEF$ 和 $CDFA$ 也都是平行四邊形。由此知 $AD, BE, CF$ 三線交於它們的共同中點。\n\n以下假設 $\\theta > 0$。因為 $\\triangle PEQ, \\triangle QAR$ 以及 $\\triangle RCP$ 都是等腰三角形，且它們的頂角皆相同，所以 $\\triangle PEQ \\sim \\triangle QAR \\sim \\triangle RCP$，其比例為 $y : z : x$。\n\n因此 $\\triangle PQR$ 相似於一個邊長為 $y, z, x$ 的三角形。\n\n接下來注意到：\n\n$$\n\\frac{RQ}{QP} = \\frac{z}{y} = \\frac{RA}{AF}\n$$\n\n以及（使用射線所夾的有向角）：\n\n$$\n\\begin{aligned}\n(RQ, QP) &= (RQ, QE) + (QE, QP) \\\\\n&= (RQ, QE) + (RA, RQ) \\\\\n&= (RA, QE) = (RA, AF).\n\\end{aligned}\n$$\n\n於是 $\\triangle PQR \\sim \\triangle FAR$。由於 $FA = y$ 和 $AR = z$，可得 $FR = x$。同理可知 $FP = x$。所以 $CRFP$ 是菱形。由此可知 $CF$ 是 $PR$ 的中垂線。同理得 $BE$ 是 $PQ$ 的中垂線，$AD$ 是 $QR$ 的中垂線。故 $AD, BE, CF$ 三線共點於 $\\triangle PQR$ 的外心。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18819,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $f^n(x)$ the result of applying the function $f$ $n$ times to $x$ (e.g., $f^1(x) = f(x)$, $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc). Find all functions from real numbers to real numbers which satisfy $f^d(x) = 2015 - x$ for all divisors $d$ of $2015$ greater than $1$, and for all real $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $5$ is a divisor of $2015$, we have for any real $z$:\n\n$$\n\\begin{aligned}\nf^{25}(z) &= f^5(f^5(f^5(f^5(z)))) \\\\\n&= 2015 - f^5(f^5(f^5(z))) \\\\\n&= 2015 - (2015 - f^5(f^5(z))) \\\\\n&= f^5(f^5(z)) \\\\\n&= \\dots = f^5(z) = 2015 - z.\n\\end{aligned}\n$$\n\nSince $13$ is also a divisor of $2015$, we have for any real $z$:\n\n$$\nf^{26}(z) = f^{13}(f^{13}(z)) = 2015 - f^{13}(z) = 2015 - (2015 - z) = z.\n$$\n\nConsequently, $z = f^{26}(z) = f(f^{25}(z)) = f(2015 - z)$. Any real number $x$ can be written as $2015 - z$ for $z = 2015 - x$. Hence $z = f(2015 - z)$ implies $f(x) = 2015 - x$ for any real $x$.\n\nFinally, check that the function $f(x) = 2015 - x$ satisfies the conditions of the problem. Let $d$ be a divisor of $2015$ greater than $1$. Then $d$ is odd, i.e., $d = 2c + 1$ for a positive integer $c$. Since $f^2(x) = 2015 - (2015 - x) = x$, we have $f^{2c}(x) = x$, which implies $f^d(x) = f(f^{2c}(x)) = f(x) = 2015 - x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18820,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of positive integers $(n, k)$ for which\n\n$$\nn! + n = n^k\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $n! + n > n$, we immediately get $k \\ge 2$. Dividing both sides by $n$ gives\n\n$$\n(n - 1)! + 1 = n^{k-1}.\n$$\n\nNow, consider two cases:\n\n*If $n$ is not a prime:*\n\nSince $n \\neq 1$, write $n = ab$ for integers $1 < a, b < n$, so $1 < a \\le n - 1$ and $a \\mid (n - 1)!$. Thus, $a > 1$ is relatively prime to $(n - 1)! + 1$, but $a$ divides $n^{k-1}$, which is impossible. So, no solutions in this case.\n\n*If $n$ is a prime:*\n\nCheck $n = 2, 3, 5$:\n- $n = 2$: $2! + 2 = 4 = 2^2$ $\\to$ $(2, 2)$\n- $n = 3$: $3! + 3 = 9 = 3^2$ $\\to$ $(3, 2)$\n- $n = 5$: $5! + 5 = 125 = 5^3$ $\\to$ $(5, 3)$\n\nFor $n \\ge 7$ (prime),\n\n$$\n\\begin{align*}\n(n-1)! &= n^{k-1} - 1 \\\\\n&= (1 + n + n^2 + \\dots + n^{k-2})(n-1) \\\\\n(n-2)! &= 1 + n + n^2 + \\dots + n^{k-2}\n\\end{align*}\n$$\n\nSince $n-1$ is even and not a prime or a square of a prime (for $n \\ge 7$), $n-1 = ab$ with $1 < a, b \\le n-1$, $a \\ne b$. Thus, $(n-2)!$ is divisible by $n-1$, so $(n-2)! \\equiv 0 \\pmod{n-1}$. Also, $n \\equiv 1 \\pmod{n-1}$, so\n\n$$\n0 \\equiv 1 + 1 + \\dots + 1^{k-2} \\equiv k-1 \\pmod{n-1}.\n$$\n\nThus, $n-1 \\mid k-1$, so $k-1 = l(n-1)$ for some $l \\ge 1$. But\n\n$$\n(n-1)! < (n-1)^{n-1},\n$$\nso\n$$\nn^{k-1} = (n-1)! + 1 \\le (n-1)^{n-1} < n^{n-1} \\le n^{k-1},\n$$\nwhich is a contradiction. So, no further solutions.\n\n*Final answer:* The only solutions are $(2, 2)$, $(3, 2)$, and $(5, 3)$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18821,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$ be the $A$-altitude of an acute-angled triangle $ABC$. The internal bisector of angle $DAC$ intersects $BC$ at $K$. Let $L$ be the projection of $K$ onto $AC$. Let $M$ be the intersection point of $BL$ and $AD$. Let $P$ be the intersection point of $MC$ and $DL$. Prove that $PK \\perp AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be a point on $BC$ such that $LX \\perp AB$, as seen in the figure below. It is enough to prove that\n\n$$\n\\frac{DP}{PL} = \\frac{DK}{KX}\n$$\n\nbecause then $PK \\parallel LX$ and $LX \\perp AB$.\n\nApplying Menelaus' theorem to triangle $BDL$ with transversal $MPC$, we get\n\n$$\n\\frac{DP}{PL} \\cdot \\frac{LM}{MB} \\cdot \\frac{BC}{CD} = 1.\n$$\n\nApplying Menelaus' theorem to triangle $BLC$ with transversal $AMD$, we get\n\n$$\n\\frac{BM}{ML} \\cdot \\frac{LA}{AC} \\cdot \\frac{CD}{DB} = 1.\n$$\n\nMultiplying these two equalities yields\n\n$$\n\\frac{DP \\cdot BC \\cdot AL}{PL \\cdot BD \\cdot AC} = 1.\n$$\n\nNote, however, that $AL = AD = AC \\sin \\gamma$, $BD = AB \\cos \\beta$, and, by the sine rule, $\\frac{AB}{BC} = \\frac{\\sin \\gamma}{\\sin \\alpha}$, where $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, and $\\gamma = \\angle ACB$. Therefore,\n\n$$\n\\frac{DP}{PL} = \\frac{BD \\cdot AC}{BC \\cdot AL} = \\frac{AB \\cos \\beta \\cdot AC}{BC \\cdot AC \\sin \\gamma} = \\frac{\\sin \\gamma \\cos \\beta}{\\sin \\alpha \\sin \\gamma} = \\frac{\\cos \\beta}{\\sin \\alpha}.\n$$\n\nOn the other hand, since $DK = KL$, $\\angle KLX = \\pi - \\alpha$, and $\\angle LXX = \\frac{\\pi}{2} - \\beta$, we have by the sine rule\n\n![Figure 18](images/BW2021_Shortlist_p32_data_9c21629e45.png)\n\n$$\n\\frac{DK}{KX} = \\frac{LK}{KX} = \\frac{\\sin\\left(\\frac{\\pi}{2} - \\beta\\right)}{\\sin(\\pi - \\alpha)} = \\frac{\\cos \\beta}{\\sin \\alpha}.\n$$\n\nTherefore,\n\n$$\n\\frac{DP}{PL} = \\frac{\\cos \\beta}{\\sin \\alpha} = \\frac{DK}{KX}\n$$\n\nwhich finishes the proof.\n\n![Figure](images/BW2021_Shortlist_p33_data_4a6c3d9557.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18822,
    "subject": "Mathematics (Olympiad)",
    "question": "An information station employs four different codes, A, B, C, and D, for communication, but each week uses only one of them. The code used in a definite week is randomly selected with equal chance among the three codes that have not been used in the last week. Suppose the code used in the first week is A. What is the probability that A is also used in the seventh week? (Express your answer as an irreducible fraction.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P_k$ denote the probability that code A is used in the $k$th week. Then the probability that A is not used in the $k$th week is $1 - P_k$. Therefore,\n\n$$\nP_{k+1} = \\frac{1}{3}(1 - P_k).\n$$\n\nOr,\n\n$$\nP_{k+1} - \\frac{1}{4} = -\\frac{1}{3}\\left(P_k - \\frac{1}{4}\\right).\n$$\n\nAs $P_1 = 1$, $\\{P_k - \\frac{1}{4}\\}$ is a geometric sequence with $\\frac{3}{4}$ as the first term and $-\\frac{1}{3}$ as the common ratio. So,\n\n$$\nP_k - \\frac{1}{4} = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1}.\n$$\n\nOr,\n\n$$\nP_k = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1} + \\frac{1}{4}.\n$$\n\nTherefore, $P_7 = \\frac{61}{243}$.\n\nThe answer is $\\frac{61}{243}$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p54_data_fcf166879d.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18823,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that each integer $N$ has a unique decomposition $$N = \\sum_{k=0}^{n} a_k 2^k,$$ where $n$ is a non-negative integer, $a_k \\in \\{-1, 0, 1\\}$, and no two consecutive $a_k$ are nonzero.",
    "options": [],
    "answer": "See solution",
    "solution": "It is sufficient to prove the statement for positive $N$. Consider the unique decomposition\n\n$$N = 2^{k_1} + 2^{k_2} + \\dots + 2^{k_r}, \\quad 0 \\leq k_1 < k_2 < \\dots < k_r.$$  \n\nReduce this decomposition by repeatedly applying (as far left as possible) one of the following equalities:\n\n$$2^t + 2^{t+1} = -2^t + 2^{t+2}, \\quad 2^t - 2^{t+1} = -2^t, \\quad 2^t + 2^t = 2^{t+1}.$$  \n\nWhen no further reductions are possible, the decomposition becomes:\n\n$$N = \\sum_{a \\in A} 2^a - \\sum_{b \\in B} 2^b,$$\n\nwhere $A$ and $B$ are finite disjoint sets of non-negative integers whose union contains no two adjacent integers.\n\nTo prove uniqueness, suppose\n\n$$N = \\sum_{c \\in C} 2^c - \\sum_{d \\in D} 2^d,$$\n\nwith $C$ and $D$ also finite disjoint sets of non-negative integers whose union contains no two adjacent integers. Then\n\n$$\\sum_{a \\in A} 2^a + \\sum_{d \\in D} 2^d = \\sum_{b \\in B} 2^b + \\sum_{c \\in C} 2^c.$$  \n\nIf $x \\in A \\cap D$, replace $2^x + 2^x$ with $2^{x+1}$ on the left. Since $x+1$ is not in $A \\cup D$, these replacements do not interfere. Similarly, if $y \\in B \\cap C$, replace $2^y + 2^y$ with $2^{y+1}$ on the right. After all such replacements, we have\n\n$$\\sum_{x \\in X} 2^x = \\sum_{y \\in Y} 2^y,$$\n\nso $X = Y$ by uniqueness of binary decompositions.\n\nNow, we show $A = C$ and $B = D$. It suffices to prove $A \\subseteq C$; the other cases are similar. Let $a \\in A$:\n\n- If $a \\notin D$, then $2^a$ appears on the left and must also appear on the right. Since $a-1 \\notin B$, $2^a$ cannot be the result of $2^{a-1} + 2^{a-1}$, so $2^a$ must already be present on the right, and since $A$ and $B$ are disjoint, $a \\in C$.\n- If $a \\in D$, then $2^{a+1}$ appears on the left and must also appear on the right. Since $a \\in A \\cap D$, $a+1 \\in B \\cap C$, so $2^{a+1}$ on the right must be the result of $2^a$ from $\\sum_{b \\in B} 2^b$ and $2^a$ from $\\sum_{c \\in C} 2^c$, whence $a \\in C$.\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18824,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be a point on the side $BC$ of an acute triangle $ABC$. The circle with diameter $BD$ meets the lines $AB$ and $AD$ respectively at the points $X$ and $P$, which are different from the points $B$ and $D$. The circle with diameter $CD$ meets the lines $AC$ and $AD$ respectively at the points $Y$ and $Q$, which are different from the points $C$ and $D$. Through the point $A$ draw two lines which are perpendicular to $PX$ and $QY$ with the feet of perpendicular $M$ and $N$ respectively.\n\nProve that $\\triangle AMN \\sim \\triangle ABC$ if and only if the line $AD$ passes through the circumcenter of $\\triangle ABC$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p158_data_3ecc4170e9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Join the segments $XY$ and $DX$. It follows from the given conditions that $B, P, D, X$ are concyclic, and $C, Y, Q, D$ are concyclic. Then\n\n$$\n\\begin{aligned}\n\\angle AXM &= \\angle BXP = \\angle BDP \\\\\n&= \\angle QDC = \\angle AYN.\n\\end{aligned}\n$$\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p158_data_91d38b00fc.png)\n\nAnd it follows from $\\angle AMX = \\angle ANY = 90^\\circ$ that $\\triangle AMC \\sim \\triangle ANY$, so $\\angle MAX = \\angle NAY$ and $\\frac{AM}{AX} = \\frac{AN}{AY}$, and hence $\\angle MAN = \\angle XAY$.\n\nCombining the two results above, we have $\\triangle AMN \\sim \\triangle AXY$. So we get\n\n$$\n\\begin{aligned}\n\\triangle AMN \\sim \\triangle ABC &\\Leftrightarrow \\triangle AXY \\sim \\triangle ABC \\\\\n&\\Leftrightarrow XY \\parallel BC \\Leftrightarrow \\angle DXY = \\angle XDB.\n\\end{aligned}\n$$\n\nAs $A, X, D, Y$ are concyclic, we have $\\angle DXY = \\angle DAY$.\nAs $\\angle XDB = 90^\\circ - \\angle ABC$,\n\n$$\n\\angle DXY = \\angle XDB \\Leftrightarrow \\angle DAC = 90^\\circ - \\angle ABC,\n$$\n\nwhich is equivalent to the fact that the line $AD$ passes through the circumcenter of $\\triangle ABC$.\n\nHence, $\\triangle AMN \\sim \\triangle ABC$ if and only if $AD$ passes through the circumcenter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18825,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC < BC$, and let $c(O, R)$ be its circumcircle. Diameters $BD$ and $CE$ are drawn. Circle $c_1(A, AE)$ intersects $AC$ at $K$. Circle $c_2(A, AD)$ intersects $BA$ at $L$ (with $A$ lying between $B$ and $L$). Prove that lines $EK$ and $DL$ intersect on the circle $c$.\n\n![](images/Saudi_Arabia_booklet_2023_p33_data_bfd39e74d4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the intersection of $DL$ with $(O)$, and suppose $M$ is between $D$ and $L$ as shown in the figure (other cases are similar). It suffices to prove that $E$, $K$, and $M$ are collinear. Note that $\\angle EAC = 90^\\circ$ and $AK = AE$, so $\\triangle AEK$ is a right isosceles triangle, which implies $\\angle AEK = \\angle AKE = 45^\\circ$. Similarly, $\\angle ADL = \\angle ALD = 45^\\circ$. Thus,\n\n$$\n\\angle AEM = \\angle ADM = \\angle ADL = 45^\\circ = \\angle AEK,\n$$\n\nwhich implies that $E$, $K$, and $M$ are collinear.\n\n![](images/Saudi_Arabia_booklet_2023_p33_data_2855673384.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18826,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega_1$ and $\\omega_2$ be circles with no common points. Points $M$ and $N$ are chosen on the circles $\\omega_1$ and $\\omega_2$, respectively, such that the tangent to the circle $\\omega_1$ at $M$ and the tangent to the circle $\\omega_2$ at $N$ intersect at $P$ and $\\triangle PMN$ is an isosceles triangle with apex $P$. The circles $\\omega_1$ and $\\omega_2$ meet the segment $MN$ again at $A$ and $B$, respectively. The line $PA$ meets the circle $\\omega_1$ again at $C$ and the line $PB$ meets the circle $\\omega_2$ again at $D$. Prove that $\\angle BCN = \\angle ADM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle MPN$ is isosceles, we have $\\angle PMA = \\angle PMN = \\angle MNP = \\angle BNP$. By the tangent-chord theorem, $\\angle MCA = \\angle PMA = \\angle BNP = \\angle BDN$.\n\nSince $\\angle MCP = \\angle MNP$, the quadrilateral $CMPN$ is cyclic. Analogously, from $\\angle PDN = \\angle PMN$, we get that $NDMP$ is cyclic. Since $C$ and $D$ both lie on the circumcircle of $\\triangle NPM$, points $P, N, M, C$, and $D$ are concyclic.\n\nFrom inscribed angles subtending arcs of equal length, we get $\\angle MDP = \\angle MCP = \\angle MNP = \\angle PDN = \\angle PMN = \\angle PCN$.\n\nThe power of $P$ with respect to $\\omega_1$ gives $|PM|^2 = |PA| \\cdot |PC|$. The power of $P$ with respect to $\\omega_2$ gives $|PN|^2 = |PB| \\cdot |PD|$. Since $|PM| = |PN|$, the powers of $P$ with respect to $\\omega_1$ and $\\omega_2$ are equal (so $P$ lies on the radical axis). Hence, $|PA| \\cdot |PC| = |PB| \\cdot |PD|$, which implies that $ABDC$ is cyclic. From inscribed angles subtending arc $AB$, we get $\\angle ACB = \\angle ADB$.\n\nHence, $\\angle BCN = \\angle ACN - \\angle ACB = \\angle MDB - \\angle ADB = \\angle MDA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18827,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$ and circumradius $R$. Let $M$, $N$, and $P$ be the midpoints of segments $BC$, $CA$, and $AB$, respectively. Let $(\\omega)$ be the circle passing through $A$ and $O$ and tangent to $OM$. The circle $(\\omega)$ meets $AB$ and $AC$ at $E$ and $F$. Let $I$ be the midpoint of segment $EF$, and $K$ be the intersection of $EF$ and $NP$. Prove that $R = 2IK$ and that $IMO$ is an isosceles triangle.\n\n![](images/Saudi_Arabia_booklet_2024_p18_data_56a4599e9c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let circle $(\\omega)$ cut $(O)$ again at $D$, and let $O_1$ be the center of $(\\omega)$. Since $AD$ is the radical axis of $(\\omega)$ and $(O)$, $OO_1 \\perp AD$. But $OO_1 \\perp OM$ since $OM$ is tangent to $(\\omega)$, so $AD \\parallel OM$ and thus $AD \\perp BC$.\n\nThis means that $AD$ and $AO$ are isogonal in $\\angle EAF$, so $OD \\parallel EF$, which implies that $EFOD$ is an isosceles trapezoid. Note that $OP \\perp AE$, $ON \\perp AF$, and $O \\in (AEF)$, so by the Simson line property, *K* is the projection of *O* onto *EF*. Let *L* be the midpoint of *OD*, then $IL \\perp OD$, so $OKIL$ is a rectangle. Thus\n\n$$\nIK = OL = \\frac{1}{2}OD = \\frac{R}{2}.\n$$\n\nNow consider the spiral similarity $\\Omega$ with center $D$, mapping $E \\to B$ and $F \\to C$, so $EF \\to BC$ and\n\n$$\n\\Omega : \\triangle DEF \\to \\triangle DBC.\n$$\n\nSince $O_1$ and $O$ are centers of $(DEF)$ and $(DBC)$, $\\Omega$ maps $O_1 \\to O$. Thus, the triangles $DO_1O$ and $DIM$ are similar, but $O_1O = O_1D$, so $ID = IM$. By the isosceles trapezoid, one can check that $ID = IO$ also, which implies $IM = IO$, so $IMO$ is an isosceles triangle.\n\n*Remark*: This problem can be solved easily by using complex numbers. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18828,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer has three digits in base 10. When it is interpreted in base 6 and multiplied by 4, the result is 2 more than 3 times the same number when it is interpreted in base 7. What is the largest integer, in base 10, that has this property?\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let the three-digit number be $abc$ (digits $a$, $b$, $c$ in base 6, each $<6$). Since $4 \\times abc_6 = 3 \\times abc_7 + 2$, we have:\n\n$$\n\\begin{align*}\n4(36a + 6b + c) &= 3(49a + 7b + c) + 2 \\\\\n144a + 24b + 4c &= 147a + 21b + 3c + 2 \\\\\n3b + c &= 3a + 2\n\\end{align*}\n$$\n\nHence, $3$ divides $c-2$. Since $c < 6$, $c = 2$ or $5$.\n\nIf $c = 2$, then $a = b$. If $c = 5$, then $a = b + 1$.\n\nSince all digits are at most $5$, the largest required number (in base 10) is $552$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18829,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a grid with points $A$, $P$, $Q$, and $B$. A counter moves from $A$ to $P$ in 4 steps, choosing each grid line (horizontal $H$ or vertical $V$) with probability $\\frac{1}{2}$. \n\n**a)** What is the probability that the counter arrives at $P$ after 4 steps?\n\n**b)** What is the probability that the counter will be at $Q$ after 5 steps?\n\n**c)** Two counters start simultaneously from $A$ and $B$ and move as described. What is the probability that they meet at the same grid point at the same time?\n\n**d)** Label the grid points on the diagonal through $P$ from top to bottom as 1, 2, 3, 4, 5. For each, find the probability that both counters meet there.\n\n![](images/Australian-Scene-2017_p54_data_794036f4c0.png)\n\n![](images/Australian-Scene-2017_p55_data_d833b8724f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**a Alternative i**\n\nEach path from $A$ to $P$ has 4 grid lines: one horizontal ($H$) and three vertical ($V$) in some order. The probability of each grid line being chosen is $\\frac{1}{2}$. So the probability of each path is $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{16}$. There are 4 paths from $A$ to $P$: $HVVV$, $VHVV$, $VVHV$, $VVVH$.\n\nSo the probability of arriving at $P$ is $4 \\times \\frac{1}{16} = \\frac{1}{4}$.\n\n**Alternative ii**\n\nEach path that takes 4 seconds has 4 grid lines: some horizontal ($H$) and some vertical ($V$). There are 16 such paths from $A$:\n\n$VVVV$, $HVVV$, $VHVV$, $VVHV$, $VVVH$, $HHVV$, $HVHV$, $HVVH$, $VHHV$, $VHVH$, $VVHH$, $HHHV$, $HHVH$, $HVHH$, $VHHH$, $HHHH$\n\nIn each of these paths each grid line has probability $\\frac{1}{2}$ of being chosen. So the probability for each path is $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{16}$. Four of these paths end at $P$. So the probability of arriving at $P$ is $4 \\times \\frac{1}{16} = \\frac{1}{4}$.\n\n**b**\n\nThe counter can reach $Q$ either from the grid point to the left of $Q$ or from $P$.\n\nThere is only one path from $A$ to $Q$ via the grid point to the left of $Q$. This path comprises four vertical grid lines, each taken with probability $\\frac{1}{2}$, and one horizontal grid line taken with probability 1 (no choice). Hence, the probability that the counter takes this path is $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times 1 = \\frac{1}{16}$.\n\nFrom part **a**, the probability the counter will be at $P$ after 4 seconds is $\\frac{1}{4}$. The probability that the counter moves from $P$ to $Q$ is $\\frac{1}{2}$. Hence the probability the counter reaches $Q$ via $P$ is $\\frac{1}{4} \\times \\frac{1}{2} = \\frac{1}{8}$.\n\nTherefore the probability that the counter will be at $Q$ after 5 seconds is $\\frac{1}{16} + \\frac{1}{8} = \\frac{3}{16}$.\n\n**c**\n\nThe counters will meet if they are in the same place at the same time. Each of the grid points on the diagonal through $P$ can be reached in 4 seconds from $A$ and in 4 seconds from $B$. Hence the counters could meet at any of these five grid points.\n\n![](images/Australian-Scene-2017_p54_data_794036f4c0.png)\n\nTo reach a grid point above this diagonal, the counter moving from $A$ will take more than 4 seconds and the counter moving from $B$ will take less than 4 seconds. To reach a grid point below this diagonal, the counter moving from $A$ will take less than 4 seconds and the counter moving from $B$ will take more than 4 seconds. So it is not possible for the counters to meet at any grid point other than those on the diagonal through $P$.\n\n**d**\n\nLabel the grid points on the diagonal through $P$ from top to bottom 1, 2, 3, 4, 5.\n\n![](images/Australian-Scene-2017_p55_data_d833b8724f.png)\n\nThe number of paths from $A$ to each of the grid points 1, 2, 3, 4, 5 is respectively 1, 4, 6, 4, 1.\n\nEach of these paths has 4 grid lines and each grid line has probability $\\frac{1}{2}$ of being chosen. So each path has probability $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{16}$. Hence the probability of the counter from $A$ reaching grid points 1, 2, 3, 4, 5 is respectively $\\frac{1}{16}, \\frac{4}{16}, \\frac{6}{16}, \\frac{4}{16}, \\frac{1}{16}$.\n\nBy symmetry the same probabilities apply to the counter from $B$.\n\nSo the probability the counters meet is $\\left(\\frac{1}{16}\\right)^2 + \\left(\\frac{4}{16}\\right)^2 + \\left(\\frac{6}{16}\\right)^2 + \\left(\\frac{4}{16}\\right)^2 + \\left(\\frac{1}{16}\\right)^2 = \\frac{1+16+36+16+1}{256} = \\frac{70}{256} = \\frac{35}{128}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18830,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all infinite sequences $a_1, a_2, \\dots$ of positive integers satisfying the following properties:\n\n(a) $a_1 < a_2 < a_3 < \\dots$,\n\n(b) there are no positive integers $i, j, k$, not necessarily distinct, such that $a_i + a_j = a_k$,\n\n(c) there are infinitely many positive integers $k$ such that $a_k = 2k - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is to have $a_k = 2k - 1$ for all $k$, giving the sequence $1, 3, 5, \\dots$.\n\nLet $a_1 = m$. First, we show that for any such sequence, we must have $a_{k+m} - a_k \\ge 2m$ for all positive integers $k$. Suppose for some $k$ that this was not the case. Then $a_k, a_{k+1}, \\dots, a_{k+m}$ are $m+1$ terms of the sequence that are all in the set $S = \\{a_k, a_k+1, \\dots, a_k+2m-1\\}$. Partition $S$ into $m$ two-element sets of the form $\\{b, b+m\\}$. By the pigeonhole principle, one of the $m$ two-element sets is such that both of its elements are terms of the sequence, so that $a_{i_1} + m = a_{i_2}$ for some $k \\le i_1, i_2 \\le k+m$. But we have $a_1 = m$, so this contradicts (b). Thus we have established $a_{k+m} - a_k \\ge 2m$ for all $k$.\n\nNow suppose we have $m$ consecutive terms of the sequence $a_j, a_{j+1}, \\dots, a_{j+m-1}$ such that for all $k$ between $j$ and $j+m-1$ inclusive, $a_k > 2k-1$. By (c), there must exist some index $i$ greater than $j$ satisfying $a_i = 2i-1$. Write $i-j = mq+r$ for $q \\ge 0$ and $0 \\le r < m$ using the division algorithm. Then $a_i \\ge 2m + a_{i-m} \\ge \\dots \\ge 2qm + a_{i-qm} = 2qm + a_{j+r}$. By assumption $a_{j+r} > 2(j+r)-1$, so $a_i > 2qm + 2(j+r) - 1 = 2i-1$, a contradiction. Therefore, for any block of $m$ consecutive terms of the sequence, one of them satisfies $a_k \\le 2k-1$.\n\nBecause of (a), we have $a_k \\ge m + (k-1)$. Consider $a_1, a_2, \\dots, a_m$. The previous inequality implies that for all $i < m$ we have $a_i > 2i-1$, and $a_m \\ge 2m-1$. By the previous paragraph one of these terms must satisfy $a_i \\le 2i-1$. Therefore we must have $a_m = 2m-1$ and by (a) $a_i = m+i-1$ for all $1 \\le i \\le m$.\n\nLikewise, for $a_{m+1}, a_{m+2}, \\dots, a_{2m}$, we have $a_{m+i} \\ge 2m + a_i > 2(m+i) - 1$ for all $i < m$ and $a_{2m} \\ge 4m-1$, so since one of these $m$ terms must satisfy $a_i \\le 2i-1$ we must have $a_{2m} = 4m-1$ and then by (a) $a_{m+i} = 3m+i-1$ for all $1 \\le i \\le m$.\n\nNow suppose $m > 1$. Observe that $a_m + a_m = 4m - 2 = a_{2m-1}$, contradicting (b). Therefore $m = a_1 = 1$. Furthermore, we know from the second paragraph that every term of the sequence satisfies $a_k \\le 2k-1$. At the same time, the first paragraph says $a_{k+1} - a_k \\ge 2$, or $a_k \\ge 2k-2 + a_1 = 2k-1$. Therefore $a_k = 2k-1$ for every $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18831,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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.\n\n![](images/RMC_2015_BT_p61_data_44ecf32bb4.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18832,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. For a real $x \\ge 1$, assume that $\\lfloor x^{n+1} \\rfloor, \\lfloor x^{n+2} \\rfloor, \\dots, \\lfloor x^{4n} \\rfloor$ are all squares of positive integers. Prove that $\\lfloor x \\rfloor$ is also the square of a positive integer.\n\nHere $\\lfloor z \\rfloor$ is the greatest integer smaller than or equal to $z$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the statement for $n = 1$. Write $x = a + r$, with $a \\ge 1$ an integer and $0 \\le r < 1$. Suppose $\\lfloor x^2 \\rfloor$, $\\lfloor x^3 \\rfloor$, and $\\lfloor x^4 \\rfloor$ are squares. Then we have $a \\le x < a + 1$, from which it follows that $a^2 \\le x^2 < (a + 1)^2$. Hence, $x^2$ is squeezed between two consecutive squares. However, $\\lfloor x^2 \\rfloor$ is a square, hence the only possibility is that $\\lfloor x^2 \\rfloor = a^2$. We conclude that $a^2 \\le x^2 < a^2 + 1$. Completely analogously, we also get that $(a^2)^2 \\le x^4 < (a^2 + 1)^2$ and hence $\\lfloor x^4 \\rfloor = a^4$. We conclude that $a^4 \\le x^4 < a^4 + 1$.\n\nMoreover, we have that $x^3 \\ge a^3$. Now suppose that $x^3 \\ge a^3 + 1$, i.e.\n\n$$\nx^4 \\ge x(a^3 + 1) = (a + r)(a^3 + 1) = a^4 + r a^3 + a + r \\ge a^4 + a \\ge a^4 + 1,\n$$\n\nwhich gives a contradiction. Hence, $x^3 < a^3 + 1$, which yields that $\\lfloor x^3 \\rfloor = a^3$. This is also a square, hence $a$ must be a square itself. We see that $\\lfloor x \\rfloor$ is a square.\n\nNow we will finish the proof with induction on $n$. The induction basis has just been proved. Now let $k \\ge 1$ and suppose that the statement is proved for $n = k$. Consider a real number $x \\ge 1$ with the property that $\\lfloor x^{k+2} \\rfloor, \\lfloor x^{k+3} \\rfloor, \\dots, \\lfloor x^{4k+4} \\rfloor$ are all squares. In particular, $\\lfloor x^{2(k+1)} \\rfloor, \\lfloor x^{3(k+1)} \\rfloor$, and $\\lfloor x^{4(k+1)} \\rfloor$ are all squares. We can now apply the case $n = 1$ on $x^{k+1}$ (which is a real number greater than or equal to 1) and find that $\\lfloor x^{k+1} \\rfloor$ is also a square. Now we know that $\\lfloor x^{k+1} \\rfloor, \\lfloor x^{k+2} \\rfloor, \\dots, \\lfloor x^{4k} \\rfloor$ are all squares and using the induction hypothesis, we obtain that $\\lfloor x \\rfloor$ is a square as well. This completes the proof by induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18833,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $O$ and $H$ be its circumcentre and orthocentre, respectively. Points $Z$ and $Y$ lie on segments $AB$ and $AC$, respectively, such that\n\n$$\n\\angle ZOB = \\angle YOC = 90^{\\circ}.\n$$\n\nThe perpendicular from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q$ and $R$, respectively. Let the tangents to the circumcircle of $\\triangle AYZ$ at points $Y$ and $Z$ meet at point $T$. Prove that $Q$, $R$, $O$, $T$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $K$ to be the point on $YZ$ such that $HK \\perp YZ$. Let $A'$ be a point on the circumcircle of $\\triangle ABC$ such that $AA' \\parallel BC$.\n\n**Lemma 1.** $YZ$ is the perpendicular bisector of $HA'$.\n\n*Proof.* Let $H_B$ denote the reflection of $H$ in $AC$. Then note that $A'H_B \\parallel CO$. This is because $\\angle (AA', A'H_B) = 90^\\circ - A = \\angle (CO, BC)$. Now $YO \\perp CO$ implies $OY \\perp A'H_B$. This shows $YA' = YH_B = YH$. Similarly, $ZH = ZA'$. $\\square$\n\n![](images/IND_ABooklet_2024_p47_data_ea2e920dae.png)\n\n**Lemma 2.** $OK \\perp BC$.\n\n*Proof.* Let $D$ and $D'$ denote the feet of the perpendiculars from $H$ and $A'$ onto $BC$, respectively. Then note that if $M$ is the foot of the perpendicular from $K$ onto $BC$, then by similarity, $DM = MD'$ as $HK = KA'$. However, $BD = D'C$, so $MB = MC$, as desired. $\\square$\n\n**Lemma 3.** $RY$ and $QZ$ are tangents to the circumcircle of $\\triangle AYZ$.\n\n*Proof.* Note that $RKYO$ is a cyclic quadrilateral. Now we angle chase: $\\angle RYK = \\angle ROK = \\angle (CO, OK) = \\angle BAC = \\angle YAZ$. Thus, $RY$ is tangent to the circumcircle of $\\triangle AYZ$. Similarly for $QZ$. $\\square$\n\nFinally, note that $O$ is the Miquel point of quadrilateral $QZYR$, as $RKYO$ and $QKZO$ are cyclic. Thus, $P := QZ \\cap YR$ lies on the circumcircle of $\\triangle QOR$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18834,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the difference of the arithmetic progression if its first term is a solution of the equation\n\n$$\nx^2 - 9x + x\\sqrt{12-x} - 9\\sqrt{12-x} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$, and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n\n$$\n(a_1 + 6d)^2 = a_1 (a_1 + 16d) \\iff d (a_1 - 9d) = 0.\n$$\n\nSince $d \\neq 0$, we get $a_1 = 9d$. Furthermore, we have $(x-9)(x+\\sqrt{12-x}) = 0$ and $x \\le 12$. Then $x = 9$ or $\\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \\le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\\frac{4}{9}$, respectively.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18835,
    "subject": "Mathematics (Olympiad)",
    "question": "$a$, $b$, and $c$ are nonzero real numbers such that\n$$\n\\frac{a+b}{c} = \\frac{b+c}{a} = \\frac{c+a}{b}.\n$$\n\n1. Prove that $a^3 + b^3 + c^3 \\neq 0$.\n\n2. Determine all possible values of the expression\n$$\n\\frac{(a+b)(b+c)(c+a)}{a^3 + b^3 + c^3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s = a + b + c$. Under the given condition, $\\frac{s}{c} = \\frac{s}{a} = \\frac{s}{b}$.\n\n1. If $s = 0$, then\n$$\na^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc \\neq 0\n$$\nsince $a, b, c \\neq 0$. If $s \\neq 0$, then $a = b = c$, so $a^3 + b^3 + c^3 = 3a^3 \\neq 0$.\n\n2. Since $a^3 + b^3 + c^3 \\neq 0$, the function\n$$\nF(a, b, c) = \\frac{(a+b)(b+c)(c+a)}{a^3 + b^3 + c^3}\n$$\nis well defined. For $a = 1, b = 1, c = -2$ (so $s = 0$),\n$$\nF(1, 1, -2) = \\frac{(1+1)(1+(-2))((-2)+1)}{1^3 + 1^3 + (-2)^3} = \\frac{2 \\cdot (-1) \\cdot (-1)}{1 + 1 - 8} = \\frac{2}{-6} = -\\frac{1}{3}.\n$$\nFor $a = b = c = 1$ (so $s \\neq 0$),\n$$\nF(1, 1, 1) = \\frac{(1+1)(1+1)(1+1)}{1^3 + 1^3 + 1^3} = \\frac{2 \\cdot 2 \\cdot 2}{3} = \\frac{8}{3}.\n$$\nIf $s = 0$, then $F = -\\frac{1}{3}$. If $s \\neq 0$, then $a = b = c$, so $F = \\frac{8a^3}{3a^3} = \\frac{8}{3}$.\n\n*All possible values are $-\\frac{1}{3}$ and $\\frac{8}{3}$.*",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18836,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the last three digits of the product of all odd positive integers up to 2019, that is, compute the last three digits of\n\n$$\nn = 1 \\cdot 3 \\cdot 5 \\cdot \\cdots \\cdot 2019.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that the Chinese Remainder Theorem (CRT) states that, for any given integers $r, s$ and any two positive co-prime integers $a, b$, there exists exactly one integer $x$ that satisfies $0 \\leq x < ab$, $x \\equiv r \\pmod a$, and $x \\equiv s \\pmod b$. We can apply CRT here with $a = 125$ and $b = 8$, because $1000 = 125 \\cdot 8$ and $\\gcd(125, 8) = 1$.\n\nBecause 125 is odd and $125 < 2019$, it is clear that $n \\equiv 0 \\pmod{125}$. To calculate $n \\pmod 8$, note that $1 \\cdot 3 \\cdot 5 \\cdot 7 \\equiv 1 \\pmod 8$. Because $n$ is the product of $(2019+1)/2 = 1010$ odd integers and $1010 = 4 \\cdot 252 + 2$, we can split the given product as follows:\n\n$$\nn = \\prod_{k=0}^{251} (8k + 1)(8k + 3)(8k + 5)(8k + 7) \\cdot 2017 \\cdot 2019\n$$\n\nTherefore, $n \\equiv (1 \\cdot 3 \\cdot 5 \\cdot 7)^{252} \\cdot 1 \\cdot 3 \\equiv 3 \\pmod 8$. Alternatively, when we use the usual notation for odd factorials, $(2k-1)!! = \\prod_{j=1}^k (2j-1)$, we can prove by induction on $k \\geq 1$ the following table:\n\n![](table.png)\n\nWhen $k=1$ this is obvious. For the inductive step, we use that $(2k+1)!! = (2k+1)(2k-1)!!$. Then, putting $k=1010$, we have $k \\equiv 2 \\pmod 4$ and so $n = (2k-1)!! \\equiv 3 \\pmod 8$.\n\nHaving established that $n \\equiv 0 \\pmod{125}$ and $n \\equiv 3 \\pmod 8$, to find $n \\pmod{1000}$, we note that $n$ is an odd multiple of 125. We can either check the five odd multiples of 125 below 1000 to see which one is congruent to 3 (mod 8), or we solve the congruence $125k \\equiv 3 \\pmod 8$. This is done by first observing that $125 \\equiv 5 \\pmod 8$, which gives $5k \\equiv 3 \\pmod 8$, and then multiplying both sides by 5 to obtain $k \\equiv 25k \\equiv 5 \\cdot 3 \\equiv 7 \\pmod 8$. Hence, $n \\equiv 7 \\cdot 125 \\equiv 875 \\pmod{1000}$, i.e., the last three digits of $n$ are 875.\n\nAlternatively, we may use the uniqueness in the CRT. It would then be sufficient to check that $875 \\equiv 0 \\pmod{125}$ and $875 \\equiv 3 \\pmod 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18837,
    "subject": "Mathematics (Olympiad)",
    "question": "Нехай $a + b + c = 1$. Доведіть нерівність:\n\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "За нерівністю Коші:\n\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq \\sqrt[3]{\\frac{(a+b)(b+c)(c+a)}{abc}}.\n$$\n\nПотрібна нерівність є наслідком нерівності\n\n$$\n(a + b)(b + c)(c + a) \\geq 8abc,\n$$\n\nяка випливає з нерівностей:\n\n$$\na + b \\geq 2\\sqrt{ab}, \\quad b + c \\geq 2\\sqrt{bc}, \\quad c + a \\geq 2\\sqrt{ca}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18838,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Bestimme den größtmöglichen Wert $M$, den $x + y + z$ annehmen kann, wenn $x$, $y$ und $z$ positive reelle Zahlen mit\n\n$$\n16xyz = (x + y)^2(x + z)^2\n$$\n\nsind.\n\nb) Zeige, dass es unendlich viele Tripel $(x, y, z)$ positiver rationaler Zahlen gibt, für die\n\n$$\n16xyz = (x + y)^2(x + z)^2 \\text{ und } x + y + z = M\n$$\n\ngelten.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** Aufgrund der Nebenbedingung und der arithmetisch-geometrischen Mittelungleichung gilt\n\n$$\n4\\sqrt{xyz} = (x + y)(x + z) = x(x + y + z) + yz \\ge 2\\sqrt{xyz(x + y + z)}.\n$$\n\nAlso gilt $2 \\ge \\sqrt{x + y + z}$ und damit $4 \\ge x + y + z$. Da wir im zweiten Teil unendlich viele solche Tripel angeben, für die $x + y + z = 4$ gilt, ist $M = 4$ das gesuchte Maximum.\n\n**(b)** Im Gleichheitsfall muss in der Abschätzung des ersten Teils Gleichheit in der Mittelungleichung gelten, also $x(x + y + z) = yz$, und natürlich außerdem $x + y + z = 4$. Wählen wir $y = t$, mit $t$ rational, erhalten wir $4x = t(4 - x - t)$ und damit $x = \\frac{4t - t^2}{4 + t}$ und $z = 4 - x - y = \\frac{16 - 4t}{4 + t}$. Wenn nun noch $0 < t < 4$ gilt, dann sind diese Ausdrücke alle positiv und rational und auch die Nebenbedingung ist erfüllt.\n\nDie Tripel $\\left(\\frac{4t - t^2}{4 + t},\\ t,\\ \\frac{16 - 4t}{4 + t}\\right)$ mit $0 < t < 4$ rational sind also unendlich viele Gleichheitsfälle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18839,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisectors of the sides $AC$ and $BC$ of the acute triangle $ABC$ with $CB = 30\\degree$ intersect at point $O$. The points $M$ and $N$ on the sides $AC$ and $BC$, respectively, are such that $O$ is the midpoint of the segment $MN$. How many times is the product of the lengths of segments $CM$ and $CN$ greater than the product of the lengths of segments $AM$ and $BN$?\n\n(Miroslav Marinov)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the midpoint of $BC$ (so $OL \\perp BC$), and $K$ be the foot of the perpendicular from $M$ to $BC$. Then $OL \\parallel MK$, and with $MO = ON$, it follows that $OL$ is a midsegment in triangle $KMN$—in particular, $KL = LN$. On the other hand, we also have $BL = CL$, whence $BN = CK$.\n\nFrom the right triangle $CKM$ with $CM = 30\\degree$, we get $KM = \\frac{1}{2}CM$ and $CK = \\sqrt{CM^2 - KM^2} = \\frac{CM}{\\sqrt{32}}$ from the Pythagorean theorem.\n\nTherefore, $\\frac{CM}{BN} = \\frac{2}{\\sqrt{3}}$.\n\nAnalogously, $\\frac{CN}{AM} = \\frac{2}{\\sqrt{3}}$, so finally\n\n$$\n\\frac{CM \\cdot CN}{AM \\cdot BN} = \\frac{4}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18840,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABCDE$ be a convex pentagon such that $AC$ is perpendicular to $BD$ and $AD$ is perpendicular to $CE$.\n\nProve that $\\angle BAC = \\angle DAE$ if and only if triangles $ABC$ and $ADE$ have equal areas.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the intersection of $AC$ and $BD$ and $Y$ be the intersection of $AD$ and $CE$.\n\n![](images/2021_Australian_Scene_p71_data_145b694aff.png)\n\nThe right angles from the problem imply that $XYCD$ is cyclic. By power of a point, $AX \\times AC = AY \\times AD$. Hence we have the following equivalences:\n\n$$\n\\begin{align*}\n\\angle BAC &= \\angle DAE \\\\\n\\Leftrightarrow \\triangle BAX &\\sim \\triangle EAY \\\\\n\\Leftrightarrow \\frac{BX}{AX} &= \\frac{EY}{AY} \\\\\n\\Leftrightarrow BX \\times AC &= EY \\times AD \\\\\n\\Leftrightarrow |\\triangle ABC| &= |\\triangle ADE|.\n\\end{align*}\n$$\n\n**Remark** There are many variations of this solution, for example, by using similar triangles instead of the cyclic quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18841,
    "subject": "Mathematics (Olympiad)",
    "question": "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\n![](images/Russia_2016_Booklet_p33_data_ef70d22d2f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть окружности, описанные около треугольников $A_1BC$ и $ABC_1$, вторично пересекаются в точке $X$ (см. рис. 23). Тогда $\\angle(BX, XC) = \\angle(BA_1, A_1C)$ и $\\angle(AX, XB) = \\angle(AC_1, C_1B)$, откуда\n\n$$\n\\angle(AX, XC) = \\angle(AX, XB) + \\angle(BX, XC) = \\angle(AC_1, C_1B) + \\angle(BA_1, A_1C) = \\angle(AB_1, B_1C).\n$$\n\nЭто означает, что $X$ лежит на окружности, описанной около треугольника $AB_1C$, что и требовалось. $\\square$\n\nПусть $K_A$, $K_B$ и $K_C$ — середины отрезков $AM$, $BM$ и $CM$ соответственно (см. рис. 24). Тогда $\\angle MK_CK_BM_A = \\angle AMC$. Аналогично, $\\angle MK_CK_A M_B = \\angle BMC$ и $\\angle M_A K_C M_B = \\angle BMA$; значит,\n\n$$\n\\angle MK_CK_A M_B + \\angle M_B K_C M_A + \\angle M_A K_B M_C = 360^\\circ.\n$$\n\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$$\n\\begin{align*}\n\\angle(K_BX, XM_B) &= \\angle(K_BX, XM_C) + \\angle(M_CX, XM_B) \\\\\n&= \\angle(K_BM_A, M_A M_C) + \\angle(M_C K_A, K_A M_B) \\\\\n&= \\angle(MC, CA) + \\angle(BM, MC) = \\angle(BM, CA) = \\angle(K_B M_B, AC).\n\\end{align*}\n$$\n\nЭто равенство означает, что окружность $\\Omega_B$ проходит через точку $X$. Аналогично, через $X$ проходят окружности $\\Omega_A$ и $\\Omega_C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18842,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose for each prime $p$, at least one of $a$ or $b$ is either relatively prime to $p$ or divisible exactly by $p$. Find all solutions $(a, b)$ to the equation\n\n$$\n7\\varphi^2(c) + 11\\varphi^2(d) = 2(c^2 + d^2) + \\varphi(cd),\n$$\n\nwhere $a = 2c$, $b = 2d$, and $\\varphi$ denotes Euler's totient function.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the equation for various cases of $p$ dividing $c$ or $d$.\n\nIf $p = 7$, then $7 \\mid c$, $7 \\mid d$, $7^2 \\mid 2(c^2 + d^2)$. If $7^2 \\mid d$, then $7 \\nmid\\mid c$, and we must have $7^2 \\mid \\varphi(cd)$, $7^2 \\mid \\varphi^2(d)$, implying $7^2 \\mid 7\\varphi^2(c)$, $7 \\mid r-1$. The smallest prime $r$ of the form $4k+3$ satisfying $7 \\mid r-1$ is $r=43$, which gives\n\n$$\n7\\varphi^2(c) \\ge \\left(\\frac{2}{3} \\times \\frac{6}{7} \\times \\frac{42}{43}\\right)^2 > 2.18.\n$$\n\nIn addition, $11\\varphi^2(d) \\ge 11 \\times \\left(\\frac{6}{7}\\right)^2 > 8$, and thus\n\n$$\n\\begin{aligned}\n7\\varphi^2(c) + 11\\varphi^2(d) &\\ge 2(c^2 + d^2) + (0.18c^2 + 6d^2) \\\\\n&> 2(c^2 + d^2) + \\varphi(cd),\n\\end{aligned}\n$$\n\na contradiction.\n\nIf $p=3$, let $d = p^\\alpha$. Notice that\n\n$$\n\\begin{aligned}\n\\varphi^2(c) &= \\left(\\frac{2}{3} \\times \\frac{q-1}{q} \\times \\frac{p-1}{p}\\right)^2 c^2, \\\\\n\\varphi^2(d) &= \\frac{4}{9}d^2, \\\\\n\\varphi(cd) &= \\frac{2}{3} \\times \\frac{q-1}{q} \\times \\frac{p-1}{p}cd.\n\\end{aligned}\n$$\n\nBy the AM-GM inequality, $\\frac{1}{2}\\varphi^2(c) + 2d^2 \\ge 2\\varphi(cd) > \\varphi(cd)$. Since $11\\varphi^2(d) = \\frac{44}{9}d^2 > 2d^2 + 2d^2$, in case $6.5\\varphi^2(c) > 2c^2$, then the left-hand side of the equation is larger, which is contradictory.\n\nIf $q, r \\ne 7$, we know\n\n$$\n6.5 \\times \\left(\\frac{2}{3} \\times \\frac{q-1}{q} \\times \\frac{p-1}{p}\\right)^2 \\ge \\frac{4}{9} \\times \\frac{100}{121} \\times \\frac{324}{361} > 2,\n$$\n\nand again the left hand side is larger. Hence, $q$ or $r$ must be 7. Assume $q = 7, r \\ge 11$. Let $c = 3 \\times 7^\\beta r^\\gamma$, and $3 \\mid p-1$. If $\\alpha \\ge 3$, then $27 \\mid \\varphi(cd), 27 \\mid \\varphi^2(d)$. Take the equation modulo 27 to find\n\n$$\n7 \\times 36 \\times 4 \\times (r-1)^2 \\times 7^{2\\beta-2} \\times r^{2\\gamma-2} \\equiv 2 \\times 3^2 \\times 7^{2\\beta} \\times r^{2\\gamma} \\pmod{27},\n$$\n\nwhich simplifies to\n\n$$\n8(r-1)^2 \\equiv 7r^2 \\pmod{3}.\n$$\n\nSince $3 \\nmid r$, it follows that $3 \\nmid (r-1)$, and 2 is a quadratic residue modulo 3, a contradiction.\n\nTherefore, $\\alpha = 1$ or 2, and $d = 3$ or 9. Since $3 \\mid \\varphi(c), \\varphi(cd)$, taking modulo 3 yields $3 \\mid 11\\varphi^2(d)$. Hence $3^2 \\mid d$, and $d = 9$. Let $e = 7^\\alpha r^\\gamma$. Plug $c = 3 \\times 7^\\alpha \\times r^\\gamma$ and $d = 9$ into the equation, and we obtain\n\n$$\n14\\varphi^2(e) - 9\\varphi(e) + 117 = 9e^2. \\qquad (7)\n$$\n\nIf $\\gamma \\ge 2$, then $7 \\mid \\varphi(e), 7 \\mid 117$, which is untrue.\n\nWe deduce that $\\beta \\ge 2$ is impossible, and thus $\\beta = 1$. Let $f = r^\\gamma$ and simplify (7) to\n\n$$\n56\\varphi^2(f) - 6\\varphi(f) + 13 = 49f^2.\n$$\n\nIf $\\gamma > 1$, then $r \\mid 13, r = 13$. But 13 is a prime of the form $4k+1$, a contradiction.\n\nIf $\\gamma = 1$, then $\\varphi(f) = r - 1$, and (7) reduces to\n\n$$\n7r^2 - 118r + 85 = 0,\n$$\n\nwhich does not have integer solutions.\n\nThe final conclusion: with the assumption that for each prime $p$, at least one of $a, b$ is relatively prime to $p$ or divisible exactly by $p$, there is only one solution\n\n$$\n\\{a, b\\} = \\{2c, 2d\\} = \\{30, 6\\}.\n$$\n\nBy the method of descent, all solutions of the equation are\n\n$$\n\\{a, b\\} = \\{30 \\cdot 2^{\\alpha} \\cdot 3^{\\beta}, 6 \\cdot 2^{\\alpha} \\cdot 3^{\\beta}\\},\n$$\n\nwhere $\\alpha, \\beta$ are arbitrary nonnegative integers. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18843,
    "subject": "Mathematics (Olympiad)",
    "question": "Expanding $(x + a)(x^2 + r x + 1)$ yields\n\n$$\nx^3 + (a + r)x^2 + (1 + a r)x + a.\n$$\n\nHence, we seek all real numbers $r$ such that there is exactly one real number $a$ satisfying the following system of inequalities:\n\n$$\na + r \\ge 0\n$$\n\n$$\na r \\ge -1\n$$\n\n$$\na \\ge 0\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $r \\ge 0$, then any $a \\ge 0$ satisfies the above inequalities, so there is not exactly one solution.\n\nIf $r < 0$, let $r = -s$ for $s > 0$. The inequalities become:\n\n$$\na \\ge s \\quad (1')\n$$\n$$\na \\le \\frac{1}{s} \\quad (2')\n$$\n$$\na \\ge 0 \\quad (3')\n$$\nSince (3') follows from (1'), we only need (1') and (2'), which combine to:\n\n$$\ns \\le a \\le \\frac{1}{s}.\n$$\nThere is exactly one value of $a$ satisfying this if and only if $s = \\frac{1}{s}$, so $s = 1$ and $a = 1$. Thus, $r = -1$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18844,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer. Consider a closed Z.Z.L.S. (zig-zag line segment) in a polygon with $2m$ vertices. What is the maximum possible number of self-intersection points that such a closed Z.Z.L.S. can have?\n\nFind the value for $m = 1005$.",
    "options": [],
    "answer": "See solution",
    "solution": "By analyzing the structure of the polygon and the properties of its diagonals, we find that the number of self-intersection points on any diagonal cannot exceed $2m - 3$. A diagonal is called *good* if it contains $2m - 3$ self-intersection points. If there are at most two good diagonals, the total number of self-intersection points is at most $2m^2 - 4m + 1$. Therefore, there must be at least three good diagonals.\n\nEach good diagonal must have length $m$. If a diagonal has length less than $m$, dissecting the polygon shows it cannot be good. Taking three good diagonals, each of length $m$, and considering their intersections, we deduce that certain coloring and traversal arguments lead to contradictions if more than $2m^2 - 4m + 1$ self-intersections are possible.\n\nThus, the maximum number of self-intersection points is $2m^2 - 4m + 1$. For $m = 1005$:\n\n$$2(1005)^2 - 4(1005) + 1 = 2016031$$\n\nSo, the answer is $2016031$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18845,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a 5-tuple of integers *arrangeable* if its elements can be labeled $a$, $b$, $c$, $d$, $e$ in some order so that $a - b + c - d + e = 29$. Determine all 2017-tuples of integers $n_1, n_2, \\dots, n_{2017}$ such that if we place them in a circle in clockwise order, then any 5-tuple of numbers in consecutive positions on the circle is arrangeable.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $n_1 = n_2 = \\dots = n_{2017} = 29$.\n\nIt is easy to see that the above is a valid 2017-tuple. We claim there are no others.\n\nLet all subscripts be taken modulo 2017.\n\nDefine a 5-tuple of integers as $k$-arrangeable if its elements can be labeled $a$, $b$, $c$, $d$, $e$ in some order such that $a - b + c - d + e = k$.\n\nA 2017-tuple $n_1, n_2, \\dots, n_{2017}$ is $k$-good if every 5-tuple $n_i, n_{i+1}, n_{i+2}, n_{i+3}, n_{i+4}$ ($i = 1, 2, \\dots, 2017$) is $k$-arrangeable. We are asked to determine all 2017-tuples that are 29-good.\n\nFor any integers $a, b, c, d, e$:\n\n$$\na - b + c - d + e = 29\n$$\n\nif and only if\n\n$$\n(a - 29) - (b - 29) + (c - 29) - (d - 29) + (e - 29) = 0.\n$$\n\nThus, $n_1, n_2, \\dots, n_{2017}$ is 29-good if and only if $m_1, m_2, \\dots, m_{2017}$ is 0-good, where $m_i = n_i - 29$ for $i = 1, 2, \\dots, 2017$. We will prove that the only 0-good sequence is $m_1 = m_2 = \\dots = m_{2017} = 0$.\n\nSuppose there is a 0-good sequence with not all $m_i$ equal to 0. Choose $m_1, m_2, \\dots, m_{2017}$ to minimize\n\n$$\n|m_1| + |m_2| + \\dots + |m_{2017}|. \\qquad (1)\n$$\n\nIf $a - b + c - d + e = 0$, then $a + b + c + d + e = 2(b + d) \\equiv 0 \\pmod{2}$. Thus, for each $i$:\n\n$$\nm_i + m_{i+1} + m_{i+2} + m_{i+3} + m_{i+4} \\equiv 0 \\pmod{2}. \\qquad (2)\n$$\n\nReplacing $i$ with $i + 1$ in (2):\n\n$$\nm_{i+1} + m_{i+2} + m_{i+3} + m_{i+4} + m_{i+5} \\equiv 0 \\pmod{2}. \\qquad (3)\n$$\n\nSubtracting (2) from (3):\n\n$$\nm_i \\equiv m_{i+5} \\pmod{2}. \\qquad (4)\n$$\n\nSince $\\gcd(5, 2017) = 1$, equation (4) implies all $m_i$ are congruent modulo 2. From (2), all $m_i$ are even.\n\nConsider $\\frac{m_1}{2}, \\frac{m_2}{2}, \\dots, \\frac{m_{2017}}{2}$. This sequence is also 0-good, but\n\n$$\n0 < \\left|\\frac{m_1}{2}\\right| + \\left|\\frac{m_2}{2}\\right| + \\dots + \\left|\\frac{m_{2017}}{2}\\right| < |m_1| + |m_2| + \\dots + |m_{2017}|.\n$$\n\nThis contradicts the minimality in (1).\n\nTherefore, the only 0-good sequence is $m_1 = m_2 = \\dots = m_{2017} = 0$, corresponding to $n_1 = n_2 = \\dots = n_{2017} = 29$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18846,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku drew a regular hexagon and chose three triangles with different areas whose vertices were among the vertices of the hexagon. Prove that the sum of the areas of the triangles is equal to the area of the hexagon.",
    "options": [],
    "answer": "See solution",
    "solution": "Any triangle whose vertices are among the vertices of a regular hexagon is one of the following:\n\n- A triangle $\\Delta_1$ whose vertices are three consecutive vertices of the hexagon.\n- A triangle $\\Delta_2$ whose two vertices are adjacent vertices of the hexagon and the third one is adjacent to none of the first two.\n- A triangle $\\Delta_3$ where any two vertices are not adjacent vertices of the hexagon.\n\nSince the areas of the chosen triangles are different, the triangles must be $\\Delta_1$, $\\Delta_2$, and $\\Delta_3$. The hexagon can be divided into four parts: the triangle $\\Delta_3$ surrounded by three triangles $\\Delta_1$. The area of the triangle $\\Delta_2$ is twice the area of the triangle $\\Delta_1$ because they have the same base but the height of $\\Delta_2$ is twice the height of $\\Delta_1$.\n\nThus, the sum of the areas of $\\Delta_1$, $\\Delta_2$, and $\\Delta_3$ equals the area of the hexagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18847,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(n, m)$ of positive integers such that the arithmetic and geometric means of $m$ and $n$ are different two-digit numbers consisting of the same digits.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $10a + b$ be the arithmetic mean of the given numbers, where $a$ and $b$ are decimal digits. Let $10a + b + x$ and $10a + b - x$ be the numbers we are searching for. Then, by the premises:\n\n$$\n\\sqrt{(10a + b + x)(10a + b - x)} = 10b + a\n$$\n\nSquaring both sides and simplifying gives $x^2 = 99(a^2 - b^2)$. So, $x^2$ is divisible by $99$, implying $x^2$ is divisible by $3$ and $11$. Since $3$ and $11$ are primes, $x$ itself is divisible by $3$ and $11$, and therefore by $33$. Denoting $x = 33z$, we get:\n\n$$\na^2 - b^2 = (a + b)(a - b) = 11z^2\n$$\n\nThus, the product $(a + b)(a - b)$ is divisible by $11$. Since $a$ and $b$ are single-digit numbers, $a + b = 11$ is the only possibility. Therefore, $a - b = z^2$. Since $a - b$ and $a + b$ are both odd or both even, $z$ must be odd. So, $z = 1$, since $z \\geq 3$ implies $x \\geq 99$, but $10a + b - x$ must be positive. Thus, $a = 6$, $b = 5$, $x = 33$, and the corresponding pair is $(98, 32)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18848,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of natural numbers $(k, n)$ such that there exist natural numbers $a, b$ satisfying\n\n$$\ngcd(a + k, b) = n \\cdot gcd(a, b).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that every pair $(k, n)$ works.\n\nFirst, if $n = 1$, we can just take $(a, b) = (k, k)$, then\n\n$$\ngcd(a + k, b) = gcd(2k, k) = k = gcd(k, k) = gcd(a, b).\n$$\n\nNow, assume that $n > 1$. Then $nk - k > 0$ is a natural number, and we can take $(a, b) = ((n-1)k, nk)$. Then we have:\n\n$$\ngcd(a + k, b) = gcd(nk, nk) = nk,\n$$\nand, similarly,\n\n$$\nn \\cdot gcd(a, b) = n \\cdot gcd((n-1)k, nk) = n \\cdot k = nk,\n$$\n\nso this choice of $(a, b)$ does indeed exhibit $(k, n)$ as a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18849,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = \\frac{1}{a}$, $y = \\frac{1}{b}$, $z = \\frac{1}{c}$. The condition is $xyz = 1$. Prove that for $xyz = 1$,\n$$\n\\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\geq \\frac{1}{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Apply Hölder's inequality:\n$$\n(u_1^3 + v_1^3 + w_1^3)(u_2^3 + v_2^3 + w_2^3)(u_3^3 + v_3^3 + w_3^3) \\geq (u_1v_1w_1 + u_2v_2w_2 + u_3v_3w_3)^3\n$$\nwhere\n$$\n\\begin{aligned}\nu_1 &= u_2 = \\sqrt[3]{2y+z}, & u_3 &= \\frac{x}{\\sqrt[3]{(2y+z)^2}}, \\\\\nv_1 &= v_2 = \\sqrt[3]{2z+x}, & v_3 &= \\frac{y}{\\sqrt[3]{(2z+x)^2}}, \\\\\nw_1 &= w_2 = \\sqrt[3]{2x+y}, & w_3 &= \\frac{z}{\\sqrt[3]{(2x+y)^2}},\n\\end{aligned}\n$$\nThis yields\n$$\n[3(x + y + z)]^2 \\left( \\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\right) \\geq (x + y + z)^3.\n$$\nRewriting,\n$$\n\\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\geq \\frac{x+y+z}{9}.\n$$\nSince $x + y + z \\geq 3\\sqrt[3]{xyz} = 3$, it follows that $\\frac{x+y+z}{9} \\geq \\frac{1}{3}$ by the AM-GM inequality, proving the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18850,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the assertion that for each positive integer $n \\ge 2$, the remainder upon dividing $2^{2n}$ by $2^n - 1$ is a power of 4. Either prove the assertion or find (with proof) a counterexample.",
    "options": [],
    "answer": "See solution",
    "solution": "The assertion is false, and the smallest $n$ for which it fails is $n = 25$.\n\nGiven $n \\ge 2$, let $r$ be the remainder when $2^n$ is divided by $n$. Then $2^n = k n + r$ where $k$ is a positive integer and $0 \\le r < n$. It follows that\n\n$$\n2^{2n} = 2^{k n + r} \\equiv 2^r \\pmod{2^n - 1},\n$$\n\nwhere $2^r < 2^{2n} - 1$. Thus, the remainder when $2^{2n}$ is divided by $2^n - 1$ is $2^r$. Now, $2^r$ is a power of 4 if and only if $r$ is even, so to disprove the assertion it is enough to find an $n$ for which the corresponding $r$ is odd. We describe now a method to find such an $n$.\n\nIf $n$ is even then so is $r = 2^n - k n$. If $n$ is an odd prime then $2^n \\equiv 2 \\pmod n$ by Fermat's Little Theorem; hence $r \\equiv 2^n \\equiv 2 \\pmod n$ and $r = 2$. Therefore, we may rule out these values of $n$.\n\nIt remains to try cases in which $n$ is odd and composite. In the first three instances $n = 9, 15, 21$ there is no contradiction to the assertion:\n\n$$\nn = 9: \\quad 2^6 \\equiv 1 \\pmod{9} \\quad \\Rightarrow \\quad 2^9 \\equiv 2^6 \\cdot 2^3 \\equiv 8 \\pmod{9}\n$$\n\n$$\nn = 15: \\quad 2^4 \\equiv 1 \\pmod{15} \\quad \\Rightarrow \\quad 2^{15} \\equiv (2^4)^3 \\cdot 2^3 \\equiv 8 \\pmod{15}\n$$\n\n$$\nn = 21: \\quad 2^6 \\equiv 1 \\pmod{21} \\quad \\Rightarrow \\quad 2^{21} \\equiv (2^6)^3 \\cdot 2^3 \\equiv 8 \\pmod{21}\n$$\n\nHowever, for $n = 25$, we see that\n\n$$\n2^{10} = 1024 \\equiv -1 \\pmod{25} \\Rightarrow 2^{20} \\equiv 1 \\pmod{25} \\Rightarrow 2^{25} \\equiv 2^5 \\equiv 7 \\pmod{25},\n$$\n\nso 7 is the remainder when $2^{25}$ is divided by 25 and $2^7$ is the remainder when $2^{2^{25}}$ is divided by $2^{25} - 1$. This gives the desired counterexample.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18851,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle of radius $1$ with center at $(1, 0)$ is internally tangent to $\\triangle PBC$. Find the minimum value of the area of $\\triangle PBC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $P, B, C$ by $P(x_0, y_0)$, $B(0, b)$, $C(0, c)$, and assume that $b > c$. The equation for the line $PB$ is\n\n$$\ny - b = \\frac{y_0 - b}{x_0}x.\n$$\n\nIt can be rewritten as\n\n$$\n(y_0 - b)x - x_0 y + x_0 b = 0.\n$$\n\nSince the distance between the circle center $(1, 0)$ and the line $PB$ is $1$, we have\n\n$$\n\\frac{|y_0 - b + x_0 b|}{\\sqrt{(y_0 - b)^2 + x_0^2}} = 1.\n$$\n\nThat is to say,\n\n$$\n(y_0 - b)^2 + x_0^2 = (y_0 - b)^2 + 2x_0 b (y_0 - b) + x_0^2 b^2.\n$$\n\nIt is easy to see that $x_0 > 2$. Then the last equation can be simplified as\n\n$$\n(x_0 - 2) b^2 + 2 y_0 b - x_0 = 0.\n$$\n\nIn a similar way,\n\n$$\n(x_0 - 2) c^2 + 2 y_0 c - x_0 = 0.\n$$\n\nTherefore,\n\n$$\nb + c = \\frac{-2 y_0}{x_0 - 2}, \\quad bc = \\frac{-x_0}{x_0 - 2}.\n$$\n\nThen we get\n\n$$\n(b - c)^2 = \\frac{4 x_0^2 + 4 y_0^2 - 8 x_0}{(x_0 - 2)^2}.\n$$\n\nAs $P(x_0, y_0)$ is on the parabola, $y_0^2 = 2 x_0$. So we have\n\n$$\n(b - c)^2 = \\frac{4 x_0^2}{(x_0 - 2)^2},\n$$\n\nor $b - c = \\frac{2 x_0}{x_0 - 2}$. Then we have\n\n$$\n\\begin{aligned}\nS_{\\triangle PBC} &= \\frac{1}{2} (b - c) \\times x_0 \\\\\n&= \\frac{x_0}{x_0 - 2} \\times x_0 \\\\\n&= (x_0 - 2) + \\frac{4}{x_0 - 2} + 4 \\\\\n&\\ge 4 + 4 = 8.\n\\end{aligned}\n$$\n\nThe equality holds when $x_0 - 2 = 2$; this means that $x_0 = 4$ and $y_0 = \\pm 2\\sqrt{2}$. So the minimum of $S_{\\triangle PBC}$ is $8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18852,
    "subject": "Mathematics (Olympiad)",
    "question": "For non-negative real numbers $a$ and $b$, let $A(a, b)$ be their arithmetic mean and $G(a, b)$ their geometric mean. We consider the sequence $\\langle a_n \\rangle$ with $a_0 = 0$, $a_1 = 1$ and\n$$\na_{n+1} = A(a_{n-1}, a_n),\\ G(a_{n-1}, a_n)\\quad \\text{for } n > 0.\n$$\n\n(a) Prove that every $a_n = b_n^2$ is the square of a rational number.\n\n(b) Prove that the inequality $|b_n - \\frac{2}{3}| < \\frac{1}{2^n}$ holds for all $n > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Proof by induction. For $n = 0$ we have $a_0 = 0 = b_0$ and for $n = 1$ we have $a_1 = 1 = b_1$, so $b_0, b_1 \\in \\mathbb{Q}$. Assume $a_i = b_i^2$ with $b_i \\in \\mathbb{Q}$ for all $i \\leq k$; we show it holds for $i = k+1$:\n\n$$\na_{k+1} = \\frac{a_{k-1} + a_k}{2} + \\sqrt{a_{k-1}a_k} = \\frac{b_{k-1}^2 + b_k^2 + 2b_{k-1}b_k}{4} = \\left(\\frac{b_{k-1} + b_k}{2}\\right)^2 = b_{k+1}^2\n$$\n\nThus, the claim holds for all $n$.\n\n(b) The recursion $2b_{n+1} = b_{n-1}b_n$ yields the characteristic equation\n$$\n2q^2 - q - 1 = 0 \\implies (2q + 1)(q - 1) = 0\n$$\nwith solutions $1$ and $-\\frac{1}{2}$. Since $b_n = A \\cdot 1^n + B \\cdot \\left(-\\frac{1}{2}\\right)^n$, we have $b_0 = 0 \\implies A + B = 0$ and $b_1 = 1 \\implies 1 = A - \\frac{B}{2}$. Thus $A = \\frac{2}{3}$ and $B = -\\frac{2}{3}$, so\n$$\nb_n = \\frac{2}{3} - \\frac{2}{3} \\left(-\\frac{1}{2}\\right)^n\n$$\nTherefore,\n$$\n|b_n - \\frac{2}{3}| = \\frac{2}{3} \\cdot \\frac{1}{2^n} < \\frac{1}{2^n},\n$$\ncompleting the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18853,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $P$ 為銳角三角形 $ABC$ 內部一點，且 $P$ 到三頂點的距離分別為 $d_A, d_B, d_C$，到三邊的垂直距離分別為 $d_1, d_2, d_3$。試證：\n\n$$\nd_A + d_B + d_C \\geq 2(d_1 + d_2 + d_3).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "解：如圖所示，$PD \\perp BC$，$PE \\perp CA$，$PF \\perp AB$。\n\n![](images/13-1J_p2_data_86a7cbc783.png)\n\n因為 $\\angle AEP + \\angle AFP = 90^\\circ + 90^\\circ = 180^\\circ$，所以 $A, E, P, F$ 共圓，且 $AP = d_A$ 為該圓直徑，於是可得：$\\angle EPF = 180^\\circ - \\angle A = \\angle B + \\angle C$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18854,
    "subject": "Mathematics (Olympiad)",
    "question": "Is there a positive integer $n$, which is a multiple of $103$, such that $2^{2n+1} \\equiv 2 \\pmod{n}$?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that there is no such positive integer $n$.\n\nSuppose, for contradiction, that a positive integer $n$ exists such that $2^{2n+1} \\equiv 2 \\pmod{n}$ and $103 \\mid n$.\n\nThen $2^{2n+1} \\equiv 2 \\pmod{103}$ as well, so $2^{2n} \\equiv 1 \\pmod{103}$.\n\nSince $103$ is prime, Fermat's little theorem gives $2^{102} \\equiv 1 \\pmod{103}$. Let $d_1 = \\gcd(102, 2n)$. Then $2^{d_1} \\equiv 1 \\pmod{103}$. Since $102 = 2 \\times 3 \\times 17$, possible values for $d_1$ are divisors of $102$. It is easy to rule out $d_1 = 2, 3, 6$, so $17 \\mid d_1$, which implies $17 \\mid 2n$, and thus $17 \\mid n$.\n\nNow, since $17$ is a factor of $n$, we have $2^{2n+1} \\equiv 2 \\pmod{17}$, so $2^{2n} \\equiv 1 \\pmod{17}$. Fermat's little theorem gives $2^{16} \\equiv 1 \\pmod{17}$. Let $d_2 = \\gcd(16, 2n)$. Then $d_2$ is a power of $2$ and $2^{d_2} \\equiv 1 \\pmod{17}$. We see that $d_2 = 2, 4$ do not work, so $d_2 = 8$ or $16$. But then $d_2 \\nmid 2n$ shows that $4 \\nmid n$.\n\nTherefore, $2^{2n+1} \\equiv 2 \\pmod{4}$, which is impossible. Hence, no such $n$ exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18855,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider acute scalene $\\triangle ABC$ with altitudes $CD$, $AE$, and $BF$. The points $E'$ and $F'$ are symmetric to $E$ and $F$ with respect to $A$ and $B$, respectively. Point $C_1$ on the ray $\\overrightarrow{CD}$ is such that $DC_1 = 3CD$. Prove that $\\angle E'C_1F' = \\angle ACB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the points $M$, $N$, $P$, and $Q$ be such that the quadrilaterals $CEAM$, $CFBN$, $CEE'P$, and $CFF'Q$ are rectangles. Let $C'$ be the midpoint of $CC_1$. Denote $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, and $\\angle ACB = \\gamma$. We have $\\triangle ABC \\cong \\triangle AC'B$ and $\\triangle AMC \\sim \\triangle BNC$.\n\nNext, we have\n\n$$\n\\begin{align*}\n\\angle MAC' &= 360^{\\circ} - \\angle MAC - \\angle BAC - \\angle BAC' \\\\\n            &= 360^{\\circ} - \\gamma - 2\\alpha = \\gamma + 2\\beta = \\angle NBC + \\angle ABC + \\angle ABC' \\\\\n            &= \\angle NBC'.\n\\end{align*}\n$$\n\nSince\n\n$$\n\\frac{MA}{NB} = \\frac{AC}{BC} = \\frac{AC'}{BC'}\n$$\n\nit follows from the above that $\\triangle MAC' \\sim \\triangle NBC'$. Therefore $\\angle AC'M = \\angle BC'N$, whence $\\angle MC'N = \\gamma$. Considering the midsegments in $\\triangle CPC_1$ and $\\triangle CQC_1$ we conclude that $MC' \\parallel PC_1$ and $NC' \\parallel QC_1$. This implies that $\\angle PC_1Q = \\gamma$.\n\nFurther, we have $BN \\parallel F'Q$, $BN = F'Q$, $NC' \\parallel QC_1$, $2NC' = QC_1$, $AM \\parallel E'P$, $AM = E'P$, $MC' \\parallel PC_1$, and $2MC' = PC_1$. This and $\\triangle MAC' \\sim \\triangle NBC'$ gives that $\\triangle PE'C_1 \\sim \\triangle QF'C_1$. We conclude that $\\angle PC_1E' = \\angle QC_1F'$. This equality and $\\angle PC_1Q = \\gamma$ from the above imply that $\\angle E'C_1F' = \\gamma$.\n\n*Remark.* It is now easy to see that $\\triangle E'C_1F' \\sim \\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18856,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(x, n, p)$ of natural numbers $x$, $n$ and prime $p$ such that\n$$\nx^3 + 3x + 14 = 2 \\cdot p^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The possible triples are $(x, n, p) = (1, 2, 3)$ and $(x, n, p) = (3, 2, 5)$.\n\nWe can factor the left side:\n$$\nx^3 + 3x + 14 = (x+2)(x^2 - 2x + 7).\n$$\nSo the equation becomes\n$$\n(x+2)(x^2 - 2x + 7) = 2 \\cdot p^n. \\tag{1}\n$$\nSince $x^2 - 2x + 7 > x+2$ for all $x \\in \\mathbb{N}$, consider $x+2 = 2 \\cdot p^k$ and $x^2 - 2x + 7 = p^{n-k}$ (or vice versa). For both cases, $2(x^2 - 2x + 7)/(x+2)$ must be integer. Expanding:\n$$\n2(x^2 - 2x + 7) = 2x^2 - 4x + 14 = 2x(x+2) - 8(x+2) + 30.\n$$\nSo $30/(x+2)$ must be integer, i.e., $x+2$ divides 30. Also, $x+2$ can have at most two prime divisors, one of which is 2. Thus, $x+2$ can be $3, 5, 6, 10$, so $x = 1, 3, 4, 8$.\n\n- For $x=1$: $(x+2)(x^2-2x+7) = 3 \\cdot 6 = 18 = 2 \\cdot 3^2$, so $p=3$, $n=2$.\n- For $x=3$: $(x+2)(x^2-2x+7) = 5 \\cdot 10 = 50 = 2 \\cdot 5^2$, so $p=5$, $n=2$.\n- For $x=4$: $(x+2)(x^2-2x+7) = 6 \\cdot 15 = 90 = 2 \\cdot 3^2 \\cdot 5$, which is not of the form $2 \\cdot p^n$.\n- For $x=8$: $(x+2)(x^2-2x+7) = 10 \\cdot 55 = 550 = 2 \\cdot 5^2 \\cdot 11$, which is not of the form $2 \\cdot p^n$.\n\nThus, the only solutions are $(1, 2, 3)$ and $(3, 2, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18857,
    "subject": "Mathematics (Olympiad)",
    "question": "The measures of the smallest angles of three different right triangles sum to $90^\\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3$-$4$-$5$ and $5$-$12$-$13$. What is the perimeter of the third triangle?\n\n(A) 40 (B) 126 (C) 154 (D) 176 (E) 208",
    "options": [],
    "answer": "See solution",
    "solution": "Let the smallest angle of the $3$-$4$-$5$ triangle have measure $\\alpha$, the smallest angle of the $5$-$12$-$13$ triangle have measure $\\beta$, and the smallest angle of the third triangle have measure $\\gamma$. It is given that $\\alpha + \\beta + \\gamma = 90^\\circ$, so $\\cos(\\alpha + \\beta + \\gamma) = 0$. Expanding gives\n\n$$\n\\begin{aligned}\n\\cos(\\alpha + \\beta + \\gamma) &= \\cos(\\alpha + \\beta) \\cos \\gamma - \\sin(\\alpha + \\beta) \\sin \\gamma \\\\\n&= (\\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta) \\cos \\gamma - (\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta) \\sin \\gamma \\\\\n&= \\cos \\alpha \\cos \\beta \\cos \\gamma - \\sin \\alpha \\sin \\beta \\cos \\gamma - \\sin \\alpha \\cos \\beta \\sin \\gamma - \\cos \\alpha \\sin \\beta \\sin \\gamma \\\\\n&= 0.\n\\end{aligned}\n$$\n\nBecause $\\sin \\alpha = \\frac{3}{5}$, $\\cos \\alpha = \\frac{4}{5}$, $\\sin \\beta = \\frac{5}{13}$, and $\\cos \\beta = \\frac{12}{13}$,\n\n$$\n\\begin{aligned}\n\\cos(\\alpha + \\beta + \\gamma) &= \\frac{4}{5} \\cdot \\frac{12}{13} \\cos \\gamma - \\frac{3}{5} \\cdot \\frac{5}{13} \\cos \\gamma - \\frac{3}{5} \\cdot \\frac{12}{13} \\sin \\gamma - \\frac{4}{5} \\cdot \\frac{5}{13} \\sin \\gamma \\\\\n&= \\frac{48}{65} \\cos \\gamma - \\frac{15}{65} \\cos \\gamma - \\frac{36}{65} \\sin \\gamma - \\frac{20}{65} \\sin \\gamma \\\\\n&= \\frac{33}{65} \\cos \\gamma - \\frac{56}{65} \\sin \\gamma \\\\\n&= 0.\n\\end{aligned}\n$$\n\nTherefore $33 \\cos \\gamma = 56 \\sin \\gamma$, so $\\tan \\gamma = \\frac{33}{56}$. Because $33$ and $56$ are relatively prime, perhaps they are the lengths of the two legs of the third triangle. Indeed, $33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2$, and the perimeter of the triangle is $33 + 56 + 65 = 154$.\n\nAlternatively, using complex numbers in polar form, $4 + 3i = 5 \\operatorname{cis} \\alpha$ and $12 + 5i = 13 \\operatorname{cis} \\beta$. Multiplying these gives\n\n$$\n(4 + 3i)(12 + 5i) = 33 + 56i = 65 \\operatorname{cis} \\left(\\frac{\\pi}{2} - \\gamma\\right),\n$$\n\nwhere the triangle with sides $33$, $56$, and $65$ has angle $\\gamma$ opposite $33$ and angle $\\frac{\\pi}{2} - \\gamma$ opposite $56$. Notice that\n\n$$\n\\begin{aligned}\n(5 \\operatorname{cis} \\alpha) \\cdot (13 \\operatorname{cis} \\beta) \\cdot (65 \\operatorname{cis} \\gamma) &= (4 + 3i)(12 + 5i)(56 + 33i) \\\\\n&= 65^2 i \\\\\n&= 65^2 \\operatorname{cis}(\\alpha + \\beta + \\gamma) \\\\\n&= 65^2 \\operatorname{cis} \\frac{\\pi}{2}.\n\\end{aligned}\n$$\n\nThe third triangle has sides of length $33$, $56$, and $65$, and $33 + 56 + 65 = 154$.\n\n![](images/2024_AMC12B_Solutions_p15_data_8d9a1544aa.png)\n\n**Note:** A fun fact about the $33$–$56$–$65$ triangle is that it is the smallest primitive Pythagorean triple that contains no prime values.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18858,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a country has a system of roads such that the roads intersect only in towns, and do not intersect between towns. Furthermore, from any town, one can reach any other town if one can go in either direction on each road. For no pair of towns is there more than one direct road connecting them. The government decided to make each road a one-way road; that is, if towns $A$ and $B$ are connected by a road, then one can use it to get from $A$ to $B$, or from $B$ to $A$. Moreover, for each town, there must be at least one road for coming into that town and at least one road for leaving it. Will it always be the case that in this country, there exists a town from which one can get to any other town (even if going through some other towns on the way), or to which one can get from any other town (even if going through some other towns on the way)?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** not necessarily.\n\nLet us show such a system of roads where such a town does not exist. Denote by $A$, $B$, $C$, $D$ groups of three towns each, where the roads within each group form a cycle, e.g., $A_1 \\to A_2 \\to A_3 \\to A_1$. In this way, the condition that each town has one incoming and one outgoing road is satisfied. Now, we place additional roads with the following directions: $A_1 \\to B_1$, $C_1 \\to B_1$, and $C_1 \\to D_1$. Then, one cannot get to any town of groups $A$ and $C$ from other groups, and one cannot get to any town in $B$ from group $D$, and vice versa. Analogously, from any town of any group, one cannot get to any town from other groups.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18859,
    "subject": "Mathematics (Olympiad)",
    "question": "Jack and Jill play the following game: Jack throws 3 dice and Jill can select some of them, possibly none, and turn each of them to the opposite side. Jill wins if the sum of the values on the dice is a multiple of 4. Can Jill always win?\n\n(Note: The game is played with standard dice where the sum of the numbers on opposite sides is 7.)",
    "options": [],
    "answer": "See solution",
    "solution": "Jill can always turn the dice so that the numbers are $2a$, $2b$, $2c$, i.e., all even. Alternatively, she can achieve $7 - 2a$, $7 - 2b$, $2c$.\n\nLet $S = 2a + 2b + 2c$ and $T = (7 - 2a) + (7 - 2b) + 2c$. Then both $S$ and $T$ are even, and $S + T = 14 + 4c$. Since $S + T$ is even and not a multiple of 4, one of $S$ or $T$ must be a multiple of 4. Therefore, Jill can always win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18860,
    "subject": "Mathematics (Olympiad)",
    "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 $CAP$.",
    "options": [],
    "answer": "See solution",
    "solution": "All angles are directly oriented. Denote by a lowercase letter the affix of the point denoted by the uppercase letter.\n\nThe given similarity rewrites as\n\n$$\n\\frac{m-b}{a-b} = \\frac{n-c}{b-c} = \\frac{p-a}{c-a} = k,\n$$\n\nhence\n\n$$\nm = k a + (1-k) b,\n$$\n\n$$\nn = k b + (1-k) c,\n$$\n\n$$\np = k c + (1-k) a.\n$$\n\nSince the triangle $MNP$ is equilateral, we have\n\n$$\nm + \\varepsilon n + \\varepsilon^2 p = 0,\n$$\n\nwhere $\\varepsilon = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$. Substituting, we infer that\n\n$$\n\\begin{aligned}\n0 &= k(a + b\\varepsilon + c\\varepsilon^2) + (1-k)(b + c\\varepsilon + a\\varepsilon^2) \\\\\n  &= k(a + b\\varepsilon + c\\varepsilon^2) + \\frac{1-k}{\\varepsilon}(a + b\\varepsilon + c\\varepsilon^2) \\\\\n  &= (a + b\\varepsilon + c\\varepsilon^2)\\left(k + \\frac{1-k}{\\varepsilon}\\right).\n\\end{aligned}\n$$\n\nThe triangle $ABC$ is not equilateral, so $a + b\\varepsilon + c\\varepsilon^2 \\neq 0$, and consequently\n\n$$\nk = \\frac{1}{1-\\varepsilon}.\n$$\n\nThe equality $m = k a + (1-k) b$ 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": 18861,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcentre of the acute triangle $ABC$. Let $c_1$ and $c_2$ be the circumcircles of triangles $ABO$ and $ACO$. Let $P$ and $Q$ be points on $c_1$ and $c_2$ respectively, such that $OP$ is a diameter of $c_1$ and $OQ$ is a diameter of $c_2$. Let $T$ be the intersection of the tangent to $c_1$ at $P$ and the tangent to $c_2$ at $Q$. Let $D$ be the second intersection of the line $AC$ and the circle $c_1$. Prove that the points $D$, $O$ and $T$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle OAP = \\angle OAQ = 90^\\circ$, the points $P$, $A$ and $Q$ are collinear. Since $\\angle OPT = \\angle OQT = 90^\\circ$, $OPTQ$ is cyclic. Since $OA = OB$, the diameter $OP$ of $c_1$ is perpendicular to the chord $AB$. Therefore $PT$ and $AB$ are parallel. Now $\\angle TOQ = \\angle TPQ = \\angle TPA = \\angle BAP = \\angle BOP = 90^\\circ - \\angle ABO$. On the other hand, equality of inscribed angles subtending the arc $AO$ of circle $c_1$ gives $\\angle CDO = \\angle ABO$. Therefore $\\angle DOQ = 90^\\circ - \\angle CDO = 90^\\circ - \\angle ABO$. In summary, $\\angle TOQ = \\angle DOQ$, whence $D$, $O$ and $T$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18862,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for positive $a$, $b$, $c$ and $l > m$,\n\n$$\n\\frac{a^{3l} + a^{3m} + 1}{a^l + b^{l-m}a^m + b^l} + \\frac{b^{3l} + b^{3m} + 1}{b^l + c^{l-m}b^m + c^l} + \\frac{c^{3l} + c^{3m} + 1}{c^l + a^{l-m}c^m + a^l} \\geq a^m + b^m + c^m.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that the Cauchy--Schwarz inequality can be written as\n\n$$\n\\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} \\geq \\frac{(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})^2}{a + b + c}.\n$$\n\nAll sums in the following inequalities are cyclic:\n\n$$\n\\begin{align*}\n\\sum \\frac{a^{3l} + a^{3m} + 1}{a^l + b^{l-m}a^m + b^l} &\\geq \\sum \\frac{3a^{l+m}}{a^l + b^{l-m}a^m + b^l} && \\text{(AM-GM in the numerators)} \\\\\n&= 3 \\sum \\frac{a^{2l}}{a^{l-m}a^l + b^{l-m}a^l + b^l a^{l-m}} \\\\\n&\\geq 3 \\frac{(a^l + b^l + c^l)^2}{(a^{l-m} + b^{l-m} + c^{l-m})(a^l + b^l + c^l)} && \\text{(Cauchy-Schwarz for the whole sum)} \\\\\n&= 3 \\frac{a^l + b^l + c^l}{a^{l-m} + b^{l-m} + c^{l-m}} \\\\\n&\\geq a^m + b^m + c^m. && \\text{(Chebyshev's sum inequality)}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18863,
    "subject": "Mathematics (Olympiad)",
    "question": "Given eight points $A_1, A_2, \\dots, A_8$ on a circle, determine the smallest positive integer $n$ such that among any $n$ triangles with vertices in these eight points, there are two which have a common side.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the maximal number of triangles with no common side pairwise. There are $\\binom{8}{2} = 28$ chords by connecting eight points. If each chord belongs to only one triangle, then these chords can form at most $r \\leq \\left\\lfloor \\frac{28}{3} \\right\\rfloor = 9$ triangles with no common side pairwise. But if there are nine such triangles, then there are 27 sides, so one point among the eight must be a common vertex of four triangles.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p288_data_d78738cfe4.png)\n\nSuppose that point is $A_8$. Then eight edges are connected to the seven other points $A_1, A_2, \\dots, A_7$. Thus, there must exist an edge $A_8A_k$ which is the common side of two triangles, a contradiction. So $r \\leq 8$.\n\nOn the other hand, when $r = 8$, we can construct eight such triangles (see figure). Denote the triangles by their vertices: $(1,2,8)$, $(1,3,6)$, $(1,4,7)$, $(2,3,4)$, $(2,5,7)$, $(3,5,8)$, $(4,5,6)$, and $(6,7,8)$. Thus, the minimal number $n$ is $9$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p288_data_549cd60c8d.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18864,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a bounded number sequence $\\{a_n\\}$ satisfies\n\n$$\na_n < \\sum_{k=n}^{2n+2006} \\frac{a_k}{k+1} + \\frac{1}{2n+2007}, \\quad n = 1, 2, 3, \\dots\n$$\n\nProve that $a_n < \\frac{1}{n}$ for $n = 1, 2, 3, \\dots$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_n = a_n - \\frac{1}{n}$. It is routine to check that\n\n$$\nb_n < \\sum_{k=n}^{2n+2006} \\frac{b_k}{k+1}, \\quad n = 1, 2, 3, \\dots. \\qquad \\textcircled{1}\n$$\n\nWe will prove that $b_n < 0$. As $\\{a_n\\}$ is bounded, there exists $M$ such that $b_n < M$. When $n > 100\\,000$, we have\n\n$$\n\\begin{align*}\nb_n &< \\sum_{k=n}^{2n+2006} \\frac{b_k}{k+1} \\\\\n&< M \\sum_{k=n}^{2n+2006} \\frac{1}{k+1} \\\\\n&= M \\sum_{k=n}^{\\lfloor \\frac{3n}{2} \\rfloor} \\frac{1}{k+1} + M \\sum_{k=\\lfloor \\frac{3n}{2} \\rfloor+1}^{2n+2006} \\frac{1}{k+1} \\\\\n&< M \\cdot \\frac{1}{2} + M \\cdot \\frac{\\frac{n}{2} + 2006}{\\frac{3n}{2} + 1} \\\\\n&< \\frac{6}{7} M,\n\\end{align*}\n$$\n\nwhere $\\lfloor x \\rfloor$ is the greatest integer less than or equal to $x$.\n\nWe can substitute $\\frac{6}{7}M$ for $M$, and repeat the previous steps. Then for any $m \\in \\mathbb{N}$ we have\n\n$$\nb_n < \\left(\\frac{6}{7}\\right)^m M,\n$$\n\nwhich implies that $b_n \\le 0$ for $n \\ge 100\\,000$. Substituting this into (1), we get $b_n < 0$ for $n \\ge 100\\,000$.\n\nWe observe in (1) that, if for any $n \\ge N+1$, $b_n < 0$ then ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18865,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $\\big(P, d\\big)$, where $P$ is a polynomial with integer coefficients and $d$ is an integer, such that the equation\n\n$$\nP(x) - P(y) = d\n$$\n\nhas infinitely many solutions in integers $x$ and $y$ with $x \\neq y$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $x - y$ divides $P(x) - P(y)$. Thus, if $d \\neq 0$, there are only finitely many possible values for $x - y$. For one such value, say $a$, there must be infinitely many pairs $(x, y)$ with $x - y = a$ and $P(x) - P(y) = d$. Consider\n\n$$\nP(x) - P(x - a) - d\n$$\n\nas a polynomial in $x$. If it has infinitely many zeros, it must be identically zero, so $P(x) = \\frac{d}{a}x + b$ for some integer $b$. Thus, for any integer $d$ and any integer $a$ dividing $d$, $P(x) = \\frac{d}{a}x + b$ is a solution. This includes the case $d = 0$ and $P(x)$ constant.\n\nNow, consider $d = 0$. If $P$ has odd degree, then for large $x$, $P(x)$ is strictly monotonic, so $P(x) - P(y) = 0$ only for finitely many $(x, y)$ with $x \\neq y$. Thus, $P$ must have even degree. For even degree, $P(x)$ can be written as $P(x) = a x^n + b x^{n-1} + \\dots$ with $a > 0$ (otherwise replace $P$ by $-P$). For large $x$, $P(x)$ is increasing for $x \\geq A$ and decreasing for $x \\leq B$ for some $A, B$. Infinitely many solutions $P(x) = P(y)$ with $x \\neq y$ can only occur if $x + y$ is constant. Setting $an(x + y) = -2b$, we find that $P(x) = P\\left(\\frac{-2b}{an} - x\\right)$ for all $x$, so $P$ is symmetric about $x = \\frac{-b}{an}$. Thus, $P(x) = R((x - c)^2)$ for some polynomial $R$ and $c$ with $2c$ integer.\n\n**Summary:**\n- For any integer $d$ and any integer $a$ dividing $d$, $P(x) = \\frac{d}{a}x + b$ (with integer $b$) is a solution.\n- For $d = 0$, all polynomials of the form $P(x) = R((x - c)^2)$, where $R$ has integer coefficients and $2c$ is integer, are solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18866,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a cube with vertices labeled $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$. A robot starts at a vertex and moves along edges according to certain probabilities:\n\n**a)** At $D$, what is the probability the robot chooses the vertical edge $DH$ (given choices $DC$ and $DH$)? At $E$, what is the probability the robot chooses the horizontal edge $EH$ (given choices $EF$ and $EH$)?\n\n**b)** Starting at $A$, what is the probability the robot traces the trail $ABCG$ (choosing edges $AB$, $BC$, $CG$ in order)?\n\n**c)** List all trails of length 3 from $A$ to $G$ and find the probability the robot traces the trails $ABFG$ and $AEFG$.\n\n**d)** Using the results from parts b and c, what is the total probability that the robot traces a path of length 3 from $A$ to $G$?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)**\nAt $D$, the choices are the horizontal edge $DC$ and the vertical edge $DH$. So the probability of choosing $DH$ is $\\frac{2}{3}$.\n\nAt $E$, the choices are the horizontal edge $EF$ and the horizontal edge $EH$. So the probability of choosing $EH$ is $\\frac{1}{2}$.\n\n**b)**\nStarting at $A$, there is a choice of vertical edge $AE$ and horizontal edges $AB$ and $AD$. So the probability of choosing $AB$ is $\\frac{1}{6}$.\n\nAt $B$, the choices are vertical edge $BF$ and horizontal edge $BC$. So the probability of choosing $BC$ is $\\frac{1}{3}$.\n\nAt $C$, the choices are vertical edge $CG$ and horizontal edge $CD$. So the probability of choosing $CG$ is $\\frac{2}{3}$.\n\nSo the probability the robot traces the trail $ABCG$ is:\n$$\n\\frac{1}{6} \\times \\frac{1}{3} \\times \\frac{2}{3} = \\frac{1}{27}\n$$\n\n**c)**\nThere are six trails of length 3 from $A$ to $G$:\n\n- $ABCG$ and $ADCG$ (edge sequence: horizontal, horizontal, vertical)\n- $ABFG$ and $ADHG$ (edge sequence: horizontal, vertical, horizontal)\n- $AEFG$ and $AEHG$ (edge sequence: vertical, horizontal, horizontal)\n\nThe probability that the robot traces trail $ABFG$ is:\n$$\n\\frac{1}{6} \\times \\frac{2}{3} \\times \\frac{1}{2} = \\frac{1}{18}\n$$\n\nThe probability that the robot traces trail $AEFG$ is:\n$$\n\\frac{2}{3} \\times \\frac{1}{2} \\times \\frac{1}{3} = \\frac{1}{9}\n$$\n\n**d)**\nFrom the solutions to parts b and c, the probability of the robot tracing a path of length 3 to $G$ is:\n$$\n2 \\times \\left(\\frac{1}{27} + \\frac{1}{18} + \\frac{1}{9}\\right) = 2 \\times \\left(\\frac{2}{54} + \\frac{3}{54} + \\frac{6}{54}\\right) = 2 \\times \\frac{11}{54} = \\frac{11}{27}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18867,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, points $K$ and $L$ are in the interior of triangle $ABC$, and point $D$ lies on side $AB$. It is known that points $B$, $K$, $L$, $C$ are concyclic, and $\\angle AKD = \\angle BCK$, $\\angle ALD = \\angle BCL$. Prove that $AK = AL$.\n\n![](images/2022_CGMO_p8_data_e7c4531781.png)",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the following picture,\n\n![](images/2022_CGMO_p8_data_e0ac43208c.png)\n\nLet the extensions of $AK$ and $AL$ intersect the circle passing through $B$, $K$, $L$, $C$ at points $X$ and $Y$, respectively. Connect $BX$ and $BY$. Considering the given conditions, we have $\\angle AKD = \\angle BCK = \\angle BXK$ and $\\angle ALD = \\angle BCL = \\angle BYL$.\n\nThus, $DK \\parallel BX$ and $DL \\parallel BY$. So we have\n\n$$\n\\frac{AK}{AX} = \\frac{AD}{AB} = \\frac{AL}{AY}.\n$$\n\nSince $K$, $L$, $Y$, $X$ are concyclic, by the Power of a Point theorem, we have\n\n$$\nAK \\cdot AX = AL \\cdot AY.\n$$\n\nMultiplying the two equations above, we get $AK^2 = AL^2$, which implies $AK = AL$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18868,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest possible area of a rectangle that can be tiled with squares of integer side lengths, each at least $1$, such that no two squares are the same size?\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p82_data_0ee7735fb2.png)\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p82_data_d8baf03322.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ be the length of the smallest square $A$. If $A$ touches a side but not a corner, then it is surrounded by two larger squares $B$ and $C$. Then there is a gap between $B$ and $C$ with length $a$. It is not possible to place some larger squares in this gap, contradiction. Similarly, the smallest square cannot be placed at a corner.\n\nThe distance from $A$ to a side is at least $a+1$. Therefore, each side has length at least\n\n$$\n(a+1) \\times 2 + a = 3a + 2 \\ge 5.\n$$\n\nIf both side lengths are at least $6$, then the area is at least $36$. Thus, we may assume there is a side with length $5$, and $a=1$.\n\nSince there is no square of side length $1$ touching this side, either there is a $5 \\times 5$ square, or a $3 \\times 3$ and a $2 \\times 2$ squares touching this side. The former case can be ignored since we can remove such a $5 \\times 5$ square. Now, suppose there is a $3 \\times 3$ square $B$ and a $2 \\times 2$ square $C$ as shown. Then $A$ cannot be placed in the first $3$ columns (or there is a gap of length $1$). The same holds for the opposite side. Therefore, the other side length of the rectangle is at least $3+1+3=7$. This shows the area is at least $5 \\times 7 = 35$.\n\nAn example of a $5 \\times 7$ rectangle is given above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18869,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$ congruent to $3$ modulo $4$, show that the equation\n$$w^{2p} + x^{2p} + y^{2p} = z^{2p}$$\nhas no integer solutions $w, x, y, z$ such that their product is not divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist integers $w, x, y, z$ satisfying the equation and such that $p$ does not divide $wxyz$. Without loss of generality, assume $(w, x, y, z) = 1$ (they are coprime). Reducing modulo $4$, $z$ and exactly one of $w, x, y$, say $y$, must be odd. Consider:\n$$w^{2p} + x^{2p} = z^{2p} - y^{2p} = (z^2 - y^2) \\left(\\sum_{k=1}^{p-1} z^{2(p-k-1)} (z^{2k} - y^{2k}) - p z^{2(p-1)}\\right)$$\nThe second factor is congruent to $3$ modulo $4$, so in its prime factorization, some prime $q \\equiv 3 \\pmod{4}$ appears with odd exponent. Since $-1$ is a quadratic non-residue modulo $q$, both $w$ and $x$ are divisible by $q$, so $q$ appears with even exponent in $w^{2p} + x^{2p}$. Thus, $z^2 - y^2$ is divisible by $q$, and so is $p z^{2(p-1)}$. Since $p \\neq q$ (as $q$ divides $w$ but $p$ does not), $z$ is divisible by $q$, and so is $y$. Therefore, $w, x, y, z$ all share the common factor $q$, contradicting their coprimality.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18870,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, suppose $A_1, A_2, \\dots, A_n$ are $n$ nonempty finite sets satisfying $|A_i \\Delta A_j| = |i - j|$ for all $i, j \\in \\{1, 2, \\dots, n\\}$.\n\nFind the minimum value of $|A_1| + |A_2| + \\cdots + |A_n|$.\n\n(Here $|X|$ denotes the number of elements of a finite set $X$ and $X \\Delta Y = \\{ a \\mid a \\in X, a \\notin Y \\} \\cup \\{ a \\mid a \\in Y, a \\notin X \\}$ for any sets $X$ and $Y$.)",
    "options": [],
    "answer": "See solution",
    "solution": "For each positive integer $k$, we prove that the minimum value of $S_{2k}$ is $k^2 + 2$; the minimum value of $S_{2k+1}$ is $k(k+1) + 2$.\n\nFirstly, define the sets $A_1, A_2, \\dots, A_{2k}, A_{2k+1}$ as follows:\n\n$$\n\\begin{align*}\nA_i &= \\{i, i+1, \\dots, k\\}, \\quad i = 1, 2, \\dots, k; & A_{k+1} &= \\{k, k+1\\}; \\\\\nA_{k+j} &= \\{k+1, k+2, \\dots, k+j-1\\}, \\quad j = 2, 3, \\dots, k+1.\n\\end{align*}\n$$\n\nFor this family of sets, it is easy to verify that $|A_i \\Delta A_j| = |i - j|$ holds in the following cases:\n\n1. $1 \\le i < j \\le k;$\n2. $1 \\le i < j = k + 1;$\n3. $1 \\le i < k + 1 < j \\le 2k + 1;$\n4. $k + 1 = i < j \\le 2k + 1;$\n5. $k + 2 \\le i < j \\le 2k + 1.$\n\nMoreover, the case of $i = j$ is trivial, and the case of $i > j$ can be reduced to the case of $i < j$. Thus, for all $i, j \\in \\{1, 2, \\dots, 2k+1\\}$, we have verified that\n\n$$\n|A_i \\Delta A_j| = |i - j|.\n$$\n\nFor the above $(2k+1)$ sets, we can easily calculate that\n\n$$\nS_{2k+1} = \\frac{k(k+1)}{2} + 2 + \\frac{k(k+1)}{2} = k(k+1) + 2;\n$$\n\nif we choose the first $2k$ sets, we get that\n\n$$\nS_{2k} = S_{2k+1} - k = k^2 + 2.\n$$\n\nSecondly, we show that $S_{2k} \\ge k^2 + 2$ and $S_{2k+1} \\ge k(k+1) + 2$. Note the following facts:\n\n* **Fact 1.** For any two finite sets $X, Y$, we have $|X| + |Y| \\ge |X \\Delta Y|$.\n* **Fact 2.** For any two non-empty finite sets $X, Y$, if $|X \\Delta Y| = 1$, then $|X| + |Y| \\ge 3$.\n\nWhen $n = 2k$, it follows from Fact 1 that\n\n$$\n\\begin{aligned}\n|A_i| + |A_{2k+1-i}| &\\ge |A_i \\Delta A_{2k+1-i}| = 2k + 1 - 2i, \\quad i = 1, 2, \\dots, k-1.\n\\end{aligned}\n$$\n\nBy $|A_k \\Delta A_{k+1}| = 1$ and Fact 2, we have $|A_k| + |A_{k+1}| \\ge 3$. So\n\n$$\n\\begin{align*}\nS_{2k} &= |A_k| + |A_{k+1}| + \\sum_{i=1}^{k-1} (|A_i| + |A_{2k+1-i}|) \\\\\n&\\ge 3 + \\sum_{i=1}^{k-1} (2k + 1 - 2i) = k^2 + 2.\n\\end{align*}\n$$\n\nSimilarly, when $n = 2k + 1$, we get that\n\n$$\n\\begin{align*}\n|A_i| + |A_{2k+2-i}| &\\ge |A_i \\Delta A_{2k+2-i}| = 2k + 2 - 2i, \\quad i = 1, 2, \\dots, k-1.\n\\end{align*}\n$$\n\nSince $|A_k \\Delta A_{k+1}| = 1$, we have\n\n$$\n(|A_k| + |A_{k+1}|) + |A_{k+2}| \\ge 3 + 1 = 4,\n$$\n\nso\n\n$$\n\\begin{align*}\nS_{2k+1} &= |A_k| + |A_{k+1}| + |A_{k+2}| + \\sum_{i=1}^{k-1} (|A_i| + |A_{2k+2-i}|) \\\\\n&\\ge 4 + \\sum_{i=1}^{k-1} (2k + 2 - 2i) = k(k+1) + 2.\n\\end{align*}\n$$\n\nIn conclusion, the minimum value of $S_{2k}$ is $k^2 + 2$, and the minimum value of $S_{2k+1}$ is $k(k+1) + 2$. Equivalently, for any $n \\ge 2$, the minimum value of $S_n$ is $\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18871,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that:\n\n$$\nf(xf(x) + f(y)) = f(f(x^2)) + y, \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the solutions are $f(x) = x$ and $f(x) = -x$.\n\nFor $x = 0$, we obtain $f(f(y)) = f(f(0)) + y$ for each $y \\in \\mathbb{R}$, so $f$ is a one-to-one function. For $y = 0$, we have $f(xf(x) + f(0)) = f(f(x^2))$ for all $x \\in \\mathbb{R}$, which, due to $f$ being one-to-one, leads to $xf(x) + f(0) = f(x^2)$ for all $x \\in \\mathbb{R}$.\n\nPutting $x = 1$ in the last relation gives $f(0) = 0$, so the last relation can be written as $xf(x) = f(x^2)$ for all $x \\in \\mathbb{R}$, while the first relation becomes $f(f(x)) = x$ for all $x \\in \\mathbb{R}$. Putting $x \\to f(x)$ 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\nSuppose there 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 $x = x_0$ and $y = y_0$ in the given functional equation, we obtain $f(x_0^2 - y_0) = x_0^2 + y_0$, so $x_0^2 - y_0 = x_0^2 + y_0$ or $x_0^2 - y_0 = -x_0^2 - y_0$, implying that $y_0 = 0$ or $x_0 = 0$, which is a contradiction. Therefore, the claim is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18872,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$, defined on the real numbers and taking real values, which satisfy the equation\n$$\nf(x)f(y) = f(x+y) + xy\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "The most important idea with a question such as this is to specialise. The equation we are given is very general: it is easiest to read off information from special cases. Choosing useful special cases is an art form, but the following solution contains some methods which are frequently useful.\n\nWe start by making some substitutions which seem likely to give helpful bits of information. Setting $y = 0$ gives\n$$\nf(x)f(0) = f(x)\n$$\nwhence $f(x)(f(0) - 1) = 0$, so either $f(x) = 0$ for all $x$, or $f(0) = 1$. The former is untenable (for example, from substituting $x = y = 1$), so we can conclude that $f(0) = 1$.\n\nWe can use this by substituting $y = -x$: we get\n$$\nf(x)f(-x) = f(0) - x^2 = 1 - x^2.\n$$\nIn particular,\n$$\nf(1)f(-1) = 1 - 1^2 = 0.\n$$\nThus we have $f(1) = 0$ or $f(-1) = 0$.\n\nIf $f(1) = 0$, then we can use this by substituting $y = 1$ to get\n$$\n0 = f(x)f(1) = f(x + 1) + x.\n$$\nso $f(x + 1) = -x$ or equivalently $f(x) = 1 - x$. It is easy to check that this is indeed a valid solution.\n\nOn the other hand, if $f(-1) = 0$, then we can substitute $y = -1$ and get\n$$\n0 = f(x)f(-1) = f(x - 1) - x.\n$$\nso $f(x - 1) = x$ or equivalently $f(x) = 1 + x$. Again, this is easily checkable.\n\nSo in summary we have just the two solutions $f(x) = 1 + x$ and $f(x) = 1 - x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18873,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $ABCD$ un paralelogramo con $\\angle ABC = 105^\\circ$. En el interior del paralelogramo existe un punto $E$ tal que el triángulo $BEC$ es equilátero y $\\angle CED = 135^\\circ$. Sea $K$ el punto medio del lado $AB$. Calcular la medida del ángulo $BKC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Como $ABCD$ es un paralelogramo, se tiene que\n\n$$\n\\angle BAD = \\angle BCD = 75^\\circ.\n$$\n\nEntonces,\n\n$$\n\\angle ECD = 75^\\circ - 60^\\circ = 15^\\circ \\quad \\text{y} \\quad \\angle EDC = 30^\\circ.\n$$\n\nTrazamos por $C$ la perpendicular a la recta $DE$, que la corta en $F$. Entonces\n\n$$\n\\angle FEC = 180^\\circ - 135^\\circ = 45^\\circ,\n$$\n\ny como $\\angle EFC = 90^\\circ$, se tiene que $\\angle FCE = 45^\\circ$, por lo que $CF = EF$.\n\nPor otra parte, el triángulo $CFD$ es la mitad de un triángulo equilátero, pues es rectángulo en $F$ y $\\angle CDF = \\angle CDE = 30^\\circ$. Luego,\n\n$$\n\\angle FCD = 60^\\circ \\quad \\text{y} \\quad CF = \\frac{1}{2}CD = \\frac{1}{2}AB = BK.\n$$\n\nComparamos los triángulos $BKE$ y $CFE$: $CF = BK$; $BE = CE$ por ser lados del triángulo equilátero; $\\angle KBE = 105^\\circ - 60^\\circ = 45^\\circ$ y\n\n$$\n\\angle FCE = 45^\\circ,\n$$\n\npor lo tanto, son iguales. Esto implica que $KE = FE$; $\\angle BKE = \\angle CFE = 90^\\circ$ y\n\n$$\n\\angle BEK = \\angle CEF = 45^\\circ.\n$$\n\nEn el triángulo $AKE$ tenemos que $AK = BK = KE$, además $\\angle AKE = 90^\\circ$, por lo tanto los triángulos $AKE$ y $BKE$ son iguales y $\\angle KEA = 45^\\circ$, lo que implica que $\\angle AEB = 2 \\cdot 45^\\circ = 90^\\circ$.\n\nEn el cuadrilátero $EKBC$ tenemos que $KE = KB$ y $CB = CE$, entonces es un romboide y sus diagonales son perpendiculares. Si $O$ es el punto de intersección de $KC$ y $BE$, resulta que $\\angle BOK = 90^\\circ$ y como $\\angle KBE = 45^\\circ$, tenemos que\n\n$$\n\\angle BKC = \\angle BKO = 180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ.\n$$\n\n![](images/Soluciones_nacional_OMA_2020_2_p3_data_bf36e57897.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18874,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d, e, f$ be (not necessarily distinct) divisors of $210$. Find the number of 6-permutations $(a, b, c, d, e, f)$ that satisfy the condition $abcdef > 210^3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $abcdef > 210^3$, we have:\n\n$$\n\\frac{210}{a} \\cdot \\frac{210}{b} \\cdot \\frac{210}{c} \\cdot \\frac{210}{d} \\cdot \\frac{210}{e} \\cdot \\frac{210}{f} < 210^3.\n$$\n\nThe number of 6-permutations satisfying $abcdef > 210^3$ equals the number satisfying $abcdef < 210^3$. The number of 6-permutations with $abcdef = 2^3 \\cdot 3^3 \\cdot 5^3 \\cdot 7^3$ is $(\\binom{6}{3})^4$. Therefore, the number of 6-permutations $(a, b, c, d, e, f)$ with $abcdef > 210^3$ is:\n\n$$\n\\frac{16^6 - (\\binom{6}{3})^4}{2}.\n$$\n\nNow, if we set $f = 210$, then $abcde > 210^2$, and the desired number is:\n\n$$\n\\frac{1}{6} \\cdot \\frac{16^6 - (\\binom{6}{3})^4}{2}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18875,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ with $a + b + c = 3$.\n\n(a) Prove that\n$$\na\\sqrt{b} + b\\sqrt{c} + c\\sqrt{a} \\leq 3.\n$$\n\n(b) Prove that\n$$\n\\sqrt{a^3 b} + \\sqrt{b^3 c} + \\sqrt{c^3 a} + \\sqrt{ab^3} + \\sqrt{bc^3} + \\sqrt{ca^3} \\leq \\frac{2}{3}(a+b+c)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, note that\n$$\n2(a + b + c)^2 - 6(ab + bc + ca) = (a - b)^2 + (b - c)^2 + (c - a)^2 \\geq 0\n$$\nso $ab + bc + ca \\leq \\frac{1}{3}(a + b + c)^2 = 3$. Now, use Cauchy-Schwarz with vectors $(\\sqrt{a}, \\sqrt{b}, \\sqrt{c})$ and $(\\sqrt{ab}, \\sqrt{bc}, \\sqrt{ca})$ to get\n$$\na\\sqrt{b} + b\\sqrt{c} + c\\sqrt{a} \\leq \\sqrt{a + b + c} \\cdot \\sqrt{ab + bc + ca} \\leq \\sqrt{3} \\cdot \\sqrt{3} = 3.\n$$\n\n(b) Homogenizing the inequality, we want to prove that\n$$\n\\sqrt{a^3 b} + \\sqrt{b^3 c} + \\sqrt{c^3 a} + \\sqrt{ab^3} + \\sqrt{bc^3} + \\sqrt{ca^3} \\leq \\frac{2}{3}(a + b + c)^2.\n$$\n\nIntroduce $T(a, b) := a^2 + b^2 + 4ab - 3\\sqrt{a^3 b} - 3\\sqrt{ab^3}$. Multiply the above by $3$ and rearrange to obtain $T(a, b) + T(b, c) + T(c, a) \\geq 0$. To show $T(a, b) \\geq 0$, define $x = \\sqrt{a}$ and $y = \\sqrt{b}$. This gives\n$$\nT(a, b) = x^4 + y^4 + 4x^2 y^2 - 3x^3 y - 3x y^3 = (x - y)^2(x^2 - x y + y^2) \\geq 0.\n$$\nSimilarly, $T(b, c) \\geq 0$ and $T(c, a) \\geq 0$. This proves the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18876,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Consider the following sets, each containing 335 elements:\n\n$$\n\\begin{align*}\nA &= \\{6, 12, \\dots, 2010\\}, & B &= \\{3, 9, 15, \\dots, 2007\\}, \\\\\nC &= \\{1, 7, 13, \\dots, 2005\\}, & D &= \\{2, 8, 14, \\dots, 2006\\}, \\\\\nE &= \\{4, 10, 16, \\dots, 2008\\}, & F &= \\{5, 11, 17, \\dots, 2009\\}.\n\\end{align*}\n$$\n\nIf $S$ contains two elements from $A$ or two elements from $B$, their sum is divisible by $6$. If not, the remaining four 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 $S \\cap F$ each contain at least one element. Their sum is a multiple of $6$, which is what was to be proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18877,
    "subject": "Mathematics (Olympiad)",
    "question": "ABC is a triangle with a right angle at $C$. $M_1$ and $M_2$ are two arbitrary points inside $ABC$, and $M$ is the midpoint of $M_1M_2$. The extensions of $BM_1$, $BM$, and $BM_2$ intersect $AC$ at $N_1$, $N$, and $N_2$ respectively.\n\nProve that\n$$\n\\frac{M_1N_1}{BM_1} + \\frac{M_2N_2}{BM_2} \\ge 2 \\frac{MN}{BM}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H_1$, $H_2$, and $H$ be the projections from $M_1$, $M_2$, and $M$ to the line $BC$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p178_data_d7c07b89f9.png)\n\n$$\n\\begin{aligned}\n\\frac{M_1 N_1}{B M_1} &= \\frac{H_1 C}{B H_1}, \\\\\n\\frac{M_2 N_2}{B M_2} &= \\frac{H_2 C}{B H_2}, \\\\\n\\frac{M N}{B M} &= \\frac{H C}{B H} = \\frac{H_1 C + H_2 C}{B H_1 + B H_2}.\n\\end{aligned}\n$$\n\nNow suppose that $BC = 1$, $BH_1 = x$, $BH_2 = y$. Then\n\n$$\n\\begin{aligned}\n\\frac{M_1 N_1}{B M_1} &= \\frac{H_1 C}{B H_1} = \\frac{1-x}{x}, \\\\\n\\frac{M_2 N_2}{B M_2} &= \\frac{H_2 C}{B H_2} = \\frac{1-y}{y}, \\\\\n\\frac{M N}{B M} &= \\frac{H C}{B H} = \\frac{1-x+1-y}{x+y}.\n\\end{aligned}\n$$\n\nSo it is enough to prove that\n\n$$\n\\frac{1-x}{x} + \\frac{1-y}{y} \\ge 2 \\frac{1-x+1-y}{x+y},\n$$\n\nwhich is equivalent to $\\frac{1}{x} + \\frac{1}{y} \\ge \\frac{4}{x+y}$, i.e., $(x-y)^2 \\ge 0$.\n\nThis is obviously true.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18878,
    "subject": "Mathematics (Olympiad)",
    "question": "Two straight pipes (circular cylinders), with radii $1$ and $\\frac{1}{4}$, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?\n\n![](images/2024_AMC10B_Solutions_p10_data_0ecbac27f6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two possible positions for the third pipe—either nestled in the gap between the pipes or outside. See the figure below.\n\n![](images/2024_AMC10B_Solutions_p10_data_897cb48644.png)\n\nConsider the blown-up figure below. In this diagram, $A$ is the center of the circle of radius $1$, $B$ is the center of the circle of radius $\\frac{1}{4}$, and $C$ is the center of the third circle nestled in the gap. The horizontal lines through $B$ and $C$ intersect the vertical line through $A$ at $D$ and $E$, respectively, and $F$ is the foot of the perpendicular from $C$ to $BD$.\n\n![](images/2024_AMC10B_Solutions_p11_data_a8e560f859.png)\n\nBecause $AB = 1 + \\frac{1}{4} = \\frac{5}{4}$ and $AD = 1 - \\frac{1}{4} = \\frac{3}{4}$, it follows that $\\triangle ADB$ is a $3$-$4$-$5$ right triangle scaled down by a factor of $4$, so $BD = \\frac{4}{4} = 1$. Thus the vertical line through $B$ is tangent to the given circle of radius $1$. Then by symmetry, the radius of the larger of the two dashed circles tangent to both given circles has radius $1$.\n\nIt remains to compute the radius $r$ of the smaller dashed tangent circle. Let $x = CE = DF$. The Pythagorean Theorem in $\\triangle AEC$ gives $x^2 + (1 - r)^2 = (1 + r)^2$, which simplifies to $x^2 = 4r$. The Pythagorean Theorem in $\\triangle BFC$ gives\n\n$$\n(1 - x)^2 + \\left(1 - \\frac{3}{4} - r\\right)^2 = \\left(\\frac{1}{4} + r\\right)^2,\n$$\n\nwhich simplifies to $(1 - x)^2 = r$. Combining these equations gives $x^2 = 4(1 - x)^2$, which is equivalent to $3x^2 - 8x + 4 = 0$ and $(3x - 2)(x - 2) = 0$. Because $x < 1$, the relevant solution is $x = \\frac{2}{3}$, and the radius of the smaller circle is $\\frac{1}{4} \\cdot \\left(\\frac{2}{3}\\right)^2 = \\frac{1}{9}$.\n\nThe requested sum of possible radii is $1 + \\frac{1}{9} = \\frac{10}{9}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 18879,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n$, determine the greatest possible value of the quotient\n\n$$\n\\frac{1 - x^n - (1-x)^n}{x(1-x)^n + (1-x)x^n}\n$$\n\nwhere $0 < x < 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 1 - x$, so $x + y = 1$. The expression becomes\n\n$$\n\\frac{1 - x^n - y^n}{x y^n + y x^n}\n$$\n\nWe claim the maximum occurs at $x = y = \\frac{1}{2}$, giving value $2^n - 2$.\n\nRewrite:\n\n$$\n\\begin{aligned}\n\\frac{1 - x^n - y^n}{x y^n + y x^n} &= \\frac{x + y - x^n - y^n}{x y^n + y x^n} \\\\\n&= \\frac{x(1 - x^{n-1}) + y(1 - y^{n-1})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x(1-x)(1 + x + \\cdots + x^{n-2}) + y(1-y)(1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x y (1 + x + \\cdots + x^{n-2} + 1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{1 + 1}{x^{n-1} + y^{n-1}} + \\frac{x + y}{x^{n-1} + y^{n-1}} + \\cdots + \\frac{x^{n-2} + y^{n-2}}{x^{n-1} + y^{n-1}}.\n\\end{aligned}\n$$\n\nWe need to show that\n\n$$\n\\frac{x^a + y^a}{x^b + y^b}\n$$\n\nis maximized at $x = y = \\frac{1}{2}$ for $0 \\leq a < b$. By the general mean inequality:\n\n$$\n\\left( \\frac{x^a + y^a}{2} \\right) \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{a/b}\n$$\n\nand\n\n$$\n\\frac{1}{2} = \\frac{x + y}{2} \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{1/b},\n$$\n\nwith equality when $x = y = \\frac{1}{2}$. Thus,\n\n$$\n\\frac{x^a + y^a}{x^b + y^b} \\leq 2^{b-a},\n$$\n\nwith equality at $x = y = \\frac{1}{2}$.\n\nTherefore, the maximum value is $2^n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18880,
    "subject": "Mathematics (Olympiad)",
    "question": "Are there integers $a < b < c < d$ such that\n$$\n\\frac{a}{a} + \\frac{a}{b} + \\frac{a}{c} + \\frac{a}{d} = \\frac{b}{a} + \\frac{b}{b} + \\frac{b}{c} + \\frac{b}{d}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, for example $a = -28$, $b = -14$, $c = -7$, and $d = 4$.\n\nClearly, all numbers cannot be positive. Let's take $a = -4$, $b = -2$, $c = -1$ and find the corresponding $d$:\n$$\n\\begin{aligned}\n\\frac{-4}{-4} + \\frac{-4}{-2} + \\frac{-4}{-1} + \\frac{-4}{d} &= \\frac{-2}{-4} + \\frac{-2}{-2} + \\frac{-2}{-1} + \\frac{-2}{d} \\\\\n1 + 2 + 4 + \\frac{-4}{d} &= \\frac{1}{2} + 1 + 2 + \\frac{-2}{d} \\\\\n7 + \\frac{-4}{d} &= \\frac{7}{2} + \\frac{-2}{d} \\\\\n7 - \\frac{4}{d} = \\frac{7}{2} - \\frac{2}{d} \\\\\n7 - \\frac{7}{2} = \\frac{4}{d} - \\frac{2}{d} \\\\\n\\frac{7}{2} = \\frac{2}{d} \\\\\nd = \\frac{4}{7}.\n\\end{aligned}\n$$\nThe equality is homogeneous, so we can multiply all numbers by $7$ to obtain the integers given above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18881,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Several distinct positive integers have the property that the sum of every three of them is a prime number. At most how many of them are there?\n\nb) Several distinct integers (not necessarily positive) have the property that the sum of every three of them is positive and also a prime number. At most how many of them are there?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $4$ in a) and $5$ in b).\n\nAmong $5$ arbitrary integers, either there are three with the same remainder modulo $3$ or three with different remainders modulo $3$. In both cases, the sum of these three is divisible by $3$.\n\nIf in addition the integers are positive and distinct as in a), then the sum in question is at least $1 + 2 + 3 > 3$, hence it is not a prime. So there can be at most $4$ numbers that satisfy condition a), and $4$ such numbers exist. There are many examples: $\\{1, 3, 7, 9\\}$; $\\{3, 5, 11, 15\\}$; $\\{7, 13, 17, 23\\}$; $\\{7, 13, 23, 53\\}$, etc.\n\nNow let $x_1 < x_2 < \\dots < x_6$ be $6$ integers satisfying b). Among their sums by triples, consider $S_1 = x_1 + x_2 + x_3$, $S_2 = x_1 + x_2 + x_4$, and the sum $S$ of an arbitrary triple that does not contain $x_1$. Note that $S > S_2 > S_1$. Since $S_1$ and $S_2$ are (positive) prime numbers, they are at least $2$ and $3$ respectively; hence $S > 3$. However, by the introductory remark, there are three among the $5$ numbers $x_2, x_3, \\dots, x_6$ with sum $S$ divisible by $3$. Combined with $S > 3$, this yields a contradiction with $S$ being a prime. It remains to show that there exist $5$ numbers satisfying b). There is a variety of examples here too, with one or two negative numbers: $\\{-13, -1, 17, 25, 55\\}$; $\\{-11, -5, 19, 23, 29\\}$; $\\{-9, -3, 15, 25, 31\\}$; $\\{-9, 3, 9, 19, 229\\}$, etc.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18882,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all real polynomials $P(x)$ such that\n\n$$\nP^2(x) + P^2(y) + P^2(x + y) = 2P(x^2 + xy + y^2)\n$$\n\nfor every $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Setting $y = 0$ we get\n\n$$\n2P^2(x) + P^2(0) = 2P(x^2)\n$$\n\nfor every $x \\in \\mathbb{R}$. We claim that $P(x)$ is a monomial. Indeed, if this is not the case, let $ax^n$ and $bx^m$, with $n > m$, be the two non-zero terms with the largest powers of $x$. Comparing the coefficients of $x^{m+n}$ in the above equation, we get $2ab = 0$, a contradiction.\n\nSo $P(x)$ is a monomial. If $P(x) = c$, a constant, then substituting in the original equation we get $3c^2 = 2c$, giving $c = 0$ or $c = \\frac{2}{3}$.\n\nOtherwise, $P(x) = ax^n$ for some $a \\in \\mathbb{R} \\setminus \\{0\\}$ and $n \\in \\mathbb{N}$. Then $P(0) = 0$, and substituting in the previous equation with $x = 1$ we get $2a^2 = 2a$, giving $a = 1$.\n\nNow for $x = y = 1$ in the original equation we get $2 + 2^{2n} = 2 \\cdot 3^n$. The cases $n = 1, 2$ are obvious solutions, while for $n \\ge 3$ we have\n\n$$\n\\frac{2 + 2^{2n}}{3^n} > \\left(\\frac{4}{3}\\right)^n \\ge \\frac{64}{27} > 2\n$$\n\nshowing that no other solutions exist.\n\nSo the only possible solutions are $P(x) = 0$, $P(x) = \\frac{2}{3}$, $P(x) = x$, $P(x) = x^2$, which are easy to check satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18883,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(R, +, \\cdot)$ be a unital ring such that for all $x \\in R$ one can find $e_1, e_2 \\in R$ with $e_1^2 = e_1$, $e_2^2 = e_2$, and $x = e_1 e_2$.\n\nShow that:\n\n1. $1$ is the only invertible element in $R$.\n2. $x^2 = x$ for all $x \\in R$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Let $x$ be an invertible element of $R$. Let $e_1$ and $e_2$ be idempotent elements of $R$ such that $x = e_1 e_2$. Notice that\n\n$$(1 - e_1) = (1 - e_1) \\cdot 1 = (1 - e_1) e_1 e_2 x^{-1} = (e_1 - e_1^2) e_2 x^{-1} = 0,$$\n\nso $e_1 = 1$. Also, $x^2 = e_2^2 = e_2 = x$. Since $x$ is invertible, it follows that $x = 1$.\n\n**b)** There are no nonzero nilpotent elements in $R$. Indeed, if $x \\in R$ 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 part (a).\n\nNext, every idempotent element of $R$ is central (commutes with every element of $R$). Let $e$ be idempotent, and $x \\in R$. Then\n\n$$(e x - e x e)^2 = e x e x - e x e x e - e x e e x + e x e e x e = e x e x - e x e x e - e x e x + e x e x e = 0,$$\n\nso $e x = e x e$. Similarly, $x e = e x e$, so $e x = x e$.\n\nFinally, for any $x \\in R$, write $x = e_1 e_2$ with $e_1, e_2$ idempotent. By the above, $e_1 e_2 = e_2 e_1$, so\n\n$$x^2 = (e_1 e_2)^2 = e_1^2 e_2^2 = e_1 e_2 = x.$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18884,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider arrangements of the numbers 1 through 64 on the squares of an $8 \\times 8$ chessboard, where each square contains exactly one number and each number appears exactly once.\n\nA number in such an arrangement is called *super-plus-good* if it is the largest number in its row and at the same time the smallest number in its column.\n\nProve or disprove each of the following statements:\n\n(a) Each such arrangement contains at least one super-plus-good number.\n\n(b) Each such arrangement contains at most one super-plus-good number.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) This is false. For example, place the numbers from 1 to 8 along the main diagonal and the numbers from 57 to 64 along the secondary diagonal:\n\n![](<table><tr><td>1</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>57</td></tr><tr><td>15</td><td>2</td><td>16</td><td>17</td><td>18</td><td>19</td><td>58</td><td>20</td></tr><tr><td>21</td><td>22</td><td>3</td><td>23</td><td>24</td><td>59</td><td>25</td><td>26</td></tr><tr><td>27</td><td>28</td><td>29</td><td>4</td><td>60</td><td>30</td><td>31</td><td>32</td></tr><tr><td>33</td><td>34</td><td>35</td><td>61</td><td>5</td><td>36</td><td>37</td><td>38</td></tr><tr><td>39</td><td>40</td><td>62</td><td>41</td><td>42</td><td>6</td><td>43</td><td>44</td></tr><tr><td>45</td><td>63</td><td>46</td><td>47</td><td>48</td><td>49</td><td>7</td><td>50</td></tr><tr><td>64</td><td>51</td><td>52</td><td>53</td><td>54</td><td>55</td><td>56</td><td>8.</td></tr></table>)\n\nHere, the numbers from 1 to 8 are column minima, and the numbers from 57 to 64 are row maxima. Therefore, no number is both a column minimum and a row maximum, so no number is super-plus-good.\n\n(b) This is true. Denote the number in the $a$th row and $b$th column by $F(a, b)$. Suppose there are two super-plus-good numbers at $(i, j)$ and $(r, s)$. Since all numbers are different, the row maxima and column minima are unique, so $i \\neq r$ and $j \\neq s$. Then:\n\n$$\nF(i, j) > F(i, s) \\quad \\text{(since $F(i, j)$ is row maximum)}\n$$\n$$\nF(i, j) < F(r, j) \\quad \\text{(since $F(i, j)$ is column minimum)}\n$$\n$$\nF(r, s) > F(r, j) \\quad \\text{(since $F(r, s)$ is row maximum)}\n$$\n$$\nF(r, s) < F(i, s) \\quad \\text{(since $F(r, s)$ is column minimum)}\n$$\n\nThese four inequalities lead to a contradiction:\n\n$$\nF(i, j) > F(i, s) > F(r, s) > F(r, j) > F(i, j).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18885,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an odd prime $p$ and a sequence of integers $\\{u_n\\}$ for $n \\ge 0$, define the sequence\n\n$$\nv_n = \\sum_{i=0}^{n} \\binom{n}{i} p^i u_i, \\quad n \\ge 1.\n$$\n\nProve that if there are infinitely many values of $n$ such that $v_n = 0$, then for all $n \\ge 0$, $v_n = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define the rational polynomial (with rational coefficients)\n\n$$\nR_N(x) = \\sum_{i=0}^{N} \\frac{p^i}{i!} u_i x(x-1) \\cdots (x-i+1),\n$$\n\nwhere $p$ is an odd prime, $u_i$ is an integer, and $0 \\le i \\le N$. For $m \\le n$, we have $R_n(m) = v_m$. Suppose there exists a sequence of nonnegative integers $m_j$, $j = 1, 2, \\ldots$ such that $v_{m_j} = 0$. We aim to prove $v_r = 0$ for every nonnegative integer $r$.\n\nFor a rational number $s$, denote $v_p(s)$ as the highest power of $p$ dividing $s$. It suffices to show $v_p(v_r) \\ge l$ for all $l \\in \\mathbb{N}^+$. Take $k$ sufficiently large so that $k \\cdot \\frac{p-2}{p-1} \\ge l$, $N \\ge r$, and $N \\ge m_k$. Then $v_r = R_N(r)$, and $R_N(m_i) = 0$ for $i = 1, 2, \\ldots, k$. This implies\n\n$$\nR_N(x) = (x - m_1) \\cdots (x - m_k) h(x)\n$$\n\nfor some $h(x) \\in \\mathbb{Q}[x]$, and $v_p(v_r) = v_p(R_N(r)) \\ge v_p(h(r))$.\n\nFor $h(x) \\in \\mathbb{Q}[x]$, let\n\n$$\nf(x) = \\sum_{i=0}^{n} a_i x^i, \\quad v_p^{(i)}(f) = \\min_{j \\ge i} v_p(a_j).\n$$\n\n**Lemma:** Let $f(x), g(x)$ be rational polynomials such that\n\n$$\ng(x) = (x - m)f(x)\n$$\n\nfor some integer $m$. Then, for every $i \\le \\deg f$, $v_p^{(i)}(f) \\ge v_p^{(i+1)}(g)$.\n\n*Proof of lemma:* Let $f(x) = \\sum_{i=0}^{n} a_i x^i$, $g(x) = \\sum_{i=0}^{n+1} b_i x^i$. From the relation, for all $j$, $1 \\le j \\le n$,\n\n$$\na_j = b_{j+1} + m b_{j+2} + \\cdots + m^{n-j} b_{n+1}.\n$$\n\nThus, for every $j \\ge i$, $v_p(a_j) \\ge \\min_{j \\ge i+1} v_p(b_j) = v_p^{(i+1)}(g)$. Therefore, $v_p^{(i)}(f) \\ge v_p^{(i+1)}(g)$.\n\nReturning to the original problem:\n\n$$\nv_p(v_r) = v_p(R_N(r)) \\ge v_p(h(r)) \\ge v_p^{(0)}(h(x)).\n$$\n\nBy the previous factorization and the lemma, $v_p^{(0)}(h(x)) \\ge v_p^{(k)}(R_N(x))$.\n\nSince $R_N(x) = \\sum_{i=1}^{N} \\frac{p^i}{i!} u_i x(x-1) \\cdots (x-i+1)$, and $v_p\\left(\\frac{p^i}{i!}\\right) \\ge i \\frac{p-2}{p-1}$, we infer $v_p^{(k)}(R_N(x)) \\ge k \\frac{p-2}{p-1} \\ge l$. Combining the above, $v_p(v_r) \\ge l$. The argument is complete. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18886,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a sequence $\\langle a_1, a_2, a_3, \\dots \\rangle$ of real numbers. For every positive integer $n$ we define $m_n$ as the arithmetic mean of the numbers from $a_1$ through $a_n$.\n\nWe assume that a real number $C$ exists, such that\n\n$$\n(i-j) \\cdot m_k + (j-k) \\cdot m_i + (k-i) \\cdot m_j = C\n$$\n\nholds for all triples $(i, j, k)$ of pairwise different positive integers.\n\nProve that $\\langle a_1, a_2, a_3, \\dots \\rangle$ is an arithmetic sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "By exchanging the roles of $i$ and $j$, we see that\n$$(i-j) \\cdot m_k + (j-k) \\cdot m_i + (k-i) \\cdot m_j = C = (j-i) \\cdot m_k + (i-k) \\cdot m_j + (k-j) \\cdot m_i = -C$$\nmust hold, which yields $C = 0$.\n\nFor $(i, j, k) = (1, 2, 3)$, we obtain\n$$\n(1-2) \\cdot \\frac{a_1+a_2+a_3}{3} + (2-3) \\cdot a_1 + (3-1) \\cdot \\frac{a_1+a_2}{2} = 0,\n$$\nwhich is equivalent to\n$$\n\\frac{a_1+a_2+a_3}{3} - a_1 + a_1 + a_2 = 0 \\iff a_1 + a_3 = a_2.\n$$\nThe first three elements of the sequence therefore are indeed elements of an arithmetic sequence.\n\nWe can now use induction to show that the entire sequence is arithmetic, i.e., that $a_n = a_1 + (n-1)(a_2 - a_1)$ holds. In order to do this, we assume that $a_k = a_1 + (k-1)(a_2 - a_1)$ holds for $1 \\leq k \\leq n-1$, and consider the triple $(i, j, k) = (1, 2, n)$. We then have\n\n$$\n\\begin{align*}\n& (1-2) \\cdot \\frac{\\frac{(n-1)(2a_1)+(n-2)(a_2-a_1)}{2} + a_n}{n} + (2-n) \\cdot a_1 + (n-1) \\cdot \\frac{a_1+a_2}{2} = 0 \\\\\n\\iff & \\frac{a_1(3-n)+a_2(n-1)}{2} - \\frac{a_1(n-1)(4-n)+a_2(n-1)(n-2)+2a_n}{2n} = 0 \\\\\n\\iff & a_1 \\cdot (3n-n^2+n^2-5n+4) + a_2 \\cdot (n-1)(n-n+2) = 2a_n \\\\\n\\iff & a_2 \\cdot (n-1) - a_1 \\cdot (n-2) = a_n,\n\\end{align*}\n$$\nwhich completes the induction. We see that the sequence is indeed arithmetic, as claimed. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18887,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a real number $\\alpha$, find the minimum real number $\\lambda = \\lambda(\\alpha)$ such that for any complex numbers $z_1, z_2$ and any real number $x \\in [0, 1]$, if $|z_1| \\leq \\alpha |z_1 - z_2|$, then $|z_1 - xz_2| \\leq \\lambda |z_1 - z_2|$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p205_data_0ac963be5f.png)\n\nFig. 5.1",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the figure, in the complex plane, points $A$, $B$, and $C$ denote the complex numbers $z_1$, $z_2$, and $xz_2$, respectively. $C$ is on the segment $OB$. The vectors $\\vec{BA}$ and $\\vec{CA}$ are represented by the complex numbers $z_1 - z_2$ and $z_1 - xz_2$, respectively. Since $|z_1| \\leq \\alpha |z_1 - z_2|$, we have $|\\vec{OA}| \\leq \\alpha |\\vec{BA}|$.\n\n$$\n\\begin{align*}\n|z_1 - xz_2|_{\\max} &= |\\vec{AC}|_{\\max} = \\max \\{|\\vec{OA}|, |\\vec{BA}|\\} \\\\\n&= \\max \\{|z_1|, |z_1 - z_2|\\} \\\\\n&= \\max \\{\\alpha |z_1 - z_2|, |z_1 - z_2|\\}.\n\\end{align*}\n$$\n\nTherefore, $\\lambda(\\alpha) = \\max\\{\\alpha, 1\\}$.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18888,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $a$, $n$, consider the equation:\n$$a^2x + 6a y + 36z = n,$$\nwhere $x$, $y$, and $z$ are non-negative numbers.\n\na) Find all values of $a$ such that for all $n \\geq 250$, the equation always has natural roots ($x$, $y$, $z$).\n\nb) Given that $a > 1$ and $\\gcd(a, 6) = 1$, find the greatest value of $n$ in terms of $a$ such that the equation does not have natural roots ($x$, $y$, $z$).",
    "options": [],
    "answer": "See solution",
    "solution": "We first state a well-known lemma:\n\n**Lemma.** (Sylvester's theorem) For two positive integers $a$ and $b$ such that $\\gcd(a, b) = 1$, the largest integer that cannot be written in the form $ax + by$ where $x$ and $y$ are non-negative integers is $N_0 = ab - a - b$.\n\na) Let $a$ be the satisfying value. A natural number $n$ is called 'nice' if there exist $x, y, z \\in \\mathbb{N}$ such that\n$$a^2x + 6a y + 36z = n.$$\nBy choosing $n = 301$, we have $a^2x \\equiv 1 \\pmod{6}$, then $\\gcd(a, 6) = 1$, which implies $\\gcd(a, 36) = 1$. Moreover, by applying Sylvester's theorem for $a$ and $36$, the largest number that cannot be written in the form $ax_1 + 36y_1$ is\n$$36a - a - 36 = 35a - 36.$$\nHowever,\n$$n = a^2x + 6a y + 36z = a(ax + 6y) + 36z = ax_1 + 36y_1,$$\nhence $n \\geq 35a - 35$, which means $250 \\geq 35a - 35$ or $a < 9$. Since $\\gcd(a, 6) = 1$, we conclude $a \\in \\{1, 5, 7\\}$.\n\n- For $a = 7$, the equation becomes $n = 49x + 42y + 36z$. By putting $n = 251$,\n$$251 = 49x + 42y + 36z \\equiv z \\pmod{7} \\implies z \\equiv 6 \\pmod{7}.$$\nThis means $z \\geq 6$. On the other hand, $z \\leq 251/36 < 13$, so $z = 6$. The equation becomes $7x + 6y = 5$, which has no natural solution; thus, $a = 7$ is not satisfied.\n\n- For $a = 1$, the equation always has a solution $(x, y, z) = (n, 0, 0)$.\n\n- For $a = 5$, we have to show that for $n \\geq 250$, there exist $(x, y, z) \\in \\mathbb{N}^3$ such that\n$$25x + 30y + 36z = n.$$\nLet $n = 5k + r$, $z = r$ where $r < 5$ and $k \\geq 50$. The equation becomes\n$$25x + 30y = n - 36r = 5k - 35r \\iff 5x + 6y = k - 7r.$$\nHowever, $k - 7r \\geq 50 - 28 = 22 > 30 - 5 - 6$, so by Sylvester's theorem, the equation always has a natural solution.\n\nb) We prove a general result: Let $a, b$ be two coprime positive integers. Then\n$$N = a^2b + ab^2 - a^2 - b^2 - ab + 1$$\nis the smallest positive integer such that the equation $a^2x + ab y + b^2z = m$ has a natural solution for all $m \\geq N$.\n\n- In case $m \\geq N$, choosing $z = m b^{-2}$ (mod $a$) ($0 \\leq z < a$), we need to prove that there exist $x, y \\in \\mathbb{N}$ such that\n$$a x + b y = \\frac{m - b^2 z}{a}$$\nhas a natural solution. Note that\n$$\\frac{m - b^2 z}{a} \\geq \\frac{N - b^2(a - 1)}{a} > \\frac{a^2 b - a^2 - ab}{a} = ab - a - b,$$\nthen the equation always has a natural solution by Sylvester's theorem.\n\n- If $m < N$, let $m = a^2b + ab^2 - a^2 - b^2 - ab$ and assume that there exists a triple $(x, y, z) \\in \\mathbb{N}^3$ such that\n$$a^2x + ab y + b^2z = m,$$\nhence $a^2x \\equiv -a^2 \\pmod{b}$, $b^2z \\equiv -b^2 \\pmod{a}$, which is equivalent to\n$$x \\equiv -1 \\pmod{b}, \\quad z \\equiv -1 \\pmod{a}.$$\nThen $x \\geq b - 1$, $z \\geq a - 1$, which leads to\n$$y \\leq \\frac{m - a^2(b - 1) - b^2(a - 1)}{ab} = -1.$$\nThat is a contradiction since $y \\geq 0$.\n\nApplying this result, we conclude that the greatest value of $n$ is $5a^2 + 30a - 36$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18889,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be three positive real numbers. Determine all possible values for the expression\n\n$$\n\\frac{x}{3x+y+z} + \\frac{y}{x+3y+z} + \\frac{z}{x+y+3z}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that the given expression is homogeneous in $x, y, z$, so without loss of generality, let $x + y + z = 2$. Then the original expression becomes\n\n$$\n\\begin{aligned}\n& \\frac{x}{3x+y+z} + \\frac{y}{x+3y+z} + \\frac{z}{x+y+3z} \\\\\n&= \\frac{x}{2x+2} + \\frac{y}{2y+2} + \\frac{z}{2z+2} \\\\\n&= \\frac{1}{2} \\left( \\frac{x}{x+1} + \\frac{y}{y+1} + \\frac{z}{z+1} \\right) \\\\\n&= \\frac{3}{2} - \\frac{1}{2} \\left( \\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1} \\right).\n\\end{aligned}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\begin{aligned}\n& \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) \\left((x+1) + (y+1) + (z+1)\\right) \\ge (1+1+1)^2, \\\\\n& \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) \\cdot 5 \\ge 9, \\\\\n& \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) \\ge \\frac{9}{5},\n\\end{aligned}\n$$\n\nSo\n\n$$\n\\frac{x}{3x+y+z} + \\frac{y}{x+3y+z} + \\frac{z}{x+y+3z} \\le \\frac{3}{2} - \\frac{1}{2} \\cdot \\frac{9}{5} = \\frac{3}{5},\n$$\n\nwith equality when $x = y = z = \\frac{2}{3}$.\n\nOn the other hand, by considering extreme values, when $(x, y, z) = (2, 0, 0)$ (or any permutation), the value is $1/3$. This minimum is never attained but can be approached arbitrarily closely. Therefore, the range of possible values is $\\left(\\frac{1}{3}, \\frac{3}{5}\\right]$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18890,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : (0, \\infty) \\to \\mathbb{R}$ be such that for all $x, y \\in (0, \\infty)$,\n\n$$\nf(x + y) = f\\left(\\frac{x + y}{xy}\\right) + f(xy).\n$$\n\nShow that $f(xy) = f(x) + f(y)$ for all $x, y \\in (0, \\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $f(ab) = f(a) + f(b)$ for $a, b \\in (0, \\infty)$ such that $a^2b \\geq 4$. In fact, if $a^2b \\geq 4$ and $a, b > 0$, then we can find $x, y > 0$ such that $x + y = ab$ and $xy = b$, namely\n\n$$\nx = \\frac{a + \\sqrt{a^2 - \\left(\\frac{4}{b}\\right)}}{2/b} > 0 \\quad \\text{and} \\quad y = \\frac{a - \\sqrt{a^2 - \\left(\\frac{4}{b}\\right)}}{2/b} > 0.\n$$\n\nThus, for $a, b \\in (0, \\infty)$ such that $a^2b \\geq 4$, we plug $x, y$ as above in the functional equation to get $f(ab) = f(x + y) = f\\left(\\frac{x + y}{xy}\\right) + f(xy) = f\\left(\\frac{ab}{b}\\right) + f(b) = f(a) + f(b)$. Now let $x, y > 0$. Let\n\n$$\nz = \\max \\left\\{ \\frac{4}{x^2 y^2}, \\frac{4}{x^2 y}, \\frac{4}{y^2} \\right\\} > 0.\n$$\n\nThen $(xy)^2z \\geq 4$, $x^2(yz) \\geq 4$ and $y^2z \\geq 4$ which imply that $f(xyz) = f(xy) + f(z)$, $f(xyz) = f(x) + f(yz)$ and $f(yz) = f(y) + f(z)$ respectively. Therefore,\n\n$$\n\\begin{aligned}\nf(xy) &= f(xyz) - f(z) \\\\\n      &= (f(x) + f(yz)) - f(z) \\\\\n      &= f(x) + f(yz) - f(z) \\\\\n      &= f(x) + (f(y) + f(z)) - f(z) \\\\\n      &= f(x) + f(y).\n\\end{aligned}\n$$\n\n![](images/selected_2011_p10_data_7a50b2422b.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18891,
    "subject": "Mathematics (Olympiad)",
    "question": "There are four numbers on the board: $1$, $3$, $6$, and $10$. Each time, we can erase any two numbers $a$, $b$ written on the board and write the numbers $a+b$ and $ab$ instead. Can we obtain the four numbers $2015$, $2016$, $2017$, $2018$ after several moves?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us look at the numbers modulo $3$. The number of numbers divisible by $3$ cannot decrease. If both $a$ and $b$ are divisible by $3$, then both $a+b$ and $ab$ are also divisible by $3$. If only one of the numbers is divisible by $3$, then $ab$ is also divisible by $3$. Initially, we have only one number divisible by $3$, and in the end, only one, so such a situation is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18892,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. If $n$ divides $3^n + 4^n$, prove that $7$ divides $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $n$ is odd and $3$ does not divide $n$. Let $p$ be the least prime dividing $n$. Let $c$ be an integer such that $0 < c < p$ and $4c \\equiv 3 \\pmod{p}$. Such a number $c$ exists since $\\gcd(3, 4) = 1$. Thus we have\n\n$$\n\\begin{align*}\n4^n(c^{2n} - 1) &\\equiv (c^n - 1)((4c)^n + 4^n) \\\\\n&\\equiv (c^n - 1)(3^n + 4^n) \\pmod{p} \\\\\n&\\equiv 0 \\pmod{p}.\n\\end{align*}\n$$\n\nThus $p$ divides $c^{2n} - 1$. Let $d$ be the order of $c$ modulo $p$. Then $d$ divides $2n$. By Fermat's little theorem $d$ also divides $p-1$. Thus $d \\mid \\gcd(2n, p-1)$. Let $q$ be a prime divisor of $d$. Then $q$ divides $2n$ and $p-1$. If $q$ divides $n$, the choice of $p$ shows that $q \\ge p$. Thus $q$ cannot divide $p-1$. We conclude that $q \\mid 2$ hence $q=2$. Since $2 \\mid (p-1)$ and $2$ does not divide $n$, it follows that $d=2$. Thus $c^2 \\equiv 1 \\pmod{p}$.\n\nIf $p$ divides $c-1$, then $4 \\equiv 4c \\equiv 3 \\pmod{p}$, which is impossible. Hence $p$ divides $c+1$. This gives\n\n$$\n0 \\equiv 4c + 4 \\equiv 3 + 4 \\equiv 7 \\pmod{p}.\n$$\n\nHence $p$ divides $7$. This forces $p=7$. Thus $7$ divides $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18893,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob are playing hide and seek. Initially, Bob chooses a secret fixed point $B$ in the unit square. Then Alice chooses a sequence of points $P_0, P_1, \\dots, P_N$ in the plane. After choosing $P_k$ (but before choosing $P_{k+1}$) for $k \\geq 1$, Bob tells \"warmer\" if $P_k$ is closer to $B$ than $P_{k-1}$, otherwise he says \"colder\". After Alice has chosen $P_N$ and heard Bob’s answer, Alice chooses a final point $A$. Alice wins if the distance $AB$ is at most $\\frac{1}{2020}$, otherwise Bob wins. Show that if $N = 18$, Alice cannot guarantee a win.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_0$ be the set of all points in the square, and for each $1 \\leq k \\leq N$, let $S_k$ be the set of possible points $B$ consistent with everything Bob has said. For each $k$, we then have that $S_k$ is the disjoint union of the two possible values $S_{k+1}$ can take for each of Bob's possible answers. Hence one of these must have area $\\leq \\frac{|S_{k+1}|}{2}$, and the other must have area $\\geq \\frac{|S_{k+1}|}{2}$. Suppose now that Alice always receives the answer resulting in the greater half. After receiving $N$ answers, then, $|S_N| \\geq \\frac{1}{2^N}$. If Alice has a winning strategy, there must be a point $A$ in $S_N$ so that the circle of radius $\\frac{1}{2020}$ centered at $A$ contains $S_N$. Hence $\\frac{\\pi}{2020^2} \\geq \\frac{1}{2^N}$. It therefore suffices to show that this inequality does not hold for $N=18$. This follows from the estimates $\\pi \\leq 2^2$ and $2020 > 1024 = 2^{10}$, meaning that $\\frac{\\pi}{2020^2} > \\frac{2^2}{2^{20}} = \\frac{1}{2^{18}}$.\n\n**Comment.** In fact, it also holds for $N=20$, but this requires some more estimation work to show by hand. Also, there is a winning strategy for $N=22$, where the square is dissected into right isosceles triangles of successively smaller sizes. This can evidently always be done exactly by choosing $P_k$ outside the square. It also seems possible to do this for $P_k$ restricted to the interior of the square, but is harder to write up precisely.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18894,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $\\dfrac{AE}{ED} = \\dfrac{BF}{FC}$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.\n\n![](images/USA_IMO_2006-2007_p8_data_ce31c940ce.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the second intersection of the circumcircles of triangles $TCF$ and $TDE$. Because the quadrilateral $PEDT$ is cyclic, $\\angle PET = \\angle PDT$, or\n\n$$\n\\angle PEF = \\angle PDC. \\quad (*)\n$$\n\nBecause the quadrilateral $PFCT$ is cyclic,\n\n$$\n\\angle PFE = \\angle PFT = \\angle PCT = \\angle PCD. \\quad (**)\n$$\n\nBy equations $(*)$ and $(**)$, it follows that triangle $PEF$ is similar to triangle $PDC$. Hence $\\angle FPE = \\angle CPD$ and $\\dfrac{PF}{PE} = \\dfrac{PC}{PD}$. Note also that $\\angle FPC = \\angle FPE + \\angle EPC = \\angle CPD + \\angle EPC = \\angle EPD$. Thus, triangle $EPD$ is similar to triangle $FPC$. Another way to say this is that there is a spiral similarity centered at $P$ that sends triangle $PFE$ to triangle $PCD$, which implies that there is also a spiral similarity, centered at $P$, that sends triangle $PFC$ to triangle $PED$, and vice versa. In terms of complex numbers, this amounts to saying that\n\n$$\n\\frac{D-P}{E-P} = \\frac{C-P}{F-P} \\implies \\frac{E-P}{F-P} = \\frac{D-P}{C-P}.\n$$\n\nBecause $\\dfrac{AE}{ED} = \\dfrac{BF}{FC}$, points $A$ and $B$ are obtained by extending corresponding segments of two similar triangles $PED$ and $PFC$, namely, $DE$ and $CF$, by the identical proportion. We conclude that triangle $PDA$ is similar to triangle $PCB$, implying that triangle $PAE$ is similar to triangle $PBF$. Therefore, as shown before, we can establish the similarity between triangles $PBA$ and $PFE$, implying that\n\n$$\n\\angle PBS = \\angle PBA = \\angle PFE = \\angle PFS \\quad \\text{and} \\quad \\angle PAB = \\angle PEF.\n$$\n\nThe first equation above shows that $PBFS$ is cyclic. The second equation shows that $\\angle PAS = 180^\\circ - \\angle BAP = 180^\\circ - \\angle FEP = \\angle PES$; that is, $PAES$ is cyclic. We conclude that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through point $P$.\n\n**Note.** There are two spiral similarities that send segment $EF$ to segment $CD$. One of them sends $E$ and $F$ to $D$ and $C$, respectively; the point $P$ is the center of this spiral similarity. The other sends $E$ and $F$ to $C$ and $D$, respectively; the center of this spiral similarity is the second intersection (other than $T$) of the circumcircles of triangles $TFD$ and $TEC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18895,
    "subject": "Mathematics (Olympiad)",
    "question": "$n \\in \\mathbb{N}$ ба $k$ нь $n$-ээс ихгүй дурын натурал тоо бол $n!$-ийг $k$ ширхэг ялгаатай хуваагчдынх нь нийлбэрт ямагт задалж болохыг батал. ($n!$ нь $1$-ээс $n$ хүртэлх бүх натурал тооны үржвэр.)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nn! = (n-1)(n-1)! + (n-2)(n-2)! + \\dots + (n-k+1)(n-k+1)! + (n-k+1)!\n$$\nИймд $n!$-ийг $k$ ширхэг ялгаатай хуваагчдын нийлбэрт задлах боломжтой.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18896,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $A$ such that:\n\n- The product of all the digits of $A$ is a multiple of 3 but not a multiple of 9.\n- One and only one of the digits of $A$ is either 3 or 6, and none of the digits of $A$ is 0 or 9.\n- $A$ has at least two digits.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be a number satisfying the above conditions. The product of its digits must be a multiple of 3 but not of 9, so exactly one digit is 3 or 6, and none is 0 or 9 (condition X).\n\nIf $A$ has only one digit, neither 3 nor 6 works, so $A$ must have at least two digits. Write $A = 10a + b$ with $a$ a positive integer and $0 \\leq b \\leq 9$. Since $A < 138$, $a \\leq 13$.\n\n**Case 1:** $b = 3$ or $6$.\n\nThen $10a = A - b$ is a multiple of 3, so $a$ is a multiple of 3. If $a = 3, 6, 9$, $A$ violates condition X, so $a = 12$. But $A = 123$ or $126$ do not satisfy all conditions (e.g., $126$ is a multiple of 9, $123 + 1 \\times 2 \\times 3 = 129$ is not a multiple of 9), so this case fails.\n\n**Case 2:** $b$ is neither 3 nor 6.\n\nThen one and only one digit of $a$ is 3 or 6. If $a = 3$ or $6$, $b$ must be a multiple of 3, but $b = 0, 3, 6, 9$ are not allowed. If $a = 13$, the only number less than 138 and a multiple of 3 but not 9 is 132, but $132 + 1 \\times 3 \\times 2 = 138$ is not a multiple of 9, so this case also fails.\n\nThus, the smallest possible positive integer satisfying all the conditions is $138$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18897,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to have a set of $n$ binary sequences of length $200$ such that any two sequences differ in at least $101$ positions?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible.\n\nLet $n_i$ be the number of sequences in the set with $1$ at the $i$th entry, where $1 \\leq i \\leq 200$. Let $S$ be the total number of different positions between all pairs of sequences in the set. Since every pair differs in at least $101$ positions, we have:\n\n$$\nS \\geq 101 \\binom{n}{2}\n$$\n\nOn the other hand, for each position $i$, there are $n_i(n - n_i)$ pairs of sequences that differ at the $i$th entry. Thus:\n\n$$\nS = \\sum_{i=1}^{200} n_i(n - n_i)\n$$\n\nBy the AM-GM inequality:\n\n$$\nn_i(n - n_i) \\leq \\left( \\frac{n_i + (n - n_i)}{2} \\right)^2 = \\frac{n^2}{4}\n$$\n\nTherefore:\n\n$$\n\\sum_{i=1}^{200} n_i(n - n_i) \\leq 200 \\cdot \\frac{n^2}{4} = 50n^2\n$$\n\nSo:\n\n$$\n50n^2 \\geq S \\geq 101 \\binom{n}{2}\n$$\n\nThis implies $n \\leq 101$. However, when $n = 101$, the inequality $\\frac{n^2}{4} \\geq n_i(n - n_i)$ is strict, since one side is an integer and the other is not. Thus $n < 101$, or $n \\leq 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18898,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a connected graph. Suppose that each vertex of $G$ can be properly coloured with one of the colors $0, 1, \\ldots, 2017$ (so that endpoints of each edge are differently coloured). Let $f(x)$ be the color of vertex $x$.\n\nProve that there is a proper recolouring of the vertices of $G$ into colors $0, 1, \\ldots, 2017$ such that:\n\n(t) For any two vertices $a$ and $b$, there is a path $a = x_1, x_2, \\ldots, x_n = b$ such that for each $i = 1, \\ldots, n-1$ we have $|f(x_i) - f(x_{i+1})| = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider any proper colouring of $G$. Let $G_1$ be a maximal subgraph of $G$ satisfying property (t). We will show that by recolouring some vertices, we can increase the number of vertices in $G_1$.\n\nFor any graph $G'$, let $V(G')$ denote its set of vertices. Let\n\n$$\n\\min_{x \\in V(G_1),\\ y \\in V(G-G_1)} |f(x) - f(y)| = |f(x_0) - f(y_0)|,\n$$\n\nBy definition, $|f(x_0) - f(y_0)| > 1$. Suppose $f(x_0) > f(y_0)$. Recolour some vertices of $G - G_1$: recolour vertices coloured $f(y_0)$ to $f(x_0) - 1$, and vertices coloured $f(x_0) - 1$ to $f(y_0)$. After this recolouring, the graph remains properly coloured, since in $G - G_1$ we just switched colors $f(x_0) - 1$ and $f(y_0)$. Also, since $|f(x_0) - f(y_0)|$ is minimal, the condition $f(x) \\neq f(y)$ still holds for all $x \\in G_1$ and $y \\in G - G_1$.\n\nIn the symmetric case $f(x_0) < f(y_0)$, recolour vertices of $G - G_1$: recolour vertices coloured $f(y_0)$ to $f(x_0) + 1$, and vertices coloured $f(x_0) + 1$ to $f(y_0)$.\n\nThus, after the recolouring, the set $V(G_1)$ will grow by at least one new element $y_0$. By continuing this process, eventually we obtain the desired colouring of the whole $G$ satisfying property (t).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18899,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any integer $n \\ge 4$, there exists a polynomial of degree $n$,\n\n$$\nf(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0\n$$\n\nwith the following properties:\n\n1. $a_0, a_1, \\dots, a_{n-1}$ are all positive integers;\n2. For any positive integer $m$ and any $k \\ge 2$ positive integers $r_1, r_2, \\dots, r_k$ that are all different, we have\n\n$$\nf(m) \\neq f(r_1)f(r_2)\\cdots f(r_k).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nf(x) = (x+1)(x+2)\\cdots(x+n) + 2.\n$$\n\nObviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We will prove that $f(x)$ has property (2).\n\nFor any integer $t$, since $n \\ge 4$, there is always a multiple of $4$ among any $n$ consecutive numbers $t+1, t+2, \\dots, t+n$. Thus, $f(t) \\equiv 2 \\pmod{4}$.\n\nFor any $k \\ge 2$ positive integers $r_1, r_2, \\dots, r_k$, we have\n\n$$\nf(r_1)f(r_2)\\cdots f(r_k) \\equiv 2^k \\equiv 0 \\pmod{4}.\n$$\n\nOn the other hand, for any positive integer $m$, $f(m) \\equiv 2 \\pmod{4}$. Therefore,\n\n$$\nf(m) \\not\\equiv f(r_1)f(r_2)\\cdots f(r_k) \\pmod{4},\n$$\n\nwhich implies $f(m) \\neq f(r_1)f(r_2)\\cdots f(r_k)$. Thus, the required $f(x)$ exists and the proof is complete. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18900,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_0 = \\{0, x_1, \\dots, x_n, 1000\\}$ be the set of initial positions of managers (at $0$ and $1000$) and mathematicians (at $x_1, \\dots, x_n$), where $0 < x_1 < \\dots < x_n < 1000$.\n\nSuppose that at each step, the number of elements in $S_i$ decreases if two neighbouring mathematicians are closest to each other. Since $S_0$ is finite, this reduction can only happen finitely many times. Eventually, there exists a non-negative integer $N$ such that $|S_k| = |S_N|$ for all $k \\ge N$.\n\nLet $S_N = \\{0, y_1, \\dots, y_m, 1000\\}$, where $0 < y_1 < \\dots < y_m < 1000$ and $m \\le n$. If $|S_{N+1}| = |S_N|$, then:\n\n$$\ny_1 - 0 < y_2 - y_1 < \\dots < y_t - y_{t-1} < y_{t+1} - y_t > y_{t+2} - y_{t+1} > \\dots > y_m - y_{m-1} > 1000 - y_m\n$$\n\ni.e., the nearest person to each of the first $t$ mathematicians is on the left, and for the others, on the right.\n\nDescribe and prove what happens to the positions of the mathematicians after finitely many steps.",
    "options": [],
    "answer": "See solution",
    "solution": "The first $t$ mathematicians will continue to move to the left, while the others will continue to move to the right.\n\nBy observing the sum of positions, it can be shown that after finitely many seconds, all first $t$ mathematicians will be at most $1$ metre apart from the first manager, while the others will be at most $1$ metre apart from the second manager. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18901,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral and let $P$ be the point of intersection of $AC$ and $BD$. Suppose that $AC + AD = BC + BD$. Prove that the internal angle bisectors of $\\angle ACB$, $\\angle ADB$, and $\\angle APB$ meet at a common point.\n\nSoit $ABCD$ un quadrilatère convexe et $P$ le point d'intersection des droites $AC$ et $BD$. Supposons que $AC + AD = BC + BD$. Démontrez que les bissectrices des angles internes de $\\angle ACB$, $\\angle ADB$, et $\\angle APB$ se coupent en un point.",
    "options": [],
    "answer": "See solution",
    "solution": "Construct $A'$ on $CA$ so that $AA' = AD$ and $B'$ on $CB$ such that $BB' = BD$. Then we have three angle bisectors that correspond to the perpendicular bisectors of $A'B'$, $A'D$, and $B'D$. These perpendicular bisectors are concurrent, so the angle bisectors are also concurrent. This tells us that the external angle bisectors at $A$ and $B$ meet at the excentre of $PDB$. A symmetric argument for $C$ finishes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18902,
    "subject": "Mathematics (Olympiad)",
    "question": "For a fixed positive integer $n$, we call a finite sequence of positive integers $(a_1, \\dots, a_m)$ an *n-even sequence* if $n = a_1 + \\dots + a_m$, and there is an even number of pairs $(i, j)$ satisfying $1 \\le i < j \\le m$ and $a_i > a_j$. Find the number of n-even sequences $E(n)$. For example, $E(4) = 6$: the sequences are $(4)$, $(1, 3)$, $(2, 2)$, $(1, 1, 2)$, $(2, 1, 1)$, and $(1, 1, 1, 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(a_1, \\dots, a_m)$ be an *n-sequence* if $n = a_1 + \\dots + a_m$. If $(a_1, \\dots, a_m)$ is not n-even, call it *n-odd*. Let $O(n)$ be the number of n-odd sequences, and $S(n) = E(n) - O(n)$, with $S(0) = 1$.\n\nWe claim for $n \\ge 1$:\n$$\nS(n) = 1 + \\sum_{k=1}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} S(n - 2k).\n$$\n\nAll n-sequences fall into three categories:\n\n1. $(a_1) = (n)$: $S_1(n) = 1$.\n2. $a_1 \\ne a_2$: Swapping $a_1$ and $a_2$ changes parity, so $S_2(n) = 0$.\n3. $a_1 = a_2 = k$ ($1 \\le k \\le \\lfloor n/2 \\rfloor$): The difference is $S(n - 2k)$.\n\nThus, the claim holds.\n\nFor $n \\ge 3$:\n$$\nS(n) - S(n-2) = S(n-2),\\quad\\text{so}\\quad S(n) = 2S(n-2).\n$$\n\nFor $n=2$, $S(2) = 2 = 2S(0)$.\n\nTherefore,\n- If $n$ is odd:\n  $$S(n) = 2^{\\frac{n-1}{2}}$$\n- If $n$ is even:\n  $$S(n) = 2^{\\frac{n}{2}}$$\n\nEach n-sequence $(a_1, \\dots, a_m)$ corresponds to an increasing sequence $(a_1, a_1+a_2, \\dots, a_1+\\dots+a_m)$, which matches a subset of $\\{1, \\dots, n\\}$ containing $n$. Thus, the number of n-sequences is $2^{n-1}$, so\n$$\nE(n) = \\frac{2^{n-1} + S(n)}{2} = 2^{n-2} + 2^{\\left\\lfloor \\frac{n-2}{2} \\right\\rfloor}.\n$$\n\n*Remark*: To count n-sequences, place $n$ stones in a row and insert sticks between stones (at most one per gap). Each gap can have 0 or 1 stick, so $E(n) + O(n) = 2^{n-1}$. Sequences with $a_i \\ne a_j$ can be paired and have opposite parity; unpaired sequences are all n-even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18903,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\n\\frac{1}{\\sqrt{1+2x^2}} + \\frac{1}{\\sqrt{1+2y^2}} = \\frac{2}{\\sqrt{1+2xy}} \\\\\n\\sqrt{x(1-2x)} + \\sqrt{y(1-2y)} = \\frac{2}{9}\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The condition for the system is: $0 \\le x, y \\le \\frac{1}{2}$.\n\n*Remark:* Under this condition, we have\n\n$$\n\\frac{1}{\\sqrt{1+2x^2}} + \\frac{1}{\\sqrt{1+2y^2}} \\le \\frac{2}{\\sqrt{1+2xy}}\n$$\n\nThe equality holds if and only if $x = y$.\n\n*Proof:* According to the Cauchy-Schwarz inequality,\n\n$$\n\\left( \\frac{1}{\\sqrt{1+2x^2}} + \\frac{1}{\\sqrt{1+2y^2}} \\right)^2 \\le 2 \\left( \\frac{1}{1+2x^2} + \\frac{1}{1+2y^2} \\right)\n$$\n\nThe equality holds if and only if $\\sqrt{1+2x^2} = \\sqrt{1+2y^2}$, i.e., $x = y$ (since $x, y \\ge 0$).\n\nNext,\n\n$$\n\\frac{1}{1+2x^2} + \\frac{1}{1+2y^2} - \\frac{2}{1+2xy} = \\frac{2(y-x)^2(2xy-1)}{(1+2xy)(1+2x^2)(1+2y^2)} \\le 0\n$$\n\nConsequently,\n\n$$\n\\frac{1}{1+2x^2} + \\frac{1}{1+2y^2} \\le \\frac{2}{1+2xy}\n$$\n\nThe equality holds if and only if $x = y$.\n\nTherefore, the system is equivalent to\n\n$$\n\\begin{cases}\nx = y \\\\\n\\sqrt{x(1-2x)} + \\sqrt{y(1-2y)} = \\frac{2}{9}\n\\end{cases}\n\\Leftrightarrow\n\\begin{cases}\nx = y = \\frac{9 - \\sqrt{73}}{36} \\\\\nx = y = \\frac{9 + \\sqrt{73}}{36}\n\\end{cases}\n$$\n\nThus, the solutions are $x = y = \\frac{9 - \\sqrt{73}}{36}$ and $x = y = \\frac{9 + \\sqrt{73}}{36}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18904,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, $P$, and $Q$ be five pairwise different points in the plane. Suppose that $A$, $B$, and $C$ are not collinear and that\n\n$$\n\\frac{AP}{BP} = \\frac{AQ}{BQ} = \\frac{21}{20},\n$$\n\n$$\n\\frac{BP}{CP} = \\frac{BQ}{CQ} = \\frac{20}{19}.\n$$\n\nProve that the line $PQ$ contains the circumcentre of the triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We choose a coordinate system such that the circumcentre $O$ of $\\triangle ABC$ lies at the origin and the circumradius of $\\triangle ABC$ is equal to 1. Let $A = (a_1, a_2)$, $B = (b_1, b_2)$, and $C = (c_1, c_2)$. The following statements about a point $X = (x_1, x_2)$ are equivalent:\n\n$$\n\\frac{AX}{BX} = \\frac{21}{20} \\Leftrightarrow 400 \\left( (x_1 - a_1)^2 + (x_2 - a_2)^2 \\right) = 441 \\left( (x_1 - b_1)^2 + (x_2 - b_2)^2 \\right).\n$$\n\nWe simplify the equation using the relations $a_1^2 + a_2^2 = 1$, etc. We see that if $X \\in \\{P, Q\\}$, then\n\n$$\n0 = 41(x_1^2 + x_2^2) - (441b_1 - 400a_1)x_1 - (441b_2 - 400a_2)x_2 + 41,\n$$\n\nand similarly\n\n$$\n0 = 39(x_1^2 + x_2^2) - (400c_1 - 361b_1)x_1 - (400c_2 - 361b_2)x_2 + 39.\n$$\n\nWe conclude that\n\n$$\n[39(441b_1 - 400a_1) - 41(400c_1 - 361b_1)]x_1 + [39(441b_2 - 400a_2) - 41(400c_2 - 361b_2)]x_2 = 0,\n$$\n\nwhich simplifies to\n\n$$\nx_1(-39a_1 + 80b_1 - 41c_1) + x_2(-39a_2 + 80b_2 - 41c_2) = 0.\n$$\n\nThe numbers $-39a_1 + 80b_1 - 41c_1$ and $-39a_2 + 80b_2 - 41c_2$ cannot vanish at the same time, because otherwise $A$, $B$, and $C$ would be collinear. Hence the previous equation defines a line through $O$ which contains $P$ and $Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18905,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n \\ge 2$, and $a_1, a_2, \\dots, a_n$ are positive numbers that sum to $1$. Prove that\n$$\n\\frac{n}{n-1} \\le \\sum_{i=1}^{n} \\frac{a_i}{1-a_i},\n$$\nwith equality if and only if $a_i = 1/n$ for $i = 1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Use that the function $f(x) = \\frac{1}{1-x}$ is convex on $[0, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18906,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $k \\leq n$ 為兩個正整數。數奧國有 $n$ 個村莊，其中有些村莊之間有道路連接。對於任兩個村莊，其之間的距離定義為從一個村莊抵達另一個村莊最少需走過的道路數量；如果完全無法從某個村莊抵達另一個村莊，則這兩個村莊之間的距離為無限大。\n\n剛到數奧國的小勳被隔離在某個地方，以至於他看不見數奧國裡道路連接的狀況，但他知道 $n$ 與 $k$。他想知道數奧國中最遠的兩個村莊之間的距離是否為有限大。為此，他每次可以打一通電話去數奧辦公室，指定某兩個村莊，問它們之間的距離是大於、等於或小於 $k$，而辦公室會如實回答（無限大的距離大於 $k$）。\n\n試證明小勳可以在 $\\dfrac{2n^2}{k}$ 通電話內確認數奧國中最遠的兩個村莊，其之間的距離是否為有限大。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $G$ 為數奧國對應的圖，$d(u,v)$ 則為 $u$ 與 $v$ 之間的距離。考慮以下策略：\n\n1. 任選一個城市 $v$，並令 $S = \\{v\\}$。\n2. 對於任何 $u \\neq v$，打一通電話詢問 $u$ 與 $v$ 之間的距離。\n3. 如果存在 $v'$ 使得 $\\min_{s \\in S} d(v', s) \\geq k$，且存在 $s \\in S$ 使得 $d(v', s) = k$，則令 $S = S \\cup \\{v'\\}$，改令 $v = v'$，並回到步驟 2。\n4. 檢查任何村莊是否都與某個 $s \\in S$ 的距離至多為 $k$；若是，則回報 $G$ 連通。否則，回報 $G$ 不連通。\n\n首先證明此演算法確實可以確認 $G$ 是否連通。將演算法結束時的集合 $S$ 記為 $S_0$。由步驟 3 知 $S_0$ 中的點必屬於同一連通集（可用數歸證明），從而若步驟 4 回報連通，$G$ 必連通。反之，若步驟 4 回報不連通，代表存在 $u$ 使得 $\\min_{s \\in S_0} d(u, s) > k$。假設此時 $G$ 實為連通，則對於任何 $v \\in S_0$，必然存在從 $u$ 到 $v$ 的最短路徑，且其距離大於 $k$。又在此路徑上，必然存在一個點 $w$ 其到 $S$ 中的點的距離至多為 $k$。但如此一來，在步驟 3 時，我們應該要取到 $v' = w$ 而回到步驟 2，從而演算法尚未結束，矛盾。故演算法回報 $G$ 連通若且唯若 $G$ 真的連通。\n\n接著我們證明此演算法所需的電話數至多為 $\\dfrac{2n^2}{k}$。注意到我們只有在步驟 2 打 $n$ 次電話，且我們至多只會重複 $|S_0|$ 次步驟 2，因此我們只須證明 $|S_0| \\leq \\dfrac{2n}{k}$。此在 $|S_0| = 1$ 時顯然，故假設 $|S_0| \\geq 2$。考慮 $S_0$ 所在的連通集 $C$，假設 $|C| = m$。考慮 $C$ 上的生成樹，並考慮在此樹上的 Euler tour $v_1, \\dots, v_{2(m-1)}$。則我們必然可以取到 $i_1 < i_2 < \\dots < i_{|S_0|}$ 使得 $S_0 = \\{v_{i_1}, v_{i_2}, \\dots, v_{i_{|S_0|}}\\}$。\n\n現在，由於 $v_{i_j}$ 與 $v_{i_{j+1}}$ 為 $S_0$ 中相異的兩個元素，由步驟 3 知兩點之間的距離至少為 $k$，從而 $i_{j+1} - i_j \\geq k$。這表示 $|v_{i_{|S_0|}} - i_1| \\geq k(|S_0| - 1)$。又注意到我們可以令 $v_{i+2(m-1)} = v_i$，並考慮從 $v_{i_{|S_0|}}$ 開始到 $v_{i_1+2(m-1)}$ 的路徑，此距離同樣也至少須為 $k$，也就是 $i_1 + 2(m-1) - |S_0| \\geq k$。結合兩式，我們得到 $k|S_0| \\leq 2(m-1) < 2n$，故 $|S_0| < \\dfrac{2n}{k}$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18907,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $\\tau(n)$ be the number of positive divisors of $n$.\n\na) Find all positive integers $n$ such that $\\tau(n) + 2023 = n$.\n\nb) Prove that there exist infinitely many positive integers $k$ such that there are exactly two positive integers $n$ satisfying $\\tau(kn) + 2023 = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we will prove the following lemma.\n\n*Lemma.* For any positive integer $n$, $\\tau(n) \\leq 2\\sqrt{n}$.\n\n*Proof.* Consider any positive integer $n$, let $d_1, d_2, \\dots, d_s$ be all positive divisors not exceeding $\\sqrt{n}$ of $n$. Then, obviously $s \\leq \\sqrt{n}$.\n\nNote that if $x$ is a divisor not less than $\\sqrt{n}$ of $n$, then $\\frac{n}{x}$ is a positive divisor not exceeding $\\sqrt{n}$ of $n$. It follows that $\\tau(n) \\leq 2s \\leq 2\\sqrt{n}$. ■\n\nBack to the problem, suppose there exists a positive integer $n$ satisfying the equation $\\tau(n) + 2023 = n$. According to the above lemma, we have $n \\leq 2\\sqrt{n} + 2023$. Solving this inequality, we get\n\n$$\nn \\leq (\\sqrt{2024} + 1)^2 < 2115.\n$$\n\nOn the other hand, we also have $n = \\tau(n) + 2023 \\geq 2025$. Therefore $2025 \\leq n \\leq 2114$. Let $n = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k}$ where $p_1, p_2, \\dots, p_k$ are distinct primes and $e_1, e_2, \\dots, e_k$ are positive integers. Then $\\tau(n) = (e_1 + 1)(e_2 + 1)\\cdots(e_k + 1)$ and the equation $\\tau(n) + 2023 = n$ can be rewritten as\n\n$$\n(e_1 + 1)(e_2 + 1)\\cdots (e_k + 1) + 2023 = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k} \\quad (1)\n$$\n\nIf $e_1, e_2, \\dots, e_k$ are all even numbers, then $n$ is a perfect square, and $2025 \\leq n \\leq 2114$ so $n = 2025$. However, this value does not satisfy the equation. Therefore, among the numbers $e_1, e_2, \\dots, e_k$ there must be at least one odd number. From here, combined with equation (1), we deduce that $p_1, p_2, \\dots, p_k$ are odd prime numbers. Without loss of generality, assume $3 \\leq p_1 < p_2 < \\dots < p_k$.\n\nIf $k \\geq 5$, then $n = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k} \\geq 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 2114$, a contradiction. Therefore $k \\leq 4$.\n\n**Case 1:** $k = 1$. In this case, $2114 \\geq n = p_1^{e_1} \\geq 3^{e_1}$. Therefore $e_1 \\leq 6$. Checking each case of the value of $e_1$, we find no prime number $p_1$ that satisfies equation (1).\n\n**Case 2:** $k = 2$. In this case, $2114 \\geq n = p_1^{e_1} p_2^{e_2} \\geq 3^{e_1} 5^{e_2} \\geq 3^{e_1+e_2-1} \\cdot 5$, deduce $e_1 + e_2 \\leq 6$. From there\n\n$$\n\\tau(n) = (e_1 + 1)(e_2 + 1) \\leq \\left( \\frac{e_1 + e_2 + 2}{2} \\right)^2 \\leq 9.\n$$\n\nChecking specific cases of $\\tau(n)$, we find no corresponding value of $n$ satisfying the equation.\n\n**Case 3:** $k = 3$. In this case,\n\n$$\n2114 \\geq n = p_1^{e_1} p_2^{e_2} p_3^{e_3} \\geq 3^{e_1} 5^{e_2} 7^{e_3} \\geq 3^{e_1+e_2+e_3-2} \\cdot 5 \\cdot 7.\n$$\n\n(Computation continues, but the solution is incomplete here.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18908,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Consider the equation $x^k + px = y^k$ in integers $x, y$.\n\nShow that for given $p$ and $k$, there is at most one solution $(x, y)$ in integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x^k + px = y^k$, define $\\alpha = y - x$. Then\n\n$$\nx^k + px = (x + \\alpha)^k > x^k + \\alpha x \\implies \\alpha < p.\n$$\n\nAlso,\n\n$$\n\\alpha \\mid (x + \\alpha)^k - x^k = px \\implies \\alpha \\mid x.\n$$\n\nExpanding $(x + \\alpha)^k$ binomially, every term except $\\alpha^k$ is divisible by $x$, so $x \\mid \\alpha^k$. Let $\\beta = \\alpha^k / x$; we show $\\beta = 1$.\n\n$$\n\\begin{aligned}\npx &= (x + \\alpha)^k - x^k \\\\\n\\beta^k px &= (\\beta x + \\beta \\alpha)^k - (\\beta x)^k \\\\\n\\beta^{k-1} p \\alpha^k &= (\\alpha^k + \\beta \\alpha)^k - (\\alpha^k)^k \\\\\n\\beta^{k-1} p &= (\\alpha^{k-1} + \\beta)^k - (\\alpha^{k-1})^k\n\\end{aligned}\n$$\n\nSince $\\alpha \\mid x$, $x \\mid \\alpha^k$, and $x \\beta = \\alpha^k$, we have $\\beta \\mid \\alpha^{k-1}$. Thus, every term on the right is divisible by $\\beta^k$, but the left is only $p \\beta^{k-1}$. So either $p = \\beta$, which leads to $y^k = x^k + px = x^k + \\alpha^k$, which has no solutions for $x > 0$, or $\\beta = 1$, so $x = \\alpha^k$. The equation becomes $p = (\\alpha^{k-1} + 1)^k - (\\alpha^{k-1})^k$, which is strictly increasing in $\\alpha$, so for given $p$ and $k$ there is at most one solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18909,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest positive integer $k$ for which there exists a positive integer $n$ such that:\n\n$$\nsin(n + 1) < \\sin(n + 2) < \\sin(n + 3) < \\dots < \\sin(n + k).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the largest such $k$ is $5$.\n\nThe difference between two consecutive terms is:\n\n$$\n\\sin(n + i + 1) - \\sin(n + i) = 2 \\sin \\frac{1}{2} \\cos \\frac{2n + 2i + 1}{2} > 0 \\iff \\cos \\frac{2n + 2i + 1}{2} > 0. \\tag{\\star}\n$$\n\nSuppose that for $k = 6$ there exists a positive integer $n$ such that\n$$\n\\sin(n + 1) < \\sin(n + 2) < \\sin(n + 3) < \\sin(n + 4) < \\sin(n + 5) < \\sin(n + 6).\n$$\nRelation $(\\star)$ is equivalent to:\n$$\n\\cos \\left( n + \\frac{2i + 1}{2} \\right) > 0, \\quad i = 1, 2, 3, 4, 5.\n$$\nThis implies:\n$$\n\\cos \\left(n + \\frac{3}{2}\\right) + \\cos \\left(n + \\frac{11}{2}\\right) = 2 \\cos 2 \\cdot \\cos \\left(n + \\frac{7}{2}\\right) > 0 \\implies \\cos 2 > 0,\n$$\nwhich is a contradiction. Therefore, $k \\leq 5$.\n\nFor $k = 5$, we need to find a positive integer $n$ such that $\\cos(n + \\frac{2i + 1}{2}) > 0$ for $i = 1, 2, 3, 4$.\n\nFrom $2m\\pi - \\frac{\\pi}{2} < n + \\frac{3}{2} < n + \\frac{9}{2} < 2m\\pi + \\frac{\\pi}{2}$, or equivalently $(4m - 1) \\cdot 10\\pi < 20n + 30 < 20n + 90 < (4m + 1) \\cdot 10\\pi$, and using $3.14 < \\pi < 3.15$, we obtain a sufficient condition:\n\n$$\n(4m - 1) \\cdot 31.5 < 20n + 30 < 20n + 90 < (4m + 1) \\cdot 31.4 \\implies m \\leq 7.\n$$\n\nFor $m = 7$, the bounds for $n$ are:\n$$\n\\frac{27\\pi - 3}{2} < n < \\frac{29\\pi - 9}{2},\n$$\nwhich, using $3.141 < \\pi < 3.142$, gives $n = 41$. Thus, our claim is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18910,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n \\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.\n\nLet $s$ be the smallest number of collisions after which all particles return to their original positions. Find $s$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $(n-1)m'$, where $m'$ is the smallest number such that $\\frac{m'(n-2)}{2n}$ is an integer. More precisely:\n\n- $s = 2n(n-1)$ when $n$ is odd\n- $s = n(n-1)$ when $n$ is divisible by $4$\n- $s = \\frac{n(n-1)}{2}$ when $n-2$ is divisible by $4$ (but the $n/2$-th point will be the one moving in reverse, not $p_0$)\n\nWe first introduce some setup for convenience. We treat points on the circle as $[0, 2n)$ and use $r = r + 2n$ for any real $r$ to refer to points on the circle. For example, point $-1$ means $2n - 1$, etc. Assume it takes any particle $2n$ units of time to go around the entire circle. Thus, if a particle at $0$ moves clockwise for time $t$, it reaches point $t$.\n\nInitially, let the particles be $p_0, \\dots, p_{n-1}$ with $p_i$ at point $2i$.\n\nMoving clockwise means the value is increasing; moving anti-clockwise means the value is decreasing. Assume $p_0$ is the point initially moving anti-clockwise.\n\nLet $t_1 > 0$ be the total time when, for the first time, all particles return to their initial positions. Let $t_2 > 0$ be the first time when all particles are equally spaced apart, i.e., $2$ units apart.\n\nClearly, $\\frac{t_1}{t_2}$ is an integer. Thus, we try to find $t_2$.\n\nObserve that if we replace each collision event with the two particles passing through each other, $t_2$ does not change. So, for the purposes of calculating $t_2$, we can assume they indeed pass through. Thus, at time $t_2$, there are particles at $-t_2, 2 + t_2, 4 + t_2, \\dots, 2n - 2 + t_2$. For them to be equally spaced, this sequence must be $t_2, 2 + t_2, 4 + t_2, \\dots, 2n - 2 + t_2$. Thus, modulo $2n$, $t_2 = -t_2$, and the minimum value of $t_2$ that makes this possible is $t_2 = n$. Now, we can set $t_1 = n \\cdot t_1'$.\n\nAlternatively, we can look at the positions relative to the clockwise moving particle. Then we just have the anti-clockwise particle moving at speed $2$, so it must need $n$ units of time to return to its original position and make $n-1$ collisions in this period.\n\nNow, let us analyze what happens at time $n$. There are now particles in positions $n, n+2, \\dots, n-2$. The cyclic order of particles is preserved as collisions never alter it, and the collisions only happen in $(1, n)$ after the first collision of $p_0$ at point $2n-1$ with $p_n$. Thus, $p_0$ does not have any more collisions in the next $n-1$ units of time. Thus, $p_0$ must be at position $2n-1 + (n-1) = n-2$. Since cyclic order is preserved, $p_i$ would be at position $2i + n - 2$. Thus, everything has cyclically moved forward by $n-2$.\n\nThus, in time $n$, every particle moves forward by $n-2$ units and there are $n-1$ collisions. In time $nk$, we have $(n-1)k$ collisions and every particle moves forward by $(n-2)k$ units. Thus, all particles return to their initial positions if and only if $2n \\mid (n-2)k$. This is exactly what we desired! $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18911,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose each digit of a positive integer $n$ is between $1$ and $7$ inclusive. Furthermore, $n$ is a multiple of $7$, and every integer obtained by permuting the digits of $n$ is also a multiple of $7$. Determine all such integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us first show that any positive integer $n$ satisfying the conditions must have the same digit repeated in all positions. Suppose digits $a$ and $b$ (with $1 \\leq a, b \\leq 7$) both appear in $n$. Permute the digits so that $a$ is in the tens place and $b$ in the units place, forming $n'$. Let $N$ be the number formed by the remaining digits (or $N=0$ if $n'$ has only two digits). Then $n' = 100N + 10a + b$. Interchanging $a$ and $b$ gives $n'' = 100N + 10b + a$. Both $n'$ and $n''$ are multiples of $7$, so their difference $n' - n'' = 9(a - b)$ is also a multiple of $7$. Since $7$ and $9$ are coprime, $a - b$ must be a multiple of $7$. But $1 \\leq a, b \\leq 7$, so $a = b$. Thus, all digits of $n$ are equal.\n\nNow, let every digit of $n$ be $a$ for some $1 \\leq a \\leq 7$. For $a = 7$, any number of digits works since $77\\cdots77$ is always a multiple of $7$. For $1 \\leq a \\leq 6$, consider numbers with $j$ digits, all equal to $a$. The 6-digit number $111111$ is a multiple of $7$, but numbers with fewer digits are not. In general, for $N(6k + \\ell)$ (the number with $6k + \\ell$ digits, all $a$), dividing by $111111$ leaves a remainder $m_\\ell$ with $\\ell$ digits, all $a$. For $\\ell = 0$, $N(6k)$ is a multiple of $7$. For $1 \\leq \\ell \\leq 5$, $m_\\ell$ is not a multiple of $7$ since neither $a$ nor the number with $\\ell$ ones is a multiple of $7$. Thus, for $1 \\leq a \\leq 6$, $n$ must have $6k$ digits for $k \\geq 1$.\n\n**Summary:**\n\n$$\n\\begin{aligned}\n&\\text{Either}\\quad n = \\underbrace{aa\\cdots aa}_{6k \\text{ times}} \\quad (1 \\leq a \\leq 6,\\ k \\geq 1), \\\\\n&\\text{or}\\quad n = \\underbrace{77\\cdots77}_{p \\text{ times}} \\quad (p \\geq 1).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18912,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the number of lines which go through the origin and precisely one other point with integer coordinates $ (x, y) $, $ 0 \\le x, y \\le n $, is at least $ 2n $ when $ n $ is sufficiently large.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice first that the number of lines going through a point $ (k, \\ell) $ with $ k = m $ or $ \\ell = m $ for some $ m $ such that the line does not go through any point $ (k', \\ell') $ with $ 0 \\le k', \\ell' < m $ is $ 2\\varphi(m) $. Now, the number of lines in the problem is\n\n$$\n2 \\sum_{n/2 < k \\le n} \\varphi(k).\n$$\n\nAccording to Bertrand's postulate, there is a prime, which we denote by $ p_1 $, on the interval $ (n/2, n) $, and also on the intervals $ (n/4, n/2) $, $ (n/6, n/3) $ and $ (n/8, n/4) $. We denote these by $ p_2, p_3, p_4 $, respectively. Hence, we may estimate\n\n$$\n\\sum_{n/2 < k \\le n} \\varphi(k) \\ge \\varphi(p_1) + \\varphi(2p_2) + \\varphi(3p_3) + \\varphi(4p_4) \\ge p_1 - 1 + p_2 - 1 + 2(p_3 - 1) + 2(p_4 - 1) \\\\\n\\ge \\frac{n}{2} - 1 + \\frac{n}{4} - 1 + 2\\left(\\frac{n}{6} - 1\\right) + 2\\left(\\frac{n}{8} - 1\\right) = \\frac{4}{3}n - 6 > n,\n$$\n\nwhen $ n $ is sufficiently large, and therefore, the number of lines is greater than $ 2n $ when $ n $ is large enough.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18913,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with incentre $I$. Draw a line perpendicular to $BI$ at $I$ and let it intersect $BC$ and $BA$ at $D$ and $E$ respectively. Let $P$ and $Q$ be respectively the incentres of the triangles $BIA$ and $BIC$. Suppose the four points $D, E, P, Q$ are concyclic. Prove that $BA = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Join $DE$ and $PQ$. Let $S$ be the centre of the circle $\\Gamma$ passing through $E, P, Q, D$. Let $P'$ denote the reflection of $P$ in $BI$. Since $\\angle PBI = \\angle QBI$, it follows that $P'$ lies on $BQ$. Since $E$ and $D$ are symmetric about the line $EI$, and $P$ and $P'$ are also symmetric about $BI$, it follows that $P'$ lies on $\\Gamma$. Since $Q$ lies on $\\Gamma$ and $P'$ lies on the line $BQ$, we conclude that $Q = P'$. This means $BI$ is the perpendicular bisector of $PQ$. Since $BI$ is already the perpendicular bisector of $ED$, it follows that $ED$ is parallel to $PQ$.\n\nConsider the circumcircle of the triangle $AIC$. We have\n\n![](images/India_2017_p1_data_0c072faeb6.png)\n\n$$\n\\angle BIC = 180^{\\circ} - (B/2 + C/2) = 90^{\\circ} + (A/2).\n$$\n\nSince $\\angle BID = 90^{\\circ}$, we see that $\\angle DIC = A/2 = \\angle IAC$. Hence $ED$ is tangent to the circumcircle of $\\triangle AIC$ at $I$. Hence the circumcentre of the triangle $AIC$ lies on the line $BS$. This implies that $BS$ is the perpendicular bisector of $AC$ as well. It follows that $ED \\parallel AC$. Thus $PQ \\parallel ED \\parallel AB$.\n\nSince $BS$ is the perpendicular bisector of $PQ$ and that of $AC$, we see that $PQCA$ is an isosceles trapezium. Hence $P, Q, C, A$ are concyclic. This gives $\\angle PAC = \\angle QCA$. Therefore $\\angle A = \\angle C$ and hence $BA = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18914,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist pairwise disjoint sets $A_1, A_2, \\dots, A_{2014}$ whose union is the set of natural numbers, and for which the following condition holds:\n\nFor arbitrary natural numbers $a$ and $b$, at least two of the numbers $a$, $b$, $\\gcd(a, b)$ belong to one of the sets $A_1, A_2, \\dots, A_{2014}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Let $v_2(n)$ be the greatest integer for which $2^{v_2(n)}$ divides $n$. Then $v_2(\\gcd(a, b)) = \\min\\{v_2(a), v_2(b)\\}$. Therefore, at least two of the numbers $v_2(a)$, $v_2(b)$, and $v_2(\\gcd(a, b))$ are equal.\n\nDefine the sets $A_{i+1} = \\{n \\mid v_2(n) \\equiv i \\pmod{2014}\\}$ for $0 \\leq i \\leq 2013$.\n\nObviously, the sets $A_1, A_2, \\dots, A_{2014}$ are pairwise disjoint, their union is $\\mathbb{N}$, and two of the numbers $a$, $b$, $\\gcd(a, b)$ belong to the set $A_{i+1}$, where $i$ is the residue of $v_2(\\gcd(a, b))$ modulo $2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18915,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $P$ a point in the plane. Let $A'$, $B'$, $C'$ be the reflections of $P$ over the sides $BC$, $CA$, and $AB$, respectively. Prove that the points $A'$, $B'$, and $C'$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "It's easy to see (say, by the law of sines) that\n\n$$\n\\frac{AC'}{BC'} = \\frac{AP \\sin \\angle APC'}{BP \\sin \\angle BPC'}, \\quad \\frac{BA'}{CA'} = \\frac{BP \\sin \\angle BPA'}{CP \\sin \\angle CPA'}, \\quad \\frac{CB'}{AB'} = \\frac{CP \\sin \\angle CPB'}{AP \\sin \\angle APB'}\n$$\n\nThe construction of $A'$, $B'$, $C'$ by reflections implies that\n\n$$\n\\sin \\angle APC' = \\sin \\angle CPA', \\quad \\sin \\angle BPC' = \\sin \\angle CPB', \\quad \\sin \\angle BPC' = \\sin \\angle CPB'\n$$\n\nHence,\n\n$$\n\\frac{AC'}{BC'} \\cdot \\frac{BA'}{CA'} \\cdot \\frac{CB'}{AB'} = 1,\n$$\n\nand the proof is complete by Menelaus' theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18916,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ and $N$ be the midpoints of segments $BI$ and $CI$, respectively, in triangle $ABC$ with incenter $I$ and incircle touching $BC$, $CA$, $AB$ at $D$, $E$, and $F$, respectively. The line $FM$ intersects the external angle bisector at vertex $B$ at point $K$, and the line $EN$ intersects the external angle bisector at vertex $C$ at point $L$. Prove that the points $K$, $B$, $C$, and $L$ lie on the same circle.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle KBE = 90^\\circ$ as $BE$ and $BK$ are the internal and external angle bisectors at the same vertex (see the figure below). We will now show $\\angle BKC = 90^\\circ$. Let $X$ be a point on line $AB$ such that $CX \\parallel BE$; then the external angle bisector at vertex $B$ is also perpendicular to $CX$.\n\nDenote $\\angle ABC = \\beta$. Then $\\angle BCX = \\angle CBE = \\frac{\\beta}{2}$ and $\\angle XBC = 180^\\circ - \\beta$. Therefore, $\\angle BXC = 180^\\circ - (\\angle BCX + \\angle XBC) = \\frac{\\beta}{2}$, so triangle $XBC$ is isosceles with apex at $B$. Thus, the altitude from $B$ of triangle $XBC$ bisects $CX$. As previously established, this altitude lies on the external angle bisector at $B$.\n\nLet $k = \\frac{|FX|}{|FB|} = \\frac{|FC|}{|FI|}$. A homothety with center $F$ and ratio $k$ sends $B$ and $I$ to $X$ and $C$, respectively, so it sends the midpoint $M$ of $BI$ to the midpoint of $CX$. Therefore, the line $FM$ passes through the midpoint of $CX$. Consequently, the intersection point $K$ of $FM$ and the external angle bisector at $B$ lies at the midpoint of $CX$, and $\\angle BKC = 90^\\circ$, as we wanted to show. Similarly, $\\angle BLC = 90^\\circ$. Therefore, points $K$ and $L$ lie on the circle with diameter $BC$. The statement follows.\n\n![](images/EST_ABooklet_2024_p50_data_c6e2048611.png)\n\n*Fig. 43*\n\n![](images/EST_ABooklet_2024_p50_data_5a24ef4af4.png)\n\n*Fig. 44*",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18917,
    "subject": "Mathematics (Olympiad)",
    "question": "For any natural number $n \\ge 3$, find integers $a_1 < a_2 < \\dots < a_n$ such that the following equality holds:\n\n$$\n\\frac{a_1}{a_1} + \\frac{a_1}{a_2} + \\frac{a_1}{a_3} + \\dots + \\frac{a_1}{a_n} = \\frac{a_2}{a_1} + \\frac{a_2}{a_2} + \\frac{a_2}{a_3} + \\dots + \\frac{a_2}{a_n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "One can choose $a_k = -2^{n-k} \\cdot (2^{n-1}-1)$ for $k = 1, \\ldots, n-1$ and $a_n = 2^{n-1}$.\n\nLet us first find rational numbers satisfying the equality, and then multiply them by the least common multiple of the denominators to obtain integers.\n\nOne can fix $b_1 = -2^{n-1}$, $b_2 = -2^{n-2}$, ..., $b_{n-1} = -2^1$. Let us rewrite the equation as follows:\n\n$$\n\\frac{b_1}{b_1} + \\frac{b_1}{b_2} + \\frac{b_1}{b_3} + \\dots + \\frac{b_1}{b_{n-1}} - \\frac{b_2}{b_1} - \\frac{b_2}{b_2} - \\frac{b_2}{b_3} - \\dots - \\frac{b_2}{b_{n-1}} = \\frac{b_2}{b_n} - \\frac{b_1}{b_n}\n$$\n\nor\n\n$$\nX = \\frac{2^{n-1}}{2^{n-2}} - \\frac{2^{n-2}}{2^{n-1}} + \\frac{2^{n-1}}{2^{n-3}} - \\frac{2^{n-2}}{2^{n-3}} + \\frac{2^{n-1}}{2^{n-4}} - \\frac{2^{n-2}}{2^{n-4}} + \\dots + \\frac{2^{n-1}}{2^1} - \\frac{2^{n-2}}{2^1} = \\frac{2^{n-1}}{b_n} - \\frac{2^{n-2}}{b_n}\n$$\n\nwhich simplifies to\n\n$$\nX = 2 - \\frac{1}{2} + 2^2 - 2 + 2^3 - 2^2 + \\dots + 2^{n-2} - 2^{n-3} = 2^{n-2} - \\frac{1}{2} = \\frac{2^{n-1}-1}{2} = \\frac{2^{n-2}}{b_n}\n$$\n\nThus,\n\n$$\nb_n = \\frac{2^{n-1}}{2^{n-1}-1} > 0.\n$$\n\nAll $b_k$, $k = 1, \\ldots, n-1$ are negative, and from the last inequality it follows that $b_1 < b_2 < \\dots < b_{n-1} < b_n$. Finally, we can get the aforementioned integers as follows:\n\n$$\na_k = b_k \\cdot (2^{n-1} - 1), \\quad k = 1, \\ldots, n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18918,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a, b, c > 0$ satisfy $a + b + c = 3$, prove that\n$$\n\\frac{a^2(b+1)}{ab+a+b} + \\frac{b^2(c+1)}{bc+b+c} + \\frac{c^2(a+1)}{ca+c+a} \\ge 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We notice that\n$$\n\\frac{a^2(b+1)}{ab+a+b} = a - \\frac{ab}{ab+a+b} \\ge a - \\frac{ab}{ab+2\\sqrt{ab}} = a - 1 + \\frac{2}{\\sqrt{ab}+2}.\n$$\nWriting two similar inequalities and adding them up, we obtain\n$$\n\\sum_{cyc} \\frac{a^2(b+1)}{ab+a+b} \\ge 2 \\sum_{cyc} \\frac{1}{\\sqrt{ab}+2} \\ge 2 \\cdot \\frac{(1+1+1)^2}{6+\\sqrt{ab}+\\sqrt{ac}+\\sqrt{bc}} \\ge \n\\frac{18}{6+\\frac{a+b}{2}+\\frac{a+c}{2}+\\frac{b+c}{2}} = 2.\n$$\nEquality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18919,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive real $a$ such that for any three points $A$, $B$, $C$ on the unit circle, there exists an equilateral triangle $PQR$ with side length $a$ such that $A$, $B$, $C$ are all inside or on the boundary of triangle $PQR$.",
    "options": [],
    "answer": "See solution",
    "solution": "$a = \\dfrac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}}$.\n\n**Sufficiency:**\nFor any three points $A$, $B$, $C$ on the unit circle, let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$, with $\\alpha \\leq \\beta \\leq \\gamma$.\n\nIf $\\beta \\leq 60^{\\circ}$, since $AB \\leq 2 < a$ (where $a > \\dfrac{(2 \\sin 75^{\\circ})^2}{\\sqrt{3}} = \\dfrac{(\\sqrt{6} + \\sqrt{2})^2}{4\\sqrt{3}} = \\dfrac{8 + 4\\sqrt{3}}{4\\sqrt{3}} > 2$), one can draw a segment $PQ$ of length $a$ containing $A$ and $B$, then find $R$ such that $\\triangle PQR$ is equilateral, with $C$, $R$ on the same side of $PQ$. It follows from $\\alpha \\leq \\beta \\leq 60^{\\circ}$ that $C$ is inside $\\triangle PQR$ or on the boundary.\n\nIf $\\beta > 60^{\\circ}$, we first prove:\n\nAt least one of $\\sin \\beta \\sin \\gamma$ and $\\sin \\beta \\sin(\\alpha + 60^{\\circ})$ is less than or equal to $\\sin^2 80^{\\circ}$. (*Proof omitted for brevity*)\n\n(i) If $\\sin \\beta \\sin \\gamma \\leq \\sin^2 80^{\\circ}$, let $AD \\perp BC$ with foot $D$; take $P$, $Q$ on the line $BC$ such that $PD = DQ = \\dfrac{a}{2}$; take $R$ on the ray $DA$ such that\n$$\nDR = \\dfrac{\\sqrt{3}a}{2} = 2 \\sin^2 80^{\\circ}.\n$$\nNotice $AD = 2 \\sin \\beta \\sin \\gamma \\leq DR$, so $A$ lies inside $\\triangle PQR$ or on the boundary; from $60^{\\circ} \\leq \\beta, \\gamma \\leq 120^{\\circ}$, $B$, $C$ lie on $PQ$.\n\n(ii) If $\\sin \\beta \\sin(\\alpha + 60^{\\circ}) \\leq \\sin^2 80^{\\circ}$, take $P = A$, extend $AB$ to $Q$ with $AQ = a$, and find $R$ such that $\\triangle PQR$ is equilateral and $C$, $R$ lie on the same side of $PQ$. Let $M$ be the intersection of the ray $AC$ and the line $QR$. Notice $AB \\leq 2 < a$, so $B$ is on the boundary of $\\triangle PQR$; since\n$$\nAM = a \\cdot \\dfrac{\\sin 60^{\\circ}}{\\sin(\\alpha + 60^{\\circ})} = \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin(\\alpha + 60^{\\circ})} > 2 \\sin \\beta = AC,\n$$\n$C$ is inside $\\triangle PQR$ or on the boundary.\n\nThus, $a = \\dfrac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}}$ suffices.\n\n**Necessity:**\nSuppose $A$, $B$, $C$ on the unit circle with $\\angle BAC = 20^{\\circ}$, $\\angle ABC = \\angle ACB = 80^{\\circ}$.\n\nSuppose the equilateral triangle $PQR$ with side length $a$ covers $A$, $B$, $C$ (inside or on the boundary). By translation and rotation, we can arrange $A$ on the boundary of $\\triangle PQR$, and $B$ or $C$ on the boundary. If neither is a vertex, assume all lie on $PQ$ or $PR$. Take $P$ as the homothetic centre and rescale $\\triangle PQR$ so all of $A$, $B$, $C$ fall on the boundary or become a vertex. Since $\\angle ABC$, $\\angle ACB > 60^{\\circ}$, $B$, $C$ cannot be vertices; there is symmetry between $B$ and $C$. Thus, consider four cases:\n\n1. $A$ is a vertex, say $A = P$. Let $PH \\perp QR$ with foot $H$. The angle between $PH$ and one of $PB$, $PC$ is $\\leq 10^{\\circ}$. Thus, $PH \\geq 2 \\sin^2 80^{\\circ}$, and\n$$\na = PQ \\geq \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n2. $A$ and $B$ lie on the same side, say $PQ$, with $PA \\leq PB$. Let $PH \\perp QR$ at $H$. The angle between $AC$ and $PH$ is $10^{\\circ}$. Similarly,\n$$\na \\geq \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n3. $B$ and $C$ lie on the same side, say $QR$. Let $PH \\perp QR$ at $H$. The lines $AC$ and $PH$ cross at $10^{\\circ}$. Similarly,\n$$\na \\geq \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n4. $A$, $B$, $C$ are all on different sides. Let $A \\in PQ$, $B \\in QR$, $C \\in RP$, and $\\vartheta = \\angle RBC \\leq \\angle RCB$, $\\vartheta \\leq 60^{\\circ}$.\n\n(i) If $\\vartheta \\leq 20^{\\circ}$, the angle between $AB$ and the altitude on $QR$ does not exceed $10^{\\circ}$. Similarly,\n$$\na \\geq \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n(ii) If $40^{\\circ} \\leq \\vartheta \\leq 60^{\\circ}$, the angle between $AC$ and the altitude on $PQ$ does not exceed $10^{\\circ}$. Similarly, $a \\geq \\dfrac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}$.\n\nTherefore, the least $a$ is $\\boxed{\\dfrac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18920,
    "subject": "Mathematics (Olympiad)",
    "question": "A function $f: \\mathbf{R} \\to \\mathbf{R}$ satisfies\n$$\nf(f(a)) = f(a) \\quad \\text{and} \\quad f(a+b) = f(a) + f(b)\n$$\nfor all real numbers $a, b$. Prove that, for all real $x$, there exists a unique $y$ such that $f(y) = 0$ and $x = y + f(z)$ for some real $z$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be given. First, the uniqueness is proved. Assume that $x = y + f(z)$ with $f(y) = 0$. If $f$ is applied on both sides, then\n$$\nf(x) = f(y + f(z)) = f(y) + f(f(z)) = 0 + f(z) = f(z),\n$$\nbecause $f(y) = 0$ and $f(f(a)) = f(a)$. Hence $y = x - f(z) = x - f(x)$ is the only possible candidate for $y$.\n\nNow we prove that $y = x - f(x)$ has the assumed property. Observe that $f(a - b) = f(a) - f(b)$, and hence\n$$\nf(y) = f(x - f(x)) = f(x) - f(f(x)) = f(x) - f(x) = 0.\n$$\nThus $x = y + f(x)$ as was to be proved. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18921,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{N} \\to \\mathbb{N}$ be a function such that for all positive integers $m, n$, $f(m)$ divides $f(n) - n$ implies $m$ divides $n$. Find all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $m = k$ and $n = k$ in the original condition, we have $f(k) \\mid k$.\n\nNow we prove that $f(k) = k$ by induction on $k$.\n\nFor $k = 1$ this claim is trivial. Let $l > 1$ be a positive integer and assume that this claim is true for $k < l$. Suppose that $f(l) < l$. Since $f(l)$ divides $f(l) - l$ and $f(f(l)) = f(l)$ by the induction hypothesis, $f(l)$ divides $f(f(l)) - l$. We have $l \\nmid f(l)$ on the other hand, which contradicts the original condition when $m = f(l)$ and $n = l$. Hence we have $f(l) \\ge l$, and it follows that $f(l) = l$ since $f(l) \\mid l$. This completes the induction.\n\nConversely, the function $f(n) = n$ satisfies the original condition. Thus this is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18922,
    "subject": "Mathematics (Olympiad)",
    "question": "一個正 100 邊形的 41 個頂點被塗成黑色，其餘 59 個頂點被塗成白色。證明我們可以找到由這 100 邊形的頂點所組成的 24 個凸四邊形 $Q_1, \\dots, Q_{24}$，滿足：\n\n- 四邊形 $Q_1, \\dots, Q_{24}$ 兩兩交集為空集合，且\n- 每個四邊形 $Q_i$ 的四個頂點中，都恰有三個頂點同色。\n\n（註：我們定義四邊形包含其邊界，故兩個共邊或共頂點的四邊形其交集非空。）",
    "options": [],
    "answer": "See solution",
    "solution": "我們稱恰有三個頂點同色的四邊形為*好四邊形*。我們只需證明以下引理，原題便立即得證：\n\n**引理：** 若一個凸 $(4k+1)$ 邊形的頂點被塗成黑白兩色，使得每個顏色都至少有 $k$ 個頂點，則我們可以從中找到 $k$ 個兩兩互斥的好四邊形。\n\n**證明：** 我們用歸納法證明。當 $k=1$ 時，由鴿籠原理知必存在一色有三個頂點，故原命題成立。接著當 $k \\ge 2$ 時，令 $b$ 和 $w$ 分別為黑白兩色的頂點數，從而有 $b, w \\ge k$ 且 $b+w = 4k+1$。不失一般性假設 $w \\ge b$，從而 $k \\le b \\le 2k$ 且 $2k+1 \\le w \\le 3k+1$。\n\n我們的目標是找到連續四個點，其中恰有三個是白的。將頂點依順時鐘方向標記為 $V_1, \\dots, V_{4k+1}$，並不失一般性假設 $V_{4k+1}$ 是黑的。考慮每四個一組的配對：\n\n$$(V_1, V_2, V_3, V_4), (V_2, V_3, V_4, V_5), \\dots, (V_{4k-3}, V_{4k-2}, V_{4k-1}, V_{4k})$$\n\n注意到以上頂點共包含 $w$ 個白點與 $b-1$ 個黑點。由於 $w > b-1$，必然存在 $(V_i, V_{i+1}, V_{i+2}, V_{i+3})$ 其中白點比黑點多：\n\n- 如果這四點中恰有三點是白的，則構成好四邊形。\n- 否則，這四點必然都是白的。此時，令 $j$ 為 $V_{i+4}, \\dots, V_{4k+1}$ 中出現的第一個黑點，則 $V_{j-3}, V_{j-2}, V_{j-1}$ 都是白的。\n\n無論如何，我們都會找到四個點 $(V_i, V_{i+1}, V_{i+2}, V_{i+3})$ 構成好四邊形。移除這四個點後，剩餘的點構成一個 $(4k-3)$ 邊形，且其包含 $b-1 \\ge k-1$ 個黑點與 $w-3 \\ge (2k+1)-3 > k-1$ 個白點，故由歸納假設知可以從中找到 $k-1$ 個互斥的好四邊形，且這些四邊形都與 $(V_i, V_{i+1}, V_{i+2}, V_{i+3})$ 互斥。得證！ $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18923,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a$, $b$, $c$, $d$ such that $a < b < c < d$, is it possible for the least common multiple of $a$ and $b$ to be greater than the least common multiple of $c$ and $d$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Yes, it is possible.\n\n**Solution:** Let $a = 8$, $b = 9$, so $\\mathrm{lcm}(a, b) = 72$. Choose $c = 10$, $d = 25$, then $\\mathrm{lcm}(c, d) = 50$. Thus, $\\mathrm{lcm}(a, b) > \\mathrm{lcm}(c, d)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18924,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist positive integers $x$ and $y$ such that\n\n$$\n11x^{5} + 33y = 13y^{5} + 31x + 2024?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to\n\n$$\n(11x^{5} - 31x) + (33y - 13y^{5}) = 2024.\n$$\n\nThe fifth power of any positive integer ends with the same digit as the number itself. Therefore, $11x^5$ ends with the same digit as $11x$, which in turn ends with the same digit as $x$. Since $31x$ also ends with the same digit, the difference $11x^5 - 31x$ ends with zero. Similarly, the difference $33y - 13y^5$ ends with zero. However, the sum of numbers ending in zero cannot end in four. Therefore, there are no integers that satisfy the given equation.\n\nNote: The differences $11x^5 - 31x$ and $33y - 13y^5$ can be either positive or negative, but this does not affect the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18925,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Suppose that $AD$ and $BE$ are its angle bisectors. Prove that $\\angle ACB = 60^\\circ$ if and only if $AE + BD = AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $I$ the incenter of the triangle $\\triangle ABC$ (see figure below). Also denote the point $D_1$, which is the symmetric reflection of $D$ with respect to $BE$. Then $DB = BD_1$ and $DI = ID_1$. Without any dependence on the given conditions, we have $D_1 \\in AB$, since in $\\triangle DBD_1$ the line $BE$ contains the altitude, so it is the bisector of $\\angle DBD_1$.\n\n![](images/Ukraine_2016_Booklet_p5_data_71eb1100c1.png)\n\nIf $\\angle ACB = 60^\\circ$, we get $\\angle AIB = 90^\\circ + \\frac{1}{2} \\angle ACB = 120^\\circ$. Then $\\angle DIB = 60^\\circ = \\angle D_1IB$, because $\\triangle DBD_1$ is isosceles. Hence, $\\angle EIA = 60^\\circ = \\angle AIB - \\angle D_1IB$. Then $\\triangle AIE \\cong \\triangle AID_1$, so $EA = AD_1$ and $AE + BD = AB$.\n\nConversely, if $AE + BD = AB$, then $EA = AD_1$, so $\\triangle AIE \\cong \\triangle AID_1$. Then $\\angle EIA = \\angle AID_1$. Since $\\triangle DBD_1$ is isosceles, we have $\\angle DIB = \\angle D_1IB$. Then $\\angle EIA + \\angle DIB = \\angle AIB = \\angle EID$, so $120^\\circ = \\angle AIB = 90^\\circ + \\frac{1}{2} \\angle ACB \\Rightarrow \\angle ACB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18926,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_n$ be positive real numbers such that $x_1 x_2 \\cdots x_n = 1$. Prove that\n\n$$\n\\sum_{i=1}^{n} \\frac{x_i^8}{(x_i^4 + x_{i+1}^4)x_{i+1}} \\geq \\frac{n}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following lemma:\n\n*Lemma.* For any positive $a$ and $b$, the following inequality holds:\n\n$$\n(a^3 + b^3)^2 \\geq 2ab(a^4 + b^4).\n$$\n\nIndeed,\n\n$$\n(a^3 + b^3)^2 - 2ab(a^4 + b^4) = (a-b)^2(a^4 - a^2b^2 + b^4) \\geq 0.\n$$\n\nNow, using the lemma, we have\n\n$$\n\\sum_{i=1}^{n} \\frac{x_i^8}{(x_i^4 + x_{i+1}^4)x_{i+1}} = \\sum_{i=1}^{n} \\frac{x_i^9}{(x_i^4 + x_{i+1}^4)x_i x_{i+1}} \\geq \\sum_{i=1}^{n} \\frac{2x_i^9}{(x_i^3 + x_{i+1}^3)^2}.\n$$\n\nNext,\n\n$$\n2 \\sum_{i=1}^{n} \\frac{(x_i^3)^3}{(x_i^3 + x_{i+1}^3)^2} \\geq 2 \\frac{\\left(\\sum_{i=1}^{n} x_i^3\\right)^3}{\\left(2 \\sum_{i=1}^{n} x_i^3\\right)^2} = \\frac{1}{2} \\sum_{i=1}^{n} x_i^3.\n$$\n\nBy the AM-GM inequality,\n\n$$\n\\sum_{i=1}^{n} x_i^3 \\geq n \\left(\\sqrt[n]{x_1 \\cdots x_n}\\right)^3 = n.\n$$\n\nTherefore,\n\n$$\n\\sum_{i=1}^{n} \\frac{x_i^8}{(x_i^4 + x_{i+1}^4)x_{i+1}} \\geq \\frac{n}{2},\n$$\n\nas was to be proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18927,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $N = \\frac{1}{n(n+1)}$ is a finite decimal fraction.",
    "options": [],
    "answer": "See solution",
    "solution": "The number $N$ is a finite decimal fraction if and only if its denominator is of the form $2^a \\cdot 5^b$, where $a, b \\in \\mathbb{N}$.\n\nSince $n$ and $n+1$ are coprime, the possible cases are:\n\n1. $n = 1$, $n+1 = 2^a \\cdot 5^b$ (obvious conclusion).\n2. $n = 5^b$, $n+1 = 2^a$.\n3. $n = 2^a$, $n+1 = 5^b$.\n\nFor case 2: $2^a = 5^b + 1$. The only solution is $a = 1$, $b = 0$, $n = 1$.\n\nFor case 3: $5^b = 2^a + 1$. The only solution is $b = 1$, $a = 2$, $n = 4$.\n\nTherefore, the sought numbers are $n = 1$ and $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18928,
    "subject": "Mathematics (Olympiad)",
    "question": "Each vertex $v$ and each edge $e$ of a graph $G$ are assigned numbers $f(v) \\in \\{1, 2\\}$ and $f(e) \\in \\{1, 2, 3\\}$, respectively. Let $S(v)$ be the sum of numbers assigned to the edges incident to $v$ plus the number $f(v)$. We say that assignment $f$ is *cool* if $S(u) \\neq S(v)$ for every pair of adjacent vertices in $G$. Prove that every graph has a cool assignment.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $v_1, v_2, \\dots, v_n$ be any ordering of the vertices of $G$. Initially, assign each vertex the number $1$, and each edge the number $2$. Imagine that there is a chip lying on each vertex, while two chips are lying on each edge. We refine this assignment to obtain a cool one by performing the following greedy procedure:\n\nIn the $i$th step, let $x_1, x_2, \\dots, x_k$ be all backward neighbors of $v_i$, and let $e_j = v_i x_j$ (for $j = 1, 2, \\dots, k$) be the corresponding backward edges. For each edge $e_j$ we have two possibilities:\n\n1. If there is only one chip on $x_j$, we may move one chip from $e_j$ to $x_j$ or do nothing.\n2. If there are two chips on $x_j$, we may move one chip from $x_j$ to $e_j$ or do nothing.\n\nNotice that none of the sums $S(x_j)$ change as a result of such actions. Also, any action on each edge may change the total sum for $v_i$ by at most one. Hence, there are $k+1$ possible values for $S(v_i)$. So, at least one combination of chips gives a sum which is different from each $S(x_j)$. We fix this combination and proceed to the next step. The proof is complete.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18929,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The case $r = 2$ is clear: any cycle of even length works.\n\nFor $r \\ge 3$, define a sequence of graphs $G_r$ as follows:\n\n- Let $G_3$ be a cycle of 7 edges (any larger odd number would also work).\n- Suppose $G_r$ is defined with $n_r$ vertices. To construct $G_{r+1}$:\n    - Take $\\displaystyle \\binom{r n_r - r + 1}{n_r}$ disjoint copies of $G_r$.\n    - Add $r n_r - r + 1$ extra vertices.\n    - Set up a one-to-one correspondence between the copies of $G_r$ and the $n_r$-element subsets of the extra vertices.\n    - For each copy of $G_r$, join it to the members of its corresponding $n_r$-element subset by $n_r$ disjoint new edges (no two share an endpoint).\n\nThe resulting graph is $G_{r+1}$. This construction ensures that no $G_r$ has a cycle of less than 6 edges.\n\nClearly, $G_3$ is 3-chromatic. If $r \\ge 3$ and $G_{r+1}$ can be colored with $r$ or fewer colors, then some $n_r$ of the extra vertices in $G_{r+1}$ must share the same color. The corresponding copy of $G_r$ would then be colored with $r-1$ or fewer colors, which by induction is impossible. Thus, no $G_r$ can be colored with fewer than $r$ colors.\n\nThis does not prove that $G_r$ is exactly $r$-chromatic, but if it is not, deleting some monochromatic classes of vertices and their incident edges yields such a graph.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18930,
    "subject": "Mathematics (Olympiad)",
    "question": "During interplanetary battles, skilled astronomers of the Earth discovered the planet Hot Dog in the Sandwich Galaxy. This planet has the shape of a convex $n$-hedron. By analyzing the spectrum of the waves emitted by this planet, scientists discovered some interesting facts about the strange way of life there. Each face of this planet is a country, each country has its own currency, and along the border between two countries, there is a *constant conversion factor* to change currencies.\n\nAll people passing through a border between two countries must change all their money into the currency of their destination, and there is no other way to change currencies. Astronomers observed that it's possible for a passenger to travel to different countries and, without spending any money, return home with an amount of money different from what they started with.\n\n% IMAGE: ![](images/2014_p18_data_dd25b4bb55.png)\n\nDuring a research project, a group of travelers was discovered who all started from the same country and owned the same amount of money. Each of them traveled a path in the form of a closed broken line and returned home! At most, how many of these travelers can have mutually different amounts of money when they are back home?\n\n*Note 1:* The only parameter is the number of countries, $n$. All other things like the conversion factors and the arrangement of the countries are variables, so your answer must be only in terms of $n$.\n\n*Note 2:* None of the travelers spend money during the journey!",
    "options": [],
    "answer": "See solution",
    "solution": "The key is that the amount of money a traveler has after completing a closed path depends on the product of the conversion factors along that path. If the traveler crosses a border from country $i$ to $j$, the conversion factor is $r_{ij}$, and the reverse is $1/r_{ij}$. For a closed path, the total conversion is the product of the conversion factors along the path.\n\nIf the path is contractible (can be shrunk to a point), the product is $1$. But for non-contractible loops, the product can differ. The set of possible different amounts corresponds to the number of independent cycles in the dual graph of the polyhedron, which is the first homology group of the surface.\n\nFor a convex $n$-hedron (a polyhedron with $n$ faces), the dual graph is a planar connected graph with $n$ vertices. The number of independent cycles is $n - 1 - (n - 2) = 1$, but more generally, the number of independent cycles is $n - 1$ (since the dual graph is a tree plus $n - 1$ edges). However, for a convex polyhedron, the maximum number of mutually different products (i.e., different final amounts) is $n - 1$.\n\n**Answer:**\n\n$$\boxed{n - 1}$$\n\nAt most $n - 1$ travelers can have mutually different amounts of money upon returning home.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18931,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be an operator $*$. Given an expression that includes this operator, one can make the following transformations:\n\n1. An expression of the form $x * (y * z)$ can be rewritten as $((1 * x) * y) * z$.\n\n2. An expression of the form $x * 1$ can be rewritten as $x$.\n\nThe transformations may be performed only on the entire expression and not on the subexpressions. For example, $(1 * 1) * (1 * 1)$ may only be rewritten using the first kind of transformation as $((1 * (1 * 1)) * 1) * 1$, but it cannot be transformed into $1 * (1 * 1)$ or $(1 * 1) * 1$ using a single step—in the latter two cases the second kind of transformation would have been applied just to the left or right subexpression of the form $1 * 1$.\n\nFor which natural numbers $n$ can the expression\n\n$$\n\\underbrace{1 * (1 * (1 * (\\cdots * (1 * 1))))}_{n\\ \\text{ones}}\n$$\n\nbe rewritten to an expression that does not include a single occurrence of the $*$ operator?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 1, 2, 3, 4$.\n\nLet's analyze the transformation process in reverse. The final result can only be $1$, which can only be obtained from the expression $1 * 1$ using the second transformation. Any intermediate expression of the form $1 * x$ can only be obtained from $(1 * x) * 1$ using the second transformation, and any intermediate result of the form $(1 * x) * y$ can only be obtained from $((1 * x) * y) * 1$ using the second transformation. Any intermediate result of the form $((1 * x) * y) * z$ can only be obtained from $x * (y * z)$ using the first transformation, because if it was obtained from $(((1 * x) * y) * z) * 1$ using the second transformation, this would require a longer expression, which cannot be represented in the form $1 * (1 * (1 * (\\cdots * (1 * 1))))$.\n\nTherefore, the end result uniquely determines all the previous expressions:\n\n$$\n\\begin{align*}\n1 &\\overset{(2)}{\\rightleftharpoons} 1 * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} (1 * 1) * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * 1) * 1) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} 1 * (1 * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} (1 * (1 * 1)) * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * (1 * 1)) * 1) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} (1 * 1) * (1 * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * 1) * (1 * 1)) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} 1 * ((1 * 1) * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} (1 * ((1 * 1) * 1)) * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * ((1 * 1) * 1)) * 1) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} ((1 * 1) * 1) * (1 * 1) \\\\\n  &\\overset{(1)}{\\rightleftharpoons} 1 * (1 * (1 * 1)) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} (1 * (1 * (1 * 1))) * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * (1 * (1 * 1)))) * 1 * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} (1 * (1 * 1)) * (1 * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * (1 * 1)) * (1 * 1)) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} (1 * 1) * ((1 * 1) * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * 1) * ((1 * 1) * 1)) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} 1 * (((1 * 1) * 1) * 1) \\\\\n  &\\overset{(2)}{\\rightleftharpoons} (1 * (((1 * 1) * 1) * 1)) * 1 \\\\\n  &\\overset{(2)}{\\rightleftharpoons} ((1 * (((1 * 1) * 1) * 1)) * 1) * 1 \\\\\n  &\\overset{(1)}{\\rightleftharpoons} (((1 * 1) * 1) * 1) * (1 * 1)\n\\end{align*}\n$$\n\nHowever, the expression $(((1 * 1) * 1) * 1) * (1 * 1)$ cannot be an intermediate result based on the earlier reasoning. Therefore, all expressions that can be transformed into $1$ are shown in the chain above. Only four of them are in the required form—for $n = 1, 2, 3, 4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18932,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube contains $n^3$ unit cubes. A line which passes through the centers of $n$ unit cubes is called \"interesting\". Does there exist $n > 1$ such that the number of \"interesting\" lines is a power of $2$?",
    "options": [],
    "answer": "See solution",
    "solution": "For a cube with side $n$, the total number of \"interesting\" lines is $3n^2 + 6n + 4$. This can be computed by observing that each \"interesting\" line can have one of three directions: parallel to an edge, parallel to the diagonal of some face, or along the main diagonal.\n\nWe want to prove that the equation $3n^2 + 6n + 4 = 2^l$ has no solutions. Let $m = n + 1$, so the equation becomes $3m^2 + 1 = 2^l$. Consider this equation modulo $8$. For $l \\geq 3$, the right-hand side is divisible by $8$, but the left-hand side can only give residues $1$, $4$, or $5$. Therefore, we only need to consider $l = 1$ and $l = 2$, which is trivial. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18933,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the number of ways to assign directions to every lattice segment contained in the region $0 \\le x, y, z \\le n$ so that all the lattice squares in the region become congruent to a given type of lattice square. Consider the following cases:\n\n(a) Type (A): Is it possible to assign directions so that all lattice squares are congruent to type (A)?\n\n(b) Type (B): How many ways are there to assign directions so that all lattice squares are congruent to type (B)?\n\n(c) Type (C): How many ways are there to assign directions so that all lattice squares are congruent to type (C)?\n\n(d) Type (D): How many ways are there to assign directions so that all lattice squares are congruent to type (D)?\n\nFind the total number of possible assignments for all cases above.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Assigning directions so that all lattice squares are congruent to type (A) is impossible. Any assignment leads to a contradiction, so there are $0$ ways.\n\n(b) For type (B), once the direction of one side of a square is chosen, the directions of the other three sides are uniquely determined. There are exactly $2$ ways to assign directions to all lattice segments so that all lattice squares are congruent to type (B).\n\n(c) For type (C), the condition is that parallel sides of a square have the same direction. Directions for lattice segments parallel to each axis and sharing the same coordinate can be assigned independently. There are $2^{3n}$ ways to assign directions in this case.\n\n(d) For type (D), the condition is that, with respect to the perpendicular axis through the center of a square, the difference between the number of sides assigned counter-clockwise and clockwise is either $-2$ or $2$. By Lemma 1, once the directions of three sides of a square are specified, the fourth is uniquely determined. By Lemma 2, if five faces of a cube are congruent to type (D), the sixth must be as well. Thus, there are $2^{(n+1)^3-1}$ ways to assign directions in this case.\n\nSumming the numbers for all cases, the total number of possible assignments is:\n\n$$2^{(n+1)^3-1} + 2^{3n} + 2$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18934,
    "subject": "Mathematics (Olympiad)",
    "question": "There are four numbers on the board: $1$, $3$, $6$, and $10$. Each time, we can erase any two numbers $a, b$ written on the board and write the numbers $a+b$ and $ab$ instead. Can we obtain the four numbers $2015$, $2016$, $2017$, and $2018$ after several moves?",
    "options": [],
    "answer": "See solution",
    "solution": "That is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18935,
    "subject": "Mathematics (Olympiad)",
    "question": "Hemos empezado la Olimpiada Matemática puntualmente a las 9:00, como he comprobado en mi reloj, que funcionaba en ese momento correctamente. Cuando he terminado, a las 13:00, he vuelto a mirar el reloj y he visto que las manecillas se habían desprendido de su eje pero manteniendo la posición en la que estaban cuando el reloj funcionaba. Curiosamente, las manecillas de las horas y de los minutos aparecían superpuestas exactamente, una sobre otra, formando un ángulo (no nulo) menor que $120^\\circ$ con el segundero. ¿A qué hora se me averió el reloj? (Dar la respuesta en horas, minutos y segundos con un error máximo de un segundo; se supone que, cuando funcionaba, las manecillas del reloj avanzaban de forma continua.)",
    "options": [],
    "answer": "See solution",
    "solution": "Si medimos el tiempo $t$ en segundos a partir de las 00:00 y los ángulos en grados, en sentido horario y a partir de la posición de las manecillas a las 00:00, tenemos que el ángulo barrido por la manecilla de las horas en el instante $t$ es $\\alpha_{hor}(t) = t/120$ y el barrido por el minutero, $\\alpha_{min}(t) = t/10$. Como ambas manecillas han aparecido superpuestas, los dos ángulos han de coincidir en el momento $t$ en que el reloj se ha averiado. El minutero ha podido dar alguna vuelta completa, por tanto debe tenerse\n\n$$\n\\frac{t}{10} = \\frac{t}{120} + 360k,\n$$\n\ndonde $k \\ge 0$ es un número entero, es decir, $t = \\frac{360 \\times 120}{11}k$. Como la avería ha sido entre las 9:00 y las 13:00, tiene que ser $9 \\leq k \\leq 12$. El ángulo para el segundero es $\\alpha_{seg}(t) = 6t$, por tanto la diferencia\n\n$$\n6t - \\frac{t}{120} = \\frac{360 \\times 719}{11}k = (360 \\times 65 + \\frac{360 \\times 4}{11})k.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18936,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two real numbers $a, b$ such that $b - a^2 > 0$, describe all matrices $A \\in M_2(\\mathbb{R})$ such that\n$$\n\\det(A^2 - 2aA + bI_2) = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $c = \\sqrt{b - a^2}$, which is real by assumption. We have\n$$\nA^2 - 2aA + bI_2 = (A - (a + ic)I_2)(A - (a - ic)I_2),\n$$\nso the characteristic polynomial of $A$ has $a + ic$ or $a - ic$ as a root. Since the polynomial has real coefficients, both are roots, so $A$ has eigenvalues $a \\pm ic$. Thus, the minimal polynomial is $x^2 - 2a x + b$, and $A^2 - 2aA + bI_2 = 0$. Therefore, $\\operatorname{tr}(A) = 2a$ and $\\det A = b$.\n\nIn conclusion, the matrices with the given property are\n$$\n\\begin{pmatrix}\na + x & y \\\\\n\\frac{a^2 - x^2 - b}{y} & a - x\n\\end{pmatrix},\n$$\nwhere $x \\in \\mathbb{R}$ and $y \\in \\mathbb{R}^* = \\mathbb{R} \\setminus \\{0\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18937,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a set $S$ of positive integers, each of which has exactly 100 decimal digits. A number in $S$ is called *bad* if it is not divisible by the sum of two (not necessarily distinct) numbers in $S$. Determine the maximum number of elements $S$ may have, under the assumption that it contains at most 10 bad numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $\\binom{19}{9} - 1$.\n\nLet $T$ be the set of bad numbers in $S$. The key idea is that each element $s$ of $S$ is a $\\mathbb{Z}_+$-linear combination of elements in $T$, with coefficients determining lattice points in a standard simplex, so they are easy to count: $s = \\sum_{t \\in T} \\alpha_{st} t$, where the $\\alpha_{st}$ are nonnegative integers such that $1 \\leq \\sum_{t \\in T} \\alpha_{st} \\leq 9$.\n\nWe prove this by induction on $S$ with the natural order. For any bad number $s \\in T$, it suffices to take $\\alpha_{ss} = 1$ and $\\alpha_{st} = 0$ for $t \\neq s$. In particular, this shows the assertion is true for $\\min S$, which is clearly bad.\n\nNow let $s \\in S \\setminus T$ and assume the assertion is true for all $r$ in $S$ with $r < s$. Since $s$ is not bad, it is divisible by the sum of two (not necessarily distinct) elements of $S$: $s = p(q + r)$, where $p$ is a positive integer and $q, r \\in S$. Clearly, $q < s$ and $r < s$, so the induction hypothesis applies, and $\\alpha_{st} = p(\\alpha_{qt} + \\alpha_{rt})$ would work if we show $\\sum_{t \\in T} \\alpha_{st} \\leq 9$ (the sum is obviously at least 1). Since each element of $S$ has exactly 100 decimal digits, $10^{100} > s = \\sum_{t \\in T} \\alpha_{st} t \\geq 10^{99} \\sum_{t \\in T} \\alpha_{st}$, so $\\sum_{t \\in T} \\alpha_{st} < 10$, i.e., at most 9.\n\nNext, we show $|S| \\leq \\binom{19}{9} - 1$. Assign each $s \\in S$ the $|T|$-tuple $(\\alpha_{st})_{t \\in T}$. This assignment is one-to-one, so $|S| \\leq |U|$, where $U$ is the set of all $|T|$-tuples $(\\alpha_t)_{t \\in T}$ of nonnegative integers such that $1 \\leq \\sum_{t \\in T} \\alpha_t \\leq 9$. Notice that $|U|$ is one less than the number of $(|T|+1)$-tuples of nonnegative integers which add up to 9. The latter is $\\binom{|T|+9}{9}$, so $|S| \\leq \\binom{|T|+9}{9} - 1 \\leq \\binom{19}{9} - 1$ for $|T| \\leq 10$.\n\nTo construct such a set, let $t_i = 10^{99} + 10^{i-1}$ for $i = 1, 2, \\dots, 10$—these are the bad numbers. The sums $\\alpha_1 t_1 + \\alpha_2 t_2 + \\dots + \\alpha_{10} t_{10}$, where the $\\alpha_i$ are nonnegative integers with $1 \\leq \\sum \\alpha_i \\leq 9$, are pairwise distinct and form a set of $\\binom{19}{9} - 1$ positive integers satisfying the conditions, with the ten $t_i$ as the bad elements.\n\n**Remark.** If we required the two elements in the definition of a bad number to be distinct, the same argument works with minor changes. In this case, at least two of the $\\alpha_{st}$ are nonzero unless $s$ is bad, and the maximum is $\\binom{19}{9} - 1 - 8 \\cdot 10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18938,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = k^2 - k + 1$ for some positive integer $k$.\n\n**a)** Show that $n = 21$ satisfies this form for $k = 5$, and construct a sequence $(i_1, i_2, i_3, i_4, i_5)$ as in the proof, for example $(1, 11, 13, 18, 19)$. Then, set $a_1 = a_2 = \\dots = a_{21} = 1$, choose 20 different primes $p_1 > p_2 > \\dots > p_{20}$, and for each $i$ from 1 to 20, multiply $a_i, a_{i+10}, a_{i+12}, a_{i+17}, a_{i+18}$ by $p_i$. Show that for each $\\ell$ from 1 to 20, the set of greatest common divisors from the problem coincides with $\\{p_1, p_2, \\dots, p_{20}\\}$.\n\n**b)** Determine whether the equation $k^2 - k + 1 = 2021$ has a positive integer solution for $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the largest of the greatest common divisors found. The equality $\\gcd(a_i, a_{i+\\ell}) = d$ is equivalent to both $a_i$ and $a_i + \\ell$ being multiples of $d$. Let $(i_1, i_2, \\dots, i_k)$ be all indices of numbers in the sequence that are multiples of $d$, in ascending order. Each pair $i_j < i_s$ yields $d$ twice: for $\\ell = s - j$ and $\\ell = j + n - s$. Thus, $d$ occurs as the greatest common divisor exactly $2\\left(\\frac{k}{2}\\right)$ times, and by assumption this number is $n-1$. Therefore, $n = 1 + 2\\left(\\frac{k}{2}\\right) = k^2 - k + 1$. The assertion is proved.\n\n**b)** The equation $k^2 - k + 1 = 2021$ is equivalent to $k(k-1) = 2020$. Since $k(k-1)$ increases for $k > 0.5$ and $44 \\cdot 45 = 1980$, $45 \\cdot 46 = 2070$, there is no positive integer $k$ such that $k(k-1) = 2020$. Thus, the answer is \"no\".\n\n**a)** For $n = 21$, $21 = 5^2 - 5 + 1$, so $k = 5$. The sequence $(1, 11, 13, 18, 19)$ can be used. Set $a_1 = a_2 = \\dots = a_{21} = 1$, choose 20 different primes $p_1 > p_2 > \\dots > p_{20}$, and for each $i$ from 1 to 20, multiply $a_i, a_{i+10}, a_{i+12}, a_{i+17}, a_{i+18}$ by $p_i$. For each $\\ell$ from 1 to 20, the set of greatest common divisors matches $\\{p_1, p_2, \\dots, p_{20}\\}$.\n\n**Note:** The statement implies $n = k^2 - k + 1$ for some positive integer $k$, and there must be a sequence $(i_1, \\dots, i_k)$ such that all pairwise differences $i_j - i_s$ give all possible nonzero remainders modulo $n$. The construction in part (a) shows these conditions are sufficient.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18939,
    "subject": "Mathematics (Olympiad)",
    "question": "Две отсечки $\\overline{AB}$ и $\\overline{CD}$ со еднакви должини лежат на иста права, така што $\\frac{1}{4}$ од нивните должини им е заедничка. Определи ја должината на тие отсечки ако растојанието меѓу нивните средни точки е 6 cm.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека со $x$ ја означиме должината на заедничкиот дел на отсечките. Тогаш, бидејќи $\\frac{1}{4}$ од должината на секоја отсечка е заедничка, целата должина на една отсечка е $4x$.\n\nСредишните точки на отсечките се оддалечени за $6$ cm. Останатиот дел од секоја отсечка (освен заедничкиот дел) е $\\frac{3}{4}$ од должината, односно $3x$. Значи, растојанието меѓу средишните точки е $3x$.\n\n$$\n3x = 6 \\implies x = 2 \\text{ cm}\n$$\n\nЗначи, должината на секоја отсечка е:\n\n$$\n4x = 4 \\cdot 2 = 8 \\text{ cm}\n$$\n\nЗатоа, должината на отсечките $\\overline{AB}$ и $\\overline{CD}$ е $8$ cm.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18940,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive real number $a$, determine the minimum value of the expression\n\n$$\n\\left( \\int_{0}^{1} f(x) \\, dx \\right)^{2} - (a + 1) \\int_{0}^{1} x^{2a} f(x) \\, dx\n$$\n\nas $f$ runs through all concave functions $f : [0, 1] \\to \\mathbb{R}$ with $f(0) = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum is $\\frac{2a - 1}{8a + 4}$, achieved for $f(x) = 1 - x$.\n\nFix a concave function $f : [0, 1] \\to \\mathbb{R}$ with $f(0) = 1$. For all $x \\in [0, 1]$,\n\n$$\nx^a f(x) + 1 - x^a = x^a f(x) + (1 - x^a) f(0) \\leq f(x^a x + (1 - x^a) \\cdot 0) = f(x^{a+1})\n$$\n\nMultiply both sides by $(a+1)x^a$ and integrate over $[0, 1]$:\n\n$$\n(a+1) \\int_{0}^{1} x^{2a} f(x) \\, dx + \\frac{a}{2a+1} \\leq \\int_{0}^{1} (a+1) x^a f(x^{a+1}) \\, dx = \\int_{0}^{1} f(x) \\, dx\n$$\n\nAlso,\n\n$$\n\\int_{0}^{1} f(x) \\, dx \\leq \\left( \\int_{0}^{1} f(x) \\, dx \\right)^2 + \\frac{1}{4}\n$$\n\nTherefore,\n\n$$\n\\left( \\int_{0}^{1} f(x) \\, dx \\right)^2 - (a+1) \\int_{0}^{1} x^{2a} f(x) \\, dx \\geq \\frac{a}{2a+1} - \\frac{1}{4} = \\frac{2a-1}{8a+4}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18941,
    "subject": "Mathematics (Olympiad)",
    "question": "The medians $AA'$, $BB'$, $CC'$ of triangle $ABC$ meet the nine-point circle at $D$, $E$, $F$, respectively. The points $L$, $M$, $N$ are the feet of the altitudes of $ABC$ ($L$ belongs to $AA'$, etc). The tangents to the nine-point circle at $D$, $E$, $F$ meet the lines $MN$, $LN$, and $LM$ at points $P$, $Q$, $R$. Show that points $P$, $Q$, and $R$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "1) Triangles $PMD$ and $PDN$ are similar (indeed, $\\widehat{PDM}$ is half inscribed, $\\widehat{DNP}$ is inscribed and both subtend the same arc in the Euler circle; moreover $\\hat{P}$ is the same in both triangles). Therefore\n$$\n\\frac{PM}{PD} = \\frac{DM}{DN} = \\frac{PD}{PN} \\implies \\frac{PM}{PN} = \\left(\\frac{DM}{DN}\\right)^2. \\quad (*)\n$$\nPoints $M$, $N$, and $B$ belong to a circle of center $A'$ and radius $a/2$. (Indeed, in triangle $A'BN$ we have $A'B = a/2$; $BN = a \\cos B$; and by the cosine law\n$$\nA'N^2 = A'B^2 + BN^2 - 2 A'B \\cdot BN \\cdot \\cos B = \\frac{a^2}{4}.\n$$\nIn triangle $A'MC$ we repeat the argument and have $A'M = a/2$).\n\nThen, since $MA' = NA' = a/2$, the angle $\\angle MDN$ has an angle bisector $DA'$, and so, if we call $T = MN \\cap AA'$, it follows that\n$$\n\\frac{DM}{DN} = \\frac{TM}{TN}.\n$$\n![](images/Spanija_b_2014_p33_data_61f196dd03.png)\nBut $MN$ is antiparallel to $BC$ with respect to $AB$ and $AC$, and so\n$$\n\\frac{TM}{TN} = \\frac{AM^2}{AN^2} = \\frac{c^2}{b^2}. \\qquad (**)\n$$\nCombining these results, we have $\\frac{PM}{PN} = \\frac{c^4}{b^4}$; in a similar way we get\n$$\n\\frac{QN}{QL} = \\frac{a^4}{c^4} \\quad \\text{and} \\quad \\frac{RL}{RM} = \\frac{b^4}{a^4},\n$$\ntherefore\n$$\n\\frac{PM \\cdot QN \\cdot RL}{PN \\cdot QL \\cdot RM} = 1.\n$$\nBy the converse of Menelaus' theorem, the points $P$, $Q$, and $R$ are collinear.\n![](images/Spanija_b_2014_p33_data_0bc9996c88.png)\n![](images/Spanija_b_2014_p33_data_14194ec819.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18942,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$ from the set of all positive integers to the same set such that, for all positive integers $a_1, \\dots, a_k$ with $k > 0$, the sum $a_1 + \\dots + a_k$ divides the sum $f(a_1) + \\dots + f(a_k)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** All functions given by $f(n) = an$, $a \\in \\mathbb{N}$.\n\n**Solution:** Suppose that $f$ is a function that satisfies the conditions of the problem. We claim that $f(n) = f(n-1) + f(1)$ for all integers $n > 1$. Indeed, for any integer $m > n$, we have $m \\mid f(n) + f(m-n)$ and $m \\mid f(n-1) + f(1) + f(m-n)$ by conditions of the problem. Hence the difference $f(n) - (f(n-1) + f(1))$ is also divisible by $m$. As $m$ was arbitrary, this implies that $f(n) - (f(n-1) + f(1))$ is divisible by an infinite number of different integers, i.e., is equal to $0$. This completes the proof of the claim.\n\nEasy induction now gives that necessarily $f(n) = n f(1)$. It remains to verify that all functions of the form $f(n) = a n$ satisfy the conditions of the problem, which is straightforward.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18943,
    "subject": "Mathematics (Olympiad)",
    "question": "a) How many possible clown costumes are there if there are 4 hats, 5 shirts, 3 pairs of pants, and 2 noses?\n\nb) How many different clown costumes can James choose if he must leave out exactly one item: either the shirt, the nose, or the hat?\n\nc) The number of possible alien costumes is 60, which equals the number of antennas times the number of masks times 10. What are the possible numbers of antennas and masks?\n\nd) An elf costume has five different items. If there are 792 possible elf costumes, what is the maximum possible number of hats?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The number of possible clown costumes is $4 \\times 5 \\times 3 \\times 2 = 120$.\n\nb) Without a shirt, the number of different costumes is $3 \\times 4 \\times 5 = 60$.\nWithout a nose, the number of different costumes is $3 \\times 4 \\times 2 = 24$.\nWithout a hat, the number of different costumes is $3 \\times 5 \\times 2 = 30$.\nSo the number of costumes James can choose is $60 + 24 + 30 = 114$.\n\nc) The number of possible alien costumes is $60 = \\text{number of antennas} \\times \\text{number of masks} \\times 10$. Therefore, $\\text{number of antennas} \\times \\text{number of masks} = 6$. So the number of antennas and number of masks are, respectively, 1 and 6, 2 and 3, 3 and 2, or 6 and 1.\n\nd) First, use division to find all the prime factors of 792. Thus $792 = 2 \\times 2 \\times 2 \\times 3 \\times 3 \\times 11$. An elf costume has five different items. So we need to express 792 as a product of 5 factors so that each is at least two and one factor is as large as possible. Thus we get $792 = 2 \\times 2 \\times 2 \\times 3 \\times 33$. So the maximum number of hats is 33.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18944,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\theta$ be an angle in the interval $(0, \\pi/2)$. Given that $\\cos \\theta$ is irrational, and that $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are both rational for some positive integer $k$, show that $\\theta = \\pi/6$.\n\n*Lemma 1*: For every positive integer $n$, there is a monic polynomial (that is, a polynomial with leading coefficient $1$) $S_n(x)$ with integer coefficients such that $S_n(2\\cos \\alpha) = 2\\cos n\\alpha$.\n\n*Proof*: We induct on $n$. The base cases $n=1$ and $n=2$ are trivial by taking $S_1(x) = x$ and $S_2(x) = x^2 - 2$. Assume the statement is true for $n \\le m$. Note that by the addition-to-product formulas, $2\\cos[(m+1)\\alpha] + 2\\cos[(m-1)\\alpha] = 4\\cos m\\alpha \\cos \\alpha$. Thus $S_{m+1}(x) = xS_m(x) - S_{m-1}(x)$ satisfies the conditions of the problem, completing the induction. $\\blacksquare$\n\n*Lemma 2*: If $\\cos \\alpha$ is rational and $\\alpha = r\\pi$ for some rational number $r$, then the possible values of $\\cos \\alpha$ are $0, \\pm 1, \\pm \\frac{1}{2}$.\n\n*Proof*: Since $r$ is rational, there exists positive integer $n$ such that $rn$ is an even integer. By lemma 1, $S_n(2\\cos \\alpha) = 2\\cos(n\\alpha) = 2\\cos(rn\\pi) = 1$; that is, $2\\cos \\alpha$ is a rational root of the monic polynomial $S_n(x)$ with integer coefficients. By Gauss' lemma, $2\\cos \\alpha$ must take integer values. Since $-1 \\le \\cos \\alpha \\le 1$, the possible values of $2\\cos \\alpha$ are $0, \\pm 1, \\pm 2$. $\\blacksquare$",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:**\n\nIf $\\cos x$ is rational, then $\\cos nx$ is rational for every positive integer $n$. This follows by induction on $n$ using the product-to-sum formula:\n\n$$\n2 \\cos n\\theta \\cos \\theta = \\cos[(n+1)\\theta] + \\cos[(n-1)\\theta].\n$$\n\nThus both $\\cos(k^2\\theta) = \\cos[k(k\\theta)]$ and $\\cos[(k^2-1)\\theta] = \\cos[(k-1)(k+1)\\theta]$ are rational. By the addition and subtraction formulas:\n\n$$\n\\cos[(k+1)\\theta] = \\cos k\\theta \\cos \\theta - \\sin k\\theta \\sin \\theta,\n$$\n$$\n\\cos(k^2\\theta) = \\cos[(k^2-1)\\theta] \\cos \\theta - \\sin[(k^2-1)\\theta] \\sin \\theta.\n$$\n\nLet $r_1 = \\cos k\\theta$, $r_2 = \\cos[(k+1)\\theta]$, $r_3 = \\cos[(k^2-1)\\theta]$, $r_4 = \\cos(k^2\\theta)$, and $x = \\cos \\theta$. Then:\n\n$$\nr_2 = r_1 x \\pm \\sqrt{(1 - r_1^2)(1 - x^2)},\n$$\n$$\nr_4 = r_3 x \\pm \\sqrt{(1 - r_3^2)(1 - x^2)}.\n$$\n\nSo:\n\n$$\n\\pm\\sqrt{(1 - r_1^2)(1 - x^2)} = r_2 - r_1 x,\n$$\n$$\n\\pm\\sqrt{(1 - r_3^2)(1 - x^2)} = r_4 - r_3 x.\n$$\n\nSquaring and subtracting gives:\n\n$$\n2(r_1 r_2 - r_3 r_4)x = r_1^2 + r_2^2 - (r_3^2 + r_4^2).\n$$\n\nSince $r_1, r_2, r_3, r_4$ are rational and $x$ is irrational, we must have $r_1 r_2 - r_3 r_4 = 0$, or:\n\n$$\n\\cos k\\theta \\cos[(k+1)\\theta] = \\cos(k^2\\theta) \\cos[(k^2-1)\\theta].\n$$\n\nBy product-to-sum formulas:\n\n$$\n\\frac{\\cos[(2k + 1)\\theta] - \\cos \\theta}{2} = \\frac{\\cos[(2k^2 - 1)\\theta] - \\cos \\theta}{2},\n$$\n\nso $\\cos[(2k + 1)\\theta] - \\cos[(2k^2 - 1)\\theta] = 0$. By sum-to-product formulas:\n\n$$\n2 \\sin[(k - k^2 + 1)\\theta] \\sin[(k^2 + k)\\theta] = 0,\n$$\n\nimplying either $(k - k^2 + 1)\\theta$ or $(k^2 + k)\\theta$ is an integer multiple of $\\pi$. Since $k$ is an integer, $\\theta = r\\pi$ for some rational $r$.\n\nBy Lemma 2 for $\\alpha = k\\theta$ and $\\alpha = (k+1)\\theta$, the possible values of $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are $0, \\pm 1, \\pm \\frac{1}{2}$. Thus both $k\\theta$ and $(k+1)\\theta$ are integer multiples of $\\frac{\\pi}{6}$. Since $0 < \\theta = k\\theta - (k-1)\\theta < \\frac{\\pi}{2}$, the only possible values of $\\theta$ are $\\frac{\\pi}{3}$ and $\\frac{\\pi}{6}$. Since $\\cos \\theta$ is irrational, $\\theta = \\frac{\\pi}{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and\n\n$$\n\\left\\{\\frac{sm}{p}\\right\\} < \\left\\{\\frac{sn}{p}\\right\\} < \\frac{s}{p}\n$$\n\nif and only if $s$ is not a divisor of $p-1$.\n\n(For $x$ a real number, let $\\lfloor x \\rfloor$ denote the greatest integer less than or equal to $x$, and let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose that $s$ is a divisor of $p-1$; write $d = (p-1)/s$. As $x$ varies among $1, 2, \\dots, p-1$, $\\left\\{sx/p\\right\\}$ takes the values $1/p, 2/p, \\dots, (p-1)/p$ once each in some order. The possible values with $\\left\\{sx/p\\right\\} < s/p$ are precisely $1/p, \\dots, (s-1)/p$. From the fact that $\\left\\{sd/p\\right\\} = (p-1)/p$, we realize that the values $\\left\\{sx/p\\right\\} = (p-1)/p, (p-2)/p, \\dots, (p-s+1)/p$ occur for\n\n$$\nx = d, 2d, \\dots, (s-1)d\n$$\n\n(which are all between $0$ and $p$), and so the values $\\left\\{sx/p\\right\\} = 1/p, 2/p, \\dots, (s-1)/p$ occur for\n\n$$\nx = p - d, p - 2d, \\dots, p - (s-1)d,\n$$\n\nrespectively. From this it is clear that $m$ and $n$ cannot exist as requested.\n\nConversely, suppose that $s$ is not a divisor of $p-1$. Put $m = \\lfloor p/s \\rfloor$; then $m$ is the smallest positive integer such that $\\left\\{ms/p\\right\\} < s/p$, and in fact $\\left\\{ms/p\\right\\} = (ms - p)/p$. However, we cannot have $\\left\\{ms/p\\right\\} = (s-1)/p$ or else $(m-1)s = p-1$, contradicting our hypothesis that $s$ does not divide $p-1$. Hence the unique $n \\in \\{1, \\dots, p-1\\}$ for which $\\left\\{nx/p\\right\\} = (s-1)/p$ has the desired properties (since the fact that $\\left\\{nx/p\\right\\} < s/p$ forces $n \\ge m$, but $m \\ne n$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18946,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M = \\max\\left\\{\\frac{a_i}{i} : 1 \\le i \\le s\\right\\}$ and fix an index $\\ell \\le s$ such that $M = \\frac{a_\\ell}{\\ell}$. Prove that $a_n = a_{n-\\ell} + a_\\ell$ for all sufficiently large $n$, specifically for all $n \\ge s^2\\ell + 2s$.",
    "options": [],
    "answer": "See solution",
    "solution": "Choose a decomposition of $a_n$ satisfying (1), (2), (3), and (4) as in the lemma. Then $n = i_1 + i_2 + \\cdots + i_k \\le sk$, so $k \\ge \\frac{n}{s} \\ge s\\ell + 2$. Suppose none of $i_3, i_4, \\ldots, i_k$ equals $\\ell$. By the pigeonhole principle, there is an index $j$ ($3 \\le j \\le s$, $j \\ne \\ell$) such that $j$ appears at least $\\ell$ times among $i_3, \\ldots, i_k$. Delete these $\\ell$ occurrences of $j$ and replace them with $j$ occurrences of $\\ell$, obtaining $(i_1, i_2, i'_3, \\ldots, i'_{k'})$. This sequence also satisfies (2), (3), and (4). Furthermore,\n\n$$\na_{i_1} + a_{i_2} + a_{i_3} + \\cdots + a_{i_k} = a_n \\ge a_{i_1} + a_{i_2} + a_{i'_3} + \\cdots + a_{i'_{k'}}$$\n\nFrom the lemma, after canceling coinciding terms, we get $\\ell a_j \\ge j a_\\ell$, so $\\frac{a_\\ell}{\\ell} \\le \\frac{a_j}{j}$. By the definition of $\\ell$, this means $\\ell a_j = j a_\\ell$. Thus,\n\n$$a_n = a_{i_1} + a_{i_2} + a_{i'_3} + \\cdots + a_{i'_{k'}}$$\n\nRearrange so $i'_{k'} = \\ell$. Thus, we have a representation of $a_n$ with $i_k = \\ell$. The indices $(i_1, \\ldots, i_{k-1})$ satisfy conditions (2), (3), and (4) with $n$ replaced by $n - \\ell$. By the lemma, $a_{n-\\ell} + a_\\ell \\ge (a_{i_1} + \\cdots + a_{i_{k-1}}) + a_\\ell = a_n$. This implies $a_n = a_{n-\\ell} + a_\\ell$ for each $n \\ge s^2\\ell + 2s$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18947,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $3$, and let $a_1, a_2, \\dots, a_n$ be nonnegative real numbers with $a_1 + a_2 + \\dots + a_n = 2$.\n\nDetermine the minimum value of\n\n$$\n\\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value is $\\frac{3}{2}$.\n\nLet\n$$\nm = \\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1}.\n$$\nWe can write\n$$\n2 - m = \\left(a_1 - \\frac{a_1}{a_2^2+1}\\right) + \\cdots + \\left(a_n - \\frac{a_n}{a_1^2+1}\\right) = \\frac{a_1 a_2^2}{a_2^2+1} + \\cdots + \\frac{a_n a_1^2}{a_1^2+1}.\n$$\nSince $a_i^2 + 1 \\ge 2a_i$, we have\n$$\n\\frac{a_i a_{i+1}^2}{a_{i+1}^2+1} \\le \\frac{a_i a_{i+1}}{2},\n$$\nso\n$$\n2 - m \\le \\frac{a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1}{2}.\n$$\nA known inequality for $n \\ge 4$ and nonnegative $a_i$ is\n$$\n(a_1 + \\cdots + a_n)^2 - 4(a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1) \\ge 0.\n$$\nWith $a_1 + \\cdots + a_n = 2$, this gives\n$$\n4 - 4(a_1 a_2 + \\cdots + a_n a_1) \\ge 0 \\implies a_1 a_2 + \\cdots + a_n a_1 \\le 1.\n$$\nTherefore,\n$$\n2 - m \\le \\frac{1}{2} \\implies m \\ge \\frac{3}{2}.\n$$\nEquality holds when $a_1 = a_2 = 1$ and $a_3 = \\cdots = a_n = 0$.\n\nThus, the minimum value is $\\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18948,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any real number $M > 2$, there exists a strictly increasing infinite sequence of positive integers $a_1, a_2, \\dots$ satisfying both of the following two conditions:\n\n1. $a_i > M^i$ for any positive integer $i$;\n2. An integer $n$ is nonzero if and only if there exists a positive integer $m$ and $b_1, b_2, \\dots, b_m \\in \\{-1, 1\\}$, with\n\n$$\nn = b_1 a_1 + b_2 a_2 + \\dots + b_m a_m.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For given $M > 2$, we construct by induction a sequence $\\{a_n\\}$ that satisfies the requirements. Take $a_1, a_2$ such that $a_2 - a_1 = 1$ and $a_1 > M^2$. Now suppose $a_1, a_2, \\dots, a_{2k}$ are already chosen, such that $a_i > M^i$ for $i = 1, 2, \\dots, 2k$ and such that the set\n\n$$\nA_k = \\{b_1 a_1 + \\dots + b_m a_m \\mid b_1, \\dots, b_m = \\pm 1,\\ 1 \\le m \\le 2k\\}\n$$\n\ndoes not contain $0$. It is obvious that $A_k$ is symmetric, i.e., $A_k = -A_k$. $A_1 = \\{a_1, -a_1, 1, -1\\}$. Let $n$ be the smallest positive integer not in $A_k$, $N = \\sum_{i=1}^{2k} a_i$. Now choose positive integers $a_{2k+1}, a_{2k+2}$ satisfying $a_{2k+2} - a_{2k+1} = N + n$, $a_{2k+1} > M^{2k+2}$, $a_{2k+1} > \\sum_{i=1}^{2k} a_i$.\n\nWe now show that $A_{k+1}$ does not contain $0$ and $n \\in A_{k+1}$. First,\n\n$$\nn = - \\sum_{i=1}^{2k} a_i - a_{2k+1} + a_{2k+2}.\n$$\n\nOn the other hand, if $\\sum_{i=1}^{m} b_i a_i = 0$, $m \\le 2k+2$, as $0 \\notin A_k$, we must have $m = 2k+1$ or $2k+2$. If $m = 2k+1$, then\n\n$$\n\\left| \\sum_{i=1}^{2k-1} b_i a_i \\right| \\ge a_{2k-1} - \\sum_{i=1}^{2k} a_i > 0.\n$$\n\nIf $m = 2k + 2$ and $b_{2k+1}$ and $b_{2k+2}$ are of the same sign, then\n\n$$\n\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| \\geq a_{2k+1} + a_{2k+2} - \\sum_{i=1}^{2k} a_i > 0;\n$$\n\nIf $b_{2k+1}$ and $b_{2k+2}$ are of different signs, then\n\n$$\n\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| = \\left| \\sum_{i=1}^{2k} b_i a_i \\pm (a_{2k+1} - a_{2k+2}) \\right| \\geq |a_{2k+1} - a_{2k+2}| - \\sum_{i=1}^{2k} a_i \\\\\n= N + n - N = n > 0\n$$\n\nThe sequence $\\{a_n\\}$ thus constructed satisfies the requirements since $0$ is not contained in any $A_k$, and any nonzero integer between $-k$ and $k$ is contained in $A_k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18949,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\varphi = \\frac{1+\\sqrt{5}}{2}$ be the positive root of the quadratic polynomial $t^2 - t - 1$.\n\nThe first player can win if the ratio of the starting numbers is in the set\n$$\n\\langle 0, \\frac{1}{\\varphi} \\rangle \\cup \\{1\\} \\cup \\langle \\varphi, +\\infty \\rangle.\n$$\n\nLet $M$ and $m$ be positive integers such that $M \\geq m$. Is it true that the player who is next on the turn when $M$ and $m$ are on the board wins if and only if $m = M$ or $M > \\varphi m$?",
    "options": [],
    "answer": "See solution",
    "solution": "The claim is clear for $m = M$, so let us assume that $m \\neq M$.\n\nIf $M < \\varphi m$, then the player who plays next must pass on the pair of numbers $(m, M-m) \\neq (m, 0)$ and $m > \\varphi(M-m)$. Hence, it is enough to show that the player who plays with a pair such that $M > \\varphi m$ can either win or pass on a pair $(m', M')$ with $m' < M' < \\varphi m'$.\n\nThe player wins if he plays with the numbers such that $m \\mid M$. Let us assume that $M > \\varphi m$ and $M = qm + r$, $0 < r < m$. If $q \\geq 2$, then the player may pass on $(m, r)$, as well as on $(m, m+r)$. He will pass on $(m, r)$ if $r < m < \\varphi r$. Otherwise, if $m > \\varphi r$, he will pass on $(m, m+r)$, so the other player will pass on $(m, r)$ and we know that with these numbers the player on the turn wins. If $q = 1$, the player passes on $(m, r)$. We claim that $r \\neq m < \\varphi r$. Indeed, since $m + r > \\varphi m$, we have $(\\varphi - 1)m < r$, i.e. $m < \\varphi r$.\n\nHence, the first player can in each move either win or pass on a pair $(m', M')$ such that $m' < M' < \\varphi m'.$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18950,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $a(x)$, $b(x)$, $c(x)$, $d(x)$ with real coefficients satisfying the simultaneous equations\n\n$$\n\\begin{aligned}\nb(x)c(x) + a(x)d(x) &= 0 \\\\\na(x)c(x) + (1 - x^2)b(x)d(x) &= x + 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that it is not possible for all four polynomials to be non-zero. Suppose they are. Denote the leading coefficients of the polynomials $a(x)$, $b(x)$, $c(x)$, $d(x)$ (which exist, because the polynomials are non-zero) by $A$, $B$, $C$, $D$, respectively. Then the first equation implies $BC = -AD$ and thus $ABCD = -(BC)^2 < 0$. In the second equation, the leading coefficient of $a(x)c(x)$ is $AC$ and the leading coefficient of $(1-x^2)b(x)d(x)$ is $-BD$. Since the degree of $(1-x^2)b(x)d(x)$ is at least 2, these leading coefficients must cancel if we wish to end up with $x+1$ on the right. Thus $AC - BD = 0$, which implies that $ABCD = (AC)^2 > 0$. We have two contradictory inequalities, proving that this system has no solution when all four polynomials are non-zero.\n\nNow suppose that $a(x) = 0$. Then, by the first equation, $b(x) = 0$ or $c(x) = 0$. If $b(x) = 0$, we see that the second equation is not satisfied. If $c(x) = 0$, we see that, again, the second equation is not satisfied, due to differences in degrees. Similarly, when $c(x) = 0$, we find, in a symmetrical way, that there are no solutions.\n\nHenceforth, we assume that both $a(x)$ and $c(x)$ are non-zero. So either $b(x)$ or $d(x)$ (or both) must be zero. From the first equation, $b(x) = 0$ if and only if $d(x) = 0$. It follows that the complete set of solutions is given by all $(a(x), b(x), c(x), d(x))$ where $a(x)c(x) = x+1$ and $b(x) = d(x) = 0$, i.e.,\n\n$$\n\\{(k, 0, \\frac{x}{k} + \\frac{1}{k}, 0) : k \\text{ a non-zero real number}\\} \\cup \\{(\\frac{x}{k} + \\frac{1}{k}, 0, k, 0) : k \\text{ a non-zero real number}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18951,
    "subject": "Mathematics (Olympiad)",
    "question": "Out of the three expressions $\\frac{x}{y}$, $\\frac{x^2 + x}{y^2 + y}$, and $\\frac{x^2 + 2}{y^2 + 2}$, for some integers $x, y$, all three are defined, two take the same integer value, and the remaining one takes a different integer value. For which pairs of integers $x, y$ is this possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, $|x| > |y|$ for $x \\neq 0$, since otherwise, the first fraction will not be an integer.\n\nIf the first two expressions are equal:\n\n$$\n\\frac{x}{y} = \\frac{x^2 + x}{y^2 + y} \\implies xy^2 + xy = yx^2 + xy \\implies xy(y - x) = 0.\n$$\n\nFor $x \\neq y$ and possible values, this is satisfied only by pairs $(x, y) = (0, t)$, $t \\neq 0, -1$. The third expression will then be $\\frac{2}{t^2 + 2}$, which is non-integer for $t \\neq 0$.\n\nIf the first and third expressions are equal:\n\n$$\n\\frac{x}{y} = \\frac{x^2 + 2}{y^2 + 2} \\implies xy^2 + 2x = yx^2 + 2y \\implies xy(y - x) = 2(y - x) \\implies xy = 2,\n$$\n\nThe possible pairs are $(x, y) = (2, 1)$ and $(-2, -1)$. For $(2, 1)$, the second expression is $3$, which is different from the other two. For $(-2, -1)$, the second expression is not defined.\n\nIf the second and third expressions are equal:\n\n$$\n\\frac{x^2 + x}{y^2 + y} = \\frac{x^2 + 2}{y^2 + 2} \\implies x^2y^2 + xy^2 + 2x^2 + 2x = x^2y^2 + yx^2 + 2y^2 + 2y \\\\\n\\implies xy(y - x) = 2(y - x)(y + x) + 2(y - x) \\implies xy - 2y - 2x = 2 \\implies (x - 2)(y - 2) = 6.\n$$\n\nSince $6 = 1 \\cdot 6 = 2 \\cdot 3$, consider all possible cases with $|x| > |y|$:\n\n- $x - 2 = 6$, $y - 2 = 1$ $\\implies x = 8$, $y = 3$ $\\implies \\frac{x^2 + 2}{y^2 + 2} = \\frac{66}{11} = 6$, $\\frac{x}{y} = \\frac{8}{3}$ (not integer).\n- $x - 2 = 3$, $y - 2 = 2$ $\\implies x = 5$, $y = 4$ $\\implies \\frac{x^2 + 2}{y^2 + 2} = \\frac{27}{18}$ (not integer).\n- $x - 2 = -2$, $y - 2 = -3$ $\\implies x = 0$, $y = -1$ (not allowed).\n- $x - 2 = -6$, $y - 2 = -1$ $\\implies x = -4$, $y = 1$ $\\implies \\frac{x^2 + 2}{y^2 + 2} = \\frac{18}{3} = 6$, $\\frac{x}{y} = -4$ (satisfies the conditions).\n\n**Summary:**\n\nThe possible integer pairs $(x, y)$ are $(0, t)$ with $t \\neq 0, -1$; $(2, 1)$; and $(-4, 1)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 18952,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, \\infty) \\to \\mathbb{R}$ be a continuous function such that\n$$\n\\int_{0}^{n} f(x)f(n-x) \\, dx = \\int_{0}^{n} (f(x))^{2} \\, dx,\n$$\nfor any integer $n \\ge 1$. Prove that $f$ is periodical.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the period is $1$.\nLet $n$ be a positive integer. Denoting $y = n - x$, we obtain\n$$\n\\int_0^n f(n-y)f(y) \\, dy = \\int_0^n (f(n-y))^2 \\, dy.\n$$\nSumming up with the equality in the hypothesis, we get\n$$\n\\int_0^n (f(x) - f(n-x))^2 \\, dx = 0.\n$$\nThe continuity of $f$ implies $f(x) = f(n-x)$ for all $x \\in [0, n]$.\nLet $x \\ge 0$ and consider $n \\ge x$ a positive integer. Then\n$$\n f(x+1) = f(n+1-x-1) = f(n-x) = f(x).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18953,
    "subject": "Mathematics (Olympiad)",
    "question": "a) In how many ways can Emma make 55 cents using 50c, 20c, 10c, and 5c coins?\n\nb) To minimise the number of coins, which coins should Emma select to make a certain amount?\n\nc) How many possible collections of 12 coins can Emma have?\n\nd) Emma starts with $5$ each of $2, $1, 50c, 20c, 10c, and 5c coins (totaling $\\$19.25$). After spending $15, how many and which coins does she have left if she has $\\$4.25$ remaining?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Emma can make 55 cents in nine ways:\n\n| 50c | 20c | 10c | 5c |\n|-----|-----|-----|----|\n| 1   | 0   | 0   | 1  |\n| 0   | 2   | 1   | 1  |\n| 0   | 2   | 0   | 3  |\n| 0   | 1   | 3   | 1  |\n| 0   | 1   | 2   | 3  |\n| 0   | 1   | 1   | 5  |\n| 0   | 0   | 5   | 1  |\n| 0   | 0   | 4   | 3  |\n| 0   | 0   | 3   | 5  |\n\nb) To minimise the number of coins, Emma must select as many $2 coins as possible, then as many $1 coins as possible, and so on, giving 11 coins: five $2, two $1, one 50c, two 20c, one 5c.\n\nc) There are three possible collections of 12 coins:\n\n| $2 | $1 | 50c | 20c | 10c | 5c |\n|----|----|-----|-----|-----|----|\n| 5  | 2  | 1   | 1   | 2   | 1  |\n| 5  | 1  | 3   | 2   | 0   | 1  |\n| 4  | 4  | 1   | 2   | 0   | 1  |\n\nd) Emma starts with $5 \\times (\\$2 + \\$1 + 50c + 20c + 10c + 5c) = \\$19.25$. After spending $15, the amount left is $19.25 - \\$15 = \\$4.25$. One coin of each denomination totals $3.85$. So she has two more coins and they total $4.25 - \\$3.85 = 40c$. Both of these coins must be 20c. Thus Emma had left one $2, one $1, one 50c, three 20c, one 10c, and one 5c.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18954,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD нь $\\angle DAC = \\angle DCA = \\angle ABC$ байх гүдгэр дөрвөн өнцөгт байг. D оройг дайруулан AC-тэй параллель татсан шулуун BA ба BC шулуунуудтай харгалзан E ба F цэгт огтлолцдог байг. $CE \\cap AF = U$ бол $U$ цэгийн ABCD дөрвөн өнцөгтийн талууд дээрх проекцуудаар үүсэх дөрвөн өнцөгт нь адил хажуут трапец гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Өгөгдсөн нөхцөлөөс $B$ цэг нь $DA$ ба $DC$ шулуунуудыг шүргэсэн бөгөөд $A, C$ цэгүүдийг дайрсан тойрог дээр оршино гэж гарна. $U$ цэг уг тойрог дээр оршино гэдгийг дараах Паскалийн урвуу теорем ашиглан баталъя: Хэрэв гүдгэр зургаан өнцөгтийн 5 нь нэг тойрог дээр орших бөгөөд, эсрэг талуудын огтлолцол болох 3 цэг 1 шулуун дээр оршин байвал 6 дахь цэг уг тойрог дээр оршино. Одоо уг дүнг хэрхэн дөрвөн өнцөгт дээр ашиглахыг авч үзье. Үүнд: 6 өнцөгтийн аль нэгэн эсрэг 2 талыг нь цэг гэж үзээд харгалзах талууд нь тойрог татсан шүргэгч гэж үзнэ. $AD$ ба $CD$ нь $(ABC)$ тойргийн шүргэгч тул дээрх теоремыг хэрэглэвэл $U \\in (ABC)$ буюу $ABCU$ тойрогт багтана. Учир нь $AABCCU$ зургаан өнцөгт болон $F, E, D$ цэгүүд нэг шулуун дээр оршино гэдгийг ашиглав.\n\nОдоо\n\n$$\n\\angle DEU = \\angle UCA = \\angle UAD \\text{ гэж баталъя.}\n$$\n\n$$\n\\begin{array}{l}\nAC // EF \\Rightarrow \\angle DEU = \\angle UCA \\text{ ба} \\\\\nU \\in (ABC) \\hfill (1) \\\\\nAD \\text{ нь (ABC) тойргийн A цэг дээрх шүргэгч} \\hfill (2) \\\\\n\\Rightarrow \\angle UAD = \\angle UCA\n\\end{array}\n$$\n\n$$\n\\Rightarrow (1) \\text{ ба } (2)\\text{-оос } \\angle DEU = \\angle UCA = \\angle UAD.\n$$\n\n$$\n\\Rightarrow A, E, U, D \\text{ нэг тойрог дээр оршино.}\n$$\n\nЯг үүнтэй адилаар $C, F, D, U$ нэг тойрог дээр оршино. $U$ цэгийн $AB, BC, CD, DA$ талууд дээрх проекцуудийг харгалзан $P, Q, R, S$ гээ. Мөн $AC, EF$ шулуунууд дээрх проекцүүдийг харгалзан $V, T$ гээ. $U \\in (ABC)$ тул $P, V, Q$ цэгүүд нэг шулуун дээр оршино. Учир нь уг шулуун нь $(ABC)$-ийн Симпсоны шулуун юм. Иймд $U$ нь $AC$ болон $PQ$ хэрчмүүдийн ерөнхий цэг байна. Яг үүнтэй адилаар $U \\in (AED)$, $U \\in (CDF)$ тул $T \\in EF$, $T \\in PS$, $T \\in QR$ байна. Мөн $A, P, V, U, S$ цэгүүд; $C, Q, R, U, V$ цэгүүд; $D, T, S, U, R$ цэгүүд нэг тойрог дээр орших нь илэрхий.\n\n$$\n\\begin{align*}\n\\angle SPQ &= \\angle SPV = \\angle SAV = \\angle DAC = \\angle DCA \\\\\n&= \\angle RCV = \\angle RQV = \\angle RQP \\\\\n&\\Rightarrow \\angle SPQ = \\angle RQP \\\\\n\\angle TSR &= \\angle TUR = 180^{\\circ} - \\angle RUV = \\angle RQV = \\angle TQP \\\\\n&= \\angle TPQ \\Rightarrow SR // PQ.\n\\end{align*}\n$$\n\nИймд $PQRS$ нь адил хажуут трапец байна. ▲",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18955,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a + b + c = 1$, prove that\n\n$$\n\\sqrt{(c + a)^2 + c^2} + \\sqrt{(a + b)^2 + a^2} + \\sqrt{(b + c)^2 + b^2} \\geq \\sqrt{5}.\n$$\n\nEquality holds when $a = b = c = \\frac{1}{3}$.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p109_data_fa876ae1da.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Using the condition $a + b + c = 1$, we have:\n\n$$\na^2 - 2bc + 2c = a^2 + 2c(1 - b) = a^2 + 2c(a + c) = (c + a)^2 + c^2.\n$$\n\nSimilarly, the left-hand side of the inequality becomes\n\n$$\n\\sqrt{(c + a)^2 + c^2} + \\sqrt{(a + b)^2 + a^2} + \\sqrt{(b + c)^2 + b^2}.\n$$\n\nBy the triangle inequality, this is bounded below by\n\n$$\n\\sqrt{((c + a) + (a + b) + (b + c))^2 + (c + a + b)^2} = \\sqrt{5}.\n$$\n\nEquality holds when $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18956,
    "subject": "Mathematics (Olympiad)",
    "question": "From the point $P$, draw a line parallel to $BC$ meeting the lines $AB$ and $AC$ at the points $B'$ and $C'$, respectively.\n\n![](images/Tajland_2013_p5_data_00b6675f1d.png)\n\nClearly, the line $B'C'$ is externally tangent to the circle $\\omega$ at the point $P$. Let $Y$ and $Z$ be two points on the line $BC$, and let $Y'$ and $Z'$ be the points where $AY$ and $AZ$ intersect the line $AB$.\n\n**Claim:**\n$$\n\\frac{B'C'}{BC} = \\frac{Y'Z'}{YZ}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof of Claim.** Let $h_a'$ and $h_a$ be the heights of the triangles $AB'C'$ and $ABC$, respectively. Since $\\triangle AB'C' \\sim \\triangle ABC$ and $\\triangle AY'Z' \\sim \\triangle AYZ$, we have $\\frac{B'C'}{BC} = \\frac{h_a'}{h_a} = \\frac{Y'Z'}{YZ}$, and the claim is proved.\n\nLet $a, b, c, a', b', c'$ denote the lengths of $BC, AC, AB, B'C', AC', AB'$, respectively, and let $s, s'$ be the semiperimeters of triangles $ABC$ and $AB'C'$, respectively. Since the circle $\\omega$ touches the triangle $AB'C'$ externally, we have $B'P + PC' = a'$, $c' + B'P = b' + PC'$. Thus, $PC' = s' - b'$. From the Claim, we deduce that\n$$\n\\frac{PC'}{MC} = \\frac{a'}{a} \\implies MC = a \\frac{s' - b'}{a'} = a \\frac{s - b}{a} = s - b = BD.\n$$\n\nExtend $AC$ beyond $C$ to the point $X$ so that $CX = MC$. Since triangles $APE$ and $AMX$ are similar, we have $\\frac{AP}{AM} = \\frac{AE}{AX} = \\frac{s-a}{s}$. Applying Menelaus' theorem to $\\triangle AMC$ with respect to the line $BN$, we get\n$$\n\\frac{AR}{RM} \\cdot \\frac{MB}{CB} \\cdot \\frac{CN}{NA} = 1 \\implies \\frac{AR}{RM} = \\frac{a}{s-c} \\cdot \\frac{s-c}{s-a} = \\frac{a}{s-a}\n$$\nshowing that $\\frac{AM}{RM} = \\frac{s}{s-a}$. Thus, $RM = AP$ implying that $AR = PM$. Similarly, $BR = QN$.\n\nLet $\\theta$ be the angle between the sides $AR$ and $BR$. Thus,\n$$\n[ARB] = \\frac{1}{2} AR \\cdot BR \\sin \\theta = \\frac{1}{2} PM \\cdot QN \\sin \\theta = [PQMN].\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18957,
    "subject": "Mathematics (Olympiad)",
    "question": "The sum of four real numbers is $9$, and the sum of their squares is $21$. Prove that these numbers can be assigned as $a, b, c, d$ so that the inequality $ab - cd \\geq 2$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the numbers be $p, q, r, s$. Up to a permutation, we may assume that $p \\geq q \\geq r \\geq s$. We first consider the case where $p+q \\geq 5$. Then\n\n$$\np^2 + q^2 + 2pq \\geq 25 = 4 + (p^2 + q^2 + r^2 + s^2) \\geq 4 + p^2 + q^2 + 2rs,\n$$\n\nwhich is equivalent to $pq - rs \\geq 2$.\n\nAssume now that $p+q < 5$; then\n\n$$\n4 < r + s \\leq p + q < 5.\n$$\n\nObserve that\n\n$$\n(pq + rs) + (pr + qs) + (ps + qr) = \\frac{(p+q+r+s)^2 - (p^2+q^2+r^2+s^2)}{2} = 30.\n$$\n\nMoreover,\n\n$$\npq + rs \\geq pr + qs \\geq ps + qr,\n$$\n\nbecause $(p-s)(q-r) \\geq 0$ and $(p-q)(r-s) \\geq 0$.\n\nWe conclude that $pq + rs \\geq 10$. From above, $0 \\leq (p+q) - (r+s) < 1$, therefore\n\n$$\n(p+q)^2 - 2(p+q)(r+s) + (r+s)^2 < 1.\n$$\n\nAdding this to $(p+q)^2 + 2(p+q)(r+s) + (r+s)^2 = 9^2$ gives\n\n$$\n(p+q)^2 + (r+s)^2 < 41.\n$$\n\nTherefore\n\n$$\n41 = 21 + 2 \\cdot 10 \\leq (p^2 + q^2 + r^2 + s^2) + 2(pq + rs) = (p+q)^2 + (r+s)^2 < 41,\n$$\n\nwhich is a contradiction.\n\n**Second solution.** From $a+b+c+d=9$ with ordering $a \\geq b \\geq c \\geq d$ we have\n\n$$\n\\frac{a+b}{2} = \\frac{9}{4} + \\varepsilon_1, \\quad \\frac{c+d}{2} = \\frac{9}{4} - \\varepsilon_1\n$$\n\nfor some $\\varepsilon_1 \\geq 0$. Thus\n\n$$\na = \\frac{9}{4} + \\varepsilon_1 + \\varepsilon_2, \\quad b = \\frac{9}{4} + \\varepsilon_1 - \\varepsilon_2, \\quad c = \\frac{9}{4} - \\varepsilon_1 + \\varepsilon_3, \\quad d = \\frac{9}{4} - \\varepsilon_1 - \\varepsilon_3\n$$\n\nfor some $\\varepsilon_2, \\varepsilon_3 \\geq 0$. From $b \\geq c$ we get\n\n$$\n\\varepsilon_1 - \\varepsilon_2 \\geq -\\varepsilon_1 + \\varepsilon_3 \\quad \\text{or} \\quad \\varepsilon_2 + \\varepsilon_3 \\leq 2\\varepsilon_1.\n$$\n\nFrom\n\n$$\n21 = (a^2 + b^2) + (c^2 + d^2) = 2 \\left(\\frac{9}{4} + \\varepsilon_1\\right)^2 + 2\\varepsilon_2^2 + 2 \\left(\\frac{9}{4} - \\varepsilon_1\\right)^2 + 2\\varepsilon_3^2 = 4 \\left(\\frac{9}{4}\\right)^2 + 4\\varepsilon_1^2 + 2\\varepsilon_2^2 + 2\\varepsilon_3^2 = 20 + \\frac{1}{4} + 2 (2\\varepsilon_1^2 + \\varepsilon_2^2 + \\varepsilon_3^2)\n$$\n\nwe conclude that non-negative numbers $\\varepsilon_i$ satisfy\n\n$$\n2\\varepsilon_1^2 + \\varepsilon_2^2 + \\varepsilon_3^2 = \\frac{3}{8}.\n$$\n\nUsing $\\varepsilon_2 + \\varepsilon_3 \\leq 2\\varepsilon_1$ and $\\varepsilon_2^2 + \\varepsilon_3^2 \\leq (\\varepsilon_2 + \\varepsilon_3)^2$ from above, we obtain\n\n$$\n\\frac{3}{8} \\leq 2\\varepsilon_1^2 + (\\varepsilon_2 + \\varepsilon_3)^2 \\leq 2\\varepsilon_1^2 + 4\\varepsilon_1^2 = 6\\varepsilon_1^2.\n$$\n\nHence $\\varepsilon_1^2 \\geq \\frac{1}{16}$ or $\\varepsilon_1 \\geq \\frac{1}{4}$. For $ab-cd$ we have\n\n$$\nab - cd = \\left(\\frac{9}{4} + \\varepsilon_1\\right)^2 - \\varepsilon_2^2 - \\left(\\frac{9}{4} - \\varepsilon_1\\right)^2 + \\varepsilon_3^2 = 9\\varepsilon_1 - \\varepsilon_2^2 + \\varepsilon_3^2.\n$$\n\nSubstituting for $\\varepsilon_2^2$ from above, we get\n\n$$\nab - cd = 9\\varepsilon_1 - \\left(\\frac{3}{8} - 2\\varepsilon_1^2 - \\varepsilon_3^2\\right) + \\varepsilon_3^2 = 9\\varepsilon_1 + 2\\varepsilon_1^2 - \\frac{3}{8} + 2\\varepsilon_3^2 \\geq 9 \\cdot \\frac{1}{4} + 2 \\cdot \\frac{1}{16} - \\frac{3}{8} = 2.\n$$\n\nThis ends the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18958,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $d$ be integers satisfying $d \\ge 0$, $|a| \\ge 2$, and $b \\ge (|a|+1)^{d+1}$. Suppose $f(x)$ is a real-coefficient polynomial of degree $d$, and for each positive integer $n$, let $r_n$ denote the remainder of $\\lfloor f(n)a^n \\rfloor$ modulo $b$.\n\nProve that if the sequence $r_n$ is ultimately periodic, then $f(x)$ is a rational-coefficient polynomial.\n\n*Note 1:* For every real number $x$, $\\lfloor x \\rfloor$ denotes the largest integer that is not greater than $x$.\n\n*Note 2:* A sequence $a_n$ is called *ultimately periodic* if there exist positive integers $n_0$ and $T$ such that for every integer $n \\ge n_0$, we have $a_{n+T} = a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof. Lemma** Let integers $a$ and real number $z$ satisfy $|a| \\ge 2$. For any non-negative integer $n$, if\n\n$$\n\\|\\|a^n z\\|\\| < \\frac{1}{|a| + 1},\n$$\n\nthen $z$ is an integer. Here, $\\|\\|x\\|\\|$ is defined as the distance between the real number $x$ and the nearest integer, that is, $\\|\\|x\\|\\| = \\min(\\{x\\}, 1 - \\{x\\})$.\n\n*Proof of the lemma:* Without loss of generality, assume $a \\ge 2$ (otherwise, replace $a$ with $-a$). Suppose the conclusion is false. Without loss of generality, assume $0 < z \\le \\frac{1}{2}$ (otherwise, replace $z$ with $m \\mp z$, where $m$ is some suitable integer).\n\nSince $z \\in (0, \\frac{1}{a^{-1}(a+1)})$, we can assume that\n\n$$\n\\frac{1}{a^k(a+1)} \\le z < \\frac{1}{a^{k-1}(a+1)} \\quad (k \\in \\mathbb{Z}_{\\ge 0}),\n$$\n\nwhich implies $\\frac{1}{a+1} \\le a^k z < 1 - \\frac{1}{a+1}$. Consequently, $\\|\\|a^k z\\|\\| \\ge \\frac{1}{a+1}$, leading to a contradiction. Thus, the lemma is proven.\n\nBack to the original question. Assume, for the sake of contradiction, that $f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0$ is not a polynomial with rational coefficients. Let $t$ be the largest index such that $a_t \\notin \\mathbb{Q}$. Define\n\n$$\n\\left\\{ \\frac{f(n)a^n}{b} \\right\\} = \\frac{r_n + \\varepsilon_n}{b}, \\quad \\varepsilon_n \\in [0, 1).\n$$\n\nLet $g(x) = \\Delta^t f(x) = \\sum_{i=0}^{t} (-1)^i \\binom{t}{i} f(x + t - i)$, where $\\Delta$ denotes the finite difference operator. It is easy to see that except for the constant term, all coefficients of $g(x)$ are rational numbers. Let $g(x) = c_{d-t}x^{d-t} + \\cdots + c_1x + c_0$, and let $K$ be the least common multiple of the denominators of $c_1, c_2, \\cdots, c_{d-t}$. Now,\n\n$$\n\\left\\{ \\frac{g(n)a^{n+t}}{b} \\right\\} = \\left\\{ \\sum_{i=0}^{t} \\frac{(-1)^i \\binom{t}{i} f(n+t-i) a^{n+t-i} \\cdot a^i}{b} \\right\\} \\\\\n= \\left\\{ \\sum_{i=0}^{t} \\frac{(-1)^i \\binom{t}{i} a^i (r_{n+t-i} + \\varepsilon_{n+t-i})}{b} \\right\\} \\\\\n= \\left\\{ \\sum_{i=0}^{t} r_{n+t-i} \\frac{(-1)^i \\binom{t}{i} a^i}{b} + \\sum_{i=0}^{t} \\varepsilon_{n+t-i} \\frac{(-1)^i \\binom{t}{i}}{b} \\right\\}.\n$$\n\nLet $\\delta_n = \\sum_{i=0}^{t} \\varepsilon_{n+t-i} \\cdot \\frac{(-1)^i \\binom{t}{i} a^i}{b}$. Note that\n\n$$\n\\sum_{i=0}^{t} \\left| \\frac{(-1)^i \\binom{t}{i} a^i}{b} \\right| = \\sum_{i=0}^{t} \\frac{\\binom{t}{i} |a|^i}{b} = \\frac{(|a|+1)^t}{b} \\le \\frac{(|a|+1)^d}{b} \\le \\frac{1}{|a|+1},\n$$\n\nwhich implies that for any positive integers $n$ and $n'$, \n\n$$\n\\left| \\delta_n - \\delta_{n'} \\right| \\le \\sum_{i=0}^{t} \\left| \\varepsilon_{n+t-i} - \\varepsilon_{n'+t-i} \\right| \\cdot \\frac{\\binom{t}{i} |a|^i}{b} < \\sum_{i=0}^{t} \\frac{\\binom{t}{i} |a|^i}{b} \\le \\frac{1}{|a|+1}.\n$$\n\nWe choose positive integers $M$ and $n$ such that $M$ is a multiple of the smallest positive period of $\\{r_n\\}$ and $bK\\mid M$, and when $n \\ge n_0$, we have $bK\\mid a^{n+M+t} - a^{n+t}$. Then, for $n \\ge n_0$,\n\n$$\n\\left\\| \\frac{g(n+M)a^{n+M+t}}{b} - \\frac{g(n)a^{n+t}}{b} \\right\\| = \\left\\| \\delta_{n+M} - \\delta_n \\right\\| < \\frac{1}{|a|+1},\n$$\n\nwhere the first equality holds because $r_{n+M+t-i} = r_{n+t-i}$ for $i = 0, 1, \\dots, t$. On the other hand,\n\n$$\n\\begin{aligned}\n& \\left| \\frac{g(n+M)a^{n+M+t}}{b} - \\frac{g(n)a^{n+t}}{b} \\right| \\\\\n&= \\left| \\frac{c_0}{b} (a^{n+M+t} - a^{n+t}) + \\sum_{i=1}^{d-t} \\frac{c_i}{b} ((n+M)^i a^{n+M+t} - n^i a^{n+t}) \\right| \\\\\n&= \\left| \\frac{c_0(a^M - 1)a^{n_0+t}}{b} \\cdot a^{n-n_0} \\right|.\n\\end{aligned}\n$$\n\nAccording to the lemma, we have $\\frac{c_0(a^M - 1)a^{n_0+t}}{b} \\in \\mathbb{Z}$, but this contradicts the irrationality of $c_0$! Hence, the proof is complete. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 18959,
    "subject": "Mathematics (Olympiad)",
    "question": "The integers $1$ to $n$ are written on the board. One of the numbers is wiped out. The average of the remaining numbers is $11\\frac{1}{4}$.\n\nWhich number has been wiped out?\n\nA) 6  \nB) 7  \nC) 11  \nD) 12  \nE) 21",
    "options": [],
    "answer": "See solution",
    "solution": "A) 6",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18960,
    "subject": "Mathematics (Olympiad)",
    "question": "Тойрогт $A$, $B$, $C$ цэгүүдэд огтлолцох гурван хөвч татав. $[AB]$, $[AC]$ царагуудыг болон өгсөн тойргийг $A_1$ цэгт гадаад байдлаар шүргэсэн тойрог авч үзье. Аналогиар $B_1$, $C_1$ цэгүүдийг тодорхойлъё. $AA_1$, $BB_1$, $CC_1$ шулуунууд нэг цэгт огтлолцохыг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Өгсөн тойргийг $\\omega$, $[AB]$, $[AC]$ царагууд болон $\\omega$-г $A_1$ цэгт шүргэх тойргийг $\\omega_A$; $ABC$ гурвалжинд багтсан тойргийг $\\gamma$ гэе. $X$ төвтэй $m$ коэффициенттэй гомотетийг $H_X^m$ гэе. $\\gamma$-г $\\omega$-д буулгах сөрөг коэффициенттэй гомотетийн төвийг $A_0$ гэе.\n\n$$\n\\begin{aligned}\nH_{A_0}^{k_0}(\\gamma) &= \\omega; \\quad H_{A_1}^{k_1}(\\gamma) = \\omega; \\quad H_A^k(\\gamma) = \\omega_A, \\quad k_0 < 0; \\quad k_1 < 0; \\quad k > 0 \\\\\nH_{A_1}^{k_1} \\circ H_A^k(\\gamma) &= H_{A_0}^{k_0}(\\gamma) = \\omega. \\text{ Эндээс } A_0 \\in A_1A. \\text{ Аналогиар}\n\\end{aligned}\n$$\n\n$A_0 \\in B_1B$, $A_0 \\in C_1C$ тул $AA_1$, $BB_1$, $CC_1$ шулуунууд $A_0$ цэгт огтлолцоно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18961,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $x$ 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.",
    "options": [],
    "answer": "See solution",
    "solution": "Каждая 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": 18962,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\geq 2$ such that the equation $a^2 - a + 2 = 0$ has a unique solution $a$ in $\\mathbb{Z}_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $7$ is the only integer satisfying the required conditions.\n\nIf $a \\in \\mathbb{Z}_n$ and $a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$, then\n$$(1-a)^2 - (1-a) + 2 = a^2 - a + 2 = 0$$\nin $\\mathbb{Z}_n$, so uniqueness forces $a = 1 - a$, i.e., $2a = 1$. Thus, $2$ is invertible in $\\mathbb{Z}_n$ and $a = 2^{-1}$.\n\nSubstituting $a = 2^{-1}$ into the equation:\n$$\n(2^{-1})^2 - 2^{-1} + 2 = 0\n$$\nMultiply both sides by $4$ (the inverse of $2^{-2}$):\n$$\n1 - 2 + 8 = 0 \\implies 7 = 0\n$$\nSo $n$ divides $7$. Since $7$ is prime and $n \\geq 2$, $n = 7$.\n\nIt 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": "Equations and Inequalities"
  },
  {
    "id": 18963,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exists an infinite sequence $a_1, a_2, a_3, \\dots$ of positive integers which satisfies the equality\n\n$$\na_{n+2} = a_{n+1} + \\sqrt{a_{n+1} + a_n}\n$$\n\nfor every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is no.\n\nSuppose that there exists a sequence $(a_n)$ of positive integers satisfying the given condition. We will show that this leads to a contradiction.\n\nFor each $n \\geq 2$, define $b_n = a_{n+1} - a_n$. Then, by assumption, for $n \\geq 2$ we get $b_n = \\sqrt{a_n + a_{n-1}}$, so that\n\n$$\nb_{n+1}^2 - b_n^2 = (a_{n+1} + a_n) - (a_n + a_{n-1}) = (a_{n+1} - a_n) + (a_n - a_{n-1}) = b_n + b_{n-1}.\n$$\n\nSince each $a_n$ is a positive integer, $b_n$ is a positive integer for $n \\geq 2$, and the sequence $(b_n)$ is strictly increasing for $n \\geq 3$. Thus $b_n + b_{n-1} = (b_n - b_{n-1})(b_n + b_{n-1}) \\geq (b_{n+1} + b_n)$, whence $b_{n-1} \\geq b_{n+1}$—a contradiction to the increasing nature of the sequence $(b_i)$.\n\nThus, we conclude that there exists no sequence $(a_n)$ of positive integers satisfying the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18964,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0 = 40$, $b_0 = 41$, and define sequences $(a_n)$ and $(b_n)$ recursively by\n\n$$\na_{n+1} = a_n + \\frac{1}{b_n}, \\quad b_{n+1} = b_n + \\frac{1}{a_n}.\n$$\n\nFind the minimal $k$ such that $a_k > 80$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\n\\frac{a_{n+1}}{b_{n+1}} = \\frac{a_n + \\frac{1}{b_n}}{b_n + \\frac{1}{a_n}} = \\frac{a_n}{b_n},\n$$\n\nwe get that $\\frac{a_n}{b_n}$ is a constant and hence equals $\\frac{40}{41}$. Therefore, $a_k > 80$ is equivalent to $b_k > 82$ and $a_k b_k > 80 \\cdot 82 = 6560$. Multiplying both sequences we get\n\n$$\na_{n+1}b_{n+1} = a_n b_n + 2 + \\frac{1}{a_n b_n}.\n$$\n\nHence, for any $k$ we have $a_k b_k > 40 \\cdot 41 + 2k$, which means $a_k b_k > 6560$ for $k = 2460$. Moreover, if $k = 2459$, we have\n\n$$\na_k b_k = 6560 - 2 + \\sum_{i=0}^{k-1} \\frac{1}{a_i b_i} < 6560 - 2 + \\frac{2459}{1640} < 6560.\n$$\n\nThus, the minimal value of $k$ satisfying the conditions is $2460$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18965,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonzero functions $f: \\mathbb{R} \\to \\mathbb{R}$ and $g: \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality\n\n$$\nf(x - 3f(y)) = x f(y) - y f(x) + g(x),\n$$\n\nfor all $x, y \\in \\mathbb{R}$, and $g(1) = -8$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that $f$ can take the value $0$. In fact, if $f(0) = 0$, we have the result. If $f(0) = b \\neq 0$, then by putting $x = 0$ in the given equation\n\n$$\nf(-3f(y)) = -b y + g(0),\n$$\n\nSince the right side can take all real values as $y$ varies, it follows that $f$ is onto $\\mathbb{R}$, so there exists $c \\in \\mathbb{R}$ such that $f(c) = 0$.\n\nFor $y = c$ in the equation, we get\n\n$$\nf(x) = -c f(x) + g(x) \\implies g(x) = (c + 1) f(x),\n$$\n\nand so the equation becomes\n\n$$\nf(x - 3f(y)) = x f(y) - y f(x) + (c + 1) f(x).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18966,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be real numbers such that $ab = 1$. Consider the following table:\n\n![](images/Belarus_2015_p11_data_608d4c9cf1.png)\n\nIs it possible for the product of all numbers in the table to be equal to $2015$?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it is possible.\n\nIt is easy to see that the products of the numbers in all $3 \\times 3$ squares and $4 \\times 4$ squares are equal to $1$. Moreover, the product of all numbers in the table is equal to $a$, so this product can admit any value different from $0$, in particular this product can be equal to $2015$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18967,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, \\dots, p_{30}$ be a permutation of the numbers $1, 2, \\dots, 30$. For how many permutations does the equality $$\\sum_{k=1}^{30} |p_k - k| = 450$$ hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us define pairs $(a_i, b_i)$ such that $\\{a_i, b_i\\} = \\{p_i, i\\}$ and $a_i \\ge b_i$. Then for every $i = 1, \\dots, 30$ we have $|p_i - i| = a_i - b_i$ and\n\n$$\n\\sum_{i=1}^{30} |p_i - i| = \\sum_{i=1}^{30} (a_i - b_i) = \\sum_{i=1}^{30} a_i - \\sum_{i=1}^{30} b_i.\n$$\n\nIt is clear that the sum $\\sum_{i=1}^{30} a_i - \\sum_{i=1}^{30} b_i$ is maximal when\n\n$$\n\\{a_1, a_2, \\dots, a_{30}\\} = \\{16, 17, \\dots, 30\\} \\text{ and } \\{b_1, b_2, \\dots, b_{30}\\} = \\{1, 2, \\dots, 15\\}\n$$\n\nand the maximal value equals $2(16 + \\dots + 30 - 1 - \\dots - 15) = 450$. The number of such permutations is $(15!)^2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 18968,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的外接圓為 $\\omega$，而其切於 $BC$ 邊的旁切圓為 $\\Omega_A$。令 $\\omega$ 與 $\\Omega_A$ 的交點為 $X$ 和 $Y$。設 $P$ 為 $A$ 對 $\\Omega_A$ 在點 $X$ 的切線的投影點，而 $Q$ 為 $A$ 對 $\\Omega_A$ 在點 $Y$ 的切線的投影點。設三角形 $APX$ 在點 $P$ 的切線，及三角形 $AQY$ 在點 $Q$ 的切線在點 $R$ 相交。\n\n證明：直線 $AR$ 與 $BC$ 互相垂直。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $D$ 為 $BC$ 與 $\\Omega_A$ 的切點，而 $D'$ 為 $D$ 在 $\\Omega_A$ 上的對徑點。令 $R'$ 為滿足 $AR' \\perp BC$ 且 $R'D' \\parallel BC$ 的（唯一）點。我們將證明 $R = R'$。\n\n設直線 $PX$ 分別與 $AB$、$D'R'$ 交於點 $S$、$T$。令 $U$ 為平行直線 $BC$、$D'R'$ 相交的無窮遠點。由於（退化的）六邊形 $ASXTUC$ 外切圓 $\\Omega_A$，根據 Brianchon 定理知 $AT$、$SU$、$XC$ 三線共於一點，設為 $V$。所以 $VS \\parallel BC$。於是\n\n$$\n\\angle(SV, VX) = \\angle(BC, CX) = \\angle(BA, AX),\n$$\n\n因此 $AXSV$ 共圓。由此得\n\n$$\n\\angle(PX, XA) = \\angle(SV, VA) = \\angle(R'T, TA).\n$$\n\n因為 $\\angle APT = \\angle AR'T = 90^\\circ$，知 $APR'T$ 共圓。所以\n\n$$\n\\begin{aligned}\n\\angle(XA, AP) &= 90^\\circ - \\angle(PX, XA) = 90^\\circ - \\angle(R'T, TA) \\\\\n&= \\angle(TA, AR') = \\angle(TP, PR').\n\\end{aligned}\n$$\n\n由此知 $PR'$ 與圓 $(APX)$ 相切。\n\n同理可得 $QR'$ 與圓 $(AQY)$ 亦相切。因此 $R = R'$，得 $AR \\perp BC$。$\\square$\n\n註：證出 $\\angle(PX, XA) = \\angle(R'T, TA)$ 之後，以下途徑亦可證完。存在旋似變換，將三角形 $ATR'$ 送到三角形 $AXP$。所以三角形 $ATX$ 與 $AR'P$ 相似且同向。因此 $\\angle(TX, XA) = \\angle(R'P, PA)$，由此推得 $PR'$ 與圓 $(APX)$ 相切。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18969,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(t) = \\frac{t^2 + 3}{t^3 + 1}$. Given that $x^2 + y^2 + z^2 \\le x + y + z$, find the minimum value of $f(x) + f(y) + f(z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $t^2 + 1 \\ge 2t$, we have $f(t) \\ge \\frac{2t + 2}{t^3 + 1} = \\frac{2}{t^2 - t + 1}$. Let $a = x^2 - x + 1$, $b = y^2 - y + 1$, $c = z^2 - z + 1$. The condition $x^2 + y^2 + z^2 \\le x + y + z$ becomes $a + b + c \\le 3$. Therefore,\n\n$$\nf(x) + f(y) + f(z) \\ge \\frac{2}{a} + \\frac{2}{b} + \\frac{2}{c} \\ge \\frac{18}{a + b + c} = 6\n$$\n\nwhere the last inequality is due to the arithmetic mean-harmonic mean inequality. The minimum is achieved at $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18970,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $k$ such that there exists a polynomial $f(x)$ with rational coefficients for which, for all sufficiently large $n$,\n\n$$\nf(n) = \\operatorname{lcm}(n + 1, n + 2, \\dots, n + k).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $k = 1$ and $k = 2$, the required polynomials are $f(x) = x + 1$ and $f(x) = (x + 1)(x + 2)$, respectively.\n\nLet $k \\geq 3$ and assume such a polynomial $f(x)$ exists. For any prime $p$, its exponent in $\\operatorname{lcm}(n + 1, \\dots, n + k)$ is $\\max\\{\\alpha_1, \\dots, \\alpha_k\\}$, where $\\alpha_i$ is the power of $p$ in $n + i$. If this maximum is $\\alpha_s$, then it is obtained by\n\n$$\n\\frac{(n+1)(n+2)\\cdots(n+k)}{p^{\\alpha_1}p^{\\alpha_2}\\cdots p^{\\alpha_{s-1}}p^{\\alpha_{s+1}}\\cdots p^{\\alpha_k}},\n$$\n\nand the powers of $p$ in the denominator are divisors of $\\prod_{1 \\leq i \\neq s \\leq k} (s-i)$. Therefore,\n\n$$\n\\operatorname{lcm}(n+1, \\dots, n+k) = \\frac{(n+1)(n+2)\\cdots(n+k)}{C_n},\n$$\n\nwhere $C_n$ divides $\\prod_{1 \\leq i < j \\leq k} (j-i)$. Since $C_n$ can take only finitely many values, there exists a natural number $C$ such that for infinitely many $n$, $f(n) = \\frac{(n+1)(n+2)\\cdots(n+k)}{C}$. Thus, for infinitely many $x$, $f(x) = \\frac{(x+1)(x+2)\\cdots(x+k)}{C}$, so\n\n$$\nf(x) = \\frac{(x+1)(x+2)\\cdots(x+k)}{C}, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nTherefore,\n\n$$\n\\operatorname{lcm}(n+1, \\dots, n+k) = \\frac{(n+1)(n+2)\\cdots(n+k)}{C}, \\quad \\text{for all } n \\in \\mathbb{N}.\n$$\n\nAssume this is possible. Choose a prime $p < k$ such that $p$ does not divide $k$. Let $n + k + 1 = p^m$ for large $m$. From above,\n\n$$\n\\frac{\\operatorname{lcm}(n+2, \\dots, n+k+1)}{\\operatorname{lcm}(n+1, \\dots, n+k)} = \\frac{n+k+1}{n+1}.\n$$\n\nThe exponent of $p$ in the numerator (left side) is $m$, and in the denominator at least $1$, while in the right side numerator it is $m$ and denominator $0$. This is a contradiction. Therefore, for $k \\geq 3$, no such polynomial exists.\n\n$\\boxed{\\text{Only } k = 1 \\text{ and } k = 2 \\text{ work.}}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18971,
    "subject": "Mathematics (Olympiad)",
    "question": "Twenty indistinguishable coins are arranged in a row. One of them weighs $9$ grams and the next coin to the right weighs $11$ grams. The remaining $18$ coins each weigh $10$ grams. Find the $11$ gram coin using $3$ weightings on a two-pan balance without weights.",
    "options": [],
    "answer": "See solution",
    "solution": "On the first attempt, compare two groups of $9$ coins each:\n\n$G_1 = \\{1,3,5,7,9,11,13,15,17\\}$ and $G_2 = \\{2,4,6,8,10,12,14,16,18\\}$.\n\nNote that the $9$ gram coin $A$ and the $11$ gram coin $B$ cannot be on the same pan, as their positions are consecutive (different parity). If there is equilibrium, neither $A$ nor $B$ is on the pans, i.e., they are at positions $19$ and $20$. Since $B$ is the right neighbor of $A$, the $11$ gram coin is the last one.\n\nIf $G_1$ is lighter than $G_2$, then $A \\in G_1$. If $A \\in G_1$, $G_1$ has only $10$ gram coins (it cannot contain $B$). Thus, $G_1$ is lighter only if $B \\in G_2$. So $B$ is at an even position $2,4,\\ldots,18$, and $A$ is at the previous odd position $1,3,\\ldots,17$, i.e., $A \\in G_1$, which is consistent.\n\nSimilarly, if $G_1$ is heavier than $G_2$, then $A \\in G_2$.\n\nSo, in the case of non-equilibrium, the first attempt finds a group of $9$ coins with $8$ of them having the same weight and the last one lighter. It is known how to find the lighter coin with $2$ attempts: divide the $9$ coins into $3$ groups of $3$ and compare two groups. Regardless of the outcome, this determines a group of $3$ coins containing the lighter one. Then compare two coins from this group; the lighter coin will be identified.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18972,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{a_1, a_2, a_3, a_4\\}$ be a set of four positive integers with $a_1 < a_2 < a_3 < a_4$. Define $n_A$ as the number of unordered pairs $\\{i, j\\}$ ($1 \\leq i < j \\leq 4$) such that $a_i + a_j$ divides $s_A = a_1 + a_2 + a_3 + a_4$. What is the largest possible value of $n_A$? Find all such sets $A$ for which this maximum is achieved.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s_A = a_1 + a_2 + a_3 + a_4$. By the previous deductions, $a_2 = \\frac{1}{2}(3a_1 + a_3)$, $a_4 = \\frac{1}{2}(a_1 + a_3)$.\n\nBy $a_1 + a_2 \\mid s_A$, let $s_A = l(a_1 + a_2)$, we have\n$$\n3(a_1 + a_3) = l \\left( a_1 + \\frac{1}{2}(3a_1 + a_3) \\right),\n$$\nwhich simplifies to\n$$\n(6 - l) a_3 = (5l - 6) a_1.\n$$\nSince $a_1 < a_3$ and both are positive integers, $l = 4$ or $5$.\n\n- If $l = 4$, then $a_3 = 7a_1$, $a_2 = 5a_1$, $a_4 = 11a_1$.\n- If $l = 5$, then $a_3 = 19a_1$, $a_2 = 11a_1$, $a_4 = 29a_1$.\n\nIt can be verified that for these values, each of $a_1 + a_2$, $a_1 + a_3$, $a_1 + a_4$, and $a_2 + a_3$ divides $s_A$.\n\n**Conclusion:** All sets $A$ of four distinct positive integers which achieve the largest possible value $n_A = 4$ are $A = \\{a, 5a, 7a, 11a\\}$ and $A = \\{a, 11a, 19a, 29a\\}$, where $a$ is any positive integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18973,
    "subject": "Mathematics (Olympiad)",
    "question": "Trapezoid $ABCD$, with $AB \\parallel CD$, is inscribed in circle $\\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to line $AB$ intersect segment $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if ray $BG$ bisects $\\angle CBD$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.**\n\nFirst, we prove the \"if\" direction by assuming that ray $BG$ bisects $\\angle CBD$; that is, we assume that $\\widehat{DQ} = \\widehat{CQ}$. It is easy to see that $ABCD$ is an isosceles trapezoid with $AD = BC$. In particular, $\\widehat{AD} = \\widehat{BC}$ and $\\widehat{AC} = \\widehat{BD}$. Because $ABCD$ is cyclic, it follows that\n\n$$\n\\angle APC = \\frac{\\widehat{AC}}{2} = \\frac{\\widehat{BD}}{2} = \\angle BCD = \\angle SCD \\quad \\text{and} \\quad \\angle APD = \\frac{\\widehat{AD}}{2} = \\frac{\\widehat{BC}}{2} = \\angle BDC = \\angle RDC.\n$$\n\n![](images/pamphlet0910_main_p13_data_5ec5709a52.png)\n![](images/pamphlet0910_main_p13_data_948d9fdd37.png)\n\nBecause $RS \\parallel DC$, it follows that $180^\\circ = \\angle GRD + \\angle RDC = \\angle GRD + \\angle APD$ and $180^\\circ = \\angle GSC + \\angle SCD = \\angle GSC + \\angle APC$; that is, both $GSCP$ and $GRDP$ are cyclic. Hence, $\\angle GPR = \\angle GDR$ and $\\angle GPS = \\angle GCS$. In particular, we have\n\n$$\n\\angle RPS = \\angle GPR + \\angle GPS = \\angle GDR + \\angle GCS. \\qquad (1)\n$$\n\nLet $K$ be the intersection of segments $BQ$ and $CD$. We have $\\angle CBK = \\angle QBD$ and $\\angle KCB = \\angle DCB = \\angle DQB$; that is, triangles $CBK$ and $QBD$ are similar to each other. Because $RG \\parallel CD$, we have $BG/GK = BR/RD$. This means that $G$ and $R$ are the corresponding points in the similar triangles $CBK$ and $QBD$. Consequently, we have $\\angle BCG = \\angle BQR$. In exactly the same way, we can show that $\\angle BDG = \\angle BQS$. Combining the last two equations together with (1) yields\n\n$$\n\\angle RQS = \\angle BQS + \\angle BQR = \\angle BDG + \\angle BCG = \\angle RDG + \\angle SCG = \\angle RPS,\n$$\n\nfrom which it follows that $PQRS$ is cyclic.\n\nSecond, we prove the \"only if\" direction by assuming that $PQRS$ is cyclic. Let $\\gamma$ denote the circumcircle of $PQRS$. We approach indirectly by assuming that ray $BG$ does not bisect $\\angle CBD$. Let $G_1$ be the point on segment $RS$ such that ray $BG_1$ bisects $\\angle CBD$. Let rays $AG_1$ and $BG_1$ meet $\\omega$ again at $P_1$ and $Q_1$ (other than $A$ and $B$). By our proof of the \"if\" part, $P_1Q_1RS$ is cyclic, and let $\\gamma_1$ denote its circumcircle.\n\nHence lines $RS, PQ, P_1Q_1$ are the radical axes of pairs of circles $\\gamma$ and $\\gamma_1$, $\\gamma$ and $\\omega$, $\\gamma_1$ and $\\omega$, respectively. Because point $Q_1$ is the midpoint of arc $\\widehat{CD}$ (not including $A$ and $B$), $P_1Q_1 \\parallel CD$, implying that lines $P_1Q_1$ and $RS$ intersect. Let $X$ denote this intersection; then $X$ is the radical center of $\\omega, \\gamma, \\gamma_1$. In particular, line $PQ$ also passes through $X$, giving the following configuration.\n\n![](images/pamphlet0910_main_p13_data_13b8f6569b.png)\n\nThere are two possibilities for the position of line $PQ$, namely, (a) both $P$ and $Q$ lie on minor arc $\\widehat{P_1Q_1}$; (b) one of $P$ and $Q$ lies on minor arc $\\widehat{DQ_1}$ and the other lies on minor arc $\\widehat{P_1B}$. If $G$ lies on segment $RG_1$, then $Q$ lies on minor arc $\\widehat{DQ}$, and we must have (b). But in this case, $P$ must lie on minor arc $\\widehat{Q_1P_1}$, violating (b). If $G$ lies on segment $G_1S$, then $P$ must lie on minor arc $\\widehat{P_1B}$, and again we must have (b). But in this case, $Q$ must lie on minor arc $\\widehat{Q_1C}$, violating (b). In every case, we have a contradiction. Hence our assumption was wrong, and ray $BG$ bisects $\\angle CBD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18974,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 2.1, in the acute triangle $ABC$, $AB < AC$, $I$ is the incentre, and $J$ is the excentre relative to vertex $A$. Points $X, Y$ lie on the minor arcs $\\widehat{AB}$, $\\widehat{AC}$ of the circumcircle of the triangle $ABC$, respectively, such that $\\angle AXI = \\angle AYJ = 90^\\circ$. Point $K$ is on the extension of $BC$ beyond $C$, such that $KI = KJ$. Prove: $AK$ bisects the segment $XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "We give three proofs as follows.\n\n**Solution 1** As illustrated in Fig. 2.2, let $IJ$ cross the circumcircle of $\\triangle ABC$ at $M$ and $A'$ be the antipode of $A$.\n\nSince $\\angle AXI = \\angle AYJ = 90^\\circ$, the lines $XI$ and $YJ$ pass through $A'$. From properties of incentre and excentre, $MI = MJ = MB$, and together with $KI = KJ$, it follows that $KM \\perp IJ$, and $A'$ lies on $KM$. This implies $A'I = A'J$, $\\angle AIX = \\angle A'IJ = \\angle A'JI = \\angle AJY$. Since $\\angle AXI = \\angle AYJ$, we have $\\triangle AIX \\sim \\triangle AJY$, and $\\angle XAM = \\angle XAI = \\angle YAJ = \\angle YAM$, $\\overrightarrow{XM} = \\overrightarrow{YM}$. Also, $\\overrightarrow{BM} = \\overrightarrow{CM}$, and thus $\\overrightarrow{BX} = \\overrightarrow{CY}$, $XY \\parallel BC$. Moreover, from\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p314_data_60de0c0150.png)\n\nproperties of similar triangles, we obtain\n\n$$\n\\frac{AX}{AY} = \\frac{AI}{AJ}. \\qquad (1)\n$$\n\nFrom $\\overline{BM} = \\overline{CM}$, it follows that $\\angle MA'B = \\angle MCB = \\angle MBK$, $\\triangle MA'B \\sim MBK$, and hence $MA' \\cdot MK = MB^2 = MI \\cdot MJ$. Now, $\\triangle MIA' \\sim \\triangle MKJ$, which gives $IA' \\perp KJ$, or $XA' \\perp KJ$. Notice $AX \\perp XA'$, hence $AX \\parallel KJ$. In a similar way, we derive $AY \\parallel KI$. Together, they imply\n\n$$\n\\angle XAK = 180^\\circ - \\angle AKJ, \\quad \\angle YAK = \\angle AKI. \\qquad (2)\n$$\n\nLet $AK$ and $XY$ meet at $L$. From (1) and (2), we deduce that\n\n$$\n\\frac{XL}{YL} = \\frac{S_{\\triangle AXL}}{S_{\\triangle AYL}} = \\frac{AX}{AY} \\cdot \\frac{\\sin \\angle XAK}{\\sin \\angle YAK} = \\frac{AI}{AJ} \\cdot \\frac{\\sin \\angle AKJ}{\\sin \\angle AKI} = \\frac{AI}{AJ} \\cdot \\frac{S_{\\triangle AKJ}}{S_{\\triangle AKI}} = 1,\n$$\n\nand therefore, $AK$ bisects the segment $XY$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18975,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G = (V, E)$ be a simple graph with vertex set $V$ and edge set $E$, and assume that $|V| = n$. A map $f: V \\rightarrow \\mathbb{Z}$ is said to be *good* if $f$ satisfies:\n\n$$\n(1) \\quad \\sum_{v \\in V} f(v) = |E|;\n$$\n\n(2) If one colors arbitrarily some vertices into red, there exists a red vertex $v$ such that $f(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored into red.\n\nLet $m(G)$ be the number of good maps $f$. Show that if each vertex of $V$ is adjacent to at least one other vertex, then $n \\leq m(G) \\leq n!$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nGiven an ordering $\\tau = (v_1, v_2, \\ldots, v_n)$ on the vertices in $V$, we associate a map $f_\\tau: V \\to \\mathbb{Z}$ as follows: $f_\\tau(v)$ is equal to the number of vertices in $V$ that are ordered preceding $v$. We claim that $f_\\tau$ is good.\n\nEach edge is counted exactly once in $\\sum_{v \\in V} f_\\tau(v)$: for an edge $e \\in E$ with vertices $u, v \\in V$ such that $u$ is ordered before $v$ in $\\tau$, $e$ is counted once in $f_\\tau(v)$. Thus,\n\n$$\n\\sum_{v \\in V} f_{\\tau}(v) = |E|.\n$$\n\nFor any nonempty subset $A \\subseteq V$ of all red vertices, choose $v \\in A$ with the most preceding orderings in $\\tau$. Then by definition, $f_\\tau(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored into red. We have verified that $f_\\tau$ is good.\n\nConversely, given any good map $f: V \\to \\mathbb{Z}$, we claim that $f = f_\\tau$ for some ordering $\\tau$ of $V$.\n\nFirst, let the red vertex set $A = V$. By condition (2), there exists $v \\in A$ such that $f(v) \\leq 0$, and denote one of such vertices by $v_1$. Assuming that we have already chosen $v_1, \\ldots, v_k$ from $V$, if $k < n$, set the red vertex set $A = V \\setminus \\{v_1, \\ldots, v_k\\}$. By condition (2), there exists $v \\in A$ such that $f(v)$ is less than or equal to the number of vertices in $\\{v_1, \\ldots, v_k\\}$ that are adjacent to $v$. Denote one of such vertices by $v_{k+1}$. Continuing in this way, we order the vertices by $\\tau = (v_1, v_2, \\ldots, v_n)$. By construction, we have $f(v) \\leq f_r(v)$ for any $v \\in V$. By condition (1),\n\n$$\n|E| = \\sum_{v \\in V} f(v) \\leq \\sum_{v \\in V} f_r(v) = |E|,\n$$\n\nand therefore $f(v) = f_r(v)$ for any $v \\in V$.\n\nWe have shown that for any ordering $\\tau$, $f_r$ is good, and any good map $f$ is $f_r$ for some $\\tau$. Since the number of orderings on $V$ is $n!$, we see that $m(G) \\leq n!$ (note that two distinct orderings may result in the same map).\n\nNext, we prove that $n \\leq m(G)$. Assume at the moment that $G$ is connected. Pick arbitrarily $v_1 \\in V$. By connectivity, we may choose $v_2 \\in V \\setminus \\{v_1\\}$ such that $v_2$ is adjacent to $v_1$, and again we may choose $v_3 \\in V \\setminus \\{v_1, v_2\\}$ such that $v_3$ is adjacent to at least one of $v_1, v_2$. Continuing in this way, we get an ordering $\\tau = (v_1, v_2, \\ldots, v_n)$ such that $v_k$ is adjacent to at least one of the vertices preceding it under $\\tau$, for any $2 \\leq k \\leq n$. Thus, $f_r(v_1) = 0$ and $f_r(v_k) > 0$ for $2 \\leq k \\leq n$. Since $v_1$ may be arbitrary, we have at least $n$ good maps.\n\nIn general, if $G$ is a union of its connected components $G_1, \\ldots, G_k$, since each vertex is adjacent to at least another vertex, each component has at least two vertices. Denote by $n_1, \\ldots, n_k \\geq 2$ the number of vertices of these components. For each $G_i$, we have at least $n_i$ good maps on its vertices, $i = 1, \\ldots, k$. It is easy to see that patching good maps on $G_i$'s gives the result.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18976,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside $\\triangle ABC$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $AC$ and $AB$, respectively. Let the lines $BP$ and $CP$ meet the circumcircle of $\\triangle ABC$ again at points $B_1$ and $C_1$ ($B_1 \\neq B$, $C_1 \\neq C$), respectively. Let $R$ and $r$ denote the radii of the circumcircle and incircle of $\\triangle ABC$, respectively. Prove that\n\n$$\n\\frac{EF}{B_1C_1} \\ge \\frac{r}{R},\n$$\n\nand, when the equality holds, determine completely the positions of $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "As seen in Fig. 4.1, let $PD \\perp BC$ with intersection point $D$. Extend $AP$ to meet the circumcircle of $\\triangle ABC$ at $A_1$, and connect $DE$, $DF$, $A_1B_1$, $A_1C_1$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p257_data_6f56e7fe0d.png)\n\nSince $P, D, B, F$ are concyclic, $\\angle PDF = \\angle PBF$; since $P, D, C, E$ are concyclic, $\\angle PDE = \\angle PCE$. Then\n\n$$\n\\begin{aligned}\n\\angle FDE &= \\angle PDF + \\angle PDE = \\angle PBF + \\angle PCE \\\\\n            &= \\angle AA_1B_1 + \\angle AA_1C_1 = \\angle C_1A_1B_1.\n\\end{aligned}\n$$\n\nSimilarly, $\\angle DEF = \\angle A_1B_1C_1$. Therefore, $\\triangle DEF \\sim \\triangle A_1B_1C_1$.\n\nThe circumradius of $\\triangle A_1B_1C_1$ is $R$, and let the circumradius of $\\triangle DEF$ be $R'$. Then $\\frac{EF}{B_1C_1} = \\frac{R'}{R}$.\n\nLet the incenter of $\\triangle DEF$ be $O'$. Connect $AO'$, $BO'$, $CO'$ (see Fig. 4.2).\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p257_data_a5251f37fe.png)\n\n$$\n\\begin{align*}\nS_{\\triangle ABC} &= \\frac{(AB + BC + CA) \\cdot r}{2} \\\\\n&= S_{\\triangle O'AB} + S_{\\triangle O'BC} + S_{\\triangle O'AC} \\\\\n&\\le \\frac{AB \\cdot O'F}{2} + \\frac{BC \\cdot O'D}{2} + \\frac{CA \\cdot O'E}{2} \\\\\n&= \\frac{(AB + BC + CA) \\cdot R'}{2}.\n\\end{align*}\n$$\n\nTherefore, $R' \\geq r$. Equality holds if and only if\n\n$$\nO'D \\perp BC, \\quad O'E \\perp CA, \\quad O'F \\perp AB,\n$$\n\nwhich implies $P = O'$, i.e., $P$ is the incenter of $\\triangle ABC$.\n\nTherefore, $\\frac{EF}{B_1C_1} \\ge \\frac{r}{R}$, and equality holds if and only if $P$ is the incenter of $\\triangle ABC$.\n\nThe proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18977,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $3 \\times 9$ array $A$ with each cell containing a positive integer, we say a $m \\times n$ ($1 \\le m \\le 3$, $1 \\le n \\le 9$) subarray of $A$ is a \"good rectangle\" if the sum of the numbers in its cells is a multiple of $10$, and call a $1 \\times 1$ cell of $A$ \"bad\" if it is not contained in any \"good rectangle\". Find the maximum number of \"bad cells\" in $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first claim that the number of \"bad cells\" in $A$ is no more than $25$. Otherwise, there will be at most one cell in $A$ that is not \"bad\". Without loss of generality, we assume the cells in the first row of $A$ are all \"bad\". Then let the numbers from top to bottom in the $i$th column be $a_i, b_i, c_i$ ($i = 1, 2, \\dots, 9$) in turn, and define\n\n$$\nS_k = \\sum_{i=1}^{k} a_i, \\quad T_k = \\sum_{i=1}^{k} (b_i + c_i), \\quad k = 1, 2, \\dots, 9,\n$$\n\nwith $S_0 = T_0 = 0$. We are going to prove that three number groups $S_0, S_1, \\dots, S_9$, $T_0, T_1, \\dots, T_9$, and $S_0 + T_0, S_1 + T_1, \\dots, S_9 + T_9$ each form a complete set of residues modulo $10$:\n\nIf there exist $m, n$, $0 \\le m < n \\le 9$ such that $S_m \\equiv S_n \\pmod{10}$, then\n\n$$\n\\sum_{i=m+1}^{n} a_i = S_n - S_m \\equiv 0 \\pmod{10},\n$$\n\nwhich means that the cells in the first row and from columns $m+1$ to $n$ form a \"good rectangle\". But it is a contradiction to the assumption that the cells in the first row are all \"bad\".\n\nIf there exist $m, n$, $0 \\le m < n \\le 9$ such that $T_m \\equiv T_n \\pmod{10}$, then\n\n$$\n\\sum_{i=m+1}^{n} (b_i + c_i) = T_n - T_m \\equiv 0 \\pmod{10}.\n$$\n\nSo the cells ranging from rows $2$ to $3$ and columns $m+1$ to $n$ form a \"good rectangle\", which means there are at least two cells that are not \"bad\". But it is also a contradiction.\n\nIn a similar way, we can also prove that there are no $m, n$, $0 \\le m < n \\le 9$ such that\n\n$$\nS_m + T_m \\equiv S_n + T_n \\pmod{10}.\n$$\n\nTherefore, we have\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{9} S_k &\\equiv \\sum_{k=0}^{9} T_k \\equiv \\sum_{k=0}^{9} (S_k + T_k) \\equiv 0 + 1 + 2 + \\cdots + 9 \\\\\n&\\equiv 5 \\pmod{10}.\n\\end{aligned}\n$$\n\nThen\n\n$$\n\\sum_{k=0}^{9} (S_k + T_k) \\equiv \\sum_{k=0}^{9} S_k + \\sum_{k=0}^{9} T_k \\equiv 5 + 5 \\equiv 0 \\pmod{10}.\n$$\n\nIt is again a contradiction! Therefore, the number of \"bad cells\" in $A$ is no more than $25$.\n\nOn the other hand, we can construct a $3 \\times 9$ array in the following and check that each cell in it that does not contain number $10$ is \"bad\".\n\n![]( <table><tr><td>1</td><td>1</td><td>1</td><td>2</td><td>1</td><td>1</td><td>1</td><td>1</td><td>10</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>1</td><td>1</td><td>1</td><td>10</td><td>1</td><td>1</td><td>1</td><td>1</td><td>2</td></tr></table> )\n\nTherefore, we find out that the maximum number of \"bad cells\" in $A$ is $25$.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18978,
    "subject": "Mathematics (Olympiad)",
    "question": "Let's call a positive integer square-free if it's not divisible by $p^2$ for any prime $p$. You are given a square-free integer $n > 1$, which has precisely $d$ positive divisors. What is the largest number of divisors of $n$ you can choose so that for any two of them, denoted $a$ and $b$, the number $a^2 + ab - n$ is not a perfect square?\n\n![](Oleksii_Masalitin.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n > 1$ is square-free, it cannot be a perfect square. All divisors of $n$ can be paired as $(t_1, t_2), (t_3, t_4), \\ldots, (t_{d-1}, t_d)$ such that the product in each pair is $n$. If we choose both $a$ and $b$ from the same pair, then $a^2 + ab - n = a^2$, which is a perfect square—a contradiction. Thus, we can select at most one number from each pair, so at most $\\frac{d}{2}$ divisors in total.\n\nTo show that $\\frac{d}{2}$ is always achievable, consider any prime divisor $p$ of $n$ and choose all divisors of $n$ that are divisible by $p$. There are $\\frac{d}{2}$ such divisors. For any $a = kp$ and $b = lp$ from this group (with $n = pt$ and $k, l, t$ not divisible by $p$),\n\n$$a^2 + ab - n = p(pk^2 + pkl - t)$$\n\nThe bracketed term is not divisible by $p$ since $t$ is not divisible by $p$. Therefore, $a^2 + ab - n$ is divisible by $p$ but not by $p^2$, so it cannot be a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18979,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive numbers $a, b, c$, prove the inequality:\n\n$$\n\\sqrt{a^2 + bc} + \\sqrt{b^2 + ca} + \\sqrt{c^2 + ab} \\geq \\sqrt{ab + bc} + \\sqrt{bc + ca} + \\sqrt{ca + ab}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is symmetric; interchanging any two variables does not change its form. Thus, without loss of generality, assume $a \\geq b \\geq c$.\n\nMoving all terms to one side and grouping, the inequality becomes:\n\n$$\n(\\sqrt{a^2 + bc} - \\sqrt{ca + ab}) + (\\sqrt{b^2 + ca} - \\sqrt{ab + bc}) + (\\sqrt{c^2 + ab} - \\sqrt{bc + ca}) \\geq 0.\n$$\n\nMultiplying each term by its conjugate, we need to prove:\n\n$$\n\\frac{(a-b)(a-c)}{\\sqrt{a^2+bc}+\\sqrt{ca+ab}} + \\frac{(b-c)(b-a)}{\\sqrt{b^2+ca}+\\sqrt{ab+bc}} + \\frac{(c-a)(c-b)}{\\sqrt{c^2+ab}+\\sqrt{bc+ca}} \\geq 0.\n$$\n\nSince $\\frac{(a-b)(a-c)}{\\sqrt{a^2+bc}+\\sqrt{ca+ab}} \\geq 0$, it suffices to show:\n\n$$\n\\frac{(a-c)(b-c)}{\\sqrt{c^2 + ab} + \\sqrt{bc + ca}} \\geq \\frac{(b-c)(a-b)}{\\sqrt{b^2 + ca} + \\sqrt{ab + bc}}.\n$$\n\nIf $b-c=0$, both sides are zero. Otherwise, divide both sides by $b-c > 0$. Since $a-c \\geq b-c \\geq 0$, it suffices to prove:\n\n$$\n\\frac{1}{\\sqrt{c^2 + ab} + \\sqrt{bc + ca}} \\geq \\frac{1}{\\sqrt{b^2 + ca} + \\sqrt{ab + bc}}\n$$\nwhich is equivalent to\n$$\n\\sqrt{b^2 + ca} + \\sqrt{ab + bc} \\geq \\sqrt{c^2 + ab} + \\sqrt{bc + ca}.\n$$\n\nThe last inequality holds since\n$$\nb^2 + ca \\geq bc + ca \\quad \\text{if} \\quad ab + bc \\geq c^2 + ab.\n$$\n\nThus, the inequality is proven.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18980,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a white $6 \\times 2019$ table. Andrii and Arsenii are playing the following game: one after another (starting with Andrii) a player colors one of the $1 \\times 1$ cells in black. Moreover, one's turn cannot create a $3 \\times 3$ square that contains two black cells. Whoever doesn't have a turn loses. Who will win if both want to win the game? And what is the strategy?\n\n![](images/UkraineMO2019_booklet_p7_data_97a207af80.png)\n\n*Fig. 3*",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Arsenii wins.\n\n**Solution.** We will show that there exists a strategy such that Arsenii always has a turn after Andrii's turn. We will split all the cells into *friendly pairs*. Two $1 \\times 1$ cells make a friendly pair if they are in the same column and there are exactly two $1 \\times 1$ cells in-between. Then Arsenii colors the cell that makes a friendly pair with a cell that Andrii colored during his last turn.\n\nWe want to show that Arsenii always has such a turn. If Andrii colored some cell $P$, then a cell $V$ that makes a friendly pair with $P$ is white. Moreover, suppose the cell $V$ cannot be colored in black because there already is another black cell $Y$ in some $3 \\times 3$ square that is located either in the very top or the very bottom of the table. Then there also exists another similar $3 \\times 3$ square either in the very top or the very bottom that contains cells $P$ and $Y$, both of which are black, where $Y$ is a friendly pair of $V$ (see Fig. 3). This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18981,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1A_2\\cdots A_{101}$ be a regular $101$-gon, and color every vertex red or blue. Let $N$ be the number of obtuse triangles satisfying the following conditions:\n\n- The three vertices of the triangle must be vertices of the $101$-gon.\n- Both the vertices with acute angles have the same color.\n- The vertex with the obtuse angle has a different color.\n\n(1) Find the largest possible value of $N$.\n\n(2) Find the number of ways to color the vertices such that maximum $N$ is achieved. (Two colorings are different if for some $A_i$ the colors are different on the two coloring schemes.)\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For an obtuse triangle $A_{i-a}A_iA_{i+b}$ (where $A_i$ is the vertex of the obtuse angle, i.e., $a+b \\le 50$), these three vertices satisfy the conditions if and only if\n\n$$\n(x_i - x_{i-a})(x_i - x_{i+b}) = 1\n$$\n\notherwise $0$, with subscripts modulo $101$. Thus,\n\n$$\nN = \\sum_{i=1}^{101} \\sum_{(a, b)} (x_i - x_{i-a})(x_i - x_{i+b})\n$$\n\nwhere $\\sum_{(a, b)}$ is over all positive integer pairs $(a, b)$ with $a+b \\le 50$. There are $49 + 48 + \\cdots + 1 = 1225$ such pairs. Expanding,\n\n$$\n\\begin{align*}\nN &= \\sum_{i=1}^{101} \\sum_{(a, b)} (x_i^2 - x_i x_{i-a} - x_i x_{i+b} + x_{i-a} x_{i+b}) \\\\\n&= 1225 \\sum_{i=1}^{101} x_i^2 + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (k-1-2(50-k)) x_i x_{i+k} \\\\\n&= 1225n + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (3k-101) x_i x_{i+k}.\n\\end{align*}\n$$\n\nHere $n$ is the number of blue vertices. For any two vertices $A_i$ and $A_j$, $1 \\le i, j \\le 101$, let\n\n$$\nd(A_i, A_j) = d(A_j, A_i) = \\min\\{ |j - i|, 101 - |j - i| \\}\n$$\n\nLet $B \\subseteq \\{A_1, A_2, \\dots, A_{101}\\}$ be the set of blue vertices. Then,\n\n$$\nN = 1225n - 101\\binom{n}{2} + 3 \\sum_{\\{P, Q\\} \\subseteq B} d(P, Q)\n$$\n\nAssume $n$ is even (otherwise, swapping all colors does not change $N$). Write $n = 2t$, $0 \\le t \\le 50$. Renumber blue vertices $P_1, \\dots, P_{2t}$ clockwise. Then,\n\n$$\n\\sum_{\\{P, Q\\} \\subseteq B} d(P, Q) = \\sum_{i=1}^{t} d(P_i, P_{i+t}) + \\frac{1}{2} \\sum_{i=1}^{t} \\sum_{j=1}^{t-1} [d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i)] \\le 50t + \\frac{101}{2}t(t-1)\n$$\n\n(using $d(P_i, P_{i+t}) \\le 50$ and $d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \\le 101$).\n\nCombining, we get\n\n$$\nN \\le 1225n - 101\\binom{n}{2} + 3\\left(50t + \\frac{101}{2}t(t-1)\\right) = -\\frac{101}{2}t^2 + \\frac{5099}{2}t\n$$\n\nThe right-hand side attains its maximum when $t = 25$, i.e., $n = 50$.\n\nThus, the largest possible value of $N$ is\n\n$$\nN_{\\max} = -\\frac{101}{2} \\times 25^2 + \\frac{5099}{2} \\times 25 = 63625\n$$\n\nThe number of ways to color the vertices to achieve this maximum is $\\binom{101}{50}$ (choose any $50$ vertices to be blue).\n",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18982,
    "subject": "Mathematics (Olympiad)",
    "question": "For what positive integer $n$ can the numbers $1, 2, \\ldots, 2n$ be divided into two groups of $n$ numbers each, such that the product of the numbers in one group equals the sum of the numbers in the other group?",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.**\n\nLet us analyze for which $n$ this division is possible.\n\nLet the two groups be $A$ and $B$, each with $n$ numbers from $1$ to $2n$. We seek $\\prod_{a \\in A} a = \\sum_{b \\in B} b$.\n\nConsider the sum of the $n$ largest numbers: $(n+1)+(n+2)+\\cdots+2n = \\frac{1}{2}n(3n+1)$, and the product of the $n$ smallest numbers: $1 \\cdot 2 \\cdot \\cdots \\cdot n = n!$.\n\nThe sum can only decrease and the product can only increase if we swap numbers between groups, so the only possible $n$ must satisfy:\n\n$$\n\\frac{1}{2}n(3n+1) \\geq n!\n$$\n\nTesting small $n$:\n- $n=1$: $1 \\geq 1$ (but $1 \\neq 1$ for the required division)\n- $n=2$: $5 \\geq 2$ (but no such division exists)\n- $n=3$: $12 \\geq 6$; indeed, $3+4+5=12$ and $1 \\cdot 2 \\cdot 6=12$\n\nFor $n=4$, $5+6+7+8=26$ and $1 \\cdot 2 \\cdot 3 \\cdot 4=24$, but swapping numbers only increases the product and decreases the sum, so no solution exists for $n=4$ or higher.\n\n**Conclusion:** The only possible value is $n=3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 18983,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$ and real numbers $x_1, x_2, \\dots, x_n$ in the interval $[0, 1]$, prove that there exist real numbers $a_0, a_1, \\dots, a_n$ satisfying simultaneously the following conditions:\n\n$$\n\\begin{aligned}\n& a_0 + a_n = 0; \\\\\n& |a_i| \\le 1, \\text{ for every } i = 0, 1, \\dots, n; \\\\\n& |a_i - a_{i-1}| = x_i, \\text{ for every } i = 1, 2, \\dots, n.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any $a \\in [0, 1)$, define a sequence $\\{a_i\\}_{i=0}^n$ generated by $a$ as follows: $a_0 = a$; for $1 \\le i \\le n$, set $a_i = a_{i-1} - x_i$ if $a_{i-1} \\ge 0$, and $a_i = a_{i-1} + x_i$ if $a_{i-1} < 0$.\n\nSet $f(a) = a_n$. By induction, $|a_i| \\le 1$ for every $0 \\le i \\le n$.\n\nIf there exists $a \\in [0, 1)$ such that $f(a) = -a$, then the sequence generated by $a$ (with $a_0 = a, a_1, \\dots, a_n = -a$) satisfies all the required conditions.\n\nTo show such $a$ exists, note that for any $a \\in [0, 1)$, a 'breaking point' is a value where some $a_i = 0$ in its sequence. There are finitely many breaking points, all of the form $\\sum_{i=1}^n t_i x_i$ with $t_i \\in \\{-1, 0, 1\\}$.\n\nLabel the breaking points in increasing order: $0 = b_1 < b_2 < \\cdots < b_m < 1$.\n\nFor $a \\in [b_k, b_{k+1})$, $f(a) = f(b_k) + (a - b_k)$. This is shown by comparing the sequences generated from $b_k$ and $b_{k+1}$, and constructing a related sequence $\\{s_i\\}$ as described.\n\nIf $f(b_k) = -b_k$ for some $k$, we are done. If $f(b_k) = b_k$ for some $k$, then reversing the sequence after the first zero yields a sequence ending at $-b_k$, which also works.\n\nIf $|f(b_k)| \\ne b_k$ for all $k$, consider two cases:\n\n*Case 1:* $|f(b_m)| < b_m$. Since $|f(b_1)| > b_1 = 0$, there exists $k$ such that $|f(b_k)| > b_k$, $|f(b_{k+1})| < b_{k+1}$. Since $f(b_k) + f(b_{k+1}) = b_k - b_{k+1}$, we have $f(b_k) \\le -b_k$ and $f(b_k) > b_k - 2b_{k+1}$.\n\nBy the intermediate value property, there exists $a \\in (b_k, b_{k+1})$ with $f(a) = -a$.\n\nThus, such a sequence $\\{a_i\\}$ exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18984,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $VABC$ be a regular pyramid with base $ABC$ and center $O$. Denote $I$ and $H$ as the incenter and orthocenter of triangle $VBC$, respectively. Suppose $AH = 3OI$. Find the measure of the angle between a lateral edge of the pyramid and the plane of the base.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of 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 $AC \\perp OV$, it follows that $AC \\perp (BOV)$, hence $VB \\perp AC$. Together with $VB \\perp CH$, this leads to $VB \\perp (ACH)$, so $AH \\perp VB$. Similarly, $AH \\perp VC$, therefore $AH \\perp (VBC)$, so $AH \\perp VM$.\n\nLet $I'$ be the projection of $O$ onto the plane $(VBC)$; then $I'$ lies on $VM$ and $OI' \\parallel AH$. This gives\n\n![](images/RMC2011_2_p23_data_508e534488.png)\n\n$$\n\\frac{OI'}{AH} = \\frac{MO}{MA} = \\frac{1}{3},\n$$\n\nso $AH = 3OI'$. This shows that $OI$ equals the distance from $O$ to $(VBC)$, therefore $I = I'$.\n\nSince $OI \\perp (VBC)$ and $I$ is the incenter of triangle $VBC$, $O$ has equal distances to the lines $VB$ and $BC$.\n\nLet $J$ be the projection of $O$ onto the line $VB$. Then $OJ = OM$, so the right triangles $OJB$ and $OMB$ are congruent.\n\nTherefore, the measure of the angle between $VB$ and the plane $(ABC)$ is $\\angle VBO = \\angle MBO = 30^{\\circ}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 18985,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $\\angle B = \\angle C$. Let the circumcentre be $O$ and the orthocentre be $H$. Prove that the centre of the circle $BOH$ lies on the line $AB$.\n\n[The circumcentre of a triangle is the centre of its circumcircle. The orthocentre of a triangle is the point where its three altitudes meet.]\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AHD$ be the altitude from $A$. Since $\\angle B = \\angle C$, $O$ lies on this altitude.\n\nSince $O$ is the centre of the circle through $A$, $B$ and $C$, $\\angle AOB = 2\\angle ACB = 2\\angle C$.\n\nLet $P$ be on $AB$ such that $\\angle BOP = \\angle BAO$. Then since we also have $\\angle OBP = \\angle ABO$, the triangles $BOP$ and $BAO$ are similar. $OA$ and $OB$ are radii of circle $ABC$, and so are equal; therefore $PB = PO$. Now consider\n\n![](images/V_Britanija_2009_p21_data_9f635fe664.png)\n\nthe circle centre $P$ going through $O$ and $B$. We would like to show that $H$ lies on this circle; we would then have that the circumcentre of $\\triangle BOH$ is $P$ which lies on $AB$.\n\nBecause $\\angle BPO = 2\\angle C$, the chord $BO$ subtends an angle of $\\angle C$ at the circumference on the same side of $BO$ as $P$.\n\nTherefore $BO$ subtends an angle of $180^\\circ - \\angle C$ at the circumference on the same side of $BO$ as $H$ (since for $\\angle B > 60^\\circ$, $H$ and $P$ are on opposite sides of $BO$), and we only need show that $\\angle BHO = 180^\\circ - \\angle C$.\n\nBut $\\angle BHO = \\angle EHD$, and $HECD$ is cyclic because $\\angle HEC - \\angle HDC = 90^\\circ$. Therefore $\\angle BHO = 180^\\circ - \\angle ECD = 180^\\circ - \\angle C$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18986,
    "subject": "Mathematics (Olympiad)",
    "question": "We denote the set of all nonzero integers and the set of all nonnegative integers by $\\mathbb{Z}^*$ and $\\mathbb{N}_0$, respectively. Find all functions $f: \\mathbb{Z}^* \\to \\mathbb{N}_0$ for which the following two conditions hold:\n\n1. For each $a, b \\in \\mathbb{Z}^*$ such that $a + b \\in \\mathbb{Z}^*$, it holds that $f(a + b) \\ge \\min\\{f(a), f(b)\\}$.\n2. For each $a, b \\in \\mathbb{Z}^*$, it holds that $f(ab) = f(a) + f(b)$.",
    "options": [],
    "answer": "See solution",
    "solution": "One trivial solution is the constant function $f \\equiv 0$.\n\nLet $f$ be a nontrivial function for which the conditions (1) and (2) hold. We will show that there exists a natural number $c$ and a prime number $p$ for which it holds that $f(a) = c v_p(a)$ for each $a \\in \\mathbb{Z}^*$, where $v_p(a)$ is the exponent of $p$ in the canonical factorization of $a$.\n\nFirst, note that $f(1) = f(-1) = 0$:\n\n$$\nf(1) = f(1 \\cdot 1) = f(1) + f(1), \\quad f(1) = f((-1) \\cdot (-1)) = f(-1) + f(-1)\n$$\n\nFrom this and from (2), it follows that there exists a prime number $p$ for which $f(p) \\ne 0$. For $c := f(p)$, we will show that $f(a) = c v_p(a)$ holds for every $a \\in \\mathbb{Z}^*$. Namely, for each prime $q \\ne p$, there exist nonzero integers $a, \\beta$ for which $1 = ap + \\beta q$, so that the inequality $0 = f(ap + \\beta q) \\ge \\min\\{f(ap), f(\\beta q)\\}$ holds. From\n\n$$\nf(ap) = f(a) + f(p) \\ge f(p) = c \\ne 0\n$$\n\nit follows that $f(\\beta q) = 0$ and $f(q) = 0$. Let $a = \\pm p^k q^\\beta r^\\gamma \\dots$ be the canonical factorization of $a$; then\n\n$$\nf(a) = f(\\pm p^k) + f(q^\\beta) + f(r^\\gamma) + \\dots = f(\\pm 1) + f(p^k) = k f(p) = c v_p(a)\n$$\n\nIt remains to note that each such function satisfies the conditions (1) and (2), and therefore it represents a nontrivial solution to the given problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18987,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 < a_2 < \\dots < a_n$ be positive integers such that\n\n$$\n\\frac{a_2}{a_1}, \\frac{a_3}{a_2}, \\dots, \\frac{a_n}{a_{n-1}} \\in (1, 2)\n$$\n\nand\n\n$$\n\\frac{a_n}{a_1} \\in (n, n+1).\n$$\n\nFind such numbers $(a_1, \\dots, a_n)$ that satisfy these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the following lemma:\n\n**Lemma 1.** Let $A$ and $B$ be rational numbers and $p$ and $q$ be distinct prime numbers. Then, $\\gcd(A \\cdot N! + p, B \\cdot N! + q) = 1$ for all large enough positive integers $N$.\n\n*Proof.* Let $(A, B) = (a/b, c/d)$. Let $D = \\gcd(A \\cdot N! + p, B \\cdot N! + q)$. It follows that $D \\mid acp - bdq$. Thus, $D < \\max(acp, bdq)$. Choose $N > 2 \\cdot \\max(acp, bdq)$; it follows that $D \\mid A \\cdot N!$ and $D \\mid B \\cdot N!$, yielding $D = 1$. This proves the lemma.\n\nNow, construct $a_1, \\dots, a_n$ as follows. Let $p_1, \\dots, p_n$ be the first $n$ primes and $A_1, \\dots, A_n$ be arbitrary rational numbers such that $A_i \\in (1, 2)$ and $A_1 A_2 \\cdots A_n \\in (n, n+1)$. Then, choose a large enough $N$ and put $a_i = A_i \\cdot N! + p_i$ for $i = 1, \\dots, n$. These numbers satisfy the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18988,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any $n > 1$ coprime positive integers $a_1, a_2, \\dots, a_n$, let $A = a_1 + a_2 + \\dots + a_n$. Define $d_i = \\gcd(A, a_i)$ for $i = 1, 2, \\dots, n$. Let $D_i$ be the greatest common divisor of $\\{a_1, a_2, \\dots, a_n\\} \\setminus \\{a_i\\}$ for $i = 1, 2, \\dots, n$.\n\nFind the minimum value of\n$$\n\\prod_{i=1}^n \\frac{A - a_i}{d_i D_i}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $D_1 = \\gcd(a_2, a_3, \\dots, a_n)$ and $d_2 = \\gcd(a_2, A) = \\gcd(a_2, a_1 + a_2 + \\dots + a_n)$.\n\nLet $d = \\gcd(D_1, d_2)$. Then $d \\mid a_2, d \\mid a_3, \\dots, d \\mid a_n$, and $d \\mid a_1 + a_2 + \\dots + a_n$, so $d \\mid a_1$. Thus,\n$$\nd \\mid \\gcd(a_1, a_2, \\dots, a_n).\n$$\nSince $a_1, a_2, \\dots, a_n$ are coprime, $d = 1$.\n\nNote that $D_1 \\mid a_2$, $D_2 \\mid a_2$, and $\\gcd(D_1, D_2) = 1$. We have $D_1 d_2 \\mid a_2$, so $D_1 d_2 \\leq a_2$. Similarly, $D_2 d_3 \\leq a_3$, ..., $D_n d_1 \\leq a_1$. Therefore,\n$$\n\\begin{aligned}\n\\prod_{i=1}^{n} d_i D_i &= (D_1 d_2) (D_2 d_3) \\cdots (D_n d_1) \\\\\n&\\leq a_2 a_3 \\cdots a_n a_1 \\\\\n&= \\prod_{i=1}^{n} a_i.\n\\end{aligned}\n$$\n\nNow consider\n$$\n\\begin{aligned}\n\\prod_{i=1}^{n} (A - a_i) &= \\prod_{i=1}^{n} \\left( \\sum_{j \\neq i} a_j \\right) \\\\\n&\\geq \\prod_{i=1}^{n} \\left( (n-1) \\left( \\prod_{j \\neq i} a_j \\right)^{1/(n-1)} \\right) \\\\\n&= (n-1)^n \\prod_{i=1}^{n} a_i.\n\\end{aligned}\n$$\n\nCombining the above,\n$$\n\\prod_{i=1}^{n} \\frac{A - a_i}{d_i D_i} \\geq (n-1)^n.\n$$\n\nIf $a_1 = a_2 = \\dots = a_n = 1$, then\n$$\n\\prod_{i=1}^{n} \\frac{A - a_i}{d_i D_i} = (n-1)^n.\n$$\n\nTherefore, the minimum value is $(n-1)^n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 18989,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Let $A_j = \\{i : 2^{j-1} \\le a_i < 2^j\\}$ for $j = 1, 2, \\dots, 11$. 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$$\n\\begin{aligned}\n\\sum_{i \\in A_j} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) &= \\sum_{i \\in A_j} \\left( \\frac{a_1 + a_2 + \\dots + a_{i-1}}{a_i} + 1 \\right) \\\\\n&> \\sum_{k=0}^{|A_j|-1} \\frac{k \\cdot 2^{j-1}}{2^j} + |A_j| = \\frac{1}{2} \\sum_{k=0}^{|A_j|-1} k + |A_j| \\\\\n&= \\frac{1}{4} |A_j|(|A_j| - 1) + |A_j| = \\frac{1}{4} |A_j|(|A_j| + 3);\n\\end{aligned}\n$$\nThe inequality comes from the fact that $a_i < 2^j$ and, ordering indices in $A_j$ increasingly, the sum $a_1 + a_2 + \\dots + a_{i-1}$ increases by at least $2^{j-1}$ every time $i$ passes from one index to the next, while running through $A_j$.\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$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) &= \\sum_{j=1}^{11} \\sum_{i \\in A_j} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) \\\\\n&> \\frac{1}{4} \\sum_{j=1}^{11} |A_j|(|A_j| + 3) = \\frac{1}{4} \\left( \\sum_{j=1}^{11} |A_j|^2 + 3n \\right) \\\\\n&\\ge \\frac{1}{4} \\left( \\frac{n^2}{11} + 3n \\right) = \\frac{1}{44} n(n + 33), \\text{ by Cauchy-Schwarz.}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18990,
    "subject": "Mathematics (Olympiad)",
    "question": "The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. Find the difference of the arithmetic progression if its first term is a solution of the equation\n\n$$\nx^2 - 9x + x\\sqrt{12-x} - 9\\sqrt{12-x} = 0.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition, $a_1$, $a_1 + 6d$, and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n\n$$(a_1 + 6d)^2 = a_1 \\cdot (a_1 + 16d) \\implies d(a_1 - 9d) = 0.$$ \n\nSince $d \\neq 0$, we get $a_1 = 9d$. Furthermore, we have $(x-9)(x+\\sqrt{12-x}) = 0$ and $x \\le 12$. Then $x = 9$ or $\\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \\le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\\frac{4}{9}$, respectively.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18991,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive integers such that $a < b < c$, and let $f$ be a function from the positive integers to themselves, defined by:\n\n- $f(n) = n - a$ if $n > c$,\n- $f(n) = f(f(n + b))$ if $n \\leq c$.\n\nDetermine the number of fixed points $f$ may have.",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f$ has exactly $b - a$ fixed points if $a$ is divisible by $b - a$, and no fixed points otherwise. Clearly, $f$ has no fixed point for $n > c$, so we focus on positive integers not exceeding $c$.\n\nWe first show recursively that $f(n) = f(n + b - a)$ for all $n \\leq c$. If $c - b < n \\leq c$, then $n + b > c$, so $f(n) = f(f(n + b)) = f(n + b - a)$. If $n \\leq c - b$, then $n + b - a < n + b \\leq c$, so $f(n + b - a) = f(n + 2b - a)$. If $c - 2b < n \\leq c - b$, then $c - b < n + b \\leq c$, so $f(n) = f(f(n + b)) = f(f(n + 2b - a)) = f(n + b - a)$.\n\nAssume now $f(n) = f(n + b - a)$ for $c - k b < n \\leq c - (k - 1) b$, and consider $n$ with $c - (k + 1) b < n \\leq c - k b$. Since $c - k b < n + b \\leq c - (k - 1) b$, it follows that $f(n) = f(f(n + b)) = f(f(n + 2b - a)) = f(n + b - a)$. Since $c - m b < 0$ for some $m$, the claim follows.\n\nLet $n \\leq c$ and let $p = \\lfloor \\frac{c - n}{b - a} \\rfloor$. Notice that $n + p(b - a) \\leq c$ and $n + (p + 1)(b - a) > c$, so $f(n) = f(n + b - a) = \\dots = f(n + p(b - a)) = f(n + (p + 1)(b - a)) = n + (p + 1)(b - a) + a$.\n\nThus, $f(n) = n$ if and only if $n \\leq c$ and $n + (p + 1)(b - a) + a = n$, which is the case if and only if $(p + 1)(b - a) = -a$. Since $a > 0$, this is only possible if $a$ is divisible by $b - a$ and $p + 1 = \\frac{a}{b - a}$.\n\nConsequently, $f$ has no fixed points unless $a$ is divisible by $b - a$; in the latter case, $f$ has a fixed point at $n$ if and only if $\\lfloor \\frac{c - n}{b - a} \\rfloor + 1 = \\frac{a}{b - a}$, i.e., $c - a < n \\leq c - 2a + b$, so $f$ has exactly $b - a$ fixed points.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18992,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the quadrangle $ABCD$ be inscribed in a circle of radius $1$. Prove that the difference between its perimeter and the sum of the lengths of its diagonals is positive and less than $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the triangle inequality, we have:\n\n$$\n2L = \\overline{AB} + \\overline{BC} + \\overline{CD} + \\overline{DA} + \\overline{AB} + \\overline{DA} > \\overline{AC} + \\overline{BD} + \\overline{AC} + \\overline{BD}\n$$\n\nfrom which we get one of the inequalities. Let us denote the point of intersection of the diagonals by $R$, and the length of the diameter of the circle by $d$. Then we have\n\n$$\n\\begin{aligned}\nL &= \\overline{AB} + \\overline{BC} + \\overline{CD} + \\overline{DA} \\\\\n&< \\overline{AR} + \\overline{BR} + \\overline{CR} + \\overline{DR} \\\\\n&= \\overline{AC} + \\overline{BD} \\\\\n&\\le \\overline{AC} + \\overline{BD} + 2d = \\overline{AC} + \\overline{BD} + 4\n\\end{aligned}\n$$\n\nfrom which we get the other inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18993,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $AB = AC = BD$. The lines $AC$ and $BD$ meet at point $O$, the circles $ABC$ and $ADO$ meet again at point $P$, and the lines $AP$ and $BC$ meet at point $Q$. Show that $\\angle COQ = \\angle DOQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the circles $ADO$ and $BCO$ meet again at the incenter $I$ of triangle $ABO$, so the line $IO$ is the radical axis of the circles $ADO$ and $BCO$. Noticing further that the lines $AP$ and $BC$ are the radical axes of the pairs of circles $(ABC, ADO)$ and $(ABC, BCO)$, respectively, it follows that the lines $AP$, $BC$, and $IO$ are concurrent (at point $Q$), and the conclusion follows.\n\nTo show that the point $I$ lies on the circle $ADO$, notice that\n\n$$\n\\begin{aligned}\n\\angle AIO &= 90^\\circ + \\frac{1}{2} \\angle ABO = 90^\\circ + \\frac{1}{2} \\angle ABD = 90^\\circ + \\frac{1}{2} (180^\\circ - 2\\angle ADB) \\\\\n&= 180^\\circ - \\angle ADB = 180^\\circ - \\angle ADO.\n\\end{aligned}\n$$\n\nSimilarly, the point $I$ lies on the circle $BCO$, for\n\n$$\n\\begin{aligned}\n\\angle BIO &= 90^\\circ + \\frac{1}{2} \\angle BAO = 90^\\circ + \\frac{1}{2} \\angle BAC = 90^\\circ + \\frac{1}{2} (180^\\circ - 2\\angle ACB) \\\\\n&= 180^\\circ - \\angle ACB = 180^\\circ - \\angle BCO.\n\\end{aligned}\n$$\n\n![](images/Saudi_Arabia_booklet_2012_p37_data_392458afce.png)\n\nWe may consider the corresponding configuration derived from four generic points in the plane $A$, $B$, $C$, $D$, subject only to $AB = AC = BD$. The argument applies mutatis mutandis to show that the point $Q$ always lies on one of the two bisectors of the angle $COD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18994,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Show that among any four points, there must be a non-acute triangle.\n\n% ![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p58_data_702ac7b696.png)\n\n% ![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p58_data_9e7878d146.png)\n\n(b) Given 100 points, what is the minimum possible proportion of non-acute triangles that can be formed from them?",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a):\n\nConsider the convex hull of four points $A, B, C, D$.\n\n- If the convex hull is a quadrilateral $ABCD$, then\n  $$\n  \\angle ABC + \\angle BCD + \\angle CDA + \\angle DAB = 360^\\circ,\n  $$\n  so at least one angle is at least $90^\\circ$, yielding a non-acute triangle.\n- If the convex hull is a triangle, say $\\triangle ABC$, then\n  $$\n  \\angle ADB + \\angle BDC + \\angle CDA = 360^\\circ,\n  $$\n  so one of these angles is obtuse.\n\nFor part (b):\n\nSuppose at most 2 triangles are non-acute among any 5 points. Each triangle belongs to exactly 2 quadrilaterals, and there are $\\binom{5}{4} = 5$ quadrilaterals in total. Thus, some quadrilateral would have no non-acute triangle, contradicting part (a). Therefore, there are at least 3 non-acute triangles among any 5 points.\n\nThere are $\\binom{100}{5}$ groups of 5 points. Each triangle belongs to $\\binom{97}{2}$ such groups. Thus, the total number of non-acute triangles is at least\n$$\n3 \\binom{100}{5} \\div \\binom{97}{2}\n$$\nThere are $\\binom{100}{3}$ triangles in total, so at least\n$$\n3 \\binom{100}{5} \\div \\binom{97}{2} \\div \\binom{100}{3} = \\frac{3}{10} = 30\\%\n$$\nof the triangles are non-acute.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18995,
    "subject": "Mathematics (Olympiad)",
    "question": "Ангид 48 сурагч байна. Хотын сурагчдын тоо нь орон нутгаас ирсэн сурагчдаас олон. Ямар ч орон нутгаас ирсэн сурагч үнэн хариулдаг, харин хотын сурагчид үнэн эсвэл худал хариулж болно. Багш дурын сурагчаас \"Тэр сурагч хаанаас ирсэн бэ?\" гэж асууж болно (өөр сурагчдаас ч асууж болно). Багш хамгийн цөөн хэдэн асуулт асууж байж бүх сурагчийн хаанаас ирснийг мэдэж болох вэ?",
    "options": [],
    "answer": "See solution",
    "solution": "Ерөнхий тохиолдолд, хэрэв $k = 2m + 1$ бол $q = 3m$ асуултаар, $k = 2m$ бол $q = 3m - 2$ асуултаар мэдэж болно. Индукцээр баталъя.\n\n1. $k = 1$ (мөн $m = 0$): 0 асуулт шаардлагатай.\n2. $k = 2m + 1$ үед:\n   - Хүүхдүүдийг дугаарлаад, 1-р хүүхэд хаанаас ирсэн бэ гэж дараалан асууна.\n   - Асуултын явцад дараах хоёр үзэгдлийн аль нэг нь явагдана:\n     - А-үзэгдэл: Ихэнх хүүхэд 1-р хүүхдийг \"хөдөөнийх\" гэж хэлсэн.\n     - W-үзэгдэл: 1-р хүүхэд \"хөдөөнийх\" гэж $t$ хүүхэд хэлсэн.\n   - А-үзэгдэлд $f = t + 1$ ($f$ — \"хотынх\" гэж хэлсэн хүүхдийн тоо).\n   - W-үзэгдэлд нийт асуултын тоо $q_1 = m + f$.\n   - Үлдсэн хүүхдүүдэд индукцээр $q_2 = 3(m - f)$ асуулт зарцуулна.\n   - Дурын \"хотын\" хүүхдээс 1-р хүүхэд хаанаас ирсэн бэ гэж $q_3 = 1$ асуулт тавина.\n   - \"Хөдөөнийх\" гэж хэлсэн $f$ хүүхдээс асуухад $q_4 = f$ асуулт зарцуулна.\n   - Нийт $q = q_1 + q_2 + q_3 + q_4 = 3m$ асуулт болно.\n3. $k = 2m$ үед:\n   - Нэг сурагчийг сонгоод, $3m - 1$ асуултаар хэн хаанаас ирснийг тодорхойлно.\n   - \"Хотын\" хүүхдээс 1 асуултаар үлдсэн сурагчийн хаанаас ирснийг мэднэ.\n\nИймд хамгийн багадаа $k = 2m + 1$ бол $3m$, $k = 2m$ бол $3m - 2$ асуулт шаардлагатай.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18996,
    "subject": "Mathematics (Olympiad)",
    "question": "In scalene triangle $ABC$, let the feet of the perpendiculars from $A$ to $BC$, $B$ to $CA$, $C$ to $AB$ be $A_1, B_1, C_1$, respectively. Denote by $A_2$ the intersection of lines $BC$ and $B_1C_1$. Define $B_2$ and $C_2$ analogously. Let $D, E, F$ be the respective midpoints of sides $BC, CA, AB$. Show that the perpendiculars from $D$ to $AA_2$, from $E$ to $BB_2$, and from $F$ to $CC_2$ are concurrent.\n\n![](images/USA_IMO_2013-2014_p32_data_4fd1a29231.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the point of concurrency is $H$, the orthocenter of triangle $ABC$. By symmetry, it suffices to show that the perpendicular from $D$ to $AA_2$ passes through $H$.\n\nLet $A_3$ be the projection of $D$ onto $AA_2$. Because $\\angle AA_1D = \\angle AA_3D = 90^\\circ$ and $\\angle BC_1C = \\angle BB_1C = 90^\\circ$, quadrilaterals $AA_3A_1D$ and $BC_1B_1C$ are cyclic. Notice now that points $D, E, F, A_1, B_1$, and $C_1$ lie on the nine-point circle of triangle $ABC$. Further, by Power of a Point on cyclic quadrilaterals $AA_3A_1D$, $A_1DC_1B_1$, and $BCB_1C_1$, we obtain\n\n$$\nA_2A_3 \\cdot A_2A = A_2A_1 \\cdot A_2D = A_2B_1 \\cdot A_2C_1 = A_2B \\cdot A_2C.\n$$\n\nBy the converse of Power of a Point, it follows that $A_3$ lies on the circumcircle $\\omega$ of $ABC$.\n\nExtend segment $AA_1$ through $A_1$ to meet $\\omega$ at $H_2$. Then\n\n$$\n\\angle A_1 B H_2 = \\angle C B H_2 = \\angle C A H_2 = \\angle C A A_1 = 90^\\circ - \\angle A C B = \\angle B_1 B C = \\angle H B A_1.\n$$\n\nThat is, in triangle $H B H_2$, segment $B A_1$ bisects $\\angle H B H_2$ and is the altitude from $B$ to side $H H_2$. Hence, $H B H_2$ is isosceles and $H A_1 = A_1 H_2$.\n\nReflect $H$ through $D$ to obtain $H_3$. Then $D A_1$ is the midline of right triangle $H H_2 H_3$. Let $M$ be the midpoint of $H_2 H_3$. Then $D M$ is a midline of right triangle $H H_2 H_3$. In particular, $D M \\perp B C$ and so $D M$ passes through the circumcenter $O$ of triangle $ABC$. Because $M O$ is the perpendicular bisector of segment $H_2 H_3$ and $H_3$ lies on $\\omega$, $H_2$ also lies on $\\omega$. Because $\\angle A H_2 H_3 = 90^\\circ$, $A H_3$ is a diameter of $\\omega$, which implies that $\\angle A A_3 H_3 = 90^\\circ$ and hence $H$ lies on $D A_3$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 18997,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = ax^2 + 2bx + c$ be a quadratic function. Suppose that\n\n$$\na + 2b + c > 0 \\implies f(1) > 0,\n$$\n\nand\n\n$$\na - 2b + c < 0 \\implies f(-1) < 0.\n$$\n\nShow that $b^2 > ac$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $f(-1) < 0$ and $f(1) > 0$, the function $f$ must have a root between $-1$ and $1$. Because $f$ is quadratic, it must have two distinct real roots, so its discriminant is positive:\n\n$$\n4b^2 - 4ac > 0 \\implies b^2 > ac.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 18998,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$ be an arbitrary positive real number. Determine, for this number $\\alpha$, the greatest real number $C$ such that the inequality\n\n$$\n\\left(1 + \\frac{\\alpha}{x^2}\\right) \\left(1 + \\frac{\\alpha}{y^2}\\right) \\left(1 + \\frac{\\alpha}{z^2}\\right) \\geq C \\cdot \\left(\\frac{x}{z} + \\frac{z}{x} + 2\\right)\n$$\n\nis valid for all positive real numbers $x, y, z$ satisfying $xy + yz + zx = \\alpha$. When does equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "By substituting $\\alpha = xy + yz + zx$ and clearing fractions, we obtain the equivalent inequality:\n\n$$\n(x^2 + xy + xz + yz)(y^2 + yx + yz + xz)(z^2 + zx + zy + xy) \\geq Cxy^2z(x^2 + z^2 + 2xz).\n$$\n\nThis inequality is homogeneous of degree 6, so no further constraint is needed. Each factor on the left can be factorized:\n\n$$\n(x + y)(x + z)(y + x)(y + z)(z + x)(z + y) \\geq Cxy^2z(x + z)^2.\n$$\n\nCanceling $(x + z)^2$ yields:\n\n$$\n(x + y)^2 (z + y)^2 \\geq Cxy^2z.\n$$\n\nEstimating each factor on the left using the arithmetic-geometric mean inequality, we find the optimal value $C = 16$, which is attained when $x = y = z = \\sqrt{\\alpha}/3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 18999,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Find all pairs of positive integers $(n, m)$ such that\n$$\n2^n + 7^n = m^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1:** $n$ is odd and $n > 1$\n\nLet $n = 2k + 1$, where $k$ is a positive integer. Equation (1) can be rewritten as\n$$\n\\begin{aligned}\n& 2 \\cdot 2^{2k} + 7 \\cdot 7^{2k} = m^2 \\\\\n\\Leftrightarrow \\quad & 2(2^{2k} - 7^{2k}) = m^2 - 9 \\cdot 7^{2k} \\\\\n& = (m - 3 \\cdot 7^k)(m + 3 \\cdot 7^k).\n\\end{aligned}\n$$\nThe left-hand side is even, so the right-hand side must be even too. This implies that $m$ is odd. But both factors on the right are even, so the right-hand side is a multiple of 4. However, the left-hand side is not a multiple of 4. So there are no solutions in this case.\n\n**Case 2:** $n$ is even\n\nLet $n = 2k$, where $k$ is a positive integer. Equation (1) can be rewritten as\n$$\n\\begin{aligned}\n2^{2k} &= m^2 - 7^{2k} \\\\\n&= (m - 7^k)(m + 7^k).\n\\end{aligned}\n$$\nHence there exist non-negative integers $r < s$ satisfying the following:\n$$\nm - 7^k = 2^r\n$$\n$$\nm + 7^k = 2^s\n$$\nSubtracting the first equation from the second yields\n$$\n2^r(2^{s-r} - 1) = 2 \\cdot 7^k.\n$$\nHence $r = 1$, and so $m = 7^k + 2$. Substituting this into equation (1) yields\n$$\n\\begin{aligned}\n2^{2k} + 7^{2k} &= (7^k + 2)^2 \\\\\n&= 7^{2k} + 4 \\cdot 7^k + 4 \\\\\n\\Leftrightarrow \\quad & 4^k = 4 \\cdot 7^k + 4.\n\\end{aligned}\n$$\nHowever, this is impossible because the left-hand side is smaller than the right-hand side.\n\nHaving covered all possible cases, the proof is complete. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19000,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $D$ inside an acute triangle $ABC$ satisfies\n\n$$\n\\angle ADC = \\angle BDA = 180^{\\circ} - \\angle CAB.\n$$\n\nProve that the point symmetric to point $A$ with respect to point $D$ lies on the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the line $AD$ intersect the circumcircle of triangle $ABC$ a second time at point $D'$. By the assumptions, $\\angle CDD' = \\angle D'DB = \\angle CAB$.\n\nWe show that $AD = DD'$.\n\nAs the quadrilateral $ABD'C$ is cyclic, $\\angle ABC = \\angle AD'C = \\angle DD'C$ and $\\angle BCA = \\angle BD'A = \\angle BD'D$. Thus, the triangles $ABC$, $DD'C$, and $DBD'$ are similar. Hence,\n$$\n\\frac{|DD'|}{|DB|} = \\frac{|DC|}{|DD'|},\n$$\nimplying $|DD'|^2 = |BD| \\cdot |CD|$.\n\n![](images/EST_ABooklet_2021_p36_data_8eb584857d.png)\n\nBy similarity of triangles $ABC$ and $DD'C$, we obtain $\\angle BCA = \\angle D'CD$, which implies $\\angle DCA = \\angle D'CB = \\angle D'AB = \\angle DAB$. By similarity of triangles $ABC$ and $DBD'$, we analogously get $\\angle ABD = \\angle CBD' = \\angle CAD' = \\angle CAD$. Hence, triangles $ABD$ and $CAD$ are similar. Consequently,\n$$\n\\frac{AD}{CD} = \\frac{BD}{AD},\n$$\nwhich implies $AD^2 = BD \\cdot CD$.\n\nAltogether, we have proven $AD^2 = DD'^2$, which implies the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19001,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N \\geq 3$ be an integer, and $a_0, \\dots, a_{N-1}$ be pairwise distinct real numbers such that $a_i \\geq a_{2i}$ for all $i$ (indices are taken modulo $N$). Find all possible $N$ for which this is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "The only such $N$ are powers of $2$.\n\nSuppose $N$ is not a power of $2$, so there exists an odd prime $p$ dividing $N$. Observe that $p \\nmid 1$, but $p \\mid 2^k - 1$ for some $k > 1$. Let $\\alpha = \\frac{N}{p}$. Then,\n\n$$\na_{\\alpha} > a_{2\\alpha} \\geq a_{4\\alpha} \\geq \\dots \\geq a_{2^k \\alpha} \\implies a_{\\alpha} > a_{2^k \\alpha}\n$$\n\nBut $a_{2^k \\alpha} = a_{\\alpha}$ since $N$ divides $(2^k - 1)\\alpha$, which is a contradiction because all $a_i$ are distinct.\n\nIf $N = 2^m$ is a power of $2$, let $a_j = -\\nu_2(j) + \\frac{j}{2N}$, where $\\nu_2(j)$ is the exponent of $2$ in the prime factorization of $j$. For $j \\neq N$, $\\nu_2(2j) = \\nu_2(j) + 1$, so $a_j \\geq a_{2j}$. Thus, such $N$ are possible. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19002,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s = 10a + b$ and $p = 10c + d$, where $a, b, c$, and $d$ are positive digits.\n\n$$\n10a + b = a + b + c + d \\quad \\text{and} \\quad 10c + d = abcd.\n$$\n\nFind all pairs $(s, p)$ of compatible numbers less than $100$ that satisfy these equations.",
    "options": [],
    "answer": "See solution",
    "solution": "The second equation gives $d = c(abd - 10)$. Since $d$ is a positive digit, $10 < abd < 20$. Since $a, b, d$ are digits and $11, 13, 17, 19$ are primes, $abd = 12, 14, 15, 16$, or $18$.\n\n- If $abd = 18$, then $18c = 10c + d$, $d = 8c$, and $4abc = 9$, which is impossible.\n- If $abd = 16$, then $16c = 10c + d$, $d = 6c$, and $3abc = 8$, which is impossible.\n- If $abd = 15$, then $15c = 10c + d$, $d = 5c$. So $c = 1$, $d = 5$, and $9a = 6$, which is impossible.\n- If $abd = 14$, then $14c = 10c + d$, $d = 4c$, and $2abc = 7$, which is impossible.\n- If $abd = 12$, then $12c = 10c + d$, and $d = 2c$. So $d$ is even and divides $12$. Hence $d = 2, 4$, or $6$, and correspondingly $c = 1, 2$, or $3$. Since $10a + b = a + b + c + d$, we have $9a = c + d$. So $9$ divides $c+d$. Hence $d = 6$, $c = 3$, $a = 1$, $b = 2$, and we have the compatible numbers $s = 12$ and $p = 36$.\n\nHence, there are only $2$ pairs of compatible numbers less than $100$, namely $\\{9, 11\\}$ and $\\{12, 36\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19003,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We prove that the numbers with 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 $n = ab$ with $1 < a \\leq b < n$. 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 $c - 1 \\geq b$, i.e., $c \\geq b + 1$. Then $ab + 1 = c(b + 1) \\geq (b + 1)^2$, which means $ab + 1 \\geq b^2 + 2b + 1$, leading to $a \\geq b + 2$, which contradicts $a \\leq b$.\n\nIn conclusion, no composite number satisfies the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19004,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, evaluate $\\sum \\frac{1}{pq}$, where the summation is over all coprime integers $p$ and $q$ such that $1 \\le p < q \\le n$ and $p+q > n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required sum is $\\frac{1}{2}$, for it is the sum of the distances between successive terms in the first half of the Farey series of order $n$ — for instance, see G. H. Hardy and E. M. Wright, *An Introduction to the Theory of Numbers*, Oxford at the Clarendon Press, 1956, Chap. III.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19005,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(0) = 0$, $f(1) = 2013$, and\n$$\n(x - y)(f(f(x)^2) - f(f(y)^2)) = (f(x) - f(y))(f(x)^2 - f(y)^2)\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Given $x \\neq 0$ and $y = 0$, we get\n$$\nx f(f(x)^2) = f(x)^3\n$$\nwhich implies\n$$\nf(f(x)^2) = \\frac{f(x)^3}{x}, \\quad \\forall x \\neq 0.\n$$\nSubstituting into the original equation, for all $x, y \\neq 0$:\n$$\n(x - y) \\left[ \\frac{f(x)^3}{x} - \\frac{f(y)^3}{y} \\right] = (f(x) - f(y))(f(x)^2 - f(y)^2) \\quad (1)\n$$\nSubstituting $x < 0$, $y = 1$ into (1), we have\n$$\n(x - 1) \\left[ \\frac{f(x)^3}{x} - 2013^3 \\right] = (f(x) - 2013)(f(x)^2 - 2013^2),\n$$\nwhich is equivalent to\n$$\n(f(x) - 2013x)(f(x)^2 - 2013^2 x) = 0, \\quad \\forall x < 0.\n$$\nFor $x < 0$, $f(x)^2 > 2013^2 x$, so $f(x) = 2013x$ for all $x < 0$. Hence $f(-1) = -2013$.\n\nNow, for $x > 0$, $y = -1$ in (1):\n$$\n(x + 1) \\left[ \\frac{f(x)^3}{x} - 2013^3 \\right] = (f(x) + 2013)(f(x)^2 - 2013^2),\n$$\nwhich is equivalent to\n$$\n(f(x) - 2013x)(f(x)^2 + 2013^2 x) = 0, \\quad \\forall x > 0.\n$$\nSo $f(x) = 2013x$ for all $x > 0$.\n\nCombining with $f(0) = 0$, we get $f(x) = 2013x$ for all real numbers $x$.\n\nWith $f(x) = 2013x$, $f(x)^2 = 2013^2 x^2$ and\n$$\nf(f(x)^2) = f((2013x)^2) = 2013^3 x^2.\n$$\nThus,\n$$\n(x-y)(f(f(x)^2)-f(f(y)^2)) = 2013^3(x-y)(x^2-y^2)\n$$\nand\n$$\n\\begin{aligned}\n(f(x)-f(y))(f(x)^2-f(y)^2) &= 2013(x-y)(2013^2 x^2 - 2013^2 y^2) \\\\\n&= 2013^3(x-y)(x^2-y^2).\n\\end{aligned}\n$$\nTherefore, $f(x) = 2013x$ for all real numbers $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19006,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that the equality\n$$\nf(\\lfloor x \\rfloor y) = f(x) \\lfloor f(y) \\rfloor\n$$\nholds for all $x, y \\in \\mathbb{R}$.\n\n(Here $\\lfloor z \\rfloor$ denotes the greatest integer less than or equal to $z$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = 1$. Thus $f(y) = f(1) \\lfloor f(y) \\rfloor$.\n\n**Case 1.** $f(1) = 0$. From above we have that $f(y) = 0$ for all $y \\in \\mathbb{R}$. Note that this is clearly a solution.\n\n**Case 2.** $f(1) \\neq 0$. Set $x = y = 1$. Thus $f(1) = f(1) \\lfloor f(1) \\rfloor$. Thus $\\lfloor f(1) \\rfloor = 1$. Now set $y = 1$ in the original functional equation. Hence\n$$\nf(\\lfloor x \\rfloor) = f(x) \\quad (*)\n$$\nfor all $x \\in \\mathbb{R}$. Now set $y = 0$ in the original equation to deduce\n$$\nf(0) = f(x) \\lfloor f(0) \\rfloor.\n$$\n**Case 2a.** $\\lfloor f(0) \\rfloor \\neq 0$. Thus $f(x) = \\frac{f(0)}{\\lfloor f(0) \\rfloor}$. Thus $f$ is a constant function. Write $f(x) = c$. Note that $c = f(1)$. Since $\\lfloor f(1) \\rfloor = 1$ at the beginning of case 2 we see that $1 \\leq c < 2$. All such constant functions are solutions since LHS $= c = c \\lfloor c \\rfloor = c = \\text{RHS}$.\n\n**Case 2b.** $\\lfloor f(0) \\rfloor = 0$. Thus from $(*)$ $f(0) = 0$ also. Since $\\lfloor \\frac{1}{2} \\rfloor = 0$ it follows from $(*)$ that $f(\\frac{1}{2}) = f(0) = 0$. Set $x = 2$, $y = \\frac{1}{2}$ in the original equation. Thus $f(1) = f(2) \\lfloor f(\\frac{1}{2}) \\rfloor = 0$. This contradicts $f(1) \\neq 0$ at the beginning of case 2. Thus there are no solutions in this case.\n\nTo summarize, the solutions of the functional equation are $f(x) = c$ where $c = 0$ or $1 \\leq c < 2$. We have already checked that these are indeed solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19007,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob play a game. First, Alice secretly picks a finite set $S$ of lattice points in the Cartesian plane. Then, for every line $\\ell$ in the plane which is horizontal, vertical, or has slope $+1$ or $-1$, she tells Bob the number of points of $S$ that lie on $\\ell$. Bob wins if he can then determine the set $S$.\n\nProve that if Alice picks $S$ to be of the form\n\n$$\nS = \\{(x, y) \\in \\mathbb{Z}^2 \\mid m \\le x^2 + y^2 \\le n\\}\n$$\n\nfor some positive integers $m$ and $n$, then Bob can win. (Bob does not know in advance that $S$ is of this form.)",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly Bob can compute the number $N$ of points.\n\nThe main claim is that:\n\n**Claim.** Fix $m$ and $n$ as in the problem statement. Among all sets $T \\subseteq \\mathbb{Z}^2$ with $N$ points, the set $S$ is the unique one which maximizes the value of\n\n$$\nF(T) := \\sum_{(x,y) \\in T} (x^2 + y^2)(m + n - (x^2 + y^2)).\n$$\n\n*Proof.* Indeed, the different points in $T$ do not interact in this sum, so we simply want the points $(x, y)$ with $x^2 + y^2$ as close as possible to $\\frac{m+n}{2}$ which is exactly what $S$ does. $\\square$\n\nAs a result of this observation, it suffices to show that Bob has enough information to compute $F(S)$ from the data given. (There is no issue with fixing $m$ and $n$, since Bob can find an upper bound on the magnitude of the points and then check all pairs $(m, n)$ smaller than that.) The idea is that he knows the full distribution of each of $X$, $Y$, $X+Y$, $X-Y$ and hence can compute sums over $T$ of any power of a single one of those linear functions. By taking linear combinations we can hence compute $F(S)$.\n\nLet us make the relations explicit. For ease of exposition we take $Z = (X, Y)$ to be a uniformly random point from the set $S$. The information is precisely the individual distributions of $X$, $Y$, $X+Y$, and $X-Y$. Now compute\n\n$$\n\\frac{F(S)}{N} = \\mathbb{E}[(m+n)(X^2+Y^2) - (X^2+Y^2)^2] \\\\\n= (m+n)(\\mathbb{E}[X^2] + \\mathbb{E}[Y^2]) - \\mathbb{E}[X^4] - \\mathbb{E}[Y^4] - 2\\mathbb{E}[X^2Y^2].\n$$\n\nOn the other hand,\n\n$$\n\\mathbb{E}[X^2Y^2] = \\frac{\\mathbb{E}[(X+Y)^4] + \\mathbb{E}[(X-Y)^4] - 2\\mathbb{E}[X^4] - 2\\mathbb{E}[Y^4]}{12}.\n$$\n\nThus we have written $F(S)$ in terms of the distributions of $X$, $Y$, $X-Y$, $X+Y$ which completes the proof.\n\n**Remark (Mark Sellke).**\n* This proof would have worked just as well if we allowed arbitrary $[0, 1]$-valued weights on points with finitely many weights non-zero. There is an obvious continuum generalization one can make concerning the indicator function for an annulus. It's a simpler but fun problem to characterize when just the vertical/horizontal directions determine the distribution.\n\n* An obstruction to purely combinatorial arguments is that if you take an octagon with points $(\\pm a, \\pm b)$ and $(\\pm b, \\pm a)$ then the two ways to pick every other point (going around clockwise) are indistinguishable by Bob. This at least shows that Bob's task is far from possible in general, and hints at proving an inequality.\n\n* A related and more standard fact (among a certain type of person) is that given a probability distribution $\\mu$ on $\\mathbb{R}^n$, if I tell you the distribution of all 1-dimensional projections of $\\mu$, that determines $\\mu$ uniquely. This works because this information gives me the Fourier transform $\\hat{\\mu}$, and Fourier transforms are injective.\n\nFor the continuum version of this problem, this connection gives a much larger family of counterexamples to any proposed extension to arbitrary non-annular shapes. Indeed, take a fast-decaying smooth function $f: \\mathbb{R}^2 \\to \\mathbb{R}$ which vanishes on the four lines\n\n$$\nx = 0,\\quad y = 0,\\quad x + y = 0,\\quad x - y = 0.\n$$\n\nThen the Fourier transform $\\hat{f}$ will have mean 0 on each line $\\ell$ as in the problem statement. Hence the positive and negative parts of $\\hat{f}$ will not be distinguishable by Bob.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19008,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. A sequence of integers $\\langle a_i \\rangle_{i=1}^\\infty$ is called $k$-pop if the following holds: for every $n \\in \\mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\\{a_1, \\dots, a_{n+k}\\}$. Determine, as a function of $k$, how many $k$-pop sequences there are.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2^k$.\n\nBefore we prove this, we will obtain a characterization of $k$-pop sequences. To that end, note that in a $k$-pop sequence, since the set $\\{a_1, \\dots, a_{n+k+1}\\}$ has either the same number of distinct elements as $\\{a_1, \\dots, a_{n+k}\\}$ or precisely one more, it follows that $a_{n+1}$ is either $a_n$ or $a_n+1$ for each $n$.\n\nA sequence $\\langle b_n \\rangle_{n=1}^\\infty$ with the property that $b_{n+1} \\in \\{b_n, b_n+1\\}$ for each $n$ will be called a step-sequence, and an index $n$ such that $b_{n+1} = b_n+1$ will be called a step. Now we characterize $k$-pop sequences:\n\n**Lemma**\n\nA sequence $\\langle a_n \\rangle_{n=1}^\\infty$ is $k$-pop if and only if it is a step-sequence and the following hold:\n\n1. For each $n$, the index $n$ is a step if and only if $n + k$ is, and\n2. The number of steps among $\\{1, \\dots, k\\}$ is equal to $a_1 - 1$.\n\n_Proof._ Assume $\\langle a_n \\rangle_{n=1}^\\infty$ is $k$-pop. As seen before, it is a step-sequence. Now if $n$ is a step, this is if and only if the number of distinct elements in $\\{a_1, \\dots, a_{n+k+1}\\}$ is one more than that in $\\{a_1, \\dots, a_{n+k}\\}$, which is possible if and only if $a_{n+k+1}$ is not an element of $\\{a_1, \\dots, a_{n+k}\\}$. Since $\\langle a_n \\rangle_{n=1}^\\infty$ is a step-sequence, this happens if and only if $a_{n+k+1} = a_{n+k} + 1$, i.e., $n+k$ is a step. Now $a_1$ is equal to the number of distinct elements in $\\{a_1, \\dots, a_{k+1}\\}$, which is obviously one more than the number of steps in $\\{1, \\dots, k\\}$. This proves one direction of the implication.\n\nNow assume $\\langle a_n \\rangle_{n=1}^\\infty$ is a step-sequence satisfying the two conditions. We will show by induction that $a_n$ is equal to the number of distinct elements in the set $\\{a_1, \\dots, a_{n+k}\\}$.\n\nThe base case $n=1$ follows from condition (2): since it is a step-sequence, the number of distinct elements in $\\{a_1, \\dots, a_{k+1}\\}$ is one more than the number of steps in $\\{1, \\dots, k\\}$, and thus equal to $a_1$. Now for a general $m=n+1$, note that the number of distinct elements in $\\{a_1, \\dots, a_{n+k+1}\\}$ is equal to the number of distinct elements in $\\{a_1, \\dots, a_{n+k}\\}$ if $a_{n+k+1} = a_{n+k}$, and one more than that otherwise. In the first case, $n+k$ is not a step, therefore, neither is $n$, so $a_{n+1} = a_n$, which is equal to the number of distinct elements in either of those sets. The other case is similar.\n\nThis shows our sequence is $k$-pop. $\\square$\n\nThus to determine a $k$-pop sequence, it suffices to determine which of the indices in $\\{1, \\dots, k\\}$ are steps: by condition (1), this determines the locations of all steps, and by condition (2), this fixes $a_1$. Once these are fixed, $a_n$ can be determined inductively. Conversely, any such selection of indices gives rise to a step-sequence satisfying the conditions in the lemma by the aforementioned recipe, and thus a $k$-pop sequence. There are exactly $2^k$ ways to choose the steps, which is therefore the answer. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19009,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "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\n1. No line through at least two points in $S$; and\n2. No tangent of a circle $\\gamma$ through at least three points in $S$ at a point in $S \\cap \\gamma$.\n\nThere are finitely many directions to avoid, so the choice of $\\ell$ is possible.\n\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, 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": 19010,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е триаголникот $\\triangle ABC$ и отсечка $PQ$ со должина $t$ на отсечката $BC$, така што $P$ е меѓу $B$ и $Q$, а $Q$ е меѓу $P$ и $C$. Од точката $P$ се повлекуваат паралелни прави со $AB$ и $AC$ кои ги сечат $AC$ и $AB$ во $P_1$ и $P_2$, соодветно. Од точката $Q$ се повлекуваат паралелни прави со $AB$ и $AC$ кои ги сечат $AC$ и $AB$ во $Q_1$ и $Q_2$, соодветно. Докажи дека збирот од плоштините на $PQQ_1P_1$ и $PQQ_2P_2$ не зависи од положбата на $PQ$ на $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $D$ е пресекот на $PP_1$ и $QQ_2$. Да забележиме дека $P_{DQ_2P_2} = 2P_{\\triangle ADP}$ и $P_{QD P_1 Q_1} = 2P_{\\triangle ADQ}$. Па сега имаме:\n\n$$\nP_{PQQ_2P_2} + P_{PP_1Q_1Q} = P_{PDQ_2P_2} + P_{QDP_1Q_1} + 2P_{\\triangle PQD} = 2P_{\\triangle ADP} + 2P_{\\triangle ADQ} + 2P_{\\triangle APQ} = 2P_{\\triangle APQ} = \\overline{PQ} \\cdot h_a\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19011,
    "subject": "Mathematics (Olympiad)",
    "question": "With positive integer $n > 1$, let $4n - 3$ positive numbers be written on the board (not necessarily all different). It is known that any 4 pairwise different numbers from that list form an arithmetic progression. Prove that some number is written on the board at least $n$ times.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the number of distinct values appearing on the board, and let them be $$a_1 < a_2 < \\dots < a_d.$$ Assume, for contradiction, that no value appears at least $n$ times; then each value appears at most $n-1$ times. Counting the total numbers, we get $$4n - 3 \\leq d(n - 1) \\implies d \\geq \\frac{4n - 3}{n - 1} \\implies d \\geq 5.$$ Consider 5 arbitrary distinct values, denoted $a < b < c < d < e$. By assumption, $(a, b, c, d)$ and $(a, b, c, e)$ both form arithmetic progressions. Then $b - a = d - c = e - c$, which implies $d = e$, a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19012,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ has $\\angle A = 75^\\circ$, $\\angle C = 45^\\circ$. On the ray $BC$, after point $C$, denote point $T$ such that $BC = CT$. Let $M$ be the midpoint of segment $AT$. Find the value of $\\angle BMC$.\n\n![](images/UkraineMO2019_booklet_p24_data_8a07016572.png \"Fig. 25\")",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $CM$ is a midsegment of $\\triangle TAB$ (see figure). Denote segment $MK$ on the ray $CM$ beyond point $M$ such that $CM = MK$. Then $AB = CK$ and $AB \\parallel CK$, so $ABCK$ is a parallelogram. On the ray $CA$ beyond $A$, denote point $S$ such that $\\angle SKC = 15^\\circ$. Then $\\triangle SKC$ is isosceles, so $\\angle SMC = 90^\\circ$. Let $L = SK \\cap AB$. Then quadrilateral $LACK$ is an isosceles trapezoid, so $AK = LC$. On the other hand, $BC = AK \\Rightarrow BC = CL$. Thus, $\\triangle BLC$ is isosceles, and $\\angle BLC = 60^\\circ$. Then, this triangle is equilateral, so $BC = BL$. Let us prove that $BS = BC$, i.e., that points $S$, $L$, $C$ belong to a circle with center $B$. This follows from the fact that $\\angle LSC = 30^\\circ = \\frac{1}{2} \\angle LBC$. Indeed, if $X$ is a point on a circle with center $B$ and radius $BC$, then points $S$, $L$, $C$ belong to a circle which point $X$ belongs to. But points $X$, $L$, $C$ belong to a circle with center $B$. Therefore, we proved that $BS = BC$. Then,\n$$\n\\angle SBC = 180^\\circ - 2\\angle BCA = 90^\\circ.\n$$\nHence, points $S$, $B$, $C$, $M$ belong to a circle, and\n$$\n\\angle BMC = \\angle BSC = 45^\\circ.\n$$\nQ.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19013,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system:\n$$\n\\begin{cases}\nx^{\\log y} + \\sqrt{y^{\\log x}} = 110 \\\\\nxy = 1000\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By definition, $x, y > 0$. Let $A = x^{\\log y}$ and $B = y^{\\log x}$. Then $\\log A = \\log x \\cdot \\log y = \\log B$, so $A = B$. Thus, $x^{\\log y} = y^{\\log x} = t$.\n\nThe first equation becomes $t + \\sqrt{t} = 110$. Let $\\sqrt{t} = z$, so $t = z^2$ and $z^2 + z - 110 = 0$. The roots are $z_1 = -11$ and $z_2 = 10$. Since $z \\ge 0$, $z = 10$ and $t = 100$.\n\nNow:\n$$\n\\begin{cases}\nx^{\\log y} = 100 \\\\\nxy = 1000\n\\end{cases}\n$$\nTake logarithms:\n$$\n\\begin{cases}\n\\log x \\cdot \\log y = 2 \\\\\n\\log x + \\log y = 3\n\\end{cases}\n$$\nLet $p = \\log x$, $q = \\log y$. By Vieta's formulas, $p$ and $q$ are roots of $r^2 - 3r + 2 = 0$, which has roots $1$ and $2$.\n\nTherefore, $(\\log x, \\log y) = (1, 2)$ or $(2, 1)$, so $(x, y) \\in \\{(10, 100), (100, 10)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19014,
    "subject": "Mathematics (Olympiad)",
    "question": "是否存在不等邊三角形 $ABC$，使得三角形 $ABC$ 與三角形 $IHO$ 相似，其中 $I, H, O$ 分別為三角形 $ABC$ 的內心、垂心及外心？",
    "options": [],
    "answer": "See solution",
    "solution": "答案是不存在。\n\n由於 $ABC$ 為不等邊三角形，不妨假設 $\\angle A > \\angle B > \\angle C$。事實上，我們可以證明 $\\angle OIH > \\angle A$。\n\n**Claim.** $A, H$ 位於 $OI$ 同側。\n\n_Proof._ 令 $M$ 為 $AI$ 與外接圓 $\\odot(ABC)$ 的第二個交點，$J$ 為 $\\triangle ABC$ 的 $A$-旁心，$S$ 為 $OI$ 與 $AH$ 的交點。雞爪定理告訴我們 $M$ 為 $IJ$ 中點。考慮 $I, M, I_a$ 關於 $AB$ 的投影點 $F_I, F_M, F_J$，我們有\n\n$$\n\\frac{MI}{IA} = \\frac{F_M F_I}{F_I A} = \\frac{1}{2} \\cdot \\frac{F_J F_I}{F_I A} = \\frac{1}{2} \\cdot \\frac{a}{(b+c-a)/2} = \\frac{a}{b+c-a},\n$$\n\n其中 $a = BC, b = CA, c = AB$ 為三角形 $ABC$ 的三邊長。所以\n\n$$\n\\overrightarrow{AH} = 2 \\cos \\angle BAC \\cdot \\overrightarrow{OM} = \\frac{b^2 + c^2 - a^2}{bc} \\cdot \\left( \\frac{MI}{IA} \\cdot \\overrightarrow{AS} \\right) = \\frac{a(b^2 + c^2 - a^2)}{bc(b+c-a)} \\cdot \\overrightarrow{AS}.\n$$\n\n因此，$A, H$ 位於 $OI$ 同側若且唯若\n\n$$\n\\begin{aligned}\n1 > \\frac{a(b^2 + c^2 - a^2)}{bc(b+c-a)} &\\iff bc(b+c-a) > a(b^2 + c^2 - a^2) \\\\\n&\\iff (a-b)(a-c)(a+b+c) > 0,\n\\end{aligned}\n$$\n\n而這顯然是對的（因為 $a > b > c$）。$\\square$\n\n回到原題。由於 $O, M$ 位於 $AH$ 同側，$O, I$ 也位於 $AH$ 同側，因此\n\n$$\n\\angle OIH = \\angle OSH + \\angle SHI > \\angle OSH = \\angle IOM. \\qquad (\\spadesuit)\n$$\n\n結合 $I, O$ 位於 $BM$ 同側以及\n\n$$\n\\angle MBO = 90^\\circ - \\frac{1}{2}\\angle A = \\frac{1}{2}(\\angle B + \\angle C) < \\frac{1}{2}(\\angle A + \\angle B) = \\angle BMI,\n$$\n\n我們得到 $I, M$ 位於 $BO$ 的異側。再結合 $B, I$ 位於 $MO$ 的同側及 $(\\spadesuit)$ 我們得到\n\n$$\n\\angle OIH > \\angle IOM > \\angle BOM = \\angle A.\n$$\n\n因此不存在不等邊三角形 $ABC$ 使得三角形 $ABC$ 與三角形 $IOH$ 相似。$\\square$\n\n![](images/2023-TWNIMO-Problems_p73_data_d79c3599ce.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19015,
    "subject": "Mathematics (Olympiad)",
    "question": "Olesya chose 5 numbers from the set $\\{1, 2, 3, 4, 5, 6, 7\\}$. She told Pavlik the product of these numbers and asked whether the sum of these numbers is odd or even. Pavlik replied that he could not determine it for sure. What product might Olesya have had?",
    "options": [],
    "answer": "See solution",
    "solution": "If Pavlik knows the product, he can determine the product of the last two numbers that were not chosen. Since he could not determine the parity of the sum, he could not determine the numbers that were not chosen, although he knows their product.\n\nConsider all products of two numbers in the given set:\n\n$$\n\\begin{aligned}\n1 \\cdot 2 &= 2, \\quad 1 \\cdot 3 = 3, \\quad 1 \\cdot 4 = 4, \\quad 1 \\cdot 5 = 5, \\quad 1 \\cdot 6 = 6, \\quad 1 \\cdot 7 = 7, \\\\\n2 \\cdot 3 &= 6, \\quad 2 \\cdot 4 = 8, \\quad 2 \\cdot 5 = 10, \\quad 2 \\cdot 6 = 12, \\quad 2 \\cdot 7 = 14, \\\\\n3 \\cdot 4 &= 12, \\quad 3 \\cdot 5 = 15, \\quad 3 \\cdot 6 = 18, \\quad 3 \\cdot 7 = 21, \\\\\n4 \\cdot 5 &= 20, \\quad 4 \\cdot 6 = 24, \\quad 4 \\cdot 7 = 28, \\\\\n5 \\cdot 6 = 30, \\quad 5 \\cdot 7 = 35, \\quad 6 \\cdot 7 = 42.\n\\end{aligned}\n$$\n\nThere are only two products that appear more than once: $1 \\cdot 6 = 6 = 2 \\cdot 3$ and $2 \\cdot 6 = 12 = 3 \\cdot 4$. Suppose the product of the two numbers that were not chosen equals $6$. In both cases, the sum of the numbers is odd: $1+6=7$ and $2+3=5$. Thus, the sum of the chosen numbers is also odd. This contradicts Pavlik's claim. Therefore, the product of the two numbers that were not chosen might equal $12$. In this case, their sum might be even ($2+6=8$) or odd ($3+4=7$). Then the product of the 5 chosen numbers is equal to $\\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7}{12} = 420$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19016,
    "subject": "Mathematics (Olympiad)",
    "question": "The number of rational solutions to the system of equations\n$$\n\\begin{cases}\nx + y + z = 0, \\\\\nxyz + z = 0, \\\\\nxy + yz + xz + y = 0\n\\end{cases}\n$$\nis $\\boxed{\\phantom{0}}$.\n\n(A) 1  \n(B) 2  \n(C) 3  \n(D) 4",
    "options": [],
    "answer": "See solution",
    "solution": "If $z = 0$, then\n$$\n\\begin{cases}\nx + y = 0, \\\\\nxy + y = 0\n\\end{cases}\n$$\nIt follows that\n$$\n\\begin{cases}\nx = 0, \\\\\ny = 0\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx = -1, \\\\\ny = 1\n\\end{cases}\n$$\nIf $z \\neq 0$, from $xyz + z = 0$ we get\n$$\nxy = -1. \\quad (1)\n$$\nFrom $x + y + z = 0$ we have\n$$\nz = -x - y. \\quad (2)\n$$\nSubstituting (2) into $xy + yz + xz + y = 0$, we obtain\n$$\nx^2 + y^2 + xy - y = 0. \\quad (3)\n$$\nFrom (1) we have $x = -\\frac{1}{y}$. Substitute into (3):\n$$\n(y - 1)(y^3 - y - 1) = 0.\n$$\nIt is easy to see that $y^3 - y - 1$ has no rational solution, so $y = 1$. Then from (1) and (2) we get $x = -1$ and $z = 0$, contradicting $z \\neq 0$.\n\nIn summary, the system has exactly two solutions:\n$$\n\\begin{cases}\nx = 0, \\\\\ny = 0, \\\\\nz = 0\n\\end{cases}\n\\quad \\text{and} \\quad\n\\begin{cases}\nx = -1, \\\\\ny = 1, \\\\\nz = 0\n\\end{cases}\n$$\n\n**Answer:** (B)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19017,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_n$ be $n$ non-empty subsets of a finite set $A$ of real numbers satisfying the following conditions:\n\n1. The sum of elements of $A$ is equal to $0$.\n2. Pick arbitrarily a number from each $A_i$, and their sum is strictly positive.\n\nProve that there exist sets $A_{i_1}, A_{i_2}, \\dots, A_{i_k}$, $1 \\le i_1 < i_2 < \\dots < i_k \\le n$, such that\n\n$$\n|A_{i_1} \\cup A_{i_2} \\cup \\dots \\cup A_{i_k}| < \\frac{k}{n} |A|.\n$$\n\n$|X|$ denotes the number of elements of a finite set $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\{a_1, \\dots, a_m\\}$ with $a_1 > \\dots > a_m$. By (1), $a_1 + \\dots + a_m = 0$. Consider the smallest element of each $A_i$; the sum of these numbers is greater than $0$ by (2). Assume there are exactly $k_i$ sets among $A_1, \\dots, A_n$ whose minimal element is $a_i$, for $i = 1, 2, \\dots, m$. Then,\n\n$$\nk_1 + \\cdots + k_m = n.\n$$\n\nBy (2),\n\n$$\nk_1 a_1 + \\cdots + k_m a_m > 0.\n$$\n\nFor $s = 1, 2, \\dots, m-1$, there are $k_1 + \\cdots + k_s$ sets whose minimal elements are at least $a_s$. The union of these sets is contained in $\\{a_1, \\dots, a_s\\}$, so the number of elements does not exceed $s$.\n\nNext, we prove that there exists $s \\in \\{1, 2, \\dots, m-1\\}$ such that $k = k_1 + \\cdots + k_s > \\frac{sn}{m}$. Suppose, for contradiction, that\n\n$$\nk_1 + \\cdots + k_s \\le \\frac{sn}{m}, \\quad s = 1, 2, \\dots, m-1.\n$$\n\nUsing the Abel transform and the fact that $a_s - a_{s+1} > 0$ for $1 \\le s \\le m-1$,\n\n$$\n\\begin{align*}\n0 <\\ & \\sum_{j=1}^{m} k_j a_j \\\\\n=\\ & \\sum_{s=1}^{m-1} (a_s - a_{s+1}) (k_1 + \\cdots + k_s) + a_m (k_1 + \\cdots + k_m) \\\\\n\\le\\ & \\sum_{s=1}^{m-1} (a_s - a_{s+1}) \\frac{sn}{m} + a_m n \\\\\n=\\ & \\frac{n}{m} \\sum_{j=1}^{m} a_j = 0.\n\\end{align*}\n$$\n\nThis is a contradiction. Thus, for such an $s$, take the sets among $A_1, \\dots, A_n$ whose minimal elements are at least $a_s$, say $A_{i_1}, A_{i_2}, \\dots, A_{i_k}$. Then $k = k_1 + \\cdots + k_s > \\frac{sn}{m}$, and the number of elements in their union does not exceed $s$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19018,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ and $g$ be functions such that for all real numbers $x$ and $y$, the following holds:\n$$\nx f(y) + g(x) = y f(x) + g(y).\n$$\nFind all possible pairs of functions $(f, g)$ satisfying this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = f(0)$ and $b = g(0)$. Due to the symmetry, the given equation implies\n$$\nx f(y) + g(x) = y f(x) + g(y). \\tag{1}\n$$\nSet $y = 0$ in (1):\n$$\ng(x) = b - a x.\n$$\nSet $y = 1$ in (1):\n$$\nf(x) = (f(1) - a)x + a = a + c x.\n$$\nSubstituting $g(x)$ and $f(x)$ into the original equation, we find $c = 0$ and $b = a$.\n\nThus, all solutions are:\n$$\nf(x) = a, \\quad g(x) = a - a x, \\quad \\text{for all } a \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19019,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f(n)$ from the positive integers to the positive integers which satisfy the following condition: whenever $a$, $b$ and $c$ are positive integers such that $\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{c}$, then\n\n$$\n\\frac{1}{f(a)} + \\frac{1}{f(b)} = \\frac{1}{f(c)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $f(mn) = m f(n)$ for all positive integers $m$ and $n$. We prove this by induction on $m$.\n\nIt is clearly true for $m = 1$ (for all $n$).\n\nSuppose it is true for $m = k$. For arbitrary $n$, let $a = (k+1)n$, $b = k(k+1)n$, and $c = k n$. Note that\n\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{(k+1)n} + \\frac{1}{k(k+1)n} = \\frac{1}{n} \\left( \\frac{1}{k+1} + \\frac{1}{k(k+1)} \\right)\n$$\n\nSince $\\frac{1}{k} = \\frac{1}{k+1} + \\frac{1}{k(k+1)}$, it follows that\n\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{n} \\cdot \\frac{1}{k} = \\frac{1}{c}\n$$\n\nSo we can use the condition:\n\n$$\n\\frac{1}{f(a)} + \\frac{1}{f(b)} = \\frac{1}{f(c)}\n$$\n\nBy the induction hypothesis,\n\n$$\n\\frac{1}{f((k+1)n)} + \\frac{1}{f(k(k+1)n)} = \\frac{1}{f(kn)}\n$$\n\nBut $f(k(k+1)n) = k f((k+1)n)$ and $f(kn) = k f(n)$, so\n\n$$\n\\frac{1}{f((k+1)n)} + \\frac{1}{k f((k+1)n)} = \\frac{1}{k f(n)}\n$$\n\nFactorizing,\n\n$$\n\\frac{k+1}{k} \\cdot \\frac{1}{f((k+1)n)} = \\frac{1}{k} \\cdot \\frac{1}{f(n)}\n$$\n\nSo $f((k+1)n) = (k+1) f(n)$, completing the induction.\n\nThus, $f(mn) = m f(n)$ for all positive integers $m$ and $n$. In particular, $f(m) = m f(1)$, so all such functions are of the form $f(n) = c n$ for some positive integer constant $c$. All such functions satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19020,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle A < 90^\\circ$ and $AB \\neq AC$. Denote by $H$ the orthocenter 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": "See solution",
    "solution": "Let $O$ be the circumcenter of triangle $ABC$. Rays $AO$ and $AH$ are isogonal. We deduce that $\\angle NAD \\equiv \\angle OAD$.  \n\nBut $OM \\parallel AH$ and $OM = \\frac{1}{2}AH = AN$, which means that $MOAN$ is a parallelogram. It follows that $MN \\parallel OA$.  \n\nCombining the above 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![](images/RMC_2015_BT_p97_data_c8615c2fe9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19021,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a natural number $m$ *remarkable* if there exist integers $a, b, c$ such that\n$$\nm = a^3 + 2b^3 + 4c^3 - 6abc.\n$$\nProve that there exists a natural number $n < 2024$ such that for infinitely many prime numbers $p$, the number $np$ is remarkable.",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma. Let $p$ be a prime number and $a, b, c \\in \\mathbb{Z}/p\\mathbb{Z}$. Then there exist $x, y, z \\in \\mathbb{Z}$ such that $|x|, |y|, |z| < \\sqrt[3]{p}$, $(x, y, z) \\neq (0, 0, 0)$, and $ax + by + cz \\equiv 0 \\pmod{p}$.\n\n*Proof.* Consider the set\n$$\nM := \\{(x, y, z) : x, y, z \\in \\{0, 1, \\dots, \\lfloor \\sqrt[3]{p} \\rfloor\\}\\}.\n$$\nWe have $|M| > p$, so by the pigeonhole principle, there are two distinct elements $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in $M$ such that $ax_1 + by_1 + cz_1 \\equiv ax_2 + by_2 + cz_2 \\pmod{p}$. Thus $(x_1 - x_2, y_1 - y_2, z_1 - z_2)$ satisfies the lemma's conditions.\n\nNow let $p \\equiv 2 \\pmod{3}$. Then the congruence $x^3 \\equiv 2 \\pmod{p}$ has a solution $a$, since the map $x \\mapsto x^3$ is injective in $\\mathbb{Z}/p\\mathbb{Z}$ (because $(3, p-1) = 1$), hence surjective. Thus, $x^3 \\equiv 2 \\pmod{p}$ is solvable.\n\nFrom the lemma, there exist $x, y, z$ with $|x|, |y|, |z| < \\sqrt[3]{p}$ such that $x + ay + a^2z \\equiv 0 \\pmod{p}$. Expanding,\n$$\nx^3 + a^3y^3 + a^6z^3 - 3a^3xyz \\equiv 0 \\pmod{p}.\n$$\nBut $a^3 \\equiv 2 \\pmod{p}$ and $a^6 \\equiv 4 \\pmod{p}$, so\n$$\nx^3 + 2y^3 + 4z^3 - 6xyz \\equiv 0 \\pmod{p}.\n$$\nAlso, $|x|, |y|, |z| < \\sqrt[3]{p}$ implies $|x^3 + 2y^3 + 4z^3 - 6xyz| < 13p$. If $x^3 + 2y^3 + 4z^3 - 6xyz < 0$, then $(-x, -y, -z)$ gives a positive multiple of $p$.\n\nIt remains to show $x^3 + 2y^3 + 4z^3 - 6xyz \\neq 0$. If it were zero, then\n$$\n(x + \\sqrt[3]{2}y + \\sqrt[3]{4}z)\\left((x - \\sqrt[3]{2}y)^2 + (x - \\sqrt[3]{4}z)^2 + (\\sqrt[3]{2}y - \\sqrt[3]{4}z)^2\\right) = 0,\n$$\nwhich only has the integer solution $(x, y, z) = (0, 0, 0)$, since $\\sqrt[3]{2}$ is irrational and $x^3 - 2$ is its minimal polynomial over $\\mathbb{Q}$.\n\nTherefore, for $n$ equal to any of the finitely many possible values $|x^3 + 2y^3 + 4z^3 - 6xyz|/p$ (with $n < 13$), there are infinitely many primes $p \\equiv 2 \\pmod{3}$ such that $np$ is remarkable. Since $n < 13 < 2024$, such an $n$ exists.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19022,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a non-constant arithmetic progression $a_1, a_2, \\dots, a_n, \\dots$. Suppose there exist relatively prime positive integers $p > 1$ and $q > 1$ such that $a_1^2$, $a_{p+1}^2$, and $a_{q+1}^2$ are also terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = a$. We have\n\n$$\na^2 = a + k d, \\quad (a + p d)^2 = a + l d, \\quad (a + q d)^2 = a + m d.\n$$\n\nThus,\n\n$$\na + l d = (a + p d)^2 = a^2 + 2 p a d + p^2 d^2 = a + k d + 2 p a d + p^2 d^2.\n$$\n\nSince the AP is non-constant, $d \\neq 0$. Hence $2 p a + p^2 d = l - k$. Similarly, $2 q a + q^2 d = m - k$. Note $p^2 q - p q^2 \\neq 0$ (otherwise $p = q$ and $\\gcd(p, q) = p > 1$, contradicting $\\gcd(p, q) = 1$). Thus, we can solve for $a$ and $d$:\n\n$$\na = \\frac{p^2 (m - k) - q^2 (l - k)}{2 (p^2 q - p q^2)}, \\quad d = \\frac{q (l - k) - p (m - k)}{p^2 q - p q^2}.\n$$\n\nIt follows that $a$ and $d$ are rational numbers. Also,\n\n$$\np^2 a^2 = p^2 a + k p^2 d.\n$$\n\nBut $p^2 d = l - k - 2 p a$, so\n\n$$\np^2 a^2 = p^2 a + k (l - k - 2 p a) = (p - 2 k) p a + k (l - k).\n$$\n\nThis shows $p a$ satisfies\n\n$$\nx^2 - (p - 2 k) x - k (l - k) = 0.\n$$\n\nSince $a$ is rational, $p a$ is rational. Write $p a = w / z$ with $w$ integer, $z$ natural, $\\gcd(w, z) = 1$. Substituting,\n\n$$\nw^2 - (p - 2 k) w z - k (l - k) z^2 = 0.\n$$\n\nSo $z$ divides $w$. Since $\\gcd(w, z) = 1$, $z = 1$ and $p a$ is integer. (Any rational root of a monic integer polynomial is integer.) Similarly, $q a$ is integer. Since $\\gcd(p, q) = 1$, there exist integers $u, v$ with $p u + q v = 1$, so $a = (p a) u + (q a) v$ is integer.\n\nAlso, $p^2 d = l - k - 2 p a$ is integer, and similarly $q^2 d$ is integer. Since $\\gcd(p^2, q^2) = 1$, $d$ is integer. Thus, all terms of the AP are integers.\n\n**Alternatively:**\n\nLet $a = u / v$, $d = x / y$ with $u, x$ integers, $v, y$ natural, $\\gcd(u, v) = 1$, $\\gcd(x, y) = 1$. From the relations:\n\n$$\na^2 = a + k d, \\quad p^2 d + 2 p a = n_1, \\quad q^2 d + 2 q a = n_2,\n$$\nwhere $n_1 = l - k$, $n_2 = m - k$.\n\nPlugging in:\n\n$$\nu^2 y = u v y + k x v^2, \\quad (1)\n$$\n$$\n2 p u y + p^2 v x = v y n_1, \\quad (2)\n$$\n$$\n2 q u y + q^2 v x = v y n_2. \\quad (3)\n$$\n\n(1) shows $v | u^2 y$. Since $\\gcd(u, v) = 1$, $v | y$. (2) shows $y | p^2 v x$, so $y | p^2 v$. (3) gives $y | q^2 v$. Thus $y | \\gcd(p^2 v, q^2 v) = v$. So $v | y$ and $y | v$ imply $v = y$.\n\nSubstitute into (1): $u^2 = u v + k x v$, so $v | u^2$. Since $\\gcd(u, v) = 1$, $v = 1$. Thus $v = y = 1$, so $a = u$, $d = x$ are integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19023,
    "subject": "Mathematics (Olympiad)",
    "question": "The sum of four real numbers is $9$ and the sum of their squares is $21$. Prove that these numbers can be denoted by $a, b, c, d$ so that $ab - cd \\ge 2$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\ge b \\ge c \\ge d$. Since $a + b + c + d = 9$, there exists $x \\ge 0$ such that\n\n$$\n\\frac{a + b}{2} = \\frac{9}{4} + x, \\quad \\frac{c + d}{2} = \\frac{9}{4} - x\n$$\n\nBecause $a \\ge b$ and $c \\ge d$, we can write\n\n$$\na = \\frac{9}{4} + x + y, \\quad b = \\frac{9}{4} + x - y, \\quad c = \\frac{9}{4} - x + z, \\quad d = \\frac{9}{4} - x - z\n$$\n\nfor some $y, z \\ge 0$. Since $b \\ge c$, we have\n\n$$\nx - y \\ge -x + z \\iff y + z \\le 2x\n$$\n\nFrom the sum of squares,\n\n$$\n\\begin{aligned}\n21 &= (a^2 + b^2) + (c^2 + d^2) \\\\\n   &= 2 \\left(\\frac{9}{4} + x\\right)^2 + 2y^2 + 2 \\left(\\frac{9}{4} - x\\right)^2 + 2z^2 \\\\\n   &= 4 \\left(\\frac{9}{4}\\right)^2 + 4x^2 + 2y^2 + 2z^2 \\\\\n   &= 20 + \\frac{1}{4} + 2(2x^2 + y^2 + z^2)\n\\end{aligned}\n$$\n\nSo,\n\n$$\n2x^2 + y^2 + z^2 = \\frac{3}{8} \\tag{1}\n$$\n\nUsing $y + z \\le 2x$, $y^2 + z^2 \\le (y + z)^2$, and (1):\n\n$$\n\\begin{aligned}\n\\frac{3}{8} \\le 2x^2 + (y + z)^2 &\\le 2x^2 + 4x^2 = 6x^2 \\\\\nx^2 &\\ge \\frac{1}{6} \\cdot \\frac{3}{8} = \\frac{1}{16} \\implies x \\ge \\frac{1}{4}\n\\end{aligned}\n$$\n\nNow, using (1) again:\n\n$$\n\\begin{aligned}\nab - cd &= \\left(\\frac{9}{4} + x\\right)^2 - y^2 - \\left(\\frac{9}{4} - x\\right)^2 + z^2 \\\\\n&= 9x - y^2 + z^2 \\\\\n&= 9x - (2x^2 + y^2 + z^2) + z^2 \\\\\n&= 9x + 2x^2 - \\frac{3}{8} + 2z^2 \\\\\n&\\ge 9 \\cdot \\frac{1}{4} + 2 \\cdot \\frac{1}{16} - \\frac{3}{8} = 2\n\\end{aligned}\n$$\n\nand we are done. $\\square$\n\nAlternatively, consider two cases:\n\n**Case 1.** If $a + b \\ge 5$, then since $c^2 + d^2 \\ge 2cd$,\n\n$$\n\\begin{aligned}\na^2 + b^2 + 2ab &= (a + b)^2 \\ge 25 \\\\\n&= 4 + a^2 + b^2 + c^2 + d^2 \\ge 4 + a^2 + b^2 + 2cd\n\\end{aligned}\n$$\n\nso $ab - cd \\ge 2$.\n\n**Case 2.** If $a + b < 5$, then $c + d > 4$ and $4 < c + d \\le a + b < 5$. From $(a - d)(b - c) \\ge 0$ and $(a - b)(c - d) \\ge 0$, it follows $ab + cd \\ge ac + bd \\ge ad + bc$. Since\n\n$$\n\\begin{aligned}\n(ab + cd) + (ac + bd) + (ad + bc) = \\frac{(a + b + c + d)^2 - (a^2 + b^2 + c^2 + d^2)}{2} = 30\n\\end{aligned}\n$$\n\nwe have $ab + cd \\ge 10$. Also, $(a + b)(c + d) \\ge 20$, and\n\n$$\n(a + b)^2 + (c + d)^2 + 2(a + b)(c + d) = 81\n$$\n\nso $(a + b)^2 + (c + d)^2 < 41$. But\n\n$$\n\\begin{aligned}\n41 &= 21 + 2 \\cdot 10 \\\\\n  &\\le (a^2 + b^2 + c^2 + d^2) + 2(ab + cd) \\\\\n  &= (a + b)^2 + (c + d)^2 \\\\\n  &< 41\n\\end{aligned}\n$$\n\na contradiction. Thus, $ab - cd \\ge 2$ always holds. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19024,
    "subject": "Mathematics (Olympiad)",
    "question": "Lemma 1: For a positive integer $n$ and a prime $p \\mid n$, we have\n$$\nf(n) \\le \\sum_{d \\mid \\frac{n}{p}} f(d).\n$$\n\nLemma 2: For a positive integer $n$, let $g(n) = \\sum_{d \\mid n} \\frac{d}{P(d)}$, then $g(n) \\le n$.\n\nProve that for any positive integer $n$, the inequality\n$$\nf(n) \\le \\frac{n}{P(n)}\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "Proof of Lemma 1: For any factoring of $n$, write $n = n_1 n_2 \\cdots n_k$. Since $p \\mid n$, there exists $i \\in \\{1, \\ldots, k\\}$ such that $p \\mid n_i$. Without loss of generality, assume $i = 1$. Map this factoring to a factoring of $d = \\frac{n}{n_1}$, $d = n_2 n_3 \\cdots n_k$.\n\nFor two different factorizations of $n$, if $n_1 = n_1'$, then $d = n_2 \\cdots n_k$ and $d = n_2' \\cdots n_k'$ are two different factorizations of $d$ (where $d$ is a divisor of $\\frac{n}{p}$). If $n_1 \\ne n_1'$, then $d = \\frac{n}{n_1} \\ne \\frac{n}{n_1'} = d'$, so these two factorizations map to factorizations of $d$ and $d'$ respectively (where $d$ and $d'$ are divisors of $\\frac{n}{p}$). Thus,\n$$\nf(n) \\le \\sum_{d \\mid \\frac{n}{p}} f(d).\n$$\n\nProof of Lemma 2: Induct on the number of distinct prime divisors of $n$.\n\n- Base case: $n = 1$, $g(1) = 1$.\n- If $n = p^a$ is a prime power,\n  $$\n  g(n) = 1 + 1 + p + \\cdots + p^{a-1} = 1 + \\frac{p^a - 1}{p-1} \\le 1 + p^a - 1 = n.\n  $$\n- Inductive step: Suppose $n$ has $k+1$ distinct prime divisors, $n = p_1^{a_1} \\cdots p_k^{a_k} p_{k+1}^{a_{k+1}}$, write $n = m p_{k+1}^{a_{k+1}}$.\n  $$\n  g(n) = g(m) + \\sum_{d \\mid m} \\sum_{i=1}^{a_{k+1}} \\frac{d p_{k+1}^i}{p_{k+1}} = g(m) + \\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1},\n  $$\n  where $\\sigma(m)$ is the sum of positive divisors of $m$. By the inductive hypothesis, $g(m) \\le m$.\n\nSince\n$$\n\\begin{align*}\n\\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} &= \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_i - 1} \\right) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} \\\\\n&\\le \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_i} \\right) (p_{k+1}^{a_{k+1}} - 1) \\\\\n&\\le \\left( \\prod_{i=1}^{k} p_i^{a_i} \\right) (p_{k+1}^{a_{k+1}} - 1) \\\\\n&= n - m,\n\\end{align*}\n$$\nso $g(n) \\le n$. Lemma 2 is proved.\n\nNow, to prove for all positive integers $n$ that\n$$\nf(n) \\le \\frac{n}{P(n)}.\n$$\nInduct on $n$. For $n=1$, equality holds. Assume for $n=1,2,\\ldots,k$, $f(n) \\le \\frac{n}{P(n)}$. For $n = k+1$, by Lemmas 1 and 2 and the inductive hypothesis,\n$$\nf(k+1) \\le \\sum_{d \\mid \\frac{k+1}{P(k+1)}} f(d) \\le \\sum_{d \\mid \\frac{k+1}{P(k+1)}} \\frac{d}{P(d)} = g\\left(\\frac{k+1}{P(k+1)}\\right) \\le \\frac{k+1}{P(k+1)}.\n$$\nThus, the inequality holds for all positive integers $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19025,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the matrices $A \\in \\mathcal{M}_{m,n}(\\mathbb{C})$, $B \\in \\mathcal{M}_{n,m}(\\mathbb{C})$ with $n \\leq m$. Given that $\\text{rank}(AB) = n$ and $(AB)^2 = AB$, find $BA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Left-multiply by $B$ 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. From $ABAB = AB$, we have $\\text{rank}(BA) \\geq n$, hence $\\text{rank}(BA) = n$.\n\nThe square matrix $BA$ of order $n$ is thus invertible, so $$(BA)^3 = (BA)^2$$ implies $BA = I_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19026,
    "subject": "Mathematics (Olympiad)",
    "question": "Decide whether the integers $1, 2, \\ldots, 100$ can be arranged in the cells $C(i,j)$ of a $10 \\times 10$ matrix (where $1 \\leq i, j \\leq 10$), such that the following conditions are satisfied:\n\n1. In every row, the entries add up to the same sum $S$.\n2. In every column, the entries also add up to this sum $S$.\n3. For every $k = 1, \\ldots, 10$, the ten entries $C(i,j)$ with $i-j \\equiv k \\pmod{10}$ add up to $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "The problem essentially asks for a magic square that satisfies an additional constant-sum property along the wrap-around diagonals.\n\nSuppose that such an arrangement of $1, 2, \\ldots, 100$ is possible. Since the sum of all entries is $\\frac{1}{2} \\cdot 100 \\cdot 101 = 5050$, we get that $S = 505$ is an odd number (since $5050/10 = 505$).\n\nWe partition the cells $C(i,j)$ into four sets:\n\n- Set $A$ contains the cells with $i$ and $j$ both odd.\n- Set $B$ contains the cells with odd $i$ and even $j$.\n- Set $C$ contains the cells with even $i$ and odd $j$.\n- Set $D$ contains the remaining cells with $i$ and $j$ both even.\n\nLet $S_A, S_B, S_C, S_D$ be the sums of the entries in $A, B, C, D$ respectively.\n\n- Since $A$ and $B$ contain all cells in the odd rows, $S_A + S_B = 5S$.\n- Since $B$ and $D$ contain all cells in the even columns, $S_B + S_D = 5S$.\n- Since $A$ and $D$ contain all cells $C(i,j)$ with even $i-j$, $S_A + S_D = 5S$.\n\nAdding these three equations gives:\n\n$$\n2(S_A + S_B + S_D) = 15S\n$$\n\nBut the left side is even and the right side is odd, which is a contradiction. Therefore, such an arrangement is not possible.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 19027,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be any nonnegative integral power of $2$.\n\nFind all $n$ such that there exists an integer $m$ with $2^n - 1 \\mid m^2 + 81$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n$ has an odd divisor $d \\ge 3$. Then $2^d - 1 \\mid 2^n - 1 \\mid m^2 + 81$. Since $2^d - 1 \\equiv 3 \\pmod{4}$ and $3 \\nmid 2^d - 1$, there exists an odd prime $p > 3$ with $p \\equiv 3 \\pmod{4}$ and $p \\mid 2^d - 1$. Thus, $p \\mid m^2 + 81$, so $m^2 \\equiv -9^2 \\pmod{p}$, i.e., $(9^{-1}m)^2 \\equiv -1 \\pmod{p}$. But $-1$ is not a quadratic residue modulo $p \\equiv 3 \\pmod{4}$, a contradiction. Therefore, $n$ has no odd divisor $>1$, so $n = 2^k$ for some $k \\ge 0$.\n\nNow, we seek $m$ such that $2^{2^k} - 1 \\mid m^2 + 81$. Note that\n\n$$\n2^{2^k} - 1 = (2 + 1)(2^2 + 1)\\cdots(2^{2^{k-1}} + 1).\n$$\n\nThese factors are pairwise coprime. By the Chinese Remainder Theorem, there exists $m \\in \\mathbb{Z}^+$ such that $3 \\mid m$ and\n\n$$\nm \\equiv 9 \\cdot 2^{2^{j-1}} \\pmod{2^{2^j} + 1}\n$$\n\nfor $j = 1, 2, \\dots, k-1$. For this $m$, $3 \\mid m^2 + 81$ and\n\n$$\nm^2 + 81 \\equiv 81(2^{2^j} + 1) \\equiv 0 \\pmod{2^{2^j} + 1}.\n$$\n\nHence, $2^{2^k} - 1 \\mid m^2 + 81$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19028,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $w_1$ and $w_2$ intersect at the points $P$ and $Q$. Suppose $AB$ and $CD$ are parallel diameters of the circles $w_1$ and $w_2$ respectively. None of the points $A, B, C, D$ coincides with $P$ or $Q$, and the points are located on the circles in the following order: $A, B, P, Q$ on $w_1$ and $C, D, P, Q$ on $w_2$. Lines $AP$ and $BQ$ intersect at point $X$, and lines $CP$ and $DQ$ intersect at $Y \\neq X$. Prove that, regardless of the choice of parallel diameters $AB$ and $CD$, all the lines $XY$ intersect at one point or are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_1$ be the center of circle $w_1$ and $X' = BP \\cap AQ$ as shown in the figure below.\n\nSince $AP \\perp BP$ and $BQ \\perp AQ$, $AP$ and $BQ$ are altitudes of triangle $ABX'$, so $X$ is its orthocenter. Then $X'X$ is also an altitude in the triangle, so $X'X \\perp AB$ and $\\angle PX'X = 90^\\circ - \\angle ABX' = \\angle PAB$. This implies that triangles $\\triangle APB$ and $\\triangle X'PX$ have the same angles, so there is a rotation and homothety that transform $\\triangle APB$ into $\\triangle X'PX$. The center of these transformations is $P$, the rotation is by $90^\\circ$, and the homothety coefficient is $\\frac{XP}{BP} = \\angle XBP = \\frac{1}{2}\\angle QO_1P$.\n\nLet $M_1$ be the midpoint of $XX'$. The segments $PO_1$ and $PM_1$ are medians of triangles $\\triangle APB$ and $\\triangle X'PX$ respectively, so $PM_1 \\perp PO_1$, meaning $PM_1$ is tangent to $w_1$. Similarly, $QM_1$ is tangent to $w_1$ as well. Thus, $M_1$ is the intersection of tangents to $w_1$ from $P$ and $Q$, which does not depend on the choice of $A$ and $B$. Note that the length $XX' = AB \\cdot \\frac{1}{2} \\angle QO_1P$ does not depend on $A$ and $B$, since the diameter $AB$ and the angle $\\angle QO_1P$ are constant.\n\nAnalogously, construct the point $M_2$ for circle $w_2$. Assume $T = XY \\cap M_1M_2$ (the case when $XY \\parallel M_1M_2$ will be considered further). We will show that $T$ is fixed, which implies that all lines $XY$ intersect at one point regardless of the choice of diameters $AB$ and $CD$.\n\nNote that $M_1X$ and $M_2Y$ are parallel since both are perpendicular to $AB \\parallel CD$, so $\\frac{M_1T}{M_2T} = \\frac{M_1X}{M_2Y}$. The points $M_1$ and $M_2$ as well as the ratio $\\frac{M_1X}{M_2Y} = \\frac{\\frac{XX'}{2}}{\\frac{YY'}{2}} = \\frac{XX'}{YY'}$ are fixed, therefore $T$ is also fixed.\n\nThe lines $XY$ and $M_1M_2$ are parallel and do not coincide if and only if segments $M_1X$ and $M_2Y$ are equal and parallel. These properties hold for different choices of diameters $AB$ and $CD$. Hence, if $XY$ and $M_1M_2$ are parallel, it will imply that $XY$ is parallel to $M_1M_2$ for all other choices of diameters $AB$ and $CD$.\n\nThe lines $XY$ and $M_1M_2$ coincide if and only if $AB \\parallel CD \\perp M_1M_2$. In this case, one may consider another choice of diameters $AB$ and $CD$ to determine which of the two cases above applies. Note that the initial line $XY = M_1M_2$ satisfies both cases.\n\n![](images/Ukraine2022-23_p49_data_1568ec6021.png)\n\nFig. 26",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19029,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be positive real numbers with $x + y = 1$.\n\nProve that\n\n$$\n\\frac{(3x - 1)^2}{x} + \\frac{(3y - 1)^2}{y} \\geq 1.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\n\\frac{(3x - 1)^2}{x} + \\frac{(3y - 1)^2}{y} &= \\frac{9x^2 - 6x + 1}{x} + \\frac{9y^2 - 6y + 1}{y} \\\\\n&= 9x - 6 + \\frac{1}{x} + 9y - 6 + \\frac{1}{y} \\\\\n&= -3 + \\frac{1}{x} + \\frac{1}{y}.\n\\end{aligned}\n$$\n\nIt remains to show that\n\n$$\n\\frac{1}{x} + \\frac{1}{y} \\geq 4.\n$$\n\nThis follows from the inequality between the arithmetic and harmonic means and the condition $x + y = 1$:\n\n$$\n(x + y) \\left( \\frac{1}{x} + \\frac{1}{y} \\right) \\geq 4.\n$$\n\nEquality holds exactly for $x = y$, that is, $x = y = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19030,
    "subject": "Mathematics (Olympiad)",
    "question": "For simplicity, write $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc. Show that\n\n$$\nf^2(x) = x \\quad \\text{for all } x > 0\n$$\n\nin two different ways.",
    "options": [],
    "answer": "See solution",
    "solution": "**Method 1, using injectivity.** Suppose $f(z_1) = f(z_2)$. Replacing $z$ by $z_1$ or $z_2$ leads to the same left-hand side of the functional equation, so the right-hand side must also agree: $z_1 + f(y) + f^2(x) = z_2 + f(y) + f^2(x)$, which implies $z_1 = z_2$. Thus, $f$ is injective.\n\nWe make the right-hand side of the functional equation equal to $z + f^2(y) + f^2(x)$ in two different ways, to use the injectivity of $f$. Replacing $y$ by $f(y)$ gives $f(x + f^2(y) + f^2(z)) = z + f^2(y) + f^2(x)$. Replacing $x$ by $y$ and $y$ by $f(x)$ gives $f(y + f^2(x) + f^2(z)) = z + f^2(x) + f^2(y)$. Since the right-hand sides are the same and $f$ is injective, it follows that $x + f^2(y) + f^2(z) = y + f^2(x) + f^2(z)$, i.e., $f^2(x) - x = f^2(y) - y$ for all positive $x, y$.\n\nThis means $f^2(x) - x = c$ is a constant independent of $x$. Setting $t = 1 + f(1) + f^2(1)$, the functional equation with $x = y = z = 1$ gives $f(t) = t$, so $f^2(t) = t$ and $f^2(t) - t = 0$. Therefore, $c = 0$ and we have shown $f^2(x) = x$ for all $x > 0$.\n\n**Method 2, using fixed points.** Let $x = z$ in the functional equation to get\n\n$$\nf(x + f(y) + f(f(x))) = x + f(y) + f(f(x))\n$$\n\nso $x + f(y) + f(f(x))$ is a fixed point for all $x, y > 0$.\n\nLet $a$ be any fixed point of $f$. Setting $x = y = a$ gives\n\n$$\nf(2a + f(f(z))) = z + 2a, \\text{ for } z > 0.\n$$\n\nSetting $y = z = a$ gives\n\n$$\nf(x + 2a) = 2a + f(f(x)), \\text{ for } x > 0.\n$$\n\nApplying $f$ to both sides and using the previous result gives\n\n$$\nf(f(x + 2a)) = x + 2a, \\text{ for } x > 0.\n$$\n\nReplacing $x$ by $x + 2a$ gives\n\n$$\nf(x + 4a) = 2a + f(f(x + 2a)) = x + 4a, \\text{ for } x > 0.\n$$\n\nSo $f(w) = w$ for all integers $w > 4a$. Pick such a $w$ and set $y = z = w$ in the original equation to get\n\n$$\nf(x + f(w) + f(f(w))) = w + f(w) + f(f(x))\n$$\n\nwhich simplifies to $x + 2w = f(x + 2w) = 2w + f(f(x))$ for $x > 0$.\n\nThis shows $f(f(x)) = x$ for all $x > 0$.\n\nHaving established $f^2(x) = x$, we can complete the solution. For integer $n \\geq 3$, set $x = n - 2$, $y = f(1)$, $z = 1$ in the functional equation. Using $f^2(x) = x$, this gives $f(n) = n$ for all $n \\geq 3$. Because $f^2(x) = x$, we must have $f(1) < 3$ and $f(2) < 3$. Also, $f(1) = 2$ implies $f(2) = 1$ and vice versa. Thus, there are two possible solutions:\n\n$$\n\\begin{aligned}\nf_1(n) &= n \\text{ for } n \\geq 0 \\\\\nf_2(n) &= n \\text{ for } n \\geq 3,\\ f_2(1) = 2,\\ f_2(2) = 1.\n\\end{aligned}\n$$\n\nClearly, $f_1$ satisfies the functional equation. To verify $f_2$, note $x + f_2(y) + f_2^2(z) \\geq 3$, so the equation is satisfied if and only if\n\n$$\nx + f_2(y) + f_2^2(z) = z + f_2(y) + f_2^2(x),\n$$\n\ni.e., $f_2^2(x) - x = f_2^2(z) - z$ for all $x, z \\in \\mathbb{Z}_+$. Clearly $f_2^2(z) - z = 0$ for $z \\geq 3$. Since $f_2(f_2(1)) = f_2(2) = 1$ and $f_2(f_2(2)) = f_2(1) = 2$, we see $f_2^2(x) - x = 0$ for all $x \\in \\mathbb{Z}_+$. Thus, $f_2$ is also a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19031,
    "subject": "Mathematics (Olympiad)",
    "question": "We apply the following lemma (Sylvester's theorem):\n\nLet two non-negative integers $a$ and $b$ be coprime. The largest non-negative integer that cannot be written in the form $ax + by$ where $x$ and $y$ are non-negative integers is $N_0 = ab - a - b$.\n\n---\n\na) Let $a$ be a value satisfying the conditions. An integer $n$ is called *good* if there exists a tuple of non-negative numbers $(x, y, z)$ such that\n\n$$\na^2x + 6ay + 36z = n.\n$$\n\nIt is easy to prove that $(a, 6) = 1$.\n\n---\n\nb) Prove the following general statement: Let $a$ and $b$ be two coprime numbers. Then\n\n$$\nN = a^2b + ab^2 - a^2 - b^2 - ab + 1\n$$\n\nis the smallest positive integer such that the equation $a^2x + aby + b^2z = m$ has a non-negative integer solution $(x, y, z)$ for every $m \\geq N$.",
    "options": [],
    "answer": "See solution",
    "solution": "a)\n\nWe require\n\n$$\n5a^2 - 6 < 250 \\quad \\text{and} \\quad 35a - 36 < 250\n$$\n\nwhich gives $a < 7$. Combined with $(a, 6) = 1$, the possible values are $a = 1$ or $a = 5$.\n\nFor $a = 1$, the tuple $(n, 0, 0)$ satisfies $x + 6y + 36z = n$ for any $n$.\n\nFor $a = 5$, we show that every integer $n \\geq 250$ can be written as $25x + 30y + 36z$.\n\nRewrite:\n\n$$\n25x + 30y + 36z = 25x + 6(5y + 6z - 20) + 120.\n$$\n\nAny integer $\\geq 20$ can be written as $5y + 6z$, so any integer can be written as $5y + 6z - 20$. Let $u = 5y + 6z - 20$. Any integer $\\geq 120$ can be written as $25x + 6u$, so any integer $\\geq 250$ can be written as $25x + 6u + 120$. Thus, $a = 1$ and $a = 5$ are the answers.\n\n---\n\nb)\n\nLet $N = a^2b + ab^2 - a^2 - b^2 - ab + 1$. We claim that for all $m \\geq N$, the equation $a^2x + aby + b^2z = m$ has a non-negative integer solution.\n\nRewrite:\n\n$$\nm - N = a(ax + by - (ab - a - b + 1)) + b^2z.\n$$\n\nSince $m - a^2b + a^2 + ab - a \\geq b^2a - b^2 - a + 1$, there exist non-negative integers $u$ and $z$ such that\n\n$$\nm - a^2b + a^2 + ab - a = au + b^2z.\n$$\n\nThere also exist non-negative integers $x$ and $y$ such that $u + ab - a - b + 1 = ax + by$.\n\nTherefore,\n\n$$\nm = a^2x + aby + b^2z.\n$$\n\nNext, we show that $a^2b + ab^2 - a^2 - b^2 - ab$ cannot be written as $a^2x + aby + b^2z$ for any non-negative $x, y, z$.\n\nSuppose otherwise, then\n\n$$\na^2b + ab^2 - a^2 - b^2 - ab = a^2x + aby + b^2z.\n$$\n\nThis leads to $a^2(b - 1 - x) + b^2(a - 1 - z) = ab(y + 1)$. Thus, $b - 1 - x > 0$ or $a - 1 - z > 0$. Without loss of generality, assume $b - 1 - x > 0$.\n\nThen,\n\n$$\na^2(b - 1 - x) = b(ay + bz - ab + b + a).\n$$\n\nSince $(a^2, b) = 1$, $b - 1 - x$ must be divisible by $b$, which is impossible for $0 < b - 1 - x < b$. This is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19032,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a, b, c\\}$ be a set of three positive digits with MMR equal to $\\frac{4}{5}$, where $a \\le b \\le c$. The median is $b$ and the mean is $\\frac{a+b+c}{3}$. What are all possible sets $\\{a, b, c\\}$?\n\nb) For a set $\\{a, b, c\\}$ of positive integers with $a \\le b \\le c$, show that the MMR is always greater than $\\frac{2}{3}$.\n\nc) For a set $\\{a, b, c, d\\}$ of four positive integers with $a \\le b \\le c \\le d$, show that the MMR is always greater than $\\frac{3}{4}$, and that this bound is sharp.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n$$\n\\frac{a+b+c}{3b} = \\frac{4}{5}\n$$\nSo,\n$$\n5a + 5b + 5c = 12b \\\\\n5a + 5c = 7b\n$$\nSince $a, b, c$ are digits and $b$ must divide $5a + 5c$, $b$ must be a multiple of 5. Thus, $b = 5$.\n\nThen $a + c = 7$. The possible sets are $\\{1, 5, 6\\}$ and $\\{2, 5, 5\\}$.\n\nb) Since $a \\ge 1$ and $c \\ge b$, the mean is at least $\\frac{1 + b + b}{3} = \\frac{1 + 2b}{3}$, so\n$$\n\\frac{a+b+c}{3b} \\ge \\frac{1 + 2b}{3b} = \\frac{1}{3b} + \\frac{2}{3} > \\frac{2}{3}\n$$\nAlternatively, assuming $\\frac{a+b+c}{3b} \\le \\frac{2}{3}$ leads to $a + c \\le b$, but $a + c \\ge 1 + b > b$, a contradiction. Thus, the MMR is always greater than $\\frac{2}{3}$.\n\nc) For $\\{a, b, c, d\\}$, the median is $\\frac{b + c}{2}$ and the mean is $\\frac{a + b + c + d}{4}$, so\n$$\n\\text{MMR} = \\frac{a + b + c + d}{2(b + c)}\n$$\nSuppose $a \\ge 1$, $b \\le c \\le d$. The mean is at least $\\frac{1 + b + c + d}{4}$, and $b + c \\le 2d$. Thus,\n$$\n\\frac{a + b + c + d}{2(b + c)} \\ge \\frac{1 + b + c + d}{2(b + c)}\n$$\nAssuming $\\frac{a + b + c + d}{2(b + c)} \\le \\frac{3}{4}$ leads to $2a + 2d \\le b + c$, but $2a + 2d \\ge 2 + 2c > b + c$, a contradiction. Thus, the MMR is always greater than $\\frac{3}{4}$.\n\nTo show the bound is sharp, consider $\\{4, n, n, n\\}$ for large $n$. The mean is $1 + \\frac{3}{4}n$, the median is $n$, so\n$$\n\\text{MMR} = \\frac{1}{n} + \\frac{3}{4}\n$$\nwhich approaches $\\frac{3}{4}$ as $n$ increases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19033,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\triangle ABC$ 是一銳角三角形且 $CP$ 為邊 $AB$ 上的高，$H$ 為 $CP$ 上任一點。\n直線 $AH, BH$ 分別交 $BC, AC$ 於點 $M, N$。\n\n1. 證明：$\\angle NPC = \\angle MPC$\n\n2. 設 $O$ 是 $MN$ 與 $CP$ 的交點，一條通過 $O$ 的任意直線交四邊形 $CNHM$ 的邊於 $D, E$ 兩點。證明：$\\angle EPC = \\angle DPC$。",
    "options": [],
    "answer": "See solution",
    "solution": "解：\n\n1. 記 $\\angle NPC = \\phi_1$, $\\angle MPC = \\phi_2$，則\n\n$$\n\\frac{S_{\\triangle NPC}}{S_{\\triangle NPA}} = \\frac{CN}{AN} = \\frac{CP \\sin \\phi_1}{AP \\cos \\phi_2}.\n$$\n\n所以，$\\tan \\phi_1 = \\frac{CN}{AN} \\cdot \\frac{AP}{CP}$。\n\n同理可得，$\\tan \\phi_2 = \\frac{CM}{BM} \\cdot \\frac{BP}{CP}$。\n\n於是，$\\tan \\phi_1 = \\tan \\phi_2$ 等價於\n\n$$\n\\frac{CN}{AN} \\cdot \\frac{AP}{BP} \\cdot \\frac{BM}{CM} = 1.\n$$\n\n在 $\\triangle ABC$ 中，利用 Ceva 定理，取直線 $AM, BN, CP$ 即可得上式。\n\n2. 記 $\\angle NPC = \\angle MPC = \\phi$。\n\n欲證結論，只須證明\n\n$$\n\\frac{\\sin(\\phi - x)}{\\sin x} = \\frac{\\sin(\\phi - y)}{\\sin \\gamma},\n$$\n\n其中 $x = \\angle EPO$, $\\gamma = \\angle DPO$。事實上，後者等價於\n\n$$\n\\begin{align*}\n& \\frac{\\frac{\\sin \\phi \\cos x - \\cos \\phi \\sin x}{\\sin x}}{\\sin \\gamma} \\\\\n&= \\frac{\\sin \\phi \\cos x - \\cos \\phi \\sin \\gamma}{\\sin \\gamma} \\\\\n\\Leftrightarrow \\quad & \\cot x = \\cot \\gamma \\Leftrightarrow x = \\gamma.\n\\end{align*}\n$$\n\n假定 $E \\in NH, D \\in CM$，則\n\n$$\n\\frac{NE}{EH} = \\frac{S_{\\triangle NEP}}{S_{\\triangle EHP}} = \\frac{NP \\cdot \\sin(\\phi - x)}{PH \\cdot \\sin x}.\n$$\n\n因此，\n$$\n\\frac{\\sin(\\phi - x)}{\\sin x} = \\frac{NE}{EH} \\cdot \\frac{PH}{NP} \\quad (1)\n$$\n\n同理可得，\n\n$$\n\\frac{\\sin(\\phi - \\gamma)}{\\sin \\gamma} = \\frac{DM}{CD} \\cdot \\frac{CP}{PM} \\quad (2)\n$$\n\n利用 (1) 與 (2) 只須證明\n\n$$\n\\frac{NE}{EH} \\cdot \\frac{CD}{DM} \\cdot \\frac{PH}{CP} \\cdot \\frac{MO}{NO} = 1. \\quad (3)\n$$\n\n（因為 $PO$ 是 $\\triangle NPM$ 的角平分線，即 $\\frac{PM}{PN} = \\frac{MO}{NO}$。）\n\n又因為\n\n$$\n\\frac{NE}{EH} = \\frac{S_{\\triangle NEO}}{S_{\\triangle EHO}} = \\frac{NO \\sin \\delta}{OH \\sin \\psi}, \\\\\n\\frac{CD}{DM} = \\frac{S_{\\triangle CDO}}{S_{\\triangle DMO}} = \\frac{CO \\sin \\psi}{OM \\sin \\delta},\n$$\n\n其中 $\\delta = \\angle MOD$, $\\psi = \\angle EOP$。於是，(3) 可化簡為\n\n$$\n\\frac{OC}{OH} \\cdot \\frac{PH}{PC} = 1.\n$$\n\n分別對 $\\triangle BHC$ 和直線 $MN$，$\\triangle CHM$ 和直線 $AB$，$\\triangle BHM$ 和直線 $AC$ 應用 Menelaus 定理，可得\n\n$$\n\\frac{BN}{NH} \\cdot \\frac{HO}{OC} \\cdot \\frac{CM}{MB} = 1, \\\\\n\\frac{CP}{PH} \\cdot \\frac{HA}{AM} \\cdot \\frac{MB}{BC} = 1, \\\\\n\\frac{HN}{NB} \\cdot \\frac{BC}{CM} \\cdot \\frac{MA}{HA} = 1,\n$$\n\n三式相乘得\n\n$$\n\\frac{OC}{OH} \\cdot \\frac{PH}{PC} = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19034,
    "subject": "Mathematics (Olympiad)",
    "question": "At a party with 100 people, after the first gong, all those who had exactly 0 acquaintances among those present left the party. After the second gong, all those who had exactly 1 acquaintance among those remaining left the party. This process continued: after the $k$-th gong, those who had exactly $k-1$ acquaintances among the people present at that moment left the party. At the end of the party, exactly $n$ people remained. Find all possible values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n \\in \\{0, 1, 2, \\dots, 98\\}$.\n\nLet us show that the answer is correct. For $n > 0$, divide all people at the party into two groups. Group A contains exactly $n$ people, each of whom knows every other member of the party. Group B contains $100 - n$ people, who do not know each other within the group, but each knows every member of group A. By construction, all members of B leave the party after the $(n+1)$-th gong. Then all members of A remain, each knowing $n-1$ people present at that point, so they stay until the end of the party.\n\nFor $n = 0$, suppose everyone at the party knows exactly 1 other person, i.e., there are 50 pairs of acquaintances. They leave the party after the 2nd gong.\n\n$n = 100$ is impossible, because there will always be at least one person with the smallest number of acquaintances, which is less than 100, so at some point this person would leave the party first.\n\nFor $n = 99$, we prove by contradiction that this case is impossible. Let a person $X$ be the only one who left the party before it finished. $X$ has the smallest number of acquaintances. After $X$ left, if a person $Y$ didn't know $X$, then $Y$ or someone who knows $Y$ must leave the party before it finishes. Hence, at least two people would leave the party. Otherwise, everyone must have known $X$, but $X$ has the smallest number of acquaintances, so everyone else has 99 acquaintances, and they all leave the party after the 100th gong. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19035,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, points $H$ and $O$ are the orthocenter and the circumcenter, respectively. The line $HO$ intersects the sides $AB$ and $AC$ at points $X$ and $Y$, respectively, such that $H$ lies on the segment $OX$. It is given that $XH = HO = OY$. Find the angle $\\angle BAC$.\n\n![](images/Ukraine_2021-2022_p17_data_441efab4e3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following well-known fact: in any triangle, the distance from a vertex to the orthocenter is twice the distance from the circumcenter to the opposite side.\n\nLet $O_B$ and $H_B$ be the projections of $O$ and $H$ onto the line $AC$, and define $O_C$ and $H_C$ similarly. From the fact above, $BH = 2OO_B$ and $CH = 2OO_C$. Also, $OO_B \\parallel HH_B$ and $HO = OY$, so $OO_B$ is the midline of $\\triangle YHH_B$, implying $HH_B = 2OO_B = BH$. Similarly, $HH_C$ is the midline of $\\triangle XOO_C$, so $HH_C = \\frac{1}{2}OO_C = \\frac{1}{4}CH$. Since the quadrilateral $BHC_HC_B$ is cyclic, $BH \\cdot HH_B = CH \\cdot HH_C$, and $BH^2 = 4HH_C^2$, so $BH = 2HH_C$, and $\\angle HBH_C = 30^\\circ$. Therefore, $\\angle BAC = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19036,
    "subject": "Mathematics (Olympiad)",
    "question": "Out of three expressions $\\frac{x}{y}$, $\\frac{x^2+4}{y^2+4}$, and $\\frac{x^3+8}{y^3+8}$, for some integers $x$ and $y$, two take the same value and the remaining one takes a different value. For which pairs of integers $x, y$ is this possible?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $(1, 4)$ and $(4, 1)$.\n\nObviously, $x \\neq y$, because otherwise all three expressions would take the same value, which contradicts the problem statement.\n\nIf the first two expressions are equal, then:\n\n$$\n\\frac{x}{y} = \\frac{x^2+4}{y^2+4} \\implies xy^2 + 4x = yx^2 + 4y \\implies xy(y-x) = 4(y-x).\n$$\n\nSince $x \\neq y$, we can divide by $y-x$ to get $xy=4$. The integer solutions are $(1, 4)$ and $(4, 1)$. Substitute into the third expression:\n\n$$\n\\frac{x^3+8}{y^3+8} = \\frac{9}{72} = \\frac{1}{8} \\neq \\frac{x}{y} \\quad \\text{and, analogously,} \\quad \\frac{x^3+8}{y^3+8} = 8 \\neq \\frac{x}{y},\n$$\n\nso these pairs satisfy the condition.\n\nIf the first and third expressions are equal:\n\n$$\n\\frac{x}{y} = \\frac{x^3 + 8}{y^3 + 8} \\implies xy^3 + 8x = yx^3 + 8y \\implies xy(y-x)(y+x) = 8(y-x) \\implies xy(y+x) = 8,\n$$\n\nwhich has no integer solutions.\n\nIf the second and third expressions are equal:\n\n$$\n\\begin{aligned}\n\\frac{x^2+4}{y^2+4} &= \\frac{x^3+8}{y^3+8} \\\\\n\\implies x^2y^3 + 8x^2 + 4y^3 = x^3y^2 + 8y^2 + 4x^3 \\\\\n\\implies x^2y^2(y-x) + 4(y-x)(y^2 + xy + x^2) = 8(y-x)(y+x) \\\\\n\\implies x^2y^2 + 4(y^2 + xy + x^2) = 8(y+x)\n\\end{aligned}\n$$\n\nFor $x, y \\geq 2$, the left side is positive and the right side is non-negative. So, $x=1$ or $y=1$. If $x=1$, then $4y+y^2 = 4(2y-y^2)+4$ or $5y^2-3y+4=0$, which is not possible for positive integer $y$. Similarly for $y=1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19037,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a group in which $a^2b = ba^2$ implies $ab = ba$.\n\n1. Show that if $G$ has $2^n$ elements, then $G$ is an abelian group.\n\n2. Exhibit an example of a nonabelian group with the given property.",
    "options": [],
    "answer": "See solution",
    "solution": "1. For $a \\in G$, let $C(a) = \\{b \\mid b \\in G \\text{ and } ab = ba\\}$. 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\n2. An example of a nonabelian group with the given property is the multiplicative group $G$ of matrices of the form\n\n$$\n\\begin{pmatrix} \\hat{1} & a & b \\\\ \\hat{0} & \\hat{1} & c \\\\ \\hat{0} & \\hat{0} & \\hat{1} \\end{pmatrix}, \\text{ with } a, b, c \\in \\mathbb{Z}_3.\n$$\n\nSince $A^3 = I_3$ for every $A \\in G$, if $A^2B = BA^2$, then $A^{-1}B = BA^{-1}$, so $AB = BA$.\n\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 part 2.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19038,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of positive integers $ (a, b) $ such that\n\n$$\n\\frac{a^2}{2ab^2 - b^3 + 1}\n$$\n\nis a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution.** (Based on work by Anders Kaseorg)\n\nRewrite the equation as $a^2/(2ab^2 - b^3 + 1) = k$, where $k$ is a positive integer. Then:\n\n$$\na^2 = 2ab^2k - b^3k + k. \\quad (*)\n$$\n\nRewrite $(*)$ as $a^2 - 2ab^2k = -b^3k + k$. Adding $b^4k^2$ to both sides completes the square:\n\n$$\n(kb^2 - a)^2 = b^4k^2 - b^3k + k,\n$$\n\nor\n\n$$\n(2kb^2 - 2a)^2 = (2b^2k)^2 - 2b(2b^2k) + 4k.\n$$\n\nCompleting the square on the right gives:\n\n$$\n(2kb^2 - 2a)^2 = (2b^2k - b)^2 + 4k - b^2.\n$$\n\nLet $x = 2kb^2 - b$ and $y = 2kb^2 - 2a$, so\n\n$$\ny^2 - x^2 = 4k - b^2.\n$$\n\nIf $4k = b^2$, then either $x = y$ or $x = -y$. In the former case, $b = 2a$; in the latter, $4kb^2 - b = 2a$, i.e., $b^4 - b = 2a$. Since $k = b^2/4$ is integer iff $b$ is even, we get the solutions\n\n$$\n\\left(\\frac{b}{2}, b\\right) \\quad \\text{and} \\quad \\left(\\frac{b^4 - b}{2}, b\\right)\n$$\n\nfor even $b$, i.e., $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ for all positive integers $t$.\n\nIf $4k < b^2$, then $y^2 < x^2$, so $y^2 \\le (x-1)^2$. Thus,\n\n$$\n4k - b^2 \\le (x - 1)^2 - x^2 = -2x + 1 = -4kb^2 + 2b + 1,\n$$\n\nor $4k(b^2 + 1) \\le b^2 + 2b + 1 < 3b^2 + 1$. Since $4k > 3$, this is a contradiction.\n\nIf $4k > b^2$, then $y^2 > x^2$, so $y^2 \\ge (x+1)^2$. Thus,\n\n$$\n4k - b^2 \\ge (x + 1)^2 - x^2 = 2x + 1 = 4kb^2 - 2b + 1,\n$$\n\nor $4k(b^2 - 1) + (b - 1)^2 \\le 0$. We must have $b = 1$ and\n\n$$\nk = \\frac{a^2}{2a} = \\frac{a}{2}.\n$$\n\nThis is integer when $a$ is even, so $(a, b) = (2t, 1)$ for all positive integers $t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19039,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x+y)f(x-y) = (f(x) + f(y))^2 - 4x^2 f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 0$:\n$$\nf(0)^2 = (f(0) + f(0))^2 - 0 = 4f(0)^2 \\implies f(0) = 0.\n$$\nLet $x = 0$:\n$$\nf(y)f(-y) = (f(0) + f(y))^2 - 0 = f(y)^2.\n$$\nSo $f(y)f(-y) = f(y)^2$, which gives $f(y) = 0$ or $f(y) = f(-y)$ for each $y$. Similarly, replacing $y$ by $-y$ shows $f$ is even or identically zero.\n\nTake $y = 1$:\n$$\nf(x+1)f(x-1) = (f(x) + f(1))^2 - 4x^2 f(1).\n$$\nAlso, swapping $x$ and $y$:\n$$\nf(1+x)f(1-x) = (f(1) + f(x))^2 - 4f(x).\n$$\nBy the evenness of $f$, comparing these gives $f(x) = c x^2$ for some $c = f(1)$.\n\nPlug $x = y = 1$ into the original equation:\n$$\nf(2)f(0) = (f(1) + f(1))^2 - 4 \\cdot 1^2 f(1) \\implies 0 = 4c^2 - 4c.\n$$\nSo $c = 0$ or $c = 1$. Thus, the solutions are $f(x) = 0$ for all $x$, or $f(x) = x^2$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19040,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$ with $AB = AC$, $M$ is the midpoint of $BC$, $H$ is the projection of $M$ onto $AB$, and $D$ is an arbitrary point on the side $AC$. Let $E$ be the intersection point of the parallel line through $B$ to $HD$ with the parallel line through $C$ to $AB$. Prove that $DM$ is the bisector of $\\angle ADE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the circle with center $M$ and radius $MH$, and let the tangent to $\\omega$ through $D$ (different from $DC$) meet the line $AB$ at $F$. It suffices to show that $E$ lies on $DF$.\n\nLet $E'$ be the intersection of $DF$ and the line through $C$ parallel to $AB$, and let $P$ be the projection of $D$ onto $MC$.\n\nLet $T$ be the tangency point of $AC$ and $\\omega$. Consider the case when $D$ lies in the segment $AT$; the case when $D$ lies in the segment $TC$ is treated analogously.\n\nSince $\\angle FBM = \\angle MCD$ and $\\angle FMB = 90^\\circ - \\angle AMF = 90^\\circ - \\frac{1}{2}\\angle ADF$ (as $M$ is the excenter of $\\triangle ADF$ opposite $A$) $= \\angle MDC$ (as $DM$ is the external angle bisector of $\\angle ADF$), the triangles $\\triangle BMF$ and $\\triangle DMC$ are similar.\n\nTherefore, we have $FH : HB = MP : PC$ (as $MH$ and $DP$ are corresponding altitudes in similar triangles) $= AD : DC$ (as $AM \\parallel DP$) $= FD : DE'$ (as $AF \\parallel CE'$). It follows that $HD \\parallel BE'$ and $E \\equiv E'$, as needed.\n\n![](images/BMO_Short_list_2014_p28_data_a4fb24e020.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19041,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all sequences $a_1, a_2, \\ldots$ of positive integers such that the expression $n a_n - m a_m + 2 a_m - 1$ is divisible by $a_n + a_m - 1$ for all $n, m \\ge 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are all sequences of the form $a_n = 1 + c(n - 1)$ for some fixed $c \\ge 0$.\n\n*Proof.*\n\nIt is clear that $a_n = 1 + c(n - 1)$ satisfies the divisibility condition. We now show there are no other solutions.\n\nConsider the expression:\n\n$$\nn - \\frac{n a_n - m a_m + 2 a_m - 1}{a_n + a_m - 1} = \\frac{n(a_m - 1) + (m - 2)a_m + 1}{a_n + a_m - 1}\n$$\n\nwhich must be an integer for all $n, m \\ge 1$. Let $c = a_2 - 1 \\ge 0$. Setting $m = 2$, we get $\\frac{c n + 1}{a_n + c}$ is a positive integer.\n\nIf $c = 0$, then $a_n = 1$ for all $n \\ge 1$. If $c \\ge 1$ and $c n + 1$ is prime, then $a_n + c \\ge 2$, so $a_n = 1 + c(n - 1)$.\n\nFinally, consider:\n\n$$\n(a_m - 1) - c \\frac{n(a_m - 1) + (m - 2)a_m + 1}{c n + a_m - c} = \\frac{a_m(a_m - 1 - c(m - 1))}{c n + a_m - c}\n$$\n\nwhich must be an integer for any $m \\ge 1$. By Dirichlet's theorem, for sufficiently large $n$, we must have $a_m = 1 + c(m - 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19042,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d, e$ ($a \\leq b \\leq c \\leq d \\leq e$) be the lengths of five given line segments. For any three line segments with lengths $x, y, z$ ($x \\leq y \\leq z$), the necessary and sufficient condition for them to form a triangle is $x + y > z$. If the triangle is acute, then $x^2 + y^2 > z^2$ must also hold.\n\nSuppose that for any choice of three line segments from the five, except for the combination $\\{a, b, e\\}$, we can form an acute triangle. Prove that we can also form a triangle with $\\{a, b, e\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By assumption, we can form an acute triangle from $\\{a, b, c\\}$ and from $\\{a, c, e\\}$. Therefore, $a^2 + b^2 > c^2$ and $a^2 + c^2 > e^2$.\n\nFrom these inequalities:\n\n$$\n(a + b)^2 = a^2 + 2ab + b^2 \\geq 2a^2 + b^2 > a^2 + c^2 > e^2,\n$$\n\nso $(a + b)^2 > e^2$, which implies $a + b > e$ (since $a, b, e > 0$). Thus, $\\{a, b, e\\}$ can also form a triangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19043,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a prime number such that both $8n^2 + 1$ and $8n^2 - 1$ are also prime numbers. Find all such $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "$n = 2$ gives $8n^2 + 1 = 33 = 3 \\cdot 11 \\notin \\mathbb{P}$, so not prime. $n = 3$ gives $8n^2 + 1 = 73 \\in \\mathbb{P}$ and $8n^2 - 1 = 71 \\in \\mathbb{P}$, both prime. For $n > 3$ and $n \\in \\mathbb{P}$, $n$ must be $3k + 1$ or $3k + 2$. If $n = 3k + 1$:\n\n$$\n8n^2 + 1 = 8(3k+1)^2 + 1 = 72k^2 + 48k + 9 = 3(24k^2 + 16k + 3)\n$$\n\nwhich is divisible by $3$, so not prime. If $n = 3k + 2$:\n\n$$\n8n^2 + 1 = 8(3k+2)^2 + 1 = 72k^2 + 96k + 33 = 3(24k^2 + 32k + 11)\n$$\n\nagain divisible by $3$, so not prime. Thus, the only solution is $n = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19044,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k = 20$. Prove that for any positive integer $n$, the number $20 \\cdot 22^n + 1$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $n \\in \\mathbb{Z}^+$,\n$$\n20 \\cdot 22^n + 1 \\equiv 20 \\cdot 1 + 1 = 21 \\equiv 0 \\pmod{21},\n$$\nso $20 \\cdot 22^n + 1$ is divisible by $21$ and thus composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19045,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations for real numbers $x$, $y$, and $z$:\n\n$$\n\\begin{aligned}\nx^2 - yz &= y - z + 1, \\\\\ny^2 - zx &= x - z + 1, \\\\\nz^2 - xy &= x - y + 1.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the second equation from the first, we obtain $x^2 - y^2 + z(x - y) = y - x$, or $(x - y)(x + y + z + 1) = 0$. This yields $x = y$ or $x + y + z = -1$. Subtracting the third equation from the second, we obtain $y^2 - z^2 + x(y - z) = y - z$, or $(y - z)(y + z + x - 1) = 0$. This yields $y = z$ or $x + y + z = 1$.\n\nWe now distinguish two cases: $x = y$ and $x \\neq y$. In the first case, we have $y \\neq z$, as otherwise we would have $x = y = z$ for which the first equation becomes $0 = 1$, a contradiction. Now it follows that $x + y + z = 1$, or $2x + z = 1$. Substituting $y = x$ and $z = 1 - 2x$ in the first equation yields $x^2 - x(1 - 2x) = x - (1 - 2x) + 1$, which can be simplified to $3x^2 - x = 3x$, or $3x^2 = 4x$. We get $x = 0$ or $x = \\frac{4}{3}$. With $x = 0$, we find $y = 0$, $z = 1$, but does not satisfy our assumption $x \\ge y \\ge z$. Thus, the only remaining possibility is $x = \\frac{4}{3}$, which gives the triple $\\left(\\frac{4}{3}, \\frac{4}{3}, -\\frac{5}{3}\\right)$. We verify that this is indeed a solution.\n\nNow consider the case $x \\neq y$. Then we have $x + y + z = -1$, hence we cannot have $x + y + z = 1$, and we see that $y = z$. Now $x + y + z = -1$ yields $x + 2z = -1$, hence $x = -1 - 2z$. Now the first equality becomes $(-1 - 2z)^2 - z^2 = 1$, which can be simplified to $3z^2 + 4z = 0$. From this, we conclude that $z = 0$ or $z = -\\frac{4}{3}$. With $z = 0$, we find $y = 0$, $x = -1$, which does not satisfy our assumption $x \\ge y \\ge z$. Hence, the only remaining possibility is $z = -\\frac{4}{3}$, and this gives rise to the triple $\\left(\\frac{5}{3}, -\\frac{4}{3}, -\\frac{4}{3}\\right)$. We verify that this is indeed a solution.\n\nBy also considering the permutations of these two solutions, we find all six solutions:\n\n- $\\left(\\frac{4}{3}, \\frac{4}{3}, -\\frac{5}{3}\\right)$\n- $\\left(\\frac{4}{3}, -\\frac{5}{3}, \\frac{4}{3}\\right)$\n- $\\left(-\\frac{5}{3}, \\frac{4}{3}, \\frac{4}{3}\\right)$\n- $\\left(\\frac{5}{3}, -\\frac{4}{3}, -\\frac{4}{3}\\right)$\n- $\\left(-\\frac{4}{3}, \\frac{5}{3}, -\\frac{4}{3}\\right)$\n- $\\left(-\\frac{4}{3}, -\\frac{4}{3}, \\frac{5}{3}\\right)$\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19046,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $ (a, b, c) $ such that\n\n$$\n\\frac{a}{2^a} = \\frac{b}{2^b} + \\frac{c}{2^c}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_n = \\frac{n}{2^n}$. For $n \\geq 2$, we have\n$$\nx_n - x_{n+1} = \\frac{n-1}{2^{n+1}} > 0,\n$$\nso\n$$\nx_1 = x_2 = \\frac{1}{2} > x_3 > x_4 > \\dots\n$$\n\nAssume without loss of generality that $b \\leq c$. From $\\frac{a}{2^a} = \\frac{b}{2^b} + \\frac{c}{2^c}$, we know $a < b \\leq c$.\n\n**Case 1:** $b = c$\n\nThen $\\frac{a}{2^a} = \\frac{b}{2^{b-1}}$, so $\\frac{b}{a} = 2^{b-a-1} \\in \\mathbb{Z}$. Let $b = ak$ with $k > 1$, then $k = 2^{ak-a-1} \\geq ak - a$, so $(a-1)(k-1) \\leq 1$. Since $k \\geq 2$, $a = 1$ or $a = 2$. In both cases, $b = 2^{b-2}$, so $b = 4$.\n\n**Case 2:** $b < c$\n\nThen\n$$\n\\frac{a}{2^a} \\leq \\frac{a+1}{2^{a+1}} + \\frac{a+2}{2^{a+2}} = \\frac{3a+4}{2^{a+2}}. \\quad (*)\n$$\nFrom this, $a \\leq 4$. Note:\n- $x_1 = x_2 = \\frac{1}{2}$\n- $x_3 = \\frac{3}{8}$\n- $x_4 = \\frac{1}{4}$\n- $x_5 = \\frac{5}{32}$\n- $x_6 = \\frac{3}{32} < \\frac{1}{8}$\n\nIf $a = 1$ or $2$, then $x_b + x_c = \\frac{1}{2}$, so $x_b > \\frac{1}{4}$, which means $b = 3$, but there is no solution for $c$.\n\nIf $a = 3$, then $x_b + x_c = \\frac{3}{8}$, so $x_b > \\frac{3}{16}$, which means $b = 4$, but again, no solution for $c$.\n\nIf $a = 4$, then equality holds in $(*)$, i.e., $b = 5$, $c = 6$, which satisfies the condition.\n\n**Summary:**\nThe solutions for $(a, b, c)$ are $(1, 4, 4)$, $(2, 4, 4)$, $(4, 5, 6)$, and $(4, 6, 5)$.\n\n![](images/2023_CGMO_p2_data_2ecb65f353.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19047,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be integers such that\n\n$$\nx^{n-1} + x^{n-2}y + \\cdots + y^{n-1} = 67.\n$$\n\nFind all possible integer solutions $(x, y, n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the divisibility properties:\n\n- $y^2 \\equiv 1 \\pmod{3}$, and $x^n = x(x^2)^k \\equiv x \\pmod{3}$, so $y^n \\equiv y \\pmod{3}$. Thus, $3 \\mid (x-y)$.\n- If $5 \\nmid x$, $5 \\nmid y$, then $x^4 \\equiv 1 \\pmod{5}$ and $y^4 \\equiv 1 \\pmod{5}$ (by Fermat's Little Theorem). For $n = 4p+1$, $x-y \\equiv 0 \\pmod{5}$; for $n = 4p+3$, $x^3 \\equiv y^3 \\pmod{5}$, which also implies $5 \\mid (x-y)$. Thus, $30 \\mid (x-y)$.\n\nSince $x^{n-1} + x^{n-2}y + \\cdots + y^{n-1} \\geq x-y$, and this sum equals $67$, if $n \\geq 5$, then $x \\leq 4$. Similarly, $y \\leq 4$, so $x = y = 0$ is the only possibility, but this does not satisfy the equation.\n\nIf $n = 3$, then $x^2 + xy + y^2 = 67$. For $x = y + 30q$, $q > 0$, the equation is impossible. If $x = y$, then $3x^2 = 67$, which has no integer solution.\n\nTherefore, there are no integer solutions $(x, y, n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19048,
    "subject": "Mathematics (Olympiad)",
    "question": "Hemos empezado la Olimpiada Matemática puntualmente a las 9:00, como he comprobado en mi reloj, que funcionaba en ese momento correctamente. Cuando he terminado, a las 13:00, he vuelto a mirar el reloj y he visto que las manecillas se habían desprendido de su eje pero manteniendo la posición en la que estaban cuando el reloj funcionaba. Curiosamente, las manecillas de las horas y de los minutos aparecían superpuestas exactamente, una sobre otra, formando un ángulo (no nulo) menor que $120^\\circ$ con el segundero. ¿A qué hora se me averió el reloj? (Dar la respuesta en horas, minutos y segundos con un error máximo de un segundo; se supone que, cuando funcionaba, las manecillas del reloj avanzaban de forma continua.)",
    "options": [],
    "answer": "See solution",
    "solution": "Si medimos el tiempo $t$ en segundos a partir de las 00:00 y los ángulos en grados, en sentido horario y a partir de la posición de las manecillas a las 00:00, tenemos que el ángulo barrido por la manecilla de las horas en el instante $t$ es $\\alpha_{hor}(t) = t/120$ y el barrido por el minutero es $\\alpha_{min}(t) = t/10$. Como ambas manecillas han aparecido superpuestas, los dos ángulos han de coincidir en el momento $t$ en que el reloj se ha averiado. El minutero ha podido dar alguna vuelta completa, por tanto debe tenerse\n\n$$\n\\frac{t}{10} = \\frac{t}{120} + 360k,\n$$\n\ndonde $k \\ge 0$ es un número entero, es decir, $t = \\frac{360 \\times 120}{11}k$. Como la avería ha sido entre las 9:00 y las 13:00, tiene que ser $9 \\leq k \\leq 12$. El ángulo para el segundero es $\\alpha_{seg}(t) = 6t$, por tanto la diferencia\n\n$$\n6t - \\frac{t}{120} = \\frac{360 \\times 719}{11}k = (360 \\times 65 + \\frac{360 \\times 4}{11})k\n$$\n\ndebe ser, salvo múltiplos de 360, un número $\\beta$ entre $-120$ y $120$. Si $k = 9$, $\\beta = (360 \\times 3)/11$, que efectivamente está en este rango. Sin embargo, si $k = 10$ o $12$, $\\beta = \\pm(360 \\times 4)/11$, que está fuera de este intervalo. El caso $k = 11$ también se excluye puesto que se tendría $\\beta = 0$ y las tres manecillas no están superpuestas. Por lo tanto, el único caso posible es $k = 9$, que corresponde al momento\n\n$$\nt = \\frac{360 \\times 120 \\times 9}{11} = 3600 \\times 9 + 60 \\times 49 + 5 + \\frac{5}{11},\n$$\n\nlo que significa que el reloj se averió a las 9:49:05.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19049,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $0 \\le x, y \\le 1$. Prove that\n\n$$\nxy + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-x), \\frac{m+n}{(m+1)n}(1-y) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$\n\nFurthermore, let $a_0, \\ldots, a_m$ and $b_0, \\ldots, b_n$ be non-negative real numbers such that $\\sum_{i=0}^m a_i = \\sum_{j=0}^n b_j = 1$. Define $S = \\max_{\\text{monotone path } \\mathcal{P}} W(\\mathcal{P})$, where $W(\\mathcal{P})$ is the weight of the path. Prove that\n\n$$\nS \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let the left-hand side of the lemma be $L$. By symmetry, assume $\\frac{m+n}{m(n+1)}(1-x) \\ge \\frac{m+n}{(m+1)n}(1-y)$, so\n\n$$\ny \\ge 1 - \\frac{(m+1)n}{m(n+1)}(1-x).\n$$\n\nSince $x \\ge 0$,\n\n$$\n\\begin{aligned}\nL &\\ge x \\left(1 - \\frac{(m+1)n}{m(n+1)}(1-x)\\right) + \\frac{m+n}{m(n+1)}(1-x) \\\\\n  &= \\frac{m+n+1}{(m+1)(n+1)} + \\frac{n}{m(m+1)(n+1)}((m+1)x - 1)^2 \\\\\n  &\\ge \\frac{m+n+1}{(m+1)(n+1)}.\n\\end{aligned}\n$$\n\nThis proves the lemma.\n\nNow, we prove the proposition by induction on $m+n$. When $m=0$ or $n=0$, the result is clear. Assume the proposition holds for $m+n < \\ell$, and consider $m+n = \\ell$ with $m, n > 0$. By homogeneity, assume $\\sum_{i=0}^m a_i = \\sum_{j=0}^n b_j = 1$, so $0 \\le a_0, b_0 \\le 1$.\n\nBy the induction hypothesis, there exists a monotone path $P_0$ from $(1,0)$ to $(m,n)$ such that\n\n$$\nW(P_0) \\ge \\frac{m+n}{m(n+1)}(1-a_0).\n$$\n\nLet $\\mathcal{P}$ be the path from $(0,0)$ to $(1,0)$ then along $P_0$, so\n\n$$\nS \\ge W(\\mathcal{P}) = a_0 b_0 + \\frac{m+n}{m(n+1)}(1-a_0).\n$$\n\nSimilarly, going from $(0,0)$ to $(0,1)$ and then using the induction hypothesis,\n\n$$\nS \\ge a_0 b_0 + \\frac{m+n}{(m+1)n}(1-b_0).\n$$\n\nCombining these and applying the lemma,\n\n$$\nS \\ge a_0 b_0 + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-a_0), \\frac{m+n}{(m+1)n}(1-b_0) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19050,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many positive integers which can't be expressed as $a^{d(a)} + b^{d(b)}$ where $a$ and $b$ are positive integers.\n\nFor a positive integer $a$, the expression $d(a)$ denotes the number of positive divisors of $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $a^{d(a)}$ is a perfect square for every positive integer $a$.\n\nIf $a$ is a perfect square, any of its powers is also a perfect square.\n\nIf $a$ is not a perfect square, the number of its positive divisors is even. We can prove this by pairing divisors of $a$ as $d$ and $\\frac{a}{d}$. A divisor $d$ won't be paired with itself unless $a = d^2$.\n\nThis proves that $d(a)$ is even and hence $a^{d(a)}$ is a perfect square for every positive integer $a$.\n\nTherefore, the given expression is always a sum of two squares. Every number of the form $4t+3$ cannot be written as a sum of two squares, because $0$ and $1$ are the only quadratic residues modulo $4$, so it is impossible for a sum of two squares to give remainder $3$ modulo $4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19051,
    "subject": "Mathematics (Olympiad)",
    "question": "In a square $ABCD$ of side length $2$, we draw lines from each vertex to the midpoints of the two opposite sides. For example, we connect $A$ to the midpoint of $BC$ and to the midpoint of $CD$. The eight resulting lines together bound an octagon inside the square (see figure). What is the area of this octagon?\n\n![](images/NLD_ABooklet_2020_p10_data_55ebb5cde0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\\frac{2}{3}$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19052,
    "subject": "Mathematics (Olympiad)",
    "question": "The areas of three semicircles are $\\frac{\\pi}{2}$, $\\frac{4\\pi}{2}$, and $\\frac{16\\pi}{2}$. Find the ratio $x : y$, where $x = \\frac{4\\pi}{2} - \\frac{\\pi}{2}$ and $y = \\frac{16\\pi}{2} - \\frac{4\\pi}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The area $x$ is $\\frac{4\\pi}{2} - \\frac{\\pi}{2} = \\frac{3\\pi}{2}$, and the area $y$ is $\\frac{16\\pi}{2} - \\frac{4\\pi}{2} = 6\\pi$. The ratio is $$\\frac{3\\pi/2}{6\\pi} = \\frac{3}{12} = \\frac{1}{4}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19053,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ for which $8^n + n$ is divisible by $2^n + n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We want all positive integers $n$ such that $2^n + n$ divides $8^n + n$.\n\nFirst, note that $8^n = (2^n)^3$, so:\n$$\n8^n + n = (2^n)^3 + n\n$$\nConsider the division:\n$$\n\\frac{8^n + n}{2^n + n}\n$$\nLet us perform polynomial division:\n\nLet $a = 2^n$, so $8^n + n = a^3 + n$ and $2^n + n = a + n$.\n\nDivide $a^3 + n$ by $a + n$:\n\n$a^3 + n = (a + n)(a^2 - a n + n^2) - (n^3 - n)$\n\nSo:\n$$\n8^n + n = (2^n + n)((2^n)^2 - n \\cdot 2^n + n^2) - (n^3 - n)\n$$\n\nTherefore, $2^n + n$ divides $8^n + n$ if and only if $2^n + n$ divides $n^3 - n$.\n\nNow, check small values of $n$:\n\n- For $n = 1$: $n^3 - n = 0$, $2^n + n = 3$, so $3$ divides $0$.\n- For $n = 2$: $n^3 - n = 6$, $2^n + n = 6$, $6$ divides $6$.\n- For $n = 3$: $n^3 - n = 24$, $2^n + n = 11$, $11$ does not divide $24$.\n- For $n = 4$: $n^3 - n = 60$, $2^n + n = 20$, $20$ divides $60$.\n- For $n = 5$: $n^3 - n = 120$, $2^n + n = 37$, $37$ does not divide $120$.\n- For $n = 6$: $n^3 - n = 210$, $2^n + n = 70$, $70$ divides $210$.\n- For $n = 7$: $n^3 - n = 336$, $2^n + n = 135$, $135$ does not divide $336$.\n- For $n = 8$: $n^3 - n = 504$, $2^n + n = 264$, $264$ does not divide $504$.\n- For $n = 9$: $n^3 - n = 720$, $2^n + n = 521$, $521$ does not divide $720$.\n\nFor $n \\geq 10$, $n^3 < 2^n$, so $n^3 - n < 2^n + n$ and division is impossible.\n\n**Conclusion:** The positive integers $n$ for which $8^n + n$ is divisible by $2^n + n$ are $n = 1, 2, 4, 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19054,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive real number $n$ and real number $x$ ($0 \\le x < n$), define\n$$\nf(n, x) = (1 - \\{x\\}) \\cdot C_{n}^{[x]} + \\{x\\} \\cdot C_{n}^{[x]+1},\n$$\nwhere $[x]$ denotes the largest integer not exceeding $x$ and $\\{x\\} = x - [x]$.\n\nIf integers $m, n \\ge 2$ satisfy\n$$\nf\\left(m, \\frac{1}{n}\\right) + f\\left(m, \\frac{2}{n}\\right) + \\dots + f\\left(m, \\frac{mn-1}{n}\\right) = 123,\n$$\nfind the value of\n$$\nf\\left(n, \\frac{1}{m}\\right) + f\\left(n, \\frac{2}{m}\\right) + \\dots + f\\left(n, \\frac{mn-1}{m}\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $k = 0, 1, \\dots, m-1$, we have\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} f\\left(m, k+\\frac{i}{n}\\right) &= C_m^k \\cdot \\sum_{i=1}^{n-1}\\left(1-\\frac{i}{n}\\right) + C_m^{k+1} \\cdot \\sum_{i=1}^{n-1} \\frac{i}{n} \\\\\n&= \\frac{n-1}{2} \\cdot \\left(C_m^k + C_m^{k+1}\\right).\n\\end{aligned}\n$$\nTherefore,\n$$\n\\begin{aligned}\n& f\\left(m, \\frac{1}{n}\\right) + f\\left(m, \\frac{2}{n}\\right) + \\cdots + f\\left(m, \\frac{mn-1}{n}\\right) \\\\\n&= \\sum_{j=1}^{m-1} C_m^j + \\sum_{k=0}^{m-1} \\sum_{i=1}^{n-1} f\\left(m, k+\\frac{i}{n}\\right) \\\\\n&= 2^m - 2 + \\frac{n-1}{2} \\cdot \\left(\\sum_{k=0}^{m-1} C_m^k + \\sum_{k=0}^{m-1} C_m^{k+1}\\right) \\\\\n&= 2^m - 2 + \\frac{n-1}{2} \\cdot (2^m - 1 + 2^m - 1) \\\\\n&= (2^m - 1)n - 1.\n\\end{aligned}\n$$\nSimilarly,\n$$\nf\\left(n, \\frac{1}{m}\\right) + f\\left(n, \\frac{2}{m}\\right) + \\cdots + f\\left(n, \\frac{mn-1}{m}\\right) = (2^{n} - 1)m - 1.\n$$\nGiven $(2^m - 1)n - 1 = 123$, so $(2^m - 1)n = 124$. Since $m \\ge 2$, $2^m - 1 \\in \\{3, 7, 15, 31, 63, 127, \\dots\\}$. $2^m - 1 = 31$ is a factor of 124 only when $m = 5$, so $n = \\frac{124}{31} = 4$.\n\nTherefore,\n$$\nf\\left(n, \\frac{1}{m}\\right) + f\\left(n, \\frac{2}{m}\\right) + \\cdots + f\\left(n, \\frac{mn-1}{m}\\right) = (2^{4} - 1) \\cdot 5 - 1 = 74.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19055,
    "subject": "Mathematics (Olympiad)",
    "question": "En un triángulo $ABC$, la bisectriz por $A$, la mediana por $B$ y la altura por $C$ son concurrentes y, además, la bisectriz por $A$ y la mediana por $B$ son perpendiculares. Si el lado $AB$ mide una unidad, ¿cuánto miden los otros dos lados?",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $P$, $M$ y $Q$ los pies de la bisectriz por $A$, la mediana por $B$ y la altura por $C$, respectivamente, que se cortan en el punto $X$. En el triángulo $ABM$, la bisectriz por $A$, $AX$, es perpendicular a $BM$ (por hipótesis la mediana y la bisectriz de $ABC$ son perpendiculares), por tanto $\\angle ABX = \\angle AMX$, es decir, $ABM$ es isósceles y $AM = AB = 1$, con lo cual $AC = 2$.\n\nSea $BP = x$ y $BQ = y$. Por el Teorema de la bisectriz, $\\frac{BP}{AB} = \\frac{PC}{AC}$, esto es $PC = 2x$. Ahora, por el Teorema de Ceva:\n\n$$\n1 = \\frac{MC}{CP} \\cdot \\frac{PB}{BQ} \\cdot \\frac{QA}{AM} = \\frac{1}{2x} \\cdot \\frac{x}{y} \\cdot \\frac{1-y}{1},\n$$\n\nde donde $y = \\frac{1}{3}$. Trazamos las perpendiculares a $AC$ por $B$ y $X$, respectivamente, con pies $R$ y $S$. Los triángulos $XQB$ y $XSM$ son congruentes (son rectángulos, $XB = XM$ por ser $AX$ la altura del triángulo isósceles $ABM$ y $XQ = XS$ por ser perpendiculares a los lados $AB$ y $AC$ desde un punto de la bisectriz), por tanto $SM = BQ = y = \\frac{1}{3}$. Por otro lado, por el Teorema de Thales, $BX = XM$ implica $RS = SM$, luego $RS = \\frac{1}{3}$ y $AR = AM - RS - SM = \\frac{1}{3}$. Finalmente, por el Teorema de Pitágoras:\n\n$$\nBC^2 = BR^2 + RC^2 = AB^2 - AR^2 + RC^2 = 1 - \\frac{1}{9} + \\left(2 - \\frac{1}{3}\\right)^2 = \\frac{11}{3}.\n$$\n\nAsí pues, los otros dos lados miden $2$ y $\\frac{\\sqrt{33}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19056,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to the sides $BC$, $CA$, and $AB$, respectively. For which triangles are two of the line segments $H_aH_b$, $H_bH_c$, and $H_cH_a$ of equal length?",
    "options": [],
    "answer": "See solution",
    "solution": "We first consider the case where no angle in $ABC$ is obtuse. If $ABC$ is right-angled with hypotenuse $AB$, then $H_a = H_b = C$, so $H_bH_c = H_cH_a$. Thus, any right-angled triangle $ABC$ has the required property. If $ABC$ is not right-angled, the triangles $AH_bB$ and $AH_aB$ certainly are.\n\n![](images/AustriaMO2012_p2_data_569cec4588.png)\n\n![](images/AustriaMO2012_p2_data_1998a45239.png)\n\nBoth $H_a$ and $H_b$ lie on the semicircle with diameter $AB$. If $H_bH_c = H_cH_a$, then $H_c$ must be the intersection of $AB$ and the bisector of $H_aH_b$, i.e., the midpoint $M_{AB}$ of $AB$. Since $CH_c$ is perpendicular to $AB$, $C$ must lie on the bisector of $AB$, so $ABC$ is isosceles.\n\nNow, suppose one angle in $ABC$ is greater than $90^\\circ$.\n\nIf $H_bH_c = H_cH_a$, we again obtain $|AC| = |BC|$ as before. If instead $H_aH_c = H_aH_b$, we have the situation in the second figure. Because of the right angles between the sides and the altitudes, each of the quadrilaterals $AH_aH_bB$, $CH_bBH_c$, and $AH_aCH_c$ is cyclic. It follows that $\\angle H_aH_bC = \\angle H_aH_bA = \\angle H_aBA = \\beta = \\angle CBH_c = \\angle CH_bH_c$, and $\\angle H_bH_cC = \\angle H_bBC = 90^\\circ - \\alpha - \\beta = \\angle H_aAH_b = \\angle H_aAC = \\angle H_aH_cC$. We see that $|H_aH_b| = |H_aH_c|$ if and only if $\\angle H_aH_bH_c = \\angle H_aH_cH_b$, which is equivalent to $2\\beta = 2 \\cdot (90^\\circ - \\alpha - \\beta)$, or $\\alpha + 2\\beta = 90^\\circ$.\n\nIn summary, exactly the right-angled triangles, the isosceles triangles, and triangles in which two angles $\\alpha$ and $\\beta$ satisfy $\\alpha + 2\\beta = 90^\\circ$ have the required property. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19057,
    "subject": "Mathematics (Olympiad)",
    "question": "是否存在無窮多個整數 $a_1, a_2, a_3, \\dots$ 及正整數 $N$，其中 $0 < a_i < 10$，使得對於所有正整數 $k > N$，\n\n$$\n\\sum_{i=1}^{k} a_i 10^{i-1}\n$$\n\n都是完全平方數？\n\nDetermine whether there exists an infinite sequence of integers $a_1, a_2, a_3, \\dots$ with $0 < a_i < 10$ and a positive integer $N$ such that, for every integer $k > N$, the number\n\n$$\n\\sum_{i=1}^{k} a_i 10^{i-1}\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "不可能。歸謬證法，假設存在這樣的無窮序列和正整數 $N$。令 $y_k = \\sum_{i=1}^{k} a_i 10^{i-1}$，則對於所有 $k > N$，存在正整數 $x_k$ 使得 $y_k = x_k^2$。\n\n1. 對所有 $n$，令 $\\gamma_n$ 為滿足 $5^{\\gamma_n}\\mid x_n$ 的最大正整數。以下證明：對於所有 $n > N$，$2\\gamma_n \\ge n$。\n\n假設存在 $n < N$ 使得 $2\\gamma_n < n$，則\n\n$$\ny_{n+1} = 10^n a_{n+1} + y_n = 5^{2\\gamma_n} \\left( 2^n 5^{n-2\\gamma_n} a_{n+1} + \\frac{y_n}{5^{2\\gamma_n}} \\right).\n$$\n\n基於 $y_n/5^{2\\gamma_n}$ 不可能被 5 整除，必然有 $\\gamma_{n+1} = \\gamma_n < n < n+1$，因此 $\\gamma_n = \\gamma_{n+1} = \\dots = \\gamma$。\n\n另一方面，對於所有 $k \\ge n$，\n\n$$\n(x_{k+1} - x_k)(x_{k+1} + x_k) = y_{k+1} - y_k = a_{k+1} 10^k.\n$$\n\n基於 $(x_{k+1} - x_k) + (x_{k+1} + x_k) = 2x_{k+1}$，因此依照 $\\gamma = \\gamma_{k+1}$ 的定義，其中必有一者不被 $5^{\\gamma+1}$ 整除。進一步地，由於 $5^k\\mid(x_{k+1} - x_k)(x_{k+1} + x_k)$，故必有一者被 $5^{k-\\gamma}$ 整除，從而\n\n$$\n5^{k-\\gamma} \\le \\max\\{x_{k+1} - x_k, x_{k+1} + x_k\\} < 2x_{k+1} = 2\\sqrt{y_{k+1}} < 2 \\times 10^{(k+1)/2},\n$$\n\n這表示 $5^{2k} < 4 \\times 5^{2\\gamma} \\times 10^{k+1}$，也就是 $(5/2)^k < 40 \\times 5^{2\\gamma}$。這對充分大的 $k$ 是不可能成立的，故矛盾。\n\n2. 現在考慮所有 $k > \\max\\{N/2, 2\\}$。由 1. 知 $2\\gamma_{2k+1} \\ge 2k+1$ 且 $2\\gamma_{2k+2} \\ge 2k+2$，故 $\\gamma_{2k+1} \\ge k+1$ 且 $\\gamma_{2k+2} \\ge k+1$。又由 $y_{2k+2} = a_{2k+2}10^{2k+1} + y_{2k+1}$ 知 $5^{2k+2}\\mid y_{2k+2} - y_{2k+1} = a_{2k+2}10^{2k+1}$，故 $5\\mid a_{2k+2} \\Rightarrow a_{2k+2} = 5$。因此\n\n$$\n(x_{2k+2} - x_{2k+1})(x_{2k+2} + x_{2k+1}) = y_{2k+2} - y_{2k+1} = 5 \\times 10^{2k+1} = 2^{2k+1}5^{2k+2}.\n$$\n\n令 $A_k = x_{2k+2}/5^{k+1}$，$B_k = x_{2k+1}/5^{k+1}$；注意到由 1.，它們必為正整數，且\n\n$$\n(A_k - B_k)(A_k + B_k) = 2^{2k+1}\n$$\n\n注意到 $a_1 \\ne 0$，因此 $A_k$ 和 $B_k$ 皆為奇數（否則 $y_{2k+2}$ 或 $y_{2k+1}$ 為 10 的倍數，不合）。這表示 $A_k - B_k$ 和 $A_k + B_k$ 中有一個不被 4 整除。從而由 (1)，$A_k - B_k = 2$ 且 $A_k + B_k = 2^{2k}$，故 $A_k = 2^{2k-1} + 1$，得\n\n$$\nx_{2k+2} = 5^{k+1}A_k = 10^{k+1}2^{k-2} + 5^{k+1} > 10^{k+1}\n$$\n\n（因 $k \\ge 2$）但這代表 $y_{2k+2} > 10^{2k+2}$，矛盾！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19058,
    "subject": "Mathematics (Olympiad)",
    "question": "How many integers $a$ satisfy the condition: for each $a$, the equation $x^3 = a x + a + 1$ with respect to $x$ has roots which are even and $|x| < 1000$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nLet $x_0 = 2n$, where $n$ is an integer and $|2n| < 1000$, so $|n| \\leq 499$. Thus, $n$ can take $2 \\times 499 + 1 = 999$ values: $n \\in \\{-499, -498, \\dots, 0, 1, \\dots, 499\\}$.\n\nSubstituting $x_0 = 2n$ into the equation gives:\n$$\nx_0^3 = a x_0 + a + 1 \\implies (2n)^3 = a (2n) + a + 1\n$$\n$$\n8n^3 = 2a n + a + 1\n$$\n$$\n8n^3 - 1 = a(2n + 1)\n$$\n$$\na = \\frac{8n^3 - 1}{2n + 1}\n$$\n\nDefine $f(n) = \\frac{8n^3 - 1}{2n + 1}$. For $n_1 \\neq n_2$, suppose $f(n_1) = f(n_2)$. Let $n_1 = \\frac{x_1}{2}$, $n_2 = \\frac{x_2}{2}$, where $x_1, x_2$ are roots of $x^3 - a x - a - 1 = 0$. Let the third root be $x_3$. By Vieta's formulas:\n$$\n\\begin{cases}\nx_3 = -(x_1 + x_2) \\\\\nx_1 x_2 + x_2 x_3 + x_3 x_1 = -a \\\\\nx_1 x_2 x_3 = a + 1\n\\end{cases}\n$$\nLet $N_1 = -(n_1^2 + n_2^2 + n_1 n_2)$, $N_2 = -n_1 n_2 (n_1 + n_2)$, then:\n$$\n\\begin{cases}\n4N_1 = -a \\\\\n8N_2 = a + 1\n\\end{cases}\n$$\nSo $4N_1 + 8N_2 = 1$, which is a contradiction.\n\nTherefore, for any $n_1 \\neq n_2$, $f(n_1) \\neq f(n_2)$, so there are exactly $999$ real numbers $a$ that satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19059,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $L$, respectively. The line $BL$ intersects the incircle of triangle $ABC$ at point $M$ ($M \\neq L$). A circle passing through point $M$ touches the lines $AB$ and $BC$ at points $P$ and $Q$, respectively, and intersects the incircle of triangle $ABC$ at point $N$ ($N \\neq M$). Prove that if $KM \\parallel AC$ then points $P$, $N$ and $L$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the circles of the problem can be obtained from each other by homothetic transformation with center $B$ since both are tangent to sides $BA$ and $BC$. Let $X$ be the other intersection point of the line $BL$ with the circumcircle of the triangle $MPQ$.\n\nThe aforementioned homothety takes point $P$ to point $K$, point $M$ to point $L$, and point $X$ to point $M$. By the homothety, $\\angle PXM = \\angle KML$. Hence $\\angle KML = \\angle KLA = \\angle LKM$.\n\nFinally,\n\n$$\n\\angle PNM + \\angle LNM = (180^\\circ - \\angle PXM) + \\angle LKM = (180^\\circ - \\angle KML) + \\angle KML = 180^\\circ.\n$$\n\nThis proves the desired claim that points $P$, $N$ and $L$ are collinear.\n\nRemark: The converse is also true: if points $P$, $N$ and $L$ are collinear then lines $KM$ and $AC$ are parallel. The proof is analogous.\n\n![](images/EST_ABooklet_2021_p20_data_a179cdc7c4.png)\n\nFig. 14",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19060,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and suppose that $3p + 10$ is the sum of the squares of six consecutive positive integers. Prove that $36 \\mid p - 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the problem, we have:\n\n$$\n3p + 10 = (n-2)^2 + (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = 6n^2 + 6n + 19.\n$$\n\nSo,\n\n$$\n3p = 6n^2 + 6n + 9,\n$$\nwhich gives\n$$\np = 2n^2 + 2n + 3 = 2n(n+1) + 3.\n$$\n\nIf either $n$ or $n+1$ is divisible by $3$, then $p$ would not be prime (since $p$ would be divisible by $3$ and greater than $3$). Thus, $n = 3k + 1$ for some integer $k$.\n\nSubstituting:\n\n$$\np = 2(3k+1)(3k+2) + 3 = 2(9k^2 + 9k + 2) + 3 = 18k^2 + 18k + 4 + 3 = 18k^2 + 18k + 7.\n$$\n\nTherefore,\n\n$$\np - 7 = 18k^2 + 18k = 18k(k+1).\n$$\n\nSince $k(k+1)$ is always even, $18k(k+1)$ is divisible by $36$. Thus, $36 \\mid p - 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19061,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers that can be expressed as\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{10}}\n$$\nwhere $a_1, a_2, \\dots, a_{10}$ are non-zero integers such that no two of them have a common factor greater than 1.",
    "options": [],
    "answer": "See solution",
    "solution": "The only integers that can be expressed in such a way are: $0$, $\\pm2$, $\\pm4$, $\\pm6$, $\\pm8$, $\\pm10$.\n\nSuppose that $\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{10}} = n$, where $\\gcd(a_i, a_j) = 1$ for all $1 \\le i < j \\le 10$. Furthermore, suppose that $p$ is a prime that divides $a_1$. Multiplying the whole equation by $a_1 a_2 \\dots a_{10}$ and using $p \\mid a_1$, we deduce that $p \\mid a_2 a_3 \\dots a_{10}$. Hence, $p$ divides $a_i$ for some $i \\ge 2$, which contradicts the fact that $\\gcd(a_1, a_i) = 1$.\n\nIt follows that $a_1 = \\pm 1$ and a similar argument shows that $a_i = \\pm 1$ for $1 \\le i \\le 10$. If $m$ of the $a_i$ are equal to $1$ and $10-m$ of the $a_i$ are equal to $-1$, then we deduce that $n = m - (10-m) = 2m - 10$. Since $m$ can be any integer from $0$ to $10$, we deduce that $n$ can be any even integer from $-10$ to $10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19062,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека равенката $x^2 + y^2 + z^2 + t^2 = 2^{2009}$, каде што $0 \\leq x \\leq y \\leq z \\leq t$, има точно едно решение во множеството на цели броеви.",
    "options": [],
    "answer": "See solution",
    "solution": "Едно решение на равенката, очигледно од нејзиниот облик, е подредената четворка $(0, 0, 2^{1004}, 2^{1004})$.\n\nЌе докажеме дека тоа е и единственото решение на равенката.\n\nНека $(x, y, z, t)$ е решение на равенката. Бидејќи квадратите на непарните природни броеви даваат остаток $1$ при делење со $8$, следува дека $x, y, z$ и $t$ се парни броеви (сите останати случаи во однос на парноста на $x, y, z, t$ може да се разгледаат поединечно, од каде ќе се добие спротивност со парноста на бројот од левата и десната страна на равенството). Затоа, $x = 2x_1$, $y = 2y_1$, $z = 2z_1$ и $t = 2t_1$, каде што $0 \\leq x_1 \\leq y_1 \\leq z_1 \\leq t_1$. Почетната равенка се трансформира во\n$$\nx_1^2 + y_1^2 + z_1^2 + t_1^2 = 2^{2007}.\n$$\nДобиената равенка е од истиот облик како и почетната, само степенот на десната страна е намален за два. Од истите причини како и претходно, заклучуваме дека $x_1 = 2x_2$, $y_1 = 2y_2$, $z_1 = 2z_2$ и $t_1 = 2t_2$, каде што $0 \\leq x_2 \\leq y_2 \\leq z_2 \\leq t_2$ и\n$$\nx_2^2 + y_2^2 + z_2^2 + t_2^2 = 2^{2005}.\n$$\nИстата постапка ќе ја повториме $1004$ пати, и добиваме дека $x = 2^{1004} a$, $y = 2^{1004} b$, $z = 2^{1004} c$ и $t = 2^{1004} d$, каде што $0 \\leq a \\leq b \\leq c \\leq d$ се цели броеви и важи $a^2 + b^2 + c^2 + d^2 = 2$. Единственото решение на последната равенка при дадените услови е подредената четворка $(0, 0, 1, 1)$. Од тука добиваме дека $(0, 0, 2^{1004}, 2^{1004})$ е единственото решение на почетната равенка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19063,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose all numbers from 1 to 80 are written on a blackboard. Two players take turns erasing any number of consecutive numbers (at least one) from the board. The player who erases the last number wins. Who has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "Jerry wins.\n\nIt is easy to see that if either 1 or 5 numbers remain on the blackboard, then the player who must move wins, but if 2 or 4 numbers remain, then this player loses. So we will solve the problem moving backward. We write all numbers from 1 to 80 and mark them with \"+\" or \"-\". If $k$ numbers remain on the blackboard before the move of a player and he can win, then we write $+k$, otherwise we write $-k$. We have the following table:\n\n| $+1$ | $-2$ | $+3$ | $-4$ | $+5$ | $-6$ | $+7$ | $+8$ | $+9$ | $+10$ | $+11$ | $+12$ | $-13$ |\n|------|------|------|------|------|------|------|------|------|-------|-------|-------|-------|\n| 1    | -    | 1    | -    | 5    | -    | 1    | 8    | 5    | 8     | 5     | 8     | -     |\n\nThe signs in the first row of the table repeat with period 13 (this can be proved by induction).\n\nSince 80 leaves remainder 2 when divided by 13, it follows that 80 and 2 are marked with the same sign, i.e. \"-\". Thus, 80 is a losing position for the starting player. To win, Jerry can erase as many numbers as indicated in the second row of the table.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19064,
    "subject": "Mathematics (Olympiad)",
    "question": "A tea cup can end up in $n$ possible positions inside a microwave, one of which is marked as \"front\".\n\nAt the start, the microwave rotates the tea cup to one of these positions. For each move, Mr. Precise enters an integer number of seconds, and the microwave decides whether to turn clockwise or counterclockwise.\n\nFor which values of $n$ can Mr. Precise ensure that, after a finite number of moves, he can take the tea cup out of the microwave precisely from the front position?",
    "options": [],
    "answer": "See solution",
    "solution": "Mr. Precise can ensure his victory when $n$ is a power of $2$.\n\nLabel the positions as $0, 1, \\ldots, n-1$, where $0$ is the front position.\n\nIf $n$ is a power of $2$, say $n = 2^k$, Mr. Precise can always enter the current position as the number of seconds. If the microwave turns the plate backwards, the tea cup will be at the front immediately. Otherwise, the position number doubles modulo $2^k$ at each turn, and after at most $k$ turns, it will be divisible by $2^k$, meaning the tea cup is at the front.\n\nIf $n = 2^k \\cdot m$ with $m > 1$ odd, the microwave can always choose a number not divisible by $m$.\n\nInitially, for example, by choosing position $1$, the tea cup is not at a position divisible by $m$. After that, suppose the tea cup is at position $p$ not divisible by $m$, and Mr. Precise enters $s$ seconds. If both $p+s$ and $p-s$ were divisible by $m$, then $m \\mid 2p$. Since $m$ is odd, this would imply $m \\mid p$, which is a contradiction.\n\nThus, only when $n$ is a power of $2$ can Mr. Precise guarantee the tea cup ends up at the front.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19065,
    "subject": "Mathematics (Olympiad)",
    "question": "21 numbers are written in a row. If $u, v, w$ are three consecutive ones, then $v = \\frac{2uw}{u+w}$. The first number is $\\frac{1}{100}$, the last one is $\\frac{1}{101}$. Find the 15th number.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $v = \\frac{2uw}{u+w}$ as $\\frac{1}{v} = \\frac{u+w}{2uw}$. This gives $\\frac{1}{v} = \\frac{1}{2} \\left( \\frac{1}{u} + \\frac{1}{w} \\right)$, or $\\frac{1}{v} - \\frac{1}{u} = \\frac{1}{w} - \\frac{1}{v}$. So, look at the sequence of reciprocals of the given numbers: $\\frac{1}{v} - \\frac{1}{u} = \\frac{1}{w} - \\frac{1}{v}$ means that consecutive reciprocals differ by the same amount, which we denote by $d$.\n\nHence, the last reciprocal $\\frac{1}{1/101} = 101$ can be obtained from the first one $\\frac{1}{1/100} = 100$ by adding $d$ 20 times. Thus, $101 = 100 + 20d$, yielding $d = \\frac{1}{20}$. To obtain the 15th reciprocal, we add $14d$ to the first one, $100$, which gives $100 + 14 \\cdot \\frac{1}{20} = \\frac{1007}{10}$. Therefore, the 15th original number is $\\frac{10}{1007}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19066,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\triangle ABC$ with angles $A$, $B$, and $C$, and corresponding sides $a$, $b$, and $c$, satisfies the equation $a \\cos B - b \\cos A = \\frac{3}{5}c$. Then the value of $\\frac{\\tan A}{\\tan B}$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "By the given condition and the Law of Cosines, we have\n\n$$\na \\cdot \\frac{c^2 + a^2 - b^2}{2ca} - b \\cdot \\frac{b^2 + c^2 - a^2}{2bc} = \\frac{3}{5}c,\n$$\n\nor $a^2 - b^2 = \\frac{3}{5}c^2$. Therefore,\n\n$$\n\\frac{\\tan A}{\\tan B} = \\frac{\\sin A \\cos B}{\\sin B \\cos A} = \\frac{a \\cdot \\frac{c^2 + a^2 - b^2}{2ca}}{b \\cdot \\frac{b^2 + c^2 - a^2}{2bc}}\n$$\n\n$$\n= \\frac{c^2 + a^2 - b^2}{b^2 + c^2 - a^2} = \\frac{\\frac{8}{5}c^2}{\\frac{2}{5}c^2} = 4.\n$$\n\nThe answer is $4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19067,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $a$, prove that $n!$ is divisible by $n^2 + n + a$ for infinitely many positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that, if some integer $n \\geq 2$ 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 $n \\geq 2$ that fits the bill.\n\nConsider the identity\n\n$$\n(n^2 + a - 1)^2 + (n^2 + a - 1) + a = (n^2 + n + a)(n^2 - n + a).\n$$\n\nSuppose now that $n!$ is divisible by $n^2 + n + a$ for some integer $n \\geq 2$. 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 above then shows $(n^2 + a - 1)!$ is divisible by $(n^2 + a - 1)^2 + (n^2 + a - 1) + a$.\n\nNext, set $n = a$ in the identity to get $(a^2 + a - 1)^2 + (a^2 + a - 1) + a = a^3(a + 2)$. If $a \\geq 3$, then $a < a + 2 < a^2 < a^2 + a - 1$, so $a^2 + a - 1 \\geq 2$ and $(a^2 + a - 1)!$ is divisible by $a^3(a + 2) = (a^2 + a - 1)^2 + (a^2 + a - 1) + a$.\n\nIf $a = 2$, set $n = 10$, and notice that $n! = 10!$ is divisible by $2 \\cdot 7 \\cdot 8 = 112 = 10^2 + 10 + 2 = n^2 + n + 2 = n^2 + n + a$.\n\nFinally, if $a = 1$, set $n = 16$, and notice that $n! = 16!$ is divisible by $3 \\cdot 7 \\cdot 13 = 273 = 16^2 + 16 + 1 = n^2 + n + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19068,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\geq 3$, prove that the diameter of a convex $n$-gon (interior and boundary) containing a disc of radius $r$ is strictly greater than $r(1 + 1/\\cos(\\pi/n))$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be a convex $n$-gon (interior and boundary) containing a disc $D$ of radius $r$, centered at some point $O$.\n\nWe show that $K$ is interior to the disc $D'$ of radius $r' = \\text{diam } K - r$, centered at $O$. To this end, let $x$ be a point of $K$, other than $O$. The point $O$ separates $x$ from one of the points, say $y$, where the line $Ox$ crosses the boundary $\\partial D$ of $D$. Consider a point $z$ of $K$ farther than $y$ from $x$, e.g., a point of $K$, other than $y$, lying on the tangent of $\\partial D$ at $y$. Then $\\text{dist}(O, x) = \\text{dist}(x, y) - \\text{dist}(O, y) = \\text{dist}(x, y) - r < \\text{dist}(x, z) - r \\leq \\text{diam } K - r = r'$, showing that $K$ is indeed interior to $D'$.\n\nNow, the lines of support along the sides of $K$ determine $n$ arcs on the boundary $\\partial D'$ of $D'$, each of length at most $2r' \\arccos(r/r')$, since $K$ is convex. By the preceding paragraph, $K$ is interior to $D'$, so an overlap of arcs occurs for each vertex of $K$. Since the $n$ arcs cover $\\partial D'$, their total length exceeds that of $\\partial D'$, so $n \\cdot 2r' \\arccos(r/r') > 2\\pi r'$. Consequently, $r/r' < \\cos(\\pi/n)$, and $\\text{diam } K = r + r' > r + r/\\cos(\\pi/n) = r(1 + 1/\\cos(\\pi/n))$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19069,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have a row of $n$ numbers, and at each step, we form a new row by grouping together equal numbers and replacing each group of $l$ equal numbers with $l$ in the next row. The process continues iteratively. What is the greatest possible number of steps this process can last, given $n$? Additionally, explain why the condition $m = 2^k - n = 2^i$ for some integer $i$ affects the process, and provide examples illustrating the maximal number of steps for small $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be the greatest integer such that $2^k \\leq n$. The process can last at most $k+1$ steps, counting the initial row. This is because, at each step, the minimal number among the remaining numbers at that stage at least doubles, and a number greater than $n$ cannot appear. If $m = 2^k - n = 2^i$ for some $i$, then a group of $2^i$ equal numbers will eventually form a group of $2^{i+1}$, shortening the process. For example, with $n = 5$, since $2^2 = 4 < 5$, the answer is at most 3 steps. If the initial row is $1, 4, 2, 2, 5$, the subsequent rows are: $1, 1, 2, 2, 1$ (1st row), $3, 3, 2, 2, 3$ (2nd row). Similar reasoning applies for $n = 2^s + 1$ or $n = 2^s + 2$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19070,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(x, y)$ of real numbers satisfying\n\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": "See solution",
    "solution": "Since $0 \\leq \\{a\\} < 1$ 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 $\\lfloor b \\rfloor = 4$.\n\nSimilarly, since $0 \\leq \\{c\\} < 1$ 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):\n$$\n\\frac{3x+2}{3} = 6 + \\frac{2}{3}, \\quad \\frac{4y+3}{4} = 4\n$$\nSo $x = 6$, $y = \\frac{13}{4}$.\n\nCase (II):\n$$\n\\frac{3x+2}{3} = 5 + \\frac{2}{3}, \\quad \\frac{4y+3}{4} = 4 + \\frac{3}{4}\n$$\nSo $x = 5$, $y = 4$.\n\nThe required pairs are $(6, \\frac{13}{4})$ and $(5, 4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19071,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the equation\n\n$$\nx^2 + xy + y^2 + 3x + 6y + 6 = 0.\n$$\n\n1. Find all pairs $(x, y)$ of integers with $x = 1$ which satisfy the equation.\n2. Find all pairs $(x, y)$ of integers which satisfy the equation.",
    "options": [],
    "answer": "See solution",
    "solution": "1. $(x, y) = (1, -2),\\ (1, -5)$\n\n2. $(x, y) = (-2, -2),\\ (-1, -4),\\ (-1, -1),\\ (1, -5),\\ (1, -2),\\ (2, -4)$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19072,
    "subject": "Mathematics (Olympiad)",
    "question": "For any 4-digit positive integer $n$, define $f(n) = (a + b)^2$, where $a$ and $b$ are the numbers formed by the first two and last two digits of $n$, respectively (leading zeroes are allowed). Find all 4-digit positive integers $n$ such that $f(n) = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a 4-digit number with first two digits $a$ and last two digits $b$, so $n = 100a + b$. We are given $f(n) = (a + b)^2 = n$.\n\nSo,\n$$\n(a + b)^2 = 100a + b\n$$\nExpanding and rearranging:\n$$\na^2 + 2ab + b^2 - 100a - b = 0\n$$\n$$\na^2 + 2ab + b^2 - 100a - b = 0\n$$\n$$\na^2 + 2ab + b^2 = 100a + b\n$$\n$$\na^2 + 2ab + b^2 - 100a - b = 0\n$$\nThis is a quadratic in $b$:\n$$\nb^2 + (2a - 1)b + (a^2 - 100a) = 0\n$$\nWe can solve for integer values of $a$ (from $10$ to $99$) and $b$ (from $0$ to $99$) such that $n$ is a 4-digit number and $b$ is a two-digit number (possibly with leading zero).\n\nBy checking all possible values, the solutions are:\n\n- $n = 2025$ ($a = 20$, $b = 25$)\n- $n = 3025$ ($a = 30$, $b = 25$)\n- $n = 9801$ ($a = 98$, $b = 01$)\n- $n = 2025$ ($a = 20$, $b = 25$) (duplicate)\n\nThus, the 4-digit numbers $n$ such that $f(n) = n$ are:\n\n$$\n\\boxed{2025,\\ 3025,\\ 9801}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19073,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a right triangular prism $A_1B_1C_1 - ABC$ with $\\angle BAC = \\frac{\\pi}{2}$ and $AB = AC = AA_1 = 1$, let $G$ and $E$ be the midpoints of $A_1B_1$ and $CC_1$ respectively; and $D$ and $F$ be variable points lying on segments $AC$ and $AB$ (not including endpoints) respectively. If $GD \\perp EF$, what is the range of the length of $DF$?\n\n(A) $\\left[\\frac{1}{\\sqrt{5}}, 1\\right)$\n\n(B) $\\left[\\frac{1}{5}, 2\\right)$\n\n(C) $[1, \\sqrt{2})$\n\n(D) $\\left[\\frac{1}{\\sqrt{5}}, \\sqrt{2}\\right)$",
    "options": [],
    "answer": "See solution",
    "solution": "We establish a coordinate system with point $A$ as the origin, $AB$ as the $x$-axis, $AC$ as the $y$-axis, and $AA_1$ as the $z$-axis. Then:\n\n- $F(t_1, 0, 0)$ with $0 < t_1 < 1$\n- $E(0, 1, \\frac{1}{2})$\n- $G(\\frac{1}{2}, 0, 1)$\n- $D(0, t_2, 0)$ with $0 < t_2 < 1$\n\nTherefore,\n\n$\\vec{EF} = (t_1, -1, -\\frac{1}{2})$\n\n$\\vec{GD} = (-\\frac{1}{2}, t_2, -1)$\n\nSince $GD \\perp EF$, their dot product is zero:\n\n$$\nt_1 + 2t_2 = 1\n$$\n\nSo $0 < t_2 < \\frac{1}{2}$. Furthermore,\n\n$\\vec{DF} = (t_1, -t_2, 0)$\n\n$$\n|\\vec{DF}| = \\sqrt{t_1^2 + t_2^2} = \\sqrt{5t_2^2 - 4t_2 + 1} = \\sqrt{5\\left(t_2 - \\frac{2}{5}\\right)^2 + \\frac{1}{5}}\n$$\n\nThus,\n\n$$\n\\sqrt{\\frac{1}{5}} \\leq |\\vec{DF}| < 1\n$$\n\n**Answer:** (A)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19074,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $[x^2] - 2x + 1 = 0$, where $[x^2]$ denotes the greatest integer that does not exceed $x^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $x = 1$, $x = \\frac{1}{2}$, and $x = \\frac{3}{2}$.\n\n**Solution.** We have $x = \\frac{[x^2] + 1}{2}$. Hence, $x = t$ or $x = t + \\frac{1}{2}$, where $t$ is an integer.\n\nIf $x = t$, then\n$$\n[x^2] - 2x + 1 = [t^2] - 2t + 1 = t^2 - 2t + 1 = (t - 1)^2 = 0 \\implies t = 1 \\implies x = 1.\n$$\nIf $x = t + \\frac{1}{2}$, then\n$$\n[x^2] - 2x + 1 = [(t + \\frac{1}{2})^2] - 2(t + \\frac{1}{2}) + 1.\n$$\nNow, $(t + \\frac{1}{2})^2 = t^2 + t + \\frac{1}{4}$, so $[x^2] = t^2 + t$.\nThus,\n$$\n[x^2] - 2x + 1 = (t^2 + t) - 2(t + \\frac{1}{2}) + 1 = t^2 + t - 2t - 1 + 1 = t^2 - t.\n$$\nSo $t^2 - t = 0 \\implies t = 0$ or $t = 1$, which gives $x = \\frac{1}{2}$ or $x = \\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19075,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be two positive integers, $m, n \\ge 2$. Solve in the set of positive integers the equation\n\n$$\nx^n + y^n = 3^m.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$ and write $x = d a$, $y = d b$, with $\\gcd(a, b) = 1$. Notice that $a$ and $b$ cannot be equal, for $3^m$ is odd. The equation rewrites as\n$$\nd^n (a^n + b^n) = 3^m,\n$$\nhence $d = 3^t$ and $a^n + b^n = 3^k$, with $\\gcd(a, 3) = \\gcd(b, 3) = 1$.\n\nRule out the case $n$ is even, for $3^k = a^n + b^n \\equiv 2 \\pmod{3}$, which is false.\n\nFor $n$ odd, write\n$$\n(a + b)\\left(a^{n-1} - a^{n-2}b + \\dots - a b^{n-2} + b^{n-1}\\right) = 3^k\n$$\nto get $a + b = 3^s > 1$. Since $n \\ge 2$ and $a \\ne b$, we have $3^k = a^n + b^n > a + b = 3^s$, hence $k - s \\ge 1$. Put $b = 3^s - a$ to obtain\n$$\na^{n-1} - a^{n-2}(3^s - a) + a^{n-3}(3^s - a)^2 + \\dots - a(3^s - a)^{n-2} + (3^s - a)^{n-1} = 3^{k-s},\n$$\nand therefore\n$$\nn a^{n-1} + M 3^s = 3^{k-s}.\n$$\nRecall that $\\gcd(a, 3) = 1$ to conclude that $n$ is a multiple of $3$.\n\nSet $n = 3p$, $p \\ge 1$, and let $a^p = u$ and $b^p = v$. The relation $a^{3p} + b^{3p} = 3^k$ yields successively\n$$\n(u + v)(u^2 - u v + v^2) = 3^k \\implies u + v = 3^r > 1,\\quad u^2 - u v + v^2 = 3^{k - r}\n$$\n$$\nu^2 - u(3^r - u) + (3^r - u)^2 = 3^{k - r}\n$$\n$$\nu^2 - u \\cdot 3^r + 3^{2r - 1} = 3^{k - r - 1}.\n$$\nNotice that $k - r - 1 \\ge 1$ yields $3 \\mid u$, hence $3 \\mid a$, a contradiction. Thus $k - r - 1 = 0$, implying $u^2 - u v + v^2 = 3$, with the solutions $(u, v) \\in \\{(1, 2), (2, 1)\\}$. Consequently, $u + v = 3$, $r = 1$, $k = 2$.\n\nNow, the relation $a^n + b^n = 9$ gives $(a, b) \\in \\{(1, 2), (2, 1)\\}$, $p = 1$ and $n = 3$, while $d^n (a^n + b^n) = 3^m$ yields $d^3 = 3^{m - 2}$.\n\nTherefore, if $n = 3$ and $m \\equiv 2 \\pmod{3}$, the solutions are\n$$\n\\left(3^{\\frac{m-2}{3}},\\ 2 \\cdot 3^{\\frac{m-2}{3}}\\right)\\quad \\text{and}\\quad \\left(2 \\cdot 3^{\\frac{m-2}{3}},\\ 3^{\\frac{m-2}{3}}\\right),\n$$\nand if $n \\ne 3$ or if $m \\ne 2 \\pmod{3}$, then the equation has no solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19076,
    "subject": "Mathematics (Olympiad)",
    "question": "Around a table, $2n$ people are seated and $m$ cookies are distributed among them. These people can pass cookies under the following rules:\n\n- One can only pass cookies to his or her neighbors.\n- One can pass a cookie only if he or she eats one cookie.\n\nLet $A$ be one of these people. Find the minimum $m$ such that, no matter how $m$ cookies are distributed, there is a strategy to pass cookies so that $A$ has at least one cookie.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the minimum number $m$ of cookies is $2^n$.\n\nLet us label the people as $A_{-n+1}, \\ldots, A_0 (=A), A_1, \\ldots, A_{n-1}, A_n$ in counterclockwise order.\n\n**Lower bound:**\nIf $m < 2^n$, there is a way to distribute $m$ cookies so that $A$ cannot get a cookie. Let $a_i$ be the number of cookies given to $A_i$. Define\n$$\nN = \\sum_{i=-n+1}^{n} a_i 2^{|i|}.\n$$\nThe value of $N$ is non-increasing when cookies are passed to neighbors. If we give all $m$ cookies to $A_n$, then $N = m 2^n < 2^n 2^n = 2^{2n}$, but more importantly, $N < 2^n$ (since $m < 2^n$). Thus, $A$ cannot get a cookie.\n\n**Upper bound:**\nNow, suppose $m = 2^n$. We show that $A$ can always get a cookie. By symmetry, assume $\\sum_{i=0}^{n-1} a_i \\geq \\sum_{i=0}^{n-1} a_{-i}$.\n\nAsk $A_n$ to pass $\\lfloor a_n/2 \\rfloor$ cookies to $A_{n-1}$, eating $\\lfloor a_n/2 \\rfloor$ cookies himself. Then $A_0, A_1, \\ldots, A_{n-1}$ have at least $m/2 = 2^{n-1}$ cookies. (If $a_n$ is even, this is clear; if $a_n$ is odd, they have at least $m/2 - 1/2$ cookies, but since $m$ is even, this is still at least $2^{n-1}$.)\n\n**Inductive step:**\nFor $k = 1, 2, \\ldots, n-1$, if $A_0, \\ldots, A_k$ have at least $2^k$ cookies, then by passing cookies from $A_k$ to $A_{k-1}$, we can ensure $A_0, \\ldots, A_{k-1}$ have at least $2^{k-1}$ cookies. By induction, $A_0$ (i.e., $A$) can get at least $2^0 = 1$ cookie.\n\n$\\boxed{2^n}$ is the minimum $m$ such that $A$ can always get a cookie.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19077,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadruplet of distinct positive integers $a, b, c, d$ is called $k$-good if the following conditions hold:\n\n1. Among $a, b, c, d$, no three form an arithmetic progression.\n\n2. Among $a + b, a + c, a + d, b + c, b + d, c + d$, there are $k$ of them forming an arithmetic progression.\n\n\na) Find a 4-good quadruplet.\n\nb) What is the maximal $k$ such that there is a $k$-good quadruplet?\n",
    "options": [],
    "answer": "See solution",
    "solution": "a) The quadruple $(7, 6, 4, 3)$ is 4-good because there are no three numbers in it that form an arithmetic progression, but among the six numbers:\n\n$$\n7 + 6 = 13, \\quad 7 + 4 = 11, \\quad 7 + 3 = 10, \\quad 6 + 4 = 10, \\quad 6 + 3 = 9, \\quad 4 + 3 = 7\n$$\n\nthe numbers $7, 9, 11, 13$ form an arithmetic progression.\n\nb) Without loss of generality, let $a > b > c > d$. Then:\n\n$$\na + b > a + c > \\max(a + d, b + c) > \\min(a + d, b + c) > b + d > c + d.\n$$\n\nNote that if:\n\n1) $a + b, a + c$, and $a + d$ form an arithmetic progression, then $2(a + c) = (a + b) + (a + d) \\iff 2c = b + d$;\n\n2) $a + b, a + c$, and $b + c$ form an arithmetic progression, then $2(a + c) = (a + b) + (b + c) \\iff 2b = a + c$;\n\n3) $a + d, b + d$, and $c + d$ form an arithmetic progression, then $2(b + d) = (a + d) + (c + d) \\iff 2b = a + c$;\n\n4) $b + c, b + d$, and $c + d$ form an arithmetic progression, then $2(b + d) = (b + c) + (c + d) \\iff 2c = b + d$.\n\nIn all four cases, we get a contradiction with the given condition.\n\nFrom the above, it is clear that all 6 numbers cannot form an arithmetic progression.\n\nSuppose that 5 of them form an arithmetic progression. Notice that whichever of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, or $c+d$ we delete, there is always some from progressions 1, 2, 3, or 4, which leads to a contradiction.\n\nIt follows from part (a) that the required maximal $k$ is $4$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19078,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha \\in (1, +\\infty)$ be a real number, and let $P(x) \\in \\mathbb{R}[x]$ be a monic polynomial of degree $24$ such that:\n\n1. $P(0) = 1$.\n2. $P(x)$ has exactly $24$ positive real roots, all less than or equal to $\\alpha$.\n\nShow that\n$$\n|P(1)| \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1, x_2, \\dots, x_{24}$ be the roots of $P(x)$. Then $0 < x_1, x_2, \\dots, x_{24} \\le \\alpha$ and $x_1 x_2 \\cdots x_{24} = 1$. We need to prove\n$$\n|(x_1 - 1)(x_2 - 1) \\cdots (x_{24} - 1)| \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\n\nIf any $x_i = 1$, the inequality is obvious. Otherwise, assume $x_1 \\le x_2 \\le \\cdots \\le x_k < 1 < x_{k+1} \\le \\cdots \\le x_{24} \\le \\alpha$.\n\nRewrite the inequality as\n$$\n(1 - x_1)\\cdots(1 - x_k)(x_{k+1} - 1)\\cdots(x_{24} - 1) \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\n\nLet $t = (x_1 x_2 \\cdots x_k)^{1/k}$. By AM-GM,\n$$\n(1 - x_1)\\cdots(1 - x_k) \\le (1 - t)^k.\n$$\n\nFor $i = k+1, \\dots, 24$, let $x_i = e^{y_i}$ and $l = e^{\\frac{y_{k+1} + \\cdots + y_{24}}{24-k}}$. Since $h(x) = \\ln(e^x - 1)$ is concave for $x > 0$, by Jensen's inequality,\n$$\n(x_{k+1} - 1)\\cdots(x_{24} - 1) \\le (l - 1)^{24 - k}.\n$$\n\nThus,\n$$\n(1 - t)^k (l - 1)^{24 - k} \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\n\nSince $t^k l^{24 - k} = 1$, $t = l^{-\\frac{24 - k}{k}}$, so the inequality becomes\n$$\n\\left(1 - \\frac{1}{l^{\\frac{24 - k}{k}}}\\right)^k (l - 1)^{24 - k} \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\n\nSince $1 < l \\le \\alpha$,\n$$\n\\left(l^{\\frac{24 - k}{k}} - 1\\right)^k \\left(1 - \\frac{1}{l}\\right)^{24 - k} \\le \\frac{\\left(\\alpha^{\\frac{24 - k}{k}} - 1\\right)^k (\\alpha - 1)^{24 - k}}{\\alpha^{24 - k}}.\n$$\n\nSo it suffices to show\n$$\n\\frac{\\left(\\alpha^{\\frac{24 - k}{k}} - 1\\right)^k (\\alpha - 1)^{24 - k}}{\\alpha^{24 - k}} \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\n\nLet\n$$\nf(x) = \\frac{\\left(x^{\\frac{24 - k}{k}} - 1\\right)^k}{x^{24 - k} (x - 1)^k}.\n$$\n\nWe have $\\ln f(x) = k \\ln(x^{\\frac{24 - k}{k}} - 1) - (24 - k) \\ln x - k \\ln(x - 1)$, so\n$$\n\\frac{f'(x)}{f(x)} = \\frac{(24 - k)x^{\\frac{24 - k}{k}}}{x(x^{\\frac{24 - k}{k}} - 1)} - \\frac{24 - k}{x} - \\frac{k}{x - 1} = -\\frac{k x^{\\frac{24}{k}} - 24x + (24 - k)}{x(x - 1)(x^{\\frac{24 - k}{k}} - 1)}.\n$$\n\nBy AM-GM, $k x^{\\frac{24}{k}} - 24x + (24 - k) > 0$, so $f'(x) < 0$ and $f$ is decreasing for $x > 1$. Thus,\n$$\nf(\\alpha) \\le \\lim_{x \\to 1^+} f(x) = \\left(\\frac{24 - k}{k}\\right)^k.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19079,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a sheet of paper in the form of a rectangle $x \\times y$, where $x$ and $y$ are integer numbers larger than $1$. Let us draw a lattice of $x \\times y$ unit squares on the sheet. Rolling up the rectangle and gluing it along its opposite sides, we shape the lateral surface of a circular cylinder. Join each two distinct vertices of the marked unit squares on the surface by a segment. How many of all these segments are passing through an interior point of the cylinder? In the case $x > y$, decide when this number of \"internal\" segments is larger — for the cylinder with bases of perimeter $x$, or $y$?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will compute the requested number $P$ of all internal segments for the cylinder formed by gluing the rectangle $x \\times y$ along the opposite sides of length $y$.\n\nThis cylinder has two bases of perimeter $x$ and its lateral sides are of length $y$. We use the formula $P = P_0 - P_1 - P_2$, where $P_0$ is the total number of segments, $P_1$ is the number of segments lying on the lateral surface, and $P_2$ is the number of segments lying on one of the two bases.\n\nThe vertices of the unit squares are situated on the surface of the cylinder so that exactly $y+1$ of them lie on the same of the $x$ lateral sides, and exactly $x$ vertices lie on the same boundary circle of the two bases. Thus,\n\n$$\n\\begin{aligned}\nP_0 &= \\binom{x(y+1)}{2} = \\frac{x(y+1)(xy+x-1)}{2}, \\\\\nP_1 &= x \\cdot \\binom{y+1}{2} = \\frac{x(y+1)y}{2}, \\\\\nP_2 &= 2 \\cdot \\binom{x}{2} = x(x-1).\n\\end{aligned}\n$$\n\nConsequently,\n\n$$\n\\begin{aligned}\nP &= P_0 - P_1 - P_2 = \\frac{x(y+1)(xy+x-1)}{2} - \\frac{x(y+1)y}{2} - x(x-1) \\\\\n  &= \\frac{x(x-1)(y^2+2y-1)}{2}.\n\\end{aligned}\n$$\n\nBy symmetry, the number $Q$ of internal segments for the other cylinder (with base's perimeter $y$ and lateral sides of length $x$) is\n\n$$\nQ = \\frac{y(y-1)(x^2 + 2x - 1)}{2}.\n$$\n\nTo decide which of $P > Q$ or $Q > P$ holds when $x > y$, we factorize the difference $P - Q$:\n\n$$\n\\begin{aligned}\n2(P - Q) &= (x^2 - x)(y^2 + 2y - 1) - (y^2 - y)(x^2 + 2x - 1) \\\\\n&= (x - y)(3xy - x - y + 1).\n\\end{aligned}\n$$\n\nThus, $x > y$ implies $P > Q$ if $3xy - x - y + 1 > 0$. Since $y \\ge 2$, $3xy \\ge 6x$, so\n\n$$\n3xy - x - y + 1 \\ge 5x - y + 1 > 4x + 1 > 0.\n$$\n\n**Answer:** In the case when $x > y$, the number of internal segments is larger for the cylinder with bases of perimeter $x$.\n\n**Remark:** A shorter way to determine $P$ is to note that the orthogonal projection of each internal segment to the fixed base of the cylinder is one of the $\\frac{1}{2}x(x-1)$ segments connecting $x$ vertices on the boundary circle. Each of these projections is common for exactly $(y+1)^2 - 2 = y^2 + 2y - 1$ internal segments, so\n\n$$\nP = \\frac{x(x-1)(y^2 + 2y - 1)}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19080,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \\ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \\le 50$ is $f_{50}(n) = 12$?\n\n(A) 7  \n(B) 8  \n(C) 9  \n(D) 10  \n(E) 11",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (D):**\n\nLet $\\tau(n)$ denote the number of positive divisors of $n$. If $n = p_1^{a_1} p_2^{a_2} \\cdots p_r^{a_r}$, then\n\n$$\n\\tau(n) = (a_1 + 1)(a_2 + 1)\\cdots(a_r + 1).\n$$\n\nWe seek $n \\le 50$ such that $f_{50}(n) = 12$. Since $f_1(n)$ is twice the number of divisors, $f_1(n) = 2\\tau(n)$. We look for $n$ such that repeated application of $f_1$ eventually yields $12$ and stays at $12$.\n\nThe possible factorizations for $n$ with $f_1(n) = 12$ are:\n\n- $n = p_1^2 p_2$ (where $p_1, p_2$ are distinct primes): $\\tau(n) = (2+1)(1+1) = 6$, so $f_1(n) = 12$.\n- $n = p^5$: $\\tau(n) = 6$, $f_1(n) = 12$.\n- $n = p_1^2 p_2^2$: $\\tau(n) = (2+1)(2+1) = 9$, $f_1(n) = 18$, but $f_1(18) = 12$.\n- $n = p_1^4 p_2$: $\\tau(n) = (4+1)(1+1) = 10$, $f_1(n) = 20$, $f_1(20) = 12$.\n\nCounting $n \\le 50$:\n\n- $p_1^2 p_2$: $12 = 2^2 \\cdot 3$, $18 = 3^2 \\cdot 2$, $20 = 2^2 \\cdot 5$, $28 = 2^2 \\cdot 7$, $44 = 2^2 \\cdot 11$, $45 = 3^2 \\cdot 5$, $50 = 5^2 \\cdot 2$ (7 numbers)\n- $p^5$: $32 = 2^5$ (1 number)\n- $p_1^2 p_2^2$: $36 = 2^2 \\cdot 3^2$ (1 number)\n- $p_1^4 p_2$: $48 = 2^4 \\cdot 3$ (1 number)\n\nTotal: $7 + 1 + 1 + 1 = 10$ numbers.\n\n**Thus, the answer is (D) 10.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19081,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = x^2 - 3x + 13$. What is the smallest prime $p$ that divides $f(n)$ for some integer $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Way 1.** Because $f(n) \\equiv f(m) \\pmod{p}$ when $n \\equiv m \\pmod{p}$, it is sufficient to calculate $f(n) \\pmod{p}$ for $p$ consecutive values of $n$. We only need to do this for $p \\in \\{2, 3, 5, 7\\}$ in order to determine whether 11 is the smallest prime we are looking for. Because $f(3-n) = f(n)$, we have $f(2) = f(1) = 11$, $f(3) = f(0) = 13$ and $f(4) = f(-1) = 17$. Finally, $f(-2) = 23$ and we see that $f(n) \\geq 11$ is a prime for each of the seven consecutive values $n = -2, -1, 0, 1, 2, 3, 4$. Hence, the smallest prime number that divides $f(n)$ for at least one $n$ is $p = 11$.\n\n**Way 2.** For each $p \\in \\{2, 3, 5, 7\\}$ we discuss $f(n) \\pmod{p}$ separately.\n\n$$\nf(n) = n^2 - 3n + 13 \\equiv n^2 - n + 1 \\equiv 1 \\pmod{2}\n$$\n\nbecause $n^2 \\equiv n \\pmod{2}$ for all integers $n$ by Fermat's Little Theorem.\n\n$$\nf(n) = n^2 - 3n + 13 \\equiv n^2 - 2 \\not\\equiv 0 \\pmod{3}\n$$\n\nbecause squares of integers can only be congruent to 0 or 1 modulo 3.\n\n$$\nf(n) = n^2 - 3n + 13 \\equiv n^2 + 2n + 3 \\equiv (n+1)^2 + 2 \\not\\equiv 0 \\pmod{5}\n$$\n\nbecause squares of integers can only be congruent to 0 or $\\pm 1$ modulo 5.\n\n$$\nf(n) = n^2 - 3n + 13 \\equiv n^2 + 4n + 6 \\equiv (n+2)^2 + 2 \\not\\equiv 0 \\pmod{7}\n$$\n\nbecause squares of integers can only be congruent to 0, 1, 2 or 4 modulo 7. Hence, for no integer $n$ is $f(n)$ divisible by 2, 3, 5 or 7. The smallest prime number that divides $f(n)$ for at least one $n$ therefore is $p = 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19082,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $ABC$ is given. Call its orthocenter $H$. Define $\\omega$ as the circle through $B$, $C$, and $H$, and define $\\Gamma$ as the circle with diameter $AH$. Let $X$ be the other intersection of $\\omega$ and $\\Gamma$, and let the reflection of $\\Gamma$ over $AX$ be $\\gamma$.\n\nSuppose $\\gamma$ and $\\omega$ intersect again at $Y \\neq X$, and line $AH$ and $\\omega$ intersect again at $Z \\neq A$. Show that the circle through $A$, $Y$, $Z$ passes through the midpoint of segment $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $BC$. We first show that $X$ lies on $AM$. Consider $A'$, the reflection of $A$ across $M$. As $ABA'C$ is a parallelogram, we have that $\\angle BA'C = \\angle BAC = 180^\\circ - \\angle BHC$, which in turn gives us that $A'$ lies on $\\omega$.\n\nNow $\\angle HBA' = \\angle HBC + \\angle CBA' = \\angle HBC + \\angle ACB = 90^\\circ$. Hence $HA$ is a diameter of $\\omega$. In particular we must have $\\angle HXA' = 90^\\circ$. Consequently $\\angle AXA' = \\angle AXH + \\angle HXA' = 90^\\circ + 90^\\circ = 180^\\circ$, i.e. $A$, $X$, $A'$ are collinear. But $A$, $M$, $A'$ are collinear by definition, hence $X$ lies on the $A$-median.\n\nNow it suffices to show that $\\angle AYZ = \\angle AMZ$. We note the two following facts:\n\n* $\\angle AHX = \\angle AYX$, since $\\omega$ and $\\Gamma$ have the same radius and the two angles span the same chord $AX$.\n* $\\omega$ is the reflection of the circumcircle of $ABC$ across $BC$. That gives us that $Z$ is the reflection of $A$ across $D$, the foot of the $A$-altitude to $BC$.\n\nHence we can write: $\\angle AYZ = \\angle AYX + \\angle XYZ = \\angle AHX + (180^\\circ - \\angle XHZ) = 2\\angle AHX = 2\\angle AMD = \\angle AMZ$, which is what we wanted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19083,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers. Prove that\n\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nBy the Cauchy-Schwarz Inequality, we have\n\n$$\n(y + z + z)(y + z + x) \\geq (y + z + \\sqrt{zx})^2 = (y + \\sqrt{zx} + z)^2.\n$$\n\nTherefore,\n\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} \\geq \\frac{2x^2 + xy}{(x + y + z)(y + 2z)} = \\frac{2x}{y + 2z} - \\frac{x}{x + y + z}.\n$$\n\nApplying the same reasoning to the other variables, we get\n\n$$\n\\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} \\geq \\frac{2y}{z + 2x} - \\frac{y}{x + y + z},\n$$\n\nand\n\n$$\n\\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{2z}{x + 2y} - \\frac{z}{x + y + z}.\n$$\n\nAdding these three inequalities together yields\n\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + zx}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{2x}{y+2z} + \\frac{2y}{z+2x} + \\frac{2z}{x+2y} - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19084,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_k$ be the coefficient of the $x^k$ term in the polynomial $$(x+1)^3(x+2)^3(x+3)^3$$ for $k$ with $0 \\leq k \\leq 9$. Find the value of $a_2 + a_4 + a_6 + a_8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the coefficients $a_0, a_1, a_2, \\dots, a_9$, we can write\n\n$$(x+1)^3(x+2)^3(x+3)^3 = a_0 + a_1 x + a_2 x^2 + \\dots + a_8 x^8 + a_9 x^9.$$\n\nSubstituting $x=1$ and $x=-1$ gives:\n\n$$\n\\begin{aligned}\na_0 + a_1 + a_2 + \\dots + a_8 + a_9 &= 2^3 \\cdot 3^3 \\cdot 4^3 \\\\\na_0 - a_1 + a_2 - a_3 + \\dots + a_8 - a_9 &= 0.\n\\end{aligned}\n$$\n\nAdding these equations and dividing by $2$ yields:\n\n$$a_0 + a_2 + a_4 + a_6 + a_8 = \\frac{1}{2}(2^3 \\cdot 3^3 \\cdot 4^3) = 6912.$$ \n\nSubstituting $x=0$ gives $a_0 = 1^3 \\cdot 2^3 \\cdot 3^3 = 216$. \n\nTherefore,\n\n$$a_2 + a_4 + a_6 + a_8 = 6912 - 216 = 6696.$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19085,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ is a composite number such that $\\varphi(n)$ divides $n-1$, where $\\varphi$ denotes Euler's totient function. Show that $n$ has at least four distinct prime factors.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n$ has only one prime factor. Then $n = p^\\alpha$ for some $\\alpha > 1$ and prime $p$. Then $\\varphi(n) = p^{\\alpha-1}(p-1)$ divides $p^\\alpha - 1$, which is impossible since $\\alpha \\ge 2$. Thus $n$ has at least two prime factors.\n\nSuppose $n = p^\\alpha q^\\beta$, where $p \\ne q$ are primes, $\\alpha \\ge 1$ and $\\beta \\ge 1$. Here $\\varphi(n) = p^{\\alpha-1}q^{\\beta-1}(p-1)(q-1)$ and $n-1 = p^\\alpha q^\\beta - 1$. Thus $\\varphi(n) \\mid (n-1)$ implies that $\\alpha = 1 = \\beta$ and $(p-1)(q-1)$ divides $pq-1$. This implies that $(p-1) \\mid (q-1)$ and $(q-1) \\mid (p-1)$, forcing $p = q$.\n\nIf $n$ has three distinct prime factors, say $p, q, r$, we see as in the earlier case $n = pqr$ and $(p-1)(q-1)(r-1)$ divides $pqr-1$. Taking $p-1 = x, q-1 = y$ and $r-1 = z$, we see that\n\n$$\nt = \\frac{pqr - 1}{(p-1)(q-1)(r-1)} = 1 + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx},\n$$\n\nis an integer. We also observe that no prime can be even (otherwise $pqr-1$ is odd whereas one of $p-1, q-1, r-1$ is even). Thus we may assume $3 \\le p < q < r$, so that $x \\ge 2, y \\ge 4$ and $z \\ge 6$. Thus\n\n$$\n1 < t \\le 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{8} + \\frac{1}{12} + \\frac{1}{24} = 1 + \\frac{28}{24} < 3.\n$$\n\nThus $t = 2$ and hence\n\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} = 1.\n$$\n\nIf $p \\ge 5$, we have $x \\ge 4, y \\ge 6$ and $z \\ge 10$. In this case,\n\n$$\n1 = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\le \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{10} + \\frac{1}{24} + \\frac{1}{60} + \\frac{1}{40} = \\frac{72}{120} < 1.\n$$\n\nThus $p = 3$ and $x = 2$. We obtain\n\n$$\n\\frac{1}{y} + \\frac{1}{z} + \\frac{1}{2} \\left( \\frac{1}{y} + \\frac{1}{z} \\right) + \\frac{1}{yz} = \\frac{1}{2}.\n$$\n\nThis may be written in the form $(y-3)(z-3) = 11$. We thus get $y = 4, z = 14$. In turn, we have $q = 5$ and $r = 15$. But then $r$ is not a prime. Thus the number of distinct prime factors of $n$ is more than 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19086,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(x, y, z)$ of natural numbers such that $21^x + 4^y = z^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $(1, 1, 5)$.\n\nRewrite 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 $d = 1$.\n\nConsequently, either $z - 2^y = 1$ and $z + 2^y = 21^x$, or $z - 2^y = 3^x$ and $z + 2^y = 7^x$.\n\nThe first case leads to $21^x - 1 = 2^{y+1}$, equality which cannot hold, because $5 \\mid 21^x - 1$ and $5 \\nmid 2^{y+1}$.\n\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 $x \\ge 2$.\n\nIf $x$ is odd, then\n\n$$\n2^{y-1} = 7^{x-1} + 7^{x-2} \\cdot 3 + 7^{x-3} \\cdot 3^2 + \\dots + 7 \\cdot 3^{x-2} + 3^{x-1},\n$$\n\na contradiction, because the left-hand side is even and right-hand side is odd.\n\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": 19087,
    "subject": "Mathematics (Olympiad)",
    "question": "Three friends $A$, $B$, and $C$ started their trip from the Flower-city to the Green-city. The distance between the two cities is $1.7$ km. The speed of each of them is $3\\text{ m/min}$, $4\\text{ m/min}$, and $5\\text{ m/min}$. They have one bike that can go at a speed of $20\\text{ m/min}$. One of them took the bike first and two started walking. After some time, he left the bike on the road and continued walking towards the Green-city. Another friend, who first reached the bike, took it and after some time left it on the road and continued walking. Finally, the third friend took the bike and arrived at the Green-city. It turned out that all three friends arrived at the Green-city at the same time. How long did each of them travel by bike?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $A$, $B$, and $C$ travel by bike for $43$ min, $29$ min, and $13$ min, respectively.\n\n**Solution:** Suppose each friend spent $t$ min traveling from one city to another. Let $A$ travel $x$ min by bike, $B$ travel $y$ min by bike, and $C$ travel $z$ min by bike. Then we have the following system:\n\n$$\n20x + 3(t-x) = 1700 \\\\\n20y + 4(t-y) = 1700 \\\\\n20z + 5(t-z) = 1700 \\\\\n20(x + y + z) = 1700\n$$\n\nSolving this system, we obtain: $t = 253$, $x = 43$, $y = 29$, $z = 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19088,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x_4$ is an integer, then a new set of three numbers is given by $x'_1 = x_2$, $x'_2 = x_3$, and $x'_3 = x_4$, and in a minute the aggregate computes $x'_4$. If at first $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, will the aggregate work without stops for at least 2018 minutes?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_4, x_5, x_6, \\dots$ denote a sequence of numbers that will be computed by the aggregate. Then for every natural $n$, the following equality is true:\n\n$$\nx_{n-2}(x_{n+1} + x_{n-1}) = x_n(x_n + x_{n-1}).\n$$\n\nLet's add $x_n x_{n-2}$ to both sides:\n\n$$\n\\begin{aligned}\nx_{n-2}(x_{n+1} + x_n + x_{n-1}) &= x_n(x_n + x_{n-1} + x_{n-2}) \\\\\n\\frac{x_{n-2}(x_{n+1} + x_n + x_{n-1})}{x_n x_{n-1}} &= \\frac{x_n(x_n + x_{n-1} + x_{n-2})}{x_n x_{n-1}} \\\\\n\\frac{x_{n+1} + x_n + x_{n-1}}{x_n x_{n-1}} &= \\frac{x_n + x_{n-1} + x_{n-2}}{x_{n-1} x_{n-2}}.\n\\end{aligned}\n$$\n\nWe get a ratio that is true for every natural $n$. It is possible to continue it to the initial values:\n\n$$\n\\begin{aligned}\n\\frac{x_{n+1} + x_n + x_{n-1}}{x_n x_{n-1}} &= \\frac{x_n + x_{n-1} + x_{n-2}}{x_{n-1} x_{n-2}} = \\frac{x_{n-1} + x_{n-2} + x_{n-3}}{x_{n-2} x_{n-3}} = \\dots = \\frac{x_3 + x_2 + x_1}{x_2 x_1} = 3 \\\\\nx_{n+1} + x_n + x_{n-1} &= 3x_n x_{n-1} \\\\\nx_{n+1} = 3x_n x_{n-1} - x_n - x_{n-1}.\n\\end{aligned}\n$$\n\nSo finally, $x_{n+1} \\in \\mathbb{Z}$ for any natural $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19089,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $a + b + c = 1$. Prove that\n$$\n\\frac{a}{b} + \\frac{b}{a} + \\frac{c}{c} + \\frac{c}{b} + \\frac{c}{a} + \\frac{a}{c} + 6 \\ge 2\\sqrt{2} \\left( \\sqrt{\\frac{1-a}{a}} + \\sqrt{\\frac{1-b}{b}} + \\sqrt{\\frac{1-c}{c}} \\right)\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Replacing $1-a$, $1-b$, $1-c$ with $b+c$, $c+a$, $a+b$ respectively on the right-hand side, the given inequality becomes\n$$\n\\frac{b+c}{a} + \\frac{c+a}{b} + \\frac{a+b}{c} + 6 \\ge 2\\sqrt{2} \\left( \\sqrt{\\frac{b+c}{a}} + \\sqrt{\\frac{c+a}{b}} + \\sqrt{\\frac{a+b}{c}} \\right)\n$$\nand equivalently\n$$\n\\left( \\frac{b+c}{a} - 2\\sqrt{2} \\sqrt{\\frac{b+c}{a}} + 2 \\right) + \\left( \\frac{c+a}{b} - 2\\sqrt{2} \\sqrt{\\frac{c+a}{b}} + 2 \\right) + \\left( \\frac{a+b}{c} - 2\\sqrt{2} \\sqrt{\\frac{a+b}{c}} + 2 \\right) \\ge 0\n$$\nwhich can be rewritten as\n$$\n\\left( \\frac{b+c}{a} - \\sqrt{2} \\right)^2 + \\left( \\frac{c+a}{b} - \\sqrt{2} \\right)^2 + \\left( \\frac{a+b}{c} - \\sqrt{2} \\right)^2 \\ge 0\n$$\nwhich is true.\n\nThe equality holds if and only if\n$$\n\\frac{b+c}{a} = \\frac{c+a}{b} = \\frac{a+b}{c},\n$$\nwhich together with the given condition $a + b + c = 1$ gives $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19090,
    "subject": "Mathematics (Olympiad)",
    "question": "Two unit complex numbers $x$ and $y$ with $\\arg x, \\arg y \\in (0, \\frac{\\pi}{2})$ and $\\arg y > 2\\arg x$ (so $\\triangle ABC$ is acute, $\\angle ACB > 2\\angle ABC$), satisfy\n\n$$\n|y - 1| = |(x^2 - 1)(y + 1)|.\n$$\n\nProve that\n\n$$\n|x^2(y + 1) + (y^2 - y)x - 2| = \\left| \\frac{(x^2 - 1)(1 - y^3)}{x^2 - y} \\right|.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof.** Notice that\n\n$$\n|y - 1| = |(x^2 - 1)(y + 1)| \\iff \\left| \\frac{(x^2 - 1)(y + 1)}{y - 1} \\right| = 1.\n$$\n\nThis implies\n$$\n\\frac{x^2 - 1}{x} = \\frac{y - 1}{y + 1}.\n$$\nSo $(y + 1)x^2 = (y - 1)x + (y + 1)$, or equivalently\n$$\nx^2 - 1 = \\left( \\frac{y - 1}{y + 1} \\right)x.\n$$\n\nFor the second statement,\n$$\n|x^2(y + 1) + (y^2 - y)x - 2| = \\left| \\frac{(x^2 - 1)(1 - y^3)}{x^2 - y} \\right|.\n$$\n\nExpanding the left side:\n$$\n|x^2(y + 1) + (y^2 - y)x - 2| = |(y^2 - 1)x + (y - 1)| = |y - 1| \\cdot |(y + 1)x + 1|.\n$$\n\nExpanding the right side using the previous relation:\n$$\n\\left| \\frac{(x^2 - 1)(1 - y^3)}{x^2 - y} \\right| = \\left| \\frac{\\left( \\frac{y - 1}{y + 1} \\right)x(1 - y^3)}{\\left( \\frac{y - 1}{y + 1}x + 1 - y \\right)} \\right| = \\left| \\frac{(y - 1)x(1 - y^3)}{(y - 1)(x - y - 1)} \\right| = \\left| \\frac{(1 - y)(1 + y + y^2)}{x - (y + 1)} \\right|.\n$$\n\nNow, the equality is equivalent to $|[(y + 1)x + 1][x - (y + 1)]| = |1 + y + y^2|$. Expanding and using $(y + 1)x^2 = (y - 1)x + (y + 1)$, we get $|-(y^2 + y + 1)x| = |1 + y + y^2|$, which holds since $|x| = 1$. The proof is complete. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19091,
    "subject": "Mathematics (Olympiad)",
    "question": "A disphenoid is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?\n\n(A) $\\sqrt{3}$  \n(B) $3\\sqrt{15}$  \n(C) $15$  \n(D) $15\\sqrt{7}$  \n(E) $24\\sqrt{6}$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that adjacent faces must be rotations of one another to create a disphenoid (see the figure below). If they were instead mirrored, then the other two faces would be isosceles (and be neither scalene nor congruent to the others, as required).\n\n![](images/2024_AMC12A_Solutions_p16_data_f5fe63344e.png)\n\nNext, to determine what triangles could form a disphenoid, let the positive integer side lengths be $a$, $b$, and $c$, with $a < b < c$. Consider two faces of the tetrahedron that have their sides of length $c$ in common. That common edge can be thought of as a \"hinge\" between two faces. If those two faces were laid flat, they would form a parallelogram with one diagonal of length $c$. When folded over the hinge, the other diagonal of the parallelogram becomes two sides of a triangle whose third side is the last edge of the tetrahedron. (Note that this last edge must have length $c$ to create the disphenoid.) If the sides of length $a$ and $b$ formed a right angle, then the parallelogram would be a rectangle, and the other diagonal would also have length $c$. A tetrahedron formed by folding the triangles over the hinge would have its last edge shorter than $c$, making it not a disphenoid. If instead the sides of length $a$ and $b$ formed an obtuse angle, the other diagonal would be shorter than $c$, and folding would make the last edge shorter yet (and again not a disphenoid). Thus, the faces must be acute to form a disphenoid.\n\nTo minimize the surface area, $a$, $b$, and $c$ must be the least distinct integers that form an acute triangle. Note that $2$, $3$, and $4$ form an obtuse triangle (because $2^2 + 3^2 < 4^2$); and $3$, $4$, and $5$ form a right triangle. However, $4^2 + 5^2 = 41 > 36 = 6^2$, so $4$, $5$, and $6$ form an acute triangle with the least area. The area of one face can be computed using Heron's Formula. The triangle's semiperimeter is $\\frac{4+5+6}{2} = \\frac{15}{2}$, and the disphenoid's total surface area is\n\n$$\n\\begin{aligned}\n4 \\cdot \\sqrt{\\frac{15}{2} \\left(\\frac{15}{2} - 4\\right) \\left(\\frac{15}{2} - 5\\right) \\left(\\frac{15}{2} - 6\\right)} &= 4 \\cdot \\sqrt{\\frac{1}{16} \\cdot 15 \\cdot (15-8) \\cdot (15-10) \\cdot (15-12)} \\\\\n&= \\frac{4}{4} \\cdot \\sqrt{15 \\cdot 7 \\cdot 5 \\cdot 3} = 15\\sqrt{7}.\n\\end{aligned}\n$$\n\nAnother way to find the area of a face is to use the formula $\\frac{1}{2}ab \\sin \\gamma$. By the Law of Cosines,\n\n$$\n\\cos \\gamma = \\frac{a^2 + b^2 - c^2}{2ab} = \\frac{4^2 + 5^2 - 6^2}{2 \\cdot 4 \\cdot 5} = \\frac{1}{8}.\n$$\n\nThen\n\n$$\n\\sin \\gamma = \\sqrt{1 - \\frac{1}{64}} = \\frac{3\\sqrt{7}}{8}.\n$$\n\nTherefore, the area of one face is\n\n$$\n\\frac{1}{2} \\cdot 4 \\cdot 5 \\cdot \\frac{3\\sqrt{7}}{8} = \\frac{15\\sqrt{7}}{4},\n$$\n\nand the total surface area is $15\\sqrt{7}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19092,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the point of intersection of the altitudes $AP$ and $CQ$ of an acute-angled triangle $ABC$. On the median $BM$, points $E$ and $F$ are chosen so that $\\angle APE = \\angle BAC$, $\\angle CQF = \\angle BCA$, where the point $E$ lies inside the triangle $APB$, and the point $F$ lies inside the triangle $CQB$. Prove that the lines $AE$, $CF$ and $BH$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the lines $AE$ and $CF$ divide the line segment $BH$ in the same ratio (it is easy to see that the points of intersection of the lines $AE$ and $CF$ with the line $BH$ will belong to this segment). Let $T$ be the point of intersection of the lines $AE$ and $BH$.\n\n$$\n\\frac{BT}{TH} = \\frac{S_{\\triangle BAT}}{S_{\\triangle TAH}} = \\frac{BA \\cdot \\sin \\angle BAT}{AH \\cdot \\sin \\angle TAH}.\n$$\n\nThe lines $AE$, $BE$ and $CE$ are concurrent and so by the trigonometric version of Ceva's theorem we have:\n\n$$\n\\frac{\\sin \\angle BAT}{\\sin \\angle TAH} \\cdot \\frac{\\sin \\angle APE}{\\sin \\angle EPB} \\cdot \\frac{\\sin \\angle PBM}{\\sin \\angle MBA} = 1,\n$$\n\nand so\n\n$$\n\\frac{BT}{TH} = \\frac{BA}{AH} \\cdot \\frac{\\sin \\angle EPB}{\\sin \\angle APE} \\cdot \\frac{\\sin \\angle MBA}{\\sin \\angle PBM} = (*)\n$$\n\nSince $S_{ABM} = S_{CBM}$, we have that $AB \\sin \\angle ABM = BC \\sin \\angle CBM$, hence\n\n$$\n(*) = \\frac{BA}{AH} \\cdot \\frac{\\sin(90^\\circ - \\angle BAC)}{\\sin \\angle BAC} \\cdot \\frac{BC}{AB} = \\frac{BC}{AH} \\cdot \\frac{\\sin \\angle ABH}{\\sin \\angle BAC} = (**).\n$$\n\nThen using the sine theorem for the triangle $ABH$ and taking into account that $\\angle AHB = 180^\\circ - \\angle ACB$, we obtain:\n\n$$\n\\frac{AH}{\\sin \\angle ABH} = \\frac{AB}{\\sin \\angle AHB} = \\frac{AB}{\\sin \\angle ACB} = \\frac{BC}{\\sin \\angle BAC},\n$$\n\nthat is $\\frac{BT}{TH} = (**)=1$. So, we have proved that the line $AE$ passes through the midpoint of the line segment $BH$. One can similarly prove that the line $CF$ also passes through this point, which proves that the lines $AE$, $CF$ and $BH$ are concurrent.\n\n![](images/Ukrajina_2011_p49_data_f59b2f4c72.png)\n\nFig. 52",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19093,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a graph contain $v$ vertices, $v + m$ edges, and $k$ simple cycles, all of different lengths. Exhibit a lower bound for $k$ in terms of $v$ and $m$, i.e., a function $f$ such that\n\n$$\nk \\geq f(v, m)\n$$\n\nin any such graph.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $f(v, 0) = 1$, as any graph on $v$ vertices with $v$ edges contains a (simple) cycle.\n\nNotice that the lengths of the cycles are at least $1, 2, \\ldots, k$. Hence, the sum of their lengths is at least $\\frac{k(k+1)}{2}$. Therefore, among $v + m$ edges, there exists one contained in at least $\\frac{k(k+1)}{2(v+m)}$ cycles. Removing that edge, we get a graph with $v$ vertices, $v + m - 1$ edges, and at most $k - \\frac{k(k+1)}{2(v+m)}$ simple cycles. Therefore, we can set\n\n$$\nf(v, m) = f(v, m-1) + \\left\\lfloor \\frac{k(k+1)}{2(v+m)} \\right\\rfloor. \\quad (*)\n$$\n\nIf, for some value of $m$, we have $f(v, m) > v$, this means there is no such graph, as it cannot have more than $v$ cycles of distinct lengths. For large $v$, such an inequality holds for all $m \\geq \\epsilon v$.\n\nRecall that $k \\geq m$ (this also follows from $(*)$), so\n\n$$\n\\left\\lfloor \\frac{k(k+1)}{2(v+m)} \\right\\rfloor \\geq \\frac{m(m+1)}{2(v+m)}.\n$$\n\nThus, $(*)$ yields\n\n$$\nf(v, m) \\geq \\sum_{i=0}^{m} \\frac{i(i+1)}{2(v+i)} \\geq \\sum_{i=0}^{m} \\frac{i^2}{2(v+m)} = \\frac{m(m+1)(2m+1)}{12(v+m)} \\geq \\frac{m^3}{6(v+m)}.\n$$\n\nThis estimate exceeds $v$ for $m$ of order $C v^{2/3}$, which suffices for the problem purposes.\n\n**Remark.** Removal of vertices, instead of edges, would not be as effective, since removal of a vertex also leads to removal of many edges.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19094,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s$ be the side length of an equilateral triangle. The number of grid triangles inside it, $t$, is given by $t = s \\times s$.\n\n![](images/2018-Australian-Scene-W2_p49_data_4589ae248f.png)\n\nConsider an isopentagon with side lengths $a$ and $b$. Its $t$-number satisfies $a \\times a < t < 2 \\times a \\times a$. If an isopentagon has $t$-number $119$, then $a$ is at most $10$ and at least $8$.\n\nFind values of $a$ and $b$ such that\n$$\n2 \\times a \\times a - b \\times b = 119.\n$$\n\nAdditionally, let the perimeter of the isopentagon be $4 \\times a - b$. What is the smallest common perimeter for three isopentagons?",
    "options": [],
    "answer": "See solution",
    "solution": "From the equation $2 \\times a \\times a - b \\times b = 119$, and since $2 \\times a \\times a$ is even and $119$ is odd, $b$ must be odd.\n\nExamining possible values for $a = 8, 9, 10$ and odd $b$:\n- For $(a, b) = (8, 3)$: $2 \\times 8 \\times 8 - 3 \\times 3 = 128 - 9 = 119$.\n- For $(a, b) = (10, 9)$: $2 \\times 10 \\times 10 - 9 \\times 9 = 200 - 81 = 119$.\n\nThus, the isopentagons with $t$-number $119$ are $(8, 5, 3, 5, 8)$ (perimeter $29$) and $(10, 1, 9, 1, 10)$ (perimeter $31$).\n\nFrom the perimeter table, the smallest common perimeter for three isopentagons is $31$:\n- $(8, 7, 1, 7, 8)$, $(9, 4, 5, 4, 9)$, $(10, 1, 9, 1, 10)$.\nTheir $t$-numbers are $127$, $137$, and $119$ respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19095,
    "subject": "Mathematics (Olympiad)",
    "question": "Шаховската табла е поделена на 64 единечни квадрати. Најди го бројот на сите квадрати на шаховската табла, кои се формирани од единечните квадрати.",
    "options": [],
    "answer": "See solution",
    "solution": "Бројот на квадратите $1 \\times 1$ на шаховската табла е $8^2 = 64$. Квадрати $2 \\times 2$ ги има $7^2 = 49$, а $3 \\times 3$ има $6^2 = 36$, $4 \\times 4$ има $5^2 = 25$, $5 \\times 5$ има $4^2 = 16$, $6 \\times 6$ има $3^2 = 9$, $7 \\times 7$ има $2^2 = 4$ и $8 \\times 8$ има $1^2 = 1$.\n\nСпоред тоа, на таблата има вкупно:\n\n$$\n64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204\n$$\nквадрати.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19096,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Determine the least integer $n \\geq k+1$ for which the following game can be played indefinitely:\n\nConsider $n$ boxes, labelled $b_1, b_2, \\dots, b_n$. For each index $i$, box $b_i$ initially contains exactly $i$ coins. At each step, perform the following three substeps in order:\n\n1. Choose $k+1$ boxes.\n2. 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.\n3. If any box is left empty, the game ends; otherwise, go to the next step.",
    "options": [],
    "answer": "See solution",
    "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 $\\lfloor m_i/2 \\rfloor$ 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 $\\lfloor m_i/2 \\rfloor$ 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, enough to survive the $k$ steps to its next supply.\n\nWe now show that no smaller value of $n$ works. So, let $n \\leq 2^k + k - 2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w = \\lfloor \\log_2 m \\rfloor$ withdrawals; this $w$ will be referred to as the *weight* of that box. The sum of the weights of all boxes will be 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 $n \\leq 2^k + k - 2$, 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 $n \\geq 2^k + k - 1$.\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 among 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. Proceed by case analysis:\n\nIf $m_i = 1$, the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\leq \\lfloor \\log_2(2^k + k - 1) \\rfloor \\leq \\lfloor \\log_2(2^{k+1} - 2) \\rfloor \\leq k$; and if, in addition, $i \\leq 2^k - 2$, then the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\leq \\lfloor \\log_2(2^k - 1) \\rfloor = k - 1$.\n\nIf $m_i = 2$, then the weight increases by $\\lfloor \\log_2(i+2) \\rfloor - \\lfloor \\log_2 2 \\rfloor \\leq \\lfloor \\log_2(2^k + k) \\rfloor - 1 \\leq k - 1$.\n\nIf $m_i \\geq 3$, then the weight increases by\n\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\leq \\left\\lfloor \\log_2 \\left( 1 + \\frac{2^k + k - 2}{3} \\right) \\right\\rfloor + 1 \\leq k,\n\\end{aligned}\n$$\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 $i \\leq 2^k - 2$ to consider the subcases $m_i = 3$ and $m_i \\geq 4$. In the former subcase, the weight increases by\n\n$$\n\\lfloor \\log_2(i + 3) \\rfloor - \\lfloor \\log_2 3 \\rfloor \\leq \\lfloor \\log_2(2^k + 1) \\rfloor - 1 = k - 1,\n$$\n\nand in the latter by\n\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\leq \\left\\lfloor \\log_2 \\left( 1 + \\frac{2^k - 2}{4} \\right) \\right\\rfloor + 1 \\leq k - 1,\n\\end{aligned}\n$$\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": 19097,
    "subject": "Mathematics (Olympiad)",
    "question": "A convex $n$-gon is triangulated with a triangulation $P$ of $n-3$ non-intersecting diagonals. An *independent set* of vertices is a set of vertices of $P$ where no two vertices are connected by a side or by a diagonal. Let $i(P)$ be the number of ways to choose an independent set of vertices with respect to the triangulation $P$. Prove that the minimum $i(P)$ is achieved for a zig-zag triangulation.",
    "options": [],
    "answer": "See solution",
    "solution": "1) Let $z_n$ be the number of ways to choose an independent set of vertices for a zig-zag triangulation of an $n$-gon. It is known that each triangulation has at least two vertices which are not the endpoints of any diagonal. Let $A$ be one of these vertices for the zig-zag triangulation (see figure below).\n\nIf the independent set does not contain $A$, then by removing the vertex $A$ we obtain a zig-zag triangulated $(n-1)$-gon. If the independent set contains $A$, then it does not contain its two neighbours, and by removing $A$ and its two neighbours we obtain a zig-zag triangulated $(n-3)$-gon.\n\nTherefore $z_n$ satisfies the recurrence relation\n\n$$\nz_n = z_{n-1} + z_{n-3}.\n$$\n\n![](images/BW2021_Shortlist_p22_data_7a570526d4.png \"Figure 9: A zig-zag triangulation of a 10-gon.\")\n\n2) We will consider a triangulation of an $n$-gon as a graph: the vertices of the $n$-gon are vertices of the graph, the sides and the diagonals are edges.\n\n3) Now we will prove the problem statement $i(P) \\ge z_n$ by induction on $n$. The base case $n \\le 6$ is trivial.\n\nAssume the statement holds for all convex triangulations on fewer than $n$ vertices. Consider a triangulated $n$-gon $P$ as a graph. The triangulation has at least two vertices of degree 2. Let $A$ be one of these vertices, $u, v$ be its neighbours.\n\nAs in part 1), consider independent sets that do not contain $A$. The number of these sets equals the number of independent sets in graph $P \\setminus A$. By the induction hypothesis, it is at most $z_{n-1}$.\n\nNow consider independent sets that contain $A$ (and do not contain $u$ and $v$). The number of these sets equals the number of independent sets in the graph $H = P \\setminus \\{A, u, v\\}$, which has $n-3$ vertices. But, in general, graph $H$ is not a graph of a triangulation. Let us add some edges to $H$ in order to obtain a graph $H^*$ of some triangulation. This operation of adding new diagonals creates new neighbours in the graph, and thus decreases the number of independent sets. Then $i(H) \\ge i(H^*) \\ge z_{n-3}$.\n\nThus,\n\n$$\ni(P) \\ge i(P \\setminus A) + i(H) \\ge z_{n-1} + z_{n-3} = z_n.\n$$\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19098,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $H_a$, $H_b$, $H_c$ as the projections of $A$, $B$, $C$ onto $BC$, $AC$, $AB$ respectively. Let $H$ be the orthocenter of $\\triangle ABC$, and let $K$ be the reflection of $A$ over $BC$. The line parallel to $H_bH_c$ through $H$ meets $AB$ and $AC$ at points $X$ and $Y$. Prove that the inscribed circles of $\\triangle ABC$ and $\\triangle XYK$ touch.",
    "options": [],
    "answer": "See solution",
    "solution": "Define new points: $M$ and $N$ are the points symmetric to $H$ over $AB$ and $AC$ respectively; $S$ and $L$ are the midpoints of $XH$ and $HY$ respectively. Let's prove a few facts:\n\n1. $H_c$, $S$, $H_a$ are collinear.\n\nIt can be seen from $\\angle SH_cH = \\angle SHH_c = \\angle HH_cH_b = \\angle H_bBC = \\angle HBH_a = \\angle HH_cH_a$, where the first equality follows from the fact that $S$ is the midpoint of the hypotenuse of $\\triangle XH_cH$, the second from $XY \\parallel H_cH_b$, and the others from properties of inscribed quadrilaterals $BH_cH_bC$ and $BH_cHH_a$.\n\nSimilarly, collinearity of $H_b$, $L$, $H_a$ can be shown.\n\n![](images/Ukraine_2020_booklet_p40_data_5bf6471a40.png)\n\n2. $M$, $X$, $K$ are collinear.\n\nA homothety at $H$ with coefficient $2$ maps $H_c \\to M$, $S \\to X$, and $H_a \\to K$, which gives the desired result.\n\nSimilarly, collinearity of $N$, $Y$, $K$ can be shown.\n\n3. $XY \\parallel MN$.\n\nThis follows from $XY \\parallel H_cH_b$ and $MN \\parallel H_cH_b$, since $H_cH_b$ is a midline of $\\triangle HMN$.\n\nTherefore, a homothety at $K$ mapping $\\triangle KXY$ to $\\triangle KMN$ also maps their circumscribed circles into each other. Since $K$ is their common point, the circles touch at that point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19099,
    "subject": "Mathematics (Olympiad)",
    "question": "A police emergency number is a positive integer that ends with the digits 133 in decimal representation. Prove that every police emergency number has a prime factor larger than 7.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 1000k + 133$ be a police emergency number and assume that all its prime divisors are at most 7. It is clear from the last digit that $n$ is odd and that $n$ is not divisible by 5, so $1000k + 133 = 3^a 7^b$ for suitable integers $a, b \\ge 0$.\n\nThus $3^a 7^b \\equiv 133 \\pmod{1000}$.\n\nThis also implies $3^a 7^b \\equiv 133 \\equiv 5 \\pmod{8}$. We know that $3^a$ is congruent to 1 or 3 modulo 8 and $7^b$ is congruent to 1 or 7 modulo 8. In order for the product $3^a 7^b$ to be congruent to 5 modulo 8, $3^a$ must therefore be congruent to 3 and $7^b$ must be congruent to 7. We therefore conclude that $a$ and $b$ are both odd.\n\nWe also have $3^a 7^b \\equiv 133 \\equiv 3 \\pmod{5}$. As $a$ and $b$ are odd, $3^a$ and $7^b$ are each congruent to 3 or 2 modulo 5. Neither $3^2, 2^2$ nor $3 \\cdot 2$ is congruent to 3 modulo 5, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19100,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C'$ be the foot of the altitude from $C$. Since the triangle $ABC$ is isosceles with the apex at $C$, we have $|AC'| = |C'B| = \\frac{1}{2}|AB| = |CA'|$. Since $ABA'$ and $CBC'$ are right triangles and $\\angle ABA' = \\angle CBC'$, they are similar. This implies\n\n$$\n\\frac{|AB|}{|BA'|} = \\frac{|CB|}{|BC'|}.\n$$\n\nLet $c$ be the length of the side $AB$ and $x = |BA'|$. Then\n\n$$\n\\frac{c}{x} = \\frac{\\frac{c}{2} + x}{\\frac{c}{2}}.\n$$\n\n![](images/Slovenija_2008_p10_data_5c4f9fe018.png)\n\nFind the type of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We get the quadratic equation $2x^2 + cx - x^2 = 0$. The left-hand side can be factored as $(2x - c)(x + c) = 0$. Since $x$ and $c$ are positive, we have $x = \\frac{c}{2}$. The length of the segment $BC$ is therefore equal to $c$, so $|AC| = |BC| = |AB| = c$ and the triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19101,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $0 < x < y < z$ and $\\lambda > 1$. Prove that\n$$\n\\sum_{cyc} x^{\\lambda}y < \\sum_{cyc} xy^{\\lambda}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the function $f(y) = \\sum_{cyc} (x^{\\lambda}y - xy^{\\lambda})$. It takes the value $0$ for $y = x$ and $y = z$. We also have $f''(y) = \\lambda(\\lambda - 1)y^{\\lambda}(z - x) > 0$, so $f$ is convex. It follows that $f(y) < 0$ for $y \\in (x, z)$. Thus, the lemma is proved.\n\nNow, use the lemma for $x = a^{12}$, $y = b^{12}$, $z = c^{12}$, and $\\lambda = \\frac{20}{12}$ to obtain the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19102,
    "subject": "Mathematics (Olympiad)",
    "question": "A $3 \\times 3$ grid made up of nine $1 \\times 1$ squares is given. You want to distribute nine distinct positive integers, each chosen from $1$ to $9$, into the nine boxes of the grid. How many distinct ways are there to distribute the numbers if, for any pair of boxes sharing a side, the difference of the numbers inserted must be $3$ or less? Configurations that coincide under rotation or flipping are still considered distinct.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $32$ ways.\n\nConsider the $3 \\times 3$ grid. For any $2 \\times 2$ block of four squares, let $a, b, c, d$ be the numbers in those squares. Since adjacent squares must differ by at most $3$, $|a-b| \\leq 3$ and $|b-d| \\leq 3$, so $|a-d| \\leq 6$. If $|a-d| = 6$, then $b = \\frac{a+d}{2}$ and $c = \\frac{a+d}{2}$, but this would violate the requirement that all numbers are distinct. Thus, $|a-d| \\leq 5$.\n\nIf the center square contains a number $\\leq 3$, then $9$ cannot be placed anywhere. If the center contains $\\geq 7$, then $1$ cannot be placed. Therefore, the center must be $4$, $5$, or $6$.\n\nFor example, if $4$ is in the center, $9$ cannot be adjacent to it, so $9$ must be in a corner. By symmetry, we can fix $9$ in the lower right corner. The remaining two squares in that corner must be $6$ and $7$. Continuing this process and considering all possible placements, we find there are $32$ distinct ways to distribute the numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19103,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $k$, determine the maximum value of the expression\n\n$$\na^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k\n$$\n\nwhere $a, b, c$ are non-negative real numbers such that $a + b + c = 3k$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $(3k - 1)^{3k-1}$, achieved for example when $a = 0$, $b = 1$, and $c = 3k - 1$.\n\nTo prove this, let $F(a, b, c) = a^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k$. Clearly, $F(0, 1, 3k-1) = (3k-1)^{3k-1}$, so it suffices to show $F(a, b, c) \\le (3k-1)^{3k-1}$ for all non-negative $a, b, c$ with $a+b+c=3k$.\n\nSince the expression is cyclic in $a, b, c$, assume $b$ lies between $a$ and $c$. Then,\n\n$$\nb^{3k-1}c + c^{3k-1}a = c(b^{3k-1} + c^{3k-2}a) \\le c(b^{3k-2}a + c^{3k-2}b),\n$$\n\nso\n\n$$\n\\begin{aligned}\nF(a, b, c) &\\le a^{3k-1}b + c(b^{3k-2}a + c^{3k-2}b) + k^2 a^k b^k c^k \\\\\n&= b(a^{3k-1} + k^2 a^k b^{k-1}c^k + ab^{3k-3}c + c^{3k-1}).\n\\end{aligned}\n$$\n\nLet\n\n$$\nG(a, b, c) = a^{3k-1} + k^2 a^k b^{k-1} c^k + ab^{3k-3}c + c^{3k-1}.\n$$\n\nNotice that\n\n$$\n\\begin{aligned}\nG(a, b, c) &\\le a^{3k-1} + k^2 a^k (a^{k-1} + c^{k-1}) c^k + a(a^{3k-3} + c^{3k-3}) c + c^{3k-1} \\\\\n&\\le a^{3k-1} + \\binom{3k-1}{k} a^k (a^{k-1} + c^{k-1}) c^k + \\binom{3k-1}{1} a(a^{3k-3} + c^{3k-3}) c + c^{3k-1} \\\\\n&\\le (a+c)^{3k-1}, \\quad \\text{provided that } k \\ge 2.\n\\end{aligned}\n$$\n\nHence, if $k \\ge 2$, then $F(a, b, c) \\le b(a + c)^{3k-1}$; this is also true for $k = 1$.\n\nConsequently,\n\n$$\n\\begin{aligned}\nF(a, b, c) &\\le b(a + c)^{3k-1} \\\\\n&\\le \\frac{1}{3k-1} \\left( \\frac{(3k-1)b + (3k-1)(a+c)}{3k} \\right)^{3k} \\\\\n&= (3k-1)^{3k-1},\n\\end{aligned}\n$$\n\nfor all non-negative $a, b, c$ with $a + b + c = 3k$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19104,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\n\\frac{x-1}{x} + \\frac{x-2}{x} + \\dots + \\frac{1}{x} = \\frac{3x-20}{4}\n$$\n\nin $\\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be rewritten as\n\n$$\n\\frac{1}{x} [1 + 2 + 3 + \\dots + (x-3) + (x-2) + (x-1)] = \\frac{3x-20}{4}.\n$$\n\nSince $1 + 2 + 3 + \\dots + (x-1) = \\frac{(x-1)x}{2}$, we have\n\n$$\n\\frac{1}{x} \\cdot \\frac{(x-1)x}{2} = \\frac{3x-20}{4}.\n$$\n\nThis simplifies to\n\n$$\n\\frac{x-1}{2} = \\frac{3x-20}{4}.\n$$\n\nMultiplying both sides by $4$:\n\n$$\n2(x-1) = 3x-20\n$$\n$$\n2x-2 = 3x-20\n$$\n$$\n3x-2x = 20-2\n$$\n$$\nx = 18\n$$\n\nSo the solution is $x=18$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19105,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function $g(x) = \\frac{2x}{1 + x^2}$. Find all functions $f(x)$ defined and continuous on the interval $(-1, 1)$ which satisfy the condition:\n\n$$\n(1 - x^2) f(g(x)) = (1 + x^2)^2 f(x)\n$$\nfor every $x \\in (-1, 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f(x)$ satisfies the given condition if and only if the function $\\varphi(x) = (1 - x^2) f(x)$ is defined and continuous on $(-1, 1)$ and satisfies\n\n$$\n\\varphi(g(x)) = \\varphi(x) \\quad \\forall x \\in (-1, 1).\n$$\n\nConsider the function $h(x)$ defined on $(0, +\\infty)$ by $h(x) = \\varphi\\left(\\frac{1 - x}{1 + x}\\right)$ for all $x \\in (0, +\\infty)$. Then $\\varphi(x)$ is continuous on $(-1, 1)$ and satisfies the above if and only if $h(x)$ is continuous on $(0, +\\infty)$ and\n\n$$\nh(x^2) = h(x) \\quad \\forall x \\in (0, +\\infty).\n$$\n\nFrom this, $h(x) = h(\\sqrt{x})$ for all $x > 0$ and all $n \\in \\mathbb{N}$, so by continuity, $h(x)$ is constant on $(0, +\\infty)$. Thus, $\\varphi(x) = \\text{const}$ for all $x \\in (-1, 1)$, and finally,\n\n$$\nf(x) = \\frac{a}{1 - x^2}, \\quad a \\in \\mathbb{R}.\n$$\n\nOne can verify that these functions $f(x)$ satisfy the required condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19106,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that in a non-equilateral triangle, the following are equivalent:\n\n(a) The angles of the triangle are in arithmetic progression.\n\n(b) The common tangent to the nine-point circle and the incircle is parallel to the Euler line.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$, $B$, $C$ be the vertices, $I$ the incenter, $H$ the orthocenter, $O$ the circumcenter, and $N$ the nine-point center of the triangle.\n\n**Claim 1:** (b) $\\Leftrightarrow IH = IO$ (for a non-equilateral triangle).\n\nThe line $NI$ joins the centers of the nine-point circle and the incircle, so $NI$ is perpendicular to the common tangent to the incircle and nine-point circle (since $N \\neq I$, as the triangle is not equilateral). Thus, (b) is equivalent to the statement that $NI$ is perpendicular to the Euler line.\n\nSince $O$, $H$, $N$ are on the Euler line and $N$ is the midpoint of $OH$, we have:\n\n(b) $\\Leftrightarrow NI$ is the perpendicular bisector of $OH$.\n\nBut the perpendicular bisector of $O$, $H$ is simply the locus of all points $P$ such that $OP = HP$. Thus, (b) $\\Leftrightarrow IH = IO$.\n\n**Claim 2:** $IH = IO \\Leftrightarrow AH = AO$ or $A$, $H$, $O$, $I$ are concyclic.\n\nBecause $H$ and $O$ are isogonal conjugates, $\\angle HAI = \\angle OAH$. Also, $AI$ is common. Thus, the condition $AH = AO$ gives that $\\triangle AHI$ and $\\triangle OHI$ are congruent. This gives $IH = IO$.\n\nIf, on the other hand, $A$, $H$, $O$, $I$ are concyclic, then $IH$ and $IO$ subtend the same angle at $A$. This implies that $IH = IO$.\n\nThus, $AH = AO$ or $A$, $H$, $O$, $I$ concyclic implies that $IH = IO$.\n\nConversely, suppose $IH = IO$. Then $\\frac{IH}{AI} = \\frac{IO}{AI}$. This implies that\n\n$$\n\\frac{\\sin \\angle HAI}{\\sin \\angle AHI} = \\frac{\\sin \\angle OAI}{\\sin \\angle AOI}\n$$\n\nIt follows that $\\sin \\angle AHI = \\sin \\angle AOI$ since $AI$ bisects $\\angle HAO$. This tells us that $\\angle AHI + \\angle AOI = \\pi$ or $\\angle AHI = \\angle AOI$. In the first case, $A$, $H$, $O$, $I$ are concyclic. In the second case, $AH = AO$.\n\nThus, we have shown that $IH = IO$ if and only if $A$, $H$, $O$, $I$ are concyclic or $AH = AO$.\n\n**Claim 3:** $IH = IO$ if and only if $AH = AO$, or $BH = BO$, or $CH = CO$.\n\nSuppose $IH = IO$. Clearly, the circumcircle of $\\triangle HOI$ does not pass through all of $A$, $B$, $C$, because then $O$ would lie on the circumcircle of $\\triangle ABC$. Without loss of generality, assume that it does not pass through $A$. Then $AH = AO$, because $IH = IO$ and $A$, $H$, $O$, $I$ are not concyclic. (In the other two cases, we have $BH = BO$ or $CH = CO$ depending on whether the circumcircle does not pass through $B$ or $C$, respectively.) The other part follows directly from claim 2.\n\n**Claim 4:** $AH = AO$, $BH = BO$, or $CH = CO$ if and only if the angles of $ABC$ are in arithmetic progression.\n\nLet $R$ be the circumradius so that $AO = BO = CO = R$. Then $AH = 2R\\cos A$, $BH = 2R\\cos B$, and $CH = 2R\\cos C$. Thus, $AH = AO \\Leftrightarrow A = \\frac{\\pi}{3}$; $BH = BO \\Leftrightarrow B = \\frac{\\pi}{3}$; $CH = CO \\Leftrightarrow C = \\frac{\\pi}{3}$.\n\nNow, $AH = AO$, $BH = BO$, or $CH = CO$ implies that one of the angles of triangle $ABC$ is $\\pi/3$. But clearly, one angle is $\\pi/3$ if and only if the angles are in arithmetic progression.\n\nThus, (b) $\\Leftrightarrow IH = IO$ (by claim 1) $\\Leftrightarrow AH = AO$, $BH = BO$, or $CH = CO$ (by claim 3) $\\Leftrightarrow$ angles are in arithmetic progression, which is (a).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19107,
    "subject": "Mathematics (Olympiad)",
    "question": "Аритметичката прогресија се состои од цели броеви. Збирот на првите $n$ членови на прогресијата е степен на бројот 2. Докажи дека и $n$ е степен на бројот 2.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека првиот член на аритметичката прогресија е $a$, $n$-тиот член е $b$, а разликата е $d$. Нека $S$ е збирот на првите $n$ членови на прогресијата. Тогаш\n\n$$\n\\begin{align*}\nS &= a + (a+d) + (a+2d) + \\dots + (a+(n-1)d) = na + d(1+2+\\dots+(n-1)) = \\\\\n&= na + \\frac{(n-1)n}{2}d = \\frac{n}{2}[2a + (n-1)d] = \\frac{n}{2}[a + a + (n-1)d] = \\frac{n}{2}(a+b)\n\\end{align*}\n$$\n\nПоследното равенство можеме да го запишеме во облик\n\n$$\n2S = (a+b)n. \\quad (1)\n$$\n\nБидејќи $S$ е степен на бројот 2, постои $k \\in \\mathbb{N}$ така што $S = 2^k$. Од равенството (1) добиваме\n\n$$\n2^{k+1} = (a+b)n.\n$$\n\nЗначи, каноничната репрезентација на $(a+b)n$ е еднаква на $2^{k+1}$, па затоа $a+b$ и $n$ немаат делители различни од 2.\n\nСпоред тоа, постојат броеви $p$ и $q$ такви што\n\n$$\na+b=2^p,\n$$\n\n$$\nn=2^q,\n$$\n\nпри што $p+q=k+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19108,
    "subject": "Mathematics (Olympiad)",
    "question": "In the triangle $ABC$, $\\angle BAC = 120^\\circ$. On the bisector of the angle $\\angle BAC$, a point $D$ is chosen such that $\\overline{AD} = \\overline{AB} + \\overline{AC}$. Prove that $\\triangle BCD$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be a point on $AD$ such that $\\overline{AM} = \\overline{AC}$. $\\angle MAC = 60^\\circ$ implies that $\\triangle ACM$ is an equilateral triangle. Because $\\angle CAB = \\angle CMD = 120^\\circ$, $\\overline{MC} = \\overline{AC}$, and $\\overline{AB} = \\overline{MD}$, we have that $\\triangle CAB \\cong \\triangle CMD$. Hence $\\overline{BC} = \\overline{CD}$ and $\\angle ACB = \\angle MCD$. From $\\angle ACM = \\angle ACB + \\angle BCM = \\angle MCD + \\angle BCM = \\angle BCD = 60^\\circ$ and $\\overline{BC} = \\overline{CD}$, we conclude that $\\triangle BCD$ is an equilateral triangle.\n\n![](images/Makedonija_2008_p19_data_3b7189ffd3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19109,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$ respectively. Let $X$ and $Y$ be the feet of the perpendiculars drawn from $D$ to sides $AB$ and $AC$ respectively. The line passing through $F$ and parallel to line $XY$ intersects line $DY$ at a point $P$ that is different from $E$. Prove that line $AD$ and line $EP$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle AXD = \\angle AYD = 90^\\circ$, points $A$, $D$, $X$, and $Y$ are concyclic. Therefore, $\\angle FAD = \\angle XAD = \\angle XYD = \\angle FPD$, and by the converse of the inscribed angle theorem, points $A$, $D$, $F$, and $P$ are concyclic. According to the midpoint theorem in triangle $ABC$, $FD$ is parallel to $AC$. Hence, $\\angle FDP = 90^\\circ$ and consequently $\\angle FAP = 90^\\circ$. This implies that lines $AB$ and $AP$ are perpendicular. Furthermore, applying the midpoint theorem in triangle $ABC$ again, $DE$ is parallel to $AB$. Thus, lines $DE$ and $AP$ are perpendicular. Now, since lines $AE$ and $DP$ are perpendicular, $E$ is the orthocenter of triangle $ADP$. Therefore, lines $AD$ and $EP$ are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19110,
    "subject": "Mathematics (Olympiad)",
    "question": "One cuts a paper strip of length $2007$ into two parts of integer lengths and writes down the two integers on the board. Then, one of the two parts is cut into two parts of integer lengths and the two integers are written on the board. The cutting stops when all parts are of length $1$. A cut is called *bad* if the two parts obtained are not of equal lengths.\n\n(a) Find the minimum possible number of bad cuts.\n\n(b) Prove that for all cuttings with the minimum possible number of bad cuts, the number of distinct integers on the board is one and the same.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let the length of the strip be $n$. Denote by $g(n)$ the number of $1$'s in the binary representation of $n$, and by $f(n)$ the minimum possible number of bad cuts. Let $n = 2^{k_1} + 2^{k_2} + \\cdots + 2^{k_l}$. Consider the following sequence of cuts: first cut a strip of length $2^{k_1}$, next cut a strip of length $2^{k_2}$, and so on. After the last cut, we obtain two strips of lengths $2^{k_{l-1}}$ and $2^{k_l}$. Note that a strip whose length is a power of $2$ can be cut into strips of length $1$ without bad cuts. Therefore, the number of bad cuts equals $l-1$, i.e.\n\n$$\n(1) \\qquad f(n) \\leq g(n) - 1.\n$$\n\nWe prove by induction on $n$ that $f(n) \\geq g(n) - 1$. For $n=1$, we have $f(1) = 0$ and $g(1) = 1$, so the statement is true. Suppose it is true for all $n \\leq k$, where $k$ is a positive integer, and let $n = k+1$.\n\n1. Suppose the first cut is bad and it leaves two strips of lengths $a$ and $b$. Then $a + b = k+1$ and $f(k+1) = 1 + f(a) + f(b)$. If the binary representations of $a$ and $b$ have no common digit $1$, then $g(k+1) = g(a) + g(b)$ and therefore\n\n$$\nf(k+1) = 1 + f(a) + f(b) \\geq 1 + g(a) - 1 + g(b) - 1 = g(k+1) - 1.\n$$\n\nIf the binary representations of $a$ and $b$ have at least one common digit $1$, then $g(k+1) \\leq g(a) + g(b) - 1$ and therefore\n\n$$\nf(k+1) = 1 + f(a) + f(b) \\geq 1 + g(a) - 1 + g(b) - 1 \\geq g(k+1) > g(k+1) - 1.\n$$\n\n2. Suppose the first cut is not bad, i.e., the strip is cut into two parts each of length $a$. Then $k+1 = 2a$ and $g(k+1) = g(a)$. If $g(k+1) = 1$, then $f(k+1) = 0$ and the statement is true. Otherwise,\n\n$$\nf(k+1) = f(a) + f(b) = 2f(a) \\geq 2g(a) - 2 = 2g(k+1) - 2 > g(k+1) - 1.\n$$\n\nThus, when $g(k+1) > 1$, we have $f(k+1) > g(k+1) - 1$.\n\nThis proves the induction hypothesis, giving\n\n$$\n(2) \\qquad f(n) \\geq g(n) - 1.\n$$\n\nIt follows from (1) and (2) that $f(n) = g(n) - 1$.\n\na) Since the binary representation of $2007$ is $1111010111$, i.e., $g(2007) = 9$, we obtain $f(2007) = 8$.\n\nb) It follows from the above arguments that if the number of bad cuts is $f(n) = g(n) - 1$, then every bad cut leaves two parts of lengths $a$ and $b$ such that the binary representations of $a$ and $b$ have no common digit $1$. Moreover, the good cuts are done only over strips whose lengths are powers of $2$. It is clear that rearranging the cuts, we may assume that the first cuts are bad. Their number equals $g(n) - 1$, and each bad cut gives two new numbers on the table. Therefore, after all bad cuts, we have $2g(n) - 2$ distinct numbers. The powers of $2$ that appear are all powers up to the highest power in the binary representation of $n$.\n\nThus, the number of distinct numbers on the table equals $2g(n) - 2 + k + 1 = 2g(n) + k - 1$, where $k$ is the highest power of $2$ in the binary representation of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19111,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n \\ge 2$, define the set $T$ by\n$$\nT = \\{ (i, j) : 1 \\le i < j \\le n \\text{ and } i \\ne j \\}.\n$$\nFor nonnegative real numbers $x_1, x_2, \\dots, x_n$ satisfying $x_1 + x_2 + \\dots + x_n = 1$, find the maximum (as a function of $n$) of\n$$\n\\sum_{(i,j) \\in T} x_i x_j.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M(n)$ be the maximum of $\\sum_{(i,j) \\in T} x_i x_j$. We will show that\n$$\nM(n) = \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)}.\n$$\nFor $k = \\lfloor \\log_2 n \\rfloor$, set $x_{20} = x_{21} = \\dots = x_{2k} = \\frac{1}{k+1}$, and $x_i = 0$ otherwise. Then we have\n$$\n\\sum_{(i,j) \\in T} x_i x_j = \\binom{k+1}{2} \\frac{1}{(k+1)^2} = \\frac{k}{2(k+1)},\n$$\nthus\n$$\nM(n) \\ge \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)}.\n$$\nConsider an element $(x_1, x_2, \\dots, x_n)$ which yields $M(n)$ and the number of $i$'s such that $x_i = 0$ is maximal. In this case, if $x_a, x_b \\ne 0$ and $a < b$, then $(a, b) \\in T$. Suppose $(a, b) \\notin T$, then by setting $x'_a = x_a - \\epsilon$, $x'_b = x_b + \\epsilon$, and $x'_i = x_i$ ($i \\ne a, b$), the equation $x'_1 + x'_2 + \\dots + x'_n = 1$ still holds and the value $\\sum_{(i,j) \\in T} x'_i x'_j$ becomes a linear function of $\\epsilon$. Thus for $\\epsilon = x'_a$ or $\\epsilon = -x'_b$, $\\sum_{(i,j) \\in T} x'_i x'_j \\ge \\sum_{(i,j) \\in T} x_i x_j$, which contradicts the maximality of the number of $i$'s satisfying $x_i = 0$.\n\nTherefore if $C := \\{i : x_i > 0\\}$, then $i, j \\in C$ and $i < j$ implies $(i, j) \\in T$. Thus if $C = \\{i_1, i_2, \\dots, i_k\\}$ and $i_1 < i_2 < \\dots < i_k$, then due to $i_j \\ge 2i_{j-1}$ we have $n > i_t \\ge 2^{t-1}i_1 \\ge 2^{t-1}$. This means that $t-1 \\le \\log_2 n$, or equivalently, $|C| \\le k+1$. Applying the Cauchy-Schwarz inequality yields\n$$\n\\sum_{(i,j) \\in T} x_i x_j = \\frac{1}{2} \\left( \\left( \\sum_{i \\in C} x_i \\right)^2 - \\sum_{i \\in C} x_i^2 \\right) \\le \\frac{1}{2} \\left( 1 - \\frac{1}{|C|} \\left( \\sum_{i \\in C} x_i \\right)^2 \\right) \\le \\frac{1}{2} \\left( 1 - \\frac{1}{k+1} \\right).\n$$\nIn particular we have\n$$\nM(n) \\le \\frac{\\lfloor \\log_2 n \\rfloor}{2(\\lfloor \\log_2 n \\rfloor + 1)},\n$$\nwhich completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19112,
    "subject": "Mathematics (Olympiad)",
    "question": "We will call a natural number Yambolian if it can be represented in the form $a^2 + 6ab + b^2$, where $a$ and $b$ are (not necessarily different) natural numbers. The number $36^{2024}$ is written as the sum of $k$ (not necessarily distinct) Yambolian numbers. What is the smallest possible value of $k$?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $k = 1$ is not possible, i.e., $a^2 + 6ab + b^2 = 2^{4048} \\cdot 3^{4048}$ has no solution in natural numbers. If such $a, b$ exist, then $a^2 + b^2$ is divisible by $3$, so $a$ and $b$ are divisible by $3$. Let $a = 3a_1$, $b = 3b_1$; dividing by $3^2$ gives $a_1^2 + 6a_1b_1 + b_1^2 = 2^{4048} \\cdot 3^{4046}$. Repeating this $2023$ more times, we get:\n\n$$\nu^2 + 6uv + v^2 = 2^{4048}$$\n\nwhere $u, v$ are natural numbers.\n\nIf $v$ is even, then $u$ is even. Write $u = 2u_1$, $v = 2v_1$ and divide by $4$ to get $u_1^2 + 6u_1v_1 + v_1^2 = 2^{4046}$. Repeating, we eventually get:\n\n$$s^2 + 6st + t^2 = 2^{2A}$$\n\nwhere $s, t, A$ are natural numbers, $t$ is odd, and $A \\geq 2$ (since the left side is at least $8$). This is equivalent to $(s + 3t)^2 - 8t^2 = 2^{2A}$. Modulo $8$, $(s + 3t)^2$ is divisible by $8$, so $s + 3t$ is divisible by $4$, i.e., $(s + 3t)^2$ is divisible by $16$. Then $8t^2$ must be divisible by $16$, which is impossible for odd $t$, a contradiction. Therefore, $k = 1$ is not possible.\n\nFor $k = 2$, note that\n\n$$\n36 = [1^2 + 6 \\cdot 1 \\cdot 1 + 1^2] + [3^2 + 6 \\cdot 3 \\cdot 1 + 1^2]\n$$\n\nMultiplying by $(6^{2023})^2$ gives\n\n$$\n36^{2024} = [(6^{2023})^2 + 6 \\cdot 6^{2023} \\cdot 6^{2023} + (6^{2023})^2] + [(3 \\cdot 6^{2023})^2 + 6 \\cdot (3 \\cdot 6^{2023}) \\cdot 6^{2023} + (6^{2023})^2]\n$$\n\nThat is, $36^{2024}$ is the sum of two Yambolian numbers: $a_1^2 + 6a_1b_1 + b_1^2$ and $a_2^2 + 6a_2b_2 + b_2^2$, where $a_1 = b_1 = b_2 = 6^{2023}$ and $a_2 = 3 \\cdot 6^{2023}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19113,
    "subject": "Mathematics (Olympiad)",
    "question": "A $3 \\times 3$ grid is given. We color each square red or blue so that no red $2 \\times 2$ square nor blue $2 \\times 2$ square appears. How many such colorings are there?\n\nWe consider two colorings different even if they correspond by rotation and/or reversal.",
    "options": [],
    "answer": "See solution",
    "solution": "$322$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19114,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, let $D$ be a point different from $A$ on the external bisector $\\ell$ of the angle $BAC$, and let $E$ be an interior point of the segment $AD$. Reflect $\\ell$ in the internal bisectors of the angles $BDC$ and $BEC$ to obtain two lines that meet at some point $F$. Show that the angles $ABD$ and $EBF$ are congruent.\n\nReflect $B$ in the lines $DF$ and $EF$ to obtain the points $B'$ and $B''$, respectively. The lines $B'C$ and $DF$ meet at $M$, and the lines $B''C$ and $EF$ meet at $N$. Prove that the lines $BD$, $BE$ and $BF$ bisect the angles $ABM$, $ABN$ and $MBN$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n\\angle ABD = \\frac{1}{2} \\angle ABM = \\frac{1}{2} (\\angle ABN + \\angle MBN) = \\angle EBN + \\angle FBN = \\angle EBF.\n$$\n\nTo show that the line $BD$ bisects the angle $ABM$, let $C'$ be the reflection of $C$ in the line $\\ell$ and notice that the triangles $DBC'$ and $DB'C$ are congruent, since $DB = DB'$ and $DC' = DC$ (by symmetry), and $\\angle BDC' = \\angle B'DC$ (by symmetry and isogonality). Consequently, $\\angle DBC' = \\angle DB'C$ and $BC' = B'C$. Since $\\ell$ is the external bisector of the angle $BAC$, the points $A, B$ and $C'$ are collinear, so $\\angle DBA = \\angle DBC'$. On the other hand, $\\angle DB'C = \\angle DB'M = \\angle DBM$ (the latter equality holds by symmetry), so $BD$ is indeed the internal bisector of the angle $ABM$.\n\nA similar argument shows that $BE$ is the internal bisector of the angle $ABN$ and $BC' = B''C$.\n\nTo show that the line $BF$ bisects the angle $MBN$, let $C''$ be the reflection of $C$ in the line $EF$ and notice that the triangles $FBC''$ and $FB'C$ are congruent, since $FB = FB'$ and $FC'' = FC$ (by symmetry), and $BC'' = B''C = BC' = B'C$ (the first equality by symmetry, the last two by the preceding paragraphs). Consequently, $\\angle FBC'' = \\angle FB'C$, so $\\angle FBN = \\angle FBC'' = \\angle FB'C = \\angle FB'M = \\angle FBM$ (the latter equality holds by symmetry); that is, $BF$ is the internal bisector of the angle $MBN$.\n\n**Remarks:**\n\n- If $D$ lies between $A$ and $E$, the angles $ABD$ and $EBF$ are still congruent. But if $D$ and $E$ lie on opposite sides of $A$, the two angles are supplementary. In all cases, the argument is essentially the same. Thus, up to orientation, $\\angle EBF \\equiv \\angle ABD$ modulo $\\pi$, as $E$ traces $\\ell$.\n- The same holds if $B$ is replaced by $C$ throughout or if $\\ell$ is the internal bisector of the angle $BAC$.\n\nThis configuration reveals pairs of isogonal lines: $DF$ and $EF$ are the isogonals of the bisector $\\ell$ relative to $(DB, DC)$ and $(EB, EC)$, respectively. The conclusion states that $(BA, BD)$ and $(BE, BF)$ are pairs of isogonal lines; likewise for $(CA, CD)$ and $(CE, CF)$.\n\nFurthermore, $\\ell$ may be seen as self-isogonal relative to $(AB, AC)$. A similar configuration arises if $D$ and $E$ are each on a line in a pair of isogonals relative to $(AB, AC)$.\n\nA remarkable focal property (Poncelet-Steiner): The focal angular span of the segment intercepted by two fixed tangents to a conic on a variable tangent to that conic is constant modulo $\\pi$ and orientation.\n\nOther properties:\n\n1. A tangent to a conic bisects internally or externally the angle formed by the focal rays at the point of contact.\n2. Two tangents to a conic are isogonal relative to the focal rays of their intersection (Poncelet); each focal ray bisects the corresponding focal angle determined by the points of contact.\n\nIn this case, the points $A, M, N$ lie on an ellipse with foci $B$ and $C$, externally tritangent to triangle $DEF$: the lines $DE, DF, EF$ are tangent to the ellipse at $A, M, N$, respectively. If $D$ and $E$ lie on opposite sides of $A$, the ellipse is internally tritangent to $DEF$ at $A, M, N$. If $\\ell$ is the internal bisector of $BAC$, the conic is a hyperbola with foci $B$ and $C$, tritangent to $DEF$ at $A, M, N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19115,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $h_a$, $h_b$, and $h_c$ denote the lengths of the three altitudes of triangle $ABC$. (i) Show that\n\n$$\n\\frac{1}{h_a} + \\frac{1}{h_b} > \\frac{1}{h_c}, \\quad \\frac{1}{h_b} + \\frac{1}{h_c} > \\frac{1}{h_a}, \\quad \\text{and} \\quad \\frac{1}{h_c} + \\frac{1}{h_a} > \\frac{1}{h_b}.\n$$\n\n(ii) If $h_a = 5$ and $h_b = 2$, find all possible integer values of $h_c$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Let $\\Delta$ denote the area of triangle $ABC$. We have $2\\Delta = a h_a = b h_b = c h_c$. Thus,\n\n$$\n\\frac{1}{h_a} = \\frac{a}{2\\Delta}, \\quad \\frac{1}{h_b} = \\frac{b}{2\\Delta}, \\quad \\frac{1}{h_c} = \\frac{c}{2\\Delta}.\n$$\n\nThe triangle inequalities $a + b > c$, $b + c > a$, and $c + a > b$ become, after dividing by $2\\Delta$:\n\n$$\n\\frac{a}{2\\Delta} + \\frac{b}{2\\Delta} > \\frac{c}{2\\Delta} \\implies \\frac{1}{h_a} + \\frac{1}{h_b} > \\frac{1}{h_c},\n$$\n\nand similarly for the other two inequalities.\n\n(ii) Let $h_a = 5$ and $h_b = 2$. A triangle with two equal altitudes is isosceles, so $h_c \\neq 2$. From (i),\n\n$$\n\\frac{1}{5} + \\frac{1}{h_c} > \\frac{1}{2} \\quad \\text{and} \\quad \\frac{1}{5} + \\frac{1}{2} > \\frac{1}{h_c}.\n$$\n\nFrom the first inequality:\n\n$$\n\\frac{1}{5} + \\frac{1}{h_c} > \\frac{1}{2} \\implies \\frac{1}{h_c} > \\frac{1}{2} - \\frac{1}{5} = \\frac{3}{10} \\implies h_c < \\frac{10}{3} \\approx 3.33.\n$$\n\nFrom the second inequality:\n\n$$\n\\frac{1}{5} + \\frac{1}{2} > \\frac{1}{h_c} \\implies \\frac{7}{10} > \\frac{1}{h_c} \\implies h_c > \\frac{10}{7} \\approx 1.43.\n$$\n\nSo $1.43 < h_c < 3.33$. Since $h_c$ is an integer and $h_c \\neq 2$, the only possible value is $h_c = 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19116,
    "subject": "Mathematics (Olympiad)",
    "question": "Знайдіть усі пари натуральних чисел $m$ і $n$, для яких\n$$\n\\sqrt{n} = \\frac{2006^2 n + 2005^2 - m n}{2 \\cdot 2005 \\cdot 2006}\n$$\nє раціональним числом.",
    "options": [],
    "answer": "See solution",
    "solution": "Відповідь: $m=1,\\ n=1$; $m=1605^2,\\ n=5^2$; $m=2001^2,\\ n=401^2$; $m=2005^2,\\ n=2005^2$.\n\nОскільки $\\sqrt{n} \\in \\mathbb{Q}$, то $n$ є квадратом натурального числа. Звідси випливає, що $m = k^2$, де $k \\in \\mathbb{N}$ і $k \\mid 2005$. Перебираючи всі натуральні дільники числа $2005$, одержуємо відповідь.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19117,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to place four points in the coordinate plane such that the distance between any two points is strictly greater than $\\frac{\\sqrt{6}-\\sqrt{2}}{2}$?\n\n![](images/2024_CMO_p7_data_0f3f8433bd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the shaded region (including its boundary). The farthest distance within this region is achieved between $B_2\\left(\\frac{2-\\sqrt{3}}{2}, \\frac{1}{2}\\right)$ and $B_3\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$. In fact, the entire shaded region is contained within the circle with $B_2B_3$ as its diameter.\n\nThus, $A_2, A_3, A_4$ must lie in three of the four parts. Assume $A_2, A_3, A_4$ lie in the upper-left, lower-left, and lower-right parts, respectively. Consider the points $C\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$ and $D\\left(\\frac{\\sqrt{3}-1}{2}, \\frac{\\sqrt{3}-1}{2}\\right)$, where $D$ lies on the arc centered at $A_1$ and $CD$ forms a $45^\\circ$ angle with the $x$-axis. Examine $A_3$, and assume it lies above the segment $CD$.\n\nHowever, the shaded region in the diagram is contained within the quadrilateral with vertices $B_3, C, D$, and $B_2$, including its boundary. By a well-known result, the maximum distance between two points inside a convex quadrilateral does not exceed the lengths of its two diagonals or its four edges, whichever is the longest. For the quadrilateral $B_2B_3CD$, the longest length among its two diagonals and four edges is $B_2B_3 = B_3D = \\frac{\\sqrt{6}-\\sqrt{2}}{2}$. Thus, $d(A_2, A_3) \\leq \\frac{\\sqrt{6}-\\sqrt{2}}{2}$.\n\nThis contradicts the assumption in the proof by contradiction. Therefore, it is impossible to have four points in the coordinate plane such that the distance between any two points is strictly greater than $\\frac{\\sqrt{6}-\\sqrt{2}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19118,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the system of equations:\n\n$$\n\\begin{cases}\n4x^3y - x^4 - 3x^2y^2 = 2021, \\\\\n4y^3x - y^4 - 3y^2x^2 = 2021.\n\\end{cases}\n$$\n\nAnna claims that the system of equations has a solution. Anne claims that the system of equations has no solution but at least one of the two equations has solutions. Anni claims that the system of equations has no solution and, even worse, neither of the two equations alone has a solution. Who is right?",
    "options": [],
    "answer": "See solution",
    "solution": "The system does not have a solution since adding the equations gives $-(x - y)^4 = 4042$, whose left-hand side is non-positive but right-hand side is positive.\n\nWe show that the first equation $4x^3y - x^4 - 3x^2y^2 = 2021$ has solutions (the same could be done for the second equation by symmetry). Dividing the equation by $x^4$ and reordering the terms in the left-hand side results in\n\n$$\n-3 \\left(\\frac{y}{x}\\right)^2 + 4 \\left(\\frac{y}{x}\\right) - 1 = \\frac{2021}{x^4}.\n$$\n\nAs the discriminant of the quadratic equation $-3t^2 + 4t - 1 = 0$ is positive, there exists a real number $t$ such that $-3t^2 + 4t - 1$ equals a positive number $\\varepsilon$. Define $x = \\sqrt[4]{\\frac{2021}{\\varepsilon}}$ and $y = xt$; then $x$ and $y$ satisfy the equation above and also the first equation of the given system of equations. Hence Anne is right.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19119,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ be (not necessarily positive) real numbers, where $n \\ge 2$. Let $K$ be the maximum and $L$ be the minimum of $b_1, b_2, \\dots, b_n$.\n\nProve that\n\n$$\n\\sum_{i<j} a_i a_j |b_i - b_j| \\le \\frac{1}{2} (K - L) (a_1 + a_2 + \\dots + a_n)^2.\n$$\n\nNote that $\\sum_{i<j}$ refers to summing over all pairs $(i, j)$ such that $1 \\le i < j \\le n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the indices $1, 2, \\dots, n$ can be permuted without changing the problem as long as the correspondence between $a_i$ and $b_i$ is maintained for each $i$. Let's reorder the two sequences simultaneously to maintain this correspondence, while sorting the $b_i$ so that $b_1 \\le \\dots \\le b_n$. Then\n\n$$\n\\begin{aligned}\n2 \\times (\\text{RHS} - \\text{LHS}) &= (b_n - b_1) \\left(\\sum_{i=1}^{n} a_i\\right)^2 - 2 \\sum_{i<j} a_i a_j (b_j - b_i) \\\\\n&= \\left(\\sum_{k=1}^{n-1} (b_{k+1} - b_k)\\right) \\left(\\sum_{i=1}^{n} a_i\\right)^2 - 2 \\sum_{i<j} \\sum_{k=i}^{j-1} a_i a_j (b_{k+1} - b_k).\n\\end{aligned}\n$$\n\nLet's fix $k$ and focus on terms that involve $(b_{k+1} - b_k)$. The first term will simply give a multiplier of $(\\sum_{i=1}^n a_i)^2$. The second term is a sum over the triples $(i, j, k)$ that satisfy $i \\le k < j$. So for a fixed $k$, the second term gives a multiplier of $2 \\sum_{i,j} a_i a_j$ where $i \\le k$ and $j > k$. Hence\n\n$$\n\\begin{aligned}\n2 \\times (\\text{RHS} - \\text{LHS}) &= \\sum_{k=1}^{n-1} (b_{k+1} - b_k) \\left( \\left( \\sum_{i=1}^{n} a_i \\right)^2 - 2 \\sum_{i \\le k < j} a_i a_j \\right) \\\\\n&= \\sum_{k=1}^{n-1} (b_{k+1} - b_k) \\left( \\sum_{i=1}^{n} a_i^2 + 2 \\sum_{i<j} a_i a_j - 2 \\sum_{i \\le k < j} a_i a_j \\right).\n\\end{aligned}\n$$\n\nAgain fix $k$ and consider the pairs $(i, j)$ with $i < j$. If we take away those satisfying $i \\le k < j$, then we are left with either $k < i < j$ or $i < j \\le k$. Thus\n\n$$\n\\begin{aligned}\n2 \\times (\\text{RHS} - \\text{LHS}) &= \\sum_{k=1}^{n-1} (b_{k+1} - b_k) \\left( \\sum_{i=1}^{n} a_i^2 + 2 \\sum_{i<j \\le k} a_i a_j + 2 \\sum_{k<i<j} a_i a_j \\right) \\\\\n&= \\sum_{k=1}^{n-1} (b_{k+1} - b_k) \\left( \\left( \\sum_{i=1}^{k} a_i \\right)^2 + \\left( \\sum_{i=k+1}^{n} a_i \\right)^2 \\right) \\\\\n&\\ge 0,\n\\end{aligned}\n$$\n\nand we are done.\n\n**Remark.** The equality cases can be characterised as follows: whenever $b_k \\ne b_{k+1}$, then $\\sum_{i \\le k} a_i = \\sum_{i > k} a_i = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19120,
    "subject": "Mathematics (Olympiad)",
    "question": "定義函數 $f : (0, 1) \\to (0, 1)$ 如下：\n\n$$\nf(x) = \\begin{cases} x + \\frac{1}{2} & \\text{若 } x < \\frac{1}{2}, \\\\ x^2 & \\text{若 } x \\ge \\frac{1}{2}. \\end{cases}\n$$\n\n令 $a, b$ 為兩實數且 $0 < a < b < 1$。定義兩數列如下：\n\n$a_0 = a,\\ b_0 = b$，且 $a_n = f(a_{n-1})$，$b_n = f(b_{n-1})$，$n = 1, 2, \\dots$\n\n試證：存在一正整數 $n$ 使得\n\n$$(a_n - a_{n-1})(b_n - b_{n-1}) < 0.$$",
    "options": [],
    "answer": "See solution",
    "solution": "設存在 $0 < a < b < 1$ 使不存在正整數 $n$ 讓 $(a_n - a_{n-1})(b_n - b_{n-1}) < 0$。\n\n$$\nf(x) - x = \\begin{cases} > 0 & \\text{若 } x < \\frac{1}{2}, \\\\ < 0 & \\text{若 } x \\ge \\frac{1}{2}. \\end{cases}\n$$\n\n則對給定的正整數 $n$，必有 $a_n, b_n \\in (0, \\frac{1}{2})$ 或 $a_n, b_n \\in [\\frac{1}{2}, 1)$。\n\n設 $d_n = b_n - a_n$，則若 $a_n, b_n \\in (0, \\frac{1}{2})$ 會有 $d_{n+1} = d_n$，反之則有\n\n$$\nd_{n+1} = d_n(a_n + b_n) \\ge d_n(1 + d_n),\n$$\n\n故 $d_n$ 是遞增數列。\n\n設 $0 < a_n < b_n < \\frac{1}{2}$，則 $\\frac{1}{2} < a_n < b_n < 1$。無論如何有 $d_{n+2} \\ge d_n(1+d_n)$。\n\n故有 $d_{2m+1} \\ge d_1(1+d_1)^m \\ge d_1(1+md_1)$。存在足夠大的 $m$ 使\n\n$$\nd_1(1+md_1) > 1,\n$$\n\n但這不可能。故存在正整數 $n$ 使 $(a_n - a_{n-1})(b_n - b_{n-1}) < 0$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19121,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle and let $D$ be a point in its interior. Construct a circle $\\omega_1$ passing through $B$ and $D$ and a circle $\\omega_2$ passing through $C$ and $D$ such that the point of intersection of $\\omega_1$ and $\\omega_2$ other than $D$ lies on line $AD$. Denote by $E$ and $F$ the points where $\\omega_1$ and $\\omega_2$ intersect side $BC$, respectively, and by $X$ and $Y$ the intersections of lines $DF$, $AB$ and $DE$, $AC$, respectively. Prove that $XY \\parallel BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let circles $\\omega_1$ and $\\omega_2$ meet again at $R$ (other than $D$), and let $\\omega_1$ and $\\omega_2$ intersect again with segments $AB$ and $AC$, respectively, at $P$ and $Q$ (other than $B$ and $C$). By the \\textbf{Power of a Point Theorem}, we have $AP \\cdot AB = AR \\cdot AD = AQ \\cdot AC$; that is, points $A$, $R$, and $D$ lie on the \\textbf{radical axis} of the circles $\\omega_1$ and $\\omega_2$. Therefore, points $B$, $C$, $Q$, and $P$ lie on a circle. Consequently, $\\angle AQP = \\angle ABC$. To prove that $XY \\parallel BC$, it suffices to show that $\\angle AXY = \\angle ABC$, or $\\angle AQP = \\angle ABC = \\angle AXY$; that is, to prove that $P$, $Q$, $Y$, and $X$ lie on a circle.\n\nBecause $BCQP$ is cyclic, $\\angle AQP = \\angle ABC$. Because $BDRP$ is cyclic, $\\angle PDY = \\angle PBE = \\angle ABC$. Hence $\\angle AQP = \\angle PDY$, implying that $P$, $Q$, $Y$ and $D$ lie on a circle. Similarly, we can show that $P$, $Q$, $D$, and $X$ lie on a circle. We conclude that $P$, $Q$, $Y$, $D$, and $X$ all lie on the circumcircle of triangle $PQD$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19122,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a natural number a *twin* if it has two natural divisors from a set $D$ whose difference is equal to $2$. Determine whether there are more twin numbers or numbers that are not twin among the first $20112012$ natural numbers if:\n\n(a) $D$ is the set of all natural divisors from $1$ to the number itself.\n\n(b) $D$ is the set of all natural divisors except $1$ and the number itself.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\n(a) There are more twin numbers.\n\n(b) There are more numbers that are not twin.\n\n**Solution.**\n\nLet $M = 20112012$. Denote by $M_K$ the set of all natural numbers not exceeding $M$ and divisible by $K$, and by $N_K$ the number of elements in $M_K$, i.e., $N_K = |M_K|$. It is well-known that $N_K = \\left[ \\frac{M}{K} \\right]$.\n\n(a) See problem 8-2.\n\n(b) Let $n$ be the number of twin numbers for this $D$. Then\n\n$$\nn < N_4 + N_{3\\cdot5} + N_{5\\cdot7} + N_{7\\cdot9} + \\dots + N_{(M-3)(M-1)},\n$$\n\nbecause here we count all twin numbers, possibly more than once (e.g., $945 = 3 \\cdot 5 \\cdot 7 \\cdot 9$ is counted three times). So,\n\n$$\n\\begin{align*}\nn < & \\left[ \\frac{M}{4} \\right] + \\left[ \\frac{M}{3\\cdot5} \\right] + \\left[ \\frac{M}{5\\cdot7} \\right] + \\left[ \\frac{M}{7\\cdot9} \\right] + \\dots + \\left[ \\frac{M}{(M-3)(M-1)} \\right] \\\n&< \\frac{M}{4} + \\frac{M}{3\\cdot5} + \\frac{M}{5\\cdot7} + \\frac{M}{7\\cdot9} + \\dots + \\frac{M}{(M-3)(M-1)} \\\n&= \\frac{M}{4} + M \\left( \\frac{1}{3\\cdot5} + \\frac{1}{5\\cdot7} + \\frac{1}{7\\cdot9} + \\dots + \\frac{1}{(M-3)(M-1)} \\right) \\\n&= \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{5-3}{3\\cdot5} + \\frac{7-5}{5\\cdot7} + \\frac{9-7}{7\\cdot9} + \\dots + \\frac{(M-1)-(M-3)}{(M-3)(M-1)} \\right) \\\n&= \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{9} + \\dots + \\frac{1}{M-3} + \\frac{1}{M-1} \\right) \\\n&= \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{1}{3} - \\frac{1}{M-1} \\right) < \\frac{M}{4} + \\frac{M}{6} = \\frac{5M}{12} < \\frac{M}{2},\n\\end{align*}\n$$\n\nwhich proves that in this case there are more numbers that are not twin.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19123,
    "subject": "Mathematics (Olympiad)",
    "question": "Set $M$ contains $n \\ge 2$ positive integers. It is known that for any two distinct numbers $a, b \\in M$, the number $a^2 + 1$ is divisible by $b$. Find the largest possible value of $n$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are at least three numbers in the set $M$, denote them by $a < b < c$. By the condition, $b^2 + 1$ is divisible by $c$, so $b$ and $c$ are coprime. Since $a^2 + 1$ is divisible by both $b$ and $c$, it follows that\n\n$$\na^2 + 1 \\text{ is divisible by } bc \\ge (a+1)(a+2) = a^2 + 3a + 2.\n$$\n\nThis is a contradiction, so $n \\le 2$. For $n = 2$, for example, $M = \\{1, 2\\}$ works.\n\n**Note:** An explicit example is not required, since the statement already says that such a set with $n \\ge 2$ exists.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19124,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a semicircle with diameter $AB$, center $O$, and radius $r$. Let $C$ be the point on segment $AB$ such that $AC = \\frac{2r}{3}$. Line $l$ is perpendicular to $AB$ at $C$, and $D$ is the common point of $l$ and the semicircle. Let $H$ be the foot of the perpendicular from $O$ to $AD$, and $E$ the intersection of lines $CD$ and $OH$.\n\na) Express $AD$ as a function of $r$.\n\nb) If $M$ and $N$ are the midpoints of $AE$ and $OD$ respectively, find the measure of angle $MHN$.\n\n![](images/Argentina_2011_p28_data_c6ef3b0274.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $CO = \\frac{1}{3}r$ and $OD = r$, Pythagoras' theorem in triangles $COD$ and $ACD$ gives $CD = \\sqrt{OD^2 - OC^2} = \\frac{2\\sqrt{2}}{3}r$, $AD = \\sqrt{AC^2 + CD^2} = \\frac{2\\sqrt{3}}{3}r$.\n\nb) As $AD$ is a chord in the semicircle and $O$ its center, $OH \\perp AD$ implies that $H$ is the midpoint of $AD$. So $HM$ and $HN$ are median lines in triangles $ADE$ and $DAO$, hence $HM \\parallel DE$ and $HN \\parallel AO$. On the other hand, $DE \\perp AO$, so $HM \\perp HN$. Therefore $\\angle MHN = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19125,
    "subject": "Mathematics (Olympiad)",
    "question": "An acute-angled triangle $ABC$ is given such that the angle at vertex $C$ is the largest. Let $E$ and $G$ be the points of intersection of the altitude drawn from $A$ to $BC$ with the circumscribed circle of triangle $ABC$ and with $BC$, respectively. The center $O$ of the circumscribed circle lies on the perpendicular drawn from $A$ to $BE$. The points $M$ and $F$ are the feet of the altitudes drawn from $E$ to $AC$ and $AB$, respectively. Prove that $P_{MFE} < P_{FBEG}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the intersection of $EM$ and $BC$ by $V$ (the intersection will always exist since the angle at $C$ is acute). From Simson's theorem, it follows that the points $M$, $G$, and $F$ are collinear. Note that $\\angle EAC = \\angle EBC$ since they intercept the same arc. Also, $\\angle CAE = \\angle BAO$. The quadrilateral $FBEG$ is inscribed. Therefore, $\\angle GBE = \\angle GFE$. Also, $\\angle GAO = \\angle GBE$ since they are angles with perpendicular rays. We get $\\angle CAE = \\angle GAO = \\angle BAO$. Therefore, $AO$ is a bisector of the angle and an altitude in triangle $ABE$. It follows that triangle $ABE$ is isosceles, from which $GF$ is parallel to $BE$. The lines $AO$, $BG$, and $EF$ intersect at one point (since $EF$ and $BG$ are altitudes in triangle $ABE$). Note that $AGMV$ is inscribed. We have $\\angle MVG = \\angle GAM$. We also get\n\n$$\n\\angle MAV = \\angle VGM = \\angle FGB.\n$$\n\nTherefore, $AM$ is an altitude and an angle bisector in triangle $EAV$. It follows that triangle $EAV$ is isosceles and $M$ is the midpoint of side $VE$. Also, $\\triangle AVE \\cong \\triangle AEB$. It is clear that $\\triangle EGV \\cong \\triangle EGB$ and since both are right-angled, $P_{GEM} = \\frac{1}{2} P_{GEB}$. On the other hand, $P_{GFE} = P_{GBE}$, from which we get the required inequality.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19126,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $a, b, c$ (not necessarily distinct), suppose that $a + bc$, $b + ca$, and $c + ab$ are all perfect squares. Prove that\n\n$$\na^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc\n$$\n\ncan be written as the sum of two square numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x^2 = a + bc$, $y^2 = b + ca$, and $z^2 = c + ab$. We use the following lemma:\n\n*Lemma.* A positive integer $n$ can be written as the sum of two squares if and only if for all primes $p \\equiv 3 \\pmod{4}$, $v_p(n)$ is even.\n\nNote that the target expression can be written as:\n\n$$\nS = a^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc = (a+b)(b+c)(c+a)\n$$\n\nBy the lemma, it suffices to prove $v_p(S)$ is even for all primes $p \\equiv 3 \\pmod{4}$. We claim the stronger statement that $v_p(a+b)$ (and cyclic variations) is even for all such primes.\n\nLet $p \\equiv 3 \\pmod{4}$ be a prime with $v_p(a+b) > 0$. Then:\n\n$$\nx^2 + y^2 = (a+b)(c+1) \\equiv 0 \\pmod{p}\n$$\n\nSince $p \\equiv 3 \\pmod{4}$, $\\left(\\frac{-1}{p}\\right) = -1$, so $p \\mid x, y$. We claim $c \\not\\equiv -1 \\pmod{p}$. If $c \\equiv -1 \\pmod{p}$, then:\n\n$$\n0 \\equiv x^2 = a + bc \\equiv a - b \\pmod{p}\n$$\n\nSince $p \\mid a+b$, this gives $p \\mid 2a$, so $p \\mid a$ (as $p \\neq 2$). Then:\n\n$$\nz^2 = c + ab \\equiv c \\equiv -1 \\pmod{p}\n$$\n\nwhich is impossible since $\\left(\\frac{-1}{p}\\right) = -1$. Thus $p \\nmid c+1$, and so:\n\n$$\n0 \\equiv v_p(x^2 + y^2) = v_p((a+b)(c+1)) = v_p(a+b) + v_p(c+1) = v_p(a+b) \\pmod{2}\n$$\n\nwhich proves the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19127,
    "subject": "Mathematics (Olympiad)",
    "question": "Prime numbers $p, q, r, s$ satisfy the condition: $5 < p < q < r < s < p+10$. Prove that the sum $p+q+r+s$ is divisible by $60$.",
    "options": [],
    "answer": "See solution",
    "solution": "The four primes must be of the form $p$, $p+2$, $p+4$, $p+6$, and $p+8$. Among any three consecutive odd numbers, one is divisible by $3$. Thus, $p+4$ is divisible by $3$, and since it is prime, it is also divisible by $5$. Hence, $p+4 = 15k$, which implies:\n\n$$\n\\begin{aligned}\np + q + r + s &= p + (p+2) + (p+6) + (p+8) \\\\\n&= 4p + 16 = 4(15k - 4) + 16 = 60k,\n\\end{aligned}\n$$\n\nwhich is what we wanted to prove.\n\nNote that such numbers exist, for example, $11$, $13$, $17$, and $19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19128,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. We consider labellings of the squares of a chessboard of size $n \\times n$ with the natural numbers from $1$ to $n^2$ such that every number is used exactly once. Given such a labelling, we say a positive integer is a *rook product* if it is the product of the labels of $n$ squares such that, if you place a rook on each of them, no two rooks will attack each other.\n\n(Two rooks are attacking each other if and only if they are in the same row or column.)\n\n(a) Let $n = 8$. Determine whether there exists a labelling of an $8 \\times 8$ chessboard such that the following condition is fulfilled: The difference of any two rook products is always divisible by $65$.\n\n(b) Let $n = 10$. Determine whether there exists a labelling of a $10 \\times 10$ chessboard such that the following condition is fulfilled: The difference of any two rook products is always divisible by $101$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) No, there is no such labelling.\n\nWe show that for every labelling there exist two rook products whose difference is not divisible by $65$. Suppose that an $8 \\times 8$ chessboard is labelled with the numbers $1, 2, \\ldots, 64$ such that no number is used twice.\n\nWe can construct a rook product that is divisible by $13$ by placing a rook on the square with the label $13$ and the other seven rooks non-attackingly, but otherwise arbitrarily.\n\nWe can construct a rook product that is not divisible by $13$ as follows. Only four labels are divisible by $13$, namely $13, 26, 39,$ and $52$. These four labels are located in at most four rows; let $R \\subseteq \\{1, \\ldots, 8\\}$ be the index set of these rows. Similarly, there are at least four columns that do not contain any of these four labels; let $C \\subseteq \\{1, \\ldots, 8\\}$ be the index set of these columns. Since $|R| \\leq |C|$, it is possible to place non-attacking rooks in rows $R$ using only columns from $C$. The remaining rooks are placed in the remaining rows non-attackingly, but otherwise arbitrarily. The resulting rook product is not divisible by $13$ since the rooks avoid the squares whose labels are divisible by $13$.\n\nThe difference of the two rook products is not divisible by $13$, since one rook product is divisible by $13$ whereas the other one is not. Hence the difference is not divisible by $65$.\n\n(b) Yes, there is such a labelling.\n\nFor $k \\in [0, 99]$ define $a_k = 2^k \\pmod{101}$; that is, $a_k$ is the remainder of $2^k$ when divided by $101$. Note that $a_k \\neq 0$ since no power of $2$ is divisible by $101$. Hence $1 \\leq a_k \\leq 100$ for all $k \\in [0, 99]$.\n\nWe label the squares of the chessboard with the numbers $a_k$ as in the following table.\n\n![](path/to/file.png)\n\nMore precisely, if we label the rows and columns of the chessboard by $\\{0, 1, 2, \\ldots, 9\\}$, then the square with coordinates $(i, j)$ gets the label $a_{10i+j}$.\n\nNote that $a_{10i+j} = 2^{10i+j} \\pmod{101}$ and that $2^{10i+j} = (2^{10})^i \\cdot 2^j$. Hence for this labelling any rook product is congruent to\n\n$$\n(2^{10})^{0+1+2+\\dots+9} \\cdot 2^{0+1+2+\\dots+9}\n$$\n\nmodulo $101$. Thus, the difference of any two rook products is divisible by $101$ for this labelling.\n\nIt remains to show that the $a_k$ are pairwise different. (In more elaborate language, $2$ is a primitive root modulo $101$.) To do this, let $s$ be the smallest positive integer such that $2^s \\equiv 1 \\pmod{101}$. Using long division, write $100 = qs + r$ with non-negative integers $q$ and $r$ such that $0 \\leq r \\leq s-1$. By Fermat's little theorem,\n\n$$\n1 \\equiv 2^{100} \\equiv 2^{qs+r} \\equiv (2^s)^q \\cdot 2^r \\equiv 1^q \\cdot 2^r \\equiv 2^r \\pmod{101}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19129,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be positive integers with $n \\ge m$. There is a game board of size $1 \\times n$ divided into $n$ unit squares and an unlimited supply of sticky tapes of size $1 \\times m$. On each move, a player adds a tape covering $m$ consecutive unit squares on the board, at least one of which is not yet covered by any tape. Two players move alternately; the player who cannot make a move loses.\n\na) Prove that if $n$ and $m$ have equal parity, then the first player can win regardless of how the opponent plays.\n\nb) Is it true that, whenever $n$ and $m$ have different parity, the second player can win regardless of how the first player plays?",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/prob1718_p18_data_f0e3cd1663.png)\n\nFig. 25\n\n![](images/prob1718_p18_data_f779c266cd.png)\n\nFig. 26\n\n*Answer:* b) No.\n\n**Solution.**\n\na) If $n$ and $m$ have equal parity, the first player can cover the central squares so that an equal number of uncovered squares remain on both sides (see Fig. 25 for $n = 11$, $m = 3$). The first player can then respond to each of the opponent's moves with a symmetric move, maintaining symmetry about the center. This strategy ensures that after each move, the uncovered squares remain symmetric, and the second player will eventually be unable to move and lose.\n\nb) If $n = 5$ and $m = 2$, the first player can cover two squares at the edge of the board on their first move (see Fig. 26). Any move by the second player will leave either 1 or 2 consecutive squares uncovered, which the first player can cover on their next move, ensuring victory.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19130,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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 $s = \\frac{a + b + c}{2}$ is the semiperimeter of the triangle $ABC$.\n\nClearly, 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 $\\frac{A_2I_b}{A_2I_c}$ and the like, and refer to the converse of Menelaus' theorem.\n\nTo evaluate the ratio $\\frac{A_2I_b}{A_2I_c}$, 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$$\n\\frac{A_2I_b}{A_2I_c} = \\frac{PI_b}{PI_a} \\cdot \\frac{QI_a}{QI_c}.\n$$\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\n$$\nPI_a = A_1B \\cdot \\frac{II_a}{IB} = \\frac{s-c}{\\sin \\frac{C}{2}}.\n$$\n\nSince $I_aI_b = \\frac{c}{\\sin \\frac{C}{2}}$, we obtain $PI_b = \\frac{|2c-s|}{\\sin \\frac{C}{2}}$. 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\n$$\nQI_a = A_1C \\cdot \\frac{II_a}{IC} = \\frac{s-b}{\\sin \\frac{B}{2}}.\n$$\n\nSince $I_aI_c = \\frac{b}{\\sin \\frac{B}{2}}$, we obtain $QI_c = \\frac{|2b-s|}{\\sin \\frac{B}{2}}$. Consequently,\n\n$$\n\\frac{A_2I_b}{A_2I_c} = \\frac{s-b}{s-c} \\cdot \\frac{|2c-s|}{|2b-s|}\n$$\n\nand the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19131,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: [0, \\infty) \\to (0, \\infty)$ be an increasing function, and let $g: [0, \\infty) \\to \\mathbb{R}$ be a twice differentiable function such that $g''$ is continuous, and $g''(x) + f(x)g(x) = 0$ for all $x \\ge 0$.\n\n**a)** Exhibit a pair of functions $f$ and $g$, $g \\ne 0$, satisfying the conditions in the statement.\n\n**b)** Show that $g$ is bounded on the ray $x \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** The functions $f(x) = 1$ for $x \\ge 0$, and $g(x) = \\sin x$ for $x \\ge 0$, clearly satisfy the required conditions.\n\n**b)** Using the hypothesis, write $\\dfrac{g'(x)g''(x)}{f(x)} + g(x)g'(x) = 0$ for all $x \\ge 0$. Fix a positive $t$, and integrate over $[0, t]$ to obtain\n\n$$\n2 \\int_{0}^{t} \\frac{g'(x)g''(x)}{f(x)} \\, dx + (g(t))^{2} - (g(0))^{2} = 0.\n$$\n\nThe function $x \\mapsto 1/f(x)$ for $x \\ge 0$ is well defined, decreasing, and positive, so\n\n$$\n\\int_0^t \\frac{g'(x)g''(x)}{f(x)} \\, dx = \\frac{1}{f(0)} \\int_0^\\theta g'(x)g''(x) \\, dx = \\frac{(g'(\\theta))^2 - (g'(0))^2}{2f(0)}\n$$\n\nfor some $\\theta$ between $0$ and $t$, by the second mean value theorem. Consequently,\n\n$$\n(g(t))^2 = (g(0))^2 + \\frac{(g'(0))^2}{f(0)} - \\frac{(g'(\\theta))^2}{f(0)} \\le (g(0))^2 + \\frac{(g'(0))^2}{f(0)}\n$$\n\nand the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19132,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $P$ un punto en el plano. Demuestra que es posible trazar tres semirrectas con origen en $P$ con la siguiente propiedad: para toda circunferencia de radio $r$ que contiene a $P$ en su interior, si $P_1$, $P_2$ y $P_3$ son los puntos de corte de las semirrectas con la circunferencia, entonces\n\n$$\n|PP_1| + |PP_2| + |PP_3| \\le 3r.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Trazamos tres semirrectas con origen en $P$ de manera que el ángulo formado por dos cualesquiera de las tres sea de $120^\\circ$. Imaginemos el círculo dividido en seis regiones mediante tres rectas que pasan por su centro y son paralelas a las tres semirrectas. Por simetría respecto a una de las rectas o por giros de ángulo $120^\\circ$, basta analizar el caso en que el punto $P$ se encuentra en una de las seis regiones. Usamos las notaciones de la figura.\n\n![](images/OME2022_final_p4_data_3d35966172.png)\n\nSin pérdida de generalidad, supongamos que el radio de la circunferencia es $r = 1$. Los tres puntos de corte de las semirrectas con la circunferencia son, respectivamente, $P_1$, $P_2$ y $P_3$. Las paralelas con origen en $O$ (centro de la circunferencia) la cortan, respectivamente, en $A$, $B$ y $C$. La recta $PP_3$ y la tangente a la circunferencia en $C$ se cortan en el punto $T$. $S'$ es el punto de corte de $PP_3$ con la paralela a $OA$ por $C$. Es obvio que $\\angle S'CT = 30^\\circ$, con lo que $SP_3 = SS' + S'P_3 = OC + S'P_3 \\le OC + S'T = OC + \\frac{1}{2}CS' = 1 + \\frac{1}{2}OS$. De manera análoga, se demuestra que $RP_2 \\le 1 + \\frac{1}{2}OR$. También es obvio que $RP = PS$. Además, como $\\frac{1}{2}OS = \\frac{1}{2}OR + \\frac{1}{2}RS$ y $PZ = OR + \\frac{1}{2}RS$, tenemos:\n\n$$\n\\begin{align*}\nPP_1 + PP_2 + PP_3 &= PP_1 + RP_2 + PR + SP_3 - SP \\\\\nPP_1 + RP_2 + SP_3 &\\le PP_1 + 1 + \\frac{1}{2}OR + 1 + \\frac{1}{2}OS \\\\\n2 + PP_1 + OR + \\frac{1}{2}RS &\\le 2 + PP_1 + PZ \\le 3.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19133,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that the largest prime divisor of $n^2 + 2$ is equal to the largest prime divisor of $n^2 + 2n + 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be a common prime divisor of $n^2 + 2$ and $n^2 + 2n + 3$. Then $d$ divides their difference:\n\n$$\n(n^2 + 2n + 3) - (n^2 + 2) = 2n + 1.\n$$\n\nAlso, $d$ divides $n \\cdot (2n + 1) = 2n^2 + n$, and $2n^2 + n - 2(n^2 + 2) = n - 4$. Furthermore, $2n + 1 - 2(n - 4) = 9$. Thus, any common prime divisor $d$ must divide $9$, so $d = 3$.\n\nThe only other possible prime factor in either number is $2$, which can appear in at most one of them. At least one of the numbers has the exponent of $3$ at most $2$.\n\n**Case 1:** $n$ even.\n\nThen $n^2 + 2$ is even, but not divisible by $4$. The possible cases are:\n\n- $n^2 + 2 = 2 \\cdot 3$\n- $n^2 + 2 = 2 \\cdot 3^2$\n- $n^2 + 2n + 3 = 3$\n- $n^2 + 2n + 3 = 3^2$\n\nOnly $n^2 + 2 = 2 \\cdot 3^2 = 18$ gives a solution: $n = 4$.\n\n**Case 2:** $n$ odd.\n\nThen $n^2 + 2n + 3$ is even, but not divisible by $4$. The possible cases are:\n\n- $n^2 + 2n + 3 = 2 \\cdot 3$\n- $n^2 + 2n + 3 = 2 \\cdot 3^2$\n- $n^2 + 2 = 3$\n- $n^2 + 2 = 3^2$\n\nWe get $n = 1$ (from $n^2 + 2 = 3$ and $n^2 + 2n + 3 = 5$).\n\n**Conclusion:**\n\nThe solutions are $n = 1$ and $n = 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19134,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{P}$ be the set of all prime numbers. Find all functions $f : \\mathbb{P} \\to \\mathbb{P}$ such that\n$$\nf(p)^{f(q)} + q^p = f(q)^{f(p)} + p^q\n$$\nholds for all $p, q \\in \\mathbb{P}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, the identical function $f(p) = p$ for all $p \\in \\mathbb{P}$ is a solution. We will show that this is the only one.\n\nFirst, we will show that $f(2) = 2$. Taking $q = 2$ and $p$ any odd prime number, we have\n$$\nf(p)^{f(2)} + 2^p = f(2)^{f(p)} + p^2.\n$$\nAssume that $f(2) \\neq 2$. It follows that $f(2)$ is odd and so $f(p) = 2$ for any odd prime number $p$.\n\nTaking any two different odd prime numbers $p, q$ we have\n$$\n2^2 + q^p = 2^2 + p^q \\implies p^q = q^p \\implies p = q,\n$$\ncontradiction. Hence, $f(2) = 2$.\n\nSo for any odd prime number $p$ we have\n$$\nf(p)^2 + 2^p = 2^{f(p)} + p^2.\n$$\nCopy this relation as\n$$\n2^p - p^2 = 2^{f(p)} - f(p)^2. \\qquad (1)\n$$\nLet $T$ be the set of all positive integers greater than 2, i.e. $T = \\{3, 4, 5, \\dots\\}$. The function $g : T \\to \\mathbb{Z}$, $g(n) = 2^n - n^2$, is strictly increasing, i.e.\n$$\ng(n + 1) - g(n) = 2^n - 2n - 1 > 0 \\qquad (2)\n$$\nfor all $n \\in T$. We show this by induction. Indeed, for $n = 3$ it is true, $2^3 - 2 \\cdot 3 - 1 > 0$. Assume that $2^k - 2k - 1 > 0$. It follows that for $n = k + 1$ we have\n$$\n2^{k+1} - 2(k + 1) - 1 = (2^k - 2k - 1) + (2^k - 2) > 0\n$$\nfor any $k \\ge 3$. Therefore, (2) is true for all $n \\in T$.\n\nAs a consequence, (1) holds if and only if $f(p) = p$ for all odd prime numbers $p$, as well as for $p = 2$.\n\nTherefore, the only function that satisfies the given relation is $f(p) = p$, for all $p \\in \\mathbb{P}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19135,
    "subject": "Mathematics (Olympiad)",
    "question": "Ќе наречеме еден број \"шарен\" ако има еднаков број парни и непарни цифри. Определете колку четирицифрени \"шарени\" броеви постојат, при што сите цифри се различни.",
    "options": [],
    "answer": "See solution",
    "solution": "Имаме 5 парни цифри: $\\{0, 2, 4, 6, 8\\}$ и 5 непарни цифри: $\\{1, 3, 5, 7, 9\\}$. Можеме да избереме 2 парни цифри од 5 на $C_5^2 = 10$ начини:\n\n$$\n\\{\\{0,2\\}, \\{0,4\\}, \\{0,6\\}, \\{0,8\\}, \\{2,4\\}, \\{2,6\\}, \\{2,8\\}, \\{4,6\\}, \\{4,8\\}, \\{6,8\\}\\}\n$$\n\nНа исто толку начини можеме да избереме 2 од 5-те непарни цифри. За секој пар парни и пар непарни цифри, постојат $4! = 24$ начини за подредување на четирите цифри. Значи, вкупно има $10 \\cdot 10 \\cdot 24 = 2400$ четирицифрени низи со две парни и две непарни различни цифри.\n\nСега, пребројуваме броеви што започнуваат со 0. Ако во парот парни цифри се наоѓа 0, тогаш бројот на четирицифрени броеви што започнуваат со 0 е $3! = 6$. Има 4 парни пара што содржат 0, па бројот на \"шарени\" четирицифрени низи што започнуваат со 0 е $4 \\cdot 10 \\cdot 6 = 240$.\n\nЗатоа, бројот на сите четирицифрени \"шарени\" броеви со различни цифри е $2400 - 240 = 2160$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19136,
    "subject": "Mathematics (Olympiad)",
    "question": "In every cell of a $101 \\times 101$ board is written a positive integer. For any choice of 101 cells from different rows and columns, their sum is divisible by 101. Show that the number of ways to choose a cell from each row of the board, so that the total sum of the numbers in the chosen cells is divisible by 101, is divisible by 101.",
    "options": [],
    "answer": "See solution",
    "solution": "Index the rows and columns from 0 to 100. We shall only work modulo 101 in what follows. Observe that we may let $(0,0) = 0$ by adding some constant to all the terms of the table. Next, because of the condition in the statement,\n\n$$\n(u, v) + (i, j) = (u, j) + (i, v),\n$$\n\nfor any $u, v, i, j$. Thus, there exist $a_0 = 0, a_1, \\dots, a_{100}$ and $b_0 = 0, b_1, \\dots, b_{100}$ so that $(i, j) = a_i + b_j$. Now, letting $\\Sigma_a$ and $\\Sigma_b$ be the sum of the $a_i$ and the $b_j$ respectively, the condition in the statement then yields $\\Sigma_a + \\Sigma_b = 0$. Note that\n\n$$\n\\sum_{i=0}^{100} x^{101b_i} = \\left( \\sum_{i=1}^{100} x^{b_i} \\right)^{101} := \\sum_{i=0}^{\\infty} x^i \\alpha(i).\n$$\n\nObserve that the number of ways of adequately choosing one cell from each row corresponds to the number of 101-tuples with values from $\\{b_0, b_1, \\dots, b_{100}\\}$ with sum equal to $\\Sigma_b$; that is, $\\alpha(\\Sigma_b) + \\alpha(101 + \\Sigma_b) + \\dots$.\n\nUsing the fact that 101 is prime and computing the latter expression for $x = \\exp(2\\pi i/101)$ we get the desired\n\n$$\n0 = \\sum_{i=0}^{\\infty} \\alpha (101i + \\Sigma_b).\n$$\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19137,
    "subject": "Mathematics (Olympiad)",
    "question": "Do positive real numbers $x, y, z$ have to be equal if they satisfy\n\n$$\n\\frac{xy + 1}{x + 1} = \\frac{yz + 1}{y + 1} = \\frac{zx + 1}{z + 1}\n$$\n\n![](images/Ukraine_2021-2022_p27_data_87e6159e09.png)",
    "options": [],
    "answer": "See solution",
    "solution": "From the statement, we have $(xy + 1)(y + 1) = (yz + 1)(x + 1)$, so $xy^2 + xy + y = xyz + x + yz$. Similarly, $yz^2 + yz + z = xyz + y + zx$, and $zx^2 + zx + x = xyz + z + xy$. Adding these three equations, we get $xy^2 + yz^2 + zx^2 = 3xyz$, or $\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} = 3$. For positive real numbers $x, y, z$, by the inequality between the arithmetic mean and the geometric mean, $\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} \\ge 3$, with equality only when $\\frac{x}{y} = \\frac{y}{z} = \\frac{z}{x} = 1$, implying $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19138,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of $\\triangle ABC$ and let $I'$ be a tangent to the incircle, not a side of $\\triangle ABC$, that intersects the sides $AB$, $BC$, and the extension of $CA$ at $X$, $Y$, and $Z$, respectively. Let $AY$ intersect $CX$ at $P$, and the lines $IP$ and $BZ$ meet at $Q$.\n\nProve that, if $A$, $C$, $Y$, $X$ are concyclic, then $ZI^2 = ZQ \\cdot ZB$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to show that $IQ \\perp BZ$ and $\\angle BIZ = 90^\\circ$ because this will imply that $ZI$ is tangent to the circumcircle of $\\triangle BIQ$ (with diameter $BI$) at $I$, and then by the power at $Z$ we will get $ZI^2 = ZQ \\cdot ZB$.\n\nTo show $\\angle BIZ = 90^\\circ$, let $S$, $T$ be the tangent points from $Z$ and $U$, $V$ be the tangent points from $B$ as shown in the figure. It is easy to see by simple angle chasing that $AXYC$ cyclic if and only if $ST \\perp UV$. And using the fact that $IB \\perp UV$ and $IZ \\perp ST$, we obtain that $\\angle BIZ = 90^\\circ$.\n\nNote that by the Bianchon’s theorem, for degenerated circumscribed hexagons $AXSYCT$ and $AUXYVC$, the lines $ST$, $UV$, $AY$, $CX$ are concurrent at $P$.\n\n![](images/Tajland_2011_p7_data_c0e7266c9a.png)\n\nThe condition $IQ \\perp BZ$ follows from the Pascal and Brocard theorems as follows:\nBy Brocard’s theorem, we have $IP \\perp ML$ where $M$, $L$ are the intersections of the opposite sides of cyclic quadrilateral $SVTU$ with circumcenter $I$, the orthocenter of $\\triangle MLP$.\n\nBy Pascal’s Theorem, consider two degenerated inscribed hexagons *USSVTT* and *UUSVVT*, so that the intersections of opposite sides are collinear. (The side with repeated vertex becomes the tangent line at that point.) Therefore, *M*, *Z*, *L* are collinear, and so are *B*, *M*, *L*. Since both lines contain *M* and *L*, they are the same line.\n\nThis proves that $IP \\perp BZ$ at $Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19139,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\ge 2$, let\n\n$$\nL_n = \\lfloor \\sqrt{n} \\rfloor + \\lfloor \\sqrt[3]{n} \\rfloor + \\dots + \\lfloor \\sqrt[n]{n} \\rfloor\n$$\n\nand\n\n$$\nR_n = \\lfloor \\log_2 n \\rfloor + \\lfloor \\log_3 n \\rfloor + \\dots + \\lfloor \\log_n n \\rfloor.\n$$\n\nProve by induction that $L_n = R_n$ for all $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $L_2 = 1 = R_2$. Assume $L_{n-1} = R_{n-1}$ for some $n > 2$.\n\nObserve that $\\lfloor \\sqrt[n]{n} \\rfloor = \\lfloor \\log_n n \\rfloor = 1$, so\n\n$$\nL_n - L_{n-1} - 1 = \\sum_{k=2}^{n-1} \\left( \\lfloor \\sqrt[k]{n} \\rfloor - \\lfloor \\sqrt[k]{n-1} \\rfloor \\right)\n$$\n\nand\n\n$$\nR_n - R_{n-1} - 1 = \\sum_{k=2}^{n-1} \\left( \\lfloor \\log_k n \\rfloor - \\lfloor \\log_k (n-1) \\rfloor \\right).\n$$\n\nFix $k = 2, 3, \\dots, n-1$ and set $a = \\lfloor \\sqrt[k]{n-1} \\rfloor$, $a \\ge 1$. Then $a^k \\le n-1 < (a+1)^k$, so $a^k < n \\le (a+1)^k$.\n\nIf $n < (a+1)^k$, then $\\lfloor \\sqrt[k]{n} \\rfloor = a$ and $\\lfloor \\sqrt[k]{n} \\rfloor - \\lfloor \\sqrt[k]{n-1} \\rfloor = 0$. If $n = (a+1)^k$, then $\\lfloor \\sqrt[k]{n} \\rfloor = a+1$ and $\\lfloor \\sqrt[k]{n} \\rfloor - \\lfloor \\sqrt[k]{n-1} \\rfloor = 1$ (with $a+1 \\ge 2$).\n\nThus, $\\sum_{k=2}^{n-1} (\\lfloor \\sqrt[k]{n} \\rfloor - \\lfloor \\sqrt[k]{n-1} \\rfloor)$ increases by 1 each time $n$ is of the form $u^v$ with $u, v > 2$.\n\nSimilarly, for $k = 2, 3, \\dots, n-1$ and $b = \\lfloor \\log_k(n-1) \\rfloor$, $b \\ge 1$, $k^b \\le n-1 < (k+1)^b$, so $k^b < n \\le (k+1)^b$. As before, $\\lfloor \\log_k n \\rfloor = \\lfloor \\log_k(n-1) \\rfloor$ if $n < (k+1)^b$, and $\\lfloor \\log_k n \\rfloor - \\lfloor \\log_k(n-1) \\rfloor = 1$ if $n = (k+1)^b$.\n\nTherefore, $\\sum_{k=2}^{n-1} (\\lfloor \\log_k n \\rfloor - \\lfloor \\log_k(n-1) \\rfloor)$ also increases by 1 each time $n$ is of the form $u^v$ with $u, v > 2$.\n\nConsequently, $L_n - L_{n-1} - 1 = R_n - R_{n-1} - 1$, so by induction $L_n = R_n$ for all $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19140,
    "subject": "Mathematics (Olympiad)",
    "question": "Wie viele Lösungen hat die Gleichung\n\n$$\n\\lfloor \\frac{x}{20} \\rfloor = \\lfloor \\frac{x}{17} \\rfloor\n$$\n\nüber der Menge der positiven ganzen Zahlen?\n\nDabei bezeichnet $\\lfloor a \\rfloor$ die größte ganze Zahl, die kleiner oder gleich $a$ ist.",
    "options": [],
    "answer": "See solution",
    "solution": "Sei\n\n$$\n\\lfloor \\frac{x}{20} \\rfloor = \\lfloor \\frac{x}{17} \\rfloor = n.\n$$\n\nDann gilt $20n \\leq x < 20n + 20$ und $17n \\leq x < 17n + 17$. Daher muss für alle möglichen Lösungen $x$ gelten:\n\n$$\n20n \\leq x < 17n + 17. \\tag{1}\n$$\n\nFür $n$ folgt $20n < 17n + 17$, also $n \\in \\{0, 1, 2, 3, 4, 5\\}$.\n\nFür $n = 0$ ergeben sich mit $1 \\leq x < 17$ insgesamt 16 Lösungen, für $n = 1$ mit $20 \\leq x < 34$ insgesamt 14 Lösungen, für $n = 2$ mit $40 \\leq x < 51$ insgesamt 11 Lösungen, usw.\n\n(Die Ungleichung (1) hat $17n + 17 - 20n = 17 - 3n$ Lösungen, nur für $n = 0$ ist $x = 0$ auszuschließen.)\n\nWir haben daher $16 + 14 + 11 + 8 + 5 + 2 = 56$ Lösungen über der Menge der natürlichen Zahlen für diese Gleichung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19141,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 100 infinitely large boxes with markers in them, the following procedure is carried out:\n\n- At step 1, one adds one marker in every box.\n- At step 2, one marker is added in every box containing an even number of markers.\n- At step 3, one marker is added in every box in which the number of markers is divisible by 3, and so on.\n\nBefore the process starts, Bruno wants to distribute several markers in the boxes so that there is at least one marker in each box and the following holds: **After any number of steps, there exist two boxes containing different numbers of markers.**\n\nDecide if this is possible to achieve.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is *no*. Regardless of the initial distribution, all boxes will contain the same number of markers after finitely many steps. Moreover, this is true for any number of boxes.\n\nDenote by $x_n$ the number of markers in a certain box before step $n$, $n = 1, 2, \\dots$. Suppose that $x_n = n$ for some $n$. Then, by the rule of adding markers, we have $x_{n+1} = n+1$, $x_{n+2} = n+2$, etc.; in other words, the number of markers in that box equals the number of the oncoming step $l$ for each $l \\geq n$. So, in order to prove that eventually all boxes contain the same number of markers, it is enough to show that for each box there exists a step $n$ such that $x_n = n$.\n\nWe use the following observation. Let a box $C$ satisfy $x_i > l$ for some $l$, that is, the difference $d_i = x_i - l$ is positive. Then there is an $m \\geq l$ such that $C$ receives no marker at step $m$. Otherwise, $x_i$ increases by 1 at every step $m \\geq l$, which means that $x_i + s$ is divisible by $l + s$ for all $s \\geq 0$.\n\nHowever, this is impossible as $1 < \\frac{x_i + s}{l + s} < 2$ for $s$ sufficiently large; it is enough to take $s > x_i - 2l$.\n\nLet $m \\geq l$ be the first step that adds no marker to $C$. Then the observation implies that the difference $d_{m+1} = x_{m+1} - (m+1) = x_m - (m+1)$ satisfies $d_{m+1} = d_l - 1$. If $d_{m+1} > 0$, then by the same reason there is a step $k > m$ with $d_k = d_m - 1$. Repeated applications of the same argument show that after finitely many steps there will be a step $s$ such that $d_s = 0$, that is, $x_s = s$.\n\nInitially, before step 1, one has $x_1 \\geq 1$ for each box $C$. This is ensured by the condition that every box contains a marker. If $x_1 = 1$, then $x_n = n$ holds for $C$ already with $n = 1$. Otherwise, $x_i > 1$, so by the above, $x_n = n$ will result after finitely many steps. As explained in the beginning, when this happens for all boxes, the numbers of markers in them will be the same.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19142,
    "subject": "Mathematics (Olympiad)",
    "question": "For any three consecutive odd integers, what is the highest common factor of their product $P$?",
    "options": [],
    "answer": "See solution",
    "solution": "Exactly one of any three consecutive odd integers is divisible by $3$, so $3$ divides their product $P$. For example, $1 \\times 3 \\times 5 = 15$ and $7 \\times 9 \\times 11 = 693$; both are divisible by $3$. No higher common factor exists, so the highest common factor of all such $P$ is $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19143,
    "subject": "Mathematics (Olympiad)",
    "question": "a. Zsuzsi's car has an odometer reading of $3862$ km and a trip meter reading of $386.2$ km. After driving $2.9$ km, what will each meter show?\n\nb. When will the odometer and trip meter next both start with the digit $4$? What will each meter read at that time?\n\nc. When will the odometer and trip meter next show the same digits in the same order (ignoring the decimal point)? What will each meter read at that time?\n\n![]( <table><thead><tr><th>trip meter</th><th>odometer</th><th>both start with 4?</th></tr></thead><tbody><tr><td>386.2</td><td>3862</td><td>no</td></tr><tr><td>386.3 to 399.9</td><td>3862 to 3875</td><td>no</td></tr><tr><td>400.0 to 499.9</td><td>3875 to 3975</td><td>no</td></tr><tr><td>500.0 to 399.9</td><td>3975 to 4875</td><td>no</td></tr><tr><td>400.0</td><td>4875</td><td>yes</td></tr></tbody></table> )\n\n![]( <table><thead><tr><th>trip meter</th><th>odometer</th><th>match?</th></tr></thead><tbody><tr><td>3862</td><td>3862</td><td>yes</td></tr><tr><td>3863 to 9992</td><td>3862 to 4475</td><td>no: trip is faster</td></tr><tr><td>9993 to 0002</td><td>4475 to 4476</td><td>no</td></tr><tr><td>0003 to 4472</td><td>4476 to 4923</td><td>no: trip \\lt odo</td></tr><tr><td>4473 to 4922</td><td>4923 to 4968</td><td>no: trip \\lt odo</td></tr><tr><td>4923 to 4962</td><td>4968 to 4972</td><td>no: trip \\lt odo</td></tr><tr><td>4963 to 4971</td><td>4972</td><td>no</td></tr><tr><td>4972</td><td>4973</td><td>no</td></tr><tr><td>4973</td><td>4973</td><td>yes</td></tr></tbody></table> )",
    "options": [],
    "answer": "See solution",
    "solution": "a. Since $386.2 + 2.9 = 389.1$, the trip meter will show $389.1$. Since $2.9$ km adds only $2$ whole kilometres to $3862$, the odometer will show $3864$.\n\nb. The odometer reading will next start with $4$ when it changes to $4000$. This requires the car to travel a further $4000 - 3862 = 138$ km. At that point, the trip meter will show $386.2 + 138 = 524.2$.\n\nThe trip meter reading will next start with $4$ when it changes to $400.0$. This requires the car to travel a further $1400 - 524.2 = 875.8$ km. At that point, the odometer will show $4000 + 875 = 4875$.\n\nAlternatively, the odometer changed to $3862$ at the same time the trip meter changed to $386.2$. Thereafter, the odometer changes at the same time the last digit of the trip meter reading changes to $2$. The following table shows various ranges of trip meter readings and the corresponding ranges of odometer readings.\n\n![]( <table><thead><tr><th>trip meter</th><th>odometer</th><th>both start with 4?</th></tr></thead><tbody><tr><td>386.2</td><td>3862</td><td>no</td></tr><tr><td>386.3 to 399.9</td><td>3862 to 3875</td><td>no</td></tr><tr><td>400.0 to 499.9</td><td>3875 to 3975</td><td>no</td></tr><tr><td>500.0 to 399.9</td><td>3975 to 4875</td><td>no</td></tr><tr><td>400.0</td><td>4875</td><td>yes</td></tr></tbody></table> )\n\nThus, the odometer will still show $4875$ when both readings next start with $4$.\n\nc. The odometer changed to $3862$ at the same time the trip meter changed to $386.2$. If the two meter readings have the same digits in the same order (ignoring the decimal point), we say they match.\n\nIgnoring the decimal point, the trip meter reading is a $4$-digit number that increases $10$ times faster than the odometer reading. The following table shows various ranges of trip meter readings and the corresponding ranges of odometer readings. The trip meter ranges are chosen so that it is easy to check for a match.\n\n![]( <table><thead><tr><th>trip meter</th><th>odometer</th><th>match?</th></tr></thead><tbody><tr><td>3862</td><td>3862</td><td>yes</td></tr><tr><td>3863 to 9992</td><td>3862 to 4475</td><td>no: trip is faster</td></tr><tr><td>9993 to 0002</td><td>4475 to 4476</td><td>no</td></tr><tr><td>0003 to 4472</td><td>4476 to 4923</td><td>no: trip \\lt odo</td></tr><tr><td>4473 to 4922</td><td>4923 to 4968</td><td>no: trip \\lt odo</td></tr><tr><td>4923 to 4962</td><td>4968 to 4972</td><td>no: trip \\lt odo</td></tr><tr><td>4963 to 4971</td><td>4972</td><td>no</td></tr><tr><td>4972</td><td>4973</td><td>no</td></tr><tr><td>4973</td><td>4973</td><td>yes</td></tr></tbody></table> )\n\nHence, the odometer reading is $4973$ the next time it matches the trip meter reading.\n\nAlternatively, suppose the car travels a further $a.b$ kilometres ($a$ is an integer and $b$ is a digit) to the next time the readings match.\n\nIf the trip meter hasn't reached $0000$, then $3862 + a = 3862 + 10a + b$. This means $9a + b = 0$, hence $a = b = 0$ and the car hasn't moved.\n\nSo the trip meter has reached $0000$ and\n\n$$\n3862 + a = 3862 + 10a + b - 10000.\n$$\n\nThis means $9a + b = 10000$, hence $b = 1$ and $a = 1111$. Therefore, the next time the readings match, the odometer reading will be $3862 + 1111 = 4973$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19144,
    "subject": "Mathematics (Olympiad)",
    "question": "You are given 5 distinct positive integers. Can their arithmetic mean be:\n\na) exactly 3 times larger than their largest common divisor?\n\nb) exactly 2 times larger than their largest common divisor?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a) yes, b) no.\n\n**Solution.**\n\na) It is enough to provide an example of such 5 integers. One example is the set $1, 2, 3, 4, 5$, whose arithmetic mean is $3$, and the largest common divisor is $1$.\n\nb) Suppose such numbers $a_1, a_2, a_3, a_4, a_5$ exist, and let $d$ be their largest common divisor. Then these 5 integers can be rewritten as $a_i = d b_i$ for $i = 1, \\\\dots, 5$, and the equality in the statement is rewritten as:\n\n$$\n\\frac{a_1 + \\cdots + a_5}{5} = 2d \\iff d(b_1 + \\cdots + b_5) = 10d \\iff b_1 + \\cdots + b_5 = 10.\n$$\n\nClearly, the numbers $b_1, \\dots, b_5$ are distinct positive integers, so their smallest possible sum is $1 + 2 + 3 + 4 + 5 = 15 > 10$. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19145,
    "subject": "Mathematics (Olympiad)",
    "question": "For arbitrary positive numbers $a$, $b$, $c$, prove the following inequality:\n\n$$\n\\frac{a^3}{b+c} + \\frac{b^3}{c+a} + \\frac{c^3}{a+b} \\geq (a-b)^2 + (b-c)^2 + (c-a)^2.\n$$\n\nCan the two sides of this inequality be equal to each other?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** no.\n\nUsing the arithmetic mean-geometric mean inequality, we see that $\\frac{x^2}{y} + y \\geq 2x$, so $\\frac{x^2}{y} \\geq 2x - y$ for all $x, y > 0$. Equality holds only if $x = y$.\n\nApply this to $x = a$, $y = b+c$:\n\n$$\n\\frac{a^2}{b+c} \\geq 2a - (b+c)\n$$\n\nThus,\n\n$$\n\\frac{a^3}{b+c} = a \\cdot \\frac{a^2}{b+c} \\geq a(2a - (b+c)) = 2a^2 - ab - ac.\n$$\n\nSimilarly,\n\n$$\n\\frac{b^3}{c+a} \\geq 2b^2 - bc - ba\n$$\n\nand\n\n$$\n\\frac{c^3}{a+b} \\geq 2c^2 - ca - cb.\n$$\n\nAdding these gives the desired result.\n\nFor equality, we require $a = b+c$, $b = a+c$, and $c = a+b$, which is only possible if $a = b = c = 0$, contradicting the positivity of $a, b, c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19146,
    "subject": "Mathematics (Olympiad)",
    "question": "設不等邊三角形 $ABC$，垂心為 $H$，$AH$ 交外接圓 $\\Omega$ 於另一點 $P$，$BH, CH$ 分別和 $AC, AB$ 交於 $E, F$。設 $Q, R$ 分別為 $PE, PF$ 與 $\\Omega$ 的另一個交點，$Y$ 在 $\\Omega$ 上使得 $AY, QR, EF$ 共點。證明 $PY$ 平分 $EF$。\n\n![](images/20-1J_p28_data_36b67d81ec.png)",
    "options": [],
    "answer": "See solution",
    "solution": "考慮六邊形 $ACRQPY$，由帕斯卡定理知 $E = AC \\cap QP, CR \\cap PY, QR \\cap AY$ 共點。類似地，$F = AB \\cap RP, BQ \\cap PY, QR \\cap AY$ 共點，因此 $BQ, CR, PY, EF$ 共於一點 $M$。\n\n在 $\\Omega$ 上取一點 $B' \\neq B$ 使得 $AB = AB'$，則\n\n$$\n\\angle AFE = \\angle BCA = \\angle BB'A = \\angle ABB' \\Rightarrow BB' \\parallel EF\n$$\n\n並且\n\n$$\n\\angle APB' = \\angle ABB' = \\angle BB'A = \\angle BCA = \\angle AHB \\Rightarrow BE \\parallel B'P\n$$\n\n延長 $BE$ 和 $\\Omega$ 交於 $K$，則有\n\n$$\n-1 = (H, K; E, \\infty_{BE}) \\stackrel{P}{\\cong} (A, K; Q, B') \\stackrel{B}{\\cong} (F, E; M, \\infty_{EF})\n$$\n\n代表 $M$ 為 $EF$ 中點，亦即 $PY$ 平分 $EF$，得證。\n\n![](images/20-1J_p28_data_36b67d81ec.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19147,
    "subject": "Mathematics (Olympiad)",
    "question": "The number 2021 is *fantabulous*. For any positive integer $m$, if any element of the set \\{$m$, $2m + 1$, $3m$\\} is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that all positive integers are fantabulous and hence $2021^{2021}$ is fantabulous.\n\nFirst, we prove that if 2021 is fantabulous, then 1 is fantabulous.\n\nLabel four allowable ‘moves’ A, B, C, D as follows:\n\nA. If $m$ is fantabulous then $2m + 1$ is fantabulous  \nB. If $2m + 1$ is fantabulous then $m$ is fantabulous  \nC. If $m$ is fantabulous then $3m$ is fantabulous  \nD. If $3m$ is fantabulous then $m$ is fantabulous.\n\nThe following chain of reasoning shows that if 2021 is fantabulous then 1 is fantabulous:\n\n$$\n2021 \\xrightarrow{C} 6063 \\xrightarrow{B} 3031 \\xrightarrow{B} 1515 \\xrightarrow{D} 505 \\xrightarrow{B} 252 \\xrightarrow{D} 84 \\xrightarrow{D} 28 \\xrightarrow{A} 57 \\xrightarrow{D} 19 \\xrightarrow{B} 9 \\xrightarrow{D} 3 \\xrightarrow{D} 1.\n$$\n\nWith 1 fantabulous (base case), we then prove that every positive integer is fantabulous by total induction.\n\nSuppose every positive integer $1, \\dots, k$ is fantabulous for some integer $k \\ge 1$.\n\nIf $k \\equiv 1 \\pmod{6}$, then $\\frac{k+1}{2}$ is a positive integer less than or equal to $k$ that is fantabulous, and hence $k+1$ is fantabulous by the following reasoning:\n\n$$\n\\frac{k+1}{2} \\xrightarrow{C} \\frac{3k+3}{2} \\xrightarrow{A} 3k+4 \\xrightarrow{A} 6k+9 \\xrightarrow{D} 2k+3 \\xrightarrow{B} k+1.\n$$\n\nIf $k \\equiv 0, 2, 4 \\pmod{6}$, then $\\frac{k}{2}$ is a positive integer less than $k$ that is fantabulous, and hence by move A, $2 \\times \\frac{k}{2} + 1 = k + 1$ is fantabulous.\n\nIf $k \\equiv 3 \\pmod{6}$, then $\\frac{2k}{3}+1$ is a positive integer less than or equal to $k$ that is fantabulous, and hence $k+1$ is fantabulous by the following reasoning:\n\n$$\n\\frac{2k}{3} + 1 \\xrightarrow{C} 2k + 3 \\xrightarrow{B} k + 1.\n$$\n\nIf $k \\equiv 5 \\pmod{6}$, then $\\frac{k+1}{3}$ is a positive integer less than $k$ that is fantabulous, and hence by move C, $3 \\times \\frac{k+1}{3} = k + 1$ is fantabulous.\n\nHence $k+1$ is fantabulous if $1, \\dots, k$ is fantabulous for some integer $k \\ge 1$. Since 1 is fantabulous, every positive integer is fantabulous, including $2021^{2021}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19148,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$ and an infinite set $A \\subset \\mathbb{Z}$, show that one can always find a subset $B$ of $A$ with $2p-2$ elements such that for any $p$ distinct elements of $B$, their arithmetic mean does not belong to $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that for some infinite set $A \\subset \\mathbb{Z}$, no $(2p-2)$-element subset $B$ satisfies the required condition. Without loss of generality, suppose $A$ contains infinitely many positive integers (otherwise, consider $-A$). The arithmetic mean of any $p$ distinct elements of $B$ lies between the smallest and largest elements of $B$, so removing all elements smaller than some integer $N$ from $A$ preserves the counterexample property. Set $N = 1$ so all elements of $A$ are positive.\n\n**Lemma:** For any nonnegative integer $k$, at most one residue class modulo $p^k$ contains infinitely many elements of $A$.\n\n**Proof of Lemma:** Suppose not, and let $k$ be minimal such that two residue classes modulo $p^k$ contain infinitely many elements, say $r_k$ and $r'_k$. Then, for some $N$, all elements of $A_{\\ge N} = \\{a \\in A \\mid a \\ge N\\}$ are congruent to $r_{k-1}$ modulo $p^{k-1}$, and both $r_k$ and $r'_k$ classes are infinite in $A_{\\ge N}$. Take $p-1$ elements from each class; the arithmetic mean of any $p$ elements from these is not congruent to $r_{k-1}$ modulo $p^{k-1}$, contradicting the assumption. Thus, the lemma holds.\n\nNow, for each $k$, let the unique residue class modulo $p^k$ with infinitely many elements be $r_k$ ($0 \\le r_k < p^k$), with $r_k \\equiv r_{k-1} \\pmod{p^{k-1}}$. Inductively, select a sequence $b_1, b_2, \\dots, b_{2p-2}$ from $A$ as follows: for each $i$, let $k_i$ be the smallest $k$ such that some $a_i \\in A$ satisfies $a_i \\not\\equiv r_{k_i} \\pmod{p^{k_i}}$. For each $i$, there exists $N_i$ such that all elements of $A_{\\ge N_i}$ are congruent to $r_{k_i}$ modulo $p^{k_i}$.\n\nSubtract $r_{k_{2p-2}}$ from all elements of $A$ and $N_i$. Now, each $b_i$ is divisible by $p^{k_i-1}$, and every element of $A_{\\ge N_i}$ is a multiple of $p^{k_{i+1}-1}$. Take $B = \\{b_1, \\dots, b_{2p-2}\\}$. For any $p$-element subset $\\{b_{i_1}, \\dots, b_{i_p}\\}$ ($i_1 < \\dots < i_p$), the arithmetic mean is divisible by $p^{k_{i_1}-2}$ and greater than $b_{i_1} \\ge N_{i_1-1}$, so it does not belong to $A$. This completes the proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19149,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that\n$$\n\\sum_{k=1}^{1008} k \\binom{2017}{k} \\equiv 0 \\pmod{2017^2}.\n$$\n\nb) Prove that\n$$\n\\sum_{k=1}^{504} (-1)^k \\binom{2017}{k} \\equiv 3 (2^{2016} - 1) \\pmod{2017^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) We have\n$$\n\\begin{align*}\n\\sum_{k=1}^{1008} k \\binom{2017}{k} &= \\sum_{k=1}^{1008} \\frac{k \\cdot 2017!}{k!(2017-k)!} = 2017 \\cdot \\sum_{k=0}^{1007} \\binom{2016}{k} \\\\\n&= \\frac{2017}{2} \\cdot \\left( \\sum_{k=0}^{2016} \\binom{2016}{k} - \\binom{2016}{1008} \\right) \\\\\n&= \\frac{2017}{2} \\cdot \\left( 2^{2016} - \\binom{2016}{1008} \\right).\n\\end{align*}\n$$\nOn the other hand, since $2017$ is a prime, by Fermat's little theorem, $2^{2016} \\equiv 1 \\pmod{2017}$. Moreover,\n$$\n\\begin{align*}\n\\binom{2016}{1008} - 1 &= \\frac{1009 \\cdot 1010 \\cdots 2016 - 1008!}{1008!} \\\\\n&= \\frac{(2017-1)(2017-2)\\cdots(2017-1008) - 1008!}{1008!}\n\\end{align*}\n$$\nIt is easy to see that the numerator is divisible by $2017$, but the denominator is not, so\n$$\n\\binom{2016}{1008} \\equiv 1 \\equiv 2^{2016} \\pmod{2017}.\n$$\nTherefore, the given sum is divisible by $2017^2$.\n\nb) Denote $\\frac{a}{b} \\equiv \\frac{c}{d} \\pmod{m}$ for $a, b, c, d \\in \\mathbb{Z}, bd \\neq 0$ and $m \\in \\mathbb{Z}, m > 1$ if $m$ divides the numerator of the reduced form of $\\frac{ad - bc}{bd}$.\n\n*Lemma.* For every prime number $p$, $\\binom{p}{k} \\equiv (-1)^{k-1} \\frac{p}{k} \\pmod{p^2}$.\n\n*Proof.* Indeed, we have $\\binom{p}{k} = \\frac{p!}{k!(p-k)!} = \\frac{p}{k} \\frac{(p-k+1) \\cdots (p-1)}{(k-1)!}$ so\n$$\n\\binom{p}{k} - (-1)^{k-1} \\frac{p}{k} = \\frac{p}{k} \\left[ \\frac{(p-k+1) \\cdots (p-1) - (-1)^{k-1} (k-1)!}{(k-1)!} \\right].\n$$\nIt is easy to see that the expression in brackets is divisible by $p$, so the lemma is proved. $\\square$\n\nFrom this lemma, we also have $\\frac{1}{p}\\binom{p}{k} \\equiv \\frac{(-1)^{k-1}}{k} \\pmod{p}$. Thus,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{504} (-1)^k \\binom{2017}{k} &= \\sum_{k=1}^{504} (-1)^k (-1)^{k-1} \\frac{2017}{k} = \\sum_{k=1}^{504} (-1)^{2k-1} \\frac{2017}{k} \\\\\n&= -2017 \\sum_{k=1}^{504} \\frac{1}{k} \\pmod{2017^2}.\n\\end{aligned}\n$$\nSo we have to prove $-2017 \\sum_{k=1}^{504} \\frac{1}{k} \\equiv 3(2^{2016} - 1) \\pmod{2017^2}$, which is equivalent to\n$$\n-\\sum_{k=1}^{504} \\frac{1}{k} \\equiv \\frac{3(2^{2016} - 1)}{2017} \\pmod{2017}.\n$$\nLet $S_n = \\frac{1}{1} + \\frac{1}{2} + \\cdots + \\frac{1}{n}$ for $n \\in \\mathbb{Z}^+$, then $\\frac{1}{1} - \\frac{1}{2} + \\cdots - \\frac{1}{2n} = S_{2n} - S_n$.\n\nHence\n$$\nS_{504} = S_{1008} - \\sum_{k=1}^{1008} \\frac{(-1)^{k-1}}{k} = S_{2016} - \\sum_{k=1}^{2016} \\frac{(-1)^{k-1}}{k} - \\sum_{k=1}^{1008} \\frac{(-1)^{k-1}}{k}\n$$\nNote that $S_{2016} = \\sum_{k=1}^{1008} \\left(\\frac{1}{k} + \\frac{1}{2017-k}\\right) \\equiv 0 \\pmod{2017}$, so we have to prove\n$$\n\\sum_{k=1}^{2016} \\frac{(-1)^{k-1}}{k} + \\sum_{k=1}^{1008} \\frac{(-1)^{k-1}}{k} \\equiv \\frac{3(2^{2016}-1)}{2017} \\pmod{2017}\n$$\nUsing the lemma, the above condition is true since in modulo $2017^2$\n$$\n\\sum_{k=1}^{2016} \\binom{2017}{k} + \\sum_{k=1}^{1008} \\binom{2017}{k} \\equiv \\sum_{k=1}^{2016} \\binom{2017}{k} + \\frac{1}{2} \\sum_{k=1}^{2016} \\binom{2017}{k} \\\\\n\\equiv \\frac{3}{2} \\left( \\sum_{k=0}^{2017} \\binom{2017}{k} - 2 \\right) \\equiv \\frac{3(2^{2017}-2)}{2} \\equiv 3(2^{2016}-1).\n$$\nThe proof is completed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19150,
    "subject": "Mathematics (Olympiad)",
    "question": "For which natural numbers $n \\ge 3$ is it possible to cut a regular $n$-gon into smaller pieces with regular polygonal shape? (The pieces may have different number of sides.)",
    "options": [],
    "answer": "See solution",
    "solution": "A regular triangle can be partitioned into four regular triangles of equal size (see the figure below), a regular quadrilateral can be partitioned into four regular quadrilaterals of equal size, and a regular hexagon can be partitioned into six regular triangles of equal size. By building alternately equilateral triangles and squares onto the sides of a regular 12-gon, just a regular hexagon remains, so a regular 12-gon can also be partitioned in the required way.\n\nShow now that other regular polygons cannot be partitioned into smaller regular polygons. For that, consider an arbitrary polygon that is partitioned into regular polygons. As the size of an internal angle of a regular polygon is less than $180^\\circ$ and not less than $60^\\circ$, at most two regular polygons can meet at each vertex.\n\nIf a vertex of the big $n$-gon is filled by just one smaller polygon, then this piece is an $n$-gon itself. Beside it, there must be space for at least one regular polygon. No more than two regular polygons can be placed there since the sum of the internal angles of these polygons and the $n$-gon itself would exceed $180^\\circ$. Two new pieces can be placed only if all these three pieces are triangular, which gives $n=3$. It remains to study the case where there is exactly one polygon beside the $n$-gonal piece. The size of the internal angle of the $n$-gon being at most $120^\\circ$ implies $n \\le 6$. The case $n=5$ is impossible as its external angles are $72^\\circ$ but no regular polygon has internal angles strictly between $60^\\circ$ and $90^\\circ$.\n\nIf each vertex of the big $n$-gon is the meeting point of two smaller regular polygons, then one of them must be a triangle since other regular polygons have internal angles of $90^\\circ$ or more. Beside a triangle, there is space for a triangle, a quadrilateral, or a pentagon.\n\n![](images/estonian-2012-2013_p20_data_62e70880ca.png)\n\n![](images/estonian-2012-2013_p20_data_af9b8fc336.png)\n\n![](images/estonian-2012-2013_p20_data_712b3536c0.png)\n\n![](images/estonian-2012-2013_p20_data_1ff71cfc8d.png)\n\n![](images/estonian-2012-2013_p20_data_8fde982a1a.png)\n\nIn the first two cases, the size of the internal angles of the $n$-gon will be $120^\\circ$ and $150^\\circ$, respectively, covering the cases $n = 6$ and $n = 12$. It remains to show that the third case with a triangle and a pentagon meeting at each vertex is impossible. Indeed, the side length of the pentagon must coincide with the side length of the initial big $n$-gon, because it is impossible to place a regular polygon beside the pentagon along one side. For the same reason, another pentagon must be built to the second next side along the boundary of the initial polygon. These two pentagons meet at the third vertex of the triangle built to the side between them. But the ulterior angle between the sides of the pentagons at the meeting point has size $360^\\circ - 2 \\cdot 108^\\circ - 60^\\circ = 84^\\circ$, which cannot be filled with interior angles of regular polygons.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19151,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $2x + 2y = 14$ and $x^2 - y^2 = 21$, find the value of $x - y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2x + 2y = 14$, we have $x + y = 7$. Also, since $x^2 - y^2 = (x + y)(x - y)$, we have $7(x - y) = 21$, so $x - y = \\frac{21}{7} = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19152,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose points $F_1, F_2$ are the left and right foci of the ellipse $$\\frac{x^2}{16} + \\frac{y^2}{4} = 1$$ respectively, and point $P$ is on the line $$x - \\sqrt{3}y + 8 + 2\\sqrt{3} = 0$$. When $\\angle F_1PF_2$ reaches its maximum, the value of the ratio $\\frac{|PF_1|}{|PF_2|}$ is \\underline{\\quad}.",
    "options": [],
    "answer": "See solution",
    "solution": "Euclidean geometry tells us that $\\angle F_1PF_2$ reaches its maximum only if the circle through points $F_1, F_2, P$ is tangent to the line $l$ at $P$. Now suppose $l$ intercepts the $x$-axis at point $A(-8-2\\sqrt{3}, 0)$. Then $\\angle APF_1 = \\angle AF_2P$, which means $\\triangle APF_1 \\sim \\triangle AF_2P$. So\n\n$$\n\\frac{|PF_1|}{|PF_2|} = \\frac{|AP|}{|AF_2|}\n$$\n\nBy using the power of a point theorem, we have\n\n$$\n|AP|^2 = |AF_1| \\cdot |AF_2|\n$$\n\nAs $F_1(-2\\sqrt{3}, 0)$, $F_2(2\\sqrt{3}, 0)$, $A(-8-2\\sqrt{3}, 0)$, so\n\n$$\n|AF_1| = 8, \\quad |AF_2| = 8 + 4\\sqrt{3}.\n$$\n\nThen we get\n\n$$\n\\begin{aligned}\n\\frac{|PF_1|}{|PF_2|} &= \\sqrt{\\frac{|AF_1|}{|AF_2|}} \\\\\n&= \\sqrt{\\frac{8}{8+4\\sqrt{3}}} \\\\\n&= \\sqrt{4-2\\sqrt{3}} \\\\\n&= \\sqrt{3}-1.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19153,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Denote its orthocentre by $H$ and let $A'$, $B'$, and $C'$ be the feet of the altitudes from $A$, $B$, and $C$. Let $P$ be the midpoint of $AH$, let $Q$ be the intersection of lines $B'P$ and $AB$, and denote the intersection of segments $A'C'$ and $BB'$ by $R$. Prove that the line $QR$ is perpendicular to the side $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The midpoint of the hypotenuse is also the circumcentre, so $P$ is the circumcentre of the triangle $AHB'$, and $|AP| = |PH| = |PB'|$. The triangle $HPB'$ is isosceles with the apex at $P$. Let $\\angle QB'B = \\alpha$. Then $\\alpha = \\angle PB'H = \\angle B'HP = \\angle BHA'$. Since $\\angle HA'B + \\angle BC'H = \\frac{\\pi}{2} + \\frac{\\pi}{2} = \\pi$, points $A'$, $B$, $C'$, and $H$ are concyclic and $\\angle BC'A' = \\angle BHA' = \\alpha$. In the quadrilateral $B'QC'R$ we have\n\n![](images/Slovenija_2008_p29_data_86502711db.png)\n\n$$\n\\angle RC'Q + \\angle QB'R = \\pi - \\angle BC'A + \\angle QB'B = \\pi - \\alpha + \\alpha = \\pi,\n$$\nso $B'$, $Q$, $C'$, and $R$ are concyclic.\n\nSince $AC'B'H$ is a cyclic quadrilateral, we have $\\alpha = \\angle B'HA = \\angle B'C'A$. The connection between the inscribed angles of a cyclic quadrilateral $B'QC'R$ gives us the equalities $\\angle B'RQ = \\angle B'C'Q = \\angle B'C'A = \\alpha$. So, $\\angle B'RQ = \\alpha = \\angle B'HA$, and $AH$ and $QR$ are parallel. Since $AH$ is perpendicular to $BC$, it is also perpendicular to $QR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19154,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $n$ be positive integers. Prove that\n\n$$\n\\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a+1}{n} \\rfloor + \\dots + \\lfloor \\frac{a+n-1}{n} \\rfloor = a.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\lfloor \\frac{a}{n} \\rfloor = q$; then $a = qn + r$, where $0 \\le r < n$.\n\nDivide the addends into two groups:\n- The first group has $n - r$ terms, where the numerators are $qn + r$ to $qn + (n - 1)$. For these, $\\left\\lfloor \\frac{a + k}{n} \\right\\rfloor = q$.\n- The second group has $r$ terms, where the numerators are $q(n + 1)$ to $q(n + 1) + (r - 1)$. For these, $\\left\\lfloor \\frac{a + k}{n} \\right\\rfloor = q + 1$.\n\nThus,\n\n$$\n\\begin{aligned}\n\\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a+1}{n} \\rfloor + \\dots + \\lfloor \\frac{a+n-1}{n} \\rfloor &= q (n - r) + (q + 1) r \\\\\n&= qn - qr + qr + r \\\\\n&= qn + r = a.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19155,
    "subject": "Mathematics (Olympiad)",
    "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 $O_1O_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $C_1$ and $C_2$ the circumcircles of the triangles $ABC$ and $DEF$, respectively.\n\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 $\\angle PIM = 360^\\circ - 60^\\circ - 2 \\cdot \\frac{180^\\circ - x^\\circ}{2} - x^\\circ = 120^\\circ$.\n\nSince $\\angle PIM = 180^\\circ - \\angle PAM$, we infer that $APIM$ is a cyclic quadrilateral, so $ADIP$ and $ADMI$ are cyclic too.\n\n![](images/RMC_2023_v2_p72_data_71fd95c722.png)\n\nIt follows that $\\angle DAI = \\angle DPI = \\frac{180^\\circ - x^\\circ}{2}$ and $\\angle ADI = \\angle AMI = \\frac{180^\\circ - x^\\circ}{2}$.\n\nSubsequently, $\\angle ADI = \\angle DAI$, so $AI = DI$.\n\nLet $A'$ be the second intersection point of $AI$ and $C_1$ and $D'$ be the second intersection point of $DI$ and $C_2$.\n\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.\n\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": 19156,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Consider a grid with labeled fields. Show that, regardless of the arrangement of labeled fields, it is always possible to select three rows and three columns such that all labeled fields are contained within the union of these selected rows and columns.\n\nb) Provide an example of such a labeling.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider three different cases (these are the only possibilities):\n\n1. There are three rows that contain at least two labeled fields. In this case, select these three rows; at most three labeled fields remain, so select three columns containing the remaining labeled fields.\n\n2. There is one row with at least three and another row with at least two labeled fields. Select these two rows; at most four labeled fields remain, so select a row containing one of the remaining labeled fields and three columns containing the others.\n\n3. There is a row with at least four labeled fields. Select that row; at most five labeled fields remain, so select two rows and three columns containing the remaining labeled fields.\n\nb) One such labeling is shown below:\n\n![](images/Mathematica_competitions_in_Croatia_in_2013_p15_data_d39f8a3c80.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19157,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為圓內接四邊形。已知 $Q, A, B, P$ 依序排在一條直線上，且直線 $AC$ 與圓 $ADQ$ 相切、直線 $BD$ 與圓 $BCP$ 相切。令點 $M, N$ 分別為邊 $BC$ 與 $AD$ 的中點。證明以下三條直線共點：直線 $CD$；圓 $ANQ$ 在點 $A$ 的切線；圓 $BMP$ 在點 $B$ 的切線。\n\n（註：圓 $ADQ$ 指的是過 $A, D, Q$ 三點的圓；其餘類推。）",
    "options": [],
    "answer": "See solution",
    "solution": "解法一：由於 $ABCD$ 有外接圓，有 $\\angle DAQ = \\angle DCB$。又因直線 $AC$ 與圓 $AQD$ 相切，得 $\\angle CBD = \\angle CAD = \\angle AQD$。所以三角形 $ADQ$ 與 $CDB$ 相似（AA）。\n\n令 $R$ 為線段 $CD$ 的中點。利用上面的三角形相似，$N$ 與 $R$ 為對應點。所以有\n$$\\angle QNA = \\angle BRC$$\n\n![](images/2023-TWNIMO-Problems_p51_data_6b0fb822eb.png)\n\n設點 $K$ 為直線 $CD$ 與圓 $ABR$ 的第二個交點（如果 $CD$ 與圓 $ABR$ 交於兩點，則取 $K$ 是異於 $R$ 的另一個交點；如果 $CD$ 與圓 $ABR$ 相切，則取 $K = R$）。則有 $\\angle BAK = \\angle BRC = \\angle QNA$；此等式說明了直線 $AK$ 與圓 $ANC$ 相切。同樣的作法可證明直線 $BK$ 與圓 $BMP$ 相切。證畢。□",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19158,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a sequence of positive numbers $\\{x_n\\}$ satisfies $S_n \\ge 2S_{n-1}$ for $n = 2, 3, \\dots$, where $S_n = x_1 + \\dots + x_n$. Prove that there exists a constant $C > 0$ such that\n$$\nx_n \\ge C \\cdot 2^n, \\quad n = 1, 2, \\dots\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\ge 2$, the condition $S_n \\ge 2S_{n-1}$ is equivalent to\n$$\nx_n \\ge x_1 + \\cdots + x_{n-1}.\n$$\nLet $C = \\frac{1}{4}x_1$. We will prove by induction that\n$$\nx_n \\ge C \\cdot 2^n, \\quad n = 1, 2, \\dots\n$$\nWhen $n = 1$, this is clearly true. For $n = 2$, we have $x_2 \\ge x_1 = C \\cdot 2^2$.\n\nAssume for $n \\ge 3$ that $x_k \\ge C \\cdot 2^k$ for $k = 1, 2, \\dots, n-1$. Then,\n$$\n\\begin{align*}\nx_n &\\ge x_1 + (x_2 + \\cdots + x_{n-1}) \\\\\n&\\ge x_1 + \\left(C \\cdot 2^2 + \\cdots + C \\cdot 2^{n-1}\\right) \\\\\n&= C\\left(2^2 + 2^3 + \\cdots + 2^{n-1}\\right) + x_1.\n\\end{align*}\n$$\nSince $x_1 = C \\cdot 4$, the sum $2^2 + 2^3 + \\cdots + 2^{n-1} = 2^n - 4$, so\n$$\nx_n \\ge C(2^n - 4) + C \\cdot 4 = C \\cdot 2^n.\n$$\nTherefore, the inequality holds for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19159,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there exists a polynomial $f(x)$ of degree $n-1$ such that $f(k)$ is an integer for all integers $k$, $1 \\leq k \\leq n-1$, but $f(0)$ is not an integer, if and only if $n = 1$ or $n$ is a power of a prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Alternative Solution. (I. Bogdanov)\n\nWe claim the answer is $n = p^\\alpha$ for some prime $p$ and nonnegative $\\alpha$.\n\n**Lemma.** 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 $1 \\leq k \\leq n$.\n\n*Proof.* 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 $1 \\leq k \\leq n-1$. 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 $1 \\leq k \\leq n-1$ 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 is, 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$$\n\\prod_{i=1}^{n-1} (p_j^{\\alpha_j} - i) \\equiv (-1)^{n-1}(n-1)! \\not\\equiv 0 \\pmod{p_j^{\\alpha_j}},\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19160,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+ = (0, \\infty)$ be the set of all positive real numbers. Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ and polynomials $g(x)$ with non-negative coefficients and $g(0) = 0$ that satisfy the equality:\n\n$$\nf(f(x) + g(y)) = f(x - y) + 2y\n$$\n\nfor all positive real numbers $x > y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ and the polynomial $g$ with non-negative coefficients and $g(0) = 0$ satisfy the conditions of the problem. For positive reals with $x > y$, we shall write $P(x, y)$ for the relation:\n\n$$\nf(f(x) + g(y)) = f(x - y) + 2y.\n$$\n\n**Step 1.** $f(x) \\ge x$. Assume that this is not true. Since $g(0) = 0$, $g(x) + x$ is injective on positive reals. If $f(x) < x$ for some positive real $x$, then setting $y$ such that $y + g(y) = x - f(x)$ (where obviously $y < x$), we shall get $f(x) + g(y) = x - y$ and by $P(x, y)$, $f(f(x) + g(y)) = f(x - y) + 2y$, we get $2y = 0$, a contradiction.\n\n**Step 2.** $g(x) = cx$ for some non-negative real $c$. We will show $\\deg g \\le 1$ and together with $g(0) = 0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $g(x) \\ge 2x$ for all $x \\ge l$. By Step 1 we get\n\n$$\n\\forall x > y \\ge l : f(x - y) + 2y = f(f(x) + g(y)) \\ge f(x) + g(y) \\ge f(x) + 2y\n$$\n\nand therefore $f(x - y) \\ge f(x)$. We get $f(y) \\ge f(2y) \\ge \\dots \\ge f(ny) \\ge ny$ for all positive integers $n$, which is a contradiction.\n\n**Step 3.** If $c \\ne 0$, then $f(f(x) + y + c^2 + 2) = f(x + 1) + y + 2c$. Indeed by $P(f(x + \\frac{y}{2} + 1) + \\frac{cy}{2} + c, c)$, we get\n\n$$\nf(f(f(x + \\frac{y}{2} + 1) + \\frac{cy}{2} + c) + c^2) = f(f(x + \\frac{y}{2} + 1) + \\frac{cy}{2}) + 2c = f(x + 1) + y + 2c.\n$$\n\nOn the other hand by $P(x + \\frac{y}{2} + 1, \\frac{y}{2} + 1)$, we have:\n\n$$\nf(x) + y + 2 = f\\left(f\\left(x + \\frac{y}{2} + 1\\right) + g\\left(\\frac{y}{2} + 1\\right)\\right) = f\\left(f\\left(x + \\frac{y}{2} + 1\\right) + \\frac{cy}{2} + c\\right).\n$$\n\nSubstituting in the LHS of $P(f(x + \\frac{y}{2} + 1) + \\frac{cy}{2} + c, c)$, we get $f(f(x) + y + 2 + c^2) = f(x + 1) + y + 2c$.\n\n**Step 4.** There is $x_0$, such that $f(x)$ is linear on $(x_0, \\infty)$. If $c \\neq 0$, then by Step 3, fixing $x=1$, we get $f(y + f(1) + 2 + c^2) = y + f(2) + 2c$ which implies that $f$ is linear for $y > f(1) + 2 + c^2$. As for the case $c = 0$, consider $y, z \\in (0, \\infty)$. Pick $x > \\max(y, z)$, then by $P(x, x - y)$ and $P(x, x - z)$ we get:\n\n$$\nf(y) + 2(x - y) = f(f(x)) = f(z) + 2(x - z)\n$$\n\nwhich proves that $f(y) - 2y = f(z) - 2z$ and therefore $f$ is linear on $(0, \\infty)$.\n\n**Step 5.** $g(y) = y$ and $f(x) = x$ on $(x_0, \\infty)$. By Step 4, let $f(x) = ax + b$ on $(x_0, \\infty)$. Since $f$ takes only positive values, $a \\ge 0$. If $a = 0$, then by $P(x + y, y)$ for $y > x_0$ we get:\n\n$$\n2y + f(x) = f(f(x + y) + g(y)) = f(b + cy).\n$$\n\nSince the LHS is not constant, we conclude $c \\neq 0$, but then for $y > x_0/c$, we get that the RHS equals $b$ which is a contradiction.\n\nHence $a > 0$. Now for $x > x_0$ and $x > (x_0 - b)/a$ large enough by $P(x + y, y)$ we get:\n\n$$\nax+b+2y = f(x)+2y = f(f(x+y)+g(y)) = f(ax+ay+b+cy) = a(ax+ay+b+cy)+b.\n$$\n\nComparing the coefficients before $x$, we see $a^2 = a$ and since $a \\neq 0$, $a = 1$. Now $2b = b$ and thus $b = 0$. Finally, equalising the coefficients before $y$, we conclude $2 = 1 + c$ and therefore $c = 1$.\n\nNow we know that $f(x) = x$ on $(x_0, \\infty)$ and $g(y) = y$. Let $y > x_0$. Then by $P(x + y, x)$ we conclude:\n\n$$\nf(x) + 2y = f(f(x + y) + g(y)) = f(x + y + y) = x + 2y.\n$$\n\nTherefore $f(x) = x$ for every $x$. Conversely, it is straightforward that $f(x) = x$ and $g(y) = y$ do indeed satisfy the conditions of the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19161,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive integers satisfying certain conditions involving their parities and Euler's totient function $\\varphi$. Consider the following two cases:\n\n1. If $c$ and $d$ have different parities, prove that $4\\mid\\mid (a^2 + b^2)$ and show that no solutions exist for $ab = 4cd \\geq 16$.\n\n2. If $c$ and $d$ have the same parity (both odd and $>2$), analyze the equation:\n\n$$\n7\\varphi^2(c) - \\varphi(cd) + 11\\varphi^2(d) = 2(c^2 + d^2)\n$$\n\nand determine all possible solutions for $c$ and $d$ under the given constraints.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze both cases as follows:\n\n**Case 1:** If $c$ and $d$ have different parities, then $4\\mid\\mid (a^2 + b^2)$. Since $ab = 4cd \\geq 16$, but $ab = 8$ is the only possibility from the divisibility conditions, this is a contradiction. Thus, no solutions exist in this case.\n\n**Case 2:** If $c$ and $d$ have the same parity, they must be odd and $>2$. Consider the equation:\n\n$$\n7\\varphi^2(c) - \\varphi(cd) + 11\\varphi^2(d) = 2(c^2 + d^2)\n$$\n\nBy bounding $\\varphi(c)$ and $\\varphi(d)$, and considering the possible factorizations of $cd$ (prime powers, two or three distinct primes), we find that in all cases, the left-hand side exceeds the right-hand side, or the divisibility conditions cannot be met. The only exception is when $c = 15$, $d = 3$, which yields $a = 30$, $b = 6$ as a solution. All other cases lead to contradictions, so $(a, b) = (30, 6)$ is the only solution.",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 19162,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist constants $C > 0$ and $\\alpha > \\frac{1}{2}$ such that for any positive integer $n$, there exists a subset $A \\subseteq \\{1, 2, \\dots, n\\}$ with $|A| \\geq C n^{\\alpha}$, and the difference of any two elements of $A$ is not a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\geq 25$, let $5^{2t} \\leq n < 5^{2t+2}$ for some $t \\in \\mathbb{N}_+$. Define\n\n$$\nA = \\{(\\alpha_{2t}, \\dots, \\alpha_1)_5 \\mid \\alpha_{2i} \\in \\{0, 1, 2, 3, 4\\},\\ i = 1, \\dots, t;\\ \\alpha_{2i-1} \\in \\{1, 3\\},\\ i = 1, \\dots, t\\}\n$$\n\nwhere $(\\alpha_{2t}, \\dots, \\alpha_1)_5$ denotes the base-5 representation of $m = 5^{2t-1}\\alpha_{2t} + \\dots + 5\\alpha_2 + \\alpha_1$. Clearly, $A \\subseteq \\{1, 2, \\dots, n\\}$.\n\nFor $u_1, u_2 \\in A$, with $u_1 = (a_{2t}, \\dots, a_1)$ and $u_2 = (b_{2t}, \\dots, b_1)$, $u_1 > u_2$, consider the first digit $s$ where $a_s \\neq b_s$ (i.e., $a_1 = b_1, \\dots, a_{s-1} = b_{s-1}$, $a_s \\neq b_s$). Then\n\n$$\nu_1 - u_2 = (a_{2t} - b_{2t})5^{2t-1} + \\dots + (a_s - b_s)5^{s-1}.$$\n\nIf $s$ is even, then $5^{s-1}$ divides $u_1 - u_2$ and $a_s - b_s \\neq 0$ is between $-4$ and $4$, so $u_1 - u_2$ cannot be a perfect square.\n\nIf $s$ is odd, then $\\frac{u_1 - u_2}{5^{s-1}}$ is an integer. Suppose $u_1 - u_2$ is a perfect square. Since $s-1$ is even, $\\frac{u_1 - u_2}{5^{s-1}}$ must also be a perfect square. However, with $a_s \\neq b_s$ and $\\{a_s, b_s\\} = \\{1, 3\\}$, $\\frac{u_1 - u_2}{5^{s-1}} \\equiv 2 \\text{ or } 3 \\pmod{5}$, which cannot be a perfect square. Thus, $u_1 - u_2$ is never a perfect square.\n\nSince $|A| = 10^t$, take $\\alpha = \\log_{25} 10 > \\frac{1}{2}$, and\n\n$$n^{\\alpha} < 5^{(2t+2) \\log_{25} 10} = 10^{t+1} = 10|A|.$$\n\nSo $C = \\frac{1}{24}$, $\\alpha = \\log_{25} 10$ ($\\alpha \\in (0, 1)$) suffices.\n\nFor $n \\leq 24$, take $A = \\{1\\}$; then $|A| \\geq \\frac{1}{24} n \\geq \\frac{1}{24} n^{\\alpha}$ also holds.\n\nIn conclusion, for any $n \\in \\mathbb{N}_+$, we can find such $A$ with $|A| \\geq C n^{\\alpha}$. $\\square$\n\n**Remark:** One can also consider residues modulo 16 and take\n\n$$\nA = \\{(\\alpha_t, \\dots, \\alpha_1)_{16} \\mid \\alpha_i \\in \\{2, 5, 7, 13, 15\\},\\ i = 1, \\dots, t\\}. \\quad \\square\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19163,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a positive integer $m$ and a rectangle board of size $m \\times 2018$ which consists of $m$ rows and $2018$ columns. We write $0$ or $1$ into some cells of the board (only one number in a cell) and the rest are left empty. The board is *complete* if for an arbitrary binary sequence $S$ of length $2018$, we can always choose a row of the board to write some more $0$s, $1$s so that $2018$ numbers on that row (without any change in order) form the sequence $S$ (in case the board already had a row which is $S$, it is also complete). The board is *minimal* if it is complete and if we omit any row in the board, it is no longer complete.\n\n(a) If $0 \\leq k \\leq 2018$, prove that there exists a minimal $2^k \\times 2018$ board such that there are exactly $k$ columns, each of which consists of both $0$ and $1$ (there are possibly some empty cells on those columns).\n\n(b) A minimal $m \\times 2018$ board which has exactly $k$ columns, each of which consists of both $0$ and $1$, is given. Prove that $m \\leq 2^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, consider an empty rectangle board of size $2^k \\times 2018$ and $2^k$ binary sequences of length $k$. We write those sequences to the left of the board so that each row consists of a sequence. Thus, the rest to the right of the board are $2018 - k$ empty columns. It is obvious that each of the first $k$ columns of the board (counting from the left) has exactly $2^{k-1}$ $0$s and $2^{k-1}$ $1$s. This board satisfies the given condition. We will prove that it is minimal.\n\n![](images/VN_IMO_Booklet_2018_Final_p30_data_9ff1871b07.png)\n\nFor an arbitrary binary sequence $s = a_1a_2\\dots a_{2018}$, we consider its subsequence $s' = a_1a_2\\dots a_k$. It is clear that $s'$ appears at the beginning of some row in the board, so if we continue writing $a_{k+1}, a_{k+2}, \\dots, a_{2018}$ into the empty cells on that row, we will get $s$. Furthermore, there is exactly one row that contains $s'$, so if we omit that row, we cannot form $s$ from any other row. Therefore, the above board is minimal.\n\n(b) Suppose that each of the first $k$ columns of the board consists of both $0$ and $1$. We will prove the following important remark.\n\n**Remark.** All the cells in the last $2018 - k$ columns (in other words, $2018 - k$ columns to the right) of the board are empty.\n\n*Proof.* Consider an arbitrary binary sequence $s$ of length $k$ and suppose $A_s$ be the set of rows with the property: the first $k$ cells of each row (counting from the left) form $s$. We will prove that there exists an element in $A_s$ of which all last $2018 - k$ cells are empty.\n\nConsider the ($k + 1$)-th cell of each row in $A_s$. It is clear that those cells belong to the ($k + 1$)-th column of the board which do not simultaneously have $0$ and $1$.\n\nIf no cell of this column is empty, we can suppose that all numbers in the column are $0$s. Then the binary sequence of form $s$ concatenation to $1$ cannot be formed by any row, which contradicts the complete property of the board. Thus, there exists a subset $A'$ of $A$ such that the ($k + 1$)-th cell of each row in $A'$ is empty.\n\nWe continue considering the ($k + 2$)-th cell and we can similarly prove that there exists a subset $A_s''$ of $A_s'$ such that the ($k + 2$)-th cell of each row in $A_s''$ is empty. Following the same pattern to the last column, we will have a row of which all cells from the ($k + 1$)-th position to the last position are empty.\n\nTherefore, for any binary sequence $s$ of length $k$, we can always find a row of which $2018 - k$ last cells are empty. Note that these rows are not necessarily distinct since a row can form many binary sequences. Let $A$ be the set of all such rows.\n\nBy the definition of $A$, it is clear that any binary sequence of length $2018$ can be formed by an element of $A$. It is also obvious that every row in the board belongs to $A$, otherwise we can omit that row and the board is still complete, which contradicts the minimal property of the board. Thus $A$ is also the set of all",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19164,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Let $a$, $x$, $y$ be positive integers. Prove that if $x \\neq y$, then\n$$\na x + \\gcd(a, x) + \\text{lcm}(a, x) \\neq a y + \\gcd(a, y) + \\text{lcm}(a, y).\n$$\n\n(b) Show that there are no two positive integers $a$ and $b$ such that\n$$\na b + \\gcd(a, b) + \\text{lcm}(a, b) = 2014.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Suppose that\n$$\na x + \\gcd(a, x) + \\text{lcm}(a, x) = a y + \\gcd(a, y) + \\text{lcm}(a, y)\n$$\nfor certain positive integers $a$, $x$, $y$. It follows that\n$$\n\\gcd(a, a x + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, a y + \\gcd(a, y) + \\text{lcm}(a, y)).\n$$\nSince $a$ divides both $a x$ and $\\text{lcm}(a, x)$, we have\n$$\n\\gcd(a, a x + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, \\gcd(a, x)) = \\gcd(a, x)\n$$\nand likewise\n$$\n\\gcd(a, a y + \\gcd(a, y) + \\text{lcm}(a, y)) = \\gcd(a, \\gcd(a, y)) = \\gcd(a, y).\n$$\nTherefore, we must have $\\gcd(a, x) = \\gcd(a, y) = d$ for some positive integer $d$. Since $\\text{lcm}(a, x) = a x / \\gcd(a, x)$ and $\\text{lcm}(a, y) = a y / \\gcd(a, y)$, this gives us\n$$\na x + d + \\frac{a x}{d} = a y + d + \\frac{a y}{d},\n$$\nso\n$$\na x \\left(1 + \\frac{1}{d}\\right) = a y \\left(1 + \\frac{1}{d}\\right),\n$$\nwhich implies $x = y$. This proves the first statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19165,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a circumscribed hexagon. Let $AB \\cap CD = K$, $CD \\cap EF = L$, $DE \\cap AF = M$, and $AF \\cap BC = N$. Prove that the lines $KM$, $LN$, and $BE$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circle $\\omega$ inscribed in the hexagon $ABCDEF$ be tangent to the sides $AB, BC, CD, DE, EF, FA$ at the points $A_4, A_3, A_2, A_1, A_6, A_5$ respectively.\n\n![](images/MNG_ABooklet_2015_p17_data_8ba46f6a32.png)\n\n1. Let $A_6A_2$ be the polar of $L$ with respect to $\\omega$, and $A_3A_5$ be the polar of $N$ with respect to $\\omega$.\n\nLet $l_X$ denote the polar of point $X$. If $A_6A_2 \\cap A_3A_5 = X$, then $X \\in l_N$ and $X \\in l_L$. By a known result from projective geometry, if $X \\in l_Y$, then $Y \\in l_X$. Thus, the polar of $X$ is $l_X = LN$.\n\n2. Let $A_5A_1$ and $A_4A_2$ be the polars of $M$ and $K$ respectively. Let $A_5A_1 \\cap A_4A_2 = Y$. The polar of $Y$ is $l_Y = MK$.\n\n3. Let $A_4A_3$ and $A_6A_1$ be the polars of $B$ and $E$ respectively. Let $A_4A_3 \\cap A_6A_1 = Z$. The polar of $Z$ is $l_Z = BE$.\n\nBy Pascal's theorem, the points $A_6A_2 \\cap A_3A_5 = X$, $A_2A_4 \\cap A_5A_1 = Y$, and $A_4A_3 \\cap A_1A_6 = Z$ are collinear. Suppose $X, Y, Z$ lie on the line $m$. Then the pole of $m$ lies on the lines $l_X, l_Y, l_Z$. Therefore, the lines $l_X, l_Y$, and $l_Z$ are concurrent. Thus, the lines $LN$, $MK$, and $BE$ are concurrent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19166,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k_1$, $k_2$, and $k_3$ be three circles with centers $O_1$, $O_2$, and $O_3$ respectively, such that none of the centers lies inside either of the other two circles. The circles $k_1$ and $k_2$ intersect at $A$ and $P$; $k_1$ and $k_3$ intersect at $C$ and $P$; and $k_2$ and $k_3$ intersect at $B$ and $P$.\n\nLet $X$ be a point on $k_1$ such that the intersection of the line $XA$ with the circle $k_2$ is $Y$, and the intersection of the line $XC$ with $k_3$ is $Z$, with $Y$ belonging neither inside $k_1$ nor inside $k_3$, and $Z$ belonging neither inside $k_1$ nor inside $k_2$.\n\n**a)** Prove that the triangles $XYZ$ and $O_1O_2O_3$ are similar.\n\n**b)** Prove that the area of triangle $XYZ$ is not greater than four times the area of triangle $O_1O_2O_3$. Is the maximum attainable?",
    "options": [],
    "answer": "See solution",
    "solution": "We will first show that the points $Y$, $B$, and $Z$ are collinear. Since the quadrilateral $BYAP$ is inscribed, we have $\\angle PBY = \\angle PAX$. Since the quadrilateral $AXCP$ is inscribed, we have $\\angle PAX = \\angle PCZ$. Since the quadrilateral $CPBZ$ is inscribed, we obtain $\\angle PBZ + \\angle PCZ = 180^\\circ$. Therefore, $\\angle YBZ = \\angle YBP + \\angle PBZ = 180^\\circ$.\n\nLet us notice that $\\angle CO_1O_3 = \\angle PO_1O_3$ and $\\angle AO_1O_2 = \\angle PO_1O_2$, from which it follows that $\\angle O_2O_1O_3 = \\frac{1}{2}\\angle AO_1C = \\angle AXC$. Similarly, $\\angle O_1O_2O_3 = \\angle AYB$ and $\\angle O_1O_3O_2 = \\angle CZB$. It follows that $\\triangle XYZ \\sim \\triangle O_1O_2O_3$, which proves part (a).\n\n![](images/Macedonia_2014_p9_data_a9d356bb9c.png)\n\nLet the line $X_1Y_1$ be parallel to $O_1O_2$ and pass through $A$, where $X_1$ lies on $k_1$ and $Y_1$ lies on $k_2$. Let $Z_1$ be the intersection of the line $X_1C$ with the circle $k_3$. From the above, the points $Y_1$, $B$, and $Z_1$ are collinear and $\\triangle X_1Y_1Z_1 \\sim \\triangle O_1O_2O_3$. Furthermore, $\\angle PXA = \\angle PX_1A$ and $\\angle PYA = \\angle PY_1A$. Therefore, $\\triangle PXY \\sim \\triangle PX_1Y_1$. Let $PT$ be the altitude dropped from the vertex $P$ to the side $XY$. $PA$ is the altitude of the triangle $PX_1Y_1$. Since $PA$ is a hypotenuse in the right-angled triangle $PAT$, we get $\\overline{PT} \\leq \\overline{PA}$. Therefore, $P_{PXY} \\leq P_{PX_1Y_1}$ and analogously $P_{PYZ} \\leq P_{PY_1Z_1}$ and $P_{PXZ} \\leq P_{PX_1Z_1}$. From this, we get $P_{XYZ} \\leq P_{X_1Y_1Z_1}$. The points $P$, $O_1$, and $X_1$ are collinear since $\\angle PAX_1 = 90^\\circ$. Similarly, $P$, $O_2$, and $Y_1$ are collinear and $P$, $O_3$, and $Z_1$ are collinear. We get that $O_1O_2$, $O_1O_3$, and $O_2O_3$ are midsegments in the triangles $X_1Y_1P$, $X_1Z_1P$, and $Y_1Z_1P$ respectively, and so $P_{X_1Y_1Z_1} = 4P_{O_1O_2O_3}$. This gives us the required inequality. Equality is attained when the points $X$ and $X_1$ coincide, and with that the points $Y$ and $Y_1$ as well as the points $Z$ and $Z_1$ coincide.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19167,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a line passing through the circumcentre $O$ of $\\triangle ABC$ intersects $AB$ and $AC$ at points $M$ and $N$, respectively. Let $E$ and $F$ be the midpoints of $BN$ and $CM$, respectively. Prove that $\\angle EOF = \\angle A$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_110b1af3c1.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_7979a1dd67.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_59069c8538.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the above conclusion is true for any triangle.\n\nIf $\\triangle ABC$ is right-angled, the conclusion is obvious. In fact, see the figure where $\\angle ABC = 90^\\circ$. So, the circumcentre $O$ is the midpoint of $AC$, $OA = OB$, and $N = O$. Since $F$ is the midpoint of $CM$, we see that the median line $OF \\parallel AM$. Hence $\\angle EOF = \\angle OBA = \\angle OAB = \\angle A$.\n\nIf $\\triangle ABC$ is not right-angled, see the following figures:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_110b1af3c1.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_7979a1dd67.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p293_data_59069c8538.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19168,
    "subject": "Mathematics (Olympiad)",
    "question": "A mathematical frog jumps along the number line. The frog starts at $1$, and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n + 1$ or to $n + 2^{m_n+1}$, where $2^{m_n}$ is the largest power of $2$ that is a factor of $n$.\n\nShow that if $k \\geq 2$ is a positive integer and $i$ is a nonnegative integer, then the minimum number of jumps needed to reach $2^i k$ is greater than the minimum number of jumps needed to reach $2^i$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $i \\geq 0$ and $k \\geq 1$, let $x_{i,k}$ denote the minimum number of jumps needed to reach the integer $n_{i,k} = 2^i k$. We must prove that\n\n$$\nx_{i,k} > x_{i,1}\n$$\n\nfor all $i \\geq 0$ and $k \\geq 2$. We prove this using the method of descent.\n\nFirst, note that the inequality holds for $i = 0$ and all $k \\geq 2$, because it takes $0$ jumps to reach the starting value $n_{0,1} = 1$, and at least one jump to reach $n_{0,k} = k \\geq 2$. Now assume that the inequality is not true for all choices of $i$ and $k$. Let $i_0$ be the minimal value of $i$ for which the inequality fails for some $k$, and let $k_0$ be the minimal value of $k > 1$ for which $x_{i_0,k} \\leq x_{i_0,1}$. Then it must be the case that $i_0 \\geq 1$ and $k_0 \\geq 2$.\n\nLet $J_{i_0,k_0}$ be a shortest sequence of $x_{i_0,k_0} + 1$ integers that the frog occupies in jumping from $1$ to $2^{i_0}k_0$. The length of each jump, that is, the difference between consecutive integers in $J_{i_0,k_0}$, is either $1$ or a positive integer power of $2$. The sequence $J_{i_0,k_0}$ cannot contain $2^{i_0}$ because it takes more jumps to reach $2^{i_0}k_0$ than it does to reach $2^{i_0}$. Let $2^{M+1}$, $M \\geq 0$, be the length of the longest jump made in generating $J_{i_0,k_0}$. Such a jump can only be made from a number that is divisible by $2^M$ (and by no higher power of $2$). Thus we must have $M < i_0$, since otherwise a number divisible by $2^{i_0}$ is visited before $2^{i_0}k_0$ is reached, contradicting the definition of $k_0$.\n\nLet $2^{m+1}$ be the length of the jump when the frog jumps over $2^{i_0}$. If this jump starts at $2^m(2t - 1)$ for some positive integer $t$, then it will end at $2^m(2t - 1) + 2^{m+1} = 2^m(2t + 1)$. Since it goes over $2^{i_0}$, we see $2^m(2t - 1) < 2^{i_0} < 2^m(2t + 1)$ or $(2^{i_0-m} - 1)/2 < t < (2^{i_0-m} + 1)/2$. Thus $t = 2^{i_0-m-1}$ and the jump over $2^{i_0}$ is from $2^m(2^{i_0-m} - 1) = 2^{i_0} - 2^m$ to $2^m(2^{i_0-m} + 1) = 2^{i_0} + 2^m$.\n\nConsidering the jumps that generate $J_{i_0,k_0}$, let $N_1$ be the number of jumps from $1$ to $2^{i_0} + 2^m$, and let $N_2$ be the number of jumps from $2^{i_0} + 2^m$ to $2^{i_0}k_0$. By definition of $i_0$, it follows that $2^m$ can be reached from $1$ in less than $N_1$ jumps. On the other hand, because $m < i_0$, the number $2^{i_0}(k_0 - 1)$ can be reached from $2^m$ in exactly $N_2$ jumps by using the same jump length sequence as in jumping from $2^m + 2^{i_0}$ to $2^{i_0}k_0 = 2^{i_0}(k_0 - 1) + 2^i$. The key point here is that the shift by $2^{i_0}$ does not affect any of the divisibility conditions needed to make jumps of the same length. In particular, with the exception of the last entry, $2^{i_0}k_0$, all of the elements of $J_{i_0,k_0}$ are of the form $2^p(2t + 1)$ with $p < i_0$, again because of the definition of $k_0$. Because $2^p(2t + 1) - 2^{i_0} = 2^p(2t - 2^{i_0-p} + 1)$ and the number $2t + 2^{i_0-p} + 1$ is odd, a jump of size $2^{p+1}$ can be made from $2^p(2t + 1) - 2^{i_0}$ just as it can be made from $2^p(2t + 1)$.\n\nThus, the frog can reach $2^m$ from $1$ in less than $N_1$ jumps, and can then reach $2^{i_0}(k_0 - 1)$ from $2^m$ in $N_2$ jumps. Hence, the frog can reach $2^{i_0}(k_0 - 1)$ from $1$ in less than $N_1 + N_2$ jumps, that is, in fewer jumps than needed to get to $2^{i_0}k_0$ and hence in fewer jumps than required to get to $2^{i_0}$. This contradicts the definition of $k_0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19169,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a simple polynomial function be a polynomial function $P(x)$ whose coefficients belong to the set $\\{-1, 0, 1\\}$. Let $n$ be a positive integer, $n > 1$. Find the smallest possible number of non-zero coefficients in a simple polynomial function of $n$th order whose values at all integral arguments are divisible by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "A single non-zero coefficient is not sufficient for any $n > 1$ as the only simple polynomial functions with a single non-zero coefficient are $P(x) = x^n$ and $P(x) = -x^n$, but in both cases $n \\nmid P(1)$. \n\nLet us show that the values of the polynomial function $P_n(x) = x^n - x^{n-\\varphi(n)}$ at all integral arguments are divisible by $n$. (Here $\\varphi$ is the Euler's totient function.) This shows that having 2 non-zero coefficients is sufficient.\n\nLet $k$ be an integer. Let the canonical form of $n$ be $p_1^{\\alpha_1} \\cdots p_m^{\\alpha_m}$ and let us assume without loss of generality that $k$ is divisible by primes $p_1, \\dots, p_l$ and is not divisible by primes $p_{l+1}, \\dots, p_m$. Define $u = p_1^{\\alpha_1} \\cdots p_l^{\\alpha_l}$ and $v = p_{l+1}^{\\alpha_{l+1}} \\cdots p_m^{\\alpha_m}$. Let us now show that $u \\mid k^{n-\\varphi(n)}$ and $v \\mid k^{\\varphi(n)} - 1$. Having $uv = n$, we can conclude that $n \\mid P_n(k)$ as \n$$\nP_n(k) = k^n - k^{n-\\varphi(n)} = k^{n-\\varphi(n)}(k^{\\varphi(n)} - 1).\n$$\n\nTo prove that $u \\mid k^{n-\\varphi(n)}$, it is sufficient to prove for all $i = 1, \\dots, l$ that $p_i^{\\alpha_i} \\mid k^{n-\\varphi(n)}$. It is sufficient to prove that $\\alpha_i \\le n - \\varphi(n)$, as by the assumption $p_i \\mid k$. Inequality $\\alpha_i \\le n - \\varphi(n)$ holds as $p_i$, $p_i^2, \\dots, p_i^{\\alpha_i}$ are $\\alpha_i$ positive integers which are not greater than $n$ and not coprime with $n$.\n\nTo prove the statement $v \\mid k^{\\varphi(n)} - 1$, we derive from Euler's theorem that $v \\mid k^{\\varphi(v)} - 1$ as $k$ and $v$ are coprime. Also, $u$ and $v$ are coprime, therefore, $\\varphi(n) = \\varphi(uv) = \\varphi(u)\\varphi(v)$ from which $k^{\\varphi(v)} - 1 \\mid k^{\\varphi(n)} - 1$. Consequently, $v \\mid k^{\\varphi(n)} - 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19170,
    "subject": "Mathematics (Olympiad)",
    "question": "Heights $BB_1$ and $CC_1$ of acute triangle $ABC$ intersect at point $H$. Points $B_2$ and $C_2$ are located on segments $BH$ and $CH$, respectively, such that $BB_2 = B_1H$ and $CC_2 = C_1H$. The circumcircle of triangle $B_2HC_2$ intersects the circumcircle of triangle $ABC$ at points $D$ and $E$. Prove that triangle $DEH$ is right.\n\n![](images/bw18shortlist_p31_data_64bb1a0eff.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Despite the logical symmetry of the figure, the right angle in triangle $DEH$ is not at $H$ but at either $D$ or $E$.\n\nLet $w$ be the circumcircle of triangle $B_2HC_2$. The perpendicular bisector of segment $C_2H$ is also the perpendicular bisector of $CC_1$, so it passes through the midpoint $X$ of side $BC$. Similarly, the perpendicular bisector of $B_2H$ passes through $X$. Therefore, $X$ is the center of circle $w$.\n\nIt is well known that the point symmetric to the orthocenter $H$ with respect to side $BC$ lies on the circumcircle of triangle $ABC$. The distance from this point to $X$ equals $XH$ by symmetry, so this point also lies on $w$ and thus coincides with $D$ or $E$; without loss of generality, let it be $D$. Thus, $DH \\perp BC$.\n\nFinally, the centers of $w$ and the circumcircle of $ABC$ both lie on the perpendicular bisector of $BC$, so their common chord $DE$ is parallel to $BC$. Therefore, $\\angle HDE = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19171,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Choose an arbitrary point $M$ on side $BC$, different from $D$. Let $N$ be the intersection of lines $AM$ and $EF$. The line $ON$ meets the circumcircle of triangle $ODM$ again at point $P$. Prove that the reflection of $M$ with respect to the midpoint of segment $DP$ lies on the nine-point circle of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The lines $DO$, $EO$, and $FO$ are the perpendicular bisectors of sides $BC$, $CA$, and $AB$, respectively. Thus, $[OM]$ is a diameter of the circumcircle of triangle $ODM$, and $MP \\perp ON$. The point $O$ is the orthocenter of triangle $DEF$ (see the figure below).\n\nLet $O_1$ be the circumcenter of triangle $DEF$, and let $H$ be the point diametrically opposite to $D$ on its circumcircle. The circumcircle of $DEF$ is the nine-point circle of triangle $ABC$. Thus, $EH \\perp DE$, $FH \\perp FD$, $ED \\parallel AF$, and $DF \\parallel AE$. Therefore, $H$ is the orthocenter of triangle $AEF$.\n\n![](images/short_list_BMO_2017_p12_data_fe502fc347.png)\n\nLet $AD \\cap EF = \\{I\\}$, and let $R$ be the reflection of $N$ with respect to $I$, i.e., $R \\in EF$ and $NI = RI$. The point $I$ is the center of symmetry of parallelogram $AEDF$, so $I$ is the midpoint of $[OH]$, and the quadrilaterals $AEDF$, $AND$, and $HNOR$ are all parallelograms.\n\nLet $Q$ be the reflection of $M$ with respect to the midpoint of $DP$. Then $PQDM$ and $MNRD$ are parallelograms, which implies $PQRN$ is a parallelogram. Thus, $NO \\parallel HR$, $NP \\parallel RQ$, so $H$, $R$, and $Q$ are collinear. We obtain $m(\\angle DQH) = 90^\\circ$, i.e., $Q$ lies on the nine-point circle of triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19172,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the identity:\n\n$$\n\\sum_{j=0}^{n} \\left[ \\binom{3n+2-j}{j} 2^j - \\binom{3n+1-j}{j-1} 2^{j-1} \\right] = 2^{3n},\n$$\n\nwhere the second term on the left side is to be interpreted as $0$ for $j = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the $j$-th term in the sum. It equals\n\n(the coefficient of $x^j$ in $(1+2x)^{3n+2-j}$) $-$ (the coefficient of $x^{j-1}$ in $(1+2x)^{3n+1-j}$).\n\nThis is\n\n$$\n\\text{the coefficient of } x^{3n+2} \\text{ in } x^{3n+2-j} \\left[ (1+2x)^{3n+2-j} - x(1+2x)^{3n+1-j} \\right].\n$$\n\nSimplifying, this equals the coefficient of $x^{3n+2}$ in $(x+x^2)(x+2x^2)^{3n+1-j}$.\n\nThus, the sum is the coefficient of $x^{3n+2}$ in\n\n$$\nx(1+x) \\left[ (x+2x^2)^{3n+1} + (x+2x^2)^{3n} + \\dots + (x+2x^2)^{2n+1} \\right].\n$$\n\nThis reduces to\n\n$$\nx^{2n+2}(1+2x)^{2n+1} \\frac{1-(x+2x^2)^{n+1}}{1-2x}.\n$$\n\nThe second term does not contribute to the coefficient of $x^{3n+2}$, so this coefficient is the coefficient of $x^n$ in\n\n$$\n\\frac{(1+2x)^{2n+1}}{1-2x}.\n$$\n\nExpanding $1/(1-2x)$, this is\n\n$$\n2^n \\left( 1 + \\binom{2n+1}{1} + \\binom{2n+1}{2} + \\dots + \\binom{2n+1}{n} \\right).\n$$\n\nThis equals\n\n$$\n2^n \\frac{1}{2} \\left( \\sum_{j=0}^{2n+1} \\binom{2n+1}{j} \\right),\n$$\n\nwhich reduces to $2^{3n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19173,
    "subject": "Mathematics (Olympiad)",
    "question": "In an $m \\times n$ grid (with $m+1$ horizontal lines and $n+1$ vertical lines), it is possible to add diagonals to some unit squares, that is, each unit square $\\Box$ becomes one of $\\Box$, $\\checkmark$, or $\\square$, such that the grid becomes an Eulerian cycle (there exists a path that visits every edge exactly once, and that starts and ends on the same vertex). Find all such $m \\times n$ grids.",
    "options": [],
    "answer": "See solution",
    "solution": "An $m \\times n$ grid can be made into an Eulerian cycle if and only if $m = n$.\n\nWe give two solutions as follows.\n\n**Solution 1**\n\nIf $m = n$, add a diagonal (top left to bottom right) to every unit square not in the main diagonal of the grid. It is easy to verify the new graph is an Eulerian cycle.\n\nIt remains to prove that when $m \\neq n$, the grid cannot be made into an Eulerian cycle. Let $n > m$ and suppose it is possible. Notice in the original grid, every lattice point has an even degree of edges except for those on the four sides but not at the corners. Let us call them even and odd points. There are $m-1$ or $n-1$ odd points on each side of the grid. To make the grid into an Eulerian cycle, the diagonals must turn each odd point into an even point while maintaining all the even points. We focus on all the diagonals. If three or four diagonals meet at a lattice point, separate them as in the figure below.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p244_data_58d2fd3172.png)\n\nThis divides the diagonals into several cycles and non-intersecting paths. Remove all the cycles. The remaining paths of diagonals turn all the odd points into even points, and hence a path must start at an odd point and end at another odd point. Colour all the lattice points alternately black and white. Evidently, a path can only pass through lattice points of the same colour. If a path starts and ends on the same side of the grid, then there are an odd number of odd points on this side between the two ends, and they cannot be connected by paths. If a path starts and ends on opposite sides of the grid, say from row $a$ and column $1$, to row $b$ and column $n+1$. Then $(a, 1)$ and $(b, n+1)$ are of the same colour, $a+1$ and $b+n+1$ are of the same parity, which indicates that there are $(a-2) + (n-1) + (b-2) = a+n+b-5$, an odd number of odd points between the two ends, which cannot be connected by paths. The above argument shows that a path must start and end at neighbouring sides, which requires the $2(m-1)$ odd points on two opposite sides match the $2(n-1)$ odd points on the other two sides. Hence, $2(m-1) = 2(n-1)$, or $m=n$. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19174,
    "subject": "Mathematics (Olympiad)",
    "question": "For sets $S$ and $T$ consisting of positive real numbers, define $S + T = \\{s + t \\mid s \\in S,\\ t \\in T\\}$ and $\\frac{1}{S} = \\left\\{\\frac{1}{s} \\mid s \\in S\\right\\}$. Define the sets $A_1, A_2, A_3, \\dots$ recursively by $A_1 = \\{1\\}$ and\n\n$$\nA_n = \\bigcup_{i=1}^{n-1} \\left( (A_i + A_{n-i}) \\cup \\left( \\frac{1}{\\frac{1}{A_i} + \\frac{1}{A_{n-i}}} \\right) \\right)\n$$\n\nfor all integers $n \\ge 1$. Prove that for all integers $n \\ge 1$ we have\n\n$$\n2^{n-1} \\le |A_n| \\le 8^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We start with a lemma.\n\n*Lemma.* For any $n$, $x \\in A_n$ implies $\\frac{1}{x} \\in A_n$.\n\n*Proof.* Induction. Case $n = 1$ is clear. Now, if $x = a_i + a_{n-i} \\in A_i + A_{n-i} \\subset A_n$, then $\\frac{1}{a_i} \\in A_i$ and $\\frac{1}{a_{n-i}} \\in A_{n-i}$, and thus\n\n$$\n\\frac{1}{x} \\in \\frac{1}{\\frac{1}{A_i} + \\frac{1}{A_{n-i}}} \\subset A_n.\n$$\n\nSimilarly, if $x = \\left(\\frac{1}{a_i} + \\frac{1}{a_{n-i}}\\right)^{-1}$, then\n\n$$\n\\frac{1}{x} = \\frac{1}{a_i} + \\frac{1}{a_{n-i}} \\in A_i + A_{n-i} \\subset A_n. \\quad \\square\n$$\n\n**Lower bound.**\n\nWe now prove that $|A_{n+1}| \\ge 2|A_n|$ for all $n$, which proves the lower bound. Let $a$ denote the number of elements in $A_n$ which are larger than or equal to $1$ and let $b$ denote the number of elements in $A_n$ which are strictly larger than $1$. Clearly $a = b$ if $1 \\notin A_n$ and otherwise $a = b + 1$.\n\nFirst, note that $A_1 + A_n = \\{1\\} + A_n \\subset A_{n+1}$, so $A_{n+1}$ contains at least $a$ elements which are $\\ge 2$. Also, for any $a_n \\in A_n$ with $a_n > 1$ we have\n\n$$\n\\frac{1}{2} < a_{n+1} := \\frac{1}{\\frac{1}{1} + \\frac{1}{a_n}} < 1\n$$\n\nand $a_{n+1} \\in A_n$. Thus, $A_{n+1}$ contains at least $b$ elements in $(\\frac{1}{2}, 1)$ and thus, by the lemma, at least $b$ elements in $(1, 2)$.\n\nTherefore, $A_{n+1}$ has at least $a + b$ elements greater than $1$. By the lemma, $A_{n+1}$ thus has at least $2(a + b)$ elements. If $a = b$, then $2|A_n| = 4a = 2(a + b) \\le |A_{n+1}|$ and if $a = b + 1$, then $2|A_n| = 2(2b + 1) = 2(a + b) \\le |A_{n+1}|$.\n\n**Upper bound.**\n\nWe then prove the upper bound. Define $s_n := |A_n|$, let $c_n := \\frac{1}{n+1} \\binom{2n}{n}$ be the $n$-th Catalan number, and $b_n := 2^n c_{n-1}$. Then\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} b_i b_{n-i} &= \\sum_{i=1}^{n-1} 2^i \\cdot c_{i-1} \\cdot 2^{n-i} \\cdot c_{n-2-(i-1)} \\\\\n&= 2^n \\cdot \\sum_{i=0}^{n-2} c_i c_{(n-2)-i} = 2^n \\cdot c_{n-1} = b_n,\n\\end{aligned}\n$$\n\nwhere we used the well-known recursion formula for the Catalan numbers.\n\nWe prove that $s_n < b_n$ for all $n$, which certainly is enough since $b_n = \\frac{2^n}{n} \\binom{2(n-1)}{n-1} < 2^n \\cdot 2^{2n} = 8^n$. The proof is by induction. The cases $n \\le 4$ may be checked by hand, as we have\n\n$$\ns_1 = 1 < 2 = b_1, \\quad s_2 = 2 < 4 = b_2, \\quad s_3 = 4 < 16 = b_3, \\quad \\text{and} \\quad s_4 = 9 < 80 = b_4.\n$$\n\nWe now assume $n \\ge 5$.\n\nWe have the trivial bound\n\n$$\ns_n \\le \\sum_{i=1}^{\\lfloor n/2 \\rfloor} 2s_i s_{n-i}.\n$$\n\nFor $n$ odd this may be written as\n\n$$\ns_n \\le \\sum_{i=1}^{n-1} s_i s_{n-i}.\n$$\n\nUse the induction hypothesis to get\n\n$$\ns_n \\le \\sum_{i=1}^{n-1} s_i s_{n-i} < \\sum_{i=1}^{n-1} b_i b_{n-i} = b_n.\n$$\n\nFor $n$ even we need more care. Note that $|A_{n/2} + A_{n/2}| \\le \\binom{s_{n/2}}{2} + s_{n/2}$ and\n\n$$\n\\left| \\frac{1}{\\frac{1}{A_{n/2}} + \\frac{1}{A_{n/2}}} \\right| \\le \\binom{s_{n/2}}{2} + s_{n/2}.\n$$\n\nTherefore, for $n$ even we have\n\n$$\ns_n \\le \\sum_{i=1}^{n/2 - 1} 2s_i s_{n-i} + s_{n/2}^2 + s_{n/2} = \\sum_{i=1}^{n-1} s_i s_{n-i} + s_{n/2}.\n$$\n\nWe now note that for $n \\ge 6$ we have, by the induction hypothesis, $s_3 s_{n-3} < (b_3 - 1) b_{n-3}$, and hence\n\n$$\n\\sum_{i=1}^{n-1} s_i s_{n-i} + s_{n/2} < \\sum_{i=1}^{n-1} b_i b_{n-i} - b_{n-3} + b_{n/2} \\le \\sum_{i=1}^{n-1} b_i b_{n-i} = b_n,\n$$\n\nas Catalan numbers and also the $b_n$ are increasing, concluding the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19175,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the distance between $A$ and $B$ be $S$ km. The speeds of the pedestrian, cyclist, and motorcyclist are $a$, $b$, and $c$ km/h, respectively, with $a < b < c$. The first meeting is between the cyclist and motorcyclist. Let the pedestrian arrive at point $D$ at the moment when the cyclist and motorcyclist meet at point $C$. It is given that $AD = \\frac{S}{6}$. Find the time at which the pedestrian, cyclist, and motorcyclist meet at point $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AD = \\frac{S}{6}$, so the time taken is $t_1 = \\frac{S}{6a}$. Since $AC + (AB + BC) = 2S$, $t_1 = \\frac{2S}{b+c}$. Equating, $\\frac{S}{6a} = \\frac{2S}{b+c}$, so $b + c = 12a$. \n\nFrom $t_1 b = \\frac{Sb}{6a}$, $AC = \\frac{Sb}{6a}$, so $CD = AC - AD = \\frac{S(b-a)}{6a}$. Given $CD = \\frac{a+c}{10}$, so $\\frac{S(b-a)}{6a} = \\frac{a+c}{10}$. \n\nTo reach $C$, the pedestrian needs $t_2 = \\frac{AC}{a} = \\frac{Sb}{6a^2}$, but also $t_2 = \\frac{2S}{a+b}$. Equating, $b(a+b) = 12a^2 \\implies b^2 + ab - 12a^2 = 0 \\implies (b+4a)(b-3a) = 0$. Since $b > 0$, $b = 3a$. From $b + c = 12a$, $c = 9a$. From the previous, $S = 3a$. Thus, $AC = \\frac{Sb}{6a} = \\frac{3a \\cdot 3a}{6a} = 1.5a$. The time for the pedestrian to reach $C$ is $t_2 = \\frac{1.5a}{a} = 1.5$ hours. Therefore, the pedestrian, cyclist, and motorcyclist meet at point $C$ at 13:30.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19176,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that\n\n$$\n\\frac{a-b}{c-d} = 2 \\quad \\text{and} \\quad \\frac{a-c}{b-d} = 3\n$$\n\nfor certain real numbers $a, b, c, d$, determine the value of\n\n$$\n\\frac{a-d}{b-c}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = c - d$ and $y = b - d$. We have\n\n$$\na - d = (c - d) + (a - c) = x + 3y\n$$\n\nand\n\n$$\na - d = (b - d) + (a - b) = y + 2x\n$$\n\nHence, $x + 3y = y + 2x$, which implies $x = 2y$.\n\nNow we get $a - d = 5y$ and\n\n$$\nb - c = (b - d) - (c - d) = y - x = -y\n$$\n\nSo\n\n$$\n\\frac{a-d}{b-c} = \\frac{5y}{-y} = -5.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19177,
    "subject": "Mathematics (Olympiad)",
    "question": "On the occasion of the 47th Mathematical Olympiad 2016, the numbers $47$ and $2016$ are written on the blackboard. Alice and Bob play the following game: Alice begins, and in turns they choose two numbers $a$ and $b$ with $a > b$ written on the blackboard, whose difference $a - b$ is not yet written on the blackboard, and write this difference additionally on the board. The game ends when no further move is possible. The winner is the player who made the last move.\n\nProve that Bob wins, no matter how they play.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the set $B$ of the numbers on the blackboard at the end of the game. It is clear that $B \\subseteq \\{1, \\dots, 2016\\}$. Let $m = \\min B$ and $n \\in B$. We claim that $m \\mid n$. Otherwise, write $n = qm + r$ with $0 < r < m$. By induction on $k$, we have $n - k m \\in B$ for $0 \\leq k \\leq q$ (because no more moves are possible, these numbers must be on the blackboard). Thus $r = n - q m \\in B$, which contradicts the minimality of $m$.\n\nWe conclude that $m \\mid 1 = \\gcd(2016, 47) \\in B$. By induction on $\\ell$, we have $n - \\ell \\in B$ for $0 \\leq \\ell \\leq 2015$. This also implies that $B = \\{1, \\dots, 2016\\}$.\n\nAs $2$ numbers had been on the blackboard at the beginning of the game, the game ends after $2014$ moves when all other numbers have been written. Therefore, Bob wins after move $2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19178,
    "subject": "Mathematics (Olympiad)",
    "question": "Two equal squares $ABCD$ and $EFGH$ are chosen such that they have exactly one common point $C$, which is the midpoint of $EF$, and the points $B$, $F$, $G$ lie on one line. Lines $BC$ and $EH$ intersect at point $K$, and lines $AC$ and $GH$ at point $M$. Denote by $L$ the midpoint of the segment $GH$, and let the line, parallel to $GH$ and passing through $K$, intersect the line $FG$ at point $N$. Prove that $CK = CL = CN$.",
    "options": [],
    "answer": "See solution",
    "solution": "It's clear that $\\triangle CFB = \\triangle CEK$, as both of them have a right angle, equal legs, and equal angles. So $CK = CB = CL$. Similarly, $CK = CN$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19179,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $n$ and $k$ with $n > k^2 > 4$, consider an $n \\times n$ grid. Any $k$ squares in distinct rows and distinct columns are called a $k$-set. Find the largest positive integer $N$ such that it is possible to choose $N$ squares of the grid and colour them in some way so that, for any coloured $k$-set, some two squares have the same colour and some two squares have different colours.",
    "options": [],
    "answer": "See solution",
    "solution": "$$N = (k - 1)^2 n.$$  \n\nChoose $(k-1)^2$ rows of the grid: colour the first $k-1$ rows in colour $c_1$; the second $k-1$ rows in colour $c_2$; \\ldots; the last $k-1$ rows in colour $c_{k-1}$. Altogether, $(k-1)^2 n$ squares are coloured. For any coloured $k$-set, since there are only $k-1$ colours, some two squares must have the same colour. On the other hand, if all the squares in this $k$-set have the same colour, then by definition they are in $k$ different rows, yet there are only $k-1$ rows in that colour, a contradiction. This implies that some two squares have different colours. Therefore, $N \\ge (k-1)^2 n$.\n\nIn a $k$-set, if all the squares have the same colour, call it a mono $k$-set; if the squares have distinct colours, call it a poly $k$-set. We assert that for any colouring of $(k-1)^2 n + 1$ squares, there must exist a mono $k$-set or a poly $k$-set. This will give $N \\le (k-1)^2 n$ and the conclusion. First, we need a lemma.\n\n**Lemma**: In an $n \\times n$ grid, among any $(m-1)n + 1$ squares, $1 \\le m \\le n$, there exists an $m$-set.\n\n**Proof of lemma**: Divide the squares of the $n \\times n$ grid into $n$ groups, such that the square in the $i$th row and $j$th column is in group $a$ ($1 \\le i, j, a \\le n$) if and only if $i - j \\equiv a \\pmod{n}$. Notice that for each group, the squares are in distinct rows and distinct columns. By the pigeonhole principle, among any $(m-1)n + 1$ squares, $m$ of them are in a group, and they form an $m$-set.\n\nFor the original problem, assume that $(k-1)^2 n + 1$ squares are coloured in a certain way. According to the lemma, there exists a $((k-1)^2 + 1)$-set, call it $A$. If there are $k-1$ or fewer colours in $A$, then by the pigeonhole principle some $k$ squares are in the same colour, and they form a mono $k$-set; if there are $k$ or more colours in $A$, then choose $k$ squares of distinct colours, and they form a poly $k$-set.\n\nTherefore, the largest $N$ is $(k-1)^2 n$.\n\n**Alternative proof of lemma**: Choose any $(m-1)n + 1$ squares and colour them black. Take $m$ rows with the most black squares: say they are row $1, 2, \\dots, m$, with $x_1, x_2, \\dots, x_m$ black squares, respectively, and the other rows have $x_{m+1}, \\dots, x_n$ black squares, respectively. If there exists $l$, $1 \\le l \\le m$, such that in the first $m$ rows, black squares of some $l$ rows are distributed in $l-1$ (or fewer) columns, say, black squares of row $1, \\dots, l$ are all in the first $l-1$ columns. Then, $x_1, \\dots, x_l \\le l-1$; for $i > m$, $x_i \\le l-1$ as well (since there are fewer black squares in those rows). Hence,\n\n$$\n\\begin{align*}\n(m-1)n + 1 &= \\sum_{i=1}^{n} x_i \\le l(l-1) + (m-l)n + (n-m)(l-1) \\\\\n&= mn - n + m - l(m+1-l) \\\\\n&\\le mn - n + m - m = (m-1)n,\n\\end{align*}\n$$\n\nwhich is a contradiction. It follows that for any $l$ of the first $m$ rows, the black squares are located in at least $l$ columns. By Hall's marriage theorem, there are $m$ black squares in distinct rows and distinct columns, which is an $m$-set. The lemma is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19180,
    "subject": "Mathematics (Olympiad)",
    "question": "Se tienen dos progresiones de números reales, una aritmética $\\{a_n\\}_{n \\in \\mathbb{N}}$ y otra geométrica $\\{g_n\\}_{n \\in \\mathbb{N}}$ no constante. Se cumple que $a_1 = g_1 \\neq 0$, $a_2 = g_2$ y $a_{10} = g_3$. Decidir, razonadamente, si para cada entero positivo $p$, existe un entero positivo $m$, tal que $g_p = a_m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $d$ y $r \\neq 1$ la diferencia y la razón, respectivamente, de las progresiones aritmética $\\{a_n\\}$ y geométrica $\\{g_n\\}$. En primer lugar, tenemos $g_1 r = g_2 = a_2 = a_1 + d = g_1 + d$, de donde $d = g_1(r - 1)$. En segundo lugar, $g_1 r^2 = g_3 = a_{10} = a_1 + 9d = g_1 + 9g_1(r - 1)$. De aquí sale $r^2 - 9r + 8 = 0$ puesto que $g_1 \\neq 0$. Las soluciones son $r = 1$ (que debemos descartar ya que la progresión geométrica no es constante) y $r = 8$, que es la razón buscada. De aquí también resulta $d = 7g_1$.\n\nSea $p$ un entero positivo cualquiera. Debemos encontrar un $m$ tal que $g_p = a_m$, es decir, $g_p = g_1 8^{p-1} = a_m = a_1 + (m-1)d = g_1 + (m-1)7g_1$, que es equivalente a $8^{p-1} + 6 = 7m$. Puesto que las potencias de 8 módulo 7 siempre son 1, resulta que $8^{p-1} + 6$ es siempre múltiplo de 7 y siempre podremos encontrar $m = \\frac{8^{p-1} + 6}{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19181,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Sally got up at 7:26 am on Monday 2 August. She went to bed that day at 8:34 pm.\n\nb) Sally was up for exactly 11 hours on 5 October. The time Sally is up decreases by 2 minutes each day. On which date was her 11-hour day?\n\nc) Sally was up for 14 hours. David was up for 24 minutes longer. How long was David up?\n\nFrom 9:21 am to 11:45 pm, how long is David up?\n\n\n\n d) For how long were Sally and David both up at the same time?\n\n% ![](images/Australian_Scene_2010_p33_data_3d14cd94ac.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) The difference between 7:26 am and 7:00 am is 26 minutes. Thus, Sally got up at 7:26 am, 26 days after Wednesday 7 July. There are 31 days in July, so the required date is $7 + 26 - 31 = 2$ August. Also, 26 days represent 3 weeks plus 5 days. So the required day is the fifth day after Wednesday, which is Monday. On this day, Sally went to bed at $9:00$ pm $- 26$ minutes $= 8:34$ pm.\n\nb) There are 14 hours between 7:00 am and 9:00 pm. The time Sally is up decreases by 2 minutes each day. Since $14 - 11 = 3$ hours $= 180$ minutes, the number of days from her 14-hour day to her 11-hour day is $180/2 = 90$. There are 31 days in July and August and 30 days in September, so her 11-hour day was $7 + 90 - 31 - 31 - 30 = 5$ October.\n\nc) Sally was up for 14 hours. David was up for 24 minutes longer.\n\n**Alternative i**\n\nTo calculate the time David was up, we use 12-hour times. From 9:21 am to 10:00 am there are 39 minutes. From 10:00 am to 12 noon there are 2 hours. From 12 noon to 11:00 pm there are 11 hours. From 11:00 pm to 11:45 pm there are 45 minutes. So the time from 9:21 am to 11:45 pm is 13 hours and $39 + 45$ minutes, that is, 14 hours and 24 minutes.\n\n**Alternative ii**\n\nThe time from 9:21 am to 9:21 pm is exactly 12 hours. The time from 9:21 pm to 11:45 pm is exactly 2 hours and 24 minutes. So the time from 9:21 am to 11:45 pm is 14 hours and 24 minutes.\n\n**Alternative iii**\n\nTo calculate the time David was up, we use 24-hour times. The time between 11:45 pm and 9:21 am is $23:45 - 09:21 = (23 - 9) \\text{ hours} + (45 - 21) \\text{ minutes} = 14 \\text{ hours and 24 minutes.}$\n\nd) They were both up at the same time for 11 hours and 39 minutes.\n\nDavid got up later than Sally and she went to bed earlier, as indicated on this time line.\n\n% ![](images/Australian_Scene_2010_p33_data_3d14cd94ac.png)\n\nThus, Sally and David were up at the same time from 9:21 am to 9:00 pm.\n\n**Alternative i**\n\nWe use 12-hour times. From 9:21 am to 10:00 am there are 39 minutes. From 10:00 am to 12 noon there are 2 hours. From 12 noon to 9:00 pm there are 9 hours. So the time from 9:21 am to 9:00 pm is 11 hours and 39 minutes.\n\n**Alternative ii**\n\nThe time from 9:21 am to 9:21 pm is exactly 12 hours. So the time from 9:21 am to 9:00 pm is 12 hours $- 21$ minutes $= 11$ hours and 39 minutes.\n\n**Alternative iii**\n\nWe use 24-hour times. The time between 9:00 pm and 9:21 am is $21:00 - 09:21 = (21 - 10)$ hours $+ (60 - 21)$ minutes $= 11$ hours and 39 minutes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19182,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x, y, z$ are real numbers satisfying\n\n$$\n\\begin{aligned}\n(x + 1)(y + 1)(z + 1) &= 3 \\\\\n(x + 2)(y + 2)(z + 2) &= -2 \\\\\n(x + 3)(y + 3)(z + 3) &= -1,\n\\end{aligned}\n$$\n\nfind the value of\n\n$$(x + 20)(y + 20)(z + 20).$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X = x + 2$, $Y = y + 2$, and $Z = z + 2$. Then the system becomes:\n\n$$\n(X - 1)(Y - 1)(Z - 1) = 3\n$$\n\n$$\nXYZ = -2\n$$\n\n$$\n(X + 1)(Y + 1)(Z + 1) = -1.\n$$\n\nExpanding $(X - 1)(Y - 1)(Z - 1)$ and $(X + 1)(Y + 1)(Z + 1)$ and substituting $XYZ = -2$, we get:\n\n$$\n\\begin{aligned}\nT_1 - T_2 &= 6 \\\\\nT_1 + T_2 &= 0,\n\\end{aligned}\n$$\n\nwhere $T_1 = X + Y + Z$ and $T_2 = XY + YZ + ZX$. Solving, $T_1 = 3$ and $T_2 = -3$.\n\nNow,\n\n$$\n\\begin{aligned}\n(x + 20)(y + 20)(z + 20) &= (X + 18)(Y + 18)(Z + 18) \\\\\n&= XYZ + 18 T_2 + 18^2 T_1 + 18^3 \\\\\n&= -2 + 18 \\times (-3) + 18^2 \\times 3 + 18^3 \\\\\n&= 6748.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19183,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying\n\n$$\nf(-f(x) - f(y)) = 1 - x - y\n$$\n\nfor all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substitute $x = y = 1$:\n\n$$\nf(-2f(1)) = -1.\n$$\n\nSubstitute $x = n$, $y = 1$:\n\n$$\nf(-f(n) - f(1)) = -n.\n$$\n\nNow substitute $x = -f(n) - f(1)$, $y = -2f(1)$:\n\n$$\nf(-f(-f(n) - f(1)) - f(-2f(1))) = 1 - (-f(n) - f(1)) - (-2f(1)).\n$$\n\nThe left side is $f(-(-n) - (-1)) = f(n + 1)$, and the right side is $1 + f(n) + f(1) + 2f(1) = f(n) + 3f(1) + 1$.\n\nLet $c = 3f(1) + 1$. Then $f(n + 1) = f(n) + c$.\n\nBy induction, $f(k) = f(0) + ck$ for all $k \\in \\mathbb{Z}$, so $f$ is linear.\n\nLet $f(x) = ax + b$. Then:\n\n$$\nf(-f(x) - f(y)) = a(-ax - b - ay - b) + b = -a^2x - a^2y - 2ab + b.\n$$\n\nSet equal to $1 - x - y$ for all $x, y$:\n\n- Coefficient of $x$: $-a^2 = -1 \\implies a = 1$ or $a = -1$.\n- If $a = -1$: $2b + b = 1 \\implies b = 1/3$ (not integer).\n- If $a = 1$: $-2b + b = 1 \\implies b = -1$.\n\nThus, the only solution is $f(x) = x - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19184,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has between 500 and 1000 students. The gymnastics teacher wants to divide the students into teams of eight for a sports day. Three students are left over. If the teacher tries to divide the students into teams of nine, again three students are left over. Also, with teams of ten students, three students will be left over. How many students attend the school?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the number of students. We have:\n\n- $n \\equiv 3 \\pmod{8}$\n- $n \\equiv 3 \\pmod{9}$\n- $n \\equiv 3 \\pmod{10}$\n- $500 < n < 1000$\n\nThe least common multiple of $8$, $9$, and $10$ is $360$. So $n \\equiv 3 \\pmod{360}$.\n\nPossible values: $363, 723$. Only $723$ is between $500$ and $1000$.\n\n**Answer:** $723$ students attend the school.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19185,
    "subject": "Mathematics (Olympiad)",
    "question": "a) For any number $A$, let $\\overleftarrow{A}$ be the reverse decimal expansion of $A$. Assume that $A = \\overline{A_nA_{n-1}\\cdots A_1}$ is a **mirror-symmetry** number with $m$ digits, all from $\\{1, 2, 3\\}$, and $A_n, A_{n-1}, \\dots, A_1$ are blocks of $A$ with number of digits $m_n, \\dots, m_1$ such that\n\n$$\nA = \\overleftarrow{A_n} \\times \\overleftarrow{A_{n-1}} \\times \\dots \\times \\overleftarrow{A_1}.\n$$\n\nNote that for all $1 \\le i \\le n$,\n\n$$\n\\overleftarrow{A}_i \\le \\underbrace{333\\cdots33}_{m_i} = \\frac{10^{m_i} - 1}{3}.\n$$\n\nOn the other hand,\n\n$$\nA \\ge \\underbrace{111\\cdots11}_{m} = \\frac{10^m - 1}{9}.\n$$\n\nb) Find all **good** numbers $m = \\overline{A_1A_2\\cdots A_n}$, where $A_i$'s are blocks of $m$, such that\n\n$$\n\\frac{m}{7} = A_1 \\times \\cdots \\times A_n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Therefore we obtain\n\n$$\n\\frac{10^m - 1}{9} \\le \\frac{10^{m_n} - 1}{3} \\times \\dots \\times \\frac{10^{m_1} - 1}{3}.\n$$\n\nIf $n \\ge 2$ we have\n\n$$\n\\begin{aligned}\n3^{n-2}(10^m - 1) &\\le (10^{m_n} - 1)(10^{m_{n-1}} - 1)\\cdots(10^{m_1} - 1) \\\\\n&< 10^{m_n} \\times 10^{m_{n-1}} \\times \\cdots \\times 10^{m_2} \\times (10^{m_1} - 1) \\\\\n&< 10^{m_n+m_{n-1}+\\cdots+m_1} - 1 = 10^m - 1.\n\\end{aligned}\n$$\n\nWhich is impossible. Therefore $n=1$. So the only possible case is when $A = \\overleftarrow{A}$, that means $A$ is a **Palindromic number** (a number that remains the same when its digits are reversed). Clearly, all Palindromic numbers with digits of $\\{1, 2, 3\\}$ satisfy the conditions.\n\nb) This part is a test of effort! Note that if we could find a **good** number $m = \\overline{A_1A_2\\cdots A_n}$ where $A_i$'s are blocks of $m$ such that\n\n$$\n\\frac{m}{7} = A_1 \\times \\cdots \\times A_n,\n$$\n\nthen $10m$ is also a **good** number because\n\n$$\n\\frac{10m}{7} = A_1 \\times \\cdots \\times \\overline{A_n 0}.\n$$\n\nTherefore $m, 10m, 100m, \\dots$ are all **good** numbers. So indeed, we just need to find a single **good** number. Now if we start to check the multiples of 7 one by one, we shall finally reach $7 \\times 45 = 315$ for which\n\n$$\n\\frac{315}{7} = 3 \\times 15.\n$$\n\nTherefore by putting $m = 315$, we can find infinitely many **good** numbers, $\\{315, 3150, 31500, \\dots\\}$.\n\n$\\blacksquare$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19186,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $m$ and $n$ is it possible to write the numbers $1, 2, \\dots, 2mn$ into the white squares of a $2m \\times 2n$ checkerboard in such a way that the sum of the numbers in every row is the same, and the sum of the numbers in every column is the same?",
    "options": [],
    "answer": "See solution",
    "solution": "For all even $m$ and $n$, except when $m = n = 2$.\n\nAll the numbers sum up to $mn(2mn+1)$. For odd $m$ this is not divisible by $2n$, breaking the equality of all column sums. Thus $m$ and likewise also $n$ cannot be odd.\n\nIn the case $m = n = 2$ we cannot write the numbers as required, because the numbers\n\n![](images/estonian-2012-2013_p8_data_4b1377e067.png)\n\nin the squares marked by * in Fig. 5 must be equal.\n\nLet us show that in the white squares of a $4 \\times 8$ checkerboard we can write the numbers $k+1, k+2, \\dots, k+8$ and $2mn - k - 7, 2mn - k - 6, \\dots, 2mn$ so that the sums of the numbers in rows are the same, and the sums of the numbers in columns are the same.\nOne possibility is the following, where $P$ stands for $2mn$:\n\n$$\n\\begin{array}{|c|c|c|c|}\n\\hline k+1 & P-k-2 & k+8 & P-k-5 \\\\\n\\hline k+2 & P-k-3 & k+7 & P-k-4 \\\\\n\\hline P-k & k+3 & P-k-7 & k+6 \\\\\n\\hline P-k-1 & k+4 & P-k-6 & k+5 \\\\\n\\hline\n\\end{array}\n$$\n\nHere the sums in the columns are $2mn + 1$ and the sums in the rows are $4mn + 2$.\n\nOne can also write the numbers $1, 2, \\dots, 12$ and $2mn - 11, 2mn - 10, \\dots, 2mn$ on a $4 \\times 12$ checkerboard in the required way, where $P$ stands for $2mn$ again:\n\n$$\n\\begin{array}{|c|c|c|c|c|c|}\n\\hline 1 & 6 & 12 & P-3 & P-4 & P-9 \\\\\n\\hline 3 & 8 & 9 & P-1 & P-6 & P-10 \\\\\n\\hline P & P-5 & P-11 & 4 & 5 & 10 \\\\\n\\hline P-2 & P-7 & P-8 & 2 & 7 & 11 \\\\\n\\hline\n\\end{array}\n$$\n\nIf one of the numbers $m$ and $n$ is even and the other is divisible by 4, then we can cover the $2m \\times 2n$ checkerboard with $4 \\times 8$ checkerboards and fill them as above, taking $k = 0, 8, \\dots, mn - 8$ in different small checkerboards. If neither $m$ nor $n$ is divisible by 4 and one of them is at least 6 then we can cover the checkerboard with one $4 \\times 12$ and $4 \\times 8$ checkerboards and in the $4 \\times 8$ checkerboards take $k = 12, 20, \\dots, mn - 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19187,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagram shows a square and two quarter-circles. Each quarter-circle is centred on a vertex of the square. If the side of the square has length $2$, then the difference between the areas of the two shaded regions is\n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p38_data_b0af50292b.png)\n\n(A) $2\\pi - 4$\n\n(B) $2\\pi$\n\n(C) $4 - \\pi$\n\n(D) $4\\pi - 2$\n\n(E) $2$",
    "options": [],
    "answer": "See solution",
    "solution": "Let the regions have areas $a$, $b$, and $c$ as shown. Then $a + b$ is a quarter circle with area $\\frac{1}{4} \\pi \\times 2^2 = \\pi$. But $a + 2b + c$ is the whole square, i.e. $4$. Now,\n\n$$a - c = 2(a + b) - (a + 2b + c) = 2\\pi - 4.$$  \n\n![](images/5213_SAMF_ANNUAL_REPORT_2016_final_p40_data_3e8a98a9e8.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19188,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest integer $n$ such that, for some $k$, $n^k$ is congruent to $\\underbrace{11\\dots1}_{2012 \\text{ ones}}$ modulo $10^{2012}$. That is,\n$$\n\\underbrace{11\\dots1}_{2012 \\text{ ones}} = \\frac{10^{2012}-1}{9}\n$$\n% IMAGE: ![](images/obm-2012_p15_data_9640d58a75.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We seek the smallest $n$ such that, for some $k$,\n$$\nn^k \\equiv \\frac{10^{2012} - 1}{9} \\pmod{10^{2012}} \\implies 9n^k \\equiv -1 \\pmod{10^{2012}}.\n$$\nThis is equivalent to\n$$\nn^{-k} \\equiv -9 \\pmod{10^{2012}}.\n$$\nBreak into prime powers:\n$$\n\\begin{cases}\nn^{-k} \\equiv -9 \\pmod{2^{2012}} \\\\\nn^{-k} \\equiv -9 \\pmod{5^{2012}}\n\\end{cases}\n$$\nFirst, modulo 8: $n$ must be odd. Since $1^2 \\equiv 3^2 \\equiv 5^2 \\equiv 7^2 \\equiv 1 \\pmod{8}$, $n \\equiv 7 \\pmod{8}$ and $-k$ is odd. Modulo 16, let $k = -m$ and $n = 8u - 1$:\n$$\n(8u - 1)^m \\equiv -1 + 8um \\pmod{16}\n$$\nSo $8um - 1 \\equiv -9 \\pmod{16} \\implies um \\equiv 1 \\pmod{2}$, so $u$ is odd. Thus $n = 8(2t + 1) - 1 \\equiv 7 \\pmod{16}$.\n\nNow modulo 5: since $-k$ is odd, and $1^3 \\equiv 1$, $2^3 \\equiv 3$, $3^3 \\equiv 2$, $4^3 \\equiv 4$ (mod 5), $n \\equiv 1 \\pmod{5}$. Combining, $n \\equiv 7 \\pmod{16}$ and $n \\equiv 1 \\pmod{5} \\implies n \\equiv 71 \\pmod{80}$, so $n \\ge 71$.\n\nWe claim $n = 71$ works. By the Lifting the Exponent lemma:\n$$\n\\nu_2(71^N - 1) = \\nu_2(71^2 - 1) + \\nu_2(N) - 1 = \\nu_2(N) + 3\n$$\n$$\n\\nu_5(71^N - 1) = \\nu_5(71 - 1) + \\nu_5(N) = \\nu_5(N) + 1\n$$\nFor $71^v \\equiv 71^w \\pmod{2^t}$, $v \\equiv w \\pmod{2^{t-3}}$, so residues are congruent to 1 or 7 mod 16, and for all $t \\ge 4$ there is $N$ such that $71^N \\equiv -9 \\pmod{2^t}$. Similarly, for $5^t$, residues are congruent to 1 mod 5, and for all $t \\ge 2$ there is $N$ such that $71^N \\equiv -9 \\pmod{5^t}$. By the Chinese remainder theorem, such $N$ exists for $t = 2012$. Thus, the smallest $n$ is $\\boxed{71}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19189,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$, $n$, $p$ be fixed positive real numbers such that $mnp = 8$. Depending on these constants, find the minimum of\n\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz\n$$\n\nwhere $x$, $y$, $z$ are arbitrary positive real numbers satisfying $xyz = 8$. When is equality attained?\n\nSolve the problem for:\n\na) $m = n = p = 2$\n\nb) arbitrary (but fixed) positive real numbers $m$, $n$, $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Using AM-GM and $xyz = 8$:\n\n$$\nx^2 + y^2 + z^2 + xy + xy + xz + xz + yz + yz \\geq 9\\sqrt{x^6 y^6 z^6} = 36.\n$$\n\nEquality holds for $x = y = z = 2$.\n\nb) Using $xyz = 8$, we can rewrite the expression:\n\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz = x^2 + \\frac{8p}{x} + y^2 + \\frac{8n}{y} + z^2 + \\frac{8m}{z}.\n$$\n\nApplying AM-GM:\n\n$$\nx^2 + \\frac{8p}{x} = x^2 + \\frac{4p}{x} + \\frac{4p}{x} \\geq 6\\sqrt{p^2}.\n$$\n\nApplying the same for $x$, $y$, $z$ and summing:\n\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz \\geq 6\\sqrt[3]{2}(\\sqrt[3]{m^2} + \\sqrt[3]{n^2} + \\sqrt[3]{p^2}).\n$$\n\nEquality holds when\n\n$$\nx = \\sqrt[3]{4p}, \\quad y = \\sqrt[3]{4n}, \\quad z = \\sqrt[3]{4m}.\n$$\n\nThus, the minimum is $6\\sqrt[3]{2}(\\sqrt[3]{m^2} + \\sqrt[3]{n^2} + \\sqrt[3]{p^2})$.\n\n*Alternative solution:* Using weighted AM-GM:\n\n$$\n\\begin{aligned}\nx^2 + y^2 + z^2 + mxy + nxz + pyz &= \\sqrt[3]{p^2} \\frac{x^2}{\\sqrt[3]{p^2}} + \\sqrt[3]{n^2} \\frac{y^2}{\\sqrt[3]{n^2}} + \\sqrt[3]{m^2} \\frac{z^2}{\\sqrt[3]{m^2}} \\\n&\\quad + 2\\sqrt[3]{m^2} \\frac{mxy}{2\\sqrt[3]{m^2}} + 2\\sqrt[3]{n^2} \\frac{nxz}{2\\sqrt[3]{n^2}} + 2\\sqrt[3]{p^2} \\frac{pyz}{2\\sqrt[3]{p^2}} \\\\\n&\\geq 6\\sqrt[3]{2}(\\sqrt[3]{m^2} + \\sqrt[3]{n^2} + \\sqrt[3]{p^2})\n\\end{aligned}\n$$\n\nEquality is achieved when $x = \\sqrt[3]{4p}$, $y = \\sqrt[3]{4n}$, $z = \\sqrt[3]{4m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19190,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a family $\\mathcal{F}$ of subsets of $X = \\{1, 2, 3, 4, 5\\}$ a *decision family* if the following properties are satisfied:\n\n1. If $A \\in \\mathcal{F}$ and $A \\subseteq B$, then $B \\in \\mathcal{F}$.\n2. $A \\in \\mathcal{F}$ if and only if $A^c \\notin \\mathcal{F}$, where $A^c = X - A$.\n\nProve that for every decision family $\\mathcal{F}$ of $\\{1, 2, 3, 4, 5\\}$, there exist nonnegative weights $w_1, w_2, \\dots, w_5$ such that\n\n$$\nA \\in \\mathcal{F} \\iff \\sum_{i \\in A} w_i > \\sum_{i \\in A^c} w_i.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the assertion, note that by property (2), the members of $\\mathcal{F}$ with at most two elements determine $\\mathcal{F}$ uniquely. Using complements, property (1) implies:\n\n1'. If $B \\notin \\mathcal{F}$ and $A \\subseteq B$, then $A \\notin \\mathcal{F}$.\n\nClearly, $\\emptyset \\notin \\mathcal{F}$, because if $\\emptyset \\in \\mathcal{F}$, then $X \\notin \\mathcal{F}$ (by (2)), but by (1'), $\\emptyset \\notin \\mathcal{F}$, a contradiction.\n\nSuppose $A, B \\in \\mathcal{F}$. If $A \\cap B = \\emptyset$, then $A \\subseteq B^c$. Since $B \\in \\mathcal{F}$, $B^c \\notin \\mathcal{F}$, so $A \\notin \\mathcal{F}$ (by (1')), a contradiction. Thus, any two members of $\\mathcal{F}$ have non-empty intersection, so $\\mathcal{F}$ has at most one single-element member.\n\nBy considering cases (up to permutations of $\\{1,2,3,4,5\\}$), the possible configurations for single- and two-element members of $\\mathcal{F}$ are:\n\n- No member of $\\mathcal{F}$ has at most two elements: set $w_1 = w_2 = w_3 = w_4 = w_5 = 1$.\n- Members with at most two elements are $\\{1,2\\}, \\{1,3\\}, \\{2,3\\}$: set $w_4 = w_5 = 0$ and $w_1, w_2, w_3$ as triangle sides, e.g., $w_1 = 3$, $w_2 = 4$, $w_3 = 5$.\n- Only $\\{1,2\\}$: set $w_1 = w_2 = a$, $w_3 = w_4 = w_5 = b$ with $2a > 3b$, $a, b > 0$, e.g., $a = 2$, $b = 1$.\n- $\\{1,2\\}$ and $\\{1,3\\}$: e.g., $(w_1, w_2, w_3, w_4, w_5) = (3, 2, 2, 1, 1)$.\n- $\\{1,2\\}, \\{1,3\\}, \\{1,4\\}$: $(w_1, w_2, w_3, w_4, w_5) = (5, 2, 2, 2, 0)$.\n- $\\{1,2\\}, \\{1,3\\}, \\{1,4\\}, \\{1,5\\}$: $(w_1, w_2, w_3, w_4, w_5) = (5, 2, 2, 2, 2)$.\n- $\\{1\\}, \\{1,2\\}, \\{1,3\\}, \\{1,4\\}, \\{1,5\\}$: $(w_1, w_2, w_3, w_4, w_5) = (5, 1, 1, 1, 1)$.\n\n*Summary Table:*\n\n![]( <table><thead><tr><th>Members of <i>F</i> that have at most two elements</th><th>w<sub>1</sub></th><th>w<sub>2</sub></th><th>w<sub>3</sub></th><th>w<sub>4</sub></th><th>w<sub>5</sub></th></tr></thead><tbody><tr><td>No such member</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>{1,2}, {1,3} and {2,3}</td><td>3</td><td>4</td><td>5</td><td>0</td><td>0</td></tr><tr><td>{1,2}</td><td>2</td><td>2</td><td>1</td><td>1</td><td>1</td></tr><tr><td>{1,2} and {1,3}</td><td>3</td><td>2</td><td>2</td><td>1</td><td>1</td></tr><tr><td>{1,2}, {1,3} and {1,4}</td><td>5</td><td>2</td><td>2</td><td>2</td><td>0</td></tr><tr><td>{1,2}, {1,3}, {1,4} and {1,5}</td><td>5</td><td>2</td><td>2</td><td>2</td><td>2</td></tr><tr><td>{1}, {1,2}, {1,3}, {1,4} and {1,5}</td><td>5</td><td>1</td><td>1</td><td>1</td><td>1</td></tr></tbody></table> )",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19191,
    "subject": "Mathematics (Olympiad)",
    "question": "試找出所有的函數 $f: \\mathbb{R} \\to \\mathbb{R}$ 滿足下列條件：\n\n1. 對於所有的 $x, y \\in \\mathbb{R}$，有 $f(1 + xy) - f(x + y) = f(x)f(y)$。\n2. $f(-1) \\neq 0$。",
    "options": [],
    "answer": "See solution",
    "solution": "唯一解是 $f(x) = x - 1$，其中 $x \\in \\mathbb{R}$。\n\n我們令 $g(x) = f(x) + 1$ 並證明 $g(x) = x$ 對所有實數 $x$ 成立。則題目的條件可寫成：\n\n$$\ng(1+xy) - g(x+y) = (g(x)-1)(g(y)-1) \\quad \\text{對所有 } x, y \\in \\mathbb{R} \\text{ 且 } g(-1) \\neq 1. \\tag{1}\n$$\n\n記 $C = g(-1) - 1 \\neq 0$。設 (1) 式中的 $y = -1$，可得：\n\n$$\ng(1-x) - g(x-1) = C(g(x) - 1). \\tag{2}\n$$\n\n令 (2) 式的 $x = 1$，得到 $C(g(1) - 1) = 0$。因此當 $C \\neq 0$ 時，$g(1) = 1$。代入 $x = 0$ 和 $x = 2$ 分別產生 $g(0) = 0$ 和 $g(2) = 2$。\n\n我們觀察到：\n\n$$\ng(x) + g(2 - x) = 2 \\quad \\text{對所有 } x \\in \\mathbb{R}. \\tag{3}\n$$\n\n$$\ng(x + 2) - g(x) = 2 \\quad \\text{對所有 } x \\in \\mathbb{R}. \\tag{4}\n$$\n\n將 (2) 中的 $x$ 替換成 $1 - x$ 可得 $g(x) - g(-x) = C(g(1-x) - 1)$，再把 $x$ 以 $-x$ 置換可得 $g(-x) - g(x) = C(g(1+x) - 1)$。兩式相加可得 $C(g(1-x) + g(1+x) - 2) = 0$。所以 $C \\neq 0$ 可得 (3)。\n\n令 $u, v$ 使得 $u + v = 1$。將 $(x, y)$ 分別以 $(u, v)$ 和 $(2 - u, 2 - v)$ 代入 (1) 式可得 $g(1 + uv) - g(1) = (g(u) - 1)(g(v) - 1)$，$g(3 + uv) - g(3) = (g(2-u)-1)(g(2-v)-1)$。利用 (3) 式我們知道最後兩個等式的右邊相等，因此 $u+v=1$ 可推得：\n\n$$\ng(uv + 3) - g(uv + 1) = g(3) - g(1).\n$$\n\n每一個 $x \\le \\frac{5}{4}$ 皆可表示成 $x = uv + 1$ 的形式，其中 $u + v = 1$（因為二次多項式 $t^2 - t + (x-1)$ 當 $x \\le \\frac{5}{4}$ 時有實根）。因此當 $x \\le \\frac{5}{4}$ 時，$g(x+2)-g(x) = g(3)-g(1)$。因為 $g(x) = x$ 成立於 $x = 0, 1, 2$，令 $x = 0$ 可得 $g(3) = 3$。如此就證明了當 $x \\le \\frac{5}{4}$ 時 (4) 成立。若 $x > \\frac{5}{4}$ 則 $-x < \\frac{5}{4}$，因此 $g(2-x)-g(-x) = 2$。另一方面，可由 (3) 得到 $g(x) = 2-g(2-x)$，$g(x+2) = 2-g(-x)$ 使得 $g(x+2) - g(x) = g(2-x) - g(-x) = 2$。所以 (4) 對所有實數 $x$ 都成立。\n\n將 (3) 式中的 $x$ 以 $-x$ 代換得到 $g(-x) + g(2+x) = 2$。由 (4) 可得 $g(x) + g(-x) = 0$ 對所有 $x$。將 $(x,y)$ 分別以 $(-x,y)$ 和 $(x,-y)$ 代入 (1) 式可得 $g(1-xy)-g(-x+y) = (g(x)+1)(1-g(y))$，$g(1-xy)-g(x-y) = (1-g(x))(g(y)+1)$。兩式相加得到 $g(1-xy) = 1-g(x)g(y)$。由 (3) 可得 $g(1+xy) = 1+g(x)g(y)$。所以原式 (1) 得到 $g(x+y) = g(x)+g(y)$ 的形式，因此 $g$ 有可加性。\n\n由可加性得到 $g(1+xy) = g(1)+g(xy) = 1+g(xy)$，因為 $g(1+xy) = 1+g(x)g(y)$ 已由上面的討論得到，所以我們可得 $g(xy) = g(x)g(y)$。特別地，$y = x$ 可得對所有實數 $x$，$g(x^2) = g(x)^2 \\ge 0$，意即當 $x \\ge 0$ 時，$g(x) \\ge 0$。因為 $g$ 有可加性且在 $[0, +\\infty)$ 有下界，$g$ 是線性的，更進一步可得對所有實數 $x$，$g(x) = g(1)x = x$。\n\n總而言之，唯一解是 $f(x) = x-1$，其中 $x \\in \\mathbb{R}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19192,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex quadrangle $ABCD$ with $|AD| = |BD| = |CD|$ and $\\angle ADB = \\angle DCA$, $\\angle CBD = \\angle BAC$, find the sizes of the angles of the quadrangle.\n\n![](images/Estonija_2010_p3_data_d6e4dfabd6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle ADB = \\angle DCA = \\alpha$ and $\\angle CBD = \\angle BAC = \\beta$ (see figure).\n\nIn triangle $DAC$, $|DA| = |DC|$, so $\\angle DAC = \\angle DCA = \\alpha$.\n\nAnalogously, in triangles $DAB$ and $DBC$, we have $\\angle DBA = \\angle DAB = \\alpha + \\beta$ and $\\angle DCB = \\angle DBC = \\beta$, respectively.\n\nSo $\\angle BCA = \\beta - \\alpha$.\n\nFrom triangle $ABC$:\n$$\\beta + \\alpha + \\beta + \\beta + \\beta - \\alpha = 180^\\circ$$\nwhich simplifies to $4\\beta = 180^\\circ$, so $\\beta = 45^\\circ$.\n\nFrom triangle $ADB$:\n$$\\alpha + \\beta + \\alpha + \\beta + \\alpha = 180^\\circ$$\nwhich simplifies to $3\\alpha = 180^\\circ - 2\\beta = 90^\\circ$, so $\\alpha = 30^\\circ$.\n\nTherefore, the sizes of the angles of quadrangle $ABCD$ are:\n- $\\angle DAB = \\alpha + \\beta = 75^\\circ$\n- $\\angle ABC = \\alpha + 2\\beta = 120^\\circ$\n- $\\angle BCD = \\beta = 45^\\circ$\n- $\\angle CDA = 360^\\circ - 75^\\circ - 120^\\circ - 45^\\circ = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19193,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a one-to-one correspondence $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ for some positive integer $n$, define four sets $A, B, C, D$ as follows:\n\n$$\nA = \\{i \\mid i > f(i)\\}\n$$\n\n$$\nB = \\{ (i,j) \\mid i < j \\le f(j) < f(i) \\text{ or } f(j) < f(i) < i < j \\}\n$$\n\n$$\nC = \\{ (i,j) \\mid i < j \\le f(i) < f(j) \\text{ or } f(i) < f(j) < i < j \\}\n$$\n\n$$\nD = \\{ (i,j) \\mid i < j \\text{ and } f(i) > f(j) \\}\n$$\n\nProve that $|A| + 2|B| + |C| = |D|$, where $|X|$ is the number of all elements in $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on the number $|D|$:\n\nFirst, if $|D| = 0$, $f$ is the identity and $A = B = C = \\emptyset$. So the statement holds for $|D| = 0$.\n\nNext, assume it holds for all bijections with $|D| < k$ for some positive integer $k$. Given $f$ with $|D_f| = k > 0$, there exists $i$ such that $f(i) > f(i+1)$. Define the function $g$ by\n\n$$\ng(j) = \\begin{cases} f(i+1), & \\text{if } j = i, \\\\ f(i), & \\text{if } j = i+1, \\\\ f(j), & \\text{otherwise.} \\end{cases}\n$$\n\nWe can define four sets $(A_g, B_g, C_g, D_g)$ by the function $g$. Since $|D_g| = |D| - 1 = k - 1 < k$, by the induction hypothesis,\n\n$$\n|A_g| + 2|B_g| + |C_g| = |D_g|. \\tag{5}\n$$\n\nNow, we consider five cases:\n\n(i) If $f(i+1) > i$ and $f(i) > i+1$, then\n\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g| + 1, |C_g| - 1, |D_g| + 1).\n$$\n\n(ii) If $f(i+1) = i$ and $f(i) > i$, then\n\n$$\n|\\{j \\mid j < i \\text{ and } i + 1 < f(j)\\}| = |\\{j \\mid f(j) < i \\text{ and } i + 1 < j\\}|\n$$\n\nyields\n\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g| + 1, |B_g|, |C_g|, |D_g| + 1).\n$$\n\n(iii) If $f(i+1) < i$ and $f(i) > i$, then\n\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g|, |C_g| + 1, |D_g| + 1)\n$$\n\nin both cases $f^{-1}(i) < i$ and $f^{-1}(i) > i$.\n\n(iv) If $f(i+1) < i$ and $f(i) = i$, then\n\n$$\n|\\{j \\mid j < i \\text{ and } i + 1 < f(j)\\}| = |\\{j \\mid f(j) < i \\text{ and } i + 1 < j\\}| + 1\n$$\n\nyields\n\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g| - 1, |B_g| + 1, |C_g|, |D_g| + 1).\n$$\n\n(v) If $f(i+1) < i$ and $f(i) < i$, then\n\n$$\n(|A_f|, |B_f|, |C_f|, |D_f|) = (|A_g|, |B_g| + 1, |C_g| - 1, |D_g| + 1).\n$$\n\nBy (5), the results in all cases complete the proof. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19194,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $E, F$ 分別為三角形 $ABC$ 的邊 $CA, AB$ 上兩點。令 $X$ 為三角形 $AEF$ 的外接圓和三角形 $ABC$ 的外接圓 $\\Gamma$ 異於 $A$ 的交點，$K$ 為三角形 $ABE$ 的外接圓和三角形 $ACF$ 的外接圓異於 $A$ 的交點。設 $AK$ 和 $\\Gamma$ 異於 $A$ 的交點為 $M$，$M$ 對於 $BC$ 的對稱點為 $N$。作 $XN$ 與 $\\Gamma$ 異於 $X$ 的交點 $S$。\n\n證明：$SM$ 平行於 $BC$。",
    "options": [],
    "answer": "See solution",
    "solution": "在 $\\odot(ABC)$ 上取一點 $S'$ 使得 $S'M$ 平行 $BC$，則 $NBS'C$ 為平行四邊形，因此 $S = S'$ 若且唯若 $XS'$ 平分 $\\overline{BC}$。注意到\n\n$$\n\\angle KBF = \\angle KFC, \\quad \\angle BFK = \\angle ECK \\implies \\triangle KBF \\sim \\triangle KEC,\n$$\n\n$$\n\\angle XBF = \\angle XCE, \\quad \\angle BFX = \\angle CEX \\implies \\triangle XBF \\sim \\triangle XCE.\n$$\n\n因此\n\n$$\n\\frac{\\overline{BS'}}{S'C} = \\frac{\\overline{BM}}{\\overline{MC}} = \\frac{\\sin \\angle BAK}{\\sin \\angle KAC} = \\frac{\\overline{BK}}{\\overline{KE}} = \\frac{\\overline{BF}}{\\overline{EC}} = \\frac{\\overline{BX}}{\\overline{XC}}.\n$$\n\n我們有\n\n$$\n\\frac{[\\triangle XBS']}{[\\triangle XS'C]} = \\frac{\\frac{1}{2} \\cdot \\overline{BX} \\cdot \\overline{BS'} \\cdot \\sin \\angle XBS'}{\\frac{1}{2} \\cdot \\overline{CX} \\cdot \\overline{CS'} \\cdot \\sin \\angle XCS'} = 1,\n$$\n\n即 $XS'$ 平分 $\\overline{BC}$。\n\n![](images/2J0410_p31_data_cc74f688b8.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19195,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square. The equilateral triangle $BCS$ is constructed on the exterior of the side $BC$. Let $N$ denote the midpoint of the line segment $AS$ and let $H$ be the midpoint of the side $CD$.\n\nProve: $\\angle NHC = 60^{\\circ}$.\n\n![](images/Austria2019_booklet_p5_data_17cc265cf2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the midpoint of $BS$ (see Figure 1). Since triangles $\\triangle SNP$ and $\\triangle SAB$ are similar with factor $2$, the segment $NP$ is parallel to $AB$ and half the length of $AB$. Therefore, $NPCH$ is a parallelogram. As $NP$ and $BC$ are orthogonal and $PC$ and $BS$ are orthogonal, we have $\\angle NPC = \\angle CBP = 60^{\\circ}$. Thus, we obtain $\\angle NHC = \\angle NPC = 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19196,
    "subject": "Mathematics (Olympiad)",
    "question": "a) What is the smallest possible value of $n$ if the largest odd and even numbers in a set of integers are both positive, and the set contains numbers of both parities?\n\nb) Given that the largest odd number in a set is $2a+1$ and the largest even number is $2b$, and all even numbers are at least $-8$ and at most $2b$, while all odd numbers are at least $-9$ and at most $2a+1$, what is the largest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The smallest possible value is $n = 3$, for example, with the numbers $-1$, $1$, and $2$.\n\nb) Let $2a+1$ be the largest odd number and $2b$ the largest even number. The number of even numbers does not exceed $b+5$, and the number of odd numbers does not exceed $a+6$. Thus,\n\n$$\n\\begin{cases}\n2a+1 \\leq b+5, \\\\\n2b \\leq a+6.\n\\end{cases}\n$$\n\nSumming gives $a+b \\leq 10$. If $a+b=9$, the total number of numbers can reach $n=19$ (e.g., $a=4$, $b=5$ with all odd numbers from $-9$ to $9$ and all even numbers from $-6$ to $10$). For $a+b \\leq 8$, $n \\leq 19$. Thus, the largest possible value is $n=19$, and this is attainable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19197,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $N$ with two digits such that $N$ equals the sum of its digits plus the cube of this sum.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the sum of the digits of $N$. Then $N = n + n^3$. Since $N$ is two digits, $n + n^3 < 100$. Checking possible values:\n\n- For $n = 1$: $1 + 1^3 = 2$\n- For $n = 2$: $2 + 8 = 10$\n- For $n = 3$: $3 + 27 = 30$\n- For $n = 4$: $4 + 64 = 68$\n- For $n = 5$: $5 + 125 = 130$ (too large)\n\nOnly $N = 10$, $30$, and $68$ are two-digit numbers. Checking their digit sums:\n\n- $10$: digits sum to $1$ ($1 + 0$), but $10 \\neq 1 + 1^3 = 2$\n- $30$: digits sum to $3$ ($3 + 0$), and $30 = 3 + 3^3 = 30$\n- $68$: digits sum to $14$ ($6 + 8$), but $68 \\neq 14 + 14^3 = 14 + 2744 = 2758$\n\nThus, the only solution is $N = 30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19198,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integer pairs $(m, n)$ that satisfy the following conditions:\n\n1. $m, n \\leq 20$\n2. $m$ and $n$ are relatively prime.\n3. $\\dfrac{5}{7} < \\dfrac{m}{n} < \\dfrac{3}{4}$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p = n - m$ and $q = n$. Finding all positive integer pairs $(m, n)$ is equivalent to finding all integer pairs $(p, q)$. Then:\n\n- Condition 1 and 3 become $1 \\leq q \\leq 20$ and $\\dfrac{1}{4} < \\dfrac{p}{q} < \\dfrac{2}{7}$.\n- Condition 2 becomes $p$ and $q$ are relatively prime.\n\nThe inequality $\\dfrac{1}{4} < \\dfrac{p}{q} < \\dfrac{2}{7}$ is equivalent to $\\dfrac{7}{2}p < q < 4p$.\n\nNow, consider possible values for $p$:\n\n- If $p \\leq 0$, $q$ doesn't exist because $\\dfrac{p}{q} \\leq 0$.\n- If $p = 1$, then $\\dfrac{7}{2} < q < 4$, so $q$ doesn't exist.\n- If $p = 2$, then $7 < q < 8$, so $q$ doesn't exist.\n- If $p = 3$, then $10.5 < q < 12$, so $q = 11$.\n- If $p = 4$, then $14 < q < 16$, so $q = 15$.\n- If $p = 5$, then $17.5 < q < 20$, so $q = 18, 19$.\n- If $p \\geq 6$, $q > 21$, which is not allowed since $q \\leq 20$.\n\nNow, check which $(p, q)$ pairs are relatively prime:\n- $(3, 11)$: $\text{gcd}(3, 11) = 1$\n- $(4, 15)$: $\text{gcd}(4, 15) = 1$\n- $(5, 18)$: $\text{gcd}(5, 18) = 1$\n- $(5, 19)$: $\text{gcd}(5, 19) = 1$\n\nConvert back to $(m, n)$:\n- $(3, 11)$: $m = n - p = 11 - 3 = 8$, $n = 11$\n- $(4, 15)$: $m = 15 - 4 = 11$, $n = 15$\n- $(5, 18)$: $m = 18 - 5 = 13$, $n = 18$\n- $(5, 19)$: $m = 19 - 5 = 14$, $n = 19$\n\nThus, all positive integer pairs $(m, n)$ are $(8, 11)$, $(11, 15)$, $(13, 18)$, $(14, 19)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19199,
    "subject": "Mathematics (Olympiad)",
    "question": "Some points are marked on a circle. Each marked point is coloured red, green, or blue. In one step, one can erase two marked points of different colours that have no marked points between them and mark a new point between the locations of the erased points with the third colour. In a final state, all marked points have the same colour, which is called the colour of the final state. Find all positive integers $n$ for which there exists an initial state of $n$ marked points with one missing colour, from which one can reach a final state of any of the three colours by applying a suitable sequence of steps.",
    "options": [],
    "answer": "See solution",
    "solution": "All even numbers $n$ greater than $2$.\n\nIf $n = 2$, then the colour of the final state is uniquely determined. We show now that required initial states are impossible for odd $n$. Note that if one colour is missing, then the numbers of marked points of the existing two colours have different parities, i.e., the difference of these numbers is odd. Each step keeps the parity of the difference of the numbers of marked points of these two colours unchanged. Hence, in every intermediate state and also in the final state, one of these two colours is represented. Consequently, a final state of the third colour is impossible.\n\nFor every even number $n > 2$, an initial state with $2$ consecutive points marked with one colour and $n-2$ points marked with another colour satisfies the conditions of the problem. Indeed, if $n > 4$, then with two symmetric steps, one can reach a similar state where the number of points marked with the more popular colour is $2$ less. Hence, it suffices to solve the case $n = 4$. In this case, making one step leads to a state with $3$ marked points, all with different colours. In order to obtain a final state of any given colour, one can replace points of the other two colours with a new point of the given colour. This completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19200,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be sets of numbers written on the blackboard. Let $|X|$ denote the number of elements in set $X$. Suppose\n\n$$\nA = \\{2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20\\}, \\quad B = \\{7, 11, 13, 17, 19\\}.\n$$\n\nEvery pair $(a, b)$ with $a \\in A$ and $b \\in B$ is relatively prime, and $|A||B| = 65$. What is the maximum possible value of $|A||B|$ for such sets $A$ and $B$?",
    "options": [],
    "answer": "See solution",
    "solution": "We are given that $|A||B| = 65$ for the example sets, and every pair $(a, b)$ with $a \\in A$, $b \\in B$ is relatively prime. We want to show that $65$ is the maximum possible value for $|A||B|$.\n\nSuppose, for contradiction, that $|A||B| \\geq 66$. By the AM-GM inequality,\n$$\n|A| + |B| \\geq 2\\sqrt{|A||B|} \\geq 2\\sqrt{66} > 16,\n$$\nso $|A| + |B| \\geq 17$. There are $19$ integers from $2$ to $20$ inclusive, so at most $2$ numbers in this range are in neither $A$ nor $B$.\n\nAmong $6, 12, 18$, at least one must be in $A$ or $B$. Suppose $6 \\in A$. Then, since multiples of the same number cannot be in both $A$ and $B$, all multiples of $2$ and $3$ must be in $A$.\n\nThe numbers from $2$ to $20$ that are not multiples of $2$ or $3$ are $5, 7, 11, 13, 17, 19$ (six numbers). Thus, $|B| \\leq 6$.\n\nIf $|B| \\leq 4$, then $|A||B| \\leq (19 - |B|)|B| \\leq 60$, which is less than $66$. So $|B| = 5$ or $6$.\n\n- If $|B| = 5$, then $B$ must contain $5$ or $7$. Then at least one of $10$ or $14$ cannot be in $A$, so $|A| \\leq 13$, and $|A||B| \\leq 65$.\n- If $|B| = 6$, then $B = \\{5, 7, 11, 13, 17, 19\\}$, so neither $10$ nor $14$ can be in $A$, so $|A| \\leq 9$, and $|A||B| \\leq 54$.\n\nIn all cases, $|A||B| \\leq 65$. Thus, $65$ is the maximum possible value.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19201,
    "subject": "Mathematics (Olympiad)",
    "question": "Dado un número entero $n$ escrito en el sistema de numeración decimal, formamos el número entero $k$ restando del número formado por las tres últimas cifras de $n$ el número formado por las cifras anteriores restantes. Demostrar que $n$ es divisible por $7$, $11$ o $13$ si y sólo si $k$ también lo es.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $A$ el número formado por las tres últimas cifras de $n$ y $B$ el número formado por las cifras anteriores. Entonces $n = 1000B + A$ y $k = A - B$. Tenemos $$n - k = 1001B = 7 \\cdot 11 \\cdot 13 B$$ y $n$ y $k$ son congruentes módulo $7$, $11$ y $13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19202,
    "subject": "Mathematics (Olympiad)",
    "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 $BE = 2 \\cdot AD$.\n\n![](images/RMC_2023_v2_p31_data_de128c9b9d.png)",
    "options": [],
    "answer": "See 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": 19203,
    "subject": "Mathematics (Olympiad)",
    "question": "The lengths of the nine edges of a regular triangular prism $ABC$-$A_1B_1C_1$ are equal. Let $P$ be the midpoint of $CC_1$, and the dihedral angle $B$-$A_1P$-$B_1$ is $\\alpha$. Then\n\n$\\sin \\alpha = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the line through segment $AB$ be the $x$-axis with the origin $O$ as the midpoint of $AB$, and let the line through $OC$ be the $y$-axis to establish a rectangular coordinate system as shown below.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p28_data_067b6102a0.png)\n\nAssume the length of each edge is $2$. Then:\n- $B(1, 0, 0)$\n- $B_1(1, 0, 2)$\n- $A_1(-1, 0, 2)$\n- $P(0, \\sqrt{3}, 1)$\n\nWe compute the relevant vectors:\n\n$$\n\\begin{aligned}\n\\overrightarrow{BA_1} &= (-2, 0, 2), \\quad \\overrightarrow{BP} = (-1, \\sqrt{3}, 1), \\\\\n\\overrightarrow{B_1A_1} &= (-2, 0, 0), \\quad \\overrightarrow{B_1P} = (-1, \\sqrt{3}, -1).\n\\end{aligned}\n$$\n\nLet $\\vec{m} = (x_1, y_1, z_1)$ and $\\vec{n} = (x_2, y_2, z_2)$ be perpendicular to $BA_1P$ and $B_1A_1P$, respectively. We have:\n\n$$\n\\begin{cases}\n\\vec{m} \\cdot \\overrightarrow{BA_1} = -2x_1 + 2z_1 = 0, \\\\\n\\vec{m} \\cdot \\overrightarrow{BP} = -x_1 + \\sqrt{3}y_1 + z_1 = 0, \\\\\n\\vec{n} \\cdot \\overrightarrow{B_1A_1} = -2x_2 = 0, \\\\\n\\vec{n} \\cdot \\overrightarrow{B_1P} = -x_2 + \\sqrt{3}y_2 - z_2 = 0.\n\\end{cases}\n$$\n\nWe can take $\\vec{m} = (1, 0, 1)$ and $\\vec{n} = (0, 1, \\sqrt{3})$.\n\nFrom $|\\vec{m} \\cdot \\vec{n}| = |\\vec{m}| \\cdot |\\vec{n}| \\cos \\alpha$, we have:\n\n$$\n\\sqrt{3} = \\sqrt{2} \\cdot 2 \\cos \\alpha \\implies \\cos \\alpha = \\frac{\\sqrt{6}}{4}.\n$$\n\nTherefore,\n\n$$\n\\sin \\alpha = \\frac{\\sqrt{10}}{4}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19204,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a $v \\times \\mu$ rectangular board, with $v \\leq \\mu$, be divided by parallel lines into $v\\mu$ unit squares. Initially, we place $N$ black tokens on some unit squares. Then, we try to fill the board entirely with black tokens using the following move:\n\n*If an empty unit square shares a side with at least two squares containing a black token, then we place a black token on that square.*\n\nFind the minimum possible value of $N$ such that, after finitely many moves, all unit squares will have a black token.",
    "options": [],
    "answer": "See solution",
    "solution": "We number the rows from $1$ to $v$ and the columns from $1$ to $\\mu$. If $v = \\mu$, the board is a $v \\times v$ square; if $v < \\mu$, the board consists of a $v \\times v$ square and a $v \\times (\\mu - v)$ rectangle. Placing tokens along the main diagonal allows us to fill the board, as illustrated below:\n\n![](images/Greece-IMO2019finalbook_p9_data_68669112d9.png)\n\nfig. 3(α)\n\n![](images/Greece-IMO2019finalbook_p9_data_46bb67d89f.png)\n\nfig. 3(β)\n\n![](images/Greece-IMO2019finalbook_p9_data_ee35f95a59.png)\n\nfig. 3(γ)\n\nFor $v = \\mu$, placing $v$ black tokens on the main diagonal suffices to fill the board after finitely many moves.\n\nFor $v < \\mu$, place $v$ tokens on the main diagonal of the $v \\times v$ square. Then:\n\n- If $\\mu - v$ is odd, place a token at $(v, v+1)$, then at $(v, v+3)$, ..., up to $(v, \\mu)$, i.e., at $(v, v+2\\kappa-1)$ for $\\kappa = 1, 2, \\dots, \\frac{\\mu - v + 1}{2}$. This requires $v + \\frac{\\mu - v + 1}{2} = \\frac{\\mu + v + 1}{2}$ tokens.\n- If $\\mu - v$ is even, place tokens at $(v, v+2), (v, v+4), \\dots, (v, \\mu)$, i.e., at $(v, v+2\\kappa)$ for $\\kappa = 1, 2, \\dots, \\frac{\\mu - v}{2}$. This requires $v + \\frac{\\mu - v}{2} = \\frac{\\mu + v}{2}$ tokens.\n\nIn both cases, the minimum number is $\\left\\lfloor \\frac{\\mu + v + 1}{2} \\right\\rfloor$.\n\nThus,\n$$\nN \\leq \\left\\lfloor \\frac{\\mu + v + 1}{2} \\right\\rfloor.\n$$\n\nTo show this is minimal, note that each initial black token can contribute at most $4$ black-and-white sides (sides shared with only one black square). The perimeter has $2(v + \\mu)$ such sides, so $4N \\geq 2(v + \\mu)$, i.e., $N \\geq \\frac{v + \\mu}{2}$. Since $N$ is integer,\n$$\nN \\geq \\left\\lfloor \\frac{\\mu + v + 1}{2} \\right\\rfloor.\n$$\n\nTherefore, the minimum possible $N$ is $\\boxed{\\left\\lfloor \\frac{\\mu + v + 1}{2} \\right\\rfloor}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19205,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xf(y) - y^2) = (y + 1)f(x - y)\n$$\nholds for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = -1$ and $x = 0$. Then $f(-1) = 0$.\n\nSuppose $f$ has another zero, i.e., there exists $a \\neq -1$ with $f(a) = 0$. Setting $x = 0$ and $y = -a$ gives $f(-a^2) = 0$. Now, choosing $y = a$, we obtain $f(x - a) = 0$. Hence, $f(x) = 0$ for all $x$, which is clearly a solution.\n\nNow assume $x = -1$ is the only zero of $f$. Setting $x = y - 1$ gives $f((y - 1)f(y) - y^2) = 0$. Thus, $(y - 1)f(y) = y^2 - 1$ for all $y$, which implies $f(y) = y + 1$ for all $y \\neq 1$. Setting $y = 2$, $x = 3$ gives $f(1) = 2$. Hence, $f(x) = x + 1$ for all $x$, which is another solution.\n\nAltogether, the only solutions are $f(x) = 0$ and $f(x) = x + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19206,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f_n(x)$ be a sequence of polynomials, where $f_0(x) = 2$, $f_1(x) = 3x$, and\n$$\nf_n(x) = 3x f_{n-1}(x) + (1 - x - 2x^2) f_{n-2}(x)\n$$\nfor all $n \\geq 2$. Determine all positive integers $n$ such that $f_n(x)$ is divisible by $x^3 - x^2 + x$.",
    "options": [],
    "answer": "See solution",
    "solution": "By direct calculation, one can obtain\n$$\nf_n(x) = (2x-1)^n + (x+1)^n\n$$\nfor all positive integers $n$. Let $Q(x) = x^3 - x^2 + x = x(x^2 - x + 1)$ and $n$ be a natural number such that $Q(x)$ divides $f_n(x)$. It is easy to see that\n$$\n(-1)^n + 1^n = f_n(0) = 0,\n$$\nso $n$ is odd.\n\nLet $R(x) = \\frac{f_n(x)}{Q(x)}$. We will show that $R(x)$ is a polynomial with integer coefficients. Assume\n$$\nR(x) = \\frac{aR_1(x)}{b},\n$$\nwhere $a, b \\in \\mathbb{Z}^+$ and $R_1(x)$ is a primitive polynomial. Hence,\n$$\nb f_n(x) = a R_1(x) Q(x).\n$$\nBy Gauss's lemma, $R_1(x)Q(x)$ is primitive, so the greatest common divisor of all the coefficients of $aR_1(x)Q(x)$ is $a$, hence $a$ is divisible by $b$. Denote $a = b \\cdot c$, then\n$$\nR(x) = cR_1(x) \\in \\mathbb{Z}[x].\n$$\nPutting $x = -2$, since $n$ is odd, we get\n$$\nQ(-2) = -14 \\mid f_n(-2) = -(5^n + 1).\n$$\nNote that $5^6 \\equiv 1 \\pmod{6}$, so we can check that the above condition is equivalent to $n = 3k$ where $k$ is an odd natural number. On the other hand, if $n = 3k$ where $k$ is odd, we have\n$$\n9Q(x) = (x+1)^3 + (2x-1)^3 \\mid f_n(x).\n$$\nIn conclusion, the answer is $n = 6p + 3$ where $p$ is a non-negative integer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19207,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a circle centred at $O$. Let $\\overline{AB}$ be a chord of that circle and $M$ its midpoint. Tangents on $k$ at points $A$ and $B$ intersect at $T$. The line $\\ell$ goes through $T$, intersects the shorter arc $\\overarc{AB}$ at the point $C$ and the longer arc $\\overarc{AB}$ at the point $D$, so that $|BC| = |BM|$.\n\nProve that the circumcentre of the triangle *ADM* is the reflection of *O* across the line *AD*.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $C$ and $D$ are on $k$, the power of the point $T$ with respect to $k$ equals $|TB|^2 = |TC| \\cdot |TD|$.\n\n![](images/Croatia_2018_p17_data_be4fc2e672.png)\n\nFurthermore, since the right-angled triangles $TBM$ and $TOB$ are similar, we have $|TB|^2 = |TM| \\cdot |TO|$. Therefore, $|TC| \\cdot |TD| = |TM| \\cdot |TO|$, i.e. the quadrilateral $CDOM$ is cyclic.\n\nLet point $C'$ be the intersection of $k$ and the line $DM$, while $\\alpha = \\angle C'MC$. Now we have\n\n$$\n\\angle C'C'M = \\angle C'C'D = \\frac{1}{2}\\angle COD = \\frac{1}{2}\\angle CMD = \\frac{1}{2}(180^\\circ - \\angle C'MC) = 90^\\circ - \\frac{1}{2}\\alpha,\n$$\n\ntherefore $\\angle MCC' = 180^\\circ - \\alpha - (90^\\circ - \\frac{1}{2}\\alpha) = 90^\\circ - \\frac{1}{2}\\alpha = \\angle CC'M$, i.e. $|C'M| = |CM|$, meaning that $C'$ is the reflection of $C$ across the line $OM$, and $|AC'| = |AM|$ holds. Let $M'$ be the point on $k$ different from $C'$ such that $|AM'| = |AC'|$. Then\n\n$$\n\\angle M'DA = \\angle C'DA = \\angle MDA.\n$$\n\nWe can conclude that triangles $MDA$ and $M'DA$ are congruent (two pairs of congruent sides, one pair of congruent angles, and both are obtuse), so $M'$ is the reflection of $M$ across the line $AD$. The fact that $O$ is the circumcentre of the triangle $ADM'$ completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19208,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, suppose $n$ numbers are chosen from the table\n\n$$\n\\begin{array}{cccc}\n0 & 1 & \\cdots & n-1 \\\\\nn & n+1 & \\cdots & 2n-1 \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n(n-1)n & (n-1)n+1 & \\cdots & n^2-1\n\\end{array}\n$$\n\nwith no two numbers from the same row or the same column. What is the maximal product of these $n$ numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "The product can be written as\n\n$$\n\\prod_{i=1}^{n} (a_i + b_{p(i)}),\n$$\n\nfor some permutation $p$ of $1, \\dots, n$, where $a_i = (i-1)n$ and $b_i = i-1$. Assume that $i + p(i) \\neq n+1$ for some $i$, and let the least such $i$ be chosen. Since $j + p(j) = n+1$ for $j < i$, then $k = p(i) < n+1-i$ and $l = p^{-1}(n+1-i) > i$. Replacing $p$ with $((n+1-i)k)p$ replaces the factor $(a_i + b_k)(a_l + b_{n+1-i})$ with $(a_i + b_{n+1-i})(a_l + b_k)$, thus increasing this factor by\n\n$$\n(a_i + b_{n+1-i})(a_l + b_k) - (a_i + b_k)(a_l + b_{n+1-i}) = (a_i - a_l)(b_k - b_{n+1-i}) > 0.\n$$\n\nFor any $p$ such that $i + p(i) \\neq n + 1$ for some $i$, the product may thus be increased by choosing another $p$. (If zero is one of the chosen numbers, any choice which avoids the zero will increase the product.) Therefore, the maximum is achieved by choosing $p(i) = n + 1 - i$, that is, taking the product along the diagonal from the upper right to the lower left corners of the table. This product is\n\n$$\n\\prod_{i=1}^{n} i(n-1) = (n-1)^n n! .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19209,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{N}$ 表示所有正整數之集合。試求所有函數 $f : \\mathbb{N} \\to \\mathbb{N}$ 滿足\n\n$$\nf(x + y f(x)) = x + f(x) f(y)\n$$\n\n對於所有正整數 $x, y$ 皆成立。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，考察原式：\n\n$$\nf(x + y f(x)) = x + f(x) f(y)\n$$\n\n嘗試代入不同值以尋找規律。\n\n1. 令 $x = 1$，則\n   $$\nf(1 + y f(1)) = 1 + f(1) f(y)$$\n   若 $f(1) = 1$，則 $f(1 + y) = 1 + f(y)$。\n   由此猜測 $f(x) = x$。\n\n2. 驗證 $f(x) = x$ 是否滿足原式：\n   $$\nf(x + y x) = x + x y = x (1 + y)$$\n   而 $f(x + y x) = x + y x$，確實相符。\n\n3. 若 $f(1) \\neq 1$，則可構造出 $a$ 使得 $(a, f(a)) = 1$，進而推導出 $f(f(y)) = f(y)$，即 $f$ 在其值域上為不動點。\n\n4. 進一步分析，若 $f$ 有最小不動點 $k > 1$，則由遞推可得 $k - 1$ 也是不動點，矛盾。\n\n因此，唯一解為\n\n$$\nf(x) = x, \\quad \\forall x \\in \\mathbb{N}$$\n\n代回原式可驗證成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19210,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers such that $5m + n$ divides $5n + m$. Prove that $m$ divides $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "There exists a positive integer $k$ such that $5n + m = k(5m + n)$, or $$(5 - k)n = (5k - 1)m.$$ The right-hand side is strictly positive, so the left-hand side must be strictly positive as well, which implies $5 - k > 0$ and so $k \\in \\{1, 2, 3, 4\\}$. \n\n- If $k = 1$, then $4n = 4m$, so $n = m$.\n- If $k = 2$, then $3n = 9m$, so $n = 3m$.\n- If $k = 3$, then $2n = 14m$, so $n = 7m$.\n- If $k = 4$, then $n = 19m$.\n\nIn each case, $m$ divides $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19211,
    "subject": "Mathematics (Olympiad)",
    "question": "證明存在正實數 $C$ 使下列敘述成立：\n\n對任意正整數 $n$，如果非負實數 $x_0, x_1, \\dots, x_{n+1}$ 滿足\n\n$$\nx_i x_{i+1} - x_{i-1}^2 \\ge 1, \\quad i = 1, \\dots, n,\n$$\n\n則 $x_0 + x_1 + \\dots + x_{n+1} > C n^{3/2}$。",
    "options": [],
    "answer": "See solution",
    "solution": "我們先引入兩個引理。\n\n**引理 1.1** 若 $a, b, c$ 為非負數且 $ab - c^2 \\ge 1$，則\n\n$$\n(a + 2b)^2 \\ge (b + 2c)^2 + 6.\n$$\n\n*證明*：\n$$(a + 2b)^2 - (b + 2c)^2 = (a - b)^2 + 2(b - c)^2 + 6(ab - c^2) \\ge 6.$$\n\n**引理 1.2**\n$$\n\\sqrt{1} + \\dots + \\sqrt{n} > \\frac{2}{3} n^{3/2}.\n$$\n\n*證明*：Bernoulli 不等式 $(1 + t)^{3/2} > 1 + \\frac{3}{2} t$ 對 $0 > t \\ge -1$（或直接驗算）給出\n\n$$\n(k-1)^{3/2} = k^{3/2} \\left(1 - \\frac{1}{k}\\right)^{3/2} > k^{3/2} \\left(1 - \\frac{3}{2k}\\right) = k^{3/2} - \\frac{3}{2} \\sqrt{k}.\n$$\n\n對 $k = 1, 2, \\dots, n$ 求和得\n\n$$\n0 > n^{3/2} - \\frac{3}{2} (\\sqrt{1} + \\dots + \\sqrt{n}).\n$$\n\n![](images/2022-TWNIMO-Problems_p55_data_dffd228cd3.png)\n\n令 $y_i := 2x_i + x_{i+1}$，$i = 0, 1, \\dots, n$。由引理 1.1 得 $y_0 \\ge 0$ 且 $y_i^2 \\ge y_{i-1}^2 + 6$，對 $i = 1, 2, \\dots, n$。由歸納法可得 $y_i \\ge \\sqrt{6i}$。利用此估計和引理 1.2，\n\n$$\n\\begin{aligned}\nx_0 + \\dots + x_{n+1} &\\ge \\frac{y_1 + \\dots + y_n}{3} \\\\\n&\\ge \\frac{\\sqrt{6}}{3} (\\sqrt{1} + \\sqrt{2} + \\dots + \\sqrt{n}) \\\\\n&> \\frac{2}{9} \\sqrt{6} n^{3/2} = \\left(\\frac{2n}{3}\\right)^{3/2}.\n\\end{aligned}\n$$\n\n因此 $C = \\left(\\frac{2}{3}\\right)^{3/2}$ 滿足所需性質。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19212,
    "subject": "Mathematics (Olympiad)",
    "question": "Assign coordinates $(i, j)$ to the squares of a board, where $i, j \\in \\{0, 1, \\dots, p^2 - 1\\}$. Consider the set\n\n$$\nS = \\{(ap + b, cp + d) : 0 \\le a, b, c, d < p \\text{ and } ac \\equiv b + d \\pmod{p}\\}.\n$$\n\nShow that $S$ contains $p^3$ elements and that no four squares from $S$ form a rectangle with sides parallel to the board.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $p^3$ elements in $S$ because for each $a, b, c \\in \\{0, 1, \\dots, p-1\\}$, there is a unique $d$ such that $ac \\equiv b + d \\pmod{p}$. \n\nSuppose four squares from $S$ form a rectangle with sides parallel to the board. Their coordinates are:\n\n$$\n(ap + b, cp + d), \\quad (ep + f, cp + d), \\quad (ap + b, gp + h), \\quad (ep + f, gp + h),\n$$\n\nwith the conditions:\n\n$$\n\\begin{cases}\nac \\equiv b + d \\pmod{p} \\\\\nec \\equiv f + d \\pmod{p} \\\\\nag \\equiv b + h \\pmod{p} \\\\\neg \\equiv f + h \\pmod{p}\n\\end{cases}\n$$\n\nSubtracting, we get $(a - e)(c - g) \\equiv 0 \\pmod{p}$. Since $p$ is prime, either $a = e$ or $c = g$. If $a = e$, then $b = f$; if $c = g$, then $d = h$. In both cases, the rectangle is degenerate. Thus, $S$ contains no unwanted rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19213,
    "subject": "Mathematics (Olympiad)",
    "question": "For some integer $n$, a set of $n^2$ magical chess pieces arrange themselves on a square $n^2 \\times n^2$ chessboard composed of $n^4$ unit squares. At a signal, the chess pieces all teleport to another square of the chessboard such that the distance between the centres of the old and new squares is $n$. The chess pieces win if, both before and after the signal, there are no two chess pieces in the same row or column. For which values of $n$ can the chess pieces win?\n\nThis problem splits into two parts: showing that it cannot be done when $n$ is odd and that it can be done when $n$ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be odd, and let the $i$th chess piece have coordinates $(x_i, y_i)$ before the signal, and $(w_i, z_i)$ afterwards (we let the bottom-left square have coordinate $(1, 1)$ and the top-right have coordinate $(n^2, n^2)$). For each $i$, $(x_i - w_i)^2 + (y_i - z_i)^2 = n^2$. Summing across all the pieces gives\n\n$$\n\\sum_{i=1}^{n^2} \\left[(x_i - w_i)^2 + (y_i - z_i)^2\\right] = \\sum_{i=1}^{n^2} n^2, \\\\\n\\text{i.e.}\\quad \\sum_{i=1}^{n^2} \\left[x_i^2 + w_i^2 + y_i^2 + z_i^2 - 2(x_i w_i + y_i z_i)\\right] = n^4.\n$$\n\nNow, as the initial $x$-coordinates take all values $1$ to $n^2$, $\\sum_{i=1}^{n^2} x_i^2 = \\sum_{i=1}^{n^2} i^2$, and likewise for $y_i, w_i$ and $z_i$. So\n\n$$\n4 \\sum_{i=1}^{n^2} i^2 - 2 \\sum_{i=1}^{n^2} (x_i w_i + y_i z_i) = n^4.\n$$\n\nHowever, the left hand side of this is even, and the right hand side is odd, so it cannot be done for $n$ odd.\n\nNow let $n$ be even, and divide the board into $n^2$ 'megasquares', each of size $n$ by $n$. We initially arrange the pieces on the top-left to bottom-right diagonal, i.e., on the diagonals of the megasquares which themselves are on the diagonal of the main grid. Numbering the megasquares on the main diagonal $1$ to $n$, we move the pieces in the odd-numbered megasquares down by $n$, and in the even-numbered ones up by $n$. As there are $n$ rows of megasquares, the last one moves up so they all stay on the board. We have in effect moved the megasquares, and there is still one in each row and column, so the final configuration meets the conditions. The diagram shows this for $n = 4$.\n\n![](images/British_2019_Booklet_p12_data_1405e330b8.png)\n\nBefore signal\n\n![](images/British_2019_Booklet_p12_data_08e21e10aa.png)\n\nAfter signal",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19214,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x, y, z$ satisfy $0 < x, y, z < 1$ and $xyz = (1-x)(1-y)(1-z)$. Show that $$\\frac{1}{4} \\le \\max\\{(1-x)y,\\ (1-y)z,\\ (1-z)x\\}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$x(1-x) \\le \\left(\\frac{x + 1 - x}{2}\\right)^2$$\nso $x(1-x) \\le \\frac{1}{4}$. Similarly,\n$$y(1-y) \\le \\frac{1}{4}, \\quad z(1-z) \\le \\frac{1}{4}.$$ \nMultiplying these inequalities and using $xyz = (1-x)(1-y)(1-z)$, we get $xyz \\le \\frac{1}{8}$. Thus, at least one of $x, y, z$ is less than or equal to $\\frac{1}{2}$. Suppose $x \\le \\frac{1}{2}$, so $1-x \\ge \\frac{1}{2}$. Assume $$\\frac{1}{4} > \\max\\{(1-x)y,\\ (1-y)z,\\ (1-z)x\\}.$$ Then $y < \\frac{1}{2}$, so $1-y > \\frac{1}{2}$, and $z < \\frac{1}{2}$, so $1-z > \\frac{1}{2}$. But then\n$$\\frac{1}{8} = \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} > xyz = (1-x)(1-y)(1-z) > \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{8}.$$ \nThis is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19215,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 100 infinitely large boxes with markers in them, the following procedure is carried out:\n\n- At step 1, add one marker in every box.\n- At step 2, add one marker in every box containing an even number of markers.\n- At step 3, add one marker in every box in which the number of markers is divisible by 3, and so on.\n\nBefore the process starts, Bruno wants to distribute several markers in the boxes so that there is at least one marker in each box and the following holds: **After any number of steps, there exist two boxes containing different numbers of markers.**\n\nDecide if this is possible to achieve.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is *no*. Regardless of the initial distribution, all boxes will contain the same number of markers after finitely many steps. Moreover, this is true for any number of boxes.\n\nDenote by $x_n$ the number of markers in a certain box before step $n$, $n = 1, 2, \\dots$. Suppose that $x_n = n$ for some $n$. Then, by the rule of adding markers, we have $x_{n+1} = n+1$, $x_{n+2} = n+2$, etc.; in other words, the number of markers in that box equals the number of the oncoming step $l$ for each $l \\geq n$. So, to prove that eventually all boxes contain the same number of markers, it is enough to show that for each box there exists a step $n$ such that $x_n = n$.\n\nWe use the following observation: Let a box $C$ satisfy $x_i > l$ for some $l$, that is, the difference $d_i = x_i - l$ is positive. Then there is an $m \\geq l$ such that $C$ receives no marker at step $m$. Otherwise, ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19216,
    "subject": "Mathematics (Olympiad)",
    "question": "Clare and Anthony are on a circular pond with 2020 lily pads. Clare jumps 201 lily pads to the right each time, while Anthony remains stationary. When is the first time Clare is five or fewer lily pads away from Anthony (counting clockwise or counterclockwise)?",
    "options": [],
    "answer": "See solution",
    "solution": "After $t$ jumps, Clare is $201t$ pads clockwise from Anthony, measured modulo 2020, so we have to solve:\n\n$$\n201t \\in \\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\\} \\pmod{2020}.\n$$\n\nWe notice that $201^2 = 40401 \\equiv 1 \\pmod{2020}$, so we can multiply both sides by 201, giving:\n\n$$\nt \\in \\{-1005, -804, -603, -402, -201, 0, 201, 402, 603, 804, 1005\\} \\pmod{2020}\n$$\n\nThe smallest $t > 0$ in this set is $t = 201$, so after 201 jumps, Clare is one pad away from Anthony, which is the first time they are five or fewer pads apart.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19217,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $\\omega$ is tangent to the side $BC$ of a triangle $ABC$ at point $T$. The side $AB$ intersects $\\omega$ at points $P$ and $R$ ($A$ is closer to $P$ than $R$); the side $AC$ intersects $\\omega$ at points $Q$ and $S$ ($A$ is closer to $Q$ than $S$). The lines $AT$, $BQ$ and $CP$ are concurrent. Prove that the lines $AT$, $BS$ and $CR$ are also concurrent.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p206_data_89ad0641ff.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By Ceva's theorem,\n\n$$\nAP \\cdot BT \\cdot CQ = BP \\cdot CT \\cdot AQ\n$$\n\nBy the power of a point,\n\n$$\nBT^2 = BR \\cdot BP\n$$\n\n$$\nCT^2 = CS \\cdot CQ\n$$\n\n$$\nAP \\cdot AR = AQ \\cdot AS\n$$\n\nCombining these, we obtain\n\n$$\nAS \\cdot BT \\cdot \\frac{CT^2}{CS} = \\frac{BT^2}{BR} \\cdot CT \\cdot AR\n$$\n\nSimplifying, we get\n\n$$\nAS \\cdot CT \\cdot BR = BT \\cdot AR \\cdot CS\n$$\n\nThus, by the converse of Ceva's theorem, the lines $AT$, $BS$, and $CR$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19218,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that if $G - v$ has a Hamiltonian cycle for each vertex $v$ in a graph $G$, but $G$ does not have a Hamiltonian cycle, then $n = |G| \\ge 10$, and produce such a graph $G$ with $n = 10$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $v$ cannot be adjacent with two consecutive vertices in a Hamiltonian cycle in $G - v$, we have $\\deg v \\le \\left\\lfloor \\frac{n-1}{2} \\right\\rfloor$. On the other hand, if $\\deg w \\le 2$ for some vertex $w$, then there is no Hamiltonian cycle in $G - v$ where $v$ is adjacent with $w$. Therefore $3 \\le \\deg v \\le \\left\\lfloor \\frac{n-1}{2} \\right\\rfloor$. In particular, $n \\ge 7$.\n\nNext, observe that if $v_1 \\to v_2 \\to \\dots \\to v_{n-1}$ is a Hamiltonian cycle in $G - v$, and if $v$ is adjacent with $v_i$ and $v_j$, $i < j$, then $v_{i-1}$ and $v_{j-1}$ cannot be adjacent with each other, as that would give a Hamiltonian cycle $v_1 \\to \\dots \\to v_{i-1} \\to v_{j-1} \\to v_{j-2} \\to \\dots \\to v_i \\to v \\to v_j \\to v_{j+1} \\to \\dots \\to v_{n-1}$ in $G$.\n\nFrom these observations, it follows that $n = 7$ and $n = 8$ are impossible. If $n = 9$, then each vertex has degree 3 or 4, and they cannot all have degree 3 by the degree sum formula. Hence, assume that $n = 9$ and there is a vertex $v_0$ with $\\deg v_0 = 4$.\n\nThen $G$ must contain the graph on the left below.\n\n![](images/Turska_2009_p11_data_2696bf80b7.png)\n\n![](images/Turska_2009_p11_data_272fa10151.png)\n\nBut then, again from the observations above, it follows that each $v_{2i}$ must be adjacent with at least—and therefore exactly—one of $v_{2i+3}$ and $v_{2i+5}$ (indices considered mod 8), and there are no other edges. Since there are edges only between vertices of different parity in the resulting graph; if we remove an odd indexed vertex, the remaining graph cannot have a Hamiltonian cycle as it has unequal numbers of odd and even indexed vertices.\n\nFinally, the following graph gives an example when $n = 10$.\n\n![](images/Turska_2009_p11_data_e3d818336a.png)\n\nIn this graph, $a \\to c \\to C \\to B \\to b \\to e \\to E \\to D \\to d \\to a$ and $A \\to B \\to C \\to c \\to e \\to b \\to d \\to D \\to E \\to A$ are Hamiltonian cycles for $G - A$ and $G - a$, respectively.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19219,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomial $f(x) = c x (x - 2)$ where $c$ is a positive real number. For any $n \\in \\mathbb{Z}^+$, let $g_n(x)$ denote the $n$-fold composition of $f$ (i.e., $g_1(x) = f(x)$, $g_2(x) = f(f(x))$, etc.). Assume that the equation $g_n(x) = 0$ has all $2^n$ solutions real.\n\n1) For $c = 5$, find, in terms of $n$, the sum of all solutions of $g_n(x)$, counting each distinct solution only once.\n\n2) Prove that $c \\geq 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "1) We prove by induction on $n$ that all solutions of $g_n(x)$ are distinct. For $n = 1$, $g_1(x) = f(x)$ has two solutions: $x = 0$ and $x = 2$. Suppose $g_n(x)$ has $2^n$ distinct solutions $x_1, x_2, \\dots, x_{2^n}$. Then,\n\n$$\ng_{n+1}(x) = g_n(f(x)) = k \\prod_{i=1}^{2^n} (f(x) - x_i)\n$$\n\nThe solution sets of $f(x) - x_i = 0$ for different $x_i$ are disjoint, and if $f(x) - x_i = 0$ has a double root, that root must be $x = 1$. However,\n\n$$\ng_1(1) = f(1) = -5, \\quad g_2(1) = f(-5) = 5(-5)(-7) = 175\n$$\n\nSo $g_n(1) > 0$ for all $n \\geq 2$, and $x = 1$ is never a solution. Thus, $g_{n+1}(x)$ also has $2^{n+1}$ distinct solutions, completing the induction.\n\nFrom the above, the solutions of $g_n(x)$ can be grouped into $2^{n-1}$ disjoint pairs, each pair summing to $2$. Therefore, the sum of all solutions is $2^n$.\n\n2) For some $d \\in \\mathbb{R}$, the equation\n\n$$\nc x (x - 2) = d \\iff x^2 - 2x = \\frac{d}{c}\n$$\n\nhas two distinct solutions if and only if $1 + \\frac{d}{c} > 0$, i.e., $c > -d$. The solutions are\n\n$$\nx = 1 \\pm \\sqrt{1 + \\frac{d}{c}}.\n$$\n\nFor $g_n(x)$ to have $2^n$ real solutions, $g_{n-1}(x)$ must have $2^{n-1}$ distinct real roots. For each root $r$, consider $f(x) = r$:\n\n- If $r < 0$, both solutions are positive.\n- If $r > 0$, the solutions have opposite signs.\n\nThe constraint $c > -r_1 = r_2 - 2$ (for $r_1 < 0 < r_2$) leads to a sequence $u_1 = 2$, $u_{n+1} = 1 + \\sqrt{1 + \\frac{u_n}{c}}$. This sequence increases and converges to $L = 2 + \\frac{1}{c}$. We require $c > u_n - 2$ for all $n$, so as $n \\to \\infty$, $c \\geq \\frac{1}{c}$, i.e., $c \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19220,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. Let $a_1, a_2, \\dots, a_n \\in \\{1, 2, 3, \\dots, 2n\\}$ be such that $\\mathrm{lcm}(a_i, a_j) > 2n$ for any $1 \\le i < j \\le n$. Prove that\n\n$$\na_1 a_2 \\dots a_n \\mid (n+1)(n+2)\\dots(2n-1)(2n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For every $i = 1, 2, \\dots, n$ let $a_i = b_i \\cdot 2^{c_i}$ where $b_i$ is odd.\n\nNote that the $b_i$'s are pairwise distinct. Indeed, if $b_i = b_j$ for some $i < j$ then one of the numbers $a_i, a_j$ divides the other one, so $\\mathrm{lcm}(a_i, a_j) = \\max(a_i, a_j) \\le 2n$ which is a contradiction.\n\nAlso, it is clear that each $b_i$ belongs to $\\{1, 3, 5, \\dots, 2n-1\\}$. Since there are exactly $n$ $b_i$'s and the set $\\{1, 3, 5, \\dots, 2n-1\\}$ has exactly $n$ elements, we have\n\n$$\n\\{b_1, b_2, \\dots, b_n\\} = \\{1, 3, 5, \\dots, 2n-1\\}.\n$$\n\nNow, for every $i$ let $d_i$ be the greatest $d$ such that $b_i \\cdot 2^d \\le 2n$. Note that $b_i \\cdot 2^{d_i} \\in \\{n+1, n+2, \\dots, 2n\\}$ as otherwise $b_i \\cdot 2^{d_i+1} \\le 2n$, contradicting maximality of $d_i$.\n\nNote that the numbers $b_i \\cdot 2^{d_i}$ are pairwise distinct (because $\\mathbb{Z}$ is a unique factorization domain). Again, we have $n$ pairwise distinct numbers $b_1 \\cdot 2^{d_1}, b_2 \\cdot 2^{d_2}, \\dots, b_n \\cdot 2^{d_n}$ belonging to the $n$-element set $\\{n+1, n+2, \\dots, 2n\\}$, hence\n\n$$\n\\{b_1 \\cdot 2^{d_1}, b_2 \\cdot 2^{d_2}, \\dots, b_n \\cdot 2^{d_n}\\} = \\{n+1, n+2, \\dots, 2n\\}.\n$$\n\nReindexing $a_i$'s if necessary, we can assume that $b_i \\cdot 2^{d_i} = n + i$ for every $i$. Clearly, $c_i \\le d_i$, so $a_i = b_i \\cdot 2^{c_i}$ for every $b_i \\cdot 2^{d_i} = n + i$. As a consequence,\n\n$$\na_1 a_2 \\dots a_n \\mid (n+1)(n+2)\\dots(n+n),\n$$\n\nas desired. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19221,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $|AB| = |BC|$. Point $E$ lies on the arc $CD$ which does not contain $A$ and $B$. The intersection of $BE$ and $CD$ is denoted by $P$, and the intersection of $AE$ and $BD$ is denoted by $Q$. Prove that $PQ \\parallel AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $|AB| = |BC|$, we have $\\angle AEB = \\angle BDC$, hence $\\angle QEP = \\angle AEB = \\angle BDC = \\angle QDP$, which yields that $QPED$ is a cyclic quadrilateral. Therefore, $\\angle QPD = \\angle QED = \\angle AED = \\angle ACD$. From this, we get that $QP$ and $AC$ are parallel. $\\square$\n\n![](images/NLD_ABooklet_2021_p24_data_b84c0addf8.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19222,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a fixed real number. Find all real numbers $b$ such that, for every real number $x$, at least one of the numbers $x^2 + a x + b$ and $x^2 - a x + b$ is non-negative.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $x^2 + a x + b$ and $x^2 - a x + b$ sum to $2x^2 + 2b$. If $b \\ge 0$, then for any real $x$, $2x^2 + 2b \\ge 0$, so at least one of the two expressions is non-negative. If $b < 0$, taking $x = 0$ gives both $x^2 + a x + b = b$ and $x^2 - a x + b = b$, which are negative. Thus, all real numbers $b \\ge 0$ satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19223,
    "subject": "Mathematics (Olympiad)",
    "question": "Find a formula for $T_n$, where\n\n$$\nT_n = 1 + 3 + 6 + \\dots + \\frac{n}{2}(n + 1)\n$$\n\nThat is, $T_n$ is the sum of the first $n$ triangular numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The $n$th triangular number is $\\frac{n}{2}(n+1)$, a quadratic polynomial in $n$. This suggests that $T_n$ is a cubic polynomial in $n$, that is, $T_n = an^3 + bn^2 + cn + d$ where $a, b, c, d$ are constants to be determined.\n\nWe have:\n\n$$\nT_1 = a + b + c + d = 1 \\quad (1)\n$$\n\n$$\nT_2 = 8a + 4b + 2c + d = 4 \\quad (2)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19224,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given positive integer $n$, let $S_n$ denote the set of all possible values that $x_n$ can obtain for different choices of numbers $a_i$, $1 \\le i \\le n$. For example:\n\n$$\nS_1 = \\left\\{\\frac{1}{2}, 2\\right\\}, \\quad S_2 = \\left\\{\\frac{1}{3}, \\frac{2}{3}, \\frac{3}{2}, 3\\right\\}, \\quad S_3 = \\left\\{\\frac{1}{4}, \\frac{2}{5}, \\frac{3}{5}, \\frac{3}{4}, \\frac{4}{3}, \\frac{5}{3}, \\frac{5}{2}, 4\\right\\}, \\dots\n$$\n\nNote that, for every $x \\in S_n$, both $x+1$ and $\\frac{1}{x+1}$ belong to $S_{n+1}$. Furthermore, one of those two numbers is smaller than 1 and the other is greater than 1. Thus, each $S_n$ consists of an even number of numbers. For a given positive integer $n$, let\n\n$$\nS_n = \\{a_1, a_2, \\dots, a_{2m}\\}, \\text{ with } a_1 < a_2 < \\dots < a_{2m}\n$$\n\nand\n\n$$\nS_{n+1} = \\{b_1, b_2, \\dots, b_{2k}\\}, \\text{ with } b_1 < b_2 < \\dots < b_{2k}.\n$$\n\nThe following statements can be proved by induction:\n\n* $m = 2^{n-1}$, $k = 2^n$ and thus $|S_{n+1}| = 2|S_n|$.\n* $a_1 = \\frac{1}{n+1}$, $a_m = \\frac{n}{n+1}$, $a_{m+1} = \\frac{n+1}{n}$, $a_{2m} = n+1$.\n* $b_1 < b_2 < \\dots < b_{2m} < 1 < b_{2m+1} < \\dots < b_{4m}$.\n* $a_i = \\frac{1}{a_{2m+1-i}}$, $b_i = \\frac{1}{b_{4m+1-i}}$.\n* $b_i = \\frac{1}{1 + a_{2m+1-i}}$, for $1 \\le i \\le 2m$.\n* $b_i = \\frac{1}{b_{2m+1-i}}$, for $1 \\le i \\le 2m$.\n\nCombining the above, we get:\n\n$$\nb_i = \\frac{a_i}{1 + a_i}, \\quad \\text{for } 1 \\le i \\le 2m. \\qquad (2)\n$$\n\nFor all integers $n \\ge 2$, prove that $a_{i+1} - a_i \\le \\frac{1}{2n-1}$ for all $1 \\le i \\le m$.\n\nFor $n = 101$, let $S_{101} = \\{c_1, c_2, \\dots, c_s\\}$, where $c_1 < c_2 < \\dots < c_s$.\n\nShow that for any $x \\in \\left[\\frac{1}{111}, \\frac{110}{111}\\right]$, there exists $c_j$ in $S_{101}$ such that $|x - c_j| \\le \\frac{1}{402}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$.\n\n**Base case ($n=2$):**\n\nFor $n=2$, $S_2 = \\left\\{\\frac{1}{3}, \\frac{2}{3}\\right\\}$ (for $m=1$), and $a_2 - a_1 = \\frac{2}{3} - \\frac{1}{3} = \\frac{1}{3} = \\frac{1}{2n-1}$, so the claim holds.\n\n**Inductive step:**\n\nAssume for $n$ that $a_{i+1} - a_i \\le \\frac{1}{2n-1}$ for all $1 \\le i \\le m$.\n\nFor $n+1$, $b_i = \\frac{a_i}{1 + a_i}$ for $1 \\le i \\le 2m$.\n\nCompute:\n$$\nb_{i+1} - b_i = \\frac{a_{i+1}}{1 + a_{i+1}} - \\frac{a_i}{1 + a_i} = \\frac{a_{i+1} - a_i}{(1 + a_i)(1 + a_{i+1})}\n$$\nBy induction, $a_{i+1} - a_i \\le \\frac{1}{2n-1}$, and $a_i, a_{i+1} \\ge \\frac{1}{n+1}$, so\n$$\nb_{i+1} - b_i \\le \\frac{\\frac{1}{2n-1}}{\\left(1 + \\frac{1}{n+1}\\right)^2} = \\frac{\\frac{1}{2n-1}}{\\left(\\frac{n+2}{n+1}\\right)^2} = \\frac{n+1}{2n-1} \\cdot \\frac{n+1}{n+2}^2\n$$\nThis is less than $\\frac{1}{2n+1}$ for $n \\ge 2$ (as shown in the original argument).\n\nFor $n=101$, $c_1 = \\frac{1}{102}$, $c_s = \\frac{101}{102}$, and $c_1 - \\frac{1}{111} < \\frac{1}{201}$, $\\frac{110}{111} - c_s < \\frac{1}{201}$.\n\nDefine $S'_n = \\{c_0, c_1, \\dots, c_s, c_{s+1}\\}$ with $c_0 = \\frac{1}{111}$, $c_{s+1} = \\frac{110}{111}$.\n\nThen $c_{i+1} - c_i \\le \\frac{1}{201}$ for $0 < i \\le s$.\n\nFor any $x \\in \\left[\\frac{1}{111}, \\frac{110}{111}\\right]$, there exists $j$ such that $c_j \\le x \\le c_{j+1}$, so $|x - c_j| \\le \\frac{1}{402}$ or $|x - c_{j+1}| \\le \\frac{1}{402}$.\n\nThus, every $x$ in the interval is within $\\frac{1}{402}$ of some $c_j$ in $S_{101}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19225,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有正整數 $x, y, z, t$ 滿足\n\n$$\nxy - zt = x + y = z + t\n$$\n\n且 $xy$ 和 $zt$ 都是完全平方數。",
    "options": [],
    "answer": "See solution",
    "solution": "無正整數解。\n\n設 $xy = a^2$ 且 $zt = c^2$。\n\n若 $x + y = z + t$ 為奇數，則 $xy$ 和 $zt$ 都是偶數，得 $xy - zt = x + y = z + t$ 也是偶數，矛盾。令 $s = \\frac{x + y}{2}$，由前可知 $s$ 是整數。令 $b = \\frac{|x - y|}{2}$，$d = \\frac{|z - t|}{2}$，則原題條件可得：\n\n$$\ns^2 = a^2 + b^2 = c^2 + d^2\n$$\n\n和\n\n$$\n2s = a^2 - c^2 = d^2 - b^2.\n$$\n\n由於上兩式中，$a, d$ 和 $b, c$ 對稱，我們只須證明上二式在 $a, s, d$ 為正整數，$b, c$ 為非負整數且不同時為零的條件下無解即可。由對稱性，不妨假設 $b \\geq c$。有 $d^2 = 2s + b^2 > c^2$，所以\n\n$$\nd^2 > \\frac{c^2 + d^2}{2} = \\frac{s^2}{2}\n$$\n\n又\n\n$$\n2s = d^2 - b^2 \\geq d^2 - (d - 2)^2 = 4(d - 1)\n$$\n\n所以有\n\n$$\n\\frac{s}{\\sqrt{2}} < d \\leq \\frac{s}{2} + 1\n$$\n\n可知 $s < 2\\sqrt{2} + 2 < 5$。因為當 $1 \\leq s \\leq 4$，$s^2$ 只能拆成 $s^2 + 0^2$ 的形式，檢查發現無解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19226,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a set $A = \\{1, 2, \\ldots, 4044\\}$. One colors 2022 numbers of them white and the rest black. For each $i \\in A$, define the weight of $i$ as the sum of the number of white numbers less than $i$ and the number of black numbers greater than $i$. For every positive integer $m$, find all positive integers $k$ such that there exists a way to color the numbers so that exactly $k$ numbers have weight $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a natural number $i$ 'good' if its weight is $m$. We will prove the following claim.\n\n**Claim.** Consider a positive integer $i \\le 4044$.\n\n(a) If there are more black numbers than white from $1$ to $i - 1$, then there exists a black number $j$ such that the numbers of black and white numbers from $j + 1$ to $i - 1$ are equal.\n\n(b) If there are more white numbers than black from $i + 1$ to $4044$, then there exists a white number $j$ such that the numbers of black and white numbers from $i + 1$ to $j - 1$ are equal.\n\n*Proof.* Clearly (a) and (b) are similar, so we only need to prove (b). Denote $f(k)$ as the difference between the numbers of black and white numbers from $i + 1$ to $i + k - 1$. It is clear that $f(0) = 0$ and\n\n$$\n|f(x) - f(x + 1)| = 1.\n$$\n\nNow we need to show that there exists $k$ such that $i + k$ is white and $f(k) = 0$. If $i + 1$ is white and $f(1) = 0$, then we can assume $i + 1$ is black, so $f(2) = 1 > 0$. Because the number of white numbers from $i + 1$ to $4044$ is more than the black numbers, $f(4044 - i) \\le 0$. Hence,\n\n$$\nf(2) = 1 > 0 \\ge f(4044 - i).\n$$\n\nWe will show that there exists $j$ such that $f(j) = 0$. If there exists the smallest natural number $t$ such that $f(t) < 0$, note that\n\n$$\n|f(t - 1) - f(t)| = 1\n$$\n\nso $f(t-1) = f(t)+1$ and $f(t-1) \\ge 0$, hence $f(t-1) = 0$. Otherwise, if $f(t) \\ge 0$ for all $t$, then $f(4044 - i) = 0$, which means there always exists a number $j$ with that property.\n\nAssume that for all $j$ with $f(j) = 0$, the number $j+1$ is always black, which means $f(j + 1) = 1$. Similarly, if there exists the smallest number $t$ such that $f(t) < 0$, then $f(t - 1) = 0$, hence $f(t) \\ge 0$ for all $t$. Thus, $f(4044 - i) = 0$, which means the numbers of black and white numbers from $i + 1$ to $4043$ are equal, so $4044$ is colored white. Otherwise, if there exists $j$ such that $f(j) = 0$ and $j + i$ is white, then $j$ satisfies the above condition. $\\square$\n\nBack to the problem, assume that $i < i'$ are two consecutive good numbers and have the same color (white). We observe that there are fewer white numbers before $i'$ than $i$ and fewer black numbers after $i'$ than $i$. Hence, the weight of $i'$ is smaller than the weight of $i$, which is a contradiction. Thus, they must have different colors.\n\nCall a natural number $i$ 'good' if it has weight $m$. We shall prove that $k$ is an even number. Indeed, let $i$ be the smallest good number and denote $a_j, b_j$ as the number of white and black numbers before and after $j$. If $j < i$ then\n\n$$\na_j + b_j \\neq a_i + b_i.\n$$\n\nIf $i$ is black, then the number of white numbers from $1$ to $i-1$ is smaller than the number of black numbers; otherwise, by lemma 1, we can find a black number $j$ such that in the interval $[j+1, i-1]$, the numbers of black and white numbers are equal, which means\n\n$$\na_i - a_j = b_i - b_j\n$$\n\nBut $j$ is also a good number, which leads to a contradiction. Note that the number of black numbers before $i$ is $2021 - b_1$, hence\n\n$$\nm = a_i + b_i < 2022 - b_i + b_i = 2022.\n$$\n\nNext, we will prove that if $s$ is good and black, then there exists $s' > s$ such that $s'$ is good and white. Because $s$ is black, the number of white numbers after $s$ is $2022 - a_s$, then\n\n$$\nb_s = m - a_s < 2022 - a_s\n$$\n\nApplying lemma 1, we get a number $s' > s$ such that $s'$ is white and the numbers of black and white numbers in $[s+1, s'-1]$ are equal. Hence,\n\n$$\na_{s'} - a_s = b_{s'} - b_s\n$$\n\nwhich means $s'$ is also a good number. Thus, if $A = \\{x_1 < x_2 < \\dots < x_k\\}$ is the set of good numbers, then two consecutive numbers have different colors by lemma 2. Note that $x_1$ is black, so $x_{2t+1}$ is black for all $t$, which is a contradiction because there must exist $k' > x_k$ such that $k'$ is good and white.\n\nSimilarly, if $i$ is white, then we can point out that $m \\ge 2022$ and for ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19227,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f$ from the set of positive integers to the set of positive integers such that, for all positive integers $a$ and $b$, there exists a non-degenerate triangle with sides of lengths\n\na, $f(b)$, and $f(b + f(a) - 1)$.\n\n(A triangle is _non-degenerate_ if its vertices are not collinear.)",
    "options": [],
    "answer": "See solution",
    "solution": "We present two solutions. Clearly, $f(x) = x$ is a solution. We will show that this is the only solution. We call an unordered triple $\\{x, y, z\\}$ of positive real numbers _triangular_ if there is a non-degenerate triangle with sides $x, y, z$. Both solutions (and in fact many others) are based on the following facts.\n\n(a) First, we show that $f(1) = 1$. Assume on the contrary that $f(1) = v_1 + 1$ for some positive integer $v_1$. Setting $a = 1$ in the given condition yields a triangular triple $\\{1, f(b), f(b+v_1)\\}$. By the triangle inequality, we must have $f(b) = f(b+v_1)$ for every positive integer $b$; that is, $f$ has a period of $v_1$. In particular, this means that $f$ is bounded. But this is impossible, because we can set $a$ to be greater than twice this bound, making $\\{a, f(b), f(b+f(a)-1)\\}$ not triangular, a contradiction. Thus, our assumption was wrong and $f(1) = 1$.\n\n(b) Second, we show that $f$ is its own inverse function; that is, $f(f(a)) = a$. Indeed, setting $b = 1$ in the given condition gives a triangular triple $\\{a, f(1) = 1, f(f(a))\\}$, from which it follows that $a = f(f(a))$.\n\n(c) Third, we claim that $f$ is injective. Indeed, if not, then there are $a_1 \\neq a_2$ with $f(a_1) = f(a_2)$. But then we have $a_1 = f(f(a_1)) = f(f(a_2)) = a_2$, violating the assumption that $a_1 \\neq a_2$.\n\nAfter establishing (a), (b), and (c), there are many possible finishes. In general, as shown in the first solution, one tries to use the triangle inequality to control the growth rate of $f$. The second solution might be more insightful: it bridges this particular triangle inequality with Freiman's theorem:\n\nLet $A$ be a set of positive integers. Then $|A+A| \\ge 2|A| - 1$, and equality holds if and only if $A$ is the set of an arithmetic progression.\n\n**Solution 1.** By (a) and (c), we know that $f(2) \\ge 2$. We assume that $f(2) = v_2+1$ for some positive integer $v_2$. Setting $a=2$ in the given condition leads to the triangular triple $\\{2, f(b), f(b+v_2)\\}$. By the triangle inequality and (c), the possible values of $f(b+v_2)$ are $f(b)+1$ and $f(b)-1$.\n\nIf $f(b+v_2) = f(b)-1$, then setting $a=2$ and $b = b+v_2$ in the given condition leads to the triangular triple $\\{2, f(b+v_2) = f(b)-1, f(b+2v_2)\\}$. By the triangle inequality and (c), the only possible value of $f(b+2v_2)$ is $f(b)-2$. Likewise, we can deduce that $f(b+3v_2) = f(b)-3$, and so on. But this is impossible, because $f$ takes values in the set of positive integers.\n\nWe conclude that $f(b + v_2) = f(b) + 1$. By a simple induction, we have $f(b + kv_2) = f(b) + k$ for positive integers $k$. In particular, setting $b = 1$ yields\n\n$$\nf(1) = 1, \\quad f(1+v_2) = 2, \\quad f(1+2v_2) = 3, \\quad \\dots, \\quad f(1+kv_2) = 1+k, \\dots \\quad (1)\n$$\n\nSetting $k = v_2$ in (1) gives\n\n$$\nf(1 + v_2^2) = 1 + v_2.\n$$\n\nBy (b), we conclude that $1+v_2^2 = f(f(1+v_2^2)) = f(1+v_2) = 2$, implying that $v_2 = 1$. Substituting $v_2 = 1$ in (1) leads to the solution $f(x) = x$.\n\n**Solution 2** (By the coordinators of this problem). We start with a lemma that is slightly stronger than Freiman's theorem. Part of this lemma could be very helpful in certain proofs of USAMO 2009 problem 2 and IMO 2000 problem 1.\n\n**Lemma 1.** Let $A, B$ be finite nonempty subsets of $\\mathbb{Z}$. Then the set $A+B = \\{a+b : a \\in A, b \\in B\\}$ has cardinality at least $|A| + |B| - 1$. Equality holds if and only if either $A$ and $B$ are arithmetic progressions with equal difference or at least one of $|A|$ or $|B|$ is equal to 1. (Here $|S|$ denotes the number of elements in $S$.)\n\n*Proof.* Let $A = \\{a_1 < a_2 < \\dots < a_{|A|}\\}$ and $B = \\{b_1 < b_2 < \\dots < b_{|B|}\\}$. The following $|A| + |B| - 1$ distinct elements, arranged in increasing order, are in $A+B$:\n\n$$\na_1 + b_1 < \\dots < a_1 + b_{|B|} < a_2 + b_{|B|} < \\dots < a_{|A|+|B|}. \\quad (2)\n$$\n\nTherefore $|A+B| \\ge |A| + |B| - 1$, establishing the inequality.\n\nNext we consider the equality case. Let $c_i$ denote the value of the $i$th element in list (2). Assume that $|A+B| = |A| + |B| - 1$ and $|A|, |B| > 1$. For any $1 < i \\le |A|$ and $1 < j \\le |B|$, consider the following list of $|A| + |B| - 1$ distinct elements in $A+B$:\n\n$$\n\\underbrace{a_1 + b_1 < \\dots < a_1 + b_{j-1}}_{j-1\\ \\text{elements}} < \\underbrace{a_2 + b_{j-1} < \\dots < a_i + b_{j-1}}_{i-1\\ \\text{elements}} < \\underbrace{a_i + b_j < \\dots < a_i + b_{|B|}}_{|B|-j+1\\ \\text{elements}} < \\underbrace{a_{i+1} + b_{|B|} < \\dots < a_{|A|} + b_{|B|}}_{|A|-i\\ \\text{elements}}.\n$$\n\nThis list must be the same as (2). Therefore $a_i+b_{j-1} = c_{i+j-2}$. Likewise $a_{i-1}+b_j = c_{i+j-2}$. Therefore $a_i + b_{j-1} = a_{i-1} + b_j$ and $a_i - a_{i-1} = b_j - b_{j-1}$. Hence, both $A$ and $B$ are arithmetic progressions with the same common difference. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19228,
    "subject": "Mathematics (Olympiad)",
    "question": "Para cada entero positivo $n$, el Banco de Ciudad del Cabo produce monedas de valor $\\frac{1}{n}$. Dada una colección finita de tales monedas (no necesariamente de distintos valores) cuyo valor total no supera $99 + \\frac{1}{2}$, demostrar que es posible separar esta colección en 100 o menos montones, de modo que el valor total de cada montón sea como máximo 1.",
    "options": [],
    "answer": "See solution",
    "solution": "Antes de empezar a particionar el conjunto de monedas, vayamos a la ventanilla del Banco de Ciudad del Cabo y realicemos los siguientes cambios:\n\n- Para cada entero $n = 2m$ par, para el que haya al menos dos monedas de valor $\\frac{1}{n}$, tomemos dos de estas monedas y cambiémoslas por una moneda de valor $\\frac{1}{m}$, hasta que no haya ningún entero par $n = 2m$ para el que haya más de una moneda de valor $\\frac{1}{n}$.\n- Para cada entero $n = 2m + 1$ impar, para el que haya al menos $2m + 1$ monedas con valor $\\frac{1}{n}$, tomemos $2m+1$ de estas monedas y cambiémoslas por una moneda de valor 1, hasta que no haya ningún entero impar $n = 2m + 1$ para el que haya más de $2m$ monedas de valor $\\frac{1}{n}$.\n\nNótese que, si podemos particionar el conjunto resultante, también podemos particionar el conjunto inicial, no teniendo para ello más que volver a la ventanilla del Banco de Ciudad del Cabo, invirtiendo los cambios realizados. Sabiendo además que siempre podemos tomar todas las monedas de valor $\\frac{1}{1}$, asignando cada una de ellas a alguno de los 100 conjuntos de la partición, vemos que nos basta con resolver el siguiente problema: particionar un conjunto $R$ de recíprocos de enteros, no necesariamente distintos y cuya suma es menor que $N - \\frac{1}{2}$, en $N$ conjuntos la suma de cada uno de los cuales es a lo sumo 1, siendo $N$ un entero, y dado que en $R$ no hay más de un recíproco de un entero par (es decir, para cada entero par $n = 2m$ existe a lo sumo un elemento en $R$ igual a $\\frac{1}{n}$), y que para cada entero impar $n = 2m + 1$, existen a lo sumo $2m$ elementos en $R$ que son los recíprocos de $\\frac{1}{n}$. Como además los elementos de $R$ que tienen valor 1 se pueden asignar cada uno a un elemento de la partición, sin afectar al problema (simplemente reduciendo el valor de $N$ y el cardinal de $R$), podemos asumir que el mayor valor de $R$ es a lo sumo $\\frac{1}{2}$. Resolveremos a continuación este problema.\n\nDefinamos conjuntos $R_1, R_2, \\dots, R_N$. Si existe una moneda de valor $\\frac{1}{2}$ (y en ese caso existiría a lo sumo una), la colocamos en $R_1$. Colocamos en $R_2$ las monedas que existan de valor $\\frac{1}{3}$ (a lo sumo dos), y de valor $\\frac{1}{4}$ (a lo sumo una), y así sucesivamente, hasta colocar en $R_N$ las monedas de valor $\\frac{1}{2N-1}$ y $\\frac{1}{2N}$. Nótese que la suma máxima de las monedas colocadas en $R_k$ es a lo sumo $\\frac{2k-2}{2k-1} + \\frac{1}{2k} = 1 - \\frac{1}{2k(2k-1)} < 1$. Coloquemos ahora las monedas restantes, de forma aleatoria, en cualquier conjunto tal que, al añadir dicha moneda, el valor no supera 1. Supongamos que en algún momento esto deja de ser posible, para una moneda de valor $\\frac{1}{n}$. Entonces, en cada conjunto la suma es mayor que $1 - \\frac{1}{n}$, para un valor total de las monedas ya asignadas superior a $N - \\frac{N}{n}$, y a su vez inferior a $N - \\frac{1}{2}$. Luego $n < 2N$, contradicción pues ya hemos colocado todas las monedas de valor mayor o igual que $\\frac{1}{2N}$ en la asignación inicial. Luego siempre podemos continuar asignando todas las monedas de valor inferior o igual a $\\frac{1}{2N+1}$, siendo por lo tanto siempre posible la partición, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19229,
    "subject": "Mathematics (Olympiad)",
    "question": "The fractional part $f(x)$ of a number $x$ is defined by\n$$\nf(x) \\in [0, 1) \\text{ and } x - f(x) \\in \\mathbb{Z}.\n$$\n\nFind all $x$ such that $f(1 - 2022x) = x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = \\frac{m}{n}$ be a rational number with $0 \\le m < n$. The condition $f(1 - 2022x) = x$ is equivalent to $1 - 2023x \\in \\mathbb{Z}$, or $\\frac{2023m}{n} \\in \\mathbb{Z}$. If we choose $n = 2023$, then for each integer $m$ with $0 \\le m < 2023$, $x = \\frac{m}{2023}$ satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19230,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_0, x_1, \\dots, x_{n_0-1}$ be integers, and let $d_1, d_2, \\dots, d_k$ be positive integers with $n_0 = d_1 > d_2 > \\dots > d_k$ and $\\gcd(d_1, d_2, \\dots, d_k) = 1$. For every integer $n \\ge n_0$, define\n\n$$\nx_n = \\left\\lfloor \\frac{x_{n-d_1} + x_{n-d_2} + \\dots + x_{n-d_k}}{k} \\right\\rfloor.\n$$\n\nShow that the sequence $\\{x_n\\}$ is eventually constant.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $x_n \\le \\max\\{x_{n-1}, \\dots, x_{n-n_0}\\}$, so $\\{x_n\\}$ is a bounded sequence. Let $X$ be the largest integer that occurs infinitely often in the sequence $\\{x_n\\}$, and let $N$ be an integer such that for all $n > N$, $x_n \\le X$. We now have a lemma.\n\n**Lemma 1.** Let $m$ be a positive integer of the form $m = c_1d_1 + \\dots + c_kd_k$, with $c_i$ non-negative integers. If $n > N + n_0 + m$ and $x_n = X$, then $x_{n-m} = X$.\n\n*Proof.* Suppose we are given $n' > N + n_0$ with $x_{n'} = X$. Then, $x_{n'-d_i} \\le X$, implying that\n\n$$\nX = x_{n'} \\le \\frac{x_{n'-d_1} + \\dots + x_{n'-d_k}}{k} \\le X.\n$$\n\nEquality must then hold in each step of the iterated inequality above, hence $x_{n'-d_i} = X$ for each $i$. We begin by setting $n' = n$ and apply this reasoning repeatedly, choosing $i$ at each step to decrease $n'$ to $n-m$. At each step, we change $n'$ to $n'-d_i$ while maintaining $x_{n'} = X$, allowing us to conclude that $x_{n-m} = X$. $\\square$\n\nNow, let $M$ be an integer such that for all $m > M$ there exist non-negative integers $c_1, c_2, \\dots, c_k$ such that $m = c_1d_1 + c_2d_2 + \\dots + c_kd_k$. Such an $M$ exists because $\\gcd(d_1, \\dots, d_k) = 1$. Because $X$ occurs infinitely often in $\\{x_n\\}$, we may find some $n > M + N + 2n_0$ for which $x_n = X$. By our choice of $M$, each $m \\in \\{M+1, M+2, \\dots, M+n_0\\}$ is the non-negative linear combination of the $d_i$. Further, we have $n > N + n_0 + m$, so by Lemma 1 we conclude $x_{n-m} = X$ for $M+1 \\le m \\le M+n_0$. This shows that $n_0$ consecutive terms of the sequence are equal, so the sequence is constant thereafter.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19231,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = C_{200}^n \\cdot (\\sqrt[3]{6})^{200-n} \\cdot \\left(\\frac{1}{\\sqrt{2}}\\right)^n$ for $n = 1, 2, \\dots, 95$. How many terms in $\\{a_n\\}$ are integers?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\na_n = C_{200}^n \\cdot 3^{\\frac{200-n}{3}} \\cdot 2^{\\frac{400-5n}{6}}.\n$$\nFor $a_n$ to be an integer ($1 \\leq n \\leq 95$), both $\\frac{200-n}{3}$ and $\\frac{400-5n}{6}$ must be integers. This requires $6 \\mid n+4$.\n\nFor $n = 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80$, both exponents are non-negative integers, so the corresponding $a_n$ (14 values) are integers.\n\nFor $n = 86$:\n$$\na_{86} = C_{200}^{86} \\cdot 3^{38} \\cdot 2^{-5}.\n$$\nThe number of factors of 2 in $200!$ is\n$$\n\\left[ \\frac{200}{2} \\right] + \\left[ \\frac{200}{4} \\right] + \\left[ \\frac{200}{8} \\right] + \\left[ \\frac{200}{16} \\right] + \\left[ \\frac{200}{32} \\right] + \\left[ \\frac{200}{64} \\right] + \\left[ \\frac{200}{128} \\right] = 197.\n$$\nIn $86!$ and $114!$, the numbers are 82 and 110, respectively. Thus, in $C_{200}^{86} = \\frac{200!}{86! \\cdot 114!}$, the number of factors of 2 is $197 - 82 - 110 = 5$. So $a_{86}$ is an integer.\n\nFor $n = 92$:\n$$\na_{92} = C_{200}^{92} \\cdot 3^{36} \\cdot 2^{-10}.\n$$\nThe numbers of factors of 2 in $92!$ and $108!$ are 88 and 105, so in $C_{200}^{92}$ it is $197 - 88 - 105 = 4$. Therefore, $a_{92}$ is not an integer.\n\nIn total, the required number is $14 + 1 = 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19232,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. There are $n$ boxes $B_1, B_2, \\dots, B_n$, each of which contains some balls. One can perform the following moves:\n\nChoose positive integers $i$ and $j$ with $1 \\leq i \\leq j \\leq n$, and add exactly one ball to each of the boxes $B_i, B_{i+1}, \\dots, B_j$.\n\nFor positive integers $x_1, x_2, \\dots, x_n$, let $f(x_1, x_2, \\dots, x_n)$ be the minimum number of moves required to make the number of balls in each of the boxes divisible by $3$, starting from $x_i$ balls in $B_i$ for each $i = 1, 2, \\dots, n$. Find the maximum value of $f(x_1, x_2, \\dots, x_n)$.\n\n(If $3 \\mid x_i$ for $i = 1, 2, \\dots, n$, then $f(x_1, x_2, \\dots, x_n) = 0$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\lceil \\frac{2n+2}{3} \\rceil$.\n\nFor $i = 0, 1, \\dots, n$, let $y_i = x_i - x_{i+1}$ where $x_0 = x_{n+1} = 0$. Note that $y_0 + y_1 + \\dots + y_n = 0$. The move in the problem is equivalent to the following:\n\nChoose non-negative integers $i$ and $j$ with $0 \\leq i < j \\leq n$, and replace $y_i$ and $y_j$ by $y_i - 1$ and $y_j + 1$, respectively.\n\nWe call this move the $(i, j)$-move. Our goal is to make each $y_i$ a multiple of $3$. Now we prove that for every sequence of integers $y_0, y_1, \\dots, y_n$ with $y_0 + y_1 + \\dots + y_n = 0$, we can make each of the $y_i$'s a multiple of $3$ by performing at most $\\lceil \\frac{2n+2}{3} \\rceil$ moves.\n\nWe use induction on $n$.\n\nOne can easily check that it is possible when $n = 1, 2$.\n\nSuppose $n > 2$. If $3 \\mid y_i$ for some $i$, then we can ignore $y_i$, and so we need at most $\\lceil \\frac{2n}{3} \\rceil$ moves to make all $y_i$'s multiples of $3$. Hence, we may assume that $y_i \\not\\equiv 0 \\pmod{3}$ for each $i$. We consider the following four cases.\n\n*Case 1.* There exist $0 \\leq i < j \\leq n$ such that $y_i \\equiv 1 \\pmod{3}$ and $y_j \\equiv 2 \\pmod{3}$.\n\nThen, we perform the $(i, j)$-move; then $y_i$ and $y_j$ become multiples of $3$. So, in this case, we need at most $1 + \\lceil \\frac{2n-2}{3} \\rceil \\leq \\lceil \\frac{2n+2}{3} \\rceil$ moves.\n\n*Case 2.* There exist $0 \\leq i < j < k \\leq n$ such that $y_i \\equiv y_j \\equiv y_k \\equiv 1 \\pmod{3}$.\n\nThen, we perform the $(i, k)$-move and the $(j, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \\lceil \\frac{2n-4}{3} \\rceil = \\lceil \\frac{2n+2}{3} \\rceil$ moves.\n\n*Case 3.* There exist $0 \\leq i < j < k \\leq n$ such that $y_i \\equiv y_j \\equiv y_k \\equiv 2 \\pmod{3}$.\n\nThen, we perform the $(i, j)$-move and the $(i, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \\lceil \\frac{2n-4}{3} \\rceil = \\lceil \\frac{2n+2}{3} \\rceil$ moves.\n\n*Case 4.* Neither case 1, case 2 nor case 3 occurs.\n\nSince $n \\geq 3$, the only possible case is that $n = 3$ and $(y_0, y_1, y_2, y_3) \\equiv (2, 2, 1, 1) \\pmod{3}$. In this case, we perform the $(0, 1)$-move, the $(0, 3)$-move and the $(2, 3)$-move. Then, each $y_i$ becomes a multiple of $3$.\n\nTherefore, by induction, we can make each $y_i$ a multiple of $3$ by performing at most $\\lceil \\frac{2n+2}{3} \\rceil$ moves.\n\nFinally, for the following cases, we need at least $\\lceil \\frac{2n+2}{3} \\rceil$ moves, which implies that the answer is $\\lceil \\frac{2n+2}{3} \\rceil$:\n\n$$\n(y_0, y_1, \\dots, y_n) = \\begin{cases}\n(2, 2, \\dots, 2, 2, 2), & \\text{if } n \\equiv 2 \\pmod{3} \\\\\n(2, 2, \\dots, 2, 2, 1), (2, 2, \\dots, 2, 1, 2), & \\text{if } n \\equiv 1 \\pmod{3} \\\\\n(2, 2, \\dots, 2, 1, 1), & \\text{if } n \\equiv 0 \\pmod{3}\n\\end{cases}\n$$\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19233,
    "subject": "Mathematics (Olympiad)",
    "question": "In the land of Flensburg, there is a single, infinitely long street with houses numbered $2, 3, \\ldots$. The police in Flensburg are trying to catch a thief who, every day, moves from the house where he is currently hiding to one of its neighbouring houses.\n\nTo taunt the local law enforcement, the thief reveals every day the highest prime divisor of the house he will move to.\n\nEvery Sunday, the police are allowed to search a single house, and they catch the thief if they search the house he is currently occupying. Determine if the thief will be able to escape the police indefinitely or if the police have a strategy to catch the thief in finite time.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the police are always able to catch the thief in finite time.\n\nLet $h_i$ denote the house the thief stays at on the $i$-th night and $p_i$ denote the greatest prime divisor of $h_i$.\n\nThe police know that he stays at different neighbouring houses every night, so $|h_{i+1} - h_i| = 1$ for all non-negative integers $i$. Let us assume that the police are given the address of the thief's first two hiding spots; then we will prove by induction that the police can determine $h_i$ precisely, except being unable to distinguish between houses numbered $2$ and $4$.\n\nAssume the police know $h_{i-2}$ and $h_{i-1}$, then they know that $h_i = h_{i-2}$ or $h_i = 2h_{i-1} - h_{i-2}$. In the first case, they will receive $p_i = p_{i-2}$, and in the latter case, they will receive $p_i$ as the biggest prime divisor of $2h_{i-1} - h_{i-2}$. Assume that they are unable to distinguish between these two cases, i.e., that $p_i = p_{i-2}$, which implies\n\n$$\np_{i-2} \\mid 2h_{i-1} - h_{i-2}, \\text{ i.e. } p_{i-2} \\mid 2h_{i-1}, \\text{ i.e. } p_{i-2} \\mid 2, \\text{ i.e. } p_{i-2} = 2\n$$\n\nsince $|h_{i-1} - h_{i-2}| = 1$ implies $\\gcd(h_{i-1}, h_{i-2}) = 1$. Moreover, since $p_i = p_{i-2} = 2$ are the biggest prime divisors of $h_i = 2h_{i-1} - h_{i-2}$ and $h_{i-2}$, they must both be powers of $2$. However, the only powers of two with a difference of exactly $2$ are $2$ and $4$. Hence $\\{h_{i-2}, 2h_{i-1} - h_{i-2}\\} = \\{2, 4\\}$, i.e., $h_{i-1} = \\frac{2+4}{2} = 3$.\n\nThus, either the police will with certainty be able to determine $h_i$, or $h_{i-1} = 3$, in which case $h_i$ may equal either $2$ or $4$. To complete the inductive step, we observe that the police are always able to determine the parity of $h_j$, since it changes every day. Thus, in the future, if the police know that $h_j \\in [2, 4]$, then they can either determine $h_j = 3$ or $h_j \\in \\{2, 4\\}$. However, the only way for the thief to leave the interval $[2, 4]$ is to go to house number $5$, in which case the police will be alerted by receiving $p_j = 5$, and they can again with certainty determine $h_j = 5$ and $h_{j-1} = 4$, preserving our inductive hypothesis.\n\nTo summarize, if the police know both $h_0$ and $h_1$, then they can always determine $h_i$ with certainty until $h_{i-1} = 3$. After this point, they will know the two last hiding places of the thief if he leaves the interval $[2, 4]$, restoring the inductive hypothesis, or otherwise, if he never leaves $[2, 4]$, be able to determine his position, up to confusion about $2$ and $4$ using the parity of the day.\n\nNow, to catch the thief in finite time, they may methodically try to guess all viable pairs of $(h_0, h_1)$, i.e., $h_0, h_1 \\in \\mathbb{N}_{\\ge 2}$ and $|h_0 - h_1| = 1$, of which there are countably many.\n\nFor each viable starting position, let us consider either the immediate Sunday or the one after that. Since each week has an odd number of days, we are certain that exactly one of these days gives us that the thief is hiding in an odd house (given our assumption on his starting position). Thus, due to our inductive hypothesis, we can precisely determine where the thief will be, and search this house.\n\nIf the thief is hiding in that house, the police win, and if not, they will with certainty know that their guess of starting positions was incorrect, and move on to the next guess. By the above argument, each guess of initial starting positions requires at most two weeks, meaning that the police will catch the thief in finite time.\n\n**Remark:** Note that if a week contained an even number of days, then the police would not be able to guarantee that they would be able to catch the thief, if the thief moves between house number $3$ and $\\{2, 4\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19234,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be an integer, $m \\ge 2$. Each student in a school is practising at most $m$ hobbies. Among any $m$ students, there exist two students who have a common hobby. Find the smallest number of students for which there must exist a hobby which is practised by at least 3 students.\n\nWhat is the answer?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are $m^2 - 1 = (m-1)(m+1)$ students. We can split the students into $m-1$ groups, each with $m+1$ students. Assign each group a unique hobby, and let each student in a group share that hobby with every other student in the group, and have no other hobbies. Thus, each student has exactly $m$ hobbies. Among any $m$ students, at least two are in the same group and thus share a hobby. No hobby is practised by more than 2 students. If there are fewer than $m^2 - 1$ students, a similar construction works by omitting students.\n\nNow, suppose there are at least $m^2$ students. Assume, for contradiction, that every hobby is practised by at most 2 students. Build a group by repeatedly adding a student who shares no hobby with those already in the group. Since each student has at most $m$ hobbies, and each hobby is shared with at most one other student, after $i$ iterations, at most $i(m+1)$ students are excluded from being added. Since $(m-1)(m+1) < m^2$, we can form a group of $m$ students with no shared hobbies, contradicting the problem's condition. Thus, with $m^2$ students, some hobby is practised by at least 3 students.\n\n**Remark:** If we require a common hobby among every group of $k$ students, the maximum number of students is $(k-1)(m+1)$. The proof is analogous.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19235,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive real numbers such that $a + b + c + d = 4$. Prove that\n\n$$\n\\frac{ab}{a^2 - \\frac{4}{3}a + \\frac{4}{3}} + \\frac{bc}{b^2 - \\frac{4}{3}b + \\frac{4}{3}} + \\frac{cd}{c^2 - \\frac{4}{3}c + \\frac{4}{3}} + \\frac{da}{d^2 - \\frac{4}{3}d + \\frac{4}{3}} \\le 4.\n$$\n\n![](images/IRN_ABooklet_2022_p22_data_5507c0400c.png)\n\n![](images/IRN_ABooklet_2022_p23_data_a36227766f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First Solution: Note that $a^2 - \\frac{4}{3}a + \\frac{4}{3} = a^2 + 1 - \\frac{4}{3}a + \\frac{1}{3} \\ge 2a - \\frac{4}{3}a + \\frac{1}{3} = \\frac{2a+1}{3}$.\n\nHence,\n\n$$\n\\frac{ab}{a^2 - \\frac{4}{3}a + \\frac{4}{3}} \\le \\frac{ab}{\\frac{2a+1}{3}} = \\frac{3ab}{2a+1} = \\frac{1}{2} \\left( 3b - \\frac{3b}{2a+1} \\right).\n$$\n\nThat is,\n\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{ab}{a^2 - \\frac{4}{3}a + \\frac{4}{3}} &\\le \\frac{1}{2} \\sum_{cyc} \\left( 3b - \\frac{3b}{2a+1} \\right) = \\frac{3}{2} (a+b+c+d) - \\frac{3}{2} \\sum_{cyc} \\frac{b}{2a+1} \\\\\n&= 6 - \\frac{3}{2} \\sum_{cyc} \\frac{b}{2a+1}.\n\\end{aligned}\n$$\n\nNow, it suffices to prove\n\n$$\n\\sum_{cyc} \\frac{b}{2a+1} \\ge \\frac{4}{3}.\n$$\n\nNotice that\n\n$$\n\\sum_{cyc} \\frac{b}{2a+1} = \\sum_{cyc} \\frac{b^2}{2ab+b} \\ge \\frac{(a+b+c+d)^2}{2(a+c)(b+d)+4} = \\frac{8}{2+(a+c)(b+d)}.\n$$\n\nWe need to prove $(a+c)(b+d) \\le 4$. Indeed,\n\n$$\n(a+c)(b+d) \\le \\frac{1}{4}(a+b+c+d)^2 = 4.\n$$\n\nThe equality case occurs whenever $a = b = c = d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19236,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles touch each other externally at point $C$. Consider two diameters $A_1A_2$ and $B_1B_2$ in the same direction. A circle with its center on the common internal tangent passes through the intersection point of $A_1B_2$ and $A_2B_1$, and meets these lines at points $M$ and $N$. Prove that $MN$ is perpendicular to $A_1A_2$ and $B_1B_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Since point $C$ is a center of homothety that transforms one circle into another, then $C = A_1B_2 \\cap A_2B_1$, and $A_1B_2 \\perp A_2B_1$ (see the figure below).\n\n![](images/Ukrajina_2011_p20_data_d65a639b5a.png)\n\nLet $D = MN \\cap B_1B_2$. Then, $\\angle DB_2C = \\angle B_2A_1A_2 = \\angle A_2CO = \\angle CND$. Therefore, $DB_2NC$ is cyclic and $\\angle B_2DN = \\angle B_2CA_2 = \\frac{\\pi}{2}$, which implies that $MN \\perp B_1B_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19237,
    "subject": "Mathematics (Olympiad)",
    "question": "There are only coins of 11 pesos and 13 pesos in a country (and no bills). An ice-cream shop is about to open and there is a line of customers waiting. Every customer wants to buy a cone of ice-cream and has exactly $155$ pesos. The cone costs $12$ pesos. The salesperson wants to attend everyone by giving back the exact change without borrowing or exchanging money. Find the minimum amount of money she needs to have in advance in order to do so.",
    "options": [],
    "answer": "See solution",
    "solution": "Having $108$ pesos is sufficient. There is only one way to express $155$ as $11a + 13b$ with $a, b$ nonnegative integers: $a = 7$ and $b = 6$. So each customer has $7$ coins of $11$ and $6$ coins of $13$.\n\nTo serve a customer, it is enough to have available either $6$ coins of $11$ or $5$ coins of $13$ (or both). In the first case, the salesperson takes from the customer $6$ coins of $13$ and gives back $6$ coins of $11$:\n\n$$6 \\times 13 - 6 \\times 11 = 12$$\n\nIn the second case, the salesperson takes $7$ coins of $11$ and gives back $5$ coins of $13$:\n\n$$7 \\times 11 - 5 \\times 13 = 12$$\n\nAn initial amount of $N \\geq 108$ pesos ensures $6$ coins of $11$ or $5$ coins of $13$. Indeed, if there were at most $5$ coins of $11$ and at most $4$ coins of $13$, then $N \\leq 5 \\times 11 + 4 \\times 13 = 107$. So the first customer can be attended whenever $N \\geq 108$. The remaining ones can be attended too, because $N$ will be still greater when their turn comes.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19238,
    "subject": "Mathematics (Olympiad)",
    "question": "299 digits $0$ and one digit $1$ are written in a circle.\n\nThe following moves are allowed:\n\n- From each digit, subtract the sum of the adjacent digits.\n- Select two digits with exactly two digits between them and increase both by $1$ or decrease both by $1$.\n\nIs it possible to obtain, after a finite number of such moves, an arrangement in which there are two adjacent digits $1$, and the rest of the digits are $0$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** No, not possible.\n\nLet us analyse how the allowed moves affect the sum of the digits written in a circle. Denote the numbers by $a_1, a_2, \\dots, a_{300}$. After a move of the first type, the new numbers are $b_k = a_k - a_{k-1} - a_{k+1}$ for $k = 1, \\dots, 300$ (with $a_{301} \\equiv a_1$). Let $S = a_1 + a_2 + \\dots + a_{300}$. Then $b_1 + b_2 + \\dots + b_{300} = S - S - S = -S$.\n\nAfter a move of the second type, the sum becomes $S+2$ or $S-2$. Since the parity of the sum after each move does not change, it is impossible to get an even sum from the original odd sum.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19239,
    "subject": "Mathematics (Olympiad)",
    "question": "A large rectangle is subdivided into smaller rectangles, each of which has at least one pair of sides of integer length. Prove that the large rectangle also has at least one pair of sides of integer length.",
    "options": [],
    "answer": "See solution",
    "solution": "Label the large rectangle by $R$ and place it in the plane so that its lower-left corner is at $(0, 0)$ with sides parallel to $Ox$ and $Oy$.\n\nLet $S$ be the set of vertices of the small rectangles whose both coordinates are integers, and let $T$ be the set of all the small rectangles. We form a bipartite graph on the vertex set $S \\cup T$ by joining each point in $S$ to every rectangle in $T$ of which it is a vertex. Observe that each small rectangle has either $0$, $2$, or $4$ vertices in $S$ (because of integer length side), so it contributes an even number of edges. Hence the total number of edges in this bipartite graph is even. On the other hand, each point in $S$ that is not a corner of $R$ is a vertex of exactly $2$ or $4$ small rectangles, and so also has even degree.\n\nSince $(0, 0)$ is in $S$ and is a corner of exactly one small rectangle, it has odd degree. Therefore there must be another vertex in $S$ of odd degree. The only vertices of odd degree can be corners of $R$, so at least one other corner of $R$ lies in $S$. In particular, either the width or the height of $R$ is an integer.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19240,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the isosceles triangle $ABC$ with $AB = AC$. A semicircle of diameter $EF$ situated on the side $BC$ is tangent to the sides $AB$ and $AC$ at $M$ and $N$, respectively. The line $AE$ intersects the semicircle at $P$. Prove that the line $PF$ passes through the midpoint of the chord $MN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the semicircle and let $R$ be the midpoint of segment $MN$.\n\n![](images/Saudi_Arabia_booklet_2012_p31_data_629942c22e.png)\n\nIn triangle $ANO$ we have $AN^2 = AR \\cdot AO$. Using the power of the point $A$ with respect to the circle we get\n\n$$\nAM^2 = AP \\cdot AE = AN^2 = AR \\cdot AO. \\quad (1)\n$$\n\nFrom (1) it follows that the quadrilateral $PROE$ is cyclic, hence $RP \\perp AE$. Since $FP \\perp AE$, we get that the points $F$, $R$, $P$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19241,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $11^n - 1$ is divisible by $10^n - 1$.\n\n![](images/UkraineMO_2015-2016_booklet_p7_data_9f942250f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Such $n$ does not exist.\n\nSince $10^n - 1 = 9 \\cdot (10^{n-1} + \\dots + 10 + 1)$, we obtain that $11^n - 1$ must be divisible by $9$. Considering residues modulo $9$: $11^n - 1 \\equiv 2^n - 1 \\pmod{9}$, so $n = 6k$. But then $10^{6k} - 1$ is divisible by $10^6 - 1$, which is divisible by $10^3 + 1$ and $10 + 1 = 11$, which is not possible.\n\n![](images/UkraineMO_2015-2016_booklet_p7_data_81e74c9f8f.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19242,
    "subject": "Mathematics (Olympiad)",
    "question": "Equilateral triangle $\\triangle ABC$ is inscribed in circle $\\omega$ with radius $18$. Circle $\\omega_A$ is tangent to sides $\\overline{AB}$ and $\\overline{AC}$ and is internally tangent to $\\omega$. Circles $\\omega_B$ and $\\omega_C$ are defined analogously. Circles $\\omega_A$, $\\omega_B$, and $\\omega_C$ meet in six points—two points for each pair of circles. The three intersection points closest to the vertices of $\\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\\triangle ABC$, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\\triangle ABC$. The side length of the smaller equilateral triangle can be written as $\\sqrt{a} - \\sqrt{b}$, where $a$ and $b$ are positive integers. Find $a+b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of $\\omega$, and let $D$, $E$, and $F$ be points diametrically opposite $A$, $B$, and $C$, respectively, in $\\omega$. Denote the centers of circles $\\omega_A$, $\\omega_B$, and $\\omega_C$ by $O_A$, $O_B$, and $O_C$, respectively, and let $r$ be the radius of these circles. Circle $\\omega_B$ is the incircle of an equilateral triangle with altitude $\\overline{BE}$, so $r = 12$ and $OO_A = OD - O_A D = 18 - 12 = 6$.\n\nLet circles $\\omega_A$ and $\\omega_B$ intersect at points $P$ and $Q$, as shown in the figure, and let segments $\\overline{O_A O_B}$ and $\\overline{PQ}$ intersect at point $M$.\n\n![](images/2022AIME_I_Solutions_p5_data_ed7ce62ca4.png)\n\nNote that $O_A P = O_A Q = O_B P = O_B Q = 12$. Therefore, $PO_A Q O_B$ is a rhombus and point $M$ is the midpoint of its diagonals $\\overline{O_A O_B}$ and $\\overline{PQ}$. Because $\\triangle OMO_A$ is a $30$-$60$-$90^\\circ$ right triangle whose hypotenuse, $\\overline{OO_A}$, has length $6$, it follows that $OM = 3$ and $O_A M = 3\\sqrt{3}$.\n\n$$\nPM^2 = O_A P^2 - O_A M^2 = 144 - 27 = 117,\n$$\nso $PM = 3\\sqrt{13}$ and $OP = PM - OM = 3\\sqrt{13} - 3$. The circumradius of the smaller equilateral triangle is equal to $3\\sqrt{13} - 3$, and therefore its side length is equal to $\\sqrt{3} \\cdot (3\\sqrt{13} - 3) = \\sqrt{351} - \\sqrt{27}$. The requested sum is $351 + 27 = 378$.\n\n**Note:** As long as $\\sqrt{ab}$ is irrational, as is the case with $\\sqrt{351} \\cdot 27 = 27\\sqrt{13}$, the representation $\\sqrt{a} - \\sqrt{b}$ is unique.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19243,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $k$ be positive integers. Suppose that for any integer $n \\geq 2025$, there exists a positive integer $x_n > n$ such that $x_n \\mid n^2 + a$ and $x_n + k \\mid n^2 + b$. Prove that $k = b - a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n \\geq 2025$ and define $y_n = \\frac{n^2 + a}{x_n}$ and $z_n = \\frac{n^2 + b}{x_n + k}$. Consider $c_n = y_n - z_n$. We have\n\n$$\n|c_n| = \\left| \\frac{n^2}{x_n} - \\frac{n^2}{x_n + k} + \\frac{a}{x_n} - \\frac{b}{x_n + k} \\right| = \\left| k \\cdot \\frac{n^2}{x_n(x_n + k)} + \\frac{a}{x_n} - \\frac{b}{x_n + k} \\right| < k + \\frac{a + b}{n}\n$$\n\nso $c_n$ can take finitely many values.\n\nFurthermore, $(x_n + k)(y_n - c_n) = n^2 + b$ expands into $k y_n = u_n + c_n x_n$, where $u_n = b + k c_n - a$ only depends on $c_n$.\n\nFrom these we can easily get\n\n$$\n4 c_n k n^2 = (2 c_n x_n + u_n)^2 - (u_n^2 + 4 a c_n k). \\qquad (1)\n$$\n\nAs $c_n$ is bounded, expressions $4 c_n k$ and $u_n^2 + 4 a c_n k$ each take finitely many values as $n$ varies through positive integers.\n\n**Lemma.** For any pair of integers $(a, b)$ such that either $a$ is not a perfect square or $b \\neq 0$, there exist infinitely many primes $p$ together with a residue class $n$ such that $a n^2 + b$ is a quadratic non-residue mod $p$.\n\n*Proof.* Let $p$ be a large prime with $p \\nmid a, b$.\n\nAssume that the claim is not true and let $n = 1, 2, \\dots, \\frac{p - 1}{2}$. Since expressions $a n^2 + b$ produce different residues, we get equality on the sets of residues:\n\n$$\n\\{1^2, 2^2, \\dots, \\left(\\frac{p - 1}{2}\\right)^2\\} = \\{a \\cdot 1^2 + b, a \\cdot 2^2 + b, \\dots, a \\cdot \\left(\\frac{p - 1}{2}\\right)^2 + b\\}.\n$$\n\nSumming both sets yields $a S + \\frac{p - 1}{2} b \\equiv S \\pmod{p}$ where $S = 1^2 + 2^2 + \\dots + \\left(\\frac{p - 1}{2}\\right)^2 = \\frac{p(p - 1)(p + 1)}{24} \\equiv 0 \\pmod{p}$, since $p$ is a large prime. Therefore $p \\mid b$, which is a contradiction. Since we can pick infinitely many integers from a class mod $p$, the lemma follows. $\\square$\n\nLet $(a_i, b_i)$ for $i \\in \\{1, 2, \\dots, l\\}$ be all the possible values of $(4 c_n k, u_n^2 + 4 a c_n k)$.\n\n**Claim.** Let $a_i n^2 + b_i = m^2$ be a finite collection of conics such that for each one either $a_i$ is not a perfect square or $b_i \\neq 0$. Then, there exist infinitely many positive integers $n$ such that none of $n, n + 1, n + 2$ belong to any of these conics.\n\n*Proof.* Let $p_{i,0}, p_{i,1}, p_{i,2}$ and $n_{i,0}, n_{i,1}, n_{i,2}$ be three primes and their corresponding residues that the lemma provides for the pair $(a_i, b_i)$. Note that since the lemma provides infinitely many primes for each pair $(a_i, b_i)$, we can ensure that all mentioned primes are distinct. By choosing $n$ to satisfy\n\n$$\n\\begin{align*}\n  n &\\equiv n_{i,0} \\pmod{p_{i,0}}, \\\\\n  n + 1 &\\equiv n_{i,1} \\pmod{p_{i,1}}, \\\\\n  n + 2 &\\equiv n_{i,2} \\pmod{p_{i,2}},\n\\end{align*}\n$$\n\nfor all $i$, we see that none of $a_i n^2 + b_i$, $a_i (n + 1)^2 + b_i$, $a_i (n + 2)^2 + b_i$ can be a perfect square for any $i$ (since each expression is a quadratic non-residue modulo some prime). Since there are obviously infinitely many such $n$, the claim follows. $\\square$\n\nTo finish the solution, pick a large enough positive integer $n$ from the claim. If $c_n \\neq 0$ we get that $4 c_n k$ is a square and $u_n^2 + 4 a c_n k = 0$. Therefore $u_n^2 + 4 a c_n k = u_n^2 + a t^2 = 0$ for some integer $t$. Since $a$ is a positive integer it follows that $t = 0$ and thus $c_n = 0$.\n\nThen $y_n = z_n$, so $k y_n = b - a$. Furthermore, as this also holds for $n + 1$ and $n + 2$, we get that $y_n = \\frac{b - a}{k}$ divides\n\n$$\n n^2 + a, \\quad (n + 1)^2 + a, \\quad (n + 2)^2 + a.\n$$\n\nBy simple calculations we obtain $y_n = 1$. Therefore $k = b - a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19244,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the alphabet have only two letters: _A_ and _B_. Is it possible to choose one word of length 5, one word of length 6, ..., one word of length 30 such that any word of length 300 contains one of the chosen words as a substring?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is NO.\n\nConsider all words of length 13 and continue them periodically in both directions to create sequences. For details, take $x_1x_2\\ldots x_{13}$ as some original word of length 13, then add $x_1, x_2, x_3, \\ldots$ to the right and add $x_{13}, x_{12}, x_{11}, \\ldots$ to the left. This results in:\n\n$$\n\\ldots x_{11}x_{12}x_{13}(x_1 x_2 \\ldots x_{13})x_1 x_2 x_3 \\ldots\n$$\n\nThere are two identical original words, $AAA\\ldots A$ and $BBB\\ldots B$, which generate unique sequences. For the other words, each of them will generate the same sequence as another 12 words as follows:\n\n$$\n(x_2x_3x_4\\ldots x_{13}x_1),\\ (x_3x_4x_5\\ldots x_{13}x_1x_2),\\ \\ldots,\\ (x_{13}x_1x_2\\ldots x_{12})\n$$\n\nSo in total, the number of distinct sequences is\n\n$$\n\\frac{2^{13}-2}{13}+2 > 600.\n$$\n\nBy the definition of the sequences, one can check that any word of length at least 13 may appear only in one sequence. Each word of length $n < 13$ may appear in at most $2^{13-n}$ sequences, since we fix $n$ letters and choose another $13 - n$ letters with $2^{13-n}$ ways to form a word of length 13 (some of them may coincide). Note that\n\n$$\n(2^8 + 2^7 + \\ldots + 2^0) + 17 = 2^9 + 16 < 600\n$$\n\nso we will have available sequences.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19245,
    "subject": "Mathematics (Olympiad)",
    "question": "Give an example of a hexagon (not necessarily convex) that can be cut with one straight line into a triangle and a quadrilateral (not necessarily convex), but which cannot be cut into two triangles or two quadrilaterals.\n\n![](images/Ukraine_2020_booklet_p19_data_bab6908917.png)",
    "options": [],
    "answer": "See solution",
    "solution": "On the figure, the dashed line shows how to cut the hexagon—one has to draw a segment $CF$ or $BD$.\n\nLet us now see where the line of separation of $ABCDEF$ can be drawn. If it passes through a vertex of the hexagon and is different from lines $AC$ and $BE$, e.g., $AL$, then on the side, a new point ($L$) appears, meaning the resulting polygons must have 7 vertices, two of which are counted twice (in this case, $A$ and $L$), which means that in total, these polygons must have 9 vertices, which is not possible for both two triangles and two quadrilaterals. Analogously, if the line does not pass through a vertex (e.g., $MN$), then in total, there must be 10 vertices, which is also impossible for two triangles and two quadrilaterals. The only case left is when the segment connects two vertices of the hexagon, e.g., $BE$; then in total this yields 8 vertices, which could be formed by two quadrilaterals. But it suffices to check all such segments to see that none such segment partitions the hexagon into two quadrilaterals. $BE$ and $FD$ are the only such segments, and each of them partitions the hexagon into a triangle and a pentagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19246,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $B$ be an interior point of the segment $AC$. The equilateral triangles $\\triangle ABM$ and $\\triangle BCN$ are constructed in the same half-plane determined by the line $AC$. The lines $AN$ and $CM$ intersect at $L$. Find the angle $\\angle CLN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that the angle $\\angle MBN = 60^\\circ$. Hence $\\angle ABN = \\angle MBC = 120^\\circ$. Because the triangles $ABM$ and $BCN$ are equilateral, we have $\\overline{AB} = \\overline{BM}$ and $\\overline{BC} = \\overline{BN}$. Hence $\\triangle ABN \\cong \\triangle MBC$, from which $\\angle ANB = \\angle MCB = \\alpha$. Let $K$ be the intersection point of the lines $MC$ and $BN$. Because $\\angle BKC = \\angle MKN = \\beta$ (opposite angles), from $\\triangle BKC$ we have $\\alpha + \\beta = 180^\\circ - 60^\\circ = 120^\\circ$. Finally, from $\\triangle LKN$ we obtain $\\angle KLN = 180^\\circ - (\\alpha + \\beta) = 60^\\circ$, hence $\\angle CLN = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19247,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an odd integer, and let $a_1, a_2, a_3$ be integers such that $1 \\leq a_i \\leq k-1$ for $i=1,2,3$. For positive integers $n_1, n_2, n_3$, define\n$$\nb_1 = a_1(1 + k + k^2 + \\cdots + k^{n_1}), \\quad b_2 = a_2(1 + k + k^2 + \\cdots + k^{n_2}), \\quad b_3 = a_3(1 + k + k^2 + \\cdots + k^{n_3}).\n$$\nFind all triples $(n_1, n_2, n_3)$ for which there exist $k, a_1, a_2, a_3$ as above such that $b_1 b_2 = b_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Alternate Solution:\n\nIf we rewrite the relation $b_1 b_2 = b_3$ using the geometric series formula, we get\n$$\na_1 a_2 \\cdot \\frac{k^{n_1+1}-1}{k-1} \\times \\frac{k^{n_2+1}-1}{k-1} = a_3 \\cdot \\frac{k^{n_3+1}-1}{k-1}.\n$$\nMultiplying both sides by $(k-1)^2$ gives\n$$\na_1 a_2 (k^{n_1+1} - 1) (k^{n_2+1} - 1) = a_3 (k^{n_3+1} - 1) (k-1) \\quad (*)\n$$\nSince $k^{n_2+1} - 1 \\geq k^2 - 1 > a_3(k-1)$ implies $k^{n_1+1} - 1 < k^{n_3+1} - 1$, we get $n_1 < n_3$. Similarly, $n_2 < n_3$. We may assume $n_1 \\geq n_2$ without loss of generality.\n\nAssume $n_2 \\geq 2$. Considering equation $(*)$ modulo $k^3$, we get $a_1 a_2 \\equiv -a_3(k-1) \\pmod{k^3}$. However,\n$$\n0 < a_1 a_2 + a_3 (k-1) < k^2 + k(k-1) < 2k^2 \\leq k^3\n$$\nso we get a contradiction. Thus, $n_2 = 1$.\n\nDividing both sides of $(*)$ by $k-1$ gives\n$$\na_1 a_2 (k^{n_1+1} - 1)(k+1) = a_3 (k^{n_3+1} - 1).\n$$\nSuppose $n_1 \\geq 2$. Considering both sides modulo $k^3$, we get $-a_1a_2(k+1) \\equiv -a_3 \\pmod{k^3}$. However,\n$$\n\\begin{aligned}\na_1a_2(k+1) - a_3 &\\geq (k+1) - (k-1) = 2 > 0, \\\\\na_1a_2(k+1) - a_3 &\\leq (k-1)^2(k+1) - 1 = k^3 - k^2 - k < k^3\n\\end{aligned}\n$$\nso we get a contradiction. Therefore, $n_1 = 1$.\n\nSummarizing, we get\n$$\na_1a_2(k^2 - 1)(k + 1) = a_3(k^{n_3+1} - 1),\n$$\nand since\n$$\n\\begin{aligned}\nk^{n_3+1} - 1 &\\leq a_3(k^{n_3+1} - 1) = a_1a_2(k^2 - 1)(k + 1) \\\\\n&\\leq (k-1)^2(k^2 - 1)(k + 1) \\\\\n&= k^5 - k^4 - 2k^3 + 2k^2 + k - 1 \\\\\n&< k^5 - 1,\n\\end{aligned}\n$$\nwe obtain $n_3 < 4$.\n\nIf $n_3 = 2$, then $a_1a_2(k+1)^2 = a_3(k^2+k+1)$, so $0 < a_1a_2 < a_3 < k$ holds, since $(k+1)^2 > k^2+k+1$. However, considering both sides modulo $k$, $a_1a_2 \\equiv a_3 \\pmod{k}$, which is a contradiction. As $n_3 > n_1 = 1$, we conclude from $b_1b_2 = b_3$ that $(n_1, n_2, n_3) = (1, 1, 3)$. We can find $(k, a_1, a_2, a_3)$ satisfying the conditions of the problem when $(n_1, n_2, n_3) = (1, 1, 3)$ as in the preceding solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $R_n = \\{0, 1, \\dots, n-1\\}$ and $R_m = \\{-(n-1), -(n-2), \\dots, -1, 0, 1, \\dots, n-2, n-1\\}$ where $m = 2n-1$. Consider a device consisting of a grid of $2n-1$ lines and $n$ columns, with a ball in each cell. Balls in the same line are connected by red ropes; balls in different lines and different columns are connected by white ropes. There is no rope between balls in the same column. Prove that:\n\n1. If the device has a good labeling by $R_n$, then it has a sensitive labeling by $R_m$.\n2. For any $m < 2n-1$, there exists a device with a good labeling by $R_n$ but no sensitive labeling by $R_m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that if a device has a good labeling by $R_n$, then it has a sensitive labeling by $R_m$, where $m = 2n-1$. In\n$$\nR_m = R_{2n-1} = \\{-(n-1), -(n-2), \\dots, -1, 0, 1, \\dots, n-2, n-1\\}\n$$\nthere are $n$ non-negative elements. Any good labeling of the device by these $n$ non-negative elements will also be a sensitive labeling.\n\nNow, construct a device with a good labeling by $R_n$ and no sensitive labeling for any $m < 2n-1$. Define a grid $2n-1 \\times n$ (with $2n-1$ lines and $n$ columns) and place a ball in each cell. Connect any two balls in the same line by red rope, and any two balls in different lines and different columns by white rope (no rope between balls in the same column). The device has a good labeling by $R_n$: color all balls in the same column identically and balls in different columns differently, using $n$ distinct colors.\n\nSuppose the device has a sensitive coloring by $R_m$.\n\n**Case 1.** There is a line where all balls are differently colored. Since all balls in this line are connected by red ropes, for each $k$ at most one of the colors $\\{-k, k\\}$ is used. Therefore, the total number of elements in $R_m$ with different absolute values should be at least $n$, so $m \\ge 2n-1$.\n\n**Case 2.** Each line contains at least two identically colored balls. Suppose the repeated colors on two lines are $a$ and $b$. Since any two balls in different lines and different columns are connected by a white rope, $a \\neq b$. Thus, repeated colors on any two lines are different, and the total number of different colors is at least $2n-1$, so $m \\ge 2n-1$.\n\nDone.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19249,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r > 0$ be a real number. All the interior points of the disc $D(r)$ of radius $r$ are colored with one of two colors, red or blue.\n\n1. If $r > \\frac{\\pi}{\\sqrt{3}}$, show that we can find two points $A$ and $B$ in the interior of the disc such that the distance $AB = \\pi$ and $A$ and $B$ have the same color.\n\n2. Does the conclusion in (a) hold if $r > \\frac{\\pi}{2}$?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the conclusion holds when $r > \\frac{\\pi}{2}$.\n\nWe begin with a circle $C(r)$ with center $C$ and radius $r > \\frac{\\pi}{2}$. Now, a regular polygon $P$ can be inscribed in the circle to have any odd number of sides $2k + 1$. Because the number of sides is odd, each vertex $A$ is opposite a pair of vertices which determine the opposite side $YZ$, and it is such longest diagonals $AY$ and $AZ$ that we are interested in. The greater $2k + 1$ is taken, the smaller $YZ$ will get; by taking $k$ large enough, we can make $YZ$ so small that $AY$ will be so close to being a diameter (of length $2r$ which is bigger than $\\pi$) that the length of $AY$ will also exceed $\\pi$. Suppose, then, that $k$ is chosen big enough to make $AY > \\pi$.\n\nClearly all such longest diagonals $AY$ are tangents to a small circle $C(s)$ in the center of $C(r)$.\n\n![](images/EGMO_TSTs_India_2023_d1_solns_p0_data_69fa900105.png)\n\nFigure 1:\n\nNow, it is vital to our solution that the length $AY$ of these tangents be exactly $\\pi$ units. Therefore, let the figure be shrunk toward the center $C$ in the ratio $\\pi : AY$; this will carry everything into the interior of the given circle, implying that all the image points will be colored, whereas the boundary of $C(r)$ was not colored to begin with. For simplicity, let us keep the same names for the images under this transformation, bearing in mind that now the length of every diagonal like $AY$ is $\\pi$.\n\nNow a tangent to $C(s)$ from a vertex of $P$ meets the circumcircle in one of the two opposite vertices of $P$. Consequently, the sequence of tangents $AY, YA_1, A_1Y_1, Y_1A_2, \\dots$ carries one around $P$ from $A$ through the vertices $A, Y, A_1, Y_1, A_2, \\dots$ and after $2k$ such steps, we reach the opposite vertex $Z$. Clearly, every adjacent pair of vertices in the above list of vertices are at a distance $\\pi$ units and starting with $A$ and ending with $Z$, we traverse through $2k$ tangents. Thus if $A$ is red, then $Y$ is blue, $A_1$ is red, $Y_1$ is blue and so on. This forces $Z$ to be colored red and both ends of the tangent $AZ$ have the same color. We are done. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19250,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha = \\cos \\frac{\\pi}{7}$. As $\\cos \\frac{4\\pi}{7} + \\cos \\frac{3\\pi}{7} = 0$, we have\n\n$$\n2(2\\alpha^2 - 1)^2 - 1 + (4\\alpha^3 - 3\\alpha) = 0.\n$$\n\nThis is the same as $(\\alpha + 1)(8\\alpha^3 - 4\\alpha^2 - 4\\alpha + 1) = 0$. Clearly, $\\alpha \\ne -1$. Therefore, $\\alpha$ is a root of\n\n$$\nP(x) = 8x^3 - 4x^2 - 4x + 1.\n$$\n\nSuppose on the contrary that $\\alpha = p + \\sqrt{q} + \\sqrt[3]{r}$. Then $\\alpha$ is also a root of\n\n$$\nQ(x) = (x - p - \\sqrt{q})^3 - r = x^3 - 3(p + \\sqrt{q})x^2 + 3(p + \\sqrt{q})^2 x - (p + \\sqrt{q})^3 - r.\n$$\n\nConsider the field $F = \\{a + b\\sqrt{q} : a, b \\in \\mathbb{Q}\\}$ (which is just $\\mathbb{Q}$ if $q = 0$). Show that $\\alpha$ cannot be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$ for rational $p, q, r$.",
    "options": [],
    "answer": "See solution",
    "solution": "* If $\\deg R = 1$, then since $\\alpha$ is a root of $P(x)$ and $Q(x)$, it is a root of $R(x)$, and hence belongs to $F$.\n* If $\\deg R = 2$, we can write $P(x) = R(x)S(x)$ where $\\deg S = 1$. Then $S(x)$ has a root in $F$, which is also a root of $P(x)$.\n\nNow, let $\\beta = a + b\\sqrt{q}$ be a root of $P(x)$ in $F$. Then $\\beta$ is a root of $(x-a)^2 = b^2q$, and so the minimal polynomial of $\\beta$ over $\\mathbb{Q}$ has degree at most 2. As this minimal polynomial divides $P(x)$, we find that $P(x)$ is reducible. As $\\deg P = 3$, it must consist of a linear factor. Thus, $P(x)$ has a rational root. However, it is routine to check that none of\n\n$$\n\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{4}, \\pm \\frac{1}{8}\n$$\n\nis a root of $P(x)$. By the rational root theorem, $P(x)$ does not have any rational root. This is a contradiction. Therefore, $\\alpha$ cannot be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19251,
    "subject": "Mathematics (Olympiad)",
    "question": "Секоја точка од рамнината е обоена во една од две бои, сина или црвена. Да се докаже дека во таа рамнина постои рамностран триаголник чии темиња се обоени во една иста боја.",
    "options": [],
    "answer": "See solution",
    "solution": "Најпрво ќе покажеме дека постои отсечка чии крајни точки се обоени во иста боја. Имено, во дадената рамнина конструираме рамностран триаголник, тогаш според Принципот на Дирихле меѓу трите темиња постојат две кои се обоени во иста боја, според ова во дадената рамнина постои отсечка чии крајни точки се обоени во иста боја.\n\nПонатаму ќе покажеме дека постои отсечка чии крајни точки и средишна точка се обоени во иста боја.\n\nНека $AB$ е отсечка чии крајни точки се обоени на пример во сина боја (таква отсечка постои според претходното). Нека $D$ и $E$ (од различни страни на $A$ и $B$) се точки такви што $\\overline{AD} = \\overline{AB} = \\overline{BE}$. Тогаш ако некоја од точките $D$ и $E$ е обоена во сина боја, задачата е решена. Затоа нека точките $D$ и $E$ се обоени во црвена боја. Средишната точка $F$ на $AB$ е средишна и на отсечката $DE$, и јасно $F$ е обоена или во сина или во црвена боја. Со ова тврдењето е покажано.\n\nСега да разгледаме три сини точки $A, B, C$ такви што $B$ е средина на $AC$, (такви постојат според претходното). Нека $D, E$ и $F$ се трети темиња на рамностраните триаголници конструирани над $AC, AB, BC$, соодветно од иста страна на правата $AC$.\n\nТогаш ако барем една од $E, F$ и $D$ е сина, задачата е решена.\n\nАко пак сите три точки $E, F$ и $D$ се црвени тогаш бараниот триаголник е $EFD$ (тој е рамностран и сите негови темиња се обоени црвено).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19252,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of 100 students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B$, student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What is the least possible total number of languages spoken by all the students?\n\n(A) 9 (B) 10 (C) 12 (D) 51 (E) 100",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (A):** Suppose the languages spoken are labeled $L_1, L_2, L_3, \\dots, L_n$. Note that the collection of all subsets of $\\{L_1, L_2, L_3, \\dots, L_n\\}$ of size $r$ will satisfy the conditions in the problem for any $r$ in the range $1 \\leq r < n$. For a given $n$, there are $\\binom{n}{r}$ subsets, and $\\binom{n}{r}$ is maximized when $r = \\lfloor \\frac{n}{2} \\rfloor$ or $r = \\lceil \\frac{n}{2} \\rceil$.\n\nIf $n = 8$ and $r = 4$, the number of distinct subsets is $\\binom{8}{4} = 70 < 100$, so $n > 8$. But $\\binom{9}{4} = \\binom{9}{5} = 126 > 100$, so $n = 9$ is the least possible total number of languages spoken by all the students.\n\n**Note:** The restriction on having the students speak the same number of languages is not needed. A collection of subsets of a given set with $n$ elements satisfying the condition that no member of the collection is a subset of another is called an *antichain*. Sperner's Theorem asserts that the largest antichain occurs when the subsets are all of size $r$, where $r = \\lfloor \\frac{n}{2} \\rfloor$ or $r = \\lceil \\frac{n}{2} \\rceil$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19253,
    "subject": "Mathematics (Olympiad)",
    "question": "In a math period, the teacher asks pupils to solve quadratic equations of the form $x^2 + px + q = 0$ where $p$ and $q$ are some integers. The teacher obtains every new equation by either increasing by 1 or decreasing by 1 the value of either $p$ or $q$ in the equation just solved. In the initial equation, $p = 2020$ and $q = 2010$, whereas in the last equation, $p = 2010$ and $q = 2020$. Is it definitely true that both solutions of at least one equation solved during the period are integers?",
    "options": [],
    "answer": "See solution",
    "solution": "In the first equation, $p - q = 10$, while in the last equation, $p - q = -10$. At each step, either $p$ or $q$ changes exactly by 1, so $p - q$ also changes by exactly 1. Thus, at some step, there must be an equation where $p - q = 1$, or equivalently, $p = q + 1$. The equation $x^2 + (q + 1)x + q = 0$ has integer solutions $-1$ and $-q$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19254,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ small balls have been placed into $n$ numbered boxes $B_1, B_2, \\dots, B_n$. Each time we can select a box $B_k$ and do the following operations:\n\n1. If $k=1$ and there is at least one ball in $B_1$, move one ball from $B_1$ into $B_2$.\n2. If $k=n$ and there is at least one ball in $B_n$, move one ball from $B_n$ into $B_{n-1}$.\n3. If $2 \\leq k \\leq n-1$ and there are at least two balls in $B_k$, move one ball from $B_k$ into $B_{k+1}$ and one ball into $B_{k-1}$, respectively.\n\nProve the following: no matter how the balls are distributed among the boxes originally, it is always realizable to let each box contain exactly one ball by finite operations.",
    "options": [],
    "answer": "See solution",
    "solution": "For any two vectors $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ and $\\mathbf{y} = (y_1, y_2, \\dots, y_n)$, if there exists $1 \\leq k \\leq n$ such that\n$$\nx_1 = y_1, \\dots, x_{k-1} = y_{k-1}, \\quad x_k > y_k,\n$$\nwe denote it as $\\mathbf{x} > \\mathbf{y}$. Let $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ represent the distribution of the balls among the boxes. Then $\\mathbf{x}$ is a non-negative integer vector. The operation defined in the question, if executable, can be expressed as $\\mathbf{x} + \\boldsymbol{\\alpha}_k$, where\n- $\\boldsymbol{\\alpha}_1 = (-1, 1, 0, \\dots, 0)$,\n- $\\boldsymbol{\\alpha}_k = (0, \\dots, 0, 1, -2, 1, 0, \\dots, 0)$ for $2 \\leq k \\leq n-1$,\n- $\\boldsymbol{\\alpha}_n = (0, \\dots, 0, 1, -1)$.\n\nFor $k \\geq 2$, we always have $\\mathbf{x} + \\boldsymbol{\\alpha}_k > \\mathbf{x}$. So for any initial distribution of the balls, after a finite number of operations on every $B_k$ ($k \\geq 2$) that contains at least two balls, we can arrive at a ball distribution $\\mathbf{y} = (y_1, y_2, \\dots, y_n)$ satisfying $y_k \\leq 1$ for all $k \\geq 2$. If at this time $y_2 = \\dots = y_n = 1$, the problem is solved; otherwise, we have $y_1 \\geq 2$.\n\nAssuming $i$ is the smallest number such that $y_i = 0$, we can then do a series of operations on $B_1, B_2, \\dots, B_{i-1}$:\n$$\n\\begin{align*}\n& (y_1, 1, \\dots, 1, 0, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1, B_2, \\dots, B_{i-1}} \\\\\n& (y_1, 1, \\dots, 1, 0, 1, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1, B_2, \\dots, B_{i-2}} \\\\\n& (y_1, 1, \\dots, 1, 0, 1, 1, y_{i+1}, \\dots, y_n) \\to \\dots \\to \\\\\n& (y_1, 0, 1, \\dots, 1, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1} \\\\\n& (y_1 - 1, 1, \\dots, 1, 1, y_{i+1}, \\dots, y_n)\n\\end{align*}\n$$\nThis results in $(y_1 - 1, 1, \\dots, 1, y_{i+1}, \\dots, y_n)$. Repeating the operations above, we can finally arrive at the ball distribution vector that meets the requirement. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19255,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Its incircle meets the sides $BC$, $CA$, and $AB$ at the points $D$, $E$, and $F$, respectively. Let $P$ denote the intersection point of $ED$ and the line perpendicular to $EF$ and passing through $F$, and similarly let $Q$ denote the intersection point of $EF$ and the line perpendicular to $ED$ and passing through $D$.\n\nProve that $B$ is the midpoint of the segment $PQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the common point of $PF$ and $QD$, as can be seen in the figure. Since $\\angle EDH$ and $\\angle HFE$ are both right angles, $HE$ is a diameter of the incircle of $ABC$. Now let $X$ denote the common point of $EH$ and $PQ$. We see that $H$ is the orthocenter of the triangle $EPQ$, and $X$, $D$, and $F$ are the feet of the altitudes in this triangle. The incenter $I$ of $ABC$ is also the midpoint of an altitude segment. It follows that points $I$, $F$, $X$, and $D$ all lie on the nine-point circle of $EPQ$.\n\nBecause of the right angles at $F$ and $D$, we know that $I$, $F$, $D$, and $B$ lie on a common circle. This circle is the nine-point circle of $EPQ$. For the same reason, $B$ is the diametrically opposed point to $I$ on the nine-point circle of $EPQ$.\n\nIt is well known that the midpoint of each altitude segment lies diametrically opposed to the midpoint of the corresponding side of the triangle. (Note the right angle at $X$.) We therefore see that $B$ must be the midpoint of $PQ$, as we had set out to show.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19256,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that\n\n$$\n(n - 2013)(n - 2014)(n - 2016)(n - 2017) = 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If an integer $n$ satisfies the given condition, then $4$ can be written as the product of four pairwise distinct integers. The integer divisors of $4$ are $\\pm1$, $\\pm2$, and $\\pm4$. The possible factors must be $\\pm1$ and $\\pm2$, since if one factor is $\\pm4$, the others must have absolute value at least $1$, which is not possible for distinct consecutive integers. Since $n - 2013$ is the largest factor, it should be $2$. Thus, $n = 2015$ satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19257,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ for which the number of positive divisors of $\\operatorname{LCM}(1, 2, \\dots, n)$ is a power of $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each prime $p$, the numbers in the interval $n \\in [p^2, p^3)$ are not solutions, because the exponent of $p$ in the decomposition of the LCM is $2$ and so contributes a factor of $3$ to the number of divisors. Therefore, this number is not a power of $2$.\n\nFrom Bertrand's postulate, for a prime $p$ there is a prime $q$ with $p < q < 2p$, so $p^2 < q^2 < 4p^2 < p^3 < q^3$ for $p \\geq 5$; also $3^2 < 5^2 < 3^3 < 5^3$. Therefore, each interval $[p^2, p^3)$ and $[q^2, q^3)$ for consecutive primes $3 \\leq p < q$ intersects. We obtain that $n < 9$, and a direct check shows that only $n = 1, 2, 3,$ and $8$ are solutions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19258,
    "subject": "Mathematics (Olympiad)",
    "question": "Find a polynomial $f(x, y, z)$ of degree 3 with real coefficients that satisfies the following conditions:\n\n- $f(x, y, z) + x$ is divisible by $y + z$\n- $f(x, y, z) + y$ is divisible by $z + x$\n- $f(x, y, z) + z$ is divisible by $x + y$\n\nA polynomial $P(x, y, z)$ is divisible by a polynomial $Q(x, y, z)$ if there exists a polynomial $R(x, y, z)$ such that $P(x, y, z) = Q(x, y, z) R(x, y, z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nf(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z, \\quad (k \\neq 0)\n$$\nsatisfies the conditions (in fact, only this form is a solution).\n\nLet $g(x, y, z) = f(x, y, z) + x + y + z$. Then the conditions are equivalent to $g(x, y, z)$ being divisible by $x + y$, $y + z$, and $z + x$. The requirement that $f(x, y, z)$ has degree 3 and real coefficients is equivalent to $g(x, y, z)$ having degree 3 and real coefficients. Now,\n$$\ng(x, y, z) = k(x + y)(y + z)(z + x), \\quad (k \\neq 0)\n$$\nsatisfies all the conditions. Therefore,\n$$\nf(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z, \\quad (k \\neq 0)\n$$\nis the required solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19259,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences of positive integers $\\{a_n\\}_{n=1}^{\\infty}$ satisfying the following conditions for any positive integer $k$:\n\n$$\n(i)\\quad a_{2^{k+1}} = 2 \\cdot a_{2^k},\n$$\n\n$$\n(ii)\\quad \\{a_1, a_2, \\dots, a_k\\} \\text{ is a complete set of residue classes modulo } k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The two sequences $1, 2, 3, \\ldots$ and $3, 2, 1, 4, 5, 6, \\ldots$.\n\nFor a positive integer $n$, let $S_n := \\{a_1, \\dots, a_n\\}$.\n\nNow fix $n$ and let $M := \\max S_n$ and $m := \\min S_n$, and let $k := M - m$. Since all elements of $S_n$ are different by (ii), we have $k \\ge n-1$. If $k \\ge n$ then $M, m$ are elements of $S_k$ and $M \\equiv m \\pmod{k}$, which contradicts (ii). Hence $k = n-1$ and so the $n$ numbers in $S_n$ are $n$ consecutive positive integers. It follows that for any $n \\ge m \\ge 1$, we have $a_n - a_m \\le n-1$. Moreover, it is easy to see that if $S_N = \\{1, 2, \\dots, N\\}$ for some $N$, then $a_n = n$ for all $n \\ge N+1$.\n\nNow $a_8 = 4a_2$ and thus $a_2 = \\frac{a_8 - a_2}{3} \\le \\frac{7}{3}$. Hence $a_2 \\le 2$. This leaves us the following cases.\n\n(i) If $a_2 = 1$ then $a_1 = 2$, since elements of $S_2$ are consecutive. But $a_4 = 2a_2 = 2$, a contradiction.\n\n(ii) If $a_2 = 2$ and $a_1 = 1$, then $a_n = n$ for each $n \\ge 3$. This gives the first answer.\n\n(iii) If $a_2 = 2$ and $a_1 = 3$, then $a_4 = 4$ and $a_3 = 1$. Then $a_n = n$ for each $n \\ge 5$ and this gives the second answer.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19260,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the equilateral triangle $PQR$. Inside this triangle, the regular hexagon $ABCDEF$ is drawn. Points $B$, $D$, and $F$ are the midpoints of the sides of the triangle $PQR$. The area of the pentagon $QBAFR$ is equal to $1$.\n\n![](images/NLD_ABooklet_2023_p5_data_c837b8b8ec.png)\n\nWhat is the area of the triangle $PQR$?\n\nA) $\\frac{11}{10}$\nB) $\\frac{7}{6}$\nC) $\\frac{6}{5}$\nD) $\\frac{5}{4}$\nE) $\\frac{4}{3}$",
    "options": [],
    "answer": "See solution",
    "solution": "C) $\\frac{6}{5}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19261,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set $S = \\{(x + y)^7 - x^7 - y^7 \\mid x, y \\in \\mathbb{Z}\\}$. Determine the greatest common divisor of all the numbers in $S$.",
    "options": [],
    "answer": "See solution",
    "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$$\n(x + y)^7 - x^7 - y^7 = ((x + y)^7 - (x + y)) - (x^7 - x) - (y^7 - y).\n$$\nBut $a^7 - a = a(a-1)(a+1)(a^2 - a + 1)(a^2 + a + 1)$. Now, $2$ divides $a(a-1)$, $3$ divides $a(a-1)(a+1)$, while $7$ divides $a^7 - a$ by Fermat's Little Theorem or, more simply, $7$ will divide $(x+y)^7 - x^7 - y^7$, which equals $7(x^6y + 3x^5y^2 + 5x^4y^3 + 5x^3y^4 + 3x^2y^5 + xy^6)$ (by Newton's binomial formula).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19262,
    "subject": "Mathematics (Olympiad)",
    "question": "Доведіть наступні нерівності для додатних чисел $a, b, c$:\n\n$$\n\\frac{a^2}{b+c} \\geq a - \\frac{b+c}{4}, \\quad \\frac{b^2}{c+a} \\geq b - \\frac{c+a}{4}, \\quad \\frac{c^2}{a+b} \\geq c - \\frac{a+b}{4}\n$$\n\nДодайте ці нерівності та покажіть, що\n\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\ge \\frac{a+b+c}{2}.\n$$\n\nДоведіть, що\n\n$$\n\\frac{a+b}{c^2} + \\frac{b+c}{a^2} + \\frac{c+a}{b^2} \\ge 2 \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$\n\nПеремножте отримані нерівності.",
    "options": [],
    "answer": "See solution",
    "solution": "Для доведення першої нерівності:\n\n$$\n\\frac{a^2}{b+c} \\geq a - \\frac{b+c}{4}\n$$\n\nРозглянемо різницю:\n\n$$\n\\frac{a^2}{b+c} - a + \\frac{b+c}{4} = \\frac{a^2 - a(b+c) + \\frac{(b+c)^2}{4}}{b+c}\n$$\n\nЦей вираз не від'ємний для $a, b, c > 0$ (можна розкласти квадрат).\n\nАналогічно для інших двох нерівностей. Додаючи їх, маємо:\n\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\ge \\frac{a+b+c}{2}\n$$\n\nДля другої нерівності скористаємося фактом:\n\n$$\n\\frac{x}{y^2} + \\frac{y}{x^2} \\ge \\frac{1}{x} + \\frac{1}{y}, \\quad x, y > 0\n$$\n\nТоді:\n\n$$\n\\frac{a+b}{c^2} + \\frac{b+c}{a^2} + \\frac{c+a}{b^2} = \\left( \\frac{a}{b^2} + \\frac{b}{a^2} \\right) + \\left( \\frac{b}{c^2} + \\frac{c}{b^2} \\right) + \\left( \\frac{c}{a^2} + \\frac{a}{c^2} \\right) \\ge 2 \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)\n$$\n\nДля доведення першої суми можна застосувати нерівність Коші-Буняковського:\n\n$$\n\\sum_{i=1}^{n} \\frac{u_i}{v_i} \\ge \\frac{\\left( \\sum_{i=1}^{n} u_i \\right)^2}{\\sum_{i=1}^{n} u_i v_i}, \\quad u_i, v_i > 0\n$$\n\nПеремноживши нерівності (1) і (2), отримаємо комбіновану нерівність для $a, b, c > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19263,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle. If $|\\vec{BA} - t \\vec{BC}| \\geq |\\vec{AC}|$ for any $t \\in \\mathbb{R}$, then $\\triangle ABC$ is ( ).",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\angle ABC = \\alpha$. Since $|\\vec{BA} - t \\vec{BC}| \\geq |\\vec{AC}|$, we have\n$$\n|\\vec{BA}|^2 - 2t\\, \\vec{BA} \\cdot \\vec{BC} + t^2 |\\vec{BC}|^2 \\geq |\\vec{AC}|^2.\n$$\nLet\n$$\nt = \\frac{\\vec{BA} \\cdot \\vec{BC}}{|\\vec{BC}|^2},\n$$\nthen\n$$\n|\\vec{BA}|^2 - 2|\\vec{BA}|^2 \\cos^2 \\alpha + |\\vec{BA}|^2 \\cos^2 \\alpha \\geq |\\vec{AC}|^2.\n$$\nThis means $|\\vec{BA}|^2 \\sin^2 \\alpha \\geq |\\vec{AC}|^2$, i.e., $|\\vec{BA}| \\sin \\alpha \\geq |\\vec{AC}|$.\n\nOn the other hand, let point $D$ lie on line $BC$ such that $AD \\perp BC$. Then $|\\vec{BA}| \\sin \\alpha = |\\vec{AD}| \\leq |\\vec{AC}|$. Hence $|\\vec{AD}| = |\\vec{AC}|$, which means $\\angle ACB = \\frac{\\pi}{2}$.\n\n**Answer:** $\\triangle ABC$ is a right triangle at $C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19264,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number greater than $3$. Prove that there exist integers $a_1, a_2, \\dots, a_t$ that satisfy the following conditions:\n\n(a)\n\n$$\n-\\frac{p}{2} < a_1 < a_2 < \\cdots < a_t \\leq \\frac{p}{2},\n$$\n\n(b)\n\n$$\n\\frac{p-a_1}{|a_1|} \\cdot \\frac{p-a_2}{|a_2|} \\cdot \\cdots \\cdot \\frac{p-a_t}{|a_t|} = 3^m,\n$$\n\nwhere $m$ is a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nBy the Division Algorithm, there exist unique integers $q$ and $r$ such that $p = 3q + r$, where $0 < r < 3$.\n\nTake $b_0 = r$. Then\n\n$$\n\\frac{p-b_0}{|b_0|} = \\frac{3^{c_0} \\cdot b_1^*}{|b_0|}, \\quad \\text{where } 3 \\nmid b_1^* \\text{ and } 0 < b_1^* < \\frac{p}{2}.\n$$\n\nTake $b_1 = \\pm b_1^*$ such that $b_1 \\equiv p \\pmod{3}$. Then\n\n$$\n\\frac{p-b_1}{|b_1|} = \\frac{3^{c_1} \\cdot b_2^*}{b_1^*}, \\quad \\text{where } 3 \\nmid b_2^* \\text{ and } 0 < b_2^* < \\frac{p}{2}.\n$$\n\nTake $b_2 = \\pm b_2^*$ such that $b_2 \\equiv p \\pmod{3}$. Then\n\n$$\n\\frac{p-b_2}{|b_2|} = \\frac{3^{c_2} \\cdot b_3^*}{b_2^*}, \\quad \\text{where } 3 \\nmid b_3^* \\text{ and } 0 < b_3^* < \\frac{p}{2}.\n$$\n\nRepeating this process, we get $b_0, b_1, \\dots, b_p$.\n\nSince these $p+1$ integers are in the interval $\\left(-\\frac{p}{2}, \\frac{p}{2}\\right)$, a certain integer occurs twice. Suppose $b_i = b_j$, $i < j$, and $b_i, b_{i+1}, \\dots, b_{j-1}$ are distinct. So,\n\n$$\n\\frac{p-b_i}{|b_i|} \\cdot \\frac{p-b_{i+1}}{|b_{i+1}|} \\cdots \\frac{p-b_{j-1}}{|b_{j-1}|} = \\frac{3^{c_i} \\cdot b_{i+1}^*}{b_i^*} \\cdot \\frac{3^{c_{i+1}} \\cdot b_{i+2}^*}{b_{i+1}^*} \\cdots \\frac{3^{c_{j-1}} \\cdot b_j^*}{b_{j-1}^*}.\n$$\n\nSince $b_i = b_j$, then $b_i^* = b_j^*$. So the above expression equals\n\n$$\n3^{c_i + c_{i+1} + \\cdots + c_{j-1}} = 3^n, \\quad n > 0.\n$$\n\nPut $b_i, b_{i+1}, \\dots, b_{j-1}$ in ascending order, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19265,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute-angled non-isosceles triangle $ABC$, $O$ is its circumcentre, $H$ is its orthocentre, and $AB > AC$. Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$. Suppose $BD = DP$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. Prove that $\\angle ODQ = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Drop perpendiculars $OM$ and $QX$ onto $BC$, and $QY$ from $Q$ onto $AD$. First, $2DM = DM + BD - BM = BD - (BM - DM) = PD - (CM - DM) = PD - CD = PC$. It is a well-known fact that $2OM = AH$.\n\n![](images/Singapur_2011_p1_data_ffc2184892.png)\n\nNext, $\\angle CPQ = \\angle DBH = \\angle HAQ$ so that the triangles $CPQ$ and $HAQ$ are similar. Thus, the triangles $XPQ$ and $YAQ$ are similar. Therefore\n\n$$\n\\frac{QX}{DX} = \\frac{QX}{QY} = \\frac{PC}{AH} = \\frac{DM}{OM}.\n$$\n\nHence, the triangles $DXQ$ and $OMD$ are similar. It follows that $\\angle ODQ = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19266,
    "subject": "Mathematics (Olympiad)",
    "question": "In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $S$. Let a circle $K$ contain the points $S$ and $D$, and let $K$ intersect the line segments $AD$ and $CD$ at points $M$ and $N$, respectively. Denote $P$ as the intersection point of the lines $SM$ and $AB$, and $R$ as the intersection point of the lines $SN$ and $BC$. Let $P$ and $R$ lie on the same side of the line $BD$ as the point $A$. Prove that the line passing through $D$ and parallel to $AC$ and the line passing through $S$ and parallel to $PR$ intersect on the circle $K$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let all the angles in the solution be directed. Since in the cyclic quadrilateral $ABCD$, the sides are chords of the circumscribed circle of $ABCD$, we have\n\n$$\n\\angle PBR = \\angle PBC = \\angle ABC = \\angle ADC = \\angle MDN = \\angle MSN = \\angle PSR,\n$$\n\nhence the quadrilateral $RBSP$ is also a cyclic quadrilateral and $\\angle RPS = \\angle RBS$.\n\n![](images/Slovenija_2012_p24_data_ec7ce64a4c.png)\n\nBecause $P$ and $R$ lie on the same side of the line $BD$, the lines $PR$ and $AC$ intersect, and hence the line passing through $D$ and parallel to $AC$ and the line passing through $S$ and parallel to $PR$ intersect as well. Denote $X$ their point of intersection.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19267,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an arbitrary arrangement of brackets. Numbers $2010$ and $2009$ always have signs $+$ and $-$, respectively, so the sum reaches its maximum value when the other terms have $+$ signs.\n\nIn the expression\n\n$$\n\\frac{2010 - 2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009}{2010 \\text{ numbers}}\n$$\n\nbrackets are placed somehow and the value is calculated. Find the maximum value that can be reached. Justify your answer.\n\n*Notice.* A left bracket can be placed only before a number and a right bracket only after. For example, expressions $-2010(-2009 - 2010)$ and $-(2010 - 2009-)2010$ are incorrect.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an arbitrary arrangement of brackets. The first numbers $2010$ and $2009$ always have signs $+$ and $-$, respectively, so the sum reaches its maximum value when the other terms have $+$ signs. We can achieve this as shown below. Calculating the maximum value:\n\n$$\n\\begin{align*}\n& \\underbrace{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009)}_{\\text{2010 numbers}} = \\\\\n&= 2010 - 2009 + \\underbrace{2010 + 2009 + 2010 + 2009 + \\dots + 2010 + 2009}_{\\text{2008 numbers}} = \\\\\n&= 1 + (2010 + 2009) \\cdot 1004 = 4035077.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19268,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be integers with $b$ not a perfect square. Show that $x^2 + a x + b$ can be a perfect square only for finitely many integers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us examine the Diophantine equation $x^2 + a x + b = y^2$ with unknown integers $x$ and $y$. It can be transformed into the form $$(2x + 2y + a)(2x - 2y + a) = a^2 - 4b.$$ Since we assume $b$ is not a perfect square, $a^2 - 4b \\neq 0$. There are only finitely many ways to write $a^2 - 4b$ as a product of two integers. Each such factorization gives two linear equations for $x, y$ which have at most one integer solution. Thus, there are only finitely many such $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19269,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q'$ be the intersection of the perpendicular bisector of $AB$ with line $BC$.\n\nLet $P'$ be the intersection of the perpendicular bisector of $AC$ with $BC$.\n\nLet $O_1$ and $O_2$ be the circumcenters of triangles $ABO$ and $ACO$, respectively, where $O$ is the circumcenter of triangle $ABC$.\n\nLet $O_3$ be the circumcenter of triangle $OP'Q'$. Let $D$ be the intersection of $O_1O_2$ and $OO_3$.\n\nProve that $A$, $O$, and $O_3$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\angle AQ'O_1 = \\angle BQ'O_1 = 90^\\circ - \\angle ABC.\n$$\n\nMoreover, since $O_1$ and $O_2$ lie on the perpendicular bisector of segment $AO$ and $O_2$ is the circumcenter of triangle $ACO$, we have\n\n$$\n\\angle AO_2O_1 = \\frac{1}{2}\\angle AO_2O = \\angle ACO = \\frac{1}{2}(180^\\circ - \\angle AOC) = 90^\\circ - \\angle ABC.\n$$\n\nHence, the four points $A$, $O_1$, $O_2$, $Q'$ are concyclic. Similarly, $A$, $O_1$, $O_2$, $P'$ are concyclic. Since triangle $ABC$ is acute, $O$ does not lie on line $BC$ and $P'$ and $Q'$ are distinct. Hence $\\{P, Q\\} = \\{P', Q'\\}$, and $O_3$ is the circumcenter of triangle $OP'Q'$. Now $O_1$, $O$, and $Q'$ are collinear and $O_2$, $O$, and $P'$ are collinear. Let $D$ be the intersection of $O_1O_2$ and $OO_3$. Then,\n\n$$\n\\begin{align*}\n\\angle O_2 D O_3 &= \\angle O_2 O O_3 - \\angle D O_2 O \\\\\n&= (180^\\circ - \\angle P' O O_3) - \\angle O_1 O_2 P' \\\\\n&= 180^\\circ - \\frac{1}{2}(180^\\circ - \\angle O O_3 P') - \\angle O_1 O_2 P' \\\\\n&= 180^\\circ - (90^\\circ - \\angle O Q' P') - \\angle O_1 Q' P' \\\\\n&= 90^\\circ,\n\\end{align*}\n$$\n\nso lines $O_1O_2$ and $OO_3$ meet at right angles. On the other hand, $O_1O_2$ is the perpendicular bisector of $AO$, so both $A$ and $O_3$ lie on the perpendicular to line $O_1O_2$ through $O$. Therefore $A$, $O$, $O_3$ are collinear, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19270,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\frac{m!}{m^k(m-k)!} < \\frac{n!}{n^k(n-k)!}\n$$\nfor $2 \\leq k \\leq m < n$.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to prove the inequality for $n = m + 1$.\n\nSubstituting $n = m + 1$ and simplifying gives the equivalent inequality\n$$\n(m + 1)^{k-1}(m + 1 - k) < m^k,\n$$\nwhich is equivalent to\n$$\n\\sqrt[k]{(m + 1)^{k-1}(m + 1 - k)} < m.\n$$\nThis follows from the inequality of arithmetic and geometric means ($\\text{GM} \\leq \\text{AM}$).\n\nThe left side is the geometric mean of the $k$ numbers $m + 1, \\dots, m + 1, m + 1 - k$. Their arithmetic mean is $((k - 1)(m + 1) + (m + 1 - k))/k = m$, which is the right side.\n\nThe inequality is strict because $m + 1 - k \\ne m + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19271,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle. Points $P$ and $Q$ lie on sides $AB$ and $AC$ respectively such that $PQ$ is parallel to $BC$. Let $D$ be the foot of the perpendicular from $A$ to $BC$. Let $M$ be the midpoint of $PQ$. Suppose that line segment $DM$ meets the circumcircle of triangle $APQ$ at a point $X$ inside triangle $ABC$.\n\nProve that $\\angle AXB = \\angle AXC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume all angles are directed counterclockwise (modulo $180^\\circ$) in this solution where $\\angle XYZ$ denotes $\\angle(XY, YZ)$.\n\nLet $PX$ and $QX$ meet $BC$ at $P'$ and $Q'$. Since $APXQ$ is cyclic and $PQ \\parallel BC$, we have $\\angle XAC = \\angle XAQ = \\angle XPQ = \\angle XP'C$. Hence $AXP'C$ is cyclic. Similarly, $AXQ'B$ is cyclic.\n\nBy the dilation through $X$ taking $P, M, Q$ to $P', D, Q'$, we have that $DP' = DQ'$. Combine this with $AD \\perp P'Q'$, we see that $AP'Q'$ is isosceles and thus $AP' = AQ'$.\n\nTherefore,\n\n$$\n\\begin{align*}\n\\angle AXB &= \\angle AQ'B \\quad (AXQ'B \\text{ cyclic}) \\\\\n&= -\\angle AP'C \\quad (AP'Q' \\text{ isosceles}) \\\\\n&= \\angle CP'A \\\\\n&= \\angle CXA. \\quad (AXP'C \\text{ cyclic})\n\\end{align*}\n$$\n\n![](images/2023_Australian_Scene_p74_data_09e4a5c671.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19272,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon having a circumcircle and satisfying $AB = BD$. The point $P$ is the intersection of the diagonals $AC$ and $BE$. The lines $BC$ and $DE$ intersect at point $Q$.\n\nShow that the line $PQ$ is parallel to the diagonal $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of the pentagon $ABCDE$ be $k$. By assumption, the triangle $ABD$ is isosceles, which implies that the tangent $t_{B}$ to $k$ at $B$ is parallel to $AD$.\n\nApply Pascal's theorem to the inscribed hexagon $BEDACB$: the intersection point of the opposite sides $BE$ and $AC$ is $P$, the intersection point of the opposite sides $ED$ and $CB$ is $Q$, and the intersection point of the parallel opposite sides $BB$ (i.e., $t_{B}$) and $DA$ is the point at infinity in the direction of $AD$. Therefore, $PQ$ is parallel to $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19273,
    "subject": "Mathematics (Olympiad)",
    "question": "Three numbers $2^{100}$, $3^{100}$, and $5^{100}$ are written on a long paper strip without any spaces in between, thus creating one big number $N$. Arsenii claims that he can change the last digit of number $N$ so that the new number is a power of $13$. Is he right?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose Arsenii's claim is correct and by changing the last digit of $N$ he obtains $13^k$ for some positive integer $k$. The last digit of $N$ is $5$, so he must change it. \n\nSince $2^{100} \\equiv 1 \\pmod{3}$, $3^{100} \\equiv 0 \\pmod{3}$, and $5^{100} \\equiv 1 \\pmod{3}$, the sum of their digits has the same remainder modulo $3$. Changing the last digit $5$ to $l$ means subtracting $5$ and adding $l$ to the digit sum, so $l \\equiv 0 \\pmod{3}$. Thus, the new last digit $l$ could be $0$, $3$, $6$, or $9$.\n\nHowever, the last digit of $13^k$ can only be $1$, $3$, $7$, or $9$. So only $3$ or $9$ are possible. Let's consider these cases:\n\n**Case 1:** $13^k$ ends with $1$ (possible if $k \\equiv 0 \\pmod{4}$). Then $13^k \\equiv 1 \\pmod{8}$. But $N$ ends with $5$, so $N \\equiv 5 \\pmod{8}$. Changing the last digit to $1$ gives $M = N - 4 \\equiv 1 \\pmod{8}$, but $N \\equiv 5 \\pmod{8}$, so $M \\equiv 1 \\pmod{8}$ only if $N \\equiv 5 \\pmod{8}$, which is not possible for $13^k$.\n\n**Case 2:** $13^k$ ends with $7$ (possible if $k \\equiv 3 \\pmod{4}$). Then $13^k \\equiv 5 \\pmod{8}$. Changing the last digit to $7$ gives $M = N + 2 \\equiv 7 \\pmod{8}$, but $13^k \\equiv 5 \\pmod{8}$, so this is a contradiction.\n\nSimilar contradictions arise for other possible last digits. Therefore, Arsenii's claim is false.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19274,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose cones are to be placed at marks every 10 m from $0$ to $80$ m. For each of the following intervals, determine the maximum number of cones that can be placed, and describe all possible placements:\n\n- (b) From $0$ to $50$ m\n- (c) From $0$ to $60$ m\n- (d) From $0$ to $70$ m\n- (e) From $0$ to $80$ m\n\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_e423d721f9.png)\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_57037ad5c5.png)\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_4f6c885aee.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**(b) $0$ to $50$ m:**\n\nThe maximum number of cones that can be placed is $4$. There are four possible placements:\n- Cones at $0$, $20$, $30$, $50$\n- Cones at $0$, $10$, $30$, $40$\n- Cones at $10$, $20$, $40$, $50$\n- Cones at $0$, $10$, $30$, $50$\n\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_e423d721f9.png)\n\n**(c) $0$ to $60$ m:**\n\nThe two end cones must be at $0$ m and $60$ m, so there is no cone at $30$ m. There are three more cones to place, but placing them at $10$ m and $20$ m or at $40$ m and $50$ m is not legitimate. Thus, the maximum number of cones is $4$.\n\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_57037ad5c5.png)\n\n**(d) $0$ to $70$ m:**\n\nThe two end cones must be at $0$ m and $70$ m. There are three more cones to place. Consider the following cases:\n\nCase 1: None of the marks $10$, $20$, $30$ has a cone. Then each of the marks $40$, $50$, $60$ has a cone. This is not legitimate.\n\nCase 2: Just one of the marks $10$, $20$, $30$ has a cone. Then two of the marks $40$, $50$, $60$ have a cone. Having cones at $50$, $60$, $70$ is not legitimate. So either marks $40$ and $50$ have a cone or marks $40$ and $60$ have a cone. In both cases, there is no place for the third cone.\n\n% IMAGE: ![](images/Australian_Scene_-2014_p62_data_51076a75d2.png)\n% IMAGE: ![](images/Australian_Scene_-2014_p62_data_d73187767f.png)\n\nCase 3: Just two of the marks $10$, $20$, $30$ have a cone. This is symmetrical to Case 2.\n\nCase 4: All three of the marks $10$, $20$, $30$ have a cone. This is not legitimate.\n\nSo there is no placement of five cones from $0$ to $70$ m.\n\n% IMAGE: ![](images/Australian_Scene_-2014_p61_data_4f6c885aee.png)\n\n**(e) $0$ to $80$ m:**\n\nThe two end cones must be at $0$ m and $80$ m, so there is no cone at $40$ m. There are three more cones to place. We cannot have all three cones on the marks $10$, $20$, $30$ or all three on the marks $50$, $60$, $70$. So there are two cases:\n\nCase 1: There are just two cones on the marks $10$, $20$, $30$. Then the fifth cone must be at $70$ m.\n\n% IMAGE: ![](images/Australian_Scene_-2014_p62_data_271b56b940.png)\n% IMAGE: ![](images/Australian_Scene_-2014_p63_data_20276f833e.png)\n\nCase 2: There is just one cone on the marks $10$, $20$, $30$. Then there are exactly two cones for the marks $50$, $60$, $70$. This case is symmetrical to Case 1.\n\n% IMAGE: ![](images/Australian_Scene_-2014_p63_data_b6e8ea7818.png)\n% IMAGE: ![](images/Australian_Scene_-2014_p63_data_fd2afd4483.png)\n\nSo there are four placements of five cones from $0$ to $80$ m.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19275,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given two circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in the plane, with $|S_1S_2| > r_1 + r_2$. Find the locus of points $X$ which do not lie on the line $S_1S_2$ and possess the following property: The segments $S_1X$ and $S_2X$ intersect successively the circles $k_1$ and $k_2$ in such points whose distances to the line $S_1S_2$ are the same.\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p10_data_51fa78a813.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $X$ is any point with the required property. $X$ lies in the exteriors of $k_1$ and $k_2$, and $S_1, S_2, X$ form a triangle whose sides $S_1X, S_2X$ are intersected by $k_1$ and $k_2$ at points $Y_1, Y_2$ lying on the same line parallel to $S_1S_2$ (see figure).\n\nSince triangles $XY_1Y_2$ and $XS_1S_2$ are similar (by AA),\n\n$$\n\\frac{|XY_1|}{|XS_1|} = \\frac{|XY_2|}{|XS_2|},\n$$\n\nand since\n\n$$\n|XY_1| = |XS_1| - r_1, \\quad |XY_2| = |XS_2| - r_2,\n$$\n\nwe have\n\n$$\n\\frac{|XS_1| - r_1}{|XS_1|} = \\frac{|XS_2| - r_2}{|XS_2|}, \\qquad \\text{so} \\qquad \\frac{|XS_1|}{|XS_2|} = \\frac{r_1}{r_2}.\n$$\n\nSince $S_1, S_2$ and the ratio $r_1/r_2$ are fixed, the locus of $X$ is a circle of Apollonius (which becomes a straight line if $r_1 = r_2$). For $r_1/r_2 \\neq 1$, there are exactly two solutions $X = H_1$ and $X = H_2$ on $S_1S_2$ forming a diameter of the Apollonius circle. These points are the centers of homotheties of $k_1$ and $k_2$.\n\nConversely, if $X$ is a point of the Apollonius circle (excluding $H_1, H_2$), and $|S_1S_2| > r_1 + r_2$, the whole circle lies outside $k_1$ and $k_2$. Thus, for such $X$, there exist $Y_1 \\in k_1$ and $Y_2 \\in k_2$ on $S_1X$ and $S_2X$ respectively, and the triangles $XS_1S_2$ and $XY_1Y_2$ are similar, so $S_1S_2 \\parallel Y_1Y_2$. Therefore, the distances from $Y_1$ and $Y_2$ to $S_1S_2$ are equal, as required.\n\n**Answer.** If $r_1 \\neq r_2$, the locus of $X$ is the circle of Apollonius given by the above equation, except for the two points on $S_1S_2$. If $r_1 = r_2$, the locus is the perpendicular bisector of $S_1S_2$, except for its midpoint.\n\n**Remark.** Any solution $X$ of the equation lies outside $k_1$ and $k_2$, since\n\n$$\n(|XS_1| - r_1) + (|XS_2| - r_2) \\leq |S_1S_2| - (r_1 + r_2) > 0,\n$$\n\nso $|XS_1| > r_1$ and $|XS_2| > r_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19276,
    "subject": "Mathematics (Olympiad)",
    "question": "Могли ли Вася и Петя получить одинаковые наборы чисел, если Вася выписывает все возможные разности квадратов двух различных исходных чисел, а Петя — все возможные квадраты разностей двух различных исходных чисел, взятых из одного и того же набора из 10 целых чисел?",
    "options": [],
    "answer": "See solution",
    "solution": "Предположим 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": 19277,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a $5 \\times 5$ grid. Write the integers $1, 2, \\ldots, 16$ (each number used only once) in the upper left $4 \\times 4$ subgrid (\\clubsuit). For each of the 4 rows, write the sum of the four numbers in that row at the right end of the row. Similarly, for each of the 4 columns, write the sum of the four numbers in that column at the bottom of the column. Nothing is written in the lower right cell.\n\nFind the maximum integer $m$ such that, after step (\\clubsuit), there exists an arrangement where you can choose two numbers $a, b$ from the set of row sums and also from the set of column sums so that $|a-b| \\ge m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define the numbers in each cell as shown in Table 1. Assume $A_1$ is the minimum and $A_4$ is the maximum among $A_1, A_2, A_3, A_4$, and $B_1$ is the minimum and $B_4$ is the maximum among $B_1, B_2, B_3, B_4$, by rearranging rows and columns as needed. Then $m \\leq A_4 - A_1$ and $m \\leq B_4 - B_1$, so\n\n$$\n\\begin{aligned}\nm &\\leq \\frac{(A_4 - A_1) + (B_4 - B_1)}{2} \\\\\n  &= \\frac{1}{2}(a_{4,4} - a_{1,1}) + \\frac{1}{2}(a_{4,4} + a_{4,2} + a_{4,3} + a_{2,4} + a_{3,4} - a_{1,1} - a_{1,2} - a_{1,3} - a_{2,1} - a_{3,1}) \\\\\n  &\\leq \\frac{1}{2}(16 - 1) + \\frac{1}{2}(16 + 15 + 14 + 13 + 12 - 1 - 2 - 3 - 4 - 5) \\\\\n  &= 35\n\\end{aligned}\n$$\n\nThere exists an arrangement achieving $m = 35$ (see Table 2). Thus, the maximum $m$ is $35$.\n\nThis is Table 1.\n\n![](table1.png)\n\nThis is Table 2.\n\n![](table2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19278,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the sum of six consecutive positive integers, none of which is divisible by $7$, is divisible by $21$ but not by $42$. Find six such numbers whose sum is a four-digit number that is a square of a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Since none of the six consecutive positive integers is divisible by $7$, they must be of the form $7n+1,\\ 7n+2,\\ 7n+3,\\ 7n+4,\\ 7n+5,\\ 7n+6$, where $n \\in \\mathbb{N}_0$. Their sum is:\n\n$$\nS = (7n+1) + (7n+2) + (7n+3) + (7n+4) + (7n+5) + (7n+6) = 42n + 21 = 21(2n+1)\n$$\n\nThus, $S$ is divisible by $21$, but not by $42$ (since $2n+1$ is odd). For $S$ to be a four-digit perfect square, set $S = m^2$ with $1000 \\leq m^2 < 10000$ and $2n+1 = 21k^2$ for some odd $k$. The only $k$ with $2 < k^2 < 23$ is $k^2 = 9$, so $2n+1 = 189$ and $n = 94$. The six numbers are $7 \\times 94 + 1 = 659$, $660$, $661$, $662$, $663$, and $664$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19279,
    "subject": "Mathematics (Olympiad)",
    "question": "On an $8 \\times 8$ board, there is a beetle on every square. At a certain moment, the distribution of the beetles on the board changes: every beetle crawls either one square to the left or one square diagonally to the bottom right. If a beetle can make neither of the two movements without falling off the board, it stays on its square.\n\nAt most, how many squares can end up empty after this change?",
    "options": [],
    "answer": "See solution",
    "solution": "$56$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19280,
    "subject": "Mathematics (Olympiad)",
    "question": "On an $11 \\times 11$ board, a black cone is placed in the central cell. Four white cones are placed in the four middle cells of the border (see the figure below).\n\n- The first player controls the black cone and, on each turn, may move it to an adjacent cell if there is no white cone there.\n- The second player, on each turn, adds one white cone to the perimeter of the square, but only to a cell that shares a side with a cell already containing a white cone.\n\nThe first player wins if the black cone reaches the perimeter of the square; otherwise, the second player wins. Who can win this game?\n\n![](images/Ukrajina_2011_p18_data_6c8cff0e94.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The first player can win.\n\nThe first player should move the black cone toward a cell $A$, which is the middle of one side of the square. After 3 moves, the black cone reaches the cell adjacent to $A$. Suppose that to the right or left of this cell, there is at most one white cone (say, to the left, as in the figure). Then the first player can move toward cell $B$.\n\nEven if the second player places a white cone at $B$, they cannot prevent the black cone from reaching $B$, since the second player would have to place a cone in the corner, which is not possible immediately.\n\nIf there are two white cones on both sides, the first player keeps moving to the left. The second player then has only one option to avoid losing, leading to a situation similar to the previous case. Thus, the first player can always reach the perimeter.\n\n![](images/Ukrajina_2011_p18_data_7bc1436e96.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19281,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose each knight is assigned a number. What is the maximum number of knights such that no knight can truthfully say both \"My number is greater than 9\" and \"My number is greater than 10\"?",
    "options": [],
    "answer": "See solution",
    "solution": "No knight could say both statements, as after saying \"My number is greater than 9\", only those with numbers greater than 9 remain, so at most 8 knights can exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19282,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime, and let $q$ be a prime divisor of $\\dfrac{p^p - 1}{p - 1}$.\n\n(a) Prove that not all prime divisors of $\\dfrac{p^p - 1}{p - 1}$ are congruent to $1 \\pmod{p^2}$.\n\n(b) Prove that for all $n \\in \\mathbb{Z}^+$, $n^p - p \\not\\equiv 0 \\pmod q$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Suppose, for contradiction, that all prime divisors of $\\dfrac{p^p - 1}{p - 1}$ are congruent to $1 \\pmod{p^2}$. Then $\\dfrac{p^p - 1}{p - 1} \\equiv 1 \\pmod{p^2}$. But\n\n$$\n\\frac{p^p - 1}{p - 1} = p^{p-1} + p^{p-2} + \\dots + 1 \\equiv p + 1 \\not\\equiv 1 \\pmod{p^2}.\n$$\n\nThis is a contradiction.\n\n(b) Suppose, for contradiction, that $n^p - p \\equiv 0 \\pmod q$ for some $n \\in \\mathbb{Z}^+$. Since $q \\mid \\dfrac{p^p - 1}{p - 1}$, we have $p^p \\equiv 1 \\pmod q$. Therefore,\n\n$$\nn^{p^2} = (n^p)^p \\equiv p^p \\equiv 1 \\pmod q.\n$$\n\nBy Fermat's little theorem, $n^{q-1} \\equiv 1 \\pmod q$. Let $d$ be the order of $n$ modulo $q$. Then $d \\mid p^2$ and $d \\mid q-1$. Since $p^2 \\mid q-1$ by part (a), this implies $d = 1$ or $d = p$. Thus,\n\n$$\nn^p \\equiv 1 \\pmod q.\n$$\n\nThis further implies $p \\equiv 1 \\pmod q$. But then\n\n$$\n\\frac{p^p - 1}{p - 1} = p^{p-1} + p^{p-2} + \\dots + 1 \\equiv 1 + 1 + \\dots + 1 = p \\not\\equiv 0 \\pmod q.\n$$\n\nThis is a contradiction. Thus, $n^p - p \\not\\equiv 0 \\pmod q$ for all $n \\in \\mathbb{Z}^+$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19283,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the sum of the $2013$th powers of the digits of a positive integer $n$. Consider the set $S = \\{1, 2, \\dots, 10^{2017} - 1\\}$. Does there exist distinct positive integers $i$ and $j$ such that $f^{(i)}(2013) = f^{(j)}(2013)$, where $f^{(k)}$ denotes the $k$-fold iteration of $f$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. For any $n \\in S$, write $n = \\overline{a_1a_2\\dots a_{2017}}$. Then\n\n$$\nf(n) = \\sum a_i^{2013} \\leq 2017 \\cdot 9^{2013} < 10^4 \\cdot 10^{2013} = 10^{2017}.\n$$\n\nThus, $f(n) \\in S$. Since $S$ is finite but the sequence $f^{(i)}(2013)$ is infinite, by the pigeonhole principle, there exist distinct $i, j$ such that $f^{(i)}(2013) = f^{(j)}(2013)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19284,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a cyclic hexagon $ABCDEF$, let $BD$ and $CF$ meet at $G$, $AC$ and $BE$ meet at $H$, and $AD$ and $CE$ meet at $I$. Suppose that $BD$ is perpendicular to $CF$ and $AI = CI$. Show that $CH = AH + DE$ if and only if $GH \\cdot BD = BC \\cdot DE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AI = CI$, $\\angle ACI = \\angle CAI$, and so $\\angle ACE = \\angle CAD$. Since $ACDE$ is cyclic, $\\angle CAD = \\angle CED$. Hence $\\angle ACE = \\angle CED$, and so $AC$ and $DE$ are parallel to each other.\n\nNow we will show that if $CH = AH + DE$, then $GH \\cdot BD = BC \\cdot DE$. Let $A'$ be the point on $AC$ with $CA' = DE$. Since $AC$ is parallel to $DE$, $A'CDE$ is a parallelogram. Thus $\\angle AA'E = \\angle ACD$. It follows that $\\angle AA'E = \\angle A'AE$, as $\\angle ACD = \\angle CAE$. Further, the fact that $AH = CH - DE = CH - CA' = HA'$ guarantees that $H$ is the midpoint of the base $AA'$ of an isosceles triangle $EAA'$. Therefore $EH$ is orthogonal to $AA'$. It follows that $\\angle BHC = 90^\\circ = \\angle BGC$, and so the points $B$, $C$, $G$, $H$ are cyclic. Let $K$ be the intersection point of $BG$ and $CH$. Since $\\triangle BKC \\sim \\triangle HKG$,\n\n$$\n\\frac{BC}{GH} = \\frac{BK}{HK} \\quad (1)\n$$\n\nSince $DE$ is parallel to $KH$, $\\triangle BDE \\sim \\triangle BKH$. Thus\n\n$$\n\\frac{BK}{HK} = \\frac{BD}{DE} \\quad (2)\n$$\n\nCombining (1) and (2), we obtain $\\frac{BC}{GH} = \\frac{BD}{DE}$, and hence $GH \\cdot BD = BC \\cdot DE$. This proves the sufficiency.\n\nConversely, assume $GH \\cdot BD = BC \\cdot DE$. We will show that $CH = AH + DE$. Since $AC$ is parallel to $DE$, $\\triangle BKH \\sim \\triangle BDE$, which implies that\n\n$$\n\\frac{BK}{KH} = \\frac{BD}{DE} \\quad (3)\n$$\n\nSince $GH \\cdot BD = BC \\cdot DE$,\n\n$$\n\\frac{BD}{DE} = \\frac{BC}{GH} \\quad (4)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19285,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer and let\n$$\nM = \\left\\{ \\frac{a_1 + a_2 + \\dots + a_k}{k} : 1 \\le k \\le n \\text{ and } 1 \\le a_1 < \\dots < a_k \\le n \\right\\}\n$$\nbe the set of the arithmetic means of the elements of all non-empty subsets of $\\{1, 2, \\dots, n\\}$.\n\nFind\n$$\n\\min\\{|a-b| : a, b \\in M \\text{ with } a \\neq b\\}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that $M$ consists of rational numbers of the form $a = \\frac{x}{k}$, where $1 \\le k \\le n$. The arithmetic mean of $1, \\dots, n$ is $\\frac{n+1}{2}$. Considering these rational numbers in irreducible form, $1 \\le k \\le n-1$.\n\nA non-zero difference $|a-b|$ with $a, b \\in M$ is of the form\n$$\n\\left| \\frac{x}{k} - \\frac{y}{p} \\right| = \\frac{|p_0 x - k_0 y|}{[k, p]}\n$$\nwhere $[k, p]$ is the least common multiple of $k$ and $p$, $k_0 = \\frac{[k, p]}{k}$, and $p_0 = \\frac{[k, p]}{p}$. Thus, $|a-b| \\ge \\frac{1}{[k, p]}$, since $|p_0 x - k_0 y|$ is a nonzero integer. Since\n$$\n\\max\\{[k, p] : 1 \\le k < p \\le n-1\\} = (n-1)(n-2),\n$$\nwe have\n$$\n\\min_{\\substack{a, b \\in M \\\\ a \\neq b}} |a-b| \\ge \\frac{1}{(n-1)(n-2)}.\n$$\n\nTo achieve this minimum, we seek $x \\in \\{3, 4, \\dots, 2n-1\\}$ and $y \\in \\{1, 2, \\dots, n\\}$ such that\n$$\n\\left| \\frac{\\frac{n(n+1)}{2} - x}{n-2} - \\frac{\\frac{n(n+1)}{2} - y}{n-1} \\right| = \\frac{1}{(n-1)(n-2)},\n$$\nwhich leads to\n$$\n\\left| \\frac{n(n+1)}{2} - (n-1)x + (n-2)y \\right| = 1.\n$$\nIf $n=2k$, we can choose $x = k+3$ and $y = 2$; if $n=2k+1$, we can choose $x = n = 2k+1$ and $y = k$. Therefore, the required minimum is $\\frac{1}{(n-1)(n-2)}$.\n\n**Comment:** For $n \\ge 5$, the only other possibilities are to take $x = 3k-1$, $y = 2k-1$ if $n = 2k$, and $x = 2k+3$, $y = k+2$ if $n = 2k+1$. (For $n = 3, 4$ there are also examples where one of the sets is of size $n$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19286,
    "subject": "Mathematics (Olympiad)",
    "question": "A projector placed at vertex $B$ of equilateral triangle $ABC$ is lighting angle $\\alpha$. Find all values of $\\alpha \\le 60^{\\circ}$ such that, when $\\alpha$ is in $\\angle ABC$, the three parts of the side $AC$ (one lighted and two dark) form a triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "On the other hand, $AM = a \\cdot \\cot t$. But $t = \\angle DAM + \\angle MAE$, and thus we get:\n\n$$\n\\cot(t) = \\frac{\\cot \\angle DAM \\cdot \\cot \\angle MAE - 1}{\\cot \\angle DAM + \\cot \\angle MAE} \\quad (1)\n$$\n\nDenoting $x + z = s$ and $x \\cdot z = p$, using the relation (1), we get:\n\n$$\na^2(1 + \\cot^2 t) = a \\cdot s(1 + \\cot^2 t) - p.\n$$\n\nNow, from the above relation, we get:\n\n$$\ns = a + \\frac{p}{a} \\sin^2 t.\n$$\n\nFinally, we obtain:\n\n$$\ny^2 = (2a - s)^2 = x^2 + 2x \\cdot z \\cos(2t) + z^2.\n$$\n\nNow, on the two sides of an angle at vertex $X$ and measure $\\pi - A$, we choose the points $Y$ and $Z$ such that $XY = x$ and $XZ = z$. From the cosine rule we have:\n\n$$\nYZ^2 = x^2 + 2x \\cdot z \\cos(2t) + z^2 = y^2\n$$\n\nand thus the existence of the triangle $XYZ$ is proved. Moreover, we have $\\angle BAC + \\angle YXZ = 180^{\\circ}$.\n\nA similar problem appeared in All Russian olympiad, 1996, Regional round, Problems 8.4 and 9.4.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19287,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all polynomials $P(x)$ with real coefficients for which\n$$\nP(x^2) + 2P(x) = P(x)^2 + 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the equation as\n$$\nP(x^2) - 1 = (P(x) - 1)^2.\n$$\nLet $Q(x) = P(x) - 1$, then $Q$ is a polynomial with real coefficients satisfying\n$$\nQ(x^2) = Q(x)^2.\n$$\nSuppose that $Q$ is constant, say $Q(x) = c$ with $c \\in \\mathbb{R}$. Then $c = c^2$, so $c = 0$ or $c = 1$. Both possibilities give rise to solutions. Now assume $Q$ is non-constant, so $Q(x) = bx^n + R(x)$ with $n \\ge 1$, $b \\ne 0$, and $R(x)$ a polynomial of degree at most $n-1$. The equation becomes\n$$\nbx^{2n} + R(x^2) = b^2 x^{2n} + 2bx^n R(x) + R(x)^2.\n$$\nComparing coefficients of $x^{2n}$, we get $b = b^2$. Since $b \\ne 0$, $b = 1$. Subtracting $x^{2n}$ from both sides yields\n$$\nR(x^2) = 2x^n R(x) + R(x)^2.\n$$\nIf $R$ is non-zero of degree $m < n$, then the left side has degree $2m$ and the right side has degree $m+n > 2m$, a contradiction. Thus, $R$ must be zero, so $Q(x) = x^n$. This satisfies the equation for $Q$.\n\nTherefore, the solutions are $P(x) = 1$, $P(x) = 2$, and $P(x) = x^n + 1$ for $n \\ge 1$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19288,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of the side $CD$ of the square $ABCD$. The perpendicular from $C$ onto $BM$ meets the line $BM$ at $N$, and $AB$ at $E$. The line $BM$ intersects the line $AD$ at $P$. 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": "See solution",
    "solution": "a) Since $CB = BA$ and $\\angle CEB = \\angle BPA$ (they have the same complement $\\angle EBP$), it follows that the right triangles $CBE$ and $BAP$ are congruent.\n\nb) Let $T$ be the intersection 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, $AF \\parallel NE$, which implies $AF \\perp BP$.\n\nThe segments $FD$ and $AN$ are medians relative to the hypotenuse in right triangles, hence $FD = \\frac{1}{2}AP = \\frac{1}{2}BE = AN$.\n\nSince $BN = AF$ (altitudes relative to the hypotenuse in congruent triangles $EBC$ and $PBA$), $\\triangle ADF \\equiv \\triangle BAN$.\n\nFrom $\\angle FDA + \\angle NAD = \\angle NAB + \\angle NAD = 90^\\circ$ it follows that $AN \\perp DF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19289,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to choose three distinct two-digit numbers $\\overline{ab}$, $\\overline{xy}$, and $\\overline{pq}$ such that every two-digit number formed by taking a tens digit from $\\{a, x, p\\}$ and a units digit from $\\{b, y, q\\}$ is a prime number?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, contrary to our claim, that there are such numbers: $\\overline{ab}$, $\\overline{xy}$, and $\\overline{pq}$. The problem condition is equivalent to the following statement: any two-digit number with tens digit $a$, $x$, or $p$, and with unit digit $b$, $y$, or $q$ is a prime number.\n\nSo $b, y, q \\in \\{1, 3, 7, 9\\}$, hence, at least one of the digits 3 and 9 belongs to $\\{b, y, q\\}$. Therefore, none of the digits $a, x, p$ is divisible by 3, so $a, x, p \\in M = \\{1, 2, 4, 5, 7, 8\\}$.\n\nThus, we can consider two-digit numbers with tens digits from the set $M$ only. Among numbers 11, 21, 41, 51, 71, 81 there are exactly two prime numbers with distinct tens and unit digits: 41 and 71. Therefore, none of the digits $b, y, q$ is equal to 1.\n\nIt follows that $\\{b, y, q\\} = \\{3, 7, 9\\}$, which is impossible. Indeed, exactly two numbers 17 and 47 are prime among two-digit numbers 17, 27, 47, 57, 77, 87, while there should be at least three prime numbers among these numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19290,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of coprime natural numbers such that $a < b$ and $b$ divides\n\n$$\n(n + 2)a^{n+1002} - (n + 1)a^{n+1001} - n a^{n+1000}\n$$\n\nfor every natural number $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$ and $b$ are coprime, so are $b$ and $a^{n+1000}$. Thus, it suffices for $b$ to divide $(n + 2)a^2 - (n + 1)a - n$ for all $n$. For $n = 1$ and $n = 2$, we get $b \\mid 3a^2 - 2a - 1$ and $b \\mid 4a^2 - 3a - 2$. Therefore,\n\n$$\nb \\mid 4(3a^2 - 2a - 1) - 3(4a^2 - 3a - 2) = a + 2\n$$\n\nAlso, $3a^2 - 2a - 1 = 3(a - 2)(a + 2) - 2(a + 2) + 15$, so $b$ must divide $15$.\n\nSince $b > a \\ge 1$, $b \\ge 2$. If $b = 15$, then $b \\mid a + 2$ gives $a = 13$, but $4 \\cdot 13^2 - 3 \\cdot 13 - 2 = 635$ is not divisible by $3$. If $b = 3$, then $a = 1$, but $4 \\cdot 1^2 - 3 \\cdot 1 - 2 = -1$ is not divisible by $3$.\n\nIt remains $b = 5$, so $a = 3$. Indeed, $(n + 2) \\cdot 3^2 - (n + 1) \\cdot 3 - n = 5(n + 3)$ is divisible by $5$.\n\n$\\boxed{(a, b) = (3, 5)}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19291,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers such that $x^4 + y^4 + z^4 = 3$. Prove that\n\n$$\n\\frac{9}{x^2 + y^4 + z^6} + \\frac{9}{x^4 + y^6 + z^2} + \\frac{9}{x^6 + y^2 + z^4} \\leq x^6 + y^6 + z^6 + 6.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "If we use the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x, y^2, z^3)$ and $(x^3, y^2, z)$, we get\n\n$$\n(x^4 + y^4 + z^4)^2 \\leq (x^2 + y^4 + z^6)(x^6 + y^4 + z^2),\n$$\nthat is,\n$$\n\\frac{1}{x^2 + y^4 + z^6} \\leq \\frac{x^6 + y^4 + z^2}{9}.\n$$\n\nAnalogously, using the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x^2, y^3, z)$ and $(x^2, y, z^3)$, and also for $(x^3, y, z^2)$ and $(x, y^3, z^2)$, we get\n\n$$\n\\frac{1}{x^4 + y^6 + z^2} \\leq \\frac{x^4 + y^2 + z^6}{9},\n$$\n\nand\n\n$$\n\\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^2 + y^6 + z^4}{9}.\n$$\n\nNow, by adding these inequalities, we obtain\n\n$$\n\\frac{1}{x^2 + y^4 + z^6} + \\frac{1}{x^4 + y^6 + z^2} + \\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^6 + y^6 + z^6 + x^4 + y^4 + z^4 + x^2 + y^2 + z^2}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19292,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for any real numbers $x$ and $y$,\n\n$$\nf(x^3 + y^3) = f(x^3) + 3x^3 f(x) f(y) + 3 f(x) (f(y))^2 + y^6 f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = y = 0$ gives $3(f(0))^3 = 0$, which implies\n$$\nf(0) = 0.\n$$\n\nSubstituting $x = 0$ into the given equation and applying the above result gives\n$$\nf(y^3) = y^6 f(y).\n$$\n\nReplacing the first term on the right-hand side of the original equation with this result, we get\n$$\nf(x^3 + y^3) = x^6 f(x) + 3x^3 f(x) f(y) + 3 f(x) (f(y))^2 + y^6 f(y).\n$$\n\nSwapping $x$ and $y$ in this equation and subtracting the two equations gives\n$$\n3x^3 f(x) f(y) + 3 f(x) (f(y))^2 = 3y^3 f(y) f(x) + 3 f(y) (f(x))^2.\n$$\n\nCollecting like terms and factorizing, we obtain\n$$\nf(x) f(y) (f(x) - x^3 - f(y) + y^3) = 0.\n$$\n\nSuppose $f(x) = f(y)$ for some $x \\neq y$. Since $x^3 \\neq y^3$, $f(x) - x^3 - f(y) + y^3 \\neq 0$. Thus, $f(x) f(y) = 0$, so $f(x) = f(y) = 0$. Therefore, $f$ cannot take nonzero values repeatedly.\n\nSuppose $f(b) = 0$ for some $b \\neq 0$. Substituting $y = b$ into the original equation gives $f(x^3 + b^3) = f(x^3)$ for all $x$. If $f(a) \\neq 0$ for some $a$, then taking $x = \\sqrt[3]{a}$ leads to a contradiction, since $b^3 \\neq 0$. Thus, $f(x) = 0$ for all $x$, which satisfies the equation.\n\nNow, consider the case where $f(x) = 0$ only when $x = 0$. Substituting $y = 1$ into the previous factorized equation gives $f(x) - x^3 - f(1) + 1 = 0$ for all $x \\neq 0$, so $f(x) = x^3 + c$ for $x \\neq 0$, where $c = f(1) - 1$. Substituting into $f(y^3) = y^6 f(y)$ and simplifying gives $c = y^6 c$ for all $y$, which is only possible if $c = 0$. Hence, $f(x) = x^3$ for all $x$, which also satisfies the original equation.\n\n**Conclusion:** The solutions are $f(x) = 0$ and $f(x) = x^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19293,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q, r,$ and $s$ be prime numbers satisfying\n\n$$\n5 < p < q < r < s < p + 10.\n$$\n\nProve that the sum of these four prime numbers is divisible by 60.",
    "options": [],
    "answer": "See solution",
    "solution": "The four prime numbers must satisfy $p > 5$ and $s < p + 10$, so they must be among the five consecutive odd numbers $p, p + 2, p + 4, p + 6,$ and $p + 8$.\n\nSince we must choose 4 out of these 5 numbers, we omit exactly one. If we omit $p$, $p + 2$, $p + 6$, or $p + 8$, then three consecutive odd numbers remain, one of which must be divisible by 3, which is impossible since all are primes greater than 5. Therefore, we must omit $p + 4$.\n\nThus, the four primes are $p$, $q = p + 2$, $r = p + 6$, and $s = p + 8$.\n\nExactly one of the five consecutive numbers $p, p + 2, p + 4, p + 6, p + 8$ is divisible by 5. Since none of the chosen numbers can be divisible by 5, $p + 4$ must be divisible by 5. So $p + 4$ is divisible by 5.\n\nNow,\n\n$$\np + q + r + s = p + (p + 2) + (p + 6) + (p + 8) = 4p + 16 = 4(p + 4)\n$$\n\nSince $p + 4$ is divisible by 5, $4(p + 4)$ is divisible by $4 \\times 5 = 20$. Also, since $p$ is odd, $p + 4$ is odd, so $4(p + 4)$ is divisible by 4, and the sum is also divisible by 3 (since among any four such primes, one is congruent to each residue mod 3). Therefore, the sum is divisible by $4 \\times 5 \\times 3 = 60$.\n\n*Remark.* The quadruple $(11, 13, 17, 19)$ shows that 60 cannot be replaced by a greater number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19294,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $k_1$ and $k_2$ intersect at points $A$ and $B$, such that $k_1$ passes through the center $O$ of circle $k_2$. The line $p$ intersects $k_1$ at points $K$ and $O$, and $k_2$ at points $L$ and $M$, with $L$ between $K$ and $O$. Let $P$ be the orthogonal projection of $L$ onto the line $AB$. Prove that the line $KP$ is parallel to the $M$-median of triangle $ABM$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p7_data_4291bafebb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $C$ be the midpoint of segment $AB$. We need to prove $MC \\parallel KP$.\n\nLet $\\alpha = \\angle BKA$. Notice that\n\n$$\n\\begin{aligned}\n\\angle BLA &= 180^\\circ - \\angle BMA = 180^\\circ - \\frac{1}{2}\\angle BOA \\\\\n&= 180^\\circ - \\frac{1}{2}(180^\\circ - \\angle BKA) = 90^\\circ + \\frac{1}{2}\\alpha\n\\end{aligned}\n$$\n\nAlso, $O$ is the midpoint of arc $\\widearc{AB}$, so $KO$ is the angle bisector of $\\angle BKA$. From this, $L$ is the incenter of triangle $ABK$.\n\nMoreover, $ML$ is a diameter of circle $k_2$, so $\\angle ABM = 90^\\circ$. Since $BL$ is the angle bisector of $\\angle ABK$, $BM$ is the exterior angle bisector of the same angle.\n\nThus, $M$ lies on both the angle bisector $KM$ and the exterior angle bisector $BM$, so $M$ is the center of the excircle of triangle $ABK$.\n\nTherefore, we need to prove that the line through the incenter $L$ of triangle $ABK$ and the point of tangency of its incircle is parallel to the line through the excircle center $M$ and the midpoint $C$ of $AB$. This is a well-known lemma, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19295,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $AB \\leq AC$ and let $P$ be an interior point on the angle bisector of $\\angle BAC$. Let $D$ and $E$ be points on the segments $PC$ and $PB$ respectively such that $\\angle PBD = \\angle PCE$. The line $BD$ meets $AC$ at $X$, and $CE$ meets $AB$ at $Y$.\n\nProve that $BX \\le CY$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use Kelly's lemma:\n\n**Lemma 1 (Kelly).** Given a triangle $ABC$. Suppose the cevians $BE$ and $CF$ are such that $\\angle CBE \\geq \\angle BCF$ and $\\angle ABE \\geq \\angle ACF$. Then $BE \\leq CF$.\n\n*Proof (from Crux):* Choose $Q$ on the segment $AE$ so that $\\angle QBE = \\angle QCF$. Let $CF$ meet $BE$, $BQ$ at $P$, $Q$ respectively. In triangle $QBC$, since $\\angle QBC \\geq \\angle QCB$, we have $QC \\geq QB$. Observe that $\\triangle QBE \\sim \\triangle QCR$, hence from $BQ \\leq CQ$ we obtain $BE \\leq CR$. Clearly $CR \\leq CF$, therefore $BE \\leq CF$. $\\square$\n\nNow we apply Kelly's lemma to our problem. We want to show that\n\n$$\n\\angle DBC \\geq \\angle ECB \\quad \\text{and} \\quad \\angle ABP \\geq \\angle ACP.\n$$\n\nThen we can apply Kelly's lemma to get $BX \\leq CY$.\n\nReflect the point $C$ about the line $AP$ to $C'$. By symmetry,\n\n$$\n\\angle ABP \\geq \\angle AC'P = \\angle ACP.\n$$\n\n$$\nBP = AP \\frac{\\sin(A/2)}{\\sin(\\angle ABP)}, \\quad CP = AP \\frac{\\sin(A/2)}{\\sin(\\angle ACP)}.\n$$\n\nThus $BP \\leq CP$. In triangle $PBC$, since $BP \\leq CP$, it follows that $\\angle PCB \\leq \\angle PBC$. Therefore,\n\n$$\n\\angle DBC \\geq \\angle ECB.\n$$\n\nThis proves the claim and the problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19296,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set of $k$ positive integers. What is the smallest value of $k$ such that there exist $n$ distinct positive integers $x_1, x_2, \\dots, x_n$ with all the sums $x_1 + x_2, x_2 + x_3, \\dots, x_{n-1} + x_n, x_n + x_1$ belonging to $A$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m_1 = x_1 + x_2, m_2 = x_2 + x_3, \\dots, m_{n-1} = x_{n-1} + x_n, m_n = x_n + x_1$.\n\nFirst, note that $m_1 \\neq m_2$, otherwise $x_1 = x_3$, which contradicts the fact that $x_i$ are distinct. Similarly, $m_i \\neq m_{i+1}$ for $i = 1, 2, \\dots, n$, where $m_{n+1} = m_1$. It follows that $k \\geq 2$.\n\nFor $k = 2$, let $A = \\{a, b\\}$, where $a \\neq b$. Then,\n\n$$\n\\begin{cases}\nx_1 + x_2 = a, \\\\\nx_2 + x_3 = b, \\\\\n\\vdots \\\\\nx_{n-1} + x_n = a, \\\\\nx_n + x_1 = b,\n\\end{cases} \\quad \\text{(if $n$ is even)}\n$$\n\nor\n\n$$\n\\begin{cases}\nx_1 + x_2 = a, \\\\\nx_2 + x_3 = b, \\\\\n\\vdots \\\\\nx_{n-1} + x_n = b, \\\\\nx_n + x_1 = a,\n\\end{cases} \\quad \\text{(if $n$ is odd)}\n$$\n\nFor the odd case, $x_n = x_2$, which is possible. For the even case,\n\n$$\n\\begin{aligned}\n\\frac{n}{2}a &= (x_1 + x_2) + (x_3 + x_4) + \\dots + (x_{n-1} + x_n) \\\\\n&= (x_2 + x_3) + (x_4 + x_5) + \\dots + (x_n + x_1) \\\\\n&= \\frac{n}{2}b,\n\\end{aligned}\n$$\n\nso $a = b$, which is impossible. Thus, $k \\geq 3$.\n\nFor $k = 3$, one can construct a valid example as follows:\n\ndefine $x_{2k-1} = k$ ($k \\geq 1$) and $x_{2k} = n+1-k$ ($k \\geq 1$). When $n$ is even,\n\n$$\nx_i + x_{i+1} = \\begin{cases} n+1, & \\text{if } i \\text{ is odd} \\\\ n+2, & \\text{if } i \\text{ is even and } i < n \\\\ \\frac{n}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n \\end{cases}\n$$\n\nWhen $n$ is odd,\n\n$$\nx_i + x_{i+1} = \\begin{cases} n+1, & \\text{if } i \\text{ is odd and } i < n \\\\ n+2, & \\text{if } i \\text{ is even} \\\\ \\frac{n-1}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n \\end{cases}\n$$\n\nTherefore, the smallest positive integer $k$ is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19297,
    "subject": "Mathematics (Olympiad)",
    "question": "Во триаголникот $ABC$, $\\angle ACB = 40^{\\circ}$. Симетралите на внатрешниот и надворешниот агол во темето $C$ ја сечат правата $AB$ во точките $D$ и $E$, така што $\\triangle CDE$ е рамнокрак. Определи ги аглите на триаголникот $ABC$.\n\n![](images/Makedonija_2009_p11_data_89f56ad702.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $CD$ е симетрала на надворешниот агол, а $CE$ на внатрешниот во темето $C$. Симетралите $CD$ и $CE$ се нормални, тоа се симетрали на два напоредни агли, $\\angle DCE = 90^{\\circ}$. Од условот $\\triangle CDE$ е рамнокрак, па следува дека тој е рамнокрак правоаголен со хипотенуза $DE$. Оттука следува дека $\\angle DEC = 45^{\\circ}$. Тој е надворешен агол за $\\triangle CEB$, т.е. $45^{\\circ} = 20^{\\circ} + \\beta$, $\\beta = 25^{\\circ}$, а аголот $\\alpha = 115^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19298,
    "subject": "Mathematics (Olympiad)",
    "question": "Write either $1$ or $-1$ in each cell of a $2n \\times 2n$ table, so that there are exactly $2n^2$ entries of each kind. Let $M$ be the minimum of the absolute values of all row sums and all column sums. Determine the largest possible value of $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "Split the table into four smaller tables of size $n \\times n$. Fill the upper left quarter with $1$s, the lower right quarter with $-1$s, and the remaining two quarters in a checkerboard pattern. If $n$ is odd, fill so that one quarter contains more $1$s than $-1$s, and the other more $-1$s than $1$s. If $n$ is even, each row and column contains either $n/2$ $1$s and $3n/2$ $-1$s, or vice versa, so $M = n$. If $n$ is odd, each row and column contains either $(n - 1)/2$ $1$s ($-1$s) and $(3n + 1)/2$ $-1$s ($1$s), or $(n + 1)/2$ $1$s ($-1$s) and $(3n - 1)/2$ $-1$s ($1$s); thus $M = n - 1$.\n\nNow, we show that $M$ cannot be larger. If a row or column contains as many $1$s as $-1$s, then $M = 0$. Otherwise, split all $4n$ rows and columns into two subsets: those with more $1$s than $-1$s, and those with more $-1$s than $1$s. One set must have at least $2n$ elements. Assume at least $2n$ rows and columns (with $k$ rows and $\\ell$ columns) have more $1$s than $-1$s. Each has at least $n + M/2$ $1$s and at most $n - M/2$ $-1$s. The total number of $1$s in these is at least\n\n$$\n(k + \\ell) \\left( n + \\frac{M}{2} \\right) - k\\ell,\n$$\n\nwhere the last term corrects for double-counting. Thus,\n\n$$\n2n^2 \\ge (k+\\ell) \\left(n + \\frac{M}{2}\\right) - k\\ell \\ge (k+\\ell) \\left(n + \\frac{M}{2}\\right) - \\frac{1}{4}(k+\\ell)^2\n$$\n\nLet $r = k + \\ell$:\n\n$$\nM \\le \\frac{2n^2 + r^2/4 - rn}{r/2} = \\frac{4n^2}{r} + \\frac{r}{2} - 2n = n - \\frac{(r-2n)(4n-r)}{2r} \\le n,\n$$\n\nsince $2n \\le r \\le 4n$. Hence $M \\le n$, which completes the proof for even $n$. If $n$ is odd, $M = n$ is impossible, since all row and column sums must be even, so $M \\le n - 1$.\n\n**Conclusion:** The largest possible value of $M$ is $n$ if $n$ is even, and $n - 1$ if $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19299,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{Z}$ be a function such that for any two positive integers $a, b$, $f(ab) = f(a) + f(b)$. Find all possible values of $f(1000)$.",
    "options": [],
    "answer": "See solution",
    "solution": "$f(1000)$ can be any integer divisible by $3$.\n\nConsider:\n\n$$\nf(1000) = f(10^3) = 3 \\cdot f(10).\n$$\n\nThus, $f(1000)$ is always a multiple of $3$.\n\nNow, for any integer $k$, define $f(a) = k \\cdot \\operatorname{ord}_2(a)$, where $\\operatorname{ord}_2(a)$ is the highest power of $2$ dividing $a$.\n\nThen,\n\n$$\nf(ab) = k \\cdot \\operatorname{ord}_2(ab) = k \\cdot (\\operatorname{ord}_2(a) + \\operatorname{ord}_2(b)) = f(a) + f(b).\n$$\n\nFor $1000 = 2^3 \\cdot 125$, $\\operatorname{ord}_2(1000) = 3$, so $f(1000) = 3k$ for any integer $k$.\n\nTherefore, all integer multiples of $3$ are possible values for $f(1000)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19300,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number.\n\na) Find all values of $a$ for which the inequality\n\n$$\nx \\log_{\\frac{1}{2}} a^4 - x^2 > 3 + 2 \\log_2 a^2\n$$\n\nhas a solution.\n\nb) Calculate the limit\n\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\log_2(a^2)$, let $2 \\log_2(a^2) = b$. The inequality becomes $x^2 + b x + 3 + b < 0$. For this quadratic to have a solution, its discriminant must be positive: $D = b^2 - 4b - 12 > 0$, which gives $b < -2$ or $b > 6$. Thus, $\\log_2(a^2) < -1$ or $\\log_2(a^2) > 3$. From logarithm properties, $a^2 < \\frac{1}{2}$ or $a^2 > 8$, with $a \\ne 0$. Therefore,\n\n$$\na \\in (-\\infty, -2\\sqrt{2}) \\cup \\left(-\\frac{\\sqrt{2}}{2}, 0\\right) \\cup \\left(0, \\frac{\\sqrt{2}}{2}\\right) \\cup (2\\sqrt{2}, \\infty).\n$$\n\nb) For $a < 0$,\n\n$$\n\\sqrt{a^2 - a + 1} + a = \\frac{(\\sqrt{a^2 - a + 1} + a)(\\sqrt{a^2 - a + 1} - a)}{\\sqrt{a^2 - a + 1} - a} = \\frac{-1 + \\frac{1}{a}}{-\\sqrt{1 - \\frac{1}{a} + \\frac{1}{a^2}} - 1}.\n$$\n\nTherefore,\n\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right) = \\frac{1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19301,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of 16 people will be partitioned into 4 indistinguishable 4-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M$, where $r$ and $M$ are positive integers and $M$ is not divisible by 3. What is $r$?\n\n(A) 5  \n(B) 6  \n(C) 7  \n(D) 8  \n(E) 9",
    "options": [],
    "answer": "See solution",
    "solution": "The 16 people can be partitioned into the 4 committees, each of size 4, in\n$$\n\\frac{16!}{(4!)^5}\n$$\nways; four of the $4!$ factors come from permuting the members of the committees and one $4!$ factor comes from permuting the committees. Then there are $4^4$ ways to choose the four chairpersons and $3^4$ ways to choose the four secretaries. This gives a total of\n$$\n\\frac{16! \\cdot 4^4 \\cdot 3^4}{(4!)^5}\n$$\nassignments. The numerator has 1 factor of 3 in each of 15, 12, 6, and 3; it has 2 factors of 3 in 9 and 4 factors in $3^4$, a total of 10. The denominator has 5 factors of 3. Thus $r = 10 - 5 = 5$.\n\nAlternatively,\n\nThere are $16 \\cdot 15$ ways to choose the chairperson and secretary of the first committee, $14 \\cdot 13$ ways for the second, $12 \\cdot 11$ for the third, and $10 \\cdot 9$ for the fourth. There are then $\\frac{8 \\cdot 7}{2} = 28$ ways to choose the remaining members of the first committee, $\\frac{6 \\cdot 5}{2} = 15$ for the second, $\\frac{4 \\cdot 3}{2} = 6$ for the third, and $\\frac{2 \\cdot 1}{2} = 1$ for the fourth. There are $4!$ ways to account for the indistinguishability of the committees. This gives a total of\n$$\n\\frac{16 \\cdot 15 \\cdot 14 \\cdot 13 \\cdot 12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 28 \\cdot 15 \\cdot 6 \\cdot 1}{4!}\n$$\nassignments. There are $1 + 1 + 2 + 1 + 1 = 6$ factors of 3 in the numerator and 1 factor of 3 in the denominator, so $r = 6 - 1 = 5$.\n\n**Note:** The number of ways to make the assignments is $54,486,432,000 = 3^5 \\cdot 224,224,000$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19302,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $x, y$ that satisfy the system:\n\n$$\n\\begin{cases}\n[x, y] + (x, y) = 2018, \\\\\n x + y = 2018,\n\\end{cases}\n$$\n\nwhere $[x, y]$ and $(x, y)$ denote the least common multiple (LCM) and greatest common divisor (GCD) of $x$ and $y$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(x, y) = (1, 2017)$, $(2, 2016)$, $(1009, 1009)$, $(2016, 2)$, and $(2017, 1)$.\n\nSince $(x, y) \\cdot [x, y] = xy$, the first equation can be rewritten as:\n\n$$\n\\frac{xy}{(x, y)} + (x, y) = 2018\n$$\n\nwhich leads to the quadratic equation in $(x, y)$:\n\n$$\n(x, y)^2 - 2018(x, y) + xy = 0.\n$$\n\nThe discriminant, using $x + y = 2018$, is:\n\n$$\nD = 2018^2 - 4xy = 2018^2 - 4x(2018 - x) = (2x - 2018)^2.\n$$\n\nThis gives $x = \\frac{2018 \\pm (2x - 2018)}{2}$, so $x_1 = x$ and $x_2 = 2018 - x = y$.\n\nThus, one of the numbers equals the GCD, so it divides the other. Let $x \\leq y$, so $y = kx$ and $(k+1)x = 2018$. The possible cases are:\n\n- $k+1 = 2018 \\Rightarrow x = 1$, $y = 2017$\n- $k+1 = 1009 \\Rightarrow x = 2$, $y = 2016$\n- $k+1 = 2 \\Rightarrow x = 1009 = y$\n\nThe symmetric pairs $(2016, 2)$ and $(2017, 1)$ also satisfy the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19303,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ players play a round-robin tournament (each pair plays one game, with possible ties). For $n=2$, the result is trivial. For $n > 2$, define a set of four players as *harmonic* if, in the games among them, no player beats all the others. Determine, for all $n \\geq 2$, the maximum number of harmonic sets of four players that can exist in such a tournament. \n\n% IMAGE: ![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, the answer is $0$.\n\nFor $n > 2$:\n\n- If $n$ is odd, the maximum number of harmonic sets is $$\\frac{n(n-3)(n^2+6n-31)}{48}.$$\n- If $n$ is even, the maximum is $$\\frac{n(n^3+3n^2-52n+108)}{48}.$$\n\n**Outline of construction and proof:**\n\nLabel the players $1$ to $n$, and let $d_i$ be the number of games the $i$-th player has won. Then $\\sum_{i=1}^n d_i = \\binom{n}{2} - n = \\frac{n(n-3)}{2}$.\n\nFor a set of four players, if one beats all others, the set is not harmonic. There cannot be two such players in a set. Thus, the number of harmonic sets is at most $\\binom{n}{4} - \\sum_{i=1}^n \\binom{d_i}{3}$.\n\nBy convexity and majorization, for odd $n$, $\\sum_{i=1}^n \\binom{d_i}{3} \\geq n \\binom{\\frac{1}{n}(n-3)}{3}$, so the maximum is $\\binom{n}{4} - n \\binom{\\frac{1}{n}(n-3)}{3} = \\frac{n(n-3)(n^2+6n-31)}{48}$.\n\nFor even $n$, $\\sum_{i=1}^n \\binom{d_i}{3} \\geq \\frac{n}{2} \\binom{\\frac{1}{n}(n-4)}{3} + \\frac{n}{2} \\binom{\\frac{1}{n}(n-2)}{3}$, so the maximum is $\\binom{n}{4} - \\frac{n}{2} \\binom{\\frac{1}{n}(n-4)}{3} - \\frac{n}{2} \\binom{\\frac{1}{n}(n-2)}{3} = \\frac{n(n^3+3n^2-52n+108)}{48}$.\n\n**Construction for odd $n$:**\n- Each player wins $\\frac{1}{2}(n-3)$ games.\n- The game between consecutive players (including $1$ and $n$) ends in a tie.\n- If $i$ and $j$ are not consecutive, with $i < j$, player $j$ beats $i$ if $i$ and $j$ have the same parity; otherwise, $i$ beats $j$.\n\nThis ensures that in any set of four players where no one beats all others, the set is harmonic.\n\n**Construction for even $n > 2$:**\n- For $1 \\leq i \\leq n$, the game between $i$ and $i+1$ (modulo $n$) ends in a tie.\n- For $1 \\leq i < j \\leq n$, player $i$ beats $j$ if $j-i$ is odd; otherwise, $j$ beats $i$.\n\nThe verification is similar to the odd $n$ case.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19304,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2018 players sitting around a round table. At the beginning of the game, we arbitrarily deal all the cards from a deck of $K$ cards to the players (some players may receive no cards). In each turn, we choose a player who draws one card from each of their two neighbours. It is only allowed to choose a player whose each neighbour holds a nonzero number of cards. The game terminates when there is no such player. Determine the largest possible value of $K$ such that, no matter how we deal the cards and how we choose the players, the game always terminates after a finite number of turns.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $K = 2017$.\n\nFor $K = 2018$, we deal 2 cards to one player, 0 cards to one of their neighbours, and 1 card to everyone else. Then in each turn, we choose the player with 0 cards:\n\n$$\n\\dots 11 \\overbrace{\\underbrace{20}_{\\sim}}^{\\sim} 11 \\dots \\rightarrow \\dots 11 \\overbrace{\\underbrace{20}_{\\sim}}^{\\sim} 11 \\dots\n$$\n\nAfter each turn, the configuration stays the same—there is one player with 2 cards, one of their neighbours with 0 cards, and all the others with 1 card (the only change is that the positions of the players with 2 and 0 cards are shifted). Therefore, we can make moves forever and the game never terminates.\n\nWhenever $K > 2018$, we can play forever using the same strategy as for $K = 2018$. We simply deal the extra cards arbitrarily and ignore them during the game.\n\nNow we will prove that for $K = 2017$ the game terminates no matter how we play. Let us call zeros the players with no cards and ones the players with exactly one card. The zeros split the other players into segments of various lengths. When two zeros sit next to each other, they form a segment with a length of 0. Also note that there is obviously at least one zero when $K = 2017$.\n\n**Lemma.** There exists a segment containing no other players than ones (possibly with a length of 0).\n\n*Proof.* If we add to each segment the zero which bounds it in the clockwise direction, then the sum of the lengths of all the segments will be 2018. There are only 2017 cards, therefore at least one segment contains fewer cards than players, which is possible only when all the players of this segment, except for the bounding zero, are ones. $\\square$\n\nLet us consider the shortest segment among the ones containing no other players than ones; the lemma assures the existence of such a segment. If we choose a zero adjacent to this segment, we shorten it by 1 (or by 2—in the special case when there is exactly one zero in the game):\n\n$$\n\\dots * \\overbrace{0111\\dots10\\dots}^{\\hat{\\hat{0}}11\\dots10\\dots} \\rightarrow \\dots * \\overbrace{2011\\dots10\\dots}^{\\hat{\\hat{2}}11\\dots10\\dots}\n$$\n\nIf we choose one of the ones inside of the shortest segment, we create two even shorter segments:\n\n$$\n\\dots 01\\dots1 \\overbrace{111}^{\\hat{\\hat{1}}} 1\\dots10\\dots \\rightarrow \\dots 01\\dots1 \\overbrace{0301}^{\\hat{\\hat{0}}11\\dots10\\dots} \\rightarrow\n$$\n\nThe length of the shortest segment could decrease only finitely many times. From the moment when it stops decreasing, we won't be able to choose any of the zeros bounding the shortest segment, nor any of the ones inside of it. This means that the game will continue on the other side of the table between the bounding zeros of the shortest segment. The neighbours of these two zeros won't be able to get any more cards, so we cannot choose them anymore. The neighbours of these neighbours will thereby be chosen at most finitely many times (at most the number of times equal to the number of cards of these neighbours), so after some time we won't be able to choose them. We can use this reasoning repeatedly. The part of the table where we still can choose players eventually decreases, which means that the game cannot last infinitely long.\n\n**Remark.** If $K = 2018$ and we give one card to every player, then after one move we would get a segment of ones bounded by two zeros. In that case, the game necessarily ends after finitely many moves (to see it we just need to use the reasoning from the solution).\n\n**Remark.** As soon as we show that the game will be played only in one part of the table bounded by two players (so no cards will ever pass some line of the table and therefore it could be thought of as a line segment), we might just use a right mono-variant to prove that the game is finite. For example, to each card we might assign its distance to one of the bounds and keep track of the sum of squares of these distances. In each move this number is decreased by\n\n$$\n(a - 1)^2 + (a + 1)^2 - 2a^2 = 2,\n$$\n\nand since it cannot be negative, the game will have to eventually end.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19305,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be the function from the set of positive integers into itself, defined by:\n\n- $f(1) = 1$\n- $f(2n) = f(n)$\n- $f(2n+1) = f(n) + f(n+1)$\n\nShow that, for any positive integer $n$, the number of positive odd integers $m$ such that $f(m) = n$ is equal to the number of positive integers less than and coprime to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "With reference to the recurrence for $f$, notice that if $n$ is a positive even integer, then $f(n) < f(n + 1)$, and if $n$ is odd, $f(n) \\geq f(n + 1)$. So $f(n) < f(n + 1)$ if and only if $n$ is even.\n\nAgain, by the recurrence for $f$, an easy induction shows $f(n)$ and $f(n + 1)$ are coprime for each positive integer $n$.\n\nDiscarding the trivial case $n = 1$, given a positive integer $n \\geq 2$, it follows that if $m$ is a positive odd integer such that $f(m) = n$, then $f(m - 1)$ is a positive integer less than and coprime to $n$.\n\nNext, we prove that for every pair of coprime positive integers $(k, n)$ there exists a unique positive integer $m$ such that $k = f(m)$ and $n = f(m + 1)$. If, in addition, $k < n$, then $m$ is even by the preceding, so $m + 1$ is a positive odd integer such that $f(m + 1) = n$ and the conclusion follows.\n\nTo prove the above claim, proceed by induction on $k + n$. The base case, $k + n = 2$ (i.e., $k = n = 1$), is clear. If $k + n > 2$, apply the induction hypothesis to the pair $(k, n - k)$ or $(k - n, n)$, according as $k < n$ or $k > n$. In the former case, $k = f(m) = f(2m)$ and $n = k + f(m + 1) = f(m) + f(m + 1) = f(2m + 1)$ for some positive integer $m$; in the latter, $n = f(m + 1) = f(2m + 2)$ and $k = f(m) + n = f(m) + f(m + 1) = f(2m + 1)$ for some positive integer $m$. This establishes the existence of the desired positive integer.\n\nTo prove uniqueness, write $k = f(m)$ and $n = f(m + 1)$ for some positive integer $m$, and consider again the two possible cases.\n\nIf $k < n$, then $m$ is even, say $m = 2m'$, where $m'$ is a positive integer, so $k = f(2m') = f(m')$ and $n - k = f(2m' + 1) - f(m') = f(m' + 1)$. The induction hypothesis applies to the pair $(k, n - k)$ to imply uniqueness of $m'$, hence uniqueness of $m$.\n\nIf $k > n$, then $m$ is odd, say $m = 2m' + 1$, where $m'$ is a non-negative integer, so $k - n = f(2m' + 1) - f(2m' + 2) = f(m') + f(m' + 1) - f(m' + 1) = f(m')$ and $n = f(2m' + 2) = f(m' + 1)$. The induction hypothesis applies now to the pair $(k - n, n)$ to imply uniqueness of $m'$, hence again uniqueness of $m$. This completes the induction step and ends the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19306,
    "subject": "Mathematics (Olympiad)",
    "question": "Examine all possible sequences $a_1, a_2, \\ldots, a_{2008}$ of non-negative integers such that $a_1 \\leq a_2 \\leq \\ldots \\leq a_{2008}$ and $a_k \\leq k-1$ for all $k$ from $1$ to $2008$. Prove that the number of such sequences exceeds:\n\na) $2^{2007}$;\n\nb) $2^{2008}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** To construct $2^{2007}$ different sequences satisfying the conditions, set $a_1 = 0$ and let each successive member be either equal to or one greater than the previous: $a_{k} = a_{k-1}$ or $a_{k} = a_{k-1} + 1$. This yields $2^{2007}$ sequences, each with $a_k \\leq k-1$. To show the total number exceeds $2^{2007}$, consider a sequence not generated this way, such as $(0, 0, 2, 2, \\ldots, 2)$, which also satisfies the conditions.\n\n**b)** Define a \"fine\" sequence as a non-decreasing sequence $a_1 \\leq a_2 \\leq \\ldots \\leq a_n$ with $a_k \\leq k-1$ for all $k$. Any fine sequence of length $n+1$ can be formed from a fine sequence of length $n$ by choosing $a_{n+1}$ such that $a_n \\leq a_{n+1} \\leq n$. Let $x_n$ be the number of fine sequences of length $n$. For each sequence, at least two choices for $a_{n+1}$ are possible: $a_n$ or $a_n + 1$, so $x_{n+m} \\geq 2^m x_n$.\n\nFor small $n$:\n- $x_1 = 1$ (only $(0)$)\n- $x_2 = 2$ ($(0,0)$ and $(0,1)$)\n- $x_3 = 5$\n- $x_4 = 14$\n\nThus, $x_{2008} \\geq 2^{2004} x_4 = 14 \\cdot 2^{2004} > 2^{2008}$.\n\nFor $x_5$, count extensions:\n- Sequences ending in $0$: add $0,1,2,3,4$ ($5$ options)\n- Ending in $1$: add $1,2,3,4$ ($4$ options, $3$ sequences)\n- Ending in $2$: add $2,3,4$ ($3$ options, $5$ sequences)\n- Ending in $3$: add $3,4$ ($2$ options, $5$ sequences)\n\nTotal: $5 \\cdot 1 + 3 \\cdot 4 + 5 \\cdot 3 + 5 \\cdot 2 = 42$, so $x_{2008} \\geq 2^{2003} x_5 = 42 \\cdot 2^{2003} > 2^{2008}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19307,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABC$ 為平面上的三角形。一圓 $\\Gamma$ 通過點 $A$，與線段 $AB$ 和 $AC$ 分別再交於點 $D$ 和 $E$，並與線段 $BC$ 交於 $F, G$ 兩點，其中 $F$ 位於 $B, G$ 兩點之間。由 $F$ 對 $\\triangle BDF$ 外接圓引切線，再由 $G$ 對 $\\triangle CEG$ 外接圓引切線，設這兩條切線交於點 $T$，且 $A, T$ 兩點相異。證明：直線 $AT$ 平行於直線 $BC$。",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $\\angle TFB = \\angle FDA$ because $FT$ is tangent to circle $BDF$, and moreover $\\angle FDA = \\angle CGA$ because quadrilateral $ADFG$ is cyclic. Similarly, $\\angle TGB = \\angle GEC$ because $GT$ is tangent to circle $CEG$, and $\\angle GEC = \\angle CFA$. Hence,\n\n$$\n\\angle TFB = \\angle CGA \\text{ and } \\angle TGB = \\angle CFA. \\qquad (1)\n$$\n\n![](images/20-2J_p4_data_273341b1c7.png)\n\nTriangles $FGA$ and $GFT$ have a common side $FG$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $AT$ is parallel to line $BFGC$. $\\square$\n\n**Comment.** Alternatively, we can prove first that $T$ lies on $\\Gamma$. For example, this can be done by showing that $\\angle AFT = \\angle AGT$ using (1). Then the statement follows as $\\angle TAF = \\angle TGF = \\angle GFA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19308,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The required maximum is $\\lfloor 2(N + 2)/3 \\rfloor$.\n\nFor 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.\n\n**REMARK.** Maximizing partitions are not necessarily unique. For instance, if $m$ is an integer greater than 1, then\n$$\n\\underbrace{2, \\dots, 2}_{m+1}, \\underbrace{1, \\dots, 1}_{m} \\quad \\text{and} \\quad 4, \\underbrace{2, \\dots, 2}_{m-1}, \\underbrace{1, \\dots, 1}_{m}.\n$$\nare both maximizing partitions of $3m + 2$ into at least three positive integers; the former is 'standard', whereas the latter is not. Similarly, if $m > 2$, then\n$$\n\\underbrace{2, \\dots, 2}_{m}, \\underbrace{1, \\dots, 1}_{m} \\quad \\text{and} \\quad 4, \\underbrace{2, \\dots, 2}_{m-1}, \\underbrace{1, \\dots, 1}_{m-2}.\n$$\nare both maximizing partitions of $3m$ into at least three positive integers; again, the former is 'standard', whereas the latter is not.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19309,
    "subject": "Mathematics (Olympiad)",
    "question": "Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?\n\n(A) 70  \n(B) 84  \n(C) 98  \n(D) 105  \n(E) 126",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ denote the required length of the ship. Assume Emily's speed is $1$ and let $v < 1$ denote the speed of the ship relative to the riverbank. When Emily walks from the back to the front, it takes 210 steps, so $210(1 - v) = L$. Similarly, when Emily walks from the front to the back of the ship, it takes 42 steps, so $42(1 + v) = L$. Solving this system yields $v = \\frac{2}{3}$ and $L = 70$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19310,
    "subject": "Mathematics (Olympiad)",
    "question": "AB is a diameter of the circle $O$. The point $C$ lies on the extension of line $AB$ produced. A line passing through $C$ intersects the circle $O$ at points $D$ and $E$. $OF$ is a diameter of the circumcircle $O_1$ of $\\triangle BOD$. Join $CF$ and extend it; its extension intersects the circle $O_1$ at $G$. Prove that points $O$, $A$, $E$, and $G$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $OF$ is a diameter of the circumcircle of $\\triangle DOB$, $OF$ is the bisector of $\\angle DOB$, so $\\angle DOB = 2\\angle DOF$. Since $\\angle DAB = \\frac{1}{2} \\angle DOB$, we have $\\angle DAB = \\angle DOF$. Since $\\angle DGF = \\angle DOF$, we obtain $\\angle DGF = \\angle DAB$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p156_data_df53526e86.png)\n\nThus, $G$, $A$, $C$, $D$ are concyclic. Hence,\n\n$$\n\\angle AGC = \\angle ADC, \\qquad ①\n$$\n\n$$\n\\angle AGC = \\angle AGO + \\angle OGF = \\angle AGO + \\frac{\\pi}{2}, \\qquad ②\n$$\n\n$$\n\\angle ADC = \\angle ADB + \\angle BDC = \\angle BDC + \\frac{\\pi}{2}. \\qquad ③\n$$\n\nCombining ①, ②, and ③ yields\n\n$$\n\\angle AGO = \\angle BDC. \\qquad ④\n$$\n\nSince $B$, $D$, $E$, $A$ are concyclic, we have\n\n$$\n\\angle BDC = \\angle EAO. \\qquad ⑤\n$$\n\nAs $OA = OE$, we obtain\n\n$$\n\\angle EAO = \\angle AEO. \\qquad ⑥\n$$\n\nCombining ④, ⑤, and ⑥ implies $\\angle AGO = \\angle AEO$.\n\nTherefore, $O$, $A$, $E$, $G$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19311,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N^*$ be the set of positive integers. Define $a_1 = 2$, and for $n = 1, 2, \\dots$,\n$$\na_{n+1} = \\min\\left\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} + \\frac{1}{\\lambda} < 1, \\lambda \\in N^*\\right\\}.\n$$\nProve that $a_{n+1} = a_n^2 - a_n + 1$ for $n = 1, 2, \\dots$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_1 = 2$, $a_2 = \\min\\left\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{\\lambda} < 1, \\lambda \\in \\mathbb{N}^*\\right\\}$,\nfrom $\\frac{1}{a_1} + \\frac{1}{\\lambda} < 1$ we have $\\frac{1}{\\lambda} < 1 - \\frac{1}{2} = \\frac{1}{2}$, $\\lambda > 2$ and so $a_2 = 3$.\n\nThis means that when $n = 1$, the conclusion is right.\n\nSuppose that for $n \\le k - 1$ ($k \\ge 2$), the conclusions are right. For $n = k$, from\n$$\na_{k+1} = \\min \\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1, \\lambda \\in \\mathbb{N}^* \\right\\}.\n$$\nAs $\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1$, i.e.\n$$\n0 < \\frac{1}{\\lambda} < 1 - \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right),\n$$\n$$\n\\lambda > \\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}}.\n$$\nNow we prove that $\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k(a_k - 1)$.\n\nBy the supposition, for $2 \\le n \\le k$, $a_n = a_{n-1}(a_{n-1}-1)+1$; then\n$$\n\\frac{1}{a_n - 1} = \\frac{1}{a_{n-1}(a_{n-1} - 1)} = \\frac{1}{a_{n-1} - 1} - \\frac{1}{a_{n-1}}.\n$$\nSo $\\frac{1}{a_{n-1}} = \\frac{1}{a_{n-1}-1} - \\frac{1}{a_n-1}$, and $\\sum_{i=2}^{k} \\frac{1}{a_{i-1}} = 1 - \\frac{1}{a_k-1}$, i.e.\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} = 1 - \\frac{1}{a_k - 1} + \\frac{1}{a_k} = 1 - \\frac{1}{a_k(a_k - 1)}\n$$\nwhich means that $\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\frac{1}{a_k}} = a_k(a_k - 1)$; then",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19312,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the value of $2 - (0 - (1 - 5))$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n2 - (0 - (1 - 5)) = 2 - (0 - (-4)) = 2 - (0 + 4) = 2 - 4 = -2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19313,
    "subject": "Mathematics (Olympiad)",
    "question": "$x, y, z \\in [0,2]$. Find the minimal value of the algebraic expression:\n\n$$\nA = \\sqrt{2 + x} + \\sqrt{2 + y} + \\sqrt{2 + z} + \\sqrt{x + y} + \\sqrt{y + z} + \\sqrt{z + x}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove first that $\\sqrt{2 + x} + \\sqrt{y + z} = \\sqrt{2 + x} + \\sqrt{4 - x} \\ge 2 + \\sqrt{2}$.\n\nIndeed, this is equivalent to\n\n$$\n2 + x + 4 - x + 2\\sqrt{(2 + x)(4 - x)} \\ge (2 + \\sqrt{2})^2\n$$\n\nwhich simplifies to\n\n$$\nx(2 - x) \\ge 0,\n$$\n\nwhich is true since $x \\in [0,2]$.\n\nSimilarly, we have $\\sqrt{2 + y} + \\sqrt{x + z} \\ge 2 + \\sqrt{2}$ and $\\sqrt{2 + z} + \\sqrt{x + y} \\ge 2 + \\sqrt{2}$.\n\nAdding all these, we get $A \\ge 6 + 3\\sqrt{2}$, and the equality holds, for example, if $x = y = 2$ and $z = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19314,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle, the feet of the altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again at point $P$ and the circumcircle of $CAB$ again at point $Q$. Prove that\n\n$$\n|MP| = |MQ|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The orthocenter $H$ of triangle $ABC$ lies on the circumcircle of triangle $CDE$, because $\\angle HDC + \\angle HEC = 90^\\circ + 90^\\circ = 180^\\circ$ (see the figure below). Let $\\alpha = \\angle BAC$; then $\\angle CHE = 90^\\circ - \\angle ECH = \\alpha$. Therefore, $\\angle MPE = 180^\\circ - \\angle CPE = 180^\\circ - \\angle CHE = 180^\\circ - \\alpha$, so points $A, M, P, E$ are concyclic. Similarly, points $B, M, P, D$ are concyclic.\n\nPoint $M$ is the circumcenter of the right triangle $ABE$. Therefore, $|ME| = |MA|$ and $\\angle MEA = \\alpha$, so $\\angle MPA = \\alpha$. Since $\\angle MQB = \\angle CQB = \\angle CAB = \\alpha$, we have $AP \\parallel BQ$. Similarly, $BP \\parallel AQ$.\n\n![](images/prob1314_p30_data_17f6bbd826.png \"Figure 35\")\n\nThus, $APBQ$ is a parallelogram with diagonals $AB$ and $PQ$. As the diagonals of a parallelogram bisect each other, the desired result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19315,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $AB = 1$, $AC = 2$, and $\\cos B = 2 \\sin C$. Then the length of side $BC$ is \\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "By the law of sines, $\\frac{\\sin B}{\\sin C} = \\frac{AC}{AB} = 2$. Thus, $\\sin B = 2 \\sin C = \\cos B$, so $\\tan B = 1$. Therefore, $B = \\frac{\\pi}{4}$.\n\nLet $BC = a > 0$. By the law of cosines:\n$$\na^2 = AB^2 + AC^2 - 2 \\cdot AB \\cdot AC \\cdot \\cos B = 1^2 + 2^2 - 2 \\cdot 1 \\cdot 2 \\cdot \\frac{\\sqrt{2}}{2} = 1 + 4 - 2\\sqrt{2}\n$$\nSo,\n$$\na = \\frac{\\sqrt{2} + \\sqrt{14}}{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19316,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^3 + f(y)) = x^2 f(x) + y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = \\lambda$. Put $x = y = 0$ in (1) to get $f(\\lambda) = 0$. Taking $y = \\lambda$, we get $f(x^3) = x^2 f(x) + \\lambda$. Put $x = \\lambda, y = 0$ in (1) and we get\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda).\n$$\nBut $x = \\lambda, y = \\lambda$ gives $f(\\lambda^3) = \\lambda$. Now put $x = \\lambda, y = \\lambda^3$ and we get\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda) + \\lambda^3.\n$$\nComparing the two, we get $\\lambda = 0$. Therefore,\n$$\nf(x^3) = x^2 f(x), \\quad f(f(x)) = x,\n$$\nfor all $x$. Now changing $y$ to $f(y)$, we get\n$$\nf(x^3 + y) = x^2 f(x) + f(y) = f(x^3) + f(y)\n$$\nfor all $x, y$. Since $x \\mapsto x^3$ is a bijection of $\\mathbb{R}$ onto $\\mathbb{R}$, we obtain\n$$\nf(x + y) = f(x) + f(y), \\quad f(f(x)) = x\n$$\nfor all $x, y$.\n\nConsider $f(x^3) = x^2 f(x)$. Replacing $x$ by $x+1$, we obtain\n$$\nf(x^3 + 3x^2 + 3x + 1) = (x^2 + 2x + 1)f(x + 1).\n$$\nUsing the additivity of $f$, we get\n$$\n\\begin{aligned}\nf(x^3) + 3f(x^2) + 3f(x) + f(1) &= (x^2 + 2x + 1)(f(x) + f(1)) \\\\\n&= x^2 f(x) + 2x f(x) + f(x) + x^2 f(1) + 2x f(1) + f(1).\n\\end{aligned}\n$$\nThis reduces to\n$$\n3f(x^2) + 2f(x) = c x^2 + 2c x + 2x f(x)\n$$\nfor all $x \\in \\mathbb{R}$. Replace $x$ by $x+1$ and expand using additivity, we get\n$$\n3f(x^2) + 6f(x) = c x^2 + 6c x + 2x f(x).\n$$\nComparing the two, we see that $f(x) = c x$. Replacing $x$ by $f(x)$, we get $x = c f(x)$. Together we get $c^2 x = x$. Hence $c^2 = 1$. Therefore $f(x) = x$ or $f(x) = -x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19317,
    "subject": "Mathematics (Olympiad)",
    "question": "Some knights are on a playground. Each knight has three properties: speed, smartness, and sightliness. For every knight, each property takes a certain integer value $x$ such that $1 \\le x \\le n$. A knight $A$ can win a knight $B$ if the speed, smartness, and sightliness of $A$ are all greater than those of $B$. It is known that none among the knights on the playground can win any of the other knights, and every two knights differ by at least one property. Find the largest possible number of knights on the playground.\n\nAnswer: $3n^2 - 3n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each knight, we can define a unique non-negative integer $x$ such that the properties of the knight are $a + x$, $b + x$, and $c + x$, where $\\min(a, b, c) = 1$. Call the vector $(a, b, c)$ the *base triple* of the knight.\n\nIf two knights had the same base triple, then either all their properties would be equal, or one of them could win the other. Both cases are excluded by the assumptions. Thus, each knight has a unique base triple, implying that there are at most as many knights as possible base triples.\n\nThe number of triples of integers $1, \\dots, n$ is $n^3$. Among them, $(n-1)^3$ triples do not contain the value $1$. Thus, the number of possible base triples is $n^3 - (n-1)^3 = 3n^2 - 3n + 1$. This number is achieved if, for every base triple, there is exactly one knight whose values of properties coincide with the components of this base triple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19318,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive real number $c$, the sequence $a_1, a_2, \\dots$ of real numbers is defined as follows.\nLet $a_1 = c$, and for $n \\ge 2$, let\n\n$$\na_n = \\sum_{i=1}^{n-1} (a_i)^{n-i+1}.\n$$\n\nFind all positive real numbers $c$ such that $a_i > a_{i+1}$ for all positive integers $i$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $c < \\frac{\\sqrt{5}-1}{2}$.\n\nTo show this is necessary, note that $a_2 = c^2$ and $a_3 = c^3 + c^4$, so if the sequence is decreasing, we have $c^2 > c^3 + c^4$, implying $c < \\frac{\\sqrt{5}-1}{2}$.\n\nIn the other direction, suppose $c$ is a positive real number with $c < \\frac{\\sqrt{5}-1}{2}$. We will apply strong induction. We have\n\n$$\n\\begin{aligned}\na_1 &= c \\\\\na_2 &= c^2 \\\\\na_3 &= c^3 + c^4\n\\end{aligned}\n$$\n\nso $a_1 > a_2$ (since $c < 1$) and $a_2 > a_3$ (since $c < \\frac{\\sqrt{5}-1}{2}$). For the inductive step, we'll show $a_n > a_{n+1}$ assuming that $n \\ge 3$ and $a_i > a_{i+1}$ for all $i < n$. We have\n\n$$\na_1^n = c^n > c^n(c + c^2) > c^n(c + c^n) = a_1^{n+1} + a_2^n.\n$$\n\nAlso, since $a_i > a_{i+1}$ for $i = 2, \\dots, n-1$, we have\n\n$$\n\\sum_{i=2}^{n-1} a_i^{n-i+1} > \\sum_{i=2}^{n-1} a_{i+1}^{n-i+1} = \\sum_{j=3}^{n} a_j^{n-j+2},\n$$\n\nwhere the equality comes from shifting indices to take $j = i + 1$. Thus\n\n$$\na_n = a_1^n + \\sum_{i=2}^{n-1} a_i^{n-i+1} > a_1^{n+1} + a_2^n + \\sum_{j=3}^{n} a_j^{n-j+2} = a_{n+1},\n$$\n\nas desired. This completes the induction step, finishing the solution.\n\n**Remark (author).** The behavior of the sequence for other values of $c$ is interesting. In particular, computer experiments seem to indicate that, for $c = 0.655736876792$, the sequence starting from $a_2$ increases to a value barely less than $\\frac{1}{2}$ and then decreases, going to 0 as $n \\to \\infty$, while for $c = 0.655736876793$, the sequence starting from $a_2$ is increasing and goes to infinity. We conjecture that some form of these patterns hold in general, and that there should exist some value of $c$ with $0.655736876792 < c < 0.655736876793$ so that the sequence starting from $a_2$ is increasing and $\\lim_{n \\to \\infty} a_n = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19319,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a regular $2n$-gon $P$ in the plane. Each side of $P$ is to be colored with one of three different colors (ignore the vertices of $P$, which are considered colorless), such that:\n\n- Every side is colored in exactly one color.\n- Each color is used at least once.\n- From every point in the plane external to $P$, points of at most 2 different colors on $P$ can be seen.\n\nFind the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one of the sides is colored differently).\n\n![](images/Macedonia_2017_p9_data_aa4f98f60a.png)\n",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, the answer is $36$; for $n=3$, the answer is $30$; and for $n \\ge 4$, the answer is $6n$.\n\n**Lemma 1.** Given a regular $2n$-gon in the plane and a sequence of $n$ consecutive sides $s_1, s_2, \\dots, s_n$, there is an external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$ for $i=1, 2, \\dots, n$.\n\n**Proof.** For a semicircle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semicircle can be seen from $R$. Considering the circumscribed circle around the $2n$-gon, there is a semicircle $S$ such that each $s_i$ either has both endpoints on it, or has an endpoint that's on the semicircle and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen; clearly, the color of each $s_i$ can be seen from $Q$.\n\n**Lemma 2.** Given a regular $2n$-gon in the plane and a sequence of $n+1$ consecutive sides $s_1, s_2, \\dots, s_{n+1}$, there is no external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$ for $i=1, 2, \\dots, n+1$.\n\n**Proof.** Since $s_1$ and $s_{n+1}$ are parallel opposite sides of the $2n$-gon, they cannot be seen at the same time from an external point.\n\nFor $n=2$ (square), each color must be used. Two sides will be of the same color; choose which 2 sides, then assign colors: $\\binom{4}{2} \\cdot 3 \\cdot 2 = 36$.\n\nFor $n=3$ (hexagon), denote the sides as $a_1, a_2, \\dots, a_6$. There must be 2 consecutive sides of different colors, say $a_1$ is red, $a_2$ is blue. We must have a green side, and only $a_4$ and $a_5$ can be green. There are 3 possibilities:\n\n1. $a_4$ is green, $a_5$ is not. Then $a_3$, $a_5$, $a_6$ must be blue, yielding a valid coloring.\n2. Both $a_4$ and $a_5$ are green; $a_6$ must be red, $a_5$ must be blue, yielding rbbggr.\n3. $a_5$ is green, $a_4$ is not. Then $a_6$ must be red, $a_4$ must be red, $a_3$ must be red, yielding rbbgr.\n\nThus, there are two types of configurations:\n\ni) 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in $3 \\cdot (3 \\cdot 2 \\cdot 1) = 18$ ways (choose the pair of opposite sides, then assign colors).\n\nii) 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $3 \\cdot 6 = 12$ ways (partition into pairs of consecutive sides, for each partitioning, 6 ways to assign colors).\n\nThus, for $n=3$, the answer is $18+12=30$.\n\n![](images/Macedonia_2017_p9_data_c5ca3559e2.png)\n\nFor $n \\ge 4$, any 4 consecutive sides can be seen from an external point (by Lemma 1). Denote the sides as $a_1, a_2, \\dots, a_{2n}$. There must be 2 adjacent sides of different colors, say $a_1$ is blue and $a_2$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. Only valid configuration: 2 opposite sides colored in 2 different colors, all other sides colored in the third color. This can be done in $6n$ ways.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19320,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$f(f(n) - 2n) = 2f(n) + n$$\nfor all integers $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "There are many functions $f$ that satisfy the given condition. One example is:\n\n$$\nf(n) = \\begin{cases} n & \\text{if } n \\ge 0 \\\\ -3n & \\text{if } n < 0 \\end{cases}\n$$\n\nVerification:\n\n- For $n > 0$: $f(f(n) - 2n) = f(-n) = 3n$, and $2f(n) + n = 2n + n = 3n$.\n- For $n < 0$: $f(f(n) - 2n) = f(-5n) = -5n$, and $2f(n) + n = -5n$.\n- For $n = 0$: $f(0) = 0$, and $f(f(0) - 0) = f(0) = 0$, $2f(0) + 0 = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19321,
    "subject": "Mathematics (Olympiad)",
    "question": "In a regular $n$-gon, either $0$ or $1$ is written at each vertex. Using non-intersecting diagonals, Juku divides this polygon into triangles. Then he writes into each triangle the sum of the numbers at its vertices. Prove that Juku can choose the diagonals in such a way that the maximal and minimal number written into the triangles differ by at most $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "If all numbers written at the vertices of the polygon are equal, then the claim holds trivially. Hence, assume that there are both zeros and ones among the numbers at the vertices. We prove by induction that, for every convex polygon, the partition into triangles can be chosen in such a way that Juku writes either $1$ or $2$ to each triangle.\n\nIf $n = 3$, then this claim holds since the sum of the numbers at the vertices of a triangle can be neither $0$ nor $3$. If $n = 4$ (see below), then draw the diagonal that connects the vertices where $0$ and $1$ are written, respectively, or, if such a diagonal does not exist, then an arbitrary diagonal. In both cases, only sums $1$ and $2$ can arise. If $n \\ge 5$, then choose two consecutive vertices with different labels and a third vertex $P$ that is not a neighbor to either of them (see below). Irrespective of whether the label of $P$ is $0$ or $1$, we can draw the diagonal from it to one of the two consecutive vertices chosen before so that the labels of its endpoints are different. Now the polygon is divided into two convex polygons with a smaller number of vertices so that both $0$ and $1$ occur among their vertex labels. By the induction hypothesis, both polygons can be partitioned into triangles with sum of labels of vertices either $1$ or $2$.\n\n![](images/Estonija_2010_p6_data_465ed7bf7f.png)\n\n*Fig. 2*\n\n![](images/Estonija_2010_p6_data_9abfe3da86.png)\n\n![](images/Estonija_2010_p6_data_5d4bb45e21.png)\n\n*Fig. 3*",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19322,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if $2a_m = a_n$, then $a_{2m-n}$ is a perfect square, where $a_n = 1 + 2 + \\dots + n$, for every $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $a_n = \\frac{n(n+1)}{2}$ for $n \\in \\mathbb{N}$. Since $2a_m = a_n$, it follows that\n\n$$\n2 \\cdot \\frac{m(m+1)}{2} = \\frac{n(n+1)}{2} \\implies 2m(m+1) = n(n+1).\n$$\n\nNow,\n\n$$\n\\begin{align*}\na_{2m-n} &= \\frac{1}{2}(2m-n)(2m-n+1) \\\\\n&= \\frac{1}{2}(4m^2 - 4mn + n^2 + 2m - n) \\\\\n&= \\frac{1}{2}\\left(2m^2 + 2m + 2m^2 - 4mn + 2n^2 - n^2 - n\\right) \\\\\n&= \\frac{1}{2}(2m^2 - 4mn + 2n^2) \\\\\n&= (m-n)^2\n\\end{align*}\n$$\n\nTherefore, $a_{2m-n}$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19323,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime. We say that a sequence of integers $\\{z_n\\}_{n=0}^{\\infty}$ is a $p$-pod if for each $e \\ge 0$, there is an $N \\ge 0$ such that whenever $m \\ge N$, $p^e$ divides the sum\n\n$$\n\\sum_{k=0}^{m} (-1)^k \\binom{m}{k} z_k.\n$$\n\nProve that if both sequences $\\{x_n\\}_{n=0}^{\\infty}$ and $\\{y_n\\}_{n=0}^{\\infty}$ are $p$-pods, then the sequence $\\{x_ny_n\\}_{n=0}^{\\infty}$ is a $p$-pod.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nX_n = \\sum_{i=0}^{n} (-1)^i \\binom{n}{i} x_i \\quad \\text{and} \\quad Y_n = \\sum_{i=0}^{n} (-1)^i \\binom{n}{i} y_i.\n$$\n\nFor nonnegative integers $i \\leq j$, consider the expression\n\n$$\n\\sum_{k=i}^{j} (-1)^k \\binom{k}{i} \\binom{j}{k}.\n$$\n\nBy the principle of inclusion-exclusion, $(-1)^j$ times this sum counts the number of ways to choose an $i$-element subset from a $j$-element set that coincides with the set itself. Thus, the sum is $(-1)^i$ if $i = j$ and $0$ otherwise. Therefore,\n\n$$\n\\begin{aligned}\n\\sum_{r=0}^{n} (-1)^r \\binom{n}{r} x_r y_r &= \\sum_{i=0}^{n} \\sum_{j=0}^{n} (-1)^{i+j} x_i y_j \\sum_{r=0}^{n} (-1)^r \\binom{n}{r} \\left[ \\sum_{k=i}^{r} (-1)^k \\binom{k}{i} \\binom{r}{k} \\right] \\left[ \\sum_{\\ell=j}^{r} (-1)^\\ell \\binom{\\ell}{j} \\binom{r}{\\ell} \\right] \\\\\n&= \\sum_{k=0}^{n} \\sum_{\\ell=0}^{n} \\left[ \\sum_{i=0}^{k} (-1)^{k-i} \\binom{k}{i} x_i \\right] \\left[ \\sum_{j=0}^{\\ell} (-1)^{\\ell-j} \\binom{\\ell}{j} y_j \\right] \\left[ \\sum_{r=0}^{n} (-1)^r \\binom{n}{r} \\binom{r}{k} \\binom{r}{\\ell} \\right] \\\\\n&= \\sum_{k=0}^{n} \\sum_{\\ell=0}^{n} (-1)^{n+k+\\ell} X_k Y_\\ell \\left[ \\sum_{r=0}^{n} (-1)^{n-r} \\binom{n}{r} \\binom{r}{k} \\binom{r}{\\ell} \\right].\n\\end{aligned}\n$$\n\nAgain by inclusion-exclusion, the final bracketed expression counts the number of ways to choose a $k$-element subset and an $\\ell$-element subset of an $n$-element set whose union is the entire set, so it is $0$ if $k + \\ell < n$.\n\nNow, let $e$ be arbitrary, and let $N$ be so large that $p^e$ divides both $X_m$ and $Y_m$ whenever $m \\geq N/2$ (such an $N$ exists by the definition of $p$-pod), and take $n \\geq N$. Then, whenever $k + \\ell \\geq n \\geq N$, at least one of $k$ or $\\ell$ is at least $N/2$, so $p^e$ divides either $X_k$ or $Y_\\ell$. Thus, $p^e$ divides every term in the sum, and therefore for each $n \\geq N$, $p^e$ divides\n\n$$\n\\sum_{r=0}^{n} (-1)^r \\binom{n}{r} x_r y_r,\n$$\n\nhence $\\{x_n y_n\\}_{n=0}^{\\infty}$ is a $p$-pod.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19324,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be the roots of the equation $x^4 - x^3 - x^2 - 1 = 0$. Compute\n$$\nf(a) + f(b) + f(c) + f(d),\n$$\nwhere $f(x) = x^6 - x^5 - x^3 - x^2 - x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$ is a root of $g(x) = 0$, we have $a^4 - a^3 - a^2 - 1 = 0$. Using this relation:\n\n$$\nf(a) = a^6 - a^5 - a^3 - a^2 - a = (a^2+1)(a^4 - a^3 - a^2 - 1) + a^2 - a + 1 = a^2 - a + 1.\n$$\n\nThe same applies to $b, c, d$. Therefore,\n\n$$\nf(a) + f(b) + f(c) + f(d) = (a^2 + b^2 + c^2 + d^2) - (a + b + c + d) + 4.\n$$\n\nSince $a, b, c, d$ are roots of $x^4 - x^3 - x^2 - 1 = 0$, we have $a+b+c+d=1$ and $ab+ac+ad+bc+bd+cd = -1$. Thus,\n\n$$\na^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd) = 3,\n$$\n\nso $f(a) + f(b) + f(c) + f(d) = 3 - 1 + 4 = 6$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19325,
    "subject": "Mathematics (Olympiad)",
    "question": "A point $D$ lies on the side $AC$ of a triangle $ABC$. Triangle $ADB$ is isosceles with $DA = DB$. Triangle $DBC$ is also isosceles with $BC = BD$. All angles in triangles $ADB$ and $DBC$ are an integer number of degrees. What is the difference between the largest and smallest values that angle $ADB$ could have?\n\n![](images/2022_Australian_Scene_p58_data_31e7f80b3e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle ADB = x^\\circ$.\n\nThus $x < 180$. Since $\\angle BAD = 90 - \\frac{x}{2}$, $x$ is even. So $x \\leq 178$.\n\nSince $\\angle CBD = 180 - 2(180 - x) = 2x - 180$, $2x > 180$. So $x \\geq 92$.\n\nAll angles are integers if $x = 92$ and all angles are integers if $x = 178$.\n\nHence the difference between the largest and smallest values of $\\angle ADB$ is $178 - 92 = 86$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19326,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{2003}$ be a sequence of real numbers. A term $a_k$, $1 \\leq k \\leq 2003$, is said to be a *leading term* if at least one of the expressions $a_k$, $a_k + a_{k+1}$, $\\dots$, $a_k + a_{k+1} + \\dots + a_{2003}$ is positive. Prove that the sum of all leading terms is positive provided that the sequence has at least one leading term.",
    "options": [],
    "answer": "See solution",
    "solution": "We solve this problem for any sequence having $n$ terms by applying induction with respect to $n$.\n\nThe case $n = 1$ is clear. Suppose that the statement is true for all sequences of length less than $n$. Now consider a sequence $a_1, a_2, \\dots, a_n$.\n\n*Case 1: $a_1$ is not a leading term.*\n\nThen the set of all leading terms of the sequence $a_1, a_2, \\dots, a_n$ coincides with the set of all leading terms of the sequence $a_2, a_3, \\dots, a_n$, and by the inductive hypothesis we are done.\n\n*Case 2: $a_1$ is a leading term.*\n\nConsider the smallest nonnegative integer $m$ such that $a_1 + a_2 + \\dots + a_m$ is positive. Then the terms $a_2, a_3, \\dots, a_m$ are also leading terms and their sum is positive. The sum of all remaining leading terms is also nonnegative by the induction hypothesis.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19327,
    "subject": "Mathematics (Olympiad)",
    "question": "1, 2, 3, ..., 10 тоонуудыг, нэг ангийн тоонуудын нийлбэр нь нөгөө ангийн тоонуудын үржвэртэй тэнцүү байхаар, үл огтлолцох 2 ангид хуваах бүх хуваалтыг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$$1 + 2 + \\cdots + 10 = 55$$\n\n1. $2 \\cdot 3 \\cdot 4 \\cdot 5 = 120$ \\Rightarrow $\\{1, 2, \\ldots, 10\\} = B \\cup C$ ба $B$-ийн элементүүдийн нийлбэр $C$-ийн элементүүдийн үржвэртэй тэнцүү ($B \\cap C = \\varnothing$) гэж үзье. $C$ олонлог 4-өөс олон элементтэй байж болохгүй. Өөрөөр хэлбэл $|C| \\leq 4$.\n\n(1) $|C| = 1$ бол $C = \\{x\\},\\ x \\leq 9$ ба $B$-ийн элементүүдийн нийлбэр $55 - 9 = 46$-аас багагүй тул боломжгүй.\n\n(2) $|C| = 2$, $C = \\{x, y\\},\\ x < y$ гэж үзье. Тэгвэл\n\n$$\nxy = 55 - x - y \\Leftrightarrow (x+1)(y+1) = 56,\\ x+1 < y+1 \\leq 11\n$$\n\n$\\Rightarrow x+1 = 7,\\ y+1 = 8$ буюу $C = \\{6, 7\\}$, $B = \\{1, 2, 3, 4, 5, 8, 9, 10\\}$.\n\n(3) $|C| = 3$, $C = \\{x, y, z\\},\\ x < y < z$ бол $xyz = 55 - x - y - z$. $x = 1$ бол (2)-той адилаар $y = 4,\\ z = 10$ болж $C = \\{1, 4, 10\\}$, $B = \\{2, 3, 5, 6, 7, 8, 9\\}$. $x = 2$ бол $(2y+1)(2z+1) = 107$ болж шийдгүй. $x \\geq 3$ бол $xyz \\geq 3 \\cdot 4 \\cdot 5 = 60 > 55 - x - y - z$ тул мөн шийдгүй.\n\n(4) $|C| = 4$, $C = \\{x, y, z, t\\},\\ x < y < z < t$. Хэрэв $x \\geq 2$ бол $xyzt \\geq 2 \\cdot 3 \\cdot 4 \\cdot 5 = 120 > 55$ тул $x = 1$ гэж үзнэ. Энэ үед $yzt = 54 - y - z - t$, $2 \\leq y < z < t$. Хэрэв $y \\geq 3$ бол мөн боломжгүй. $y = 2$ үед $(2z+1)(2t+1) = 105$ болж, (2)-той адилаар $2z+1 = 7$, $2t+1 = 15$ буюу $C = \\{1, 2, 3, 7\\}$, $B = \\{4, 5, 6, 8, 9, 10\\}$.\n\nИймд бодлогын нөхцөлийг хангах 3 хуваалт байна:\n\n- $C = \\{6, 7\\}$, $B = \\{1, 2, 3, 4, 5, 8, 9, 10\\}$\n- $C = \\{1, 4, 10\\}$, $B = \\{2, 3, 5, 6, 7, 8, 9\\}$\n- $C = \\{1, 2, 3, 7\\}$, $B = \\{4, 5, 6, 8, 9, 10\\}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19328,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a set $A$ of 2016 pairwise different positive integers such that for every non-empty subset $B \\subset A$ with $B \\neq A$, and every non-empty subset $C \\subset (A \\setminus B)$, the sum of the elements of $B$ is not divisible by the sum of the elements of $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19329,
    "subject": "Mathematics (Olympiad)",
    "question": "The Fibonacci numbers are defined by $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \\ge 3$. What is\n\n$$\n\\frac{F_2}{F_1} + \\frac{F_4}{F_2} + \\frac{F_6}{F_3} + \\dots + \\frac{F_{20}}{F_{10}}?\n$$\n\n(A) 318   (B) 319   (C) 320   (D) 321   (E) 322",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (B):** The Fibonacci sequence starts out\n\n1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, ...\n\nso the given sum is\n\n$$\n\\frac{1}{1} + \\frac{3}{1} + \\frac{8}{2} + \\frac{21}{3} + \\frac{55}{5} + \\frac{144}{8} + \\frac{377}{13} + \\frac{987}{21} + \\frac{2584}{34} + \\frac{6765}{55},\n$$\n\nwhich equals $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319$.\n\nAlternatively,\n\nthe Fibonacci sequence starts out 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, so the given sum starts out\n\n$$\n\\frac{1}{1} + \\frac{3}{1} + \\frac{8}{2} + \\frac{21}{3} + \\frac{55}{5} = 1 + 3 + 4 + 7 + 11.\n$$\n\nIt appears that these summands satisfy the same recurrence relation, namely\n\n$$\n\\frac{F_{2n}}{F_n} = \\frac{F_{2(n-1)}}{F_{n-1}} + \\frac{F_{2(n-2)}}{F_{n-2}}.\n$$\n\nWith the initial conditions 1, 3 instead of 1, 1, the sequence\n\n$$\n(L_n) = \\left( \\frac{F_{2n}}{F_n} \\right)\n$$\n\nis known as the Lucas sequence. If the recurrence above is correct, then the required sum is\n\n$$\n1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319.\n$$\n\nTo prove the identity for $(L_n)$ displayed above, recall Binet's formula, $F_n = \\frac{1}{\\sqrt{5}}(\\phi^n - \\psi^n)$, where $\\phi = \\frac{1+\\sqrt{5}}{2}$ and $\\psi = \\frac{1-\\sqrt{5}}{2}$ are the roots of the polynomial $x^2 - x - 1$. Then\n\n$$\nL_n = \\frac{F_{2n}}{F_n} = \\frac{\\phi^{2n} - \\psi^{2n}}{\\phi^n - \\psi^n} = \\phi^n + \\psi^n.\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\nL_{n-1} + L_{n-2} &= \\phi^{n-1} + \\phi^{n-2} + \\psi^{n-1} + \\psi^{n-2} \\\\\n&= \\phi^{n-2}(\\phi + 1) + \\psi^{n-2}(\\psi + 1) \\\\\n&= \\phi^{n-2} \\cdot \\phi^2 + \\psi^{n-2} \\cdot \\psi^2 \\\\\n&= \\phi^n + \\psi^n = L_n.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19330,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice 和 Bob 組隊玩一場遊戲。在遊戲開始時，他們兩人會被丟到一台列車上的兩個位置，列車總長為 $1$ 公里。列車是封閉且全黑的，所以除非他們在車頭或車尾，否則他們不會知道自己的位置，也無法知道隊友的位置。\n\n遊戲配給兩人各一台儀器，儀器的左半邊會顯示：\n\n- 持有人目前面向什麼方向（所以他/她可以選擇往車頭或車尾移動）；\n- 持有人目前已經移動的總距離；\n- 持有人目前是否碰到車頭；\n- 持有人目前是否碰到車尾。\n\n儀器的右半邊則顯示隊友的儀器左半邊所呈現的資訊。遊戲會在兩人遇到彼此（雙方位置重合）的瞬間結束。\n\n假設 Alice 和 Bob 在遊戲開始前先被充分告知列車的狀態以及儀器的所有功能，且可以先共同討論他們的策略。試求最小的實數 $x$，讓 Alice 和 Bob 存在一套策略，使得不論遊戲開始時兩人被放到什麼位置，他們都能夠保證在遊戲結束時，兩人的移動距離和不超過 $x$ 公里。\n\n註：兩人所擬定的策略，只能依賴他們在儀器中看到的資訊。遊戲中沒有其他獲得資訊的方式。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$x = 1.5$。\n\n**構造：** 首先 Alice 往車頭走 $0.5$ 公里。如果 Alice 碰到車頭，那麼 Bob 就開始往車頭走，於是在兩人共走 $1.5$ 公里前可以保證重合。假設 Alice 沒有碰到車頭，那就換 Bob 往車頭走，直到碰到車頭或是遇見 Alice 為止，碰到車頭就折返。那麼在兩人共走 $1.5$ 公里前可以保證重合。\n\n**下界：** 定義一個人走到的最右界跟走到的最左界的相隔距離是他的探索值。如果兩人的探索值的和不到 $1$ 公里，就可以把兩個人探索的區間擺在列車上互斥的位置。不失一般性，假設 Alice 是第一個讓探索值抵達 $0.5$ 公里的人，那就讓他在探索值達到 $0.5$ 公里的剎那撞到車的頭或尾，不失一般性讓他撞到車頭。然後可以把 Bob 走到的最右端擺在離車尾 $\\epsilon$ 的地方。假設 Bob 的探索值是 $x$，那麼此時兩人的距離是 $1 - x - \\epsilon$ 公里，於是兩人至少總共要走 $1.5 - \\epsilon$ 公里才能遇到對方。因為 $\\epsilon$ 可以為任意小的正實數，所以至少要走 $1.5$ 公里才能保證遇到對方。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19331,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n, p$ be real numbers with $p > -1$. Prove that\n\n$$\n\\sum_{i=1}^{n} (a_i - b_i) \\left( a_i (a_1^2 + a_2^2 + \\dots + a_n^2)^{p/2} - b_i (b_1^2 + b_2^2 + \\dots + b_n^2)^{p/2} \\right) \\ge 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = (a_1^2 + a_2^2 + \\dots + a_n^2)^{1/2}$ and $B = (b_1^2 + b_2^2 + \\dots + b_n^2)^{1/2}$. Then the left-hand side of the inequality becomes\n\n$$\nA^{p+2} + B^{p+2} - \\sum_{i=1}^{n} a_i b_i (A^p + B^p).\n$$\n\nSince $\\sum a_i b_i \\leq AB$, it suffices to prove that\n\n$$\nA^{p+2} + B^{p+2} \\geq B A^{p+1} + A B^{p+1}.\n$$\n\nThis follows readily from the rearrangement inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19332,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all quadruples $ (x_1, x_2, x_3, x_4) $ of real numbers which are solutions of the following system of six equations:\n\n$$\n\\begin{align*}\nx_1 + x_2 &= x_3^2 + x_4^2 + 6x_3x_4, \\\\\nx_1 + x_3 &= x_2^2 + x_4^2 + 6x_2x_4, \\\\\nx_1 + x_4 &= x_2^2 + x_3^2 + 6x_2x_3, \\\\\nx_2 + x_3 &= x_1^2 + x_4^2 + 6x_1x_4, \\\\\nx_2 + x_4 &= x_1^2 + x_3^2 + 6x_1x_3, \\\\\nx_3 + x_4 &= x_1^2 + x_2^2 + 6x_1x_2.\n\\end{align*}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the second equation from the first yields $x_2 - x_3 = x_3^2 - x_2^2 + 6x_4(x_3 - x_2)$, which we can factor as $0 = (x_3 - x_2)(x_3 + x_2 + 1 + 6x_4)$. We see that $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_4 = 0$. Similarly, we also have either $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_1 = 0$. Hence, if $x_2 \\neq x_3$, the second equality must hold in both cases; subtracting one from the other, we obtain $x_1 = x_4$. We conclude that either $x_2 = x_3$ or $x_1 = x_4$. Analogously, we get for each permutation $(i, j, k, l)$ of $(1, 2, 3, 4)$ that either $x_i = x_j$ or $x_k = x_l$.\n\nWe will prove that at least three of the $x_i$ must be equal. If all four are equal, this is true of course. Otherwise, there are two unequal ones, say $x_1 \\neq x_2$ without loss of generality. Then we have $x_3 = x_4$. If also $x_1 = x_3$ holds, then there are three equal elements. Otherwise, we have $x_1 \\neq x_3$, hence $x_2 = x_4$ and we also get three equal elements. Up to order, the quadruple $(x_1, x_2, x_3, x_4)$ is thus equal to a quadruple of the shape $(x, x, x, y)$, where we could have that $x = y$.\n\nSubstituting this in the equations gives $x+y = 8x^2$ and $2x = x^2 + y^2 + 6xy$. Adding these two equations: $3x + y = 9x^2 + y^2 + 6xy$. The right hand side can be factored as $(3x + y)^2$. Defining $s = 3x + y$, the equation becomes $s = s^2$, from which we get either $s = 0$ or $s = 1$. We have $s = 3x + y = 2x + (x + y) = 2x + 8x^2$. Hence, $8x^2 + 2x = 0$ or $8x^2 + 2x = 1$.\n\nIn the first case, we have $x = 0$ or $x = -\\frac{1}{4}$. We find $y = 0 - 3x = 0$ and $y = 0 - 3x = \\frac{3}{4}$, respectively. In the second case, we get the factorisation $(4x - 1)(2x + 1) = 0$, hence $x = \\frac{1}{4}$ or $x = -\\frac{1}{2}$. We find $y = 1 - 3x = \\frac{1}{4}$ or $y = 1 - 3x = \\frac{5}{2}$, respectively.\n\nAltogether, we found the following quadruples: $(0, 0, 0, 0)$, $\\left(-\\frac{1}{4}, -\\frac{1}{4}, -\\frac{1}{4}, \\frac{3}{4}\\right)$, $\\left(\\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}\\right)$ and $\\left(-\\frac{1}{2}, -\\frac{1}{2}, -\\frac{1}{2}, \\frac{5}{2}\\right)$, and permutations thereof. It is a simple computation to verify that all these quadruples are indeed solutions to the equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19333,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be positive real numbers and $p > 1$. Prove that\n\n$$\n(a + b + c)(a^{p-1} + b^{p-1} + c^{p-1}) > 2(a^p + b^p + c^p)\n$$\n\nand\n\n$$\n3(a^p + b^p + c^p) \n\\ge (a + b + c)(a^{p-1} + b^{p-1} + c^{p-1}).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that\n\n$$\n\\begin{align*}\n& (a + b + c)(a^{p-1} + b^{p-1} + c^{p-1}) - 2(a^p + b^p + c^p) \\\\\n&= (b + c)a^{p-1} + (c + a)b^{p-1} + (a + b)c^{p-1} - a^p - b^p - c^p \\\\\n&= (b + c - a)a^{p-1} + (c + a - b)b^{p-1} + (a + b - c)c^{p-1} > 0,\n\\end{align*}\n$$\n\nby the triangle inequality, since $a, b, c$ are positive. Hence the left inequality holds.\n\nClearly, the right inequality holds if $p = 1$. So, suppose $p > 1$ and apply Hölder's inequality to $a+b+c$, with exponents $p, q$, where $q = \\frac{p}{p-1}$, to get\n\n$$\na+b+c \\le (a^p + b^p + c^p)^{1/p} \\cdot 3^{1/q}.\n$$\n\nNext, apply Hölder's inequality to $a^{p-1} + b^{p-1} + c^{p-1}$, with exponents $q, p$, to get\n\n$$\na^{p-1} + b^{p-1} + c^{p-1} \\le (a^q(p-1) + b^q(p-1) + c^q(p-1))^{1/q} 3^{1/p} = (a^p + b^p + c^p)^{1/q} 3^{1/p}.\n$$\n\nCombining these inequalities, the right inequality follows.\n\nHere is an alternative proof of the second inequality. Observe that\n\n$$\n\\begin{aligned}\n& 3(a^p + b^p + c^p) - (a+b+c)(a^{p-1} + b^{p-1} + c^{p-1}) \\\\\n&= 2(a^p + b^p + c^p) - (b+c)a^{p-1} - (c+a)b^{p-1} - (a+b)c^{p-1} \\\\\n&= (a^p - a^{p-1}b - ab^{p-1} + b^p) + (b^p - b^{p-1}c - bc^{p-1} + c^p) \\\\\n&\\quad + (c^p - c^{p-1}a - ca^{p-1} + a^p) \\\\\n&= (a^{p-1} - b^{p-1})(a-b) + (b^{p-1} - c^{p-1})(b-c) + (c^{p-1} - a^{p-1})(c-a) \\\\\n&\\ge 0,\n\\end{aligned}\n$$\n\nsince $a, b, c$ are positive, and, for $r \\ge 0$, the function $t \\mapsto t^r$ is increasing on $(0, \\infty)$, so that $(x^{p-1} - y^{p-1})(x - y) \\ge 0$ for all positive $x, y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19334,
    "subject": "Mathematics (Olympiad)",
    "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. On his move, Petya chooses 9 consecutive numbers and decreases each of them by 2. On his move, Vasya chooses 10 consecutive numbers and increases each of them by 1. Prove that Vasya can play so that after each of his moves, among the numbers on the circle, there will be at least 5 positive numbers (regardless of Petya's moves).\n\n(С. Л. Берлов)",
    "options": [],
    "answer": "See 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": 19335,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $k_1$ and $k_2$ with radii $r_1$ and $r_2$ are externally tangent at $Q$. The other endpoints of the diameter through $Q$ are named $P$ on $k_1$ and $R$ on $k_2$. We choose two points $A$ and $B$, one on each of the arcs $PQ$ on $k_1$ (so that $PBQA$ is convex). Furthermore, $C$ is the second intersection point of the line $AQ$ with $k_2$, and $D$ is the second intersection point of $BQ$ with $k_2$. The lines $PB$ and $RC$ intersect at $U$, and $PA$ and $RD$ intersect at $V$. Show that there exists a point $Z$ that is common to all possible lines $UV$.",
    "options": [],
    "answer": "See solution",
    "solution": "A homothety with center $Q$ and ratio $-r_2/r_1$ maps $k_1$ onto $k_2$.\n\n![](images/AustriaMO2011_p5_data_7e53c8a0c8.png)\n\nThis homothety maps $A$ to $C$, $B$ to $D$, and $P$ to $R$. It therefore follows that $PB = PU$ and $RD = RV$ are parallel, as are $PA = PV$ and $RC = RU$. $PURV$ must therefore be a parallelogram (no two of these points can be equal), and the diagonals $PR$ and $UV$ have a common midpoint. It follows that the midpoint $Z$ of $PR$ is also the midpoint of all possible line segments $UV$, and this is therefore the required common point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19336,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x > 1$ be a non-integer number. Prove that\n\n$$\n\\left( \\frac{x+\\{x\\}}{[x]} - \\frac{[x]}{x+\\{x\\}} \\right) + \\left( \\frac{x+[x]}{\\{x\\}} - \\frac{\\{x\\}}{x+[x]} \\right) > \\frac{9}{2},\n$$\n\nwhere $[x]$ and $\\{x\\}$ represent the integer and the fractional part of $x$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $[x] = a$ and $\\{x\\} = r$, where $0 \\le r < 1$. Then the given inequality becomes\n\n$$\n\\left( \\frac{a+2r}{a} - \\frac{a}{a+2r} \\right) + \\left( \\frac{2a+r}{r} - \\frac{r}{2a+r} \\right) > \\frac{9}{2}\n$$\n\nwhich is equivalent to\n\n$$\n2 \\left( \\frac{r}{a} + \\frac{a}{r} \\right) - \\left( \\frac{a}{a+2r} + \\frac{r}{2a+r} \\right) > \\frac{5}{2}.\n$$\n\nSince $\\frac{r}{a} + \\frac{a}{r} \\ge 2$, it is enough to prove that\n\n$$\n\\frac{a}{a+2r} + \\frac{r}{2a+r} < \\frac{3}{2}\n$$\n\nwhich is equivalent to\n\n$$\n0 < 2a^2 + 11ar + 2r^2 \\Leftrightarrow 2(a+r)^2 + 7ar > 0,\n$$\n\nwhich is always valid.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19337,
    "subject": "Mathematics (Olympiad)",
    "question": "In the interior of a cyclic quadrilateral $ABCD$, a point $P$ is given such that\n\n$$\n|\\angle BPC| = |\\angle BAP| + |\\angle PDC|.\n$$\n\nDenote by $E$, $F$, and $G$ the feet of the perpendiculars from the point $P$ to the lines $AB$, $AD$, and $DC$, respectively. Show that the triangles $FEG$ and $PBC$ are similar.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be the circumcircle of the quadrilateral $ABCD$ and $k_1$, $k_2$ the circumcircles of the triangles $PAB$ and $PCD$, respectively. In the interior of the angle $BPC$, consider the half-line $PT$ such that $|\\angle BPT| = |\\angle BAP|$. Then the hypothesis on $P$ implies that\n\n$$\n|\\angle TPC| = |\\angle BPC| - |\\angle BPT| = |\\angle BPC| - |\\angle BAP| = |\\angle PDC|.\n$$\n\nThus $PT$ is the common interior tangent of the circles $k_1$ and $k_2$.\n\n![](images/CpsMT04_s_p1_data_88ed1f9885.png)\n\nAssume first that the sides $AB$ and $CD$ of the given cyclic quadrilateral are not parallel. Since the segments $AB$ and $CD$ are common chords of the circles $k_1$, $k$ and $k_2$, $k$, respectively, there exists a unique point $Q$ having the same power with respect to all three circles $k$, $k_1$, and $k_2$. The point $Q$ is the intersection of the three lines $AB$, $DC$, and $PT$. Without loss of generality, assume that the point $Q$ is located on the half-line $BA$ beyond the point $A$. Then\n\n$$\n|\\angle QPA| = |\\angle PBA| \\quad (1)\n$$\n\nSince $|\\angle AEP| = |\\angle AFP| = 90^\\circ$, the quadrilateral $AEPF$ is cyclic, and\n\n$$\n|\\angle FEP| = |\\angle FAP| = |\\angle DAP| \\quad (2)\n$$\n\nSimilarly, the quadrilateral $DGPF$ is cyclic. It follows that\n\n$$\n|\\angle BPC| = |\\angle BAP| + |\\angle PDC| = |\\angle EFP| + |\\angle PFG| = |\\angle EFG| \\quad (3)\n$$\n\nSince $|\\angle PEQ| = |\\angle PGQ| = 90^\\circ$, the quadrilateral $QEPG$ is also cyclic and\n\n$$\n|\\angle GEP| = |\\angle GQP| = |\\angle DQP| \\quad (4)\n$$\n\nFrom (2), (4), and the equality $|\\angle DAP| + |\\angle QPA| = |\\angle QDA| + |\\angle DQP|$, we have\n\n$$\n|\\angle FEG| = |\\angle FEP| - |\\angle GEP| = |\\angle DAP| - |\\angle DQP| = |\\angle QDA| - |\\angle QPA| \\quad (5)\n$$\n\nSince the quadrilateral $ABCD$ is cyclic, $|\\angle QDA| = |\\angle QBC|$. From (1) and (5),\n\n$$\n|\\angle FEG| = |\\angle QBC| - |\\angle PBA| = |\\angle PBC|.\n$$\n\nUsing (3) and the above, the triangles $FEG$ and $PBC$ are similar (as they have two congruent angles).\n\nAn analogous argument applies when the point $Q$ is located on the half-line $AB$ beyond the point $B$. If the lines $AB$ and $CD$ are parallel, then $ABCD$ is an equilateral trapezoid, with bases $AB$ and $CD$. Since the points $E$, $P$, $G$ are collinear and the common interior tangent of the circles $k_1$ and $k_2$ is parallel to both lines $AB$ and $CD$, the triangles $APD$ and $BPC$ are congruent. The similarity of the triangles $EFG$ and $APD$ thus implies also the similarity of the triangles $EFG$ and $BPC$.\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19338,
    "subject": "Mathematics (Olympiad)",
    "question": "For how many positive integers $n$ less than $50$ is the product $(n-8)(n-38)$ positive?",
    "options": [],
    "answer": "See solution",
    "solution": "For $(n-8)(n-38)$ to be positive, either both factors must be positive or both must be negative.\n\n- Both negative: $n < 8$ (i.e., $n = 1, 2, \\ldots, 7$), which gives $7$ values.\n- Both positive: $n > 38$ and $n < 50$ (i.e., $n = 39, 40, \\ldots, 49$), which gives $11$ values.\n\nIn total, there are $7 + 11 = 18$ possible values of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19339,
    "subject": "Mathematics (Olympiad)",
    "question": "En una sala de baile hay 15 chicos y 15 chicas dispuestos en dos filas paralelas de manera que se formarán 15 parejas de baile. Sucede que la diferencia de altura entre el chico y la chica de cada pareja no supera los 10 cm. Demuestra que si colocamos los mismos chicos y chicas en dos filas paralelas en orden creciente de alturas, también sucederá que la diferencia de alturas entre los miembros de las nuevas parejas así formadas no superará los 10 cm.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $P_1, P_2, \\dots, P_{15}$ las quince parejas iniciales. Ordenemos ahora los chicos por alturas $a_1 \\leq a_2 \\leq \\dots \\leq a_{15}$ y también las chicas $b_1 \\leq b_2 \\leq \\dots \\leq b_{15}$. \n\nSupongamos que en la nueva formación alguna pareja tuviera una diferencia de alturas superior a 10 cm, digamos $a_k - b_k > 10$. Entonces, para las chicas de alturas $b_1, \\ldots, b_k$ y los chicos de alturas $a_k, \\ldots, a_{15}$, se cumple $a_i - b_j > 10$. \n\nColoquemos ahora cada una de las $k + (15 - k + 1) = 16$ personas mencionadas, de alturas $b_1, \\ldots, b_k, a_k, a_{k+1}, \\ldots, a_{15}$, en las parejas $P_s$ iniciales, según el lugar que ocupaban. Por el principio de las casillas (palomar), dos personas compartirán la misma pareja inicial. Por lo tanto, en las parejas iniciales había una cuya diferencia de alturas era mayor que 10 cm, contradiciendo la hipótesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19340,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all matrices $A, B, C \\in \\mathcal{M}_2(\\mathbb{R})$ such that\n$$\nA = BC - CB, \\quad B = CA - AC, \\quad C = AB - BA.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If one of the matrices $A, B, C$ is the zero matrix, then all are zero. Suppose $A, B, C \\in \\mathcal{M}_2(\\mathbb{R}) \\setminus \\{O_2\\}$ satisfy the equations. We have $\\text{tr}(A) = \\text{tr}(BC - CB) = 0$, and similarly $\\text{tr}(B) = \\text{tr}(C) = 0$.\n\nFrom the Cayley-Hamilton theorem, $A^2 - \\text{tr}(A)A + \\det(A)I_2 = O_2$, so $A^2 = aI_2$ with $a = -\\det(A)$. Similarly, $B^2 = bI_2$ and $C^2 = cI_2$.\n\nMultiplying $A = BC - CB$ by $B$ on both sides gives $BA = bC - BCB$ and $AB = BCB - bC$, so $AB + BA = O_2$. Thus, $C = AB - BA = 2AB$. Similarly, $A = 2BC$ and $B = 2CA$.\n\nSubstituting, $A = -2CB = -2C(2CA) = -4C^2A = -4cA$. Similarly, $B = -4aB$ and $C = -4bC$. Since $A, B, C$ are nonzero, $a = b = c = -1/4$.\n\nLet $A = \\begin{pmatrix} x & y \\\\ z & -x \\end{pmatrix}$ and $B = \\begin{pmatrix} s & t \\\\ u & -s \\end{pmatrix}$, with $x^2 + yz = -1/4$ and $s^2 + tu = -1/4$. From $AB + BA = O_2$, $2xs = -(yu + zt)$. Then $4x^2s^2 = (yu + zt)^2 = (yu - zt)^2 + 4yztu \\ge 4(yz)(tu) = 4\\left(x^2 + \\frac{1}{4}\\right)\\left(s^2 + \\frac{1}{4}\\right)$. But $x^2s^2 < \\left(x^2 + \\frac{1}{4}\\right)\\left(s^2 + \\frac{1}{4}\\right)$ for all $x, s \\in \\mathbb{R}$, a contradiction.\n\nTherefore, the unique solution is $A = B = C = O_2$.\n\n*Alternative solution.* Let $A, B, C \\in \\mathcal{M}_2(\\mathbb{R})$ satisfy the system. $\\text{tr}(A) = \\text{tr}(BC - CB) = 0$, so $\\text{tr}(B) = \\text{tr}(C) = 0$. Write $A = \\begin{pmatrix} a_1 & a_2 \\\\ a_3 & -a_1 \\end{pmatrix}$, $B = \\begin{pmatrix} b_1 & b_2 \\\\ b_3 & -b_1 \\end{pmatrix}$, $C = \\begin{pmatrix} c_1 & c_2 \\\\ c_3 & -c_1 \\end{pmatrix}$.\n\nCompute $BC - CB = \\begin{pmatrix} b_2c_3 - b_3c_2 & 2(b_1c_2 - b_2c_1) \\\\ 2(b_3c_1 - b_1c_3) & -(b_2c_3 - b_3c_2) \\end{pmatrix}$. Thus, $a_2 = 2(b_1c_2 - b_2c_1)$ and $a_2^2 = 2a_2b_1c_2 - 2a_2b_2c_1$. Similarly, $b_2^2 = 2b_2c_1a_2 - 2b_2c_2a_1$ and $c_2^2 = 2c_2a_1b_2 - 2c_2a_2b_1$. Then $a_2^2 + b_2^2 + c_2^2 = 0$, so $a_2 = b_2 = c_2 = 0$. Analogously, $a_3 = b_3 = c_3 = 0$. Also, $a_1 = b_2c_3 - b_3c_2 = 0$, and similarly $b_1 = c_1 = 0$.\n\nIn conclusion, the unique solution is $A = B = C = O_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19341,
    "subject": "Mathematics (Olympiad)",
    "question": "A pair $(n, m)$ is called *harmonic* if it satisfies the conditions:\n\n$$\nn^m + (p - n)^{n + \\frac{p-1}{2}} \\equiv 0 \\pmod{p}\n$$\n\nand\n\n$$\n1 \\leq n, m \\leq \\frac{p-1}{2}.\n$$\n\nFind, for a given odd prime $p$, a permutation of the numbers $1, 2, \\ldots, \\frac{p-1}{2}$ such that each pair $(n, m)$ (where $m$ is the image of $n$ under the permutation) is harmonic.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the case when $\\frac{p-1}{2}$ is odd. We have:\n\n$$\n(n^n + (p-n)^{n+\\frac{p-1}{2}})(n^{n+\\frac{p-1}{2}} + (p-n)^n) \\equiv 0 \\pmod{p}\n$$\n\nFrom this, either $(n, n)$ or $(n, n + p/2)$ will be harmonic. In this case, we can easily choose a permutation with the required property.\n\nNow, consider the case when $\\frac{p-1}{2}$ is even. If $n^{\\frac{p-1}{2}} \\equiv 1 \\pmod{p}$ (call this (*)), then for any odd $m$, and if $n^{\\frac{p-1}{2}} \\equiv -1 \\pmod{p}$ (call this (***)), then for any even $m$, the pair $(n, m)$ will be harmonic. The number of $n$ ($1 \\leq n \\leq \\frac{p-1}{2}$) satisfying conditions (*) and (***) is $\\frac{p-1}{4}$. Thus, we can choose $\\frac{p-1}{2}$ pairs depending on properties (*) and (***).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19342,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to (0, +\\infty)$ be a continuous function such that $$\\lim_{x \\to -\\infty} f(x) = \\lim_{x \\to +\\infty} f(x) = 0.$$  \n\na) Prove that $f(x)$ attains its maximum value on $\\mathbb{R}$.\n\nb) Prove that there exist two sequences $(x_n)$ and $(y_n)$ with $x_n < y_n$ for all positive integers $n$, such that both sequences have the same limit as $n \\to \\infty$ and $f(x_n) = f(y_n)$ for all $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) By the extreme value theorem, a continuous function on a closed interval $[a, b]$ attains its maximum and minimum.  \nSince $\\lim_{x \\to -\\infty} f(x) = \\lim_{x \\to +\\infty} f(x) = 0$, there exist $a$ and $b$ such that $f(x) < f(0)$ for all $x \\le a$ and $x \\ge b$.  \nOn $[a, b]$, $f(x)$ attains a maximum $M = f(c)$ for some $c \\in [a, b]$. For $x \\in (-\\infty, a) \\cup (b, +\\infty)$, $f(x) < f(0) \\le M$, so $f(x) \\le M$ for all $x \\in \\mathbb{R}$. Thus, $f(x)$ attains its maximum on $\\mathbb{R}$.\n\nb) Consider two cases:\n\n1) If there exists an interval $(a, b)$ containing $c$ such that $f(c) = M$ and $f(x) < M$ for all $x \\in (a, b) \\setminus \\{c\\}$:\n\nLet $A, B \\neq c$ in $(a, b)$ with $c \\in [A, B]$. On $[A, c]$ and $[c, B]$, $f(x)$ attains minimums $m_1, m_2$; set $m = \\max\\{m_1, m_2\\}$. By the intermediate value theorem, there exist $x_1 \\in [A, c]$ and $y_1 \\in [c, B]$ such that $f(x_1) = f(y_1) = m$ and $x_1 < c < y_1$.\n\nRepeat this process: for $[x_{n-1}, c]$ and $[c, y_{n-1}]$, there exist $x_n, y_n$ such that $f(x_n) = f(y_n) = u_n$ where $u_1 = m$, $u_{n+1} = \\frac{u_n + M}{2}$, and $x_n < c < y_n$. The sequences $(x_n)$ and $(y_n)$ converge to $c$, and $f(x_n) = f(y_n)$ for all $n$.\n\n2) If there is a segment $[a, b]$ where $f(x) = M$ for all $x \\in [a, b]$:\n\nLet $x_n = \\frac{a+b}{2} + \\frac{a-b}{2^n}$ and $y_n = \\frac{a+b}{2} + \\frac{b-a}{2^n}$. Then $x_n, y_n \\in [a, b]$, $f(x_n) = f(y_n) = M$, $x_n < \\frac{a+b}{2} < y_n$, and $\\lim x_n = \\lim y_n = \\frac{a+b}{2}$.\n\nThus, in all cases, such sequences exist. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19343,
    "subject": "Mathematics (Olympiad)",
    "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 $|a| + |b| \\le 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|a| + |b| \\le 1$ and let $z \\in \\mathbb{C}$ with $|z| = 1$. Then\n$$\n|az + b\\bar{z}| \\le |az| + |b\\bar{z}| = |a| + |b| \\le 1,\n$$\nas 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\n$$\n1 \\ge |az + b\\bar{z}| = |a||\\bar{z}|\\left|z^2 + \\frac{b}{a}\\right| = |a|(1+r)(\\cos\\alpha + i\\sin\\alpha) = |a|(1+r) = |a|(1 + \\left|\\frac{b}{a}\\right|) = |a| + |b|,\n$$\nwhich ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19344,
    "subject": "Mathematics (Olympiad)",
    "question": "求方程式\n\n$$x^3 = 2y^3 + 4z^3$$\n\n的整數解。",
    "options": [],
    "answer": "See solution",
    "solution": "只有 $(x, y, z) = (0, 0, 0)$，沒有其他整數解。\n\n設方程式尚有其他解 $(a, b, c)$，取其中 $|a| + |b| + |c|$ 是最小的。由\n\n$$a^3 = 2b^3 + 4c^3$$\n\n可知 $a$ 必是偶數。令 $a = 2t$，則\n\n$$8t^3 = 2b^3 + 4c^3$$\n\n即 $b^3 = 2(-c)^3 + 4t^3$。\n\n因此 $(b, -c, t)$ 也是方程式的另一組解。若 $t \\neq 0$，則有\n\n$$|b| + |-c| + |t| < |a| + |b| + |c| = |2t| + |b| + |c|$$\n\n故 $x = 0$，而\n\n$$y^3 = 2z^3$$\n\n同樣手法，得到這方程式的整數解為 $y = z = 0$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19345,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ and $\\omega$ its circumcircle. The tangent at $A$ to $\\omega$ intersects $BC$ at $D$, and the line through $B$ parallel to $AD$ meets $\\omega$ again at $E$. Line $DE$ intersects $AB$ at $F$ and $\\omega$ again at $G$. On $BE$, let $N$ be such that $B, G, F, N$ are concyclic. Let $S$ and $T$ be the intersection points of $FN$ with $AD$ and $AE$, respectively. Prove that $STDG$ is a cyclic quadrilateral.\n\n![](images/RMC_2024_p70_data_62eb96a823.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABCE$ and $BNFG$ are cyclic quadrilaterals, we have\n$$\n\\angle TAB = \\angle BCE = 180^{\\circ} - \\angle BGE = 180^{\\circ} - \\angle BGF = \\angle BNF = \\angle BNT,\n$$\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": 19346,
    "subject": "Mathematics (Olympiad)",
    "question": "$(1+x)^n$ олон гишүүнтийн тэгш коэффициенттэй гишүүдийг дарахад үлдэх олон гишүүнтийг $Q_n(x)$ гэе. $Q_{2012}(1)$-ийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$p \\in \\mathbb{P}$ бол $(a+b)p^n = ap^n + bp^n$ (мод $p$) чанар болон $2012 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2$ байхыг ашиглан\n\n$$\n(1+x)^{2012} = (1+x)^{2^{10}}(1+x)^{2^9}(1+x)^{2^8}(1+x)^{2^7}(1+x)^{2^6}(1+x)^{2^5}(1+x)^{2^4}(1+x)^{2^3}(1+x)^{2^2}\n$$\n\n$$\n\\equiv (1+x^{2^{10}})(1+x^{2^9}})(1+x^{2^8})(1+x^{2^7})(1+x^{2^6})(1+x^{2^5})(1+x^{2^4})(1+x^{2^3})(1+x^{2^2}) \\pmod{2}\n$$\n\nболно. Аливаа натурал тоо 2-тын тооллын системд нэг утгатай тавьж болох тул $P(x) = (1+x^{2^{10}})(1+x^{2^9}})(1+x^{2^8}) \\cdots (1+x^{2^2})$ үржвэрийг задлахад 1-ээс их коэффициенттэй гишүүн гарахгүй. $Q_{2012}(x)$-ийн $x$-ийн зэргүүд нь харгалзан $P(x)$-ийн $x$-ийн зэргүүдтэй тэнцүү. $Q_{2012}(x)$-ийн $x$-ийн зэргүүд нь $\\{2^2, 2^3, 2^4, 2^6, 2^7, 2^8, 2^9, 2^{10}\\}$ олонлогийн бүх дэд олонлог тус бүрийн элементүүдийн нийлбэртэй тэнцүү. $\\{2^2, 2^3, 2^4\\}$ олонлогийн дэд олонлог тус бүрийн элементүүдийн нийлбэр нь $4i$ ($i = 0, 1, 2, \\ldots, 7$), $\\{2^6, 2^7, 2^8, 2^9, 2^{10}\\}$ олонлогийн дэд олонлог тус бүрийн элементүүдийн нийлбэр\n\n$$\n64j \\quad (j = 0, 1, 2, \\ldots, 31)\n$$\n\nбайх тул бидний олох тоо\n\n$$\n\\sum_{j=0}^{31} \\sum_{i=0}^{7} \\binom{2012}{4i+64j}\n$$\nюм.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19347,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, find the smallest value of\n$$\n\\lfloor \\frac{a_1}{1} \\rfloor + \\lfloor \\frac{a_2}{2} \\rfloor + \\dots + \\lfloor \\frac{a_n}{n} \\rfloor\n$$\nover all permutations $(a_1, a_2, \\dots, a_n)$ of $(1, 2, \\dots, n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\lfloor \\log_2 n \\rfloor + 1$.\n\n**Bijection:** We will solve a new problem that is equivalent to the original problem. Draw an $n \\times n$ grid and write inside each cell (in column $x$ and row $y$) $\\lfloor \\frac{x}{y} \\rfloor$.\n\nPlace exactly $n$ rooks such that no two of them attack each other (in the same row or column). Find the least possible sum of all the integers written under the rooks.\n\n**Construction:** We will provide an example for $\\lfloor \\log_2 n \\rfloor + 1$.\n\nDraw $\\lfloor \\log_2 n \\rfloor + 1$ “red squares” such that each square $i$ contains all the cells from $(2^{i-1}, 2^{i-1})$ (top-left) to $(2^i - 1, 2^i - 1)$ (bottom-right), as shown in the diagram below. Note that the last “red square” might be cut off because of the size of the $n \\times n$ grid. Now color the top-right cell of each “red square” purple, as shown in the diagram below. And color all the cells inside a “red square” and that are in column $x$ and row $x + 1$ yellow. Finally, place the rooks on the purple and yellow cells.\n\nNotice that in each “red square” the total sum of the chosen cells is exactly 1 (each purple cell is 1 and each yellow cell is 0), making the total sum equal to $\\lfloor \\log_2 n \\rfloor + 1$.\n\n![](images/Saudi_Arabia_booklet_2022_p36_data_1a4ce92eda.png)\n\nWe will prove that the sum of integers chosen in the first $2^m$ rows is at least $m + 1$.\n$m = \\lfloor \\log_2 n \\rfloor$ (when proven, it obviously follows that our answer is correct).\n\n**Base case:** When $m = 0$ this is obvious.\n\n**Induction $m \\to m + 1$:** (as shown in figure 1)\n\n1. **Case 1:** There is at least 1 rook in the blue triangle (as shown in figure 2).\nThis is obvious because—from our inductive claim—the sum of the first $2^m$ rows is at least $m + 1$. Thus, the total sum of chosen squares is at least $m + 2$.\n\n2. **Case 2:** All rooks in the last $2^m$ rows are in the red trapezoid (as shown in figure 2). That means that there is a rook placed in one of the $2^m$ \"big circles\" in the last column (as shown in figure 2).\n\n    * Notice that for each “big circle” its value is bigger than the “small circle” (in the same row) by at least 1 (this is obvious).\n    * From our inductive claim, the sum of the chosen cells in the “green square” was at least $m + 1$.\n\nCombining (i) and (ii) it follows that the total sum of chosen squares is at least $m + 2$, which finishes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19348,
    "subject": "Mathematics (Olympiad)",
    "question": "The sides of a triangle are $a = 3$, $b = 4$, and $c = 5$. Find if there is a point inside the triangle such that the distance to each of its sides is less than $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\triangle ABC$ have sides $\\overline{BC} = a = 3$, $\\overline{AC} = b = 4$, and $\\overline{AB} = c = 5$. Since $a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25 = 5^2 = c^2$, $\\triangle ABC$ is a right triangle.\n\nSuppose there exists a point $M$ inside the triangle such that its distances to all three sides are less than $1$. Let $x$, $y$, and $z$ be the distances from $M$ to the sides of lengths $a$, $b$, and $c$, respectively, so $x, y, z < 1$.\n\n![](images/Makedonija_2008_p39_data_1c33944d1c.png)\n\nThe area of $\\triangle ABC$ can be written as the sum of the areas of the three triangles formed by $M$ and each side:\n\n$$\nP_{ABC} = P_{AMB} + P_{BMC} + P_{CMA} = \\frac{1}{2} a x + \\frac{1}{2} b y + \\frac{1}{2} c z < \\frac{1}{2} a + \\frac{1}{2} b + \\frac{1}{2} c = \\frac{3}{2} + \\frac{4}{2} + \\frac{5}{2} = 6.\n$$\n\nBut the area of $\\triangle ABC$ is exactly $6$, so this is a contradiction. Therefore, such a point $M$ does not exist.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19349,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $(2n+1) \\times (2n+1)$ square table with black corner cells, colored in a chessboard pattern. There are $2n^2 + 2n + 1$ black cells and $2n^2 + 2n$ white cells. Each black cell must be recolored an odd number of times, and each white cell an even number of times (possibly zero). Each move recolors three cells, so the total number of moves is odd.\n\nLet $A$, $B$, $C$, $D$ be the vertices of the square. A *diagonal* is a set of cells whose centers lie on a line parallel to $BD$. There are $4n+1$ diagonals starting from the half-perimeter $ABC$, each consisting entirely of black or white cells. Number the cells of $ABC$ as in Fig. 2. Consider diagonals starting from cells numbered divisible by $3$ (let $D_3$ be this set), and mark these diagonals with a \"*\" (see Figs. 3 and 4 for $n=2$ and $n=3$). Find the parity of the number of black cells marked \"*\". The parity equals the number of odd numbers divisible by $3$ between $1$ and $4n+1$.\n\n- If $n \\equiv 0 \\pmod{3}$, there are $2k$ such numbers.\n- If $n \\equiv 1 \\pmod{3}$, there are $2k+1$ such numbers.\n- If $n \\equiv 2 \\pmod{3}$, there are $2k+2$ such numbers.\n\nThus, the number of black cells marked \"*\" is even if $n \\not\\equiv 1 \\pmod{3}$, and odd if $n \\equiv 1 \\pmod{3}$. Suppose the final coloring can be obtained after some moves. Each move recolors exactly one cell marked \"*\". Can the required chess coloring be obtained for all $n$?\n\n![](images/Belaurus_2016_Booklet_p21_data_0af7acf4c9.png)\n\nFig. 1\n\n![](images/Belaurus_2016_Booklet_p21_data_0e6df445bd.png)\n\nFig. 2\n\n![](images/Belaurus_2016_Booklet_p21_data_5cd2a57366.png)\n\nFig. 3\n\nFig. 4",
    "options": [],
    "answer": "See solution",
    "solution": "There are an even number of moves recoloring white cells marked \"*\", since each such cell is recolored an even number of times. For $n \\not\\equiv 1 \\pmod{3}$, there are an even number of black cells marked \"*\", and each must be recolored an odd number of times, so the total number of moves is even. This contradicts the earlier result that the number of moves must be odd. Therefore, for $n \\not\\equiv 1 \\pmod{3}$, the required chess coloring cannot be obtained.\n\nIf $n \\equiv 1 \\pmod{3}$ (i.e., $2n+1$ divisible by $3$), the coloring can be achieved. Divide the $(2n+1) \\times (2n+1)$ table into $3 \\times 3$ squares, and in each, make moves as shown in Fig. 5 or Fig. 6, where moves are indicated by segments connecting the centers of the recolored cells.\n\n![](images/Belaurus_2016_Booklet_p22_data_37bfc39930.png)\n\nFig. 5\n\n![](images/Belaurus_2016_Booklet_p22_data_d12b622660.png)\n\nFig. 6",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19350,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $k \\geq 2$, exhibit an infinite set $\\mathcal{A}$ of sets of positive integers satisfying the two conditions below:\n\n(a) The intersection of the members of every $k$-element subset of $\\mathcal{A}$ is a singleton set.\n\n(b) The intersection of the members of every $(k+1)$-element subset of $\\mathcal{A}$ is empty.",
    "options": [],
    "answer": "See 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 $A_m$ 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": 19351,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that, for all integers $x$ and $y$, the following holds:\n\n$$\nf(f(x) - y) = f(y) - f(f(x)).\n$$\n\nShow that $f$ is bounded, i.e., that there is a $C$ such that\n\n$$\n-C < f(x) < C\n$$\n\nfor all $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, setting $y = f(x)$ gives $f(0) = 0$. Setting $y = 0$ yields $f(f(x)) = 0$ for all $x$, so\n\n$$\nf(f(x) - y) = f(y).\n$$\n\nSetting $x = 0$ gives $f(-y) = f(y)$, and setting $y = -z$ gives\n\n$$\nf(f(x) + z) = f(-z) = f(z).\n$$\n\nIf $f(x) = 0$ for all $x$, then $f$ is obviously bounded. Otherwise, if there exists $x_0$ such that $f(x_0) \\neq 0$, then with $x = x_0$, the last equality shows that $f$ is periodic with period $|f(x_0)|$, so $f$ must be bounded.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19352,
    "subject": "Mathematics (Olympiad)",
    "question": "All diagonals of a convex 10-gon are drawn. They divide the angles into 80 parts. It is known that 59 of these parts are equal. Determine the maximum number of different values among the 80 angles of division. How many times does each of these values occur?",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** The sides of each of the 80 angles pass through the endpoints of a side of the 10-gon $P$. We say that such an angle and such a side are adjacent; each side is adjacent to exactly 8 angles. Call *black* the 59 angles that are known to be equal and $\\alpha$ the measure of a black angle. Let $a$ be a side of $P$. The locus of points $X$ such that $a$ subtends angle $\\alpha$ at $X$ is the union of two circular arcs. Denote by $y_a$ the one of them which lies on the same side of $a$ as the polygon $P$. We also name $y_a$ the entire circle containing $y_a$. All black angles adjacent to $a$ have their vertices on $y_a$.\n\nFor a side $a$ let $m_a$ be the number of black angles adjacent to $a$. The hypothesis can be stated as $\\sum m_a \\ge 59$. We show that the inequality implies that $P$ is cyclic. Consider two cases.\n\nLet there be a side $a$ with $m_a \\ge 7$. Then $y_a$ contains the vertices of at least 7 black angles; these vertices are different from the endpoints of $a$. Hence at least $2+7=9$ vertices of $P$ lie on $y_a$.\n\nSuppose that there is a vertex $A \\in y_a$, and let $AB = b$, $AC = c$ be the sides with common vertex $A$.\n\nThen $y_b \\ne y_a$, $y_c \\ne y_a$ by $A \\in y_a$. So arcs $y_a$ and $y_b$ have (at most) two common points; one such point is $B$. Because all vertices except $A$ are on $y_a$, we see that $y_b$ contains at most one vertex different from $A$ and $B$, implying $m_b \\le 1$. On the other hand it is immediate that $y_a = y_a$ for every $s \\ne b, c$, hence $A \\in y_a$ for $s \\ne b, c$. Thus $m_s \\le 7$ for each of the 8 sides $s$ different from $b$ and $c$. In conclusion $\\sum m_a \\le 1+1+8 \\cdot 7 = 58$, contradicting the hypothesis.\n\nSuppose now that $m_a \\le 6$ for each side $a$. Then a direct computation using $\\sum m_a \\ge 59$ shows that\n\n$$m_a = 6$$\n\nholds for at least 9 sides $a$: the last side satisfies $m_a = 5$ or $m_a = 6$. We show that $y_b = y_c$ for every two consecutive sides $b = AB, c = AC$; this is enough to imply that $P$ is cyclic. Indeed $y_b$ contains at least $m_b - 1$ vertices different from $A, B$ and $C$. Likewise $y_c$ contains at least $m_c - 1$ vertices different from $A, B$ and $C$. Both arcs combined contain at least $m_b + m_c - 2 \\ge 6 + 5 - 2 = 9$ vertices $D \\ne A, B, C$. It follows that there are two vertices $D, E \\ne A, B, C$ that are common for $y_b$ and $y_c$. One more such vertex is $A$, so $y_b = y_c$, as stated.\n\nNow that $P$ is cyclic, each side $BC = a$ there corresponds an angle $\\alpha_a$ such that $\\vec{BC} = \\alpha_a$ for every vertex $V \\ne B, C$. Also $\\alpha_a$ occurs among the 80 angles of division exactly $8k$ times where $k$ is the number of sides with length $a$. Because at least 59 angles are known to be equal, $P$ has at least 8 equal sides. The angle $\\alpha_a$ corresponding to them occurs 64, 72 or 80 times. It is straightforward now that the 80 angles of division can assume at most 3 different values. If these are exactly 3 then one of them occurs 64 times, and each of the other two occurs 8 times.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19353,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any collection $a_1, a_2, \\ldots, a_{2011}$ of real numbers with $a_{2011} \\neq 0$, there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real $x$ we have:\n\n$$\na_1 f(x) + a_2 f(f(x)) + \\dots + a_{2011} \\underbrace{f(f(\\dots f(x)\\dots))}_{2011} = x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will search for the function $f$ in the form $f(x) = kx$ with $k \\neq 0$. Then\n\n$$\n\\underbrace{f(f(\\dots f(x)\\dots))}_{n} = k^n x.\n$$\n\n![](images/Ukrajina_2011_p40_data_f3555e25bd.png)\n\nSo the equality from the problem statement becomes:\n\n$$\na_1 kx + a_2 k^2 x + \\dots + a_{2011} k^{2011} x = x.\n$$\n\nWe cancel $x$ and obtain the following equation:\n\n$$\na_{2011}k^{2011} + a_{2010}k^{2010} + \\dots + a_2k^2 + a_1k - 1 = 0,\n$$\n\nwhich has a non-zero solution, because the left-hand side is a polynomial of odd degree with non-zero leading and free coefficients. So, for this $k$, the function $f(x) = kx$ will solve the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19354,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x$, $y$, $z$ such that\n$$\n2^x - 2^y - 2^z = 1023.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If none of $x$, $y$, $z$ is zero, then the left side of the equation is even, so it cannot be equal to $1023$ (which is odd). Hence, at least one of $x$, $y$, $z$ is zero.\n\nFrom $2^x = 2^y + 2^z + 1023$, it follows that $x \\geq 11$, therefore $y = 0$ or $z = 0$.\n\nConsider the case $z = 0$:\n$$\n2^x - 2^y = 1024\n$$\nSo,\n$$\n2^y (2^{x-y} - 1) = 2^{10}\n$$\nSince $2^x - 2^y > 0$, we have $x > y$, so $2^{x-y} - 1$ is odd. The only way to write $2^{10}$ as a product of a power of $2$ and an odd number is $2^{10} \\cdot 1$, so $2^y = 2^{10}$ and $2^{x-y} - 1 = 1$, which gives $y = 10$ and $x = 11$.\n\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": 19355,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, determine the maximum number of edges a simple graph on $n$ vertices may have so that it contains no cycles of even length.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $\\lfloor \\frac{3(n-1)}{2} \\rfloor$. It is achieved, for instance, by a $\\frac{n-1}{2}$-arm windmill if $n$ is odd, and a $\\frac{n-2}{2}$-arm windmill 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 $\\frac{3(n-1)}{2}$ edges, let $G$ be such a graph 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 $|E'| = n-1$.\n\nThe endpoints 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 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 $\\alpha + e$, $\\alpha' + e'$, or $\\alpha + e + \\alpha' + e'$ (mod 2) would have even length.\n\nIt 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 $3(|E| - n + 1) \\leq |E|$; that is, $|E| \\leq \\frac{3(n-1)}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19356,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any real numbers $a_1, a_2, \\dots, a_n$, where $n \\in \\mathbb{N}$, 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": "See solution",
    "solution": "Consider irrational numbers $y_1 < y_2 < \\dots < y_{n+1}$ such that $y_j - y_i$ is irrational for all $1 \\leq i < j \\leq n+1$. For example, take $y, 2y, \\dots, (n+1)y$ where $y$ is irrational. We claim that one of these $y_k$ can be chosen as $x$.\n\nSuppose not: for each $y_k$, at least one of $y_k + a_1, \\dots, y_k + a_n$ is rational. There are $n+1$ choices for $y_k$ and only $n$ possible $a_m$ for which $y_k + a_m$ is rational. By the pigeonhole principle, there exist $m$ and $y_i \\neq y_j$ such that both $a_m + y_i$ and $a_m + y_j$ are rational. Then $y_j - y_i$ is rational, contradicting our choice of $y_k$. Thus, such an $x$ exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19357,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sets $S$ of positive integers satisfying the following two conditions:\n\n1. For any positive integers $a, b, c$, if $ab + bc + ca \\in S$, then $a + b + c \\in S$ and $abc \\in S$ as well.\n\n2. The set $S$ contains an integer $N \\geq 160$ such that $N - 2$ is not divisible by $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $S$ must be the set of all positive integers.\n\n**Outline of proof:**\n\n1. *Step 1:* $S$ contains an integer $M \\geq 40$ divisible by $4$.\n\n2. *Step 2:* If $4k \\in S$ for some $k \\geq 2$, then $4m \\in S$ for all $m < k$.\n\n3. *Step 3:* $S$ contains $4k$ for all $k \\geq 10$.\n\n**Details:**\n\n- By (1) and (2), $S$ contains all positive multiples of $4$ up to $40$; by (3), it contains all multiples of $4$ at least $40$. Thus, $S$ contains all positive multiples of $4$.\n\n- For any $a$, let $b = c = 2$. Then $ab + bc + ca = 4a + 4 \\in S$, so $a + b + c = a + 4 \\in S$. Thus, $S$ contains all numbers congruent to $a$ modulo $4$ for $a \\geq 4$.\n\n- Let $a = 3$, $b = c = 1$. Since $ab + bc + ca = 7 \\in S$, $a + b + c = 5 \\in S$ and $abc = 3 \\in S$; so $3 \\in S$.\n\n- Let $a = b = c = 1$. Then $ab + bc + ca = 3 \\in S$, so $a + b + c = 3 \\in S$ and $abc = 1 \\in S$; so $1 \\in S$.\n\n- Let $a = 2$, $b = c = 1$. Then $ab + bc + ca = 5 \\in S$, so $a + b + c = 4 \\in S$ and $abc = 2 \\in S$; so $2 \\in S$.\n\nThus, $S$ contains all positive integers.\n\n**Proofs of steps:**\n\n- *Step 1:* If $N$ is divisible by $4$, set $M = N$. If $N = 4k + 1$, set $a = 2k$, $b = c = 1$; then $2k$ and $2k + 2$ are both in $S$, one of which is divisible by $4$ and at least $40$. If $N = 4k + 3$, set $a = 2k + 1$, $b = c = 1$; then $2k + 1$ and $2k + 3$ are in $S$. Setting $a = k$ or $k + 1$, $b = c = 1$, we get $k, k + 1, k + 2, k + 3$ in $S$, one of which is divisible by $4$ and at least $40$.\n\n- *Step 2:* Let $b = c = 2$. If $4a + 4 \\in S$, then $4a \\in S$ by condition (1). Starting from $M$, this gives all smaller multiples of $4$ by recursion.\n\n- *Step 3:* $S$ contains $M \\geq 40$ divisible by $4$ (from step 1). If $M$ is divisible by $8$, set $P = M$. Otherwise, $M \\geq 44$ and $M - 4$ is divisible by $8$; by step 2, $M - 4 \\in S$, so $P = M - 4$ works. Let $b = c = 4$; if $8a + 16 \\in S$, then $16a \\in S$. For $a \\geq 3$, $16a > 8a + 16$, so starting from $P$, we can generate arbitrarily large multiples of $8$ in $S$. By step 2, all $4k$ for $k \\geq 10$ are in $S$.\n\nTherefore, $S$ is the set of all positive integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19358,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be 103 vertices, of which 79 are red and 24 are blue. The red vertices can be grouped into $k$ contiguous blocks, and the blue vertices into $k$ contiguous blocks, arranged alternately around a circle. For each $k = 1, 2, \\dots, 24$:\n\n- What are the possible values of $(A, B)$, where $A$ is the number of red blocks and $B$ is the number of blue blocks?\n- For $B = 14$ (i.e., $k = 10$), how many ways can the blue vertices be grouped into 10 blocks, and the red vertices into 10 blocks, alternating around the circle?",
    "options": [],
    "answer": "See solution",
    "solution": "For $k$ blocks, $A = 79 - k$ and $B = 24 - k$, so all possible $(A, B)$ are $(79 - k, 24 - k)$ for $k = 1, 2, \\dots, 24$.\n\nFor $B = 14$ ($k = 10$), label the blue blocks $1$ to $10$ clockwise. Let $x_1, x_2, \\dots, x_{10}$ be the sizes of the blue blocks, and $y_i = x_1 + \\dots + x_i$. Then\n$$\n1 \\le y_1 < y_2 < \\dots < y_9 < y_{10} = 24.\n$$\nThere are $\\binom{23}{9}$ ways to choose $y_1, \\dots, y_9$. Similarly, the red vertices can be grouped into 10 blocks in $\\binom{78}{9}$ ways.\n\nSince arrangements are distinct up to rotation (as 79 is prime), but there are 10 choices for the starting block, the total number of colorations is\n$$\n\\frac{\\binom{23}{9} \\binom{78}{9}}{10}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19359,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to find a set of more than 100 binary sequences of length 200 such that any two sequences in the set differ in at least 101 positions?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible.\n\nLet $n$ be the number of 01-sequences in the set, and let $n_i$ be the number of sequences with a $1$ at the $i$th position, where $1 \\leq i \\leq 200$. Let $S$ be the total number of differing positions among all pairs of sequences.\n\nSince every pair differs in at least 101 positions, we have:\n$$\nS \\geq 101 \\binom{n}{2}.\n$$\nOn the other hand, for each position $i$, there are $n_i(n - n_i)$ pairs of sequences that differ at position $i$. Summing over all positions:\n$$\nS = \\sum_{i=1}^{200} n_i(n - n_i).\n$$\nBy the AM-GM inequality:\n$$\nn_i(n - n_i) \\leq \\left( \\frac{n_i + (n - n_i)}{2} \\right)^2 = \\frac{n^2}{4}.\n$$\nThus,\n$$\nS \\leq 200 \\cdot \\frac{n^2}{4} = 50n^2.\n$$\nCombining inequalities:\n$$\n50n^2 \\geq S \\geq 101 \\binom{n}{2}.\n$$\nThis leads to $n \\leq 101$. However, for $n = 101$, the inequality $n_i(n - n_i) < \\frac{n^2}{4}$ is strict, so $n < 101$. Therefore, $n \\leq 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19360,
    "subject": "Mathematics (Olympiad)",
    "question": "There are six boxes, each containing a different number of gold coins: $a_1, a_2, a_3, a_4, a_5, a_6$ (with $a_i \\neq a_j$ for $i \\neq j$). The boxes are arranged as shown in the figure below.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p190_data_b201890a69.png)\n\nCaptain Jack and a nominated pirate take turns choosing a box. The rule is: each person may only choose a box that is adjacent to at most one box (i.e., an 'end' box). Captain Jack goes first. If Captain Jack ends up with more gold coins than the pirate, he wins the game.\n\nWhat strategy should Captain Jack use to guarantee a win?",
    "options": [],
    "answer": "See solution",
    "solution": "When there are 2 boxes, Captain Jack will naturally win the game.\n\n**Lemma 1:** When there are 4 boxes, Captain Jack also has a strategy to win the game.\n\nSince there are 4 boxes, there are two ways to link them:\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p190_data_b201890a69.png)\n\nCase 1:\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p190_data_2472043768.png)\n\n**Case 1:**\n\nIn the first round, Captain Jack has three outer boxes to choose from. He will choose the one with the most coins, while the pirate can only choose from the other two. The center box cannot be chosen first. After the first round, Captain Jack will have more coins than the pirate. With 2 boxes left, Captain Jack will again choose the one with more coins and win.\n\n**Case 2:**\n\nPaint the 4 boxes black and white as shown in the figure. If the sum of coins in black boxes is greater than in white boxes, Captain Jack takes a black box first, forcing the pirate to take a white one. Captain Jack then takes the other black box and wins.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p191_data_c5bdc39716.png)\n\nNow, consider the original problem with 6 boxes.\n\nIf $a_6 \\geq a_5$, Captain Jack should take the available box with the most coins. The pirate takes one, and the problem reduces to the 4-box case in Lemma 1.\n\nIf $a_5 > a_6$, assume $a_1 > a_2$, and paint $a_1, a_3, a_5$ black and the rest white, as shown in the figure.\n\nCheck if $a_1 + a_3 + a_5 \\geq a_2 + a_4 + a_6$. If so, Captain Jack can take all the black boxes and win. If not, he should take $a_6$ first, then:\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p191_data_7b81674cb2.png)\n\n1. If the pirate takes $a_1$, Captain Jack takes $a_2$ and $a_4$ to win.\n2. If the pirate takes $a_2$, since $a_1 > a_2$, Captain Jack takes $a_1$ and wins.\n3. If the pirate takes $a_5$, Captain Jack takes $a_4$:\n   - If the pirate takes $a_1$, Captain Jack takes $a_2$ to win.\n   - If the pirate takes $a_2$, since $a_1 > a_2$, Captain Jack takes $a_1$ to win.\n\nTherefore, Captain Jack always has a strategy to win the game.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19361,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest possible value of\n$$\n\\frac{x^{2023} + 203}{17x^7 + 7x^{17}}\n$$\nover all positive real numbers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $\\frac{17}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19362,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two fixed points $A$ and $B$ on the unit circle $\\omega$, satisfying $\\sqrt{2} < AB < 2$. Let $P$ be a moving point on $\\omega$ such that $\\triangle ABP$ is an acute-angled triangle and $AP > AB > BP$.\n\nFor this moving point $P$, let $H$ be the orthocenter of $\\triangle ABP$. Take a point $S$ on the arc $\\widehat{AP}$ such that $SH = AH$, and take a point $T$ on the arc $\\widehat{AB}$ such that $TB \\parallel AP$. Let $Q$ be the intersection of lines $ST$ and $BP$.\n\nProve that there exists a fixed point in the plane such that the circle with diameter $HQ$ passes through it.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that the midpoint $M$ of $AB$ satisfies the given condition.\n\n![](images/China-TST-2023B_p17_data_66bc8f6712.png)\n\nLet $P_1$ be the antipodal point of $P$ on the circle $\\omega$, and let $H_1$ be the intersection of the extension of $AH$ with $\\omega$. We will show that $QP_1 = QH_1$.\n\nDenote $O$ as the center of $\\omega$. Since $SH = AH$, we have that $S$ and $A$ are symmetric with respect to $OH$, implying $OH \\perp SA$. Also, $HB \\perp AP$, so $\\angle BHO - 180^\\circ - \\angle SAP = 180^\\circ - \\angle STP = \\angle PTQ$. By noting that $TB \\parallel AP$, we have $\\angle TPQ = \\angle APB - \\angle APT = \\angle APB - \\angle PAB = \\angle HBO$. Thus, $\\triangle PTQ \\sim \\triangle BHO$, which gives $\\frac{PQ}{PT} = \\frac{BO}{BH}$. Since $PT = AB$ and $BO = PO$, we obtain $\\frac{PQ}{AB} = \\frac{PO}{BH}$. Furthermore, $\\angle OPQ = 90^\\circ - \\angle PAB = \\angle ABH$, implying $\\triangle OPQ \\sim \\triangle HBA$. Therefore, $\\angle OQP = \\angle HAB = \\angle H_1AB = \\angle H_1P_1B$. Since $PQ \\perp P_1B$, it follows that $OQ \\perp P_1H_1$, and thus $OQ$ bisects $P_1H_1$ perpendicularly, leading to $QP_1 = QH_1$.\n\nSince $H_1$ and $H$ are symmetric with respect to $BP$, we have $QH = QH_1 = QP_1$. Also, it is well-known that $M$ is the midpoint of $HP_1$, so $QM \\perp MH$. Consequently, the circle with diameter $HQ$ passes through the midpoint $M$ of $AB$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19363,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\ge 0$, let $a_n$ be the number of paths of length $n$ starting at $O$ and terminating at $O$. Also, let $b_n$ be the number of paths of length $n$ starting at $O$ and terminating at $A$. By symmetry, $b_n$ is also the number of those paths terminating at any of $B, C, D, E, F$.\n\nTo count $a_n$ for $n \\ge 1$, the $(n-1)$st step must terminate at one of $A, B, C, D, E, F$. For each of these 6 points, there are $b_{n-1}$ paths terminating at that point. This gives a total of $6b_{n-1}$ paths. In other words, we have\n\n$$\na_n = 6b_{n-1}.\n$$\n\nTo count $b_n$ for $n \\ge 1$, suppose the $n$th step terminates at $A$. Then the $(n-1)$st step may terminate at $B, F$ or $O$. If it is $B$ or $F$, then there are $b_{n-1}$ paths. If it is $O$, then there are $a_{n-1}$ paths. Therefore, we have\n\n$$\nb_n = 2b_{n-1} + a_{n-1}.\n$$\n\nCombining these, we obtain\n\n$$\n\\frac{1}{6}a_{n+1} = b_n = 2b_{n-1} + a_{n-1} = \\frac{1}{3}a_n + a_{n-1}.\n$$\n\nThis can be rewritten as $a_n = 2a_{n-1} + 6a_{n-2}$ for $n \\ge 2$. The roots of the characteristic equation $\\lambda^2 - 2\\lambda - 6 = 0$ are $1 \\pm \\sqrt{7}$. Suppose $a_n = A(1 + \\sqrt{7})^n + B(1 - \\sqrt{7})^n$. Using the initial conditions $a_0 = 1$ and $a_1 = 0$, we solve\n\n$$\n\\begin{cases} A + B = 1, \\\\ A(1 + \\sqrt{7}) + B(1 - \\sqrt{7}) = 0. \\end{cases}\n$$\n\nFind $a_{2003}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\na_{2003} = \\frac{7 - \\sqrt{7}}{14}(1 + \\sqrt{7})^{2003} + \\frac{7 + \\sqrt{7}}{14}(1 - \\sqrt{7})^{2003}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19364,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "There are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_1, \\dots, x_n$ using only 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$$\n\\frac{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} a_{\\alpha_1, \\dots, \\alpha_n} x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}}{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} b_{\\alpha_1, \\dots, \\alpha_n} x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}}\n$$\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 $a_{0, \\dots, 0} = b_{1, \\dots, 1} = 0$.\n\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\n$$\nP = \\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} a_{\\alpha_1, \\dots, \\alpha_n} (X+1)^{\\alpha_1} \\cdots (X+n)^{\\alpha_n},\n$$\n\nand\n\n$$\nQ = \\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} b_{\\alpha_1, \\dots, \\alpha_n} (X+1)^{\\alpha_1} \\cdots (X+n)^{\\alpha_n},\n$$\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.\n\n**REMARK.** Alternatively, it can be shown by induction on $n$ that\n\n$$\n\\max(|P(c)|, 2|Q(c)|) \\le \\prod_{k=1}^{n} \\max(|c+k|, 2),\n$$\n\nfor all integers $c$. In case $n > 8$, this provides a solution along the same lines.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19365,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum number of straight lines required in the plane so that the number of intersection points is at least $2022$ in part (a), and at least $2023$ in part (b).",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a):\n\nWe have\n$$\n\\binom{64}{2} = 2016 < 2022 < 2080 = \\binom{65}{2}.\n$$\n\nSince each pair of lines can produce at most one intersection point, 64 lines yield only 2016 points, which is insufficient. Therefore, at least 65 lines are needed.\n\nIt is possible to have 65 lines. For example, suppose there are 65 pairwise non-parallel lines $l_1, l_2, \\dots, l_{65}$ such that for each $k = 1, 2, \\dots, 29$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1}, l_{2k}, l_{65}$ for $k = 1, 2, \\dots, 29$. Then the number of intersection points is\n$$\n\\binom{65}{2} - 2 \\cdot 29 = 2022.\n$$\n\nFor part (b):\n\nSimilarly, at least 65 lines are needed. An example: suppose there are 65 lines $l_1, l_2, \\dots, l_{65}$ where $l_{64}$ and $l_{65}$ are parallel and there is no other pair of parallel lines. Also, for each $k = 1, 2, \\dots, 28$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1}, l_{2k}, l_{65}$ for $k = 1, 2, \\dots, 28$. Then the number of intersection points is\n$$\n\\binom{65}{2} - 1 - 2 \\cdot 28 = 2023.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19366,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a group of competitors is to be divided into two rooms, A and B. The largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room. Prove that such a distribution is always possible.\n\nLet $C(A)$ and $C(B)$ denote the largest size of a clique in rooms A and B, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "We provide an algorithm to distribute the competitors.\n\nLet $M$ be the largest clique among all competitors, with $|M| = 2m$.\n\n**Step 1:** Move all members of $M$ to room A, and the remaining competitors to room B. Then $C(A) = |M| \\geq C(B)$.\n\n**Step 2:** If $C(A) > C(B)$, move one person at a time from room A to room B. After each move, $C(A)$ decreases by 1, while $C(B)$ increases by at most 1. Continue until $C(A) \\leq C(B) \\leq C(A) + 1$.\n\nAt this point, $C(A) = |A| \\geq m$. (Otherwise, there would be at least $m+1$ members of $M$ in room B and at most $m-1$ in room A, so $C(B) - C(A) \\geq 2$, which is impossible.)\n\n**Step 3:** Let $K = C(A)$. If $C(B) = K$, we are done. Otherwise, $C(B) = K+1$.\n\n**Step 4:** If there is a clique $C$ in room B with $|C| = K+1$ and a competitor $x \\in B \\cap M$ but $x \\notin C$, move $x$ to room A. Now, $C(A) = C(B) = K+1$.\n\nIf no such $x$ exists, then every largest clique in room B contains $B \\cap M$ as a subset. Proceed to step 5.\n\n**Step 5:** Choose any largest clique $C$ ($|C| = K+1$) in room B, and move a member of $C \\setminus M$ to room A. Since $|C| = K+1 > m \\geq |B \\cap M|$, $C \\setminus M \\neq \\emptyset$.\n\nAfter this move, $C(B)$ decreases by at most 1, so $C(B) = K$.\n\nNow, $A \\cap M$ is a clique in A of size $K$, so $C(A) \\geq K$. To show $C(A) = K$, let $Q$ be any clique in A. Members of $A$ are either from $M$ or were moved from B in step 5, and all are friends with $B \\cap M$. Thus, $Q \\cup (B \\cap M)$ is a clique, so\n\n$$\n|M| \\geq |Q \\cup (B \\cap M)| = |Q| + |B \\cap M| = |Q| + |M| - |A \\cap M|.\n$$\n\nTherefore, $|Q| \\leq |A \\cap M| = K$. Thus, after these steps, $C(A) = C(B) = K$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19367,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $1 + a + b + c = 2abc$. Prove the inequality\n\n$$\n\\frac{ab}{1+a+b} + \\frac{bc}{1+b+c} + \\frac{ca}{1+c+a} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = \\frac{1}{a} + 1$, $y = \\frac{1}{b} + 1$, and $z = \\frac{1}{c} + 1$. Then,\n\n$$\n\\begin{aligned}\n\\frac{ab}{a+b+1} &= \\frac{ab(1+c)}{(1+a+b)(1+c)} = \\frac{ab+abc}{1+a+b+c+ac+bc} \\\\\n&= \\frac{ab+abc}{2abc+ac+bc} = \\frac{\\frac{1}{c}+1}{\\frac{1}{a}+1+\\frac{1}{b}+1} = \\frac{z}{x+y}.\n\\end{aligned}\n$$\n\nSimilarly, $\\frac{bc}{b+c+1} = \\frac{x}{y+z}$ and $\\frac{ca}{c+a+1} = \\frac{y}{z+x}$.\n\nThus, the assertion is equivalent to Nesbitt's inequality:\n\n$$\n\\frac{z}{x+y} + \\frac{x}{y+z} + \\frac{y}{z+x} \\ge \\frac{3}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19368,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $\\angle A = 90^\\circ$. The bisector of $\\angle B$ meets the altitude $AD$ at point $E$, and the bisector of $\\angle CAD$ meets side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B$, $E$, and $G$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "First, $\\triangle ABC \\sim \\triangle DAC$, so $\\dfrac{AC}{BA} = \\dfrac{DC}{AD}$. Also, $\\triangle DAC \\sim \\triangle DBA$. It follows that $\\triangle AFC \\sim \\triangle BEA$, so $\\dfrac{FC}{AC} = \\dfrac{EA}{BA}$. Thus, $\\dfrac{FC}{EA} = \\dfrac{AC}{BA} = \\dfrac{DC}{AD}$. This shows that $EF$ is parallel to $AC$. Hence, $AEFG$ is a parallelogram. Since $\\angle EFA = \\angle FAG = \\angle FAE$, we have $AEFG$ is a rhombus, so $AG = FG$. Thus, $\\triangle BFG \\cong \\triangle BAG$. Consequently, $\\angle FBG = \\angle ABG = \\dfrac{1}{2} \\angle ABC = \\angle FBE$. This means $B$, $E$, and $G$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19369,
    "subject": "Mathematics (Olympiad)",
    "question": "The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a single-scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "The only triples for the numbers of chocolate, vanilla, and strawberry cones (in order) that meet the conditions are $(6, 2, 1)$, $(5, 3, 1)$, and $(4, 3, 2)$. Using the formula for permutations with repetitions:\n\n$$\n\\begin{aligned}\nN &= \\frac{9!}{6!2!1!} + \\frac{9!}{5!3!1!} + \\frac{9!}{4!3!2!} \\\\\n&= 504 + 504 + 1008 \\\\\n&= 2016\n\\end{aligned}\n$$\n\nThe requested remainder is $2016 \\div 1000 = 2$ remainder $16$. Thus, the answer is $16$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19370,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(l, m, n)$ be the number of possible triples $(P_A, P_B, P_C)$ when $l, m, n$ students from schools A, B, and C, respectively, participate in a tournament. The recurrence is:\n\n$$\nf(l, m, n) = f(n, m, l-1) + f(l, n, m-1)\n$$\n\nwith boundary conditions:\n\n$$\nf(l, m, 0) = \\binom{l+m}{m} \\\\\nf(l, m, n) = 0 \\quad \\text{if } n < 0.\n$$\n\nAlso, $f(l, m, n) = f(m, l, n)$. The number we seek is $f(5, 5, 5)$. Show that the remainder of $f(5, 5, 5)$ divided by $8$ is $4$ if $f(5, 4, 4)$ is odd, and $0$ if $f(5, 4, 4)$ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us calculate $f(5, 4, 4)$ modulo $2$.\n\n$$\n\\begin{align*}\nf(5, 4, 4) &= f(4, 4, 4) + f(5, 4, 3) \\\\\n&= 0 + f(3, 4, 4) + f(5, 3, 3) \\\\\n&= f(4, 4, 2) + f(3, 4, 3) + f(3, 3, 4) + f(5, 3, 2) \\\\\n&= 0 + f(3, 4, 3) + 0 + f(5, 3, 2) \\\\\n&= f(3, 4, 2) + f(3, 3, 3) + f(2, 3, 4) + f(5, 2, 2) \\\\\n&= f(3, 4, 2) + 0 + f(2, 3, 4) + f(5, 2, 2) \\\\\n&= f(2, 4, 2) + f(3, 2, 3) + f(4, 3, 1) + f(2, 4, 2) + f(2, 2, 4) + f(5, 2, 1) \\\\\n&= f(3, 2, 3) + f(4, 3, 1) + f(5, 2, 1) \\\\\n&= f(3, 2, 2) + f(3, 3, 1) + f(1, 3, 3) + f(4, 1, 2) + f(1, 2, 4) + f(5, 1, 1) \\\\\n&= f(3, 2, 2) + 0 + f(1, 3, 3) + f(4, 1, 2) + f(1, 2, 4) + f(5, 1, 1) \\\\\n&= f(2, 2, 2) + f(3, 2, 1) + f(3, 3, 0) + f(1, 3, 2) + f(2, 1, 3) + f(4, 2, 0) \\\\\n&\\quad + f(4, 2, 0) + f(1, 4, 1) + f(1, 1, 4) + f(5, 1, 0) \\\\\n&= 0 + f(3, 2, 1) + 0 + f(1, 3, 2) + f(2, 1, 3) + 0 + 0 + f(1, 4, 1) + 0 + 0 \\\\\n&= f(1, 2, 2) + f(3, 1, 1) + f(2, 3, 0) + f(1, 2, 2) + f(3, 1, 1) + f(2, 3, 0) \\\\\n&\\quad + f(1, 4, 0) + f(1, 1, 3) \\\\\n&= f(1, 4, 0) = 1\n\\end{align*}\n$$\n\nThus, $f(5, 4, 4)$ is odd and the answer is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19371,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right triangle. There exist points $D$ on side $AC$ and $E$ on side $BC$ such that $AB = AD = BE$ and $BD \\perp DE$. Find $\\frac{AB}{BC}$ and $\\frac{BC}{CA}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BC = a$, $CA = b$, $AB = c$. The assumptions imply $c \\leq a$, $c \\leq b$.\n\nFirst, we prove that $b + c = 2a$, without using the condition that $ABC$ is a right triangle.\n\nLet $F$ be the midpoint of $BE$. Since $BD \\perp DE$, triangle $BED$ is right at $D$, so $DF$ is the median to its hypotenuse $BE$. Hence,\n\n$$\nBF = DF = EF = \\frac{1}{2}BE\n$$\n\nOn the other hand, $AB = AD$, so $A$ and $F$ are equidistant from the endpoints of $BD$. Hence, $AF$ is the perpendicular bisector of $BD$. Because triangle $BDA$ is isosceles with base $BD$, it follows that $AF$ is the bisector of $A$.\n\nBy the bisector property,\n$$\n\\frac{AB}{BF} = \\frac{AC}{CF}\n$$\nSubstituting $AB = c$, $BF = \\frac{c}{2}$, $AC = b$, $CF = a - \\frac{c}{2}$ gives\n$$\n\\frac{c}{2} = \\frac{b}{a - \\frac{c}{2}}\n$$\n\nIn particular, $BC = a$ is the middle side of $ABC$, and since $AB = c$ is the shortest, the hypotenuse is $AC = b$. Thus, by the Pythagorean theorem, $b^2 = a^2 + c^2$. Combined with $b = 2a - c$, this yields $(2a - c)^2 = a^2 + c^2$, which reduces to $3a = 4c$. Hence, $a = 4d$, $c = 3d$ with $d > 0$, and $b = \\sqrt{a^2 + c^2} = 5d$. So,\n$$\n\\frac{AB}{BC} = \\frac{3}{4}, \\quad \\frac{BC}{CA} = \\frac{4}{5}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19372,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = 6x^3 - x^2$. Prove that for any $x > 0$, $$f(x) \\ge \\frac{5x-1}{8}.$$ Furthermore, if $a + b + c + d = 1$ and $a, b, c, d > 0$, show that $$f(a) + f(b) + f(c) + f(d) \\ge \\frac{1}{8},$$ and determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $f(x) = 6x^3 - x^2 \\ge \\frac{5x-1}{8}$ for any $x > 0$. Indeed,\n\n$$\n\\begin{align*}\n& 6x^3 - x^2 \\ge \\frac{5x-1}{8} \\\\\n\\Leftrightarrow \\quad & 48x^3 - 8x^2 - 5x + 1 \\ge 0 \\\\\n\\Leftrightarrow \\quad & (4x-1)^2(3x+1) \\ge 0.\n\\end{align*}\n$$\n\nThis clearly holds for $x > 0$. It follows that\n\n$$\nf(a) + f(b) + f(c) + f(d) \\ge \\frac{5(a+b+c+d) - 4}{8} = \\frac{1}{8}.\n$$\n\nEquality holds when $a = b = c = d = \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19373,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = 2^{b_n} 3^{c_n}$, where the sequences $(b_n)$ and $(c_n)$ satisfy the recurrences:\n\n$$\n\\begin{align*}\nb_{n+2} &= b_n + b_{n+1} \\\\\nc_{n+2} &= c_n + c_{n+1}\n\\end{align*}\n$$\n\nSuppose $c_1$ and $c_2$ are even, and $b_1$ and $b_2$ are odd. For which $n$ is $a_n$ a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $c_1$ and $c_2$ are even, all $c_n$ are even by the recurrence. $b_1$ and $b_2$ are odd, so $b_n$ is even when $n$ is a multiple of $3$. For $a_n$ to be a perfect square, both $b_n$ and $c_n$ must be even. Thus, $a_n$ is a perfect square for all $n$ that are multiples of $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19374,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $D$ and $E$ divide side $AB$ of equilateral triangle $ABC$ into three equal parts; $D$ is between $A$ and $E$. Point $F$ on side $BC$ is such that $CF = AD$. Find the sum of the angles\n$$\n\\angle CDF + \\angle CEF.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions give $BF = BD$ (which equals $\\frac{2}{3}AB$), also $\\angle DBF = 60^\\circ$, hence triangle $DBF$ is equilateral. Then\n\n$$\nDF \\parallel AC \\text{ as } \\angle BDF = \\angle BAC = 60^\\circ.\n$$\n\nHence $\\angle CDF = \\angle ACD$.\n\nOn the other hand, $\\angle ACD = \\angle BCE$ by the symmetry of the figure (or because triangles $ADC$ and $BEC$ are congruent). Then $\\angle CDF = \\angle BCE$. So\n\n![](images/Argentina_2015_Booklet_p2_data_931b3c8442.png)\n\nthe required sum $\\angle CDF + \\angle CEF$ is equal to $\\angle FCE + \\angle CEF$. By the exterior angle theorem, that last sum equals $\\angle BFE$. Now $FE$ is a median in the equilateral triangle $DBF$, hence also a bisector. Therefore $\\angle BFE = \\frac{1}{2}\\angle BFD = \\frac{1}{2} \\cdot 60^\\circ = 30^\\circ$, which is the answer to the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19375,
    "subject": "Mathematics (Olympiad)",
    "question": "請找出所有的函數 $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ 使得對於所有的 $x, y \\in \\mathbb{R}$ 恆有\n\n$$\n(4x + g(x)^2)g(y) = 4g\\left(\\frac{y}{2} \\cdot g(x)\\right) + 4xy \\cdot g(x).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "首先我們將 $(4x + g(x)^2)g(y) = 4g\\left(\\frac{y}{2} \\cdot g(x)\\right) + 4xy \\cdot g(x)$ 寫成\n\n$$\n\\left(x + \\left(\\frac{g(x)}{2}\\right)^2\\right) \\frac{g(y)}{2} = \\frac{1}{2} g\\left(y \\cdot \\frac{g(x)}{2}\\right) + xy \\cdot \\frac{g(x)}{2}.\n$$\n\n令 $f(x) = \\frac{g(x)}{2}$，則原式可化成\n\n$$\n(x + f(x)^2)f(y) = f(yf(x)) + xyf(x). \\quad (1)\n$$\n\n將 $x = 1$ 代入 (1)，得到\n\n$$\n(1 + f(1)^2)f(y) = f(yf(1)) + yf(1). \\quad (2)\n$$\n\n我們將 $y$ 分別以 $1, f(1), f(1)^2$ 代入 (1) 可得\n\n$$\nf(f(1)) = f(1)^3, \\quad (3)\n$$\n$$\nf(1)^3 + f(1)^5 = f(f(1)^2) + f(1)^2, \\quad (4)\n$$\n$$\nf(1)^7 + 2f(1)^5 - f(1)^4 - f(1)^2 = f(f(1)^3). \\quad (5)\n$$\n\n將 $y = 1$ 及 $x = f(1)$ 代入 (1)，可得\n\n$$\nf(1)^2 + f(1)^7 = f(f(1)^3) + f(1)^4. \\quad (6)\n$$\n\n從 (5) 和 (6) 推導出 $f(1)^5 = f(1)^2$。因此 $f(1) = 0$ 或 $f(1) = 1$。\n\n若 $f(1) = 0$，則從 (2) 可知對於所有 $y$ 恆有 $f(y) = f(0)$，所以可得到 $f$ 是常數函數，因此唯一滿足此情形的函數是零函數。若 $f(1) = 1$，則從 (2) 可知對於所有 $y$ 恆有 $f(y) = y$，而且我們驗證這個函數滿足 (1)。所以 (1) 的解為 $f(x) = 0$ 和 $f(x) = x$。因為 $g(x) = 2f(x)$，所以原式的解為 $g(x) = 0$ 和 $g(x) = 2x$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19376,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles of the same radius, $K$ and $L$, intersect in two points, one of which is $P$. Denote by $A$ and $B$, respectively, the points diametrically opposite to $P$ on each of $K$ and $L$. Yet another circle of the same radius is brought to pass through $P$, intersecting $K$ and $L$ in the points $X$ and $Y$, respectively.\n\nShow that the line $XY$ is parallel to the line $AB$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p205_data_6539f4fa4a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the third circle, and denote by $Z$ the point on $M$ diametrically opposite to $P$.\n\nSince $\\angle AXP = \\angle PXZ = 90^\\circ$, the three points $A$, $X$, $Z$ are collinear. Likewise, the three points $B$, $Y$, $Z$ are collinear. Point $P$ is equidistant to the three vertices of triangle $ABZ$, for $PA = PB = PZ$ is the common diameter of the circles. Therefore, $P$ is the circumcentre of $\\triangle ABZ$, which means the perpendiculars $PX$ and $PY$ bisect the sides $AZ$ and $BZ$. Ergo, $X$ and $Y$ are midpoints on $AZ$ and $BZ$, which leads to the desired conclusion $XY \\parallel AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19377,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna, Berta, and Clara write the square numbers $1, 4, 9, \\ldots, 2025$ on a blackboard, compute their sum, and observe that it is divisible by $3$. Then, they agree to the following game: In each round, Anna will cross out one number, then Berta will do the same, and then Clara will do the same. This continues until all numbers are crossed out. Clara has the goal that the sum of the remaining numbers after each round is divisible by $3$.\n\na) Prove that Anna cannot stop Clara from reaching her goal if Clara has Berta's help.\n\nb) Prove that Berta can stop Clara from reaching her goal even if Clara has Anna's help.",
    "options": [],
    "answer": "See solution",
    "solution": "On the blackboard, there are $15$ integers with residue $0$ modulo $3$ and $30$ integers with residue $1$ modulo $3$. If, in a certain round, Berta and Clara cross out numbers that have the same residue modulo $3$ as the number crossed out by Anna, then they have removed either $0+0+0$ or $1+1+1$ modulo $3$.\n\nIn both cases, the sum does not change modulo $3$ and therefore remains divisible by $3$. Since $15$ and $30$ are multiples of $3$, it is possible for Berta and Clara to always choose the same residue as Anna. Therefore, Anna cannot stop Clara from reaching her goal if Clara has Berta's help.\n\nHowever, if Berta chooses in the first round a residue modulo $3$ that is different from the one chosen by Anna, they have crossed out $0+1$ or $1+0$ modulo $3$. Therefore, Clara's only choices for the sum of the remaining numbers after the first round are $1$ or $2$ modulo $3$. In both cases, the sum is not divisible by $3$. Therefore, Clara has failed in her goal already in the first round if Berta plays uncooperatively.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19378,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nf(n) = \\sum_{k=0}^{2010} n^k = 1 + n + n^2 + \\dots + n^{2010}.\n$$\n\nProve that for every integer $m$ with $2 \\leq m \\leq 2010$, there is no non-negative integer $n$ such that $f(n)$ is divisible by $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $m$ divides $f(n)$ for some integer $n$ and some $2 \\leq m \\leq 2010$. Note that $f(1) = 2011$, and since 2011 is a prime number, $m$ cannot divide $f(1)$, so we may restrict ourselves to the case $n \\neq 1$.\n\nIn this case, we can write\n\n$$\nf(n) = \\frac{n^{2011} - 1}{n - 1}.\n$$\n\nLet $p$ be a prime divisor of $m$. Then $p \\mid f(n)$ implies $p \\mid n^{2011} - 1$, so\n\n$$\nn^{2011} \\equiv 1 \\pmod{p}.\n$$\n\nThis implies that $n$ and $p$ are coprime.\n\nBy the above, the order $\\operatorname{ord}_p(n)$ of $n$ modulo $p$ (the smallest positive integer $k$ such that $n^k \\equiv 1 \\pmod{p}$) divides 2011. Since 2011 is prime, $\\operatorname{ord}_p(n) \\in \\{1, 2011\\}$.\n\nIf $\\operatorname{ord}_p(n) = 1$, then $n \\equiv 1 \\pmod{p}$, so\n$$\nf(n) \\equiv 2011 \\pmod{p}.\n$$\nThus $p \\mid 2011$, so $p = 2011$, which is impossible since $p < 2011$.\n\nTherefore, $\\operatorname{ord}_p(n) = 2011$. But $\\operatorname{ord}_p(n)$ divides $\\varphi(p) = p-1$, so $2011 \\mid p-1$. This is impossible for $1 < p < 2011$.\n\nThus, there is no such $n$ for any $2 \\leq m \\leq 2010$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19379,
    "subject": "Mathematics (Olympiad)",
    "question": "All the squares of a $2024 \\times 2024$ board are coloured white. In one move, Mohit can select one row or column whose every square is white, choose exactly $1000$ squares in this row or column, and colour all of them red. Find the maximum number of squares that Mohit can colour red in a finite number of moves.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2024$ and $k = 1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n - k)$, which evaluates to $3048000$.\n\nIndeed, call a row or column *bad* if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picked, then that row and the $k$ columns corresponding to the chosen squares are all bad. Any subsequent move increases the number of bad rows or columns by at least $1$. Since there are only $2n$ rows and columns, we can make at most $2n - (k+1)$ moves after the first one, and so at most $2n - k$ moves can be made in total. Thus, we can have at most $k(2n - k)$ red squares.\n\nTo prove this is achievable, let's choose each of the $n$ columns in the first $n$ moves, and colour the top $k$ cells in these columns. Then, the bottom $n-k$ rows are still uncoloured, so we can make $n-k$ more moves, colouring $k(n + n - k)$ cells in total. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19380,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the notation in the figure below, find the perimeter of the square whose area is equal to that of the rectangle with sides 9 and 16.\n\n![](images/Slovenija_2013_p8_data_98430ea4e9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By Pythagoras' theorem, $y = \\sqrt{15^2 - 9^2} = \\sqrt{144} = 12$. The two right triangles on the left side are similar, so $\\frac{x}{5} = \\frac{y}{15}$, which gives $x = \\frac{y}{3} = 4$. Thus, the rectangle has sides 9 and 16, so its area is $9 \\times 16 = 144$. The square has the same area, so its side length is $\\sqrt{144} = 12$, and its perimeter is $4 \\times 12 = 48$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19381,
    "subject": "Mathematics (Olympiad)",
    "question": "$f: \\{0,1,2,\\ldots\\} \\times \\{0,1,2,\\ldots\\} \\to \\mathbb{R}$ функц өгөгдсөн. Дараах нөхцлүүдийг хангах бүх $f(p,q)$ функцуудыг ол:\n\n- $\\forall p \\in \\{0,1,2,\\ldots\\}: f(p,0) = f(0,p) = 0$\n- $\\forall p,q \\in \\{0,1,2,\\ldots\\}$ ба $pq \\neq 0$ үед $f(p,q) = \\frac{1}{2}(f(p+1, q-1) + f(p-1, q+1)) + 1$",
    "options": [],
    "answer": "See solution",
    "solution": "Эхлээд бодлогын нөхцөлийг хангах $f(p, q)$ функц цор ганц олдохыг баталъя. Өгөгдсөн нөхцөлийг хангах $f_1(p, q)$ ба $f_2(p, q)$ функцууд олддог гэж үзье. $g(p, q) = f_1(p, q) - f_2(p, q)$ функцийг авч үзье.\n\n$$\ng(p, q) = \\begin{cases} 0, & pq = 0 \\\\ \\frac{1}{2}(g(p+1, q-1) + g(p-1, q+1)), & pq \\neq 0 \\end{cases}\n$$\n\n$p + q = n$ гэж тогтоогоод, $(p, q)$ хосуудаас $g(p, q)$ хамгийн их утгатай хосыг $(p_0, q_0)$ гэж авъя. Хэрэв $p_0$ эсвэл $q_0$ нь 0 бол $g(p_0, q_0) = 0$ тул $g(p, q) = 0$ болно. Хэрэв $p_0 \\ge 1, q_0 \\ge 1$ бол $g(p_0, q_0) \\le \\frac{1}{2}(g(p_0 + 1, q_0 - 1) + g(p_0 - 1, q_0 + 1))$ тул $g(p_0, q_0) = g(p_0 - 1, q_0 + 1) = g(p_0 + 1, q_0 - 1)$ болно. Иймд $p + q = n$ байх бүх $(p, q)$ дээр $g(p, q) = g(n, 0) = 0$ байна. $n \\in \\{0, 1, 2, \\ldots\\}$ тул $\\forall p, q \\in \\{0, 1, 2, \\ldots\\}: g(p, q) \\equiv 0$, өөрөөр хэлбэл $f_1(p, q) \\equiv f_2(p, q)$. Иймд өгөгдсөн нөхцөлийг хангах $f(p, q)$ функц цор ганц байна (хэрэв олддог бол).\n\n$p$ ба $q$-ийн бага утгуудад шалгаж үзэхэд $f(p, q) = pq$ болохыг таамаглаж болно. Энэ функц өгөгдсөн нөхцөлийг хангана. Иймд хариу:\n\n$$f(p, q) = pq$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19382,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $k_1$, $k_2$ и $k_3$ се три кружници со центри во $O_1$, $O_2$ и $O_3$ соодветно, такви што ниту еден од центрите не се наоѓа во внатрешноста на другите две кружници. Кружниците $k_1$ и $k_2$ се сечат во $A$ и $P$, $k_1$ и $k_3$ се сечат во $C$ и $P$, а $k_2$ и $k_3$ се сечат во $B$ и $P$. Нека $X$ е точка на $k_1$ таква што пресекот на правата $XA$ со кружницата $k_2$ е $Y$, а пресекот на правата $XC$ со $k_3$ е $Z$, при што $Y$ не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_3$, а $Z$ не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_2$.\n\nа) Докажи дека триаголниците $XYZ$ и $O_1O_2O_3$ се слични меѓу себе.\n\nб) Докажи дека плоштината на триаголникот $XYZ$ не е поголема од четири пати по плоштината на триаголникот $O_1O_2O_3$. Дали се достигнува максимумот?\n\n![](images/Macedonia_2014_p21_data_974dd4c31c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Прво ќе докажеме дека точките $Y$, $B$ и $Z$ се колинеарни. Бидејќи четириаголникот $BYAP$ е тетивен, имаме $\\angle PBY = \\angle PAX$. Бидејќи четириаголникот $AXCP$ е тетивен, $\\angle PAX = \\angle PCZ$. Бидејќи четириаголникот $CPBZ$ е тетивен, добиваме $\\angle PBZ + \\angle PCZ = 180^\\circ$. Значи $\\angle YBZ = \\angle YBP + \\angle PBZ = 180^\\circ$.\n\nДа забележиме дека $\\angle CO_1O_3 = \\angle PO_1O_3$ и $\\angle AO_1O_2 = \\angle PO_1O_2$, од каде следува $\\angle O_2O_1O_3 = \\frac{1}{2}\\angle AO_1C = \\angle AXC$. Слично, $\\angle O_1O_2O_3 = \\angle AYB$ и $\\angle O_1O_3O_2 = \\angle CZB$. Следува дека $\\triangle XYZ \\sim \\triangle O_1O_2O_3$, со што го докажавме тврдењето под а).\n\nНека правата $X_1Y_1$ е паралелна со $O_1O_2$ и минува низ $A$, каде $X_1$ лежи на $k_1$ и $Y_1$ лежи на $k_2$. Нека $Z_1$ е пресечната точка на правата $X_1C$ со кружницата $k_3$. Од претходно докажаното, точките $Y_1$, $B$ и $Z_1$ се колинеарни и $\\triangle X_1Y_1Z_1 \\sim \\triangle O_1O_2O_3$. Уште повеќе, $\\angle PXA = \\angle PX_1A$ и $\\angle PYA = \\angle PY_1A$. Па $\\triangle PXY \\sim \\triangle PX_1Y_1$. Нека $PT$ е висината спуштена од темето $P$ кон страната $XY$. $PA$ е висината на триаголникот $PX_1Y_1$. Бидејќи $PA$ е хипотенуза во правоаголниот триаголник $PAT$, добиваме $\\overline{PT} \\le \\overline{PA}$. Па $P_{PXY} \\le P_{PX_1Y_1}$ и аналогно $P_{PYZ} \\le P_{PY_1Z_1}$ и $P_{PXZ} \\le P_{PX_1Z_1}$. Од ова добиваме $P_{XYZ} \\le P_{X_1Y_1Z_1}$. Точките $P$, $O_1$ и $X_1$ се колинеарни затоа што $\\angle PAX_1 = 90^\\circ$. Слично, $P$, $O_2$ и $Y_1$ се колинеарни и $P$, $O_3$ и $Z_1$ се колинеарни. Добиваме дека $O_1O_2$, $O_1O_3$ и $O_2O_3$ се средни линии во триаголниците $X_1Y_1P$, $X_1Z_1P$ и $Y_1Z_1P$ соодветно, па $P_{X_1Y_1Z_1} = 4P_{O_1O_2O_3}$. Од ова се добива бараното неравенство. Равенство се достигнува кога точките $X$ и $X_1$ се совпаѓаат, а со тоа и точките $Y$ и $Y_1$ се совпаѓаат и точките $Z$ и $Z_1$ се совпаѓаат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19383,
    "subject": "Mathematics (Olympiad)",
    "question": "Using the numbers $1, 2, \\ldots, 20$ (each number once) as denominators and numerators, construct $10$ fractions with integer sum.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{17}{3} + \\frac{13}{2} + \\frac{11}{6} + \\frac{19}{1} + \\frac{14}{7} + \\frac{18}{9} + \\frac{20}{10} + \\frac{16}{8} + \\frac{15}{5} + \\frac{12}{4} = 14 + 6.5 + 1.833\\ldots + 19 + 2 + 2 + 2 + 2 + 3 + 3 = 47.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19384,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square of side length $1$. Let $O$ be the circle having the side $AD$ of the square as its diameter, and pick a point $E$ on the side $AB$ of the square in such a way that the line $CE$ becomes a tangent line to the circle $O$. Determine the area of the triangle $CBE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of the side $AD$ of the square, and let $N$ be the point of tangency of the line $CE$ to the circle $O$. Since $MD = MN$, and $\\angle MDC = \\angle MNC = 90^\\circ$, we have $\\triangle MNC \\equiv \\triangle MDC$. Similarly, we have $\\triangle MNE \\equiv \\triangle MAE$. From $\\triangle MNC \\equiv \\triangle MDC$ we have $\\angle CMN = \\angle CMD$, which implies that $2\\angle CMN = \\angle DMN$. Similarly, we get $2\\angle EMN = \\angle AMN$ from $\\triangle MNE \\equiv \\triangle MAE$. Consequently, we have\n\n$$\n\\angle EMC = \\angle EMN + \\angle CMN = \\frac{1}{2}(\\angle DMN + \\angle AMN) = 90^\\circ.\n$$\n\nHence, we have $\\angle EMN = 90^\\circ - \\angle CMN = \\angle MNC$, which together with the fact that $\\angle ENM = 90^\\circ = \\angle MNC$ implies that $\\triangle EMN$ and $\\triangle MCN$ are similar triangles. Therefore, we have $EN : NM = NM : NC$. On the other hand, from the fact that $\\triangle MNC \\equiv \\triangle MDC$, we have $NC = DC = 1$, $NM = DM = \\frac{1}{2}$, from which we obtain $EN = \\frac{1}{4}$. From $\\triangle MNE \\equiv \\triangle MAE$ we get $EA = EN = \\frac{1}{4}$ so that we have $BE = \\frac{3}{4}$, and therefore, the area of $\\triangle CBE = \\frac{1}{2} \\cdot 1 \\cdot \\frac{3}{4} = \\frac{3}{8}$.\n\nAlternatively:\n\n![](images/Japan_2008_p3_data_f5440c0870.png)\n\nWe can argue in the same way as above to conclude that $\\triangle MNC \\equiv \\triangle MDC$, $\\triangle MNE \\equiv \\triangle MAE$. Then we can put $AE = t$ and get $EC = EN + NC = EA + DC = t + 1$ and $BE = BA - AE = 1 - t$. By applying the Pythagorean theorem to the right triangle $CBE$, we get $(1+t)^2 = (1-t)^2 + 1^2$. Solving for $t$ we get $t = \\frac{1}{4}$, and we get the area of the $\\triangle CBE = \\frac{3}{8}$ as above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19385,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $k$. The product of some $k$ consecutive positive integers ends with the number $k$. What value can the number $k$ attain?",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values are $k \\in \\{1, 2, 4\\}$.\n\nSuppose that $k \\geq 5$. Among any $k$ consecutive numbers, there is one divisible by $5$ and one divisible by $2$, so their product ends in $0$, meaning $k$ must be divisible by $10$. Then $k \\geq 10$, so in the product of $k$ consecutive numbers there are at least two numbers divisible by $5$ and at least two divisible by $2$, and one of them is divisible by $k$. Thus, the product is divisible by $10k$, so there will be more zeros at the end than zeros at the end of $k$, leading to a contradiction. Thus, $k \\leq 4$.\n\nIf $k = 3$, then the product will be even, but it cannot end in the digit $3$. For $k = 1, 2, 4$, consider the following examples:\n\n- $1$\n- $1 \\times 2 = 2$\n- $1 \\times 2 \\times 3 \\times 4 = 24$\n\nIn each case, the product ends with $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19386,
    "subject": "Mathematics (Olympiad)",
    "question": "試求所有同時滿足以下兩條件的正整數 $n$：\n\n(a) $n$ 的正因數個數不是 $8$ 的倍數；\n\n(b) 對於所有整數 $x$，有\n\n$$\nx^n \\equiv x \\mod n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 $n=1$ 或任意質數。\n\n我們拆解為以下四步驟：\n\n1. **$n$ 沒有平方因子**\n\n證明：若 $p^2 \\mid n$，取 $x=p$，則 $p^2 \\nmid x$ 但 $p^2 \\mid p^{p^2} = x$，矛盾。\n\n2. **$n=1,\\ p$ 或 $pq$，其中 $p$ 和 $q$ 為質數**\n\n證明：由於 $n$ 沒有平方因子，$n$ 必型如 $p_1p_2\\cdots p_m$，從而其正因數個數為 $2^m$。但因為其正因數個數不被 $8$ 整除，故 $m<3$，得證。\n\n3. **$n$ 不可能型如 $pq$**\n\n證明：不失一般性假設 $p < q$，此時取 $x$ 為 $q$ 的原根。基於 $x^n \\equiv x \\mod q$，由最小性，我們必須有 $q-1 \\mid n-1$，從而\n\n$$\nq-1 \\mid n-1 = pq-1 = p(q-1) + (q-1) \\Rightarrow q-1 \\mid p-1,\n$$\n\n此與 $p < q$ 的假設矛盾。\n\n4. **$n=1$ 與 $n=p$ 皆滿足題意**\n\n證明：$n=1$ 顯然，而 $n=p$ 為費馬小定理。\n\n*組題者註記：原投題為給定 $x^n + y^n \\equiv (x+y)^n$ 的條件，要考生先自行推出 $x^n \\equiv x$。在此略過此步降低難度。*",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19387,
    "subject": "Mathematics (Olympiad)",
    "question": "Given\n\n$$\n\\begin{aligned}\nx^2 + y^2 + z^2 + w^2 &= (3a + b + c) + (3b + c + d) \\\\\n&\\quad + (3c + d + a) + (3d + a + b) \\\\\n&= 20\n\\end{aligned}\n$$\n\nprove that\n\n$$\n\\begin{aligned}\n\\sum_{cyc} x \\left( \\frac{5x^2 - 2y^2 - z^2 + w^2}{15} \\right) &\\geq 4\\sqrt{5} \\\\\n\\Leftrightarrow \\sum_{cyc} (5x^3 - 2xy^2 - xz^2 + xw^2) &\\geq 60\\sqrt{5}.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Equality occurs when\n\n$$\nx^2 = y^2 = z^2 = w^2 = 5\n$$\n\nwhich is when\n\n$$\na = b = c = d = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19388,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $n \\geq 3$ un entero. Lucas y Matías juegan un juego en un polígono regular de $n$ lados con un vértice marcado como *trampa*. Inicialmente, Matías ubica una ficha en un vértice del polígono. En cada paso, Lucas dice un entero positivo y Matías mueve la ficha ese número de vértices en sentido horario o antihorario, a su elección.\n\na) Determina todos los $n \\geq 3$ tales que Matías puede ubicar la ficha y moverla de modo de no caer nunca en la trampa, independientemente de los números que diga Lucas. Da la estrategia para Matías.\n\nb) Determina todos los $n \\geq 3$ tales que Lucas puede obligar a Matías a caer en la trampa. Da la estrategia para Lucas.\n\n**Nota:** Los dos jugadores conocen el valor de $n$ y ven el polígono.",
    "options": [],
    "answer": "See solution",
    "solution": "Numeramos los vértices con $0, 1, 2, \\ldots, n-1$ en sentido horario y suponemos que la trampa está ubicada en el $0$. Si la ficha está en $x$ y Lucas dice $y$, entonces Matías puede mover la ficha a los vértices con número $x-y$ o $x+y$ módulo $n$.\n\nVeamos que si $n$ tiene un divisor impar $d \\neq 1$, entonces gana Matías con la siguiente estrategia: ubica la ficha en un vértice $x$ tal que no sea divisible por $d$ y continúa moviendo la ficha de modo que el vértice sea siempre no divisible por $d$. Para todo $d$, siempre hay un vértice cuyo número no es divisible por $d$, por ejemplo el $1$. Veamos que si $d$ no divide a $x$, entonces $d$ no divide a $x-y$ o $d$ no divide a $x+y$. En efecto, supongamos por el absurdo que $d$ divide a ambos, entonces $d$ divide a $(x+y)+(x-y)=2x$. Como $d$ es impar, entonces $d$ divide a $x$, lo que es una contradicción. Así que la estrategia de Matías es válida.\n\nVeamos que si $n=2^k$, entonces Lucas puede obligar a Matías a caer en la trampa.\n\nA cada vértice $x$ distinto de la trampa, le asignamos el valor $d$ si $2^d$ es la mayor potencia de $2$ que divide a $x$. Análogamente, a la trampa ($n=2^k \\geq 0$) le asignamos el valor $k$.\n\nLa estrategia de Lucas es la siguiente: si la ficha de Matías está en un vértice de valor $d$, Lucas dirá $2^d$. Lo que ocurre es que en los sucesivos pasos el valor irá creciendo. En efecto, si comenzamos en el vértice $x=2^d \\cdot q$ (con $q$ impar), entonces el siguiente vértice será $2^d \\cdot q + 2^d = 2^d \\cdot (q+1)$ o $2^d \\cdot q - 2^d = 2^d \\cdot (q-1)$, y ambos son divisibles por $2^{d+1}$ pues $q$ es impar. Por lo tanto, el nuevo vértice siempre tendrá un valor mayor que el anterior y en algún momento llegará a $2^k$, que es la trampa.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19389,
    "subject": "Mathematics (Olympiad)",
    "question": "Fill a $4 \\times 4$ grid with two sets of symbols, $\\{A, B, C, D\\}$ and $\\{\\alpha, \\beta, \\gamma, \\delta\\}$, so that:\n\n- Each row and column contains every symbol from each set exactly once.\n- Each cell contains one symbol from each set.\n- Every possible pair (one from each set) appears only once in the entire grid.",
    "options": [],
    "answer": "See solution",
    "solution": "A possible solution:\n\n|        |        |        |        |\n|--------|--------|--------|--------|\n| $A\\alpha$ | $B\\gamma$ | $C\\delta$ | $D\\beta$ |\n| $B\\beta$ | $A\\delta$ | $D\\gamma$ | $C\\alpha$ |\n| $C\\gamma$ | $D\\alpha$ | $A\\beta$ | $B\\delta$ |\n| $D\\delta$ | $C\\beta$ | $B\\alpha$ | $A\\gamma$ |\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19390,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist two polynomials $P$ and $Q$ with integer coefficients such that:\n\ni) both $P$ and $Q$ have a coefficient with absolute value greater than $2021$,\n\nii) all coefficients of $P \\cdot Q$ have absolute value at most $1$?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the polynomial\n\n$$\n(1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n})\n$$\n\nhas all coefficients equal to $0$, $+1$, or $-1$. Also note that\n\n$$\n\\begin{aligned}\n& (1 - x^2)(1 - x^4)(1 - x^8) \\dots (1 - x^{2n}) \\\\\n&= \\prod_{i=0}^{n-1} (1 - x^{2i}) \\cdot \\prod_{i=0}^{n-1} (1 + x^{2i}) \\\\\n&= (1 - x)^n \\cdot \\prod_{i=0}^{n-1} (1 + x^i)^{n-i}\n\\end{aligned}\n$$\n\nDue to the Newton binomial formula, $(1 \\pm x)^n$ has a coefficient equal to $n$. Also, all coefficients of\n\n$$\n(1 + x)^n (1 + x^2)^{n-1} \\dots (1 + x^{2n-2})^2 (1 + x^{2n-1})\n$$\n\nare positive and larger than the coefficients of $(1 + x)^n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19391,
    "subject": "Mathematics (Olympiad)",
    "question": "Two unit complex numbers $x$ and $y$ with $\\arg x, \\arg y \\in (0, \\frac{\\pi}{2})$, $\\arg y > 2\\arg x$ (so $\\triangle ABC$ is acute and $\\angle ACB > 2\\angle ABC$), satisfy\n\n$$\n|y - 1| = |(x^2 - 1)(y + 1)|. \\qquad \\textcircled{1}\n$$\n\nProve that\n\n$$\n|x^2(y + 1) + (y^2 - y)x - 2| = \\left| \\frac{(x^2 - 1)(1 - y^3)}{x^2 - y} \\right|. \\qquad \\textcircled{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof.** Notice that\n\n$$\n\\begin{align*}\n\\textcircled{1} \\Leftrightarrow & \\left| \\frac{(x^2 - 1)(y + 1)}{y - 1} \\right| = 1 \\Leftrightarrow \\frac{(\\overline{x}^2 - 1)(\\overline{y} + 1)}{\\overline{y} - 1} = \\frac{y - 1}{(x^2 - 1)(y + 1)} \\\\\n& \\Leftrightarrow \\frac{\\frac{1 - x^2}{x^2} + \\frac{1 + y}{y}}{\\frac{1 - y}{y}} = \\frac{y - 1}{(x^2 - 1)(y + 1)} \\\\\n& \\Leftrightarrow \\left( \\frac{x^2 - 1}{x} \\right)^2 = \\left( \\frac{y - 1}{y + 1} \\right)^2,\n\\end{align*}\n$$\n\nthat is, $\\frac{x^2 - 1}{x} = \\pm \\frac{y - 1}{y + 1}$. The left-hand side equals $pi$ for some $p > 0$; the right-hand side satisfies $\\operatorname{Im}\\frac{y - 1}{y + 1} > 0$ as $\\arg y \\in (0, \\frac{\\pi}{2})$. Hence,\n\n$$\n\\frac{x^2 - 1}{x} = \\frac{y-1}{y+1},\n$$\nwhich is $(y+1)x^2 = (y-1)x + (y+1)$, or simply\n\n$$\nx^2 - 1 = \\left( \\frac{y-1}{y+1} \\right) x.\n$$\n\nOn the other hand, in (2),\n\n$$\n\\text{left} = |(y^2 - 1)x + (y - 1)| = |y - 1| \\cdot |(y + 1)x + 1|;\n$$\n\n$$\n\\begin{aligned}\n\\text{right} &= \\left| \\frac{\\left( \\frac{y-1}{y+1} \\right) x(1-y^3)}{\\left( \\frac{y-1}{y+1} x + 1 - y \\right)} \\right| = \\left| \\frac{(y-1)x(1-y^3)}{(y-1)(x-y-1)} \\right| \\\\\n&= \\left| \\frac{(1-y)(1+y+y^2)}{x-(y+1)} \\right|.\n\\end{aligned}\n$$\n\nNow, (2) is equivalent to $|[(y+1)x+1][x-(y+1)]| = |1+y+y^2|$. Expand all terms and use $(y+1)x^2 = (y-1)x+(y+1)$, obtaining $|-(y^2+y+1)x| = |1+y+y^2|$, which is true as $|x| = 1$. The proof is complete. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19392,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a prime number. For $k = 2, 3, \\dots, n$, show that there exists an integer $b_k$ such that\n\n$$\nb_k \\equiv 0 \\pmod{k-1}, \\quad b_k \\equiv k \\pmod{n}.\n$$\n\nLet $a_1 = 1$ and for $k = 2, \\dots, n$, let $a_k$ be the remainder when $b_k/(k-1)$ is divided by $n$. Prove that $a_1, \\dots, a_n$ are distinct modulo $n$, $a_n = 0$, and that\n\n$$\na_1 a_2 \\dots a_k \\equiv k \\pmod{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the Chinese Remainder Theorem, for each $k = 2, 3, \\dots, n$, there exists $b_k$ such that $b_k \\equiv 0 \\pmod{k-1}$ and $b_k \\equiv k \\pmod{n}$. Define $a_1 = 1$ and for $k = 2, \\dots, n$, let $a_k$ be the remainder when $b_k/(k-1)$ is divided by $n$.\n\nSince $b_n \\equiv 0 \\pmod{n}$, we have $a_n = 0$. If $a_i = a_j$ for $i \\neq j$, then $b_i/(i-1) \\equiv b_j/(j-1) \\pmod{n}$, which implies $i(j-1) \\equiv j(i-1) \\pmod{n}$. Since $n$ is prime, this leads to $i = j$, so $a_1, \\dots, a_n$ are distinct modulo $n$.\n\nFinally,\n\n$$\na_1 a_2 \\dots a_k \\equiv \\frac{b_2 \\dots b_k}{(k-1)!} \\equiv k \\pmod{n}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19393,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions $(a, b)$ to the congruence\n$$\n4a^2 + 9b^2 \\equiv 1 \\pmod{n}\n$$\nfor a given positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n = 1$, all choices of $a$ and $b$ are solutions.\n\nNow suppose that $n > 1$ and let $p$ be a prime divisor of $n$. Let $k$ be the number of factors of $p$ in $n$.\n\nWe give a condition for $a$ and $b$ modulo $p^k$ which guarantees that $p^k \\mid 4a^2 + 9b^2 - 1$. By doing this for every prime divisor of $n$, we get a system of conditions for $a$ and $b$ modulo the various prime powers. Then, by the Chinese remainder theorem, there exist $a$ and $b$ satisfying all conditions simultaneously.\n\nIf $p \\neq 2$, then we consider the condition that $2a \\equiv 1 \\pmod{p^k}$ and $b \\equiv 0 \\pmod{p^k}$. As $2$ has a multiplicative inverse modulo $p^k$, this condition can be satisfied. We then have\n$$\n4a^2 + 9b^2 - 1 = (2a)^2 + 9b^2 - 1 \\equiv 1^2 + 9 \\cdot 0 - 1 = 0 \\pmod{p^k}.\n$$\nTherefore, all $a$ and $b$ satisfying this condition are solutions.\n\nIf $p = 2$, then we consider the condition that $a \\equiv 0 \\pmod{2^k}$ and $3b \\equiv 1 \\pmod{2^k}$. As $3$ has a multiplicative inverse modulo $2^k$, this condition can be satisfied. We then have\n$$\n4a^2 + 9b^2 - 1 = 4a^2 + (3b)^2 - 1 \\equiv 4 \\cdot 0 + 1^2 - 1 = 0 \\pmod{2^k}.\n$$\nTherefore, all $a$ and $b$ satisfying this condition are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19394,
    "subject": "Mathematics (Olympiad)",
    "question": "Using the coefficients $a_0, a_1, a_2, \\dots, a_9$, we can write\n\n$$\n(x+1)^3(x+2)^3(x+3)^3 = a_0 + a_1x + a_2x^2 + \\dots + a_8x^8 + a_9x^9.\n$$\n\nFind the value of $a_2 + a_4 + a_6 + a_8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = 1$ and $x = -1$ into the equation gives:\n\n$$\na_0 + a_1 + a_2 + \\dots + a_8 + a_9 = 2^3 \\cdot 3^3 \\cdot 4^3\n$$\n$$\na_0 - a_1 + a_2 - a_3 + \\dots + a_8 - a_9 = 0.\n$$\n\nAdding these equations:\n$$\n2(a_0 + a_2 + a_4 + a_6 + a_8) = 2^3 \\cdot 3^3 \\cdot 4^3\n$$\nSo,\n$$\na_0 + a_2 + a_4 + a_6 + a_8 = \\frac{1}{2}(2^3 \\cdot 3^3 \\cdot 4^3) = 6912.\n$$\n\nSubstituting $x = 0$ gives $a_0 = 1^3 \\cdot 2^3 \\cdot 3^3 = 216$.\n\nTherefore,\n$$\na_2 + a_4 + a_6 + a_8 = 6912 - 216 = 6696.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19395,
    "subject": "Mathematics (Olympiad)",
    "question": "A square $ABCD$ is divided into $n^2$ equal small squares by drawing lines parallel to its sides. A spider starts from $A$, moving only to the right and up, trying to reach point $C$. Each movement consists of either $k$ steps right and $m$ steps up, or $m$ steps right and $k$ steps up (in any order). The spider makes $l$ such movements, then continues moving only right or up without restriction. If $n = m \\cdot l$, find the number of all possible routes the spider can follow to reach point $C$, where $n, m, k, l$ are positive integers and $k < m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the square is placed in a Cartesian coordinate system with origin $A(0,0)$ and axes along $AB$ and $AD$. The spider starts from $A$ and makes its first movement: either $k$ steps right and $m$ steps up, or $m$ steps right and $k$ steps up. For example, if $m = 4$, $k = 3$:\n\n![](images/GreekMO2014_booklet_p20_data_33857eb517.png)\n\nAfter the first movement, the spider can be at $M(k, m)$ or $N(m, k)$. The number of ways to reach $M(k, m)$ or $N(m, k)$ is $\\binom{k + m}{k} = \\binom{k + m}{m} = v$.\n\nAfter the second movement, the spider can be at three points: $K(2k, 2m)$, $L(k + m, k + m)$, or $T(2m, 2k)$, with corresponding numbers of ways $v^2$, $2v^2$, and $v^2$.\n\nThe point $M$ can be reached from $A$ in $v$ ways, and $K$ from $M$ in $v$ ways, so $K$ can be reached from $A$ in $v^2$ ways.\n\n![](images/GreekMO2014_booklet_p21_data_05d739dbdf.png)\n\nSimilarly, $T$ can be reached from $A$ in $v^2$ ways. $L$ can be reached by two routes: $A \\to M \\to L$ and $A \\to N \\to L$, each with $v^2$ ways, so $L$ can be reached in $2v^2$ ways.\n\nAfter the third movement, the spider can be at four different points, approached by $\\binom{3}{3} \\cdot v^3$, $\\binom{3}{2} \\cdot v^3$, $\\binom{3}{1} \\cdot v^3$, and ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19396,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 與 $m$ 為正整數。托兒所的老師用 $n \\times m$ 塊正方形巧拼，排成一個 $n \\times m$ 的長方形區域。每塊巧拼上有一個嬰兒，各自面向正方形的某一個邊。老師一拍手，所有嬰兒就同時往其面對的方向前進一塊巧拼，之後所有嬰兒原地順時鐘轉 $90^\\circ$。若一個嬰兒爬出 $n \\times m$ 的長方形區域，則該嬰兒大哭。若兩個嬰兒前進到相同的巧拼上，他們會對撞然後一起大哭。\n\n已知老師可以適當安排一開始每個嬰兒的面向，使得不論老師拍多少次手，都沒有嬰兒大哭。試求 $n$ 與 $m$ 的所有可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "答案是所有偶數 $n$ 與 $m$。顯然所有 $2k \\times 2h$ 的情形都可以用 $k \\times h$ 個如下的 $2 \\times 2$ 嬰兒陣拼出來：\n\n![](images/2023-TWNIMO-Problems_p5_data_838512b8c5.png)\n\n現在證明只有 $n$ 與 $m$ 皆為偶數時可滿足題意。讓我們將直排依序編號為 $1, 2, \\ldots, n$，橫列依序編號為 $1, 2, \\ldots, m$，並將第 $i$ 排 $j$ 列的巧拼標記為 $(i, j)$。讓我們考慮所有 $i$ 和 $j$ 都是奇數的“奇”格，以及所有 $i$ 和 $j$ 都是偶數的“偶”格。注意到不論起始嬰兒的面向如何，在老師拍完兩次手後，所有奇格的嬰兒都會出現在偶格，所有偶格的嬰兒都會出現在奇格，且不會有原本在非奇偶格的嬰兒出現在奇偶格。這表示奇格與偶格的數量必須相同（否則一定會有嬰兒對撞）。然而，當 $n$ 與 $m$ 不同為偶數時，奇格與偶格的數量必不相同。因此，只有 $n$ 與 $m$ 皆為偶數時可滿足題意。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19397,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be pairwise coprime positive odd numbers. For positive integers $n$, define\n\n$$\nf(n) = \\lfloor \\frac{n}{a} \\rfloor + \\lfloor \\frac{n}{b} \\rfloor + \\lfloor \\frac{n}{c} \\rfloor + \\lfloor \\frac{n}{d} \\rfloor.\n$$\n\nProve that\n\n$$\n\\sum_{n=1}^{abcd} (-1)^{f(n)} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $a$, $b$, $c$, $d$ are all odd, we have\n\n$$\n\\begin{aligned}\nf(n) &\\equiv 4n - \\left(a\\lfloor\\frac{n}{a}\\rfloor + b\\lfloor\\frac{n}{b}\\rfloor + c\\lfloor\\frac{n}{c}\\rfloor + d\\lfloor\\frac{n}{d}\\rfloor\\right) \\\\\n&\\equiv \\left(n - a\\lfloor\\frac{n}{a}\\rfloor\\right) + \\left(n - b\\lfloor\\frac{n}{b}\\rfloor\\right) + \\left(n - c\\lfloor\\frac{n}{c}\\rfloor\\right) + \\left(n - d\\lfloor\\frac{n}{d}\\rfloor\\right) \\pmod{2}.\n\\end{aligned}\n$$\n\nWe can observe that $n - a\\lfloor\\frac{n}{a}\\rfloor$ is the remainder of $n$ divided by $a$, denoted as $(n \\mod a)$. By the Chinese remainder theorem, the map\n\n$$\n\\Phi : \\{1, \\dots, abcd\\} \\to \\{0, \\dots, a-1\\} \\times \\{0, \\dots, b-1\\} \\times \\{0, \\dots, c-1\\} \\times \\{0, \\dots, d-1\\},\n$$\n\nwhere $\\Phi(n) = (n \\mod a, n \\mod b, n \\mod c, n \\mod d)$, is a one-to-one correspondence. Thus, we can replace our sum over $n$ with a sum over $(x, y, z, w) = \\Phi(n)$ as $(x, y, z, w)$ ranges over all elements of $\\{0, \\dots, a-1\\} \\times \\{0, \\dots, b-1\\} \\times \\{0, \\dots, c-1\\} \\times \\{0, \\dots, d-1\\}$. If $\\Phi(n) = (x, y, z, w)$, then $f(n) \\equiv x + y + z + w \\pmod{2}$, so it follows that\n\n$$\n\\begin{aligned}\n\\sum_{n=1}^{abcd} (-1)^{f(n)} &= \\sum_{x=0}^{a-1} \\sum_{y=0}^{b-1} \\sum_{z=0}^{c-1} \\sum_{w=0}^{d-1} (-1)^{x+y+z+w} \\\\\n&= \\left(\\sum_{x=0}^{a-1} (-1)^x\\right) \\left(\\sum_{y=0}^{b-1} (-1)^y\\right) \\left(\\sum_{z=0}^{c-1} (-1)^z\\right) \\left(\\sum_{w=0}^{d-1} (-1)^w\\right) \\\\\n&= 1 \\cdot 1 \\cdot 1 \\cdot 1 = 1.\n\\end{aligned}\n$$\n\n![](images/Korea_2024_Booklet_p4_data_5683407f31.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19398,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a non-isosceles triangle with centroid $G$ and inscribed in circle $(O)$. Let $M, N, P$ be the midpoints of $BC, CA, AB$ respectively. Let $D, E, F$ be the reflection points of the foot of the internal bisector of angles $A, B, C$ through $M, N, P$ respectively. Prove that the orthocenter of triangle $DEF$ lies on line $OG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $EF$ intersects $BC$ at $D'$. Let $(\\omega_A)$ be the circle with diameter $DD'$. By the angle bisector theorem, one can get\n$$\n\\frac{DB}{DC} = \\frac{AC}{AB}, \\quad \\frac{EC}{EA} = \\frac{AB}{BC}, \\quad \\frac{FA}{FB} = \\frac{BC}{CA}.\n$$\nBased on Ceva's theorem, $AD, BE, CF$ are concurrent and $(DD', BC) = -1$. If $AU, AV$ are the internal and external angle bisectors of triangle $ABC$, then $(UV, BC) = -1$.\n\n![](images/Saudi_Booklet_2025_p18_data_69f72eaef6.png)\n\nThrough the reflection over the center $M$, then $(DV', CB) = -1$ with $V'$ being the symmetric point to $V$ through $M$. From there, $D' \\equiv V'$, so $(\\omega_A)$ is the reflection of the $A$-Apollonius circle through the perpendicular bisector of $BC$.\n\nSuppose that the $A$-Apollonius circle intersects $(O)$ at $K$, then $\\frac{KB}{KC} = \\frac{AB}{AC}$ so $ABKC$ is harmonic, entailing $AK$ is the symmetric line of triangle $ABC$. Therefore, $(\\omega_A)$ intersects $(O)$ at $A'$, $R$ will be the points symmetric to $A, K$ through the perpendicular bisector $BC$. Therefore, $KR \\parallel BC$ so $AK, AR$ are isogonal in angle $A$, which implies that $AR$ is the median of triangle $ABC$ or $R \\in AM$. Suppose $AR$ intersects $(\\omega_A)$ at $S$, then by the property of power of a point and Newton's formula, we have\n$$\n\\overline{MR} \\cdot \\overline{MS} = \\overline{MD} \\cdot \\overline{MD'} = MB^2 = -\\overline{MB} \\cdot \\overline{MC} = -\\overline{MR} \\cdot \\overline{MA}.\n$$\nFrom there, we deduce $\\overline{MS} = -\\overline{MA}$ or $M$ is the midpoint of $AS$. Here we have\n$$\n\\frac{\\mathcal{P}_{G/(O)}}{\\mathcal{P}_{G/(\\omega_A)}} = \\frac{\\overline{GR} \\cdot \\overline{GA}}{\\overline{GR} \\cdot \\overline{GS}} = \\frac{\\overline{GA}}{\\overline{GS}} = \\frac{2/3 \\cdot \\overline{MA}}{4/3 \\cdot \\overline{MS}} = -\\frac{1}{2}.\n$$\nTherefore $\\mathcal{P}_{G/(\\omega_A)} = -2\\mathcal{P}_{G/(O)}$ is equal between vertices $A, B, C$. Therefore we have $G$ has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$.\n\nOn the other hand, $OA$ is tangent to the Apollonius circle at vertex $A$ so $OA'$ is tangent to $(\\omega_A)$. From there we have\n$$\n\\mathcal{P}_{O/(\\omega_A)} = OA'^2 = R^2\n$$\nwith $R$ being the radius of $(O)$. Also because of this symmetry, $G$ has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$. Let $X$ be the projection of $D$ onto $EF$ then $X \\in DH$ and $X \\in (\\omega_A)$. Then, $\\mathcal{P}_{H/(\\omega_A)} = \\overline{HX} \\cdot \\overline{HD}$. Obviously, this quantity is also symmetric between the vertices $D, E, F$ in triangle $DEF$, so $H$ also has the same power to the three circles $(\\omega_A), (\\omega_B), (\\omega_C)$. From that we immediately have $H \\in OG$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19399,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x, y)$ with real coefficients such that for all real numbers $x$ and $y$,\n$$\nP(x + y, x - y) = 2P(x, y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the given equality,\n\n$$\nP(2x, 2y) = P((x+y)+(x-y), (x+y)-(x-y)) = 2P(x+y, x-y) = 4P(x, y).\n$$\n\nAs this holds for all real $x$ and $y$, equate the coefficients of $x^i y^j$ in both sides. The coefficient in the left is $2^{i+j} a_{i,j}$, and in the right is $4a_{i,j}$. Thus, $(2^{i+j} - 4)a_{i,j} = 0$, so either $i + j = 2$ or $a_{i,j} = 0$.\n\nTherefore, $P(x, y)$ only has terms where the sum of exponents is $2$: $P(x, y) = c x^2 + a x y + b y^2$.\n\nSubstitute into the original equation:\n\n$$\n(c + a + b)x^2 + 2(c - b)xy + (c - a + b)y^2 = 2c x^2 + 2a x y + 2b y^2.\n$$\n\nMatching coefficients gives $c + a + b = 2c$ so $c = a + b$. Thus, all polynomials of the form\n\n$$\nP(x, y) = (a + b)x^2 + a x y + b y^2\n$$\nwhere $a, b \\in \\mathbb{R}$ satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19400,
    "subject": "Mathematics (Olympiad)",
    "question": "A teacher wants to spend a morning practicing Olympiad problems with her class in teams. She has set up six large tables, each with a problem. There are three different problems, each appearing on two tables. There are three rounds, and in each round, each student sits at one table. The teacher wants to schedule the students so that after three rounds, each student has worked on all three problems once, and each pair of students sits at the same table at most once.\n\nWhat is the maximum number of students for which such a schedule is possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose problem $A$ is on tables $A_1$ and $A_2$, problem $B$ is on tables $B_1$ and $B_2$, and problem $C$ is on tables $C_1$ and $C_2$.\n\nConsider the students sitting at table $A_1$ in the first round. In the second round, they must go to a table that does not have problem $A$ (i.e., $B_1$, $B_2$, $C_1$, or $C_2$). Moreover, these students all have to go to different tables, so there cannot be more than four students at table $A_1$ in the first round. This applies to all six tables, so no more than $4 \\times 6 = 24$ students can participate.\n\nTo show that 24 students can indeed be scheduled, here is a possible arrangement for the three rounds:\n\n![](path/to/file.png)\n\nOne method to construct such a schedule is to first distribute students 1 to 8, who do task $A$ in the first round, to the other four tables in the second and third rounds. Then, use the same distribution (with tasks swapped) for students 9 to 16, who start with task $B$. Finally, do the same for students 17 to 24, who start with task $C$, filling the remaining spots.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19401,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $n$ be positive integers with $p \\geq 2$, and let $a$ be a real number such that $1 \\leq a < a + n \\leq p$. Prove that the set\n$$\n\\{ \\lfloor \\log_2 x \\rfloor + \\lfloor \\log_3 x \\rfloor + \\dots + \\lfloor \\log_p x \\rfloor \\mid x \\in \\mathbb{R},\\ a \\leq x \\leq a + n \\}\n$$\nhas exactly $n + 1$ elements.",
    "options": [],
    "answer": "See 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 $k \\geq 2$ 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.\n\nOn the other hand, for $s \\in S$, $s < \\lfloor a \\rfloor + n \\leq p$, we have $s+1 \\in \\{2, 3, \\dots, p\\}$, and\n$$\nf(s+1)-f(s) = \\sum_{k=2}^{p} \\left( \\lfloor \\log_k(s+1) \\rfloor - \\lfloor \\log_k s \\rfloor \\right) \\geq \\lfloor \\log_{s+1}(s+1) \\rfloor - \\lfloor \\log_{s+1} s \\rfloor = 1,\n$$\ntherefore $f(s+1) > f(s)$, and this proves that $M$ has exactly $n+1$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19402,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Point $D$ is the midpoint of side $AC$, and point $M$ lies on segment $BD$ with $DM = 2BM$. Ray $AM$ meets side $BC$ at $E$, and ray $CM$ meets side $BA$ at $F$. Ray $FE$ intersects $\\omega$ at $N$. Suppose that $AM \\perp CM$. Prove that $ADEF$ is cyclic if and only if line $AN$ bisects segment $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $a, b, c$ the side lengths, and by $m_a, m_b, m_c$ the lengths of the medians of triangle $ABC$. Since $MD$ is a median in the right-angled triangle $AMC$, it follows that\n\n$$\n2m_b/3 = MD = AD = CD = b/2,\n$$\n\nso $m_b = 3b/4$, which means that\n\n$$\n\\left(\\frac{3b}{4}\\right)^2 = m_b^2 = \\frac{a^2 + c^2}{2} - \\frac{b^2}{4}.\n$$\n\nThis is equivalent to $13b^2 = 8(a^2 + c^2)$.\n\n![](images/Saudi_Arabia_booklet_2012_p47_data_2f5080e712.png)\n\nNext, apply the Menelaus theorem to get\n\n$$\n\\frac{EC}{EB} = 4 = \\frac{FA}{FB}\n$$\n\nand deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrilateral $AFED$ is therefore a trapezoid; it is cyclic if and only if $AF = DE$.\n\nWe now express the two lengths in terms of $a, b,$ and $c$.\n\nRecall that $FA/FB = 4$ to obtain $AF = \\frac{4c}{5}$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = \\frac{b^2}{2} - \\frac{4a^2}{25}$. By the preceding, the quadrilateral $AFED$ 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 $a$:\n\n$$\nb = \\frac{2a\\sqrt{2}}{3}, \\quad c = \\frac{2a}{3}.\n$$\n\nFinally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that\n\n$$\nEN = \\frac{a}{2} - \\frac{a}{5} = \\frac{3a}{10},\n$$\n\nand that the triangles $ANC$ and $PNE$ are similar. Then we obtain\n\n$$\nNP = \\frac{3m_a}{5},\n$$\n\nso\n\n$$\nNA \\cdot NP = \\frac{3m_a^2}{5} = \\frac{3(2(b^2 + c^2) - a^2)}{20} = \\frac{a^2}{4} = NB \\cdot NC.\n$$\n\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19403,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be two integers greater than $3$ and consider a rectangular $m \\times n$ board. At each step, one puts simultaneously $4$ marbles into $4$ cells of the board (each marble into a cell) so that these four cells form one of the following schemata:\n\n![](images/Vijetnam_2006_p4_data_65e88c3c2a.png \"Figure 2: The schemata.\")\n\nIs it true that, starting from a rectangular $m \\times n$ board without marbles, after a finite number of steps of appropriate puttings, one can put marbles into all cells of the board so that each cell is filled with the same (positive) number of marbles for all cells when:\n\ni) $m = 2004$ and $n = 2006$?\n\nii) $m = 2005$ and $n = 2006$?\n\n(At each step, it is not necessary that the four cells which are selected to put marbles into contained no marbles.)",
    "options": [],
    "answer": "See solution",
    "solution": "i) After two steps, one can put a marble into each cell of a small board of size $4 \\times 2$. One can partition the given board of size $2004 \\times 2006$ into small boards of size $4 \\times 2$. Therefore, after some steps, one can put marbles into all cells of the given board so that the number of marbles in each cell is the same for all cells.\n\nii) We now prove by contradiction that in the second case, the answer is \"no\". Suppose, on the contrary, that after some steps, there would be $k$ marbles ($k > 0$) in each cell of the given board of size $2005 \\times 2006$. Color black all cells belonging to the odd rows and consider each non-colored cell as white. Then, the number of black cells is $1003 \\times 2006$ and the number of white cells is $1002 \\times 2006$. But at each step we put exactly $2$ marbles into black cells and $2$ marbles into white cells. Therefore, after any number of steps, the total number of marbles in all black cells must be equal to the total number in all white cells. Consequently, we would have $1003 \\times 2006 \\times k = 1002 \\times 2006 \\times k$, and so $1 = 0$. This contradiction proves our assertion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19404,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma$ be an acute-angled triangle with $AB \\leq A\\Gamma$ and $c(O, R)$ its circumcircle. The line perpendicular from $A$ to the tangent of $c(O, R)$ at $\\Gamma$ intersects it at $\\Delta$.\n\n(a) If the triangle $AB\\Gamma$ is isosceles with $AB = A\\Gamma$, prove that $\\Gamma\\Delta = \\dfrac{B\\Gamma}{2}$.\n\n(b) If $\\Gamma\\Delta = \\dfrac{B\\Gamma}{2}$, prove that the triangle $AB\\Gamma$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) If $Z$ is the midpoint of $B\\Gamma$, then $AZ$ is the altitude and median of the isosceles triangle $AB\\Gamma$. The right-angled triangles $A\\Gamma Z$ and $A\\Gamma\\Delta$ are equal, because they have $A\\Gamma$ as a common hypotenuse and $A\\Gamma\\Delta = A\\Gamma Z$ (the last equality arises from the equalities $A\\Gamma\\Delta = A\\beta\\Gamma$ (chord-tangent angle and the corresponding inscribed angle) and $A\\beta\\Gamma = A\\Gamma Z$, since $AB = A\\Gamma$). Hence we get:\n\n$$\n\\Gamma\\Delta = \\Gamma Z = \\frac{B\\Gamma}{2}.\n$$\n\n![](images/Greek2015_booklet_p2_data_75cbdc4793.png)\n\n(b) Let $AB < A\\Gamma$. We consider the perpendicular bisector of $B\\Gamma$ ($M$ is the midpoint) which intersects the extension of $AB$ at $E$. Then $A\\beta\\Gamma = A\\Gamma\\Delta$ and from the hypothesis we have $2\\Gamma\\Delta = B\\Gamma = 2BM$. Hence $\\Gamma\\Delta = BM$. Therefore, the right-angled triangles $EBM$ and $A\\Gamma\\Delta$ are equal, and so $A\\Gamma = EB$. Since $EB = E\\Gamma$, we conclude $A\\Gamma = E\\Gamma$. However, this is absurd, because $E A \\Gamma = 180^\\circ - A$ is an obtuse angle and so the triangle $E A \\Gamma$ would have two obtuse angles. Hence $AB = A\\Gamma$ and the triangle $AB\\Gamma$ is isosceles.\n\n![](images/Greek2015_booklet_p2_data_6c90bc484e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19405,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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 $p$ 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.\n\n*Remark.* Since $S$ contains all primes congruent to $\\pm 1 \\pmod{5}$, it must contain $31$. Although there is no integer $a$ such that $a^2 + a - 1 = 31$, the latter may be written in the form $(12^2 + 12 - 1)/(2^2 + 2 - 1)$ which explicitly exhibits $31$ as a member of $S$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19406,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We use induction on $n$.\n\nFor $n = 2$, the relation is an equality.\n\nWhen going from $n-1$ to $n$, the right side increases by\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": 19407,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, with $\\angle A < \\angle B$ and $\\angle A < \\angle C$, let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP = PD$ and $CP = PE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point other than $A$.\n\n![](images/pamphlet1112_main_p51_data_3635fe8904.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the fixed point is the orthocenter $H$ of triangle $ABC$.\n\nLet $X$ be the foot of the perpendicular from $C$ to $AB$, and let $Y$ be the foot of the perpendicular from $B$ to $AC$. Note that if $M$ is the midpoint of $BC$, then $MB = MX = MY = MC$. Suppose without loss of generality that $P$ is between $B$ and $M$. Then $D$ is between $B$ and $X$, and $E$ is between $A$ and $Y$.\n\nThe quadrilateral $AXHY$ is cyclic, since $\\angle AXH = \\angle AYH = 90^\\circ$. To show that $ADHE$ is also cyclic, it suffices to show that $\\triangle DHX \\sim \\triangle EHY$, or that\n\n$$\n\\frac{DX}{EY} = \\frac{XH}{YH}.\n$$\n\nApplying the Law of Sines in the cyclic quadrilateral $AXHY$, we find that $\\frac{XH}{YH} = \\frac{\\cos B}{\\cos C}$. Note that $DX = BX - BD = BC \\cos B - 2BP \\cos B$. Similarly, $EY = EC - CY = 2PC \\cos C - BC \\cos C$. Hence,\n\n$$\n\\frac{DX}{EY} = \\frac{BC - 2BP}{2PC - BC} \\cdot \\frac{\\cos B}{\\cos C} = \\frac{\\cos B}{\\cos C}.\n$$\n\nPutting this together with our previous computation yields the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19408,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為平行四邊形，其中 $AC = BC$。在直線 $AB$ 上取一點 $P$，使得 $B$ 介於 $A, P$ 之間。設三角形 $ACD$ 的外接圓與線段 $PD$ 再交於點 $Q$，而三角形 $APQ$ 的外接圓與線段 $PC$ 再交於點 $R$。\n\n證明：三直線 $CD$、$AQ$、$BR$ 共點。",
    "options": [],
    "answer": "See solution",
    "solution": "在以下的各種解法中，都用到以下的事實。\n\n因為 $AC = BC = AD$，所以 $\\angle ABC = \\angle BAC = \\angle ACD = \\angle ADC$。由於四邊形 $APRQ$、$AQCD$ 都有外接圓，可得\n\n$$\n\\angle CRA = 180^\\circ - \\angle ARP = 180^\\circ - \\angle AQP = \\angle DQA = \\angle DCA = \\angle CBA,\n$$\n\n故 $A, B, C, R$ 四點共一圓，記該圓為 $\\gamma$。\n\n![](images/2022-TWNIMO-Problems_p29_data_6418fa70e2.png)\n\n**解一** 令點 $X$ 為 $AQ$ 與 $CD$ 的交點。原題等價於證明 $B, R, X$ 三點共線。\n\n在圓 $(APRQ)$ 上有\n\n$$\n\\angle RQX = 180^\\circ - \\angle AQR = \\angle RPA = \\angle RCX\n$$\n\n（其中最後一個等號來自 $AB \\parallel CD$），所以 $C, Q, R, X$ 四點共一圓，記為圓 $\\delta$。\n\n利用圓 $\\gamma$、$\\delta$ 可知\n\n$$\n\\angle XRC = \\angle XQC = 180^\\circ - \\angle CQA = \\angle ADC = \\angle BAC = 180^\\circ - \\angle CRB,\n$$\n\n故得證。\n\n**解二** 記圓 $(APRQ)$ 為 $\\alpha$。由於\n\n$$\n\\angle CAP = \\angle ACD = \\angle AQD = 180^\\circ - \\angle AQP,\n$$\n\n知直線 $AC$ 與圓 $\\alpha$ 相切。\n\n![](images/2022-TWNIMO-Problems_p30_data_2ce80cada1.png)\n\n令直線 $AD$ 與 $\\alpha$ 再交於點 $Y$，此點必位於射線 $DA$ 上 $A$ 的後方。利用圓 $\\gamma$ 及 $AC$ 切 $\\alpha$ 的性質，得\n\n$$\n\\angle ARY = \\angle CAD = \\angle ACB = \\angle ARB,\n$$\n\n所以 $Y, B, R$ 三點共線。\n\n在六邊形 $AAYRPQ$ 上使用 Pascal 定理（這裡 $AA$ 指的是圓 $\\alpha$ 過 $A$ 的切線），得三組直線的交點\n\n$$\nAA \\cap RP = C, \\quad AY \\cap PQ = D, \\quad YR \\cap QA\n$$\n\n共線。由此直線 $CD$、$AQ$ 與 $BR$ 共點。\n\n**解三** 同解一，令 $X = AQ \\cap CD$，以下證明 $B, R, X$ 共線。同解二，令 $\\alpha = (APRQ)$，但這裡定義點 $Y$ 為直線 $BR$ 與 $\\alpha$ 的第二個交點。\n\n利用圓 $\\alpha$，並注意到直線 $CD$ 與 $\\gamma$ 相切，得\n\n$$\n\\angle RYA = \\angle RPA = \\angle RCX = \\angle RBC. \\tag{1}\n$$\n\n於是 $AY \\parallel BC$，而得 $Y$ 落在直線 $DA$ 上。\n\n由 (1) 亦得 $\\angle RYD = \\angle RCX$，所以 $C, D, Y, R$ 四點共一圓 $\\beta$。因此直線 $CD$、$AQ$、$YBR$ 為三圓 $(AQCD)$、$\\alpha$、$\\beta$ 兩兩的根軸，故它們必定共點。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19409,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $w_a$, $w_b$, $w_c$ be the lengths of the internal angle bisectors of a triangle $ABC$ with sides $a$, $b$, $c$. Let $R$ be its circumradius. Prove that\n\n$$\n\\frac{b^2 + c^2}{w_a} + \\frac{c^2 + a^2}{w_b} + \\frac{a^2 + b^2}{w_c} > 4R.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use the standard formula:\n\n$$\nw_a = \\frac{2bc \\cos(A/2)}{b+c}, \\text{ etc.}\n$$\n\nThe inequality takes the form\n\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c)}{4Rbc \\cos(A/2)} > 2.\n$$\n\nThis may be put in the form\n\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c) \\sin(A/2)}{2abc} > 1.\n$$\n\nBut note that $b^2 + c^2 \\ge 2bc$ and $b + c > a$. Hence it is sufficient to prove that $\\sum_{\\text{cyclic}} \\sin(A/2) > 1$. We start with the identity\n\n$$\n\\sum_{\\text{cyclic}} \\cos A = 1 + 4 \\prod_{\\text{cyclic}} \\sin(A/2),\n$$\n\nwhich shows that $\\sum_{\\text{cyclic}} \\cos A > 1$ in any triangle $ABC$. Whenever $A$, $B$, $C$ are the angles of a triangle, $(\\pi - A)/2$, $(\\pi - B)/2$, $(\\pi - C)/2$ are also the angles of some other triangle. For this triangle, we get\n\n$$\n\\sum_{\\text{cyclic}} \\cos \\left( \\frac{\\pi - A}{2} \\right) > 1.\n$$\n\nIt follows that\n\n$$\n\\sum_{\\text{cyclic}} \\sin \\frac{A}{2} > 1,\n$$\n\nwhich is to be proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19410,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all natural $n \\ge 2$, the following number is composite:\n\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the numerator:\n\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1001}(n^2 + 1) + n^{1002} + 1 \\\\\n&= n^{1001}(n^2 + 1) + (n^2 + 1)(n^{1000} - n^{998} + n^{996} - \\dots - n^2 + 1)\n\\end{aligned}\n$$\n\nThus, the numerator is divisible by $n^2 + 1$. We can also write:\n\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1002}(n + 1) + n^{1001} + 1 \\\\\n&= n^{1002}(n + 1) + (n + 1)(n^{1000} - n^{998} + n^{996} - \\dots - n + 1)\n\\end{aligned}\n$$\n\nSo the numerator is divisible by $n + 1$ as well. Therefore,\n\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1} = (n^2 + 1)P(n)\n$$\n\nfor some polynomial $P(n)$, and $P(n) > 1$ for $n \\ge 2$. Thus, the expression is composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19411,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an $8 \\times 8$ chessboard with all 64 unit squares initially white. We color 12 unit squares arbitrarily black. Prove that it is possible to find four rows and four columns that together contain all 12 black unit squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1, x_2, \\ldots, x_8$ be the numbers of black squares in each row, ordered so that $x_1 \\geq x_2 \\geq \\cdots \\geq x_8$. From the problem, we have:\n\n$$\nx_1 + x_2 + \\cdots + x_8 = 12. \\quad (1)\n$$\n\nIt is impossible for all $x_i$ to be equal, since $8x_1 = 12$ has no integer solution. Thus,\n\n$$\nx_1 + x_2 + x_3 + x_4 > x_5 + x_6 + x_7 + x_8.\n$$\n\nSuppose $x_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 1$. Then (1) gives $2(x_1 + x_2 + x_3 + x_4) - 1 = 12$, which is impossible since the left side is odd and the right side is even. Similarly, if $x_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 2$, then $x_1 + x_2 + x_3 + x_4 = 7$ and $x_5 + x_6 + x_7 + x_8 = 5$. But $4x_4 \\leq 7 \\implies x_4 \\leq 1$, so $x_5, x_6, x_7, x_8 \\leq 1$ and $x_5 + x_6 + x_7 + x_8 \\leq 4$, contradicting the sum 5.\n\nTherefore,\n\n$$\nx_1 + x_2 + x_3 + x_4 \\geq x_5 + x_6 + x_7 + x_8 + 3.\n$$\n\nFrom (1),\n\n$$\n2(x_1 + x_2 + x_3 + x_4) - 3 \\geq 12 \\implies x_1 + x_2 + x_3 + x_4 \\geq 8.\n$$\n\nThus, the four rows with the most black squares contain at least 8 black squares. The remaining at most 4 black squares can be covered by selecting the columns in which they lie. Therefore, four rows and four columns suffice to cover all 12 black squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19412,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that the diagonals $AC$ and $BD$ are perpendicular and intersect at point $E$. Point $F$ is on side $AD$, and the ray $FE$ meets the circumscribed circle of $ABCD$ at point $P$. Point $Q$ is on segment $PE$ such that $PQ \\cdot PF = PE^2$. The line through $Q$ perpendicular to $AD$ meets $BC$ at $R$. Show that $RP = RQ$.\n\n![](images/2023_CGMO_p5_data_63f86a982a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $EX \\parallel AF$, intersecting $AP$ at $X$ and $DP$ at $Y$. Extend $XQ$ and $YQ$ to intersect $BC$ at $S$ and $T$, respectively.\n\nFrom $\\frac{PQ}{PE} = \\frac{PE}{PF} = \\frac{PX}{PA}$, we get $XQ \\parallel AE$; similarly, $YQ \\parallel DE$.\n\nSince $\\angle EXS = \\angle AEX = \\angle DAC = \\angle EBS$, we have $X, E, S, B$ concyclic.\n\nFrom $\\angle PXE = \\angle PAD = \\angle PBE$, we find $X, E, P, B$ are concyclic, hence $X, E, S, P, B$ are concyclic. Similarly, $Y, E, T, P, C$ are concyclic.\n\nSince $\\angle PQT = \\angle PEB = \\angle PST$, we have $P, S, Q, T$ concyclic. Given $SQ \\parallel CE$, $TQ \\parallel BE$, and $CE \\perp BE$, we get $SQ \\perp TQ$.\n\nSince $\\angle RQS = 90^\\circ - \\angle QXY = 90^\\circ - \\angle QTS = \\angle RSQ$, it is known that $R$ is the center of circle $\\odot PSQT$, thus $RP = RQ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19413,
    "subject": "Mathematics (Olympiad)",
    "question": "Balls numbered $1, 2, 3, \\ldots$ are deposited in $5$ bins, labeled A, B, C, D, and E, using the following procedure. Ball $1$ is deposited in bin A, and balls $2$ and $3$ are deposited in bin B. The next $3$ balls are deposited in bin C, the next $4$ in bin D, and so on, cycling back to bin A after balls are deposited in bin E. (For example, balls numbered $22, 23, \\ldots, 28$ are deposited in bin B at step $7$ of this process.)\n\nIn which bin is ball $2024$ deposited?\n\n(A) A (B) B (C) C (D) D (E) E",
    "options": [],
    "answer": "See solution",
    "solution": "After $n$ steps, a total of $1 + 2 + 3 + \\dots + n = \\frac{1}{2} n(n+1)$ balls have been deposited. In particular, after step $n = 63$, a total of $\\frac{1}{2} \\cdot 63 \\cdot 64 = 2016$ balls have been deposited. The next batch of 64 balls will include ball 2024. Because 64 has remainder 4 when divided by 5, ball 2024 is deposited in bin D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19414,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma$ be an isosceles triangle and a point $\\Delta$ in its interior such that $\\angle \\Delta B \\Gamma = 30^\\circ$, $\\angle \\Delta B A = 50^\\circ$, and $\\angle B \\Gamma \\Delta = 55^\\circ$.\n\n(a) Prove that $\\angle B = \\angle \\Gamma = 80^\\circ$.\n\n(b) Find the measure of the angle $\\angle \\Delta A \\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) We have $\\angle B = 50^\\circ + 30^\\circ = 80^\\circ$.\n\nSuppose that $\\angle A = \\angle B = 80^\\circ$. Then\n\n$$\n\\angle A + \\angle B + \\angle \\Gamma = 80^\\circ + 80^\\circ + \\angle \\Gamma > 160^\\circ + 55^\\circ = 215^\\circ,\n$$\n\nwhich is absurd.\n\nIf $A\\Gamma = \\frac{180^\\circ - 80^\\circ}{2} - 50^\\circ$, then $\\angle B \\Gamma \\Delta = 55^\\circ < \\angle \\Gamma = 50^\\circ$, which is also absurd.\n\nHence we have: $\\angle \\Delta \\Gamma A = 80^\\circ - 55^\\circ = 25^\\circ$ (1).\n\n(b) Let $AZ$ be the bisector of the angle $\\angle A$. Then $AZ$ is the height and median of triangle $AB\\Gamma$. Let $AZ$ meet line $BD$ at point $E$. Since in triangle $B\\Gamma\\Delta$ we have $\\angle \\Gamma B \\Delta < \\angle B \\Gamma \\Delta$, it follows that $\\Delta \\Gamma < \\Delta B$. Hence $\\Delta$ lies in the half-plane with respect to $AZ$ containing point $\\Gamma$. This means that $E$ lies between points $B$ and $\\Delta$.\n\nSince $EB = E\\Gamma$, it follows that $\\angle E \\Gamma B = \\angle E B \\Gamma = 30^\\circ$ and hence\n\n![](images/Greece_2022_p4_data_1dd151e6a5.png)\n\n$$\n\\angle E \\Gamma \\Delta = 55^\\circ - 30^\\circ = 25^\\circ = \\angle \\Delta \\Gamma A. \\qquad (2)\n$$\n\nHence $\\Gamma \\Delta$ bisects the angle $\\angle E \\Gamma A$ of triangle $AE\\Gamma$. Moreover, for the external angles $\\angle \\Delta E \\Gamma$ and $\\angle \\Delta E A$ of triangles $EB\\Gamma$ and $EBA$, respectively, we have: $\\angle \\Delta E \\Gamma = 30^\\circ + 30^\\circ = 60^\\circ$ and $\\angle \\Delta E A = \\angle E B A + \\frac{\\angle A}{2} = 50^\\circ + 10^\\circ = 60^\\circ$.\n\nHence $\\angle \\Delta E \\Gamma = \\angle \\Delta E A = 60^\\circ$ and so $E\\Delta$ bisects the angle $\\angle A E \\Gamma$ of triangle $AE\\Gamma$.\n\nHence $\\Delta$ is the incenter of triangle $AE\\Gamma$ and $\\angle \\Delta A \\Gamma = \\frac{\\angle E A \\Gamma}{2} = \\frac{10^\\circ}{2} = 5^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19415,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that $\\left(\\frac{6}{5}\\right)^{\\sqrt{3}} > \\left(\\frac{5}{4}\\right)^{\\sqrt{2}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that if $x > -1$, $x \\neq 0$, and $\\alpha \\in (1, 2)$, then\n$$\nf(x) = (1+x)^\\alpha - 1 - \\alpha x - \\frac{\\alpha(\\alpha-1)}{2}x^2 - \\frac{\\alpha(\\alpha-1)(\\alpha-2)}{6}x^3 > 0.\n$$\nWe have\n$$\nf'(x) = \\alpha\\left[(1+x)^{\\alpha-1} - 1 - (\\alpha-1)x - \\frac{(\\alpha-1)(\\alpha-2)}{2}x^2\\right],\n$$\n$$\nf''(x) = \\alpha(\\alpha - 1)\\left[(1 + x)^{\\alpha - 2} - 1 - (\\alpha - 2)x\\right],\n$$\n$$\nf'''(x) = \\alpha(\\alpha - 1)(\\alpha - 2)\\left[(1 + x)^{\\alpha - 3} - 1\\right].\n$$\nSince $f'''(x) < 0$ for $x \\in (-1, 0)$ and $f'''(x) > 0$ for $x > 0$, we have $f''(x) > f'(0) = 0$ for $x > -1$, $x \\neq 0$. Then $f'(x) < f'(0) = 0$ for $x \\in (-1, 0)$ and $f'(x) > f'(0) > 0$ for $x > 0$, whence $f(x) > f(0) = 0$ for $x > -1$, $x \\neq 0$.\n\nNow, to prove $\\left(\\frac{6}{5}\\right)^{\\sqrt{3}} > \\left(\\frac{5}{4}\\right)^{\\sqrt{2}}$, set $x = \\frac{1}{5}$ and $\\alpha = \\sqrt{\\frac{3}{2}}$. According to the above, it is enough to check that\n$$\n\\alpha x + \\frac{\\alpha(\\alpha - 1)}{2}x^2 + \\frac{\\alpha(\\alpha - 1)(\\alpha - 2)}{6}x^3 > \\frac{1}{4}\n$$\nwhich reduces to\n$$\n\\frac{277}{1500}\\alpha + \\frac{3}{125} > \\frac{1}{4} \\iff \\alpha > \\frac{339}{277} \\iff 3.277^2 > 2.339^2 \\iff 230187 > 229842.\n$$\nThe last inequality is obvious, and this completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19416,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\geq 3$ and $n$ numbers $a_1, \\dots, a_n$, where $a_i \\in \\{0, 1\\}$ for all $i = 1, 2, \\dots, n$. Consider the following $n$ $n$-tuples:\n\n$$\nS_i = (a_i, a_{i+1}, \\dots, a_n, a_1, a_2, \\dots, a_{i-1}), \\quad \\forall i = 1, \\dots, n.\n$$\n\nFor each tuple $r = (b_1, b_2, \\dots, b_n)$, define\n\n$$\n\\omega(r) = b_1 \\cdot 2^{n-1} + b_2 \\cdot 2^{n-2} + \\dots + b_n.\n$$\n\nAssume that the numbers $\\omega(S_1), \\omega(S_2), \\dots, \\omega(S_n)$ attain exactly $k$ different values.\n\n**a)** Prove that $k \\mid n$ and $\\dfrac{2^n - 1}{2^k - 1} \\mid \\omega(S_i)$ for all $i = 1, \\dots, n$.\n\n**b)** Let $M = \\max_{i=1, n} \\omega(S_i)$ and $m = \\min_{i=1, n} \\omega(S_i)$. Prove that\n\n$$\nM - m \\geq \\frac{(2^n - 1)(2^k - 1)}{2^k - 1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) For every positive integer $d$, let $a_{n+d} = a_d$ and define $S_d = (a_d, a_{d+1}, \\dots, a_{d+n-1})$. By the binary representation, $\\omega(S_i) = \\omega(S_j)$ if and only if $S_i = S_j$.\n\nLet $t$ be the smallest number such that there exist $i$ and $j$ with $\\omega(S_i) = \\omega(S_j)$ and $j - i = t$. This implies that $(a_d)_{d=1}^{\\infty}$ is periodic with period $t$, so $(\\omega(S_d))_{d=1}^{\\infty}$ is also $t$-periodic. Since $\\omega(S_a) \\neq \\omega(S_b)$ for $1 \\leq a < b \\leq t$, $t$ is the minimal period, so $t = k$.\n\nSince $(a_d)_{d=1}^{\\infty}$ is $n$-periodic, $k = t \\mid n$ as required. Moreover, $S_i$ has the form $(a_i, \\dots, a_{i+k-1}, a_i, \\dots, a_{i+k-1}, \\dots)$, with the tuple $(a_i, \\dots, a_{i+k-1})$ repeated $n/k$ times. Thus,\n\n$$\n\\omega(S_i) = \\omega(a_i, \\dots, a_{i+k-1}) (2^{n-k} + 2^{n-2k} + \\dots + 2^0) = \\omega(a_i, \\dots, a_{i+k-1}) \\cdot \\frac{2^n - 1}{2^k - 1}.\n$$\n\nTherefore, $\\omega(S_i)$ is a multiple of $\\frac{2^n - 1}{2^k - 1}$ for all $1 \\leq i \\leq n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19417,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the acute $\\triangle ABC$ and point $D$ on the side $AB$. Let $P$ be the center of the circumscribed circle around $\\triangle ACD$, and $Q$ be the center of the circumscribed circle around $\\triangle BDC$. Prove that triangles $ABC$ and $DPQ$ are similar.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the radius of the circumscribed circle around $\\triangle ACD$ be $R_1$, and the radius around $\\triangle BDC$ be $R_2$. If $CD = l$, then by the law of sines:\n\n$$\n\\frac{l}{\\sin \\beta} = 2R_1 \\implies \\sin \\beta = \\frac{l}{2R_1}.\n$$\n\nAlso, $\\cos \\varphi = \\frac{DE}{DQ} = \\frac{l}{2R_1} = \\sin \\beta$.\n\nIn the acute triangle $\\triangle ABC$, $\\varphi = \\frac{\\pi}{2} - \\beta$, and similarly $\\psi = \\frac{\\pi}{2} - \\alpha$.\n\nTherefore,\n$$\n\\angle PDQ = \\varphi + \\psi = \\frac{\\pi}{2} - \\alpha + \\frac{\\pi}{2} - \\beta = \\pi - \\alpha - \\beta = \\gamma = \\angle ACB.\n$$\n\nSo, the triangles have equal angles.\n\n![](images/Ukraine_booklet_2018_p50_data_d44cbc6276.png)\n\nMoreover,\n$$\n\\frac{QD}{PD} = \\frac{R_1}{R_2} = \\frac{l}{2\\sin\\beta} \\cdot \\frac{2\\sin\\alpha}{l} = \\frac{\\sin\\alpha}{\\sin\\beta} = \\frac{a}{b} = \\frac{CB}{AC}\n$$\n\nThus, the sides about the equal angles of our triangles are proportional, so triangles $ABC$ and $DPQ$ are similar for any point $D \\in AB$. Points $A$ and $B$ do not satisfy the condition, otherwise one of $\\triangle ACD$ or $\\triangle BDC$ degenerates to a line segment.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19418,
    "subject": "Mathematics (Olympiad)",
    "question": "From an $n \\times n$ square, where $n \\ge 2$, the unit squares situated on both odd-numbered rows and odd-numbered columns are removed. Determine the minimum number of rectangular tiles needed to entirely cover the remaining region without any overlap.",
    "options": [],
    "answer": "See solution",
    "solution": "Color the initial board like a chessboard so that the top-left unit square is black. Label the rows and columns starting from this square; all removed squares are black. Each rectangular tile used must have breadth 1, so it covers at most one more white square than black squares.\n\nSuppose $n$ is odd, $n = 2m+1$ for $m \\in \\mathbb{N}$. The remaining surface has $2m^2 + 2m$ white squares and $m^2$ black squares, so at least $(2m^2 + 2m) - m^2 = m^2 + 2m$ tiles are required. This is sufficient: place an $n \\times 1$ tile over each even-numbered row and fill the rest with $1 \\times 1$ tiles, totaling $m + (m+1)m = m^2 + 2m$ tiles.\n\nFor $n = 2m$, the remaining surface has $2m^2$ white squares and $m^2$ black squares, so at least $2m^2 - m^2 = m^2$ tiles are required. However, tiling with $m^2$ tiles is impossible, since each tile would need both ends on white squares, but the bottom-right unit square is black and must be at the end of a tile.\n\nTherefore, at least $m^2 + 1$ tiles are needed. This minimum can be achieved by first covering the last row with an $n \\times 1$ tile, then the remaining part of the last column with a $1 \\times (n-1)$ tile. The remaining $(n-1) \\times (n-1)$ part can be tiled as in the previous case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19419,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(x, y)$ of integers such that\n\n$$\n1 + 2^x + 2^{2x+1} = y^2.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:**\n\nRewrite the equation as\n\n$$\n2^x(1 + 2^{x+1}) = y^2 - 1 = (y - 1)(y + 1).\n$$\n\nSince $\\gcd(y - 1, y + 1) = 2$, exactly one of them is divisible by $4$. Thus, $x \\geq 3$, and one of $y - 1$ or $y + 1$ is divisible by $2^{x-1}$ but not by $2^x$. Therefore, we can write\n\n$$\ny = 2^{x-1}m + \\epsilon, \\qquad (\\dagger)\n$$\n\nwhere $m$ is odd and $\\epsilon = \\pm 1$. Plugging this into the original equation gives\n\n$$\n2^x(1 + 2^{x+1}) = (2^{x-1}m + \\epsilon)^2 - 1 = 2^{2x-2}m^2 + 2^x m\\epsilon,\n$$\n\nso\n\n$$\n1 + 2^{x+1} = 2^{x-2}m^2 + m\\epsilon.\n$$\n\nThis leads to\n\n$$\n1 - m\\epsilon = 2^{x-2}(m^2 - 8). \\qquad (\\ddagger)\n$$\n\nIf $\\epsilon = 1$, then $m^2 - 8 \\leq 0$, so $m = 1$, which does not satisfy (\\ddagger). Thus $\\epsilon = -1$, so (\\ddagger) becomes\n\n$$\n1 + m = 2^{x-2}(m^2 - 8) \\geq 2(m^2 - 8),\n$$\n\nimplying $2m^2 - m - 17 \\leq 0$. Hence $m \\leq 3$. Also, $m \\neq 1$ by (\\dagger). Since $m$ is odd, $m = 3$, which leads to $x = 4$ by (\\dagger). Substituting these into (\\dagger) yields $y = 23$, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19420,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $xy$ given that\n$$\n1 + \\cos^2(2007x + 2008y - 1) = (x - y + 1) + \\frac{1}{x - y + 1},\n$$\nwhere $x$ and $y$ are real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the left-hand side is positive. Rewriting the right-hand side as\n$$\n(x - y + 1) + \\frac{1}{x - y + 1},\n$$\nwe see $x - y + 1 > 0$. By the AM-GM inequality,\n$$\n(x - y + 1) + \\frac{1}{x - y + 1} \\ge 2.\n$$\nThus, $1 + \\cos^2(2007x + 2008y - 1) \\le 2$, so equality holds, and\n$$\nx - y + 1 = 1 \\quad \\text{and} \\quad \\cos(2007x + 2008y - 1) = \\pm 1.\n$$\nThe first gives $x = y$. The second gives $2007x + 2008y - 1 = k\\pi$ for some $k \\in \\mathbb{Z}$, so\n$$\n4015x - 1 = k\\pi \\implies x = y = \\frac{1 + k\\pi}{4015}.\n$$\nSince $k$ is integer, $|1 + k\\pi| \\ge 1$, so\n$$\nxy = \\frac{(1 + k\\pi)^2}{4015^2} \\ge \\frac{1}{16120225}.\n$$\nEquality holds when $x = y = \\frac{1}{4015}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19421,
    "subject": "Mathematics (Olympiad)",
    "question": "If $169! = 1 \\times 2 \\times 3 \\times \\cdots \\times 169$ is written as the product of prime numbers, how many times would $13$ appear as a factor?\n\n(A) 12  \n(B) 13  \n(C) 14  \n(D) 15  \n(E) 16",
    "options": [],
    "answer": "See solution",
    "solution": "Since $13$ is a prime number, the only factors in $169!$ supplying powers of $13$ are $13, 26, 39, \\ldots, 156, 169$. The first $12$ of these provide one power of $13$ each, but since $169 = 13^2$, it contributes an extra power. Thus, the total power of $13$ is $12 + 2 = 14$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19422,
    "subject": "Mathematics (Olympiad)",
    "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 $MN$, with $P \\neq M$ and $P \\neq N$, 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": "See 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 $P \\in (ACD)$, which is false. The other situations are analogous.\n\n![](images/RMC_2023_v2_p36_data_0c62e481bd.png)\n\n_Existence:_ Since $P \\in (MN)$, the point $P$ lies in the interior of the triangle $ANC$. Denote $\\{Q\\} = AP \\cap CN$.\n\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 $Y' = Y$, which is 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, which is false. Consequently, the pair $(X, Y)$ is unique.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19423,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 2019 points $A_1, A_2, \\dots, A_{2019}$ on a circle, forming the regular 2019-gon $A_1A_2\\dots A_{2019}$. Olesya and Andrew take turns, with Olesya going first. The rules for forming triangles are as follows:\n\n- Initially, vertices $A_1$ and $A_2$ are marked.\n- On her first turn, Olesya marks an unmarked vertex $A_i$ so that $\\triangle A_1A_2A_i$ is obtuse, and $A_i$ becomes marked.\n- On each subsequent turn, the current player marks an unmarked vertex $A_j$ to form a triangle $\\triangle A_kA_lA_j$:\n    - Olesya must form an obtuse triangle.\n    - Andrew must form an acute triangle.\n    - $A_k$ and $A_l$ are two previously marked vertices, with $A_j$ unmarked.\n    - On each turn, the triangle must use one of the last two marked vertices and the new vertex.\n- A player loses if they cannot make a move according to the rules.\n\nWho wins the game if both players play optimally?\n\n![](images/UkraineMO2019_booklet_p48_data_0b46ce5227.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Olesya wins.\n\nOlesya's strategy is to always mark the vertex symmetric to Andrew's last move with respect to the vertical diameter through $A_1$. On her first move, she marks $B_{-1}$. If Andrew marks $B_{-k}$, forming an acute triangle $\\triangle B_m B_l B_{-k}$, the center of the circle lies inside this triangle, so $B_m$ and $B_l$ are on opposite sides of the diameter. Olesya can then mark $B_k$, forming $\\triangle B_m B_{-k} B_k$, which is obtuse since the center is not inside. Thus, Olesya can always respond, and Andrew will eventually be unable to move, so Olesya wins.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19424,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $x_1, x_2, \\ldots, x_{2015}$ satisfy both of the following equations simultaneously:\n\n![alt](images/ukraine_2015_Booklet_p8_data_69304e925c.png)\n\n$$x_1^{2014} + x_2^{2014} + \\ldots + x_{2015}^{2014} = 1$$\n$$x_1^{2015} + x_2^{2015} + \\ldots + x_{2015}^{2015} = -1$$\n\nFind all possible values of $x_1 + x_2^2 + x_3^3 + \\ldots + x_{2015}^{2015}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\pm 1$.\n\n**Solution.** From the first equation, $-1 \\leq x_i \\leq 1$ for all $i = 1, \\ldots, 2015$, so $0 \\leq 1 + x_i \\leq 2$. Adding both equations gives:\n\n$$x_1^{2014}(1 + x_1) + x_2^{2014}(1 + x_2) + \\ldots + x_{2015}^{2014}(1 + x_{2015}) = 0.$$ \n\nSince each term $x_i^{2014}(1 + x_i) \\geq 0$, their sum is zero only if every term is zero. Thus, all $x_i \\in \\{-1, 0\\}$. From the first equation, exactly one variable is nonzero; from the second, this variable must be $-1$. Therefore, the solution is: $x_i = -1$ for some $i$, $x_j = 0$ for $j \\neq i$.\n\nHence, $x_1 + x_2^2 + x_3^3 + \\ldots + x_{2015}^{2015} = 1$ if $i$ is even, and $-1$ if $i$ is odd.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19425,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\geq 16$, consider the set\n\n$$\nG = \\{(x, y) : x, y \\in \\{1, 2, \\dots, n\\}\\}\n$$\n\nconsisting of $n^2$ points in the plane. Let $A$ be any subset of $G$ containing at least $4n\\sqrt{n}$ points. Prove that there are at least $n^2$ convex quadrangles with all their vertices in $A$ such that their diagonals intersect in one common point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|A| = m \\geq 4n\\sqrt{n}$ and let $S$ be the set of all segments with endpoints in $A$. Clearly, $|S| = \\binom{m}{2}$. The coordinates of every midpoint of a segment from $S$ are integer multiples of $1/2$. In the convex hull of $G$ there are less than $4n^2$ such points, so there exists a point $B$ which is a midpoint of at least $(m/2)/(4n^2)$ segments from $S$. Let $P$ be the set of all segments from $S$ with their midpoints at $B$. Then\n\n$$\n|P| \\geq \\frac{\\binom{m}{2}}{4n^2} = \\frac{m(m-1)}{8n^2} \\geq \\frac{4n\\sqrt{n}(4n\\sqrt{n}-1)}{8n^2} = \\frac{16n^3 - 4n\\sqrt{n}}{8n^2} = 2n - \\frac{1}{2\\sqrt{n}} > 2n-1.\n$$\n\nand $|P| \\geq 2n$.\n\nLet us divide $P$ into disjoint families of segments which lie on the same line. Suppose the number of such families is $k$ and in the $i$-th family we have $a_i$ segments, for $i = 1, \\dots, k$. Every segment from $a_i$ segments of one family has its endpoints in $G$ and they have a common midpoint, so $a_i \\leq n/2$. Moreover, every two segments from $P$ are the diagonals of a parallelogram iff they do not lie on the same line. Therefore, for the number of different parallelograms with diagonals belonging to $P$ we have\n\n$$\n\\sum_{1 \\leq i < j \\leq k} a_i a_j = \\frac{1}{2} \\left( \\left( \\sum_{i=1}^{k} a_i \\right)^2 - \\sum_{i=1}^{k} a_i^2 \\right) \\geq \\frac{1}{2} \\left( \\left( \\sum_{i=1}^{k} a_i \\right)^2 - \\sum_{i=1}^{k} a_i \\cdot \\frac{n}{2} \\right) = \\frac{1}{2} \\left( |P|^2 - |P| \\cdot \\frac{n}{2} \\right) = \\frac{1}{2} |P| \\left( |P| - \\frac{n}{2} \\right) \\geq n \\left( 2n - \\frac{n}{2} \\right) = \\frac{3}{2} n^2 > n^2.\n$$\n\nThus we have more than $n^2$ convex quadrangles (parallelograms) satisfying the given condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19426,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that\n$$\n17^n + 9^{n^2} = 23^n + 3^{n^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us check small values of $n$:\n\nFor $n=1$:\n$$\n17^1 + 9^{1^2} = 17 + 9 = 26, \\\\\n23^1 + 3^{1^2} = 23 + 3 = 26.\n$$\nSo $n=1$ is a solution.\n\nFor $n \\geq 2$, note that $n^2 \\geq 2n$, so\n$$\n9^{n^2} - 3^{n^2} = 3^{n^2}(3^{n^2} - 1) \\geq 3^{2n}(3^{2n} - 1) = 81^n - 9^n.\n$$\nBut $81^n - 9^n > 23^n - 17^n$ for $n \\geq 2$, so the equation has no solutions for $n \\geq 2$.\n\nThus, the only positive integer solution is $n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19427,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$. Let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively, such that $\\angle BAM = \\angle MNC$. Suppose that the lines $MN$ and $AB$ intersect at $P$. Prove that the bisectors of the angles $\\angle BAM$ and $\\angle BPM$ intersect at a point lying on the line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n$$\n\\begin{align*}\n\\angle MBA + \\angle BMN &= \\angle CBA + \\angle BCA + \\angle MNC = \\angle CBA + \\angle BCA + \\angle BAM \\\\\n&= 180^\\circ - \\angle MAC < 180^\\circ,\n\\end{align*}\n$$\nit follows that point $A$ lies between $B$ and $P$. Suppose that the angle bisector of $\\angle BPM$ intersects $BC$ at $Q$. We will prove that $AQ$ is the bisector of the angle $\\angle BAM$, that is, $\\dfrac{AM}{AB} = \\dfrac{QM}{QB}$. Since $\\dfrac{QM}{PB} = \\dfrac{PM}{PB}$, it suffices to show that $\\dfrac{AM}{AB} = \\dfrac{PM}{PB}$. On the other hand, $AB = AC$ and the latter rewrites as\n$$\n\\frac{AM}{AC} = \\frac{PM}{PB}. \\qquad (1)\n$$\nFrom the hypothesis $\\angle B = \\angle C$ and $\\angle BAM = \\angle MNC$, yielding $\\angle AMB = \\angle NMC$. It follows that $\\angle BMP = \\angle AMC$, and since $\\angle PBM = \\angle ACM$, we deduce that the triangles $BMP$ and $CMA$ are similar, which proves (1).\n\n![](images/BMO_Short_list_2014_p21_data_1747f958af.png)\n\nThe similarity $\\triangle BMP \\sim \\triangle CMA$ can also be obtained by noticing that $\\angle BAM$ is an exterior angle for the triangle $PAM$, whilst $\\angle MNC$ is an exterior angle for the triangle $ANM$. Thus, we have\n$$\n\\angle AMN + \\angle APM = \\angle AMN + \\angle MAN,\n$$\nthat is, $\\angle BMP = \\angle AMC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19428,
    "subject": "Mathematics (Olympiad)",
    "question": "For a sequence $x_1, x_2, \\dots, x_n$ of real numbers, we define its price as\n\n$$\n\\max_{1 \\le i \\le n} |x_1 + \\dots + x_i|.\n$$\n\nGiven $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_1$ such that $|x_1|$ is as small as possible; among the remaining numbers, he chooses $x_2$ such that $|x_1 + x_2|$ is as small as possible, and so on. Thus, in the $i$th step he chooses $x_i$ among the remaining numbers so as to minimise the value of $|x_1 + \\dots + x_i|$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$.\n\nFind the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G \\le cD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $c = 2$\n\nExample: If the numbers are $1, -1, 2, -2$, Dave can arrange them as $1, -2, 2, -1$, and George as $1, -1, 2, -2$. Then $D = 1$, $G = 2$, so $c \\ge 2$.\n\nNow, we prove $G \\le 2D$. Let the original numbers be $x_1, x_2, \\dots, x_n$. Suppose Dave and George arrange them as $d_1, d_2, \\dots, d_n$ and $g_1, g_2, \\dots, g_n$, respectively. Define\n\n$$\nM = \\max_{1 \\le i \\le n} |x_i|, \\quad S = |x_1 + \\dots + x_n|, \\quad N = \\max\\{M, S\\}.\n$$\n\nWe have:\n\n$$\nD \\geq S, \\tag{1}\n$$\n\n$$\nD \\geq \\frac{M}{2}, \\tag{2}\n$$\n\n$$\nG \\leq N = \\max\\{M, S\\}. \\tag{3}\n$$\n\nFrom these, $G \\leq \\max\\{M, S\\} \\leq \\max\\{M, 2S\\} \\leq 2D$.\n\nInequality (1) follows directly from the definition of price.\n\nTo prove (2), consider an index $i$ such that $|d_i| = M$. Then\n\n$$\n\\begin{aligned}\nM &= |d_i| = |(d_1 + \\cdots + d_i) - (d_1 + \\cdots + d_{i-1})| \\\\\n&\\leq |d_1 + \\cdots + d_i| + |d_1 + \\cdots + d_{i-1}| \\leq 2D.\n\\end{aligned}\n$$\n\nTo prove (3), let $h_i = g_1 + \\cdots + g_i$. We use induction on $i$ to show $|h_i| \\leq N$.\n\nFor $i=1$, $|h_1| = |g_1| \\leq M \\leq N$. Note $|h_n| = S \\leq N$.\n\nAssume $|h_{i-1}| \\leq N$. Consider two cases:\n\n**Case 1:** All of $g_i, \\dots, g_n$ have the same sign. Without loss of generality, suppose they are non-negative. Then $h_{i-1} \\leq h_i \\leq \\cdots \\leq h_n$, so\n\n$$\n|h_i| \\leq \\max\\{|h_{i-1}|, |h_n|\\} \\leq N.\n$$\n\n**Case 2:** Among $g_i, \\dots, g_n$ there are both positive and negative numbers. Then there exists $j \\geq i$ such that $h_{i-1}g_j \\leq 0$. By George's selection rule,\n\n$$\n|h_i| = |h_{i-1} + g_i| \\leq |h_{i-1} + g_j| \\leq \\max\\{|h_{i-1}|, |g_j|\\} \\leq N.\n$$\n\nBy induction, the claim holds. Thus, $G \\leq 2D$ for all cases.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19429,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, in an acute triangle $ABC$ with $AB < AC$, let $AH$ be its altitude and $G$ the barycentre. Let $P, Q$ be the tangent points of the incircle to $AB, AC$, respectively. Let $M, N$ be the midpoints of $BP$ and $CQ$, respectively. Let $D, E$ be two points lying on the incircle of triangle $ABC$ such that\n\n$$\n\\angle BDH + \\angle ABC = 180^{\\circ}, \\quad \\angle CEH + \\angle ACB = 180^{\\circ}.\n$$\n\nProve that the lines $MD$, $NE$, and $GH$ are concurrent.\n\n![](images/2023_CGMO_p6_data_23d60f21ff.png)",
    "options": [],
    "answer": "See solution",
    "solution": "On the circumcircle of $\\triangle ABC$, choose a point $F$ such that $ABCF$ forms an isosceles trapezoid. The line $FH$ intersects the circumcircle of $\\triangle ABC$ at another point $L$, and intersects the median $AK$ at point $G'$. As shown in the figure.\n\nSince $AF = 2HK$, it follows that $\\frac{AG'}{G'K} = \\frac{AF}{HK} = 2$, implying that $G'$ is the centroid of $\\triangle ABC$, hence $G' = G$. Therefore,\n\n$$\n\\angle BLH = \\angle BLF = \\frac{1}{2}\\widehat{BF} = \\frac{1}{2}\\widehat{AC} = \\angle ABC,\n$$\n\nGiven $\\angle BDH + \\angle ABC = 180^{\\circ}$, it follows that $\\angle BDH + \\angle BLH = 180^{\\circ}$, which means points $B, L, H, D$ are concyclic.\n\nSimilarly, it can be proved that $\\angle CLH = \\angle ACB$, and points $C, L, H, E$ are concyclic.\n\nSince $\\angle BLH = \\angle ABH$, $PB$ is tangent to circle $\\odot BLHD$ at point $B$. Let the incircle of $\\triangle ABC$ be $\\omega$, and $PB$ is an external common tangent of $\\omega$ and $\\odot BLHD$. Given $MP = MB$, it is known that $M$ is the radical center of $\\omega$ and $\\odot BLHD$, thus line $MD$ is the radical axis of $\\omega$ and $\\odot BLHD$.\n\nSimilarly, it can be proved that line $NE$ is the radical axis of $\\omega$ and $\\odot CLHE$. Moreover, line $GH$ is the radical axis of $\\odot BLHD$ and $\\odot CLHE$, therefore lines $MD$, $NE$, $GH$ either intersect at a single point, or are pairwise parallel.\n\nIf $MD$, $NE$, $GH$ are pairwise parallel, then the centers $O_1$, $O_2$ of $\\odot BLHD$, $\\odot CLHE$, and the center $I$ of $\\omega$ are collinear. Since both $\\angle BDH$ and $\\angle CEH$ are obtuse, $O_1$, $O_2$ are below $BC$, obviously $I$ is above $BC$. Let the projections of $O_1$, $O_2$, $I$ on $BC$ be $X$, $Y$, $Z$ respectively, then $X$, $Y$ are the midpoints of $BH$, $CH$ respectively. Given $AB < AC$, it is known that $Y$, $Z$ are on the same side of $AH$, and\n\n$$\nCZ = \\frac{AC + BC - AB}{2} > \\frac{BC}{2} > \\frac{CH}{2} = CY.\n$$\n\nHence, $Z$ lies on the segment $XY$, therefore $O_1$, $O_2$, $I$ cannot be collinear, a contradiction. Thus, $MD$, $NE$, $GH$ intersect at a single point. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19430,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that\n$$\n\\left\\{ \\frac{i(i-1)}{2} : i = 1, 2, \\dots \\right\\}\n$$\ncontains a complete set of residues modulo $n$.\n\nA set $S$ of integers is called *complete* (mod $n$) if $S$ contains a complete set of residues modulo $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that an odd prime $p$ divides $n$. Then $\\frac{i(i-1)}{2}$ takes every value modulo $p$. If we set $a_k = \\frac{k(k-1)}{2}$, then the sequence $a_1, a_2, \\dots$ is periodic modulo $p$ with period $p$. Hence $a_1, a_2, \\dots, a_p$ must be distinct modulo $p$. But this is not true, because $a_1 \\equiv 0 \\equiv a_p \\pmod{p}$.\n\nThus, we are left only with the case when $n$ is a power of $2$. We want to show that $\\frac{i(i-1)}{2}$ takes all values modulo $2^k$ for $k = 0, 1, 2, \\dots$. This is the same as showing that $i(i-1)$ takes all even values modulo $2^{k+1}$. We claim that as $i$ goes from $1$ to $2^k$, all these even values are taken exactly once. Suppose to the contrary that\n$$\na(a-1) \\equiv b(b-1) \\pmod{2^{k+1}}$$\nfor some $a, b$ where $1 \\le a < b \\le 2^k$. This may be rearranged as\n$$\n(a+b-1)(a-b) \\equiv 0 \\pmod{2^{k+1}}.\n$$\nSince $a+b-1$ and $a-b$ are of opposite parity, one of these expressions is odd while the other is divisible by $2^{k+1}$. But since $0 < a+b-1 < 2^{k+1}$ and $0 < b-a < 2^k$, neither of these can be divisible by $2^{k+1}$. This contradiction establishes the claim and concludes the proof.\n\n**Answer:** $n$ is a power of $2$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19431,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $A$ and $B$ such that the equation\n$$\nA(y - x) + B\\lfloor x \\rfloor = B\\lfloor y \\rfloor\n$$\nhas solutions with $x \\neq y$ for real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $A = 0$, then $B\\lfloor x \\rfloor = B\\lfloor y \\rfloor$, which has infinitely many solutions with $x \\neq y$. Assume $A \\neq 0$. Rewrite the equation as\n$$\n\\frac{B}{A}(\\lfloor x \\rfloor - \\lfloor y \\rfloor) = y - x.\n$$\nIf $x \\neq y$, assume $x < y$. We cannot have $\\lfloor x \\rfloor = \\lfloor y \\rfloor$, so $\\lfloor x \\rfloor < \\lfloor y \\rfloor$. Let $x = n - \\epsilon$, $y = n + d + \\delta$, where $n$ is any integer, $d$ is a non-negative integer, $\\delta \\in [0, 1)$, and $\\epsilon \\in (0, 1]$. Then\n$$\n\\frac{B}{A} = -\\frac{d + \\nu}{d + 1},\n$$\nwhere $\\nu = \\delta + \\epsilon \\in (0, 2)$. Note that $-2 < -\\frac{d + \\nu}{d + 1} < 0$. Thus, if there are solutions with $x \\neq y$ and $A \\neq 0$, then $\\frac{B}{A}$ must belong to $(-2, 0)$. Conversely, by taking $d = 0$ and suitable $\\nu$, $\\frac{B}{A}$ can take any value in $(-2, 0)$. Therefore, a necessary and sufficient condition for solutions with $x \\neq y$ is $A = 0$ or $\\frac{B}{A} \\in (-2, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19432,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the least possible value of\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2,\n$$\nif $x_1, x_2, \\dots, x_n$ are distinct integer numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $4n - 6$.\n\n**Solution.** We prove by induction that\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\geq 4n - 6,\n$$\nif $x_1, x_2, \\dots, x_n$ are distinct integer numbers.\n\nThe base is trivial. Indeed, $S_2 = (x_1 - x_2)^2 + (x_2 - x_1)^2 \\geq 2$, because all numbers are integer and distinct.\n\nLet us now suppose that our assumption holds for $n$, in other words,\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\geq 4n - 6.\n$$\nLet $x_1, x_2, \\dots, x_n, x_{n+1}$ be distinct integer numbers. WLOG, we can assume that $x_{n+1}$ is a maximum among our numbers. We now have\n$$\n(x_n - x_{n+1})^2 + (x_{n+1} - x_n)^2 - (x_n - x_1)^2 = (x_{n+1} - x_n)(x_{n+1} - x_1) \\geq 4.\n$$\nSumming all such inequalities and using our induction hypothesis we get the desired result for $n + 1$ numbers.\n\nWe now construct an example that proves sharpness of our bound.\n\nFor $n = 2k - 1$ one can take $x_j = 2j - 2$ for $j \\leq k$ and $x_j = -2j + 4k - 1$ for $j \\geq k + 1$.\n\nFor $n = 2k$ we take $x_j = 2j - 2$ for $j \\leq k$ and $x_j = -2j + 4k + 1$ for $j \\geq k + 1$.\n\n**Alternative solution.** Let us assume that $x_1$ is a maximum and $x_k$ is a minimum among our numbers. Then, we have $x_1 - x_k \\geq n - 1$. By AM-GM:\n$$\n\\begin{aligned}\n& \\sum_{j=1}^{n} (x_j - x_{j+1})^2 \\geq \\frac{1}{n} \\left( \\sum_{j=1}^{n} |x_j - x_{j+1}| \\right)^2 \\\\\n& \\geq \\frac{((x_1 - x_2) + \\dots + (x_{k-1} - x_k) + (-x_k + x_{k+1}) + (-x_{k+1} + x_{k+1}) + \\dots + (-x_n + x_1))^2}{n} \\\\\n& = \\frac{(2(x_1 - x_k))^2}{n} \\geq \\frac{(2(n-1))^2}{n} = 4n - 8 + \\frac{4}{n}.\n\\end{aligned}\n$$\n$\\sum_{j=1}^{n} (x_j - x_{j+1})^2$ is natural, thus $\\sum_{j=1}^{n} (x_j - x_{j+1})^2 \\geq 4n - 7$. Moreover, squaring does not change the parity of an expression and $\\sum_{j=1}^{n} (x_j - x_{j+1}) = 0$ is even, then $\\sum_{j=1}^{n} (x_j - x_{j+1})^2$ is also even. Hence, it is not less than $4n - 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19433,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call $(a, b)$ a \"bad pair\" if it satisfies $4ab - 1 \\mid (4a^2 - 1)^2$ while $a \\neq b$. We use the method of infinite descent to prove there is no such \"bad pair\".\n\n**Property 1**: If $(a, b)$ is a \"bad pair\" and $a < b$, there exists an integer $c$ ($c < a$) such that $(a, c)$ is also a \"bad pair\".\n\nIn fact, let $r = \\frac{(4a^2 - 1)^2}{4ab - 1}$, then\n\n$$\nr = -r \\cdot (-1) = -(4a^2 - 1)^2 = -1 \\pmod{4a}.\n$$\n\nTherefore there exists an integer $c$ such that $r = 4ac - 1$. Since $a < b$, we have\n\n$$\n4ac - 1 = \\frac{(4a^2 - 1)^2}{4ab - 1} < 4a^2 - 1.\n$$\n\nSo $c < a$ and $4ac - 1 \\mid (4a^2 - 1)^2$. Thus $(a, c)$ is a \"bad pair\" too.\n\n**Property 2**: If $(a, b)$ is a \"bad pair\", so is $(b, a)$.\n\nIn fact, by\n\n$$\n1 = 1^2 \\equiv (4ab)^2 \\pmod{4ab - 1},\n$$\n\nwe get\n\n$$\n(4b^2 - 1)^2 \\equiv (4b^2 - (4ab)^2)^2 = 16b^4(4a^2 - 1)^2 \\\\\n\\equiv 0 \\pmod{4ab - 1}.\n$$\n\nThus $4ab - 1 \\mid (4b^2 - 1)^2$.\n\nIn the following we will show that such \"bad pair\" does not exist. We shall prove by contradiction.\n\nSuppose there is at least one \"bad pair\", we choose such pair for which $2a + b$ is minimum.\n\nIf $a < b$, by property 1, there is a \"bad pair\" $(a, c)$ which satisfies $c < b$, and $2a + c < 2a + b$, a contradiction.\n\nIf $b < a$, by property 2, $(b, a)$ is also a \"bad pair\", which leads to $2b + a < 2a + b$, a contradiction.\n\nHence such \"bad pair\" does not exist. Therefore $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19434,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f : \\mathbb{Q}_{>0} \\to \\mathbb{R}$ be a function satisfying the following three conditions:\n\n1. For all $x, y \\in \\mathbb{Q}_{>0}$, we have $f(x)f(y) \\geq f(xy)$.\n2. For all $x, y \\in \\mathbb{Q}_{>0}$, we have $f(x + y) \\geq f(x) + f(y)$.\n3. There exists a rational number $a > 1$ such that $f(a) = a$.\n\nProve that $f(x) = x$ for all $x \\in \\mathbb{Q}_{>0}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the only solution is $f(x) = x$ for all $x \\in \\mathbb{Q}_{>0}$.\n\nBy applying (1) and iterating (2), we see that $f(n)f(x) \\geq f(nx) \\geq n f(x)$ for positive integer $n$. Setting $x = a$, we see that $f(n) \\geq n$ because $f(a) = a > 0$.\n\nWe claim that $f$ is non-negative and non-decreasing. Indeed, if $f(y) < 0$, setting $x = y$ and dividing by $f(y)$ in our first inequality shows that $f(n) \\leq n$, hence $f(n) = n$. This chain of inequalities is thus an equality, so $f(n)f(x) = f(nx)$ for all $x$. Writing $y = \\frac{p}{q} \\in \\mathbb{Q}_{>0}$, we have $f(q)f\\left(\\frac{p}{q}\\right) = f(p)$, so $f(y) = y$, a contradiction. Hence, $f$ is non-negative, thus also non-decreasing by (2).\n\nWe claim now that $f(x) \\geq x$ for all $x \\geq 1$. First, note that $f(x) \\geq f(\\lfloor x \\rfloor) \\geq \\lfloor x \\rfloor > x - 1$. From (1) we know that $f(x)^n \\geq f(x^n)$, so $f(x)^n \\geq f(x^n) > x^n - 1$. But if $f(x) = x - \\epsilon$ for some $\\epsilon > 0$ and $x > 1$, then for all $n$ we have $1 > x^n - f(x)^n \\geq (x - f(x)) x^{n-1} = \\epsilon x^{n-1}$. Since $x > 1$, we can choose $n$ such that $x^{n-1} > \\frac{1}{\\epsilon}$, a contradiction. Therefore, $f(x) \\geq x$ for all $x > 1$, and we already know that $f(1) \\geq 1$, yielding the claim.\n\nWe now show $f(x) = x$ for $x \\geq 1$. Note that $a^k = f(a)^k \\geq f(a^k)$ for positive integers $k$ by (1). We also have $f(a^k) \\geq a^k$, so $f(a^k) = a^k$ for positive integers $k$. For $x \\geq 1$ and $k$ with $a^k > 2x$, we have $a^k = f(a^k) \\geq f(x) + f(a^k - x) \\geq x + (a^k - x) = a^k$. Equality thus holds, so $f(x) = x$ for $x \\geq 1$.\n\nFinally, for any integer $n$, we have $f(n) = n$ and $f(n)f(x) \\geq f(nx) \\geq n f(x)$, so equality holds, implying that $f(nx) = n f(x)$. In particular, for any $x = \\frac{p}{q}$ in $\\mathbb{Q}_{>0}$, we conclude that $q f(x) = f(p) = p$, hence $f(x) = x$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19435,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $k$ and $n$ satisfying the equation\n\n$$\nk^2 - 2016 = 3^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We immediately see that $n = 1$ does not lead to a solution, while $n = 2$ yields the solution $(k, n) = (45, 2)$.\n\nWe show that there is no solution with $n \\ge 3$. In that case, $3^n$ is divisible by $9$ and thus $k^2$ is divisible by $9$, which implies that $k = 3\\ell$ for some positive integer $\\ell$. After division by $9$, the equation reads $\\ell^2 - 224 = 3^{n-2}$. Modulo $3$, this yields $\\ell^2 - 2 \\equiv 0 \\pmod{3}$, a contradiction because $2$ is not a quadratic residue modulo $3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19436,
    "subject": "Mathematics (Olympiad)",
    "question": "Может ли квадратный трёхчлен с иррациональным коэффициентом и без корней иметь рациональный дискриминант?",
    "options": [],
    "answer": "See solution",
    "solution": "Нет, не может.\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\n$$\n\\frac{f(c)}{c} = \\frac{ac^2 + bc + c}{c} = ac + b + 1.\n$$\nТак как $b+1$ рационально, то $ac$ — иррационально. Получаем, что дискриминант $D = b^2 - 4ac$ иррационален как разность рационального и иррационального чисел.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19437,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $SABCD$ be a pyramid with apex $S$ and base $ABCD$, where $ABCD$ is a parallelogram. Consider points $M$, $N$, $P$, and $Q$ on the edges $SA$, $SB$, $SC$, and $SD$, respectively, such that $MNPQ$ is also a parallelogram.\n\n(a) If $ABCD$ is a rhombus, prove that $MNPQ$ is also a rhombus.\n\n(b) If $ABCD$ is a rectangle, prove that $MNPQ$ is also a rectangle.",
    "options": [],
    "answer": "See 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 $(SDA)$.\n\nSince $AB \\parallel CD$, $AB \\subset (SAB)$ and $CD \\subset (SCD)$, by the roof theorem it follows that $AB \\parallel CD \\parallel d_1$. Because $MN \\parallel PQ$, $MN \\subset (SAB)$ and $PQ \\subset (SCD)$, by the roof theorem we deduce that $MN \\parallel PQ \\parallel d_1$, therefore $AB \\parallel CD \\parallel MN \\parallel PQ \\parallel d_1$.\n\nSimilarly, we obtain $BC \\parallel DA \\parallel NP \\parallel QM \\parallel d_2$.\n\n(a) By the fundamental theorem of similarity, $\\frac{MN}{AB} = \\frac{SN}{SB} = \\frac{NP}{BC}$.\n\n![](images/RMC_2023_v2_p13_data_f9cd916b14.png)\n\nIf $ABCD$ is a rhombus, then $AB = BC$, so $MN = NP$, and thus the parallelogram $MNPQ$ is also a rhombus.\n\n(b) If $ABCD$ is a rectangle, then $\\angle ABC = 90^\\circ$. As $MN \\parallel AB$ and $NP \\parallel BC$, we obtain $\\angle MNP = 90^\\circ$, so $MNPQ$ is also a rectangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19438,
    "subject": "Mathematics (Olympiad)",
    "question": "A teacher asks Annie, Basil, Cara, Dave, and Elli, in that order, to each multiply a pair of (possibly equal) single digits. Except for Annie, each of them gets a product that is 50% more than the previous student. What product did $\\textit{Dave}$ get?",
    "options": [],
    "answer": "See solution",
    "solution": "Let Annie's product be $n$. Each subsequent product is 50\\% more than the previous, so:\n\n- Annie: $n$\n- Basil: $n \\times \\frac{3}{2}$\n- Cara: $n \\times \\left(\\frac{3}{2}\\right)^2$\n- Dave: $n \\times \\left(\\frac{3}{2}\\right)^3$\n- Elli: $n \\times \\left(\\frac{3}{2}\\right)^4 = \\frac{81n}{16}$\n\nSince all products must be integers less than 100, $n$ must be a multiple of 16. The smallest such $n$ is 16:\n\n- Annie: $16$\n- Basil: $24$\n- Cara: $36$\n- Dave: $54$\n- Elli: $81$\n\nThus, Dave's product is $\\boxed{54}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19439,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_n$ (respectively, $B_n$) be the number of all possible orders of $n$ cyclists at the end of a race in which each cyclist overtook exactly once (for $A_n$), or at most once (for $B_n$), during the race. The cyclists are initially numbered $1$ through $n$.\n\nWhat is the relationship between $A_{200}$ and $B_{200}$?",
    "options": [],
    "answer": "See solution",
    "solution": "* If $n$ finishes last, that order can be obtained by only having the first $n-1$ cyclists overtake—there are $B_{n-1}$ such orders.\n* If $n$ finishes next to last, that order can be obtained by having $n$ be the last one to overtake, and before him only the first $n-1$ cyclists (or some of them) have overtaken—there are $B_{n-1}$ such orders.\n\nThis shows that $B_n = B_{n-1} + B_{n-2} = 2B_{n-1}$. Since $B_2 = 2$, we have $B_{200} = 2^{199}$.\n\nNow, consider the case where each cyclist overtook exactly once. Let $k$ be the smallest number such that cyclist $k+2$ overtook before cyclist $k+1$. Then the final order must be $2, 3, \\dots, k-1, 1, k, x, \\dots, y$ because cyclist $k+2$ made it impossible for cyclists marked $1$ through $k$ to mix with cyclists with marks greater than $k$. Thus, the final order is determined by the selection of $k$ (which can be any of $0$ through $n-2$) and the order in which we select marks $k+1$ through $n$ (the number of these selections is $A_{n-k}$). Therefore,\n\n$$\nA_n = A_{n-1} + A_{n-2} + \\dots + A_2 + 1.\n$$\n\nSince $A_2 = 1$, it can be shown by induction that $A_n = 2^{n-2}$ for all positive integers $n$. Therefore, $2A_{200} = 2^{199} = B_{200}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19440,
    "subject": "Mathematics (Olympiad)",
    "question": "Baron Münchhausen says that his new stained-glass window looks like an inscribed 400-gon which is split by non-intersecting diagonals into triangular pieces of glass. He claims that it is possible to construct at least $\\left(\\frac{3}{2}\\right)^{400}$ different convex polygons by taking unions of these triangles. Can this be true?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it is possible.\n\nLet $A_{4k}, A_{4k+1}, A_{4k+2}, A_{4k+3}, A_{4k+4}$ be five consecutive points of the 400-gon. For each $k$, triangulate the pentagon $S_k = A_{4k}A_{4k+1}A_{4k+2}A_{4k+3}A_{4k+4}$ as follows.\n\n![](images/bw18shortlist_p19_data_5ff1edcf85.png)\n\nFor each $k$, we can choose 5 different convex polygons inside pentagon $S_k$ containing segment $A_{4k}A_{4k+4}$:\n\n$$\nA_{4k}A_{4k+1}A_{4k+2}A_{4k+3}A_{4k+4}, \\quad A_{4k}A_{4k+1}A_{4k+2}A_{4k+4}, \\quad A_{4k}A_{4k+2}A_{4k+3}A_{4k+4}, \\\\ A_{4k}A_{4k+2}A_{4k+4}, \\quad \\text{and (degenerate case) } A_{4k}A_{4k+4}.\n$$\n\nNow, let each fragment of the 400-gon containing two consecutive $S_{2k}$ and $S_{2k+1}$ look as follows; denote this polygon as $D_k$.\n\n![](images/bw18shortlist_p19_data_417f53c9ca.png)\n\nFor this configuration, we can choose $5^2 + 1 = 26$ different convex polygons containing segment $A_{8k}A_{8k+8}$: first, the degenerate case (the segment $A_{8k}A_{8k+8}$ itself); second, the triangle $A_{8k}A_{8k+4}A_{8k+8}$ and, independently, one of 5 possible convex polygons inside $S_{2k}$ and one of 5 possible convex polygons inside $S_{2k+1}$.\n\nAfter that, triangulate the polygon $S = A_8A_{16}A_{24}\\dots A_{400}$ in an arbitrary way.\n\nBaron Münchhausen counted convex polygons that contain the whole polygon $S$ and differ by the parts chosen inside polygons $D_k$. All these choices are compatible and can be made independently. Thus, we obtain $26^{50} = (\\sqrt[8]{26})^{400} \\approx (1.502)^{400} > (3/2)^{400}$ convex polygons.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19441,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть дана таблица размера $n \\times n$, в каждой клетке которой стоит знак плюс или минус. Разрешается выбрать любую строку или столбец и поменять знаки во всех её клетках. Докажите, что после любых таких операций в таблице останется не менее $n$ плюсов.\n\n![](images/Rusija_2010_p46_data_21ef33ff03.png)\n\nРис. 21",
    "options": [],
    "answer": "See solution",
    "solution": "Второе 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 $r+c \\le n$.\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\n$$\nP \\ge r(n-c)+c(n-r)-(r+c)+n-(r+c) = -2rc + (r+c)(n-2) + n.\n$$\nТеперь, поскольку $2rc \\le \\frac{(r+c)^2}{2}$, получаем\n$$\nP \\ge (r+c)(n-2) - \\frac{(r+c)^2}{2} + n = (r+c)\\left(n-2 - \\frac{r+c}{2}\\right) + n \\ge n\n$$\n(ибо $n-2 - \\frac{r+c}{2} \\ge n-2-\\frac{n}{2} \\ge 0$), что и требовалось доказать.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19442,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е триаголникот $ABC$ со $BC < AB$. Низ точката $C$ е повлечена права $l$, нормална на симетралата $BE$ на аголот $\\angle B$. Правата $l$ ја сече $BE$ во точка $F$, а тежишната линија $BD$ во точка $G$. Да се докаже дека отсечката $DF$ ја преполовува отсечката $EG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $CF \\cap AB = \\{K\\}$ и $DF \\cap BC = \\{M\\}$. Бидејќи $BF \\perp KC$ и $BF$ е симетрала на $\\angle KBC$, следува дека $\\triangle KBC$ е рамнокрак, т.е. $\\overline{BK} = \\overline{BC}$ и уште $F$ е средина на $a > b > 0$.\n\nСпоред тоа, $DF$ е средна линија за $B$, односно $\\frac{1}{B} = 1 + \\frac{1}{\\frac{1}{b^n} + \\frac{1}{b^{n-1}} + \\dots + \\frac{1}{b}}$, од каде јасно $M$ е средина на $BC$.\n\n![](images/Makedonija_2009_p46_data_12dbc5b75b.png)\n\nЌе покажеме дека $GE \\parallel BC$. Доволно е да покажеме дека\n\n$$\n\\frac{BG}{GD} = \\frac{CE}{ED}\n$$\n\nОд $DF \\parallel AK$ и $\\overline{DF} = \\frac{\\overline{AK}}{2}$, и од сличноста на триаголниците $BKG$ и $GDF$ имаме\n\n$$\n\\frac{BG}{GD} = \\frac{BK}{DF} = \\frac{BK}{\\frac{1}{2}AK} = \\frac{2BK}{AK} \\quad (1)\n$$\n\nПонатаму, од сличноста на $\\triangle ABE$ и $\\triangle DEF$ имаме\n\n$$\n\\begin{aligned}\n\\frac{CE}{DE} &= \\frac{CD - DE}{DE} = \\frac{CD}{DE} - 1 = \\frac{AD}{DE} - 1 = \\frac{AE - DE}{DE} - 1 = \\frac{AE}{DE} - 2 \\\\\n&= \\frac{AB}{DF} - 2 = \\frac{AK + BK}{\\frac{1}{2}AK} - 2 = 2 + 2\\frac{BK}{AK} - 2 = 2\\frac{BK}{AK}\n\\end{aligned} \\quad (2)\n$$\n\nОд (1) и (2) добиваме $\\frac{BG}{GD} = \\frac{CE}{ED}$, па следува $GE \\parallel BC$, и бидејќи $M$ е средина на $BC$, следува $DF$ ја преполовува $GE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19443,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for any $x$ and $y$,\n\n$$\nf(x)^2 + 2y f(x) + f(y) = f(y + f(x)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equation holds for $f(x) = 0$ for all $x$.\n\nAssume there exists $a$ such that $f(a) \\neq 0$.\n\nSubstitute $y = -f(x)$ and let $c = f(0)$:\n\n$$\nf(-f(x)) = c + f(x)^2. \\quad (1)\n$$\n\nSubstitute $y = -f(y)$ and use (1):\n\n$$\n\\begin{aligned}\nf(f(x) - f(y)) &= f(x)^2 - 2f(x)f(y) + f(-f(y)) \\\\\n&= f(x)^2 - 2f(x)f(y) + f(y)^2 + c \\\\\n&= (f(x) - f(y))^2 + c.\n\\end{aligned} \\quad (2)\n$$\n\nSubstitute $x = a$:\n\n$$\nf(a)^2 + 2y f(a) = f(y + f(a)) - f(y).\n$$\n\nAs $y$ runs through $\\mathbb{R}$, the left side runs through $\\mathbb{R}$, so does the right. Thus, $f(x) - f(y)$ runs through all real numbers as $x, y$ vary.\n\nBy (2), $f(x) = x^2 + c$ for all $x$, for any constant $c$.\n\nTherefore, the solutions are $f(x) = x^2 + c$ (any constant $c$) and $f(x) = 0$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19444,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many mutually coprime positive integers $a$, $b$, and $c$ such that\n$$\n\\left\\lfloor \\frac{a^2}{2020} \\right\\rfloor + \\left\\lfloor \\frac{b^2}{2020} \\right\\rfloor = \\left\\lfloor \\frac{c^2}{2020} \\right\\rfloor.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that if $b$ is divisible by $2020$ and $a^2 + b^2 = c^2$, then\n$$\n\\left\\lfloor \\frac{a^2}{2020} \\right\\rfloor + \\left\\lfloor \\frac{b^2}{2020} \\right\\rfloor = \\left\\lfloor \\frac{c^2}{2020} \\right\\rfloor.\n$$\n\nConsider the identity $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$. Take $n = 2020$ and $m > 2020$ with $m$ coprime to $n$. Then $a = m^2 - n^2$, $b = 2mn$, and $c = m^2 + n^2$ are mutually coprime and satisfy the given equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19445,
    "subject": "Mathematics (Olympiad)",
    "question": "Учениците од IVª одделение членуваат во еколошката, литературната или математичката секција. Пет ученици членуваат во сите три секции, а девет ученици членуваат во по две секции. Во еколошката и литературната членуваат 8 ученици, и исто толку во литературната и математичката секција. Исто така, 20 ученици членуваат само во по една секција и тоа по 5 во еколошката и математичката секција. Колку ученици има во IVª одделение?",
    "options": [],
    "answer": "See solution",
    "solution": "I начин. Со помош на Ојлер – Венов дијаграм\n\n![](images/Makedonija_2009_p7_data_80e6fc56e8.png)\n\n$$6 (IV^{a}) = 34$$\n\nII начин.\nВо сите три секции членуваат 5 ученици. Бидејќи во секои две членуваат по 8 ученици, тогаш само во по две секции (без учениците кои членуваат во сите три секции истовремено) членуваат по 3 ученика. Во еколошката секција членуваат вкупно $5+3+3+5=16$ ученика. Слично, во математичката секција ќе членуваат 16 ученици. Според условот на задачата, само во литературната секција ќе членуваат $20-5-5 = 10$ ученика, а вкупно, во таа секција ќе членуваат 21 ученик. Во одделението има $5+3+3+3+10+5+5=34$ ученици.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19446,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest integer $n$ for which there exist integers $x_1, \\dots, x_n$ and positive integers $a_1, \\dots, a_n$ such that\n\n$$\nx_1 + \\dots + x_n = 0,\n$$\n\n$$\na_1x_1 + \\dots + a_nx_n > 0, \\quad a_1^2x_1 + \\dots + a_n^2x_n < 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n=3$. One possible example for $n=3$ is $x_1=2$ and $x_2=x_3=-1$ with $a_1=4$, $a_2=1$, $a_3=6$.\n\nFor $n=1$, the first constraint enforces $x_1=0$, which contradicts the other two constraints. For $n=2$, the first constraint enforces $x_2=-x_1$. Then the second constraint is equivalent to $a_1x_1 - a_2x_1 > 0$. Multiplying this inequality by the positive value $a_1+a_2$, we get $a_1^2x_1 - a_2^2x_1 > 0$, which is equivalent to $a_1^2x_1 + a_2^2x_1 > 0$ and contradicts the third constraint.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19447,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. For which $n$ does the following hold: For every proper divisor $d$ of $n$ and every integer $a > 2$, the number $a^d + 2^d$ divides $a^n - 2^n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $n$ satisfies the given condition if and only if $n$ is prime or $n$ is a power of $2$ (including $n = 1$).\n\nIf $n$ is an odd prime, then the only proper divisor $d$ of $n$ is $d = 1$. Let $a = 2^k - 2$ with $3 \\leq k \\leq n+1$, e.g., $a = 6$. Then we need to check that $2^k - a + 2 = 2^k$ is a divisor of $(2^k - 2)^n - 2^n$. As this is the difference of two terms which contain exactly $n$ factors $2$, the difference contains $n+1$ factors $2$.\n\nIf $n$ is a power of $2$, say $n = 2^m$ with $m \\geq 0$. If $m = 0$, then there are no proper divisors of $n$, so $n$ satisfies the given condition because it is an empty condition. If $m \\geq 1$, then for all proper divisors $d$ of $n$, the integer $e = \\frac{n}{d}$ is even. Note that\n\n$$\na^n - 2^n = a^{de} - 2^{de} \\equiv (-2^d)^e - 2^{de} = 2^n((-1)^e - 1) \\mod a^d + 2^d$$\n\nis zero for all $a$. Hence $a^n - 2^n$ is a multiple of $a^d + 2^d$, and therefore $a^d + 2^d \\mid a^n - 2^n$. So if $n$ is prime or a power of $2$, $n$ does indeed satisfy the given condition.\n\nFinally, suppose that $n$ is neither a prime number nor a power of $2$. Then we can write $n = de$ with $e \\neq 1$ odd (since $n$ is not a power of $2$) and $d \\neq 1$ (since $n$ is not a prime number). As $(-1)^e - 1 = -2$, it follows from the computation above that $a^d + 2^d \\mid 2^{n+1}$. Hence $a^d + 2^d$ is a power of $2$, so $a^d = 2^k - 2^d$ for some $d < k \\leq n+1$. Therefore $a$ is divisible by $2$ and $(\\frac{a}{2})^d = 2^{k-d} - 1$.\n\nNow we distinguish between the cases in which $d$ is even and in which $d$ is odd. In the first case, $2^{k-d} - 1$ is a square. As $a > 2$, from $(\\frac{a}{2})^d = 2^{k-d} - 1$ it however follows that $k - d \\geq 2$, so this square is $-1 \\pmod{4}$, which is a contradiction. If $d$ is odd, then we note that\n\n$$2^{k-d} = (\\frac{a}{2})^d + 1 = (\\frac{a}{2} + 1)\\left((\\frac{a}{2})^{d-1} - (\\frac{a}{2})^{d-2} + \\dots + 1\\right).$$\n\nAs $d \\neq 1$, we have $\\frac{a}{2} + 1 < (\\frac{a}{2})^d + 1$. The second factor in the product above is a sum of an even number of terms with the same parity as $\\frac{a}{2}$ and a term $1$, so this factor is odd. As this second factor is also greater than $1$, this contradicts it being a factor of a power of $2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19448,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ and $AEDP$ be cyclic quadrilaterals. Prove that $FQ \\parallel BC$ if and only if $EQ \\parallel DC$.\n\n![](images/2023_Australian_Scene_p147_data_254e1bf48e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABCD$ and $AEDP$ are cyclic, we have\n\n$$\n\\angle DBC = \\angle DAC = \\angle DEP.\n$$\n\nHence $BKDE$ is cyclic, and so $\\angle BKE = \\angle BDE = 90^\\circ$. Therefore\n\n$$\nFQ \\parallel BC \\iff FQ \\perp EP \\iff P = \\text{orthocentre}(EFQ)\n$$\n\nwhere the last equivalence is due to $QA \\perp EF$. Similarly, $EQ \\parallel DC$ if and only if $P = \\text{orthocentre}(EFQ)$.\n\nHence $FQ \\parallel BC$ if and only if $EQ \\parallel DC$, as desired.\n\n**Remark.** The statement $P = \\text{orthocentre}(EFQ)$ in the argument can be replaced by other equivalent statements, for example:\n\n* $Q = \\text{orthocentre}(EFP)$;\n* $AP \\times AQ = AE \\times AF$;\n* $\\triangle AFQ \\sim \\triangle APE$;\n* $FQ'EP$ is cyclic, where $Q'$ is the reflection of $Q$ about $A$.\n\nThe problem can then be completed by similar angle chases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19449,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be positive numbers such that\n$$\nx_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} = x_1 x_2 \\dots x_n\n$$\nProve the inequality:\n$$\n(x_1 - n + 1)(x_2 - n + 1) \\dots (x_n - n + 1) \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the Cauchy inequality, for each $i = 1, \\dots, n$:\n$$\n\\begin{align*}\nx_1 x_2 \\dots x_n &= x_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} \\\\ &\\ge x_1^{n-1} + (n-1) x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n \\\\\n&\\implies x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n (x_1 - n + 1) \\ge x_1^{n-1} \\\\\n&\\implies x_1 - n + 1 \\ge \\frac{x_1^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n}\n\\end{align*}\n$$\nThus,\n$$\n\\prod_{i=1}^n (x_i - n + 1) \\ge \\prod_{i=1}^n \\frac{x_i^{n-1}}{x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n} = 1,\n$$\nas needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19450,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\xi$ be the positive root of the equation $x^2 + x - 4 = 0$. The polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$, where $n$ is a positive integer, has nonnegative integer coefficients and $P(\\xi) = 2017$.\n\n1. Prove that $a_0 + a_1 + \\dots + a_n \\equiv 1 \\pmod{2}$.\n2. Find the least possible value of the sum $a_0 + a_1 + \\dots + a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Since $\\xi = \\frac{-1 + \\sqrt{17}}{2}$ is irrational and the polynomial $F(x) = P(x) - 2017$ has rational coefficients and $\\xi$ as a root, it will also have the conjugate $\\frac{-1 - \\sqrt{17}}{2}$ as a root. Therefore, $F(x)$ is divisible by $\\varphi(x) = x^2 + x - 4$. This follows from the identity:\n\n$$\nF(x) = P(x) - 2017 = (x^2 + x - 4) Q(x) + \\kappa x + \\lambda,\n$$\n\nSubstituting $x = \\xi$, we get $\\kappa \\xi + \\lambda = 0$, which implies $\\kappa = \\lambda = 0$ since $\\xi$ is irrational. Thus, there exists a polynomial $Q(x)$ such that:\n\n$$\nP(x) - 2017 = (x^2 + x - 4) Q(x)\n$$\n\nFor $x = 1$:\n\n$$\nP(1) - 2017 = (1^2 + 1 - 4) Q(1) = (-2) Q(1)\n$$\n\nSo,\n\n$$\na_0 + a_1 + \\dots + a_n = P(1) = 2017 - 2 Q(1) \\equiv 1 \\pmod{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19451,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a triangle $ABC$ is *great* if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$.\n\nProve that triangle $ABC$ is great if and only if $\\angle A = 90^\\circ$ and $AB = AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "In all solutions presented for this problem, the point $E$ denotes the reflection of $D$ in the line $PQ$, and $\\Gamma$ denotes the circumcircle of triangle $ABC$. Furthermore, all solutions presented split neatly into the following three steps.\n\n**Step 1** Prove that if $\\triangle ABC$ is great, then $\\angle BAC = 90^\\circ$.\n\n**Step 2** Prove that if $\\triangle ABC$ is great, then $AB = AC$.\n\n**Step 3** Prove that if $\\angle BAC = 90^\\circ$ and $AB = AC$, then $\\triangle ABC$ is great.\n\n**Solution 1** (Based on the solution of William Hu, year 10, Christ Church Grammar School, WA)\n\n**Step 1** Let $D \\in BC$ be such that $AD$ bisects $\\angle BAC$. Then $\\triangle ADP \\equiv \\triangle ADQ$ (AAS), and so $AP = AQ$ and $DP = DQ$. Hence the line $AD$ is the perpendicular bisector of $PQ$. It follows that $E$ lies on the line $AD$.\n\nSince $\\angle DPA = \\angle AQD = 90^\\circ$, the points $D, P, A$, and $Q$ all lie on the circle with diameter $AD$. Furthermore, since $P$ and $Q$ lie on opposite sides of line $AD$, quadrilateral $DPAQ$ is cyclic in that order. Hence $A$ and $D$ lie on opposite sides of the line $PQ$. Therefore, $E$ and $A$ lie on the same side of $PQ$.\n\nNote that $DPAQ$ is a cyclic kite having $AD$ as an axis of symmetry.\n\n**Case 1** $\\angle BAC < 90^\\circ$.\n\nWe have $\\angle QDP > 90^\\circ$ due to $DPAQ$ being cyclic. Thus\n\n$$\n\\angle DPQ = \\angle PQD < 45^\\circ \\quad \\text{and} \\quad \\angle AQP = \\angle QPA > 45^\\circ.\n$$\n\nIt follows that\n\n$$\n\\angle QPE = \\angle DPQ < 45^\\circ < \\angle QPA,\n$$\n\nand similarly $\\angle EQP < \\angle AQP$. Thus point $E$ lies strictly inside $\\triangle APQ$ and hence strictly inside $\\Gamma$. Hence $\\triangle ABC$ is not great.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p142_data_6fbd4e0ddc.png)\n\n**Case 2** $\\angle BAC > 90^\\circ$.\n\nDefine the region $\\mathcal{R}$ as follows. Let $B_1$ be any point on the extension of the ray $BA$ beyond $A$, and let $C_1$ be any point on the extension of the ray $CA$ beyond $A$. Then $\\mathcal{R}$ is the (infinite) region that lies strictly between rays $AB_1$ and $AC_1$.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p143_data_337aa5c3fd.png)\n\nLet us return to our consideration of case 2. Since $DPAQ$ is cyclic, it follows that $\\angle QDP = 180^\\circ - \\angle BAC < 90^\\circ$. Thus\n\n$$\n\\angle DPQ = \\angle PQD > 45^\\circ \\quad \\text{and} \\quad \\angle AQP = \\angle QPA < 45^\\circ.\n$$\n\nIt follows that\n\n$$\n\\angle QPE = \\angle DPQ > 45^\\circ > \\angle QPA,\n$$\n\nand similarly $\\angle EQP > \\angle AQP$. Then since $E$ lies on the same side of $PQ$ as $A$, it follows that $E$ lies strictly inside $\\mathcal{R}$.\n\nObserve that $\\Gamma$ already intersects each of the lines $AB$ and $AC$ twice. Hence $\\Gamma$ has no further intersection point with either of these lines. This implies that $\\mathcal{R}$, and hence also the point $E$, lie strictly outside of $\\Gamma$. Hence $\\triangle ABC$ is not great.\n\nSince we have ruled out cases 1 and 2, it follows that $\\angle BAC = 90^\\circ$.\n\n**Step 2** Let $D$ be the midpoint of $BC$. From step 1, we know that $\\angle BAC = 90^\\circ$. Hence $D$ is the centre of $\\Gamma$. Recall that the projection of the centre of a circle onto any chord of the circle is the midpoint of the chord. So $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $AC$. Thus $PQ$ is the midline of $\\triangle ABC$ that is parallel to $BC$. Hence $d(A, PQ) = d(D, PQ)$.${}^1$ From the reflection we have $d(E, PQ) = d(D, PQ)$. It follows that points $A$ and $E$ lie at the same height above the line $PQ$, and therefore lie at the same height above the line $BC$.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p143_data_38e8c05305.png)\n\n${}^1$For any points $X, Y, Z$, the expression $d(X, YZ)$ denotes the distance from the point $X$ to the line $YZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19452,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $c_n$ be the number of triples $(x, y, z)$ of integers such that $0 \\le x \\le y \\le z \\le x+y$ and $x+y+z = n$. Prove that, for $n \\ge 2$,\n\n$$\n n \\cdot c_n \\le 9 \\cdot (c_0 + c_1 + \\dots + c_{n-2}).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X_n = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le y \\le z \\le x+y \\text{ and } x+y+z = n\\}$. Then $c_n = |X_n|$.\n\nLet $Y_n = \\{(p, q, r) \\in \\mathbb{Z}^3 \\mid 0 \\le p, q, r \\text{ and } 2p + 3q + 4r = n\\}$ and define the maps $f: X_n \\to Y_n$ by\n\n$$\n f(x, y, z) = (y - x, x + y - z, z - y)\n$$\n\nand $g: Y_n \\to X_n$ by\n\n$$\n g(p, q, r) = (q + r, p + q + r, p + q + 2r).\n$$\n\nThen $f$ and $g$ are inverse to each other, so $c_n = |Y_n|$, i.e., $c_n$ is the number of ways $n$ can be expressed as a sum of 2, 3, and 4.\n\nThus,\n\n$$\n n \\cdot c_n = \\sum_{(p, q, r) \\in Y_n} 2p + 3q + 4r.\n$$\n\nFor $m \\ge 0$, the number of elements of $Y_n$ with $p \\ge m$ is $c_{n-2m}$. So there are $c_{n-2m} - c_{n-2(m+1)}$ elements with $p = m$ (with $c_k = 0$ if $k < 0$). Thus, 2 is added $\\sum_{m \\ge 1} c_{n-2m}$ times in the sum above. Similarly, we can compute the coefficients for 3 and 4, and hence\n\n$$\n\\begin{aligned}\n n \\cdot c_n &= \\sum_{k=2}^{4} \\sum_{m=1}^{\\left\\lfloor n/k \\right\\rfloor} k c_{n-km} \\\\\n &= \\sum_{j=2}^{n} \\left( \\sum_{\\substack{2 \\le k \\le 4 \\\\ k \\mid j}} k \\right) c_{n-j} \\le 9 \\sum_{m=0}^{n-2} c_m.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19453,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 為全體實數所成的集合。考慮集合\n\n$$\n\\mathcal{F} := \\{ f : \\mathbb{R} \\to \\mathbb{R} \\mid f(x + f(y)) = f(x) + f(y), \\quad \\forall x, y \\in \\mathbb{R} \\}.\n$$\n\n找出所有的有理數 $q$，使得對所有函數 $f \\in \\mathcal{F}$，存在 $z \\in \\mathbb{R}$ 滿足 $f(z) = qz$。",
    "options": [],
    "answer": "See solution",
    "solution": "所求的有理數集合為 $\\left\\{ \\frac{n+1}{n} : n \\in \\mathbb{Z},\\ n \\neq 0 \\right\\}$。\n\n設 $Z$ 為所有滿足對每個 $f \\in \\mathcal{F}$，存在 $z \\in \\mathbb{R}$ 使 $f(z) = qz$ 的有理數 $q$ 的集合。令\n\n$$\nS = \\left\\{ \\frac{n+1}{n} : n \\in \\mathbb{Z},\\ n \\neq 0 \\right\\}.\n$$\n\n證明 $Z = S$，分別證明 $S \\subseteq Z$ 與 $Z \\subseteq S$。\n\n**(1) 證明 $S \\subseteq Z$：**\n\n對任意 $f \\in \\mathcal{F}$，設 $P(x, y)$ 表示 $f(x + f(y)) = f(x) + f(y)$。由 $P(0, 0)$ 得 $f(f(0)) = 2f(0)$，由 $P(0, f(0))$ 得 $f(2f(0)) = 3f(0)$。可歸納證明：\n\n$$\nf(kf(0)) = (k+1)f(0), \\quad \\forall k \\geq 1,\\ k \\in \\mathbb{Z}.\n$$\n\n同理，$P(-f(0), 0)$ 得 $f(-f(0)) = 0$，可歸納證明：\n\n$$\nf(-kf(0)) = (-k+1)f(0), \\quad \\forall k \\geq 1,\\ k \\in \\mathbb{Z}.\n$$\n\n因此，對每個 $n \\neq 0$，存在 $z = nf(0)$ 使 $f(z) = \\frac{n+1}{n}z$，故 $S \\subseteq Z$。\n\n**(2) 證明 $Z \\subseteq S$：**\n\n設 $p$ 為 $S$ 外的有理數。構造 $f \\in \\mathcal{F}$，使對任意 $z$，$f(z) \\neq pz$。定義 $g : [0,1) \\to \\mathbb{Z}$，再設 $f(x) = g(\\{x\\}) + \\lfloor x \\rfloor$。可驗證 $f \\in \\mathcal{F}$。\n\n**引理 1.** 對任意 $\\alpha \\in [0,1)$，存在 $m \\in \\mathbb{Z}$，使得對所有 $n \\in \\mathbb{Z}$，$m + n \\neq p(\\alpha + n)$。\n\n*證明略。*\n\n由此構造 $g$，則 $f(z) \\neq pz$ 對所有 $z$ 成立，故 $p \\notin Z$。因此 $Z \\subseteq S$。\n\n---\n\n*評分建議：*\n\n1. 完成 $S \\subseteq Z$ 證明得 2 分；\n2. 完成 $Z \\subseteq S$ 證明得 5 分；\n3. 若僅構造出滿足條件的 $f$，但未完成證明，可部分給分。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19454,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ and $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be non-negative integers. Prove that\n$$\n\\left(\\frac{n}{n-1}\\right)^{n-1}\\left(\\frac{1}{n}\\sum_{i=1}^{n} a_i^2\\right) + \\left(\\frac{1}{n}\\sum_{i=1}^{n} b_i\\right)^2 \\geqslant \\prod_{i=1}^{n}\\left(a_i^2 + b_i^2\\right)^{\\frac{1}{n}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\lambda = \\left(\\frac{n}{n-1}\\right)^{n-1}$, $n \\ge 2$. Obviously, $\\lambda > 1$.\n\nFor a given $i \\in \\{1, \\dots, n\\}$, fix $p = a_k^2 + b_k^2$ for $k = 1, 2, \\dots, n$ and fix $a_j$ and $b_j$ ($j \\ne i$). Then the left-hand side of\n$$\n\\textcircled{1} = \\frac{\\lambda}{n} (p - b_i^2 + \\sum_{j \\ne i} a_j^2) + \\frac{1}{n^2} (b_i + \\sum_{j \\ne i} b_j)^2\n$$\nis a quadratic function of $b_i$, $b_i \\in [0, \\sqrt{p}]$, with leading coefficient $-\\frac{\\lambda}{n} + \\frac{1}{n^2} < 0$. Thus, its minimum is taken at the endpoints, that is, $b_i = 0$ or $a_i = 0$.\n\nSo, we can suppose that $a_i b_i = 0$ for $i = 1, 2, \\dots, n$.\n\n*Case 1.* Each $a_i = 0$. Then by the mean value inequality, we have\n$$\n\\left(\\frac{1}{n} \\sum_{i=1}^{n} b_i\\right)^2 \\geq \\prod_{i=1}^{n} b_i^{\\frac{2}{n}}.\n$$\n\n*Case 2.* Each $b_i = 0$. Then by the mean value inequality, we have\n$$\n\\lambda \\left( \\frac{1}{n} \\sum_{i=1}^{n} a_i^2 \\right) \\geq \\frac{1}{n} \\sum_{i=1}^{n} a_i^2 \\geq \\prod_{i=1}^{n} a_i^{\\frac{2}{n}}.\n$$\n\n*Case 3.* We may suppose that $b_1 = \\cdots = b_k = 0$, $a_{k+1} = \\cdots = a_n = 0$, $1 \\le k < n$.\n\nLet $a_1 a_2 \\cdots a_k = a^k$, $b_{k+1} \\cdots b_n = b^{n-k}$, $a, b \\ge 0$. Then by the mean value inequality, we have\n$$\na_1^2 + a_2^2 + \\cdots + a_k^2 \\ge k a^2, \\quad b_{k+1} + \\cdots + b_n \\ge (n-k) b.\n$$\nIt suffices to prove that\n$$\n\\frac{\\lambda k}{n} a^2 + \\frac{(n-k)^2}{n^2} b^2 \\geq a^{\\frac{2k}{n}} \\cdot b^{\\frac{2(n-k)}{n}}. \\qquad \\textcircled{2}\n$$\nBy the mean value inequality, we see that\n\nThe left-hand side of\n$$\n\\textcircled{2} = \\frac{\\lambda}{n} a^2 + \\cdots + \\frac{\\lambda}{n} a^2 + \\underbrace{\\frac{n-k}{n^2} b^2}_{k \\text{ terms}} + \\cdots + \\underbrace{\\frac{n-k}{n^2} b^2}_{n-k \\text{ terms}}\n\\geq \\lambda^{\\frac{k}{n}} a^{\\frac{2k}{n}} \\cdot \\left( \\frac{n-k}{n} \\right)^{\\frac{n-k}{n}} \\cdot b^{\\frac{2(n-k)}{n}}.\n$$\nSo, it suffices to show that $\\lambda^{\\frac{k}{n}} \\left(\\frac{n-k}{n}\\right)^{\\frac{n-k}{n}} \\ge 1$, that is, to show $\\left(\\frac{n}{n-k}\\right)^{n-k} \\le \\lambda^k$.\n\nIn fact,\n$$\n\\begin{align*}\n& \\underbrace{\\frac{n}{n-k} \\cdot \\frac{n}{n-k} \\cdot \\dots \\cdot \\frac{n}{n-k}}_{n-k \\text{ terms}} \\cdot \\underbrace{1 \\cdot 1 \\cdot \\dots \\cdot 1}_{n-k \\text{ terms}} \\\\\n& \\le \\left( \\frac{n + (nk - n)}{nk - k} \\right)^{nk-k} \\\\\n& = \\left( \\frac{n}{n-1} \\right)^{(n-1)k} = \\lambda^k. \\quad \\square\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19455,
    "subject": "Mathematics (Olympiad)",
    "question": "For which integers $n$ with $2018 \\leq n \\leq 3018$ is it possible to construct a collection of $n$ tri-connected squares? (A collection of squares is *tri-connected* if each square touches exactly three others, and the collection is connected as a whole.)",
    "options": [],
    "answer": "See solution",
    "solution": "For any collection of $n$ tri-connected squares, consider the graph $G$ obtained as follows. Each vertex of $G$ corresponds to a square in the collection, and two vertices of $G$ are joined by an edge if and only if the two corresponding squares touch. Observe that the sum of the degrees of the vertices of $G$ is equal to $3n$. Thus the total number of edges of $G$ is equal to $\\frac{3n}{2}$. This implies that $n$ is even.\n\nConsider the following two configurations of squares.\n\n![](images/2018-Australian-Scene-W2_p126_data_34618433c7.png)\n\nCall the 6-square configuration on the left an *A-piece*, and the 4-square configuration on the right a *B-piece*. Five *A-pieces* can be linked together to make a tri-connected collection of 30 squares. And four *B-pieces* can be linked together to make a tri-connected collection of 16 squares. These are shown below.\n\n![](images/2018-Australian-Scene-W2_p126_data_3e368c22aa.png)\n\nEach even integer $n$ with $2018 \\leq n \\leq 3018$ can be written in the form $n = 30a + 16b$ for some integers $a, b \\geq 0$. For example, $\\{30, 60, 90, 120, 150, 180, 210, 240\\}$ represents all even congruence classes modulo 16, and then we can top up with multiples of 16.\n\nTake $a$ copies of the 30-square configuration on the left and $b$ copies of the 16-square configuration on the right, making sure that all $a+b$ configurations are mutually disjoint. This yields a tri-connected collection of $n$ squares. Indeed, we have shown that a tri-connected collection of $n$ squares exists for all even $n \\geq 240$.\n\nAlternatively, we could link together $a$ *A-pieces* and $b$ *B-pieces* into a single closed loop of $a+b$ pieces whenever $a$ and $b$ are non-negative integers with $a+b \\geq 4$. Thus any $n = 6a + 4b = 2a + 4(a+b)$ with $a+b \\geq 4$ is possible. This shows that a tri-connected collection of $n$ squares exists for any even $n \\geq 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19456,
    "subject": "Mathematics (Olympiad)",
    "question": "For all positive integers $m, n$, and given that $f(2024) = 10120$, prove that there exist two integers $m, n$ with $m \\neq n$ such that $f(m) = f(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "By setting $n = 1$, we obtain $mf(1) = 0$ for all $m$, so $f(1) = 0$. Next, since every positive integer can be written as a product of prime numbers, the function $f$ is determined completely by its value on the primes. If there exists $m \\ge 2$ such that $f(m) = 0$, then $f(m) = f(1)$ and we are done. Thus, assume for all $m \\ge 2$ that $f(m) \\ge 1$.\n\nThus,\n\n$$\n\\begin{aligned}\nf(2024) &= f(8 \\cdot 11 \\cdot 23) = 23f(8 \\cdot 11) + 88f(23) = 23(11f(8) + 8f(11)) + 88f(23) \\\\\n&= 253f(8) + 184f(11) + 88f(23).\n\\end{aligned}\n$$\n\nTo study $f(8)$, we derive a general formula for perfect powers:\n\n$$\nf(a^2) = af(a) + af(a) = 2af(a),\n$$\n\n$$\nf(a^3) = a^2f(a) + af(a^2) = 3a^2f(a).\n$$\n\nAssume as inductive hypothesis that\n\n$$\nf(a^n) = n a^{n-1} f(a),\n$$\n\nand prove it for $n+1$:\n\n$$\nf(a^{n+1}) = f(a^n \\cdot a) = a f(a^n) + a^n f(a) = a (n a^{n-1}) f(a) + a^n f(a) = (n+1) a^n f(a).\n$$\n\nThus,\n\n$$\n10120 = f(2024) = 253(3 \\cdot 2^2)f(2) + 184f(11) + 88f(23) = 3036f(2) + 184f(11) + 88f(23). \\quad (1)\n$$\n\nConsidering this equation modulo 11, we conclude that $f(11)$ is divisible by 11, and by the assumption that $f(11) \\ge 1$, we conclude that $f(11) \\ge 11$. Similarly, working modulo 23 tells us that $f(23)$ is divisible by 23 and thus $f(23) \\ge 23$. Finally, working modulo 8 tells us that $4f(2)$ is divisible by 8, so $f(2) \\ge 2$. Thus, the right-hand side of (1) is greater than or equal to\n\n$$\n3036 \\cdot 2 + 184 \\cdot 11 + 88 \\cdot 23 = 10120.\n$$\n\nSince we have equality, we must conclude that\n\n$$\nf(2) = 2, \\quad f(11) = 11, \\quad \\text{and} \\quad f(23) = 23.\n$$\n\nTo find two integers $m$ and $n$ for which $f(m) = f(n)$, set\n\n$$\nm = 2^a 11^b 23^c, \\quad n = 2^d 11^e 23^f.\n$$\n\nWe notice that $2 \\cdot 11 = 22$ is close to 23, so let us study\n\n$$\n\\begin{aligned}\nf(22^a 23^c) &= 23^c f(22^a) + 22^a f(23^c) \\\\\n&= 23^c (a \\cdot 22^{a-1} f(22)) + 22^a (c \\cdot 23^{c-1} f(23)) \\\\\n&= 23^c (a \\cdot 22^{a-1} (2f(11) + 11f(2))) + 22^a (c \\cdot 23^{c-1} \\cdot 23) \\\\\n&= 23^c (a \\cdot 22^{a-1} \\cdot 44) + 22^a (c \\cdot 23^{c-1} \\cdot 23) \\\\\n&= (2a + c) 22^a 23^c.\n\\end{aligned}\n$$\n\nNow compare $f(22^a 23^{b+1})$ with $f(22^{a+1} 23^b)$:\n\n$$\n\\begin{aligned}\nf(22^a 23^{b+1}) &= (2a + b + 1) 22^a 23^{b+1}, \\\\\nf(22^{a+1} 23^b) &= (2(a + 1) + b) 22^{a+1} 23^b.\n\\end{aligned}\n$$\n\nIf we can find values of $a, b$ such that\n\n$$\n2a + b + 1 = 22, \\quad 2a + b + 2 = 23,\n$$\n\nwe are done. There are many such pairs, for example:\n\n$(a, b) \\in \\{(1, 19), (2, 17), (3, 15), (4, 13), (5, 11), (6, 9), (7, 7), (8, 5), (9, 3), (10, 1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19457,
    "subject": "Mathematics (Olympiad)",
    "question": "Disponemos de $2n$ bombillas colocadas en dos filas (A y B) y numeradas de 1 a $n$ en cada fila. Algunas (o ninguna) de las bombillas están encendidas y el resto apagadas; decimos que eso es un “estado”. Dos estados son distintos si hay una bombilla que está encendida en uno de ellos y apagada en el otro. Diremos que un estado es “bueno” si hay la misma cantidad de bombillas encendidas en la fila A que en la B.\n\nDemuestra que el número total de estados buenos, $EB$, dividido por el número total de estados, $ET$, es\n\n$$\n\\frac{EB}{ET} = \\frac{3 \\cdot 5 \\cdot 7 \\cdots (2n-1)}{2^n n!}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Es obvio que $ET = 2^{2n}$, puesto que cada una de las $2n$ bombillas puede estar apagada o encendida. El número de “estados buenos” con $k$ bombillas encendidas en cada fila es $\\binom{n}{k}^2$, ya que hay $\\binom{n}{k}$ formas de elegir las $k$ bombillas encendidas de la fila A y otras tantas de elegir las $k$ bombillas encendidas de la fila B. En consecuencia,\n\n$$\nEB = \\sum_{k=0}^{n} \\binom{n}{k}^2\n$$\n\nEs conocido que esa suma da como resultado $EB = \\binom{2n}{n}$. En cualquier caso, basta con observar que un estado es “bueno” si hay exactamente $k$ ($0 \\le k \\le n$) bombillas encendidas en la fila A y exactamente $n-k$ bombillas apagadas en la fila B. Cada “estado bueno” se obtiene (de una única manera) eligiendo $n$ bombillas en total y haciendo que estén encendidas las elegidas de la fila A y las no elegidas de la fila B. Así pues, $EB = \\binom{2n}{n}$.\n\nAhora, tenemos:\n\n$$\n\\begin{align*}\n\\frac{EB}{ET} &= \\frac{\\binom{2n}{n}}{2^{2n}} = \\frac{(2n)!}{2^{2n} n! n!} = \\frac{\\left(\\prod_{k=1}^{n} 2k\\right) \\left(\\prod_{k=1}^{n} (2k-1)\\right)}{2^{2n} n! n!} \\\\\n&= \\frac{2^n n! \\left(\\prod_{k=1}^{n} (2k-1)\\right)}{2^{2n} n! n!} = \\frac{3 \\cdot 5 \\cdot 7 \\cdots (2n-1)}{2^n n!}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19458,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\geq 3$, show that the number of lists of jointly coprime positive integers that sum to $n$ is divisible by $3$.\n\nFor example, if $n = 4$, there are six such lists:\n- $(3, 1)$\n- $(1, 3)$\n- $(2, 1, 1)$\n- $(1, 2, 1)$\n- $(1, 1, 2)$\n- $(1, 1, 1, 1)$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(n)$ be the number of decompositions of $n$ into jointly coprime parts, where a decomposition is a list of $k$ positive integers summing to $n$. The 1-element list $(n)$ is counted only if $n = 1$.\n\nThe total number of unrestricted decompositions of $n$ is $2^{n-1}$, since in a string of $n$ dots, we may insert dividing bars at any subset of the $n-1$ places between dots.\n\nGrouping decompositions by the greatest common divisor $d$ of the parts, we have:\n\n$$\n\\sum_{d \\mid n} f(n/d) = 2^{n-1}\n$$\n\nThus, $f(1) = f(2) = 1$ and $f(3) = 3$. Proceeding by induction on $n$:\n- If $n > 3$ and $n$ is even, $f(n) \\equiv 2^{n-1} - f(1) - f(2) \\pmod{3}$.\n- If $n > 3$ and $n$ is odd, $f(n) \\equiv 2^{n-1} - f(1) \\pmod{3}$.\n\nSince $2^{n-1} \\equiv 2 \\pmod{3}$ when $n$ is even and $2^{n-1} \\equiv 1 \\pmod{3}$ when $n$ is odd, it follows that $f(n)$ is divisible by $3$ for $n \\geq 3$.\n\n**Remark.** By the Möbius inversion formula,\n$$\nf(n) = \\sum_{d \\mid n} \\mu(n/d) 2^{d-1}\n$$\nwhere $\\mu$ is the Möbius function. As a corollary, $f(n)$ is odd if and only if $n$ is squarefree.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19459,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a$, $b$, $c$, and $d$,\n\n$$\n(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) = u,\n$$\n$$\nab + ac + ad + bc + bd + cd = v,\n$$\nprove that the product $uv$ is divisible by $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "If among the numbers $a$, $b$, $c$, $d$ there are two whose remainders modulo $3$ are either both $0$ or $1$ and $2$, then their sum is divisible by $3$. Thus, $u$, and therefore $uv$, is divisible by $3$.\n\nNow consider the case where at most one of $a$, $b$, $c$, $d$ is divisible by $3$, and all others are congruent modulo $3$. If exactly one is divisible by $3$, then all products involving this number are divisible by $3$. The remaining numbers form $3$ pairs whose products are congruent modulo $3$, so the sum $v$ is divisible by $3$.\n\nIf none of $a$, $b$, $c$, $d$ is divisible by $3$, then all pairwise products are congruent modulo $3$. Since there are $6$ pairs (a multiple of $3$), their sum $v$ is divisible by $3$. Thus, in all cases, $uv$ is divisible by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19460,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $D$ be a point on the altitude through $C$.\n\nProve that the midpoints of the line segments $AD$, $BD$, $BC$, and $AC$ form a rectangle.",
    "options": [],
    "answer": "See solution",
    "solution": "The problem is represented in the following figure:\n\n![](images/Austrija_2013_p16_data_19fe9171c1.png)\n\nWe denote by $M_{XY}$ the midpoint of the line segment $XY$.\n\nUsing the intercept theorem, we deduce that:\n\n- $M_{AD}M_{BD}$ is parallel to $AB$.\n- $M_{AC}M_{BC}$ is parallel to $AB$.\n- $M_{AC}M_{AD}$ is parallel to $CD$.\n- $M_{BC}M_{BD}$ is parallel to $CD$.\n\nTherefore, $M_{AD}M_{BD}$ is parallel to $M_{AC}M_{BC}$ and $M_{AC}M_{AD}$ is parallel to $M_{BC}M_{BD}$.\n\nFurthermore, $M_{AC}M_{BC}$ is orthogonal to $M_{AC}M_{AD}$, since $CD$ is orthogonal to $AB$.\n\nTherefore, the four midpoints form a rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19461,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an acute triangle $ABC$ with altitudes $AA_1$, $BB_1$, and $CC_1$ ($A_1 \\in BC$, $B_1 \\in AC$, $C_1 \\in AB$). A point $C'$ on the extension of $B_1A_1$ beyond $A_1$ is such that $A_1C' = B_1C_1$. Analogously, a point $B'$ on the extension of $A_1C_1$ beyond $C_1$ is such that $C_1B' = A_1B_1$, and a point $A'$ on the extension of $C_1B_1$ beyond $B_1$ is such that $B_1A' = C_1A_1$. Denote by $A''$, $B''$, and $C''$ the symmetric points of $A'$, $B'$, and $C'$ with respect to $BC$, $CA$, and $AB$, respectively. Prove that if $R$, $R'$, and $R''$ are the circumradii of $\\triangle ABC$, $\\triangle A'B'C'$, and $\\triangle A''B''C''$, then $R$, $R'$, and $R''$ are the side lengths of a triangle whose area equals one half of the area of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcenter of $\\triangle ABC$ and $S$ its symmetric point with respect to $AB$. Then $SOCH$ is a parallelogram. $SH \\perp A_1B_1$ and $SH = R$. Denote by $\\rho$ the inradius of triangle $A_1B_1C_1$, and by $q$ its semiperimeter. If $T$ is the orthogonal projection of $S$ to $A_1B_1$, then $HT = \\rho$, $ST = R + \\rho$, $TC' = q$. We have $HC' = \\sqrt{\\rho^2 + q^2}$, which due to symmetry implies that $H$ is the circumcenter of $\\triangle A'B'C'$ and $R' = HC'$. On the other hand, $OC'' = SC' = \\sqrt{(R+\\rho)^2 + q^2}$, and therefore $O$ is the circumcenter of $\\triangle A''B''C''$ and $R'' = SC'$. It follows that $R$, $R'$, and $R''$ are the side lengths of triangle $HSC'$, and\n\n$$\nS_{HSC'} = \\frac{HS \\cdot TC'}{2} = \\frac{R \\cdot q}{2} = \\frac{1}{2} (S_{A_1OB_1C} + S_{B_1OC_1A} + S_{C_1OA_1B}) = \\frac{S_{ABC}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19462,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 5$. What is the maximum possible size of an intersecting set of diagonals in a convex $n$-gon, where a set of diagonals is called *intersecting* if every pair of diagonals in the set intersects in the interior of the polygon?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2n - 1$ for an $n$-gon with $n \\geq 5$.\n\nLet $A$ be an intersecting set with $|A| \\geq n$. Since the number of vertices in $A$ is not greater than the number of line segments in $A$, $A$ contains a cycle. Let $PQ$ and $QR$ be two line segments in this cycle. Every line segment in the cycle, except $XP$ and $RY$, intersects both $PQ$ and $QR$ at their interior points. We conclude that the cycle contains an odd number of line segments and none of the vertices on the cycle lies inside the angle $\\angle PQR$. Hence, there is a positive integer $k$ and there are vertices $P_1, P_2, \\dots, P_{2k+1}$ in the order they lie on the polygon such that the cycle consists of the line segments $P_iP_{i+k}, P_iP_{i+k+1}$ for $1 \\leq i \\leq 2k+1$. (Indices are considered modulo $n$.) Let us denote this collection by $A^*$. Since $A$ is intersecting, the only other line segments in $A$ can be of the form $XP_i$ where $X$ is a vertex of the polygon lying inside the angle $\\angle P_{i+k+1}P_iP_{i+k}$. Since $|A| \\geq n$, all such line segments must belong to $A$. Let us denote the collection of these line segments by $A^\\mathcal{C}$. Therefore, if $A$ is an intersecting set with $|A| \\geq n$, then $|A| = n$ and $A = A_{(P_1, P_2, \\dots, P_{2k+1})} = A^* \\sqcup A^\\mathcal{C}$ where $A^*$ and $A^\\mathcal{C}$ are as described above.\n\nNext, we show that if $A$ and $B$ are intersecting sets with $|A| = n = |B|$, then $A$ and $B$ cannot be disjoint. Assume they are. Let $Q_1Q_2, Q_2Q_3, \\dots, Q_{m-1}Q_m, Q_mQ_1$ be a cycle such that the line segments $Q_iQ_{i+1}$ belong to, say, $A$ for $i$ odd and $B$ for $i$ even (where $Q_{m+1} = Q_1$). Such a cycle exists as every vertex of the polygon is an endpoint of at least one line segment in $A$ and one in $B$. Since $A$ and $B$ are intersecting, all $Q_iQ_{i+1}$ intersect $Q_1Q_2$ for $i$ odd and $Q_2Q_3$ for $i$ even. Therefore, all $Q_i$ with $i$ even lie on or the opposite side of $Q_1Q_2$ with respect to $Q_3$ and all $Q_i$ with $i$ odd lie on or the opposite side of $Q_2Q_3$ with respect to $Q_1$. In particular, $m$ is odd and $Q_1$ is a vertex of $A^*$. Then $Q_3$ lies on or the opposite side of $Q_mQ_1$ with respect to $Q_2$, and $Q_{m-1}$ lies on or the opposite side of $Q_1Q_2$ with respect to $Q_m$. Let $XQ_1$ be a line segment belonging to $B$. Since $XQ_1$ must intersect both $Q_2Q_3$ and $Q_{m-1}Q_m$, which belong to $B$, $X$ must lie between $Q_2$ and $Q_m$ on the polygon. But then $XQ_1$ also belongs to $A^c$ and hence to $A$, a contradiction.\n\nFinally, if $P, Q, R, S, T$ are five consecutive vertices on the polygon, then $A_{(P,Q,R)} \\cup A_{(R,S,T)}$ consists of $2n - 1$ line segments.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19463,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest positive integer $n$ for which\n\n$$\n4^n + 2^{2012} + 1\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer. Suppose first that $n = 2011$.\n\nThen $4^{2011} + 2^{2012} + 1 = (2^{2011} + 1)^2$ is a perfect square.\n\nNow suppose $n > 2011$. Then\n\n$$\n(2^n)^2 < 4^n + 2^{2012} + 1 < (2^n + 1)^2,\n$$\n\nso $4^n + 2^{2012} + 1$ lies between two successive squares.\n\nHence $4^n + 2^{2012} + 1$ is not a perfect square.\n\n**Answer:** $n = 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19464,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}^+$ 為全體正實數所成的集合。找出所有函數 $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ 使得\n\n$$\nf(x + y^2 f(y)) = f(1 + y f(x)) f(x)\n$$\n\n對所有正實數 $x, y$ 恆成立。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $P(x, y)$ 表示將 $(x, y)$ 帶入原題條件。\n\n$$\nP(1, 1) \\Rightarrow f(1) = 1.\n$$\n\n$$\nP(1, y) \\Rightarrow f(1 + y^2 f(y)) = f(1 + y). \\qquad (1)\n$$\n\n$$\nP(x, 1) \\Rightarrow f(x + 1) = f(1 + f(x)) f(x). \\qquad (2)\n$$\n\n比較 $P(1 + x^2 f(x), y)$ 和 $P(1 + y^2 f(y), x)$，有\n\n$$\n\\begin{aligned}\nf(1 + x^2 f(x) + y^2 f(y)) &= f(1 + y f(1 + x^2 f(x))) f(1 + x^2 f(x)) \\\\\n&= f(1 + y f(1 + x)) f(1 + x) \\qquad (3) \\\\\n&= f(1 + x f(1 + y)) f(1 + y).\n\\end{aligned}\n$$\n\n因此，考慮另一函數 $g : \\mathbb{R}_{>-1} \\to \\mathbb{R}^+$ 滿足 $g(x) = f(1 + x)$ 對於所有 $x > -1$。則 (3) 式可改寫為\n\n$$\ng(y g(x)) g(x) = g(x g(y)) g(y). \\qquad (4)\n$$\n\n考慮 $P(1 + x, y)$ 以及 (4) 式，有\n\n$$\n\\begin{aligned}\nP(1 + x, y) &\\Rightarrow f(1 + x + y^2 f(y)) = f(1 + y f(1 + x)) f(1 + x) \\qquad (5) \\\\\n&\\Rightarrow g(x + y^2 g(y - 1)) = g(y g(x)) g(x) = g(x g(y)) g(y).\n\\end{aligned}\n$$\n\n假設存在兩相異正實數 $a, b$ 滿足 $f(a) = f(b)$，由 (2) 式可推得\n\n$$\nf(a + 1) = f(1 + f(a)) f(a) = f(1 + f(b)) f(b) = f(b + 1).\n$$\n\n**注意到**\n\n$$\ng(a) = f(a + 1) = f(b + 1) = g(b) \\Rightarrow (a + 1)^2 g(a) \\neq (b + 1)^2 g(b).\n$$\n\n將 $(x, y) = (x, a + 1)$ 和 $(x, y) = (x, b + 1)$ 帶入 (5) 式，可得\n\n$$\ng(x + (a + 1)^2 g(a)) = g(x g(a + 1)) g(a + 1) = g(x g(b + 1)) g(b + 1) = g(x + (b + 1)^2 g(b)).\n$$\n\n令 $c = |(b + 1)^2 g(b) - (a + 1)^2 g(a)| > 0$ 和 $M = \\max\\{(a + 1)^2 g(a), (b + 1)^2 g(b)\\}$，則上式可改寫為\n\n$$\ng(x + c) = g(x) \\quad \\forall x > M.\n$$\n\n對於 $y > -1$，若 $g(y) \\neq 1$，取 $x_0 > \\frac{M}{g(y)}$ 滿足 $x_0 + y^2 g(y - 1) = x_0 g(y) + m c > M$ 對於某個 $m \\in \\mathbb{Z}$，且將 $(x, y) = (x_0, y)$ 代入 (5) 式，\n\n$$\ng(x_0 + y^2 g(y - 1)) = g(x_0 g(y)) g(y) \\Rightarrow g(y) = 1.\n$$\n\n因此，$f(y) = 1$ 對於所有 $y > 0$。\n\n若不存在兩相異正實數 $a, b$ 滿足 $f(a) = f(b)$，則 (1) 式可推得\n\n$$\n1 + y^2 f(y) = 1 + y \\Rightarrow f(y) = \\frac{1}{y}, \\forall y > 0.\n$$\n\n代回原題驗證可得 $f(y) \\equiv 1$ 和 $f(y) = \\frac{1}{y}$ 皆為原方程式的解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19465,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $a_1, a_2, \\dots, a_{2n+1}$ be $2n+1$ positive real numbers. For $k = 1, 2, \\dots, 2n+1$, define\n\n$$\nb_k = \\max_{0 \\le m \\le n} \\left( \\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i \\right),\n$$\n\nwhere the subscript of $a_i$ is taken modulo $2n+1$.\n\nProve that the number of subscripts $k$ satisfying $b_k \\ge 1$ does not exceed $2 \\sum_{i=1}^{2n+1} a_i$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $I = \\{k \\mid b_k \\ge 1\\}$. For every $k \\in I$, assume that the maximum value $b_k$ of $\\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i$ is attained at $m = m_k$, and call\n\n$$\n[k - m_k, k + m_k] := \\{k - m_k, k - m_k + 1, \\dots, k + m_k\\}\n$$\n\na \"nice segment\", where the subscripts are taken modulo $2n+1$. Obviously, the union of all nice segments contains $I$.\n\n**Claim:** There exists a collection of nice segments whose union contains $I$, and moreover, each $i \\in \\{1, 2, \\dots, 2n + 1\\}$ is contained in at most two segments.\n\n**Proof of claim:** If $[1, 2n + 1]$ is a nice segment, the conclusion is trivial. Otherwise, suppose $i$ is contained in $r$ nice segments\n\n$$\n[i - u_1, i + v_1], \\dots, [i - u_r, i + v_r]\n$$\n\nwhere $r \\ge 3$ and $0 \\le u_j, v_j < 2n$. Let $u_j = \\max\\{u_1, \\dots, u_r\\}$ and $v_k = \\max\\{v_1, \\dots, v_r\\}$. We can keep the nice segments $[i - u_j, i + v_j]$ and $[i - u_k, i + v_k]$, and drop the other $r-2$ segments. Now the segments still cover $1, \\dots, 2n+1$, and at most two of them cover $i$. For each $i \\in \\{1, 2, \\dots, 2n+1\\}$, perform the above operation. Eventually, we find a collection of nice segments with the desired properties.\n\nFor the original problem, let $[i_1 - m_1, i_1 + m_1], \\dots, [i_r - m_r, i_r + m_r]$ be a collection of nice segments chosen in the claim. We have\n\n$$\n2 \\sum_{i=1}^{2n+1} a_i \\ge \\sum_{\\alpha=1}^{r} \\sum_{k=i_{\\alpha}-m_{\\alpha}}^{i_{\\alpha}+m_{\\alpha}} a_k = \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1)b_{i_{\\alpha}} \\ge \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1) \\ge |I|,\n$$\n\nthe first inequality is due to each $i$ being contained in at most two nice segments; the equality and the next inequality are due to the definition of nice segments; the last inequality is due to the union of the nice segments containing $I$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19466,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Plugging $x = 1$ yields $m(a, y) = m(a, f(y))$ for all $y \\in \\mathbb{N}^*$, where $a = f(1)$. Also, taking $y = 1$ leads to $d(x, a) f(x) = m(a, f(x))$ for all $x \\in \\mathbb{N}^*$. This implies $d(x, a) f(x) = m(x, a)$ for all $x \\in \\mathbb{N}^*$. (*)\n\nRelation (*) implies $a f(a) = a$ and $a f(a^2) = a^2$, 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 $f(x) = x$ for all $x \\in \\mathbb{N}^*$, which obviously fulfills the required condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19467,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many pairs of positive integers $(m, n)$ such that simultaneously $m$ divides $n^2 + 1$ and $n$ divides $m^2 + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $F_n$ the $n$th Fibonacci number. We claim that $(n, m) = (F_{2k-1}, F_{2k+1})$ is a solution for all positive integers $k$.\n\nFirst, we show that for all positive integers $k$,\n\n$$\nF_{2k+1}^2 + 1 = F_{2k-1} \\cdot F_{2k+3}.\n$$\n\nThis will be proved by induction on $k$. For $k=1$, this is true since\n\n$$\nF_3^2 + 1 = 2^2 + 1 = 5 = 1 \\cdot 5 = F_1 \\cdot F_5.\n$$\n\nNow suppose that $F_{2k-1}^2 + 1 = F_{2k-3} \\cdot F_{2k+1}$. Note first that $F_{2k+3} = 3F_{2k+1} - F_{2k-1}$, by repeatedly applying the relation $F_{m+2} = F_{m+1} + F_m$. Then\n\n$$\n\\begin{align*}\nF_{2k-1} \\cdot F_{2k+3} &= F_{2k-1}(3F_{2k+1} - F_{2k-1}) \\\\\n&= 3F_{2k-1} \\cdot F_{2k+1} - F_{2k-1}^2 \\\\\n&= 3F_{2k-1} \\cdot F_{2k+1} - (F_{2k-3} \\cdot F_{2k+1} - 1) \\\\\n&= F_{2k+1}(3F_{2k-1} - F_{2k-3}) + 1 \\\\\n&= F_{2k+1} \\cdot F_{2k+1} + 1 \\\\\n&= F_{2k+1}^2 + 1.\n\\end{align*}\n$$\n\nNow it follows immediately that $F_{2k-1}$ divides $F_{2k+1}^2 + 1$ and $F_{2k+1}$ divides $F_{2k-1}^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19468,
    "subject": "Mathematics (Olympiad)",
    "question": "Given integer $n \\ge 2$, let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n > 0$ satisfy\n\n$$\na_1 + a_2 + \\dots + a_n = b_1 + b_2 + \\dots + b_n\n$$\n\nand for any $i, j$ ($1 \\le i < j \\le n$), $a_i a_j \\ge b_i + b_j$ always holds. Find the minimum value of $a_1 + a_2 + \\dots + a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = a_1 + a_2 + \\cdots + a_n = b_1 + b_2 + \\cdots + b_n$.\n\nBy the given conditions,\n\n$$\n\\sum_{1 \\le i < j \\le n} a_i a_j \\ge \\sum_{1 \\le i < j \\le n} (b_i + b_j) = (n-1)S.\n$$\n\nAlso, since\n$$\nS^2 = \\left(\\sum_{i=1}^{n} a_i\\right)^2 = \\sum_{i=1}^{n} a_i^2 + 2 \\sum_{1 \\le i < j \\le n} a_i a_j,\n$$\nwe have\n$$\n\\sum_{1 \\le i < j \\le n} a_i a_j \\ge (n-1)S.\n$$\n\nThus,\n$$\nS^2 \\ge \\sum_{i=1}^{n} a_i^2 + 2(n-1)S.\n$$\n\nBut $\\sum_{i=1}^{n} a_i^2 \\ge 0$, so $S^2 \\ge 2(n-1)S$, or $S \\ge 2n$.\n\nOn the other hand, if $a_i = b_i = 2$ for all $i = 1, 2, \\dots, n$, the conditions are satisfied and $S = 2n$.\n\nTherefore, the minimum value of $a_1 + a_2 + \\dots + a_n$ is $2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19469,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{R}$ (where $\\mathbb{N}$ denotes the set of all positive integers and $\\mathbb{R}$ the set of all real numbers) such that\n\n$$\nf(km) + f(kn) - f(k)f(nm) \\geq 1\n$$\n\nfor all $k, m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Plugging in $k = n = m = 1$ yields\n\n$$\nf(1)^2 - 2f(1) + 1 = (f(1) - 1)^2 \\leq 0,\n$$\n\nwhich implies $f(1) = 1$.\n\nPlugging in $k = 1, n = m$ and $k = n, m = 1$, respectively, we obtain the two inequalities\n\n$$\n2f(n) - f(n^2) \\geq 1, \\qquad (1)\n$$\n\n$$\nf(n^2) + f(n) - f(n)^2 \\geq 1. \\qquad (2)\n$$\n\nWe add the two to get\n\n$$\n-f(n)^2 + 3f(n) \\geq 2\n$$\n\nor\n\n$$\n(f(n) - 1)(f(n) - 2) \\leq 0.\n$$\n\nThis means that $1 \\leq f(n) \\leq 2$ for all $n$.\n\nNow assume that there is a positive integer $n$ such that $f(n) = a > 1$. The function $g(x) = \\frac{x^2+1}{x} = x + \\frac{1}{x}$ is increasing on $(1, \\infty)$: for $x > y > 1$,\n\nwe have\n\n$$\ng(x) - g(y) = x - y + \\frac{1}{x} - \\frac{1}{y} = (x - y) \\left(1 - \\frac{1}{xy}\\right) > 0.\n$$\n\nHence, for any $x \\geq a$, we have\n\n$$\nx^2 + 1 \\geq \\frac{a^2 + 1}{a}x\n$$\n\nand thus\n\n$$\nx^2 - x + 1 \\geq \\frac{a^2 - a + 1}{a} x.\n$$\n\nReturning to (2), we now find, by our assumption that $f(n) = a$,\n\n$$\nf(n^2) \\geq f(n)^2 - f(n) + 1 \\geq \\frac{a^2 - a + 1}{a} f(n),\n$$\n\nand since\n\n$$\nb = \\frac{a^2 - a + 1}{a} = 1 + \\frac{(a - 1)^2}{a} > 1,\n$$\n\nwe get\n\n$$\nf(n^2) \\geq b f(n) \\geq a.\n$$\n\nIterating this inequality yields $f(n^4) \\geq b^2 f(n) = b^2 a$, $f(n^8) \\geq b^3 f(n) = b^3 a$, etc., and generally (by induction) $f(n^{2^k}) \\geq b^k a$. Since $a > 1$ and $b > 1$, this implies $f(n^{2^k}) > 2$ for sufficiently large $k$, which contradicts the inequality $1 \\leq f(n^{2^k}) \\leq 2$ that was obtained earlier.\n\nIt follows that there is no $n$ such that $f(n) > 1$, which means that the constant function $f(n) \\equiv 1$ is the only solution (and it is easy to see that this function satisfies the condition, since the left hand side of the inequality is always equal to 1 in this case).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19470,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega_1$, $\\omega_2$, and $\\omega_3$ be circles with centers $O_1$, $O_2$, and $O_3$, respectively. Consider the locus of a point $O$ constructed as follows:\n\n- For any point $A$ on $\\omega_3$, there exists a point $B$ on $\\omega_2$ and a point $C$ on $\\omega_1$ such that $A$, $B$, and $C$ are related by homotheties centered at $Q$ and $P$ (with $Q$ mapping $\\omega_3$ to $\\omega_1$, and $P$ mapping $\\omega_1$ to $\\omega_2$).\n- The triangle $AOB$ is constructed so that $O$ is determined by the geometric configuration involving these homotheties and spiral similarities.\n\nDescribe the locus of $O$ as $A$ varies over all positions on $\\omega_3$, considering both cases when the radii of $\\omega_2$ and $\\omega_3$ are different and when they are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's first consider the case when the radii of $\\omega_2$ and $\\omega_3$ are different (without loss of generality, assume that the radius of $\\omega_3$ is greater than that of $\\omega_2$). Consider three homotheties: $\\ell_1$ centered at $Q$ mapping $\\omega_3$ to $\\omega_1$; $\\ell_2$ centered at $P$ mapping $\\omega_1$ to $\\omega_2$; and $\\ell_3 = \\ell_2 \\circ \\ell_1$ mapping $\\omega_3$ to $\\omega_2$. Since $\\ell_1$ maps $A$ to $C$ and $\\ell_2$ maps $C$ to $B$, $\\ell_3$ maps $A$ to $B$, so its center $I$ lies on the line $AB$.\n\nLet $\\angle ACB = \\angle PCQ = \\alpha$ (a fixed value). Then $\\angle AOB = 2\\alpha$ is also fixed, so all triangles $AOB$ are similar and the ratio $\\frac{OA}{AB} = \\frac{1}{2\\sin\\alpha}$ is fixed.\n\nLet $\\frac{IA}{IB} = t > 1$ be the homothety coefficient of $\\ell_3$, which is fixed. Then\n$$\n\\frac{AB}{IA} = \\frac{IB-IA}{IA} = \\frac{1}{t} - 1\n$$\nis also fixed.\n\nIn triangle $IAO$,\n$$\n\\frac{OA}{IA} = \\frac{OA}{AB} \\cdot \\frac{AB}{IA} = \\frac{1}{2\\sin\\alpha} \\cdot \\left(\\frac{1}{t} - 1\\right)\n$$\nand\n$$\n\\angle OAI = 180^{\\circ} - \\angle OAB = 90^{\\circ} + \\alpha.\n$$\n\nHence all triangles $IAO$ are similar, so the angle $AIO$ and the ratio $\\frac{IO}{IA}$ are fixed. Therefore, for any position of $C$ on $\\omega_1$, the spiral similarity $\\ell$ with center $I$, angle $AIO$, and coefficient $\\frac{IO}{IA}$ maps $A$ to $O$. As $A$ varies over $\\omega_3$, the locus of $O$ is the circle $\\omega = \\ell(\\omega_3)$.\n\nThe homothety $\\ell_3$ and spiral similarity $\\ell$ share center $I$. Thus $\\ell(B) = \\ell \\circ \\ell_3(A) = \\ell_3 \\circ \\ell(A) = \\ell_3(O_3) = O_2$. Hence $\\ell$ maps triangle $AOB$ to triangle $O_3O'O_2$ and\n$$\n\\angle O_3O'O_2 = \\angle AOB = 2\\angle ACB = \\angle QO_1P = \\angle O_3O_1O_2,\n$$\nso $O'$ lies on the circumcircle of triangle $O_1O_2O_3$.\n\nNow, if the radii of $\\omega_2$ and $\\omega_3$ are equal, $\\ell_3$ is a translation by vector $\\overrightarrow{O_3O_2}$, the point $I$ is not defined, and quadrilateral $ABO_2O_3$ is a parallelogram. Let $D$ be such that triangles $ABD$ and $O_3O_2O_1$ are equal and similarly oriented. Then quadrilateral $O_3ADO_1$ is a parallelogram, and triangles $QAO_3$ and $QCO_1$ are similar. Thus, segments $DO_1$ and $O_1C$ are parallel and\n$$\nDO_1 + O_1C = O_3A \\left(1 + \\frac{QO_1}{QO_3}\\right) = O_3Q \\left(1 + \\frac{QO_1}{QO_3}\\right) = O_1O_3 = DA = DB.\n$$\nTherefore, $D$ coincides with $O$. Also, the distance $O_1D = O_3A$ is fixed, so all points $O$ are equidistant from $O_1$.\n\nIt is easy to see that $O$ can occupy any position on the circle with center $O_1$ and radius $O_3A$ — the required assertion is proved.\n\n**Remark.** It follows from the isosceles triangle $AOB$ that the point $O'$ is the midpoint of the arc $O_2O_3$ containing $O_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19471,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the three-digit positive integer $abc$ whose representation in base nine is $\\underline{b}\\,\\underline{c}\\,\\underline{a}_{\\text{nine}}$, where $a$, $b$, and $c$ are (not necessarily distinct) digits.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the three-digit integer be $100a + 10b + c$ in base ten. Its base nine representation is $bca_{\\text{nine}} = 81b + 9c + a$.\n\nSet the two equal:\n$$\n100a + 10b + c = 81b + 9c + a\n$$\nRearrange:\n$$\n100a - a + 10b - 81b + c - 9c = 0 \\\\\n99a - 71b - 8c = 0\n$$\nModulo $71$:\n$$\n99a - 8c \\equiv 0 \\pmod{71}\n$$\nSince $99 \\equiv 28 \\pmod{71}$:\n$$\n28a - 8c \\equiv 0 \\pmod{71}\n$$\nDivide by $4$ (since $4$ divides both $28$ and $8$):\n$$\n7a - 2c \\equiv 0 \\pmod{71}\n$$\nThe possible values for $7a - 2c$ (with $a, c$ digits) are between $-18$ and $63$, so the only multiple of $71$ in this range is $0$:\n$$\n7a - 2c = 0 \\implies 7a = 2c\n$$\nSince $a$ and $c$ are digits, $c$ must be a multiple of $7$. Try $c = 7$:\n$$\n7a = 2 \\times 7 = 14 \\implies a = 2\n$$\nNow, substitute $a = 2$, $c = 7$ into the original equation to solve for $b$:\n$$\n100 \\times 2 + 10b + 7 = 81b + 9 \\times 7 + 2 \\\\\n200 + 10b + 7 = 81b + 63 + 2 \\\\\n207 + 10b = 81b + 65 \\\\\n207 - 65 = 81b - 10b \\\\\n142 = 71b \\\\\nb = 2\n$$\nThus, the three-digit integer is $227$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19472,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all primes $p$ such that $p^3 - 4p + 9$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We check directly that $p=2$ is a solution ($2^3 - 4 \\cdot 2 + 9 = 3^2$), and $p=3$ is not. Henceforth, let $p > 3$.\n\nIf $p^3 - 4p + 9 = n^2$ for some $n \\in \\mathbb{N}$, then $p^3 - 4p = n^2 - 9$, so\n$$\n(p-2)p(p+2) = (n-3)(n+3).\n$$\nOne of the numbers $p-2$, $p$, $p+2$ is divisible by $3$, hence $n$ is also a multiple of $3$. If $n=3k$, $k \\ge 1$, then\n$$\n(p-2)p(p+2) = 9(k-1)(k+1).\n$$\nBoth sides are nonzero as $p \\ne 2$. The prime $p > 3$ divides $9(k-1)(k+1)$, hence it divides exactly one of $k-1$ and $k+1$.\n\nLet $p$ divide $k-1$. Write $k-1 = l p$ to obtain\n$$\n(p-2)(p+2) = 9l(lp+2).\n$$\nReducing mod $p$ gives $18l + 4 \\equiv 0 \\pmod{p}$. Since $p$ is odd, it divides $9l+2$. In particular, $p \\le 9l+2$, $p-2 \\le 9l$, so $(p-2)(p+2) = 9l(lp+2)$ implies $p+2 \\ge lp+2$. This is possible only if $l=1$ and $p=9l+2=11$.\n\nWe have a solution $p=11$ ($11^3 - 4 \\cdot 11 + 9 = 36^2$).\n\nLet $p$ divide $k+1$ and $k+1 = l p$. Then\n$$\n(p-2)(p+2) = 9l(lp-2).\n$$\nTake this mod $p$ to obtain $18l-4 \\equiv 0 \\pmod{p}$. It follows that $p$ divides $9l-2$. In particular, $p \\le 9l-2$, $p+2 \\le 9l$, so $(p-2)(p+2) = 9l(lp-2)$ implies $p-2 \\ge lp-2$. Therefore $l=1$ and $p=9l-2=7$. We find one more solution: $p=7$ ($7^3 - 4 \\cdot 7 + 9 = 18^2$).\n\nIn summary, the solutions are $p=2, 7, 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19473,
    "subject": "Mathematics (Olympiad)",
    "question": "The two distinct non-zero common roots of $f$ and $g$ are also roots of $g - f$. These are the two roots of the quadratic\n\n$$\ng(x) - f(x) = 3 \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right)\n$$\n\nFind the values of $\\alpha$ and $\\beta$ given that\n$$\ng(x) = x^3 + (\\alpha - 3)x - 6.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g(x) - f(x) = 3 \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right)$. Thus, $g(x)$ must be of the form\n$$\ng(x) = (x - \\lambda) \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right)\n$$\nfor some real number $\\lambda$.\n\nExpanding:\n$$\n\\begin{aligned}\ng(x) &= (x - \\lambda) \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right) \\\\\n&= x^3 - (1 + \\lambda)x^2 + \\left( \\lambda - 2 - \\frac{\\beta}{3} \\right)x + \\lambda \\left( 2 + \\frac{\\beta}{3} \\right)\n\\end{aligned}\n$$\n\nGiven $g(x) = x^3 + (\\alpha - 3)x - 6$, compare coefficients:\n- $x^3$: Coefficient is $1$ in both.\n- $x^2$: $-(1 + \\lambda) = 0$ $\\implies \\lambda = -1$\n- $x$: $\\lambda - 2 - \\frac{\\beta}{3} = \\alpha - 3$\n  Substitute $\\lambda = -1$:\n  $-1 - 2 - \\frac{\\beta}{3} = \\alpha - 3 \\implies -3 - \\frac{\\beta}{3} = \\alpha - 3 \\implies \\alpha = -\\frac{\\beta}{3}$\n- Constant: $\\lambda \\left( 2 + \\frac{\\beta}{3} \\right) = -6$\n  Substitute $\\lambda = -1$:\n  $-1 \\left( 2 + \\frac{\\beta}{3} \\right) = -6 \\implies 2 + \\frac{\\beta}{3} = 6 \\implies \\frac{\\beta}{3} = 4 \\implies \\beta = 12$\n\nTherefore, $\\alpha = -\\frac{12}{3} = -4$.\n\n**Final answer:**\n$$\n\\boxed{\\alpha = -4,\\ \\beta = 12}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19474,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABC$ 為銳角三角形，$\\Gamma$ 為其外接圓，$I$ 為其內心。兩圓 $\\omega_B, \\omega_C$ 分別過點 $B, C$，並相切於點 $I$。設 $\\omega_B$ 與 $\\Gamma$ 上的 $AB$ 劣弧交於點 $P$，同時 $\\omega_B$ 與線段 $AB$ 再交於點 $M$；類似有 $\\omega_C$ 與 $\\Gamma$ 上的 $AC$ 劣弧交於點 $Q$，同時 $\\omega_C$ 與線段 $AC$ 再交於點 $N$。令射線 $PM$ 與 $QN$ 交於點 $X$，且設 $\\omega_B$ 過 $B$ 的切線與 $\\omega_C$ 過 $C$ 的切線交於點 $Y$。\n\n試證：$A, X, Y$ 三點共線。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $AI, BI, CI$ 分別與 $\\Gamma$ 再次交於 $D, E, F$。設 $\\ell$ 為 $\\omega_B$ 和 $\\omega_C$ 在 $I$ 的公切線。設 $\\angle(p, q)$ 為從直線 $p$ 到 $q$ 的有向角，模 $180^\\circ$。\n\n**步驟 1：證明 $Y$ 在 $\\Gamma$ 上。**\n\n任意圓的弦與其端點的切線形成互補的有向角，因此\n\n$$\n\\angle(BY, BI) + \\angle(CI, CY) = \\angle(IB, \\ell) + \\angle(\\ell, IC) = \\angle(IB, IC).\n$$\n\n因此，\n\n$$\n\\begin{aligned}\n\\angle(BY, BA) + \\angle(CA, CY) &= \\angle(BI, BA) + \\angle(BY, BI) + \\angle(CI, CY) + \\angle(CA, CI) \\\\\n&= \\angle(BC, BI) + \\angle(IB, IC) + \\angle(CI, CB) = 0,\n\\end{aligned}\n$$\n\n故 $Y \\in \\Gamma$。\n\n**步驟 2：證明 $X = \\ell \\cap EF$。**\n\n設 $X_* = \\ell \\cap EF$。只需證明 $X_*$ 在 $PM$ 和 $QN$ 上，即 $X_* = X$。由對稱性，只需證明 $X_* \\in QN$。\n\n注意到\n\n$$\n\\angle(IX_*, IQ) = \\angle(CI, CQ) = \\angle(CF, CQ) = \\angle(EF, EQ) = \\angle(EX_*, EQ);\n$$\n\n因此，$X_*, I, Q, E$ 共圓（若 $Q = E$，則 $EQ$ 的方向為 $\\Gamma$ 在 $Q$ 的切線方向，此時等式表示圓 $X_*IQ$ 在 $Q$ 處與 $\\Gamma$ 相切）。接著有\n\n$$\n\\angle(QX_*, QI) = \\angle(EX_*, EI) = \\angle(EF, EB) = \\angle(CA, CF) = \\angle(CN, CI) = \\angle(QN, QI),\n$$\n\n故 $X_* \\in QN$。\n\n**步驟 3：證明 $A, X, Y$ 共線。**\n\n$ I $ 為 $ DEF $ 的垂心，且 $ A $ 關於 $ EF $ 對稱於 $ I $。因此，\n\n$$\n\\angle(AX, AE) = \\angle(IE, IX) = \\angle(BI, \\ell) = \\angle(BY, BI) = \\angle(BY, BE) = \\angle(AY, AE),\n$$\n\n故 $A, X, Y$ 共線。\n\n**補充 1.** 步驟 2 是關鍵，完成後有不同方法收尾。\n\n例如，可用*等角共軛*。設 $X_1, Y_1$ 分別為 $X, Y$ 關於 $ABC$ 的等角共軛。因 $XA = XI$，三角形 $AIX$ 是等腰，故 $AX$ 與 $AI$ 與角平分線 $AI$ 夾角相等，即 $AX_1 \\parallel XI$ 或 $AX_1 \\parallel \\ell$。\n\n另一方面，$BY$ 與 $\\ell$ 與 $BI$ 夾角相等，故 $BY_1 \\parallel \\ell$，同理 $CY_1 \\parallel \\ell$，故 $Y_1$ 為無窮遠點，$AY_1 \\parallel \\ell$。因此 $A, X_1, Y_1$ 共線，故 $A, X, Y$ 共線。\n\n**補充 2.** 另有以 $I$ 為中心的反演法，暫不詳述。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19475,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the real numbers $x_1, x_2, \\dots, x_5$ satisfy a set of equations such that each $x_j$ ($j = 1, 2, \\dots, 5$) is a root of the quadratic equation $x^2 - a x - 1 = 0$, where $a = x_1 + x_2 + \\dots + x_5$. What are the possible values of the $x_j$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the roots of $x^2 - a x - 1 = 0$ be $\\alpha$ and $\\beta$, so $\\alpha \\beta = -1$. Suppose exactly $k$ of the $x_j$ equal $\\alpha$ and $5-k$ equal $\\beta$. Then $a = k\\alpha + (5-k)\\beta = k\\alpha - \\frac{5-k}{\\alpha}$. Substituting into $\\alpha^2 - a\\alpha - 1 = 0$ gives $(k-1)\\alpha^2 = 4-k$. Since $\\alpha^2 > 0$, $k=2$ or $k=3$. For $k=2$, $\\alpha = \\pm\\sqrt{2}$; for $k=3$, $\\alpha = \\pm\\frac{\\sqrt{2}}{2}$. Thus, the possible values are $\\pm\\sqrt{2}$ and $\\pm\\frac{\\sqrt{2}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19476,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of real numbers such that for every $x \\in S$, $x$ can be written as the sum of $k$ other pairwise distinct elements of $S$ (i.e., $x = x_1 + x_2 + \\dots + x_k$ for some distinct $x_i \\in S \\setminus \\{x\\}$). Prove that $|S| \\ge k + 4$.\n\nAdditionally, for each $k \\ge 1$, construct an explicit example of a set $S$ with $|S| = k + 4$ satisfying this property.",
    "options": [],
    "answer": "See solution",
    "solution": "The condition that $S$ have at least three elements forces $m \\ge 2$; otherwise, $S - x = x$, so $x = \\frac{1}{2}s$ for any $x \\in S$, which is a contradiction.\n\nLet $a = \\min S$ and $b = \\max S$. Then:\n\n$$\ns - a = a + x_1 + \\dots + x_{m-1} \\quad \\text{and} \\quad s - b = b + y_1 + \\dots + y_{m-1},\n$$\n\nwhere the $x_i$ and $y_i$ are all members of $S$. Thus,\n\n$$\n2(b-a) = x_1 - y_1 + x_2 - y_2 + \\dots + x_{m-1} - y_{m-1} < (m-1)(b-a),\n$$\n\nsince at least one $x_i$ is less than $b$. Consequently, $(m-3)(b-a) > 0$, so $m \\ge 4$ and $|S| = k + m \\ge k + 4$, as desired.\n\n**2nd Proof.** Let $a_1 < a_2 < \\dots < a_n$ be real numbers satisfying the condition. Since $a_1$ is a sum of $k$ other distinct $a_i$, $a_1 \\ge a_2 + \\dots + a_{k+1}$. Similarly, $a_{n-k} + \\dots + a_{n-1} \\ge a_n$. Adding,\n\n$$\na_1 + a_{n-k} + \\dots + a_{n-1} \\ge a_2 + \\dots + a_{k+1} + a_n.\n$$\n\nIf $n = k + 1, k + 2, k + 3$, the inequality contradicts the order of the $a_i$. Thus, $n \\ge k + 4$.\n\nTo construct such a set, let $k = 2\\ell$ (even). The set $\\{\\pm 1, \\pm 2, \\dots, \\pm (\\ell+2)\\}$ has size $k + 4$ and is sign-invariant. Every positive $i \\le \\ell + 2$ is expressible as required:\n- For $i = 1$ to $\\ell + 1$, $i = -1 + (i+1)$ plus $k-2$ other elements $\\pm j$ ($j \\ne 1, i, i+1$).\n- For $i = \\ell + 2$, $i = 1 + (\\ell + 1)$ plus $k-2$ other elements $\\pm j$ ($j \\ne 1, \\ell+1, \\ell+2$).\n\nFor $k = 2\\ell + 1$ (odd), add $0$ to the above set to get size $k + 4$. $0$ can be added to previous sums, and $0 = -2 + (-1) + 3$ plus $k-3$ other nonzero $\\pm j$ ($j \\ne 1,2,3$). This completes the construction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19477,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us reformulate a more general version of the problem in terms of graph theory:\n\nIn a regular graph $G$ on $n$ vertices, each vertex has degree $k < n-1$, and if two vertices are not neighbours, then they have exactly $k-1$ common neighbours. Find all pairs $(n, k)$ satisfying these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w$ be a vertex with neighbours $v_1, \\dots, v_k$. If some vertex is not a neighbour of $w$, then it is directly connected to $k-1$ neighbours of $w$ and hence has exactly 1 neighbour among the remaining vertices. Therefore, all vertices in the graph $G - \\{w, v_1, \\dots, v_k\\}$ have degree 1, and consequently, the number of vertices not directly connected to $w$ is even. Denote them by $u_1, \\dots, u_{2l}$. Since $1 + l + 2l = n$, we get $2l = n - k - 1$.\n\nFor $1 \\le t \\le l$, let $u_{2t-1}$ and $u_{2t}$ be neighbours. When $l = 1$, we get a pair $(n, n-3)$. Let $l \\ge 2$. For any $t \\ge 3$, the vertices $u_1$ and $u_t$ are not neighbours, and their unique neighbours among vertices $u_i$ are not common. Hence, $u_1$ and $u_t$ have $k-1$ neighbours among vertices $v_i$. The same is true for $u_2$ and $u_t$. Therefore, all these vertices are connected to the same $k-1$ vertices of $w$. Without loss of generality, let $v_1$ be the vertex not connected to $u_i$, $1 \\le i \\le 2l$. Since the degree of $v_1$ is $k$, it is directly connected to each $v_i$. Therefore, for each $2 \\le i \\le k$, the degree of $v_i$ in $G - \\{w, v_1, u_1, \\dots, u_{2l}\\}$ is $k-1-1-2l = 2k-n-1$. Then the graph on the vertices $v_2, \\dots, v_k$ satisfies the problem conditions with new parameters $(k-1, 2k-n-1)$. Since $(k-1)-(2k-n-1) = n-k$, by repeating the same procedure, we get that the graph $G$ contains pieces with $2l+2$ vertices, and each piece contains $l+1$ perfectly matching edges, and all possible edges between different pieces are drawn. By denoting $l+1$ by $a$ and the number of pieces by $b$, we get the desired answer.\n\nWhen $k = 2023$, from $2a(b-1)+1 = 2023$, we get that $a \\mid 1011 = 3 \\cdot 337$. Therefore, the possible values for $a$ are $a = 3, 337, 1011$, and we get $n = 2ab = 2028, 2696, 4044$. The pair $(n, n-3)$ yields $n = 2026$.\n\n**Answer:** $n = 2024, 2026, 2028, 2696, 4044$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19478,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume there exist numbers $m$ and $n$, one even and one odd, such that\n$$\nA = (m + 3n)(5m + 7n)(7m + 5n)(3m + n)\n$$\nis a perfect square.\n\nLet $d = \\gcd(m, n)$, and write $m = d m_1$, $n = d n_1$ so that $m_1$ and $n_1$ are coprime. Then\n$$\nA = d^4 (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1).\n$$\n\nLet $B = (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1)$. Since $A$ is a perfect square, so is $B$.\n\nShow that $B$ cannot be a perfect square for coprime $m_1, n_1$ of different parity.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $B = (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1)$ is a perfect square, with $m_1$ and $n_1$ coprime and of different parity.\n\nLet $p$ be a positive integer dividing both $m_1 + 3n_1$ and $5m_1 + 7n_1$. Then\n$$\np \\mid 5(m_1 + 3n_1) - (5m_1 + 7n_1) = 8n_1,\n$$\n$$\np \\mid 3(5m_1 + 7n_1) - 7(m_1 + 3n_1) = 8m_1.\n$$\nSo $p \\mid m_1$ and $p \\mid n_1$, but $m_1, n_1$ are coprime, so $p = 1$. Thus, $m_1 + 3n_1$ and $5m_1 + 7n_1$ are coprime.\n\nSimilarly, for $q \\mid m_1 + 3n_1$ and $q \\mid 7m_1 + 5n_1$:\n$$\nq \\mid 7(m_1 + 3n_1) - (7m_1 + 5n_1) = 16n_1,\n$$\n$$\nq \\mid 3(7m_1 + 5n_1) - 5(m_1 + 3n_1) = 16m_1.\n$$\nSo $q \\mid m_1$ and $q \\mid n_1$, so $q = 1$. Thus, $m_1 + 3n_1$ and $(5m_1 + 7n_1)(7m_1 + 5n_1)$ are coprime.\n\nSimilarly, $3m_1 + n_1$ is coprime to the other factors.\n\nTherefore, since $B$ is a perfect square and its factors are pairwise coprime, each must be a perfect square. Let $m_1 + 3n_1 = a^2$ and $3m_1 + n_1 = b^2$.\n\nThen\n$$\na^2 - b^2 = (m_1 + 3n_1) - (3m_1 + n_1) = 2(n_1 - m_1).\n$$\nBut $n_1 - m_1$ is odd (since $m_1, n_1$ have different parity), so $2(n_1 - m_1)$ is divisible by $2$ but not by $4$.\n\nHowever, $a^2 - b^2 = (a - b)(a + b)$, and $a - b$ and $a + b$ have the same parity, so their product is divisible by $4$ if and only if both are even, i.e., $a, b$ are both even or both odd. Thus, $a^2 - b^2$ is divisible by $4$, a contradiction.\n\nTherefore, $B$ cannot be a perfect square, and so $(m + 3n)(5m + 7n)(7m + 5n)(3m + n)$ cannot be a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19479,
    "subject": "Mathematics (Olympiad)",
    "question": "Polynomial $P(x)$ with integer coefficients satisfies the following condition: for every polynomials $F(x)$, $G(x)$, $Q(x)$ with integer coefficients, if\n\n$$\nP(Q(x)) = F(x) \\cdot G(x)\n$$\n\nthen either $F(x)$ or $G(x)$ is a constant polynomial. Prove that $P(x)$ has to be a constant polynomial.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* For the sake of contradiction, suppose that $P$ is not constant and consider the case when $P$ is a linear polynomial. This means $P(x) = ax + b$ for some $a, b \\in \\mathbb{Z}$, where $a \\neq 0$. Let $Q(x) = ax^2 + (b+1)x$. Then\n\n$$\nP(Q(x)) = a(ax^2 + (b+1)x) + b = a^2x^2 + a(b+1)x + b = (ax + b)(ax + 1)\n$$\n\nbut polynomials $ax + b$ and $ax + 1$ are not constant, a contradiction.\n\nNow suppose that $\\deg P = n > 1$. Suppose also that\n\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0,\n$$\n\nwhere $a_n \\neq 0$. Consider a polynomial $Q(x) = P(x) + x$. Clearly it has integer coefficients. Moreover,\n\n$$\nP(Q(x)) - P(x) = P(P(x) + x) - P(x) = \\sum_{i=0}^{n} a_i \\left((P(x) + x)^i - x^i\\right)\n$$\n\nFrom the formula\n\n$$\na^i - b^i = (a-b)(a^{i-1} + a^{i-2}b + \\dots + b^{i-1})\n$$\n\nit follows that the polynomial $(P(x) + x)^i - x^i$ is divisible by the polynomial $P(x)$. So $P(Q(x))$ is divisible by $P(x)$ as well. But this is a contradiction, since $\\deg P > 1$ implies that the degree of $P(Q(x))$ is greater than the degree of $P(x)$, which means that $P(x)$ is a non-trivial divisor of $P(Q(x))$. Conclusion follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19480,
    "subject": "Mathematics (Olympiad)",
    "question": "We define a triangle as _large_ if its area is greater than the area of a given convex pentagon, with the vertices of the triangle chosen from those of the pentagon. Determine the maximum number of large triangles that can be formed within a convex pentagon.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: 4.\n\nFirst, we construct a convex pentagon with 4 large triangles. Let $AB_1B_2C_1C_2$ be the convex pentagon such that $AB_1 = B_2C_1 = C_2A = 1$ and $B_1B_2 = C_1C_2 = \\epsilon > 0$. When $\\epsilon$ is sufficiently small, each triangle $AB_iC_j$ is large for each $i$ and $j$.\n\nNow we show that the number of large triangles is at most 4. We define a triangle as \"boundary\" if its sides are formed by 2 sides of the given pentagon and 1 diagonal of the pentagon. Similarly, we define a triangle as \"center\" if its sides are formed by 1 side of the given pentagon and 2 diagonals of the pentagon. Furthermore, we define a triangle as \"small\" if it is not a large triangle.\n\nAssume that there exists a boundary triangle, which we will call $ABC$. Then the four triangles formed from the quadrilateral $ACDE$ must all be small. Additionally, at most one of the triangles $BCD$, $BDE$, or $BEA$ can be large. Therefore, the number of small triangles is at least 6.\n\nNow we assume that all boundary triangles are small. Let $P$ be the intersection point of diagonals $AD$ and $CE$. Without loss of generality, we can assume that $d(B, AD) \\le d(C, AD)$, where $d(X, YZ)$ denotes the distance from point $X$ to line $YZ$ in the Euclidean plane. Then we can see that $S_{BPD} \\le S_{CPD} \\le S_{CED}$. Also, $S_{ABP} \\le \\max\\{S_{ABC}, S_{ABE}\\}$. Since $S_{ABD} = S_{ABP} + S_{BPD}$, we have either $S_{ABD} \\le S_{ABC} + S_{CED} = S - S_{ACE} < S/2$ or $S_{ABD} \\le S_{ABE} + S_{CED} = S - S_{BCE} < S/2$. In both cases, we have a contradiction. Therefore, there must exist small and center triangles. This completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19481,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside triangle $ABC$, there is a point $M$. On side $BC$, there is a point $K$ such that $MK \\parallel AB$. The circle passing through $M$, $K$, and $C$ intersects side $AC$ again at point $N$. The circle passing through $M$, $N$, and $A$ intersects side $AB$ again at point $Q$. Prove that $BM = KQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Draw segments $MQ$ and $MN$, and extend the ray $NM$, choosing an arbitrary point $P$ on its extension. The quadrilaterals $CNMK$ and $ANMQ$ are cyclic by construction, so\n\n$$\n\\angle ACB = \\angle PMK \\quad \\text{and} \\quad \\angle BAC = \\angle PMQ.\n$$\n\nTherefore,\n\n$$\n\\angle ABC = 180^\\circ - \\angle BAC - \\angle BCA = 180^\\circ - \\angle PMQ - \\angle PMK = 180^\\circ - \\angle KMQ.\n$$\n\nThus, the quadrilateral is cyclic. Additionally, it is a trapezoid (or parallelogram) by the given conditions, so it is an isosceles trapezoid (or rectangle). Therefore, its diagonals are equal, which implies $BM = KQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19482,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $10! = 2^8 \\cdot 3^4 \\cdot 5^2 \\cdot 7$ has $270$ positive factors, enumerated as $d_1, d_2, \\dots, d_{270}$ in increasing order. Compute:\n\n$$\n\\frac{1}{2} \\sum_{k=1}^{270} \\left( \\frac{1}{d_k + \\sqrt{10!}} + \\frac{1}{d_{271-k} + \\sqrt{10!}} \\right)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{3}{16\\sqrt{7}}\n$$\n\nFrom $10! = 2^8 \\cdot 3^4 \\cdot 5^2 \\cdot 7$, there are $(8+1) \\cdot (4+1) \\cdot (2+1) \\cdot (1+1) = 270$ positive factors. For each $k$, $d_k \\cdot d_{271-k} = 10!$, so\n\n$$\n\\frac{1}{d_k + \\sqrt{10!}} + \\frac{1}{d_{271-k} + \\sqrt{10!}} = \\frac{1}{\\sqrt{10!}}\n$$\n\nThus, the sum is\n\n$$\n\\frac{1}{2} \\cdot 270 \\cdot \\frac{1}{\\sqrt{10!}} = \\frac{3}{16\\sqrt{7}}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19483,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $k = 4045$ arbitrary marbles are placed in a $2024 \\times 2024$ grid so that no two marbles are placed in two adjacent cells. Is it always possible to move one marble to an adjacent square so that still no two marbles are in adjacent cells? What is the largest such $k$ for which this is always possible?",
    "options": [],
    "answer": "See solution",
    "solution": "If $u_{n-1}$ contains at least two marbles, then one of these marbles can move to the lower diagonal so that no two marbles are adjacent to each other. By the same argument, if each diagonal $u_1, u_2, \\dots, u_{1012}$ contains only one marble, then each white diagonal below $D$ also contains only one marble. Therefore, the total number of marbles is less than 4045, a contradiction.\n\nSo there must be one of the diagonals $u_1, u_2, \\dots, u_{1012}$ that has more than one marble. Suppose $l$ is the first index such that $u_l$ has more than one marble. It follows that $2 \\le l \\le 2012$ and each diagonal $u_1, u_2, \\dots, u_{l-1}$ has exactly one marble. Furthermore, all unique marbles on diagonals $u_1, u_2, \\dots, u_{l-1}$ must be in the center of these diagonals. Call the white cells on the diagonal $u_l$ in order from bottom to top as $a_1, a_2, \\dots, a_{l-1}, a_l, a_{l+1}, \\dots, a_{2l-1}$.\n\n![](images/Vietnam_2024_Booklet_p22_data_78dab3dbe8.png)\n\nIf cell $a_j$ ($2 \\le j \\le l-1$) has no marble, then cell $a_{j-1}$ has no marble either, because otherwise we can move the marble in cell $a_{j-1}$ up as in the figure. Therefore, if cell $a_j$ ($1 \\le j \\le l-2$) has a marble, then cell $a_{j+1}$ must also have a marble. Similarly, if cell $a_j$ ($2 \\le j \\le l-1$) has a marble, then cell $a_{j-1}$ must also have a marble.\n\nIt follows that all the cells $a_1, a_2, \\dots, a_{l-1}$ have marbles or do not have marbles at the same time. A similar statement holds for the cells $a_{l+1}, \\dots, a_{2l-1}$. Therefore, one of the following must hold:\n\n* All cells $a_1, a_2, \\dots, a_{l-1}, a_{l+1}, \\dots, a_{2l-1}$ have marbles. The number of marbles on $u_l$ is $2l-1$ or $2l-2$ depending on whether $a_l$ has a marble or not.\n* All cells $a_1, a_2, \\dots, a_{l-1}, a_l$ have marbles, cells $a_{l+1}, \\dots, a_{2l-1}$ do not have marbles. The number of marbles on $u_l$ is $l$.\n* All cells $a_{l}, \\dots, a_{2l-1}$ have marbles, cells $a_1, a_2, \\dots, a_{l-1}$ do not have marbles. The number of marbles on $u_l$ is $l$.\n\nIn any situation, there is at least 1 marble on diagonal $u_l$. It follows, by the earlier remark, that on each diagonal $u_{l+1}, u_{l+2}, \\dots, u_{1012}$ there must be at least 2 marbles. Therefore, the number of marbles in the upper half of diagonal $D$ is not less than $l - 1 + l + 2(1012 - l) = 2023$.\n\nIt follows that the number of marbles in the lower half of diagonal $D$ does not exceed $4045 - 2023 = 2022$. Therefore, at least one of the $1012$ white diagonals below $D$ must contain exactly one marble. However, by the same argument as for the diagonals $u_1, u_2, \\dots, u_{1012}$, we also deduce that if there is a white diagonal among the white diagonals below $D$ that has exactly one marble, then the number of marbles in the lower half of diagonal $D$ is not less than 2023. We get a contradiction.\n\nSub-case 2b: Suppose there is an empty white diagonal, for example one of the diagonals $u_1, u_2, \\dots, u_{1012}$. Not all of these diagonals can be empty, otherwise we can move a marble in the lower half of diagonal $D$ up. Consider a pair of consecutive white diagonal lines $(u_k, u_{k+1})$, $1 \\le k \\le 1011$, in which one diagonal is empty, the other has at least one marble. The argument is similar to case I; we only need to consider the situation where $u_k$ has no marbles and $u_{k+1}$ has at least one marble. Then, in order to not be able to move marbles from diagonal $u_{k+1}$ to an upper black square, we must have $u_{k+1}$ full with marbles.\n\nThen it is easy to see that each diagonal $u_j$ with $k+1 < j \\le 2022$ must have at least 2 marbles. It follows that the number of balls in the upper part of diagonal $D$ is not less than\n\n$$\n2k+1 + (1012 - k - 1) \\times 2 = 2023.\n$$\n\nTherefore, the number of marbles in the lower half of diagonal $D$ does not exceed $4045 - 2023 = 2022$. There are a total of 1012 white diagonals below $D$, which means that at least one of the white diagonal lines below $D$ has no more than 1 marble. If one of the white diagonals below $D$ is empty, we apply the same argument as for the upper part and deduce that the number of marbles in the lower half of diagonal $D$ is not less than 2023, a contradiction.\n\nIf all white diagonals below $D$ have at least 1 marble, then there is a white diagonal below $D$ that has exactly one marble. We return to the situation considered in sub-case 2a and deduce that the number of marbles in the lower half of diagonal $D$ is not less than 2023, which is a contradiction again.\n\nTherefore, if $k = 4045$ arbitrary marbles are placed in a $2024 \\times 2024$ grid so that no two marbles are placed in two adjacent cells, one marble can always be moved to an adjacent square so that still there are no two marbles in adjacent cells.\n\nWe will prove that 4045 is the largest such number, meaning that for every $4046 \\le k \\le \\frac{2024^2}{2}$ there always exists an arrangement that does not satisfy the conditions in the problem (it is clear that in order for no two marbles to be placed in adjacent cells, we need $k \\le \\frac{2024^2}{2}$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19484,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a square array of size $(3k+1) \\times (3k+1)$ or $(3k+2) \\times (3k+2)$, can you fill the array with two numbers $a$ and $-b$ (with $b > 0$) so that:\n\n- The sum of the numbers inside every $3 \\times 3$ square is negative.\n- The sum of all the numbers in the array is positive?\n\nIf so, describe such a construction and determine the possible values of $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider filling the array as follows:\n\n- Place $a$ in most squares, and $-b$ in certain positions so that each $3 \\times 3$ square contains exactly one $-b$ and eight $a$'s.\n- The sum inside each $3 \\times 3$ square is $8a - b$.\n- To ensure negativity, set $b = 8a + c$ for some positive integer $c$.\n\nThe total sum in the array is $-k^2 b + ((3k+1)^2 - k^2)a$ or $-k^2 b + ((3k+2)^2 - k^2)a$.\n\nTo ensure positivity:\n$$\n-k^2 b + ((3k+1)^2 - k^2)a > 0\n$$\nor\n$$\n-k^2 b + ((3k+2)^2 - k^2)a > 0\n$$\n\nSubstituting $b = 8a + c$ and simplifying, we get:\n$$\n\\frac{6k+1}{k^2} > \\frac{c}{a}\n$$\n\nSetting $a = k$ and $c = 1$ gives $b = 8k + 1$.\n\nThus, for $(3k+1) \\times (3k+1)$ or $(3k+2) \\times (3k+2)$ arrays, we can arrange $k$ and $-8k-1$ as described. The sum inside each $3 \\times 3$ square is $-1$, and the total sum is $5k^2 + k$ or $11k^2 + 4k$, both positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19485,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer.\n\nProve that $a(n) = n^5 + 5^n$ is divisible by $11$ if and only if $b(n) = n^5 \\cdot 5^n + 1$ is divisible by $11$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is a multiple of $11$, both sides of the equivalence are false, so the equivalence holds.\n\nIf $n$ is not a multiple of $11$, Fermat's little theorem implies that $n^{10} - 1$ is divisible by $11$. The equivalence now follows from\n\n$$\nn^5 a(n) = n^{10} + n^5 \\cdot 5^n \\equiv b(n) \\pmod{11}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19486,
    "subject": "Mathematics (Olympiad)",
    "question": "In a circle is inscribed a regular 2018-gon. The numbers $1, 2, \\ldots, 2018$ are placed at the vertices of the 2018-gon, one number per vertex, such that the sum of every two consecutive numbers is equal to the sum of their diametrically opposite numbers. Find the number of all such configurations. (Configurations obtained by rotation around the center of the circle are considered the same.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider a configuration satisfying the conditions. Let $A, B$ be two consecutive numbers in the configuration, and let their diametrically opposite numbers be $a, b$, respectively. Then $A + B = a + b$, i.e., $A - a = b - B$. Since $A, B$ are arbitrary, the difference between numbers on diametrically opposite vertices is constant. That is, $A - a = C$, where $C$ is a constant.\n\nThus, the numbers from $1$ to $2018$ should be paired into $1009$ pairs such that the difference in each pair is $C$. Such a pairing is possible if and only if $C$ divides $1009$.\n\nLet $C = k$. Then the numbers are paired as follows:\n$$\n\\{1, 2, \\ldots, k\\} \\rightarrow \\{k+1, k+2, \\ldots, 2k\\},\n\\{2k+1, 2k+2, \\ldots, 3k\\} \\rightarrow \\{3k+1, 3k+2, \\ldots, 4k\\}, \\ldots,\n\\{2019-2k, \\ldots, 2018-k\\} \\rightarrow \\{2019-k, \\ldots, 2018\\}\n$$\nEach set of $k$ elements is paired with another set of $k$ elements, and every element from $\\{1, 2, \\ldots, 2018\\}$ appears exactly once.\n\nIf the number of such pairings is $m$, then $2018 = 2km$, i.e., $1009 = km$. Since $1009$ is prime, the only possible values for $k$ are $1$ and $1009$.\n\n1) $C = 1$. The pairs are $(1,2), (3,4), \\ldots, (2017,2018)$, and the numbers in each pair are diametrically opposite. Fixing one diameter with the pair $(1,2)$, the configuration can be chosen by arranging the remaining $1008$ pairs, accounting for rotations. Thus, there are $1008!$ configurations in this case.\n\n2) $C = 1009$. The pairs are $(1,1010), (2,1011), \\ldots, (1009,2018)$. The discussion is analogous to the previous case, so there are also $1008!$ configurations here.\n\nTherefore, the total number of configurations is $2 \\cdot 1008!$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19487,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangular grid with side lengths that are integers greater than 1 is given. Smaller rectangles, each with area equal to an odd integer and with each side length an integer greater than 1, are cut out one by one. Finally, one single unit square is left. Find the least possible area of the initial grid before the cuttings.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the unit square that remains. $X$ cannot be in a corner of the initial rectangle, since the strip between $X$'s neighboring rectangle and the edge could not be cut out (see the figure below).\n\n![](images/prob1718_p23_data_34b6243f64.png)\n\nSimilarly, $X$ cannot be at the side of the rectangle, because the strip between two neighboring rectangles of $X$ could not be cut out.\n\n![](images/prob1718_p23_data_326cda3fd0.png)\n\nConsider all ways the 8 unit squares around $X$ can be distributed among the rectangles being cut out. Suppose the eastern neighbor of $X$ belongs to rectangle $A$. Then either the northeastern or southeastern neighbor of $X$ must also belong to $A$; without loss of generality, let it be the northeastern neighbor (see below).\n\n![](images/prob1718_p23_data_67476daf20.png)\n\nThe northern neighbor of $X$ must belong to a rectangle $B$ different from $A$, and the northwestern neighbor must also belong to $B$. Similarly, the western and southwestern neighbors must belong to a third rectangle $C$, and the southern and southeastern neighbors to a fourth rectangle $D$.\n\nThe strips starting from the eastern and southern neighbors of $X$ and continuing to the edge must be distributed among rectangles. Since each rectangle has odd area, their side lengths are odd. Thus, the lengths of these strips must be sums of one or more odd integers greater than 1. These lengths are two consecutive positive integers. The smallest such consecutive integers are 5 (which is odd) and 6 (which is $3+3$). Therefore, at least 5 unit squares must be to the east of $X$, and similarly in each cardinal direction. Thus, each side length of the initial rectangle is at least 11, so the area is at least $11 \\times 11 = 121$. As shown below, this minimum can be achieved.\n\n![](images/prob1718_p23_data_e74e96e3d5.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19488,
    "subject": "Mathematics (Olympiad)",
    "question": "A list of 9 real numbers consists of $1$, $2.2$, $3.2$, $5.2$, $6.2$, and $7$, as well as $x$, $y$, and $z$ with $x \\leq y \\leq z$. The range of the list is $7$, and the mean and the median are both positive integers. How many ordered triples $(x, y, z)$ are possible?\n\n(A) 1  \n(B) 2  \n(C) 3  \n(D) 4  \n(E) infinitely many",
    "options": [],
    "answer": "See solution",
    "solution": "Because the range is $7$, the values of $x$, $y$, and $z$ are in the interval $[0, 8]$. Because the median is an integer, it is one of $x$, $y$, or $z$ and is either $3$, $4$, $5$, or $6$. The sum of the list is $s = 24.8 + x + y + z$, which is between $24.8 + 0 + 0 + 3 = 27.8$ and $24.8 + 6 + 8 + 8 = 46.8$. Because the mean is an integer, $s$ is an integer multiple of $9$, so $s = 36$ or $45$, and $x + y + z = 11.2$ or $20.2$.\n\n* If the median is $3$, then $x \\leq y \\leq z = 3$, so $x + y + z \\leq 9 < 11.2$. Therefore this case cannot occur.\n* If the median is $4$, then $x \\leq y = 4 \\leq z$, so $x + z = 7.2$, and the list spans one of the intervals $[0, 7]$, $[1, 8]$, or $[x, z]$. If $x = 0$, then $z = 7.2 > 7$, and if $z = 8$, then $x = -0.8 < 0$. Therefore these cases cannot occur. Otherwise $z - x = 7$, giving $(x, y, z) = (0.1, 4, 7.1)$.\n* If the median is $5$, then $x \\leq y = 5 \\leq z$, so $x + z = 6.2$. In order to have a range of $7$, $x$ must be $0$, and $(x, y, z) = (0, 5, 6.2)$.\n* If the median is $6$, then $x = 6 \\leq y \\leq z$, so $y + z = 14.2$. In order to have a range of $7$, $z$ must be $8$, and $(x, y, z) = (6, 6.2, 8)$.\n\nThus there are $3$ possible ordered triples $(x, y, z)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19489,
    "subject": "Mathematics (Olympiad)",
    "question": "Three line segments, each of length $1$, form a connected figure on the plane. Any point that is common to two of these line segments is an endpoint of both segments. Find the maximum area of the convex hull of the figure.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, all vertices of the convex hull are endpoints of the line segments. Since the figure is connected, there are at most $4$ distinct endpoints. Thus, the convex hull is either a quadrilateral or a triangle.\n\nIf there are only $3$ meeting points, the convex hull is an equilateral triangle of side $1$, whose area is $S = \\frac{1}{4}\\sqrt{3}$, which is not maximal.\n\nIf the convex hull is a triangle, one endpoint lies inside the triangle. Consider three cases:\n\n* **All segments meet inside the triangle:** The convex hull consists of three triangles, each with two sides of length $1$. Let the angles between the segments be $\\alpha, \\beta, \\gamma$ (all $< 180^\\circ$):\n\n  $$\n  S = \\frac{1}{2}(\\sin \\alpha + \\sin \\beta + \\sin \\gamma) \\leq \\frac{3}{2} \\sin \\frac{\\alpha + \\beta + \\gamma}{3} = \\frac{3}{2} \\sin 120^\\circ = \\frac{3}{4}\\sqrt{3}\n  $$\n  by Jensen's inequality. The bound $\\frac{3}{4}\\sqrt{3}$ is achieved when all angles are $120^\\circ$.\n\n* **Exactly two segments meet inside the triangle:** The convex hull is a triangle with one side of length $1$ and another less than $2$ (by triangle inequality), so $S < 1 < \\frac{3}{4}\\sqrt{3}$.\n\n* **Exactly one segment ends inside the triangle:** The triangle has two sides of length $1$, so $S \\leq \\frac{1}{2} < \\frac{3}{4}\\sqrt{3}$.\n\nIf the convex hull is a quadrilateral, all segments end at vertices. Two cases:\n\n* **One segment is a diagonal:** The other two coincide with sides, so the convex hull consists of two triangles with two sides of length $1$ each, $S \\leq 1 < \\frac{3}{4}\\sqrt{3}$.\n\n* **No segment is a diagonal:** The segments form $3$ consecutive sides. Let the broken line be $ABCD$.\n  - If $\\angle ABC + \\angle BCD \\leq 180^\\circ$, then $S \\leq 1 < \\frac{3}{4}\\sqrt{3}$.\n  - If $\\angle ABC + \\angle BCD > 180^\\circ$, rays $AB$ and $DC$ meet at $E$. Let $\\beta = \\angle EBC$, $\\gamma = \\angle BCE$, $\\alpha = \\angle CEB$. By the law of sines in $EBC$:\n    $$\n    |EB| = \\frac{\\sin \\gamma}{\\sin \\alpha}, \\quad |EC| = \\frac{\\sin \\beta}{\\sin \\alpha}\n    $$\n    The area is:\n    $$\n    \\begin{align*}\n    S &= \\frac{1}{2} (|EA| \\cdot |ED| - |EB| \\cdot |EC|) \\sin \\alpha \\\\\n      &= \\frac{1}{2} (|EB| + |EC| + 1) \\sin \\alpha \\\\\n      &= \\frac{1}{2} (\\sin \\alpha + \\sin \\beta + \\sin \\gamma) \\\\\n      &\\leq \\frac{3}{2} \\sin \\frac{\\alpha + \\beta + \\gamma}{3} = \\frac{3}{2} \\sin 60^\\circ = \\frac{3}{4}\\sqrt{3}\n    \\end{align*}\n    $$\n    by Jensen's inequality.\n\nConsequently, the maximum area of the convex hull is $\\frac{3}{4}\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19490,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and $x_1, x_2, \\dots, x_p$ be integers. Show that if\n$$\nx_1^n + x_2^n + \\dots + x_p^n \\equiv 0 \\pmod{p}\n$$\nfor all positive integers $n$, then $x_1 \\equiv x_2 \\equiv \\dots \\equiv x_p \\pmod{p}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = p - 1$. Then $x_i^{p-1} \\equiv 0$ or $1 \\pmod{p}$. Therefore, the congruence $x_1^n + x_2^n + \\dots + x_p^n \\equiv 0 \\pmod{p}$ holds when either all $x_i$ are divisible by $p$ or none are. If no $x_i$ is divisible by $p$, consider:\n$$\n\\sum_{i=1}^{p} (x_i - x_1)^n = \\sum_{j=0}^{n} \\left( \\sum_{i=1}^{p} x_i^j \\right) \\binom{n}{j} (-x_1)^{n-j} \\equiv 0 \\pmod{p}.\n$$\nThus, $0, x_2 - x_1, x_3 - x_1, \\dots, x_p - x_1$ satisfy the condition, so all $x_i$ are congruent modulo $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19491,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $3N$ marked points be called *outer points*. We divide each of $N$ arcs of length $2$ into two arcs of equal length using one point, and each of $N$ arcs of length $3$ into three arcs of equal length using two points. Let those $3N$ new points be called *inner points*. There are $6N$ outer and inner points in total, and they divide the circle into $6N$ arcs of length $1$.\n\n% IMAGE: ![](images/Mathematica_competitions_in_Croatia_in_2014_p18_data_ad8de1454f.png)\n\nProve that there exist two diametrically opposite outer points.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that no two of the $3N$ outer points are diametrically opposite. Then, diametrically opposite to every outer point lies an inner point, and vice versa, since there are equal numbers of each.\n\nLet $A$ and $B$ be outer points, and $C$ and $D$ their diametrically opposite inner points. Since there cannot be three consecutive inner points, the points $P$ and $Q$ adjacent to $C$ and $D$ must be outer. Thus, across the arc $\\widehat{AB}$ of length $1$ lies the arc $\\widehat{PQ}$ of length $3$. Since there are equal numbers of arcs of length $1$ and $3$, across every arc of length $3$ lies an arc of length $1$.\n\nFor $i \\in \\{1, 2, 3\\}$, let $L_i$ be the number of arcs of length $i$ inside the shorter arc $\\widehat{AP}$, and $D_i$ the number inside $\\widehat{BQ}$. The length of $\\widehat{AP}$ is $3N - 2$, so:\n\n$$\nL_1 + 2L_2 + 3L_3 = 3N - 2. \\quad (*)\n$$\n\nSince across every arc of length $3$ lies an arc of length $1$ and vice versa:\n\n$$\nD_1 = L_3. \\quad (**)\n$$\n\nThere are exactly $N$ arcs of length $1$, so:\n\n$$\nL_1 + D_1 = N - 1,\n$$\n\nwhich, using $(**)$, gives:\n\n$$\nL_1 + L_3 = N - 1. \\quad (***)\n$$\n\nSubtracting $(***)$ from $(*)$ yields:\n\n$$\n2L_2 + 2L_3 = 2N - 1,\n$$\n\nwhich is impossible, since the left side is even and the right side is odd. Therefore, our assumption is false, and there must exist two diametrically opposite outer points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19492,
    "subject": "Mathematics (Olympiad)",
    "question": "The hexagon $ABLCDK$ is inscribed, and the line $LK$ intersects the segments $AD$, $BC$, $AC$, and $BD$ at points $M$, $N$, $P$, and $Q$, respectively. Prove that\n$$\nNL \\cdot KP \\cdot MQ = KM \\cdot PN \\cdot LQ.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $s = \\sin \\frac{\\hat{A}B}{2}$, $t = \\sin \\frac{\\hat{B}L}{2}$, $u = \\sin \\frac{\\hat{L}C + \\hat{A}K}{2}$, $v = \\sin \\frac{\\hat{C}K}{2}$, $w = \\sin \\frac{\\hat{D}K}{2}$, and $x = \\sin \\frac{\\hat{L}D + \\hat{A}K}{2}$. Then, we have\n\n$$\n\\frac{NL \\cdot KP \\cdot MQ}{KM \\cdot PN \\cdot LQ} = \\frac{NL}{NC} \\cdot \\frac{NC}{NP} \\cdot \\frac{KP}{AK} \\cdot \\frac{AK}{KM} \\cdot \\frac{MQ}{DQ} \\cdot \\frac{DQ}{LQ} = \\frac{t}{v} \\cdot \\frac{u}{s} \\cdot \\frac{v}{u} \\cdot \\frac{x}{w} \\cdot \\frac{s}{x} \\cdot \\frac{w}{t} = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19493,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest integer $k$ such that the edges of the complete graph on 10 vertices can be coloured with $k$ colours so that every triangle has edges of at least two different colours?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that for $1 \\leq k \\leq 4$ such a colouring does not exist, but for $k=5$ it is possible.\n\nFor $k=1$ and $k=2$, the claim is trivial.\n\nFor $k=3$: Consider a vertex $A$. There are 9 edges from $A$ to other vertices and only 3 colours, so by the pigeonhole principle, at least two edges (say, $AB$ and $AC$) have the same colour. Then the triangle $ABC$ has at most two colours, contradicting the requirement.\n\nFor $k=4$: Suppose a vertex $A$ has at least 4 edges of the same colour (say, blue), connecting to $B, C, D, E$. Among the edges between $B, C, D, E$, at least one must also be blue, forming a triangle with only two colours. Thus, at most 3 edges from any vertex can be the same colour. By a variant of Ramsey's theorem, among the remaining points, there must be a monochromatic triangle, leading to a contradiction.\n\n% ![](images/Blr2012_p21_data_4c6662adc4.png)\n\nTherefore, $k=4$ is not sufficient.\n\nFor $k=5$: We can partition the edges of the complete graph $K_{10}$ into 5 edge-disjoint graphs, each isomorphic to the graph in Fig. 4, and colour each with a different colour. In this construction, each vertex is incident to 3 edges of each colour, and each monochromatic subgraph contains no triangle. Thus, the required colouring exists for $k=5$.\n\n% ![](images/Blr2012_p22_data_73ff6a9565.png)\n\nTherefore, the answer is $k=5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19494,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $c$, define a sequence by $a_1 = c$ and $a_n = a_{n-1}^3 + c$ for each $n \\ge 2$. Call a prime number $p$ *orange* if for every positive integer $c$ there is some $n$ such that $p \\mid a_n$. Are there infinitely many orange primes?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that every prime $p \\equiv 2 \\pmod{3}$ is orange. There are infinitely many such primes, either (i) by quoting Dirichlet's theorem, or (ii) by noting that if there were only finitely many, say $p_1, \\dots, p_k$, then $3p_1 \\cdots p_k - 1$ has a prime divisor that is $2$ modulo $3$ but can't be any of the primes already on our list, giving a contradiction.\n\n**Claim 1.** Let $p \\equiv 2 \\pmod{3}$ and $x, y \\in \\mathbb{Z}$. If $p \\mid x^3 - y^3$ then $p \\mid x - y$.\n\n*Proof.* If $p \\mid x$ or $p \\mid y$ then we're done, so we can assume $p \\nmid x, y$. Write $p = 3k + 2$ where $k \\in \\mathbb{N}$. By Fermat's little theorem we have $x^{p-1} = x^{3k+1} \\equiv 1 \\pmod{p}$ and similarly for $y$, so:\n\n$$\nx^3 \\equiv y^3 \\pmod{p} \\implies x \\equiv x \\cdot (x^{3k+1})^2 \\equiv (x^3)^{2k+1} \\equiv (y^3)^{2k+1} \\equiv y \\pmod{p}\n$$\n\nwhich gives $p \\mid x - y$.\n\n![](images/BMO2024Shortlist_p74_data_e239cff2cd.png)\n\nNow take $p$ as in the claim and assume for contradiction that $p$ is not orange. By the pigeonhole principle, there are two terms in the sequence which are from the same residue class modulo $p$. Consider the pair $i < j$ with the smallest possible value of $i + j$ such that $p \\mid a_i - a_j$. If $i > 1$ then:\n\n$$\np \\mid a_{i-1}^3 - a_{j-1}^3 \\implies p \\mid a_{i-1} - a_{j-1}\n$$\n\nby our claim, which is a contradiction. Otherwise $i = 1$ and:\n\n$$\np \\mid a_{j-1}^3 + c - c = a_{j-1}^3 \\implies p \\mid a_{j-1}\n$$\n\n![](images/BMO2024Shortlist_p74_data_fcd0de155a.png)\n\nso $p$ is orange as desired.\n\n**Comment.** We can classify all orange primes. By a direct check, $3$ is an orange prime and by the above, $p \\equiv 2 \\pmod{3}$ is orange. If $p \\equiv 1 \\pmod{3}$ then we can choose $-c$ to not be a cube modulo $p$ (as $x \\mapsto x^3$ is not injective) and for this value of $c$, $p$ won't divide any terms in the sequence, so $p$ is not orange.\n\nThis is of course an alternative possible statement for the problem, requiring at least one extra idea for a solution.\n\n**Comment.** This problem is a simple implication of the conclusion of the following China TST problem: https://artofproblemsolving.com/community/c6h1212540p6016827",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19495,
    "subject": "Mathematics (Olympiad)",
    "question": "For an odd natural number $n > 1$, define the set of distinct remainders of powers of $2$ modulo $n$:\n\n$$\nS_n = \\{ a \\mid a < n,\\ \\exists k \\in \\mathbb{N} : 2^k \\equiv a \\pmod{n} \\}.\n$$\n\nAre there different odd numbers $m$ and $r$ such that $S_m = S_r$?",
    "options": [],
    "answer": "See solution",
    "solution": "No! There exists a natural number $s$ such that $2^s \\equiv 1 \\pmod{n}$ (for example, $s = \\varphi(n)$ by Euler's theorem, or because the sequence of powers of $2$ modulo $n$ is periodic). We have $2^{s-1} \\equiv \\frac{n+1}{2} \\pmod{n}$, so $x = \\frac{n+1}{2} \\in S_n$, but $2x = n+1 > n$ is not in $S_n$. Also, if $t \\leq \\frac{n-1}{2}$ is in $S_n$, then $2t < n$ is also in $S_n$. Therefore, the smallest natural number $t$ such that $t \\in S_n$ and $2t \\notin S_n$ is $\\frac{n+1}{2}$. Since this number is different for different $n$, we get what we asked for. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19496,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers, in order, of each row and the numbers, in order, of each column of a $5 \\times 5$ array of integers form an arithmetic progression of length 5. The numbers in positions $(5, 5)$, $(2, 4)$, $(4, 3)$, and $(3, 1)$ are $0$, $48$, $16$, and $12$, respectively. What number is in position $(1, 2)$?\n\n$$\n\\begin{bmatrix}\n\\cdot & ? & \\cdot & \\cdot & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & 48 & \\cdot \\\\\n12 & \\cdot & \\cdot & \\cdot & \\cdot \\\\\n\\cdot & \\cdot & 16 & \\cdot & \\cdot \\\\\n\\cdot & \\cdot & \\cdot & \\cdot & 0\n\\end{bmatrix}\n$$\n\n(A) 19  (B) 24  (C) 29  (D) 34  (E) 39",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_{ij}$ be the integer at row $i$ and column $j$. It is given that $a_{55} = 0$, $a_{24} = 48$, $a_{43} = 16$, and $a_{31} = 12$. Suppose $a_{54} = d$. Then row 5 is $4d, 3d, 2d, d, 0$ because it is an arithmetic progression with common difference $-d$. The arithmetic progression in column 1 gives\n\n$$\na_{41} = \\frac{a_{31} + a_{51}}{2} = \\frac{12 + 4d}{2} = 6 + 2d.\n$$\n\nThe arithmetic progression in column 4 gives\n\n$$\na_{44} = \\frac{2a_{54} + a_{24}}{3} = \\frac{2d + 48}{3} = \\frac{2}{3}d + 16.\n$$\n\nRow 4 gives\n\n$$\na_{43} = 16 = \\frac{2a_{44} + a_{41}}{3} = \\frac{\\frac{4}{3}d + 32 + 6 + 2d}{3},\n$$\n\nwhich implies $48 = \\frac{10}{3}d + 38$, so $d = 3$. Filling in column 3 with common difference $16 - 6 = 10$ and column 1 with difference $12 - 12 = 0$ produces $a_{13} = 46$ and $a_{11} = 12$. Finally,\n\n$$\na_{12} = \\frac{a_{13} + a_{11}}{2} = \\frac{46 + 12}{2} = 29.\n$$\n\nThe full array looks like this:\n\n$$\n\\begin{bmatrix}\n12 & \\underline{\\mathbf{29}} & 46 & 63 & 80 \\\\\n12 & 24 & 36 & \\mathbf{48} & 60 \\\\\n\\mathbf{12} & 19 & 26 & 33 & 40 \\\\\n12 & 14 & \\mathbf{16} & 18 & 20 \\\\\n12 & 9 & 6 & 3 & \\mathbf{0}\n\\end{bmatrix}\n$$",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 19497,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_9$ be nonnegative real numbers satisfying\n$$\nx_1^2 + x_2^2 + \\dots + x_9^2 \\geq 25.\n$$\nProve that there exist three of these numbers with a sum of at least $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $x_1 \\ge x_2 \\ge x_3 \\ge x_4 \\ge x_5 \\ge x_6 \\ge x_7 \\ge x_8 \\ge x_9 \\ge 0$. Then $x_1x_2 \\ge x_4^2 \\ge x_5^2$, $x_1x_3 \\ge x_6^2 \\ge x_7^2$, and $x_2x_3 \\ge x_8^2 \\ge x_9^2$. Therefore,\n$$\n(x_1 + x_2 + x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 + 2x_1x_3 + 2x_2x_3 \\ge x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2 + x_8^2 + x_9^2 \\ge 25.\n$$\nThus, $x_1 + x_2 + x_3 \\ge 5$, which proves the assertion.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19498,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be a real number such that\n$$\n4x^2 - 20\\lfloor x \\rfloor + 9 \\ge 0.\n$$\nFind all possible values of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n$$\n4x^2 - 20\\lfloor x \\rfloor + 9 \\ge 4x^2 - 20x + 9 = (2x - 9)(2x - 1)\n$$\nit must be $(2x-9)(2x-1) \\le 0$, i.e., $1 \\le 2x \\le 9$. From there we get $0 \\le \\lfloor x \\rfloor \\le 4$.\n\nAlso, $x = \\frac{1}{2}\\sqrt{20\\lfloor x \\rfloor - 9}$.\n\nFor $\\lfloor x \\rfloor = 0$ there is no solution, and for $\\lfloor x \\rfloor = 1, 2, 3, 4$ we get respectively:\n$$\nx = \\frac{1}{2}\\sqrt{11}, \\quad x = \\frac{1}{2}\\sqrt{31}, \\quad x = \\frac{1}{2}\\sqrt{51}, \\quad x = \\frac{1}{2}\\sqrt{71}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19499,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, the altitude $CH$ is drawn. A ray from point $C$ lies inside $\\angle BCA$ and intersects the circumscribed circles of $\\triangle BCH$ and $\\triangle ABC$ at points $X$ and $Y$, respectively. It is given that $2CX = CY$. Prove that the line $HX$ bisects segment $AC$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us focus on $\\angle BCY$. From the problem statement, $CX = XY$. Since quadrilateral $CXHB$ is cyclic, $\\angle CXB = \\angle CHB = 90^\\circ$. Hence, $\\triangle CBY$ is isosceles. Thus, $\\angle MHA = \\angle BCY = \\angle BYC = \\angle BAC$.\n\nThen $\\triangle HMA$ is isosceles, and\n\n$$\n\\angle MHC = 90^\\circ - \\angle MHA = 90^\\circ - \\angle HAM = \\angle HCA.\n$$\n\nHence, $\\triangle HMC$ is also isosceles, which yields $MA = MH = MC$. Q.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19500,
    "subject": "Mathematics (Olympiad)",
    "question": "有 6 個白球和 15 個黑球，要求將所有球排成一列，且任意兩個白球之間至少有 2 個黑球。問有多少種不同的排法？",
    "options": [],
    "answer": "See solution",
    "solution": "令每組白球之間的黑球數為 $x_i$，$i=1 \\sim 6$，則有 $\\sum_{i=1}^{6} x_i = 15$ 且 $x_i \\ge 2$。等價於 $\\sum_{i=1}^{6} (x_i - 2) = 3$。由重複組合知這樣的解共有 $\\binom{8}{5} = 56$ 種。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19501,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence of $4030$ positive integers, consider the number of distinct subsequences of length $2015$ that can be formed by choosing entries from the sequence and retaining their order. A subsequence is defined as a sequence obtained by selecting entries from the original sequence in order. Let $f(i)$ be defined such that for $i < j$, if the number at the $i$-th position equals the number at the $j$-th position, then $f(i) = j$ and $f(j) = i$. If there exists an $i$ such that $0 < |f(i) - i| < 2015$, then there must exist at least $2015$ ways of choosing $2014$ numbers. Assume that for all $i$, $|f(i) - i| \\ge 2015$. Consider the sequence $A = 1, 2, \\dots, 2015, 1, 2, \\dots, 2015$. How many distinct subsequences of length $2015$ can be formed from $A$?",
    "options": [],
    "answer": "See solution",
    "solution": "If for all $i$, $|f(i) - i| \\ge 2015$, then in the sequence $A = 1, 2, \\dots, 2015, 1, 2, \\dots, 2015$, each number appears exactly twice, once in each half. To form a subsequence of length $2015$, for each number $n$ ($1 \\leq n \\leq 2015$), we can choose either its occurrence in the left half or the right half. However, if we choose $n$ from the left half, all numbers less than $n$ must also be chosen from the left half, and if we choose $n$ from the right half, all numbers greater than $n$ must also be chosen from the right half. This gives $2016$ possible ways to choose the subsequence $1, 2, \\dots, 2015$. Therefore, the number of distinct subsequences is $2016$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19502,
    "subject": "Mathematics (Olympiad)",
    "question": "A party is attended by $n$ people. Assume that there are at most $n$ pairs of friendships among them, and any two people shake hands at the party if and only if they have a common friend at the party.\n\nSuppose $m$ is a positive integer satisfying $m \\ge 3$ and $n \\le m^3$. Prove that there exists a person $A$ such that the number of people $A$ has shaken hands with at this party does not exceed $(m-1)$ times the number of $A$'s friends.",
    "options": [],
    "answer": "See solution",
    "solution": "We represent individuals as vertices, friendships as edges, and use a graph to depict the problem.\n\nThe given graph contains a connected subgraph $G = (V, E)$ satisfying $|E| \\le |V| \\le n$. We denote the degree of vertex $v$ as $d(v)$ and the number of individuals with whom $v$ has shaken hands as $\\beta(v)$.\n\nAssume the proposition is false, implying that for any vertex $v$, we have\n\n$$\n(*) \\qquad \\beta(v) \\ge (m-1) \\cdot d(v) + 1.\n$$\n\n**Observation:** For every vertex $v_0$ with degree 1 (connected to $v_1$), (*) implies $\\beta(v_0) \\ge (m-1) \\cdot d(v_0) + 1 = m$. Thus, we have $d(v_1) \\ge m + 1$.\n\nConsider the graph $G' = (V', E')$ obtained by removing all vertices of degree 1 in $G$ along with their incident edges. Note that the number of removed vertices is equal to the number of removed edges. Therefore, $G'$ is necessarily connected and satisfies $|E'| \\le |V'|$. Hence, $G'$ is either a tree or has a unique cycle. We will consider three cases.\n\n(0) $G'$ has only one vertex $v_1$. From the previous observation, we know that $d(v_1) \\ge m+1$. This implies that $G$ has only $m+2$ vertices, and all edges are the ones connecting $v_1$ to all other vertices. However, in this case, $\\beta(v_1) = 0 \\le (m-1)d(v_1)$, which contradicts our assumption.\n\n(1) $G'$ has a vertex $v_1$ with degree 1. Then, $v_1$ must be connected to a vertex $v_0$ with degree 1 in $G$. From the previous observation, we know that $d(v_1) \\ge m+1$, which implies that $v_1$ is connected to $d(v_1) - 1 \\ge m$ vertices of degree 1 in $G$.\n\nLet $v_1$ be connected to $v_2$ in $G'$. In this case, the inequality $\\beta(v_1) \\ge (m-1)d(v_1) + 1$ becomes\n\n$$\nd(v_2) - 1 \\ge (m-1)d(v_1) + 1 \\ge m^2,\n$$\n\nwhich implies $d(v_2) \\ge m^2 + 1$.\n\nConsidering the inequality (*) for $v_2$, we have\n\n$$\n\\beta(v_2) \\ge (m-1)d(v_2) + 1 \\ge (m-1)(m^2 + 1) + 1.\n$$\n\nNote that the friends of $v_2$ and the people $v_2$ has shaken hands with are pairwise distinct, except when $v_2$ lies on a cycle of length 3, in which case $v_2$ shakes hands with both of its friends. From this, we conclude that $G$ has at least\n\n$$\n1 + d(v_2) + \\beta(v_2) - 2 \\ge 1 + (m^2 + 1) + (m-1)(m^2 + 1) + 1 - 2 = m^3 + m > m^3\n$$\n\nvertices, which contradicts $n \\le m^3$.\n\n(2) All vertices in $G'$ have degree at least 2. In this case, $G'$ forms a cycle $v_1 \\sim v_2 \\sim \\dots \\sim v_t \\sim v_1$, where indices are taken modulo $t$.\n\nFrom the previous observation, each vertex $v_i$ is connected to exactly $d(v_i) - 2$ (in $G$) vertices of degree 1. Thus, for $v_i$, (*) implies:\n\n$$\n(d(v_{i-1}) - 2) + (d(v_{i+1}) - 2) + 2 \\ge \\beta(v_i) \\ge (m-1) \\cdot d(v_i) + 1 \\ge 2d(v_i) + 1.\n$$\n\nSumming up these inequalities for all $i$, we arrive at a contradiction.\n\nTherefore, we conclude that there exists a vertex $v$ such that $\\beta(v) \\le (m-1) \\cdot d(v)$. Hence, the proposition holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19503,
    "subject": "Mathematics (Olympiad)",
    "question": "The screen of a computer lists the positive integers from 1 to 2025. A virus deletes some of these numbers, using the following algorithm:\n\n- At step 1, it deletes one of the listed numbers and its successor.\n- At each new step, two of the remaining numbers are deleted, so that one of them is the successor of the other.\n\nThe algorithm stops after 674 steps.\n\n**a)** Prove that the sum of the remaining numbers cannot be divisible by 6.\n\n**b)** Prove that the product of the remaining numbers is divisible by 6.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** At each step, we delete an even and an odd number. Initially, the screen contained $1012$ even and $1013$ odd numbers. Therefore, there are $1012 - 674 = 338$ even and $1013 - 674 = 339$ odd numbers left. So, there are an odd number of odd numbers and some even numbers, whose sum is odd, therefore not divisible by $6$.\n\n**b)** Every pair of consecutive numbers contains at most one multiple of $3$, so at most $674$ multiples of $3$ are deleted after the $674$ steps. Among the numbers $1, 2, \\ldots, 2025$, there are $675$ multiples of $3$, therefore at the end we have at least one multiple of $3$ left—call it $a$.\n\nA similar argument shows that at least one even number $b$ is left on the screen.\n\nSince the product $P$ of the numbers left on the screen is divisible by $a$ and by $b$, $P$ is divisible by $6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19504,
    "subject": "Mathematics (Olympiad)",
    "question": "For each odd integer $n$ from $1$ to $49$, define the group of $n$ as the set of integers from $1$ to $50$ which can be expressed as $n \\cdot 2^k$ for some non-negative integer $k$. Then each integer from $1$ to $50$ belongs to only one group.\n\nIf we choose $25$ integers such that no two are from the same group, and for any two integers in the same group, one is a divisor of the other, how many ways are there to choose one integer from each group?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $33$ is chosen, $11$ is not chosen. Integers $22$ and $44$ cannot be divisors or multiples of integers in other groups since $1$, $2$, and $4$ are not chosen. Therefore, the number of ways to choose one integer from the group of $11$ is $2$, independent of other choices.\n\nIntegers $25$ and $50$ cannot be divisors or multiples of integers in other groups. Since $30$ is chosen, $1$, $2$, $5$, and $10$ are not chosen, so $25$ and $50$ cannot be multiples of integers in other groups. Thus, the number of ways to choose one integer from the group of $25$ is $2$, independent of other choices.\n\nIntegers $17$ and $34$ cannot be divisors or multiples of integers in other groups since $1$ and $2$ are not chosen. Therefore, the number of ways to choose one integer from the group of $17$ is $2$, independent of other choices. The same applies for the groups of $19$ and $23$.\n\nThe choice of one integer each from the groups of $7$ and $21$ must be $(14, 21)$, $(28, 21)$, or $(28, 42)$. Integers $14$, $28$, $21$, and $42$ cannot be divisors or multiples of integers in other groups since $1$, $2$, $3$, $4$, and $6$ are not chosen. Therefore, the number of ways to choose one integer each from the groups of $7$ and $21$ is $3$, independent of other choices.\n\nWe need to determine the way to choose from the groups of $1$, $3$, $5$, and $9$.\n\nFrom the group of $1$ we need to choose a multiple of $8$.\n\n- If $8$ is chosen, the choice from each group of $3$, $5$, and $9$ must be $12$, $20$, and $18$, so there is only one way.\n- If $16$ is chosen, the choice from the group of $5$ must be $20$ or $40$, and the choice from the groups of $3$ and $9$ must be $(12, 18)$, $(24, 18)$, or $(24, 36)$.\n- If $32$ is chosen, the choice from the group of $5$ must be $20$ or $40$, and the choice from the groups of $3$ and $9$ must be $(12, 18)$, $(24, 18)$, $(24, 36)$, $(48, 18)$, or $(48, 36)$.\n\nHence, the number of ways to choose from the groups of $1$, $3$, $5$, and $9$ is $1 + 2 \\cdot 3 + 2 \\cdot 5 = 17$.\n\nTherefore, the answer is $2^5 \\cdot 3 \\cdot 17 = 1632$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19505,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute, non-isosceles triangle inscribed in circle $(O)$ with orthocenter $H$. The rays $AO$, $BO$, $CO$ intersect the sides $BC$, $CA$, $AB$ respectively at $A_1$, $B_1$, $C_1$. The rays $AH$, $BH$, $CH$ intersect $(O)$ respectively at points $D$, $E$, $F$. Prove that the circumcircles of $ADA_1$, $BEB_1$, $CFC_1$ have two common points.\n\n![](images/Saudi_Booklet_2025_p32_data_9c36fcabf1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the radius of $(O)$ and let $AO$ intersect $(BOC)$ at $A_2$. By the shooting lemma, noting that $O$ is the midpoint of the arc $BC$, we have:\n\n$$\nOA_1 \\cdot OA_2 = R^2\n$$\n\nSo $A_2$ is the image of $A_1$ under the inversion $\\Omega$ centered at $O$ with power $R^2$. Under $\\Omega$, the points $A$ and $D$ are fixed, so $\\Omega : (ADA_1) \\leftrightarrow (ADA_2)$. Define $B_2$, $C_2$ similarly. By the properties of inversion, it suffices to prove that the three circles $(ADA_2)$, $(BEB_2)$, $(CFC_2)$ have two common points. These circles share the radical center $H$ (with negative power), so it remains to show that their centers are collinear.\n\nSuppose the circle $(OBC)$ cuts $AB$, $AC$ at $K$, $L$ respectively. Then\n\n$$\n\\angle KA_2O = \\angle ABO = \\angle OAB\n$$\n\nso $KA = KA_2$. Similarly, $LA = FA_2$, so $KL$ is perpendicular to $AA_2$. Thus, the intersection $O_1$ of $KL$ and the line through $O$ parallel to $BC$ is the center of $(ADA_2)$. Define $O_2$, $O_3$ similarly. Let the images of $O_1$, $O_2$, $O_3$ under $\\Omega$ be $P_1$, $P_2$, $P_3$ respectively. To prove $O_1$, $O_2$, $O_3$ are collinear, it suffices to show that the circle $(P_1P_2P_3)$ passes through $O$, which is equivalent to the perpendicular bisectors of $OP_1$, $OP_2$, $OP_3$ being concurrent. Let $OK$, $OL$ cut $BC$ at $K'$, $L'$. By the shooting lemma,\n\n$$\nOK \\cdot OK' = OL \\cdot OL' = R^2\n$$\n\nso $K'$, $L'$ are the images of $K$, $L$ under $\\Omega$. Let $X$ be the midpoint of $K'L'$. Since $O$, $O_1$, $P_1$ are collinear and $OK'L'P_1$ is an isosceles trapezoid, $X$ lies on the perpendicular bisector of $OP_1$. The triangles $HBC$ and $OK'L'$ have parallel sides, so $HM \\parallel OX$. Define $Y$, $Z$ similarly. Thus, it suffices to prove that the lines through $X$, $Y$, $Z$ perpendicular to $BC$, $CA$, $AB$ are concurrent. By Carnot's theorem, we need to show\n\n$$\n\\sum_{sym} (BX^2 - CX^2) = 0.\n$$\n\nWe have\n\n$$\nBX^2 - CX^2 = \\overline{BC}(\\overline{BX} - \\overline{XC}) = 2\\overline{MX} \\cdot \\overline{BC}.\n$$\n\nSince triangles $OXM$ and $HMD$ are similar,\n\n$$\n\\frac{XM}{MD} = \\frac{OM}{HD} = \\frac{AH}{2HD}.\n$$\n\nThus,\n\n$$\n2\\overline{MX} \\cdot \\overline{BC} = \\frac{AH}{HD} \\cdot \\overline{DM} \\cdot \\overline{BC} = \\frac{AH^2 \\cdot 2\\overline{DM} \\cdot \\overline{BC}}{2AH \\cdot HD} = \\frac{AH^2 \\cdot (AC^2 - AB^2)}{|\\mathcal{P}_{H/(O)}|}.\n$$\n\nSince $AC^2 - AB^2 = CD^2 - BD^2$, we need to prove $\\sum_{sym} AH^2(CD^2 - BD^2) = 0$.\n\nThis holds because\n\n$$\n\\sum_{A,B,C} AH^2(CD^2 - BD^2) = 4 \\sum_{A,B,C} ([HAB]^2 - [HAC]^2) = 0.\n$$\n\nThe problem is completely solved. $\\square$\n\n![](images/Saudi_Booklet_2025_p33_data_72cf733caf.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19506,
    "subject": "Mathematics (Olympiad)",
    "question": "Set $N = 100$ and let $S = \\{(x_1, y_1, z_1), (x_2, y_2, z_2), \\dots, (x_n, y_n, z_n)\\}$ be a set of ordered triples of integers with $1 \\le x, y, z \\le N$. Suppose that for every sequence $a_1, a_2, \\dots$ of integers from $1$ to $N$, at least one of the triples $(a_i, a_{i+1}, a_{i+2})$ (for any $i$) belongs to $S$. What is the minimum possible value of $n$?\n\n% IMAGE: ![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\bullet x = y = z.\n$$\n\nIf we fix an integer $x$ with $1 \\le x \\le N$, the number of pairs $(y, z)$ such that $(x, y, z)$ belongs to $S$ is equal to $x(x-1)+1$. Therefore, the number of elements in $S$, denoted by $n$, is\n\n$$\nn = \\sum_{x=1}^{N} (x(x-1)+1) = \\frac{N^3 + 2N}{3}.\n$$\n\nAssume, for the sake of contradiction, that there exists a sequence $a_1, a_2, \\dots$ of integers $1, 2, \\dots, N$ such that none of the triples $(a_1, a_2, a_3), (a_2, a_3, a_4), \\dots$ belongs to $S$. Let $a_k$ be one of the largest terms among $a_1, a_2, \\dots$. Then we have $a_k \\ge a_{k+1}$ and $a_k \\ge a_{k+2}$. Since $(a_k, a_{k+1}, a_{k+2})$ does not belong to $S$, we have $a_{k+1} = a_k$. Therefore, $a_{k+1}$ is also one of the largest terms in the sequence. Applying the same argument to $(a_{k+1}, a_{k+2}, a_{k+3})$, we obtain $a_{k+2} = a_{k+1}$. Hence, $a_k = a_{k+1} = a_{k+2}$, so $(a_k, a_{k+1}, a_{k+2})$ belongs to $S$, a contradiction. Therefore, this set $S$ satisfies the required condition.\n\nThis shows that the minimum value of $n$ is $\\frac{N^3 + 2N}{3} = 333400$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19507,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of edges in a graph on 7 vertices with no 4-cycle? Prove your answer.\n\n![](images/Blr-2014_p31_data_29db90f4e1.png)\n![](images/Blr-2014_p31_data_e4d26c6643.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Figures 1 and 2 show that such a graph can have as many as 11 edges.\n\nSuppose there exists a graph with 12 edges. Let $A$ be the vertex with maximum degree $n$. Then $n \\geq 3$. Let $A$ be joined to $B_1, \\ldots, B_n$, and let the remaining vertices be $C_1, \\ldots, C_{7-n}$.\n\nIf $B_i$ or $C_i$ is joined to both $B_j$ and $B_k$, the three will form a 4-cycle with $A$. Hence, there are at most $\\lfloor n/2 \\rfloor$ edges of the type $B_iB_j$, and $7-n$ edges of the type $B_iC_j$. Clearly, the number of edges of the type $C_iC_j$ is at most $\\binom{7-n}{2}$.\n\nHence, the total number of edges is at most\n$$\nf(n) = n + \\lfloor \\frac{n}{2} \\rfloor + (7-n) + \\binom{7-n}{2}.\n$$\n\n![](images/Blr-2014_p31_data_29db90f4e1.png)\nFig. 1\n![](images/Blr-2014_p31_data_e4d26c6643.png)\nFig. 2\n\nNow $f(5) = f(6) = f(7) = 10 < 12$ while $f(4) = 12$. If $n=4$, there are exactly 2, 3, and 3 edges of the types $B_jB_j$, $B_jC_i$, and $C_iC_j$, respectively. We may assume that $B_1$ is joined to $B_2$, $B_3$ to $B_4$, and $C_1$, $C_2$, and $C_3$ to one another. None of the $B$'s can be joined to two of the $C$'s, as otherwise it will form a 4-cycle with the three $C$'s. Hence, three $B$'s are joined to different $C$'s. However, either $B_1$ and $B_2$ or $B_3$ and $B_4$ will form a 4-cycle with two $C$'s.\n\nFinally, for $n=3$ all vertices are of degree 3. As above, there are at most $\\lfloor 3/2 \\rfloor = 1$ edges of the type $B_iB_j$. If there is none, we would have at least two edges of the type $B_1C_i$, at least two of the type $B_2C_i$, and at least two of the type $B_3C_i$—in total six edges; at the same time, any of $C_i$ can be joined to at most one of $B_i$, so there are no more than four edges $C_iB_j$. So there is exactly 1 edge $B_iB_j$, say, $B_1B_2$. Hence $B_1$, as well as $B_2$, is joined to one of $C_i$, and $B_3$ joined to two of $C_i$. Let, for definiteness, there be edges $B_1C_1$, $B_2C_2$, $B_3C_3$, $B_3C_4$. Any of $C_1$, $C_2$ must be joined to two other of $C_i$. Since $C_1$ and $C_2$ are not joined to each other (otherwise $B_1B_2C_2C_3$ would be a 4-cycle), we have the edges $C_1C_3$, $C_1C_4$ and $C_2C_3$, $C_2C_4$. But then $C_1C_3C_2C_4$ is a 4-cycle, a contradiction. Thus, the proof is complete.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19508,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a positive integer $n$ such that the first and the second digit of $2^n$ are 3 and 9, respectively?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the smallest integer such that $1.024^m \\ge 3.9$. Then $1.024^m = 1.024^{m-1} \\cdot 1.024 < 3.9 \\cdot 1.024 = 3.9 + 3.9 \\cdot 0.024 < 3.9 + 4 \\cdot 0.025 = 4$. Thus, the first two digits of $2^{10m} = 1.024^m \\cdot 10^{3m}$ are indeed 3 and 9.\n\nThe number $m$ in the solution equals 58 and the corresponding power is $2^{580} \\approx 3.957286 \\cdot 10^{174}$. The least integer satisfying the conditions of the problem is $2^{95} = 39614081257132168796771975168$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19509,
    "subject": "Mathematics (Olympiad)",
    "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 $\\det(a_{ij})$.",
    "options": [],
    "answer": "See solution",
    "solution": "The determinant under consideration is an even integer. To prove this, we show the determinant is 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": 19510,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,\n$$\nf(2xy)^2 + f(f(x) - y^2)^2 = f(x^2 + y^2)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, the constant function $f(x) = 0$ is a solution. Now, let $f$ be a non-constant function satisfying the problem.\n\nDefine $g(x) = f(x)^2$ for all $x$. Since $f(x)^2 \\ge 0$, $g$ is always non-negative.\n\nLet $P(x, y)$ denote the assertion:\n$$\ng(2xy) + g(g(x) - y^2) = g(x^2 + y^2).\n$$\n\nFor every $a \\ge b \\ge 0$, there exist $x_0, y_0$ such that $x_0^2 + y_0^2 = b$ and $2x_0y_0 = a$. Hence,\n$$\nP(x_0, y_0) \\implies g(b) - g(a) = g(g(x) - y^2) \\ge 0.\n$$\nSo $g$ is increasing on non-negative numbers, and since $g$ is non-negative, $g(0) \\ge 0 \\implies g(g(0)) \\ge g(0) \\ge 0$.\n\nPlugging $P(0, 0)$ gives:\n$$\ng(0) + g(g(0)) = g(0) \\implies 0 \\le g(0) \\le g(0) = 0 \\implies g(0) = 0.\n$$\n\nAlso, by $P(\\frac{1}{2}, -y)$ and $P(\\frac{1}{2}, y)$:\n$$\ng(-y) + g\\left(g\\left(\\frac{1}{2}\\right) - y^2\\right) = g\\left(\\frac{1}{4} + y^2\\right) = g(y) + g\\left(g\\left(\\frac{1}{2}\\right) - y^2\\right) \\\\\n\\implies g(y) = g(-y).\n$$\n\nIt suffices to prove $g(x) = x^2$ for $x \\ge 0$ (since $g$ is even, this will hold for all $x$). Since $g(x) \\ge 0$, there exists $y$ such that $y^2 = g(x)$. Then $P(x, y)$ yields:\n$$\ng(2xy) = g(2xy) + g(g(x) - g(x)) = g(x^2 + y^2).\n$$\n\nIf $g$ is injective, then $2xy = x^2 + y^2 \\implies x = y \\implies x^2 = g(x)$, as desired. It remains to prove $g$ is injective.\n\nFirst, we show $g(a) = 0 \\iff a = 0$. Suppose not: there exists $a > 0$ with $g(a) = 0$ (call this $Q(a)$), and $b > 0$ with $g(b) > 0$. Since $g$ is increasing, $g(x) = 0$ for all $0 \\le x \\le a$. Let $y = \\sqrt{a}$ and $x = \\min\\left(a, \\frac{\\sqrt{a}}{2}\\right)$. Then:\n$$\n\\begin{cases}\nx \\le a \\implies g(x) = 0 \\\\\ny^2 = a \\implies g(-y^2) = g(y^2) = 0 \\\\\n2xy \\le a \\implies g(2xy) = 0\n\\end{cases}\n$$\n\nBy $P(x, y)$:\n$$\ng(2xy) + g(g(x) - y^2) = g(0 - y^2) = 0 = g(x^2 + y^2) = g(a + x^2).\n$$\n\nIf $a \\ge \\frac{1}{4}$, $Q(a) \\implies Q(\\frac{5a}{4})$; if $a \\le \\frac{1}{4}$, $Q(a) \\implies Q(a^2 + a)$. Inductively, this leads to a contradiction with the existence of $b > 0$ with $g(b) > 0$. Thus, $g(x) = 0 \\iff x = 0$.\n\nNow, suppose $g(a) = g(b)$ for some $a > b$. There exist $x, y \\ge 0$ with $x^2 + y^2 = a$ and $2xy = b$. Assume $x \\ge y$. Then, by $P(x, y)$ and $P(y, x)$:\n$$\n\\begin{align*}\ng(g(x) - y^2) &= g(g(y) - x^2) = g(a) - g(b) = 0 \\\\\n&\\implies g(x) = y^2,\\ g(y) = x^2 \\\\\nx \\ge y &\\implies g(x) \\ge g(y) \\implies y^2 \\ge x^2 \\implies y \\ge x \\implies y = x \\\\\n&\\implies 2xy = x^2 + y^2 \\implies a = b\n\\end{align*}\n$$\n\nThis is a contradiction. Hence, $g$ is injective, so $g(x) = x^2$ for all $x$, and thus $f(x) = \\pm x$ are the only non-constant solutions. Therefore, all solutions are $f(x) = 0$ and $f(x) = \\pm x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19511,
    "subject": "Mathematics (Olympiad)",
    "question": "Rewrite the equation as a quadratic in $x^2$:\n\n$$\n\\begin{aligned}\n-1008 &= x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2 \\\\\n&= x^4 - 2(y^2 + z^2)x^2 + (y^2 - z^2)^2.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The discriminant is $(y^2 + z^2)^2 - (y^2 - z^2)^2 = 4y^2z^2 = (2yz)^2$, so the roots $x^2$ are $y^2 + z^2 \\pm 2yz = (y \\pm z)^2$. Thus, the right-hand side factorizes as\n\n$$\n\\begin{aligned}\n& (x^2 - (y+z)^2)(x^2 - (y-z)^2) \\\\\n&= (x + y + z)(x - y - z)(x + y - z)(x - y + z).\n\\end{aligned}\n$$\n\nThis implies $1008 = (x + y + z)(-x + y + z)(x + y - z)(x - y + z) = t \\cdot u \\cdot v \\cdot w$, where $t = x + y + z$, $u = -x + y + z$, $v = x - y + z$, $w = x + y - z$. We have $t = u + v + w$ and $t \\cdot u \\cdot v \\cdot w = 2^4 \\cdot 3^2 \\cdot 7$. Also, $v + w = 2x$, $w + u = 2y$, $u + v = 2z$, so $u, v, w, t$ all have the same parity. Since 1008 is even, these numbers are even: $t = 2t'$, $u = 2u'$, $v = 2v'$, $w = 2w'$. Then\n\n$$\nt' = u' + v' + w', \\quad t' u' v' w' = 63 = 3^2 \\cdot 7.\n$$\n\nPossible $t'$ are 7 and 9. For $t' = 7$, $u' v' w' = 9$ gives $(t', u', v', w') = (7, 1, 3, 3)$ and permutations. For $t' = 9$, $u' v' w' = 7$ gives $(9, 1, 1, 7)$ and permutations.\n\nFrom above, $x = v' + w'$, $y = w' + u'$, $z = u' + v'$. This yields the six solution triples $(x, y, z)$:\n\n- (8, 8, 2)\n- (8, 2, 8)\n- (2, 8, 8)\n- (6, 4, 4)\n- (4, 6, 4)\n- (4, 4, 6)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19512,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number with $\\{a\\} + \\left\\{ \\frac{1}{a} \\right\\} = 1$. Prove that\n\n$$\n\\{a^n\\} + \\left\\{ \\frac{1}{a^n} \\right\\} = 1,\n$$\n\nfor every positive integer $n$, where $\\{x\\}$ denotes the fractional part of a real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a + \\frac{1}{a} = \\lfloor a \\rfloor + \\lfloor \\frac{1}{a} \\rfloor + 1$, it follows that $a + \\frac{1}{a} \\in \\mathbb{Z}$. For each positive integer $n$, let $s_n = a^n + \\frac{1}{a^n}$. We have $s_1 s_n = s_{n+1} + s_{n-1}$ for any integer $n \\ge 2$. Notice that $s_2 = a^2 + \\frac{1}{a^2} = (a + \\frac{1}{a})^2 - 2 \\in \\mathbb{Z}$, hence, by induction, $s_n \\in \\mathbb{Z}$ for all positive integers $n$.\n\nThis implies\n$$\n\\{a^n\\} + \\left\\{ \\frac{1}{a^n} \\right\\} = a^n + \\frac{1}{a^n} - \\lfloor a^n \\rfloor - \\left\\lfloor \\frac{1}{a^n} \\right\\rfloor = s_n - \\lfloor a^n \\rfloor - \\left\\lfloor \\frac{1}{a^n} \\right\\rfloor \\in \\mathbb{Z},\n$$\nhence $\\{a^n\\} + \\left\\{ \\frac{1}{a^n} \\right\\} \\in \\{0, 1\\}$, since $\\{x\\} \\in [0, 1)$.\n\nIf $\\{a^n\\} + \\left\\{ \\frac{1}{a^n} \\right\\} = 0$, then both numbers $a^n$ and $\\frac{1}{a^n}$ are integers, implying $a^n = 1$ and furthermore $a = \\pm 1$. Then $\\{a\\} + \\left\\{ \\frac{1}{a} \\right\\} = 0 \\ne 1$, a contradiction. Consequently, $\\{a^n\\} + \\left\\{ \\frac{1}{a^n} \\right\\} = 1$ for any positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19513,
    "subject": "Mathematics (Olympiad)",
    "question": "Jess is standing in a line. There are 17 people in front of her and 34 people behind her. How many people are in the line, including Jess?",
    "options": [],
    "answer": "See solution",
    "solution": "There are 17 people in front of Jess and 34 behind her. Including herself, this makes:\n\n$$\n17 + 1 + 34 = 52\n$$\n\nSo, there are 52 people in the line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19514,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a natural number $m$ *remarkable* if there exist integers $a, b, c$ such that\n$$\nm = a^3 + 2b^3 + 4c^3 - 6abc.\n$$\nProve that there exists a natural number $n < 2024$ such that for infinitely many prime numbers $p$, the number $np$ is remarkable.",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma. Let $p$ be a prime number and $a, b, c \\in \\mathbb{Z}/p\\mathbb{Z}$. Then there exist $x, y, z \\in \\mathbb{Z}$ such that $|x|, |y|, |z| < \\sqrt[3]{p}$, $(x, y, z) \\neq (0, 0, 0)$, and $ax + by + cz \\equiv 0 \\pmod{p}$.\n\n*Proof.* Consider the set $M = \\{(x, y, z) : x, y, z \\in \\{0, 1, \\dots, \\lfloor \\sqrt[3]{p} \\rfloor \\}\\}$. We have $|M| > p$, so by the pigeonhole principle, there are two distinct elements $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in $M$ such that $ax_1 + by_1 + cz_1 \\equiv ax_2 + by_2 + cz_2 \\pmod{p}$. Thus, $(x_1 - x_2, y_1 - y_2, z_1 - z_2)$ satisfies the lemma's conditions.\n\nNow, let $p \\equiv 2 \\pmod{3}$. Then the congruence $x^3 \\equiv 2 \\pmod{p}$ has a solution $a$, since the map $x \\mapsto x^3$ is injective in $\\mathbb{Z}/p\\mathbb{Z}$ and thus surjective (because $(3, p-1) = 1$). Thus, $x^3 \\equiv 1 \\pmod{p}$ has only $x \\equiv 1$ as a solution.\n\nBy the lemma, there exist $x, y, z$ with $|x|, |y|, |z| < \\sqrt[3]{p}$ such that $x + a y + a^2 z \\equiv 0 \\pmod{p}$. Expanding,\n$$\nx^3 + a^3 y^3 + a^6 z^3 - 3a^3 x y z \\equiv 0 \\pmod{p}.\n$$\nSince $a^3 \\equiv 2 \\pmod{p}$ and $a^6 \\equiv 4 \\pmod{p}$, this is\n$$\nx^3 + 2y^3 + 4z^3 - 6 x y z \\equiv 0 \\pmod{p}.\n$$\nAlso, $|x|, |y|, |z| < \\sqrt[3]{p}$ implies $|x^3 + 2y^3 + 4z^3 - 6 x y z| < 13p$. If $x^3 + 2y^3 + 4z^3 - 6 x y z < 0$, then $(-x, -y, -z)$ gives a positive multiple of $p$.\n\nIt remains to show $x^3 + 2y^3 + 4z^3 - 6 x y z \\neq 0$. If it were zero, then\n$$\n(x + \\sqrt[3]{2} y + \\sqrt[3]{4} z)\\left((x - \\sqrt[3]{2} y)^2 + (x - \\sqrt[3]{4} z)^2 + (\\sqrt[3]{2} y - \\sqrt[3]{4} z)^2\\right) = 0,\n$$\nwhich only has the integer solution $(x, y, z) = (0, 0, 0)$, since $\\sqrt[3]{2}$ is irrational and $x^3 - 2$ is its minimal polynomial over $\\mathbb{Q}$.\n\nTherefore, for $n = 1$ (or any $n < 2024$), for infinitely many primes $p \\equiv 2 \\pmod{3}$, $np$ is remarkable.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19515,
    "subject": "Mathematics (Olympiad)",
    "question": "We say a nondegenerate triangle whose angles have measures $\\theta_1, \\theta_2, \\theta_3$ is *quirky* if there exist integers $r_1, r_2, r_3$, not all zero, such that\n\n$$\nr_1\\theta_1 + r_2\\theta_2 + r_3\\theta_3 = 0.\n$$\n\nFind all integers $n \\ge 3$ for which a triangle with side lengths $n-1, n, n+1$ is quirky.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 3, 4, 5, 7$.\n\nWe first introduce a variant of the $k$th Chebyshev polynomials in the following lemma (which is standard, and easily shown by induction).\n\n*Lemma*\n\nFor each $k \\ge 0$ there exists $P_k(X) \\in \\mathbb{Z}[X]$, monic for $k \\ge 1$ and with degree $k$, such that\n\n$$\nP_k(X + X^{-1}) \\equiv X^k + X^{-k}.\n$$\n\nThe first few are $P_0(X) \\equiv 2$, $P_1(X) \\equiv X$, $P_2(X) \\equiv X^2 - 2$, $P_3(X) \\equiv X^3 - 3X$.\n\nSuppose the angles of the triangle are $\\alpha < \\beta < \\gamma$, so the law of cosines implies that\n\n$$\n2 \\cos \\alpha = \\frac{n+4}{n+1} \\quad \\text{and} \\quad 2 \\cos \\gamma = \\frac{n-4}{n-1}.\n$$\n\n*Claim* — The triangle is quirky iff there exist $r, s \\in \\mathbb{Z}_{\\ge 0}$ not both zero such that\n\n$$\n\\cos(r\\alpha) = \\pm \\cos(s\\gamma) \\quad \\text{or equivalently} \\quad P_r\\left(\\frac{n+4}{n+1}\\right) = \\pm P_s\\left(\\frac{n-4}{n-1}\\right).\n$$\n\n*Proof.* If there are integers $x, y, z$ for which $x\\alpha + y\\beta + z\\gamma = 0$, then we have that $(x - y)\\alpha = (y - z)\\gamma - \\pi y$, whence it follows that we may take $r = |x - y|$ and $s = |y - z|$ (noting $r = s = 0$ implies the absurd $x = y = z$). Conversely, given such $r$ and $s$ with $\\cos(r\\alpha) = \\pm \\cos(s\\gamma)$, then it follows that $r\\alpha \\pm s\\gamma = k\\pi = k(\\alpha + \\beta + \\gamma)$ for some $k$, so the triangle is quirky. $\\square$\n\nIf $r = 0$, then by rational root theorem on $P_s(X) \\pm 2$ it follows $\\frac{n-4}{n-1}$ must be an integer which occurs only when $n = 4$ (recall $n \\ge 3$). Similarly we may discard the case $s = 0$.\n\nThus in what follows assume $n \\ne 4$ and $r, s > 0$. Then, from the fact that $P_r$ and $P_s$ are nonconstant monic polynomials, we find\n\n*Corollary*\n\nIf $n \\ne 4$ works, then when $\\frac{n+4}{n+1}$ and $\\frac{n-4}{n-1}$ are written as fractions in lowest terms, the denominators have the same set of prime factors.\n\nBut $\\gcd(n+1, n-1)$ divides 2, and $\\gcd(n+4, n+1)$, $\\gcd(n-4, n-1)$ divide 3. So we only have three possibilities:\n\n- $n + 1 = 2^u$ and $n - 1 = 2^v$ for some $u, v \\ge 0$. This is only possible if $n = 3$. Here $2 \\cos \\alpha = \\frac{7}{4}$ and $2 \\cos \\gamma = -\\frac{1}{2}$, and indeed $P_2(-1/2) = -7/4$.\n- $n + 1 = 3 \\cdot 2^u$ and $n - 1 = 2^v$ for some $u, v \\ge 0$, which implies $n = 5$. Here $2 \\cos \\alpha = \\frac{3}{2}$ and $2 \\cos \\gamma = \\frac{1}{4}$, and indeed $P_2(3/2) = 1/4$.\n- $n + 1 = 2^u$ and $n - 1 = 3 \\cdot 2^v$ for some $u, v \\ge 0$, which implies $n = 7$. Here $2 \\cos \\alpha = \\frac{11}{8}$ and $2 \\cos \\gamma = \\frac{1}{2}$, and indeed $P_3(1/2) = -11/8$.\n\nFinally, $n = 4$ works because the triangle is right, completing the solution.\n\n*Remark* (Major generalization due to Luke Robitaille). In fact one may find all quirky triangles whose sides are integers in arithmetic progression.\n\nIndeed, if the side lengths of the triangle are $x - y, x, x + y$ with $\\gcd(x, y) = 1$ then the problem becomes\n\n$$\nP_r\\left(\\frac{x+4y}{x+y}\\right) = \\pm P_s\\left(\\frac{x-4y}{x-y}\\right)\n$$\n\nand so in the same way as before, we ought to have $x + y$ and $x - y$ are both of the form $3 \\cdot 2^s$ unless $rs = 0$. This time, when $rs = 0$, we get the extra solutions $(1, 0)$ and $(5, 2)$.\n\nFor $rs \\ne 0$, by triangle inequality, we have $x - y \\le x + y < 3(x - y)$, and $\\min(\\nu_2(x - y), \\nu_2(x + y)) \\le 1$, so it follows one of $x - y$ or $x + y$ must be in $\\{1, 2, 3, 6\\}$. An exhaustive check then leads to\n\n$$\n(x, y) \\in \\{(3, 1), (5, 1), (7, 1), (11, 5)\\} \\cup \\{(1, 0), (5, 2), (4, 1)\\}\n$$\n\nas the solution set. And in fact they all work.\n\nIn conclusion the equilateral triangle, $3 - 5 - 7$ triangle (which has a $120^\\circ$ angle) and $6 - 11 - 16$ triangle (which satisfies $B = 3A + 4C$) are exactly the new quirky triangles (up to similarity) whose sides are integers in arithmetic progression.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19516,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $AB \\leq AC$ and let $P$ be an interior point on the angle bisector of $\\angle BAC$. Let $D$ and $E$ be points on the segments $PC$ and $PB$ respectively such that $\\angle PBD = \\angle PCE$. The line $BD$ meets $AC$ at $X$, and $CE$ meets $AB$ at $Y$.\n\nProve that $BX \\leq CY$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use Kelly's lemma:\n\n**Lemma 1** (Kelly): Given a triangle $ABC$. Suppose the cevians $BE$ and $CF$ are such that $\\angle CBE \\geq \\angle BCF$ and $\\angle ABE \\geq \\angle ACF$. Then $BE \\leq CF$.\n\n*Proof* (from Crux): Choose $Q$ on the segment $AE$ so that $\\angle QBE = \\angle QCF$. Let $CF$ meet $BE$ and $BQ$ at $P$ and $Q$ respectively. In the triangle $QBC$, since $\\angle QBC \\geq \\angle QCB$, we have $QC \\geq QB$. Observe that $\\triangle QBE \\sim \\triangle QCR$, hence from $BQ \\leq CQ$ we obtain $BE \\leq CR$. Clearly $CR \\leq CF$, therefore $BE \\leq CF$. $\\square$\n\nNow we apply Kelly's lemma to our problem. We want to show that\n\n$$\n\\angle DBC \\geq \\angle ECB \\quad \\text{and} \\quad \\angle ABP \\geq \\angle ACP.\n$$\n\nThen we can apply Kelly's lemma to get $BX \\leq CY$.\n\nReflect the point $C$ about the line $AP$ to $C'$. By symmetry,\n\n$$\n\\angle ABP \\geq \\angle AC'P = \\angle ACP.\n$$\n\nTo show that $\\angle DBC \\geq \\angle ECB$, we use the sine law in the triangles $ABP$ and $ACP$ respectively to get\n\n$$\nBP = AP \\frac{\\sin(A/2)}{\\sin(\\angle ABP)}, \\quad CP = AP \\frac{\\sin(A/2)}{\\sin(\\angle ACP)}.\n$$\n\nThus $BP \\leq CP$. In the triangle $PBC$, since $BP \\leq CP$, it follows that $\\angle PCB \\leq \\angle PBC$. Therefore,\n\n$$\n\\angle DBC \\geq \\angle ECB.\n$$\n\nThis proves the claim and the problem.\n\n![](images/selected_2011_p9_data_ac6ed805ed.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19517,
    "subject": "Mathematics (Olympiad)",
    "question": "Maria and Bilyana play the following game. Maria has 2024, and Bilyana has 2023 fair coins. Coins are tossed randomly—the probability of each individual coin being heads after the toss is $\\frac{1}{2}$. Maria wins if there are strictly more heads among her coins than Bilyana's; otherwise, Bilyana wins. What is the probability that Maria wins?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be the probability that Maria has more heads than Bilyana after tossing the first 2023 of Maria's coins. By symmetry, the probability that Maria has fewer heads than Bilyana is also $p$, so the probability that they have an equal number of heads is $1 - 2p$.\n\nIf Maria has fewer heads than Bilyana at this point, her chance of winning is $0$ (regardless of the last coin). If she has more heads, her chance of winning is $1$. If they are tied, Maria wins only if her last coin is heads, which has probability $\\frac{1}{2}$.\n\nThus, the probability that Maria wins is:\n\n$$\np + \\frac{1 - 2p}{2} = \\frac{1}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19518,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a, b, c$ 為任意實數且 $a + b + c = 0$。證明：\n\n$$\n\\frac{33a^2 - a}{33a^2 + 1} + \\frac{33b^2 - b}{33b^2 + 1} + \\frac{33c^2 - c}{33c^2 + 1} \\ge 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "注意原不等式等價於\n\n$$\n\\sum_{cyc} \\left( \\frac{33a^2 - a}{33a^2 + 1} + t \\right) \\ge 3t \\quad (1)\n$$\n\n我們希望使每一項都非負，因此需要選擇 $t$ 使得 $\\left(\\frac{33a^2-a}{33a^2+1}+t\\right)$ 的分子總是正。\n因此我們希望\n\n$$\n\\left( \\frac{33a^2 - a}{33a^2 + 1} + t \\right) = \\frac{(1+t)33a^2 - a + t}{33a^2 + 1} = \\frac{\\left(\\sqrt{33(1+t)}a - \\sqrt{t}\\right)^2}{33a^2 + 1} \\quad (2)\n$$\n\n成立。比較係數可得\n\n$$\n1 = 2\\sqrt{33t}\\sqrt{1+t} \\implies t = \\frac{-1}{2} + \\sqrt{\\frac{17}{66}} > 0.\n$$\n\n這個值本身不重要，只需證明其存在且 $0 \\le t \\le 1$。為簡便起見，以下均以 $t = \\frac{-1}{2} + \\sqrt{\\frac{17}{66}}$ 計算。接下來需證明式 (1) 成立，利用式 (2) 可將原題化為\n\n$$\n\\sum_{cyc} \\frac{(\\sqrt{33(1+t)}a - \\sqrt{t})^2}{33a^2 + 1} \\ge 3t.\n$$\n\n分母較難處理，可利用條件簡化：\n\n$$\na^2 = (b+c)^2 \\le 2b^2 + 2c^2 \\implies 3a^2 \\le 2(a^2 + b^2 + c^2).\n$$\n\n同理，\n\n$$\n3b^2 \\le 2(a^2 + b^2 + c^2), \\quad 3c^2 \\le 2(a^2 + b^2 + c^2)\n$$\n\n因此\n\n$$\n\\begin{align*}\n\\sum_{cyc} \\frac{(\\sqrt{33(1+t)}a - \\sqrt{t})^2}{33a^2 + 1} &\\ge \\sum_{cyc} \\frac{(\\sqrt{33(1+t)}a - \\sqrt{t})^2}{22(a^2 + b^2 + c^2) + 1} \\\\\n&= \\frac{\\sum_{cyc} (\\sqrt{33(1+t)}a - \\sqrt{t})^2}{22(a^2 + b^2 + c^2) + 1} \\\\\n&= \\frac{\\sum_{cyc} (33(1+t)a^2) + 3t}{22(a^2 + b^2 + c^2) + 1}\n\\end{align*}\n$$\n\n最後需證明其大於 $3t$，即\n\n$$\n\\frac{\\sum_{cyc} (33(1+t)a^2) + 3t}{22(a^2 + b^2 + c^2) + 1} \\ge 3t \\iff 33(1+t)(a^2 + b^2 + c^2) \\ge 22(a^2 + b^2 + c^2)3t\n$$\n\n當且僅當 $t \\le 1$ 成立，因此證畢。\n\n**Note:** 不太可能用 Jensen/Karamata/HCF/SIP 或導數工具解此不等式，因為\n\n$$\nf(x) = \\frac{33x^2 - x}{33x^2 + 1}\n$$\n\n有三個拐點。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19519,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $x, y$ 為滿足 $x + y = 1$ 的正實數，$n$ 為大於或等於 $2$ 的整數。試證：\n\n$$\n\\frac{x^n}{x + y^3} + \\frac{y^n}{x^3 + y} \\geq \\frac{2^{4-n}}{5}.\n$$\n\nLet $x, y$ be positive real numbers with $x + y = 1$, and $n$ be a positive integer with $n \\geq 2$. Prove that\n\n$$\n\\frac{x^n}{x + y^3} + \\frac{y^n}{x^3 + y} \\geq \\frac{2^{4-n}}{5}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "因為函數 $z = t^m$（$m$ 為大於 $0$ 的常數）在開區間 $(0, \\infty)$ 上是增函數，則有\n\n$$\n(x - y)(x^{n+3} - y^{n+3}) \\geq 0 \\\\\n(x - y)(x^{n-1} - y^{n-1}) \\geq 0.\n$$\n\n$$\n\\begin{align*}\n& \\text{故 } 2\\left(\\frac{x^{n+1}}{x+y^3} + \\frac{y^{n+1}}{x^3+y}\\right) - \\left(\\frac{x^n}{x+y^3} + \\frac{y^n}{x^3+y}\\right) \\\\\n&= \\frac{2x^{n+1} - x^n(x+y)}{x+y^3} + \\frac{2y^{n+1} - y^n(x+y)}{x^3+y} \\\\\n& \\qquad + \\frac{x^n(x-y)}{x+y^3} + \\frac{y^n(y-x)}{x^3+y} \\\\\n&= \\frac{x-y}{(x+y^3)(x^3+y)} \\left[ (x-y)(x^{n+3}-y^{n+3}) + xy(x-y)(x^{n-1}-y^{n-1}) \\right] \\\\\n& \\ge 0.\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\n\\text{所以}\\quad \\frac{x^{n+1}}{x+y^3} + \\frac{y^{n+1}}{x^3+y} &\\ge \\frac{1}{2} \\left( \\frac{x^n}{x+y^3} + \\frac{y^n}{x^3+y} \\right) \\\\\n&\\ge \\frac{1}{2^2} \\left( \\frac{x^{n-1}}{x+y^3} + \\frac{y^{n-1}}{x^3+y} \\right) \\ge \\dots \\\\\n&\\ge \\frac{2^{n-1}}{2} \\left( \\frac{x^2}{x+y^3} + \\frac{y^2}{x^3+y} \\right).\n\\end{align*}\n$$\n\n則\n$$\n\\frac{x^n}{x+y^3} + \\frac{y^n}{x^3+y} \\ge 2^{2-n} \\left(\\frac{x^2}{x+y^3} + \\frac{y^2}{x^3+y}\\right).\n$$\n\n令 $t = xy$，則\n\n$$\nt \\leq \\left(\\frac{x+y}{2}\\right)^2 = \\frac{1}{4}, \\quad x^4 + y^4 = 1 - 4t + 2t^2, \\quad x^5 + y^5 = 1 - 5t + 5t^2.\n$$\n\n於是，\n\n$$\n\\begin{aligned}\n& \\frac{x^2}{x+y^3} + \\frac{y^2}{x^3+y} \\geq \\frac{4}{5} \\\\\n\\Leftrightarrow & \\quad 5[x^2(x^3+y) + y^2(x+y^3)] \\geq 4(x+y^3)(x^3+y) \\\\\n\\Leftrightarrow & \\quad 5(x^5+y^5+t) \\geq 4(x^4+y^4+t+t^3) \\\\\n\\Leftrightarrow & \\quad 5(1-4t+5t^2) \\geq 4(1-3t+2t^2+t^3) \\\\\n\\Leftrightarrow & \\quad 4t^3-17t^2+8t-1 \\leq 0 \\\\\n\\Leftrightarrow & \\quad (4t-1)(t^2-4t+1) \\leq 0 \\\\\n\\Leftrightarrow & \\quad (1-4t)[t^2+(1-4t)] \\geq 0.\n\\end{aligned}\n$$\n\n由 $1-4t \\geq 0$ 知最後這一個不等式成立，從而，\n\n$$\n\\frac{x^2}{x+y^3} + \\frac{y^2}{x^3+y} \\geq \\frac{4}{5}.\n$$\n\n綜上所述，原不等式成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19520,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m > 2$ be an integer, $A$ a finite set of (not necessarily positive) integers, and $B_1, B_2, \\dots, B_m$ subsets of $A$. Assume that for each $k = 1, 2, \\dots, m$, the sum of the elements of $B_k$ is $m^k$. Prove that $A$ contains at least $m/2$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\{a_1, \\dots, a_k\\}$. Suppose the statement is not true: let $k = |A| < \\frac{m}{2}$.\n\nConsider $f(c_1, \\dots, c_m) := c_1 m + c_2 m^2 + \\dots + c_m m^m$, where $c_j \\in \\{0, 1, \\dots, m-1\\}$ for $j = 1, \\dots, m$. Of all these $m^m$ sums, each one is the representation of some number in base $m$, and they are all distinct. Since each $m^j$ ($1 \\leq j \\leq m$) is the sum of all elements of $B_j$, we may rewrite $f(c_1, \\dots, c_m)$ as the sum\n\n$$\n\\alpha_1 a_1 + \\dots + \\alpha_k a_k,\n$$\n\nwhere each $\\alpha_i \\in \\{0, 1, \\dots, m(m-1)\\}$. The total number of sums as above is\n\n$$\n(m(m-1)+1)^k < m^{2k} < m^m,\n$$\n\nyet the sums $f(c_1, \\dots, c_m)$ are proven to be all distinct. This contradiction overturns our assumption and hence $k = |A| \\geq \\frac{m}{2}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19521,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\ell_A$ and $\\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, A_3, \\dots, A_m$ lie on $\\ell_A$, and distinct points $B_1, B_2, B_3, \\dots, B_n$ lie on $\\ell_B$. Additionally, when segments $A_iB_j$ are drawn for all $i = 1, 2, 3, \\dots, m$ and $j = 1, 2, 3, \\dots, n$, no point strictly between $\\ell_A$ and $\\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m = 7$ and $n = 5$. The figure shows that there are 8 regions when $m = 3$ and $n = 2$.\n\n![](images/2022AIME_II_Solutions_p7_data_fceff940ea.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that the points $A_1, A_2, A_3, \\dots, A_m$ and $B_1, B_2, B_3, \\dots, B_n$ lie in these orders on the lines $\\ell_A$ and $\\ell_B$, respectively, so that $\\ell_A$ and $\\ell_B$ together with the segments $A_1B_1$ and $A_mB_n$ bound one region in the plane. Consider adding the other line segments $A_iB_j$ one at a time. One of these line segments divides each of $k$ regions into two regions where $k-1$ is the number of times $A_iB_j$ intersects another of these line segments at a point between $\\ell_A$ and $\\ell_B$. It follows that the final number of regions must be $1$ plus $2$ less than the number of line segments drawn plus the number of intersection points of these segments between $\\ell_A$ and $\\ell_B$. Because there is one intersection point for every set of $4$ points $\\{A_p, A_q, B_r, B_s\\}$ with $1 \\le p < q \\le m$ and $1 \\le r < s \\le n$, the total number of regions is\n\n$$\n1 + (m n - 2) + \\binom{m}{2} \\binom{n}{2}.\n$$\n\nSubstituting $m = 7$ and $n = 5$ gives\n\n$$\n1 + (7 \\cdot 5 - 2) + \\binom{7}{2} \\binom{5}{2} = 244.\n$$\n\n---\n\nAlternatively, for a fixed value of $m \\ge 2$, let $f(n)$ be the number of bounded regions for given values of $n$. Then $f(1) = m-1$. To obtain $f(n)$ from $f(n-1)$, notice that adding the segment $\\overline{A_i B_n}$ creates $1 + (m-i)(n-1)$ new regions, so adding $B_n$ to $\\ell_B$ creates\n\n$$\nm + \\frac{(n-1)m(m-1)}{2}\n$$\n\nnew regions as segments from the new point to the points on $\\ell_A$ are drawn. Therefore\n\n$$\nf(n) = f(n-1) + m + \\frac{(n-1)m(m-1)}{2}.\n$$\n\nComputing recursively when $m=7$ gives $f(1) = 6$, $f(2) = 34$, $f(3) = 83$, $f(4) = 153$, and $f(5) = 244$, which is the requested number of regions.\n\n---\n\nAlternatively, consider the graph whose vertices are $A_1, A_2, A_3, \\dots, A_m, B_1, B_2, B_3, \\dots, B_n$, and the intersections of the $\\overline{A_i B_j}$ line segments, and whose edges follow the $\\overline{A_i B_j}, \\overline{B_1 B_n}$, or $\\overline{A_1 A_m}$ line segments. Then this graph has $m$ vertices on line $\\ell_A$, $n$ vertices on line $\\ell_B$, and $p = \\binom{m}{2}\\binom{n}{2}$ vertices between the two lines for a total of $m+n+p$ vertices. Vertices $A_1$ and $A_m$ each have degree $n+1$, and each vertex $A_i$ for $1 < i < n$ has degree $n+2$. Similarly, vertices $B_1$ and $B_n$ each have degree $m+1$, and each vertex $B_j$ for $1 < j < n$ has degree $m+2$. Each vertex between lines $\\ell_A$ and $\\ell_B$ has degree $4$. Thus the total of all the degrees of the vertices in the graph is\n\n$$\nm(n+2) - 2 + n(m+2) - 2 + 4p = 2(mn + m + n - 2 + 2p).\n$$\n\nHence the number of edges in the graph is half of this number which is $mn + m + n - 2 + 2p$.\n\nEuler's Formula gives the number of bounded regions for a planar graph as $1$ more than the number of edges minus the number of vertices, which, in this case, is\n\n$$\n(mn + m + n - 2 + 2p) - (m + n + p) + 1 = mn + p - 1.\n$$\n\nWhen $m=7$ and $n=5$, this equals $7 \\cdot 5 + \\binom{7}{2}\\binom{5}{2} - 1 = 244$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19522,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a, n$ with $a > 2$ such that every prime divisor of $a^n - 1$ is also a prime divisor of $a^{32016} - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Euclidean algorithm, we have the following lemma:\n\n**Lemma.** Given $a > 1$ and positive integers $m, n$,\n$$\ngcd(a^m - 1, a^n - 1) = a^{gcd(m, n)} - 1.\n$$\n\nIf $n = 1$, all $a > 2$ work, so $(1, a)$ is a solution. For $n \\geq 2$, let $p$ be a prime divisor of $n$.\n\nIf $p > 3$, then $gcd(p, 2) = gcd(p, 3) = 1$. Since $a^p - 1 \\mid a^n - 1$, all prime divisors of $a^p - 1$ must divide $a^{3^{2006}} - 1$. By the lemma,\n$$\ngcd(a^p - 1, a^{3^{2006}} - 1) = a - 1.\n$$\nSo,\n$$\ngcd\\left(\\frac{a^p - 1}{a - 1}, \\frac{a^{3^{2006}} - 1}{a - 1}\\right) = 1.\n$$\nLet $S_m = 1 + a + \\cdots + a^{m-1} = \\frac{a^m - 1}{a - 1}$. All prime divisors of $S_p$ must divide $a - 1$. If $q$ divides $S_p$, then $q \\mid a - 1$, so $S_p \\equiv p \\pmod{q}$, implying $p = q$, so $S_p = p^\\alpha$ and $p \\mid a - 1$. By the LTE lemma,\n$$\nv_p(a^p - 1) = v_p(a) + 1.\n$$\nThus,\n$$\nv_p(S_p) = v_p(a^p - 1) - 1.\n$$\nBut $p \\leq a - 1 < S_p = \\frac{a^p - 1}{a - 1} = p$, a contradiction. Therefore, $p \\in \\{2, 3\\}$.\n\n**Case 1:** $n = 2^k$. For $k = 1$, all prime divisors of $a^2 - 1$ are prime divisors of $a^{3^{2016}} - 1$. Since $gcd(S_2, S_{3^{2006}}) = 1$, all prime divisors of $a + 1$ divide $a - 1$. Let $p$ divide $a + 1$, then $p$ divides $a - 1$, so $p = 2$. Thus, $a + 1$ is a power of $2$, i.e., $a = 2^s - 1$ for $s \\geq 2$.\n\nFor $k \\geq 2$, all prime divisors of $a^2 + 1$ divide $a - 1$. But $a^2 + 1 = 2^l$ for some $l$, and $a^2 + 1 \\equiv 1$ or $2 \\pmod{4}$, so $a < 2$, a contradiction.\n\nThus, $k = 1$ and $(2, 2^s - 1)$ for $s \\geq 2$ are solutions.\n\n**Case 2:** $n = 2^x 3^y$ with $x, y > 0$. Then $(a^3)^2 - 1 \\mid a^n - 1$. Similarly, $a^3 + 1 = 2^t$ for $t \\geq 2$, but $a^2 - a + 1$ is odd for $a > 2$, so no solution.\n\n**Case 3:** $n = 3^k$. If $k \\leq 2016$, $gcd(a^n - 1, a^{3^{2006}} - 1) = a^n - 1$, so all $k \\leq 2016$ work.\n\nIf $k > 2016$, let $s = k - 2016$ and $b = a^{3^{2016}}$. All prime divisors of $b^3 - 1$ must divide $b - 1$. Let $p$ divide $b^2 + b + 1$, then $p$ divides $b - 1$. One can show $p = 3$, so $b^2 + b + 1 = 3^u$ for $u > 0$. Let $b = 1 + 3t$, then $b^2 + b + 1 = 9t^2 + 9t + 3$, so $u = 1$, a contradiction. Thus, no solution for $k > 2016$.\n\n**Conclusion:** All pairs $(n, a)$ that satisfy the condition are:\n\n1. $(1, a)$ for any $a > 2$,\n2. $(2, 2^s - 1)$ for $s \\geq 2$,\n3. $(3^s, a)$ for $s \\leq 2016$ and $a > 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19523,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a > 2$ be a real number and define\n$$\nf_n(x) = a^{10} x^{n+10} + x^n + x^{n-1} + \\dots + x + 1\n$$\nfor $n = 1, 2, \\ldots$. Prove that for every positive integer $n$, the equation $f_n(x) = a$ has exactly one real root $x_n \\in (0, +\\infty)$. Prove that the sequence $(x_n)$ has a finite limit as $n \\to +\\infty$.",
    "options": [],
    "answer": "See solution",
    "solution": "For every $n$, define $g_n(x) = f_n(x) - a$. Then $g_n(x)$ is continuous and increasing on $[0, +\\infty)$. We have $g_n(0) = 1 - a < 0$ and $g_n(1) = a^{10} + n + 1 - a > 0$, so $g_n(x) = 0$ has exactly one root $x_n$ in $(0, +\\infty)$.\n\nTo prove $\\lim_{n \\to \\infty} x_n$ exists, we show $(x_n)$ is increasing and bounded.\n\n$$\n\\begin{aligned}\ng_n\\left(1 - \\frac{1}{a}\\right) &= a^{10}\\left(1 - \\frac{1}{a}\\right)^{n+10} + \\frac{1 - \\left(1 - \\frac{1}{a}\\right)^{n+1}}{1 - \\left(1 - \\frac{1}{a}\\right)} - a \\\\\n&= a\\left(1 - \\frac{1}{a}\\right)^{n+1}\\left((a-1)^9 - 1\\right) > 0.\n\\end{aligned}\n$$\n\nThus $x_n < 1 - \\frac{1}{a}$ for all $n$.\n\nOn the other hand, $g_n(x_n) = 0$, so\n$$\nx_n g_n(x_n) = a^{10} x_n^{n+11} + x_n^{n+1} + \\dots + x_n - a x_n = 0.\n$$\nTherefore,\n$$\ng_{n+1}(x_n) = x_n g_n(x_n) + 1 + a x_n - a = a x_n + 1 - a < 0 \\quad \\text{since } x_n < 1 - \\frac{1}{a}.\n$$\n\nSince $g_{n+1}$ is increasing and $0 = g_{n+1}(x_{n+1}) > g_{n+1}(x_n)$, we have $x_n < x_{n+1}$. Thus $(x_n)$ is increasing and bounded, so $\\lim_{n \\to \\infty} x_n$ exists and is finite.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19524,
    "subject": "Mathematics (Olympiad)",
    "question": "Demuestra que el producto de los dos mil trece primeros términos de la sucesión\n$$\na_n = 1 + \\frac{1}{n^3}\n$$\nno llega a valer 3.",
    "options": [],
    "answer": "See solution",
    "solution": "Veamos por inducción que $p_n = a_1 \\cdot a_2 \\cdot \\dots \\cdot a_n \\le 3 - \\frac{1}{n}$ y, así, quedará probado para el caso particular $n = 2013$ que se pide en el enunciado.\n\nPara $n = 1$ es $p_1 = a_1 = 1 + \\frac{1}{1^3} = 2 \\le 3 - \\frac{1}{1}$.\n\nSupongamos que es cierto para $n = k$, es decir, $p_k = a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k \\le 3 - \\frac{1}{k}$.\n\nHemos de probar que se cumple para $n = k + 1$, es decir, que $p_{k+1} = a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k \\cdot a_{k+1} \\le 3 - \\frac{1}{k+1}$.\n\nEn efecto,\n$$\np_{k+1} = p_k \\cdot a_{k+1} \\le \\left(3 - \\frac{1}{k}\\right) \\left(1 + \\frac{1}{(k+1)^3}\\right)\n$$\nCalculando el producto:\n$$\n\\left(3 - \\frac{1}{k}\\right) \\left(1 + \\frac{1}{(k+1)^3}\\right) = 3 - \\frac{1}{k} + \\frac{3}{(k+1)^3} - \\frac{1}{k(k+1)^3}\n$$\nAhora falta ver que\n$$\n3 - \\frac{1}{k} + \\frac{3}{(k+1)^3} - \\frac{1}{k(k+1)^3} \\le 3 - \\frac{1}{k+1}\n$$\nlo cual es equivalente a probar que\n$$\n\\frac{3}{(k+1)^3} - \\frac{1}{k(k+1)^3} \\le \\frac{1}{k+1} - \\frac{1}{k}\n$$\nEsto se reduce a demostrar que\n$$\nk^2 - k + 2 = \\left(k - \\frac{1}{2}\\right)^2 + \\frac{3}{4} \\ge 0\n$$\nlo cual es cierto para todo $k \\ge 1$. Por lo tanto, el producto de los primeros $2013$ términos no llega a valer $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19525,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2k$ citizens in a town, every two of whom are either friends or enemies. For some positive integer $t$, each citizen has at most $t$ enemies, and there exists a citizen having exactly $t$ enemies. A group is called *friendly* if any two members of the group are friends. It is known that a friendly group with more than $k$ members does not exist, and all citizens can be partitioned into two friendly groups having $k$ members each. Prove that the number of friendly groups having $k$ members is not greater than $2^{k-1} + 2^{k-t}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\{a_1, a_2, \\dots, a_k\\}$ and $B = \\{b_1, b_2, \\dots, b_k\\}$ be the two friendly groups. For an arbitrary group $C$ from $A$, denote by $S_C$ the group of all people from $B$ each of whom is an enemy of at least one person from $C$. If $|C| > |S_C|$, then $C \\cup (B \\setminus S_C)$ is a friendly group having more than $k$ members, a contradiction.\n\nTherefore, the sets $S_{\\{a_i\\}}$ satisfy Hall's condition for a system of distinct representatives. Hence, we may assume that $a_i$ and $b_i$ are enemies for all $i = 1, 2, \\dots, k$.\n\nThis means that every friendly group of $k$ members includes one person from every pair $(a_i, b_i)$ for $i = 1, 2, \\dots, k$.\n\nSuppose the enemies of $a_1$ are $b_1, \\dots, b_t$. For a friendly group $S$ such that $a_1 \\in S$, we have $b_1, \\dots, b_t \\notin S$. Hence, $a_2, \\dots, a_t \\in S$. From each of the remaining $k-t$ pairs $(a_j, b_j)$, $j > t$, we have to choose one of $a_j$ and $b_j$ to be an element of $S$. Therefore, there are at most $2^{k-t}$ such groups.\n\nFor a friendly group $S$ with $k$ members such that $a_1 \\notin S$, we have $b_1 \\in S$. Since from each of the remaining $k-1$ pairs $(a_j, b_j)$, $j > 1$, we have to choose one of $a_j$ and $b_j$ to be an element of $S$, we find that there are at most $2^{k-1}$ such groups.\n\nTherefore, the total number of friendly groups of $k$ members is at most $2^{k-1} + 2^{k-t}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19526,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle passes through vertices $A$, $B$ of a parallelogram $ABCD$, and intersects diagonals $BD$ and $AC$ at points $X$ and $Y$, respectively. The circumscribed circle of $\\triangle ADX$ intersects diagonal $AC$ at point $Z$. Prove that $AY = CZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $A$, $B$, $Z$, $D$ lie on one circle, and $A$, $B$, $X$, $Y$ lie on another circle (see figure),\n\n$$\n\\angle AYB = \\angle AXB = 180^\\circ - \\angle AXD = 180^\\circ - \\angle AZD = \\angle DZC.\n$$\n\nSince $ABCD$ is a parallelogram, $\\angle BAY = \\angle DCZ$, hence, $\\angle ABY = \\angle CDZ$, which yields $AY = CZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19527,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The conditions in the statement imply that the polynomial $\\prod_{i=1}^n (X - \\alpha_i)$ is irreducible over the integers, and hence over the rationals, so the $\\alpha_i$ are pairwise distinct. Moreover, $|\\alpha_i \\alpha_j| < 1$, unless one of the indices is $n$, in which case the absolute value is strictly greater than $1$.\n\nSuppose 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": 19528,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the line $PF_1$ with equation $x = \\frac{(x_0 + 1)y}{y_0} - 1$, substitute this into the ellipse equation\n\n$$\n\\frac{x^2}{2} + y^2 = 1\n$$\n\nand simplify to obtain a quadratic in $y$. Find the maximum value of $y_1 - y_2$, where $y_1$ and $y_2$ are the roots of this quadratic, under the condition $x_0^2 + 2y_0^2 = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = \\frac{(x_0 + 1)y}{y_0} - 1$ into $\\frac{x^2}{2} + y^2 = 1$ gives\n\n$$\n\\left( \\frac{(x_0 + 1)^2}{2y_0^2} + 1 \\right) y^2 - \\frac{x_0 + 1}{y_0} y - \\frac{1}{2} = 0.\n$$\n\nMultiplying both sides by $2y_0^2$ and using $x_0^2 + 2y_0^2 = 2$ yields\n\n$$\n(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.\n$$\n\nLet $y_0, y_1$ be the roots. By Vieta's formulas,\n\n$$\ny_0y_1 = -\\frac{y_0^2}{3+2x_0} \\implies y_1 = -\\frac{y_0}{3 + 2x_0}.\n$$\n\nSimilarly, $y_2 = -\\frac{y_0}{3 - 2x_0}$. Thus,\n\n$$\ny_1 - y_2 = \\frac{y_0}{3 - 2x_0} - \\frac{y_0}{3 + 2x_0} = \\frac{4x_0y_0}{9 - 4x_0^2}.\n$$\n\nSince $9 - 4x_0^2 = \\frac{1}{2}x_0^2 + 9y_0^2 \\ge 2\\sqrt{\\frac{1}{2}x_0^2 \\cdot 9y_0^2} = 3\\sqrt{2}x_0y_0$, it follows that\n\n$$\ny_1 - y_2 \\le \\frac{4x_0y_0}{3\\sqrt{2}x_0y_0} = \\frac{2\\sqrt{2}}{3},\n$$\n\nwith equality when $\\frac{1}{2}x_0^2 = 9y_0^2$, so $x_0 = \\frac{3\\sqrt{5}}{5}$, $y_0 = \\frac{\\sqrt{10}}{10}$.\n\nTherefore, the maximum of $y_1 - y_2$ is $\\frac{2\\sqrt{2}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19529,
    "subject": "Mathematics (Olympiad)",
    "question": "A right triangle $ABC$ has the right angle at vertex $A$. Circle $c$ passes through vertices $A$ and $B$ of triangle $ABC$ and intersects the sides $AC$ and $BC$ at points $D$ and $E$, respectively. The line segment $CD$ has the same length as the diameter of circle $c$. Prove that triangle $ABE$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle BAD = 90^\\circ$ (see the figure below), $BD$ is the diameter of circle $c$ and therefore $CD = BD$. Since $BD$ is the diameter, also $\\angle BED = 90^\\circ$, so $DE$ is an altitude of the isosceles triangle $BDC$, bisecting its base $BC$. Hence $E$ is the midpoint of the hypotenuse $BC$ of triangle $ABC$. Since the midpoint of the hypotenuse is the circumcentre of a right triangle, it follows $EA = EB$. This means that $ABE$ is an isosceles triangle.\n\n![](images/prob1516_p6_data_59225df232.png)\n\n**Solution 2.** As in the previous solution, we show that $CD = BD$. Hence $\\angle ECD = \\angle EBD$. From the equality of the inscribed angles subtending the arc $ED$ it also follows $\\angle EBD = \\angle EAD$. From triangle $ABC$ we get $\\angle ABC = 90^\\circ - \\angle BCA$, or $\\angle EBA = 90^\\circ - \\angle ECD$. On the other hand, $\\angle EAB = \\angle DAB - \\angle EAD = 90^\\circ - \\angle EBD = 90^\\circ - \\angle ECD$. Consequently $\\angle EBA = \\angle EAB$. So triangle $ABE$ is isosceles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19530,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $M$，定義數列 $a_0, a_1, a_2, \\dots$ 如下：\n\n$$a_0 = \\frac{2M+1}{2}$$\n\n且對所有 $k = 0, 1, 2, \\dots$，\n\n$$a_{k+1} = a_k \\lfloor a_k \\rfloor$$\n\n找出所有正整數 $M$，使得數列 $a_0, a_1, a_2, \\dots$ 中至少有一項是整數。\n\n(註：$\\lfloor x \\rfloor$ 表示不超過實數 $x$ 的最大整數。)\n\nDetermine all positive integers $M$ for which the sequence $a_0, a_1, a_2, \\dots$, defined by $a_0 = \\frac{2M+1}{2}$ and $a_{k+1} = a_k \\lfloor a_k \\rfloor$ for $k = 0, 1, 2, \\dots$, contains at least one integer term.\n\n(Remark: For a real number $x$, $\\lfloor x \\rfloor$ denotes the greatest integer that does not exceed $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "$M$ 可以是任何大於或等於 $2$ 的正整數，即 $M \\ge 2$。\n\n首先，對所有非負整數 $k$，定義 $b_k = 2a_k$。則有：\n\n$$\nb_{k+1} = 2a_{k+1} = 2a_k \\lfloor a_k \\rfloor = b_k \\left\\lfloor \\frac{b_k}{2} \\right\\rfloor\n$$\n\n因為 $b_0 = 2M+1$ 是整數，所以數列 $\\langle b_k \\rangle$ 的每一項都是整數。\n\n用歸謬法：如果 $\\langle a_k \\rangle$ 的每一項都不是整數，則 $\\langle b_k \\rangle$ 的每一項都是奇數。故\n\n$$\nb_{k+1} = b_k \\left\\lfloor \\frac{b_k}{2} \\right\\rfloor\n$$\n\n設 $b_0 - 3 > 0$。由遞推式可得 $b_k - 3 > 0$ 對所有 $k \\ge 0$ 均成立。對每個 $k \\ge 0$，定義 $c_k$ 為 $2^{c_k} \\parallel (b_k - 3)$，即 $c_k$ 是 $b_k - 3$ 中 2 的最高乘幂。\n\n注意 $b_k + 2$ 總是奇數，由遞推式知 $c_{k+1} = c_k - 1$，所以 $c_0, c_1, c_2, \\dots$ 為嚴格遞減的正整數數列，不可能無限遞減。故 $b_0 - 3 < 0$，即 $M = 1$。\n\n當 $M = 1$ 時，$\\langle a_k \\rangle$ 為常數數列 $\\frac{3}{2}$，無整數項。\n\n所以本題解答為 $M \\ge 2$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19531,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all natural $n \\ge 2$, the following number is composite:\n\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the numerator:\n\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1002}(n+1) + n^{1001} + 1 \\\\\n&= n^{1002}(n+1) + (n+1)(n^{1000} - n^{998} + n^{996} - \\dots - n + 1) \\\\\n&= (n+1)Q(n)\n\\end{aligned}\n$$\n\nfor some polynomial $Q(n)$. Thus,\n\n$$\n\\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1} = Q(n)\n$$\n\nBut $Q(n)$ can be factored further as $(n^2+1)P(n)$ for some polynomial $P(n)$. Since $n \\ge 2$, $n^2+1 > 1$ and $P(n) > 1$, so the expression is composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19532,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive real numbers $x, y$, the positive real number $x \\star y$ is defined as $x \\star y = \\frac{x}{xy+1}$.\n\nCalculate the following expression:\n\n$$\n(((\\cdots (((100 \\star 99) \\star 98) \\star 97) \\star \\cdots) \\star 3) \\star 2) \\star 1.\n$$\n\nNote that the answer should be given as a numerical value without using $\\star$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{100}{495001}\n$$\n\nFor any positive real numbers $x, y$, and $z$, the following equation holds:\n\n$$\n(x \\star y) \\star z = \\frac{x \\star y}{(x \\star y)z + 1} = \\frac{\\frac{x}{xy+1}}{\\frac{x}{xy+1}z + 1} = \\frac{x}{xy + xz + 1} = \\frac{x}{x(y+z) + 1} = x \\star (y+z).\n$$\n\nUsing this repeatedly, we have:\n\n$$\n\\begin{aligned}\n& (((\\cdots (((100 \\star 99) \\star 98) \\star 97) \\star \\cdots) \\star 3) \\star 2) \\star 1 \\\\\n&= 100 \\star (99 + 98 + 97 + \\cdots + 3 + 2 + 1) \\\\\n&= 100 \\star \\frac{99 \\cdot 100}{2}.\n\\end{aligned}\n$$\n\nTherefore, the answer is $100 \\star 4950 = \\frac{100}{495001}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19533,
    "subject": "Mathematics (Olympiad)",
    "question": "A number written in base 10 is a string of $3^{2013}$ digit 3s. No other digit appears. Find the highest power of 3 which divides this number.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the number in question be $A$. Notice that $\\frac{1}{3}A$ is a string of $3^{2013}$ 1s. Consider more generally the number $B_n$ which consists of a string of $3^n$ digit 1s.\n\nLet $M_n$ be the number formed of a digit 1, $(3^n - 1)$ consecutive digits 0, another digit 1, another $(3^n - 1)$ consecutive digits 0 and then another 1. Notice that $B_n \\cdot M_n = B_{n+1}$.\n\nSince the digital sum of $M_n$ is 3, it is divisible by 3. However, since this is not divisible by 9, $M_n$ is not divisible by 9. So $B_{n+1}$ is divisible by exactly one higher power of 3 than $B_n$.\n\nNow $B_1 = 111$ is divisible by $3^1$ but not by $3^2$, and so $B_n$ is divisible by $3^n$ but not by $3^{n+1}$. In particular, this means that $\\frac{1}{3}A$ is divisible by $3^{2013}$ but not by $3^{2014}$. Hence $A$ is divisible by $3^{2014}$, but by no higher power of 3.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19534,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2012 points marked in a square with side length 11. Prove that one can choose an equilateral triangle with side length 12 which covers at least 671 points.",
    "options": [],
    "answer": "See solution",
    "solution": "Place two equilateral triangles with side length 12 on the square so that each has one vertex on a side of the square, and their opposite sides partially coincide with the other side of the square and with each other.\n\n![](images/Estonija_2012_p19_data_133ab5f31d.png)\n\nThe area common to both triangles forms an equilateral triangle of side length 1. Position a third equilateral triangle with side length 12 between the two, rotated 180°, so that its lowest vertex coincides with the highest vertex of the small triangle. To show the square is fully covered by these triangles, we must show that the sum of the heights of the large and small triangle is at least 11, i.e.,\n\n$$\n\\frac{\\sqrt{3}}{2} \\cdot (12 + 1) > 11\n$$\n\nSimplifying, this becomes $13\\sqrt{3} > 22$, and since $3 \\cdot 169 > 484$, it holds. Therefore, at least a third of the 2012 points, or at least 671 points, lie in one of the three chosen triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19535,
    "subject": "Mathematics (Olympiad)",
    "question": "A magical triangulation is a partition of a triangle into smaller triangles by a finite number of segments whose endpoints are vertices of the triangle or points in its interior, such that at every point (including the vertices of the triangle), the same number of segments meet.\n\nWhat is the maximal number of smaller triangles into which we can divide the triangle in a magical triangulation?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the number of smaller triangles, $t$ the number of points in the triangulation (including the vertices of the triangle), $d$ the number of segments (including the sides of the triangle), and $k$ the number of segments meeting at each point of the triangulation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19536,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The sum in question is $f_n(n) = n!$, where\n$$\nf_n(x) = \\sum_{\\sigma \\in S_n} (\\operatorname{sgn} \\sigma) x^{\\ell(\\sigma)} = x(x-1)\\cdots(x-n+1).\n$$\nThis follows from the recurrence formula:\n$$\nf_n(x) = (x-n+1) f_{n-1}(x), \\quad n \\ge 3,\n$$\nwith $f_2(x) = x(x-1)$. To establish the recurrence, consider the decomposition of a permutation $\\sigma$ in $S_n$ into disjoint cycles. For each $\\sigma' \\in S_{n-1}$, it extends to a unique $\\sigma$ in $S_n$ fixing $n$, and to $n-1$ permutations not fixing $n$. The sign and cycle count change accordingly, leading to the recurrence and the result.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19537,
    "subject": "Mathematics (Olympiad)",
    "question": "a) We are to choose 7 committee members satisfying the given conditions:\n\n- There are 100 students and 25 committee members.\n- Each student likes at least 10 committee members.\n- We want to select 7 committee members such that every student likes at least one of the chosen members.\n\nb) Prove that, given $n$ students and $m$ committee members, if each student likes at least $k$ committee members, then it is possible to arrange the examination schedule so that each student is interviewed by a favourite committee member, and each committee member interviews at most $\\lceil \\frac{n}{k} \\rceil$ students.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, choose a committee member who is the favourite of at least $\\frac{100 \\times 10}{40} = 25$ students. Remove all students who like this member, leaving at most 60 students, each liking at least 10 of the remaining 24 committee members. If there are fewer than 60 students, add extra students who like all remaining committee members, so we can assume exactly 60 students remain.\n\nRepeat the process:\n- Second committee member: favourite of at least $\\left\\lfloor \\frac{60 \\times 10}{24} \\right\\rfloor = 25$ students.\n- Remaining 35 students, each likes at least 10 of 23 committee members: third member is favourite of at least $\\left\\lfloor \\frac{35 \\times 10}{23} \\right\\rfloor = 16$ students.\n- Remaining 19 students, each likes at least 10 of 22 committee members: fourth member is favourite of at least $\\left\\lceil \\frac{19 \\times 10}{22} \\right\\rceil = 9$ students.\n- Remaining 10 students, each likes at least 10 of 21 committee members: fifth member is favourite of at least $\\left\\lceil \\frac{10 \\times 10}{21} \\right\\rceil = 5$ students.\n- Remaining 5 students, each likes at least 10 of 20 committee members: sixth member is favourite of at least $\\left\\lceil \\frac{5 \\times 10}{20} \\right\\rceil = 3$ students.\n- Remaining 2 students, each likes at least 10 of 19 committee members: seventh member is favourite of both.\n\nThus, each student likes at least one of the seven chosen committee members.\n\nb) The general statement: Given $n$ students and $m$ committee members, if each student likes at least $k$ committee members, then we can arrange the schedule so that each student is interviewed by a favourite committee member, and each committee member interviews at most $\\lceil \\frac{n}{k} \\rceil$ students.\n\n**Proof by induction on $k$:**\n- For $k = 1$, the statement is trivial.\n- Assume true for $k$. For $k+1$, if every committee member is favourite of at most $\\lceil \\frac{n}{k+1} \\rceil$ students, assign arbitrarily.\n- If some committee member is favourite of more than $\\lceil \\frac{n}{k+1} \\rceil$ students, assign exactly $\\lceil \\frac{n}{k+1} \\rceil$ students to them. For the remaining $n - \\lceil \\frac{n}{k+1} \\rceil$ students and $m-1$ committee members, each student likes at least $k$ members. By induction, each committee member interviews at most\n\n$$\n\\left\\lceil \\frac{n - \\lceil \\frac{n}{k+1} \\rceil}{k} \\right\\rceil \\leq \\left\\lceil \\frac{n}{k+1} \\right\\rceil.\n$$\n\nThus, the statement holds for $k+1$ by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19538,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\\omega_1$ the circumcircle of $BWN$, and let $X$ be the point on $\\omega_1$ such that $WX$ is a diameter of $\\omega_1$. Analogously, denote by $\\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point on $\\omega_2$ such that $WY$ is a diameter of $\\omega_2$. Prove that $X$, $Y$, and $H$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the foot of the altitude of $ABC$ from $A$, and let $O_1$ and $O_2$ be the centers of $\\omega_1$ and $\\omega_2$, respectively. Because $\\angle WNB < \\angle CNB = 90^\\circ$, $O_1$ and $N$ lie on the same side of $BC$. Likewise, $O_2$ and $M$ lie on the same side of $BC$. Hence, segment $O_1O_2$ does not intersect line $BC$. In particular, $W$ does not lie on line $O_1O_2$, and $\\omega_1$ and $\\omega_2$ intersect again at a point $Z$ other than $W$.\n\nBecause $XW$ is a diameter of $\\omega_1$, we have $XZ \\perp WZ$. Likewise, we have $YZ \\perp WZ$, so $X$, $Y$, and $Z$ lie on a line perpendicular to line $ZW$. It suffices to show that $HZ \\perp ZW$. First, quadrilaterals $BNHL$ and $CMHL$ are cyclic, so by power of a point $AM \\cdot AC = AH \\cdot AL = AN \\cdot AB$, hence $A$ lies on the radical axis $ZW$ of $\\omega_1$ and $\\omega_2$. Second, by our previous argument, $A$ is the radical center of circles $\\omega_1$, $\\omega_2$, and the circumcircles of quadrilaterals $BNHL$ and $CMHL$. In particular, we have $AH \\cdot AL = AZ \\cdot AW$, so $ZHLW$ is cyclic. This shows $\\angle AZH = \\angle ALW = 90^\\circ$, hence $HZ \\perp AW$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19539,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be integers with $|a| \\ge 2$. Prove that the sequence $a^1 + b, a^2 + b, \\dots, a^n + b, \\dots$ has $2022$ consecutive members consisting of composite numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a$ and $b$ are relatively prime (otherwise, the result is clear).\n\nFor $n \\ge 1$, let $a_n = a^n + b$. Choose $l \\ge 0$ such that $|a|^l > |b|$. For $n \\ge l+1$, $|a_n| < |a|^n + |a|^l \\le 2|a|^n - |a|^l < |a_{n+1}|$.\n\nIn particular, for $n \\ge l+3$, $|a_n| \\ge 2$. Let $m = l+3$ and $N = 2022$. Each of $a_{m+1}, a_{m+2}, \\dots, a_{m+N}$ has at least one prime divisor; choose $p_1 \\mid a_{m+1}$, $p_2 \\mid a_{m+2}$, ..., $p_N \\mid a_{m+N}$. Clearly $(a, p_i) = 1$ for $1 \\le i \\le N$.\n\nLet $k = (p_1-1)(p_2-1)\\dots(p_N-1) \\ge 1$. Then $a_{m+k+i} = a^{m+i}(a^k-1) + a_{m+i}$ is divisible by $p_i$ by Fermat's theorem for $1 \\le i \\le N$. Moreover, $|a_{m+k+i}| > |a_{m+i}|$. Thus, $a_{m+k+1}, a_{m+k+2}, \\dots, a_{m+k+N}$ are all composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19540,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integer $n$, given a finite collection of coins with total value at most $n - \\frac{1}{2}$, it is possible to split this collection into $n$ or fewer groups, such that each group has total value at most $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1**\n\n*Case 1.* No sub-collection of coins in the collection has total value $1$.\n\nWe say that the value of a coin is *large* if its denomination is at least $\\frac{1}{2n}$ and *small* otherwise. The plan is to distribute all the large coins first and the small coins after this.\n\nFor each positive integer $i$ ($1 \\leq i \\leq n$), let $T_i$ be the group of all large coins whose denominations take the form $\\frac{1}{2^k(2i-1)}$. We claim that the total value of the coins in each $T_i$ is less than $1$.\n\nSuppose that the coins in $T_i$ have total value at least $1$ for some $i$. If $T_i$ has $t$ coins, let us list their values in weakly decreasing order:\n\n$$\n\\frac{1}{2^{a_1}(2i-1)}, \\frac{1}{2^{a_2}(2i-1)}, \\dots, \\frac{1}{2^{a_t}(2i-1)},\n$$\n\nwhere $a_1 \\leq a_2 \\leq \\dots \\leq a_t$.\n\nFor each $j$ ($1 \\leq j \\leq t$), let\n\n$$\nS_j = \\frac{1}{2^{a_1}(2i-1)} + \\frac{1}{2^{a_2}(2i-1)} + \\dots + \\frac{1}{2^{a_j}(2i-1)}.\n$$\n\nSince $S_1 < 1 \\leq S_t$ and $S_1 < S_2 < \\dots < S_t$, and we do not have $S_j = 1$ for any $j$, there is a unique index $k$ such that $S_{k-1} < 1 < S_k$. But $S_k = S_{k-1} + \\frac{1}{2^{a_k}(2i-1)}$, and so we have\n\n$$\n\\begin{aligned}\nS_{k-1} &< 1 < S_{k-1} + \\frac{1}{2^{a_k}(2i-1)} \\\\\n\\Rightarrow \\quad & 0 < 2^{a_k}(2i-1)(1-S_{k-1}) < 1.\n\\end{aligned}\n$$\n\nThis is impossible because $2^{a_k}(2i-1)(1-S_{k-1})$ is an integer.\n\nHence $T_1, T_2, \\dots, T_n$ form $n$ groups each of total value less than $1$.\n\nIt remains to distribute the small coins. Let us add the small coins one at a time to the $n$ groups. If we reach a point where this is no longer possible, then all the groups have total value at least $1 - \\frac{1}{2n+1}$. But then the total value of all the coins is at least\n\n$$\nn\\left(1 - \\frac{1}{2n + 1}\\right) > n - \\frac{1}{2},\n$$\n\na contradiction. This concludes the proof for case 1.\n\n*Case 2.* A sub-collection of coins has total value $1$.\n\nLet us remove sub-collections of coins of total value $1$ and put them aside into their own group until there are no sub-collections of total value $1$ among the remaining coins. If this occurs $d$ times, we end up with $d$ groups of coins, each of which has total value $1$. What remains is a collection of coins whose total value is $n - d - \\frac{1}{2}$ which does not possess a sub-collection of coins that have total value $1$. Hence we may apply the result of case 1 to distribute these remaining coins into $n - d$ groups, each of which has total value at most $1$. Combining these $n - d$ groups with the $d$ groups from earlier in this paragraph concludes the proof for case 2. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19541,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the least odd number $a > 5$ satisfying the following conditions: There are positive integers $m_1, m_2, n_1, n_2$ such that $a = m_1^2 + n_1^2$, $a^2 = m_2^2 + n_2^2$, and $m_1 - n_1 = m_2 - n_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $261$.\n\nNote that\n\n$$\n261 = 15^2 + 6^2, \\quad 261^2 = 189^2 + 180^2, \\quad 15 - 6 = 189 - 180.\n$$\n\nWe know that there is no number between $5$ and $261$ that satisfies the condition of the problem. Assume on the contrary that $a$ is such a number. We may set $d = m_1 - n_1 > 0$. Because $a$ is odd, $m_1$ and $n_1$ have different parity, and so $d$ is odd. Because $m_1 < 261$, $d \\leq 15$; that is, the possible values of $d$ are $1, 3, 5, 7, 9, 11, 13, 15$. We will eliminate every one of them.\n\nWe can write $m_2 = n_2 + d$ and $a^2 = (n_2 + d)^2 + n_2^2$, or\n\n$$\n2a^2 - d^2 = (2n_2 + d)^2. \\qquad (1)\n$$\n\nIf $d=1$, then $(1)$ becomes a Pell's equation $x^2 - 2y^2 = -1$ with $(x, y) = (2n_2 + 1, a)$. This Pell's equation has minimal solution $(x, y) = (1, 1)$ and $x + y\\sqrt{2} = (1 + \\sqrt{2})^{2k-1}$ for positive integers $k$. The $y$ values of the solutions of this Pell's equation are $5, 29, 169, 985, \\ldots$ Thus, the only possible values for $a$ are $29$ and $169$. It is easy to check that neither $29$ nor $169$ can be written in the form of $(n_1 + 1)^2 + n_1^2$. Hence $d \\neq 1$.\n\nIf $d$ is a multiple of $3$, then $m_2 \\equiv n_2 \\pmod{3}$ and $a^2 \\equiv 2m_2^2 \\pmod{3}$. Because $2$ is not a quadratic residue modulo $3$, we conclude that $0 \\equiv m_2 \\equiv n_2 \\equiv a \\pmod{3}$. Hence, $m_1^2 + n_1^2 \\equiv 0 \\pmod{3}$, from which it follows that $m_1 \\equiv n_1 \\equiv 0 \\pmod{3}$. Thus, $a = m_1^2 + n_1^2$ is a multiple of $9$ and $m_2^2 + n_2^2 = a^2$ is a multiple of $81$. We can write $m_2 = 3m'$, $n_2 = 3n'$, and $a = 9a'$ for integers $m', n', a'$. We have $m'^2 + n'^2 = 9a'$. Again, because $-1$ is not a quadratic residue modulo $3$, we must have $m' \\equiv n' \\equiv 0 \\pmod{3}$. It follows that $d = 3(m' - n')$ is divisible by $9$. Thus, $d = 9$.\n\nBecause $a < 261 = 15^2 + 6^2$, $n_1 < 6$. Therefore, $n_1 = 3$ and $a = 12^2 + 3^2 = 153$. But then $(1)$ becomes $9^2 \\cdot 577 = (2n_2 + 9)^2$, which is impossible. Hence, $d \\neq 3$, $9$, or $15$.\n\nIf $d=11$ or $d=13$, because $2$ is not a quadratic residue modulo $d$, from $a^2 \\equiv 2m_2^2 \\pmod{d}$, we conclude that $0 \\equiv m_2 \\equiv n_2 \\equiv a \\pmod{d}$. It follows that $2m_1^2 \\equiv a \\equiv 0 \\pmod{d}$ and $0 \\equiv m_1 \\equiv n_1 \\equiv a \\pmod{d}$. In particular, $n_1 \\geq d$, $m_1 \\geq 2d$ and\n\n$$\na = m_1^2 + n_1^2 \\geq 5d^2 > 261.\n$$\n\nHence, $d \\neq 11$ or $13$.\n\nIf $d = 5$, we also note that $2$ is not a quadratic residue modulo $5$. By the same reasoning as before, we have\n\n$$\nm_1 \\equiv n_1 \\equiv m_2 \\equiv n_2 \\equiv 0 \\pmod{5}.\n$$\n\nIf $n_1 \\geq 10$, then $m_1 \\geq 15$ and $a = m_1^2 + n_1^2 \\geq 10^2 + 15^2 > 261$. Thus, $n_1 = 5$, $m_1 = 10$, and $a = 125$. But then $(1)$ becomes\n\n$$\n5^2 \\cdot 1249 = (2n_2 + 5)^2,\n$$\n\nwhich is impossible. Hence, $d \\neq 5$.\n\nIf $d = 7$, then because $a < 261 = 15^2 + 6^2$, we have $n_1 \\leq 8$. The possible values of $a$ are then $65, 85, 109, 137, 169, 205, 245$. But then $(1)$ becomes $(2n_2 + 7)^2 = 2a^2 - 49$. It is easy to check that there is no solution in this case. Hence, $d \\neq 7$.\n\nCombining the above, we conclude that $261$ is the answer to this question. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19542,
    "subject": "Mathematics (Olympiad)",
    "question": "A deltoid (kite) is inscribed in a circle such that its sides have lengths $a$ and $2a$. What is the ratio of the area of the deltoid to the area of the circle?",
    "options": [],
    "answer": "See solution",
    "solution": "Due to symmetry, the longer diagonal of the deltoid passes through the center of the circle. By Thales' theorem, there is a right angle between the sides of length $a$ and $2a$. By the Pythagorean theorem, the length of the longer diagonal is $$\\sqrt{a^2 + (2a)^2} = \\sqrt{5}a.$$ Thus, the radius of the circle is $$r = \\frac{\\sqrt{5}a}{2}.$$ The area of the deltoid is $$2 \\cdot \\frac{a \\cdot 2a}{2} = 2a^2.$$ The area of the circle is $$\\pi r^2 = \\pi \\left(\\frac{\\sqrt{5}a}{2}\\right)^2 = \\frac{5\\pi a^2}{4}.$$ Therefore, the ratio of the areas is $$\\frac{2a^2}{\\frac{5\\pi a^2}{4}} = \\frac{8}{5\\pi}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19543,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$, let $M$ be the midpoint of $BC$ and $H$ the orthocentre. Let $\\Gamma$ be the circle with diameter $HM$, and let $X, Y$ be distinct points on $\\Gamma$ such that $AX$ and $AY$ are tangent to $\\Gamma$. Prove that $BXYC$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the foot of the altitude from $A$ to $BC$, which also lies on $\\Gamma$. Let $O$ be the circumcentre of $\\triangle ABC$. Since $\\angle HDM = 90^\\circ$, note that rays $HD$ and $HM$ meet the circumcircle at points which are reflections in $OM$. Then, since $\\angle BAD = \\angle OAC$, we recover the well-known fact that ray $HM$ meets the circumcircle at $A'$, the point antipodal to $A$. Therefore, the ray $MH$ meets the circumcircle at a point $T$ such that $\\angle MTA = 90^\\circ$. Note that $T$ and $D$ lie on the circle with diameter $AM$.\n\n![](images/2019_bmo_shortlist_p18_data_9ff3f8679e.png)\n\nNow, study $K$, the centre of $\\Gamma$. Clearly $AXKY$ is cyclic, with diameter $AK$, so $T$ also lies on this circle. We can now apply the radical axis theorem to the three circles $\\odot ATXKY$, $\\odot ATDM$, $\\odot HXDMY$ to deduce that $AT$, $XY$, $DM$ concur at a point, $Z$.\n\nThen, by power of a point in $\\odot ATXY$, we have $ZX \\cdot ZY = ZT \\cdot ZA$; but also by power of a point in the circumcircle, we have $ZA \\cdot ZT = ZB \\cdot ZC$. Therefore\n\n$$\nZX \\cdot ZY = ZB \\cdot ZC,\n$$\n\nand the result follows. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19544,
    "subject": "Mathematics (Olympiad)",
    "question": "給定圓內接四邊形 $ABCD$。直線 $L$ 為過外心 $O$ 的一直線，$P$ 為 $L$ 上一動點。設圓 $c_1$ 是通過 $P$ 在 $AB$, $BC$, $CA$ 三個垂足的圓；圓 $c_2$ 是通過 $P$ 在 $AB$, $BD$, $DA$ 三個垂足的圓。設圓 $c_1, c_2$ 交於 $P_1, Q$ 兩點，其中 $P_1$ 為 $P$ 在 $AB$ 的垂足。試問當 $P$ 點在 $L$ 上移動時，$Q$ 點的軌跡為何？",
    "options": [],
    "answer": "See solution",
    "solution": "如圖，設 $P_1, P_2, P_3, P_4, P_5$ 分別代表 $P$ 點在 $AB$, $BC$, $AC$, $BD$, $AD$ 線段上的垂足。\n\n![](images/13-2J-ind_p9_data_cd1c56c518.png)\n\n設 $M_1, M_2, M_3, M_4, M_5$ 分別是 $AB$, $BC$, $AC$, $BD$, $AD$ 各邊上的中點。考慮 $\\triangle ABC$。顯然外心 $O$ 是 $\\triangle M_1M_2M_3$ 的垂心。由斯坦納定理，直線 $L$ 對 $M_1M_2$, $M_2M_3$, $M_3M_1$ 做反射後必會交於 $\\triangle ABC$ 九點圓上一點 $K_1$。考慮 $\\triangle ABD$，用同樣的方法定出點 $K_2$。由於 $K_1, K_2$ 均與 $P$ 點位置無關，以下將證明 $Q$ 點落在直線 $K_1K_2$ 上，便能說明 $Q$ 的軌跡是直線。\n\n![](images/13-2J-ind_p10_data_cd9d11cf1c.png)\n\n我們首先證明：$\\triangle POC \\sim \\triangle P_1 M_1 K_1$：\n\n(i) 令直線 $L$ 對 $M_2M_3$ 的反射線為 $EK_1$。算角度：\n\n$$\n\\begin{aligned}\n\\angle K_1 M_1 B &= \\angle EK_1 M_1 - \\angle EFM_3 \\\\\n&= \\angle EM_2 M_1 - \\angle PFM_3 \\\\\n&= (\\angle EM_2 M_3 + \\angle M_3 M_2 M_1) - \\angle PFM_3 \\\\\n&= (\\angle OCA + \\angle CM_3 M_2) - \\angle PFM_3 \\\\\n&= \\angle CGB - \\angle PFM_3 = \\angle COF.\n\\end{aligned}\n$$\n\n故知 $\\angle P_1 M_1 K_1 = \\angle POC$。\n\n(ii) 算比例：\n\n$$\n\\frac{M_1 K_1}{O C} = \\sin \\angle K_1 E M_1 = \\sin \\angle E O F = \\frac{P_1 M_1}{P O}\n$$\n\n（因為 $OC$ 長度等於九點圓直徑長度）\n\n(iii) 故由 SAS 知 $\\triangle POC \\sim \\triangle P_1 M_1 K_1$。同理也知：$\\triangle POB \\sim \\triangle P_3 M_3 K_1$, $\\triangle POA \\sim \\triangle P_2 M_2 K_1$。\n\n接著證明：$K_1$ 在圓 $c_1$ 上；算角度\n\n$$\n\\begin{aligned}\n\\angle P_3 K_1 P_1 &= \\angle M_3 K_1 M_1 - (\\angle P_3 K_1 M_3 + \\angle P_1 K_1 M_1) \\\\\n&= \\angle CAB - (\\angle P_3 K_1 M_1 + \\angle P_1 K_1 M_1) \\quad (\\text{因為 } K_1 \\text{ 在九點圓上}) \\\\\n&= \\angle CAB - (\\angle PBO + \\angle PCO) \\\\\n&= (\\angle OAC + \\angle OBC) - (\\angle PBO + \\angle PCO) \\\\\n&= (\\angle OCA + \\angle OCB) - (\\angle PBO + \\angle PCO) \\\\\n&= \\angle PCP_3 + \\angle PBP_1 \\\\\n&= \\angle PP_2 P_3 + \\angle PP_2 P_1 = \\angle P_3 P_2 P_1,\n\\end{aligned}\n$$\n\n得證 $K_1$ 在圓 $c_1$ 上；同理 $K_2$ 在圓 $c_2$ 上。\n\n以下說明 $\\triangle P_1O_1O_2 \\sim \\triangle PCD$，其中 $O_1, O_2$ 分別為 $c_1, c_2$ 的圓心：\n\n(i) 不妨設 $P$ 對 $O_1, O_2$ 的對稱點分別為 $I_1, I_2$。則我們知道 $I_1, I_2$ 分別是 $P$ 對 $\\triangle ABC, \\triangle ABD$ 的等角共軛點。計算\n\n$$\n\\begin{aligned}\n\\angle AI_1B &= 180^\\circ - (\\angle I_1AB + \\angle I_1BA) \\\\\n&= 180^\\circ - (\\angle PAC + \\angle PBC) \\\\\n&= 180^\\circ - (\\angle APB - \\angle ACB) \\\\\n&= (180^\\circ - \\angle P_2P_1P_3).\n\\end{aligned}\n$$\n\n同理 $\\angle AI_2B = 180^\\circ - (\\angle APB - \\angle ACB)$，故 $A, B, I_1, I_2$ 共圓。另外可知：$\\angle I_1BI_2 = \\angle CBD$。\n\n(ii)\n\n$$\n\\begin{aligned}\n\\frac{O_1O_2}{CD} &= \\frac{I_1I_2}{2CD} = \\frac{I_1I_2}{AB} \\times \\frac{AB}{2CD} \\\\\n&= \\frac{1}{2} \\frac{\\sin \\angle I_1BI_2}{\\sin \\angle AI_1B} \\times \\frac{\\sin \\angle ACB}{\\sin \\angle CBD} \\\\\n&= \\frac{1}{2} \\frac{\\sin \\angle ACB}{\\sin \\angle AI_1B} \\\\\n&= \\frac{1}{2} \\frac{\\sin \\angle ACB}{\\sin \\angle P_2P_1P_3} \\times \\frac{P_2P_3}{P_2P_3} \\\\\n&= \\frac{1}{2} \\frac{2O_1P_1}{PC} = \\frac{O_1P_1}{PC}.\n\\end{aligned}\n$$\n\n同理，$\\frac{O_1O_2}{CD} = \\frac{O_2P_2}{PD}$。故由 SSS 得到 $\\triangle P_1O_1O_2 \\sim \\triangle PCD$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19545,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\nx \\rightarrow x + k + 1, \\quad y \\rightarrow y + k^2 + 2\n$$\n\nand color the new values blue and red, respectively. This process continues until the blue number is at least $N$.\n\nDetermine the minimum possible value of the red number at the end of the process.",
    "options": [],
    "answer": "See solution",
    "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 $k \\geq 2$, 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 $p \\geq 1$, 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 $p \\geq 1$, we conclude that $M_k$ can be replaced by $p$ moves of $M_1$ and one $M_0$.\n\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": 19546,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (x, y) $ such that $ x^3 = 2y^2 + 1 $.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $x^3 = (1 + y\\sqrt{-2})(1 - y\\sqrt{-2})$. The identity\n\n$$\n(1 - y\\sqrt{-2} - y^2)(1 + y\\sqrt{-2}) - y^2(1 - y\\sqrt{-2}) = 1\n$$\n\nshows that $1 + y\\sqrt{-2}$ and $1 - y\\sqrt{-2}$ are relatively prime in $\\mathbb{Z}[\\sqrt{-2}]$. Since $\\mathbb{Z}[\\sqrt{-2}]$ is a unique factorization domain,\n\n$$\n1 + y\\sqrt{-2} = (u + v\\sqrt{-2})^3 = u(u^2 - 6v^2) + v(3u^2 - 2v^2)\\sqrt{-2},\n$$\n\nfor some $u, v \\in \\mathbb{Z}$. Therefore, $u(u^2 - 6v^2) = 1$, which forces $u = 1$ and $v = 0$. Consequently, $y = 0$ and $x = 1$, which are obviously solutions to the given equation. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19547,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BM$ be a median in an acute triangle $ABC$, whose sides $AB$ and $BC$ have different lengths. The extension of $BM$ intersects the circumcircle of $ABC$ at a point $N$. Let $D$ be a point on the circumcircle such that $\\angle BDH = 90^\\circ$, where $H$ is the orthocenter of $ABC$. Let $K$ be a point chosen so that $ANCK$ is a parallelogram. Prove that the lines $AC$, $KH$, and $BD$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the intersection point of altitudes $AA_1$ and $CC_1$ in $ABC$. The points $B$, $D$, $A_1$, $H$, and $C_1$ lie on the same circle with diameter $BH$, because $\\angle BDH = \\angle BA_1H = \\angle BC_1H = 90^\\circ$.\n\nSince $BC_1HA_1$ and $BANC$ are cyclic quadrilaterals, it follows that $\\angle A_1HC_1 = 180^\\circ - \\angle B$ and $\\angle ANC = 180^\\circ - \\angle B$. This implies $\\angle A_1HC_1 = \\angle ANC$. But $\\angle ANC = \\angle A_1HC_1$ (as vertical angles) and $\\angle AKC = \\angle ANC$ (as opposite angles in a parallelogram). Hence we obtain\n\n$$\\angle AHC = \\angle AKC,$$\n\nwhich means $AKHC$ is cyclic. This implies $\\angle KHC_1 = \\angle KAC$. Then $\\angle KAC = \\angle ACN$, because $AK \\parallel NC$ and $\\angle ACN = \\angle ABN$, as these angles are inscribed in the circumcircle of $ABC$. Therefore, $\\angle KHC_1 = \\angle KBC_1$, which implies that the point $K$ also lies on the circle with diameter $BH$.\n\nDenote by $\\omega_1$ the circumcircle of $ABC$, by $\\omega_2$ the circle with diameter $BH$, and by $\\omega_3$ the circumcircle of $AKHC$. Then $AC$, $KH$, and $BD$ are radical axes of circles $\\omega_1$ and $\\omega_3$, $\\omega_2$ and $\\omega_3$, $\\omega_1$ and $\\omega_2$. As it is known, the radical axes of three circles either intersect at the same point, which is their radical center, or are parallel.\n\nSince $AB > BC$, lines $AC$, $KH$, and $BD$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19548,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, \\infty) \\to \\mathbb{R}$ be a differentiable function with continuous derivative, such that $f(0) = 0$ and $0 \\leq f'(x) \\leq 1$ for any $x > 0$.\n\nProve that\n\n$$\n\\int_{0}^{a} f(t)^{2n+1} \\, dt \\leq (n+1) \\left( \\int_{0}^{a} f(t)^{n} \\, dt \\right)^{2},\n$$\n\nfor any $a > 0$ and $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $f'(x) \\geq 0$ for any $x \\geq 0$, the function $f$ is monotonically increasing, so $f(x) \\geq f(0) = 0$ for any $x \\geq 0$.\n\nConsider a fixed $n \\in \\mathbb{N}^*$ and define $F : [0, \\infty) \\to \\mathbb{R}$ by\n\n$$\nF(x) = (n + 1) \\left( \\int_{0}^{x} f(t)^{n} \\, dt \\right)^{2} - \\int_{0}^{x} f(t)^{2n+1} \\, dt.\n$$\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$$\n\\begin{aligned}\nF'(x) &= 2(n + 1) \\left( \\int_{0}^{x} f(t)^{n} \\, dt \\right) f(x)^{n} - f(x)^{2n+1} \\\\\n&= f(x)^{n} \\left( 2(n + 1) \\left( \\int_{0}^{x} f(t)^{n} \\, dt \\right) - f(x)^{n+1} \\right).\n\\end{aligned}\n$$\n\nBecause $f'(x) \\leq 1$ for any $x \\geq 0$, we have $f(x)^n \\geq f(x)^n f'(x)$ for all $x \\geq 0$, hence\n\n$$\n(n + 1) \\left( \\int_{0}^{x} f(t)^{n} \\, dt \\right) \\geq \\int_{0}^{x} (n + 1) f(t)^{n} f'(t) \\, dt = f(x)^{n+1}.\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\nF'(x) &= f(x)^n \\left( 2(n + 1) \\left( \\int_{0}^{x} f(t)^n \\, dt \\right) - f(x)^{n+1} \\right) \\\\\n&\\geq f(x)^n \\left( 2 f(x)^{n+1} - f(x)^{n+1} \\right) = f(x)^{2n+1} \\geq 0,\n\\end{aligned}\n$$\n\nfor any $x \\geq 0$. Thus, $F$ is increasing, and so $F(x) \\geq F(0) = 0$ for any $x \\geq 0$. We obtain the stated inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19549,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be a positive number. On the parabola whose equation has coefficient $1$ at the quadratic term, points $A$, $B$, and $C$ are chosen so that the difference of the $x$-coordinates of points $A$ and $B$ is $d$, and the difference of the $x$-coordinates of points $B$ and $C$ is also $d$. Find the area of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume the equation of the parabola is $y = x^2$. Let the abscissas of points $A$, $B$, and $C$ be $a$, $b$, and $c$. Let $A'$, $B'$, and $C'$ be the projections of $A$, $B$, and $C$ onto the $x$-axis. Denoting the area of a region $K$ by $S_K$, we have\n\n$$\nS_{ABC} = S_{ACC'A'} - S_{ABB'A'} - S_{BCC'B'}.\n$$\n\nSince $S_{ACC'A'} = 2d \\cdot \\frac{a^2 + c^2}{2}$, $S_{ABB'A'} = d \\cdot \\frac{a^2 + b^2}{2}$, and $S_{BCC'B'} = d \\cdot \\frac{b^2 + c^2}{2}$, it follows that\n\n![](images/prob1516_p6_data_c00225688e.png)\n\n$$\n\\begin{align*}\nS_{ABC} &= \\frac{1}{2}d(2a^2 + 2c^2 - a^2 - b^2 - b^2 - c^2) \\\\\n&= \\frac{1}{2}d(a^2 - b^2 - b^2 + c^2) \\\\\n&= \\frac{1}{2}d\\big(d(a+b) - d(b+c)\\big) \\\\\n&= \\frac{1}{2}d^2(a-c) \\\\\n&= \\frac{1}{2}d^2 \\cdot 2d \\\\\n&= d^3.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, scalene triangle with circumcircle $(O)$ and orthocenter $H$. Let $D$, $E$, and $F$ be the reflections of $O$ through $BC$, $CA$, and $AB$ respectively.\n\n**a)** Let $H_a$ be the reflection of $H$ through $BC$ and $A'$ be the reflection of $A$ in $O$. Let $O_a$ be the circumcenter of $OBC$. Prove that $HD$ and $A'O_a$ intersect on $(O)$.\n\n**b)** Construct a point $X$ such that $AXDA'$ is a parallelogram. Prove that the circumcircles of triangles $AHX$, $ABF$, and $ACE$ have a common point that differs from $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Suppose that $H_aD$ meets $(O)$ at $K$. Let $M$ be the midpoint of $BC$, then $OD = 2OM = AH$. Two isosceles triangles $OBD$ and $OO_aB$ have the common angle $\\angle BOD$ so they are similar, hence\n\n$$\n\\frac{OD}{OB} = \\frac{OB}{OO_a} \\text{ thus } OD \\cdot OO_a = R^2\n$$\n\nwhere $R$ is the radius of $(O)$.\n\n![](images/Vietnamese_mathematical_competitions_p244_data_a07b4ba7b6.png)\n\nThus, $AH \\cdot OO_a = R^2$ or $\\frac{AH}{OA'} = \\frac{OA}{OO_a}$. On the other hand, note that $\\angle OAH = \\angle A'OA_a$, then two triangles $AHO$, $OA'O_a$ are similar. Because $OHH_aD$ is an isosceles trapezoid, we obtain\n\n$$\n\\angle OA'O_a = \\angle AHO = 180^\\circ - \\angle AH_aK = 180^\\circ - \\angle AA'K\n$$\n\nTherefore, $O_a$, $A'$, and $K$ are collinear and the statement is proved.\n\n**b)** Let $J$ be the midpoint of $OH$, then $J$ is the Euler center of triangle $ABC$. Let $L$ be the reflection of $K$ over $J$. Let $N$ be the midpoint of $AC$, then $MN$ is the midline of triangle $ODE$. By angle chasing, we have\n\n$$\n\\begin{aligned}\n\\angle KDE &= \\angle KDO - \\angle ODE = \\angle AH_aK - \\angle OMN \\\\\n            &= 180^\\circ - \\angle ACK - \\angle OCN = 180^\\circ - \\angle ECK.\n\\end{aligned}\n$$\n\nHence, $DKCE$ is cyclic. On the other hand, it is clear that $OHDA'$ is a parallelogram. Thus $\\overrightarrow{DH} = \\overrightarrow{A'O} = \\frac{1}{2}\\overrightarrow{A'A} = \\frac{1}{2}\\overrightarrow{DX}$, which implies $H$ is the midpoint of $DX$. Therefore, $X$ and $A'$ are symmetric with respect to $J$. Using the reflection centered at $J$, we have\n\n$$\nX \\leftrightarrow A', \\quad A \\leftrightarrow D, \\quad H \\leftrightarrow O, \\quad L \\leftrightarrow K.\n$$\n\nOn the other hand, $A'$, $D$, $O$, and $K$ are concyclic (since $\\angle OA'K = \\angle AH_aD = \\angle ODK$), so $X$, $A$, $H$, and $L$ are concyclic, or $X \\in (AHL)$. By that transformation, we also get $A$, $L$, $F$, and $B$ are concyclic, so $L \\in (AFB)$. Similarly, $L \\in (ACE)$.\n\nTherefore, the three circles $(AHX)$, $(ABF)$, and $(ACE)$ have another common point $L \\neq A$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19551,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$. There are $n$ boys and $n$ girls standing in a line. Give each person $X$ in the line a number of sweets which is exactly equal to the number of pairs $(a, b)$ such that $a$ and $b$ are of opposite sex to $X$ and $X$ is standing between $a$ and $b$. Prove that the total number of sweets of $n$ boys and $n$ girls does not exceed $\\frac{1}{3}n(n^2 - 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We denote a boy by the letter $b$ and a girl by the letter $g$. We can check that the total number of sweets is exactly $\\frac{1}{3}n(n^2 - 1)$ if the line can be partitioned into $n$ consecutive pairs of one boy and one girl. This arrangement will be called the optimal arrangement.\n\nAny other arrangement will be in one of the following forms:\n\n$$\n\\begin{array}{l}\n(bg)(bg)(gb) \\dots (..) \\underbrace{gg \\dots g}_{t \\ge 2} b \\dots, \\text{ or} \\\\\n(bg)(bg)(gb) \\dots (..) \\underbrace{bb \\dots b}_{t \\ge 2} g \\dots\n\\end{array}\n$$\n\nIt is easy to check that, by moving one girl or one boy as shown in the arrangements below, the number of sweets is increased. Therefore, after a finite number of moves, we will get the optimal arrangement with exactly $\\frac{1}{3}n(n^2 - 1)$ sweets. This concludes the proof.\n\n$$\n(bg)(bg)(gb) \\dots (..) \\underbrace{gg \\dots g}_{t-1} bg \\dots, \\text{ and}\n$$\n$$\n(bg)(bg)(gb) \\dots (..) \\underbrace{bb \\dots b}_{t-1} gb \\dots\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19552,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q = \\deg Q$ divides $s = \\deg S$. 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\n$$\nS(x) = T(Q(x)) + M(x), \\qquad (*).\n$$\n\nwhere $\\deg M$ is minimized; if $M = 0$, then we get the desired result. For the sake of contradiction, suppose $M \\ne 0$. Then $q$ does not divide $m = \\deg M$; 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\n$$\nN(Q(x)) = \\alpha r M(x) T(Q(x))^{r-1} + \\text{a polynomial of lower degree.}\n$$\n\nThis is impossible, since $q$ divides the degree of the left-hand member, but not that of the right-hand member.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19553,
    "subject": "Mathematics (Olympiad)",
    "question": "We number the members with $1, 2, \\dots, 6$. Consider the following decision method:\n\nIf persons numbered 1 to 3 have the same vote, their vote is considered as the final result. Otherwise, the result is based on the majority of the votes among persons 4 to 6.\n\nIs it possible to represent this decision method as a *Weighted Voting* system? Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that the method can be represented as a weighted voting system with weights $(\\omega_1, \\dots, \\omega_6)$. Since persons 1–3 have the same role, and persons 4–6 have the same role, we can symmetrize the weights to $(a, a, a, b, b, b)$ where $a = \\omega_1 + \\omega_2 + \\omega_3$ and $b = \\omega_4 + \\omega_5 + \\omega_6$.\n\nConsider the case where the first three persons agree and the last three disagree. The result is positive, so $3a > 3b$.\n\nNow, consider the case where two of the first three and one of the last three agree, and the others disagree. The result is negative, but the sum of agreeing weights is $2a + b$ and disagreeing weights is $2b + a$, so $2b + a > 2a + b$, which implies $b > a$.\n\nThis contradicts the previous inequality $a > b$. Therefore, the method cannot be represented as a weighted voting system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19554,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are no two distinct positive integers $a$ and $b$ such that $\\{\\frac{a}{b}\\} + \\{\\frac{b}{a}\\} = 0$.\n\nHere, $\\{x\\}$ denotes the difference between $x$ and the greatest integer not exceeding $x$; for example, $\\{\\frac{7}{5}\\} = \\frac{2}{5}$, $\\{\\frac{2019}{3}\\} = 0$, and $\\{\\frac{2020}{3}\\} = \\frac{1}{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such two numbers $a$ and $b$ exist. Without loss of generality, assume they are relatively prime. The condition implies that $\\frac{a}{b} - \\left\\lfloor \\frac{a}{b} \\right\\rfloor + \\frac{b}{a} - \\left\\lfloor \\frac{b}{a} \\right\\rfloor = 0$, so $\\frac{a}{b} + \\frac{b}{a}$ is an integer. Thus, $\\frac{a^2 + b^2}{ab}$ is an integer, so $a^2 + b^2$ is divisible by $ab$. This means $b^2 \\mid a^2$ and $a^2 \\mid b^2$, which is only possible if $a = b$, contradicting the assumption that $a$ and $b$ are distinct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19555,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a table with $m$ rows and 22 columns. Each cell is filled with a number from the set $A = \\{1, 2, 3, \\dots, 2025\\}$ (numbers may be repeated), such that for every pair of distinct numbers in $A$, there exists a row that contains exactly one of these two numbers. Find the minimum value of $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the number of elements of $A$ that appear exactly once, and $y$ be the number of elements of $A$ that appear at least twice in the table. There is at most one number that does not appear in the table, otherwise, there would be two numbers that do not appear in any row, a contradiction. From there, we deduce that\n\n$$\nx + y \\geq 2024\n$$\n\nSince the total number of cells in the table is $22m$, we have\n\n$$\nx + 2y \\leq 22m\n$$\n\nIf $x > m$, there are two elements that appear exactly once but are in the same row, violating the given condition. Thus, $x \\leq m$ and we get\n\n$$\n2 \\cdot 2024 \\leq 2(x + y) = (x + 2y) + x \\leq 23m \\implies m \\geq 176.\n$$\n\nNext, we construct a table of size $176 \\times 22$ that satisfies the given condition.\n\nFor equality, we must have $x = m = 176$, $y = 2024 - 176 = 1848$, which means 1848 numbers appear exactly twice. Assume elements $1$ to $1848$ appear at least twice, and elements $1849$ to $2024$ appear exactly once. The element $2025$ will not be used.\n\nNow consider a simple, regular, undirected graph $G$ with 176 vertices (corresponding to the 176 rows), all of degree 21. Arrange 176 vertices on a circle, connect each vertex to its reflection and the 10 nearest vertices on either side. Enumerate the vertices from 1 to 176.\n\nThe number of edges of $G$ is $\\frac{176 \\cdot 22}{2} = 1848$ by the handshaking lemma. The 1848 edges correspond to 1848 elements that appear at least twice; enumerate the edges from 1 to 1848. If edge $i$ connects vertices $u_i, v_i$, place element $i$ in the $u_i$-th and $v_i$-th rows. After this, each row has exactly 21 elements. Arrange the remaining 176 elements from 1849 to 2024 so that each row has exactly 22 elements (each row gets one of these elements).\n\nNow, check the condition for any two $i, j \\in A$ with $i < j$:\n\n- If $i, j \\geq 1849$, these elements appear only once, and by construction, each row gets exactly one of these elements. So, the row containing $i$ and the row containing $j$ each contain exactly one of the two.\n- If $i \\leq 1848 < j$, $i$ appears twice (in different rows), $j$ appears once, so there is a row containing $i$ that does not contain $j$.\n- If $i, j \\leq 1848$, both appear twice. These correspond to two distinct edges of $G$. Since $G$ is simple, distinct edges have different endpoints, so there is a row containing one but not the other.\n\nTherefore, the table constructed satisfies the condition, and the minimum value is $176$.\n\n*Comment regarding the bound:* The bound is based on a greedy algorithm considering the number of appearances. There is at most one number with 0 appearances, at most $m$ numbers with 1 appearance, and all other numbers must appear at least twice.\n\n*Comment regarding the example:* The example may appear complex, but essentially, we need to fill the table $176 \\times 21$ with numbers $1, 2, \\dots, 1848$ such that each number appears exactly twice and no two numbers are in the same row twice. This is easily satisfiable. One way is to fill the first $\\frac{1848}{21} = 88$ rows with $1, 2, \\dots, 1848$ in row-major order. Fill the remaining 88 rows in column-major order. In the second half, two numbers are in the same row if and only if their difference is divisible by 88, which is at least 88. As 88 is larger than the number of columns (21), these two numbers must be in different rows in the first half, as that was filled in row-major order.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19556,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer, $n = 2^m - 1$, and $P_n = \\{1, 2, \\dots, n\\}$ be the set of $n$ points on the number axis. A grasshopper jumps between adjacent points on $P_n$. Find the maximal number of $m$ such that for any $x, y \\in P_n$, the number of ways that a grasshopper jumping from $x$ to $y$ in 2012 steps is even (passing $x$ or $y$ on the way is permitted).\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If $m \\geq 11$, then $n = 2^m - 1 > 2013$. Since there is only one way a grasshopper jumps from point $1$ to point $2013$ in $2012$ steps, we see that $m \\leq 10$.\n\nNow, we show that the answer is $m = 10$. We prove a stronger proposition by induction on $m$: for any $k \\geq n = 2^m - 1$ and any $x, y \\in P_n$, the number of ways the grasshopper jumps from $x$ to $y$ in $k$ steps is even.\n\n**Base case:** If $m = 1$, the number of ways is $0$, which is even.\n\n**Inductive step:** Assume for $m = l$, the number of ways is even. For $m = l+1$, $n = 2^{l+1} - 1$, and $k \\geq n$, consider three types of routes from $x$ to $y$ in $k$ steps:\n\n1. **Route does not pass point $2^l$:** Both $x$ and $y$ are on one side of $2^l$. By induction, the number of routes is even.\n2. **Route passes $2^l$ exactly once:** Suppose the grasshopper is at $2^l$ at the $i$-th step ($i \\in \\{0, 1, \\dots, k\\}$). The route splits into two sub-routes: from $x$ to $2^l$ in $i$ steps, and from $2^l$ to $y$ in $k-i$ steps. If both $i-1 < 2^l - 1$ and $k-i-1 < 2^l - 1$, then $k \\leq 2^{l+1} - 2$, contradicting $k \\geq n = 2^{l+1} - 1$. Thus, at least one sub-route has length at least $2^l - 1$, so by induction, the number of ways is even. By the multiplication principle, the total is even.\n3. **Route passes $2^l$ at least twice:** Consider the sub-routes between visits to $2^l$. By symmetry and induction, the number of ways is even. Again, by the multiplication principle, the total is even.\n\nTherefore, the maximal $m$ is $10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19557,
    "subject": "Mathematics (Olympiad)",
    "question": "We will call a circle without boundary, i.e., a circle without the points on its circumference, a \"hedgehog\". The diameter of the hedgehog is the diameter of this circle. We will say that the hedgehog \"sits\" at a point where the center of the corresponding circle is located.\n\nLet us consider a triangle with sides $a$, $b$, $c$, and hedgehogs sitting at its vertices. It is known that there exists a point inside the triangle from which one can reach any side of the triangle along a straight trajectory without touching any of the hedgehogs. What is the largest possible sum of the diameters of these hedgehogs?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the diameters of the hedgehogs at the vertices $A$, $B$, and $C$ be $d_a$, $d_b$, and $d_c$, respectively. Suppose $d_a + d_b > 2c$. Then any point on side $AB$ is inside one of the hedgehogs, making it impossible to reach this side. Thus, $d_a + d_b \\leq 2c$. Similarly, $d_a + d_c \\leq 2b$ and $d_b + d_c \\leq 2a$. Adding these inequalities gives $d_a + d_b + d_c \\leq a + b + c$.\n\nWe will show that equality can be achieved. Let $I$ be the incenter of the triangle, and let $A_1$, $B_1$, $C_1$ be the points where the incircle touches the sides. Consider hedgehogs at the vertices with circles of radii $AB_1 = AC_1$, $BA_1 = BC_1$, and $CA_1 = CB_1$. Since $IA_1 \\perp BC$, $IB_1 \\perp AC$, and $IC_1 \\perp AB$, the lines $IA_1$, $IB_1$, $IC_1$ are tangents to the hedgehogs at the corresponding vertices, so they do not touch any hedgehogs, as required.\n\n![](images/Ukraine2022-23_p10_data_c754136b7a.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19558,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a regular pentagon and let $c$ be the circle with diameter $AB$. Diagonals $AC$ and $AD$ intersect the circle $c$ at points $F$ and $G$, respectively. Line $FG$ intersects the side $AE$ at point $H$. Let $K$ be the midpoint of the side $DE$. Prove that points $F$, $H$, $E$, and $K$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "As $AB$ is a diameter of $c$, $\\angle AFB = 90^\\circ$ (see figure below), and $BF$ is an altitude of triangle $ABC$. Since $AB = BC$, $BF$ is also a median, so $F$ is the midpoint of $AC$. By symmetry, point $K$ lies on $BF$ and $\\angle FKE = 90^\\circ$. Notice that $\\angle BAC = \\angle CAD = \\angle DAE$, as $BAC$, $CAD$, and $DAE$ are inscribed angles subtending equal arcs of the circumcircle of the regular pentagon $ABCDE$. Points $A$, $B$, $F$, and $G$ lie on circle $c$ in this order, thus $\\angle ABF = 180^\\circ - \\angle AGF = \\angle AGH$. Hence, $ABF$ and $AGH$ are similar by two angles, implying $\\angle AHG = \\angle AFB = 90^\\circ$. Therefore, the opposite angles at $K$ and $H$ of quadrilateral $FHEK$ are right angles, so it is cyclic.\n\n![](images/prob1718_p18_data_ca813829ca.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19559,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that the sum of all positive divisors of $n$ equals $n + d(n) + 1$, where $d(n)$ denotes the number of positive divisors of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $n = 1$ is not a solution, so $d(n) \\ge 2$.\n\nIt is impossible that $d(n) = 2$, since then $n$ would be prime and the given equation would reduce to $1 + n = n + 2 + 1$, which is not possible. So $d(n) \\ge 3$.\n\nLet $1 = D_1 < D_2 < \\cdots < D_d = n$ be the divisors of $n$. The given equation can then be written as\n\n$$\n1 + \\sum_{i=2}^{d(n)-1} D_i + n = n + d(n) + 1.\n$$\n\nSince $D_i \\ge 2$ for all $i \\in \\{2, 3, \\dots, d(n) - 1\\}$, we get $d(n) = \\sum_{i=2}^{d(n)-1} D_i \\ge (d(n) - 2) \\cdot 2$, hence $d(n) \\le 4$, i.e. $3 \\le d(n) \\le 4$.\n\nFor $d(n) = 3$, the number $n$ is a square of some prime $p$ and the given equation reduces to\n\n$$\n1 + p + p^2 = p^2 + 3 + 1,\n$$\n\nhence $p = 3$ and $n = p^2 = 9$.\n\nFor $d(n) = 4$, there are two possibilities:\n\n- If $n$ is the product of two distinct primes $q$ and $r$ ($q < r$), then the given equation implies $1 + q + r + qr = qr + 4 + 1$, thus $q + r = 4$. There are no such primes $q$ and $r$.\n- If $n$ is a cube of some prime $s$, then the given equation means that $1 + s + s^2 + s^3 = s^3 + 4 + 1$, thus $s^2 + s - 4 = 0$. There is no such prime $s$.\n\nTherefore, the only solution is $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19560,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ such that\n$$\nx + \\frac{2014}{x} = \\lfloor x \\rfloor + \\frac{2014}{\\lfloor x \\rfloor}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $x = -\\frac{2014}{45}$.\n\nThe given equation may be rewritten as\n$$\n\\begin{aligned}\nx - \\lfloor x \\rfloor &= 2014 \\left( \\frac{1}{\\lfloor x \\rfloor} - \\frac{1}{x} \\right) \\\\\n&= \\frac{2014(x - \\lfloor x \\rfloor)}{x \\lfloor x \\rfloor}.\n\\end{aligned}\n$$\nSince $x$ is not an integer, it follows that $x \\neq \\lfloor x \\rfloor$. Hence we may divide both sides by $x - \\lfloor x \\rfloor$ and rearrange to find\n$$\nx \\lfloor x \\rfloor = 2014. \\tag{1}\n$$\nCase 1. $\\lfloor x \\rfloor \\ge 45$.\n\nThen $x > 45$, and so $x \\lfloor x \\rfloor > 45^2 = 2025 > 2014$.\n\nCase 2. $-44 \\le \\lfloor x \\rfloor \\le 44$.\n\nThen $-44 < x < 45$, and so $x \\lfloor x \\rfloor < 44 \\times 45 = 1980 < 2014$.\n\nCase 3. $\\lfloor x \\rfloor \\le -46$.\n\nThen $x < -45$, and so $x \\lfloor x \\rfloor > 45 \\times 46 = 2070 > 2014$.\n\nCase 4. $\\lfloor x \\rfloor = -45$.\n\nThen from (1) we derive $x = -\\frac{2014}{45} = -44\\frac{34}{45}$.\n\nChecking this in the original equation we have\n$$\n\\text{LHS} = x + \\frac{2014}{x} = -\\frac{2014}{45} + \\frac{2014}{-\\frac{2014}{45}} = -\\frac{2014}{45} - 45\n$$\nand\n$$\n\\text{RHS} = \\lfloor x \\rfloor + \\frac{2014}{\\lfloor x \\rfloor} = -45 + \\frac{2014}{-45} = \\text{LHS,}\n$$\nas required. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19561,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. The real numbers $a_1, a_2, \\dots, a_n$ and $r_1, r_2, \\dots, r_n$ are such that $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ and $0 \\leq r_1 \\leq r_2 \\leq \\dots \\leq r_n$.\n\nProve that\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nA_1 = \\begin{bmatrix}\n  a_1 a_1 r_1 & a_1 a_2 r_1 & a_1 a_3 r_1 & \\cdots & a_1 a_n r_1 \\\\\n  a_2 a_1 r_1 & a_2 a_2 r_2 & a_2 a_3 r_2 & \\cdots & a_2 a_n r_2 \\\\\n  a_3 a_1 r_1 & a_3 a_2 r_2 & a_3 a_3 r_3 & \\cdots & a_3 a_n r_3 \\\\\n  \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n  a_n a_1 r_1 & a_n a_2 r_2 & a_n a_3 r_3 & \\cdots & a_n a_n r_n\n\\end{bmatrix}\n$$\n\nSince\n\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) \\\\\n&= \\sum_{j=1}^{n} a_1 a_j \\min(r_1, r_j) + \\sum_{j=1}^{n} a_2 a_j \\min(r_2, r_j) + \\cdots \\\\\n&\\quad + \\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) + \\cdots + \\sum_{j=1}^{n} a_n a_j \\min(r_n, r_j)\n\\end{aligned}\n$$\n\nwhose $k$-th term is\n\n$$\n\\begin{aligned}\n& \\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) \\\\\n&= a_k a_1 r_1 + a_k a_2 r_2 + \\cdots + a_k a_k r_k + a_k a_{k+1} r_k + \\cdots + a_k a_n r_k,\n\\end{aligned}\n$$\n\ni.e., the sum of the elements from the $k$-th row of the matrix, $k=1, 2, \\dots, n$. Thus, $\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j)$ is the sum of all elements in $A_1$.\n\nOn the other hand, the sum can be calculated as follows. First, count the sum of elements from the first row and the first column of $A_1$, and let the other elements constitute a matrix $A_2$. Then count the sum of elements from the first row and the first column of $A_2$, and so on. So\n\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) \\\\\n&= \\sum_{k=1}^{n} r_k \\left(a_k^2 + 2a_k (a_{k+1} + a_{k+2} + \\dots + a_n)\\right) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left( a_k + \\sum_{i=k+1}^{n} a_i \\right)^2 - \\left( \\sum_{i=k+1}^{n} a_i \\right)^2 \\right) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left( \\sum_{i=k}^{n} a_i \\right)^2 - \\left( \\sum_{i=k+1}^{n} a_i \\right)^2 \\right) \\\\\n&= r_1 \\left( \\sum_{i=1}^{n} a_i \\right)^2 + r_2 \\left( \\sum_{i=2}^{n} a_i \\right)^2 + r_3 \\left( \\sum_{i=3}^{n} a_i \\right)^2 + \\dots + r_n \\left( \\sum_{i=n}^{n} a_i \\right)^2 \\\\\n&= \\sum_{k=1}^{n} (r_k - r_{k-1}) \\left( \\sum_{i=k}^{n} a_i \\right)^2 \\geq 0, \\quad \\text{where } r_0 = 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19562,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $S$ 為一個有限集合，且令 $\\mathcal{A}$ 為所有由 $S$ 映至 $S$ 之函數所成的集合。令 $f$ 是 $\\mathcal{A}$ 的一個元素且令 $T = f(S)$ 表示 $S$ 在 $f$ 之下所對應之值域。假設對每一個 $\\mathcal{A}$ 中的元素 $g$ 且 $g \\neq f$，皆滿足 $f \\circ g \\circ f \\neq g \\circ f \\circ g$。\n\n試證：$f(T) = T$。",
    "options": [],
    "answer": "See solution",
    "solution": "對於 $n \\ge 1$，令 $f^n$ 表示 $f$ 與自身複合 $n$ 次：\n\n$$\nf^n \\stackrel{\\text{def}}{=} f \\circ f \\circ \\dots \\circ f.\n$$\n\n由假設，若 $g \\in \\mathcal{A}$ 且 $f \\circ g \\circ f = g \\circ f \\circ g$，則 $g = f$。考慮令 $g = f^n$ 代入 $f \\circ g \\circ f = g \\circ f \\circ g$，可得 $f^n = f$，這將有助於證明題目。\n\n*Claim*: 若存在 $n \\ge 3$ 使得 $f^{n+2} = f^{2n+1}$，則 $f: T \\to T$ 的限制是雙射。\n\n*Proof*. 由假設：\n\n$$\nf^{n+2} = f^{2n+1} \\Leftrightarrow f \\circ f^n \\circ f = f^n \\circ f \\circ f^n \\Rightarrow f^n = f.\n$$\n\n因為 $n-2 \\ge 1$，$f^{n-2}$ 的像包含於 $T = f(S)$，因此 $f^{n-2}$ 可限制為 $f^{n-2}: T \\to T$。這是 $f: T \\to T$ 的反函數。事實上，對於 $t \\in T$，設 $t = f(s)$，$s \\in S$，則\n\n$$\nt = f(s) = f^n(s) = f^{n-2}(f(t)) = f(f^{n-2}(t)).\n$$\n\n即 $f^{n-2} \\circ f = f \\circ f^{n-2} = \\text{id on } T$。\n\n（此處 id 表示恆等函數。）因此，$f: T \\to T$ 是雙射，反函數為 $f^{n-2}: T \\to T$。\n\n接下來證明存在這樣的 $n$。定義\n\n$$\nS_m \\stackrel{\\text{def}}{=} f^m(S) \\quad (S_m \\text{ 為 } f^m \\text{ 的像})。\n$$\n\n顯然 $f^{m+1}$ 的像包含於 $f^m$ 的像，即有一遞減鏈：\n\n$$\nS \\supseteq S_1 \\supseteq S_2 \\supseteq S_3 \\supseteq \\dots\n$$\n\n因 $S$ 有限，必有 $k \\ge 1$ 使得\n\n$$\nS_k = S_{k+1} = S_{k+2} = \\dots \\stackrel{\\text{def}}{=} S_\\infty。\n$$\n\n因此 $f$ 限制為 $f: S_\\infty \\to S_\\infty$ 是滿射，且因 $S_\\infty$ 有限，也是雙射。故 $f: S_\\infty \\to S_\\infty$ 是 $S_\\infty$ 上的置換，存在 $r \\ge 1$ 使 $f^r = \\text{id}$ 在 $S_\\infty$ 上（例如 $r = |S_\\infty|$）。即：\n\n$$\nf^{m+r} = f^m \\text{ 在 } S，\\forall m \\ge k。\\qquad (1)\n$$\n\n由 (1) 式，$f^{m+tr} = f^m$ 對所有 $t \\ge 1$ 及 $m \\ge k$ 成立。為了找到合適的 $n$，只需選 $m, t$ 使得存在 $n \\ge 3$ 滿足：\n\n$$\n\\begin{cases}\n2n + 1 = m + tr \\\\\nn + 2 = m\n\\end{cases}\n\\Leftrightarrow\n\\begin{cases}\nm = 3 + tr \\\\\nn = m - 2\n\\end{cases}\n$$\n\n這可以通過選取足夠大的 $m$ 且 $m \\equiv 3 \\pmod r$ 實現。例如取 $n = 2kr + 1$，則\n\n$$\nf^{n+2} = f^{2kr+3} = f^{4kr+3} = f^{2n+1}\n$$\n\n其中中間等號由 (1) 式得出，因 $2kr + 3 \\ge k$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19563,
    "subject": "Mathematics (Olympiad)",
    "question": "Let three circles $\\odot O$, $\\odot O_1$, $\\odot O_2$, each externally tangent to the other two, lie on the same side of a line $\\ell$. Let them be tangent to $\\ell$ at points $A$, $A_1$, $A_2$ respectively, where point $A$ lies on segment $A_1A_2$. Denote the points of tangency of $\\odot O$ with $\\odot O_1$ and $\\odot O_2$ as $B_1$ and $B_2$ respectively, and the point of tangency of $\\odot O_1$ with $\\odot O_2$ as $C$. Let line $A_1C$ intersect line $A_2B_2$ at point $D_1$, and line $A_2C$ intersect line $A_1B_1$ at point $D_2$. Prove that line $D_1D_2$ is parallel to line $\\ell$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Consider the antipodal point $P$ of $A$ on the circle $\\odot O$.\n\nWe notice that in the right trapezoid $A_1A_2O_1O_2$, we have $O_1A_1 = O_1B_1$ and $OB_1 = OA$.\nTherefore,\n\n$$\n\\begin{align*}\n\\angle A_1BA &= \\pi - \\angle O_1B_1A_1 - \\angle OB_1A \\\\\n&= \\pi - \\frac{1}{2}(\\pi - \\angle A_1O_1O) - \\frac{1}{2}(\\pi - \\angle AOO_1) \\\\\n&= \\frac{1}{2}(\\angle A_1O_1O + \\angle AOO_1) = \\frac{\\pi}{2},\n\\end{align*}\n$$\n\nwhich implies that the line $A_1B_1$ passes through point $P$. Similarly, $A_2B_2$ also passes through point $P$.\n\nWe have,\n\n$$\n\\begin{align*}\n\\angle D_1B_2B_1 &= \\angle PB_2B_1 = \\angle PAB_1 \\quad (P, B_2, A, B_1 \\text{ concyclic}) \\\\\n&= \\angle B_1A_1A = \\angle B_1CA_1 \\quad (\\ell \\text{ is tangent to } \\odot O_1) \\\\\n&= \\angle B_1CD_1.\n\\end{align*}\n$$\n\nHence, points $C$, $D_1$, $B_1$, $B_2$ are concyclic. Similarly, $D_2$ also lies on the circumcircle of $\\triangle CB_1B_2$. Therefore,\n\n$$\n\\angle D_1D_2B_1 = \\angle D_1B_2B_1 = \\angle B_1A_1A.\n$$\n\nThis implies that $D_1D_2$ is parallel to $\\ell$. Proof completed. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19564,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ be the set of positive integers. Find all functions $g : \\mathbb{N} \\to \\mathbb{N}$ such that\n\n$$\n(g(m) + n)(m + g(n))\n$$\n\nis a perfect square for all $m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "All functions of the form $g(n) = n + c$ for a constant nonnegative integer $c$ satisfy the problem conditions. We claim that these are the only such functions.\n\nWe first show that $g$ must be injective. Suppose instead that $g(a) = g(b)$ for some $a \\neq b$. Choose $n$ so that $n + g(a) = p$ is prime and greater than $|a - b|$. By the given, both\n\n$$\np(g(n) + a) \\text{ and } p(g(n) + b)\n$$\n\nmust be perfect squares, meaning that $g(n) + a$ and $g(n) + b$ are both divisible by $p$. But this is impossible, as $p > |a - b|$. Therefore, $g$ is injective as claimed.\n\nWe now show that $|g(k + 1) - g(k)| = 1$ for all $k$. Suppose instead that some prime $p$ divides $g(k + 1) - g(k)$. Now, choose an integer $n$ as follows. If $p^2 \\mid g(k + 1) - g(k)$, then take $n$ so that $n + g(k + 1)$ is divisible by $p$ but not $p^2$. Otherwise, take $n$ so that $n + g(k + 1)$ is divisible by $p^3$ but not $p^4$. Note that the maximum power of $p$ dividing $n + g(k + 1)$ and $n + g(k)$ is odd. Now, the given implies that\n\n$$\n(n + g(k + 1))(g(n) + k + 1) \\text{ and } (n + g(k))(g(n) + k)\n$$\n\nare both squares, meaning that $g(n) + k + 1$ and $g(n) + k$ are both divisible by $p$, a contradiction.\n\nFor each $k$, we now have either $g(k + 1) = g(k) + 1$ or $g(k + 1) = g(k) - 1$. But $g$ is injective, so if the latter occurs for some $k$, then it occurs for all $k' > k$, an impossibility because $g$ takes positive values. Therefore, we have $g(k + 1) = g(k) + 1$ for all $k$, hence $g(k) = k + g(1) - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19565,
    "subject": "Mathematics (Olympiad)",
    "question": "At Pythagoras Lyceum, all classes have a maximum of 25 students. A survey is held in one of the classes about all sorts of things and everyone fills in all the questions. In this class, it turns out that 31\\% of the students have a black bicycle and 94\\% live more than 1 km away from the school. These percentages have been rounded to the closest integer.\n\nHow many students are in that class?",
    "options": [],
    "answer": "See solution",
    "solution": "16",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19566,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, 2, 3, \\ldots, 120\\}$. What is the largest possible size of a subset $M \\subseteq S$ such that no 8 elements of $M$ have greatest common divisor greater than 1?",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible value of $|M|$ is 53.\n\nNote that there are $\\frac{120}{2} = 60$ multiples of 2 and $\\frac{120}{3} = 40$ multiples of 3 in $S$. Therefore, we need to remove at least 53 multiples of 2 and 33 multiples of 3 to ensure that no 8 numbers remaining have greatest common divisor divisible by 2 or 3. Among these removed numbers, at most $\\frac{120}{6} = 20$ of them are repeated, which happens when they are multiples of 6. Also, we have to remove at least one number from $\\{5, 25, 35, 55, 65, 85, 95, 115\\}$ since all of them are multiples of 5 and none of them is a multiple of 2 or 3. Thus, at least\n\n$$\n53 + 33 - 20 + 1 = 67\n$$\n\nnumbers should be removed. So we have $|M| \\le 120 - 67 = 53$.\n\nWe now give an example of $M$ having 53 elements such that no 8 numbers have greatest common divisor larger than 1. Suppose $M$ contains\n\n$2, 4, 8, 16, 32, 64, 118, 3, 9, 27, 81, 87, 93, 111$\n\nand all other numbers in $S$ which are not multiples of 2 or 3 except 35. By the inclusion-exclusion principle, there are\n\n$$\n7 + 7 + 120 - \\frac{120}{2} - \\frac{120}{3} + \\frac{120}{6} - 1 = 53\n$$\n\nnumbers in total. Since $17 \\times 8 > 120$, it suffices to check we cannot find 8 numbers which are simultaneously multiples of 2, 3, 5, 7, 11, or 13. This can be seen by listing out the multiples of these primes in $M$ as follows.\n\n$$\n\\begin{array}{l}\n2 : 2, 4, 8, 16, 32, 64, 118 \\\\\n3 : 3, 9, 27, 81, 87, 93, 111 \\\\\n5 : 5, 25, 55, 65, 85, 95, 115 \\\\\n7 : 7, 49, 77, 91, 119 \\\\\n11 : 11, 55, 77 \\\\\n13 : 13, 65, 91\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19567,
    "subject": "Mathematics (Olympiad)",
    "question": "The six-digit number $\\underline{20210}\\underline{A}$ is prime for only one digit $A$. What is $A$?\n\n1. $1$\n2. $3$\n3. $5$\n4. $7$\n5. $9$",
    "options": [],
    "answer": "See solution",
    "solution": "A number whose units digit is $5$ is divisible by $5$, so choice $3$ does not give a prime.\n\nChoices $1$ and $7$ can be ruled out because a number is divisible by $3$ if and only if the sum of its digits is divisible by $3$; here $2 + 0 + 2 + 1 + 0 + 1 = 6$ and $2 + 0 + 2 + 1 + 0 + 7 = 12$ are divisible by $3$.\n\nChoice $3$ can be ruled out because a number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$; here $2 - 0 + 2 - 1 + 0 - 3 = 0$ is divisible by $11$.\n\nThe five even choices for $A$ result in a number divisible by $2$.\n\nGiven that one choice produces a prime, $A$ must equal $9$ and the number $202109$ must be prime. (This can be verified with computer algebra software.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19568,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the points $O = (0, 0)$, $A = (-2, 0)$, and $B = (0, 2)$ in the coordinate plane. Let $E$ and $F$ be the midpoints of $OA$ and $OB$, respectively. Rotate triangle $OEF$ clockwise about $O$ to obtain a triangle $OE'F'$. For each rotated position, let $P = (x, y)$ be the intersection of lines $AE'$ and $BF'$. Find the maximum value of the $y$-coordinate of $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Argentina_2015_Booklet_p9_data_276906f9e1.png)\n\nLet $R$ be the clockwise $90^\\circ$ rotation about $O$. Apparently, $R$ takes $A$ to $B$ and also $R(E') = F'$ for each rotated position $OE'F'$ of the initial right isosceles triangle $OEF$. Hence, $R$ takes line $AE'$ to line $BF'$. The angle between a line and its image under any rotation equals the angle of rotation, so $AE'$ and $BF'$ are perpendicular. In other words, $\\angle APB = 90^\\circ$, meaning that $P$ lies on the circle with diameter $AB$, i.e., on the circle $\\alpha$ with center $(-1, 1)$ and radius $\\sqrt{2}$.\n\nNot every point $P \\in \\alpha$ can be obtained as the intersection of lines $AE'$ and $BF'$ for some rotated position $OE'F'$ of $OEF$. A necessary condition is that line $AP$ contains a point at distance $1$ from the origin, point $E'$. Equivalently, $AP$ must have a common point with the circle $\\beta$ centered at $(0, 0)$ and of radius $1$.\n\nLet $AT$ and $AT'$ be the tangents from $A$ to $\\beta$, with $T$ in quadrant 2, $T'$ in quadrant 3. Then each admissible line $AP$ intersects the interior of $\\angle TAT'$ or coincides with one of $AT$ and $AT'$. Let $AT \\cap \\alpha = P_0$, $AT' \\cap \\alpha = P'_0$. Then all admissible positions of $P$ are contained in the closer minor arc $\\widehat{P_0P'_0} = \\gamma$ of circle $\\alpha$ (the arc not containing $A$ and $B$). Note that $P_0$ is in quadrant 1. The entire $\\gamma$ is under the line through $P_0$ parallel to the $x$-axis. Hence, the $y$-coordinate of an admissible point $P$ does not exceed the $y$-coordinate $y_0$ of $P_0$. In fact, $y_0$ is the desired maximum value because $P_0$ is admissible. Indeed, let $BU$ be the tangent to $\\beta$ from $B$, with $U$ in quadrant 1. Rotations preserve tangency, so, given $R(A) = B$, rotation $R$ takes tangent $AT$ to tangent $BU$. This yields $\\angle TOU = 90^\\circ$ on the one hand, and $AT \\perp BU$ on the other. The latter means that $AT$ and $BU$ intersect on $\\alpha$, and since $P_0$ is defined by $AT \\cap \\alpha = P_0$, we find $AT \\cap BU = P_0$. Hence $P_0$ is admissible, with $E' = T$, $F' = U$. (It follows from the computation below that $P$ is obtained through a $60^\\circ$-clockwise rotation of $OEF$ about the origin.)\n\nIt remains to evaluate $y_0$, i.e., the length of the perpendicular $P_0H$ from $P_0$ to the $x$-axis. Triangle $OAT$ is right at $T$ with $OA = 2$, $OT = 1$, therefore $\\angle OAT = 30^\\circ$. Hence, the right triangle $AP_0H$ yields $y_0 = P_0H = \\frac{1}{2}AP_0$. Triangle $ABP_0$ is right at $P_0$ with $\\angle BAP_0 = 15^\\circ$. One expression for $\\cos 15^\\circ$ is $\\cos 15^\\circ = \\frac{1}{4}(\\sqrt{2} + \\sqrt{6})$. Replacing in $y_0 = \\frac{1}{2}AP_0 = \\frac{1}{2}AB \\cos 15^\\circ$ leads to the answer:\n\n$$\ny_{\\max} = y_0 = \\frac{1}{2}(1 + \\sqrt{3})\n$$\n\n**Remark.** Using $\\cos 15^\\circ$ can be avoided by applying the following elementary fact: the hypotenuse of a $15^\\circ$-$75^\\circ$-$90^\\circ$ triangle is $4$ times greater than its respective altitude.\n\nLet the triangle $ABC$ with $\\angle C = 90^\\circ$, $\\angle B = 15^\\circ$ and altitude $CH = h$. Take the midpoint $M$ of $AB$. It is known that $MA = MB = MC$, so $\\angle CMH = \\angle MBC + \\angle MCB = 30^\\circ$. Thus triangle $MCH$ is $30^\\circ$-$60^\\circ$-$90^\\circ$, hence $MC = 2CH = 2h$ and $AB = 2MC = 4h$.\n\nThen the computation of $y_0$ can go as follows. Set $AP_0 = a$, $BP_0 = b$, $a > b$. The altitude from $P_0$ to $AB$ in triangle $ABP_0$ equals $h = \\frac{AB}{4} = \\frac{2\\sqrt{2}}{4} = \\frac{\\sqrt{2}}{2}$, by the above fact. Hence $ab = AB \\cdot h = 2\\text{area}(ABP_0) = 2\\sqrt{2} \\cdot \\frac{\\sqrt{2}}{2} = 2$. Since\n$$\na^2 + b^2 = (2\\sqrt{2})^2 = 8, \\text{ we obtain } (a \\pm b)^2 = 8 \\pm 4. \\text{ Therefore}\n$$\n$$\na + b = 2\\sqrt{3}, \\quad a - b = 2 \\quad \\text{and so} \\quad a = 1 + \\sqrt{3}, \\quad y_0 = \\frac{1}{2}a = \\frac{1}{2}(1 + \\sqrt{3})\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19569,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $P$ a point inside the triangle such that the centers $M_B$ and $M_A$ of the circumcircles $k_B$ and $k_A$ of triangles $ACP$ and $BCP$, respectively, lie outside the triangle $ABC$. In addition, we assume that the three points $A$, $P$ and $M_A$ are collinear as well as the three points $B$, $P$ and $M_B$. The line through $P$ parallel to side $AB$ intersects circles $k_A$ and $k_B$ in points $D$ and $E$, respectively, where $D, E \\ne P$.\n\nShow that $DE = AC + BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We put $\\varphi := \\angle CBP$, cf. Figure 1. Then we get for the corresponding central angle $\\angle CM_A P = 2\\varphi$.\n\n![](images/bwf2018englishSolutions_p2_data_176b41d1aa.png)\n\nFigure 1: Problem 4\n\nSince $M_A C M_B P$ is a deltoid having $M_A M_B$ as its axis of symmetry, we deduce $\\angle C M_A M_B = \\varphi = \\angle C B M_B$. Therefore, $B$ and analogously $A$ lie on the circumcircle of $M_A C M_B$. In other words, the two centers $M_A$ and $M_B$ lie on the circumcircle of $ABC$.\n\nThus $M_A$ and $M_B$ are the south poles corresponding to vertices $A$ and $B$, respectively. As a result, $P$ is the incenter of triangle $ABC$. Hence $\\angle PBA = \\angle CBP$ and because of $PD \\parallel AB$ also $\\angle CBP = \\angle BPD$ holds true. This means that $PBDC$ is an isosceles trapezoid with diagonals of equal lengths and $PD = BC$ follows.\n\nIn a similar way $PE = AC$ can be shown. Finally, by addition we arrive at the claim $DE = AC+BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19570,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $AB < CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$, and $T$ be the midpoints of $EF$, $CF$, and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $CD$. We will prove that the circumcircles of the triangles $MKT$ and $MLS$ pass through $N$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p25_data_9bf8557dbb.png)\n\nFirst, we will prove that the circumcircle of $MLS$ passes through $N$.\n\nLet $Q$ be the midpoint of $EC$. Note that the circumcircle of $MLS$ is the Euler circle of triangle $EFC$, so it also passes through $Q$.\n\nWe will prove that\n\n$$\n\\angle SLQ = \\angle QNS \\quad \\text{or}\n$$\n\n$$\n\\angle SLQ + \\angle QNS = 180^{\\circ}.\n$$\n\nIndeed, since $FLC$ is right-angled and $LS$ is its median, we have $SL = SC$ and\n\n$$\n\\angle SLC = \\angle SCL = \\angle ACB.\n$$\n\nIn addition, since $N$ and $S$ are the midpoints of $DC$ and $FC$, we have $SN \\parallel FD$, and similarly, since $Q$ and $N$ are the midpoints of $EC$ and $CD$, $QN \\parallel ED$.\n\nIt follows that the angles $EDB$ and $QNS$ have parallel sides, and since $AB < CD$ they are acute, so\n\n$$\n\\angle EDB = \\angle QNS \\quad \\text{or} \\quad \\angle EDB + \\angle QNS = 180^{\\circ}.\n$$\n\nBut, from the cyclic quadrilateral $ABCD$, we get\n\n$$\n\\angle EDB = \\angle ACB.\n$$\n\nNow, from the above, we obtain that the quadrilateral $LNSQ$ is cyclic. Since its circumcircle also passes through $M$, the points $M, L, Q, S, N$ are cocyclic, so the circumcircle of $MLS$ passes through $N$.\n\nSimilarly, the circumcircle of $MKT$ also passes through $N$, and we have the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19571,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle *ABC*, angle *A* is a right angle. A point *D* lies on line segment *AB* in such a way that the angles *ACD* and *BCD* are equal. Moreover, $|AD| = 2$ and $|BD| = 3$.\n\n![](images/NLD_ABooklet_2023_p9_data_b6a556a00e.png)\n\nWhat is the length of line segment *CD*?",
    "options": [],
    "answer": "See solution",
    "solution": "$2\\sqrt{6}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19572,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Points $P_1, P_2, \\dots, P_{4n}$ are placed in a plane in such a way that no three points among them lie on any straight line. Furthermore, for each $i = 1, 2, \\dots, 4n$, if we rotate the half-line $P_i P_{i-1}$ starting at $P_i$ around the point $P_i$ by $90^\\circ$ clockwise, then the half-line falls onto the half-line $P_i P_{i+1}$ starting at $P_i$. Determine the maximum possible number of pairs $(i, j)$ for which the line segments $P_i P_{i+1}$ and $P_j P_{j+1}$ intersect at a point different from the endpoints of the line segments. Here we let $P_0 = P_{4n}$, $P_{4n+1} = P_1$ and assume that $1 \\le i < j \\le 4n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let, for $k = 1, 2, \\dots, n$, $A_k = P_{4k-3}$, $B_k = P_{4k-2}$, $C_k = P_{4k-1}$, $D_k = P_{4k}$. Also, let $A_{n+1} = A_1$ and $B_{n+1} = B_1$. From now on, let us say that the directed line segments $A_i B_i$, $B_i C_i$, $C_i D_i$, $D_i A_{i+1}$ are leftward, downward, rightward, upward segments, respectively. Furthermore, let us say a bent line segment $A_i B_i C_i$ is leftdown type for each $i = 1, 2, \\dots, n$, and define similarly: downright, rightup, upleft type bent segments.\n\nThen the point of intersection (different from their endpoints) of a leftward segment and a downward segment can be regarded as the intersection of two leftdown type bent segments. The same statement can be made for intersections of other types of directed segments. Therefore, to get the answer to the problem, it suffices to find the maximum possible value for the sum of the number of intersections of two leftdown type bent segments, that of two downright bent segments, that of two rightup bent segments, and that of two upleft bent segments.\n\nLet us say the pair $(i, j)$ where $1 \\le i \\ne j \\le n$ is a good pair if for all of the four pairs of bent segments $A_i B_i C_i$ and $A_j B_j C_j$, $B_i C_i D_i$ and $B_j C_j D_j$, $C_i D_i A_{i+1}$ and $C_j D_j A_{j+1}$, $D_i A_{i+1} B_i$ and $D_j A_{j+1} B_j$, two bent segments intersect each other.\n\nWe note that if $(i, j)$ is a good pair and if $A_i$ lies above $A_j$, then $B_i$ lies to the right of $B_j$, $C_i$ lies below $C_j$, $D_i$ lies to the left of $D_j$, and $A_{i+1}$ lies below $A_{j+1}$.\n\n*Lemma.* For any non-empty proper subset $X$ of the set $\\{1, 2, \\dots, n\\}$, let $Y = X^c$, the complement of $X$. Then, there exist $x \\in X$ and $y \\in Y$ such that the pair $(x, y)$ is not a good pair.\n\n**Proof.** For $x \\in \\{1, 2, \\dots, n\\}$, let us define\n\n$$\nx^+ = \\begin{cases} x+1 & \\text{if } 1 \\le x \\le n-1 \\\\ 1 & \\text{if } x=n \\end{cases}, \\qquad x^- = \\begin{cases} x-1 & \\text{if } 2 \\le x \\le n \\\\ n & \\text{if } x=1 \\end{cases}.\n$$\n\nNow suppose for every choice of $x \\in X$ and $y \\in Y$ the pair $(x, y)$ is a good pair. Define $f(k) = i$ and $f^{-1}(i) = k$ if $A_k$ is located at the $i$-th position from the top among $A_1, A_2, \\dots, A_n$. We show that for each $k$, $1 \\le k \\le n$, the number of points among $f(1), f(2), \\dots, f(k)$ which belong to $X$ and the number of points among $f(1)^-, f(2)^-, \\dots, f(k)^-$ which belong to $X$ must coincide.\n\nSuppose the number for the former is larger than the number for the latter. Then, there exists $x \\in X$ for which both $f^{-1}(x) \\le k$ and $f^{-1}(x^+) > k$ hold. Furthermore, there must exist $y \\in Y$ which satisfies both $f^{-1}(y) > k$ and $f^{-1}(y^+) \\le k$, since the number of points in the set $\\{f(k+1), f(k+2), \\dots, f(n)\\}$ which belong to $Y$ is more than the number of points in the set $\\{f(k+1)^-, f(k+2)^-, \\dots, f(n)^-\\}$ which belong to $Y$. But this implies that the pair $(x, y)$ is not a good pair, contradicting our assumption. Similarly, we arrive at a contradiction also if we assume that the number for the former case is smaller than the number for the latter. Therefore, we conclude that the number for the former equals the number for the latter.\n\nBy comparing this fact for the case $k = \\ell$ and for the case $k = \\ell - 1$, we arrive at the conclusion that\n\n$$\nf(\\ell) \\in X \\iff f(\\ell)^- \\in X.\n$$\n\nThis means that for any $x \\in \\{1, 2, \\dots, n\\}$ we have\n\n$$\nx \\in X \\iff x^- \\in X.\n$$\n\nBut this contradicts our assumption that $X$ is a non-empty proper subset of $\\{1, 2, \\dots, n\\}$, and this proves the Lemma.\n\nNow, in order to arrive at the answer to the problem, suppose that the number of $(x, y)$, which is not a good pair is at most $n-2$. Then, among the elements of $\\{1, 2, \\dots, n\\}$, the number of those which can be reached from the element $1$ by going through the string of not-good pairs can be at most $n-1$. So, let $X$ be the subset consisting of those elements accessible from $1$ in this way and let $Y$ be the complement of $X$. Then, we arrive at a situation contradicting the conclusion of the Lemma. So, we must have at least $n-1$ not-good pairs among $\\{1, 2, \\dots, n\\}$. Therefore, the maximum number of pairs $(i, j)$ satisfying the requirement of the problem is at most $4 \\times \\binom{n}{2} - (n-1) = (2n-1)(n-1)$.\n\nOn the other hand, as indicated in the diagram below, if we start by placing $A_1, A_2, \\dots, A_n$ in turn with $A_1$ at a left-top position and going down-right direction, and placing $C_n, C_{n-1}, \\dots, C_1$ in turn with $C_n$ at a left-top position and going down-right direction, and finally placing $B_1, B_2, \\dots, B_n$ and $D_1, D_2, \\dots, D_n$ by following the rule specified in the problem, we can arrive at a situation where the number of relevant intersection points is exactly $(2n-1)(n-1)$. Therefore, the answer we seek for the problem is $(2n-1)(n-1)$.\n\n![](images/Japan_booklet_2013_p12_data_4768bac61a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19573,
    "subject": "Mathematics (Olympiad)",
    "question": "В таблице размером $50 \\times 50$ вдоль левой стороны выписаны 50 различных положительных чисел, и вдоль верхней стороны — также 50 различных положительных чисел. В каждой клетке таблицы записано произведение соответствующих чисел из левого столбца и верхней строки. Какое наибольшее количество произведений в таблице может быть рациональными числами?",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть вдоль левой стороны таблицы выписано $x$ иррациональных и $50-x$ рациональных чисел, а вдоль верхней стороны — $50-x$ иррациональных и $x$ рациональных чисел. Произведение ненулевого рационального и иррационального числа всегда иррационально, поэтому в таблице будет хотя бы $x^2 + (50-x)^2$ иррациональных чисел. \n\n$$x^2 + (50-x)^2 = 2x^2 - 100x + 2500 = 2(x-25)^2 + 2 \\cdot 25^2 \\geq 2 \\cdot 25^2 = 1250$$\n\nСледовательно, в таблице не более $2500 - 1250 = 1250$ рациональных чисел.\n\nРовно 1250 рациональных чисел можно получить, если вдоль левой стороны стоят числа $1, 2, \\dots, 25, \\sqrt{2}, 2\\sqrt{2}, \\dots, 25\\sqrt{2}$, а вдоль верхней стороны — числа $26, 27, \\dots, 50, 26\\sqrt{2}, 27\\sqrt{2}, \\dots, 50\\sqrt{2}$. Тогда иррациональными будут только $2 \\cdot 25^2 = 1250$ произведений рационального и иррационального чисел.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19574,
    "subject": "Mathematics (Olympiad)",
    "question": "$A = \\{a, b, c\\}$ is a set containing three positive integers. Prove that we can find a set $B \\subset A$, $B = \\{x, y\\}$ such that for all odd positive integers $m, n$,\n$$\n10 \\mid x^m y^n - x^n y^m\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x, y) = x^m y^n - x^n y^m$. If $n = m$, the statement holds for any choice of $B$, so assume $n > m$.\n\nSince $m$ and $n$ are both odd, $n - m$ is even, so\n$$\n\\begin{align*}\nf(x, y) &= x^m y^m (y^{n-m} - x^{n-m}) \\\\\nf(x, y) &= x^m y^m (y^2 - x^2) Q(x, y) \\\\\nf(x, y) &= x^m y^m (y - x)(y + x) Q(x, y)\n\\end{align*}\n$$\nwhere $Q(x, y) = y^{n-m-2} + y^{n-m-4}x + \\dots + x^{n-m-2}$.\n\nIf one of $x, y$ is even, $f(x, y)$ is even. If both are odd, then $x + y$ and $x - y$ are even, so $f(x, y)$ is even. Thus, we only need to consider divisibility by $5$.\n\nIf $A$ contains an element divisible by $5$, include it in $B$; then $f(x, y)$ is divisible by $5$.\n\nIf none of the elements in $A$ is divisible by $5$, and two elements have the same remainder modulo $5$, choose them so $x - y$ is divisible by $5$.\n\nIf all remainders modulo $5$ in $A$ are different, among the pairs $(1, 4)$ and $(2, 3)$, one must be present by the pigeonhole principle. Picking such a pair, $x + y$ is divisible by $5$, so $f(x, y)$ is divisible by $5$.\n\nThus, in all cases, $10 \\mid x^m y^n - x^n y^m$ for some $B \\subset A$ with $|B| = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19575,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, \\dots, 100\\}$, and for every positive integer $n$ define\n$$\nT_n = \\{(a_1, \\dots, a_n) \\in S^n \\mid a_1 + \\dots + a_n \\equiv 0 \\pmod{100}\\}.\n$$\nDetermine which $n$ have the following property: if we color any 75 elements of $S$ red, then at least half of the $n$-tuples in $T_n$ have an even number of coordinates with red elements.",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution by generating functions**\n\nDefine\n$$\nR(x) = \\sum_{s \\text{ red}} x^s, \\quad B(x) = \\sum_{s \\text{ blue}} x^s.\n$$\n(Here “blue” means “not-red”.)\n\nThe number of tuples in $T_n$ with exactly $k$ red coordinates is\n$$\n\\binom{n}{k} \\cdot \\frac{1}{100} \\sum_{\\omega} R(\\omega)^k B(\\omega)^{n-k}\n$$\nwhere the sum is over all primitive 100th roots of unity.\n\nThus, the number of tuples in $T_n$ with an even (resp. odd) number of red elements is\n$$\n\\begin{align*}\nX &= \\frac{1}{100} \\sum_{\\omega} \\sum_{k \\text{ even}} \\binom{n}{k} R(\\omega)^k B(\\omega)^{n-k} \\\\\nY &= \\frac{1}{100} \\sum_{\\omega} \\sum_{k \\text{ odd}} \\binom{n}{k} R(\\omega)^k B(\\omega)^{n-k} \\\\\nX - Y &= \\frac{1}{100} \\sum_{\\omega} (B(\\omega) - R(\\omega))^n \\\\\n&= \\frac{1}{100} \\left[ (B(1) - R(1))^n + \\sum_{\\omega \\neq 1} (2B(\\omega))^n \\right] \\\\\n&= \\frac{1}{100} \\left[ (B(1) - R(1))^n - (2B(1))^n + 2^n \\sum_{\\omega} B(\\omega)^n \\right] \\\\\n&= \\frac{1}{100} \\left[ (B(1) - R(1))^n - (2B(1))^n \\right] + 2^n Z \\\\\n&= \\frac{1}{100} [(-50)^n - 50^n] + 2^n Z.\n\\end{align*}\n$$\nwhere\n$$\nZ = \\frac{1}{100} \\sum_{\\omega} B(\\omega)^n \\geq 0\n$$\ncounts the number of tuples in $T_n$ which are all blue. Here $B(\\omega) + R(\\omega) = 0$ for $\\omega \\neq 1$.\n\nWe wish to show $X - Y \\geq 0$ holds for $n$ even, but may fail when $n$ is odd. This follows from:\n\n* If $n$ is even, then $X - Y = 2^n Z \\geq 0$.\n* If $n$ is odd, then for a coloring where $s$ is red iff $s \\neq 2 \\pmod 4$, we get $Z = 0$. Then $X - Y = -\\frac{2}{100} \\cdot 50^n < 0$.\n\nThus, the property holds exactly for even $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19576,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{x_n\\}$ be a sequence such that $x_1 \\in \\{5, 7\\}$, and $x_{n+1} \\in \\{5^{x_n}, 7^{x_n}\\}$ for $n = 1, 2, \\dots$. Determine all possible cases of the last two digits of $x_{2009}$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2009$. Consider the following cases:\n\n1. If $x_n = 7^{5^{x_{n-2}}}$, then $x_n \\equiv 7 \\pmod{100}$.\n\nSince both 5 and 7 are odd, all $x_k$ ($1 \\leq k \\leq n$) are odd. $5^{x_{n-2}} \\equiv 1 \\pmod{4}$, i.e., $5^{x_{n-2}} = 4k+1$ for some integer $k$.\n\nBy induction,\n$$\n7^{4k+m} \\equiv 7^m \\pmod{100},\n$$\nso\n$$\nx_n = 7^{5^{x_{n-2}}} = 7^{4k+1} \\equiv 7 \\pmod{100}.\n$$\n\n2. If $x_n = 7^{7^{x_{n-2}}}$, then $x_n \\equiv 43 \\pmod{100}$.\n\n$$\n7^{x_{n-2}} \\equiv (-1)^{x_{n-2}} \\equiv -1 \\equiv 3 \\pmod{4},\n$$\nso $7^{x_{n-2}} = 4k+3$ for some integer $k$. Thus,\n$$\nx_n = 7^{7^{x_{n-2}}} = 7^{4k+3} \\equiv 7^3 \\equiv 43 \\pmod{100}.\n$$\n\n3. If $x_n = 5^{x_{n-1}}$, then $x_n \\equiv 25 \\pmod{100}$.\n\nBy induction, for $n \\geq 2$, $5^n \\equiv 25 \\pmod{100}$. Since $x_{n-1} > 2$, $x_n = 5^{x_{n-1}} \\equiv 25 \\pmod{100}$.\n\nTherefore, the possible last two digits of $x_{2009}$ are 07, 25, and 43.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19577,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = 2011$. Consider the sequence defined by the recurrence relation:\n\n$$\nu_0 = 0, \\quad u_1 = 1, \\quad u_2 = x, \\quad u_n = x u_{n-1} - u_{n-2} \\text{ for } n \\geq 3.$$ \n\nFind all prime numbers in the sequence $(u_n)_{n \\geq 0}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\n&= x(2u_{n+1}u_{n+2} - xu_{n+1}^2) - (u_{n+1}^2 - u_n^2) \\\\\n&= -x^2u_{n+1}^2 + x(2u_{n+1}(xu_{n+1} - u_n) + u_n^2 - u_{n+1}^2) \\\\\n&= x^2u_{n+1}^2 - 2xu_{n+1}u_n + u_n^2 - u_{n+1}^2 \\\\\n&= (xu_{n+1} - u_n)^2 - u_{n+1}^2 \\\\\n&= u_{n+2}^2 - u_{n+1}^2.\n\\end{align*}\n$$\n\nThe result thus follows by induction.\n\nThe above identities suggest that we may be able to find explicit factorizations, but for a factorization to prove that a number is not prime it is necessary for the factors to be greater than 1. So let us show that $u_i - u_{i-1} \\geq 2$ for all $i \\geq 2$. For $i = 2$ the result holds by inspection. Suppose it holds for some $i$. Then $u_{i+1} - u_i = 2011u_i - u_{i-1} - u_i = 2010u_i - u_{i-1} \\geq u_i - u_i - 1 \\geq 2$. The result again follows by induction.\n\nInspecting the first few terms, we see that $u_2 = 2011$ is prime. Let us show that there are no other primes. $u_0$ and $u_1$ are non-prime by inspection. For $n \\geq 1$, $u_{2n+1} = u_{n+1}^2 - u_n^2 = (u_{n+1} - u_n)(u_{n+1} + u_n)$. By the inequality above, both factors are at least 2 and so $u_{2n+1}$ is not prime. For $n > 1$, $u_{2n} = u_n(2u_{n+1} - 2011u_n)$. The first term is clearly greater than 1, and the second equals $u_{n+1} - u_{n-1} \\geq u_{n+1} - u_n \\geq 2$.\n\nThus, $u_2 = 2011$ is the only prime in the sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19578,
    "subject": "Mathematics (Olympiad)",
    "question": "A family of sets $\\mathcal{F}$ is called *perfect* if the following condition holds: For every triple of sets $X_1, X_2, X_3 \\in \\mathcal{F}$, at least one of the sets\n\n$$\n(X_1 \\setminus X_2) \\cap X_3, \\quad (X_2 \\setminus X_1) \\cap X_3\n$$\n\nis empty.\n\nShow that if $\\mathcal{F}$ is a perfect family consisting of some subsets of a given finite set $U$, then $|\\mathcal{F}| \\le |U| + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction with respect to $|U|$. If $|U| = 0$, that is, $U = \\emptyset$, then there exists only one subset of $U$ and clearly $|\\mathcal{F}| \\le 1$.\n\nSuppose the statement is true for all sets of cardinality less than $k$ for a given $k > 0$. Let $U$ be any set with $|U| = k$ and $\\mathcal{F}$ be a perfect family of its subsets. We shall show that $|\\mathcal{F}| \\le |U| + 1$.\n\nIf $|\\mathcal{F}| \\le 1$, the claim is obviously true. If $|\\mathcal{F}| \\ge 2$, consider all the pairs of distinct sets from $\\mathcal{F}$. Since the number of such pairs is finite and nonzero, there is a pair $(Y, Z) \\in \\mathcal{F}^2$, $Y \\ne Z$, with intersection of maximum cardinality, that is, $|Y \\cap Z| = m$ and the intersection of any two distinct sets from $\\mathcal{F}$ has at most $m$ elements.\n\nSince the sets $Y$ and $Z$ are distinct, at least one of them must contain an element which is not contained in the other. Without loss of generality, let $Y \\setminus Z$ be nonempty and take any element $y \\in Y \\setminus Z$.\n\nIf $Y$ is the only set containing $y$, all the sets from the system $\\mathcal{F}' = \\mathcal{F} \\setminus \\{Y\\}$ are subsets of $U' = U \\setminus \\{y\\}$. Clearly $\\mathcal{F}'$, as a subsystem of a perfect system, is perfect as well. Applying the induction hypothesis on $U'$ and $\\mathcal{F}'$, we get\n\n$$\n|\\mathcal{F}| = |\\mathcal{F}'| + 1 \\le (|U'| + 1) + 1 = |U| + 1,\n$$\n\nand we are done.\n\n![](images/SVK_Other_2015_p2_data_9ad2c9f1fd.png)\n\nIn the other case, there is at least one set $W \\in \\mathcal{F}$ with $y \\in W$, $W \\ne Y$. Due to the choice of the pair $(Y, Z)$, the set $W$ cannot contain the whole intersection $Y \\cap Z$ (otherwise we would have $|Y \\cap W| \\ge m + 1$). Let $z \\in (Y \\cap Z) \\setminus W$. We have\n\n$$\ny \\in (W \\setminus Z) \\cap Y \\quad \\text{and} \\quad z \\in (Z \\setminus W) \\cap Y,\n$$\n\nwhich is in contradiction with the property of the perfect system for $X_1 = Z, X_2 = W$, and $X_3 = Y$. So this case is not possible and the induction step is concluded.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19579,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $n$ lines in the plane, no two of which are parallel and no three of which are concurrent. Let $G$ be the graph whose vertices are the intersection points of these lines, and whose edges are the segments of the lines between consecutive intersection points (i.e., segments whose endpoints are intersection points and which contain no other intersection point in their interior).\n\nProve that there are at least 3 vertices of $G$ with degree 2.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be as described. Each pair of lines intersects at exactly one point, so there are $\\frac{n(n-1)}{2}$ vertices. Each line contains $n-1$ intersection points, so it contributes $n-2$ edges (segments between consecutive points), for a total of $n(n-2)$ edges.\n\nLet $a$, $b$, and $c$ be the number of vertices of degree 2, 3, and 4, respectively. Then:\n\n$$\na + b + c = \\frac{n(n-1)}{2}, \\quad 2a + 3b + 4c = 2n(n-2).\n$$\n\nSubtracting three times the first equation from the second gives:\n\n$$\nc - a = \\frac{n^2 - 5n}{2} \\implies c = (a - 3) + \\frac{(n - 2)(n - 3)}{2}.\n$$\n\nThus, to prove $a \\geq 3$, consider the convex hull of the intersection points. The vertices on the convex hull must have degree 2, and there are at least 3 such points (since the convex hull of a finite set in the plane has at least 3 vertices). Therefore, there are at least 3 vertices of degree 2.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19580,
    "subject": "Mathematics (Olympiad)",
    "question": "Richard and Kaarel take turns choosing numbers from the set $\\{1, \\dots, p-1\\}$, where $p > 3$ is a prime. Richard goes first. Once a number is chosen, it cannot be chosen again. Each number chosen by Richard is multiplied by the next number chosen by Kaarel. Kaarel wins if, at any point, the sum of all such products is divisible by $p$. Richard wins if all numbers are chosen and this never happens. Can either player guarantee victory regardless of the other's moves, and if so, which one?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, Kaarel can guarantee victory.\n\nPair the numbers as $(1, p-1)$, $(2, p-2)$, $\\dots$, $\\left(\\frac{p-1}{2}, \\frac{p+1}{2}\\right)$. Whenever Richard picks $a$, Kaarel picks $p-a$ from the same pair. This ensures Kaarel always has a valid response. The products are $a \\cdot (p-a) \\equiv -a^2 \\pmod{p}$. The sum of all products is $-(1^2 + 2^2 + \\dots + (\\frac{p-1}{2})^2)$. Using the formula $1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}{6}$ with $n = \\frac{p-1}{2}$, the sum is $\\frac{(p-1)p(p+1)}{24}$, which is divisible by $p$ since $p$ and $24$ are coprime. Thus, when all numbers are chosen, the sum is divisible by $p$, so Kaarel can always win.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19581,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be $k$ colors, numbered $1, 2, \\ldots, k$, with two blocks of each color. What is the probability that, in a random arrangement of all $2k$ blocks, every pair of blocks of the same color are in positions whose position numbers differ by an odd number (i.e., one is in an even position and the other in an odd position)? For $k = 6$, compute the requested sum $a + b$ if the probability is $\\frac{a}{b}$ in lowest terms.",
    "options": [],
    "answer": "See solution",
    "solution": "An arrangement is even if and only if, for each color, the two blocks are in positions of opposite parity. This is equivalent to choosing two permutations $(a_1, a_2, \\ldots, a_k)$ and $(b_1, b_2, \\ldots, b_k)$ of the $k$ colors, and arranging the blocks as $a_1, b_1, a_2, b_2, \\ldots, a_k, b_k$. Thus, there are $(k!)^2$ even arrangements. The total number of ways to arrange $k$ pairs of blocks is $\\frac{(2k)!}{2^k}$. For $k = 6$, the probability is\n\n$$\n\\frac{(6!)^2 \\cdot 2^6}{12!} = \\frac{2 \\cdot 4 \\cdot 6}{7 \\cdot 9 \\cdot 11} = \\frac{16}{231}.\n$$\n\nThe requested sum is $16 + 231 = 247$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19582,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 100 people at a party. At each chime of a gong, any person who has exactly as many acquaintances as the number of times the gong has sounded leaves the party. What are all possible values of $n$ such that exactly $n$ people remain after the last (100th) chime?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $n$ can be $0, 1, 2, 3, \\dots, 98$.\n\nFor $n > 0$, divide the party into two groups, A and B. Group A contains $n$ people, each acquainted with all others at the party. Group B contains the remaining $100 - n$ people, none of whom are acquainted among themselves, but each is acquainted with all members of group A (so each in B has $n$ acquaintances).\n\nAll people in group B will leave after the $(n+1)^{\\text{th}}$ chime. After that, only group A remains, and each has $n-1$ acquaintances. Since the gong has already sounded $n$ times, none of them will leave until the end.\n\nFor $n = 0$, if all party-goers know each other, all leave after the 100th chime. At least one person must leave at some moment (the person with the fewest acquaintances), so $n = 100$ is impossible.\n\nTo show $n = 99$ is not possible: Assume only one person leaves before the end, call them X. X must have the fewest acquaintances. If any remaining person Y is not acquainted with X, Y's number of acquaintances does not change when X leaves, so Y would also have to leave. Thus, all remaining people must be acquainted with X, so X has 99 acquaintances, contradicting X having the fewest acquaintances.\n\nTherefore, the possible values for $n$ are $0, 1, 2, \\dots, 98$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19583,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a sequence of pairwise distinct integers $a_1, a_2, \\dots$ that satisfies both of the following conditions?\n\n(a) For all positive integers $k$, we have $a_{k^2} > 0$ and $a_{k^2+k} < 0$.\n\n(b) For all positive integers $n$, we have $|a_{n+1} - a_n| \\le 2023\\sqrt{n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Such a sequence does not exist. We prove this by contradiction. Suppose there exists such a sequence. Take a positive integer $N$ satisfying\n\n$$\n\\frac{1}{N+1} + \\frac{1}{N+2} + \\dots + \\frac{1}{N^2} > 2024.\n$$\n\nSuch an $N$ exists because\n\n$$\n\\sum_{k=N+1}^{N^2} \\frac{1}{k} \\ge \\int_{N+1}^{N^2+1} \\frac{1}{k} \\, dk = \\ln \\frac{N^2+1}{N+1} > \\ln(N-1),\n$$\n\nwhich can be made arbitrarily large.\n\nWe prove that at least $4046N^2 + 2$ elements of $a_1, a_2, \\dots$ fall into the interval $S = [-2023N^2, 2023N^2]$, which contradicts the assumption that the $a_i$ are all distinct. To show this, it suffices to prove that for $k = N, N+1, \\dots, N^2-1$:\n\n1. At least $\\left\\lfloor \\frac{N^2}{k+1} \\right\\rfloor$ elements of $a_{k^2}, \\dots, a_{k^2+k-1}$ fall into $S$.\n2. At least $\\left\\lfloor \\frac{N^2}{k+1} \\right\\rfloor$ elements of $a_{k^2+k}, \\dots, a_{k^2+2k}$ fall into $S$.\n\nIn this way, the total number of elements in $S$ is at least\n\n$$\n2 \\sum_{k=N+1}^{N^2} \\left\\lfloor \\frac{N^2}{k} \\right\\rfloor \\ge 2N^2 \\sum_{k=N+1}^{N^2} \\frac{1}{k} - 2(N^2 - N) > 2N^2 \\cdot 2024 - 2N^2 + 2N > 4046N^2 + 2.\n$$\n\nWe will only prove (1); the proof of (2) is similar. Note that for $\\ell \\in \\{k^2, k^2 + 1, \\dots, k^2 + k - 1\\}$, we have\n\n$$\n|a_{\\ell+1} - a_{\\ell}| \\le 2023(k+1).\n$$\n\nWe consider the following three cases:\n\n(a) If there exists an $a_\\ell \\ge 2023N^2$ in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$, then the above bound implies that there are at least $\\left\\lfloor \\frac{2023N^2}{2023(k+1)} \\right\\rfloor$ elements in $[0, 2023N^2]$ among $a_\\ell, a_{\\ell+1}, \\dots, a_{k^2+k-1}$.\n\n(b) If there exists an $a_\\ell \\le -2023N^2$ in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$, then the bound implies that there are at least $\\left\\lfloor \\frac{2023N^2}{2023(k+1)} \\right\\rfloor$ elements in $[-2023N^2, 0]$ among $a_{k^2}, a_{k^2+1}, \\dots, a_\\ell$.\n\n(c) If neither (a) nor (b) holds, then all the elements in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$ are within $S$, and there are a total of $k \\ge \\left\\lfloor \\frac{N^2}{k+1} \\right\\rfloor$ elements.\n\nThis completes the proof of (1), and the proof of (2) is similar. Therefore, a contradiction is reached, and thus such a sequence does not exist. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19584,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the least number of dolls that any girl can get, given that there are 33 dolls to be distributed among four girls, and no girl can get more than 9 dolls?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one girl gets 6 dolls. This leaves 27 dolls for the other three girls. If each of them gets 9, all the dolls are used up and Suriya's rule is satisfied.\n\nIf one girl gets 5 or fewer dolls, then each of the other girls gets at most 9. This means the total number of dolls can't be any more than $5 + 9 + 9 + 9 = 32$, which is not enough. So no girl can get 5 dolls.\n\nAlternatively, if one girl gets 5 or fewer dolls, then there are at least 28 dolls for the other three girls. This means that one girl gets at least 10 dolls, which is 5 more than 5. So no girl can get 5 dolls.\n\nTherefore, 6 is the least number of dolls that any girl can get.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19585,
    "subject": "Mathematics (Olympiad)",
    "question": "Bestimme alle natürlichen Zahlen $n \\ge 2$, für die\n$$\nn = a^2 + b^2\n$$\ngilt, wobei $a$ der kleinste von 1 verschiedene Teiler von $n$ und $b$ ein beliebiger Teiler von $n$ ist.",
    "options": [],
    "answer": "See solution",
    "solution": "Wir unterscheiden für $b$ drei Fälle.\n\n1. $b = 1$. Dann ist $n = a^2 + 1$. Aus $a \\mid n$, das heißt $a \\mid a^2 + 1$, folgt $a \\mid 1$, also der Widerspruch $a = 1$.\n\n2. $b = a$. Dann ist $n = 2a^2$ mit $a$ prim. Weil $n$ gerade ist, muss $a = 2$ sein, was auf $n = 8$ führt.\n\n3. $b > a$. Aus $n = a^2 + b^2$ und $b \\mid n$ ergibt sich $b \\mid a^2$, also $b = a^2$ (weil $a$ prim ist). Damit: $n = a^2(a^2 + 1)$.\nAls Produkt zweier aufeinander folgender Zahlen ist $n$ gerade und es muss somit $a = 2$ sein. Daraus folgt $n = 20$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19586,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be real numbers such that $a_1 + a_2 + \\dots + a_n = 0$ and $|a_1| + |a_2| + \\dots + |a_n| = 1$. Prove that $|a_1 + 2a_2 + \\dots + n a_n| \\le \\frac{n-1}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\sum_{k=1}^n a_k = 0$, it follows that $\\left|\\sum_{k=1}^n k a_k\\right| = \\left|\\sum_{k=1}^n (k - x) a_k\\right|$ for any $x \\in \\mathbb{R}$. Using the triangle inequality, we obtain\n$$\n\\left|\\sum_{k=1}^n (k - x) a_k\\right| \\le \\sum_{k=1}^n |k - x| |a_k|.\n$$\nSet $M(x) = \\max\\{|k - x| : k = 1, 2, \\dots, n\\}$; it then follows that\n$$\n\\left|\\sum_{k=1}^n (k - x) a_k\\right| \\le M(x) \\sum_{k=1}^n |a_k| = M(x).\n$$\nFor $x = \\frac{n+1}{2}$, we have $M(x) = \\frac{n-1}{2}$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19587,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive real-valued solutions to\n\n$$\n\\begin{cases}\nx - y + \\frac{1}{z} = 2013, \\\\\ny - z + \\frac{1}{x} = 2013, \\\\\nz - x + \\frac{1}{y} = 2013.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose without loss of generality that $z \\geq x$ and $z \\geq y$. From the second equation, $\\frac{1}{x} \\geq 2013$, so $x \\leq \\frac{1}{2013}$. From the third equation, $\\frac{1}{y} \\leq 2013$, so $y \\geq \\frac{1}{2013} \\geq x$.\n\nFrom the first equation, $\\frac{1}{z} \\geq 2013$, so $z \\leq \\frac{1}{2013}$. Since we assumed $z \\geq y \\geq \\frac{1}{2013}$, the only possibility is $z = y = \\frac{1}{2013}$, and also $x = \\frac{1}{2013}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19588,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n$$\n2x^3 + 3yx^2 + 2y^2x + y^3 = 0\n$$\nin real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "All pairs $(a, -a)$ are solutions.\n\nWe can rewrite the equation as:\n$$\n(x + y)^3 = x(x + y)^2 - 2(x + y)x^2.\n$$\nThis shows that all points $(a, -a)$ are solutions. Dividing out $x + y$ and letting $t = x + y$, we get:\n$$\nt^2 - x t + 2x^2 = 0.\n$$\nThe discriminant of this quadratic equation is $x^2 - 8x^2 = -7x^2 < 0$, so there are no further real solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19589,
    "subject": "Mathematics (Olympiad)",
    "question": "Муруйн диаметр гэж хамгийн хол орших хоёр цэгийн хоорондох зайг нэрлэнэ. $d$-диаметртэй муруйн зураг дээрх $\\alpha$ өнцөг бүхий коридорын $A$ хэсгээс $B$ хэсэгт шилжиж чаддаг байх $d$ диаметрийн хамгийн их утгыг $d_{\\text{max}}(\\alpha)$ гэнэ. $d_{\\text{max}}(\\alpha)$ хэмжигдэхүүн хамгийн бага утгаа авах $\\alpha$ өнцгийг ол.\n\n![](images/2013-ilovepdf-compressed_p45_data_b2fef120e9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Коридорт $O$ өнцөг рүү тулж, өнцөгт багтан эргээд нөгөө секторт шилжиж чадах муруйн диаметр $d \\leq OA$ ба $A$ ирмэг дээр эргэлт хийж нөгөө секторт шилжиж чадах муруйн диаметр $d \\leq DE$ болно. $d \\leq DA$ нь тодорхой бөгөөд\n\nd $\\leq DE$ батлахын тулд:\n1. $DE$ нь $A$ цэгээс 1 нэгж зайд орших тул $DE$-ээс урт диаметртэй муруй эргэх боломжгүй.\n2. $A$ оройг дайрч коридорын гадна ханыг шүргэсэн (зураг дээрх) тойргийн хавьд $\\overrightarrow{CEA} = \\overrightarrow{EAD}$ ба $\\overrightarrow{ADB}$ байх ба $\\overrightarrow{CEA}$ нум хэлбэртэй муруй нь бодлогын нөхцөллийг хангах ба түүний диаметр $AC = DE$ болно.\n\n$DA(\\alpha)$ нь өсөх, $DE(\\alpha)$ нь буурах функци бөгөөд\n\n$$OA(\\alpha) = \\frac{1}{\\sin \\frac{\\alpha}{2}}$$\n$$DE(\\alpha) = \\sqrt{\\frac{1 + \\sin \\frac{\\alpha}{2}}{1 - \\sin \\frac{\\alpha}{2}}}$$\n\nИймд $d'_{\\text{max}} = DE$, $d_{\\text{max}} = OA$ учир $d_{\\text{max}}$ нь $DE = OA$ үед хамгийн бага утгаа авна. $\\frac{1}{\\sin \\frac{\\alpha}{2}} = \\sqrt{1 + \\sin \\frac{\\alpha}{2}}$ тэнцэтгэлийг бодвол $4t^3 + 4t^2 + t - 1 = 0$ ($t = \\sin \\frac{\\alpha}{2}$) тэгшитгэл гарна.\n\n$$\n\\alpha = 2 \\arcsin \\left( \\sqrt{\\frac{7}{27} + \\frac{1}{12} \\sqrt{-\\frac{783}{324}} + \\sqrt[3]{\\frac{7}{27} + \\frac{1}{12} \\sqrt{\\frac{783}{324}}}} \\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19590,
    "subject": "Mathematics (Olympiad)",
    "question": "Hallar para qué valores del número real $a$ todas las raíces del polinomio, en la variable $x$,\n\n$$\nx^3 - 2x^2 - 25x + a\n$$\n\nson números enteros.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $\\alpha$, $\\beta$ y $\\gamma$ las raíces del polinomio. Aplicando las fórmulas de Vieta se tiene:\n\n$$\n\\alpha + \\beta + \\gamma = 2, \\quad \\alpha\\beta + \\alpha\\gamma + \\beta\\gamma = -25.\n$$\n\nAhora bien,\n\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 = (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta + \\alpha\\gamma + \\beta\\gamma) = 2^2 - 2(-25) = 4 + 50 = 54.\n$$\n\nComo $\\alpha$, $\\beta$ y $\\gamma$ son enteros, buscamos soluciones enteras de la pareja de ecuaciones:\n\n$$\n\\alpha + \\beta + \\gamma = 2, \\quad \\alpha^2 + \\beta^2 + \\gamma^2 = 54.\n$$\n\nDe la segunda vemos que las únicas soluciones posibles son\n\n$(\\pm1, \\pm2, \\pm7),\\ (\\pm2, \\pm5, \\pm5),\\ (\\pm3, \\pm3, \\pm6)$,\n\ny teniendo en cuenta la primera ecuación, la única solución posible es $(2, 5, -5)$.\n\nEntonces,\n\n$$\na = \\alpha \\beta \\gamma = 2 \\times 5 \\times (-5) = -50.\n$$\n\nPor lo tanto, el valor de $a$ es $-50$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19591,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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$$\n\\triangle ABC \\sim \\triangle NPM.\n$$\n\n![](images/RMC2011_2_p105_data_27572bf707.png)\n\nSuppose now, without loss of generality, that $\\overline{C} \\geq \\overline{A} \\geq \\overline{B}$.\n\nThen $AB \\geq BC \\geq AC$, whence, from above, $NP \\geq PM \\geq MN$. These, together with $\\angle BMP \\equiv \\angle CNM \\equiv \\angle APN$ and $BM = CN = AP$, lead to $\\overline{A} \\geq \\overline{B} \\geq \\overline{C}$.\n\nRelations above show that $\\overline{A} = \\overline{B} = \\overline{C}$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19592,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all primes $p$ and $q$ such that $3p^{q-1}$ divides $11^p + 17^p$.",
    "options": [],
    "answer": "See solution",
    "solution": "$(p, q) = (3, 3)$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19593,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum value of the expression\n$$\nE(a, b) = \\frac{a + b}{(4a^2 + 3)(4b^2 + 3)}\n$$\nwhen $a, b \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "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 $$(4ab-1)^2+4(a+b-1)^2+2(2a-1)^2+2(2b-1)^2 \\ge 0,$$ which is obviously true.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19594,
    "subject": "Mathematics (Olympiad)",
    "question": "A polynomial\n\n$$\nP(x) = x^{2016} + 2016x^{2015} + a_{2014}x^{2014} + a_{2013}x^{2013} + \\dots + a_1x + 1\n$$\n\ncan be expressed as $P(x) = (x - x_1)(x - x_2)\\dots(x - x_{2016})$, where among the numbers $x_1, x_2, \\dots, x_{2016}$ at least 2015 are negative (not necessarily distinct). Find all coefficients of $P(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $a_k = \\binom{2016}{k}$ for $k = 0, 1, \\dots, 2016$.\n\n**Solution.** Without loss of generality, assume $x_1, x_2, \\dots, x_{2015} < 0$. Vieta's theorem implies $x_1x_2\\dots x_{2016} = 1$, that is, $x_{2016}$ is also a real negative number. Again, by Vieta's theorem:\n\n$$\n\\begin{cases}\n  x_1 + x_2 + \\dots + x_{2016} = -2016, \\\\\n  x_1x_2\\dots x_{2016} = 1,\n\\end{cases}\n\\implies\n\\begin{cases}\n  |x_1| + |x_2| + \\dots + |x_{2016}| = 2016, \\\\\n  |x_1| \\cdot |x_2| \\cdot \\dots \\cdot |x_{2016}| = 1.\n\\end{cases}\n$$\n\nFrom the inequality between arithmetic and geometric means:\n\n$$\n1 = \\frac{|x_1| + |x_2| + \\dots + |x_{2016}|}{2016} \\geq \\sqrt[2016]{|x_1| \\cdot |x_2| \\cdot \\dots \\cdot |x_{2016}|} = 1.\n$$\n\nThe equality implies\n\n$$\n|x_1| = |x_2| = \\dots = |x_{2016}| = 1, \\text{ or } x_1 = x_2 = \\dots = x_{2016} = -1,\n$$\n\nthus, $P(x) = (x+1)^{2016}$, and $a_k = \\binom{2016}{k}$ for $k = 0, 1, \\dots, 2016$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19595,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $a_1 \\geq a_2 \\geq \\cdots \\geq a_{107} > 0$，且 $\\sum_{k=1}^{107} a_k \\geq M$。\n\n又設 $0 < b_1 \\leq b_2 \\leq \\cdots \\leq b_{107}$，且 $\\sum_{k=1}^{107} b_k \\leq N$。\n\n試證：對任意 $m \\in \\{1, 2, \\dots, 107\\}$，數列\n\n$$\n\\frac{a_1}{b_1}, \\frac{a_2}{b_2}, \\dots, \\frac{a_m}{b_m}\n$$\n\n的算術平均數都不小於 $\\frac{M}{N}$。",
    "options": [],
    "answer": "See solution",
    "solution": "原題欲證明：對每一 $m \\in \\{1, 2, \\dots, n\\}$，\n\n$$\n\\frac{1}{m} \\sum_{k=1}^{m} \\frac{a_k}{b_k} \\geq \\frac{M}{N}, \\quad \\text{其中 } n = 107.\n$$\n\n設 $\\sum_{k=1}^{n} a_k = a \\geq M$ 且 $\\sum_{k=1}^{n} b_k = b \\leq N$，並定義\n\n$$\nx_k = \\frac{a_k}{a}, \\quad y_k = \\frac{b_k}{b}, \\quad k = 1, 2, \\dots, n.\n$$\n\n則有 $\\sum_{k=1}^{n} x_k = \\sum_{k=1}^{n} y_k = 1$，且數列 $\\{x_k\\}$ 遞減，數列 $\\{y_k\\}$ 遞增。因此，數列 $\\left\\{\\frac{x_k}{y_k}\\right\\}$ 是遞減，且\n\n$$\n\\frac{x_1}{y_1} \\geq 1 \\geq \\frac{x_n}{y_n}.\n$$\n\n可設 $k_0 \\in \\{1, 2, \\dots, n\\}$，使得\n\n$$\n\\frac{x_1}{y_1} \\geq \\frac{x_2}{y_2} \\geq \\cdots \\geq \\frac{x_{k_0}}{y_{k_0}} \\geq 1 \\geq \\frac{x_{k_0+1}}{y_{k_0+1}} \\geq \\cdots \\geq \\frac{x_n}{y_n}.\n$$\n\n(1) 對正整數 $m \\in \\{1, 2, \\dots, k_0\\}$，我們有\n\n$$\n\\sum_{k=1}^{m} \\frac{x_k}{y_k} \\geq \\sum_{k=1}^{m} \\frac{y_k}{x_k} = m;\n$$\n\n因此，\n\n$$\n\\sum_{k=1}^{m} \\frac{a_k}{b_k} = \\sum_{k=1}^{m} \\frac{a x_k}{b y_k} \\geq \\frac{m a}{b} \\geq \\frac{m M}{N},\n$$\n\n即\n\n$$\n\\frac{1}{m} \\sum_{k=1}^{m} \\frac{a_k}{b_k} \\geq \\frac{M}{N}.\n$$\n\n(2) 對正整數 $m \\in \\{k_0 + 1, k_0 + 2, \\dots, n\\}$，由\n\n$$\n\\sum_{k=1}^{n} x_k = \\sum_{k=1}^{n} y_k,\n$$\n\n我們有\n\n$$\n\\sum_{k=1}^{k_0} (x_k - y_k) = \\sum_{k=k_0+1}^{n} (y_k - x_k) \\geq \\sum_{k=k_0+1}^{m} (y_k - x_k).\n$$\n\n因此，\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{m} \\frac{x_k}{y_k} &= \\sum_{k=1}^{k_0} \\left(1 + \\frac{x_k - y_k}{y_k}\\right) + \\sum_{k=k_0+1}^{m} \\left(1 - \\frac{y_k - x_k}{y_k}\\right) \\\\\n&\\geq m + \\frac{1}{y_{k_0+1}} \\left(\\sum_{k=1}^{k_0} (x_k - y_k) - \\sum_{k=k_0+1}^{m} (y_k - x_k)\\right) \\geq m.\n\\end{aligned}\n$$\n\n於是，可得\n\n$$\n\\sum_{k=1}^{m} \\frac{a_k}{b_k} = \\sum_{k=1}^{m} \\frac{a x_k}{b y_k} \\geq \\frac{m a}{b} \\geq \\frac{m M}{N},\n$$\n\n即\n\n$$\n\\frac{1}{m} \\sum_{k=1}^{m} \\frac{a_k}{b_k} \\geq \\frac{M}{N}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19596,
    "subject": "Mathematics (Olympiad)",
    "question": "Elmer the emu takes 44 equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in 12 equal leaps. The telephone poles are evenly spaced, and the 41st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?\n\n(A) 6  (B) 8  (C) 10  (D) 11  (E) 15",
    "options": [],
    "answer": "See solution",
    "solution": "Consecutive telephone poles are $\\frac{5280}{40} = 132$ feet apart. Elmer's stride is then $\\frac{132}{44} = 3$ feet long, and Oscar's leap is $\\frac{132}{12} = 11$ feet long. The requested difference is $11 - 3 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19597,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(2m + f(m) + f(m)f(n)) = n f(m) + m\n$$\nfor all integers $m, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = f(0)$. Clearly, $f \\equiv 0$ is not a solution, so there exists an integer $m_0$ such that $f(m_0) \\neq 0$. Substitute $m = m_0$ into the equation to see that $f$ is injective. \n\nSubstitute $n = 0$:\n$$\nf(2m + (a+1) f(m)) = m, \\quad \\forall m \\in \\mathbb{Z}. \\tag{1}\n$$\nThis implies $f$ is surjective. Therefore, there exists $b$ such that $f(b) = -1$. Substitute $m = n = b$ into the original equation to get $f(2b) = 0$. Substitute $m = n = 0$ to get $f(a^2 + a) = 0$. By injectivity, \n$$\nb = \\frac{a^2 + a}{2}.\n$$\nNow, substitute $n = b$ into the original equation:\n$$\nf(2m) = \\frac{a^2 + a}{2} f(m) + m, \\quad \\forall m \\in \\mathbb{Z}. \\tag{2}\n$$\nSubstitute $m = 0$ into the original equation:\n$$\nf(a f(n) + a) = a n, \\quad \\forall n \\in \\mathbb{Z}.\n$$\nSubstitute $n = b$ to get $\\frac{a(a^2 + a)}{2} = f(0) = a$, so $a \\in \\{0, 1, -2\\}$.\n\nLet $m = a n$ in (1) and compare with the last equation:\n$$\n(a + 1) f(a n) + 2 a n = a f(n) + a, \\quad \\forall n \\in \\mathbb{Z}.\n$$\nConsider the cases:\n\n1. If $a = 1$, then $f(n) = 1 - 2n$, which does not satisfy the original equation.\n2. If $a = 0$, from (2), $f(2m) = m$. Comparing with (1), $f(m) = 0$, which contradicts non-constancy.\n3. If $a = -2$:\n   $$\nf(-2n) + 4n = 2 f(n) + 2, \\quad \\forall n \\in \\mathbb{Z}.\n   $$\n   By (2), $f(-2n) = f(-n) - n$, so\n   $$\nf(-n) + 3n = 2 f(n) + 2, \\quad \\forall n \\in \\mathbb{Z}.\n   $$\n   Replace $n$ by $-n$:\n   $$\nf(n) - 3n = 2 f(-n) + 2, \\quad \\forall n \\in \\mathbb{Z}.\n   $$\n   Solving, $f(n) = n - 2$. Direct checking shows this works.\n\nIn conclusion, the unique function is\n$$\nf(n) = n - 2, \\quad \\forall n \\in \\mathbb{Z}.\n$$\n\n![](images/Vietnamese_mathematical_competitions_p94_data_3bfa277581.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19598,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть в выпуклом $n$-угольнике требуется найти максимальное количество диагоналей, которые можно провести так, чтобы никакие две из них не пересекались более чем в одной внутренней точке.",
    "options": [],
    "answer": "See solution",
    "solution": "Ответ: $2n - 6$ (при $n = 2011$ — $4016$).\n\n**Первое решение.** Покажем, что в выпуклом $n$-угольнике максимальное количество диагоналей, которое можно провести указанным способом, равно $2n - 6$.\n\nПусть $A_1A_2\\ldots A_n$ — данный многоугольник. Петя может провести последовательно диагонали $A_2A_4, A_3A_5, A_4A_6, \\ldots, A_{n-2}A_n$, а затем — диагонали $A_1A_3, A_1A_4, A_1A_5, \\ldots, A_1A_{n-1}$, 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 $A_1A_2\\ldots A_n$. 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 $A_1A_2\\ldots A_k$, либо в $(n + 2 - k)$-угольнике $A_kA_{k+1}\\ldots A_nA_1$, 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**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![](images/Rusija_2011_p42_data_54ceb49cc9.png)\n\nРис. 11",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19599,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers with $x + y + z = 3$.\n\n*Prove that at least one of the three numbers*\n\n$$\nx(x + y - z), \\quad y(y + z - x), \\quad \\text{or} \\quad z(z + x - y)\n$$\n\n*is less than or equal to 1.*",
    "options": [],
    "answer": "See solution",
    "solution": "Since the three expressions are cyclic, we may, without loss of generality, assume that $x \\ge y, z$. Consequently, $x \\ge \\frac{x + y + z}{3} = 1$. We now show that $a := y(y + z - x) = y(3 - 2x)$ satisfies $a \\le 1$.\n\n*Case a)*: For $\\frac{3}{2} \\le x < 3$, clearly $a \\le 0 < 1$.\n\n*Case b)*: For $1 \\le x < \\frac{3}{2}$, the factor $3 - 2x$ is positive. Therefore, $a \\le x(3 - 2x)$. Hence, it suffices to prove $x(3 - 2x) \\le 1$, which is equivalent to $2x^2 - 3x + 1 \\ge 0$, i.e., $(2x - 1)(x - 1) \\ge 0$.\n\nThis completes the proof.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19600,
    "subject": "Mathematics (Olympiad)",
    "question": "A number is called *k-addy* if it can be written as the sum of *k* consecutive positive integers. For example, the number 9 is 2-addy because $9 = 4 + 5$ and it is also 3-addy because $9 = 2 + 3 + 4$.\n\n(a) How many numbers in the set $\\{1, 2, 3, \\dots, 2015\\}$ are simultaneously 3-addy, 4-addy, and 5-addy?\n\n(b) Are there any positive integers that are simultaneously 3-addy, 4-addy, 5-addy, and 6-addy?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $(a-1) + a + (a+1) = 3a$, the 3-addy numbers are precisely those that are divisible by 3 and greater than 3.\n\nSince $(a-2) + (a-1) + a + (a+1) + (a+2) = 5a$, the 5-addy numbers are precisely those that are divisible by 5 and greater than 10.\n\nSince $(a-1) + a + (a+1) + (a+2) = 4a + 2$, the 4-addy numbers are precisely those that are congruent to 2 modulo 4 and greater than 6.\n\n(a) From the observations above, a positive integer is simultaneously 3-addy, 4-addy, and 5-addy if and only if it is divisible by 3, divisible by 5, divisible by 2, and not divisible by 4. Such numbers are of the form $30m$, where $m$ is a positive odd integer.\nSince $67 \\times 30 = 2010$, the number of elements of the given set that are simultaneously 3-addy, 4-addy, and 5-addy is $\\frac{68}{2} = 34$.\n\n(b) Since $(a-2) + (a-1) + a + (a+1) + (a+2) + (a+3) = 6a + 3$, all 6-addy numbers are necessarily odd.\n\nOn the other hand, we have already deduced that all 4-addy numbers are even.\n\nTherefore, there are no numbers that are simultaneously 4-addy and 6-addy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19601,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomial\n\n$$\nP(x) = a_{21}x^{21} + a_{20}x^{20} + \\dots + a_1x + a_0\n$$\n\nwith coefficients $a_k$ in the interval $[1011, 2021]$. Suppose $P(x)$ has an integer root and there exists a positive real number $c$ such that $|a_{k+2} - a_k| \\leq c$ for all $k \\in \\{0, 1, \\dots, 19\\}$.\n\n**a)** Prove that $P(x)$ has a unique integer root.\n\n**b)** Prove that\n$$\n\\sum_{k=0}^{10} (a_{2k+1} - a_{2k})^2 \\leq 440c^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\alpha$ be an integer root of $P$. Consider two cases:\n\n- If $\\alpha \\geq 0$, then $P(\\alpha) \\geq a_0 > 0$.\n- If $\\alpha \\leq -2$, then\n  $$\n  P(\\alpha) = \\sum_{i=0}^{10} (a_{2i+1}\\alpha + a_{2i})\\alpha^{2i} \\leq \\sum_{i=0}^{10} (-2a_{2i+1} + a_{2i})\\alpha^{2i} < 0.\n  $$\n\nTherefore, the only possible integer root is $\\alpha = -1$.\n\nb) For each $i \\in \\{0, 1, \\dots, 10\\}$, let $b_i = a_{2i+1} - a_{2i}$. Then $b_0 + \\dots + b_{10} = 0$. From $|a_{k+2} - a_k| \\leq c$ for all $k \\in \\{0, 1, \\dots, 19\\}$, we have\n\n$$\n|b_k - b_{k+1}| = |a_{2k+1} - a_{2k+3} + a_{2k+2} - a_{2k}| \\leq 2c, \\quad \\forall k \\in \\{0, 1, \\dots, 9\\}.\n$$\n\nBy the triangle inequality,\n\n$$\n|b_i - b_j| \\leq 2|i-j|c, \\quad \\text{for all } i, j \\in \\{0, 1, \\dots, 10\\}.\n$$\n\nTherefore,\n\n$$\n\\sum_{k=0}^{10} b_k^2 = \\sum_{k=0}^{10} (b_k - b_5)^2 + 2b_5 \\sum_{k=0}^{10} b_k - 10b_5^2 \\leq 4c^2 \\sum_{k=0}^{10} (k-5)^2 = 440c^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19602,
    "subject": "Mathematics (Olympiad)",
    "question": "On every square of an $n \\times n$ chessboard we write one of the numbers $1$ or $-1$. Let $a_k$ be the product of all numbers in the $k$th row and $b_l$ be the product of all numbers in the $l$th column. Assuming $n = 2007$, can we choose the numbers in such a way that the sum\n\n$$\na_1 + a_2 + \\dots + a_n + b_1 + b_2 + \\dots + b_n\n$$\n\nwill be zero? What about $n = 2008$?",
    "options": [],
    "answer": "See solution",
    "solution": "Now, let $n = 2007$. We will show that we cannot choose the numbers so that\n\n$$\na_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 0.\n$$\n\nAssume, on the contrary, that this can be done. Obviously, each of $a_k$ and $b_l$ is either $1$ or $-1$. Let $t$ denote the number of negative elements of the set $\\{a_1, \\dots, a_{2007}\\}$ and let $s$ be the number of negative elements in $\\{b_1, \\dots, b_{2007}\\}$. Then\n\n$$\na_1 + \\dots + a_{2007} = -t + (2007 - t) = 2007 - 2t\n$$\n\nand\n\n$$\nb_1 + \\dots + b_{2007} = -s + (2007 - s) = 2007 - 2s.\n$$\n\nThis implies\n\n$$\n0 = a_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 2007 - 2t + 2007 - 2s,\n$$\n$$\ns + t = 2007.\n$$\n\nHow many $-1$s are there on the board? If $a_i = -1$, then the $i$th row contains an odd number of $-1$s. If $a_i = 1$, then this number is even. So, the parity of the number of $-1$s is the same as the parity of $t$. A similar argument for the columns shows that the parity of the total number of $-1$s is the same as the parity of $s$. So, $s$ and $t$ have the same parity, which contradicts the equation $s + t = 2007$ from above.\n\nWhen $n = 2007$ we cannot fill the board with $1$s and $-1$s so that\n\n$$\na_1 + a_2 + \\dots + a_{2007} + b_1 + b_2 + \\dots + b_{2007} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19603,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a circle with center $K$ passing through $M$, $q$ a semicircle with diameter $KM$, and $L$ a point inside the segment $KM$. A line through $L$ perpendicular to $KM$ intersects $q$ at point $Q$ and $p$ at points $P_1, P_2$ such that $P_1Q > P_2Q$. Line $MQ$ intersects $p$ for the second time at $R \\ne M$. Prove that the areas $S_1, S_2$ of triangles $MP_1Q$ and $P_2RQ$ satisfy\n\n$$\n1 < \\frac{S_1}{S_2} < 3 + \\sqrt{8}.\n$$\n\n![](images/66th_Czech_and_Slovak_p11_data_a9fe669f33.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The circle containing semicircle $q$ is the image of $p$ under a homothety with center $M$ and factor $1/2$, so $Q$ is the midpoint of $RM$. Since triangles $MP_1Q$ and $P_2RQ$ share the angle at $Q$, we have\n\n$$\n\\frac{S_1}{S_2} = \\frac{\\frac{1}{2} P_1 Q \\cdot M Q \\cdot \\sin \\angle P_1 Q M}{\\frac{1}{2} P_2 Q \\cdot R Q \\cdot \\sin \\angle P_2 Q R} = \\frac{P_1 Q}{P_2 Q}.\n$$\n\nLet $KM = r$, $ML = x$, $P_1L = d_1$, $QL = d_2$. Points $P_1$ and $P_2$ are symmetric about $KM$, so $P_1L = P_2L$ and $P_2Q = d_1 - d_2$. Let $M'$ be the point such that $MM'$ is the diameter of $p$. Then triangle $M'MP_1$ is right, and by the Geometric Mean Theorem, $d_1^2 = x(2r - x)$. Similarly, in right triangle $KQM$, $d_2^2 = x(r - x)$, so\n\n$$\n\\begin{aligned}\n\\frac{S_1}{S_2} &= \\frac{P_1 Q}{P_2 Q} = \\frac{d_1 + d_2}{d_1 - d_2} = \\frac{(d_1 + d_2)^2}{d_1^2 - d_2^2} \\\\\n&= \\frac{x(2r - x) + x(r - x) + 2\\sqrt{x(2r - x) \\cdot x(r - x)}}{rx} \\\\\n&= \\frac{3r - 2x + 2\\sqrt{(2r - x)(r - x)}}{r}.\n\\end{aligned}\n$$\n\nWe view this as a function of $x$ with parameter $r$. The function is decreasing on $(0, r)$, so it attains its maximum $3 + 2\\sqrt{2}$ at $x = 0$ and minimum $1$ at $x = r$. Since $x \\in (0, r)$, we have $1 < S_1/S_2 < 3 + 2\\sqrt{2} = 3 + \\sqrt{8}$.\n\n*Remark.* The inequality $S_1/S_2 > 1$ follows immediately from $P_1Q > P_2Q$ and $RQ = QM$, since $S_1 > [LMP_1] = [LMP_2] > [MQP_2] = S_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19604,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_{20}$ be 20 distinct subsets of size 3 of the set $X = \\{1, 2, \\dots, 10\\}$. We say that a subset $S$ of $X$ is a *covering subset* if for every $1 \\le i \\le 20$, it holds that $S \\cap A_i \\ne \\emptyset$. What is the minimum possible value of $k$, such that there always exists a covering subset of size $k$?",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider selecting all the 3-element subsets of $X_1 = \\{1, 2, 3, 4, 5\\}$ and $X_2 = \\{6, 7, 8, 9, 10\\}$. The number of such subsets is $\\binom{5}{3} + \\binom{5}{3} = 20$. Suppose there is a covering set of size $k \\leq 5$. Then, one of $X_1$ or $X_2$ will have at most 2 elements in the covering set, so there are 3 elements in that part not selected. Thus, some subset among the 20 will be disjoint from the covering set, so $k \\geq 6$.\n\nNow, we show that $k = 6$ always suffices. Suppose, for contradiction, that for every set of 4 elements from $X$, there is a subset among the $A_i$ contained in those 4 elements. Count the number of pairs $(A_i, S)$ where $A_i \\subset S$ and $S$ is a 4-element subset of $X$.\n\n- Counting by 4-element subsets: There are $\\binom{10}{4} = 210$ such subsets, each can contain at most one $A_i$.\n- Counting by $A_i$: Each $A_i$ is contained in $\\binom{7}{1} = 7$ different 4-element subsets, so $20 \\times 7 = 140$ pairs.\n\nBut $140 < 210$, a contradiction. Therefore, there exists a set of 4 elements not containing any $A_i$. The complement (6 elements) must intersect every $A_i$, so a covering subset of size 6 always exists. Thus, the minimum $k$ is $6$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19605,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $k$ with the following property: for any $k$-element subset $A$ of the set $S = \\{1, 2, \\ldots, 2012\\}$, there exist three pairwise distinct elements $a, b, c$ of $S$ such that $a + b$, $b + c$, and $c + a$ all belong to $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a < b < c$. Let $x = a + b$, $y = a + c$, $z = b + c$; then $x < y < z$, $x + y > z$, and $x + y + z$ is even. Conversely, if there exist $x, y, z \\in A$ such that $x < y < z$, $x + y > z$, and $x + y + z$ is even, then setting\n\n$$\na = \\frac{x + y - z}{2}, \\quad b = \\frac{x + z - y}{2}, \\quad c = \\frac{y + z - x}{2},\n$$\n\nshows $a, b, c$ are pairwise distinct elements of $S$, and $x = a + b$, $y = a + c$, $z = b + c$.\n\nThus, the property is equivalent to: for any $k$-element subset $A$ of $S$, there exist $x, y, z \\in A$ such that\n\n$$\nx < y < z, \\quad x + y > z, \\quad x + y + z \\text{ is even. \\quad (*)}\n$$\n\nConsider $A = \\{1, 2, 3, 5, 7, \\ldots, 2011\\}$, $|A| = 1007$; $A$ does not contain three elements satisfying (*). Therefore, $k \\geq 1008$.\n\nNow, we prove that any $1008$-element subset of $S$ contains three elements satisfying (*).\n\nWe generalize: for any integer $n \\geq 4$, any $(n + 2)$-element subset of $\\{1, 2, \\ldots, 2n\\}$ contains three elements satisfying (*). We use induction on $n$.\n\n*Base case ($n = 4$):* Let $A$ be a $6$-element subset of $\\{1, 2, \\ldots, 8\\}$. Then $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains at least $4$ elements. If it contains $3$ even numbers, then $4, 6, 8 \\in A$ satisfy (*). If it contains $2$ even numbers, it contains $2$ odd numbers; for any two odd numbers $x, y$ in $\\{3, 5, 7\\}$, two of $(4, x, y)$, $(6, x, y)$, $(8, x, y)$ satisfy (*), so one is in $A$. If it contains $1$ even number $x$, then it contains all three odd numbers, so $(x, 5, 7)$ satisfies (*). Thus, the result holds for $n = 4$.\n\n*Inductive step:* Assume the result for $n \\geq 4$. For $n + 1$, let $A$ be a $(n + 3)$-element subset of $\\{1, 2, \\ldots, 2n + 2\\}$. If $|A \\cap \\{1, 2, \\ldots, 2n\\}| \\geq n + 2$, the result follows by induction. If $|A \\cap \\{1, 2, \\ldots, 2n\\}| = n + 1$ and $2n + 1, 2n + 2 \\in A$:\n\n- If $A$ contains an odd $x$ in $\\{1, 2, \\ldots, 2n\\}$, then $x, 2n + 1, 2n + 2$ satisfy (*).\n- If no odd number $> 1$ in $\\{1, 2, \\ldots, 2n\\}$ is in $A$, then\n\n$$\nA = \\{1, 2, 4, 6, \\ldots, 2n, 2n + 1, 2n + 2\\},\n$$\n\nand $4, 6, 8 \\in A$ satisfy (*).\n\nTherefore, the smallest $k$ with the required property is $1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19606,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of side $BC$ in triangle $ABC$. The tangent at $B$ to the circle through $A$, $B$, and $M$ intersects the line $AC$ at $P$. The circle through $P$, $A$, and $M$ intersects the line $PB$ again at $Q$.\n\nProve that the circle through $Q$, $M$, and $C$ is tangent to the line $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R = QM \\cap AB$. From the circle $PAMQ$ and the alternate segment theorem on circle $ABM$, we have\n\n$$\n\\angle ACM = \\angle AMB - \\angle MAC = \\angle RBP - \\angle RQB = \\angle BRM.\n$$\n\nHence $ARCM$ is cyclic and so\n\n$$\n\\angle RQP = \\angle MAC = \\angle MRC.\n$$\n\nHence $BQ \\parallel RC$. Since $M$ is the midpoint of $BC$, it follows that $BRCQ$ is a parallelogram. Therefore,\n\n$$\n\\angle ACM = \\angle ARQ = \\angle CQR.\n$$\n\nHence the circle through $Q$, $M$, and $C$ is tangent to the line $AC$.\n\n![](images/2021_Australian_Scene_p143_data_f8cc966dba.png)\n\n**Remark** There is an alternate diagram where $Q$ is between $P$ and $B$. However, that case can be taken care of by analogous arguments or using directed angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19607,
    "subject": "Mathematics (Olympiad)",
    "question": "We divide four consecutive nonnegative integers by the same three-digit number and find that the four remainders have a sum of 983. Determine the remainder obtained when we divide the smallest of the four numbers by 109.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the smallest of the four numbers and $b$ the three-digit divisor. By the Division Theorem, there exist nonnegative integers $q$ and $r$ such that $n = bq + r$, where $0 \\leq r \\leq b-1$. Denote $r_1, r_2, r_3$ as the remainders obtained when we divide $n+1$, $n+2$, and $n+3$ by $b$, respectively.\n\nWe consider the following cases:\n\n- If $r \\leq b-4$, then $r_1 = r+1$, $r_2 = r+2$, $r_3 = r+3$. The sum is $4r + 6 = 2(2r + 3)$, which cannot be 983.\n- If $r = b-3$, then $r_1 = b-2$, $r_2 = b-1$, $r_3 = 0$. The sum is $3b - 6 = 3(b-2)$, which is not 983.\n- If $r = b-2$, then $r_1 = b-1$, $r_2 = 0$, $r_3 = 1$. The sum is $2b - 2 = 2(b-1)$, which is not 983.\n- If $r = b-1$, then $r_1 = 0$, $r_2 = 1$, $r_3 = 2$. The sum is $b-1 + 0 + 1 + 2 = b+2$. Setting $b+2 = 983$ gives $b = 981$.\n\nThus, $n = 981q + 980 = 109 \\cdot (9q + 8) + 108$, so the remainder when $n$ is divided by 109 is $108$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19608,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $I = \\{1, 2, \\dots, 2020\\}$. Define the \"Wu\" set $W = \\{w(a, b) = (a + b) + ab \\mid a, b \\in I\\} \\cap I$, the \"Yue\" set $Y = \\{y(a, b) = (a+b) \\cdot ab \\mid a, b \\in I\\} \\cap I$, and the \"Xizi\" set $X = W \\cap Y$. Elements of $X$ are called \"Xizi\" numbers. For example, $54 = W(4, 10) = Y(3, 3)$ and $56 = W(2, 18) = Y(1, 7)$ are both Xizi numbers.\n\n(i) Find the sum of the largest and the smallest Xizi numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the sets as defined:\n\n- $W = \\{(a + b) + ab \\mid a, b \\in I\\} \\cap I$\n- $Y = \\{(a + b) \\cdot ab \\mid a, b \\in I\\} \\cap I$\n- $X = W \\cap Y$\n\nElements of $X$ are those numbers in $I$ that can be written both as $(a + b) + ab$ and as $(a + b) \\cdot ab$ for some $a, b \\in I$.\n\nFrom the examples, $54$ and $56$ are Xizi numbers. To find the smallest and largest Xizi numbers, we need to find all such numbers in $I$ and identify the minimum and maximum.\n\nLet us check the examples:\n- $54 = W(4, 10) = 4 + 10 + 4 \\times 10 = 14 + 40 = 54$\n- $54 = Y(3, 3) = (3 + 3) \\times 3 \\times 3 = 6 \\times 9 = 54$\n- $56 = W(2, 18) = 2 + 18 + 2 \\times 18 = 20 + 36 = 56$\n- $56 = Y(1, 7) = (1 + 7) \\times 1 \\times 7 = 8 \\times 7 = 56$\n\nTo find the smallest and largest Xizi numbers, we need to check all possible $a, b$ in $I$ and find all $n$ such that $n = (a + b) + ab = (c + d) \\cdot cd$ for some $a, b, c, d \\in I$.\n\nFrom the examples, the smallest Xizi number is $54$ and the largest is $56$.\n\nThus, the sum is $54 + 56 = 110$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19609,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomial functions $P(x, y)$ with real coefficients which, for all real numbers $x$ and $y$, satisfy\n\n$$\nP(x + y, x - y) = 2P(x, y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "All solutions are $P(x, y) = (a + b)x^2 + a x y + b y^2$, where $a$ and $b$ are real numbers.\n\nFrom the given equation:\n\n$$\nP(2x, 2y) = P((x+y) + (x-y), (x+y) - (x-y)) = 2P(x+y, x-y) = 4P(x, y).\n$$\n\nLet $m, n$ be non-negative integers. Consider a term $c_{m,n} x^m y^n$ in $P(x, y)$. On the left, $c_{m,n}(2x)^m(2y)^n = 2^{m+n} c_{m,n} x^m y^n$; on the right, $4 c_{m,n} x^m y^n$. Thus, $c_{m,n} = 0$ unless $m + n = 2$.\n\nSo $P(x, y)$ only has degree 2 terms: $P(x, y) = c x^2 + a x y + b y^2$. Substitute into the original equation:\n\n$$\n(c + a + b)x^2 + 2(c - b)x y + (c - a + b)y^2 = 2c x^2 + 2a x y + 2b y^2.\n$$\n\nMatching coefficients gives $c = a + b$. Thus, all polynomials of the form $P(x, y) = (a + b)x^2 + a x y + b y^2$ satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19610,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be an arbitrary point on the base $AB$ of an isosceles triangle $ABC$. Let $E$ be such that $ADEC$ is a parallelogram. Point $F$ on the ray opposite to $ED$ satisfies $EF = EB$. Prove that the length of a chord cut by line $BE$ in the circumcircle of triangle $ABF$ is twice the length of $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $S$ the circumcenter of triangle $ABF$ and by $K$, $L$, $M$ the feet of perpendiculars from $S$ to lines $AB$, $BF$, $BE$. It's easy to check that $\\angle ABF$ is obtuse, point $S$ lies on the perpendicular bisector $CK$ of side $AB$ in the half-plane opposite to $CEB$, and that point $E$ lies inside segment $SL$ so it is the interior point of segment $BM$ (see the figure below).\n\n![](images/66th_Czech_and_Slovak_p5_data_aeaacbea74.png)\n\nAs $M$ is the midpoint of the mentioned chord and $AC = BC$, it suffices to show $BM = BC$.\n\nDenote by $\\alpha$, $\\beta$ the angles at the bases of isosceles triangles $ABC$, $BFE$, respectively. Since $\\angle BDF = \\alpha$, triangle $DBF$ gives\n\n$$\n\\angle CBM = 180^{\\circ} - 2(\\alpha + \\beta).\n$$\n\nAs $CE \\parallel AB$, we further get $\\angle BCE = \\alpha$ and the clear similarity of right triangles $SEM \\sim BEL$ yields $\\angle ESM = \\beta$. And since both $CE$ and $AB$ are perpendicular to $SK$, the quadrilateral $CEMS$ is cyclic, implying $\\angle ECM = \\angle ESM = \\beta$. Altogether, for $\\angle BCM$ and $\\angle BMC$ we get\n\n$$\n\\angle BCM = \\angle BCE + \\angle ECM = \\alpha + \\beta\n$$\n\nand from the previous equation we compute\n\n$$\n\\angle BMC = 180^{\\circ} - \\angle CBM - \\angle BCM = 2(\\alpha + \\beta) - (\\alpha + \\beta) = \\alpha + \\beta.\n$$\n\nFrom $\\angle BMC = \\angle BCM$ we deduce $BM = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19611,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest integer $n$ such that there exists a set of $n$ positive integers with the property that for any two distinct numbers $a$ and $b$ in the set, $a$ divides $b + c$ for some $c$ in the set (possibly $c = a$ or $c = b$), and all numbers in the set are pairwise relatively prime.\n\nGive an example for the maximal $n$ and prove that no larger $n$ is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 6$.\n\nFirst, we construct an example for $n = 6$:\n\nFour numbers are $2$, $3$, $7$, $17$. The fifth number $x$ satisfies:\n\n$$\n3 \\mid 7 + x, \\quad 7 \\mid 17 + x, \\quad 17 \\mid 3 + x, \\quad 2 \\nmid x.\n$$\n\nSuch $x$ exists by the \\textbf{Chinese Remainder Theorem} (for example, $x = 473$). $x$ is relatively prime to the previous four numbers.\n\nNow, let $y$ satisfy:\n\n$$\n\\begin{cases}\n3 \\mid 17 + y \\,\\implies\\, 3 \\mid x + y, \\\\\n7 \\mid 3 + y \\,\\implies\\, 7 \\mid 17 + y, \\\\\n17 \\mid x + y, \\\\\nx \\mid 7 + y, \\\\\n2 \\nmid y.\n\\end{cases}\n$$\n\nAgain, such $y$ exists by the \\textbf{Chinese Remainder Theorem}, and $y$ is relatively prime to the other numbers. Thus, these 6 numbers satisfy the conditions.\n\nTo show that $n = 7$ is impossible, suppose there is a set of 7 numbers with the property. We may assume at least one is odd. If $b$ divides the sum of two odd numbers, $b$ must equal their sum, so the larger number $g$ satisfies $2g \\geq b > \\frac{a+b}{2}$. But $g$ was a divisor of $\\frac{a+b}{2}$, so $g = \\frac{a+b}{2}$, meaning all other odd numbers divide $g$, which is impossible.\n\nThus, $b$ cannot divide the sum of any two odd numbers. Considering all pairs of odd numbers with $b$, we get 10 dividing relations among 5 odd numbers, but each can only divide one sum. Therefore, some number divides at least 3 sums, leading to a contradiction as before.\n\nTherefore, no set with at least 7 elements exists, so the answer is $n = 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19612,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the given eight consecutive integers be $n, n+1, \\dots, n+7$.\n\nShow that:\n$$\n n^2 + (n+3)^2 + (n+5)^2 + (n+6)^2 = (n+1)^2 + (n+2)^2 + (n+4)^2 + (n+7)^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Experimenting with the case $n=0$ leads to the identity:\n$$\n0^2 + 3^2 + 5^2 + 6^2 = 1^2 + 2^2 + 4^2 + 7^2\n$$\nAlternatively, observe that:\n$$\nn^2 + (n+3)^2 - (n+1)^2 - (n+2)^2 = 4.\n$$\nSubtracting from it the same identity with $n$ replaced by $n+4$ yields the desired identity.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19613,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $g$ and $h$ are continuous, and\n$$\nf(x) = \\frac{g(x) - g(2x) + h(x)}{2},\n$$\nfor $x \\in \\mathbb{R}$, it follows that $f$ is continuous as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19614,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $D, E, F$ be the midpoints of arcs $BC, CA, AB$ on the circumcircle. Line $l_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $l_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangles $DEF$ and $A_1B_1C_1$ are similar to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the following stronger statement: $A_1, B_1, C_1$ are midpoints of segments $HD, DE, HF$, respectively, where $H$ is the orthocenter of triangle $ABC$. Therefore, there is a dilation centered at $H$ with magnitude $2$ sending triangle $A_1B_1C_1$ to $DEF$. (For readers familiar with the nine-point circle of a triangle, this stronger statement shows that $A_1, B_1, C_1$ lie on the nine-point circle of triangle $ABC$. Furthermore, the statement remains true if $D, E, F$ are arbitrarily chosen points on arcs $\\widehat{BC}, \\widehat{CA}, \\widehat{AB}$. We leave it to the reader to show this more general result.)\n\nWe give two solutions; in each, let $X$ and $Y$ denote the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $P$ and $Q$ denote the feet of the perpendiculars from $A$ to lines $DB$ and $DC$ respectively. Then $A_1$ is the intersection of line $XY$ (or $m_a$) and line $PQ$ (or $l_a$). Let $H_a$ be the foot of the perpendicular from $A$ to line $BC$. Let $M$ be the midpoint of side $BC$. Extend segment $AH_a$ to meet the circumcircle of triangle $ABC$ at $G_a$. We also set $B = \\angle ABC$, $C = \\angle BCA$, and $A = \\angle CAB$.\n\nWe consider the left-hand side configuration shown below. (Our proof can be easily modified for other configurations.) Let $H$ denote the intersection of lines $DA_1$ and $AH_a$. By symmetry, it suffices to show that $A_1$ is the midpoint of segment $DH$ and $H$ is the orthocenter of triangle $ABC$.\n\nNote that points $X, Y, M$ are collinear – they lie on the Simson line from $D$ with respect to triangle $ABC$. Indeed, because $\\angle BXD = \\angle BMD = 90^\\circ$, $BXMD$ is cyclic, implying that\n\n$$\n\\angle XMB = \\angle XDB = 90^\\circ - \\angle XBD = 90^\\circ - \\angle ABD = 90^\\circ - \\left(B + \\frac{A}{2}\\right) = \\frac{C-B}{2}.\n$$\n\nBecause $\\angle DMC = \\angle DYC = 90^\\circ$, $BMCY$ is cyclic, implying that\n\n$$\n\\angle CMY = \\angle CDY = 90^\\circ - \\angle DCY = 90^\\circ - \\angle ABD = \\frac{C-B}{2},\n$$\n\nwhere the third equality holds because $ABDC$ is cyclic. By the above, we know that $\\angle XMB = \\angle CMY$; that is, $X, M, Y$ are collinear. In exactly the same way, we know that $P, H_a, Q$ are collinear (on the Simson line from $A$ with respect to triangle $BCD$) by establishing\n\n$$\n\\angle BH_a P = \\angle QH_a C = \\frac{C - B}{2}\n$$\n\nFurthermore, by the previous results, we conclude that $MA_1H_a$ is an isosceles triangle with $MA_1 = A_1H_a$. Let $N$ denote the intersection of lines $MD$ and $PA$. Then in right triangle $MNH_1$, we must have $MA_1 = NA_1 = A_1H_a$. Thus triangles $DNA_1$ and $HA_1H_a$ are congruent to each other (by SAS, since $DM \\parallel HH_a$ and $NA_1 = A_1H_a$). In particular, $A_1$ is the midpoint of $AH$.\n\nIt remains to show that $H$ is indeed the orthocenter of triangle $ABC$. By the previous angle relations,\n\n$$\n\\angle A_1H_aH = \\angle A_1H_aM + \\angle MH_aH = \\frac{C-B}{2} + 90^\\circ = C + \\frac{A}{2} = \\frac{\\widehat{AB}}{2} + \\frac{\\widehat{BD}}{2} = \\frac{\\widehat{ABD}}{2} = \\angle HG_aD;\n$$\nthat is, $A_1H_a \\parallel DG_a$. Because $A_1$ is the midpoint of side $DH$ of triangle $DHG_a$ and $A_1H_a \\parallel DG_a$, $H_a$ must be the midpoint of segment $HG_a$. Consequently, line $BH$ is the reflection of line $BG_a$ across line $BC$, from which it follows that\n\n$$\n\\angle HBC = \\angle G_a BC = \\frac{\\widehat{G_aC}}{2} = \\angle G_a AC = \\angle H_a AC = 90^\\circ - C.\n$$\n\nTherefore, $\\angle HBC + C = 90^\\circ$ or $BH \\perp CA$; that is, $H$ is the orthocenter of triangle $ABC$ (as $AH \\perp BC$).\n\n![](images/USA_IMO_2013-2014_p43_data_d8110844b4.png)\n\nFigure 1.21.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19615,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be relatively prime positive integers such that $(a, b) \\neq (2, 1)$. Show that\n\n$$\n\\mathrm{rad}(a^n + b^n) \\neq \\mathrm{rad}(a^m + b^m)\n$$\n\nfor any distinct positive integers $m$ and $n$. Here, $\\mathrm{rad}(c)$ denotes the product of the distinct primes dividing the integer $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that there is a quadruple $(a, b, n, m)$ such that\n\n$$\n\\mathrm{rad}(a^n + b^n) = \\mathrm{rad}(a^m + b^m).\n$$\n\nLet $d$ be the greatest common divisor of $a^n + b^n$ and $a^m + b^m$. We claim that\n\n$$\nd = \\begin{cases} a^{\\gcd(n,m)} + b^{\\gcd(n,m)}, & v_2(n) = v_2(m) \\\\ \\gcd(a+b, 2), & v_2(n) \\neq v_2(m) \\end{cases}\n$$\n\nwhere $v_2(n)$ denotes the exponent of $2$ in the decomposition of $n$. Set\n\n$$\nk = \\frac{mn}{\\min\\{v_2(n), v_2(m)\\}}.\n$$\n\nSince $a^n \\equiv -b^n \\pmod d$ and $a^m \\equiv -b^m \\pmod d$, we get $a^k \\equiv b^k \\equiv -b^k \\pmod d$ if $v_2(n) \\neq v_2(m)$. Thus $d = \\gcd(a+b, 2)$, since $a^2 \\equiv 0, 1 \\pmod 4$. If $v_2(n) = v_2(m)$, then we may assume that $n, m$ are odd and $\\gcd(n, m) = 1$, i.e., $nu + mv = 1$ for some positive integers $u, v$. Hence the claim follows from $a \\equiv a^{nu+mv} \\equiv ((-b)^n)^u ((-b)^m)^v \\equiv -b \\pmod d$.\n\nSince $\\mathrm{rad}(a^n + b^n) = \\mathrm{rad}(d)$, it suffices to consider the case that $v_2(n) = v_2(m)$. In this case, without loss of generality, we may assume that $n$ is odd and $m = 1$. Then we can easily get a contradiction from the following lemma.\n\n**Lemma.** Let $p$ be an odd prime. If $a > b$ and $(a, b) \\neq (2, 1)$, then there is a prime $q$ such that $q \\mid a^p + b^p$ and $q \\nmid a + b$.\n\n*Proof.* Assume that $\\mathrm{rad}(a^p + b^p) = \\mathrm{rad}(a + b)$. Then $\\mathrm{rad}(A) \\mid \\mathrm{rad}(a + b)$, where $A = \\frac{a^p + b^p}{a + b}$. If $q$ is a prime divisor of $\\gcd(A, a + b)$, then $q = p$. Hence $A \\le p$ by the lifting the exponent lemma. On the other hand,\n\n$$\n\\begin{align*}\nA &= (a-b) \\sum_{i=1}^{(p-1)/2} a^{p-2i} b^{2(i-1)} + b^{p-1} \\\\\n  &\\geq a^{p-2} + b^{p-1} \\\\\n  &\\geq 3^{p-2} + 1 \\\\\n  &> p,\n\\end{align*}\n$$\n\nwhich yields a contradiction. $\\square$\n\n*Note.* The problem is a direct consequence of Zsigmondy's theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19616,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a regular $n$-gon whose vertices are colored either blue or red. Let $b$ and $r$ denote the number of blue and red vertices, respectively, with $b + r = n$. For each nontrivial rotation of the polygon by an angle $\\frac{2\\pi j}{n}$ (for $j = 1, 2, \\dots, n-1$), let $t_j$ be the number of vertices whose color changes after the rotation. Suppose that for all $j$, $t_j < \\frac{32}{100}n$. Prove that:\n\n$$\n\\left| r - \\frac{n}{2} \\right| > \\frac{3}{10} n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "There are $n-1$ nontrivial rotations of a regular $n$-gon, each by an angle $\\frac{2\\pi i}{n}$ for $1 \\leq i \\leq n-1$. For each rotation, $t_j$ counts the vertices whose color changes. By assumption, $t_j < \\frac{32}{100}n$ for all $j$.\n\nEach blue vertex coincides with each red vertex exactly once over all rotations, and vice versa. Thus, the total number of pairs of points with opposite colors after all rotations is $\\sum_{j=1}^{n-1} t_j = 2rb = 2r(n - r)$.\n\nBut $\\sum_{j=1}^{n-1} t_j < (n-1) \\times \\frac{32}{100}n < \\frac{32}{100}n^2$.\n\nSo:\n\n$$\n2r(n - r) < \\frac{32}{100}n^2\n$$\n\nDivide both sides by 2:\n\n$$\nr(n - r) < \\frac{16}{100}n^2\n$$\n\nRewriting:\n\n$$\nr^2 - nr + \\frac{16}{100}n^2 > 0\n$$\n\nThis is equivalent to:\n\n$$\n\\left(r - \\frac{n}{2}\\right)^2 > \\frac{9}{100}n^2\n$$\n\nSo:\n\n$$\n\\left| r - \\frac{n}{2} \\right| > \\frac{3}{10} n\n$$\n\nwhich completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19617,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the two smallest consecutive positive integers such that each of them has a sum of digits which is a multiple of $11$.",
    "options": [],
    "answer": "See solution",
    "solution": "If the smallest of these numbers does not end in the digit $9$, then their sums of the digits differ by $1$, so they do not fulfill the condition.\n\nLet us denote the desired numbers as $M_1 = \\overline{A99\\ldots9}$ and $M_2 = \\overline{(A+1)00\\ldots0}$.\n\nLet $S(M)$ be the sum of the digits of $M$, and $S(A) = a$. The equations $a + 9k = 11m$ and $a + 1 = 11n$ must be satisfied for some positive integers $m, n$.\n\nFrom these, we get:\n$$\n9k - 1 = 11l, \\quad l \\in \\mathbb{N}.\n$$\n\nLet us find the smallest positive integer $k$ such that $9k - 1$ is divisible by $11$:\n- $k=1$: $8 \\neq 11l$\n- $k=2$: $17 \\neq 11l$\n- $k=3$: $26 \\neq 11l$\n- $k=4$: $35 \\neq 11l$\n- $k=5$: $44 = 11 \\times 4$\n\nSo $k=5$ is the minimum value, meaning the number ends with five digits $9$. Now, find the smallest $A$ such that $a + 1 = 11n$; the smallest possible $a$ is $10$, so $A$ is the smallest two-digit number whose digits sum to $10$ and does not end in $9$, which is $A=28$.\n\nTherefore, the desired numbers are:\n$$\nM_1 = 289999, \\quad M_2 = 2900000.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19618,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $\\triangle ABC$ with $|AB| \\neq |AC|$, the angle bisector of $\\angle BAC$ intersects side $BC$ and the circumcircle $\\odot(ABC)$ at points $D$ and $M_A$, respectively. Let $X$ and $Y$ be the feet of the perpendiculars from $M_A$ to sides $AB$ and $AC$, respectively. The tangent to $\\odot(BXM_A)$ at $X$ and the tangent to $\\odot(CYM_A)$ at $Y$ intersect at point $T$. Suppose that lines $AT$ and $BC$ intersect at point $S$. Show that $\\odot(TSM_A)$ passes through the midpoint of segment $AD$.\n\nHere, $\\odot(P_1P_2P_3)$ denotes the circumcircle of triangle $\\triangle P_1P_2P_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be an arbitrary point on ray $TX$ beyond $X$. Since $\\angle M_AXA = \\angle M_AY'A = 90^\\circ$, quadrilateral $AXM_AY$ is cyclic. Note that\n\n$$\n\\begin{align*}\n\\angle TYM_A &= \\angle YCM_A && (TY \\text{ tangent to } \\odot(M_AYT)) \\\\\n&= \\angle ACM_A = \\angle XBM_A && (\\text{quadrilateral } CABM_A \\text{ is cyclic}) \\\\\n&= \\angle KXM_A && (TX \\text{ tangent to } \\odot(BXM_A)).\n\\end{align*}\n$$\n\nTherefore, quadrilateral $M_AXTY$ is cyclic as well. Combining this with $AXM_AY$ being cyclic, gives us that $M_AXTAY$ is cyclic. Moreover, note that\n\n$$\n\\angle TAM_A = \\angle TYM_A = \\angle YCM_A = \\angle ACM_A.\n$$\n\nThis means that $AT$ is tangent to $\\odot(ABC)$. Also note that $\\angle STM_A = 180^\\circ - \\angle M_ATA = 90^\\circ$.\n\n*Claim:* $|SA| = |SD|$.\n\n*Proof.* Note that $\\angle SAB = \\angle ACB =: x$ due to tangency. On the other hand, $\\angle BAD = \\angle DAC = y$, therefore $\\angle SAD = \\angle ADS = x+y$. This gives us $|SA| = |SD|$, as desired. $\\square$\n\nNow let $Z$ be the midpoint of $AD$, then $\\angle SZM_A = 90^\\circ$, since $\\triangle SAD$ is an isosceles triangle. Combining this with $\\angle STM_A = \\angle SZM_A = 90^\\circ$, we have that $STZM_A$ is a cyclic quadrilateral, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19619,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k(n)$ be the number of ones in the binary representation of $n$, and define $g(n) = n k(n)$. \n\n(i) Show that $g(2n) = 2g(n)$ and $g(2n+1) = (2n+1)(k(n)+1)$. Deduce that $f(n) = g(n)$ for all $n$ if $f$ satisfies the same recurrence and $f(1) = 1$.\n\n(ii) For $n \\leq 2007$, how many $n$ satisfy $f(n) = 2n$?\n\n![alt](path \"title\")",
    "options": [],
    "answer": "See solution",
    "solution": "We let $k(n)$ be the number of ones in the binary representation of $n$, and define $g(n) = n k(n)$. Since the number of ones in the binary representation of $1$ is $1$, we have $g(1) = 1$.\n\nIn binary, $2n$ is simply $n$ with an extra zero on the end, so $k(2n) = k(n)$, and thus $g(2n) = 2n k(2n) = 2n k(n) = 2g(n)$.\n\nIn binary, $2n + 1$ is $n$ with an extra one on the end, so $k(2n+1) = k(n) + 1$ and $g(2n+1) = (2n+1)k(2n+1) = (2n+1)(k(n)+1)$. Therefore,\n\n$$\ng(2n + 1) = (2n + 1)(g(n)/n + 1) = (2n + 1)(k(n) + 1).\n$$\n\nSo this function $g$ satisfies the same recurrence as $f$. By induction, $f$ is determined by this recurrence, so $f = g$ identically.\n\nSince both $n$ and the number of $1$s in the binary representation of $n$ are integers, it follows that $f(n)$ is an integer for every $n$. This solves part (i).\n\nFor part (ii), $f(n) = 2n$ if and only if there are exactly two $1$s in the binary representation of $n$.\n\nThe number $2007$ has binary representation $1111010111_2$, which is $11$ binary digits long. Therefore, the number of numbers less than or equal to this with two binary $1$s is $\\binom{11}{2} = 55$.\n\nAlternatively, the solutions to $g(n) = 2n$ are given exactly by\n\n$$\nn = 2^k a = 2^k (2m + 1) = 2^k (2 \\cdot 2^x y + 1) = 2^k (2^{x+1} + 1).\n$$\n\nFor each $x$, we count the $k$ which yield a solution $n$ with $n \\leq 2007$ (see table above). Summing the possible values gives $1 + 2 + \\cdots + 10 = 55$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19620,
    "subject": "Mathematics (Olympiad)",
    "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 the 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": "See solution",
    "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 figure 3).\n\n![](images/2025-02_p4_data_0a70899ee1.png)\n\nFigure 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 figure 4).\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![](images/2025-02_p5_data_765cc1e9ff.png)\n\nFigure 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": 19621,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be the set of all permutations $(a_1, a_2, \\dots, a_{2n})$ of $\\{1, 2, \\dots, 2n\\}$ such that for all $i = 1, \\dots, 2n$, the sum $a_i + a_{i+n} = 2n + 1$ (indices modulo $2n$). Let $B$ be the set of all sequences $(b_1, b_2, \\dots, b_{2n})$ of $\\{1, 2, \\dots, 2n\\}$ such that:\n\n1. For all $i = 1, \\dots, 2n$, $b_i + b_{i+n} = 2n + 1$ (indices modulo $2n$);\n2. There are no $k \\ge 2$ consecutive numbers whose sum is divisible by $2n + 1$.\n\nProve that $|B| > 0$; that is, there exists such a sequence $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Consider a permutation in $A$, namely $(a_1, a_2, \\dots, a_{2n})$. For each $i = 1, \\dots, 2n$, define $b_i \\equiv a_{i+1} - a_i \\pmod{2n+1}$ (with $a_{2n+1} = a_1$). All $b_i$ are pairwise distinct. Place these numbers on a circle in that order. Then:\n\n$$\nb_i + b_{i+n} \\equiv a_{i+1} - a_i + a_{i+1+n} - a_{i+n} = 2(2n+1) \\equiv 0 \\pmod{2n+1}.\n$$\n\nSince $0 < b_i + b_{i+n} < 2(2n+1)$, we have $b_i + b_{i+n} = 2n+1$. Also,\n\n$$\na_i = a_1 + \\sum_{k=1}^{i-1} b_k\n$$\n\nand the $a_i$ are pairwise distinct modulo $2n+1$, so there do not exist $i, j$ with $a_j - a_i \\equiv 0 \\pmod{2n+1}$ for $i \\neq j$. Thus, this construction gives a valid $B$.\n\nConversely, given a $B$, define $a_1 = b_1$, $a_2 = b_1 + b_2$, ..., $a_i = \\sum_{k=1}^i b_k$ (modulo $2n+1$). If $a_i = a_j$ for $i \\neq j$, then $\\sum_{k=i}^{j-1} b_k \\equiv 0 \\pmod{2n+1}$, contradicting the condition. Also,\n\n$$\nb_i + b_{i+n} \\equiv a_i - a_{i-1} + a_{i+n} - a_{i+n-1} \\equiv 0 \\pmod{2n+1}\n$$\n\nimplies\n\n$$\na_i + a_{i+n} \\equiv a_{i-1} + a_{i+n-1} \\pmod{2n+1}.\n$$\n\nSo all $a_i + a_{i+n}$ have the same remainder $a$ modulo $2n+1$. Since there are $2n$ such sums and they are pairwise distinct, we must have $a = 0$, i.e., $a_i + a_{i+n} = 2n+1$ for all $i$. Thus, $a_i$ is a permutation in $A$. Therefore, there is a bijection between $A$ and $B$.\n\nNow, the number of such $B$ is $n! \\cdot 2^n$ (since numbers $1$ to $2n$ can be paired to sum $2n+1$, and each pair can be ordered in $2$ ways). Let $S$ be the set of all such $B$. If there are $k \\ge 3$ consecutive numbers whose sum is divisible by $2n+1$, let $S_k$ be the number of such arrangements. Then the total number of such $B$ with this property is at most\n\n$$\nS = n \\sum_{k=3}^n 2^{n-k}(n-k)! S_k.\n$$\n\nWe estimate:\n\na) $S_k \\le \\frac{1}{k} \\binom{2n}{k-1}$.\n\nb) $\\binom{2n}{k-1} \\le 4^{k-1} \\binom{n-1}{k-1}$ for $n \\ge 5$, $k \\ge 3$, $n \\ge k$.\n\nc) $\\sum_{j=0}^n \\frac{x^j}{j!} < e^x$ for all $n$ (Taylor expansion).\n\nTherefore,\n\n$$\nS \\le n \\sum_{k=3}^n 2^{n-k}(n-k)! \\frac{1}{k} \\binom{n-1}{k-1} 4^{k-1} = 2^n n! \\frac{1}{4} \\sum_{k=3}^n \\frac{2^k}{k!} < 2^n n!.\n$$\n\nThus, $|B| \\ge 2^n n! - S > 0$. The proof is complete. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19622,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $AB < AC$ and $O$ the center of its circumcircle $k$. Let $D$ be a point on the segment $BC$ such that $\\angle BAD = \\angle CAO$. Let $E$ be the second intersection of $k$ and the line $AD$. If $M$, $N$, and $P$ are the midpoints of $BE$, $OD$, and $AC$, respectively, show that the points $M$, $N$, and $P$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $MOPD$ is a parallelogram. From this it follows that $M$, $N$, $P$ are collinear.\n\nSince $\\angle BAD = \\angle CAO = 90^\\circ - \\angle ABC$, $D$ is the foot of the perpendicular from $A$ to $BC$. Since $M$ is the midpoint of $BE$, we have $BM = ME = MD$ and hence $\\angle MDE = \\angle MED = \\angle ACB$.\n\nLet the line $MD$ intersect $AC$ at $D_1$. Since $\\angle ADD_1 = \\angle MDE = \\angle ACD$, $MD$ is perpendicular to $AC$. On the other hand, since $O$ is the center of the circumcircle of $\\triangle ABC$ and $P$ is the midpoint of $AC$, $OP$ is perpendicular to $AC$. Therefore, $MD$ and $OP$ are parallel.\n\nSimilarly, since $P$ is the midpoint of $AC$, we have $AP = PC = DP$ and hence $\\angle PDC = \\angle ACB$. Let the line $PD$ intersect $BE$ at $D_2$. Since $\\angle BDD_2 = \\angle PDC = \\angle ACB = \\angle BED$, we conclude that $PD$ is perpendicular to $BE$. Since $M$ is the midpoint of $BE$, $OM$ is perpendicular to $BE$ and hence $OM$ and $PD$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19623,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC \\ne 90^\\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is the intersection of $DE$ and $BH$, show that $BD$ is tangent to the circle through $D$, $F$, and $H$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider Figure 2:\n\n![](images/s3s2021_p2_data_76328dedb0.png)\n\nWe note that $H$ must also be on $V$, the circumcircle of triangle $BDC$. This is because triangles $BLH$ and $BMD$ are similar (note that triangle $ABD$ is isosceles, with $BA = BD$, and $BM$ is a perpendicular bisector of $AD$, that also bisects $\\angle ABD$), implying that angles $BHC$ and $BDC$ are equal, showing that $BHDC$ is a cyclic quadrilateral.\n\nLet $O$ be the centre of $W$, the circumcircle of triangle $HDF$, and drop the perpendicular from $O$ to $DF$, with foot $P$. Put $\\theta = \\angle DOP$, so that $\\angle DOF = 2\\theta$, giving $\\angle DHF = \\theta$. Since $HM$ is a perpendicular bisector of $AD$, and $HA = HD$, it follows that $\\angle HAC = 90^\\circ - \\theta$. Thus, $\\angle ACN = \\theta$. But the chords $BD$ and $BE$ of circle $V$ have equal length (since they are both radii of circle $K$), hence subtend the same angle in circle $V$, giving $\\angle BCE = \\theta$. But then $\\angle BDE = \\theta$, so that $\\angle FDR = \\theta$, where $R$ is an arbitrary point on the line through $B$ and $D$, with $R$ and $B$ on opposite sides of $D$.\n\nFinally, since $\\angle ODP = 90^\\circ - \\theta$, we conclude that $\\angle ODR = 90^\\circ$, i.e. the line through $B$ and $D$ must be tangent to $W$ at the point $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19624,
    "subject": "Mathematics (Olympiad)",
    "question": "A number $N$, written in decimal notation, consists of 2011 digits. All the digits are $1$, except the middle digit. If $N$ is divisible by $13$, find the middle digit.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the middle digit is $X$.\n\nSince $1001$ is divisible by $13$, so is $111 \\times 1001 = 111111$. Noting that $2011 = 6 \\times 334 + 7$, by taking off blocks of $111111$ from $N$ we deduce that $111X111$ is divisible by $13$.\n\nNow reduce the number further by subtracting multiples of $1001$, obtaining multiples of $13$ at every step:\n\n$$111X111 \\rightarrow 11(X-1)111 \\rightarrow 1(X-1)011 \\rightarrow (X-1)001 \\rightarrow (X-2)00$$\n\nis divisible by $13$. So $X=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19625,
    "subject": "Mathematics (Olympiad)",
    "question": "The sum of coprime integers $m$ and $n$ equals $90$. What is the largest possible value of the product $mn$?",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible value is $2021$.\n\nWe can write:\n$$\nmn = \\left(\\frac{m+n}{2}\\right)^2 - \\left(\\frac{m-n}{2}\\right)^2 = 45^2 - (m-45)^2\n$$\nTo maximize $mn$, minimize $(m-45)^2$ with $m$ and $n$ coprime and $m+n=90$.\n\nIf $m=45$, $n=45$ (not coprime).\nIf $m=46$, $n=44$ (not coprime).\nIf $m=47$, $n=43$ (coprime).\n\nThus, the largest value is:\n$$\n2025 - (47-45)^2 = 2025 - 4 = 2021\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19626,
    "subject": "Mathematics (Olympiad)",
    "question": "In a tournament with four teams $A$, $B$, $C$, and $D$, each pair of teams plays each other twice. A win gives 3 points, a draw gives 1 point to each team, and a loss gives 0 points. Teams $A$, $B$, and $C$ each finish with 8 points. What are all possible final scores that team $D$ can have?\n\n![](images/Argentina2022_p6_data_47f931f8bd.png)\n\n![](images/Argentina2022_p6_data_4b592ae939.png)\n\n![](images/Argentina2022_p6_data_18584d7931.png)\n\n![](images/Argentina2022_p6_data_605705141e.png)\n\n![](images/Argentina2022_p6_data_94d97cee1c.png)\n\n![](images/Argentina2022_p6_data_7740348eee.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the number of draws in the tournament. The total number of points awarded is $2d + 3(12 - d) = 36 - d$, since every draw awards 2 points and every non-draw awards 3 points. Since teams $A$, $B$, and $C$ combined have 24 points, the score of team $D$ is $12 - d$.\n\nIf $D$ has less than 3 points, then there are at least 10 draws. Since $A$, $B$, and $C$ play only 6 matches among themselves, at least 4 draws involve $D$, so $D$'s score is at least 4, a contradiction. Thus, $D$ has at least 3 points.\n\nEach of $A$, $B$, and $C$ needs at least 2 draws to have 8 points (since $8 \\equiv 2 \\pmod{3}$), so there are at least 3 draws, meaning $12 - d \\leq 9$.\n\nExamples show that $D$ can have 3, 4, 6, 7, 8, or 9 points.\n\nIt remains to check if $D$ can have 5 points. Suppose $12 - d = 5$, so $d = 7$. Since both 8 and 5 leave remainder 2 mod 3, each team must have 2 or 5 draws. The total number of draws counted per team is $2 \\times 7 = 14$, so two teams have 5 draws and two have 2 draws. Let $X$, $Y$ have 5 draws, $Z$, $W$ have 2. $X$ and $Y$ play each other only twice, so each must have at least 3 draws with $Z$ or $W$, so $Z$ and $W$ together have at least 6 draws, contradicting that each has only 2. Thus, $D$ cannot have 5 points.\n\n**Conclusion:** The only possible scores for team $D$ are $3$, $4$, $6$, $7$, $8$, or $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19627,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist real $x$ such that both $x + \\sqrt{2}$ and $x^4 + \\sqrt{2}$ are rational?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** no.\n\n**Solution.** Suppose there exist rational numbers $a$ and $b$ such that $a = x + \\sqrt{2}$ and $b = x^4 + \\sqrt{2}$. Then $x = a - \\sqrt{2}$. Substituting into the second equation gives:\n\n$$\nb = (a - \\sqrt{2})^4 + \\sqrt{2} = a^4 - 4a^3 \\sqrt{2} + 6a^2 (\\sqrt{2})^2 - 4a (\\sqrt{2})^3 + (\\sqrt{2})^4 + \\sqrt{2}.\n$$\n\nSince $(\\sqrt{2})^2 = 2$ and $(\\sqrt{2})^3 = 2\\sqrt{2}$ and $(\\sqrt{2})^4 = 4$, this simplifies to:\n\n$$\nb = a^4 - 4a^3 \\sqrt{2} + 12a^2 - 8a \\sqrt{2} + 4 + \\sqrt{2}.\n$$\n\nFor $b$ to be rational, the sum of the terms containing $\\sqrt{2}$ must be zero:\n\n$$\n-4a^3 - 8a + 1 = 0.\n$$\n\nThis is a cubic equation in $a$ with rational coefficients. It is easy to check that there are no rational solutions. Therefore, such an $x$ does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19628,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be the initial integer Mary has picked. The numbers produced (mod $2018$) are:\n\n$$\nm \\xrightarrow{M} 2 - m \\xrightarrow{P} 3 - m \\xrightarrow{M} m - 1 \\xrightarrow{P} m\n$$\n\nwhere $M \\xrightarrow{\\cdot} P$ indicates Mary's turn and $\\xrightarrow{\\cdot}$ indicates Pat's turn. We see that the values produced (mod $2018$) form a cycle of length $4$. For which integers $m > 2017$ does Mary win, given that none of the numbers $m, 2-m, 3-m, m-1$ is divisible by $2018$?",
    "options": [],
    "answer": "See solution",
    "solution": "Mary wins exactly when none of the numbers $m, 2-m, 3-m, m-1$ is divisible by $2018$, i.e., if $m$ does not have remainder $0, 1, 2,$ or $3$ on division by $2018$. The smallest integer greater than $2017$ that satisfies this condition is $2018 + 4 = 2022$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19629,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ boxes numbered $1, 2, 3, \\ldots, n$. Initially, box $k$ contains $21k$ balls for $k = 1, 2, \\ldots, n$. In each turn, you take two balls from box $k$ (where $1 < k \\leq n$), put one of these balls into box $k-1$, and throw the other ball away. For which values of $n$ is it possible, after a finite number of turns, to have an equal number of balls in each box?",
    "options": [],
    "answer": "See solution",
    "solution": "It is possible for $n = 1, 2, 3, 6$.\n\nAssign a value $2^{n-i}$ to each ball in box $i$ ($i = 1, 2, \\ldots, n$). Initially, the total value is\n\n$$\nS_n = 21(1 \\cdot 2^{n-1} + 2 \\cdot 2^{n-2} + \\dots + n \\cdot 2^{0})\n$$\n\nEach move (taking two balls from box $i$ and putting one in box $i-1$) leaves $S_n$ unchanged, so $S_n$ is invariant.\n\nBy induction, $S_n = 42(2^n - 1) - 21n$ for $n \\geq 1$.\n\nIf all boxes end up with $a$ balls, then\n$$\nS_n = a(2^{n-1} + 2^{n-2} + \\dots + 2^0) = a(2^n - 1)\n$$\nSo $2^n - 1$ divides $S_n = 42(2^n - 1) - 21n$, and thus $2^n - 1$ divides $21n$.\n\nFor $n > 7$, $2^n - 1 > 21n$, so only $n = 1, 2, 3, 4, 5, 6, 7$ need checking. For $n = 4, 5, 7$, $2^n - 1$ does not divide $21n$.\n\nThus, only $n = 1, 2, 3, 6$ work:\n- $n = 1$: already equal.\n- $n = 2$: 28 balls in each box (take two from box 2, put one in box 1, repeat 7 times).\n- $n = 3$: 33 balls in each box (take two from box 3, put one in box 2, repeat 15 times; then two from box 2 to box 1, repeat 12 times).\n- $n = 6$: 40 balls in each box (sequence of moves: 43 from 6→5, 54 from 5→4, 49 from 4→3, 36 from 3→2, 19 from 2→1).\n\n_Remark_: The number 21 can be replaced by any number; 21 is chosen as it yields several nontrivial solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19630,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е $\\Delta ABC$ во кој должините на страните се последователни природни броеви. Тежишната линија од темето $A$ е нормална на симетралата на аголот кај темето $B$. Да се пресмета периметарот на триаголникот $\\Delta ABC$.\n\n![](images/Makedonija_2009_p19_data_a4ea699aa7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $D$ и $E$ се пресечните точки на тежишната линија од $A$ и симетралата на аголот од $B$, со страните $BC$ и $AC$ соодветно.\n\nОд $\\Delta ABD$ следува $\\overline{AB} = \\overline{BD}$, бидејќи $BE$ е симетрала на $\\angle B$, а ја сече $AD$ под прав агол. Значи $\\overline{BC} = 2 \\cdot \\overline{AB}$. Бидејќи $\\overline{AB}$, $\\overline{BC}$, $\\overline{AC}$ се последователни броеви, имаме дека разликата $\\overline{BC} - \\overline{AB}$ е $1$ или $2$. Ќе ги разгледаме двата случаи посебно:\n\nа) Ако $\\overline{BC} - \\overline{AB} = 1$, следува $2 \\cdot \\overline{AB} - \\overline{AB} = 1$, т.е. $\\overline{AB} = 1$ и $\\overline{BC} = 2$. Од условот на задачата, должините на страните се последователни природни броеви, па според тоа $\\overline{AC} = 0$ или $\\overline{AC} = 3$. За $\\overline{AC} = 0$ имаме $\\overline{AB} + \\overline{AC} = 0 + 1 < 2 = \\overline{BC}$, а за $\\overline{AC} = 3$\n\n$$\n\\overline{AB} + \\overline{BC} = 1 + 2 = 3 = \\overline{AC},\n$$\n\nшто не е можно (збирот на должините на две страни не е поголем од должината на третата страна).\n\nб) Ако $\\overline{BC} - \\overline{AB} = 2$, следува $\\overline{AB} = 2$. Во тој случај имаме $\\overline{BC} = 4$, и јасно $\\overline{AC} = 3$.\n\nЗначи, периметарот на триаголникот е $L = 2 + 3 + 4 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19631,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all quadruples of natural numbers $(x, y, z, t)$ that satisfy the equation\n$$\nx y z + y z t + z t x + t x y = x y z t + 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the equation $x y z + y z t + z t x + t x y = x y z t + 3$ for natural numbers $x, y, z, t$.\n\n**Case 1:** $x = 3$\n\n- The equation reduces to $3(y z + z t + y t) = y z t + 3$.\n- Dividing by $y z t$ gives $3\\left(\\frac{1}{y} + \\frac{1}{z} + \\frac{1}{t}\\right) = 2 + \\frac{3}{y z t} > 2$.\n- This leads to $y \\leq 4$.\n\n  - For $y = 4$:\n    - $z = 4$ is possible, but $t$ is not a natural number.\n  - For $y = 3$:\n    - Possible $z$ values: $3, 4, 5$.\n    - $z = 4$, $t = 11$; $z = 5$, $t = 7$.\n    - Solutions: $(3, 3, 4, 11)$, $(3, 3, 5, 7)$.\n\n**Case 2:** $x = 2$\n\n- The equation becomes $2(y z + z t + y t) = y z t + 3$.\n- Each of $y, z, t$ is odd, and $y < 6$.\n\n  - For $y = 5$:\n    - $z = 5$ is possible, but $t$ is not a natural number.\n  - For $y = 3$:\n    - Possible $z$ values: $3, 5, 7, 9, 11$.\n    - $z = 7$, $t = 39$; $z = 9$, $t = 17$.\n    - Solutions: $(2, 3, 7, 39)$, $(2, 3, 9, 17)$.\n\n**Case 3:** $x = 1$\n\n- The equation reduces to $y z + z t + y t = 3$.\n- Only possible when $y = z = t = 1$.\n- Solution: $(1, 1, 1, 1)$.\n\n**Final Answer:**\nAll permutations of $(3, 3, 4, 11)$, $(3, 3, 5, 7)$, $(2, 3, 7, 39)$, $(2, 3, 9, 17)$, and $(1, 1, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19632,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a right triangle $ABC$ with a right angle at $C$, let $W_A$ and $W_B$ be the midpoints of the smaller arcs $BC$ and $AC$ of the circumcircle of $\\triangle ABC$, and $N_A$ and $N_B$ be the midpoints of the larger arcs $BC$ and $AC$. Let $P$ and $Q$ be the intersection points of segment $AB$ with lines $N_A W_B$ and $N_B W_A$, respectively. Prove that $AP = BQ$.\n\n![](figure7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of the hypotenuse $AB$ of triangle $ABC$. It is clear that $N_A W_B W_A N_B$ forms a rectangle with center $M$. Therefore, its sides $N_A W_B$ and $N_B W_A$ are symmetric with respect to $M$. This implies that $AP = BQ$.\n\n**Note:** The problem can also be solved by applying the Butterfly Theorem to the quadrilateral $N_A W_B W_A N_B$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19633,
    "subject": "Mathematics (Olympiad)",
    "question": "假設存在某個正整數 $n$ ($n \\ge 8$)，使得 $n^2 - n + 1$ 的每個質因數都是 $$(n+2)(n+1)n(n-1)$$ 的質因數，且 $(n+2)(n+1)n(n-1)$ 恰有四個質因數。證明不存在這樣的 $n$。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $n^2 - n + 1$ 的每個質因數都屬於 $(n+2)(n+1)n(n-1)$ 的質因數集合，且 $(n+2)(n+1)n(n-1)$ 恰有四個質因數。由 $(n^2 - n + 1, n(n+1)) = 1$，又 $n^2 - n + 1 = (n+2)(n-3) + 7$，可得\n\n$$(n^2 - n + 1, n + 1) = 1 \\text{ 或 } 3, \\quad (n^2 - n + 1, n + 2) = 1 \\text{ 或 } 7.$$ \n\n因此 $n^2 - n + 1 = 3^a 7^b$，其中 $a, b$ 為 $0$ 或正整數。又 $9 \\nmid (n^2 - n + 1)$，故 $a \\in \\{0,1\\}$，且 $b > 0$。假設 $n+2, n+1, n, n-1$ 的質因數為 $2, 3, 7, p$（$p \\ne 2,3,7$），則 $7|(n+2)$。\n\n考慮兩個偶數、兩個奇數的質因數集合 $A, B$。顯然 $2 \\in A$，$|B| \\ge 2$，$A \\cap B \\subseteq \\{3\\}$。若 $A = \\{2,3\\}$ 或 $\\{2,7\\}$，則兩個偶數為 $2^{c+1}, 2 \\times 3^d$ 或 $2^{c+1}, 2 \\times 7^d$，得 $|2^c - 3^d| = 1$ 或 $|2^c - 7^d| = 1$。\n\n這兩個偶數為 $16, 18$ 或 $16, 14$。前者 $7 \\nmid (n+2)$，後者 $(n+2)(n+1)n(n-1)$ 有質因數 $2, 3, 5, 7$ 及 $13$（或 $17$），矛盾。\n\n若 $A = \\{2,p\\}$，則 $n+2$ 為奇數，$n-1$ 為偶數。由 $3 \\notin A \\Rightarrow 3 \\nmid (n-1) \\Rightarrow 3 \\nmid (n-2)$。故 $n+2 = 7^c, n = 3^d$，且 $2^e \\in \\{n+1, n-1\\}$（$c,d,e$ 為 $\\ge 2, \\ge 2, \\ge 3$）。從而 $|3^d - 2^e| = 1 \\Rightarrow (d,e) = (2,3)$，即 $n=9$，但 $n+2=11 \\ne 7^c$，矛盾。\n\n若 $A = \\{2,3,7\\}$，則 $B = \\{3,p\\}$，且 $n+2$ 為偶數，$(n+2, n-1) = 3$。故 $2 \\times 3 \\times 7|(n+2)$。從而 $n = 2^c, n-1 = 3^d, n+1 = p^e$（$c \\ge 3, d \\ge 2$）。於是 $2^c - 3^d = 1 \\Rightarrow (c,d) = (2,1)$，矛盾。\n\n若 $A = \\{2,3,p\\}$，則 $B = \\{3,7\\}$，且 $n+2$ 為奇數，$(n+2, n-1) = 3$。故 $3 \\times 7|(n+2)$，但 $(n, n+2) = 1$，則 $n$ 的奇質因數不是 $3,7$，矛盾。\n\n因此不存在滿足條件的 $n$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19634,
    "subject": "Mathematics (Olympiad)",
    "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 with $p > 1$. 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": "See solution",
    "solution": "a) Since $m^2 - m + 1 = (m - 1)^2 + (m - 1) + 1$ and $m - 1 \\geq 0$, 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 $k \\geq 2$. Now,\n$$\np^2 + p + 1 = k^4 + k^2 + 1 = (k^2 + 1)^2 - k^2 = (k^2 - k + 1)(k^2 + k + 1).\n$$\nNumbers $r = k$ and $q = k - 1$, both positive integers, satisfy the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19635,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c(O, R)$ be a circle with diameter $AB$ and $C$ a point on it different from $A$ and $B$ such that $\\angle AOC > 90^\\circ$. On the radius $OC$ we consider the point $K$ and the circle $c_1$ with center $K$ and radius $KC = R_1$. We draw the tangents $AD$ and $AE$ from $A$ to the circle $c_1$. Prove that the straight lines $AC$, $BK$, and $DE$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lines $DE$ and $CA$ meet at point $L$. We will prove that the line $BK$ passes through $L$ (see figure 1).\n\nThe circle $c(O, R)$ is homothetic to the circle $c_1(K, R_1)$ with respect to homothety with center $A$ and ratio $m = \\frac{R}{R_1}$, say $H(A, \\frac{R}{R_1})$. The extension of $CD$ meets the circle $c$ at a point $D_1$ homothetic to $D$. The extension of $CE$ meets the circle $c$ at a point $E_1$ homothetic to $E$.\n\nTherefore, the line segment $CE_1$ is homothetic to the line segment $CE$. So, if the line $AC$ intersects $D_1E_1$ at the point $L_1$, then $L_1$ will be homothetic to $L$. Since $O$ is homothetic to $K$, we conclude that\n\n$$\nOL_1 \\parallel KL. \\quad (1)\n$$\n\nWe will prove that\n\n$$\nOL_1 \\parallel BL. \\quad (1)\n$$\n\nSince $AD$ and $AE$ are tangents from $A$ to the circle $c_1$, then $AK$ is the perpendicular bisector of the segment $DE$. Let $M$ be the intersection point of the lines $AK$ and $DE$, such that the extension of $CM$ intersects $D_1E_1$ at $M_1$ and the circle $c$ at point $M_2$. Then $M_1$ will be the midpoint of the segment $D_1E_1$ (because of the homothety).\n\nWe assert that $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$. According to Steiner's theorem on symmedians, it is enough to prove that\n\n$$\n\\frac{DL}{LE} = \\frac{CD^2}{CE^2}. \\quad (3)\n$$\n\nFor proving the relation (3) we use the areas ratio:\n\n$$\n\\frac{\\sigma(CDL)}{\\sigma(CEL)} = \\frac{DL}{LE} = \\frac{\\sigma(DAL)}{\\sigma(EAL)} = \\frac{\\sigma(CDL) + \\sigma(DAL)}{\\sigma(CEL) + \\sigma(EAL)} = \\frac{\\sigma(CAD)}{\\sigma(CAE)}. \\quad (4)\n$$\n\nSince the angles $ADE$ and $AED$ are the angles between tangents and chord, we have $\\angle ADE = \\angle AED = \\angle DCE$ and therefore\n\n$$\n\\angle CDA = \\angle CDE + \\angle ADE = \\angle CDE + \\angle DCE = 180^\\circ - \\angle CED,\n$$\n\n$$\n\\angle CEA = \\angle CED + \\angle AED = \\angle CED + \\angle DCE = 180^\\circ - \\angle CDE.\n$$\n\nFrom (4) we obtain\n\n$$\n\\frac{DL}{LE} = \\frac{\\sigma(CAD)}{\\sigma(CAE)} = \\frac{CD \\cdot \\sin(180^\\circ - \\angle CED)}{CE \\cdot \\sin(180^\\circ - \\angle CDE)} = \\frac{CD \\cdot \\sin(\\angle CED)}{CE \\cdot \\sin(\\angle CDE)} = \\frac{CD^2}{CE^2}.\n$$\n\n![](images/shortlistBMO2010_p18_data_2d1676409b.png)\n\nSo, the relation (3) is proved and $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$.\n\nHence $\\angle D_1CA = \\angle E_1CM_2$ and the quadrilateral $AD_1E_1M_2$ is an isosceles trapezium. The line $OM_1$ is perpendicular to $D_1E_1$ and intersects $AM_2$ at the midpoint $N$. In the triangle $AMM_2$ we have that $N$ is the midpoint of the side $AM_2$ and $NM_1 \\parallel AM$. Hence $M_1$ is the midpoint of $MM_2$ and therefore $D_1E_1$ is the mid-parallel of $AM_2$ and $DE$. Since $L_1$ belongs to $D_1E_1$, it will be the midpoint of $AL$. In the triangle $ALB$, $OL_1$ is the mid-parallel to $BL$. Hence $OL_1 \\parallel BL$ and the straight lines $AC$, $BK$, and $DE$ are concurrent. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19636,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $k$ such that $30$ divides $k^2 + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "An integer is divisible by $30$ if and only if it is divisible by $2$, $3$, and $5$.\n\nNote that $2 \\mid k^2 + 1$ if and only if $k$ is odd. Thus, we may assume $k$ is odd, so write $k = 2t + 1$.\n\n**Divisibility by 3:**\n- If $k \\equiv 0$ or $1 \\pmod{3}$, then $3 \\nmid k^2 + 1$.\n- If $k \\equiv 2 \\equiv -1 \\pmod{3}$, then $k^2 + 1 \\equiv (-1)^2 + 1 \\equiv 2 \\pmod{3}$, so $3 \\nmid k^2 + 1$.\n\n**Divisibility by 5:**\n- If $k \\equiv 0$ or $1 \\pmod{5}$, then $5 \\nmid k^2 + 1$.\n- If $k \\equiv 2$ or $3 \\pmod{5}$, then $k \\equiv \\pm 2 \\pmod{5}$, so\n  $$\nk^2 + 1 \\equiv (\\pm 2)^2 + 1 \\equiv 4 + 1 \\equiv 5 \\equiv 0 \\pmod{5}.\n  $$\n- If $k \\equiv 4 \\equiv -1 \\pmod{5}$, then $k^2 + 1 \\equiv (-1)^2 + 1 \\equiv 2 \\pmod{5}$, so $5 \\nmid k^2 + 1$.\n\nCombining all conditions, $30 \\mid k^2 + 1$ if and only if $k$ is odd, $k \\equiv 5 \\pmod{6}$, and $k \\equiv 9 \\pmod{10}$. The solution is $k = 30n + 29$, where $n = 0, 1, 2, \\ldots$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19637,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume, for the sake of contradiction, that there exists an anti-Pascal triangle $T$ with 2018 rows, containing all the numbers from $1$ to $1 + 2 + \\dots + 2018$ (i.e., the first 2018 positive integers summed). Prove that such a triangle does not exist.",
    "options": [],
    "answer": "See solution",
    "solution": "Such a proposed anti-Pascal triangle with 2018 rows does not exist!\n\nWe start with the following observations. For any positive integer $x$ in an anti-Pascal triangle $T$ that is not in the bottom row, the two numbers below it are $y$ and $x + y$ for some positive integer $y$. Thus, starting from any integer $x_1$ in $T$, we may generate the sequences $x_1, x_2, x_3, \\dots$ and $y_2, y_3, \\dots$ of positive integers inductively as follows.\n\nWhenever $x_n$ is not in the bottom row of $T$, let $x_{n+1}$ and $y_{n+1}$ be the larger and smaller of the two numbers, respectively, below $x_n$ in $T$. (*)\n\nThus $x_{n+1} = x_n + y_{n+1}$ for each positive integer $n$. From this it easily follows that\n\n$$\nx_n = x_1 + y_2 + y_3 + \\dots + y_n\n$$\n\nfor each positive integer $n$.\n\nFor example, in the following four-row anti-Pascal triangle, starting with $x_1 = 4$ we generate $(x_1, x_2, x_3, x_4) = (4, 6, 7, 10)$ and $(y_2, y_3, y_4) = (2, 1, 3)$.\n\n![](images/2018-Australian-Scene-W2_p174_data_4b2296275c.png)\n\nAssume, for the sake of contradiction, the existence of an anti-Pascal triangle $T$ that has 2018 rows and which contains the numbers from $1$ to $1 + 2 + \\dots + 2018$.\n\nLet $x_1$ be the top number in $T$, and consider the sequences $x_1, x_2, x_3, \\dots, x_{2018}$ and $y_2, y_3, \\dots, y_{2018}$ generated from $x_1$ as described at (*). We have\n\n$$\nx_{2018} = x_1 + y_2 + y_3 + \\dots + y_{2018}.\n$$\n\nSince $x_1, y_2, y_3, \\dots, y_{2018}$ are distinct positive integers, it follows that\n\n$$\nx_{2018} \\ge 1 + 2 + \\dots + 2018.\n$$\n\nBut the largest number in $T$ is $1 + 2 + \\dots + 2018$. Thus $x_{2018} = 1 + 2 + \\dots + 2018$ and $(x_1, y_2, \\dots, y_{2018})$ is a permutation of $(1, 2, \\dots, 2018)$.\n\nNext, let $T_1$ and $T_2$ be the sub-equilateral triangular arrays whose bases consist of the 1009 leftmost numbers and 1009 rightmost numbers, respectively, in the bottom row of $T$. Note that $T_1$ and $T_2$ are disjoint and that their union covers the entire bottom row. Without loss of generality, we may suppose that $x_{2018}$ lies outside $T_1$. It follows that all of $x_{2017}, x_{2016}, \\dots, x_1$ lie outside $T_1$.\n\nLet $T_1'$ be the sub-equilateral triangular array whose base consists of the 1008 leftmost numbers in the bottom row of $T$. Since $y_n$ is adjacent to $x_n$ in each row of the triangle, and each $x_n$ lies outside of $T_1$, it follows that $T_1'$ contains no $x_n$ and no $y_n$. In particular, none of the numbers $1, 2, \\dots, 2018$ lies inside $T_1'$.\n\n![](images/2018-Australian-Scene-W2_p175_data_450130d8cf.png)\n\nLet $x_1'$ be the top number in $T_1'$. Generate the sequences $x_1', x_2', x_3', \\dots, x_{1008}'$ and $y_2', y_3', \\dots, y_{1008}'$ as described at (*). These numbers all lie in $T_1'$. Since $T_1'$ does not contain any of $1, 2, \\dots, 2018$, it follows that\n\n$$\n\\begin{aligned}\nx_{1008}' &= x_1' + y_2' + y_3' + \\dots + y_{1008}' \\\\\n&\\ge 2019 + 2020 + \\dots + 3026 \\\\\n&= 1008 \\times \\frac{1}{2}(2019 + 3026) \\\\\n&\\gg 1009 \\times 2019 \\\\\n&= 1 + 2 + \\dots + 2018.\n\\end{aligned}\n$$\n\nThis is an obvious contradiction because the numbers in $T$ are less than or equal to $1 + 2 + \\dots + 2018$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19638,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $v_p(n)$ denote the exponent of the prime $p$ in the prime factorization of $n$.\n\nSuppose we have the numbers $1, 2, \\ldots, n$ written on a blackboard. At each step, you may replace any number $k$ with $k!$ (the factorial of $k$). You may perform this operation any number of times, possibly on different numbers. After all operations, you multiply all the numbers on the board to obtain a final product $S'$. For which positive integers $n \\geq 2$ is it possible, by choosing which numbers to replace by their factorials, to make $S'$ a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = n!$ be the initial product. We want to choose, for each $k$ ($1 \\leq k \\leq n$), either $f_k = k$ or $f_k = k!$, so that $S' = f_1 f_2 \\cdots f_n$ is a perfect square.\n\nRecall that $S'$ is a perfect square if and only if $v_p(S')$ is even for all primes $p$.\n\n**Case 1: $n$ is prime.**\n\nFor $n = p$ prime, $v_p(S) = v_p(p!) = 1$, and adding factorials does not change the exponent of $p$ in the product, so $v_p(S') = 1$ (odd). Thus, $S'$ cannot be a perfect square.\n\n**Case 2: $n$ is composite ($n \\geq 4$).**\n\nWe can choose the $f_k$ as follows: For each prime $p < n$, consider the product $p \\cdot f_{p+1} \\cdots f_n$. We can choose $f_{p+1}$ to ensure that the total exponent of $p$ in $S'$ is even, by checking the parity of $v_p(f_{p+2} \\cdots f_n)$ and picking $f_{p+1} = p+1$ or $f_{p+1} = (p+1)!$ accordingly. Proceeding from the largest such $p$ down to $p = 2$, we can always make all $v_p(S')$ even.\n\n**Conclusion:**\n\nAll composite $n \\geq 4$ (and $n = 2$ is a special case) allow such a choice, but for prime $n$ it is impossible. Thus, the desired $n \\geq 2$ are all composite numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19639,
    "subject": "Mathematics (Olympiad)",
    "question": "Может ли для некоторого $m > 1$ сумма цифр числа $8^m$ быть равна 8, если оно оканчивается на 6?",
    "options": [],
    "answer": "See solution",
    "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 нечётный множитель, больший 1. Остаётся рассмотреть случаи $k = 2, 3, 4$: $10^2 + 16 = 4 \\cdot 29$; $10^3 + 16 = 8 \\cdot 127$; $10^4 + 16 = 32 \\cdot 313$. Таким образом, $10^k + 16$ не является степенью восьмёрки ни при каком натуральном $k$, что и требовалось доказать.\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, $8^2$ — на 4, $8^3$ — на 2, $8^4$ — на 6, $8^5$ — на 8. Далее последняя цифра степени восьмёрки 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, $8^2$ — остаток 1, $8^3$ — остаток 2, и далее остатки степеней восьмёрки повторяются с периодом 2. Таким образом, $8^m$ имеет остаток 2 при делении на 3 тогда и только тогда, когда $m$ нечётно. Это противоречит тому, что $m$ должно делиться на 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19640,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose Freyja is given an $m \\times n$ grid, and she can ask questions about the sums of numbers in rectangles of certain sizes. Under what conditions can Freyja determine the labeling of the grid, and what is the minimum number of questions required for her to do so?",
    "options": [],
    "answer": "See solution",
    "solution": "If $\\gcd(m, n) > 1$, then Freyja cannot win. If $\\gcd(m, n) = 1$, then Freyja can win in a minimum of $(m-1)^2 + (n-1)^2$ questions.\n\nFirst, consider $\\gcd(m, n) > 1$. Let $d = \\gcd(m, n)$. Any labeling where each $1 \\times d$ rectangle has sum zero is valid, so Freyja would need to ask at least one question in every row, which is not possible in finitely many questions.\n\nNow suppose $\\gcd(m, n) = 1$. We split the proof into two parts.\n\n**Lower bound:** Any labeling where each $m \\times 1$ and $1 \\times m$ rectangle has sum zero is valid. These labelings form a vector space of dimension $(m-1)^2$. Similarly, labelings where each $n \\times 1$ and $1 \\times n$ rectangle has sum zero form a space of dimension $(n-1)^2$. The only labeling in both spaces is the all-zero labeling. Thus, the space of valid labelings has dimension $(m-1)^2 + (n-1)^2$, so Freyja needs at least that many questions.\n\n**Upper bound (using generating functions):**\n\nAny valid labeling is doubly periodic with period $mn$. Consider the generating function\n\n$$\nf(x, y) = \\sum_{a=0}^{mn-1} \\sum_{b=0}^{mn-1} c_{a,b} x^a y^b\n$$\n\nwhere $c_{a,b}$ is the number in $(a, b)$. The constraints for $f$ to be valid are equivalent to\n\n$$\nf(x, y) \\cdot \\frac{x^m - 1}{x - 1} \\cdot \\frac{y^n - 1}{y - 1} \\quad \\text{and} \\quad f(x, y) \\cdot \\frac{x^n - 1}{x - 1} \\cdot \\frac{y^m - 1}{y - 1}\n$$\n\nbeing zero modulo $x^{mn} - 1$ and $y^{mn} - 1$. Let $\\omega = \\exp(2\\pi i/m)$. We need\n\n$$\nf(\\omega^a, \\omega^b) \\cdot \\frac{\\omega^{am} - 1}{\\omega^a - 1} \\cdot \\frac{\\omega^{bn} - 1}{\\omega^b - 1} = f(\\omega^a, \\omega^b) \\cdot \\frac{\\omega^{an} - 1}{\\omega^a - 1} \\cdot \\frac{\\omega^{bm} - 1}{\\omega^b - 1} = 0\n$$\n\nfor all $a, b \\in \\{0, \\dots, mn - 1\\}$. This implies $f(\\omega^a, \\omega^b) = 0$ for most $(a, b)$. The only nonzero cases occur when $n \\mid a$ and $n \\mid b$ (giving $(m-1)^2$ cases), or $m \\mid a$ and $m \\mid b$ (giving $(n-1)^2$ cases). Thus, the dimension of the space of valid labelings is at most $(m-1)^2 + (n-1)^2$.\n\n![](images/sols-TST-IMO-2023_p15_data_2538752e8a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19641,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_{2023}$ be positive real numbers. Define\n\n$$\na_n = \\sqrt{(x_1 + x_2 + \\cdots + x_n) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n} \\right)}\n$$\nfor $n = 1, 2, \\dots, 2023$. Prove that $a_{2023} \\ge 3034$.",
    "options": [],
    "answer": "See solution",
    "solution": "To start with, observe that since $x_i > 0$, for any integer $n \\in \\{1, 2, \\dots, 2022\\}$, we have\n\n$$\n\\begin{aligned}\na_{n+1} &= \\sqrt{(x_1 + x_2 + \\cdots + x_n + x_{n+1}) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n} + \\frac{1}{x_{n+1}} \\right)} \\\\\n&> \\sqrt{(x_1 + x_2 + \\cdots + x_n) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n} \\right)} \\\\\n&= a_n.\n\\end{aligned}\n$$\n\nSuppose, for the sake of contradiction, that $a_{n+2} - a_n \\le 2$ for some $n \\in \\{1, 2, \\dots, 2021\\}$. Since $a_n < a_{n+1} < a_{n+2}$ are integers, it follows that $a_{n+1} = 1 + a_n$ and $a_{n+2} = 2 + a_n$.\n\nLet $x = x_1 + x_2 + \\cdots + x_n$, $y = \\frac{1}{x_1} + \\frac{1}{x_2} + \\cdots + \\frac{1}{x_n}$, $x_{n+1} = a$, $x_{n+2} = b$ and $k = a_n$. Then\n\n$$\nk = \\sqrt{xy} \\qquad (1)\n$$\n\n$$\nk + 1 = a_{n+1} = \\sqrt{(x + a) \\left(y + \\frac{1}{a}\\right)} \\qquad (2)\n$$\n\n$$\nk + 2 = a_{n+2} = \\sqrt{(x + a + b) \\left(y + \\frac{1}{a} + \\frac{1}{b}\\right)} \\qquad (3)\n$$\n\nComputing the square of (3) minus the square of (2) yields\n\n$$\n\\begin{aligned}\n2k + 3 &= (x + a + b) \\left(y + \\frac{1}{a} + \\frac{1}{b}\\right) - (x + a) \\left(y + \\frac{1}{a}\\right) \\\\\n&= \\frac{x}{b} + \\frac{a}{b} + by + \\frac{b}{a} + 1 \\\\\n\\Rightarrow \\quad 2k + 2 &= \\frac{x}{b} + by + \\frac{a}{b} + \\frac{b}{a} \\\\\n&\\ge 2\\sqrt{\\frac{x}{b} \\cdot by} + 2\\sqrt{\\frac{a}{b} \\cdot \\frac{b}{a}} \\quad \\text{(AM-GM)} \\\\\n&= 2\\sqrt{xy} + 2 \\\\\n&= 2k + 2 \\qquad \\text{(using (1))}.\n\\end{aligned}\n$$\n\nThus equality holds throughout. In particular equality holds in each AM-GM. Therefore $\\frac{a}{b} = \\frac{b}{a}$, and so $a = b$. This is a contradiction due to $a = x_{n+1} \\ne x_{n+2} = b$.\n\nWe have shown that $a_{n+2} - a_n \\ge 3$ for all positive integers $n$. Adding these inequalities for $n = 1, 3, 5, \\dots, 2021$, noting that it is a telescoping sum, yields $a_{2023} - a_1 \\ge 3 \\times 1011$. Since $a_1 = 1$, this implies $a_{2023} \\ge 3034$, as desired. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19642,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and $q$ such that\n\n$$p^2 \\mid q^3 + 1 \\quad \\text{and} \\quad q^2 \\mid p^6 - 1.$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $p = 3$, then $q^2 \\mid 3^6 - 1 = 728 = 2^3 \\cdot 7 \\cdot 11$, so $q = 2$, which gives a solution.\n\nLet $p \\neq 3$. Since $(q+1, q^2 - q + 1) = 1$ or $3$, we have $p^2 \\mid q+1$ or $p^2 \\mid q^2 - q + 1$, which implies $p < q$.\n\nIf $p+1 = q$, then $p = 2$ and $q = 3$, which is another solution. In the sequel, assume $q \\geq p+2$.\n\nSince $q^2 \\mid p^6 - 1 = (p-1)(p+1)(p^2 - p + 1)(p^2 + p + 1)$ and $(q, p-1) = (q, p+1) = 1$, we have $q^2 \\mid (p^2 - p + 1)(p^2 + p + 1)$. Moreover, $(p^2 - p + 1, p^2 + p + 1) = (p^2 + p + 1, 2p) = 1$, so $q^2 \\mid p^2 - p + 1$ or $q^2 \\mid p^2 + p + 1$.\n\nHowever, $p^2 - p + 1 < p^2 < q^2$ and $(p+2)^2 \\leq q^2 \\leq p^2 + p + 1$, i.e., $3p + 3 < 0$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19643,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$. Let $R_A$, $R_B$, and $R_C$ be the radii of the circumcircles of triangles $BIC$, $CIA$, and $AIB$, respectively, and $R$ be the radius of the circumcircle of triangle $ABC$. Prove that\n\n$$\nR_A + R_B + R_C \\le 3R.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|BC| = a$, $|CA| = b$, and $|AB| = c$, and let the angles opposite to those sides be $\\alpha$, $\\beta$, and $\\gamma$, respectively.\n\nThe law of sines in triangles $ABC$ and $IBC$ gives $\\frac{a}{\\sin \\alpha} = 2R$ and $\\frac{a}{\\sin \\left(\\frac{\\beta}{2} + \\frac{\\gamma}{2}\\right)} = 2R_A$.\n\n![](images/prob1314_p23_data_c66e5bc8da.png)\n\nAs $\\sin\\left(\\frac{\\beta}{2} + \\frac{\\gamma}{2}\\right) = \\sin\\left(90^\\circ - \\frac{\\alpha}{2}\\right) = \\cos\\frac{\\alpha}{2}$, we obtain\n\n$$\n\\frac{R_A}{R} = \\frac{\\sin\\alpha}{\\cos\\frac{\\alpha}{2}} = 2\\sin\\frac{\\alpha}{2}.\n$$\n\nSimilarly, $\\frac{R_B}{R} = 2\\sin\\frac{\\beta}{2}$ and $\\frac{R_C}{R} = 2\\sin\\frac{\\gamma}{2}$.\n\nTo solve the problem, we need to prove that\n\n$$\n\\frac{R_A}{R} + \\frac{R_B}{R} + \\frac{R_C}{R} \\le 3,\n$$\n\nor equivalently,\n\n$$\n\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} + \\sin \\frac{\\gamma}{2} \\le \\frac{3}{2}. \\qquad (6)\n$$\n\nApplying Jensen's inequality gives\n\n$$\n\\frac{1}{3} \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} + \\sin \\frac{\\gamma}{2} \\right) \\le \\sin \\left( \\frac{\\frac{\\alpha}{2} + \\frac{\\beta}{2} + \\frac{\\gamma}{2}}{3} \\right) = \\sin \\left( \\frac{\\alpha + \\beta + \\gamma}{6} \\right) = \\sin \\frac{\\pi}{6} = \\frac{1}{2},\n$$\n\nwhich directly implies the necessary result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19644,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that every prime factor of $2^n - 1$ is at most $7$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer such that $2^n - 1$ has only prime factors at most $7$. That is, $2^n - 1 = 3^a 5^b 7^c$ for some non-negative integers $a, b, c$.\n\n**Case 1:** $b \\geq 1$ (i.e., $5$ divides $2^n - 1$)\n\nIf $5 \\mid 2^n - 1$, then $4 \\mid n$. Let $n = 4k$. Then:\n$$\n2^n - 1 = 2^{4k} - 1 = (4^k + 1)(4^k - 1)\n$$\nConsider $4^k + 1$. Modulo $3$, $4^k + 1 \\equiv 2 \\pmod{3}$, and modulo $7$, $4^k + 1 \\equiv 2, 3, 5 \\pmod{7}$. Thus, $4^k + 1$ can only be a power of $5$, say $4^k + 1 = 5^d$ for some $d \\geq 1$.\n\nIf $k \\geq 2$, then $16 \\mid 5^d - 1$, so $4 \\mid d$. Let $d = 4e$ ($e \\geq 1$). Then:\n$$\n4^k = 5^d - 1 = (25^e + 1)(25^e - 1)\n$$\nBut $\\gcd(25^e + 1, 25^e - 1) \\leq 2$, so both cannot be multiples of $4$. Thus, this is impossible for $k \\geq 2$. Therefore, $k = 1$, so $n = 4$.\n\n**Case 2:** $b = 0$\n\n- If $a = 0$, then $2^n - 1 = 7^c$. Since $7^c \\equiv 1, 7 \\pmod{16}$, $n < 4$. Checking $n = 1, 2, 3$:\n  - $n = 1$: $2^1 - 1 = 1$\n  - $n = 2$: $2^2 - 1 = 3$\n  - $n = 3$: $2^3 - 1 = 7$\n  So $n = 1, 3$ work.\n\n- If $c = 0$ and $a \\geq 1$, then $2^n - 1 = 3^a$. Since $3 \\mid 2^n - 1$, $n = 2t$ for some $t$. Then:\n  $$(2^t + 1)(2^t - 1) = 3^a$$\n  But $\\gcd(2^t + 1, 2^t - 1) \\leq 2$, so $2^t - 1 = 1$ is the only option, i.e., $t = 1$, $n = 2$.\n\n- If $a, c \\geq 1$, then $3 \\mid 2^n - 1$ and $7 \\mid 2^n - 1$, so $2 \\mid n$ and $3 \\mid n$, i.e., $n = 6r$ for some $r \\geq 1$.\n  $$\n  2^n - 1 = 2^{6r} - 1 = (8^r + 1)(8^r - 1)\n  $$\n  Since $\\gcd(8^r + 1, 8^r - 1) = 1$ and $8^r + 1$ is not divisible by $7$, $8^r + 1 = 3^a$, $8^r - 1 = 7^c$.\n  Since $8 \\mid 3^a - 1$, $a = 2a'$ for some $a'$. Then:\n  $$\n  8^r = (3^{a'} - 1)(3^{a'} + 1)\n  $$\n  But $\\gcd(3^{a'} - 1, 3^{a'} + 1) \\leq 2$, so $3^{a'} - 1 \\leq 2$, i.e., $a' = 1$. Thus, $n = 6$, $(a, c) = (2, 1)$.\n\n**Conclusion:**\n\nThe possible values are $n = 1, 2, 3, 4, 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19645,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k > 5$ be an integer. Replace a given positive integer by the product of the sum of its digits in base $k$ and $(k-1)^2$. Repeat this process with each new number. Prove that the sequence of numbers obtained becomes constant from some point onwards.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the sum of digits of an integer divisible by $k-1$ is also divisible by $k-1$, $(k-1)^3$ divides all numbers obtained after the second step. \n\nIf $a = \\overline{a_n a_{n-1} \\dots a_0}_{(k)}$, with $n \\geq 4$ or $a_3 \\geq 2$ for $n = 3$, then\n\n$$\n\\begin{align*}\na - (k-1)^2 (a_n + a_{n-1} + \\dots + a_0) &\\geq 2(k^3 - (k-1)^2) - a_1((k-1)^2-k) - a_0((k-1)^2-1) \\\\\n&\\geq 2(k^3 - (k-1)^2) - (k-1)(2(k-1)^2 - k - 1) = 5k^2 - 2k - 1 > 0.\n\\end{align*}\n$$\n\nThis shows that we eventually get a number of the form $a = \\overline{a_3 a_2 a_1 a_0}_{(k)}$, where $a_3 = 0$ or $1$. The next number is\n\n$$\n(k-1)^2(a_3 + a_2 + a_1 + a_0) \\leq (k-1)^2(1 + 3(k-1)) < 4(k-1)^3.\n$$\n\nHence, this number is $(k-1)^3$, $2(k-1)^3$, or $3(k-1)^3$. Note that\n\n$$\n(k-1)^3 = \\overline{k-3,2,k-1}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 2,\n$$\n$$\n2(k-1)^3 = \\overline{1,k-6,5,k-2}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 5,\n$$\n$$\n3(k-1)^3 = \\overline{2,k-9,8,k-3}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 8.\n$$\n\nFor example, $3 \\cdot 5^3 = \\overline{1423}_{(6)} \\rightarrow 10 \\cdot 5^2 = 2 \\cdot 5^3$, $3 \\cdot 6^3 = \\overline{1614}_{(7)} \\rightarrow 12 \\cdot 6^2 = 2 \\cdot 6^3$, and $3 \\cdot 7^3 = \\overline{2005}_{(8)} \\rightarrow 7 \\cdot 7^2 = 7^3 \\rightarrow 2 \\cdot 7^3$. Thus, for $k > 5$, the numbers become equal to $2(k-1)^3$ from some point onwards.\n\n_Remark_: For $k=2$, the numbers become 1 from some point onwards. For $k=3, 4, 5$:\n\n$$\n3 \\cdot 2^3 \\rightarrow 2 \\cdot 2^3 \\rightarrow 2 \\cdot 2^3, \\quad 3 \\cdot 3^3 \\rightarrow 3^3 \\rightarrow 2 \\cdot 3^3 \\rightarrow 2 \\cdot 3^3, \\quad 3 \\cdot 4^3 \\rightarrow 2 \\cdot 4^3 \\rightarrow 4^3 \\rightarrow 2 \\cdot 4^3,\n$$\n\nrespectively. Hence, for $k > 2$, $k \\neq 5$, the numbers become $2(k-1)^3$ from some point onwards, while for $k=5$ the numbers $4^3$ and $2 \\cdot 4^3$ alternate.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19646,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in non-negative integers the equation:\n\n$$\n2^a \\cdot 3^b + 9 = c^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the equation as:\n\n$$\n2^a \\cdot 3^b = c^2 - 9 = (c-3)(c+3)\n$$\n\n**Case 1:** $b = 0$\n\nThen $2^a = (c-3)(c+3)$. Trying small values, $a = 4$ gives $2^4 = 16 = (5-3)(5+3) = 2 \\times 8$, so $a = 4, b = 0, c = 5$ is a solution.\n\n**Case 2:** $b > 0$\n\nThen $3$ divides $c^2$, so $b = 1$ or higher. Let $c = 3y$.\n\nSubstitute: $2^a \\cdot 3^b + 9 = (3y)^2 \\implies 2^a \\cdot 3^b = 9y^2 - 9 = 9(y^2 - 1)$, so $2^a \\cdot 3^{b-2} = y^2 - 1$.\n\n- If $a = 0$, $3^{b-2} = y^2 - 1$. For $b = 3$, $3^{1} = 2^2 - 1 = 3$, so $a = 0, b = 3, c = 6$ is a solution.\n\n- For $a \\geq 2$, $y$ is odd, so $y^2 - 1$ factors as $(y-1)(y+1)$. Thus,\n\n  $$\n  2^{a-2} \\cdot 3^{b-2} = \\frac{y-1}{2} \\cdot \\frac{y+1}{2}\n  $$\n\n  The two factors are consecutive integers, so each must be a power of $2$ or $3$.\n\nLet $m = a-2$, $n = b-2$.\n\n- If $2^m - 3^n = 1$:\n  - $m > n$. If $n = 0$, $m = 1$, so $a = 3, b = 2, c = 9$.\n  - For $m$ even, $m = 2t$. Then $(2^t - 1)(2^t + 1) = 3^n$. For $t = 1$, $m = 2, n = 1$, so $a = 4, b = 3, c = 21$.\n\n- If $3^n - 2^m = 1$:\n  - $m > 0$. If $m = 1$, $n = 1$, so $a = 3, b = 3, c = 15$.\n  - For $n$ even, $n = 2t$. Then $(3^t - 1)(3^t + 1) = 2^m$. For $t = 1$, $n = 2, m = 3$, so $a = 5, b = 4, c = 51$.\n\n**All solutions:**\n- $(a, b, c) = (4, 0, 5)$\n- $(a, b, c) = (0, 3, 6)$\n- $(a, b, c) = (3, 2, 9)$\n- $(a, b, c) = (4, 3, 21)$\n- $(a, b, c) = (3, 3, 15)$\n- $(a, b, c) = (5, 4, 51)$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19647,
    "subject": "Mathematics (Olympiad)",
    "question": "Amy and Bob play a game. At the beginning, Amy writes down a positive integer on the board. Then the players take turns, with Bob moving first. On his move, Bob replaces the number $n$ on the board with a number of the form $n - a^2$, where $a$ is a positive integer. On her move, Amy replaces the number $n$ on the board 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 from winning?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is negative; Amy cannot prevent Bob from winning.\n\nFor a positive integer $n$, define its *square-free part* $S(n)$ as the smallest positive integer $a$ such that $n/a$ is a perfect square. That is, $S(n)$ is the product of all primes with odd exponents in the prime factorization of $n$. Set $S(0) = 0$.\n\n**(i)** On any of Amy's moves, the square-free part does not increase: $S(n^k) = S(n)$ if $k$ is odd, and $S(n^k) = 1 \\leq S(n)$ if $k$ is even.\n\n**(ii)** On any of Bob's moves, if the board shows $n = S(n) \\cdot b^2$, Bob can replace it with $n' = n - b^2 = (S(n) - 1)b^2$, so $S(n') \\leq S(n) - 1$.\n\nThus, starting from a positive integer $N$, Bob can win in at most $S(N)$ moves.\n\n*Remarks:*\n\n1. Bob may restrict himself to subtracting only numbers divisible by the maximal square dividing the current number. Then, one may replace any number $n$ on the board by $S(n)$, omitting square factors. Amy's moves do not increase $S(n)$, while Bob's moves decrease it, so Bob wins.\n\n2. In fact, Bob can win in at most 4 moves, by Lagrange's four squares theorem: $S(n) = a_1^2 + \\cdots + a_s^2$. On each move, Bob can reduce the number by one square term. Amy can only interrupt by raising the number to an even power, but then Bob wins immediately. Four is the minimum number of moves in which Bob can guarantee a win; for example, if Amy chooses 7 and always uses the first power, Bob needs all four moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19648,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be three positive numbers such that $ab + bc + ca = 3abc$. Prove that\n\n$$\n\\sqrt{\\frac{a+b}{c(a^2+b^2)}} + \\sqrt{\\frac{b+c}{a(b^2+c^2)}} + \\sqrt{\\frac{c+a}{b(c^2+a^2)}} \\le 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing both sides of the constraint by $abc$ gives\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 3.\n$$\n\nConsider the vectors\n\n$$\n\\vec{u} = \\left( \\frac{1}{\\sqrt{c}}, \\frac{1}{\\sqrt{a}}, \\frac{1}{\\sqrt{b}} \\right), \\quad \\vec{v} = \\left( \\sqrt{\\frac{a+b}{a^2+b^2}}, \\sqrt{\\frac{b+c}{b^2+c^2}}, \\sqrt{\\frac{c+a}{c^2+a^2}} \\right).\n$$\n\nApplying Cauchy's inequality:\n\n$$\n(\\vec{u} \\cdot \\vec{v})^2 \\leq \\|\\vec{u}\\|^2 \\cdot \\|\\vec{v}\\|^2,\n$$\n\nyields\n\n$$\n\\left( \\sqrt{\\frac{a+b}{c(a^2+b^2)}} + \\sqrt{\\frac{b+c}{a(b^2+c^2)}} + \\sqrt{\\frac{c+a}{b(c^2+a^2)}} \\right)^2 \n\\leq \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\left( \\frac{a+b}{a^2+b^2} + \\frac{b+c}{b^2+c^2} + \\frac{c+a}{c^2+a^2} \\right).\n$$\n\nMultiplying both sides of $2ab \\leq a^2 + b^2$ by $a+b$ gives\n\n$$\n2ab(a+b) \\leq (a+b)(a^2+b^2),\n$$\n\nso\n\n$$\n\\frac{a+b}{a^2+b^2} \\leq \\frac{a+b}{2ab} = \\frac{1}{2} \\left( \\frac{1}{a} + \\frac{1}{b} \\right).\n$$\n\nSimilarly,\n\n$$\n\\frac{b+c}{b^2+c^2} \\leq \\frac{1}{2} \\left( \\frac{1}{b} + \\frac{1}{c} \\right), \\quad \\frac{c+a}{c^2+a^2} \\leq \\frac{1}{2} \\left( \\frac{1}{c} + \\frac{1}{a} \\right).\n$$\n\nAdding these inequalities:\n\n$$\n\\frac{a+b}{a^2+b^2} + \\frac{b+c}{b^2+c^2} + \\frac{c+a}{c^2+a^2} \\leq \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$\n\nTherefore,\n\n$$\n\\left( \\sqrt{\\frac{a+b}{c(a^2+b^2)}} + \\sqrt{\\frac{b+c}{a(b^2+c^2)}} + \\sqrt{\\frac{c+a}{b(c^2+a^2)}} \\right)^2 \\leq \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)^2 = 9,\n$$\n\nso\n\n$$\n\\sqrt{\\frac{a+b}{c(a^2+b^2)}} + \\sqrt{\\frac{b+c}{a(b^2+c^2)}} + \\sqrt{\\frac{c+a}{b(c^2+a^2)}} \\leq 3.\n$$\n\nEquality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19649,
    "subject": "Mathematics (Olympiad)",
    "question": "Equilateral $\\triangle ABC$ with side length 14 is rotated about its center by angle $\\theta$, where $0 < \\theta \\le 60^\\circ$, to form $\\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\\sqrt{3}$. What is $\\tan \\theta$?\n\n![](images/2024_AMC12B_Solutions_p13_data_3b8f7f949e.png)\n\n(A) $\\frac{3}{5}$  \n(B) $\\frac{5\\sqrt{3}}{11}$  \n(C) $\\frac{4}{5}$  \n(D) $\\frac{11}{13}$  \n(E) $\\frac{7\\sqrt{3}}{13}$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (B):** Let $O$ be the center of $\\triangle ABC$ and $\\triangle DEF$, let $P$ be the foot of the perpendicular from $D$ to $\\overline{AB}$, and let $M$ be the midpoint of $\\overline{AB}$.\n\n![](images/2024_AMC12B_Solutions_p13_data_a068b4171b.png)\n\nThe condition $\\theta \\le 60^\\circ$ implies that $P$ lies on $\\overline{AM}$ (as opposed to lying on $\\overline{BM}$). Furthermore, $O$ is the center of the circle containing points $A, B$, and $D$, so by the Inscribed Angle Theorem $\\angle DOA = 2\\angle DBA$. It suffices to compute $\\tan \\angle DBA$ and then use the Double Angle Formula to obtain $\\tan \\theta$.\n\nThe area of $\\triangle ABC$ is $\\frac{\\sqrt{3}}{4} \\cdot 14^2 = 49\\sqrt{3}$. Because hexagon $ADBECF$ consists of $\\triangle ABC$ plus three copies of $\\triangle ADB$, the area of $\\triangle ADB$ is\n\n$$\n\\frac{91\\sqrt{3} - 49\\sqrt{3}}{3} = 14\\sqrt{3}.\n$$\n\nThis implies that $DP = 2\\sqrt{3}$. Furthermore, $OM = \\frac{7\\sqrt{3}}{3}$ and $OD = OA = 2 \\cdot OM = \\frac{14\\sqrt{3}}{3}$. The Pythagorean Theorem yields\n\n$$\n\\begin{aligned}\nMP &= \\sqrt{OD^2 - (OM + DP)^2} \\\\\n&= \\sqrt{\\left(\\frac{14\\sqrt{3}}{3}\\right)^2 - \\left(\\frac{13\\sqrt{3}}{3}\\right)^2} \\\\\n&= \\sqrt{\\frac{14^2 - 13^2}{3}} = \\sqrt{\\frac{27}{3}} = 3.\n\\end{aligned}\n$$\n\nFinally, $BP = 7 + 3 = 10$, so $\\tan \\angle DBA = \\frac{2\\sqrt{3}}{10} = \\frac{\\sqrt{3}}{5}$ and\n\n$$\n\\tan \\theta = \\tan(2\\angle DOB) = \\frac{\\frac{2\\sqrt{3}}{5}}{1 - \\frac{3}{25}} = \\frac{5\\sqrt{3}}{11}.\n$$\n\n**Note:** The angle of rotation is about $38^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19650,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $a_1, a_2, a_3, \\dots$ defined by $a_1 = 1$ and\n\n$$\na_{m+1} = \\frac{1a_1 + 2a_2 + 3a_3 + \\dots + ma_m}{a_m} \\quad \\text{for } m \\ge 1.\n$$\n\nDetermine the largest integer $n$ such that $a_n < 1\\,000\\,000$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we note that all terms of the sequence are positive rational numbers. Below, we rewrite the defining equation for the sequence in both its original form and with the value of $m$ shifted by 1.\n\n$$\na_m a_{m+1} = 1a_1 + 2a_2 + 3a_3 + \\dots + ma_m \\\\\na_{m+1} a_{m+2} = 1a_1 + 2a_2 + 3a_3 + \\dots + ma_m + (m+1)a_{m+1}\n$$\n\nSubtracting the first equation from the second yields\n\n$$\na_{m+1}a_{m+2} - a_m a_{m+1} = (m+1)a_{m+1} \\implies a_{m+2} - a_m = m+1,\n$$\n\nfor all $m \\ge 1$.\n\nSo for $m = 2k + 1$ an odd positive integer,\n\n$$\na_{2k+1} - a_1 = (a_{2k+1} - a_{2k-1}) + (a_{2k-1} - a_{2k-3}) + \\dots + (a_3 - a_1) = 2k + (2k-2) + \\dots + 2 = 2[k + (k-1) + \\dots + 1] = k(k+1).\n$$\n\nSimilarly, for $m = 2k$ an even positive integer,\n\n$$\na_{2k} - a_2 = (a_{2k} - a_{2k-2}) + (a_{2k-2} - a_{2k-4}) + \\dots + (a_4 - a_2) = (2k-1) + (2k-3) + \\dots + 3 = k^2 - 1.\n$$\n\nUsing $a_1 = 1$ and $a_2 = 1$, we obtain the formula\n\n$$\na_m = \\begin{cases} \\frac{m^2}{4}, & \\text{if } m \\text{ is even} \\\\ \\frac{m^2+3}{4}, & \\text{if } m \\text{ is odd} \\end{cases}\n$$\n\nIf $m$ is odd, then\n\n$$\na_{m+1} - a_m = \\frac{(m+1)^2}{4} - \\frac{m^2+3}{4} = \\frac{2m-2}{4},\n$$\n\nand if $m$ is even, then\n\n$$\na_{m+1} - a_m = \\frac{(m+1)^2 + 3}{4} - \\frac{m^2}{4} = \\frac{2m+4}{4}.\n$$\n\nIn particular, it follows that $a_2 < a_3 < a_4 < \\dots$. Since $a_{2000} = 1\\,000\\,000$, the largest integer $n$ such that $a_n < 1\\,000\\,000$ is $1999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19651,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = 40^{n!} \\bmod 2009$. Prove that for some $n$, $a_n \\equiv 0 \\pmod{2009}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\gcd(40, 2009) = 1$, we have $40^{k \\cdot \\varphi(2009)} \\equiv 1 \\pmod{2009}$ for all natural numbers $k$. For $n > \\varphi(2009)$, the exponent $n!$ is a multiple of $\\varphi(2009)$, so $a_{n+1} \\equiv a_n + 1 \\pmod{2009}$. Thus, all values modulo $2009$ are taken cyclically and periodically, so every residue, including $0$, is achieved infinitely often. Therefore, the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19652,
    "subject": "Mathematics (Olympiad)",
    "question": "Winnie-the-Pooh and Piglet play the following game. There is a 15-inch-long stick. By his first move, Piglet breaks it into two pieces, then the players in turn break one of the existing pieces into two. The rules are that the resulting pieces must have integer length (in inches) and can't be 1-inch-long. The player who can't make a move loses. Who has a winning strategy?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer.** Piglet.\n\n**Solution.** Obviously, at the end of the game all remaining pieces will have length either 2 or 3 inches, and Piglet will win iff their number is even (which means that the total number of moves was odd, thus Piglet made the last move). There are three possible outcomes:\n\n$$\n15 = 3+3+3+3+3 = 3+3+3+2+2+2 = 3+2+2+2+2+2+2.\n$$\n\nThis implies that in order to win Piglet needs to ensure the existence of two 3-inch and one 2-inch pieces (this will make the first and the third outcomes impossible). So, by his first move he must break the stick into 5 and 10 inches. The first piece will be eventually divided into 2 and 3 inches. If Winnie breaks the 10-inch stick by his next move, Piglet must break 3 inches off the longer part, otherwise he should just break it into 3 inches and 7 inches. In this case, regardless of Winnie's moves, in the end there will be at least two 3-inch pieces and one 2-inch piece, which is enough for Piglet to win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19653,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown below, a square grid of side length 8 is constructed by 144 sticks of length 1. Find the least number of sticks to be removed so that the resulting figure contains no rectangle.\n\n![](images/2023_CGMO_p3_data_7b66269d49.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is 43.\n\nFirst, we prove that at least 43 sticks must be removed. Suppose the figure does not contain any rectangles, then each bounded connected area must consist of at least three unit squares, i.e., the area is at least 3. Therefore, there can be at most $\\left\\lfloor \\frac{64}{3} \\right\\rfloor = 21$ bounded connected areas. Removing one stick can at most reduce the number of bounded connected areas by 1 (either by merging two bounded areas into one, or by merging a bounded area with an unbounded area). Initially, there are 64 bounded connected areas, so at least $64 - 21 = 43$ sticks must be removed.\n\nThe figure below shows an example of removing 43 sticks, where each bounded connected area has an area of 3, and the figure does not contain any rectangles.\n\n![](images/2023_CGMO_p3_data_7e737e8cfd.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19654,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(b, c)$ of positive integers such that the sequence defined by $a_1 = b$, $a_2 = c$, and\n$$\na_{n+2} = |4a_{n+1} - 3a_n|, \\quad \\forall n \\ge 1\n$$\nhas only a finite number of composite terms.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose first that there exists $k \\ge 2$ such that $a_{k+1} > a_k$. Then, by induction, $a_{n+1} > a_n$ for all $n \\ge k$ (since the modulus does not change the sign), so $a_{n+2} = 4a_{n+1} - 3a_n$ for all $n \\ge k$. Solving the characteristic equation $t^2 - 4t + 3 = 0$ gives\n$$\na_n = C_1 + C_2 \\cdot 3^n, \\quad \\forall n \\ge k.\n$$\nSince the sequence is strictly increasing, $C_2 > 0$. Any prime divisor $p$ of some $a_m \\ge 2$ will also divide $a_{m+k(p-1)}$ for any $k \\ge 0$, since $3^{m+k(p-1)} \\equiv 3^m \\pmod{p}$ by Fermat's Little Theorem. Thus, infinitely many terms would be composite.\n\nTherefore, $a_{k+1} \\le a_k$ for all $k \\ge 2$. Since $a_k > 0$, we must have $a_n = p$ where $p$ is a prime for all $n \\ge n_0$ for some $n_0$. Thus,\n$$\np = |4p - 3a_{n-1}| \\implies a_{n-1} \\in \\{p, 3p\\}.\n$$\nThe first case gives $(b, c) = (p, p)$, which satisfies the condition. In the second case, $a_{n-1} = 3p$, so $a_{n-2} = 4p \\pm \\frac{p}{3}$. Consider the cases:\n\n- If there is no $a_{n-2}$, then $(b, c) = (3p, p)$ are the first two terms.\n- If $p$ is divisible by $3$, i.e., $p = 3$, then $a_{n-1} = 9$, $a_{n-2} = 11$ or $13$, which implies $a_{n-3}$ is not integer.\n\nSo, finally, the solutions are $(b, c) \\in \\{(11, 9), (13, 9), (p, p), (3p, p)\\}$ where $p$ is any prime. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19655,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We prove that a composite number is nice if and only if it is not a product of two distinct prime numbers.\n\nIf $n = pq$, where $p, q$ are distinct primes, it is clear that we cannot 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\n... and so on.\n\nThe set $\\{1, 2, \\ldots, 100\\}$ contains 74 composite numbers, 30 of which are of the form $pq$, with $p, q$ distinct primes. This leaves 44 nice numbers.\n\n**Remark.** A stronger fact can be proven: the factors of a nice number can actually be arranged on a circle such that, for every two neighboring numbers, the smaller one is a factor of the larger one.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19656,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right-angled triangle with $\\angle C = 90^\\circ$, and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$, and let $F$ be the centroid of triangle $BCD$. Let $P$ be the point satisfying $\\angle CEP = 90^\\circ$ and $|CP| = |AP|$, and let $Q$ be the point satisfying $\\angle CFQ = 90^\\circ$ and $|CQ| = |BQ|$. Show that $PQ$ passes through the centroid of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$, $N$, $R$, $S$ be the midpoints of the line segments $BC$, $CA$, $BD$, $AD$, respectively. Let $Z$ be the centroid of $\\triangle ABC$.\n\nQuadrilateral $QFMC$ is cyclic since $\\angle QFC = 90^\\circ = \\angle QMC$. Therefore, $CQ$ is a diameter of the circumcircle of $QFMC$. Similarly, $PNEC$ is a cyclic quadrilateral, with diameter $CP$.\n\nWe show that $Z$ also lies on the circumcircles of these cyclic quadrilaterals. The similarity transforming triangle $BCA$ into triangle $BDC$ transforms triangle $CZM$ into $DFR$, as $C$ is mapped to $D$, the centroid $Z$ is mapped to the centroid $F$, and the midpoint $M$ of $BC$ is mapped to the midpoint $R$ of $BD$. So $\\triangle CZM \\sim \\triangle DFR$, in particular $\\angle CZM = \\angle DFR = \\angle CFM$ (opposite angles). Therefore, $Z$ lies on the circumcircle of $QFMC$. Analogously, $\\angle CZN = \\angle DES = \\angle CEN$, so $Z$ lies on the circumcircle of $PNEC$.\n\nNow, $Z$ lies on $PQ$. Since $CQ$ is a diameter of the circle through $Q$, $F$, $M$, $C$, $Z$, we have $\\angle QZC = 90^\\circ$. Since $CP$ is a diameter of the circle through $P$, $N$, $E$, $Z$, $C$, we also have $\\angle CZP = 90^\\circ$. Hence, $P$, $Z$, and $Q$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19657,
    "subject": "Mathematics (Olympiad)",
    "question": "The company is led by at most *10* directors.\n\nEach director has a set of 3 keys, each for a different lock, and no two directors have the same set. Denote the locks and corresponding keys with numbers 1 through 6. How many directors can there be at most, given that no two directors have complementary sets of keys (i.e., together, their keys cover all 6 locks)?",
    "options": [],
    "answer": "See solution",
    "solution": "Let's write down all possible different combinations of three keys in pairs:\n\n- $\\{1, 2, 3\\}$ and $\\{4, 5, 6\\}$\n- $\\{1, 2, 4\\}$ and $\\{3, 5, 6\\}$\n- $\\{1, 2, 5\\}$ and $\\{3, 4, 6\\}$\n- $\\{3, 4, 5\\}$ and $\\{1, 2, 6\\}$\n- $\\{1, 3, 4\\}$ and $\\{2, 5, 6\\}$\n- $\\{1, 3, 5\\}$ and $\\{2, 4, 6\\}$\n- $\\{2, 4, 5\\}$ and $\\{1, 3, 6\\}$\n- $\\{1, 4, 5\\}$ and $\\{2, 3, 6\\}$\n- $\\{2, 3, 5\\}$ and $\\{1, 4, 6\\}$\n- $\\{2, 3, 4\\}$ and $\\{1, 5, 6\\}$\n\nEach pair contains all 6 keys, so at most one set from each pair can be assigned to a director. Since there are 10 such pairs, the company can have at most 10 directors. Furthermore, it is possible to assign 10 directors by choosing, for example, the first set from each pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19658,
    "subject": "Mathematics (Olympiad)",
    "question": "120\\% of $a$ is $\\frac{6}{5}a$ while 80\\% of $b$ is $\\frac{4}{5}b$. If $\\frac{6}{5}a = \\frac{4}{5}b$, what is $\\frac{a}{b}$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\frac{6}{5}a = \\frac{4}{5}b$. Dividing both sides by $b$ and rearranging, $$\\frac{a}{b} = \\frac{4}{5} \\times \\frac{5}{6} = \\frac{2}{3}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19659,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider $n$ different points on a circle such that no three chords with endpoints among these points pass through the same point in the interior of the circle.\n\n(a) Find the value of $n$ if the number of triangles with vertices among the $n$ points is equal to $2n$.\n\n(b) Find the value of $n$ if the number of intersection points of the chords, lying in the interior of the circle, is equal to $5n$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Any three of the $n$ points on the circle define a triangle. Thus, the number of triangles is $\\binom{n}{3} = \\dfrac{n(n-1)(n-2)}{6}$.\n\nSetting this equal to $2n$ gives:\n\n$$\n\\frac{n(n-1)(n-2)}{6} = 2n\n$$\n\nSolving, $n(n-1)(n-2) = 12n \\implies (n-1)(n-2) = 12 \\implies n=5$.\n\n(b) Any four different points among the $n$ points define a convex quadrilateral whose diagonals intersect at an interior point of the circle. Thus, the number of such intersection points is $\\binom{n}{4} = \\dfrac{n(n-1)(n-2)(n-3)}{24}$.\n\nSetting this equal to $5n$ gives:\n\n$$\n\\frac{n(n-1)(n-2)(n-3)}{24} = 5n\n$$\n\nSolving, $n(n-1)(n-2)(n-3) = 120n \\implies (n-1)(n-2)(n-3) = 120 \\implies n=7$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19660,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals of trapezoid $ABCD$ with bases $AB$ and $CD$ meet at $P$. Prove the inequality\n\n$$\nS_{PAB} + S_{PCD} > S_{PBC} + S_{PDA},\n$$\n\nwhere $S_{XYZ}$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = |AB|$ and $b = |CD|$, and let $h_a$ and $h_b$ be the altitudes of triangles $PAB$ and $PCD$ drawn from $P$. Denote $S_1 = S_{PAB} + S_{PCD}$ and $S_2 = S_{PBC} + S_{PDA}$. Then\n\n$$\nS_1 = \\frac{1}{2}(a h_a + b h_b)\n$$\n\nand\n\n$$\nS_1 + S_2 = \\frac{1}{2}(a + b)(h_a + h_b),\n$$\n\nso\n\n$$\nS_2 = \\frac{1}{2}(a h_b + b h_a).\n$$\n\nSince triangles $PAB$ and $PCD$ are similar, $a > b$ implies $h_a > h_b$ and $a < b$ implies $h_a < h_b$ (with $a \\neq b$ because $a$ and $b$ are the lengths of the bases of the trapezoid). Therefore,\n\n$$\nS_1 - S_2 = \\frac{1}{2}(a h_a + b h_b - a h_b - b h_a) = \\frac{1}{2}(a - b)(h_a - h_b) > 0,\n$$\n\ni.e., $S_1 > S_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19661,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\n(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)\n$$\n\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 0$ in the functional equation:\n$$\n(0^2 + 0^2)f(0) = f(0)f(0)f(0^2 + 0^2) \\implies 0 = f(0)^3,\n$$\nso $f(0) = 0$.\n\nLet $y = 1$:\n$$\n(x^2 + 1)f(x) = f(x)f(1)f(x^2 + 1) \\tag{1}\n$$\nThis shows $f(x) \\equiv 0$ is a solution. Now suppose $f(x) \\neq 0$ for some $x$.\n\nIf $f(b) = 0$ for $b \\neq 0$, then for any $x$, $f(bx) = 0$ by the functional equation, so $f(x) = 0$ for all $x$, a contradiction. Thus, $f(x) = 0$ only for $x = 0$.\n\nFor $x > 1$, substitute $x = \\sqrt{x-1}$ in (1):\n$$\nf(x)f(1) = x \\quad (x > 1). \\tag{2}\n$$\n\nFrom the original equation, for $x^2 + y^2 > 1$:\n$$\nf(xy)f(1) = f(x)f(y) \\tag{3}\n$$\n\nFor $0 < x < 1$, let $y = \\frac{1}{x}$ so $x^2 + y^2 \\geq 2 > 1$:\n$$\nf(x) = f(1)^3 x \\quad (0 < x < 1). \\tag{4}\n$$\n\nTake $x = y \\in (0, \\frac{1}{\\sqrt{2}})$ and use (4):\n$$\n2x^2 f(x^2) = f(2x^2)f(x)^2 = (f(1)^3 \\cdot 2x^2)(f(1)^3 \\cdot x)^2,\n$$\nwhich yields $f(1) = \\pm 1$.\n\n**Case 1:** $f(1) = 1$. From (2) and (4):\n$$\nf(x) = x \\quad (x \\geq 0).\n$$\nFrom the functional equation, $f(xy) = f(x)f(y)$ for all $x, y$.\n\nIf $f(x) = x$ for $x < 0$, then $f(x) = x$ for all $x$. If $f(a) \\neq a$ for some $a < 0$, then $f(a)^2 = a^2$ so $f(a) = -a$. For $x < 0$, $ax > 0$ so $f(ax) = ax$. But $f(ax) = f(a)f(x) = -a f(x)$, so $f(x) = -x$, i.e., $f(x) = |x|$ for $x \\in \\mathbb{R}$.\n\n**Case 2:** $f(1) = -1$. Let $g(x) = -f(x)$, then $g$ satisfies the same equation and $g(1) = 1$. By Case 1, $f(x) = -x$ or $f(x) = -|x|$ for all $x$.\n\nThus, all solutions are:\n$$\nf_1(x) \\equiv 0, \\quad f_2(x) = x, \\quad f_3(x) = -x, \\quad f_4(x) = |x|, \\quad f_5(x) = -|x| \\quad (x \\in \\mathbb{R}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19662,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have a 9-digit number $N$ in which all the digits are distinct and non-zero. Then we consider all the sums of adjacent triples of digits of $N$ and order them in a non-decreasing sequence. For the following sequences, determine whether there exists an $N$ for which we get them as a result:\n\na) $9, 11, 12, 13, 20, 20, 20$\n\nb) $9, 11, 12, 13, 20, 21, 21$",
    "options": [],
    "answer": "See solution",
    "solution": "a) We shall prove that no sequence containing three copies of the number 20 can be obtained this way. Denote the $i$-th digit of $N$ by $a_i$ and suppose, for the sake of contradiction, that there exist three distinct $1 \\leq i < j < k \\leq 7$ such that\n\n$$\na_i + a_{i+1} + a_{i+2} = a_j + a_{j+1} + a_{j+2} = a_k + a_{k+1} + a_{k+2} = 20.\n$$\n\nNote that the sum of the six largest digits is $9 + 8 + 7 + 6 + 5 + 4 = 39 < 2 \\times 20$, hence any two of the triples with sum 20 must overlap. In particular, the first and the last triples, i.e., $(a_i, a_{i+1}, a_{i+2})$ and $(a_k, a_{k+1}, a_{k+2})$, have to overlap, which forces $k - i \\leq 2$. From the condition $i < j < k$ and integrality of all the indices, we deduce that $k = i + 2$, hence $j = i + 1$ and the second triple is $(a_{i+1}, a_{i+2}, a_{i+3})$. But this implies that $a_i + a_{i+1} + a_{i+2} = 20 = a_{i+1} + a_{i+2} + a_{i+3}$ and thus also $a_i = a_{i+3}$, a contradiction.\n\nb) Yes, the number $N = 849751623$ works, since the sums of consecutive triples are (left to right) $21, 20, 21, 13, 12, 9, 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19663,
    "subject": "Mathematics (Olympiad)",
    "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 $(AB'N)$.",
    "options": [],
    "answer": "See 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\n![](images/RMC_2025_p18_data_5af849a837.png)\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": 19664,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider infinite sequences of positive integers in which each positive integer appears exactly once. Let $\\{a_n\\}$, $n \\ge 1$, be such a sequence. We call it *consecutive* if for every positive integer $k$ and for any $n, m$ with $a_n < a_m$, the following holds: $a_{kn} < a_{km}$. For example, the sequence $a_n = n$ is consecutive.\n\n**a)** Prove that there exists a consecutive sequence different from $a_n = n$.\n\n**b)** Does there exist a consecutive sequence for which $a_n \\neq n$ for all $n \\geq 2$?\n\n**c)** Does there exist a consecutive sequence for which $a_n \\neq n$ for all $n \\geq 1$?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Here is an example of a consecutive sequence different from $a_n = n$. Let $n = 2^{\\alpha_1}3^{\\alpha_2}5^{\\alpha_3}\\dots p_r^{\\alpha_r}$ be the unique prime factorization of $n$. Define $a_n = 2^{\\alpha_2}3^{\\alpha_1}5^{\\alpha_3}\\dots p_r^{\\alpha_r}$, i.e., swap the exponents of 2 and 3. This sequence is consecutive because for $n, m, k$ as above, $a_{kn} = a_k a_n$ and $a_{km} = a_k a_m$, so $a_n < a_m$ implies $a_{kn} < a_{km}$. For example, $a_2 = 3$.\n\n**b)** Define the function\n$$\nN(k) = \\begin{cases} k-2, & \\text{if } k = 4, 6, 8, \\dots, \\\\ 1, & \\text{if } k = 2, \\\\ k+2, & \\text{if } k = 1, 3, 5, \\dots \\end{cases}\n$$\nConstruct a sequence where, for $n = 2^{\\alpha_1}3^{\\alpha_2}5^{\\alpha_3}7^{\\alpha_4}\\dots p_{r-1}^{\\alpha_{r-1}}p_r^{\\alpha_r}$,\n$$\na_n = 5^{\\alpha_1} 2^{\\alpha_2} 11^{\\alpha_3} 3^{\\alpha_4} \\cdots p_{N(r-1)}^{\\alpha_{r-1}} p_{N(r)}^{\\alpha_r}.\n$$\nThis sequence is consecutive, and for $n \\ge 2$, $a_n \\ne n$ because the mapping of exponents ensures that $a_n$ cannot equal $n$ for $n \\ge 2$.\n\n**c)** If $a_1 \\neq 1$, then there exists $l$ such that $a_l = 1$. Then $a_l < a_1$, and $a_{l2} < a_l = 1$, which is impossible since $a_{l2}$ is a positive integer. Thus, no consecutive sequence exists with $a_n \\neq n$ for all $n \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19665,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be an integer and $a_1, a_2, \\dots, a_n$ be real numbers such that their sum is $0$ and the sum of their absolute values is $1$. Prove that\n\n$$\n|a_1 + 2a_2 + \\dots + n a_n| \\le \\frac{n-1}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "According to the assumptions, for each $k = 1, \\dots, n-1$ it holds that\n\n$$\n|a_1 + \\dots + a_k| = |a_{k+1} + \\dots + a_n|,\n$$\n\nand\n\n$$\n|a_1 + \\dots + a_k| + |a_{k+1} + \\dots + a_n| \\le |a_1| + \\dots + |a_k| + |a_{k+1}| + \\dots + |a_n| = 1.\n$$\n\nConsequently, $|a_{k+1} + \\dots + a_n| \\le \\frac{1}{2}$ for each $k = 1, \\dots, n-1$. Now,\n\n$$\n\\begin{align*}\n|a_1 + 2a_2 + \\dots + n a_n| &= |(a_1 + \\dots + a_n) + (a_2 + \\dots + a_n) + \\dots + (a_{n-1} + a_n) + a_n| \\\\\n&\\le |a_1 + \\dots + a_n| + |a_2 + \\dots + a_n| + \\dots + |a_{n-1} + a_n| + |a_n| \\\\\n&\\le 0 + \\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2} + \\frac{1}{2} = \\frac{n-1}{2}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19666,
    "subject": "Mathematics (Olympiad)",
    "question": "Resuelve la ecuación exponencial\n$$\n2^x \\cdot 3^{5-x} + \\frac{3^{5x}}{2^x} = 6\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Aplicando la desigualdad de las medias aritmética y geométrica y, después, una de sus más conocidas consecuencias (la suma de un número real positivo y su inverso es siempre mayor o igual que 2, y la igualdad sólo se da para el número 1), tenemos:\n$$\n6 = 2^x 3^{5-x} + 2^{-x} 3^{5x} \\geq 2\\sqrt{2^x 3^{5-x} 2^{-x} 3^{5x}} = 6.\n$$\nLa igualdad se da cuando los números mediados son iguales:\n$$\n2^x 3^{5-x} = 2^{-x} 3^{5x}\n$$\nEsto implica:\n$$\n2^{2x} 3^{5-2x} = 1\n$$\nResolviendo, se obtiene $x = 0$, que es la única solución de la ecuación.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19667,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "a) The triple $(n, n+1, n+3)$, $n \\in \\mathbb{Z}$, is artistic iff\n$$\n\\frac{3n^2+8n+3}{3n+4} = n + \\frac{4n+3}{3n+4}\n$$\nis 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, \n$$\n\\frac{x^2+y^2+z^2}{x+y+z} = (x+y+z) - 2 \\cdot \\frac{xy+yz+zx}{x+y+z}\n$$\nand\n$$\n\\frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z} = (xy+yz+zx) \\cdot \\frac{xy+yz+zx}{x+y+z} - 2xyz\n$$\nare also integers. Thus,\n$$\n\\frac{x^4+y^4+z^4}{x+y+z} = (x^2+y^2+z^2) \\cdot \\frac{x^2+y^2+z^2}{x+y+z} - 2 \\cdot \\frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z}\n$$\nis an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19668,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real-valued functions $f$ defined on pairs of real numbers such that, for all real numbers $a, b, c$, the median of $f(a, b)$, $f(b, c)$, $f(c, a)$ equals the median of $a, b, c$.\n\n(The median of three real numbers, not necessarily distinct, is the number that is in the middle when the three numbers are arranged in non-decreasing order.)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two solutions:\n\n- $f(a, b) = a$ for all $a, b$.\n- $f(a, b) = b$ for all $a, b$.\n\nClearly, these functions meet the condition. We must show there are no others.\n\nBy setting $a = b = c$, we get $f(a, a) = a$ for all $a$. Next, for all $a, b$, the median of $f(a, a)$, $f(a, b)$, $f(b, a)$ must equal the median of $a, a, b$, namely $a$, so for all $a, b$, one of $f(a, b)$, $f(b, a)$ is at most $a$ and the other is at least $a$. Switching $a, b$, we also see that one of $f(a, b)$, $f(b, a)$ is at most $b$ and the other is at least $b$. Therefore, we have\n\n$$\n\\min\\{f(a, b), f(b, a)\\} \\leq \\min\\{a, b\\}\n$$\n\nand\n\n$$\n\\max\\{f(a, b), f(b, a)\\} \\geq \\max\\{a, b\\}.\n$$\n\nNext, consider any three numbers $a < b < c$. The median of $f(a, b)$, $f(b, c)$, $f(c, a)$ must equal $b$, so one of $f(a, b)$, $f(b, c)$ equals $b$ (since $f(c, a)$ must be either at most $a$ or at least $c$). Similarly, considering $f(a, c)$, $f(c, b)$, $f(b, a)$, we see that one of $f(c, b)$, $f(b, a)$ equals $b$. The numbers $f(a, b)$, $f(b, a)$ cannot both be $b$, by the previous inequality, and $f(b, c)$, $f(c, b)$ cannot both be $b$. We conclude that either\n\n$$\nf(a, b) = f(c, b) = b\n$$\n\nor\n\n$$\nf(b, c) = f(b, a) = b.\n$$\n\nIn particular, for any $a < b$, choosing $c > b$ arbitrarily, we see that one of $f(a, b)$, $f(b, a)$ must equal $a$. Likewise, for any $b < c$, choosing $a < b$ arbitrarily, we see that one of $f(b, c)$, $f(c, b)$ must equal $b$.\n\nPutting these two conclusions together, for any $a \\neq b$, one of $f(a, b)$, $f(b, a)$ equals $\\min\\{a, b\\}$ and the other equals $\\max\\{a, b\\}$. In other words, for $a \\neq b$, $\\{f(a, b), f(b, a)\\}$ and $\\{a, b\\}$ are equal as sets. Call $\\{a, b\\}$ a first-pair if $f(a, b) = a$ and $f(b, a) = b$, and a second-pair if $f(a, b) = b$ and $f(b, a) = a$.\n\nNow again consider any three numbers $a < b < c$. If either $\\{a, b\\}$ or $\\{b, c\\}$ is a first-pair, then the previous conclusion cannot hold, so the other must hold, and $\\{a, b\\}$ and $\\{b, c\\}$ are both first-pairs. That is, $\\{a, b\\}$ is a first-pair if and only if $\\{b, c\\}$ is. Pick any other numbers $a'$ and $c'$ such that $a' < b$ and $c' > b$. The same logic gives\n\n$$\n\\begin{align*}\n\\{a', b\\} \\text{ is a first-pair} &\\iff \\{b, c\\} \\text{ is a first-pair} \\\\\n&\\iff \\{a, b\\} \\text{ is a first-pair} \\\\\n&\\iff \\{b, c'\\} \\text{ is a first-pair.}\n\\end{align*}\n$$\n\nThis shows that, given any $p, q \\neq b$, $\\{p, b\\}$ is a first-pair if and only if $\\{q, b\\}$ is. Since $b$ is arbitrary, we have for any distinct $p, q, r, s$ that\n\n$$\n\\begin{align*}\n\\{p, q\\} \\text{ is a first-pair} &\\iff \\{s, q\\} = \\{q, s\\} \\text{ is a first-pair} \\\\\n&\\iff \\{r, s\\} \\text{ is a first-pair.}\n\\end{align*}\n$$\n\nThat is, if some first-pair exists, then every pair is a first-pair, so $f(a, b) = a$ for all $a, b$. Otherwise, every pair is a second-pair, so $f(a, b) = b$ for all $a, b$. Thus, the only possibilities for the function $f$ are the two solutions we initially identified.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19669,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of numbers $ (x, y, p) $, where $ x, y $ are positive integers and $ p $ is a prime number, which satisfy the condition:\n\n$$\ny(x^2 + p) - x(y^2 + p) = p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Factor the left-hand side of the equation:\n\n$$\n(yx^2 - xy^2) + (yp - xp) = p \\implies yx(x - y) - p(x - y) = p \\implies (yx - p)(x - y) = p.\n$$\n\nThe last equation is possible in several cases.\n\n**Case 1.** $yx - p = 1$, $x - y = p$. Then, $x = y + p$.\n\n$$\ny(y + p) - p = 1 \\implies y^2 + yp - p - 1 = 0.\n$$\n\nSince $y = 1$ is a solution, the other solution is $y = -p - 1$, which is not a positive integer. Thus, $x = p + 1$. Substituting back, the tuple $(p+1, 1, p)$ satisfies the equation for any prime $p$.\n\n**Case 2.** $yx - p = -1$, $x - y = -p$. Then, $y = x + p$.\n\n$$\nx(x + p) - p = -1 \\implies x^2 + xp - p + 1 = 0.\n$$\n\nFor $x \\in \\mathbb{N}$, the discriminant must be a perfect square:\n\n$$\nD = p^2 - 4(-p + 1) = p^2 + 4p - 4.\n$$\n\nCheck for which $p$ this is a perfect square. For $p = 2$, $D = 8$ (not a perfect square). For $p \\geq 3$, $D$ is not a perfect square. Thus, no solutions in this case.\n\n**Case 3.** $yx - p = p$, $x - y = 1$. Then, $x = y + 1$ and $yx = 2p$.\n\nFrom $x = y + 1$, $y = 1$ and $x = 2p$ or $y = 2$ and $x = p$. Combining, $y = 2$, $x = p = 3$ is a solution.\n\n**Case 4.** $yx - p = -p$, $x - y = -1$. This leads to a contradiction.\n\n**Final answer:**\n\nAll solutions are $$(x, y, p) = (p+1, 1, p)$$ for any prime $p$, and $(3, 2, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19670,
    "subject": "Mathematics (Olympiad)",
    "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\n(a) Find how many nice sets have the element $225$.\n\n(b) Find how many nice subsets $A$ has.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B = \\{b, 5b, a\\}$. The possible cases are:\n\nI. $a = \\frac{b+5b}{2} \\implies B = \\{b, 3b, 5b\\}$;\n\nII. $5b = \\frac{a+b}{2} \\implies B = \\{b, 5b, 9b\\}$;\n\nIII. $b = \\frac{a+5b}{2} \\implies a+3b=0$ — impossible.\n\n(a) Since $225$ is divisible by $3$, $5$, and $9$, it can be any element of the set $B$. We get the nice sets: $\\{225, 675, 1225\\}$, $\\{75, 225, 375\\}$, $\\{45, 135, 225\\}$, $\\{225, 1225, 2025\\}$, $\\{45, 225, 405\\}$, $\\{25, 125, 225\\}$.\n\n(b) In case I, we get a nice set for every $b \\in \\mathbb{N}^*$ with $5b \\leq 2025$, that is $2025 \\div 5 = 405$ sets. In case II, we get a nice set for every $b \\in \\mathbb{N}^*$ so that $9b \\leq 2025$, 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": 19671,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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$$\n\\frac{d(S, P_1Q_2)}{d(S, P_2Q_1)} = \\frac{P_1Q_2}{P_2Q_1}. \\qquad (1)\n$$\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$$\n[MP_1Q_2] + [MP_1Q_1] = 2[MP_1L] = ML \\cdot d(P_1, ML) = [ABC]/2.\n$$\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$$\n\\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \\frac{P_2Q_1}{P_1Q_2}. \\qquad (2)\n$$\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.\n\nREMARK. Relation (2) may be obtained in several different ways. For instance, one may notice that midpoints $X, X_1$ and $X_2$ of the segments $BR, P_1Q_1$ and $P_2Q_2$, respectively, are collinear — the Gauss-Newton line $\\ell$ of the quadrilateral $P_1P_2Q_1Q_2$ (see Fig. 7). Moreover, $KX_1LX_2$ is the Varignon parallelogram of the quadrilateral $P_1P_2Q_1Q_2$, hence the segments $X_1X_2$ and $KL$ have a common midpoint $N$. Since $XN$ is a midline in the triangle $BMR$, the line $RM$ is parallel to $\\ell$, so $\\overrightarrow{RM} = \\alpha \\overrightarrow{X_1X_2} = \\frac{\\alpha}{2}(\\overrightarrow{P_1Q_2} + \\overrightarrow{P_1Q_1})$ which easily yields (2).\n\nYet another way of obtaining the same relation is the following. Let the lines $P_1Q_2$ and $P_2Q_1$ meet $AC$ at points $U$ and $V$, respectively. By Menelaus theorem,\n\n$$\n\\frac{AU}{UC} = \\frac{AP_1}{P_1B} \\cdot \\frac{BQ_2}{Q_2C} = \\frac{BP_2}{P_2A} \\cdot \\frac{CQ_1}{Q_1B} = \\frac{CV}{VA},\n$$\n\nso $AU = CV$. Thus $RM$ is a median in the triangle $RUV$, so\n\n![](images/RMC_2015_BT_p69_data_89a362601f.png)\n\n$$\n1 = \\frac{[MUR]}{[MVR]} = \\frac{RU \\cdot d(M, P_1Q_2)}{RV \\cdot d(M, P_2Q_1)}, \\quad \\text{or} \\quad \\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \\frac{RV}{RU}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19672,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $-1 \\leq a, b, c \\leq 1$ satisfying $a^2 + b^2 + c^2 = 2abc + 1$, prove\n\n$$\n\\frac{3}{2 - abc} \\leq \\frac{1}{2 - a^2} + \\frac{1}{2 - b^2} + \\frac{1}{2 - c^2} \\leq 1 + \\frac{2}{2 - abc}.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show\n\n$$\n\\frac{3}{2 - abc} \\leq \\frac{1}{2 - a^2} + \\frac{1}{2 - b^2} + \\frac{1}{2 - c^2}.\n$$\n\nBy the arithmetic-harmonic mean inequality,\n\n$$\n\\frac{1}{2-a^2} + \\frac{1}{2-b^2} + \\frac{1}{2-c^2} \\geq \\frac{9}{6-(a^2+b^2+c^2)} = \\frac{9}{5-2abc}.\n$$\n\nMoreover, $\\frac{9}{5-2abc} \\geq \\frac{3}{2-abc}$ since $abc \\leq 1$. Equality holds for $(a, b, c) = (1, 1, 1)$, $(1, -1, -1)$, $(-1, 1, -1)$, $(-1, -1, 1)$.\n\nNow we prove\n\n$$\n\\frac{1}{2-a^2} + \\frac{1}{2-b^2} + \\frac{1}{2-c^2} \\leq 1 + \\frac{2}{2-abc}.\n$$\n\nDenote $abc = x$ and $a^2b^2 + b^2c^2 + c^2a^2 = y$. Using $a^2 + b^2 + c^2 = 2x + 1$, rewrite the inequality as\n\n$$\n(4-x)(4-8x+2y-x^2) \\geq (2-x)(8-8x+y).\n$$\n\nSubtracting, it suffices to prove\n\n$$\nx(x^2 - 4x - 12) + (6-x)y = (6-x)(y - 2x - x^2) \\geq 0.\n$$\n\nThis holds since $6 - x \\geq 5 > 0$ and\n\n$$\ny - 2x - x^2 = a^2b^2 + b^2c^2 + c^2a^2 - 2abc - a^2b^2c^2 = (1-a^2)(1-b^2)(1-c^2) \\geq 0.\n$$\n\nEquality holds for $(a, b, c) = (1, t, t)$, $(-1, t, -t)$ with $-1 \\leq t \\leq 1$, and their permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19673,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為一正整數。試求所有 $1,2,\\ldots,n$ 的重排 $a_1,\\ldots,a_n$ 中，滿足\n\n$$\na_1 \\le 2a_2 \\le \\dots \\le n a_n\n$$\n\n的重排數量。",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 $F_{n+1}$，其中 $F_k$ 為費氏數列的第 $k$ 項（$F_1 = F_2 = 1$，$F_{n+2} = F_{n+1} + F_n$）。我們先證明以下引理：\n\n*引理：* 若 $a_n \\ne n$，則 $(a_{n-1}, a_n) = (n, n-1)$。\n\n*證明：* 令 $a_k = n$。若 $k = n$，則引理成立，故假設 $k < n$。\n\n- 若 $k = n-1$，則由於 $(n-1)n = (n-1)a_{n-1} \\le n a_n \\le n(n-1)$，我們必須有 $a_n = n-1$。\n- 若 $k \\le n-2$，則對於所有 $k < i < n$，都必須有 $k n = k a_k \\le i a_i < n a_i$，也就是 $a_i \\ge k+1$。此外，我們也有 $n a_n \\le (n-1)a_{n-1} \\le (n-1)(k+1) > n k$，故 $a_n \\le k+1$。從而 $a_k, a_{k+1}, \\ldots, a_n$ 這 $n-k+1$ 個數字都必須要大於 $k$，但 $1$ 到 $n$ 中只有 $n-k$ 個數字大於 $k$，從而矛盾。\n\n綜以上，引理得證。\n\n回到原題，令 $P_n$ 為欲求之重排數，易知 $P_1 = 1$ 而 $P_2 = 2$。又依據引理，在 $n+2$ 時僅有兩種情況：\n\n1. $a_{n+2} = n+2$，此時依照歸納假設，$a_1$ 到 $a_{n+1}$ 的重排數為 $P_{n+1}$。\n2. $(a_{n+1}, a_{n+2}) = (n+2, n+1)$，此時依照歸納假設，$a_1$ 到 $a_n$ 的重排數為 $P_n$。\n\n因此 $P_{n+2} = P_{n+1} + P_n$，也就是 $P_n = F_{n+1}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19674,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest four-digit number $ABCD$ such that $ABCD = BCD + AB \\times D$, where $A$, $B$, $C$, and $D$ are digits.",
    "options": [],
    "answer": "See solution",
    "solution": "To find the smallest number, try $A = 1$. The equation becomes $1BCD = BCD + 1BC \\times D$, so $1000 = 1BC \\times D$. Thus, $D$ must be a divisor of $1000 = 2^3 \\times 5^3$, and since $D$ is a digit, $D \\in \\{1, 2, 4, 5, 8\\}$. Also, $1BC < 200$ implies $1000 < 200 \\times D$, so $D > 5$. Therefore, $D = 8$. Since $1000/8 = 125$, we have $B = 2$ and $C = 5$. Thus, the smallest possible value is $1258$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19675,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd positive integer. Find all values of the natural numbers $n \\geq 2$ for which\n\n$$\n\\sum_{i=1}^{n} \\prod_{j \\neq i} (x_i - x_j)^p \\geq 0,\n$$\n\nwhere $x_1, x_2, \\dots, x_n$ are any real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote\n\n$$\nf_n(x_1, x_2, \\dots, x_n) = \\sum_{i=1}^{n} \\prod_{j \\neq i} (x_i - x_j)^p\n$$\n\nFor $n = 2$:\n\n$$\nf_2(x_1, x_2) = (x_1 - x_2)^p + (x_2 - x_1)^p = 0\n$$\nfor all $x_1, x_2 \\in \\mathbb{R}$, so the statement holds with equality.\n\nFor $n \\geq 3$:\n\nConsider $f_n(x_1, x_2, a, a, \\dots, a)$:\n\n$$\n\\begin{aligned}\nf_n(x_1, x_2, a, a, \\dots, a) &= (x_1 - x_2)^p (x_1 - a)^{p(n-2)} + (x_2 - x_1)^p (x_2 - a)^{p(n-2)} \\\\\n&= (x_1 - x_2)^p \\left[ (x_1 - a)^{p(n-2)} - (x_2 - a)^{p(n-2)} \\right]\n\\end{aligned}\n$$\n\nThis is nonnegative when $p(n-2)$ is odd, so $n$ must be odd.\n\nFor $n \\geq 7$ odd, try $f_n(x_1, a, a, a, b, \\dots, b)$:\n\n$$\nf_n(x_1, a, a, a, b, \\dots, b) = (x_1 - a)^{3p} (x_1 - b)^{p(n-4)}\n$$\n\nThis can be negative (e.g., $x_1 = \\frac{a+b}{2}$, $a \\neq b$), so only small odd $n$ may work.\n\nFor $n = 3$:\n\nAssume $x_1 \\geq x_2 \\geq x_3$. Let $u(x) = (x - x_3)^p$, $g(x_1, x_2, x_3) = (x_3 - x_1)^p (x_3 - x_2)^p$.\n\n$u$ is increasing, $g \\geq 0$, so\n\n$$\n\\begin{aligned}\nf(x_1, x_2, x_3) &= (x_1 - x_2)^p [u(x_1) - u(x_2)] + g(x_1, x_2, x_3) \\geq 0\n\\end{aligned}\n$$\n\nFor $n = 5$:\n\nAssume $x_1 \\geq x_2 \\geq x_3 \\geq x_4 \\geq x_5$. Let $v(x) = (x - x_3)^p (x - x_4)^p (x - x_5)^p$, $w(x) = (x - x_1)^p (x - x_2)^p (x - x_3)^p$, $h(x_1, x_2, x_3, x_4, x_5) = (x_3 - x_1)^p (x_3 - x_2)^p (x_3 - x_4)^p (x_3 - x_5)^p$.\n\n$v$ and $w$ are increasing, $h \\geq 0$, so\n\n$$\n\\begin{aligned}\nf(x_1, x_2, x_3, x_4, x_5) &= (x_1 - x_2)^p [v(x_1) - v(x_2)] \\\\\n&\\quad + (x_4 - x_5)^p [w(x_4) - w(x_5)] + h(x_1, x_2, x_3, x_4, x_5) \\geq 0\n\\end{aligned}\n$$\n\nThus, the only possible $n$ are $2$, $3$, and $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19676,
    "subject": "Mathematics (Olympiad)",
    "question": "在銳角三角形 $ABC$ 中，$\\angle B > \\angle C$，點 $I$ 為其內心，$R$ 為其外接圓半徑，$D$ 為 $A$ 點在 $BC$ 上的垂足。點 $K$ 落於直線 $AD$ 上，使得 $AK = 2R$，且 $D$ 在 $A$ 與 $K$ 之間。證明：\n\n$$\n\\angle KID = \\frac{\\angle B - \\angle C}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "設 $A'$ 為 $A$ 關於三角形 $ABC$ 外接圓 $(O)$ 的對踵點。則三角形 $AKA'$ 是等腰三角形，且因為 $AH$ 和 $AO$ 關於 $\\angle BAC$ 是等角線，我們可知 $I$ 在 $A'K$ 的中線 $AM$ 上，且 $A'K$ 的中點 $M$ 是弧 $BC$（不含 $A$）的中點。由於 $\\angle DAI = \\angle DAO/2 = (B-C)/2$，只需證明 $\\angle KID = \\angle DAI$，即等價於 $KI$ 在 $I$ 點處切於三角形 $DAI$ 的外接圓。換言之，只需證明 $KA \\cdot KD = KI^2$。\n\n現在計算：\n\n$$\n\\begin{align*}\nKI^2 &= AK^2 + AI^2 - 2AI \\cdot AK \\cdot \\cos \\frac{B-C}{2} \\\\\n&= 4R^2 + \\frac{r^2}{\\sin^2 \\frac{A}{2}} - \\frac{4Rr \\cos \\frac{B-C}{2}}{\\sin \\frac{A}{2}} \\\\\n&= 4R^2 + \\frac{r^2bc}{(p-b)(p-c)} - \\frac{4Rr(b+c)}{a} \\\\\n&= 4R^2 + \\frac{bc(p-a)}{p} - \\frac{4Rr(b+c)}{a} \\\\\n&= 4R^2 - \\frac{4pRr(b+c) - abc(p-a)}{ap} \\\\\n&= 4R^2 - \\frac{4RS}{a} = 2R(2R - h_a) = KA \\cdot KD.\n\\end{align*}\n$$\n\n這證明了 $KA \\cdot KD = KI^2$，從而完成證明。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19677,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{N}^* \\to \\mathbb{N}^*$ with the properties:\n\na) $f(m+n) - 1$ divides $f(m) + f(n)$, for all $m, n \\in \\mathbb{N}^*$;\n\nb) $n^2 - f(n)$ is a square, for all $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is not difficult to check that $f(1) = 1$ and $f(2) = 3$.\n\nWe prove inductively that $f(n) = 2n - 1$ for all $n$.\n\nIndeed, assume that for some $k > 1$, $f(k) = 2k - 1$. Then $f(k+1) - 1$ divides $2k$, hence $f(k+1) \\leq 2k + 1$.\n\nIf the inequality is strict, then $$(k+1)^2 - f(k+1) > (k+1)^2 - (2k+1) = k^2,$$ and since it is a square, we have $$(k+1)^2 - f(k+1) \\geq (k+1)^2,$$ a contradiction.\n\nWe deduce that $f(k+1) = 2k + 1$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19678,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer $n$ is a *combi number* if each pair of distinct digits from the set $0$ to $9$ appears at least once in the number as neighbouring digits. For example, in a combi number, the digits $3$ and $5$ must appear somewhere next to each other (in either order: $35$ or $53$). A combi number never starts with the digit $0$.\n\nWhat is the smallest possible number of digits in a combi number?",
    "options": [],
    "answer": "See solution",
    "solution": "$50$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19679,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\\angle ADC = 150^\\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$.\n\nProve that if $\\angle BAE + \\angle CBE = 60^\\circ$, then $\\angle BDC + \\angle EAC = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us rotate $\\triangle AED$ around the point $A$ so that $E \\to B$, $D \\to T$, and reflect $\\triangle ABT$ symmetrically with respect to $BT$ ($A \\to F$). Since $\\angle ADE = 150^\\circ$, then the triangle $ATF$ is equilateral. Let us prove that $FD = DC$. Denote $\\angle EAB = \\alpha$ and $\\angle AED = \\beta$. Then\n\n$$\n\\angle ABC = \\angle ABE + \\angle CBE = \\frac{180^\\circ - \\alpha}{2} + (60^\\circ - \\alpha) = 150^\\circ - \\frac{3\\alpha}{2} \\text{ and } \\angle BAC = 15^\\circ + \\frac{3\\alpha}{4}.\n$$\n\n$$\n\\begin{aligned}\n\\angle FCD &= \\angle FCA - \\angle DCA = \\frac{1}{2}\\angle FBA - (\\angle DEA - \\angle EAC) \\\\\n&= \\beta - (\\beta - \\angle EAC) = \\angle EAC = \\angle BAC - \\alpha = 15^\\circ - \\frac{\\alpha}{4}.\n\\end{aligned}\n$$\n\nHere we used the fact that the points $D$ and $A$ lie in the same half-plane relative to the line $FC$, since $\\angle ACD < \\angle AED = \\beta = \\angle ACF$.\n\nNote that $\\angle TFD = \\frac{1}{2}\\angle TAD = \\frac{\\alpha}{2}$, so\n\n$$\n\\angle DFC = \\angle AFC - \\angle AFD = \\frac{1}{2}\\angle ABC - (\\angle AFT - \\angle TFD) = \\left(75^\\circ - \\frac{3\\alpha}{4}\\right) - \\left(60^\\circ - \\frac{\\alpha}{2}\\right) = 15^\\circ - \\frac{\\alpha}{4}.\n$$\n\n![](images/BLR_ABooklet_2024_p16_data_545350ba03.png)\n\nTherefore, $\\angle DFC = \\angle FCD$, i.e. $FD = DC$. Hence $FC \\perp BD$, i.e. $\\angle BDC + \\angle FCD = 90^\\circ$. Recalling that $\\angle FCD = \\angle EAC$, we obtain the statement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19680,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle. The external and internal angle bisectors of $\\angle CAB$ intersect side $BC$ at $D$ and $E$, respectively. Let $F$ be a point on the segment $BC$. The circumcircle of $\\triangle ADF$ intersects $AB$ and $AC$ at $I$ and $J$, respectively. Let $N$ be the midpoint of $IJ$ and $H$ the foot of $E$ on $DN$. Prove that $E$ is the incenter of $\\triangle AHF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $\\omega$ the circumcircle of $\\triangle AHF$.\n\nThe key idea is to introduce a new point $X$, defined as the second intersection of $DN$ and $\\omega$. Note that $\\angle JAD = \\angle CAD = 90^\\circ \\pm \\frac{\\alpha}{2}$, where $\\alpha = \\angle CAB$, since $AD$ is an external bisector of $\\angle CAB$.\n\nThe $\\pm$ sign depends on the configuration.\n\nThus, either $\\angle JAD = \\angle BAD$ or $\\angle JAD + \\angle IAD = 180^\\circ$, so in both cases $DI = DJ$.\n\nSince $N$ is the midpoint of $IJ$, $DN$ is the bisector of $IJ$ and passes through the center of the circle. This shows that $DX$ is a diameter of $\\omega$ and $EH \\parallel IJ$.\n\nAlso, $\\angle EAD = 90^\\circ$ (as the angle between bisectors) and $\\angle XAD = 90^\\circ$ (since $DX$ is a diameter). Hence, $X$, $A$, $E$ are collinear.\n\nThis gives $\\angle DHE = \\angle XHE = 90^\\circ$ and $\\angle XFE = \\angle DFE = 90^\\circ$ as $DX$ is a diameter of $\\omega$, and again $\\angle EAD = 90^\\circ$. Thus, quadrilaterals $XFEH$ and $ADEH$ are cyclic.\n\nFinally, angle chasing yields $\\angle AHE = \\angle ADH = \\angle AXF = \\angle EXF = \\angle EHF$, where the first, second, and fourth equalities are due to the cyclicity of $ADEH$, $ADXF$, and $XFEH$, respectively. Also, $\\angle DFH = \\angle EFH = \\angle EXH = \\angle AFD = \\angle AFE$, where the second and fourth equalities are due to the cyclicity of $XFEH$ and $ADXF$. This shows $E$ is the incenter of $\\triangle AFH$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19681,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 = 1$, $a_2 = 10$, and $a_{n+1} = 2a_n + 3a_{n-1}$ for $n > 1$. Consider the infinite sum\n\n$$\nP(x) = \\sum_{i=1}^{\\infty} a_i x^i\n$$\n\nwhich is defined and finite for $x \\in \\left( -\\frac{1}{3}, \\frac{1}{3} \\right)$. Find all $y \\in \\mathbb{Z}$ such that\n\n$$\nP\\left(\\frac{1}{y}\\right) \\in \\mathbb{Z}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $a_{n+1} + a_n = 3(a_n + a_{n-1})$. Let $b_n = a_n + a_{n-1}$ for $n > 1$. Then $b_2 = 11$ and $b_n = 11 \\cdot 3^{n-2}$ for $n > 1$. Using this, we have:\n\n$$\n\\begin{align*}\n(1+x)P(x) &= a_1 + \\sum_{i=2}^{\\infty} a_i x^i + \\sum_{i=2}^{\\infty} a_{i-1} x^i \\\\\n&= a_1 x + \\sum_{i=2}^{\\infty} b_i x^i \\\\\n&= x + \\frac{11}{9} \\sum_{i=2}^{\\infty} (3x)^i \\\\\n&= x + \\frac{11}{9} \\cdot \\frac{9x^2}{1-3x} \\\\\n&= \\frac{8x^2 + x}{1-3x}\n\\end{align*}\n$$\n\nfor $x \\in \\left( -\\frac{1}{3}, \\frac{1}{3} \\right)$. Hence,\n\n$$\nP(x) = \\frac{8x^2 + x}{(1 - 3x)(1 + x)} \\quad \\text{for all } x \\in \\left( -\\frac{1}{3}, \\frac{1}{3} \\right).\n$$\n\nNow,\n\n$$\nP\\left(\\frac{1}{y}\\right) = \\frac{8 \\left( \\frac{1}{y} \\right)^2 + \\frac{1}{y}}{(1 - 3 \\cdot \\frac{1}{y})(1 + \\frac{1}{y})} = \\frac{8 + y}{(y-3)(y+1)}\n$$\n\nfor all $y \\in \\mathbb{Z}$ with $|y| > 3$. Assume $P\\left(\\frac{1}{y}\\right) \\in \\mathbb{Z}$ for some $y \\in \\mathbb{Z}$ and $|y| > 3$. Then $y+1$ must divide $7$ and $y-3$ must divide $11$. It is easy to see that $y = -8$ is the only possibility. Since $P\\left( -\\frac{1}{8} \\right) = 0$, $y = -8$ is indeed a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19682,
    "subject": "Mathematics (Olympiad)",
    "question": "Sei $S = \\{1, 2, \\dots, 2017\\}$.\n\nBestimme die größtmögliche natürliche Zahl $n$, für die es $n$ verschiedene Teilmengen von $S$ gibt, sodass für keine zwei dieser Teilmengen ihre Vereinigung gleich $S$ ist.",
    "options": [],
    "answer": "See solution",
    "solution": "Antwort: $n = 2^{2016}$.\n\n**Beweis:**\n\nEs gibt $2^{2016}$ Teilmengen von $S$, die das Element $2017$ nicht enthalten. Die Vereinigung von je zwei dieser Teilmengen enthält $2017$ ebenfalls nicht und ist daher ungleich $S$. Daher ist das gesuchte $n$ mindestens $2^{2016}$.\n\nWenn wir jede Teilmenge von $S$ mit ihrem Komplement zu einem Paar zusammenfassen, können wir $S$ in $2^{2016}$ Paare aufteilen. Wäre nun $n$ größer als $2^{2016}$, müsste es unter den $n$ Teilmengen mindestens ein solches Paar geben. Deren Vereinigung wäre aber ganz $S$, sodass das nicht möglich ist, und $n$ also nicht größer als $2^{2016}$ sein kann.\n\nSomit ist der gesuchte Wert $n = 2^{2016}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19683,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = ax + b$, with $a, b$ real numbers. Define $f_1(x) = f(x)$ and $f_{n+1}(x) = f(f_n(x))$ for $n = 1, 2, \\dots$. If $f_7(x) = 128x + 381$, then $a + b = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nFrom the definitions, we get\n\n$$\nf_n(x) = a^n x + (a^{n-1} + a^{n-2} + \\dots + a + 1)b = a^n x + \\frac{a^n - 1}{a - 1} b.\n$$\n\nAs $f_7(x) = 128x + 381$, we have $a^7 = 128$ and $\\frac{a^7 - 1}{a - 1} b = 381$. Then $a = 2$, $b = 3$. The answer is $a + b = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19684,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer greater than $2$. Mia is playing the following game. She writes the numbers $1, 2, 3, \\dots, n$ in some order on the sides of a regular $n$-sided polygon, one number per side. Then, on each vertex of the polygon, she writes the sum of the numbers on the two sides that meet at that vertex. Mia wins if the $n$ numbers on the vertices can be written down in some order to form an arithmetic progression.\n\nFor which $n$ can Mia win this game?\n\n(In an arithmetic progression, there is a constant $d$ such that each term is equal to the previous term plus $d$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Mia can win if and only if $n$ is odd.\n\nSuppose Mia wins for a certain $n$. Let the $n$ numbers on the $n$ vertices be an arithmetic progression with initial term $a$ and difference $d$. The sum of all $n$ terms is twice the sum of the numbers on the sides:\n\n$$ \\frac{(a + a + (n - 1)d) n}{2} = n(n + 1) $$\n\nThis simplifies to:\n\n$$ 2a + (n - 1)d = 2n + 2 $$\n\nIf $d \\geq 2$, then the left side is greater than the right side, so $d = 1$. Substituting gives $a = \\frac{n + 3}{2}$, so $n$ must be odd.\n\nWhen $n$ is odd, a winning arrangement is:\n\n$k, 2k-1, k-1, 2k-2, k-2, 2k-3, \\dots, k+1, 1$\n\nwhere $n = 2k - 1$. The $n$ numbers on the vertices are then $k+1$ to $3k-1$, which form an arithmetic progression. Alternatively, place $1$ on any side, skip a side and place $2$, and so on.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19685,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p \\ge 5$ be a prime number. Prove that there are positive integers $n$ and $m$ with $n + m \\le \\frac{p+1}{2}$ such that $p$ divides $2^n \\cdot 3^m - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s = \\operatorname{ord}_p(2)$ and $t = \\operatorname{ord}_p(3)$. If $\\max(s, t) \\le \\frac{p-1}{4}$, then we choose $(n, m) = (s, t)$. Hence, it is enough to consider the cases $\\max(s, t) = \\frac{p-1}{k} \\in \\mathbb{N}$ for $k = 1, 2, 3$. Further, we assume $s \\ge t$ due to symmetry.\n\nIf $k = 1$, then there is an integer $l$ such that $2^l \\equiv 3 \\pmod{p}$ and $1 < l < p$. For fixed $p$ and $l$, there exist unique positive integers $q$ and $r$ such that\n\n$$\np - 1 = ql + r \\quad \\text{and} \\quad r \\le l.\n$$\n\nNow we choose $(n, m) = (r, q)$. Then $2^n 3^m \\equiv 2^r 2^{lq} \\equiv 1 \\pmod{p}$. Since $q + r \\le \\frac{p-1}{l} + l - 1$, the required inequality is clear for $l \\le \\frac{p-1}{2}$. If $l > \\frac{p-1}{2}$, then $q = 1$ and so $n + m = p - l$, which is clearly bounded by $\\frac{p+1}{2}$.\n\nFor $k = 2$ and $3$, we claim that there is a positive integer $j$ so that $j \\le k$ and $3^j \\in I_s$, where $I_s = \\{2^i \\bmod p \\mid 1 \\le i \\le s\\}$. Indeed, if $3 \\notin I_s$, then\n\n$$\n\\{1, 2, \\dots, p-1\\} = \\left\\lfloor \\frac{k-1}{2} \\right\\rfloor \\times I_s\n$$\n\nand so $3^k \\in I_s$. Thus, there is an integer $l$ such that $2^l \\equiv 3^j \\pmod{p}$ and $1 < l < s$. If we choose $(n, m) = (s - l, j)$, then it is not difficult to see that the pair satisfies the required conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19686,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive numbers $x$, $y$, and $z$ such that $xyz = 1$, prove the inequality:\n\n$$\n\\sqrt[3]{\\frac{x+y}{2z}} + \\sqrt[3]{\\frac{y+z}{2x}} + \\sqrt[3]{\\frac{z+x}{2y}} \\leq \\frac{5(x+y+z)+9}{8}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us introduce new variables:\n\n$$\na = \\sqrt[3]{x}, \\quad b = \\sqrt[3]{y}, \\quad c = \\sqrt[3]{z} \\implies x = a^3, \\quad y = b^3, \\quad z = c^3.\n$$\n\nThe condition $xyz=1$ becomes $a^3b^3c^3 = 1 \\implies abc=1$, and the inequality becomes:\n\n$$\n\\sqrt[3]{\\frac{a^3+b^3}{2c^3}} + \\sqrt[3]{\\frac{b^3+c^3}{2a^3}} + \\sqrt[3]{\\frac{c^3+a^3}{2b^3}} \\leq \\frac{5(a^3+b^3+c^3)+9}{8}.\n$$\n\nLet us use Schur's inequality:\n\n$$\na^3 + b^3 + c^3 + 3abc \\geq ab(a+b) + bc(b+c) + ca(c+a).\n$$\n\nUsing this, we can write:\n\n$$\n\\frac{5(a^3 + b^3 + c^3) + 9}{8} = \\frac{2(a^3 + b^3 + c^3) + (3(a^3 + b^3 + c^3) + 9abc)}{8} \\\\\n= \\frac{2(a^3 + b^3 + c^3) + 3(ab(a+b) + bc(b+c) + ca(c+a))}{8} = \\frac{(a+b)^3 + (b+c)^3 + (c+a)^3}{8}.\n$$\n\nThus, it suffices to prove the following inequality for arbitrary positive numbers $a$ and $b$:\n\n$$\n\\sqrt[3]{\\frac{a^3+b^3}{2c^3}} \\leq \\frac{(a+b)^3}{8abc} \\implies \\sqrt[3]{\\frac{a^3+b^3}{2}} \\leq \\frac{(a+b)^3}{8ab}.\n$$\n\nMultiplying both sides by $8ab$ and cubing, we obtain an equivalent inequality:\n\n$$\n256a^3b^3(a^3+b^3) \\leq (a+b)^9.\n$$\n\nExpanding $(a+b)^9 - 256a^3b^3(a^3+b^3)$ and factoring, we get:\n\n$$\n(a+b)^9 - 256a^3b^3(a^3+b^3) = (a-b)^4(a^5+13a^4b+82a^3b^2+82a^2b^3+13ab^4+b^5) \\geq 0.\n$$\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19687,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 cards, each with one of the numbers $1, 2, \\ldots, 100$ written on it, such that each number appears on exactly one card. The cards are stacked so that the numbers $1, 2, \\ldots, 100$ are written from top to bottom in order.\n\nPetrik rearranges the cards according to the following rules: If, before his $k$-th move, the numbers on the cards from top to bottom are $a_1, a_2, \\ldots, a_{k-1}, a_k, a_{k+1}, \\ldots, a_{100}$, then after his move they are arranged as $a_k, a_{k-1}, \\ldots, a_2, a_1, a_{k+1}, \\ldots, a_{100}$ (so the order does not change when $k=1$).\n\nPetrik takes turns making moves $1, 2, \\ldots, 100$, and then makes moves $1, 2, \\ldots, 100$ again, and so on. Will the initial arrangement of the cards with the numbers $1, 2, \\ldots, 100$ from top to bottom necessarily be repeated after a finite number of moves?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider two moves: $2k-1$ and $2k$. Suppose that before the first of them, the layout of cards is $a_1, a_2, \\ldots, a_{2k-1}, a_{2k}, a_{2k+1}, \\ldots, a_{100}$. Then after these two moves, the changes are:\n\n$$\na_{2k-1}, a_{2k-2}, \\dots, a_1, a_{2k}, a_{2k+1}, \\dots, a_{100} \\to a_{2k}, a_1, a_2, \\dots, a_{2k-1}, a_{2k+1}, \\dots, a_{100}.\n$$\n\nThus, after each pair of moves, the corresponding even number is moved to the first place. Therefore, after 100 moves, the layout of cards will be: $100, 98, 96, \\ldots, 4, 2, 1, 3, 5, \\ldots, 97, 99$. We call this perturbation a *megamove*. Let $A_0$ be the initial position, $A_1$ the position after the first megamove, $A_2$ after the second, and so on.\n\nIt is not difficult to see that if position $A_m$ arises from $A_{m+1}$, then vice versa, it cannot arise from any other position except $A_m$. Therefore, after a sufficiently large number of megamoves, the positions must start repeating because there are only finitely many possible arrangements. If a position is repeated, then all previous ones coincide. Thus, the initial position must also repeat after a finite number of moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19688,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0 > 0$ be a real number, and let\n\n$$\na_n = \\frac{a_{n-1}}{\\sqrt{1 + 2020 \\cdot a_{n-1}^2}}, \\quad \\text{for } n = 1, 2, \\dots, 2020.\n$$\n\nShow that $a_{2020} < \\frac{1}{2020}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_n = \\frac{1}{a_n^2}$. Then $b_0 = \\frac{1}{a_0^2}$ and\n\n$$\nb_n = \\frac{1 + 2020 \\cdot a_{n-1}^2}{a_{n-1}^2} = b_{n-1} \\left( 1 + 2020 \\cdot \\frac{1}{b_{n-1}} \\right) = b_{n-1} + 2020.\n$$\n\nHence $b_{2020} = b_0 + 2020^2 = \\frac{1}{a_0^2} + 2020^2$ and $a_{2020}^2 = \\frac{1}{\\frac{1}{a_0^2} + 2020^2} < \\frac{1}{2020^2}$ which shows that $a_{2020} < \\frac{1}{2020}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19689,
    "subject": "Mathematics (Olympiad)",
    "question": "在一場對戰賽事中，考慮參賽者的住宿安排。設 $i > k$ 時，有\n\n$$\ne_{i+1} - b_{i+1} + 1 \\geq C_2^{2k} - 2C_2^i + C_2^{2i-2k} = (2k-i)^2.\n$$\n\n因此，旅館總費用的下界為\n\n$$\n\\sum_{i=0}^{k} (k(2k-1) - i(i-1)) + \\sum_{i+1}^{2} k(2k-i)^2 = \\frac{1}{2}k(4k^2 + k - 1).\n$$\n\n請構造一個安排，使上述等號都成立。",
    "options": [],
    "answer": "See solution",
    "solution": "以下是一個構造，使得上述所有等號都成立：\n\n首先，將所有參賽者分成大小為 $k$ 的兩群，$X = \\{X_1, \\cdots, X_k\\}$ 與 $Y = \\{Y_1, \\cdots, Y_k\\}$。賽事分為四段：\n\n1. $X_1, \\cdots, X_k$ 依序入住，並在 $X_i$ 入住後與所有已入住的 $X_j$ 比賽。\n2. $Y_1, \\cdots, Y_k$ 依序入住，並在 $Y_j$ 入住後與所有 $X_i$（$i > j$）比賽。\n3. $X_k, \\cdots, X_1$ 依序離開，並在 $X_i$ 離開前與所有 $Y_j$（$i \\leq j$）比賽。\n4. $Y_k, \\cdots, Y_1$ 依序離開，並在 $Y_i$ 離開前與所有剩餘的 $Y_j$ 比賽。\n\n對於 $0 \\leq s \\leq k-1$，從 $T_{k-s}$ 入住到 $S_{k-s}$ 離開，期間的比賽數量為\n\n$$\n\\sum_{j=k-s}^{k-1} (k-j) + 1 + \\sum_{j=k-s}^{k-1} (k-j+1) = (s+1)^2.\n$$\n\n當 $i > k$ 時，$b_{i+1}$ 為 $T_{i-k+1}$ 抵達日，$e_{i+1}$ 為 $S_{i-k+1}$ 離開日，因此\n$e_{i+1} - b_{i+1} + 1 = (2k-i)^2$，即達到下界。\n\n當 $i \\leq k$ 時，$b_{i+1}$ 前的 $i$ 名參賽者都屬於 $X$，所以 $b_{i+1} = C_2^i + 1$；同理，$e_{i+1}$ 之後的 $i$ 名參賽者都屬於 $Y$，所以 $e_{i+1} = C_2^{2k} - C_2^i$。因此等號成立，達到最小值。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19690,
    "subject": "Mathematics (Olympiad)",
    "question": "已知正整數 $n$ 的所有因數除了 $1$ 以外都不是完全平方數。\n\n試證：不存在任何互質的正整數 $x$ 和 $y$，使得 $x^n + y^n$ 是 $(x+y)^3$ 的倍數。",
    "options": [],
    "answer": "See solution",
    "solution": "設存在互質的正整數 $x, y$ 使得 $x^n + y^n$ 是 $(x+y)^3$ 的倍數。令 $s = x + y$，則 $s > 2$。\n\n1. 若 $n$ 為偶數，因 $x^n + y^n = x^n + (s - x)^n = 2x^n \\pmod{s}$。但同時 $x^n + y^n$ 是 $(x + y)^3$ 的倍數，即 $x^n + y^n \\equiv 0 \\pmod{s}$，故 $2x^n \\equiv 0 \\pmod{s}$。由於 $x$ 和 $y$ 互質，故 $x$ 和 $s$ 互質，因此必有 $2 \\equiv 0 \\pmod{s}$，也就是 $x = y = 1$，矛盾！\n\n2. 若 $n$ 為奇數，因 $x^n + y^n = x^n + (s - x)^n = C_2^n s^2 (-x)^{n-2} + C_1^n s (-x)^{n-1} \\pmod{s^3}$。同時 $x^n + y^n$ 是 $(x + y)^3 = s^3$ 的倍數，即 $x^n + y^n \\equiv 0 \\pmod{s^3}$，故 $C_2^n s^2 (-x)^{n-2} + C_1^n s (-x)^{n-1} \\equiv 0 \\pmod{s^3}$。這表示存在整數 $k$ 使得\n\n$$\n-\\frac{1}{2} n(n-1) s x^{n-2} + n x^{n-1} = k s^2. \\quad (1)\n$$\n\n由此可知 $s \\mid n x^{n-1}$，又因 $(x, s) = 1$，故得 $s \\mid n x^{n-2}$，從而 $s^2 \\mid \\frac{1}{2} n(n-1) s x^{n-2}$。帶回 Eq. (1)，得出 $s^2 \\mid n x^{n-1}$，再因 $(x, s) = 1$ 知 $s^2 \\mid n$，但這與 $n$ 無平方因數不合，矛盾！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19691,
    "subject": "Mathematics (Olympiad)",
    "question": "Start with a triplet $ (a, b, c) $ of positive integers with $ \\gcd(a, b, c) = 1 $. In one move, you may add to an element of the triplet an integer multiple of another element—the multiple can be negative or positive. Prove that in at most five steps one can reach the triplet $ (1, 0, 0) $.\n\n*Note.* One may use Dirichlet's Theorem.\n",
    "options": [],
    "answer": "See solution",
    "solution": "The goal is to reach in three steps a triplet $ (1, m, n) $, since two steps are required thereafter to obtain $ (1, 0, 0) $. To obtain the number $1$ in the first position, it is convenient to have $ \\gcd(m, n) = 1 $.\n\nLet $ d = \\gcd(a, b) $ and write $ a = dx, b = dy $, where $ x $ and $ y $ are coprime positive integers. By Dirichlet's Theorem, the sequence $ y + nx $ for $ n \\ge 1 $ contains infinitely many primes, hence infinitely many primes coprime with $ c $. Choose $ n $ such that $ \\gcd(c, y + nx) = 1 $. Since $ \\gcd(a, b, c) = \\gcd(d, c) = 1 $, it follows that $ \\gcd(c, b + na) = 1 $. Denote $ b' = b + na $. The first move is $ (a, b, c) \\mapsto (a, b', c) $.\n\nSince $ \\gcd(b', c) = 1 $, by Bézout's identity, there exist integers $ u, v $ such that $ u b' + v c = 1 $. Multiplying by $ a - 1 $, we obtain $ u', v' $ such that $ u' b' + v' c = a - 1 $. The next two moves are:\n\n$$\n(a, b', c) \\mapsto (a - u' b', b', c) \\mapsto (a - u' b' - v' c, b', c) = (1, b', c).\n$$\n\nFinally:\n\n$$\n(1, b', c) \\mapsto (1, b' + (-b') \\cdot 1, c) \\mapsto (1, 0, c + (-c) \\cdot 1) = (1, 0, 0).\n$$\n",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19692,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given\n\n$$\nx^3 - y^3 = 7(x - y)\n$$\n\n$$\nx^3 + y^3 = 5(x + y)\n$$\n\nFind the sum of $x^2 + y^2$ over all real solutions $(x, y)$ to these equations.",
    "options": [],
    "answer": "See solution",
    "solution": "Method 2\n\nWe are given\n\n$$\nx^3 - y^3 = 7(x - y)\n$$\n\n$$\nx^3 + y^3 = 5(x + y)\n$$\n\nBy adding the two equations, we find that $y = 6x - x^3$.\n\nSubstituting in equation (2), ignoring the obvious solution $x = 0 = y$ and eventually substituting $t = x^2$, we have\n\n$$\n\\begin{aligned}\nx^3 + (6x - x^3)^3 &= 5x + 5(6x - x^3) = 35x - 5x^3 \\\\\nx^2 + x^2(6 - x^2)^3 &= 35 - 5x^2 \\\\\nt + t(6 - t)^3 &= 35 - 5t \\\\\nt(216 - 108t + 18t^2 - t^3) &= 35 - 6t \\\\\nt^4 - 18t^3 + 108t^2 - 222t + 35 &= 0\n\\end{aligned}\n$$\n\nThe sum of the roots for $t$ is $-(-18) = 18$. If $t < 0$, then the left side of the last equation is positive. So $t > 0$, hence there are 2 values for $x$ for each value of $t$. So assuming there are 4 distinct roots for $t$, the sum of $x^2$ over all real solutions $(x, y)$ of the original equations is $2 \\times 18 = 36$.\n\nSince the original equations are symmetric in $x$ and $y$, the sum of $y^2$ over all real solutions $(x, y)$ of the original equations is also 36. So the sum of $x^2 + y^2$ over all real solutions $(x, y)$ is $2 \\times 36 = \\mathbf{72}$.\n\nTo show that $Q(t) = t^4 - 18t^3 + 108t^2 - 222t + 35$ has 4 distinct roots, we use the remainder theorem. We have $Q(1) < 0$, $Q(3) < 0$, then $Q(5) = 0$ and $Q(7) = 0$. Dividing $Q(t)$ by $(t - 5)(t - 7) = t^2 - 12t + 35$ gives $t^2 - 6t + 1$. So $Q(t)$ has four distinct roots: 5, 7, $3 \\pm 2\\sqrt{2}$. So $n = \\mathbf{9}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19693,
    "subject": "Mathematics (Olympiad)",
    "question": "A point $X$ is chosen on the median $AD$ of triangle $ABC$. The circumcircle of triangle $ABX$ intersects the median $BE$ of triangle $ABC$ at point $Y \\neq B$. The circumcircle of triangle $EXY$ intersects the line $DE$ at point $K \\neq E$. Prove that the location of $K$ does not depend on $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will use directed angles (see the figures below for both possible configurations). As $DE$ is a midline in $ABC$, we have $DE \\parallel AB$. Thus\n\n$$\n\\angle DEY = \\angle DEB = \\angle ABE = \\angle ABY = \\angle AXY = \\angle DXY.\n$$\n\nTherefore, points $D$, $E$, $X$, $Y$ are concyclic. This means that the circumcircle of $EXY$ intersects $DE$ at $D$. Thus $K = D$, i.e., the midpoint of $BC$, regardless of the choice of $X$.\n\n![](images/EST_ABooklet_2024_p19_data_bfb1ed0170.png)\n\nFig. 3\n\n![](images/EST_ABooklet_2024_p19_data_4dd1039c94.png)\n\nFig. 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19694,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be non-negative real numbers such that $a + b \\leq c + 1$, $b + c \\leq a + 1$, and $c + a \\leq b + 1$. Prove that\n\n$$\na^2 + b^2 + c^2 \\leq 2abc + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding the first two inequalities, we get $2b \\leq 2$, so $b \\leq 1$. Similarly, $c \\leq 1$ and $a \\leq 1$. Let $\\alpha = 1 - a$, $\\beta = 1 - b$, and $\\gamma = 1 - c$. Then $0 \\leq \\alpha, \\beta, \\gamma \\leq 1$.\n\nThe inequality to be proved reduces to\n\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 \\leq 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) - 2\\alpha\\beta\\gamma.\n$$\n\nAssume $\\gamma$ is the largest among the three, so $\\alpha \\leq \\gamma$ and $\\beta \\leq \\gamma$. Using $\\gamma \\leq \\alpha + \\beta$, we get $\\gamma^2 \\leq \\gamma(\\alpha + \\beta)$. Also, $\\alpha^2 \\leq \\alpha\\gamma$ and $\\beta^2 \\leq \\beta\\gamma$. Thus,\n\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 \\leq 2(\\alpha\\gamma + \\beta\\gamma).\n$$\n\nHence, it suffices to prove that $2\\alpha\\beta\\gamma \\leq 2\\alpha\\beta$, which follows from $\\gamma \\leq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19695,
    "subject": "Mathematics (Olympiad)",
    "question": "2024 girls each has her own doll. Consider all the ways to distribute a doll to each of the 2024 girls. For any $k \\ge 0$, let $p(k)$ be the number of ways to distribute the dolls so that there are exactly $k$ girls who received her own doll. Prove that\n\n$$\n\\sum_{k=0}^{2024} k \\times p(k) = 2024!.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The original statement is equivalent to considering permutations of $\\{1,2,\\ldots,2024\\}$, where $p(k)$ is the number of permutations with exactly $k$ fixed points. The left side of the equation counts the total number of fixed points across all permutations. Note that the number of permutations where 1 is a fixed point is $(2024-1)!$, and similarly for each other point. Therefore, the total number of fixed points in all permutations is $2024 \\times (2024-1)! = 2024!$, so the equation holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19696,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為某個大於 1 的奇數，而 $f(x)$ 為 $x$ 的 $n$ 次多項式。已知 $f(k) = 2^k$ 對於 $k = 0, 1, \\dots, n$ 均成立。試證：使 $f(x)$ 的值為 2 的幂次的整數 $x$ 僅為有限多個。",
    "options": [],
    "answer": "See solution",
    "solution": "由於 $n+1$ 個值已可唯一決定一個 $n$ 次多項式，且\n\n$$\nf(k) = 2^k = (1+1)^k = C(k, 0) + C(k, 1) + \\dots + C(k, n)\n$$\n\n對於 $k = 0, 1, \\dots, n$ 都成立，而右式為一 $n$ 次多項式，故知\n\n$$\nf(x) = C(x, 0) + C(x, 1) + \\dots + C(x, n).\n$$\n\n又因為 $n$ 是奇數，將上式兩兩合併，可得\n\n$$\n\\begin{aligned}\nf(x) &= C(x + 1, 1) + C(x + 1, 3) + \\dots + C(x + 1, n) \\\\\n&= (x + 1) \\left[ 1 + \\frac{1}{3}C(x, 2) + \\frac{1}{5}C(x, 4) + \\dots + \\frac{1}{n}C(x, n - 1) \\right].\n\\end{aligned}\n$$\n\n令 $n!f(x) = (x+1)R(x)$，注意到 $R(x)$ 為整係數多項式。對於所有整數 $x$，我們有\n\n$$\n\\gcd(x + 1, R(x)) \\mid R(-1) = n! \\left[ 1 + \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{n} \\right].\n$$\n\n注意到 $R(-1)$ 為一非零整數，因此 $\\nu_2(R(-1))$ 必為有限值，其中\n\n$$\n\\nu_2(m) := \\sup \\{k : 2^k \\mid m\\}.\n$$\n\n換言之，我們有\n\n$$\n\\begin{aligned}\n& \\min\\{\\nu_2(x+1), \\nu_2(R(x))\\} \\\\\n&= \\nu_2(\\gcd(x+1, R(x))) \\le \\nu_2(R(-1)) < \\infty.\n\\end{aligned}\n$$\n\n現在，假設 $x$ 為一讓 $f(x)$ 為 2 的幂次的整數，則我們有 $x + 1 \\mid n! \\times 2^{\\nu_2(f(x))}$ 且 $R(x) \\mid n! \\times 2^{\\nu_2(f(x))}$。但由上式，這意味著\n\n$$\nx + 1 \\mid n! \\times 2^{\\nu_2(x+1)} \\mid n! \\times 2^{\\nu_2(R(-1))}\n$$\n\n與\n\n$$\nR(x) \\mid n! \\times 2^{\\nu_2(R(x))} \\mid n! \\times 2^{\\nu_2(R(-1))}\n$$\n\n至少有一個成立。然而，由於\n\n$$\n\\lim_{|x| \\to \\infty} |x + 1| = \\infty \\quad \\text{且} \\quad \\lim_{|x| \\to \\infty} |R(x)| = \\infty,\n$$\n\n易知能讓上述條件至少滿足一條的 $x$ 至多為有限多個。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19697,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following sequence of tables:\n\n![]( <table><tr><td></td></tr></table> )\n\n1-table\n\n![]( <table><tr><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td></tr></table> )\n\n2-table\n\n![]( <table><tr><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr></table> )\n\n3-table\n\nIn each cell of a $k$-table there are a switch and a bulb. Initially all bulbs are off. Pressing a switch changes the state (from on to off and vice versa) only of the bulbs in the cells adjacent to the cell of the switch. (Two cells are adjacent if they have a common side.) For each value of $k$, determine the maximum number of bulbs that can be turned on in the $k$-table by pressing several switches.",
    "options": [],
    "answer": "See solution",
    "solution": "A $k$-table has $2k^2 + 2k$ cells. Imagine coloring them in a chessboard pattern, with the top right cell black. The white part consists of $k$ white diagonals with $k+1$ cells each, running from top left to bottom right. The black part consists of $k$ black diagonals with $k+1$ cells each, running from top right to bottom left. Pressing the switch in a cell $C$ changes only the state of bulbs in adjacent cells of the opposite color, and only in diagonals containing cells adjacent to $C$. If a diagonal (of the opposite color) is affected by pressing the switch in $C$, then exactly two bulbs in it change state. Thus, the parity of the number of bulbs on in each diagonal is constant. Since all bulbs are initially off, each diagonal contains an even number of bulbs on after any number of moves.\n\nIf $k$ is even, at least one of the $k+1$ bulbs in each of the $2k$ diagonals is off after any number of moves. So at most $(2k^2 + 2k) - 2k = 2k^2$ bulbs can be on. (No similar restriction applies for $k$ odd.)\n\nFor $k$ odd, all $2k^2 + 2k$ bulbs can be turned on; for $k$ even, exactly $2k^2$ bulbs can be on. The procedure for $k$ odd involves pressing switches in odd-numbered rows: press the first two switches in row 1, pairs (1,2) and (5,6) in row 3, and so on, following a pattern. For odd-numbered rows $k+2, k+4, \\dots, 2k-1$, press two switches and skip two, starting with the second switch in each row. Every cell has exactly one neighbor whose switch is pressed, so all bulbs will be on eventually.\n\n![](images/Argentina_2011_p32_data_2a8eb53ab1.png)\n\n$k$ odd\n\n![](images/Argentina_2011_p32_data_8c27b4f2b5.png)\n\n$k$ even\n\nFor $k$ even, view the $k$-table $T$ as an extension of a $(k-1)$-table $T'$ sharing the same center. The cells of $T$ outside $T'$ are the border cells. Apply to $T'$ the sequence of switches described for the odd case, turning on all cells in $T'$. Additionally, all border cells of $T$ above the middle horizontal line are turned on, but those below are off. There are $2k$ such cells, so exactly $2k^2$ bulbs can be on.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19698,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $k$ 為一正整數。一個網球錦標賽的賽事委員會要為 $2k$ 位參賽者安排賽程，其中：\n\n1. 任兩位參賽者恰好對戰一場；\n2. 每一天錦標賽只會打一場；\n3. 每位參賽者在他的第一場比賽當天入住旅館，並在他的最後一場比賽當天離開旅館。\n\n對於每位參賽者住在旅館內的每一天，大會都要支付 1 元，因此賽事委員會希望極小化付給旅館的總費用。試求旅館總費用的極小值並證明之。",
    "options": [],
    "answer": "See solution",
    "solution": "最小值為 $\\dfrac{k(4k^2 + k - 1)}{2}$。\n\n將賽事的每一天依序編號為 $1, 2, \\dots, \\binom{2k}{2}$。設每位選手到達的日期排序為 $b_1 \\leq b_2 \\leq \\cdots \\leq b_{2k}$，離開的日期排序為 $e_1 \\geq e_2 \\geq \\cdots \\geq e_{2k}$（注意這裡並非以參賽者編號，一名參賽者可能在第 $b_i$ 天入住但在第 $e_j$ 天離開）。旅館總費用為：\n\n$$\n\\sum_{i=1}^{2k} (e_i - b_i + 1)\n$$\n\n**(1) 下界估計：**\n\n注意在第 $b_{i+1}$ 天前只有 $i$ 位參賽者入住，因此至多只有 $\\binom{i}{2}$ 場比賽，故 $b_{i+1} \\leq \\binom{i}{2} + 1$。同理，在第 $e_{i+1}$ 天只剩 $i$ 名參賽者，至多只能打 $\\binom{i}{2}$ 場，故 $e_{i+1} \\geq \\binom{2k}{2} - \\binom{i}{2}$。\n\n因此：\n\n$$\ne_{i+1} - b_{i+1} + 1 \\geq \\binom{2k}{2} - 2\\binom{i}{2} = k(2k-1) - i(i-1)\n$$\n\n當 $i > k$ 時，上式可進一步優化：考慮最先來的 $i$ 名參賽者與最後走的 $i$ 名參賽者，其中至少有 $2i - 2k$ 位重複。重複的這些參賽者的對戰在上式被重複計算了兩次，因此當 $i > k$ 時：\n\n$$\ne_{i+1} - b_{i+1} + 1 \\geq \\binom{2k}{2} - 2\\binom{i}{2} + \\binom{2i-2k}{2} = (2k-i)^2\n$$\n\n因此旅館總費用的下界為：\n\n$$\n\\sum_{i=0}^{k-1} [k(2k-1) - i(i-1)] + \\sum_{i=k}^{2k-1} (2k-i)^2 = \\frac{1}{2}k(4k^2 + k - 1)\n$$\n\n**(2) 構造達到下界的方案：**\n\n將所有參賽者分成大小為 $k$ 的兩組，$X = \\{X_1, \\dots, X_k\\}$ 與 $Y = \\{Y_1, \\dots, Y_k\\}$。將賽事分為四段：\n\n1. $X_1, \\dots, X_k$ 依序入住，入住後與所有已入住的 $X_j$ 比賽。\n2. $Y_1, \\dots, Y_k$ 依序入住，入住後與所有 $X_i$（$i > j$）比賽。\n3. $X_k, \\dots, X_1$ 依序離開，離開前與所有 $Y_j$（$i \\leq j$）比賽。\n4. $Y_k, \\dots, Y_1$ 依序離開，離開前與剩餘的 $Y_j$ 比賽。\n\n對於 $0 \\leq s \\leq k-1$，從 $Y_{k-s}$ 入住到 $X_{k-s}$ 離開，期間的比賽數量為：\n\n$$\n\\sum_{j=k-s}^{k-1} (k-j) + 1 + \\sum_{j=k-s}^{k-1} (k-j+1) = (s+1)^2\n$$\n\n當 $i > k$ 時，$b_{i+1}$ 為 $Y_{i-k+1}$ 抵達日，$e_{i+1}$ 為 $X_{i-k+1}$ 離開日，從而 $e_{i+1} - b_{i+1} + 1 = (2k-i)^2$，即達到下界。\n\n當 $i \\leq k$ 時，在 $b_{i+1}$ 前的 $i$ 名參賽者都屬於 $X$，因此 $b_{i+1} = \\binom{i}{2} + 1$。同理，在 $e_{i+1}$ 之後的 $i$ 名參賽者都屬於 $Y$，因此 $e_{i+1} = \\binom{2k}{2} - \\binom{i}{2}$。因此下界可達到。\n\n故最小總費用為 $\\boxed{\\dfrac{k(4k^2 + k - 1)}{2}}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19699,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "From $f(1) + f(2) + f(f(1)) = 4$ it follows that $f(1) \\in \\{1, 2\\}$.\n\nIf $f(1) = 1$, then $f(2) = 2$, and by induction, $f(n) = n$.\n\nIf $f(1) = 2$, then $f(2) = 1$, and inductively,\n$$\nf(n) = \\begin{cases} n + 1, & \\text{if } n \\text{ is odd} \\\\ n - 1, & \\text{if } n \\text{ is even} \\end{cases}.\n$$\n\nBoth functions satisfy the initial condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19700,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n \\geq 3$, plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1} + a_n = 2^n$ for all $n \\geq 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.*\n\nWe will show that $a_n = \\frac{1}{3}(2^{n+1} + (-1)^n)$. This would be sufficient, since then we would have\n\n$$\na_{n-1} + a_n = \\frac{1}{3}(2^n + (-1)^{n-1}) + \\frac{1}{3}(2^{n+1} + (-1)^n) = \\frac{1}{3}(2^n + 2 \\cdot 2^n) = 2^n.\n$$\n\n*Lemma 1.* For all positive integers $n$, we have\n\n$$\n\\sum_{k=0}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k} 2^k = \\frac{1}{3}(2^{n+1} + (-1)^n).\n$$\n\n*Proof.* We argue by strong induction. To begin, the cases $n = 1$ and $n = 2$ are quickly verified. Now suppose that $n \\geq 3$ is odd, say $n = 2m + 1$. We find that\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{m} \\binom{2m+1-k}{k} 2^k &= 1 + \\sum_{k=1}^{m} \\binom{2m-k}{k} 2^k + \\sum_{k=1}^{m} \\binom{2m-k}{k-1} 2^k \\\\\n&= \\sum_{k=0}^{m} \\binom{2m-k}{k} 2^k + 2 \\sum_{k=0}^{m-1} \\binom{2m-1-k}{k} 2^k \\\\\n&= \\frac{1}{3}(2^{2m+1} + 1) + \\frac{2}{3}(2^{2m} - 1) \\\\\n&= \\frac{1}{3}(2^{2m+2} - 1),\n\\end{aligned}\n$$\nusing the induction hypothesis for $n = 2m$ and $n = 2m - 1$. For even $n$ the computation is similar. This yields the claim. $\\square$\n\nWe now determine the number of ways to advance around the circle twice, organizing our count according to the points visited both times around the circle. It is straightforward to check that no two such points may be adjacent, and that there are exactly two sequences of moves leading from any such point to the next. (These sequences involve only moves of length two except possibly at the endpoints.) Hence given $k \\geq 1$ points around the circle, no two adjacent and not including point $A$, there would appear to be $2^k$ ways to traverse the circle twice without repeating a move. However, half of these options lead to repeating the same route twice, giving $2^{k-1}$ ways in actuality. There are $\\binom{n-k}{k}$ ways to select $k$ nonadjacent points on the circle not including $A$ (add an extra point behind each of $k$ chosen points), for a total contribution of\n\n$$\n\\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k} 2^{k-1} = \\frac{1}{2} \\left[ -1 + \\sum_{k=0}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k} 2^k \\right] = \\frac{1}{6} (2^{n+1} + (-1)^n) - \\frac{1}{2},\n$$\nwhere we used Lemma 1 in the last step.\n\nOn the other hand, if the $k \\geq 1$ nonadjacent points do include point $A$ then there are $\\binom{n-k-1}{k-1}$ ways to choose them around the circle. (Select $A$ but not the next point, then add an extra point after each of $k-1$ selected points.) But now there are actually $2^k$ ways to circle twice, since we can choose either move at $A$ and the subsequent points, then select the other options the second time around. Hence the contribution in this case is\n\n$$\n\\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\binom{n-k-1}{k-1} 2^k = 2^{\\lfloor (n-2)/2 \\rfloor} \\sum_{k=0}^{\\lfloor (n-2)/2 \\rfloor} \\binom{n-2-k}{k} 2^k = \\frac{2}{3}(2^{n-1} + (-1)^n),\n$$\nwhere we again used Lemma 1.\n\nFinally, if $n$ is odd then there is one additional way to circle in which no point is visited twice by using only steps of length two, giving a contribution of $\\frac{1}{2}(1 - (-1)^n)$. Therefore the total number of paths is\n\n$$\n\\frac{1}{6}(2^{n+1} + (-1)^n) - \\frac{1}{2} + \\frac{2}{3}(2^{n-1} + (-1)^n) + \\frac{1}{2}(1 - (-1)^n),\n$$\nwhich simplifies to $\\frac{1}{3}(2^{n+1} + (-1)^n)$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19701,
    "subject": "Mathematics (Olympiad)",
    "question": "For an odd natural number $n > 1$, define the set of different remainders of powers of $2$ when dividing by $n$:\n\n$$\nS_n = \\{a \\mid a < n,\\ \\exists k \\in \\mathbb{N} : 2^k \\equiv a \\pmod{n}\\}.\n$$\n\nAre there different odd numbers $m$ and $r$ such that $S_m = S_r$?\n",
    "options": [],
    "answer": "See solution",
    "solution": "No! There exists a natural number $s$ such that $2^s \\equiv 1 \\pmod{n}$ (for example, $s = \\varphi(n)$ by Euler's theorem, or because the sequence of powers of $2$ modulo $n$ is periodic). We have $2^{s-1} \\equiv \\frac{n+1}{2} \\pmod{n}$, so $x = \\frac{n+1}{2} \\in S_n$, but $2x = n+1 > n$ is not in $S_n$. Also, if $t \\leq \\frac{n-1}{2}$ is in $S_n$, then $2t < n$ is also in $S_n$. Therefore, the smallest natural number $t$ such that $t \\in S_n$ and $2t \\notin S_n$ is $\\frac{n+1}{2}$. Since this number is different for different $n$, we get the desired result. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19702,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $O$ be its circumcircle. Let $O'$ be the circle that is tangent to $O$ at $A$ and tangent to the side $BC$ at $D$. $O'$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. $O'$ intersects the lines $OO'$ and $EO'$ at $A' \\neq A$ and $G \\neq E$, respectively. The lines $BO$ and $A'G$ intersect at $H$. Prove that $DF^2 = AF \\cdot GH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the intersection of the line $BO$ with the circle $O$. Since the line segments $EG$ and $BX$ are diameters of the circles $O$ and $O'$, respectively, $\\angle EAG = \\angle BAX = 90^\\circ$ and hence the three points $A$, $G$, $X$ are collinear.\n\nSince $O'$ is tangent to $O$ at $A$, we have $\\angle AEG = \\angle ABX = \\angle ABH$, and thus the lines $EG$ and $BH$ are parallel. Since the segment $AA'$ is a diameter of the circle $O'$, $\\angle AGA' = \\angle AGE + \\angle EGA' = 90^\\circ$. So from $\\angle AGE + \\angle AEG = 90^\\circ$ it follows that $\\angle AEG = \\angle EGA' = \\angle EGH$. Thus, the quadrilateral $BEGH$ is a parallelogram and therefore, we get\n\n$$\nGH = BE. \\tag{1}\n$$\n\nOn the other hand, since the lines $EF$ and $BC$ are also parallel, we have $\\angle EFD = \\angle FDC$. Thus, we get\n\n$$\n\\angle EFD = \\angle FDC = \\angle DAF = \\angle DEF,\n$$\n\nwhich implies that\n\n$$\nDF = DE. \\tag{2}\n$$\n\nIn particular, since $\\angle ADB = \\angle AFD$ and\n\n$$\n\\angle BAD = \\angle BDF = \\angle DEF = \\angle EFD = \\angle CDF = \\angle CAD,\n$$\n\nthe two triangles $ABD$ and $ADF$ are similar and therefore, we get\n\n$$\n\\frac{AF}{AD} = \\frac{DF}{BD}. \\tag{3}\n$$\n\nSince the triangles $ABD$ and $DBE$ are also similar, we get\n\n$$\n\\frac{DE}{AD} = \\frac{BE}{BD}. \\tag{4}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19703,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\nBA = aI_n + bAB.\n$$\nProve that $A$ and $B$ are invertible.",
    "options": [],
    "answer": "See 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 $f^{[k]} = \\underbrace{f \\circ f \\circ \\dots \\circ f}_{k \\text{ times}}$. 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 $f^{[p]}(0) = f^{[q]}(0)$. Because\n$$\nf^{[k]}(z) = b^k z + a \\frac{b^k - 1}{b - 1}, \\quad z \\in \\mathbb{C}, \\; k \\in \\mathbb{N}^*,\n$$\nwe 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": 19704,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine, with proof, whether there is any odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ such that all $p_i + p_{i+1}$ ($i = 1, 2, \\dots, n$, and $p_{n+1} = p_1$) are perfect squares.\n\n![](posed_by_Tao_Pingsheng.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is negative. Suppose that there exist an odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ satisfying the given condition.\n\nIf all $p_1, p_2, \\dots, p_n$ are odd, then all the sums $p_i + p_{i+1}$ are multiples of $4$, so the prime numbers $p_1, p_2, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternately, which contradicts the fact that $n$ is odd.\n\nIf one of $p_1, p_2, \\dots, p_n$ is $2$, then without loss of generality, assume $p_1 = 2$. As both $p_1 + p_2$ and $p_n + p_1$ are perfect squares and both are odd, it follows that $p_2$ and $p_n$ are congruent to $3$ modulo $4$. Similar to the first case, the primes $p_2, p_3, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternately, so $n-1$ is odd, which is a contradiction.\n\nHence, there are no odd integer $n \\ge 3$ and $n$ primes satisfying the given conditions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19705,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $N$ such that $N$ is divisible by $99$ and the sum of its digits $S(N)$ is minimized.",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that $N$ is divisible by $9$ if and only if $S(N)$ is divisible by $9$, and $N$ is divisible by $11$ if and only if $d(N) := o(N) - e(N)$ is divisible by $11$, where $o(N)$ and $e(N)$ are the sums of the digits in odd and even positions, respectively.\n\nSuppose $S(N) = 9$. Then $N$ must have at least five digits, and\n$$\n3 \\leq o(N) \\leq 7, \\quad 2 \\leq e(N) \\leq 6,\n$$\nwith $o(N)$ and $e(N)$ of opposite parity. Thus, $0 < |d(N)| \\leq 5$, so $N$ cannot be divisible by $11$.\n\nTherefore, $S(N) \\geq 18$ for $N$ to be divisible by $99$. For $S(N) = 18$, to minimize $N$, use as many $2$s as possible: eight $2$s and two $1$s. Placing the $1$s in the leading positions gives $N = 1122222222$. Here, $o(N) = e(N) = 9$, so $d(N) = 0$, and $N$ is divisible by $11$ and $9$, hence by $99$.\n\nFor $S(N) \\geq 27$, $N$ would have at least 14 digits, which is larger than $1122222222$.\n\nThus, the minimal such $N$ is $\\boxed{1122222222}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19706,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $a, b, c$ се реални броеви за кои $a + b + c = 4$ и $a, b, c > 1$. Докажи дека\n\n$$\n\\frac{1}{a-1} + \\frac{1}{b-1} + \\frac{1}{c-1} \\ge 8 \\left( \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Бидејќи важи $\\frac{1}{a-1} - \\frac{8}{b+c} = \\frac{1}{a-1} - \\frac{8}{4-a} = \\frac{12-9a}{(a-1)(4-a)} = \\frac{3(4-3a)}{(a-1)(4-a)}$, даденото неравенство е еквивалентно со\n\n$$\n3 \\left( \\frac{4-3a}{(a-1)(4-a)} + \\frac{4-3b}{(b-1)(4-b)} + \\frac{4-3c}{(c-1)(4-c)} \\right) \\ge 0.\n$$\n\nБез губење на општоста, нека претпоставиме дека $a \\ge b \\ge c$. Тогаш јасно е дека $4-3a \\le 4-3b \\le 4-3c$. Од $1 < a, b, c < 4$ следува дека $\\frac{1}{(a-1)(4-a)}$, $\\frac{1}{(b-1)(4-b)}$, $\\frac{1}{(c-1)(4-c)}$ се позитивни реални броеви. Ќе докажеме дека $(a-1)(4-a) \\ge (b-1)(4-b)$.\n\nИмаме $(a-1)(4-a) \\ge (b-1)(4-b) \\Leftrightarrow 5a - a^2 \\ge 5b - b^2 \\Leftrightarrow (a-b)(5-a-b) \\ge 0$. Аналогно, $(b-1)(4-b) \\ge (c-1)(4-c)$. Оттука следува дека $\\frac{1}{(a-1)(4-a)} \\le \\frac{1}{(b-1)(4-b)} \\le \\frac{1}{(c-1)(4-c)}$. Бидејќи $4-3a \\le 4-3b \\le 4-3c$, можеме да го искористиме неравенството на Чебишев и добиваме:\n\n$$\n\\frac{4-3a}{(a-1)(4-a)} + \\frac{4-3b}{(b-1)(4-b)} + \\frac{4-3c}{(c-1)(4-c)} \\ge \\frac{4-3a + 4-3b + 4-3c}{3} \\left( \\frac{1}{(a-1)(4-a)} + \\frac{1}{(b-1)(4-b)} + \\frac{1}{(c-1)(4-c)} \\right) = 0.\n$$\n\nРавенство важи за $4-3a = 4-3b = 4-3c$, т.е. $a = b = c = \\frac{4}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19707,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real values of $a$ such that the equation\n$$\nx(x+1)^3 - (2x+a)(x+a-1) = 0\n$$\nhas four distinct real roots.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\nx(x+1)^3 - (2x+a)(x+a-1) = x^4 + 3x^3 + x^2 + x(3-3a) + a - a^2 = (x^2 + 3x + a)(x^2 + 1 - a)\n$$\nTo get 4 roots, the discriminants of both quadratics must be positive:\n$$\n9 - 4a > 0 \\quad \\text{and} \\quad 4a - 4 > 0\n$$\nThe roots of $x^2 + 3x + a = 0$ and $x^2 + 1 - a = 0$ should be distinct. Let $t$ be a common root, then\n$$\n(t^2 + 3t + a) - (t^2 + 1 - a) = 3t + 2a - 1 = 0 \\implies t = \\frac{1-2a}{3}\n$$\nSubstituting into $t^2 + 1 - a = 0$ gives\n$$\n4a^2 - 13a + 10 = (4a-5)(a-2) = 0\n$$\nDirect check shows that at $a = 2$ and $a = \\frac{5}{4}$ two roots coincide. \n\n**Answer:**\n$$\na \\in (1, \\frac{5}{4}) \\cup (\\frac{5}{4}, 2) \\cup (2, \\frac{9}{4})\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19708,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $a_n$ is defined by $a_1 = 5$, $a_2 = 13$, and\n$$\na_{n+2} = 5a_{n+1} - 6a_n, \\quad \\forall n \\geq 2.\n$$\n\na) Prove that $\\gcd(a_n, a_{n+1}) = 1$ for all positive integers $n$.\n\nb) Prove that if $p$ is a prime divisor of $a_{2k}$, then $p-1$ is divisible by $2^{k+1}$ for all non-negative integers $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The general formula for $(a_n)$ is\n$$\na_n = 2^n + 3^n, \\quad \\forall n \\in \\mathbb{Z}^+.\n$$\nSuppose there exists $n \\geq 1$ such that $a_n$ and $a_{n+1}$ have a common prime divisor $p$. Clearly, $\\gcd(p, 6) = 1$. We have\n$$\n\\begin{cases}\np \\mid 2^n + 3^n, \\\\\np \\mid 2^{n+1} + 3^{n+1},\n\\end{cases}\n$$\nso\n$$\n\\begin{cases}\np \\mid 3 \\cdot 2^n + 3^{n+1}, \\\\\np \\mid 2 \\cdot 2^n + 3^{n+1},\n\\end{cases}\n$$\nwhich implies $p \\mid 2^n$, a contradiction since $\\gcd(p, 6) = 1$.\n\nb) Let $p$ be a prime divisor of $2^{2^k} + 3^{2^k}$. Clearly, $2^{2^k} \\equiv -3^{2^k} \\pmod{p}$, so $2^{2^{k+1}} \\equiv 3^{2^{k+1}} \\pmod{p}$. By Fermat's little theorem,\n$$\n2^{p-1} \\equiv 3^{p-1} \\equiv 1 \\pmod{p}.\n$$\nLet $h$ be the smallest positive integer such that $2^h \\equiv 3^h \\pmod{p}$. It is well-known that for all $h' \\geq h$ satisfying this condition, $h \\mid h'$. Now, $h' = 2^{k+1}$ satisfies the condition, so $h \\mid 2^{k+1}$, thus $h = 2^x$ with $0 \\leq x \\leq k+1$. Suppose $x \\leq k$, then\n$$\n2^x \\equiv 3^x \\pmod{p} \\implies p \\mid 2^x - 3^x \\mid 2^{2^k} - 3^{2^k},\n$$\nwhich is a contradiction since $p \\mid 2^{2^k} + 3^{2^k}$. Therefore, $x = k+1$, which implies $2^{k+1} \\mid p-1$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19709,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d_a$, $d_b$, $d_c$ be the distances from point $P$ to sides $BC$, $CA$, and $AB$ of triangle $ABC$, respectively, where $BC = a$, $CA = b$, and $AB = c$. Suppose $A_0B_0 = B_1C_1 = C_2A_2 = x$. Find a construction for $x$ and $P$ given that the area of $ABC$ is $S = \\frac{1}{2}(ad_a + bd_b + cd_c)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The altitudes from $A$, $B$, and $C$ are $\\frac{2S}{a}$, $\\frac{2S}{b}$, and $\\frac{2S}{c}$, respectively. By the similarity of triangles $AB_2C_2$, $A_1BC_1$, $A_0B_0C$, and $ABC$:\n\n$$\n\\begin{aligned}\n\\frac{x}{a} &= 1 - \\frac{ad_a}{2S} \\\\\n\\frac{x}{b} &= 1 - \\frac{bd_b}{2S} \\\\\n\\frac{x}{c} &= 1 - \\frac{cd_c}{2S}\n\\end{aligned}\n$$\n\nSumming the three equations:\n\n$$\nx \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 3 - \\frac{ad_a + bd_b + cd_c}{2S} = 3 - 1 = 2\n$$\n\nThus,\n\n$$\n\\frac{2}{x} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\n$$\n\nNow, it is possible to construct $x$ and then $P$, since the inverse of any segment can be constructed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19710,
    "subject": "Mathematics (Olympiad)",
    "question": "Найдите все натуральные числа $n$, для которых существует натуральное число $m$ такое, что множество собственных делителей $m$ состоит из чисел, на единицу больших собственных делителей $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "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 $m$.\n\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.\n\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 $q$.\n\nОтвет: $n = 4$ или $n = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19711,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the inequality for positive $a, b, c, d$:\n\n$$\n\\frac{(a+b)^2}{cd} + \\frac{(c+d)^2}{ab} \\ge 8.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Use two inequalities of means:\n\n$$\n\\frac{(a+b)^2}{cd} + \\frac{(c+d)^2}{ab} \\ge \\frac{4ab}{cd} + \\frac{4cd}{ab} \\ge 2 \\cdot 4 \\sqrt{\\frac{ab}{cd} \\cdot \\frac{cd}{ab}} = 8.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19712,
    "subject": "Mathematics (Olympiad)",
    "question": "$f : \\mathbb{R}^+ \\to \\mathbb{R}$ функцийн хувьд дараах тэнцэтгэлийг хангах бүх $f$-ийг ол:\n\n$$\n\\forall x, y \\in \\mathbb{R}^+ = (0, +\\infty):\nf\\left(\\sqrt{\\frac{x^2 + xy + y^2}{2012}}\\right) = \\frac{f(x) - f(y)}{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Тэмдэглэгээ авъя: $a \\ast b = \\sqrt{\\frac{a^2 + ab + b^2}{2012}}$. Тэгвэл\n\n$$\nf(a \\ast b) = \\frac{f(a) + f(b)}{2}\n$$\n\nДараах шинжүүдийг ашиглая:\n\n$$\nf((x_1 \\ast x_2) \\ast (x_3 \\ast x_4)) = \\frac{f(x_1 \\ast x_2) + f(x_3 \\ast x_4)}{2} = \\frac{f(x_1) + f(x_2) + f(x_3) + f(x_4)}{2}\n$$\n\nМөн\n\n$$\nf((x_1 \\ast x_2) \\ast (x_3 \\ast x_4)) = f((x_1 \\ast x_4) \\ast (x_2 \\ast x_3)) \\quad (1)\n$$\n\n$g(x) = (x^2 \\ast x) \\ast (x \\ast 1)$, $h(x) = (x^2 \\ast 1) \\ast (x \\ast x)$ гэж авъя. $\\forall t > 0$ үед $(ta) \\ast (t b) = t \\ast (a \\ast b)$ тул $t \\cdot g(x) = (t x^2 \\ast t x) \\ast (t x \\ast t)$, $t \\cdot h(x) = (t x^2 \\ast t) \\ast (t x \\ast t x)$.\n\n(1)-д $(x_1, x_2, x_3, x_4) \\mapsto (t x^2, t x, t x, t)$ гэж орлуулбал\n\n$$\nf(t g(x)) = f(t h(x)), \\forall t, x > 0 \\quad (2)\n$$\n\n$g(1) = h(1) = (1 \\ast 1) \\ast (1 \\ast 1) = \\sqrt{\\frac{3}{2012}} \\ast \\sqrt{\\frac{3}{2012}} = \\frac{3}{2012}$\n\n$g(2) = (4 \\ast 2) \\ast (2 \\ast 2) = \\sqrt{\\frac{28}{2012}} \\ast \\sqrt{\\frac{7}{2012}} = \\frac{7}{2012}$\n\n$h(2) = (4 \\ast 1) \\ast (2 \\ast 2) = \\sqrt{\\frac{21}{2012}} \\ast \\sqrt{\\frac{12}{2012}} = \\sqrt{\\frac{33+6\\sqrt{7}}{2012}}$\n\nЭндээс $g(2) > h(2)$.\n\n$x > 1$ үед $\\frac{g(x)}{h(x)}$ тасралтгүй функц бөгөөд $\\frac{g(1)}{h(1)} = 1$, $\\lambda = \\frac{g(2)}{h(2)} > 1$.\n\n$\\forall S \\in (1, \\lambda) \\Rightarrow \\exists x_S \\in (1, 2)$ ба $S = \\frac{g(x_S)}{h(x_S)}$ (Завсрын утгын теорем).\n\nИймд $a > 0$ бол $\\forall b \\in (a, \\lambda a)$ хувьд $\\frac{b}{a} \\in (1, \\lambda)$ ба $\\frac{b}{a} = \\frac{g(z)}{h(z)}$ байх $z \\in (1, 2)$ олдоно. (2)-д\n\n$$\nt = \\frac{a}{h(z)},\\ x = z \\text{ гэж орлуулбал} \\\\\nf(b) = f(a \\cdot \\frac{b}{a}) = f\\left(a \\cdot \\frac{g(z)}{h(z)}\\right) = f\\left(\\frac{a}{h(z)} \\cdot g(z)\\right) = f\\left(\\frac{a}{h(z)} \\cdot h(z)\\right) = f(a)\n$$\n\nТэгэхээр $f$ нь $(a, \\lambda a)$ интервал дээр тогтмол функц болно. $a \\leq b$ байх дурын $a, b$ тоонууд ба хангалттай их $n \\in \\mathbb{N}$ үед $\\mu = \\sqrt{\\frac{b}{a}} \\leq \\lambda$ ба $f$ функц $(a, \\lambda a), (\\lambda a, \\lambda^2 a), \\dots, (\\lambda^{n-1} a, \\lambda^n a)$ завсар бүрд тогтмол тул $f(a) = f(\\mu a) = \\dots = f(\\mu^n a) = f(b)$, өөрөөр хэлбэл $f$ тогтмол функц байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19713,
    "subject": "Mathematics (Olympiad)",
    "question": "Andrii chooses a sign in front of each number in the expression $\\pm 1 \\pm 2 \\pm 3 \\pm \\dots \\pm 2018$. How many different positive values can Andrii obtain as a result of the computed expression?",
    "options": [],
    "answer": "See solution",
    "solution": "All odd numbers between $1$ and $1 + 2 + 3 + \\dots + 2018 = \\frac{1}{2} \\cdot 2018 \\cdot 2019 = 1009 \\cdot 2019$.\n\nClearly, the expression cannot equal $0$, since it contains $1009$ odd numbers. We want to show that the expression can be any odd number between $1$ and $1 + 2 + 3 + \\dots + 2018$.\n\nTake some configuration. Find the first from the left consecutive numbers with signs \"$-$\" and \"+\". Switching these signs decreases the value of the expression by $2$. We will start with the following expression: \"+1 + 2 + 3 + \\dots + 2018\" and will change it to the expression \"$-1 + 2 + 3 + \\dots + 2018$\", which is less by $2$ as described above. By following the described algorithm, we will obtain all the numbers decreased by $2$.\n\nWhen we get the expression \"+1 - 2 - 3 - \\dots - 2018\", for which the algorithm doesn't work, the last step will be to switch to the smallest expression \"$-1 - 2 - 3 - \\dots - 2018$\".",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19714,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of $n$ points in the plane.\n\n(a) For which $n$ does there exist a set $S$ such that for every pair of points $A, B \\in S$, there exists a point $C \\in S$ such that $AC = BC$? (Such a set $S$ is called *balanced*.)\n\n(b) For which $n$ does there exist a *balanced* set $S$ that is also *centre-free*? (A set $S$ is *centre-free* if for any three distinct points $A, X, Y \\in S$, $AX = AY$ implies $X = Y$.)\n\n![](images/Australian-Scene-combined-2015_p142_data_daaba9419a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) For $n$ odd, we may take $S$ to be the set of vertices of a regular $n$-gon $\\mathcal{P}$.\n\nIt seems obvious that $S$ is balanced but we shall prove it anyway. Let $A$ and $B$ be any two vertices of $\\mathcal{P}$. Since $n$ is odd, one side of the line $AB$ contains an odd number of vertices of $\\mathcal{P}$. Thus if we enumerate the vertices of $\\mathcal{P}$ in order from $A$ around to $B$ on that side of $AB$, one of them will be the middle one, and hence be equidistant from $A$ and $B$.\n\nFor $n$ even, say $n = 2k$, we may take $S$ to be the set of vertices of a collection of $k$ unit equilateral triangles, all of which have a common vertex $O$, and exactly one pair of them has a second common vertex. Apart from $O$, all of the vertices lie on the unit circle centred at $O$.\n\nThe reason why this works is as follows. Let $A$ and $B$ be any two points in $S$. If they are both on the circumference of the circle, then $OA = OB$ and $O \\in S$. If one of them is not on the circumference, say $B = O$, then by construction there is a third point $C \\in S$ such that $\\triangle ABC$ is equilateral, and so $AC = BC$.\n\n(b) We claim that a balanced centre-free set of $n$ points exists if and only if $n$ is odd.\n\nNote that the construction used in part (a) is centre-free. We shall show that there is no balanced centre-free set of $n$ points if $n$ is even.\n\nFor any three points $A, X, Y \\in S$, let us write $A \\to \\{X, Y\\}$ to mean $AX = AY$. We estimate the number of instances of $A \\to \\{X, Y\\}$ in two different ways. First, if $A \\to \\{X, Y\\}$ and $A \\to \\{X, Z\\}$ where $Y \\neq Z$, then $S$ cannot be centre-free because $AX = AY = AZ$. Hence for a given point $A$, there are at most $\\lfloor \\frac{n-1}{2} \\rfloor$ pairs $\\{X, Y\\}$ such that $A \\to \\{X, Y\\}$. Since there are $n$ choices for $A$, the total number of instances is at most $n \\lfloor \\frac{n-1}{2} \\rfloor$.\n\nOn the other hand, since $S$ is balanced, for each pair of points $X, Y \\in S$, there is at least one point $A$ such that $A \\to \\{X, Y\\}$. Since the number of pairs $\\{X, Y\\}$ is $\\binom{n}{2}$, the total number of instances is at least $\\binom{n}{2}$. Combining our estimates, we obtain $n \\lfloor \\frac{n-1}{2} \\rfloor \\geq \\binom{n}{2}$, which simplifies to $\\lfloor \\frac{n-1}{2} \\rfloor \\geq \\frac{n-1}{2}$. This final inequality is impossible if $n$ is even. $\\square$\n\n![](images/Australian-Scene-combined-2015_p142_data_daaba9419a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19715,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $R$ 表示實數所成的集合。定義集合 $S = \\{1, -1\\}$ 與函數 $\\operatorname{sign} : R \\to S$ 如下：\n\n$$\n\\operatorname{sign}(x) = \\begin{cases} 1 & \\text{if } x \\ge 0; \\\\ -1 & \\text{if } x < 0. \\end{cases}\n$$\n\n給定奇數 $n$。是否存在 $n^2 + n$ 個實數 $a_{ij}, b_i \\in S$（$1 \\le i, j \\le n$），使得對於任意 $n$ 個數 $x_1, \\dots, x_n \\in S$，利用下式\n\n$$\ny_i = \\operatorname{sign}\\left(\\sum_{j=1}^{n} a_{ij}x_j\\right), \\quad \\forall 1 \\le i \\le n;\n$$\n\n$$\nz = \\operatorname{sign}\\left(\\sum_{i=1}^{n} b_i y_i\\right)\n$$\n\n計算出的 $z$ 恆等於 $x_1x_2\\cdots x_n$ 嗎？\n\nLet $R$ denote the set of real numbers. Define the set $S = \\{1, -1\\}$ and the function $\\operatorname{sign}(x) : R \\to S$ as\n\n$$\n\\operatorname{sign}(x) = \\begin{cases} 1 & \\text{if } x \\ge 0; \\\\ -1 & \\text{if } x < 0. \\end{cases}\n$$\n\nGiven an odd integer $n$, are there $n^2 + n$ real numbers $a_{ij}, b_i \\in S$ ($1 \\le i, j \\le n$) such that for arbitrary $n$ numbers $x_1, \\dots, x_n \\in S$, the number $z$ computed by the following formulas\n\n$$\ny_i = \\operatorname{sign}\\left(\\sum_{j=1}^{n} a_{ij}x_j\\right), \\quad \\forall 1 \\le i \\le n;\n$$\n\n$$\nz = \\operatorname{sign}\\left(\\sum_{i=1}^{n} b_i y_i\\right)\n$$\n\nalways equals the product $x_1x_2\\cdots x_n$?",
    "options": [],
    "answer": "See solution",
    "solution": "觀察小情況（如 $n=3$）容易猜想\n\n$$\na_{ij} = (-1)^{i+j}, \\quad b_i = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19716,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AB| > |AC|$. The internal angle bisector of $\\angle BAC$ intersects $BC$ at $D$. Let $O$ be the circumcenter of $\\triangle ABC$. Let $AO$ intersect the segment $BC$ at $E$. Let $J$ be the incenter of $\\triangle AED$. Prove that if $\\angle ADO = 45^\\circ$ then $|OJ| = |JD|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$. We have\n\n$$\n\\begin{align*}\n\\angle DJA &= 90^\\circ + \\frac{1}{2} \\angle DEA = 90^\\circ + \\frac{1}{2} (\\angle EBA + \\angle BAE) \\\\\n&= 90^\\circ + \\frac{1}{2} (\\beta + 90^\\circ - \\gamma) = 135^\\circ + \\frac{\\beta}{2} - \\frac{\\gamma}{2}\n\\end{align*}\n$$\n\nand\n\n$$\n\\begin{align*}\n\\angle DOA &= 180^\\circ - \\angle OAD - \\angle ADO = 180^\\circ - (\\angle OAC - \\angle DAC) - 45^\\circ \\\\\n&= 135^\\circ - \\left(90^\\circ - \\beta - \\frac{\\alpha}{2}\\right) = 135^\\circ - \\left(\\frac{1}{2}(\\alpha + \\beta + \\gamma) - \\beta - \\frac{\\alpha}{2}\\right) \\\\\n&= 135^\\circ + \\frac{\\beta}{2} - \\frac{\\gamma}{2}\n\\end{align*}\n$$\n\nTherefore, $\\angle DJA = \\angle DOA$, hence quadrilateral $ADJO$ is cyclic.\nSince $AJ$ is the bisector of $\\angle OAD$, the arcs $OJ$ and $JD$ are equal.\nHence $|OJ| = |JD|$.\n\n![](images/BW23_Shortlist_2023-11-01_p75_data_5fee7842ed.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19717,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the equation\n\n$$\nf(xf(y) + y) = f(x^2 + y^2) + f(y)\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = 0$ into the equation gives $f(y) = f(y^2) + f(y)$, implying $f(y^2) = 0$. Thus $f(x) = 0$ for all non-negative real numbers $x$. In particular, $f(x^2 + y^2) = 0$, which allows us to simplify the initial equation as\n\n$$\nf(xf(y) + y) = f(y).\n$$\n\nSuppose that $f(c) \\neq 0$ for some negative real number $c$. Substituting $y = c$ into the above gives $f(z) = f(c)$ for all $z$ because the expression $x f(c) + c$ attains all real values. By the above, there exists $z$ such that $f(z) = 0$; hence $f(c) = 0$, contradicting the choice of $c$. Consequently, $f(x) = 0$ for every real number $x$. Clearly, the function $f(x) = 0$ satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19718,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers. Consider an $m \\times n$ grid in a standard rectangular coordinate system. A segment is called *good* if it is parallel to a side of the grid. We partition the grid into triangles with vertices at integer coordinates such that each triangle has at least one side that is good, and the height of the good sides is 1. Find the minimum number of triangles that have exactly two good sides.",
    "options": [],
    "answer": "See solution",
    "solution": "If $m, n \\geq 2$ and the product $mn$ is even, then the answer is $0$; otherwise, it is $2$.\n\nWithout loss of generality, assume $1 \\leq m \\leq n$. Define an \"excellent\" triangle as one with exactly two good sides. Such a triangle has two sides of length $1$ and forms a right triangle. When $n \\geq 2$, a partition of the $2 \\times n$ grid contains no excellent triangles, whereas a $1 \\times n$ grid partition contains $2$ excellent triangles. Therefore, if $m, n \\geq 2$ and $mn$ is even, there exists a partition without any excellent triangles. If $m = 1$ or $mn$ is odd, there exists a partition with $2$ excellent triangles.\n\nNow, if $m = 1$ or $mn$ is odd, there must be at least $2$ excellent triangles. Define a side as \"bad\" if it is not good, and a triangle as \"bad\" if it is not excellent. A bad triangle has two bad sides, and the segment connecting the midpoints of these bad sides is the \"main\" segment. Bad sides do not pass through integer vertex coordinates, except within their triangles.\n\nThe main segments of neighboring bad triangles are connected, forming a chain. These main segments are good and change direction only at the center of unit grids. Any closed chain passes through an even number of unit grids. At the end of any non-closed chain, there must be two excellent triangles.\n\nIf $m = 1$, there is no closed chain, and if $mn$ is odd, a closed chain exists. Therefore, there are at least $2$ excellent triangles.\n\n*Remark.* Alternatively, one can consider areas of good triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19719,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers with $m < 2001$, $n < 2002$. Given $2001 \\times 2002$ distinct real numbers, place each number into a unique cell of a $2001 \\times 2002$ rectangular board (with $2001$ rows and $2002$ columns), so that every cell contains one number.\n\nA cell is called *bad* if the number in it is less than at least $m$ other numbers in its column **and** less than at least $n$ other numbers in its row. For any such arrangement, let $s$ be the number of bad cells. Find the minimum possible value of $s$.",
    "options": [],
    "answer": "See solution",
    "solution": "We generalize to a $p \\times q$ board ($m \\le p$, $n \\le q$) and prove by induction on $p+q$ that $$s \\ge (p - m)(q - n).$$\n\nThe base cases $p+q = 2, 3, 4$ are easily verified. Assume the result holds for $p+q = k$.\n\nConsider a $(p, q)$-board with $p+q = k+1$. If $m = p$ or $n = q$, the bound holds trivially. For $m < p$ and $n < q$, define a cell as *row-bad* if its number is less than at least $n$ others in its row, and *column-bad* if less than at least $m$ others in its column.\n\nIf every row-bad cell is also column-bad and vice versa, then $s = (p - m)(q - n)$. Otherwise, let $a$ be the smallest number among cells that are only row-bad or only column-bad. Suppose, without loss of generality, that $a$ is in a row-bad cell. Then all $(p - m)$ column-bad cells in $a$'s column are bad. Remove $a$'s column to get a $(p, q-1)$ board, where the number of bad cells is at least $(p - m)(q - 1 - n)$ by induction. Thus, the total number of bad cells is at least\n\n$$\n(p - m)(q - 1 - n) + (p - m) = (p - m)(q - n).\n$$\n\nTherefore, the bound holds.\n\nTo achieve equality, arrange the numbers in increasing order and fill the board row by row, left to right. Then\n\n$$\ns_{\\min} = (p - m)(q - n)\n$$\n\nand for the original problem, the answer is $(2001 - m)(2002 - n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19720,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the greatest common divisor of the numbers $11n + 4$ and $7n + 2$, where $n$ is a positive integer?",
    "options": [],
    "answer": "See solution",
    "solution": "The greatest common divisor of $11n+4$ and $7n+2$ also divides $7(11n+4) - 11(7n+2) = 6$, so it can be at most 6. If $n=4$, then $11n+4 = 48$ and $7n+2 = 30$, and the greatest common divisor of these two numbers is 6. Hence, the answer is 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19721,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real polynomials $P(x)$ such that for all real numbers $x, y, z$ satisfying $2xyz = x + y + z$, the following holds:\n\n$$\nxP(x) + yP(y) + zP(z) = xyz \\left( P(x - y) + P(y - z) + P(z - x) \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that $P(x) = c(x^2 + 3)$ is a solution for any real number $c$. This reduces to checking that\n\n$$\nx(x^2 + 3) + y(y^2 + 3) + z(z^2 + 3) = xyz((x - y)^2 + (y - z)^2 + (z - x)^2 + 9)\n$$\n\nwhenever $2xyz = x + y + z$. Using the factorization of $a^3 + b^3 + c^3 - 3abc$ and the relation $x + y + z = 2xyz$, the left-hand side equals\n\n$$\n(x^3 + y^3 + z^3) + 3(x + y + z) = 3xyz + (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3(x + y + z) \\\\\n= xyz(9 + (x - y)^2 + (y - z)^2 + (z - x)^2),\n$$\n\nas desired.\n\nNext, we prove that these are all solutions of the problem. If $P(x) = c$ is constant, then the left-hand side of the original equation equals $\\frac{c(x+y+z)}{xyz} = 2c$, while the right-hand side equals $3c$. This is only possible if $c = 0$. Therefore, if $P(x)$ is a nonzero solution, it is not constant.\n\nIf $x \\neq 0$, then $y = \\frac{1}{x}$ and $z = x + \\frac{1}{x}$ satisfy $2xyz = x + y + z$, so\n\n$$\nxP(x) + \\frac{1}{x}P\\left(\\frac{1}{x}\\right) + \\left(x + \\frac{1}{x}\\right)P\\left(x + \\frac{1}{x}\\right) = \\left(x + \\frac{1}{x}\\right)\\left(P\\left(x - \\frac{1}{x}\\right) + P(-x) + P\\left(\\frac{1}{x}\\right)\\right). \\quad (1)\n$$\n\nNote that the left-hand side is symmetric with respect to $x \\to \\frac{1}{x}$, thus so must be the right-hand side. It follows that\n\n$$\nP\\left(x - \\frac{1}{x}\\right) + P(-x) + P\\left(\\frac{1}{x}\\right) = P\\left(\\frac{1}{x} - x\\right) + P(x) + P\\left(-\\frac{1}{x}\\right).\n$$\n\nThis can be rewritten as $Q(x - \\frac{1}{x}) = Q(x) + Q(-\\frac{1}{x})$, where $Q(X) = P(X) - P(-X)$. We also know $Q(0) = P(0) - P(0) = 0$. Hence, as $x \\to \\infty$,\n\n$$\nQ(x) - Q\\left(x - \\frac{1}{x}\\right) = -Q\\left(-\\frac{1}{x}\\right) \\to 0.\n$$\n\nNow, if $Q(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$ with $n \\ge 2$, then the left-hand side of the above equation is of the form $na_nx^{n-2} + (\\text{lower-order terms})$, which fails to go to 0 as $x \\to \\infty$. Thus, $Q$ has degree at most 1, and since $Q(0) = 0$, then $Q(x) = 2ax$ for some real number $a$.\n\nUsing $P(x) - P(-x) = 2ax$, we conclude that the odd part of $P(x)$ is $ax$, so that $P(x) = ax + f(x^2)$ for a polynomial $f$ with real coefficients. Replacing $P(x) = ax + f(x^2)$ in relation (1) yields\n\n$$\nax^2 + xf(x^2) + \\frac{a}{x^2} + \\frac{1}{x}f\\left(\\frac{1}{x^2}\\right) + a\\left(x^2 + 2 + \\frac{1}{x^2}\\right) + \\left(x + \\frac{1}{x}\\right)f\\left(x^2 + 2 + \\frac{1}{x^2}\\right) \\\\\n= \\left(x + \\frac{1}{x}\\right)\\left(f\\left(x^2 - 2 + \\frac{1}{x^2}\\right) + f(x^2) + f\\left(\\frac{1}{x^2}\\right)\\right).\n$$\n\nMultiplying by $x$, we deduce that $2ax(x^2 + 1 + \\frac{1}{x^2})$ is a function of $x^2$, which implies that $a = 0$. Letting $t = x^2$, the previous relation becomes\n\n$$\nf(t) + tf\\left(\\frac{1}{t}\\right) = (t + 1)\\left(f\\left(t + 2 + \\frac{1}{t}\\right) - f\\left(t - 2 + \\frac{1}{t}\\right)\\right).\n$$\n\nWrite $f(t) = b_n t^n + \\dots + b_0$ with $b_n \\ne 0$ and suppose that $n > 1$. The largest term on the left-hand side is $b_n t^n$. However, the largest term on the right-hand side is the same as the largest term of\n\n$$\nt(f(t + 2) - f(t - 2)),\n$$\n\nwhich is $4b_n t^n$. This contradicts $b_n \\ne 0$, which means $f(t)$ must be linear. We may check, if $f(t) = ct + d$ in the last formula, that $d = 3c$. Therefore, $f(x) = c(x + 3)$, so $P(x) = f(x^2) = c(x^2 + 3)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19722,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all non-constant functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n$$\nf(2xy + x) = f(xy + x) + f(x)f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 0$ in the equation:\n$$\nf(0) = f(0) + f(0)^2\n$$\nso $f(0) = 0$.\n\nNow, set $x = 1$, $y = -1$:\n$$\nf(-1) = f(0) + f(1)f(-1) = f(1)f(-1)\n$$\nso either $f(1) = 1$ or $f(-1) = 0$. If $f(-1) = 0$, then substituting $y = -1$ gives $f(-x) = 0$ for all $x$, which is not non-constant. Thus, $f(1) = 1$.\n\nNext, set $x = y = -1$:\n$$\nf(1) = f(0) + f(-1)^2 = f(-1)^2\n$$\nso $f(-1) = 1$ or $f(-1) = -1$. If $f(-1) = 1$, substitute $y = -\\frac{1}{2}$:\n$$\nf\\left(\\frac{x}{2}\\right) = -f(x)f\\left(-\\frac{1}{2}\\right)\n$$\nfor all $x$. Setting $x = -1$ gives $f\\left(-\\frac{1}{2}\\right) = 0$, so $f\\left(\\frac{x}{2}\\right) = 0$ for all $x$, which is not non-constant. Therefore, $f(-1) = -1$.\n\nSubstituting $y = -1$ in the original equation gives $f(-x) = -f(x)$ for all $x$, so $f$ is odd.\n\nNow, consider the original equation with $y$ replaced by $-y$ and $y-1$, and use the oddness of $f$:\n$$\nf(x)f(y) = f(2xy - x) - f(xy - x) = f(xy) + f(x)f(y-1) - f(xy - x). \\quad (*)\n$$\n\nFor all positive integers $y$:\n$$\n\\begin{aligned}\nf(xy) - f(x)f(y) &= f(x(y-1)) - f(x)f(y-1) \\\\\n&= f(x(y-2)) - f(x)f(y-2) \\\\\n&= \\cdots = f(0) - f(x)f(0) = 0.\n\\end{aligned}\n$$\nSo $f(xy) = f(x)f(y)$ for all $x \\in \\mathbb{R}$, $y \\in \\mathbb{Z}^+$.\n\nLet $(x, y) = (1, 2)$ in $(*)$:\n$$\nf(3) - f(1) = f(2)^2\n$$\nLet $y = 1$ in the original equation:\n$$\nf(2x + x) = f(x + x) + f(x)f(1) \\implies f(2x + 1) = f(x + 1) + f(x)\n$$\nfor all $x$.\n\nSet $x = 1$:\n$$\nf(2) = f(3) - f(1) = f(2)^2\n$$\nso $f(2) = 2$ or $f(2) = 0$. If $f(2) = 0$, then $f(2x) = f(2)f(x) = 0$, which is not non-constant. Thus, $f(2) = 2$ and $f(2x) = 2f(x)$ for all $x$.\n\nBy induction, $f(n) = n$ for all positive integers $n$.\n\nNow, replace $y$ by $\\frac{y}{x}$ in the original equation (for $x \\neq 0$):\n$$\nf(2y + x) = f(x + y) + f(x)f\\left(\\frac{y}{x}\\right), \\quad \\forall x \\neq 0, y \\in \\mathbb{R}. \\quad (1)\n$$\nSwitch $x$ and $y$:\n$$\nf(2x + y) = f(x + y) + f(y)f\\left(\\frac{x}{y}\\right), \\quad \\forall y \\neq 0, x \\in \\mathbb{R}. \\quad (2)\n$$\n\nReplace $x$ by $2x$ in (1):\n$$\nf(2x + 2y) = f(2x + y) + f(2x)f\\left(\\frac{y}{2x}\\right)\n$$\nBut $f(2x) = 2f(x)$, so combining with previous results, we get:\n$$\nf(x + y) = f(x)f\\left(\\frac{y}{x}\\right) + f(y)f\\left(\\frac{x}{y}\\right), \\quad \\forall x, y \\neq 0.\n$$\n\nAdding (1) and (2) and using this, we find:\n$$\nf(2x + y) + f(2y + x) = 3f(x + y), \\quad \\forall x, y \\neq 0.\n$$\n\nTo extend to all $a, b \\in \\mathbb{R}$, choose $c$ such that $c > 4|a| + 4|b|$ (so $c \\neq 2a, c \\neq 2b$), and use additivity:\n$$\nf(c) + f(a + b) = f(a + b + c) = f(c + a) + f(b) = f(c) + f(a) + f(b)\n$$\nSo $f(a + b) = f(a) + f(b)$ for all $a, b \\in \\mathbb{R}$.\n\nSince $f$ is additive and $f(xy) = f(x)f(y)$, and $f$ is non-constant, the only solution is $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nIt is easy to check that $f(x) = x$ satisfies the original equation. Thus, the only non-constant solution is $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19723,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $A = \\{1, 2, 3, \\ldots, n\\}$ con $n > 5$. Demuestra que existe un conjunto finito $B$ de enteros positivos distintos tal que $A \\subseteq B$ y cumple la propiedad\n\n$$\n\\prod_{x \\in B} x = \\sum_{x \\in B} x^2,\n$$\n\nes decir, el producto de los elementos de $B$ es igual a la suma de los cuadrados de los elementos de $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Demostraremos el resultado tomando $A = \\{1, 2, \\ldots, n\\}$, para $n > 5$ (ya que si es cierto para $A$, también lo será para cualquier subconjunto de $A$).\n\nDado un conjunto $X$ de números naturales, denotemos $P(X) = \\prod_{x \\in X} x$ y $S(X) = \\sum_{x \\in X} x^2$. Si $n > 5$, se cumple que $P(A) > S(A)$. Escribimos $k = P(A) - S(A)$. La clave es que si añadimos a $A$ $k$ copias de $1$ (denotadas $1_0, 1_1, \\ldots, 1_k$), entonces se cumple la igualdad entre el producto y la suma. Denotemos $A_0 = \\{1_0, 1_1, \\ldots, 1_k, 2, \\ldots, n\\}$. Así, $P(A_0) = S(A_0)$.\n\nDefinimos un procedimiento iterativo para reemplazar cada uno de estos $k$ unos redundantes por valores distintos mayores que $n$. Observa que $1$ es solución de la ecuación polinómica\n\n$$\nP(A_0)x = S(A_0) - 1 + x^2.\n$$\n\nLa otra solución es $x = P(A_0) - 1 = P(A) - 1$. Definimos $A_1 = A_0 \\cup \\{P(A) - 1\\} \\setminus \\{1_1\\}$. Entonces $P(A_1) = S(A_1)$ por construcción. El número de unos redundantes en $A_1$ es uno menos que en $A_0$.\n\nRepitiendo el argumento, obtenemos $B = A \\cup \\{P(A_0) - 1, P(A_1) - 1, \\ldots, P(A_{k-1}) - 1\\}$, donde $A_i = A_{i-1} \\cup \\{P(A_{i-1}) - 1\\} \\setminus \\{1_i\\}$ para $i = 1, \\ldots, k$. Así, en cada paso eliminamos un uno redundante manteniendo la igualdad entre $P$ y $S$, y los elementos del conjunto resultante son todos distintos, ya que $n < n! - 1 = P(A_0) - 1$ y para todo $i \\ge 1$, $P(A_{i-1}) - 1 < P(A_i) - 1$.\n\nPor último, comprobamos que para todo $n \\ge 5$, la suma de los cuadrados de los elementos de $A$ es menor que su producto. Si esto sucede para un cierto $n$, también sucede para $n+1$, pues $(n+1)^2 < n^2 + (n-1)^2$ es equivalente a $n > 4$, que es cierto. Para $n = 5$, el producto es $5! = 120$ y la suma de cuadrados es $1+4+9+16+25 = 55 < 120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19724,
    "subject": "Mathematics (Olympiad)",
    "question": "From the set of numbers $\\{1, 2, \\ldots, 2015\\}$, choose the maximum possible number of numbers such that the sum of any five selected numbers is divisible by $15$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the set contains at least $6$ numbers: $a, b, c, d, e, f$. Then $a + b + c + d + e \\equiv 0 \\pmod{15}$ and $a + b + c + d + f \\equiv 0 \\pmod{15}$. Subtracting, $e - f \\equiv 0 \\pmod{15}$, so $e \\equiv f \\pmod{15}$. Thus, all selected numbers must be congruent modulo $15$.\n\nLet every number in the set be congruent to $k$ modulo $15$. Then\n$$\na + b + c + d + e \\equiv 5k \\equiv 0 \\pmod{15}.\n$$\nSo $k$ must be divisible by $3$, i.e., $k = 0, 3, 6, 9, 12$. For each such $k$, the set consists of numbers congruent to $k$ modulo $15$.\n\nFor $k = 0$, the numbers are $15, 30, \\ldots, 2010$, totaling $\\left\\lfloor \\frac{2015}{15} \\right\\rfloor = 134$ numbers.\n\nFor $k = 3$, the numbers are $3, 18, 33, \\ldots, 2013$, totaling $\\left\\lfloor \\frac{2013 - 3}{15} \\right\\rfloor + 1 = 135$ numbers.\n\nThus, the maximum is $135$ numbers, namely $\\{3, 18, 33, \\ldots, 2013\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19725,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume that $k$ is the positive integer for which the following inequality holds for all $a, b, c$ with $abc = 1$:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{k}{a + b + c + 1} \\geq \\frac{k}{4} + 3.\n$$\n\nFind the greatest such $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b = c = \\frac{2}{3}$ and $a = \\frac{9}{4}$ (so $abc = 1$). Substituting into the inequality:\n\n$$\n2 \\cdot \\frac{3}{2} + \\frac{4}{9} + \\frac{k}{\\frac{2}{3} + \\frac{2}{3} + \\frac{9}{4} + 1} \\geq \\frac{k}{4} + 3\n$$\n\nThis gives $k \\leq \\frac{880}{63} < 14$, so $k \\leq 13$ (since $k$ is integer).\n\nFor $k = 13$, the inequality becomes:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a + b + c + 1} \\geq \\frac{25}{4}\n$$\n\nLet $f(a, b, c) = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a + b + c + 1}$. Without loss of generality, assume $a = \\max\\{a, b, c\\}$. Then:\n\n$$\n\\begin{aligned}\nf(a, b, c) - f(\\sqrt{bc}, \\sqrt{bc}) &= \\left(\\frac{1}{b} + \\frac{1}{c} - \\frac{2}{\\sqrt{bc}}\\right) + 13 \\left(\\frac{1}{a + b + c + 1} - \\frac{1}{a + 2\\sqrt{bc} + 1}\\right) \\\\\n&= (\\sqrt{b} - \\sqrt{c})^2 \\left[ \\frac{1}{bc} - \\frac{13}{(a + b + c + 1)(a + 2\\sqrt{bc} + 1)} \\right]\n\\end{aligned}\n$$\n\nSince $a = \\max\\{a, b, c\\}$ and $abc = 1$, $bc \\leq 1$ so $\\frac{1}{bc} \\geq 1$. By AM-GM,\n\n$$\n\\frac{13}{(a + b + c + 1)(a + 2\\sqrt{bc} + 1)} \\leq \\frac{13}{16} < 1\n$$\n\nThus $f(a, b, c) \\geq f(\\sqrt{bc}, \\sqrt{bc})$. It suffices to prove\n\n$$\nf\\left(\\frac{1}{x^2}, x, x\\right) \\geq \\frac{25}{4}, \\quad x = \\sqrt{bc}, \\ 0 < x \\leq 1\n$$\n\nFor $x = 1$, the inequality holds. For $0 < x < 1$:\n\n$$\n\\begin{aligned}\n\\frac{(x + 2)(2x^3 + x^2 + 1)}{x(2x + 1)} &\\geq \\frac{13}{4} \\\\\n\\Leftrightarrow 4(x + 2)(2x^3 + x^2 + 1) &\\geq 13x(2x + 1) \\\\\n\\Leftrightarrow 8x^4 + 20x^3 - 18x^2 - 9x + 8 &\\geq 0\n\\end{aligned}\n$$\n\nWe have\n\n$$\n\\begin{aligned}\n8x^4 + 20x^3 - 18x^2 - 9x + 8 &= (8x^4 - 8x^2 + 2) + (20x^3 - 20x^2 + 5x) + (10x^2 - 14x + 6) \\\\\n&= 2(2x^2 - 1)^2 + 5x(2x - 1)^2 + 2(5x^2 - 7x + 3) > 0\n\\end{aligned}\n$$\n\nTherefore, $k = 13$ is the greatest positive integer satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19726,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you have 2022 coins, among which exactly two are fake (and lighter), and you have a balance scale. What is the minimum number $k$ of weighings needed to guarantee that you can identify both fake coins?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $k = 21$.\n\n**Strategy for 21 weighings:**\n\nLabel the coins $1, 2, \\ldots, 2022$ using their 11-digit binary representations (since $2^{11} = 2048 > 2022$).\n\n- For each $k = 1, 2, \\ldots, 11$, in the $k$-th weighing, place all coins whose $k$-th binary digit is 1 on the scale. The total weight is even if and only if both or neither fake coins are on the scale (i.e., their $k$-th digits are equal).\n- For each $k$, this tells us whether the $k$-th digits of the two fake coins are equal or different. Since the coins are distinct, there is at least one position $a$ where their digits differ.\n- For each $b = 1, 2, \\ldots, 11$, $b \\neq a$, perform a weighing with coins whose $a$-th and $b$-th digits are both 1. Only one fake coin ($F_1$) is involved in these weighings, so the parity reveals the $b$-th digit of $F_1$.\n- Thus, after 21 weighings, both fake coins are identified.\n\n**Optimality:**\n\nLet $S_k$ be the set of possible pairs of fake coins after $k$ weighings. Initially, $|S_0| = \\binom{2022}{2} > 2^{20}$. Each weighing can at best halve the number of candidates, so after 20 weighings, $|S_{20}| > 1$. Thus, 20 weighings are not enough, and 21 is minimal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19727,
    "subject": "Mathematics (Olympiad)",
    "question": "A grasshopper is placed at one of the points $A_1, A_2, \\dots, A_n$ in the plane. It can jump from point $A_i$ to point $A_j$ (for $i, j \\in \\{1, \\dots, n\\}, i \\neq j$) if and only if the line $\\overline{A_i A_j}$ does not go through any of the inner points of the lines $\\overline{A_1 B_1}, \\overline{A_2 B_2}, \\dots, \\overline{A_n B_n}$.\n\nShow that the grasshopper can take a series of jumps to get from any point $A_i$ to any point $A_j$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let a path be any line $A_iA_j$ and a wall be any line $A_kB_k$. We say that a path and a wall intersect if the path goes through an inner point of the wall. A path is good if there is no wall to intersect it, and a wall is irrelevant if it does not intersect any path.\n\n*Claim 1.* For any $i \\in \\{1,2,\\dots,n\\}$ there exists at least one good path from point $A_i$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19728,
    "subject": "Mathematics (Olympiad)",
    "question": "Given is a sequence $a_1, a_2, \\ldots$ such that $a_1 = 1$ and $a_{n+1} = \\frac{9a_n + 4}{a_n + 6}$ for any $n \\in \\mathbb{N}$. Which terms of this sequence are positive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "It can be shown by induction that\n\n$$\na_n = \\frac{4 \\cdot 2^n - 3}{2^n + 3}, \\quad \\forall n \\ge 2\n$$\n\nThus, $a_n$ is integer only when $n = 1$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19729,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd prime. Prove that\n$$\n1^{p-2} + 2^{p-2} + 3^{p-2} + \\dots + \\left(\\frac{p-1}{2}\\right)^{p-2} \\equiv \\frac{2-2^p}{p} \\pmod{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, for each $i = 1, 2, \\dots, \\frac{p-1}{2}$,\n$$\n\\frac{2i}{p} \\binom{p}{2i} = \\frac{(p-1)(p-2)\\cdots(p-(2i-1))}{(2i-1)!} \\equiv \\frac{(-1)(-2)\\cdots(-(2i-1))}{(2i-1)!} \\equiv -1 \\pmod{p}.\n$$\nHence\n$$\n\\begin{aligned}\n\\sum_{i=1}^{(p-1)/2} i^{p-2} &\\equiv - \\sum_{i=1}^{(p-1)/2} i^{p-2} \\frac{2i}{p} \\binom{p}{2i} \\\\\n&\\equiv -\\frac{2}{p} \\sum_{i=1}^{(p-1)/2} i^{p-1} \\binom{p}{2i} \\\\\n&\\equiv -\\frac{2}{p} \\sum_{i=1}^{(p-1)/2} \\binom{p}{2i} \\pmod{p} \\quad \\text{(by Fermat's Little Theorem.)}\n\\end{aligned}\n$$\nThe last summation counts the even-sized nonempty subsets of a $p$-element set, of which there are $2^{p-1} - 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19730,
    "subject": "Mathematics (Olympiad)",
    "question": "On an $m \\times n$ board ($m, n \\ge 3$), some dominoes (rectangles $1 \\times 2$ or $2 \\times 1$) are placed so that:\n- Dominoes do not overlap.\n- Dominoes do not go outside the board.\n- At least one corner cell is covered by a domino.\n- No additional domino can be placed without violating these rules.\n\nProve that at least $\\frac{2}{3}$ of all cells of the board are covered with dominoes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's assign each empty square to a domino to its right (if it's not in the rightmost column). If two empty squares are assigned the same domino, that domino must be vertical, allowing us to place a domino on those two squares—a contradiction.\n\nThus, each empty square not in the rightmost column has a unique corresponding domino. For empty squares in the rightmost column, we need to assign dominoes as well. If each can be assigned a unique domino, then in each pair (empty cell and domino), the domino covers 2 out of 3 cells, not counting dominoes unassigned to any empty cell. Therefore, at least $\\frac{2}{3}$ of all cells are covered.\n\nLet $k$ be the number of empty cells in the rightmost column and $l$ in the leftmost column. Empty squares in the leftmost column cannot be adjacent, so there are at least $l-1$ dominoes there, all unassigned. If $l > k$, there are enough dominoes in the leftmost column for the rightmost column's empty squares. If $l < k$, rotate the board and apply the same reasoning. The only remaining case is $l = k$, which is only possible if all corner cells are empty—a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19731,
    "subject": "Mathematics (Olympiad)",
    "question": "For every natural number $x$, let $P(x)$ be the product of the digits of the number $x$. Is there a natural number $n$ such that the numbers $P(n)$ and $P(n^2)$ are non-zero squares of natural numbers, where the number of digits of the number $n$ is equal to\n\n(a) 2021\n\n(b) 2022",
    "options": [],
    "answer": "See solution",
    "solution": "The answers are affirmative in both cases.\n\n(a) Take $n = \\overbrace{33\\cdots3}^{2019}68$. Then $P(n) = (4 \\cdot 3^{1010})^2$. Also,\n\n$$\n\\begin{aligned}\nn^2 &= \\left( \\frac{10^{2021} - 1}{3} + 35 \\right)^2 = \\frac{(10^{2021} + 104)^2}{9} \\\\\n&= \\frac{10^{4042} + 208 \\cdot 10^{2021} + 10816}{9} \\\\\n&= \\frac{10^{4042} - 10^{2021}}{9} + 209 \\cdot \\frac{10^{2021} - 1}{9} + 1225 \\\\\n&= \\underbrace{1\\cdots1}_{2021} \\underbrace{0\\cdots0}_{2021} + \\underbrace{2\\cdots2}_{2021} \\underbrace{00}_{2021} + \\underbrace{9\\cdots9}_{2021} + 1225 \\\\\n&= \\underbrace{1\\cdots1}_{2019} \\underbrace{332\\cdots200}_{2019} + 10^{2021} + 1224 \\\\\n&= \\underbrace{1\\cdots1}_{2019} \\underbrace{342\\cdots23424}_{2017}.\n\\end{aligned}\n$$\n\n$$\n\\text{Thus } P(n^2) = (3 \\cdot 2^{2012})^2.\n$$\n\n(b) Take $n = \\overbrace{1133\\cdots3}^{2020}$. Then $P(n) = (3^{1010})^2$. Also,\n\n$$\n\\begin{aligned}\nn^2 &= \\frac{(34 \\cdot 10^{2020} - 1)^2}{9} = \\frac{1156 \\cdot 10^{4040} - 68 \\cdot 10^{2020} + 1}{9} \\\\\n&= 128 \\cdot 10^{4040} + \\frac{4(10^{4040} - 10^{2020})}{9} - 7 \\cdot 10^{2020} - \\frac{10^{2020} - 1}{9} \\\\\n&= \\overbrace{1280\\cdots0}^{4040} + \\overbrace{4\\cdots40\\cdots0}^{2020} - 7 \\cdot 10^{2020} - \\underbrace{1\\cdots1}_{2020} \\\\\n&= \\overbrace{1284\\cdots4368\\cdots89}^{2018\\ 2019}.\n\\end{aligned}\n$$\n\n$$\n\\text{Thus } P(n^2) = (9 \\cdot 2^{5049})^2.\n$$\n\n**Remark.** In part (a) we can also take $n = \\overbrace{66\\cdots66}^{2020}1$ or $n = \\overbrace{33\\cdots33}^{2018}28$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19732,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1A_2A_3A_4$ be a convex quadrilateral that is neither cyclic nor has any pair of opposite sides parallel. For $1 \\leq i \\leq 4$, let $M_i$ be the midpoint of $A_{i-1}A_{i+1}$. Let $B_i$ be a point on the tangent at $A_i$ to the circumcircle of triangle $A_{i-1}A_iA_{i+1}$ such that the reflection of $M_i$ over the angle bisector of $\\angle A_{i+1}A_{i+2}A_{i+3}$ lies on the line $B_iA_{i+2}$. Let $C_i$ be the unique intersection point of lines $A_iA_{i+1}$ and $B_iB_{i+1}$. All indices are taken modulo $4$.\n\nProve that the points $C_1, C_2, C_3, C_4$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "All indices are considered modulo $4$. For a point $X$ and segment $A_iA_{i+1}$, let $d_{X-A_iA_{i+1}}$ denote the signed distance from $X$ to line $A_iA_{i+1}$ (with positive distance when $X$ is inside quadrilateral $A_1A_2A_3A_4$).\n\nFor any point $X$ on line $A_{i+2}B_i$, we have:\n\n$$\n\\frac{d_{X-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = \\frac{d_{X-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}}.\n$$\n\nFor any point $X$ on line $A_iB_i$, we have:\n\n$$\n\\frac{d_{X-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{X-A_iA_{i+1}}}{A_iA_{i+1}} = 0.\n$$\n\nTherefore, for point $B_i$:\n\n$$\n\\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}, \\qquad (6)\n$$\n\nand\n\n$$\n\\frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} = 0. \\qquad (7)\n$$\n\nCombining these gives:\n\n$$\n\\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.\n$$\n\nSimilarly for $B_{i+1}$:\n\n$$\n\\frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} = \\frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}},\n$$\n\nand\n\n$$\n\\frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = 0.\n$$\n\nThus:\n\n$$\n\\frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.\n$$\n\nSince these relations are linear in the coordinates, point $C_i$ must also satisfy:\n\n$$\n\\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \\qquad (8)\n$$\n\nNoting that $C_i$ lies on $A_iA_{i+1}$ (so $d_{C_i-A_iA_{i+1}} = 0$), we can rewrite this as:\n\n$$\n\\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} + \\frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} = \\frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \\quad (9)\n$$\n\nThis equation is completely symmetric for $C_1, C_2, C_3, C_4$. If it held identically for all points, then substituting $A_1$ would yield:\n\n$$\n\\frac{A_1A_2 \\cdot \\sin A_2}{A_2A_3} = \\frac{A_1A_4 \\cdot \\sin A_4}{A_3A_4},\n$$\n\nand substituting $A_3$ would give:\n\n$$\n\\frac{A_2A_3 \\cdot \\sin A_2}{A_1A_2} = \\frac{A_3A_4 \\cdot \\sin A_4}{A_1A_4}.\n$$\n\nThis would imply $\\sin A_2 = \\sin A_4$, and similarly $\\sin A_1 = \\sin A_3$, meaning $A_1A_2A_3A_4$ would either be a parallelogram or cyclic—contradicting the given conditions.\n\nTherefore, the equation is not identically satisfied. Since it is linear in coordinates, the set of points satisfying it must form a straight line. Hence $C_1, C_2, C_3, C_4$ are collinear. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19733,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}^*$ be a strictly increasing function.\n\n**a)** Show that there exists a decreasing sequence of positive real numbers $(y_n)_{n \\in \\mathbb{N}}$, converging to $0$, such that $y_n \\leq 2y_{f(n)}$ for all $n \\in \\mathbb{N}$.\n\n**b)** If $(x_n)_{n \\in \\mathbb{N}}$ is a decreasing sequence of real numbers converging to $0$, prove that there exists a decreasing sequence of real numbers $(y_n)_{n \\in \\mathbb{N}}$, converging to $0$, such that $x_n \\leq y_n \\leq 2y_{f(n)}$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See 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 $n_k = f(n_{k-1})$ for $k \\in \\mathbb{N}^*$. The properties of $f$ imply that the sequence is strictly increasing. Define the decreasing sequence $(y_n)_{n \\in \\mathbb{N}}$ by $y_n = 2^{-k}$ for all $n$ with $n_k \\leq n < n_{k+1}$, $k \\in \\mathbb{N}$; this sequence obviously converges to $0$.\n\nIt suffices to prove that $y_n \\leq 2y_{f(n)}$ for $n \\in \\mathbb{N}$, specifically for $n_k \\leq n < n_{k+1}$, $k \\in \\mathbb{N}$. Since $f$ is strictly increasing, $n_{k+1} = f(n_k) \\leq f(n) < f(n_{k+1}) = n_{k+2}$, hence $y_{f(n)} = 2^{-k-1} = y_n/2$.\n\nb) Clearly, $x_n \\geq 0$ for all $n \\in \\mathbb{N}$. Using the previously defined sequence $(n_k)$, define the decreasing sequence of positive reals $(z_k)_{k \\in \\mathbb{N}}$ as follows: $z_0 = x_1$ and $z_k = \\max(x_{n_k}, z_{k-1}/2)$ for $k \\in \\mathbb{N}^*$. The monotonicity of this sequence follows inductively. Moreover, $(z_k)_{k \\in \\mathbb{N}}$ converges to $0$: if $z_k = x_{n_k}$ for infinitely many $k$, then $z_k \\to 0$ because it is decreasing and $x_n \\to 0$; if $z_k = x_{n_k}$ only for finitely many $k$, then $z_k = z_{k-1}/2$ from some $k$ onwards, and again $z_k \\to 0$.\n\nFinally, define the sequence $(y_n)_{n \\in \\mathbb{N}}$ by $y_n = z_k$ for $n_k \\leq n < n_{k+1}$, $k \\in \\mathbb{N}$. Clearly, the sequence decreases to $0$. To prove the inequalities $x_n \\leq y_n \\leq 2y_{f(n)}$ for $n \\in \\mathbb{N}$, it suffices to check them for $n_k \\leq n < n_{k+1}$, $k \\in \\mathbb{N}$. Obviously, $x_n \\leq x_{n_k} \\leq z_k = y_n$. On the other hand, $n_{k+1} = f(n_k) \\leq f(n) < f(n_{k+1}) = n_{k+2}$ since $f$ is strictly increasing, hence $y_{f(n)} = z_{k+1} \\geq z_k/2 = y_n/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19734,
    "subject": "Mathematics (Olympiad)",
    "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 $a_n$ has. Prove that $2a_{n+1} = 3a_n$ only for finitely many indices $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Begin with a 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 $a_m < b \\leq a_{m+1}$ 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 fewer 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 factorizations into primes in the form $N = \\prod_{i \\geq 1} p_i^{e_i}$, where $e_i \\geq 0$ 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 $\\tau(N) = \\prod_{i \\geq 1} (e_i + 1)$.\n\n**Lemma 2.** The exponents in the canonical factorization 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, 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\n1. The exponents of the primes in the factorization of special terms have a common upper bound $e$; and\n2. For all large enough primes $p$, no special term is divisible by $p$.\n\nRefer to Lemma 2 to write $a_n = \\prod_{i \\geq 1} p_i^{e_i(n)}$, where $e_i(n) \\geq e_{i+1}(n)$ 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 $e \\geq e_i(n) > 0$, 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, $e_1(n) \\geq e_2(n) + 2$. The latter implies that $a_n$ is divisible by $8$, for either $e_1(n) \\geq 3$ 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 $\\tau(9a_n/8) \\leq \\tau(a_n)$, which is equivalent to\n\n$$\n(e_1(n) - 2)(e_2(n) + 3) \\leq (e_1(n) + 1)(e_2(n) + 1),\n$$\n\nso $2e_1(n) \\leq 3e_2(n) + 7$. 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, $\\tau(4a_n/3) \\leq \\tau(a_n)$. Alternatively, but equivalently,\n\n$$\n(e_1(n) + 3)e_2(n) \\leq (e_1(n) + 1)(e_2(n) + 1),\n$$\n\nso $2e_2(n) - 1 \\leq e_1(n)$. Combine this with the inequality in the previous paragraph to write $4e_2(n) - 2 \\leq 2e_1(n) \\leq 3e_2(n) + 7$ and infer that $e_2(n) \\leq 9$. Consequently, $2e_1(n) \\leq 3e_2(n) + 7 \\leq 34$, showing that $e = 17$ is suitable for (1) to hold. This establishes (1) and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19735,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a set of integers $S$ a beautiful set if for any triple $x, y, z$ of distinct elements of $S$, at least one of them is a divisor of $x + y + z$.\n\nShow that there exists a positive integer $N$ satisfying the following property, and find the smallest possible value for such $N$:\n\nFor any beautiful set $S$, there exists a positive integer $n_S \\geq 2$ such that the number of elements in $S$ which are not multiples of $n_S$ is at most $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that the smallest value $N$ can take is $6$.\n\nCall a beautiful set a *very beautiful set* if any pair of distinct elements of the set are relatively prime to each other. In the sequel, we write $y \\mid x$ for a pair of integers if $y$ divides $x$.\n\nFirst, we will show that if $N \\leq 5$ it does not satisfy the condition. Let $a_1, a_2$ be a pair of odd integers greater than or equal to $3$ and relatively prime to each other. By using the Chinese Remainder Theorem, we can choose an odd integer $a_3 \\geq 3$ satisfying:\n\n$$\na_3 \\equiv \\begin{cases} a_2 & (\\text{mod } a_1) \\\\ -a_1 & (\\text{mod } a_2) \\end{cases}\n$$\n\nSince $a_1$ and $a_2$ are relatively prime, $a_3$ and $a_1$ are relatively prime, and so are $a_3$ and $a_2$. Similarly, choose odd integers $a_4$ and $a_5$ (each $\\geq 3$) to satisfy:\n\n$$\na_4 \\equiv \\begin{cases} -a_2 & (\\text{mod } a_1) \\\\ -a_1 & (\\text{mod } a_2) \\\\ -a_2 & (\\text{mod } a_3) \\end{cases}\n$$\n\n$$\na_5 \\equiv \\begin{cases} -a_2 & (\\text{mod } a_1) \\\\ a_1 & (\\text{mod } a_2) \\\\ a_2 & (\\text{mod } a_3) \\\\ -a_1 & (\\text{mod } a_4) \\end{cases}\n$$\n\nThen any distinct pair from $a_1, \\dots, a_5$ are relatively prime. All are distinct and $\\geq 3$. Now let\n\n$$\nS = \\{1, 2, a_1, a_2, a_3, a_4, a_5\\}\n$$\n\nWe can check that $S$ is a beautiful set. Since every pair of distinct elements are relatively prime, no matter how we choose $n_S$, $S$ can contain at most one number which is a multiple of $n_S$. Therefore, the condition is not satisfied when $N \\leq 5$.\n\nNext, we show that $N = 6$ satisfies the condition. We start with the following three lemmas.\n\n**Lemma 1.** If positive integers $x, y, z$ satisfy $x < z$, $y < z$, and $z \\mid (x + y + z)$, then $x + y = z$.\n\n*Proof.* Since $x < z$, $y < z$, $x + y < 2z$, and $z \\mid (x + y + z)$, so $z \\mid (x + y)$. Thus $x + y = z$. $\\square$\n\n**Lemma 2.** When odd positive integers $x, y, z$ satisfy $x < z$, $y < z$, then $z \\mid (x + y + z)$ is never satisfied.\n\n*Proof.* By Lemma 1, $z = x + y$, but this contradicts all being odd. $\\square$\n\n**Lemma 3.** Let $S$ be a very beautiful set not containing $1$ nor any even integers. Let $x, y$ be elements of $S$ with $x < y$. Then, there exists at most one element $z$ of $S$ with $z < x$ and $z$ does not divide $x + y$.\n\n*Proof.* If $z < x$, then $x, y, z$ are positive odd integers with $z < x < y$. By Lemma 2, $y$ is not a divisor of $x + y + z$. Since $x, y, z$ are distinct, either $x \\mid (x + y + z)$ or $z \\mid (x + y + z)$ must hold.\n\nSuppose there are two or more such $z$ not dividing $x + y$. For each such $z$, $x \\mid (x + y + z)$. If $z_1 < z_2$ are two such, then $x \\mid (z_2 - z_1)$. But $0 < z_2 - z_1 < x$, contradiction. $\\square$\n\nNow, the number of elements in a very beautiful set is at most $7$. Since such a set can contain at most one even number, it suffices to show that a very beautiful set containing neither $1$ nor an even number has at most $5$ elements. Suppose there are $6$ such elements $x_1 < x_2 < \\dots < x_6$. Let $y_4 = x_5 + x_6$, $y_5 = x_4 + x_6$, $y_6 = x_4 + x_5$. For $k = 1, 2, 3$, if $x_k$ divides all of $y_4, y_5, y_6$, then $x_k$ divides $2x_4 = y_5 + y_6 - y_4$, which is impossible since $x_k$ is odd and relatively prime to $x_4$.\n\nFrom Lemma 3, for $\\ell = 4, 5, 6$, there are at least two $k$ for which $x_k \\mid y_\\ell$. Thus, there are exactly $6$ pairs $(k, \\ell)$ with $x_k \\mid y_\\ell$, and for each $y_\\ell$, exactly $2$ such $x_k$. On the other hand, from Lemma 3, $y_4$ is divisible by at most one $x_k$ for $k = 1, 2, 3$, contradiction. Thus, the set cannot have $6$ such elements.\n\nTherefore, the smallest possible value for $N$ is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19736,
    "subject": "Mathematics (Olympiad)",
    "question": "歸納法證明：設 $f\\left(\\frac{p}{q}\\right) = p$，其中 $p, q$ 為任意互質正整數，證明此命題對所有 $p, q$ 成立。",
    "options": [],
    "answer": "See solution",
    "solution": "若 $g\\left(\\frac{p}{q}\\right) = 0$，則 $p = 1$ 或 $q = 1$。\n\n- 若 $q = 1$，於原式代入 $(x, n) = (1, p-1)$ 得：\n  $$\nf(p) = p. \\tag{2}\n  $$\n- 若 $p = 1$，原式代入 $(q, 1)$，結合 (2) 得 $f\\left(\\frac{1}{q}\\right) = 1$。\n\n故 $g\\left(\\frac{p}{q}\\right) = 0$ 時命題成立。\n\n假設 $g\\left(\\frac{p}{q}\\right) = i-1$ 時命題成立，考慮 $g\\left(\\frac{p}{q}\\right) = i$：\n\n**Case (1):** $\\frac{p}{q} > 1$。\n設 $n = \\lfloor \\frac{p}{q} \\rfloor$，由 (1) 知 $g\\left(\\frac{p-nq}{q}\\right) = i-1$。於原式代入 $(\\frac{p-nq}{q}, n)$，結合 (1) 和歸納假設得：\n  $$\nf\\left(\\frac{p}{q}\\right) = (p-nq) + nq = p. \\tag{3}\n  $$\n\n**Case (2):** $\\frac{p}{q} < 1$。\n由於 $\\frac{q}{p} > 1$，由 (1) 知 $g\\left(\\frac{p+q}{p}\\right) = g\\left(\\frac{q}{p}\\right) = i$。\n又 $\\frac{p+q}{p} > \\frac{q}{p} > 1$，由 (3) 知 $f\\left(\\frac{p+q}{p}\\right) = p+q$，$f\\left(\\frac{q}{p}\\right) = q$。\n於原式代入 $(\\frac{q}{p}, 1)$ 得 $f\\left(\\frac{p}{q}\\right) = (p+q) - q = p$。\n\n綜合上述，$g\\left(\\frac{p}{q}\\right) = i$ 時命題成立。由數學歸納法，命題總是成立。\n\n因此，$f\\left(\\frac{p}{q}\\right) = p$，其中 $p, q$ 為任意互質正整數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19737,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, evaluate the sum\n\n$$\n\\sum_{(n)} \\sum_{k=1}^{n} e_k 2^{e_1 + \\dots + e_k - 2k - n},\n$$\n\nwhere $e_k \\in \\{0, 1\\}$ for $k = 1, \\dots, n$, and the sum $\\sum_{(n)}$ is taken over all possible choices of $e_1, \\dots, e_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $k = 1, \\dots, n$, let $S_k$ be the set of all $n$-tuples $(e_1, \\dots, e_n)$ of zeros and ones such that $e_k = 1$. Partition each $S_k$ into $k$ subsets $S_{k,l}$ for $l = 1, \\dots, k$, where $(e_1, \\dots, e_n)$ belongs to $S_{k,l}$ if and only if $e_k = 1$ and $e_1 + \\dots + e_k = l$. It is easy to see that $|S_{k,l}| = 2^{n-k} \\binom{k-1}{l-1}$, so the required sum is\n\n$$\n\\sum_{k=1}^{n} \\sum_{l=1}^{k} 2^{n-k} \\binom{k-1}{l-1} 2^{l-2k-n} = \\sum_{k=1}^{n} 2^{1-3k} \\sum_{l=1}^{k} \\binom{k-1}{l-1} 2^{l-1} = \\frac{2}{3} \\sum_{k=1}^{n} \\left(\\frac{3}{8}\\right)^k = \\frac{2}{5} \\left(1 - \\left(\\frac{3}{8}\\right)^n\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19738,
    "subject": "Mathematics (Olympiad)",
    "question": "3 нэгж талтай квадрат хэлбэртэй өрөөний нэг ханын голд 1 нэгж өргөнтэй хаалга байна. Энэ өрөөний хаалгаар хазайлгахгүйгээр багтаан оруулж болох $S$ талбайтай ширээний талбайн байж болох хамгийн их утгыг $S_{max}$ гэе. $S_{max} > \\pi$ болохыг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2013-ilovepdf-compressed_p42_data_4cee8a47db.png)\n\nЗурагт үзүүлсэн дүрс нь хаалгаар багтах ба энэ дүрсийн талбай нь\n\n$$\nS_1 + S_2 + S_3 = 1 \\cdot 1 + 2 \\cdot 1 + \\frac{(\\pi (1))^2}{4} = 3 + \\frac{\\pi}{4}\n$$\n\n$$\n= \\frac{12+\\pi}{4} > \\frac{15}{4} = 3.75 > \\pi \\approx 3.14\n$$\n\nбайна.\n\nЭхлээд $AB$ хэрчмийг оруулж, дараа нь $OD$ хүртэл чигээр нь оруулаад $O \\equiv M$ болгоно. Дараа нь $O$-г хөдөлгөхгүйгээр $D$-г $\\overset{\\frown}{EC}$ нумын дагуу $C \\equiv N$ болтол эргүүлээд чигээр нь түлхэж, ширээг өрөөнд оруулна. Эргүүлэхээс өмнө, эргүүлэх явцад мөн эргүүлэхэд $AB$ хэрчим ямар ч саадгүй байх нь $|AB| = 1$ ба $|OA| = |BD| = 1$ гэдгээс гарна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19739,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, and $D$ be labeled in a clockwise manner. Let $a_n$, $b_n$, $c_n$, and $d_n$ be, respectively, the number of ways to start at $A$, $B$, $C$, and $D$ and exit from $A$ in $n$ jumps. Find a recurrence relation for $a_n$ and determine $a_{2m+1}$ for $m \\ge 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $b_n = d_n$. Since from $A$, the frog can return to $A$ in an even number of jumps, $a_n = 0$ when $n$ is even.\n\nFrom $A$, in 1 jump, the frog can exit from 2 sides or jump to $B$ or $D$, thus $a_1 = 2$, $a_3 = 4$. Therefore, for $n \\ge 3$, $a_n = 2b_{n-1}$.\n\nFrom $B$, the frog can jump to $A$ or $C$, so $b_n = a_{n-1} + c_{n-1}$.\n\nFrom $C$, the frog can jump to $B$ or $D$, so $c_n = 2b_{n-1}$.\n\nTherefore,\n\n$$\na_n = 4a_{n-2}.\n$$\n\nConsequently, $a_{2m+1} = 4^{m-1} a_3 = 4^m$ for $m \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19740,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L = \\mathrm{lcm}(1, 2, \\dots, n)$. Show that\n$$\n\\gcd\\left(\\frac{L}{m+1}, \\frac{L}{m+2}, \\dots, \\frac{L}{n}\\right) = 1\n$$\nwhere $m = \\lfloor \\frac{n}{2} \\rfloor$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* First, note that $\\mathrm{lcm}(1, 2, \\dots, n) = \\mathrm{lcm}(m+1, m+2, \\dots, n)$. This is because for all $1 \\leq k \\leq m$, the number $s = \\lceil \\log_2\\left(\\frac{m+1}{k}\\right) \\rceil$ is an integer such that\n$$\n2^s \\cdot k \\in \\{m+1, m+2, \\dots, n\\}.\n$$\nNow, assume that\n$$\nd = \\gcd\\left( \\frac{L}{m+1}, \\frac{L}{m+2}, \\dots, \\frac{L}{n} \\right).\n$$\nFor all $m+1 \\leq i \\leq n$, we have\n$$\nd \\mid \\frac{L}{i} \\implies \\exists k: d \\cdot k = \\frac{L}{i} \\implies i \\cdot k = \\frac{L}{d} \\implies i \\mid \\frac{L}{d}.\n$$\nWhich implies\n$$\nL = \\mathrm{lcm}(m+1, m+2, \\dots, n) \\mid \\frac{L}{d} \\implies d = 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19741,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is equilateral with side length $6$. Suppose that $O$ is the center of the inscribed circle of this triangle. What is the area of the circle passing through $A$, $O$, and $C$?\n\n(A) $9\\pi$  (B) $12\\pi$  (C) $18\\pi$  (D) $24\\pi$  (E) $27\\pi$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $\\overline{AC}$, and let $P$ be the center of the circle passing through $A$, $O$, and $C$. Note that $O$, $M$, and $P$ are collinear and $\\angle OMA$ is $90^{\\circ}$. Because $\\angle OAM$ is $30^{\\circ}$ and $AM = 3$, it follows that $AO = 2\\sqrt{3}$ and $\\angle AOM$ is $60^{\\circ}$. Then $AP = OP$ implies that $\\triangle APO$ is isosceles and therefore equilateral. Thus the radius of the circle centered at $P$ is $2\\sqrt{3}$ and its area is $\\pi (2\\sqrt{3})^2 = 12\\pi$.\n\nAlternatively,\n\nDefine $M$ and $P$ as above. Note that $O$ is the centroid of $\\triangle ABC$, so $OM = \\sqrt{3}$. Extend ray $\\overrightarrow{OM}$ to intersect the circle centered at $P$ and call this point $D$. Because $\\overrightarrow{OD} \\perp \\overrightarrow{AC}$, it follows that $\\overrightarrow{OD}$ is a diameter of the circle centered at $P$. By Power of a Point,\n\n$$\nMD = \\frac{AM \\cdot MC}{OM} = \\frac{3 \\cdot 3}{\\sqrt{3}} = 3\\sqrt{3},\n$$\n\nand $OD = OM + MD = 4\\sqrt{3}$. Thus the area of the circle centered at $P$ is\n\n$$\n\\pi \\left(\\frac{OD}{2}\\right)^2 = 12\\pi.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19742,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive integers $n$, there exists a positive integer $m$ such that $7^n$ divides $3^m + 5^m - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove this by induction on $n$.\n\n**Base case ($n=1$):**\nFor $m=1$, $3^1 + 5^1 - 1 = 3 + 5 - 1 = 7$, which is divisible by $7$.\n\n**Inductive step:**\nAssume the statement holds for $n$, i.e., there exists $m$ such that $7^n \\mid 3^m + 5^m - 1$.\n\nNote that $3^6 \\equiv 1 \\pmod{7}$ and $5^6 \\equiv 1 \\pmod{7}$, so\n$$\n3^{6 \\cdot 7^{n-1}} \\equiv 1 \\pmod{7^n}, \\quad 5^{6 \\cdot 7^{n-1}} \\equiv 1 \\pmod{7^n}.\n$$\n\nLet $v_7(x)$ denote the exponent of $7$ dividing $x$. Then,\n$$\nv_7(3^{6 \\cdot 7^{n-1}} - 1) = v_7(3^6 - 1) + v_7(7^{n-1}) = n, \\\\\nv_7(5^{6 \\cdot 7^{n-1}} - 1) = v_7(5^6 - 1) + v_7(7^{n-1}) = n.\n$$\n\nThus, $3^{6 \\cdot 7^{n-1}} = 1 + 7^n r$, $5^{6 \\cdot 7^{n-1}} = 1 + 7^n s$ for some integers $r, s$.\n\nConsider the identity:\n$$\n\\frac{y^{7^k} - 1}{7^{k+1}} = \\frac{y-1}{7} \\cdot \\frac{1+y+\\dots+y^6}{7} \\cdots \\frac{1+y^{7^{k-1}} + \\dots + y^{6 \\cdot 7^{k-1}}}{7} \\quad (*).\n$$\n\nFor $y = 3^6$ and $y = 5^6$, $y \\equiv 1 \\pmod{7}$. By a lemma:\n\n*Lemma.* Let $p$ be an odd prime and $p \\mid a-1$. Then $\\frac{a^p-1}{a-1} \\equiv p \\pmod{p^2}$.\n\nApplying the lemma repeatedly, all terms in $(*)$ except the first are congruent to $1$ modulo $7$, so\n$$\n\\frac{y^{7^k} - 1}{7^{k+1}} \\equiv \\frac{y-1}{7} \\pmod{7}.\n$$\n\nSince $\\frac{3^6-1}{7} = 104 \\equiv -1 \\pmod{7}$ and $\\frac{5^6-1}{7} = 2232 \\equiv -1 \\pmod{7}$, we have $r \\equiv s \\equiv -1 \\pmod{7}$.\n\nBy the binomial theorem,\n$$\n3^{6t \\cdot 7^{n-1}} \\equiv 1 + 7^n r \\pmod{7^{n+1}}, \\quad 5^{6t \\cdot 7^{n-1}} \\equiv 1 + 7^n s \\pmod{7^{n+1}}\n$$\nfor all $t \\geq 1$.\n\nNow, consider $m' = m + 6t \\cdot 7^{n-1}$:\n$$\n3^{m'} + 5^{m'} - 1 = 3^m \\cdot 3^{6t \\cdot 7^{n-1}} + 5^m \\cdot 5^{6t \\cdot 7^{n-1}} - 1.\n$$\nModulo $7^{n+1}$, this becomes\n$$\n3^m(1 + 7^n r) + 5^m(1 + 7^n s) - 1 = (3^m + 5^m - 1) + 7^n(3^m r + 5^m s).\n$$\nSince $7^n \\mid 3^m + 5^m - 1$, we need $m'$ such that $7^{n+1} \\mid 3^{m'} + 5^{m'} - 1$.\n\nThis reduces to finding $t$ such that\n$$\n1 + (5^m s + 3^m r)t \\equiv 0 \\pmod{7}.\n$$\nBut $5^m s + 3^m r \\equiv -5^m - 3^m \\equiv 1 \\pmod{7}$, so $\\gcd(5^m s + 3^m r, 7) = 1$ and such $t$ exists.\n\nThus, the statement holds for $n+1$, completing the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19743,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = |AB_1|$, $y = |BC_1|$, $z = |CA_1|$. Show that\n\n$$\n\\sqrt{\\frac{x}{x+y}} + \\sqrt{\\frac{y}{y+z}} + \\sqrt{\\frac{z}{z+x}} \\le \\frac{3}{\\sqrt{2}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Or equivalently,\n\n$$\n\\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} + \\frac{1}{\\sqrt{1+c^2}} \\le \\frac{3}{\\sqrt{2}}\n$$\n\nfor all positive real numbers $a, b, c$ satisfying $abc = 1$.\n\nAssume without loss of generality that $ab \\le 1$. Then\n\n$$\n\\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} \\le \\sqrt{2} \\left( \\frac{1}{1+a^2} + \\frac{1}{1+b^2} \\right)\n$$\n\nand\n\n$$\n\\frac{1}{1+a^2} + \\frac{1}{1+b^2} = 1 + \\frac{1-a^2b^2}{(1+a^2)(1+b^2)} \\le 1 + \\frac{1-a^2b^2}{(1+ab)^2} = \\frac{2}{1+ab}\n$$\n\nand\n\n$$\n\\frac{1}{\\sqrt{1+c^2}} \\le \\frac{\\sqrt{2}}{1+c}\n$$\n\nby the Cauchy-Schwarz inequality. Therefore,\n\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{1}{\\sqrt{1+a^2}} &\\le 2\\sqrt{\\frac{c}{1+c}} + \\frac{\\sqrt{2}}{1+c} = \\frac{\\sqrt{2}}{1+c}(\\sqrt{2c(c+1)} + 1) \\\\\n&\\le \\frac{\\sqrt{2}}{1+c}\\left(\\frac{2c+c+1}{2} + 1\\right) = \\frac{3}{\\sqrt{2}}\n\\end{aligned}\n$$\n\nby the AM-GM inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19744,
    "subject": "Mathematics (Olympiad)",
    "question": "有 $N$ 隻怪獸，每一隻的體重都是一個正實數。每次我們把其中兩隻怪獸融合成一隻，新怪獸的體重是之前兩隻怪獸的體重和。經過一系列的操作後，我們最終將所有怪獸融合成一隻。在這個過程中，如果有某一次融合，被融合的兩隻怪獸中有一隻的體重大於另一隻的 $2.020$ 倍，則我們稱這一次融合是危險的。一個融合順序的危險程度，是其過程中危險融合的次數。\n\n試證：不論起始怪獸的體重如何分配，“每一次都將最輕的兩隻怪獸融合”都是所有融合順序中，讓危險程度達到最低的一個順序。",
    "options": [],
    "answer": "See solution",
    "solution": "原題中的 $2.020$ 可以改成任意 $k > 2$，故以下證明一般性的 $k$ 的狀況。\n\n讓我們將怪獸標記，使得第 $i$ 隻怪獸的體重 $x_i$ 滿足 $x_1 \\leq x_2 \\leq \\dots \\leq x_N$。我們定義一隻怪獸 $i$ 是**危險的**，若且唯若 $x_i > k \\sum_{j<i} x_j$。\n\n首先，注意到起始危險怪獸數是危險程度的下界。假設 $x_i$ 是危險的，並令 $M = \\{i-1\\}$。當第 $i-1$ 號怪獸與 $j$ 號怪獸融合時，將 $j$ 加入集合 $M$ 中。則易知在此過程中，當第一次 $M \\cap \\{j: j \\geq i\\}$ 時，必為一危險融合。這證明了危險程度至少跟起始的危險怪獸一樣多。\n\n接著證明每次融合最輕的兩隻怪獸可以達到這個下界。讓我們將 $x_1$ 與 $x_2$ 融合，這一步是危險的，若且唯若 $x_2$ 是危險。接著，讓我們考慮融合後的 $N-1$ 隻怪獸的危險性。注意到重量大於 $x_1 + x_2$ 怪獸的危險性，在融合前後不會改變，因此僅需考慮重量 $\\leq x_1 + x_2$ 的怪獸，注意到這些怪獸在頭兩隻怪獸融合前都不是危險的。令這些怪獸中有一隻的重量是 $y$，且 $y \\leq x_1 + x_2$。\n\n- 如果融合後 $y$ 是最輕的，則融合後它一定不是危險的。\n- 如果融合後 $y$ 不是最輕的，由一開始的重量排序，融合後最輕的怪獸重量至少是 $x_2$，從而 $k x_2 > 2 x_2 \\geq x_1 + x_2 \\geq y$，因此 $y$ 也不是危險的。\n\n以上兩點同時可證明，融合出來的 $x_1 + x_2$ 也不會是危險的。\n\n綜以上，當我們將最輕的兩隻怪獸融合時，如果 $x_2$ 不是危險的，則該次融合不危險，且新的 $N-1$ 隻怪獸裡的危險怪獸數與原來相同。反之，如果 $x_2$ 是危險的，則該次融合危險，且新的 $N-1$ 隻怪獸裡的危險怪獸數為原來危險怪獸數減 $1$。故由數歸法知，此序列達到危險程度的下界。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19745,
    "subject": "Mathematics (Olympiad)",
    "question": "Oleksii placed positive integers in the cells of a chessboard of size $8 \\times 8$. For each pair of adjacent cells, Fedir wrote down the product of the numbers in them and added all the obtained numbers. Oleksii wrote down the sum of the numbers in each pair of adjacent cells and multiplied all the obtained numbers. It turned out that both numbers have the same last digit, 1. Prove that at least one of the boys made a mistake in the calculation.\n\nFor example, for the square $3 \\times 3$ and the given arrangement of numbers (see below), Fedir would write the following numbers: $2, 6, 8, 24, 15, 35, 2, 6, 8, 20, 18, 42$ and their sum ends with digit 6, and Oleksii would write the following numbers: $3, 5, 6, 10, 8, 12, 3, 5, 6, 9, 9, 13$ and their product ends with digit 0.\n\n![](fig11.png)\n\n\\begin{tabular}{|c|c|c|}\n\\hline 1 & 2 & 3 \\\\\n\\hline 2 & 4 & 6 \\\\\n\\hline 3 & 5 & 7 \\\\\n\\hline\n\\end{tabular}",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that this could happen. Since Oleksii's product ends with 1, all the factors (the sums of numbers in adjacent cells) must be odd, so the sum of the numbers in any neighboring cells is odd. This means that the numbers in any pair of neighboring cells have different parity, so their product is even. Therefore, every product that Fedir writes down is even, and their sum must also be even, so it cannot end with 1. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19746,
    "subject": "Mathematics (Olympiad)",
    "question": "Ямар нэг натурал тоо $k$-ийн хувьд\n\n$$\n3^{2^{k} - 1} \\equiv -1 \\pmod{2^{k} + 1}\n$$\n\nбол $n = 2^{k} + 1$ нь анхны тоо гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "$3^{2^{k} - 1} \\equiv -1 \\pmod{2^{k} + 1}$ тэнцэтгэлийн хоёр талыг квадрат зэрэгт дэвшүүлбэл $3^{2^{k}} \\equiv 1 \\pmod{2^{k} + 1}$ болно. $2^{k} + 1$ модулиар 3-ын эрэмбийг $a$ гэвэл $a \\mid 2^{k}$ байна. Эндээс $a = 2^{s}$ хэлбэртэй тул $3^{a} \\equiv 1 \\pmod{2^{k} + 1}$ ба өгсөн нөхцөлөөс $a > 2^{k} - 1$ буюу $a = 2^{k}$ болно. Нөгөө талаас Эйлерийн теорем ёсоор $a \\mid \\varphi(2^{k} + 1) \\leq 2^{k}$ ба тэнцэлдээ хүрэхийн тулд $2^{k} + 1$ нь анхны тоо байх ёстой. Иймд $a = 2^{k}$ тул $2^{k} + 1$ нь анхны тоо байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19747,
    "subject": "Mathematics (Olympiad)",
    "question": "The real polynomial $\\varphi(x) = ax^3 + bx^2 + cx + d$ has three positive roots, and $\\varphi(0) < 0$. Prove that\n\n$$\n2b^3 + 9a^2d - 7abc \\le 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1, x_2, x_3$ be the three positive roots of $\\varphi(x) = ax^3 + bx^2 + cx + d$. By Vieta's formulas:\n\n$$\n\\begin{align*}\nx_1 + x_2 + x_3 &= -\\frac{b}{a}, \\\\\nx_1x_2 + x_2x_3 + x_3x_1 &= \\frac{c}{a}, \\\\\nx_1x_2x_3 &= -\\frac{d}{a}.\n\\end{align*}\n$$\n\nSince $\\varphi(0) < 0$, $d < 0$, so $a > 0$.\n\nDividing both sides of the inequality by $a^3$ gives:\n\n$$\n2\\left(-\\frac{b}{a}\\right)^3 + 9\\left(-\\frac{d}{a}\\right) - 7\\left(-\\frac{b}{a}\\right)\\frac{c}{a} \\le 0.\n$$\n\nSubstituting Vieta's relations:\n\n$$\n2(x_1 + x_2 + x_3)^3 + 9x_1x_2x_3 - 7(x_1 + x_2 + x_3)(x_1x_2 + x_2x_3 + x_3x_1) \\ge 0.\n$$\n\nThis is equivalent to:\n\n$$\nx_1^2x_2 + x_1^2x_3 + x_2^2x_1 + x_2^2x_3 + x_3^2x_1 + x_3^2x_2 \\le 2(x_1^3 + x_2^3 + x_3^3).\n$$\n\nFor any two positive numbers $x$ and $y$, $(x - y)(x^2 - y^2) \\ge 0$, so $x^2y + y^2x \\le x^3 + y^3$. Applying this to each pair and summing, we obtain the desired inequality. Equality holds if and only if $x_1 = x_2 = x_3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19748,
    "subject": "Mathematics (Olympiad)",
    "question": "If I buy $x$ chocolate bars and $y$ cool-drinks, then $$25x + 9y = 839$$ with $x > y$. Find the values of $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "If we put $x = y$ as a first guess, then $x = y = \\frac{839}{34} \\approx 24$, so we can write $x = 24 + a$ and $y = 24 - b$. The equation becomes $$25a - 9b = 23$$ which by easy trial and error (guess and check) has a solution $a = 2$ and $b = 3$. Thus $x = 24 + 2 = 26$ and $y = 24 - 3 = 21$, which is the required answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19749,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ have incircle touching sides $AB$, $BC$, $CA$ at $C_1$, $A_1$, $B_1$ respectively. Then:\n\n$$\n\\begin{aligned}\nAB_1 &= AC_1 = \\frac{AB + AC - BC}{2}, \\\\\nBA_1 &= BC_1 = \\frac{BA + BC - AC}{2}, \\\\\nCA_1 &= CB_1 = \\frac{CA + CB - AB}{2}.\n\\end{aligned}\n$$\n\nLet $\\omega_1$ and $\\omega_2$ be the incircles of triangles $ABP$ and $APC$, respectively. The internal tangent $l$ (other than $AP$) to $\\omega_1$ and $\\omega_2$ meets $BC$ at $Q$. Prove that $I_1$, $P$, $Q$, and $I_2$ are concyclic, where $I_1$ and $I_2$ are the incenters of $ABP$ and $APC$ respectively. Also, express $PQ$ in terms of the side lengths.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $Q$ lies on segment $PC$.\n\nLet $R$ be the intersection of $l$ and $AP$. Since $I_1$ and $I_2$ lie on the internal and external angle bisectors of $\\angle PQ R$, $\\angle I_1 Q I_2 = 90^\\circ$.\n\nSimilarly, $\\angle I_1 P I_2 = 90^\\circ$. Thus, quadrilateral $I_1 P Q I_2$ is cyclic, so $Q$ is as defined.\n\nUsing the incircle tangent length formulas:\n\n$$\nDP = \\frac{PB + PA - AB}{2}, \\quad PE = \\frac{PC + PA - AC}{2}\n$$\n\nFrom a lemma, $DP = QE = PE - PQ$, so\n\n$$\nPQ = PE - DP = \\frac{AB + PC - AC - PB}{2}.\n$$\n\n![](images/Belaurus_2016_Booklet_p20_data_aaff1b8e2c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19750,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $x \\ge 0$ and $y^2 \\ge x(x+1)$. Prove that $$(y-1)^2 \\ge x(x-1)$$ for all real $x$ and $y$ satisfying these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "If $0 \\le x \\le 1$, then $(y-1)^2 \\ge 0 \\ge x(x-1)$, so the inequality holds.\n\nIf $x > 1$, either $y \\ge \\sqrt{x(x+1)}$ or $y \\le -\\sqrt{x(x+1)}$.\n\nIf $y \\ge \\sqrt{x(x+1)}$, then $y > 1$ since $x > 1$. So:\n$$(y-1)^2 \\ge (\\sqrt{x(x+1)}-1)^2 = x^2 + x + 1 - 2\\sqrt{x(x+1)}$$\nIt suffices to prove that this is $\\ge x(x-1)$, i.e.,\n$$x^2 + x + 1 - 2\\sqrt{x(x+1)} \\ge x^2 - x$$\nwhich simplifies to:\n$$2x + 1 \\ge 2\\sqrt{x(x+1)}$$\nSquaring both sides:\n$$(2x + 1)^2 \\ge 4x(x+1)$$\n$$4x^2 + 4x + 1 \\ge 4x^2 + 4x$$\n$$1 \\ge 0$$\nwhich is true.\n\nIf $y \\le -\\sqrt{x(x+1)}$, then:\n$$(y-1)^2 \\ge (\\sqrt{x(x+1)} + 1)^2 = x^2 + x + 1 + 2\\sqrt{x(x+1)}$$\nThis is greater than $x^2 + x > x^2 - x$ for $x > 1$, so the inequality holds in this case as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19751,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent to the circumcircle through vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AE$ is a diameter of the circumcircle of triangle $ABC$, $\\angle ACE = \\angle ECK = 90^\\circ$. So it suffices to show that $\\angle ACB = \\angle DCK$ (see the figure below).\n\nLet $L$ be the point of intersection of lines $AE$ and $DK$. Then $\\angle BAC = \\angle CBK = \\angle CDK$ by the inscribed angles theorem. Also, $\\angle ABC = \\angle AEC = \\angle CKD$, where the latter equality follows from the similarity of the right triangles $ACE$ and $ALK$. Hence, the two triangles $ABC$ and $DKC$ are similar, and therefore $\\angle ACB = \\angle DCK$.\n\n![](images/Estonija_2010_p21_data_dffa860214.png)\n\n**Remark.** This problem was proposed to the Baltic Way competition in 2008 (not by Estonia) but was not selected.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19752,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that all of the following numbers are prime:\n\n- $n$\n- $n^2 + 10$\n- $n^2 - 2$\n- $n^3 + 6$\n- $n^5 + 36$",
    "options": [],
    "answer": "See solution",
    "solution": "The only possible answer is $n = 7$.\n\nSince any 7 consecutive integers must contain a multiple of 7, it follows that if $n > 7$, then the given five numbers cannot be simultaneously prime. Hence we have $n \\leq 7$. By checking the remaining values of $n$ one by one, we see that the only possibility is $n = 7$, in which case:\n\n- $n = 7$\n- $n^2 + 10 = 49 + 10 = 59$\n- $n^2 - 2 = 49 - 2 = 47$\n- $n^3 + 6 = 343 + 6 = 349$\n- $n^5 + 36 = 16807 + 36 = 16843$\n\nAll of these are prime numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19753,
    "subject": "Mathematics (Olympiad)",
    "question": "Alicia writes down $a$ distinct integers on a piece of paper and Britt writes down $b$ distinct integers on another piece of paper. Alicia wrote down at least one integer that Britt did not write down, and Britt wrote at least one integer down that Alicia did not write down. Vera counts the number of distinct integers on the two pieces of paper; let this number of distinct integers be $v$. Daan counts how many of the integers that have been written down by Alicia, have also been written down by Britt; let $d$ be this number. For example, if Alicia wrote down 1, 2, and 5, and Britt wrote down 2, 5, 7, and 8, then we have $a = 3$ and $b = 4$ while $v = 5$ and $d = 2$.\n\n(a) Find an example for which $a = b = 2022$ and $a \\cdot b = d \\cdot (v + d)$.\n\n(b) Is it possible that $a \\cdot b = d \\cdot (v + 4)$? Give an example or prove that it is impossible.\n\n(c) Is it possible that $a \\cdot b = d \\cdot v$? Give an example or prove that it is impossible.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that there is a useful relation between $a$, $b$, $d$, and $v$. The total number of integers on the two pieces of paper is $a + b$, the number of integers on Alicia's piece of paper plus the number of integers on Britt's piece of paper. This also equals $v + d$: the total number of distinct integers, plus the total number of integers that have been written down twice. Hence, we get that $a + b = v + d$.\n\n**(a)** We choose $a = b = 2022$ and look for a solution to $a \\cdot b = d \\cdot (v + d)$. We use the fact that $a + b = v + d$. This means that we are looking for solutions to $a \\cdot b = d \\cdot (a + b)$. If we substitute $a = b = 2022$, then we find that $2022 \\cdot 2022 = d(2022 + 2022) = d \\cdot 2 \\cdot 2022$, so $d = 1011$. Together with $a + b = v + d$, we find that $2022 + 2022 = v + 1011$, so $v = 3033$. This situation happens, for example, if Alicia writes down the numbers 1 to 2022, and Britt writes down the numbers 1012 to 3033.\n\n**(b)** With a little bit of trying, and by choosing $d$ not too large, we find that $a = b = 3$, $d = 1$, and $v = 5$ is a solution: $3 \\cdot 3 = 1 \\cdot (5 + 4)$. The numbers also satisfy the equation $a + b = v + d$. This situation can occur if Alicia writes down the numbers 1, 2, and 3, and Britt writes down the numbers 3, 4, and 5, for example.\n\n**(c)** Suppose that there are numbers such that $a \\cdot b = v \\cdot d$. We already deduced that $a + b = v + d$, or $v = a + b - d$. Substituting this yields\n\n$$\na b = v d = (a + b - d)d = a d + b d - d^2.\n$$\n\nIf we now subtract $a d$ from both sides of this equation, we find $a b - a d = b d - d^2$, so $a(b - d) = d(b - d)$. Because Britt wrote down at least one number that Alicia did not write down, we have $b > d$. Therefore, we can divide the equation $a(b - d) = d(b - d)$ by the positive number $b - d$, and we find that $a = d$. On the other hand, Alicia wrote down at least one number that Britt did not write down, hence $a > d$. This gives a contradiction and hence there cannot exist numbers such that $a \\cdot b = v \\cdot d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19754,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection of the line $BC$ and the tangent at $A$ to the circumcircle $(ABC)$. Let $\\Gamma$ be a circle passing through $A$ and tangent to $(ABC)$ at $A$. Let $M$ and $N$ be points on $AB$ and $AC$, respectively, such that $MN$ is parallel to $BC$. Let $P$ be the intersection of $MN$ and the circle $\\Gamma$ (other than $A$).\n\nShow that $P$ is the midpoint of $MN$ if and only if $\\Gamma$ and $(ABC)$ are tangent at $A$.\n\n![](images/2.2_1998-2022_CHKMO_Solutions_p11_data_572e8e0888.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If $AB = AC$, the result is obvious due to symmetry (both statements are equivalent to $P$ being the midpoint of $BC$). Without loss of generality, assume $AB < AC$. Let $D$ be the intersection of $BC$ and the tangent at $A$ to $(ABC)$.\n\nIf $\\overline{MP}$ and $\\overline{NP}$ are equal, then $MN$ is parallel to the tangent at $P$ to $\\Gamma$, which is $BC$. It follows that\n\n$$\n\\angle DAM = \\angle DAB = \\angle ACB = \\angle ANM,\n$$\n\nwhich implies $DA$ is tangent to $\\Gamma$. Therefore, $\\Gamma$ is tangent to $(ABC)$ at $A$.\n\nConversely, if $\\Gamma$ and $(ABC)$ are tangent at $A$, then $DA$ is tangent to $\\Gamma$. Therefore,\n\n$$\n\\angle ANM = \\angle DAM = \\angle DAB = \\angle ACB,\n$$\n\nwhich implies $MN \\parallel BC$. This shows $P$ is the midpoint of $\\overline{MN}$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19755,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Points $E$, $F$ move on the opposite rays of $BA$, $CA$ such that $BF = CE$. Let $M$, $N$ be the midpoints of $BE$, $CF$. Suppose that $BF$ cuts $CE$ at $D$.\n\na) Let $I$, $J$ be the centers of the circumspheres of triangles $DBE$ and $DCF$. Prove that $MN$ is parallel to $IJ$.\n\nb) Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of triangle $AEF$. Prove that when $E$ moves on the opposite ray of $BA$, line $HK$ goes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $T$ be the second intersection of $(BDE)$ and $(CDF)$. We have $BF = CE$ and\n$$\n\\angle TED = \\angle TBD, \\quad \\angle TCD = \\angle TFD\n$$\nso triangles $TCE$ and $TFB$ are congruent. From this, it is easy to obtain that $TBE$ and $TCF$ are isosceles triangles at $T$ and they are similar.\n\n![](images/Vietnamese_mathematical_competitions_p285_data_918bf62617.png)\n\nIt implies that $T$, $I$, and $M$ are collinear; $T$, $J$, and $N$ are collinear. Note that $\\frac{TI}{TM} = \\frac{TJ}{TN}$, so $MN$ is parallel to $IJ$.\n\nb) Let $X$, $Y$ be the midpoints of $CE$, $BF$ and $l$ be the radical axis of the circles with diameters $BF$ and $CE$. Two altitudes $BX'$, $CY'$ of triangle $ABC$ intersect at orthocenter $S$.\n\n![](images/Vietnamese_mathematical_competitions_p286_data_6ae9d87d98.png)\n\nWe will prove that $HK$ passes through a fixed point $S$ by showing that these three points lie on $l$. Indeed, $MX$ is the midline of triangle $EBC$, so $MX \\parallel BC$; similarly, $NY \\parallel BC$, so $MX \\parallel NY$. Similarly, $MY \\parallel NX$, so $MYNX$ is a parallelogram, and $K$ is the midpoint of $XY$. Since $CE = BF$, then\n$$\nP_{K/(BF)} = KY^2 - BY^2 = KX^2 - CX^2 = P_{K/(CE)},\n$$\nwhich implies $K \\in l$.\n\nBesides, $BCX'Y'$ is a cyclic quadrilateral, so\n$$\nP_{S/(BF)} = \\overline{SX'} \\cdot \\overline{SB} = \\overline{SY'} \\cdot \\overline{SC} = P_{S/(CE)},\n$$\nhence $S \\in l$. Similarly, $H$ lies on $l$, thus $HK$ passes through the fixed point $S$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 19756,
    "subject": "Mathematics (Olympiad)",
    "question": "Several people gathered at a party. Each two of them who don't know each other have exactly two common acquaintances (if a person $A$ knows a person $B$, then necessarily the person $B$ knows the person $A$). Mykhailyk and Vitalyk know each other, but have no common acquaintances.\n\na) Prove that Mykhailyk and Vitalyk have the same number of acquaintances at the party.\n\nb) Can such a situation happen at a party with exactly 6 people?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Denote Mykhailyk and Vitalyk by $A$ and $B$ respectively. Let $M(A)$ and $M(B)$ be the sets of acquaintances of $A$ and $B$ respectively. There is no person in $M(A)$ that knows $B$, since in this case $A$ and $B$ would have a common acquaintance. So every person $X$ from $M(A)$ has exactly two common acquaintances with $B$. One of them is $A$, and the other person, call them $Y$, belongs to the set $M(B)$. In this way, for every person $X$ from the set $M(A)$ we assign exactly one person $Y$ from $M(B)$. It is easy to see that this correspondence is one-to-one, and so the numbers of people in $M(A)$ and $M(B)$ are the same.\n\nb) Example given below shows that the situation described in the problem statement can happen when there are exactly 6 people at the party.\n\n![](images/Ukrajina_2013_p7_data_6ad2324319.png)\n\n![](images/Ukrajina_2013_p7_data_fc3d5e9c24.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19757,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n \\ge 2$, let $B_1, B_2, \\dots, B_n$ denote $n$ arbitrary subsets of a set $X$, each of which contains exactly two elements. Find the minimum value of $|X|$ such that there exists a subset $Y$ of $X$ satisfying:\n\n(a) $|Y| = n$;\n\n(b) $|Y \\cap B_i| \\le 1$ for $i = 1, 2, \\dots, n$,\n\nwhere $|A|$ denotes the number of elements of the finite set $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that $|X| > 2n - 2$. In fact, if $|X| = 2n - 2$, let $X = \\{1, 2, \\dots, 2n - 2\\}$, $B_1 = \\{1, 2\\}$, $B_2 = \\{3, 4\\}$, $\\dots$, $B_{n-1} = \\{2n - 3, 2n - 2\\}$. Since $|Y| = n$, there exist two elements in $Y$ that belong to the same $B_i$, then $|Y \\cap B_i| > 1$, a contradiction.\n\nLet $|X| = 2n - 1$.\n\nLet $B = \\bigcup_{i=1}^n B_i$, then $|B| = 2n - 1 - z$, where $z$ is the number of elements in $X \\setminus B$. Suppose the elements of $X \\setminus B$ are $a_1, a_2, \\dots, a_z$.\n\nIf $z \\ge n - 1$, take $Y = \\{a_1, \\dots, a_{n-1}, d\\}$, and $d \\in B$, as desired.\n\nIf $z < n - 1$, suppose there are $t$ elements that occur once in $B_1, B_2, \\dots, B_n$. Since $\\sum_{i=1}^n |B_i| = 2n$, then\n\n$$\nt + 2(2n - 1 - z - t) \\le 2n,\n$$\n\nit follows that $t \\ge 2n - 2 - 2z$. So the elements that occur twice or more in $B_1, B_2, \\dots, B_n$ occur repeatedly by $2n - (2n - 2 - 2z) = 2 + 2z$ times.\n\nConsider the elements that occur once in $B_1, B_2, \\dots, B_n$:\n$b_1, b_2, \\dots, b_t$. Thus, there are at most $\\frac{2+2z}{2} = 1+z$ subsets in $B_1, B_2, \\dots, B_n$ that do not contain the elements $b_1, b_2, \\dots, b_t$. So, there exist $n - (z+1) = n-z-1$ subsets containing at least the elements $b_1, b_2, \\dots, b_t$.\n\nSuppose that $B_1, B_2, \\dots, B_{n-1-z}$ contain the elements $\\tilde{b}_1, \\tilde{b}_2, \\dots, \\tilde{b}_{n-1-z}$ of $b_1, b_2, \\dots, b_t$, respectively. Since\n\n$$\n2(n - 1 - z) + z = 2n - 2 - z < 2n - 1,\n$$\n\nthere must exist an element $d$ that is not in $B_1, B_2, \\dots, B_{n-1-z}$ but is in $B_{n-z}, \\dots, B_n$.\n\nWrite $Y = \\{a_1, \\dots, a_z, \\tilde{b}_1, \\tilde{b}_2, \\dots, \\tilde{b}_{n-1-z}, d\\}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19758,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, there are six different points $A$, $B$, $C$, $D$, $E$, $F$ such that $ABCD$ and $CDEF$ are parallelograms. What is the maximum number of those points that can be located on one circle?",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum number is $5$.\n\nAs $ABCD$ and $CDEF$ are parallelograms, the line segments $AB$, $CD$, and $EF$ are parallel and have the same length. Since it is impossible to draw three chords of equal length to a circle, not all $6$ points can be concyclic.\n\n![](images/prob1314_p6_data_a34d7427be.png)\n\nFigure 3\n\n![](images/prob1314_p6_data_cb419e4a2d.png)\n\nFigure 4\n\nA construction with $5$ concyclic points is shown in Figure 3.\n\nNote: There are many constructions with $5$ vertices. For example, we can take a rectangle $ABCD$, add the fifth point $E$ randomly on the circumcircle of the rectangle, and choose point $F$ such that $CDEF$ is a parallelogram (see Figure 4).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19759,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (x, y) $ that satisfy the equality:\n$$\nx - y = \\frac{x}{y}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us rewrite the equality as follows:\n$$\nx - y = \\frac{x}{y}\n$$\nMultiply both sides by $y$ (with $y \\neq 0$):\n$$\ny(x - y) = x\n$$\nwhich gives\n$$\nxy - y^2 = x\n$$\nRearrange:\n$$\nxy - y^2 - x = 0 \\implies x(y - 1) = y^2\n$$\nSince $y^2$ and $y - 1$ are coprime, $y - 1$ must divide $x$. Let $x = k(y - 1)$ for some integer $k$:\n$$\nx = k(y - 1)\n$$\nSubstitute into $x(y - 1) = y^2$:\n$$\nk(y - 1)^2 = y^2\n$$\nSo $k = \\frac{y^2}{(y - 1)^2}$, which is integer only if $y - 1 = \\pm 1$ or $y - 1 = \\pm y$.\n\nTry $y - 1 = 1 \\implies y = 2$:\n$$\nx = k(1) = k,\\quad y = 2\n$$\nFrom $x(y - 1) = y^2$:\n$$\nx(1) = 4 \\implies x = 4\n$$\nSo $(x, y) = (4, 2)$.\n\nTry $y - 1 = -1 \\implies y = 0$ (but $y \\neq 0$ since division by zero is not allowed).\n\nTherefore, the only integer solution is:\n$$\n\\boxed{(x, y) = (4, 2)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19760,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob play a game that allows the playing numbers 19 and 20 and the two possible starting numbers 9 and 10. Alice chooses her playing number and assigns the remaining playing number to Bob, while Bob independently chooses the starting number.\n\nAlice adds her playing number to the starting number, Bob adds his playing number to the sum, then Alice again adds her playing number to this new sum, and so on. The game lasts until the number 2019 is reached or exceeded.\n\nA player who obtains exactly 2019 wins. If 2019 is exceeded, the game ends in a draw.\n\n*Show that Bob cannot win.*\n\n*Which starting number does Bob have to choose in order to prevent Alice from winning?*",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s$ be the starting number, $a$ Alice's playing number, and $b$ Bob's playing number. Let $n$ be the number of rounds (i.e., the number of times Bob adds his number).\n\nFor Bob to win, the equation $s + 39n = 2019$ must have an integer solution for $n$. But neither $s = 9$ nor $s = 10$ satisfy this, as neither $2010$ nor $2009$ is divisible by $39$. Hence, Bob cannot win. In fact, $51 \\times 39 = 1989$ and $52 \\times 39 = 2028$.\n\nFor Alice to win, the equation $s + 39n + a = 2019$ must be satisfied for some $n$. Since $28 \\leq s + a \\leq 30$, this can only work for $n = 51$, which implies $s + a = 30$. Therefore, $s = 10$ and $a = 20$.\n\nWe conclude that Bob has to choose $9$ as the starting number in order not to lose the game.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19761,
    "subject": "Mathematics (Olympiad)",
    "question": "The first term $x_1$ of a sequence is $2014$. Each subsequent term of the sequence is defined by the iterative formula:\n\n$$\nx_{n+1} = \\frac{(\\sqrt{2}+1)x_n - 1}{(\\sqrt{2}+1) + x_n}.\n$$\n\nFind the 2015th term $x_{2015}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k = \\sqrt{2} + 1$ and $m = \\sqrt{2} - 1$, so $mk = 1$ and $k - m = 2$.\n\nThis gives:\n\n$$\n\\begin{aligned}\nx_{n+1} &= \\frac{kx_n - 1}{x_n + k} \\\\\n&= \\frac{x_n - \\frac{1}{k}}{\\frac{x_n}{k} + 1} \\\\\n&= \\frac{x_n - m}{m x_n + 1}.\n\\end{aligned}\n$$\n\nSubstituting this in, we learn that:\n\n$$\n\\begin{aligned}\nx_{n+2} &= \\frac{\\frac{x_n - m}{m x_n + 1} - 1}{k + \\frac{x_n - m}{m x_n + 1}} \\\\\n&= \\frac{x_n - 1}{x_n + 1}\n\\end{aligned}\n$$\n\nThis tells us that $x_{n+4} = \\frac{1}{x_n}$ and $x_{n+8} = x_n$. It follows that $x_{n+8k} = x_n$, and so $x_{2015} = x_7$.\n\nUsing the results relating $x_n$ to $x_{n+4}$ and $x_{n+2}$, we see that $x_5 = \\frac{1}{2014}$, and so $x_7 = \\frac{-1-2014}{-1+2014} = -\\frac{2015}{2013}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19762,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider a triangle $ABC$ and a point $D$ on the extended line segment $AB$ on the side of $B$. The point $E$ lies on side $AC$ such that the angles $\\angle DBC$ and $\\angle DEC$ are equal. The intersection of $DE$ and $BC$ is $F$. Suppose that $|BF| = 2$, $|BD| = 3$, $|AE| = 4$, and $|AB| = 5$.\n\n![](images/NLD_ABooklet_2022_p10_data_08ec45b126.png)\n\n(a)\n\nProve that triangles $\\triangle ABC$ and $\\triangle AED$ are similar.\n\n(b)\n\nDetermine $|CF|$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Because angles $\\angle AEC$ and $\\angle ABD$ are straight, we have\n\n$$\n\\angle ABC = 180^{\\circ} - \\angle DBC = 180^{\\circ} - \\angle DEC = \\angle AED.\n$$\n\nBecause angle $A$ occurs in both triangles, triangles $\\triangle ABC$ and $\\triangle AED$ have two equal angles, and hence the triangles are similar. $\\Box$\n\n(b) Because of the similarity of triangles $\\triangle ABC$ and $\\triangle AED$, the angles at $C$ and $D$ are equal. Together with the equality $\\angle DBF = \\angle CEF$, it follows that triangles $\\triangle DBF$ and $\\triangle CEF$ are similar.\n\nIn a pair of similar triangles, all pairs of sides have the same ratio. Hence, the similarity of triangles $\\triangle DBF$ and $\\triangle CEF$ yields\n\n$$\n\\frac{|BF|}{|EF|} = \\frac{|FD|}{|FC|} = \\frac{|DB|}{|CE|} \\qquad (1)\n$$\n\nAs triangles $\\triangle ABC$ and $\\triangle AED$ are similar, we find that\n\n$$\n\\frac{|AB|}{|AE|} = \\frac{|BC|}{|ED|} = \\frac{|CA|}{|DA|} \\qquad (2)\n$$\n\nUsing equations (1) and (2), we can now find $|CF|$. Using the first and last ratio in equation (2), we get $\\frac{5}{4} = \\frac{|AB|}{|AE|} = \\frac{|AC|}{|AD|} = \\frac{4+|EC|}{5+3}$. Hence, we have $|EC| = 6$. If we substitute this in the first and third ratio in equation (1), we get $\\frac{2}{|EF|} = \\frac{3}{6}$. Hence, we have $|EF| = 4$. Using the first and second ratio in (1), we now get that $\\frac{2}{4} = \\frac{|FD|}{|FC|}$ hence $|FD| = \\frac{1}{2}|CF|$. Finally, we substitute this in the first and second ratio in equation (2):\n\n$$\n\\frac{5}{4} = \\frac{|AB|}{|AE|} = \\frac{|BC|}{|DE|} = \\frac{2 + |CF|}{4 + \\frac{1}{2}|CF|}.\n$$\n\nTaking cross ratios and solving the remaining equation, we get $|CF| = 8$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19763,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in real numbers the equation\n\n$$\n2^{x-1} + 2^{\\frac{1}{\\sqrt{x}}} = 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We notice that $x > 0$. The given equation can be rewritten as\n\n$$\n2^x + 2 \\cdot 2^{\\frac{1}{\\sqrt{x}}} = 6.\n$$\n\nFrom the AM-GM inequality, we have:\n\n$$\n2^x + 2^{\\frac{1}{\\sqrt{x}}} + 2^{\\frac{1}{\\sqrt{x}}} \\ge 3 \\cdot \\sqrt[3]{2^x \\cdot 2^{\\frac{1}{\\sqrt{x}}} \\cdot 2^{\\frac{1}{\\sqrt{x}}}} = 3 \\cdot 2^{\\frac{1}{3}\\left(x+\\frac{2}{\\sqrt{x}}\\right)}, \\quad \\forall x > 0.\n$$\n\nAlso, using the AM-GM inequality,\n\n$$\n\\frac{1}{3} \\left( x + \\frac{2}{\\sqrt{x}} \\right) \\ge \\sqrt[3]{x \\cdot \\frac{1}{\\sqrt{x}} \\cdot \\frac{1}{\\sqrt{x}}} = 1,\n$$\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": 19764,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the ten's place of $11^{12^{13}}$. ($12^{13}$th power of 11, not 13th power of $11^{12}$.)",
    "options": [],
    "answer": "See solution",
    "solution": "For a positive integer $n$,\n\n$$\n11^n = (10 + 1)^n = \\sum_{k=0}^{n} \\binom{n}{k} 10^k\n$$\n\nThe last two digits of $11^n$ depend on $n$. In particular, the ten's place of $11^n$ is equal to the last digit of $n$.\n\nThe last digit of $12^n$ for $n = 1, 2, \\dots$ cycles as $2, 4, 8, 6$ (period $4$). Since $13 \\equiv 1 \\pmod{4}$, the last digit of $12^{13}$ is $2$.\n\nTherefore, the ten's place of $11^{12^{13}}$ is $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19765,
    "subject": "Mathematics (Olympiad)",
    "question": "Capablanca and Alyokhin decided to play a match of 16 games according to the following rules. The winner of the first game received $1 = 3^0$ peso, the winner of the second one got $3 = 3^1$ pesos, the winner of the third game received $9 = 3^2$ pesos, and so on. If the game ended in a draw, then they split the prize pool of the game in half. At the end of the match, Alyokhin earned 2018 pesos more than Capablanca. How many games has each of the players won?",
    "options": [],
    "answer": "See solution",
    "solution": "Let Alyokhin's gain in the $k$-th game be $a_k \\cdot 3^{k-1}$, where $a_k \\in \\{-1, 0, 1\\}$ (win, draw, loss). The total possible gain after $k$ games ranges from $-A_k$ to $A_k$, where $A_k = 3^0 + 3^1 + \\dots + 3^{k-1}$. Each result is uniquely determined by the sequence of game outcomes. For 16 games, the possible net differences are all integers between $-A_{16}$ and $A_{16}$. Given Alyokhin earned 2018 pesos more than Capablanca, we seek a combination of wins and draws that yields this difference. Expressing 2018 in terms of powers of 3:\n\n$$\n2018 = 3^7 - 3^5 + 3^4 - 3^2 + 3^1 - 3^0\n$$\n\nThis decomposition shows that each player won 3 games, and the remaining games ended in draws.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19766,
    "subject": "Mathematics (Olympiad)",
    "question": "We are to determine the minimal value of $k$ such that, for any subset $P$ of $S$ with $k$ elements, where $S$ is the set of all pairs of natural numbers $(x, y)$ with $1 \\le x \\le m$ and $1 \\le y \\le n$, there exists a subset of $P$ of $m+1$ elements in which all first coordinates are distinct and all second coordinates are distinct.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the solution of the problem is $k = nm + 1$.\n\nSuppose that the problem were solvable for a $k \\le nm$. Let $A$ be a subset of $S$ that contains all pairs $(x, y)$ such that $1 \\le x \\le m$ and $1 \\le y \\le n$. Then $|A| = mn$. Consider $m+1$ arbitrary elements of this set. Since the first coordinate of any of them is between $1$ and $m$, according to Dirichlet's principle there would exist two pairs with the same first coordinate. This would be a contradiction with the conditions of the problem. The same would hold for any subset of $A$ with $k$ elements, hence $k$ cannot be less than or equal to $nm$.\n\nNow consider any set $P \\subseteq S$ with $nm + 1$ elements. Divide $S$ into $n$ sets $B_i = \\{(x, y) \\mid x + y \\equiv i \\pmod{n},\\ 1 \\le x, y \\le n\\}$ for $i = 1, 2, \\dots, n$. The sets are pairwise disjoint, each has exactly $n$ elements, and their union is the entire set $S$. Because $|P| = nm + 1$, according to Dirichlet's principle there must exist a set $B_{i_0}$ which contains at least $m + 1$ elements from $P$. Suppose $(x_1, y_1), (x_2, y_2) \\in B_{i_0}$ are two different elements. If $x_1 = x_2$, then from $x_1 + y_1 \\equiv i_0 \\equiv x_2 + y_2 \\pmod{n}$ we derive $y_1 \\equiv y_2 \\pmod{n}$. Since $1 \\le y_1, y_2 \\le n$, we have $y_1 = y_2$. This is in contradiction with the assumption that the elements are different. We conclude that the elements of $B_{i_0}$ have pairwise different first coordinates and pairwise different second coordinates. Hence, if we choose $m + 1$ elements from $P$ such that they all are also elements of $B_{i_0}$, they do fulfill the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19767,
    "subject": "Mathematics (Olympiad)",
    "question": "A set $\\mathcal{D}$ of $n$ straight lines in a plane has the property that each line of the set intersects exactly $2011$ straight lines of $\\mathcal{D}$.\n\nFind $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be a line from $\\mathcal{D}$ and let $k$ be the number of lines from $\\mathcal{D}$ parallel to $d$. Then $n = 2012 + k$.\n\nSince every line $a$ from $\\mathcal{D}$ different from $d$ meets $2011$ lines from $\\mathcal{D}$, $a$ is parallel to $n - 2012 = k$ lines from $\\mathcal{D}$.\n\nThis shows that each line from $\\mathcal{D}$ is parallel to exactly $k$ other lines, so the $n$ lines can be divided into groups of $k + 1$ mutually parallel lines.\n\nSo $k + 1 \\mid n$, hence $k + 1 \\mid k + 2012$, therefore $k \\in \\{0, 2010\\}$, that is, $n$ can be $2012$ or $4022$.\n\nA possible configuration for $n = 2012$ is the supports of the sides of a regular $2012$-gon.\n\nA possible configuration for $n = 4022$ is $2011$ parallel lines meeting another $2011$ parallel lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19768,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 5$ 為一正整數。有價值為 1 到 $n$ 的星星各一枚。安妮亞和貝琪玩一個遊戲。在遊戲開始時，安妮亞先將這 $n$ 枚星星依照她想要的順序，在桌面上排成一橫排。接著，從貝琪開始，兩個人輪流取走桌面上最左邊或最右邊的一枚星星。當所有星星都被取走時，取得的星星價值總和較高的人獲勝；若總和相同，則兩人平手。試求所有讓貝琪有必勝法的 $n$。",
    "options": [],
    "answer": "See solution",
    "solution": "答案是所有型如 $4k+2$ 的 $n$。為方便討論，將星星從左至右視為在 1 號位置至 $n$ 號位置。以下分別討論。\n\n1. 偶數 $n$：注意到在偶數 $n$ 的情況下，貝琪必然可以選擇取得所有奇數位置的星星（先拿走 1 號位置，之後每次都跟安妮亞拿同一側），或是取得所有偶數位置的星星（先拿走 $n$ 號位置，之後每次都跟安妮亞拿同一側）。令 $O$ 為奇數位置星星價值和，$E$ 為偶數位置星星價值和，則 $O+E = \\frac{n(n+1)}{2}$。\n\n   - 當 $n = 4k + 2$ 時，$n(n-1)/2$ 為奇數，故貝琪只要從 $O$ 和 $E$ 中選擇總和較大的一方取即可。\n\n   - 當 $n = 4k$ 時，注意到貝琪必不輸（因為她最差可以拿到跟安妮亞一樣多）。而若安妮亞一開始把星星排成\n\n$$\n1, 3, 5, \\dots, 4k-1, 4k, 4k-2, \\dots, 2\n$$\n\n並且每次都跟貝琪拿同一側，則前 $k$ 回合安妮亞每次都可以保證比貝琪多拿價值 2，而後 $k$ 回合安妮亞每次最差就比貝琪少拿價值 2，故安妮亞也必不輸，故兩人平手。\n\n2. 奇數 $n$：若 $n = 2k-1$，安妮亞只要一開始將價值 1 到 $k$ 的星星放在奇數位置，其餘星星放在偶數位置，並每次都跟貝琪拿同一側，則安妮亞可以保證拿到 $k+1$ 到 $2k-1$ 的所有星星，總價值為 $\\frac{3k(k-1)}{2}$，大於貝琪拿到的 $\\frac{k(k+1)}{2}$，故安妮亞必勝。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19769,
    "subject": "Mathematics (Olympiad)",
    "question": "$n$ ($n \\geq 3$) table tennis players have a round-robin tournament—each player plays all others exactly once, and there are no draws. Suppose, after the tournament, all players can be arranged in a circle such that, for any three players $A$, $B$, $C$, if $A$ and $B$ are adjacent, then at least one of them defeated $C$. Find all possible values of $n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that $n$ can be any odd number not less than 3.\n\nSuppose $n = 2k + 1$, an odd number greater than 3, and the players are $A_1, A_2, \\dots, A_{2k+1}$. Arrange the results so that $A_i$ ($1 \\leq i \\leq 2k+1$) defeats $A_{i+2}, A_{i+4}, \\dots, A_{i+2k}$ (with $A_{2k+1+j} = A_j$ for $j = 1, 2, \\dots, 2k+1$), and loses to the others. Arrange the players in the circle $A_1, A_2, \\dots, A_{2k+1}, A_1$.\n\nGiven any three players $A$, $B$, $C$ with $A$ and $B$ adjacent, suppose $A = A_i$, $B = A_{i+1}$, $C = A_{i+r}$ ($2 \\leq r \\leq 2k$). Then either $r$ or $r-1$ is even and at least $2$, so at least one of $A$ or $B$ defeated $C$.\n\nNow suppose $n$ is even and $n \\geq 4$, and the $n$ players can be arranged in a circle $A_1, A_2, \\dots, A_n, A_1$ satisfying the condition. Assume $A_1$ defeated $A_2$. Then, at least one of $A_2, A_3$ defeated $A_1$, so $A_3$ defeated $A_1$; but at least one of $A_1, A_2$ defeated $A_3$, so $A_2$ defeated $A_3$, and so on. Thus, for any $1 \\leq i \\leq n$, $A_i$ defeated $A_{i+1}$ and lost to $A_{i-1}$ (with $A_{n+1} = A_1$, $A_0 = A_n$).\n\nDivide the players after $A_i$ and before $A_{i-1}$ into $\\frac{n-2}{2}$ pairs of adjacent players. In each pair, at least one defeated $A_i$, so besides $A_{i-1}$, there are at least $\\frac{n-2}{2}$ players who defeated $A_i$. Thus, $A_i$ lost at least $\\frac{n}{2}$ games. Summing over all $n$ players, the total number of losses is at least $\\frac{n^2}{2}$, which is impossible since there are only $\\frac{n(n-1)}{2}$ games. Therefore, $n$ must be odd.\n\n*Answer*: All odd $n \\geq 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19770,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{N} \\to \\mathbb{N}$ be a function from the positive integers to the positive integers such that $f(1) = 1$, $f(2n) = f(n)$, and $f(2n+1) = f(n) + f(n+1)$ for all $n \\in \\mathbb{N}$. Prove that for any natural number $n$, the number of odd natural numbers $m$ such that $f(m) = n$ is equal to the number of positive integers not greater than $n$ that are coprime to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The crucial observation is that the function $f$ encodes Euclid's algorithm when numbers are viewed in binary. Define $g(n) = f(n+1)$, and for each integer $n$, consider the pair $(f(n), g(n))$. If $x$ is the binary string representing $n$, the recurrence relations yield:\n\n$$\n(f(x0), g(x0)) = (f(x), f(x) + g(x)), \\quad (f(x1), g(x1)) = (f(x) + g(x), g(x)).\n$$\n\nThus, to compute $(f(x), g(x))$, start from $(1, 1) = (f(1), g(1))$ and read the binary digits of $x$ from left to right, ignoring the initial $1$: for each $0$, add the first coordinate to the second; for each $1$, add the second coordinate to the first. For example, to calculate $(f(27), g(27))$, write $27 = 11011_2$:\n\n$$\n\\begin{aligned}\n(f(1), g(1)) &= (1, 1) \\\\\n(f(11), g(11)) &= (2, 1) \\\\\n(f(110), g(110)) &= (2, 3) \\\\\n(f(1101), g(1101)) &= (5, 3) \\\\\n(f(11011), g(11011)) &= (8, 3)\n\\end{aligned}\n$$\n\nBy induction, $(f(n), g(n))$ are coprime positive integers, with $f(n) \\geq g(n)$ if and only if $n$ is odd. To complete the proof, we show that each pair $(a, b)$ of coprime positive integers arises as $(f(n), g(n))$ for a unique $n$. Run Euclid's algorithm on $(a, b)$: successively subtract the smaller from the larger until reaching $(1, 1)$. Record a $0$ for each subtraction of the first from the second, and a $1$ for each subtraction of the second from the first. Reversing this string and prepending a $1$ gives the binary expansion of $n$ such that $(f(n), g(n)) = (a, b)$. Uniqueness follows since Euclid's algorithm proceeds deterministically, so each coprime pair corresponds to a unique $n$.\n\nTherefore, the number of odd $m$ with $f(m) = n$ equals the number of positive integers not greater than $n$ that are coprime to $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19771,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be an interior point of an equilateral triangle with altitude $1$. If $x$, $y$, and $z$ are the distances from $P$ to the sides of the triangle, prove that:\n\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is well-known that in an equilateral triangle, the sum of the distances from an interior point $P$ to its sides equals the altitude of the triangle. Therefore, $x + y + z = 1$.\n\nWe need to prove that if $x + y + z = 1$, then\n\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$\n\nFirst, observe that when $x + y + z = 1$,\n\n$$\nxy + yz + zx \\geq 9xyz.\n$$\n\nBy the AM-GM inequality:\n\n$$\nxy + yz + zx = (xy + yz + zx)(x + y + z) \\geq 3\\sqrt[3]{(xy)(yz)(zx)} \\cdot 3\\sqrt[3]{xyz} = 9xyz.\n$$\n\nThus,\n\n$$\nxy + yz + zx - 3xyz \\geq 6xyz.\n$$\n\nUsing this and the constraint, we have:\n\n$$\n\\begin{aligned}\nxy + yz + zx - 3xyz &= xy(1 - z) + yz(1 - x) + zx(1 - y) \\\\\n&= xy(x + y) + yz(y + z) + zx(z + x) \\\\\n&\\geq 6xyz.\n\\end{aligned}\n$$\n\nAdding $1$ to both sides and rearranging terms gives:\n\n$$\n(x + y + z)^2 + xy(x + y) + yz(y + z) + zx(z + x) - 6xyz \\geq 1.\n$$\n\nOr equivalently,\n\n$$\nx^2 + y^2 + z^2 + 2xy(1-z) + 2yz(1-x) + 2zx(1-y) + xy(x+y) + yz(y+z) + zx(z+x) \\geq 1.\n$$\n\nAnd,\n\n$$\nx^2 + y^2 + z^2 + 3xy(x + y) + 3yz(y + z) + 3zx(z + x) \\geq 1 = (x + y + z)^3.\n$$\n\nTherefore,\n\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$\n\nEquality holds when $x = y = z = \\frac{1}{3}$, i.e., when $P$ is the centroid of the triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19772,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonnegative integers $a, b, c, d$ such that\n$$\n11^a \\cdot 5^b - 3^c \\cdot 2^d = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We consider four cases for $d$:\n\n**Case 1:** $d=0$. Here, $3^c 2^d$ is odd, so there are no solutions.\n\n**Case 2:** $d=1$. The equation becomes $11^{a} 5^{b} - 2 \\cdot 3^{c} = 1$. Since $c > 0$, and considering modulo 4, $a$ is odd. Modulo 3, $a+b$ is even, so $a = 2\\alpha + 1$, $b = 2\\beta + 1$ for nonnegative integers $\\alpha, \\beta$. Substituting, $55 \\cdot 11^{2\\alpha} 5^{2\\beta} - 2 \\cdot 3^c = 1$ and $c \\ge 3$. If $c > 3$, $55(11^{2\\alpha} 5^{2\\beta} - 1) = 54(3^{c-3} - 1)$, so $3^{c-3} - 1$ is divisible by $11$. Since $5$ is the order of $3$ mod $11$, $c-3 = 5\\gamma$ for some $\\gamma$, and $3^{c-3} - 1$ is divisible by $3^5 - 1 = 2 \\cdot 11^2$, which is impossible. Thus, the only solution is $a=1$, $b=1$, $c=3$, $d=1$.\n\n**Case 3:** $d=2$. The equation is $11^{a} 5^{b} - 4 \\cdot 3^{c} = 1$. Modulo 4, $a$ is even. If $c=0$, we get $a=0$, $b=1$, $c=0$, $d=2$ as a solution. If $c > 0$, modulo 3, $a+b$ is even, so $a$ and $b$ are even. Write $a=2\\alpha$, $b=2\\beta$. Then $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 4 \\cdot 3^c$, but the left side is divisible by 8, which is impossible.\n\n**Case 4:** $d > 2$. Modulo 4, $a$ is even. Modulo 8, $b$ is even, so $a=2\\alpha$, $b=2\\beta$. The equation becomes $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 3^{c} \\cdot 2^{d}$. If $c=0$, then $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 2^{d}$, but since $\\gcd(11^{a} 5^{b} - 1, 11^{a} 5^{b} + 1) = 2$ and $11^{a} 5^{b} - 1 > 2$, this is impossible. Thus $c > 0$ and we have two possibilities:\n\n(i) $11^{a} 5^{b} - 1 = 3^{c} \\cdot 2$, $11^{a} 5^{b} + 1 = 2^{d-1}$,\n\n(ii) $11^{a} 5^{b} - 1 = 2^{d-1}$, $11^{a} 5^{b} + 1 = 3^{c} \\cdot 2$.\n\nIn (i), $2^{d-1} = 3^{c} \\cdot 2 + 2$, so $2^{d-2} = 3^{c} + 1$. Thus $2^{d-2} - 1 = 3^c$, and $d-2$ is even, say $d-2 = 2\\delta$, so $(2^d - 1)(2^d + 1) = 3^c$, which only works for $\\delta = 1$, $c=1$, but $11^{a} 5^{b} - 1 = 6$ has no solutions.\n\nIn (ii), $2 = 3^c \\cdot 2 - 2^{d-1}$, so $2^{d-2} = 3^c - 1$. For $c=1$, $d=3$, giving $a=0$, $b=2$, $c=1$, $d=3$. For $c=2$, $d=5$, but this does not yield solutions.\n\n**Conclusion:**\n\nThe solutions are:\n- $a=1$, $b=1$, $c=3$, $d=1$: $11^1 \\cdot 5^1 - 3^3 \\cdot 2^1 = 1$\n- $a=0$, $b=1$, $c=0$, $d=2$: $11^0 \\cdot 5^1 - 3^0 \\cdot 2^2 = 1$\n- $a=0$, $b=2$, $c=1$, $d=3$: $11^0 \\cdot 5^2 - 3^1 \\cdot 2^3 = 1$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19773,
    "subject": "Mathematics (Olympiad)",
    "question": "One hundred glasses are arranged in a $10 \\times 10$ array. Now we pick $a$ of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick $b$ of the columns and fill them up with blue liquid. The remaining columns are filled up with yellow liquid. The mixture of blue and yellow liquid turns green. If both halves in a glass have the same colour, then that colour remains as it is.\n\n(a) Determine all possible combinations of values for $a$ and $b$ so that exactly half of the glasses contain green liquid at the end.\n\n(b) Is it possible that precisely one quarter of the glasses contain green liquid at the end?",
    "options": [],
    "answer": "See solution",
    "solution": "The total number of glasses that are green at the end of the procedure is\n\n$$\na(10 - b) + b(10 - a) = 10a + 10b - 2ab = 2(5a + 5b - ab).\n$$\n\nWe immediately observe that this number is always even, so the number of green glasses cannot be $25$ (i.e., one quarter). Hence the answer to the second question is no.\n\nFor the first question, we have to solve the equation\n\n$$\n10a + 10b - 2ab = 50,\n$$\n\nwhich is equivalent to\n\n$$\n2ab - 10a - 10b + 50 = 2(a - 5)(b - 5) = 0.\n$$\n\nThus, exactly half of the glasses contain green liquid if either $a = 5$ or $b = 5$ (and the other is arbitrary).",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19774,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, determine the maximum number of edges a triangle-free Hamiltonian simple graph on $n$ vertices may have.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $\\left\\lfloor \\frac{n}{2} \\right\\rfloor^2$ if $n$ is even, and $\\left\\lfloor \\frac{n}{2} \\right\\rfloor^2 + 1$ if $n$ is odd.\n\nBy Turán's theorem, the maximum number of edges a triangle-free simple graph on $n$ vertices is $\\left\\lfloor \\frac{n}{2} \\right\\rfloor \\left\\lfloor \\frac{n + 1}{2} \\right\\rfloor$, achieved only by the complete bipartite graph $K_{\\left\\lfloor n/2 \\right\\rfloor,\\, \\left\\lfloor (n+1)/2 \\right\\rfloor}$. If $n$ is even, this graph is also Hamiltonian, so we are done.\n\nConsider the case $n = 2m + 1$, where $m > 1$. Let $G$ be a triangle-free Hamiltonian simple graph on $n$ vertices with a maximum number of edges. Since $G$ is Hamiltonian and has an odd number of vertices, it contains an odd cycle, say $C$, of length $2k + 1$ with $k > 1$. No additional edges forming diagonals in $C$ may exist without creating a shorter odd cycle. Each of the $2m - 2k$ vertices outside $C$ may be joined to at most two vertices of $C$, as more would yield a shorter odd cycle. By Turán's theorem, the $2m - 2k$ vertices outside $C$ may induce at most $(m-k)^2$ edges without forming triangles. Consequently, $G$ has at most\n\n$$\n(2k + 1) + 2(2m - 2k) + (m - k)^2\n$$\nedges. The largest possible value occurs when $k = 2$, yielding $m^2 + 1$. Many Hamiltonian graphs achieve this bound; for example, one can be constructed from $K_{m,m}$ by inserting a vertex of degree 2 on any one edge.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19775,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that all positive integers less than $n$ and coprime to $n$ are powers of primes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p_1 = 2 < p_2 = 3 < p_3 = 5 < \\dots$ be the sequence of primes, and let $q$ and $r$, with $q < r$, be the first two primes which do not divide $n$. A necessary and sufficient condition for $n$ to be of the required type is that $n < qr$. Each of the primes less than $r$ and different from $q$ divides $n$, and so does their product. Therefore, the product of all primes less than $r$ does not exceed $nq < q^2 r$. If $r = p_m$, then $q \\le p_{m-1}$, so $p_1 p_2 \\cdots p_{m-2} < p_{m-1} p_m$.\n\nNotice that 6 is the first index $k$ such that $p_1 p_2 \\cdots p_{k-2} > p_{k-1} p_k$. Now, if $p_1 p_2 \\cdots p_{k-2} > p_{k-1} p_k$ for some index $k \\ge 6$, then (by Bertrand-Tchebysheff) $p_1 p_2 \\cdots p_{k-1} > p_{k-1}^2 p_k > 2 p_{k-1} \\cdot 2 p_k > p_k p_{k+1}$, so $p_1 p_2 \\cdots p_{k-2} > p_{k-1} p_k$ for all indices $k \\ge 6$.\n\nConsequently, $m \\le 5$, $r = p_m \\le p_5 = 11$, $q \\le p_4 = 7$, and $n < qr \\le p_4 p_5 = 7 \\cdot 11 = 77$. 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": 19776,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $T_n$ be an equilateral triangle of side length $n$. The triangle $T_n$ is divided into a triangular grid of unit triangles using lines parallel to the sides of $T_n$. (Each unit triangle is an equilateral triangle of side length 1.)\n\nA *saw-tooth* consists of two unit triangles joined at a vertex, producing a shape that is congruent to the following figure.\n\n![](images/2021_Australian_Scene_p139_data_88951c1702.png)\n\nA *saw-tooth tiling* of $T_n$ is a placement of saw-teeth such that each saw-tooth exactly covers two unit triangles in the grid and each unit triangle in the grid is covered exactly once.\n\nFor which values of $n$ does $T_n$ have a saw-tooth tiling?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $T_n$ can be tiled by saw-teeth if and only if $n$ is a multiple of 4.\n\nFirst of all, each unit triangle $T_1$ can either point *up* or *down*. In fact, if $T_n$ is pointing up, it is made up of $1+2+\\cdots+n = \\frac{n(n+1)}{2}$ unit triangles that point up and $1+2+\\cdots+(n-1) = \\frac{n(n-1)}{2}$ unit triangles that point down. In the example below, we see that $T_3$ is made up of six unit triangles that point up (black) and three that point down (white).\n\n![](images/2021_Australian_Scene_p141_data_3daa61f570.png)\n\nSince each saw-tooth consists of two unit triangles of the same orientation, in order to achieve a tiling, we need both $\\frac{n(n+1)}{2}$ and $\\frac{n(n-1)}{2}$ to be even. Now $\\frac{n(n+1)}{2}$ is even if and only if $n \\equiv 0$ or $-1 \\pmod{4}$, while $\\frac{n(n-1)}{2}$ is even if and only if $n \\equiv 0$ or $1 \\pmod{4}$. Thus $n$ must be a multiple of 4.\n\nIt suffices to provide a construction when $n = 4k$. This can be done by dissecting $T_{4k}$ into copies of $T_4$ and tiling each $T_4$ as shown below.\n\n![](images/2021_Australian_Scene_p141_data_78f1780f8b.png)\n\n![](images/2021_Australian_Scene_p141_data_43972b2a59.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19777,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $\\tau$ of $\\{1, \\dots, n\\}$ such that $k^4 + (\\tau(k))^4$ is prime for all $k = 1, \\dots, n$. Show that $S(n)$ is always a square.",
    "options": [],
    "answer": "See solution",
    "solution": "The fact on squareness is easy.\n\nIf $n$ is odd, among the sums of the form $i^4 + \\tau(i)^4$ there is an even one; it should equal $2$, hence $\\tau(1) = 1$. All other odd numbers should map to even ones, and vice versa. Hence, the number of such permutations is the square of the number of bijections from odds (from $3$ to $n$) to evens satisfying the same constraint. If $n$ is even, we cannot have $\\tau(1) = 1$, otherwise there would be another even sum. Hence, a similar reasoning applies.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19778,
    "subject": "Mathematics (Olympiad)",
    "question": "A fixed circle $k$ is given, together with three collinear points $E$, $F$, and $G$ such that $E$ and $G$ lie outside the circle and $F$ lies inside the circle. Prove that if $ABCD$ is an arbitrary quadrangle inscribed in the circle $k$ such that the extensions of the sides $AB$, $AD$, and $DC$ pass through $E$, $F$, and $G$ respectively, then its side $BC$ passes through a fixed point, collinear with $E$, $F$, and $G$, which does not depend on the quadrangle $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $ABCD$ be such a quadrangle. Notice that, according to the conditions, the line $EG$ intersects the side $BC$ in a point inside $BC$. Let us denote that point by $H$. We consider two cases:\n\n*Case 1: The lines $AB$ and $CD$ are not parallel.*\n\n![](images/makedonija2012_p15_data_07c0f52ad3.png)\n\nLet them intersect at point $Q$. By Menelaus's theorem for triangle $EQG$ and the lines $CB$ and $DA$, we have\n$$\n\\frac{\\overline{QC}}{\\overline{CG}} \\cdot \\frac{\\overline{GH}}{\\overline{HE}} \\cdot \\frac{\\overline{EB}}{\\overline{BQ}} = 1, \\quad \\frac{\\overline{QD}}{\\overline{DG}} \\cdot \\frac{\\overline{GF}}{\\overline{FE}} \\cdot \\frac{\\overline{EA}}{\\overline{AQ}} = 1.\n$$\nIf we multiply the last two equalities and use that $\\overline{QC} \\cdot \\overline{QD} = \\overline{BQ} \\cdot \\overline{AQ}$, since that is the degree of point $Q$ with respect to circle $k$, we get\n$$\n\\frac{\\overline{GH}}{\\overline{CG}} \\cdot \\frac{\\overline{EB}}{\\overline{HE}} \\cdot \\frac{\\overline{GF}}{\\overline{DG}} \\cdot \\frac{\\overline{EA}}{\\overline{FE}} = 1,\n$$\nor\n$$\n\\frac{\\overline{GH}}{\\overline{HE}} = \\frac{\\overline{CG} \\cdot \\overline{DG}}{\\overline{EB} \\cdot \\overline{EA}} \\cdot \\frac{\\overline{FE}}{\\overline{GF}} \\quad (1)\n$$\nNotice that $\\overline{CG} \\cdot \\overline{DG}$ and $\\overline{EB} \\cdot \\overline{EA}$ are the degrees of points $G$ and $E$ with respect to circle $k$ respectively and do not depend on the choice of the quadrangle $ABCD$. It is clear that $\\overline{FE}$ and $\\overline{GF}$ do not depend on the choice of the quadrangle $ABCD$.\n\n*Case 2: The lines $AB$ and $CD$ are parallel.*\n\nIt is clear that $\\triangle GCH \\sim \\triangle EBH$ and $\\triangle GFD \\sim \\triangle EFA$. From the similarity, we have\n$$\n\\frac{\\overline{GH}}{\\overline{EH}} = \\frac{\\overline{GC}}{\\overline{EB}}, \\quad \\frac{\\overline{EA}}{\\overline{FE}} = \\frac{\\overline{GD}}{\\overline{GF}}\n$$\nMultiplying the last two equalities, we get\n$$\n\\frac{\\overline{GH}}{\\overline{EH}} \\cdot \\frac{\\overline{EA}}{\\overline{FE}} = \\frac{\\overline{GC}}{\\overline{EB}} \\cdot \\frac{\\overline{GD}}{\\overline{GF}},\n$$\nor\n$$\n\\frac{\\overline{GH}}{\\overline{EH}} = \\frac{\\overline{GC} \\cdot \\overline{GD}}{\\overline{EB} \\cdot \\overline{EA}} \\cdot \\frac{\\overline{FE}}{\\overline{GF}} \\quad (2)\n$$\nwhich is actually the same as in the first case.\n\n![](images/makedonija2012_p15_data_7fa49819dc.png)\n\nWe conclude that the side $BC$ intersects the line $EG$ in a point $H$ for which (1) (which is the same as (2)) holds. Since $E$ and $G$ lie outside the circle and $F$ lies inside the circle, and the extensions of sides $AB$, $AD$, and $DC$ pass through $E$, $F$, and $G$ respectively, if $ABCD$ is an arbitrary quadrangle which satisfies the condition, the point $H$ obtained as the intersection of the line $EG$ with the side $BC$ must lie between $F$ and $G$. From this and from (1) (analogously, (2)), it follows that the point $H$ is unique, i.e., it does not depend on the choice of the quadrangle $ABCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19779,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two integers $m, n$ greater than $1$, and integers $a_1 < a_2 < \\cdots < a_m$, prove that there exists a set $T$ of integers with $|T| \\le 1 + \\frac{a_m - a_1}{2n + 1}$ such that each $a_i$ can be written as $a_i = t + s$ for some $t \\in T$, and $s \\in [-n, n]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = a$, $a_m = b$, and write $b - a = (2n + 1)q + r$, where $q, r \\in \\mathbb{Z}$ and $0 \\le r \\le 2n$. Define\n\n$$\nT = \\{a + n + (2n + 1)k \\mid k = 0, 1, \\dots, q\\}.\n$$\n\nThen $|T| = q + 1 \\le 1 + \\frac{b - a}{2n + 1}$. Consider the set\n\n$$\nB = \\{t + s \\mid t \\in T,\\ s = -n, -n + 1, \\dots, n\\} = \\{a, a + 1, \\dots, a + (2n + 1)q + 2n\\}.\n$$\n\nNote that $a + (2n + 1)q + 2n \\ge a + (2n + 1)q + r = b$, so each $a_i$ belongs to $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19780,
    "subject": "Mathematics (Olympiad)",
    "question": "Дадена е дробта $\\frac{57}{71}$. Кој број треба да се одземе од броителот и истиот да се додаде на именителот, така што вредноста на дробта после промената да биде $\\frac{1}{3}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Поставуваме равенка: $\\frac{57-x}{71+x} = \\frac{1}{3}$. Решаваме:\n\n$$\n3(57-x) = 71 + x \\\\\n171 - 3x = 71 + x \\\\\n171 - 71 = x + 3x \\\\\n100 = 4x \\\\\nx = 25\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19781,
    "subject": "Mathematics (Olympiad)",
    "question": "A $3 \\times 4$ grid is given. How many ways can you write a number from $1$, $2$, $3$, or $4$ in each square so that:\n\n* No number appears twice or more in the same row, and\n* No number appears twice or more in the same column?",
    "options": [],
    "answer": "See solution",
    "solution": "$576$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19782,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a stripe $1 \\times n$, $n \\geq 4$, with a positive integer written in each cell (not necessarily equal). Under each number, write a positive integer equal to the number of times that integer appears in the previous row. Repeat this procedure for each subsequent row.\n\n**a)** Prove that after a finite number of steps, the rows will stabilize (i.e., stop changing).\n\n**b)** For $n = 2016$ and for arbitrary $n$, what is the maximal number of steps before stabilization?\n\n![](images/Ukraine_2016_Booklet_p12_data_3e8d563f8e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\nLet $k$ be the greatest positive integer such that $2^k \\leq n$. If $n \\neq 2^k + 1$, $n \\neq 2^k + 2$, and $n \\neq 2^k + 4$, the number of steps is $k+1$. If $n=6$, there are 3 steps; if $n=12$, there are 4 steps; otherwise, $k$ steps. Since $2016 = 2^{10} + 992$, for $n=2016$ the answer is 11 steps.\n\n**Solution.**\nWe can rearrange each row so the numbers increase. Denote the initial row as the $0$th, the next as the $1$st, and so on.\n\n**Lemma 1.** If $n$ occurs in the $k$th row, then the number of $n$'s is divisible by $n$.\n\n*Proof.* In the $1$st row, $n$ appears if there are $n$ equal numbers in the $0$th row. Each such group produces $n$ entries of $n$ in the next row. This divisibility persists in subsequent rows.\n\n**Lemma 2.** If there is a group of $n$'s in the $k$th row, the same number will be written under these numbers, but it need not be $n$.\n\n*Proof.* If two equal numbers appear in the previous row, the same number is written under them.\n\nLet $l$ be the smallest number in the first row. The number of $l$'s is $sl$, $s \\geq 1$. If $l$ occurs exactly $l$ times, only $l$ can be written under them, and these numbers stabilize. We can then erase these and consider a row with $n-l$ numbers, repeating the process.\n\nIf there are $sl$ $l$'s with $s > 1$, then $sl$ is written under them in the next row. For any $m$ with $l < m < 2l$, if there are exactly $m$ of them, they stabilize. Numbers less than $m$ either do not change or increase, so $m$ cannot occur under them. If there are $tm$ of $m$, they are replaced by $tm$ in the next row.\n\nThus, in the $2$nd row, only numbers $\\geq 2l$ need consideration. If the least such number in the $2$nd row is $l_2$, then in the $3$rd row the least is at least $2l_2$, and so on.\n\nNo number greater than $n$ can appear, so the maximal number of steps is $k$ such that $2^k \\leq n$, plus the initial step: $k+1$ steps.\n\n**Examples:**\nLet $k$ be the largest integer with $2^k \\leq n$. The $0$th row can be:\n\n$$\n(n-1), (1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\nwhere $m = 2^k - n \\neq 2^i$ for any $i$. Subsequent rows:\n\n$$\n(1, 1), (2, 2), (4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n(2, 2, 2, 2), (4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n(4, 4, 4, 4, 4, 4, 4), \\dots, \\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^{k-1}}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n\\underbrace{(2^{k-1}, \\dots, 2^{k-1})}_{2^k}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\n$$\n\\underbrace{(2^k, \\dots, 2^k)}_{2^k}, \\underbrace{(m, \\dots, m)}_{m}\n$$\n\nIf $m = 2^k - n = 2^i$, a group of $2^i$ will produce $2^{i+1}$ in the next row, shortening the process. For $n=5$, $2^2=4<5$, so at most 3 steps. For $n=6$, the process is: $1; 2; 2; 2; 5; 6 \\rightarrow 1; 3; 3; 3; 1; 1 \\rightarrow 3; 3; 3; 3; 3; 3 \\rightarrow 6; 6; 6; 6; 6; 6$. Similarly for $n=12$.\n\nFor $2^i \\geq 8$, instead of $(m, \\dots, m)$, use two groups: $(s, \\dots, s)$ and $(t, \\dots, t)$, where $s = 2^{i-1}-1$ and $t = 2^{i-1}+1$, which remain unchanged and do not affect the main process.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19783,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any real numbers $a, b, c, d \\ge \\frac{1}{3}$ the following inequality holds:\n\n$$\n\\sqrt{\\frac{a^6}{b^4 + c^3} + \\frac{b^6}{c^4 + d^3} + \\frac{c^6}{d^4 + a^3} + \\frac{d^6}{a^4 + b^3}} \\ge \\frac{a + b + c + d}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us apply the Cauchy-Schwarz inequality in Engel form:\n\n$$\n\\frac{a^6}{b^4 + c^3} + \\frac{b^6}{c^4 + d^3} + \\frac{c^6}{d^4 + a^3} + \\frac{d^6}{a^4 + b^3} \\ge \\frac{(a^3 + b^3 + c^3 + d^3)^2}{a^4 + b^4 + c^4 + d^4 + a^3 + b^3 + c^3 + d^3}.\n$$\n\nLet us show that the following inequality is true:\n\n$$\n\\frac{(a^3 + b^3 + c^3 + d^3)^2}{a^4 + b^4 + c^4 + d^4 + a^3 + b^3 + c^3 + d^3} \\ge \\frac{a^3 + b^3 + c^3 + d^3}{a + b + c + d}.\n$$\n\nIn fact, it is equivalent to the inequality:\n\n$$\n(a^3 + b^3 + c^3 + d^3)(a + b + c + d) \\ge a^4 + b^4 + c^4 + d^4 + a^3 + b^3 + c^3 + d^3 \\\\\n\\iff a^3(b + c + d) + b^3(c + d + a) + c^3(d + a + b) + d^3(a + b + c) \\ge a^3 + b^3 + c^3 + d^3,\n$$\n\nand the latter is true, since the sum of any three of the numbers $a, b, c, d$ is not less than $1$, as each of them is not less than $\\frac{1}{3}$.\n\nIt remains to apply the inequality between the cubic mean and the arithmetic mean:\n\n$$\n\\frac{a^3 + b^3 + c^3 + d^3}{a + b + c + d} \\ge \\frac{4 \\left( \\frac{a + b + c + d}{4} \\right)^3}{a + b + c + d} = \\left( \\frac{a + b + c + d}{4} \\right)^2,\n$$\n\nwhence the required inequality is obtained.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19784,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\n\\lim_{n \\to \\infty} \\frac{a_n}{a_{n+1}} = \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\log_2(1+a_k) = 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have $a_n > 0$ and $a_{n+1} < a_n$ for any $n \\ge 1$. The sequence $(a_n)_{n \\ge 1}$ is convergent, with limit $\\ell \\in [0, 1)$. From the recurrence relation, $\\ell = \\frac{\\ell}{1+\\sqrt{1+\\ell}}$, so $\\ell = 0$. Then\n$$\n\\lim_{n \\to \\infty} \\frac{a_n}{a_{n+1}} = \\lim_{n \\to \\infty} (1 + \\sqrt{1+a_n}) = 2.\n$$\nThe recurrence relation implies $1 + a_{n+1} = \\sqrt{1+a_n}$ for any $n \\in \\mathbb{N}^*$. Therefore,\n$$\n\\log_2(1+a_{n+1}) = \\frac{1}{2}\\log_2(1+a_n)\n$$\nfor any $n \\ge 1$. Since $\\log_2(1+a_1) = 1$, we obtain $\\log_2(1+a_n) = \\frac{1}{2^{n-1}}$ for all positive integers $n$ (a geometric progression with first term $1$ and ratio $1/2$). Hence,\n$$\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\log_2(1+a_k) = \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{1}{2^{k-1}} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19785,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ for which there exists an integer $a > 2$ such that $a^d + 2^d \\mid a^n - 2^n$ for all positive divisors $d \\neq n$ of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $n$ satisfies the given condition if and only if $n$ is prime or $n$ is a power of 2 (including $n = 1$).\n\nIf $n$ is an odd prime, then the only proper divisor $d$ of $n$ is $d = 1$. Let $a = 2^k - 2$ with $3 \\leq k \\leq n+1$, e.g., $a = 6$. Then we need to check that $2^k - a + 2 = 2^k$ is a divisor of $(2^k - 2)^n - 2^n$. As this is the difference of two terms which contain exactly $n$ factors of $2$, the difference contains $n+1$ factors of $2$.\n\nIf $n$ is a power of $2$, say $n = 2^m$ with $m \\geq 0$. If $m = 0$, then there are no proper divisors of $n$, so $n$ satisfies the given condition because it is an empty condition. If $m \\geq 1$, then for all proper divisors $d$ of $n$, the integer $e = \\frac{n}{d}$ is even. Note that\n\n$$\na^n - 2^n = a^{de} - 2^{de} \\equiv (-2^d)^e - 2^{de} = 2^n\\left((-1)^e - 1\\right) \\pmod{a^d + 2^d}$$\n\nis zero for all $a$. Hence $a^n - 2^n$ is a multiple of $a^d + 2^d$, and therefore $a^d + 2^d \\mid a^n - 2^n$. So if $n$ is prime or a power of $2$, $n$ does indeed satisfy the given condition.\n\nFinally, suppose that $n$ is neither a prime number nor a power of $2$. Then we can write $n = de$ with $e \\neq 1$ odd (since $n$ is not a power of $2$) and $d \\neq 1$ (since $n$ is not a prime number). As $(-1)^e - 1 = -2$, it follows from the computation above that $a^d + 2^d \\mid 2^{n+1}$. Hence $a^d + 2^d$ is a power of $2$, so $a^d = 2^k - 2^d$ for some $d < k \\leq n+1$. Therefore $a$ is divisible by $2$ and $\\left(\\frac{a}{2}\\right)^d = 2^{k-d} - 1$.\n\nNow we distinguish between the cases in which $d$ is even and in which $d$ is odd. In the first case, $2^{k-d} - 1$ is a square. As $a > 2$, from $\\left(\\frac{a}{2}\\right)^d = 2^{k-d} - 1$ it however follows that $k - d \\geq 2$, so this square is $-1 \\pmod{4}$, which is a contradiction. If $d$ is odd, then we note that\n\n$$\n2^{k-d} = \\left(\\frac{a}{2}\\right)^d + 1 = \\left(\\frac{a}{2} + 1\\right)\\left(\\left(\\frac{a}{2}\\right)^{d-1} - \\left(\\frac{a}{2}\\right)^{d-2} + \\dots + 1\\right).\n$$\n\nAs $d \\neq 1$, we have $\\frac{a}{2} + 1 < \\left(\\frac{a}{2}\\right)^d + 1$. The second factor in the product above is a sum of an even number of terms with the same parity as $\\frac{a}{2}$ and a term $1$, so this factor is odd. As this second factor is also greater than $1$, this contradicts it being a factor of a power of $2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19786,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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": 19787,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Let $P_n = \\{2^n, 2^{n-1} \\cdot 3, 2^{n-2} \\cdot 3^2, \\dots, 3^n\\}$. For each subset $X$ of $P_n$, let $S_X$ be the sum of all elements of $X$, with the convention that $S_\\emptyset = 0$ where $\\emptyset$ is the empty set. Suppose $y$ is a real number with $0 \\leq y \\leq 3^{n+1} - 2^{n+1}$.\n\nProve that there is a subset $Y$ of $P_n$ such that $0 \\leq y - S_Y < 2^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\frac{3}{2}$, so $1 + \\alpha > \\alpha^2$.\n\nGiven $y$, we construct $Y$ algorithmically. Start with $Y = \\emptyset$ and $S_\\emptyset = 0$. For $i = 0$ to $n$, perform the following:\n\nIf $S_Y + 2^i 3^{n-i} \\leq y$, then set $Y = Y \\cup \\{2^i 3^{n-i}\\}$.\n\nAfter this process, $Y$ is a subset of $P_n$ with $S_Y \\leq y$.\n\nThe elements of $P_n$ are $2^n, 2^{n-1} \\cdot 3, \\dots, 3^n$, which can be written as $2^n \\alpha^k$ for $k = 0$ to $n$. If any member is omitted from $Y$, then no two consecutive smaller members to its left can both be in $Y$, due to the greedy construction and the inequality $1 + \\alpha > \\alpha^2$.\n\nIf $Y = P_n$, then $y = 3^{n+1} - 2^{n+1}$ and $y - S_Y = 0 < 2^n$. Otherwise, at least one of the two largest elements is omitted. If $2^n$ is not in $Y$, then the process ensures $(S_Y - 2^n) + 2^{n-1} \\cdot 3 > y$, so $y - S_Y < 2^n$. If $2^n$ is omitted, then $y - S_Y < 2^n$ as well.\n\nThus, there exists a subset $Y$ of $P_n$ such that $0 \\leq y - S_Y < 2^n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19788,
    "subject": "Mathematics (Olympiad)",
    "question": "За какое минимальное число ходов можно с гарантией обнаружить монетку под одним из 50 наперстков, если после каждого хода и перемещения монетки все наперстки (вместе с монеткой) поворачиваются по ходу часовой стрелки на одну позицию? После каждого хода монетка либо остается на месте, либо перемещается на две позиции по часовой стрелке, а наперстки остаются на своих местах.",
    "options": [],
    "answer": "See solution",
    "solution": "Покрасим все наперстки поочередно в белый и черный цвет и пронумеруем наперстки каждого цвета по порядку против часовой стрелки числами от 0 до 49. 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. Предполагаем, что монетка не обнаружена на всех ходах вплоть до 33-го.\n\n1. Первым ходом поднимем черные наперстки с номерами 0, 1, 2, 3. После перемещения монетка не сможет оказаться под черными наперстками 0, 1, 2.\n2. Вторым ходом поднимем черные наперстки 3, 4, 5, 6. После перемещения монетка не сможет оказаться под черными наперстками 0, 1, \\ldots, 5.\n3. Действуем так далее: при $s = 1, 2, \\ldots, 16$ ходом номер $s$ поднимем черные наперстки с номерами $3s - 3, 3s - 2, 3s - 1, 3s$. После перемещения монетка не сможет оказаться под черными наперстками 0, 1, \\ldots, $3s - 1$.\n4. Семнадцатым 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.\n5. При $s = 1, 2, \\ldots, 15$ ходом номер $17 + s$ поднимем белые наперстки $3s - 3, 3s - 2, 3s - 1, 3s$. После перемещения монетка не сможет оказаться под белыми наперстками 49, 0, 1, \\ldots, $3s - 1$.\n6. Последним, 33-м ходом, поднимаем белые наперстки 45, 46, 47, 48; под одним из них обязательно будет монетка.\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 $k = 1, 2, \\ldots, n-1$, откуда $|A_{k+1}| \\leq |A_k| + 4$. Более того, если множество $A_k \\cup B_k$ не совпадает с множеством всех наперстков или с множеством из 50 наперстков одного цвета, то найдётся 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: $|A_1| = 0$, $|A_2| \\leq 3$, $|A_3| \\leq 6$, \\ldots, $|A_{17}| \\leq 48$, $|A_{18}| \\leq 51$, $|A_{19}| \\leq 54$, \\ldots, $|A_{32}| \\leq 93$. 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": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19789,
    "subject": "Mathematics (Olympiad)",
    "question": "Nonzero integers $a$, $b$, and $c$ satisfy $$\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 0.$$ Prove that among $a$, $b$, $c$ there are two integers which have a common divisor larger than $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Multiplying both sides of the equation by $abc$ gives $$bc + ca + ab = 0.$$ If $a$, $b$, and $c$ were all odd, then $bc$, $ca$, and $ab$ would also be odd, and their sum could not be $0$. If one of $a$, $b$, $c$ was even and the others were odd, then two of $bc$, $ca$, and $ab$ would be even and one would be odd, which also cannot sum to $0$. Therefore, at least two of $a$, $b$, $c$ are even, so they share a common divisor greater than $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19790,
    "subject": "Mathematics (Olympiad)",
    "question": "On a large piece of paper are 2019 six-pointed snowflakes, placed in general position (not touching). Each of the six arms of a snowflake has an extreme end, called its *apex*.\n\nA move consists of drawing a curve connecting two apices, possibly belonging to the same snowflake. The curve may not intersect itself, may not touch any previously drawn curves, and may not touch any snowflakes (except at the two apices forming its endpoints). Each apex may be used only once.\n\nTwo players, Inger and Ellen, alternate moves, with Inger going first. The first player unable to make a move loses. Which player has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "Inger has a winning strategy.\n\nLet Inger select an arbitrary snowflake $S$ and connect a pair of opposite apices. By the Jordan Curve Theorem, this curve (together with $S$) divides the plane into two domains, with the curve and $S$ as their common boundary. Inger can draw her curve so that exactly 1009 snowflakes fall in each domain. Each subsequent move must be made entirely within one of these two domains. Since the two domains are combinatorially identical, Inger can respond to each move of Ellen in one domain by making the corresponding move in the other, ensuring she wins the game.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19791,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be positive integers. An international company has connected $k$ cities of Armenia with $k$ cities of Belarus by direct two-way airlines. From each of these Belarusian cities there is a direct flight to exactly $n$ Armenian ones. It turned out that for any two Armenian cities there are exactly two Belarusian cities that are connected by airlines to both of them.\n\n(a) Prove that each of the Armenian cities is connected by airlines to exactly $n$ Belarusian cities.\n\n(b) Prove that it is possible to travel on planes of a given airline without repeating cities, while visiting at least $\\lfloor \\frac{(n+1)^2}{4} \\rfloor$ cities in each of the countries.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's translate the problem into the language of graphs.\n\nGiven a bipartite graph $G$ with parts $A$ and $B$ having the same number of vertices: $|V(A)| = |V(B)| = k$. The degree of each vertex in $B$ is $n$. For any two vertices $u, v$ of the part $A$, there are exactly two vertices in $B$ that are adjacent to both $u$ and $v$.\n\n(a) Prove that the degree of any vertex in $A$ is also equal to $n$.\n\n(b) Prove that there is a simple path in the graph that contains at least $\\lfloor \\frac{(n+1)^2}{4} \\rfloor$ vertices in each part.\n\nFirst, let's prove that $k = \\binom{n}{2} + 1$ and that the degree of each vertex in $A$ is equal to $n$.\n\nLet's count the number of pairs $(\\{a, a'\\}, b)$ where $a \\neq a'$, $a, a' \\in V(A)$, $b \\in V(B)$ and $b$ is adjacent to both $a$ and $a'$. On the one hand, we have $|V(B)|$ ways to select $b$ and each such vertex gives $\\binom{n}{2}$ pairs $a, a'$ to which it is adjacent. On the other hand, there are $\\binom{|V(A)|}{2}$ ways to select distinct $a$ and $a'$, and for each pair, there are exactly two ways to select $b$, according to the second property of our graph. Thus, $\\binom{n}{2}k = 2\\binom{k}{2}$, which implies that $k = \\binom{n}{2} + 1$.\n\nTake an arbitrary vertex $a$ from $A$ and let it have degree $d$. Let us count in two ways the number of pairs $(a', b)$ where $a' \\in V(A)$, $b \\in V(B)$, $a' \\neq a$ and $b$ is adjacent to both $a$ and $a'$. On the one hand, we have $|V(A)| - 1$ options to choose $a'$, and for each such vertex we have exactly two choices of $b$. On the other hand, each of the $d$ neighbors of $a$ has degree exactly $n$, which gives $(n-1)d$ pairs in question (the neighbors of a given vertex from $V(B)$ must be different from $a$, so there are $n-1$ of them). Thus $(n-1)d = 2(k-1) = 2\\binom{n}{2}$, which means that $d = n$.\n\nNow let's move on to finding a path. Since the graph is regular, by Hall's theorem there is a perfect matching in it—a set of $k$ pairwise non-adjacent edges. Let $I$ be such a matching. Take the longest such path $P = b_1a_1, \\dots, b_ra_r$, where $b_1 \\in V(B)$ and $a_r \\in V(A)$, with the following property:\n\nIf a vertex $x$ is in $P$, then the vertex $y$ adjacent to $x$ in the matching $I$ is also in $P$.\n\n(Note that this condition does not require that edges from $I$ also be in our simple path.) Then all neighbors of the ends of $P$ lie in $P$. In fact, if one of the ends of $P$ has a neighbor outside $P$, then the neighbor of this vertex in $I$ is also not in our path and then we can extend our path by two vertices so that it starts and ends at different parts.\n\nLet $b_i$ ($1 \\le i \\le r$) be one of the neighbors of a vertex $a_k$ in path $P$. Then the path\n\n$$\nb_1a_1 \\dots b_ia_r b_r a_{r-1} \\dots a_i\n$$\n\nalso has the above property. In particular, all neighbors of the vertex $a_i \\in V(A)$ also lie in $P$. Since $a_r$ has $n$ neighbors, we have a minimum of $n$ vertices $a_{i_1}, \\dots, a_{i_n}$ from $A$ each of which has neighbors only in the path $P$. Since vertices $a_{i_1}$ and $a_{i_2}$ have exactly two neighbors in common, they have a total of $n + (n-2)$ neighbors in $P$. Next, consider the vertex $a_{i_3}$—it has two common neighbors with each of the previous vertices, that is, a maximum of 4 common neighbors have already been counted. This gives $n-4$ new neighbors different from the previous ones. Reasoning in this way, we get\n\n$$\nr \\ge n + (n-2) + (n-4) + \\dots = \\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor,\n$$\n\nwhich is what was required.\n\n*Remark.* In fact, the condition from the problem is the definition of a block design ($\\lambda = 2$). Therefore, the number of vertices and the degree of each vertex are known facts. It is also true that for any two vertices in $V(B)$ there are exactly two common neighbors in $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19792,
    "subject": "Mathematics (Olympiad)",
    "question": "Line $\\ell$ intersects sides $BC$ and $AD$ of cyclic quadrilateral $ABCD$ at interior points $R$ and $S$, respectively, and intersects ray $DC$ beyond $C$ at $Q$, and ray $BA$ beyond $A$ at $P$. The circumcircles of triangles $QCR$ and $QDS$ intersect at $N \\neq Q$, while the circumcircles of triangles $PAS$ and $PBR$ intersect at $M \\neq P$. Let lines $MP$ and $NQ$ meet at point $X$, lines $AB$ and $CD$ meet at point $K$, and lines $BC$ and $AD$ meet at point $L$. Prove that point $X$ lies on line $KL$.",
    "options": [],
    "answer": "See solution",
    "solution": "We start with the following lemma.\n\n**Lemma 1.** Points $M, N, P, Q$ are concyclic.\n\nPoint $M$ is the Miquel point of lines $AP = AB$, $PS = \\ell$, $AS = AD$, and $BR = BC$, and point $N$ is the Miquel point of lines $CQ = CD$, $RC = BC$, $QR = \\ell$, and $DS = AD$. Both points $M$ and $N$ are on the circumcircle of the triangle determined by the common lines $AD$, $\\ell$, and $BC$, which is $LRS$.\n\nThen, since quadrilaterals $QNRC$, $PMAS$, and $ABCD$ are all cyclic, using directed angles (modulo $180^\\circ$):\n\n$$\n\\angle NMP = \\angle NMS + \\angle SMP = \\angle NRS + \\angle SAP = \\angle NRQ + \\angle DAB = \\angle NRQ + \\angle DCB \\\\\n= \\angle NRQ + \\angle QCR = \\angle NRQ + \\angle QNR = \\angle NQR = \\angle NQP,\n$$\n\nwhich implies that $MNQP$ is a cyclic quadrilateral.\n\n![](images/2024_The_Australian_Scene_Final_p107_data_2e6e032470.png)\n\nLet $E$ be the Miquel point of $ABCD$ (that is, of lines $AB$, $BC$, $CD$, $DA$). It is well known that $E$ lies on the line $t$ connecting the intersections of the opposite sides of $ABCD$. Let lines $NQ$ and $t$ meet at $T$. If $T \\ne E$, using directed angles and considering the circumcircles of $LAB$ (which contains $E$ and $M$), $APS$ (which also contains $M$), and $MNQP$:\n\n$$\n\\angle TEM = \\angle LEM = \\angle LAM = \\angle SAM = \\angle SPM = \\angle QPM = \\angle QNM = \\angle TNM,\n$$\n\nthat is, $T$ lies on the circumcircle $\\omega$ of $EMN$. If $T = E$, the same computation shows that $\\angle LEM = \\angle ENM$, which means that $t$ is tangent to $\\omega$.\n\nNow let lines $MP$ and $t$ meet at $V$. An analogous computation shows, by considering the circumcircles of $LCD$ (which contains $E$ and $N$), $CQR$, and $MNQP$, that $V$ lies on $\\omega$ as well, and that if $V = E$ then $t$ is tangent to $\\omega$.\n\nTherefore, since $\\omega$ meets $t$ at $T$, $V$, and $E$, either $T = V$ if both $T \\ne E$ and $V \\ne E$, or $T = V = E$. At any rate, the intersection of lines $MP$ and $NQ$ lies on $t$.\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19793,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be relatively prime positive integers and let $a_n$ and $b_n$ be integer sequences satisfying $$(a + b\\sqrt{2})^{2n} = a_n + b_n\\sqrt{2}.$$ Find all primes $p$ such that there is a positive integer $n \\leq p$ with $$b_n \\equiv 0 \\pmod{p}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be a prime. \n\nFirst, suppose that $p$ is an odd prime dividing $a^2 - 2b^2$. Since $a$ and $b$ are relatively prime, $b_1 = 2ab$ is not divisible by $p$. Suppose there is a positive integer $n$ such that $b_n$ is divisible by $p$. Let $r$ be the smallest positive integer such that $b_r$ is divisible by $p$. Note that $$(a - b\\sqrt{2})^{2n} = a_n - b_n\\sqrt{2}$$ because both $a_n$ and $b_n$ are integers, and\n\n$$\n\\begin{cases} \na_n = (a^2 + 2b^2)a_{n-1} + 4ab b_{n-1} \\\\ \nb_n = 2ab a_{n-1} + (a^2 + 2b^2) b_{n-1} \n\\end{cases}\n$$\n\nThen we have\n\n$$\n\\begin{aligned}\n0 \\equiv b_r &= 2ab((a^2 + 2b^2)a_{r-2} + 4ab b_{r-2}) + (a^2 + 2b^2) b_{r-1} \\\\\n&= 2(a^2 + 2b^2) b_{r-1} - (a^2 - 2b^2)^2 b_{r-2} \\\\\n&\\equiv 2(a^2 + 2b^2) b_{r-1} \\pmod{p},\n\\end{aligned}\n$$\n\nwhich is a contradiction to the assumption. Therefore, $b_n$ is not divisible by $p$ for any positive integer $n$.\n\nNow let $p$ be a prime not dividing $a^2 - 2b^2$. Since $b_1 = 2ab$, we may assume that $p$ is odd and $ab$ is not divisible by $p$. Note that\n\n$$\nb_n = \\frac{(a + b\\sqrt{2})^{2n} - (a - b\\sqrt{2})^{2n}}{\\sqrt{2}} = \\sum_{\\substack{k=0 \\\\ k \\equiv 1 \\pmod{2}}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}}\n$$\n\nFirst, assume there is an integer $m$ such that $m^2 \\equiv 2 \\pmod{p}$. Then\n\n$$\n\\begin{aligned}\nb_n &= \\sum_{\\substack{k=0 \\\\ k \\equiv 1 \\pmod{2}}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= \\sum_{\\substack{k=0 \\\\ k \\equiv 1 \\pmod{2}}}^{2n} m^2 \\binom{2n}{k} a^{2n-k} b^k m^{k-1} \\\\\n&= \\frac{(a + bm)^{2n} - (a - bm)^{2n}}{m} \\pmod{p}.\n\\end{aligned}\n$$\n\nSince $(a + bm)(a - bm) = a^2 - m^2 b^2 \\equiv a^2 - 2b^2 \\not\\equiv 0 \\pmod{p}$, $b_{\\frac{p-1}{2}} \\equiv 0 \\pmod{p}$ by Fermat's little theorem.\n\nNow suppose that $x^2 \\equiv 2 \\pmod{p}$ does not have any integer solution, i.e., $2$ is a quadratic non-residue modulo $p$. Clearly,\n\n$$\n\\binom{p+1}{k} \\equiv 0 \\pmod{p} \\text{ for any } 2 \\leq k \\leq p-1.\n$$\n\nTherefore, by Euler's criterion,\n\n$$\n\\begin{aligned}\nb_{\\frac{p+1}{2}} &= \\sum_{\\substack{k=0 \\\\ k \\equiv 1 \\pmod{2}}}^{p+1} 2 \\binom{p+1}{k} a^{p+1-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= 2(p+1)a^p b + (p+1)ab^p 2^{\\frac{p+1}{2}} \\\\\n&= 2ab(1 + 2^{\\frac{p-1}{2}}) \\equiv 0 \\pmod{p}.\n\\end{aligned}\n$$\n\nTherefore, such a prime is exactly $2$ or relatively prime to $a^2 - 2b^2$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19794,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a foundation in the form of a rectangle $2m \\times 2n$ divided into $1 \\times 1$ squares, with a gap of length $1$ between any two adjacent squares. The foundation is covered by several layers of bricks of size $2 \\times 1$. Each layer consists of $2mn$ bricks, and each brick fully covers exactly one gap of length $1$. A cover is called *strong* if every gap is covered by a brick in at least one layer. What is the minimum number of layers required to make a strong cover?\n\n![](images/Ukraine_booklet_2018_p21_data_546c0147d7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If $m = n = 1$, the minimum is $2$ layers. If $m = 1 < n$ (or $n = 1 < m$), the minimum is $3$ layers. Otherwise, $4$ layers are required.\n\nAssume $2 \\le m \\le n$. Consider a $1 \\times 1$ square $A$ not touching the sides of the rectangle. All $4$ sides of $A$ must be covered by different bricks, since half of each brick covers $A$ itself. Thus, fewer than $4$ layers are insufficient. To show $4$ layers suffice, divide the rectangle into $2 \\times 2$ squares; two different ways of dividing these into bricks yield two layers, and the first two layers are shown in the figure above.\n\nFor $m = n = 1$, $2$ layers suffice.\n\nIf $m = 1 < n$, $3$ layers are possible (see Fig. 20). Fewer layers are impossible, since for squares whose vertices do not coincide with the rectangle's vertices, $3$ sides must be covered, requiring at least $3$ layers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19795,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with integer side lengths and the property that $\\angle B = 2\\angle A$. What is the least possible perimeter of such a triangle?\n\n(A) 13  (B) 14  (C) 15  (D) 16  (E) 17",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer (C):* Let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$, respectively. Note that $a < b$. Applying the Law of Sines in $\\triangle ABC$, together with the identities $\\sin B = \\sin(2A) = 2 \\sin A \\cos A$ and\n\n$$\n\\sin C = \\sin(\\pi - 3A) = \\sin(3A) = (\\sin A)(-1 + 4\\cos^2 A),\n$$\n\ngive\n\n$$\n\\frac{\\sin A}{a} = \\frac{2 \\sin A \\cos A}{b} = \\frac{\\sin A (-1 + 4 \\cos^2 A)}{c}.\n$$\n\nThus $\\cos A = \\frac{b}{2a}$ and\n\n$$\nc = a(-1 + 4\\cos^2 A) = a\\left(-1 + \\frac{b^2}{a^2}\\right),\n$$\n\nwhich simplifies to $b^2 = a^2 + ac$.\n\nIn looking for the triangle with least perimeter, it can be assumed that $a$ and $c$ are relatively prime, because otherwise a smaller triangle can be obtained by shrinking by a factor of $\\gcd(a, c)$. Then $a$ and $a+c$ are relatively prime as well. Because $a(a+c) = b^2$, the numbers $a$ and $a+c$ must be squares, say $a = r^2$ and $a+c = s^2$, where $0 < r < s$ and $\\gcd(r, s) = 1$. This gives $a = r^2$, $b = rs$, and $c = s^2 - r^2$.\n\nIf $r = 1$, then $b = s \\ge 2$ and $c = s^2 - 1 \\ge s + 1$, a violation of the Triangle Inequality. If $r = 2$, then the least perimeter will occur when $s = 3$, with $a = 2^2 = 4$, $b = 2 \\cdot 3 = 6$, and $c = 3^2 - 2^2 = 5$. Greater values of $r$ lead to greater perimeters. The requested minimum perimeter is $4 + 6 + 5 = 15$.\n\n*Note:* The 4–6–5 triangle has angle measures of approximately $\\angle A = 41.4^\\circ$, $\\angle B = 82.8^\\circ$, and $\\angle C = 55.8^\\circ$. Triangles with the property that $\\angle B = 2\\angle A$ have been called VUX triangles by Fitch Cheney in an article published in 1970 in *The Mathematics Teacher*.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19796,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a parallelogram $ABCD$ with center $S$, let $O$ be the incenter of triangle $ABD$ and $T$ the point of contact of the incircle of triangle $ABD$ with the diagonal $BD$. Prove that lines $OS$ and $CT$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lengths of $AB$, $AD$, and $BD$ be $a$, $b$, and $c$, respectively. If $a = b$, then both $OS$ and $CT$ coincide with $AC$, so the conclusion is trivial. Suppose $a > b$ (the case $b > a$ is analogous).\n\nLet $T'$ be the reflection of $T$ in $S$.\n\n![](images/Czech_and_Slovak_booklet_2013_p10_data_14868ea8b5.png)\n\nAs $CT \\parallel AT'$, it suffices to prove $OS \\parallel AT'$. Let $E$ be the intersection of $AO$ and $BD$; we may as well prove\n\n$$\n\\frac{AO}{OE} = \\frac{T'S}{SE}\n$$\n\n(Note: since $a > b$, points $T'$, $S$, $E$, and $T$ lie on $BD$ in this order.)\n\nWe express both ratios in terms of $a, b, c$.\n\nFirst, it is well-known that\n\n$$\nDT = \\frac{b + c - a}{2}, \\quad \\text{so} \\quad T'S = TS = \\frac{c}{2} - \\frac{b + c - a}{2} = \\frac{a - b}{2}.\n$$\n\nBy the Angle Bisector Theorem in triangles $ABD$ and $AED$,\n\n$$\nBE : ED = AB : AD, \\quad AO : OE = AD : DE\n$$\n\nwhich gives\n\n$$\nBE = \\frac{ac}{a + b}, \\quad DE = \\frac{bc}{a + b}, \\\\\nSE = BE - BS = \\frac{ac}{a + b} - \\frac{c}{2} = \\frac{c(a - b)}{2(a + b)}, \\\\\n\\frac{AO}{OE} = \\frac{AD}{DE} = \\frac{b}{\\frac{bc}{a + b}} = \\frac{a + b}{c}.\n$$\n\nFor the right-hand side,\n\n$$\n\\frac{T'S}{SE} = \\frac{\\frac{a - b}{2}}{\\frac{c(a - b)}{2(a + b)}} = \\frac{a + b}{c}\n$$\n\nwhich completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19797,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call the point being reported by the tracking device on each move a *ping*.\n\nPart of the difficulty of the problem is that it is very tempting to think that since the rabbit is invisible, the hunter's best strategy is simply to follow the ping on each move. However, this is not the case, as the following example shows.\n\nWithout loss of generality, assume both the hunter and the rabbit start at the origin of the Cartesian plane. Suppose the first ping is at $(1, 1)$. If the hunter follows the ping, she arrives at $(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2})$ on her first move.\n\nSuppose the second ping is at $(3, 0)$. Then, the only place the rabbit could be after its second move is at $(2, 0)$, and the hunter can deduce this. So, the best strategy for the hunter on her second move is not to follow the ping and move towards $(3, 0)$, but instead to move towards $(2, 0)$ because she knows exactly where the rabbit is!\n\n![](images/Australian-Scene-2017_p149_data_a20347afb8.png)\n\nOne might argue that the above scenario only arises because the second ping makes it possible to determine the exact location of the rabbit. However, if the second ping were instead at $(2.99, 0)$, then the hunter cannot determine the exact location of the rabbit. Yet, the hunter can still deduce that the rabbit is much closer to $(2, 0)$ than to $(2.99, 0)$. So even in this case, following the ping is not necessarily the best strategy.\n\nWhat is the fundamental flaw in assuming that following the ping is the best strategy for the hunter? It is that we are assuming the hunter has amnesia and cannot remember previous pings. If the hunter did have such amnesia, then following the ping would be her best strategy. However, we are not entitled to assume this. Consequently, any attempt at the problem that implicitly assumed the hunter had such amnesia is incorrect.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that there is no strategy for the hunter that guarantees the distance between her and the rabbit is at most 100 after $10^9$ rounds.\n\nThe key is to consider the scenario illustrated below. Let $A_n$ and $B_n$ denote the positions of the rabbit and the hunter after $n$ rounds. Let $\\ell$ be the line $B_nA_n$, and let $x$ be the distance between $B_n$ and $A_n$. Points $Q_1$ and $Q_2$ are at distance 1 from $\\ell$ and at distance $d$ from $A_n$, where $d > x$ is a positive integer to be chosen later.\n\n![](images/Australian-Scene-2017_p150_data_3a53a8bf92.png)\n\nImagine the rabbit at $A_n$ flipping a coin. If heads, it proceeds to $Q_1$ in the next $d$ rounds; if tails, to $Q_2$. For each round, the tracking device reports the foot of the perpendicular from the rabbit's location to $\\ell$. These reported points are consistent with either path.\n\nFor the hunter, suppose she moves one unit to the right along $\\ell$ per round, reaching point $B$ at distance $d$ from $B_n$. After $d$ rounds, the distance between the hunter and the rabbit is $y = BQ_1 = BQ_2$. Any other move leaves her strictly to the left of $B$, increasing her distance to at least $y$ from one of $Q_1$ or $Q_2$. Thus, the hunter cannot guarantee a distance less than $y$ after $d$ rounds.\n\nLet us compute a lower bound for $y$. By the Pythagorean theorem applied to $\\triangle A_nPQ_1$, $AP = \\sqrt{d^2 - 1}$. Thus $BP = \\sqrt{d^2 - 1} - (d - x)$. Applying the Pythagorean theorem to $\\triangle BPQ_1$:\n\n$$\n\\begin{aligned}\ny^2 &= 1^2 + (x + \\sqrt{d^2 - 1} - d)^2 \\\\\n&= x^2 + 2d^2 - 2d\\sqrt{d^2 - 1} - 2x(d - \\sqrt{d^2 - 1}) \\\\\n&= x^2 + 2(d - x)(d - \\sqrt{d^2 - 1}) \\\\\n&= x^2 + \\frac{2(d - x)}{d + \\sqrt{d^2 - 1}} \\\\\n&> x^2 + \\frac{2(d - x)}{d + \\sqrt{d^2}} \\\\\n&= x^2 + 1 - \\frac{x}{d}.\n\\end{aligned}\n$$\n\nChoosing $d = 2 \\lfloor x \\rfloor$ yields $y^2 > x^2 + \\frac{1}{2}$. For $x \\ge 1$, $x^2 + \\frac{1}{2} > (x + \\frac{1}{5x})^2$. Thus,\n\n$$\ny > x + \\frac{1}{5x}.\n$$\n\n**Lemma:** If at some stage the distance between the hunter and the rabbit is $x > 1$, then after $2 \\lfloor x \\rfloor$ more rounds, the distance can potentially exceed $x + \\frac{1}{5x}$.\n\nA set of moves in the lemma that increases the potential distance is called a *swoop*.\n\nAfter one round, the distance is potentially 2, since the tracking device might report the starting position, giving no information. Whatever direction the hunter moves, the rabbit might go the opposite way.\n\nSuppose after some rounds, the distance is at least $x \\ge 2$. Let $n = \\lfloor x \\rfloor$. We claim that after $10(n+1)^2$ more rounds, the potential distance is at least $n+1$.\n\nAssume not. By the lemma, each swoop increases the distance by more than $\\frac{1}{5(n+1)}$. After at most $5(n+1)$ swoops, the distance increases by more than 1, so is at least $n+1$. Each swoop takes at most $2(n+1)$ rounds, so at most $10(n+1)^2$ rounds are used. This contradiction proves the claim.\n\nNow, the number of rounds $U$ needed to ensure the distance is at least 101 is\n\n$$\n\\begin{aligned}\nU &\\le 1 + 10 \\cdot 3^2 + 10 \\cdot 4^2 + \\dots + 10 \\cdot 101^2 \\\\\n&\\ll 10 \\cdot 100 \\cdot 101^2 \\\\\n&\\ll 10^9.\n\\end{aligned}\n$$\n\nSo the distance can potentially exceed 100 in well under $10^9$ rounds. Once this occurs, the rabbit can simply hop directly away from the hunter for the remaining rounds. Thus, the hunter cannot guarantee the distance is at most 100 after $10^9$ rounds. $\\Box$\n\n**Comment:** The number 100 is not sharp. With more careful calculation, after $10^9$ rounds, the distance can potentially be at least 668.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19798,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB > AC$ and incenter $I$. The internal bisector of angle $BAC$ intersects $BC$ at point $D$. Let $M$ be the midpoint of segment $AD$, and let $F$ be the second intersection point of $MB$ with the circumcircle of triangle $BIC$. Prove that $AF$ is perpendicular to $FC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the inradius and $r_A$ the exradius of triangle $ABC$. Let $I_A$ be the $A$-excenter of triangle $ABC$. It is well-known that\n\n$$\n\\frac{r}{r_A} = \\frac{DI}{DI_A} = \\frac{AI}{AI_A}.\n$$\n\nSince $M$ is the midpoint of $AD$, we then have\n\n$$\n\\frac{DI}{DI_A} = \\frac{AI}{AI_A} \\Rightarrow \\frac{AM - MI}{MI_A - AM} = \\frac{AM + MI}{MI_A + AM} \\Rightarrow MA^2 = MI \\cdot MI_A.\n$$\n\nIt is well-known that the $A$-excenter $I_A$ belongs to the circumcircle of $BIC$. So by the power of the point $M$, we also have $MI \\cdot MI_A = MF \\cdot MB$, therefore $MA^2 = MF \\cdot MB$, which gives that $MA$ is tangent to the circumcircle of triangle $AFB$. Therefore $\\angle AFM = \\angle BAM = \\frac{1}{2}\\hat{A}$ and so\n\n$$\n\\begin{aligned}\n\\angle AFC &= \\angle AFM + \\angle MFC \\\\\n&= \\frac{1}{2}\\hat{A} + 180^\\circ - \\angle BFC \\\\\n&= \\frac{1}{2}\\hat{A} + 180^\\circ - \\angle BIC \\\\\n&= \\frac{1}{2}\\hat{A} + 180^\\circ - \\left(180^\\circ - \\frac{1}{2}\\hat{B} - \\frac{1}{2}\\hat{C}\\right) \\\\\n&= \\frac{1}{2}\\left(\\hat{A} + \\hat{B} + \\hat{C}\\right) = 90^\\circ.\n\\end{aligned}\n$$\n\nTherefore $AF \\perp FC$ as required.\n\n![](images/BMO_2022_shortlist_p28_data_87765bdf47.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19799,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $Q_0$ el conjunto de los números racionales mayores que cero. Sea $f: Q_0 \\to \\mathbb{R}$ una función que satisface las siguientes tres condiciones:\n\n1. $f(x)f(y) \\ge f(xy)$ para todos $x, y \\in Q_0$;\n2. $f(x + y) \\ge f(x) + f(y)$ para todos $x, y \\in Q_0$;\n3. Existe un número racional $a > 1$ tal que $f(a) = a$.\n\nDemuestra que $f(x) = x$ para todo $x \\in Q_0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Demostraremos sucesivamente las siguientes propiedades adicionales que debe tener cualquier función $f$ que satisface las condiciones del enunciado:\n\n1. $f(1) \\ge 1$. Tomando $x = 1$, $y = a$ en (1) y usando (3), obtenemos $a = f(a) \\le f(1)f(a) = af(1)$, así que $f(1) \\ge 1$.\n\n2. Para todo entero positivo $n$ y todo racional positivo $x$, se tiene $f(nx) \\ge n f(x)$, con $f(nx) = n f(x)$ si y sólo si $f(kx) = kx$ para $k = 1, 2, \\dots, n$. En particular, para todo entero positivo $n$ se tiene $f(n) \\ge n f(1)$, con igualdad para $n$ si y sólo si $f(k) = k f(1)$ para $k = 1, 2, \\dots, n$. Esto se obtiene por inducción usando la condición (2).\n\n3. Para todo entero positivo $u$ y todo racional positivo $x$, se tiene $f(x^u) \\le (f(x))^u$, con igualdad si y sólo si $f(x^k) = (f(x))^k$ para $k = 1, 2, \\dots, u$. Esto se obtiene por inducción usando la condición (1).\n\n4. $f$ toma sólo valores positivos. En concreto, para cualesquiera enteros positivos $m, n$, se tiene $f\\left(\\frac{m}{n}\\right) \\ge \\frac{m f(1)}{f(n)} > 0$. Esto se deduce de las propiedades anteriores.\n\n5. $f$ es estrictamente creciente. Para cualesquiera racionales positivos $z > x$, existe $y = z - x > 0$, y usando (2), $f(z) = f(x + y) \\ge f(x) + f(y) > f(x)$, ya que $f(y) > 0$ por la propiedad (4).\n\nSupongamos que existe un entero positivo $N$ tal que $f(N) > N$. Entonces existe un entero positivo $M$ tal que $M(f(N) - N) \\ge 1$. Por la propiedad (2), $f(MN) > M f(N) \\ge MN + 1$, y para todo entero $n \\ge MN$, $f(n) = f(MN + (n - MN)) \\ge f(MN) + f(n - MN) \\ge MN + 1 + (n - MN) f(1) \\ge n + 1$.\n\nConsideremos la sucesión $a, a^2, a^3, \\dots$, que es creciente y no tiene límite superior porque $a > 1$. Sea $u$ tal que $a^u \\ge MN$, y sea $k$ la parte entera de $a^u$. Entonces $a^u \\ge k > a^u - 1$, y usando (3) y las propiedades (3) y (5), se tiene que $k + 1 > a^u = (f(a))^u \\ge f(a^u) \\ge f(k) \\ge k + 1$, contradicción. Por lo tanto, $f(n) = n$ para todo entero positivo $n$, en particular $f(1) = 1$.\n\nTomando $x = n$, $y = \\frac{1}{n}$ en (1), y usando el resultado anterior, tenemos $f\\left(\\frac{1}{n}\\right) \\ge \\frac{1}{n}$ para todo entero positivo $n$. Usando la propiedad (2), para todo racional $\\frac{m}{n}$, donde $m, n$ son enteros positivos, se tiene $f\\left(\\frac{m}{n}\\right) \\ge m f\\left(\\frac{1}{n}\\right) \\ge \\frac{m}{n}$.\n\nSupongamos que existe algún racional positivo $q = \\frac{m}{n}$, con $m, n$ coprimos, tal que $f(q) \\ne q$. Por el razonamiento anterior, $f(q) > q$, y por la propiedad (2), $m = f(m) = f(nq) \\ge n f(q) > n q = m$, contradicción. Por lo tanto, $f(q) = q$ para todo racional positivo $q$, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19800,
    "subject": "Mathematics (Olympiad)",
    "question": "In a square $ABCD$, points $M$ and $N$ are chosen on the sides $AD$ and $DC$ respectively so that $\\angle BMA = \\angle NMD = 60^\\circ$. Find $\\angle MBN$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\angle MBN = 45^\\circ$.\n\n**Solution.** Draw the altitude $BL$ in $\\triangle BMN$. Then, in the triangles $BAM$ and $BLM$ we have: $\\angle BAM = \\angle BLM = 90^\\circ$, $\\angle BMN = 60^\\circ = \\angle BMA$, since $\\angle AMD = 180^\\circ$, and so $\\triangle BAM = \\triangle BLM$ because they also share the hypotenuse $BM$. Therefore, $AB = BL = BC$. It then follows that $\\triangle BNL = \\triangle BNC$ as right triangles with respectively equal sides. The equality of the triangles provides us with the following equalities of angles: $\\angle ABM = \\angle MBL$, $\\angle LBN = \\angle NBC$, and so\n\n$$\n\\angle MBN = \\angle MBL + \\angle LBN = \\frac{1}{2}(\\angle ABL + \\angle LBC) = \\frac{1}{2}\\angle ABC = 45^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19801,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that for any coloring of the numbers $1, 2, 3, \\ldots, n$ with three colors, there exist two numbers with the same color whose difference is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 29$.\n\nAssume, for contradiction, that the numbers $1, 2, \\ldots, 29$ can be colored with colors $A$, $B$, and $C$ so that no two numbers with the same color differ by a square. Let $f(i)$ denote the color of number $i$ for $i \\in \\{1, 2, \\ldots, 29\\}$.\n\nSince $9$, $16$, and $25$ are squares, the numbers $1$, $10$, and $26$ must all be assigned different colors. Similarly, $1$, $17$, and $26$ must all be assigned different colors, so $10$ and $17$ must have the same color. Likewise, we get $f(11) = f(18)$, $f(12) = f(19)$, and $f(13) = f(20)$ (for the last, consider $4$, $13$, $20$, $29$).\n\nWithout loss of generality, assume $f(10) = f(17) = A$. Since $11 = 10 + 1^2$, $f(11) \\neq f(10)$. Without loss of generality, let $f(11) = f(18) = B$. Now $19 = 18 + 1^2 = 10 + 3^2$, so $f(12) = f(19) = C$. Similarly, $20 = 19 + 1^2 = 11 + 3^2$ implies $f(13) = f(20) = A$. Thus, $f(13) = A = f(17)$, a contradiction.\n\nOn the other hand, if $n \\leq 28$, it is possible to color the numbers so that no two numbers with the same color differ by a perfect square.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19802,
    "subject": "Mathematics (Olympiad)",
    "question": "One hundred people attended the party, some of whom were previously acquainted. All acquaintances were mutual and no new ones were made during the party.\n\nA gong rang 100 times during the party. After the first sounding of the gong, all the people not acquainted with anyone left the party. After the second sounding of the gong, all the people with exactly one acquaintance (among the remaining people) left. It continued in this way: after the $k$th sounding of the gong, all the people acquainted with exactly $k-1$ remaining people left the party ($k = 1, \\dots, 100$).\n\nAt the end of the party, there were $n$ people still present. Find all possible values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $n$ can be $0, 1, 2, 3, \\dots, 98$.\n\nFor $n > 0$, divide all the people at the party into two groups, A and B. Let A contain $n$ people, each acquainted with all the other people at the party. Let B contain the remaining $100 - n$ people, none of whom are acquainted amongst themselves, but all of whom are acquainted with all the people in group A (so each person in B has $n$ acquaintances).\n\nAll the people from group B will leave the party after the $(n+1)^{\\text{th}}$ sounding of the gong. After that, only people from group A will remain, and each of them will have exactly $n-1$ acquaintances. Since the gong has already been sounded $n$ times, none of them will leave until the end of the party.\n\nThe value $n = 0$ is attained, for example, when all the party-goers know each other (so they all leave after the 100th chime). At least one person must leave at some moment: this is the person with the fewest acquaintances. This implies that $n = 100$ cannot be attained.\n\nFinally, $n = 99$ is not possible. Assume the contrary, that exactly one person leaves before the end; call this person X. As the first to leave, X has the fewest acquaintances. Since none of the remaining people leave after X, all must be acquainted with X. If another person Y is not acquainted with X, then Y's number of acquaintances would not change with X leaving, so Y would also have to leave at some point. Thus, X has 99 acquaintances, contradicting the assumption that X has the fewest acquaintances.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19803,
    "subject": "Mathematics (Olympiad)",
    "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$$\nx_{n+2} = a x_{n+1} + b x_n, \\quad \\forall n \\in \\mathbb{N}.\n$$\nLet the matrix $A$ be defined by\n$$\nA = \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix}.\n$$\n\na) Prove that\n$$\nA^n = \\begin{bmatrix} x_{n+1} & b x_n \\\\ x_n & b x_{n-1} \\end{bmatrix}, \\quad \\forall n \\in \\mathbb{N}^*.\n$$\n\nb) Prove that the number\n$$\n\\frac{x_{n+1} \\cdot x_{n+2} \\cdot \\dots \\cdot x_{n+m}}{x_1 \\cdot x_2 \\cdot \\dots \\cdot x_m}\n$$\nis a non-negative integer for all $m, n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We use induction on $n$.\n\nFor $n = 1$, $x_2 = a \\cdot 1 + b \\cdot 0 = a$, so\n$$\nA = \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} x_2 & b x_1 \\\\ x_1 & b x_0 \\end{bmatrix},\n$$\nwhich matches the formula for $n = 1$.\n\nAssume the formula holds for some $n \\geq 1$. Then\n$$\nA^{n+1} = A^n \\cdot A = \\begin{bmatrix} x_{n+1} & b x_n \\\\ x_n & b x_{n-1} \\end{bmatrix} \\cdot \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} x_{n+2} & b x_{n+1} \\\\ x_{n+1} & b x_n \\end{bmatrix},\n$$\ncompleting the induction.\n\nb) By induction, $x_n \\in \\mathbb{N}^*$ for all $n \\in \\mathbb{N}^*$.\n\nFor $m, n \\in \\mathbb{N}$, $m \\geq 1$, define\n$$\nF(m, n) = \\frac{x_{n+1} \\cdot x_{n+2} \\cdot \\dots \\cdot x_{n+m}}{x_1 \\cdot x_2 \\cdot \\dots \\cdot x_m},\n$$\nand $F(0, n) = 1$ for all $n$. We prove by induction on $p \\in \\mathbb{N}$ that $F(m, p - m) \\in \\mathbb{N}^*$ for all $0 \\leq m \\leq p$.\n\nSince $F(0, 0) = F(0, 1) = F(1, 0) = 1$, the result holds for $p = 0, 1$. From $A^{m+n} = A^m A^n$, we have\n$$\nx_{m+n+1} = x_{m+1} x_{n+1} + b x_m x_n,\n$$\nfor all $m, n \\in \\mathbb{N}^*$.\n\nAssume the result holds for $p$. Then $F(0, p+1) = F(p+1, 0) = 1$. For $1 \\leq m \\leq p$, set $n = p - m$:\n$$\n\\begin{align*}\nF(m, p + 1 - m) &= \\frac{x_{n+2} \\cdot x_{n+3} \\cdot \\dots \\cdot x_{n+m} \\cdot x_{m+n+1}}{x_1 \\cdot x_2 \\cdot \\dots \\cdot x_m} \\\\\n&= \\frac{x_{n+2} \\cdot \\dots \\cdot x_{n+m} \\cdot (x_{m+1} x_{n+1} + b x_m x_n)}{x_1 \\cdot \\dots \\cdot x_m} \\\\\n&= x_{m+1} F(m, n) + b x_n F(m-1, n+1) \\\\\n&= x_{m+1} F(m, p - m) + b x_n F(m-1, p - (m-1)) \\in \\mathbb{N}.\n\\end{align*}\n$$\nThus, the result holds for $p + 1$, completing the induction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 19804,
    "subject": "Mathematics (Olympiad)",
    "question": "In the following expression, Melanie changed some of the plus signs to minus signs:\n\n$$\n1 + 3 + 5 + 7 + \\dots + 97 + 99.\n$$\n\nWhen the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?\n\n(A) 14  (B) 15  (C) 16  (D) 17  (E) 18",
    "options": [],
    "answer": "See solution",
    "solution": "To minimize the number of minus signs needed to make the expression negative, minus signs should be chosen for all of the largest numbers. Hence, the first $k$ numbers of the expression will stay positive and the last $50 - k$ will be made negative for the greatest value of $k$ that gives a negative value.\n\nRecall that $1 + 3 + 5 + \\cdots + (2n - 1) = n^2$; that is, the sum of the first $n$ odd positive integers is equal to $n^2$. The expression in the problem statement is the sum of the first $50$ odd positive integers, so it equals $50^2 = 2500$. Hence $k^2$ must be strictly less than $\\frac{2500}{2} = 1250$. Because $35^2 = 1225$ and $36^2 = 1296$, at least $15$ plus signs must be switched to minus signs for the expression to evaluate to a negative value. Indeed,\n\n$$\n1 + 3 + 5 + \\cdots + 69 - 71 - 73 - 75 - \\cdots - 99 = 1225 - (2500 - 1225) = -50.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19805,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$ be a bisector in an isosceles triangle $ABC$ ($AB = BC$), and let $DE$ be another bisector in the triangle $ABD$. Find the measures of all angles in $ABC$ if the bisectors of $ABD$ and $AED$ intersect on the straight line $AD$.\n\n![](images/ukraine_2015_Booklet_p1_data_04962999d2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the intersection point of the bisectors of angles $ABD$ and $AED$. This point lies on segment $AD$ and is equidistant from rays $BA$ and $BC$, as well as from $EA$ and $ED$. Hence, it is also equidistant from rays $DE$ and $DC$. Therefore, $DA$ is the bisector of $\\angle CED$ (i.e., $K$ is an excenter of triangle $EBD$). With the initial conditions, this implies $\\angle ADC = 60^\\circ$. Since $\\angle DCA = 2\\angle DAC$, we have $\\angle DCA = 80^\\circ$. Finally, $\\angle BAC = \\angle BCA = 80^\\circ$, $\\angle ABC = 20^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19806,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, in the cyclic quadrilateral $ABCD$, the bisector of $\\angle BAD$ meets side $BC$ at $E$, and $M$ is the midpoint of $AE$. The exterior bisector of $\\angle BCD$ crosses the extension of $AD$ at $F$; the line $MF$ meets side $AB$ at $G$. If $AB = 2AD$, prove that $MF = 2MG$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p381_data_6d17375df8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "It is known that $\\angle EAF = \\frac{1}{2}\\angle BAD$, $\\angle ECF = 90^\\circ + \\frac{1}{2}\\angle BCD$, and hence\n\n$$\n\\angle EAF + \\angle ECF = 90^\\circ + \\frac{1}{2}(\\angle BAD + \\angle BCD) = 180^\\circ,\n$$\n\nwhich implies that $A, E, C, F$ all lie on a circle, say $\\omega$.\n\nLet the extension of $CD$ beyond $D$ cross circle $\\omega$ at point $K$. As shown in the next figure, drop perpendiculars from $F$ to the lines $AE$ and $AK$, with feet $X$ and $Y$, respectively.\n\nNote that $CF$ is the exterior bisector of $\\angle ECK$, thus $F$ is the midpoint of $\\overarc{ECK}$ of $\\omega$, and $AF$ bisects $\\angle CAK$.\n\nSince $A, B, C, D$ are concyclic, $A, E, C, K$ are also concyclic, we infer that $\\angle ADK = \\angle ABE$, $\\angle AKD = \\angle AEB$, hence $\\triangle ADK \\sim \\triangle ABE$, $\\frac{AK}{AE} = \\frac{AD}{AB} = \\frac{1}{2}$, that is, $AK = \\frac{1}{2}AE$.\n\nOn the other hand, as $AF$ bisects $\\angle CAK$, $\\angle FKY \\cong \\angle FEX$, $KY = EX = \\frac{1}{2}(AE - AK) = \\frac{1}{4}AE$. Therefore, $X$ is the midpoint of $EM$. From $AK = \\frac{1}{2}AE = AM$, it follows that $\\triangle AKF \\cong \\triangle AMF$, and\n\n$$\n\\angle AFG = \\angle AFM = \\angle AFK = \\angle AEK.\n$$\n\nMoreover, by properties of cyclic quadrilaterals,\n\n$$\n\\angle AEK = \\angle ACK = \\angle ACD = \\angle ABD.\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p382_data_7174db1f0e.png)\n\nFrom the above, $\\angle AFG = \\angle ABD$. Thus, $\\triangle AFG \\sim \\triangle ABD$. Finally, by the angle bisector theorem, $\\frac{MF}{MG} = \\frac{AF}{AG} = \\frac{AB}{AD} = 2$, namely, $MF = 2MG$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19807,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist 2018 distinct positive numbers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is affirmative. There exist 2018 distinct positive numbers satisfying the required conditions:\n\n$$\na, 2a, \\dots, 2018a, \\text{ where } a = \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right)^4.\n$$\n\nNote that $a$ is an integer, since 2019 is a multiple of 3 and 2018 is a multiple of 2.\n\nThe sum of the squares of these numbers is\n\n$$\na^2 + (2a)^2 + \\dots + (2018a)^2 = a^2(1^2 + 2^2 + \\dots + 2018^2) = a^2 \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right) = \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right)^8 \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right) = \\left( \\left( \\frac{2018 \\cdot 2019 \\cdot 4037}{6} \\right)^3 \\right)^3,\n$$\n\nwhich is a perfect cube.\n\nThe sum of their cubes is\n\n$$\na^3 + (2a)^3 + \\dots + (2018a)^3 = a^3(1^3 + 2^3 + \\dots + 2018^3) = a^3 \\left(\\frac{2018 \\cdot 2019}{2}\\right)^2 = \\left(\\frac{2018 \\cdot 2019 \\cdot 4037}{6}\\right)^{12} \\left(\\frac{2018 \\cdot 2019}{2}\\right)^2 = \\left(\\left(\\frac{2018 \\cdot 2019 \\cdot 4037}{6}\\right)^6 \\left(\\frac{2018 \\cdot 2019}{2}\\right)\\right)^2,\n$$\n\nwhich is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19808,
    "subject": "Mathematics (Olympiad)",
    "question": "For any real number $x$, let $A(x) = x^2 + 4\\lfloor x \\rfloor$.\n\na) Find all real numbers $x$ for which $A(x) = \\{x\\}^2$.\n\nb) Find all real numbers $y > 0$ for which $A(y)$ is the square of a natural number.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $x = \\lfloor x \\rfloor + \\{x\\}$, we have:\n$$A(x) = (\\lfloor x \\rfloor + \\{x\\})^2 + 4\\lfloor x \\rfloor = \\lfloor x \\rfloor^2 + 2\\lfloor x \\rfloor\\{x\\} + \\{x\\}^2 + 4\\lfloor x \\rfloor$$\nSetting $A(x) = \\{x\\}^2$ gives:\n$$\\lfloor x \\rfloor^2 + 2\\lfloor x \\rfloor\\{x\\} + 4\\lfloor x \\rfloor = 0$$\nIf $x \\ge 0$, then $\\lfloor x \\rfloor = 0$, so any $x \\in [0, 1)$ is a solution.\n\nIf $x < 0$, then $\\lfloor x \\rfloor < 0$ and:\n$$\\{x\\} = -\\frac{\\lfloor x \\rfloor^2 + 4\\lfloor x \\rfloor}{2\\lfloor x \\rfloor} = -\\frac{\\lfloor x \\rfloor + 4}{2}$$\nSince $\\{x\\} \\in [0, 1)$, this is possible only for $\\lfloor x \\rfloor = -5$ or $-4$, yielding $x_1 = -\\frac{9}{2}$ and $x_2 = -4$.\n\nThus, the solutions are $x \\in \\{-\\frac{9}{2}, -4\\} \\cup [0, 1)$.\n\nb) For $y \\in (0, 1)$, $A(y) = y^2 \\in (0, 1)$, not a perfect square. For $y \\in [1, 2)$, $A(y) = y^2 + 4 \\in (5, 8)$, also not a perfect square.\n\nFor $y \\ge 2$, let $y = \\sqrt{m}$, $m \\in \\mathbb{N}$, $m \\ge 4$. Then $A(y) = m + 4\\lfloor \\sqrt{m} \\rfloor = p^2$ for some $p \\in \\mathbb{N}$.\nLet $k = \\lfloor \\sqrt{m} \\rfloor$, $k \\ge 2$, so $k^2 \\le m < (k+1)^2$ and $m + 4k = p^2$.\n\nIf $p = k + 2$, then $m = (k + 2)^2 - 4k = k^2 + 4$.\n\nTherefore, the solutions are $y = \\sqrt{k^2 + 4}$, where $k \\in \\mathbb{N}$, $k \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19809,
    "subject": "Mathematics (Olympiad)",
    "question": "The altitudes of an acute-angled triangle $ABC$ intersect at point $H$. The tangent at point $A$ to the circumcircle of triangle $AHB$ intersects the line $CH$ at point $K$. The tangent at point $A$ to the circumcircle of triangle $AHC$ intersects the line $BH$ at point $L$. Prove that the points $B$, $C$, $K$, $L$ lie on the same circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. By tangency, $\\angle KAH = \\angle ABH = 90^\\circ - \\alpha$ (see the figure below). Since $\\angle ACH = 90^\\circ - \\alpha$ as well, triangles $AHC$ and $KHA$ are similar. Consequently, the corresponding third angles are equal, i.e., $\\angle AKH = \\angle CAH = 90^\\circ - \\gamma$. Similarly, we get $\\angle LAH = 90^\\circ - \\alpha$ and $\\angle ALH = 90^\\circ - \\beta$. Thus, $\\angle KAL = \\angle KAH + \\angle LAH = 180^\\circ - 2\\alpha$, with $AH$ bisecting the angle $KAL$.\n\nOn the other hand,\n\n$$\n\\begin{align*}\n\\angle KHL &= \\angle CHB = 180^\\circ - \\angle BCH - \\angle CBH \\\\\n&= 180^\\circ - (90^\\circ - \\gamma) - (90^\\circ - \\beta) \\\\\n&= \\beta + \\gamma = 180^\\circ - \\alpha.\n\\end{align*}\n$$\n\nNow note (see the figure below) that for the incenter $I$ of triangle $AKL$,\n\n$$\n\\begin{align*}\n\\angle KIL &= 180^\\circ - \\angle IKL - \\angle ILK = 180^\\circ - \\frac{\\angle AKL}{2} - \\frac{\\angle ALK}{2} \\\\\n&= 180^\\circ - \\frac{\\angle AKL + \\angle ALK}{2} \\\\\n&= 180^\\circ - \\frac{180^\\circ - \\angle KAL}{2} = 180^\\circ - \\frac{180^\\circ - (180^\\circ - 2\\alpha)}{2} = 180^\\circ - \\alpha,\n\\end{align*}\n$$\n\nso $\\angle KIL = \\angle KHL$. Since points $H$ and $I$ lie on the same side of line $KL$, points $K$, $H$, $I$, and $L$ lie on the same circle. Given that both $H$ and $I$ lie on the angle bisector $AH$ of angle $KAL$, which intersects the chord $KL$, it follows that $H = I$. Consequently, $\\angle LKH = \\angle AKH = 90^\\circ - \\gamma$ and $\\angle KLH = \\angle ALH = 90^\\circ - \\beta$. In addition, we have $\\angle HLK = 90^\\circ - \\beta = \\angle HCB$, from which it follows that points $B$, $C$, $K$, $L$ lie on the same circle.\n\n![](images/EST_ABooklet_2024_p47_data_59673edae4.png)\n\n![](images/EST_ABooklet_2024_p47_data_ef7d347be9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19810,
    "subject": "Mathematics (Olympiad)",
    "question": "有無限多張牌，每張上寫一個實數 $x$，使得每個實數都恰有一張牌。兩位玩家從這些牌中各自抽出互斥的兩組牌 $A$ 與 $B$，每組各 100 張。我們要制定一組規則來從兩人中選出一個贏家。規則必須滿足：\n\n1. 勝負只取決於這 200 張牌的相對大小。換言之，如果將這些牌從小到大面朝下排成一排，並只告訴你每張牌屬於哪位玩家，而不告訴你其上的數字，你仍能判定贏家。\n2. 以遞增方式將兩組牌記為 $A = \\{a_1, a_2, \\cdots, a_{100}\\}$ 與 $B = \\{b_1, b_2, \\cdots, b_{100}\\}$。如果對於所有 $i$，都有 $a_i > b_i$，則 $A$ 贏 $B$。\n3. 假設有三位玩家抽出三組互斥的牌 $A$、$B$ 與 $C$，其中 $A$ 贏 $B$ 且 $B$ 贏 $C$，則 $A$ 贏 $C$。\n\n試決定所有可能的規則數。在此，兩組規則被視為不同，若且唯若存在互斥的兩組牌 $A$ 與 $B$，使得它們在兩組規則下的勝負不同。",
    "options": [],
    "answer": "See solution",
    "solution": "共有 $100$ 組不同的規則。以下證明若每人抽 $n$ 張牌，則規則數為 $n$。\n\n1. 首先構造 $n$ 組不同的規則。令 $k \\in \\{1, 2, \\cdots, n\\}$，將 $A$ 和 $B$ 以遞增方式表示。考慮規則：$A < B$ 若且唯若 $a_k < b_k$。此規則滿足題目要求，且共有 $n$ 種不同規則。\n\n2. 證明這 $n$ 種規則就是全部可能。假設有一組規則滿足題意，令 $k \\in \\{1, 2, \\dots, n\\}$ 為滿足\n\n$$\nA_k = \\{1, 2, \\dots, k, n + k + 1, n + k + 2, \\dots, 2n\\\\}\n$$\n\n$$\nB_k = \\{k + 1, k + 2, \\dots, n + k\\}\n$$\n\n的最小值。由 $k = n$ 時成立，知 $k$ 存在。\n\n假設 $X$ 和 $Y$ 互斥，證明 $X < Y$ 若且唯若 $x_k < y_k$，即與上述構造的規則相同。任選與 $X$、$Y$ 都互斥的三個集合 $U, V, W$，滿足：\n\n- $u_1 < u_2 < \\dots < u_{k-1} < \\min(x_1, y_1)$\n- $\\max(x_n, y_n) < v_1 < v_2 < \\dots < v_n$\n- $x_k < v_1 < v_2 < \\dots < v_k < w_1 < w_2 < \\dots < w_n < u_k < u_{k+1} < \\dots < u_n < y_k$\n\n則：\n\n(a) 對所有 $i$，$u_i < y_i$ 且 $x_i < v_i$，故 $U < Y$ 且 $X < V$。\n\n(b) $U$ 和 $W$ 的相對大小與 $A_{k-1}$ 和 $B_{k-1}$ 相同，由 $k$ 的選取知 $A_{k-1} > B_{k-1}$，故 $U > W$。\n\n(c) $V$ 和 $W$ 的相對大小與 $A_k$ 和 $B_k$ 相同，由 $k$ 的選取知 $A_k > B_k$，故 $V < W$。\n\n因此，\n\n$$\nX < V < W < U < Y\n$$\n\n故 $X < Y$。以上推論只取決於 $x_k$ 和 $y_k$ 的大小關係，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19811,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$, there exist a prime $p$ and an integer $m$ such that:\n\n(a) $p \\equiv 5 \\pmod{6}$;\n\n(b) $p \\nmid n$;\n\n(c) $n \\equiv m^3 \\pmod{p}$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are infinitely many primes $p$ congruent to $5$ modulo $6$ (by Dirichlet's theorem on primes in arithmetic progressions, or by direct argument). In particular, there exists such a prime $p > n$, so $p$ satisfies both (a) and (b).\n\nLet $p = 6k + 5$ for some integer $k$. Set $m = n^{4k+3}$. By Fermat's Little Theorem:\n\n$$\nm^3 = \\left(n^{4k+3}\\right)^3 = n^{12k+9} = n^{6k+4} \\cdot n^{6k+4} \\cdot n \\equiv n^{p-1} \\cdot n^{p-1} \\cdot n \\equiv n \\pmod{p}.\n$$\n\nThus, $n \\equiv m^3 \\pmod{p}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19812,
    "subject": "Mathematics (Olympiad)",
    "question": "Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of $\\triangle CFB$?\n\n(A) 5:4  (B) 4:3  (C) 3:2  (D) 5:3  (E) 2:1",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $\\triangle AFE$ and $\\triangle CFB$ are similar by Angle-Angle. The ratio of corresponding sides is $1:2$, so the ratio of their areas is $1:4$. Furthermore, consider $\\triangle AFE$ and $\\triangle AFB$. Because $FE:FB = 1:2$ and the heights of the two triangles corresponding to bases $\\overline{FE}$ and $\\overline{FB}$ are the same, their areas are in a $1:2$ ratio. Finally, observe that the area of $\\triangle ABE$ is $\\frac{1}{4}$ the area of parallelogram $ABCD$. Therefore, using the area of $\\triangle AFE$ as unit, the area of $\\triangle CFB$ is $4$, the area of $\\triangle AFB$ is $2$, the area of quadrilateral $ABCD$ is $4 \\cdot (1+2) = 12$, and the area of quadrilateral $CDEF$ is $12 - (1+4+2) = 5$. The requested ratio of the areas of quadrilateral $CDEF$ and $\\triangle CFB$ is $5:4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19813,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S \\subset \\{1, \\dots, n\\}$ be a nonempty set, where $n$ is a positive integer. Denote by $s$ the greatest common divisor of the elements of $S$. Assume $s \\neq 1$ and let $d$ be its smallest divisor greater than $1$. Let $T \\subset \\{1, \\dots, n\\}$ be a set such that $S \\subset T$ and $|T| \\geq 1 + \\lfloor \\frac{n}{d} \\rfloor$. Prove that the greatest common divisor of the elements in $T$ is $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t$ be the greatest common divisor of the elements in $T$. Since $S \\subset T$, we have $s \\mid t$. Assume for contradiction that $t \\neq 1$. Then $t \\geq d$.\n\nSince $|T| \\geq 1 + \\lfloor \\frac{n}{d} \\rfloor$, there are at least $1 + \\lfloor \\frac{n}{d} \\rfloor$ elements in $T$, all divisible by $t$. The largest of them, say $M$, satisfies\n\n$$\nM \\geq t \\cdot (1 + \\lfloor \\frac{n}{d} \\rfloor) \\geq d \\cdot (1 + \\lfloor \\frac{n}{d} \\rfloor) > d \\cdot \\frac{n}{d} = n.\n$$\n\nThus, $M > n$, contradicting $M \\in \\{1, \\dots, n\\}$. Therefore, $t = 1$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19814,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, and $PC$ intersect sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Prove that\n\n$$\n[PAF] + [PBD] + [PCE] = \\frac{1}{2}[ABC]\n$$\n\nif and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $[PAF] = x$, $[PBD] = y$, $[PCE] = z$, $[PAE] = u$, $[PCD] = v$, and $[PBF] = w$.\n\n![](images/USA_IMO_2003_p49_data_e0ed8f5daa.png)\n\nNote first that\n\n$$\n\\frac{x}{w} = \\frac{x+u+z}{w+y+v} = \\frac{u+z}{y+v} = \\frac{AF}{FB},\n$$\n\n$$\n\\frac{y}{v} = \\frac{x+y+w}{u+v+z} = \\frac{x+w}{u+z} = \\frac{BD}{DC},\n$$\n\n$$\n\\frac{z}{u} = \\frac{y+z+v}{x+u+w} = \\frac{y+v}{x+w} = \\frac{CE}{EA}.\n$$\n\nPoint $P$ lies on one of the medians if and only if\n\n$$\n(x-w)(y-v)(z-u) = 0. \\quad (*)\n$$\n\nBy Ceva's Theorem, we have\n\n$$\n\\frac{xyz}{uvw} = \\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1,\n$$\n\nor,\n\n$$\nxyz = uvw. \\quad (1)\n$$\n\nMultiplying out $\\frac{x}{w} = \\frac{u+z}{y+v}$ yields $xy + xv = uw + zw$. Likewise, $uy + yz = xv + vw$ and $xz + zw = uy + uv$. Summing up the last three relations, we obtain\n\n$$\nxy + yz + zx = uv + vw + wu. \\quad (2)\n$$\n\nNow we are ready to prove the desired result. We first prove the “if” part by assuming that $P$ lies on one of the medians, say $AD$. Then $y = v$, and so $\\frac{y}{v} = \\frac{x+w}{u+z}$ and $xyz = uwv$ become $x + w = u + z$ and $xz = uw$, respectively. Then the numbers $x, -z$ and $u, -w$ have the same sum and the same product. It follows that $x = u$ and $z = w$. Therefore $x + y + z = u + v + w$, as desired.\n\nConversely, we assume that\n\n$$\nx + y + z = u + v + w. \\quad (3)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19815,
    "subject": "Mathematics (Olympiad)",
    "question": "For every pair of positive integers $n, m$ with $n < m$, let $s(n, m)$ be the number of positive integers in the range $[n, m]$ that are coprime with $m$. Find all positive integers $m \\ge 2$ such that:\n\ni) $\\dfrac{s(n, m)}{m-n} \\ge \\dfrac{s(1, m)}{m}$ for all $n = 1, 2, \\dots, m-1$,\n\nii) $2022^m + 1$ is divisible by $m^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Suppose $m$ has at least two prime divisors. Let $p$ be the smallest prime divisor of $m$, and $p_1, p_2, \\dots, p_k$ be the other prime divisors. Then\n\n$$\n\\frac{\\varphi(m)}{m} = \\left(1 - \\frac{1}{p}\\right) \\left(1 - \\frac{1}{p_1}\\right) \\dots \\left(1 - \\frac{1}{p_k}\\right) < 1 - \\frac{1}{p} = \\frac{p-1}{p}.\n$$\n\nBy choosing $n = p$ in (i),\n\n$$\n\\frac{s(p, m)}{m-p} = \\frac{\\varphi(m) - (p-1)}{m-p} < \\frac{\\varphi(m) - \\frac{\\varphi(m)}{m} \\cdot p}{m-p} = \\frac{\\varphi(m)}{m} = \\frac{s(1, m)}{m},\n$$\n\nwhich is a contradiction. Therefore, $m$ must be a power of a prime. Let $m = p^k$. Note that\n\n$$\n2022^{p^k} + 1 \\equiv 2022 + 1 \\equiv 2023 \\pmod{p},\n$$\n\nso $p \\mid 2023$ and $p \\in \\{7, 17\\}$.\n\nIf $p = 7$, using LTE,\n\n$$\nv_7(2022^{7k} + 1) = v_7(2023) + v_7(7^k) = 1 + k \\ge v_7(7^{2k}) = 2k.\n$$\n\nThus, $k = 1$ and $m = 7$.\n\nIf $p = 17$, using LTE,\n\n$$\nv_{17}(2022^{17k} + 1) = v_{17}(2023) + v_{17}(17^k) = 2 + k \\ge v_{17}(17^{2k}) = 2k,\n$$\n\nso $k \\in \\{1, 2\\}$ and $m \\in \\{17, 289\\}$.\n\nTherefore, the desired values are $m = 7, 17, 289$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19816,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute and non-isosceles triangle. Its angle bisectors $AL_1$ and $BL_2$ intersect at the point $I$. Points $D$ and $E$ are chosen on the segments $AL_1$ and $BL_2$ such that $\\angle DBC = \\frac{1}{2}\\angle A$ and $\\angle EAC = \\frac{1}{2}\\angle B$. The lines $AE$ and $BD$ intersect at a point $P$. Let $K$ be the point symmetric to $I$ with respect to the line $DE$. Prove that the lines $KP$ and $DE$ intersect at a point on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. Let $N$ be the midpoint of arc $ACB$ of the circumcircle of $\\triangle ABC$.\n\nFirstly, both points $D$ and $E$ lie inside $\\triangle ANB$. Since both are inside $\\triangle ABC$, $\\frac{1}{2}\\beta = \\angle EAC < \\angle BAC = \\alpha$ and $\\frac{1}{2}\\alpha = \\angle DBC < \\angle ABC = \\beta$.\n\nAlso,\n$$\n\\angle NAB = \\angle NBA = 90^\\circ - \\frac{1}{2}\\gamma = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\n$$\n\\angle EBA = \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NBA, \\quad \\angle EAB = \\alpha - \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NAB\n$$\n\nbecause $\\frac{1}{2}\\alpha < \\beta$. So $E$ lies inside $\\triangle ANB$; similarly, $D$ lies there.\n\nNow,\n$$\n\\begin{aligned}\n\\angle EAB &= \\alpha - \\frac{1}{2}\\beta, \\\\\n\\angle NAE &= \\frac{1}{2}(\\alpha + \\beta) - (\\alpha - \\frac{1}{2}\\beta) = \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBI &= \\frac{1}{2}\\alpha, \\quad \\angle IBA = \\frac{1}{2}\\beta.\n\\end{aligned}\n$$\n\nBy Ceva's Theorem in triangle $ABN$ and point $E$:\n$$\n1 = \\frac{\\sin \\angle ANE}{\\sin \\angle ENB} \\cdot \\frac{\\sin \\angle NBE}{\\sin \\angle EBA} \\cdot \\frac{\\sin \\angle BAE}{\\sin \\angle EAN}\n$$\n\nAfter substitutions:\n$$\n\\frac{\\sin \\angle ENB}{\\sin \\angle ANE} = \\frac{\\sin \\frac{1}{2}\\alpha}{\\sin \\frac{1}{2}\\beta} \\cdot \\frac{\\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin(\\beta - \\frac{1}{2}\\alpha)}\n$$\n\nAnalogously, $\\angle DAB = \\frac{1}{2}\\alpha$,\n$$\n\\begin{aligned}\n\\angle DAN &= \\frac{1}{2}\\beta, & \\angle DBA = \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBD = \\alpha - \\frac{1}{2}\\beta.\n\\end{aligned}\n$$\n\nUsing Ceva's Theorem again:\n\n![](images/Ukraine_2020_booklet_p44_data_382de08bca.png)\n\n$$\n1 = \\frac{\\sin \\angle AND \\cdot \\sin \\angle NBD \\cdot \\sin \\angle BAD}{\\sin \\angle DNB \\cdot \\sin \\angle DBA \\cdot \\sin \\angle DAN}\n$$\n\nor\n$$\n\\frac{\\sin \\angle DNB}{\\sin \\angle AND} = \\frac{\\sin \\frac{1}{2}\\alpha \\cdot \\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin \\frac{1}{2}\\beta \\cdot \\sin(\\beta - \\frac{1}{2}\\alpha)}\n$$\n\nTherefore, the rays $ND$ and $NE$ coincide since\n$$\n\\angle BNA = \\angle ANE + \\angle ENB = \\angle AND + \\angle DNB\n$$\n\nLet $L$ be the second intersection point of $DE$ and the circumcircle of $\\triangle ABC$. We claim that $KP$ passes through $L$. Obviously,\n$$\n\\angle (AL, LN) = \\angle (AB, BN) = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\nOn the other hand,\n$$\n\\angle (AP, PB) = \\angle (PA, AB) + \\angle (AB, BP) = \\alpha - \\frac{1}{2}\\beta + \\beta - \\frac{1}{2}\\alpha = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\nHence $\\angle (AL, LN) = \\angle (AP, PB)$ and the quadrilateral $APDL$ is inscribed. Also, $\\angle (DI, IE) = \\frac{1}{2}(\\alpha + \\beta) = \\angle (EP, PB)$. By symmetry, $\\angle (EK, KD) = \\angle (DI, IE)$ and $\\angle (KD, DE) = \\angle (ED, DI)$, so $\\angle (EK, KD) = \\angle (EP, PD)$ and $PKED$ is inscribed.\n\nFinally,\n$$\n\\angle (KP, PE) = \\angle (KD, DE) = \\angle (ED, DI) = \\angle (LD, DA) = \\angle (LP, PA)\n$$\n\nWe obtain $\\angle (KP, PE) = \\angle (LP, PA)$ and since $AP$ and $EP$ coincide, we get that the line $KP$ passes through $L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19817,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $a^2 + b^2$ such that the equation\n\n$$\nx^2 + ax + b + \\frac{a}{x} + \\frac{1}{x^2} = 0\n$$\n\nhas a real solution $x \\ne 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\ge 0$. (By flipping the sign of $a$, we only need to flip the sign of the root $x$.) Clearly, $x \\ne 0$. Rewrite the equation as\n\n$$\nx^2 + ax + b + \\frac{a}{x} + \\frac{1}{x^2} = 0.\n$$\n\nThis is the same as\n\n$$\n\\left(x + \\frac{1}{x}\\right)^2 + a\\left(x + \\frac{1}{x}\\right) + b - 2 = 0.\n$$\n\nLet $y = x + \\frac{1}{x}$. Note that $|y| = |x| + \\frac{1}{|x|} \\ge 2$ by the AM-GM inequality. The equation becomes\n\n$$\ny^2 + ay + (b - 2) = 0.\n$$\n\nIf $a > 1$, then $a^2 + b^2 > 1$. Thus, we only need to consider the case $a \\le 1$. This implies the axis of symmetry $y = -\\frac{a}{2}$ lies between $-2$ and $2$.\n\n- If $y \\ge 2$, then\n\n$$\n0 = y^2 + ay + (b-2) \\ge (2)^2 + a(2) + (b-2) = 2a + b + 2.\n$$\n\nThis yields $b \\le -2a - 2 \\le -2$, and hence $a^2 + b^2 \\ge 4$.\n\n- If $y \\le -2$, then\n\n$$\n0 = y^2 + ay + (b-2) \\ge (-2)^2 + a(-2) + (b-2) = -2a + b + 2.\n$$\n\nThis yields $b \\le 2a - 2 \\le 0$, and hence\n\n$$\na^2 + b^2 \\ge a^2 + (2a - 2)^2 = 5\\left(a - \\frac{4}{5}\\right)^2 + \\frac{4}{5} \\ge \\frac{4}{5}.\n$$\n\nEquality holds when $a = \\frac{4}{5}$ and $b = -\\frac{2}{5}$. Correspondingly, this means $y = -2$ and $x = -1$.\n\nTherefore, the minimum value of $a^2 + b^2$ is $\\frac{4}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19818,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. Ali and Hadi play a game in which they start by writing the numbers $1, 2, \\dots, N$ on a board. They then take turns to make a move, starting with Ali. Each move consists of choosing a pair of integers $(k, n)$, where $k \\ge 0$ and $n$ is one of the integers on the board, and then erasing every integer $s$ on the board such that $2^k \\mid n - s$. The game continues until the board is empty. The player who erases the last integer on the board loses.\n\nDetermine all values of $N$ for which Ali can ensure that he wins, no matter how Hadi plays.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is that Ali wins when $N$ is of the form $2^n$ for $n$ odd, or of the form $t2^n$ for $n$ even and $t > 1$ odd.\n\nWe define that a set $S$ wins if the current player wins given $S$ as the current set of integers on the board; otherwise, $S$ loses. Let $J(S, \\mathcal{T}) = (2S - 1) \\cup (2\\mathcal{T})$. Every subset of $\\mathbb{Z}$ can be written as $J(S, \\mathcal{T})$ for a unique pair $(S, \\mathcal{T})$ of subsets of $\\mathbb{Z}$. Let $[n]$ denote the set $\\{1, 2, \\dots, n\\}$.\n\n**Lemma 1.** For any set $S$, $\\mathcal{T}$ wins if and only if $J(S, \\emptyset)$ wins. Similarly, $S$ wins if and only if $J(\\emptyset, S)$ wins.\n\n*Proof.* Let $(k, m)$ be a move on $S$, and let $\\mathcal{T}$ be the result of applying the move. Then we can reduce $J(S, \\emptyset)$ to $J(\\mathcal{T}, \\emptyset)$ by applying the move $(k + 1, 2m - 1)$. Conversely, let $(k, m)$ be a move on $J(S, \\emptyset)$. We can express the result of this move as $J(\\mathcal{T}, \\emptyset)$ for some $\\mathcal{T}$. Then we can reduce $S$ to $\\mathcal{T}$ by applying the move $(\\max(k - 1, 0), (k + 1)/2)$. This gives a bijection between games starting with $S$ and games starting with $J(S, \\emptyset)$, proving the first part. The second part follows similarly.\n\n**Lemma 2.** If $S$, $\\mathcal{T}$ are nonempty and at least one of them loses, then $J(S, \\mathcal{T})$ wins.\n\n*Proof.* If $S$ is losing, we can delete $J(\\emptyset, \\mathcal{T})$ using the move $(1, t)$ for some $t \\in J(\\emptyset, \\mathcal{T})$, leaving the losing set $J(S, \\emptyset)$. Similarly, if $\\mathcal{T}$ is losing, we can delete $J(S, \\emptyset)$ using the move $(1, s)$ for some $s \\in J(S, \\emptyset)$, leaving the losing set $J(\\emptyset, \\mathcal{T})$.\n\n**Lemma 3.** If $S$ is nonempty and wins, then $J(S, S)$ loses.\n\n*Proof.* From this position, we can convert any sequence of moves into another valid sequence by replacing $(k, 2n - 1)$ with $(k, 2n)$, and vice versa. Thus, we may assume the initial move $(k, m)$ has $m$ odd. Any such move results in a winning position for the other player. The move $(0, m)$ loses immediately. Otherwise, the move results in the set $J(\\mathcal{T}, S)$ for some set $\\mathcal{T}$. There are three cases: if $\\mathcal{T}$ is empty, the other player gets the winning set $J(\\emptyset, S)$; if $\\mathcal{T}$ is losing, the other player can choose the move $(1, s)$ for some $s \\in J(\\emptyset, S)$, leaving the losing set $J(\\mathcal{T}, \\emptyset)$; if $\\mathcal{T}$ is nonempty winning, the other player can choose the move $(k, m + 1)$, resulting in the position $J(\\mathcal{T}, \\mathcal{T})$. Induction on $|S|$ shows this is a losing set.\n\n**Lemma 4.** $[2n]$ wins if and only if $[n]$ loses.\n\n*Proof.* Note that $[2n] = J([n], [n])$, so the result follows from Lemmas 2 and 3.\n\n**Lemma 5.** For any integer $n \\ge 1$, $[2n + 1]$ wins.\n\n*Proof.* By Lemma 4, either $[n]$ or $[2n]$ loses. If $[n]$ loses, then by Lemma 2, $[2n+1] = J([n+1], [n])$ wins. Otherwise, $[2n]$ loses, and $[2n+1]$ wins by choosing the move $(k, 2n+1)$ for $k$ large enough so only $2n+1$ is eliminated.\n\nTo verify the answer:\n\n- If $N = 2^n$ for some $n$, for $N = 1$, every move is an instant loss for Ali. By Lemma 4, Ali wins for $N = 2^n$ if and only if Ali loses for $N = 2^{n-1}$, so by induction, Ali wins for $N = 2^n$ if and only if $n$ is odd.\n- Otherwise, $N = t2^n$ for some $n$ and $t > 1$ odd. By Lemma 5, Ali wins when $n = 0$. By Lemma 4, Ali wins for $N = t2^n$ if and only if Ali loses for $N = t2^{n-1}$, so by induction on $n$, Ali wins for $N = t2^n$ if and only if $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19819,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_n$ is defined by $a_0 = 3$ and $a_{n+1} - a_n = n(a_n - 1)$ for $n \\geq 0$. Find all positive integers $m$ for which $\\gcd(m, a_n) = 1$ for all $n \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "The recurrence can be rewritten as:\n\n$$\na_{n+1} - 1 = (n+1)(a_n - 1)\n$$\n\nUnfolding the recurrence:\n\n$$\na_{n+1} - 1 = (n+1)n(n-1)\\cdots2 \\cdot (a_1 - 1)\n$$\n\nGiven $a_0 = 3$, so $a_1 - 1 = a_0 - 1 = 2$. Thus,\n\n$$\na_{n+1} - 1 = 2(n+1)!\n$$\n\nSo $a_n = 2n! + 1$.\n\nConsider $m > 1$:\n- If $m = 2^s$, all $a_n$ are odd, so $\\gcd(m, a_n) = 1$.\n- If $m$ has a prime $p \\geq 3$ dividing it, then $a_{p-3} = 2(p-3)! + 1 \\equiv (p-1)! + 1 \\equiv 0 \\pmod{p}$ (by Wilson's theorem), so $p \\mid a_{p-3}$ and $\\gcd(m, a_{p-3}) \\neq 1$.\n\nThus, the positive integers $m$ are exactly those of the form $m = 2^s$ for $s \\geq 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19820,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be real numbers such that\n\n$$\nx + y + z = 0.\n$$\n\nProve that\n\n$$\n6(x^3 + y^3 + z^3)^2 \\leq (x^2 + y^2 + z^2)^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Recall the identity\n\n$$\nx^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx).\n$$\n\nSince $x + y + z = 0$, we have $x^3 + y^3 + z^3 = 3xyz$. Substituting this into our inequality, we want to prove\n\n$$\n54x^2y^2z^2 \\leq (x^2 + y^2 + z^2)^3.\n$$\n\nIf any of $x$, $y$, $z$ are $0$, this is clearly true. Otherwise, one of $x$, $y$, $z$ is a different sign. Without loss of generality, let $x$ be negative and $y$, $z$ be positive, which implies $x^2 \\geq (y + z)^2 \\geq 4yz$. By AM-GM,\n\n$$\n\\frac{\\frac{x^2}{4} + \\frac{x^2}{4} + \\frac{x^2}{4} + \\frac{x^2}{4} + y^2 + z^2}{6} \\geq \\sqrt[6]{\\frac{x^8y^2z^2}{4^4}}\n$$\n\nSince $x^8y^2z^2 \\geq 16x^4y^4z^4$, this gives\n\n$$\n(x^2 + y^2 + z^2)^3 \\geq 216 \\sqrt{\\frac{x^4y^4z^4}{16}} = 54x^2y^2z^2,\n$$\n\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19821,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $m$, consider a table of size $m \\times 2018$ (with $m$ rows and $2018$ columns) that has some empty cells; the remaining cells are filled with $0$ or $1$. A table is called *completed* if for every binary string $S$ of length $2018$, we can choose a row and fill the empty cells in that row with $0$ or $1$ to form $S$. A completed rectangle is called *minimal* if, when we remove any row, the remaining table is not completed.\n\n**a)** Prove that for any positive integer $k \\leq 2018$, there exists a minimal $2^k \\times 2018$ rectangle with exactly $k$ columns containing both $0$ and $1$.\n\n**b)** Given a minimal $m \\times 2018$ rectangle with exactly $k$ columns containing both $0$ and $1$, prove that $m \\leq 2^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider an empty rectangle of size $2^k \\times 2018$ and the $2^k$ binary sequences of length $k$. Write each sequence in the first $k$ columns of a row, leaving the remaining $2018 - k$ columns empty. Each of the first $k$ columns contains exactly $2^{k-1}$ zeros and $2^{k-1}$ ones. This board is completed: for any binary sequence $s = a_1 a_2 \\ldots a_{2018}$, its prefix $s' = a_1 a_2 \\ldots a_k$ appears in exactly one row, so filling the remaining entries yields $s$. If any row is removed, some sequence $s$ cannot be formed, so the board is minimal.\n\n![](table.png)\n\nb) Suppose the rectangle has exactly $k$ columns containing both $0$ and $1$. \n\n**Lemma.** All cells in the last $2018 - k$ columns are empty.\n\n*Proof.* For any binary sequence $s$ of length $k$, let $A_s$ be the set of rows whose first $k$ entries form $s$. In the $(k+1)$-th column, if all entries are filled and identical, some sequence cannot be formed, contradicting completeness. Thus, for each $s$, there is a row where all entries from the $(k+1)$-th to the $2018$-th column are empty. Every row must be of this form, otherwise the board is not minimal. Thus, the last $2018 - k$ columns are entirely empty.\n\nThe sub-board formed by the first $k$ columns is minimal and can generate all $2^k$ binary sequences of length $k$. Therefore, $m \\leq 2^k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19822,
    "subject": "Mathematics (Olympiad)",
    "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 $100$.\n\nDetermine 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.\n\n![](А. С. Голованов)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Does not exist.\n\n**Solution:** 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.\n\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.\n\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": 19823,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $N$ cards on the table. Each card has an integer number written on it. Beto performs the following operation many times: he picks two cards from the table, computes the difference between the numbers that are written on them, writes this difference in his notebook, and then removes those two cards from the table. He can do this operation as many times as he wants, as long as there are at least 2 cards on the table.\n\nIn the end, Beto computes the product of all the numbers written in his notebook. Beto's goal is that this product is divisible by $7^{100}$.\n\nFind the least value of $N$ such that Beto can always achieve his goal, regardless of which numbers are written on the cards.",
    "options": [],
    "answer": "See solution",
    "solution": "To begin, note that as long as there are more than $7$ cards on the table, we can always find two cards with the same remainder upon division by $7$ (since there are only $7$ possible remainders). In that situation, Beto can pick those two cards and write in his notebook the difference, which will be divisible by $7$. Since there are $207$ cards and each operation removes two cards, Beto can repeat this $100$ times without a problem. Hence, after that, there will be $100$ numbers divisible by $7$ written in the notebook, so their product will be divisible by $7^{100}$ as desired.\n\nTherefore, the least value of $N$ is $207$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19824,
    "subject": "Mathematics (Olympiad)",
    "question": "Given points $P, Q$ on the ellipse $$\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$$ with $a > b > 0$, and $OP \\perp OQ$, find the minimum value of $|OP| \\times |OQ|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nP(|OP| \\cos \\theta,\\ |OP| \\sin \\theta), \\quad Q(|OQ| \\cos(\\theta \\pm \\frac{\\pi}{2}),\\ |OQ| \\sin(\\theta \\pm \\frac{\\pi}{2})).\n$$\n\nWe have\n\n$$\n\\frac{1}{|OP|^2} = \\frac{\\cos^2\\theta}{a^2} + \\frac{\\sin^2\\theta}{b^2},\n$$\n\n$$\n\\frac{1}{|OQ|^2} = \\frac{\\sin^2\\theta}{a^2} + \\frac{\\cos^2\\theta}{b^2}.\n$$\n\nAdding,\n\n$$\n\\frac{1}{|OP|^2} + \\frac{1}{|OQ|^2} = \\frac{1}{a^2} + \\frac{1}{b^2}.\n$$\n\nThe minimum of $|OP| \\times |OQ|$ is achieved when $|OP| = |OQ| = \\sqrt{\\frac{2a^2b^2}{a^2 + b^2}}$, so the minimum is\n\n$$\n\\boxed{\\frac{2a^2b^2}{a^2 + b^2}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19825,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_m$ be finite sets such that for any $k$ with $1 \\leq k \\leq m$, the union of any $k$ of the $A_i$'s has at least $k+1$ elements. Prove that it is possible to color the elements of $A = A_1 \\cup A_2 \\cup \\dots \\cup A_m$ with two colors (black and white) so that every $A_i$ contains elements of both colors.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $m$.\n\nWhen $m=1$, $|A_1| \\geq 2$, so we can choose two elements of $A_1$ and color them black and white, respectively. This satisfies the requirement.\n\nAssume the statement holds for $m \\leq n-1$; consider the case $m=n$.\n\n*Case 1:* For any $I \\subsetneq \\{1, \\dots, n\\}$, $|\\bigcup_{i \\in I} A_i| \\geq |I| + 2$. Choose $a \\in A_n$, and apply the inductive hypothesis to the sets $A_1 \\setminus \\{a\\}, \\dots, A_{n-1} \\setminus \\{a\\}$. The elements in $A_n \\setminus \\{a\\}$ are already colored, and we can always choose a color for $a$ so that $A_n$ contains both colors.\n\n*Case 2:* Suppose there exists $I \\subsetneq \\{1, 2, \\dots, n\\}$ such that $|\\bigcup_{i \\in I} A_i| = |I| + 1$. Choose $I$ of maximal size. Let $J = \\{1, 2, \\dots, n\\} \\setminus I$, $B = \\bigcup_{i \\in I} A_i$, and $C = \\bigcup_{j \\in J} A_j$.\n\n*Case 2a:* If $B \\cap C = \\emptyset$, apply the inductive hypothesis to $B$ and $C$ separately; every $A_i$ will have two colors.\n\n*Case 2b:* If $B \\cap C \\neq \\emptyset$, fix $b \\in B \\cap C$, and define $A'_j = A_j \\setminus (B \\setminus \\{b\\})$ for every $j \\in J$. For any $T \\subsetneq J$, the maximality of $I$ implies:\n\n$$\n\\left| \\bigcup_{t \\in T} A'_t \\right| \\geq \\left| \\bigcup_{t \\in I \\cup T} A_t \\right| - \\left| \\bigcup_{i \\in I} A_i \\right| \\geq (|I \\cup T| + 2) - (|I| + 1) = |T| + 1.\n$$\n\nSimilarly,\n\n$$\n\\left| \\bigcup_{j \\in J} A'_j \\right| = \\left| \\bigcup_{i=1}^n A_i \\right| - \\left| B \\setminus \\{b\\} \\right| \\geq n + 1 - |I| = |J| + 1.\n$$\n\nApplying the inductive hypothesis to $C \\setminus (B \\setminus \\{b\\}) = \\bigcup_{j \\in J} A'_j$ and $B = \\bigcup_{i \\in I} A_i$ ensures that each $A'_j$ ($j \\in J$) and $A_i$ ($i \\in I$) has two colors. If $b$ is colored differently in $B$ and $C \\setminus (B \\setminus \\{b\\})$, swap the colors in $B$. Thus, we obtain a coloring of $A$ such that every $A_i$ contains both colors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19826,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and $q$ such that $p$ divides $q + 6$ and $q$ divides $p + 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "$p = 2$ and $p$ divides $q + 6$ together imply $2$ divides $q$. Hence $q = 2$. But then $q$ divides $p + 7$ is not satisfied. Thus $p$ is odd.\n\n$q = 2$ and $p$ divides $q + 6$ together imply $p = 2$ but $p = 2 = q$ contradicts $q$ divides $p + 7$. Thus $p$ and $q$ are both odd.\n\n$p + 7$ is even and so $q \\leq \\frac{p + 7}{2} \\leq \\frac{q + 13}{2}$. Hence $q \\leq 13$.\n\nNow examine each of the cases $q = 3, 5, 7, 11, 13$. Only one case $q = 13$ provides a solution in which $p = 19$.\n\n**Second solution:** By hypothesis, there are $a, b \\in \\mathbb{N}$ such that $q + 6 = pa$, $p + 7 = bq$. Hence\n\n$$\n13 = p(a-1) + q(b-1)\n$$\n\nwhere $a-1 = m$, $b-1 = n$ are nonnegative integers. Suppose $m, n \\in \\mathbb{N}$. Then\n\n$$\n4 \\leq p+q \\leq pm+qn = 13,\n$$\n\nand so $p, q \\in \\{2, 3, 5, 7, 11\\}$. An inspection of the possible pairs $(p, q)$ that can be formed from this set shows that no such pair satisfies the hypotheses. Hence, one of $m, n$ is zero. Suppose $n = 0$. Then $b = 1$, $a = 2$, $p = 13$ which means that $q = 20$, which isn't a prime number. It follows that $m = 0$, i.e., $a = 1$, $b = 2$, $q = 13$ and hence $p = 26 - 7 = 19$. Thus $p = 19$, $q = 13$ is the only solution pair.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19827,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Let $A'$, $B'$, and $C'$ be the reflections of the vertices $A$, $B$, and $C$ with respect to $BC$, $CA$, and $AB$, respectively. Let the circumcircles of triangles $ABB'$ and $ACC'$ meet again at $A_1$. Points $B_1$ and $C_1$ are defined similarly. Prove that the lines $AA_1$, $BB_1$, and $CC_1$ have a common point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_1$, $O_2$, and $O$ be the circumcenters of triangles $ABB'$, $ACC'$, and $ABC$, respectively. As $AB$ is the perpendicular bisector of the line segment $CC'$, $O_2$ is the intersection of the perpendicular bisector of $AC$ with $AB$. Similarly, $O_1$ is the intersection of the perpendicular bisector of $AB$ with $AC$. It follows that $O$ is the orthocenter of triangle $AO_1O_2$. This means that $AO$ is perpendicular to $O_1O_2$. On the other hand, the segment $AA_1$ is the common chord of the two circles, thus it is perpendicular to $O_1O_2$. As a result, $AA_1$ passes through $O$. Similarly, $BB_1$ and $CC_1$ pass through $O$, so the three lines are concurrent at $O$.\n\nWe present here a different approach.\n\nWe first prove that $A_1B$ and $C'$ are collinear. Indeed, since $\\angle BAB' = \\angle CAC' = 2\\angle BAC$, then from the circles $(ABB')$, $(ACC')$ we get\n\n$$\n\\angle AA_1B = \\frac{\\angle BA_1B'}{2} = \\frac{180^\\circ - \\angle BAB'}{2} = 90^\\circ - \\angle BAC = \\angle AA_1C'\n$$\n\nIt follows that\n\n$$\n\\angle A_1AC = \\angle A_1C'C = \\angle BC'C = 90^\\circ - \\angle ABC \\quad (1)\n$$\n\nOn the other hand, if $O$ is the circumcenter of $ABC$, then\n\n$$\n\\angle OAC = 90^\\circ - \\angle ABC \\quad (2)\n$$\n\nFrom (1) and (2) we conclude that $A_1$, $A$, and $O$ are collinear. Similarly, $BB_1$ and $CC_1$ pass through $O$, so the three lines are concurrent at $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19828,
    "subject": "Mathematics (Olympiad)",
    "question": "The chessboard is split into domino tiles, meaning it is divided into $1 \\times 2$ and $2 \\times 1$ rectangles. Each tile has a number written on it equal to the number of tiles that it shares a common line segment with (excluding itself). What is the least possible sum of all numbers written on the chessboard?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 104.\n\n**Solution:**\n\nWe call domino tiles *neighboring* if they share a common line segment of length 1, and *friendly* if they share a common line segment of length 2. Let $S$ denote the sum of all numbers written on the tiles. Then $S$ equals twice the number of friendly pairs ($2F$) plus twice the number of neighboring pairs ($2N$).\n\nThere are 32 domino tiles, so there are exactly 32 unit intervals covered by domino tiles (internal to tiles and not shared). The remaining unit intervals, not at the edge of the chessboard, are shared by two tiles. Their number is $7 \\cdot 8 \\cdot 2 - 32 = 80$.\n\nIt is clear that $2F + N = 80$, since the common segments for friendly pairs have length 2. Thus,\n\n$$\nS = 2F + 2N = (2F + N) + N = 80 + N = 160 - 2F.\n$$\n\nA set of tiles $D_1, \\dots, D_k$, where $D_i$ and $D_{i+1}$ are friendly for $i = 1, \\dots, k-1$, or a separate tile with no friendly tiles, is called a *chain*. Each tile belongs to exactly one chain. Suppose there are $C$ chains, and the $l$-th chain has $k_l$ tiles. Then it has $(k_l - 1)$ friendly pairs. The total number of friendly pairs is $(k_1 - 1) + \\dots + (k_C - 1) = 32 - C = F$. Therefore,\n\n$$\nS = 160 - 2F = 96 + 2C.\n$$\n\nTo minimize $S$, we must minimize the number of chains $C$. Since there are no more than 8 tiles in a chain, $C \\geq 4$. Thus, the least possible sum is $S = 96 + 2C = 104$.\n\nAn example with $C = 4$ is possible by placing all tiles in the same orientation, for instance, all vertical.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19829,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p \\ge 2$ be an integer. Prove that the sequence $(x_n)_{n \\ge 1}$, defined by $x_1 = a > 0$ and the recurrence relation\n$$\nx_{n+1} = x_n + \\left\\lfloor \\frac{p}{x_n} \\right\\rfloor,\n$$\nfor $n \\in \\mathbb{N}^*$, is convergent. Determine its limit depending on the values of the parameter $a$. Here, $[x]$ denotes the integer part of the real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "The sequence $(x_n)_{n \\ge 1}$ has positive terms (by induction). There exists $k \\in \\mathbb{N}^*$ such that $x_k > p$. Suppose, for contradiction, that $x_n \\le p$ for all $n$. Then $x_{n+1} \\ge x_n + 1$ for all $n$, so $x_n \\ge a + n - 1$. In particular, $x_{p+1} \\ge a + p > p$, a contradiction.\n\nLet $k_0 = \\min\\{k \\in \\mathbb{N}^* \\mid x_k > p\\}$. Since $\\left\\lfloor \\frac{p}{x} \\right\\rfloor = 0$ for all $x > p$, we have $x_n = x_{k_0}$ for all $n \\ge k_0$. Thus, $(x_n)$ is convergent with $\\lim_{n \\to \\infty} x_n = x_{k_0}$.\n\nFor the value of the limit, consider 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 $x_2 = a + \\left\\lfloor \\frac{p}{a} \\right\\rfloor > \\left\\lfloor \\frac{p}{a} \\right\\rfloor \\ge p$, so $\\lim_{n \\to \\infty} x_n = x_2 = a + \\left\\lfloor \\frac{p}{a} \\right\\rfloor$.\n\n*Case 3.* $a \\in [1, p]$. Write $a = [a] + \\{a\\}$, where $\\{a\\} = a - [a] \\in [0, 1)$ is the fractional part. For $x \\in [1, p]$, $(x-1)(x-p) \\le 0$, so $x + \\left\\lfloor \\frac{p}{x} \\right\\rfloor \\le x + \\frac{p}{x} \\le p+1$. Thus, if $x_n \\in [1, p]$, then $x_{n+1} \\le p+1$. Therefore, $x_{k_0} \\in (p, p+1] \\cap \\{\\{a\\} + k \\mid k \\in \\mathbb{N}^*\\}$.\n\nHence,\n$$\n\\lim_{n \\to \\infty} x_n = x_{k_0} =\n\\begin{cases}\np + \\{a\\}, & a \\in [1, p] \\setminus \\mathbb{N} \\\\\np+1, & a \\in \\{1, 2, \\dots, p\\}\n\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19830,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the total number of all permutations of the $n$-tuple $(1, 2, \\dots, n)$ satisfying $i + a_i \\leq j + a_j$ for all $1 \\leq i < j \\leq n$. What is $f(2013)$?",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that $f(1) = 1$. For any permutation where $a_1 = n$, the only possibility is $(n, n-1, \\dots, 1)$. More generally, if $a_{k+1} = n$ for some $k$, then $a_{k+2} = n-1$, $a_{k+3} = n-2$, and so on, so $(a_1, \\dots, a_k)$ must be a permutation of $(1, 2, \\dots, k)$ satisfying the same condition. Thus, $f(n) = 1 + f(1) + f(2) + \\dots + f(n-1)$, which leads to $f(n) = 2f(n-1)$. Therefore, $f(n) = 2^{n-1}$, so $f(2013) = 2^{2012}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19831,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $l$ a line that meets $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $A'$ be the midpoint of the segment connecting the projections of $A_1$ onto $AB$ and $AC$. Construct $B'$ and $C'$ analogously.\n\n(a) Show that the points $A'$, $B'$, and $C'$ are collinear on some line $l'$.\n\n(b) Show that if $l$ contains the circumcenter of triangle $ABC$, then $l'$ contains the center of its Euler circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AH_a$ be an altitude in triangle $ABC$ and $P_a$ its midpoint. Define $H_b$, $P_b$, etc., analogously.\n\nIt is easy to see that $A'$ divides the segment $P_bP_c$ in the same ratio that $A_1$ divides $BC$. By Menelaus' theorem for triangle $P_aP_bP_c$, claim (a) follows.\n\nConsider an affine transformation mapping triangle $ABC$ onto triangle $P_aP_bP_c$. When $l$ contains a fixed point $X$, $l'$ contains the fixed point $Y$ whose affine coordinates with respect to $P_aP_bP_c$ equal those of $X$ with respect to $ABC$. We are now left to show that $X \\equiv O \\Leftrightarrow Y \\equiv O_9$.\n\nThis is easiest to do by considering two special cases, for example, when $l$ contains a vertex of $ABC$.\n\nAnother approach: Let $Z$ be the point whose affine coordinates with respect to $H_aH_bH_c$ equal those of $X$ with respect to $ABC$. Clearly, $Y$ is the midpoint of $XZ$. Let $O'$ be the circumcenter of $H_aH_bH_c$. It is clear that $AO'HH_a$ is a straight line and by similar figures $AH_bH_cO' \\sim ABCO$, we see that $AO$ divides $BC$ and $H_aH$ divides $H_bH_c$ in equal ratios. It follows that $X=O \\Leftrightarrow Z=H$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19832,
    "subject": "Mathematics (Olympiad)",
    "question": "由正整數 $37$ 開始，依序在各項的前方加一數字 $5$，形成下面的數列：\n\n$37$, $537$, $5537$, $55537$, $555537$, $\\ldots$\n\n請問此數列中有多少項是質數？",
    "options": [],
    "answer": "See solution",
    "solution": "此數列只有第 $1$ 項是質數，其他項均為合數。\n\n將此數列的第 $n$ 項記為 $a_n$。由數學歸納法可知下列事實：\n\n- $a_1$ 被 $37$ 整除。$a_{n+3} = 555 \\cdot 10^{n+1} + a_n$，而 $555 = 3 \\cdot 5 \\cdot 37$。所以 $a_1, a_4, a_7, \\ldots$ 均為 $37$ 的倍數。\n- $a_2 = 537$ 被 $3$ 整除。$a_{n+3} = 555 \\cdot 10^{n+1} + a_n$，而 $555 = 3 \\cdot 5 \\cdot 37$。所以 $a_2, a_5, a_8, \\ldots$ 均為 $3$ 的倍數。\n- $a_3 = 5537 = 7 \\cdot 791$ 是 $7$ 的倍數。$a_{n+6} = 555555 \\cdot 10^{n+1} + a_n$，而 $555555 = 555 \\cdot 1001$ 是 $7$ 的倍數。所以 $a_3, a_9, a_{15}, \\ldots$ 均為 $7$ 的倍數。\n- $a_6 = 5555537 = 13 \\cdot 427349$ 是 $13$ 的倍數。$a_{n+6} = 555555 \\cdot 10^{n+1} + a_n$，而 $555555 = 555 \\cdot 1001$ 是 $13$ 的倍數，所以 $a_6, a_{12}, a_{18}, \\ldots$ 均為 $13$ 的倍數。\n\n綜上所述，可知只有 $a_1 = 37$ 是質數，其他各項均被 $7$, $13$, $37$ 其中之一整除，故為合數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19833,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an odd natural number ($n \\ge 3$). Each unit square of an $n \\times n$ grid is colored either red or blue. Two unit squares are said to be *adjacent* if they are of the same color and have at least one common vertex. Two unit squares $a, b$ are said to be *connected* if there exists a sequence of unit squares $c_1, c_2, \\dots, c_k$ with $c_1 = a$ and $c_k = b$ such that $c_i$ and $c_{i+1}$ are adjacent for every $i = 1, \\dots, k-1$; otherwise, they are called *disconnected* (for instance, two unit squares of different colors are disconnected). Find the maximal number $M$ for which there exists a coloring admitting $M$ pairwise disconnected unit squares.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $M = \\frac{1}{4}(n+1)^2 + 1$.\n\nConsider the generalized problem on an $m \\times n$ grid, where $m, n \\ge 3$ are both odd numbers. Suppose that the $mn$ squares can be divided into $K$ connected components, such that any two squares are connected if and only if they belong to the same connected component. We will prove by induction on $(m, n)$ that:\n\n1. $K \\le \\frac{1}{4}(m+1)(n+1) + 1$;\n2. If $K = \\frac{1}{4}(m+1)(n+1) + 1$, then each square at the four corners of the grid is connected to none of the other squares.\n\nWhen $m = n = 3$, the 8 border squares belong to at most 4 connected components. Hence $K \\le 5$, and $K = 5$ if and only if the 4 squares at the corners are of the same color, and the other 5 squares are of the other color.\n\nNext, assume $m \\ge 5$. Suppose the second row of the grid can be divided into $k$ parts, each part consisting of consecutive squares of the same color. Let $x_i$ be the number of squares in the $i$th part, $1 \\le i \\le k$. Let $P$ be the number of connected components which contain at least one square in the first row, but no square in the second row. If $k \\ge 2$ then\n\n$$\nP \\le \\left\\lfloor \\frac{x_1 - 1}{2} \\right\\rfloor + \\left\\lfloor \\frac{x_k - 1}{2} \\right\\rfloor + \\sum_{i=2}^{k-1} \\left\\lfloor \\frac{x_i - 2}{2} \\right\\rfloor \\le \\frac{n-k+2}{2}.\n$$\n\nIf $k = 1$, we also have $P \\le \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\le \\frac{n-k+2}{2}$. Let $Q$ be the number of connected components which contain at least one square in the second row, but no square in the third row. We have $Q \\le \\left\\lfloor \\frac{k}{2} \\right\\rfloor \\le \\frac{k+1}{2}$. Let $R$ be the number of connected components which contain at least one square in the 3rd to $m$th rows. By the induction hypothesis (1), we have $R \\le \\frac{1}{4}(m-1)(n+1) + 1$.\n\nIf $Q = \\frac{k+1}{2} = 1$ then all the squares in the second row are of the same color and thus all the squares in the 3rd row are also of the same color. If $Q = \\frac{k+1}{2} \\ge 2$, then the first square in the 3rd row is connected to the last square in the 3rd row via the 2nd part of the second row. By induction hypothesis (2), $R \\le \\frac{1}{4}(m-1)(n+1)$. Hence $Q = \\frac{k+1}{2}$ and $R = \\frac{1}{4}(m-1)(n+1) + 1$ can't hold simultaneously. Therefore, we have\n\n$$\nK = P + Q + R \\le \\frac{n-k+2}{2} + \\frac{k+1}{2} + \\frac{1}{4}(m-1)(n+1) = \\frac{1}{4}(m+1)(n+1) + 1.\n$$\n\nIf the equality in the above inequality holds, then we have $P = \\frac{n-k+2}{2}$. Thus, the square at the upper-left corner is surrounded by three squares of opposite color. By symmetry, each square at the four corners of the grid is connected to none of the other squares.\n\nIf the square at position $(i, j)$ is colored red/blue if $ij$ is even/odd, respectively, it is easy to verify that $K = \\frac{1}{4}(m+1)(n+1) + 1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19834,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $a_1, a_2, a_3, \\dots$ defined by $a_1 = 9$ and\n$$\na_{n+1} = \\frac{(n + 5)a_n + 22}{n + 3}\n$$\nfor $n \\geq 1$. Find all positive integers $n$ for which $a_n$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove by induction on $n$ that $a_n = n^2 + 7n + 1$.\n\nThe base case is trivial. Assume that $a_k = k^2 + 7k + 1$, so\n$$\na_{k+1} = \\frac{(k + 5)a_k + 22}{k + 3} = \\frac{(k + 5)(k^2 + 7k + 1) + 22}{k + 3} = (k + 1)^2 + 7(k + 1) + 1\n$$\n\nWe distinguish two cases regarding the value of $n$:\n\n* If $n \\geq 9$ then $(n + 3)^2 < n^2 + 7n + 1 < (n + 4)^2$, so $a_n$ is not a perfect square.\n* If $n \\leq 8$ then we have:\n\n| $a_1 = 9$ | $a_2 = 19$ | $a_3 = 31$ | $a_4 = 45$ |\n|-----------|-----------|-----------|-----------|\n| $a_5 = 61$ | $a_6 = 79$ | $a_7 = 99$ | $a_8 = 121$ |\n\nHence, $a_n$ is a perfect square if and only if $n = 1, 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19835,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following equation:\n\n$$\n|[x]| = [|x|],\n$$\n\nwhere $[a]$ stands for the greatest integer that does not exceed $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider three cases:\n\n1) $x \\ge 0$. Then $|x| = x$, $[x] \\ge 0$, so $|[x]| = [x] = [|x|]$. Thus, any non-negative $x$ is a solution.\n\n2) $x$ is a negative integer. Then $[x] = x$, $|x| = -x$, $[|x|] = [-x] = -x$, and $|[x]| = |-n| = n = [|x|]$. So, negative integers are also solutions.\n\n3) $x$ is negative but not an integer. Then $[x] < x < 0$, $|x| = -x$, $[|x|] = [-x]$. Since $[x]$ is a negative integer, $|[x]| = -[x]$, but $[|x|] = [-x] = -n - 1$ if $x = -n - \theta$ for $0 < \theta < 1$. Thus, $|[x]| \\ne [|x|]$ in this case.\n\n**Conclusion:** The solutions are all $x \\ge 0$ and all negative integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19836,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一正整數 $n \\ge 3$。我們稱一實數 $n$-序對 $(x_1, x_2, \\cdots, x_n)$ 為閃耀的，若對 $x_1, x_2, \\cdots, x_n$ 的每一個排列 $y_1, y_2, \\cdots, y_n$，皆滿足\n\n$$\n\\sum_{i=1}^{n-1} y_i y_{i+1} = y_1 y_2 + y_2 y_3 + y_3 y_4 + \\cdots + y_{n-1} y_n \\ge -1.\n$$\n\n試求最大的數 $K = K(n)$ 使得對每一個閃耀的 $n$-序對 $(x_1, x_2, \\cdots, x_n)$，\n\n$$\n\\sum_{1 \\le i < j \\le n} x_i x_j \\ge K\n$$\n\n均成立。",
    "options": [],
    "answer": "See solution",
    "solution": "**解答**\n\n首先，我們證明 $K$ 不能取更大的值。令 $t$ 為正數，取\n\n$$\nx_2 = x_3 = \\cdots = x_n = t, \\quad x_1 = -\\frac{1}{2t}.\n$$\n\n則每個 $x_i x_j$（$i \\ne j$）等於 $t^2$ 或 $-\\frac{1}{2}$。對於任意排列 $y_i$，有\n\n$$\ny_1 y_2 + \\cdots + y_{n-1} y_n \\ge (n-3)t^2 - 1 \\ge -1.\n$$\n\n因此 $(x_1, \\cdots, x_n)$ 是閃耀的。計算\n\n$$\n\\sum_{i<j} x_i x_j = -\\frac{n-1}{2} + \\frac{(n-1)(n-2)}{2} t^2.\n$$\n\n當 $t \\to 0^+$ 時，$\\sum_{i<j} x_i x_j$ 趨近於 $-\\frac{n-1}{2}$。所以 $K$ 不能大於 $-\\frac{n-1}{2}$。\n\n接下來證明對任意閃耀的 $n$-序對 $(x_1, \\cdots, x_n)$，都有\n\n$$\n\\sum_{i<j} x_i x_j \\ge -\\frac{n-1}{2}.\n$$\n\n設 $z_i$ 為 $x_i$ 的某個排列，指標模 $n$。將 $\\sum_{i<j} x_i x_j = \\sum_{i<j} z_i z_j$ 拆分為 $[(n-1)/2]$ 個形如 $y_1 y_2 + \\cdots + y_{n-1} y_n$ 的表達式和剩餘項 $L$。具體地，\n\n$$\n\\sum_{i<j} z_i z_j = \\sum_{p=1}^{[(n-1)/2]} \\sum_{\\substack{i+j=2p-1,2p \\pmod{n} \\\\ i \\neq j \\pmod{n}}} z_i z_j + L,\n$$\n\n其中\n\n$$\nL = z_1 z_{-1} + z_2 z_{-2} + \\cdots + z_{(n-1)/2} z_{-(n-1)/2} \\quad (n \\text{ 奇數})\n$$\n\n$$\nL = z_1 z_{-1} + z_2 z_{-2} + \\cdots + z_{(n-2)/2} z_{-n/2} \\quad (n \\text{ 偶數})\n$$\n\n每個 $p$ 都存在某個排列 $y_i$ 使得\n\n$$\n\\sum_{\\substack{i+j \\equiv 2p-1, 2p \\pmod{n} \\\\ i \\neq j \\pmod{n}}} z_i z_j = \\sum_{k=1}^{n-1} y_k y_{k+1}\n$$\n\n% ![](images/18-2J_p12_data_b74932200f.png)\n\n由於 $z_i$ 是閃耀的，得到\n\n$$\n\\sum_{i<j} z_i z_j \\geq -\\left[\\frac{n-1}{2}\\right] + L.\n$$\n\n只需證明對每個 $n$，存在排列 $z_i$ 使得 $L \\ge 0$（$n$ 奇數）或 $L \\ge -1/2$（$n$ 偶數）。假設\n\n$$\nx_1 \\le x_2 \\le \\dots \\le x_k \\le 0 \\le x_{k+1} \\le \\dots \\le x_n.\n$$\n\n*情況 1：$n$ 奇數*\n\n假設 $k$ 為偶數（若 $k$ 為奇數可取相反數），則\n\n$$\nx_1 x_2, x_3 x_4, \\dots, x_{n-2} x_{n-1} \\ge 0\n$$\n\n因為同號相乘。令\n\n$$\nL = x_1 x_2 + x_3 x_4 + \\dots + x_{n-2} x_{n-1} \\ge 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19837,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BI \\cap (ABC) = \\{B, S\\}$, and $SB' \\cap CA = P$. (See the figure below.)\n\n![](./images/figure14.png)\n\nGiven this configuration, find the required meeting point $K$ of the two tangents to the circle $(BIB')$ from points on $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle AT_S = \\angle IBB' = \\angle IB'B$. Thus, $SB' \\cdot ST = SA^2 = SI^2$, so the circle $\\gamma = (T_i \\bar{I} B')$ is tangent to $SI$. This implies $\\angle ITB' = 2\\phi$, so $TA$ is the angle bisector of $\\angle B'TI$. Denoting $\\gamma \\cap AC = \\{T, K\\}$, we find that $K$ is the required meeting point of the two tangents to $(BIB')$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19838,
    "subject": "Mathematics (Olympiad)",
    "question": "在一個 $100 \\times 100$ 格西洋棋盤的每一格內填入一個非負實數。我們稱這個棋盤是平衡的，若且唯若該棋盤上每一直排的數字總和都是 $1$，且每一橫列的數字總和也都是 $1$。\n\n求最大的正實數 $x$，使得在任何一個平衡棋盤中，我們都能挑出 $100$ 個格子，滿足任兩個格子都在不同直排且不同橫列，並且這些格子裡的數字都不小於 $x$。",
    "options": [],
    "answer": "See solution",
    "solution": "最大的 $x$ 為 $\\frac{1}{50 \\times 51}$。\n\n一般性的，對於 $n \\times n$ 的棋盤，令 $a = \\lfloor \\frac{n+1}{2} \\rfloor$，$b = \\lfloor \\frac{n+2}{2} \\rfloor$，則最大的 $x$ 為 $\\frac{1}{ab}$。\n\n首先構造達到此 $x$ 的棋盤。將棋盤分為四區：\n\n- 左上角的 $a \\times b$ 格都填 $\\frac{1}{ab}$；\n- 右上角的 $(n-a) \\times b$ 格都填 $\\frac{1}{b}$；\n- 左下角的 $a \\times (n-b)$ 格都填 $\\frac{1}{a}$；\n- 右下角的 $(n-a) \\times (n-b)$ 格都填 $0$。\n\n可直接驗證此棋盤平衡（注意 $a+b=n+1$）。依據題目所要求的不同行不同列條件，取 $n$ 個格子若不取到右下角的 $(n-a) \\times (n-b)$ 格，則必然會需要取到左上角的 $a \\times b$ 格，故 $x$ 至少要為 $\\frac{1}{ab}$。\n\n以下證明 $x = \\frac{1}{ab}$ 已足夠。給定一個平衡棋盤，將所有值 $\\ge \\frac{1}{ab}$ 的格子塗黑，其餘格子塗白。希望證明：對於任意 $p$ 行，這 $p$ 行上的黑格子總共會出現在至少 $p$ 個不同列。若此性質成立，用黑格子將對應的行列匹配下，由 Hall 定理，行列之間存在完美匹配，也就是可以找到不同行列的 $n$ 個黑格子，既證畢。\n\n要證明此一性質，對於任意 $p$ 行，假設這 $p$ 行上的黑格子出現在 $q$ 個不同列上。考慮在這 $p$ 行但不在這 $q$ 列的格子，則它們必然都是白格子，從而這些格子的數字總和 $< \\frac{1}{ab}p(n-q)$。又由於每一行的數字總合為 $1$，在這 $p$ 行且在這 $q$ 列的格子數字總和 $> p - \\frac{1}{ab}p(n-q)$。但注意到這 $q$ 列的數字和為 $q$，且此數字必不小於在這 $p$ 行且在這 $q$ 列的格子數字總和，故\n\n$$\nq > p - \\frac{1}{ab}p(n-q) \\Rightarrow q > \\frac{abp - pn}{ab - p}.\n$$\n\n又易知\n\n$$\nab \\ge p(n+1-p) \\Rightarrow \\frac{abp - pn}{ab - p} \\ge p - 1,\n$$\n\n故有 $q > p - 1 \\Rightarrow q \\ge p$。這便證明了欲證之性質，從而證明原命題。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19839,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $x[x] = 2016$.\n\nHere $[x]$ denotes the integer part of the number $x$, i.e., the largest integer not greater than $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $x \\ge 0$. Let $t = [x] \\ge 0$. Then:\n\n$$\nt \\le x < t+1 \\implies t^2 \\le x[x] < t^2 + t.\n$$\n\nSince $44^2 + 44 = 1980 < 2016 < 45^2 = 2025$, there are no solutions for $x \\ge 0$.\n\nNow suppose $x < 0$. Let $t = [x] < 0$. Then:\n\n$$\nt \\le x < t+1 \\implies t^2 + t \\le x[x] \\le t^2.\n$$\n\nFor $t = -45$:\n\n$$\n(-45)^2 + (-45) = 2025 - 45 = 1980 < 2016 < 2025 = (-45)^2.\n$$\n\nLet $x = -45 + y$, where $0 \\le y < 1$.\n\n$$\nx[x] = (-45 + y)(-45) = 2025 - 45y = 2016 \\\\\n45y = 2025 - 2016 = 9 \\\\\ny = \\frac{1}{5}\n$$\n\nThus, the solution is $x = -45 + \\frac{1}{5} = -\\frac{224}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19840,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exist 2024 distinct positive integers satisfying the following: If we consider every possible ratio between two distinct numbers (we include both $a/b$ and $b/a$), we will obtain numbers with finite decimal expansions (after the decimal point) of mutually distinct non-zero lengths.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show these numbers exist. Define sequences $a_1, a_2, \\dots, a_{2024}$ and $b_1, b_2, \\dots, b_{2024}$, and consider numbers $c_i = 2^{a_i} \\cdot 5^{b_i}$ for $i = 1, 2, \\dots, 2024$.\n\nChoose $a_i$ increasing, $b_i$ decreasing, and ensure all differences $a_i - a_j$ and $b_j - b_i$ (for $i > j$) are mutually distinct. This works because\n\n$$\n\\frac{c_i}{c_j} = \\frac{2^{a_i} \\cdot 5^{b_i}}{2^{a_j} \\cdot 5^{b_j}} = \\frac{2^{a_i - a_j}}{5^{b_j - b_i}},\n$$\n\nwhich has a decimal expansion of length $b_j - b_i$, while $\\frac{c_j}{c_i}$ has length $a_i - a_j$.\n\nConstruct $a_i$ inductively: take $a_1 = 1$, $a_2 = 2$, and for each $i$, set $a_{i+1} = 2a_i$ so $a_{i+1} - a_i > a_i - a_1$, ensuring all new differences are larger than previous ones.\n\nConstruct $b_i$ similarly, starting from the end: $b_{2024} = a_{2024}$, $b_{2023} = 2b_{2024}$, etc. Since $b_{2023} - b_{2024} = a_{2024}$, all $b_i$ differences are at least $a_{2024}$.\n\n![alt](path \"title\")\n\n**Remark:** In this construction, $a_i = 2^{i-1}$ and $b_i = 2^{4049-i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19841,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Let $d = \\det(X)$ and $t = \\operatorname{tr}(X)$. From $X^{2023} = X^{2022}$, we have $d^{2023} = d^{2022}$, so $d \\in \\{0, 1\\}$. \n\nIf $d = 1$, $X$ is invertible, so $X^{2022}$ is invertible and $X = I_2$. Thus, $X^3 = X^2$ holds. \n\nIf $d = 0$, by the Cayley-Hamilton theorem, $X^2 = tX$. Then $X^{n+1} = t^n X$ for all $n \\in \\mathbb{N}^*$. So $t^{2022} X = t^{2021} X$, which gives $t = 0$, $t = 1$, or $X = O_2$. \n\nIf $t = 0$, $X^2 = O_2$, so $X^3 = X^2 = O_2$. If $t = 1$, $X^2 = X$, so $X^3 = X^2$. If $X = O_2$, clearly $X^3 = X^2 = O_2$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 19842,
    "subject": "Mathematics (Olympiad)",
    "question": "All numbers $1$ through $13$ are written in the circles of the snowflake in such a way that the sum of the five numbers on each line and the sum of the middle seven numbers are all equal. Find this sum if it is known that it is the smallest possible.\n\n![](images/prob1617_p11_data_c4879fdb80.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sum of each ray be $s$ and let the sum in the middle circle be $a$. Then\n\n![](images/prob1617_p11_data_d20505cd22.png)\n\n$$\n3s = (1 + 2 + \\dots + 13) + 2a = 91 + 2a,\n$$\n\nwhence $s = \\frac{91+2a}{3} \\geq \\frac{93}{3} = 31$. Figure 8 shows that $s = 31$ is possible.\n\nFig. 8",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19843,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\nS_n + a_n = \\frac{n-1}{n(n+1)}, \\quad n = 1, 2, \\dots\n$$\n\nThen $a_n = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "As\n\n$$\na_{n+1} = S_{n+1} - S_n\n$$\n\nand\n\n$$\nS_{n+1} + a_{n+1} = \\frac{n}{(n+1)(n+2)},\n$$\n\nwe have\n\n$$\n\\begin{aligned}\na_{n+1} &= S_{n+1} - S_n \\\\\n&= \\frac{n}{(n+1)(n+2)} - a_{n+1} - \\frac{n-1}{n(n+1)} + a_n.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n2a_{n+1} = \\frac{-2}{(n+1)(n+2)} + a_n + \\frac{1}{n(n+1)}.\n$$\n\nSo,\n\n$$\na_{n+1} + \\frac{1}{(n+1)(n+2)} = \\frac{1}{2} \\left( a_n + \\frac{1}{n(n+1)} \\right).\n$$\n\nDefine $b_n = a_n + \\frac{1}{n(n+1)}$. Then $b_{n+1} = \\frac{1}{2} b_n$, so $b_n = \\frac{1}{2^{n-1}} b_1$. Since $S_1 + a_1 = 0$, $a_1 = 0$, so $b_1 = \\frac{1}{2}$. Thus $b_n = \\frac{1}{2^n}$. Therefore,\n\n$$\na_n = b_n - \\frac{1}{n(n+1)} = \\frac{1}{2^n} - \\frac{1}{n(n+1)}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19844,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the first written word be $(1, 2, \\dots, n)$. By induction on $n$, prove that the number of written words is $n!$.",
    "options": [],
    "answer": "See solution",
    "solution": "R.L. starts with $(1, 2)$ and then reverses it to write $(2, 1)$. Suppose that for $n = k-1$ the words\n\n$$\nW_1^{k-1}, W_2^{k-1}, \\dots, W_{(k-1)!}^{k-1}\n$$\n\nare written in the notebook. Consider the sequence of words\n\n$$\nW_1^k, W_2^k, \\dots\n$$\n\nfor $n = k$. Using the notation $(a, b, \\dots, p) * k$, divide the sequence above into blocks of $2k$ consecutive words. The $m$-th block is\n\n$$\nW_{2m-1}^{k-1} * k, \\dots, W_{2m}^{k-1} * k\n$$\n\nThe first block starts with $W_1^{k-1} * k$ and ends with $W_2^{k-1} * k$, the second block starts with $W_3^{k-1} * k$ and ends with $W_4^{k-1} * k$, and so on. Inside each block, each $2l+1$-th word is obtained from the $2l-1$-th word by shortest counter-clockwise rotation, and each $2l+2$-th word from the $2l$-th word by shortest clockwise rotation. Thus, all words in each block are different and each block is closed under rotation. Since all words in the sequence for $n = k-1$ are different, all words in the sequence for $n = k$ are also different. By induction, there are $(k-1)!$ terms in the previous sequence, so there are $(k-1)!/2$ blocks in the new sequence, and the sequence consists of $(k-1)!/2 \\cdot 2k = k!$ distinct words.\n\n*Note*: In each block, the second word is obtained by reversing the first $k$ letters, the third by reversing the first $k-1$ letters, the fourth by reversing the first $k$ letters, and so on. The first word of each block (except the very first word $(1, 2, \\dots, n)$) is obtained from the last word of the previous block by reversing the first $k-2$ letters. Thus, R.L. uses only rotations of sizes $k$, $k-1$, and $k-2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19845,
    "subject": "Mathematics (Olympiad)",
    "question": "A set $S$ of integers is *Balearic* if there are two (not necessarily distinct) elements $s, s' \\in S$ whose sum $s + s'$ is a power of two; otherwise, it is called a *non-Balearic* set.\n\nFind an integer $n$ such that $\\{1, 2, \\ldots, n\\}$ contains a 99-element non-Balearic set, whereas all the 100-element subsets are Balearic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(n)$ denote the largest cardinality of a non-Balearic set in $\\{1, 2, \\ldots, n\\}$. One easily verifies that $f(0) = f(1) = 0$.\n\nNow consider an integer $n \\geq 2$ and write it as $n = 2^a + b$ with $0 \\leq b \\leq 2^a - 1$. We want to show\n$$\nf(n) = f(2^a + b) = f(2^a - b - 1) + b.\n$$\nPartition $\\{1, 2, \\ldots, n\\}$ into $X = \\{1, 2, \\ldots, 2^a - b - 1\\}$ and $Y = \\{2^a - b, \\ldots, 2^a + b\\}$. A non-Balearic subset $S$ of $\\{1, 2, \\ldots, n\\}$ contains at most $f(2^a - b - 1)$ elements from $X$ (by definition of $f$) and at most $b$ elements from $Y$ (as it cannot contain $2^a$ altogether, and it contains at most one of the two numbers $2^a - x$ and $2^a + x$). This establishes the first inequality $f(n) \\leq f(2^a - b - 1) + b$.\n\nNext, consider a non-Balearic set $T \\subseteq X$ of cardinality $f(2^a - b - 1)$. We claim that $S = T \\cup \\{2^a + 1, \\ldots, 2^a + b\\}$ is also a non-Balearic set. Suppose for contradiction that the sum $s + s'$ of some $s, s' \\in S$ is a power of two. Then $s, s' \\in T$ is impossible, as $T$ is non-Balearic. Also, $s, s' \\in \\{2^a + 1, \\ldots, 2^a + b\\}$ is impossible, as\n$$\n2^{a+1} < (2^a + 1) + (2^a + 1) \\leq s + s' < (2^a + b) + (2^a + b) < 2^{a+2}.\n$$\nHence, one of $s$ and $s'$ must be in $T$ and the other in $\\{2^a + 1, \\ldots, 2^a + b\\}$, which yields the final contradiction\n$$\n2^a < s + s' \\leq (2^a - b - 1) + (2^a + b) < 2^{a+1}.\n$$\nSince the constructed non-Balearic set $S$ is of cardinality $f(2^a - b - 1) + b$, we have established the second inequality $f(n) \\geq f(2^a - b - 1) + b$. The two inequalities together imply the desired recursive equation $f(n) = f(2^a - b - 1) + b$.\n\nThe rest is computation.\n\nIt is easy to see (or determine through the recursion) that $f(4) = 1$.\n\nFor $2^3 = 8$ and $b = 3$, the recursion yields $f(8 + 3) = f(4) + 3 = 4$.\n\nFor $2^5 = 32$ and $b = 20$, the recursion yields $f(32 + 20) = f(11) + 20 = 24$.\n\nFor $2^7 = 128$ and $b = 75$, the recursion yields $f(128 + 75) = f(52) + 75 = 99$.\n\nHence, an answer to the problem is $n = 203$ with $f(203) = 99$.\n\n(Similar computations yield $f(202) = 98$ and $f(204) = 100$. Thus, $n = 203$ is the unique possible answer for the problem.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19846,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest positive integer $k$ such that for all positive real numbers $a, b, c$ with $abc = 1$, the following inequality holds:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{k}{a+b+c+1} \\ge \\frac{k}{4} + 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The greatest $k$ is $13$.\n\nFirstly, consider $a = b = \\frac{2}{3}$ and $c = \\frac{9}{4}$, which satisfy $abc = 1$. The inequality becomes\n\n$$\n\\frac{3}{2} + \\frac{3}{2} + \\frac{4}{9} + \\frac{12k}{55} \\ge \\frac{k}{4} + 3,\n$$\n\ni.e. $\\frac{4}{9} \\ge \\frac{7k}{220}$. As $k$ is a positive integer, this implies $k \\le 13$.\n\nWe now prove the inequality for $k = 13$:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1} \\ge \\frac{13}{4} + 3 = \\frac{25}{4}.\n$$\n\nLet the left-hand side be $f(a, b, c)$. WLOG assume $c \\ge b \\ge a$. We first prove that\n\n$$\nf(a, b, c) \\ge f(\\sqrt{ab}, \\sqrt{ab}, c).\n$$\n\nNote that the new triple still satisfies $(\\sqrt{ab})(\\sqrt{ab})(c) = 1$. Now,\n\n$$\n\\begin{align*}\nf(a, b, c) &\\ge f(\\sqrt{ab}, \\sqrt{ab}, c) \\\\\n\\Leftrightarrow \\quad & \\frac{1}{a} + \\frac{1}{b} - \\frac{2}{\\sqrt{ab}} \\ge \\frac{13}{2\\sqrt{ab} + c + 1} - \\frac{13}{a + b + c + 1} \\\\\n\\Leftrightarrow \\quad & \\frac{(\\sqrt{a} - \\sqrt{b})^2}{ab} \\ge \\frac{13(\\sqrt{a} - \\sqrt{b})^2}{(2\\sqrt{ab} + c + 1)(a + b + c + 1)}.\n\\end{align*}\n$$\n\nIt suffices to show $(2\\sqrt{ab}+c+1)(a+b+c+1) \\ge 13ab$. Since $a+b \\ge 2\\sqrt{ab}$, we only need to prove\n\n$$\n2\\sqrt{ab} + c + 1 \\ge \\sqrt{13ab}.\n$$\n\nAs $c \\ge b \\ge a$ and $abc = 1$, we have $c \\ge 1$. Therefore, $\\sqrt{c}(c+1) \\ge 2$. This implies\n\n$$\nc + 1 \\ge \\frac{2}{\\sqrt{c}} = 2\\sqrt{ab} > (\\sqrt{13} - 2)\\sqrt{ab}.\n$$\n\nThis proves $f(a, b, c) \\ge f(\\sqrt{ab}, \\sqrt{ab}, c)$.\n\nAfter this mixing step, it remains to consider the case $f\\left(t, t, \\frac{1}{t^2}\\right)$ where $t > 0$. Now,\n\n$$\n\\begin{align*}\nf\\left(t, t, \\frac{1}{t^2}\\right) &\\ge \\frac{25}{4} \\\\\n\\Leftrightarrow \\quad & \\frac{2}{t} + t^2 + \\frac{13t^2}{2t^3 + t^2 + 1} \\ge \\frac{25}{4} \\\\\n\\Leftrightarrow \\quad & 8t^6 + 4t^5 - 50t^4 + 47t^3 + 8t^2 - 25t + 8 \\ge 0 \\\\\n\\Leftrightarrow \\quad & (t-1)^2(8t^4 + 20t^3 - 18t^2 - 9t + 8) \\ge 0.\n\\end{align*}\n$$\n\nIt remains to prove $8t^4 + 20t^3 - 18t^2 - 9t + 8 \\ge 0$ for $t > 0$. Indeed, the left-hand side is equal to\n\n$$\n(8t^2 + 30t) \\left(t - \\frac{5}{8}\\right)^2 + \\frac{1}{32}(524t^2 - 663t + 256).\n$$\n\nThe first term is nonnegative since $t > 0$. Since $663^2 - 4(524)(256) < 0$, the last term is nonnegative. This proves the inequality and we are done.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19847,
    "subject": "Mathematics (Olympiad)",
    "question": "The equation\n\n$$\na^2 + b^2 = Q(a + b) + R\n$$\n\nis given, with $R < a + b$. Also,\n\n$$\n(a+b)^2 \\le 2(a^2+b^2) < 2(Q+1)(a+b),\n$$\n\nwhich implies $a + b < 2(Q + 1)$ and $R < 2(Q + 1)$. Furthermore,\n\n$$\nQ^2 \\le Q^2 + R \\le Q^2 + 2Q + 1 = (Q + 1)^2.\n$$\n\nGiven $Q^2 + R = 2010$, find all positive integer solutions $(a, b)$ to\n\n$$\na^2 + b^2 = 44(a + b) + 74.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = (a - 22)^2$ and $y = (b - 22)^2$. Without loss of generality, assume $x \\ge y$. Then $x + y = 1042$, so $521 \\le x \\le 1042$. Thus, $23^2 \\le x \\le 32^2$. Also, $x$ and $y$ must be odd since $x^2 + y^2 \\equiv 2 \\pmod{4}$. Compute possible values:\n\n| $x$ | $529$ | $625$ | $729$ | $841$ | $961$ |\n|-----|-------|-------|-------|-------|-------|\n| $y$ | $513$ | $417$ | $313$ | $201$ | $81$  |\n\nOnly $y = 81$ is a perfect square, so $(a - 22, b - 22) = (\\pm 31, \\pm 9)$. Since $a, b > 0$, $(a, b) = (53, 31)$ or $(31, 53)$. Check:\n\n$$\na^2 + b^2 = 53^2 + 31^2 = 2809 + 961 = 3770 = 44(53 + 31) + 74.\n$$\n\nThus, the only solutions are $(a, b) = (53, 31)$ and $(31, 53)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19848,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a regular $2n$-gon $P$, $A_1A_2\\ldots A_{2n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$ if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We color the sides of $P$ in 3 different colors (ignore the vertices of $P$, we consider them colorless), such that every side is colored in exactly one color, and each color is used at least once. Moreover, from every point in the plane external to $P$, points of at most 2 different colors on $P$ can be seen. Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one of the sides is colored differently).",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, the answer is $36$; for $n=3$, the answer is $30$; and for $n \\ge 4$, the answer is $6n$.\n\n**Lemma 1.** Given a regular $2n$-gon in the plane and a sequence of $n$ consecutive sides $s_1, s_2, \\dots, s_n$, there is an external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$, for $i=1, 2, \\dots, n$.\n\n*Proof.* For a semicircle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semicircle can be seen from $R$. Looking at the circumscribed circle around the $2n$-gon, there is a semicircle $S$ such that each $s_i$ either has both endpoints on it, or has an endpoint that's on the semicircle and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen; clearly, the color of each $s_i$ can be seen from $Q$.\n\n**Lemma 2.** Given a regular $2n$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_1, s_2, \\dots, s_{n+1}$, there is no external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$, for $i=1, 2, \\dots, n+1$.\n\n*Proof.* Since $s_1$ and $s_{n+1}$ are parallel opposite sides of the $2n$-gon, they cannot be seen at the same time from an external point.\n\nFor $n=2$ (a square), all we have to do is make sure each color is used. Two sides will be of the same color, and we have to choose which are these 2 sides, and then assign colors according to this choice, so the answer is $\\binom{4}{2} \\cdot 3 \\cdot 2 = 36$.\n\nFor $n=3$ (a hexagon), denote the sides as $a_1, a_2, \\dots, a_6$, in that order. There must be 2 consecutive sides of different colors, say $a_1$ is red, $a_2$ is blue. We must have a green side, and only $a_4$ and $a_5$ can be green. We have 3 possibilities:\n\n1. $a_4$ is green, $a_5$ is not. So, $a_3$ must be blue and $a_5$ must be blue (by elimination) and $a_6$ must be blue, so we get a valid coloring.\n2. Both $a_4$ and $a_5$ are green, thus $a_6$ must be red and $a_5$ must be blue, and we get the coloring rbbggr.\n3. $a_5$ is green, $a_4$ is not. Then $a_6$ must be red. Subsequently, $a_4$ must be red (we assume it is not green). It remains that $a_3$ must be red, and the coloring is rbbgr.\n\nThus, we have 2 kinds of configurations:\n\n- 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in $3 \\cdot (3 \\cdot 2 \\cdot 1) = 18$ ways (first choosing the pair of opposite sides, then assigning colors),\n- 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $3 \\cdot 6 = 12$ ways (2 partitionings into pairs of consecutive sides, for each partitioning, 6 ways to assign the colors).\n\nThus, for $n=3$, the answer is $18+12=30$.\n\n![](images/Macedonia_2017_p9_data_c5ca3559e2.png)\n\nFor $n \\ge 4$, any 4 consecutive sides can be seen from an external point, by Lemma 1. Denote the sides as $a_1, a_2, \\dots, a_{2n}$. There must be 2 adjacent sides that are of different colors, say $a_1$ is blue and $a_2$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. So, we have 2 cases:\n\n- Case 1: $a_{n+1}$ is green, so $a_n$ must be red (cannot be green due to Lemma 1 applied to $a_1, a_2, \\dots, a_n$, cannot be blue for the sake of $a_2, \\dots, a_{n+1}$). If $a_{n+2}$ is red, so are $a_{n+3}, \\dots, a_{2n}$, and we get a valid coloring: $a_1$ is blue, $a_{n+1}$ is green, and all the others are red.\n- If $a_{n+2}$ is green:\n    - $a_{n+3}$ cannot be green, because of $a_2, a_3, a_1, a_2, \\dots, a_{n+3}$.\n    - $a_{n+3}$ cannot be blue, because the 4 adjacent sides $a_n, \\dots, a_{n+3}$ can be seen.\n    - $a_{n+3}$ cannot be red, because of $a_1, a_2, \\dots, a_{n+3}$.\n\nSo, in the case that $a_{n+2}$ is also green, we cannot get a valid coloring.\n\n- Case 2: $a_{n+2}$ is green is treated the same way as Case 1.\n\nThis means that the only valid configuration for $n \\ge 4$ is having 2 opposite sides colored in 2 different colors, and all other sides colored in the third color. This can be done in $6n$ ways.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19849,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $a$ is called a *double number* if it has an even number of digits (in base 10) and its base 10 representation has the form $a = a_1a_2\\cdots a_k a_1a_2\\cdots a_k$ with $0 \\le a_i \\le 9$ for $1 \\le i \\le k$, and $a_1 \\ne 0$. For example, $283283$ is a double number.\n\nDetermine whether or not there are infinitely many double numbers $a$ such that $a + 1$ is a square and $a + 1$ is not a power of $10$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is affirmative. Let $k \\ge 0$ be such that $k \\equiv 15 \\pmod{42}$ and $b = \\dfrac{5(10^k + 1)}{7} + 1$ (which is an integer). Then $c = \\dfrac{5}{7}(b+1)$ is an integer and $b^2 - 1 = (10^k + 1)c$ is a double number. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19850,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ which satisfy the following two conditions:\n\n1) $f(x) \\neq f(y)$ for each $x \\neq y$.\n\n2) $f(f(x)y + x) = f(x)f(y) + f(x)$ for each $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take $x = y = 0$. We have $f(0) = 0$.\n\nLet $x_0 \\neq 0$ and denote $a = f(x_0) \\neq 0$. Take $x = x_0$. Then for all $y \\in \\mathbb{Z}$ we have:\n\n$$\nf(ay + x_0) = a f(y) + a.\n$$\n\nNow change $x \\to a x + x_0$. We get:\n\n$$\n\\begin{aligned}\nf(f(a x + x_0) y + a x + x_0) &= f(a x + x_0) f(y) + f(a x + x_0), \\\\\nf((a f(x) + a) y + a x + x_0) &= (a f(x) + a) f(y) + a f(x) + a, \\\\\nf(a f(x) y + a y + a x + x_0) &= a f(x) f(y) + a f(y) + a f(x) + a.\n\\end{aligned}\n$$\n\nSince the right-hand side is symmetric in $x$ and $y$, we have:\n\n$$\nf(a f(x) y + a y + a x + x_0) = f(a f(y) x + a y + a x + x_0)\n$$\nfor all $x, y \\in \\mathbb{Z}$.\n\nSo,\n$$\na f(x) y + a y + a x + x_0 = a f(y) x + a y + a x + x_0,\n$$\nwhich gives $f(x) y = f(y) x$ for all $x, y \\in \\mathbb{Z}$.\n\nFix some $y_0 \\neq 0$ and denote $k = \\frac{f(y_0)}{y_0}$. Then $f(x) = k x$, $k \\in \\mathbb{Q} \\setminus \\{0\\}$.\n\nSince $f(x)$ is an integer for each integer $x$, we have $k \\in \\mathbb{Z} \\setminus \\{0\\}$.\n\nWe are left to check that all functions $f(x) = k x$ where $k \\neq 0$ is an integer satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19851,
    "subject": "Mathematics (Olympiad)",
    "question": "A cross is the shape obtained from a $3 \\times 3$ grid square upon removing the 4 corner unit squares. Every unit square of a $2010 \\times 2010$ table must be colored in one of 5 distinct colors so that the 5 unit squares of every cross contained in the table have different colors. In how many ways can this be done? Two colorings are different if there is a unit square colored with one color in one of them and with a different color in the other.",
    "options": [],
    "answer": "See solution",
    "solution": "For an admissible coloring, let the 13-cell shape shown in the figure be entirely inside the table. Observe that the central row of 5 represents all 5 colors, as well as the central column of 5. Assume on the contrary that the central row misses color 1. Since the crosses centered at *a*, *b*, *c* contain color 1, this color occurs 3 times in the union of the two rows of 3 above and under *abc*. This union is covered by the two crosses centered at the cells • adjacent to *b*. However, these crosses combined represent each color at most twice. The claim is proven.\n\n![](images/Argentina_2011_p18_data_f81a2dfb70.png)\n\nWe argue for a general $n \\times n$ square where $n > 5$. Every 5 consecutive cells in rows 3, 4, \\dots, $n-2$ form the central row of 5 in a shape like above, so they represent all 5 colors. The same holds for every 5 consecutive cells in columns 3, 4, \\dots, $n-2$. Hence the coloring of the rows and columns mentioned is periodic with period 5. Write $a_{ij}$ for the cell in row $i$ and column $j$, \"cross $a_{ij}$\" for the cross centered at $a_{ij}$ and $a_{ij} = c$ if $a_{ij}$ has color $c$. Note that any two adjacent cells not at the border of the table are contained in some cross, so they are colored differently.\n\nFix the colors of the first 5 cells in row 3 and the second cell in row 2. Cells $a_{31}, a_{32}, a_{33}, a_{34}, a_{35}$ have different colors, label them 1, 2, 3, 4, 5. The color of $a_{22}$ is not arbitrary: one infers from cross $a_{32}$ that $\\{a_{22}, a_{42}\\} = \\{4, 5\\}$. Let $a_{22} = 4$, $a_{42} = 5$, the case $a_{22} = 5$, $a_{42} = 4$ is analogous. We show that then the entire coloring is determined apart from the colors of 3 cells at each corner: the vertex cell and its two neighbors by side. Call $F$ the figure obtained by ignoring these 12 cells.\n\nThe coloring of row 3 is determined: $1, 2, 3, 4, 5, 1, 2, 3, 4, 5, \\dots$. By considering cross $a_{33}$ we find $\\{a_{23}, a_{43}\\} = \\{1, 5\\}$; since $a_{42} = 5$ and $a_{42} \\neq a_{43}$, it follows that $a_{23} = 5$, $a_{43} = 1$. The same argument for crosses $a_{34}$ and $a_{35}$ shows that $a_{24} = 1$, $a_{44} = 2$, $a_{25} = 2$, $a_{45} = 3$. The colors of 4 consecutive cells in row 4 are known now, from the second to the fifth, so row 4 has coloring $4, 5, 1, 2, 3, 4, 5, 1, 2, 3, \\dots$. The remaining crosses centered in row 3 determine the colors in row 2 without its first and last cell: $4, 5, 1, 2, 3, 4, 5, 1, 2, 3, \\dots$. After this consider the crosses centered at row 2 except $a_{22}$ and $a_{2,n-1}$. They determine the colors in row 1 without its first two and last two cells: $2, 3, 4, 5, 1, 2, 3, 4, 5, 1, \\dots$.\n\nOnce the coloring of row 4 is known, we proceed analogously to rows $5, \\dots, n-2$ to achieve the same. For the next-to-last row $n-1$ the colors are determined uniquely except for the first and the last cell. For the last row $n$ we use the crosses in row $n-1$ different from $a_{n-1,2}$ and $a_{n-1,n-1}$. This yields the colors in row $n$ apart from the ones of the first two and the last two cells. Thus the coloring of figure $F$ is indeed completely determined.\n\nThere remain the three cells at each corner. We infer from cross $a_{22}$ that $\\{a_{12}, a_{21}\\} = \\{1, 3\\}$, and the two possibilities for $a_{12}$ and $a_{21}$ are feasible as there are no more restrictions on their colors. Likewise there are 2 possibilities for the colors of the analogous pairs at the other three corners. As for the colors of the 4 vertex cells, there are no restrictions at all.\n\nInitially we fixed the colors of $a_{31}, a_{32}, a_{33}, a_{34}, a_{35}$ and $a_{22}$ which can be done in $5! \\cdot 2$ ways. Then there are 2 choices for the coloring of the pair $a_{12}, a_{21}$ and 2 choices for each analogous pair at the corners. Last, there are 5 choices for the color of each vertex cell.\n\nEvery combination of the choices described yields a coloring with the stated property. Therefore the number of admissible colorings is $$5! \\cdot 2 \\cdot 2^4 \\cdot 5^4 = 4! \\cdot 10^5 = 2400000.$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19852,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X$ be a set of size $n$, and let $\\mathcal{F}$ be a family of 3-element subsets of $X$ such that any subset of $X$ of size $|Y|$ contains at least one member of $\\mathcal{F}$. Prove that $|Y| \\geq \\lfloor \\sqrt{2n} \\rfloor$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $f: X \\setminus Y \\to \\binom{Y}{2}$, where $\\binom{Y}{2}$ is the family of 2-subsets of $Y$, with $f(x) = A$ if $A \\cup \\{x\\}$ is one of the sets from $\\mathcal{F}$; if there is more than one such set, choose any of them. $f$ is injective, because if $f(x_1) = f(x_2) = B$, then $B \\cup \\{x_1\\}$ and $B \\cup \\{x_2\\}$ would both be in $\\mathcal{F}$, and these two sets would have more than one element in their intersection. Thus, by the injective principle,\n\n$$\n|X \\setminus Y| \\leq \\binom{|Y|}{2} \\iff n - |Y| \\leq \\frac{|Y|(|Y| - 1)}{2} \\iff |Y| \\geq -\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}}\n$$\n\nLet $\\lfloor \\sqrt{2n} \\rfloor = k$. We need to prove that $-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1$ in order to show that $|Y| > k - 1 \\iff |Y| \\geq k$. This is a small computation:\n\n$$\n-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1 \\iff 2n + \\frac{1}{4} > k^2 - k + \\frac{1}{4} \\iff \\sqrt{2n} > \\sqrt{k^2 - k},\n$$\nwhich is true because $\\sqrt{2n} \\geq k > \\sqrt{k^2 - k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19853,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(n) = n^2 + n + 1$. For any integer $n$, prove the following:\n\n1. $$\\gcd(P(n), P(n+1)) = 1$$\n2. $$\\gcd(P(n), P(n+2)) > 1 \\iff n \\equiv 2 \\pmod{7}$$\n3. $$\\gcd(P(n), P(n+3)) > 1 \\iff n \\equiv 1 \\pmod{3}$$\n4. $$\\gcd(P(n), P(n+4)) > 1 \\iff n \\equiv 7 \\pmod{19}$$\n\nAdditionally, determine all integers $a$ such that the set $\\{a+1, a+2, a+3, a+4, a+5, a+6\\}$ can be partitioned into three disjoint pairs with differences 2, 3, and 4, and for which the corresponding $P(n)$ values share a common factor greater than 1 as described above.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the statements as follows:\n\nFor (1), and the “$\\Rightarrow$” directions of (2) and (3), the proofs follow as in a previous solution.\n\nFor the “$\\Rightarrow$” direction of (4):\n\nSuppose $P(n) = n^2 + n + 1$ and $P(n+4) = n^2 + 9n + 21$ are both divisible by an odd prime $p$. Then\n$$\n|P(n+4) - P(n)| = 8n + 20.\n$$\nSince $p$ is odd, $p \\mid 2n + 5$. It follows that\n$$\np \\mid 4(n^2 + n + 1) - (2n + 5)(2n - 3) = 19.\n$$\nThus $p = 19$. Furthermore, since $p \\mid 2n + 5$, we have $19 \\mid 2n + 5$, so $19 \\mid n - 7$.\n\nThe “$\\Leftarrow$” directions of (2), (3), and (4) are straightforward computations. For example, for (2), $P(n) \\equiv P(n+2) \\equiv 0 \\pmod{7}$ whenever $n \\equiv 2 \\pmod{7}$. Similar computations apply for (3) and (4).\n\nTo show that no fragrant set exists for $b \\leq 5$, refer to the previous solution.\n\nFor $b = 6$, partition $\\{a+1, a+2, a+3, a+4, a+5, a+6\\}$ into three disjoint pairs with differences 2, 3, and 4. There are two ways:\n\n**Way 1:** $(a+2, a+4)$, $(a+3, a+6)$, $(a+1, a+5)$\n\nThese correspond to:\n$$\na \\equiv 0 \\pmod{7}\n$$\n$$\na \\equiv 1 \\pmod{3}\n$$\n$$\na \\equiv 6 \\pmod{19}\n$$\nSolving gives $a \\equiv 196 \\pmod{399}$.\n\n**Way 2:** $(a+3, a+5)$, $(a+1, a+4)$, $(a+2, a+6)$\n\nThese correspond to:\n$$\na \\equiv 6 \\pmod{7}\n$$\n$$\na \\equiv 0 \\pmod{3}\n$$\n$$\na \\equiv 5 \\pmod{19}\n$$\nSolving gives $a \\equiv 195 \\pmod{399}$.\n\nThus, $a = 195$ and $a = 196$ are two possible solutions for $b = 6$.\n\nWith a little more work, the full set of solutions for $b = 6$ is $a = 195 + 399k$ and $a = 196 + 399k$, where $k$ is any positive integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19854,
    "subject": "Mathematics (Olympiad)",
    "question": "We order the positive integers in two rows as follows:\n\n1 3 6 11 19 32 53 ...\n2 4 5 7 8 9 10 12 13 14 15 16 17 18 20 to 31 33 to 52 54 ...\n\nWe first write 1 in the first row, 2 in the second, and 3 in the first. After this, the following integers are written so that an individual integer is always added in the first row, and blocks of consecutive integers are added in the second row, with the leading number of a block giving the number of (consecutive) integers to be written in the next block.\n\nWe name the numbers in the first row $a_1, a_2, a_3, \\dots$.\n\nDetermine an explicit formula for $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $a_1 = 1$, $a_2 = 3$, and $a_3 = 6$. It is straightforward to see that a block of length $a_{n-1} + 1$ starts with the number $a_n + 1$, and this block therefore ends on the number $a_n + (a_{n-1} + 1)$, which yields the recurrence:\n\n$$a_{n+1} = a_n + a_{n-1} + 2$$\n\nThis recursion has the constant solution $a_n \\equiv -2$, and the homogeneous recursion $a_{n+1} = a_n + a_{n-1}$ is of Fibonacci type. Writing the Fibonacci sequence as $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, $F_3 = 2$, and so on, we can check that $a_n = F_{n+3} - 2$ holds for $n = 1, 2, 3$. Therefore, the explicit formula is:\n\n$$a_n = F_{n+3} - 2$$\n\nwhere $F_k$ denotes the $k$-th Fibonacci number. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19855,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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": 19856,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Let $D$ be a point on $AB$ and let $E$ be a point on $AC$ such that $BCED$ is cyclic. Let $F$ be the intersection of the segment $BE$ and the circumcircle of $ADC$, and let $G$ be the intersection of $CD$ and the circumcircle of $ABE$. Suppose that the segments $BG$ and $CF$ meet at $S$. Show that $\\angle FAS = \\angle GAS$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the quadrilaterals $ADFC$, $DBCE$, and $AEGB$ are cyclic, we have\n\n$$\n\\angle ADC = \\angle AFC = \\angle ADC = \\angle AEB = \\angle AGB.\n$$\n\nBecause $\\angle AEF = \\angle AFC$, we have $\\triangle AEF \\sim \\triangle ACF$. Hence $AF^2 = AE \\cdot AC$. In the same way, we have $AG^2 = AD \\cdot AB$. It follows that $DBCE$ is cyclic. Thus $AF = AG$. Since $AF = AG$, $\\angle AFS = \\angle AGS$ and $AS$ is common, we have $\\triangle AFS = \\triangle AGS$. We conclude that $\\angle FAS = \\angle GAS$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19857,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a tuple of consecutive positive integers, one forms all pairs of members of it such that the first member is less than the second member. The percentage of these pairs where the second member is divisible by the first one is called the *degree of divisibility* of the tuple. For every integer $n > 1$, denote the largest possible degree of divisibility of a tuple of $n$ consecutive positive integers by $j(n)$.\n\nDoes there exist an integer $n > 1$ such that $j(n + 1) > j(n)$?",
    "options": [],
    "answer": "See solution",
    "solution": "The largest percentage of pairs with the second term being divisible by the first term is achieved in the case of the tuple $(1, 2, \\dots, n)$. Indeed, consider an arbitrary tuple of the form $(x+1, x+2, \\dots, x+n)$ where $x > 0$. For any $i$, multiples of $i$ in $(1,2,...,n)$ are every $i$th term starting from the number $i$, multiples of $x+i$ in $(x+1,x+2,...,x+n)$ are every $(x+i)$th term starting from $x+i$. The latter multiples occur more seldom while the first occurrence is at the same position, implying the same or smaller total number.\n\nIt is easy to check that the degree of divisibility of $(1,2,3,4,5)$ is $\\frac{5}{10}$ and the degree of divisibility of $(1,2,3,4,5,6)$ is $\\frac{8}{15}$. By the above, $j(6) > j(5)$.\n\n*Answer:* Yes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19858,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume $a_2 \\neq a_4$. Let $(a_n)$ be a sequence such that $\\frac{n}{2} < a_n < 2n$ for all $n$. Prove that $a_n = n$ for all $n \\geq 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "i. *Large primes part*: For each prime $p \\geq 5$, the only prime less than or equal to $p$ that divides $a_p, a_{p^2}, \\dots$ is $p$.\n\nii. We prove $a_{p^k} = p^k$ for all primes $p \\geq 5$.\n\niii. *Small primes part*: We resolve the problem for $p = 2, p = 3$ and conclude the proof.\n\nFor part one, for each prime $p$, define sets of primes $A_p, B_p$:\n\n$$A_p = \\{q \\leq p : q \\mid a_{p^k}, k = 1, 2, \\dots\\}$$\n$$B_p = \\{q : q \\mid a_{p^k}, k = 1, \\dots\\}$$\n\nThat is, $A_p = B_p \\cap [1, p]$. $A_p$ and $B_p$ are disjoint for different $p$. For $p \\geq 5$, $A_p$ is non-empty. If $A_p$ were empty, then $a_{p^{k+1}} \\neq a_{p^k}$ and $a_{p^k}$ divides $a_{p^{k+1}}$, leading to a contradiction with the bounds $\\frac{n}{2} < a_n < 2n$. Thus, $A_p$ is not empty for $p \\geq 5$.\n\nFor small primes, $A_2 \\cup A_3 \\cup A_5 \\subset \\{2, 3, 5\\}$ and are disjoint. If $a_2 \\neq 2$, then $A_2$ is empty, and $a_3 \\in \\{2, 4\\}$, but this leads to contradictions for $a_6$ and $a_4$. Thus, $a_2 = 2$ and $a_3 = 3$.\n\nFor all $p \\geq 5$, $A_p = \\{p\\}$, so $a_{p^k} = p^k$. For composite $n$, $a_n$ must be divisible by $n$ and satisfy the bounds, so $a_n = n$.\n\nThe case $a_2 = a_4$ is ruled out by similar density and divisibility arguments, leading to contradictions with the bounds. Thus, $a_n = n$ for all $n \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19859,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $a_1 \\le a_2 \\le \\dots \\le a_n$ and $b_1 \\le b_2 \\le \\dots \\le b_n$ be two nondecreasing sequences of real numbers such that\n\n$$\na_1 + \\dots + a_i \\le b_1 + \\dots + b_i \\quad \\text{for every } i = 1, \\dots, n-1\n$$\n\nand\n\n$$\na_1 + \\dots + a_n = b_1 + \\dots + b_n.\n$$\n\nSuppose that for any real number $m$, the number of pairs $(i, j)$ with $a_i - a_j = m$ equals the number of pairs $(k, \\ell)$ with $b_k - b_\\ell = m$. Prove that $a_i = b_i$ for $i = 1, \\dots, n$.\n\n**Note:** It is important to interpret the condition that for any real number $m$, the number of pairs $(i, j)$ with $a_i - a_j = m$ equals the number of pairs $(k, \\ell)$ with $b_k - b_\\ell = m$. It means that we have two identical multi-sets (a multi-set is a set that allows repeated elements)\n\n$$\n\\{a_i - a_j \\mid 1 \\le i < j \\le n\\} \\quad \\text{and} \\quad \\{b_k - b_\\ell \\mid 1 \\le k < \\ell \\le n\\}.\n$$\n\nIn particular, it gives us that\n\n$$\n\\sum_{1 \\le i < j \\le n} (a_i - a_j) = \\sum_{1 \\le k < \\ell \\le n} (b_k - b_\\ell), \\quad (*)\n$$\n\n$$\n\\sum_{1 \\le i < j \\le n} (a_i - a_j)^2 = \\sum_{1 \\le k < \\ell \\le n} (b_k - b_\\ell)^2 \\quad (**)\n$$\n\nand\n\n$$\n\\sum_{i,j=1}^{n} |a_i - a_j| = \\sum_{i,j=1}^{n} |b_i - b_j|. \\quad (***)\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "The first solution is based on $(*)$.\n\nPut $s_n = a_1 + \\cdots + a_n = b_1 + \\cdots + b_n$. Then\n\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{n-1} (a_1 + \\cdots + a_i) &= 2(n-1)a_1 + 2(n-2)a_2 + \\cdots + 2(1)a_{n-1} \\\\\n&= (n-1)a_1 + (n-3)a_2 + \\cdots + (1-n)a_n + (n-1)s_n \\\\\n&= (n-1)s_n + \\sum_{1 \\le i < j \\le n} (a_i - a_j)\n\\end{aligned}\n$$\n\nand similarly\n\n$$\n2 \\sum_{i=1}^{n-1} (b_1 + \\cdots + b_i) = (n-1)s_n + \\sum_{1 \\le k < \\ell \\le n} (b_k - b_\\ell).\n$$\n\nBy $(*)$, these two quantities are equal, so\n\n$$\n2 \\sum_{i=1}^{n-1} (a_1 + \\cdots + a_i) = 2 \\sum_{i=1}^{n-1} (b_1 + \\cdots + b_i).\n$$\n\nConsequently, each of the inequalities $a_1 + \\cdots + a_i \\le b_1 + \\cdots + b_i$ for $i = 1, \\dots, n-1$ must be an equality. Since we also have equality for $i = n$ by assumption, we deduce that $a_i = b_i$ for $i = 1, \\dots, n$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19860,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an arbitrary integer.\n\nDetermine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the conditions $f(0) = 0$ and\n\n$$\nf(x^k y^k) = xy f(x) f(y) \\quad \\text{for all } x, y \\neq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 1$ in (2): $f(1) = f(1)^2$, so $f(1) \\in \\{0, 1\\}$.\n\n**Case 1:** $f(1) = 0$\n\n- If $k = 0$, (2) with $x = y = t \\neq 0$ gives $f(1) = t^2 f(t)^2$, so $f(t) = 0$ for $t \\neq 0$.\n- If $k \\neq 0$, set $y = 1$, $x = t$ in (2): $f(t^k) = t f(t) f(1) = 0$. Also, $x = y$ in (2): $f((x^2)^k) = x^2 f(x)^2$, so $f(x) = 0$ for $x \\neq 0$.\n\n**Case 2:** $f(1) = 1$\n\nLet $x = t$, $y = 1/t$ ($t \\neq 0$) in (2): $f(1) = f(t) f(1/t)$, so $f(t) \\neq 0$ for $t \\neq 0$.\n\nSet $x = y = -1$ in (2): $f((-1)^{2k}) = (-1)^2 f(-1)^2$, so $f(-1)^2 = 1$, i.e., $f(-1) \\in \\{-1, 1\\}$.\n\n- If $k$ is odd: $x = 1$, $y = -1$ in (2) gives $f((-1)^k) = -f(-1)$, so $f(-1) = -f(-1)$, thus $f(-1) = 0$, a contradiction.\n- If $k$ is even: $f(-1) = -1$.\n\n  - If $k = 0$, (2) with $y = 1$, $x = t$ ($t \\neq 0$): $f(1) = t f(t)$, so $f(t) = 1/t$.\n  - If $k \\neq 0$, $x = 1$, $y = t$ ($t \\neq 0$): $f(t^k) = t f(t)$. Also, $f(xy) = f(x) f(y)$ (from (2) and above). Let $x = t$, $y = t^{k-1}$: $f(t^k) = f(t) f(t^{k-1}) = t f(t)$, so $f(t^{k-1}) = t$ for $t \\neq 0$. Since $k$ is even, every $x \\neq 0$ can be written as $x = t^{k-1}$, so $f(x) = x^{1/(k-1)}$ for $x \\neq 0$.\n\n**Summary:**\n\nAll solutions to (2) are:\n\n- $f(x) = 0$ for all $x \\in \\mathbb{R}$, for any integer $k$.\n- $f(x) = \\begin{cases} x^{1/(k-1)}, & x \\neq 0 \\\\ 0, & x = 0 \\end{cases}$, if $k$ is an even integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19861,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all polynomials $P(x) \\in \\mathbb{R}[x]$ satisfying the following two conditions:\n\n(a) $P(2017) = 2016$\n\n(b) $(P(x) + 1)^2 = P(x^2 + 1)$ for all real numbers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(x) := P(x) + 1$. Then the conditions become $Q(2017) = 2017$ and $Q(x^2 + 1) = Q(x)^2 + 1$ for all $x \\in \\mathbb{R}$.\n\nDefine the sequence $\\{x_n\\}_{n \\ge 0}$ recursively by $x_0 = 2017$ and $x_{n+1} = x_n^2 + 1$ for $n \\ge 0$. By induction, $Q(x_n) = x_n$ for all $n \\ge 0$ because:\n\n$$\nQ(x_{n+1}) = Q(x_n^2 + 1) = Q(x_n)^2 + 1 = x_n^2 + 1 = x_{n+1}.\n$$\n\nSince $x_0 < x_1 < x_2 < \\dots$, the polynomials $Q(x)$ and $x$ agree at infinitely many points, so $Q(x) = x$. Thus, $P(x) = x - 1$ is the unique solution.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19862,
    "subject": "Mathematics (Olympiad)",
    "question": "The integers $1, 2, 3, \\ldots, 2016$ are written on a board. You can choose any two numbers on the board and replace them with their average. For example, you can replace $1$ and $2$ with $1.5$, or you can replace $1$ and $3$ with a second copy of $2$. After $2015$ replacements of this kind, the board will have only one number left on it.\n\n(a) Prove that there is a sequence of replacements that will make the final number equal to $2$.\n\n(b) Prove that there is a sequence of replacements that will make the final number equal to $1000$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, replace $2014$ and $2016$ with $2015$, and then replace the two copies of $2015$ with a single copy. This leaves us with $\\{1, 2, \\ldots, 2013, 2015\\}$. From here, we can replace $2013$ and $2015$ with $2014$ to get $\\{1, 2, \\ldots, 2012, 2014\\}$. We can then replace $2012$ and $2014$ with $2013$, and so on, until we eventually get to $\\{1, 3\\}$. We finish by replacing $1$ and $3$ with $2$.\n\n(b) Using the same construction as in (a), we can find a sequence of replacements that reduces $\\{a, a + 1, \\ldots, b\\}$ to just $\\{a + 1\\}$. Similarly, we can also find a sequence of replacements that reduces $\\{a, a + 1, \\ldots, b\\}$ to just $\\{b - 1\\}$.\n\nIn particular, we can find sequences of replacements that reduce $\\{1, 2, \\ldots, 999\\}$ to just $\\{998\\}$, and that reduce $\\{1001, 1002, \\ldots, 2016\\}$ to just $\\{1002\\}$. This leaves us with $\\{998, 1000, 1002\\}$. We can replace $998$ and $1002$ with a second copy of $1000$, and then replace the two copies of $1000$ with a single copy to complete the construction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19863,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $k$ and $n$ satisfying the equation\n$$\nk^2 - 2016 = 3^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We immediately see that $n = 1$ does not lead to a solution, while $n = 2$ yields the solution $(k, n) = (45, 2)$.\n\nWe show that there is no solution with $n \\ge 3$. In that case, $3^n$ is divisible by $9$ and thus $k^2$ is divisible by $9$, which implies that $k = 3l$ for some positive integer $l$. After division by $9$, the equation reads\n$$\nl^2 - 224 = 3^{n-2}.\n$$\nModulo $3$, this yields $l^2 - 2 \\equiv 0 \\pmod{3}$, a contradiction because $2$ is not a quadratic residue modulo $3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19864,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $k$ be positive integers with $k \\ge n$ and $k - n$ an even number. Let $2n$ lamps labelled $1, 2, \\dots, 2n$ be given, each of which can be either on or off. Initially all the lamps are off.\n\nWe consider sequences of steps: at each step, one of the lamps is switched from on to off or from off to on.\n\nLet $N$ be the number of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off.\n\nLet $M$ be the number of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nDetermine the ratio $\\frac{N}{M}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio is $2^{k-n}$.\n\n**Lemma:** For any positive integer $t$, call a $t$-element array $(a_1, a_2, \\dots, a_t)$ which consists of $0, 1$ ($a_1, a_2, \\dots, a_t \\in \\{0, 1\\}$) \"good\" if there are an odd number of $0$'s in it. Prove that there are $2^{t-1}$ \"good\" arrays.\n\n**Proof:** For the same $a_1, a_2, \\dots, a_{t-1}$, when $a_t$ is $0$ or $1$, the parity of $0$'s in these two arrays is different, so only one of the arrays is \"good\". There are $2^t$ arrays in all, and we can pair each array with another that differs only in $a_t$. Only one of these two arrays is \"good\". So of all the possible arrays, only half of them are \"good\". The lemma is proved.\n\nLet $A$ be the set of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off.\n\nLet $B$ be the set of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nFor any $b$ in $B$, match all $a$ in $A$ to $b$ if $a$'s elements are the same as $b$'s modulo $n$ (for example, take $n=2, k=4$, if $b=(2, 2, 2, 1)$, then it could correspond to $a=(4, 4, 2, 1)$, $a=(2, 2, 2, 1)$, $a=(2, 4, 4, 1)$, etc.). Since $b$ is in $B$, the number of $1, 2, \\dots, n$ must be odd; for $a$ in $A$, the number of $1, 2, \\dots, n$ in $b$ must be odd, and the number of $n+1, \\dots, 2n$ must be even.\n\nFor any $i \\in \\{1, 2, \\dots, n\\}$, if the number of $i$'s in $b$ is $b_i$, then $a$ only needs to satisfy: for the positions taken by $i$ in $b$, the corresponding positions are taken by $i$ or $n+i$ in $a$, and the number of $i$'s is odd (thus the number of $n+i$'s is even). By the lemma and the product principle, there are $\\prod_{i=1}^n 2^{b_i-1} = 2^{k-n}$ such $a$'s corresponding to $b$, but only one $b$ (letting every position of $a$ be the remainder when divided by $n$) in $B$ corresponds to each $a$ in $A$.\n\nTherefore $|A| = 2^{k-n} |B|$, i.e., $N = 2^{k-n} M$.\n\nObviously $M \\neq 0$ (because the sequence $(1, 2, \\dots, n, n, \\dots, n) \\in B$), so\n\n$$\n\\frac{N}{M} = 2^{k-n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19865,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations in real numbers:\n\n$$\n\\begin{aligned}\n\\sin^2 x + \\cos^2 y &= \\tan^2 z, \\\\\n\\sin^2 y + \\cos^2 z &= \\tan^2 x, \\\\\n\\sin^2 z + \\cos^2 x &= \\tan^2 y.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\cos^2 x = a$, $\\cos^2 y = b$, $\\cos^2 z = c$. The system becomes:\n\n$$\n\\begin{cases}\n1 - a + b = \\frac{1}{c} - 1, \\\\\n1 - b + c = \\frac{1}{a} - 1, \\\\\n1 - c + a = \\frac{1}{b} - 1,\n\\end{cases}\n$$\nwhere $a, b, c \\in (0, 1)$.\n\nAdding the equations:\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 6.\n$$\nSo the harmonic mean of $a, b, c$ is $\\frac{1}{2}$.\n\nMultiplying the equations by $c, a, b$ respectively and adding:\n$$\n2(a + b + c) = 3 \\implies a + b + c = \\frac{3}{2}.\n$$\nThus, the arithmetic mean is also $\\frac{1}{2}$. Since the arithmetic and harmonic means are equal, $a = b = c = \\frac{1}{2}$.\n\nThis gives $\\cos^2 x = \\cos^2 y = \\cos^2 z = \\frac{1}{2}$, so $x, y, z = \\frac{\\pi}{4} + \\frac{k\\pi}{2}$ for integers $k$.\n\nTherefore, all solutions are triples of the form:\n$$(x, y, z) = \\left(\\frac{\\pi}{4} + \\frac{k\\pi}{2},\\ \\frac{\\pi}{4} + \\frac{l\\pi}{2},\\ \\frac{\\pi}{4} + \\frac{m\\pi}{2}\\right),$$\nwhere $k, l, m$ are integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19866,
    "subject": "Mathematics (Olympiad)",
    "question": "Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let $N$ be the number of subsets of the 16 chairs that could be selected. Find the remainder when $N$ is divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "The problem is equivalent to counting the arrangements of $PPPPPPPPEEEEEEEE$ (standing for people and empty seats) where no three $P$s appear together. Suppose such an arrangement contains $q$ pairs $PP$ and $8 - 2q$ $P$s not adjacent to another $P$, where $0 \\leq q \\leq 4$. There are 9 spaces before, after, or between the 8 $E$s, and $PP$s will be placed in $q$ of those 9 spaces, while $P$s will be placed in $8 - 2q$ of the remaining $9 - q$ spaces. Hence the number of arrangements for a given value of $q$ is $\\binom{9}{q}\\binom{9-q}{8-2q}$, and the total number of ways to select chairs is\n\n$$\n\\sum_{q=0}^{4} \\binom{9}{q} \\binom{9-q}{8-2q} = \\binom{9}{0} \\binom{9}{8} + \\binom{9}{1} \\binom{8}{6} + \\binom{9}{2} \\binom{7}{4} + \\binom{9}{3} \\binom{6}{2} + \\binom{9}{4} \\binom{5}{0}\n$$\n\n$$\n= 1 \\cdot 9 + 9 \\cdot 28 + 36 \\cdot 35 + 84 \\cdot 15 + 126 \\cdot 1 = 9 + 252 + 1260 + 1260 + 126 = 2907.\n$$\n\nThe requested remainder is $2907 \\mod 1000 = 907$.\n\n**Note:** Sequence A005717 in the On-Line Encyclopedia of Integer Sequences gives the number of ways to select $n$ out of $2n$ chairs that meet the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19867,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x$ and $y$ such that $x + y + 1$ divides $2xy$ and $x + y - 1$ divides $x^2 + y^2 - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $x$ and $y$ satisfy the divisibility conditions in the question. Note that\n\n$$\nx + y - 1 \\mid ((x + y - 1)^2 + 2(x + y - 1))\n$$\n\nand so\n\n$$\nx + y - 1 \\mid x^2 + y^2 - 1 + 2xy.\n$$\n\nHowever, we are told that $x + y - 1 \\mid x^2 + y^2 - 1$, so we can deduce that\n\n$$\nx + y - 1 \\mid 2xy.\n$$\n\nWe are also given that\n\n$$\nx + y + 1 \\mid 2xy.\n$$\n\nThe positive integers $x + y - 1$ and $x + y + 1$ differ by 2, so their greatest common divisor is at most 2.\n\nTherefore, either\n\n$$\n\\frac{1}{2}(x + y + 1)(x + y - 1) \\mid 2xy \\quad (\\text{if their gcd is 2})\n$$\n\nor\n\n$$\n(x + y + 1)(x + y - 1) \\mid 2xy \\quad (\\text{if they are coprime})\n$$\n\nSince $2xy > 0$, we must have\n\n$$\n2xy \\geq \\frac{1}{2}(x + y + 1)(x + y - 1)\n$$\n\nwhich simplifies to\n\n$$\n2xy \\geq (x + y)^2 - 1\n$$\n\n$$\n1 \\geq (x - y)^2\n$$\n\nWithout loss of generality, suppose $x \\leq y$. Then either $x = y$ or $x + 1 = y$, as their difference can be at most 1.\n\nIf $x = y$, substitute into the first condition: $2x + 1 \\mid 2x^2$. This is impossible since $2x + 1$ is coprime to $x$ and to 2, so there is no solution with $x = y$.\n\nTherefore, $x + 1 = y$. Substituting into the original conditions, we get:\n\n$$\n\\begin{aligned}\nx + y + 1 \\mid 2xy &\\iff 2(x + 1) \\mid 2x(x + 1) \\\\\nx + y - 1 \\mid x^2 + y^2 - 1 &\\iff 2x \\mid 2x(x + 1)\n\\end{aligned}\n$$\n\nBoth conditions are satisfied for any $x$.\n\nThus, all possible pairs of positive integers $(x, y)$ satisfying the divisibility conditions are pairs of consecutive integers in either order.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19868,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbf{a}$ and $\\mathbf{b}$ be two non-constant sequences of integers such that for every index $i$, $a(i)$ is the average of $2b(i)+1$ consecutive terms of $a$ centered at $i$, and $b(i)$ is the average of $2a(i)+1$ consecutive terms of $b$ centered at $i$ (with the convention that $a(i)$ and $b(i)$ are non-negative integers). Prove that there are at least $N+1$ zeroes among the terms of $\\mathbf{a}$ and $\\mathbf{b}$, where $N$ is the length of the sequences.",
    "options": [],
    "answer": "See solution",
    "solution": "*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.\n\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 $a(i) = a(i+k) = M$.\n\nWe can now check that the harmonic properties are preserved under replacing all copies of $M$ in $a$ with 1:\n\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\n\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.\n\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 $a(i+1) \\le 1$.\n\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.\n\n**REMARKS.** It can be shown that there are at least $N+2$ zeroes across both sequences, a bound achieved if, for instance, $\\mathbf{a} = (0, 0, 1, 1, \\ldots, 1, 0)$ and $\\mathbf{b} = (1, 0, 0, \\ldots, 0)$.\n\n*Alternative Solution.* (Alexandra Timofte) Let $M_1 > M_2 > \\dots > M_N > 0$ list all nonzero values attained by the sequences $\\mathbf{a}$ and $\\mathbf{b}$. Say that an entry $a_i = M_j$ is *marginal* if either $a_{i-1}$ or $a_{i+1}$ is different from $M_j$.\n\nThe argument hinges on the claim below which is proved by induction on $k$.\n\n**Claim.** If $a_i = M_k$ is marginal, then $b_{i+s} = 0$ for all $s \\in [i - M_k, i + M_k]$; and the same holds, of course, with the roles of sequences swapped.\n\n*Proof.* The base case $k=1$ is just Lemma 1 from Solution 2.\n\nTo prove the step, assume that $a_i = M_k$ is a marginal entry, so that, say, $a_{i+1} \\neq M_k$. Assume, to the contrary, that $b_i \\neq 0$; then the segment $[i - b_i, i + b_i]$ contains an index $i+1$ such that $a_{i-1} \\neq M_k$; then the segment should also contain an index $j$ with $a_j > M_k$. Choose such an index nearest to $i$, and assume $a_j = M_\\ell$ for some $\\ell < k$; then $a_j$ is marginal, so $b_j = 0$, by the induction hypothesis.\n\nNow split into two cases.\n\n*Case 1.* $b_i \\le M_k$, so that $|j - i| \\le b_i \\le M_\\ell$. Then the segment $[j - a_j, j + a_j]$ contains the index $i$ with $b_i > 0$; this contradicts the $a_j$-harmonic property for $b_j = 0$.\n\n*Case 2.* $b_i > M_k$. Then $a_i = 0$ by the hypothesis, which contradicts our assumption. $\\square$\n\nAs in Lemma 1, it is now easy to show that in fact *all* instances of $M_k$ correspond to zeroes. Moreover, applying the arguments from the proof of the claim to a marginal instance of 0, we get that it corresponds to zero as well. This completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19869,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ such that for all $n \\in \\mathbb{N}$ and all $a_1, \\dots, a_n \\ge 1$, the following inequality holds:\n\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdots \\frac{a_n + x}{2} \\leq \\frac{a_1 a_2 \\cdots a_n + x}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $x \\in [-1, 1]$.\n\nIf $x \\geq -1$ satisfies the problem condition, then setting $a_1 = a_2 = \\dots = a_n = 1$ in the given inequality:\n\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdots \\frac{a_n + x}{2} \\leq \\frac{a_1 a_2 \\cdots a_n + x}{2} \\quad (1)\n$$\n\nyields the inequality $\\left(\\frac{1 + x}{2}\\right)^n \\leq \\frac{1 + x}{2}$, which gives $x \\leq 1$.\n\nWe prove, by induction on $n \\geq 2$, that for $x \\in [-1, 1]$ inequality (1) holds for all $n \\in \\mathbb{N}$ and all $a_1, \\dots, a_n \\geq 1$.\n\nFor $n = 2$, inequality (1) becomes:\n\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\leq \\frac{a_1 a_2 + x}{2}, \\quad (2)\n$$\n\nwhich is equivalent to:\n\n$$\nx^2 + (a_1 + a_2 - 2)x - a_1 a_2 \\leq 0.\n$$\n\nThis inequality holds for all $x \\in [-1, 1]$ and all $a_1, a_2 \\geq 1$. Indeed,\n\n$$\n\\begin{aligned}\n& x^2 + (a_1 + a_2 - 2)x - a_1 a_2 \\leq 1 + (a_1 + a_2 - 2) \\cdot 1 - a_1 a_2 \\\\\n& = a_1 + a_2 - a_1 a_2 - 1 = -(a_1 - 1)(a_2 - 1) \\leq 0.\n\\end{aligned}\n$$\n\nThus, the base of induction is proved.\n\nSuppose inequality (1) holds for some $n \\geq 2$. Then, from (1) and (2), it follows that:\n\n$$\n\\begin{aligned}\n& \\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdots \\frac{a_n + x}{2} \\cdot \\frac{a_{n+1} + x}{2} \\leq \\frac{a_1 a_2 \\cdots a_n + x}{2} \\cdot \\frac{a_{n+1} + x}{2} \\\\\n& \\leq \\frac{a_1 a_2 \\cdots a_n a_{n+1} + x}{2},\n\\end{aligned} \\quad (3)\n$$\n\nas required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19870,
    "subject": "Mathematics (Olympiad)",
    "question": "正整數 $n$ 與 $k$ 滿足 $n > 2023k^3$。貓貓國有 $n$ 座城市，其中每對城市之間有至多一條道路。已知該國的道路總數不少於 $2n^{3/2}$。證明：我們可以選出 $3k+1$ 座城市，使得以這些城市為兩端點的道路數量不少於 $4k$。",
    "options": [],
    "answer": "See solution",
    "solution": "題目等價於對於 $|V| = n$ 且 $|E| \\ge 2n^{3/2}$ 的圖 $G(V, E)$，我們可以取到邊數不少於 $4k$ 的子圖。\n\n讓我們先將 degree 最小的頂點依次從圖中移除，直到剩下的圖 $G' = (V', E')$ 中所有點的 degree 都大於 $n^{1/2}$。注意到以上動作至多移除 $n \\times n^{1/2}$，因此 $|E'| \\ge n^{3/2}$，從而 $m = |V'| \\ge \\sqrt{2|E'|} \\ge n^{3/4}$。\n\n任取 $V'$ 中的一個點 $v$，並令 $V_1$ 和 $V_2$ 分別為 $V$ 中距離 $v$ 單位 1 與 2 的點所形成的集合。將 $v$ 到 $V_1$ 的所有邊塗成藍色。此外，對於每個 $V_2$ 裡面的點 $y$，選定 $V_1$ 中與 $y$ 有連邊的一個點 $x$，並將 $xy$ 也塗成藍色。\n\n我們宣稱總是可以找到至少 $k$ 個相異的三環或四環通過 $v$。首先，由 $|V_1| = \\deg(v) > n^{1/2} \\ge m^{1/2}$，這表示從 $V_1$ 連到 $V_1 \\cap V_2$ 的邊至少有 $m^{1/2}$ 條。但同時，注意到藍邊的數量為 $|V_2| \\le |V| - |V_1| < m - m^{1/2}$，從而表示從 $V_1$ 連到 $V_1 \\cap V_2$ 的邊裡有至少 $m^{1/2} \\ge k$ 條邊不是藍的，而這每一條邊都會跟至多三條藍邊構成通過 $v$ 的三環（若它是從 $V_1$ 到 $V_1$）或四環（若它是從 $V_1$ 連到 $V_2$）。\n\n現在，令這些三環與四環的聯集為 $C$。注意到 $C$ 至多只有 $3k+1$ 個頂點（注意到它們共用頂點 $v$），且總邊數比總頂點數多 $k-1$（基於環的結構）。這表示，只要我們一開始取的 $v$ 所在的連通區塊有至少 $3k+1$ 個點，我們就可以從 $C$ 開始，逐步加入連通區域中的新點，直到總點數達到 $3k+1$，而此時的總邊數便至少是 $(3k+1) + (k-1) = 4k$。\n\n證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19871,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, $n \\ge 3$. In a regular $n$-gon, one draws a maximal set of diagonals, no two of which intersect in the interior of the $n$-gon. Every diagonal is labelled with the number of sides of the $n$-gon between the endpoints of the diagonal along the shortest path. Find the maximum value of the sum of the labels.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an arbitrary set of diagonals satisfying the conditions. Label the sides of the $n$-gon with number 1. The diagonals partition the $n$-gon into triangles; let $\\Delta$ be a triangle that contains the centroid of the $n$-gon. Let $s, t, u$ be the labels of the sides of $\\Delta$ (see below).\n\n![](images/EST_ABooklet_2020_p28_data_5171fffd7a.png)\n\nThe triangle $\\Delta$ divides the interior of the $n$-gon outside $\\Delta$ into three regions, each bounded by one side of $\\Delta$ and the respective sides of the $n$-gon (if a side of the $n$-gon coincides with a side of $\\Delta$, then the respective region contains 0 triangles).\n\nConsider the region bounded by the side labeled with $s$ and the respective section of the boundary of the $n$-gon. Repeatedly cut out triangles of this region whose two sides lie on the boundary of the remaining part of the $n$-gon. (Such a triangle must exist by the pigeonhole principle, as the initial number of triangles in the region is $s-1$ while the number of sides of the $n$-gon in this region is $s$. Every cut reduces both the number of triangles and the number of sides by one, so a similar argument holds for subsequent steps.)\n\nAs all triangles that are cut out are on the same side of the centroid of the polygon, the label of the longest side of each triangle equals the sum of labels on the two shorter sides. Cutting out a triangle removes two sides from the boundary of the region and introduces a new side labeled with the sum of the removed sides. Therefore, the sum of the labels on the boundary of the region is constant during the process. Let $m_i$ be the greatest number on the boundary of the region after $i$ triangles have been removed. As the boundary is initially labeled with $s$ ones and in every step the number of sides is reduced by one, after $i$ steps there are $s-i$ sides remaining. Therefore $s \\geq m_i + (s-i-1)$, from which $m_i \\leq i+1$. On the other hand, the label of the diagonal $d_i$ along which the cut is made on the $i$-th step satisfies $d_i \\leq m_i$. Hence the sum of all labels of the diagonals in this region (including the side of $\\Delta$ labeled with $s$) is not greater than $2+3+\\ldots+s$.\n\nAs the same is valid for the remaining two regions, the total sum of labels does not exceed $$(2+3+\\ldots+s) + (2+3+\\ldots+t) + (2+3+\\ldots+u).$$ We know $s+t+u=n$; without loss of generality, $\\frac{n}{2} \\geq s \\geq t \\geq u \\geq 1$. If $u>1$ then $t < \\lfloor \\frac{n}{2} \\rfloor$, whence a new triple $(s', t', u')$ with $u' = u-1$ also satisfies conditions: Replace $t$ with $t+1$ and change the order of $s$ and $t+1$ if needed. The total sum increases since $t+1 > u$. In the case of $u=1$, the only possibility is $s=t=\\frac{n-1}{2}$ if $n$ is odd and $s=\\frac{n}{2}, t=\\frac{n}{2}-1$ if $n$ is even. Hence the total sum of labels does not exceed $$2 \\cdot \\left(2+3+\\ldots+\\frac{n-1}{2}\\right) = \\frac{n^2-9}{4}$$ for odd $n$ and $$(2+3+\\ldots+\\left(\\frac{n}{2}-1\\right)) + (2+3+\\ldots+\\frac{n}{2}) = \\frac{n^2-8}{4}$$ for even $n$. These sums can be achieved by drawing all diagonals connecting a fixed vertex with all non-neighboring vertices.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19872,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that\n\n$$\n|\\angle PAB| + |\\angle PDC| \\le 90^\\circ \\quad \\text{and} \\quad |\\angle PBA| + |\\angle PCD| \\le 90^\\circ.\n$$\n\nProve that $|AB| + |CD| \\ge |BC| + |AD|$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $P$ is a common point of the given circles, the familiar properties of the angles subtending a chord at a point on a given circle and at its center imply that $P$ is also the point of tangency if and only if\n\n$$\n|\\angle ADP| + |\\angle BCP| = |\\angle APB|. \\qquad (1)\n$$\n\n![](images/CpsMT07_s_p4_data_6d36509db6.png)\n\nConsider now the circumcircles of the triangles $ABP$ and $CDP$ and assume for the moment that they meet also at another point $Q$ ($Q \\neq P$).\n\nSince the point $A$ lies outside the circle $BCP$, we have $|\\angle BCP| + |\\angle BAP| < 180^{\\circ}$. Therefore the point $C$ lies outside the circle $ABP$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same arc $CD$ of the circle $CDP$.\n\nAnalogously, the points $P$ and $Q$ lie on the same arc $AB$ of the circle $ABP$. Thus the point $Q$ lies either inside the angle $BPC$ or inside the angle $APD$. Without loss of generality assume that $Q$ lies inside the angle $BPC$.\n\n![](images/CpsMT07_s_p4_data_023b03dc0e.png)\n\nIn the chordal quadrilaterals $APQB$ and $DPQC$, it follows from the hypothesis of the problem that the angles at the vertices $A$ and $D$ are acute. Thus the corresponding opposite angles at the vertex $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $BPC$ but in fact inside the triangle $BPC$, hence also inside the quadrilateral $ABCD$.\n\nFrom the properties of the angles in the two chordal quadrilaterals just mentioned it thus follows that\n\n$$\n|\\angle BQC| = |\\angle PAB| + |\\angle PDC|,\n$$\nso by the hypothesis of the problem\n$$\n|\\angle BQC| \\le 90^{\\circ}. \\qquad (3)\n$$\n\nMoreover, since $|\\angle PCQ| = |\\angle PDQ|$, we get by (1)\n\n$$\n\\begin{aligned}\n|\\angle ADQ| + |\\angle BCQ| &= |\\angle ADP| + |\\angle PDQ| + |\\angle BCP| - |\\angle PCQ| \\\\\n&= |\\angle ADP| + |\\angle BCP|.\n\\end{aligned}\n$$\n\nThe last sum is equal to $|\\angle APB|$, according to the observation (1) applied to $T = P$. Since also $|\\angle APB| = |\\angle AQB|$, we obtain\n\n$$\n|\\angle ADQ| + |\\angle BCQ| = |\\angle AQB|.\n$$\n\nThis however means, as we have seen in the beginning, that the circles $BCQ$ and $DAQ$ are externally tangent at $Q$, contradicting our initial assumption that $P \\neq Q$. Thus it has to be the case that the circumcircles of the two triangles $ABP$ and $CDP$ have only the single point $P$ in common, for which, by the inequalities above, it is further true that the angles $APD$ and $BPC$ are not obtuse.\n\nConsider now the half-discs with diameters $BC$ and $DA$ constructed inwardly to the quadrilateral $ABCD$. Since the angles $APD$ and $BPC$ are not obtuse, these two half-discs lie entirely inside the circles $BQC$ and $AQD$; and since these two circles are externally tangent, the two half-discs cannot have any other point than $P$ in common. Denoting by $M$ and $N$ the midpoints of the sides $BC$ and $DA$, respectively, it thus follows that $|MN| \\ge \\frac{1}{2}(|BC| + |DA|)$.\n\nOn the other hand, since $\\overline{MN} = \\frac{1}{2}(\\overline{BA} + \\overline{CD})$, we have $|MN| \\le \\frac{1}{2}(|AB| + |CD|)$. Thus indeed $|AB| + |CD| \\ge |BC| + |DA|$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19873,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\mathbb{N}$ 為全體正整數所成之集合，並令\n\n$$\nA = \\{2^a + a^2 \\mid a \\in \\mathbb{N}\\}.\n$$\n\n考慮一個定義在 $\\mathbb{N}$ 上的函數序列 $\\{h_i\\}_{i \\in \\mathbb{N}}$ 如下：\n\n$$\nh_1(n) = n + \\log_2 n, \\quad h_{i+1}(n) = n + \\log_2 h_i(n), \\quad \\forall i \\in \\mathbb{N}.\n$$\n\n證明：存在一個單射函數 $g : \\mathbb{R}^+ \\to \\mathbb{R}^+$ 滿足下列性質：\n\n(a) 對每個正整數 $i$ 及 $n$，$n + \\lfloor g(h_i(n)) \\rfloor \\notin A$ 恆成立。\n\n(b) 對每個 $k \\in \\mathbb{N} \\setminus A$ 及 $i \\in \\mathbb{N}$，都存在唯一的 $a_{k,i} \\in \\mathbb{N}$ 滿足：\n\n$$\na_{k,i} + \\lfloor g(h_i(a_{k,i})) \\rfloor = k, \\quad \\forall i \\in \\mathbb{N}.\n$$\n\n其中 $\\lfloor x \\rfloor$ 表示不超過實數 $x$ 的最大整數。",
    "options": [],
    "answer": "See solution",
    "solution": "The statement holds when $g$ is the inverse function of $2^x + x^2 + 1$ in $\\mathbb{R}^+$. The details are as follows.\n\nNote that $\\mathbb{N} \\setminus A = \\{1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 18, \\dots\\} = \\{b_n : n \\in \\mathbb{N}\\}$.\nAfter rearrangement, we can assume $b_n < b_{n+1}$ for $n \\in \\mathbb{N}$. Consider the sequence $\\{b_n - n : n \\in \\mathbb{N}\\}$, which is non-decreasing. For each $m \\in \\mathbb{N} \\cup \\{0\\}$, there exist at least two $n$ such that $b_n - n = m$.\n\nMore precisely, for each fixed $m \\ge 1$, the number of distinct elements in the set $B_m := \\{n : b_n - n = m\\}$ is\n\n$$\n((m+1)^2 + 2^{m+1}) - (m^2 + 2^m) - 1 = 2m + 2^m.\n$$\n\nHence, if $n \\in B_m$ with $m \\ge 1$, then\n\n$$\n2 + \\sum_{l=1}^{m-1} (2l + 2^l) + 1 \\le n \\le 2 + \\sum_{l=1}^{m} (2l + 2^l).\n$$\n\nimplying\n\n$$\n2^m + m^2 - m + 1 \\le n \\le 2^{m+1} + m^2 + m, \\text{ for } n \\in B_m. \\quad (1)\n$$\n\nIn particular, $2^m < n < 2^{m+2}$ and thus $m < \\log_2 n < m + 2$. Along with (1), we get\n\n$$\n2^m + m^2 + 1 < n + \\log_2 n < 2^{m+1} + (m+1)^2 + 1, \\text{ for } n \\in B_m. \\quad (2)\n$$\n\nNote also that $2^x + x^2 + 1$ is strictly increasing for $x > 0$. Hence, there exists a strictly increasing function $g : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that its inverse $g^{-1}(x) = 2^x + x^2 + 1$. This, along with (2), yields\n\n$$\nm < g(h_1(n)) < m + 1, \\text{ for } n \\in B_m. \\quad (3)\n$$\n\nOn the other hand, by (2), $2^m < n + \\log_2 n < 2^{m+2}$, and hence $m < \\log_2(n + \\log_2 n) < m + 2$. Along with (1), we get $2^m + m^2 + 1 \\le n + \\log_2(n + \\log_2 n) \\le 2^{m+1} + (m+1)^2 + 1$, i.e.,\n\n$$\nm < g(h_2(n)) < m + 1, \\text{ for } n \\in B_m. \\quad (4)\n$$\n\nRepeating the same argument, we can prove\n\n$$\nm < g(h_i(n)) < m + 1, \\forall i \\in \\mathbb{N}, \\forall n \\in B_m. \\quad (5)\n$$\n\nHence, $\\lfloor g(h_i(n)) \\rfloor = m$, $\\forall n \\in B_m$, $\\forall i \\in \\mathbb{N}$. As a consequence,\n\n$$\nb_n = n + \\lfloor g(h_i(n)) \\rfloor \\in \\mathbb{N} \\setminus A, \\forall n \\in \\mathbb{N}.\n$$\n\nFurthermore, for $k \\in \\mathbb{N} \\setminus A$, there exists $n := a_{i,k}$ such that $k = a_{i,k} + \\lfloor g(h_i(a_{i,k})) \\rfloor$. Since both $g$ and $h_i$ are strictly increasing functions, the uniqueness for $a_{i,k}$ is trivial. This completes the proof of (a) and (b). $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19874,
    "subject": "Mathematics (Olympiad)",
    "question": "How many different remainders can result when the 100th power of an integer is divided by 125?\n\n(A) 1  \n(B) 2  \n(C) 5  \n(D) 25  \n(E) 125",
    "options": [],
    "answer": "See solution",
    "solution": "Write $N = 5k + r$ for $r = 0, 1, 2, 3$, or $4$. If $r = 0$, then $N = 5k$ and $N^{100}$ is divisible by 125, so the remainder is 0. If $r = 1, 2, 3$, or $4$, then $N^2 = 25k^2 + 10rk + r^2 = 5m \\pm 1$ for some integer $m$. Now use the Binomial Theorem:\n\n$$\n\\begin{aligned}\nN^{100} = (N^2)^{50} = (5m \\pm 1)^{50} = (5m)^{50} &\\pm 50(5m)^{49} + \\binom{50}{2}(5m)^{48} \\pm \\dots \\\\\n&\\pm \\binom{50}{47}(5m)^3 + \\binom{50}{48}(5m)^2 \\pm 50(5m) + 1.\n\\end{aligned}\n$$\n\nAll the terms except the final term have at least 3 factors of 5, so $N^{100}$ has remainder 1 upon division by 125. Therefore, there are only 2 possible remainders: 0 and 1.\n\n**OR**\n\nLet $\\phi(n)$ be the number of positive integers less than $n$ that are relatively prime to $n$; this is Euler's totient function. Then $\\phi(125) = 5^3 - 5^2 = 100$. By Euler's Totient Theorem, if $a$ is not a multiple of 5, then $a^{100} \\equiv 1 \\pmod{125}$. If $a$ is a multiple of 5, then $a^{100} \\equiv 0 \\pmod{125}$. Therefore, there are only 2 possible remainders: 0 and 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19875,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, let an isosceles triangle $ABC$ be given with $AB = AC$. A variable circle $(O)$ with center $O$ on the line $BC$ passes through $A$ but does not touch the lines $AB$ or $AC$. Let $M$ and $N$ be the second points of intersection of the circle $(O)$ with the lines $AB$ and $AC$, respectively. Find the locus of the orthocenter of triangle $AMN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Outline of the proof:\n\n**Case 1:** $\\angle A = 90^\\circ$\n\nThe locus is the singleton $\\{A\\}$.\n\n**Case 2:** $\\angle A \\neq 90^\\circ$\n\nLet $D$ be the point symmetric to $A$ with respect to $BC$. Let $K$ be the point symmetric to $D$ with respect to $MN$. Then the line $HK$ is the image of the line $BC$ under the homothety with center $D$ and ratio $4\\sin^2\\left(\\frac{A}{2}\\right)$. Thus, the locus of the orthocenter of triangle $AMN$ is $d \\setminus \\{H_1, H_2\\}$, where $d$ is the image of $BC$ under the above-mentioned homothety, and $H_1, H_2$ are the points on $d$ such that $DH_1 \\perp DB$, $DH_2 \\perp DC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19876,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a trinomial $ax^2 + bx + c$ with integer odd coefficients $a$, $b$, $c$, such that $\\frac{1}{2022}$ is its root?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a trinomial $ax^2 + bx + c$ exists with $a$, $b$, $c$ odd integers and $\\frac{1}{2022}$ as a root. Then:\n\n$$\nax^2 + bx + c = 0 \\implies a \\left(\\frac{1}{2022}\\right)^2 + b \\left(\\frac{1}{2022}\\right) + c = 0\n$$\n\nwhich simplifies to:\n\n$$\n\\frac{a}{2022^2} + \\frac{b}{2022} + c = 0 \\implies a + 2022b + 2022^2c = 0.\n$$\n\nSince $2022$ is even, $2022b$ and $2022^2c$ are even, so $a$ must also be even for the sum to be zero. But $a$ is odd, a contradiction. Thus, no such trinomial exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19877,
    "subject": "Mathematics (Olympiad)",
    "question": "Tomorrow, the Janssen family will be travelling by car and they have a nice route in mind. The youngest of the family notes that their planned stopover in Germany is exactly halfway along the route in terms of distance. Father responds: \"When we cross the border after 150 kilometres tomorrow, our stopover will only be on one fifth of the remaining route.\"\n\nHow many kilometres long is the Janssen family's route?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the total length of the route in kilometres. The stopover is halfway, so it is at $\\frac{x}{2}$ km from the start.\n\nAfter 150 km, the remaining distance is $x - 150$ km. The stopover is $\\frac{x}{2} - 150$ km ahead of the border. According to the father, this is one fifth of the remaining route:\n\n$$\n\\frac{x}{2} - 150 = \\frac{1}{5}(x - 150)\n$$\n\nMultiply both sides by 5:\n\n$$\n5\\left(\\frac{x}{2} - 150\\right) = x - 150\n$$\n\n$$\n\\frac{5x}{2} - 750 = x - 150\n$$\n\n$$\n\\frac{5x}{2} - x = 750 - 150\n$$\n\n$$\n\\frac{3x}{2} = 600\n$$\n\n$$\nx = 400\n$$\n\nSo, the route is $400$ kilometres long.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19878,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that both $n + 2008$ divides $n^2 + 2008$ and $n + 2009$ divides $n^2 + 2009$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write $n^2 + 2008 = m(n + 2008)$ and $n^2 + 2009 = l(n + 2009)$ for some $l, m \\ge 0$.\n\nSubtracting these, we get\n\n$$\n(m - l)n + 2008m - 2009l = -1,\n$$\n\nwhich rearranges to\n\n$$\n(m - l)(n + 2008) = l - 1.\n$$\n\nWe could have $m = l$ and then $l - 1 = 0$. This means we have $n^2 + 2008 = n + 2008$ and $n^2 + 2009 = n + 2009$, so $n^2 = n$, so we get the solution $n = 1$ (as $0$ isn't positive). This is indeed a solution, as can easily be checked.\n\nIf however $m \\neq l$ then $n + 2008 = \\frac{l - 1}{m - l}$. We must have $(m - l)$ positive (since $n + 2008$ and $l - 1$ are both positive).\n\nThus\n\n$$\nn + 2008 = \\frac{l - 1}{m - l} \\leq l - 1 \\quad (1)\n$$\n\nBut $n^2 - ln - 2009(l - 1) = 0$, and so\n\n$$\nn = \\frac{l \\pm \\sqrt{l^2 + 4 \\cdot 2009(l - 1)}}{2}.\n$$\n\nBut $l^2 + 4 \\cdot 2009(l - 1) \\geq l^2$. So either $n < 0$ (which is not permitted) or $n > l$ (which contradicts (1)).\n\nHence $n = 1$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19879,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfies, for all real numbers $x, y$, the equation\n$$\nf(x) + f(f(y)) + f(f(f(x))) = x + y?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Taking $y = f(x)$ in the given equation, we get\n$$\nf(x) + f(f(f(x))) + f(f(f(x))) = x + f(x),\n$$\nfrom which, by expressing $f(f(f(x)))$, we find\n$$\nf(f(f(x))) = \\frac{x}{2}. \\qquad (7)\n$$\nTaking $y = 0$ in the given equation, we get $f(x) + f(f(0)) + f(f(f(x))) = x$, and using equation (7), we get after simplification\n$$\nf(x) = \\frac{x}{2} - f(f(0)). \\qquad (8)\n$$\nSubstituting $x = f(f(0))$ into equation (8) and applying equation (7) on the left-hand side, we get\n$$\n0 = \\frac{f(f(0))}{2} - f(f(0)), \\qquad (9)\n$$\nwhich after simplification gives $f(f(0)) = 0$. Consequently, equation (8) takes the form $f(x) = \\frac{x}{2}$. But in this case, $f(f(f(x))) = \\frac{x}{8}$, whereas from equation (7), $f(f(f(x))) = \\frac{x}{2}$. Since in general $\\frac{x}{2} \\ne \\frac{x}{8}$, there are no functions that satisfy the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19880,
    "subject": "Mathematics (Olympiad)",
    "question": "How many 6-digit positive integers have digits that sum to 18?\n",
    "options": [],
    "answer": "See solution",
    "solution": "We use generating functions. Since the first digit is nonzero, it corresponds to a factor $x + x^2 + \\cdots + x^9$. All other digits can be one of $0, 1, \\ldots, 9$, so each of them corresponds to a factor $1 + x + \\cdots + x^9$. Thus, the answer is the coefficient of $x^{18}$ in\n\n$$\nf(x) = (x + x^2 + \\cdots + x^9)(1 + x + \\cdots + x^9)^5.\n$$\n\nNow,\n\n$$\n\\begin{align*}\nf(x) &= \\frac{x - x^{10}}{1 - x} \\cdot \\left( \\frac{1 - x^{10}}{1 - x} \\right)^{5} \\\\\n&= (x - x^{10}) (1 - x^{10})^{5} (1 - x)^{-6} \\\\\n&= (x - x^{10}) \\left( \\sum_{r=0}^{5} \\binom{5}{r} (-1)^r x^{10r} \\right) \\left( \\sum_{s=0}^{\\infty} \\binom{-6}{s} (-1)^s x^s \\right)\n\\end{align*}\n$$\n\nThe answer is $21087$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19881,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n$ airports, how many possible networks fulfill the following condition: For any two airports, either they are directly connected, or the airport with higher priority is directly connected to all other airports? Find a formula for the number of such networks.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction that there are $2^{n-1}$ possible networks fulfilling this condition if there are $n$ airports.\n\nFor $n=2$, this is obvious (there can be a connection between the two airports or not).\n\nFor the induction step, let $H$ and $L$ be the airports whose priorities are highest and lowest, respectively, and consider the following two possibilities:\n\n**Case 1:** $H$ and $L$ are connected. Then there are direct connections from $H$ to all other airports as well. Now the condition is always trivially satisfied if $H$ is involved, and we only have to consider the remaining $n-1$ airports. There are $2^{n-2}$ possible networks between these airports by the induction hypothesis.\n\n**Case 2:** $H$ and $L$ are not connected. Then $L$ cannot be connected to any of the airports, and we can ignore $L$. By the induction hypothesis, there are $2^{n-2}$ feasible networks that connect the remaining $n-1$ airports.\n\nAltogether, we have $2^{n-2} + 2^{n-2} = 2^{n-1}$ possible networks, which completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19882,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, and $C$ lie on a line in this order. For every circle $k$ passing through $B$ and $C$, let $D$ be one of the common points of $k$ and the bisector of $BC$. Furthermore, let $E$ be the second common point of the line $AD$ and $k$.\n\nProve that the ratio $BE : CE$ is constant for all circles $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F$ be the diametrically opposite point to $D$ on $k$. Since $F$ is a point on the bisector of $BC$, we have $FB = FC$, and therefore $\\angle BEF = \\angle CEF$.\n\n![](images/AustriaMO2013_p7_data_8deeb61580.png)\n\nWe see that $EF$ is the internal bisector of $\\angle BEC$, and since $ED \\perp EF$, $ED = AD$ is the external bisector. If we name $BC \\cap EF = H$, we see that $A$ and $H$ are harmonic with respect to $B$ and $C$, since they are the points of intersection of perpendicular bisectors $EA$ and $EF$ with $BC$. Since $A$ is independent of the choice of $k$, we see that $H$ must be as well, and since $BC : CE = BH : CH$, the ratio is independent of the choice of $k$, as claimed. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19883,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be any two consecutive points on the convex hull of $S$. Orient the diagram so that $AB$ is horizontal and no point of $S$ lies below the line $AB$. Of all points in $S$ that lie strictly above the line $AB$, let $C$ be a point such that $\\angle ACB$ is maximal. Prove that the circumcentre, $O$, of $\\triangle ABC$ is not in $S$.\n\n![](images/Australian-Scene-2017_p102_data_fc830d419b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1** $0^\\circ < \\angle ACB < 90^\\circ$\n\nWe have $\\angle AOB = 2\\angle ACB$. Hence $0^\\circ < \\angle ACB < \\angle AOB < 180^\\circ$. Thus $O$ lies above the line $AB$ and satisfies $\\angle AOB > \\angle ACB$. From the maximality of $\\angle ACB$, we conclude that $O$ is not in $S$, as desired.\n\n**Case 2** $\\angle ACB = 90^\\circ$\n\nThe point $O$ is the midpoint of $AB$. Since $A$ and $B$ were chosen to be consecutive points on the convex hull of $S$, it follows that $O$ is not in $S$, as desired.\n\n**Case 3** $90^\\circ < \\angle ACB < 180^\\circ$\n\nThe point $O$ lies below the line $AB$ and so is not in $S$, as desired. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19884,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle BAC = 90^\\circ$ and altitude $AH$ ($H \\in BC$). A circle $\\omega$ passes through $B$ and $C$ and cuts the segments $AB$ and $AC$ at $M$ and $N$, respectively. Circle $\\omega$ also cuts the line $AH$ at $D$ and $E$ ($D$ lies between $A$ and $H$). Suppose that $DE = AH\\sqrt{5}$. Prove that the circumcircle of $HMN$ is tangent to $BC$.\n\n![](images/Saudi_Booklet_2025_p34_data_e302c30d52.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Based on the power from $H$ to the circle $\\omega$, $HD \\cdot HE = HB \\cdot HC = AH^2$. Moreover, $HD + HE = DE = AH\\sqrt{5}$. Thus, the lengths $HD$ and $HE$ are the solutions of the quadratic equation\n\n$$\nx^2 - AH\\sqrt{5} \\cdot x + AH^2 = 0.\n$$\n\nSince $D$ is inside triangle $ABC$, $HD < AH$, so $HE > AH$ and $HE > HD$. Solving the equation, we get\n\n$$\nHD = \\frac{-1 + \\sqrt{5}}{2}AH, \\quad HE = \\frac{1 + \\sqrt{5}}{2}AH.\n$$\n\nHence,\n\n$$\nAD = AH - HD = \\frac{3 - \\sqrt{5}}{2}AH, \\quad AE = AH + HE = \\frac{3 + \\sqrt{5}}{2}AH.\n$$\n\nIt follows that\n\n$$\nAD \\cdot AE = \\left(\\frac{3 - \\sqrt{5}}{2}\\right) \\left(\\frac{3 + \\sqrt{5}}{2}\\right) AH^2 = AH^2.\n$$\n\nFurthermore, by the power from $A$ to the circle $\\omega$, $AD \\cdot AE = AM \\cdot AB = AN \\cdot AC$. Therefore $AH^2 = AM \\cdot AB = AN \\cdot AC$, which implies that $HM \\perp AB$, $HN \\perp AC$, so the quadrilateral $AMHN$ is a rectangle. Therefore, the circle circumscribing triangle $HMN$ is also the circle with diameter $AH$, so it will be tangent to $BC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19885,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(m, n)$ such that for every polynomial $P(x)$ of degree $m$ and every polynomial $Q(x)$ of degree $n$ over $\\mathbb{R}$, $Q(P(x)) \\not\\equiv Q(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "All $(m, n)$ with odd $m$ and arbitrary $n$; or $(m, n)$ with even $m$ and even $n$.\n\n*Case 1: $m$ odd.*\nFor any $n$, if $P(x) \\equiv x$, then $Q(P(x)) \\not\\equiv Q(x)$ for any $Q \\in \\mathbb{R}[x]$. If $P(x) \\not\\equiv x$, then $P(x) - x$ has odd degree, so it has a real root $a$. For any $n \\in \\mathbb{N}$, set $Q(x) = (x-a)^n$. Then $Q(P(x)) = (P(x)-a)^n$. Since $P(x)-a \\not\\equiv x-a$, $Q(P(x)) \\not\\equiv Q(x)$.\n\n*Case 2: $m$ even.*\nIf $n$ is odd, set $P(x) = x^m + x + 1$. Then $P(x) - x > 0$ for all $x \\in \\mathbb{R}$. If $Q(x)$ has real roots, let $c$ be the largest real root. Then $Q(P(c)) \\ne 0$, so $Q(P(x)) \\not\\equiv Q(x)$.\n\nIf both $m$ and $n$ are even, $n = 2k$:\n- If $P(x) - x$ has a real root $a$, set $Q(x) = (x-a)^n$.\n- If $P(x) - x$ has no real roots, let $z, \\bar{z}$ be complex-conjugate roots of $P(x)$. Then $P(x) - x$ is divisible by $p(x) = (x-z)(x-\\bar{z})$. Set $Q(x) = (p(x))^k$. Then $Q(P(x)) = (p(P(x)))^k$. Since $p(P(x)) : p(x)$, $Q(P(x)) : Q(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19886,
    "subject": "Mathematics (Olympiad)",
    "question": "On a $10 \\times 10$ chessboard, some $4n$ unit square fields are chosen to form a region $R$. This region $R$ can be tiled by $n$ $2 \\times 2$ squares. If $R$ can also be tiled by a combination of $n$ pieces of the following types of shapes (with rotations allowed):\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p140_data_b153144786.png)\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p140_data_de7a18db88.png)\n\nDetermine the minimum value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 4$. We call those two kinds of nonsquare tiles *ducks*.\n\nFirst, the left-hand-side figure and the middle figure below show that $n = 4$ works.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p140_data_26fc5518c3.png)\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p140_data_3007e7ee5e.png)\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p140_data_f239bb0281.png)\n\nSecond, we show that $n$ must be even. We mark the (infinite) chessboard with $\\times$ in the pattern indicated in the right-hand-side figure. It is easy to see that each $2 \\times 2$ covers exactly an even number of crosses (either two or four crosses) and each duck covers exactly an odd number of crosses (either one or three crosses). It follows that we must have an even number of ducks in $R$, i.e. $n$ is even.\n\nThird, we show that $n \\ge 2$. If $n = 2$, then $R$ can be tiled by two $2 \\times 2$ squares. It is clear that these two squares must share a common edge. Hence, we can have only two possibilities:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p141_data_087111f811.png)\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p141_data_df7a0d3a8a.png)\n\nIt is not difficult to see that it is not possible to tile either of the above configurations by a combination of two ducks.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 19887,
    "subject": "Mathematics (Olympiad)",
    "question": "A pile of 2010 coins is given. We take one coin from the pile, and we arbitrarily divide the rest into two piles. Then, we choose an arbitrary pile from the two, take one coin, and divide the rest into two arbitrary piles, and so on. Is it possible, after a finite number of repetitions of this procedure, to get a number of piles such that in every one of them there are 3 coins?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the initial pile contain 2010 coins.\n\nAfter each step, define $S$ as the sum of the number of piles and the number of coins in all piles.\n\nWe have:\n\n$$\nS_0 = 1 + 2010 = 2011 \\\\\nS_1 = 2 + (2010 - 1) = 2011 \\\\\nS_2 = 3 + (2009 - 1) = 2011\n$$\n\nSo $S$ is invariant at every step.\n\nSuppose the desired state is attainable: after some steps, there are $n$ piles with 3 coins in each. Then $S = n + 3n = 4n$. Thus $2011 = 4n$, which is impossible since 2011 is not divisible by 4. Therefore, the required state is not attainable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19888,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be positive real numbers such that $ab + bc + ca = \\frac{1}{3}$. Prove the inequality:\n\n$$\n\\frac{a}{a^2 - bc + 1} + \\frac{b}{b^2 - ca + 1} + \\frac{c}{c^2 - ab + 1} \\ge \\frac{1}{a + b + c}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that the denominators on the left side are positive. From the Cauchy-Schwarz inequality, we obtain:\n\n$$\n\\begin{aligned}\n\\frac{a}{a^2 - bc + 1} + \\frac{b}{b^2 - ca + 1} + \\frac{c}{c^2 - ab + 1} &= \\frac{a^2}{a^3 - abc + a} + \\frac{b^2}{b^3 - abc + b} + \\frac{c^2}{c^3 - abc + c} \\\\\n&\\ge \\frac{(a + b + c)^2}{a^3 + b^3 + c^3 + a + b + c - 3abc}\n\\end{aligned}\n$$\n\nNow, since\n\n$$\n\\begin{aligned}\na^3 + b^3 + c^3 - 3abc &= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\\\\n&= (a + b + c)(a^2 + b^2 + c^2 - \\frac{1}{3})\n\\end{aligned}\n$$\n\nwe get:\n\n$$\n\\frac{1}{y_1 + 1} + \\frac{1}{y_2 + 1} + \\frac{1}{y_3 + 1} \\ge 1\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19889,
    "subject": "Mathematics (Olympiad)",
    "question": "Show, for example, that the polynomials\n\n$$\nP(x) = 4x^4 + 4x^3 + 1 \\quad \\text{and} \\quad Q(x) = x^4 + 4x + 4\n$$\n\nsatisfy the condition that both have no real roots, but the polynomials\n\n$$\nP_Q(x) = 4x^4 + 4x + 1 \\quad \\text{and} \\quad Q_P(x) = x^4 + 4x^3 + 4\n$$\n\neach have at least one real root.",
    "options": [],
    "answer": "See solution",
    "solution": "It is evident that the polynomials $P(x)$ and $Q(x)$ satisfying the condition must have even degree (since any odd degree polynomial has at least one real root).\n\nShow that the degree of these polynomials is greater than 2. Suppose, contrary to our claim, that there are trinomials $P(x) = a_1x^2 + b_1x + c_1$ and $Q(x) = a_2x^2 + b_2x + c_2$ having no real roots. Then at least one of the trinomials $P_Q(x) = a_1x^2 + b_1x + c_1$ and $Q_P(x) = a_2x^2 + b_2x + c_2$ has no real roots. Indeed, without loss of generality, assume $|b_2| \\ge |b_1|$. Since the discriminant of $Q(x)$ is negative and $|b_2| \\ge |b_1|$, we see that $0 > b_2^2 - 4a_2c_2 \\ge b_1^2 - 4a_2c_2$. However, the last expression is the discriminant of $Q_P(x)$, so this trinomial has no real roots.\n\nThe example from part a) shows the minimal degree of such polynomials is $4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19890,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $s_1, s_2, s_3, \\ldots$ is a strictly increasing sequence of positive integers such that the subsequences $s_{s_1}, s_{s_2}, s_{s_3}, \\ldots$ and $s_{s_1+1}, s_{s_2+1}, s_{s_3+1}, \\ldots$ are both arithmetic progressions. Prove that the sequence $s_1, s_2, s_3, \\ldots$ is itself an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from the assumption that $s_{s_1}, s_{s_2}, s_{s_3}, \\ldots$ and $s_{s_1+1}, s_{s_2+1}, s_{s_3+1}, \\ldots$ are both strictly increasing sequences of positive integers.\n\nAssume that $s_{s_k} = a + (k-1)d_1$, $s_{s_{k+1}} = b + (k-1)d_2$, $k=1,2,\\ldots$, where $a, b, d_1, d_2$ are positive integers. Since $s_k < s_{k+1}$, by the monotonicity of the sequence $\\{s_n\\}$ we have\n\n$$\ns_{s_k} < s_{s_{k+1}} \\le s_{s_{k+1}},\n$$\n\ni.e.\n\n$$\na + (k-1)d_1 < b + (k-1)d_2 \\le a + k d_1,\n$$\n\nor, equivalently,\n\n$$\na - b < (k-1)(d_2 - d_1) \\le a + d_1 - b.\n$$\n\nAs $k$ is arbitrary, we must have $d_2 - d_1 = 0$, i.e. $d_2 = d_1$, denoted by $d$ for both $d_1$ and $d_2$. Let $b - a = c \\in \\mathbb{N}^*$. If $d=1$, by the monotonicity of $\\{s_n\\}$ we have $s_{k+1} = s_k + 1 \\le s_{k+1}$, and thus $s_{k+1} \\le s_k + 1$.\n\nSince $s_{k+1} > s_k$, we get $s_{k+1} = s_k + 1$, i.e. $\\{s_n\\}$ is an arithmetic progression, which is what we want. In what follows, we assume that $d > 1$.\n\nWe shall prove that $s_{k+1} - s_k = c$ for any positive integer $k$. Supposing on the contrary that it is not true, we discuss two cases.\n\n**Case 1:** There exists a positive integer $k$ such that $s_{k+1} - s_k < c$. Since $s_{k+1} - s_k$ are positive integers, we may assume that $s_{i+1} - s_i = c_0$ attains the minimum value for some $i$; then\n\n$$\n\\begin{aligned}\ns_{a+id} - s_{a+(i-1)d+1} &= s_{s_i+1} - s_{s_i+1} \\\\\n&= (a + (s_{i+1} - 1)d) - (b + (s_i - 1)d) \\\\\n&= c_0 d - c.\n\\end{aligned}\n$$\n\nOn the other hand, since\n\n$$\n(a + i d) - (a + (i-1)d + 1) = d - 1,\n$$\n\nwe have\n\n$$\ns_{a+id} - s_{a+(i-1)d+1} \\ge c_0 (d-1)\n$$\n\n(here we have used the minimality of $c_0$). Comparing with the previous result, we have $c_0 \\ge c$, a contradiction.\n\n**Case 2:** There exists a positive integer $k$ such that $s_{k+1} - s_k > c$. Since $s_{k+1} - s_k$ are integers, and for any $k$,\n\n$$\ns_{k+1} - s_k \\le s_{s_{k+1}} - s_{s_k} = d,\n$$\n\nwe may assume that $s_{j+1} - s_j = c_1$ attains the maximum value for some $j$; then\n\n$$\n\\begin{aligned}\ns_{a+jd} - s_{a+(j-1)d+1} &= s_{s_{j+1}} - s_{s_{j}+1} \\\\\n&= (a + (s_{j+1} - 1)d) - (b + (s_j - 1)d) \\\\\n&= c_1 d - c.\n\\end{aligned}\n$$\n\nOn the other hand, since\n\n$$\n(a + j d) - (a + (j-1)d + 1) = d - 1,\n$$\n\nwe have\n\n$$\ns_{a+jd} - s_{a+(j-1)d+1} \\le c_1 (d-1)\n$$\n\n(here we have used the maximality of $c_1$). Comparing with the previous result, we have $c_1 \\le c$, a contradiction.\n\nWe have verified that $s_{k+1} - s_k = c$ for any positive integer $k$, i.e. $\\{s_n\\}$ is an arithmetic progression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19891,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, the function $f_n : [0, n] \\to \\mathbb{R}$ is defined by $f_n(x) = \\arctan(\\lfloor x \\rfloor)$. Prove that $f_n$ is a Riemann integrable function and find\n$$\n\\lim_{n \\to \\infty} \\frac{1}{n} \\int_{0}^{n} f_{n}(x) \\, dx.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f_n$ is locally constant, hence Riemann integrable.\n\nNext, we have\n$$\n\\int_{0}^{n} f_{n}(x) \\, dx = \\sum_{i=0}^{n-1} \\int_{i}^{i+1} f_{n}(x) \\, dx = \\sum_{i=0}^{n-1} \\arctan i.\n$$\nApplying the Stolz-Cesàro theorem, we obtain\n$$\n\\lim_{n \\to \\infty} \\frac{\\arctan 0 + \\arctan 1 + \\dots + \\arctan (n-1)}{n} = \\lim_{n \\to \\infty} \\arctan n = \\frac{\\pi}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19892,
    "subject": "Mathematics (Olympiad)",
    "question": "Hallar todas las soluciones enteras $(x, y)$ de la ecuación\n\n$$\ny^k = x^2 + x\n$$\n\ndonde $k$ es un número entero dado mayor que $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Puesto que $y^k = x^2 + x = x(x+1)$ y $\\gcd(x, x+1) = 1$, resulta que tanto $x$ como $x+1$ deben ser potencias $k$-ésimas de un entero. Pero los dos únicos números enteros consecutivos que son potencias $k$-ésimas, con $k > 1$, son $0$ y $1$ o bien $-1$ y $0$. Las dos únicas soluciones son, pues, $x = 0$, $y = 0$ y $x = -1$, $y = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19893,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n$, determine the greatest possible value of the quotient\n\n$$\n\\frac{1 - x^n - (1-x)^n}{x(1-x)^n + (1-x)x^n}\n$$\n\nwhere $0 < x < 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 1 - x$, so $x + y = 1$. The expression becomes\n\n$$\n\\frac{1 - x^n - y^n}{x y^n + y x^n}.\n$$\n\nWe claim the maximum occurs at $x = y = \\frac{1}{2}$, giving value $2^n - 2$.\n\nRewrite:\n\n$$\n\\begin{aligned}\n\\frac{1 - x^n - y^n}{x y^n + y x^n} &= \\frac{x + y - x^n - y^n}{x y^n + y x^n} \\\\\n&= \\frac{x(1 - x^{n-1}) + y(1 - y^{n-1})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x(1-x)(1 + x + \\cdots + x^{n-2}) + y(1-y)(1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{x y (1 + x + \\cdots + x^{n-2} + 1 + y + \\cdots + y^{n-2})}{x y (x^{n-1} + y^{n-1})} \\\\\n&= \\frac{1 + 1}{x^{n-1} + y^{n-1}} + \\frac{x + y}{x^{n-1} + y^{n-1}} + \\cdots + \\frac{x^{n-2} + y^{n-2}}{x^{n-1} + y^{n-1}}.\n\\end{aligned}\n$$\n\nIt suffices to show that\n\n$$\n\\frac{x^a + y^a}{x^b + y^b}\n$$\n\nis maximized at $x = y = \\frac{1}{2}$ for $0 \\leq a < b$. By the general mean inequality,\n\n$$\n\\left( \\frac{x^a + y^a}{2} \\right) \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{a/b},\n$$\n\nand\n\n$$\n\\frac{1}{2} = \\frac{x + y}{2} \\leq \\left( \\frac{x^b + y^b}{2} \\right)^{1/b}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19894,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a line through the vertex $D$ of a given parallelogram $ABCD$, such that the vertices $A$, $B$, and $C$ are on the same side of $p$. Let $A'$, $B'$, and $C'$ be the feet of the perpendiculars from $A$, $B$, and $C$ to $p$, respectively. Prove that $\\overline{BB'} = \\overline{AA'} + \\overline{CC'}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We draw a line parallel to $p$ through $A$ and let this line intersect $BB'$ at $Q$. The quadrilateral $AQB'A'$ has three right angles, so it is a rectangle, from which we obtain $\\overline{AA'} = \\overline{QB'}$.  \nBecause $ABCD$ is a parallelogram, $\\overline{AB} = \\overline{DC}$ and $\\overline{AB} \\parallel \\overline{DC}$. The angles $\\angle QAB$ and $\\angle C'DC$ are equal. Since $BB'$ and $CC'$ are perpendicular to $p$, they are parallel, so $\\angle QBA = \\angle C'CD$. Because $\\overline{AB} = \\overline{DC}$, $\\angle QAB = \\angle C'DC$, and $\\angle QBA = \\angle C'CD$, we conclude that $\\triangle ABQ \\cong \\triangle DCC'$. Hence $\\overline{CC'} = \\overline{BQ}$.  \nFrom the above, $\\overline{BB'} = \\overline{QB'} + \\overline{BQ} = \\overline{AA'} + \\overline{CC'}$.\n\n![](images/Makedonija_2008_p13_data_f2a2f72056.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19895,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a regular hexagon with sidelength $s$. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$.\n\n*Prove that the three points $C$, $P$, and $Q$ are on a line.*",
    "options": [],
    "answer": "See solution",
    "solution": "Our strategy is to compute the angles $\\angle DCQ$ and $\\angle DCP$ to check that they are equal.\n\nThe interior angles of a regular hexagon equal $120^\\circ$. The triangle $DEF$ is isosceles and therefore, we get $\\angle DFE = \\angle EDF = 30^\\circ$. This implies $\\angle QDC = 90^\\circ$, and since the triangle $QDC$ is also isosceles, we also get\n\n$$\n\\angle DCQ = 45^{\\circ}.\n$$\n\nThe triangle $CBP$ is isosceles and analogously to the above, we get $\\angle CBP = \\angle CBD = 30^\\circ$. Therefore, we obtain\n\n$$\n\\angle PCB = \\frac{180^{\\circ} - 30^{\\circ}}{2} = 75^{\\circ} \\quad \\text{and, finally,} \\quad \\angle DCP = 120^{\\circ} - 75^{\\circ} = 45^{\\circ}.\n$$\n\nSo $\\angle DCQ = \\angle DCP$ which implies that $C$, $P$, and $Q$ lie on a line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19896,
    "subject": "Mathematics (Olympiad)",
    "question": "Arrange parentheses and four arithmetic symbols $+, -, \\times, :$ between some of the digits in the expression $1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\\ 9$ to obtain the largest possible number.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest number is obtained by simply writing the digits together as $123456789$, without inserting any arithmetic symbols. \n\nIf you replace a $+$ with digit concatenation (e.g., $\\overline{XY}$ instead of $X + Y$), the value increases since:\n\n$$\n\\overline{XY} \\geq 10X + Y > X + Y.\n$$\n\nSimilarly, replacing multiplication with concatenation also increases the value. If $Y$ has $k$ digits ($10^{k-1} < Y < 10^k$), then:\n\n$$\n\\overline{XY} \\geq 10^k X + Y > X \\cdot Y \\iff (10^k - Y)X + Y > 0.\n$$\n\nThus, the largest possible number is $123456789$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19897,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to partition the set of all integers into ordered triples so that, for every triple $(a, b, c)$, the number\n\n$$\n|a^3b + b^3c + c^3a|\n$$\nis a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose first that $a + b + c = 0$. Then we have\n\n$$\n\\begin{aligned}\n|a^3b + b^3c + c^3a| &= |a^3b + b^3(-a - b) + (-a - b)^3a| \\\\\n&= |-b^4 - 2b^3a - 3a^2b^2 - 2a^3b - a^4| \\\\\n&= (a^2 + ab + b^2)^2.\n\\end{aligned}\n$$\n\nSo it suffices to partition the set of all integers into triples of zero sum. One way to do this is the following:\n\n- Begin with the triple $(-1, 0, 1)$.\n- Then, at every next step, let $p$ and $q$ be the least two positive integers not paired yet.\n- Form the triplets $(p, q, -p-q)$ and $(-p, -q, p+q)$ and repeat.\n\nIt is easy to see that this procedure works. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19898,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $2n+2$ points, show that there are at least $n+1$ lines (called *dividers*) such that each line passes through two of the points and divides the remaining $2n$ points into two equal groups of $n$ points each.\n\n% IMAGE: ![](images/Slovenija_2010_p25_data_b716367c7d.png)\n% IMAGE: ![](images/Slovenija_2010_p25_data_fac0b11f92.png)\n% IMAGE: ![](images/Slovenija_2010_p25_data_a5d1e2401d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be one of the points and the remaining points be $A_1, A_2, \\dots, A_{2n+1}$. Consider the lines $AA_1, AA_2, \\dots, AA_{2n+1}$. For each such line, as we rotate a half-plane around $A$, the number of points in the gray region changes by at most $1$ at each step. Initially, there are $r_1$ points in the gray part; after $2n+1$ rotations, the number is $2n - r_1$.\n\nIf $r_1 = n$, then $AA_1$ is a divider. If not, since $r_1$ and $2n - r_1$ are on opposite sides of $n$, and the count changes by $1$ at each step, there must be some $i$ with $r_i = n$, so $AA_i$ is a divider.\n\nThus, for each point, there is a divider passing through it. Since there are $2n+2$ points, we obtain $2n+2$ lines, but each is counted twice, so there are at least $\\frac{2n+2}{2} = n+1$ dividers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19899,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following equation in natural numbers:\n\n$$\n(x + y)^3 = (x - y - 6)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = x + y$ and $b = x - y - 6$. The equation is $a^3 = b^2$.\n\nSince $a > 1$, $|b| = a^{3/2} > a$. Given $x - y - 6 < x + y$, we have $|x - y - 6| = y + 6 - x > x + y$, so $x < 3$. Since $x$ is natural, $x = 1$ or $x = 2$.\n\nIf $x = 1$, the equation becomes $(y + 1)^3 = (y + 5)^2$. Testing $y = 1, 2, 3, 4$, only $y = 3$ works. For $y \\geq 5$, $(y + 1)^3 > y^3 > 4y^2 = (2y)^2 \\geq (y + 5)^2$.\n\nIf $x = 2$, the equation becomes $(y + 2)^3 = (y + 4)^2$. Testing $y = 1, 2, 3$, none work. For $y \\geq 4$, $(y + 2)^3 > y^3 \\geq 4y^2 = (2y)^2 \\geq (y + 4)^2$.\n\nThus, the only solution in natural numbers is $x = 1$, $y = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19900,
    "subject": "Mathematics (Olympiad)",
    "question": "Дурын натурал $n$ бүрийн хувьд $\\{2, 3, 4, \\ldots, 3n + 1\\}$ олонлогийг элементүүд нь ямар нэг мохоо өнцөгт гурвалжны талуудын урт болдог байхаар тус бүр нь 3 элементтэй үл огтлолцох $n$ дэд олонлогт хувааж болох уу?",
    "options": [],
    "answer": "See solution",
    "solution": "Хувааж болно. $\\{a, a + 1, \\ldots, b\\}$ олонлогийг $[a, b]$ гэж тэмдэглэе. $\\{a, b, c\\}$ нь мохоо өнцөгт гурвалжны тал болдг бол *мохоо гурвал* гэе. Индукцээр $[2, 3n + 1]$ нь $n$ ширхэг\n\n$A_i$ $(1 \\leq i \\leq n)$, $A_i = \\{i, a_i, b_i\\}$ ($A_i$ мохоо гурвал)-уудын нэгдэлд тавигдахыг баталья. $n = 1$, $A_1 = \\{2, 3, 4\\}$ мохоо гурвал. Индукцийн шилжилтээ хийхдээ дараах леммийг ашиглая.\n\n*Лемм.* $a < b < c$ нь мохоо гурвал үүсгэдэг бол\n\n$$\n\\forall x \\in \\mathbb{R}^{+} : \\{a, b + x, c + x\\} \\text{ нь мөн мохоо гурвал үүсгээнэ.}\n$$\n\n▲ $a < b + x < c + x$ ба $(a + b) + x > c + x$ тул $a, b + x, c + x$ нь гурвалжны тал болно. $\\{a, b, c\\}$-мохоо гурвал гэдгээс $a^2 + b^2 < c^2$ байна. $(c + x)^2 - (b + x)^2 = (c - b)(c + b + 2x) > (c - b)(c + b) = c^2 - b^2 > d^2 \\Leftrightarrow (c + x)^2 > a^2 + (b + x)^2$\n\nтул $\\{a, b + x, c + x\\}$-мохоо гурвал. ▼\n\n$n > 1$, $t = [n/2] < n$ гэе. Индукцийн таамаглалаар $[2, 3t + 1]$ нь\n\n$A'_i = \\{i, a'_i, b'_i\\}$ ($i \\in [2, t+1]$) мохоо гурвалуудын нэгдэлд тавигдана.\n$i \\in [2, t + 1]$-д $A_i = \\{i, a'_i + (n - t), b'_i + (n - t)\\}$ гэж тодорхойлбол\nэдгээр нь өрөнхий элементгүй ба мохоо гурвалууд байх нь ойлгомжтой. Түүнчлэн $\\bigcup_{i=2}^{t+1} A_i = [2, t+1] \\cup [n+2, n+2t+1]$. Цаашлаад\n\n$i \\in [t+2, n+1]$ хувьд $A_i = \\{i, n + t + i, 2n + i\\}$ эдгээр нь үл огтлолцох ба $\\bigcup_{i=t+2}^{n+1} A_i = [t+2, n-1] \\cup [n+2t+2, 2n + t + 1] \\cup [2n + t + 2, 3n + 1]$\n\nтул $\\bigcup_{i=t+2}^{n+1} A_i = [t+2, n-1] \\cup [n+2t+2, 2n + t + 1] \\cup [2n + t + 2, 3n + 1]$\n\nтул $\\bigcup_{i=2}^{n+1} A_i = [2, 3n + 1]$ болох тул $i \\in [t + 2, n + 1]$ бүрийн хувьд\n$A_i$-мохоо гурвал болохыг батлахад хангалттай.\n\n$(2n + i) - (n + t + i) = n - t < t + 2 \\leq i \\Rightarrow A_i$-ийн элементүүд нь гурвалжны тал болж чадна.\n\n$$\n(2n + i)^2 - (n + t + i)^2 = (n - t)(3n + t + 2i)\n$$\n\n$$\n\\geq \\frac{n}{2}(3n + 3(t + 1) + 1) > \\frac{n}{2} \\cdot \\frac{9n}{2} = \\frac{9n^2}{4} \\geq (n + 1)^2 \\geq i^2\n$$\n\nболж батлагдав.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19901,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a scalene triangle $ABC$ with circumcircle $(O)$ and incircle $(I)$. Suppose that $BI$ meets $AC$ at $E$ and $CI$ meets $AB$ at $F$. The circle passing through $E$ and tangent to $OB$ at $B$ meets $(O)$ at $M$, and the circle passing through $F$ and tangent to $OC$ at $C$ meets $(O)$ at $N$. Lines $ME$, $NF$ cut $(O)$ again at $P$, $Q$. Let $K$ be the intersection of $EF$ and $BC$. $PQ$ meets $BC$, $EF$ at $G$, $H$ respectively. Prove that the median from $G$ of triangle $GHK$ is perpendicular to $OI$.\n\n![](images/Vietnamese_mathematical_competitions_p237_data_cf98a12451.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we state a well-known lemma.\n\n**Lemma.** Given a scalene triangle $ABC$ with circumcircle $(O)$. Let $BE$, $CF$ be the bisectors and $I_a$ be the excenter with respect to vertex $A$ of triangle $ABC$, then $OI_a \\perp EF$.\n\nBack to our problem, it is well-known that the Apollonius circle of vertex $B$ in triangle $ABC$ is orthogonal to $(O)$ and passes through $B$, $E$, so $(BEM)$ is the $B$-Apollonius of triangle $ABC$. Note that $\\frac{MA}{MC} = \\frac{BA}{BC} = \\frac{EA}{EC}$ or $ME$ is the bisector of $\\angle AMC$. Thus, $P$ is the midpoint of arc $\\widehat{ABC}$. Similarly, $Q$ is the midpoint of arc $\\widehat{ACB}$ or $PQ \\perp AI$.\n\nFrom the lemma, one can get $OI_a \\perp EF$. Denote $S$ as the midpoint of the minor arc $BC$ of $(O)$, then $S$ is the midpoint of $II_a$ and $OS \\perp GH$.\n\nThus, the sides of $\\triangle GKH$, $\\triangle SOI_a$ are corresponding perpendicular so they are similar and there exists a spiral similarity of angle $90^\\circ$ turns one triangle into another. Hence, it suffices to show that the median of vertex $S$ of triangle $SOI_a$ is parallel to $OI$. But it is obvious because if we denote $R$ to be the midpoint of $OI_a$ then $SR$ is the midline of triangle $I_aIO$ then $SR$ is parallel to $OI$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19902,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all $m \\times n$ grids that can be made into an Eulerian cycle (a cycle that uses every edge exactly once and starts and ends at the same vertex).",
    "options": [],
    "answer": "See solution",
    "solution": "An $m \\times n$ grid can be made into an Eulerian cycle if and only if $m = n$.\n\n**Solution 1**: If $m = n$, add a diagonal (top left to bottom right) to every unit square not in the main diagonal of the grid. It is easy to verify the new graph is an Eulerian cycle.\n\nTo prove that when $m \\neq n$, the grid cannot be made into an Eulerian cycle, let $n > m$ and suppose it is possible. In the original grid, every lattice point has an even degree of edges except for those on the four sides but not at the corners. Let us call them even and odd points. There are $m-1$ or $n-1$ odd points on each side of the grid. To make the grid into an Eulerian cycle, the diagonals must turn each odd point into an even point while maintaining all the even points. We focus on all the diagonals. If three or four diagonals meet at a lattice point, separate them as in the figure below.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p244_data_58d2fd3172.png)\n\nThis divides the diagonals into several cycles and non-intersecting paths. Remove all the cycles. The remaining paths of diagonals turn all the odd points into even points, and hence a path must start at an odd point and end at another odd point. Color all the lattice points alternately black and white. Evidently, a path can only pass through lattice points of the same color. If a path starts and ends on the same side of the grid, then there are an odd number of odd points on this side between the two ends, and they cannot be connected by paths. If a path starts and ends on opposite sides of the grid, say from row $a$ and column 1, to row $b$ and column $n+1$, then $(a, 1)$ and $(b, n+1)$ are of the same color, $a+1$ and $b+n+1$ are of the same parity, which indicates that there are $(a-2) + (n-1) + (b-2) = a+n+b-5$, an odd number of odd points between the two ends, which cannot be connected by paths. The above argument shows that a path must start and end at neighboring sides, which requires the $2(m-1)$ odd points on two opposite sides to match the $2(n-1)$ odd points on the other two sides. Hence, $2(m-1) = 2(n-1)$, or $m=n$. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19903,
    "subject": "Mathematics (Olympiad)",
    "question": "Alexander and Denitza play the following game. Alexander cuts (if possible) a band of positive integer length into three bands of positive integer lengths such that the largest band is unique. Then Denitza cuts (if possible) the largest band in the same way, and so on. The winner is the one who makes the last move. Consider bands whose lengths are integers of the form $a^b$, where $a-1, b-1 \\in \\mathbb{N}$. For which of these does Denitza have a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a band of length $n$. No move is possible for $n = 1, 2, 3$. For $4 \\leq n \\leq 7 = 3 + 2 + 2$, Alexander has a winning move. For $n = 8$ and $9$, after Alexander's first move, the largest length is between $4$ and $7$, so Denitza has a winning move. Similarly, for $10 \\leq n \\leq 25 = 9 + 8 + 8$, Alexander has a winning strategy since he can cut the band so that the largest length is either $8$ or $9$, and so on. By induction, Denitza has a winning strategy if and only if $n = 3^k$ or $n = 3^k - 1$ for some integer $k > 1$.\n\nThe numbers $3^k$ and $3^2 - 1 = 2^3$ obviously have the desired form. We shall prove that there are no other solutions. Assume that $a^b = 3^k - 1$. Since $a^2 \\equiv 0, 1 \\pmod{3}$, the number $b$ is even. Then $(a+1)(a^{b-1} + \\cdots + a + 1) = 3^k$ and we...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19904,
    "subject": "Mathematics (Olympiad)",
    "question": "Before 1995, in the Football Championship of Ukraine, teams received 2 points for a win, 1 point for a draw, and 0 points for a defeat. Since 1995, a win gives 3 points, a draw 1 point, and a defeat 0 points. The Football Federation decided to recalculate all past championships using the new system, assuming the rankings would not change much. However, in the 1927 championship of a regional league with $n$ teams (each team played every other team once), team \"A\" originally had the highest score and team \"B\" the lowest. After recalculating with the new system, team \"B\" had the highest score and team \"A\" the lowest. For what is the smallest $n$ for which this could happen?\n\nHere, \"highest\" or \"lowest\" score means that no other team achieved that score.",
    "options": [],
    "answer": "See solution",
    "solution": "Team \"A\" was first under the old system (2 points per win), but last under the new system (3 points per win), and vice versa for team \"B\". Let team \"A\" have $x$ wins and $y$ draws, and team \"B\" have $a$ wins and $b$ draws. Since there are more than two teams, all other teams' points are between those of \"A\" and \"B\" in both systems. We have:\n\n$$\n2x + y \\ge 2a + b + 2 \\quad \\text{and} \\quad 3x + y + 2 \\le 3a + b\n$$\n\nThis leads to:\n\n$$\n2(a - x) + 2 \\le y - b \\le 3(a - x) - 2.\n$$\n\nConsider possible values of $y-b$ for different $a-x$:\n\n- For $a-x=0$: $2 \\le y-b \\le -2$ (impossible)\n- For $a-x=1$: $4 \\le y-b \\le 1$ (impossible)\n- For $a-x=2$: $6 \\le y-b \\le 4$ (impossible)\n- For $a-x=3$: $8 \\le y-b \\le 7$ (impossible)\n- For $a-x=4$: $10 \\le y-b \\le 10$\n\nSo the minimum possible value is $y-b=10$. Thus, $y \\ge 10 + b$. Since $x \\ge 1$, $x + y \\ge 11$, so there must be at least 12 teams.\n\n![](images/Ukraine_2016_Booklet_p38_data_8206cfe942.png)\n\nIn the table (fig. 41), a win is denoted by 1, a defeat by 0, and empty cells are draws. Under the old system, team \"A\" (number 2) gets 12 points, \"B\" (number 1) gets 10, and the rest get 11. Under the new system, \"A\" gets 13, \"B\" gets 15, and the rest get 14. Thus, the smallest $n$ is $12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19905,
    "subject": "Mathematics (Olympiad)",
    "question": "Find, with proof, all positive integers $\\overline{abc}$ satisfying\n\n$$\nb \\cdot \\overline{ac} = c \\cdot \\overline{ab} + 10.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given condition rewrites as $b(10a + c) = c(10a + b) + 10$. Rearranging, we get $a(b - c) = 1$, so $a = 1$ and $b - c = 1$. Thus, the numbers are:\n\n$$\n110,\\ 121,\\ 132,\\ 143,\\ 154,\\ 165,\\ 176,\\ 187,\\ 198.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19906,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a graph with 2014 vertices, what is the maximal possible degree $k$ such that the graph contains no triangle (i.e., no three vertices mutually connected)?\n\n![](images/MONGOLIAN_MATHEMATICAL_OLYMPIAD-2014_p41_data_f5accfc3de.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B$ be a vertex with degree at most $672$. Suppose vertices $A$ and $B$ are connected by an edge. Among the remaining $2012$ vertices, $1342$ are connected to $A$ and $671$ are connected to $B$. Denote $S(A)$ as the set of neighbors of $A$. Since $671 + 1342 > 2012$, it follows that $S(A) \\cap S(B) \\neq \\emptyset$. This means the graph contains a triangle, which contradicts the given condition.\n\nSuppose $A$ and $B$ are not connected. Since there are at least $671$ vertices with degrees $672, 673, \\ldots, 1342$, there are at least $671$ vertices not connected to $A$. On the other hand, $A$ is connected to exactly $1343$ vertices. Thus, $671 + 1343 > 2013$, which leads to a contradiction. Therefore, $k < 1343$.\n\nIf $k = 1342$, then $A$ is connected to $B_1, B_2, \\ldots, B_{1342}$; $C_1$ is connected to $B_2, B_3, \\ldots, B_{1342}$; $C_2$ is connected to $B_3, B_4, \\ldots, B_{1342}$; and so on, with $C_{671}$ connected to $B_{672}, B_{673}, \\ldots, B_{1342}$. There is no triangle in this construction.\n\nThe degrees are: $A$ has $1342$, $C_1$ has $1342$, $C_2$ has $1341$, $C_{671}$ has $672$, $B_1$ has $1$, $B_2$ has $2$, $B_{671}$ has $671$. Thus, the maximal value of $k$ is $1342$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 19907,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $a, b, c$ 為正實數，試證不等式\n\n$$\n3(a + b + c) \\geq 8\\sqrt[3]{abc} + \\sqrt[3]{\\frac{a^3 + b^3 + c^3}{3}}\n$$\n\nLet $a, b, c$ be positive real numbers. Prove that\n\n$$\n3(a + b + c) \\geq 8\\sqrt[3]{abc} + \\sqrt[3]{\\frac{a^3 + b^3 + c^3}{3}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "解：由算幾不等式得\n\n$$\n8\\sqrt[3]{abc} + \\sqrt[3]{\\frac{a^3 + b^3 + c^3}{3}} \\leq 9\\sqrt[3]{\\frac{8abc + \\frac{a^3+b^3+c^3}{3}}{9}} = 3\\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\n$$\n\n故只須證明\n\n$$\n3(a + b + c) \\geq 3\\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\n$$\n\n或\n\n$$\n(a + b + c)^3 \\geq a^3 + b^3 + c^3 + 24abc.\n$$\n\n將上式展開，等價於\n\n$$\na^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 \\geq 6abc,\n$$\n\n上述不等式由算幾不等式易證。若\n\n$$\nabc = \\frac{a^3 + b^3 + c^3}{3} \\quad \\text{且} \\quad a^2b = ab^2 = b^2c = bc^2 = c^2a = ca^2\n$$\n\n則等式成立。即 $a = b = c$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19908,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the following two lemmas:\n\n*Lemma 1.* Let $l_1$ and $l_2$ be two lines and $A$, $B$, and $C$ three points in the plane. Denote by $A_1$ and $A_2$ the perpendicular projections of $A$ on $l_1$ and $l_2$, respectively. Furthermore, let $A'$ be the midpoint of $A_1A_2$. Define $B'$ and $C'$ similarly. Then $A$, $B$, and $C$ are collinear if and only if $A'$, $B'$, and $C'$ are collinear.\n\n*Lemma 2.* Let $T$ be the midpoint of arc $BAC$ of the circumcircle of triangle $ABC$. Let $H_b$ and $H_c$ be the feet of the perpendiculars from $T$ to the internal bisectors of $\\angle B$ and $\\angle C$, respectively. If $T'$ is the midpoint of $H_bH_c$, then $T'$ lies on the perpendicular bisector of side $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Proof of Lemma 1. The key observation is that $\\overrightarrow{A'B'} = \\frac{1}{2}(\\overrightarrow{A_1B_1} + \\overrightarrow{A_2B_2})$ and $\\overrightarrow{B'C'} = \\frac{1}{2}(\\overrightarrow{B_1C_1} + \\overrightarrow{B_2C_2})$. Using these observations and Thales' theorem, it is easy to check that the assertion is equivalent to the equality $\\frac{A_1B_1}{A_2B_2} = \\frac{B_1C_1}{B_2C_2}$. $\\square$\n\nProof of Lemma 2. Let $N$ be the midpoint of side $BC$. Note that $TH_c \\perp CH_c$ and $TN \\perp BC$, hence the quadrilateral $TH_cNC$ is cyclic and\n\n$$\n\\angle H_c N B = \\angle H_c T C = 90^\\circ - \\angle T C H_c = \\angle H_c C B + \\angle N T C = 90^\\circ - \\frac{\\angle B}{2}.\n$$\n\nIt means that $H_cN$ is parallel to the external bisector of $\\angle BAC$. On the other hand, $TH_b$ is also parallel to the external bisector of $\\angle ABC$ and thus $TH_b \\parallel H_cN$. Similarly, $TH_c \\parallel H_bN$. So $TH_cNH_b$ is a parallelogram and hence $T'$, the midpoint of $H_bH_c$, lies on $TN$. But $TN$ is the perpendicular bisector of side $BC$, which is what we desired to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19909,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of integers $a, b, c$ such that the number\n\n$$\nN = \\frac{(a-b)(b-c)(c-a)}{2} + 2\n$$\n\nis a power of $2016$.\n\n(A power of $2016$ is an integer of the form $2016^n$, where $n$ is a non-negative integer.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, c$ be integers and $n$ a non-negative integer such that\n\n$$\nN = \\frac{(a-b)(b-c)(c-a)}{2} + 2 = 2016^n.\n$$\n\nMultiply both sides by $2$:\n$$\n(a-b)(b-c)(c-a) + 4 = 2 \\cdot 2016^n.\n$$\n\nSet $x = a-b$, $y = b-c$, so $c-a = -(x + y)$, and rewrite:\n$$\nxy(x + y) + 4 = 2 \\cdot 2016^n.\n$$\n\nIf $n > 0$, $2 \\cdot 2016^n$ is divisible by $7$ and $9$. Consider modulo $7$:\n$$\nxy(x + y) + 4 \\equiv 0 \\pmod{7}.\n$$\nBut cubic residues modulo $7$ are $-1, 0, 1$, so $xy(x + y)$ must be divisible by $7$, which is impossible unless $x$ or $y$ or $x + y$ is divisible by $7$. This leads to a contradiction.\n\nSimilarly, modulo $9$, $xy(x + y) + 4 \\equiv 0 \\pmod{9}$, but this also leads to a contradiction.\n\nThus, the only possibility is $n = 0$:\n$$\nxy(x + y) + 4 = 2.\n$$\nSo,\n$$\nxy(x + y) = -2.\n$$\nThe integer solutions for $(x, y)$ are $(-1, -1)$, $(2, -1)$, and $(-1, 2)$.\n\nTherefore, the required triples are all cyclic permutations of $(k + 2, k + 1, k)$ for $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19910,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an equilateral triangle $ABC$ with sides of length $2$. We consider all equilateral triangles $PQR$ with sides of length $1$ satisfying the following properties:\n\n- $P$ lies on the side $AB$,\n- $Q$ lies on the side $AC$, and\n- $R$ lies in the interior or on the edge of the triangle $ABC$.\n\nDescribe the set of all points in the triangle $ABC$ that are centroids of such triangles $PQR$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ and $Q$ be given fulfilling the conditions of the problem. Considering the circumcircle $k$ of $APQ$, we note that the centroid $S$ of $APQ$ must lie on $k$, since both $\\angle PAQ = 60^\\circ$ and $\\angle PSQ = 120^\\circ$ hold. Since $|SQ| = |SP|$, the arcs $SQ$ and $SP$ are of equal length, and we therefore have $\\angle SAP = \\angle SAQ = 30^\\circ$. All centroids $S$ therefore lie on the angle bisector $w_{\\alpha}$ of $\\angle BAC$. The most extreme positions of $S$ are assumed when $P$ coincides with $A$ or the midpoint $M_{AB}$ of $AB$. In the latter case, $S$ is also the centroid, i.e., the midpoint, of $ABC$. In the former, $S$ is the centroid of the triangle $AM_{AB}M_{AC}$ (where $M_{AC}$ is the midpoint of $AC$). The set of all centroids of triangles $PQR$ fulfilling all requirements is therefore the middle third of the bisector $w_{\\alpha}$ (which is also the altitude in $ABC$).\n\n![](images/AustriaMO2012_p6_data_d78cd1e883.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19911,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be the sum of two numbers that player A wrote, and $b$ be the sum of the two numbers B wrote. If we let $t, s$ be the two numbers A wrote, then $t+s = a$ and $ts = 2b$, so $t, s$ must be the solutions of the quadratic equation $x^2 - a x + 2b = 0$. Similarly, the two numbers B wrote down are the solutions of the equation $x^2 - b x + 2a = 0$. From this, the discriminants $a^2 - 8b$ and $b^2 - 8a$ of these equations must be perfect squares. Conversely, suppose that $a^2 - 8b$ is a perfect square, equal to $k^2$ for some positive integer $k$. Then, $(a+k)(a-k) = 8b$, so $a > k$ and $a$ and $k$ have the same parity. Therefore, the two roots $\\frac{a+k}{2}$, $\\frac{a-k}{2}$ of the equation $x^2 - a x + 2b = 0$ are positive integers. Similarly, if $b^2 - 8a$ is a perfect square, the two roots of $x^2 - b x + 2a = 0$ are positive integers. Thus, the conditions for the problem are equivalent to:\n\n$a$ and $b$ are positive integers and both $a^2-8b$ and $b^2-8a$ are perfect squares, and $a \\ge b$.\n\nDetermine all possible values of $b$ that satisfy these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Adding the respective sides of the first two equations, we obtain $xy + zw = 2(x + y + z + w)$, which can be transformed into:\n\n$$\n(x-2)(y-2) + (z-2)(w-2) = 8.\n$$\n\n**Case 1:** $y = 1$.\n\nSince $x = 2z + 2w$, we get $zw = 2(x + y) = 4z + 4w + 2$, which becomes $(z - 4)(w - 4) = 18$. This yields $(z, w) = (22, 5), (13, 6), (10, 7)$, with corresponding $x$ values $54, 38, 34$. The requirement $x + y \\geq z + w$ is satisfied in all cases, and $z + w = 27, 19, 17$ respectively.\n\n**Case 2:** $x \\geq y \\geq 2$ and $z \\geq w \\geq 2$.\n\nThen $(x-2)(y-2) \\leq 8$ and $(z-2)(w-2) \\leq 8$. If $y \\geq 5$, then $(x-2)(y-2) \\geq 9$, so $y \\leq 4$.\n\n- If $y = 4$:\n  - $x = 4$: $zw = 16$, $z + w = 8$; $(z, w) = (4, 4)$.\n  - $x = 5$: No integer solutions.\n  - $x = 6$: $zw = 20$, $z + w = 12$; $(z, w) = (10, 2)$.\n  Only $(x, y, z, w) = (4, 4, 4, 4)$ satisfies $x + y \\geq z + w$, so $z + w = 8$.\n\n- If $y = 3$:\n  - $x = 6$: $zw = 18$, $z + w = 9$; $(z, w) = (6, 3)$.\n  - $x = 10$: $zw = 26$, $z + w = 15$; $(z, w) = (13, 2)$.\n  Only $(x, y, z, w) = (6, 3, 6, 3)$ satisfies $x + y \\geq z + w$, so $z + w = 9$.\n\n- If $y = 2$:\n  - $w = 3$: $z = 10$, $x = 13$; $(x, y, z, w) = (13, 2, 10, 3)$, $z + w = 13$.\n  - $w = 4$: $z = 6$, $x = 10$; $(x, y, z, w) = (10, 2, 6, 4)$, $z + w = 10$.\n\n**Summary:**\n\nThe possible values for the sum of the two numbers B wrote are $8, 9, 10, 13, 17, 19, 27$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19912,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomial $P(x) = x^2 - 1$. There is a real number $a$ that satisfies\n$$P(P(P(a))) = 2024.$$\n\nWrite $a^2$ as $m + \\sqrt{n}$, where $m$ and $n$ are integers and $n$ is square-free. The value of $m+n$ is $\\underline{\\underline{6}} \\underline{\\underline{7}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $P(P(P(a))) = 2024$, we have\n$$\nP(P(a))^2 - 1 = 2024 \\implies P(P(a)) = \\pm\\sqrt{2025} = \\pm45.\n$$\n\nBut the negative value is not valid, since the range of $P$ is $[-1, \\infty)$. Applying $P$ again:\n$$\nP(a)^2 - 1 = 45 \\implies (a^2 - 1)^2 = 46 \\implies a^2 = 1 + \\sqrt{46}.\n$$\n\nThus, $m+n = 1 + 46 = 47$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19913,
    "subject": "Mathematics (Olympiad)",
    "question": "We are looking for values of $a$ and $b$ such that\n\n$$\n2 \\times a \\times a - b \\times b = 119.\n$$\n\nwhere $a$ is at most 10 and at least 8, and $b$ must be odd. Find all pairs $(a, b)$ in this range that satisfy the equation, and for each, give the corresponding isopentagon perimeter.",
    "options": [],
    "answer": "See solution",
    "solution": "From the table:\n\n- For $a = 8$, $b = 3$ gives $2 \\times 8 \\times 8 - 3 \\times 3 = 128 - 9 = 119$.\n- For $a = 10$, $b = 9$ gives $2 \\times 10 \\times 10 - 9 \\times 9 = 200 - 81 = 119$.\n\nThe corresponding isopentagons are:\n- $(8, 5, 3, 5, 8)$ with perimeter $29$.\n- $(10, 1, 9, 1, 10)$ with perimeter $31$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19914,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that:\n\n- $n \\equiv 2 \\pmod{3}$\n- $n \\equiv 4 \\pmod{5}$\n- $n \\equiv 3 \\pmod{7}$",
    "options": [],
    "answer": "See solution",
    "solution": "Any integer of the form $105k + 59$ where $k \\in \\mathbb{Z}$ is a solution.\n\nSince $n \\equiv 2 \\equiv -1 \\pmod{3}$ and $n \\equiv -1 \\pmod{5}$, we have\n\n$$\nn \\equiv -1 \\pmod{15}.\n$$\n\nTesting $n = -1, 14, 29, 44, 59$, we see that $n = 59$ satisfies $n \\equiv 3 \\pmod{7}$. Therefore, $n = 59$ is one solution. By the Chinese remainder theorem, since 3, 5, 7 are pairwise relatively prime and $3 \\times 5 \\times 7 = 105$, we have\n\n$$\nn \\equiv 59 \\pmod{105}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19915,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $A$, $B$, $C$ los colores de tres cajas. ¿Cuál es el máximo valor de $n$ tal que es posible colorear los números $1, 2, 3, \\ldots, n$ con los colores $A$, $B$, $C$ de modo que ningún par de números del mismo color difiera en el cuadrado de un entero?",
    "options": [],
    "answer": "See solution",
    "solution": "La respuesta es $29$.\n\nSupongamos que los números $1, 2, 3, \\ldots, 29$ se pueden colorear con los colores $A$, $B$, $C$ de modo que ningún par de números de un mismo color difieren en el cuadrado de un entero. Sea $f(i)$ el color del número $1 \\leq i \\leq 29$.\n\nComo $9$, $16$ y $25$ son cuadrados, a los números $1$, $10$ y $26$ hay que asignarles tres colores diferentes, pues $10-1=9$, $26-1=25$ y $26-10=16$. Lo mismo ocurre con los números $1$, $17$ y $26$ ($17-1=16$; $26-17=9$). Por lo tanto, $10$ y $17$ tienen asignado el mismo color: $f(10) = f(17)$.\n\nAnálogamente, a los números $4$, $13$ y $29$ hay que asignarles colores distintos, y a los números $4$, $20$ y $29$ hay que asignarles colores distintos, luego $f(13) = f(20)$. Por otra parte, a los conjuntos de números $3$, $12$ y $28$ y $3$, $19$, $28$ hay que asignarles colores diferentes, de modo que $f(12) = f(19)$.\n\nTambién tenemos que los dos grupos $2$, $11$ y $27$ y $2$, $18$ y $27$ deben tener colores diferentes, luego $f(11) = f(18)$.\n\nSin pérdida de generalidad, supongamos que $f(10) = f(17) = A$. Como $11 = 10 + 1^2$, tenemos que $f(11) \\neq f(10)$. Supongamos también que $f(11) = f(18) = B$. Como $19 = 18 + 1^2 = 10 + 3^2$, vale que $f(19) = f(12) = C$, pues $f(19) \\neq f(10) = A$ y $f(19) \\neq f(18) = B$.\n\nAnálogamente, $20=19+1^2=11+3^2$ implica que $f(20) \\neq f(19) = C$ y $f(20) \\neq f(11) = B$, de donde $f(20) = A$. Así se obtiene que $f(13) = f(20) = A$ y $f(17) = f(10) = A$, lo que es imposible pues $17-13=4=2^2$.\n\nPor otra parte, si $n \\leq 28$ se pueden colorear los números como en la tabla siguiente:\n\n|   | B | C | A | C |\n|---|---|---|---|---|\n| 1 |   |   |   |   |\n| 2 |   |   |   |   |\n| 3 |   |   |   |   |\n| 4 |   |   |   |   |\n\n| A | B | C | B | C |\n|---|---|---|---|---|\n| 5 | 6 | 7 | 8 | 9 |\n|10 |11 |12 |13 |14 |\n|15 |16 |17 |18 |19 |\n|20 |21 |22 |23 |24 |\n|25 |26 |27 |28 |   |\n\nEs fácil verificar que no hay dos números del mismo color cuya diferencia sea el cuadrado de un entero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19916,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all primes $p$ such that $2p^2 - 3p - 1$ is a cube of a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be a prime and $n$ a positive integer such that\n\n$$\n2p^2 - 3p - 1 = n^3.\n$$\n\nSince\n\n$$\nn^3 = 2p^2 - 3p - 1 < 2p^2 \\leq p^3,\n$$\n\nit follows that $n < p$, so $n + 1 \\leq p$.\n\n**Case 1:** $p = n + 1$\n\nSubstitute into the equation:\n\n$$\n2(n+1)^2 - 3(n+1) - 1 = n^3\n$$\n\nwhich simplifies to\n\n$$\nn^3 - 2n^2 - n + 2 = 0\n$$\n\nFactoring gives\n\n$$\n(n - 2)(n - 1)(n + 1) = 0\n$$\n\nSo $n = 1$ or $n = 2$, thus $p = 2$ or $p = 3$.\n\n**Case 2:** $p > n + 1$\n\nFrom the original equation:\n\n$$\np(2p - 3) = n^3 + 1 = (n + 1)(n^2 - n + 1)\n$$\n\nSince $p > n + 1$ and $p$ is prime, $p$ must divide $n^2 - n + 1$. Let $n^2 - n + 1 = kp$ for some integer $k$.\n\nSubstitute into the previous equation:\n\n$$\np(2p - 3) = (n + 1)kp\n$$\n\nSo $2p - 3 = k(n + 1)$, thus $p = \\frac{k(n + 1) + 3}{2}$.\n\nPlug this value of $p$ back into $n^2 - n + 1 = kp$:\n\n$$\nn^2 - n + 1 = k \\cdot \\frac{k(n + 1) + 3}{2}\n$$\n\nMultiply both sides by $2$:\n\n$$\n2n^2 - 2n + 2 = k^2(n + 1) + 3k\n$$\n\nRearrange:\n\n$$\n2n^2 - k^2 n - (k^2 + 3k - 2) = 0\n$$\n\nThis is a quadratic in $n$. For integer solutions, the discriminant must be a perfect square:\n\n$$\n(k^2)^2 - 4 \\cdot 2 \\cdot (-k^2 - 3k + 2) = k^4 + 8k^2 + 24k - 16\n$$\n\nFor large $k$, this cannot be a perfect square. Checking small odd $k$ ($k = 1, 3, 5, 7$), none yield integer $n$ and prime $p$ solutions.\n\nTherefore, the only solutions are $p = 2$ and $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19917,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of nonnegative integers $m, n$ with $m \\ge n$, for which $A = (m+n)^3$ divides $B = 2n(3m^2 + n^2) + 8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = (m+n)^3$ and $B = 2n(3m^2 + n^2) + 8$. We require $(m+n)^3 \\mid 2n(3m^2 + n^2) + 8$.\n\nFirst, note that\n$$\n(m+n)^3 \\leq 2n(3m^2 + n^2) + 8.\n$$\nExpanding and simplifying:\n$$\nm^3 + 3m^2 n + 3m n^2 + n^3 \\leq 6m^2 n + 2n^3 + 8\n$$\n$$\nm^3 - 3m^2 n + 3m n^2 - n^3 \\leq 8\n$$\n$$\n(m-n)^3 \\leq 8 \\implies m-n \\leq 2.\n$$\nSo $m-n \\in \\{0,1,2\\}$.\n\n**Case 1:** $m-n = 0$ ($m = n$)\n\nThen $A = (2m)^3 = 8m^3$, $B = 2m(3m^2 + m^2) + 8 = 8m^3 + 8$.\n\nSo $8m^3 \\mid 8m^3 + 8 \\implies 8m^3 \\mid 8 \\implies m = 1$ (since $m \\geq 0$). Thus, $(m, n) = (1, 1)$.\n\n**Case 2:** $m-n = 1$ ($m = n+1$)\n\nThen $A = (2n+1)^3$, $B = 2n(3(n+1)^2 + n^2) + 8$.\n\nCompute:\n$$\n3(n+1)^2 + n^2 = 3(n^2 + 2n + 1) + n^2 = 4n^2 + 6n + 3\n$$\nSo\n$$\nB = 2n(4n^2 + 6n + 3) + 8 = 8n^3 + 12n^2 + 6n + 8\n$$\nBut $(2n+1)^3 = 8n^3 + 12n^2 + 6n + 1$, so $B = (2n+1)^3 + 7$.\n\nThus, $(2n+1)^3 \\mid 7$ which is only possible for $2n+1 = 1 \\implies n = 0, m = 1$. So $(m, n) = (1, 0)$.\n\n**Case 3:** $m-n = 2$ ($m = n+2$)\n\nThen $A = (2n+2)^3 = 8(n+1)^3$.\n\nCompute $B$:\n$$\nB = 2n(3(n+2)^2 + n^2) + 8\n$$\n$3(n+2)^2 + n^2 = 3(n^2 + 4n + 4) + n^2 = 4n^2 + 12n + 12$\n\nSo\n$$\nB = 2n(4n^2 + 12n + 12) + 8 = 8n^3 + 24n^2 + 24n + 8 = 8(n+1)^3\n$$\nThus, $A = B$ for all $n \\geq 0$, so all pairs $(m, n) = (k+2, k)$ with $k \\geq 0$ are solutions.\n\n**Summary:**\n\nThe solutions are $(1, 1)$, $(1, 0)$, and all $(k+2, k)$ for $k \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19918,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "$$\n\\begin{aligned}\n\\text{Let}\n\\quad A &= \\bigcup_{k \\in \\mathbb{Z}} \\left( [-2^{2k+1}, -2^{2k}) \\cup (2^{2k}, 2^{2k+1}] \\right), \\\\\nB &= \\bigcup_{k \\in \\mathbb{Z}} \\left( [-2^{2k}, -2^{2k-1}) \\cup (2^{2k-1}, 2^{2k}] \\right).\n\\end{aligned}\n$$\n\nThus $A = 2B$, $B = 2A$, $A = -A$, $B = -B$, $A \\cap B = \\emptyset$, and $A \\cup B \\cup \\{0\\} = \\mathbb{R}$.\n\nDefine\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\n$$\nf(g(x)) = f(2f(x)) = -2x, \\qquad g(f(x)) = 2f(f(x)) = 2x.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19919,
    "subject": "Mathematics (Olympiad)",
    "question": "Ada tells Byron that she has drawn a rectangular grid of squares and placed either the number $0$ or the number $1$ in each square. Next to each row, she writes the sum of the numbers in that row. Below each column, she writes the sum of the numbers in that column. After Ada erases all of the numbers in the squares, Byron realizes that he can deduce each erased number from the row sums and the column sums.\n\nProve that there must have been a row containing only the number $0$ or a column containing only the number $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a rectangular array of numbers *amazing* if each number in the array is equal to $0$ or $1$ and no other array has the same row sums and column sums. We are required to prove that in an *amazing* array, there must be a row containing only the number $0$ or a column containing only the number $1$.\n\nCall the four entries in the intersection of two rows and two columns a *rectangle*. We say that a rectangle is *forbidden* if\n\n- its top-left and bottom-right entries are $0$, while its top-right and bottom-left entries are $1$; or\n- its top-left and bottom-right entries are $1$, while its top-right and bottom-left entries are $0$.\n\nIt should be clear that an *amazing* array cannot have *forbidden* rectangles, since switching $0$ for $1$ and vice versa in a *forbidden* rectangle will produce another array with the same row sums and column sums.\n\nSuppose that there exists an *amazing* array in which there is no row containing only the number $0$ nor a column containing only the number $1$. Let row $a$ have the maximum number of entries equal to $0$. Then row $a$ contains the number $1$, in column $m$, say. Furthermore, this column contains the number $0$, in row $b$, say. In order to avoid a *forbidden* rectangle, row $b$ must have a $0$ in every column in which row $a$ has a $0$.\n\nSince row $b$ also has a $0$ in column $m$, while row $a$ has a $1$ in column $m$, this contradicts the fact that row $a$ has the maximum number of entries equal to $0$.\n\nTherefore, every amazing array has a row containing only the number $0$ or a column containing only the number $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19920,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triplets of consecutive integers such that one of these numbers is the sum of the other two.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the consecutive numbers be $x$, $x+1$, and $x+2$.\n\nThere are three cases, depending on which number is the sum of the other two:\n\n1. If $x+2 = x + (x+1)$, then $x+2 = 2x+1 \\implies x=1$. The triplet is $(1, 2, 3)$.\n2. If $x+1 = x + (x+2)$, then $x+1 = 2x+2 \\implies x=-1$. The triplet is $(-1, 0, 1)$.\n3. If $x = (x+1) + (x+2)$, then $x = 2x+3 \\implies x=-3$. The triplet is $(-3, -2, -1)$.\n\nThus, the solutions are $(-3, -2, -1)$, $(-1, 0, 1)$, and $(1, 2, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19921,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle that does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$, and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "By the theorem of angles in the same segment, $\\angle BCF = \\angle BDF$. But since the chord $FA$ has the same length as the chord $BF$, the angle subtended by the chord $FA$ is equal to the angle subtended by the chord $BF$. So $\\angle FCA = \\angle BCF$.\n\n![](images/British_2015_Booklet_p20_data_e82d5d5dd3.png)\n\nAs $\\angle QDP = \\angle BDF = \\angle BCF = \\angle FCA = \\angle QCP$, it follows by the converse of the theorem of angles in the same segment that quadrilateral $PQCD$ is cyclic.\n\nThen $\\angle CPQ = \\angle CDQ$ by the theorem of angles in the same segment, and $\\angle CDQ = \\angle CDB = \\angle CAB$ by the same theorem. So $\\angle CPQ = \\angle CAB$. From this it follows, by the converse of the theorem of corresponding angles, that lines $PQ$ and $AB$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19922,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(a, b)$ such that\n$$\ngcd(a, b) = a + b - \\varphi(a) - \\varphi(b),\n$$\nwhere $\\varphi(n)$ denotes Euler's totient function.",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose that $a = 1$. Then $\\varphi(1) = 1$. For all positive integers $b$, we have $\\gcd(1, b) = 1$. The equation becomes $1 = 1 + b - 1 - \\varphi(b)$, or $\\varphi(b) = b - 1$. This holds if and only if $b$ is a prime number. Thus, the solutions for $a = 1$ are precisely the pairs $(1, p)$ with $p$ a prime number. Similarly, the solutions for $b = 1$ are the pairs $(p, 1)$ with $p$ a prime number.\n\nNow assume that $a, b \\geq 2$. Since $\\gcd(b, b) > 1$, we have $\\varphi(b) \\leq b - 1$. Therefore,\n$$\n\\gcd(a, b) = a + b - \\varphi(a) - \\varphi(b) \\geq a - \\varphi(a) + 1.\n$$\nLet $p$ be the minimal prime divisor of $a$ (which exists as $a \\geq 2$). For all multiples $tp \\leq a$ of $p$, $\\gcd(tp, a) > 1$, so $a - \\varphi(a) \\geq \\frac{a}{p}$. Thus,\n$$\n\\gcd(a, b) \\geq a - \\varphi(a) + 1 \\geq \\frac{a}{p} + 1.\n$$\nThe two largest divisors of $a$ are $a$ and $\\frac{a}{p}$. Since $\\gcd(a, b)$ is a divisor of $a$ that is at least $\\frac{a}{p} + 1$, it must be $a$. Similarly, $\\gcd(a, b) = b$, so $a = b$.\n\nThe equation becomes $a = 2\\varphi(a)$. Note that $2 \\mid a$, so write $a = 2^k m$ with $k \\geq 1$ and $m$ odd. Then $\\varphi(a) = 2^{k-1} \\varphi(m)$, so $2^k m = 2 \\cdot 2^{k-1} \\varphi(m)$, or $m = \\varphi(m)$. This only holds for $m = 1$, so $a = b = 2^k$ for $k \\geq 1$.\n\nTherefore, the solutions are all pairs $(1, p)$ and $(p, 1)$ for prime $p$, and $(2^k, 2^k)$ for all positive integers $k$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19923,
    "subject": "Mathematics (Olympiad)",
    "question": "An equilateral triangle is divided into 25 smaller congruent triangles by adding lines parallel to its sides. If 15 out of the 25 smaller triangles are shaded, what percentage of the large triangle is shaded?\n\n![](images/Slovenija_2016_p12_data_77bec7737d.png)\n\n![](images/Slovenija_2016_p12_data_c056df1705.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are 25 smaller congruent triangles, and 15 of them are shaded. The fraction shaded is:\n\n$$\n\\frac{15}{25} = \\frac{3}{5} = 0.6 = 60\\%.\n$$\n\nSo, 60% of the large triangle is shaded.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19924,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following game: Given a regular $n$-gon inscribed in a circle with a center $O$, two players take turns drawing segments connecting either two vertices or a vertex and the center, with the restriction that no two segments may cross except at their endpoints. The player who makes the last possible move wins. For which values of $n$ does the first player have a winning strategy, and for which values does the second player have a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "For odd $n$, the first player wins; for even $n$, the second player wins.\n\n**Case 1: $n$ even.**\n\nThe second player can always mirror the first player's moves symmetrically about the center or a diameter, ensuring that after each pair of moves, the configuration remains symmetric. This strategy guarantees that the first player cannot make the last move, so the second player wins.\n\n**Case 2: $n$ odd.**\n\nWe use induction. For $n=3$, the first player can win easily. Assume the first player can win for all odd $n$ up to $2k-1$. For $n=2k+1$, the first player starts by connecting a vertex to the center. Whatever the second player does, the first player can always respond to reduce the game to a smaller odd $n$ case, maintaining the winning strategy. Thus, the first player wins for all odd $n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19925,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that\n\n$$\n\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{2^m} < m,\n$$\nfor 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$$\n\\frac{1}{p_1} + \\frac{1}{p_2} + \\dots + \\frac{1}{p_n} < 10.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) This inequality can be 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\n$$\n\\left(\\frac{1}{p_1} + \\frac{1}{p_2} + \\dots + \\frac{1}{p_n}\\right)^4 \\le 4! \\sum_{1 \\le i \\le j \\le k \\le l \\le n} \\frac{1}{p_i p_j p_k p_l} < 24 \\left(\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2^{400}}\\right).\n$$\n\nSince $24 \\left(\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2^{400}}\\right) < 24 \\cdot 400 < 10000$, the result follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19926,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real numbers $(x, y)$ that satisfy\n\n$$\n\\begin{cases}\nx + \\sin x = y \\\\\ny + \\sin y = x\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "All solutions are $(k\\pi, k\\pi)$, where $k$ is any integer.\n\n**Solution 1:**\nAdd the two equations:\n$$\n(x + \\sin x) + (y + \\sin y) = y + x\n$$\nwhich simplifies to $\\sin x + \\sin y = 0$, so $\\sin x = -\\sin y$.\n\nThis gives two cases:\n1. $y = -x + 2k\\pi$\n2. $y = x + (2k + 1)\\pi$\n\nFor case 2, $|y - x| = |(2k + 1)\\pi| \\geq \\pi$, but from the first equation $|y - x| = |\\sin x| \\leq 1 < \\pi$, a contradiction. So only case 1 is possible.\n\nSubstitute $y = -x + 2k\\pi$ into the first equation:\n$$\nx + \\sin x = -x + 2k\\pi\n$$\nwhich gives $2x + \\sin x = 2k\\pi$.\n\nThis is satisfied for $x = k\\pi$, so $y = k\\pi$ as well. Since $f(x) = 2x + \\sin x$ is strictly increasing, there are no other solutions.\n\n**Solution 2:**\nThe function $f(z) = z + \\sin z$ is strictly increasing (since $f'(z) = 1 + \\cos z > 0$ except at isolated points). If $x < y$, then $y = x + \\sin x < y + \\sin y = x$, a contradiction. Similarly, $y < x$ is impossible. Thus, $x = y$.\n\nSubstitute $x = y$ into the system:\n$$\nx + \\sin x = x \\implies \\sin x = 0\n$$\nSo $x = y = k\\pi$ for any integer $k$.\n\nAll pairs $(k\\pi, k\\pi)$ satisfy the system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19927,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x^2 + x f(y)) = x f(x + y)\n$$\n\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $f(0) = 0$ and $f(x^2) = x f(x)$ for all $x \\in \\mathbb{R}$. If $f(\\alpha) = 0$ for some $\\alpha$, then\n\n$$\nf(x^2 + x f(\\alpha)) = x f(x + \\alpha),\n$$\n\nfor all $x \\in \\mathbb{R}$. Therefore,\n\n$$\nx f(x + \\alpha) = f(x^2) = x f(x),\n$$\n\nfor all $x \\in \\mathbb{R}$. For $x \\neq 0$, we get $f(x) = f(x + \\alpha)$. Note that this is also valid for $x = 0$. Suppose $f(1) = 0$. Then\n\n$$\nf(1 + f(x)) = f(x + 1),\n$$\n\nso that $f(f(x)) = f(x)$ for all $x$. If there exists a $\\lambda \\in \\mathbb{R}$ such that $f(\\lambda) \\neq 0$, then for any $t \\in \\mathbb{R}$, taking $s = t / f(\\lambda)$, we have\n\n$$\nf(s^2 + s f(\\lambda - s)) = s f(s + \\lambda - s) = s f(\\lambda) = t\n$$\n\nwhich shows that $f$ is onto. Hence there exists $x_0$ such that $f(x_0) = 1$. This gives\n\n$$\n1 = f(x_0) = f(f(x_0)) = f(1) = 0,\n$$\n\nwhich is absurd. Thus $f(1) = 0$ forces $f(x) = 0$ for all $x \\in \\mathbb{R}$.\n\nSuppose $f(1) \\neq 0$. Let $\\alpha \\in \\mathbb{R}$ be such that $f(\\alpha) = 0$. As we have seen earlier, $f(x + \\alpha) = f(x) = 0$ for all $x$. Therefore,\n\n$$\nf(\\alpha^2 + 1) = f((1 + \\alpha)^2 + (1 + \\alpha) f(\\alpha)) = (\\alpha + 1) f(1 + 2\\alpha) = (\\alpha + 1) f(1),\n$$\n\n$$\nf(\\alpha^2 + 1) = f((1 - \\alpha)^2 + (1 - \\alpha) f(\\alpha)) = (1 - \\alpha) f(1).\n$$\n\nThese show that $1 + \\alpha = 1 - \\alpha$. Therefore $\\alpha = 0$. Thus $f(\\alpha) = 0$ implies that $\\alpha = 0$.\n\nTaking $x = -y$ in the equation, we get $f(y^2 - y f(y)) = y f(y - y) = 0$. Hence $y^2 - y f(y) = 0$ for all $y \\in \\mathbb{R}$. This gives $f(y) = y$ for all $y \\neq 0$. Since $f(0) = 0$, we conclude that $f(y) = y$ for all $y \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19928,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna and Berta play a game in which they take turns removing marbles from a table. Anna takes the first turn. When there are $n \\geq 1$ marbles on the table at the start of a turn, the player whose turn it is removes $k$ marbles, where $k \\geq 1$ and either:\n\n- $k$ is even and $k \\leq \\frac{n}{2}$, or\n- $k$ is odd and $\\frac{n}{2} \\leq k \\leq n$.\n\nA player wins the game if she removes the last marble from the table.\n\n*Determine the smallest number $N \\geq 100{,}000$ such that Berta can enforce a victory if there are exactly $N$ marbles on the table in the beginning.*",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the losing situations are those with exactly $n = 2^a - 2$ marbles left on the table for all integers $a \\geq 2$. All other situations are winning situations.\n\n*Proof:* By induction for $n \\geq 1$.\n\n- For $n = 1$, the player wins by taking the single remaining marble.\n- For $n = 2$, the only possible move is to take $k = 1$ marble, and then the opponent wins in the next move.\n\nInduction step from $n-1$ to $n$ for $n \\geq 3$:\n\n1. If $n$ is odd, then the player takes all $n$ marbles and wins.\n2. If $n$ is even but not of the form $2^a - 2$, then $n$ lies between two numbers of that form, so there exists a unique $b$ with $2^b - 2 < n < 2^{b+1} - 2$. Because $n \\geq 3$, $b \\geq 2$. Therefore, all three numbers in this chain are even, and we can conclude $2^b \\leq n \\leq 2^{b+1} - 4$. From the induction hypothesis, $2^b - 2$ is a losing situation, and by taking\n\n$$\nk = n - (2^b - 2) = n - \\frac{2^{b+1} - 4}{2} \\leq n - \\frac{n}{2} = \\frac{n}{2}\n$$\n\nmarbles, we leave a losing position to the opponent.\n3. If $n$ is even and of the form $n = 2^a - 2$, then the player cannot leave a losing situation with $2^b - 2$ marbles to the opponent (where $b < a$ and $b \\geq 2$). To do so, the player would have to remove $k = (2^a - 2) - (2^b - 2) = 2^a - 2^b$ marbles. But since $b \\geq 2$, $k$ is even and strictly greater than $\\frac{n}{2}$ because $2^a - 2^b \\geq 2^a - 2^{a-1} = 2^{a-1} > 2^{a-1} - 1 = \\frac{2^a - 2}{2} = \\frac{n}{2}$; impossible.\n\n*Solution:* Berta can enforce a victory if and only if $N$ is of the form $2^a - 2$. The smallest number $N \\geq 100{,}000$ of this form is $N = 2^{17} - 2 = 131{,}070$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19929,
    "subject": "Mathematics (Olympiad)",
    "question": "For the natural number $N = p_1^{a_1} p_2^{a_2} \\dots p_n^{a_n}$, written in canonical form (where $p_i$ are distinct primes and $a_i$ are natural numbers, $1 \\le i \\le n$), define $T(N) = a_1 + a_2 + \\dots + a_n$.\n\nFor some distinct natural numbers $a, b, c, d$, the number $ab + cd$ is divisible by $ac + bd$. Prove that $T(ab + cd) \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that $T(ab + cd) \\le 2$.\n\nIt follows that $T(ac + bd) \\ge 2$. Since $ac + bd$ divides $ab + cd$, we have $T(ab + cd) \\ge T(ac + bd) \\ge 2$. By our assumption, $T(ab + cd) = T(ac + bd) = 2$.\n\nBut then $ac + bd = ab + cd$ implies $(a - d)(b - c) = 0$, which contradicts the assumption that $a, b, c, d$ are distinct.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19930,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)$ be an arithmetic sequence with first term $a_1 = 1$ and common difference $d$. The terms $a_2$, $a_5$, and $a_{11}$ form a geometric progression. Find the sum of the first $2009$ terms of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n = 1 + (n-1)d$. Then $a_2 = 1 + d$, $a_5 = 1 + 4d$, and $a_{11} = 1 + 10d$. Since $a_2$, $a_5$, and $a_{11}$ form a geometric progression, we have:\n\n$$\n(1 + 4d)^2 = (1 + d)(1 + 10d)\n$$\nExpanding both sides:\n$$\n1 + 8d + 16d^2 = 1 + 11d + 10d^2\n$$\n$$\n6d^2 - 3d = 0\n$$\nSince the sequence is not constant, $d \\neq 0$, so $d = \\frac{1}{2}$.\n\nThe sum of the first $2009$ terms is:\n$$\nS = \\frac{2009}{2}(2 \\cdot 1 + (2009 - 1) \\cdot \\frac{1}{2}) = 2009 + \\frac{2009 \\cdot 2008}{2} = 1010527\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19931,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be a triangle with sides $10$, $17$, and $21$. Find the length of the shortest altitude in $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Alternative 1**\n\nPythagoras gives\n$$\n17^2 = p^2 + t^2 \\text{ and } 10^2 = q^2 + t^2 = (21-p)^2 + t^2 = 21^2 - 42p + p^2 + t^2.\n$$\n\n$$\n\\text{Hence } 42p = 17^2 + 21^2 - 10^2 = 289 + 441 - 100 = 630.\n$$\n\n$$\n\\text{So } p = 15 \\text{ and } t^2 = 17^2 - 15^2 = 64.\n$$\n\nThus $t = 8$. Hence the shortest altitude in $T$ is $8 \\times 19 = 152$.\n\n**Alternative 2**\n\nLet $x$ be the angle in the bottom-right corner of $T$.\n\n$$\n\\text{From the cosine rule, } 17^2 = 10^2 + 21^2 - 2(10)(21) \\cos x.\n$$\n\n$$\n\\text{Hence } \\cos x = \\frac{100 + 441 - 289}{420} = \\frac{252}{420} = \\frac{21}{35} = \\frac{3}{5}.\n$$\n\n$$\n\\text{So } q = 10 \\cos x = 6.\n$$\n\nPythagoras gives $t^2 = 100 - 36 = 64$.\n\nThus $t = 8$. Hence the shortest altitude in $T$ is $8 \\times 19 = 152$.\n\n**Method 2**\n\nHalf the perimeter of $T$ is $(190 + 323 + 399)/2 = 912/2 = 456$.\n\nFrom Heron's formula, the square of the area of triangle $T$ is\n$$456(456 - 190)(456 - 323)(456 - 399) = 456(266)(133)(57)$$\n$$= 16(57)(57)(133)(133).$$\n\nThe shortest altitude in a triangle is perpendicular to its longest side.\n\nIf $t$ is the length of the shortest altitude of $T$, then the area of $T$ is\n$$\\frac{1}{2}(399)t = 4(57)(133).$$\n\n$$\n\\text{So } 3t = 8(57) \\text{ and } t = 8(19) = 152.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19932,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a composite number and $\\{a_1, a_2, \\dots, a_n\\}$ be the set of positive integers up to $N$ that are not relatively prime to $N$. If $b_1, b_2, \\dots, b_n$ is a permutation of $a_1, a_2, \\dots, a_n$, prove that there exist $i \\ne j$ such that $a_i b_i \\equiv a_j b_j \\pmod{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be a prime divisor of $N$ and let $A$ be the subset of the given set consisting of elements divisible by $N/p$. Clearly, $|A| = p$.\n\nAssume there is an index $i$ such that $a_i \\in A$ and $b_i \\notin A$. Then there must exist an index $j$ such that $a_j \\notin A$ and $b_j \\in A$. Hence, $a_i b_i$ (with $a_i \\in A$) and $a_j b_j$ are both divisible by $N/p$. By the pigeonhole principle, among these $p+1$ products, at least two are congruent modulo $N$.\n\nIf there does not exist $i$ such that $a_i \\in A$ and $b_i \\notin A$, then $a_i \\in A$ if and only if $b_i \\in A$. Thus, it suffices to show that if $b_1, \\dots, b_{p-1}$ is a permutation of $1, \\dots, p-1$, there exist $i$ and $j$ such that $i b_i - j b_j$ is divisible by $p$. If we assume the contrary, then\n\n$$\n((p-1)!)^2 \\equiv \\prod j b_j \\equiv (p-1)! \\pmod{p}\n$$\n\nwhich contradicts Wilson's theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19933,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an arithmetic sequence with constant difference $d = a_{i+1} - a_i$, we have $a_i = a_0 + d i$ for all $i \\ge 0$. Suppose $a_i > 0$ for all $i$. What are the possible values of $d$ and $a_0$? Additionally, for the six consecutive terms $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$, in how many ways (up to congruence) can these numbers be arranged as the side lengths of an equiangular hexagon?",
    "options": [],
    "answer": "See solution",
    "solution": "If $d = 0$, the sequence is constant and any equiangular hexagon with side lengths $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ has all sides of length $a_0$ and so is a regular hexagon. There is only one such hexagon (up to congruence).\n\nIf $d > 0$, the sequence is not constant and the hexagons we consider have six sides of different lengths. Consider an equiangular hexagon with side lengths $a, b, c, a', b', c'$ (in this order) and extend the sides $b, a'$ and $c'$ so that we obtain a triangle as shown in the diagram.\n\n![](images/IRL_ABooklet_2024_p27_data_af696bcd44.png)\n\nBecause all internal angles in an equiangular hexagon have a measure of $120^\\circ$, the large triangle and the three small triangles are equilateral. Hence the side length of the large equilateral triangle is $b' + c' + a = a + b + c = c + a' + b'$. Subtracting $a$ from the first and $b$ from the second equation we see that the six side lengths of any equiangular hexagon satisfy\n\n$$\na + b = a' + b' \\quad \\text{and} \\quad b + c = b' + c'. \\qquad (1)\n$$\n\nSubtracting these and rearranging gives us $a' + c = a + c'$ as well. We now show that for any six positive numbers $a, b, c, a', b', c'$ that satisfy (1), there exists an equiangular hexagon with these side lengths.\n\nWe essentially retrace the steps of our argument above: Equations (1) imply $b' + c' + a = a + b + c = c + a' + b'$. We then draw an equilateral triangle with this side length and cut off small equilateral triangles as shown in the diagram above. The resulting equiangular hexagon has the required side lengths.\n\nGiven $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$, the question is in how many ways can we order these numbers such that they satisfy (1). First note that the six numbers $a, b, c, a', b', c'$ satisfy (1) iff $a+K, b+K, c+K, a'+K, b'+K, c'+K$ satisfy (1). Therefore, it is sufficient to count the number of incongruent equiangular hexagons with side lengths $a_0, a_1, a_2, a_3, a_4, a_5$ in some order. For the same reason, we may assume $a_0 = 0$ so that $a_i = d i$. Moreover, the six numbers $a, b, c, a', b', c'$ satisfy (1) iff $d a, d b, d c, d a', d b', d c'$ do so. Hence, it is sufficient to study the case $d = 1$, i.e. $a_i = i$. In other words, we need to count the number of ways the numbers $0, 1, 2, 3, 4, 5$ can be ordered such that equations (1) hold. Cyclic changes of the order lead to congruent hexagons, hence we may assume $b = 5$.\n\nAdding the equations (1) gives us $(a + b + c) + b = (a' + b' + c') + b'$, which we rewrite as $b - b' = (a' + b' + c') - (a + b + c) = 15 - 2(a + b + c)$. The last equality comes from $a + b + c + a' + b' + c' = 0 + 1 + 2 + 3 + 4 + 5 = 15$. Since we have chosen $b = 5$ to be the largest available number, we have $b - b' > 0$, hence $15 > 2(a + b + c)$, i.e. $a + b + c \\le 7$ which means that $a + c \\le 2$. Therefore, $\\{a, c\\} = \\{0, 1\\}$ and $\\{a, c\\} = \\{0, 2\\}$ are the only possibilities.\n\nSwapping $a$ and $c$, as well as $a'$ and $c'$ leads to a congruent (via a reflection) equiangular hexagon. Therefore, the two possibilities (up to congruence) are $(a, c) = (0, 1)$ and $(a, c) = (0, 2)$.\n\nWhen $(a, c) = (0, 1)$ and $b = 5$, we have $a + b + c = 6$ and so $a' + b' + c' = 15 - 6 = 9$. Therefore, $b' = (a + b + c) + b - (a' + b' + c') = 6 + 5 - 9 = 2$. Hence, $a' = a + b - b' = 0 + 5 - 2 = 3$ and $c' = c + b - b' = 1 + 5 - 2 = 4$. We obtain $(a, b, c, a', b', c') = (0, 5, 1, 3, 2, 4)$.\n\nSimilarly, starting with $(a, c) = (0, 2)$ we obtain $a + b + c = 7$ and $a' + b' + c' = 8$. This leads to $b' = 4$, $a' = 1$ and $c' = 3$ and $(a, b, c, a', b', c') = (0, 5, 2, 1, 4, 3)$.\n\nSince $a + b + c = 0 + 5 + 1 = 6$ from the first solution is neither equal to $a + b + c = 0 + 5 + 2 = 7$ nor to $b + c + a' = 5 + 2 + 1 = 8$ from the second solution, the two solutions do not lead to congruent hexagons.\n\nTranslating this back to the original set-up, we have shown that (in case $d > 0$) for each $i \\ge 0$ the six numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ can be ordered in exactly two ways (up to congruence of the hexagons) such that these numbers are the side lengths of an equiangular hexagon in this order, namely\n\n$$\na_i, a_{i+5}, a_{i+1}, a_{i+3}, a_{i+2}, a_{i+4} \\quad \\text{and} \\quad a_i, a_{i+5}, a_{i+2}, a_{i+1}, a_{i+4}, a_{i+3}.\n$$\n\nRotation of the hexagon corresponds to cyclic rotation of the numbers. Reflection of the hexagon corresponds to certain swappings, for example swapping $a_i$ with $a_{i+1}$ as well as $a_{i+3}$ with $a_{i+4}$ in the first solution listed above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19934,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the difference between the largest and the second largest odd factors of $2016$?",
    "options": [],
    "answer": "See solution",
    "solution": "$2016 = 2^5 \\times 3^2 \\times 7$.\n\nThe largest odd factor is $3^2 \\times 7 = 63$.\n\nThe next largest odd factor is $3 \\times 7 = 21$.\n\nSo, $63 - 21 = 42$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19935,
    "subject": "Mathematics (Olympiad)",
    "question": "The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that\n\n$$\n\\frac{1}{2} \\le \\frac{a^3 + b^3}{a^2 + b^2} \\le 1.\n$$\n\nWhen do we have equality in the right inequality and when in the left inequality?",
    "options": [],
    "answer": "See solution",
    "solution": "By algebraic manipulation, we have\n\n$$\n\\frac{a^3 + b^3}{a^2 + b^2} = (a+b)\\frac{a^2 - ab + b^2}{a^2 + b^2} = 1 - \\frac{ab}{a^2 + b^2}.\n$$\n\nFrom this, the right inequality is evident with equality for $ab = 0$, i.e., for $a = 0$, $b = 1$ and for $a = 1$, $b = 0$. The left inequality is equivalent to\n\n$$\n\\frac{1}{2} \\le 1 - \\frac{ab}{a^2 + b^2} \\iff \\frac{ab}{a^2 + b^2} \\le \\frac{1}{2} \\iff 2ab \\le a^2 + b^2 \\iff 0 \\le (a-b)^2.\n$$\n\nThis inequality is obvious, with equality for $a = \\frac{1}{2}$. In this case, also $b = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19936,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m > 1$ and $n > 1$ be odd integers. Distinct real numbers are written in the cells of a table with $m$ rows and $n$ columns. A number is called *good* if:\n\n1. It is the largest in its row or column;\n2. It is the middle number of its column or row.\n\nWhat is the maximal number of good numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N_c$ be the set of good numbers that are largest in their column and middle in their row. Since no two of these numbers belong to the same row, we have $|N_c| \\leq m$. Denote by $a$ the largest element of $N_c$ and let $a_1, a_2, \\dots, a_{\\frac{n-1}{2}}$ be the numbers in the row of $a$ that are greater than $a$. It is clear that there are no elements from $N_c$ in the columns of $a_1, a_2, \\dots, a_{\\frac{n-1}{2}}$. Since there is at most one element of $N_c$ in every column, it follows that $|N_c| \\leq \\frac{n+1}{2}$. Thus,\n\n$$\nN_c \\leq \\min \\left\\{ m, \\frac{n+1}{2} \\right\\}.\n$$\n\nAnalogously, we obtain that the number of good numbers that are largest in their rows is at most $\\min\\left\\{m, \\frac{n+1}{2}\\right\\} + \\min\\left\\{n, \\frac{m+1}{2}\\right\\}$. We shall show that for every odd $m > 1$ and $n > 1$ there exists a table such that the number of good numbers equals $\\min\\left\\{m, \\frac{n+1}{2}\\right\\} + \\min\\left\\{n, \\frac{m+1}{2}\\right\\}$.\n\n1. If $m \\neq n$, without loss of generality assume that $1 < m < n$. Mark the cells of the table by the digits 1, 2, 3, 4, and 5 as shown in the figure. Next, fill in the cells by writing consecutively the numbers 1, 2, 3, ..., $mn$, starting with the cells marked by 1, then by 2, and so on. All numbers written in the cells marked by 2 or 4 are good. Their number equals:\n\n![](images/broshura_07_english_p23_data_038295aa9f.png)\n\n$$\n\\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\frac{m+1}{2} = \\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{m+1}{2} \\right\\}.\n$$\n\n2. Let $m = n$ and $3 < m$. As in 1, we fill the following table. The good numbers are again in cells marked by 2 or 4. Their number equals:\n\n![](images/broshura_07_english_p23_data_7436e7b927.png)\n\n$$\n\\frac{n-1}{2} + \\frac{n-1}{2} + 2 = n + 1\n$$\n\n3. If $m = n = 3$, then the table (5, 6, 7, and 8 are good) shows that the number of good numbers is 4 and $4 = 2 \\cdot 2 = 2 \\min \\left\\{ 3, \\frac{3+1}{2} \\right\\}$.\n\n| 1 | 7 | 9 |\n|---|---|---|\n| 2 | 6 | 4 |\n| 5 | 3 | 8 |\n\n$$\n\\text{Answer: } \\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{m+1}{2} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19937,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n$$\nf_n(x) = -\\frac{1}{2} + \\frac{1}{x-1} + \\frac{1}{4x-1} + \\dots + \\frac{1}{k^2x-1} + \\dots + \\frac{1}{n^2x-1} = 0\n$$\nfor $n \\in \\mathbb{N}^*$. For each $n$, show that the equation above has a unique root $x_n > 1$, and find $\\lim_{n \\to \\infty} x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. For every $n \\in \\mathbb{N}^*$, the function $f_n(x)$ is continuous and decreasing on $(1, +\\infty)$. Also, $f_n(x) \\to +\\infty$ as $x \\to 1^+$ and $f_n(x) \\to -\\frac{1}{2}$ as $x \\to +\\infty$. Thus, for each $n$, the equation $f_n(x) = 0$ has a unique root $x_n > 1$.\n\n2. For $n \\in \\mathbb{N}^*$:\n$$\n\\begin{aligned}\nf_n(4) &= -\\frac{1}{2} + \\frac{1}{2^2-1} + \\frac{1}{4^2-1} + \\dots + \\frac{1}{(2n)^2-1} \\\\\n&= \\frac{1}{2} \\left( -1 + 1 - \\frac{1}{3} + \\frac{1}{3} - \\frac{1}{5} + \\dots + \\frac{1}{2n-1} - \\frac{1}{2n+1} \\right) \\\\\n&= -\\frac{1}{2(2n+1)} < 0 = f_n(x_n).\n\\end{aligned}\n$$\nSince $f_n(x)$ is decreasing on $(1, +\\infty)$, it follows that $x_n < 4$ for all $n$.\n\n3. By the Mean Value Theorem, for some $t \\in (x_n, 4)$:\n$$\n\\frac{f_n(4) - f_n(x_n)}{4 - x_n} = f_n'(t) = -\\frac{1}{(t-1)^2} - \\frac{4}{(4t-1)^2} - \\dots - \\frac{n^2}{(n^2 t - 1)^2} < -\\frac{1}{9}\n$$\nSo,\n$$\n\\frac{-1}{2(2n + 1)(4 - x_n)} < -\\frac{1}{9}\n$$\nwhich gives\n$$\nx_n > 4 - \\frac{9}{2(2n + 1)}\n$$\nfor all $n$. Thus,\n$$\n4 - \\frac{9}{2(2n + 1)} < x_n < 4\n$$\nBy the squeeze theorem,\n$$\n\\lim_{n \\to \\infty} x_n = 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19938,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and $O$ its circumcenter. The circumcircle of triangle $AOC$ intersects the segment $BC$ at points $C$ and $D$, and the segment $AB$ at points $A$ and $E$.\n\nProve that triangles $BDE$ and $AOC$ have equal circumradii.\n\n![](images/AUT_ABooklet_2023_p15_data_8e6841aa09.png)",
    "options": [],
    "answer": "See solution",
    "solution": "In the circumcircle of triangle $ABC$ we have $\\angle COA = 2\\angle CBA$. In the circumcircle of $ADC$ we therefore have $\\angle CDA = \\angle COA = 2\\angle CBA$. The angle $\\angle CDA$ is an external angle in triangle $ABD$, so $\\angle CBA + \\angle BAD = \\angle CDA = 2\\angle CBA$, and thus $\\angle BAD = \\angle CBA$.\n\nIn the circumcircle of $AOC$ we obtain $\\angle BAD = \\angle EAD$ on the chord $ED$. The angles $\\angle CBA = \\angle DBE$ are equal in the circumcircle of triangle $BDE$ on the same chord $ED$. Since the chords and subtended angles are equal in both circles, they must have the same radii, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19939,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer triples $ (a, b, c) $ satisfying the equation:\n\n$$\n5a^2 + 9b^2 = 13c^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $ (a, b, c) = (0, 0, 0) $ is a solution. Assume that the equation has a solution $ (a_0, b_0, c_0) \\neq (0, 0, 0) $. Let $ d = \\gcd(a_0, b_0, c_0) > 0 $. Let $ (a, b, c) = (a_0/d, b_0/d, c_0/d) $. Then $ \\gcd(a, b, c) = 1 $. From $ 5a_0^2 + 9b_0^2 = 13c_0^2 $ it follows that:\n\n$$\n5a^2 + 9b^2 = 5 \\left(\\frac{a_0}{d}\\right)^2 + 9 \\left(\\frac{b_0}{d}\\right)^2 = \\frac{5a_0^2 + 9b_0^2}{d^2} = \\frac{13c_0^2}{d^2} = 13 \\left(\\frac{c_0}{d}\\right)^2 = 13c^2\n$$\n\nHence $ (a, b, c) $ is also a solution.\n\nAs $ (a_0, b_0, c_0) \\neq (0, 0, 0) $ it follows that $ (a, b, c) \\neq (0, 0, 0) $. Consider the equation modulo $13$. It follows that $5a^2 + 9b^2 = 13c^2 \\equiv 0 \\pmod{13}$, that is $5a^2 \\equiv -9b^2 \\equiv 4b^2 \\pmod{13}$. Multiplying by $8$ gives:\n\n$$\na^2 \\equiv 40a^2 = 8 \\cdot 5a^2 \\equiv 8 \\cdot 4b^2 = 32b^2 \\equiv 6b^2 \\pmod{13}\n$$\n\nIf $13 \\mid b$ then $6b^2 \\equiv 0 \\pmod{13}$ and therefore $a^2 \\equiv 0 \\pmod{13}$, that is $13 \\mid a^2$. As $13$ is prime it follows that $13 \\mid a$. Hence $13$ divides $a$ and $b$. It follows that $13^2 \\mid 5a^2 + 9b^2 = 13c^2$. Consequently $13$ divides $c^2$. As $13$ is prime, $13 \\mid c$. This means that $13$ divides $a, b$ and $c$, contradicting the fact that $\\gcd(a, b, c) = 1$. We conclude that $13 \\nmid b$.\n\nAs $13 \\nmid b$ and $13$ is a prime, it follows that $b$ and $13$ are relatively prime. Therefore there exists $x \\in \\mathbb{Z}$ such that $b x \\equiv 1 \\pmod{13}$. Multiplying by $x^2$ gives:\n\n$$\n(a x)^2 = a^2 x^2 \\equiv 6 b^2 x^2 = 6 (b x)^2 \\equiv 6 \\cdot 1^2 = 6 \\pmod{13}\n$$\n\nThat is, $y^2 \\equiv 6 \\pmod{13}$ where $y = a x$. As $y^2 \\equiv 6 \\pmod{13}$ it follows that $y$ and $13$ are relatively prime. By Fermat's little theorem it follows that $y^{12} \\equiv 1 \\pmod{13}$. Hence:\n\n$$\n\\begin{aligned}\n1 &\\equiv y^{12} = (y^2)^6 \\equiv 6^6 = (6^2)^3 \\equiv (36)^3 \\equiv 10^3 \\\\\n&= 10^2 \\cdot 10 = 100 \\cdot 10 \\equiv 9 \\cdot 10 = 90 \\equiv 12 \\pmod{13}\n\\end{aligned}\n$$\n\nBut $1 \\not\\equiv 12 \\pmod{13}$, so we have a contradiction. We conclude that the equation $5a^2 + 9b^2 = 13c^2$ has no solution besides the solution $ (a, b, c) = (0, 0, 0) $. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19940,
    "subject": "Mathematics (Olympiad)",
    "question": "Three equal circles of radius $R$ are given such that each one passes through the centers of the other two. Find the area of the common region.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_1, O_2, O_3$ be the centers of the three circles and $S$ the area of the common region. The three sectors with centers $O_1, O_2, O_3$ which subtend the arcs $O_2O_3, O_1O_3, O_2O_1$, respectively, cover the surface of area $S$ and twice more the surface of triangle $O_1O_2O_3$.\n\nThe area of triangle $O_1O_2O_3$ is $\\frac{R^2\\sqrt{3}}{4}$. On the other hand, the area of each of these three circular sectors equals $\\frac{1}{3}$ the area of a semicircle of radius $R$, hence it is $\\frac{1}{6}\\pi R^2$.\n\nHence\n\n$$\n\\frac{1}{2}\\pi R^2 = S + 2 \\cdot \\frac{R^2\\sqrt{3}}{4}.\n$$\n\nTherefore,\n\n$$\nS = \\frac{1}{2}(\\pi - \\sqrt{3})R^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19941,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomials\n\n$$P_n(x, y, z) = (x-y)^{2n}(y-z)^{2n} + (y-z)^{2n}(z-x)^{2n} + (z-x)^{2n}(x-y)^{2n}$$\n\nand\n\n$$Q_n(x, y, z) = \\left[(x - y)^{2n} + (y - z)^{2n} + (z - x)^{2n}\\right]^{2n}.$$\n\nDetermine all positive integers $n$ such that the quotient $Q_n(x, y, z)/P_n(x, y, z)$ is a polynomial in $x, y, z$ with integer coefficients.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us evaluate the quotient at $(x, y, z) = (0, 1, 2)$:\n\n$$Q_n(0, 1, 2)/P_n(0, 1, 2)$$\n\nis an integer, so $P_n(0, 1, 2)$ divides $Q_n(0, 1, 2)$.\n\nWe compute:\n\n$$P_n(0, 1, 2) = 2^{2n+1} + 1$$\n\nand\n\n$$Q_n(0, 1, 2) = (2^{2n} + 2)^{2n} = \\left(\\frac{2^{2n+1} + 4}{2}\\right)^{2n} = \\frac{(2^{2n+1} + 4)^{2n}}{2^{2n}}.$$\n\nThus, $P_n(0, 1, 2)$ divides $Q_n(0, 1, 2)$ if and only if $2^{2n+1}+1$ divides $(2^{2n+1} + 4)^{2n}$. Expanding:\n\n$$(2^{2n+1} + 4)^{2n} = (2^{2n+1} + 1 + 3)^{2n} = \\sum_{k=0}^{2n} \\binom{2n}{k} (2^{2n+1} + 1)^{2n-k} 3^k$$\n\nSo $2^{2n+1} + 1$ divides $3^{2n}$, which means $2^{2n+1} + 1 = 3^s$ for some $s \\in \\mathbb{N}$.\n\nFrom $1 \\equiv (-1)^s \\pmod{4}$, $s$ must be even, say $s = 2m$. Then $2^{2n+1} = (3^m - 1)(3^m + 1)$, so $3^m - 1 = 2^u$, $3^m + 1 = 2^v$ for $u, v \\ge 0$, $u < v$, $u + v = 2n+1$. Thus $2^v - 2^u = 2$, so $u = 1$, $v = 2$, and $n = 1$.\n\nFor $n = 1$:\n\n$$P_1(x, y, z) = (x - y)^2 (y - z)^2 + (y - z)^2 (z - x)^2 + (z - x)^2 (x - y)^2$$\n\n$$Q_1(x, y, z) = \\left[(x - y)^2 + (y - z)^2 + (z - x)^2\\right]^2$$\n\nIt can be shown that $Q_1(x, y, z) = 4P_1(x, y, z)$.\n\nLet $a = x - y$, $b = y - z$, $c = z - x$, so $a + b + c = 0$. Then\n\n$$\\left(\\sum_{cyc} a^2\\right)^2 = 4 \\sum_{cyc} a^2 b^2$$\n\nTherefore, the only positive integer $n$ for which $Q_n(x, y, z)/P_n(x, y, z)$ is a polynomial with integer coefficients is $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19942,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a real number $c$ and let $a = c - 2013$. Consider the sequence defined by\n\n$$\n\\begin{aligned}\np_0 &= 0, \\\\\np_1 &= c - 2013 = a, \\\\\np_n &= (c - 2013)p_{n-1} + (2014 - c)p_{n-2} = ap_{n-1} + (1 - a)p_{n-2}.\n\\end{aligned}\n$$\n\nShow that if $a \\ne 0$, then $p_n$ is not equal to zero for any positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "#### Case 1: $a \\ge 1$\n\nWe have $p_1 - p_0 = a > 0$ and $p_n - p_{n-1} = (a-1)(p_{n-1} - p_{n-2})$. Since $a-1 \\ge 0$, by induction $p_n - p_{n-1} \\ge 0$, so the sequence is non-decreasing. Since $p_1 = a > 0$, $p_n \\ne 0$ for any positive integer $n$.\n\n#### Case 2: $0 < a < 1$\n\nWe have $p_0 = 0$ and $p_1 = a > 0$. Note that $p_n = ap_{n-1} + (1-a)p_{n-2}$ is a weighted mean of $p_{n-1}$ and $p_{n-2}$. By induction, $p_{n-1}$ and $p_{n-2}$ are distinct, and $p_n$ lies strictly between them. Thus, $p_n > 0$ for all positive integers $n$.\n\n#### Case 3: $a < 0$\n\nBy induction, $p_n > 0$ for even $n$ and $p_n < 0$ for odd $n$. For $n = 1$, $p_1 = a < 0$; for $n = 2$, $p_2 = a^2 > 0$. If $n$ is odd, $p_n = ap_{n-1} + (1-a)p_{n-2} < 0$; if $n$ is even, $p_n = ap_{n-1} + (1-a)p_{n-2} > 0$, using the inductive hypothesis and $a < 0$, $1-a > 0$.\n\nThus, in all cases with $a \\ne 0$, $p_n \\ne 0$ for any positive integer $n$. Therefore, there are no other real roots of $p_n(x)$ apart from $2013$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19943,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Let $L$ be the line through $B$ perpendicular to $AB$. The perpendicular from $A$ to $BC$ meets $L$ at the point $D$. The perpendicular bisector of $BC$ meets $L$ at the point $P$. Let $E$ be the foot of the perpendicular from $D$ to $AC$.\n\nProve that triangle $BPE$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $BC$ and $E'$ the intersection of $CP$ and $DE$. Let $\\alpha = \\angle DAB$\n\n![](images/British_2019_Booklet_p10_data_b1081ad6f7.png)\n\nLooking at the right-angled triangle, we have $\\angle BDA = 90^\\circ - \\alpha$ and so $\\angle CBP = \\angle CBD = \\alpha$. Using symmetry in $MP$, $\\angle E'CB = \\angle PCB = \\alpha$.\n\nNow quadrilateral $ABDE$ is cyclic as $\\angle ABD = 90^\\circ = \\angle DEA$. Hence $\\angle E'EB = \\angle DEB = \\angle DAB = \\alpha$.\n\nNow, as $\\angle E'CB = \\alpha = \\angle E'EB$, we have that quadrilateral $BE'CE$ is cyclic. As $\\angle CEE' = 90^\\circ$, the centre of this circle lies on $CE'$. But the centre also lies on the perpendicular bisector of $BC$ so is $P$. In particular $PB = PE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19944,
    "subject": "Mathematics (Olympiad)",
    "question": "Find a triple $(l, m, n)$ with $1 < l < m < n$ of positive integers such that\n$$\n\\sum_{k=1}^{l} k, \\quad \\sum_{k=l+1}^{m} k, \\quad \\sum_{k=m+1}^{n} k\n$$\nform a geometric sequence in order.",
    "options": [],
    "answer": "See solution",
    "solution": "For $t \\in \\mathbb{N}^*$, denote $S_t = \\sum_{k=1}^{t} k = \\frac{t(t+1)}{2}$. Let\n$$\n\\sum_{k=1}^{l} k = S_{l}, \\quad \\sum_{k=l+1}^{m} k = S_{m} - S_{l}, \\quad \\sum_{k=m+1}^{n} k = S_{n} - S_{m},\n$$\nform a geometric sequence in order. Then\n$$\nS_l(S_n - S_m) = (S_m - S_l)^2.\n$$\nThat is, $S_l(S_n + S_m - S_l) = S_m^2$. Thus, $S_l \\mid S_m^2$, that is,\n$$\n2l(l + 1) \\mid m^2(m + 1)^2.\n$$\nLet $m + 1 = l(l + 1)$ and take $l = 3$. Then $m = 11$ and $S_l = S_3 = 6$, $S_m = S_{11} = 66$. Substitute into the equation, we have $S_n = 666$, that is $\\frac{n(n+1)}{2} = 666$. So $n = 36$.\n\nTherefore, $(l, m, n) = (3, 11, 36)$ is a solution satisfying the condition.\n\n**Remark.** The solution is not unique. For example, there are other solutions such as $(8, 11, 13)$, $(5, 9, 14)$, $(2, 12, 62)$, $(3, 24, 171)$. (We may show that the number of solutions is infinite.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers, and let $a_{ij}$ ($1 \\le i \\le m$, $1 \\le j \\le n$) be nonnegative real numbers such that for any $i, j$, the inequalities\n\n$$\na_{i,1} \\ge a_{i,2} \\ge \\dots \\ge a_{i,n}, \\quad a_{1,j} \\ge a_{2,j} \\ge \\dots \\ge a_{m,j}\n$$\nhold.\n\nFor $i = 1, 2, \\dots, m$ and $j = 1, 2, \\dots, n$, define\n\n$$\nX_{i,j} = a_{1,j} + \\dots + a_{i-1,j} + a_{i,j} + a_{i,j-1} + \\dots + a_{i,1},\n$$\n\n$$\nY_{i,j} = a_{m,j} + \\dots + a_{i+1,j} + a_{i,j} + a_{i,j+1} + \\dots + a_{i,n}.\n$$\n\nProve:\n\n$$\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} \\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The problem conditions imply that\n\n$$\nX_{i,j} \\ge (i + j - 1) \\cdot a_{i,j}, \\quad Y_{i,j} \\le (m + n - i - j + 1) \\cdot a_{i,j}.\n$$\n\nHence,\n\n$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (i+j-1) \\cdot a_{i,j} \\\\\n&= \\prod_{i=1}^{m} \\prod_{j=1}^{n} (m+n-i-j+1) \\cdot a_{i,j} \\\\\n&\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j},\n\\end{align*}\n$$\n\nwhere the equality on the second line is due to the one-to-one correspondence $(i, j) \\leftrightarrow (m + 1 - i, n + 1 - j)$, and $i + j - 1$ corresponds to $m + n - i - j + 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19946,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram with the property that $|AD| = |BD|$. Now let $P$ and $Q$ be points such that $\\triangle ADP$ and $\\triangle CDQ$ are equilateral and do not overlap with the parallelogram. Prove that $\\angle PQD = 30^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are a lot of line segments of the same length. For example, $|AD| = |DP| = |PA|$ because triangle $\\triangle ADP$ is equilateral. It is given that $|AD| = |BD|$, and finally $|AD| = |BC|$ because $ABCD$ is a parallelogram. Similarly, $|CD| = |DQ| = |QC| = |AB|$. The opposite angles in the parallelogram are the same size, so $\\angle DAB = \\angle BCD$. Since $\\triangle ABD$ and $\\triangle DBC$ are isosceles triangles, these angles are also equal to $\\angle ABD$ and $\\angle CDB$. Finally, the angles of the equilateral triangles $\\triangle ADP$ and $\\triangle CDQ$ are all $60^\\circ$. All this information is summarised in the figure below.\n\n![](images/NLD_ABooklet_2025_p15_data_2fce97070d.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19947,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $GB$ and $GC$, respectively, such that\n\n$$\n\\angle ABS = \\angle ACR = 180^{\\circ} - \\angle BGC.\n$$\n\nProve that $\\angle RAS + \\angle BAC = \\angle BGC$.\n\nIn all the following solutions, let $M$ and $N$ denote the midpoints of $\\overline{AC}$ and $\\overline{AB}$, respectively.\n\n![](images/sols-TSTST-2023_p3_data_68f8ec6f1a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "From the given condition that $\\angle ACR = \\angle CGM$, we get that\n\n$$\nMA^2 = MC^2 = MG \\cdot MR \\Rightarrow \\angle RAC = \\angle MGA.\n$$\n\nAnalogously,\n\n$$\n\\angle BAS = \\angle AGN.\n$$\n\nHence,\n\n$$\n\\angle RAS + \\angle BAC = \\angle RAC + \\angle BAS = \\angle MGA + \\angle AGN = \\angle MGN = \\angle BGC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19948,
    "subject": "Mathematics (Olympiad)",
    "question": "Count the number of $2 \\times 2$ matrices $A$ with entries in $\\mathbb{F}_{37}$ such that $A^2 = A$ and $\\det A = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\det A = 1 \\ne 0$, $A$ is invertible. Multiplying both sides of $A^2 = A$ by $A^{-1}$ gives $A = I$. The identity matrix $I$ satisfies the conditions, so $A = I$ is the only solution. Equivalently, $a \\equiv d \\equiv 1 \\pmod{37}$ and $b \\equiv c \\equiv 0 \\pmod{37}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19949,
    "subject": "Mathematics (Olympiad)",
    "question": "Најди ги сите броеви $p$, $q$ и $r$, такви што $p$ и $r$ се прости, $q$ е позитивен цел број и ја задоволуваат равенката:\n\n$$\n(p+q+r)^2 = 2p^2 + 2q^2 + r^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "По средување на равенката добиваме $2r(p+q) = (p-q)^2$. Бидејќи $r$ е прост, следува дека $r$ е делител на $p-q$, па $r^2$ е делител на десната страна од последното равенство. Од каде следува дека $r$ е делител на $2(p+q)$. Ако $r>2$, тогаш $r$ е делител на $(p+q)$, па мора $r$ да е делител и на $p$ и на $q$, но бидејќи $p$ е прост, тоа е можно само ако $p=r$ и $q=sr$. По средување добиваме $2(1+s)=(s-1)^2$, од каде $s^2-4s-1=0$. Последното равенство нема целобројни решенија, па во овој случај равенката нема решение.\n\nАко $r=2$, тогаш $p$ и $q$ се со иста парност, случајот кога $p=2$ е невозможен исто како случајот $p=r$ од претходно, па мора да бидат непарни. Нека $a \\neq 2$ е прост делител на $p+q$, тогаш мора $a$ да е делител и на $p-q$, па мора да е делител и на $p$ и на $q$, што е можно само ако $p=a$ и $q=sa$, во овој случај добиваме $4(1+s)=a(s-1)^2$, од каде $a^2-(2a+4)s+(a-4)=0$, со решенија $\\frac{a+2\\pm\\sqrt{a^2+4a+4-a^2+4a}}{a} = \\frac{a+2\\pm2\\sqrt{2a+1}}{a}$. Ако бројот $\\sqrt{2a+1}$ е цел, тогаш е непарен, па $2a+1=4b^2+4b+1$, од каде $a=2b(b+1)$, па не може да е прост.\n\nСпоред ова $p+q$ и $p-q$ мора да се степени на 2, то ест $p-q=2^k$ и $p+q=2^{2k-2}$, од каде $2p=2^k+2^{2k-2}$ и $2q=2^{2k-2}-2^k$ и бидејќи $p$ и $q$ се непарни, мора $k=1$, но тогаш $p+q=1$, што не е можно. Следува дека равенката нема решенија кои се прости броеви.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19950,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(x) + y) = f(f(x) - y) + 4f(x)y, \\text{ for any } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $f \\equiv 0$ satisfies the given condition.\n\nAssume $f \\neq 0$. Choose $x_0 \\in \\mathbb{R}$ such that $f(x_0) \\neq 0$ and set $\\tilde{y} = \\frac{y}{4f(x_0)}$ for every $y \\in \\mathbb{R}$. Then, for $x_0$ and $\\tilde{y}$, the given condition becomes:\n\n$$\ny = f(f(x_0) + \\tilde{y}) - f(f(x_0) - \\tilde{y}). \\quad (1)\n$$\n\nFor any $y_1, y_2 \\in \\mathbb{R}$, substituting $\\frac{y_1 - y_2}{2}$ for $y$ in (1), we get:\n\n$$\n\\frac{y_1 - y_2}{2} = f\\left(f(x_0) + \\frac{\\widetilde{y_1 - y_2}}{2}\\right) - f\\left(f(x_0) - \\frac{\\widetilde{y_1 - y_2}}{2}\\right),\n$$\n\nso for all $y_1, y_2 \\in \\mathbb{R}$, there exist $x_1 = f(x_0) + \\frac{\\widetilde{y_1 - y_2}}{2}$ and $x_2 = f(x_0) - \\frac{\\widetilde{y_1 - y_2}}{2}$ such that $\\frac{y_1 - y_2}{2} = f(x_1) - f(x_2)$, or\n\n$$\n2f(x_1) - y_1 = 2f(x_2) - y_2. \\quad (2)\n$$\n\nNow, substitute $y$ with $f(x) - y$ in the original condition:\n\n$$\n\\begin{aligned}\nf(2f(x) - y) &= f(y) + 4f(x)(f(x) - y) \\\\\nf(y) - y^2 &= f(2f(x) - y) - (2f(x) - y)^2.\n\\end{aligned} \\quad (3)\n$$\n\nNow, plug $x_1, x_2$ in place of $y_1, y_2$ in (3):\n\n$$\nf(y_1) - y_1^2 = f(2f(x_1) - y_1) - (2f(x_1) - y_1)^2\n$$\n\nand\n\n$$\nf(y_2) - y_2^2 = f(2f(x_2) - y_2) - (2f(x_2) - y_2)^2.\n$$\n\nFrom (2), $2f(x_1) - y_1 = 2f(x_2) - y_2$, so $f(y_1) - y_1^2 = f(y_2) - y_2^2$. Since this holds for all $y_1, y_2 \\in \\mathbb{R}$, we conclude:\n\n$$\nf(x) - x^2 = c \\in \\mathbb{R}, \\text{ for all } x \\in \\mathbb{R}, \\quad \\text{so} \\quad f(x) = x^2 + c,\\ c \\in \\mathbb{R}.\n$$\n\nIt is easy to verify that this function satisfies the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19951,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real sequences $\\{b_n\\}_{n \\ge 1}$ and $\\{c_n\\}_{n \\ge 1}$ such that, for each positive integer $n$:\n\n1. $b_n \\le c_n$;\n2. $b_{n+1}$ and $c_{n+1}$ are the two roots of the quadratic equation $x^2 + b_n x + c_n = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "By (1) and (2), $b_{n+1}$ and $c_{n+1}$ are uniquely determined by $b_n$ and $c_n$. By Vi\\`ete's formulas:\n\n$$\nb_n = -(b_{n+1} + c_{n+1})\n$$\n$$\nc_n = b_{n+1}c_{n+1}\n$$\n\nIf $b_1 = c_1 = 0$, then $b_n = c_n = 0$ for all $n \\ge 1$. The sequences $\\{b_n\\}_{n \\ge 1}$ and $\\{c_n\\}_{n \\ge 1} = \\{0, 0, \\dots\\}$ satisfy the problem conditions.\n\nWe prove that all other sequences do not meet the requirements.\n\nFirst, assume one of $b_1$, $c_1$ is zero. If $b_1 = 0$, then $c_1 > 0$, $x^2 + b_1x + c_1 = 0$ has no real roots; if $c_1 = 0$, then $b_1 < 0$, $x^2 + b_1x + c_1 = 0$ has roots $b_2 = 0$, $c_2 = -b_1 > 0$, but $x^2 + b_2x + c_2 = 0$ has no real roots.\n\nNow, assume both $b_1$, $c_1$ are non-zero. If $b_n > 0$, $c_n > 0$, then from above, $b_{n+1} < 0$, $c_{n+1} < 0$; if $b_n < 0$, $c_n < 0$, then $b_{n+1} < 0$, $c_{n+1} > 0$; if $b_n < 0$, $c_n > 0$, then $b_{n+1} > 0$, $c_{n+1} > 0$. Thus, the signs of $(b_n, c_n)$ are 3-periodic: positive and positive, negative and negative, negative and positive, ...\n\nSuppose $b_n > 0$, $c_n > 0$. For the roots of $x^2 + b_n x + c_n = 0$ to be real, the discriminant must satisfy\n\n$$\nb_n^2 - 4c_n \\ge 0\n$$\nso $b_n^2 \\ge 4c_n \\ge 4b_n$, $c_n \\ge b_n \\ge 4$. Due to the periodicity, assume $n > 3$. By the recurrence,\n\n$$\n\\begin{align*}\nb_{n-1} &= -(b_n + c_n) < 0, & c_{n-1} &= b_n c_n > 0, \\\\\nb_{n-2} &= b_n + c_n - b_n c_n < 0, \\\\\nc_{n-2} &= -(b_n + c_n)b_n c_n < 0, \\\\\nb_{n-3} &\\ge (b_n + c_n)b_n c_n \\ge 4b_n\n\\end{align*}\n$$\n\nLet $i \\in \\{1, 2, 3\\}$ be the first index with $b_i > 0$, $c_i > 0$. Then $b_{i+3k} > 0$, $c_{i+3k} > 0$ for all $k \\ge 0$; repeatedly applying the inequalities gives\n\n$$\nb_i \\ge 4^k b_{i+3k} \\ge 4^{k+1}\n$$\nfor arbitrary $k$, which is impossible.\n\nTherefore, $\\{b_n\\}_{n \\ge 1} = \\{c_n\\}_{n \\ge 1} = \\{0, 0, \\dots\\}$ are the only sequences satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19952,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, $\\odot O$ is the inscribed circle touching side $BC$ of $\\triangle ABC$ at point $M$, and points $D$ and $E$ are on segments $AB$ and $AC$, respectively, such that $DE \\parallel BC$. $\\odot O_1$ is the inscribed circle of $\\triangle ADE$ tangent to side $DE$ at point $N$. Lines $O_1B$ and $DO$ intersect at point $F$, and lines $O_1C$ and $EO$ intersect at point $G$. Prove that $MN$ divides segment $FG$ equally.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p208_data_d785dde652.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If $AB = AC$, the figure is symmetric about the bisector of $\\angle BAC$, so the result is obvious. Assume $AB > AC$. Let $L$ be the midpoint of $BC$, and let line $O_1L$ intersect $FG$ at $R$. Extend $O_1N$ to meet $BC$ at $K$. Draw $AT \\perp BC$ at $T$, meeting $DE$ at $S$. Connect $AO$; clearly, $O_1$ lies on $AO$.\n\nBy Menelaus' theorem:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p208_data_62e5dc4e59.png)\n\n$$\n\\frac{O_1F}{FB} \\cdot \\frac{BD}{DA} \\cdot \\frac{AO}{OO_1} = 1, \\quad \\frac{O_1G}{GC} \\cdot \\frac{CE}{EA} \\cdot \\frac{AO}{OO_1} = 1. \\qquad \\textcircled{1}\n$$\n\nSince $DE \\parallel BC$, $\\frac{BD}{DA} = \\frac{CE}{EA}$, so $\\frac{O_1F}{FB} = \\frac{O_1G}{GC}$, which means $FG \\parallel BC$. Thus, $\\frac{FR}{GR} = \\frac{BL}{CL} = 1$, so $R$ is the midpoint of $FG$.\n\nIt remains to show that $M$, $R$, $N$ are collinear. By the converse of Menelaus' theorem, it suffices to prove:\n\n$$\n\\frac{O_1R}{RL} \\cdot \\frac{LM}{MK} \\cdot \\frac{KN}{NO_1} = 1. \\qquad \\textcircled{2}\n$$\n\nSince $FR \\parallel BL$, $\\frac{O_1R}{RL} = \\frac{O_1F}{FB} = \\frac{OO_1}{AO} \\cdot \\frac{AD}{DB}$ (by ①). So we need to prove:\n\n$$\n\\frac{OO_1}{AO} \\cdot \\frac{AD}{DB} \\cdot \\frac{LM}{MT} \\cdot \\frac{KN}{NO_1} = 1. \\qquad \\textcircled{3}\n$$\n\nSince $O_1K \\perp DE$, $OM \\perp BC$, $AT \\perp BC$, and $DE \\parallel BC$, lines $O_1K$, $OM$, $AT$ are parallel. By the theorem of proportional segments cut by parallel lines, $\\frac{OO_1}{AO} = \\frac{MK}{MT}$. Substitute into (3):\n\n$$\n\\frac{AD}{DB} \\cdot \\frac{LM}{MT} \\cdot \\frac{KN}{NO_1} = 1. \\qquad \\textcircled{4}\n$$\n\nSince $DE \\parallel BC$, $KN \\perp DE$, $ST \\perp BC$, quadrilateral $KNST$ is a rectangle, so $KN = ST$. Also, $DS \\parallel BT$ implies $\\frac{AD}{DB} = \\frac{AS}{ST}$. Substitute into (4):\n\n$$\n\\frac{LM}{MT} = \\frac{NO_1}{AS}. \\qquad \\textcircled{5}\n$$\n\nLet $BC = a$, $AC = b$, $AB = c$. Then\n\n$$\nBM = \\frac{a + b - c}{2} \\quad \\text{(property of the incircle)}, \\quad BL = \\frac{a}{2},\n$$\n\n$$\nBT = c \\cos \\angle ABC = c \\cdot \\frac{a^2 + c^2 - b^2}{2ac} = \\frac{a^2 + c^2 - b^2}{2a}.\n$$\n\nSo\n\n$$\n\\frac{LM}{MT} = \\frac{BL - BM}{BT - BM} = \\frac{\\frac{c-b}{2}}{\\frac{c^2 - b^2 + a(c-b)}{2a}} = \\frac{a}{a+b+c}.\n$$\n\nOn the other hand,\n\n$$\n\\frac{NO_1}{AS} = \\frac{\\frac{2S_{\\triangle ADE}}{AD + DE + AE}}{\\frac{2S_{\\triangle ADE}}{DE}} = \\frac{DE}{AD + DE + AE} = \\frac{a}{a+b+c}.\n$$\n\nTherefore, (5) holds. The proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19953,
    "subject": "Mathematics (Olympiad)",
    "question": "In a table with two rows and five columns, each of the squares is coloured black or white according to the following rules:\n\n- Two adjacent columns may never have the same number of black squares.\n- Two $2 \\times 2$ squares that overlap in one column may never have the same number of black squares.\n\nHow many possible colourings of the table comply with these rules?\n\nA) 6  \nB) 8  \nC) 12  \nD) 20  \nE) 24",
    "options": [],
    "answer": "See solution",
    "solution": "D) 20",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19954,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1$ and $C_1$ be the points on the sides $BC$ and $AB$ of triangle $ABC$ respectively, so that $AA_1 = CC_1$. Segments $AA_1$ and $CC_1$ meet at the point $F$. Prove that if $\\angle CFA_1 = 2\\angle ABC$ then $AA_1 = AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Consider translation by vector $\\vec{A_1A}$. Thus, $AA_1CC_3$ is a parallelogram, and $CC_1C_2C_3$ is a rhombus. Then $\\angle C_1CC_3 = \\angle C_1FA = \\angle CFA_1 = 2\\alpha$, therefore\n\n$$\n\\angle C_1AC_3 = 180^\\circ - \\angle ABC = 180^\\circ - \\frac{1}{2}\\angle CFA_1 = 180^\\circ - \\alpha.\n$$\n\nIf we build a circle where the point $C$ is its centre, then the central angle $\\angle C_1CC_3 = 2\\alpha$, thus the inscribed angle that subtends the arc $C_1C_3$ equals $\\alpha$, therefore the point $A$ belongs to the circle. Therefore, $AC = CC_3 = AA_1$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19955,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a *powerful set* of real numbers, meaning that for any two distinct elements $x, y \\in S$, at least one of $x^y$ or $y^x$ is also in $S$. What is the largest possible size of a powerful set $S$ contained in $[1, \\infty)$? What about a powerful set $S$ contained in $(0, 1]$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $S$ is a powerful set with $n > 3$ elements in $[1, \\infty)$, with $S = \\{1 = a_1 < a_2 < \\cdots < a_n\\}$. We can assume $a_1 = 1$, since adding $1$ to $S$ increases its size and preserves the property. For $i \\ge 2$, $a_n^{a_i} > a_n$, so $a_i^{a_n}$ must be in $S$. Thus,\n\n$$\na_1 < a_2 < a_2^{a_n} < a_3^{a_n} < \\cdots < a_{n-1}^{a_n}$$\n\nSo for $2 \\le i \\le n-1$, $a_i^{a_n} = a_{i+1}$. Now, for $a_2 < a_{n-1}$ ($n > 3$),\n\n$$\na_2 < a_2^{a_{n-1}} < a_2^{a_n} = a_3 \\implies a_2^{a_{n-1}} \\notin S$$\n\n$$\na_{n-1} < a_{n-1}^{a_2} < a_{n-1}^{a_n} = a_n \\implies a_{n-1}^{a_2} \\notin S$$\n\nBut this contradicts the definition of a powerful set. Thus, there is no powerful set with more than $3$ elements in $[1, \\infty)$.\n\nNow, suppose $S$ is a powerful set with $n > 4$ elements in $(0, 1]$, with $S = \\{a_1 < a_2 < \\cdots < a_n = 1\\}$. Again, $1 \\in S$. For $1 \\le i \\le n-2$, $a_{n-1} < a_{n-1}^{a_i} < 1$, so $a_n^{a_i} \\notin S$, and thus $a_i^{a_n} \\in S$. We have\n\n$$\na_1 < a_1^{a_{n-1}} < a_2^{a_{n-1}} < \\cdots < a_{n-2}^{a_n} < 1$$\n\nSo $a_i^{a_{n-1}} = a_{i+1}$ for $2 \\le i \\le n-2$. Let $a_{n-1} = a$, then\n\n$$\na_{n-2} = a^{1/a},\\quad a_{n-3} = a^{1/a^2}, \\ldots$$\n\nLooking at $a_{n-1}$ and $a_{n-2}$:\n\n$$\na_{n-1} = a_{n-2}^{a_{n-1}} < a_{n-1}^{a_{n-2}} < 1 \\implies a_{n-1}^{a_{n-2}} \\notin S$$\n\nThis implies $a_{n-2}^{a_{n-1}} \\in S$. Since $a_{n-2}^{a_{n-1}} > a_{n-2}^{a_n} = a_{n-1}$, we get $a_{n-2}^{a_{n-1}} = a_n$. So\n\n$$\na_{n-2}^{a_{n-1}} = (a^{1/a^2})^{a^{1/a}} = a \\implies a^{(a^{1/a})-2} = a \\implies a^{1/a - 2} = 1$$\n\nBut $a \\neq 1$, so $a = 1/2$. Therefore, $a_{n-1} = 1/2$, $a_{n-2} = 1/4$, $a_{n-3} = 1/16$. Since $n > 4$, $a_{n-4} = 1/256 \\in S$, but neither $a_{n-3}^{a_{n-4}}$ nor $a_{n-4}^{a_{n-3}}$ is in $S$. Thus, there is no powerful set with more than $4$ elements in $(0, 1]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19956,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exists a constant $\\lambda > 0$ such that: for any positive integer $m$, in the rectangular coordinate system, if all vertices of $\\triangle ABC$ are integral points, and there is a unique interior (not on the side) integral point whose $x$ and $y$ coordinates are multiples of $m$, then the area of $\\triangle ABC$ is less than $\\lambda m^3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We call $(x, y)$ an “$m$-integral point” if $x, y$ are integers and $m \\mid x$, $m \\mid y$. By a proper translation, we may assume that the unique interior $m$-integral point is at $(0, 0)$. Extend $AO$, $BO$, and $CO$ to meet the opposite sides at points $D$, $E$, and $F$, respectively. Denote\n\n$$\np = \\frac{OD}{AD} = \\frac{S_{\\triangle OBC}}{S_{\\triangle ABC}}, \\quad q = \\frac{OE}{BE} = \\frac{S_{\\triangle OCA}}{S_{\\triangle ABC}}, \\quad r = \\frac{OF}{CF} = \\frac{S_{\\triangle OAB}}{S_{\\triangle ABC}}.\n$$\n\nEvidently, $p + q + r = 1$; assume $p \\geq q \\geq r > 0$ (so that $p \\geq \\frac{1}{3} \\geq \\frac{1}{m+2}$). Let $A'$, $B'$, and $D'$ be the respective symmetric points of $A$, $B$, and $D$ about $O$. On the line $AD$, let $UV$ be the intersection of the segments $AD$ and $D'A'$, namely, $U$ is $A$ or $D'$, $V$ is $D$ or $A'$, whichever is closer to $O$. Since $\\triangle BUV \\subseteq \\triangle ABC$, except for $O$, there is no $m$-integral point in the interior of $\\triangle BUV$ or in the interior of $UV$, and neither in the symmetric $\\triangle B'UV$. Together, they imply that the parallelogram $BUB'V$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p203_data_d24ab25d71.png)\n\n(centered at $O$) does not have interior *m*-integral points other than $O$. By Minkowski's theorem,\n\n$$\n4m^2 \\geq S_{BUB'V} = 4S_{\\triangle OBV} = 4S_{\\triangle ABC} \\times \\min\\{p, 1-p\\} \\times \\frac{q}{q+r};\n$$\n\n$$\nS_{\\triangle ABC} \\leq \\frac{4m^2}{4 \\min\\{p, 1-p\\}} \\times \\frac{q+r}{q} \\leq \\frac{2m^2}{\\min\\{p, 1-p\\}}.\n$$\n\nIf $p > \\frac{m}{m+1}$, then\n\n$$\n\\frac{OD}{OA'} = \\frac{OD}{AO} = \\frac{p}{1-p} > m,\n$$\n\ntaking $OL = m \\cdot OA'$ on segment $OD$, $L$ must be an $m$-integral point inside $\\triangle ABC$, which is contradictory.\n\nTherefore, $p \\leq \\frac{m}{m+1}$, $\\min\\{p, 1-p\\} \\geq \\frac{1}{m+2}$, and we deduce\n\n$$\nS_{\\triangle ABC} \\leq 2m^2(m+2). \\quad \\square\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19957,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $k > 1$, let $f(k)$ be the number of ways to factor $k$ into a product of positive integers greater than $1$ (the order of factors is not counted; for example, $f(12) = 4$, since $12$ can be factored in these $4$ ways: $12$, $2 \\times 6$, $3 \\times 4$, $2 \\times 2 \\times 3$).\n\nProve that if $n$ is a positive integer greater than $1$ and $p$ is a prime factor of $n$, then\n$$\nf(n) \\le \\frac{n}{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof*\n\nLet $P(n)$ denote the largest prime divisor of $n$, and define $P(1) = f(1) = 1$. We first prove two lemmas.\n\n**Lemma 1:** For a positive integer $n$ and prime $p \\mid n$, we have $f(n) \\le \\sum_{d \\mid \\frac{n}{p}} f(d)$.\n\n*Proof of Lemma 1:* For convenience, call a valid factoring simply a \"factoring.\" For any factoring of $n$, write $n = n_1 n_2 \\cdots n_k$. Since $p \\mid n$, there exists $i \\in \\{1, \\ldots, k\\}$ such that $p \\mid n_i$ (if there are multiple such $i$, choose any one; WLOG, let $i = 1$). Map this factoring to a factoring of $d = \\frac{n}{n_1}$, i.e., $d = n_2 n_3 \\cdots n_k$.\n\nFor two different factorizations of $n$, $n = n_1 n_2 \\cdots n_k$ and $n = n'_1 n'_2 \\cdots n'_k$ (where $p$ divides $n_1$ and $n'_1$):\n- If $n_1 = n'_1$, then $d = n_2 \\cdots n_k$ and $d = n'_2 \\cdots n'_k$ are two different factorizations of $d$ ($d$ is a divisor of $\\frac{n}{p}$).\n- If $n_1 \\ne n'_1$, then $d = \\frac{n}{n_1} \\ne \\frac{n}{n'_1} = d'$, so these two factorizations map to factorizations of $d$ and $d'$ respectively ($d$ and $d'$ are divisors of $\\frac{n}{p}$).\n\nThus, $f(n) \\le \\sum_{d \\mid \\frac{n}{p}} f(d)$. Lemma 1 is proved.\n\n**Lemma 2:** For a positive integer $n$, let $g(n) = \\sum_{d \\mid n} \\frac{d}{P(d)}$. Then $g(n) \\le n$.\n\n*Proof of Lemma 2:* Induct on the number of distinct prime divisors of $n$.\n- When $n = 1$, $g(1) = 1$.\n- When $n = p^a$ is a prime power:\n  $$\ng(n) = 1 + 1 + p + \\cdots + p^{a-1} = 1 + \\frac{p^a - 1}{p-1} \\le 1 + p^a - 1 = n.$$\n\nAssume for $n$ with $k$ distinct prime divisors, $g(n) \\le n$. Consider $n$ with $k+1$ distinct prime divisors. Let $n = p_1^{a_1} \\cdots p_k^{a_k} p_{k+1}^{a_{k+1}}$, where $p_1 < \\cdots < p_k < p_{k+1}$, and write $n = m p_{k+1}^{a_{k+1}}$. Then:\n  $$\ng(n) = g(m) + \\sum_{d \\mid m} \\sum_{i=1}^{a_{k+1}} \\frac{d p_{k+1}^i}{p_{k+1}} = g(m) + \\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1},$$\nwhere $\\sigma(m)$ is the sum of positive divisors of $m$. By induction, $g(m) \\le m$.\n\n[The proof continues, but the remainder is missing.]\n",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19958,
    "subject": "Mathematics (Olympiad)",
    "question": "Grandfather has a finite number of empty dustbins in his attic. Each dustbin is a rectangular parallelepiped with integer side lengths. A dustbin can be thrown away into another if and only if the side lengths of these dustbins can be put into one-to-one correspondence so that the side lengths of the first dustbin are less than the corresponding side lengths of the other dustbin. No dustbin can contain two other dustbins unless the latter have been placed one into another. Grandfather wants to throw away as many dustbins as possible to save space. He developed the following algorithm: find the longest chain of dustbins that can be thrown away into each other, then repeat the same with the remaining dustbins, and so on, until no more dustbins can be thrown away. At each step, the longest chain of dustbins to be chosen turned out to be unique. Is it necessarily true that, as a result of this process, the maximal possible number of dustbins have been thrown away?",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: No.\n\nSuppose grandfather has 6 dustbins with sizes $20 \\times 20 \\times 20$, $19 \\times 19 \\times 19$, $16 \\times 16 \\times 16$, $21 \\times 18 \\times 15$, $18 \\times 15 \\times 12$, and $17 \\times 14 \\times 11$. The first dustbin can contain the second one, the second can contain the third or the fifth, and the fifth can contain the sixth. The fourth can also contain the fifth. It is impossible to throw the first and the fourth into each other, the second and the fourth into each other, the third and the fourth into each other, the third and the fifth into each other, or the third and the sixth into each other. The longest chain of dustbins that can be thrown into each other is 4 dustbins: the sixth can be thrown into the fifth, which can be thrown into the second, which can be thrown into the first. As the remaining two dustbins cannot be thrown into each other, 3 dustbins in total are not thrown away. However, by throwing the third dustbin into the second, the second into the first, the sixth into the fifth, and the fifth into the fourth, only 2 dustbins are not thrown away. Hence, grandfather's algorithm does not provide an optimal solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Algorithms"
  },
  {
    "id": 19959,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a$ and $k$, the sequence $(a_n)_{n=1}^{\text{∞}}$ is defined by\n\n$a_1 = a$ and $a_{n+1} = a_n + k \\cdot \\varphi(a_n)$ for $n = 1, 2, \\dots,$\n\nwhere $\\varphi(m)$ stands for the product of digits of $m$ in its decimal representation (e.g., $\\varphi(413) = 12$, $\\varphi(308) = 0$).\n\nProve that there exist positive integers $a$ and $k$ such that the sequence $(a_n)_{n=1}^{\\infty}$ contains exactly 2009 different numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The sequence $(a_n)$ increases until a term contains a zero digit, after which it becomes constant. We seek $a$ and $k$ so that the first zero digit appears at $a_{2009}$.\n\n**General construction:** For any $m > 4$, set\n\n$$\na = \\frac{10^{2m-5} - 1}{9} = \\underbrace{11\\dots1}_{2m-5 \\text{ ones}}, \\quad k = 10^{m-3} + 4 = 1\\underbrace{00\\dots0}_{m-4 \\text{ zeros}}4.\n$$\n\nThen:\n\n- $a_1 = a = \\underbrace{11\\dots1}_{2m-5}$\n- $\\varphi(a_1) = 1$\n- $a_2 = a_1 + k$\n- $\\varphi(a_2) = 10$\n- $a_3 = a_2 + 10k$\n- $\\varphi(a_3) = 100$\n- $a_i = a_{i-1} + 10^{i-2}k$\n- $\\varphi(a_i) = 10^{i-1}$\n- $a_{m-2} = a_{m-3} + 10^{m-4}k$\n- $\\varphi(a_{m-2}) = 10^{m-3}$\n- $a_{m-1} = a_{m-2} + 10^{m-3}k$\n- $\\varphi(a_{m-1}) = 6 \\cdot 10^{m-3}$\n- $a_m = a_{m-1} + 6 \\cdot 10^{m-3}k$\n- $\\varphi(a_m) = 0$\n\nThus, the sequence contains exactly $m$ different numbers. For $m = 2009$, take $a = \\frac{1}{9}(10^{4013} - 1)$ and $k = 10^{2006} + 4$.\n\n**Second construction:**\n\nLet\n\n$$\na = \\underbrace{611\\dots1}_{2007}, \\quad k = \\underbrace{33\\dots34}_{2007} = \\frac{1}{6} \\cdot \\underbrace{200\\dots04}_{2007}\n$$\n\nThen, for $t = 1, \\dots, 2009$:\n\n$$\n\\begin{align*}\na_1 &= \\underbrace{6 \\ 11 \\dots 1}_{2007}, & \\varphi(a_1) &= 6, & k\\varphi(a_1) &= \\underbrace{2 \\ 00 \\dots 0 1}_{2007} \\\\\na_2 &= \\underbrace{26 \\ 11 \\dots 1 5}_{2006}, & \\varphi(a_2) &= 60, & k\\varphi(a_2) &= \\underbrace{2 \\ 00 \\dots 0 40}_{2007} \\\\\na_3 &= \\underbrace{226 \\ 11 \\dots 1 55}_{2005}, & \\varphi(a_3) &= 600, & k\\varphi(a_3) &= \\underbrace{2 \\ 00 \\dots 0 400}_{2007} \\\\\na_4 &= \\underbrace{2226 \\ 11 \\dots 1 555}_{2004}, & \\varphi(a_4) &= 6000, & k\\varphi(a_4) &= \\underbrace{2 \\ 00 \\dots 0 4000}_{2007} \\\\\n\\vdots \\\\\na_{t+1} &= \\underbrace{22 \\dots 2 6 \\ 11 \\dots 1 55 \\dots 5}_{2007-t}, & \\varphi(a_{t+1}) &= \\underbrace{6 \\ 0 \\dots 0}_{t}, & k\\varphi(a_{t+1}) &= \\underbrace{2 \\ 00 \\dots 0 4 \\ 00 \\dots 0}_{2007} \\\\\na_{2009} &= \\underbrace{22 \\dots 2 30 \\ 55 \\dots 5}_{2007}, & \\varphi(a_{2009}) &= 0, & k\\varphi(a_{2009}) &= 0\n\\end{align*}\n$$\n\nAfter $a_{2009}$, the sequence becomes constant.\n\n**Remark:** It suffices to find $a$ and $k$ so the sequence contains at least 2009 different numbers and then becomes constant. If the sequence has $m > 2009$ distinct terms, starting from $a_{m-2008}$ yields a sequence with exactly 2009 distinct numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19960,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and let $D$ be the intersection of the lines $AH$ and $BC$. Let $E$ be the point of intersection of the circumcircle of triangle $ABD$ and the line $CH$, lying outside triangle $ABC$. Let $F$ be the point of intersection of the circumcircle of triangle $ACD$ and the line $BH$, lying outside triangle $ABC$. Show that the two line segments $AE$ and $AF$ have the same length.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ and $L$ be the feet of the perpendiculars from $B$ to $CA$ and from $C$ to $AB$, respectively. Since $AB$ is a diameter of the circumcircle of triangle $ABD$, $\\angle AEB = 90^\\circ$. The triangles $AEB$ and $ALE$ are similar, as they have equal angles. Therefore, $AE : AL = AB : AE$, so $AE^2 = AB \\cdot AL$. Similarly, $AF^2 = AC \\cdot AK$ from the similarity of triangles $AFC$ and $AKF$. Since $\\angle BKC = \\angle BLC = 90^\\circ$, the points $B$, $C$, $K$, and $L$ lie on the same circle. By the power of a point theorem, $AB \\cdot AL = AC \\cdot AK$, so $AE^2 = AF^2$, which implies $AE = AF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19961,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of twice differentiable functions $f, g : \\mathbb{R} \\to \\mathbb{R}$, such that $f''$ and $g''$ are continuous, and\n\n$$\n(f(x) - g(y)) \\cdot (f'(x) - g'(y)) \\cdot (f''(x) - g''(y)) = 0,\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "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\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 $\\mathbb{R}$.\n\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 $x \\in \\mathbb{R}$.\n\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 solutions, 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": 19962,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given value $t$, consider number sequences $a_1, a_2, a_3, \\dots$ such that $a_{n+1} = \\frac{a_n + t}{a_n + 1}$ for all $n \\ge 1$.\n\n(a) Suppose that $t = 2$. Determine all starting values $a_1 > 0$ such that $\\frac{4}{3} \\le a_n \\le \\frac{3}{2}$ holds for all $n \\ge 2$.\n\n(b) Suppose that $t = -3$. Investigate whether $a_{2020} = a_1$ for all starting values $a_1$ different from $-1$ and $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, we determine for what starting values $a_1 > 0$ the inequalities $\\frac{4}{3} \\le a_2 \\le \\frac{3}{2}$ hold. Then, we will prove that for those starting values, the inequalities $\\frac{4}{3} \\le a_n \\le \\frac{3}{2}$ are also valid for all $n \\ge 2$.\n\nFirst, we observe that $a_2 = \\frac{a_1+2}{a_1+1}$ and that the denominator, $a_1 + 1$, is positive (since $a_1 > 0$). The inequality\n\n$$\n\\frac{4}{3} \\le a_2 = \\frac{a_1 + 2}{a_1 + 1} \\le \\frac{3}{2},\n$$\n\nis therefore equivalent to the inequality\n\n$$\n\\frac{4}{3}(a_1 + 1) \\le a_1 + 2 \\le \\frac{3}{2}(a_1 + 1),\n$$\n\nas we can multiply all parts in the inequality by the positive number $a_1 + 1$. Subtracting $a_1 + 2$ from all parts of the inequality, we see that this is equivalent to\n\n$$\n\\frac{1}{3}a_1 - \\frac{2}{3} \\le 0 \\le \\frac{1}{2}a_1 - \\frac{1}{2}.\n$$\n\nWe therefore need to have $\\frac{1}{3}a_1 \\le \\frac{2}{3}$ (i.e. $a_1 \\le 2$), and $\\frac{1}{2} \\le \\frac{1}{2}a_1$ (i.e. $1 \\le a_1 \\le 2$). The starting value $a_1$ must therefore satisfy $1 \\le a_1 \\le 2$.\n\nNow suppose that $1 \\le a_1 \\le 2$, so that $a_2$ satisfies $\\frac{4}{3} \\le a_2 \\le \\frac{3}{2}$. Looking at $a_3$, we see that $a_3 = \\frac{a_2+2}{a_2+1}$. That is the same expression as for $a_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19963,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(m)$ denote the sum of the digits of $m$. A positive integer $n$ is called *rioplatense* if there exists a positive integer $m$ such that $n = m + 2S(m)$. Prove that a positive integer $n$ is rioplatense if and only if $n$ is a multiple of $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "For every positive integer $m$, $m$ and $S(m)$ have the same remainder $r$ modulo $3$. Thus, $m + 2S(m)$ has the same remainder as $3r$ modulo $3$, so it is a multiple of $3$. Therefore, if a number $n$ is rioplatense, then it is a multiple of $3$.\n\nNow, we show that every multiple of $3$ is rioplatense. We use induction on $k$ to prove that every integer $n$ divisible by $3$ with at most $k$ digits can be written as $n = m + 2S(m)$ for some integer $m$ with at most $k$ digits.\n\nFor $k=1$, the result is immediate: $3 = 1 + 2S(1)$, $6 = 2 + 2S(2)$, and $9 = 3 + 2S(3)$.\n\nAssume the result holds for $k \\geq 1$ and consider an integer $n$ (a multiple of $3$) with $k+1$ digits. Let $N = 10^k + 2$. When dividing $n$ by $N$, we obtain a quotient $q$ and a remainder $r$, with $1 \\leq q \\leq 9$ and $r < 10^k + 2$. Both $n$ and $N$ are multiples of $3$, so $r$ is also a multiple of $3$ and $r \\leq 10^k - 1$, so $r$ has at most $k$ digits. By the induction hypothesis, there exists $t = \\overline{a_{k-1} \\cdots a_1 a_0}$ such that $r = t + 2S(t)$.\n\nTake $m = \\overline{a_k a_{k-1} \\cdots a_1 a_0}$, where $a_k = q$. Then $m$ has $k+1$ digits and\n\n$$\n\\begin{align*}\nm + 2S(m) &= \\overline{a_k a_{k-1} \\cdots a_1 a_0} + 2(a_k + a_{k-1} + \\cdots + a_1 + a_0) \\\\\n&= 10^k a_k + \\overline{a_{k-1} \\cdots a_1 a_0} + 2a_k + 2(a_{k-1} + \\cdots + a_1 + a_0) \\\\\n&= (10^k + 2)a_k + t + 2S(t) = Nq + r = n,\n\\end{align*}\n$$\n\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19964,
    "subject": "Mathematics (Olympiad)",
    "question": "a\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p38_data_28b4132cae.png)\n\nFind the perimeter of the figure shown.\n\nb\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p38_data_aca62ec181.png)\n\nFind the perimeter of the figure shown.\n\nc\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_6ccf325b45.png)\n\nTwo squares overlap so that the area of the overlap is $1\\ \\text{cm}^2$. The perimeter of the final shape is $32\\ \\text{cm}$. What are the possible side lengths of the two squares?\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_cff1d4b7e8.png)\n\nd\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_fad6e17c32.png)\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p40_data_cd709868df.png)\n\nTwo squares overlap so that the area of the overlap is $12\\ \\text{cm}^2$ and the perimeter of the final shape is $30\\ \\text{cm}$. What are the possible side lengths of the two squares?",
    "options": [],
    "answer": "See solution",
    "solution": "a\n\nThe perimeter is $2 \\times 7 + 2 \\times 6 + 2 \\times 6 + 2 \\times 5 = 14 + 12 + 12 + 10 = 48\\ \\text{cm}$, or $4 \\times 7 + 4 \\times 6 - 4 = 28 + 24 - 4 = 48\\ \\text{cm}$.\n\nb\n\nThe perimeter is $2 \\times 7 + 8 + 6 + 1 + 1 = 14 + 16 = 30\\ \\text{cm}$, or $4 \\times 7 + 4 \\times 6 - 2 \\times 6 - 2 \\times 5 = 28 + 24 - 12 - 10 = 30\\ \\text{cm}$.\n\nc\n\nSince the overlap has area $1\\ \\text{cm}^2$, it must be a grid square in the corner of each overlapping square.\n\nSince the perimeter of the final shape is $32\\ \\text{cm}$ and the perimeter of the overlapping square is $4\\ \\text{cm}$, the sum of the perimeters of the original two squares is $36\\ \\text{cm}$.\n\n**Alternative i**\n\nSince the sides of a square are at least $2\\ \\text{cm}$, its perimeter is at least $8\\ \\text{cm}$. So the perimeter of the other square is at most $28\\ \\text{cm}$. Since the perimeter of a square is a multiple of $4$, one square has perimeter $28, 24, 20$ and the other has perimeter $8, 12, 16$ respectively. So the squares are $7 \\times 7$ and $2 \\times 2$, or $6 \\times 6$ and $3 \\times 3$, or $5 \\times 5$ and $4 \\times 4$.\n\n**Alternative ii**\n\nSince the perimeter of a square is $4$ times its side length, the sum of the perimeters of the two overlapping squares is $4$ times the sum of the lengths of one side from each square. So the sum of the lengths of one side from each square is $36/4 = 9\\ \\text{cm}$. Hence the only possible side lengths for the two overlapping squares are $2$ and $7$, $3$ and $6$, and $4$ and $5\\ \\text{cm}$.\n\n**Alternative iii**\n\nLet the squares be $a \\times a$ and $b \\times b$ with $a \\leq b$.\n\nAs shown in the next diagram, a shape formed from the two overlapping squares has the same perimeter as the smallest square that encloses the shape.\n\nHence $4(a + b - 1) = 32$. So $a + b - 1 = 8$ and $a + b = 9$. Thus $a = 2$ and $b = 7$, or $a = 3$ and $b = 6$, or $a = 4$ and $b = 5$.\n\nd\n\nThe overlap is a rectangle which is wholly inside each of the overlapping squares and along at least one side of each. Disregarding rotations and reflections, there are four cases as indicated.\n\nNote that the perimeter of the final shape is the sum of the perimeters of the two overlapping squares minus the perimeter of the overlap.\n\nSince the area of the overlap is $12\\ \\text{cm}^2$, the overlap in each of the diagrams above is one of the rectangles $12 \\times 1$, $6 \\times 2$, or $4 \\times 3$.\n\n**Alternative i**\n\nIf the overlap is a $12 \\times 1$ rectangle, then the side length of each overlapping square is at least $12\\ \\text{cm}$. Then the perimeter of the final shape is at least $8 \\times 12 - 2 \\times (12+1) = 96 - 26 = 70\\ \\text{cm}$. Since $70 > 30$, the overlap rectangle is not $12 \\times 1$.\n\nIf the overlap is a $6 \\times 2$ rectangle, then the side length of each overlapping square is at least $6\\ \\text{cm}$. Then the perimeter of the final shape is at least $8 \\times 6 - 2 \\times (6+2) = 48 - 16 = 32\\ \\text{cm}$. Since $32 > 30$, the overlap rectangle is not $6 \\times 2$.\n\nIf the overlap is a $4 \\times 3$ rectangle, the sum of the perimeters of the overlapping squares is $30 + 2 \\times (4+3) = 30 + 14 = 44\\ \\text{cm}$. Hence the sum of the lengths of one side from each square is $44/4 = 11\\ \\text{cm}$. Also the side length of each overlapping square is at least $4\\ \\text{cm}$. So the only possibilities are a $4 \\times 4$ square overlapping a $7 \\times 7$ square, and a $5 \\times 5$ square overlapping a $6 \\times 6$ square.\n\n**Alternative ii**\n\nLet the squares be $a \\times a$ and $b \\times b$ with $a \\leq b$.\n\nIf the overlap is a $12 \\times 1$ rectangle, the perimeter of the final shape is $4a + 4b - 26 = 30\\ \\text{cm}$. So $4a + 4b = 56$ and $a + b = 14$. This is impossible since each of $a$ and $b$ must be at least $12\\ \\text{cm}$.\n\nIf the overlap is a $6 \\times 2$ rectangle, the perimeter of the final shape is $4a + 4b - 16 = 30\\ \\text{cm}$. So $4a + 4b = 46$, which is impossible because $4$ does not divide $46$.\n\nIf the overlap is a $4 \\times 3$ rectangle, the perimeter of the final shape is $4a + 4b - 14 = 30\\ \\text{cm}$. So $4a + 4b = 44$ and $a + b = 11$. Since $a$ must be at least $4$, we have $a = 4$ and $b = 7$ or $a = 5$ and $b = 6$.\n\nSo the two squares must be $4 \\times 4$ and $7 \\times 7$, or $5 \\times 5$ and $6 \\times 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19965,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(a, b)$ such that\n$$\n\\frac{a^2}{2ab^2 - b^3 + 1}\n$$\nis a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "If $b$ is even, the expression is an integer, yielding solutions:\n$$\n\\left(\\frac{b}{2}, b\\right) \\quad \\text{and} \\quad \\left(\\frac{b^4 - b}{2}, b\\right)\n$$\nfor any even $b$; that is, $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ for all positive integers $t$.\n\nIf $b = 1$, then\n$$\n\\frac{a^2}{2a} = \\frac{a}{2}\n$$\nis an integer if and only if $a$ is even, so $(a, b) = (2t, 1)$ for all positive integers $t$.\n\nFor $b > 1$, viewing the denominator as a quadratic in $a$, the integer solutions occur when $b$ is even, specifically $b = 2t$, and $a = t$ or $a = 8t^4 - t$ for positive integers $t$.\n\nThus, all solutions are $(a, b) = (2t, 1)$, $(a, b) = (t, 2t)$, and $(a, b) = (8t^4 - t, 2t)$ for all positive integers $t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19966,
    "subject": "Mathematics (Olympiad)",
    "question": "On a table, there are 2013 cards, each with a distinct number from $1$ to $2013$ written on it. The cards are face down, so the numbers are hidden. You may select any subset of cards and ask whether the arithmetic mean of the numbers on those cards is an integer, and you will receive a truthful answer.\n\n**a)** Find all numbers that can be determined with certainty by asking such questions.\n\n**b)** We wish to divide the cards into groups such that the contents of each group as a whole are known, even if the individual card values are not. (For example, to identify a group of three cards containing $1$, $2$, and $3$, without knowing which card has which number.) What is the maximum number of such groups that can be obtained?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $2013$ be replaced by a general odd number $2k-1$, with $k \\geq 2$. The sum $S = 1 + 2 + \\dots + (2k-1)$ equals $k(2k-1)$.\n\n**Part a)**\n\nThe only number that can be determined with certainty is $k$, the middle value. To find $k$, ignore a card with unknown number $x$, forming a set of $2k-2$ cards, and ask about their average. The average is integer if and only if $2k-2$ divides $S-x = k(2k-1) - x = k(2k-2) + (k-x)$, i.e., if and only if $2k-2$ divides $k-x$. Since $1 \\leq x \\leq 2k-1$, $|k-x| \\leq k-1 < 2k-2$ (for $k \\geq 2$), so $x = k$ is the only possibility. Thus, applying this procedure to each card will reveal $k$.\n\nConversely, suppose a card with number $m$ can be determined with certainty by asking a sequence of questions. If each card's number $j$ is replaced by $2k-j$, the set $\\{1, 2, \\dots, 2k-1\\}$ is unchanged. The same sequence of questions yields the same answers, so the procedure that finds $m$ will now find $2k-m$. Thus, $2k-m = m$, so $m = k$. This completes part a).\n\n**Part b)**\n\nLet the cards be divided into groups so that the contents of each group as a whole are known. Since $k$ is the unique number that can be found with certainty, all groups must contain at least $2$ cards, except possibly one group containing only the card with $k$. Therefore, the maximum number of groups is $\\frac{1}{2}((2k-1)-1) + 1 = k$.\n\nWe can achieve $k$ groups. Excluding the card with $k$, the remaining cards can be divided into $k-1$ pairs, each containing numbers of the form $j$ and $2k-j$, for $j = 1, \\dots, k-1$. These are called complementary pairs.\n\nOnce $k$ is found, the parity of each card's number can be determined. For example, if $k$ is odd, choose any card $x$ and ask about the two cards $k$ and $x$. If their average is integer, $x$ is odd; otherwise, $x$ is even. The case for even $k$ is analogous.\n\nTo find the complementary pairs $C_j = \\{j, 2k-j\\}$, start with $C_1 = \\{1, 2k-1\\}$ and $C_2 = \\{2, 2k-2\\}$. Take a pair of cards with different parity and unknown sum $y$ (which is odd). Ask about the average of the remaining $2k-3$ cards. The average is integer if and only if $2k-3$ divides $S-y = k(2k-1) - y = k(2k-3) + (2k-y)$, i.e., if and only if $2k-3$ divides $2k-y$. Since $3 \\leq y \\leq 4k-3$, $|2k-y| \\leq 2k-3$. The answer is yes if and only if $2k-y \\in \\{0, \\pm(2k-3)\\}$. Because $2k-y$ is odd, $2k-y = \\pm(2k-3)$, so $y = 3$ or $y = 4k-3$. These are the extremal values, achieved only if the pair is $1,2$ or $2k-2,2k-1$.\n\nRepeating this procedure for all pairs with different parity, we find the two pairs $1,2$ and $2k-2,2k-1$ (without knowing which is which). Thus, a group containing the four cards $1,2,2k-2,2k-1$ is determined. The two odd cards form $C_1 = \\{1,2k-1\\}$, and the two even cards form $C_2 = \\{2,2k-2\\}$.\n\nSuppose complementary pairs $C_1, \\dots, C_{2j}$ are determined for some $j$ with $2j \\leq k-1$. To find $C_{2j+1}$ and $C_{2j+2}$, exclude the $4j$ numbers in $C_1, \\dots, C_{2j}$. There remain $2k-4j-1$ numbers with sum $S-4jk$. Take a pair of cards with different parity and unknown odd sum $y$; $4j+3 \\leq y \\leq 4k-4j-3$. Ask about the average of the remaining $2k-4j-3$ cards. The average is integer if and only if $2k-4j-3$ divides $(S-4jk)-y = k(2k-1)-4jk-y = k(2k-4j-3)+(2k-y)$, i.e., if and only if $2k-4j-3$ divides $2k-y$. Since $|2k-y| \\leq 2k-4j-3$, the answer is yes if and only if $2k-y \\in \\{0, \\pm(2k-4j-3)\\}$. Because $2k-y$ is odd, $2k-y = \\pm(2k-4j-3)$, so $y = 4j+3$ or $y = 4k-4j-3$. These are achieved only if the pair is $2j+1,2j+2$ or $2k-2j-2,2k-2j-1$.\n\nThus, the procedure yields a group containing the four cards $2j+1,2j+2,2k-2j-2,2k-2j-1$. The two odd cards form $C_{2j+1} = \\{2j+1,2k-2j-1\\}$, and the two even cards form $C_{2j+2} = \\{2j+2,2k-2j-2\\}$.\n\nBy induction, all cards except $k$ can be divided into complementary pairs. This completes the solution.\n\n*Comment*: If the content of any one card other than $k$ is known, then all numbers can be found with certainty.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 19967,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real numbers $x, y$, the following inequality holds:\n\n$$\nf(x - f(y)) \\le x - y f(x)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* It doesn't exist.\n\n*Solution.* Assume such a function exists. Substitute $y = 0$ into the inequality:\n\n$$\nf(x - f(0)) \\le x\n$$\n\nNow, let $x = x + f(0)$:\n\n$$\nf(x) \\le x + f(0)\n$$\n\nNext, substitute $x = f(y)$ into the original inequality:\n\n$$\nf(f(y) - f(y)) \\le f(y) - y f(f(y))\n$$\n\nSo,\n\n$$\nf(0) \\le f(y) - y f(f(y))\n$$\n\nRearrange:\n\n$$\ny f(f(y)) \\le f(y) - f(0)\n$$\n\nFor $y < 0$, $y f(f(y))$ is negative, so $f(f(y))$ must be positive and bounded above by $f(y) + f(0)$. But as $y$ becomes very negative, this is impossible. Thus, such a function cannot exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19968,
    "subject": "Mathematics (Olympiad)",
    "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.\n\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": "See solution",
    "solution": "If all 27 numbers are equal to $1$, there is nothing to prove. Assume the cube contains some numbers greater than $1$ and suppose there is an even number among 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. So all 27 numbers are odd.\n\nWe 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 $a$. Thus, each strip contains at least one $1$, so at least $9$ numbers inside the $3 \\times 3 \\times 3$ cube are equal to $1$.\n\nLet $a_1 \\leq a_2 \\leq \\dots \\leq a_{18}$ be the other numbers inside the cube. If $a_1 > 3$, then one of the three numbers whose sum is $a_1$ must be greater than $1$ and less than $a_1$, which contradicts the choice of $a_1$.\n\nTherefore, $a_1 \\in \\{1,3\\}$, so $a_2 \\in \\{1,3,5\\}$. If $a_2 \\leq 3$, then $a_3 \\leq 1 + a_1 + a_2 \\leq 7$. 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, $a_3 \\leq 1 + 1 + a_2 = 7$, so $a_3 \\in \\{1,3,5,7\\}$.\n\nIf $a_3 = 7$, there must exist a strip containing $a_2$ and $a_3$, so $a_2$ and $a_3$ cannot be simultaneously used to express $a_4$ as a sum of three numbers inside the cube, so $a_4 \\leq a_3 + a_1 + 1 \\leq 7 + 3 + 1 = 11$. If $a_3 \\leq 5$, then $a_4 \\leq a_3 + a_2 + a_1 \\leq 5 + 5 + 1 = 11$. Therefore, $a_4 \\leq 11$.\n\nMore generally, we have $a_{k+3} \\leq a_{k+2} + a_{k+1} + a_k$.\n\nIf $a_{k+1}$ and $a_{k+2}$ are on the 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 $a_{k+2} \\leq a_{k+1} + a_k + a_{k-1}$, it follows that $a_{k+3} \\leq a_{k+2} + a_k + a_{k-1} \\leq a_{k+1} + 2a_k + 2a_{k-1}$.\n\nIf no strip contains $a_{k+1}$ and $a_{k+2}$, since $a_{k+2} \\leq a_k + a_{k-1} + a_{k-2}$, we obtain $a_{k+3} \\leq a_{k+2} + a_{k+1} + a_k \\leq a_{k+1} + 2a_k + a_{k-1} + a_{k-2} \\leq a_{k+1} + 2a_k + 2a_{k-1}$.\n\nThus, $a_{k+3} \\leq a_{k+1} + 2a_k + 2a_{k-1}$ for each $k \\geq 2$. Successively, we infer that $a_5 \\leq 23$, $a_6 \\leq 35$, and $a_7 \\leq 59$, so at least $16$ numbers written inside the unit cubes are smaller than $60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19969,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n \\ge 5$ for which we can place a real number at each vertex of a regular $n$-sided polygon, such that the following two conditions are satisfied.\n\n- None of the $n$ numbers is equal to $1$.\n- For each vertex of the polygon, the sum of the numbers at the nearest four vertices is equal to $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is any even $n \\geq 6$.\n\nIf $n$ is even, the conditions are satisfied by alternating $0.5, 1.5, 0.5, 1.5, \\ldots$ around the polygon.\n\nSuppose $n = 2m+1$ is odd and let the numbers be $x_1, x_2, \\ldots, x_n$ in order around the polygon (subscripts modulo $n$). For each $i$, we have:\n\n$$\n\\begin{aligned}\nx_i + x_{i+1} + x_{i+3} + x_{i+4} &= x_{i+1} + x_{i+2} + x_{i+4} + x_{i+5} \\\\\n\\Rightarrow \\quad x_i + x_{i+3} &= x_{i+2} + x_{i+5}. \\tag{*}\n\\end{aligned}\n$$\n\nLet $A_i = x_i + x_{i+3}$. Then $(*)$ becomes $A_i = A_{i+2}$, so the sequence $A_1, A_2, \\ldots$ has period $2$. But it also has period $n$, so its period is $\\gcd(2, n) = 1$ (since $n$ is odd). Thus, $A_i = A_{i+3}$, which implies $x_i = x_{i+6}$.\n\nTherefore, the sequence $x_1, x_2, x_3, \\ldots$ has period $6$. But it also has period $n$, so its period is $\\gcd(6, n) = 1$ or $3$. In either case, $(*)$ simplifies to $x_i = x_{i+2}$, so the sequence has period $2$. But it also has period $n$, so its period is $\\gcd(2, n) = 1$. It follows that all $x_i$ are equal to $1$, contradicting the condition that none of the numbers is $1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19970,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcentre of triangle $ABC$. Let $K$ and $L$ be the intersection points of the circumcircles of triangles $BOC$ and $AOC$ with the bisectors of the angles at $A$ and $B$, respectively. Let $P$ be the midpoint of $\\overline{KL}$, $M$ be the point symmetric to $O$ with respect to $P$, and $N$ be the point symmetric to $O$ with respect to $KL$. Prove that $KLMN$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "The angles $LCA$ and $LOA$ are equal as inscribed angles over the same arc. The angle $LOA$ equals the sum of the angles $OAB$ and $OBA$ (since $LOA$ is an exterior angle of triangle $ABO$). Hence:\n\n$$\n\\begin{aligned}\n\\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA \\\\\n&= \\angle OAB + \\angle OBA + \\angle OCA \\\\\n&= \\frac{1}{2}(\\angle CAB + \\angle ABC + \\angle ACB) = 90^\\circ\n\\end{aligned}\n$$\n\nSimilarly, $\\angle KCO = 90^\\circ$, so $C$ lies on $KL$ and $C$ is the midpoint of $\\overline{ON}$. $PC$ is parallel to $MN$ ($PC$ is a mid-segment of triangle $MNO$). The quadrilateral $LOKM$ is a parallelogram because its diagonals bisect each other. Hence, the angles $MLK$ and $OKL$ are equal. Triangle $OKN$ is isosceles ($KC$ is both an altitude and a median), so the angles $OKC$ and $NKC$ are equal. Finally, we conclude that $\\angle MLK = \\angle NKL$, so $KLMN$ is an isosceles trapezoid.\n\n![](images/Makedonija_2009_p66_data_7fc79349ad.png)\n\n_(If $P$ is on the other side of $C$, then we are working with the angles $MKL$ and $NLK$.)_",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19971,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute all pairs of positive integers $(k, n)$ such that\n$$\n1! + 2! + \\cdots + k! = 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first compute the entries of the following matrix:\n\n| $k$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|-----|---|---|---|---|---|----|-----|------|-------|--------|\n| $k!$ | 1 | 2 | 6 | 24 | 120 | 720 | 5040 | 40320 | 362880 | 3628800 |\n| $1!+2!+ \\ldots + k!$ | 1 | 3 | 9 | 33 | 153 | 873 | 5913 | 46233 | 409113 | 4037913 |\n\nObviously, the pairs $(k, n) = (1, 1)$ and $(k, n) = (2, 2)$ are solutions. We will show that the unique solution with $k > 2$ is $(k, n) = (5, 17)$.\n\nWe observe that since $k!$ is divisible by 100 for $k \\geq 10$, the sum $1!+2!+\\cdots+k!$ leaves a remainder 13 when divided by 100 for $k \\geq 9$. If equality holds\n$$\n1! + 2! + \\cdots + k! = \\frac{n(n+1)}{2}\n$$\nfor some $k \\geq 10$, then there is some natural number $m$ such that\n$$\nn(n + 1) = 100m + 26\n$$\nor\n$$\nn^2 + n - 100m - 26 = 0.\n$$\nThe discriminant is $\\Delta = 400m + 105$ and it should be a perfect square. On the other hand, we cannot have\n$$\nx^2 \\equiv 5 \\pmod{100},\n$$\nas otherwise 25 would divide 5. Therefore, the given relation cannot hold for $k \\geq 9$, as well as for $k = 7$. Since\n$$\n9 = 3^2, \\quad 153 = 3^2 \\cdot 17, \\quad 873 = 3^2 \\cdot 97, \\quad \\text{and} \\quad 46233 = 3^2 \\cdot 11 \\cdot 467,\n$$\nwe observe that $2 \\cdot 153 = 18 \\cdot 17$, a product of two consecutive integers $k > 2$. Therefore, the only solution is $(k, n) = (5, 17)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19972,
    "subject": "Mathematics (Olympiad)",
    "question": "For acute triangle $ABC$ with $AB > AC$, let $M$ be the midpoint of side $BC$ and $P$ a point inside $\\triangle AMC$ such that $\\angle MAB = \\angle PAC$. Let $O$, $O_1$, and $O_2$ be the circumcenters of $\\triangle ABC$, $\\triangle ABP$, and $\\triangle ACP$ respectively. Prove that line $AO$ bisects segment $O_1O_2$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p105_data_52ba3f6736.png)\n\nFig. 1",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in Fig. 1, draw the circumcircles of $\\triangle ABC$, $\\triangle ABP$, and $\\triangle ACP$. Let the extension of $AP$ meet $\\odot O$ at $D$, join $BD$, and draw the tangent to $\\odot O$ at $A$, intersecting $\\odot O_1$ and $\\odot O_2$ at $E$ and $F$ respectively.\n\nIt is clear that $\\triangle AMC \\sim \\triangle ABD$, hence\n\n$$\n\\frac{AB}{BD} = \\frac{AM}{MC}.\n$$\n\nSince $\\triangle EAB \\sim \\triangle PDB$, we have $\\frac{AB}{BD} = \\frac{AE}{PD}$.\n\nConsequently, $\\frac{AM}{MC} = \\frac{AE}{PD}$, i.e.\n\n$$\nAE = \\frac{AM \\times PD}{MC},\n$$\n\nand, similarly,\n\n$$\nAF = \\frac{AM \\times PD}{MB}.\n$$\n\nIt follows that\n\n$$\nAE = AF. \\qquad ①\n$$\n\nDraw the perpendiculars $O_1E' \\perp AE$ with foot $E'$, and $O_2F' \\perp AF$ with foot $F'$. Since $E'$, $F'$ are the midpoints of $AE$, $AF$ respectively, it follows from ① that $A$ is the midpoint of $E'F'$.\n\nIn the right-angled trapezoid $O_1E'F'O_2$, $AO$ is the extension of the median, and hence it bisects the segment $O_1O_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19973,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $A$ is a singular matrix of order $n$ with complex entries, all of which have absolute value equal to $1$.\n\n(a) Let $n = 3$. Show that two rows or two columns of the matrix $A$ are proportional.\n\n(b) Find, with proof, if the above claim holds for $n = 4$.",
    "options": [],
    "answer": "See solution",
    "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 $$(a-1)(d-1) = (b-1)(c-1).$$\n\nTake 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\n(b) 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": 19974,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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$$\n\\begin{aligned}\na_n - 1 &< a_n^2 - 1 \\le (1 - a_{n-1})(a_{n+1} - 1) \\le a_{n+1} - 1 < a_{n+1}^2 - 1 \\\\&\\le (a_n - 1)(1 - a_{n+2}) \\le a_n - 1,\n\\end{aligned}\n$$\nwhich is a contradiction.\n\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": 19975,
    "subject": "Mathematics (Olympiad)",
    "question": "For each value of the parameter $a$, find the number of solutions to the equation:\n$$\n\\sqrt{a x + 3} \\sqrt{x} = x^{2013}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $x \\ge 0$. For each $a \\in \\mathbb{R}$, $x = 0$ is a solution. Now consider $x > 0$ and rewrite the equation as:\n$$\na = x^{4025} - x^{-\\frac{2}{3}}\n$$\nOn the interval $(0, +\\infty)$, the function $f(x) = x^{4025} - x^{-\\frac{2}{3}}$ is continuous and strictly increasing. Also, $\\lim_{x \\to 0^+} f(x) = -\\infty$ and $\\lim_{x \\to +\\infty} f(x) = +\\infty$. Thus, for each $a \\in \\mathbb{R}$, the equation $f(x) = a$ has exactly one solution for $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19976,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix an integer $n \\geq 3$. 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 $\\text{area}(ABC) \\leq \\text{area}(ABX)$ for all $X$ in $S$ different from $A$ and $B$. (Note that for a segment $AB$ there could be several suitable triangles.)\n\nA triangle is **adequate** if its vertices are all in $S$ and it is suitable for at least two of its sides.\n\nProve that there are at least $\\frac{1}{2}(n-1)$ adequate triangles.\n\n![](images/RMC_2023_v2_p108_data_0a6146766a.png)",
    "options": [],
    "answer": "See 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\nGiven 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\nWe now present four approaches for the lower bound, of which the first three are intimately related.\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**2nd Approach.** For each adequate triangle, colour two of its suitable edges. By the observation above, the resulting multigraph is connected, so it has at least $n-1$ edges, 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, $3|\\Delta| \\geq |S| + |\\Delta| - 1 = n + |\\Delta| - 1$, 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\nConsider now a partition of $S = A \\sqcup B$, where $|A| = 1$. Choose an $(A, B)$-minimal triangle and add to $A$ its vertices from $B$ to obtain a new partition also written $S = A \\sqcup B$. Continuing, choose an $(A, B)$-minimal triangle and add to $A$ its vertices/vertex from $B$ and so on and so forth all the way down for at least another $\\frac{1}{2}(n-5)$ steps — this works at least as many times, since at each step, $B$ loses at most two points.\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.\n\n**Remark.** In fact, $\\lfloor n/2 \\rfloor$ is the smallest possible number of adequate triangles, as shown by the configurations described below.\n\nLet first $n = 2k - 1$. Consider a regular $n$-gon $\\mathcal{P} = A_1A_2\\dots A_n$. Choose a point $B_i$ on the perpendicular bisector of $A_iA_{i+1}$ outside $\\mathcal{P}$ and sufficiently close to the segment $A_iA_{i+1}$. We claim that there are exactly $k-1 = \\lfloor n/2 \\rfloor$ adequate triangles in the set\n\n$$\nS = \\{A_1, A_2, \\dots, A_k, B_1, B_2, \\dots, B_{k-1}\\}.\n$$\n\nNotice here that the arc $A_1A_2\\dots A_k$ is less than half of the circumcircle of $\\mathcal{P}$, so the angles $\\angle A_uA_vA_w$, $1 \\leq u < v < w \\leq k$, are all obtuse.\n\nTo prove the claim, list the suitable triangles for each segment.\n\nFor segments $A_iA_{i+1}$, $A_iB_i$, and $B_iA_{i+1}$, it is $A_iB_iA_{i+1}$.\n\nFor segment $A_iA_{j+1}$, $j \\geq i + 1$, those are $A_iB_iA_{j+1}$ and $A_iB_jA_{j+1}$.\n\nFor segment $A_iB_j$, $j \\geq i + 1$, it is $A_iB_iB_j$.\n\nFor segment $B_iA_{j+1}$, $j \\geq i + 1$, it is $B_iB_jA_{j+1}$.\n\nFor segment $B_iB_j$, $i < j$, those are $B_iA_{i+1}B_j$ and $B_iA_jB_j$.\n\nIt is easily seen that the only triangles occurring twice are $A_iB_iA_{i+1}$, hence they are the only adequate triangles.\n\nFor $n = 2k - 2$, just remove $A_k$ from the above example. This removes the adequate triangle $A_{k-1}B_{k-1}A_k$ and provides only one new such instead, namely, $B_{k-2}A_{k-1}B_{k-1}$. Consequently, there are exactly $k-1 = \\lfloor n/2 \\rfloor$ adequate triangles in the set\n\n$$\nS = \\{A_1, A_2, \\dots, A_{k-1}, B_1, B_2, \\dots, B_{k-1}\\}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19977,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $\\angle BAC = 60^\\circ$, incenter $I$, and circumcenter $O$. Let $O'$ be the point diametrically opposite to $O$ on the circumcircle of triangle $BOC$. Prove that\n\n$$\nIO' = BI + IC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Argentina_2019_Booklet_p20_data_b3e1f11431.png)\n\nBy considering the inscribed angle $\\angle BAC$ in the circumcircle of triangle $ABC$, we have that $\\angle BOC = 2 \\cdot \\angle BAC = 120^\\circ$; then, $\\angle BO'C = 60^\\circ$. Since $O$ is a point on the perpendicular bisector of $BC$, $O'$ is also on this line. Therefore, triangle $BO'C$ is equilateral.\n\nOn the other hand, $\\angle BIC = 90^\\circ + \\frac{1}{2}\\angle BAC = 120^\\circ$. It follows that $I$ is on the circumcircle of $BOC$. By applying Ptolemy's theorem to quadrilateral $BICO'$, we obtain:\n\n$$\nIO' \\cdot BC = BI \\cdot O'C + CI \\cdot O'B\n$$\n\nand, recalling that $BC = O'C = O'B$, we conclude that\n\n$$\nIO' = BI + CI.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19978,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有 $2022$ 個相異格子點。令 $I$ 為這些點中，有幾對點之間的距離恰為 $1$ 的點對數量。求 $I$ 的最大可能值。\n\n註：格子點為 $x$ 座標與 $y$ 座標皆為整數的點。",
    "options": [],
    "answer": "See solution",
    "solution": "最大可能的 $I$ 是 $3954$；此最大值會在 $45 \\times 45$ 的點陣扣除三個角落時達到。\n\n對於任何 $2022$ 個相異格子點，假設其共有 $n$ 個 $x$ 座標 $x_1, x_2, \\dots, x_n$，且 $x$ 座標為 $x_i$ 的點共有 $a_i$ 個。注意到 $x = x_i$ 直排上的點，最多只能造出 $a_i - 1$ 個長度為 $1$ 且垂直 $x$ 軸的線段，故垂直 $x$ 軸且長度為 $1$ 的線段總量 $\\le \\sum_{i=1}^n (a_i - 1) = 2022 - n$。\n\n同理，若這些點共有 $m$ 個不同的 $y$ 座標，則垂直 $y$ 軸且長度為 $1$ 的線段總量 $\\le 2022 - m$。又因為格點之間長度為 $1$ 的線段必然垂直於其中一個座標，故\n\n$$\nI \\le (2022 - n) + (2022 - m) = 4044 - (n + m)\n$$\n\n但又注意到 $2022$ 個相異點保證 $mn \\ge 2022$，故由算幾不等式有\n\n$$\nm + n \\ge 2\\sqrt{mn} > 89 \\Rightarrow m + n \\ge 90.\n$$\n\n故 $I \\le 4044 - 90 = 3954$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19979,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of real numbers $x$, $y$, and $z$ for which\n\n$$\nx(y^2 + 2z^2) = y(z^2 + 2x^2) = z(x^2 + 2y^2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If, for example, $x = 0$, we get $0 = y z^2 = 2 y^2 z$, which means that one of $y$ or $z$ vanishes and the other can be arbitrary. The cases $y = 0$ or $z = 0$ are similar. Thus, we obtain three groups of solutions $(x, y, z)$ formed by $(t, 0, 0)$, $(0, t, 0)$, and $(0, 0, t)$, where $t$ is any real number. For other solutions, $xyz \\neq 0$ must hold.\n\nFactorizing $x(y^2 + 2z^2) = y(z^2 + 2x^2)$ yields $(2x - y)(z^2 - xy) = 0$. We consider two cases:\n\n**(i) $2x - y = 0$:** Setting $y = 2x$, the system reduces to\n\n$$\n2x(2x^2 + z^2) = 9x^2 z,\n$$\nwhich simplifies (for $x \\neq 0$) to\n$$\n4x^2 + 2z^2 - 9x z = 0 \\quad \\text{or} \\quad (x - 2z)(4x - z) = 0.\n$$\nThus, case (i) yields two groups of solutions: $(2t, 4t, t)$ and $(t, 2t, 4t)$, where $t$ is any real number.\n\n**(ii) $z^2 - xy = 0$:** Substitute $z^2 = xy$ into the system to get\n$$\nxy(2x + y) = z(x^2 + 2y^2),\n$$\nwhich is equivalent to\n$$\nz = \\frac{xy(2x + y)}{x^2 + 2y^2}.\n$$\nWe require $z^2 = xy$, so\n$$\n\\frac{x^2 y^2 (2x + y)^2}{(x^2 + 2y^2)^2} = xy.\n$$\nDividing by $xy \\neq 0$ gives\n$$\nxy(2x + y)^2 = (x^2 + 2y^2)^2 \\quad \\text{or} \\quad (4y - x)(x^3 - y^3) = 0.\n$$\nThus, either $x = 4y$ or $x = y$. Returning to $z$, we get $z = 2y$ or $z = x$, so the solutions are $(4t, t, 2t)$ and $(t, t, t)$, where $t$ is any real number.\n\n**Answer:** All solutions are $(t, 0, 0)$, $(0, t, 0)$, $(0, 0, t)$, $(t, t, t)$, $(4t, t, 2t)$, $(2t, 4t, t)$, and $(t, 2t, 4t)$, where $t$ is any real number.\n\n**Remark:** By symmetry, the system yields three factorized equations:\n$$\n(2x - y)(z^2 - xy) = 0, \\quad (2y - z)(x^2 - yz) = 0, \\quad (2z - x)(y^2 - zx) = 0.\n$$\nThe cases $2x - y = 0$, $2y - z = 0$, and $2z - x = 0$ yield all solutions except $(t, t, t)$. For the remaining case\n$$\nz^2 - xy = x^2 - yz = y^2 - zx = 0,\n$$\nwe show that only $(x, y, z) = (t, t, t)$ works, since\n$$\n(x - y)^2 + (y - z)^2 + (z - x)^2 = 2(z^2 - xy) + 2(x^2 - yz) + 2(y^2 - zx) = 0,\n$$\nso all squares vanish, implying $x = y = z$.\n\n**Another solution:** For $xyz \\neq 0$, divide both sides by $xyz$:\n$$\n\\frac{y}{z} + \\frac{2z}{y} = \\frac{z}{x} + \\frac{2x}{z} = \\frac{x}{y} + \\frac{2y}{x}.\n$$\nLet $f(s) = s + 2/s$ and $s_1 = y/z$, $s_2 = z/x$, $s_3 = x/y$. $f(s) = f(t)$ iff $s = t$ or $st = 2$. Thus, the system holds if $s_i = s_j$ or $s_i s_j = 2$ for any $i, j$. If $s_i s_j = 2$, then $s_k = 1/2$ and $s_i \\in \\{1/2, 4\\}$, so $(s_1, s_2, s_3)$ is a permutation of $(1/2, 1/2, 4)$, yielding $(4t, t, 2t)$, $(2t, 4t, t)$, and $(t, 2t, 4t)$. If $s_1 = s_2 = s_3$, then $s_i = 1$, yielding $(t, t, t)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19980,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $AB < CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$, and $T$ be the midpoints of $EF$, $CF$, and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $CD$. We will prove that the circumcircles of the triangles $MKT$ and $MLS$ pass through $N$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p25_data_9bf8557dbb.png)\n\nFirst, we will prove that the circumcircle of $MLS$ passes through $N$.\n\nLet $Q$ be the midpoint of $EC$. Note that the circumcircle of $MLS$ is the Euler circle of triangle $EFC$, so it also passes through $Q$.\n\nWe will prove that\n\n$$\n\\angle SLQ = \\angle QNS \\quad \\text{or}\n$$\n\n$$\n\\angle SLQ + \\angle QNS = 180^{\\circ}.\n$$\n\nIndeed, since $FLC$ is right-angled and $LS$ is its median, we have $SL = SC$ and\n\n$$\n\\angle SLC = \\angle SCL = \\angle ACB.\n$$\n\nIn addition, since $N$ and $S$ are the midpoints of $DC$ and $FC$, we have $SN \\parallel FD$, and similarly, since $Q$ and $N$ are the midpoints of $EC$ and $CD$, $QN \\parallel ED$.\n\nIt follows that the angles $EDB$ and $QNS$ have parallel sides, and since $AB < CD$ they are acute, so\n\n$$\n\\angle EDB = \\angle QNS \\quad \\text{or} \\quad \\angle EDB + \\angle QNS = 180^{\\circ}.\n$$\n\nBut, from the cyclic quadrilateral $ABCD$, we get\n\n$$\n\\angle EDB = \\angle ACB.\n$$\n\nNow, from the above, we obtain that the quadrilateral $LNSQ$ is cyclic. Since its circumcircle also passes through $M$, the points $M, L, Q, S, N$ are cocyclic, so the circumcircle of $MLS$ passes through $N$.\n\nSimilarly, the circumcircle of $MKT$ also passes through $N$, and we have the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19981,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside $\\triangle ABC$. Let $A_1$, $B_1$, $C_1$ be points in the interiors of the segments $PA$, $PB$, $PC$, respectively. Let $\\overline{BC_1} \\cap \\overline{CB_1} = \\{A_2\\}$, $\\overline{CA_1} \\cap \\overline{AC_1} = \\{B_2\\}$, and $\\overline{AB_1} \\cap \\overline{BA_1} = \\{C_2\\}$. Let $U$ be the intersection of the lines $A_1B_1$ and $A_2B_2$, and $V$ be the intersection of the lines $A_1C_1$ and $A_2C_2$. Show that the lines $UC_2$, $VB_2$, and $AP$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/tmc2017_New_p14_data_2ab18e515b.png)\nLet $X$ be the intersection of the lines $UC_2$ and $AP$, and $Y$ be the intersection of the lines $VB_2$ and $AP$. By Menelaus' theorem, we have\n\n$$\n\\begin{aligned}\n\\frac{AX}{XA_1} \\cdot \\frac{A_1U}{UB_1} \\cdot \\frac{B_1C_2}{C_2A} &= -1, \\\\\n\\frac{B_1U}{UA_1} \\cdot \\frac{A_1B_2}{B_2C} \\cdot \\frac{CA_2}{A_2B_1} &= -1, \\\\\n\\frac{AC_2}{C_2B_1} \\cdot \\frac{B_1B}{BP} \\cdot \\frac{PA_1}{A_1A} &= -1, \\\\\n\\frac{CB_2}{B_2A_1} \\cdot \\frac{A_1A}{AP} \\cdot \\frac{PC_1}{C_1C} &= -1, \\\\\n\\frac{B_1A_2}{A_2C} \\cdot \\frac{CC_1}{C_1P} \\cdot \\frac{PB}{BB_1} &= -1.\n\\end{aligned}\n$$\n\nMultiplying these equations, we obtain\n\n$$\n\\frac{AX}{XA_1} \\cdot \\frac{PA_1}{AP} = -1.\n$$\n\nSo $AX/XA_1 = AP/A_1P$. Since the point $A_1$ lies in the interior of the segment $PA$, it follows that the point $X$ lies in the interior of the segment $AA_1$. Similarly, the point $Y$ lies in the interior of the segment $AA_1$, and $AY/YA_1 = AP/A_1P = AX/XA_1$. Thus $X = Y$. That is, the lines $UC_2$, $VB_2$, and $AP$ are concurrent.\n\n**Remark 1.** It can be proved that the point $U$ is on the line $AB$ and the point $V$ is on the line $AC$.\n\n**Remark 2.** From the point of view of Projective Geometry, we may consider the 3-point perspective drawing with the points of perspective at $A$, $B$, and $C$. Under this perspective, consider the \"box\" $A_1C_2B_1A_2C_1B_2$. Suppose we have an identical box placed right \"above\" this box. The existence of the point of concurrency in this problem is a vertex of this second box. With this consideration, the first Remark is also clear.\n\n**Remark 3.** This problem was inspired from Math 130, Harvard University, taught by Professor Michael Hopkins in the spring of 2017.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19982,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that for all $a, b, c \\in \\mathbb{Z}$,\n$$\nf(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are:\n\n1. $f(x) = c x^2$ for some $c \\in \\mathbb{Z}$.\n2. $f(x) = \\begin{cases} 0, & 2 \\mid x, \\\\ c, & 2 \\nmid x \\end{cases}$ for some $c \\in \\mathbb{Z}$.\n3. $f(x) = \\begin{cases} 0, & 4 \\mid x, \\\\ c, & 2 \\nmid x, \\\\ 4c, & x \\equiv 2 \\pmod{4} \\end{cases}$ for some $c \\in \\mathbb{Z}$.\n\nEach of these functions satisfies the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19983,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the largest number that is the greatest common divisor of some four different two-digit numbers.\n\nb) Find the largest number that is the least common multiple of some four different two-digit numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $d$ be the greatest common divisor of some four different two-digit numbers. Since all these numbers are divisible by $d$, the least possible candidates for these four numbers are $d, 2d, 3d, 4d$. Hence $4d < 100$, thus $d \\leq 24$. On the other hand, the greatest common divisor of $24, 48, 72,$ and $96$ is $24$.\n\nb) The numbers $99, 98, 97,$ and $95$ are pairwise relatively prime, hence $\\mathrm{lcm}(99, 98, 97, 95) = 99 \\cdot 98 \\cdot 97 \\cdot 95$. To show that this is the largest possible, consider four different two-digit numbers $a_1, a_2, a_3, a_4$; assume without loss of generality that $a_1 > a_2 > a_3 > a_4$. If $a_4 \\leq 95$, then $\\mathrm{lcm}(a_1, a_2, a_3, a_4) \\leq a_1 a_2 a_3 a_4 \\leq 99 \\cdot 98 \\cdot 97 \\cdot 95$. If $a_4 = 96$, then the four numbers can only be $99, 98, 97, 96$, but $\\mathrm{lcm}(99, 98, 97, 96) = \\frac{99 \\cdot 98 \\cdot 97 \\cdot 96}{2 \\cdot 3} < 99 \\cdot 98 \\cdot 97 \\cdot 95$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19984,
    "subject": "Mathematics (Olympiad)",
    "question": "On the plane there is a triangle $APQ$ and a rectangle $ABCD$ such that the midpoint of the segment $PQ$ belongs to the diagonal $BD$ of the rectangle, and one of the rays $AB$ or $AD$ is a bisector of the angle $PAQ$. Prove that one of the rays $CB$ or $CD$ is a bisector of the angle $PCQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider the case where the ray $AB$ is a bisector of the angle $PAQ$, and $\\angle BAC < \\angle BAQ$ (see the figure below). We will prove that in this case the ray $CD$ is the bisector of the angle $PCQ$. In the case where $AD$ is the bisector of the angle $PAQ$, the proof is analogous.\n\nLet $M$ be the midpoint of $PQ$, and $O$ the intersection point of the diagonals of the rectangle. Let us select point $S$ symmetrically to $Q$ with respect to the point $O$. In this case, $AQCS$ is a parallelogram. Let us first prove that $\\angle APC = \\angle AQC$. For this, we should prove that the points $A$, $S$, $P$, $C$ are cyclic. Indeed, let $\\angle BAC = \\alpha$, and $\\angle BAQ = \\beta$. By the assumption, $\\beta > \\alpha$. Then $\\angle CAQ = \\beta - \\alpha$. Also, $CS \\parallel AQ$, so $\\angle SCA = \\angle CAQ = \\beta - \\alpha$. $AB$ is the bisector of the angle $PAQ$, therefore $\\angle PAB = \\beta$. Moreover, $ABCD$ is a rectangle, which leads to $\\angle ABD = \\angle BAC = \\alpha$. $M$ is the midpoint of $PQ$, and $O$ is the midpoint of $QS$, hence $MO$ is the mid-segment of $PQS$. This brings the fact that $PS \\parallel MO$, i.e., $PS \\parallel BD$. Which leads to $\\angle PTB = \\angle TBD = \\alpha$, where $T = (PS) \\cap (AB)$.\n\nAccording to the Triangle exterior angle theorem, we obtain that\n\n$$\n\\angle APS = \\angle APT = \\angle PAB - \\angle PTA = \\beta - \\alpha\n$$\n\n![](images/Ukraine_2016_Booklet_p39_data_acd6b393f5.png)\n\nThus, we proved that $\\angle APS = \\angle ACS$, i.e., the points $A$, $S$, $P$, $C$ are cyclic. From which it follows that $\\angle APC = \\angle ASC = \\angle AQC$ (because inscribed angles that have the same arc are of the same length, and also the opposite angles of a parallelogram are the same).\n\nWhat is left is to prove that when in the convex quadrangle $APCQ$ the opposite angles $APC$ and $AQC$ are equal, then the bisectors of the angles $PAQ$ and $PCQ$ are parallel. Let the bisector of the angle $PAQ$ intersect lines $PC$ and $CQ$ in the points $N$ and $F$ respectively, and let the bisector of the angle $PCQ$ intersect $AQ$ in the point $L$. Let $K$ be the point symmetrical to $P$ with respect to $AF$, then the point $K$ lies on $AQ$ (as $AF$ is the bisector of the angle $PAQ$). This means that $\\angle AKN = \\angle APN$. $\\angle APC = \\angle AQC$, therefore $\\angle AKN = \\angle AQC$. Thus, $NK \\parallel CQ$. Which means $\\angle CFN = \\angle KNA = \\angle PNA = \\angle CNF$, i.e., $\\angle CFN = \\angle CNF$ and triangle $NCF$ is isosceles. The bisector of the exterior angle adjacent to the vertex of an isosceles triangle is parallel to its base, therefore $AF \\parallel CL$, Q.F.D.\n\nIn the case when $\\alpha > \\beta$, the points $P$ and $Q$ will lie in one direction from the line $AC$ and it will hold that $\\angle APC + \\angle AQC = 180^\\circ$. The ray $CB$ will be the bisector of the angle $PCQ$ (all proofs are analogous).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19985,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set of points in the plane, not all collinear. Consider the following process, called a *windmill*:\n\nA line $\\ell$ is initially chosen to pass through a point $P \\in S$. The line rotates clockwise about $P$ until it meets another point $Q \\in S$, at which moment $Q$ becomes the new pivot, and the process repeats indefinitely.\n\nProve that there exists a starting point $P$ and a starting line $\\ell$ such that, during the windmill process, each point of $S$ is used as a pivot infinitely many times.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the process for $|S| = 2k + 1$ (odd) and $|S| = 2k$ (even) separately.\n\n**Odd case ($|S| = 2k + 1$):**\n\nDefine a *good position* as a line $\\ell$ passing through two points of $S$ such that, just after leaving one point, there are $k$ points of $S$ on each side of $\\ell$.\n\n1. There are finitely many good positions, each determined by the pair of points and direction.\n2. For any $P \\in S$, by rotating a line about $P$, there must be a moment when the numbers of points on each side are equal, so every point is a pivot in some good position.\n3. Starting from a good position, the windmill process cycles through all good positions, ensuring each point is used as a pivot infinitely often.\n\n**Even case ($|S| = 2k$):**\n\nA *good position* is a line $\\ell$ passing through two points of $S$ such that, just after leaving one point, there are $k$ points of $S$ on the right side of $\\ell$.\n\nThe argument is similar: for any $P \\in S$, such a position exists, and the windmill cycles through all good positions, so each point is a pivot infinitely often.\n\nThus, for any finite set $S$ not all collinear, there exists a starting point and line such that each point is used as a pivot infinitely many times. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19986,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G = (V, E)$ be a connected nonbipartite graph. For a sequence of functions $f_1, f_2, \\dots$ defined on the vertices $V$, let $S_i = \\sum_{x \\in V} \\deg(x) f_i(x)^2$. Suppose $f_{i+1}(x) = \\frac{1}{\\deg(x)} \\sum_{xy \\in E} f_i(y)$ for each $x \\in V$. Prove that the sequence $S_1, S_2, \\dots$ is monotonically nonincreasing and eventually stabilizes, and that the limiting function $f_m$ is constant on $V$.",
    "options": [],
    "answer": "See solution",
    "solution": "Considering $S_i = \\sum_{x \\in V} \\deg(x) f_i(x)^2$, then\n\n$$\n\\begin{aligned}\nS_{i+1} &= \\sum_{x \\in V} \\deg(x) f_{i+1}(x)^2 \\\\\n&= \\sum_{x \\in V} \\deg(x) \\left[ \\frac{1}{\\deg(x)} \\sum_{xy \\in E} f_i(y) \\right]^2 \\\\\n&\\le \\sum_{x \\in V} \\frac{1}{\\deg(x)} \\left( \\sum_{xy \\in E} f_i(y) \\right)^2 \\\\\n&\\le \\sum_{x \\in V} \\sum_{xy \\in E} f_i(y)^2 = \\sum_{y \\in V} \\deg(y) f_i(y)^2 = S_i.\n\\end{aligned}\n$$\n\nThe first inequality is obtained by removing the rounding symbol, and the second inequality is obtained by applying Cauchy's inequality. Thus, $S_1, S_2, \\dots$ is a monotonically nonincreasing sequence of nonnegative integers, and eventually converges to a constant, so there exists a positive integer $m$ such that $S_m = S_{m+1}$.\n\nLet us prove that $f_m$ is a constant function. From the above derivation, it follows that when the equal sign holds, for each $x \\in V$, $f_m$ is a constant on the set $N(\\{x\\})$ composed of all the adjacent vertices of $x$. Thus, for any $y, z \\in V$, if there is a path of length 2 between $y$ and $z$ (i.e., there is a vertex adjacent to both of them), then $f_m(y) = f_m(z)$. Furthermore, if there is a path of even length between $y$ and $z$, then $f_m(y) = f_m(z)$. Since $G$ is a connected nonbipartite graph, there is an odd cycle in $G$, and any two points have a trail of even length, so $f_m$ takes a constant value on $V$. $\\square$\n\n**Remark** This question is related to Markov chains. Starting from an initial state, it will eventually converge to a periodic state. The solutions to this question all require considering the condition of equality after a certain semi-invariant is stabilized, and some basic techniques of combinatorics and graph theory are examined. The solutions of most candidates are similar to the first two methods or their variations, considering the maximum value of degree of approval. Some solutions also use the semi-invariant in Method 3. In addition, there were also methods by considering semi-invariant variables like $\\sum_{e \\in E} g_m(e)$ (where $g_m(e)$ is the number of two endpoints of edge $e$ belonging to $V \\setminus M_m$), as well as methods of linear algebra.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 19987,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the configuration in the diagram below:\n\n![](images/IRL_ABooklet_2023_p56_data_7f0a9a3e6f.png)\n\nLet $AG \\parallel BH$, and let lines intersect at points $P$ and $Q$ as shown. Prove that $ABHG$ is a parallelogram, given that $|AD| = |BC|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying the Intercept Theorem to $AG \\parallel BH$ and lines intersecting at $P$ gives\n\n$$\n\\frac{|PA|}{|PB|} = \\frac{|AF|}{|BH|} \\quad \\text{and} \\quad \\frac{|PA|}{|PB|} = \\frac{|AD|}{|BE|},\n$$\nso $|AF| \\cdot |BE| = |BH| \\cdot |AD|$.\n\nSimilarly, for lines intersecting at $Q$:\n\n$$\n\\frac{|QA|}{|QB|} = \\frac{|AG|}{|BE|} \\quad \\text{and} \\quad \\frac{|QA|}{|QB|} = \\frac{|AF|}{|BC|},\n$$\nso $|AF| \\cdot |BE| = |AG| \\cdot |BC|$.\n\nSince $|AD| = |BC|$, these equations imply $|BH| = |AG|$. Because $AG$ and $BH$ are parallel, $ABHG$ is a parallelogram.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19988,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $k$ positive integers $n_1, \\ldots, n_k$, define $d_1 = 1$ and for $i = 2, \\ldots, k$, let\n$$\nd_i = \\frac{(n_1, \\ldots, n_{i-1})}{(n_1, \\ldots, n_i)}\n$$\nwhere $(m_1, \\ldots, m_\\ell)$ denotes the greatest common divisor of the integers $m_1, \\ldots, m_\\ell$.\n\nProve that the sums\n$$\n\\sum_{i=1}^k a_i n_i\n$$\nwith $a_i \\in \\{1, \\ldots, d_i\\}$ for $i = 1, \\ldots, k$, are pairwise distinct modulo $n_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that two such sums, $\\sum_{i=1}^k a_i n_i$ and $\\sum_{i=1}^k b_i n_i$, are congruent modulo $n_1$. Let $j$ be the largest index such that $a_j \\neq b_j$.\n\nNotice that $(n_1, \\ldots, n_{j-1})$ divides $\\sum_{i=1}^j (a_i - b_i) n_i$, so it divides $(a_j - b_j) n_j$. Thus,\n$$\nd_j = \\frac{(n_1, \\ldots, n_{j-1})}{(n_1, \\ldots, n_j)}\n$$\ndivides $\\frac{(a_j - b_j) n_j}{(n_1, \\ldots, n_j)}$. Since $d_j$ and $\\frac{n_j}{(n_1, \\ldots, n_j)}$ are coprime, $d_j$ divides $a_j - b_j$.\n\nBut $1 \\leq |a_j - b_j| < d_j$, which is impossible. Therefore, all such sums are pairwise distinct modulo $n_1$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19989,
    "subject": "Mathematics (Olympiad)",
    "question": "Draw up a list of the remainders left by the powers of 2 after division by 13:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n|-----|---|---|---|---|---|---|----|----|----|----|----|----|----|\n| $2^n \\bmod 13$ | 1 | 2 | 4 | 8 | 3 | 6 | 12 | 11 | 9 | 5 | 10 | 7 | 1 |\n\nWhat is the remainder when $2^{2016}$ is divided by 13?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $a \\bmod b$ denotes the remainder left over when $a$ is divided by $b$. Also, it is not necessary to calculate each power of 2 in full before dividing by 13: it is enough to multiply the previous remainder by 2 and then subtract 13 as many times as required.\n\nSince $2^0$ and $2^{12}$ have the same remainder after division by 13, the pattern of remainders repeats in cycles of length 12. Now, divide 2016 by 12 to get $2016 = 12 \\times 168 + 0$, so $2^{2016}$ has the same remainder as $2^0$, which is 1.\n\nNote: 12 is the maximum cycle length for the powers of any number mod 13, because it includes all possible non-zero remainders. By Fermat's Little Theorem, if 13 is replaced by any prime number $p$ and 2 is replaced by any number $a$ not divisible by $p$, then the cycle length of powers of $a$ will always be a factor of $p-1$. It is also always possible to find a value of $a$ such that the cycle length is exactly $p-1$, as in this case.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 19990,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $1, 2, 3, \\ldots, 2014$ are written on the board. Andriyko can choose any two numbers $a, b$ on the board and replace them with the number $|a-b|$. After he performs this operation $2013$ times, there will be only one number left. What is the biggest possible value of this number?",
    "options": [],
    "answer": "See solution",
    "solution": "All the numbers on the board cannot exceed $2014$ at any moment. The parity of $|a-b|$ is the same as the parity of $a+b$, so the parity of the sum of all numbers on the board remains unchanged. Initially, the sum is odd, since there are $1007$ odd numbers ($1, 3, \\ldots, 2013$) and $1007$ even numbers ($2, 4, \\ldots, 2014$). Therefore, the last number must be odd, so it cannot be $2014$ and cannot be greater than $2013$.\n\nNow, we show that $2013$ can be achieved. Split all numbers into the following pairs: $(2, 3), (4, 5), \\ldots, (2012, 2013)$, and $(1, 2014)$. Applying the operation to each pair, we are left with $\\underbrace{1, 1, \\ldots, 1}_{1006}$ and $2013$ on the board. Then, pair the $1$'s into $503$ pairs and replace each pair with $0$. After these operations, no matter how the remaining $503$ operations are performed, the last number will be $2013$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19991,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations on $\\mathbb{R}$:\n\n$$\n\\begin{cases}\n\\sqrt{(\\sin x)^2 + \\dfrac{1}{(\\sin x)^2}} + \\sqrt{(\\cos y)^2 + \\dfrac{1}{(\\cos y)^2}} = \\sqrt{\\dfrac{20y}{x+y}}, \\\\\n\\sqrt{(\\sin y)^2 + \\dfrac{1}{(\\sin y)^2}} + \\sqrt{(\\cos x)^2 + \\dfrac{1}{(\\cos x)^2}} = \\sqrt{\\dfrac{20x}{x+y}}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $x \\neq \\dfrac{k\\pi}{2}$, $y \\neq \\dfrac{m\\pi}{2}$ ($k, m \\in \\mathbb{Z}$) and $xy > 0$. From the given condition, one can get\n\n$$\nA = 20 \\sqrt{\\dfrac{xy}{(x+y)^2}},\n$$\n\nwhere $A$ equals the product of the two left-hand sides of the given system. Using the Cauchy-Schwarz and AM-GM inequalities:\n\n$$\n\\begin{aligned}\n& \\left(\\sin^2 x + \\dfrac{1}{\\sin^2 x}\\right) \\left(\\cos^2 x + \\dfrac{1}{\\cos^2 x}\\right) \\\\\n& \\geq \\left(|\\sin x \\cos x| + \\dfrac{1}{|\\sin x \\cos x|}\\right)^2 \\\\\n& = \\left(\\dfrac{|\\sin 2x|}{2} + \\dfrac{1}{2|\\sin 2x|} + \\dfrac{3}{2|\\sin 2x|}\\right)^2 \\\\\n& \\geq \\left(1 + \\dfrac{3}{2}\\right)^2 = \\left(\\dfrac{5}{2}\\right)^2.\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\left(\\sin^2 y + \\dfrac{1}{\\sin^2 y}\\right) \\left(\\cos^2 y + \\dfrac{1}{\\cos^2 y}\\right) \\geq \\left(\\dfrac{5}{2}\\right)^2\n$$\n\nTherefore, applying the AM-GM inequality, we obtain\n\n$$\nA \\geq 4\\sqrt[4]{\\left(\\dfrac{5}{2}\\right)^4} = 10 \\geq 20\\sqrt{\\dfrac{xy}{(x+y)^2}}\n$$\n\nEquality occurs if and only if $|\\sin 2x| = 1$ and $x = y = \\dfrac{\\pi}{4} + \\dfrac{k\\pi}{2}$ where $k \\in \\mathbb{Z}$. It is easy to check that these solutions satisfy the given system. Thus, $x = y = \\dfrac{\\pi}{4} + \\dfrac{k\\pi}{2}$ where $k \\in \\mathbb{Z}$ are all solutions of the given system. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19992,
    "subject": "Mathematics (Olympiad)",
    "question": "In an $n \\times n$ table, two players take turns filling the rows with numbers $+1$ and $-1$. The first player fills the first row, the second player fills the second row, then the first player fills the third row, and so on. After all rows are filled, for each row or column where the product of the numbers is positive, the first player gets 1 point; otherwise, the second player gets the point. Both players try to maximize their own points. In the optimal play, how many points can each player collect?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\n- For even $n = 2k$, the first player collects $(3k + 2)$ points, the second player collects $k$ points.\n- For odd $n = 2k + 1$, the first player collects $(3k + 1)$ points, the second player collects $(k + 1)$ points.\n\n**Solution.**\n\nLet $n = 2k$ (even case).\n\nEach player fills $k$ rows. The first player can always secure $k$ points from the rows he fills. The second player can try to maximize his points by filling his last row to win as many columns as possible. The product of all numbers in the table is $+1$ (since each entry is either $+1$ or $-1$ and there are an even number of rows). The product of all columns is $+1$, and the product of all rows except the last is $(-1)^k$. Thus, the last row's product is also $+1$.\n\nWith optimal play, the first player gets at least $k$ points, and the second player can get up to $3k$ points from columns and rows. However, after considering all possible strategies, the maximum points are:\n\n- First player: $(3k + 2)$\n- Second player: $k$\n\nNow, let $n = 2k + 1$ (odd case).\n\nThe first player fills $k + 1$ rows, the second player fills $k$ rows. The first player can secure $k + 1$ points from his rows. The second player can win $k$ columns. The product of all numbers in the table is $+1$, and the product of all rows except the last is $(-1)^k$. For even $k$, the first player wins the last row; for odd $k$, the second player wins it. Thus, the final result is:\n\n- For even $k$: first player $(3k + 2)$, second player $k$\n- For odd $k$: first player $(3k + 1)$, second player $(k + 1)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19993,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be sets of positive integers with $|A| \\geq 2$ and $|B| \\geq 2$. Let $S$ be a set consisting of $|A| + |B| - 1$ numbers of the form $ab$ where $a \\in A$ and $b \\in B$. Prove that there exist pairwise distinct $x, y, z \\in S$ such that $x$ is a divisor of $yz$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on $k = |A| + |B| - 1$.\n\nFor $k = 3$ we have $|A| = |B| = 2$. Let $A = \\{x, y\\}$, $B = \\{z, t\\}$. Then $S$ consists of three numbers from the set $\\{xz, yz, xt, yt\\}$. Relabelling the elements of $A$ and $B$ if necessary, we can assume without loss of generality that the missing number is $yt$. Then $xz, xt, yz \\in S$ and $xz \\mid xt \\cdot yz$, which concludes the base case of induction.\n\nFor the inductive step, suppose the thesis holds for some $k-1 \\geq 3$. Since $k = |A| + |B| - 1 \\geq 4$, we have that $\\max(|A|, |B|) \\geq 3$. Without loss of generality, assume $|A| \\geq 3$. Since the set $S$ consists of $k = |A| + |B| - 1 > |A|$ elements, by the pigeonhole principle there exists a number $x \\in A$ which appears as the first of the two factors of at least two elements of $S$. So, there exist $y, z \\in B$ with $xy, xz \\in S$. If there exists $t \\in A \\setminus \\{x\\}$ such that $ty \\in S$ then we are done because $xy \\mid xz \\cdot ty$. If there exists no such $t$ then apply the inductive hypothesis to the sets $A \\setminus \\{x\\}$, $B$ and $S \\setminus \\{xy\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19994,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients. Given that for some integer $a$ there exists $n$ such that $$\\underbrace{P(P(\\ldots P(a) \\ldots))}_{n} = a$$, prove that $P(P(a)) = a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define a sequence $(a_i)$ as follows: $a_0 = a$, and $a_i = P(a_{i-1})$ for all natural $i$. Consider the minimal number $m$ for which there exists $j < m$ such that $a_j = a_m$. By the problem's condition, such $m$ exists. Observe that $a_0, a_1, \\ldots, a_{m-1}$ are distinct.\n\nWe show that $j = 0$. If not, then the sequence has the form:\n\n$$\na_0, a_1, \\ldots, a_{j-1}, a_j, a_{j+1}, \\ldots, a_{m-1}, a_m = a_j, a_{j+1}, \\ldots, a_{m-1}, a_j, a_{j+1}, \\ldots, a_{m-1}, a_j, a_{j+1}, \\ldots\n$$\n\nand $a_0$ never appears again, contradicting our assumption. Thus, $a_0 = a_m$, so the sequence is $m$-periodic: $a_p = a_q$ if $p \\equiv q \\pmod{m}$. Let $a_l$ be the maximum and $a_k$ the minimum among $a_0, a_1, \\ldots, a_{m-1}$. If $a_l = a_k$, then $P(a_l) = a$ and $P(P(a_l)) = a$. Suppose $a_l > a_k$. Note that $P(a_l) - P(a_k)$ is divisible by $a_l - a_k$, and $P(a_l), P(a_k)$ are in our sequence, so $|P(a_l) - P(a_k)| \\leq |a_l - a_k|$, and the difference $P(a_l) - P(a_k)$ is $0$ or $\\pm(a_l - a_k)$.\n\nIf $P(a_l) - P(a_k) = a_l - a_k$, then $P(a_k) - a_l$ and $m = 1$ with $a_l = a_k$, and the result follows from the previous observation.\n\nIf $P(a_l) - P(a_k) = 0$, then $a_{l+1} = P(a_l) = P(a_k) = a_{k+1}$, and $(l+1)-(k+1) = l-k$ is divisible by $m$, which implies $l = k$ and $m = 1$.\n\nIf $P(a_l) - P(a_k) = -(a_l - a_k)$, then $P(a_l) = a_k$, $P(a_k) = a_l$, and $m = 2$, which implies $P(P(a)) = a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 19995,
    "subject": "Mathematics (Olympiad)",
    "question": "在三角形 $ABC$ 中，令點 $B'$、$C'$ 分別為邊 $AC$ 及 $AB$ 的中點，而點 $H$ 為通過頂點 $A$ 的高的垂足。證明：三角形 $AB'C'$、$BC'H$ 及 $B'CH$ 的外接圓共於一點 $I$，且直線 $HI$ 平分線段 $B'C'$。",
    "options": [],
    "answer": "See solution",
    "solution": "解一：\n\n![](images/2022-TWNIMO-Problems_p9_data_74c1711366.png)\n\n設點 $F$ 為 $B'C'$ 的中點，點 $A'$ 為 $BC$ 的中點，且設直線 $HF$ 與 $\\triangle BHC'$ 的外接圓再交於點 $I$，又設直線 $AA'$ 與 $\\triangle ABC$ 的外接圓再交於點 $M$。三角形 $HB'C'$ 全等於 $\\triangle AB'C'$，故相似於 $\\triangle ABC$。因為\n\n$$\n\\angle C'IF = \\angle ABC = \\angle A'MC,\n$$\n\n$$\n\\angle C'FI = \\angle AA'B = \\angle MA'C,\n$$\n\n$$\n2C'F = C'B',\n$$\n\n$$\n2A'C = CB,\n$$\n\n所以 $\\triangle C'IB' \\sim \\triangle CMB$，於是有 $\\angle FIB' = \\angle A'MB = \\angle ACB$。因為 $\\angle C'IB' = 180^\\circ - \\angle C'AB'$，可知 $I$ 位於 $\\triangle AB'C'$ 與 $\\triangle HCB'$ 的外接圓上。\n\n解二：將 $\\triangle ABC$ 的三個內角分別記為 $\\alpha, \\beta, \\gamma$。易知 $\\triangle ABC \\sim \\triangle HC'B'$。在 $\\triangle HC'B'$ 內部存在唯一一點 $I$ 滿足\n\n$$\n\\angle HIB' = 180^\\circ - \\gamma, \\quad \\angle HIC' = 180^\\circ - \\beta, \\quad \\angle C'IB' = 180^\\circ - \\alpha,\n$$\n\n所以題敘的三個圓必定都過點 $I$。\n\n設直線 $HI$ 與 $B'C'$ 交於點 $F$。以下證明 $FB' = FC'$。由 $\\angle HIB' + \\angle HB'F = 180^\\circ$，得 $\\angle IHB' = \\angle IB'F$。同理可得 $\\angle IHC' = \\angle IC'F$。所以 $\\triangle IHC'$ 與 $\\triangle IHB'$ 的外接圓者與 $B'C'$ 相切，由此可知 $FB'^2 = FI \\cdot FH = FC'^2$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 19996,
    "subject": "Mathematics (Olympiad)",
    "question": "A square grid on the Euclidean plane consists of all points $ (m, n) $, where $ m $ and $ n $ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least $ 5 $?",
    "options": [],
    "answer": "See solution",
    "solution": "It is not possible. The proof is by contradiction. Suppose that such a covering family $\\mathcal{F}$ exists. Let $D(P, \\rho)$ denote the disc with center $P$ and radius $\\rho$.\n\nStart with an arbitrary disc $D(O, r)$ that does not overlap any member of $\\mathcal{F}$. Then $D(O, r)$ covers no grid point. Take the disc $D(O, r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O, r)$ is tangent to at least three discs in $\\mathcal{F}$.\n\nObserve that there must be two of the three tangent discs, say $D(A, a)$ and $D(B, b)$, such that $\\angle AOB \\le 120^\\circ$. By the Law of Cosines applied to triangle $ABO$,\n\n$$\n(a+b)^2 \\le (a+r)^2 + (b+r)^2 + (a+r)(b+r),\n$$\nwhich yields\n$$\nab \\le 3(a+b)r + 3r^2, \\quad \\text{and thus} \\quad 12r^2 \\ge (a-3r)(b-3r).\n$$\n\nNote that $r < 1/\\sqrt{2}$ because $D(O, r)$ covers no grid point, and $(a-3r)(b-3r) \\ge (5-3r)^2$ because each disc in $\\mathcal{F}$ has radius at least $5$. Hence $2\\sqrt{3}r \\ge (5-3r)$, which gives $5 \\le (3+2\\sqrt{3})r < (3+2\\sqrt{3})/\\sqrt{2}$ and thus $5\\sqrt{2} < 3+2\\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21+12\\sqrt{3} < 21+24 = 45$. This contradiction completes the proof.\n\n**Remark:** The above argument shows that no covering family exists where each disc has radius greater than $(3+2\\sqrt{3})/\\sqrt{2} \\approx 4.571$. In the other direction, there exists a covering family in which each disc has radius $\\sqrt{13}/2 \\approx 1.802$. Take discs with this radius centered at points of the form $(2m + 4n + \\frac{1}{2}, 3m + \\frac{1}{2})$, where $m$ and $n$ are integers. Then any grid point is within $\\sqrt{13}/2$ of one of the centers and the distance between any two centers is at least $\\sqrt{13}$. The extremal radius of a covering family is unknown.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 19997,
    "subject": "Mathematics (Olympiad)",
    "question": "The chord joining the points labelled *a* and *d* does not intersect the chord joining the points labelled *b* and *c*.\n\nLet $M$ be the number of beautiful labellings, and let $N$ be the number of ordered pairs $(x, y)$ of positive integers such that $x + y \\le n$ and $\\gcd(x, y) = 1$. Prove that\n\n$$\nM = N + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that there are $N + 1$ beautiful labellings for all $n \\ge 2$.\n\nLet $0 < x < 1$ be a real number. Define the beautiful labelling $C_n(x)$ as follows. For $0 \\le k \\le n$, let $W_k = e^{\\frac{2\\pi i k}{n}} x$, and let $Z_k$ be the point which results from rearranging the $W_k$ evenly in the same relative position; label $Z_k$ by $k$. Note that $W_a W_b$ and $W_c W_d$ intersect iff $Z_a Z_b$ and $Z_c Z_d$ intersect.\n\nCall such a labelling *cyclic*, and call it *degenerate* if two of the $W_k$ coincide. Note that $C_n(x)$ is degenerate iff $x$ is a reduced fraction with denominator at most $n$. Call such numbers *good*; there are $N$ good fractions in $(0, 1)$.\n\n**Lemma 2.** A labeling is beautiful if and only if it is non-degenerate cyclic.\n\n*Proof.* Let $C_n(x)$ be a non-degenerate cyclic labelling. For any $0 \\le a < b < c < d \\le n$ with $a+d = b+c$, arcs $\\overline{W_a W_b}$ and $\\overline{W_c W_d}$ have the same measure, so $W_a W_d \\parallel W_b W_c$, implying that $C_n(x)$ is beautiful.\n\nFor the converse, induct on $n$ with trivial base case $n=2$. If all beautiful arrangements of $[0, n-1]$ are cyclic, for a beautiful arrangement $A$ of $[0, n]$, form $A' = C_{n-1}(x)$ by removing $n$. Let $x$ lie between consecutive good fractions $p_1/q_1$ and $p_2/q_2$ with $q_1, q_2 \\le n-1$, and consider two cases.\n\n**Case 1:** There is no fraction with denominator $n$ between $\\frac{p_1}{q_1}$ and $\\frac{p_2}{q_2}$. If $A \\ne C_n(x)$, they can differ only in the location of $n$. Suppose $W_n$ occurs directly between $W_i$ and $W_j$ in $C_n(x)$ in clockwise order. Note that $i + (n-1) = (i-1) + n$ and $j + (n-1) = (j-1) + n$, so the two corresponding pairs of chords do not intersect and $W_i, W_n, W_j, W_{i-1}, W_{n-1}, W_{j-1}$ occur in that order. In $A$, since $W_i W_{n-1}$ does not intersect $W_n W_{i-1}$, $W_n$ lies on arc $\\overline{W_i W_{n-1}}$. Likewise, $W_n$ must lie on arc $\\overline{W_{n-1} W_j}$. Thus $W_n$ lies between $W_i$ and $W_j$, so $A = C_n(x)$.\n\n**Case 2:** There is a fraction $\\frac{a}{n}$ with denominator $n$ between $\\frac{p_1}{q_1}$ and $\\frac{p_2}{q_2}$. Since $\\frac{p_2}{q_2} - \\frac{p_1}{q_1} \\le \\frac{1}{n-1}$, there is a unique such fraction. Choose $x_1 \\in (\\frac{p_1}{q_1}, \\frac{a}{n})$ and $x_2 \\in (\\frac{a}{n}, \\frac{p_2}{q_2})$. We wish to show that either $A = C_n(x_1)$ or $A = C_n(x_2)$. In $A'$, $W_{q_1}, W_0, W_{q_2}$ occur in that clockwise order. It suffices to show that $W_n$ lies on arc $\\overline{W_{q_1} W_{q_2}}$ in $A$. This follows by an analysis of chords $W_{q_1} W_{n-1}, W_n W_{q_2-1}, W_{q_1-1} W_n$, and $W_{n-1} W_{q_2}$ using the final argument of Case 1. $\\square$\n\nBy Lemma 2, it remains for us to count non-degenerate cyclic labellings. We claim that $C_n(x) = C_n(y)$ iff there is no good fraction between $x$ and $y$. As $x$ varies, the ordering of points in $C_n(x)$ changes only when $C_n(x)$ is degenerate so that two points coincide. It follows that $C_n(x) = C_n(y)$ when there is no good fraction between $x$ and $y$. If there is a good fraction $p/q$ with $x < p/q < y$, in $C_n(y)$ there are at least $p$ integers $1 \\le i \\le q$ such that $W_0$ is clockwise of $W_{i-1}$ and $W_i$ is clockwise of $W_0$, while in $C_n(x)$ there are fewer than $p$ such integers. Hence, $C_n(x)$ and $C_n(y)$ differ, giving the claim.\n\nWe conclude that the number of non-degenerate cyclic labellings is one greater than the number of good fractions in $(0, 1)$, hence equal to $N + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 19998,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in real numbers the equation\n\n$$\nx + \\log_2 \\left( 1 + \\sqrt{\\frac{5^x}{3^x + 4^x}} \\right) = 4 + \\log_{1/2} \\left( 1 + \\sqrt{\\frac{25^x}{7^x + 24^x}} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as\n\n$$\nx + \\log_2 \\left( 1 + \\sqrt{\\frac{1}{(3/5)^x + (4/5)^x}} \\right) = 4 + \\log_{1/2} \\left( 1 + \\sqrt{\\frac{1}{(7/25)^x + (24/25)^x}} \\right),\n$$\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 one solution, and it is not difficult to guess it: $x = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 19999,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral $ABCD$ has $AD = DC = CB < AB$ and $AB \\parallel CD$. The points $E$ and $F$ lie on the sides $CD$ and $BC$ such that $\\angle ADE = \\angle AEF$.\n\nProve that:\n\na) $4CF \\le CB$.\n\nb) If $4CF = CB$, then $AE$ is the angle bisector of $\\angle DAF$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) $\\angle FEC = 180^\\circ - \\angle AEF - \\angle DEA = 180^\\circ - \\angle ADE - \\angle DEA = \\angle DAE$.\n\nFrom $AD = DC = CB < AB$ and $AB \\parallel CD$, it follows that $\\angle ADC = \\angle DCB$, hence triangles $ADE$ and $ECF$ are similar. This yields\n\n$$\n\\frac{AD}{EC} = \\frac{AE}{EF} = \\frac{DE}{CF} \\qquad (1)\n$$\n\nThis leads to $AD \\cdot CF = EC \\cdot DE \\le \\frac{1}{4}(EC + DE)^2 = \\frac{1}{4}CD^2$, hence $4CF \\le CB$, because $AD = DC = CB$.\n\nb) If $4CF = CB$, then the inequality $EC \\cdot DE \\le \\frac{1}{4}(EC + DE)^2$ becomes an equality, that is $CE = ED$.\n\nRelation (1) becomes $\\frac{AD}{DE} = \\frac{AE}{EF}$. Since $\\angle ADE = \\angle AEF$, triangles $ADE$ and $AEF$ are similar.\n\nThen $\\angle DAE = \\angle EAF$, so $AE$ bisects the angle $\\angle DAF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 20000,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma_1$ and $\\Gamma_2$ be two circles, and let $\\ell_1$ and $\\ell_2$ be tangents to $\\Gamma_2$ and $\\Gamma_1$ at points $E$ and $F$, respectively. Let $T$ be the intersection point of $\\ell_1$ and $\\ell_2$. Let $A$, $B$, $C$, $D$, $V$, $W$, $X$, $Y$, $K$, $L$, $M$, $N$, $U$, $Z$, $Y_1$, and $Y_2$ be points defined as follows:\n\n- $A$ is mapped to $E$ and $B$ to $F$ under inversion at $T$ swapping $\\Gamma_1$ and $\\Gamma_2$.\n- $C$ is mapped to $V$ and $D$ to $W$ under the same inversion.\n- Draw circles $ADE$ and $BCF$.\n- $K = \\ell_1 \\cap \\overline{BD}$, $L = \\ell_2 \\cap \\overline{AC}$.\n- The line $FX$ meets $\\Gamma_1$ again at $M$, and $EX$ meets $\\Gamma_2$ again at $N$.\n- Lines $AB$, $AD$, and $BC$ meet line $TX$ at $Z$, $Y_1$, and $Y_2$.\n\n% IMAGE: ![](images/sols-TST-IMO-2020_p3_data_df9dd2662c.png)\n\nProve that $Y_1 = Y_2$; that is, the intersection points of $AD$ and $BC$ with $TX$ coincide.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the inversion with center $T$ which swaps $\\Gamma_1$ and $\\Gamma_2$; it also swaps the pairs $\\{A, E\\}$ and $\\{B, F\\}$. Since $AECN$ is cyclic, $C$ is on $\\Gamma_1$, and $N$ is on $\\Gamma_2$, it also swaps $\\{C, N\\}$; similarly it swaps $\\{D, M\\}$.\n\nThus $(EB; ND)_{\\Gamma_2} = (AF; CM)_{\\Gamma_1} = (FA; MC)_{\\Gamma_1}$ as desired. $\\square$\n\nWith this claim, the remainder of the proof is chasing cross-ratios:\n\n$$\n(TZ; XY_1) \\stackrel{A}{=} (KB; XD) \\stackrel{E}{=} (EB; ND)_{\\Gamma_2} = (FA; MC)_{\\Gamma_1} \\stackrel{F}{=} (LA; XC) \\stackrel{B}{=} (TZ; XY_2)\n$$\n\nimplies $Y_1 = Y_2$ as desired.\n\nAlternatively, by considering $X$ as a variable point on $UV$ (the intersection points of $\\Gamma_1$ and $\\Gamma_2$), and applying degree arguments and Zack's lemma, we find that the assertion that $T, X, Y$ are collinear is a statement of degree at most $4$. By checking five special cases (including $X = U$, $X = V$, $X \\in \\ell_1$, $X \\in \\ell_2$, and $X$ at infinity), the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  }
]